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8500 | https://tamucc-ir.tdl.org/server/api/core/bitstreams/50b89337-54cc-48ce-b0fd-137b9c2dde42/content | http: // www.aimspress.com /journal /Math AIMS Mathematics, 6(1): 518–531. DOI:10.3934 /math.2021031 Received: 8 October 2020 Accepted: 11 October 2020 Published: 19 October 2020
Research article
Existence of a unique solution to an elliptic partial di ff erential equation when the average value is known
Diane Denny ∗
Department of Mathematics and Statistics, Texas A&M University-Corpus Christi, 6300 Ocean Drive, Corpus Christi, TX 78412, USA Correspondence: Email: diane.denny@tamucc.edu; Tel: +13618253485; Fax: +13618255789.
Abstract: The purpose of this paper is to prove the existence of a unique classical solution u(x) to the quasilinear elliptic partial di ff erential equation ∇ · (a(u)∇u) = f for x ∈ Ω, which satisfies the condition that the average value 1
|Ω|
∫
Ω
ud x = u0, where u0 is a given constant and 1
|Ω|
∫
Ω
f d x = 0. Periodic boundary conditions will be used. That is, we choose for our spatial domain the N-dimensional torus
TN , where N = 2 or N = 3. The key to the proof lies in obtaining a priori estimates for u.
Keywords: elliptic; existence; uniqueness
Mathematics Subject Classification: 35A01
Introduction
In this paper, we consider the existence of a unique, classical solution u(x) to the quasilinear elliptic equation
∇ · (a(u)∇u) = f (1.1) for x ∈ Ω, which satisfies the condition that the average value 1
|Ω|
∫
Ω
ud x = u0 (1.2) where u0 is a given constant and 1
|Ω|
∫
Ω
f d x = 0 . Periodic boundary conditions will be used. That is, we choose for our spatial domain the N-dimensional torus TN , where N = 2 or N = 3. The purpose of this paper is to prove the existence of a unique classical solution u to (1.1), (1.2). The proof of the existence theorem uses the method of successive approximations in which an iteration scheme, based on solving a linearized version of Eq (1.1), will be defined and then convergence of the sequence of approximating solutions to a unique solution satisfying the quasilinear equation will be 519
proven. The key to the proof lies in obtaining a priori estimates for u. To the best of our knowledge, no other researcher has proven the existence and uniqueness of the solution to this partial di ff erential equation when the given data is the average value of the solution. The paper is organized as follows. The main result, Theorem 2.1, is presented and proven in the next section. The existence of a solution to the linearized equation used in the iteration scheme is proven in Appendix A. Appendix B presents lemmas supporting the proof of the theorem.
Existence theorem
We will be working with the Sobolev space Hs(Ω) (where s ≥ 0 is an integer) of real-valued functions in L2(Ω) whose distribution derivatives up to order s are in L2(Ω). The norm is ‖u‖2
s
=
∑0≤| α|≤ s
∫
Ω
|Dαu|2dx. We are using the standard multi-index notation. We define |F|r,G0 = max {
∣∣∣∣d j Fdu j (u∗)
∣∣∣∣ :
u∗ ∈ G0, 0 ≤ j ≤ r}, where F is a function of u and where G0 ⊂ R is a closed, bounded interval. Also, we let both ∇u and Du denote the gradient of u. And Ck(Ω) is the set of real-valued functions having all derivatives of order ≤ k continuous in Ω (where k = integer ≥ 0 or k = ∞). The purpose of this paper is to prove the following theorem:
Theorem 2.1. Let a be a smooth, positive function of u. Let f ∈ H2(Ω) and let 1
|Ω|
∫
Ω
f d x = 0. Let the domain Ω = TN , the N-dimensional torus, where N = 2 or N = 3.There exists a constant C 1 which depends only on N, Ω such that if
1(min u∗∈G0 a(u∗)) 4
∣∣∣∣da du
∣∣∣∣20,G0
‖∇ f ‖20 ≤ C1
and if
∣∣∣∣d2adu 2
∣∣∣∣0,G0
≤ 1(min u∗∈G0 a(u∗))
∣∣∣∣da du
∣∣∣∣20,G0
(2.1)
where G 0 ⊂ R is a closed, bounded interval, then there exists a unique solution u ∈ C2(Ω) to the equation
∇ · (a(u)∇u) = f (2.2)
which satisfies the condition that the average value
1
|Ω|
∫
Ω
ud x = u0 (2.3)
where u 0 is a given constant. Proof.
We begin by using the following change of variables:
v = ( a0
‖∇ f ‖0
)ub(v) = ( 1
a0
)a(‖∇ f ‖0
a0
v)
AIMS Mathematics Volume 6, Issue 1, 518–531. 520
g = ( 1
‖∇ f ‖0
) f (2.4) where the constant a0 = min u∗∈G0 a(u∗) and G0 ⊂ R is a closed, bounded interval. Under this change of variables the equation (2.2) becomes
∇ · (b(v)∇v) = g (2.5) And under this change of variables, (2.3) becomes 1
|Ω|
∫
Ω
vd x = v0 = a0
‖∇ f ‖0
u0 (2.6) We fix closed, bounded intervals G0 ⊂ R and G1 ⊂ R by defining G0 = {u∗ ∈ R : |u∗ − u0|L∞ ≤
R‖∇ f‖0
a0
} and G1 = {v∗ ∈ R : |v∗ − v0|L∞ ≤ R}, where R is a constant to be defined later. We will prove that v(x) ∈ G1 for x ∈ Ω. It follows that u(x) ∈ G0 for x ∈ Ω.We will construct the solution of (2.5), (2.6) through an iteration scheme. To define the iteration scheme, we will let the sequence of approximate solutions be {vk}. Set the initial iterate v0 = v0. For
k = 0, 1, 2, . . . , construct vk+1 from the previous iterate vk by solving the linear equation
∇ · (b(vk)∇vk+1) = g (2.7) which satisfies the condition that the average value 1
|Ω|
∫
Ω
vk+1dx = v0 (2.8) and using periodic boundary conditions. The existence of a unique solution vk+1 ∈ C2(Ω) to the linear equation (2.7) for fixed k which satisfies (2.8) is proven in Appendix A. Lemmas supporting the proof are presented in Appendix B. We proceed now to prove convergence of the iterates as k → ∞ to a unique, classical solution v of (2.5), (2.6), which therefore produces a unique, classical solution u = ‖∇ f ‖0
a0
v of (2.2), (2.3). We begin by proving the following proposition:
Proposition 1. Assume that the hypotheses of Theorem 2.1 hold. Then there exist constants C 2, C 3,and R such that the following inequalities hold for k = 1, 2, 3 . . . :
‖∇ vk‖22 ≤ C2 (2.9)
‖vk‖24 ≤ C3 (2.10)
|vk − v0|L∞ ≤ R (2.11)
‖∇ (vk+1 − vk)‖20 ≤ (12
)k
C2 (2.12)
where the constants C 2, R depend on N and Ω, and where the constant C 3 depends on R, u 0, a 0,
‖∇ f ‖0, ‖∇ f ‖1,
∣∣∣∣da du
∣∣∣∣2,G0
, N, and Ω. From (2.11) it follows that v k(x) ∈ G1 for x ∈ Ω and for k = 1, 2, 3 . . . .
AIMS Mathematics Volume 6, Issue 1, 518–531. 521
Proof. The proof is by induction on k. We prove in Lemma B.2 in Appendix B that if vk satisfies (2.9) and (2.11), then vk+1 satisfies (2.9) and (2.10). See Lemma B.2 in Appendix B for the detailed proof. It only remains to prove inequalities (2.11) for vk+1 − v0 and (2.12) for ∇(vk+1 − vk). In the estimates below, we will let C denote a generic constant whose value may change from one relation to the next.
Estimate for |vk+1 − v0|L∞ :
Lemma B.2 in Appendix B presents the proof that ‖∇ vk+1‖22 ≤ C2. Then by using standard Sobolev space inequalities we obtain the inequality:
|vk+1 − v0|L∞ ≤ C‖vk+1 − v0‖2
≤ C‖∇ (vk+1 − v0)‖1
= C‖∇ vk+1‖1
≤ C √C2
= R
where the constants C and C2 depend on Ω, N. Here we used the fact that |vk+1 − v0|L∞ ≤ C‖vk+1 − v0‖2
by Sobolev’s Lemma. Since 1
|Ω|
∫
Ω
vk+1dx = v0 by (2.8), it follows that vk+1 − v0 is a zero-mean function and ‖vk+1 − v0‖0 ≤ C‖∇ (vk+1 − v0)‖0 by Poincar´ e’s inequality. Therefore ‖vk+1 − v0‖2 ≤ C‖∇ (vk+1 − v0)‖1.We define R = C √C2. Then inequality (2.11) of Proposition 1 holds for vk+1 − v0.
Estimate for ‖∇ (vk+1 − vk)‖20:
From successive iterates of Eq (2.7) we obtain the following:
∇ · (b(vk)∇(vk+1 − vk)) = ∇ · (b(vk)∇vk+1) − ∇ · (b(vk)∇vk)
= g − ∇ · (( b(vk) − b(vk−1)) ∇vk) − ∇ · (b(vk−1)∇vk)
= −∇ · (( b(vk) − b(vk−1)) ∇vk) (2.13) In the estimates that follow, we use the notation ( h1, h2) = ∫
Ω
h1h2dx for the L2 inner product of functions h1, h2. Note that vk − vk−1 is a zero-mean function because vk − vk−1 = (vk − v0) − (vk−1 − v0)and vk − v0, vk−1 − v0 are zero-mean functions by successive iterates of (2.8). We define the constant b0 = min v∗∈G1 b(v∗), where G1 ⊂ R is a closed, bounded interval. Note that
b0 = 1 by the definition of the function b in (2.4). Then integration by parts and using Eq (2.13) yields
‖∇ (vk+1 − vk)‖20 = (∇(vk+1 − vk), ∇(vk+1 − vk))
≤ 1
b0
(b(vk)∇(vk+1 − vk), ∇(vk+1 − vk))
= − 1
b0
(∇ · (b(vk)∇(vk+1 − vk)) , (vk+1 − vk))
= 1
b0
(∇ · (( b(vk) − b(vk−1)) ∇vk), (vk+1 − vk))
= − 1
b0
(( b(vk) − b(vk−1)) ∇vk, ∇(vk+1 − vk))
≤ 1
b0
‖(b(vk) − b(vk−1)) ∇vk‖0‖∇ (vk+1 − vk)) ‖0
AIMS Mathematics Volume 6, Issue 1, 518–531. 522
≤ 1
b0
∣∣∣∣db dv
∣∣∣∣0,G1
‖vk − vk−1‖0|∇ vk|L∞ ‖∇ (vk+1 − vk)‖0
≤ C( 1
b0
)2∣∣∣∣db dv
∣∣∣∣20,G1
‖∇ (vk − vk−1)‖20‖∇ vk‖22
12‖∇ (vk+1 − vk)‖20 (2.14) where C is a constant which depends on N, Ω. Here we used the fact that |∇ vk|L∞ ≤ C‖∇ vk‖2 by Sobolev’s Lemma. And we used Poincar´ e’s inequality to obtain ‖vk − vk−1‖0 ≤ C‖∇ (vk − vk−1)‖0, since
vk − vk−1 is a zero-mean function. Using the facts that 1
b0
= 1 and that
∣∣∣∣db dv
∣∣∣∣20,G1
= 1
a40
‖∇ f ‖20
∣∣∣∣da du
∣∣∣∣20,G0
≤ C1 by the definition of b(v) in (2.4) and by the statement of the theorem, and using the fact that ‖∇ vk‖22 ≤ C2 by the induction hypothesis, we obtain from re-arranging terms in (2.14) the inequality
‖∇ (vk+1 − vk)‖20 ≤ C( 1
b0
)2∣∣∣∣db dv
∣∣∣∣20,G1
‖∇ vk‖22‖∇ (vk − vk−1)‖20
≤ CC 1C2‖∇ (vk − vk−1)‖20
≤ 12‖∇ (vk − vk−1)‖20
(2.15) where we define the constant C1 to be su ffi ciently small so that CC 1C2 ≤ 12 . And the constants C, C1,
C2 depend on N, Ω.By repeatedly applying inequality (2.15) it follows that
‖∇ (vk+1 − vk)‖20 ≤ (12
)k‖∇ (v1 − v0)‖20
= (12
)k‖∇ v1‖20
≤ (12
)kC2 (2.16) where the initial iterate v0 = v0, which is a constant, and where ‖∇ v1‖20 ≤ ‖∇ v1‖22 ≤ C2 by Lemma B.2 in Appendix B. Therefore inequality (2.12) of Proposition 1 holds for ∇(vk+1 − vk). This completes the proof of Proposition 1.
We now complete the proof of Theorem 2.1. By (2.16), ‖∇ (vk+1 − vk)‖0 → 0 as k → ∞ . By Poincar´ e’s inequality, ‖vk+1 − vk‖20 ≤ C‖∇ (vk+1 − vk)‖20. It follows that ‖vk+1 − vk‖0 → 0 as k → ∞ . We next use the standard interpolation inequality ‖vk+1 − vk‖r ≤ C‖vk+1 − vk‖β
0
‖vk+1 − vk‖1−β
4
, where β = 4−r
4
,and 0 < r < 4. Then since ‖vk+1 − vk‖24 ≤ C(‖vk+1‖24 + ‖vk‖24
) ≤ CC 3 by (2.10) in Proposition 1, it follows that ‖vk+1 − vk‖r → 0 as k → ∞ for 0 < r < 4. Therefore there exists v ∈ Hr(Ω), where r < 4, such that ‖vk − v‖r → 0 as k → ∞ . The fact that
v ∈ H4(Ω) can be deduced using boundedness in high norm and a standard compactness argument (see, for example, Embid , Majda ). Sobolev’s Lemma implies that v ∈ C2(Ω). From Lemma A.1 in Appendix A, vk+1 ∈ C2(Ω) is a solution of the linear equation ∇ · (b(vk)∇vk+1) =
g for each k ≥ 0, and vk+1 satisfies the condition that 1
|Ω|
∫
Ω
vk+1dx = v0. It follows that v is a classical
AIMS Mathematics Volume 6, Issue 1, 518–531. 523
solution of the equation ∇ · (b(v)∇v) = g, and v satisfies the condition that 1
|Ω|
∫
Ω
vd x = v0. The uniqueness of the solution follows by a standard proof using estimates similar to the estimates used in the proof of inequality (2.12). Therefore, there exists a unique classical solution u = (‖∇ f ‖0
a0
)v of
∇ · (a(u)∇u) = f which satisfies the condition that 1
|Ω|
∫
Ω
ud x = u0. This completes the proof of the theorem.
Conclusion
We have proven that if 1(min u∗∈G0 a(u∗)) 4
∣∣∣∣da du
∣∣∣∣20,G0
‖∇ f ‖20 ≤ C1
and if
∣∣∣∣d2adu 2
∣∣∣∣0,G0
≤ 1(min u∗∈G0 a(u∗))
∣∣∣∣da du
∣∣∣∣20,G0
where G0 ⊂ R is a closed, bounded interval and where the constant C1 depends on N, Ω, then there exists a unique solution u ∈ C2(Ω) to the equation
∇ · (a(u)∇u) = f
which satisfies the condition that the average value 1
|Ω|
∫
Ω
ud x = u0
where u0 is a given constant, under periodic boundary conditions. We remark that in the trivial case in which ∇ f = 0 (and therefore f = 0), it follows that u = u0 is the unique solution.
Conflict of interest
The author confirms that there is no conflict of interest.
References
D. Denny, Existence of a unique solution to a quasilinear elliptic equation, J. Math. Anal. Appl. ,
380 (2011), 653–668. 2. P. Embid, On the Reactive and Non-di ff usive Equations for Zero Mach Number Flow, Commun. Part. Di ff er. Eq. , 14 (1989), 1249–1281. 3. L. Evans, Partial Di ff erential Equations , Graduate Studies in Mathematics 19 , American Mathematical Society, Providence, Rhode Island, 1998. 4. D. Gilbarg, N. S. Trudinger, Elliptic Partial Di ff erential Equations of Second Order , Springer-Verlag, Berlin, 1983.
AIMS Mathematics Volume 6, Issue 1, 518–531. 524
S. Klainerman, A. Majda, Singular limits of quasilinear hyperbolic systems with large parameters and the incompressible limit of compressible fluids, Comm. Pure Appl. Math. , 34 (1981), 481–524. 6. A. Majda, Compressible Fluid Flow and Systems of Conservation Laws in Several Space Variables ,Springer-Verlag: New York, 1984. 7. J. Moser, A rapidly convergent iteration method and non-linear di ff erential equations, Ann. Scuola Norm. Sup., Pisa , 20 (1966), 265–315.
A. Existence for the linear equation
In this appendix, we present the proof of the existence of a unique, classical solution to the linear problem (2.7), (2.8).
Lemma A.1. Let b be a smooth positive function of w. Let w ∈ C2(Ω), let g ∈ H2(Ω), and let
1
|Ω|
∫
Ω
gd x = 0. Let the domain Ω = TN , the N-dimensional torus, where N = 2 or N = 3. Then there exists a unique solution v ∈ C2(Ω) of the equation
∇ · (b(w)∇v) = g (A.1)
which satisfies the condition
1
|Ω|
∫
Ω
vd x = v0 (A.2)
where v 0 is a given constant. Proof.
We define the zero-mean function
v = v − 1
|Ω|
∫
Ω
vd x (A.3) The existence of a unique zero-mean solution v ∈ C2(Ω) to equation (A.1) under periodic boundary conditions is a well-known result from the standard theory of elliptic equations (see, e.g., Embid , Evans , Gilbarg and Trudinger ). It follows that the function v defined by
v(x) = v(x) + v0 (A.4) is the unique solution to equation (A.1) which satisfies the condition (A.2) that 1
|Ω|
∫
Ω
vd x = v0.This completes the proof of the lemma.
B. A priori estimates
In this appendix, we present lemmas supporting the proof of the theorem. We begin by listing several standard Sobolev space inequalities.
AIMS Mathematics Volume 6, Issue 1, 518–531. 525
Lemma B.1. (Standard Sobolev Space Inequalities)
(a) Let b be a smooth function of w, and let w (x) be a continuous function such that w (x) ∈ G1 for
x ∈ Ω where G 1 ⊂ R is a closed, bounded interval. And let w ∈ Hr+1(Ω) where r ≥ 0.Then
‖D(b(w)) ‖2
r
≤ C
∣∣∣∣ db dw
∣∣∣∣2
r,¯G1
(1 + |w|L∞ )2r‖∇ w‖2
r
(B.1)
where | db dw |r,G1 = max {
∣∣∣∣ d j+1bdw j+1 (w∗)
∣∣∣∣ : w∗ ∈ G1, 0 ≤ j ≤ r}. And the constant C depends on r, N, Ω.(b) If f ∈ Hn(Ω), where Ω ⊂ RN , and r = βm + (1 − β)n, with 0 ≤ β ≤ 1 and m < n, then
‖ f ‖r ≤ C‖ f ‖β
m
‖ f ‖1−β
n
(B.2)
Here C is a constant which depends on m, n, N, Ω.(c) If f ∈ Hs0 (Ω) where Ω ⊂ RN , N = 2 or N = 3, and s 0 = [ N
2
] + 1, then
| f |L∞ ≤ C‖ f ‖s0 (B.3)
Here C is a constant which depends on N, Ω.(d) If D f ∈ Hr1 (Ω), g ∈ Hr−1(Ω), where r ≥ 1 and where r 1 = max {r − 1, s0} and s 0 = [ N
2
] + 1, then:
‖Dα( f g ) − f D αg‖0 ≤ C‖D f ‖r1 ‖g‖r−1, (B.4)
where |α| = r and where the constant C depends on r, N, Ω.
These inequalities are well-known. Proofs may be found, for example, in , . These inequalities also appear in , .
Lemma B.2. Let the function w ∈ C2(Ω) satisfy (2.9), (2.11) in Proposition 1 and let the hypotheses in the statement of Theorem 2.1 hold. Let b be a smooth, positive function of w. Let g ∈ H2(Ω) and let
1
|Ω|
∫
Ω
gd x = 0. Let (2.4) define the functions b, g. Let the domain Ω = TN , the N-dimensional torus, where N = 2 or N = 3.Let v be the solution from Lemma A.1 in Appendix A of
∇ · (b(w)∇v) = g (B.5)
which satisfies the condition 1
|Ω|
∫
Ω
vd x = v0, (B.6)
where v 0 is a given constant. Then ∇v and v satisfy the following inequalities:
‖∇ v‖22 ≤ C2
‖v‖24 ≤ C3
where the constant C 2 depends on N and Ω and where the constant C 3 depends on R, u 0, a 0, ‖∇ f ‖0,
‖∇ f ‖1,
∣∣∣∣da du
∣∣∣∣2,G0
, N, and Ω.
AIMS Mathematics Volume 6, Issue 1, 518–531. 526
Proof.
In the estimates below, we will let C denote a generic constant whose value may change from one relation to the next. We use the notation ( h1, h2) = ∫
Ω
h1h2dx for the L2 inner product of two functions
h1, h2. And we use the notation hα = Dαh for di ff erentiation with a multi-index α.
Estimate for ‖∇ v‖20:
Using integration by parts and then substituting equation (B.5) yields
‖∇ v‖20 = (∇v, ∇v)
≤ 1
b0
(b(w)∇v, ∇v)
= − 1
b0
(∇ · (b(w)∇v), v − 1
|Ω|
∫
Ω
vd x)
= − 1
b0
(g, v − 1
|Ω|
∫
Ω
vd x)
≤ 1
b0
‖g‖0‖v − 1
|Ω|
∫
Ω
vd x‖0
≤ Cb0
‖∇ g‖0‖∇ v‖0
= C‖∇ v‖0 (B.7) where b0 = min w∗∈G1 b(w∗) = 1 by definition of the function b, and ‖∇ g‖0 = 1 by definition of the function g. Here we used the fact that g and v − 1
|Ω|
∫
Ω
vd x are zero-mean functions and we used Poincar´ e’s inequality for a zero-mean function h, namely ‖h‖0 ≤ C‖∇ h‖0. The constant C depends on
N, Ω.It follows that
‖∇ v‖20 ≤ ˜C (B.8) where the generic constant ˜ C depends on N, Ω.
Estimate for ‖∇ v‖21: To begin, let |α| ≥ 1. Using integration by parts and then substituting equation (B.5) yields
‖∇ vα‖20 = (∇vα, ∇vα)
≤ 1
b0
(b(w)∇vα, ∇vα)
= 1
b0
(( b(w)∇v)α, ∇vα) − 1
b0
((b(w)∇v)α − b(w)∇vα, ∇vα
)
= − 1
b0
(∇ · (b(w)∇v)α, vα) − 1
b0
((b(w)∇v)α − b(w)∇vα, ∇vα
)
= − 1
b0
(gα, vα) − 1
b0
((b(w)∇v)α − b(w)∇vα, ∇vα
) (B.9) where b0 = min w∗∈G1 b(w∗). If |α| = 1 in (B.9) then
‖∇ vα‖20 ≤ − 1
b0
(gα, vα) − 1
b0
((b(w)∇v)α − b(w)∇vα, ∇vα
)
AIMS Mathematics Volume 6, Issue 1, 518–531.
527
= − 1
b0
(gα, vα) − 1
b0
(b(w)α∇v, ∇vα)
≤ ( 1
b0
)‖gα‖0‖vα‖0 + ( 1
b0
)‖b(w)α∇v‖0‖∇ vα‖0
≤ 12
( 1
b0
)2‖gα‖20 + 12‖vα‖20 + ( 1
b0
)‖ db dw wα∇v‖0‖∇ vα‖0
≤ 12
( 1
b0
)2‖gα‖20 + 12‖vα‖20
12
( 1
b0
)2∣∣∣∣ db dw
∣∣∣∣20,G1
‖wα∇v‖20 + 12‖∇ vα‖20 (B.10) Re-arranging the terms in (B.10) and adding the resulting inequality over |α| = 1 yields
∑
|α|=1
‖∇ vα‖20 ≤ ( 1
b0
)2‖∇ g‖20 + ‖∇ v‖20 + ( 1
b0
)2∣∣∣∣ db dw
∣∣∣∣20,G1
|∇ w|2
L∞
‖∇ v‖20
≤ 1 + ˜C + C
∣∣∣∣ db dw
∣∣∣∣20,G1
‖∇ w‖22 ˜C
≤ 1 + ˜C + CC 1C2 ˜C (B.11) where the generic constants C, ˜ C depend on N, Ω. Here we used the facts that 1
b0
= 1, ‖∇ g‖0 = 1, and
‖∇ w‖22 ≤ C2. And ‖∇ v‖20 ≤ ˜C from (B.8). And we used the fact that
∣∣∣∣ db dw
∣∣∣∣20,G1
= (‖∇ f ‖20
a40
)∣ ∣∣∣da du
∣∣∣∣20,G0
≤ C1 by definition of the function b in (2.4) and by the statement of Theorem 2.1. From (B.8), (B.11) it follows that
‖∇ v‖21 =
∑
0≤| α|≤ 1
‖∇ vα‖20 = ‖∇ v‖20 +
∑
|α|=1
‖∇ vα‖20
≤ 1 + 2 ˜ C + CC 1C2 ˜C (B.12)
Estimate for ‖∇ v‖22:
Letting |α| = 2 in inequality (B.9) and then using integration by parts with |γ| = 1 produces
‖∇ vα‖20 ≤ − 1
b0
(gα, vα) − 1
b0
((b(w)∇v)α − b(w)∇vα, ∇vα
)
= ( 1
b0
)(gα−γ, vα+γ) − 1
b0
(b(w)α∇v, ∇vα
)
− 1
b0
(b(w)γ∇vα−γ, ∇vα
) − 1
b0
(b(w)α−γ∇vγ, ∇vα
)
= ( 1
b0
)(gα−γ, vα+γ) − 1
b0
(( d2bdw 2 wα−γwγ)∇v, ∇vα
) − 1
b0
(( db dw wα)∇v, ∇vα
)
− 1
b0
( db dw wγ∇vα−γ, ∇vα) − 1
b0
( db dw wα−γ∇vγ, ∇vα)
≤ ( 1
b0
)‖gα−γ‖0‖vα+γ‖0 + ( 1
b0
)∣∣∣∣ d2bdw 2
∣∣∣∣0,G1
|wα−γ|L∞ |wγ|L∞ ‖∇ v‖0‖∇ vα‖0
AIMS Mathematics Volume 6, Issue 1, 518–531. 528
( 1
b0
)∣∣∣∣ db dw
∣∣∣∣0,G1
‖wα‖0|∇ v|L∞ ‖∇ vα‖0
( 1
b0
)∣∣∣∣ db dw
∣∣∣∣0,G1
|wγ|L∞ ‖∇ vα−γ‖0‖∇ vα‖0
( 1
b0
)∣∣∣∣ db dw
∣∣∣∣0,G1
|wα−γ|L∞ ‖∇ vγ‖0‖∇ vα‖0
≤ C
( 1
b0
)2‖gα−γ‖20 + ‖∇ vα‖20
14
( 1
b0
)2∣∣∣∣ d2bdw 2
∣∣∣∣20,G1
|wα−γ|2
L∞
|wγ|2
L∞
‖∇ v‖20 + ‖∇ vα‖20
14
( 1
b0
)2∣∣∣∣ db dw
∣∣∣∣20,G1
‖wα‖20|∇ v|2
L∞
‖∇ vα‖20
14
( 1
b0
)2∣∣∣∣ db dw
∣∣∣∣20,G1
|wγ|2
L∞
‖∇ vα−γ‖20 + ‖∇ vα‖20
14
( 1
b0
)2∣∣∣∣ db dw
∣∣∣∣20,G1
|wα−γ|2
L∞
‖∇ vγ‖20 + ‖∇ vα‖20
≤ C
( 1
b0
)2‖gα−γ‖20 + C
( 1
b0
)2∣∣∣∣ d2bdw 2
∣∣∣∣20,G1
‖wα−γ‖22‖wγ‖22‖∇ v‖20
C
( 1
b0
)2∣∣∣∣ db dw
∣∣∣∣20,G1
‖Dγwα−γ‖20‖∇ v‖22 + C
( 1
b0
)2∣∣∣∣ db dw
∣∣∣∣20,G1
‖wγ‖22‖∇ vα−γ‖20
C
( 1
b0
)2∣∣∣∣ db dw
∣∣∣∣20,G1
‖wα−γ‖22‖∇ vγ‖20 + 5‖∇ vα‖20
(B.13) where we used Cauchy’s inequality with and we define = 110 . We also used Sobolev’s Lemma, i.e.,
|h|L∞ ≤ C‖h‖2.Re-arranging terms in (B.13), and then adding the resulting inequality over |α| = 2 and |γ| = 1, produces
∑
|α|=2
‖∇ vα‖20 ≤ C( 1
b0
)2‖∇ g‖20 + C( 1
b0
)2∣∣∣∣ d2bdw 2
∣∣∣∣20,G1
‖∇ w‖42‖∇ v‖20
C( 1
b0
)2∣∣∣∣ db dw
∣∣∣∣20,G1
‖∇ w‖21‖∇ v‖22 + C( 1
b0
)2∣∣∣∣ db dw
∣∣∣∣20,G1
‖∇ w‖22‖∇ v‖21
≤ C( 1
b0
)2‖∇ g‖20 + C( 1
b0
)2∣∣∣∣ d2bdw 2
∣∣∣∣20,G1
‖∇ w‖42‖∇ v‖22
C( 1
b0
)2∣∣∣∣ db dw
∣∣∣∣20,G1
‖∇ w‖22‖∇ v‖22
≤ C + (CC 21C22 + CC 1C2)‖∇ v‖22
≤ C + CC 1C22‖∇ v‖22 (B.14) where we can assume that C1 < 1 and that C2 > 1. Here we used the fact that ‖∇ w‖22 ≤ C2. And we used the fact that
∣∣∣∣ db dw
∣∣∣∣20,G1
=(‖∇ f ‖20
a40
)∣ ∣∣∣da du
∣∣∣∣20,G0
≤ C1. And we used the fact that
∣∣∣∣ d2bdw 2
∣∣∣∣20,G1
= (‖∇ f ‖40
a60
)∣ ∣∣∣d2adu 2
∣∣∣∣20,G0
AIMS Mathematics Volume 6, Issue 1, 518–531. 529
≤ (‖∇ f ‖40
a80
)∣ ∣∣∣da du
∣∣∣∣40,G0
≤ C21 by the definition of b(v) in (2.4) and by the statement of the theorem. And we used the facts that ‖∇ g‖0 = 1 and that 1
b0
= 1. From (B.14) and from inequality (B.12) for ‖∇ v‖21, it follows that
‖∇ v‖22 =
∑
0≤| α|≤ 2
‖∇ vα‖20
= ‖∇ v‖21 +
∑
|α|=2
‖∇ vα‖20
≤ 1 + 2 ˜ C + CC 1C2 ˜C + C + CC 1C22‖∇ v‖22
≤ 1 + 2 ˜ C + 12 + C + 12‖∇ v‖22 (B.15) where the generic constants C, ˜ C depend on N, Ω, and where C1 is su ffi ciently small so that CC 1C2 ˜C ≤
12
and so that CC 1C22 ≤ 12 .Re-arranging terms in (B.15) yields
‖∇ v‖22 ≤ 4 ˜ C + C
= C2 (B.16) where we define C2 = 4 ˜ C + C, and where the constant C2 depends on N, Ω.
Estimate for ‖∇ v‖23:
Letting |α| = 3 in inequality (B.9) and then using integration by parts with |γ| = 1 produces
‖∇ vα‖20 ≤ − 1
b0
(gα, vα) − 1
b0
((( b(w)∇v)α − b(w)∇vα), ∇vα)
= 1
b0
(gα−γ, vα+γ) − 1
b0
((( b(w)∇v)α − b(w)∇vα), ∇vα)
≤ ( 1
b0
)‖gα−γ‖0‖vα+γ‖0 + ( 1
b0
)‖(b(w)∇v)α − b(w)∇vα‖0‖∇ vα‖0
≤ ( 1
b0
)‖gα−γ‖0‖vα+γ‖0 + C( 1
b0
)‖Db ‖2‖∇ v‖2‖∇ vα‖0
≤ C
( 1
b0
)2‖gα−γ‖20 + ‖∇ vα‖20 + C
( 1
b0
)2‖Db ‖22‖∇ v‖22
‖∇ vα‖20 (B.17) where = 14 and where we used the Sobolev space inequality (B.4) from Lemma B.1 with r = |α| = 3and r1 = 2. Re-arranging the terms in (B.17) and then adding the resulting inequality over |α| = 3 and |γ| = 1yields
∑
|α|=3
‖∇ vα‖20 ≤ C( 1
b0
)2‖g‖22 + C( 1
b0
)2‖Db ‖22‖∇ v‖22
AIMS Mathematics Volume 6, Issue 1, 518–531. 530
≤ C( 1
b0
)2‖g‖22 + C( 1
b0
)2∣∣∣∣ db dw
∣∣∣∣22,G1
(1 + |w|L∞ )4‖∇ w‖22‖∇ v‖22
= C( 1
b0
)2‖g‖22 + C( 1
b0
)2∣∣∣∣ db dw
∣∣∣∣22,G1
(1 + |w − v0 + v0|L∞ )4‖∇ w‖22‖∇ v‖22
≤ C( 1
b0
)2‖g‖22 + C( 1
b0
)2∣∣∣∣ db dw
∣∣∣∣22,G1
(1 + |w − v0|L∞ + |v0|)4‖∇ w‖22‖∇ v‖22
≤ C‖∇ g‖21 + C
∣∣∣∣ db dw
∣∣∣∣22,G1
(1 + R + |v0|)4C2‖∇ v‖22
(B.18) where we used the Sobolev space inequality (B.1) from Lemma B.1 for ‖Db ‖22. We also used the facts that |w − v0|L∞ ≤ R, ‖∇ w‖22 ≤ C2, and b0 = 1. And we used the fact that g is a zero-mean function, so that ‖g‖0 ≤ C‖∇ g‖0 by Poincar´ e’s inequality. From (B.18) and from inequality (B.16) for ‖∇ v‖22, it follows that
‖∇ v‖23 =
∑
0≤| α|≤ 3
‖∇ vα‖20
=
∑
0≤| α|≤ 2
‖∇ vα‖20 +
∑
|α|=3
‖∇ vα‖20
= ‖∇ v‖22 +
∑
|α|=3
‖∇ vα‖20
≤ C2 + C‖∇ g‖21
C
∣∣∣∣ db dw
∣∣∣∣22,G1
(1 + R + |v0|)4C22 (B.19)
Estimate for ‖v‖24:
From (B.19) it follows that
‖v‖24 = ‖v − v0 + v0‖24
≤ C‖v − v0‖24 + C‖v0‖24
≤ C‖∇ (v − v0)‖23 + C|v0|2|Ω|
= C‖∇ v‖23 + C|v0|2|Ω|≤ CC 2 + C‖∇ g‖21 + C
∣∣∣∣ db dw
∣∣∣∣22,G1
(1 + R + |v0|)4C22
C|v0|2|Ω|
= CC 2 + C ‖∇ f ‖21
‖∇ f ‖20
C(‖∇ f ‖20
a40
)∣ ∣∣∣da du
∣∣∣∣22,G0
(1 + R + ( a0
‖∇ f ‖0
)|u0|)4C22
C( a0
‖∇ f ‖0
)2
|u0|2|Ω|
= C3 (B.20)
AIMS Mathematics Volume 6, Issue 1, 518–531. 531
Here we used Poincar´ e’s inequality for the zero-mean function v−v0, where the constant v0 = 1
|Ω|
∫
Ω
vd x.And we used the definition of the functions v, b, g from (2.4). This completes the proof of the lemma.
c© 2021 the Author(s), licensee AIMS Press. This is an open access article distributed under the terms of the Creative Commons Attribution License (http: // creativecommons.org /licenses /by /4.0) |
8501 | https://seer.cancer.gov/statistics-network/explorer/application.html?site=600&data_type=4&graph_type=5&compareBy=stage&chk_stage_104=104&chk_stage_105=105&chk_stage_106=106&chk_stage_107=107&series=9&sex=1&race=1&age_range=1&advopt_precision=1&advopt_show_ci=on&advopt_display=2 | SEERExplorer Application
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[x] All Cancer Sites Combined - [x] Anus, Anal Canal & Anorectum - [x] Bones and Joints - [x] Brain and Other Nervous System
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[x] Diffuse Astrocytoma and Anaplastic Astrocytoma - [x] Glioblastoma - [x] Other Glioma - [x] Embryonal Tumors - [x] Meningioma - [x] Other Tumors of Brain and ONS
[x] Non-malignant Brain and Other Nervous System
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[x] Non-malignant Meningioma - [x] Non-malignant Tumors of the Cranial and Paraspinal Nerves
[x] Female Breast 1
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[x] HR+/HER2- (Luminal A) 1 - [x] HR-/HER2- (Triple Negative) 1 - [x] HR+/HER2+ (Luminal B) 1 - [x] HR-/HER2+ (HER2-enriched) 1
[x] Cervix Uteri 1 - [x] Chronic myeloproliferative disorders (CMD) - [x] Colon and Rectum (including Appendix)
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[x] Colon (including Appendix) - [x] Rectum
[x] Colon and Rectum (excluding Appendix) - [x] Corpus and Uterus, NOS 1
[x] Esophagus
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[x] Adenocarcinoma - [x] Squamous Cell Carcinoma
[x] Eye and Orbit - [x] Gallbladder - [x] Hodgkin Lymphoma - [x] Kaposi Sarcoma - [x] Kidney and Renal Pelvis
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[x] Kidney - [x] Renal Pelvis
[x] Larynx - [x] Leukemia
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[x] Acute Lymphocytic Leukemia (ALL) - [x] Chronic Lymphocytic Leukemia (CLL) - [x] Acute Myeloid Leukemia (AML) - [x] Chronic Myeloid Leukemia (CML) - [x] Acute Monocytic Leukemia (AML-M5) - [x] Chronic Myelomonocytic Leukemia (CMML)
[x] Liver and Intrahepatic Bile Duct - [x] Lung and Bronchus
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[x] Adenocarcinoma - [x] Large Cell Carcinoma - [x] Small cell carcinoma - [x] Squamous Cell Carcinoma
[x] Melanoma of the Skin - [x] Mesothelioma - [x] Myelodysplastic syndromes (MDS)
[x] Myeloma - [x] Non-Hodgkin Lymphoma - [x] Oral Cavity and Pharynx
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[x] Lip - [x] Tongue - [x] Salivary Gland - [x] Floor of Mouth - [x] Gum & other mouth - [x] Oropharynx & Tonsil
[x] Ovary 1 - [x] Pancreas - [x] Prostate 2 - [x] Small Intestine - [x] Soft Tissue including Heart - [x] Stomach - [x] Testis 2 - [x] Thyroid
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1 Female only.
2 Male only.
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Adenocarcinoma of the Esophagus
SEER 5-Year Relative Survival Rates, 2015-2021
By Stage at Diagnosis, Both Sexes, All Races / Ethnicities, All Ages
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| | Stage at Diagnosis | 5-Year Relative Survival (%) | Lower 95% C.I. | Upper 95% C.I. |
--- --- |
| Stage at Diagnosis | 5-Year Relative Survival (%) | Lower 95% C.I. | Upper 95% C.I. |
| Localized | 53.5 | 51.5 | 55.5 |
| Regional | 29.3 | 28.0 | 30.7 |
| Distant | 5.0 | 4.4 | 5.6 |
| Unstaged | 15.9 | 13.9 | 18.1 |
| |
Recent 5 and 10 Year Trends
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Data Source
SEER Incidence Data, November 2024 Submission (1975-2022), SEER 21 registries (excluding Illinois).
Expected Survival Life Tables by Socio-Economic Standards.
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The five-year survival rates are calculated using monthly intervals.
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For more details on SEER race/ethnicity groupings, please see Race and Hispanic Ethnicity Changes.
Incidence data for Hispanics and Non-Hispanics are based on the NAACCR Hispanic Latino Identification Algorithm (NHIA).
Rates for American Indians/Alaska Natives only include cases that are in a Purchased/Referred Care Delivery Area (PRCDA).
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See SEERExplorer Cancer Site Definitions for details about the cancer site coding used for SEER Incidence data.
Stage at diagnosis is calculated using the Combined Summary Stage (2004+) recode, created from SEER Combined Summary Stage 2000 (2004-2017) & Derived Summary Stage 2018 (2018+). For more information, see the SEER documentation.
Suggested Citation
SEERExplorer: An interactive website for SEER cancer statistics [Internet]. Surveillance Research Program, National Cancer Institute; 2025 Jul 2. [cited 2025 Jul 14]. Available from: Data source(s): SEER Incidence Data, November 2024 Submission (1975-2022), SEER 21 registries (excluding Illinois). Expected Survival Life Tables by Socio-Economic Standards.
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8502 | https://www.groundedinthebible.com/p/god-sovereign | Characteristics of God
9 Ways That God is Sovereign
Characteristics of God
Karen Hoffman
Aug 29, 2021
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What comes to mind when you think of the word “sovereign”? Do you think of the rulers of the world—kings and queens, presidents, prime ministers? Or maybe heads of companies—CEOs, presidents, founders, chairmen? Whoever these leaders are, they usually are in charge of making the most important decisions, and they have a pretty good idea about what goes on in most parts of their nation or company.
And then there’s God. He’s in charge of all of creation. That’s a lot more than just a nation or a company! And he not only has a “pretty good idea” about what’s going on, he knows exactly what is going on in every corner of the earth and beyond. He doesn’t just have a 5-year plan, he has a plan for all eternity.
So what characteristics does God need to be sovereign over all creation? What characteristics support his ability to rule over everything? Let’s briefly look at a few. Links are provided for each verse if you would like to see more context for the verse.
God is Sovereign. God owns the world and everything in it. He controls the seasons, the rulers, and all life events.
Psalm 103:19 – The Lord has established his throne in heaven, and his kingdom rules over all.
Deuteronomy 10:14 – To the Lord your God belong the heavens, even the highest heavens, the earth and everything in it.
God is Creator. He made everything that exists, from the smallest bacteria to the biggest whale in the ocean. From the smallest flower to the largest star in the sky.
Colossians 1:16-17 – For in him all things were created: things in heaven and on earth, visible and invisible, whether thrones or powers or rulers or authorities; all things have been created through him and for him. He is before all things, and in him all things hold together.
Nehemiah 9:6 – You alone are the Lord. You made the heavens, even the highest heavens, and all their starry host, the earth and all that is on it, the seas and all that is in them. You give life to everything, and the multitudes of heaven worship you.
God is the Sustainer of Life. As creator, God holds the power of life and death for every living thing, from plants and animals to humans.
Deuteronomy 32:39 – “See now that I myself am he! There is no god besides me. I put to death and I bring to life, I have wounded and I will heal, and no one can deliver out of my hand.”
Job 14:5 – A person’s days are determined; you have decreed the number of his months and have set limits he cannot exceed.
God is Eternal. Even though God created everything, God himself is not a created being. He has no beginning and no end.
Psalm 90:2 – Before the mountains were born or you brought forth the whole world, from everlasting to everlasting you are God.
Revelation 22:13 – I am the Alpha and the Omega, the First and the Last, the Beginning and the End.
God is Omnipotent. God is all-powerful. Sometimes he is even called God Almighty. He can do anything. He can perform miracles, which shows his power to change creation to his will.
Matthew 19:26 – Jesus looked at them and said, “With man this is impossible, but with God all things are possible.”
Hebrews 1:3 – The Son is the radiance of God’s glory and the exact representation of his being, sustaining all things by his powerful word. After he had provided purification for sins, he sat down at the right hand of the Majesty in heaven.
God is Omniscient. God knows everything. There’s nothing in the past, present, or future that is unknown to God.
Psalm 147:4-5 – He determines the number of the stars and calls them each by name. Great is our Lord and mighty in power; his understanding has no limit.
Hebrews 4:13 – Nothing in all creation is hidden from God’s sight. Everything is uncovered and laid bare before the eyes of him to whom we must give account.
God is Omnipresent. God is everywhere, and he sees everything. There’s nowhere on this earth you can go to escape God.
Jeremiah 23:23-24 – “Am I only a God nearby,” declares the Lord, “and not a God far away? Who can hide in secret places so that I cannot see them?” declares the Lord. “Do not I fill heaven and earth?” declares the Lord.
Psalm 139:7-10 – Where can I go from your Spirit? Where can I flee from your presence? If I go up to the heavens, you are there; if I make my bed in the depths, you are there. If I rise on the wings of the dawn, if I settle on the far side of the sea, even there your hand will guide me, your right hand will hold me fast.
God is Unchanging. God is the same yesterday, today, and forever.
Malachi 3:6a – I the Lord do not change.
1 Samuel 15:29 – He who is the Glory of Israel does not lie or change his mind; for he is not a human being, that he should change his mind.
God has No Equal. No human or other living creature can ever have the status that God has as sovereign over everything. No other god will ever be as great as the one true God.
Psalm 40:5 – Many, Lord my God, are the wonders you have done, the things you planned for us. None can compare with you; were I to speak and tell of your deeds, they would be too many to declare.
Isaiah 45:5-6 – I am the Lord, and there is no other; apart from me there is no God. I will strengthen you, though you have not acknowledged me, so that from the rising of the sun to the place of its setting people may know there is none besides me. I am the Lord, and there is no other.
Pretty amazing, right? And we’re just getting started! The verses listed here are just a small taste of what the Bible says to support these characteristics of God. We’ll cover each of these characteristics in more depth over time.
The verses listed here are just a small taste of what the Bible says to support these characteristics of God. If you’d like to study more about God’s sovereignty over the next month, I’d encourage you to download and print the premium resource that goes along with this study. It is a “Bible Bingo” with 25 Scripture passages that provide further evidence for the characteristics of God we discussed above.
Click here to download the God is Sovereign Bible bingo card: Original / Printer-friendly
Click here to see all premium resources for this series.
Are there any other characteristics of God that support his sovereignty over all? I’d love to hear your thoughts in the comments! It might even turn into a Bible study one day.
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Frances Hoffman
Sep 27, 2021
Read this study today. It is great to have all these verses together. God is so amazing! Thanks for sharing.
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8503 | https://bestpractice.bmj.com/topics/en-us/398?locale=fa | Skull fractures - Symptoms, diagnosis and treatment | BMJ Best Practice US
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Skull fractures
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Last reviewed: 28 Aug 2025
Last updated: 18 Jun 2025
Summary
Skull fractures commonly result from a fall, a traffic accident, or an assault.
Skull fractures may be linear or comminuted with multiple fracture lines, may be located on the cranial vault or in the basilar skull, may have a varying degree of depression or elevation, and can be open or closed. Open fractures communicate with the skin through a wound, a sinus, the ear, or the oropharynx.
Computed tomography, with thin axial cuts, remains the imaging modality of choice. With basilar skull fractures, 3D reconstructions are useful.
May be associated with other significant injuries, most importantly intracranial hemorrhage.
For isolated skull fractures, treatment is primarily conservative.
Surgical intervention is determined not by the fracture per se but by the extent of the associated intracranial pathology, cranial nerve deficit, or cerebrospinal fluid leak.
Definition
Skull fracture refers to a fracture of one or more bones of the cranial vault or skull base. Skull fractures are categorized according to the appearance, location, degree of depression, and if they are open or closed. Open fractures communicate with the skin through a wound, a sinus, the ear, or the oropharynx. Skull fractures may be linear or comminuted; comminuted fractures are complex with multiple fracture lines.[Figure caption and citation for the preceding image starts]:Linear parietal fracture without depressionFrom the teaching collection of Demetrios Demetriades; used with permission [Citation ends].[Figure caption and citation for the preceding image starts]:Axial CT scan demonstrating an open nondepressed linear skull fracture (arrow) associated with pneumocephalus (circle)From the teaching collection of Demetrios Demetriades; used with permission [Citation ends].[Figure caption and citation for the preceding image starts]:Axial CT scan showing nondepressed linear skull fracture (arrow) of the skull base involving the foramen magnum. This injury pattern is concerning for associated spinal fracture, cord injury, and blunt cerebrovascular injuryFrom the teaching collection of Demetrios Demetriades; used with permission [Citation ends].[Figure caption and citation for the preceding image starts]:Comminuted nondepressed fractureFrom the teaching collection of Demetrios Demetriades; used with permission [Citation ends].[Figure caption and citation for the preceding image starts]:Comminuted depressed skull fracture with pneumocephalusFrom the teaching collection of Demetrios Demetriades; used with permission [Citation ends].[Figure caption and citation for the preceding image starts]:Comminuted depressed fracture of the frontal sinus with air, fluid, and bone fragments in frontal sinus and pneumocephalus; level of depression greater than width of cortexFrom the teaching collection of Demetrios Demetriades; used with permission [Citation ends].[Figure caption and citation for the preceding image starts]:Depressed skull fracture: level of depression equal to thickness of cortexFrom the teaching collection of Demetrios Demetriades; used with permission [Citation ends].[Figure caption and citation for the preceding image starts]:Axial CT scan demonstrating open elevated linear skull fracture (large arrow). Note the air in the soft tissues (small arrow), the small amount of pneumocephalus associated with the fracture (circle), and that the level of elevation of the bone fragment is significantly more than the thickness of the bony tableFrom the teaching collection of Demetrios Demetriades; used with permission [Citation ends].[Figure caption and citation for the preceding image starts]:Sagittal CT images of an open, comminuted, depressed skull fracture. Note the associated pneumocephalus (small arrows). The level of depression is greater than the bony table and there are several bone fragments impacted below the inner cortex of the opposing bone (large arrow). Despite lack of underlying associated brain injury this fracture required operative debridement and elevation of the bone fragments. See also the corresponding coronal CT imageFrom the teaching collection of Demetrios Demetriades; used with permission [Citation ends].[Figure caption and citation for the preceding image starts]:Coronal CT of an open, comminuted, depressed skull fracture. The level of depression is greater than the bony table and there are several bone fragments impacted below the inner cortex of the opposing bone (large arrow). Despite lack of underlying associated brain injury this fracture required operative debridement and elevation of the bone fragments. See also the corresponding sagittal CT imageFrom the teaching collection of Demetrios Demetriades; used with permission [Citation ends].
History and exam
Key diagnostic factors
open fracture
palpable discrepancy in bone contour
Battle sign
periorbital ecchymosis
bloody otorrhea
cerebrospinal fluid rhinorrhea
facial paralysis, nystagmus, or paresthesia
Full details
Other diagnostic factors
evidence of trauma
cranial pain or headache
nausea
altered mental state/loss of consciousness
abnormal pupillary reflexes
hearing loss
Full details
Risk factors
fall from height
motor vehicle accident
assault resulting in head trauma
gunshots to the head
male sex
Full details
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Diagnostic tests
1st tests to order
cranial CT
Full details
Tests to consider
beta-2 transferrin assay
MRI brain
MR angiography
cranial ultrasound
plain skull x-ray
CT scan cervical spine
skeletal survey
CT angiogram
CT venogram
Full details
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Treatment algorithm
ACUTE
closed nondepressed fracture
closed depressed fracture
open fracture
ONGOING
persistent cranial nerve injury or CSF leakage
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Contributors
Expert advisers
VIEW ALL Expert advisers
Demetrios Demetriades, MD, PhD, FACS
Professor of Surgery
Director
Division of Trauma and Surgical Intensive Care
LAC+USC Trauma Center
Keck School of Medicine at USC
University of Southern California
Los Angeles
CA
Disclosures
DD declares that he has no competing interests.
Leslie Kobayashi, MD, FACS
Professor of Surgery
Division of Trauma, Surgical Critical Care and Burns
University of California San Diego
San Diego
CA
Disclosures
LK declares that she has no competing interests.
Peer reviewers
VIEW ALL Peer reviewers
Prof Sherard Austin Tatum, MD
Professor Otolaryngology and Pediatrics
SUNY Upstate Medical University
Syracuse
New York
Disclosures
SAT declares that he has no competing interests.
Tunji Lasoye, FRCS, FCEM, MA, Med Ed
Consultant and Honorary Senior Lecturer in Emergency Medicine
Clinical Lead
Emergency Department
Director of Medical Education
King's College Hospital
London
UK
Disclosures
TL declares that he has no competing interests.
Peer reviewer acknowledgements
BMJ Best Practice topics are updated on a rolling basis in line with developments in evidence and guidance. The peer reviewers listed here have reviewed the content at least once during the history of the topic.
Disclosures
Peer reviewer affiliations and disclosures pertain to the time of the review.
References
Our in-house evidence and editorial teams collaborate with international expert contributors and peer reviewers to ensure that we provide access to the most clinically relevant information possible.
Key articles
Expert Panel on Neurological Imaging: Shih RY, Burns J, et al. ACR Appropriateness Criteria® head trauma: 2021 update. J Am Coll Radiol. 2021 May;18(5S):S13-36.Full textAbstract
Reference articles
A full list of sources referenced in this topic is available to users with access to all of BMJ Best Practice.
Differentials
Intracranial hemorrhage
Suture lines in children
Cephalhematoma
More Differentials
Guidelines
ACR Appropriateness Criteria: head trauma
ACR Appropriateness Criteria: head trauma - child
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8504 | https://pubmed.ncbi.nlm.nih.gov/7634848/ | Pseudotumor cerebri after treatment with tetracycline and isotretinoin for acne - PubMed
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. 1995 Mar;55(3):165-8.
Pseudotumor cerebri after treatment with tetracycline and isotretinoin for acne
A G Lee1
Affiliations Expand
Affiliation
1 Department of Ophthalmology, Cullen Eye Institute, Baylor College of Medicine, Houston, Texas 77030, USA.
PMID: 7634848
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Pseudotumor cerebri after treatment with tetracycline and isotretinoin for acne
A G Lee. Cutis.1995 Mar.
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. 1995 Mar;55(3):165-8.
Author
A G Lee1
Affiliation
1 Department of Ophthalmology, Cullen Eye Institute, Baylor College of Medicine, Houston, Texas 77030, USA.
PMID: 7634848
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Tetracyclines and isotretinoin are widely used treatments for patients with acne. Although generally safe, the use of these agents has been associated with pseudotumor cerebri and combination therapy with these agents may increase the risk for pseudotumor cerebri. A 14-year-old boy presented with headaches and bilateral visual loss secondary to papilledema. He had been treated with tetracycline and isotretinoin for acne for three weeks prior to presentation and was subsequently diagnosed as having pseudotumor cerebri. He required long-term medical therapy and eventually underwent bilateral optic nerve sheath decompression. The literature regarding pseudotumor cerebri in association with tetracyclines and isotretinoin treatment for acne is reviewed. Dermatologists should be aware of the risk of pseudotumor cerebri in patients receiving tetracycline or isotretinoin treatment for acne and should be particularly cautious about using both agents simultaneously.
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Pseudotumor cerebri caused by isotretinoin.Roytman M, Frumkin A, Bohn TG.Roytman M, et al.Cutis. 1988 Nov;42(5):399-400.Cutis. 1988.PMID: 2973971
Pseudotumor cerebri--a complication of tetracycline treatment of acne.Pierog SH, Al-Salihi FL, Cinotti D.Pierog SH, et al.J Adolesc Health Care. 1986 Mar;7(2):139-40. doi: 10.1016/s0197-0070(86)80010-9.J Adolesc Health Care. 1986.PMID: 2937759
Safe Use of Therapeutic-Dose Oral Isotretinoin in Patients With a History of Pseudotumor Cerebri.Tintle SJ, Harper JC, Webster GF, Kim GK, Thiboutot DM.Tintle SJ, et al.JAMA Dermatol. 2016 May 1;152(5):582-4. doi: 10.1001/jamadermatol.2015.3447.JAMA Dermatol. 2016.PMID: 26580859 No abstract available.
Idiopathic intracranial hypertension associated with tetracycline use in fraternal twins: case reports and review.Gardner K, Cox T, Digre KB.Gardner K, et al.Neurology. 1995 Jan;45(1):6-10. doi: 10.1212/wnl.45.1.6.Neurology. 1995.PMID: 7824136 Review.
Tetracycline-, Doxycycline-, Minocycline-Induced Pseudotumor Cerebri and Esophageal Perforation.Angelette AL, Rando LL, Wadhwa RD, Barras AA, Delacroix BM, Talbot NC, Ahmadzadeh S, Shekoohi S, Cornett EM, Kaye AM, Kaye AD.Angelette AL, et al.Adv Ther. 2023 Apr;40(4):1366-1378. doi: 10.1007/s12325-023-02435-y. Epub 2023 Feb 10.Adv Ther. 2023.PMID: 36763302 Review.
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8505 | https://www.youtube.com/watch?v=k7oAl3AiV2I | Domain and Range in Interval Notation
Kyle Ritter
988 subscribers
1522 likes
Description
354083 views
Posted: 23 Nov 2012
Transcript:
Intro in this uh video we're going to talk about interval notation uh interval notation is a way that we can write um sets of numbers or ranges and domains in a certain uh format so that's called notation and there are lots of different notations and this video we'll just focus on interval notation so interval notation uses two types of symbols we have parenthesis or we have brackets parentheses deal with values that are open so if you think about your graph if you have an open circle on a graph which does not include that number then we're going to use parentheses when we talk about that number when we have a closed set or a closed value meaning that we have a filled in so Circle then we're going to use brackets when we talk about that value um so let's give it a shot here Graphing let me bring up a graph for this video I'm going to focus on quadratics and talk about domain range and interval notation so um I'm also going to use a specific form of quadratics called Vertex or standard form um so if I have a quadratic that looks like this then it's a lot easier to graph in this form a little bit faster so that's why I'm going to use it um we could expand this we could foil this and combine some like terms to put it into standard form but I'm not going to bother so I'll leave that up to you if you'd like uh in this form our vertex is really easy to find because it's given to us in the H and K form so if this is not familiar to you I would suggest watching some previous videos that I created on vertex form or standard form of a quadratic so our H value is always the opposite of what it looks like there so it's going to be positive2 our K value is going to be positive 3 so the vertex of our Parabola is going to be at 23 and then we can kind of uh plug in some values here just to get a couple more points on our graph if I plug in three 3 - 2 is 1 1 2 is 1 + 3 is 4 if I plug in one I should have a symmetry point right here but let's just double check plug in one I get 1 - 2 is netive 1 1^2 is 1 and 1 + 3 is 4 so that worked out now if I plug in four let's see what happens 4 - 2 is 2 2^ 2 is 4 and 4 + 3 is 7 seven and again I'm just going to use my definition of symmetry there and know that if I plug in zero I should get seven as well uh this is always a pattern for parabolas it's always up one over one to the right up one to the left one from this point it's always up three and over one up three over one um but only if this a value here is at positive one okay so we got our Parabola here let's talk about the domain domain values are our possible X values well domain for parabas is always going to be uh really no limit to the left this arm while going very slowly to the left will eventually go forever to the left so we know that uh it will always be negative Infinity to the left and likewise to the right this arm will always stretch eventually to positive Infinity to the right that's horrible try that one more time so there's our domain negative Infinity to positive infinity and so the way that we write that in interval notation specifically because we cannot contain Infinity whether it's negative or positive we're going to have to use that open symbol and so we'd say open parentheses negative Infinity to positive Infinity close parentheses I'm sorry open on the other side so we we have two open symbols we can't include or uh tie down or in some way capture Infinity so we always got to leave them open so there's our domain and interval notation that's all it is let's talk about the range uh we always start with the most negative value when we're talking about domain and range so for the most negative or less value for range would be the bottom or the minimum of the graph uh so we know that our least Y value is going to be three and we know it included three so it must be a closed three so we use our bracket three and then what's our limit on the top is there well no because these are going to Forever go up so it's going to be a positive infinity and we're going to use a parentheses always for Infinity so there's our range in interval notation let's do another one see if you got it here let's see what happens when a is negative one let's do x + 2^ 2 - 1 I guess all right so we know our vertex is going to be at the H value of -2 and K value of1 so -2 1 let's see if our pattern holds up um of going up one over one so if I plug in negative one here 1 + 2 is POS 1 POS 1 2 is POS 1 -1 oops I already made a mistake I know what I did I wasn't paying attention it's why the XY table always helps to check uh I know I'm going to be right here so again let's check sorry - 1 + 2 is 1 1 2 is 1 1 1 is 1 and- 1 - 1 is -2 all right so so if we use symmetry we know that's going to line up right there and then supposedly if we go down this time three and to the right one we should get another Point well let's check it out if I plug in zero should get5 out of it 0 + 2 is 2 2^ 2 is 4 4 -1 is -4 and4 - 1 is5 so that worked and we'll use symmetry to plot our other point and now the reason I made a mistake and why I should have been thinking it's because I made a negative here anytime a is negative we know we have an upside down Parabola and that's why I should have been counting the reverse way there all right so now we got our upside down Parabola we know our vertex is a maximum now so that's going to change our range slightly let's start with domain an interval notation again like we said all parabas are going to have a domain of negative Infinity to positive Infinity in and we use our parentheses range is the only thing that's different this time we're going to start with our lowest value well now that we have an upside down Parabola our lowest value is going to be parenthesis of negative Infinity can't contain Infinity but it will have a maximum will have a top range value and that top range value is ne1 and it includes negative 1 it's a closed value so use a bracket all right so that's basically it domain and range for parabas is somewhat simple we can do one more just to Check check make sure you got it try and do this one on your own as much as possible pause the video and then check your work when you're done um let's do uh x + 5^ 2ar + 3 all right see if you got it see if you can list the domain and range in interval notation I know that my vertex is going to be at5 positive3 and know I'm going to open up because a is positive and if I use my pattern I know that my Parabola should look like this you can double check your values if you want using your XY table so my domain again every single time for parabas negative Infinity to positive Infinity with parentheses range we're going to have a minimum this time so we know we're going to use a bracket at the minimum value the minimum value is three and it's going to go up for Infinity so there's our domain and range and interval notation hopefully you got it good luck on your other problems writing no domain and range and interval notation I'll continue with this notation and other uh functions um in the next unit we'll be dealing with uh root functions so you'll see this uh notation come up again to describe their domain and range as well as a brand brand new one in that unit as well called set builder notation so look forward to that |
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This is a great activity for an adult to do with children or for a reader to do on their own. Keep the corresponding counters in their own containers for easy access. Read word problem, child uses counters to make addition word problem and solve. Comes with 12 word problems and 12 sets of 10 counters.
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8507 | https://www.youtube.com/watch?v=PCOvjDW7zcY&pp=0gcJCf8Ao7VqN5tD | How to Solve Inequality with Reciprocals Graphical and Algebraic Approach
Anil Kumar
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Description
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Posted: 16 Oct 2018
globalmathinstiute #anilkumarmath Introduction to Inequalities:
Polynomial Inequalities: #graph_equation_relation #GCSE_functions #Precalculus_anilkumar
1 comments
Transcript:
I wanna go Kumar now let's move forward and learn how to solve inequalities involving reciprocal functions here is an example we are going to solve 2 over X plus 3 greater than 29 over X I'll show you two different ways of doing it the first method will kind of use a graph to plot the graph I'll take the constant terms on one side so leaving 3 here we could write our inequality as 3 is greater than 29 over X minus 2 over X and that is 3 is greater than 27 over X now we can plot both these graphs and graphically see the solution so let's say that is our coordinate system 27 or X will be a graph which is kind of one over X vertically stretched so here is a very approximate diagram okay and definitely X is not equal to zero we have a vertical asymptote right there now the ground for three should be something like this y equals to three let me just sketch a line here let's say this is y equals to three now from the graph we have a clear solution which is we want this line to be above the graph of reciprocal function that is we have to find what this point is once we find this point then we know that line is above the graph on the right side of this point and from 0 onwards to the left side so that becomes the solution is that clear now when are they equal so we can solve for it when is 3 equal to 27 over X well that gives us x equals to 27 over 3 which is 9 so this point is 9 for us once we know this is 9 we can see that the solution is what solution is from minus infinity to 0 Union from 9 to plus infinity right 9 to plus infinity is that clear so this is a very straightforward method how graphically we may not get very accurate results so here we have the second method which we use values so we'll have a table of intervals so we'll kind of analyze and then find the solution now in the second method what do we do we bring all the terms to one side so so let's take this three on the right side so we get 0 is greater than 27 over X minus 3 now we could write this as 0 is greater than common denominator is X we get 27 minus 3 and we can rewrite this factor the numerator I mean I'm sorry 3x X is the common denominator so 3 is common we get 9 minus x over X so we have a rational function on the right side do you see that we want that to be less than 0 we want that to be less than 0 so we'll analyze this particular function now so what we will do here is kind of make table with zeros and undefined values at x equals to 9 we have a 0 right so so we have a 0 at x equals to 9 when this factor is 0 and then this factor is undefined at x equals to 0 we have undefined so we'll analyze the rational function on the intervals which are divided by zeros and the vertical asymptotes so this interval is from minus infinity to 0 and then we have from 0 to 9 and from 9 to infinity let us take test points in these intervals so from 0 to minus infinity I could take minus 1 as a test point here I could take one as a test and ten right now what do we test we'll test these two factors so we can test the value for X and also the factor 3 times 9 minus X right as far as X is concerned for a negative value it will be negative for positive it will be positive now 9 minus X is going to be negative for negative 1 I mean positive if I substitute X is negative 1 so it becomes positive for one also it is positive 9 minus 1 is 8 but for 10 it is going to be negative but when we divide these two what do we get that is what we try to analyze when you divide positive by negative you do get negative right positive by positive is positive and this would be negative when we say zero is greater than this we are looking for a solution where rational function is negative which is these intervals correct so these intervals are between minus infinity to 0 and from 9 to infinity just as we saw earlier okay so we get exactly the same solution from this table right so we get our answer as from minus infinity to 0 Union from 9 to infinity is that clear now these are two very good methods to apply for situations where we have rational functions as shown here now if it is specified a particular method you have to go for that otherwise you can choose any one of these methods I hope these steps are absolutely clear and you understand how we have done it feel free to write your comments and share your views if you like and subscribe to my videos that great thanks for watching and all the best |
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Linear Equations Unit Plan
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mathclasswithsass
This is a unit plan for a linear equations unit. It includes the common core standards, essential questions, key student learning objectives, sequence of key learning activities, texts to be used, and assessments and tasks.
8th - 9th
Math
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Absolute Value Functions Unit Lesson Plan
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Secondary Math Solutions
This is a free lesson plan on Absolute Value Functions. This lesson plan covers the following topics & TEKS but could be used for anyone looking for an Absolute Value Functions Unit Lesson Plan. graph the function f(x) = |x| and when applicable analyze the key attributes such as domain, range, intercepts, symmetries and maximum and minimum given an interval (2A.2A)analyze the effect on the graphs f(x) = |x| when f(x) is replaced by af(x), f(bx), f(x - c) and f(x) + d for specific positive a
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Also included in: Algebra 2 Curriculum - Texas TEKS Aligned
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Writing Linear Equations in Slope Intercept Form y=mx+b (a mini unit)
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Building Linear Function from a Pattern, Explicit and Recursive Linear Equations
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Lindsey Henderson
I use these guided notes in conjunction with the Mathematics Vision Project Secondary Math 1 curriculum, Getting Ready chapter 1 lesson 1. Topics this covers are: linear growth, rates of change, table of values, constructing a linear equation given a pattern, graphing points, and slope of a line.
I have the students work in small groups or pairs and explore the linear pattern together.
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HSF-BF.A.1
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7th grade math unit plan: Linear Relationships
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This unit can be used to teach 7th grade Expressions and Equations: Linear Relationships. It contains Common Core standards, assessments, learning activities, enduring understandings and essential questions.
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3.3 Point-slope equation (Unit: Linear, Course: Algebra 1)
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Alissa Berger
This is a typical, 50-minute lesson plan to fit a traditional Algebra 1 classroom. It is downloadable as a PowerPoint, but it has been formatted so that you may print 2 slides per page to create a handout, or worksheet, for students. This lesson is all about how to write "point-slope" within the unit of "linear functions." I make sure to target our focus vocabulary of "equations" while showing three types of equations that can be found with linear functions. Students like this comparison
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Algebra 1 - Linear Equations - Complete Unit - FREE SAMPLE
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This is a sample of our Linear Equations Unit and includes the table of contents for the full unit, the standards alignment chart, a sample calendar, one quiz and one lesson (including teacher notes).
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3.2 Slope-intercept equation (Unit: Linear, Course: Algebra 1)
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This is a typical, 50-minute lesson plan to fit a traditional Algebra 1 classroom. It is downloadable as a PowerPoint, but it has been formatted so that you may print 2 slides per page to create a handout, or worksheet, for students. This lesson is all about how to write "slope-intercept equations" within the unit of "linear functions." I make sure to target our focus vocabulary of "slope-intercept". Over time, this will be compared to the 2 alternate linear equations to develop reasoning
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Graphing Linear Equations Review
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Solving Linear, Quadratic, Radical, and Absolute Value Equations Handouts
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Algebra, Algebra 2, Math
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Algebra Performance Task: Multistep Equations with Tables and Graphs
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Systems Unit Lesson Plan for Algebra 2
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Secondary Math Solutions
This is a free lesson plan for an Algebra 2 unit on Systems. This lesson plan covers the following topics and TEKS: solving 2 and 3 variable systems of equations (2A.3A, 2A.3B)solving 2 and 3 variable systems of inequalities (2A.3E, 2A.3F, 2A.3G)solving linear-quadratic systems (2A.3A, 2A.3C, 2A.3D)using systems to write a quadratic equation given 3 points on its graph (2A.4A)To make your life easier, you can purchase all of the resources mentioned in the lesson plan here: Systems Unit Bundl
9th - 12th
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Also included in: Algebra 2 Curriculum - Texas TEKS Aligned
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Multiple Representations of Linear Relationships FREEBIE
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Eugenia's Learning Tools
I save this activity for the end of our linear relationships unit. For the task, students must create a poster showing a linear relationship represented with a verbal description, table, graph, and equation. This is a one page template to help your students create a draft, helpful in planning out their student created poster! More display posters✽ Converting Decimals, Fractions, and Percent Math Posters✽ Integer Operations Rules Math PostersBest sellers that may help you out during the year! -
5th - 8th
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Systems of Equations Unit Lesson Plan for Algebra 1
Created by
Secondary Math Solutions
This is a free lesson plan for an Algebra 1 unit on Solving Systems of Linear Equations. This is aligned to the Texas TEKS (A2I, A3F, A3G, A5C) but could be used for anyone looking for a Systems of Equations lesson plan for Algebra 1 students. Need more unit plans for Algebra 1? I've got you covered! -- > Algebra 1 Scope & Sequence
8th - 9th
Algebra, Math
Also included in: Systems of Equations Unit Bundle - Algebra 1 Curriculum
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Systems of Equations Puzzle Math Activity
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Beyond the Worksheet with Lindsay Gould
This puzzle has students match up a linear system, an equation that shows the system in a substituted format and the solution to the system. Puzzle must be pre-cut as the format it is in currently is the answer key! This puzzle can be found in my Algebra 1 Linear Systems Unit, Algebra 1 Linear Systems Digital Notebook and in Unit 5 - Linear Systems of Equations of my Pre-Algebra Curriculum!Licensing Terms :By purchasing this product, you own a license for one teacher only for personal use
7th - 8th
Math
Also included in: 8th Grade Math Curriculum Supplemental Activities Bundle
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Unit 3 Algebra 1 Lesson Plans
Created by
Justin Seabolt
These lesson plans align with the Holt McDougal Larson Algebra 1 textbook (Florida edition). With this lesson plan, it covers all of Unit 3 on Solving Linear Equations (except for solving percent problems as it is not covered on our state exam).
Includes:
3-4 weeks of lesson plans
Gradual Release model (I do, We do, You do)
Shows alignment with state standards (Florida only)
Unit Essential Question
Lesson Essential Questions for each concept
Key Vocabulary
Description of Activating and Summa
8th - 12th
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Algebra 2 Functions Test Review guide with answer key
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In Depth No Prep Math Lessons Plans
This is an example of a review guide for a test complete with an answer key. This unit covered
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• Solving Absolute Value equations
• Defining quadratic equations
• Solve quadratic equations by factoring, completion of squares, square roots, quadratic formula
• Using the discriminant in identifying types of zeroes for a quadratic equation
• Identify and solve polynomial equations
• Simplify and solve rat
10th - 12th
Algebra, Algebra 2, Math
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Solving Systems of Equations by Elimination - Perfect 10 Activity - FREEBIE
Created by
Tater Taught by Jen Tate
In this no-prep, self-checking, high school math activity, students will practice solving 2-variable systems of equations (simultaneous linear equations) using elimination. How does this Perfect 10 activity work?Students will solve 5 systems of linear equations using elimination. They will then cancel out the x- and y-values of the coordinate answer in the large box at the top of the paper. If done correctly, they will have crossed out every number, 1 through 10 (and gotten a Perfect 10). When
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8509 | https://www.youtube.com/watch?v=b0xC3JlILKE | Sum of First N Cubes Equals Square of 1+2+3+...+n | Number Theory
Wrath of Math
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Posted: 10 May 2020
The sum 1^3 + 2^3 + 3^3 + ... + n^3 is equal to (1+2+...+n)^2. Amazing! In today's number theory video lesson, we'll prove this wonderful equality using - you guessed it - induction!
The sum of cubes and the sum of squares are cool, but the sum of cubes and the square of a sum is an even cooler pair!
Here are two proofs for the sum of the first n natural numbers: and
Lesson on induction proofs:
I hope you find this video helpful, and be sure to ask any questions down in the comments!
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here's one of those beautiful hard to believe but easy to prove number theory results the sum of the first n cubes 1 cubed plus 2 cubed and so on up to M cubed is equal to the square of the sum of the first n natural numbers 1 plus 2 and so on up to N squared it's almost too beautiful that math should work out like this but lo and behold this is true and we're going to go over how to prove it in today's wrath of math lesson you can probably guess this is going to be a proof by induction if you're not familiar with proof by induction check out my lesson on the topic I'll leave a link to it in the description I'll also leave a link to a proof of this result which we're going to use in our proof today this result states that the sum of the first n natural numbers 1 plus 2 plus all the way up to N is equal to that natural number n that last natural number in the sum times the next natural number n plus 1/2 so this is going to come in super handy and will really be the key to our proof and then we'll just have to do a little bit of algebra and we'll be done all right let's get into it you probably know where to start an induction proof starts with the basis step I'll just write basis the basis step is where we prove our result is true for the first number of interest in this case that's the first natural number as it often is N equals 1 and as is usual the basis case is pretty trivial 1 cubed is indeed equal to 1 squared so we see that the sum of the first cube which is just 1 cubed is equal to the square of the sum of the first one natural numbers that's just 1 squared so this is just 1 equals 1 the basis case checks out so that's it that's all we have to do for the basis case show that the result is true for N equals 1 quickly let's just check out N equals 2 since it's a little more interesting we expect that the sum of the first 2 cubes 1 cubed plus 2 cubed should be equal to the square of the sum of the first two natural numbers 1 plus 2 squared 1 cubed is 1 2 cubed is 8 so that's 1 plus eight and then we have 1 plus 2 which is 3 ^ 2 and look at that 9 equals 9 it checks out for N equals 2 as well that's not necessary for the proof but it's definitely more interesting let's go ahead and jump into the induction step for the induction step of our induction proof remember that all we do is first assume the result is true for some natural number K this is called the induction hypothesis of course this is a valid assumption to make because in the basis case we proved that indeed our result is true for some natural number in particular it's true for the first natural number then in our induction step all we need to do to complete the proof is show that if our result is true for some natural number K then it must also be true for the next natural number K plus 1 I want to quickly point something out that will help us see where we're going in this proof and this is often a helpful thing to do remember the theorem we mentioned at the beginning of the lesson for an expression that's equal to the sum of the first n natural numbers in this case we've got the sum of the first K plus 1 natural numbers squared applying that theorem tells us that this is equal to the last natural number in the sum which is K plus 1 multiplied by the next natural number which would be K plus 2 divided by 2 and then all squared so if we could show that the sum of the first K plus 1 cubes is equal to this then we could apply that same theorem about the sum of the first and natural numbers to go the other way and to show that this sum is equal to this so keep that in mind that's where we're going and I'm just going to shrink this a little bit and set it aside for us so here's how we begin we want to show that the sum of the first K plus 1 cubes is equal to this expression here because we know that's equal to the square of the sum of the first K plus 1 natural numbers so where can we start well we know that the sum of the first K plus 1 cubes is equal to the sum of the first K cubes plus K plus one cubed and that's helpful because we know the sum of the first K cubes is equal to that the sum of the first K natural numbers squared so this is all equal to 1 plus 2 all the way up to K squared and then we've got to add that cube of k plus 1 remember it's just substitution to get this equality we know the sum of the first K cubes is equal to this that was our induction hypothesis so we just did some substitution then since we've got the sum of the first K natural numbers let's go ahead and apply that theorem that we've mentioned that the sum of the first K natural numbers is equal to the last natural in the sum K multiplied by the next natural number k plus 1 divided by 2 and then since this is getting squared we've got to square this expression and of course we've got K plus 1 cubed still on the end now let's go ahead and carry out this squaring operation so that we can work with everything a bit more squaring what's here in the brackets gives us K squared multiplied by K plus 1 squared all divided by 2 squared which is equal to 4 and then again plus K plus 1 cubed now let's just bring our orange expression down so we can keep the goal in sight so we know where we are headed let's rewrite this sum in one fraction in order to give K plus 1 cubed a denominator of 4 we of course just have to multiply it by 4 over 4 and so this is equal to K squared multiplied by K plus 1 squared plus 4 times K plus 1 cubed all over our common denominator of 4 now in order for the expression that we've got our sites on to be a bit more useful let's also carry out the exponentiation in that expression because that's going to look a bit closer to what we have here we'll go ahead and zoom in and do that so carrying out this squared operation we have K plus 1 squared times K plus 2 squared divided by 2 squared which of course is 4 all right now we we are very close notice that in the expression we want to get to we've got a factor of k plus 1 squared additionally in the sum that were currently at we've also got a factor of k plus 1 squared in both terms this K plus 1 cubed could be rewritten as K plus 1 squared times K plus 1 so let's do that so this is equal to K squared multiplied by K plus 1 squared plus 4 times k plus 1 squared times k plus 1 again that's just rewriting k plus 1 cubed as k plus 1 squared times k plus 1 and of course we've got a denominator of 4 all right now what do you think we should do well certainly it would be a good idea to factor k plus 1 squared out of both of these terms in the sum that'll get us one step closer to our goal which has a factor of K plus 1 squared so factoring out k plus 1 squared gives us of course k plus 1 squared multiplied by k squared that's that factor of K squared there plus 4 times k plus 1 that's what's left over here on the right and then again can't forget divide by 4 all right we've got two more steps perhaps you can see exactly where this is going let's distribute the 4 through this term of K plus 1 so that will give us K plus 1 squared multiplied by K squared plus distributing 4 through K plus 1 gives us plus 4 k plus 4 and again all of this is being divided by let me straighten that out all of this is being divided by 4 and look at that folks you're not going to get a much more obvious term to factor than that k squared plus 4 k plus 4 is equal to k plus 2 squared so we can rewrite this as k plus 1 squared multiplied by k plus 2 squared divided by 4 look at that just like we wanted let's point out the big move again we rewrote k squared for K plus 4 as K plus 2 squared because K times K will give us the K squared K times 2 plus 2 times K will give us the for K and 2 times 2 will give us the 4 so these two things are equal it's possible you've forgotten why we wanted to get an expression like this in the first place if so let me remind you all the factors we've got here are squares so now we're going to go in the other direction and take out the square exponent by that I mean we're going to rewrite this as K plus 1 multiplied by K plus 2 divided by 2 all squared remember we want to show that this is equal to the sum of the first K natural numbers squared so we know we need something like a bunch of stuff in parentheses to the power of 2 and that's exactly where we are and now my friends what is this equal to just remember that classic theorem the sum of the first n natural numbers is equal to that last natural number in the sum n times the next natural number n plus 1 divided by 2 so then what do you notice about this here well this must be equal to the sum of the first K plus 1 natural numbers because we've got the last number in that sum K plus 1 times the next natural number K plus 2 divided by 2 so by applying this theorem this is equal to 1 plus 2 plus all the way up to k plus k plus 1 all squared and again there are links in the description in case you haven't seen a proof of this beautiful result we're using but that my friends concludes the proof because this is the square of the sum of the first K plus 1 natural numbers and remember what was at the beginning of all of these equalities well it was the sum of the first K plus 1 cubes so we just completed the induction proof we first showed that our result is true for the first natural number then we showed that if it's true for some natural it must also be true for the next natural number thus we have now proven that the sum of the first n cubes is equal to the square of the sum of the first n natural numbers beautiful so I hope this video helped you understand this induction proof let me know in the comments if you have any questions need anything clarified or have any other video requests thank you very much for watching I'll see you next time and be sure to subscribe to the swankiest math lessons [Music] [Applause] [Music] |
8510 | https://cstheory.stackexchange.com/questions/10542/when-are-two-algorithms-said-to-be-similar | Stack Exchange Network
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When are two algorithms said to be "similar"?
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I do not work in theory, but my work requires reading (and understanding) theory papers every once in a while. Once I understand a (set of) results, I discuss these results with people I work with, most of whom do not work in theory as well. During one of such discussions, the following question came up:
When does one say that two given algorithms are "similar"?
What do I mean by "similar"? Let us say that two algorithms are said to be similar if you can make either of the following claims in a paper without confusing/annoying any reviewer (better definitions welcomed):
Claim 1. "Algorithm $A$, which is similar to algorithm $B$, also solves problem $X$"
Claim 2. "Our algorithm is similar to Algorithm $C$"
Let me make it slightly more specific. Suppose we are working with graph algorithms. First some necessary conditions for the two algorithms to be similar:
They must be solving the same problem.
They must have the same high level intuitive idea.
For instance, talking about graph traversal, breadth-first and depth-first traversal satisfy the above two conditions; for shortest-path computations, breadth-first and Dijkstra's algorithm satisfy the above two conditions (on unweighted graphs, of course); etc.
Are these also sufficient conditions? More specifically, suppose two algorithms satisfy the necessary conditions to be similar. Would you indeed call them similar, if
they have different asymptotic performance?
for a special class of graphs, one algorithm requires $\Omega(n)$ time while the other requires $O(n^{1/3})$ time?
they have different terminating conditions? (recall, they are solving the same problem)
the pre-processing step is different in the two algorithms?
the memory complexity is different in the two algorithms?
Edit: The question is clearly very context dependent and is subjective. I was hoping that the above five conditions, however, will allow getting some suggestions. I am happy to further modify the question and give more details, if needed to get an answer. Thanks!
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edited Mar 6, 2012 at 7:13
RachitRachit
asked Mar 6, 2012 at 6:06
RachitRachit
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$\begingroup$ it really depends on the context. For example, for certain sequential algorithms, DFS and BFS are very different and one might not even work. In parallel settings, DFS (or at least one variant) is P-complete, whereas BFS is "easy in parallel". $\endgroup$
Suresh Venkat
– Suresh Venkat
2012-03-06 06:37:33 +00:00
Commented Mar 6, 2012 at 6:37
$\begingroup$ @SureshVenkat -- I agree that the question is very context dependent. In the interest of not starting a debate, I refrained from taking names of "the two algorithms" at the risk of sounding vague :-) $\endgroup$
Rachit
– Rachit
2012-03-06 06:44:21 +00:00
Commented Mar 6, 2012 at 6:44
4
$\begingroup$ The problem is that there is close and there is close. There's a way of thinking of the multiplicative-weight-update method as "essentially a binary search", but in the wrong context this would sound insane. FWIW, in all of your cases above I can imagine declaring the two algorithms to be different. $\endgroup$
Suresh Venkat
– Suresh Venkat
2012-03-06 06:49:37 +00:00
Commented Mar 6, 2012 at 6:49
1
$\begingroup$ This question seems too subjective to me. You're basically asking for a definition of "similar", when no canonical definition exists. $\endgroup$
Joe
– Joe
2012-03-06 07:04:08 +00:00
Commented Mar 6, 2012 at 7:04
1
$\begingroup$ Somewhat related: cstheory.stackexchange.com/questions/9409/… $\endgroup$
Radu GRIGore
– Radu GRIGore
2012-03-06 11:43:17 +00:00
Commented Mar 6, 2012 at 11:43
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It is a tough problem to give even a coherent definition of "Algorithm A is similar to Algorithm B". For one, I don't think that "they must be solving the same problem" is a necessary condition. Often when one says in a paper that "the algorithm $A$ of Theorem $2$ is similar to the algorithm $B$ in Theorem $1$", the algorithm $A$ is actually solving a different problem than that of $B$, but has some minor modifications to handle the new problem.
Even trying to determine what it means for two algorithms to be the same is an interesting and difficult problem. See the paper "When are two algorithms the same?"
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answered Mar 6, 2012 at 7:18
Ryan WilliamsRyan Williams
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More often than not, it means "I don't want to write out Algorithm B in detail, because all the interesting details are nearly identical to those in Algorithm A, and I don't want to go over the 10-page limit, and anyway the submission deadline is in three hours."
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answered Mar 6, 2012 at 8:01
JeffεJeffε
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If you mean "similar" in the colloquial sense, I think JeffE's answer captures what some people mean.
In a technical sense though, it depends on what you care about. If asymptotic time complexity is all you care about, the difference between recursion and iteration may not matter. If computatability is all you care about, the difference between a counter variable and a one-symbol stack do not matter.
To compare algorithms, a first step would be to make the notion of equivalence precise. Intuitively, let $A$ be the space of algorithms and $M$ be a space of mathematical objects and $\mathit{sem}: A \to M$ be a function encoding that $\mathit{sem}(P)$ is the meaning of algorithm $P$. The space $M$ could contain anything ranging from the number of variables in your algorithm, to its state-graph or it's time complexity. I don't believe there is an absolute notion of what $M$ can be. Given $M$ though, we can say two algorithms are equivalent if $\mathit{sem}(P)$ equals $\mathit{sem}(Q)$. Let me add that I think each of the five criteria you mentioned can be formalised mathematically in this manner.
If we want to talk about an algorithm being more general than another (or an algorithm refining another), I would endow $M$ with more structure. Imagine that $(M, \sqsubseteq)$ is a partially ordered set and the order $x \sqsubseteq y$ encodes that $x$ is a more defined object than $y$. For example, if $M$ contains sets of traces of an algorithm and $\sqsubseteq$ is set inclusion, $\mathit{sem}(P) \sqsubseteq \mathit{sem}(Q)$ means that every trace of $P$ is a trace of $Q$. We can interpret this as saying that $P$ is more deterministic than $Q$.
Next, we could ask if it's possible to quantify how close two algorithms are. In this case, I would imagine that $M$ has to be endowed with a metric. Then, we can measure the distance between the mathematical objects that two algorithms represent. Further possibilities are to map algorithms to measure spaces or probability spaces and compare them using other criteria.
More generally, I would ask - what do you care about (in intuitive terms), what are the mathematical objects representing these intuitive properties, how can I map from algorithms to these objects, and what is the structure of this space? I would also ask if the space of objects enjoys enough structure to admit a notion of similarity. This is the approach I would take coming from a programming language semantics perspective. I'm not sure if you find this approach appealing, given the vastly different cultures of thought in computer science.
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answered Mar 7, 2012 at 7:57
Vijay DVijay D
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Along the lines of Jeff's answer, two algorithms are similar if the author of one of them expects that the author of the other one might be reviewing her paper.
But joking aside, in the theory community, I would say that what problem algorithm A is solving is rather tangental to whether it is "similar" to algorithm B, which might be solving a completely different problem. A is similar to B if it "works" because of the same main theoretical idea. For example, is the main idea in both algorithms that you can project the data into a much lower dimensional space, preserve norms with the Johnson-Lindenstrauss lemma, and then do a brute-force search? Then your algorithm is similar to other algorithms that do this, no matter what problem you are solving. There are some small number of heavy-duty algorithmic techniques that can be used to solve a wide variety of problems, and I would think that these techniques form the centroids of many sets of "similar" algorithms.
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answered Mar 8, 2012 at 12:55
Aaron RothAaron Roth
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Very interesting question, and very nice paper Ryan!
I do definitely agree with the idea that making an assessment on the overall similarity between algorithms is mainly a subjective value judgement. While from a technical point of view there are a number of features that are closely observed to decide upon the similarity of algorithms, in the end, it is also a matter of personal taste. I will try to provide a description of the importance of both sides of the same coin while referring to the particular points of your question:
From a technical point of view:
Ryan already pointed out that both algorithms must solve the same problem. One could go even further and generalize this notion by saying that it is usually enough to prove that there is a polynomial transformation of the same instance that is understoodable by algorithm A so that algorithm B can handle it. However, this would be actually very weak. I do prefer to think of the similarity in a stronger sense.
However, I would never expect for two equivalent algorithms to have the same intuitive idea ---though, again, this is a definition which ain't easy to capture. More than that, it is often the case that algorithms that are deemed to be similar do not follow the main rationale. Consider for example some sorting algorithms which however, originated in different ways following different ideas. As an extreme example consider genetic algorithms which are usually considered by the mathematical community just as stochastic processes (and therefore they are equivalent in their view) which are then modeled and analyzed in quite a different way.
Moreover, I would even generalize this notion to say that other technicalities such as the termination condition or the pre-processing stage do not matter often. But this is not always the case. See for example Dijkstra's Algorithm versus Uniform Cost Search or a Case Against Dijkstra's Algorithm. Both algorithms are so close that most people do not tell the difference, yet the differences (being technical) were very important for the author of that paper. Much the same happens with the pre-processing step. In case you are familiar with the $N$-puzzle, then observe that an A$^$-like search algorithm using the Manhattan distance or $(N^2-1)$ additive pattern databases would actually expand the same number of nodes in exactly the same order, and that makes both algorithms (and their heuristics) to be strictly equivalent in a very strong sense, whereas the first approach has no pre-processing and the second one has a significant overhead before starting to solve a particular instance. However, as soon as your Pattern Databases consider more simulatenous interactions there is a huge gap in performance between them, so that they are definitely different ideas/algorithms.
As a matter of fact, I think that most people would judge algorithms for their purpose and performance. Therefore, asymptotic performance is a good metric to reason about the similarity between programs. However, bear in mind that this performance is not necessarily the typical case so that if two algorithms have the same asymptotic performance but behave differently in practice, then you would probably conclude that they are different. The strong evidence in this regard would be that both algorithms have the same performance both on time and memory (and this, as Suresh said, makes DFS and BFS to look different). In case this assertion does not sound convincing to you, please refer to the excellent (and very recommended book): Programming the Universe by Seth Lloyd. In page 189 he refers to a list with more than 30 measures of complexity that can be used to regard algorithms as being different.
So what makes algorithms to be similar/different? In my view (and this is purely speculative), the main difference is about what they suggest to you. Many, many (many!) algorithms differ just in a few technicalities when serving to the same purpose so that the typical case is different for different ranges of the input. However, the greatest of all differences is (to my eye) what they suggest to you. Algorithms have different capabilities and therefore their own strengths and weaknesses. If two algorithms look like being the same but might be extended in different ways to cope with different cases then I would conclude that they are different. Often, however, two algorithms do look much the same so that you would regard them to be the same ... until someone arrives making a key distinction and suddenly, they are completely different!
Sorry, my response was in the end so long ...
Cheers,
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answered Mar 7, 2012 at 9:44
Carlos Linares LópezCarlos Linares López
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$\begingroup$ Actually, Ryan suggested that it is not necessary for both algorithms to solve the same problem. $\endgroup$
Jeffε
– Jeffε
2012-05-10 06:21:05 +00:00
Commented May 10, 2012 at 6:21
$\begingroup$ True! I was just collecting my opinions in this regard, but you are definitely right! $\endgroup$
Carlos Linares López
– Carlos Linares López
2012-05-12 12:17:29 +00:00
Commented May 12, 2012 at 12:17
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Any mention of similarity without defining a similarity metric is not well-defined. There are many ways in which two algorithms can be similar:
Quicksort and Mergesort solve very similar problems, but they use different algorithms to do so. They have similar algorithmic complexity (although their worst-case performance and memory usage can vary). Quicksort and Mergesort are both similar to Bubblesort,however Bubblesort has very different performance metrics. If you ignore complexity statistics Quicksort, Mergesort and Bubblesort are all in the same equivalence class. However, if you care at all about algorithmic complexity, then Quicksort and Mergesort are much more similar to each other than either is to Bubblesort.
Smith-Waterman dynamic programming and HMM-sequence comparison attempt to solve the problem of aligning two sequences. However, they take different inputs. Smith-Waterman takes two sequences as input, and HMM-sequence comparisons take a HMM and a sequence as input. Both output sequence alignments. In terms of motivating ideas, both of these are similar to Levenshtein's edit distance, but only at a very high level.
Here are some criteria by which two algorithms might be called similar:
Input/output types
Algorithmic/Memory Complexity
Assumptions about types of inputs (eg only positive numbers or floating point stability)
Nested relationships (eg some algorithms are special cases of others)
The critical decision about the meaning of similarity remains. Sometimes you care about the complexity of an algorithm, sometimes you don't. As the definition of similarity depends on the context of the discussion, the term "similar algorithm" isn't well-defined.
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answered Mar 13, 2012 at 2:50
James ThompsonJames Thompson
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8511 | https://www.engineersedge.com/heat_transfer/convection.htm | Engineers Edge utilizes cookies to enable essential site functionality, and targeted advertising. To learn more, see our Privacy Policy.
Convective Heat Transfer Convection Equation and Calculator
Convective Heat Transfer Convection Equation and Calculator
Heat Transfer Engineering
Thermodynamics
Convection of Known Surface Area Calculator
Convective heat transfer , often referred to simply as convection , is the transfer of heat from one place to another by the movement of fluids . Convection is usually the dominant form of heat transfer in liquids and gases. Although often discussed as a distinct method of heat transfer, convective heat transfer involves the combined processes of conduction (heat diffusion) and advection (heat transfer by bulk fluid flow).
The convective heat transfer coefficient (h), defines, in part, the heat transfer due to convection.
The convective heat transfer coefficient is sometimes referred to as a film coefficient and represents the thermal resistance of a relatively stagnant layer of fluid between a heat transfer surface and the fluid medium. Common units used to measure the convective heat transfer coefficient are:
1 W/(m2 K) = 0.85984 kcal/(h m2 ° C) = 0.1761 Btu/(ft2 h ° F)
1 kcal/(h m2 ° C) = 1.163 W/(m2 K) = 0.205 Btu/(ft2 h ° F)
Btu/hr - ft2 - °F = 5.678 W/(m2 K) = 4.882 kcal/(h m2 ° C)
Convection involves the transfer of heat by the motion and mixing of "macroscopic" portions of a fluid (that is, the flow of a fluid past a solid boundary). The term natural convection is used if this motion and mixing is caused by density variations resulting from temperature differences within the fluid. The term forced convection is used if this motion and mixing is caused by an outside force, such as a pump. The transfer of heat from a hot water radiator to a room is an example of heat transfer by natural convection. The transfer of heat from the surface of a heat exchanger to the bulk of a fluid being pumped through the heat exchanger is an example of forced convection.
Heat transfer by convection is more difficult to analyze than heat transfer by conduction because no single property of the heat transfer medium, such as thermal conductivity, can be defined to describe the mechanism. Heat transfer by convection varies from situation to situation (upon the fluid flow conditions), and it is frequently coupled with the mode of fluid flow. In practice, analysis of heat transfer by convection is treated empirically (by direct observation).
Convection heat transfer is treated empirically because of the factors that affect the stagnant film thickness:
Fluid velocity
Fluid viscosity
Heat flux
Surface roughness
Type of flow (single-phase/two-phase)
Convection involves the transfer of heat between a surface at a given temperature (T1) and fluid at a bulk temperature (Tb). The exact definition of the bulk temperature (Tb) varies depending on the details of the situation. For flow adjacent to a hot or cold surface, Tb is the temperature of the fluid "far" from the surface. For boiling or condensation, Tbis the saturation temperature of the fluid. For flow in a pipe, Tb is the average temperature measured at a particular crosssection of the pipe.
The basic relationship for heat transfer by convection has the same form as that for heat transfer by conduction:
Q = h · A · ΔT
where
Q = rate of heat transfer (Btu/hr)
h = convective heat transfer coefficient (Btu/hr-ft2 - °F)
A = surface area for heat transfer (ft2)
ΔT = temperature differnece (°F)
or
Q = hc · A · (Ts - Ta)
where
Q = heat transferred per unit time (W)
A = heat transfer area of the surface (m2)
hc = convective heat transfer coefficient of the process (W/(m2 K) or W/(m2 ° C))
Ts = Temperature surface
Ta = Temperature air
The convective heat transfer coefficient (h) is dependent upon the physical properties of the fluid and the physical situation. Typically, the convective heat transfer coefficient for laminar flow is relatively low compared to the convective heat transfer coefficient for turbulent flow. This is due to turbulent flow having a thinner stagnant fluid film layer on the heat transfer surface. Values of h have been measured and tabulated for the commonly encountered fluids and flow situations occurring during heat transfer by convection.
Example calculations:
A 22 foot uninsulated steam line crosses a room. The outer diameter of the steam line is 18 in. and the outer surface temperature is 280oF. The convective heat transfer coefficient for the air is 18 Btu/hr-ft 2 - oF.
Calculate the heat transfer rate from the pipe into the room if the room temperature is 72 oF.
Solution
Q = h · A · ΔT
Q = h ( 2 · π · r · L ) ΔT
Q = ( 18 Btu / ( hr-ft2-°F) · 2 · ( 3.14157) · ( 0.75 ft) · ( 22 ft ) · ( 280°F - 72°F)
Q = 3.88 x 105 Btu/hr
Many applications involving convective heat transfer take place within pipes, tubes, or some similar cylindrical device. In such circumstances, the surface area of heat transfer normally given in the convection equation ( ) varies as heat passes through the cylinder. In addition, the temperature difference existing between the inside and the outside of the pipe, as well as the temperature differences along the pipe, necessitates the use of some average temperature value in order to analyze the problem. This average temperature difference is called the log mean temperature difference (LMTD), described earlier.
It is the temperature difference at one end of the heat exchanger minus the temperature difference at the other end of the heat exchanger, divided by the natural logarithm of the ratio of these two temperature differences. The above definition for LMTD involves two important assumptions: (1) the fluid specific heats do not vary significantly with temperature, and (2) the convection heat transfer coefficients are relatively constant throughout the heat exchanger.
Related:
Convective Heat Transfer Coefficients Table Chart
Extruded ( Finned )Heat Sink Radiation Formulae
Convection of Known Surface Area Calculator
Heat Loss from a Pipe
Heat Sink Convection with Fins Calculator
Heat Radiation of a Surface Calculator
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8512 | https://d1yqpar94jqbqm.cloudfront.net/ecs/TXOER_Geo_M02_T01_L03_TIG.pdf | LESSON 3: Into the Ring • 315A Into the Ring Constructing an Inscribed Regular Polygon 3 Lesson Overview Students learn how to construct three regular polygons: regular hexagons, squares, and equilateral triangles. They start by constructing a regular hexagon inscribed in a circle using two different methods. First, they duplicate 60° angles to create six equilateral triangles sharing the center of the circle as a vertex. Secondly, they construct six adjacent congruent chords the same length as the radius of the circle around the circumference of the circle to inscribe a regular hexagon. Students then construct a square inscribed in a circle using two different methods. First, they use patty paper to rotate a right triangle to create four congruent right triangles sharing the center of the circle as the vertex of the right angle. Secondly, they construct the diagonals of a square by constructing perpendicular diameters of a circle and then connect the endpoints. Finally, students analyze Worked Examples that demonstrate how to bisect an angle using patty paper and a compass and straightedge. Students construct both an equilateral triangle given a side length and an equilateral triangle inscribed in a circle. They demonstrate their construction skills by constructing inscribed angles, a 75° angle, and a regular octagon inscribed in a circle.
Geometry Logical Argument and Constructions (5) The student uses constructions to validate conjectures about geometric figures. The student is expected to: (B) construct congruent segments, congruent angles, a segment bisector, an angle bisector, perpendicular lines, the perpendicular bisector of a line segment, and a line parallel to a given line through a point not on a line using a compass and a straightedge. (C) use the constructions of congruent segments, congruent angles, angle bisectors, and perpendicular bisectors to make conjectures about geometric relationships.
ELPS 1.A, 1.C, 1.D, 1.E, 2.C, 2.D, 2.G, 2.H, 2.I, 3.A, 3.B, 3.C, 3.D, 3.F, 4.A, 4.B, 4.C, 4.D, 4.F, 4.K, 5.E MATERIALS Compasses Patty paper Straightedges ©2020 Carnegie Learning, Inc. Created on behalf of the Texas Education Agency. This work is subject to a CC BY-NC 4.0 license.
315B • TOPIC 1: Composing and Decomposing Shapes Essential Ideas • Constructions can be used to duplicate a given angle.
• A 60° angle can be constructed by creating an equilateral triangle within a circle.
• A regular hexagon can be inscribed in a circle by duplicating 60° angles to create six equilateral triangles sharing the center of the circle as a vertex.
• A regular hexagon can be inscribed in of a circle by constructing six adjacent congruent chords the same length as the radius of the circle around the circumference of the circle.
• When a square is inscribed in a circle, a segment that is a diagonal of the square is also a diameter of the circle. • An angle bisector is a line, segment, or ray that is drawn through the vertex of an angle and divides the angle into two congruent angles. Angle bisectors can be constructed using patty paper or a compass and straightedge.
• Both an equilateral triangle with a given side length and an equilateral triangle inscribed in a circle can be created using construction tools.
• The central angle of a circle is twice the measure of an inscribed angle that intercepts the same arc of the circle.
• Constructions can be used to verify geometric theorems.
LESSON 3: Into the Ring • 315C Lesson Structure and Pacing: 3 Days Day 1 Engage Getting Started: Duped Students construct an equilateral triangle in a circle by duplicating a segment and using the fact that the radii of a circle are congruent.
Develop Activity 3.1: Duplicating an Angle Students duplicate an angle and construct an angle that is twice the measure of a given angle. Students also revisit the construction from the Getting Started to construct a regular hexagon inscribed in a circle by duplicating 60° angles to create six equilateral triangles sharing the center of the circle as a vertex. They compare the processes used to duplicate an angle and duplicate a line segment.
Day 2 Activity 3.2: Constructing an Inscribed Hexagon Students construct a regular hexagon inscribed in a circle by creating six adjacent congruent chords, each the same length as the radius of the circle. They also determine the measure of each interior angle of a regular hexagon.
Activity 3.3: Constructing an Inscribed Square Students construct a square inscribed in a circle using patty paper and rigid motion transformations. They use the Pythagorean Theorem to determine the side lengths of the square based on a radius, r. Students then use the converse of the Pythagorean Theorem to determine that each angle of the square has a measure of 90°. They explain that when a square is inscribed in a circle, the diagonals of the square are also diameters of the circle in which it is inscribed. Finally, students use the fact that diagonals of a square are perpendicular to construct a square inscribed in a circle.
Activity 3.4: Bisecting an Angle Students analyze Worked Examples that demonstrate how to bisect an angle using patty paper and using a compass and straightedge. Students then bisect angles and use repeated reasoning to construct angles that are one-fourth and one-eighth the measure of a given angle.
Day 3 Activity 3.5: Constructing and Inscribing an Equilateral Triangle Students determine that given three line segments of the same length, an equilateral triangle can always be constructed. They use this reasoning and their knowledge of constructions to construct an equilateral triangle using a compass and straightedge. They then show two different ways to construct an inscribed equilateral triangle in a circle and describe the process they used in each construction.
Demonstrate Talk the Talk: Playing the Angles Students are given a circle with a central angle and its intercepted arc. Within the circle, they draw two inscribed angles that have the same intercepted arc as the central angle. Students then conjecture about the relationship between the measure of a central angle and the measure of any inscribed angle in the same circle that intercepts the same arc. Next, students use the constructions they have learned to construct a 75° angle and inscribe a regular octagon in a given circle.
315D • TOPIC 1: Composing and Decomposing Shapes Getting Started: Duped Facilitation Notes In this activity, students construct an equilateral triangle in a circle by duplicating a segment and using the fact that the radii of a circle are congruent.
Have students work with a partner or in a group to complete Question 1. Share responses as a class.
Questions to ask • Is there more than one possible location for point C?
• What are the possible locations for point C?
• How did you determine the correct length for ¯ BC ?
• How is ¯ BC related to the circle?
• If the triangle is equilateral, is the triangle also equiangular?
• If the triangle is equiangular, what is the measure of each interior angle?
Differentiation strategy To assist all students, highlight the importance of this activity. Have students mark the three sides congruent, label each angle in the triangle with its measure, and make a side note that this is a strategy to construct a 60° angle. Students may need to refer to this strategy to complete other constructions.
Summary A 60° central angle can be constructed without using measuring tools by creating an equilateral triangle within a circle.
Activity 3.1 Duplicating an Angle Facilitation Notes In this activity, students duplicate an angle and construct an angle that is twice the measure of a given angle. Students also revisit the construction from the Getting Started activity to construct a regular hexagon inscribed in a circle by duplicating 60° angles to create six equilateral triangles sharing the center of the circle as a vertex. They compare the processes used to duplicate an angle and duplicate a line segment.
DEVELOP ENGAGE LESSON 3: Into the Ring • 315E Differentiation strategies To assist all students, • Allow them to use patty paper or a compass and straightedge to complete constructions, unless the specific tools to be used are noted in the question.
• Provide options for placing patty paper constructions in the textbook. Suggest students either staple their patty paper in the book or transfer the solution from their patty paper to the diagram in the textbook.
Have students work with a partner or in a group to complete Questions 1 through 3. Share responses as a class.
Differentiation strategy To assist all students in writing concise explanations of how they performed constructions, suggest that they label points of intersection in their diagrams. That way, they can refer to locations by the letter of a point rather than a description of the location of the point. You may want to model this practice in the discussion of Question 2 if you find that this strategy is not intuitive to students. See the answer to Question 2 as an example.
As students work look for • Missing construction arcs. Students may forget to make the first arc to “standardize” the spread of the angle at a fixed distance from the vertex. If so, explain the importance of this step.
• Strategies to deal with the orientation of the angle in Question 2. Are students unaffected? Do they turn their book so the original angle had a horizontal base? Do they draw a starter line with a horizontal base or a base parallel to the bottom ray of the original angle?
Questions to ask • What is the first step?
• How did you decide the initial setting for the compass? Does it make a difference?
• Does the compass setting have to be the same for the arc on the original angle and your construction?
• Where did you place the compass point to create the first arc?
• Why do you think you need to construct that first arc?
• Did you need to change the setting for the compass in the next step? Why?
• Why is the new compass setting so important?
• Will everyone use the same setting? Why not?
• To create the other arc, where do you place the compass point?
• In Question 2, did you change your method at all because the orientation of the angle was different than in Question 1? If so, what did you do differently?
315F • TOPIC 1: Composing and Decomposing Shapes • After you duplicated / A, what ray did you use as a new starter line? Could you have used the other side of angle A as a new starter line?
• Are the same construction tools used to duplicate an angle and a line segment?
• How is duplicating an angle more complex than duplicating a line segment?
Have students work with a partner or in a group to complete Question 4. Share responses as a class.
Questions to ask • When first duplicating / A, which side of the angle did you use as a starter line?
• Did you construct a new first arc each time, or did you construct one concentric circle with a smaller radius to construct all of the angles?
• What is the name of a six-sided polygon?
• Is the hexagon you created equilateral? Equiangular? How do you know?
• What is the measure of each angle of the regular hexagon?
• What makes this hexagon an inscribed hexagon?
Summary A regular hexagon can be inscribed in a circle by duplicating 60° angles to create six equilateral triangles sharing the center of the circle as a vertex.
Activity 3.2 Constructing an Inscribed Hexagon Facilitation Notes In this activity, students construct a regular hexagon inscribed in a circle by creating six adjacent congruent chords, each the same length as the radius of the circle. They also determine the measure of each interior angle of a regular hexagon.
Ask a student to read the introduction aloud. Discuss as a class.
Questions to ask • What are you going to be constructing in this activity?
• How is this construction of an inscribed hexagon going to be different than the construction of an inscribed hexagon you completed in the previous activity?
• Do you think this construction will be easier or more difficult than the previous construction? Why?
LESSON 3: Into the Ring • 315G Have students work with a partner or in a group to complete Question 1. Share responses as a class.
Questions to ask • What is a chord?
• What is the first step in this process?
• Where is the radius of circle A? How did you duplicate it?
• Where did you locate the first chord around the circle? Does it make a difference where you place it?
• Do you think the radius of a circle will always map onto the circumference of the circle exactly six times? Do you suppose this happens with any size circle?
• Is the figure formed by connecting the endpoints of the chords a regular hexagon? How do you know?
Differentiation strategies To extend the activity, • Have students use what they know about chords and perpendicular bisectors to determine the exact center of the circle in the example diagram shown.
• Ask students to draw several circles of different sizes and show that the radius of each circle will always map onto the circumference of the circle to create exactly six congruent chords.
Have students work with a partner or in a group to complete Questions 2 and 3. Share responses as a class.
Misconception Students may misunderstand what is meant by interior angles. They may think the term interior angles refers to the six central angles formed by six equilateral triangles and exterior angles refers to the six angles of the hexagon with vertices at the circumference of the circle. If that is the case, clarify the difference between interior angles of the hexagon and exterior angles of the hexagon.
Questions to ask • If all of the interior angles are the same measure, what is the measure of each angle?
• How can you use the information from the Getting Started to determine the measure of each interior angle of the regular hexagon?
• By drawing the radii to each of the endpoints of the six chords, six triangles are formed. Are these triangles congruent?
• Are the six congruent triangles formed by connecting the radii to each of the endpoints of the six chords equilateral triangles? How do you know?
• Where are the interior angles of a hexagon?
• Why is the measure of each interior angle of the regular hexagon 120° rather than 60°?
315H • TOPIC 1: Composing and Decomposing Shapes Differentiation strategy To extend the activity, have students construct other figures by expanding upon their construction of the hexagon.
• If every other vertex of the hexagon is connected to the center of the circle, a two-dimensional representation of a cube is created.
• If the arcs to create the chord lengths are extended inside of the circle, a petal pattern is created. Summary A regular hexagon can be inscribed in a circle by creating six adjacent congruent chords that are the same length as the radius of the circle.
Activity 3.3 Constructing an Inscribed Square Facilitation Notes In this activity, students construct a square inscribed in a circle using patty paper and rigid motion transformations. They use the Pythagorean Theorem to determine the side lengths of the square based on a radius, r. Students then use the converse of the Pythagorean Theorem to determine that each angle of the square has a measure of 90°. They explain that when a square is inscribed in a circle, the diagonals of the square are also diameters of the circle in which it is inscribed. Finally, students use the fact that diagonals of a square are perpendicular to construct a square inscribed in a circle.
Ask a student to read the introduction aloud. Discuss as a class.
Have students work with a partner or in a group to complete Questions 1 and 2. Share responses as a class.
As students work, look for Correct placement of the labels for vertices D and E. They will be referenced in another question.
LESSON 3: Into the Ring • 315I Questions to ask • Is / B a central angle or an inscribed angle?
• How do you know that ¯ BA and ¯ BC are congruent?
• Is each central angle created by the rotation of nABC also a right angle? How do you know?
• What properties regarding circles can you use to verify that when nABC is rotated, its vertices still lie on the circle?
Have students work with a partner or in a group to complete Questions 3 and 4. Share responses as a class.
Misconceptions Students sometimes have difficulty recognizing the different relationships that a single segment may have within a complex diagram.
• Students may not recognize that the diameters of the circles are also diagonals of the square, especially due to their horizontal and vertical orientation.
• When students are verifying that the angles of the square are each 90°, they may not recognize that a side of the square is also a hypotenuse of a right triangle.
A C B 2x x x 2 x 2 Questions to ask • Outline one of the triangles you are using to answer Question 3, part (a). How can you use the Pythagorean Theorem to determine the side lengths of the square?
• Outline one of the triangles you are using to answer Question 3, part (b). How can you use the Converse of the Pythagorean Theorem to determine that each angle of the square has a measure of 90°?
• How can you use the fact that the sum of the angles of a triangle is 180° to determine that each angle of the square has a measure of 90°?
Have students work with a partner or in a group to complete Question 5. Share responses as a class.
As students work, look for Whether patty paper or a compass and straightedge are used to complete the construction.
315J • TOPIC 1: Composing and Decomposing Shapes Questions to ask • How many different diameters can be drawn in circle O?
• How did you know that the second diameter you drew was perpendicular to the first diameter?
• How do you know this method works?
• Do you prefer to use transformations or diameters to inscribe a square inside a circle?
Summary A square can be inscribed in a circle by using the fact that the diagonals of a square are perpendicular to one another.
Activity 3.4 Bisecting an Angle Facilitation Notes In this activity, the term angle bisector is defined. Students analyze Worked Examples that demonstrate how to bisect an angle using patty paper and using a compass and straightedge. Students then bisect angles and use repeated reasoning to construct angles that are one-fourth and one-eighth the measure of a given angle.
Have students work with a partner or in a group to read the introduction, analyze the Worked Example, and complete Question 1. Share responses as a class.
Misconception Students may associate the size of an angle with the length of the sides forming the angle. The size of an angle is a degree measure and it is determined by the spread between the sides of the angle; the length of the sides have no bearing on the size of the angle. Provide examples to contradict students’ misunderstanding.
Questions to ask • How is an angle bisector different than a segment bisector?
• What do an angle bisector and a segment bisector have in common?
• How is an angle bisector different than a perpendicular bisector?
• Would the process change if the angle was an obtuse angle? A right angle?
LESSON 3: Into the Ring • 315K • Do all angles have an angle bisector?
• How many creases could Angela make through the vertex?
• Do all creases through the vertex divide the angle into two congruent angles?
• Can an angle have more than one bisector? Why not?
Have students work with a partner or in a group to analyze the Worked Example following Question 1, and complete Questions 2 through 4. Share responses as a class.
Misconception Students may assume that the compass must remain in the same position throughout all steps of this construction. The construction will still work if the compass is changed after step 1 to perform step 2, provided the new setting is used twice in step 2 and allows the arcs to intersect. All possible arc intersections will lie on the angle bisector.
Questions to ask • How do you know what setting to use on the compass to create the first arc?
• Is important to maintain the same compass setting from step 1 to step 2?
• If the compass setting is different in step 2 than it was in the previous step, will the construction still work?
• To bisect an angle, do you prefer construction tools or patty paper? Why?
• How many times will the bisection construction need to be performed if the goal is to create an angle that is one-fourth the measure of the given angle?
• How many times will the bisection construction need to be performed if the goal is to create an angle that is one-eighth the measure of the given angle?
• What other fractions of an angle can you construct?
Differentiation strategy To extend the activity, have students create an angle 3 __ 4 the measure of the given angle and an angle that is 5 __ 2 times the measure of the given angle, then describe the strategy used to create the angles.
Summary An angle bisector is a line, segment, or ray that is drawn through the vertex of an angle and divides the angle into two congruent angles. Angle bisectors can be constructed using patty paper or a compass and straightedge.
315L • TOPIC 1: Composing and Decomposing Shapes Activity 3.5 Constructing and Inscribing an Equilateral Triangle Facilitation Notes In this activity, students determine that given three line segments of the same length, an equilateral triangle can always be constructed. They use this reasoning and their knowledge of constructions to construct an equilateral triangle using a compass and straightedge. They then show two different ways to construct an inscribed equilateral triangle in a circle and describe the process they used in each construction.
Have students work with a partner or in a group to complete Question 1. Share responses as a class.
As students work, look for The use of patty paper or a compass and straightedge to attempt to construct an equilateral triangle. The purpose of this question is for students to understand that given three line segments of equal length, a triangle can always be constructed. Share explanations and construction attempts; however, do not provide instruction regarding how to construct an equilateral triangle at this time. Differentiation strategy To extend the activity, discuss how the Triangle Inequality Theorem could support Sophie’s conclusion. The Triangle Inequality Theorem states that the sum of the lengths of any two sides of a triangle is greater than the length of the third side.
Questions to ask • How is this situation different from the Getting Started?
• Do you think you can construct an equilateral triangle without using a circle and its radii to support your construction?
• How do you know how much to slant the sides so that they connect at their endpoints to make a triangle?
• How can you use patty paper to support your conclusion?
• Were you able to construct an equilateral triangle using a compass and straightedge? If so, explain your steps.
• What is the measure of each interior angle of an equilateral triangle? How do you know?
Have students work with a partner or in a group to complete Question 2. Share responses as a class.
LESSON 3: Into the Ring • 315M Differentiation strategies • To scaffold support for using the space to create the triangle provided below the horizontal segment, suggest that students trace the segment and complete the construction on a piece of patty paper in the orientation they prefer.
• To assist all students, if the construction using a compass and straightedge is not shared by a student in the class discussion, show that strategy. Then, question students as to why this method works. See the answer to Question 2 for the diagram and strategy.
Questions to ask • Were the steps any different when you had to construct your triangle below the horizontal line? Why or why not?
• Did you perform the construction using patty paper or a compass and straightedge?
• How did you construct an equilateral triangle using patty paper?
• How did you construct an equilateral triangle using a compass and straightedge?
• Did you place the given line segment in a circle in order to complete your construction? If so, how was that helpful?
Have students work with a partner or in a group to complete Question 3. Share responses as a class.
Differentiation strategies • To scaffold support, suggest that students construct a hexagon inscribed in a circle using six chords around the circumference of the circle. Then, ask them to modify that procedure to create a triangle inscribed in a circle.
• To assist all students, make sure they are familiar with at least one method of constructing an equilateral triangle inscribed in a circle. Developing a second method may be a challenge for many students. See the answers for possible methods.
Questions to ask • How can you use what you know about constructing a hexagon inscribed in a circle to support your thinking?
• What made constructing the equilateral triangle inscribed in a circle more difficult than constructing the equilateral triangle in Question 2?
• Where did you get stuck? What additional information did you need?
• Is there any other information you can use from the relationships you know exist in the diagram?
• Is there any way you can use the six equilateral triangles you constructed around the radius of a circle to create a regular hexagon to help in this construction?
315N • TOPIC 1: Composing and Decomposing Shapes Summary Given three line segments of the same length, an equilateral triangle can always be constructed. An equilateral triangle can be inscribed in a circle using construction tools and modifying the procedure used for inscribing a hexagon in a circle.
Talk the Talk: Playing the Angles Facilitation Notes In this activity, students are given a circle with a central angle and its intercepted arc. Within the circle, they draw two inscribed angles that have the same intercepted arc as the central angle. Students then conjecture about the relationship between the measure of a central angle and the measure of any inscribed angle in the same circle that intercepts the same arc. Next, students use the constructions they have learned to construct a 75° angle and inscribe a regular octagon in a given circle.
Have students work with a partner or in a group to complete Question 1.
Share responses as a class.
Differentiation strategies To assist all students, have them extend and outline each angle with a different colored pencil to see that they intercept the same arc.
Questions to ask • What is meant by an intercepted arc?
• Can two different inscribed angles in the same circle have the same intercepted arc?
• If two inscribed angles in the same circle intercept the same arc, will those angles have the same measure? How can you tell?
• Can two different central angles in the same circle have the same intercepted arc?
• Which angle appears larger, the central angle or the inscribed angle?
• Why does it make sense that the angle that has a vertex closer to the intercepted arc is larger than the angle that has a vertex further from the intercepted arc?
• Does that relationship exist in others’ diagrams?
• Do you think the relationship you noticed occurs in any circle when a central angle and an inscribed angle intercept the same arc? Why or why not?
Have students work with a partner or in a group to complete Question 2. Share responses as a class.
DEMONSTRATE LESSON 3: Into the Ring • 315O Differentiation strategy To scaffold support, suggest that students create a sketch of their strategy before completing it with construction tools.
As students work, look for Unnecessary duplication of angles. Students may think they have to construct a 75° angle that is isolated from all other angles. The 75° angle they construct may be included as part of a 120° angle. Students just need to identify the 75° angle using vertices that have been labeled.
Questions to ask • Why did you begin by constructing an equilateral triangle?
• Why are two adjacent 60° angles needed?
• How can a 60° angle be used to determine a 75° angle?
• How could a bisection construction performed on a 60° angle be helpful?
• If you performed a double bisection construction on a 60° angle, what would be the measure of the smallest angle?
• Does it matter which 30° angle you bisect?
• How can a 15° angle be used to determine a 75° angle?
Have students work with a partner or in a group to complete Question 3.
Share responses as a class.
As students work, look for Different strategies to identify the center of the circle, such as: • Constructing a diameter of the circle and its perpendicular bisector.
• Tracing the circle on patty paper and using folds to determine the center of the circle.
Differentiation strategies To assist all students, discuss how the different strategies to identify the center of the circle are related. For example, • Using the diameter of a circle and its perpendicular bisector is a special case of using the intersection of the perpendicular bisectors of two chords; this is true because the diameter of a circle is a chord that passes through the center of the circle.
• Using patty paper to determine the center of the circle is related to constructing a diameter and its perpendicular bisector.
Questions to ask • How did you locate the center of the circle?
• What is another way to identify the center of the circle?
• How can you use the method of constructing a square inscribed in a circle, specifically using the diagonals of a square, to determine the center of the circle?
• How can the perpendicular bisectors of two chords be used to locate the center of a circle?
315P • TOPIC 1: Composing and Decomposing Shapes • Why did you divide the circle into four equal parts?
• What construction divides each of the four right angles into two equivalent angles?
• How did you use the fact that you have eight equivalent angles to construct a regular octagon inscribed in a circle?
• What is the measure of each interior angle of an octagon? What calculation did you use to identify this angle measure?
Differentiation strategies To extend the activity, • Ask students what other angles they can construct using a combination of 90° angles and 60° angles, along with angle bisectors and multiples of angle measures. Then, have them construct some of those angles.
• Have students construct a regular dodecagon inscribed in a circle.
Summary Within a circle, an relationship exists between the measure of a central angle and the measure of any inscribed angle that intercepts the same arc of the circle. Constructions of 90° angles, 60° angles, and angle bisectors can be used to construct angles of other measures.
LESSON 3: Into the Ring • 315 Warm Up Answers 1. Sample answer. Using a reflection across ¯ CH , ¯ DE > ¯ AB .
2. Sample answer. Using the center of rotation as the center of the star, by a clockwise rotation of ( 360 _ 5 ) ° , /HIJ > /FGH .
3. Sample answer. Using the center of rotation as the center of the star, by a counterclockwise rotation of ( 360 ___ 5 ) ° , /IHG > /AJI .
LESSON 3: Into the Ring • 315 3 Into the Ring Constructing an Inscribed Regular Polygon Learning Goals • Use construction tools to duplicate an angle.
• Construct a regular hexagon inscribed in a circle using a variety of strategies.
• Construct a square inscribed in a circle using patty paper and using a compass and straightedge.
• Use construction tools to bisect an angle.
• Construct an equilateral triangle using a compass and straightedge.
• Construct an equilateral triangle inscribed in a circle.
You have used construction tools to construct figures in a plane. How can you construct a regular polygon inscribed in a circle?
Key Terms • inscribed polygon • angle bisector Warm Up Use patty paper to identify a geometric figure that is congruent to each figure in the diagram. Describe the transformation(s) you used.
A B C D E F G H J I 1. ¯ AB 2. /FGH 3. /AJI Geo_M02_T01_L03_Student Lesson.indd 315 Geo_M02_T01_L03_Student Lesson.indd 315 03/12/22 11:45 AM 03/12/22 11:45 AM 316 • TOPIC 1: Composing and Decomposing Shapes Answers 1a. A B C 1b. A B C The length of ¯ AC is equal to the radius of circle A.
1c. Triangle ABC is an equilateral triangle because both ¯ AB and ¯ AC are radii, and ¯ BC is congruent to ¯ AB . All three side lengths are equal.
1d. The angle measures of n ABC are each 60°.
ELL Tip One term that is used throughout the lesson is duplicate. Determine whether students are familiar with the term. Note that duplicate is used as a noun, an adjective, and a verb. Define it as a noun as one of two or more identical things. Define it as a verb as to make an exact copy of something. Define it as an adjective as describing something that is exactly like something else. Read aloud the first sentence in the section, “You can use your construction skills to duplicate a line segment inside a circle.” Discuss the use of duplicate in the context of the sentence as a verb, which means to make an exact copy of a line segment inside a circle.
316 • TOPIC 1: Composing and Decomposing Shapes GETTING STARTED Duped You can use your construction skills to duplicate a line segment inside a circle.
1. Consider circle A.
A B a. Construct ¯ BC congruent to ¯ AB in circle A so that point C lies on circle A.
b. Connect points A and C. What do you know about the length of ¯ AC ?
c. What type of triangle is △ABC? Explain your reasoning.
d. What do you know about the angle measures of △ABC?
Geo_M02_T01_L03_Student Lesson.indd 316 Geo_M02_T01_L03_Student Lesson.indd 316 03/12/22 11:45 AM 03/12/22 11:45 AM LESSON 3: Into the Ring • 317 Answer 1. Check students’ diagrams.
2. First, I labeled /A as /BAC. Then I duplicated /BAC and labeled it as /B’A’C’. Then, I used ¯ A’B’ as my starter line, duplicated /BAC again, and labeled it as /D’A’C’. Angle D’A’C’ is twice the measure of /BAC.
LESSON 3: Into the Ring • 317 Duplicating an Angle ACTIVIT Y 3.1 Not only can you copy a line segment using construction, remember that you can also copy an angle.
B 1. Duplicate angle B. Verify with patty paper.
2. Construct an angle that is twice the measure of ∠A. Then explain how you performed the construction.
A Geo_M02_T01_L03_Student Lesson.indd 317 Geo_M02_T01_L03_Student Lesson.indd 317 03/12/22 11:45 AM 03/12/22 11:45 AM 318 • TOPIC 1: Composing and Decomposing Shapes Answers 3. When duplicating a line segment and an angle, the same tools are used. However, when duplicating a line segment, I have to make only one mark. When duplicating an angle, I must make several marks because I must duplicate both rays of the angle.
4. The six triangles compose a regular hexagon.
318 • TOPIC 1: Composing and Decomposing Shapes 3. How is duplicating an angle similar to duplicating a line segment? How is it different?
4. In the Getting Started, you constructed an equilateral triangle ABC. Duplicate the 60° angle shown to construct 5 additional equilateral triangles that share the center of the circle as a vertex. What do you notice about the polygon you have constructed?
A B C Geo_M02_T01_L03_Student Lesson.indd 318 Geo_M02_T01_L03_Student Lesson.indd 318 03/12/22 11:45 AM 03/12/22 11:45 AM LESSON 3: Into the Ring • 319 ELL Tip Assess students’ prior knowledge of the term equidistant. If students are unfamiliar, break the term down into two parts: equal and distance. Remind students of common synonyms for equal, which include the same, identical, alike, and equivalent. Note that distant in the context of the term equidistant is used to describe the amount of space between two objects. Read aloud Question 1, part (b) in the activity, and remind students that endpoints of chords that are equidistant from each other are endpoints that have the exact same distance from each other.
Answers 1a–c. Check students’ diagrams.
2. I can use a compass and straightedge to duplicate one angle from my diagram onto patty paper, then use the patty paper to verify that the angles of the hexagon are congruent.
3. The measures of the interior angles of a regular hexagon are 120°.
LESSON 3: Into the Ring • 319 Constructing an Inscribed Hexagon ACTIVIT Y 3.2 In the Getting Started, you created an equilateral triangle by duplicating the length of the radius to create a chord. You can use what you know about copying a line segment to construct a regular hexagon inscribed in a circle using a compass and straightedge.
1. Use circle A to complete your construction.
a. Draw and copy the radius of circle A.
b. Create 6 chords around the circumference of circle A so that they are congruent to the radius and their endpoints are equidistant from each other.
c. Connect consecutive endpoints of the chords.
The figure you constructed is a regular hexagon inscribed in a circle.
2. How can you copy an angle to verify that the inscribed hexagon is a regular hexagon?
3. What do you know about the measures of the interior angles of a regular hexagon?
Remember: A regular polygon is a polygon with all sides congruent and all angles congruent.
An inscribed polygon is a polygon drawn inside another polygon or circle in which all the vertices of the interior polygon lie on the outer figure.
A Geo_M02_T01_L03_Student Lesson.indd 319 Geo_M02_T01_L03_Student Lesson.indd 319 03/12/22 11:45 AM 03/12/22 11:45 AM 320 • TOPIC 1: Composing and Decomposing Shapes Answers 1a. Central angle 1b. Each line segment is a radius.
1c. Chord 2. The rotation of the triangle creates an additional triangle with two sides that are radii and one side that is a chord. The endpoints of every radius lie on the circle.
320 • TOPIC 1: Composing and Decomposing Shapes In a previous activity, you constructed a square given a side length. Let’s now consider how to inscribe a square inside a circle. One way to create a square is to use transformations.
B C A 1. Circle B contains a right triangle, △ABC, with points A and C on the circle and point B at the center. Describe each element using a term associated with a circle.
a. ∠B b. ¯ BC and ¯ BA c. ¯ AC 2. Use patty paper to rotate the right triangle to create square ACDE. How do you know all 4 vertices lie on the circle?
Constructing an Inscribed Square ACTIVIT Y 3.3 Geo_M02_T01_L03_Student Lesson.indd 320 Geo_M02_T01_L03_Student Lesson.indd 320 03/12/22 11:45 AM 03/12/22 11:45 AM LESSON 3: Into the Ring • 321 Answers 3a. Each side of the square is the hypotenuse of a right triangle with leg lengths equal to the length of the radius of the circle, r. Using the Pythagorean Theorem, each side length is equal to x √ __ 2 .
3b. The Converse of the Pythagorean Theorem says that if the sum of the squares of the leg lengths is equal to the square of the hypotenuse length, then the triangle is a right triangle. The diameter of the circle is equal to 2x. Each side length is equal to x √ __ 2 . So the diameter squared, (2x)2, or 4x2, is equal to the sum of the squares of two side lengths. ( √ __ 2 ? x)2 1 ( √ __ 2 ? x)2 5 2x2 1 2x2 5 4x2.
4a. Segments AD and CE are both diameters of the circle and diagonals of the inscribed square.
4b. Segments AD and CE are perpendicular to each other.
LESSON 3: Into the Ring • 321 3. Verify that you created a square inscribed in circle B. To do this, you need to show that all side lengths of the square are equal and each interior angle measures 90°.
a. Show that all side lengths are equal. Suppose BC = x . Determine each side length of the square. Label each length.
b. Verify that each interior angle measure of the square is 90°.
4. Consider ¯ AD and ¯ CE .
a. Describe each line segment in terms of the circle and in terms of the square.
b. Describe the relationship between ¯ AD and ¯ CE .
Remember: Recall that the Pythagorean Theorem says that a2 1 b2 5 c2 .
b c a The Converse of the Pythagorean Theorem says that if a2 1 b2 5 c2 , then the triangle is a right triangle.
Geo_M02_T01_L03_Student Lesson.indd 321 Geo_M02_T01_L03_Student Lesson.indd 321 03/12/22 11:45 AM 03/12/22 11:45 AM 322 • TOPIC 1: Composing and Decomposing Shapes Answer 5. Check students’ constructions.
322 • TOPIC 1: Composing and Decomposing Shapes 5. Use diagonals to construct a square inscribed in circle O. Describe your process.
O Is an inscribed square a cyclic quadrilateral?
Think about: Geo_M02_T01_L03_Student Lesson.indd 322 Geo_M02_T01_L03_Student Lesson.indd 322 03/12/22 11:45 AM 03/12/22 11:45 AM LESSON 3: Into the Ring • 323 Answer 1. No. Angela is not correct. Just because the crease goes through the vertex does not mean it bisects the angle. To bisect the angle, the crease must divide the angle into two angles of equal measure.
LESSON 3: Into the Ring • 323 Bisecting an Angle ACTIVIT Y 3.4 Just as line segments can be bisected, angles can be bisected too. If a ray is drawn through the vertex of an angle and divides the angle into two angles of equal measure, or two congruent angles, this ray is called an angle bisector. The construction used to create an angle bisector is called bisecting an angle.
One way to bisect an angle is using patty paper.
Worked Example Draw an angle on the patty paper.
Fold the patty paper so that the rays lie on top of each other. The fold should pass through the vertex of the angle.
Open the patty paper. The crease represents the angle bisector.
Lines and line segments can also bisect an angle.
1. Angela states that as long as the crease goes through the vertex, it is an angle bisector. Is she correct? Why or why not?
Geo_M02_T01_L03_Student Lesson.indd 323 Geo_M02_T01_L03_Student Lesson.indd 323 03/12/22 11:45 AM 03/12/22 11:45 AM 324 • TOPIC 1: Composing and Decomposing Shapes Answers 2. A C D B 3. B H C D P First, bisect /H to form two angles, /BHP and /PHC, that are each one-half the measure of /H. Then, bisect either /BHP or /PHC to form two angles that are each one-fourth the measure of /BHC. /PHD and /DHC are each one-fourth the measure of /BHC.
4. I must bisect one of the angles that are one-fourth the measure of /BHC. The resulting pair of angles will both be one-eighth the measure of /BHC.
324 • TOPIC 1: Composing and Decomposing Shapes You can also bisect an angle using a compass and a straightedge.
2. Construct the bisector of ∠A. Use patty paper to verify your construction.
3. Construct an angle that is one-fourth the measure of ∠H. Explain how you performed the construction.
4. Describe how to construct an angle that is one-eighth the measure of ∠H from Question 3.
Worked Example Construct an Arc Place the compass at C. Construct an arc that intersects both sides of the angle. Label the intersections A and B.
Construct Another Arc Place the compass at A. Construct an arc. Then, place the compass point at B. Using the same radius, construct another arc.
Construct a Ray Label the intersection of the two arcs D. Use a straightedge to construct a ray through C and D. Ray CD bisects /C.
C B A C B A C B D A A H Geo_M02_T01_L03_Student Lesson.indd 324 Geo_M02_T01_L03_Student Lesson.indd 324 03/12/22 11:45 AM 03/12/22 11:45 AM LESSON 3: Into the Ring • 325 Answers 1. Sophie is correct. Roberto didn’t realize that three congruent segments can always be connected at their endpoints to create an equilateral triangle. According to the Triangle Inequality Theorem, if the sides lengths are x, x, and x, a triangle can be constructed because 2x . x.
2. Sample answers. I set my compass to the length of ¯ AB . I placed my compass point on point A and made and arc, and then on point B and made an arc. I labeled the intersection of the two arcs as point C and connected point A and point B to point C. All 3 sides of the triangle have the same length. A B C I used patty paper and traced two additional line segments. I then turned the patty paper so that the endpoints of the line segments connected to make a triangle. I then traced the triangle on a single piece of patty paper. I repeated the process in the Getting Started. I used the side length as my radius, ¯ AB , and constructed a circle. I created a chord, ¯ BC , the same length as the radius. Then, I connected the other endpoint of my chord to create another radius, ¯ AC , and completed the triangle.
LESSON 3: Into the Ring • 325 In this activity, you will construct an equilateral triangle. To perform the construction, you will use only a compass and straightedge and rely on the basic geometric constructions you have learned, such as duplicating a line segment.
2. Construct an equilateral triangle using the given side length. Then, describe the steps you performed for the construction.
Constructing and Inscribing an Equilateral Triangle ACTIVIT Y 3.5 1. Sophie claims that she can construct an equilateral triangle by constructing and then duplicating a line segment three times and having the endpoints of all three line segments intersect. Roberto thinks that Sophie’s method will not result in an equilateral triangle. Is Sophie correct? Explain why Sophie’s reasoning or Roberto’s reasoning is incorrect.
Remember: An equilateral triangle has three congruent sides and three congruent angles.
Geo_M02_T01_L03_Student Lesson.indd 325 Geo_M02_T01_L03_Student Lesson.indd 325 03/12/22 11:45 AM 03/12/22 11:45 AM 326 • TOPIC 1: Composing and Decomposing Shapes Answers 3. Sample answers. I used the fact that all sides of an equilateral triangle have equal lengths. I set my compass to the length of the radius and placed 6 points on the circumference of the circle as if I was creating a regular hexagon. Then, I connected every other point to create an equilateral triangle. Triangle XYZ is an equilateral triangle. Y Z Q X I used the fact that all angles of an equilateral triangle have a measure of 60°. First I created two equilateral triangles, nABP and nPBC, that shared a common side, using the radius as the side lengths. Then, I took the two adjacent 60° angles, /ABP and /PBC, and bisected each of them to create a new 60° angle, /GBH, with the center of the circle, Point P, on the bisector of the angle. I then connected point G and point H to create the third side of the triangle. Triangle BGH is an inscribed equilateral triangle. G H P B A C I constructed diameter ¯ AC . I determined the midpoint, point M, of radius ¯ QC . I then constructed a perpendicular bisector through point M, creating chord ¯ DE . I connected point D and point E to the endpoint, A, of the diameter to create the other two sides of the triangle. Triangle ADE is an equilateral triangle. A C Q M D E 326 • TOPIC 1: Composing and Decomposing Shapes An inscribed equilateral triangle is an equilateral triangle inside a circle with all of its vertices touching the circle.
3. Show two different ways to construct an inscribed equilateral triangle using each circle given. Explain your process for each construction.
P Q How can you use the process for constructing an inscribed regular hexagon to inscribe an equilateral triangle?
Think about: Geo_M02_T01_L03_Student Lesson.indd 326 Geo_M02_T01_L03_Student Lesson.indd 326 03/12/22 11:45 AM 03/12/22 11:45 AM LESSON 3: Into the Ring • 327 Answers 1a. Sample answer.
O C A B D 1b. Sample answer. Conjecture: Within a circle, the measure of the central angle of a circle is twice the measure of an inscribed angle which intercepts the same arc of the circle.
ELL Tip Students are familiar with the term intercepts referring to the points where a line crosses over an axis on a coordinate plane. For example, “Identify the x- and y-intercepts of the line shown on the graph.” Remind students that intercepts in verb form means to cross over something. Refer to the scenario, “… the measure of an inscribed angle which intercepts the same arc of the circle.” Help students visualize the meaning of intercepts using the diagram. You may also want to compare the meanings of intercepts and intersects.
LESSON 3: Into the Ring • 327 NOTES TALK the TALK Playing the Angles Within a circle, an interesting relationship exists between the measure of a central angle and the measure of any inscribed angle that intercepts the same arc of the circle.
1. Consider Circle O.
O C A a. Draw two different inscribed angles that intercept the same arc as the one intercepted by ∠COA.
b. Use patty paper to analyze your diagram. Make a conjecture regarding a relationship you notice.
Ask yourself: What do an inscribed angle and an inscribed polygon have in common?
Geo_M02_T01_L03_Student Lesson.indd 327 Geo_M02_T01_L03_Student Lesson.indd 327 03/12/22 11:45 AM 03/12/22 11:45 AM 328 • TOPIC 1: Composing and Decomposing Shapes X=35p7.173 Y=-2p1.933 Scale 66.68% Answers 2. A F E D B C I started with ¯ AB , constructed the vertices for an equilateral triangle, nABC, but connected only two sides to create 60° angle, /CBA. I duplicated the 60° angle to create adjacent angle, /ABD. I bisected /ABD to create 30° angle, /ABE. Then, I bisected /ABE to create angle 15° angle, /ABF. angle CBF is a 75° angle.
3. A C Z E Y B W D X I drew diameter ¯ BC . Then, I constructed perpendicular bisector ¯ DE . I bisected /DAC and extended the angle bisector to also bisect /BAE. I then bisected /DAB and extended the angle bisector to also bisect /CAE. I then connected the eight chords on the circumference of the circle to inscribe a regular octagon.
328 • TOPIC 1: Composing and Decomposing Shapes NOTES 2. Use the constructions you have learned to construct an angle with a measure of 75° using only a compass and straightedge. Summarize your process.
You have constructed a regular quadrilateral (square), a regular hexagon, and a regular triangle (equilateral triangle) inscribed in a circle. What other regular polygons can you inscribe in a circle?
3. Inscribe a regular octagon in the circle provided, if possible.
Geo_M02_T01_L03_Student Lesson.indd 328 Geo_M02_T01_L03_Student Lesson.indd 328 03/12/22 11:45 AM 03/12/22 11:45 AM |
8513 | https://www.korpisworld.com/Mathematics/Calculus%20Maximus/NOTES/NOTES%2005.5%20Partial%20Fractions%20&%20Logistic%20Growth.pdf | Calculus Maximus Notes5.5: Partial Fractions & Logistic Growth Page 1 of 6 §5.5—Partial Fractions & Logistic Growth Many things that grow exponentially cannot continue to do so indefinitely. This is a good thing. Imagine if human population growth went unchecked: we’d have people covering every square inch of the Earth, then the Galaxy, then the Universe. Imagine trying to find a parking place then! After a while, things that start off growing exponentially begin to compete for resources like food, water, money, and parking spaces. The growth begins to taper off as it approaches some carrying capacity of the system. In this case, the growth rate is not only proportional to the current value, but also how far the current value is from the carrying capacity. Imagine a rumor spreading throughout a school of 2000 students. The rate at which the rumor spreads is directly proportional to BOTH the students who have heard the rumor AND the students who have yet to hear the rumor as the number of people hearing the rumor approaches 2000. The curve for the spread of the rumor might look like something shown below. This type of curve and growth is called Logistic Growth. For quantities, y, that grow logistically with a carrying capacity of y L = , we can state the relation mathematically the following way: ( ) dy ky L y dt = − Solving this differential equation requires an new integration technique called Integration by Partial Fraction Decomposition. Example 1: Add then simplify by finding a common denominator: 5 2 2 1 3 x x − − − Calculus Maximus Notes5.5: Partial Fractions & Logistic Growth Page 2 of 6 Example 2: Evaluate 2 13 2 7 3 x dx x x − − + ∫ On the AP Exam, you’ll be expected to decompose rational expressions involving non-repeating, linear factors in the denominator. For cases as this one, there is a very slick method you can use that was developed by Oliver “No So Much On The” Heaviside, called the “Heaviside Cover-Up” Method. Here’s an example we’ll do under the watchful eye of Mr. Heaviside. Example 3: Evaluate 2 5 2 x dx x x + + − ∫ Calculus Maximus Notes5.5: Partial Fractions & Logistic Growth Page 3 of 6 In order for partial fraction decomposition to work, the degree of the denominator must be greater than that of the numerator. When it’s not we’ll use our tried and true method of long dividing first. Example 4: Evaluate 4 2 3 1 1 x dx x + − ∫ Back to Logistic Growth. Example 5: Solve the differential equation ( ) dy ky L y dt = − Calculus Maximus Notes5.5: Partial Fractions & Logistic Growth Page 4 of 6 Memorize the solution: Logistic Growth If ( ) dy ky L y dt = − , then 1 Lkt L y Ce− = + Think “Lice Minus Licked.” It’s gross, but it works. Notice the prominence of our carrying capacity value L in each form of the equation. This is very important, especially when asked for the limit at infinity OR when asked to find the y-value when the y-values are increasing most rapidly (i.e. the inflection value: 2 L y = ). Example 6: (Calculator) The population of Alaska since from 1900 to 2000 can be modeled by the following logistic equation. 0.0516 895598 ( ) 1 71.57 t P t e− = + where P is the population and t years after 1900, with 0 t = corresponding to 1900. a) What is the predicted population of Alaska in 2020? b) How fast was the population of Alaska changing in 1920? In 1940? In 1999? c) When was Alaska growing the fastest, and what was the population then? d) What information does the equation tell us about the population of Alaska in the long run? − Calculus Maximus Notes5.5: Partial Fractions & Logistic Growth Page 5 of 6 The AP exam loves to ask questions that require you to recognize the parameters of logistic growth for either the equation or the differential equation written in a DIFFERERNT FORMAT. This requires you to manipulate the equation to fit one of the two standard forms below: ( ) 1 Lkt dy L ky L y y dt Ce− = − → = + Example 7: The growth rate of a population P of bears in a newly established wildlife preserve is modeled by the differential equation ( ) 0.008 100 dP P P dt = − , where t measured in years. a) What is the carrying capacity for bears in this wildlife preserve? b) What is the bear population when the population is growing the fastest? c) What is the rate of change of population when it is growing the fastest? Example 8: Suppose that a population develops according to the logistic differential equation 2 0.2 0.002 dP P P dt = − , where t is measured in weeks, 0 t ≥ . a) If (0) 5 P = , what is lim ( ) t P t →∞ ? b) If (0) 60 P = , what is lim ( ) t P t →∞ ? c) If (0) 120 P = , what is lim ( ) t P t →∞ ? d) Sketch the solution curves for a), b), and c). Which one has an inflection point? e) Which solutions are strictly increasing? decreasing? Calculus Maximus Notes5.5: Partial Fractions & Logistic Growth Page 6 of 6 Example 8: The rate at which the flu spreads through a community is modeled by the logistic differential equation ( ) 0.001 3000 dP P P dt = − , where t is measured in days, 0 t ≥ . a) If (0) 50 P = , solve for P as a function of t. b) Use your solution to a) to find the size of the population when 2 t = days. c) Use your solution to a) to find the number of days that have occurred when the flu is spreading the fastest. |
8514 | https://www.gauthmath.com/solution/1835092942989393/x-square-root-of-x-3-which-inequality-can-be-used-to-find-the-domain-of-fx-x-3-0 | Solved: (x)=sqrt(x-3) , which inequality can be used to find the domain of f(x) ? x-3≥ 0 sqrt(x-3) [Math]
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Question
(x)=sqrt(x-3) , which inequality can be used to find the domain of f(x) ? x-3≥ 0 sqrt(x-3)≥ 0 x-3≤ 0 sqrt(x-3)≤ 0
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Answer
The answer is Option 1: $$x-3\geq 0$$x−3≥0
Explanation
Option 1: $$x-3\geq 0$$x−3≥0
For the function $$f(x) = \sqrt{x-3}$$f(x)=x−3 to be defined in the real number system, the expression inside the square root must be non-negative. This means that $$x-3$$x−3 must be greater than or equal to 0. So Option 1 is correct.
Option 2: $$\sqrt {x-3}\geq 0$$x−3≥0
While it is true that the square root of any non-negative number is always greater than or equal to 0, this inequality does not help us find the domain of the function. It is a property of the square root function itself, not a condition to determine the possible values of $$x$$x
Option 3: $$x-3\leq 0$$x−3≤0
This inequality implies that $$x-3$$x−3 is less than or equal to 0, which means $$x$$x is less than or equal to 3. This would result in a negative number inside the square root, which is not allowed for real-valued functions.
Option 4: $$\sqrt {x-3}\leq 0$$x−3≤0
Since the square root of a real number is always non-negative, the only way for $$\sqrt{x-3}$$x−3 to be less than or equal to 0 is if it is equal to 0. This would imply $$x-3 = 0$$x−3=0, so $$x = 3$$x=3. However, this inequality does not define the entire domain of the function, only a single point.
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8515 | https://www.pw.live/school-prep/exams/consecutive-integers-formula | Consecutive Integers Formula: Even, Odd and Solved Examples
Learn the concise formula for calculating consecutive integers and solve mathematical sequences effortlessly. In essence, consecutive integers are whole numbers that follow one another in a sequence. Find examples and explanations here.
Ranvijay Singh5 Oct, 2023
In the realm of mathematics, you frequently encounter problem statements that task you with determining two or more
Consecutive Integers
when their sum or difference is provided. Often, these problem statements specify whether these consecutive numbers should be odd or even. To tackle these problems, you typically introduce a variable to represent one of the integers and then express the others as consecutive integers. Let me illustrate this with an example.
Suppose you need to identify two consecutive integers whose sum equals 89. How do you approach this problem? Initially, you assign a variable, say 'x,' to one of the unknown integers. Subsequently, since the problem calls for two consecutive integers, the integer immediately following 'x' would be represented as (x + 1). In accordance with the problem, the sum of 'x' and (x + 1) should equal 89. This is expressed in the form of an equation: x + (x + 1) = 89. Solving this equation yields 'x' as 44 and the subsequent integer (x + 1) as 45, and together they sum up to 89.
Similarly, the formula for consecutive integers finds application in various mathematical problems. In this section, we will explore some of these formulas.
What are Consecutive Integers?
Whenever we need to enumerate or arrange items in a specific order, consecutive integers come into play. In essence, consecutive integers are whole numbers that follow one another in a sequence, with a consistent fixed difference between them. For instance, when examining the sequence of natural numbers, such as 1, 2, 3, 4, 5, and 6, we observe that each integer is precisely one unit apart from the next. Likewise, we can construct lists of consecutive even integers, consecutive odd integers, and various other arrangements. It's essential to note that the difference between consecutive integers remains constant, and as they are integers, they can take on positive, negative, or zero values, but they do not encompass fractions or decimals.
Consecutive Integer Formula
When dealing with integers, the consecutive integers following a given integer 'n' are typically expressed as (n + 1) and (n + 2). For instance, if we take 'n' to be 1, the consecutive integers are calculated as (1 + 1) and (1 + 2), resulting in 2 and 3.
Therefore, the general formula for representing consecutive integers is:
n, n + 1, n + 2, n + 3, ...
Even Consecutive Integer Formula
In the realm of mathematics, we represent an even integer as 2n. When we have 2n as an even integer, the next two consecutive even integers can be expressed as (2n + 2) and (2n + 4). For instance, if we take 2n to be 4, which is indeed an even integer, we can calculate its consecutive even integers as (4 + 2) and (4 + 4), resulting in 6 and 8.
Thus, the formula for representing consecutive even integers can be written as:
2n, 2n + 2, 2n + 4, 2n + 6, ...
It's important to note that the difference between two even consecutive integers in this sequence is always 2, ensuring that they remain even integers throughout.
Also Check –
Ratio and proportion Formula
Odd Consecutive Integer Formula
In the realm of mathematics, an odd integer is typically represented as 2n + 1. When we start with 2n + 1 as an odd integer, the next two consecutive odd integers can be expressed as (2n + 3) and (2n + 5). For instance, if we consider 2n + 1 to be 7, which is indeed an odd integer, we can calculate its consecutive odd integers as (7 + 2) and (7 + 4), resulting in 9 and 11.
As a result, the formula for representing consecutive odd integers can be written as:
2n + 1, 2n + 3, 2n + 5, 2n + 7, ...
It's important to note that the difference between two consecutive odd integers in this sequence is always 2, ensuring that they remain odd integers throughout.
Also Check –
Line and Angles Formula
Solved Examples
Question: Determine three consecutive integers that sum up to 76.
Solution:
Let's use 'n' to represent the first integer. Therefore, the next three consecutive integers would be n + 1, n + 2, and n + 3.
Now, we want their sum to equal 76:
n + (n + 1) + (n + 2) + (n + 3) = 76
Combine like terms:
4n + 6 = 76
Subtract 6 from both sides:
4n = 70
Divide by 4:
n = 70 / 4
n = 17.5
Since 'n' must be an integer, these consecutive integers won't work to sum up to 76.
Also Check –
Introduction to Graph Formula
Question: Find three consecutive even integers that start from -8.
Solution:
Let's represent -8 as 2n since it's an even integer. The next three consecutive even integers would be 2n + 2, 2n + 4, and 2n + 6.
So, we have:
2n + 2 = -8
2n + 4 = -6
2n + 6 = -4
Now, let's find the value of 'n':
2n + 2 = -8
Subtract 2 from both sides:
2n = -8 - 2
2n = -10
Divide by 2:
n = -10 / 2
n = -5
Now, we can find the three consecutive even integers:
2n + 2 = 2(-5) + 2 = -10 + 2 = -8
2n + 4 = 2(-5) + 4 = -10 + 4 = -6
2n + 6 = 2(-5) + 6 = -10 + 6 = -4
Therefore, we have -8, -6, and -4 as the three consecutive even integers starting from -8.
Consecutive Integers Formula FAQs
Q1. What are consecutive integers?
Ans. Consecutive integers are a sequence of numbers where each number is followed by the next number without any gaps. They differ by a constant value of 1. For example, 1, 2, 3, 4, and so on are consecutive integers.
Q2. How do you find consecutive integers when given a starting point?
Ans. To find consecutive integers starting from a given integer 'n,' you simply add 1 to 'n' to get the next consecutive integer, and so on. The formula for consecutive integers is n, n + 1, n + 2, n + 3, and so forth.
Q3. How can consecutive integers be used to solve problems?
Ans. Consecutive integers are commonly used in various mathematical problems, especially those involving arithmetic sequences. They are used to represent quantities that increase or decrease in a consistent pattern, making problem-solving more structured and straightforward.
Q4. Can consecutive integers be both positive and negative?
Ans. Yes, consecutive integers can be both positive and negative. If you start with a positive or negative integer, the consecutive integers will follow the pattern accordingly. For example, consecutive positive integers are 1, 2, 3, 4, and so on, while consecutive negative integers are -1, -2, -3, -4, and so forth.
Q5. Are there specific formulas for consecutive even and odd integers?
Ans. Yes, there are formulas for consecutive even and odd integers. Consecutive even integers can be represented as 2n, 2n + 2, 2n + 4, and so on. Consecutive odd integers can be represented as 2n + 1, 2n + 3, 2n + 5, and so forth, where 'n' is an integer.
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8516 | https://reference.wolfram.com/language/ref/ArcCosh.html | Wolfram Language & System Documentation Center
ArcCosh
See Also
Cosh
ArcSinh
ArcCos
TrigToExp
TrigExpand
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Hyperbolic Functions
GPU Computing
Inverse Functions
Elementary Functions
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GPU Computing with Apple
Tech Notes
Elementary Transcendental Functions
See Also
Cosh
ArcSinh
ArcCos
TrigToExp
TrigExpand
Related Guides
Hyperbolic Functions
GPU Computing
Inverse Functions
Elementary Functions
GPU Computing with NVIDIA
GPU Computing with Apple
Tech Notes
Elementary Transcendental Functions
ArcCosh[z]
gives the inverse hyperbolic cosine of the complex number .
Details
Background & Context
Examples
Basic Examples
Scope
Numerical Evaluation
Specific Values
Visualization
Function Properties
Differentiation
Integration
Series Expansions
Function Identities and Simplifications
Function Representations
Applications
Properties & Relations
Possible Issues
Neat Examples
See Also
Tech Notes
Related Guides
Related Links
History
Cite this Page
ArcCosh ✖ ArcCosh
ArcCosh[z]
✖
ArcCosh[z]
gives the inverse hyperbolic cosine of the complex number .
Details
Mathematical function, suitable for both symbolic and numerical manipulation.
For certain special arguments, ArcCosh automatically evaluates to exact values.
ArcCosh can be evaluated to arbitrary numerical precision.
ArcCosh automatically threads over lists.
ArcCosh[z] has a branch cut discontinuity in the complex plane running from to .
ArcCosh can be used with Interval and CenteredInterval objects. »
Background & Context
ArcCosh is the inverse hyperbolic cosine function. For a real number , ArcCosh[x] represents the hyperbolic angle measure such that .
ArcCosh automatically threads over lists. For certain special arguments, ArcCosh automatically evaluates to exact values. When given exact numeric expressions as arguments, ArcCosh may be evaluated to arbitrary numeric precision. Operations useful for manipulation of symbolic expressions involving ArcCosh include FunctionExpand, TrigToExp, TrigExpand, Simplify, and FullSimplify.
ArcCosh is defined for complex argument by . ArcCosh[z] has a branch cut discontinuity in the complex plane.
Related mathematical functions include Cosh, ArcSinh, and ArcCos.
Examples
open all
close all
Basic Examples (5)Summary of the most common use cases
Evaluate numerically:
In:=1
✖
Out=1
In:=2
✖
Out=2
Plot over a subset of the reals:
In:=1
✖
Out=1
Plot over a subset of the complexes:
In:=1
✖
Out=1
Asymptotic expansion at Infinity:
In:=1
✖
Out=1
Asymptotic expansion at a singular point:
In:=1
✖
Out=1
Scope (41)Survey of the scope of standard use cases
Numerical Evaluation (6)
Evaluate numerically:
In:=1
✖
Out=1
Evaluate to high precision:
In:=1
✖
Out=1
The precision of the output tracks the precision of the input:
In:=2
✖
Out=2
Evaluate for complex arguments:
In:=1
✖
Out=1
Evaluate ArcCosh efficiently at high precision:
In:=1
✖
Out=1
In:=2
✖
Out=2
Compute worst-case guaranteed intervals using Interval and CenteredInterval objects:
In:=1
✖
Out=1
In:=2
✖
Out=2
In:=3
✖
Out=3
Or compute average-case statistical intervals using Around:
In:=4
✖
Out=4
Compute the elementwise values of an array:
In:=1
✖
Out=1
Or compute the matrix ArcCosh function using MatrixFunction:
In:=2
✖
Out=2
Specific Values (4)
Values of ArcCosh at fixed points:
In:=1
✖
Out=1
Values at infinity:
In:=1
✖
Out=1
In:=2
✖
Out=2
Zero of ArcCosh:
In:=1
✖
Out=1
Find the value of satisfying equation :
In:=1
✖
In:=2
✖
Out=2
Substitute in the value:
In:=3
✖
Out=3
Visualize the result:
In:=4
✖
Out=4
Visualization (3)
Plot the ArcCosh function:
In:=1
✖
Out=1
Plot the real part of :
In:=1
✖
Out=1
Plot the imaginary part of :
In:=2
✖
Out=2
Polar plot with :
In:=1
✖
Out=1
Function Properties (10)
ArcCosh is defined for all real values greater than or equal to 1:
In:=1
✖
Out=1
Complex domain is the whole plane:
In:=2
✖
Out=2
ArcCosh achieves all real values greater than or equal to 0:
In:=1
✖
Out=1
Function range for arguments from the complex domain:
In:=2
✖
Out=2
ArcCosh is not an analytic function:
In:=1
✖
Out=1
Nor is it meromorphic:
In:=2
✖
Out=2
ArcCosh is increasing over its real domain:
In:=1
✖
Out=1
ArcCosh is injective:
In:=1
✖
Out=1
In:=2
✖
Out=2
ArcCosh is not surjective:
In:=1
✖
Out=1
In:=2
✖
Out=2
ArcCosh is non-negative over its real domain:
In:=1
✖
Out=1
It has both singularity and discontinuity in (-∞,1]:
In:=1
✖
Out=1
In:=2
✖
Out=2
ArcCosh is concave over its real domain:
In:=1
✖
Out=1
TraditionalForm formatting:
In:=1
✖
Differentiation (3)
First derivative:
In:=1
✖
Out=1
Higher derivatives:
In:=1
✖
Out=1
In:=2
✖
Out=2
Formula for the derivative:
In:=1
✖
Out=1
Integration (3)
Indefinite integral of ArcCosh:
In:=1
✖
Out=1
Definite integral of ArcCosh over an interval outside of the real domain is imaginary:
In:=1
✖
Out=1
More integrals:
In:=1
✖
Out=1
In:=2
✖
Out=2
Series Expansions (4)
Find the Taylor expansion using Series:
In:=1
✖
Out=1
Plot the first three approximations for ArcCosh around :
In:=2
✖
Out=2
General term in the series expansion of ArcCosh around :
In:=1
✖
Out=1
Find series expansions at branch points and branch cuts:
In:=1
✖
Out=1
In:=2
✖
Out=2
ArcCosh can be applied to power series:
In:=1
✖
Out=1
Function Identities and Simplifications (3)
Simplify expressions involving ArcCosh:
In:=1
✖
Out=1
Use TrigToExp to express through logarithms and square roots:
In:=1
✖
Out=1
Convert back:
In:=2
✖
Out=2
Expand assuming real variables and :
In:=1
✖
Out=1
Function Representations (5)
Represent using ArcSech:
In:=1
✖
Out=1
Representation through inverse Jacobi functions:
In:=1
✖
Out=1
In:=2
✖
Out=2
Represent using Hypergeometric2F1:
In:=1
✖
Out=1
ArcCosh can be represented in terms of MeijerG:
In:=1
✖
Out=1
In:=2
✖
Out=2
ArcCosh can be represented as a DifferentialRoot:
In:=1
✖
Out=1
Applications (4)Sample problems that can be solved with this function
Find the rapidity of a boost that makes the energy of a body twice its rest energy:
In:=1
✖
Out=1
Fraction of the speed of light required:
In:=2
✖
Out=2
Width at the base of the inverted catenary arch of Gateway Arch in St. Louis, Missouri, in feet:
In:=1
✖
Out=1
Plot the real and imaginary part of ArcCosh:
In:=1
✖
Out=1
Solve a differential equation:
In:=1
✖
Out=1
Properties & Relations (5)Properties of the function, and connections to other functions
Compositions with the inverse function might need PowerExpand to simplify to an identity:
In:=1
✖
Out=1
In:=2
✖
Out=2
Alternatively, use additional assumptions:
In:=3
✖
Out=3
This shows the branch cuts of the ArcCosh function:
In:=1
✖
Out=1
Solve an inverse trigonometric equation:
In:=1
✖
Out=1
In:=2
✖
Out=2
Solve for zeros:
In:=1
✖
Out=1
Solve the differential equation satisfied by ArcCosh:
In:=1
✖
Out=1
Verify it is satisfied by ArcCosh:
In:=2
✖
Out=2
In:=3
✖
Out=3
Possible Issues (2)Common pitfalls and unexpected behavior
Generically :
In:=1
✖
Out=1
In:=2
✖
Out=2
On branch cuts, machine-precision inputs can give numerically wrong answers:
In:=1
✖
Out=1
In:=2
✖
Out=2
Neat Examples (1)Surprising or curious use cases
This shows the branch cuts of the ArcCosh power function:
In:=1
✖
Out=1
Wolfram Research (1988), ArcCosh, Wolfram Language function, (updated 2021).
✖
Wolfram Research (1988), ArcCosh, Wolfram Language function, (updated 2021).
Text
Wolfram Research (1988), ArcCosh, Wolfram Language function, (updated 2021).
✖
Wolfram Research (1988), ArcCosh, Wolfram Language function, (updated 2021).
CMS
Wolfram Language. 1988. "ArcCosh." Wolfram Language & System Documentation Center. Wolfram Research. Last Modified 2021.
✖
Wolfram Language. 1988. "ArcCosh." Wolfram Language & System Documentation Center. Wolfram Research. Last Modified 2021.
APA
Wolfram Language. (1988). ArcCosh. Wolfram Language & System Documentation Center. Retrieved from
✖
Wolfram Language. (1988). ArcCosh. Wolfram Language & System Documentation Center. Retrieved from
BibTeX
@misc{reference.wolfram_2025_arccosh, author="Wolfram Research", title="{ArcCosh}", year="2021", howpublished="\url{ note=[Accessed: 22-August-2025]}
✖
@misc{reference.wolfram_2025_arccosh, author="Wolfram Research", title="{ArcCosh}", year="2021", howpublished="\url{ note=[Accessed: 22-August-2025]}
BibLaTeX
@online{reference.wolfram_2025_arccosh, organization={Wolfram Research}, title={ArcCosh}, year={2021}, url={ note=[Accessed: 22-August-2025]}
✖
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8517 | https://www.unitconverters.net/energy/j-to-cal.htm | Convert J to cal
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Convert J to cal
Please provide values below to convert joule [J] to calorie (th) [cal (th)], or vice versa.
From:joule
To:calorie (th)
Joule
Definition: A joule (symbol: J) is a derived unit of energy in the International System of Units (SI). It is defined as the energy transferred to an object when a one newton force is applied to the object in the direction of its motion through a distance of one meter. It is also defined as the energy dissipated as heat when a 1 ampere electric current passes through a resistance of one ohm in the course of one second. It has a number of representations both in SI base units as well as other SI units such that:
J = kg·m 2/s 2 = N·m = Pa·m 3 = W·s = C·V
History/origin: The unit, joule, is named after James Prescott Joule, an English physicist and mathematician who helped develop the Kelvin scale. He also discovered the relationship between heat and mechanical work, leading to the law of conservation of energy, and subsequently, the first law of thermodynamics.
Current use: As an SI derived unit, the joule is used within a variety of scientific contexts. Practical examples of energy measurement using joules include the energy required to lift objects, the energy released when objects fall, the heat required to raise temperature, and the kinetic energy of moving objects.
One of the dimensional representations of a joule is the N·m (Newton-meter), which is equivalent to the SI unit for torque. These units however, are different, and should be considered as such. Even though the joule is algebraically equal to the N·m, the N·m should not be used to represent the joule whenever possible, to avoid confusion with torque.
calorie
Definition: A calorie (symbol: cal) is a unit of energy defined as the amount of energy required to increase the temperature of one gram of water by one °C. This is referred to as the small calorie or the gram calorie, and is equal to 4.184 joules, the SI (International System of Units) unit of energy.
A large calorie (symbol: Cal), also known as a kilogram calorie (symbol: Cal), is technically a kilocalorie (symbol: kcal), the equivalent of 1000 small calories, but is also sometimes referred to as simply "Calorie." Large calories are usually used for labeling foods, and as such, is known as the food calorie.
History/origin: The term calorie comes from the Latin word "calor," which means heat. It was first defined as a unit of heat energy in 1824 by Nicolas Clement and appeared in French and English dictionaries between 1841 and 1867. The large calorie was introduced to the American public later, in 1887, by Wilbur Olin Atwater.
Current use: Although the calorie can be used with SI, since the official adoption of SI in 1960, the calorie is considered obsolete. Despite this, the large calorie is still widely used as a unit of food energy, alongside, or in place of the SI unit of food energy, the kilojoule. The use of a capital "C" in Calorie is intended to denote the use of kilocalories rather than calories denoting a single calorie. This is not often understood however, resulting in some confusion when foods are only labeled as Calories rather than kilocalories.
Calories are also used within scientific contexts, such as chemistry. In these contexts, the term being referenced is most often the small calorie, though measurements are often reported in kilocalories.
Joule to Calorie (th) Conversion Table
| Joule [J] | Calorie (th) [cal (th)] |
--- |
| 0.01 J | 0.0023900574 cal (th) |
| 0.1 J | 0.0239005736 cal (th) |
| 1 J | 0.2390057361 cal (th) |
| 2 J | 0.4780114723 cal (th) |
| 3 J | 0.7170172084 cal (th) |
| 5 J | 1.1950286807 cal (th) |
| 10 J | 2.3900573614 cal (th) |
| 20 J | 4.7801147228 cal (th) |
| 50 J | 11.9502868069 cal (th) |
| 100 J | 23.9005736138 cal (th) |
| 1000 J | 239.0057361377 cal (th) |
How to Convert Joule to Calorie (th)
1 J = 0.2390057361 cal (th)
1 cal (th) = 4.184 J
Example: convert 15 J to cal (th):
15 J = 15 × 0.2390057361 cal (th) = 3.5850860421 cal (th)
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8519 | https://pmc.ncbi.nlm.nih.gov/articles/PMC5264509/ | Photosynthesis - PMC
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Essays Biochem
. 2016 Oct 26;60(3):255–273. doi: 10.1042/EBC20160016
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Photosynthesis
Matthew P Johnson
Matthew P Johnson
1 Department of Molecular Biology and Biotechnology, University of Sheffield, Firth Court, Western Bank, Sheffield S10 2TN, U.K.
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1 Department of Molecular Biology and Biotechnology, University of Sheffield, Firth Court, Western Bank, Sheffield S10 2TN, U.K.
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Correspondence: Matthew P. Johnson (email matt.johnson@sheffield.ac.uk).
Received 2016 Jan 14; Revised 2016 Jul 22; Accepted 2016 Jul 26; Issue date 2016 Oct 31.
© 2016 The Author(s)
This is an open access article published by Portland Press Limited on behalf of the Biochemical Society and distributed under the Creative Commons Attribution Licence 4.0 (CC BY).
PMC Copyright notice
PMCID: PMC5264509 PMID: 27784776
This article has been corrected. See Essays Biochem. 2017 Oct 31;61(4):429.
See "Editorial Note: Photosynthesis" in volume 65 on page 405.
Abstract
Photosynthesis sustains virtually all life on planet Earth providing the oxygen we breathe and the food we eat; it forms the basis of global food chains and meets the majority of humankind's current energy needs through fossilized photosynthetic fuels. The process of photosynthesis in plants is based on two reactions that are carried out by separate parts of the chloroplast. The light reactions occur in the chloroplast thylakoid membrane and involve the splitting of water into oxygen, protons and electrons. The protons and electrons are then transferred through the thylakoid membrane to create the energy storage molecules adenosine triphosphate (ATP) and nicotinomide–adenine dinucleotide phosphate (NADPH). The ATP and NADPH are then utilized by the enzymes of the Calvin–Benson cycle (the dark reactions), which converts CO 2 into carbohydrate in the chloroplast stroma. The basic principles of solar energy capture, energy, electron and proton transfer and the biochemical basis of carbon fixation are explained and their significance is discussed.
Keywords: membrane, photosynthesis, thylakoid
An overview of photosynthesis
Introduction
Photosynthesis is the ultimate source of all of humankind's food and oxygen, whereas fossilized photosynthetic fuels provide ∼87% of the world's energy. It is the biochemical process that sustains the biosphere as the basis for the food chain. The oxygen produced as a by-product of photosynthesis allowed the formation of the ozone layer, the evolution of aerobic respiration and thus complex multicellular life.
Oxygenic photosynthesis involves the conversion of water and CO 2 into complex organic molecules such as carbohydrates and oxygen. Photosynthesis may be split into the ‘light’ and ‘dark’ reactions. In the light reactions, water is split using light into oxygen, protons and electrons, and in the dark reactions, the protons and electrons are used to reduce CO 2 to carbohydrate (given here by the general formula CH 2 O). The two processes can be summarized thus:
Light reactions:
Dark reactions:
Overall:
The positive sign of the standard free energy change of the reaction (Δ G°) given above means that the reaction requires energy (an endergonic reaction). The energy required is provided by absorbed solar energy, which is converted into the chemical bond energy of the products (Box 1).
Box 1. Standard free energy change.
Photosynthesis converts ∼200 billion tonnes of CO 2 into complex organic compounds annually and produces ∼140 billion tonnes of oxygen into the atmosphere. By facilitating conversion of solar energy into chemical energy, photosynthesis acts as the primary energy input into the global food chain. Nearly all living organisms use the complex organic compounds derived from photosynthesis as a source of energy. The breakdown of these organic compounds occurs via the process of aerobic respiration, which of course also requires the oxygen produced by photosynthesis.
Unlike photosynthesis, aerobic respiration is an exergonic process (negative Δ G°) with the energy released being used by the organism to power biosynthetic processes that allow growth and renewal, mechanical work (such as muscle contraction or flagella rotation) and facilitating changes in chemical concentrations within the cell (e.g. accumulation of nutrients and expulsion of waste). The use of exergonic reactions to power endergonic ones associated with biosynthesis and housekeeping in biological organisms such that the overall free energy change is negative is known as ‘coupling’.
Photosynthesis and respiration are thus seemingly the reverse of one another, with the important caveat that both oxygen formation during photosynthesis and its utilization during respiration result in its liberation or incorporation respectively into water rather than CO 2. In addition, glucose is one of several possible products of photosynthesis with amino acids and lipids also being synthesized rapidly from the primary photosynthetic products.
The consideration of photosynthesis and respiration as opposing processes helps us to appreciate their role in shaping our environment. The fixation of CO 2 by photosynthesis and its release during breakdown of organic molecules during respiration, decay and combustion of organic matter and fossil fuels can be visualized as the global carbon cycle (Figure 1).
Figure 1. The global carbon cycle.
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The relationship between respiration, photosynthesis and global CO 2 and O 2 levels.
At present, this cycle may be considered to be in a state of imbalance due to the burning of fossil fuels (fossilized photosynthesis), which is increasing the proportion of CO 2 entering the Earth's atmosphere, leading to the so-called ‘greenhouse effect’ and human-made climate change.
Oxygenic photosynthesis is thought to have evolved only once during Earth's history in the cyanobacteria. All other organisms, such as plants, algae and diatoms, which perform oxygenic photosynthesis actually do so via cyanobacterial endosymbionts or ‘chloroplasts’. An endosymbiotoic event between an ancestral eukaryotic cell and a cyanobacterium that gave rise to plants is estimated to have occurred ∼1.5 billion years ago. Free-living cyanobacteria still exist today and are responsible for ∼50% of the world's photosynthesis. Cyanobacteria themselves are thought to have evolved from simpler photosynthetic bacteria that use either organic or inorganic compounds such a hydrogen sulfide as a source of electrons rather than water and thus do not produce oxygen.
The site of photosynthesis in plants
In land plants, the principal organs of photosynthesis are the leaves (Figure 2A). Leaves have evolved to expose the largest possible area of green tissue to light and entry of CO 2 to the leaf is controlled by small holes in the lower epidermis called stomata (Figure 2B). The size of the stomatal openings is variable and regulated by a pair of guard cells, which respond to the turgor pressure (water content) of the leaf, thus when the leaf is hydrated, the stomata can open to allow CO 2 in. In contrast, when water is scarce, the guard cells lose turgor pressure and close, preventing the escape of water from the leaf via transpiration.
Figure 2. Location of the photosynthetic machinery.
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(A) The model plant Arabidopsis thaliana. (B) Basic structure of a leaf shown in cross-section. Chloroplasts are shown as green dots within the cells. (C) An electron micrograph of an Arabidopsis chloroplast within the leaf. (D) Close-up region of the chloroplast showing the stacked structure of the thylakoid membrane.
Within the green tissue of the leaf (mainly the mesophyll) each cell (∼100 μm in length) contains ∼100 chloroplasts (2–3 μm in length), the tiny organelles where photosynthesis takes place. The chloroplast has a complex structure (Figure 2C, D) with two outer membranes (the envelope), which are colourless and do not participate in photosynthesis, enclosing an aqueous space (the stroma) wherein sits a third membrane known as the thylakoid, which in turn encloses a single continuous aqueous space called the lumen.
The light reactions of photosynthesis involve light-driven electron and proton transfers, which occur in the thylakoid membrane, whereas the dark reactions involve the fixation of CO 2 into carbohydrate, via the Calvin–Benson cycle, which occurs in the stroma (Figure 3). The light reactions involve electron transfer from water to NADP+ to form NADPH and these reactions are coupled to proton transfers that lead to the phosphorylation of adenosine diphosphate (ADP) into ATP. The Calvin–Benson cycle uses ATP and NADPH to convert CO 2 into carbohydrates (Figure 3), regenerating ADP and NADP+. The light and dark reactions are therefore mutually dependent on one another.
Figure 3. Division of labour within the chloroplast.
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The light reactions of photosynthesis take place in the thylakoid membrane, whereas the dark reactions are located in the chloroplast stroma.
Photosynthetic electron and proton transfer chain
The light-driven electron transfer reactions of photosynthesis begin with the splitting of water by Photosystem II (PSII). PSII is a chlorophyll–protein complex embedded in the thylakoid membrane that uses light to oxidize water to oxygen and reduce the electron acceptor plastoquinone to plastoquinol. Plastoquinol in turn carries the electrons derived from water to another thylakoid-embedded protein complex called cytochrome b 6 f (cyt b 6 f). cyt b 6 f oxidizes plastoquinol to plastoquinone and reduces a small water-soluble electron carrier protein plastocyanin, which resides in the lumen. A second light-driven reaction is then carried out by another chlorophyll protein complex called Photosystem I (PSI). PSI oxidizes plastocyanin and reduces another soluble electron carrier protein ferredoxin that resides in the stroma. Ferredoxin can then be used by the ferredoxin–NADP+ reductase (FNR) enzyme to reduce NADP+ to NADPH. This scheme is known as the linear electron transfer pathway or Z-scheme (Figure 4).
Figure 4. The photosynthetic electron and proton transfer chain.
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The linear electron transfer pathway from water to NADP+ to form NADPH results in the formation of a proton gradient across the thylakoid membrane that is used by the ATP synthase enzyme to make ATP.
The Z-scheme, so-called since it resembles the letter ‘Z’ when turned on its side (Figure 5), thus shows how the electrons move from the water–oxygen couple (+820 mV) via a chain of redox carriers to NADP+/NADPH (−320 mV) during photosynthetic electron transfer. Generally, electrons are transferred from redox couples with low potentials (good reductants) to those with higher potentials (good oxidants) (e.g. during respiratory electron transfer in mitochondria) since this process is exergonic (see Box 2). However, photosynthetic electron transfer also involves two endergonic steps, which occur at PSII and at PSI and require an energy input in the form of light. The light energy is used to excite an electron within a chlorophyll molecule residing in PSII or PSI to a higher energy level; this excited chlorophyll is then able to reduce the subsequent acceptors in the chain. The oxidized chlorophyll is then reduced by water in the case of PSII and plastocyanin in the case of PSI.
Figure 5. Z-scheme of photosynthetic electron transfer.
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The main components of the linear electron transfer pathway are shown on a scale of redox potential to illustrate how two separate inputs of light energy at PSI and PSII result in the endergonic transfer of electrons from water to NADP+.
Box 2. Relationship between redox potentials and standard free energy changes.
The water-splitting reaction at PSII and plastoquinol oxidation at cyt b 6 f result in the release of protons into the lumen, resulting in a build-up of protons in this compartment relative to the stroma. The difference in the proton concentration between the two sides of the membrane is called a proton gradient. The proton gradient is a store of free energy (similar to a gradient of ions in a battery) that is utilized by a molecular mechanical motor ATP synthase, which resides in the thylakoid membrane (Figure 4). The ATP synthase allows the protons to move down their concentration gradient from the lumen (high H+ concentration) to the stroma (low H+ concentration). This exergonic reaction is used to power the endergonic synthesis of ATP from ADP and inorganic phosphate (P i). This process of photophosphorylation is thus essentially similar to oxidative phosphorylation, which occurs in the inner mitochondrial membrane during respiration.
An alternative electron transfer pathway exists in plants and algae, known as cyclic electron flow. Cyclic electron flow involves the recycling of electrons from ferredoxin to plastoquinone, with the result that there is no net production of NADPH; however, since protons are still transferred into the lumen by oxidation of plastoquinol by cyt b 6 f, ATP can still be formed. Thus photosynthetic organisms can control the ratio of NADPH/ATP to meet metabolic need by controlling the relative amounts of cyclic and linear electron transfer.
How the photosystems work
Light absorption by pigments
Photosynthesis begins with the absorption of light by pigments molecules located in the thylakoid membrane. The most well-known of these is chlorophyll, but there are also carotenoids and, in cyanobacteria and some algae, bilins. These pigments all have in common within their chemical structures an alternating series of carbon single and double bonds, which form a conjugated system π–electron system (Figure 6).
Figure 6. Major photosynthetic pigments in plants.
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The chemical structures of the chlorophyll and carotenoid pigments present in the thylakoid membrane. Note the presence in each of a conjugated system of carbon–carbon double bonds that is responsible for light absorption.
The variety of pigments present within each type of photosynthetic organism reflects the light environment in which it lives; plants on land contain chlorophylls a and b and carotenoids such as β-carotene, lutein, zeaxanthin, violaxanthin, antheraxanthin and neoxanthin (Figure 6). The chlorophylls absorb blue and red light and so appear green in colour, whereas carotenoids absorb light only in the blue and so appear yellow/red (Figure 7), colours more obvious in the autumn as chlorophyll is the first pigment to be broken down in decaying leaves.
Figure 7. Basic absorption spectra of the major chlorophyll and carotenoid pigments found in plants.
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Chlorophylls absorb light energy in the red and blue part of the visible spectrum, whereas carotenoids only absorb light in the blue/green.
Light, or electromagnetic radiation, has the properties of both a wave and a stream of particles (light quanta). Each quantum of light contains a discrete amount of energy that can be calculated by multiplying Planck's constant, h (6.626×10−34 J·s) by ν, the frequency of the radiation in cycles per second (s−1):
The frequency (ν) of the light and so its energy varies with its colour, thus blue photons (∼450 nm) are more energetic than red photons (∼650 nm). The frequency (ν) and wavelength (λ) of light are related by:
where c is the velocity of light (3.0×10 8 m·s−1), and the energy of a particular wavelength (λ) of light is given by:
Thus 1 mol of 680 nm photons of red light has an energy of 176 kJ·mol−1.
The electrons within the delocalized π system of the pigment have the ability to jump up from the lowest occupied molecular orbital (ground state) to higher unoccupied molecular electron orbitals (excited states) via the absorption of specific wavelengths of light in the visible range (400–725 nm). Chlorophyll has two excited states known as S 1 and S 2 and, upon interaction of the molecule with a photon of light, one of its π electrons is promoted from the ground state (S 0) to an excited state, a process taking just 10−15 s (Figure 8). The energy gap between the S 0 and S 1 states is spanned by the energy provided by a red photon (∼600–700 nm), whereas the energy gap between the S 0 and S 2 states is larger and therefore requires a more energetic (shorter wavelength, higher frequency) blue photon (∼400–500 nm) to span the energy gap.
Figure 8. Jablonski diagram of chlorophyll showing the possible fates of the S 1 and S 2 excited states and timescales of the transitions involved.
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Photons with slightly different energies (colours) excite each of the vibrational substates of each excited state (as shown by variation in the size and colour of the arrows).
Upon excitation, the electron in the S 2 state quickly undergoes losses of energy as heat through molecular vibration and undergoes conversion into the energy of the S 1 state by a process called internal conversion. Internal conversion occurs on a timescale of 10−12 s. The energy of a blue photon is thus rapidly degraded to that of a red photon. Excitation of the molecule with a red photon would lead to promotion of an electron to the S 1 state directly. Once the electron resides in the S 1 state, it is lower in energy and thus stable on a somewhat longer timescale (10−9 s). The energy of the excited electron in the S 1 state can have one of several fates: it could return to the ground state (S 0) by emission of the energy as a photon of light (fluorescence), or it could be lost as heat due to internal conversion between S 1 and S 0. Alternatively, if another chlorophyll is nearby, a process known as excitation energy transfer (EET) can result in the non-radiative exchange of energy between the two molecules (Figure 9). For this to occur, the two chlorophylls must be close by (<7 nm), have a specific orientation with respect to one another, and excited state energies that overlap (are resonant) with one another. If these conditions are met, the energy is exchanged, resulting in a mirror S 0→S 1 transition in the acceptor molecule and a S 1→S 0 transition in the other.
Figure 9. Basic mechanism of excitation energy transfer between chlorophyll molecules.
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Two chlorophyll molecules with resonant S 1 states undergo a mirror transition resulting in the non-radiative transfer of excitation energy between them.
Light-harvesting complexes
In photosynthetic systems, chlorophylls and carotenoids are found attached to membrane-embedded proteins known as light-harvesting complexes (LHCs). Through careful binding and orientation of the pigment molecules, absorbed energy can be transferred among them by EET. Each pigment is bound to the protein by a series of non-covalent bonding interactions (such as, hydrogen bonds, van der Waals interactions, hydrophobic interaction and co-ordination bonds between lone pair electrons of residues such as histidine in the protein and the Mg 2+ ion in chlorophyll); the protein structure is such that each bound pigment experiences a slightly different environment in terms of the surrounding amino acid side chains, lipids, etc., meaning that the S 1 and S 2 energy levels are shifted in energy with respect to that of other neighbouring pigment molecules. The effect is to create a range of pigment energies that act to ‘funnel’ the energy on to the lowest-energy pigments in the LHC by EET.
Reaction centres
A photosystem consists of numerous LHCs that form an antenna of hundreds of pigment molecules. The antenna pigments act to collect and concentrate excitation energy and transfer it towards a ‘special pair’ of chlorophyll molecules that reside in the reaction centre (RC) (Figure 10). Unlike the antenna pigments, the special pair of chlorophylls are ‘redox-active’ in the sense that they can return to the ground state (S 0) by the transfer of the electron residing in the S 1 excited state (Chl) to another species. This process is known as charge separation and result in formation of an oxidized special pair (Chl+) and a reduced acceptor (A−). The acceptor in PSII is plastoquinone and in PSI it is ferredoxin. If the RC is to go on functioning, the electron deficiency on the special pair must be made good, in PSII the electron donor is water and in PSI it is plastocyanin.
Figure 10. Basic structure of a photosystem.
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Light energy is captured by the antenna pigments and transferred to the special pair of RC chlorophylls which undergo a redox reaction leading to reduction of an acceptor molecule. The oxidized special pair is regenerated by an electron donor.
It is worth asking why photosynthetic organisms bother to have a large antenna of pigments serving an RC rather than more numerous RCs. The answer lies in the fact that the special pair of chlorophylls alone have a rather small spatial and spectral cross-section, meaning that there is a limit to the amount of light they can efficiently absorb. The amount of light they can practically absorb is around two orders of magnitude smaller than their maximum possible turnover rate, Thus LHCs act to increase the spatial (hundreds of pigments) and spectral (several types of pigments with different light absorption characteristics) cross-section of the RC special pair ensuring that its turnover rate runs much closer to capacity.
Photosystem II
PSII is a light-driven water–plastoquinone oxidoreductase and is the only enzyme in Nature that is capable of performing the difficult chemistry of splitting water into protons, electrons and oxygen (Figure 11). In principle, water is an extremely poor electron donor since the redox potential of the water–oxygen couple is +820 mV. PSII uses light energy to excite a special pair of chlorophylls, known as P680 due to their 680 nm absorption peak in the red part of the spectrum. P680 undergoes charge separation that results in the formation of an extremely oxidizing species P680+ which has a redox potential of +1200 mV, sufficient to oxidize water. Nonetheless, since water splitting involves four electron chemistry and charge separation only involves transfer of one electron, four separate charge separations (turnovers of PSII) are required to drive formation of one molecule of O 2 from two molecules of water. The initial electron donation to generate the P680 from P680+ is therefore provided by a cluster of manganese ions within the oxygen-evolving complex (OEC), which is attached to the lumen side of PSII (Figure 12). Manganese is a transition metal that can exist in a range of oxidation states from +1 to +5 and thus accumulates the positive charges derived from each light-driven turnover of P680. Progressive extraction of electrons from the manganese cluster is driven by the oxidation of P680 within PSII by light and is known as the S-state cycle (Figure 12). After the fourth turnover of P680, sufficient positive charge is built up in the manganese cluster to permit the splitting of water into electrons, which regenerate the original state of the manganese cluster, protons, which are released into the lumen and contribute to the proton gradient used for ATP synthesis, and the by-product O 2. Thus charge separation at P680 provides the thermodynamic driving force, whereas the manganese cluster acts as a catalyst for the water-splitting reaction.
Figure 11. Basic structure of the PSII–LHCII supercomplex from spinach.
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The organization of PSII and its light-harvesting antenna. Protein is shown in grey, with chlorophylls in green and carotenoids in orange. Drawn from PDB code 3JCU
Figure 12. S-state cycle of water oxidation by the manganese cluster (shown as circles with roman numerals representing the manganese ion oxidation states) within the PSII oxygen-evolving complex.
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Progressive extraction of electrons from the manganese cluster is driven by the oxidation of P680 within PSII by light. Each of the electrons given up by the cluster is eventually repaid at the S 4 to S 0 transition when molecular oxygen (O 2) is formed. The protons extracted from water during the process are deposited into the lumen and contribute to the protonmotive force.
The electrons yielded by P680 following charge separation are not passed directly to plastoquinone, but rather via another acceptor called pheophytin, a porphyrin molecule lacking the central magnesium ion as in chlorophyll. Plastoquinone reduction to plastoquinol requires two electrons and thus two molecules of plastoquinol are formed per O 2 molecule evolved by PSII. Two protons are also taken up upon formation of plastoquinol and these are derived from the stroma. PSII is found within the thylakoid membrane of plants as a dimeric RC complex surrounded by a peripheral antenna of six minor monomeric antenna LHC complexes and two to eight trimeric LHC complexes, which together form a PSII–LHCII supercomplex (Figure 11).
Photosystem I
PSI is a light-driven plastocyanin–ferredoxin oxidoreductase (Figure 13). In PSI, the special pair of chlorophylls are known as P700 due to their 700 nm absorption peak in the red part of the spectrum. P700 is an extremely strong reductant that is able to reduce ferredoxin which has a redox potential of −450 mV (and is thus is, in principle, a poor electron acceptor). Reduced ferredoxin is then used to generate NADPH for the Calvin–Benson cycle at a separate complex known as FNR. The electron from P700 is donated via another chlorophyll molecule and a bound quinone to a series of iron–sulfur clusters at the stromal side of the complex, whereupon the electron is donated to ferredoxin. The P700 species is regenerated form P700+ via donation of an electron from the soluble electron carrier protein plastocyanin.
Figure 13. Basic structure of the PSI–LHCI supercomplex from pea.
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The organization of PSI and its light-harvesting antenna. Protein is shown in grey, with chlorophylls in green and carotenoids in orange. Drawn from PDB code 4XK8.
PSI is found within the thylakoid membrane as a monomeric RC surrounded on one side by four LHC complexes known as LHCI. The PSI–LHCI supercomplex is found mainly in the unstacked regions of the thylakoid membrane (Figure 13).
Other electron transfer chain components
Plastoquinone/plastoquinol
Plastoquinone is a small lipophilic electron carrier molecule that resides within the thylakoid membrane and carries two electrons and two protons from PSII to the cyt b 6 f complex. It has a very similar structure to that of the molecule ubiquinone (coenzyme Q 10) in the mitochondrial inner membrane.
Cytochrome b 6 f complex
The cyt b 6 f complex is a plastoquinol–plastocyanin oxidoreductase and possess a similar structure to that of the cytochrome bc 1 complex (complex III) in mitochondria (Figure 14A). As with Complex III, cyt b 6 f exists as a dimer in the membrane and carries out both the oxidation and reduction of quinones via the so-called Q-cycle. The Q-cycle (Figure 14B) involves oxidation of one plastoquinol molecule at the Qp site of the complex, both protons from this molecule are deposited in the lumen and contribute to the proton gradient for ATP synthesis. The two electrons, however, have different fates. The first is transferred via an iron–sulfur cluster and a haem cofactor to the soluble electron carrier plastocyanin (see below). The second electron derived from plastoquinol is passed via two separate haem cofactors to another molecule of plastoquinone bound to a separate site (Qn) on the complex, thus reducing it to a semiquinone. When a second plastoquinol molecule is oxidized at Qp, a second molecule of plastocyanin is reduced and two further protons are deposited in the lumen. The second electron reduces the semiquinone at the Qn site which, concomitant with uptake of two protons from the stroma, causes its reduction to plastoquinol. Thus for each pair of plastoquinol molecules oxidized by the complex, one is regenerated, yet all four protons are deposited into the lumen. The Q-cycle thus doubles the number of protons transferred from the stroma to the lumen per plastoquinol molecule oxidized.
Figure 14. Cytochrome b 6 f complex.
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(A) Structure drawn from PDB code 1Q90. (B) The protonmotive Q-cycle showing how electrons from plastoquinol are passed to both plastocyanin and plastoquinone, doubling the protons deposited in the lumen for every plastoquinol molecule oxidized by the complex.
Plastocyanin
Plastocyanin is a small soluble electron carrier protein that resides in the thylakoid lumen. The active site of the plastocyanin protein binds a copper ion, which cycles between the Cu 2+ and Cu+ oxidation states following its oxidation by PSI and reduction by cyt b 6 f respectively.
Ferredoxin
Ferredoxin is a small soluble electron carrier protein that resides in the chloroplast stroma. The active site of the ferredoxin protein binds an iron–sulfur cluster, which cycles between the Fe 2+ and Fe 3+ oxidation states following its reduction by PSI and oxidation by the FNR complex respectively.
Ferredoxin–NADP+ reductase
The FNR complex is found in both soluble and thylakoid membrane-bound forms. The complex binds a flavin–adenine dinucleotide (FAD) cofactor at its active site, which accepts two electrons from two molecules of ferredoxin before using them reduce NADP+ to NADPH.
ATP synthase
The ATP synthase enzyme is responsible for making ATP from ADP and P i; this endergonic reaction is powered by the energy contained within the protonmotive force. According to the structure, 4.67 H+ are required for every ATP molecule synthesized by the chloroplast ATP synthase. The enzyme is a rotary motor which contains two domains: the membrane-spanning F O portion which conducts protons from the lumen to the stroma, and the F 1 catalytic domain that couples this exergonic proton movement to ATP synthesis.
Membrane stacking and the regulation of photosynthesis
Within the thylakoid membrane, PSII–LHCII supercomplexes are packed together into domains known as the grana, which associate with one another to form grana stacks. PSI and ATP synthase are excluded from these stacked PSII–LHCII regions by steric constraints and thus PSII and PSI are segregated in the thylakoid membrane between the stacked and unstacked regions (Figure 15). The cyt b 6 f complex, in contrast, is evenly distributed throughout the grana and stromal lamellae. The evolutionary advantage of membrane stacking is believed to be a higher efficiency of electron transport by preventing the fast energy trap PSI from ‘stealing’ excitation energy from the slower trap PSII, a phenomenon known as spillover. Another possible advantage of membrane stacking in thylakoids may be the segregation of the linear and cyclic electron transfer pathways, which might otherwise compete to reduce plastoquinone. In this view, PSII, cyt b 6 f and a sub-fraction of PSI closest to the grana is involved in linear flow, whereas PSI and cyt b 6 f in the stromal lamellae participates in cyclic flow. The cyclic electron transfer pathway recycles electrons from ferredoxin back to plastoquinone and thus allows protonmotive force generation (and ATP synthesis) without net NADPH production. Cyclic electron transfer thereby provides the additional ATP required for the Calvin–Benson cycle (see below).
Figure 15. Lateral heterogeneity in thylakoid membrane organization.
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(A) Electron micrograph of the thylakoid membrane showing stacked grana and unstacked stromal lamellae regions. (B) Model showing the distribution of the major complexes of photosynthetic electron and proton transfer between the stacked grana and unstacked stromal lamellae regions.
‘Dark’ reactions: the Calvin–Benson cycle
CO 2 is fixed into carbohydrate via the Calvin–Benson cycle in plants, which consumes the ATP and NADPH produced during the light reactions and thus in turn regenerates ADP, P i and NADP+. In the first step of the Calvin–Benson cycle (Figure 16), CO 2 is combined with a 5-carbon (5C) sugar, ribulose 1,5-bisphosphate in a reaction catalysed by the enzyme ribulose-1,5-bisphosphate carboxylase/oxygenase (Rubisco). The reaction forms an unstable 6C intermediate that immediately splits into two molecules of 3-phosphoglycerate. 3-Phosphoglycerate is first phosphorylated by 3-phosphoglycerate kinase using ATP to form 1,3-bisphosphoglycerate. 1,3-Bisphosphoglycerate is then reduced by glyceraldehyde 3-phosphate dehydrogenase using NADPH to form glyceraldehyde 3-phosphate (GAP, a triose or 3C sugar) in reactions, which are the reverse of glycolysis. For every three CO 2 molecules initially combined with ribulose 1,5-bisphopshate, six molecules of GAP are produced by the subsequent steps. However only one of these six molecules can be considered as a product of the Calvin–Benson cycle since the remaining five are required to regenerate ribulose 1,5-bisphosphate in a complex series of reactions that also require ATP. The one molecule of GAP that is produced for each turn of the cycle can be quickly converted by a range of metabolic pathways into amino acids, lipids or sugars such as glucose. Glucose in turn may be stored as the polymer starch as large granules within chloroplasts.
Figure 16. The Calvin–Benson cycle.
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Overview of the biochemical pathway for the fixation of CO 2 into carbohydrate in plants.
A complex biochemical ‘dance’ (Figure 16) is then involved in the regeneration of three ribulose 1,5-bisphosphate (5C) from the remaining five GAP (3C) molecules. The regeneration begins with the conversion of two molecules of GAP into dihydroxyacetone phosphate (DHAP) by triose phosphate isomerase; one of the DHAP molecules is the combined with another GAP molecule to make fructose 1,6-bisphosphate (6C) by aldolase. The fructose 1,6-bisphosphate is then dephosphorylated by fructose-1,6-bisphosphatase to yield fructose 6-phosphate (6C) and releasing P i. Two carbons are then removed from fructose 6-phosphate by transketolase, generating erythrose 4-phosphate (4C); the two carbons are transferred to another molecule of GAP generating xylulose 5-phosphate (5C). Another DHAP molecule, formed from GAP by triose phosphate isomerase is then combined with the erythrose 4-phosphate by aldolase to form sedoheptulose 1,7-bisphosphate (7C). Sedoheptulose 1,7-bisphosphate is then dephosphorylated to sedoheptulose 7-phosphate (7C) by sedoheptulose-1,7-bisphosphatase releasing P i. Sedoheptulose 7-phosphate has two carbons removed by transketolase to produce ribose 5-phosphate (5C) and the two carbons are transferred to another GAP molecule producing another xylulose 5-phosphate (5C). Ribose 5-phosphate and the two molecules of xylulose 5-phosphate (5C) are then converted by phosphopentose isomerase to three molecules of ribulose 5-phosphate (5C). The three ribulose 5-phosphate molecules are then phosphorylated using three ATP by phosphoribulokinase to regenerate three ribulose 1,5-bisphosphate (5C).
Overall the synthesis of 1 mol of GAP requires 9 mol of ATP and 6 mol of NADPH, a required ratio of 1.5 ATP/NADPH. Linear electron transfer is generally thought to supply ATP/NADPH in a ratio of 1.28 (assuming an H+/ATP ratio of 4.67) with the shortfall of ATP believed to be provided by cyclic electron transfer reactions. Since the product of the Calvin cycle is GAP (a 3C sugar) the pathway is often referred to as C 3 photosynthesis and plants that utilize it are called C 3 plants and include many of the world's major crops such as rice, wheat and potato.
Many of the enzymes involved in the Calvin–Benson cycle (e.g. transketolase, glyceraldehyde-3-phosphate dehydrogenase and aldolase) are also involved in the glycolysis pathway of carbohydrate degradation and their activity must therefore be carefully regulated to avoid futile cycling when light is present, i.e. the unwanted degradation of carbohydrate. The regulation of the Calvin–Benson cycle enzymes is achieved by the activity of the light reactions, which modify the environment of the dark reactions (i.e. the stroma). Proton gradient formation across the thylakoid membrane during the light reactions increases the pH and also increases the Mg 2+ concentration in the stroma (as Mg 2+ flows out of the lumen as H+ flows in to compensate for the influx of positive charges). In addition, by reducing ferredoxin and NADP+, PSI changes the redox state of the stroma, which is sensed by the regulatory protein thioredoxin. Thioredoxin, pH and Mg 2+ concentration play a key role in regulating the activity of the Calvin–Benson cycle enzymes, ensuring the activity of the light and dark reactions is closely co-ordinated.
Rubisco
It is noteworthy that, despite the complexity of the dark reactions outlined above, the carbon fixation step itself (i.e. the incorporation of CO 2 into carbohydrate) is carried out by a single enzyme, Rubisco. Rubisco is a large multisubunit soluble protein complex found in the chloroplast stroma. The complex consists of eight large (56 kDa) subunits, which contain both catalytic and regulatory domains, and eight small subunits (14 kDa), which enhance the catalytic function of the L subunits (Figure 17A). The carboxylation reaction carried out by Rubisco is highly exergonic (Δ G°=−51.9 kJ·mol-1), yet kinetically very slow (just 3 s−1) and begins with the protonation of ribulose 1,5-bisphosphate to form an enediolate intermediate which can be combined with CO 2 to form an unstable 6C intermediate that is quickly hydrolysed to yield two 3C 3-phosphoglycerate molecules. The active site in the Rubisco enzyme contains a key lysine residue, which reacts with another (non-substrate) molecule of CO 2 to form a carbamate anion that is then able to bind Mg 2+. The Mg 2+ in the active site is essential for the catalytic function of Rubisco, playing a key role in binding ribulose 1,5-bisphosphate and activating it such that it readily reacts with CO 2.. Rubisco activity is co-ordinated with that of the light reactions since carbamate formation requires both high Mg 2+ concentration and alkaline conditions, which are provided by the light-driven changes in the stromal environment discussed above (Figure 17B).
Figure 17. Rubisco.
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(A) Structure of the Rubisco enzyme (the large subunits are shown in blue and the small subunits in green); four of each type of subunit are visible in the image. Drawn from PDB code 1RXO. (B) Activation of the lysine residue within the active site of Rubisco occurs via elevated stromal pH and Mg 2+ concentration as a result of the activity of the light reactions.
In addition to carboxylation, Rubisco also catalyses a competitive oxygenation reaction, known as photorespiration, that results in the combination of ribulose 1,5-bisphosphate with O 2 rather than CO 2. In the oxygenation reaction, one rather than two molecules of 3-phosphoglycerate and one molecule of a 2C sugar known as phosphoglycolate are produced by Rubisco. The phosphoglycolate must be converted in a series of reactions that regenerate one molecule of 3-phosphoglycerate and one molecule of CO 2. These reactions consume additional ATP and thus result in an energy loss to the plant. Although the oxygenation reaction of Rubisco is much less favourable than the carboxylation reaction, the relatively high concentration of O 2 in the leaf (250 μM) compared with CO 2 (10 μM) means that a significant amount of photorespiration is always occurring. Under normal conditions, the ratio of carboxylation to oxygenation is between 3:1 and 4:1. However, this ratio can be decreased with increasing temperature due to decreased CO 2 concentration in the leaf, a decrease in the affinity of Rubisco for CO 2 compared with O 2 and an increase in the maximum rate of the oxygenation reaction compared with the carboxylation reaction. The inefficiencies of the Rubisco enzyme mean that plants must produce it in very large amounts (∼30–50% of total soluble protein in a spinach leaf) to achieve the maximal photosynthetic rate.
CO 2-concentrating mechanisms
To counter photorespiration, plants, algae and cyanobacteria have evolved different CO 2-concentrating mechanisms CCMs that aim to increase the concentration of CO 2 relative to O 2 in the vicinity of Rubisco. One such CCM is C 4 photosynthesis that is found in plants such as maize, sugar cane and savanna grasses. C 4 plants show a specialized leaf anatomy: Kranz anatomy (Figure 18). Kranz, German for wreath, refers to a bundle sheath of cells that surrounds the central vein within the leaf, which in turn are surrounded by the mesophyll cells. The mesophyll cells in such leaves are rich in the enzyme phosphoenolpyruvate (PEP) carboxylase, which fixes CO 2 into a 4C carboxylic acid: oxaloaceatate. The oxaloacetate formed by the mesophyll cells is reduced using NADPH to malate, another 4C acid: malate. The malate is then exported from the mesophyll cells to the bundle sheath cells, where it is decarboxylated to pyruvate thus regenerating NADPH and CO 2. The CO 2 is then utilized by Rubisco in the Calvin cycle. The pyruvate is in turn returned to the mesophyll cells where it is phosphorylated using ATP to reform PEP (Figure 19). The advantage of C 4 photosynthesis is that CO 2 accumulates at a very high concentration in the bundle sheath cells that is then sufficient to allow Rubisco to operate efficiently.
Figure 18. Diagram of a C 4 plant leaf showing Kranz anatomy.
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Figure 19. The C 4 pathway (NADP+–malic enzyme type) for fixation of CO 2.
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Plants growing in hot, bright and dry conditions inevitably have to have their stomata closed for large parts of the day to avoid excessive water loss and wilting. The net result is that the internal CO 2 concentration in the leaf is very low, meaning that C 3 photosynthesis is not possible. To counter this limitation, another CCM is found in succulent plants such as cacti. The Crassulaceae fix CO 2 into malate during the day via PEP carboxylase, store it within the vacuole of the plant cell at night and then release it within their tissues by day to be fixed via normal C 3 photosynthesis. This is termed crassulacean acid metabolism (CAM).
Acknowledgments
I thank Professor Colin Osborne (University of Sheffield, Sheffield, U.K.) for useful discussions on the article, Dr Dan Canniffe (Penn State University, Pennsylvania, PA, U.S.A.) for providing pure pigment spectra and Dr P.J. Weaire (Kingston University, Kingston-upon-Thames, U.K.) for his original Photosynthesis BASC article (1994) on which this essay is partly based.
Abbreviations
ADP
adenosine diphosphate
ATP
adenosine triphosphate
CH 2 O
carbohydrate
cyt b 6 f
cytochrome b 6 f
DHAP
dihydroxyacetone phosphate
EET
excitation energy transfer
FNR
ferredoxin–NADP+ reductase
GAP
glyceraldehyde 3-phosphate
LHC
light-harvesting complex
NADPH
nicotinomide–adenine dinucleotide phosphate
PEP
phosphoenolpyruvate
P i
inorganic phosphate
PSI
Photosystem I
PSII
Photosystem II
RC
reaction centre
Rubisco
ribulose-1,5-bisphosphate carboxylase/oxygenase
This article is a reviewed, revised and updated version of the following ‘Biochemistry Across the School Curriculum’ (BASC) booklet: Weaire, P.J. (1994) Photosynthesis. For further information and to provide feedback on this or any other Biochemical Society education resource, please contact education@biochemistry.org. For further information on other Biochemical Society publications, please visit www.biochemistry.org/publications.
Competing Interests
The Author declares that there are no competing interests associated with this article.
Recommended reading and key publications
Blankenship R.E. Early evolution of photosynthesis. Plant Physiol. 2010;154:434–438. doi: 10.1104/pp.110.161687. [DOI] [PMC free article] [PubMed] [Google Scholar]
Blankenship R.E. Molecular Mechanisms of Photosynthesis. Chichester: Wiley–Blackwell Publishing; 2014. [Google Scholar]
Nelson N., Ben Shem A. The complex architecture of oxygenic photosynthesis. Nat. Rev. 2004;5:1–12. doi: 10.1038/nrm1525. [DOI] [PubMed] [Google Scholar]
Raines C. The Calvin cycle revisited. Photosynth. Res. 2003;75:1–10. doi: 10.1023/A:1022421515027. [DOI] [PubMed] [Google Scholar]
Ruban A.V. Evolution under the sun: optimizing light harvesting in photosynthesis. J. Exp. Bot. 2015;66:7–23. doi: 10.1093/jxb/eru400. [DOI] [PubMed] [Google Scholar]
Sage R.F. The evolution of C4 photosynthesis. New Phytol. 2004;161:341–370. doi: 10.1111/j.1469-8137.2004.00974.x. [DOI] [PubMed] [Google Scholar]
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How the photosystems work
Other electron transfer chain components
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8520 | https://www.media4math.com/LessonPlans/ApplyingRatiosGr6 | Skip to content
Lesson Plan: Applying Ratios and Proportional Reasoning
Lesson Summary
In this lesson, students will apply their understanding of ratios and proportional reasoning to solve complex, real-world problems. They will utilize various representations, such as tables, graphs, and equations, to model and analyze situations involving proportional relationships. Through engaging activities and practical examples, students will enhance their problem-solving skills and ability to communicate mathematical reasoning effectively.
Lesson Objectives
Apply ratio and proportional reasoning to solve complex real-world problems
Use multiple representations to model and solve problems
Common Core Standards
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
Prerequisite Skills
Understanding of ratios, rates, and proportional relationships
Problem-solving skills
Key Vocabulary
Proportion: An equation stating that two ratios are equivalent. For example, if (\frac{a}{b} = \frac{c}{d}), then (a), (b), (c), and (d) are in proportion.
Multimedia Resource: Definition
Constant of Proportionality: The constant value ((k)) that relates two proportional quantities, expressed as (y = kx). It represents the rate at which one quantity changes relative to another.
Linear Relationship: A relationship between two variables where the rate of change is constant, resulting in a straight-line graph. In the context of proportional relationships, the line passes through the origin.
Multimedia Resources
A collection of definitions on the topic of ratios, proportions, and percents:
A student tutorial slide show on definitions on the topic of ratios, proportions, and percents:
Warm Up Activities
Choose from one or more activities.
Activity 1: Review of Equivalent Ratios
Present the following multi-step real-world problem involving ratios:
"A recipe for 12 muffins requires 2 cups of flour and 3/4 cup of sugar. If you want to make 30 muffins, how much flour and sugar will you need?"
Solution using a table of equivalent ratios
Create a table with columns for muffins, flour, and sugar.
Fill in the given information for 12 muffins.
Extend the table to show equivalent ratios for 24 and 36 muffins.
Use the table to solve for 30 muffins.
| Muffins | Flour (cups) | Sugar (cups) |
---
| 12 | 2 | 3/4 |
| 24 | 4 | 1 1/2 |
| 36 | 6 | 2 1/4 |
Calculation for 30 muffins
Observe that 30 is between 24 and 36 muffins.
Find the scale factor: 30 ÷ 12 = 2.5
Multiply the original amounts by 2.5: Flour: 2 cups × 2.5 = 5 cups Sugar: 3/4 cup × 2.5 = 1 7/8 cups
Answer: For 30 muffins, you need 5 cups of flour and 1 7/8 cups of sugar.
Activity 2: Review of Tape Diagrams
Objective: Help students visualize ratios and proportional relationships using tape diagrams.
Instructions:
Draw a tape diagram on the board to represent the ratio 2:3. Explain that each segment represents one part of the ratio.
Provide a real-world scenario: "For every 2 cups of flour, 3 cups of sugar are needed in a recipe. If we use 4 cups of flour, how much sugar is needed?"
Have students extend the tape diagram to show an equivalent ratio of 4:6.
Encourage students to use tape diagrams to model other proportional relationships, such as the ratio of boys to girls in a classroom.
Activity 3: Review of Double Number Lines
Objective: Reinforce proportional reasoning by using double number lines with whole-number alignments.
Instructions:
Draw a double number line with two corresponding scales: one for the number of pencils in a pack and one for the total cost.
Use the example: "A store sells packs of pencils at a rate of 3 pencils for $2."
Label the number of pencils on the top line (3, 6, 9, 12, 15) and the cost on the bottom line ($2, $4, $6, $8, $10).
Ask students: "If a customer buys 12 pencils, how much will they pay?"
Discuss how the double number line helps visualize proportional relationships and find missing values.
This activity provides a clear and practical example of using double number lines to solve proportional problems while keeping numbers whole and easy to work with.
Teach
Demonstrate how to approach complex problems using various representations and strategies, such as tables, graphs, and equations. Emphasize the importance of clear communication in problem-solving, encouraging students to explain their reasoning.
Present the following real-world problems and their solutions:
1. Art Application: Color Mixing
Problem: An artist mixes yellow and blue paint in a 3:2 ratio to create green. If they need 25 ounces of green paint, how much yellow and blue paint should they use?
Solution using equivalent ratios:Set up the proportion: 3 + 2 = 5 parts total25 ÷ 5 = 5 ounces per partYellow: 3 × 5 = 15 ouncesBlue: 2 × 5 = 10 ounces
Answer: The artist needs 15 ounces of yellow paint and 10 ounces of blue paint.
2. Business Application: Sales Commission
Problem: A salesperson earns a 5% commission on all sales. Last month, their total sales were \$45,000. This month, they aim to earn \$3,000 in commission. What should their total sales be?
Solution using an equation
Set up a percent equation:
Sales • 5% = Commission
We know the commission and need to calculate the sales. Sales is the unknown:
x • 5% = 3000
Solve for x:
x = 3000 ÷ 0.05
x = 60,000
Answer: The salesperson needs \$60,000 in total sales to earn \$3,000 in commission.
3. Science Application: Chemical Dilution
Problem: A chemist needs to dilute a solution that is 90% acid. The diluted solution should be 40% acid and have a volume of 225 mL. How much of the original solution and water should be used?
This slide show provides a detailed solution to this problem:
Emphasize how these different solution techniques (tape diagrams, equivalence tables, and equivalent ratios) can be applied to various real-world problems. Encourage students to choose the method that works best for them in each situation.
Additional Examples
Example 1: Using Equivalent Ratios
Problem: A bakery uses 4 cups of flour to make 6 loaves of bread. How many cups of flour are needed to make 18 loaves?
Solution:
Set up the given ratio: [ \frac{4 \text{ cups}}{6 \text{ loaves}} ]
Find an equivalent ratio with 18 loaves by scaling up:
Multiply both terms by 3: [ \frac{4 \times 3}{6 \times 3} = \frac{12}{18} ]
Final Answer: The bakery needs 12 cups of flour to make 18 loaves.
Example 2: Solving with Proportions
Problem: A school has 24 teachers for 480 students. If the school expands to 600 students, how many teachers should it have to maintain the same ratio?
Solution:
Set up a proportion: [ \frac{24}{480} = \frac{x}{600} ]
Cross multiply: [ 24 \times 600 = 480 \times x ] [ 14400 = 480x ]
Divide by 480: [ x = \frac{14400}{480} = 30 ]
Final Answer: The school needs 30 teachers for 600 students.
Example 3: Solving with Tape Diagrams
Problem: A fruit stand sells 3 apples for every 2 oranges. If the stand has 12 apples, how many oranges should they have?
Solution:
Draw a tape diagram with two sections: one for apples and one for oranges.
Since the ratio is 3 apples to 2 oranges, divide the apple section into 3 equal parts and the orange section into 2 equal parts.
Label the tape diagram for the apples as 3, 6, 9, and 12 (in increments of 3).
Label the corresponding orange sections as 2, 4, 6, and 8.
Since 12 apples correspond to 8 oranges in the diagram, the fruit stand should have 8 oranges.
Example 4: Solving with Double Number Lines
Problem: A grocery store sells 2 pounds of bananas for $3. How much would 8 pounds of bananas cost?
Solution:
Draw a double number line with two scales:
Top scale: Pounds of bananas (2, 4, 6, 8).
Bottom scale: Price ($3, $6, $9, $12).
Identify the cost of 8 pounds by following the pattern.
Since 8 pounds aligns with $12, the total cost is $12.
Review
Lesson Summary
In this lesson, students applied their understanding of ratios and proportions to solve real-world problems using different mathematical representations. They explored:
Equivalent Ratios: Finding missing values in a proportional relationship by scaling up or down.
Proportions: Setting up and solving equations that express two equal ratios.
Tape Diagrams: Visual models that represent ratios and help in solving problems step by step.
Double Number Lines: A structured way to visualize proportional relationships and find missing values.
By practicing these methods, students developed flexibility in problem-solving and strengthened their ability to analyze proportional relationships in various contexts.
Key Vocabulary
Ratio: A comparison between two quantities, often written as a fraction, using a colon (3:4), or in words ("3 to 4").
Equivalent Ratios: Ratios that express the same relationship but are scaled versions of one another (e.g., 2:3 is equivalent to 4:6).
Proportion: An equation that shows two ratios are equal, such as ( \frac{a}{b} = \frac{c}{d} ).
Constant of Proportionality: The fixed value ( k ) that relates two variables in a proportional relationship, where ( y = kx ).
Tape Diagram: A visual model that represents ratios using bar segments.
Double Number Line: A pair of aligned number lines used to compare proportional relationships.
Activities
Students work in groups to solve complex ratio and proportion problems, presenting their solutions to the class. Encourage the use of tables of equivalent ratios and other representations. Use problems from as a starting point, adapting them to be more challenging.
Additional Worked-Out Examples
Example 1: Using Equivalent Ratios
Problem: A recipe calls for 2 cups of sugar for every 5 cups of flour. How much sugar is needed for a batch that uses 20 cups of flour?
Solution:
Set up the equivalent ratio: [ \frac{2}{5} = \frac{x}{20} ]
Find the scale factor: ( 20 \div 5 = 4 ).
Multiply the sugar by the same factor: ( 2 \times 4 = 8 ).
Final Answer: 8 cups of sugar are needed.
Example 2: Solving with Proportions
Problem: A car travels 180 miles in 3 hours. At the same speed, how far can it travel in 5 hours?
Solution:
Set up a proportion: [ \frac{180}{3} = \frac{x}{5} ]
Cross multiply: [ 180 \times 5 = 3x ] [ 900 = 3x ]
Divide by 3: [ x = 300 ]
Final Answer: The car can travel 300 miles in 5 hours.
Example 3: Using a Tape Diagram
Problem: At a pet store, the ratio of cats to dogs is 5:2. If there are 15 cats, how many dogs are there?
Solution:
Draw a tape diagram with 5 equal sections for cats and 2 equal sections for dogs.
Label the sections for cats as 5, 10, and 15.
Since each section represents 3, the corresponding sections for dogs are 2, 4, and 6.
Final Answer: The pet store has 6 dogs.
Example 4: Using a Double Number Line
Problem: A juice company sells 3 bottles of juice for $5. How much would 12 bottles cost?
Solution:
Draw a double number line with:
Top scale: Number of bottles (3, 6, 9, 12).
Bottom scale: Price ($5, $10, $15, $20).
Identify the cost of 12 bottles from the pattern.
Final Answer: The total cost is $20.
Quiz
Answer the following questions.
A car travels 210 miles on 7 gallons of gas. How many miles can it travel on 12 gallons?
In a school, the ratio of boys to girls is 5:6. If there are 132 students in total, how many boys are there?
A recipe calls for 2 1/4 cups of flour for 9 servings. How much flour is needed for 15 servings?
If you earn \$720 for working 40 hours, what is your hourly wage?
A map has a scale of 1 inch : 25 miles. If two cities are 7.5 inches apart on the map, what is their actual distance?
In a bag of marbles, the ratio of red to blue marbles is 3:5. If there are 24 red marbles, how many blue marbles are there?
A store offers a 15% discount on all items. If an item originally cost $80, what is its discounted price?
If 3 pounds of apples cost $4.50, how much would 7 pounds cost?
A rectangle has a length-to-width ratio of 4:3. If its perimeter is 70 inches, what are its dimensions?
In a survey, 3 out of every 8 people preferred brand A. If 600 people were surveyed, how many preferred brand A?
Answer Key
360 miles
60 boys
3 3/4 cups
\$18/hr
187.5 miles
40 blue marbles
$68
$10.50
Length: 20 inches, Width: 15 inches
225 people
Purchase the lesson plan bundle. Click here. |
8521 | https://pmc.ncbi.nlm.nih.gov/articles/PMC4214538/ | Nephrotic Syndrome: Components, Connections, and Angiopoietin-Like 4–Related Therapeutics - PMC
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J Am Soc Nephrol
. 2014 May 22;25(11):2393–2398. doi: 10.1681/ASN.2014030267
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Nephrotic Syndrome: Components, Connections, and Angiopoietin-Like 4–Related Therapeutics
Camille Macé
Camille Macé
1 Glomerular Disease Therapeutics Laboratory, University of Alabama at Birmingham, Birmingham, Alabama
Find articles by Camille Macé
1, Sumant S Chugh
Sumant S Chugh
1 Glomerular Disease Therapeutics Laboratory, University of Alabama at Birmingham, Birmingham, Alabama
Find articles by Sumant S Chugh
1,✉
Author information
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1 Glomerular Disease Therapeutics Laboratory, University of Alabama at Birmingham, Birmingham, Alabama
✉
Correspondence: Dr. Sumant S. Chugh, University of Alabama at Birmingham Division of Nephrology, THT 611L, 1900 University Boulevard, Birmingham, AL 35294. Email: chugh@uab.edu
✉
Corresponding author.
Issue date 2014 Nov.
Copyright © 2014 by the American Society of Nephrology
PMC Copyright notice
PMCID: PMC4214538 PMID: 24854282
Abstract
Nephrotic syndrome is recognized by the presence of proteinuria in excess of 3.5 g/24 h along with hypoalbuminemia, edema, hyperlipidemia (hypertriglyceridemia and hypercholesterolemia), and lipiduria. Each component has been investigated individually over the past four decades with some success. Studies published recently have started unraveling the molecular basis of proteinuria and its relationship with other components. We now have improved understanding of the threshold for nephrotic-range proteinuria and the pathogenesis of hypertriglyceridemia. These studies reveal that modifying sialylation of the soluble glycoprotein angiopoietin-like 4 or changing key amino acids in its sequence can be used successfully to treat proteinuria. Treatment strategies on the basis of fundamental relationships among different components of nephrotic syndrome use naturally occurring pathways and have great potential for future development into clinically relevant therapeutic agents.
Keywords: podocyte, nephrotic syndrome, proteinuria, lipids, FSGS, diabetic nephropathy
Nephrotic syndrome is a hallmark of glomerular disease and characterized by the presence of proteinuria in excess of 3.5 g/24 h, hypoalbuminemia, and variable amounts of hyperlipidemia (hypertriglyceridemia and hypercholesterolemia), lipiduria, and edema1 (Figure 1). In children, nephrotic-range proteinuria is defined by urinary protein excretion rates >40 mg/h per meter 2. Patients with primary glomerular diseases (e.g., minimal change disease [MCD], FSGS, and membranous nephropathy) and systemic disorders (e.g., diabetes mellitus, systemic lupus erythematosis, and amyloidosis) can present with nephrotic syndrome. The principle driving force in nephrotic syndrome is proteinuria, because other components develop only after proteinuria reaches a certain threshold. Substantial research effort has been committed to understanding the pathogenesis of each individual component. The purpose of this review is to discuss in broad terms what is understood about each component and then consider what remains to be investigated. Next, molecular connections, if known, between proteinuria and the other components will be discussed. Finally, novel therapeutic strategies derived from the study of nephrotic syndrome will be outlined.
Figure 1.
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The nephrotic syndrome tree is shown. The trunk depicts increasing proteinuria, and the branches represent other components that appear when proteinuria crosses the nephrotic-range threshold.
Proteinuria
Several genes expressed in podocytes have now been directly or indirectly implicated in the pathogenesis of proteinuria. They can be classified as slit diaphragm-related, cell matrix interface–related, cytoskeleton-related, podocyte surface proteins, transcriptional factors, and podocyte-secreted proteins.2 Many of the structural proteins among these categories were discovered when screening for mutations in patients with disease (e.g., nephrin for congenital nephrotic syndrome of the Finnish type3 and podocin for FSGS4). These genes are best viewed as disease-related genes, with proteinuria being a major component of the associated disease. Others were discovered through mutagenesis studies in mice (e.g., neph1).5,6 In most cases, these genes do not single-handedly explain the development of proteinuria. Transcriptional factors, like WT1, came into light from a combination of genetics7 and animal model8 studies, whereas the role of ZHX proteins was discovered through differential gene expression studies in animal models.2,9 Among podocyte-secreted proteins implicated in human disease, the roles of angiopoietin-like 4 (Angptl4) and vascular endothelial growth factor in proteinuria were established through the study of experimental models.10–12 A hyposialylated form of Angptl4 secreted from podocytes is directly implicated in the pathogenesis of proteinuria in MCD and accounts for most of the cardinal manifestations of this disease, including glucocorticoid sensitivity, selective proteinuria, loss of glomerular basement membrane (GBM) charge, and classic morphologic changes.1,11 Despite demonstration that loss of GBM charge in MCD is caused by binding of Angptl4, it is not known whether complex interaction of this protein with heparan sulfate proteoglycans and other GBM proteins or actual loss of GBM charge is responsible for proteinuria. Vascular endothelial growth factor has been implicated in human thrombotic microangiopathy.13 The added dimension of podocyte-secreted proteins is their ability to reach out to binding partners on the surface of glomerular endothelial cells and potentially participate in feedback loops within the glomerulus.14 In addition to podocyte-secreted proteins, high plasma levels of the soluble urokinase receptor are being investigated for their role in the pathogenesis of FSGS.15
A new dimension recently added to this field is the systemic response to proteinuria when it reaches nephrotic range. A circulating sialylated form of Angptl4 secreted predominantly from adipose tissue, skeletal muscle, and heart reduces proteinuria, at least in part, by binding to glomerular endothelial α v β 5 integrin.12 Thematically, it opens up a new area of investigation to study how other organs reduce established proteinuria by secreting proteins into the circulation. Angptl4 is the first of several such proteins that are likely to be identified in the future.
Edema
The onset of edema in nephrotic syndrome occupies a clinical spectrum that varies from subacute to acute onset in many patients with FSGS or membranous nephropathy, explosive onset (often overnight) in MCD, and complete absence in many patients with HIV-related collapsing glomerulopathy. From a pathophysiology standpoint, edema requires a combination of hypoalbuminemia, renal salt retention, and increased peripheral capillary permeability, because there are numerous clinical situations involving a single component that are not associated with edema. The variability of edema in different clinical situations may be directly related to differences between these pathogenic components. The best studied aspect and indeed, the primary target of diuretic therapy is renal tubular salt retention.16 In order of diuretic efficacy, transporters in the thick ascending loop of Henle, distal tubule, collecting duct, and proximal tubule have been targeted to reduce edema. Recent studies suggest that the proteolytic activation of collecting duct epithelial sodium channel may be mediated by plasmin converted from filtered plasminogen.17,18 Treatment of edema with albumin infusions is generally not practiced, except in selected cases of refractory anasarca. Perhaps the least understood part of edema is increased peripheral capillary permeability, and no current therapy targets this aspect. This area has tremendous potential for investigation, and mechanistic studies may reveal useful clues for treating patients with refractory edema. Investigating the explosive onset of edema in MCD could provide insight into additional molecular mechanisms in nephrotic syndrome, because there is a potential for glomerular or peripherally secreted proteins in the pathogenesis of this phenomenon.
Hypercholesterolemia
Nephrotic patients have elevated total and LDL cholesterol levels, largely related to an acquired LDL receptor deficiency, which limits the removal of cholesterol-rich LDL particles from the circulation.19 This reduction in uptake of extracellular cholesterol stimulates cholesterol biosynthesis by upregulating hepatic 3-hydroxy-3-methyl glutaryl-CoA reductase expression and activity in the nephrotic liver. Increased hepatic activity of Acyl-CoA cholesterol acyltransferase-2, an enzyme responsible for esterification of cholesterol, is also noted. Presence of normal LDL receptor mRNA expression in the nephrotic liver suggests a post-transcriptional etiology. A recent study in experimental animals suggests that increased hepatic degradation of the LDL receptor by proprotein convertase subtilisin/kexin type 9 and inducible degrader of the LDL receptor may be involved.20 Also, urinary loss of plasma proteins like lecithin–cholesterol acyltransferase may also contribute to hypercholesterolemia. However, the precise sequence of events and the molecular relationship of these changes with proteinuria remain unknown.
Hypertriglyceridemia
Hypertriglyceridemia results from impaired clearance of triglycerides in very low-density lipoprotein and chylomicrons because of inactivation of endothelium-bound lipoprotein lipase (LPL) activity by the circulating glycoprotein Angptl4, which reduces the conversion of circulating triglycerides to free fatty acids (FFAs).12 Circulating sialylated Angptl4 is mostly secreted from skeletal muscle, adipose tissue, and heart to reduce proteinuria through glomerular endothelial binding, but it also causes hypertriglyceridemia as a side effect. Interaction of Angptl4 with LPL converts the active dimeric form of this protein into inactive monomers. Both dimers and monomers of LPL are lost in the urine in nephrotic syndrome. Proteinuria and hypertriglyceridemia are linked by two negative feedback loops, which are discussed below.
Hypoalbuminemia
Hypoalbuminemia results from urinary losses of albumin during proteinuria, insufficient compensation by hepatic synthesis, and perhaps, increased albumin catabolism. The major enigma in the pathogenesis of hypoalbuminemia is the inability of the nephrotic liver to increase albumin synthesis to compensate for urinary losses, although a normal liver synthesizes 12–14 g albumin/d and can increase production 3-fold in times of demand.21 Whereas a lot of importance was traditionally placed on reduced plasma oncotic pressure resulting from hypoalbuminemia, it has now become clear that changes in FFA binding to albumin are equally important.12 Plasma FFAs are noncovalently bound to albumin through six high-affinity sites and several low-affinity sites.14 During proteinuria, patients lose, for unclear reasons, albumin with lower FFA content, resulting in the accumulation of albumin with higher FFA content.12,22,23 This loss, along with the development of hypoalbuminemia, results in a high plasma FFA-to-albumin ratio, which in turn, drives FFA uptake in skeletal muscle, heart, and adipose tissue.12 This event is primarily responsible for the development of hypertriglyceridemia and as discussed below, an attempt of these organs to reduce proteinuria through the secretion of Angptl4 into the circulation.
Lipiduria
Lipiduria is thought to be secondary to hyperlipidemia, and it mostly results from filtration of HDL particles because of their relatively small size.24 These lipids are found in oval fat bodies (sloughed tubular cells with lipids), fatty casts, or free-floating lipid globules. Components that have high amounts of esterified cholesterol have a Maltese cross appearance under polarized light.
Molecular Links between Proteinuria and Other Components of Nephrotic Syndrome
There are large gaps of knowledge in our understanding of the molecular relationship between proteinuria, the primary driver in nephrotic syndrome, and most of the other components. Only the link between proteinuria and hypertriglyceridemia has been clearly elucidated.12,14 This relationship is strongly influenced by the link between FFA and albumin. FFAs are a major fuel source for skeletal muscle and heart, and they are stored in adipose tissue as triglycerides. Under normal conditions, FFA uptake in these organs relies on a combination of FFA released by LPL-mediated hydrolysis of circulating triglyceride and albumin-bound FFAs that are derived from diet (medium-chain FFA) or lipolysis of stored triglyceride in adipose tissue in the fasting state (Figure 2). These organs also have high expression for LPL, Angptl4, and peroxisome proliferator-activated receptor (PPAR) family members, which regulate Angptl4 expression in response to FFA uptake. In nephrotic syndrome, there is urinary loss of albumin with low FFA content, leading to retention of albumin with high FFA content, which alters the FFA uptake balance in favor of albumin-bound FFA. Increase in uptake of albumin-bound FFA induces local Angptl4 upregulation, likely through PPARs, which then inactivates LPL in the same tissues, leading to reduction in the formation of FFA from triglycerides and consequently, hypertriglyceridemia. This change in balance between two sources of FFA can be viewed as a product of a local negative feedback loop (Figure 3). However, this local feedback loop is subservient to a much larger systemic negative feedback loop to reduce proteinuria. Indeed, Angptl4 released from the skeletal muscle, heart, and adipose tissue into the circulation also binds to glomerular endothelial α v β 5 integrin to reduce proteinuria. Patients with all causes of nephrotic syndrome studied have elevated plasma Angptl4 levels.12 Studies in diabetic rats using low doses of recombinant human Angptl4 suggest a lower threshold to participate in the systemic rather than the local feedback loop. Overall, it seems that the local feedback loop reduces the effectiveness of the systemic feedback loop by limiting the extent of Angptl4 upregulation. Overall, the antiproteinuric and hypertriglyceridemia-inducing effects of Angptl4 are dependent on high uptake of albumin-bound FFA in peripheral organs and indirectly dependent on disproportionate retention of albumin with high FFA content in nephrotic syndrome.
Figure 2.
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Schematic illustration of the two sources of FFA available for uptake by skeletal muscle, heart, and adipose tissue in the normal and nephrotic state. Green shows normal conditions, and red illustrates changes in nephrotic syndrome. The balance shifts significantly to albumin-bound FFA because of retention of albumin with high FFA content in nephrotic syndrome. Angptl4 secreted from these organs reduces the conversion of triglycerides to FFA by inactivating LPL, thereby reducing use of triglycerides and resulting in hypertriglyceridemia.
Figure 3.
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Schematic illustration of negative feedback loops in the link between proteinuria, hypoalbuminemia, and hypertriglyceridemia mediated by Angptl4 and FFA. Plasma FFAs are noncovalently bound to albumin. Because of the preferential loss of albumin with low FFA content in nephrotic syndrome, there is a relative increase in circulating albumin with higher FFA content. Because glomerular disease progresses to severe proteinuria, hypoalbuminemia develops, and the combination of high albumin FFA content and lower plasma albumin levels increases the FFA-to-albumin ratio. It promotes entry of FFA into skeletal muscle, heart, and adipose tissue, which causes upregulation of Angptl4 at least partially mediated by PPARs. Angptl4 secreted from these organs participates in two feedback loops. In the systemic loop, it binds to glomerular endothelial α v β 5 integrin and reduces proteinuria. In a local loop, it inactivates LPL in the same organs from which it is secreted to reduce the uptake of FFA, thereby curtailing a stimulus for its own upregulation. Reproduced from reference 14, with permission.
Molecular Basis for Nephrotic-Range Proteinuria Threshold
Patients do not start manifesting the other components of nephrotic syndrome until they cross the nephrotic-range proteinuria threshold, which is usually defined as about 3.5 g/d in adults, but it is likely to be quite variable between different individuals and within the same individual among different components. Until recently, the molecular basis for this threshold was not known. It is, at present, only possible to explain the nephrotic threshold in the context of hypertriglyceridemia.12 Hypertriglyceridemia is dependent on the retention of albumin with high FFA content, resulting in a plasma FFA-to-albumin ratio that is high enough to induce upregulation of Angptl4 expression in skeletal muscle, heart, and adipose tissue. Studies in experimental animals reveal that, during mild proteinuria, plasma FFA-to-albumin ratio, plasma Angptl4 levels, peripheral organ Angptl4, and PPAR mRNA expression are similar to control nonproteinuric animals. During severe proteinuria, all these parameters are significantly elevated, suggesting that the threshold for nephrotic-range proteinuria, in the context of hypertriglyceridemia, correlates with the downstream effects of an increased plasma FFA-to-albumin ratio.
Therapeutic Strategies Developed from the Study of Nephrotic Syndrome
A central role played by Angptl4 in nephrotic syndrome makes it suitable as a therapeutic agent as well as a target.14,25 An important lesson learned from Angptl4 is its “Jekyll and Hyde” effect, because the same protein can have very different biologic effects on the basis of the presence or absence of sialic acid residues and the compartment in which it is secreted. Because circulating sialylated Angptl4 has an antiproteinuric effect, it is conceivable that injecting recombinant human Angptl4 or increasing the secretion of native Angptl4 from heart, skeletal muscle, and adipose tissue using PPAR agonists, glucocorticoids, FFA supplements, or β-adrenergic agonists could be of potential benefit in reducing proteinuria through the systemic feedback loop. However, these interventions run the risk of aggravating hypertriglyceridemia through the local feedback loop and could also reduce FFA uptake in the heart and skeletal muscle to below a threshold for organ dysfunction. In addition, all the drugs mentioned above have multiple side effects. Mutating Angptl4 at amino acid 40 or 39 to reduce its interaction with LPL bypasses the local feedback loop and allows for mutant recombinant human Angptl4 to very significantly reduce proteinuria in nephrotic animals with FSGS or diabetic nephropathy without affecting plasma triglyceride levels.12 This strategy is a novel futuristic treatment for all etiologies of proteinuria and nephrotic syndrome, especially if studies on the long-term administration of mutant human Angptl4 currently in progress also show improvement in the progression of CKD.
It is also clear that podocyte-secreted hyposialylated Angptl4 mediates proteinuria in MCD1,11 and also contributes to proteinuria in diabetic nephropathy.14 The effects of hyposialylated Angptl4 are most likely related to its binding to the GBM,11 but adverse consequences of glomerular endothelial binding are also possible.12 Converting hyposialylated Angptl4 to sialylated protein using N-acetyl-d-manosamine, a precursor of sialic acid that can be taken up and stored in podocytes, very significantly reduces proteinuria and has the potential for use in small maintenance doses to prevent relapse in MCD and as maintenance therapy for diabetic nephropathy.14
Future Interventions on the Basis of the Study of Nephrotic Syndrome
The proteinuria–hypertriglyceridemia relationship shows that other organs in the body make an attempt to reduce proteinuria by secreting specific proteins. This area of investigation would benefit from additional clinical studies in the future. Perhaps other unresolved or partially resolved components of nephrotic syndrome may also result from a similar systemic response involving other putative proteins. Identifying and modifying this circulating glomerulophilic proteome may hold the key to developing additional novel therapies for proteinuric kidney disease in the future.
Disclosures
S.S.C. is Founder, President, and Chief Executive Officer of GDTHERAPY LLC and filed patents related to the use of Angptl4 mutants (PCT/US2011/039255) and precursors of sialic acid, including N-acetyl-d-manosamine (PCT/US2011/039058), for the treatment of nephrotic syndrome. He may benefit financially from these patents in the future. C.M. declares no competing financial interests.
Acknowledgments
This work was supported by National Institutes of Health Grants T32-DK007545 (to C.M.), R01-DK077073 (to S.S.C.), R01-DK090035 (to S.S.C.), and R01-DK101637 (to S.S.C.).
Footnotes
Published online ahead of print. Publication date available at www.jasn.org.
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Abstract
Proteinuria
Edema
Hypercholesterolemia
Hypertriglyceridemia
Hypoalbuminemia
Lipiduria
Molecular Links between Proteinuria and Other Components of Nephrotic Syndrome
Molecular Basis for Nephrotic-Range Proteinuria Threshold
Therapeutic Strategies Developed from the Study of Nephrotic Syndrome
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Back to Top |
8522 | https://www.chess.com/forum/view/general/social-media-and-development | The Chess Board
I dont have the answer to this question.
is there a relationship between the 4 center squares of a board, and the 4 corners?
If you take a center square and it's corresponding corner (d4 and a1, for example), it forms a 4x4 chess board with the proper 'white on the right.'
If you fold the four center squares in half diagonally, you get two triangles and a diamond of opposite colors. If you fold them in half vertically or horizontally, you get a grey blob. :( (The same would apply for the whole chess board, but on a larger scale.)
A white pawn on e4 or d4 can reach either a8 or h8, but a white pawn on e5 can no longer reach a8, and a white pawn on d5 can no longer reach h8.
There are lots of other strange things we could make up. Are you getting at something in particular?
None of those squares are protected by any peice in the starting position?
right.. i just thought of it. sorry, please dont feel compelled to solve it.
Thank you!
The real question is whether there's any way to get those six minutes of my life back.
i doubt it... i think i am getting close to solving this one.
Well, the only thing that video triggered for me that might apply to the chess board is that if you similarly number the squares of the chess board the sum of the numbers on the four corners (1+8+57+64=130) is the same as the sum of the numbers on the center squares (28+29+36+37=130). This is actually true of any four squares that form a larger square the the corners on those two diagonals. The video itself was nonsense though.
The problem is that this isn't something that's an attribute of the chess board, but rather of any sequential NxN array of numbers. The coordinates of a chess board aren't even referenced in this way....
thank you for sharing that info with me.. interesting..
how about this to think about.. the relationship... well if you were to set a number of 1 or 8 starting in any diagonal no matter what point and number you begin with the center squares will always be 4... 4+4+4+4
we could say that the center is a microcosm of the board. the 4 center squares, with the 4 corners.
this leads us to another number which is the number 16.
if you were to add 4 4 4 4 in the center that gives you 16.
if you were to devide the board in 4 that gives you 16 squares in each direction
if you devide the board in any direction or form you get 16 squares.
there are 16 pieces for white, and 16 pieces for black.
the last thing i could think of, which is one of the first i did when thinking baout this, was that if you imagine the chessboard being lifted by the center squares, you get a shape. i dont know the name of that shape.
interesting?!
The 4/16/64 relationship happens to work out because all are powers of two. the board is 8x8 (8 = 2³) so this should really come as no surprise. Pure coincidence.
I don't really follow you on the shape that results from picking the board up by the center four squares though -- assuming the squares themselves are rigid, depending the relative rigidity of the vertices the board will fold one of two ways: Either leaving the d and e files facing skyward or the 4th and 5th ranks.
if you control the d5 square as white does that mean you could gain an initiative on the queen side? if you gain control of the e5 square as whites does that mean you could own the kingside!?!?!?!
thessseeeeeeeeee are the quessstttiooonnsssss that you musttttt assskkkkkkkkkk (check out my latest video)
I specialize in useless information.
this is too advanced. i agree.
There are a couple of lessons in Chess Mentor involving the concept of
"Corresponding Squares" which are fascinating, although I do not fully grasp
the concept yet.
Perhaps these center squares correspond in a way with the corner squares
in a way nobody has solved yet, and this is why computers are now so good at
chess.
In one of my blogs I wrote about the idea of "Visualizing the chessboard as
a circle instead of a square", something I have been experimenting with now.
Perhaps there is a paradox, by controlling the d5 square white gets an
initiative on the kingside, not the queenside, and we just don't see why yet.
Since computers are doing so well at chess now it seems logical that humans
have much to learn about this mysterious enigma known as chess.
This question is either very, very deep or completely pointless. I haven't decided which one yet.
I dont have the answer to this question.
is there a relationship between the 4 center squares of a board, and the 4 corners?
I know of a relationship between the 4 center squares and the 4 corners. If you arrange knights such that they form their own squares, only one center square and one corner square may be occupied. The first diagram shows an arrangement of knights which uses e4 and h8. The involved center square and corner square must be of opposite color. Thus e4 can correspond to either h8 or a1. There are 10 possible arrangements of knights because there are two arrangements which do not touch either the center or the corners. The second diagram shows an arrangement of knights which neither touches the center nor the corners.
thats beautiful ;)
There are a couple of lessons in Chess Mentor involving the concept of
"Corresponding Squares" which are fascinating, although I do not fully grasp
the concept yet.
Perhaps these center squares correspond in a way with the corner squares
in a way nobody has solved yet, and this is why computers are now so good at
chess.
In one of my blogs I wrote about the idea of "Visualizing the chessboard as
a circle instead of a square", something I have been experimenting with now.
Perhaps there is a paradox, by controlling the d5 square white gets an
initiative on the kingside, not the queenside, and we just don't see why yet.
Since computers are doing so well at chess now it seems logical that humans
have much to learn about this mysterious enigma known as chess.
corresponding sqaures... difficult! |
8523 | https://introcs.cs.princeton.edu/java/52turing/ | Intro to Programming
1. Elements of Programming
1.1 Your First Program
1.2 Built-in Types of Data
1.3 Conditionals and Loops
1.4 Arrays
1.5 Input and Output
1.6 Case Study: PageRank
2. Functions
2.1 Static Methods
2.2 Libraries and Clients
2.3 Recursion
2.4 Case Study: Percolation
3. OOP
3.1 Using Data Types
3.2 Creating Data Types
3.3 Designing Data Types
3.4 Case Study: N-Body
4. Data Structures
4.1 Performance
4.2 Sorting and Searching
4.3 Stacks and Queues
4.4 Symbol Tables
4.5 Case Study: Small World
Computer Science
5. Theory of Computing
5.1 Formal Languages
5.2 Turing Machines
5.3 Universality
5.4 Computability
5.5 Intractability
9.9 Cryptography
6. A Computing Machine
6.1 Representing Info
6.2 TOY Machine
6.3 TOY Programming
6.4 TOY Virtual Machine
7. Building a Computer
7.1 Boolean Logic
7.2 Basic Circuit Model
7.3 Combinational Circuits
7.4 Sequential Circuits
7.5 Digital Devices
Beyond
8. Systems
8.1 Library Programming
8.2 Compilers
8.3 Operating Systems
8.4 Networking
8.5 Applications Systems
9. Scientific Computation
9.1 Floating Point
9.2 Symbolic Methods
9.3 Numerical Integration
9.4 Differential Equations
9.5 Linear Algebra
9.6 Optimization
9.7 Data Analysis
9.8 Simulation
Related Booksites
| | |
--- |
| | |
Web Resources
+ FAQ
+ Data
+ Code
+ Errata
+ Lectures
+ Appendices
- A. Operator Precedence
- B. Writing Clear Code
- C. Glossary
- D. TOY Cheatsheet
- E. Matlab
+ Coursera MOOCs
- Programming with a Purpose
- Algorithms, Theory, Machines
+ Certificate Courses
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5.2 Turing Machines
This section under major construction.
Turing machine.
The Turing machine is one of the most beautiful and intriguing intellectual discoveries of the 20th century. Turing machine is a simple and useful abstract model of computation (and digital computers) that is general enough to embody any computer program. It forms the foundation of theoretical computer science. Because of its simple description and behavior, it is amenable to mathematical analysis. This analysis has led to a deeper understanding of digital computers and computation, including the revelation that there are some computational problems that cannot be solved on computers at all, no matter how fast the processor, or how much memory is available.
Turing machine simulator. This is a graphical Turing machine simulator that was written in Java by Tom Ventimiglia under the supervision of Bob Sedgewick and Kevin Wayne.
Executable jar (turing.jar). To execute, type java -jar turing.jar from the command line.
OS X app (Turing.zip). To execute, double click Turing.zip file to unzip. Double click Turing.app to launch.
You are welcome to inspect and modify the source code for your own use.
Components. Alan Turing sought to describe the most primitive model of a mechanical device that had the same basic capabilities as a human "computer." In his epoch making 1936 paper, Alan Turing introduced an abstract machine, which would later come to be known as a Turing machine. The machine consists of the following components:
The ticker-tape stores the input, the intermediate results, and the output. The tape is one arbitrarily long strip, divided into cells. Each cell stores one of a finite alphabet of symbols. In the example below, we use a 4 character alphabet consisting of 0, 1, A, X, and #.
The tape head of the Turing machine scans the tape one cell at a time. We refer to the cell being scanned as the active cell and the symbol it contains as the input symbol. At each time step, the tape head reads the input symbol, and leaves it either unchanged or overwrites it with a new symbol. At the end of each time step, the tape head moves one position to the left or right. We highlight the active cell in yellow. In the example below, the A is replaced with an X and the tape head moves one cell to the left.
The control unit is the analog of the CPU in modern day microprocessors. It consists of a state transition diagram, which is a finite table of instructions that specifies exactly what action the machine takes at each step. Each state represents one of the possible configurations of the machine. Depending on its current state and input symbol, the Turing machine overwrites the input symbol with a new symbol and moves to a new state. Each transition connects one state, say s, to another state, say t, and is labeled with two symbols, say A and X: this means that if the Turing machine is in state s and the input symbol is A, then it overwrite the A with an X and transitions to state t. Each state is labeled with one of five designations: L (left), R (right), Y (yes), N (no), or H (halt). Upon entering a state, the Turing machine either moves its tape head or halts according to the state's designation. Below is an illustration of the state transition diagram for a machine with four states.
Execution. Initially the Turing machine starts in one distinguished state called the start state, and the tape head points to one distinguished cell called the start cell. There is at most one possible transition corresponding to each combination of state and input symbol; thus, the actions of the machine are completely determined in advance. (If there is no possible transition from a state with some input symbol, then the Turing machine remains in the same state and does not overwrite the input symbol.) Each step in a Turing machine proceeds as follows:
Read the input symbol from the active cell.
Look up the transition rule associated with the current state and input symbol.
Overwrite the input symbol with the new symbol.
Change the current state according to the transition rule.
Shift the tape head one cell to the left or right, according to the new state's designation.
These steps are repeated until the current state is labeled H for halt, Y (in which case the machine answers yes) or N (in which case the machine answers no). It is possible that the machine runs forever without ever reaching one of these terminating states.
Computation must allow repetitive actions - do action A over and over until a certain condition is met. This amounts to staying in a state (and moving the tape head left or right) until a certain condition is met. Computation must also allow adaptive actions - if a certain condition is met, do action A; otherwise do action B. This is captured by state transitions according to the contents of the tape head at a particular location.
An example: unary to binary conversion. We consider the 4 state Turing machine illustrated below. The current state and input symbol are highlighted in yellow. We trace its execution.
Since the input symbol is A, the Turing machine follows the appropriate transition arrow leaving the current state - the one labeled A : X. The Turing machine overwrites the input symbol with an X, changes state to the bottom right state, and moves the tape head one position to the left (since the new state is labeled with L). The illustration below shows the Turing machine at the end of this first step.
Since the input symbol is now #, the Turing machine follows the appropriate transition arrow leaving the current state -- the one labeled # : 1. This overwrites the current cell with a 1, changes the state back to the bottom left state, and moves the tape head one position to the right (since the new state is labeled with R).
Here are the contents of the tape after the next several steps.
(Errata: in the fourth row, the highlighted cell should contain a # instead of a 1.)
Once all of the As are overwritten with Xs, the Turing machine erases all of the Xs (overwriting them with #s).
What it does and why it works. The Turing machine described above converts from unary to binary. That is, if the input consists of n consecutive A's, then the Turing machine prints the number n in binary to the left of sequence of A's (and overwrites the A's with X's). In the example above, the input consists of 6 A's and the Turing machine writes the binary number 110 to the tape.
To describe how this is accomplished, we first review an algorithm for incrementing a binary integer by 1: scan the bits from right to left, changing 1's to 0's until you see a 0. Then change the 0 to a 1.
The Turing machine repeatedly knocks off one A at a time and increments the binary number. Our Turing machine mimics this strategy. The initial state seeks out the next A, overwrites it with an X, and then transitions to the Increment state. The Increment state increments the binary integer by one (leaving the X's alone, changing 1's to 0's, until seeing a 0 or #, which it changes to a 1), and then transitions back to the Initial state. When all of the A's are overwritten with X's, the Cleanup state replaces all of the X's with #'s, and the transitions to the Halt state.
Turing machine implementation in Java.
We encapsulate each of the main Turing machine components (tape, transition, control) using good OOP principles.
The tape. Program Tape.java is an ADT that represents an unbounded Turing machine tape. It supports the following operations: move tape head left, move tape head right, read the symbol in the current cell, and write a symbol to the current cell. To implement it, we use two stacks (one to store all of the symbols to the left of the tape head, and one to the right). To print out the contents of the tape, we print out the reverse of the first stack, the current element, then the second stack.
The states. Each state has a name and a type (halt, left, right, accept, or reject).
The transitions. Each transition has the name of the initial state, the name of the final state, and the symbol to be written to the tape.
The Turing machine. We implement a Turing machine as a tape, a symbol table of states, and a symbol table of transitions.
Non-terminating Turing machines. From a theoretical standpoint, we are primarily concerned with machines that perform finite computations and then halt. However, many practical applications involve programs that are designed never to terminate (operating system, air traffic control system, nuclear reactor control system) or produce an infinite amount of output (web browser, program that computes the digits of π = 3.1415...). The Turing machine model of computation extends to handle such non-terminating situations as well.
Turing machines connect physics and mathematics (Turing's original motivation, thermodynamics of computation).
Exercises
What does the following Turing machine do when started with the given tape ...
Binary adder.
Binary counter.
Binary palindrome.
Unary multiplication.
Equal number of a's and b's.
Multiple of 3 or 7.
Balanced parentheses.
Power-of-2.
String compare.
Unary-to-binary. How many steps does the 3-state unary-to-binary Turing machine make to convert N to binary? Answer: proportional to N^2.
Unary-to-binary. Design a 6-state unary-to-binary Turing machine that converts the unary number N to binary in time proportional to N log N. Hint: cross out every other A. If the number of A's is odd, write a 1; otherwise write a 0. Repeat with the uncrossed out A's remaining.
Swap two cells on a Turing machine. Use the state to encode the temporary symbol.
Hex-to-binary. Design a Turing machine that converts from hexadecimal to binary.
Comparator. Create a Turing machine that takes as input two binary integers, separate by a # character, and accepts the input if the first string is strictly less than the second. How many steps does the Turing machine in the previous question take to compare two N-bit integers? (Each step is one move of the tape head.)
Efficient comparator. Create a comparator that runs in time polynomial in N.
Bitwise OR. Create a Turing machine that computes the bitwise OR of its two binary inputs of length N.
Creative Exercises
Doubling. Write a Turing machine that transform an input consisting of k consecutive 1's to an input that consists of 2k consecutive 1's. (unary multiplication by 2) Hint: write two 1's on the left, and delete one 1 on the right.
Copying. Write a Turing machine that transform an input consisting of 0's and 1's instead two copies of the original input, separated by the symbol #.
Langtons Ant. Write a program LangtonsAnt.java that simulates a two dimensional Turing machine known as Langton's Ant, and animate the results using Turtle graphics.
Turmites. Create some other two dimensional Turing machines or Turmites that produce interesting patterns.
Turing tape. Write a program Tape.java that implement a one-dimensional Turing tape. The tape consists of a sequence of cells, each of which stores an integer (that is initialized to 0). At any instant, there is a tape head which points to one of the cells. Support the following interface methods: moveLeft() to move the tape head one cell to the left, moveRight() to move the tape head one cell to the right, look() to return the contents of the active cell, and write(int a) to change the contents of the active cell to a. Hint: use an int for the active cell, and two stacks for the left and right parts of the tape.
Turing machine simulator. Write a program TuringMachine.java that simulates a Turing machine. Design your program as follows: Tape.java, State.java, Transition.java.
Collatz Turing machine. Design a Turing machine that takes as input the binary representation of a binary integer and repeatedly divides it by 2 (if even) or multiplies by 3 and adds 1 (if odd) until it equals 1. Famous open conjecture that this machine will terminates for inputs.
Last modified on June 30, 2017. Copyright © 2000–2025 Robert Sedgewick and Kevin Wayne. All rights reserved. |
8524 | https://artofproblemsolving.com/community/c3045h1057033_some_geometry_before_inmo?srsltid=AfmBOorMUPCgx-CrioAdhuP2MFVkQk4ZGObAWhBTYH3MAr8dZDMzWbxy | MemoryFromMyHighSchool : Some Geometry before INMO
Community » Blogs » MemoryFromMyHighSchool » Some Geometry before INMO
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MemoryFromMyHighSchool ======================
Some Geometry before INMO
by TripteshBiswas, Jan 31, 2015, 8:41 AM
L et be an acute-angled triangle with an dD enote the circumcentre by and the orthocentre by and let meet at and at
Prove that
L et be an acute-angled triangle whose inscribed circle touches and at and respectively. Let and be the points of intersection of the bisectors of the angles and with the line and let be the midpoint of . Prove that the triangle is equilateral if and only if
Let be the incenter of
are concyclic.
are concyclic.
So is the center of the circle passing through
So is equilateral
Consider a triangle with . Let be the foot of the altitude from .Circle touches the line segment at point , the altitude at point and the circumcircle of at point Prove that points are collinear and .
Let be the circumcircle of and
It is easy to see
Hence
Hence are collinear.
are concyclic
L et be a trapezoid with parallel sides . Points and lie on the line segments and , respectively, so that . Suppose that there are points and on the line segment satisfying and . Prove that the points ,, and concyclic.
Proposed by Vyacheslev Yasinskiy, Ukraine
Let and
Then
Let be a point on such that
Then
are concyclic.
are concyclic.
This post has been edited 26 times. Last edited by TripteshBiswas, Feb 20, 2015, 7:32 AM
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8525 | https://www.gauthmath.com/solution/1751035623748613/Solve-by-completing-the-square-r-2-4r-3-0-Write-your-answers-as-integers-proper- | Solved: Solve by completing the square. r^2-4r+3=0 Write your answers as integers, proper or impro [Math]
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Solve by completing the square. r^2-4r+3=0 Write your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth. r=□ or r= □ /□
Show transcript
Gauth AI Solution
Answer
$$r=3$$r=3 o $$r=1$$r=1
Explanation
Mueve la constante al lado derecho de la ecuación:
$$r^{2} - 4r = -3$$r 2−4 r=−3
Completa el cuadrado en el lado izquierdo. Toma la mitad del coeficiente del término $$r$$r, eleva al cuadrado y suma a ambos lados de la ecuación. La mitad de $$-4$$−4 es $$-2$$−2, y $$(-2)^{2} = 4$$(−2)2=4
$$r^{2} - 4r + 4 = -3 + 4$$r 2−4 r+4=−3+4
Factoriza el lado izquierdo como un cuadrado perfecto y simplifica el lado derecho.
$$(r - 2)^{2} = 1$$(r−2)2=1
Toma la raíz cuadrada de ambos lados. Recuerda incluir ambos signos.
$$r - 2 = \pm 1$$r−2=±1
Resuelve para $$r$$r
$$r = 2 \pm 1$$r=2±1
Encuentra las dos soluciones.
$$r = 2 + 1$$r=2+1 o $$r = 2 - 1$$r=2−1
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8526 | https://worldmentalcalculation.com/mental-cube-roots-algorithm/ | Skip to content
Mental Cube Roots Algorithm
There are many methods to mentally calculate cube roots (for numbers that are not an exact cube). In this article we explain a quick method that works on any number, and is similar to the method described for mental calculation of square roots.
Introduction:
This method is easier to understand if we first look at the answer squared. As an example, we will find the cube root of 397758. We can write this as:
3√397758 = ab.cdef… = 10a + 1b + 0.1c + 0.01d + 0.001e + 0.0001f + …
Therefore, by forming the cube of the left-hand and right-hand sides:
397758 =
1000 a³
100 3a²b
10 [3a²c + 3ab²]
1 [3a²d + 6abc + b³]
0.1 [3a²e + 6abd + 3ac² + 3b²c]
0.01 [et cetera]
Notice that the numbers become generally smaller and less significant as we read from the top to the bottom, because of the factors 1000, 100, 10, etc. Therefore we can start by calculating a, subtracting the top line and using the second line to calculate b, subtracting the second line and using the third line to calculate c, etc.
Method:
Step 1:
Calculate the first digit of the answer. Here, it is 7, because:
70³ = 343000 ≤ 397758 < 512000 = 80³
Step 2:
The quantity 3a² appears a lot in the explanation above, so we will calculate this now to help us later:
3a² = 377 = 147
Step 3:
We have used the first line of the explanation (1000 a³) to calculate the first digit a, so we can subtract this first line. The first line only uses multiples of 1000, so we can simplify the mental work by using only the largest digits:
397 – 343 = 54
We will next use the second line of the algorithm, which uses multiples of 100, so now we do need to consider the next (fourth) digit of the number 397758. From the explanation, we now have:
547 = 3a²b + [all the other lines, but these are small numbers that we will ignore until later]
To calculate b, we need to divide by 3a², and we know that in this example, 3a² = 147.
547 / 147 = 3 remainder 106
So b = 3, and the answer so far is 73.
Step 4:
We have finished with one line and now move onto the next. Bring the next digit from the original number 397758 onto the remainder:
106 –> 1065
Then the next (third) line of the explanation gives us:
1065 = 3a²c + 3ab² + [all the other lines]
We know a=7 and b=3, so we can subtract the term 3ab² = 3 7 3 3 = 189:
1065 – 189 = 3a² c + [all the other lines]
876 = 147 c + [all the other lines]
876 / 147 = 5 (remainder 141)
So c = 5, and the answer so far is 73.5…
Step 5:
Repeat step 4 for every new digit you wish to calculate. In summary, for every new digit:
multiply the remainder by 10, and add the next digit from the original question
subtract all terms from the next line of the explanation, except the term with the new unknown letter
divide the result by 3a². The answer is the next digit of the answer, and the remainder will be used to calculate the next digit.
Sometimes it is necessary to choose a smaller number for the answer so that the (extra-large) remainder will be large enough for subtracting the terms in the following step. See immediately below for an example of this.
To illustrate, here is a continuation of the current example:
141 10 + 8 = 1418
6abc = 6735 = 630; 1418 – 630 = 788; b³ = 27; 788 – 27 = 761
761 / 147 = 4 (remainder 173)
Note that although 761 / 147 = 5 (remainder 26), this would result in 260 – 630 – 525 – 135 < 0 while calculating the next digit! Therefore we take a smaller dividend (4) to obtain a large remainder. Unfortunately, this happens very frequently during this algorithm.
Back to the correct answer:
173 10 + 0 = 1730 (+0, because the original question was 397758.0…)
6abd = 6734 = 504; 1730 – 504 = 1226; 3ac² = 3755 = 525; 1226 – 525 = 701; 3b²c = 3335 = 135; 701 – 135 = 566
566 / 147 = 3 (remainder 125)
This gives us an answer of:
3√397758 = 73.543 (actual answer: 73.542712…)
Summary:
This algorithm is significantly more cumbersome than the equivalent for square roots, and requires some intuition to choose the size needed for the remainders.
Alternative methods—which generalize to fourth-, fifth-, and deeper roots, include:
a quicker estimation method that provides about 3 digits of accuracy for cube roots.
a more advanced method involving logarithms.
For practice with cube roots and deeper roots, you can use the Pegasus training software listed on this website.
Inexact cube roots of 6-digit integers was the challenge task in the 2012 Mental Calculation World Cup. Results and challenge questions are here.
If you have any questions or would like specialist coaching from myself, you’re welcome to contact me here. |
8527 | https://www.chegg.com/homework-help/questions-and-answers/prove-statement-give-counterexample-show-false-nonnegative-real-numbers-b-square-root-b-sq-q120317123 | Solved prove the statement or give a counterexample to show | Chegg.com
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prove the statement or give a counterexample to show that is false for all nonnegative real numbers a and b square root a+b = square root a + square root b
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Question: prove the statement or give a counterexample to show that is false for all nonnegative real numbers a and b square root a+b = square root a + square root b
prove the statement or give a counterexample to show that is false for all nonnegative real numbers a and b square root a+b = square root a + square root b
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8528 | https://www.expii.com/t/solve-inequalities-with-positive-multiplication-or-division-examples-4271 | Expii
Solve Inequalities with Positive Multiplication or Division — Examples - Expii
Solving an inequality with multiplication or division of a positive number does not flip the inequality symbol.
Explanations (4)
Alex Federspiel
Video
9
(Video) Solving 1-Step Inequalities
by Friendly Math 101
This video by Friendly Math 101 shows you how to solve one-step inequalities.
Summary
Solving inequalities is almost the exact same as solving equations. You apply inverse operations to both sides in order to isolate the variable. Instead of having the variable equal one number, with inequalities the variable will be a range of numbers.
Let's look at the last example, p5≥−3. In order to undo the division of 5 we multiply by 5 on both sides.
p5≥−3p5×5≥−3×5p≥−15
This means that x can be any number greater than −15 including −15 itself. We represent this on a number line by drawing a closed circle on the −15 which tells us that it can be −15. Then an arrow to the right which tells us that it can be values above −15.
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Solve Inequalities with Addition or Subtraction — Examples
Solve Inequalities with Negative Multiplication or Division — Examples
Transitive Property — Definition & Examples
Equation Word Problems — Examples & Practice
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Caroline K
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Dividing by a Positive to Solve an Inequality
Dividing by a positive to solve an inequality is the same as dividing by a positive to solve an equation. Just remember to divide both sides by the same quantity, then simplify! The inequality sign will stay the same the entire time.
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Solving Inequalities: Division Property
Much like with normal equations, you can use division to reduce an inequality to a simpler form. Specifically, the division property of inequalities is:
When dividing an inequality by a negative number, flip the inequality symbol. When dividing by a positive number, keep the symbol the same.
When c is positive:
Let's say that c is positive and a is greater than b. That means that we know that:
c>0a>b
If we divide a and b by c here, their sign and their relationship stays the same. So, the term with a in it is greater than the term with b in it:
ac>bc
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Multiplying by a Positive to Solve an Inequality
When we multiply by a positive to solve an inequality, it's the same as multiplying by a positive to solve a regular equation! Just remember to multiply both sides by the same quantity, then simplify! The inequality sign will stay the same the entire time.
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8529 | https://www.youtube.com/watch?v=eSsfMMTLrxg | Factorising Cubic Expressions - 4 different ways
AlphaMath'
1200 subscribers
17 likes
Description
722 views
Posted: 8 Jun 2022
4 Different ways to factorise Cubic Expressions...
1) Table Method - finds x solutions
2) Grouping & Common factor/bracket
3) Synthetic Division
4) K-method
1 comments
Transcript:
in this video i'm going to show you how to factorize cubic expressions and i'm going to take you through four different methods starting from the easiest and then working our way up we'll start with the calculator slash table method we literally just use our calculator and the table in our calculator to factorize the polynomials so looking at this first example x cubed minus 5x squared minus 8x plus 12 is equal to zero so you want to start by turning on your calculator go to mode and then press three for the table mode your calculator should show you an fx equation so fx is equal to and there you just punch in the expression that is shown here so alpha close bracket gives you the x and you cube that and then just continue to punch in the entire function then when you've done so press equal to if your calculator shows a gx equation just press equal to to skip that part because we're only going to be dealing with this one for now and then it asks you where we're going to start we like to start at negative 9. i always try and make my domain as big as possible and then we're going to end at 9 and we're going to have a step of one okay so a table should appear with x and y values now you need to look for the y values that is zero because remember an x intercept is where a y value is zero okay so here i got a y value of zero and that is where x is negative two okay so a solution for this equation is where x is equal to negative two which means that the factor for this equation is x plus two okay let's find all the other factors this is a cubic expression so it should have three factors so we keep going down again i hit zero y is zero where x is one okay so if x is equal to one that means i have another factor of x minus one notice the brackets and the x-intercepts or the the roots the solutions are always going to be opposite signs if it's x equal to negative 2 our bracket is x plus 2. if it's x equal to 1 our bracket is x minus one so let's find a last x-intercept and that is where x is six where x is six y is zero so that's another solution as a factor we write x minus six okay so this is how you would factorize that expression and if the question was for us to solve x we would simply just leave it as these answers here at the bottom okay let's have a look at the second example start by punching in the function into your calculator once you are done press equal to again we're gonna skip gx we'll start at negative nine end at nine and keep our step one okay now we need to look for the y values that are zero and then once we find that we look at what that x value is this is checking for your x intercepts okay so here's one x intercept that's where x is negative two then y is zero so i'll just write that down where x is negative two so our first factor is x plus two let's keep going we should find three right there is another one where x is equal to three so i write x minus three and then we still need to find a third factor but in this case the calculator only shows me these two so how do i find in my third factor well this is one way to do it we'll start off by saying x times x times what will give us x cubed and the answer is x x times x times x give us x cubed then we look at this last value which is 18 so we're going to say 2 times negative 3 times what gives us the 18. and again the answer is negative 3 because 2 times negative 3 times negative 3 gives you positive 18. okay so the reason the calculator never showed you this third factor is because they are actually the same thing okay so the calculator won't show you this twice it's just going to show it once but we managed to find it out using some case and check okay moving on to the second method method number two is either to take out a common factor or to group the terms now even though this equation looks very similar to the one that we previously had it's only the c that is different instead of 18 it's now 12 this equation is not easily solved by using the calculator method as you can see i punched the function in my calculator and once i press equal to to follow all the same steps that i followed before i only managed to find one factor and that is where x is equal to four or uh i guess an x-intercept um so the factor is x minus 4 because it's always opposite of what the root is so that's the only factor i was able to find so this method this calculator method won't work for solving this equation what i now can do is group the first two terms and take out a common factor so if i take out x squared as a common factor that's my highest common factor i am left with an x minus four okay then i look at my next two terms i'll take out negative three and that gives me x minus four notice i have a common bracket so i'll write the x squared minus three with my common bracket x minus four and there we go there's it all factorized now this bracket over here if or since this is an equation let's just write the solutions for this equation i'll do it over here we already know that x is equal to four for this one but for that one x will be equal to plus or minus the square root of three because you take three across and then the square root on both sides for the next one we'll have to take out a common factor all of these terms has an x in it so i can take out a common factor of x this is the highest common factor i can remove from the equation now i just have a trinomial so i can factorize that trinomial it's going to be 3x and x that's the factors of 3x squared then the factors of 7 since it's a prime number it's just 7 and 1. notice when i multiply 3 with 7 i get 21 and 1 with one i just get one so 21 minus one is 20. okay this should get the minus and that should get the plus again if i just were to write out the solutions for this equation the first is x equal to zero that's because we took out x as a common factor the next is this bracket negative one over three so negative a third and for this bracket seven okay we would most likely be able to find these two answers using the calculator method but then you'd have to try something to figure out that one this next method is called synthetic division but to start of this method you first need to find at least one factor using some trial and error so again i'm just going to make use of the table in our calculator to find one x value which is going to give us uh the factor for this equation my calculator will actually give me all the solutions for this equation but i'm just going to find one and then show you guys how to do synthetic division because for the next one we wouldn't be able to find all the solutions from the calculator so let's punch it in the calculator first so here's my table i need to go down look for my x-intercepts right the first x-intercept is where x is equal to negative three so like i said you just need one x-intercept that's all um i'm going to write it as a factor so that's x plus three okay so now this is how you do the synthetic method you put the number um the the solution which is negative three you put that on this side and then the coefficients for all the numbers or for all the um terms we are going to fill in here on top so it's the 1 that's the negative 4 it's the negative 11 and 30. okay you just fold that in the top row then the one you just bring it down and now you multiply the negative three with this one which gives us negative three then you add that gives us negative seven then you multiply again so negative 3 times negative 37 that gives you 21 and negative 11 plus 21 is equal to 10 and then negative 3 times 10 is negative 30 and 30 minus 30 is zero you will always notice that this cancels so these numbers here are the coefficients for the trinomial that goes here so one x squared minus seven x plus ten then all we need to do is factorize the trinomial so the factors of ten to give us the seven is five and two negative five and negative two okay so therefore you can now read off your x um intercepts or your x solutions it's negative three it's two and it's five okay and we did it via synthetic division let's have a look at this one we'll start off by punching the equation into our calculator so it's x cubed minus x squared minus x plus 10 equal to follow the same steps and we get our table look for all the x intercepts that you can find and the only one that i can find is negative two which means we have a factor of x minus two okay so now i'm going to apply the synthetic division method okay so negative 2 goes over here then i write the coefficients of these terms 1 negative 1 negative one again and ten okay we bring this one down then we multiply negative two times one is negative two negative one plus negative two gives us negative three negative 2 times negative 3 is 6. negative 1 plus 6 is 5 negative 2 times 5 is negative 10 10 minus 10 is zero so again we end up with zero these numbers here are the coefficients for the trinomial so that is x squared minus three x plus five cool now this trinomial can't be factorized also if you make use of the quadratic formula there are no solutions for x for this trinomial so this is the only solution for the cubic expression for these last two examples i'm going to be using the k method and so we are first going to start off by finding at least one x intercept so let me again just make use of my calculator so over here i got the table for this expression um i start at negative nine in that nine let's check for our x-intercepts so negative one is my x-intercept i'm going to write that as a factor so that is x plus one okay and now we know that the next bracket has to be a trinomial okay x times what number year what term here will give us a 2 x cubed well it's x times 2 x squared when we multiply these two things we get the 2x cubed then to find this last value here positive 1 times what gives us 3 it has to be 1 times positive 3. okay so now to find the middle term i'm going to call that k or let's call it kx okay because k is the coefficient of x now to work out what k is we are going to look at the 5x squared in the middle over here okay so if you multiply this this is minus sorry this is positive 1 times 2 x squared that gives us positive 2x squared and also if we multiply the x with the kx that's a positive k x squared these two terms once you add them up is supposed to give you the negative 5x squared okay so it's 2 x squared plus k x squared and that's supposed to give us negative 5 x squared this must mean then that k x squared is equal to negative 7 x squared if i take that across and then if i divide by k sorry if i divide by x squared k is equal to 7 or negative 7. now i'm going to substitute k with negative 7 and then factorize the trinomial if k is equal to negative 7 i'll factorize this trinomial as two x and x that's the factors of two x squared three is a prime number so it's just three and one two times three is six one times one is one one plus six is seven so both signs should be the same it's negative and negative and here we can write down the solutions if this were to be an equation we'd write down the solutions as x is equal to negative 1 or x is equal to 1 over 2 positive and x is equal to positive 3. these two we found on the calculator but we had to do a little magic to find this one so you know the drill already punch the function into your calculator start at negative nine end at nine look for your x-intercepts uh the first x-intercept i get is negative four okay so i'm going to use that if x is equal to negative 4 that means x plus 4 is my factor okay i'm going to use the k method again x times x squared gives us x cubed then 4 times 4 is equal to 16. now to find this middle term i'm going to label that as kx now there's no x squared term here which means that the coefficient of that x squared term is zero okay so uh if we multiply 4 with x squared we get 4x squared and if we multiply x with the kx that gives us kx squared now there's a 0x squared here so we know that 4 x squared plus k x squared is equal to zero x squared or just zero i'll just leave it as zero um therefore k x squared is equal to negative four x squared if i take that across and divide by x squared on both sides k is equal to negative four let's substitute that here and factorize further now factorizing this trinomial further is easy because that's just a perfect square the factors of 4 to give us 4 is 2 and 2 both brackets should be negative so here's the solutions for x if this equation was equal to zero um x is equal to negative four and x is equal to two both these answers you could have found using the calculator and using the first method that we learned in the beginning you could have actually just figured out this second one but i just thought i'd show you guys all four methods so you have multiple tools in your arsenal to factorize these cubic expressions |
8530 | https://www.cuemath.com/ncert-solutions/ncert-solutions-class-7-maths-chapter-14-symmetry/ | NCERT Solutions Class 7 Maths Chapter 14 Symmetry
The NCERT Solutions for Class 7 maths Chapter 14 Symmetry explains the concept of symmetry that is almost everywhere in our surroundings, in nature, and in real-life scenarios as well. Symmetric patterns are seen throughout nature, such as the growth and development of living organisms, flowers, trees, and leaves, as well as many other natural occurrences.
Symmetry can also be created in architecture, as well as art and music. Similarly, mathematics is also governed by certain symmetries or patterns that repeat themselves over and over again. These NCERT solutions Class 7 maths Chapter 14
starts with recalling the concept of line symmetry that means a line, over which if the figure is folded, its parts will coincide. This line of symmetry is also known as the axis.
The chapter talks about the symmetry of regular polygons. The students will discover interesting facts about them, including that the regular polygons have multiple lines of symmetry. More such facts can be read in the Class 7 maths NCERT solutions Class 7 maths Chapter 14 Symmetry, which is given below and also find some of these in the exercises given below.
NCERT Solutions for Class 7 Maths Chapter 14 PDF
Symmetrical shapes are pleasing to the eye and convey qualities of balance and order. No wonder that a symmetrical shape such as a square or rectangle is often used as the basis for architectural design, hence making it an important concept for the students to study. The pdf file of the NCERT solutions Class 7 maths Chapter 14 representing various sections can be downloaded from the links below:
☛ Download Class 7 Maths NCERT Solutions Chapter 14 Symmetry
NCERT Class 7 Maths Chapter 14 Download PDF
NCERT Solutions for Class 7 Maths Chapter 14 Symmetry
Symmetry is one of the distinguishing features of life - it is one of the fundamental properties that give rise to a wide variety of phenomena, from fractals to crystals and hurricanes. It doesn't matter what field you're in: you'll find symmetries everywhere, from the mechanics of waves and turbulence to the economic theories behind monetary systems.
For these reasons, learning about symmetry is an absolute must for anyone interested in science or engineering. Hence, the students must make use of the NCERT Solutions Class 7 Maths Chapter 14 Symmetry, the section-wise exercise classification of which is given below:
☛ Download Class 7 Maths Chapter 14 NCERT Book
Topics Covered: Line and Rotational symmetry, order of rotational symmetry, symmetry of regular polygons, terms like the center of rotation, angle of rotation, also, the two kinds of rotation like clockwise and anticlockwise have been covered in detail in the Class 7 maths NCERT solutions Chapter 14.
Total Questions: Class 7 Maths Chapter 14 Symmetry Chapter 14 has 19 questions in all, 12 of which are very easy, 4 involve drawing the figures in order to complete their symmetry, while the rest 3 can be classified as moderately difficult.
List of Formulas in NCERT Solutions Class 7 Maths Chapter 14
The concept of symmetry entails certain criteria to be fulfilled in order for a figure to be called symmetrical. The students need to have dedicated observational skills and focus, in order to analyze and categorize the given figure as symmetrical while applying the knowledge of these rules. Hence, these criteria or logics are important to remember. Some of them are given below.
Important Questions for Class 7 Maths NCERT Solutions Chapter 14
| CBSE Important Questions for Class 7 Maths Chapter 14 Exercise 14.1 |
| Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 10 |
| CBSE Important Questions for Class 7 Maths Chapter 14 Exercise 14.2 |
| Question 1 Question 2 |
| CBSE Important Questions for Class 7 Maths Chapter 14 Exercise 14.3 |
| Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 |
NCERT Solutions for Class 7 Maths Video Chapter 14
| NCERT Class 7 Maths Videos for Chapter 14 | |
--- |
| Video Solutions for Class 7 Maths Exercise 14.1 | |
| Exercise 14.1 Question 1 | Exercise 14.1 Question 6 |
| Exercise 14.1 Question 2 | Exercise 14.1 Question 7 |
| Exercise 14.1 Question 3 | Exercise 14.1 Question 8 |
| Exercise 14.1 Question 4 | Exercise 14.1 Question 9 |
| Exercise 14.1 Question 5 | Exercise 14.1 Question 10 |
| Video Solutions for Class 7 Maths Exercise 14.2 | |
| Exercise 14.2 Question 1 | Exercise 14.2 Question 2 |
| Video Solutions for Class 7 Maths Exercise 14.3 | |
| Exercise 14.3 Question 1 | Exercise 14.3 Question 5 |
| Exercise 14.3 Question 2 | Exercise 14.3 Question 6 |
| Exercise 14.3 Question 3 | Exercise 14.3 Question 7 |
| Exercise 14.3 Question 4 | |
FAQs on NCERT Solutions Class 7 Maths Chapter 14
Why are NCERT Solutions Class 7 Maths Chapter 14 Important?
The NCERT Solutions Class 7 Maths Chapter 14 has all the details about the topic of symmetry in a comprehensive manner. All the related terms have been covered explicitly with the help of examples to make sure the students understand their practicality. The team at NCERT does full justice to all the concepts they wish to teach in an efficient manner. The solutions given to every problem are very accurate and explanatory making this a very important resource for the students.
Do I Need to Practice all Questions Provided in NCERT Solutions Class 7 Maths Symmetry?
The examples and the exercise questions in the NCERT books encompass all the concerned aspects of symmetry, solving which will result in a quick revision of the concepts which the students have read so far. Hence, it is essential that the students practice all the problems to explore these concepts further. While some problems require the students to draw diagrams as a clarification of the question, other problems will require the students to make use of their logical and analytical skills, which will be advantageous for them.
How Many Questions are there in Class 7 Maths NCERT Solutions Chapter 14 Symmetry?
The NCERT Solutions Class 7 Maths Chapter 14 Symmetry has summed up the knowledge of symmetry in 19 questions. The students will find 12 relatively easy to solve, 4 are moderately easy while 3 would require some logical thinking taking a longer time.
What are the Important Formulas in NCERT Solutions Class 7 Maths Chapter 14?
The topic of symmetry is based on the concept of balance which is central to the understanding of how things are ordered in time, space, and form. There are certain rules and logics related to symmetry, and if the object or figure follows these criteria, it can be termed symmetrical. Some of these criteria are as follows: A figure can be called symmetrical only if, when folded on its line of symmetry, both parts of its images coincide exactly. Also, if we take the case of regular polygons, which have equal sides, their lines of symmetry are as many in number as the number of sides that they possess.
How CBSE Students can utilize NCERT Solutions Class 7 Maths Chapter 14 effectively?
Since the CBSE board recommends studying from the NCERT Solutions Class 7 Maths Chapter 14 hence, students must make sure to practice all the solved examples in the book as well as the exercise questions. They should pay attention to how the statements are framed in the solved examples, which will give them an idea as to how a problem should be solved in the exams by presenting proper logic step by step. Thus, consistent practice will help the students utilize this important learning resource more effectively.
Why Should I Practice NCERT Solutions Class 7 Maths Symmetry Chapter 14?
The concept of symmetry has been very well explained in the NCERT Solutions Class 7 Maths Symmetry Chapter 14 in simple language through easy practical activities and examples. The expert scholars at NCERT have combined the chapter with relevant important facts and pointers, which are essential for the students to understand. The objectives have been properly explained in the proper sequence; hence, the students must make use of the knowledge presented in this chapter and practice the concepts to strengthen their foundation of geometry.
| | |
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| NCERT Solutions Class 7 Maths Chapters | |
| Chapter 1 Integers | Chapter 2 Fractions and Decimals |
| Chapter 3 Data Handling | Chapter 4 Simple Equations |
| Chapter 5 Lines and Angles | Chapter 6 The Triangles and its Properties |
| Chapter 7 Congruence of Triangles | Chapter 8 Comparing Quantities |
| Chapter 9 Rational Numbers | Chapter 10 Practical Geometry |
| Chapter 11 Perimeter and Area | Chapter 12 Algebraic Expressions |
| Chapter 13 Exponents and Powers | Chapter 15 Visualising Solid Shapes | |
8531 | https://webbook.nist.gov/cgi/cbook.cgi?ID=C64175&Type=TFREEZE | Ethanol
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Ethanol
Formula: C 2 H 6 O
Molecular weight: 46.0684
IUPAC Standard InChI:InChI=1S/C2H6O/c1-2-3/h3H,2H2,1H3 Copy
IUPAC Standard InChIKey:LFQSCWFLJHTTHZ-UHFFFAOYSA-N Copy
CAS Registry Number: 64-17-5
Chemical structure:
This structure is also available as a 2d Mol file or as a computed3d SD file
View 3d structure (requires JavaScript / HTML 5)
Isotopologues:
[2H6]ethanol
Ethanol-d1
C2H3D3O
Ethanol-d5
Ethanol-1,1-d2
Other names: Ethyl alcohol; Alcohol; Alcohol anhydrous; Algrain; Anhydrol; Denatured ethanol; Ethyl hydrate; Ethyl hydroxide; Jaysol; Jaysol S; Methylcarbinol; SD Alchol 23-hydrogen; Tecsol; C2H5OH; Absolute ethanol; Cologne spirit; Fermentation alcohol; Grain alcohol; Molasses alcohol; Potato alcohol; Aethanol; Aethylalkohol; Alcohol, dehydrated; Alcool ethylique; Alcool etilico; Alkohol; Cologne spirits; Denatured alcohol CD-10; Denatured alcohol CD-5; Denatured alcohol CD-5a; Denatured alcohol SD-1; Denatured alcohol SD-13a; Denatured alcohol SD-17; Denatured alcohol SD-23a; Denatured alcohol SD-28; Denatured alcohol SD-3a; Denatured alcohol SD-30; Denatured alcohol SD-39b; Denatured alcohol SD-39c; Denatured alcohol SD-40m; Etanolo; Ethanol 200 proof; Ethyl alc; Etylowy alkohol; EtOH; NCI-C03134; Spirits of wine; Spirt; Alkoholu etylowego; Ethyl alcohol anhydrous; SD alcohol 23-hydrogen; UN 1170; Tecsol C; Alcare Hand Degermer; Absolute alcohol; Denatured alcohol; Ethanol, silent spirit; Ethylol; Punctilious ethyl alcohol; SD 3A
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Normal melting point
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Normal melting point
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Data compilation copyright by the U.S. Secretary of Commerce on behalf of the U.S.A. All rights reserved.
Data compiled as indicated in comments:
DH - Eugene S. Domalski and Elizabeth D. Hearing
TRC - Thermodynamics Research Center, NIST Boulder Laboratories, Chris Muzny director
| T fus (K) | Reference | Comment |
---
| 159.015 | Hwa and Ziegler, 1966 | DH |
| 158.75 | Anonymous, 1958 | TRC |
| 158.66 | Dreisbach and Martin, 1949 | Uncertainty assigned by TRC = 0.2 K; TRC |
| 158.35 | Tschamler, Richter, et al., 1949 | Uncertainty assigned by TRC = 0.6 K; TRC |
| 159. | Yoshida, 1944 | Uncertainty assigned by TRC = 2. K; TRC |
| 159. | Timmermans, 1935 | Uncertainty assigned by TRC = 2. K; TRC |
| 158.65 | Biltz, Fischer, et al., 1930 | Uncertainty assigned by TRC = 1. K; TRC |
| 157.93 | Mitsukuri, 1926 | Uncertainty assigned by TRC = 0.5 K; TRC |
| 159. | Maass and Boomer, 1922 | Uncertainty assigned by TRC = 2. K; may not be measured by authors; TRC |
| 158.90 | Timmermans, 1911 | Uncertainty assigned by TRC = 0.3 K; TRC |
| 156. | Timmermans, 1907 | Uncertainty assigned by TRC = 2. K; TRC |
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References
Go To:Top, Normal melting point, Notes
Data compilation copyright by the U.S. Secretary of Commerce on behalf of the U.S.A. All rights reserved.
Hwa and Ziegler, 1966
Hwa, S.C.P.; Ziegler, W.T., Temperature dependence of excess thermodynamic properties of ethanol-methylcyclohexane and ethanol-toluene systems, J. Phys. Chem., 1966, 70(8), 2572-2593. [all data]
Anonymous, 1958
Anonymous, X., Am. Pet. Inst. Res. Proj. 50, 1958, Unpublished, 1958. [all data]
Dreisbach and Martin, 1949
Dreisbach, R.R.; Martin, R.A., Physical Data on Some Organic Compounds, Ind. Eng. Chem., 1949, 41, 2875-8. [all data]
Tschamler, Richter, et al., 1949
Tschamler, H.; Richter, E.; Wettig, F., Mixtures of Primry Aliphatic Alcohols with Chlorex and Other Organic Substances. Binary Liquid Mixtures XII., Monatsh. Chem., 1949, 80, 749. [all data]
Yoshida, 1944
Yoshida, U., Structural relaxation of amorphous solids and the cybotactic structure of super-cooled liquids, Mem. Coll. Sci., Univ. Kyoto, Ser. A, 1944, 24, 135. [all data]
Timmermans, 1935
Timmermans, J., Researches in Stoichiometry. I. The Heat of Fusion of Organic Compounds., Bull. Soc. Chim. Belg., 1935, 44, 17-40. [all data]
Biltz, Fischer, et al., 1930
Biltz, W.; Fischer, W.; Wunnenberg, E., Molecular and Atomic Volumes. The Volume Requirements of Crystalline Organic Compounds and Low Temperatures, Z. Phys. Chem., Abt. A, 1930, 151, 13-55. [all data]
Mitsukuri, 1926
Mitsukuri, S., The Heats of Fusion of SOme ORganic Solvents Whose Melting Points Are Relatively Low, Bull. Chem. Soc. Jpn., 1926, 1, 30-4. [all data]
Maass and Boomer, 1922
Maass, O.; Boomer, E.H., Vapor Densities at Low Pressures and Over and Extended Temperature Range. I. The Properties of Ethylene Oxide Compared to Oxygen Compounds of Similar Molecular Weight, J. Am. Chem. Soc., 1922, 44, 8, 1709-1728, . [all data]
Timmermans, 1911
Timmermans, J., Researches on the freezing point of organic liquid compounds, Bull. Soc. Chim. Belg., 1911, 25, 300. [all data]
Timmermans, 1907
Timmermans, J., New research on the density of liquids below zero, Bull. Soc. Chim. Belg., 1907, 21, 395. [all data]
Notes
Go To:Top, Normal melting point, References
Symbols used in this document:
T fus Fusion (melting) point
Data from NIST Standard Reference Database 69: NIST Chemistry WebBook
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8532 | https://vc.bridgew.edu/cgi/viewcontent.cgi?article=1240&context=honors_proj | Bridgewater State University Bridgewater State University
Virtual Commons - Bridgewater State University Virtual Commons - Bridgewater State University
Honors Program Theses and Projects Undergraduate Honors Program
5-9-2017
Extremal Graph Theory: Turán’s Theorem Extremal Graph Theory: Turán’s Theorem
Vincent Vascimini
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Vascimini, Vincent. (2017). Extremal Graph Theory: Turán’s Theorem. In
BSU Honors Program Theses and
Projects. Item 234. Available at:
Copyright © 2017 Vincent Vascimini
This item is available as part of Virtual Commons, the open-access institutional repository of Bridgewater State
University, Bridgewater, Massachusetts. Extremal Graph Theory: Turán ’s Theorem
Vincent Vascimini
Submitted in Partial Completion of the
Requirements for Departmental Honors in Mathematics
Bridgewater State University
May 9, 2017
Dr. Shannon Lockard, Thesis Director
Dr. John Pike, Committee Member
Dr. Stephen Flood, Committee Member Extremal Graph Theory: Tur´ an’s Theorem
Vincent Vascimini May 9, 2017
1 History of Tur´ an’s Theorem
Extremal graph theory is a branch of graph theory that involves finding the largest or smallest graph with certain properties. Extremal graph theory began with this question : if we have a graph of order n, what is the maximum size allowed so that it does not contain a triangle? A graph G = ( V, E ) is a set of vertices V together with a set of edges E. A triangle is simply a graph with 3 vertices and 3 edges. We call the number of vertices of a graph G the order of
G, and the number of edges in G the size of G. Consider a graph G of order
n = 4 and size m = 4 as in Figure 1.
12
34
Figure 1: Graph G of order n = 4 and size m = 4. This graph is a cycle of order 4, denoted C4. Clearly G does not contain a triangle. However, if we were to add one more edge to G, then no matter where we attempt to put the edge we are guaranteed to create a triangle (see Figure 2).
12
34
12
34
Figure 2: Graphs of order n = 4 and size m = 5 that contain a triangle. In fact, in any graph with 4 vertices and 5 edges we are forced to have a triangle. This is easier to see for a graph of relatively small order. In this case there are only six different labeled graphs with 4 vertices and 5 edges, as seen in Figure 3. 11 2
34
12
34
12
34
12
34
12
34
12
34
Figure 3: All labeled graphs of order n = 4 and size m = 5. Therefore, we see that for a graph of order n = 4, the maximum number of edges so that the graph does not have to contain a triangle is m = 4. This question was answered in general by Mantel, a Dutch mathematician, in 1907 . Mantel gave a lower bound on the number of edges in a graph so that it must contain a triangle.
Theorem 1.1. (Mantel, ) If G is a graph of order n ≥ 3 and size m >
⌊ n2
4
⌋
,then G contains a triangle. Notice that for the graph G in Figure 1, the number of edges m = 4 is not greater than
⌊ 42
4
⌋
= 4. However, if we add one more edge to have 5 edges, then 5 is indeed greater than 4, so by Theorem 1.1 we are guaranteed to have a triangle as we saw in Figure 2. Similarly, consider a graph H of order n = 5. First, note that
⌊ n2
4
⌋
=
⌊ 52
4
⌋
= 6. We see that it is possible to draw a graph H without a triangle using 6 edges, like the graph on the left in Figure 4, but impossible with m = 7 edges, which is indeed greater than
⌊ 52
4
⌋
, as seen by the graph on the right in Figure 4.
1
2
3
4
51
2
3
4
5
Figure 4: Graph H on the left that does not contain a triangle, and graph H′
on the right that does contain a triangle. Of course, mathematicians have built upon Mantel’s work and more results have been given. Some of these results guarantee whether or a not a graph of a specific size contains a triangle as a subgraph. For example, it has been shown that for a graph of even order n = 2 and size m =2 for some positive integer
, the only graph with no triangle as a subgraph is K,. Additionally, a 2graph of odd order n = 2 + 1 and size m ≥ 2 + + 1 must contain a triangle . It is important to note that a triangle is a complete graph on p = 3 vertices.
Definition 1.1. A graph G is complete if every two distinct vertices of G are adjacent, or if there is an edge between every two distinct vertices. The complete graph with p vertices is denoted Kp.Indeed, a triangle is K3. Some examples of complete graphs can be seen in Figure 5.
1
2
31
23
4
1
2
3
4
5
1
2
3
4
5
6
Figure 5: Complete graphs K3, K4, K5, and K6 respectively.
Definition 1.2. A graph G = ( V, E ) is regular if the vertices of G all have the same degree. If deg( v) = r for every vertex v ∈ V (G), then G is called an
r-regular graph. Notice that the vertices in each of the graphs in Figure 5 all have the same degree. Respectively, these graphs are 2-regular, 3-regular, 4-regular, and 5-regular. Theorem 1.1 provides us with parameters regarding graphs containing K3
as a subgraph .
Definition 1.3. A graph H = ( VH , E H ) is a subgraph of G = ( VG, E G), denoted
H ⊆ G, if VH ⊆ VG and EH ⊆ EG.Naturally, the next question to ask is: what is the maximum number of edges in a graph with n vertices that does not contain a complete subgraph with p
vertices? That is, what is the maximum number of edges in a graph of order
n that does not contain Kp as a subgraph? This generalization is precisely the question P´ al Tur´ an was interested in answering. P´ al Tur´ an was a Hungarian mathematician born on August 28, 1910 in Bu-dapest. While he was primarily a probabilistic and analytical number theorist, his renowned work in graph theory is considered to be the birth of extremal graph theory . Tur´ an’s Theorem is a major result in extremal graph theory that gives an upper bound on the number of edges in a graph with no Kp as a subgraph. First we have a related definition.
Definition 1.4. A p-clique is a subset of p vertices of a graph G = ( V, E ) such that each of the p vertices are adjacent. That is, a p-clique is a subgraph of G
that is isomorphic to Kp.3Now we state Tur´ an’s Theorem.
Tur´ an’s Theorem - Version 1. If a graph G = ( V, E ) on n vertices has no p-clique, p ≥ 2, then
|E| ≤
(
1 − 1
p − 1
) n2
2 . (1) Being such an important theorem in extremal graph theory, there are many different methods to go about proving Tur´ an’s Theorem. We will continue by presenting a few of these proofs, explaining the logic and tools used in them, and bringing attention to notable differences between them.
2 A Probabilistic Proof
This first proof, as adapted from a proof in Chapter 40 of , uses tools from probability theory. The proof outline is as follows. We first prove a lemma about the order of a complete subgraph of a graph G, which will later be useful in proving Tur´ an’s Theorem. To prove the lemma we take a random permutation
of vertices of G and extract a subset of vertices from the permutation in such a way that this subset forms a clique.
Definition 2.1. A permutation of a list, or set, is a reordering of the elements of the list. Then we find the expected value of the random variable assigned to keep track of the order of this set and compare it to the order of a largest clique, which completes the proof of the lemma.
Definition 2.2. A random variable X is measurable function on the sample space of an experiment which attaches numbers to outcomes.
Definition 2.3. The expected value of a discrete random variable X is the sum of all values of X times the probability that each value occurs. That is,
E(X) =
n
∑
k=1
k · P (X = k).
We are then able to deduce the result in Tur´ an’s Theorem using the Cauchy-Schwarz inequality to give a needed bound, Theorem 2.1, and the lemma that follows. The Cauchy-Schwarz inequality states that for any sequences {ai} and {bi}
we have ( n∑
i=1
aibi
)2
≤
( n∑
i=1
a2
i
)( n∑
i=1
b2
i
)
. (2) The following theorem is often called the First Theorem of Graph Theory, as it easily relates the degrees of the vertices of a graph to the size of the graph . 4Theorem 2.1. If G = ( V, E ) is a graph of order n and di is the degree of vertex
vi for 1 ≤ i ≤ n, then n∑
i=1
di = 2 |E|
Proof. Each edge in G is counted twice when adding the degrees of each vertex. Thus, the sum of the degrees is twice the number of edges as claimed. We will now prove the lemma.
Lemma 2.2. Let ω(G) be the order of a largest clique in a graph G = ( V, E )and let di be the degree of a vertex vi in G. Then
ω(G) ≥
n
∑
i=1
1
n − di
.
Proof. Let G = ( V, E ) be a graph on n vertices. Let π be any permutation of the set {1, 2, ..., n }. Then π induces an ordering vπ(1) , v π(2) , ..., v π(n) on the vertex set V (G). Using this notation, a vertex vπ(k) is listed in the kth position in π, and the position a vertex vj in π is given by π−1(j). Consider the subset
Cπ of vertices of G defined by Cπ = {vk ∈ V : vk ∼ vj for all j with π−1(j) <π−1(k)}. Note that vπ(1) is always in Cπ by construction. Let Π be chosen uniformly at random from all n! permutations. Then CΠ is a subset of vertices of G such that all vertices in CΠ are adjacent. Hence these vertices form a clique. Now let
Xi =
{
1, vi ∈ CΠ
0, vi /∈ CΠ
be the indicator of the event that vi ∈ CΠ.Let
X = |CΠ| =
n
∑
i=1
Xi,
be a random variable corresponding to CΠ that tracks the order of CΠ.We are interested in the probability that a vertex vi is in CΠ. A vertex
vi has di neighbors. There are ( ndi
) ways to locate the di neighbors amongst all n vertices. Then, there are ( di)! ways to arrange the di neighbors amongst themselves. Notice that if any of the n − di non-neighbors are listed before vi
in Π, then vi would not be adjacent to all previous vertices hence vi would not be in CΠ. Finally, there are ( n − di − 1)! ways to order the non-neighbors with
vi listed first. Thus the number of permutations π for which vi is in Cπ is
( ndi
)
· (di)! · (n − di − 1)! = n!( di)!( n − di − 1)! (n − di)!( di)! = n!
n − di
.
5As each of the n! permutations is equally likely, we have
P (vi ∈ CΠ) = 1
n! · n!
n − di
= 1
n − di
.
It follows that
P (vi /∈ CΠ) = 1 − 1
n − di
.
Thus by linearity of expectation we have
E[X] = E
[ n∑
i=1
Xi
]
=
n
∑
i=1
E[Xi]=
n
∑
i=1
1 · P (vi ∈ CΠ) + 0 · P (vi /∈ CΠ)=
n
∑
i=1
1 ·
( 1
n − di
)
0 ·
(
1 − 1
n − di
)
=
n
∑
i=1
1
n − di
As the average cannot exceed the maximum, we see that a largest clique has order
ω(G) ≥
n
∑
i=1
1
n − di
. (3) Thus the lemma is proved. Now to prove Tur´ an’s Theorem - Version 1.
Proof. To prove Tur´ an’s Theorem we will make use of the Cauchy-Schwarz inequality in (2). Let ai = √n − di and bi = 1√n−di
. Substituting into (2), we have ( n∑
i=1
aibi
)2
=
( n∑
i=1
√n − di · 1
√n − di
)2
=
( n∑
i=1
1
)2
= n2
and ( n∑
i=1
a2
i
)( n∑
i=1
b2
i
)
=
( n∑
i=1
n − di
)( n∑
i=1
1
n − di
)
.
6Thus
n2 ≤
( n∑
i=1
n − di
)( n∑
i=1
1
n − di
)
≤
( n∑
i=1
n − di
)
ω(G), by Lemma 2.2
≤
( n∑
i=1
n − di
)
(p − 1) , since G has no p-clique, ω(G) ≤ p − 1=
(
n2 − 2|E|
)
(p − 1) , by linearity and Theorem 2.1 .
Hence
n2 ≤
(
n2 − 2|E|
)
(p − 1) .
Solving for |E|, we see
|E| ≤
(
1 − 1
p − 1
) n2
2 .
Thus Tur´ an’s Theorem is proved.
3 A Combinatorial Proof
This next proof involves taking a more combinatorial approach, and is also adapted from a proof in Chapter 40 of . The proof outline is as follows. We first assume a graph G has the most possible edges. We use the assumption in Tur´ an’s Theorem - Version 1 to obtain a ( p − 1)-clique as a subgraph and call those vertices the set A. We then count the number of edges in A, the number of edges between the remaining vertices of G, and finally the number of edges between the vertices of A and the remaining vertices of G. We use these three quantities to bound the total number of edges in G above by the bound given in Tur´ an’s Theorem. In the following proof, we will also use the fact that the number of edges in a complete graph of order n is (n
2
) = n(n−1) 2 .Now let us prove Tur´ an’s Theorem.
Proof. We will prove Tur´ an’s Theorem - Version 1 by strong induction on n.For our base case, we will show that the theorem holds for all n < p . To begin, suppose a graph G on n vertices has no p-clique. Note that since n < p ,we have n ≤ p − 1, hence −1
n
≤ −1
p−1
. Since G has no p-clique and n < p , we see 7that G has at most (n
2
) edges. Thus the number of edges in G is at most
|E| ≤
(n
2
)
= n(n − 1) 2=
(
1 − 1
n
) n2
2
≤
(
1 − 1
p − 1
) n2
2 .
Thus the base case holds. For our induction hypothesis, suppose that for some k ≥ p, a graph G =(V, E ) of order , where 1 ≤ ≤ k, with no p-clique has at most
(
1 − 1
p − 1
) `2
2edges. We want to show that for a graph G = ( V, E ) on k + 1 vertices with no
p-clique, we have
|E| ≤
(
1 − 1
p − 1
) (k + 1) 2
2 .
Let G be a graph on k + 1 vertices with no p-clique, p ≥ 2. Since we are interested in finding an upper bound on the number of edges in G, suppose G
has maximal edges. Since G has maximal edges and no p-clique, G must have a ( p − 1)-clique, otherwise we could add more edges. Let A be the subset of vertices that form a ( p−1)-clique. Since A is complete, each of the p−1 vertices are adjacent to the remaining p − 2 vertices. Thus the number of edges in A,denoted eA, is
eA =
(p − 12
)
= (p − 1)( p − 2) 2 . (4) Let B = V \ A be the set containing the remaining vertices in G. Then B has (k + 1) − (p − 1) = k − p + 2 vertices. Since B has no p-clique, by our induction assumption we have
eB ≤
(
1 − 1
p − 1
) (k − p + 2) 2
2 . (5) The number of edges in G is now given by eG = eA + eB + eA,B , where eA,B is the number of edges between A and B. To find eA,B , consider a vertex vj in B.If vj is adjacent to p − 1 vertices in A, then G would have a p-clique which is a contradiction. So vj is adjacent to at most p − 2 vertices in A, for any vj ∈ B.Hence
eA,B ≤ (k − p + 2)( p − 2) . (6) 8Combining the quantities in (4), (5), and (6) gives the number of edges in G as
eG = eA + eB + eA,B
≤ (p − 1)( p − 2) 2 +
(
1 − 1
p − 1
) (k − p + 2) 2
2 + ( k − p + 2)( p − 2) = p − 22( p − 1)
(
(p − 1) 2 + ( k − p + 2) 2 + 2( p − 1)( k − p + 2)
)
= p − 22( p − 1)
(
(p − 1) + ( k − p + 2)
)2
= p − 22( p − 1) (k + 1) 2
=
(
1 − 1
p − 1
) (k + 1) 2
2 .
Thus
|E| ≤
(
1 − 1
p − 1
) (k + 1) 2
2 ,
which completes the proof of Tur´ an’s Theorem.
4 Graphs of Maximum Size
Upon initial inspection, the bound given in Version 1 of Tur´ an’s Theorem seems unforeseen. To better understand bounds in graph theory, it often helps to look at different types and families of graphs. Of course, there are many different types and families of graphs, each of which have their own interesting properties and usefulness. We will proceed by discussing a particular family of graphs and discover how the upper bound in Version 1 was obtained. First we will turn our attention to k-partite graphs.
Definition 4.1. A graph G is k-partite if V (G) can be partitioned into k subsets
V1, V 2, ...V k, called partite sets , such that if uv ∈ E(G), then u and v belong to different partite sets. We also have complete k-partite graphs .
Definition 4.2. A k-partite graph G is a complete k-partite graph if every two vertices in different partite sets are joined by an edge. For each partite set Vi, if
|Vi| = ni for 1 ≤ i ≤ k, then the complete k-partite graph is denoted Kn1,n 2,..,n k .Some complete k-partite graphs can be seen in Figure 6. The first and second graphs in Figure 6 are 2-partite, which are typically called bipartite graphs, and the third graph is 4-partite. While working to find certain graphs of maximum size, Tur´ an focused on a particular family of finite graphs, namely Tur´ an graphs .9Figure 6: Complete partite graphs K1,4, K2,3, and K1,1,1,3 respectively.
Definition 4.3. The Tur´ an graph Tn,k is the complete k-partite graph of order
n, the cardinalities of whose partite sets differ by at most one. The Tur´ an graph Tn,k has n vertices and k partite sets. Since the cardinal-ities of each partite set differ by at most one, each partite set will have either ⌊ nk
⌋ or ⌈ nk
⌉ vertices. If k divides n, then each partite set will have the same cardinality and Tn,k will be regular. If k does not divide n and r is the remain-der when n is divided by k, then exactly r of the partite sets of Tn,k will have cardinality ⌈ nk
⌉ and k − r partite sets will have cardinality ⌊ nk
⌋ . Since Tn,k
is complete k-partite, every vertex x will be adjacent to every vertex y unless x
and y are in the same partite set. Thus if U1, U 2, ..., U k are the partite sets of
Tn,k , then every vertex in Ui is adjacent to every vertex in Uj , for i 6 = j.As an example, consider the Tur´ an graph T10 ,3. To construct T10 ,3, we note that each of the three partite sets will have either ⌊ 10 3
⌋ = 3 or ⌈ 10 3
⌉ = 4 vertices. Since 3 does not divide 10, exactly 1 of the partite sets will have ⌈ 10 3
⌉ = 4 vertices, and 2 partite sets will have ⌊ 10 3
⌋ = 3 vertices. We then draw an edge from each vertex to every other vertex not in the same partite set. Figure 7 shows two isomorphic representations of T10 ,3.
Figure 7: The Tur´ an graph T10 ,3.10 Specifically, the graph on the right in Figure 7 illustrates the three partite sets vertically, each consisting of 4 vertices, 3 vertices, and 3 vertices respectively. Additionally, Figure 8 shows some other Tur´ an graphs.
Figure 8: Tur´ an graphs T5,3, T6,5, and T7,4 respectively. Let us compute the size of a Tur´ an graph Tn,p −1 with maximal edges. When
p − 1 divides n, there are ( np−1
)2 edges between any two partite sets. Also, there are (p−12
) ways to choose any two different partite sets from all p − 1 partite sets. Thus the total number of edges in Tn,p −1 when p − 1 divides n is
(p − 12
)( np − 1
)2
. (7) Notice that since Tn,p −1 has p − 1 partite sets, it cannot contain a p-clique. Recalling Tur´ an’s Theorem - Version 1, the maximum number of edges in this graph is bounded above by
|E| ≤
(
1 − 1
p − 1
) n2
2 .
In fact, from (7) we have that for Tn,p −1,
|E| ≤
(
1 − 1
p − 1
) n2
2 =
(p − 12
)( np − 1
)2
.
Therefore, we see now that Tur´ an graphs actually attain the maximum number of edges without some complete subgraph. We observe that when p − 1 divides
n, the number of edges in Tn,p −1 is equal to the upper bound given in Version 1. Correspondingly when p − 1 does not divide n, the number of edges in Tn,p −1
is strictly less than this upper bound. As an example, consider a graph F of order n = 5 such that F does not contain a 4-clique. There are many ways to construct F so that it does not contain K4 as a subgraph, such as the graphs in Figure 9. 11 1
2
3
45
1
2
3
45
1
2
3
45
Figure 9: Graphs of order n = 5 that do not contain a K4 as a subgraph. We see that these graphs have size m = 5, m = 7, and m = 5 respectively. Tur´ an’s Theorem - Version 1 says that for any graph on n = 5 vertices with no 4-clique, the number of edges is bounded above by
(
1 − 14 − 1
) 52
2 = 25 3 = 8 .33 .
Hence the maximum number of edges possible in such a graph is m = 8. Notice that the graphs in Figure 9 all have less than 8 edges, so these graphs do satisfy the bound given in Version 1. The question now becomes: can we construct a graph so that it actually attains the maximum of m = 8 edges? Indeed as we saw above, the Tur´ an graph T5,3 is a graph with no 4-clique that attains the maximum number of edges. Figure 8 shows T5,3 as the leftmost graph. Notice that T5,3 has m = 8 edges, which is more than the graphs in Figure 9 and, in fact, the maximum number of possible edges.
5 A Graph Theoretic Proof
Tur´ an graphs are graphs whose sizes meet the upper bound given by Version 1 of Tur´ an’s Theorem. These graphs lead us to a second version that we will prove in this section.
Tur´ an’s Theorem - Version 2. The largest graph with n vertices that con-tains no subgraph isomorphic to Kk+1 is a complete k-partite graph Kn1,n 2,...,n k
with n = n1 + n2 + ... + nk and |ni − nj | ≤ 1 for all i, j ∈ { 1, 2, ..., k } with i 6 = j.The final proof is adapted from a proof in Chapter 4 of . We will need the following lemmas and terminology.
Definition 5.1. A graph H is an induced subgraph of a graph G provided if
u, v ∈ V (H) and uv ∈ E(G), then uv ∈ E(H).
Lemma 5.1. The graph of maximum size with n vertices and k partite sets is a complete k-partite graph Kn1,n 2,...,n k with n = n1 + n2 + ... + nk and
|ni − nj | ≤ 1. 12 Proof. Let G be a k-partite graph of maximum size with partite sets Vi for 1 ≤ i ≤ k. Let ni be the number of vertices in the partite set Vi. Adding the vertices in each partite set, we have n = n1 + n2 + ... + nk. By assumption,
G has maximal edges, so every pair of vertices in different partite sets must be adjacent. Thus by definition, G is the complete k-partite graph Kn1,n 2,...,n k .Denote the number of edges in a k-partite graph Km1,m 2,...,m k by
e(m1, m 2, ..., m k).
Consider the graph H obtained by removing the edges between the vertices in V1
and V2. Notice that H is a ( k−1)-partite graph with partite sets U1, U 2, ..., U k−1
where U1 = V1 ∪ V2 and Ui = Vi+1 for 2 ≤ i ≤ k − 1. Therefore, H =
Kn1+n2,n 3,...,n k . Then the number of edges in H is
eH = e(n1 + n2, n 3, ..., n k).
When comparing H to G, note that H does not have the edges between V1 and
V2. Thus the number of edges in G is
eG = n1n2 + eH = n1n2 + e(n1 + n2, n 3, ..., n k). (8) Suppose by way of contradiction that the number of vertices in two of the partite sets of G differs by more than 1. Without loss of generality, suppose
n1 ≥ n2 + 2. Consider a new graph ˆG with partite sets W1, W 2, ..., W k such that ˆG = Kn1−1,n 2+1 ,n 3,...,n k . Notice that W1 in ˆG has 1 less vertex than V1 in G,and W2 in ˆG has 1 more vertex than V2 in G. Since n1 ≥ n2 + 2, the number of vertices in W1 and W2 in ˆG differ by at least 2. From (8), the number of edges in ˆG is
e ˆG = e(n1 − 1, n 2 + 1 , n 3, ..., n k)= ( n1 − 1)( n2 + 1) + e(( n1 − 1 + n2 + 1) , n 3, ..., n k).
Then the number of edges in ˆG minus the number of edges in G is
e ˆG − eG = ( n1 − 1)( n2 + 1) + e(( n1 − 1 + n2 + 1) , n 3, ..., n k)
− (n1n2 + e(n1 + n2, n 3, ..., n k)) = ( n1 − 1)( n2 + 1) − n1n2
= n1 − n2 − 1.
Since n1 ≥ n2 + 2, we have that n1 − n2 − 1 ≥ 1. Thus
e ˆG − eG = n1 − n2 − 1 ≥ 1.
Hence when G has two partite sets such that n1 ≥ n2 + 2, there exists a graph, namely ˆG, with at least one more edge than G. Therefore G cannot have the maximum number of edges when the number of vertices in any two partite sets differs by more than 1, a contradiction. Thus for any two partite sets Vi, V j of the graph G = Kn1,n 2,...,n k , we have |ni −nj | ≤ 1. This completes the proof. 13 Lemma 5.2. If G is a graph on n vertices that contains no Kk+1 , then there is a k-partite graph H with the same vertex set as G such that deg G(z) ≤ deg H (z)for every vertex z of G.
Proof. This proof is by strong induction on k.For our base case, let k = 2. Let G = ( V, E G) be a graph on n vertices that does not contain a K3. We want to show there is a bipartite graph H = ( V, E H )with the same vertex set as G such that deg G(z) ≤ deg H (z)for every vertex z in G.Let ∆( G) = max {deg( v) : v ∈ V (G)} and choose a vertex x in G such that deg G(x) = ∆( G). Let W be the set of vertices adjacent to x. Note that any two vertices in W cannot be adjacent, otherwise G would contain a K3. Also,
W only contains vertices adjacent to x, so x is in the set of vertices V \ W .Define a new graph H on the vertices of G such that H is a complete bipartite graph with partite sets W and V \ W . Note that H only has the same vertex set as G, and not necessarily the same edge set as G. We want to show that deg G(z) ≤ deg H (z) for every vertex z in G.Consider a vertex z in V (G) = V (H). First, suppose z ∈ W . Note that no two vertices in W are adjacent in G or H. Then since z is adjacent to all vertices in V \ W in H we have deg H (z) = n − | W | ≥ deg G(z).
Now suppose z ∈ V \ W . Then deg H (z) = deg H (x), since W consists of all vertices adjacent to x
= deg G(x), since x is adjacent to all vertices in W in the graph G
≥ deg G(z), since we chose x to have maximal degree in G.
Thus there is a bipartite graph H with the same vertex set as G such that deg G(z) ≤ deg H (z)for every vertex z in G. Therefore the base case holds. For our induction hypothesis, suppose that for a graph G of order n with no
K+1 as a subgraph, for all 1 ≤ < k , there exists an `-partite graph H with the same vertex set as G such that deg G(z) ≤ deg H (z) for every vertex z in G.We want to show that for a graph G of order n with no Kk+1 as a subgraph that there exists a k-partite graph H such that deg G(z) ≤ deg H (z) for every vertex z in G.Let G = ( V, E ) be a graph on n vertices with no Kk+1 as a subgraph. Let
x be a vertex in V with maximal degree. Let W be the set of vertices adjacent to x.14 Let G0 be the subgraph induced by the vertices in W . Then G0 cannot contain Kk as a subgraph, otherwise G would contain a Kk+1 as a subgraph since x is adjacent to all vertices in W . By our induction hypothesis, there exists a ( k − 1)-partite graph H0 such that deg G0 (z) ≤ deg H0 (z) (9) for every vertex z in W .Define a new graph H on the vertices of G where the EH includes all the edges in H0 and the edges between every vertex in W and in V \ W . First, consider a vertex z in V \ W . Recall that x is also in V \ W . Then we have deg G(z) ≤ deg G(x)since the vertex x was chosen with maximal degree in G. Then deg G(x) = deg H (x) = deg H (z)since all vertices in W in H are adjacent to all vertices in V \ W in H, and
deg G(x) = |W |. Thus for every vertex z in V \ W , we have deg G(z) ≤ deg H (z) (10) since all vertices in W are adjacent to all vertices in V \ W in H.Now consider a vertex z in W . Then deg G(z) ≤ deg G0 (z) + |V \ W |
since G0 does not contain the vertices in V \ W . Then, adding |V \ W | to (9) we have deg G0 (z) + |V \ W | ≤ deg H0 (z) + |V \ W |.
Notice deg H0 (z) + |V \ W | = deg H (z),
because H was constructed by connecting every vertex in W to every vertex in
V \ W , in addition to keep all the adjacencies in H0. Thus deg G(z) ≤ deg H (z) (11) for every vertex z in W . Now by (10) and (11), we have that deg G(z) ≤ deg H (z)for every vertex z in V . Thus we have shown that there is a k-partite graph
H such that deg G(z) ≤ deg H (z) for every vertex z in G, hence the lemma is proved. Now to prove Tur´ an’s Theorem - Version 2. 15 Tur´ an’s Theorem - Version 2. The largest graph with n vertices that con-tains no subgraph isomorphic to Kk+1 is a complete k-partite graph Kn1,n 2,...,n k
with n = n1 + n2 + ... + nk and |ni − nj | ≤ 1 for all i, j ∈ { 1, 2, ..., k } with i 6 = j.
Proof. Let G be a graph on n vertices that does not contain a subgraph iso-morphic to Kk+1 . By Lemma 5.2, we have that there exists a k-partite graph
H with the same vertex set as G such that deg G(z) ≤ deg H (z) for every ver-tex z in G. Therefore the size of H is greater than or equal than the size of
G. By Lemma 5.1, we have that the largest k-partite graph with n vertices is the complete k-partite graph Kn1,n 2,...,n k with |ni − nj | ≤ 1. Thus the largest graph with n vertices with no Kk+1 as a subgraph is a complete k-partite graph
Kn1,...,n k with n1 + n2 + ... + nk = n and |ni − nj | ≤ 1. Hence Tur´ an’s Theorem is proved. Indeed, these two versions of Tur´ an’s Theorem are equivalent. Version 1 gives an upper bound on the number of edges in a graph without some p-clique. In section 4 we found the size of Tur´ an graphs Tn,p −1 to meet this bound when
p − 1 divides n, or to fall just below this bound when p − 1 does not divide
n. Version 2 says that Tur´ an graphs are the largest graphs that satisfy these conditions. It has been shown in section 11.2 in that Tur´ an graphs are in fact the unique graphs of maximum size that satisfy Tur´ an’s Theorem. Finally we have seen three very different proof of Tur´ an’s Theorem. In the probabilistic proof we relied on the expected value of the order of a largest clique in a graph to deduce the upper bound for the number of edges. In the combinatorial proof we cleverly counted the number of edges in any graph without some p-clique to arrive at that same upper bound. In the graph theoretic proof we utilized the structure of Tur´ an graphs to show that these graphs attain the maximum number of edges. Of course, there are more than just three ways to prove Tur´ an’s Theorem, and more than just two ways to state it. The proofs presented here used various techniques from distinct areas of mathematics. Each of them has their own practicality and elegance in extremal graph theory. It remains the reader’s decision to choose which proof of Tur´ an’s Theorem is most favorable.
References
M. Aigner and G. M. Ziegler, Proofs from THE BOOK , Springer-Verlag Berlin Heidelberg, 2014. B. Bollob´ as, Extremal Graph Theory , Dover Publications, Inc., Mineola, 2004. G. Chartrand and P. Zhang, A First Course in Graph Theory , Dover Pub-lications, Inc., Mineola, 2012. N. Hartsfield and G. Ringel, Pearls in Graph Theory , Dover Publications, Inc., Mineola, 2003. 16 A. Llad´ o, An Introduction to Extremal Graph Theory, Accessed 11 September 2016. 17 |
8533 | https://www.facebook.com/CivilEngineerings.Co/posts/activate-your-mind-and-solve-this-math-by-pemdas-rule/1047779870715584/ | Activate your mind and solve this Math... - Civil Engineering | Facebook
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Arturo Kabigting
15 - is the answer
15 - 10 + 5 × 2 = ?
15 - 10 = 5
5 + 10 = 15 ✓
30w
321
View all 144 replies
Buddy Smith
Put it in the calculator it is 15
30w
85
View all 22 replies
Mike Mundo
Correct answer is 20
The pemdas answer is 15
Boy o boy is our education system screwed up
30w
151
View all 157 replies
Minaxi Dhar
This answer is -5
30w
237
View all 88 replies
MikeandBea Cress
It is not new math to get 15. Order of operations has been around since 1800's and the principles behind it for much longer than that. Believe it or not, you Were taught this in elementary school if you went to school.
15-10+5×2
5×2=10
15-0=5
5+10=15
Multiply/divide before add/subtract. Add/subtract are equally ranked, so it is whichever comes first, left to right. Multiply/divide likewise are equally ranked.
30w
55
View all 18 replies
Sadiq Rabiyat
20 is the right answer
30w
101
View all 23 replies
Murugesan
Apply bodmas rule.
Answer is -5.
30w
67
View all 22 replies
Rubi Das
Answer is 15
30w
98
View 1 reply
Stephen Wells
15 - 10 + ( 5 x 2 ) = 15
30w
134
View all 45 replies
Lori Dinneweth Beerli
we're at meeting or some event...there's a table of 15 people & two tables of 5 people. before the end of the event, 10 people leave the table of 15, how many total people are still at the event/meeting?
30w
38
View all 27 replies |
8534 | https://www.youtube.com/watch?v=Qvh-DgPGZVk | Expand and simplify trinomial square (a + b + c)^2 = a^2+b^2+c^2+2ab+2ac+2bc
Anil Kumar
402000 subscribers
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Posted: 12 Oct 2017
Squares of Trinomial:
Distributive Property:
For Guidance Contact : anil.anilkhandelwal@gmail.com
YouTube Channel:
distributiveproperty_applications #mcr3u_functions #degreeofPolynomial #polynomialdegree #mhf4u_polynomials #polynomialexpressions #GCSE_functions #sat_math2
37 comments
Transcript:
I'm Malcolm are sharing with you a question from my student and I hope it's solution helps you all the question is to expand and simplify a plus B plus C whole square now to do this what you can do is you can do a plus B plus C times a plus B plus C right multiply and expand this is one way of doing it perfectly fine the other way could be we could apply the formula which is which is let me write here let's say if we have X plus y whole square we could write x square plus 2xy plus y square right so you could treat this as x + c sy expand and then again expand so two levels you can do so either way you're going to get the same result so let me do both ways in this particular video so first we'll multiply with a all the three terms correct and then with B and C so how many terms will you get we get 3 3 and 3 that means 9 terms multiplying by a gives me a square plus a B plus AC now let's multiply by B now if you multiply B and a you get be a but you could write this as a B right since you know a B is same as as be a red miracles because of the commutative property right okay so I'll prefer to write in the order ABC since it becomes simpler to add them add like terms you get the idea perfect so b x is a be B times B is B Square and this is a good practice B times C is BC this is in the same order so I just maintain now with C see I change order instead of AC I mean CA I'm writing AC is it okay then see with B I'm reading B C and then C square perfect so what do you notice here you notice that we have three square terms so we could write this as a square plus B square plus C square so we have taken care of those square terms and now we will take care of terms like a B a B right so there are two abs similarly you find two ACS do you see and then you find to be seized perfect so that takes care of all the terms so we can now write here to a B plus to AC plus to PC does it make sense so you could do like this and you get your expression and that's a easy way to expand a plus B plus C whole square so the result is you get a square plus B square plus C square and you also get 2 a B to a C and to BC okay now let us do the alternate method sometimes that may be the requirement correct so so we'll write this time a plus B plus C whole square we'll use the formula so we'll treat this as our first term X and this is y perfect so that is how we are going to do okay so what I'm trying to say here is that let me write like this I am using a plus B as one term plus C whole square you get the idea perfect so that is what I'm trying to do so and applying this formula a square plus X square plus 2xy plus y square so let me rewrite this a plus B whole square that is the first term plus two times a plus B times C right product of these two twice is there okay and then plus C square plus C square now we again have a plus B whole square which I could expand so I get a square plus 2 a B plus B Square let me expand these terms to C times a B so we get 2 AC + 2 BC and here we get plus C square is it okay so again you see we have got these three terms I could Club them together a square plus B square plus C square and we have the other three terms we don't even have to Club them so we have 2a b plus 2a c plus 2b c there seems to be complicated to start with however it turns out to be simpler if you really see it right anyway so these are two ways you could always do it you could always group them in different way also your wish try it out and then see what you get I hope you appreciate it feel free to subscribe to my videos and write your comments thank you and all the best |
8535 | https://openstax.org/books/calculus-volume-3/pages/chapter-3 | Skip to ContentGo to accessibility pageKeyboard shortcuts menu
Calculus Volume 3
Chapter 3
Calculus Volume 3Chapter 3
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Checkpoint
3.1
The domain of is all real numbers.
3.3
3.4
3.5
3.6
3.7
3.8
3.9
so
3.10
or Substituting this into gives
3.11
3.12
3.13
At the point the curvature is equal to 4. Therefore, the radius of the osculating circle is
A graph of this function appears next:
The vertex of this parabola is located at the point Furthermore, the center of the osculating circle is directly above the vertex. Therefore, the coordinates of the center are The equation of the osculating circle is
3.14
The units for velocity and speed are feet per second, and the units for acceleration are feet per second squared.
3.15
3.16
967.15 m
3.17
Section 3.1 Exercises
1.
3.
5.
a. b. c. Yes, the limit as t approaches is equal to d.
7.
a. b. c. Yes
9.
11.
13.
The limit does not exist because the limit of as t approaches infinity does not exist.
15.
where k is an integer
17.
where n is an integer
19.
21.
All t such that
23.
a variation of the cube-root function
25.
a circle centered at with radius 3, and a counterclockwise orientation
27.
29.
Find a vector-valued function that traces out the given curve in the indicated direction.
31.
For left to right, where t increases
33.
35.
37.
39.
One possibility is By increasing the coefficient of t in the third component, the number of turning points will increase.
Section 3.2 Exercises
41.
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53.
55.
57.
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Undefined or infinite
67.
To show orthogonality, note that
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The last statement implies that the velocity and acceleration are perpendicular or orthogonal.
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at
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Arc-length function: r as a parameter of s:
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The maximum value of the curvature occurs at
133.
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137.
139.
The curvature approaches zero.
141.
and
143.
145.
147.
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151.
The curvature is decreasing over this interval.
153.
Section 3.4 Exercises
155.
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speed =
163.
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167.
169.
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44.185 sec
175.
sec
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88.37 sec
179.
The range is approximately 886.29 m.
181.
m/sec
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10.94 km/sec
201.
Review Exercises
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False,
205.
False, it is
207.
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unit tangent vector:
215.
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221.
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225.
m/sec, m/sec; at m, m/sec, m/sec2, and m/sec
227.
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© Jul 24, 2025 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University. |
8536 | https://pmc.ncbi.nlm.nih.gov/articles/PMC6430423/ | Morphological characteristics and microstructure of kidney stones using synchrotron radiation μCT reveal the mechanism of crystal growth and aggregation in mixed stones - PMC
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Morphological characteristics and microstructure of kidney stones using synchrotron radiation μCT reveal the mechanism of crystal growth and aggregation in mixed stones
Muhammed A P Manzoor
Muhammed A P Manzoor
1 Yenepoya Research Centre, Yenepoya (Deemed to be University), Mangalore, Karnataka, India
2 Department of Urology, Yenepoya Medical College, Yenepoya (Deemed to be University), Mangalore, Karnataka, India
Conceptualization, Data curation, Formal analysis, Methodology, Writing – original draft
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1,2,¤, Ashish K Agrawal
Ashish K Agrawal
3 Technical Physics Division, Bhabha Atomic Research Centre, Indore-Mumbai, India
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3, Balwant Singh
Balwant Singh
3 Technical Physics Division, Bhabha Atomic Research Centre, Indore-Mumbai, India
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3, M Mujeeburahiman
M Mujeeburahiman
2 Department of Urology, Yenepoya Medical College, Yenepoya (Deemed to be University), Mangalore, Karnataka, India
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2, Punchappady-Devasya Rekha
Punchappady-Devasya Rekha
1 Yenepoya Research Centre, Yenepoya (Deemed to be University), Mangalore, Karnataka, India
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Editor: Yogendra Kumar Mishra 4
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1 Yenepoya Research Centre, Yenepoya (Deemed to be University), Mangalore, Karnataka, India
2 Department of Urology, Yenepoya Medical College, Yenepoya (Deemed to be University), Mangalore, Karnataka, India
3 Technical Physics Division, Bhabha Atomic Research Centre, Indore-Mumbai, India
4 Institute of Materials Science, GERMANY
Competing Interests:The authors have declared that no competing interests exist.
¤
Current address: ICAR–Indian Institute of Spices Research, Kozhikode, Kerala, India
✉
E-mail: rekhapd@hotmail.com, dydirectoryrc@yenepoya.edu.in
Roles
Muhammed A P Manzoor: Conceptualization, Data curation, Formal analysis, Methodology, Writing – original draft
Ashish K Agrawal: Conceptualization, Formal analysis, Investigation, Resources, Validation, Writing – review & editing
Balwant Singh: Data curation, Formal analysis, Methodology, Writing – review & editing
M Mujeeburahiman: Conceptualization, Investigation, Resources, Supervision, Writing – review & editing
Punchappady-Devasya Rekha: Conceptualization, Resources, Supervision, Validation, Writing – review & editing
Yogendra Kumar Mishra: Editor
Received 2018 Nov 14; Accepted 2019 Mar 5; Collection date 2019.
© 2019 Manzoor et al
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PMCID: PMC6430423 PMID: 30901364
Abstract
Understanding the mechanisms of kidney stone formation, development patterns and associated pathological features are gaining importance due to an increase in the prevalence of the disease and diversity in the presentation of the stone composition. Based on the microstructural characteristics of kidney stones, it may be possible to explain the differences in the pathogenesis of pure and mixed types of stones. In this study, the microstructure and distribution of mineral components of kidney stones of different mineralogy (pure and mixed types) were analyzed. The intact stones removed from patients were investigated using synchrotron radiation X-ray computed microtomography (SR-μCT) and the tomography slice images were reconstructed representing the density and structure distribution at various elevation planes. Infrared (IR) spectroscopes, X-ray diffraction (XRD) and scanning electron microscopy (SEM) were used to confirm the bulk mineral composition in the thin section stones. Observations revealed differences in the micro-morphology of the kidney stones with similar composition in the internal 3-D structure. Calcium oxalate monohydrate stones showed well-organised layering patterns, while uric acid stones showed lower absorption signals with homogenous inner structure. Distinct mineral phases in the mixed types were identified based on the differential absorption rates. The 3-D quantitative analysis of internal porosity and spatial variation between nine different types of stones were compared. The diversity among the microstructure of similar and different types of stones shows that the stone formation is complex and may be governed by both physiological and micro-environmental factors. These factors may predispose a few towards crystal aggregation and stone growth, while, in others the crystals may not establish stable attachment and/or growth.
Introduction
Kidney stone disease is caused by deposition of mineralized crystals in the renal calyces and pelvis, and these depositions transform into stone(s) in a series of events. The stones formed are managed with surgical procedures or medical interventions; however, prevention of recurrence requires an understanding of the stone composition for medical management that is specific for the stone mineral type [1, 2]. Crystal retention and growth are governed by various factors such as hyaluronic acid, osteopontin, composition of the renal tubular epithelial cell surfaces along with superstauration [3, 4]. However, differences in the interplay of these factors and other intrinsic physiological and metabolic factors may govern the crystallization process and determine the compositions .
The microstructure of the stone matrix varies among the stone types and is determined by the mineral composition. The compositional analysis reveals the mineral types responsible for the stone formation, and the ultra-structural investigation of kidney stone matrix provide additional details that are crucial links to the pathogenesis . Certain stones especially composed of calcium oxalate are attached to ‘plaque’, while others often form in free solution in the renal collection system. Under each of these conditions the nucleation and aggregation process differ significantly allowing the stone to aggregate and grow under different conditions leading to heterogeneity in the physical and chemical architecture. During the progression of stone genesis, extensive and repeated dissolution can also occur and may alter the stone matrix at ultra-structural level . The renal inflammatory injury induced by cell-crystal reaction plays an important role in the formation of intra-renal calcium oxalate crystals .
The diversity in the lithogenesis process itself can significantly influence the stone internal structure and composition. Further, the mineral distribution pattern in the stone matrix is defined by the mineral accumulation process during the crystal growth that is governed by the availability and or super-saturation of the species. Various advanced techniques are used to study the kidney stone composition and ultrastructure [7, 9]. In a study, detailed characterization of Randall’s plaque using μCT provided details on the mechanism of stone formation; especially the role of apatite in stone formation within the interstitium as a denser portion present within the plaque . A high diversity of mixed stones among the patients indicates that the nucleation process can be random events that can favour the co-aggregation of other urine crystals and facilitate the stone retention. Hence, it is important to understand the relative distribution of the different mineral types in the kidney stones using technology that is more precise and accurate.
Imaging using synchrotron radiation X-ray computed microtomography (SR-μCT) due to higher photon flux in parallel beam morphology as compared with conventional μCT provides higher resolution details in 3-D . Kaiser et al. initially used SR-μCT to study the microstructure and mineralogy of kidney stone with 3-D features and internal composition of kidney stones. This study showed the presence of apatite as concentric and continuous layers in certain stones especially in calcium oxalate stones. However, the study was limited to fragmented kidney stones and could not represent the actual growth pattern characteristics. Hence in this study, we employed SR-μCT to study the texture, mineral deposition pattern, 3-D quantitative analysis of porosity and spatial variation of intact kidney stones representing major mineral types.
Methods
Ethics statement
All the procedures involving human participants were approved by the Yenepoya University Institutional Ethics Committee (YUEC.022/16) and Yenepoya Research Centre Scientific Review Board (YRCSRB034/17). The study was performed in accordance with the ethical standards of the institutional and/or national research committee and with the 1964 Helsinki declaration and written informed consent was obtained from all the participants.
Sample and specimen
Surgically removed kidney stones (n = 22) were collected from the patients with symptomatic stone diseases. All the patients underwent baseline assessment, including detailed medical history and complete blood and urine analysis and non-contrast computerized tomography (NCCT). Each patient had his or her kidney stone(s) removed following the common clinical practice (laparoscopic, percutaneous nephrolithotomy or open stone surgery). The intact stones removed were washed and dried prior to SR-μCT investigations.
Analytical procedures used in the blood analysis
Blood (5 mL) was collected in vacutainer from each patient using Ethylene diamine tetraacetic acid as a preservative. Blood sugar was estimated by glucose oxidase-hydrogen peroxide (Trinder) method. Serum calcium and electrolytes such as sodium, potassium, and chloride were estimated by Direct-ISE (VITROS 5600 Integrated System Ortho-Clinical Diagnostics NJ, US). ESR was assessed by the photometry method. The 24 h urine pH was measured using a digital pH meter.
SR-μCT scans and 3-D reconstruction
The experiments were performed at the X-ray imaging beamline (BL-04) on Indus-2 synchrotron source at Raja Ramanna Centre for Advanced Technology, (RRCAT, India). The experimental setup for X-ray μCT consisted of motorized precision translation stages x, y and z and a rotating stage [12, 13]. For data acquisition, the kidney stone samples were mounted over the sample holding chuck. The energy of the incident X-ray beam was optimized for the samples depending on their thicknesses and approximate composition (24–30 keV). The effective voxel size was 2.25 micron. The samples were rotated about their axis in the angular range of 0–180° with step size 0.2° and a total of 901 radiographic projections were collected for each sample.
Image reconstruction and analysis
Tomography slice images were reconstructed using filtered back projection method to represent the density and structure distribution at various elevation planes in the samples. The cross-sectional slice images were stacked together, volume rendered to show stone microstructure in 3-D and to highlight the features. Quantitative analysis of the SR-μCT data on grey values and porosity variations was carried out using ImageJ software. Porosity was calculated from the images using the Otsu method of threshold followed by noise removal using a median filter . Grey value calculation was done using average grey value for every 10 slices while propagating from top to bottom.
Compositional analysis using XRD, FT-IR spectroscopy, and FESEM
Following the SR-μCT studies, the stone was subjected to chemical composition analysis using XRD, FT-IR spectroscopy and Field Emission Scanning Electron Microscopy (FESEM). The XRD patterns were recorded with a laboratory diffractometer (Rigaku MiniFlex 600) using Cu-Kα radiation (λ = 1.5406 Å). The diffraction patterns were registered within the 2θ angle range from 10 to 50° and from the diffraction pattern the crystalline phases were identified [15, 16]. The mineral compositions were assessed using ATR-FTIR (Shimadzu IR Prestige-21), in the frequency range of 4000–400 cm-1 at 4 cm-1 resolution [17, 18]. The FESEM was used to study the microstructure and morphology (Carl Zeiss, Germany).
Statistical analyses
Statistical analysis was performed using SPSS, Version 22.0. (IBM Corp). Results of categorical data were summarized using frequencies and percentages. Continuous variables were reported as mean ± standard deviation. One-way ANOVA was used to compare means of quantitative variables across the stone types. Significance tests were two-sided and the value of p<0.05 was considered statistically significant.
Results
Among all the 22 stones tested, 16 (72.7%) were derived from males, and 6 (27.3%) were from females. Summary of patient demographics is shown in Table 1. The stones represented 9 mineral types namely; calcium oxalate monohydrate (COM), uric acid, struvite, COM-apatite mixed, struvite-apatite mixed, COM-uric acid mixed, COM-calcium oxalate dihydrate (COD) mixed, COM-COD-apatite and struvite-COM-apatite mixed. Majority of the patients showed blood biochemistries in the normal ranges; the clinical characteristics such as blood and urine biochemistry are given in S1 Table. The XRD pattern and FESEM images of representative samples are shown in Fig 1.
Table 1. Patient demographics and stone mineral composition.
| Sample ID | Age | Sex | Location | Side | Recurrence | Mineral |
:--- :--- :---
| KS-1, KS-2 | 40 | M | Kidney | Right | No | COM |
| KS-3 | 64 | M | Kidney | Right | No | Uric acid |
| KS-4 | 31 | F | Kidney | Right | No | Struvite- apatite mixed |
| KS-5 | 39 | F | PUJ | Right | No | COM-uric acid mixed |
| KS-6, KS-8 KS-16 | 34 | M | Kidney | Bilateral | No | COM |
| KS-7 | 64 | M | PUJ | Right | No | Struvite—COM-apatite mixed |
| KS-9 | 37 | M | Kidney | Left | No | COM-apatite mixed |
| KS-10, KS-11 | 53 | F | Kidney | Left | Yes | Struvite |
| KS-12 | 50 | M | Kidney | Bilateral | Yes | COM-COD-apatite |
| KS-13 | 65 | F | Kidney | Right | Yes | COM -uric acid mixed |
| KS-14 | 34 | M | Ureter | Right | No | COM-COD mixed |
| KS-15 | 65 | M | kidney | Left | Yes | Struvite-apatite mixed |
| KS-17 | 24 | M | Ureter | Right | No | Uric acid |
| KS-18, KS-19 | 50 | M | PUJ | Right | No | COM-COD-apatite mixed |
| KS-20 | 42 | M | Ureter | Right | Yes | COM-COD-apatite mixed |
| KS-21 | 60 | F | Kidney | Left | No | Struvite |
| KS-22 | 34 | M | Kidney | Left | No | COM-apatite mixed |
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PUJ- Pelvi-ureteric junction, COM-Calcium oxalate monohydrate, COD-Calcium oxalate dihydrate
Fig 1. Representative XRD and FESEM images of kidney stones used in the present study.
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(a) KS-8 and KS-10 COM and struvite crystals showing highest peak intensity at (040) plane. KS-17 Uric acid crystal showing preferred texture growth along (121) plane. (b-d) Pure stones viewed by conventional photography and FESEM. (b) COM appeared as reticulate type of appearance and boundaries of small crystallites. (c) Struvite crystals showed small crystals with hemimorphic morphology. (d) Uric acid crystals showed agglomerates of plate like crystals with clusters of small crystals (scale 10 μm).
Micro-tomography of pure stone types: SR- μCT
The COM stones exhibited visible, shell-like texture with a radial pattern and concentric layers of organization. The images with mineral density and grey-scale architecture are given in Fig 2. The central denser region of high X-ray attenuation values was consistent with the core of apatite nucleus surrounded by well organized outer layers. COM stones showed multiple layers of organisation with visible internal cracks (Fig 2B & 2C). The porosity was 0.23% (S2 Table). Similar microstructure in different stones derived from the same patient was observed. However, the microstructure varied in samples derived from different patients (S1 Fig). Fig 3 shows a reconstructed tomographic slice image, the internal 3-D morphology and the histograms indicate the differences in the density between the apatite core and outer layers. The tomographic slice image of different sections of the stones showing mineral precipitation and accumulation of minerals at different elevation planes is given in S2 Fig.
Fig 2. Pure COM stone micro-morphology.
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(a) A reconstructed μCT slice showing concentric layers of organization. (b) A typical SR-μCT image slice showing visible, radial pattern with concentric layers of organization showing a prominent ring artefact. Arrow mark shows central denser region of high X-ray attenuation values consistent with apatite nucleus. (c) SR-μCT slice showing multiple layers of organisation. Arrow mark shows the presence of visible internal cracks.
Fig 3. Micro-morphology of COM showing tomographic slice image and the internal 3-D morphology.
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(a) Maximum-intensity projection of the stone that displays the internal 3-D morphology. Inset showing the conventional photography. (b) Histogram and profile lines measured for outer layer and internal apatite nucleus showing the difference of density in the two regions. (c & d) 3-D reconstruction of the stone that displays the internal 3-D morphology showing a central denser nucleation point and the presence of apatite nucleus. Arrow marks show occasional layers of apatite laid down in the COM.
The struvite stones consisted of layered pattern and exhibited void spaces with cracks inwards from the outer surfaces (Fig 4A). The apatite content in the struvite specimen was conspicuous with increased X-ray absorption. The porosity of struvite stones was 7.9% (S2 Table). The uric acid stones showed lower absorption signals compared to other stone types. These stones were less porous (4.36%) and the inner structure displayed multinucleation points with homogenous texture (Fig 4B). However, the outer surface of the uric acid stones showed heterogeneous regions with distinct surface roughness containing small cavities and sharp edges (Fig 4C).
Fig 4. Micro-morphology of pure stone types.
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(a) SR-μCT image slice of struvite stone showing a pattern of two separate layers with void spaces, micro and macro pores (arrow). The thin white lines showing the apatite. (b) SR-μCT image slice of uric acid stone showing homogenous texture. (c) 3-D surface rendering with slice across the image stack of uric acid stone.
Micro-tomography of mixed stone types
The mixed types of stones comprised a more variable crystalline growth pattern, having more irregular and diverse shape and structure (Fig 5). The mixed stones with apatite and COM presented unorganized distribution of the mineral components in the stone matrix. All the mixed stones having apatite component showed relatively higher absorption signals and particles of apatite crystals (Fig 5A & 5B). The COM-uric acid mixed stones displayed intricate structure,calcium oxalate crystal components having relatively higher absorption and weak scattering signal and the opposite was observed in uric acid crystals (Fig 5C & 5D). These stones exhibited a relatively visible radial pattern of core structures. The struvite-apatite mixed stones showed comparatively higher X-ray attenuation compared to other stone types (Fig 5E & 5F). The COM-COD-apatite system, the calcium oxalate component exhibited uniform microtomography, while the apatite showed a relatively denser structure and X-ray attenuation (Fig 5G & 5H). In the case of COM-COD and struvite-COM-apatite mixed stones, apatite showed a relatively denser structure compared to calcium oxalate (S3 Fig).
Fig 5. Micro-morphology and 3-D reconstruction of the representative mixed stone types.
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(a & b) COD-apatite mixed stones. Apatite found along with COD in an unorganized manner. The apatite mixed stones showed relatively high absorption signals and articles of embedded crystals (green). (c & d) COM-uric acid mixed stones. The COM crystal components having relatively higher absorption (green) are directly inverse to the relatively weak absorption observed in uric acid crystals (red). (e & f) Struvite-apatite mixed showing two mineral types by distinct X-ray attenuation; struvite (red), apatite (or carbapatite) (green) and porous air cavities (blue). (g & h) COD-COM-apatite mixed. These stones exhibited visible internal micro and macro pores and showed three distinct layers with varying density and porosity. Arrow indicates apatite nucleus.
Porosity and grey value variation in pure and mixed stone types
Porosity was calculated based on the percentage area of pores in the samples after binarization using standard methods. We found a distinct variation in the porosity among different types of kidney stones as well as stones from different patients in the same group (S2 Table). The purity of the stones was evident from the uniformity in the shade of grey distribution in segmented tomography slices. Among the stone types, the least porosity was found in pure stones and highest porosity was observed in mixed stones (S2 Table)
Discussion
The ex vivo imaging is often used to build data on stones of different morphologies obtained from diverse patient populations. In this study, we compared 22 kidney stones with 9 different compositions (3 pure and 6 mixed types) using SR-μCT. We observed variation in the micro-tomography among the similar type of stones as well as among different composition. The mixed composition stones displayed high heterogeneity in the 3-D internal morphology. This heterogeneity may be due to the fact that crystallization is influenced by many factors like the concentration of ions, their chelators and ionic strength . Similarly, we observed variations in the porosity among different types of kidney stones as well as stones from different patients in the same group that may indicate the mineral concentration or rate of crystal accumulation during the growth. Highly porous stones like COD and struvite are susceptible to shock wave lithotripsy (SWL) compared with less porous stones such as COM. The porosity variation among mixed type stones may pose a challenge during SWL.
Calcium oxalate, struvite and apatite stones exhibit relatively high absorption signals, while directly inverse observations were found in case of the uric acid stone, which is in accordance with previously described chemical and morphological stone properties . Keeping the operational conditions similar (conditions of source size, beam energy and magnification) for the analysis of all stone types in the same facility the variations observed in the grey distribution is attributed to the inherent mineralogical and morphological variations of different types of the stones. We used the original greyscale for the identification of different stones and this imaging system can be correlated to non-contrast helical CT in human .
Pure COM stones exhibited a radial pattern and concentric layers of organization with apatite in the nidus [22, 23]. Most of the calcium oxalate stones generally grow from Randall’s plaque having apatite core. The reconstructed tomographic slice image of pure COM suggests that the layers could have been formed by organic material being laid down first and crystals forming within the matrix . Moreover, the urinary constituents like citrate and osteopontin can involve in shape modification and modulation of crystal habits as well the rate of crystal aggregation . It has been shown that the formation of COD crystals is mostly associated with a high urinary calcium concentration and COM crystallization is associated with high oxalate concentration, thus supporting the concept of calcium-dependence of COD and of oxalate-dependence of COM .
Uric acid stones are known to grow in a layer-wise manner as concentric rings around a crystallite core [22, 27]; however, in our cases, it was not observed. It has been shown in earlier reports that the crystal size of uric acid stones is significantly different between male and female patients . Previous studies have highlighted the difficulty in interpretation of struvite stone by CT mainly due to the presence of other mineral components . However, struvite can be distinguished in μCT from other stone types due to its characteristic internal structure and radiation attenuation properties. According to a previous investigation, bacterial imprints can also appear on kidney stones with small nanocrystals, such as carbonated apatite than with large nanocrystals, such as struvite .
Among the kidney stones, a relative high proportion and diversity of mixed type stones are now commonly reported . The mechanism and sequence of events during the mixed stone growth is highly diverse and yet to be fully explored. The mixed mineral phases present distinct radiographic attenuation values corresponding to each mineral composition in the mixed stone matrix and this enables the visualization of crystal structures and stone mineralogy . For example, in COM-apatite mixed stones, the apatite found along with COM was arranged in an unorganized manner unlike the apatite core in the pure COM. Though, the crystallization process cannot be easily outlined; it may be formed by multiple apatite nucleation interrupting the continuous COM crystal growth resulting in this mixed morphology. It can be possible to explain with the theory of nascent crystal growth from the ductal plugs, with aggressive growth of two different minerals. The process may be governed by the urinary pH, and rate of calcium and oxalate excretion [5, 32]. Due to the chemical interaction and relative concentration of mineral species distributed in the urine, the stone formation can be modified with the incorporation of other minerals present in the urine to a mixed-component complex stone. Studies have shown the co-aggregation effect of amorphous calcium phosphate to promote the formation of large calcium oxalate complexes forming COM stable nuclei. And these clusters aggregate around calcium phosphate that provides multiple sites for nucleation and growth, resulting in multiple COM crystals encapsulating the calcium phosphate . This may be a long process mediated by the biochemical characteristics of the urine and physiological process of the patient. Similar observations were also made showing large porous non-homogenous regions throughout the samples with fragmented particles accumulated in micro-CT by Miller et al. . The presence of uric acid in calcium oxalate stones has been commonly observed and may be a result of the similar crystal lattices present in both crystal types. It is also shown that uric acid binding proteins that bind to uric acid can act as a bridge to bind calcium oxalate to uric acid during the stone formation .
Based on the 3-D distribution pattern of the mixed composition in the stone matrix, it is evident that incorporation of different mineral compositions may be dependent on the chemical interaction, changes in the pH and supersaturation of species. The relative distribution of the species may determine the kinetics and thermodynamic aspects of the crystallization process. Mass transfer of a solute from supersaturated liquid solution to a solid crystalline and mixing, also have an effect on crystal purity and morphology. During the initial aggregation stages, the stone exterior may exhibit solute exchange allowing the binding of biological and non-biological entities that can attract various stone constituents, and help in the adherence of smaller crystals to form aggregates.This process will help in aggregation and/or binding of crystal phases. The retained crystals will eventually continue to accumulate the mineral with a periodic interruption in the process by the precipitation of organic material leading to complex structures. Some of the accumulated minerals may get degraded due to dissolution and form micro-porous structures. Similarly, the organic components over the long retention time may get degraded and lost leading to the porous structure. The minerals in the stone may also interact with urine constituents resulting in concentration or leaching as represented by density variation in the stone microstructure. These processes may also encompass the phenomenon of secondary nucleation of newly formed crystals on the surface of those already formed. Our SR-μCT images confirm the law of superposition (i.e., older layers at the bottom and younger layers at the top) in the mixed stones . The porosity in these stones can vary depending on the distribution of other elements. Hence, SR-μCT can be used to easily distinguish the compositions of each mineral and quantitatively calculate the percentages of the total volume of the stone samples based on the radiographic attenuation values of each segmented planes. These parameters can be used to establish the stone growth rate and estimate the age of the stone. These require both in vitro and in vivo parallel studies in comparison with clinically received samples.
Conclusion
SR-μCT technique holds potential for better analysis of kidney stones especially their micro-structure, 3-D quantitative analysis of porosity and its spatial variation, thereby providing useful information for the understanding the disease. The micro-morphology of the kidney stones of similar composition showed diversity in the internal 3-D structure indicating different mechanisms of stone initiation and growth. Mixed stones can be formed due to the interaction between the two mineral phases or may also be influenced by the biological residues released due to the damage caused by the stone retention in the tissues. With further efforts, the SR-μCT technique may also be able to provide solution for estimating the age of the stone and the mineralization process.
Supporting information
S1 Table. The blood and urine parameters of the patients with kidney stone.
(DOCX)
Click here for additional data file. (16.8KB, docx)
S2 Table. Porosity value of pure and mixed types of kidney stones.
(DOCX)
Click here for additional data file. (14.9KB, docx)
S1 Fig. Micro-morphology of pure calcium oxalate monohydrate.
(a & b) Different stones obtained from same patient (KS 8 and KS 16). (c & d) Different stones obtained from same patient (KS 1 and KS 2).
(TIF)
Click here for additional data file. (9.1MB, tif)
S2 Fig. The tomographic slice image of different sections of the calcium oxalate monohydrate stone showing mineral precipitation and accumulation of minerals at different elevation plane.
(i–xx) Different sections of the stones showing mineral precipitation and accumulation of minerals at different elevation plane. Arrows indicates the nucleus.
(TIF)
Click here for additional data file. (93.7MB, tif)
S3 Fig. Micro-tomography of mixed kidney stones.
(a) COM-COD mixed showing uniform micro-tomography and (b) COM-struvite-apatite mixed stones. Apatite showing comparatively denser structure (arrow).
(TIF)
Click here for additional data file. (2.5MB, tif)
Acknowledgments
The authors acknowledge the Yenepoya University for the seed grant for carrying out this study and are thankful to the technical staff at Raja Ramanna Centre for Advanced Technology (RRCAT), Indore, for providing the synchrotron beam, the necessary support and infrastructure to carry out this research and their active support during data acquisition and analysis.
Data Availability
All relevant data are within the paper and its Supporting Information files.
Funding Statement
The authors acknowledge the Yenepoya University for the seed grant for carrying out this study. The funder had no role in study design, data collection and analysis, decision to publish, or preparation of the manuscript.
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Associated Data
This section collects any data citations, data availability statements, or supplementary materials included in this article.
Supplementary Materials
S1 Table. The blood and urine parameters of the patients with kidney stone.
(DOCX)
Click here for additional data file. (16.8KB, docx)
S2 Table. Porosity value of pure and mixed types of kidney stones.
(DOCX)
Click here for additional data file. (14.9KB, docx)
S1 Fig. Micro-morphology of pure calcium oxalate monohydrate.
(a & b) Different stones obtained from same patient (KS 8 and KS 16). (c & d) Different stones obtained from same patient (KS 1 and KS 2).
(TIF)
Click here for additional data file. (9.1MB, tif)
S2 Fig. The tomographic slice image of different sections of the calcium oxalate monohydrate stone showing mineral precipitation and accumulation of minerals at different elevation plane.
(i–xx) Different sections of the stones showing mineral precipitation and accumulation of minerals at different elevation plane. Arrows indicates the nucleus.
(TIF)
Click here for additional data file. (93.7MB, tif)
S3 Fig. Micro-tomography of mixed kidney stones.
(a) COM-COD mixed showing uniform micro-tomography and (b) COM-struvite-apatite mixed stones. Apatite showing comparatively denser structure (arrow).
(TIF)
Click here for additional data file. (2.5MB, tif)
Data Availability Statement
All relevant data are within the paper and its Supporting Information files.
Articles from PLoS ONE are provided here courtesy of PLOS
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8537 | https://storage-cdn.labflow.com/data/filedir/ff/ba/ffba85f7e9d2c62e22cf69f0f7002e5daa0542eb | Published Time: Wed, 26 Mar 2025 13:33:14 GMT
Experiment 3:
The Molar Volume of Gases
Purpose
The purpose of this experiment is to determine the molar volume of a gas.
Learning Objectives
Quantitatively measure the volume of oxygen gas produced from the decomposition of hydrogen per-oxide. Calculate the molar volume of oxygen gas at standard temperature and pressure using the ideal gas law and Dalton’s law of partial pressures.
Principles
Molar Volume of an Ideal Gas
The molar volume is defined as the volume of one mole of a substance, or the volume per mole of a substance,
𝑉/𝑛 . The molar volume of a gas varies with temperature and pressure as the gas expands and compresses. We can calculate the molar volume of a gas from its density and molar mass at a given temperature and pressure.
𝑉 𝑛 = molar mass (g/mol) density (g/L) = molar volume (L/mol)
Table 3.1: Molar Volumes of Gases at 0 °C and 1 atm
Gas Density (g/L) Molar Mass (g/mol) Molar Volume (L/mol)
CO 1.250 28.010 22.406 N2 1.250 28.013 22.403 He 0.178 4.003 22.429 H2 0.0899 2.016 22.429 CO 2 1.977 44.010 22.262 HCl 1.639 36.461 22.244 O2 1.429 31.999 22.392
Laboratory Manual Prepared by Catalyst Education, LLC for the University of California at Santa Barbara, Department of Chemistry. Copyright 2023 by Petra Van Koppen; University of California, Santa Barbara
3.1 The Molar Volume of Gases
Table 3.1 clearly shows that for a variety of gases, at standard temperature and pressure (STP), the molar volume is very near 22.4 L/mol. In other words, at 0 °C and 1 atm, one mole of a gas occupies approximately 22.4 L. Two moles of gas would occupy 44.8 L. This is known as Avogadro’s Law : For a gas at constant temperature and pressure the volume is directly proportional to the number of moles of gas, n.
𝑉 ∝ 𝑛 when pressure and temperature are constant The relationship between molar volume and pressure and temperature are described in Boyle’s law and Charles’ law as follows.
Boyle’s Law : The volume of a fixed quantity of gas at a fixed temperature is inversely proportional to pressure. As the volume increases the pressure decreases.
𝑃 ∝ (1/𝑉) when temperature and the amount of gas are constant
Charles’ Law : For a fixed quantity of gas at constant pressure, the volume increases as the temperature increases. A plot of volume versus temperature is a straight line and the intercept of this plot, when T is measured in ° C, is
−273.15 °C. This is why absolute zero of temperature is defined as 0 K = −273.15 °C.
T (in Kelvin) = T (in ° C) – 273.15.
𝑉 ∝ 𝑇 when pressure and the amount of gas are constant Combining the gas laws of Boyle, Charles, and Avogadro, the ideal gas law is derived. The general equation given for the ideal gas law is an empirical law which holds approximately for all gases near atmospheric pressure and becomes increasingly accurate at low pressures.
𝑃𝑉 = 𝑛𝑅𝑇 Ideal Gas Law The temperature, T, is the absolute temperature in units of Kelvin. R is the universal gas constant that has the same value for all gases. R can be estimated from the observation that one mole of any gas at 0 °C and 1 atm occupies approximately 22.4 L.
𝑅 = 𝑃𝑉 𝑛𝑇 = (1.0 atm )(22.4 L)(1 mol )(273 K) = 0.08206 L atm mol K The numerical value of R depends on the units chosen for pressure and volume. Commonly used values of the gas constant, obtained through simple unit conversions, include the following:
R = 0.08206 L atm mol −1 K−1 R = 1.987 cal mol −1 K−1
R = 8.314 J mol −1 K−1 R = 8.314 kg m2 sec −2 mol −1 K−1
Laboratory Manual Prepared by Catalyst Education, LLC for the University of California at Santa Barbara, Department of Chemistry. Copyright 2023 by Petra Van Koppen; University of California, Santa Barbara
3.2 The Molar Volume of Gases
In this experiment you will determine the molar volume of oxygen at standard temperature and pressure (STP;
0 °C and 1 atm) and compare this value to the molar volume of 22.414 L/mol for an ideal gas at STP. To determine the molar volume of oxygen, the decomposition reaction of hydrogen peroxide, H 2O2, to produce water and oxygen will be used.
2 H 2O2 (aq ) catalyst
−−−→ 2 H 2O (l) + O2 (g)
This reaction is slow and a catalyst is needed to speed up the reaction. The catalyst used in this experiment is FeCl 3; it speeds up the reaction without being consumed in the reaction itself. Because mass is always conserved in a reaction, the difference between the weight of reactants before the reaction takes place and the weight of products after the reaction goes to completion equals the mass of oxygen gas evolved. By measuring the volume of O 2 produced at a known temperature and pressure, we can calculate the molar volume of O 2 in L/mol (the volume of O 2 divided by the moles of O 2 produced). The volume of oxygen gas produced in this reaction corresponds to the volume of water displaced by the gas. We must take into account that when any gas in a closed container is collected over liquid water, or is exposed to water, the water contributes to the total vapor pressure. Water evaporates until a saturated vapor results, that is, until opposing rates of evaporation and condensation of water molecules at the liquid surface reach a balance. According to Dalton’s law of partial pressures, each gas exerts its own pressure regardless of the presence of other gases. In the collection of oxygen gas the total pressure is the sum of the partial pressure of oxygen and water.
𝑃 TOTAL = 𝑃 O2 + 𝑃 H2O
Thus, to determine the pressure of oxygen gas, a correction for the vapor pressure of water ( 𝑃 H2O) must be made.
𝑃 O2 = 𝑃 TOTAL − 𝑃 H2O
The state of a gas is defined by its pressure, volume, temperature, and the number of moles. For the change in state from 𝑃 1, 𝑉 1, 𝑇 1, 𝑛 1 to 𝑃 2, 𝑉 2, 𝑇 2, 𝑛 2 the following equations hold:
𝑃 1𝑉 1
𝑛 1𝑇 1
= 𝑃 2𝑉 2
𝑛 2𝑇 2
𝑃 1𝑉 1
𝑇 1
= 𝑃 2𝑉 2
𝑇 2
at constant 𝑛
This equation can be rearranged to: 𝑉 2 = 𝑉 1 ( 𝑃 1
𝑃 2
) ( 𝑇 2
𝑇 1
) at constant 𝑛
Thus, if we measure the volume of a gas to be 𝑉 1, at temperature 𝑇 1, and pressure 𝑃 1, the volume of the gas ( 𝑉 2) can be calculated at any other temperature ( 𝑇 2) and pressure ( 𝑃 2). In this experiment, you will measure the volume of oxygen and hydrogen at the experimental temperature and pressure and you will then calculate the volume of each gas at STP ( 0 °C and 1 atm).
Laboratory Manual Prepared by Catalyst Education, LLC for the University of California at Santa Barbara, Department of Chemistry. Copyright 2023 by Petra Van Koppen; University of California, Santa Barbara
3.3 The Molar Volume of Gases
Procedure
Equipment
Extra small test tubes, 12 × 75 mm, are supplied for the FeCl 3 catalyst. These test tubes must be returned after class.
Chemicals
Deionized water Hydrogen peroxide solution, 3% Iron(III) chloride solution, 3 M
SAFETY
WEAR SAFETY GLASSES If you spill any chemicals on your skin, flush with water immediately.
Assemble the apparatus as shown in Figure 3.1. Clean the glassware with soap and water, use deionized water (DI water) as the final rinse. For flask B use a 500 mL Erlenmeyer flask or a 500 mL filter flask. You may have one in your drawer. If not, borrow one from the stock room. Remove the yellow side arm from the filter flask and put a cork #3 in the hole on the side of the flask. Corks are provided in a bin along with the rubber stoppers and hoses.
Figure 3.1: Note the position of the clamp—it is near the top of flask B.
Obtain an extra small test tube (12 × 75 mm) provided for this experiment. Test to make sure the small test tube remains standing at an angle in the 250 mL flask. If it drops completely horizontal inside the flask, obtain another flask or you can secure a metal paper clip to the open end of the test tube to prevent this.
Laboratory Manual Prepared by Catalyst Education, LLC for the University of California at Santa Barbara, Department of Chemistry. Copyright 2023 by Petra Van Koppen; University of California, Santa Barbara
3.4 The Molar Volume of Gases
Measure 20 mL of 3% H 2O2 solution in a graduated cylinder and pour it into a 250 mL Erlenmeyer flask (flask A in the figure). Measure and pour approximately 4.5 mL of 3 M FeCl 3 into a 12 × 75 mm test tube (a very small test tube provided for this experiment). Holding flask A at an angle, carefully slide the test tube into the flask or lower the test tube with a pair of tongs. 4. Weigh flask A and its contents (before you weigh flask A, make sure it is dry on the outside). Record your data in Report Table 3.1 in the Report Sheet. 5. Fill the 500 mL Erlenmeyer flask B to the neck with tap water. Fill the 600 mL beaker C about one-third full with tap water. Disconnect the rubber stopper at point A and attach a pipette bulb. With the pinch clamp open, force air into rubber tube AB by squeezing the pipette bulb, forcing water from flask B into the beaker C. Raise and lower beaker C to move water back and forth through rubber tube BC to remove all air bubbles. With water halfway up the neck of flask B, reconnect the stopper at point A. Push down firmly on the stopper on flask A and on the stopper on flask B. 6. With the pinch clamp open, test the apparatus for leaks, as follows. Raise beaker C as high as possible without removing tubing from the beaker. The water level in flask B should move a little and then remain fixed. If the water level continues to change, a leak is present. Do not proceed until the leak is fixed. 7. Equalize the pressures inside and outside flask B by raising the beaker until the water level in the beaker and in flask B are the same. While one student holds the beaker to keep the water levels the same, another student closes the pinch clamp, positioned near the stopper on flask B. Pour out all the water in beaker C. Replace the tube in the beaker and open the pinch clamp. A little water will flow out and should be retained in the beaker. 8. Fill a trough half full with water. 9. Check to make sure the pinch clamp is open. Tip flask A carefully so that the FeCl 3 catalyst spills out of the test tube into the hydrogen peroxide solution (avoid getting the solution on the stopper). Place the flask in the water trough. Swirl the flask to mix the reactants. Note and record any changes that you see. As the oxygen is released, water is forced from flask B into beaker C. Feel the bottom of flask A. Is heat released or absorbed in the reaction? Note the level of the water in beaker C. Swirl the flask some more. When the reaction stops, and the water level remains constant, you can assume the decomposition of the hydrogen peroxide is complete.
Laboratory Manual Prepared by Catalyst Education, LLC for the University of California at Santa Barbara, Department of Chemistry. Copyright 2023 by Petra Van Koppen; University of California, Santa Barbara
3.5 The Molar Volume of Gases
Let the solution sit for 15 minutes to allow the temperature of the solutions in flasks A and B to equilibrate to room temperature. 11. Raise beaker C until the water levels in flask B and beaker C are equal, at which point another student closes the pinch clamp on tube BC. Measure the volume of oxygen produced by carefully pouring the water in beaker C into a graduated cylinder. Record the volume of water displaced by the oxygen gas. 12. To measure the temperature, use a digital thermometer. Remove the cap. Press the “on” button and check that the display reads ° C rather than Fahrenheit. To measure the temperature, the ° C display should not blink.
NOTE: When the ° C display blinks, the thermometer is in hold position and will not record the temperature measured or any changes in temperature. If the ° C display blinks, press the hold button to stop the blinking. When the ° C display does not blink, the ther-mometer is set to measure the temperature. 13. To measure the temperature of the gas produced in the reaction, loosen the stopper on flask A and insert a thermometer into the solution. Measure and record the temperature of the solution in flask A. Similarly, measure and record the temperature of the solution in flask B. The average of these two temperatures rep-resents the temperature of the oxygen gas. 14. Disconnect flask A at point A, dry the outside of flask A, weigh and record the flask and its contents. 15. Obtain the barometric pressure from your instructor. You can also obtain the value on the web: weather.com. For the local barometric pressure enter the name of the city or the zip code. 16. All waste is disposed of in the appropriately labeled waste container located in the hood. 17. Return the small test tubes provided for this experiment.
Waste Disposal
Dispose of all waste in the appropriately labeled waste container in the hood.
Laboratory Manual Prepared by Catalyst Education, LLC for the University of California at Santa Barbara, Department of Chemistry. Copyright 2023 by Petra Van Koppen; University of California, Santa Barbara
3.6 Report Sheet:
The Molar Volume of Gases
Name: Section: Date:
Report Table 3.1: Decomposition of H 2O2 Using FeCl 3 as a Catalyst Measurements
Mass of flask A and its contents before reaction (g) Mass of flask A and its contents after reaction (g) Temperature of the gas in flask A (° C) (which equals to the temperature of the solution in flask A) Temperature of the gas in flask B (° C) (which equals to the temperature of the solution in flask B) Average temperature of the gas (° C) and (K) Volume of oxygen gas collected (mL) Barometric pressure (atm) Water vapor pressure at the average temperature of the gas (atm) (see Appendix, interpolate the value) See Appendix Table 3.3 at the end of this experiment. Show your calculation of the water vapor pressure, including interpolation, below Report Table 3.2.
Calculations
Completed calculations must be shown to the instructor before you leave the lab.
Report Table 3.2: Molar Volume of Oxygen Calculations
Mass of oxygen (g) Moles of oxygen Average temperature of gas (absolute, K) Pressure of oxygen, 𝑃 O2 (atm) Volume of O 2 at STP (calculate this from your data) (L) Molar volume of O 2 at STP (L/mol) % error The mass of oxygen gas is very small. Keep 4 decimal places throughout calculations. Compare the molar volume of O 2 (g) at STP that you obtained to the actual value, 22.4 L/mol.
Laboratory Manual Prepared by Catalyst Education, LLC for the University of California at Santa Barbara, Department of Chemistry. Copyright 2023 by Petra Van Koppen; University of California, Santa Barbara
3.1 The Molar Volume of Gases Report Sheet
CALCULATIONS: Show your calculations, including UNITS throughout your calculations, for Average Tem-perature; Water vapor pressure at the average temperature; Mass of Oxygen gas produced; Moles of Oxygen gas, O2 (g); Pressure of O 2 (g); Volume of O 2 (g) at STP; Molar Volume of O 2 (g) at STP (compare your value to the actual value, 22.4 L/mol).
Laboratory Manual Prepared by Catalyst Education, LLC for the University of California at Santa Barbara, Department of Chemistry. Copyright 2023 by Petra Van Koppen; University of California, Santa Barbara
3.2 The Molar Volume of Gases Report Sheet
Review Questions
Explain the function of FeCl 3 in the decomposition of H 2O2. If we would have used 5.0 mL of 3 M FeCl 3
rather than 4.5 mL of 3 M FeCl 3, would it have made any difference in the amount of oxygen gas collected? Explain why or why not. 2. In this experiment, some oxygen gas will dissolve in water. How will this affect the molar volume calculated? Would the molar volume be too high or too low? 3. In step 14, after the reaction has gone to completion, if the outside of flask A was wet when you weighed it, how would this affect your results? Would this cause the measured molar volume of oxygen gas to be higher or lower? Explain why. 4. If 0.65 g of a gas occupies 275 mL at 26 °C and 1.10 atm, calculate the molar mass of the gas. 5. The density of a gas is 0.980 g/L at 25 °C and 365 torr. Calculate the molar mass of the gas.
Laboratory Manual Prepared by Catalyst Education, LLC for the University of California at Santa Barbara, Department of Chemistry. Copyright 2023 by Petra Van Koppen; University of California, Santa Barbara
3.3 The Molar Volume of Gases Report Sheet
The density of dry air is 1.2929 g/L at STP. Calculate the average molar mass of air. Is this a reasonable value? Explain why or why not. 7. In this experiment, after the reaction went to completion, we let the solutions sit to equilibrate the temper-ature in flasks A and B. If you did not do this and the gas temperature was too high (in the reaction flask A) when you measured the volume of water displaced in the reaction, would the molar volume calculated be too high or too low? Explain why. 8. Under what conditions do real gases behave most like ideal gases? Does the air in a room at room temperature (22 °C) and pressure (1 atm) behave ideally? 9. The molar volume for CO 2 and HCl are 22.262 L/mol and 22.244 L/mol, respectively. Why are these molar volumes slightly less than the ideal molar volume, 22.4 L/mol? 10. Hydrogen gas can be produced from the reaction of calcium hydride and water.
CaH 2 (s) + 2 H 2O (l) − −−→ 2 H 2 (g) + Ca (OH )2 (aq )
How many grams of calcium hydride are needed to produce 2.5 L of hydrogen gas, collected over water at
26 °C and 760. torr total pressure?
Laboratory Manual Prepared by Catalyst Education, LLC for the University of California at Santa Barbara, Department of Chemistry. Copyright 2023 by Petra Van Koppen; University of California, Santa Barbara
3.4 The Molar Volume of Gases Report Sheet
A series of measurements are made in order to determine the molar mass of an unknown gas. First, a large flask is evacuated and found to weigh 134.567 g. It is then filled with the gas to a pressure of 735 torr at
31 °C and reweighed; its mass is now 137.456 g. Finally, the flask is filled with water at 31 °C and found to weigh 1067.9 g. (The density of the water at this temperature is 0.997 g/mL.) Assuming that the ideal gas law equation applies, calculate the molar mass of the unknown gas. 12. A study of the effects of certain gases on plant growth requires a synthetic atmosphere composed of 1.5 mol percent CO 2, 18.0 mol percent O 2, and 80.5 mol percent Ar. a. Calculate the partial pressure of O 2 in the mixture if the total pressure of the atmosphere is to be 745 torr. b. If this atmosphere is to be held in a 120 L space at 295 K, how many moles of O 2 are needed? 13. Many compounds used in the fragrance industry are derived from plant extracts. One step in identifying a desired compound is the determination of its molar mass. The volatile organic compound geraniol is a component of oil of roses. The mass density of the vapor at 260 °C and 103 Torr is 0.480 g/L. What is the molar mass of geraniol?
Laboratory Manual Prepared by Catalyst Education, LLC for the University of California at Santa Barbara, Department of Chemistry. Copyright 2023 by Petra Van Koppen; University of California, Santa Barbara
3.5 The Molar Volume of Gases Report Sheet
Lab Report 3
For this experiment, the lab report should summarize the data and results in your own words. Write the discussion section in paragraph form and discuss the following: Explain what was measured to determine the molar volume of O 2 (g). Include the balanced equation for the reaction after you introduce the reaction. Indicate the catalyst that was used and explain its purpose. Explain how you determined the volume of O 2 (g) at STP with the equation you used. Explain how you determined the molar volume of O 2 (g) at STP. From your experimental data, state the molar volume of O 2 gas at STP that you determined (including units); state the accepted value at STP, 22.4 L/mol, and the percent error in the experimental molar volume relative to 22.4 L/mol.
Sources of Error: Review how the experiment was done and look at the data/results to state possible sources of error and how these errors affect the results: 1. State the experimental molar volume of O 2 (g) at STP and indicate if it is higher or lower than the accepted molar volume of O 2 (g), 22.4 L/mol, at STP. 2. State possible sources of error in determining the moles of O 2 (g) and explain if a given error caused your experimental molar volume at STP to be higher or lower than the accepted value at STP. When weighing the flask after the reaction, if the flask was not completely dry on the outside, explain how this would affect the mass, the moles and the molar volume of O 2 (g). 3. State possible sources of error in determining the volume of O 2 (g) and explain if a given error caused your experimental molar volume at STP to be higher or lower than the accepted value at STP. Explain the affect of oxygen dissolved in water on the volume of water displaced and on the experimen-tal molar volume of O 2 (g). Explain how a difference in temperature affects the pressure of oxygen gas, the volume of water dis-placed and the experimental molar volume of O 2 (g). 4. Indicate which source of error is consistent with the experimental molar volume at STP relative to the known value. For the conclusion, restate the objective and purpose of the experiment. Summarize major findings and how you would fix any of the errors that may have occurred. The original data recorded in your laboratory notebook must be scanned and uploaded as part of your lab report.
Preparation for Experiment 4
Before lab you MUST: 1. Read Experiment 4 procedure 2. Watch the two videos posted on LabFlow 3. Complete the PRE-LAB quiz 4. Complete the PRE-LAB Assignment
Laboratory Manual Prepared by Catalyst Education, LLC for the University of California at Santa Barbara, Department of Chemistry. Copyright 2023 by Petra Van Koppen; University of California, Santa Barbara
3.6 The Molar Volume of Gases Report Sheet
Appendix
Table 3.3: Vapor Pressure of Water at Selected Temperatures
Degrees Centigrade Vapor Pressure (torr) Degrees Centigrade Vapor Pressure (torr)
–10 2.149 30 31.824 –5 3.163 35 42.175 0 4.570 40 55.320 5 6.543 45 71.881 10 9.209 50 92.510 15 12.788 55 118.04 16 13.634 60 149.38 17 14.530 65 187.54 18 15.477 70 233.7 19 16.477 75 289.1 20 17.535 80 355.1 21 18.650 85 433.6 22 19.827 90 525.8 23 21.068 95 633.9 24 22.377 100 760.00 25 23.756 125 1740 26 25.209 150 3570 27 26.739 175 6694 28 28.349 200 11659 29 30.043 300 64432
Laboratory Manual Prepared by Catalyst Education, LLC for the University of California at Santa Barbara, Department of Chemistry. Copyright 2023 by Petra Van Koppen; University of California, Santa Barbara
3.7 |
8538 | https://www.ncbi.nlm.nih.gov/books/NBK538287/ | Sideroblastic Anemia - StatPearls - NCBI Bookshelf
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StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
StatPearls [Internet].
Show details
Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
Search term
Sideroblastic Anemia
Damilola Ashorobi; Anahat Kaur; Anil Chhabra.
Author Information and Affiliations
Authors
Damilola Ashorobi 1; Anahat Kaur 2; Anil Chhabra 3.
Affiliations
1 Nassau University Medical Center
2 Charleston Area Medical Center
3 Mary Black Health System
Last Update: December 11, 2024.
Go to:
Continuing Education Activity
Sideroblastic anemia is characterized by impaired iron utilization during erythropoiesis, leading to ring sideroblasts in the bone marrow. These ring sideroblasts are erythroid precursors with iron-laden mitochondria encircling the nucleus. The condition can be hereditary or acquired, with hereditary forms presenting as either microcytic or macrocytic anemia, depending on the specific genetic mutation and its effect on red blood cell maturation and hemoglobin synthesis. Sideroblastic anemia is associated with normal to elevated iron levels, unlike iron deficiency anemia. A notable subtype, refractory anemia with ring sideroblasts (RARS), falls under the myelodysplastic syndrome (MDS) spectrum and is characterized by significant ring sideroblasts and resistance to standard anemia treatments. Although RARS often follows an indolent course with isolated anemia, some cases may progress to more severe forms of MDS or acute myeloid leukemia.
This course provides healthcare professionals with a comprehensive understanding of the pathophysiology, diagnostic criteria, and treatment advancements in sideroblastic anemia and RARS. Participants learn to identify key genetic mutations, analyze bone marrow morphology, and assess iron metabolism’s role in disease progression. Emphasizing a multidisciplinary approach, the course highlights the collaborative efforts of hematologists, nutritionists, primary care clinicians, and mental health professionals in delivering holistic, patient-centered care. This interprofessional collaboration improves diagnostic accuracy, enhances personalized treatment plans, and addresses patients' broader nutritional, psychological, and hematologic needs, ultimately leading to better clinical outcomes and quality of life.
Objectives:
Identify the clinical and laboratory features of sideroblastic anemia and refractory anemia with ring sideroblasts, including ring sideroblasts and abnormal iron metabolism.
Evaluate and differentiate sideroblastic and refractory anemia with ring sideroblasts from other types of anemia, particularly iron deficiency anemia, by assessing iron levels, red blood cell morphology, and other laboratory findings.
Differentiate between sideroblastic anemia, iron deficiency anemia, and other microcytic or normochromic anemias based on iron levels, red blood cell morphology, and genetic markers.
Collaborate with an interprofessional team, including hematologists, genetic counselors, dietitians, and primary care providers, to deliver comprehensive care for patients with sideroblastic anemia.
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Introduction
Sideroblastic anemia is a rare type that results from abnormal utilization of iron during erythropoiesis. There are different forms of sideroblastic anemia, and all forms are defined by the presence of ring sideroblasts in the bone marrow. Ring sideroblasts are erythroid precursors containing deposits of non-heme iron in mitochondria, forming a ring-like distribution around the nucleus.The iron-formed ring covers at least one-third of the nucleus rim (seeImages.Sideroblastic Anemia and Ringed Sideroblasts).
The hemoglobin molecule contains a crucial prosthetic group called heme, integral to its oxygen-carrying function. Structurally, heme consists of a porphyrin ring formed by 4 pyrrole rings connected by methine bridges at the alpha positions, with a central iron atom coordinated within the ring. This iron center enables hemoglobin to bind and transport oxygen efficiently to tissues throughout the body. Other functions of heme apart from the formation of hemoglobin include gas sensing, signal transduction, biological clock, circadian rhythm, and microRNA processing.
Sideroblastic anemia results from abnormal erythropoiesis during heme production.Eighty-five percent of heme is produced in the cytoplasm and mitochondria of the erythroblast cells, while the remainder is produced in hepatocytes. In the Shemin pathway, also known as the heme biosynthesis pathway, 8 enzymes catalyze sequential reactions for heme synthesis. These enzymes include aminolevulinic acid synthase, porphobilinogen synthase, porphobilinogen deaminase, uroporphyrinogen III synthase, uroporphyrinogen decarboxylase, coproporphyrinogen oxidase, protoporphyrinogen oxidase, and ferrochelatase.
There are 2 forms of sideroblastic anemia: hereditary and acquired. Hereditary sideroblastic anemia primarily causes microcytic anemia due to defective heme synthesis, but certain mutations can lead to normocytic or, in rare cases, macrocytic anemia.Unlike iron deficiency anemia, where there is a depletion of iron stores, patients with sideroblastic anemia have normal or high iron levels. Other microcytic anemias include thalassemia and anemia of chronic disease.
Refractory anemia with ring sideroblasts (RARS) is an acquired form of sideroblastic anemia that is classified as a type of myelodysplastic syndrome (MDS), which is characterized by anemia and the presence of 15% or more ring sideroblasts in the marrow. Generally, the patient presents with normochromic, normocytic anemia with overall erythroid hyperplasia as the body attempts to compensate for ineffective erythropoiesis.The hemoglobin level is generally 9 to 12 g/dL, though lower levels may be present. The red blood cells may show dimorphism (with both hypochromic and normochromic populations) and variable degrees of dyserythropoietic and occasional megaloblastic features. Granulopoiesis and megakaryocytopoiesis are normal. Platelet and neutrophil counts can vary but are typically normal or mildly increased.
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Etiology
The hereditary form of sideroblastic anemia is caused by mutations in genes that are involved in heme synthesis, iron-sulfur cluster biogenesis, or mitochondrial metabolism.For heme synthesis, the gene defect is on the X chromosome, leading to mutations in the aminolevulinate synthase (ALAS2), adenosine triphosphate-binding cassette B7 (ABCB7) or glutaredoxin 5 (GRLX5) enzymes. Other congenital causes include mutations in the mitochondrial transporter (SLC25A38), thiamine transporter (SLC19A2), RNA modifying enzyme pseudouridine synthase (PUS1), mitochondrial tyrosyl-tRNA synthase (YARS2), and mitochondrial DNA deletions.
The most common cause of hereditary sideroblastic anemia is the X-linked type caused by a mutation of the gene-forming ALAS2 enzyme.The ALAS2, SLC25A38, GLRX5, HSPA9, ABCB7 genes when mutated cause microcytic anemia while the mitochondrial genes (SLC19A2, PUS1, YARS2, TRNT1) when mutated cause macrocytic anemia.In other words, the X-linked mutations cause microcytic anemia, and mitochondrial DNA deletions cause macrocytic anemia.
Acquired sideroblastic anemias may arise from primary or secondary causes. Primary causes include clonal sideroblastic anemias, bone marrow stem cell disorders categorized under the broader groups of MDS and myeloproliferative neoplasms (MPN).RARS is now classified under myelodysplastic syndrome with ring sideroblasts (MDS-RS), and RARS with thrombocytosis is now classified as a myelodysplastic, myeloproliferative neoplasm with ring sideroblasts and thrombocytosis (MDS/MPN-RS-T).The morphology observed during diagnosis is critical, significantly influencing the disease's progression.
Most cases of RARS evolve in 1 of 2 ways: either they become a type of low-risk MDS with a normal lifespan, or they progress to a higher-grade MDS or acute myeloid leukemia. Most RARS cases fall into the first category, with only 7% to 10% of patients progressing to the more severe forms. Secondary causes of sideroblastic anemia are copper deficiency (or zinc toxicity), drugs (such as isoniazid, chloramphenicol, or linezolid), and excessive alcohol use.Sideroblastic anemia secondary to copper deficiency, certain medications, or excessive alcohol consumption can be completely reversed once the underlying cause is eliminated.
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Epidemiology
Sideroblastic anemia is considered a rare disease.By definition, rare diseases affect fewer than 200,000 people in the United States. Due to the low incidence and prevalence, researchers do not have definite statistical data on its epidemiology. The disorder affects a wide age range, from infants with congenital forms to middle-aged and older individuals.
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Pathophysiology
Understanding the pathophysiology of sideroblastic anemia, heme biosynthesis, and the consequences of defective enzyme/transport genes is important. The first step in heme synthesis involves the transport of glycine into the mitochondria by the SLC25A38 transporter, which combines with succinyl-coenzyme A to form aminolevulinic acid (ALA). This reaction is catalyzed by the enzyme aminolevulinic acid synthase 2 (ALAS2). Once synthesized, ALA is transported from the mitochondria to the cytosol. A defect in ALAS2, which impairs ALA formation, leads to heme deficiency and the development of sideroblastic anemia. Additionally, mutations in the ABCB7 and GLRX5 genes cause heme deficiency through distinct pathways, as outlined below. Other gene mutations, with diverse functions, also contribute to the development of sideroblastic anemia.
Hereditary sideroblastic anemia arises from defects in genes encoding enzymes involved in heme biosynthesis, located both on autosomal chromosomes and within the mitochondria.The diseases caused by these gene mutations are described as non-syndromic and syndromic. Non-syndromic disorders include X-linked sideroblastic anemia, also known as sideroblastic anemia 1 (SIDBA1), SIDBA2, SIDBA3, and SIDBA4. In contrast, syndromic diseases include X-linked sideroblastic anemia with ataxia; Pearson marrow-pancreas syndrome; thiamine-responsive megaloblastic anemia; myopathy, lactic acidosis, and sideroblastic anemia; sideroblastic anemia with immunodeficiency, fevers, and developmental delay; and NDUFB11 deficiency.The list below outlines different types of sideroblastic anemia and their corresponding pathophysiology. All of these types are involved in heme synthesis either directly or indirectly.
Non-syndromic conditions include:
SIDBA1 has been reported to be caused by various mutations, including missense, nonsense mutations, and mutations at promoter or enhancer regions.
All these mutations show an X-linked inheritance pattern. However, ALAS2 has also been reported in females due to acquired unbalanced lyonization causing heterozygous ALAS2 mutation.
A mutation in SLC25A38 causes SIDBA2. Various mutations have been reported, including nonsense, frameshift, and missense mutations.
As mentioned above, SLC25A38 is a mitochondrial glycine transporter that transports glycine into the mitochondria. Glycine is required to form ALA when joined with succinyl-coenzyme A. When mutated, this formation is limited, impeding the formation of heme.
SIDBA3 is a result of a GLRX5 mutation. This type is rare, as only 2 affected families have been reported. This condition is caused by a homozygous mutation that interferes with the splicing of GLRX5 messenger RNA, which reduces the function of GLRX5, which produces an iron-sulfur cluster in the mitochondria.
A mutation of HSPA9, the mitochondrial chaperone protein, causes SIDBA4. As in previous types, various mutations occur, including missense, nonsense, frameshift, and in-frame deletion mutations. The absence of HSPA9 in erythroid cell lines inhibits erythroid cell differentiation.
Syndromic conditions are:
X-linked sideroblastic anemia with ataxia (XLSA/A) is caused by a mutation in the adenosine triphosphate-binding cassette transporter (ABCB7). ABCB7 is involved in iron-sulfur cluster biogenesis, which helps transport iron to the cytosol during heme production. A missense mutation in this gene would lead to X-linked sideroblastic anemia with cerebellar ataxia.
Pearson marrow-pancreas syndrome is due to a deletion of mitochondria DNA. The mechanism causing sideroblastic anemia is unknown, but it was reported that the deletion of the mitochondria DNA causes a defect in the respiratory chain of the mitochondria, causing anemia.This defect leads to refractory sideroblastic anemia and exocrine pancreatic insufficiency.
Thiamine-responsive megaloblastic anemia is caused by a mutation of the SLC19A2 gene, which aids in thiamine transport. Thiamine is needed for the production of succinyl-coenzyme A. A defect in this protein will lead to megaloblastic anemia, diabetes mellitus, and deafness.
Myopathy, lactic acidosis, and sideroblastic anemia are due to a mutation of PUS1, which helps translate the respiratory chain in the mitochondria (eg, a missense mutation). The reason it causes sideroblastic anemia is unknown.
Myopathy, lactic acidosis, and sideroblastic anemia are due to a mutation in the YAR2 gene, which encodes mitochondrial tyrosyl transfer RNA synthase. The mutation of this enzyme will deplete the quantity of this enzyme. The cause of sideroblastic anemia is also unknown with this type.
Sideroblastic anemia with immunodeficiency, fevers, and developmental delay is due to a mutation of the TRNT1 gene, which encodes CCA-adding transfer RNA nucleotidyltransferase. The pathology of sideroblastic anemia is unknown.
NDUFB11 deficiency is due to a deletion of 3 nucleotides in the NDUFB11 gene, leading to phenylalanine deletion in the NDUFB protein, causing normocytic sideroblastic anemia and lactic acidosis.
Clonal hematologic disorders, including MDS/MPN variants, are a primary cause of acquired sideroblastic anemia and are classified as lower-risk MDS. These disorders are the most commonly encountered form of sideroblastic anemia in clinical practice, typically affecting individuals older than 40. The MDS/MPN variants associated explicitly with ring sideroblasts include: MDS with ring sideroblasts and single lineage dysplasia (MDS-RS-SLD), previously known as RARS; MDS with ring sideroblasts and multilineage dysplasia (MDS-RS-MLD), formerly called refractory cytopenia with multilineage dysplasia and ring sideroblasts (RCMD-RS); and MDS/MPN with ring sideroblasts and thrombocytosis (MDS/MPN-RS), previously referred to as refractory anemia with ring sideroblasts and thrombocytosis (RARS-T).
Acquired missense mutations in SF3B1, which encodes a component of the messenger RNA spliceosome, are present in up to 85% of patients with the above MDS/MPN variants.SF3B1 is inserted into the spliceosome by arranging a pre-splicing complex to become a type of U2 small nuclear ribonucleoprotein.The function of this gene is to help join the U2 snRNP to the pre-mRNA, and it also helps in the formation of an intermolecular helix between the 5' end of the U2 and the 3' end of the U6 small nuclear RNAs.The detection of SFB3B1 mutations confirms the diagnosis of an MDS/MPN variant, is associated with a more favorable clinical outcome, and helps rule out X-linked sideroblastic anemia in female patients.
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Histopathology
Siderocytes on peripheral blood smear are considered highly characteristic of sideroblastic anemia. These are hypochromic red blood cells with prominent coarse basophilic granules visible on the Wright-Giemsa stain. The granules consist of residual iron-laden mitochondria and are known as Pappenheimer bodies. Siderocytes are the mature red blood cell equivalent of the ring sideroblasts observed in the bone marrow.
If a bone marrow assay is ordered for worsening anemia or progressively low neutrophil counts, it often reveals a hypocellular picture with dysplastic changes or even erythroid hyperplasia.Prussian blue staining of the bone marrow aspirate smear will show the typical ring sideroblasts when viewed under the microscope (see Image. Ring Sideroblasts). This Prussian blue staining is called the Perls reaction, and there should be a minimum of 5 granules surrounding one-third of the nuclear diameter.Granulopoiesis is often left-shifted, but full maturation still occurs. Bone marrow is often stained for iron and will show numerous iron-laden macrophages, which are not appreciated as well on the smear. Dyserythropoietic features, including a left shift in erythroid precursors, megaloblastoid changes, and binuclearity, are frequently observed in MDS variants but are less common in other types of sideroblastic anemia. Vacuolization of erythroid and myeloid precursors is also commonly seen in copper deficiency, and there may be hyperplasia of lymphoid precursor cells.
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History and Physical
The presentation of patients with sideroblastic anemia includes the common symptoms of anemia, such as fatigue, malaise, shortness of breath, palpitations, and headache. Clinicians must review all potential social exposures, alcohol use, work-related exposures, and previous surgical procedures. A history of gastrointestinal surgery, enteral or parenteral nutrition without copper supplementation, or excessive zinc intake may suggest copper deficiency. Evaluating for any past medical history of cancer, familial hematologic malignancies, and other iatrogenic causes is also essential.
This condition requires a thorough physical examination of the patient, including a head-to-toe examination of mucous membranes, a conjunctivae examination for pallor, a cardiopulmonary examination, an evaluation of fingers for changes in nail beds secondary to anemia, and an evaluation for alternative causes of anemia. Physical examination may reveal conjunctival pallor, pale skin, and tachycardia. Some affected individuals may have bronze-colored skin due to iron overload.In some cases, especially with severe chronic forms or in associated MDS, mild hepatosplenomegaly may be present due to extramedullary hematopoiesis. Patients with syndromic hereditary sideroblastic anemia may present with uncontrolled diabetes and deafness, such as the thiamine-responsive megaloblastic anemia type. Hereditary types will be seen in younger patients with a family history, while acquired sideroblastic anemia typically occurs in older patients with possible myelodysplastic syndrome.
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Evaluation
Sideroblastic anemia is primarily a laboratory diagnosis. A complete blood panel with differential chemistries and an evaluation of vitamin and mineral deficiencies (B12, folate, and copper) are necessary. The mean corpuscular volume (MCV) is a useful parameter for differentiating among types of sideroblastic anemia. The anemia is typically microcytic (low MCV) in most congenital subtypes, and nearly all acquired forms—including MDS, copper deficiency, and most medication-induced cases—tend to be normocytic or macrocytic (normal to high MCV).
Iron studies should be conducted to assess the extent of iron overload, which will help determine the necessity for phlebotomy or iron chelation therapy and ongoing monitoring. In individuals with lab-confirmed iron overload, the severity of iron accumulation in tissues is most accurately evaluated through magnetic resonance imaging or liver biopsy. Ring sideroblasts diagnose sideroblastic anemia in the bone marrow. The red blood cells in the peripheral blood smear that contain these iron inclusions are called siderocytes.
When ring sideroblasts are confirmed using the Perls reaction, the patient's bone marrow should be checked for dysplasia and SF3B1 mutations; if present, the patient most likely has a clonal hematologic disorder such as MDS/MPN variants. If absent, the patient either has congenital or secondary acquired sideroblastic anemia.Genetic testing should also be considered if secondary acquired sideroblastic anemia has been ruled out and the cause of the sideroblastic anemia is unknown.
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Treatment / Management
Managing sideroblastic anemia depends on its severity. Patients with a mild or asymptomatic presentation can follow up in the outpatient clinic. Oral pyridoxine 50 to 100 mg/day has been proven to partially or completely correct anemia for patients diagnosed with X-linked sideroblastic anemia.Luspatercept, a drug that promotes erythroid maturation, could be beneficial in treating congenital sideroblastic anemias characterized by significantly ineffective erythropoiesis.Thiamine is used to treat thiamine-responsive megaloblastic anemia.
For patients who are not responsive to pyridoxine, blood transfusion is indicated if they have severe anemia.Iron chelation must be considered for patients who require chronic transfusion to avoid iron overload. Starting deferoxamine or oral chelators when serum ferritin is more than 1000 ng/L is recommended.Iron overload, if left untreated, can result in unresponsiveness of pyridoxine. Therefore, it is also recommended that patients with normal hemoglobin levels after pyridoxine should have phlebotomy as a treatment for iron overload.Iron overload treatment is critical as studies have shown an improvement in anemia due to low iron levels and better responsiveness to pyridoxine.
Diabetes can develop in patients with syndromic congenital sideroblastic anemia, and tight glycemic control should be encouraged in such patients. Hypoglycemia should also be addressed if blood sugar is not well controlled. For secondary acquired sideroblastic anemia caused by known drugs or toxins, such drugs should be discontinued and avoided. Since it is acquired, the patient’s anemia will improve after the removal of the drug. For isoniazid, the anemia can also be reversed by administering large doses of pyridoxine. Copper should be replenished in a patient with a copper deficiency through dietary intake.
Symptomatic sideroblastic anemia in MDS/MPN variants is treated similarly to low-risk MDS. Therapies previously used include erythropoiesis-stimulating agents, granulocyte colony-stimulating factor (G-CSF), and hypomethylating agents (HMAs) such as azacytidine and decitabine, all of which are potential options to reduce the need for red blood cell transfusions.Though some patients may be successful on the watch-and-wait regimen, HMAs are frequently used after the failure of conservative management with erythropoiesis-stimulating agents and G-CSF. Administration of 75 mg/m 2 azacitidine for 7 consecutive days with a 28-day cycle subcutaneously or intravenously is an option for patients older than 20 who meet the criteria for RARS and whose neutrophil counts are less than 1x10 9/L.
Emerging treatments include luspatercept, an agent that stimulates late-stage erythropoiesis. In the randomized, double-blind MEDALIST trial, luspatercept significantly reduced the need for red blood cell transfusions and improved hematologic outcomes in transfusion-dependent patients with low-risk MDS with ring sideroblasts.Luspatercept can be used as a first-line treatment or after trying an erythropoiesis-stimulating agent.
Another newer treatment option is imetelstat, a telomerase inhibitor that showed promising results in the phase 3 IMerge trial and was approved in 2024 for transfusion-dependent low to intermediate-risk MDS. This can be used in patients who are refractory to erythropoiesis-stimulating agents.For patients with MDS/MPN-RS-T, aspirin therapy is recommended if the patient has an additional JAK2V617F mutation due to an increased risk of thrombosis.Often, the underlying causes of the sideroblastic anemia require therapeutic action before any condition improvement ensues.
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Differential Diagnosis
The differential diagnosis of sideroblastic anemia includes non-clonal and clonal disorders and other more common causes of anemia.
Non-clonal disorders include:
Congenital conditions comprising nonsyndromic and syndromic diseases
Secondary acquired conditions, such as copper deficiency, zinc overload, lead poisoning, and isoniazid use.
Clonal disorders include conditions include:
Myelodysplastic syndromes with ring sideroblasts
Myelodysplastic syndromes with ring sideroblasts, with single lineage dysplasia
Myelodysplastic syndromes with ring sideroblasts, with multilineage dysplasia
Myelodysplastic/myeloproliferative neoplasm with ring sideroblasts and thrombocytosis.
Other more common causes of anemia which should be considered in the differential diagnosis are:
Iron deficiency anemia
Thalassemia
Anemia of chronic disease
Lead poisoning
Blood loss.
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Prognosis
The prognosis for patients with sideroblastic anemia differs depending on the underlying cause. In secondary acquired sideroblastic anemia, the prognosis is generally favorable once the causative drugs or toxins are discontinued. In X-linked sideroblastic anemia, appropriate pyridoxine supplementation and management of iron overload can significantly improve outcomes. However, the prognosis may be poor if these conditions are not adequately addressed.
The prognosis of acquired sideroblastic anemia due to MDS/MPN variants varies depending on the underlying genetic and clinical characteristics. These disorders are typically seen in older adults and often involve mutations in the SF3B1 gene, associated with favorable outcomes. However, the prognosis remains influenced by the presence of other co-occurring mutations and the risk of progression to acute myeloid leukemia. The level of blast counts and response to potential treatments will often determine the overall survival of these patients.
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Complications
Sideroblastic anemia caused by mutations in XLSA, GLRX5, and SLC25A38 has been reported to cause heaptic and systemic iron overload.Iron concentrates in the liver and can eventually lead to fibrosis and cirrhosis, just like in hereditary hemochromatosis. In rare situations, patients may also experience hepatomegaly, cardiomyopathy, and endocrine dysfunction, particularly when iron accumulation becomes excessive over time.
Acquired sideroblastic anemia due to MDS/MPN variants is linked with severe anemia, which can cause fatigue, weakness, and exercise intolerance. Leukopenia can result in frequent infection, and thrombocytopenia can lead to easy bruising and bleeding. Additionally, there is an increased risk of transforming into acute myeloid leukemia,especially in cases with high-risk mutations. Vascular and thrombotic complications are also very prevalent in these patients.
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Deterrence and Patient Education
Deterrence and patient education are critical components in managing patients with sideroblastic anemia. Educating patients and caregivers about the disease, including its triggers and complications, is essential for effective prevention and management. Clinicians should stress the necessity of avoiding drugs that can induce sideroblastic anemia. Patients should be informed about the importance of avoiding substances that can exacerbate the condition, such as alcohol, lead, and certain medications like isoniazid, which can contribute to the development of sideroblastic anemia.
Additionally, those with iron overload must understand the need to limit their intake of iron-rich foods, particularly red meat, to prevent further complications. Regular follow-up with clinicians, adherence to prescribed treatments like pyridoxine and iron chelators, and maintaining a balanced diet rich in essential vitamins and minerals are all vital for improving patient outcomes.More clinical research is needed to better understand these diseases' prognosis and epidemiology. Through comprehensive education and proactive management strategies, patients can better manage their condition and reduce the risk of severe complications associated with sideroblastic anemia.
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Enhancing Healthcare Team Outcomes
Sideroblastic anemia and RARS are rare diseases. When patients are seen in primary care clinics with anemia, most are assumed to have iron deficiency anemia, which is significantly more common. However, clinicians must consider the possibility of less common differentials, such as sideroblastic anemia and RARS, in the appropriate clinical scenario. Primary care physicians, advanced clinicians, and internists are responsible for recognizing these diseases early for proper patient treatment and best outcomes.
Patients who develop RARS may be at risk for subsequent progression to acute myeloid leukemia. Therefore, communication between the pathologist and hematologist/primary care physician is essential. Once a case of RARS is considered in the differential diagnosis, a careful evaluation of the peripheral smear with iron staining performed is in order, as well as ancillary analysis for SF3B1 mutations. Treatment will often be determined by the clinician shortly after the review of the peripheral blood smear.
Patients with sideroblastic anemia and RARS benefit significantly from an interprofessional team, as this collaborative approach ensures comprehensive care.Patients with sideroblastic anemia require the expertise of diverse healthcare professionals, including hematologists, pathologists, primary care clinicians, advanced clinicians, pharmacists, nutritionists, genetic counselors, and mental health specialists. Hematologists and pathologists provide specialized knowledge for diagnosing and treating patients with anemia, while primary care and advanced clinicians coordinate overall patient care and monitor for associated conditions.
Pharmacists play a key role in managing medications and monitoring for adverse events. Nutritionists can address dietary deficiencies that may exacerbate symptoms, and genetic counselors offer insights into hereditary aspects of the disease. Mental health specialists support patients in coping with the emotional and psychological impacts of chronic illness.The collaboration among the interprofessional healthcare team ensures a holistic approach to patient care, addressing not only the direct effects of these diseases but also the broader health and well-being of the patient. This integrated approach improves care management, enhances patient quality of life, and can potentially slow disease progression.
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Figure
Sideroblastic Anemia. There are different forms of sideroblastic anemia, and all forms are defined by the presence of ring sideroblasts in the bone marrow, shown in the image. Contributed by S Bhimji, MD
Figure
Ring Sideroblasts. This image shows abnormal deposition of iron in the mitochondria of red cell precursors, forming a ring around the nucleus. Paulo Henrique Orlandi Mourao,Pubic Domain, via Wikimedia Commons
Figure
Ringed Sideroblasts. This image shows ringed sideroblasts in dark pink. Contributed by S Bhimji, MD
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Disclosure:Damilola Ashorobi declares no relevant financial relationships with ineligible companies.
Disclosure:Anahat Kaur declares no relevant financial relationships with ineligible companies.
Disclosure:Anil Chhabra declares no relevant financial relationships with ineligible companies.
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8539 | https://www.thoughtco.com/the-ph-and-pka-relationship-603643 | pH, pKa, and the Henderson-Hasselbalch Equation
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How to Convert pH to pKa
The Henderson-Hasselbalch Equation
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By
Anne Marie Helmenstine, Ph.D.
Anne Marie Helmenstine, Ph.D.
Chemistry Expert
Ph.D., Biomedical Sciences, University of Tennessee at Knoxville
B.A., Physics and Mathematics, Hastings College
Dr. Helmenstine holds a Ph.D. in biomedical sciences and is a science writer, educator, and consultant. She has taught science courses at the high school, college, and graduate levels.
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Updated on May 25, 2024
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The pH of a solution is a measure of the concentration of hydrogen ions. pKa (acid dissociation constant) and pH are related, but pKa is more specific in that it helps you predict what a molecule will do at a specific pH.
Essentially, pKa tells you what the pH needs to be in order for a chemical species to donate or accept a proton.
The relationship between pH and pKa is described by the Henderson-Hasselbalch equation.1
Henderson-Hasselbalch Equation: pH to pKa
The pKa is the pH value at which a chemical species will accept or donate a proton.
The lower the pKa, the stronger the acid and the greater the ability to donate a proton in aqueous solution.
The Henderson-Hasselbalch equation relates pKa and pH. However, it is only an approximation and should not be used for concentrated solutions, extremely low pH acids, or high pH bases.
What Do pH and pKa Measurements Tell Us?
Once you have pH or pKa values, you know certain things about a solution and how it compares with other solutions:
The lower the pH, the higher the concentration of hydrogen ions [H+].
The lower the pKa, the stronger the acid and the greater its ability to donate protons.
The pH depends on the concentration of the solution. This is important because it means a weak acid could actually have a lower pH than a diluted strong acid. For example, concentrated vinegar (acetic acid, which is a weak acid) could have a lower pH than a dilute solution of hydrochloric acid (a strong acid).
On the other hand, the pKa value is constant for each type of molecule. It is unaffected by concentration.
Even a chemical ordinarily considered a base can have a pKa value because the terms "acids" and "bases" simply refer to whether a species will give up protons (acid) or remove them (base). For example, if you have a base Y with a pKa of 13, it will accept protons and form YH, but when the pH exceeds 13, YH will be deprotonated and become Y. Because Y removes protons at a pH greater than the pH of neutral water (seven), it is considered a base.
Relating pH and pKa With the Henderson-Hasselbalch Equation
If you know either pH or pKa, you can solve for the other value using an approximation called the Henderson-Hasselbalch equation:
pH = pKa+ log ([conjugate base]/[weak acid])
pH = pka+log ([A-]/[HA])
pH is the sum of the pKa value and the log of the concentration of the conjugate base divided by the concentration of the weak acid.
At half the equivalence point:
pH = pKa
It's worth noting that sometimes this equation is written for the K a value rather than pKa, so you should know the relationship:
pKa = -logK a
Assumptions for the Henderson-Hasselbalch Equation
The reason the Henderson-Hasselbalch equation is an approximation is because it takes water chemistry out of the equation. This works when water is the solvent and is present in a very large proportion to the [H+] and acid-conjugate base. You shouldn't try to apply the approximation for concentrated solutions. Use the approximation only when the following conditions are met:2 3
−1<log ([A−]/[HA])<1
The molarity of buffers should be 100 times greater than that of the acid ionization constant K a.
Only use strong acids or strong bases if the pKa values fall between five and nine.
Example pKa and pH Problem
Find [H+] for a solution of 0.225 M NaNO 2 and 1.0 M HNO 2. The K a value (from a table) of HNO 2 is 5.6 x 10-4.
pKa=−log K a=−log(7.4×10−4)=3.14
pH = pka + log ([A-]/[HA])
pH=pKa+log([NO 2-]/[HNO 2])
pH=3.14+log(1/0.225)
pH=3.14+0.648=3.788
[H+]=10−pH=10−3.788=1.6×10−4
View Article Sources
de Levie, Robert. “The Henderson-Hasselbalch Equation: Its History and Limitations.” Journal of Chemical Education, 2003.
Hasselbalch, K. A. "Die Berechnung der Wasserstoffzahl des Blutes aus der freien und gebundenen Kohlensäure desselben, und die Sauerstoffbindung des Blutes als Funktion der Wasserstoffzahl." Biochemische Zeitschrift, 1917, pp.112–144.
Henderson, Lawrence J. "Concerning the relationship between the strength of acids and their capacity to preserve neutrality." American Journal of Physiology-Legacy Content, vol. 21, no. 2, Feb. 1908, pp. 173–179.
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Helmenstine, Anne Marie, Ph.D. "How to Convert pH to pKa." ThoughtCo, May. 25, 2024, thoughtco.com/the-ph-and-pka-relationship-603643.Helmenstine, Anne Marie, Ph.D. (2024, May 25). How to Convert pH to pKa. Retrieved from Helmenstine, Anne Marie, Ph.D. "How to Convert pH to pKa." ThoughtCo. (accessed September 28, 2025).
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8540 | https://www.doubtnut.com/qna/53748666 | The line mx+ny=1 passes through the points (1,2)and(2,1). What is the value of m ?
1
3
12
13
The correct Answer is:D
To find the value of m for the line mx+ny=1 that passes through the points (1,2) and (2,1), we can follow these steps: Step 1: Substitute the first point into the equation We start by substituting the first point (1,2) into the equation mx+ny=1. m(1)+n(2)=1 This simplifies to: m+2n=1(Equation 1) Step 2: Substitute the second point into the equation Next, we substitute the second point (2,1) into the same equation. m(2)+n(1)=1 This simplifies to: 2m+n=1(Equation 2) Step 3: Solve the system of equations Now we have a system of two equations: 1. m+2n=1 (Equation 1) 2. 2m+n=1 (Equation 2) We can solve these equations simultaneously. Let's express n from Equation 1: 2n=1−m⟹n=1−m2(Substituting into Equation 2) Now, substitute n into Equation 2: 2m+(1−m2)=1 Step 4: Clear the fraction To eliminate the fraction, multiply the entire equation by 2: 4m+(1−m)=2 This simplifies to: 4m+1−m=2 Step 5: Combine like terms Combine the terms involving m: 3m+1=2 Step 6: Isolate m Now, isolate m: 3m=2−1 3m=1 m=13 Final Answer Thus, the value of m is: 13 ---
To find the value of m for the line mx+ny=1 that passes through the points (1,2) and (2,1), we can follow these steps:
Step 1: Substitute the first point into the equation
We start by substituting the first point (1,2) into the equation mx+ny=1.
m(1)+n(2)=1
This simplifies to:
m+2n=1(Equation 1)
Step 2: Substitute the second point into the equation
Next, we substitute the second point (2,1) into the same equation.
m(2)+n(1)=1
This simplifies to:
2m+n=1(Equation 2)
Step 3: Solve the system of equations
Now we have a system of two equations:
m+2n=1 (Equation 1)
2m+n=1 (Equation 2)
We can solve these equations simultaneously. Let's express n from Equation 1:
2n=1−m⟹n=1−m2(Substituting into Equation 2)
Now, substitute n into Equation 2:
2m+(1−m2)=1
Step 4: Clear the fraction
To eliminate the fraction, multiply the entire equation by 2:
4m+(1−m)=2
This simplifies to:
4m+1−m=2
Step 5: Combine like terms
Combine the terms involving m:
3m+1=2
Step 6: Isolate m
Now, isolate m:
3m=2−1
3m=1
m=13
Final Answer
Thus, the value of m is:
13
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To check if two or more vectors are Coplanar.
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I have written my own understand for this. Please correct me if I’m right or wrong.
To get two or more vectors to be coplanar , I have noticed that there is a kind of series which forms between the coordinates.
For example : (1,1,1) , (2,2,2) , (3,3,3) .
Similarly , when I made it a little difficult. I still got the same correct result.
If the Q is like : If vector 2 i+2 j−2 k 2 i+2 j−2 k , 5 i+y j+k 5 i+y j+k and −i+2 j+2 k−i+2 j+2 k. Check if the vectors are co planar and then find the value of y? Then , my approach can be a little too time consuming .
I would like to know if you have any different kind of approach to these kind of questions.
Maybe intuitively or using formula of vectors . Anything.
Thank you.
vectors
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edited Aug 8, 2021 at 8:14
S.M.TS.M.T
asked Aug 8, 2021 at 8:01
S.M.TS.M.T
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Hint:
If three vectors are coplanar, cross product of any two is perpendicular to the third one.
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answered Aug 8, 2021 at 8:19
DatBoiDatBoi
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A little nudge in the right direction; -What property does two or more vectors that lie in the same plane have in common?
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answered Aug 8, 2021 at 8:07
ElectromagneticSwindleElectromagneticSwindle
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Parallel to each other. Therefore , a.b = ab always .S.M.T –S.M.T 2021-08-08 08:13:21 +00:00 Commented Aug 8, 2021 at 8:13
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Let four points are given. (When you are given three points, there always exist a plane such that the points are on the plane. I think 'three vectors are coplanar' means that origin and the given three points are on one plane.)
Subtract one vector to every given vectors (translation doesn't harm the coplanarity). Say this set of vectors are S={0,(a 1,a 2,a 3),(b 1,b 2,b 3),(c 1,c 2,c 3)}S={0,(a 1,a 2,a 3),(b 1,b 2,b 3),(c 1,c 2,c 3)}. Then, there exist a plane π:{α x+β y+γ z=0}π:{α x+β y+γ z=0} such that S⊂π S⊂π. The equation of π π is homogeneous (which means the right hand side of its defining equation is 0) since 0∈S 0∈S. This implies ⎛⎝⎜a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3⎞⎠⎟⎛⎝⎜α β γ⎞⎠⎟=0(a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3)(α β γ)=0 has a nontrivial solution. This implies that ⎛⎝⎜a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3⎞⎠⎟(a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3) is singular. So what to check is the singularity of this matrix.
If you were given m+1 m+1 points for m≥3 m≥3, then the matrix which we want to check the singularity would be m m by 3 3 matrix.
To check the singularity, finding determinant is one way, but it is computationally heavy and not extended to the situation of that five or more points were given. (you can find the determinants of 3 3 by 3 3 submatrices. It is somewhat similar to the situation of choosing all 4 4-tuples of points and find the coplanarity) Another way to do this is finding the RREF (row reduced echelon form) of this matrix. Refer to any basic linear algebra textbook.
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answered Aug 8, 2021 at 9:23
dust05dust05
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2 i+2 j−2 k 2 i+2 j−2 k, 5 i+y j+k 5 i+𝑦 j+k and −i+2 j+2 k−i+2 j+2 k are coplanar (with origin, the fourth point) if and only if ⎡⎣⎢2 5−1 2 y 2−2 1 2⎤⎦⎥[2 2−2 5 y 1−1 2 2] is singular. Its determinant = 2 y−46 2 y−46, so y=23 y=23 gives the coplanarity.dust05 –dust05 2021-08-08 09:26:40 +00:00 Commented Aug 8, 2021 at 9:26
Thank you. @dust05 S.M.T –S.M.T 2021-08-08 10:06:29 +00:00 Commented Aug 8, 2021 at 10:06
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Put k stones on a n m grid, to maximize the number of rectangles for which there is one stone in each corner
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Put k stones on a n m grid, each stone should be on an intersection point of the lines of grid.Try to find a way to put the stones, in order to maximize the number of rectangles for which there is one stone in each corner. Output the number.
for example, if k <= 3, the answer is 0, meaning no such rectangle; if k is 4 and n, m >= 2, the answer is 1; more examples:
(n, m, k, answer): (3, 3, 8, 5), (4, 5, 13, 18), (7, 14, 86, 1398)
k is between 0 and n m. n, m are positive integers less than 30000
ps:This is actually a problem in Microsoft-Beauty-of-Programming qualification round(But you may not be able to find it since it is held in China and I translate it to English myself.) pss:I have made some progress. It can be proved that to get the answer, searching through all possible Ferrers diagrams is enough, but the complexity is exponential with regard to k.
EDIT: (by Dukeling)
A visualization of (3, 3, 8, 5), with the rectangles indicated in different colours.
As you noticed, it's actually a (n-1) (m-1) grid, there's another interpretation using a n m grid where the stones are placed inside the cells, but then you'll need to add an additional constraint that rectangles can't be width / height 1.
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1 I don't really understand how you got those answers. Can you explain exactly where you place the stones and why you get that many rectangles? What is "a conjunction of the grid"?IVlad –IVlad 2013-04-12 09:16:17 +00:00 Commented Apr 12, 2013 at 9:16
@IVlad I made an edit which should make things clearer.Bernhard Barker –Bernhard Barker 2013-04-12 10:09:51 +00:00 Commented Apr 12, 2013 at 10:09
And your question is?Klas Lindbäck –Klas Lindbäck 2013-04-12 10:37:38 +00:00 Commented Apr 12, 2013 at 10:37
1 What have you tried? Have you considered the densest case, i.e. computed the number of rectangles for a full mn grid with stones at every crossing? How many rects do you get when you remove one of these stones? Does it matter which one you remove? What would be a good choice to remove a seconds stone? How about a third? Does removing two stones together suggest a different set of stones than when you remove first one and then the other? Can you detect a pattern? Can you describe it? Can you show that it is indeed optimal? This might be better posted at the Math SE, due to MathJax typesetting.MvG –MvG 2013-04-12 11:03:55 +00:00 Commented Apr 12, 2013 at 11:03
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From a programming perspective this suggests a search that takes advantage of the symmetries of the rectangle to reduce search space. What follows is something of an extended hint.
As the OP points out, a naive implementation would check all possible k-subsets of nodes, a size of C(nm,k).
The amount of search space reduction depends on how symmetries are exploited. The square has symmetries of reflection and rotation, so for n=m there's an 8-fold symmetry group. If say n < m, then the lesser amount of symmetry gives a 4-fold group.
A typical approach is to organize the possible k-subsets by a lexicographic ordering, so that a potential configuration is skipped when it's equivalent to one of earlier appearance in that ordering.
But there are additional "wrap-around" symmetries to be exploited. Suppose the last row of the grid is moved to the top (along with any assignment of stones to its nodes). This transformation preserves the count of the 4-stone rectangles (though the exact sizes of those rectangles will differ).
Indeed transposing two rows or two columns preserves the counts of 4-stone rectangles. Once you have that insight, can you see how to parameterize the search space more efficiently?
Added: Even though it's more of a math insight than programming, consider the number of 4-stone rectangles provided by a "full subrectangle", say r x c if rc < k. Consider the incremental number of extra rectangles provided by one more stone; by two more stones.
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edited Apr 12, 2013 at 18:34
answered Apr 12, 2013 at 12:59
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Does communications with spacecraft red shift the further away they are?
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As Voyager 1 and 2 proceed further away do the radio waves red shift (change frequency)?
If so, then don't we have to compensate for that red shift by modulating the frequency so that it arrives at the expected frequency at the destination?
Theoretical Example: if the spacecraft is expecting 5.0 GHz then doesn't Earth have to transmit at 5.2 GHz to allow for the red shift?
Or am I confusing issues?
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edited Apr 20, 2016 at 14:04
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10 Red Shift is an artefact of velocity not distance.Brian Tompsett - 汤莱恩 –Brian Tompsett - 汤莱恩 2016-04-20 13:23:53 +00:00 Commented Apr 20, 2016 at 13:23
4 The question is valid and a bit deeper than meets the eye. Be careful with quick answers unless you know your cosmology!user12102 –user12102 2016-04-20 18:31:09 +00:00 Commented Apr 20, 2016 at 18:31
3 Red-shift as a function of distance applies to stars (Hubble's Law), not to man-made objects. Due to the (more-or-less) even expansion of the universe, more distant stars are moving proportionally faster away from us than closer stars. But none of this applies to man-made objects, since they are way too close, and were not formed from the expanding matter of the early universe.Reversed Engineer –Reversed Engineer 2016-04-20 19:33:01 +00:00 Commented Apr 20, 2016 at 19:33
@DaveBoltman Well, you are looking at millihertz of Doppler shift due to expansion of the universe for the distances and frequencies involved in the OP's question. Valid, but totally overshadowed by the simple fact that the probes are still carrying a great deal of velocity relative to Earth after being cut loose from the boosters and their gravity slingshot maneuvers...user –user 2016-04-20 19:47:33 +00:00 Commented Apr 20, 2016 at 19:47
1 @DaveBoltman "doesn't apply because it's too close" or even "...way too close" is not usually how these things go. It applies. It's small. BOTH are true.user12102 –user12102 2016-04-20 20:08:36 +00:00 Commented Apr 20, 2016 at 20:08
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No, at the distances we have sent any spacecraft or probe to, there is no need to compensate for redshift (the Doppler effect) because a spacecraft is far away. Doppler shift is an effect of relative velocity, not distance.
The Doppler effect is defined such that the frequency f f as observed by a receiver moving at velocity v r v r relative to a source moving at velocity v s v s and where the source is transmitting at frequency f 0 f 0 is
f=(c+v r c+v s)f 0 f=(c+v r c+v s)f 0
In the case of radio transmissions, c c is the speed of light. v r v r and v s v s are measured in some inertial (that's basically a fancy word for "non-moving") reference frame.
For a stationary sender or receiver, simply set v s=0 v s=0 or v r=0 v r=0 respectively. From this it is obvious how, when both sender and receiver are stationary (in an inertial reference frame), the transmitted frequency equals the received frequency.
As you can see, for a given type of transmission (such as a radio transmission), only the relative velocities of the sender and receiver and the transmission frequency affect the received frequency.
This effect is caused by the fact that when the sender is moving toward the receiver, each wave front is closer to its neighboring wave fronts than merely its frequency would warrant. Likewise, when the two are moving away from each other, each wave front is farther away from its neighbors. Colloquially, the transmission is being "pushed together" or "dragged apart" by the relative movement of the sender and receiver. This is observed as a change in frequency at the receiver.
The distance between the two do not play a role in this effect, and thus we do not need to compensate for Doppler shift merely because an object is far away. We do however have to compensate for Doppler shift because the object is moving relative to us. Of course, because spacecraft generally move at relatively high speeds relative to an Earth ground station, in spacecraft communications we often do need to pay attention to Doppler shift, regardless of whether they are close to or far from Earth.
For example, if the spacecraft is moving away from the Sun at 15 km/s (not unreasonable at all for interplanetary missions) and Earth happens to be moving in the opposite direction at 30 km/s (which it does once a year), and the spacecraft is transmitting on 5 GHz, the frequency as received on Earth will be
f=(300 000 000−15 000 300 000 000+30 000)×5 000 000 000≈4 999 250 075 f=(300 000 000−15 000 300 000 000+30 000)×5 000 000 000≈4 999 250 075
or approximately 4999.25 MHz instead of 5000.00 MHz. Conversely, if we want the spacecraft to receive our transmission on 5000.00 MHz, then we need to transmit at approximately 5000.75 MHz. If we didn't compensate for Doppler shift, we would need a receiver bandpass filter bandwidth of nearly a megahertz; enough to cram in two or three broadcast FM stations.
It's also worth keeping in mind that this is an issue not just for deep space missions. The math is a bit more involved because of the trigonometry to calculate the absolute value of the velocity and if you want an absolute value also compensating for the planet's rotation, but the exact same principles apply when a low orbit spacecraft is communicating with a ground station or for that matter with a spacecraft in a different orbit.
If we were to send spacecraft to other galaxies or even far-away stars, then Doppler shift due to the expansion of the universe might start to actually matter. However, for as long as it can be argued whether we have even exited our own solar system, the relative velocities caused by expansion of the universe are of magnitudes that are completely irrelevant for practical radio communications. The expansion of the universe is given as Hubble's constantH 0 H 0, and the most recent estimate for its value is 67 800±770 67 800±770 m/s per megaparsec. The Voyagers are about 134 and 110 AU away, respectively, which works out to 6.497×10−10 6.497×10−10 and 5.333×10−10 5.333×10−10 megaparsec. If my math is right, taking Voyager 1 because it is farther away, this means an observed expansion rate of about 0.00004405 m/s (44 µm/s) corresponding to a received frequency of
(300 000 000 300 000 000+0.00004405)×5 000 000 000=4 999 999 999.999265833(300 000 000 300 000 000+0.00004405)×5 000 000 000=4 999 999 999.999265833
or a Doppler shift of about 0.001 Hz when the transmitted frequency is exactly 5 GHz. In other words, valid in theory but completely negligible in practice.
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edited Apr 20, 2016 at 19:52
answered Apr 20, 2016 at 15:30
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I realise that theoretically we need to compensate, but is it ever actually done in reality? Are today's spacecraft moving fast enough (still only a fraction of c c, the driving part of that equation) that it's a real issue?James Thorpe –James Thorpe 2016-04-20 15:36:13 +00:00 Commented Apr 20, 2016 at 15:36
2 What about doppler shift caused by the expansion of space? Does that pose a significant enough effect to require adjusting?gandalf3 –gandalf3 2016-04-20 17:54:03 +00:00 Commented Apr 20, 2016 at 17:54
1 @gandalf3 has a point - there is a 'red shift' due to the fact that every point is moving away from every other point (or at least so say smart people). The answer may be "it's really small and so no you don't need to compensate for it explicitly" but the answer is probably NOT "no, because it doesn't happen." Somebody just go look up the Hubble constant for us!user12102 –user12102 2016-04-20 18:37:15 +00:00 Commented Apr 20, 2016 at 18:37
2 i did. Question is about red shift. You mix the term together with Doppler shift and create confusion. They are two totally different things. Why don't you remove everything about Doppler shift. The question was worded well.user12102 –user12102 2016-04-20 19:56:51 +00:00 Commented Apr 20, 2016 at 19:56
1 @uhoh Redshift is the Doppler effect applied to light waves. Perhaps what you're trying to express is the difference between redshift due to velocity relative to the Earth, and redshift due to the expansion of the Universe? The original question is vague about which they're asking about, perhaps the OP doesn't know either. This answer properly covered both.Schwern –Schwern 2016-04-20 22:24:56 +00:00 Commented Apr 20, 2016 at 22:24
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Contrary to the other answers, distance from the earth does does cause/influence red-shift (albeit indirectly). The universe is expanding away from us, and the further away a point in space is, the faster the expansion is at that point. The expansion rate is rather regular and predictable at macroscopic scales, so scientists actually use red-shifts of hydrogen emission spectra for distant stars to gauge how far stars and galaxies are from us.
If you had a spacecraft very, very far away, the universe's relative expansion to us would cause signals between us and the craft to be redshifted.
The expansion rate of our universe is usually given as Hubble's constant -- H 0=67.15(k m/s)/M p c H 0=67.15(k m/s)/M p c. We can calculate the velocity of the universe's expansion at a given distance as H 0 H 0 the distance away.
Plugging in a distance like the distance to the closest star in wolfram alpha, we get that between us and Alpha Centauri, the universe is expanding at a rate of 0.2 miles per hour, or 0.3 kilometers per hour.
This is probably much lower than the actual relative velocity between us and Alpha Centurai just due to the fact that we're moving in different directions! So redshift will not be a significant factor if we ever decide to communicate to Alpha Centauri -- our actual relative velocities due to our relative motion will be more significant.
At intergalactic differences things might start getting a bit more noticeable. Between us and andromeda, the universe is expanding at a rate of about 53 km/s, about 120 million miles per hour, or about one ten thousandths of the speed of light. This is still not really enough to be terribly significant, I believe, but it does give a good scale as to how distance necessarily affects relative velocity and, indirectly, redshift.
EDIT A comment was made that reminded me that while expansion is homogeneous at the macroscopic level, the metric expansion of space might behave as homogeneously at sub-lightyear levels. I'm not familiar enough with this topic to give a definitive answer as to what models could be useful here.
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edited Apr 20, 2016 at 20:39
answered Apr 20, 2016 at 18:16
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Kudos for math! ...and quantitative answering!!!user12102 –user12102 2016-04-20 18:40:46 +00:00 Commented Apr 20, 2016 at 18:40
check my math - I think that's 1 meter/sec at 4.4E+14 kilometers. At Neptune (4.5 billion km) that's an apparent (real?) velocity of 1.0E-05 meters/sec. For a 2GHz signal from earth received and rebroadcast (for doppler sounding) that's a shift of 0.001 Hz. Small, but real!, (real small).user12102 –user12102 2016-04-20 18:54:11 +00:00 Commented Apr 20, 2016 at 18:54
@uhoh Looks like your math is correct; see my answer.user –user 2016-04-20 19:44:24 +00:00 Commented Apr 20, 2016 at 19:44
4 From Wikipedia: "However, the (expansion) model is valid only on large scales (roughly the scale of galaxy clusters and above). At smaller scales matter has become bound together under the influence of gravitational attraction and such things do not expand at the metric expansion rate as the universe ages. " IMO that means you wouldn't get a redshift from a probe at Alpha Centauri.Hobbes –Hobbes 2016-04-20 20:03:56 +00:00 Commented Apr 20, 2016 at 20:03
1 @Hobbes that's about "the expansion model". It says that in gravitationally bound clusters of "star stuff" and other mass, their motion does not track the large scale expansion - roughly speaking because of their mutual gravitational interaction. It doesn't mean necessarily that space isn't expanding, it does mean (among other things) that if you want to measure the Hubble constant by measuring motion of "stuff", better choose points far enough apart that they are not appreciably interacting. Copy/pasting paragraphs about cosmology is not a good idea.user12102 –user12102 2016-04-20 20:17:15 +00:00 Commented Apr 20, 2016 at 20:17
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Let me add to @Michael's answer by saying that this is also very much a factor when communicating with earth-orbiting satellites. For instance, the ISS signal bounces off a geostationary communications satellite before it (the signal) gets to the ground. If I recall correctly, these two spacecraft can differ in velocity up to 7 km/s. Not only can this cause either red or blue shift, but the Doppler shift varies dynamically during a single communication event. So Doppler handling is an important feature in the modems used to process these communications.
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answered Apr 20, 2016 at 16:32
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3 There is an interesting story about how the Huygens mission was doomed, because they had forgot to think about Doppler, as they found out during the 7 year voyage to Saturn! They saved the mission at the expense of some rocket fuel by optimizing the trajectory for minimum Doppler effect.user2394284 –user2394284 2016-04-20 18:40:59 +00:00 Commented Apr 20, 2016 at 18:40
@user2394284 wow, AWESOME read!Magic Octopus Urn –Magic Octopus Urn 2018-09-22 00:19:02 +00:00 Commented Sep 22, 2018 at 0:19
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Yes and no. You have to take into account the time dilation (as in GPS clocks). Maybe the comm equipment does allow for some minor frequency change and it just accepts it still working. However as the gravity does not change in linear fashion, I think the biggest changes will be in LEO/GO. You have to calculate the time speed difference between "there" and "here". For LEO or similar (GPS orbits) it's about ~38us per day. Diffrent time speed is visible in microwave radiation as redshift/blueshift. Not "speed" of the wave.
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answered Sep 21, 2018 at 21:45
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I think I see where you are going with the answer but see events like thespacereview.com/article/306/1 which make some of your assertions questionable. If going down this line some maths on how time dilation would impact(or not) something like the equipment on the Parker solar probe would be interesting noting that it is not just clock time but the frequency of the radio equipment that will be impacted.GremlinWranger –GremlinWranger 2018-09-22 00:04:34 +00:00 Commented Sep 22, 2018 at 0:04
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8544 | https://libguides.mnhs.org/directories/telephone | Telephone - Directories - LibGuides at Minnesota Historical Society Library
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Telephone directories, although not originally created for this purpose, can assist in locating where a person was living in a particular time and place. Telephone directories are published by telephone companies and includealphabetical listings of residential and business phone numbers and street addresses in the white pages, and a listing of businesses arranged by subject in the yellow pages.
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8545 | https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:poly-factor/x2ec2f6f830c9fb89:factor-w-structure/v/id-quad-patterns | Identifying quadratic patterns (video) | Khan Academy
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Algebra 2
Course: Algebra 2>Unit 3
Lesson 5: Factoring using structure
Identifying quadratic patterns
Identify quadratic patterns
Factorization with substitution
Factorization with substitution
Factoring using the perfect square pattern
Factoring using the difference of squares pattern
Factor polynomials using structure
Math>
Algebra 2>
Polynomial factorization>
Factoring using structure
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Identifying quadratic patterns
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Factoring using structure involves recognizing patterns in expressions, like the difference of squares or perfect square trinomials, and using these patterns to break down complex expressions into simpler ones. As an example, you’ll determine whether the polynomial 9x⁸+6x⁴y+y² can be factored using the perfect square pattern or the difference of squares pattern (or neither).
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soratotokuo 5 years ago Posted 5 years ago. Direct link to soratotokuo's post “On answers, A and B would...” more On answers, A and B wouldn't (U-V)^2 be equal to (U+V)(U-V)?
and why is "(U+V)^2 or (U-V)^2" aren't they different?
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Bradley Reynolds 5 years ago Posted 5 years ago. Direct link to Bradley Reynolds's post “(U-V)^2 is not equal to (...” more (U-V)^2 is not equal to (U+V)(U-V). (U+V)(U-V) evaluates to U^2 - V^2, while (U-V)^2 evaluates to U^2 - 2UV + V^2.
For the second part of your question, yes (U+V)^2 and (U-V)^2 are different things but the question is simply looking for similar patterns so they group the two together.
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Andy 5 years ago Posted 5 years ago. Direct link to Andy's post “For the Identifying Quad...” more For the Identifying Quadratic Patterns, I watched the video, did the practice, I just can't get to proficient. I am massively confused with the topic. Can anyone help me?
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happypelican 5 years ago Posted 5 years ago. Direct link to happypelican's post “These patterns are common...” more These patterns are common ways to factor quadratics. (U+V)^2 or (U-V)^2 are the factorizations of perfect square trinomials. You use them anytime the expression is in the pattern U^2+2UV+V^2 or U^2-2UV+V^2.
For example: x^2+2x+1 uses the (U+V)^2 pattern because it factors into (x+1)^2, where U=x and V=1.
(U+V)(U-V) is always used when the expression is in the pattern U^2-V^2 because that's what it expands into.
Ex: 9x^2-16y^2 uses the (U+V)(U-V) pattern since it factors into (3x+4y)(3x-4y), where U=3x and V=4y.
Hope this clears things up a little bit! and just as a note, a and b are often used instead of U and V.
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dustin.carter a year ago Posted a year ago. Direct link to dustin.carter's post “if i had two dollars and ...” more if i had two dollars and i used one how much would i have left
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Logar a year ago Posted a year ago. Direct link to Logar's post “more than my life savings” more more than my life savings
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Michelle Banks 2 years ago Posted 2 years ago. Direct link to Michelle Banks 's post “i jus aced the exam witho...” more i jus aced the exam without doing any work
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John Buz 2 years ago Posted 2 years ago. Direct link to John Buz's post “What are some real-life a...” more What are some real-life applications for the currently useless quadratic patterns
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Tianyue Ma 2 years ago Posted 2 years ago. Direct link to Tianyue Ma's post “In case someone walks up ...” more In case someone walks up to you on the street.
1 comment Comment on Tianyue Ma's post “In case someone walks up ...”
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louisaandgreta 2 years ago Posted 2 years ago. Direct link to louisaandgreta's post “From 5:00, I rewrote the ...” more From 5:00 , I rewrote the expression in the standard form:
-10x^3 +49
And assigned U^2 = -10x^3
And took the square root to find U. I found this:
U= √(10x) ix
Is this correct?
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Charlotte 2 years ago Posted 2 years ago. Direct link to Charlotte's post “I suppose it is correct s...” more I suppose it is correct so great and a good use of knowledge! But to make it complete, for this question it says that U and V are either constant integers or single-variable expressions, so i and radical sign better not to be involved. In addition, for (U+V)(U-V)formula, you'd better remain U^2-V^2, and this will help you figure it out. That's my opinion with it, so reference only :)
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Addie Jones 7 months ago Posted 7 months ago. Direct link to Addie Jones's post “I get how to recognize p...” more I get how to recognize patterns but IDK how to use them..... is this covered in a later lesson?
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Quantum Cat 7 months ago Posted 7 months ago. Direct link to Quantum Cat's post “Yes, Perfect Squares: htt...” more Yes,
Perfect Squares:
Difference of Squares:
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SeaFoam 3 months ago Posted 3 months ago. Direct link to SeaFoam's post “If you listen closely you...” more If you listen closely you can hear Sal's phone binging with a notification...I guess that shows that he's a real person, not just some artificial intelligence teaching me things for my whole entire life....
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dhloreth 2 months ago Posted 2 months ago. Direct link to dhloreth's post “Maybe that’s what he want...” more Maybe that’s what he wants you to think 🤔
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altonmarshall1 a year ago Posted a year ago. Direct link to altonmarshall1's post “How do you check answers ...” more How do you check answers like this after factoring?
2x(x+3)(x-2)(x+2)
I know how to get to a conclusion like this, and get the right answer but I like to be able to check after the fact to be entirely sure it is correct. It's difficult because you cannot FOIL three expressions next to each other properly, and just multiplying the outlying 2x with everything doesn't give the original problem either.
I guess I'm kind of having the issue of knowing it is right but not really knowing why since I can't replicate it by retracing the steps in reverse order.
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TheReal3A a year ago Posted a year ago. Direct link to TheReal3A's post “I would focus on pairs of...” more I would focus on pairs of 2 groups at a time. FOIL is the same as: Multiply every term in the first group by every term in the second group, then add them all up (comes from distributive property, always works).
2x(x+3)(x-2)(x+2)
(x-2)(x+2) = x^2 - 2x + 2x - 4 = x^2 - 4 (can also use difference of 2 squares formula)
=> 2x(x+3)(x^2 - 4)
(x+3)(x^2 - 4) = x^3 + 3x^2 - 4x - 12
=> 2x(x^3 + 3x^2 - 4x - 12)
2x^4 + 6x^3 - 8x^2 - 24x
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deer hunter 5 5 years ago Posted 5 years ago. Direct link to deer hunter 5's post “At 1:40, Sal mentions the...” more At 1:40 , Sal mentions the whole 2UV thing. He includes x and y in 2UV. In the next example, he says at 2:45 that because there is a y, (U+V)^2 or (U-V)^2 is incorrect. What am I to do?
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A/V 5 years ago Posted 5 years ago. Direct link to A/V's post “Realizing that in the sec...” more Realizing that in the second example that both squared binomial rules do not apply, you should put that you cannot use either of such formulas.
If you also are unsure on why you cannot use the formula,
Note that as a perfect squared binomial, V=5.
If V=5, then the product should be 4x^6+10x^3+25.
However! You will notice that there is an extra y in the 2UV term. Since V=5 and not = 5y, then you will realize that the polynomial in the problem is exempted from those formulas.
Hopefully that helps !
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Video transcript
[Instructor] We're told that we wanna factor the following expression and they ask us which pattern can we use to factor the expression? And U and V are either constant integers or single-variable expressions. So we'll do this one together and then we'll have a few more examples where I'll encourage you to pause the video. So when they're talking about patterns they're really saying, "Hey, can we say "that some of these can generally form "a pattern that matches what we have here "and then we can use that pattern to factor "it into one of these forms?" What do I mean by that? Well, let's just imagine something like U plus V squared. We've squared binomials in the past. This is going to be equal to U squared plus two times the product of these terms so two U V and then plus V squared. Now, when you look at this polynomial right over here, it actually has this form if you look at it carefully. How can it have this form? Well, if we view U squared as nine X to the eighth then that means that U is, and let me write it as a capital U, U is equal to three X to the fourth. 'Cause notice if you square this you're gonna get nine X to the eighth. So this right over here is U squared. And if we said that V squared is equal to Y squared, so if this is capital V squared then that means that V is equal to Y. And then this would have to be two times U V. Is it? Well, see if I multiply U times V I get three X to the fourth Y and then two times that is indeed six X to the fourth Y. So this right over here is two U V. So notice this polynomial, this higher-degree polynomial can be expressed in this pattern, which means it can be factored this way. So when they say which pattern can we use to factor this expression, well, I would use a pattern for U plus V squared. So I would go with that choice right over there. Let's do a few more examples. So here once again we're told the same thing. We're given a different expression and they're asking us what pattern can we use to factor the expression? So I have these three terms here. It looks like maybe I could use, I can see a perfect square here. Let's see if that works. If this is U squared, if this is U squared then that means that U is going to be equal to two X to the third power. And if this is V squared then that means that V is equal to five. Now is this equal to two times U V? Well, let's see, two times U V would be equal to, well, you're not gonna have any Y in it so this is not going to be two U V. So this actually is not fitting the perfect square pattern. So we could rule this out. And both of these are perfect squares of some form. One just has a, I guess you'd say adding V. The other one is subtracting V. This right over here, if I were to multiply this out, this is going to be equal to, this is a difference of squares and we've seen this before. This is U squared minus V squared. So you wouldn't have a three-term polynomial like that. So we could rule that one out. So I would pick that we can't use any of the patterns. Let's do yet another example. And I encourage you, pause the video and see if you can work this one out on your own. So the same idea, they wanna factor the following expression. And this one essentially has two terms. We have a term here and we have a term here. They both look like they are the square of something and we have a difference of squares. So this is making me feel pretty good about this pattern but let's see if that works out. Remember, U plus V times U minus V is equal to U squared minus V squared. So if this is equal to U squared then that means that capital U is equal to six X squared. That works. And if this is equal to V squared, well, that means that V is equal to Y plus three. So this is fitting this pattern right over here. And they're just asking us to say what pattern can we use to factor the expression. They're not asking us to actually factor it. So we'll just pick this choice. But once you identify the pattern it's actually pretty straightforward to factor it because if you say this is just going to factor into U plus V times U minus V, well, U plus V is going to be six X squared plus V plus Y plus three times U minus V. U is six X squared. Minus V is minus, we could write minus Y plus three or we could distribute the negative sign. But either way this might make it a little bit clearer what we just did. We used a pattern to factor this higher-degree polynomial which is essentially just a difference of squares. Let's do one last example. So here once again we wanna factor an expression. Which pattern can we use? Pause the video. All right, so we have two terms here. So it looks like it might be a difference of squares if we set U is equal to seven and then this would be U U squared. But then what can we square to get 10 X to the third power? Remember, we wanna have integer exponents here. And the square root of 10 X to the third power, if I were to take the square root of 10 X to the third power it would be something a little bit involved like the square root of 10 times X times the square root of X to the third power and I'm not going to get an integer exponent here. So it doesn't look like I can express this as V squared. So I would go with that we can't use any of the patterns. And we're done.
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8546 | https://www.degruyterbrill.com/document/doi/10.1351/pac199870061157/html?srsltid=AfmBOooMo0Q03P3IntJqdcKciCFgw2iKsgYJkOn5yPo8HGXZkK6Fd-7z | Pure &App/. Chern., Vol. 70, No. 6, pp. 1157-1162, 1998. Printed in Great Britain.
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Water-stabilized plasma generators
Milan Hrabovsky
Institute of Plasma Physics, Academy of Sciences of CR, I82 21 Prague 8, Czech Republic Abstract: Plasma torches with liquid-stabilized arcs provide an alternative to commonly used sources of thermal plasmas based on gas-stabilized arcs or RF discharges. Relatively long arc columns can be stabilized by liquid walls under very low mass flow rates through the torch chamber. This leads to high values of plasma temperature and enthalpy. In water-stabilized torches, oxygen-hydrogen plasma jet is produced with extremely high plasma enthalpy and flow velocity. Plasma torches with water-stabilized arc provide special performance characteristics in some plasma processing applications like plasma spraying or waste treatment. Physical processes which determine properties of generated plasma are discussed and basic characteristics of water plasma torch are presented in the paper.
INTRODUCTION
Thermal plasmas are commonly generated in inductively coupled discharges or in electric arcs which are stabilized by flowing gas. Typically, inductively coupled plasma torches are characterized by plasma temperatures from 6 000 K to 10 000 K. Averaged temperatures in torches with non-transferred electric arcs with gas stabilization are somewhat higher, usually in the range from 8 000 K to 14 000 K. Average plasma enthalpies, determined as ratio of the usefbl power of plasma generator to the flow rate of plasma forming gas, vary from 1 to 100 MJ/kg. Further increase of plasma temperatures and enthalpies is limited by the fact that flowing gas protects the arc chamber walls from thermal overloading and thus a minimum possible gas flow rate exists for given arc power. Higher thermal loading is possible if the walls are created by liquid and an arc is stabilized by wall evaporation. Thus, liquid-stabilized arcs can be utilized as sources of thermal plasmas with high temperatures and enthalpies. The electric arc with the stabilization of arc column by water vortex was first described more than seventy years ago by Gerdien and Lotz (ref. 1, 2). Basic experimental investigations of the water stabilized arcs were performed in the fifties. Maecker et al. (ref. 3, 4) measured basic electric characteristics of the arc and studied effect of the length and diameter of the stabilizing channel. Several investigators measured arc plasma temperatures using methods of emission spectroscopy (ref. 5, 6, 7). All authors reported very high plasma temperatures in the arc column with maximum about 50 000 K in the centerline position. The principle of arc stabilization by water vortex was utilized in the plasma torch designed for plasma spraying and cutting (ref. 8). Despite of the potential of achieving extreme performance characteristics in plasma processing applications the development of liquid-stabilized torches has been limited, especially due to their more complex structure and due to lack of understanding of physical processes in the arc which is necessary for improvement of the torch design. This paper presents results of investigation of water-stabilized plasma torches which has been established several years ago with the aim of better understanding of arc processes, determination of characteristics of generated plasma jet and further development of plasma torch design. Differences of water and gas stabilized generators are discussed and experimental characteristics of arc and of the generated plasma jet are presented.
1157 1158 M. HRAEOVSK?
ARC COLUMN STABILIZED BY WATER VORTEX
Principle of stabilization of arc bv liauid The schematic picture of a water-stabilized arc is shown in Fig. 1. An arc is ignited in the center of water Yortex which is created in an arc chamber with tangential injecliun of water. Water flows over the segmenl which determines inner diamcter of the vortex into the exhaust slot. Energy dissipatcd in the conducting arc core by Joule heating is transportcd to the inner surface of water vortex by radiation, heat conduction and turbulent transfer. Evaporation of water and heating and ionization of vapour are principal mechanisms that produce arc plasma. Evaporation rate m is determined by a fraction or total power reaching the water surface. The other sheath between the arc column and water surface where it causes heating and ionization of vapour. Power loss Q is given by a part of energy which is transferred into thc water volume. tangential water water inlet outlet nozzle part of transcened energy is absorbed in a vapour
i w
cathod
,.
Fig. ISchematics of water-stabilized
arc Effect of proDerIies of Dlasma gas and DrinciDle of arc stabiliidtion on arc characteristics Effect ofmaterial properties of plasma medium and torch chamber dimensions on arc characteristics can be analyzed on the basis of integral energy balance equation of cylindrical arc column (ref. 9) where p is plasma density. v, axial velocity, h enthalpy, CT electric conductivity, k thermal conductivity, 7
temperature, &, net emission coefficient reprcsenting power loss due to'radiation and E electric field intensity. A is cross section of the arc chamber and R its radius. Quantities averaged over the cross section
A are defined by the equation x = 7j2mXdr . Simple equation can be obtained if derivatives in
I R
d o- ~~
B(pv,hA) pv,hA s
equation (1) are approximated as ~~ - and(k$) =[$) =-- where heat
R : r=K r=R Bz L
i-
flux potential S is defined as S = k dT and L is arc length. Enthalpy h(R) of vapor flowing from the wall into the arc chamber corresponds to the boiling temperature of water TB and can be put equal to zero
h(R)=h(Tn)=O. Following two equations for electric field intensity and arc current can be then derived
TO
( 3 )
I = R G / + 2 s S + 4 r 2 R 2-&,,
L
RK
00
where arc cnrrentl = E I2modr= zR2GE and total mass flow rate G = !2mpv,dr = zR2x
01998 IUPAC. Pure&AppliedChernistw70.1157-1162 Water-stabilized plasma generators 1159
Plasma medium
The same equations can be written for arcs in plasma torches with gas stabilization, where m=O. On the right hand sides of the equations (2) and (3) are physical quantities dependent on temperature, geometrical design parameters L , R and total mass flow rate G. Thus, relations between electric characteristics of the arc I, E and averaged plasma temperature can be evaluated from the equations for given values of the ratio
arc current arc povier mass flow rate G/L mean enthalpy m.temperature
[A] [kW] [&I [kg/s.m] [MJ/kg] [K]
GIL and for given arc chamber radius R.
16000 1II
I
water water r
212000-b
2 8000-
i
E0L-
o 4000 -.-
5
300 54 0.20 0.004 157 13 750 600 133 0.33 0.006 272 16 200
2
GIL
= 0.1 k g k m 8000 12000 16000 20000 plasma temperature [Kl
Fig. 2Electric field intensity in the arc column with radius R=3
mm for different ratios of gas flow rate G to the arc length L for arc in water (full line), nitrogen (dash-and-dot line) and argon (dash line).
-ml i?i
2~ 2Et8 P
m
5C
al
-!IE+%
m
P
-
OE+O 0200 400 600 800 1000
arc current [A]
Fig. 3Dependence
of mean plasma enthalpy on the arc current for different ratios of gas flow rate G to the arc length L for arc in water (full line), nitrogen (dash-and-dot line) and argon (dash line), R = 3 mm. Figs. 2 and 3 show curves calculated from (2) and (3) for three plasma mediums - argon. nitrogen and water. 1,TE values of fl), h(T) and S(T) were determined using computer :ode ADEP (ref. IO) values of net emission coefficient E,, were taken from (ref. 11, 12). It can be seen th.?t stabilization of arc by water leads t r high electric field intensities and thus to high arc powers, and a:.io to high plasma .nthalpies. However. principle difference of water-stabilized and g-s-si ibilized arcs ‘ollows from the eftsct of ratio
G/L on arc characteristics. Due to possibility of higher theiinal loading of the wall in water-stabilized plasma torches an arc cd:i be stabilized with very low values of ratio GIL. For water torches, values G/L of order 0.001 kg/m.s are typical while for gas-stabilized torches minimum possible values G/L are more than one order higher. Thus, very high plasma enthalpy and temperature can be achieved in water-stabilized torches. This is illustrated in Table 1, where typical parameters for several plasma spraying torches are presented. Values of mean plasma enthalpy and temperature for gas-stabilized torches were estimated assuming 60% torch efficiency and using thermodynamic properties of plasma gas calculated on the basis of computer code ADEP (ref. lo). Ar/H1 Ar/H2 (33110 slpm)
N2IH2
(235194 slpni) (65/3 slpir ) 750 44 1.93 0.15 13.5 12 100 500 25 0.98 0.08 15.3 10 800 500 200 5.0 0.1 24 6 200
01998 IUPAC, Pure &Applied Chemistry70, 1157-1162
1160
.-C5
0 0.10:
t
!n
0.01:
M. HRABOVSK?
II 2 1 Y downstream of the torch exit are presented in Fig. 5 (ref. 15). Plasma temperature close to 28 000 K
_ _ _ _ - - - - - - - - - - - -- c
-2-
- - - - - - - - - - - - - and velocity 7 kmls were found in the - experiments. Coefficients characterizing balance of radial transfer of energy in the arc chamber are shown in Fig. 6. The coefficients were evaluated from the series of experiments on water stabilized torch
-[-]1 - W. = 0.57 !
2 - ~ . = 0 . 8 7
I
- (ref. 14). It can be seen that 57% - 87% of arc I I power is transferred in radial direction, about
. - . - . - - - . - . _ _ _ ;--.-.1.
- - i - - - - -- - - - .p, - --.-I
CHARACTERISTICS OF WATER-STABILIZED PLASMA TORCH Characteristics of arc
This paragraph presents results of experiments performed with water stabilized plasma torch designed for power range from 90 to 200 kW (ref. 14, 15). The stabilizing chamber of the torch was created by three sections with tangential water injection. The sections were separated by segments determining diameter of water vortex, the exhaust slots were at the both ends of the stabilizing chamber. The cathode was made of a small piece of zirconium fixed in copper rod, the anode in the form of rotating internally cooled cooper disc was positioned outside of the arc chamber 2 mm downstream of the nozzle exit. The length of the stabilized part of arc column was 5 5 mm, the inner diameter of water vortex was 7 mm, the diameter of exit nozzle was 6 mm. Total flow rate of water through the chamber was 30 l/min.
28
P 24
CI 0I600 Avelocity
12 87
0.0 0.5 1.0 1.5 2.0 2.5 3.0
radial coordinate [mm]
5 i ? 8
4 :
32
Fig. 4 Power balance of water-stabilized plasma torch for several distances d of surface of anode from the axis of arc. Fig. 5 Radial Profiles of Plasma temperame and velocity 2 mm d o w n s t m n ofthe nozzle exit for d = 5 mm.
01998 IUPAC, Pure 8tAppliedChernistry70, 1157-1162 Water-stabilized plasma generators 1161
6
Properties of generated Dlasma iet
,4
, -- 't + I = 3 0 0 A 4
Basic characteristics of plasma jet measured at the torch exit are shown in Table 2. The jet is characterized by very high plasma temperatures, enthalpies and flow velocities. High flow velocity combined with very low ratio S of plasma density to the density of ambient air leads to intensive turbulent mixing of flowing plasma with ambient gas. Characteristic frequencies of the production of turbulent eddies due to entrainment of cold gas into jet given in the table were determined from power spectra of light emitted from the jet (ref. 16). High characteristic frequencies are related to high flow velocities. The non-dimensional frequencies given by Strouhal number St=f.D/v, are close to the values found in jets produced in gas-stabilized torches with substantially lower flow velocities.
3 -2 -1 -O - --1
-
-2
-
-3
-
TABLE 2. Torch operating parameters and jet exit conditions. 600 arc current (A) 300 400 500 power input (kW) 84 106.8 139 176
I-__l
mass flow rate ( g / s )
mean temperature (K) centerline temperature (K) mean enthalpy (MJ/kg) mean velocity (m/s) centerline velocity ( d s )
mean density (kg/m3) centerline density (kg/m3) mean density ratio Scharacteristic frequency (kHz)
Strouhal number St Reynolds number Re Machnumber M_ _ I lll__ I I_ 1 ~ 1 1
0.204 13 750 19 000 157 1736 2 494 4 . 1 5 ~ 1 O 3
1 . 9 2 ~ 1 0 ~ 0.0034 52
0.18
473 0.317 .lllll.-l___""
0.272 0.285 0.325 14 500 15 400 16 200 23 000 26 200 27 200 185 230 272 2 635 3 247 4 230 4 407 5 649 7 054 3 . 6 4 ~ 10-3 3.1x 10" 2 . 7 2 ~ 10-3 1 . 2 3 ~ 1 0 . ~ 0 . 9 8 ~ 1 0 - ~ 0 . 9 2 ~ 1 0 ' ~ 0.0030 0.0026 0.0023 68 96 118 0.15 0.18 0.17 786 1140 1770 0.445 0.505 0.617 ~ ll._.l"." .... " _ I ",. . .
The jet is fully turbulent and it is characterized by a presence of large scale structures with very high gradients of density (ref. 17). Intensive turbulent mixing causes rapid decrease of temperature and velocity along the jet. Isotherms of plasma jet generated at arc current 400 A are shown in Fig. 7, development of centerline velocity along the jet is shown in Fig. 8. The temperatures were obtained by series of spectroscopic measurements (ref. 15), the velocities were determined by electric probe diagnostics (ref. 18). Despite of their rapid decrease with increasing distance from the torch exit, both the temperature and the velocity are high in the jet.
I=400 A. P=107 kW 84III
t
1Fig. 7 Isotherms of plasma jet generated in water torch at arc current 400 A and arc power 107 kW. Fig. 8plasma jet. Development of centerline velocity along the
01998 IUPAC, Pure &Applied Chemistry70, 1157-1162 1162 M. HRABOVSK?
CONCLUSIONS
Principle of stabilization of arcs by liquid wall can be utilized for stabilization of long arc columns at very low flow rates of plasma gas. Thus, plasma jets with high temperatures can be generated in torches with liquid stabilization. In water plasma torches the oxygen-hydrogen plasma is produced, high content of hydrogen results in high enthalpy and sound velocity of plasma. Water-stabilized plasma torches thus generate plasma jet with extremely high plasma temperature and flow velocity. Basic parameter which determine arc and plasma characteristics is mass flow rate through the arc chamber. This parameter can not be controlled independently as it is determined by power balance of radial transfer of energy in the arc chamber. In water plasma torch the energy spent for evaporation is substantially lower than the energy absorbed in the produced vapour. This is principle cause of low mass flow rates and high temperatures found in experiments. High velocities in combination with low plasma densities lead to intensive interaction of plasma jet with surrounding atmosphere and to high level of turbulence. The mixing of plasma with surrounding cold gas is characterized by very low time constants. High enthalpy and flow velocity of generated plasma, high level of turbulence and short time constants of mixing result in special performance characteristics in some plasma processing applications. Especially high throughputs of treated material and high process rates are characteristic for water plasma torches. At present the only industrial scale application of water-stabilized plasma torch is plasma spraying. Powder throughputs almost one order higher than for common gas stabilized torches can be achieved (ref. 19). Therefore water-stabilized systems are used for large-area coatings, for production of self-supporting ceramic parts and for powder processing. Application for destruction of liquid waste is investigated (ref. 20, 21). Besides high level of turbulence, high rate of mixing process and high plasma enthalpy also chemical composition of plasma is advantageous for this application.
ACKNOWLEDGMENTS
The author would like to thank to the Grant Agency of Czech Republic for the support of this work under the projects No. 102/95/0592 and No. 106/96/K245.
REFERENCES
H. Gerdien, A. Lotz. Wiss. VeroffentlichungenSiemenswerk 2, 489 (1922). 2. H. Gerdien, A. Lotz. Z. Tech. Phys. 4, 157 (1923). 3. H. Maecker. 2 . t Phys. 129, 108-122 (1951). 4.
F. Bumhorn, H. Maecker. Zf Phys. 129, 369-376 (1951). 5. R. W. Larentz. Z.f Phys. 129, 343-364 (1951). 6. R. Weiss. 2,f Phys. 138, 170-182 (1954).
7 . F.
Bumhorn, H. Maecker, T. Peters. Z.$ Phys. 131, 28-40 (1951). 8. B.Gross, B. Grycz, K. Miklossy. Plasma Technology, Iliffe Books Ltd., London (1968). 9. B.W. Swanson. In Current Interruption ofHigh-Voltage Network (K.Ragaller, ed.), pp. 137-179. Plenum Press, New York -
London (1978). 10. ADEP
- Data bank and computer code. LMCTS URA 320 CNRS, Universitt? de Limoges. 1 1 . H. Riad, J.J. Gonzales, A. Gleizes. Proc. oflSPC 12, pp. 1731-1736. Minneapolis, Aug. 21-25, 1995. 12. V. Aubrecht, B. Gross, J. Phys. D: Appl. Phys. 27,95100 (1994). 13. G . Irons. Proc. of Workshop on Industrial Applications ofplasma Chemistry, Vol. B, pp. 53-65. Minneapolis, Aug. 25-26, 14. M. Hrabovslj, M. Konrad, V Kopeclj, V. Sember. IEEE Trans. on Plasma Science, in press. 15. M. Hrabovslj, V Kopeclj, V. Sember. In Heat and Mass Transfer under Plasma Conditions (P. Fauchais, M. Boulos, 16. M. Hrabovsw, M. Konrad, V Kopecw, J. Hlina. . Proc. ofZSPC 12, pp. 1627-1632. Minneapolis, Aug. 21-25, 1995. 17. M. Hrabovslj, M. Konrad, V Kopeclj, J. Kravarik, P. KubeS. Proc. of 171h Symp. on Plasma Physics and Technology, pp. 18. M. Hrabovslj, M. Konrad, V Kopeclj. In Heat and Mass Transfer under Plasma Conditions (P, Fauchais, M. Boulos, 19. PChraska, M. Hrabovslj. Proc. of lnt. Thermal Spray Conf & Exhib., pp. 81-85. Orlando, USA, May 28-June 5, 1992. 20. V. Broiek, M. Hrabovslj, V. Kopecw. Proc oflSPC 13, pp.1735-1739. Beijing, August 18-22, 1997. 21. V. BroZek, M. Hrabovsw, V. Kopecw. Journ. ofHigh Temp. Mat. Process, in press. 1995. J. van der Mullen, ed.), pp. 91-98. Begell House, New York-Wallingford (1994). 179-181. Prague, June 13-16 1995. |
8547 | https://math.libretexts.org/Bookshelves/Algebra/Elementary_Algebra_(LibreTexts)/07%3A_Rational_Expressions_and_Equations/7.02%3A_Multiplying_and_Dividing_Rational_Expressions | Skip to main content
7.2: Multiplying and Dividing Rational Expressions
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: Aug 18, 2025
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7.1: Simplifying Rational Expressions
7.3: Adding and Subtracting Rational Expressions
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Learning Objectives
Multiply rational expressions.
Divide rational expressions.
Multiply and divide rational
functions
.
Multiplying Rational Expressions
When multiplying fractions, we can multiply the numerators and denominators together and then reduce, as illustrated:
Multiplying rational expressions is performed in a similar manner. For example,
In general, given polynomials , , , and , where and , we have
In this section, assume that all variable expressions in the denominator are nonzero unless otherwise stated.
Example
Multiply:
Solution:
Multiply numerators and denominators and then cancel common
factors
.
Answer:
Example
Multiply:
Solution:
Leave the product in factored form and cancel the common
factors
.
Answer:
Example
Multiply:
Solution:
Leave the polynomials in the numerator and denominator factored so that we can cancel the
factors
. In other words, do not apply the
distributive property
.
Answer:
Typically, rational expressions will not be given in factored form. In this case, first factor all numerators and denominators completely. Next, multiply and cancel any common
factors
, if there are any.
Example
Multiply:
Solution
Factor the denominator as a
difference of squares
. Then multiply and cancel.
Keep in mind that 1 is always a factor; so when the entire numerator cancels out, make sure to write the factor 1.
Answer:
Example
Multiply:
Solution:
It is a best practice to leave the final answer in factored form.
Answer:
Example
Multiply:
Solution:
The
trinomial
in the numerator has a negative
leading coefficient
. Recall that it is a best practice to first factor out a and then factor the resulting
trinomial
.
Answer:
Example
Multiply:
Solution:
We replace with so that we can cancel this factor.
Answer:
Exercise
Multiply:
Answer
Dividing Rational Expressions
To divide two fractions, we multiply by the
reciprocal
of the
divisor
, as illustrated:
Dividing rational expressions is performed in a similar manner. For example,
In general, given polynomials P, Q, R, and S, where , , and , we have
Example
Divide:
Solution:
First, multiply by the
reciprocal
of the
divisor
and then cancel.
Answer:
Example
Divide:
Solution:
After multiplying by the
reciprocal
of the
divisor
, factor and cancel.
Answer:
Example
Divide:
Solution:
Begin by multiplying by the
reciprocal
of the
divisor
. After doing so, factor and cancel.
Answer:
Example
Divide:
Solution:
Just as we do with fractions, think of the
divisor
as an
algebraic fraction
over 1.
Answer:
Exercise
Divide:
Answer
Multiplying and Dividing Rational Functions
The product and
quotient
of two rational
functions
can be
simplified
using the techniques described in this section. The
restrictions
to the domain of a product consist of the
restrictions
of each function.
Example
Calculate and determine the
restrictions
to the domain.
Solution:
In this case, the domain of consists of all
real numbers
except 0, and the domain of consists of all
real numbers
except . Therefore, the domain of the product consists of all
real numbers
except 0 and . Multiply the
functions
and then simplify the result.
Answer:
, where
The
restrictions
to the domain of a
quotient
will consist of the
restrictions
of each function as well as the
restrictions
on the
reciprocal
of the
divisor
.
Example
Calculate and determine the
restrictions
.
Solution:
In this case, the domain of consists of all
real numbers
except 3 and 8, and the domain of consists all
real numbers
except 3. In addition, the
reciprocal
of has a restriction of −8. Therefore, the domain of this
quotient
consists of all
real numbers
except 3, 8, and −8.
Answer:
, where
Key Takeaways
After multiplying rational expressions, factor both the numerator and denominator and then cancel common
factors
. Make note of the
restrictions
to the domain. The values that give a value of 0 in the denominator are the
restrictions
.
To divide rational expressions, multiply by the
reciprocal
of the
divisor
.
The
restrictions
to the domain of a product consist of the
restrictions
to the domain of each factor.
The
restrictions
to the domain of a
quotient
consist of the
restrictions
to the domain of each rational expression as well as the
restrictions
on the
reciprocal
of the
divisor
.
Exercise Multiplying Rational Expressions
Multiply. (Assume all denominators are nonzero.)
Answer
: 1.
3.
5.
7.
9.
11.
13.
15.
17.
19.
21.
23.
25.
Exercise Dividing Rational Expressions
Divide. (Assume all denominators are nonzero.)
Answer
: 1.
3.
5.
7.
9.
11.
13.
15.
17.
19.
21.
23.
25.
27.
29.
31.
Exercise Dividing Rational Expressions
Recall that multiplication and division are to be performed in the
they appear from left to right. Simplify the following.
Answer
: 1.
3.
5.
Exercise Multiplying and Dividing Rational Functions
Calculate and determine the
restrictions
to the domain.
and
and
and
and
and
and
Answer
: 1.
3.
5.
Exercise Multiplying and Dividing Rational Functions
Calculate and state the
restrictions
.
and
and
and
and
and
and
Answer
: 1.
3.
5.
Exercise Discussion Board Topics
In the history of fractions, who is credited for the first use of the fraction bar?
How did the ancient Egyptians use fractions?
Explain why is a restriction to .
Answer
: 1. Answer may vary
3. Answer may vary
7.1: Simplifying Rational Expressions
7.3: Adding and Subtracting Rational Expressions |
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Area of Parallelogram: Definition, Formula And Worksheets
Ever seen a slanted rectangle and wondered, “Is that still a rectangle?” It might be a parallelogram—a clever four-sided shape with secrets hidden in its angles. In this guide, we’ll break down what a parallelogram is, how to find its base and height, and how to calculate its area with confidence.
Looking for extra practice? Scroll to the bottom to download free printable worksheets!
Definition of parallelogram
A parallelogram is defined as a quadrilateral (a four-sided polygon) where both pairs of opposite sides are parallel.
Now, when it comes to finding the area, two parts of a parallelogram matter most: the base and the height.
But wait — how do we know which side is the base?
If the parallelogram is sitting flat, the bottom edge is usually the base.
If it’s tilted or slanted, any side can be used as the base — as long as we know how to find the correct height for it!
The height is the straight-up distance from the base to the opposite side — like a ladder leaning against a wall. It must be perpendicular (forms a 90° angle) to the base.
Sometimes, the height falls outside the parallelogram. That’s totally okay! Just imagine dropping a straight line from the top down to the base. As long as it’s at a right angle, it counts as the height.
So whether the base is flat or slanted, the rule is the same: Height must be perpendicular to the base, even if it’s outside the shape!
Once you’ve got the base and the height, you’re all set to find the area.
Parallelogram Area Formula
We often think of rectangles when we hear the word “area.” But how do we find the area of a parallelogram? Let’s break it down with a trick!
Here is a parallelogram. If you cut off a triangle from one side and slide it to the other, you can form a rectangle! This neat transformation doesn’t change the area.
That’s why the area of a parallelogram uses the same idea as the area of a rectangle:
Area the Parallelogram = base × height
Important tip: The height is not the side of the parallelogram—unless that side forms a right angle with the base! Think of height as the straight-up distance from the base to the top, like a ladder standing tall and perfectly upright.
Example: Applying the Parallelogram Area Formula
Example 1
Problem:
What is the area of this parallelogram?
Answer:
base = 12, height = 8
Area = base × height = 12 × 8 = 96
Example 2
Problem:
The area of this parallelogram is 60. If the base is the side with length 12, what is the corresponding height?
Answer:
Area = 60, base = 12
Area = base × height
60 = 12 × height
height = 60 ÷ 12 = 5
Want more practice?
Download Free Worksheets Now
Wrapping Up the Parallelogram Area Formula
Parallelograms may look like they’re leaning, but their math stands tall and strong! By remembering the base and the perpendicular height, you can find the area of any parallelogram — even if it’s wearing a fancy disguise as a rhombus, rectangle, or square.
To help you see how parallelograms fit into the big picture of geometry, here’s a quick summary of the area formulas for some common shapes:
| Shape | Figure | Area Formula | Special Notes |
--- --- |
| Square | Area = side² | All four sides are equal |
| Rectangle | Area = length × width | Opposite sides are equal |
| Parallelogram | Area = base × height | Opposite sides are parallel |
| Trapezoid | Area = ½ × (base₁ + base₂) × height | One pair of parallel sides |
Additional Math Topics for Grade 6 – with Free Worksheets
Area of Trapezoid
Area of Triangles
How to Divide Fractions
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Published On: June 29, 2025
Post Content
Definition of parallelogram
Parallelogram Area Formula
Example: Applying the Parallelogram Area Formula
Example 1
Example 2
Wrapping Up the Parallelogram Area Formula
Additional Math Topics for Grade 6 – with Free Worksheets
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8549 | https://pages.uoregon.edu/haowang/teaching/425_FALL2008/425LN3.pdf | HW MATH425/525 Lecture Notes 1 Law of Total Probability Given a sequence of mutually exclusive events S1, S2, · · · , Sn. If event A ⊂∪n i=1Si and P(Si) > 0, then P(A) = P(S1)P(A|S1) + · · · + P(Sn)P(A|Sn) Proof: Since (A ∩Si) ∩(A ∩Sj) = φ for any i ̸= j, this means that the sequence {(A ∩Si), i = 1, · · · , n} are mu-tually exclusive, according to the second condition of the definition of probability, we have P(A) = P(A ∩S1) + · · · + P(A ∩Sn).
Since P(A ∩Si) = P(Si)P(A|Si) for i = 1, · · · , n, the law of total probability is proved.
Example 7.1 A fair coin is flipped. If a head turns up, a fair die is tossed; if a tail turns up, two fair dice are tossed. What is the probability of the event B that the sum of the appearing number(s) is equal to 6?
Solution: Since P(H) = 1 2 and P(T) = 1 2, and P(B|H) = 1 6 P(B|T) = 5 36 HW MATH425/525 Lecture Notes 2 {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}. Therefore, P(B) = P(H)P(B|H) + P(T)P(B|T) = 1 2 × 1 6 + 1 2 × 5 36 = 0.15.
Bayes’ Rule Given a sequence of mutually exclusive events S1, S2, · · · , Sn and P(Si) > 0. If event A ⊂∪n i=1Si, then P(Si|A) = P(Si)P(A|Si) Pn k=1 P(Sk)P(A|Sk) Proof: According to the definition of conditional prob-ability, we have P(Si|A) = P(Si ∩A) P(A) .
Since P(Si ∩A) = P(Si)P(A|Si), and according to the law of total probability, P(A) = P(S1)P(A|S1) + · · · + P(Sn)P(A|Sn), this proves the Bayes’ rule.
Example 7.2 A man takes either a bus or the subway to work with probabilities 0.3 and 0.7 , respectively.
When he takes the bus, he is late 30% of the days.
When he takes the subway, he is late 20% of the days.
HW MATH425/525 Lecture Notes 3 If the man is late for work on a particular day, what is the probability that he took the bus?
Solution: Define B = { The man takes a bus }, S = { The man takes the subway }, and L = { The man is late on the day }. Then, P(B) = 0.3, P(S) = 0.7, P(L|B) = 0.3, P(L|S) = 0.2.
By Bayes’ rule, we have P(B|L) = P(B)P(L|B) P(S)P(L|S) + P(B)P(L|B) = 0.3 × 0.3 0.7 × 0.2 + 0.3 × 0.3 = 0.09 0.23 = 0.3913.
Example 7.3 To evaluate the effectiveness of a screen-ing procedure, we will evaluate the probability of a false negative or a false positive using the following no-tation: T + : The test is positive and indicate that the person has the disease.
T −: The test is negative and indicate that the person does not have the disease.
Dc : The person really does not have the disease.
D : The person really has the disease.
According to the test results, we found that the sensi-tivity of the test has following conditional probabilities: P(T +|D) = 0.98, HW MATH425/525 Lecture Notes 4 and P(T −|Dc) = 0.99.
If the proportion of the general population infected with this disease is 2 per million, what is (a) the probability of a false positive, P(Dc|T +) ?
(b) the probability of a false negative, P(D|T −) ?
Solution: From the given information, we know the fol-lowing: P(D) = 0.000002, P(Dc) = 0.999998 P(T +|D) = 0.98 P(T −|D) = 0.02 P(T +|Dc) = 0.01 P(T −|Dc) = 0.99 (a) From Bayes’ Rule, P(Dc|T +) = P(Dc ∩T +) P(T +) = P(Dc)P(T +|Dc) P(Dc)P(T +|Dc) + P(D)P(T +|D) Therefore, P(Dc|T +) = 0.00999998 0.01000194 = 0.999804038 HW MATH425/525 Lecture Notes 5 (b) Using a similar calculation, P(D|T −) = P(D ∩T −) P(T −) = P(D)P(T −|D) P(Dc)P(T −|Dc) + P(D)P(T −|D) Therefore, P(D|T −) = 0.00000004 0.98999806 = 0.00000004 Hence, the probability of a false positive is near 1 and very likely, while the probability of a false negative is quite small and very unlikely.
Example 7.4 If men constitute 47% of the population and tell the truth 78% of the time, while women tell the truth 63% of the time, what is the probability that a person selected at random will answer a question truthfully?
Solution: Define B = {The person interviewd answers truthfully } A = { The person interviewed is a man } According to the law of total probability, we have P(B) = P(A)P(B|A) + P(Ac)P(B|Ac) = (0.47)(0.78) + (0.53)(0.63) = 0.70 HW MATH425/525 Lecture Notes 6 Example 7.4 A worker-operated machine produces a defective item with probability 0.01 if the worker fol-lows the machine’s operating instructions exactly, and with probability 0.03 if he does not .
If the worker follows the instructions 90% of time, what proportion of all items produced by the machine will be defective?
Given that a defective item is produced, what is the conditional probability of the event that the worker exactly follows the machine operating instructions?
Solution: Define D : Machine produces a defective item.
F : Worker follows instructions.
Then, we have following information: P(D|F) = 0.01 P(F) = 0.9 P(D|F c) = 0.03 P(F c) = 0.1 According to the law of total probability, we have P(D) = P(D|F)P(F) + P(D|F c)P(F c) = 0.01(0.9) + 0.03(0.1) = 0.012.
According to the Bayes’ Rule, we have P(F|D) = P(F)P(D|F) P(D) = 0.9(0.01) 0.012 = 0.75 P(F c|D) = 0.25 HW MATH425/525 Lecture Notes 7 Example8.3 Many companies are testing prospective employees for drug use with the intent of improving efficiency and reducing accidents. Suppose a company uses a test that is 98% accuracy to identify a user or a nonuser. To reduce the chance of error, two indepen-dent tests are required for each applicant. What are the probabilities of following events?
(a) A nonuser fails both tests.
(b) A drug user is detected.
(c) A drug user is not detected.
Solution: Define P1 := {The first test is positive}; P2 := {The second test is positive}; N1 := {The first test is negative}; N2 := {The second test is negative}.
D := { An applicant is a drug user}.
Dc := { An applicant is not a drug user}.
We know that P(P1|D) = 0.98, P(P2|D) = 0.98, P(N1|Dc) = 0.98 and P(N2|Dc) = 0.98. Then, P(N1|D) = 1 −0.98 = 0.02, P(N2|D) = 0.02, P(P1|Dc) = 0.02 and P(P2|Dc) = 0.02.
(a) P(P1 ∩P2|Dc) = P(P1|Dc)P(P2|Dc) = (0.02)(0.02) = 0.0004 HW MATH425/525 Lecture Notes 8 (b) P((P1 ∩N2) ∪(N1 ∩P2) ∪(P1 ∩P2)|D) = P((P1 ∩N2)|D) + P((N1 ∩P2)|D) + P((P1 ∩P2)|D) = P(P1|D)P(N2|D) + P(N1|D)P(P2|D) + P(P1|D)P(P2|D) = (0.98)(0.02) + (0.02)(0.98) + (0.98)(0.98) = 0.9996 (c) P(N1 ∩N2|D) = P(N1|D)P(N2|D) = (0.02)(0.02) = 0.0004 HW MATH425/525 Lecture Notes 9 Example As items come to the end of a production line, an inspector chooses items to undergo a complete in-spection. Of all items produced 20% are defective. 50% of all defective items go through a complete inspection, and 30% of all good items go through a complete in-spection. Given that an item is completely inspected, what is the probability that it is defective?
Solution: Define D : An item defective.
C : An item is completely inspected.
Then, we have following information: P(C|D) = 50% P(D) = 20% P(C|Dc) = 30% P(Dc) = 80% According to the Bayes’ Rule, we have P(D|C) = P(D)P(C|D) P(D)P(C|D) + P(Dc)P(C|Dc) = 0.2(0.5) 0.2(0.5) + 0.8(0.3) = 10 34 Discrete Random Variables and Their Distribution In order to use the existing mathematical tools to find out the probabilities of events, mean, variance, and so on, statisticians figured out a nice way as follows: We associate each simple event with a real number. For HW MATH425/525 Lecture Notes 10 example, in the coin tossing experiment, we define: Λ = { 1, 0 } ⊂R = {all the real numbers} ↑ ↑ Ω= {head, tail } This is called a transformation. Once transformed, we can use the existing mathematical tools. For our con-venience, in this section we denote the sample space by Ω= {ω1, ω2, · · · , ωn}, in which each ωi is a simple event.
Definition 9.1 Consider a random experiment with sample space Ω= {ω1, ω2, · · · , ωn}. A function X : Ω→R which assigns to each simple event ω ∈Ωone and only one real number X(ω) = x ∈R, is called a random variable.
Λ := {xi : xi = X(ωi), ωi ∈Ω} is called the range space of the random variable (r.v.) X.
Definition 9.2 If Λ is a finite set or a countable set, then X is called a discrete random variable. Given a discrete random variable X with range space Λ = {x1,· · · , xn}. For each xi ∈Λ, we define p(xi) = P({ω : X(ω) = xi}).
Then, p(x) is called the discrete density function of r.v.
X and the following table is called a discrete probability distribution of r.v. X.
HW MATH425/525 Lecture Notes 11 The distribution table of r.v X x x1 x2 · · · xn p(x) p(x1) p(x2) · · · p(xn) Example 9.1 Consider a random experiment that con-sists of tossing two fair coins successively and let X be equal to the number of heads observed. Find its sam-ple space, range space, and distribution.
Solution: The sample space Ω= {ω1 = {TT}, ω2 = {HT}, ω3 = {TH}, ω4 = {HH}} and Λ := {x1 = 0, x2 = 1, x3 = 2}.
p(x1) = p(0) = P(X = 0) = P({TT}) = 1 4, p(x2) = p(1) = P(X = 1) = P({TH, HT}) = P({TH}) +P({HT}) = 2 4, p(x3) = p(2) = P(X = 2) = P({HH}) = 1 4.
Therefore, The distribution table of r.v X x 0 1 2 p(x) 1/4 1/2 1/4 Remark: (a) 0 ≤p(xi) ≤1, HW MATH425/525 Lecture Notes 12 (b) P xi∈Λ p(xi) = P(Ω) = 1 Definition 9.3 Let X be a discrete r.v. with range space Λ = {x1,· · · , xn} and distribution p(xi), i = 1, · · · , n. The expected value or mean of X is defined by µ = E(X) = X xi∈Λ xip(xi).
The variance of X is defined by σ2 = E[(X −µ)2] = X xi∈Λ (xi −µ)2p(xi).
The standard deviation of X is defined as σ = √ σ2.
Example 9.2 A company has five applicants for two po-sitions: two women and three men. Suppose that the five applicants are equally qualified and no preference is given for choosing either gender.
Let X equal the number of women chosen to fill the two positions.
(a) Find the probability distribution of X; (b) Find the mean and standard deviation of X.
Example 9.2 A company has five applicants for two po-sitions: two women and three men. Suppose that the five applicants are equally qualified and no preference is given for choosing either gender.
Let X equal the number of women chosen to fill the two positions.
(a) Find the probability distribution of X; (b) Find the mean and standard deviation of X.
HW MATH425/525 Lecture Notes 13 Solution: (a). Define short notations as follows: M1: Candidate man one; M2: Candidate man two; M3: Candidate man three; W1: Candidate woman one; W2: Candidate woman two.
The sample space Ω= {ω1 = {M1M2}, ω2 = {M1M3}, ω3 = {M2M3}, ω4 = {W1M1}, ω5 = {W1M2}, ω6 = {W1M3}, ω7 = {W2M1}, ω8 = {W2M2}, ω9 = {W2M3}, ω10 = {W1W2}} and the range space Λ := {x1 = 0, x2 = 1, x3 = 2}.
p(x1) = p(0) = P(X = 0) = P({ω1, ω2, ω3}) = P({{M1M2}, {M1M3}, {M2M3}}) = 3 10, p(x2) = p(1) = P(X = 1) = P({ω4, ω5, ω6, ω7, ω8, ω9}) = P({{W1M1}, {W1M2}, {W1M3}, {W2M1}, {W2M2}, {W2M3}}) = 6 10, HW MATH425/525 Lecture Notes 14 p(x3) = p(2) = P(X = 2) = P({ω10 = {W1W2}}) = 1 10.
Therefore, The distribution table of r.v X x 0 1 2 p(x) 3/10 6/10 1/10 (b) µ = E(X) = P xi∈Λ xip(xi) = 0(3/10) + 1(6/10) + 2(1/10) = 8/10.
The variance of X is σ2 = E[(X −µ)2] = P xi∈Λ(xi −µ)2p(xi) = (0 −8/10)2(3/10) + (1 −8/10)2(6/10)+ (2 −8/10)2(1/10) = 0.36 and σ = √ σ2 = 0.6 Example 9.3 A jar contains four coins: a nickel, a dime, a quarter, and a half-dollar. Three coins are randomly selected from the jar.
Let X be equal to the total amount drawn.
a. List all the simple events in the sample space Ωand find the range space Λ.
b. Find the probability P(X ≥0.5 dollar).
HW MATH425/525 Lecture Notes 15 c. Find the probability distribution of X.
Solution: a. Denote: N: nickel; D: dime; Q: quarter; H: half-dollar.
and E1 = (NDQ),E2 = (NDH), E3 = (NQH), E4 = (DQH).
Then, Ω= {E1, E2, E3, E4} and Λ := {x1 = 0.4, x2 = 0.65, x3 = 0.80, x4 = 0.85}.
b. The simple event along with their monetary values follow: E1 = NDQ = $0.4 E2 = NDH = $0.65 E3 = NQH = $0.80 E4 = DQH = $0.85 P(X ≥0.5 dollar) = P(E2) + P(E3) + P(E4) = 1 4 + 1 4 + 1 4 = 3 4 c. We have p(x1) = p(0.4) = P(X = 0.4) = P({NDQ}) = 1 4, p(x2) = p(0.65) = P(X = 0.65) = P({NDH}) = 1 4, p(x3) = p(0.80) = P(X = 0.80) = P({NQH}) = 1 4.
HW MATH425/525 Lecture Notes 16 p(x4) = p(0.85) = P(X = 0.85) = P({DQH}) = 1 4.
Therefore, The distribution table of r.v X x 0.4 0.65 0.80 0.85 p(x) 1/4 1/4 1/4 1/4 Example 9.4 A student prepares for a quiz by studying a list of ten problems. She only can solve six of them.
For the quiz, the instructor selects five questions at random from the list of ten. Let X be the number of questions she can solve.
a Find total number of simple events in the sample space and the range space.
b Find the probability distribution of X.
Solution: a.
The total number of simple events is the sample space is equal to C10 5 = 252.
Then, Λ := {x1 = 1, x2 = 2, x3 = 3, x4 = 4, x5 = 5}.
b. We have p(x1) = p(1) = P(X = 1) == C6 1C4 4 252 , p(x2) = p(2) = P(X = 2) == C6 2C4 3 252 , HW MATH425/525 Lecture Notes 17 p(x3) = p(3) = P(X = 3) == C6 3C4 2 252 , p(x4) = p(4) = P(X = 4) == C6 4C4 1 252 , p(x5) = p(5) = P(X = 5) == C6 5C4 0 252 , Therefore, The distribution table of r.v X x 1 2 3 4 5 p(x) C6 1C4 4 252 C6 2C4 3 252 C6 3C4 2 252 C6 4C4 1 252 C6 5C4 0 252 Example 9.5 In a pocket there are 3 black and 2 white balls. Balls are identical except their colors. We randomly draw a ball and observe its color two times successively, with replacement. We define that X is the number of black balls observed.
a Find the sample space and the range space.
b Find the probability P(X ≥1).
c Find the probability distribution of X.
Solution: a. Denote: B: observe a black ball; W: observe a white ball; and E1 = (BB), E2 = (BW), E3 = (WB), E4 = (WW), HW MATH425/525 Lecture Notes 18 Then, Ω= {E1, E2, E3, E4} and Λ := {x1 = 0, x2 = 1, x3 = 2}.
b.
P(X ≥1) = 1 −P(X = 0) = 1 −P({WW}) = 1 −P(W)P(W) = 1 −(2 5)(2 5) = 1 −4 25 = 21 25 c. We have p(x1) = p(0) = P(X = 0) = P({WW}) = 4 25, p(x2) = p(1) = P(X = 1) = P({BW} ∪{WB}) = P({BW}) + P({WB}) = 2P({W})P({B}) = 2(2 5)(3 5) = 12 25, p(x3) = p(2) = P(X = 2) = P({BB}) = P({B})P({B}) = (3 5)(3 5) = 9 25, Therefore, The distribution table of r.v X x 0 1 2 p(x) 4/25 12/25 9/25 The Binomial Probability Distribution Definition 10.1 A binomial experiment is one that has following three characteristics: (1) The experiment consists of n ordered, independent, HW MATH425/525 Lecture Notes 19 identical trials.
(2) Each trial has two possible outcomes: success A, failure ¯ A.
(3) 0 < P(A) = p < 1 and P( ¯ A) = 1 −p = q.
Remark: In binomial experiment, if p ̸= q, then simple events are not equally likely. Therefore, it is not a clas-sical probability model.
Example10.1 Consider a binomial experiment of flip-ping a biased coin three times successively. Let A be the event of observing a head and ¯ A be the event of observing a tail.
Suppose that P(A) = 1 3 P( ¯ A) = 2 3 Then Ω= {AAA, ¯ AAA, A ¯ AA, AA ¯ A, ¯ A ¯ AA, ¯ AA ¯ A, A ¯ A ¯ A, ¯ A ¯ A ¯ A} P(AAA) = P(A)P(A)P(A) = 1 27 P(A ¯ AA) = P(A)P( ¯ A)P(A) = (1 3)(2 3)(1 3) = 2 27 Therefore, it is not a classical probability model.
Let X be number of heads observed. Find the proba-bility P(X = 2).
P(X = 2) = P({AA ¯ A}, {A ¯ AA}, { ¯ AAA}) P({AA ¯ A}) = P(A)P(A)P( ¯ A) = ppq HW MATH425/525 Lecture Notes 20 P({ ¯ AAA}) = P(A)P(A)P( ¯ A) = ppq P({A ¯ AA}) = P(A)P(A)P( ¯ A) = ppq P(X = 2) = C3 2ppq = C3 2p2q3−2 Generally, consider a binomial experiment with n tri-als. let X be the number of successes in the n trials.
What is the probability of event {X = k}. Then, P(X = k) = Cn k pkqn−k (For Cn k , Consider there are n positions and how many ways to choose n −k positions to put bars on tops of these positions.) Definition 10.2 A binomial experiment consists of n ordered, independent, identical trials with probability of success p and probability of failure q = 1 −p on each trial. Let X be the number of successes in the n trials.
Then, The distribution table of r.v X x 0 1 2 · · · n p(x) Cn 0 p0qn Cn 1 p1qn−1 Cn 2 p2qn−2 · · · Cn npnq0 is called the binomial distribution. Its mean µ = np, its variance σ2 = npq, its standard deviation σ = √npq Example 9.6 In a pocket there are 3 black and 2 white balls. Balls HW MATH425/525 Lecture Notes 21 are identical except their colors. We randomly draw a ball and observe its color three times successively, with replacement. We define that X is the number of black balls observed.
a Find the sample space and the range space.
b Find the probability P(X ≥1).
c Find the probability distribution of X.
Solution: a. Denote: B: observe a black ball; W: observe a white ball; and E1 = (BBB), E2 = (BBW), E3 = (BWB), E4 = (BWW), E5 = (WBB), E6 = (WBW), E7 = (WWB), E8 = (WWW).
Then, Ω= {E1, E2, E3, E4, E5, E6, E7, E8} and Λ := {x1 = 0, x2 = 1, x3 = 2, x4 = 3}.
b.
P(X ≥1) = 1 −P(X = 0) = 1 −P({WWW}) (0.1) = 1 −P(W)P(W)P(W) = 1 −(2 5)(2 5)(2 5) = 1 −8 125 = 117 125 c. We have p(x1) = p(0) = P(X = 0) = P({WWW}) = 8 125, HW MATH425/525 Lecture Notes 22 p(x2) = p(1) = P(X = 1) = P({BWW} ∪{WBW} ∪{WWB}) = P({BWW}) + P({WBW}) + P({WWB}) = 3P({W})P({W})P({B}) = 3(2 5)(2 5)(3 5) = 36 125, p(x3) = p(2) = P(X = 2) = P({WBB} ∪{BWB} ∪{BBW}) = P({WBB}) + P({BWB}) + P({BBW}) = 3P({B})P({B})P({W}) = 3(3 5)(3 5)(2 5) = 54 125, p(x4) = p(3) = P(X = 3) = P({BBB}) = P({B})P({B})P({B}) = (3 5)(3 5)(3 5) = 27 125.
Therefore, The distribution table of r.v X x 0 1 2 3 p(x) 8/125 36/125 54/125 27/125 Example 10.1 Suppose that a family will certainly have 5 children, but each child being a boy or girl is totally uncertain with equal probability. Let X be the number of boys this family will have. Find the distribution of X and the probability P(X ≥1).
Example10.2 Consider a binomial experiment of flip-ping a biased coin twenty times successively. Let A be HW MATH425/525 Lecture Notes 23 the event of observing a head and ¯ A be the event of observing a tail.
Suppose that P(A) = 1 5 P( ¯ A) = 4 5 Let X be number of heads observed in the twenty flip-pings. (a) Find the probability P(X = 8). (b) Find its mean and variance. (c) Find the probability of event (X > 2).
solution: (a) According to the binomial distribution formula, we have P(X = 8) = C20 8 (1 5) 8 (4 5) 20−8 .
We have P(X ≤8) = 0.990 and P(X ≤7) = 0.968 according to the table 1 (n = 20, k = 8, k = 7, p = 0.2 ) on page 684.
P(X = 8) = P(X ≤8) −P(X ≤7) = 0.022 .
(b) Its mean is equal to µ = 20(1/5) = 4 and its variance is equal to σ2 = 20(1/5)(4/5) = 16 5 (c) P(X > 2) = 1 −P(X ≤2) = 1 −0.206 = 0.794 Example10.3 Consider an experiment of randomly draw-ing a chip 8 times successively with replacement from a box containing two black chips and three white chips.
HW MATH425/525 Lecture Notes 24 The chips are identical except their colors. Let X be the number of black chips observed in the 8 drawings.
(a) Find the probability P(X = 5). (b) Find its mean and variance. (c) Find the probability of event (X > 2).
solution: (a) According to the binomial distribution formula, we have P(X = 5) = C8 5(2 5) 5 (3 5) 8−5 .
We have P(X ≤5) = 0.950 and P(X ≤4) = 0.826 according to the table 1 (n = 8, k = 5, k = 4, p = 0.4 ) on page 681.
P(X = 5) = P(X ≤5) −P(X ≤4) = 0.124 .
(b) Its mean is equal to µ = 8(2/5) = 16/5 and its variance is equal to σ2 = 8(2/5)(3/5) = 48 25 (c) P(X > 2) = 1 −P(X ≤2) = 1 −0.315 = 0.685 Example: A factory employs several thousand work-ers, of whom 30% are Hispanic. If the 15 members of the union executive committee were chosen from the workers at random,let X be the number of Hispanics on the committee.
(a) Find P(X = 3).
(b) Find P(X ≤3).
HW MATH425/525 Lecture Notes 25 Example: Suppose that early statewide election re-turns indicate totals of 33, 000 votes for candidate A versus 27, 000 for candidate B, and that these early re-turns can be regarded as a random sample selected from the population of all 10, 000, 000 eligible voters in the state. Let X be the number of votes for A.
(a) If the statewide vote will be split 50 −50, find the expected number of votes for A in the sample of 60, 000 early returns.
(b) Find the standard deviation of X.
(c) Find the probability that (X > 40, 000).
Solution: (a) µ = np = 60, 000 × 0.5 = 30, 000.
(b) σ = √npq = √15, 000 = 122.47449 (c) P(X > 40, 000) = 1 −P40,000 i=0 C60,000 i (0.5)i(0.5)60,000−i |
8550 | https://vencru.com/blog/understanding-retail-inventory-method-pros-and-cons/ | Understanding Retail Inventory Method: Pros and Cons - Vencru
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Understanding Retail Inventory Method: Pros and Cons
August 24, 2023
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Inventory management is a crucial aspect of running any business that involves selling goods. To make informed decisions about inventory, businesses need to understand inventory costing methods, one of which is the retail inventory method. In this blog post, we will discuss the method, its advantages and disadvantages, and alternatives to it. We will also discuss how Vencru can help retailers perform inventory accounting using weighted average costs.
What is Inventory Costing?
Inventory costing is the process of determining the value of goods that a business has in its possession. It involves assigning a value to each unit of inventory and keeping track of the total value of inventory.
What is Retail Inventory Method?
This method of inventory costing that is commonly used in the retail industry. It involves calculating the value of inventory by using the ratio of cost to retail price. In other words, the retail inventory method determines the cost of goods sold and the value of inventory by using the retail price of goods.
This is important because it provides businesses with a real-time valuation of their inventory. This can help businesses make informed decisions about pricing, ordering, and inventory management.
How does retail inventory method work?
This works by calculating the cost of goods sold as a percentage of retail sales. This percentage is known as the cost-to-retail percentage. The cost-to-retail percentage is calculated by dividing the cost of goods by the retail price of goods.
First, the cost-to-retail percentage is calculated.
Then, this percentage is applied to the total retail value of inventory.
Finally, the cost value of inventory is determined.
Calculation of average cost to retail percentage
The average cost to retail percentage is calculated by dividing the cost of goods sold by the retail price of goods sold. For example, if the cost of goods sold is $50,000 and the retail price of goods sold is $100,000, the average cost to retail percentage is 50%.
Advantages of Retail Inventory Method
Provides real-time inventory valuation:This provides businesses with a real-time valuation of their inventory. This can help businesses make informed decisions about pricing, ordering, and inventory management.
Enables businesses to identify and address inventory issues quickly:Since the retail inventory method provides a real-time inventory valuation, it enables businesses to identify and address inventory issues quickly. For example, if a business notices that the cost-to-retail percentage is increasing, it may indicate inventory shrinkage or overpricing.
Facilitates effective decision-making:The real-time inventory valuation provided by the retail inventory method can help businesses make informed decisions about pricing, ordering, and inventory management. This can lead to increased profitability and better business outcomes.
Disadvantages of Retail Inventory Method
Requires a lot of record-keeping and calculations: This requires a lot of record-keeping and calculations. Businesses must keep track of retail prices, cost of goods sold, and other inventory-related data to use this method effectively.
Can be time-consuming and labor-intensive:Due to the record-keeping and calculations required, the retail inventory method can be time-consuming and labor-intensive. Businesses need to allocate sufficient resources to perform these tasks.
May not be suitable for all businesses: This may not be suitable for all businesses. For example, businesses with a low sales volume may find it difficult to use this method effectively.
Alternatives to Retail Inventory Method
There are several alternative inventory costing methods to the inventory valuation method, and it’s important to understand their advantages and disadvantages to determine which method works best for your business.
First-In, First-Out (FIFO)
The FIFO method assumes that the oldest inventory is sold first, and the newest inventory remains in stock. This method is ideal for businesses selling perishable goods. The older inventory is likely to spoil or become outdated.
This method guarantees that the oldest inventory is sold first. FIFO also provides a better representation of the current inventory value when prices are rising.
Last-In, First-Out (LIFO)
The LIFO method assumes that the newest inventory is sold first, and the oldest inventory remains in stock. This method is ideal for businesses that sell non-perishable goods. Such goods do not decrease in value over time. Therefore, older inventory does not have a major effect on the overall value of the inventory.
LIFO is advantageous for businesses during times of inflation. This is because it reduces taxable income. It achieves this by pairing the most recent, and thus higher-priced, inventory with the income from sales.
Weighted Average Cost
The weighted average cost method calculates the average cost of all inventory items.
This is done by dividing the total cost of goods for sale by the total units available. This method can benefit businesses with many inventory items or frequent price changes, as it smooths out price fluctuations over time and provides a stable average cost per unit.
How to Implement Retail Inventory Method
Implementing this involves several steps, including:
Choosing a software program or system to perform the necessary calculations and record-keeping.
Establishing a regular schedule for inventory counts and adjustments.
Determining the cost and retail value of inventory items.
Calculating the average cost to retail percentage.
Recording and tracking inventory purchases, sales, and adjustments.
It’s also important to consider factors such as shrinkage, seasonality, and customer demand when implementing the retail inventory method.
By following best practices and ensuring accuracy in inventory tracking and valuation, businesses can effectively utilize the retail inventory method to manage inventory and make informed decisions.
Conclusion
Inventory costing methods are important for inventory management. It is important to understand the advantages and disadvantages of each method. This will help businesses to decide which method is the best for their needs. While the retail inventory method has advantages and disadvantages, it can provide real-time inventory valuation and facilitate effective decision-making.
Businesses can use a software program like Vencru to manage their inventory with the weighted average cost method. This method offers a steady average cost per unit. It is beneficial for businesses that have frequent price changes or a large number of inventory items.
Related: Retail Chart of Accounts, Inventory accounting: Key terms and valuation methods for retail, All about the retail inventory method
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StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
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Type I and Type II Errors and Statistical Power
Jacob Shreffler; Martin R. Huecker.
Author Information and Affiliations
Authors
Jacob Shreffler 1; Martin R. Huecker 2.
Affiliations
1 University of Louisville School of Medicine
2 University of Louisville
Last Update: March 13, 2023.
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Definition/Introduction
Healthcare professionals, when determining the impact of patient interventions in clinical studies or research endeavors that provide evidence for clinical practice, must distinguish well-designed studies with valid results from studies with research design or statistical flaws. This topic helps providers determine the likelihood of type I or type II errors and judge the adequacy of statistical power (see Table. Type I and Type II Errors and Statistical Power). Then, one can decide whether or not the evidence provided should be implemented in practice or used to guide future studies.
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Issues of Concern
Understanding the concepts discussed in this topic allows healthcare providers to accurately and thoroughly assess the results and validity of medical research. Without understanding type I and II errors and power analysis, clinicians could make poor clinical decisions without evidence to support them.
Type I and Type II Errors
Type I and Type II errors can lead to confusion as providers assess medical literature. A vignette that illustrates the errors is The Boy Who Cried Wolf. First, the citizens commit a type I error by believing there is a wolf when there is not. Second, the citizens commit a type II error by believing there is no wolf when there is one. A type I error occurs when, in research, we reject the null hypothesis and erroneously state that the study found significant differences when there was no difference. In other words, it is equivalent to saying that the groups or variables differ when, in fact, they do not or have false positives.
Here is a sample research hypothesis: Drug 23 will significantly reduce symptoms associated with Disease A compared to Drug 22. If we were to state that Drug 23 significantly reduced symptoms of Disease A compared to Drug 22 when it did not, this would be a type I error. Committing a type I error can be very grave in specific scenarios. For example, if we did move ahead with Drug 23 based on our research findings, even though there was no difference between groups, and the drug costs significantly more money for patients or has more side effects, then we would raise healthcare costs, cause iatrogenic harm, and not improve clinical outcomes. If a p-value is used to examine type I error, the lower the p-value, the lower the likelihood of the type I error to occur.
A type II error occurs when we declare no differences or associations between study groups when, in fact, there were.As with type I errors, type II errors in certain cause problems. Picture an example with a new, less invasive surgical technique developed and tested compared to the more invasive standard care. Researchers would seek to show no differences between patients receiving the 2 treatment methods in health outcomes (noninferiority study). If, however, the less invasive procedure resulted in less favorable health outcomes, it would be a severe error. Table 1 provides a depiction of type I and type II errors. (See Type I and Type II Errors and Statistical Power Table 1)
Power
A concept closely aligned to type II error is statistical power. Statistical power is a crucial part of the research process that is most valuable in the design and planning phases of studies, though it requires assessment when interpreting results. Power is the ability to reject a null hypothesis that is indeed false correctly.Unfortunately, many studies lack sufficient power and should be presented as having inconclusive findings.Power is the probability of a study to make correct decisions or detect an effect when one exists.
The power of a statistical test depends on the significance level set by the researcher, the sample size, and the effect size or the extent to which the groups differ based on treatment.Statistical power is critical for healthcare providers to decide how many patients to enroll in clinical studies.Power is strongly associated with sample size; when the sample size is large, power is generally not an issue.Thus, when conducting a study with a low sample size and, ultimately, low power, researchers should be aware of the likelihood of a type II error. The greater the N within a study, the more likely a researcher will reject the null hypothesis. The concern with this approach is that a very large sample could show a statistically significant finding due to the ability to detect small differences in the dataset; thus, using p values alone based on a large sample can be troublesome.
It is essential to recognize that power can be deemed adequate with a smaller sample if the effect size is large.What is an acceptable level of power? Many researchers agree that a power of 80% or higher is credible enough to determine the effects of research studies.Ultimately, studies with lower power find fewer true effects than studies with higher power; thus, clinicians should be aware of the likelihood of a power issue resulting in a type II error.Unfortunately, many researchers and providers who assess medical literature do not scrutinize power analyses. Studies with low power may inhibit future work as they cannot detect actual effects with variables; this could lead to potential impacts remaining undiscovered or noted as ineffective when they may be.
Medical researchers should invest time conducting power analyses to distinguish a difference or association sufficiently.Luckily, there are many tables of power values and statistical software packages that can help determine study power and guide researchers in study design and analysis. If choosing statistical software to calculate power, the following are necessary for entry: the predetermined alpha level, proposed sample size, and effect size the investigator(s) aims to detect.By utilizing power calculations on the front end, researchers can determine an adequate sample size to compute effect and determine, based on statistical findings, that sufficient power was observed.
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Clinical Significance
By limiting type I and type II errors, healthcare providers can ensure patients' decisions based on research outputs are safe.Additionally, while power analysis can be time-consuming, making inferences on low-powered studies can be inaccurate and irresponsible. By utilizing adequately designed studies, balancing the likelihood of type I and type II errors, and understanding power, providers, and researchers can determine which studies are clinically significant and should be implemented into practice.
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Nursing, Allied Health, and Interprofessional Team Interventions
All physicians, nurses, pharmacists, and other healthcare professionals should strive to understand the concepts of Type I and II errors and power. These individuals should maintain the ability to review and incorporate new literature for evidence-based and safe care. They also more effectively work in teams with other professionals.
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Review Questions
Access free multiple choice questions on this topic.
Comment on this article.
Figure
Type I and Type II Errors and Statistical Power Contributed by M Huecker, MD and J Shreffler, PhD
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References
1.
Lieberman MD, Cunningham WA. Type I and Type II error concerns in fMRI research: re-balancing the scale. Soc Cogn Affect Neurosci. 2009 Dec;4(4):423-8. [PMC free article: PMC2799956] [PubMed: 20035017]
2.
Singh G. A shift from significance test to hypothesis test through power analysis in medical research. J Postgrad Med. 2006 Apr-Jun;52(2):148-50. [PubMed: 16679686]
3.
Bezeau S, Graves R. Statistical power and effect sizes of clinical neuropsychology research. J Clin Exp Neuropsychol. 2001 Jun;23(3):399-406. [PubMed: 11419453]
4.
Keen HI, Pile K, Hill CL. The prevalence of underpowered randomized clinical trials in rheumatology. J Rheumatol. 2005 Nov;32(11):2083-8. [PubMed: 16265683]
5.
Gaskin CJ, Happell B. Power, effects, confidence, and significance: an investigation of statistical practices in nursing research. Int J Nurs Stud. 2014 May;51(5):795-806. [PubMed: 24207028]
6.
Algina J, Olejnik S. Conducting power analyses for ANOVA and ANCOVA in between-subjects designs. Eval Health Prof. 2003 Sep;26(3):288-314. [PubMed: 12971201]
7.
Ueki C, Sakaguchi G. Importance of Awareness of Type II Error. Ann Thorac Surg. 2018 Jan;105(1):333. [PubMed: 29233341]
8.
Leon AC. Multiplicity-adjusted sample size requirements: a strategy to maintain statistical power with Bonferroni adjustments. J Clin Psychiatry. 2004 Nov;65(11):1511-4. [PubMed: 15554764]
Disclosure:Jacob Shreffler declares no relevant financial relationships with ineligible companies.
Disclosure:Martin Huecker declares no relevant financial relationships with ineligible companies.
Definition/Introduction
Issues of Concern
Clinical Significance
Nursing, Allied Health, and Interprofessional Team Interventions
Review Questions
References
Copyright © 2025, StatPearls Publishing LLC.
This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( which permits others to distribute the work, provided that the article is not altered or used commercially. You are not required to obtain permission to distribute this article, provided that you credit the author and journal.
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Black Bear Pass, between Silverton and Telluride, CO. If you rent a jeep, you must sign a paper stating you will not take it over Black Bear. They say you must be crazy to drive it, so who better to ride with, than my husband, Cuz! At one point as I was video taping, he had to remind me to breath. He'd been going out west the past 18 years, how was I to know it was only his 2nd time on the pass! At the top, Black Bear sits at 12,800' & overlooks Telluride nestled in the valley 4000' below you.
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8553 | https://www.tiger-algebra.com/drill/4x-3y-2=0/ | Copyright Ⓒ 2013-2025
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1.1 Solve 4x-3y-2 = 0
Tiger recognizes that we have here an equation of a straight line. Such an equation is usually written y=mx+b ("y=mx+c" in the UK).
"y=mx+b" is the formula of a straight line drawn on Cartesian coordinate system in which "y" is the vertical axis and "x" the horizontal axis.
In this formula :
y tells us how far up the line goes
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m is the Slope or Gradient i.e. how steep the line is
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The X and Y intercepts and the Slope are called the line properties. We shall now graph the line 4x-3y-2 = 0 and calculate its properties
Graph of a Straight Line :
Calculate the Y-Intercept :
Notice that when x = 0 the value of y is 2/-3 so this line "cuts" the y axis at y=-0.66667
Calculate the X-Intercept :
When y = 0 the value of x is 1/2 Our line therefore "cuts" the x axis at x= 0.50000
Calculate the Slope :
Slope is defined as the change in y divided by the change in x. We note that for x=0, the value of y is -0.667 and for x=2.000, the value of y is 2.000. So, for a change of 2.000 in x (The change in x is sometimes referred to as "RUN") we get a change of 2.000 - (-0.667) = 2.667 in y. (The change in y is sometimes referred to as "RISE" and the Slope is m = RISE / RUN)
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Law of Sines
The Law of Sines is a useful identity in a triangle, which, along with the law of cosines and the law of tangents can be used to determine sides and angles. The law of sines can also be used to determine the circumradius, another useful function.
Contents
[hide]
1 Statement
2 Proof
2.1 Method 1
2.2 Method 2
3 Method 3
4 Problems
4.1 Introductory
4.2 Intermediate
4.3 Olympiad
5 See Also
Statement
In triangle , where is the side opposite to , opposite to , opposite to , and where is the circumradius:
Proof
Method 1
In the diagram above, point is the circumcenter of . Point is on such that is perpendicular to . Since , and . But making . We can use simple trigonometry in right triangle to find that
The same holds for and , thus establishing the identity.
Method 2
This method only works to prove the regular (and not extended) Law of Sines.
The formula for the area of a triangle is .
Since it doesn't matter which sides are chosen as , , and , the following equality holds:
Assuming the triangle in question is nondegenerate, . Multiplying the equation by yields:
Method 3
We can circumvent some of the work in Method 1 by setting up the circle in a different way. Let be a diameter and be the center of the circle, and let be on . Furthermore, let , and let , , and .
We have that is a right angle, as is a diameter. Therefore, , so, rearranging, we have , or . Likewise, . Finally, we observe that , so evidently . Combining all three equalities,
Problems
Introductory
If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?
(Source)
Intermediate
Triangle has sides , , and of length 43, 13, and 48, respectively. Let be the circlecircumscribed around and let be the intersection of and the perpendicular bisector of that is not on the same side of as . The length of can be expressed as , where and are positive integers and is not divisible by the square of any prime. Find the greatest integer less than or equal to .
(Source)
Olympiad
Let be a convex quadrilateral with , , and let be the intersection point of its diagonals. Prove that if and only if .
(Source)
See Also
Trigonometry
Trigonometric identities
Geometry
Law of Cosines
Retrieved from "
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8555 | https://www.som.org.uk/sites/som.org.uk/files/NICE_guideline_menopause.pdf | Menopause: diagnosis and management Menopause: diagnosis and management NICE guideline Published: 12 November 2015 nice.org.uk/guidance/ng23 © NICE 2017. All rights reserved. Subject to Notice of rights (
Y Your responsibility our responsibility The recommendations in this guideline represent the view of NICE, arrived at after careful consideration of the evidence available. When exercising their judgement, professionals and practitioners are expected to take this guideline fully into account, alongside the individual needs, preferences and values of their patients or the people using their service. It is not mandatory to apply the recommendations, and the guideline does not override the responsibility to make decisions appropriate to the circumstances of the individual, in consultation with them and their families and carers or guardian.
Local commissioners and providers of healthcare have a responsibility to enable the guideline to be applied when individual professionals and people using services wish to use it. They should do so in the context of local and national priorities for funding and developing services, and in light of their duties to have due regard to the need to eliminate unlawful discrimination, to advance equality of opportunity and to reduce health inequalities. Nothing in this guideline should be interpreted in a way that would be inconsistent with complying with those duties.
Commissioners and providers have a responsibility to promote an environmentally sustainable health and care system and should assess and reduce the environmental impact of implementing NICE recommendations wherever possible.
Menopause: diagnosis and management (NG23) © NICE 2017. All rights reserved. Subject to Notice of rights (
Page 2 of 31 Contents Contents Overview ................................................................................................................................................................................ 4 Who is it for? ......................................................................................................................................................................................
4 Recommendations.............................................................................................................................................................. 5 1.1 Individualised care ...................................................................................................................................................................
5 1.2 Diagnosis of perimenopause and menopause ..............................................................................................................
5 1.3 Information and advice ..........................................................................................................................................................
6 1.4 Managing short-term menopausal symptoms..............................................................................................................
7 1.5 Long-term benefits and risks of hormone replacement therapy........................................................................... 11 1.6 Diagnosing and managing premature ovarian insufficiency ................................................................................... 17 Terms used in this guideline......................................................................................................................................................... 19 Menopause implementation: getting started..........................................................................................................21 The challenge: stopping the use of follicle-stimulating hormone tests to diagnose menopause in women aged over 45 years........................................................................................................................................................................... 21 The challenge: communicating the long-term benefits and risks of hormone replacement therapy............. 22 The challenge: providing enough specialist services ......................................................................................................... 24 Need more help?............................................................................................................................................................................... 25 Context....................................................................................................................................................................................26 Recommendations for research....................................................................................................................................28 1 Women with breast cancer ...................................................................................................................................................... 28 2 Effects of HRT on breast cancer risk .................................................................................................................................... 28 3 Effects of HRT on venous thromboembolism risk .......................................................................................................... 29 4 Effects of HRT on dementia risk............................................................................................................................................. 29 5 Premature ovarian insufficiency............................................................................................................................................ 30 Menopause: diagnosis and management (NG23) © NICE 2017. All rights reserved. Subject to Notice of rights (
Page 3 of 31 This guideline is the basis of QS143.
Ov Overview erview This guideline covers the diagnosis and management of menopause, including in women who have premature ovarian insufficiency. The guideline aims to improve the consistency of support and information provided to women in menopause.
Who is it for?
Healthcare professionals who care for women in menopause.
Women in menopause, and their families and carers.
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Page 4 of 31 Recommendations Recommendations People have the right to be involved in discussions and make informed decisions about their care, as described in your care.
Making decisions using NICE guidelines explains how we use words to show the strength of our recommendations, and has information about safeguarding, consent and prescribing medicines (including 'off-label' use).
1.1 Individualised care 1.1.1 Adopt an individualised approach at all stages of diagnosis, investigation and management of menopause. Follow recommendations in the NICE guideline on patient experience in adult NHS services.
1.2 Diagnosis of perimenopause and menopause 1.2.1 Diagnose the following without laboratory tests in otherwise healthy women aged over 45 years with menopausal symptoms: perimenopause based on vasomotor symptoms and irregular periods menopause in women who have not had a period for at least 12 months and are not using hormonal contraception menopause based on symptoms in women without a uterus.
1.2.2 Take into account that it can be difficult to diagnose menopause in women who are taking hormonal treatments, for example for the treatment of heavy periods.
1.2.3 Do not use the following laboratory and imaging tests to diagnose perimenopause or menopause in women aged over 45 years: anti-Müllerian hormone inhibin A inhibin B oestradiol Menopause: diagnosis and management (NG23) © NICE 2017. All rights reserved. Subject to Notice of rights (
Page 5 of 31 antral follicle count ovarian volume.
1.2.4 Do not use a serum follicle-stimulating hormone (FSH) test to diagnose menopause in women using combined oestrogen and progestogen contraception or high-dose progestogen.
1.2.5 Consider using a FSH test to diagnose menopause only: in women aged 40 to 45 years with menopausal symptoms, including a change in their menstrual cycle in women aged under 40 years in whom menopause is suspected (see also section 1.6).
1.3 Information and advice 1.3.1 Give information to menopausal women and their family members or carers (as appropriate) that includes: an explanation of the stages of menopause common symptoms (see recommendation 1.3.2) and diagnosis lifestyle changes and interventions that could help general health and wellbeing benefits and risks of treatments for menopausal symptoms long-term health implications of menopause.
1.3.2 Explain to women that as well as a change in their menstrual cycle they may experience a variety of symptoms associated with menopause, including: vasomotor symptoms (for example, hot flushes and sweats) musculoskeletal symptoms (for example, joint and muscle pain) effects on mood (for example, low mood) urogenital symptoms (for example, vaginal dryness) sexual difficulties (for example, low sexual desire).
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Page 6 of 31 1.3.3 Give information to menopausal women and their family members or carers (as appropriate) about the following types of treatment for menopausal symptoms: hormonal, for example hormone replacement therapy (HRT) non-hormonal, for example clonidine non-pharmaceutical, for example cognitive behavioural therapy (CBT).
1.3.4 Give information on menopause in different ways to help encourage women to discuss their symptoms and needs.
1.3.5 Give information about contraception to women who are in the perimenopausal and postmenopausal phase. See guidance from the Faculty of Sexual & Reproductive Healthcare on contraception for women aged over 40 years.
1.3.6 Offer women who are likely to go through menopause as a result of medical or surgical treatment (including women with cancer, at high risk of hormone-sensitive cancer or having gynaecological surgery) support and: information about menopause and fertility before they have their treatment referral to a healthcare professional with expertise in menopause.
1.4 Managing short-term menopausal symptoms The recommendations in this section are not intended for women with premature ovarian insufficiency (see recommendations 1.6.6 to 1.6.8 for management of premature ovarian insufficiency).
1.4.1 Adapt a woman's treatment as needed, based on her changing symptoms.
V Vasomotor symptoms asomotor symptoms 1.4.2 Offer women HRT for vasomotor symptoms after discussing with them the short-term (up to 5 years) and longer-term benefits and risks. Offer a choice of preparations as follows: oestrogen and progestogen to women with a uterus Menopause: diagnosis and management (NG23) © NICE 2017. All rights reserved. Subject to Notice of rights (
Page 7 of 31 oestrogen alone to women without a uterus.
1.4.3 Do not routinely offer selective serotonin reuptake inhibitors (SSRIs), serotonin and norepinephrine reuptake inhibitors (SNRIs) or clonidine as first-line treatment for vasomotor symptoms alone.
1.4.4 Explain to women that there is some evidence that isoflavones or black cohosh may relieve vasomotor symptoms. However, explain that: multiple preparations are available and their safety is uncertain different preparations may vary interactions with other medicines have been reported.
Psy Psychological symptoms chological symptoms 1.4.5 Consider HRT to alleviate low mood that arises as a result of the menopause.
1.4.6 Consider CBT to alleviate low mood or anxiety that arise as a result of the menopause.
1.4.7 Ensure that menopausal women and healthcare professionals involved in their care understand that there is no clear evidence for SSRIs or SNRIs to ease low mood in menopausal women who have not been diagnosed with depression (see the NICE guideline on depression in adults).
Altered se Altered sexual function xual function 1.4.8 Consider testosterone supplementation for menopausal women with low sexual desire if HRT alone is not effective.
Urogenital atroph Urogenital atrophy y 1.4.9 Offer vaginal oestrogen to women with urogenital atrophy (including those on systemic HRT) and continue treatment for as long as needed to relieve symptoms.
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Page 8 of 31 1.4.10 Consider vaginal oestrogen for women with urogenital atrophy in whom systemic HRT is contraindicated, after seeking advice from a healthcare professional with expertise in menopause.
1.4.11 If vaginal oestrogen does not relieve symptoms of urogenital atrophy, consider increasing the dose after seeking advice from a healthcare professional with expertise in menopause.
1.4.12 Explain to women with urogenital atrophy that: symptoms often come back when treatment is stopped adverse effects from vaginal oestrogen are very rare they should report unscheduled vaginal bleeding to their GP.
1.4.13 Advise women with vaginal dryness that moisturisers and lubricants can be used alone or in addition to vaginal oestrogen.
1.4.14 Do not offer routine monitoring of endometrial thickness during treatment for urogenital atrophy.
Complementary ther Complementary therapies and unregulated prepar apies and unregulated preparations ations 1.4.15 Explain to women that the efficacy and safety of unregulated compounded bioidentical hormones are unknown.
1.4.16 Explain to women who wish to try complementary therapies that the quality, purity and constituents of products may be unknown.
1.4.17 Advise women with a history of, or at high risk of, breast cancer that, although there is some evidence that St John's wort may be of benefit in the relief of vasomotor symptoms, there is uncertainty about: appropriate doses persistence of effect variation in the nature and potency of preparations Menopause: diagnosis and management (NG23) © NICE 2017. All rights reserved. Subject to Notice of rights (
Page 9 of 31 potential serious interactions with other drugs (including tamoxifen, anticoagulants and anticonvulsants).
Re Review and referr view and referral al 1.4.18 Discuss with women the importance of keeping up to date with nationally recommended health screening.
1.4.19 Review each treatment for short-term menopausal symptoms: at 3 months to assess efficacy and tolerability annually thereafter unless there are clinical indications for an earlier review (such as treatment ineffectiveness, side effects or adverse events).
1.4.20 Refer women to a healthcare professional with expertise in menopause if treatments do not improve their menopausal symptoms or they have ongoing troublesome side effects.
1.4.21 Consider referring women to a healthcare professional with expertise in menopause if: they have menopausal symptoms and contraindications to HRT or or there is uncertainty about the most suitable treatment options for their menopausal symptoms.
Starting and stopping HR Starting and stopping HRT T 1.4.22 Explain to women with a uterus that unscheduled vaginal bleeding is a common side effect of HRT within the first 3 months of treatment but should be reported at the 3-month review appointment, or promptly if it occurs after the first 3 months (see recommendations on endometrial cancer in the NICE guideline on suspected cancer).
1.4.23 Offer women who are stopping HRT a choice of gradually reducing or immediately stopping treatment.
1.4.24 Explain to women that: Menopause: diagnosis and management (NG23) © NICE 2017. All rights reserved. Subject to Notice of rights (
Page 10 of 31 gradually reducing HRT may limit recurrence of symptoms in the short term gradually reducing or immediately stopping HRT makes no difference to their symptoms in the longer term.
W Women with, or at high risk of omen with, or at high risk of, breast cancer , breast cancer 1.4.25 For advice on the treatment of menopausal symptoms in women with breast cancer or at high risk of breast cancer, see section 1.13 of the NICE guideline on early and locally advanced breast cancer and section 1.7 of the NICE guideline on familial breast cancer.
1.4.26 Offer menopausal women with, or at high risk of, breast cancer: information on all available treatment options information that the SSRIs paroxetine and fluoxetine should not be offered to women with breast cancer who are taking tamoxifen referral to a healthcare professional with expertise in menopause.
1.5 Long-term benefits and risks of hormone replacement therapy V Venous thromboembolism enous thromboembolism 1.5.1 Explain to women that: the risk of venous thromboembolism (VTE) is increased by oral HRT compared with baseline population risk the risk of VTE associated with HRT is greater for oral than transdermal preparations the risk associated with transdermal HRT given at standard therapeutic doses is no greater than baseline population risk.
1.5.2 Consider transdermal rather than oral HRT for menopausal women who are at increased risk of VTE, including those with a BMI over 30 kg/m 2.
1.5.3 Consider referring menopausal women at high risk of VTE (for example, those with a strong family history of VTE or a hereditary thrombophilia) to a haematologist for assessment before considering HRT.
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Page 11 of 31 Cardio Cardiovascular disease vascular disease 1.5.4 Ensure that menopausal women and healthcare professionals involved in their care understand that HRT: does not increase cardiovascular disease risk when started in women aged under 60 years does not affect the risk of dying from cardiovascular disease.
1.5.5 Be aware that the presence of cardiovascular risk factors is not a contraindication to HRT as long as they are optimally managed.
1.5.6 Using tables 1 and 2, explain to women that: the baseline risk of coronary heart disease and stroke for women around menopausal age varies from one woman to another according to the presence of cardiovascular risk factors HRT with oestrogen alone is associated with no, or reduced, risk of coronary heart disease HRT with oestrogen and progestogen is associated with little or no increase in the risk of coronary heart disease.
1.5.7 Explain to women that taking oral (but not transdermal) oestrogen is associated with a small increase in the risk of stroke. Also explain that the baseline population risk of stroke in women aged under 60 years is very low (see table 2).
T Table 1 Absolute r able 1 Absolute rates of coronary heart disease for different types of HR ates of coronary heart disease for different types of HRT T compared with no HR compared with no HRT ( T (or placebo or placebo), different dur ), different durations of HR ations of HRT use and time since T use and time since stopping HR stopping HRT for menopausal women T for menopausal women Difference in coronary heart disease incidence per 1000 Difference in coronary heart disease incidence per 1000 menopausal women o menopausal women ov ver 7.5 y er 7.5 years (95% confidence interval) ears (95% confidence interval) (baseline population risk in the UK o (baseline population risk in the UK ov ver 7.5 y er 7.5 years: 26.3 per ears: 26.3 per 1000 1000 1) ) Menopause: diagnosis and management (NG23) © NICE 2017. All rights reserved. Subject to Notice of rights (
Page 12 of 31 Current HRT users Treatment duration <5 years Treatment duration 5–10 years >5 years since stopping treatment RCT estimate 2 6 fewer (-10 to 1) No available data No available data 6 fewer (-9 to -2) Women on oestrogen alone Observational estimate 3 6 fewer (-9 to -3) No available data No available data No available data RCT estimate 2 5 more (-3 to 18) No available data No available data 4 more (-1 to 11) Women on oestrogen + progestogen Observational estimate 3 No available data No available data No available data No available data HRT, hormone replacement therapy; RCT, randomised controlled trial For full source references, see Appendix M in the full guideline.
1 Results from Weiner 2008 were used for the baseline population risk estimation.
2 For women aged 50–59 years at entry to the RCT.
3 Observational estimates are based on cohort studies with several thousand women.
T Table 2 Absolute r able 2 Absolute rates of strok ates of stroke for different types of HR e for different types of HRT compared with no HR T compared with no HRT ( T (or or placebo placebo), different dur ), different durations of HR ations of HRT use and time since stopping HR T use and time since stopping HRT for T for menopausal women menopausal women Difference in strok Difference in stroke incidence per 1000 menopausal women e incidence per 1000 menopausal women o ov ver 7.5 y er 7.5 years (95% confidence interval) (baseline population ears (95% confidence interval) (baseline population risk in the UK o risk in the UK ov ver 7.5 y er 7.5 years: 11.3 per 1000 ears: 11.3 per 1000 1) Current HRT users Treatment duration <5 years Treatment duration 5–10 years >5 years since stopping treatment RCT estimate 2 0 (-5 to 10) No available data No available data 1 more (-4 to 9) Women on oestrogen alone Observational estimate 3 3 more (-1 to 8) No available data No available data No available data Menopause: diagnosis and management (NG23) © NICE 2017. All rights reserved. Subject to Notice of rights (
Page 13 of 31 RCT estimate 2 6 more (-2 to 21) No available data No available data 4 more (-1 to 13) Women on oestrogen + progestogen Observational estimate 3 4 more (1 to 7) No available data No available data No available data HRT, hormone replacement therapy; RCT, randomised controlled trial For full source references, see Appendix M in the full guideline.
1 Results from Weiner 2008 were used for the baseline population risk estimation.
2 For women aged 50–59 years at entry to the RCT.
3Observational estimates are based on cohort studies with several thousand women.
T Type 2 diabetes ype 2 diabetes 1.5.8 Explain to women that taking HRT (either orally or transdermally) is not associated with an increased risk of developing type 2 diabetes.
1.5.9 Ensure that women with type 2 diabetes and all healthcare professionals involved in their care are aware that HRT is not generally associated with an adverse effect on blood glucose control.
1.5.10 Consider HRT for menopausal symptoms in women with type 2 diabetes after taking comorbidities into account and seeking specialist advice if needed.
Breast cancer Breast cancer 1.5.11 Using table 3, explain to women around the age of natural menopause that: the baseline risk of breast cancer for women around menopausal age varies from one woman to another according to the presence of underlying risk factors HRT with oestrogen alone is associated with little or no change in the risk of breast cancer HRT with oestrogen and progestogen can be associated with an increase in the risk of breast cancer any increase in the risk of breast cancer is related to treatment duration and reduces after stopping HRT.
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Page 14 of 31 T Table 3 Absolute r able 3 Absolute rates of breast cancer for different types of HR ates of breast cancer for different types of HRT compared with no T compared with no HR HRT ( T (or placebo or placebo), different dur ), different durations of HR ations of HRT use and time since stopping HR T use and time since stopping HRT for T for menopausal women menopausal women Difference in breast cancer incidence per 1000 menopausal Difference in breast cancer incidence per 1000 menopausal women o women ov ver 7.5 y er 7.5 years (95% confidence interval) (baseline ears (95% confidence interval) (baseline population risk in the UK o population risk in the UK ov ver 7.5 y er 7.5 years: 22.48 per 1000 ears: 22.48 per 1000 1 1) ) Current HRT users Treatment duration <5 years Treatment duration 5–10 years >5 years since stopping treatment RCT estimate 2 4 fewer (-11 to 8) No available data No available data 5 fewer (-11 to 2) Women on oestrogen alone Observational estimate 3 6 more (1 to 12) 4 4 more (1 to 9) 5 more (-1 to 14) 5 fewer (-9 to -1) RCT estimate 2 5 more (-4 to 36) No available data No available data 8 more (1 to 17) Women on oestrogen + progestogen Observational estimate 3 17 more (14 to 20) 12 more (6 to 19) 21 more (9 to 37) 9 fewer (-16 to 13) 5 HRT, hormone replacement therapy; RCT, randomised controlled trial For full source references, see Appendix M in the full guideline.
1 Office for National Statistics (2010) breast cancer incidence statistics.
2 For women aged 50–59 years at entry to the RCT.
3Observational estimates are based on cohort studies with several thousand women.
4 Evidence on observational estimate demonstrated very serious heterogeneity without plausible explanation by subgroup analysis.
5 Evidence on observational estimate demonstrated very serious imprecision in the estimate of effect.
Osteoporosis Osteoporosis 1.5.12 Give women advice on bone health and discuss these issues at review appointments (see the NICE guideline on osteoporosis: assessing the risk of fragility fracture).
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Page 15 of 31 1.5.13 Using table 4, explain to women that the baseline population risk of fragility fracture for women around menopausal age in the UK is low and varies from one woman to another.
1.5.14 Using table 4, explain to women that their risk of fragility fracture is decreased while taking HRT and that this benefit: is maintained during treatment but decreases once treatment stops may continue for longer in women who take HRT for longer.
T Table 4 Absolute r able 4 Absolute rates of an ates of any fr y fragility fr agility fracture for HR acture for HRT compared with no HR T compared with no HRT ( T (or or placebo placebo), different dur ), different durations of HR ations of HRT use and time since stopping HR T use and time since stopping HRT for T for menopausal women menopausal women Difference in an Difference in any fr y fragility fr agility fracture incidence per 1000 menopausal acture incidence per 1000 menopausal women (95% confidence interval) (see footnotes for information on women (95% confidence interval) (see footnotes for information on baseline population risk and length of follow-up time o baseline population risk and length of follow-up time ov ver which er which absolute risk difference is calculated) absolute risk difference is calculated) Current HRT users Treatment duration <5 years Treatment duration 5–10 years >5 years since stopping treatment RCT estimate 1 23 fewer (-10 to -33) 3 25 fewer (-9 to -37) 4 No available data No available data Women on any HRT Observational estimate 2 16 fewer (-15 to -18) 5 15 fewer (-11 to -17) 5 18 fewer (-15 to -20) 5 2 more (-19 to 27) 6 Menopause: diagnosis and management (NG23) © NICE 2017. All rights reserved. Subject to Notice of rights (
Page 16 of 31 HRT, hormone replacement therapy; RCT, randomised controlled trial For full source references, see Appendix M in the full guideline.
Absolute risks calculated by using the baseline population risk in the control arm for each analysis, following GRADE methodology.
1 For women aged 50–59 years at entry to the RCT.
2 Observational estimate is based on cohort studies with several thousand women.
3 Baseline population risk = 69 per 1000 women (follow-up: 3.43 years).
4 Baseline population risk = 78 per 1000 women (follow-up: 3.71 years).
5 Baseline population risk = 15.4 per 1000 women (follow-up: 2.8 years).
6 Baseline population risk = 106 per 1000 women (follow-up: 5 years).
Dementia Dementia 1.5.15 Explain to menopausal women that the likelihood of HRT affecting their risk of dementia is unknown.
L Loss of muscle mass and strength oss of muscle mass and strength 1.5.16 Explain to women that: there is limited evidence suggesting that HRT may improve muscle mass and strength muscle mass and strength is maintained through, and is important for, activities of daily living.
1.6 Diagnosing and managing premature ovarian insufficiency Diagnosing premature o Diagnosing premature ovarian insufficiency varian insufficiency 1.6.1 Take into account the woman's clinical history (for example, previous medical or surgical treatment) and family history when diagnosing premature ovarian insufficiency.
1.6.2 Diagnose premature ovarian insufficiency in women aged under 40 years based on: menopausal symptoms, including no or infrequent periods (taking into account whether the woman has a uterus) and and Menopause: diagnosis and management (NG23) © NICE 2017. All rights reserved. Subject to Notice of rights (
Page 17 of 31 elevated FSH levels on 2 blood samples taken 4–6 weeks apart.
1.6.3 Do not diagnose premature ovarian insufficiency on the basis of a single blood test.
1.6.4 Do not routinely use anti-Müllerian hormone testing to diagnose premature ovarian insufficiency.
1.6.5 If there is doubt about the diagnosis of premature ovarian insufficiency, refer the woman to a specialist with expertise in menopause or reproductive medicine.
Managing premature o Managing premature ovarian insufficiency varian insufficiency 1.6.6 Offer sex steroid replacement with a choice of HRT or a combined hormonal contraceptive to women with premature ovarian insufficiency, unless contraindicated (for example, in women with hormone-sensitive cancer).
1.6.7 Explain to women with premature ovarian insufficiency: the importance of starting hormonal treatment either with HRT or a combined hormonal contraceptive and continuing treatment until at least the age of natural menopause (unless contraindicated) that the baseline population risk of diseases such as breast cancer and cardiovascular disease increases with age and is very low in women aged under 40 that HRT may have a beneficial effect on blood pressure when compared with a combined oral contraceptive that both HRT and combined oral contraceptives offer bone protection that HRT is not a contraceptive.
1.6.8 Give women with premature ovarian insufficiency and contraindications to hormonal treatments advice, including on bone and cardiovascular health, and symptom management.
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Page 18 of 31 1.6.9 Consider referring women with premature ovarian insufficiency to healthcare professionals who have the relevant experience to help them manage all aspects of physical and psychosocial health related to their condition.
Terms used in this guideline Compounded bioidentical hormones Compounded bioidentical hormones Unregulated plant-derived hormonal combinations similar or identical to human hormones that are compounded by pharmacies to the specification of the prescriber.
F Fr ragility fr agility fracture acture Fractures that result from mechanical forces that would not ordinarily result in fracture (such as a fall from a standing height or less). Reduced bone density is a major risk factor for fragility fractures, which occur most commonly in the spine, hip and wrist.
L Low mood ow mood Mild depressive symptoms that impair quality of life but are usually intermittent and often associated with hormonal fluctuations in perimenopause.
Menopause Menopause A biological stage in a woman's life that occurs when she stops menstruating and reaches the end of her natural reproductive life. Usually it is defined as having occurred when a woman has not had a period for 12 consecutive months (for women reaching menopause naturally).
The changes associated with menopause occur when the ovaries stop maturing eggs and secreting oestrogen and progesterone.
Menopausal women Menopausal women This includes women in perimenopause and postmenopause.
P Perimenopause erimenopause The time in which a woman has irregular cycles of ovulation and menstruation leading up to menopause and continuing until 12 months after her final period. The perimenopause is also known as the menopausal transition or climacteric.
P Postmenopause ostmenopause The time after menopause has occurred, starting when a woman has not had a period for 12 consecutive months.
Premature o Premature ovarian insufficiency varian insufficiency Menopause occurring before the age of 40 years (also known as premature ovarian failure or premature menopause). It can occur naturally or as a result of medical or surgical treatment.
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Page 19 of 31 Urogenital atroph Urogenital atrophy y Thinning and shrinking of the tissues of the vulva, vagina, urethra and bladder caused by oestrogen deficiency. This results in multiple symptoms such as vaginal dryness, vaginal irritation, a frequent need to urinate and urinary tract infections.
V Vasomotor symptoms asomotor symptoms Menopausal symptoms such as hot flushes and night sweats caused by constriction and dilatation of blood vessels in the skin that can lead to a sudden increase in blood flow to allow heat loss. These symptoms can have a major impact on activities of daily living.
You can also see this guideline in the NICE pathway on menopause.
To find out what NICE has said on topics related to this guideline, see our web page on menopause.
At the time of publication (November 2015), testosterone did not have a UK marketing authorisation for this indication in women. The prescriber should follow relevant professional guidance, taking full responsibility for the decision. Informed consent should be obtained and documented. See the General Medical Council's Prescribing guidance: prescribing unlicensed medicines for further information.
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Page 20 of 31 Menopause implementation: getting started Menopause implementation: getting started This section highlights 3 areas of the menopause guideline that could have a big impact on practice and be challenging to implement, along with the reasons why we are proposing change in these areas (given in the box at the start of each area). We identified these with the help of stakeholders and Guideline Development Group members (see section 9.4 of the manual). The section also gives information on resources to help with implementation.
The challenge: stopping the use of follicle-stimulating hormone tests to diagnose menopause in women aged over 45 years See recommendation 1.2.5.
The follicle-stimulating hormone (FSH) test is often performed unnecessarily in women aged over 45 years. The evidence underpinning this guideline highlights that hormonal tests should not routinely be used in the diagnosis of menopause and that FSH tests should not be used in women aged over 45 years. This is because FSH fluctuates considerably over short periods of time during the years leading up to menopause and so blood levels are not a helpful addition to what is a clinical diagnosis. If a woman is aged over 45 years and has not had a period for at least 12 months, or has vasomotor symptoms and irregular periods (or just symptoms if she doesn't have a uterus), this is adequate information to diagnose menopause and perimenopause respectively. In younger women, FSH tests should not be used to diagnose menopause in those taking combined oestrogen and progestogen contraception or high-dose progestogen because these affect FSH measurements.
Carrying out this test in this group of women does not improve menopause management and so this is an area of care where considerable savings could be made through disinvestment.
Raising a Raising awareness of the need to change pr wareness of the need to change practice actice There may be staff in primary care services who do not know that FSH tests should not be carried out in women aged over 45 years.
To raise awareness, clinical commissioning groups, practice managers and lead GPs could: Use newsletters, bulletins and education events to help ensure that GPs and other practice staff (in particular practice nurses) are aware of this change in practice and that they understand the evidence underpinning this recommendation.
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Page 21 of 31 Add a prompt to electronic requesting systems which remind primary care staff that this test should not be requested for women aged over 45.
Refer GPs to NICE's clinical knowledge summary for menopause.
Use the NICE costing report and template to estimate the local savings that can be made. A sample calculation using this template showed that savings of £16,500 could be made for a population of 100,000.
Use the baseline assessment tool to establish current practice in requesting tests and carry out clinical audit so this can be monitored and improved.
Also, laboratory staff and managers could: Engage with their local GP practices. For example, in NHS Lothian a GP/laboratory liaison group meets regularly every 2 months and holds an annual update meeting to which all GPs are invited. This continuing professional development (CPD) accredited meeting provides a good opportunity to promote changes in practice.
Encourage GPs to stop requesting FSH tests for women aged over 45 years by drawing attention to the fact that this test is unlikely to be informative and is not recommended. Lab Tests Online UK and the UK National External Quality Assessment Service (UK NEQAS) are also raising awareness of the new NICE guidance.
The challenge: communicating the long-term benefits and risks of hormone replacement therapy See section 1.5.
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Page 22 of 31 It is important to provide information on the benefits and risks of hormone replacement therapy (HRT) to help women make an informed choice about which treatment to use for menopausal symptoms. Media reports about HRT have not always been accurate, so providing healthcare professionals and women with a robust source of information is vital. Before publication of this guideline there was no consensus about the long-term benefits and risks of HRT. Although the Women's Health Initiative found that HRT prevented osteoporotic fractures and colon cancer, it initially reported that HRT increased the risk of having a cardiovascular event as well as the incidence of breast cancer. However, the association between HRT and cardiovascular disease has since been disputed and the results show that the risk varies in accordance with individual factors. One of the aims of this guideline is to help GPs and other healthcare professionals to be more confident in prescribing HRT and women more confident in taking it. A knowledge gap among some GPs and other healthcare professionals could mean that they are reluctant to prescribe HRT because they overestimate the risks and contraindications, and underestimate the impact of menopausal symptoms on a woman's quality of life.
Impro Improving knowledge among healthcare professionals ving knowledge among healthcare professionals There is a need to improve knowledge about the long-term benefits and risks of HRT. No other treatment has been shown to be as effective as HRT for menopausal symptoms, though the balance of risks and benefits varies among women. Healthcare professionals need to be in a position to be able to support women to make an informed decision about individual benefits and risks of HRT.
NICE is working with the Royal College of Obstetricians and Gynaecologists to ensure that management of menopause, including the benefits and risks of HRT, is covered within the core curriculum. This includes supporting the update and promotion of the advanced training specialist module on menopause and the subspecialty training in reproductive medicine. We are also working with the Faculty of Sexual & Reproductive Healthcare (FSRH) to highlight the menopause special skills theory course and the basic and advanced special skills module.
To improve knowledge, clinical commissioning group prescribing leads could: Help to develop formularies of good HRT prescribing for GPs. This could be done with input from GPs with a specialist interest in menopause and interested consultant gynaecologists.
Use briefings and newsletters to help disseminate prescribing knowledge on HRT.
Also, GPs could: Menopause: diagnosis and management (NG23) © NICE 2017. All rights reserved. Subject to Notice of rights (
Page 23 of 31 Set up local meetings or teaching sessions (particularly those GPs with a specialist interest) to target interested GPs in each practice who can then take the information back to their partners.
Use the recommendations in section 1.5 of the guideline and the HRT section of NICE's clinical knowledge summary to ensure that they are informed of the actual long-term benefits and risks of HRT for each individual and are not basing decisions on perceived knowledge.
Use materials such as NICE's information for the public, NHS choices and the Women's Health Concern leaflet to help support women to make informed decisions when advising them about HRT.
Link with the menopause specialist in their area for advice. This could be by telephone or email about specific cases and/or through training delivered by the menopause specialist.
Complete the basic or advanced FSRH special skills course. Practice nurses may also complete this training.
The challenge: providing enough specialist services See guideline recommendations.
The number of women aged over 45 years in the UK has been steadily increasing and will continue to rise. The associated increase in the number of women going through menopause is expected to result in more new referrals to secondary care of both women needing short-term symptom control and those with associated long-term health issues. There is currently a lack of specialist services and their availability varies nationally. Throughout this guideline there are recommendations to refer certain women to a healthcare professional with expertise in menopause. Currently, there may not be enough services nationally to refer these women to.
Re Reviewing and redesigning local service pro viewing and redesigning local service provision vision In order to address variation and potential gaps in service provision, local health services may need to review, map and redesign local service provision.
To do this, commissioners and clinical commissioning groups could: Use Menopause UK's national menopause map as a starting point. This highlights variations in practice and the lack of overall provision.
Menopause: diagnosis and management (NG23) © NICE 2017. All rights reserved. Subject to Notice of rights (
Page 24 of 31 Clarify current referral routes and communicate them if they are effective.
Identify lead clinicians to drive a change in service provision if a gap is identified. Ideally all clinical commissioning groups should have a GP with a specialist interest or a community gynaecologist who could do this.
Establish whether current referrals are appropriate. These may be to secondary care (hospitals), community services or a GP with a specialist interest and will vary according to local facilities. Ideally, services should be provided by a dedicated menopause clinic.
Confirm that care is provided by a healthcare professional with expertise in menopause (for example, women with breast cancer should have access to a specialist menopause clinic or professional but often receive treatment for menopause from their oncologist who may not have the appropriate training).
Consider the feasibility of providing dedicated menopause support by setting up clinics within current gynaecology services.
Menopause clinics may be multispecialist and so jointly led by a nurse consultant and a consultant ensuring that when a member of staff is unavailable the clinic may still run.
Establish regional menopause clinics if services are unable to have their own.
Use the learning from examples of practice where successful services have been set up to help.
For example, a primary care service in Essex manages specialist clinic waiting lists through an established agreement whereby a GP with specialist interest accepts emails or written requests from all GPs within a clinical commissioning group. These requests are answered once a week. A specialist service in London has set up a helpline that receives calls outside of clinic times and can allow women to be given support and advice without the need for a clinic appointment.
Need more help?
Further resources are available from NICE that may help to support implementation: uptake data about guideline recommendations and quality standard measures the British Menopause Society provides information to healthcare professionals about menopause.
Menopause: diagnosis and management (NG23) © NICE 2017. All rights reserved. Subject to Notice of rights (
Page 25 of 31 Conte Context xt Menopause is when a woman stops having periods as she reaches the end of her natural reproductive life. This is not usually abrupt, but a gradual process during which women experience perimenopause before reaching postmenopause. The average age of menopause in the UK is 51.
However, this varies widely and 1 in 100 women experience premature ovarian insufficiency (menopause occurring before the age of 40 years).
Oestrogen depletion associated with menopause causes irregular periods and has many other effects on the body. The most common symptoms are hot flushes and night sweats. Other symptoms include mood changes, memory and concentration loss, vaginal dryness, a lack of interest in sex, headaches, and joint and muscle stiffness. Quality of life may be severely affected.
Most women (8 out of 10) experience some symptoms, typically lasting about 4 years after the last period, but continuing for up to 12 years in about 10% of women. Prolonged lack of oestrogen affects the bones and cardiovascular system and postmenopausal women are at increased risk of a number of long-term conditions, such as osteoporosis.
Around a million women in the UK use treatment for their menopausal symptoms. The advice and support available is variable, and use of hormone replacement therapy (HRT) – a highly successful treatment for common symptoms of menopause – varies with socioeconomic and cultural factors.
The number of prescriptions for HRT almost halved after the publication of 2 large studies: the Women's Health Initiative (2002) and the Million Women Study (2003). These studies focused on the use of HRT in chronic disease prevention and potential long-term risks rather than considering the benefits in terms of symptom relief. One of the aims of this NICE guideline was to clarify the balance of benefits and risks of HRT use for both women and their healthcare providers.
This guideline addresses the diagnosis and management of menopause. It covers women in perimenopause and postmenopause, and the particular needs of women with premature ovarian insufficiency and women with hormone-sensitive cancer (for example, breast cancer). The guideline concentrates on the clinical management of menopause-related symptoms, considers both pharmaceutical and non-pharmaceutical treatments, includes a health economic analysis, and reviews the benefits and adverse effects of HRT. It applies to all settings in which NHS services are provided.
At the time of publication (November 2015), the MHRA is consulting with marketing authorisation holders on amending the existing warning about the risk of ovarian cancer in the Summary of Product Characteristics (SPC) information for HRT products. The current core SPC Menopause: diagnosis and management (NG23) © NICE 2017. All rights reserved. Subject to Notice of rights (
Page 26 of 31 states that long-term use of oestrogen-only and combined oestrogen-progestogen HRT has been associated with a slightly increased risk of ovarian cancer.
Menopause: diagnosis and management (NG23) © NICE 2017. All rights reserved. Subject to Notice of rights (
Page 27 of 31 Recommendations for research Recommendations for research The Guideline Development Group has made the following recommendations for research. The full set of research recommendations is detailed in the full guideline.
1 Women with breast cancer What is the safety and effectiveness of alternatives to systemic HRT as treatments for menopausal symptoms in women who have had treatment for breast cancer?
Wh Why this is important y this is important Women with a history of breast cancer are rarely offered hormonal treatment for menopausal symptoms but the available alternatives are less effective. There is limited evidence from randomised controlled trials on the safety and effectiveness of options such as non-hormonal treatments, ospemifene, conjugated equine estrogen/bazedoxifene (CEE/BZA) or local vaginal oestrogen for menopausal symptoms in women who have had treatment for breast cancer. There is insufficient evidence on the efficacy and safety of non-pharmaceutical treatments in women with breast cancer and other hormone-sensitive conditions. Randomised controlled trials or large cohort studies are needed to understand the effects of these treatments in women with breast cancer, and to investigate if there is a difference in breast cancer recurrence, mortality and tumour aggression with different types of treatment.
2 Effects of HRT on breast cancer risk What is the difference in the risk of breast cancer in menopausal women on HRT with progesterone, progestogen or selective oestrogen receptor modulators?
Wh Why this is important y this is important Fear of breast cancer deters many women from taking HRT, even in the presence of debilitating menopausal symptoms. There is a lack of evidence from randomised controlled trials directly comparing the risk of breast cancer in menopausal women on HRT with progesterone, progestogen or selective oestrogen receptor modulators. There is a need for a national registry of women with breast cancer.
Menopause: diagnosis and management (NG23) © NICE 2017. All rights reserved. Subject to Notice of rights (
Page 28 of 31 Optimising the risk–benefit profile of HRT will potentially reduce morbidity and mortality from breast cancer in women who need HRT over the long term because of continuing menopausal symptoms.
3 Effects of HRT on venous thromboembolism risk How does the preparation of HRT affect the risk of venous thromboembolism (VTE)?
Wh Why this is important y this is important An increase in the risk of VTE (deep vein thrombosis [DVT] or pulmonary embolism [PE]) is a significant side effect of HRT, particularly because PEs can be fatal. This risk appears to be greater with oral than transdermal HRT. DVT risk increases with age and BMI, among other risk factors.
The progestogen component of HRT may also influence the risk of a DVT, which may be greater with androgenic synthetic progestogens than natural progesterone (but findings from observational studies need confirmation). Most women in the UK take oral HRT comprising oestrogen combined with a synthetic progestogen, and the use of progesterone is less common.
Randomised controlled trials are needed to compare oral with transdermal HRT, and HRT containing different types of progestogens. These trials should measure coagulation factors and the incidence of VTE in women at increased risk of VTE for whom transdermal oestrogen is indicated.
4 Effects of HRT on dementia risk What are the effects of early HRT use on the risk of dementia?
Wh Why this is important y this is important Concern about the prospect of dementia in older age is increasing and any beneficial effect on the future risk of dementia will be important to women who are considering using HRT. There is a need for good-quality observational studies controlling for the effect of important confounders on how early HRT use affects dementia risk in women with early natural menopause, including women with premature ovarian insufficiency.
Menopause: diagnosis and management (NG23) © NICE 2017. All rights reserved. Subject to Notice of rights (
Page 29 of 31 5 Premature ovarian insufficiency What are the main clinical manifestations of premature ovarian insufficiency and the short- and long-term impact of the most common therapeutic interventions?
Wh Why this is important y this is important Women with premature ovarian insufficiency can experience the effects of menopause for most of their adult life. This can lead to reduced quality of life and an increased risk of osteoporosis, cardiovascular disease and possibly dementia. There is uncertainty about the diagnosis, time course and management of premature ovarian insufficiency. For example, it is possible that different interventions produce different outcomes in terms of quality of life, and bone, cardiovascular and brain protection. Combined oral contraceptives are often prescribed when this might not be the best treatment in terms of quality of life and preservation of bone density and cardiovascular health. Short- and long-term outcomes of HRT versus combined hormonal contraceptives in women with premature ovarian insufficiency therefore need to be investigated.
Development of a collaborative premature ovarian insufficiency registry would allow the collection of high-quality demographic, biobank (genomic) and clinical data in order to clarify: the diagnosis and presentation of premature ovarian insufficiency the impact of therapeutic interventions such as combined hormonal contraceptives, HRT and androgens the long-term impact of premature ovarian insufficiency on bone density and fracture, and cardiovascular and cognitive health the long-term risk of cancer, which can be determined by linking with relevant cancer and mortality registries.
ISBN: 978-1-4731-1525-5 Menopause: diagnosis and management (NG23) © NICE 2017. All rights reserved. Subject to Notice of rights (
Page 30 of 31 Accreditation Menopause: diagnosis and management (NG23) © NICE 2017. All rights reserved. Subject to Notice of rights (
Page 31 of 31 |
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Altitudes of a tetrahedron
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$\begingroup$
Consider the following scenario.
The position vectors of the vertices A, B and C , of a tetrahedron are $(1,1,1),(1,0,0), (3,0,0)$ respectively. The altitude from the vertex D to the opposite face $\triangle ABC$ meets the median line through A of the $\triangle ABC$ at point E. The length of side AD is $4$ and volume of the tetrahedron is $\frac{2\sqrt{3}}{3}$.
I am thinking that we get a unique tetrahedron here and thus a unique position for E
But using the fact that E divides AD in ratio say $k:1$ and that $\triangle AED$ is a right triangle. I am getting two values of $k$.
Is one of them an extra solution? If yes, how do I reject the extra one? If no, can you tell how two positions are possible.
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edited Apr 6, 2021 at 17:25
TheBestMagician
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ishwar b bishwar b b
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$\endgroup$
6
$\begingroup$ Welcome to MSE. Please use MathJax to format your posts. To begin with, surround math expressions (including numbers) with $ signs and use _ for subscripts. $x_1$ comes out as $x_1$. $\endgroup$
saulspatz
– saulspatz
2021-04-06 16:44:16 +00:00
Commented Apr 6, 2021 at 16:44
$\begingroup$ I will keep that in mind $\endgroup$
ishwar b b
– ishwar b b
2021-04-06 16:55:30 +00:00
Commented Apr 6, 2021 at 16:55
$\begingroup$ It is not clear what the tetrahedron volume is. That is why you need to use mathjax. Is it $\frac{2}{3\sqrt3}$ or $\frac{2\sqrt3}{3}$? Also what is that you are trying to find? Can you write that clearly? Is it just the altitude from $D$ to $\triangle ABC$? $\endgroup$
Math Lover
– Math Lover
2021-04-06 17:01:49 +00:00
Commented Apr 6, 2021 at 17:01
3
$\begingroup$ if you think of $ABC$ plane, the vertex $D$ has two possible positions relative to the plane $\endgroup$
Vasili
– Vasili
2021-04-06 17:43:47 +00:00
Commented Apr 6, 2021 at 17:43
$\begingroup$ $AE^2=AD^2-DE^2$. But $AD=4$ and $DE=\ $altitude$\ =\sqrt6$. Hence $AE^2=10$. Where is the other solution? $\endgroup$
Intelligenti pauca
– Intelligenti pauca
2021-04-07 15:49:05 +00:00
Commented Apr 7, 2021 at 15:49
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What is meant by celibate?
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Dilip Bhatt (Dr.)
Ph.D. in English Language and Literature&English (language), Sardar Patel University (Graduated 2005) · Author has 3.6K answers and 4.9M answer views
·5y
Related
What is the antonym of celibate?
bad
cluttered
complex
complicated
corrupt
evil
imprecise
impure
intricate
sinful
stained
unchaste
defiled
dirty
lewd
unchaste
wanton
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9 2
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How can I be a celibate?
What are the benefits of celibacy?
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Does being a celibate mean I have no sexuality?
What are the differences between celibate and non-celibate people?
Ramakrishnan Annamalai
Author has 20.7K answers and 2.8M answer views
·5y
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What is the antonym of celibate?
Celibate(v).
Opposite:
Marry, materialize.
Don’t be solitary and marry to experience the material life.
Upvote ·
9 3
Ithamar Paraguassu Ramos
Works at Movies · Author has 3.5K answers and 6M answer views
·6y
Basically that you don't marry.
Some people confuses with abstinence, that means not having sex,
You can have sex and be in celibate.
Upvote ·
9 2
Sadasiva S
Author has 3.9K answers and 1.9M answer views
·5y
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What is the antonym of celibate?
celibate - noun , adjective.
A person who abstains from sexual relations; not married.
chaste; virgin; pure; virtuous.
ANTONYMS:
active
promiscuous
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Ashish Singh Rajput
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What is the antonym of celibate?
abstaining from marriage and sexual relations, typically for religious reasons.
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9 1
Prof Saroj Kumar Tripathi
Author has 3.6K answers and 4.7M answer views
·5y
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What is the meaning of “Celibacy” in Hindi?
THE MEANING OF THE ENGLISH WORD “CELIBACY” : IN HINDI , IT DENOTES : 1. Bramhmacharya. 2. Kaumaryavrat. 3 . Abivahit Aur Sharirik Sambandhon Sey Parhej Rakhneywala , Prayah Dharmik Karnon Sey.
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Karen Davis
Former Translator and Analyst, Report Writer, Instructor at Federal Government of the United States (1974–2016) · Author has 5.9K answers and 696K answer views
·2y
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What is the correct word to use: celibate or cannot?
Depends. Do you want to say “unmarried and chaste" or do you want to say “is unable to"? One's an adjective and one's a verb, so they can't be swapped syntacticly.
Of course, if you're looking for an insult, I suppose it doesn't matter.
Upvote ·
Kalkala Vadyappa Venkataramana
HRD and Soft Skills Trainer at Freelancing (2001–present)
·6y
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Is celibant a correct word verses celibate?
Celibate is the correct answer.
Upvote ·
9 1
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How can I be a celibate?
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Why would one choose to be celibate before marriage?
Does being a celibate mean I have no sexuality?
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How can I be a celibate?
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What are the differences between celibate and non-celibate people?
Autistics: If you are celibate is it by choice?
Has anyone ever decided to become celibate?
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A hypothesis is a testable prediction about the relationship between variables, often formulated based on existing knowledge or observations. It serves as a foundation for scientific investigation, guiding the research process by suggesting possible outcomes and allowing researchers to design experiments to validate or refute it. A well-structured hypothesis is crucial in the scientific method, as it directs experimentation and helps in drawing conclusions from the results.
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A hypothesis must be specific and measurable, allowing for clear testing through observation or experimentation.
The formulation of a hypothesis typically involves reviewing existing literature to identify gaps in knowledge that can be explored.
Hypotheses can be classified into two types: null hypothesis (which states there is no effect or relationship) and alternative hypothesis (which suggests there is an effect or relationship).
A hypothesis should lead to predictions that can be tested, enabling scientists to gather data to support or reject it based on empirical evidence.
The process of testing a hypothesis is iterative; if results do not support the initial hypothesis, researchers may refine it or develop a new one based on their findings.
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How does formulating a hypothesis contribute to the overall scientific method?
Formulating a hypothesis is a critical step in the scientific method because it provides a clear direction for research. It allows scientists to make predictions about what they expect to find, which guides the design of experiments. A good hypothesis enables researchers to focus on specific variables and relationships, facilitating systematic investigation and helping them draw meaningful conclusions from their work.
Discuss how hypotheses can evolve during the course of scientific research and why this is important.
Hypotheses can evolve based on new evidence gathered during research, which is important for scientific progress. As experiments yield results, scientists may find that their initial predictions were incorrect or incomplete. By refining hypotheses or creating new ones, researchers adapt their approach to align with emerging data, ensuring that their investigations remain relevant and scientifically rigorous. This flexibility is essential for developing a deeper understanding of complex phenomena.
Evaluate the role of hypotheses in establishing theories within the context of scientific inquiry.
Hypotheses play a foundational role in establishing theories within scientific inquiry by serving as initial predictions that are rigorously tested through experiments. When multiple hypotheses related to a particular phenomenon are consistently supported by empirical evidence, they can contribute to a broader understanding, leading to the development of a comprehensive theory. Theories represent well-established explanations that integrate various validated hypotheses, illustrating how they collectively enhance our knowledge of natural processes and relationships.
Related terms
Variable:A factor or condition that can change in an experiment, influencing the outcome and potentially affecting the relationship being tested.
Theory:A well-substantiated explanation of an aspect of the natural world, formed after many hypotheses have been tested and supported by a significant body of evidence.
Experiment:A controlled procedure carried out to test a hypothesis, where variables are manipulated to observe their effects on other variables.
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8559 | https://experimentationlab.berkeley.edu/sites/default/files/Books/Kittel_Introduction%20to%20Solid%20State%20Physics-Eighth%20Edition.pdf | i r . EIGHTH EDlTlGlv Y - - -- "- .- 7 >. t ; ' 1 d' - lnfroduction to Solid State Physics '- t n L - - -- CHARLES KITTEL Name Symbol Name Symbol Name Symbol Actinium Aluminum Americium Antimony Argon Arsenic Astatine Barium Berkelium Beryllium Bismuth Boron Bromine Cadmium Calcium Californium Carbon Cerium Cesium Chlorine Chromium Cobalt Capper Curium Dysprosium Einsteinium Erbium Europium Fermium Fluorine Francium Gadolinium Gallium Germanium Gold Hafnium Helium Holmium Hydrogen Indium Iodine Iridium Iron Krypton Lanthanum Lawrencium Lead Lithium Lutetium Magnesium Manganese Mendelevium Mercury Molybdenum Neodymium Neon Neptunium Nickel Niobium Nitrogen Nobelium Osmium Oxygen Palladium Phosphorus Platinum Plutonium Poloni~~m Potassium Praseodymium Promethium Protactinium Radium Radon Rhenium Rhodium Rubidium Ruthenium Samarium Scandium Selenium Silicon Silver Sodium Strontium Sulfur Tantalum Technetium Tellurium Terbium Thallium Thorium Thulium Tin Titanium Tungsten Uranium Vanadium Xenon Ytterbium Yttrium Zinc Zirconium - H' Periodic Table, with the Outer Electron Configurations of Neutral Is Atoms in Their Ground States L i :, Be' The notation used to descrilx the electronic configuratior~ o f atoms 8 " ' N' OX FY Nel0 ;nrd ions is discussed in all textl,nokc uf introdoctory atomic physics. The letters s, p , d, . . . signifj. flectrorrs having nrlrital angular 2y " ' tnomentum 0, 1, 2, . . . in units fi; the rruml,er to the left of the Zs22p 2 ~ ~ 2 ~ ' 25122~1 2~~211' 2sZ2pi 2'2p0 ~ ~ > i ~ ~ $ 2 letter dcwotrs the principal quantum nurnl~er of one c~rl)it, and the ~ 1 4 3 s i t 4 PI-S'~ c l n 7 A ~ ~ X superscript to the right denotes the nrlrnher of electn)rrs in the r~rl)it. 3s 3s' 3s23p 3s23pZ 3s23p:' 3s3p4 3s13pi 3S23p6 K'Y cato sex ~ i ' z ~ 2 ' $ Co" Ni" C u ' Y Zn:" Ga3' Ger' A s 3 3 Se71 B r 3 " r 3 @ XeS4 4d 4d2 4d4 4d5 4d6 4d7 4dX 4dIU 4dn1' 4dlU Ss 5s' ssz 5sZ 58 5s 5s 5~ 5s - 5s 5sZ 5s25p 5S'5p2 5s25p7 5s25p' 5Y'5p5 5s25p6 CeSX p r S Y N d 6 O Pm61 SmlZ EUI? Gd61 Tb6i DYC6 HOIl 6 1 6 8 Tm69 YblD LUIl 6d 4f2 4 f 3 4f4 4f5 4f6 4f' 4 f 4fX 4f1° 4fl3 4fI2 4f13 4f14 4f14 7s 7S2 , 7s2 5d 5d 5d \ 6 s X 6 x 2 6s' 6s' 6s' 6s2 63% 6sZ 6sZ 6y2 Cs2 6s2 6s2 6s1 -------------- T h W Pa81 "82 NPII pU94 A m 9 5 C m 9 6 Bk91 Cf98 E .
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2 Lr'Yi - Sf2 sf" S f " f 6 sf' Sf' 6d2 6d 6d 6d 7sZ 7s2 7sZ 7s2 7s2 7sZ 7 S 1 - - -. Introduction to E I G H T H E D I T I O N Charles Kittel Pr($essor Enwritus L'nitiersity of Cal$c~nlia, Berkeley B ! t Chapter 18, Nanostructures, was written by I Professor Paul McEuen of Cornell University. John Wiley & Sons, Inc EXECUTIVE EDITOR Stuart Johnson SENIOR PRODUCTION EDITOR Patricia McFadden SENIOR MARKETING MANAGER Robert Smith DESIGN DIRECTOR Madclyn Lesure SENIOR MEDIA EDITOR Martin Batey PRODUCTION MANAGEMENT Suzanne Ingrao/lngrao Associates This book was set in 10112 New Caledonia by GGS Books Services, Atlantic Highlands and printrd and bound by hfalloy Litt~ugraphing. Tlt: cover was printed by Phoenix Color This book is printed on acid free paper a , Copyright 6 2 2005 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any fonn or hy any means, electronic, mechanical, photucopying, recording, scanning or otbenvise, execpt as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc. 222 Rosewood Drive. Danvers, M A 01993, (97R)i50-8400. fax (978)646-8600. Requests to the Publisher for permission should be addressed to thc Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hobaken, NJ 07030-5774, (201)748-6011, fax (201)748-6008. To order books or for customer service please, call 1-800~CALL WILEY (225-5945). Library ojCongress Cataloging in Publication Data: Kittcl, Charles. Introduction to solid state physics 1 Charles Kitte1.-8th cd. p. cm. ISBN 0-471-41526-X 1. Solid state physics. 1, Title. ~ 7 6 . K 5 2005 5304+>dc22 ISBNV-471-41526-X WIE ISBN 0-471-68057-5 Printed in the United States of America 1 0 9 8 7 6 5 4 About the Author Charles Kittel did his undergraduate work in physics at M.1.T and at the Cavendish Laboratory of Cambridge University He received his Ph.D. from the University of Wisconsin. He worked in the solid state group at Bell Laboratories, along with Bardeen and Shockley, leaving to start the theoretical solid state physics group at Berkeley in 1951. His research has been largely in magnetism and in semiconductors. In magnetism he developed the theories of ferromagnetic and antiferromagnetic resonance and the theory of single ferromagnetic domains, and extended the Bloch theory of magnons. In semi- conductor physics he participated in the first cyclotron and plasma resonance experiments and extended the results to the theory of impurity states and to electron-hole drops. He has been awarded three Guggenheim fellowships, the Oliver Buckley Prize for Solid State Physics, and, for contributions to teaching, the Oersted Medal of the American Association of Physics Teachers. He is a member of the National Academy of Science and of the American Academy of Arts and Sciences. Preface This book is the eighth edition of an elementary text on solid state/ condensed matter physics fur seniors and beginning grad~rate students of the physical sciences, chemistry, and engineering. In the years since the first edi- tion was pnhlished the field has devcloped \,igoronsly, and there are notable applications. The challenge to the author has been to treat significant new areas while maintaining the introductory level of the text. It would be a pity to present such a physical, tactile field as an exercise in formalism. At the first editic~n in 1953 superconductivity was not lmderstood; Fermi snrfaces in metals were beginning to he explored and cyclotron resonance in semiconductors had just been observed; ferrites and permanent magnets were beginning to be understood; only a few physicists then believed in the reality of spin waves. Nanophysics was forty years off. In other fields, the structure of DNA was determined and the drift of continents on the Earth was demon- strated. It was a great time to be in Science, as it is now. I have tried with the successivt: editions of lSSY to introduce new generations to the same excitement. There are several changes from the seventh edition, as well as rnucll clarification: An important chapter has been added on nanophysics, contributed by an active worker in the field, Professor Paul L. McEuen of Cornell University Nanophysics is the science of materials with one, two, or three small dimen- sions, where "small" means (nanometer 10-%m). This field is the most excit- ing and vigorous addition to solid state science in the last ten years. The text makes use of the simplificati(~ns made possible hy the nniversal availability of computers. Bibliographies and references have been nearly eliminated because simple computer searches using keywords on a search engine slreh as Google will quickly generate many useful and rnore recent references. As an cxamplc of what can ho dons on the Web, explore the entry No lack of honor is in- tended by the omissions of early or traditiorral references to the workers who first worked on the problems of the solid state. The order nf the chapters has been changed: superconducti\ity and magnetism appear earlier, thereby making it easier to arrange an interesting one-semester course. The crystallographic notation conforms with current usage in physics. Im- portant equations in the body of the text are repeated in SI and CGS-Gaussian units, where these differ, except where a single indicated substitution will translate frnm CGS to SI. The dual usage in this book has been found helpful and acceptable. Tables arc in conventional units. The symbol e denotes the charge on the proton and is positive. The notation (18) refers to Equation 18 of the current chapter, but (3.18) refers to Equation 18 of Chapter 3. A caret (^) over a vector denotes a nnit vector. Few of the problems are exactly easy: Most were devised to carry forward the subject of the chapter. With few exceptions, the problems are those of the original sixth and seventh editions. The notation QTS refers to my Quantum Theory of Solirls, with solutions by C. Y. Fong; TP refers to Thermal Physics, with H. Kroemer. This edition owes much to detailed reviews of the entire text by Professor Paul L. McEuen of Cornell University and Professor Roger Lewis of Wollongong University in Australia. They helped make the book much easier to read and un- derstand. However, I must assume responsibility for the close relation of the text to the earlier editions, Many credits for suggestions, reviews, and photographs are given in the prefaces to earlier editions. I have a great debt to Stuart Johnson, my publisher at Wiley; Suzanne Ingrao, my editor; and Barbara Bell, my per- sonal assistant. Corrections and suggestions will be gratefully received and may be ad- dressed to the author by rmail to kittelQberke1ey.edu. The Instructor's Manual is available for download at: \m.wiley.coml collegelkittel. Charles Kittel Contents E CHAPTER 1: CRYSTAL STRUCTURE 1 Periodic Array of Atoms Lattice Translation Vectors Basis and the Crystal S t ~ c t u r e Primitive Lattice Cell Fundamental Types of Lattices Two-Dimensional Lattice Types Three-Dimensional Lattice Types Index Systems for Crystal Planes Simple Crystal Structures Sodium Chloride Structure Cesium Chloride Structure Hexagonal Close-Packed Structure (hcp) Diamond Structure Cubic Zinc Sulfide Structure Direct Imaging of Atomic Structure Nonideal Crystal Structures Random Stacldng and Polytypism Crystal Structure Data Summary Problems CHAPTER 2: WAVE DIFFRACTION AND THE RECIPROCAL LATTICE Diffraction of Waves by Crystals Bragg Law Scattered Wave Amplitude Fourier Analysis Reciprocal Lattice Vectors Diffraction Conditions Laue Equations Brillonin Zones Reciprocal Lattice to sc Lattice Reciprocal Lattice to hcc Lattice Reciprocal Lattice to fcc Lattice Fourier Analysis of the Basis Structure Factor of the bcc Lattice Structure factor of the fcc Lattice Atomic Form Factor Summary Problems Crystals of Inert Gases Van der Wads-London Interaction Repulsive Interaction Equilibrium Lattice Constants Cohesive Energy Ionic Crystals Electrostatic or Madelung Energy Evaluation of the Madelung Constant Covalent Crystals Metals Hydrogen Bonds Atomic Radii Ionic Crystal Radii Analysis of Elastic Strains Dilation Stress Components Elastic Compliance and Stiffness Constants Elastic Energy Density Elastic Stiffness Constants of Cubic Crystals Bulk Modulus and Compressibility Elastic Waves in Cubic Crstals Waves in the [I001 Direction Waves in the [I101 Direction Summary Problems CIIAPTER 4: PHONONS I. CRYSTAL VIBRATIONS Vibrations of Crystals with Monatomic Basis First Brillouin Zone Group Velocity Long Wavelength Limit Derivation of Force Constants from Experiment Two Atoms per Primitive Basis Quantization of Elastic Waves Phonon Momentum Inelastic Scattering by Phonons Summary Problems CHAFTER 5: PHONONS 11. THERMAL PROPERTIES Phonon Heat Capacity Planck Distribution Normal Mode Enumeration Density of States in One Dimension Density of States in Three Dimensions Debye Model for Density of States Debye Law Einstein Model of the Density of States General Result for D(w) Anharmonic Crystal Interactions Thermal Expansion Thermal Conductivity Thermal Resistivity of Phonon Gas Umklapp Processes Imperfecions Problems CHAPTER 6: FREE ELECTRON FERMI GAS Energy Levels in One Dimension Effect of Temperature on the Fermi- Dirac Distribution Free Electron Gas in Three Dimensions Heat Capacity of the Electron Gas Experimental Heat Capacity of Metals Heavy Fermions Electrical Conductivity and Ohm's Law Experimental Electrical Resistivity of Metals Umklapp Scattering Motion in Magnetic Fields Hall Effect Thermal Conductivity of Metals Ratio of Thermal to Electrical Conductivity Problems CHAPTER 7: ENERGY BANDS Nearly Free Electron Model Origin of the Energy Gap Magnitude of the Energy Gap Bloch Functions Kronig-Penney Model Wave Equation of Electron in a Periodic Potential Restatement of the Bloch Theorem Crystal Momentum of an Electron Solution of the Central Equation Kronig-Penney Model in Reciprocal Space Empty Lattice Approximation Approximate Solution Near a Zone Boundary Number of Orbitals in a Band Metals and Insr~lators Summary Problems CHAPTER 8: SEMICONDUCTOR CRYSTALS Band Cap Equations of Motion Physical Derivation of i & = F Holes Effective Mass Physical Interpretation of the Effective Mass Effective Masses in Semiconductors Silicon and Germanium Intrinsic Carrier Concentration Intrinsic Mobility Impurity Conductivity Donor States Acceptor States Thermal Ionization of Donors and Acceptors Thermoelectric Effects Semimetals Superlattices Bloch Oscillator Zener Tunneling Summary Problems C ~ E R 9: FERMI SURFACES AND METALS Reduced Zone Scheme Periodic Zone Scheme Construction of Fermi Surfaces Nearly Free Electrons Electron Orbits, Hole Orbits, and Open Orbits Calculation of Energy Bands Tight Binding Method of Energy Bands Wigner-Seitz Method Cohesive Energy Pseudopotential Methods Experimental Methods in Fermi Surface Studies Quantization of Orbits in a Magnetic Field De Haas-van Alphen Effect Extremal Orbits Fermi Surface of Copper Magnetic Breakdown Summary Problems CHAPTER 10: SUPERCONDUCTIVITY Experimental Survey Occurrence of Superconductivity Destruction of Superconductivity of Magnetic Fields Meissner Effect Heat Capacity Energy Gap Microwave and Infrared Properties Isotope Effect Theoretical Survey Thermodynamics of the Superconducting Transition London Equation Coherence Length BCS Theory of Superconductivity BCS Ground State Flux Quantization in a Superconducting Ring Duration of Persistent Currents Type I1 Superconductors Vortex State Estimation of H,, and H,, Single Particle Tunneling Josephson Superconductor Tunneling Dc Josephson Effect Ac Josephson Effect Macroscopic Quantum Interference High-Temperature Superconductors Summary Problems Reference Langevin Diamagnetism Equation Quantum Theory of Diamagnetism of Mononuclear Systems Paramagnetism Quantum Theory of Paramagnetism Rare Earth Ions Hund Rules Iron Group Ions Clystal Field Splitting Quenching of the Orbital Angular Momentum Spectroscopic Splitting Factor Van Vleck Temperature-Independent Paramagnetism Cooling by Isentropic Demagnetization Nuclear Demagnetization Paramagnetic Susceptibility of Conduction Electrons Summary Problems CHAPTER 12: FERROMAGNETISM AND ANTIFERROMAGNETISM Ferromagnetic Order Curie Point and the Exchange Integral Temperature Dependence of the Saturation Magnetization Saturation Magnetization at Absolute Zero Magnons Quantization of Spin Waves Thermal Excitation of Magnons Neutron Magnetic Scattering Ferrimagnetic Order Curie Temperature and Susceptibility of Ferrimagnets Iron Garnets Antiferromagnetic Order Susceptibility Below the NBel Temperature Antiferromagnetic Magnons Ferromagnetic Domains Anisotropy Energy Transition Region between Domains Origin of Domains Coercivity and Hysteresis Single Domain Particles Geomagnetism and Biomagnetism Magnetic Force Microscopy Summary Problems CHAPTER 13: MAGNETIC RESONANCE Nuclear Magnetic Resonance Equations of Motion Line Width Motional Narrowing Hyperfine Splitting Examples: Paramagnetic Point Defects F Centers in Alkali Halides Donor Atoms in Silicon Knight Shift Nuclear Quadmpole Resonance Ferromagnetic Resonance Shape Effects in FMR Spin Wave Resonance Antiferromagnetic Resonance Electron Paramagnetic Resonance Exchange Narrowing Zero-field Splitting Principle of Maser Action Three-Level Maser Lasers Summary Problems CHAPTER 14: PLASMONS, POLARITONS, AND POLARONS Dielectric Function of the Electron Gas Definitions of the Dielectric Function Plasma Optics Dispersion Relation for Electromagnetic Waves Transverse Optical Modes in a Plasma Transparency of Metals in the Ultraviolet Longitudinal Plasma Oscillations Plasmons Electrostatic Screening Screened Coulomb Potential Pseudopotential Component U(0) Matt Metal-Insulator Transition Screening and Phonons in Metals Polaritons LST Relation Electron-Electron Interaction Femi Liquid Electron-Electron Collisions Electron-Pbonon Interaction: Polarons Peierls Instability of Linear Metals Summary Problems CHAPTER 15: OPTICAL PROCESSES AND EXCITONS Optical Reflectance Kramers-Kronig Relations Mathematical Note Example: Conductivity of collisionless Electron Gas Electronic Interband Transitions Excitons Frenkel Excitons Alkali Halides Molecular Crystals Weakly Bound (Molt-Wannier) Excitons Exciton Condensation into Electron-Hole Drops (EHD) Raman Effects in Crystals Electron Spectroscopy with X-Rays Energy Loss of Fast Particles in a Solid Summary Problems CHAPTER 16: DIELECTRICS AND FERROELECTRICS Maxwell Equations Polarization Macroscopic Electric Field Depolarization Field, E, Local Electric Field at an Atom Lorentz Field, E , Field of Dipoles Inside Cavity, E3 Dielectric Constant and Polarizability Electronic Polarizability Classical Theory of Electronic Polarizability Stmctural Phase Transitions Ferroelectric Crystals Classification of Ferroelectric Crystals Displacive Transitions Soft Optical Phonons Landau Theory of the Phase Transition Second-Order Transition First-Order Transition Antiferroelectricity Ferroelectric Domains Piezoelectricity Summaq Problems CHAPTER 17: SURFACE AND INTERFACE PHYSICS Reconstruction and Relaxation Surface Crystallography Reflection High-Energy Electron Diffraction Surface Electronic Structure Work Function Thermionic Emission Surface States Tangential Surface Transport Magnetoresistance in a Two-Dimensional Channel Integral Quantized Hall Effect (IQHE) IQHE in Real Systems Fractional Quantized Hall Effect (FQHE) p-n Junctions Rectification Solar Cells and Photovoltaic Detectors Schottky Barrier Heterostructures n-N Heterojunction Semiconductor Lasers Light-Emitting Diodes Problems Imaging Techniques for Nanostructures Electron Microscopy Optical Microscopy Scanning Tunneling Microscopy Atomic Force Microscopy EIectronic Structure of I D Systems One-Dimensional Subbands Spectroscopy of Van Hove Singularities 1D Metals - Coluomb Interactions and Lattice Copnlings Electrical Transport in 1D Conductance Quantization and the Larldauer Formula Two Barriers in Series-resonant Tunneling Incoherent Addition and Ohm's Law Localization Voltage Probes and the Buttiker-Landauer Formalism Electronic Structure of O D Systems Quantized Energy Levels Semiconductor Nanocrystals Metallic Dots Discrete Charge States Electrical Transport in OD Coulomb Oscillations Spin, Mott Insulators, and the Kondo Effect Cooper Pairing in Superconducting Dots Vibrational and Thermal Properties of Nanostructures Quantized Vibrational Modes Transverse Vibrations Heat Capacity and Thermal Transport Summary Problems Diffraction Pattern Monatomic Amorphous Materials Radial Distribution Function Structure of Vitreous Silica, SiO, Glasses Viscosity and the Hopping Rate Amorphous Ferromagnets Amorphous Semiconductors Low Energy Excitations in Amorphous Solids Heat Capacity Calculation Thermal Conductivity Fiber Optics Rayleigh Attenuation Problems Lattice Vacancies Diffusion Metals Color Centers F Centers Other Centers in Alkali Halides Problems Shear Strength of Single Crystals Slip Dislocations Burgers Vectors Stress Fields of Dislocations Low-angle Grain Boundaries Dislocation Densities Dislocation Multiplication and Slip Strength of Alloys Dislocations and Crystal Growth Whiskers Hardness of Materials Problems General Considerations Substitutional Solid Solutions- Hume-Rothery Rules Order-Disorder Transformation Elementary Theoly of Order Phase Diagrams Eutectics Transition Metal Alloys Electrical Conductivity Kondo Effect Problems APPENDIXA: TEMPERATURE DEPENDENCE OF THE REFLECTION LINES 641 APPENDIX B: EWALD CALCULATION OF LATTICE SUMS 644 Ewald-Kornfeld Method for Lattice Sums for Dipole Arrays 647 APPENDIX C: QUANTIZATION OF ELASTIC WAVES: PHONONS Phonon Coordinates Creation and Annihilation Operators APPENDIX F: BOLTZMANN TRANSPORT EQUATION Particle Diffusion Classical Distribution Fermi-Dirac Distribution Electrical Conductivity APPENDIX G: VECTOR POTENTIAL, FIELD MOMENTUM, AND GAUGE TRANSFORMATIONS Lagrangian Equations of Motion Derivation of the Hamiltonian Field Momentum Gauge Transformation Gauge in the London Equation Crystal Structure PERIODIC ARRAYS OF ATOMS Lattice translation vectors Basis and the crystal structure Primitive lattice cell FUNDAMENTAL TYPES OF LATTICES Two-dimensional lattice types Three-dimensional lattice types INDEX SYSTEM FOR CRYSTAL PLANES SIMPLE CRYSTAL STRUCTURES Sodium chloride structure Cesium chloride structure Hexagonal close-packed structure Diamond structure Cubic zinc s d d e structure DIRECT IMAGING OF ATOMIC STRUCTURE 18 NONIDEAL CRYSTAL STRUCTURES 18 Random stacking and polytypism 19 CRYSTAL STRUCTURE DATA 19 SUMMARY 22 PROBLEMS 22 1. Tetrahedral angles 2. Indices of planes 3. Hcp structure UNITS: 1 A = 1 angstrom = 10-'cm = O.lnm = 10-"m. (c) Figure 1 Relation of the external form of crystals to the form of tlre elementary building hlackr. The building blocks are identical in (a) and (b), but Merent c ~ s t a l faces ;Ire developed. ( c ) Cleaving a crystal of rocksalt. CHAPTER 1: CRYSTAL STRUCTURE PERIODIC ARRAYS O F ATOMS The serious study of solid state physics began with the discovery of x-ray diffraction by crystals and the publication of a series of simple calculations of the properties of crystals and of electrons in crystals. Why crystalline solids rather than nonclystalline solids? The important electronic properties of solids are best expressed in crystals. Thus the properties of the most important semi- conductors depend on the crystalline structure of the host, essentially because electrons have short wavelength components that respond dramatically to the regular periodic atomic order of the specimen. Noncrystalline materials, no- tably glasses, are important for optical propagation because light waves have a longer wavelength than electrons and see an average over the order, and not the less regular local order itself. We start the book with crystals. A crystal is formed by adding atoms in a constant environment, usually in a solution. Possibly the first crystal you ever saw was a natural quartz crystal grown in a slow geological process from a sili- cate solution in hot water under pressure. The crystal form develops as identical building blocks are added continuously. Figure 1 shows an idealized picture of the growth process, as imagined two centuries ago. The building blocks here are atoms or groups of atoms. The crystal thus formed is a three-dimensional periodic array of identical building blocks, apart from any imperfections and impurities that may accidentally be included or built into the structure. The original experimental evidence for the periodicity of the structure rests on the discovery by mineralogists that the index numbers that define the orientations of the faces of a crystal are exact integers. This evidence was sup- ported by the discovery in 1912 of x-ray diffraction by crystals, when Laue de- veloped the theory of x-ray diffraction by a periodic array, and his coworkers reported the first experimental observation of x-ray diffraction by crystals. The importance of x-rays for this task is that they are waves and have a wave- length comparable with the length of a building block of the structnre. Such analysis can also be done with neutron diffraction and with electron diffraction, hut x-rays are usually the tool of choice. The diffraction work proved decisively that crystals are built of a periodic array of atoms or groups of atoms. With an established atomic model of a crys- tal, physicists could think much further, and the development of quantum the- ory was of great importance to the birth of solid state physics. Related studies have been extended to noncrystalline solids and to quantum fluids. The wider field is h o w n as condensed matter physics and is one of the largest and most vigorous areas of physics. An ideal crystal is constn~cted by the infinite repetition of idenbcal groups of atoms (Fig 2) A group is called the basis. The set of mathematical points to which the basls is attached is called the lattice The lattice in three dimensions may be defined by three translabon vectors a,, a,, a,, such that the arrange- ment of atoms in the crystal looks the same when viewed from the point r as when viewed from every polnt r' translated by an mtegral multiple of the a's: Here u,, u,, u, are arhitraryintegers. The set of points r' defined by (1) for all u,, u,, u, defines the lattice. The lattice is said to be primitive if any two points from which the atomic arrangement looks the same always satisfy (1) with a suitable choice of the in- tegers ui. This statement defines the primitive translation vectors a,. There is no cell of smaller volume than a, . a, x a, that can serve as a building block for the crystal structure. We often use the primitive translation vectors to de- fine the crystal axes, which form three adjacent edges of the primitive paral- lelepiped. Nonprimitive axes are often used as crystal axes when they have a simple relation to the symmetry of the structure. Figure 2 The crystal srmchre is formed by the addition af the basis (b) to evely lattice point of the space laisice (a). By looking at ( c ) , one oan recognize the basis and then one can abstract the space lattice. It does not matter where the basis is put in relation to a lattice point. 1 Crystal Struchrre 5 Basis and the Crystal Structure The basis of the crystal structure can be idendled once the crystal mes have been chosen Figure 2 shows how a crystal is made by adding a basis to every lamce pomnt-of course the lattice points are just mathematical con- structions. Every bas~s in a given crystal is dentical to every other ~n composi- tmn, arrangement, and orientation The number of atoms in the basis may be one, or it may be more than one. The position of the center of an atom3 of the basis relahve to the associated lattice point is We may arrange the origin, wl~ich we have called the associated lattice point, so that 0 5 x,, yj, zj 5 1. (b) (c) Fieure 3a Latttce nomts of a soam lnmce m two &mensrons AU ~ a r s of vector? a , , a , are trans- twice the area of a primitive c e l l . Figare 3b Primitive ccll "fa space lattice in three dimensions Figure 3c Suppose these points are identical atoms: Sketchin on the figure a set of lattice points, s choice ofprimitive axes, aprimitivc cell, and the basis of atoms associatedwith alattice point. F i p e 4 A primitive c e l l may also be chosen fol- lowing this procedure: (1) draw lines to connect a given l a t t i c e point to all nearby lattice points: (2) at the midpoint and normal to these lines, draw new l i n e s or planes. The smallest volume enclosed in this way is the Wigner-Seitz primitive cell. All space may be filled by these cells, j u s t as by the ceh of Fig. 3. Primitive Lattice Cell The parallelepiped defined by primitive axes al, a,, a , is called a primitive cell (Fig. 3b). A primitive cell is a t).pe of cell or unit cell. (The adjective unit is superfluous and not needed ) A cell will fill all space by the repetition of suit- able crystal b-anslatlon operationq. A primitive cell is a m~nimum-volume cell. There are many ways of choosing the prnnitive axes and primitive cell for a given lattice. The number of atoms m a primitive cell or primtive basis is alwap the same for a given crystal smllcture There is always one lattice point per primitive cell. If the primitwe cell is a parallelepiped with lattice po~uts at each of the eight corners, each lattice point is shared among e~ght cells, so that the total number of lattice points in the cell is one: 8 X = 1. The volume of a parallelepiped wrth axes a,, %, a3 is V, = (a, - as X a , 1 , (3) by elementary vector analysis. The basis associated wrth a primitive cellis cdued a prim~tive bans. No basis contans fewer atoms than a pnmibve basis contains. Another way of choosing a primitive cell is shown in Fig. 4. This is hown to physicists as a Wigner-Seitz cell. FUNDAMENTAL TYPES OF LATTICES Crystal lattices can be carried or mapped into themselves by the lattice translations T and by vanous other symmetzy operattons. A typical symmetry operation is that of rotation about an axis that passes through a lattice point. Lattices can be found such that one., two-, three., four., and sixfold rotation axes cany the lattice into itself, corresponhng to rotatrons by ZT, 2 ~ 1 2 , 2 d 3 , 2 ~ 1 4 , and 2 ~ 1 6 radians and by integral multiples of these rotations. The rota- tion axes are denoted by the symbols 1,2, 3, 4, and 6. We cannot find a lattice that goes into itself under other rotations, such as by 2.d7 rachans or 2 ~ / 5 rachans. A single molecule properly designed can have any degree of rotational symmetry, but an Infinite periodic lattice cannot. We can make a crystal from molecules that individually have a fivefold rotation ms, but we should not expect the latbce to have a fivefold rotation ms. In Fig. 5 we show what happens if we try to construct a penodic latbce havlng fivefold Figure 5 A fivefold tuis of symmetry can- not exist in a periodic lattice because it is not ~ossible to fa thc area of a plane with a mmected m a y of pentagons. We can, however, fill all the area oTa plane with just two distina designs of "tiles" or elemeutary potysms. (c) (e) Figure 6 (a) A plane of symmetry to the faces of a cube. (b) A diagonal plane of s)rmmehy in a cube. (c) The three tetrad xes of a cube. (d) Thc four t r i d axes of a cube. (e) The six diad axes of a cube. symmetry: the pentagons do not fit together to fdl all space, shoulng that we can- not combmne fwefold point symmetrywith the requed translational penodicity. By lattice point group we mean the collection of symmetry operations which, apphed about a lathce pomt, carry the lattice into itself The possible ro- tations have been listed. We can have mirror reflecbons m about a plane through y p p by -r The symmetry axes and symmetry planes of a cube are shown in Fig 6. Two-Dimensional Lattice Types The lattice in Fig. 3a was dram for arbitrary al and as. A general lattice such as this is known as an oblique lattice and is invariant only under rotation of .rr and 27r ahout any lattice point. But special lattices of the oblique type can he invariant under rotation of 2 ~ 1 3 , 2 ~ 1 4 , or 2 d 6 , or under mirror reflection. We mist impose restrictive conditions on a, and a% if we want to construct a lat- tice that will he invariant under one or more of these new operations. There are four distinct types of restriction, and each leads to what we may call a special lattice type. Thus there are five distinct lattice types in two dimensions, the oblique lattice and the four special lattices shown in Fig. 7. Bravais lattice is the common phrase for a distinct lattice.%e; we say that there are five Bravdrs lattices in two dimensions. Figure 7 Four special lattices in twodirnmsions, 1 Crystal St-ture Three-Dimensional Lattice Types The point symmetry groups in three dimensions require the 14 different lattice types listed in Table 1. The general lattice is triclinic, and there are 13 special lattices. These are grouped for convenience into systems classified according to seven types of cells, which are triclinic, monoclinic, orthorbom- bic, tetragonal, cubic, higonal, and hexagonal. The division into systems is expressed in the table in terms of the adal relations that describe the cells. The cells in Fig. 8 are conventional cells: of these only the sc is a primitive cell. Often a nonprimitive cell has a more obvious relation with the point symmetry operations than has a primitive cell. There are three lattices in the cubic system: the simple cubic (sc) lattice, the body-centered cubic (hcc) lattice, and the face-centered cubic (fcc) lattice. Table 1 The 14 latlioe types in three dimensions Numhrr of Restr~chons on cunvenhonal System lathces cell axes and angles Triclimc Monoclinic Orthorhomb~c Cubic Trigonal Hexagonal Figure 8 The cubic space lattices. The cells s h m are the conventional cells. a Lattice points per cell 1 2 4 Volume. primitive cell a3 ZQ 1 3 x a 1 3 Lattice points per unit volume l/a3 2/aR 4/a3 Numher of nearest neighbors 6 8 12 Nearest-neighbor distance a 3u2a/2 = 0.866a a/2'" 0.707a Number of second neighbors 12 6 6 Second neighbor distance 2'% a a Packing fractionn ZW &V5 i d 5 =0.524 =O.B80 =0.740 "The packing fraction is the manirnum proportion of the available volume that can be filled with hard spheres. Figure 10 Pnmibve translation vectors of the body- ~ i ~ , , , . ~ 9 ~ ~ d ~ .
~ ~ ~ ~ ~ ~ ~ d c,lbic lattice, shouing a centered cubic lattice; these vectors connect the lattice prilnitive ~h~ rimifive she%,,,, is a rhonl+,o point at the origin to lattice points at the body centers. hedron , , f edge ; 2 3 a, and the angle behen adja. The primitive ceU is obtained on completing the rhody- cent edges is 109"ZR'. huhedron. In terms of the cube edge a, the primitive translation vectors are al=ia(i+i-2) ; a s = $ ~ ( - i + j + L i ) . a3=;a(%-f+i) Ifere i , j, i are the Cartesian untt vectors The charactenstics of the three cubic lattxes are summanzed in Table 2. A prirmhve cell of the bcc lattice is shown in Fig. 9, and the primtive tran~lation vectors are shown in Fig. 10 The primitive translabon vectors of the fcc lathce are shown In Fig. 11. Primitive cells by definition contain only one lattice point, but the conventional bcc cell contains two lattice points, and the fcc cell contains four lattice points. I Crystal Stwctun 11 Figore 11 The rhombohedra1 primitive cell of the hce-centered Figore 12 Relation of the plimitive cell cubic clystal. The primitive translation vectors a,, a , , connen in the hexagonal system (heavy lines) to the lattice point at the origin with lattice points at the face centers, a prism of hexagonal symmew. Here As drawn, the primitive vectors are: rr,=o,#a,. The angles between& axes are 6 0 ' The pos~hon of a point in a cell is spec~fied by (2) in terms of the atomic coordinates x , y, z. Here each coordinate is a fraction of the axla1 length a,, a,, a, in the direction of the coordinate axis, with the origin taken at one corner of the cell Thus the coorhnate? of the body center of a cell are ;$$, and the face 1 1 1 1 centers include iiO, 0,s; 5 % . In the hexagonal system the primitive cell is a right prism based on a rhomhu3 with an included angle of 120". F~gure 12 shows the relabonship of the rhombic cell to a hexagonal prism. INDEX SYSTEM FOR CRYSTAL PLANES The orientahon of a crystal plane is determined by three points in the plane, provided they are not collinear. If each point lay on a different crystal axis, the plane could be specdied by giving the coordinates of the points in terms of the latbce constants a,, a,, a3. However, it turns out to be more useful for structure analysis to specify the orientation of a plane by the indices deter- mined by the followlug rules (Fig. 13). Find the intercepts on the axes in terms of the lattice constants a,, a , , a,. The axes may be those of a primitive or nonprimibve cell. Figure I3 This plane intercepts the a , , +, a , axes at 3a,, Za,, Z a , The recrprocals of these numbers , I & are 5, ., . The smallest three mte- gers havlng the same rabo are 2, 3, 3, and thus the m&ces of the plane are (233) I Figure 14 Inlces oElmportant lanes in a cubx ctystal The plane (200) is ~ a r d e l to (100) and to (100) Take the reciprocals of these numbers and then reduce to three integers having the same ratio, usually the smallest three integers. The result, en- closed in parentheses (hkl), is called the index of the plane. For the plane whose intercepts are 4,1,2, the reciprocals are $, 1, and $: the smallest three integers having the same ratio are (142). For an intercept at infin- ity, the corresponding index is zero. The indices of some important plades in a cubic crystal are illustrated by Fig. 14. The indices (hkl) may denote a single plane or a set of parallel planes. If a plane cuts an axis on the negative side of the origin, the corresponding index is negative, indicated hy placing a minus sign 1 Crystal Structure 13 above the index: (hkl). The cube faces of a cubic crystal are (100). (OlO), (OOl), (TOO), (o~o), and (001). Planes equivalent by symmeq may he denoted by curly brackets (braces) around indices; the set of cube faces is {100}. When we speak of the (200) plane we mean a plane parallel to (100) but cutting the a, axis at i n . The indices [uvw] of a direction in a clystal are the set of the smallest inte- gers that have the ratio of the components of a vector in the desired direction, referred to the axes. The a, axis is the [loo] direction; the -a, axis is the [o~o] direction. In cubic crystals the direction [hkl] is perpendicular to a plane (hkl) having the same indices, but this is not generally true in other crystal systems. SIMPLE CRYSTAL STRUCTURES We di~cuss simple crystal structures of general interest the sohum chlo- nde, cesium chloride, hexagonal close-packed, hamond and cuh~c zinc sulfide structures. Sodium Chloride Structure The sohum chloride, NaCI, structure 1 s shown in Figs. 1.5 and 16. The lattice is face-centered cublc: tile basis conslsts of one Na+ Ion and one C 1 ion Figure 15 We may construct the soditzrn chloride cvstal shuchlre by arranging Naf and C 1 ions alter- nately at the lattice points of a simple cubic lattice. In the ctysral each ion is surrounded by six nearest neigh- bors of the opposite charge. The space lattice is fcc, and the basis has one C I ion at 000 w d one Na' L ? L . , , . The 'figre shows one conventional cubi The ionic diameters here are reduced in relation cell in order to darify the spatial arrangement. .
.
. c cell. Fii tothe ?m P. : pre 16 M d e r than th Singer) ulll cirlnride ns. (Courtes The sodium iuns are y of A. N. Holden and Pigure 17 Na1ui.d c~?stals of lead snlfide. PbS. whir11 lias the NaCl crystal stmcturc. (Phutag~apl~ by R. Burl?aon.) Figure 18 The cesium chloride cqstal struehne. The space lattice is silnple cubic, and &the basis has one C s t ion at 000 and one C 1 ion at i. separated by one-half the body diagonal of a umt cube There are four units of NaCl ~n each unit cube, with atoms in tlie pos~tions Each atom has as nearest neighbors six atoms of the opposire kind. Represen- tative crystals having the ~ k l arrangement include those in the following table. The cube edge a is given in angstroms; 1 if -- cm lo-'' m 0.1 nm. Figure 17 is a photograph of crystals of lead sulfide (PbS) from Joplin, Missouri. The Joplin specimens form in beautiful cuhes. Cesium Chloride Stnccture The cesium chloride structure is shown in Fig. 18. There is one molecule per primitive cell, with atoms at the comers 000 and body-centered positions l i l --- , , . of the simple cubic space lattice. Each atom may he viewed as at the center Figure 19 A close-packed layer of spheres is shown, with centers at points marked A. A second and identical layer of spheres can he placed on top of this, above and patallel to the plane of the drawing, with centen over the points marked B. There are two choices for a third layer. It can go in over A or over C. Ifit goes in over A, the sequence is AEABAB . . . and the structure is hexagonal close-packed. If the third layer goes in over C, the sequence is ABCABCABC . . . and the Struchlre is face-centered cubic. A B Figure 20 The hexagonal close-packed structure. c The atom positions in t h i s smcture do not constitute a soace lattice. The mace lattice is sim~le hexa~onal I wik a basis of hvo ;dentical atoms asiociatedwith A each latttce nomt. The lamce parameters o and c are inhcated, where o is i n the basal plane and c is the rnagmtude of the an? a , of R g 12 of a cube of atoms of the opposite kind, $0 that the number of nearest neigh- bors or coordination number is eight. Hexagonal Close-Packed Structure (hcp) There are an infinite number of ways of arranging identical spheres in a regular array that maximuzes the packing fraction (Fig. 19) One 1s the face- centered cub^ structure; another is the hexagonal close-packed structure (Fig. 20). The fraction of the total volume occupled by the spheres is 0.74 for both structures. No structure, regilar or not, has denserpaclung. Figure 21 The primitive cell h a s a, = h , with an included angle of 120". The c axis (or a,) is normal to t h e plane of a, and a,. The ideal hcp stmohre has c = 1.633 a, The two atoms of one basis are shown as solid circles. One atom of the bais is at the ori- gin; the other atom is at $Qb, which m e a n s at the position r = $a, + +a, + $ a , . Spheres are arranged in a single closest-packed layer A by placmg each sphere in contact with SIX others in a plane. This layer may serve as either the basal plane of an hcp structure or the (111) plane of the fcc shcture. A sec- ond s~milar layer B may be added by placlng each sphere of B in contact with three spheres of the bottom layer, as in Figs. 19-21. A third layer C may be added In two ways. We obtam the fcc structure if the spheres of the third layer are added over the holes in the first layer that are not occupied by B We obtain the hcp structure when the spheres in the third layer are placed directly over the centers of the spheres in the first layer. The number of nearest-nelghbor atoms is 12 for both hcp and fcc stnlc- tures. If the b~nding energy (or free energy) depended only on the number of nearest-neighbor bonds per atom, there would be no difference in energy between the fcc and hcp structures. Diamond Structure The diamond structure is the structwe of the semiconductors silicon and germanium and is related to the structure of several important semicondnctor binary compouncLs. The space lattice of damond 1s face-centered cubic. The primitive basis of the diamond structure has two identical atoms at coordinates 000 and 2 ; ; assoc~ated mth each point of the fcc latt~ce, as shown in Fig. 22. Because the convenbonal unit cube of the fcc lattice contains 4 latbce points, it follows that the conven~onal unit cube of the dlamond structure contams 2 X 4 = 8 atoms. There is no way to choose a primitive cell such that the basis of diamond contains only one atom. Figure 22 Atomic positions in the cubic cell uf the diamond Figure 23 Crystal structure of diamond, st~uch~rr projected on a cub? face; fiacticms denote height showingthetetrahedralbondarrangement. above the hasp in units of a cubc edge. The paints at 0 and $ are on the fcc lattice those at and are on a similar lattice displvcerl along the body diagonal by one-fourth of its lengh. With a fcc space lattice, the basis consists of mia identical atoms at 000 0 d i i ; . The tetrahedral bonding characteristic of the diamond structure is shown in Fig. 23. Each atom has 4 nearest neighbors and 12 next nearest neighbors. The diamond structure is relatively empty: the maximum proportion of the available volume which may he filled by hard spheres is only 0.34, which is 46 percent of the f11ling factor for a closest-packed stmcture such as fcc or hcp. The diamond structure is an example of the directional covalent bonding found in column IV of the periodic table of elements. Carbon, silicon, germa- nium, and tin can crystallize in the diamond structure, with lattice constants n = 3.567, 5.430, 5.658, and 6.49 A, respectively. Here a is the edge of the conventional cubic cell. Cubic Zinc Sulfide Structure The diamond structure may be viewed as twn fcc structures displaced from each other by one-quarter of a body diagonal. The cubic zinc sulfide (zinc blende) structure results when Zn atoms are placed on one fcc lattice and S atoms on the other fcc lattice, as in Fig. 24. The conventional cell is a cube. The coordinates of the Zn atoms are 000; 0;;; $0;; $ $0; the coordinates of the 1 1 1 1 3 3 3 1 3 3 5 1 S atoms are 444; , 44; 2 j j ; 4 4 4. The lattice is fcc. There are four molecules '5- ZnS per conventional cell. About each atom there are four equally distafft atoms of the opposite kind arranged at the comers of a regular tetrahed~w ..- Figure 24 sulfide Crystal structure of cubic zinc The diamond structure allows a center-of-inversion symmetry operation at the midpoint of every hne between nearest-neighbor atoms. The inversion operation carries an atom at r into an atom at -r. The cubic ZnS struc- ture does not have inversion symmetry. Examples of the cubic zinc sulfide structure are The close equality of the lattice constants of several pairs, notably (Al, Ga)P and (Al, Ga)As, makes possible the construction of sem~conductor hetemjunc- tions (Chapter 19). DIRECT IMAGING OF ATOMIC STRUCTURE Direct images of crystal structure have been produced by transmission electron microscopy. Perhaps the most beaubful Images are produced by scan- ning tunneling microscopy; in STM (Chapter 19) one exploits the large vana- tions in quantum tunneling as a function of the height of a fine metal tip above the surface of a crystal. The image of Fig 25 was produced m t h ~ ~ way. An STM method has been developed that will assemble single atoms Into an orga- nized layer nanometer structure on a crystal substrate. NONIDEAL CRYSTAL STRUCTURES l",v - ., . ,. ; .
: t ..,. ; > The ideal crystal of classical crystallographers is formed by the periodic 7 ' ~ repetition of identical units in space. But no general proof bas been given that I Crystal Structure 19 Figure 25 A scanning tunneling microscope image of atorns on a (111) surface of fcc plat- inum at 4 K. The nearest-neighbor spacing is 2.78 A. (Photo courtesy of D. M. Eigler, IHM Rrerarch Divi~irn.) the ideal crystal is the state of minimum energy of identical atoms at the tem- perature of absolute zero. At finite temperatures this is likely not to be true. We give a further example here. Random Stacking and Polytypism :d planes I - . The fcc and hcp structures are made up of close-pack< 3f atoms. The structures differ in the stacking sequence of the planes, fcc having the se- quence ABCABC . . . and hcp having the sequence ABABAB . . . . Structures are h o w n in which the stacking sequence of close-packed planes is random. This is known as random stacking and may be thought of as crystalline in two dimensions and noncrystalline or glasslike in the third. Polytypism is characterized by a stacking sequence with a long repeat unit along the s t a c h g axis. The hest known example is zinc sulfide, ZnS, in which more than 150 polytypes have been identified, with the lnngest period- icity being 360 layers. Another example is silicon carbide, Sic, which occurs with more than 45 stacking sequences of the close-packed layers. The polytype of SiC known as 393R has a primitive cell with a = 3.079 A and c = 989.6 A. The longest primitive cell observed for Sic has a repeat distance of 594 layers. A given sequence is repeated many times within a single crystal. The mecha- nism that induces such long-range crystallographic order is not a la force, but arises from spiral steps due to dislocations in the growtl (Chapter 20). ~ng-range I nucleus CRYSTAL STRUCTURE DATA In Table 3 we hst the more common crystal stmctureq and lattlce structures of the elements Values of the atomic concentration and the density are glven in Table 4. Many dements occur m several crystal structures and transform from Table 3 Crystal structures of the elements .rhe data given are at morn temperature for the most common form, Y. the stated temperature in deg K. (Inorganic Cr)-stal Stlucture Database Table 4 Density and atomic concentration The data are given at atmospheric pressure and mom temperature, or at the m z g ~ m r ~ ~ ~ m ~ e a m ~ s ~ m ~ ~ e x s m ~ r s ~ v v more stable. SUMMARY A lattice is an array of points related by the lattice translation operator T = ula, f uza2 f u,a3, where u,, us, u3 are integers and a,, a,, a, are the crystal axes. . To form a crystal we attach to every lattice point an identical basis composed of s atoms at the positions r , = xja, + y,a, + zja3, withj = 1,2, . . . , s. Here x, y, z may be selected to have values between 0 and 1. . The axes a,, a, a3 are primitive for the minimum cell volume la,. as X a,( for which the crystal can be constructed from a lattice translation operator T and a basis at every lattice point. Problems 1. Tetrahedral angles. The angles behveen the tetrahedral bonds of diamond are the same as the angles between the body diagonals of a cube, as in Fig. 10. Use elemen- tary vector analysis to find the value of the angle. 2. Indices o f planes. Consider the planes with indices (100) and (001); the lattice is fcc, and the indices refer to the conventional cubic cell. What are the indices of these planes when referred to the primitive axes of Fig. II? 3. Hcp structum. Show that the c/a ratio for an ideal hexagonal close-packed struc- ture is (:)IR = 1.633. If c/a is significantly larger than this value, the crystal structure may be thought of as composed of planes of closely packed atoms, the planes being loosely stacked. Wave Dvfraction and the Reciprocal Lattice DIFFRACTION OF WAVES B Y CRYSTALS 25 The Bragg law 25 SCATTERED WAVE AMPLITUDE 26 Fourier analysis 27 Reciprocal lattice vectors 29 Diffraction conditions 30 Laue equations 33 BRILLOUIN ZONES 32 Reciprocal lattice to sc lattice 34 Reciprocal lattice to hcc lattice 36 Reciprocal lattice to fcc lattice 37 FOURIER ANALYSIS OF THE BASIS 39 Structure factor of the hcc lattice 40 Structure factor of the fcc lattice 40 Atomic form factor 41 SUMMARY 43 PROBLEMS 43 1 . Interplanar separation 43 2. Hexagonal space lattice 44 3. Volume of Brillouin zone 44 4. Width of diffraction maximum 44 5. Structure factor of diamond 45 6. Form factor of atomic hydrogen 45 7. Diatomic line 45 Figure 1 Wavelength versus parti- cle energy, for photons, neutrons, and electrons Photon energy, keV Neuhon energy, 0 01 eV Elech-an energy, 100 PV Figure 2 Derivation of thr Bragg equation 2rl sin 8 = nA; here d is tile spacing of pardel atomic planes and Z m is the difference in phase between retlections from successive planer. The reflecting planes have nothing to do with the surface planes botlnding the particular specimen. CHAPTER 2: WAVE DIFFRACTION AND THE RECIPROCAL LATTICE DIFFRACTION OF WAVES BY CRYSTALS The Bragg law We study crystal structure through the diffraction of photons, neutrons, and electrons (Fig. 1). The diffraction depends on the crystal structure and on the wavelength. At optical wavelengths such as 5000 A, the superposition of the waves scattered elastically by the individual atoms of a crystal results in or- dinary optical refraction. When the wavelength of the radiation is comparable with or smaller than the lattice constant, we may find diffracted beams in directions quite different from the incident direction. W. L. Bragg presented a simple explanation of the diffracted beams from a crystal. The Bragg derivation is simple but is convincing only because it repro- duces the correct result. Suppose that the incident waves are reflected specu- larly from parallel planes of atoms in the crystal, with each plane reflecting only a very small fraction of the radiation, like a lightly silvered mirror. In specular (mirrorlike) reflection the angle of incidence is equal to the angle of reflection. The diffracted beams are found when the reflections from parallel planes of atoms interfere constructively, as in Fig. 2. We treat elastic scatter- ing, in which the energy of the x-ray is not changed on reflection. Consider parallel lattice planes spaced d apart. The radiation is incident in the plane of the paper. The path difference for rays reflected from adjacent planes is 2d sin 0, where 0 is measured from the plane. Constructive interfer- ence of the radiation from successive planes occurs when the path difference is an integral number n of wavelengths A, so that This is the Bragg law, which can he satisfied only for wavelength A 5 2d. Although the reflection from each plane is specular, for only certain values of 0 will the reflections from all periodic parallel planes add up in phase to give a strong reflected beam. If each plane were perfectly reflecting, only the first plane of a parallel set would see the radiation, and any wavelength would be re- flected. But each plane reflects to of the incident radiation, so that lo3 to lo5 planes may contribute to the formation of the Bragg-reflected beam in a perfect crystal. Reflection by a single plane of atoms is treated in Chapter 17 on surface physics. The Bragg law is a consequence of the periodicity of the lattice. Notice that the law does not referto the composition of the basis of atoms associated Incident bean / h m x-ray tube 01 r?a&or Bragg angle 0 -Undniated mmponmts of m m beam Figure 3 Sketch of a monochromator which by Bragg reflection selects a narrow spectrum of x-ray or neubon wavelengths tiom a broad spectrum incident heam. The upperpart of the f i p o shows the analysis (obtained by reflection from a secnnd crystal) of the purity of a 1.16 A beam of neutrons frum a calcil~m fluoride crystal monochromator (After G. Bacon.) Figure 4 X-ray diffractometer recording of powdered silicon, silowing a countor recording of the hffracted beams. (Courtesy of W. Parrisll.) with every lattice point We shall see, however, that the composition of the bas~s determmes the relative intens~ty of the various orders of diffraction (denoted by n above) from a given set of parallel planes Bragg reflection from a single crystal is $horn in Fig 3 and from a powder m Fig 4. SCATTERED WAVE AMPLITUDE The Bragg derivation of the hffraction condrtion (1) gives a neat state- ment of the condihon for the ~onstructive interference of waves scattered from the lamce points. We need a deeper analysis to determine the scattering 2 Reciprocal Lottics 27 intensity from the basis of atoms, whch means from the spahal distribution of electrons within each cell. Fozrrier Analysis We have seen that a crystal 1s ~nvariant under any translation of the form T = ulal + uza, + u3a3, where ul, u,, u, are integers and a,, as, a3 are the cr~stal axes. Any local physical property of the crystal, such a? the charge concentra- tion, electron number density, or magnetic moment density IS invanant under T. What is most lmportanf to us here 1s that the electron number den+ n(r) is a periodic function of r, w~th periods al, a,, a, in the direchons of the three crys- tal axes, respechvely. Thus Such penohcity creates an ideal situation for Fourier analys~s The most inter- estmg propemes of crystals are directly related to the Fourier components of the electron dens~ty We cons~der first a funchon n(x) in one bmension with period a in the brection x. We expand n(x) in a Fourier series of sines and cosines: where the p are positive integers and Cp, Sp are real constants, caned the Fourier coefficients of the expansion. The factor 27rIa in the arguments en- sures that n(z) has the period a: We say that 2rrpIa is a point in the reciprocal lattice or Fourier space of the crystal. In one dimension these points lie on a line. The reciprocal lattice points tell us the allowed terms in the Fourier series (4) or (5). A term is al- lowed if it is consistent with the periodicity of the crystal, as in Fig. 5; other Figure 5 perlad a, 4 1 1 2~ 2 1 1 4 1 1 may appt o o ( I J T n(x) = 21 riodic function. It is convenient to write the series (4) in the compact form where the sum is over all integers p: positive, negative, and zero. The coeffi- cients np now are complex numbers. To ensure that n(x) is a real function, we require nlp=nP , (6) for then the sum of the terms in p and - p is real. The asterisk on nl, denotes the complex conjugate of n -, . With q = 2~px/a, the sum of the terms in p and -p in (5) is real if (6) is satisfied. The sum is n,(cos q + i sin q) + n_,(cos q - i sin q) = (np + n_,)cos 9 f i(nP - n+in 9 , (7) which in turn is equal to the real function 2Re(np} cos q - 21m{nP} sin q (8) if (6) is satisfied. Here Re(nP) and Im(np] are real and denote the real and imaginary parts of n,,. Thus the number density n(x) is a real function, as desired. The extension of the Fourier analysis to periodic functions n(r) in three dimensions is straightfonvard. We must find a set of vectors G such that n(r)=): nc exp(iG. r) G (9) is invariant under all crystal translations T that leave the crystal invariant. It will be shown below that the set of Fourier coefficients n, determines the x-ray scattering amplitude. Inversion of Fourier Series. We now show that the Fourier coefficient n,, in the series (5) is given by n, =a-' dr n(x) exp(-i2npxla) . I (10) Substitute (5) in (10) to obtain . , % , . . . n, = a-' z nss loa dx expb27r(pr - p)x/a] . (11) P 2 Rsciprncal Lattice 29 If p' # p the valuie of the integral is because p' - p is an integer and exp[i2.ir(integer)] = 1. For the term p' = p the integrand is exp(i0) = 1, and the value of the integral is a, so that np = a-'npa = n,, which is an identity, so that (10) is an identity. As in (lo), the inversion of (9) gives ~ Here V, is the volume of a cell of the c~ystal. Reciprocal Lattice Vectors To proceed further with the Fourier analysis of the electron concentration we must find the vectors G of the Fourier sum Znr exp(iG . r) as In (9). There is a powerful, somewhat abstract procedure for doing this. The procedure forms the theoretical basis for much of solid state physics, where Fourier analysis is the order of the day. I We construct the axis vector? b,, b,, b3 of the reciprocal lattice: The factors 2.rr are not used by crystallographer? but are convenient in solid state physics. If al, a,, a, are primrtive vectors of the crystal lattice, then hl, h,, b3 are primitive vectors of the reciprocal lattice. Each vector defined by (13) is orthogonal to two axis vectors of the crystal lattice Thus h,, b,, bS have the property where6, = l i f i =jandS!, = O i f z i j . Points in the reciprocal lattice are mapped by the set of vectors where vl, v2, u2 are integers. A vector G of &is fonn is a reciprocal l a t t i c e vector. The vector? Gin the Fourier senes (9) are just the rec~procal lattice vectors (15), for then the Fourier senes reprcsentahon of the electron density has the desired io- varmce under any crystal translabon T = ulal + ups, + u,a,. From (9). The argument of the exponential has the form 2ni times an integer, because v,u, + u,u, + u3u, is an integer, being the sum of products of integers. Thus by (9) we have the desired invariance, n(r + T) = n(r) = 2 nc eq(iG . r). Every crystal strl~cture has two lattices associated with it, the crystal lattice and the reciprocal lattice. A diffraction pattern of a crystal is, as we shdl show, a map of the reciprocal lattice of the crystal. A microscope image, if it could be resolved on a fine enough scale, is a map of the crystal structure in real space. The two lattices are related by the definitions (13). Thus when we rotate a crys- taI in a holder, we rotate both the direct lattice and the reciprocal lattice. Vectors in the direct lattice have the dimensions of [length]; vectors in the reciprocal lattice have the dimensions of [I/length]. The reciprocal lattice is a lattice in the Fourier space associated with the crystal. The term is motivated below. Wavevectors are always drawn in Fourier space, so that every position in Fourier space may have a meaning as a description of a wave, but there is a special significance to the points defined by the set of G's associated with a crystal structure. Diffraction Conditions Theorem. The set of reciprocal lattice vectors G detemnnes the possible x-ray reflections We see in Flg 6 that the difference ~n phase factors is exp[i(k - k') . r] between beams scattered from volume elements r apart The wavevectors of the incoming and outgomg beams are k and k ' We suppose that the amplitude Figure 6 The difFerence inpath length oftheincident wave kat thepoints 0, r is rsin v, and the difference in phase angle is (2msin q)/A, which is equal to k . r. For the diffracted wave the dif- ference in phase angle is k ' . r. The total difference in phase angle is (k - k ' ) . r, and the wave scattered from rN at r has the pbase factor exp[i(k - k') . rl relative to the wave scattered from a vohlme element at the origin 0. stitution we can write (22) as This particular expression is often used as the condition for diffraction. Equation (23) is another statement of the Bragg condition (1). The result of Problem 1 is fbat the spacing djhkl) between parallel lattice planes that are normal to the direction G = hb, + kb, + lb, is djhkl) = ~T/IGI. Thus the result 2k . G = G2 may be written as 2(2m/A) sin 0 = Z?rld(hkl) , or 2d(hkl) sm 0 = A . Here B is the angle between the incident beam and the crystal plane. The integers hkl that define G are not necessarily den tical with the in- dices of an actual crystal plane, because the hkl may contain a common factor n, whereas in the defiuition of the indices m Chapter 1 the common factor has been ehminated. We thus obtain the Bragg result. where d is the spacing between adjacent parallel planes with indices htn, kln, l/n. h u e Equations The original result (21) of diffraction theory, namely that Ak = G, may be expressed in another way to give what are called the Laue equations These are valuable because of their geometrical representation. Take the scalar prod- uct of both Ak and G successively withal, a,, a , . From (14) and (15) we get These equations have a simple geometrical interpretation. The first equation a, . Ak = 2ml tells us that Ak lies on a certam cone about the direction of a,. The second equation tells us that Ak lies on a cone about a , as well, and the third equation requlres that Ak lies on a cone about a3. Thus, at a reflection Ak must satisfy all three equations: it must lie at the common line of mtersec- tion of three cones, which is a severe condition that can be satisfied only by systematic sweeplng or searching in wavelength or crystal orientation-or by sheer accident. A beauhful construction, the Ewald construction, 1s exhibited in Fig. 8. This helps us visualize the nature of the accident that must occur in order to satisfy the diffraction condition in three dimensions. Figure 8 The points on ills tight-hand side are reciprocal-latticr points of the crystal. The vector k is drawn in the dircctioi> of the incident x-ray hcam, and the migin in chosen such that k term- nates at any reciprocal lattice point. We h a w a sphere of radius k = 2wlA about the oligin of k. A diffracted beam will he formed if this sphere intersects any other point in the reciprocal lattice. The sphere as d r a m intercepts a p i n t connected with the end of k by a reciprocal lattice vedor G. The diffracted x-ray beam is in the direction k ' = k + G. The angle B is the Bragg angle of Fig. 2. This constructionis due to P. P Ewald. BRZLLOUIN ZONES Brillouin gave the statement of the diffraction condition that is most widely used in solid state physics, which means in the description of electron energy band theory and of the elementary excitations of other kinds. A Brillouin zone is defined as a Wigner-Seitz primitive cell in the reciprocal lat- tice. (Tlie construction in the direct lattice was shown in Fig. 1.4.) The Brillouin zone gives a vivid geometrical interpretation of the diffraction condi- tion 2k . G = GZof Eq. (23). We divide both sides by 4 to obtain We now work in reciprocal space, the space of the k's and G's. Select a vector G from the origin to a reciprocal lattice point. Construct a plane normal to this vector G at its midpoint. This plane forms a part of a zone boundary (Fig. 9a). An x-ray beam in the crystal will he diffracted if its wavevector k has the magnitude and direction required by (26). The diffracted beam will then be in the direction k - G, as we see from (19) with Ak = -G. 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Figure 10 Conshuction of the fitst Brilloain zone for an oblique lattioe in two dimensions. We first draw a number of vectors from 0 to nearby . points in the reciprocal lattice. Next we construct lines perpendicular to these vectors at their mid- paints. Tbe smallest endosed area is the first Btil- lauin zone. k = -' k = ' Figure 11 Crystal and reciprocal lattices in one dimension. The basis vector in the reciprocal lat- tice k b, of length equal to ZrIa. The shortest reciprocal latticevectors from the origin are b and -b The perpendicular bisectors of these vectors form the boundaries of the first BriUouin zone. The boundaries are at k = 2wIa. Here 9, i i are orthogonal vectors of unit length. The volume of the cell is al . a, X a3 = a3. The primitive translationvectoi's of the reciprocal lattice are found from the standard prescription (13): - 2 ) ; b, = (27r/a)9 ; b, = (2?r/a)Z . (27b) Here the reciprocal lattice is itself a simple cubic lattice, now of lattice constant 2da. Figme 12 Pnm~hve bmls vectors of the body centered mbrc lattice Figure 13 Ftrst Brilloum zone of the body- centered cublc lathce The fipre a a regular rbomb~c dodecahedron The boundaries of the first Brillduin zones are the planes normal to the six reciprocal lattice vectors ?b,, ?bz, Cb, at their rmdpoints: -ti bl = ?(?ria)% ; 2 b = a ; 2; b, = ? ( d a ) i . (28) The six planes bound a cube of edge 2w/a and of volume ( 2 ~ 1 ~ ) ~ ; this cube is the first Brillouin zone of the sc crystal lattice. Reciprocal Lattice to bcc Lattice The primitive translation vectors of the bcc lattice (Fig. 12) are ax=$.(-%+f+2) ; % = ; a ( ; - f + i ) ; a3=$a(%+f-2) , (29) where a 1s the side of the conventional cube and 4 f, 2 are orthogonal unit vectors parallel to the cube edges The volume of the primtive cell is V = la,.s,xa31=;a3 . (30) The prim~tive translations of the reciprocal lattice are defined by (13). We have, using (28), b = ( 2 a ( + 2) ; b = ( 2 a + 2) ; b, = (2da)(%+ 9) . (31) Note by cornpanson with Fig. 14 (p. 37) that these are just the primitive vectors of an fcc lattice, so that an fcc lattice is the reciprocal lattice of the bcc lattice. The general rec~procal lattice vector is, for integral u,, o,, o,, G = vlbl + nabz + v,b3 = (2.rr/a)[(uz + u,)% + (0, + 03)f + (vl + v2)21 . (32) 2 Reciprocal L d t i ~ e 37 3 Figure 14 Pnm~tlve basrs vectors of the face-centered cttblc lathce The shortest G's are the following 12 vectors, where all choices of sign are independent. One primitive cell of the reciprocal Iattice is the parallelepiped described by the b,, bb b3 defined by (31). The volume of this cell in reciprocal space is b, . b2 X b, = 2(2~/a)~. The cell contains one reciprocal lattice point, because each of the eight corner points is shared among eight parallelepipeds. Each parallelepiped contains one-eighth of each of eight comer points (see Fig. 12). Another primitive cell is the central (Wigner-Seitz) cell of the reciprocal lattice which is the first Brillouin zone. Each such cell contains one lattice point at the central point of the cell. This zone (for the hcc lattice) is bounded by the planes normal to the 12 vectors of Eq. (33) at their midpoints. The zone is a regular 12-faced solid, a rhombic dodecahedron, as shown in Fig. 13. Reciprocal Lattice to fec Lattice The primitive translation vectors of the fcc lattice of Fig. 14 are The volume of the primitive cell is 'I' 4wla Figore 15 Brlllouin zones of the face-centered cubic lattice. The cells are i n reciprocal space, and the reciprocal lattice i s body centered. The primrtlve translabon vectors of the lattice reciprocal to the fcc lattice are These are pnmltlve translation vectors of a bcc lattice, so that the bcc lattice is reciprocal to the fcc lattice The volume of the primibve cell of the reciprocal la&= is 4 ( 2 ? r l ~ ) ~ The shorte~t G's are the eight vectors: The boundaries of the ceneal cell in the reciprocal latt~ce are determined for the most part by the e~ght planes normal to these vectors at them midpoints. But the corners of the octahedron thus formed are cut by the planes that are the perpendicular b~qectors of s i x other rec~procal lattice vectors Note that (Zda)(G) 1s a reciprocal lattice vector because it is equal to b, + b, The first Brillonin zone is the smallest bounded volume about the or~gin, the truncated octahedron shown m Fig 15. The six planes bound a cube of edge &la and (before truncation) of voh~me ( 4 ~ r I a ) ~ . 2 Recipmed Lattice 39 FOURIER ANALYSIS OF THE BASIS When the diffraction condition A k = G of Eq. (21) is satisfied, the scatter- ing amplitude (18) for a crystal of N cells may be written as The quantity S, is called the structure factor and is defined as an integral over a slngle cell, with r = 0 at one corner. Often it is useful to write the electron concentration n(r) as the super- position of electron concentration functions nl associated with each atom J of the cell. If r, 1 s the vector to the center of atom j, then the function nl(r - rl) defines the contribution of that atom to the electron concentration at r. The total electron concentration at r due to all atoms m the smgle cell is the sum over the s atoms of the basis. The decornposifion of njr) is not unique, for we cannot always say how much charge density IS assoc~ated mth each atom. This is not an important difficulty. The structure factor defined by (39) may now be written as integrals over the s atoms of a cell. where p = r - q . We now define the atomic form factor as integrated over all space. If nl(p) is an atomic property,fi is an atomic property. We combine (4l)and (42) to obtain the structure factor of the basis in the form S, = z f , e q ( - z ~ . r , ) . (43) J The usual form of this result follows on writing for atomj: 5 = xlal + ylaz + zla3 , so that (43) becomes The structure factor S need not be real because the scattered intensity will involve S'S, where S' is the complex conjugate of S so that S'S is real. Structure Factor of the bcc Lattice The bcc basis referred to the cnbic cell has identical atoms at x, = yl = z, = 0 and at x2 = ys = z2 = 2 . Thus (46) becomes where f 1s the form factor of an atom. The value of S is zero whenever the exponential has the value -1, which IS whenever the argument is +rr X (odd mteger) Thus we have S = 0 when u, + v2 + u, = odd integer ; S = 2 f when u, + u2 + u3 = even integer Metallic sod~um has a bcc structure. The diffraction pattern does not con- tainlines such as (loo), (300), (Ill), or (221), butlines such as (ZOO), (110). and (222) wdl be present: here the indices (olueu3) are referred to a cubic cell. What is the physical interpretahon of the result that the (100) reflection vanishes? The (100) reflection normally occurs when reflections from the planes that hound the cubic cell differ in phase by 27r. In the bcc lattice there is an inter- veningplane (Fig. 16) of atoms, labeled the second plane in the figure, which is equal ~n scattering power to the other planes. Situated midway between them, it gives a reflection retarded m phase by 7r with respect to the first plane, thereby canceling the contribution from that plane The cancellahon of the (100) reflection occurs m the bcc lattice because the planes are identical in composition. A similar cancellation can easily be found in the hcp structure. Structure Factor of the fcc Lattice The basis of the fcc structure referred to the cubic cell has identical atoms at 000; G; +@; S O . Thus (46) becomes 2 Reciprocal Lanice 41 Figurc 16 Explanation uf t l ~ r absence of a (100) reflection from a body-centered cnbic lattice. The plrasr difference between s~lccessivo planes is w, so that the reflected amplitude from two adjacent planes is 1 + a-'" = 1 - 1 = 0. If all mndices are even mtegers, S = 4f; similarly if all indices are odd mntegers. But if only one of the Integers is even, two of the exponents will be odd multi- ples of -zw and Swill vamrh. If only one of the integers is odd, the same argu- ment applies and S will also van~sh. Thus in the fcc lathce no reflections can occur for wlnch the Indices are partly even and partly odd. The point is heautifnlly illustrated by Fig 17 both KC1 and KBr have an fcc lattice, but n(r) for KC1 simulates an sc lathce because the K+ and C 1 ions have equal numbers of electrons. Atomic Form Factor In the expression (46) for the structure factor, there occurs the quanbtyf;, which is a measure of the scattenng power of theyth atom in the unit cell. The value off involve? the number and distribut~on of atomic electrons, and the wavelength and angle of scattering of the radiation. We now give a classical calculation of the scatter~ng factor. The scattered rad~abon from a single atom takes account of interference effects withln the atom We defined the form factor in (42): with the integral extended over the electron concentration associated with a single atom. Let r make an angle ru with G; then G . r = Gr cos a. If the elec- tron dntribution is spherically symmetnc about the origin, then fi = 2 1 1 J dr P d(cos a) n,(r) exp-iGr cos a) Figure 17 Comparison of x-ray reflections horn KC1 and KBr powders. In KC1 the numbers of electrons of K t and CT ions are equal. The scatteliog ampli- tudes ill(+) and f(C1-) me almost exactly equal, so that the crystal looks to a-rays as i f it were a monatomic simple cc~bic laKim of lattice constant a12. Only even integers occnr in the reflection indices when ttiese are based an n cuhic lam= of lattice can- stant a. In KBr the form factor of R r is qc"tc differ- ent to that of K. and all reflections of the fcc lattice are presxnt. (Courtesy of R. van Norristrand.1 I I I I I I (200)' KBr (220) A 7 S O 0 70" 60" 50" 40" 30" 20' -28 after integration over d(cos a) between -1 and 1. Thus the form factor is given by If the same total electron density were concentrated at r = 0, only Gr = 0 would contribute to theintegrand. In thls limit (sin Gr)lGr = 1, and f j = 4 7 1 J dr nl(r)P = Z , (51) the number of atomic electrons Therefore f is the ratio of the radiation ampli- tude ~cattered by the actual electron distnbubon in an atom to that scattered by one electron localized at a point. In the fonvard direction G = 0, and f reduces again to the value Z The overall electron distribution m a solid as seen m x-ray diffraction is fairly close to that of the appropriate free atoms Th~s statement does not mean that the outermost or valence electrons are not red~stnbuted somewhat m forming the solid: it means only that the x-ray reflection intensities are represented well by the free atom values of the form factors and are not vely sensitive to small rehstributions of the electrons. SUMMARY . Various statements of the Bragg condition: 2dsinO=nA ; A k = G ; 2 k . G = G 2 . Lane conditions: al.Ak=2m, ; s,.Ak=2mz ; a,.Ak=2m3 . The primitive translation vectors of the reciprocal lattice are Here al, az, a, are the primitive translation vectors of the crystal lattice. . A reciprocal lattice vector has the form G = o,b, + 02bz + o,b, , where ol, o,, u, are integers or zero. The scattered amplitude in the direction k' = k + Ak = k + G is propor. tional to the geometrical structure factor: where j runs over the s atoms of the basis, and& is the atomic form factor (49) of the jth atom of the basis. The expression on the right-hand side is written for a reflection (u,u~,), for which G = o,b, + 02b2 + 03b3. . Any function invariant under a lattice translation T may be expanded in a Fourier series of the form The first Brillouin zone is the Wigner-Seitz primitive cell of the reciprocal lattice. Only waves whose wavevector k drawn from the origin terminates on a surface of the Brillouin zone can be diffracted by the crystal. . Crystal lattice First Brillouin zone Simple cubic Cube Body-centered cubic Rhombic dodecahedron (Fig. 13) Face-centered cubic Truncated octahedron (Fig. 15) Problems 1. Interplanor separation. Consider a plane hkl in a crystal lattice. (a) Prove that the reciprocal lattice vector G = hb, + kb2 + lb3 is perpendicular to this plane. (h) Prove that the distance between two adjacent parallel planes of the lattice is d(hkl) = 27r//GI. (c) Show for a simple cubic lattice that d2 = aP/(h2 + k2 + 1'). a, = (3"a/Z)i + (a/2)9 ; q = -(31na/2)i + (a/2)9 ; a, = & . (a) Show that the volume of the primitive cell is (3"V2)a2c. (b) Show that the primitive translations of the reciprocal lattice are b, = (25~/3~a)i + (2nlag ; b, = - ( 2 ~ r / 3 ~ a ) i + (2nIa)j. ; b, = (2vIc)i , so that the lattice is its own reciprocal, hut with a rotation of axes. (c) Describe and sketch the first Brillouin zone of the hexagonal space lattice 3. Volume of Brillowin zone. Show that the volume of the first Brillouin zone is ( 2 ~ ) ~ i V ~ . where V, is the volume of a clystal primitive cell. Hint: The volume of a Brillouin zone is equal to the volume of the primitive parallelepiped in Fourier space. Recall the vector identity (c X a) X (a X b) = (c . a X b)a . 4. Width of diffraction maximum. We suppose that in a linear crystal there are identical point scattering centers at every lattice point p_ = ma, where m is an inte- ger. By analogy with (20), the total scattered radiation amplitude will be proportional to F = 2 exp[-ima . Ak]. The sum over M lattice points is 1 - exp[-iM(a. Ak] F = 1 - exp-i(a. Ak)] by the use of the series (a) The scattered intensity is proportional to IFP. Show that IFP-PF= sin2$ M(a . Ak) sin2 (a . Ak) (b) We know that a diffraction maximum appears when a . Ak = 2nh, where h is an integer. We change Ak slightly and define E in a . Ak = 2nh + E such that gives the position of the first zero in s i n i ~ ( a . Ak). Show that E = 2n/M, so that the width of the diffraction maximum is proportional to 1/M and can be extremely narrow for macroscopic values of M. The same result holds true for a three-dimensional crystal. 5. Structure factor of diamond. The crystal structure of diamond is described in Chapter 1. The basis consists of eight atoms if the cell is taken as the conventional cube. (a) Find the structure factor S of this basis. (b) Find the zeros of S and show that the allowed reflections of the diamond structure satisfy v , + o, + o, = 4n, where all indices are even and n is any integer, or else all indices are odd (Fig. 18). (Notice that h, k, I may be written for o,, u,, u, and this is often done.) 6. Form factor of atomic hydrogen. For the hydrogen atom in its ground state, the number density is n(r) = (ma;)-' exp(-2r/%), where a , , is the Bohr radius. Show that the form factor is f , = 16/(4 + G24)'. 2 Reciprocal lattice 45 Counter position 28 Figure 18 Neutron diffraction pattern for powdered diamond. (After G. Bacon.) i 7. Diatomic line. Consider a Line of atoms ABAB . . AB, with an A-B bond length of ;a. The form factors are fA, fa for atoms A, B, respectively. The incident beam of x-rays is perpendicular to the Line of atoms. (a) Show that the interference condition is n A = a cos 0, where 0 is the angle between the diffracted beam and the line of atoms. (b) Show that the intensity of the diffracted beam is proportional to If, - f a r for n odd, and to 1 f , + fB 1% for n even. (c) Explain what happens iffA =fa. Crystal Binding and Elastic Constants CRYSTALS OF INERT GASES 49 Van der Waals-London Interaction 53 Repulsive Interaction 56 Equilibrium Lattice Constants 58 Cohesive Energy 59 IONIC CRYSTALS 60 Electrostatic or Madelung Energy 60 Evaluation of the Madelung Constant 64 COVALENT CRYSTALS 67 METALS 69 HYDROGEN BONDS 70 ATOMIC RADII 70 Ionic Crystal Radii 72 ANALYSIS O F ELASTIC STRAINS 73 Dilation 75 Stress Components 75 ELASTIC COMPLIANCE AND STIFFNESS CONSTANTS 77 Elastic Energy Density 77 Elastic Stiffness Constants of Cubic Crystals 78 Bulk Modulus and Compressibility 80 ELASTIC WAVES IN CUBIC CRYSTALS 80 Waves in the [loo] Direction 81 Waves in the [I101 Direction 82 SUMMARY 85 PROBLEMS 85 1. Quantum solid 85 2. Cohesive energy of bce and fcc neon 86 3. Solid molecular hydrogen 86 7. Divalent ionic crystals 8. Young's modulus and Poisson's ratio 9. Longitudinal wave velocity 1 0 . Transverse wave velocity 11. Effective shear constant 12. Determinantal approach 13. General propagation direction 14. Stability criteria Figure 1 Thc principal types of crystalline binding. In (a) neutral atoms with closed electron shells arc bound together weakly by the van der Wads forces associated with fluctuations in the charge distributions. In (b) electrons are transferred from the alkali atoms to the halogen atoms, and the restdting ions are held together by attractive cloctrostatic forces between the positive and negative ions. In ( c ) the valence electrons are taken away from each alkali atom to form a commu- nal electron sea in which the positive ions are dispersed. In (d) the neutral atoms are bo~md to- gether by the overlapping parts of their electron disttib~~tions. CHAPTER 3 : CRYSTAL BINDING AND ELASTIC CONSTANTS In this chapter we are concerned with the question: What holds a crystal together? The attractive electrostatic interaction between the negative charges of the electrons and the positive charges of the nuclei is entirely responsible for the cohesion of solids. Magnetic forces have only a weak effect on cohe- sion, and gravitational forces are negligible. Specialized terms categorize dis- tinctive situations: exchange energy, van der Wads forces, and covalent bonds. The observed differences between the forms of condensed matter are caused in the final analysis by differences in the distribution of the outermost elec- : trons and the ion cores (Fig. 1). The cohesive energy of a crystal is defined as the energy that must be added to the crystal to separate its components into neutral free atoms at rest, : at infinite separation, with the same electronic configuration. The term lattice I energy is used in the discussion of ionic crystals and is defined as the energy j that must be added to the crystal to separate its component ions into free ions at rest at infinite separation. Values of the cohesive energy of the crystalline elements are given in Table 1. Notice the wide variation in cohesive energy between different columns of the periodic table. The inert gas crystals are weakly bound, with cohesive energies less than a few percent of the cohesive energies of the ele- 1 ments in the C, Si, Ge . . . column. The alkali metal crystals have intermediate ' values of the cohesive energy, The transition element metals (in the middle f columns) are quite strongly bound. The melting temperatures (Table 2) and 1 bulk modulii (Table 3) vary roughly as the cohesive energies. CRYSTALS OF INERT GASES The inert gases form the simplest crystals. The electron distribution is very close to that of the free atoms. Their properties at absolute zero are sum- marized in Table 4. The crystals are transparent insulators, weakly hound, with low melting temperatures. The atoms have very high ionization energies (see Table 5). The outermost electron shells of the atoms are completely filled, and the distribution of electron charge in the free atom is spherically symmetric. In the crystal the inert gas atoms pack together as closely as possible1: the 'Zer~-~oint motion of the atoms (Idnetic energy at absolute zero) is a quantum effect that plays a dominant role in He3 and He4. They do not solidify at zero pressure even at absolute zero temp- erature. The average fluctuation at 0 K of a He atom fmm its equilibrium position is of the order of ' 30 to 40 percent of the nearest-neighbor distance. The heavier the atom, the less important the zero- point effects. If we omit zero-point motion, we calculate a molar volume of 9 cm3 mol-' for solid . helium, as compared with the obsemd values of 27.5 and 36.8 cm3 mol-' for liquid He4 and liquid He3, respectively. Table 2 Melting points, in K. (After R H Larnoreaux) m~sssss~~~m~E~Esvw#~m~swsvwe~~s~~~v~~w#w~Ems~ss Li 4 5 3 .
7 Na 3 7 1 .
0 2365 Be 1562 Mg 922 B C N O F N e 63.15 K 3 3 6 .
3 Rb 3 1 2 .
6 Cs 3 0 1 6 Fr 2 4 .
5 6 5 4 .
3 6 Ca 1 1 1 3 Sr 1 0 4 2 Ba 1 0 0 2 Ra 973 53.48 Ar 8 3 .
8 1 Th 2 0 3 1 Al 9 3 3 .
5 Ni 1 7 2 8 Sc 1814 S 3 8 8 .
4 Cr 2 1 3 3 Y 1 8 0 1 La 1 1 9 4 Ac 1 3 -9 4 CI 1 7 2 2 SI 1687 Cu 1 3 5 8 Ti 1 9 4 6 Pa la8 P w 317 r 8 6 3 V 2202 I I I I I I I I I I I I I 1 Zn 6927 Co 1 7 7 0 Mn 1 5 2 0 Z r 2 1 2 6 U 1 4 0 8 Am 1 4 4 9 Fe 1 8 1 1 Es Ga 3 0 2 .
9 Np 910 Cm 1 6 1 3 Nb 2750 Pu 913 Fm Ge 1 2 1 1 Mo 2 8 9 5 Hf Ta 2504 3293 Bk 1 5 6 2 Rh Pd Ag Cd In Sn Sb Te I Xe 2 2 3 6 1827 1 2 3 5 5 9 4 3 4298 505 1 903 9 7 2 2 .
7 3 8 6 7 1 6 1 .
4 Tc 2477 W 3695 Cf Md As 1 0 8 9 Ru 2527 Ir 2720 Re 3 4 5 9 I I I I I I I I I I I I I I \ 0 s 3 3 0 6 Ce 1 0 7 2 No Lw Kr 1158 Se 494 Br 2 6 5 .
9 Pr 1 2 0 5 Nd 1 2 9 0 Eu 1 0 9 1 Ho 1745 Pm Gd 1 5 8 7 Sm 1 3 4 6 Er 1797 Tb 1 6 3 2 Dy 1 6 8 4 Tm 1 8 2 0 Yb 1 0 9 8 Lu 1 9 3 8 Table 3 Isothermal bulk madulii and cnrnpressibilities at mom temperature U I N After K. Cschneidner, Jr., Solid State Physics 16, 275-426 (1964), several data are from F. Birch, in Handbook o f physical constants, Geological Soci- ety of America Memoir 97, 107-173 (1966). Original references should be consulted when values are needed for research purposes. Values in paren- .theses are estimates. Letters in parentheses refer to the crystal form. Let- ters in brackets refer to the temperature: [a] = 77 K; [b] = 273 K; [c] = 1 K; [dl = 4 K; [el = 81 K. ~llCib3JIIs25#5~~13ir#ifil:fII~?:?:EIi:~IZIZ3i?#i?llI~?1?t~ll#~#IJ~W~ii~f~ - I ?E81 : : T I .... B,ik rnooL,Ls in un,ts l o " dyn/cm2 or 10" NIW ComDie~s:Wm, n w h y 10 'zcmbaye o r 10. ' nlLlP 3 Crystal Binding 53 Table 4 Properties of i n e r t gas crystals (Extrapolated to 0 K and zero pressure) Parameters in Experimental Lennard-Jones Nearest- Ionization cohesive potential, Eq. 10 neighbor potential energy distance, Melting of free 6, 0, in k kT/mol eVIatom point, K atom, eV in 10-''erg i n k He (liquid at zero pressure) 24.58 14 2.56 Ne 3.13 1.88 0.02 24.56 21.56 50 2.74 A r 3.76 7.74 0.080 83.81 15.76 167 3.40 Kr 4.01 11.2 0.116 115.8 14.00 225 3.65 Xe 4.35 16.0 0.17 161.4 12.13 320 3.98 crystal structures (Fig. 2) are all cubic close-packed (fcc), except He3 and He4. What holds an inert gas crystal together? The electron distribution in the crystal is not significantly distorted from the electron distribution around the free atoms because not much energy is available to distort the free atom charge distributions. The cohesive energy of an atom in the crystal is only 1 percent or less of the ionization energy of an atomic electron. Part of this distortion gives the van der Wads interaction. Van der Waals-London Interaction Consider two identical inert gas atoms at a separation R large in compari- son with the radii of the atoms. What interactions exist between the two neu- tral atoms? If the charge distributions on the atoms were rigid, the interaction between atoms would be zero, because the electrostatic potential of a spheri- cal distribution of electronic charge is canceled outside a neutral atom by the electrostatic potential of the charge on the nucleus. Then the inert gas atoms could show no cohesion and could not condense. But the atoms induce dipole moments in each other, and the induced moments cause an attractive interac- tion between the atoms. As a model, we consider two identical linear harmonic oscillators 1 and 2 separated by R. Each oscillator bears charges 5 e with separations 1-1 and x2, as in Fig. 3. The particles oscillate along the x axis. Let p, andp, denote the momenta. The force constant is C. Then the hamiltonian of the unperturbed system is Each uncoupled oscillator is assumed to have the frequency o, of the strongest optical absorption line of the atom. Thus C = mog. Table 5 Ionization energies 3 Crystal Binding 55 Figure 2 Cubic cl~sc-~x~cked (fcc) crystal structure uf the imrt ~ R S C F Ne, AL Kc and Xe. The lat. tice parameters of thc cubic cells arc 4.46, 5.31, 5.64, and 613 A, r~spectivcly, at 4 4. Figme 3 Coordinates of the two oscillators Let XI be the coulomb interaction energy of the two oscillators. The geometlyis shown in the fipre. The internuclear coordinate is R. Then in the approximation lx, 1, ix, / R we expand (2) to obtain in lowest order: The total hamiltonian with the approximate form (3) for XI can he diago- nalized by the normal mode transformation The subscripts s and a denote symmetric and antisymmetric modes of motion. Further, we have the momenta p,, p, associated with the two modes: The total hamiltonian 'Xo + 'X, after the transformations (5) and (6) is The two frequencies of the coupled oscillators are found by inspection of (7) to be with w, given by (C/m)l'Z. In (8) we have expanded the square root. The zero point energy of the system is $fi(w, + w,); because of the interac- tion the sum is lowered from the uncoupled value 2 . $firno by This attractive interaction varies as the minus sixth power of the separation of the two oscillators. This is called the van der Wads interaction, known also as the London in- teraction or the induced dipole-dipole interaction. It is the principal attractive interaction in crystals of inert gases and also in crystals of many organic mole- cules. The interaction is a quantum effect, in the sense that AU + 0 as fi - 0. Thus the zero point energy of the system is lowered by the dipole-dipole cou- pling of Eq. (3). The van der Waals interaction does not depend for its exis- tence on any overlap of the charge densities of the two atoms. An approximate value of the constant A in (9) for identical atoms is given by fiw,a2, where fiw, is the energy of the strongest optical absorption line and a is the electronic p~larizabilit~ (Chapter 15). Repulsive Interaction As the two atoms are brought together, their charge distributions gradually overlap (Fig. 4), thereby changing the electrostatic energy of the system. At sufficiently close separations the overlap energy is repulsive, in large part he- cause of the Pauli exclusion principle. The elementary statement of the principle is that two electrons cannot have all their quantum numbers equal. When the charge distributions of two atoms overlap, there is a tendency for electrons from atom B to occupy in part states of atom A already occupied by electrons of atom A, and vice versa. 3 Crystal Binding 57 Figure 4 Electronic charge distribu- tions overlap as atoms approach. The solid circles denote the nuclei. 0 Total (a) energy: e l -n 7 8 98 eV lstlsl Total spin zero Total elearon energy -59 38 eV 1st IS? ISTZST Total spin one Figure 5 The effect of Pauli principle on the repulsive energy: in an extreme example, two hydro- gen atoms are pushed together until the protons are almost in contact. The energy of the electron system alone can be taken from observations on atomic He, which has two electrons. In (a) the elec- trons have antipardel spins and t h e Pauli principle has no effect: the electrons are bound by 78.98 eV In (b) the spins are parallel: the Pauli principle forces the promotion of an electron from a 1s l . orbital of H to a 2s f orbital of He. The electrons now are bound by -59.38 eV, less than (a) by 19.60 eV This is the amount by which the Pauli principle has increased the repulsion. We have omitted the repulsive coulomb energy of the two protons, which is the same in both (a) and (b). The Pauli principle prevents multiple occupancy, and electron distribu- tions of atoms with closed shells can overlap only if accompanied by the partial promotion of electrons to unoccupied high energy states of the atoms. Thus the electron overlap increases the total energyof the system and gives a repul- sive contribution to the interaction. An extreme example in which the overlap is complete is shown in Fig. 5. We make no attempt here to evaluate the repulsive interaction2 from first principles. Experimental data on the inert gases can be fitted well by an empirical repulsive potential of the form B/R12, where B is a positive constant, when used 'The overlap energy naturally depends on the radial distribution of charge about each atom. The mathematical calculation is always complicated even if the charge distribution is known. Nu+ Figure 6 Form of the Lemard-Jones potential (10) which describes the interaction of hm, inert gas atoms. The minimum orcurs at Nu = 2 " e 1.12. Notice how steep the curve is inside the minimum, and how Bat it is outside the minimum. The value of U at the minimum is -E; and U = 0 at R = u. together with a long-range attractive potential of the form of (9). The constants A and B are empirical parameters determined from independent measurements made in the gas phase; the data used include the virial coefficients and the viscos- ity. It is usual to write the total potential energy of two atoms at separation R as where E and u are the new parameters, with 4eu6 = A and ~ E U " = B. The potential (10) is h o w n as the Lennard-Jones potential, Fig. 6. The force between the two atoms is given by -dU/dR. Values of E and u given in Table 4 can be obtained from gas-phase data, so that calculations on properties of the solid do not involve disposable parameters. Other empirical forms for the repulsive interaction are widely used, in par- ticular the exponential form A exp(-Wp), where p is a measure of the range of the interaction. This is generally as easy to handle analytically as the inverse power law form. Equilibrium Lattice Constants If we neglect the kinetic energy of the inert gas atoms, the cohesive en- ergy of an inert gas crystal is given by summing the Lennard-Jones potential (10) over all pairs of atoms in the crystal. If there are N atoms in the crystal, the total potential energy is 3 Crystal Binding 59 where pYR is the distance between reference atom i and any other atom j, ex- pressed in terms of the nearest-neighbor distance R. The factor ; occurs with the N to compensate for counting twice each pair of atoms. The summations in (11) have been evaluated, and for the fcc structure There are 12 nearest-neighbor sites in the fcc structure; we see that the series are rapidly converging and have values not far from 12. The nearest neighbors contribute most of the interaction energy of inert gas crystals. The corre- sponding sums for the hcp structure are 12.13229 and 14.45489. If we take Ut,, in (11) as the total energy of the crystal, the equilibrium value R, is given by requiring that U,,, be a minimum with respect to variations in the nearest-neighbor distance R: whence the same for all elements with an fcc structure. The observed values of Rdu, using the independently determined values of u given in Table 4, are: The agreement with (14) is remarkable. The slight departure of R o l u for the lighter atoms from the universal value 1.09 ~redicted for inert gases can he ex- plained by zero-point quantum effects. From measurements on the gas phase we have predicted the lattice constant of the clystal. Cohesive Energy The cohesive energy of inert gas crystals at absolute zero and at zero pres- sure is obtained by substituting (12) and (14) in (11): and, at R = R,, U,,,(&) = -(2.15)(m€) , (16) the same for all inert gases. This is the calculated cohesive energy when the atoms are at rest. Quantum-mechanical corrections act to reduce the binding by 28, 10, 6, and 4 percent of Eq. (16) for Ne, Ar, Kr, and Xe, respectively. stand the origin of the quantum correction by consideration of a simple model in which an atom is confined by fixed boundaries. If the particle has the quan- tum wavelength A, where A is determined by the boundaries, then the particle has kinetic energy p2/2M = ( h l ~ ) ~ l Z M with the de Broglie relation p = hlA for the connection between the momentum and the wavelength of a particle. On this model the quantum zero-point correction to the energy is inversely pro- portional to the mass. The final calculated cohesive energies agree with the ex- perimental values of Table 4 within 1 to 7percent. One consequence of the quantum kinetic energy is that a crystal of the iso- tope Ne2" is observed to have a larger lattice constant than a crystal of NeZ2. The higher quantum kinetic energy of the lighter isotope expands the lattice because the kinetic energy is reduced by expansion. The observed lattice constants (extrapolated to absolute zero from 2.5 K) are NeZ0, 4.4644 A; NeZ2, 4.4559 A. IONIC CRYSTALS Ionic crystals are made up of positive and negative ions. The ionic bond results from the electrostatic interaction of oppositely charged ions. Two com- mon crystal structures found for ionic crystals, the sodium chloride and the ce- sium chloride structures, were shown in Chapter 1. The electronic configurations of all ions of a simple ionic crystal corre- spond to closed electronic shells, as in the inert gas atoms. In lithium fluoride the configuration of the neutral atoms are, according to the periodic table in the front endpapers of this book, Li: ls22s, F: ls22s22p5. The singly charged ions have the configurations Li+:ls2, F-: ls22s22p6, as for helium and neon, re- spectively. Inert gas atoms have closed shells, and the charge distributions are spherically symmetric. We expect that the charge distributions on each ion in an ionic crystal will have approximately spherical symmetry, with some distor- tion near the region of contact with neighboring atoms. This picture is con- firmed by x-ray studies of electron distributions (Fig. 7). A quick estimate suggests that we are not misguided in looking to electro- static interactions for a large part of the binding energy of an ionic crystal. The distance between a positive ion and the nearest negative ion in crystallirle sodium chloride is 2.81 X lo-' cm, and the attractive coulomb part of the potential energy of the two ions by themselves is 5.1 eV. This value may be compared (Fig. 8) with the experimental value of 7.9 eV per molecular unit for the lattice energy of crystalline NaCl with respect to separated Naf and CI- ions. We now calculate the energy more closely Electrostatic or Madelung Energy The long-range interaction between ions with charge ?q is the electrostatic interaction +q2/r, attractive between ions of opposite charge and repulsive 3 Crystal Binding 61 Figure 7 Electron density distribution in the base plane of NaCI, after x-ray sbdies by G. Schoknecht. The numbers an the contours @ve the relative electron concentration. Elechon I G a &ty Figure 8 The energyper molecule unit of a crys- tal of sodium chloride is (7.9 - 5.1 + 3.6) = 6.4 eV lower than the energy of separated neutral atoms. The lattice energy with respect to separated ions 4 7 o a r is 7.9 eV oer molecule unit. All values on the fie- I eneW tton affmity are given in Table 6. between ions of the same charge. The ions arrange themselves in whatever c~ys- tal structnre gives the strongest attractive interaction compatible with the repul- sive interaction at short distances between ion cores. The repulsive interactions between ions with inert gas configurations are similar to those between inert gas atoms. The van der Weals part of the attractive interaction in ionic crystals makes a relatively small contribution to the cohesive energy in ionic crystals, of the order of 1 or 2 percent. The main contribution to the binding energy of ionic crystals is electrostatic and is called the Madelung energy. Atom Electron affinity enerw eV Atom Electron affinity energy eV Source: H. Hotop and W C. Lineberger, J. Phys. Chem. Ref. Data 4,539 (1975). If U, is the interaction energy between ions i and j, we define a sum U which includes all interactions involving the ion i: u , = Z ' u , , / (17 where the summation includes all ions exceptj = i. We suppose that U,, may b written as the sum of a central field repulsive potential of the form A expi-rlp), where A and p are empirical parameters, and a coulomb potential kq2/r. Thus where the + sign is taken for the like charges and the - sign for unlike charges In SI units the coulomb interaction is ?q2/4mor; we write this section in CGS units in which the coulomb interaction is ?q2/r. The repulsive term describes the fact that each ion resists overlap with th electron distributions of neighboring ions. We treat the strength A and range p as constants to be determined from observed values of the lattice constant an compressibility; we have used the exponential form of the empirical repulsiv potential rather than the R-l2 form used for the inert gases. The change i made because it may give a better representation of the repulsive interaction For the ions, we do not have gas-phase data available to permit the indepen dent determination of A and p. We note that p is a measure of the range of th repulsive interaction; when r = p, the repulsive interaction is reduced to e-of the value at r = 0. In the NaCl strnctnre the value of U, does not depend on whether th reference ion i is a positive or a negative ion. The sum in (17) can he arrange to converge rapidly, so that its value will not depend on the site of the referenc ion in the crystal, as long as it is not near the surface. We neglect surface effect 3 Crystal Binding 63 and write the total lattice energy U,,, of a crystal composed of N molecules or 2N ions as U , = NU,. Here N, rather than 2N, occurs because we must count each pair of interactions only once or each bond only once. The total lattice en- ergy is defined as the energy required to separate the crystal into individual -. ions at an infinite distance apart. It is convenient again to introduce quantities p, such that r, = p,R, where R is the nearest-neighbor separation in the crystal. If we include the repulsive interaction only among nearest neighbors, we have (CGS) q2 A exp-Wp) - - (nearest neighbors) R u , , = (19) +- - (otherwise). P,- R Thus where z is the number of nearest neighbors of any ion and (21) The sum should include the nearest-neighbor contribution, which is just z. The (2) sign is discussed just before (25). The value of the Madelung constant is of central importance in the theory of an ionic crystal. Methods for its calcu- lation are discussed next. At the equilibrium separation dU,,ldR = 0, so that This determines the equilibrium separation R, if the parameters p, A of the re- pulsive interaction are known. For SI, replace q2 by q2/4'7r€,. The total lattice energy of the crystal of 2N ions at their equilibrium sepa- ration R, may he written, using (20) and (23), as The term -Naq2/R, is the Madelung energy. We shall find that p is of the order of O.lRo, so that the repulsive interaction has a very short range. Figure 9 Line of ions of alternating signs, with &stance H between ions. Evaluation of the Madelung Constant The first calculation of the coulomb energy constant a was made by Madelung. A powerful general method for lattice sum calculations was devel- oped by Ewald and is developed in Appendix B. Computers are now used for the calculations. The definition of the Madelung constant a is, by (21), For (20) to give a stable crystal it is necessary that a be positive. If we take the reference ion as a negative charge, the plus sign will apply to positive ions and the minus sign to negative ions. An equivalent definition is where 5 is the distance of thejth ion from the reference ion and R is the near- est-neighbor distance. The value given for a will depend on whether it is defined in terms of the nearest-neighbor distance R or in terms of the lattice parameter a or in terms of some other relevant length. As an example, we compute the Madelung constant for the infinite line of ions of alternating sign in Fig. 9. Pick a negative ion as reference ion, and let R denote the &stance between adjacent ions. Then the factor 2 occurs because there are two ions, one to the right and one to the left, at equal distances r,. We sum this series by the expansion Thus the Madelung constant for the one-dimensional chainis a = 2 In 2. 3 Crystal Binding 65 Figure 10 Energy per molecule of KC1 clystal, showing Madelung (coulomb) and repulsive contribut~ons. I 14 12 10 8 6 I 4 I % 2 - .5 i 0- 1u !
2 -2- 1 4 ! -6 -8 I -10 -12 -14 In three dimensions the series presents greater difficulty. It is not possible to write down the successive terms by a casual inspection. More important, the series will not converge unless the successive terms in the se- ries are arranged so that the contributions from the positive and negative terms nearly cancel. Typical values of the Madelung constant are listed below, based on unit charges and referred to the nearest-neighbor distance: - - - (2.4 x lo4) exp(-RI0.30) eV - - - 1 2 3 4 s 6 R, i n 1 ~ c m Equiliblium - position - - (2521R) eV - - - The Madelung and repulsive contributions to the binding of a KC1 crystal are shown in Fig. 10. Properties of alkali halide crystals having the sodium chloride structure are given in Table 7. The calculated values of the lattice en- ergy are in exceedingly good agreement with the observed values. Table 7 Properties of alkali halide crystals with the NaCl structure All values (except those in square brackets) at room temperature and atmospheric pressure, with no correction for changes in R, and absolute zero. Values in square brackets at absolute zero temperature and zero pressure, from private communication by L. Brewer. Nearest- Repulsive Repulsive neighbor Bulk modulus B, energy range Lattice energy comp separatip in 10" dyn/cmP parameter parameter to free ions, in kcal/ R, in A or loLo ~ / m % %A, in 10F erg p, in A Experimental LiF 2.014 6.71 0.296 0.291 242.3[246.81 LiCl 2.570 2.98 0.490 0.330 198.9[201.8] LiBr 2.751 2.38 0.591 0.340 189.8 LiI 3.000 (1.71) 0.599 0.366 177.7 NaF 2.317 4.65 0.641 0.290 214.4[217.91 NaCl 2.820 2.40 1.05 0.321 182.6[185.3] NaBr 2.989 1.99 1.33 0.328 173.6[174.31 NaI 3.237 1.51 1.58 0.345 163.2[162.3] KF 2.674 3.05 1.31 0.298 189.8[194.5] KC1 3.147 1.74 2.05 0.326 165.8[169.51 KBr 3.298 1.48 2.30 0.336 158.5[159.3] KI 3.533 1.17 2.85 0.348 149.9[151.1] RbF 2.815 2.62 1.78 0.301 181.4 RbCl 3.291 1.56 3.19 0.323 159.3 RbBr 3.445 1.30 3.03 0.338 152.6 RbI 3.671 1.06 3.99 0.348 144.9 Data from various tables by M. P. Tosi, Solid State Physics 16, 1 (1964). 3 Crystol Binding 67 Figure 11 Calculatrd valence electron concentration in germanium. The numbers on the con- tours give the electron concentration per primitive cell, with four valence electrons per atom (eight electrons per primitive cell). Note the high concentration midway along the Ce-Ge bond, as we expect for covalent honding. (After J. R. Chelikowsband M. L. Cohen.) COVALENT CRYSTALS The covalent bond is the classical electron pair or homopolar bond of chemistry, particularly of organic chemistry. It is a strong bond: the bond be- tween two carbon atoms in diamond with respect to separated neutral atoms is comparable with the bond strength in ionic crystals. The covalent bond is usually formed from two electrons, one from each atom participating in the bond. The electrons forming the bond tend to be partly localized in the region between the two atoms joined by the bond. The spins of the two electrons in the bond are antiparallel. The covalent bond has strong directional properties (Fig. 11). Thus car- bon, silicon, and germanium have the diamond structure, with atoms joined to four nearest neighbors at tetrahedral angles, even though this arrangement gives a low filling of space, 0.34 of the available space, compared with 0.74 for a close-packed structure. The tetrahedral bond allows only four nearest neigb- bors, whereas a close-packed structure has 12. We should not overemphasize the similarity of the bonding of carbon and silicon. Carbon gives biology, but silicon gives geology and semiconductor technology. The binding of molecular hydrogen is a simple example of a covalent bond. The strongest binding (Fig. 12) occurs when the spins of the two electrons are antiparallel. The binding depends on the relative spin orientation not because there are strong magnetic dipole forces between the spins, but because the Pauli principle modifies the distribution of charge according to the spin orientation. This spin-dependent coulomb energy is called the exchange interaction. .paw.roj ale spuoq aq) uaqM pau~8a.I m q l arom Junome ue i\a p sai!nba~ ale)s pun018 a v m o ~ j uo!) -omold s q ~ ',~zsz,s~ uoge~n%guoa quorlaaIa aql 04 palomo~d aq JSJIJ xsnur moJe U O ~ J R ~ ay? spuoq xuapoa jo ma~sks ppayeyal e m ~ o j o~ .,dz,s~,s~ s ! uoqlea jo uoge~&rjuoa uoJzaaIa ayL .depano a%eya q ? ! ~ paaeposse uo~l3e1aau1 ai\rpeq$e ue aAeq U E ~ (a~dmexa roj) sluaurala asaq, snql pue 'sIIays paII!j 01 ~ a a d s a ~ yl suosaa~a Inoj ~3x1 a 3 pue '!s '3 s1uamaIa aqL .13 n ! neyl .IV u! 1a2uo1~s s ! uog3era) -u ! ai!s~nda~ aqx J E ~ J os 'IIaqs aqz %u!ng 'qs seq moxe .IV aqx put. [lays dz apl u ! suo1~3a1a ai\g seq mo?e 13 aql leq) s ! ZIV pue 213 uaaMJaq aaua~ajjp ayA .I alqeL u ! uan@ sa!%.raua ai!sayoa aqx a1eduro3 osp irv pqos u ! ~ v j o (v g ~ . z ) aauelslp 3!mo>eIam! aqj qw 213 30 (y 2) qGua1 puoq aq) aleduo3 .;auoqs aq 1 1 " puoq a 9 pus saxeas B ~ a u a q%!q 02 snoqaaIa jo uognxpxa anoy)% pa3ep -ourmo33e aq ut.3 d e p a ~ o uo.r)aaIa 'pa119 )on ale sI[ays aqajI q a q s paIIg y p ~ smole uaawaq uoqaeJa)u! a ~ ~ s ~ n d a r 2uo1as e sail!% aId13upd ynea aqA ' S puv v sams aq, roj sann rnoj U 0 3 X q paauasa~dar SF a2mqajo hrsnap a q ~ .sn!ds lallemdgua xoj (ams alquss aq,) s poe 'lunoa -3s o~n! a~d!auud uo!snpxa !pea ayl B u ~ x 'so~ds uorjaala p l l ~ ~ e d 103 qnsax aq s ! v isag!suap azmqa mole ssxj qrw uogv~na~ea prssep e o x srajar N aiuna aqL .8urpu!q o a spuodsa~ro3 Xaraua anrz~3au y .smole lsrznau pa%~1adas 0% pa11a3ar (%) uaBo~pXq relnoalom $0 Wlaua 2 ; ~ axn%rs 3 C~ystol Binding 69 Table 8 Fractional ionic character of bonds in binary crystals - ~ Fractional Fractional Clvstal ionic character Clystal ionic character Si Sic Ge Z n o ZnS ZnSe ZnTe CdO CdS CdSe CdTe InP InAs InSb LiF NaCI RbF ~ -- - - ~ ~ After J. C. Phillips, Bonds and bands in semiconductors. There is a continuous range of crystals between the ionic and the covalent limits. It is often important to estimate the extent a given bond is ionic or cova- lent. A semiempirical theory of the fractional ionic or covalent character of a bond in a dielectric crystal has been developed with considerable success by J. C. Phillips, Table 8. METALS Metals are characterized by high electrical conductivity, and a large num- ber of electrons in a metal are free to move about, usually one or two per atom. The electrons available to move about are called conduction electrons. The valence electrons of the atom become the conduction electrons of the metal. In some metals the interaction of the ion cores with the conduction elec- trons always makes a large contribution to the binding energy, but the charac- teristic feature of metallic binding is the lowering of the energy of the valence electrons in the metal as compared with the free atom. The binding energy of an alkali metal crystal is considerably less than that of an alkali halide c~ystal: the bond formed by a conduction electron is not very strong. The interatomic distances are relatively large in the alkali metals because the kinetic energy of the conduction electrons is lower at large interatomic distances. This leads to weak binding. Metals tend to crystallize in relatively Figure 13 The hydrogen difluoride ion HF4 is stabilized by a hydrogen bond. The sketch is of an extreme model of the bond, extreme in the sense that the proton is shown bare of electrons close packed structures: hcp, fcc, bcc, and some other closely related structures, and not in loosely-packed structures such as diamond. In the transition metals there is additional binding from inner electron shells. Transition metals and the metals immediately following them in the periodic table have large d-elec~on shells and are characterized by high binding energy HYDROGEN BONDS Because neutral hydrogen has only one electron, it should form a covalent bond with only one other atom. It is known, however, that under certain condi- tions an atom of hydrogen is attracted by rather strong forces to two atoms, thus forming a hydrogen bond between them, with a bond energy of the order of 0.1 eV It is believed that the hydrogen bond is largely ionic in charac- ter, being formed only between the most electronegative atoms, particularly F, 0, and N. In the extreme ionic form of the hydrogen bond, the hydrogen atom loses its electron to another atom in the molecule; the bare proton forms the hydrogen bond. The atoms adjacent to the proton are so close that more than two of them would get in each other's way; thus the hydrogen bond connects only twc atoms (Fig. 13). The hydrogen bond is an important part of the interaction between HzO molecules and is responsible together with the electrostatic attraction of the electric dipole moments for the strihng physical properties of water and ice. It is important in certain ferroelectric crystals and in DNA. ATOMIC RADII Distances between atoms in crystals can be measured very accurately by x-ray diffraction, often to 1 part in lo5. Can we say that the observed distance between atoms may be assigned partly to atom A and partly to atom B? Can a definite meaning be assigned to the radius of an atom or an ion, irrespective of the nature and composition of the crystal? Strictly, the answer is no. The charge distribution around an atom is not limited by a rigid spherical boundary. Nonetheless, the concept of an atomic from the additive properties of the atomic radii. Further, the electronic config- uration of the constituent atoms often can be inferred by comparison of mea- sured and predicted values of the lattice constants. To make predictions of lattice constants it is convenient to assign (Table 9) sets of self-consistent radii to various types of bonds: one set for ionic crystals with the constituent ions 6-coordinated in inert gas closed-shell configura- tions, another set for the ions in tetrahedrally-coordinated structures, and an- other set for 12-coordinated (close-packed) metals. The predicted self-consistent radii of the cation Na+ and the anion F- as given in Table 9 would lead to 0.97 k + 1.36 A = 2.33 A for the interatomic separation in the crystal NaF, as compared with the observed 2.32 A. This agreement is much better than if we assume atomic (neutral) configurations for Na and F, for this would lead to 2.58 A for the interatomic separation in the crystal. The latter value is $(n.n. distance in metallic Na+ interatomic distance in gaseous F,). The interatomic distance between C atoms in diamond is 1.54 A; one-half of this is 0.77 k. In silicon, which has the same crystal structure, one-half the interatomic distance is 1.17 A. In Sic each atom is surrounded by four atoms of the opposite kind. If we add the C and Si radii just given, we predict 1.94 k for the length of the C-Si bond, in fair agreement with the 1.89 observed for the bond length. This is the kind of agreement (a few percent) that we shall find in using tables of atomic radii. Ionic Crystal Radii Table 9 gives the ionic crystal radii in inert gas configurations for &fold coordination. The ionic radii can be used in conjunction with Table 10. Let us Table 10 Use of tbe standard radii of ions given in Table 9 The interionic distance D is represented by D , = RC + RA + A,, for ionic c~ystals, where N is the coordination number of the cation (positive ion), R, and R, are the stan- dard radii of the cation and anion, and A, is a correction for coordination number. Room temperature. (After Zachariasen.) 3 Clystol Binding 73 consider BaTiO, with a lattice constant of 4.004 A at room temperature. Each Ba++ ion has 12 nearest 0-- ions, so that the coordination number is 12 and the correction A,, of Table 10 applies. If we suppose that the strncture is determined by the Ba-0 contacts, we have Dl, = 1.35 + 1.40 + 0.19 = 2.94 A or a = 4.16 A; if the Ti-0 contact determines the structure, we have D, = 0.68 + 1.40 = 2.08 or a = 4.16 A. The actual lattice constant is somewhat smaller than the estimates and may perhaps suggest that the bonding is not purely ionic, hut is partly covalent. ANALYSIS OF ELASTIC STRAINS We consider the elastic properties of a crystal viewed as a homogeneous continuous medium rather than as a periodic array of atoms. The continuum approximation is usually valid for elastic waves of wavelengths A longer than 10-6cm, which means for frequencies below 10" or 10" Hz. Some of the ma- terial below looks complicated because of the unavoidable multiplicity of sub- scripts on the symbols. The basic physical ideas are simple: we use Hooke's law and Newton's second law. Hooke's law states that in an elastic solid the strain is directly proportional to the stress. The law applies to small strains only. We say that we are in the nonlinear region when the strains are so large that Hooke's law is no longer satisfied. We specify the strain in terms of the components e , , e , , , em, exY, e , , e,, which are defined below. We treat infinitesimal strains only. We shall not distinguish in our notation between isothermal (constant temperature) and adiabatic (constant entropy) deformations. The small differences between the isothermal and adiabatic elastic constants are not often of importance at room temperature and below. We imagine that three orthogonal vectors ir,j., i of unit length are emhed- ded securely in the unstrained solid, as shown in Fig. 14. After a small uniform deformation of the solid has taken place, the axes are distorted in orientation and in length. In a uniform deformation each primitive cell of the crystal is deformed in the same way. The new axes x', y', z' may he written in terms of the old axes: The coefficients eaB define the deformation; they are dimensionless and have values < 1 if the strain is small. The original axes were of unit length, hut the new axes will not necessarily be of unit length. For example, Figure 14 Coordinate axes for the description of the state of strain; the orthogonal unit axes in the unstrained state (a) are deformed in the strained state (b). whence x' - 1 + E, +. . .. The fractional changes of length of the i, 9, and i axes are e,, eyy, E , , , respectively, to the first order. What is the effect of the deformation (26) on an atom originally at r = x i + yf + zi? The origin is taken at some other atom. If the deformation is uniform, then after deformation the point will he at the position r' = xx' + yy' + z z ' . This is obviously correct if we choose the 2 axis such that r = xi; then r' = xx' by definition of x'. The displacement R of the deforma- tion is defined by or, from (26), This may be written in a more general form by introducing u, u, w such that the displacement is given by If the deformation is nonuniform we must relate u, v, w to the local strains. We take the origin of r close to the region of interest; then comparison of (28) and (29) gives, by Taylor series expansion of R using R(0) = 0, 3 C ~ y ~ t a l Binding 75 It is usual to work with coefficients enP rather than E , ~ We define the strain components e,, eYY, em by the relations using (30). The other strain components e,, e,, e, are defined in terms of the changes in angle between the axes: using (26) we may define We may replace the - signs by = signs if we neglect terms of order 2. The six dimensionless coefficients eaP(=epm) completely define the strain. Dilation The fractional increase of volume associated with a deformation is called the dilation. The dilation is negative for hydrostatic pressure. The unit cube of edges i, 9, i has a volume after deformation of by virtue of a well-hown result for the volume of a parallelepiped having edges x', y', z'. From (26) we have Products of two strain components have been neglected. The dilation 8 is then given by Stress Components The force acting on a unit area in the solid is defined as the stress. There are nine stress components: &, Xy, Xz, Yx, Yy, Y , , Z,, Zy, Zz. The capital letter indicates the direction of the force, and the subscript indicates the normal to the plane to which the force is applied. In Fig. 15 the stress component XI .ssa~uo!suarur~ ale pue s@uaI jo so;rel ale sxuauoduro3 uplas aqJ .arnnlon fun lad &aua .to sale :,Tun lad aaloj jo suo!suam!p aqz aneq s:,uauodu103 ssalas .hX 'xk '=A 'zZ '" ‘2 se uaye:, aqXem s?uauodrno3 ssans xuapuadapn! q s aqL (SF) . 6 ' " A = % ' ' x = 2 . Z = ' h W v S M O n O J $1 .o~az aq lsnm anblo3 aqa ~2q? amaq pux 'qsluen uo!jelalaaae le1nZu-e a q ~ :,ey:, u o ~ r ~ u o a aq$ (91 u ! m) aqn3 Xre$uama~a ue 0% 2uj1dde X q q s 03 au!u uroq paanpal s ! s~uauodmo3 ssar:,s luapuadapu! jo laqurnu aqL .uogaal!p fi ayl u ! saq purrou asoqM auqd e jo earn l ~ n e o x uogaanp x aq:, u ! paqdde aalof B s-lnasa-lda~ liX ~uauodmo3 ssa~ls aq3 :uo!$aa~~ x aq:, u ! saq purJou asoqM aue~d e jo eam :,Tun e o x uo~aanp x aqq m paqdde axoj e s~uasalda~ .
6 x = ' ~ j ; oraz asp s : u:8uo a p anoqe an6103 [eaoa a q ~ , .says!uPA aaroj [eaol aqJ .maz os[e s ! uo:$aa~rp li aq u: samoj aqajo mns a q ~ .
o r a 7 . s r ooqaaqp x aq> U : Saxoj aql jo u m s a q ~ .% = rnnpqr[:nba a!ms m Lpoq e xqf aeq% noqvrxsuomaa 91 .uoqaarrp fi arfl u ! sa![ prnzou asoq amid ejo earn a:un e 0% uo:%Jarrp % aqa m pagdda sg % : U O R J ~ ~ x ayr n! sat1 prnrau asaqk aosld e ja earn vun v 0% uowarlp r , x aqj u: paqddv aaroj e sly auanodrnoa s s a q pr a ~ n a ! ~ I 3 Crystal Binding ELASTIC COMPLIANCE AND STIFFNESS CONSTANTS Hooke's law states that for sufficiently small deformations the strain is di- rectly proportional to the stress, so that the strain components are linear func- tions of the stress components: The quantities S,,, Slz . . . are called elastic compliance constants or elastic constants; the quantities Cll, C , , , . . . are called the elastic stiffness constants or moduli of elasticity. The S's have the dimensions of [area]/ [force] or [volume]/[energyl. The CS have the dimensions of [forcel/[area] or [energyl/[volume]. Elastic Energy Density The 36 constants in (37) or in (38) may be reduced in number by several considerations. The elastic energy density U is a quadratic function of the strains, in the approximation of Hooke's law (recall the expression for the energy of a stretched spring). Thus we may write where the indices 1 through 6 are defined as: The C's are related to the C's of (38), as in (42) below. p potential energy. Consider the stress X , applied to one face of a unit cube, the opposite face being held at rest: au -au- - l 6 - & = - = - - C,,e, + 2 2 (CIB + Cpl)ep de, de, 8 =z Note that only the combination $(Ea8 + eBa) enters the stress-strain relations. It follows that the elastic stiffness constants are symmetrical: Thus the thirty-six elastic stiffness constants are reduced to twenty-one. Elastic Stiffness Constants o f Cubic Crystab The number of independent elastic stiffness constants is reduced further if the crystal possesses symmetry elements. We now show that in cubic crystals there are only three independent stiffness constants. We assert that the elastic energy density of a cubic crystal is and that no other quadratic terns occur; that is, do not occur. The minimum symmetry requirement for a cubic structure is the exis- tence of four three-fold rotation axes. The axes are in the [ I l l ] and equivalent directions (Fig. 17). The effect of a rotation of 2 ~ 1 3 about these four axes is to interchange the x, y, z axes according to the schemes according to the axis chosen. Under the first of these schemes, for example, and similarly for the other terms in parentheses in (43). Thus (43) is invariant under the operations considered. But each of the terms exhibited in (44) is odd in one or more indices. A rotation in the set (45) can be found which will 3 Crystal Binding 79 Figure 17 Rotation by 2 ~ 1 3 about the axis marked 3 changes r + y; y + z: andz + x . change the sign of the term, because e,, = -e,(y,, for example. Thus the terms (44) are not invariant under the required operations. It remains to venfy that the numetical factors in (43) are correct. By (41), aulae,, = & = Clle, + cl,(ey, + e,) . (46) The appearance of C,,e, agrees with (38). On further comparison, we see that Cl2=Cl3; c L 4 = C 1 5 = C 1 6 = o . (47) Further, from (43), on comparison with (38) we have Thus from (43) we find that the array of values of the elastic stiffness constants is reduced for a cubic crystal to the matrix These relations follow on evaluating the inverse matrix to (50) Bulk Modulus and Compressibility Consider the uniform dilation e, = e, = e, = iS. For this deformation the energy density (43) of a cubic crystal is We may define the bulk modulus B by the relation which is equivalent to the definition -Vdp/dV. For a cubic crystal, The compressibility K is defined as K = 11B. Values of B and K are given in Table 3. ELASTIC WAVES IN CUBIC CRYSTALS By consideling as in Figs. 18 and 19 the forces acting on an element of volume in the crystal we obtain the equation of motion in the x direction azu a& a% ax, P-=-+- +-; at2 ax ay az here p is the density and u is the displacement in the x direction. There are similar equations for the y and 2 directions. From (38) and (50) it follows that for a cubic crystal here the x , y, z directions are parallel to the cube edges. Using the definitions (31) and (32) of the strain components we have where u, v , w are the components of the displacement R as defined by (29). 3 Crystal Binding 81 Figure 18 Cube of volume h r Ay hz acted on by a stress -&(XI on the face at r, and ax, &(x + hr) = Z(x) + h r on the parallel face at r + h. The net force is ( 2 h r ) A y hz. Other forces in the x direction anse from the vanahon across the cube of the stresses& and&, whch are not s h m The net x component ofthe force on the cube is The force equals the mass of the cube times the component of the acceleration in the r direction. The mass i s p 4.z Ay 4 s and the acceleration is a2u/af?. Figure 19 If springs A and B are stretched equally. the black between them experiences no net force. This illustrates the fact that a uniform stress & in a solid does not give a net force on a "01- ume element If the spring at B is stretched more than the spring at A, the block between them will be accelerated by the farce X,(B) - &(A). The corresponding equations of motion for a2~/at2 and aZw/at2 are found directly from (57a) by symmetly: We now look for simple special solutions of these equations. Waves in the [loo] Direction One solution of (57a) is given by a longitudinal wave where u is the x component of the particle displacement. Both the wavevector and the particle motion are along the x cube edge. Here K = 27r/A is the thus the velocity olK of a longitudinal wave in the [loo] direction is Consider a transverse or shear wave with the wavevector along the x cube edge and with the particle displacement o in they direction: o = v, exp [ i ( k - ot)] . (61) On substitution in (57h) this gives the dispersion relation thus the velocity o/K of a transverse wave in the [loo] direction is 0 , = (c&p)'" . (63) The identical velocity is obtained if the particle displacement is in the z direc- tion. Thus for K parallel to [I001 the two independent shear waves have equal velocities. This is not true for K in a general direction in the crystal. Waves in the [I101 Direction There is a special interest in waves that propagate in a face diagonal direc- tion of a cubic crystal, because the three elastic constants can be found simply from the three propagation velocities in this direction. Consider a shear wave that propagates in the xy plane with particle dis- placement w in the z direction whence (32c) gives independent of propagation direction in the plane. Consider other waves that propagate in the xy plane with particle motion in the xy plane: let u = u, exp [ i ( G + K y y - ot)] ; o = v, exp [ i ( Q + K y y - wt)] . (66) From (57a) and (57b), 3 Crystal Binding 83 This pair of equations has a particularly simple solution for a wave in the [I101 direction, for which & = K , , = . The condition for a solution is that the determinant of the coefficients of u and o in (67) should equal zero: This equation has the roots The first root describes a longitudinal wave; the second root describes a 1 shear wave. How do we determine the direction of paficle displacement? The j first root when substituted into the upper equation of (67) gives whence the displacement components satisfy u = o. Thus the particle dis- placement is along [I101 and parallel to the K vector (Fig. 20). The second root of (44) when substituted into the upper equation of (67) gives whence u = -0. The particle displacement is along [IiO] and perpendicular to the K vector. Selected values of the adiabatic elastic stiffness constants of cubic crystals at low temperatures and at room temperature are given in Table 11. Notice the general tendency for the elastic constants to decrease as the temperature is in- creased. Further values at room temperature alone are given in Table 12. , Wave in [lo01 direction Wave in LllOI direction Wave in [I111 direction L:C11 L : k(Cl1 + C12 + 2C&j L : $C,, + ZC12 + 4C4) T : C 4 TI : C , T:f(C,,-C12+ C,) T2:;(cll-cl%) Figure 20 Effective elastic constants for the three modes of elastic waves in the principal propa- g a t i o n directions in cubic crystals. The hvo transverse modes are degenerate for propagation in the [loo] and [Ill] directions. The values given at 0 K were obtained by extrapolation of measurements carried out down to 4 K. The table was compiled with the assistance of Professor Charles S. Smith. Stiffness constants, in 10" dyne/cmZ (10"~lrn') Crystal G I C I S C , Temperature, K Density, g/crn3 Table 12 Adiabatic elastic stiffness constants of several cubic crystals at room temperature or 300 K Stiffness constants, in 10" dynelcmP or 10" N/m2 Diamond Na Li Ge Si Gash InSb MgO NaCl 3 Crystal Binding 85 There are three normal modes of wave motion in a clystal for a given magnitude and direction of the wavevector K. In general, the polarizations (directions of article displacement) of these modes are not exactly parallel or perpendicular to K. In the special propagation directions [loo], [Ill], and [I101 of a cubic crystal two of the three modes for a given K are such that the article motion is exactly transverse to K and in the third mode the motion is exactly longitudinal (parallel to K). The analysis is much simpler in these special directions than in general directions. SUMMARY Crystals of inert gas atoms are bound by the van der Waals interaction (in- duced dipole-dipole interaction), and this varies with distance as 1/R6. The repulsive interaction between atoms arises generally from the electro- static repulsion of overlapping charge distributions and the Pauli principle, which compels overlapping electrons of parallel spin to enter orbitals of higher enera. . Ionic crystals are hound by the electrostatic attraction of charged ions of opposite sign. The electrostatic energy of a structure of 2N inns of charge ?q is where a is the Madelung constant and R is the distance between nearest neighbors. . Metals are hound by the reduction in the hnetic energy of the valence elec- trons in the metal as compared with the free atom. A covalent bond is characterized by the overlap of charge distributions of antiparallel electron spin. The Pauli contribution to the repulsion is reduced for antiparallel spins, and this makes possible a greater degree of overlap. The overlapping electrons hind their associated ion cores by electrostatic attraction. Problems 1. Quantum solid. In a qnantum solid the dominant repulsive energy is the zero- point energy of the atoms. Consider a crude one-dimensional model of crystalline He4 with each He atom confined to a line segment of length L. In the ground state the wave function within each segment is taken as a half wavelength of a free pari- cle. Find the zero-point kinetic energy per particle. 0.958). The lattice sums for the bcc structures are 3. Solid molecular hydrogen. For Hz one finds from measurements on the gas that the Lennard-Jones parameters are r = 50 X erg and u = 2.96 A. Find the cohesive energy in kJ per mole of H,; do the calculation for an fcc structure. Treat each H, molecule as a sphere. The observed value of the cohesive energy is 0.751 kJ/mol, much less than we calculated; thus, quantum corrections must be very important. 4. Possibility of ionic cryatab R+R-. Imagine a crystal that exploits for binding the coulomb attraction of the positive and negative ions of the same atom or molecule R. This is believed to occur with certain organic molecules, but it is not found when R is a single atom. Use the data in Tables 5 and 6 to evaluate the stability of such a form of Na in the NaCl structure relative to normal metallic sodium. Evalu- ate the energy at the observed interatomic distance in metallic sodium, and use 0.78 eV as the electron affinity of Na. 5. Linear ionic crystal. Consider a line of 2N ions of alternating charge ?q with a repulsive potential energy AIR" hehveen nearest neighbors. (a) Show that at the equilibrium separation (h) Let the clystal be compressed so that &+&(I - 6). Show that the work done in compressing a unit length of the crystal has the leading term $a2, where To obtain the results in SI, replace q2 by q 2 / 4 ~ 6 . Note: We should not expect to ob- tain this result from the expression for U(Ro), hut we must use the complete expres- sion for U(R). 6. Cubic ZnS structure. Using A and p from Table 7 and the Madelung constants given in the tea, calculate the cohesive energy of KC1 in the cubic ZnS structure described in Chapter 1 . Compare with the value calculated for KC1 in the NaCl structure. 7. Divalent ionic crystab. Barium oxide has the NaCl structure. Estimate the cohesive energies per molecule of the hypothetical crystals BaiO- and Batto-- referred to separated neutral atoms. The observed nearest-neighbor internuclear distance is & = 2.76 A; the first and second ionization potentials of Ba are 5.19 and 9.96 eV; and the electron affinities of the first and second electrons added to the neutral oxygen atom are 1.5 and -9.0 eV. The first electron affinity of the 3 Crystal Binding 87 Undeformed body 7 Tension , Figure 21 Youngs modulus is defined as stresslstrain for a ten- sile stress acting i n one direction, with the specimen sides left Figure 22 This deformation is compounded free. Poisson's ratio is defined as (h/1o)/(SI/l) for this situation. from the two shears e , = -e,. neutral oxygen atom is the energy released in the reaction 0 + e + 0-. The sec- ond electmn affinity is the energy released in the reaction 0- f e + 0--. Which valence state do you predict will occur? Assume R , , is the same for both forms, and neglect the repulsive energy. 8. Young's modulus and Poisson's ratio. A cubic crystal is subject to tension in the [loo] direction. Find expressions in terns of the elastic stiffnesses for Young's modulus and Poisson's ratio as defined in Fig. 21. 9. Longitudinal wave velocity. Show that the velocity of a longitudinal wave in the [ I l l ] direction of a cubic crystal is given by o , = [$(C,,+ 2C,, + 4~,)/~]. Hint: For such a wave u = v = w. Let u = u8'~'+yfzJAe~'"l, and use Eq. (57a). 10. Transverse wave velocity. Show that the velocity of transverse waves in the [ I l l ] direction of a cubic crystal is given by v, = [+(c,,- C,, + C,lplE. Hint: See Problem 9. 11. Effective shear constant. Show that the shear constant :(c,, -CIS) in a cubic clystal is defined by setting e, = -eyy = $e and all other strains equal to zero, as in Fig. 22. Hint: Consider the energy density (43); look for a C' such that U = ;C'eP. 12. Determinantal approach. It is known that an R-dimensional square matrix with all elements equal to unity has roots R and 0, with the R occurring once and the zero occurring R - 1 times. If all elements have the value p, then the roots are Rp and 0. (a) Show that if the diagonal elements are q and all other elements are p, then there is one root equal to ( R - l)p + q and R - 1 roots equal to q - p. (b) Show from the elastic equation (57) for a wave in the [ I l l ] direction of a cubic clystal that the determinantal equation which gives o2 as a function of K is where q = $P(c,, + 2C,) and p = $ ?
( c , , + C,). This expresses the condition that three linear homogeneous algebraic equations for the three displacement 13. General propagation direction. (a) By substitution in (57) find the determinan- tal equation which expresses the condition that the displacement R ( ~ ) = [u,$ + yo? + ~ $ 1 exp li(K . r - wt)l be a solution of the elastic wave equations in a cubic crystal. (b) The sum of the roots of a determinantal equation is equal to the sum of the diagonal elements at<. Show from part (a) that the sum of the squares of the three elastic warre velocities in any direction in a cubic crystal is equal to (CII + 2C,)lp. Recall that v : = oz/lC. 14. Stability criteria. The criterion that a cubic crystal with one atom in the primi- tive c e l l be stable against small homogeneous deformations is that the energy den- sity (43) be positive for all cornhinations of strain components. What restrictions are thereby imposed on the elastic stiffness constants? (In mathematical language the problem is to find the conditions that a real symmetric quadratic form should be positive definite. The solution is given in books on algebra; see also Korn and Korn, Mathematical Handbook, McGraw-Hill, 1961, Sec. 13.5-6.) Ans. C , , > 0, C,, > 0, Cf, - C t > 0, and C,, + Z C , , > 0 . For an example of the instability which results when C,, - C,,, see L. R. Testardi et al., Phys. Rev. Letters 15, 250 (1965). Phonons I. Crystal Vibrations VIBRATIONS O F CRYSTALS WITH MONATOMIC BASIS 91 First Brillonin zone 93 Group velocity 94 Long wavelength limit 94 Derivation of force constants from experiment 94 TWO ATOMS PER PRIMITIVE BASIS QUANTIZATION OF ELASTIC WAVES PHONON MOMENTUM INELASTIC SCATTERING BY PHONONS SUMMARY PROBLEMS k t 1. Monatomic linear lattice 2. Continuum wave equation 3. Basis of two unlike atoms 4. Kohn anomaly 5. Diatomic chain 6. Atomic vibrations in a metal 7. Soft phonon modes Chapter 5 treats the thermal properties of phonons. Figure 1 Important elementary excitations in solids. - - -+I+ - - Figure 2 (Dashed lines) Planes of atoms when in equilibrium. (Solid lines) Planes of atoms when displaced a s for a longitudi- nal wave. The coordinate u measures the displacement of the planes. Figure 3 Planes of atoms as displaced during passage of a transverse wave. Name Electron Photon Phonon Plasmon Magnon P u h n Exciton Field - Elechomagnetic -wave Elastic wave Collective electron wave M~etization wave Elechon + elastic deformation Polarization wave CHAPTER 4: PHONONS I. CRYSTAL VIBRATIONS VIBRATIONS OF CRYSTALS WITH MONATOMIC BASIS Consider the elastic vibrations of a crystal with one atom in the primitive cell. We want to find the frequency of an elastic wave in terms of the wavevec- tor that describes the wave and in terms of the elastic constants. The mathematical solution is simplest in the [loo], , and [Ill] propa- gation directions in cubic crystals. These are the directions of the cube edge, face diagonal, and body diagonal. When a wave propagates along one of these directions, entire planes of atoms move in phase with displacements either b parallel or perpendicular to the direction of the wavevector. We can describe with a single coordinate u, the displacement of the planes from its equilibrium position. The problem is now one dimensional. For each wavevector there are i three modes as solutions for us, one of longitudinal polarization (Fig. 2) and i two of transverse polarization (Fig. 3). k We assume that the elastic response of the crystal is a linear function of / the forces. That is equivalent to the assumption that the elastic energy is a I quadratic function of the relative displacement of any two points in the crystal. Terms in the energy that are linear in the displacements will vanish in 1 equilibrium-see the minimum in Fig. 3.6. Cubic and higher-order terms may i be neglected for sufficiently small elastic deformations. ! 1 We assume that the force on the planes caused by the displacement of the plane s + p is proportional to the difference us+, -us of their displacements. i For brevity we consider only nearest-neighbor interactions, with p = 21. The 1 total force on s from planes s + 1: i F, = C(u,+, - us) + C(u,-, - u,) . (1) This expression is linear in the displacements and is of the form of Hooke's law. The constant C is the force constant between nearest-neighbor planes and will differ for longitudinal and transverse waves. It is convenient hereafter to regard C as defined for one atom of the plane, so that F, is the force on one atom in the planes. The equation of motion of an atom in the planes is where M is the mass of an atom. We look for solutions with a l l displacements having the time dependence exp(-iot). Then dZu,ldt2 = -ozua, and (2) becomes u,,, = u exp(isKa) exp(+ iKa) , (4) where a is the spacing between planes and K is the wavevector. The value to use for a will depend on the direction of K. With (4), we have from (3): -02Mu exp(iKa) = Cu{exp[i(s + l)Ka]+ exp[i(s - I)&] - 2 exp(isKa)J . (5) We cancel u exp(isKa) from both sides, to leave With the identity 2 cos Ka = exp(iKa) + exp(-i&), we have the dispersion relation w(K). oz = (ZC/M)(l- cos Ka) . (7) The boundary of the first Brillouiu zone lies at K = +da. We show from (7) that the slope of o versus K is zero at the zone boundary: do2/dK = (2CaIM) sin Ka = 0 (8) at K = ?&a, for here sin Ka = sin (km) = 0 . The special significance of phonon wavevectors that lie on the zone boundary is developed in (12) below. By a trigonometric identity, (7) may be written as A plot of o versus K is given in Fig. 4. Figure 4 Plot of o versus K . The region of K l/n o r A B o corresponds to the contin- uum approximation; here o is directly proportional to K. 4 Phonons I. Crystal Vibratim 93 First Brillouin Zone What range of K is physically significant for elastic waves? Only those in the first Brillouin zone. From (4) the ratio of the displacements of two succes- sive planes is given by The range -.rr to +.rr for the phase Ka covers all independent values of the exponential. The range of independent values of K is specified by This range is the first Brillouin zone of the linear lattice, as defined in Chapter 2. The extreme values are G, = ? d a . Values of K outside of the first Brillouin zone (Fig. 5) merely reproduce lattice motions described by values within the limits ?ria. We may treat a value of K outside these limits by subtracting the integral multiple of 2.rrla that will give a wavevector inside these limits. Suppose K lies out- side the first zone, but a related wavevector K' defined K' = K - 2mla lies within the first zone, where n is an integer. Then the displacement ratio (10) becomes because exp(i2m) = 1. Thus the displacement can always be described by a wavevector within the fust zone. We note that 2mla is a reciprocal lattice vec- tor because 2 d a is a reciprocal lattice vector. Thus by subtraction of an appro- priate reciprocal lattice vector from K, we always obtain an equivalent wavevector in the first zone. At the boundaries K,, = -C.rrla of the Brillouin zone the solution u, = u exp(isKa) does not represent a traveling wave, but a standing wave. At the zone boundaries sK,,ua = ?ST, whence i Figure 5 The wave represented by the solid curve conveys no information not given by the dashed curve. Only wavelengths longer than 2n are needed to represent the ,notion. g g neither to the right nor to the left. This situation is equivalent to Bragg reflection of x-rays: when the B condition is satisfied a traveling wave cannot propagate in a lattice, through successive reflections back and forth, a standing wave is set up. The critical value K,, = +m/a found here satisfies the Bragg cond 2d sin 0 = nA: we have 0 = $m, d = a, K = 2m/A, n = 1, so that A = 2a. W x-rays it is possible to haven equal to other integers besides unity because amplitude of the electromagnetic wave has a meaning in the space betw atoms, hut the displacement amplitude of an elastic wave usually has a m ing only at the atoms themselves. Group Velocity The transmission velocity of a wave packet is the group velocity, give va = do/dK , or the gradient of the frequency with respect to K. This is the velocity of en propagation in the medium. With the particular dispersion relation (9), the group velocity (Fig. 6) i vg = ( c ~ ~ / M ) ~ cos $ Ka . This is zero at the edge of the zone where K = r/a. Here the wave is a stan wave, as in (12), and we expect zero net transmission velocity for a standing w Long Wavelength Limit When Ka < 1 we expand cos Ka - I - ; ( K a ) ' , so that the dispersion r tion (7) becomes w2 = (C/M)@a2 . The result that the frequency is directly proportional to the wavevector in long wavelength limit is equivalent to the statement that the velocity of so is independent of frequency in this limit. Thus v = olK, exactly as in the tinuum theory of elastic waves-in the continuum limit Ka < 1. Derivation of Force Constants from Experiment In metals the effective forces may be of quite long range and are car from ion to ion through the conduction electron sea. Interactions have b found between planes of atoms separated by as many as 20 planes. We can m a statement about the range of the forces from the observed experime 4 Phomm I. Crystal Vibration8 95 Figure 6 Group velocity u , versus K for model of Fig. 4. At the zone boundary K = wla the group velocity is zero. dispersion relation for w. The generalization of the dispersion relation (7) top nearest planes is easily found to be w2 = (21M) c. Cp(l - cos pKa) . (16a) p>n We solve for the interplanar force constants C , by multiplying both sides by cos rKa, where r is an integer, and integrating over the range of indepen- dent values of K: via MIIT'" dK w: cos rKa = 2~ C p L dK (1 - cos pKa) ms rKa WIO P>O li/o The integral vanishes except for p = r. Thus ,./a CP = - dK w$ cos ,./a gives the force constant at range pa, for a structure with a monatomic basis TWO ATOMS PER PRIMITIVE BASIS The phonon dispersion relation shows new features in crystals with two or more atoms per primitive basis. Consider, for example, the NaCl or diamond structures, with two atoms in the primitive cell. For each polarization mode in a given propagation direction the dispersion relation w versus K develops two branches, known as the acoustical and optical branches, as in Fig. 7. We have longitudinal LA and transverse acoustical TA modes, and longitudinal LO and transverse optical TO modes. If there are p atoms in the primitive cell, there are 3p branches to the dis- persion relation: 3 acoustical branches and 3p - 3 optical branches. Thus ger- manium (Fig. 8a) and KBr (Fig. Sh), each with two atoms in a primitive cell, have six branches: one LA, one LO, two TA, and two TO. Figure 7 Optical and acoustical branches of the dis- pbonon branch persion relation far a diatomic linear lattice, showing the limiting frequencies at K = 0 and K = K., = vta. ?r K - The l a t t i c e constant is a. a Kt&,, in [Ill] direction Figure 8a Pbonon dispersion relations in the I1111 direction in germanium at 80 K. The huo TA phonon branches are horizontal at the zone boundary position, & , = (2/a)(+$+). The LO and TO branches coincide at K = 0 ; this also is a consequence of the crystal symmetry of Ge. The results were obtained with neutron inelastic scattering by G. Nilsson and G. Nelin. 0 Kt&,, in I1111 d i d o n Figure 8b Dispersion curves in the [Ill] direction in KBr at 90 K, after A. D. B. Woods, B. N. Bmckhouse, R. A. Cowley, and W. Cochran. The extrapolation to K = 0 of the TO, LO branches are called mr, mL. The numerology of the branches follows from the number of degrees of free- dom of the atoms. With p atoms in the primitive cell and N primitive cells, there are pN atoms. Each atom has three degrees of freedom, one for each of the x, y, z directions, mahng a total of 3pN degrees of freedom for the crystal. The number of allowed K values in a single branch is just N for one Brillouin zone.' Thus the 'We show in Chapter 5 by application of periodic boundq conditions to the modes of the crystal of volume V that there is one K value in the volume (2w)VV in Fourier space. The volume of a Brillouin zone is (Zn)'N, where V . is the volume of a crystal primitive c e l l . Thus the number of allowed Kvalues in a Brillouin zone is VN., which is just N, t h e number ofprimitive c e l l s in the crystal. 4 Phonona I. Crystal Vibrations 97 Figure 9 A diatomic ctystal structure with masses M,, Mz connected by force constant C be- tween adjacent planes. The displacements of atoms M I are denoted by u , -, , u,, u,,,, . . . , and of atoms M, by 0,-,, v., v,,,. The repeat &stance is a in the direction o f the wavevector K. The atoms are shown in their undisplaced positions. LA and the two TA branches have a total of 3N modes, thereby accounting for 3N of the total degrees of freedom. The remaining (3p - 3)N degrees of freedom are accommodated by the optical branches. We consider a cubic clystal where atoms of mass MI lie on one set of planes and atoms of mass M, lie on planes interleaved between those of the first set (Fig. 9). It is not essential that the masses be different, but either the force con- stants or the masses will be different if the two atoms of the basis are in non- equivalent sites. Let a denote the repeat distance of the lattice in the direction normal to the lattice planes considered. We treat waves that propagate in a symmetry direction such that a single plane contains only a single type of ion; such directions are [ I l l ] in the NaCl structure and [loo] in the CsCl structure. We write the equations of motion under the assumption that each plane interacts only with its nearest-neighbor planes and that the force constants are identical between all pairs of nearest-neighbor planes. We refer to Fig. 9 to obtain We look for a solution in the form of a traveling wave, now with different amplitudes u, u on alternate planes: We define n in Fig. 9 as the distance between nearest identical planes, not nearest-neighbor planes. On substitution of (19) in (18) we have the coefficients of the unknowns u, o vanishes: or M,M204 - 2C(M1 + M2)02 + 2C2(1 - cos Ka) = 0 . (22) We can solve this equation exactly for w2, but it is simpler to examine the limiting cases Xn < 1 and Ka = +TI at the zone boundary. For small Ka we have cos Ka E 1 - K2a2 + . . . , and the two roots are (optical branch) ; 02= - ;c K2a2 MI + MZ (acoustical branch) The extent of the first Brillouin zone is -v/a 5 K 5 d a , where a is the repeat distance of the lattice. At K , = ?r/a the roots are The dependence of o on K is shown in Fig. 7 for M, > M2. The particle displacements in the transverse acoustical (TA) and trans- verse optical (TO) branches are shown in Fig. 10. For the optical branch at K = 0 we find, on substitution of (23) in (201, The atoms vibrate against each other, hut their center of Inass is fured. If the two atoms cany opposite charges, as in Fig. 10, we may excite a motion of this Figure 1 0 Transverse optical and transverse amustical waves in a di- atomic linear lattice, illustrated by the particle dqlacements far the two modes at the same wavelength. Acoustical mode 4 Phonons I. Crystal Vibrations 99 type with the electric field of a light wave, so that the branch is called the opti- cal branch. At a general K the ratio ulu will be complex, as follows from either of the equations (20). Another solution for the amplitude ratio at small K is u = u, obtained as the K = 0 limit of (24). The atoms (and their center of mass) move together, as in long wavelength acoustical vibrations, whence the term acoustical branch. Wavelike solutions do not exist for certain frequencies, here between (2C/M,)'" and (2C/M,)'". This is a characteristic feature of elastic waves in polyatomic lattices. There is a frequency gap at the boundary K,, = ? ~ / a of the first Brillouin zone. QUANTIZATON OF ELASTIC WAVES i The energy of a lattice vibration is quantized. The quantum of energy is j called a phonon in analogy with the photon of the electromagnetic wave. The ; energy of an elastic mode of angular frequency o is L when the mode is excited to quantum number n; that is, when the mode is occu- pied by n phonons. The term $ fiw is the zero point energy of the mode. It occurs for both phonons and photons as a consequence of their equivalence to a quan- tum harmonic oscillator of frequency w, for which the energy eigenvalues are also (n + i)fi~. The quantum theory of phonons is developed in Appendix C. We can quantize the mean square phonon amplitude. Consider the stand- ing wave mode of amplitude Here u is the displacement of a volume element from its equilibrium position at x in the crystal. The energy in the mode, as in any harmonic oscillator, is half kinetic energy and half potential energy, when averaged over time. The kinetic energy density is 2 p(&lat)2, where p is the mass density. In a crystal of volume V, the volume integral of the kinetic energy is ipVo2u; sin2&. The time aver- age kinetic energy is because = i. The square of the amplitude of the mode is This relates the displacement in a given mode to the phonon occupancy n of the mode. What is the sign of o ? The equations of motion such as (2) are equations for oZ, and if this is positive then w can have either sign, + or -. But the energy of a phonon must be positive, so it is conventional and suitable to vie o as positive. If the crystal structure is unstable, then o2 will be negative and will be imaginary. PHONON MOMENTUM A phonon of wavevector K will interact with particles such as photon neutrons, and electrons as if it had a momentum hK. However, a phonon do not carry physical momentum. The reason that phonons on a lattice do not carry momentum is that phonon coordinate (except for K = 0) involves relative coordinates of th atoms. Thus in an Hz molecule the internuclear vibrational coordinate rl -is a relative coordinate and does not carry linear momentum; the center mass coordinate $(rl + r2) corresponds to the uniform mode K = 0 and c carry linear momentum. In crystals there exist wavevector selection rules for allowed transitio between quantum states. We saw in Chapter 2 that the elastic scattering of x-ray photon by a crystal is governed by the wavevector selection rule where G is a vector in the reciprocal lattice, k is the wavevector of the incide photon, and k' is the wavevector of the scattered photon. In the reflectio process the crystal as a whole will recoil with momentum -hG, but this un form mode momentum is rarely considered explicitly. Equation (30) is an example of the rule that the total wavevector of inte acting waves is conserved in a periodic lattice, with the possible addition of reciprocal lattice vector G. The true momentum of the whole system always rigorously conserved. If the scattering of the photon is inelastic, with th creation of a phonon of wavevector K, then the wavevector selection ru becomes If a phonon K is absorbed in the process, we have instead the relation Relations (31) and (32) are the natural extensions of (30). INELASTIC SCAWERING BY PHONONS Phonon dispersion relations o(K) are most often determined experime tally by the inelastic scattering of neutrons with the emission or absorption of phonon. A neutron sees the crystal lattice chiefly by interaction with the nucl 4 Phonons I. Crystal Vibrations 101 of the atoms. The kinematics tttering of a neutron beam by a crystal lattice are described by the g c evector selection m1 and by the requirement of conservation of energy ; the wavevector of the phonon created (+) or absorbed (-) in the process, and G is any reciprocal lattice vector. 1 (on we choose G such that K lies in the first Brillouin zone. Wavevector, in units Snla Figure 1 1 The dispersion curves of sodium far ~honons propagating in the , , and [Ill] directions at 90 K, as determined hy inelastic scattering of neutrons, by Woods, Brockhouse, March and Bowers. i Figure 12 .
4 triple ads neutron spectrometer at Bruoklravm. (Coxirtesy oCB. If. Grier.) of the neutron. The momentum p is given by hk, where k is the wavevecto the neutron. Thus h2k2/2M, is the kinetic energy of the incident neutron. I is the wavevector of the scattered neutron, the energy of the scattered neut is fi2k'2/2M,. The statement of conservation of energy is where h o is the energy of the phonon created (+) or absorbed (-) in process. To determine the dispersion relation using (33) and (34) it is necessar the experiment to find the energy gain or loss of the scattered neutrons function of the scattering direction k - k ' . Results for germanium and KBr given in Fig. 8; results for sodium are given in Fig. 11. A spectrometer used phonon studies is shown in Fig. 12. SUMMARY The quantum unit of a crystal vibration is a phonon. If the angular quency is o, the energy of the phonon is fio. When a phonon of wavevector K is created by the inelastic scattering o photon or neutron from wavevector k to k', the wavevector selection rule governs the process is k = k l + K + G , where G is a reciprocal lattice vector. All elastic waves can be described by wavevectors that lie within the f Brillouin zone in reciprocal space. If there are p atoms in the primitive cell, the phonon dispersion relation have 3 acoustical phonon branches and 3p - 3 optical phonon branches. Problems 1. Monatomic linear lattice. Consider a longitudinal wave u , = u cos(mt - sKa) which propagates in a monatomic linear lattice of atoms of mass M, spacing a, nearest-neighbor interaction C. (a) Show that the total energy of the wave is where s runs over all atoms (h) By substitution of u, in this expression, show that the time-average total energy per atom is where in the last step we have used the dispersion relation (9) for this problem 2. Continuum wave equation. Show that for long wavelengths the equation of mo- tion (2) reduces to the continuum elastic wave equation where o is the velocity of sound 3. Basis oftwo unlike atom. For the problem treated by (18) to (26), find the am- ~litude ratios ulv for the two branches at & , = ria. Show that at this value of K the two lattices act as if decoupled: one lattice remains at rest while the other lat- tice moves. 4. Kohn anomaly. We suppose that the interplanar force constant C, between planes s and s + p is of the form sin pk,a C, =A- Pa where A and k, are constants and p runs over all integers. Such a form is expected in metals. Use this and Eq. (16a) to find an expression for 0% and also for do2/JK. Prove that JwZ/aK is infinite when K = k , . Thus a plot of wZ versus K or of o versus K has a vertical tangent at k,: there is a kink at k, in the phonon dispersion relation o(K). 5. Diatomic chain. Consider the normal modes of a linear chain in which the force constants between nearest-neighbor atoms are alternately C and 10C. Let the masses he equal, and let the nearest-neighbor separation be aI2. Find o(K) at K = 0 and K = &a. Sketch in the dispersion relation by eye. This problem simu- lates a crystal of diatomic molecriles such as H,. 6. Atomic vibrations in a metal. Consider point ions of mass M and charge e im- mersed in a uniform sea of conduction electrons. The ions are imagined to be in stable equilibrium when at regular lattice points. If one ion is displaced a small dis- tance r from its equilibrium position, the restoring force is largely due to the elec- tnc charge within the sphere of radius r centered at the equilibrium position. Take the number density of ions (or of conduction electrons) as 3/4?rR3, which defines R. (a) Show that the frequency of a single ion set into oscillation is o = (e2/MR3)1'e. (b) Estimate the value of this frequency for sodium, roughly. (c) From (a), (b), and some common sense, estimate the order of magnitude of the velocity of sound in the metal. ' 7 . Soft phonon modes. Consider a Line of ions of equal mass but alternating in charge, with e, = e(- 1)P as the charge on the pth ion. The interatomic potential is 'This problem is rather difficult. Phonons II. Thermal Properties PHONON HEAT CAPACITY Planck distribution Normal mode enumeration Density of states in one dimension Density of states in three dimensions Debye model for density of states Debye T3 law Einstein model of the density of states General result for D(o) ANHARMONIC CRYSTAL INTERACTIONS Thermal expansion THERMAL CONDUCTMTY Thermal resistivity of phonon gas Umklapp processes Imperfections PROBLEMS 1. Singularity in density of states 2. Rms thermal dilation of crystal cell 3 . Zero point lattice displacement and strain 4. Heat capacity of layer lattice 5. Griineisen constant Figure 1 Plot of Planck distribution function. At high temperatures the occupancy of a state approximately linear in the temperature. The function (n) + b, which is not plotted, approach the dashed line as asymptote at high temperatures. 1 0 6 We discuss the heat capacity of a phonon gas and then the effects of anharmonic lattice interactions on the phonons and on the crystal. PHONON HEAT CAPACITY By heat capacity we shall usually mean the heat capacity at constant vol- ume, which is more fundamental than the heat capacity at constant pressure, which is what the experiments determine.' The heat capacity at constant vol- ume is defined as Cv = (dU/dT), where U is the energy and T the temperature. The contribution of the phonons to the heat capacity of a crystal is called the lattice heat capacity and is denoted by C,.,. The total energy of the phonons at a temperature T(= k,T) in a crystal may he written as the sum of the energies over all phonon modes, here indexed by the wavevector K and polarization index p: Ui, = 2 2 U,, = z zcn,NJJio, 2 K v K P (1) 1 (n) = errp(ho/.r) - l ' (2) where the (...) denotes the average in thermal equilibrium. A graph of (n) is Planck Distribution Consider a set of identical harmonic oscillators in thermal equilibrium. The ratio of the number of oscillators in their (n + 1)th quantum state of exci- tation to the number in the nth quantum state is N,,+,IN. = exp(-fio/~) , 7= kBT , (3) 'A thermodynamic relation gives Cp - C, = 902BVT, where a is the temperature coefficient of linear expansion, V the volume, and B the bulk modulus. The fractional difference between C , and C, is usually small in solids and often may be neglected. As T- 0 we see that C,+Cv, pro- vided a and B are constant. We see that the average excitation quantum number of an oscillator i z s exp-shw/~) (n) = ' z eT(-sfiolr) The summations in (5) are with x = exp(-ftwl~). Thus we may rewrite (5) as the Plauck distribution: x - (n) = -- - 1 1 - x exp(fw/7) - 1 Nomal Mode Enumeration The energy of a collection of oscillators of frequencies on;, in th equilibrium is found from (1) and (2): It is usually convenient to replace the summation over K by an integral pose that the crystal has DP(o)do modes of a given polarization p in th quency range o to o +do. Then the energy is The lattice heat capacity is found by differentiation with respect to tem ture. Let x = h o / ~ = ho/kBT: then 8U/aT gives x2 exp x ~ ~ = k , ~ I d o ~ , , ( o ) p (expx - ' The central prohlem is to find D(w), the number of modes per un quency range. This function is called the density of modes or, more often sity of states. Density of States in One Dimension Consider the boundary value prohlem for vibrations of a one-dimen line (Fig. 2) of length L carrying N + 1 particles at separation a. We su 5 Phonons 11. Thermal Properties 109 Figure 2 Elastic line of N + I atoms, with N = 10, fur boundary conditions that the end atoms s = 0 and s = 10 are k c d . The particle displacements in the normal modes for eitl~cr longitu&~d or transverse displacrme~~ts are of the form u, sin sKa. This form is antomatically zero at the atom at the ends = 0 , and we choose K to make the displacement zero at the ends = 10 Figure 3 Thc boundary condition sin sKa = O for s = 10 can be satisfied by choosing K = .rr/lOa, ZdIOa, . . ., S.rr/lOa, where 10a is the length L of the line. The present figure is in K space. The dots are not atoms but are the allowed valucs of K. Of the N + 1 particles on the line, only N - 1 are allowcd to move, and their most general motion car1 be expressed in terms of the N - 1 al- lowed vali~es of K. This quantization of K has nothing to do with qnantnm mechanics but follows classically from the boondaryconditions that tlre cnd atoms be fixed. that the s = 0 and s = N at the ends of the line are held fixed. Each norrrlal vibrational modc of polarization p has the form of a standing wave, where u , is the displacement of the particle s: v, = 4 0 ) exp-io,,+,t) sin sKtl , (11) wtiere wKl, is related to K by the appropriate dispersion relation. As in Fig. 3, thc wavevector K is restricted by the fixed-end boundary con- ditions to the values The solution for K = n/L has u, a sin (s.rra/L) (13) and vanishes for s = 0 and s = N as required. The solution for K = NT/L = d a = K,,,, has u, sin ST; this permits no 111otioll of any atom, because sinsz- vanishes at each atom. Thus there are N - 1 allowed independent values of K in (12). This number is equal to the number of particles allowed to Inove. Each allowed value of K is associated with a sta~ldi~ig wave. For the one-dimensional line there is one mode for each iriterval AK = T/L, so that the number of modes per unit range of K is LIT for K 5 d a , and 0 for K > rrla. Therc are three polarizations p for each value of K: in one dimension two of these are transverse and one longitudinal. In three dimensions the polariza- tions are this simple only for wavevectors in ccrtain special crystal directions. Another device for enumerating modes is equally valid. We consider the medium as unbounded, hut require that the solutions be periodic over a large X q uo!suam!p auo u ! uan@ s ! m $8 mp u ! mp (m)Q sapom jo Jaqmnu aqL .uo~ez!nqod u a ~ @ e JOJ a2uer Xauanbalj %run ad sapour jo Jaqmnu aql '(m)a ~ o u q 02 paau aM '9 .%!a U ! paXelj~od s ! aqfel leuo!suamrp-ow e u ! uogenfs aqL .asWaq%o 0 pm: 'n/" 5 o/&- loj "zf7 s ! x jo a z u e ~ ?Fun ad sapour 30 Jaqmnu aya 'suo!$!puoa hepunoq xpouad I O ~ .X jo s a n p anrssaaans uaawaq 7/uz = [eiuaau: aq q ~ w '>I jo s a n p snuim pue s n ~ d q ~ o q MOU aneq am anq ' ( ~ 1 ) X q uang s v (mo?e apqom lad auo) sapour jo raqmnu awes a p sai!2 uo!lelamnua 30 poqjam s!q~ 1 aJe XJO SanleA pamo~p aq$ [(tXm - qs))]dxa (0)n = "n / uognps a A E M 8u!uun~ ayj u~ .maasks azlel e ~ o j $aadsa~ pguassa ! Xm: u! ma~qord aq$ jo s3!sXqd aqa a2u~qa JOU saop (s pue p .s%j) suog!pu03 I Lepunoq xpouad jo poqaaur ayL .(7 + os)n = (m)n aeq~ os '7 aam:~s!p .(sq)dxa '(p/su~!~)daa '(~/.w!~)dua '(p/sul~)&a 'I 0% leuo!xrudord mom qxs aql jo s%uauza3e~ds!p q~ 'sapom p a ~ a l p xq3!a am amy) snql :(s'~!-)dra o x pwuap! s ! (s+)dxa asnwaq uo!xn[os a@u!s 8 Xluo luanardal ? / u ~ i slu!od legads a u apom imaj!un aql s ! uogn[os o = x 1 q ~ .7 q@ua~jo an![ r, uo smom 8 = N hrarpouadjo a3q -$el rean![ u 03 pa!ldde suog!puoa Xrepunoq n!pouad 103 x zonanannfijo s a n p p a ~ o w I : -x (91) b x u ! SP '6.qm.g pm 'DN/Y~F ' D N , U + + 'ON/UZT 'O = x q w sapam % $ !
a aql 0% pea1 p p ~ uogrpuoa Xrppunoq arpouad ay) '(q.w)dxa rmoj xald -moa aqj o ! sapom ay) uaqq peq ah31 -g I!J jo uoq ~ ~ 0 3 Xrepunoq pua-pax9 aql roj se Xluaa 'apg.zsd lad apom p a ~ o p auo o x qmal uogrpuoa i(repunaq n p ~ a d aqJ snq~, .sapsred iqZ!s aw rqt sapom p a ~ o l p qS!a 30 p o l v XU& sapom aqsw pm a - ayr qjaq 103 p s m p are x 30 sanpn mqjo aaryr a q ~ 0 = " 2 " U !
S = ( ~ / O ' L ~ S ) " I S asnnaaq '-03 au!soa ayr zoj Xpo Su!uvam e scq cqjfi~ anpn a q ~ .
O = ~ Y ) S "S asnEJaq rmoj au!s ayr ~ a j ssa14u!mam A o = x anph a u ' ~ w 8 pm 'cq1~9 '%/fie 'cq/sz 'O ale X J ~ SanleA luapuadapm pa~ollp a g = = N'OJ .YZ 30 aldqpm @a"! m aq lsnm 0s 's [@ loj ' n = '+h leql E ! uoqpoa hpunoq ayr Bou am jo Xlnlpou -ad pgamoaS ay) Xg .sapom mapuadapu! are asav : q s saa 10 q o u!s uuoj ayr jo aq l p s mom ja ' n ~uama3eldsrp aql apom pmmu e u1 .Buuds Dqscla dq paaauuoa j! alqpso ma s a 1 -d a u .Sup m p ~ ! a e uo apqs 4 paupzxsuoa salngmd , q 1ap!suo3 p ;u&~g 5 Phonons 11. Themol Properties 111 Figure 6 Allowed values in Fourier space of the phonon wavevector K for a square lattice of lat- tice constant a, with periodic boundary conditions applied over a square of side L = 10o. The uni- form mode is marked with a cross. There is one allowrd value of K per area (271/l&~)~ = (ZwIL)', so that within the circle of area 7iKi the smoothed number of allowed paints is ITK'(L/ZW)'. We can obtain the group velocity doldK from the dispersion relation o versus K. There is a singularity in Dl(o)whenever the dispersion relation w(K) is hori- zontal; that is, whenever the group velocity is zero. Density of States in Three Dimensions We apply periodic boundary conditions over N3 primitive cells within a cube of side L, so that K is determined by the condition whence Therefore, there is one allowed value of K per volume (25~lL)~ in K space, or allowed values of K per unlt volume of K space, for each polarization and for each branch. The volume of the specimen is V = L3. The total number of modes with wavevector less than K is found from (18) to he (L125~)~ times the volume of a sphere of radius K. Thus N = (L/25~)~(4?ik"/3) (19) D(W) = d / d = ( v I C / 2 d ) ( d l d w ) . (2 Debye Model for Density o f States In the Debye approximation the velocity of sound is taken as constant fo each polarization type, as it would be for a classical elastic continuum. The di persion relation is written as w = u K , (2 with v the constant velocity of sound. The density of states (20) becomes If there are N primitive cells in the specimen, the total number of acousti phonon modes is N. A cutoff frequency oD is determined by (19) as To this frequency there corresponds a cutoff wavevector in K space: On the Debye model we do not allow modes of wavevector larger than K,. Th number of modes with K 5 K, exhausts the number of degrees of freedom of monatomic lattice. The thermal energy (9) is given by for each polarization type. For brevity we assume that the phonon velocity independent of the polarization, so that we multiply by the factor 3 to obtain where x = ho/r --= fi.wlk,T and xD = hwulk,T = B I T . (2 This defines the Debye temperature 0 in terms of w, defined by (23 We may express 0 as 5 Phomn8 11. Thermal Properties 113 Figure 7 Heat capacity C, of a solid, according to the Debye approximation. The vertical scale is in J mol-' K-I. The holizuntal scale is the temperature normalized to the Debye temperature 0. The re- gion of the T3 law is below 0.18. The asymptotic value at high values of TI0 is 24.943 J mol-' deg-'. Figure 8 Heat capacity of silicon and germa- nium. Note the decrcase at low temperatures. To convert a value in caVmol-K to Jlmol-K, Temperature, K multiply by 4.186. so that the total phonon energy is where N is the number of atoms in the specimen and XD = BIT. The heat capacity is found most easily by differentiating the middle ex- pression of (26) with respect to temperature. Then The Debye heat capacity is plotted in Fig. 7. At T P 0 the heat capacity ap- proaches the classical value of 3Nkn. Measured values for silicon and germa- nium are plotted in Fig. 8. y p y pp y g pp limit go to infinity. We have where the sum over s-4 is found in standard tables. Thus U - 37r4Nk,P/503 f T G 8, and which is the Dehye T3 approximation. Experimental results for argon are plo ted in Fig. 9. At sufficiently low temperature the T3 approximation is quite good; that when only long wavelength acoustic modes are thermally excited. These are ju the modes that may be treated as an elastic continuum with macroscopic elas constants. The energy of the short wavelength modes (for which this approxim tion fails) is too high for them to he populated significantly at low temperature We understand the T3 result by a simple argument (Fig. 10). Only tho lattice modes having h o < kBT will be excited to any appreciable extent a low temperature T. The excitation of these modes will he approximately clas cal, each with an energy close to k,T, according to Fig. 1. Of the allowed volume in K space, the fraction occupied by the excit modes is of the order of do^)^ or (KT/KD)3, where KT is a "thermal" waveve tor defined such that hvK, = k,T and K, is the Debye cutoff wavevector. Th the fraction occupied is (T/O)3 of the total volume in K space. There are of t order of 3N(T/8)3excited modes, each having energy kBT. The energy -3Nk,T(T/O)3, and the heat capacity is -12NkB(T/O)3. For actual crystals the temperatures at which the T3 approximation hol are quite low. It may be necessary to be below T = 8/50 to get reasonably pu T3 behavior. Selected values of 8 are given in Table 1. Note, for example, in the alk metals that the heavier atoms have the lowest 8>, because the velocity sound decreases as the density increases. Einstein Model of the Density of States Consider N oscillators of the same frequency o, and in one dimensio The Einstein density of states is D(o) = N6(o - w,), where the delta functi is centered at ow The thermal energy of the system is Nho U = N(n)ho = e " " / ' , with o now written in place of o,, for convenience Figure 9 Low temperature heat capacity of solid argon, plotted against T3. In this temperature region the experimental results are in excellent agreement with the Debye T3 law with B = 92.0 K. (Conrtesy of L. Finegold and N. E. Phillips.) Figure 10 To obtain a qualitative explanation of the Debye T3 law, we suppose that all phonon modes of wavevector less than K , have the classical thermal energy k,T and that modes between K, and the Debye cutoff K , are not excited at all. Of the 3N possible modes, the fraction excited is (KdKDJ1 = (T/O)3, because this is the ratio of the volume of the inner sphere to the outer sphere. Tne enerais U - k,T . 3N(T@, and the heat capacity is C , = JU/aT= 12NkB(T/B)3. 5 Phonons 11. T h e m 1 Properties 0 no, Figure 11 Comparison of values of the heat capacity of diamond w i t h values calcu- hted on the earliest quantum (Einstein) model, using the characteristic temperature & = W k , = 1320 K. To convert to Jlmol-deg, multiply by 4.186. The heat capacity of the oscillators is C v - - (;gv - =Nk, f:y(e6iy - , (34) as plotted in Fig. 11. This expresses the Einstein (1907) result for the contribu- tion of N identical oscillators to the heat capacity of a solid. In three dimensions N is replaced by 3N, there being three modes per oscillator. The high tempera- ture limit of Cv becomes 3Nk8, which is known as the Dnlong and Petit value. At low temperatures (34) decreases as exp(-fiw/~), whereas the experi- mental form of the phonon contribution is known to he T3as accounted for by the Debye model treated above. The Einstein model, however, is often used to approximate the optical phonon part of the phonon spectrum. General Result for D(m) We want to find a general expression for D(w), the number of states per unit frequency range, given the phonon dispersion relation o(K). The number of d- lowed values of K for which the phonon frequency is between o and w + dw is Mw) dw = ($ Ishe" BK . (35) where the integral is extended over the volume of the shell in K space hounded by the two surfaces on which the phonon frequency is constant, one surface on which the frequency is w and the other on which the frequency is o + dw. The real problem is to evaluate the volume of this shell. We let dS, denote an element of area (Fig. 12) on the surface in K space of the selected constant Figure 12 Element of area dS, on a constant frequency surface in K space. The volume between -two surfaces of constant frequency at wand w + dw is equal to J dS,do/lV,wl. frequency w. The element of volume between the constant frequency surfaces w and w + dw is a right cylinder of base dS, and altitude dK,, SO that J = J ~ S J K ~ . shell Here dKL is the perpendicular distance (Fig. 13) between the surface w con- stant and the surface w + dw constant. The value of dK, will vary from one point to another on the surface. The gradient of w, which is VKw, is also normal to the surface w constant, and the quantity is the difference in frequency between the two surfaces connected by dKk Thus the element of the volume is where vg = lVKwl is the magnitude of the group velocity of a phonon. For (35) we have We divide both sides by dw and write V = L3 for the volume of the crystal: the result for the density of states is 5 Phonons 11. Thelma1 Properties Surface o + dw = constant Figure 13 Tlre quantity dK, is the perpendicular distance between two constant frequency surfaccs in K space, one at frequency o and the other at frequency o + dw. (a) (b) Figure 14 Density of states as a function of frequency for (a) the Debye solid and (b) an actual crystal structure. The specbum for the crystal starts as o2 for small o, but discontinuities develop at singular points. Thc integral is taken over the area of the surface o constant, in K space. The result refers to a single branch of the dispersion relation. We can use this re- sult also in electro~l band theory. There is a special interest in the contribution to D(w) frorn points at which the group velocity is zero. Such critical points produce singularitics (known as Van Hove singnlarities) in the distribution function (Fig. 14). ANHARMONIC CRYSTAL INTERACTIONS The theory of lattice vibrations disciissed thus far has been limited in the potential energy to terms quadratic in the interatomic displacements. This is the harmonic theory; among its consequences are: Two lattice waves do not interact; a single wave docs not decay or change form with time. There is no thermal expansion. Adiabatic and isothermal elastic constants are equal. The elastic constants are independent of pressure and temperature. The heat capacity becomes constant at high temperatures T > 8. tions may bc attributed to the neglect of anharmonic (higher than quadratic terms in the interatomic displacements. We discuss some of the simpler as pects of anharnionic effects. Beautiful demonstrations of anharmonic effects are the experiments o thc interaction of two pllonons to poduce a third phonon at a frequenc w3 = wl + 0~ Three-phonon processes are caused by third-order terms in th lattice potential energy. The physics of the phonon interaction can be state simply: the presence of one phonon canses a periodic elastic strain whid (through the anharmonic interaction) modulates in space and time the elasti constant of the crystal. A second phonon perceives the modulation of the elas tic constant and thereupon is scattered to produce a third phonon, just as from a moving three-dimensional grating. Thermal Expansion We may understand thermal expansion by considering for a classical osci lator the ellect of anharmonic terms in the potential energy on the mean scpa ration of a pair of atoms at a temperature T . We take the potential energy of th atoms at a displacement x from their equilibrium separation at absolute zero as with c, g, andf all positive. The term in x3 represents the asymmetry of th mutual repulsion of the atoms and the term in x4 represents the softening of th vibration at large amplitudes. The ~ninimum at x = 0 is not an absolute mini mum, hut for small oscillations the form is an adequate representation of an in teratomic potential. We calculate the average displacement by using the Boltzmann distribu tion function, which weights the possible values of x according to thei thermodynamic probability with p = l/k,T. For displacements such that the anharmonic terms in th energy are small in comparison with k,T, we may expand the integrands as whence the thermal expansion is 3 t z (x) = -kRT 4cZ Temperature, in K 5 Phonons ZI. T h s m l Properties 121 Figure 15 Lattice constant of solid argon as a funaion of temperature. in the classical region. Note that in (38) we have left a2 in the exponential, but we have expanded exp(pgx3 + pfi4) s 1 + pgx3 + pfi4 + . . .. Measurements of the lattice constant of solid argon are shown in Fig. 15. The slope of the curve is proportional to the thermal expansion coefficient. The expansion coefficient vanishes as T+ 0, as we expect from Problem 5. In lowest order the thermal expansion does not involve the symmetric termfi4 in U(x), but only the antisymmetric term gx3. THERMAL CONDUCTMTY The thermal conductivity coefficient K of a solid is defined with respect to the steady-state flow of heat down a long rod with a temperature gradient dT/&: where jL, is the flux of thermal energy, or the energy transmitted across unit area per unit time. This form implies that the process of thermal energy transfer is a random process. The energy does not simply enter one end of the specimen and pro- ceed directly (hallistically) in a straight path to the other end, but diffuses through the specimen, suffering frequent collisions. If the energy were propa- gated directly through the specimen without deflection, then the expression for the thermal flux would not depend on the temperature gradient, but only on the difference in temperaturc AT between the ends of the specimen, re- gardless of the Tength of the specimen. The random nature of the conductivity process brings the temperature qadient and, as we shall see, a mean free path into the expression for the thermal flux. The e's obtained in this way refer to umklapp processes.] 'Parallel to optic axis. From the kinetic theory of gases we find below thc tbllowing expression for the thermal conductivity: K = ;cue , (42) where C is the heat capacity per unit volu~ne, v is the average particle velocity, and Z is the mean free path of a prticle between collisions. This result was ap- plied first by Debye to describe thermal conductivity in dielectric solids, with C as the heat capacity of the phonons, o the phonon velocity, and e the phonon mean free path. Several representative values of the mean free path are given in Table 2. We give the elementary kinetic theory which leads to (42). The flux of par- ticles in the x direction is in(lozl), where n is the concentration of molec~iles in equilibrium there is a flux of equal magnitnde in the opposite direction. The (. . .) denote average valuc. If c is thc heat capacity of a particle, then in moving frurn a region at local temperatine T + AT to a region at local temperature 2' a particle will give up energy c AT. Now AT between the ends of a free path of the particle is given hy where T is the average time between collisions. The net flnx of energy (from both senses of the particle flux) is therefore dT 1 dT j,, = - n ( d ) c r = - z n ( v 2 ) c ~ - . dx dx If, as for phonons, u is constant, we may write (43) as - -k dl- u - ~ e - ; dx with e = OT and C = nc. Thns K = $cut. 5 Phononn Z I . Thermal Properties Thermal Resistivity o f Phonon Gas The phonon mean free path t ! is determined principally by two processes, geometrical scattering and scattering by other phonons. If the forces between atoms were purely harmonic, there would be no mechanism for collisions be- tween different phonons, and the mean free path wolild be limited solely by collisions of a phonon with the crystal boundary, and by lattice imperfections. There are situations where these effects are dominant. With anharmonic lattice interactions, there is a coupling between differ- ent phonons which limits the value of the mean free path. The exact states of the anharmonic system are no longer like pure phonons. The theory of the effect of anharmonic coupling on thermal resistivity pre- dicts that C is proportional to l/T at high temperatures, in agreement with many experiments. We can understand this dependence in terms of the nnm- ber of phonons with which a given phonon can interact: at high temperature the total number of excited phonons is proportional to T. The collision fre- quency of a given phonon should be proportional to the number of phonons with which it can collide, whence e 1/T. To define a thermal conductivity there must exist mechanisms in the crys- tal whereby the distribution of phonons may be brought locally into thermal equilibrium. Without such mechanisms we may not speak of the phonons at one end of the crystal as being in thermal equilibrium at a temperature T, and those at the other end in equilibrium at T , . It is not sufficient to have only a way of limiting the mean free path, but there must also be a way of establishing a local thermal equilibrium distribu- tion of phonons. Phonon collisions with a static imperfection or a crystal boundary will not by themselves establish thermal equilibrium, because such collisions do not change the energy of individual phonons: the frequency o2 of the scattered phonon is equal to the frequency o, of the incident phonon. It is rather remarkable also that a three-phonon collision process will not establish equilibrium, but for a subtle reason: the total momentum of the phonon gas is not changed by such a collision. An equilibrium distribution of phonons at a temperature T can move down the crystal with a drift velocity which is not disturbed by three-phonon collisions of the form (45). For such collisions the phonon momentum is conserved, because on collision the change in J is K3 - K2 - K1 = 0. Here nK is the number of phonons having wavevector K. For a distribution with J + 0, collisions such as (45) are incapable of es- tablishing complete thermal cquilihrium because they leave J unchanged. If Figure 16a Flow of gas molecules in a state of drifting equilibrium down a long open tube wit frictionless walls. Elastic collision processes among the gas molecules do not change the momen tum or energy flux of the gas because in each collision the velocity of the center of mass of the col liding particles and their energy remain unchanged. Thus energy is transported from left to righ without being driven by a temperature gradient. Therefore the thermal resistivity is zero and th thermal conductivity is infinite. Figure 16b The usual definition of thermal conductivity in a gas refers to a situation where n mass flow is permitted. Here the tube is closed at both ends, preventing the escape or entrance o molecules. With a temperature gradient the colliding pairs with above-average center of mass ve locities will tend to be directed to the right, those with below-average velocities will tend to he di rected to the left. A slight concentration gradient, high on the right, will be set up to enable th net mass transport to be zero while allowing a net energy transport from the hot to the cold end. Figure 16c In a crystal we may arrange to create phonons chiefly at one end, as by illuminatin the left end with a lamp. From that end there will be a net flux of phonons toward the right end o the crystal. If only N processes (K, + K, = K,) occur, the phonon f l u x is unchanged in momentum on collision and some phonon flux will persist down the length of the crystal. On arrival o phonons at the right end we can arrange in principle to convert most of their energy to radiation thereby creating a sink for the phonons. Just as in (a) the thermal resistivity is zero. we start a distribution of hot phonons down a rod with J # 0, the distribution will propagate down the rod with J unchanged. Therefore there is no therma resistance. The problem as illustrated in Fig. 16 is like that of the collisions be tween molecules of a gas in a straight tube with frictionless walls. 5 Phonons 11. Thermal Propedies 125 Figure 16d In U processes there is a large net change in phonon momentum in each collision I event. An initial net phonon f l u will rapidly decay as we move to the right. The ends may act as I I sources and sinks. Net energy transport under a temperature gradient occurs as in (b). Figure 17 (a) Normal K, + K, = K, and (b) umklapp K, + K, = K3 + G phonon collision processes in a two-dimensional square lattice. The square in each figure represents the first Brillouin zone in the phonon K space; this zone contains all the possible independent values of the phonon wavevector. Vectors K with arrowheads at the center of the zone represent phonons absorbed in the collision process; those with arrowheads away from the center of the zone repre- sent phonons emitted in the collision. We see in (b) that in the umklapp process the direction of the x-component of the phonon flux has been reversed. The reciprocal lattice vector G as shown is of length 2 d a , where a is the lattice constant of the crystal lattice, and is parallel to the K, axis. For all processes, N or U , energy must be conserved, so that o, + w, = o , . Umklapp Processes The important three-phonon processes that cause thermal resistivity are not of the form K1 + K2 = K3 in which K is conserved, but are of the form where G is a reciprocal lattice vector (Fig. 17). These processes, discovered by Peierls, are called umklapp processes. We recall that G may occur in all mo- mentum conservation laws in crystals. In all allowed processes of the form of (46) and (47), energy is conserved. vector. Such processes are always possible in periodic lattices. The argument is particularly strong for phonons: the only meaningful phonon K's lie in the first Brillouin zone, so that any longer K produced in a collision must he brought back into the first zone by addition of a G. A collision of two phonons both with a negative valiie of K, can by an umklapp process (G # O), create a phonon with positive K,. Umklapp processes are also called U processes. Collisions in wliidi G = O are called normal processes or N processes. At high temperatures T > 0 all phonon modes are excited because kBT > &om. A substantial proportion of all phonon collisions will then he U processes, with the attendant high momentum change in the collision. In this regime we can estimate the thermal resistivity without particular distinction between Nand U processes; by the earlier argument about nonlinear effects we expect to find a lattice thermal resistivity T at high temperatures. The energy of phonons K,, K, suitable for umklapp to occur is of the order of ikB8, because each of thc phonons 1 and 2 mnst have wavevectors of the order of ;G in order for the collision (47) to be possible. If both phonons have low K, and therefore low energy, there is no way to get from their collision a phonon of wavevector outside the first zone. The uniklapp process must con- serve energy, just as for the normal process. At low temperatures the number of suitable phonons of the high energy ;kB0 rcquired may he expected to vary roughly as exp(-8/2T), according to the Boltzmann factor. The exponential form is in good agreement with experiment. In summary, the phonon mean free path which enters (42) is the mean free path for urnklapp collisions be- tween phonons and not for all collisions between phonons. Geometrical effects may also be in~portant in limiting the mean free path. We must consider scattering by crystal boundaries, the distribution of isotopic masses in natural chemical elements, chemical impurities, lattice imperfec- tions, and amorphous structiires. When at low temperatures the mean free path t becomes comparable with the width of the test specimen, the value o f t is limited by the width, and the thermal conductivity becomes a function of the dimensions of the specimen. This effect was discovered by de Haas and Biermasz. The abrupt decrease in thermal conductivity of pure crystals at low temperatures is caused by the size effect. At low tcmperatiires the umklapp process becomes ineffective in limiting the thermal conductivity, and the size effect becomes dominamt, as shown in Fig. 18. One would expect then that the pho~~om niean free path would be con- stant and of the order of the diameter U of the specimen, so that K=CvD . (48) 5 Phonons IZ. Thermal Properties Figure 18 Thermal conductivity of a highly pr~rified crystal of sodium fluo- ride, after 11. E. Jackson, C. T. Walker, and T. F. McNelly. 50 - 20 L4 + k lo 0 9 .c 5 x 6 ' S 5 2 Figure 19 Isotope effect on thermal -3 conduction in germanium, amor~nting to 8 a factor of three at the conductivity a maximum. The enriched specimen is 96 f percent Ge74, natural germanium i s 20 8 Q.5 percent Ge7", 27 percent Gei2, 8 percent ~ e " , 37 percent Ge", and 8 percent 0.2 6eV6. Below 5 K the enriched specimen has K = 0.06 T3, which agrees well with 0.1 Casimirh theory for thennal resistance 1 2 5 10 20 50 100 200 500 caused by bonndaly scattering. (After Te~nperaturc, in K T. H. Geballe and 6. W. IIull.) The only temperature-dependent term on the right is C, the heat capacity, which varies as T%t low temperatures. We expect the thermal conductivity to vary as T h t low temperatures. The s i z e effect enters whenever the phonon mean free pat11 becomes comparahle with the diameter of the specimen. Dielectric crystals may have thermal conductivities as high as metals. Syn- thetic sapphire (A1,0,) has one of the highest values of the conductivity: nearly 200 W c m K-' at 30 K. The maximum of the thermal conductivity in sapphire is greater than the maximum of 100 W cm-' K-' in copper. Metallic gallium, however, has a co~lductivity of 845 W cm-' K-' at 1.8 K. The electronic contri- bution to the thermal conductivity of metals is treated in Chaptcr 6. In an otherwise perfect crystal, the distribution of isotopes of the chemical elements often provides an important mechanism for phonon scattering. The random distribution of isotopic mass disturbs the periodicity of the density as seen by an elastic wave. In some substances scattering of phonons by isotopes is comparable in importance to scattering by other phonons. Kcsults for gcr- manium are show1 in Fig. 19. Enhanced thcrmal condncti\lty has been oh- served also in isotopically pure silicon and diamond; the latter has device importance as a heat sink for laser sources. Problems 1. Singularity in density of ntatex. (a) Frnm the dispersion relation rlcrivcd in Chap- ter 4 for a irroiratoinic linear lattice nf N atnms with nearest-neighhnr interactions, show that t l ~ r density of modes is where w , is the maximum frequenc).. (b) Suppose that an optical phonon branch has the form w(K) = w,-AK2, near K = 0 in three dimensions. Show that D(w) = ( L / ~ ' T ) ~ ( ~ ' T / A " ' ) ( ~ - w)lJ2 for w < w , and D(w) = 0 for o > o,. Here the density o C modes is discontinuous. 2. Rma thennal dilation of crystal cell. (a) Estimate for 300 K the root mean squarc thcrmal dilation Am' for a primitive cell of sodium. Take the bulk modulus as 7 X 10" crg ~ m - ~ . Note that the Debye temperature 158 K is less than 300 K, so that thc thcrmal cncrg is of the order of k,T. (b) Use this result to estimate the root mean square thcrmal fluctuation Aala of the lattice parameter. 3 . Zero point lattice displacement and atrain. (a) In the Debye approximation, show that the mean square displacement of an atom at absolute zero is (RZ) = 3hw38dp3, where v is the velocity of sound. Start from the result (4.29) summed over the independent lattice modes: (R? )= (h/2pV)Zw-'. We have included a factor of to go from mean square amplitude to mean square displacement. (b) Show that Zw-' and (d) diverge for a one-dimensional lattice, but that the mean square strain is finite. Consider ((dkva~)? = i2ICui as the mean square strain, and show that it is equal to fiw%/4h4N03 Tor a linc of i \ r atoms each of mass M , counting longitudinal modes only The divergence of R2 is not significant for any physical measurement. 4. Heat capacity of layer lattice. (a) Consider a dielectric crystal made up of layers of atoms, with rigid coupling between layers so that the motion of the atoms is restricted to the plane of the layer. Show that the phonon heat capacity in the Debye approximation in the low temperature limit is proportional to ' P . 5 Phononn 11. Thermal Properties (b) Suppnse instead, as in many layer structures, that adjacent layers are very weakly bound to each other. What form would you expect the phonon heat capacity to ap- proach at extremely low temperatures? '5. Griineinen constant. (a) Show that the free energy of a phonon mode of fre- quency o is kgT In [2 sinh (ho/2k,T)]. It is necessary to retai~~ the zero-point energy iho to obtain this result. (h) If A is the fractional volurrre change, then the free en- ergy of the crystal may be written as F(A, T) = ~BA' + k , ~ x In [ 2 sinh (fiwK/2kBT)] where B is the bulk modulus. Assume that the volume deparrdence of o ~ ( is Swlw = -yA, whcrc y is known as the Criineisen constant. If y is taken as indepcn- dent nf the mode K, show that P is a minimum with respect to A whcn BA = yZiho cot11 (fiw/2kBT), and show that this may be written in terrrrs of the thermal energy density as A = yU(T)/B. (c) Show that on the Debye model y = -a In Old In V. Note: Many approximations are involved in this theory: the result (a) is valid only if o is in- dependent of tcmperature; y may be quite different for differerrt modes. h his pmblcm is solnewhat difficult. Free Electron Fermi Gas ENERGY LEVELS IN ONE DIMENSION EFFECT OF TEMPERATURE ON THE FERMI-DIRAC DISTRIBUTION FREE ELECTRON GAS IN TIIREE DIMENSIONS HEAT CAPACITY OF THE ELECTRON GAS Experimental heat capacity of metals Heavy fermions ELECTRICAL CONDUCTIVITY AND OHM'S LAW Experimental electrical resistivity of metals Umklapp scattering MOTION IN MAGNETIC FIELDS Hall effect THERMAL CONDUCTIVITY OF METALS Ratio of thermal to electrical conductivity PROBLEMS 1. Kinetic energy of eleclron gas 2. Pressure and bulk modulus of an elcctron gas 3. Chemical potential in two dimensions 4. Fermi gases in astrophysics 5. Liquid He3 6. Frequency dependence of the electrical conductivity 7. Dynamic magnetoconductivity tensor for frcc electrons 8. Cohesive energy of free electron Fermi gas 9. Static magnetoconductivity tensor 10. Maximum surface resistance Figure 1 Schematic model of a crystal of sodium metal. The atomic cores are Na' ions: they are immersed in a sea of conduction electrons. The conduction electrons are derived from the 3s valence electrons of the free atoms. The atomic cores contain 10 electrons in the configuration ls22s22p~ In an alkali metal the atomic cores occupy a relatively small part (-15 percent) of the total volume of the crystal, hut in a nohle metal (Cu, Ag, Au) the atomic cores are relatively larger and may he in contact with each other. The common crystal structure at room temperature is hcc for the alkali metals and fcc for the nohle metals. In a theory which has given results like these, there must certainly be a great deal of truth. H . A. Lorentz We can understand many physical properties of metals, and not only of the simple metals, in terms of the free electron model. According to this rnodel, die valence electrons of the constituent atoms becorne coriduction electrons and move about freely through the volurrie of the metal. Even in metals for which the free electron model works best, the charge distribution of thc conduction electrons reflects the strong electrostatic potential of the ion cores. The utility of the free electron model is greatest for properties that depend essentially on the kinetic properties of the conduction electrons. The interaction of the conduction clcctrons with the ions of the lattice is treated in the next chapter. Thc simplest metals are the alkali metals-lithium, sodium, potassium, cesium, and rubidium. In a free atom of sodium tlie valence electron is in a 3s state; in the metal this electror~ becomes a conduction electron in the 3 s conduction band. A r~ionovalent crystal which contains N atoms will have N conduction electrons and N positive ion cores. Thc Nat ion core contains 10 electrons that occupy the Is, 29, and 2p shells of the free ion, with a spatial distribution that is csscntially the same when in the metal as in the free ion. The ion cores fill only about 15 percent of the volume of a sodiurri crystal, as in Fig. 1. The radius of the free Na+ ion is 0.98 A, whereas one-half of the nearest-neighbor distance of the rr~etal is 1.83 A. The interpretation of metallic properties in terms of the motion of free electrons was developed long before the invention of quantum mechanics. The classical theory had several conspicuous successes, notably the derivation of tlie form of Ohm's law and the relation between the electrical and thermal conduc- tivity. The classical theory fails to explain the heat capacity and the magnetic susceptibility of the conduction electrons. (These are not failures of the free electron model, but failures of the classical Maxwell distribution function.) There is a further difficulty with the classical model. From many types of experiments it is clear that a conduction electron in a metal can move freely in a straight path over many atomic distances, undeflected by collisions with other cond~~ction electrons or by collisions with the atom cores. In a very pure specimen at low temperatures, the mean free path rnay be as long as 10' inter- atomic spacings (rnore tivan 1 cm). Why is condensed matter so transparent to conduction electrons? The ' answer to the question contains two parts: (a) A conduction electron is not deflected by ion cores arranged on a periodic lattice because mattcr waves can propagate freely in a periodic structure, as a consequencc of the mathematics treated in thc following chapter. (b) A conduction elrctron is scattered only in- frequently hy other conductio~i electrons. This property is a consequence of the Pauli exclusion principle. By a free electron Fermi gas, we shall mean a gas of free electroris subject to thc Pa111i principle. ENERGY LEVELS IN ONE DIMENSION Consider a free electron gas in one dimension, taki~~g account of quantum theory and of the Pauli principle. An electron of maqs m is confined to a length L by infinite harriers (Fig. 2). The wavefunction $,(x) of the electron is a solu- tion of the Schrodinger equation X+ = E+; with the neglect of potential cnergy we have X = p2/2m, where p is the momentum. In quantum theory p may be represented by the operator -i?i dldx, so that where t , is the e n c r a of the electron in the orbital. We use thc term orbital to denote a solution of the wave equation for a system of only one electron. The term allows us to distinguish between an exact quantum state of the wave equation of a system of N interacting elec- trons and an approxirrlate quantum state which we construct by assigning the N electrons to N different orbitals, where each orbital is a solution of a wave equation for one electron. The orbital model is exact only if there are no inter- actions between electrons. The boundary conditions are cL,(O) = 0; $,,(L) = 0, as imposed by the infi- riite potential energy barriers. They are satisfied if the wavefunction is sir~elike with an integral number n of half-wavelengths between 0 and L: where A is a constant. \Ve see that (2) is a solution of (1), because whence the energy E, is given by We want to accommodate N electrons on the linc. According to the Pauli exclusion principle, no two electrons can have all their quantum numbers 6 Free Electron Fermi Gas 135 Energy levels Wavefunctions, A = Z L relative scale V w - g ---- ----- ---- 3 2 IN : Figure 2 First three energy levels and wave- + .- 9 G functions of a free electron of mass m confined 9 C. to a line of length L. The energy levels arc la- .- beled according to the quantum number n $ 4 ------- ------- 2 3 which gives thc liu~nber of half-wavelengths in r W A =2L the wavefunction. The wavelengths are indi- --------------- cated on the wavefunctions. The energy E, of the level of quantum number n is equal to x - (h'/~m)(n/21,)~. identical. That is, each orbital can be occupied by at most one electron. This applies to electrons in atoms, molecules, or solids. In a linear solid the quantum numbers of a conduction electron orbital are n and m,, where n is any positive integer and the magnetic qnanti~m number m, = + : , according to spin orientation. A pair of orbitals labeled by the quan- tum number n can accomlnodate two electrons, one with spin up and one with spin down. If there are six electrons, then in the ground state of the system the filled orbitals are thosc given in the table: Electroll glectron n nccupancy n o~rupancy More than one orbital may have the same energy. The number of orbitals with the saIrle energy is called the degeneracy. Let nF denote thc topmost filled energy level, where we start filling the levels from the bottom (n = 1) and continue filling higher levels with elec- trons until all N electrons are accommodated. It is convenient to suppose that N is an even number. The condition enF = N determines nF, the value of n for the uppermost filled level. The Fermi energy eF is defined as the energy of the topmost filled level in the ground state of the N electron system. By (3) with n = n, we have in one dimension: EFFECT OF TEMPERATURE ON THE FERMI-DIRAC DISTRIBUTION The ground state is the state of the N electron system at absolute zero. What happens as the temperature is increased? This is a standard problem in elementary statistical mechanics, and thc sohition is given by the Fermi-Dirac distribution function (Appendix D and TP, Chapter 7). The kinetic cncrgy of the electron gas increases as the temperature is in- creased: some energy levels are occupied which were vacant at absolute zero, and some levels are vacant which were occupied at absolute zero (Fig. 3). Thc Fermi-Dirac distribution gives the probability that an orbital at energy E will be occupied in art ideal electron gas in thermal cq~iilihrium: The quantity p is a function of the temperature; p is to be chosen for the particular problcm in siich a way that the total number of in the system comcs out correctly-that is, equal to N . At absolute zero = E ~ , because in the limit T + 0 the functionf(e) changes discontinuously from the value 1 (filled) to the value 0 (empty) at = cF = p. At all iemperatures f j ~ ) i s equal to when E = p, for then the denominator of (5) has the valuc 2. 6/kB, in units of 1@ K Figure 3 Femi-Dirac distrihutiorr function (5) at the valious labelled temperah~res, for T, - cl/kB = 50,000 K. The results apply to a gas in three di~ne~lsions. The total number of parti- cles is constant, independent of temperature. The chemical potential p at each te~nperaturc may be read off the graph as the energy at whichj = 0.5. 6 Free Electron Fermi Gas 137 The quantity y is the chemical potential (TP, Chapter 5), and we see that at absolute zero the chemical potential is eqml to the Fermi energy, de- fined as the energy of the topmost filled orbital at absolute zero. The high energy tail of the distrihi~tion is that part for which 6 - y 9 k,T; here the exponential term is dominant in the denominator of (5), so that f(e) - exp[(p - <)/k,T]. This lim~t is called the Boltzmann or Maxwell distribution. FREE ELECTRON GAS IN THREE DIMENSIONS The free-particle Schrodinger equation in three dimensions is If the electrons are confined to a cube of edge J,, the wavefunction is the standing wave $,,(r) = A sin ( m n ~ / L ) sin (m,y/L) sin (m,z/L) , (7) where n,, fly, 11, are positive integers. The origin is at one corner of the cubc. It is convenient to introduce wavefiinctions that satisfy periodic boundary conditions, as we did for phonons in Chapter 5. We now require the wavefunc- tions to be periodic in 1 , y, z with period L. Thus and si~liilarly for thc y and z coordinates. Wavefunctions satisfying the free- particle Schrodinger equation and the periodicity condition are of the form of a traveling plane wave: $k(r) = exp (ik . r) r provided that the components of the wavevector k satisfy and similarly fork,, and k , . Any component of k of the form 2 n d L will satisfy thc periodicity coridition over a Icngth L, where n is a positive or negativr integer. The com- ponents of k are the quantum nurnhers of the prohlem, along with the quantum number m, for the spin direction. We confirm that these values of k, satisfy (8), for On substituting (9) in (6) we have the energy ek of the orbital with wavevector k: The magnitude k of the wavevector is related to the wavelength h by k = 2?rlh. The linear momentum p may be represented in quantum mechanics by the operator p = -ifiV, whence for the orbital (9) so that the plane wave $ k is an eigenfunction of the linear momentum with the eigenvalue f i k . The particle velocity in the orbital k is given by v = f i k l m . In the ground state of a system of N free electrons, the occupied orbitals may be represented as points inside a sphere in k space. The energy at the sur- face of the sphere is the Fermi energy; the wavevectors at the Fermi surface have a magnitude k, such that (Fig. 4): From (10) we see that there is one allowed wavevector-that is, one dis- tinct triplet of quantum numbers k,, k , , , k,-for the volume element ( 2 7 r / ~ ) ~ of k space. Thus in the sphere of volume 4?rk23 the total number of orbitals is where the factor 2 on the left comes from the two allowed values of the spin quantum number for each allowed value of k. Then (15) gives which depends only on the particle concentration. Figure 4 In the ground state of a system of N free electrons the occupied orbitals of the system fill a sphere of radius k , where EF = fL2k,22m is the energy of an electron having a wavevector k , . Table 1 Calculated free electron Fermi surface parameters for metals at room temperature (Except for Na, K, Rh, Cs at 5 K and Li at 78 K) Fermi Elcciron Radius'" Fermi Fermi Fer~ni terriperature concentration, parameter \vdvevectur, vc1ocit)i energy, & - ~ ~ / k V;llmw Metal in C I I I - ~ r. in cm-' in cm sC1 in eV in dee K - -- "The d~mens~onless radlus parameter IS defined as r,, = ~,la,, where a" is the first Bohr radlus and r, IS the radlus of a ~ p h e r r that contrlns one electron Energy, + Using (14) and (l6), Figure 5 Density of single-particle states as a func- tion of energy, for a free electron gas in three dimen- sions. The dashed curve represents the density f ( E , T)D(E) of filled orbitals at a finite temperature, but such that k,T is small in comparison with E , . The shaded area represents the filled orbitals at absolute zero. The average energy is increased when the tem- perature is increased from 0 to T, for electrons are thermally excited from region 1 to region 2. This relates the Fermi energy to the electron concentration NN. The electron velocity vF at the Fermi surface is Calculated values of k , , v,, and E , are given in Table 1 for selected metals; also given are values of the quantity TF which is defined as ~,/k,. (The quantity TF has nothing to do with the temperature of the electron gas!) We now find an expression for the number of orbitals per unit energy range, D(E), called the density of states.' We use (17) to obtain the total number of orbitals of energy SE: so that the density of states (Fig. 5) is 'Strictly, D(E) is the density of one-particle states, or density of orbitals. 6 Free Electron Fermi G a a 141 This result may he expressed more simply by comparing (19) and (20) to ohtain at E Within a factor of the ordcr of unity, the number of orbitals per unit energy range at the Fermi energy is the total number of conduction electrons divided by the Fermi energy, just as we would expect. HEAT CAPACITY OF THE ELECTRON GAS The question that caused the greatest difficulty in the early development of the electron theory of metals concerns the heat capacity of the conduction : electrons. Classical statistical mechanics predicts that a free particle should : have a heat capacity of k,, where k, is the Boltzrnann constant. If N atoms i each give one valence electron to the electron gas, and the electrons arc freely i mobile, then the electronic contribution to the heat capacity shonld be ;h'k,, i just as for the atoms of a monatomic gas. But the observed electronic contribu- tion at room temperature is usually less than 0.01 of this value. i This important discrepancy distracted the early workers, such as Lorentz: [ How can the electrons participate in electrical conduction processes as if they ' were mobile, while not contributing to the heat capacity? The question was / answcrcd only upon the discovery of the Pauli exclusion principle and the Fermi distribution function. Fermi found the correct result and he wrote, "One recognizes that the specific heat vanishes at absolute zero and that at low temperatures it is proportional to the absolute temperatnre." When we heat the specimen from absohite zero, not every electron gains 1 an energy -kBT as expectcd clawically, but only those electrons in orbitals / within an energy range k,T of the Fermi level are excited thermally, as in ! i Fig. 5. This gives an immediate qualitative solution to the problem of the heat I capacity of the conduction electron gas. If N is the total number of electrons, ! i only a fraction of the order of TITF can he excited thermally at temperature T, [ because only these lie within an energy rangc of the order of kBT of the top of I the energy distribution. Each of these NT/Tr clcctrons ha5 a thermal energy of the order of kBT. The total electronic thermal kinetic energy U is of the order of The electronic heat capacity is given by and is directly proportional to T, in agreement with the experimental results discussed in the following section. At room temperature CeI is smaller than the classical value Nk, by a factor of the order of 0.01 or less, for TF -5 X 104K. We now derive a quantitative expression for the electronic heat capacity valid at low temperatures kllT 4 eF. The increase AU = U(T) - 17(0) in the total energy (Fig. 5) of a system of W electrons when heated from 0 to T is Here f(c) is the Fer~ni-Dirac function (5): and D(c) is the number of orbitals per unit energy range. i17e multiply the identity by eF to obtain We use (26) to rcwritc (24) as The first integral on the right-hand side of (27) gives the energy needed to take electrons from eF to the orbitals of energy l > c ~ , a i d the second integral gives the energy needed to bring the electrons to C , fro~n orbitals below c,. Both contributions to the energy are positive. The product f ( ~ ) D ( r ) d e in the first intcgral of (27) is the number of electrons elevated to orhitals in the energy range d~ at an energy C . The factor [l - . f ( ~ ) ] in the second integral is the probability that an electron has been removed from an orbital E . The function AU is plotted in Fig. 6. The heat capacity of the electron gas is f o u ~ ~ d on differentiating AU with respect to T. Tlie orily temperature-depe~idellt term in (27) is f(r), whence we can group terms to obtain At the terriperatures of interest in metals, r/tF < 0.01, and we see from Fig. 3 that ( C - r F ) dYdT has large positive peaks at energies near c,. It is a 6 Free Electron Fenni Gas 143 Figure 6 Temperature dependence of the energy of a noninteracting fermion gas in three dimensions. The energy is plotted in normal- ized form as AUINE,, where N is the number of electrons. The temperature is plotted as kBTkp. Region of degenerate quantum gas Figure 7 Plot of the chemical potential p versus temperature as k,T for a gas of noointeracting fermions m three dimensions. For convenience in plotting, the units of p and k,T are 0 . 7 6 3 ~ ~ good approximation to evaluate the density of states D(E) at E, and take it outside of the integral: Examination of the graphs in Figs. 7 and 8 of the variation of the chemical potential p with T suggests that when kgT < eF we ignore the temperature I 1.00 P - EF Figure 8 Variation with temperature of the chemical potential p, for free electrorr Fermi gases in unc and three dimensions. In common metals T / E ~ = 0.01 at room temperature, so that p is closely equal to E,. 0.95 1 I These curves were calculated from series expansions 0 0.1 0.2 of the integral for the number of particles in the - r - system. t~ dependence of the chemical potential p in the Fermi-Dirac distribution func- tion and replace p by the constant E,. W e have then, with T = kBT, We set X = (E - E&T , (31) and it follows from (29) and (30) that We may safely replace the lower limit by -w because the factor ex in the inte- grand is already negligible at r = e F / 7 if we are concerned with low tempera- tures such that E ~ / T - 100 or more. The integral in (32) then becomes whence the heat capacity of an electron gas is From (21) we have for a free electron gas, with kRTF -- eF. Thus (34) becomes 6 Free Electron Fermi Gas 145 Recall that although T, is called the Fermi temperature, it is not the electron temperature, but only a convenier~t reference notation. Experimental Heat Capacity o f Metals At temperatures much below both the Debye temperature 0 and the Fermi temperature T,, the heat capacity of metals may be written as the sum of elcctron and phonon contributions: C = yT + AT3, where y and A are con- stants characteristic of the material. The electronic term is linear in T and is dominant at sufficiently low temperatures. It is convenient to exhibit the ex- perimental values of C as a plot of CIT versus p: for then the points should lie on a straight line with slope A and Intercept 7. Such a plot for potassium is shown In Fig. 9. Observed values of y, called the Sommerfeld paramctcr, are pven in Table 2. Thr ohscrved values of the coefficient y are of the expected magnitude, , but often do not agree very closely with the value calculated for free electrons of mass m by use of (17) and (34). It is coIrlmon practice to express the ratio of the observed to the free electron values of the electronic hcat capacity as a ratio of a thermal effective mass nLth to the electron mass m, where mm IS de- 1 fined by the relation This form arises in a natural way because eF is i~iversely proportional to the mass of the electron, whence y a m. Values of the ratio are given in Table 2. The departure from unity involves three scparate effects: The interaction of thc conduction electrons with the periodic potential of the rigid crystal lattice. The effective mass of an electron in this potential is called the hand effective mass. Figure Y Experimental heat capacity values for potassi~lm, plotted a s C/T versus T2. (Aftcr W. H. Lien and N. E. Phillips.) 6 Free Electron Fermi Gas 147 The interaction of the conduction electrons with phonons. An electron tends to polarize or distort the lattice in its neighborhood, so that the rnov- ing electron tries to drag nearby ions along, thereby increasing the effective mass of the electron. The interaction of the conduction electrons with themselves. A moving elec- tron causes an inertial reaction in the surrounding electron gas, thereby in- creasing the effective mass of the electron. Heavy Fermions. Several metallic compoiinds have been discovered that have enormous values, two or three orders of magnitude higher than usual, of the elec- tronic heat capacity constant y. The heavy fermion compourids include UBe13, CcAI,, and CeCu,Si,. It has been suggested that f electrons in these compounds may have inertial masses as high as 1000 rn, because of the weak overlap of wave- functions off electrons on neighboring ions (see Chapter 9, "tight binding"). ELECTRICAL CONDUCTIVITY AND OHM'S LAW The momentum of a free electron is related to the wavevector by m v = hk. In an electric field E and magnetic field B the force F on an electron of charge -e is -e[E + (1Ic)v X B ] , so that Newton's second law of motion becomes In the absence of collisions the Fermi sphere (Fig. 10) moves in k space at a uniform rate by a constant applied electric field. We integrate (39) with B = 0 to obtain k(t) - k(0) = -eEt/h . (40) If the force F = -eE is applied at tirne t = 0 to an electron gas that fills the Fermi sphere centered at the origin of k space, then at a later time t the sphere will be displaced to a new center at Sk = -eEtlh . (41) Notice that the Fcrmi sphere is displaced as a whole because every electron is displaced by the same 6k. Because of collisions of electrons with impurities, lattice imperfections, and phonons, the displaced sphere may be maintained in a steady state in an electric field. If the collisio~i time is r, the displacement of the Fermi sphere in the steady state is given by (41) with t = 7 . The incremental velocity is v =$%ldm = -eErlm. If in a constant clectric field E there are n electrons of charge q = -e per unit volume, the electric current density is j = nqv = ne2rE/m . (42) This is Ohm's law. Fermi sphere k Figure 10 (a) The Ferrni sphere encloses the occupied electron orbitals i n k space in the ground state of the electron gas. The net momentum is zero, because for every orbital k there is an occu- pied orbital at -k. (b) Under the illfluc~rcc of a constant force F acting fir a time interval t eveT orbital has its k vector increased hy Sk = Ftlfi. This is equivalent to a displacement of the whole Fermi sphere by 6k. The total momentum is ATfiSk, if there are N electrons present. The applica- tion oT the force incrcascs tlrc clrcrgy of tile system by N(fi6k)2/2m. The electrical conducti\lty u is defined by j = uE, so by (42) The electrical resistivity p is defined as the reciprocal of the conductivity, so that I I p = mn/ne27 . (44) Values of the electrical conductivity and resistivity of the elements are given in Table 3. In Gaussian units u has the dimensions of frequency. It is easy to understand thc result (43) for the conductivity of a Fermi gas. We expect the charge transported to he proportional to thc chargc dcnsity ne; the factor e/m enters (43) because the acceleration in a given electric field is proportional to e and inversely proportional to the mass m. The time T describes the free time during whidi the field acts on the carrier. Closely the same result for the electrical conductivity is obtained for a classical (Mawwellian) gas of elec- trons, as realized at low carrier concentration in many semico~lductor problems. Experimental Electrical Resistivity o f Metals The electrical resistivity of most metals is dominated at room te~nperature (300 K) hy collisions of thc conduction electrons with lattice phonons and at (a) (b) Figure 11 Electrical resistivity in most metals arises from collisions of electrons with irregulari- ties in the lattice, as in (a) by phonons and in (h) by impurities and vacant lattice sites. liquid helium temperature (4 K) by collisions with impurity atoms and me- chanical imperfections in the lattice (Fig. 11). The rates of these collisions are often independent to a good approximation, so that if the electric field were switched off the momentum distribution would relax back to its ground state with the net relaxation rate where rL and T, are the collision times for scattering by phonons and by imper- fections, respectively. The net resistivity is given by P = P L + P i , (46) where pL is the resistivity caused by the thermal phonons, and p, is the resistiv- ity caused by scattering of the electron waves by static defects that disturb the periodicity of the lattice. Often pL is independent of the number of defects when their concentration is small, and often p, is independent of temperature. This empirical observation expresses Matthiessen's rule, which is convenient in analyzing experimental data (Fig. 12). The residual resistivity, p,(O), is the extrapolated resistivity at 0 K because p, vanishes as T + 0. The lattice resistivity, pL(T) = p - p,(O), is the same for different specimens of a metal, even though p,(O) may itself vary widely. The resistivity ratio of a specimen is usually defined as the ratio of its resistivity at room temperature to its residual resistivity. It is a convenient approximate in- dicator of sample purity: for many materials an impurity in solid solution cre- ates a residual resistivity of about 1 pohm-cm (1 X ohm-cm) per atomic percent of impurity. A copper specimen with a resistivity ratio of 1000 will have a residual resistivity of 1.7 x pohm-cm, corresponding to an 6 Free Electron Permi Gas 151 Figure 12 Resistance of potassium below 20 K, as measured on two specirrrells by D. MacDonald and K. Mendelssohn. The differe~~t i~iterceyts at 0 K are attributed to different concentrations of impurities arrd static imperfections in the two specimens. impurity concentration of about 20 ppm. In exceptionally pure specimens the resistivity ratio may be as high as 10" whereas in some alloys (e.g., manganin) it is as low as 1.1. It is possible to obtain crystals of copper so pure that their conducticity at liquid helium temperatures (4 K) is nearly 10' times that at room temperature; for these conditio~~s T = 2 X s at 4 K. The mean free path t of a conduc- tion electron is defined as where uF is the velocity at the Fermi surface, because all collisions involve only electrons near the Fermi surface. From Table 1 we have c~ = 1.57 X 10' cm s-' for Cu, thus the mean free path is ((4 K) = 0.3 cm. Mean free paths as long as 10 cm have been observed in very pure metals in the liquid helium tempera- ture range. The temperature-dependent part of the electrical resistivity is proportional to the rate at which an electron collides with thermal phonons and thermal elec- trons. The collision rate with phonons is proportional to the co~rcentration of thermal phonons. One simple limit is at te~r~peratures over the Debye tempera- ture 0: here the phonon co~rcentration is proportional to the temperature T, so that p a T for T > 8. A sketch of the theory is given in Appendix J. Umklapp Scattering Umklapp scattering of electrons by phonons (Chapter 5) accounts for most of the electrical resistivity of metals at low tenrperatures. These are electron-phonon scattering processes in which a reciprocal lattice vector G is involved, so that electron momentum change in thc process may be much larger Figure 13 Two Fermi spheres in adjacent . . - zones: a construction to show the role of phonon umklapp processes in electrical resistivity. 40 than in a normal electron-phonon scattering process at low temperatures. (In an umklapp process the wavevector of one particle may be "flipped over.") Consider a section perpendicular to [loo] through two adjacent Brillouin zones in bcc potassium, with the equivalent Fermi spheres inscribed within each (Fig. 13). The lower half of the figure shows the normal electron-phonon collision k' = k + q, while the upper half shows a possible scattering process k' = k + q + G involving the same phonon and terminating outside the first Brillouin zone, at the point A. This point is exactly equivalent to the point A' inside the orignal zone, where AA' is a reciprocal lattice vector G. This scat- tering is an umklapp process, in analogy to phonons. Such collisions are strong scatterers because the scattering angle can be close to T. When the Fermi surface does not intersect the zone boundary, there is some minimum phonon wavevector q, for umklapp scattering. At low enough temperatures the number of phonons available for umklapp scattering falls as exp(-OdT), where 0, is a characteristic temperature calculable from the geometry of the Fermi surface inside the Brillouin zone. For a spherical Fermi surface with one electron orbital per atom inside the bcc Brillouin zone, one shows by geometry that q, = 0.267 k,. The experimental data (Fig. 12) for potassium have the expected exponen- tial form with 6, = 23 K compared with the Debye O = 91 K. At the very low- est temperatures (below about 2 K in potassium) the number of umklapp processes is negligible and the lattice resistivity is then caused only by small angle scattering, which is the normal (not umklapp) scattering. MOTION IN MAGNETIC FIELDS By the arguments of (39) and (41) we are led to the equation of motion for the displacement 6k of a Fermi sphere of particles acted on by a force F and by friction as represented by collisions at a rate 11~: 6 Free Electron Fermi Gas 153 The free particle acceleration term is (M/dt) 6k and the effect of collisions (the friction) is represented by UWT, where T is the collision time. Consider now the motion of the system in a uniform magnetic field B. The Lorentz force on an electron is If mv = fi6k, then the equation of motion is An important situation is the following: let a static magnetic field B lie along the z axis. Then the component equations of motion are The results in SI are obtained by replacing c by 1. In the steady state in a static electric field the time derivatives are zero, so that the drift velocity is where w, = eBlmc is the cyclotron frequency, as discussed in Chapter 8 for cyclotron resonance in semiconductors. Hall Eflect The Hall field is the electric field developed across two faces of a conduc- tor, in the direction j x B, when a current j flows across a magnetic field B. Consider a rod-shaped specimen in a longitudinal electric field E, and a trans- verse magnetic field, as in Fig. 14. If current cannot flow out of the rod in the y direction we must have 8uy = 0. From (52) this is possible only if there is a transverse electric field Section + + + + + + + + perpendicular to B axis; drift velocitv Figure 14 The standard geometry for the Hall effect: a rod-shaped specimen of rectangular cross-section is placed in a magnetic field EL, as in (a). An electric field E, applied across the end electrodes causes an electric current density j, to flow down the rod. The drift velocity of the negatively-charged electrons immediately after the electric field is applied as shown in (b). The deflection in the -y direction is caused by the magnetic field. Electrons accumulate on one face of the rod and a positive ion excess is established on the opposite face until, as in (c), the trans- verse electric field (Hall field) just cancels the Lorentz force due to the magnetic field. The quantity defined by is called the Hall coefficient. To evaluate it on our simple model we use j, = ne27E/m and obtain This is negative for free electrons, fore is positive by definition. 6 Free Electron Fermi Gas Table 4 Comparison of observed Hall coefficients with free electron theory ~ ~ ' wavr method at 4 K are by J. M. Goodman. The values of the carrier concentratioh n are from I Table 1.4 except for Na, K, Al, In. where Goodman's vahles are nsed. To corrvert tlrc valuc oTR, in CGS nnits to the value in volt-cn~/amp-gauss, ~nultipl~ by 9 x 10"; to convert A, in CGS to conv. -1.89 1 electron - 1.48 -2.619 1 electron 2.603 conv. -2.3 1 electron -4.944 con^^. 4 . 7 1 electron -6.04 1 electron 0 . 8 2 1 electron 1 . 1 9 1 clcctron -1.18 - - -0.92 - - +1.136 I hole +I.135 + 1.774 1 hole +1.780 - - - - conw -6000. - - The lower the carrier concesltration, the greater the magnitude of the Hall coefficient. Measuring RH is all important way of measuring the carrier concentration. Note: The symbol RH denotes the Hall coefficient (54), but the same sysnbol is sometimes used with a different meaning, that of Hall resis- tance in two-dimensional problems. The simple result (55) follows from the assusnption that all relaxation times are equal, independent of the velocity of the electron. A numerical fac- tor of order unity enters if the relaxation time is a function of the velocity. The expression becomes somewhat more complicated if both electrons and holes contribute to the conductivity. In Table 4 observed values of the IIall coefficient are compared with val- ues calculated from the carrier concentration. The most accurate Ineasure- ments are made by the method of helicon resonance which is treated as a problem in Chapter 14. The accurate values for sodium and potassium arc in excellent agreement with values calculated for one cond~~ction electron per atom, using (55). Notice, however, the experimental values for the trivalent elerrlents aluminum and indiu~n: these agree with values calculated for one positive charge carrier per atom and thus disagree in magnitude and sign with values calculated for the expected three negative charge carriers. The problem of an apparent positive sign for the charge carriers arises also for Be and As, as seen in the table. The anomaly of the sign was explained by Peierls (1928). The motion of carriers of apparent positive sign, which Heisenberg later called "holes," cannot be explaitled by a free electron gas, but finds a natural explanation in terms of the energy band theory to be devclopcd in Chapters 7-9. Band theory also accounts for thc occiirrence of very large values of the Hall coefficient, as for As, Sh, and Bi. THERMAL CONDUCTIVITY OF METALS In Chapter Fj we found an expression K = ;Cut for thr thermal cond~ictiv- ity of particles of velocity v , heat capacity C per nnit vohlme, and mean free path t ?
. The thermal conductivity of a Fermi gas follows from (36) for the heat capacity, and with E, = :mu; : 2 nkZT &=-.L. u,.Z =- 2nkiT.r 3 mu: 3m Here 4 = V ~ T ; the electron concentration is n, and T is the collision time. Do the electrons or the phonons carry the greater part of the heat current in a metal? In pure metals the electronic contribution is dominant at all tem- peratures. In impure metals or in disordered alloys, the electron mean free path is rednced by collisions with impurities, and the phonon contribution may be comparable with the electronic contribution. Ratio of Thermal to Electrical Conductivity The Wiedemann-Franz law states that for metals at not too low tcmper- atures the ratio of the thermal conductivity to the electrical cond~lctlvlty is directly proportional to the temperature, ulth the value of the constant of proportionaky independent of the particular metal. This result was important in the history of the theory of metals, for it supported the picture of an electron gas as the carrier of charge and energy. It can be explained by using (43) for u and (56) for K: 6 Free Electron Fermi Gas 157 Table 5 Experimental Lorenz numbers L x lO%att-ohmldeg2 L X l0"vatt-ohm/de$ Metal 0°C 100°C Metal 0°C 1 0 0 ° C The Lorenz number L is defined as and according to (57) shollld have the value This re~r~arkable result involves neither n nor m. Experimental values of L at 0°C and at 100°C as given in Table 5 are in good agreement with (59). Problems 1. Kinetic energy ofelectron gas. Show that the kinetic energy of a three-dimensional g a s of N free electrons at 0 K is U , , = ~ N E , . (60) 2. Pressure and bulk modulus of an electron gas. (a) Derive a relation connecting the pressure and volume of an electron gas at 0 K. Hint: Use the result of Problem 1 and the relation hctween E, and electron concentration. The result may be writ- ten as p - $(~J,/v). (b) Show that the bulk modulus B = -V(apl;tV) of an electron gas at 0 K is B = $13 = 10Ud9V. (c) Estimate for potassium, using Table 1, the value of the electron gas contribution to B. 3. Chemical potential in two dimensions. Show that the chemical potential of a Fermi gas ill two dimensions is given by: for n electrolls per unit area. Note: The density of orbitals of a free electron gav in two dimensions is independent of enera: D(e) = m./7Tf12, per unit area of specimen. 4. Fermi gases in astrophysics. (a) Given = 2 x g for the mass of thc Son, estimate the numbcr of electrons in the Sun. In a white dwarf star this number nf electrons may be ionized and contained in a sphere of radius 2 % 10' cm; find the Fermi energy of the clectrnns in electron volts. (b) The energr. of an electron in the relativistic limit E S mc%s related to the wavevector as t = pc = hkc. Show that the Fermi energy in this limit is E P = J2C (N/v)"~, roughly. (c) If the abovc numher of electrons were container1 within a pulsar of radius 10 km, show that thc Fermi en- ergy would bc =loX eV. This value explains why pulsars are believed to be cnrnposed largely of neutrons rather than of protons and electrons, for the energy rclease in the reaction n + p + e- is only 0.8 x lo6 eV, which is not large enough to enable many electrons to lnrm a Ferrrii sea. The neutron decay proceeds only until the electron concentration h~rilds up enough to create a Fermi level of 0.8 % lO%\7, at wlricl~ point the neutron, proton, and electron concentrations are in equiliblilnn. 5. Liquid He". The atom He3 has spin and is a fermion. Thc dcnsity nf liquid He" is 0.081 g cm-'' near absolute zero. Calculate the Fermi encrgy E, and the Ferrni temperature TF. 6. Frequency dependence of the electrical conductivity. Use the equation m(du/dt + t ; / ~ ) = -eE for the electron drift velocity v to show that the conductivity at frcq~lenc~ w is '7. Dynamic magnetoconductivity tensor for free electrons. A metal with a concen- tration n of frec clcctrons of charge -e is in a static m a ~ e t i c field 84. The clcctric current density in the xy plane is related to the electric field by Assumc that the freqirency w 9 w, and w l / ~ , where w, = eB/mc and T is the collision time. (a) Solve the drift velocity- equation (51) to find thc compnne~rts of the magnetncnnductivity tensor: where W; = 4me2/m. (b) Note from a Maxwell cqnatinn that the dielectric func- tion tensor of the medium is related to the conductivity tensor as E = 1 + i ( 4 - d ~ ) ~ . Coi~sider an electromagnetic wave with wavcvcctnr k = kg. Show that the disper- sion relation for this wave in the medium is At a given frequency there are two modes nf propagation with different wavevec- tors and different velocities. The two modes correspond to circularly polarized '~llis problem is somewhat difficult. 6 h e Electron Fenni Gas wdves. Because a linearly polarized wave can he dccomposed into two circularly I~olarized waves, it follows that the plane of polarization of a linearly polarized wave will be rotated by the magnetic field. '8. Cohesive energy of free electron Fenni gas. We define the dimensionless length r,, as r,/a,, where r, is the radius of a spherc that contains one electron, and oH is thc bohr radius h2/e2m. (a) Show that the average kinetic energyper elec- trori in a frcc electron Fermi gas at 0 K is 2.21/<, where the energy is expressed in $bergs, with 1 Ry = me4Ah! (b) Show that the co~~lnmb energy of a point posi- tive charge e interacting with the uniform electron distribution of one electron in the volnme of radius r, is 3e2/2r0, or -3/rs in rydbergs. (c) Show that the coulornh self-encrg of the electron distribution in the sphcre is 3e2/5r,,, or 6/5r, in rydbergs. (d) The sum of (b) and (c) gives -1.80/rS fnr thc total coulomb energy per electron. Show that the equilibrium value of r, is 2.45. Will such a metal be stable with respect to separated I1 atoms? 9. Static magnetoconductivity tensor. For the drift velocity theory of (51), show that the static cnrreut dcnsity can be written in matrix fonir as In the high magnetic field limit of W,T 1, show that In this limit a , , = 0, to order l / w < ~ . llie quantity uy, is called the Hall conductivity. 10. Maximum surface resistance. Consider a square sheet of side L, thickness d, and electrical resistivity p. The resistance measured hetwccn opposite edges of the sheet is called the surface resistance: RSq = pL/lrl = pld, which is independent of the area L%f the sheet. (R,,, is called the resistance per square and is expressed in ohrrls per square, because pld has the dimensions of ohms.) If we express p by (44), then R,, = m/nde2r. Suppose now that the rrrininrilm value of the collision time is deter~rrined by scattering from the surfaces of the sheet, so that r = dlu,, where t i , is the Fernri velocity. Thus the maximum surface resistivity is RSq = rnok/nd2e2. Show for a rnonatomic metal sheet one atom in thickness that Rq = ii/e2 = 4.1 k a . 'This problem is somewhat ddficult Energy Bands NEARLY FREE ELECTRON MODEL Origin of the energy gap Magnitude of the energy gap BLOCH FUNCTIONS KRONIG-PENNEY MODEL W A V E EQUATION OF ELECTRON IN A PERIODIC POTENTIAL Restatement of the Bloch theorem Crystal momentum of an electron Solution of the central equation Kronig-Penney model in reciprocal space Empty lattice approximation Approximate solution near a zone boundary NUMBER OF ORBITALS IN A BAND Metals and insulators SUMMARY PROBLEMS 1. Square lattice, free electron energies 2. Free electron energies in reduced zone 3. Kronig-Penney model 4. Potential energy in the diamond structure 5. Complex wavevectors in the energy gap 6. Square lattice Energy Insulator Metal Semimetal Sem~conductor Sem~conductor Figure 1 Schematic electron occupancy of allowed energy bands for an insulator, metal, semi- metal, and semiconductor. The vertical extent of the boxes indicates the allowed energy regions; the shaded areas indicate the regions filled with electrons. In a semimetal (such as bismuth) one band is almost filled and another band is nearly empty at absolute zero, but a pure semiconduc- tor (such as silicon) becomes an insulator at absolute zero. The left of the two semiconductors shown is at a finite temperature, with carriers excited thermally. The other semiconductor is electron-deficient because of impurities. CHAPTER 7: ENERGY BANDS When I started to think about it, I felt that the main problem was to explain how the electrons could meak b y all the ions in a metal.. . . By straight Fourier analysis I found to my delight that the wave differed from the plane wave of free electrons only by a periodic modulation. F . Bloch The free electron model of metals gives us good insight into the heat capacity, thermal conductivity, electrical conductivity, magnetic susceptibility, and e l e c t r o ~ a m i c s of mctals. But the model fails to hclp 1 1 s with other large questions: the distinction between metals, semimetals, semiconductors, and insulators; the occnrrence of positive values of the Hall coefficient; the rela- tion of conduction electrons in the metal to the valence electrons of free atoms; and many transport properties, partic~llarly magnetotransport. We need a less nayve theory, and fortullately it tnms out that alniost any simple attempt to improve upon the free electron model is enormously profitable. The difference between a good conductor and a good insulator is striking. The electrical resistivity of a pure metal may be as low as 10-lo ohm-cm at a temperature of 1 K, apart from the possibility of superconductivity. The resis- tivity of a good insnlator may be as high as 10" ohm-cm. This range of lo3' may be the widest of any comrrloll physical property of solids. Evcry solid contains electrons. Thr important question for electrical con- ductibity is how the electrons respond to an applied electric field. We shall see that electrons in crystals are arranged in energy bands (Fig. 1) separated by regions in energy for which no wavelike electron orbitals exist. Such forbidden regions are called energy gaps or band gaps, and resnlt from the interaction of the co~lduction electron waves with the ion cores of the crystal. The crystal behaves as an insulator if the allowed energy bands are either filled or cmpty, for then no electrons can move in an electric field. The crystal behaves as a metal if one or more bands are partly filled, say between 10 and 90 percent filled. The crystal is a semiconductor or a semimetal if one or two bands are slightly filled or slightly empty. To understand thc difference between insulators and conductors, we rnust extend the free clectron model to take account of the periodic lattice of the solid. The possibility of a band gap is the most important new property that emerges. \'c shall encounter other quite remarkable properties of electrons in crys- tals. For example, they respond to applied electric or magnetic fields as if the electrons were endowed with an effective mass m, which may be larger or smaller than the frec clectron mass, or may even bc negative. Electrons in crystals respond to applied fields as if endowed with negative or positive charges, -e or +e, and herein lies the explanation of the negative and positive values of the Hall coefficient. NEARLY FREE ELECTRON MODEL On the free electron model the allowed energy values are distributed es- sentially continuously from zero to infinity. We saw in Chapter 6 that where, for periodic boundary conditions over a cube of side L, The free electron wavefunctions are of the form they represent running waves and carry momentum p = fik. The band structure of a crystal can often be explained by the nearly free electron model for which the band electrons are treated as perturbed only weakly by the periodic potential of the ion cores. This model answers almost all the qualitative questions about the behavior of electrons in metals. We know that Bragg reflection is a characteristic feature of wave propaga- tion in crystals. Bragg reflection of electron waves in crystals is the cause of energy gaps. (At Bragg reflection wavelike solutions of the Schrodinger equa- tion do not exist, as in Fig. 2.) These energy gaps are of decisive significance in determining whether a solid is an insulator or a conductor. We explain physically the origin of energy gaps in the simple problem of a linear solid of lattice constant a. The low energy portions of the band structure - Forbidden band 1 % k "7 k Figure 2 (a) Plot of energy E versus wavevector k for a free electron. (b) Plot of energy versus wavevector for an electron in a monatomic linear lattice of lattice constant a. The energy gap E, shown is associated with the first Bragg reflection at k = ? d a ; other gaps are found at higher energies at k n d a , for integral values of n. 7 Energy Ban& are shown qualitatively in Fig. 2, in (a) for entirely free electrons and in (h) for electrons that are nearly free, but with an energy gap at k = f n/a. The Bragg condition (k + G)' = kZ for diffraction of a wave of wavevector k becomes in one dirrrension where C = 2xn/u is a reciprocal lattice vector and n is an integer. The first re- flections and the first energy gap occur at k = + d u . The region in k space be- tween -ria and n/a is the first Brillouin zone of this lattice. Other energy gaps occur for other values of the integer n. The wavefimctions at k = ?n/a are not thc traveling waves exp(i~x1a) or exp(-i~x/a) of free electrons. At these special values of k the wavefunctions are made up of equal parts of waves traveling to the right and to the left. UThen the Bragg reflection condition k = +n/a is satisfied by the wavevector, a wave traveling to the right is Bragg-reflected to travel to the left, and vice versa. Each subsequent Bragg reflection will reverse the direction of travel of the wave. A wave that travels neither to the right nor to the left is a standing wave: it doesn't go anjwhere. Thc time-independent state is represented by standing waves. We can form two different standing waves from the two traveling waves so that the standing waves are +(+) = exp(inx/a) + exp(-inx/a) = 2 cos (nx/a) ; +(- ) = exp(i~x/u) - exp(-inx/a) = 2i sin (nxla) . (5) The standing waves are labeled (+) or (-) according to whether or not they change sign when -x is substituted for x. Both standing waves are composed of equal parts of right- and left-directed traveling waves. Origin o f the Energy Gap The two standing waves $(+) and +(-) pile up electrons at different regions, and therefore the two waves have different values of the potential energy in the field of the ions of the lattice. This is the origin of the energy gap. The probability density p of a particle is $+ = I+12. For a pure traveling wave exp(ikx), we have p = exp(-ikx) exp(ikx) = 1, so that the charge density is constant. The charge density is not constant for linear combinations of plane waves. Consider the standing wave $(+) in (5); for this we have This function piles up electrons (negative charge) on the positive ions centered at x = 0, a, ea, . . . in Fig. 3, where the potential energy is lowest. U, potential energy I p, probability density eling wave x Figure 3 (a) Variation of potential energy of a conduction electron in the field of the ion cores of a linear lattice. (b) Distribution of probability density p in the lattice for I$(-)I2 a sinZ m l a ; I$(+)I2 = cosZ m l a ; and for a traveling wave. The wavefunction $(+) piles up electronic charge on the cores of the positive ions, thereby lowering the potential energy in comparison with the average potential energy seen by a traveling wave. The wavefnnction $(-) piles up charge in the region between the ions, thereby raising the potential energy in comparison with that seen by a traveling wave. This figure is the key to understanding the origin of the energy gap. Figure 3a pictures the variation of the electrostatic potential energy of a conduction electron in the field of the positive ion cores. The ion cores bear a net positive charge because the atoms are ionized in the metal, with the va- lence electrons taken off to form the conduction band. The potential energy of an electron in the field of a positive ion is negative, so that the force between them is attractive. For the other standing wave $(- ) the probability density is which concentrates electrons away from the ion cores. In Fig. 3b we show the electron concentration for the standing waves $(+), $(-), and for a travel- ing wave. When we calculate the average or expectation values of the potential energy over these three charge distributions, we find that the potential energy of p(+) is lower than that of the traveling wave, whereas the potential energy of p ( - ) is higher than the traveling wave. We have an energy gap of width E, if 7 Energy Bands 167 the energies of p ( - ) and p ( + ) differ by Eg. Just below the energy gap at points A in Fig. 2 the w-avefunction is +(+), and just above the gap at points B the wavefunction is $(-). Magnitude o f the Energy Gap The wavefunctions at the Brillouin zone boundar). k = ?r/a are fi cos m / a and ~ sin mla, normalized over unit length or line. Let 11s suppose that the potential energ). of an electron in the crystal at point x is U(x) = U cos 2 d a . The first-order enerD difference between the two standing wave states is We see that the gap is equal to the Fourier component of the crystal potential. BLOCH FUNCTIONS F. Bloch proved the important theorem that the solutioris of the Schrochnger equation for a periodic potential must be of a special form: where uk(r) ha5 the period of the crystal lattice with uk(r) = llk(' + T). Here T is a translation vector of the lattice. The result (7) expresses the Bloch theorerri: The eigenfunctions of the wave equation for a periodic potential are the prodr~ct of a plane wave exp(ik . r ) times a function uk(r) with the periodicity of the crystal lattice. A one-electron wavefunction of the for~n (7) is called a Rloch function and can be decomposed into a sum of traveling waves, as we see later. Bloch func- tions can he assembled into wave packets to represent electrons that propa- gate freely through the potential field of the ion cores. We give now a restricted proof of the Bloch theorem, valid \vlren +k is nondegenerate; that is, when there is no other wavefunction with the samc energy and wavevector as &. The general case will be treated later. \?7c con- sider N identical lattice points on a ring of le~lgth Nu. The potcntial energy is periodic in a, with U(x) = U(x + sa), where s is an integer. Let us be guided by tlie symmetry of the ring to look for solutions of the wave equation such that where C is a constant. Then, on going once around the ring, +(x + Nu) = +(x) = C " +(x) , because $(x) must be single-valued. It follows that C is one of the N roots ol unity, or C = exp(i2nslN) ; s = 0, 1,2, . . . , N - I . (9) We use (9) to see that satisfies (8), provided that uL(x) has the periodicity a, so that uk(x) = uk(x + a). This is the Bloch result (7). KRONIG-PENNEY MODEL A periodic for which the wave equation can he solved in terms of elementary functions is the square-well array of Fig. 4. The wave equation is where U(x) is the potential energy and is the energy eigenvalue. In the region 0 < x < a in which U = 0, the eigenfunction is a linear combination, of plane waves traveling to the right and to the left, with energy E = h2P/2rn . (13) In the region -b < x < 0 within the barrier the solution is of the form fi = CeQ" + ~ e - ~ " , (14) with U, - G = h2Q2/2m . (1s) Figure 4 Square-well periodic potential as introduced by Kronig and Penney. i a + b ) 41 0 a a c b x - 7 Energy Bands 169 \'t: want the complcte solution to have the Bloch form (7). Thus the solu- tion in the region n < x < a + b must be related to the solution (14) in the region -b < r < 0 by the Bloch theorem: wl~ich serves to dcfine the wavevector k used as an index to label the solution. The constants A, B, C, D are chosen so that $ and d+idx are continuous at x = O and x = a. These are the usual quantum mechanical boundary condi- tions in problems that involve squarc potential wells. At x = 0, with Q lrom (14). At x = a, with the use of (16) for $(a) under the barrier in terms of $(-h), ~ ~ i f i + = (ce-Q" + neQb) eik(a+b) ; (19) i~(&'fi - B~-'&) = Q ( c ~ - Q ~ - neQb) e'kt"+b) , (20) The lour cqnations (17) to (20) have a solution only if the determinant of the coefficients ofA, B, C, D vanishes, yielding [(Q" K~)/ZQK] sinl~ Qb sin Ka + cosh Qb cos Ka = cos k(a + b) . (214 It is rather tedious to obtain this equation. The result is simplified if we represent the potential by the periodic delta function obtained when we pass to the limit b = 0 and Uo = m in such a way that Qzba12 = Y, a finite quantity. In this limit Q & K and Qb 4 1. Then (21a) reduces to (P/Ka)sin Ka + cos Ka = cos ka . (2lb) The ranges of K for which this equation has solutio~ls are plotted in Fig. 5, for the case P = 3 d 2 . The corresponhng values of the energy are plotted in Fig. 6. Note thc cnergy gaps at the zone boundaries. The wavevector k of the Bloch function is the important index, uot the K in (12), which is related to the cnergy by (13). A treatment of this problem in wavevector space is given later in this chapter. WAVE EQUATlON OF ELECTRON IN A PERIODIC POTENTIAL We considered in Fig. 3 the approximate fonu we expect for the sohltion of the Schrodinger equation if the wavevector is at a zone houndq, as at k = 2min. We treat in detail the wave equation for a general potential, at general E t (PI&) sin Ka + cos Ka Figure 5 Plot of the function (PIKa) sin Ka + cos Ka, for P = 3 ~ 1 2 , The allowed values of the energy e are given by those ranges of Ka = ( 2 r n ~ f f i ' ) ~ a for which the function lies between 51. For'other values of the energy there are no traveling wave or Bloch-like solutions to the wave equation, so that forbidden gaps in the energy spectrum are formed. Figure 6 Plot of energyvs. wavenumber for the Kronig-Penney potential, with P = 3 ~ 1 2 . Notice the energy gaps at ka = W , ZW, 377. . . . values of k. Let U(x) denote the potential energy of an electron in a linear lattice of lattice constant a. We know that the potential energy is invariant under a crys- tal lattice translation: U(x) = U(x + a). A function invariant under a crystal lattice translation may be expanded as a Fourier series in the reciprocal lattice vectors G. We write the Fourier series for the potential energy as The values of the coefficients UG for actual crystal potentials tend to decrease rapidly with increasing magnitude of G. For a bare coulomb potential U , decreases as 1/G2. We want the poteutial energy U(x) to be a real function: U(x) = 2 UG(eiG" + e-'") = 2 IIc cos Gx G > O G>O 7 Energy Bands For convenience we have assumed that the crystal is symmetric about x = 0 and that UO = 0. The wave equation of an electron in the crystal is X = e+, where X is the hamiltonian and is the energy eigenvahe. The solutions + are called eigen- functions or orbitals or Bloch functions. Explicitly, the wave equation is Equation (24) is written in the one-electron approximation in which the orbital $(x) describes the motion of one electron in the potential of the ion cores and in the averagc potential of the other conduction electrons. The wavefunction $(x) may be expressed as a Fouricr series summed over all values of the wavevector permitted by the boundary conditions, so that where k is real. (We could equally well write the index k as a subscript on C, as in Ck. ) The set of values of k has the form 2 ~ n l L , because these vallles satisfy periodic boundary conditions over length L. Here n is any integer, positive or negative. We do not assume, nor is it generally tnic, that ( x ) itself is periodic in the fundamental lattice translation a. Thc translational properties of ( x ) are determined by the Bloch theorem (7). Not all wavevectors of the sct 2m/L enter the Fourier expansion of any one Bloch function. If one particular wavevector k is contained in a @, then all other wavevectors in the Fourier expansion of this @ will have the form k + G, where G is any reciprocal lattice vector. V7e prove this result in (29) below. Wc can label a wavefunction t , b that contains a component k as 4k or, eqnally well, as & + , , because if k enters the Fourier expansion then k + G may enter. The wavevectors k + C rnnning over G are a restricted subsct of the set ZmlL, as shown in Fig. 7. We shall usually choose as a label for the Bloch function that k which lies within the first Brillouin zone. When other conventions are used, we shall say so. This situation differs from the phonon problcrn for a monatomic lattice where there are no comporle~lts of the ion motion outside the first zone. The electron problen~ is like the x-ray diffraction problem because like the electron wavefunctior~ the electromagnetic field exists everywhere within the crystal and not o111y at the ions. Figure 7 The lower points represent values of the wavevector k = 2m/L allowed by the periodic houndary condition on thc wavefunction over a ring of circumfcrcnce L composed of 20 primitive cells. The allowed valiies continue to i m. The uppcr points represent the first few wavevcctors which may enter into the Fourier expansiorl of a wavefunction (XI, starting from a paltic~~lar wavevector k = k, = -8(2~r/L). Thc: shortest reciprocal lattice vector is 2wIa = 20(2~7/L). To solve the wave equation, substitute (25) in (24) to obtain a set of linear algebraic cquations for the Fourier coefficients. The kinetic energy term is and the potential energy term is The wave eqi~ation is obtained as the sum: Each Fourier component must have the same coefficient on both sides of the equation. Thus we have the central equation with the notation Equation (27) is a useful form of the wave equation in a periodic lattice, although unfamiliar because a set of algebraic equations has taken the place of the usual differential equation (24). The set appears unpleasant and formida- ble because there are, in principle, an infinite number of C(k - G) to be de- termined. In practice a small number will often suffice, pcrhaps two or four. It takes some experience to appreciate the practical advantages of the algebraic approach. 7 Energy Bands Rentatement of the Bloch Theorem Once we determine the C's from (27), the wavcfiinction (25) is given as +bk(x) = 2 C(k - G) ei(k '" , C which may he rearranged as with the definition uk(x) = 2 C(k - G) e-ic7 G Because uk(x) is a Fourier series over the reciprocal latticc vectors, it is in- variant under a crystal lattice translati011 T, so that uk(x) = uk(x + T). We verify this directly by evaluating uk(x + T): Because exp(-iGT) = 1 by (2.17), it follows that uL(x + T) = uk(x), thereby establishing the periodicity of uk. This is an alternate and exact proof of the Bloch theorem and is valid even when the $rk are degenerate. Crystal Momentum of an Electron What is the significance of the wavevector k used to label the Bloch func- tion? It has several properties: Under a crystal lattice translation which carries r to r + T we have because uk(r + T) = uk(r). Thus exp(ik T) is the phase factor by which a Bloch function is multiplied when we make a crystal lattice translation T. If the lattice potential vanishes, the central equation (27) reduces to (Ak - c)C(k) = 0, so that all C(k - G) are zero exccpt C(k), and thus uk(r) is constant. We have I J ~ ( ~ ) = eik.r, just as for a free electron. (This assumes we have had the foresight to pick the "right" k as the label. For many pur- poses other choices of k, differing by a reciprocal lattice vector, will be more convenient.) The quantity k enters in the conservation laws that govern collision processes in crystals. (The conservation laws are really selection rules for transitions.) Thus fik is called the crystal momentum of an electron. If an electron k absorbs in a collision a phonon of wavevector q, the selection rule is k + q = k' + G. In this process the electron is scattered from a state k to a state kt, with G a reciprocal lattice vector. Any arbitrariness in labeling the Bloch func- tions can be absorbed in the G without changing the physics of the process. Solution of the Central Equation The central equation (27), represents a set of simultaneous linear equations that connect the coefficients C(k - G) for all reciprocal lattice vectors G. It is a set because there are as many cquations as there are coefficients C. These equations are consistent if the determinant of the coefficients vanishes. Let us write out the eqnations for an explicit problem. \Ve let g denote the shortest G. We suppose that the potential energy U(x) contains only a single Fourier component Up = Kg, denoted by U. Then a hlock of the determinant of the coefficients is given by: To see this, write out five successive equations of the set (31). The determi- nant in principle is infinite in extent, but it will often be sufficient to set equal to zero the portion we have shown. At a given k, each root E or ek lies on a different energy band, except in case of coincidence. The solntion of the determinant (32) gives a set of energy eigenvalues enk, where n is an index for ordering the energies and k is the wavevector that labels Ck. Most often k will be taken in the first zone, to reduce possible confusion in the labeling. If we chose a k different from the original by some reciprocal lattice vector, we would have obtained the same set of equations in a different order-but having the same energy spectrum. Kronig-Penney Mock1 in Reciprocal Space As an example of the use of the central equation (31) for a problem that is exactly solvable, we use the Kronig-Penney model of a periodic delta-filnction potential: U(z) = 2 UG cos G x = AGE 6(x - sa) , G>O where A is a constant and a the lattice spacing. The sum is over all integers s between 0 and lla. The boundary conditions are periodic over a ring of unit 7 Energy Bands 175 length, which lrieans over l/o atoms. Thus the Fourier coefficients of the potential are U, = 1 dx WGI) cos Gx = dr S(x - so) cos GGI (34) All U , are equal for the delta-function potential. We write the central equation with k as the Bloch index. Thus (31) becomes where Ar = fi2k'/2m and the sum is over all integers n. We want to solve (35) for ~ ( k ) . \Ve define f(k) = Z c ( k - 2m/a) , (36) so that (35) becomes Because the suru (36) is over all coefficients C, we have, for any n; f (k) = f (k - 2 m / a ) . (38) This relation lets us write C(k - 2m/cr) = - (2mA/ri2) f(k)[(k - 2 n ~ r / a ) ~ - 2fn</ri2)]-' . (39) We sum but11 sides over all n to obtain, using (36) and cancelling f(k) from both sides, (fi2/2m~) = -Z [(k - 2rmla)" ( 2 d h 2 ) ] - ' . (40) The sum can be carried out with the help of the standard relation After trigonometric manipulations in which we use relations for the difference of two cotangents and the of two sines, the sum in (40) bccomes a2 sin Ka 4Ka(cos ka - cos Ka) ' where we write: l ? = 2me/7i2 as in (13). The final result for (40) is (m~n~/2fi')(Ka)- ' sin K u + cos Ka = cos ka , (43) which agrees with the Kronig-Penney result (21b) with P written for m ~ a ~ / 2 i i ' . Empty Lattice Approximation Actual band structures are usually exhibited as plots of energy versus wavevector in the first Brillo~~in zonc. When wavevectors happen to be given outside the first zone, they are carried hack into the first zone by subtracting a suitable reciprocal lattice vector. Such a translation can always be found. The operation is helpful in visualization. When band energies are approximated fairly well by free electron ener- gies et = fi212/2m, it is advisable to start a calculation by carrying the free elec- tron energies hack into thc first zone. The procedure is simple enough once one gets the hang of it. We look for a G such that a k' in the first zone satisfies where k is unrestricted and is the true free electron wavevector in the empty lattice. (Once the plane wave is niodulated by the lattice, there is no single "true" wavevector for the state I//.) If we drop the prime on k' as unnecessary baggage, the free electron energy can always be written as with k in the first zone and G allowed to run over the appropriate reciprocal lattice points. We consider as an example the low-lying free electron bands of a simple cubic lattice. Sl~ppose we want to exhibit the energy as a function of k in the [loo] direction. For convenience, choosc units such that 6212m = 1. We sllow several low-lying bands in this empty lattice approximation with their energies ~ ( 0 0 0 ) at k = 0 and e(k,OO) along the k, axis in the first zonc: - -- Band 1 000 0 k," 2,3 100,100 (k, 2 2v/a)" 4,5,6,7 010,0i0,001,00i ( Z ~ / U ) ~ + ( 2 ~ 1 ~ ) ~ 8,9,10,11 - ~lo,lol,lTo,loi 2(2rr/a)' (k, + ~ T / < I ) ~ + ( ~ T / < L ) ~ 12,13,14,15 110,~0l,fi0,i0i 2(27r/n)' (k, - ~ T / ( L ) ~ + (2da)' 16,17,18,19 0 1 1 , ~ ~ 1 , 0 1 ~ , 0 i i 2(2rrl[~)' kz + 2(2~/a)' 7 Energy Bands Figure 8 Low-lying free electron energy bands of the empty sc lattice, as tra~isfornied to the first Brillouin zone and plotted vs. (k,OO). The free electron energy is fi2(k + G)'/Zrn, where the G's are given in the second coli~mn of the table. Thc bold curvcs are in the first Urillouin zone, with w i n 5 k, 5 mlu. Energy bands drawn in this way are said to be in the rednced wne sche~ne. These free electron bands are plotted in Fig. 8. It is a good exercise to plot the same bands for k parallel to the [Ill] direction of wwevector space. Approximate Solution Near a Zone Boundary \Vc suppose that the Fourier components LrG of the potential energy are small in comparison with the kinetic energy of a free electron at the zone boundary. We first consider a wavevector exactly at the zone boundary at :G, that is, at T/U. Here so that at the zone boundary the kinetic energy of the two component waves k = ?$G are equal. If c($G) is an important coefficient in the orbital (29) at the zone boundary, then c(-;G) is also an important coefficient. This result also follows from the discussion of (5). We retain only those equations in thc central equation that contain both coefficients c ( ~ G ) and C(-:G), and neglect all other coefficients. One equation of (31) becomes, with k = : G and h = fi2(&2)'/2rn, Another equation of (31) becomes. with k = ;G: These two equations have nontrivial solutions for the two coefficients if the energy E satisfies whence The energy has two roots, one lower than the free electron kinetic energy by U , and one higher by U. Thus the potential energy 2U cos Gx has created an energy gap 2U at the zone boundary. The ratio of the C's may be found from either (44) or (45): where the last step uses (47). Thus the Fourier expansion of $(x) at the zone boundary has the two solutions These orbitals are identical to (5). One solution gives the wavefunction at the bottom of the encrgy gap; the other gives the wavefunction at the top of the gap. Which solution has the lower energy depends on the sign of U. We now solve for orbitals with wavevcctor k near the zone boundary iG. We nse the same two-component approximation, now with a wavefunction of the form $(x) = C(k) e"" + C(k - G ) ei(k-Gb . (49) As directed by the central eqnation (31), we solve the pair of equations 7 Energy Bands with hk defined as k 2 k 2 m . These equations have a solution if the energy r satisfies u - E = o U hi-c I Pk - whencc 6 ' - E ( A ~ - ~ + hLj + hk-Ghk-@ = 0 The energy has two roots: and each root describes an energy band, plotted in Fig. 9. It is convenient to expand the energy in terms of a quantity K (thc mark over the K is called a tilde), which rrieasures the difference k - k - ;G in wavevector between k and the zone boundary: in the region k2G1(1/2rn < 1U(. Here h = (fi2/2m)(k G)' as beforc. Writing the two zone boundary roots of (47) as E ( ? ) , we nlay write (51) as Figure 9 Solutions of (50) in the periodic zone scheme, in the region near a boundaly of the first Brillouin zone. Thc units are such that U = -0.45, G = 2, and fLZ/m = 1. The ficc electron curve is drawn for comparison. The coergy gap at the zone houndaryis 0.90. The value of U has deliberately heen chosen large for this illi~stration, too large for the hvo-terrn approximation to be accurate. / ~ i r s t I hand I Figure 10 Ratio of the coefficients i11 (x) = C(k) exp(ikx) + C(k - 6) exp[i(k - G)x] as calcu- lated rrzar the boundary of the first Rrillouin zone. One component dominates as we move away from the boundary These are the roots for the energy when the wavevector is very close to the zone boundary at f ~ . Note the quadratic dependence of the energy on the wavevector K. For I ! negative, the solution E ( - ) correspondq to the nppcr of the two bands, and e(+) to the lower of the two bands. The two C's are plotted in Fig. 10. NUMBER OF ORBITALS IN A BAND Consider a linear crystal constructed of an even number N of primitive cells of lattice constant a. In order to count states we apply periodic boundary conditions to the wavefunctions over the length of the crystal. The allowcd values of the electron wavevector k in the first Brillouin zone are given by (2): We cut the series off at Nn/L = n/a, for this is the zone boundary. The point -Nn/L = -nla is not to be counted as an independent point because it is connected by a reciprocal lattice vector with nla. The total number of points is exactly N, the number of primitive cells. Each primitive cell contributes exactly one independent value of k to each energy band. This result carries over into three dimensions. With account taken of the two independent orientations of the electron spin, there 7 Energy Bands are 2N independent orbitals in each energy band. If there is a single atom of valence, one in each primitive cell, the band can be half filled with electrons. If each atom contributes two valence electrons to the band, the band can be exactly filled. If there are two atoms of valence, one in each prim- itive cell, the band can also be exactly filled. Metals and Insulators If the valence electrons exactly fill one or more bands, leaving others empty, the crystal will be an insulator. An external electric field will not cause current flow in an insulator. (We suppose that the electric field is not strong enough to disrupt the electronic structure.) Provided that a filled band is sepa- rated by an energy gap from the next higher band, there is no continuous way to change the total momentum of the electrons if every accessible state is filled. Nothing changes when the field is applied. This is quite unlike the situa- tion for free electrons for which k increases uniformly in a field (Chapter 6). A crystal can be an insulator only if the number of valence electrons in a primitive cell of the crystal is an even integer. (An exception must be made for electrons in tightly bound inner shells which cannot be treated by band theory.) If a crystal has an even number of valence electrons per primitive cell, it is necessary to consider whether or not the bands overlap in energy. If the bands overlap in energy, then instead of one filled band giving an insulator, we can have two partly filled bands giving a metal (Fig. 11). The alkali metals and the noble metals have one valence electron per primitive cell, so that they have to be metals. The alkaline earth metals have two valence electrons per primitive cell; they could be insulators, but the bands overlap in energy to give metals, but not very good metals. Diamond, silicon, and germanium each have two atoms of valence four, so that there are Figure 11 Occupied states and band structures giving (a) an insulator, (b) a metal or a semimetal because of band overlap, and (c) a metal because of electron concentration. In (b) the overlap need not occur along the same directions in the Brillouin zone. If the overlap is small, with rela- tively few states involved, we speak of a semimetal. eight valence electrons per prir~~itive cell; the bands do not overlap, and thc pure crystals are insulators at absolute zero. SUMMARY The solutions of the wave equation in a periodic lattice are of the Bloch form i,bk(r) = uk(r), where uk(r) is invariant under a cystal lattice translation. There are repons of energy for which no Bloch furictiorl solutions of the wave equation exist (see Proble~ri 5). These energies form forbidden regions in which the wavefu~ictio~is are damped in space and the values of thc k's are complex, as pictured in Fig. 12. The existence of forbiddcn rcgions of energy is prerequisite to the existence of insulators. Energy bands may often be approximated by one or two plane waves: for example, I/J~(?~) = c(k)eikx + C(k - G)c'(~-'" near the zone bounday at :G. The number of orbitals in a band is 2N, where N is the number of primitive cells in the specimen. Problems 1. Square lattice, free electron energies. (a) Show for a simple square lattice (two dimensions) that the kinetic energy of a free electron at a corner of the first zone is higher than that of an clcctron at midpoint of a side face of the zone by a factor of 2. (b) What is the cnrresponding factor for a simple cubic lattice (three dimensions)? (c) What hearing might the result of (b) have on the conductivity of divalent metals? 2. Free electron energies in reduced zone. Consider the free electron energy bands of an fcc crystal lattice in the a~>proxiniatinn of an clnpty lattice, but in the reduced zone scheme in which all k ' s are transformed to lic in the first Brillouin zone. Plot roughly in the [ill] direction the energies of all bands up to six times the lowest band energy at the zone boundary at k = i~.rr/a)(;, fr, i). Let this be the unit of en- ergy This problem shows why band edges need not necessarily be at the zone cen- ter. Several of the degeneracies (hand crossings) will he removcd when account is taken of the crystal potential. 3. Kronig-Penney model. (a) For the delta-filnctinn potential and with P + 1, find at k = 0 the energy of the lowest energy band. (h) Fnr the same problem find the band gap at k = d a . 4. Potential energy in the diamond structure. (a) Show that for the diamond struc- ture the Fourier component U , of the cystal pote~ltial seen an electron is cqual to zero for G = 2A, where A is a basis vector in the reciprocal lattice referred to the conventional cubic cell. (b) Show that in the usual first-order approximation to the solutions of the wave equation in a periodic lattice the energy gap vanishes at the zone boundary plane normal to the end of the vector A. 7 Energy Bands 183 Real part of klG Figure 12 In the energy gap there exist solutions of the wave equation for complex values of the wavevector. At the buu~~dary uf the first zone thc real part of the wavevector is +c. The imaginaly part of k in the gap is plotted in the approximation of two plane waves, for U = 0.01 hPG2/2nr. In an infinite unbounded crystal the wavevector must be real, or else the amplitude will increase with- ol~t limit. But on a surface or at a ~ U I I L ~ ~ U I I there can exist solutions with complex wavevector. '5. Complex wavevectors in the energy gap. Find an expression for the imaginary p a t of the wavevector in the energy gap at the boundaly of the first Brillouin zone, in the approximation that led to Eq. (46). Give the result for the Im(k) at the center of the energy gap. The r e s ~ ~ l t for small Im(k) is The fur111 as plotted irr Fig. 12 is of impnrtance in the theory of Zener tunneling from one band to another in the presence of a strong electric ficld. 6. Square lattice. Consider a square lattice in two dimensiom with the crystal potential Apply the central equation to find approximately the energy gap at the comer point (via, ria) of the Brillouin zone. It will suffice to solve a 2 X 2 dctcrminantal equation. h his problem is somewhat difficnlt. Semiconductor Crystals BAND GAP EQUATIONS OF MOTION Physical derivation of hk = F Holes Effective mass Physical interpretation of the effective maw Effective masses in semiconductors Silicon and germanium INTRINSIC CARRIER CONCENTRATION Intrinsic mohility IMPURITY CONDUCTIVITY Donor states Acceptor states Thermal ionization of donors and acceptors THERMOELECTRIC EFFECTS SEMIMETALS SUPERLATTICES Bloch oscillator Zener tunneling SUMMARY PROBLEMS 1. Impurity orbits 2. Ionization of donors NOTE: The discussion of carrier orbits in applied fields is continued in Chapter 9. 3. Hall effect with two carrier types 218 4. Cyclotron resonance for a spheroidal energy surface 219 5. Magnetoresistance with two carrier types 219 Figure 1 Carrier concentrations for metals, semimetals, and semiconductors. The semiconductor range may be extended upward by increasing the impurity concentration, and the range can be ex- tended downward to merge eventually with the insulator range. CHAPTER 8: SEMICONDUCTOR CRYSTALS Carrier conccntrations representative of metals, semimetals, and semicon- ductors arc shown in Fig. 1. Semiconductors are generally classified by their clcctrical resistivity at room temperature, with values in the range of lo-' to 10"hm-cm, and strongly dependent " 1 1 temperature. At absolutc zcro a pnre, perfect crystal of most serr~icor~ductors will be an insulator, if we arbitrarily de- fine an insulator as having a resistivity above 1014 ohm-cm. Uevices based on semiconductors include transistors, switches, diodes, photovoltaic cclls, detectors, and thermistors. These may be used as single circuit elemcnts or as components of integrated circuits. \Ye discuss in this chapter the central physical features of the classical semiconductor crystals, particularly silicon, germanium, and galliuni arsenide. Some useful no~r~enclature: the se~niconductor componnds of chemical formula AB, where A is a trivalent elemcnt and B is a pentavalent element, are called II1-V (three-five) compounds. Examples are indium antimonide and galliuni arsenide. il'herc A is divalent and B is hexavalent, the compound is called a 11-VJ compo~lnd; examples are zinc sulfide and cadmium sulfide. Silicon and germanium are sometimes called chamond-h-pe semiconductors, because they have the crystal structure of diamond. Diarriond itself is more an insulator rather than a semiconductor. Silicon carbide SiC is a IV-IV compound. A highly purified se~r~ico~iductor exhibits intrinsic conductivity, as distin- guisl~ed fro111 the impurity conductivity of lcss pnre specimens. In the intrln- sic temperature range thc clcctrical properties of a semiconductor are not essentially modificd by impnrities in the crystal. An electronic band scheme leading to intrinsic conductivity is indicated in Fig. 2. The coriduction band is vacant at absolute zero and is separated by a m energy gap Eg from the filled valence band. The band gap is the difference in energy betwccn the lowest point of the conduction band and the highest point of the valence band. The lowest point in the conduction band is called the conduction band edge; the highest point in the valencc band is called the valence band edge. As thc temperatlire is increased, electrons are thermally excited from the valence band to the conduction band (Fig. 3). Both the electrons in the con- duction band and the vacant orbitals or holes left behind in the valcnce band contribute to the electrical conductivity. BAND GAP Thc intrinsic condiictivity and intrinsic carrier concentrations are largely controlled by Edk,T, the ratio of the band gap to the temperature. When this ratio is large, the concentration of intrinsic carriers will be low and the - . , . . . . . . . , . . , I V.wnnt r o n d u ~ o ~ ~ hand P 3 I Forbidden band Figure 2 Band scheme for intrinsic conductivity in a semiconductor. At 0 K the conductivity is zero because all states in the valence band are filled and all states in the conduction band are va- cant. As the temperature is increased, electrons are thermally excited from the valence band to the conduction band, where they become mobile. Such carriers are called 'intrinsic." Temperature, K Temperature, K (a) (b) Figure 3 Intrinsic electron concentration as a function of temperature for (a) germanium and (b) silicon. Under intrinsic conditions the hole concentration is equal to the electron concentra- tion. The intrinsic concentration at a given temperature is higher in Ge than in Si because the energy gap is narrower in Ge (0.66 eV) than in Si (1 1 1 eV). (After W. C. Dnnlap.) conductivity will be low. Band gaps of representative semiconductors are given in Table 1. The best values of the band gap are obtained by optical absorption. In a direct absorption process the threshold of continuous optical ab- sorption at frequency wg measures the band gap Eg = hwg as shown in Figs. 4a and 5a. A photon is absorbed by the crystal with the creation of an electron and a hole. In the indirect absorption process in Figs. 4b and 5b the minimum energy gap of the band structure involves electrons and holes separated by a 8 Semiconductor Crystals 189 CRYSTAL WITH DIRECT GAP CRYSTAL WITH INDIRECT GAP Onset of indirect fiwg E~ + fin E,.~, Photon energy Q o -+ Photon energy fio -+ (a) (h) Figure 4 Optical absorption in pure insulators at absolute zero. In (a) the threshold determines the energy gap as E,:= nop. In (h) the optical absorption is weaker near the threshold: at Qo = E, + fin a photon is absorbed with the creation of three particles: a free electron, a free hole, and a phonon of energy ha. In (b) the energy E , , , marks the threshold for the creation of a free electron and a free hole, with no phonon involved. Such a transition is called vertical; it is similar to the direct transition in (a). These plots do not show absorption lines that sometimes are seen lying just to the low energy side of the threshold. Such lines are due to the creation of a hound electron-hole pair, called an exciton. Conduhon band edge band edge /Valence band edge Valence hand edge Figure 5 In (a) the lowest point of the conduction hand occurs at the same value of k as the highest point of the valence band. A direct optical transition is drawn vertically with no significant change of k, because the absorbed photon has a very small wavevector. The threshold frequency og for absorp- tion by the direct transition determines the energy gap E, = fiw,. The indirect transition in (h) in- volves both aphoton and aphonon because the hand edges of the conduction and valence bands are widely separated in k space. The threshold energy for the indirect process in (h) is greater than the true band gap. The absorption threshold for the indirect transition between the hand edges is at fiw = E , + fin, where n is the frequency of an emitted phonon of wavevector K - -kc At higher temperatures phonons are already present; if a phonon is absorbed along with a photon, the thresh- old energy is fiw = Eg - fin. Note: The figure shows only the threshold transitions. Transitions occur generally between almost all points of the two hands for which the wavevectors and energy can he conserved. Table 1 Energy gap between the valence and conduction bands (i = indirect gap; d = direct gap) E, e V - - - - E,, eV - - - Crystal Gap 0 K 300 K C~ystal Gap 0 K 300 K Diamond i j.4 Si i 1.17 1.11 Ge i 0.744 0.66 aSn d 0.00 0.00 InSb d 0.23 0.17 1114s d 0.43 0.36 InP d 1.42 1.27 Gap i 2.32 2.25 GaAs d 1.52 1.43 GaSb d 0.81 0.68 AlSb i 1.65 1.6 SiC(11rx) i 3.0 Tc d 0.33 HgTea d 0 . 3 0 PbS d 0.286 PbSe i 0.165 PhTr i 0.190 CdS d 2.582 CdSe d 1.840 C dTe d 1.607 SnTe d 0.3 Co2O d 2.172 'HgTe is a semimetal: the bands overlap substantial wavevector kc. Here a direct photon transition at the energy of the minimum gap cannot satisfy the requirement of conservation of wavevector, because photon wavevectors are negligible at the energy range of interest. But if a phonon of wa\~evector K and frequency Cl is created in the process, then we can have as required hy the conservation laws. The phonon energy fin will generally he much less than E,: a phonon even of high wavevector is an easily accessible source of crystal momentum because the phonon energes are characteristi- cally small (-0.01 to 0.03 eV) in comparison with the energy gap. If the tem- perature is high ellough that the necessary p l ~ o ~ ~ o ~ l is already tl~errr~ally excited in the crystal, it is possible also to have a photon absorption process in which thc phonon is absorbcd. The band gap may also he dediiced from the temperatnre dependence of the conductivity or of the carrier concentration in the intrinsic range. The carrier concentration is obtained from measurements of the Hall voltage (Chapter 6), sometimes supplemented by conductivity measurements. Optical ~neasurements determine whether the gap is direct or indirect. The band edges in Ge and in Si are connected by indirect transitions; the band edges in InSb and GaAs are connected by a direct transition (Fig. 6). The gap in aSn is 8 Semiconductor Crystals Figure 6 Optical absorption in pure indium antimonide, InSb. The transition is direct because both ronrlnction and valencc band edges are at the center of the Brilloilin zone, k = 0. Noticc the sharp threshold. (After 6. W Coheli and H. Y. Fan.) direct and is exactly zero; HgTe and IIgSe are semi~netals and have negative gaps-the conduction and valence bands overlap. EQUATIONS OF MOTION We derive the equation of motion of an electron in an energy band. We look at the notion of a wave packet in an applied electric field. Suppose that the wave packet is made up of wavefunctions assembled near a particular wavevector k. The group velocity by definition is cr = dddk. The frequency as- sociated with a wavefiinction of energy E by quantum theoly is o = d i i , and so The eflects of the crystal on the electron rnotion are contained in the disper- sion relation ~ ( k ) . The work S E done on the electron by the electric field E in the time interval 6t is S E = -eEvg 6t . We observe that S E = (de/dk)Sk = fivg 6k , using (1). On comparing (2) with (3) we have whence fidkldt = -eE. We may write (4) in terms of the external force F as This is an important relation: in a crystal fidkldt is equal to the external force on the electron. In free space d(mv)/dt is equal to the force. We have not over- thrown Newton's second law of motion: the electron in the crystal is subject to forces from the crystal lattice as well as from external sources. The force term in (5) also includes the electric field and the Lorentz force on an electron in a magnetic field, under ordinary conditions where the mag- netic field is not so strong that it breaks down the band structure. Thus the equation of motion of an electron of group velocity v in a constant magnetic field B is dk (CGS) fi- = -sv x B ; dt where the right-hand side of each equation is the Lorentz force on the electron. With the group velocity v = C1gradp, the rate of change of the wavevector is dk e (CGS) - = -- Vkc X B ; dt fizc where now both sides of the equation refer to the coordinates in k space. We see from the vector cross-product in (7) that in a magnetic field an electron moves in k space in a direction normal to the direction of the gra- dient of the energy E , so that the electron moves on a surface of constant energy. The value of the projection kB of k on B is constant during the motion. The motion in k space is on a plane normal to the direction of B, and the orbit is defined by the intersection of this plane with a surface of constant energy. 8 Semiconductor Crystals 193 Physical Deriuation of hk = F \'t! cons~der the Bloch eigenfunction $k belonging to thc cncrg): cigcn- value ek and wavevector k: The expectation valuc of the momentum of an electron in the Bloch state k is using Z IC(k + 6 ) j2 = 1. We examine the transfer of momentum between the electron and the lat- tice when the state k of the electron is changed to k + Ak by the application of an external force. We imagine an insulating crystal electrostatically neutral except for a single electron in the state k or an othemise empty band. We suppose that a wreak external force is applied for a time interval such that the total impulsc givcn to the entire crystal system is J = SF dt. If the cond~lction electron were free (mh = m), the total momentum imparted to the crystal system by the impulse would appear in the change of momentum of the conduction electron: The neutral cqystal suffers no net interaction with the electric field, either directly or indirectly through the free electron. If the conduction electron interacts with thc pcriodic potential of the crys- tal lattice, wc must have From the result (9) for pel we have Ap,, = fi,Ak + hG[(VkIC(k + G)I2) . Ak] . G (12) The change Apl,, in the lattice ~rlornentunr resulting from the change of state of the electron may be derived by an elementary physical consideration. An electron reflected by the lattice transfers momentum to thc lattice. If an incident electron with plane wave componcnt of momentum hk is reflected with momant~~m h(k + G), the lattice acquires the momentum -hG, as re- quired by momentum conservation. The momentum transfer to the lattice when the state $k goes over to $k+hk is Aplat = -fix G[(Vk/C(k + G)I2 . Ak] , G (13) because the portion Vk C(k + 6) 1' . A k of each individual component of the initial state is reflected during the state change Ak. The total momentum change is therefore Apa + Ap~,t = J = fiAk , (15) exactly as for free electrons, Eq. (10). Thus from the definition of J, we have /dt = F , (16) derived in (5) by a different method. A rigorous derivation of (16) by an en- tirely different method is given in Appendix E. Holes The properties of vacant orbitals in an othenvise filled band are important in semiconductor physics and in solid state electronics. Vacant orbitals in a band are commonly called holes, and without holes there would be no transis- tors. A hole acts in applied electric and magnetic fields as if it has a positive charge +e. The reason is given in five stcps in the boxes that follow. 1. k,,= -k, . 0 7 ) The total wavevector of the electrons in a filled band is zero: Zk = 0, where the sum is over all states in a Brillouin zone. This result follows from the geometrical syrn~rletry of the Brillouin zone: every fundamental lattice type has symmetly under the inversion operation r+ -r about any lattice point; it follows that the Brillouin none of the latticc also has inversion symmetry. If the band is filled all pairs of orhitals k and -k are filled, and the total wavevector is zero. If an electron is missing from an orbital of wavevector k,, the total wavevector of the system is -k, and is attributed to the hole. This result is surprising: the electron is missing fro~ri k , and tllt: position of the hole is usually indicated graphically as situated at ke, as in Fig. 7. But the true wavevector of the hole is -k,, which is the wavevector of the point G if the hole is at E. The wavevector -k, cntcrs into sclcction rtilcs for photon absorption. The hole is an alternate description of a band with one missing elec- tron, and we either say that the hole has wavevector k , or that the band with one missing electron has total wavevector -k,. 8 Semiconductor Crystals 195 Figure 7 Absorption of a photon of energy fio and negligible wavevector takes an electron from E in the filled valence band to Q in the conduction band. If k , was the wavevector of the electron at E, it becomes the wavevector of the electron at Q. The total wavevector of the valence band after the absorption is -+, and this is the wavevector we must ascribe to the hole if we describe the valence band as occupied by one hole. Thus kh = -k,; the wavevector of the hole is the same as the wavevector of the electron which remains at 6. For the entire system the total wavevector after the absorption of the photon is k , + = 0, so that the total wavevector is unchanged by the absorption of the photon and the creation of a free electron and free hole. 2. = - ~ , ( k , ) . (18) Here the zero of energy of the valence band is at the top of the band. The lower in the band the missing electron lies, the higher the energy of the system. The energy of the hole is opposite in sign to the energy of the missing electron, because it takes more work to remove an electron from a low orbital than from a high orbital. Thus if the band is symmet- ric,' e,(k,) = €@(-kg) = -eh(-k)= -ch(kh). We construct in Fig. 8 a band scheme to represent the properties of a hole. This hole band is a helpful representation because it appears right side up. 3. vh = v, . (19) The velocity of the hole is equal to the velocity of the missing electron. From Fig. 8 we see that Veh(kh) = Ve,(ke), SO that vh(kh) = v,(k,). 'Bands are always symmetric under the inversion k + -k if the spin-orbit interaction is neglected. Even with spin-orbit interaction, bands are always symmetric if the crystal structure permits the inversion operation. Without a center of symmetry, but with spin-orbit interaction, the bands are symmetric if we compare subbands for which the spin direction is reversed. ~ ( k , T) = E(-k, J). See QTS, Chapter 9. Hole band constructed k Figure 8 The upper half of the figure shows the hole band that simulates the dynamics of a hole, constructed by inversion of the valence band in the origin. The wavevector and energy of the hole are equal, but opposite in sign, to the wavevector and energy of the empty electron orbital in the va- lence band. We do not show the disposition of the electron removed from the valence band at k,. 4. rnh = -me . (20) We show below that the effective mass is inversely proportional to the curvature d2e/dk2, and for the hole band this has the opposite sign to that for an electron in the valence band. Near the top of the valence band m, is negative, so that mh is positive. mc, 1 5. f i - = e ( E + ? v h X B ) dt . (21) This comes from the equation of motion (CGS) mc, fi- = - 1 dt e(E X B) (22) that applies to the missing electron when we substitute -kh for k , and vh for v,. The equation of motion for a hole is that of a particle of positive charge e. The positive charge is consistent with the electric current carried by the valence band of Fig. 9: the current is carried by the unpaired electron in the orbital 6: j = (-e)v(G) =(-e)[-v(E)] = ev(E) , (23) which is just the current of a positive charge moving with the velocity as- cribed to the missing electron at E. The current is shown in Fig. 10. 8 Semiconductor Crystals 197 Figure 9 (a) At t = 0 all states are filled except F at the top of the band; the velocity o, is zero at F because deldk, = 0. (b) An electric field E, is applied in the +x direction. The force on the elec- trons is in the -k, direction and all electrons make transitions together in the -k, direction, mov- ing the hole to the state E. (c) After a further interval the electrons move farther along ink space and the hole is now at D Figure 10 Motion of electrons in the conduction band and holes in the valence hand in the electric field E. The hole and electron drift velocities are in opposite directions, but their electric currents are in the same direction, the direction of the electric field. Effective Mass When we look at the energy-wavevector relation E = (h2/2m)k2 for free electrons, we see that the coefficient of k2 determines the curvature of E versus k. Turned about, we can say that llm, the reciprocal mass, determines the cur- vature. For electrons in a band there can be regions of unusually high curva- ture near the band gap at the zone boundary, as we see from the solutions in Chapter 7 of the wave equation near the zone boundary. If the energy gap is small in comparison with the free electron energy A at the boundary, the cur- vature is enhanced by the factor MEg. In semiconductors the band width, which is like the free electron energy, is of the order of 20 eV, while the band gap is of the order of 0.2 to 2 eV. Thus the reciprocal mass is enhanced by a factor 10 to 100, and the effective mass is reduced to 0.1-0.01 of the free electron mass. These values apply near the band gap; as we go away from the gap the curvatures and the masses are likely to approach those of free electrons. To summarize the solutions of Chapter 7 for U positive, an electron near the lower edge of the second band has an energy that may be written as E(K) = E, + (h212m,)l? ; m,lm = l/[(W/U)-11 . (24) Here K is the wavevector measured from the zone boundary, and me denotes the effective mass of the electron near the edge of the second band. An elac- tron near the top of the first band has the energy The curvature and hence the mass will bc ncgativc ncar thc top of the first band, hut we have introduced a mimls sign into (25) in order that the symbol m, for the hole mass will have a positive value-see (20) above. The crystal does not weigh any less if the effective mass of a carrier is less than the free electron mass, nor is Newton's secorld law violated for the crystal taken us n whole, ions plus carriers. The important point is that an electron in a periodic potential is accelerated relative to the lattice in an applicd clcctric or magnetic field as if the mass of the electron were equal to an effective mass which we now define. We differentiate the result (1) for the group velocity to obtain hie know from (5) that dkldt = Ffi, whence If we identify fi2/(d2~/dk2) as a I I ~ ~ S S , the11 (27) ~ S S U I I I ~ S the for111 of Newton's second law. \Ve define the effective mass m by It is easy to generalize this to take account of an anisotropic clcctron cn- ergy surface, as for electrons in Si or Ge. Me introduce the components of the reciprocal effective mass tensor where p, v are Cartesian coordinates. Physical Interpretation ofthe Effectke Mass How can an clrctron of mass m whcn put into a clystal respond to applicd fields as if the mass were m'? It is helpfill to think of the process of Bragg re- flection of electron waves in a lattice. Consider the weak interaction approxi- mation treated in Chapter 7. Near the bottom of the lower band the orbital is represented quite adequately by a plar~e wave exp(ikx) with rrionielitn~ri hk; the wave component exp[i(k - G)r] with mome~ltum h(k-G) is sulall and 8 Semiconductor Crystals 199 Beam v Figure 11 Explanation of negative effective masses which occur near, but below, a Brillouin zone boundary. In (a) the energy of the electron beam incident on a thin crystal is slightly too low to sat- isfy the condition for Bragg reflection and the beam is transmitted through the crystal. The appli- cation of a small voltage across the grid may, as in (b), cause the Bragg condition to be satisfied, and the electron beam will then he reflected from the appropriate set of crystal planes. increases only slowly as k is increased, and in this regon m = m. An increase in the reflected component exp[i(k - G)x] as k is increased represents mo- mentum transfer to the electron from the lattice. Near the boundary the reflected component is quite large; at the bound- ary it becomes equal in amplitude to the forward component, at which point the eigenfunctions are standing waves, rather than running waves. Here the momentum component h(- k G ) cancels the momentum component fi($ G). A single electron in an energy band may have positive or negative effective mass: the states of positive effective mass occur near the bottom of a band be- cause positive effective mass means that the band has upward curvature (d2eldk2 is positive). States of negative effective mass occur near the top of the band. A negative effective mass means that on going from state k to state k + Ak, the momentum transfer to the lattice from the electron is larger than the momentum transfer from the applied force to the electron. Although k is increased by A k by the applied electric field, the approach to Bragg reflection can gve an overall decrease in the forward momentum of the electron; when this happens the effective mass is negative (Fig. 11). As we proceed in the second band away from the boundary, the amplitude of exp[i(k - G)x] decreases rapidly and m assumes a small positive value. Here the increase in electron velocity resulting from a given external impulse is larger than that which a free electron would experience. The lattice makes up the difference through the reduced recoil it experiences when the ampli- tude of exp[i(k - G)x] is diminished. If the energy in a band depends only slightly on k, then the effective mass will be very large. That is, mlm % - 1 when d2eldk2 is very small. The tight- binding approximation discussed in Chapter 9 gives quick insight into the for- mation of narrow bands. If the wavefunctions centered on neighboring atoms overlap very little, then the overlap integral is small; the width of the band narrow, and the effective mass large. The overlap of wavefunctions centered on neighboring atoms is small for the inner or core electrons. The 4 f electrons of the rare earth metals, for example, overlap very little. EfJkctive Masses in Semiconductors In many semiconductors it has been possible to determine by cyclotron resonance the effective masses of carriers in the conduction and valence bands near the band edges. The determination of the energy surface is equivalent to a determination of the effective mass tensor (29). Cyclotron resonance in a semiconductor is carried out with centimeter wave or millimeter wave radia- tion at low carrier concentration. The current carriers are accelerated in helical orbits about the axis of a static magnetic field. The angular rotation frequency w, is eB (CGS) w = - mc ' where m is the appropriate cyclotron effective mass. Resonant absorption of energy from an rf electric field perpendicular to the static magnetic field (Fig. 12) occurs when the rf frequency is equal to the cyclotron frequency. Holes and electrons rotate in opposite senses in a magnetic field. We consider the experiment for m/m = 0.1. At f, = 24 GHz, or w, = 1.5 X 10" s-', we have B = 860 G at resonance. The line width is determined by the collision relaxation time T , and to obtain a distinctive resonance it is necessary that wcr 3 1. The mean free path must be long enough to permit the average carrier to get one radian around a circle between collisions. The re- quirements are met with the use of higher frequency radiation and higher magnetic fields, with high purity crystals in liquid helium. In direct-gap semiconductors with band edges at the center of the Bril- louin zone, the bands have the structure shown in Fig. 13. The conduction band edge is spherical with the effective mass mo: , , . , , , . . . , . . . . . . . ... ' 8 . . A 8 (shtiel - Figure 12 Arrangcmcnt of fields in Ef a cvclotron rrso~~ancr rxprrllrtrr!t 111 a sr~~~icor~ductor. Tl~r srl!,t. oS the circulation is opposite for electrons and holes. 8 Semiconductor Crystals 201 ~ ~ l i t . ~ f f h ~ l ~ ~ Figure 13 Simplified view of the band edge structure of a direct-gap I semiconductor. Table 2 Effective masses of electrons and holes in direct-gap semiconductors Electron Heavy hole Light hole Split-off hole Spin-orhit Crystal m..Jm A. eV InSb 0.015 0.39 0.021 (0.11) 0.83 InAs 0.026 0.41 0.025 0.08 0.43 InP 0.073 0.4 (0.078) (0.15) 0.11 GaSb 0.047 0.3 0.06 (0.14) 0.80 GaAb 0.066 0.5 0.082 0.17 0.34 Cu,O 0.99 - 0.58 0.69 0.13 refcrred to the valence hand edge. The valence bands are characteristically threefold near the edge, with the heavy hole hh and light hole lh bands degen- erate at the center, and a band soh split off by the spin-orbit splitting A: Values of the mass parameters are given in Table 2. The forms (32) are only approximate, because even close to k = O the heavy and light hole hands are not spherical-see the discussion below for Ge and Si. The perturbation theory of band edgcs (Problem 9.8) suggests that the electron effective mass should be proportional to the band gap, approximately, for a direct gap crystal. We use Tables 1 and 2 to find the fairly constant values mJ(mEg) = 0.063, 0.060, and 0.051 in (eV)-' for the series InSb, InAs, and In!?, in agreement with this suggestion. Silicon and Germanium The conduction amd valer~ce bands of ger~rlaniunl are shown in Fig. 14, based on a combination of theoretical and experime~ital results. The valence band edge in both Si and Ge is at k = 0 and is derivcd from p,, and pllz states of the frce atoms, as is clear from the tight-hinding approximation (Chapter 9) to the wa\~efiinctions. The p , level is fourfold degenerate as in the atom; the four states corre- spond to m , values & and & $. The p , , level is doubly degenerate, with mJ = ? i. The p31z states are higher in energy tllan tlie p , , states; the energy difference A is a measure of the spin-orbit interaction. The valence band edges are not simple. Holes ncar thc band edge are characterized by two cffrctivc masses, light and heavy These arise from the two hands formed from the p,, level of the atom. There is also a band formed from the p , , level, split off from the p,,, level by the spin-orbit interaction. The energy surfaces are not spherical, but warped (QTS, p. 271): ~ ( k ) = ~k~ % [FI2k4 + C2(k:k; + k$: + kfki)J112 (33) The choice of sign distiriguishes the two Inasses. The split-off band has ~ ( k ) = -A + Ak2. The experiments give, in units h2/29ra, Si: A = -4.29 ; IBI = 0.68 ; ICI = 4.87 ; A = 0.044 eV Ge: A = -13.38 ; IBI = 8.48 ; ICI = 13.15 ; A = 0.29eV Roughly, the light and heavy holes in germanium have masses 0.043 m and 0.34 m; in silicon 0.16 n L and 0.52 in; in diamond 0.7 rn and 2.12 rn. The conduction band edges in Ge are at the equivalent points L of the Brillouin zone, Fig. 15a. Each band edge has a spheroidal energy surfacc ori- ented along a (111) crystal axis, with a longitudinal mass ml = 1.59 nl and a transverse mass m, = 0.082 m. For a static magnetic field at an angle 0 with the longitudinal axis of a spheroid, the effective cyclotron mass m, is Results for Ce are shown in Fig. 16. 1 1 1 silicor~ the conduction band edges are spheroids oriented along the equivalent (100) directions in the Brillouin zone, with mass parameters ml = 0.92 m and m, = 0.19 m, as in Fig. 17a. The hand edges lie along the lines laheled A in the zone of Fig. 15a, a little way in from the boundary points X. In GaAs we have A = -6.98, B = -4.5, I C I = 6.2, A = 0.341 eV. The band structure is shown in Fig. 1%. It has a mrect band gap with an isotropic conduction electron mass of 0.067 in. Figure 14 Calculated band structure of germanium, after C. Y. Fong. The general features are in good agreement with experiment. The four valence bands are shown in gray The fine structure of the valence band edge is caused by spin-orbit splitting. The energy gap is indirect; the conduction band edge is at the point (2.rr/a)(: i). The constant energy surfaces around this point are ellipsoidal. Figure 15 Sta~~dard labels of the symmetry points and awes of the Rrillouin zones of the fcc and hcc lattices. The zone centers are T. In (a) the boundary point at (2m/a)(100) is X; the houndary point at (2w/a)(; if) is L; the line 4 runs between I' and X. In (b) the corresponding sy~ribals are H, P, and A. Figure 16 Effective cyclotron mass of electrons in germa- nium at 4 K for magnetic field directions in a (110) planc. Thcrc are four independent m a s spheroids in Ge, one along each [Ill] axis, but viewed in the (110) plane hvo spheroids always appear equivalent. (After Drcssclhaus, Kip, and Kittel.) .4ngle in degrees in (110) plane from axis 8 Semiconductor Crystals 205 (a) Figure 17a Constant energy ellipsoids for electrons in silicon, drawn for mllm, = 5. 4 3 2 % 1 C . 8 O B W -1 -2 3 4 L r x (b) Figure 17b Band structure of GaAs, after S. 6. Louie. INTRINSIC CARRIER CONCENTRATION We want the concentration of intrinsic carriers as a function of tempera- ture, in terms of the band gap. We do the calculation for simple parabolic band edges. We first calculate in terms of the chemical potential p the number of electrons excited to the conduction band at temperature T . In semiconductor physics p is called the Fermi level. At the temperatures of interest we may suppose for the conduction band of a semiconductor that E - jt k,T, so that the Fermi-Dirac distribution function reduces to This is the probability that a conduction electron orbital is occupied, in an approximation valid when f, < 1. The energy of an electron in the conduction band is where E, is the energy at the conduction band edge, as in Fig. 18. Here me is the effective mass of an electron. Thus from (6.20) the density of states at E is The concentration of electrons in the conduction band is whicl~ integrates to give The probleln is solved for when y is known. It is useful to calculate the equilibrium concentration of holes p. The distribution functionfi, for lloles is rclatcd to the electron distribution functionf, byfh = 1 -f,, because a hole is the absence of an electron. Iic: have provided (y - E ) % k,T. If the holes near the top of the valence band behave as particles with effcctivc mass mh, the density of hole states is given by where E, is the energy at the valence band edge. Proceeding as in (38) we obtain for the concentration p of holes in the valence band. Wc multiply together the expressions for n and p to obtain the equilibrium relation, with the energy gap E, = E, - E, as in Fig. 18, This useful result does not involve the Ferrrii level p. At 300 K the value of rrp is 2.10 X 10'%m-" 2.89 x 10'%m-" and 6.53 X 10'\1n-" for the actual band structures or Si, Ge, and GaAs, respectively. Lic haw nowhcre assumed in the derivation that the material is intrinsic: the result holds for impnrity ionization as well. The only asslimption made is that the distance of the Fermi level from the edge of both bands is large in comparison with kBT. A simple kinetic argument shows why the product n p is constant at a given te~nperature. Suppose that the equilibrium population of electrons and lloles 8 Semiconductor Crystals 207 Figure 18 Energy scale for statistical calcula- tions. The Fermi distribution function is shown on the same scale, for a temperature kgT < Eg. The Fermi level p is taken to lie well within the band gap, as for an intrinsic semiconductor. If E = p, then f = i. is maintained by black-body photon radiation at temperature T. The photons generate electron-hole pairs at a rate A(T), while B(T)np is the rate of the re- combination reaction e + h = photon. Then dnldt = A(T) - B(T)np = dpldt . (44) In equilibrium dnldt = 0, dpldt = 0, whence np = A(T)IB(T). Because the product of the electron and hole concentrations is a constant independent of impurity concentration at a given temperature, the introduction of a small proportion of a suitable impurity to increase n, say, must decrease p. This result is important in practice-we can reduce the total canier concentra- tion n + p in an impure crystal, sometimes enormously, by the controlled intro- duction of suitable impurities. Such a reduction is called compensation. In an intrinsic semiconductor the number of electrons is equal to the number of holes, because the thermal excitation of an electron leaves behind a hole in the valence band. Thus from (43) we have, letting the subscript i de- note intrinsic and E, = E, - E,, The intrinsic carrier concentration depends exponentially on Ep12kBT, where Eg is the energy gap. We set (39) equal to (42) to obtain, for the Fermi level as measured from the top of the valence band, If m,, = m,, then p = Eg and the Fermi level is in the middle of the forbid- den gap. Intrinsic Mobility The mobility is the magnitude of the drift velocity of a charge carrier per unit electric field: p = ~ u I / E . (48) The mobility is defined to he positive for both electrons and holes, although their drift velocities are opposite in a given field. By writing pe or p,, with subscripts for the electron or hole mobility we can avoid any confusion be- tween p as the chemical potential and as the mobility. The electrical conductivity is the sum of the electron and hole contributions: where n and p are the concentrations of electrons and holes. In Chapter 6 the drift velocity of a charge q was found to be u = q~E/m, whence where T is the collision time. The mobilities depend o m temperature as a modest power law. The tem- perature dependence of the conductivity in the intrinsic region will be dominated by the exponential dependence exp(-Epk,T) of the carrier con- centration, Eq. (45). Table 3 gives experimental values of the mobility at room temperature. The mobility in SI units is expressed in m2N-s and is lo-' of the mobility in practical units. For most substances the values quoted are limited by the scat- tering of carriers by thermal phonons. The hole mohilities typically are smaller than the electron mobilities because of the occurrence of band degeneracy at the valence band edge at the zone center, thereby making possible interband scattering processes that reduce the mobility considerably. Table 3 Carrier mobilities at room temperature, in cm2N-s Crystal Electrons Hohs Crystal Electrons Holm Diamond Si Ge InSb InAs InP A 1 As AlSb GaAs GaSb PbS PhSe PbTe AgCl KBr (100 K) SIC 8 Semiconductor Crynials 209 In some crystals, particularly in ionic crystals, the holes are essentially immobile and get about only by thermally-activated hopping from ion to ion. The principal cause of this "self-trapping" is the lattice distortion associated with the Jahn-Teller cf'fect of degenerate states. The orbital degeneracy neces- sary for self-trapping is much more frequent for holcs than for electrons. There is a tendency for crystals with small cncrgy gaps at direct barid edges to havc high values of the electron mobility. Small gaps lead to small effcctive masses, which favor high mobilities. The highest mobility observed in a bulk semiconductor is 5 X lo6 cm2117-s in PbTe at 4 K, where the gap is 0.19 eV. IMPURITY CONDUCTIVITY Certain impurities and imperfections drastically affect the rlectrical prop- erties of a se~niconductor. The addition of boron to silicon in the proportion of 1 boron atom to lo5 silicon atoms increases the conductivity of pure silicorl at room temperature by a factor of 1 0 ' . I 1 1 a componnd semiconductor a stoichio- ~netric deficiency of one constituent will act as an impurity; such semiconduc- tors are known as deficit semiconductors. The deliberate additioil of impuri- ties to a semiconductor is called doping. We consider the affect of impurities in silicon and germanium. These ele- ments crystallize in the diamond structure. Eacli atom lorms f'oiir covalent bonds, one with each of its nearest neighbors, corresponding to the chemical valence four. II an impurity atom of valence five, such as phosphorus, arsenic, or antimony, is slibstituted in the lattice in place of a normal atom, there will be one valcnce electron from the impurity atom left over after the four cova- lent bonds are establislied with the ncarest neighbors, that is, after the impii- rity atom has been acco~nmodated in the structure wit11 as little disturbance as possible. Impurity atoms that can give up an electron are called donors. Donor States. The structure in Fig. 19 has a positive charge on the impurity atom (which has lost one electron). Lattice constant studies have verified that the pentavalent impurities enter the lattice by substitution for normal atorns, and not in interstitial positions. The crystal as a wholc remains neutral because the electron remains in the crystal. The extra electron moves in the coulomb potential e / ~ r of the impurity ion, where E in a co\~alent crystal is the static dielectric constant of the medium. The factor l/e takes account of the reduction in the coulomb force between charges caused by thc electronic polarization of the mcdi~im. This treatment is valid for orbits large in compariso~l with the distance between atoms, aid for slow motions of the electron such that thc orbital frequency is low in comparison with the frequency wg corrcsponding to the energy gap. These conditions are satisfied quite well in Ge and Si by the donor electron of P, As, or Sb. Figure 19 Charges associated with an arsenic ~mpurity atom in silicon. Arsenic has five valence electrons, but silicon has only four valence electrons. Thus four electrons on the arsenic form tetra- hedral covalent bonds similar to silicon, and the fifth electron is available for conduction. The arsenic atom is called a donor because when ionized it donates an electron to the conduction band. We estimate the ionization energy of the donor impurity. The Bohr theory of the hydrogen atom may be modified to take into account the dielectric constant of the medium and the effective mass of an electron in the periodic potential of the crystal. The ionization energy of atomic hydrogen is -e4m/2fi2 in CGS and -e4m/2(4wc0fi)' in SI. In the semiconductor with dielectric constant E we replace e2 by e% and m by the effective mass me to obtain e4% - (13.6 m e ) ev ; (CGS) Ed =-- -- 2c2A2 € 2 as the donor ionization energy of the semiconductor. The Bohr radius of the ground state of hydrogen is ti2/me2 in CGS or 4wc0A2/me2 in SI. Thus the Bohr radius of the donor is €ti2 - ( 0 . 5 3 ~ ) ; (CGS) ad = -- - - m,e2 m$m The application of impurity state theory to germanium and silicon is com- plicated by the anisotropic effective mass of the conduction electrons. But the dielectric constant has the more important effect on the donor energy because it enters as the square, whereas the effective mass enters only as the first power. To obtain a general impression of the impurity levels we use me = 0.1 m for electrons in germanium and m, = 0.2 m in silicon. The static dielectric constant is given in Table 4. The ionization energy of the free hydrogen atom is 13.6 eV. For germanium the donor ionization energy Ed on our model is 5 meV, reduced with respect to hydrogen by the factor m$me2 = 4 x The corresponding result for silicon is 20 meV. Calculations using the correct 8 Semiconductor Cqstah 211 Table 4 Static relative dielectric constant of semiconductors Crystal Crystal E Diamond Si c:c I11SI-r InAs InP GaS b GaAs AlAs AlSb Sic cu,o Table 5 Donor ionization energies Ed of pentavalent impurities in germanium and silicon, in meV anisotropic niass tensor predict 9.05 meV for germanium and 29.8 meV for silico~i. Observed values of donor ionizatioli energics in Si and Ge are given in Table 5. In GaAs donors have Ed = 6 meV. The radius of the first Bohr orbit is increased by em/rr~, over the value 0.53 A For the free hydrogen atom. Thc corresponding radius is (160)(0.53) = 80 in germanium and (60)(0.,53) = 30 A in silicon. These arc large radii, so that donor orbits overlap at relatively low donor concentrations, compared to the number of host atoms. With appreciable orbit overlap, an "impurity band" is formed from the donor states: see the discussion of the metal-insulator tran- sition in Chaptcr 14. The semiconductor can conduct in the impurity band by electrons hop- pi~lg from donor to donor. The process of impurity band conduction sets in at lowcr donor concentratiori levels if there are also some acceptor atoms pre- sent, so that some of the donors are always ionized. It is easier for a donor electron to hop to an ionized (unoccupied) donor than to an occupied donor atom, in order that two electrons will not have to occnpy the same site during charge transport. Acceptor States. A hole may be bound to a trivalent impurity in germanium or silicon (Fig. 20), just as an electron is hound to a pentavalent impurity. Trivalent impurities such as B, Al, Ga, and In are called acceptors because they accept electrons from thc valence band i 1 1 order to complete the covalent bonds with neighbor atoms, leaving holes in the band. Figure 20 Boron has only three valence electrons; it can complete its tetrahedral bonds only by taking an electron from a Si-Si bond, leaving behind a hole in the silicon valence band. The positive hole is then available for conduction. The boron atom is called an acceptor because when ionized it accepts an electron from the valence band. At 0 K the hole is hound. Table 6 Acceptor ionization energies E, of trivalent impurities in germanium and silicon, in meV When an acceptor is ionized a hole is freed, which requires an input of energy. On the usual energy band diagram, an electron rises when it gains energy, whereas a hole sinks in gaining energy. Experimental ionization energies of acceptors in germanium and silicon are given in Table 6. The Bohr model applies qualitatively for holes just as for electrons, but the degeneracy at the top of the valence band complicates the effective mass problem. The tables show that donor and acceptor ionization energies in Si are com- parable with k,T at room temperature (26 meV), so that the thermal ionization of donors and acceptors is important in the electrical conductivity of silicon at room temperature. If donor atoms are present in considerably greater num- bers than acceptors, the thermal ionization of donors will release electrons into the conduction band. The conductivity of the specimen then will be con- trolled by electrons (negative charges), and the material is said to be n type. If acceptors are dominant, holes will be released into the valence band and the conductivity will be controlled by holes (positive charges): the mater- ial is p type. The sign of the Hall voltage (6.53) is a rough test for n or p type. 8 Semiconductor Crystaln 213 A-type annealed 1000/T Figure 21 Temperature dependence of the free carrier concentration in ultrapure Ge, after R. N. Hall. The net cuncentration of electrically active irnpuritics is 2 X 10" ~ r n - ~ , as determined by Hall coefficient measurements. The rapid onset of intrinsic excitation as tlic temperature is in- creasrd is evident at low values of 1/T. The carrier corrccntration is closely constant between 20 K and 200 K. Another handy laboratory test is the sign of the thermoelectric potential, dis- cussed below. The numbers of holes and electrons arc equal in the intrinsic regime. The intrinsic electron concentration ni at 300 K is 1.7 X loi3 cm-3 in germanium and 4.6 X 10' cm-3 in silicon. Thc electrical resistivity of intrinsic material is 43 ohm-cm for germanium and 2.6 X 10' ohm-crn for silicon. Germanium has 4.42 x 10" atoms per c m ! The pi~rification of Ge has been carried further than any other element. The concentration of the comIrlon electrically active impurities-the shallow donor and acceptor impurities-has been reduced below 1 impurity atom in 10" Ge atoms (Fig. 21). For example, the concentration of I ' in Ge can be reduced below 4 X 10'' cm-! There are irnpuritics (H, 0; Si, C) whose conccntrations in Ge cannot usually be reduced below 10"- loL4 ~ r n - ~ , but these do not affect elec- trical measure~nents and therefore may be hard to detect. Thermal Ionization of Donors and Acceptors The calcl~lation of the equilibrium concentration of conduction electrons from ionized donors is identical with thc standard calculatio~l in statistical me- chanics of the thermal ionization of hydrogen atoms (TP, p. 369). If there are no acceptors present, the result in the low temperature limit kBT 4 Ed is Electron concentration, cm-3 Figure 22 Electrical conductivity and hole concentration p calculated as a function of electron co~ice~~tration n for a semicondnctor at a temperature such that np = l V U c m P The conductivity is symmetrical about n = 10'" cm-! For n > lO"', the specimen is n type; for n < lo"', it is p hFe. We have taken p, = ph, for the niobilities. with no = 2 ( r n J ~ , T / 2 d ~ ) ~ / ~ ; here Nd is the concentration of donors. To obtain (53) we apply the laws of chemical eqililibria to the concentration ratio [e][@]/[Nd], and then set [Nil = [el = n. Identical reslllts hold for acceptors, under the assumption of no donor atoms. If the donor and acceptor concentrations are comparable, affairs are com- plicated and the equations are solved by numerical methods. However, the law of mass action (43) requires the n p product to be constant at a given tempera- ture. An excess of donors will increase the electron concentration and de- crease the hole concentration; the sum n + p will increase. The conductivity will increase as n + p if the mobilities are equal, as in Fig. 22. THERMOELECTRIC EFFECTS Consider a selniconductor maintained at a constant temperature while an electric field drives through it an electric current density j,. If the current is carried only by electrons, the charge flux is where p, is the electron mobility. The average energy transported by an elec- tron is referred to the Fermi level p, 8 Semiconductor Crystals 215 where E, is the energy at the conduction band cdge. R7e refer the energy to the Fermi level because different conductors in contact have the saIne Fermi level. The energy flux that acco~npanies the charge flux is The Peltier coefficient I I is defined hy j, = IIjy; or the energy carried per unit charge. For electrons, II,= - ( ~ , - p + $ k ~ ~ ) l e (56) and is negative because the energy flux is opposite to the charge flux. For holes jq = p e p ~ E ; j ~ , = p(p - E , + $kBT)phE , (57) where E , is thc energy at the valence band edge. Thus and is positive. Equations (56) and (58) are the result of our simple drift veloc- ity t b e o ~ ~ ; a trcatment by the Boltzn~ann transport cqliation gives minor nu- merical difkrence~.~ The absolute thermoelectric power Q is defined from the open circuit electric field created by a temperature gradient: E = Q grad T . (59) The Peltier coefficient I I is related to the thcrmoelectric power Q by This is thc famous Kelvin relation of irreversible thermodjmamics. A measure- ment of the sign of the voltage across a scrniconductor specirr~en, one end of which is heated, is a rough and ready uray to tell if the speci~nen is n typc or p type (Fig. 23). SEMIMETALS In semimetals the conduction band edge is very slightly lower in energy than the valence band edge. A small overlap in energy of the cor~duction and valence bands leads to small concentration of holes in the valence band and of electrons in the conduction band (Tahle 7). Three of the semimetals, arsenic, antimony, and bismuth, are in group V of the periodic table. Their atoms associate in pairs in the crystal lattice, with two ions and ten valence electrons per primitive cell. The even number of valence electrons 'A si111ple discussion of Uoltzmann transport theory is given in Appendix I?.'. Figure 23 Peltier coclficient of n and p silicon as a function uf ternperaturc. Above 600 K the spec- imens act as intlirrsic scmiconduc- tors. The curves are ralci~lated and tlre points are experimentd. (After T H. Gehalle and G. \V, Hull.) Table 7 Electron and hole concentrations in semimetals Semimetal n,, in ern? Arsenic Antirrrony Bismuth Graphite could allow these elements to be insulators. Like semiconductors, the serni- metals may be doped with suitable impurities to vary the relative numbers of holes and electrons. Their concentrations may also be varied with pressure, for the band edge overlap varies with pressure. SUPERLATTICES Consider a multilayer crystal of alternating thin layers of different composi- tions. Coherent layers on a nanometer thickness scale may be deposited by moleciilar-beam epitaxy or metal-organic vapor deposition, thus building up a sriperperio&c structure on a large scale. Systems of alternate lay.ers of GaAs and GaAlAs have been studied to 50 periods or more, wit11 lattice spacing A of per- haps 5 nm (50 A). A superperiodic crystal potential arises from the sulperperiodic structure and acts on the conduction electrons and holes to create new (small) Brillouin zoncs and mini energy bands superposed on the hand structures of the 8 Semiconductor Crystals 217 constitnent layers. Here we treat the motion of an electron in a superlatticc in an applied electric field. Bloch Oscillator Consider a collisionless electrori in a periodic lattice in one dimension, with motion normal to the planes of the superlattice. The equation of motion in a constant electric field parallel to k is fidkldt = -eE or, for motion across a Brillouin zone with reciprocal lattice vector G = 27r/A, we have f L G = fi2?r/A = eET, where T is the period of the motion. Thc Bloch frequency of the motion is o , = 2v/T = eEA/fi. The electron accelerates from k = 0 towarcl the zonc hoiinda~y; when it reaches k = ?r/A it reappears (as by an Unrklapp proccss) at the zone boundaly at the identical point -dA, using the argument of Chapter 2. \Ve consider the motion in a rnodel systcm in real space. We suppose that the clcctron lies in a simple energy band of width 6,: The velocity in k-space (momentum space) is v = fi-'de/tlk = (AedfL) sin k A , (62) and the position or the clectron in real space, with the initial condition z = 0 at t = 0, is given by z = $0 dt = Jclk v(k)(dtldk) = (Aedfi) Jdk(-fileE) sin kA =(-~~leE)(coa k A - 1) = (-e,leE)(cos(-eEAtlTc) -1) . (63) This confirins that the Bloch oscillation frequency in real space is w, = eEAfi. The motion in the periodic lattice is quite different from the motion in free space, for which the acceleration is constant. Zener Tunneling Thus far we have considered the effect of the electrostatic potential -eEz (or -eEnA) on onc energy band; the potential tilts the urhole band. Higher bands will also he tilted similarly, creating the possibility of crossing between ladder levcls of different bands. The interaction hctween different band levels at thc same energy opens the possibility for an electron in one band at n to cross to another band at n'. This field-indnced interband tunneling is an example of Zener breakdown, met most often at a single junction as in the Zener diode. SUMMARY The motion of a wave packet centered at wavevector k is described by F = f&dt, where F is the applied force. The motion in real space is ob- tained from the group vclocity vg = fi"Vkc(k). The smaller the energy gap, the smaller is the effective mass Im ncar thc A crystal with one hole has one empty electron state in an otherwise filled bald. The properties of the hole are those of the N - 1 electrons in this band. (a) If the electron is lnissilig from the state of wavevector k,, then the wavevector of the hole is k,, = -k,. (h) The rate of change of kh In an applied ficld rcquirrs thc assignmrnt of a positive charge to the hole: eh = e = -e,. (c) If u, is the velocity an electron would have in the state k,, then the veloc- ity to be ascribed to the hole of wavevector kh = - k , is uh = u,. (d) The energy of the hole referred to zero for a filled band is positive and is eh(kh)= - 4 , ) . (e) The effective mass of a hole is opposite to the effective mass of an elec- tron at the same point on thc e n e r g hand: mh = -me. Problems 1. Impurity orbits. Indium antilnonidc has Eg = 0.23 eV; rlielectric constant E = 18; electron cffcctivc mass 7n, = 0.015 m. (:alcnlate (a) the donor ioniratiorr enerc; (b) thc radius of thc ground state orbit. (c) At what miriimilnr donor corrcer~tratior~ will appreciable overlap effects hetween the orbits of acljacerrt impurity atorrls occur? This overlap tends to prodllce an in~pnrity band-a havrd of energy levels which permit cond~~ctivity presi~mabl~ by a hopI>ing rr~ecl~ariisrn in url~id~ electroris iiiove froin one inipurity site to a neighboring ionized impurity site. 2. Ionization of donors. Irr a particular semiconductor there are 1013 donors/cm3 wit11 a 1 1 iomizatio~r energy Ed of 1 meV and an effective mass 0.01 m. (a) Estimate the coricrrrtratio~l of curlduction electrons at 4 K. (b) What is the value of the Hall coeff- icelit? Assurne no acceptor atoms are present and that Eg % kgT. 3. Hall effect with two carrier types. Assuming concentration n; p; relaxation times T,, .rjb; and masses m,, mi,, show that the Hall coefficient in the drift velocity approxi- mation is where b = p,/CLI, is the mobility ratio. In the derivation neglect terms of order B2. In SI we drop the c. Hint: In the presence of a longitudinal electric field, find the transverse electric field such that the transverse current vanishes. The algebra may seem tedious, but the result is worth the trouble. Use (6.64), but for two carrier types; neglect (w,:~)' in comparison with w,:~. 8 Semiconductor Crystals 219 4 . Cyclotron resonance for a spheroidal energy surface. Consider the energy surface whcrc m, is the transverse mass parameter and rrLl is the longitudinal mass parame- ter. A surface on which c(k) is constant will be a spheroid. Use the equation of mo- tion (6). with v = fi- lVkc. to show that w, = eBl(mlm,)'"c when the static magnetic field B lics in the xy plane. This result agrees with (34) when 0 = d 2 . The rcsult is in CGS: to obtain SI, omit the c. 5. Magnetoresistance with two carrier types. Problerri 6.9 shows thal in the drift velocity approximation the motion of charge carriers in electric and magnetic ficlds does not lead to transverse magnetoresistance, The result is different with two car- rier types. Considcr a conductor with a concentration n of electro~is of effective n~ws me anrl rclaxation time 7,; and a concentration p of holes of effective rnws 7nh and relaxation time rh. Treat the limit of \rely strong magnetic fields, w , ~ 1. (a) Slrow in this limit that uyr = (n - p)ec/B. (h) Show that the Hall field is given by, with Q = wc7, which vanishes if n = p. (c) Show that the effective conductivity in the x direction is If n = p, u B-'. If 71 + p, u saturatcs in strong fields; that is, it approaches a limit independent ofB as B + m. Fermi Surfaces and Metals Reduced zone scheme 223 Periodic zone scheme 225 CONSTRUCTION OF FERMI SURFACES 226 Nearly free electrons 228 ELECTRON ORBITS, HOLE ORBITS, AND OPEN ORBITS 230 CALCULATION OF ENERGY BANDS 232 Tight binding method for e n e r e bands 232 Wigner-Seitz method 236 Cohesive e n e r g 237 Pseudopotential methods 239 EXPERIMENTAL METHODS IN FERMI SURFACE STUDIES 242 Quantization of orbits in a magnetic field 242 De Haas-van Alphen effect 244 Extremal orbits 248 Fermi surface of copper 249 Example: Fermi surface of gold 249 Magnetic breakdown 251 SUMMARY 252 PROBLEMS 252 1. Brillouin zones of rectangular lattice 252 2. Brillouin zone, rectangular lattice 252 3. Hexagonal close-packed structure 252 4. Brillouin zones of two-dimensional divalent metal 253 5. Open orbits 253 6. Cohesive energy for a square well potential 253 7. De Haas-van Alphen period of potassium 253 8. Band edge structure on k . p perturbation theory 253 9. Wannier functions 254 10. Open orbits and magnetoresistance 254 11. Landau levels 254 Zone 3 Copper Nu~~~innrn Figure 1 Free electron Fermi surfaces for fcc metals with one (Cu) and three (Al) valence elec trons per primitive cell. The Fermi surfacc shown for copper has been deformed from a sphert to agree wit11 the qerimental resnlts. The second ;.one of aluminum is nearly half-filled wit1 electrons. (A. R. Mackintosh.) CHAPTER 9: FERMI SURFACES AND METALS Few people would define a metal as "a solid with a Fermi surface." This may nevertheless be the most meaningful definition of a metal one can give today; it represents a profound advance in the understanding of why metals behave as they do. The concept of the Fermi surface, as developed by quantum physics, provides a pre- cise explanation of the main physical properh'es of metals. A. R. Mackintosh Thc Fermi surface is the surface of constant energy E~ in k space. The Fcrmi surface separates the unfilled orbitals from the filled orbitals, at absolute zero. The electrical properties of thc metal are determined by the volume and shape of the Fermi surfacc, hecaiise the current is due to changes in the occupancy of states near thc Fcrmi sllrface. The shape may be vcry intricate as viewed in the reduced zone scheme below and yet havc a simple interpretation when reconstructed to lie near the surfacc of a sphere. Lie exhibit in Fig. 1 the free electron Ferrni surfaces con- striicted for two metals that have the face-centered cubic crystal strtictiire: copper, with one valence electron, and aluminum, with three. The free elec- tron Fermi surfaces were developed from spheres of radi~is k , determined by the valence electron concentration. The wrface for copper is deformed by in- teraction with the lattice. How do we construct these surfaces from a sphere? The constructions rcquire the reduced and also the periodic zone schemes. Reduced Zone Scheme It is always possible to select the wavevector index k of any Rloch function to lie within the first Brilloui~i zone. The procedurc is known as mapping the band in the reduced zone scheme. If we encounter a Bloch function written as Jik.(r) = ei""ut.(r), with k' outside the first zone, as in Fig. 2, we may always find a suitable reciprocal lat- tice vector G such that k = k' + G lies within the first Brillouin zone. Then where uk(r) = e-iC'rt~k,(~). Rnth e-'G'r and uw(r) are periodic in the crystal lat- tice, so uk(r) is also, whence 4i(r) is of the Bloch form. Even with free electrons it is useful to work in the reduced zone scheme, as in Fig. 3. Any energy tk. for k' outside the first zone is equal to an e k in the first zone, where k = k' + G. Thus we necd solve for the energy only in the Figure 2 First Brillouin zone of a square lattice of side a. The wavevector k' can he carried into the first zone by forming k ' + G. The wavevector at a point A on the zone boundary is carried by G to the point A' on the opposite boundq of the same zone. Shall we count both A and A' as lying in the first zone? Because they can be connected by a reciprocal lattice vector, we count them as one identical point in the zone. Figure 3 Energy-wavevector relation q = &'k2/2rn for \ free electrons as drawn in the reduced zone scheme. \ This construction often gives a useful idea of the over- \ all appearance of the hand structure of a crystal. The \ branch AC if displaced by -2'7r/a gives the usual free \ electron c u m for negative k, as suggested by the \ \ dashed curve. The branch A'C if displaced by 2 d a \ gives the usual curve for positive k. A crystal potential \ \ U(x) will introduce hand gaps at the edges of the zone A' A (as at A and A') and at the center of the zone (as at C ) . The point C when viewed in the extended zone scheme falls at the edges of the second zone. The 7r -- 0 ' 7 r overall width and gross features of the hand structure - a are often indicated properly by such free electron k - bands in the reduced zone scheme. I - First Bdlouin zone first Brillouin zone, for each band. An energy band is a single branch of the el, versus k surface. In the reduced zone scheme we may find different energies at the same value of the wavevector. Each different energy characterizes a dif- ferent band. Two bands are shown in Fig. 3. Two wavefunctions at the same k but of different energies will be inde- pendent of each other: the wavefunctions will be made up of different combi- nations of the plane wave components exp[i(k + G) . r] in the expansion of (7.29). Because the values of the coefficients C(k + G) will differ for the 9 Femi Sutfaces and Metals 225 different bands, we should add a symbol, say n, to the C's to serve as a band index: C,(k + G). Thus the Bloch function for a state of wavevector k in the band n can be written as Periodic Zone Scheme We can repeat a given Brillouin zone periodically through all of wavevec- tor space. To repeat a zone, we translate the zone by a reciprocal lattice vector. If we can translate a band from other zones into the first zone, we can translate a band in the first zone into every other zone. In this scheme the energy ek of a band is a periodic function in the reciprocal lattice: Here E ~ + ~ is understood to refer to the same energy band as EL. 0 k - Figure 4 Three energy bands of a linear lattice lotted in (a) the extended (Brillonin), (b) reduced, and (c) periodic zone schemes. The result of this construction is known as the periodic zone scheme. The periodic property of thr energy also can be seen easily from the central equation (7.27). Consider for example an energy band of a simple cubic lattice as calcu- lated in the tight-binding approximation in (13) below: ~k = -a - 2 y (COS kxa + cos k,a + cos k,a) , (3) where a and y are constants. One reciprocal lattice vector of the sc lattice is G = (2~rla)i; if we add this vector to k the only change in ( 3 ) is cos kp -+ cos (k, + 2.rrla)a = cos (k,a + 27r) , but this is identically equal to cos k,a. The energy is unchanged when the wavevector is increased by a reciprocal lattice vector, so that the energy is a periodic function of the wavevector. Thrce different zone schemes are useful (Fig. 4): The extended zone scheme in which different bands are drawn in differ- ent zones in wavevector space. The reduced zone scheme in which all bands are drawn in the first Brillouin zone. The periodic zone scheme in which every hand is drawn in every zone. CONSTRUCTION OF FERMI SURFACES We consider in Fig. 5 the analysis for a square lattice. The equation of the zone boundaries is 2k.G + G~ = 0 and is satisfied if k terminates on the plane normal to G at thc midpoint of G. The first Brillouin zone of the square lattice is the area enclosed by the perpendicular bisectors of GI and of the three reci- procal lattice vectors equivalent by symmetry to G, in Fig. 5a. These four reci- procal lattice vectors are 2(2?r/a)& and ?(2.da)$. The second zone is constructed from G2 and the three vectors equivalent to it by syrn~netry, and similarly for the third zone. The pieces of the second and third zones are drawn in Fig. 5b. To determine the boundaries of some zones we have to co~lsider sets of several nonequivalent reciprocal lattice vectors. Thus the boundaries of sec- tion 3, of the third zone are formed from the perpendicular hisectors of three G's, namely (27r/a)k; ( 4 d a ) k ; and (27r/a)(& + 4). The free electron Fermi surface for an arbitrary electron concentration is shown in Fig. 6. It is inconvenient to have sections of the Fermi surface that belong to the same zone appear detached from one another. The detachment can be repaired by a transformation to the reduced zone scheme. We take the triangle labeled 2, and move it by a reciprocal lattice vector G = -(2?r/a)& such that the triangle reappears in the area of the first 9 Femi Surfaces and Metals 227 Figure 5 (a) Construction in k space of the first three Brillouin zones of a square lattice. The three shortest forms of the reciprocal lattice vectors are indicated as G,, G,, and G3. The lines drawn are the perpendicular bisectors of these G's. (b) On constructing all lines equivalent by symmetly to the three lines in (a) we obtain the regions in k space which form the first three Brillouin zones. The numbers denote the zone to which the regions belong; the numbers here are ordered according to the length of the vector G involved in the construction of the outer boundary of the region. Figure 6 Brillouin zones of a square lattice in two dimensions. The circle shown is a surface of constant energy for free electrons; it will be the Fermi surface for some particular value of the electron concentra- tion. The total area of the filled region ink space de- pends only on the electron concentration and is inde- pendent of the interaction of the electrons with the lattice. The shape of the Fermi surface depends on the lattice interaction, and the shape will not be an exact circle in an actual lattice. The labels within the sections of the second and third zones refer to Fig. 7. 1st zone 2nd zone 3rd zone Figure 7 Mapping of the first, second, and third Brillouin zones in the reduced zone scheme. The sections of the second zone in Fig. 6 are put together into a square by translation through an appropriate reciprocal lattice vector. A different G is needed for each piece of a zone. 1st zone 2nd zone 3rd zone Figure 8 The free electron Fermi surface of Fig. 6, as viewed in the reduced zone scheme. The shaded areas represent occupied electron states. Parts of the Fermi surface fall in the second, third, and fourth zones. The fourth zone is not shown. The first zone is entirely occupied. Figure 9 The Ferm~ surface in the thlrd zone as drawn m the peno&c zone scheme. The fiwre was - constmcted by repeating the third zone of Fig. 8. I I I I I I Brillouin zone (Fig. 7). Other reciprocal lattice vectors will shift the triangles Z6, 2,, 2d to other parts of the first zone, completing the mapping of the second zone into the reduced zone scheme. The parts of the Fermi surface falling in the second zone are now connected, as shown in Fig. 8. A third zone is assembled into a square in Fig. 8, but the parts of the Fermi surface still appear disconnected. When we look at it in the periodic zone scheme (Fig. 9), the Fermi surface forms a lattice of rosettes. Nearly Free Electrons How do we go from Fermi surfaces for free electrons to Fermi surfaces for nearly free electrons? We can make approximate constructions freehand by the use of four facts: The interaction of the electron with the periodic potential of the crystal creates energy gaps at the zone boundaries. Almost always the Fermi surface will intersect zone boundaries perpendicu- larly. 9 Femi Surfaces and Metals 2nd zone 3rd zone Figure 10 Qualitative impression of the effect of a weak periodic crystal potential on the Fermi surface of Fig. 8. At one point on each Fermi surface we have shown the vector gradk€. In the sec- ond zone the energy increases toward the interior of the figure, and in the third zone the energy increases toward the exterior. The shaded regions are filled with electrons and are lower in energy than the unshaded regions. We shall see that a Fermi surface like that of the third zone is elec- tronlike, whereas one like that of the second zone is holelike. Figure 11 Hanison construction of free elec- tron Fermi surfaces on the second, third, and fourth zones for a square lattice. The Fermi surface encloses the entire first zone, which therefore is filled with electrons. The crystal potential will round out sharp comers in the Fermi surfaces. The total volume enclosed by the Fermi surface depends only on the electron concentration and is independent of the details of the lattice interaction. We cannot make quantitative statements without calculation, but qualitatively we expect the Fermi surfaces in the second and third zones of Fig. 8 to be changed as shown in Fig. 10. Freehand impressions of the Fermi surfaces derived from free electron surfaces are useful. Fermi surfaces for free electrons are constructed by a pro- cedure credited to Harrison, Fig. 11. The reciprocal lattice points are deter- mined, and a free electron sphere of radius appropriate to the electron concentration is drawn around each point. Any point in k space that lies within at least one sphere corresponds to an occupied state in the first zone. Points within at least two spheres correspond to occupied states in the second zone, and similarly for points in three or more spheres. We said earlier that the alkali metals are the simplest metals, with weak in- teractions between the conduction electrons and the lattice. Because the alkalis have only one valence electron per atom, the first Brillouin zone bound- aries are distant from the approximately spherical Fermi surface that fills one- half of the volume of the zone. It is known by calculation and experiment that the Fermi surface of Na is closely spherical, and that the Fermi surface for Cs is deformed by perhaps 10 percent from a sphere. The divalent metals Be and Mg also have weak lattice interactions and nearly spherical Fermi surfaces. But because they have two valence electrons each, the Fermi surface encloses twice the volume in k space as for the alkalis. That is, the volume enclosed by the Fermi surface is exactly equal to that of a zone, but because the surface is spherical it extends out of the first zone and into the second zone. ELECTRON ORBITS, HOLE ORBITS, AND OPEN ORBITS We saw in Eq. (8.7) that electrons in a static magnetic field move on a curve of constant energy on a plane normal to B. An electron on the Fermi surface will move in a curve on the Fermi surface, because this is a surface of constant energy. Three types of orbits in a magnetic field are shown in Fig. 12. The closed orbits of (a) and (b) are traversed in opposite senses. Because particles of opposite charge circulate in a magnetic field in opposite senses, we say that one orbit is electronlike and the other orbit is holelike. Electrons in holelike orbits move in a magnetic field as if endowed with a positive charge. This is consistent with the treatment of holes in Chapter 8. In (c) the orbit is not closed: the particle on reaching the zone boundary at A is instantly folded back to B, where B is equivalent to B' because Hole orbit Electron orbit Ooen orbits Figure 12 Motion in a magnetic field of the wavevector of an electron on the Fermi surface, in (a) and (b) for Fermi surfaces topologically equivalent to those of Fig. 10. In (a) the wavevector moves around the orbit in a clockwise direction; in (b) the wavevector moves around the orbit in a counter-clockwise direction. The direction in (b) is what we expect for a free electron of charge -e; the smaller k values have the lower energy, so that the filled electron states lie inside the Fermi surface. We call the orbit in (b) electronlike. The sense of the motion in a magnetic field is opposite in (a) to that in (b), so that we refer to the orbit in (a) as holelike. A hole moves as a par- ticle of positive charge e. In (c) for a rectangular zone we show the motion on an open orbit in the periodic zone scheme. An open orbit is topologically intermediate between a hole orbit and an electron orbit. 9 Femi Surfaces and Metals Figure 13 (a) Vacant states at the comers of an almost-filled band, drawn in the reduced zone scheme. (b) In the periodic zone scheme the various parts of the Fermi surface are con- nected. Each circle forms a bolelike orbit. The different circles are entirely equivalent to each other, and the density of states is that of a single circle. (The orbits need not he true cir- cles: for the lattice shown it is only required that the orbits have fourfold symmetry.) Figure 14 Vacant states near the top of an almost filled band in a two- dimensional crystal. This figure is equivalent to Fig. 12a. Figure I5 Constant energy surface in the Brillouin zone of a simple cubic lattice, for the assumed energy band E~ = -a - 2y(cos k,a + cos k,a + cos k,a). (a) Constant energy surface E = -a. The filled volume contains one electron per primitive cell. (b) The same surface exhibited in the periodic zone scheme. The connectivity of the orbits is clearly shown. Can you find electron, hole, and open orbits for motion in a magnetic field BP? (A. Sommerfeld and H. A. Bethe.) they are connected by a reciprocal lattice vector. Such an orbit is called an open orbit. Open orbits have an important effect on the magnetoresistance. Vacant orbitals near the top of an otherwise filled band give rise to hole- like orbits, as in Figs. 13 and 14. A view of a possible energy surface in three dimensions is given in Fig. 15. Orbits that enclose filled states are electron orbits. Orbits that en- close empty states are hole orbits. Orbits that move from zone to zone without closing are open orbits. CALCULATION OF ENERGY BANDS \Vigner and Seitz, who in 1933 performed the first serious band calcula- tions, refer to afternoons spcnt 0 1 1 thc mani~al desk calc~~lators of those days, l~sing one afternoon for a trial wavefunction. Here we limit ourselves to three introductory methods: the tight-binding method, useful for interpolation; tlie Wiper-Seitz method, useful for the visualization and ur~derstar~dirig of the alkali metals; and the pseudopoter~tial method, utilizing the general theory of Chapter 7, which shuws tlie simplicity of many problems. Tight Binding Method for Energy Bands Lct 1 1 s start with nei~tral separated atoms and watch the changes in the atomic energy levels as the charge distributions of adjacent atoms overlap when the atoms are brought together to forrri a crystal. Consider twu hydrogen atoms, each wit11 a11 electron in the Is ground state. The wavefunctio~ls $,4, i,bB on tlie separated atoms are show~l in Fig. 16a. As the atoms are brought together, their wavcfunctions ovcrlap. 14Jc con- sider the two combinations clr, ? i,bB. Each cornhination shares an electron with thc hvo protons, hut an electron in the state i , b A + $ , will have a some- what lower energy than in the state J/, - I/J~. In I ) + i,bB the electron spends part of the time in the region rriidway between the two protons, and in this region it is in the attractive potential of both protons at once, thereby increasir~g tlie binding energy. I 1 1 i,bA - i,bB the density \,anishes midway between the nuclei; an extra binding docs not appear. (b? (c! Figure 16 (a) Schematic drawing of wavefunctions of electrons on two hydrogen atoms at largr separatio11. (b! Gruu~~d state waucfunctiorr at closcr separation. (c) Excited state wavefunction. 9 Fermi Surfaces and Metals 233 0 1 2 3 4 5 Nearest-neighhor distan~x, in Bohr radii Figure 17 The Is band of a ring of 20 hydrogen atoms; the one-electron energies are calculated in the tight-binding approxi- nlation with the nearest-neighbor overlap integral of Ey. (9). .4s two atoms arc brought together. two separated energy levels are formed for each lcvel of the isolated atom. For N atoms, N orbitals are formed for each orhital of the isolated atom (Fig. 17). As free atoms are brought together, the coulomb interaction between the aton1 cores and the electron splits the energy levcls, spreading them into bands. Each state of given quantum number of the free atom is spread in the cYstal into a band of energies. The width of the hand is proportional to the strength of the overlap interaction between neighboring atoms. There will also hc hands formed from p, d, . . . states (I = 1, 2, . . .) of the lree atoms. States degenerate in the free atom will form different bands. Each will not have the same energy as any other band over any substantial range of the wavevector. Bands nlay coincide in energy at ccrtain values of k in the Brillouin zone. The approximation that starts out from the wavefunctions of the free atoms is krlown as the tight-binding approximation or the LCAO (linear co~ribi~~ation of atomic orbitals) approximation. The approximation is quite good for the inner electrons of atoms: hut it is not often a good description of the conduction clec- trans themselves. It is used to describe approximately the d bands of the transi- tion metals and the valence bands of diamondlike and inert gm crystals. Suppose that the ground state of an electron moving in the potential U(r) of an isolated atom is p(r), an s state. The treatment of bands arising from degenerate (p, d, . . .) atomic levels is more complicated. If the influence of one atom on anothcr is small, we obtain an approximate wavefunction for one electron in the whole crystal by talang where the sum is over all lattice points. We assume the primitive hasis contains one atom. This function is of the Bloch form (7.7) if Cki = N-li2 e , which gives, for a crystal of N atoms, We prove (5) is of the Bloch form. Consider a translation T connecting two lattice points: exactly the Bloch condition. We find the first-order energy by calculating the diagonal matrix elements of the hamiltonian of the crystal: where q, - ~ ( r - r,). Writing p , = r,,, - rj, We now neglect all integrals in (8) except those on the same atom and those between nearest neighbors connected by p. We write $ dV cpe(r)Hq(r) = -a ; J " dV p(r - p)Hq(r) = - y ; (9) and we have the first-order energy, provided (klk) = 1: The dependence of the overlap energy y on the interatomic separation p can be evaluated explicitly for two hydrogen atoms in Is states. In rydberg energy units, Ry = me4/2fi2, we have where no= fi2/nlnle? The overlap energy decreases exponentially with the separation. 9 Fermi Surfaces and Metals 235 For a simple cubic structure the nearest-neighbor atoms are at so that (10) becomes ck = -a - ~ ~ ( C O S kp + cos kya + cos kg) . (13) Thus the energies are confined to a band of width 127. The weaker the over- lap, the narrower is the energy band. A constant energy surface is shown in Fig. 15. For ka << 1, ck = -a - 67 + yk2a2. The effective mass is m = fi2/2ya2. When the overlap integral y is small, the band is narrow and the effec- tive mass is high. We considered one orbital of each free atom and obtained one band EL. The number of orbitals in the band that corresponds to a nondegenerate atomic level is 2N, for N atoms. We see this directly: values of k within the first Brillouin zone define independent wavefunctions. The simple cubic zone has -ria < k, < via, etc. The zone volume is 8n3/a3. The number of orbitals (counting both spin orientations) per unit volume of k space is V / 4 g , SO that the number of orbitals is 2vla3. Here V is the volume of the crystal, and l/a3 is the number of atoms per unit volume. Thus there are 2N orbitals. For the fcc structure with eight nearest neighbors, For the fcc structure with 12 nearest neighbors, ek = -a - ~ ~ ( C O S i kya cos i k,a + cos $ k,a cos i kp + cos ;kg cos i k,a) . (15) A constant energy surface is shown in Fig. 18. Figure 18 A constant energy surface of an fcc crystal structure, in the nearest-neighbor tight-binding approx- imation. The surface shown has E = -a + 21yl. Wigner-Seitz Method Wigner and Seitz showed that for tlie alkali metals there is no inconsis- tency between the electron wa\d'unctions of frcc atoms and the nearly free electron model of the band structure of a crystal. Over most of a band the energy may depend on the wavevector nearly as for a free electron. However the Bloch wavefunction, unlike a plane wave, will pile up charge on the posi- tive ion cores as in the atomic wavefunction. A Blocli functiori satisfies the wave equation With p = -ih grad, we have p e l k 'uk(r) = R k e'kruk(r) + elhpuk(r) , p2ezhuk(r) = (hk)2e'"uk(r) + efk'(2hk .p)uk(r) + eZhp2uk(r) , thus tlie wave equation (16) Irlay be written as an equation for uk: At k = 0 we have go = uo(r), where u,(r) has the periodicity of the lattice, see: the ion cores, and near them will look like the wavefunction of the free atom. It is much easier to find a solution at k = 0 than at a general k, because at k = 0 a nondegenerate solution will have the full syrnmetry of Cir(r), that is, ol the crystal. We can then use uo(r) to construct the approximate solution This is of the Bloch form, but u, is not an exact solution of (17): it is a soliltion only if we drop the term in k.p. Often this term is treated as a perturbation, a: in Problem 8. The k.p perturbation theory developed there is especially useful in finding the effective mass m at a band edge. Because it takes account of the ion core potential the function (18) is a much better approximation than a plane wave to the correct wavefunction The energy of the approximate solution depends on k as (hI~)~/2rn, exactly as for the plane wave, even though the modulation represented by uO(r) may be very strong. Because u, is a solution of the function (18) has the e n c r o expectation valne E,+ (h2k2/27n). The function u,,(r) oftcn will give 11s a good picture of the charge distribution within a cell. 9 F e m i Surfaces and Metals 237 0 1 2 3 4 r (Bohr units) Figure 19 Radial u~avefimctions for the 3y orbital of free sodium atom and for the 3s conduction band in sodium metal. The wavefunctions, which are nnt normalized here, are found by integrat- iug the Sclirodingcr equation for an electron in the potential urell of an Na' ion core. For the free atom the wavefunction i s integrated subject to the usual Sctlrudinger bou~idary condition +(r) + 0 as r + m; the energy eigenvalue is -5.15 eV. The wa\refunction for wavevector k = 0 in the metal is subject to the U'igner-Seitz boundary condition that d+/dr = 0 when r is midway between neighboring atoms; the energy of this orhital is 8 . 2 eV, considerably lower tl~an fur the free atom. The orbitals at the zone boundaly are not filled in sodium; their energy is +2.7 eV (After E. Wigner and F. Seitz.) W'igner and Seitz developed a simple and fairly accurate method of calcu- lating u,,(r). Figure 19 shows the Wigner-Seitz wavefunction for k = 0 in the 3s conduction band of metallic sodium. The function is practically constant over 0.9 of the atomic volume. To the extent that the solutions for higher k may be approximaterl by exp(ik . r)u,(r), the wavefunctions in the conduction hand will he similar to plane waves over most of the atomic volume, but in- crease markedly and oscillate within the ion core. Cohesive Energy. The stability of the simple metals with respect to free atoms is caused by the loweri~~g of the energy of the Bloch orbital with k = 0 in the crystal compared to the ground valence orbital of the free atom. The effect is illustrated in Fig. 19 for sodium and in Fig. 20 for a linear periodic - potential of attractive square wells. The ground orbital energy is much lower (because of lower kinetic energy) at the actual spacing in the metal than for isolated atoms. A decrease in ground orbikal energy will increase the binding. The decrease in ground orbital energy is a consequence of the change in the boundary condi- tion on the wavefu~lction: The Schrodinger boundary condition for the free atom is $(r) 4 0 as r 4 a. In thr crystal the k = 0 wavefunction u,(r) has the symmetry of the lattice and is symmetric about r = 0. To have this, the normal derivative of $ must vanish across every plane midway between adjacent atoms. .f e , I I w n 1 2 3 bla + Figure 20 Ground orbital (k = 0) energy for an electron in a periodic square well potential of depth I U o l = 2h21m2. The energy is lowered as the wells come closer together. Here a is held con- stant and b is varied. Large bla corresponds to separated atoms. (Courtesy of C. Y. Fong.) Figure 21 Cohesive energy of sodium metal is the dif- ference between the average energy of an electron in the metal (-6.3 eV) and the ground state energy (-5.15 eV) of the valence 3s electron in the free atom, referred to an Na+ ion plus free electron at infinite separation. Metal 5.15 eV Cohesive energy ---- ------ 4 , 3 e V -8.2 eV k = 0 state In a spherical approximation to the shape of the smallest Wigner-Seitz cell we use the Wigner-Seitz boundary condition where ro is the radius of a sphere equal in volume to a primitive cell of the lat- tice. In sodium, r - 3.95 Bohr units, or 2.08 A; the half distance to a nearest O . - neighbor is 1.86 A. The spherical approximation is not bad for fcc and bcc structures. The boundary condition allows the ground orbital wavefunction to have much less curvature than the free atom boundary condition. Much less curvature means much less kinetic energy. In sodium the other filled orbitals in the conduction band can be repre- sented in a rough approximation by wavefunctions of the form (18), with 9 Permi Surfaces and Metals 239 Tlie Fer~ni energy is 3.1 eV, from Table 6.1. The average kinetic energy per electron is 0.6 of the Fermi energy, or 1.9 eV. Bccause co = -8.2 eV at k = 0, the average clcctron cncrgy is ( E ~ ) = -8.2 + 1.9 = -6.3 eV, compared with -5.15 eV for the valence electron of the free atom, Fig. 21. \Ve therefore estimate that sodium metal is stable by about 1.1 eV with respect to the free atom. This result agrees well with the experimental value 1.13 eV. Pseudopotential Methods Conduction electron wavefimctions are usually smoothly varying in the re- gion between the ion cores, but have a complicated nodal structure in the re- gion of the cores. This behavior is illustrated by the ground orbital of sodium, Fig. 19. It is helpful to view the nodes in the conduction electron wavefunction in the core regon as created by the requirement that the function be ortho- g o ~ ~ a l to the wavefunctions of the core electrons. This all comes out of the Schrodinger equation, hut we can see that we nced the flexibility of two nodes in the 3 s conduction orbital of La in order to be orthogonal both to the 1s core orbital w i t h no nodes and the 2 7 core orhital with one node. Oiitside the core the potential energy that acts on the conduction electron is relatively weak: the potential energy is only the coulomb ~otential of the singly-charged positive ion cores and is reduced markedly by the electrostatic screer~ing of tlie other conduction electrons, Chapter 14. In this outer region the conduction electron wavefunctions are as smoothly varying as plane waves. If the conduction orbitals in this outer rcgion arc approximately plane wavcs, thc cncrgy must dcpend on the wavevector approximately as ek = fi2k2/2m as for free electrons. But how do we treat the conduction orbitals in the core region where the orbitals are not at all like plane waves? What goes on in the core is largely irrelevant to the dependence of on k. Recall that we can calculate tlie energy by applying the liamiltonian operator to an orbital at any point in space. Applied in the outer region, this opcration will give an energy nearly cqual to thc frcc electron energy. This argument leads nati~rally to the idea that we might replace the actual potential energy (and filled shells) in the core region by an effective potential energy' that gives the same wavefunctions outside the core as are given by the actual ion cores. It is startling to find that tlie effective potential or 'J. C. Phillips and L. Klci~i~nan, Phys. Rcv. 116, 287 (1959); E. Antoncik, J. Phys. Chem. Solids 10, 314 (1959). The ~eneral theory of pseudopotentials is disci~ssed by B. J. Anstin, V . Heine, and L. J. Sham, Phys. Rev. 127; 276 (1962); see also Vol. 24 of Solid state physics. The utility of the empty cure model has beer1 known fur niany years: it gocs back to E. Fcrmi, Nuovo Cimento 2, 157 (1934); H. Hellmann, Acta Physiochimica URSS 1, 91.3 (1935); and H. Hellmann a d TV, K~assatotschkin, J. Chem. Phys. 4, 324 (1936), who wrote "Since the field of the ion determined in thls way runs a rather flat course, it is sofficicnt in thc first approximation to set the valence electron in the lattice equal to a plane wave." pseudopotential that satisfies this requirement is nearly zero. This co~lclusion about pseudopotentials is supported by a large amount of empirical experience as well as by theoretical arguments. The result is referred to as the cancella- tion theorem. The pseudopotential for a problem is not unique nor exact, but it may be very good. On the Empty Core Model (ECM) we can even take the uxi- screened pseudopotential to be zero inside some radius 8,: 0 , f o r r < R , ; U(r) = -e2/r , for r > Re . This potential should now be screened as described in Chapter 10. Each com- ponent U(K) of U(r) is to be divided by the dielectric constant e(K) of the electron gas. If, just as an exampIe, we use the Thomas-Fermi dielectric func- tion (14.33), we obtain the screened pseudopotential plotted in Fig. 22a. Figure 22a Pseudopotential for metallic sodium, based on the empty core model and screened by thc Thomas-Fermi dielectric function. The calculations were made for an empty core radius R, = I.fiha,, where a, is the Bohr radiuu, and for a screening paranictcr k,a, = 0.79. The dashed curve shows the assumed unscreened potential, as from (21). The dotted cnrve is the actual potential of tlrc ion core; other v-dues of U(r) are -50.4, -11.6, and -4.6, for r = 0.15, 0.4, and 0.7, respectively. Th~ls the act~ial potential of the ion (chosen to fit the energy Ic\,els oi the free atom) is vely much larger than the pseudopotential, over 200 times larger at r = 0.15. 4 .
2 r, in uruts of Bohr radii + n I 3 4 0.6 C 3 + .
; -1- a .B - 3 -2 .... .... - : Ionic poteutid 9 Fenni Surfaces and Metab 241 6 3 61 6 2 A I I U'avevector k Figure 22b A typical reciprocal space pqeodopotential. Values of U(k) tirr warevec- tors equal to the reciprocal lattice vectors, 6, arc indicated by the dots. For very small k the potential apprnaches (-213) times the Fermi energy, which is the screened-ion limit for 111etals. (After M. L. Cohcn.) The pseudopotential as drawn is much wcaker than the tnie potential, hut the pseudopotential was adjusted so that the wavefunction in the outer region is ncarly identical to that for the true potential. In the language of scattering theory, we adjust the phase shifts of the pseudopotential to match those of the true potential. Calculation of the band structure depe~ids only 0 1 1 the Fourier compo- nents of t l ~ e pseudopotential at the reciprocal lattice vectors. Usually only a few values of the coefficients U(G) are needed to get a good band structure: see the L1(G) in Fig. 22b. These coefficients arc sometimes calc~~lated from modcl potentials, and sometimes they are obtained from fits of tentative band stnicti~res to the results of optical measurements. Good values of U ( 0 ) can be estimated from first principles; it is shown in (14.43) that for a screened coulornb potential U(0) = -gtF. In the re~narkably successful Elnpirical Pseudopotential Method (EPM) the band structure is calculated using a few coefficients U(G) deduced from theoretical fits to measurements of the optical reflectance and absorption of crystals, as discusscd in Chapter 15. Charge density maps can be plotted from the wavefimctions generated by the EPM-see Fig. 3.11. The results are in excellent agreement with x-ray diffraction determinations; such maps give an understanding of the bonding and have great predictive value for proposed new structures and conlpounds. The EPM values of the coemcients U(G) often are additive in the contri- butions of the several types of ions that are prcscnt. Thus it may he possible to construct thc C'(G) for entirely new structures, starting from results on known structures. Further, the pressure dependence of a band structure may be de- ternlined when it is possible to estimate from the form of the U(r) curve the dependence of U(G) on srnall variations of G. It is often possible to calculate band structures, cohesive energy, lattice constants, and bulk moduli from first principles. T n such ah initio pscudo- potential calculations the basic inputs are the crystal structure type and the atomic number, along with well-tested theoretical approximations to exchange energy terms. This is not the same as calculating from atomic number alone, but it is the most reasonable basis for a first-principles calculation. The results of Yin and Cohen are compared with experiment in the table that follows. Lattice Cohesive Bulk modulus constant (A) energy (eV) (Mbar) Silicon Calculated 5.45 Experimental 5.43 Germanium Calculated 5.66 Experimental 5.65 Diamond Calculated 3.60 Experimental 3.57 EXPERIMENTAL METHODS IN FERMI SURFACE STUDIES Powerful experimental methods have been developed for the determina- tion of Fermi surfaces. The methods include magnetoresistance, anomalous skin effect, cyclotron resonance, magneto-acoustic geometric effects, the Shubnikow-de Haas effect, and the de Haas-van Alphan effect. Further infor- mation on the momentum distribution is given by positron annihilation, Compton scattering, and thc Kohn cffcct. We propose to study one method rather thoronghly. All the methods are useful, but need detailed theoretical analysis. \Ve select the de Haas-van Alphen effect because it exhibits very well the characteristic periodicity in 1/B of the properties of a metal in a uniform magnetic field. Quantization of Orbits in a Magnetic Field The momentum p of a particle in a magnetic Geld is the sum (Appendix 6) of two parts, the kinetic rnomcntum pldn = mv = fik and the potential momcn- tnm or field momentum pfield = qA/c, where q is the charge. The vector poten- tial is related to the magnetic field by B = curl A. The total momentum is In SI the factor c-' is omitted 9 F e m i Surfaces and Metals 243 Following the semiclassical approach of Onsager and Lifslutz, we assume that the orbits in a magnetic field are quantized by the Bohr-Sommerfeld rclation when n is an integer and y is a phase correction that for free electrons has the vallle i. Then The equation of motion of a particle of charge q in a magnetic field is We integrate with respect to time to give apart from an additive constant which does not contribute to the final result. Thus one of the path integrals in (24) is where @ is the magnetic flux contained within the orbit in real space. We have used the geometrical result that f r X dr = 2 X (area enclosed by the orhit) The other path integral in (24) is by the Stokes theorem. Here dais the area element in real space. The momen- tum path integral is the sum of (25b) and (25c): It follows that the orbit of an electron is quantized in such a way that the flux through it is @, = (n + y)(Z&/e) . (27) The flux llnit 2&/e = 4.14 X gauss crnZ or Tm2. In the de Haas-van Alphen effect discussed below we need the area of the orbit in wavevector space. We obtained in (27) the flux through the orbit in real space By (25a) we know that a line element Ar in the plane normal to B is rclatcd to A k by Ar = (fic/~R)Ak, so that thr area S, in k spacr is rclatcd to the area A, of the orbit in r ypace by A,, = (L/~-.B)~S,, (28) It follow^ that from (27), \,hence the area of an orbit in k space will satisly In Ferrni surface experirrients we may be interested in the increment A B for which two successive orbits, and n + 1, have the same area in k space on the Fermi surlace. The areas are equal when from (30). We ha-e the important result that equal increments of 1/B repro- duce similar orbits-this periodcity in 1IB is a striking feature of the magneto- oscillatory effects in metals at low temperatures: resistivity, susceptibility heat capacity. The population of orbits on or near the Fermi surface oscillates as B is var- ied, causing a wide variety of eflects. From the period of the oscillation we reconstruct thc Fcrmi surf'acc. Thc result (SO) is indcpcndcnt of thr galigc of the vector potrntial uscd in thc cxprcssion (22) for momcntnm; that is, p is not gauge invariant, but S, is. Gauge invariance is discussed further in Chapter 10 and in Appendix G. De Haas-van Alphen Effect The de Raas-van Alphen effect is the oscillation of the magnetic moment of a metal as a function of the static magnetic field intensity. The effect can be observed in pure specimens at low temperatures in strong magnetic fields: we do not want the quantizatio~l of the electron orbits to be blurred by collisions, and we do not want the population oscillations to be averaged out by thermal population of adjacent orbits. The analysis of thc dHv.4 cff'cct is givcn tbr absolutc zcro in Fig. 23. The electron spin is neglected. The treatment is given for a two-dimensional (2D) system; in 3D we need only multiply the 2D wavefunction by plane wave factors exp(ik,z), where the magnetic field is parallel to the z axis. The area of an orbit in k,, k,, space is quantized as i 1 1 (30). Tl~e area between successive orbits is AS = S, - = 2veB/fic . (32) 9 Femi Surfaces and Metals 245 (a) (h) (4 (d) (4 Figure 23 Explanation of the de Haas-van Alphen effect for a free electron gas in two dimen- sions in a magnetic field. The filled orbitals of the Fermi sea in the absence of a magnetic field are shaded in a and d. The energy levels in a magnetic field are shown in b, c, and e. In b the field has a value B, such that the total energy of the electrons is the same as in the absence of a magnetic field: as many electrons have their energy raised as lowered by the orbital quantization in the mag- netic field B,. When we increase the field to B, the total electron energy is increased, because the uppermost electrons have their energy raised. In e for field B, the energy is again equal to that for the field B = 0. The total energy is a minimum at points such as B,, B,, B,, . . . , and a maximum near points such as B2, B,, . . . . The area in k space occupied by a single orbital is (2rlL)', neglecting spin, for a square specimen of side L. Using (32) we find that the number of free electron orbitals that coalesce in a single magnetic level is where p = e~'/2&, as in Fig. 24. Such a magnetic level is called a Landau level. The dependence of the Fermi level on B is dramatic. For a system of N electrons at absolute zero the Landau levels are entirely filled up to a magnetic quantum number we identify by s, where s is a positive integer. Orbitals at the next higher levels + 1 will be partly filled to the extent needed to accommo- date the electrons. The Fermi level will lie in the Landau level s + 1 if there are electrons in this level; as the magnetic field is increased the electrons move to lower levels. When s + 1 is vacated, the Fermi level moves down abruptly to the next lower levels. The electron transfer to lower Landau levels can occur because their degeneracy D increases as B is increased, as shown in Fig. 25. As B is (a) (b) Figure 24 (a) Allowed electron orbitals in two dimensions in absence of a magnetlc field. (b) In a magnetic field the points which represent the orbitals of free electrons may he viewed as re- stricted to circles in the former kAY plane. The successive circles correspond to successive values of the quantum number n in the energy (n - f )ko,. The area between successive circles is The angular position of the points has no significance. The number of orbitals on a circle is con- stant and is equal to the area between successive circles times the number of orbitals per unit area in (a), or (2?reB/fic)(L/2~)~ = L2eB/2dc, neglecting electron spin. Figure 25 (a) The heavy line gives the number of particles in levels which are completely occu- pied in a magnetic field B, for a two-dimensional system with N = 50 and p = 0.50. The shaded area gives the number of particles in levels partially occupied. The value of s denotes the quantum number of the highest level whlch is completely filled. Thus at B = 40 we have s = 2; the levels n = 1 and n = 2 are filled and there are 10 particles in the level n = 3. At B = 50 the level n = 3 is empty. (b) The periodicity in 1/B is evident when the same points are plotted against 1/B. increased there occur values of B at which the quantum number of the upper- most filled level decreases abruptly by unity. At the critical magnetic fields labeled B, no level is partly occupied at absolute zero, so that 9 Fermi Sulfaces and Metals 247 Figure 26 The upper curve is the total electronic energy versus 1IB. The oscillations in the en- ergy U may be detected by measurement of the magnetic moment, given by - a U l a ~ . The thermal and transport properties of the metal also oscillate as successive orbital levels cut through the Fermi level when the field is increased. The shaded region in the figure gives the contribution to the energy from levels that are only partly filled. The parameters for the figure are the same as for Fig 25, and we have taken the units of B such that B = fiw,. The number of filled levels times the degeneracy at B, must equal the number of electrons N. To show the periodicity of the energy as B is varied, we use the result that the energy of the Landau level of magnetic quantum number n is E, = ( n - ;)Tim,, where w, = eBlmc is the cyclotron frequency. The result for E , follows from the analogy between the cyclotron resonance orbits and the simple harmonic oscillator, but now we have found it convenient to start counting at n = 1 instead of at n = 0. The total energy of the electrons in levels that are fully occupied is where D is the number of electrons in each level. The total energy of the electrons in the partly occupied levels + 1 is where s D is the number of electrons in the lower filled levels. The total energy of the N electrons is the sum of (35) and (36), as in Fig. 26. The magnetic moment p of a system at absolute zero is given by p = -8UIaB. The moment here is an oscillatory function of 1/B, Fig. 27. This os- " cillatory magnetic moment of the Fermi gas at low temperatures is the de ~aas-van ~ l p h e n effect. From (31) we see that the oscillations occur at equal intervals of 1/B such that Figure 27 At absolute zero the magnetic moment is given by -BUBB. The energy plotted in Fig. 26 leads to the magnetic moment shown here, an oscillatoly function of 1IB. In impure speci- mens the oscillations are smudged out in part because the energy levels are no longer sharply defined. A' Magnetic - Figure 28 The orbits in the section AA' are ex- field tremal orbits: the cyclotron period is roughly con- stant over a reasonable section of the Fermi surface. Other sections such as BB' have orbits that vary in period along the section. a where S is the extremal area (see below) of the Fermi surface normal to the di- rection of B. From measurements of A(l/B), we deduce the corresponding ex- tremal areas S: thereby much can be inferred about the shape and size of the Fermi surface. Extremal Orbits. One point in the interpretation of the dHvA effect is sub- tle. For a Fermi surface of general shape the sections at different values of k, will have different periods. Here k, is the component of k along the direction of the magnetic field. The response will be the sum of contributions from all sections or all orbits. But the dominant response of the system comes from or- bits whose periods are stationary with respect to small changes in k , . Such 9 Fermi Sulfaces and Metals 249 orbits are called extremal orbits. Thus, in Fig. 28 the section AA' dominates the obsewed cyclotron period. The argument can be put in mathematical form, but we do not give the proof here (QTS, p. 223). Essentially it is a question of phase cancellation; the contributions of different nonextremal orbits cancel, but near the extrema thc phase varies only slowly and there is a net signal fronr these orbits. Sharp resonances are obtained even from complicated Fermi surfaces because the experiment selects the extermal orbits. Fermi Surface of Copper. The Fermi surface of copper (Fig. 29) is distinctly nonspherical: eight necks make contact with thc hexagonal faces of the first Brillouin zone of the fcc lattice. The electron concentration in a monovalent metal with an fcc structure is n = 4/a3; there are four electrons in a cube of volume a3. Thc radius of a free electron Fermi sphere is and the diameter is 9.801~. The shortest distance across the Brillouin zone (the distance between hexagonal faces) is ( 2 ~ / a ) ( 3 ) ~ ' ~ = 10.88/a, somewlrat larger than the diameter of the free electron sphere. The sphere does not touch the zone boundary, but we know that the presence of a zone boundary tends to lower the band energy near the boundary. Thus it is plausible that the Fermi surface should neck ont to meet the closest (hexagonal) faces of the zone (Figs. 18 and 29). The square faces of thc zone are more distant, with separation 12.57/a, and the Fermi surface does not neck out to meet these faces. EXAMPLE: Fermi Surface o f Gold. In gold for quite a wide range of field directions Shoenberg finds the magnetic moment has a period of 2 x lO-\auss-'. This period corresponds to an extremal orbit of area From Tahlc 6.1, we have k , = 1.2 X 10' cm-' for a free electron Fermi sphere for gold, or an cxtrcmal area of 4.5 X 1016 cm-', in general agreement with the experi- mental value. Thc actual periods reported by Shoenberg are 2.05 x lo-' gauss-' and 1.95 X I()-' gauss-1. In thc [Ill] direction in A u a large period of 6 X lo-@ gauss-' is also found; tlir corresponding orbital area is 1.6 X l O I 5 cm-% This is the "neck orbit N. Another extrernal orbit, the "dog's bone," is shown in Fig. 30; its area in A u is about 0.4 of the belly area. Experimental results are shown in Fig. 31. To do the example in SI, drop c from the relation for S and use as the period 2 X tesla-'. Figure 29 Fermi surface of copper, after Pippard. The Figure 30 Dog's bone orbit of an electron on Brillouin zone of the fcc structure is the truncated octa- the Fermi surface of copper or gold in a mag- hedron derived in Chapter 2. The Fermi surface makes netic field. This orbit is classified as holelike be- contact with the boundary at the center of the hexagonal cause the energy increases toward the interior of faces of the zone, in the [Ill] directions in k space. Two the orbit. "belly" extremal orbits are shown, denoted by B; the extremal "neck" orbit is denoted by N. 45.0 kG 45.5 kG 46.0 k~ Figure 31 De Haas-van Alphen effect in gold with B 1 1 . The oscillation is from the dog\ bone orbit of Fig. 30. The signal is related to the second derivative of the magnetic moment with respect to field. The results were obtained by a field modulation technique in a high-homogeneity superconducting solenoid at about 1.2 K. (Courtesy of I. M. Templeton.) The free electron Fermi sphere of aluminum fills the first zone entirely and has a large overlap into the second and third zones, Fig. 1. The third zone Fermi surface is quite complicated, even though it is just made up of certain pieces of the surface of the free electron sphere. The free electron model also gives small pockets of holes in the fourth zone, but when the lattice potential is 9 Fenni Surfaces and Metals 251 Figure 32 Multiply-connected hole surface of magnesium in bands 1 and 2; according to L. M. Falicuv. (Drawing by Marta Puebla.) High magnetic field Weak magnetic field (a) (b) Figure 33 Brwakdown of band structure by a strong magnetic field. Brillouin zone boundaries are the light lines. The free electron orhits (a) in a strorlg field changc connectitity in a weak field (b) to become open orbits in the first hand and electron orbits in the second band. Both bauds are mapped together. taken into account these empty out, the electrons being added to the third zone. The general features of the predicted Fermi surface of aliiminum are quite well verified by experiment. Figure 32 shows part of the free electron Ferrr~i surface of magnesium. Magnetic Breakdown. Electrons in slifficiently high magnetic fields will movc in frcc particle orbits, the circular cyclotron orbits of Fig. 33a. Here the magnetic forces are dominant, and the lattice is a slight perturbation. In this limit the classification of the orbitals into bands may have little impor- tance. However, we know that at low magnetic fields the motion is described hy (8.7) with the band structure ek that obtains in the abscncc of a magnetic field. The eventual breakdown of this description as the magnetic field is in- creased is called magnetic hreakdown. The passage to strong magnetic fields may drastically change the connectivity of the orbits, as in the figure. The onset of magnetic breakdown will be revealed by physical properties such as magnetoresistance that depend sensitively on the connectivity. The condition for magnetic breakdown is that hw,~, > Ei, approximately. Here e, is the free electron Fermi energy and E, is the energy gap. This condition is much milder, especially in metals with small gaps, than the nalve condition that the mag- netic splitting Rw, exceed the gap. Small gaps may be found in hcp metals where the gap acrvss the hexagonal face of the zone would be zero except for a small splitting introduced by the spin-orbit interaction. In Mg the splitting is of the order of eV; for this gap and E~ - 10 eV the breakdown condition is hw, > 1 0 5 eV, or B > 1000 C,. SUMMARY A Fermi surface is the surface in k space of constant energy equal to E,. The Fer~ni surface separates filled states from empty states at absolute zero. The form of the Fermi surface is usually exhibited best in the reduced zone scheme, but the connectivity of the surfaces is clearest in the periodic zone scheme. An energy band is a single branch of the GI, versus k surface The cohesion of simple metals is accounted for by the lowering of energy of the k = 0 conduction band orbital when the boundary conditions on the wavefunction are changed from Schrodinger to Wigncr-Seitz. The periodicity in the de Haas-van Alphen effect measures thc rxtremal cross-section area S in k space of the Fermi silrface, the cross section being taken perpendicular to B: Problems 1. Brillouin zones of rectangular lattice. Make a plot of the first two Brillouin zones of a primitive rectangular two-dimensional lattice with axes a, b = 3a. 2. Brillouin zone, rectangular lattice. A two-dimensional metal has one atom of vdency one in a simple rectangular primitive cell a = 2 6; b = 4 A. (a) Draw the first Brillouin zone. Give its dimensions, in cm-'. (b) Calculate the radius of the free electron Fermi sphere, in cm-'. (c) Draw this sphere to scalc on a drawing of the first Brillouin zone. Make another sketch to show thc first fcw periods of thc free electron band in the periodic zonc schcmc, for both the Grst and sccond cn- ergy bands. Assume there is a small energy gap at thc zone boundaly. 3. Hexagonal close-packed structure. Consider the Grst Brillnuin zone of a crystal with a simple hexagonal lattice in three dimensions with lattice constants a and c. Let G, denote the shortest reciprocal lattice vector parallel to the c axis of the 9 Fenni Surfaces and Metals 253 crystal lattice. (a) Show that for a hexagonal-close-packed crystal stn~ctllre the Fourier corrrponent UiG,) of the crystal potential U[r) is zero. (b) Is U(2G,) also zero? (c) \%'Ililly is it possihle in principle to obtain an insulator made up of divalent atoms at the lattice points or a simple hexagonal lattice? (d) Why is it not possihle to obtain an irrsulator made up of monovalent atoms in a hexagonal-close-packed structure? 4. Brillouin zones of two-dimensional divalent metal. A two-dimensional metal in the form of a square lattice has two conduction electrons per atom. In the d- most free electron approxirriation. sketch carefully the electron and hole energy surfaces. For the electror~s chnosr a zonc scheme such that the Fermi surface is shown as closed. 5. Open orbits. An open orbit in a monovalent tetragonal metal connects opposite faces of the boundary of a Brillonin zonc. The faces are separated by G = 2 X 10%m '. A magnetic field B = 10'' ganss = lo-' tesla is normal to the plane of the open orbit. (a) What is the order of magnitude of the period of the motion in k space? Take c = 10' cm/sec. (b) Descrihc in real space the motion of an electron on this orbit in the presence of tlie magnetic field. 6. Cohesive energy for a square well potentiul. (a) Find an expression for the hinrling energy of an electron in one dimension in a single square well of depth U, and width a. (This is the standard first problem in elerrlerltary quantum mechan- ics.) Assume that the solution is symmetric about the rrridpoint of the wcl. (b) Find a nirmcrical result for the binding energy in terms of Uo for the special case IUo1 = 2K4nm%nd conlpare with the appropriate limit of Fig. 20. In this limit of widely separated \vclls the band width goes to zero, so the energy fork = 0 is the same as the enrrm for any other k in the lowest energy band. Other bands are formed frorr~ tlir excited states of the well, in this limit. 7 . De Hans-van Alphen period of potassium. (a) Calculate the period A(1IB) ex- pected for potassium or1 d1c frce electron model. (b) What is the area in real space of the extrslrral orhit, for B = 10 kG = 1 T? The same period applies to os- cillations in the electrical resistiviky, )sown as the Shubnikow-de Haas effect. '8. Band edge structure on k - p perturbation theory. Consider a nondegrrrerate orbital $nk at k = 0 1 1 1 the band n of a cubic crystal. Use second-order perturba- t~on theory to find the re~ult where the surrl is over a 1 1 othcr orbitals ICrJk at k = 0. The effective mass at this point is 'This problem is surrlewhat difficult. The mass at the conduction band edge in a narrow gap semicond~~ctor is often dominated by the effect of the valence band edge, whcnce where the sum is over the valcnce hands; Eg is the energy gap. For given matrix elements, small gaps lead to small masses. 9. Wannier functions. Thc Wannier functions of a band are defined in terms of the Bloch functions of thc same band hy ~ ( r - r,,) = N- In Z e x p - ik . r,) $k(r) ; k ( 4 2 ) where r, is a lattice point. (a) Prove that Wannier functions about different lattice points n,m are orthogonal: This orthogonality property makes the functions often of greater use than atomic orbitals centered on different lattice sites, because the latter are not generally or- thogonal. (b) The Wannier functions are peaked around the lattice sites. Show that for $k = N-'" elkx U&X) the W7annier {unction is sin v(x - x,)/a w(x - x") = u0(x) r(x - x,J\a for N atoins on a line of lattice constant a 10. Open orbits and magnetoresistance. We considered the transverse magncto- resistance of free electrons in Problem 6.9 and of electrons and holes in Problcm 8.5. In some crystals the magnetoresistance saturates except in spccial crystal nri- entations. An open orbit carries current only in a single direction in the plane normal to the magnetic field; such carriers are not deflected by thc ficld. In the arrangement of Fig. 6.14, let the open orbits be parallel to k,; in real space these orbits carry current parallel to the y axis. Let a?,, = su0 bc the cond~lctivity of the open orbits; this defines the constant s. The magnetncond~~ctivity tensor iri the high field limit w,r B 1 is Q-2 -0-1 0 uo(Qil ; : ) with Q = w,r. (a) Show that the Hall field is E, = -EJsQ. (b) Show that the ef- fectivc rcsistivity in the x directiorr is p = (Q2/uo)(s/s + l), so that the resistivity does not saturate, bnt increases as BZ. 11. Landau leoeln. The vector potential of a uniform magnetic field Bi is A = -By% in the Landau gauge. The hamiltonian of a free electron without spin is 9 Fenni SurJaces and Metolx 255 We will look for an eigenfilnction of the wave equation H$ = E(J irr tlra form $ = x(y) exp[i(k~ + k,z)l (a) Sho\v that ~ ( y ) satisfies the equation where o, = eBlmc and yo = cfik,/eB. (b) Show that this is the wave equation of a harmonic oscillator with frequency o, , where 6 , = (n + $)fro, + fizk:/2rn . Superconductivity EXPERIMENTAL SURVEY Occurrence of superconductivity Destruction of superconductivity by magnetic fields Meissner effect IIeat capacity Energy gap Microwave and infrared properties Isotope effect THEORETlCAL SURVEY Thermodynamics of the ~upercouducting transition London equation Coherence length BCS theory of superconductivity BCS ground state Flux quantization in a superconducting ring Duration of persistent currents Type I1 superconductors Vortex state Estimation of H,, and H,, Single particle tunneling Josephson superconductur tunneling Dc Josephson effect Ac Josephson effect Macroscopic quantum interference HIGH-TEMPERATURE SUPERCONDUCTORS NOTATION: 1 1 1 this chapter B, denotes the applied magnetic field. 111 the CGS system the critical value B,, of the applied field will he denoted by the syn~bol H, in accordance with the custom of workers in superconductivity Valucs of B,, are given in gauss in CGS units and in teslas in ST units, with 1 T = lo4 G. In SI we have B,,, = p a r . SUMMARY PROBLEMS 1. Magnetic field penetration in a plate 2. Critical field of thin films 3. Two-fluid model of a superconductor 4. Structure of a vortex 5. London penetration depth 6. Diffraction effect of Josephson junction 7. Meissner effect in sphere REFERENCE APPENDICES RELEVANT TO SUPERCONDUCTIVITY H Cooper Pairs I Ginzburg-Landau Equation J Electron-Phonon Collisions Figure 1 Resistance in ohms of a specimen of rnerculyversus absolute temperature. This plot by Kamerlingh Onnes marked the discover). of superconductivity. CIIAPTER 10: SUPERCONDUCTIVITY The electrical resistivity of many metals and alloys drops suddenly to zero when the speci~neri is cooled to a sufficiently low temperature, often a temper- ature in tlle liquid helium range. This phenomenon, called superconductivity, was observed first by Kamerlingh Onnes in Leideli in 1911, three years after he first liquificd helium. At a critical temperature T, the spccimen undergoes a phase transition from a state of nur~nal electrical resistivity to a supercon- ducting state, Fig. I. Superconductivity is now very well understood. It is a field with many practical and theoretical aspects. Thc length of this chapter and the relevant appendices reflect the richness and silhtleties of the field. EXPERIMENTAL SURVEY In the superconducting state the dc electrical resistivity is zero, or so close to zero that persistent clectrical currents have been observed to flow without attenuation in supcrcondi~cting rings for more than a year, until at last the ex- perinlentalist wearied of the experiment. Thc decay of supercurrents in a solenoid was studied hy File and Mills using precision nuclear magnetic resonance methods to measure the magnetic field associated with the supercurrent. They concluded that the decay ti~ne of the supercurrent is not less than 100.000 years. We estimate the decay time below. In some superconducting materials, particularly those used for super- corlducting magnets, finite decay times are observed because of an irrevcrsihle redistribution of magnetic flux in the material. The magnetic properties exhibited by superconductors are as dramatic as their clectrical properties. The magnetic propertics cannot be accounted for hy the assumption that a superconductor is a normal conductor with zero elec- trical resistivity. It is an experimental fact that a hulk superconductor in a weak magnetic field will act as a perlect diamagnet, with zero magnetic induction in the inte- rior. When a specimen is placed in a magnetic field and is then cooled through the transition temperature for superconductivity, the magnctic flnx originally prcscnt is ejected from the specimen. This is called thc Meissner effect. The sequence of events is shown in Fig. 2. The unique magnetic properties of su- perconductors are central to the characterization of the superconducting state. The supercvnducti~lg state is an ordered state of the conduction electrons of the ~netal. The order is in the formation of loosely associated pairs of elec- trons. The electrons are ordered at temperatures below the transition temper- ature, and they are disordered above the transition temperature. Figure 2 Meissner effect in a superconducting sphere cooled in a constant applied magnetic field; on passing below the transition temperature the lines of induction B are ejected from the sphere. The nature and origin of the ordering was explained by Bardeen, Cooper, and Schrieffer.' In the present chapter we develop as far as we can in an ele- mentary way the physics of the superconducting state. We shall also discuss the basic physics of the materials used for superconducting magnets, but not their technology. Appendices H and I give deeper treatments of the super- conducting state. Occurrence of Superconductivity Superconductivity occurs in many metallic elements of the periodic system and also in alloys, intermetallic compounds, and doped semiconductors. The range of transition temperatures best confirmed at present extends from 90.0 K for the compound YBa2Cu,0, to below 0.001 K for the element Rh. Several f-band superconductors, also known as "exotic superconductors," are listed in Chapter 6. Several materials become superconducting only under high pres- sure; for example, Si has a superconducting form at 165 kbar, with T, = 8.3 K. The elements known to be superconducting are displayed in Table 1, for zero pressure. Will every nonmagnetic metallic element become a superconductor at sufficiently low temperatures? We do not know. In experimental searches for superconductors with ultralow transition temperatures it is important to eliminate from the specimen even trace quantities of foreign paramagnetic elements, because they can lower the transition temperature severely. One part of Fe in lo4 will destroy the superconductivity of Mo, which when pure has T, = 0.92 K; and 1 at. percent of gadolinium lowers the transition temper- ature of lanthanum from 5.6 K to 0.6 K. Nonmagnetic impurities have no very marked effect on the transition temperature. The transition temperatures of a number of interesting superconducting compounds are listed in Table 2. Several organic compounds show superconductivity at fairly low temperatures. 'J. Bardeen, L. N. Cooper, and J. R. Schrieffer, Phys. Rev. 106,162 (1957); 108,1175 (1957). Table 2 Superconductivity of selected compounds Compo~~nd T,, in K Compound T,, in K Nb,Sn Nb,Ge Nb,AI NbN C60 Figure 3 Experimental threshold curves of the critical field H,(T) versus temperature for sc\reral su- perconductors. .%specimen is super- conducting bclow the curve and normal above the C I I ~ V . Temperature, in K Destruction of Superconductivity by Magnetic Fields A sufficiently strong magnetic field will destroy superconductivity. The tt~reshold or critical value of the applied magnetic field for the destruction of supercondl~ctivity is denoted by H,(T) and is a function of the temperaturc. At the critical temperature the critical field is zero: H,(T,) = 0. The variation of the critical field with temperature for several superconducting elements is shown in Fig. 3. The threshold curves separate the superconducting state in the lower left of the figure from the normal state in the upper right. Note: We should denote the critical value of the applied magnetic field as B,,, hut this is not common practice among workers in superconductivity. In the CGS system we shall al- ways understand that H, - B,,, and in thc SI we have H,. - B,Jpo The s p h o l B, denotes the applied rrlagnetic field. Meissner Effect Meissner and Ochsenfeld (1933) found that if a superconductor is cooled in a rr~agnetic field to helow the transition temperature, then at the transition the lines of induction B are pushed out (Fig. 2). The Meissner effect sho~vs that a bulk superconductor behaves as if B = 0 inside the specimen. 10 Superconductivity 263 We obtain a particularly useful form of this result if we limit ourselves to long thin specimens with long axes parallel to R,: now the demagnet~zing field contrihution (see Chapter 16) to B will bc negligible, whence:" The resillt B = 0 cannot be derived from the characterization of a super- condr~ctor as a medium of zero resistivity. From Ohm's law, E = pj, we see that if the resistivity p goes to zero while j is held finite, then E must be zero. By a Maxwell equation clBldi is proportional to curl E, so that zero resistivity im- plies dB/& = 0, but not B = 0. This argument is not entirely transparent, but the result predicts that the flux through the metal cannot changc on cooling through the transition. The Meissner effect suggests that perfect diamagnet- ism is an essential property of the supercondi~cting state. \'ire expect another differe~~ce between a superconductor and a perfect condirctor, defined as a cor~ductor in which the electrons have an infinite srrean free path. W e n the problem is solved in detail, it turns out that a perfect conductor when placed in a magnetic field cannot produce a perrrlarlent eddy current screen: the field will penetrate about 1 cm in an hour.:' The ~nag~letization curve expected for a supercorlductor under the condi- tions or the Mcissner-Ochsenfeld experiment is sketched in Fjg. 4a. This ap- plies quantitatively to a specimen in the forrn of a long solid cylinder placed in a longitudinal magnetic field. Pure speci~nens of many materials exhibit this behavior; they are called type I superconductors or, formerly, soft super- conductors. The values of H, are always too low for type I supercunductors to have application i 1 1 coils for supcrcond~~cting magnets. Other materials exhibit a magnetization curve of the form of Fig. 4b and are known as type 11 superconductors. They terid to be alloys (as in Fig. 5a) or transition metals with high values of the electrical resistivity in the normal state: that is, thc electronic mean free path in the normal state is short. We shall see later why the mean free path is involved in the "magnctiwtion" of superconductors. Tjye I1 superconductors have supercondi~cting electrical properties up to a field denoted by H,,. Behveen the lower critical field H,, and the upper criti- cal field H,, the flux density R # 0 and the Meissner effect is said to be incom- plete. The value of H,, may be 100 times or more higher (Fig. Sb) than 'Diarrragnctism, the magnetization LU, and the magnetic susceptibility are defined in Chapter 14. The magnitude of the apparent diamagnetic susceptibility of hulk s~lpercur~ductors is vely much larger than in typiral diamagnetic substa~lces. In (I), M is the magnetization eql~ivalent to the superco~~ductirig currents in the specimen. 3A. B. Pippard. Dynamics o f condvclion electrons, Gordon and Breach, 1965. 11,: Applicd mapctic field B, + (a! 41 H, 4 2 Applied rnag~etic field B, - Figure 4 (a) Magnetization versus applied magnetic field for a bulk snpercooductor exhibiting a complete Meissner effect (perfect diamagnetism). A superconductor with this behavior is callcd a type I ruperconductor. Above the critical field H, the specimen is a normal condl~ctor and the mag- netization is too small to be seen on this scale. Xote that minus 4vBf is plotted on the vertical scale: the negative value olM corresponds to damagnclism. (b) Supcrconducti~~g lrlagnetizatio~l cuwe of- a type I1 superconductor The flnx starts to penetrate the specimen at a field H,, lower than the thermodynamic critical field f f ' . The specimen is in a vortex state behveen H,, and H,,; and it has superconducting electrical propcrtics up to H,,. Ahovc H,, the speci~rren is a nor~~lal conductor ill every respect, except for possible sllrfacr effects. For given H, the area under the magnetization curve is the same for a type I1 superconductor as for a type 1. (CGS units in all parts of this figure.) the value of the critical field H,, calculated from the thermodynamics of the transition. In the region benveen HC1 and H,2 the superconductor is threaded by flux lines and is said to be in the vortex state. A field H,, of 410 k c (41 tes- las) has been attained in an alloy of Nb, Al, and Ge at the boiling point of he- lium, and 540 kG (54 tcslas) has been reported for PbMo6S8. Commercial solenoids wonnd with a hard superconductor producc high steady fields over 100 kG. A "hard supercondi~ctor" is a type I1 snperconduc- tor with a large magnetic hysteresis, usually induced by mechanical treatment. Such rnaterials have an important medical application in magnetic resonance imaging (MKI). Heat Capacity In all superconductors the entropy decreases markedly on cooling below the critical temperature T,. Measurements for aluminum are plotted in Fig. 6. The decrease in entropy between the ~lorrnal state and the superconducting statc tclls us that the superconducting state is more ordered than the normal state, for the entropy is a mcasurc of thc disorder of a system. Some or all of the electrons thermally excited in the normal state are ordered in the snpercon- ducting state. The change in entropy is small, in aluminum of the order of lo-" kg per atom. The small entropy change must mean that only a small fraction (of the order of of the conductiorl electrons participate in the transition to the ordered superconducting state. The free energies of normal and supercon- ducting states are compared in Fig. 7. 10 Superconductivity 265 Applied magnetic field B, in gauss Figure 5a S~~perconductimg ~~ragnctization curves of annealed polycrystalline lead and lead- indium alloys at 4.2 K. (A) lead; (R) lead-2.08 u. percent indium; (C) lead-8.23 wt. percent indinm; (D) lad-20.4 wl, percent indium. (After Livingston.) Temperature, K Figure 5b Strong magnetic fields are within thc capability of certain Type I1 materials. Temperature, K Figure 6 E~~tropy S of aluminnm in the l~nrrnal and superconducting states as a function of the temperature. The entropy is lower in the superconducting state because thc clcctrons are murc ur- dered here than in the normal statc. At any tenlyerature below the critical temperature Tc the speci- men car1 be put in the normal state hy application of a magnetic field stronger than the critical field. The heat capacity of gallium is plotted in Fig. 8: (a) compares the normal and superconducting states; (b) shows that the electronic contribution to the heat capacity in the superconducting state is an exponential form with an argn- ment proportiond to - 1/T, suggestive of excitation of electrons across an en- ergy gap. An energy gap (Fig. 9) is a characteristic, but not universal, feature of the supcrconducting state. The gap is accou~lted for by the Bardeen- Cooper-Schrieffer (RCS) thcory of superconductivity (see Appe~ldix H). Energy Gap The energy gap of superconductors is of entirely different origin and na- ture than the energy gap of insulators. In an insulator the energy gap is caused by the electron-lattice interaction, Chapter 7. This interaction ties the electrons to the lattice. In a superconductor the i~nportarlt interaction is the electron- electron interaction which orders the electrons in k space with respect to t l ~ e Fermi gas of electrons. The argument of the exponential factor in the clcctronic heat capacity of a superconductor is found to be -E&2kBT and not -Edk,T. This has been learnt from corriparison with optical and electron tunneling determinations of the gap E,. Values of the gap in several superconductors are given in Table 3. The transition in zero magnetic field fro~n the superconducting state to the normal statc is observed to be a second-order phase transition. At a -0.1 - -0.2 - 0 . 3 - .C- -0.6 - 9 ; -0.7- a , 4 . 8 - i r 4 . 9 - 1 . 0 - -1.1 - -1.2 - I I I I I I I I I I I I I I 0 0.5 1.0 1.5 Ternperatue, K Figure 7 Experilllcntal values of the fiec energy as a fimction of temperatnre fur alurninum in the ~u~crconducting state and in the normal state. Bclow the transition tc~nperature T, = 1.180 K the free energy is lower in the silperconducting state. The two curves merge at thc transition tem- perature, so that the phase transition is second order (there is 1 1 " latent heat of transition at T,). The curve F, is measured in zero magnetic field, and F> is measured in a magnetic field snfiicient to put the specirncn in the normal state. (Courtesy of U. E. Phillips.) I I I I I I I I I I I I I I . . 1.5 - Gallium I - OB, = 200G I - 'B,=O .
& .
' t - & S " I Tc G .
4 - B I I I _Y' 1.0- I I I I C/T = 0.596 + 0.0568 T' - - # - I - L I L 1 l L 1 l l l l l l 0 0.5 1.0 T 2 , K~ T</T (a) (b) Figure 8 (a) The heat capacity of gallium in the norrnal and snpercondlcti~~g states. The normal state (which is restored by a 200 G field) has electronic, lattice, and (at low tempenrturcs) nuclear rluddrupole contrihntions. In (b) the electronic part C , , of the heat capacitj' in the soper~vnduct- ing state is plotted on a log scale versus T,fl: the cxpo~~ential dependence on 1IT is evident. Here y = 0.60 mJ mol-I deg-'. (After N. E. PlliUips.) Superconductor (h Figure 9 (a) Conduction band in the normal state: (h) enerby gap at the Fermi level in the super- conducting state. Electrons in excited states ahove the gap hehave as normal electrons in rf Fields: they cause resistance; at dc they are shorted out hy the superconducting electrons. The gap Ez is exaggerated in the figure: tn~ically l$ - lo-' E,;. Table 3 Energy gaps in superconductors, at T = 0 second-order transition there is no latent heat, but there is a discontinuity in the heat capacity, evident in Fig. 8a. Further, the energy gap decreases contin- uously to zero as the temperature is increased to the transition temperature T,, as in Fig. 10. A first-order transition would be characterized by a latent heat and by a discontinuity in the energy gap. Microwave and Infrared Properties The existence of an energy gap means that photons of energy less than the gap energy are not absorbed. Nearly all the photons incident are reflected as for any metal because of the impedance mismatch at the boundary between vacuum and metal, but for a very thin (-20 A) film more photons are transrnit- ted in the superconducting state than in the normal state. Figure 10 Reduced values of the observed energy gap E,(T)/E,(O) as a functioil of the reduced temperature T/T,, after Tomsend and Sutton. The solid curve is drawn for the BCS theory For photon energies less than the encrgy gap, the resistivity of a supercon- ductor va~~islles at absolute zero. At T 4 T, the resistance in the superconduct- ing state has a sharp threshold at the gap energy. Photons of lower energy see a resistanceless sl~rface. Photons of higher energy than t l ~ e energy gap see a re- sistancc that approaches that of the normal state because such photons causc transitions to unoccupied normal energy levels above the gap. As the temperature is increased not only does thc gap decrease in energy, but the resistivity for photons with energy bclow the gap energy no longer van- ishes, except at zero frequency. At zcro frequency the superconducting elec- trons short-circuit any normal electrons that have been ther~irally excited above the gap. At finite frequencies the inertia of t l ~ e superco~lducting elec- trons prevents them from completely screerlirlg the electric field, so that ther- mally excited normal electrons now can absorb energy (Problem 3). Isotope Effect It ha5 been observed that the critical temperature of supcrcondi~ctors varies with isotopic mass. I 1 1 mercury T, varies from 4.185 K to 4.146 K as the average atomic mass A4 varies from 199.5 to 203.4 atomic mass units. The tran- sition terr~perature changes smoothly when we mix different isotopes of the same element. The experimental resillts within each series of isotopes may be fitted by a relation of the form MOT, = constant . (2) Observed values of cu are given in Table 4 Tahle 4 Tsutope effect in superconductors Esperirnental values of w in MaT, = constant, where A4 is the isotopic mass. Substance Substance From the dependence of T, on the isotopic mass we learn that lattice vibrations and hence electron-lattice interactions arc deeply involved in super- conductivity This was a fundamental discovery: there is no other rcason for the superconductirlg transition temperature to depend on the number of new trons in the nucleus. Thc orignal BCS model gave the result ?: eDebvr a M-'I2 , so that cu = in (2), bnt the inclusion of coulomb interactions between the electrons changes the relation. Nothing is sacrcd about a = i. The absence of an isotope effect in Ru and Zr has been accounted for in tcrms of the electron band struc- ture of these metals. THEORETICAL SURVEY A theoretical understandir~g of the phenomena associated with supercon- dnctivity has been reached in several ways. Certain results follow directly from thermodynamics. Many important results can be described by p~~enomenolog- ical equations: the London eqnations and the Landau-Ginzburg equations (Appendix 1). A successful quantum theory of s~iperconductivity was given by Bardeen, Cooper, and Schrieffer, and has provided the basis for snhscqucnt work. Josepl~son and Anderson discovered the importance of the phase of the superconducting wavefunction. Thermodynamics of the Superconducting Transition The transition between the normal and superconducting states is thermo- dynamically reversible, just as tlie transition between liquid and vapor phases of a substance is reversible. Tbus we may apply therrnody~~arnics to the transi- tion, and we thereby obtain an expression for the entropy difference between normal and supercond~~cting states in tcrms of the critical field curve H, ver- sus T. This is analogous to the vapor pressnre eq~iation for thc liquid-gas coexistence cunTe (TP, Chapter 10). 10 Superconductivity 271 We treat a type I superconductor with a complete Meissner effect. so that B = 0 inside the superconductor. We shall see that the critical field H, is a quan- titative measure of the free energy difference between the superconducting and normal states at constant temperature. The symbol H, will always refer to a bulk specimen, never to a thin film. For type I1 superconductors, H, is understood to be the thermodynamic critical field related to the stabilization free energy. The stabilization free energy of the superconducting state with respect to the normal state can be determined by calorimetric or magnetic measure- ments. In the calorimetric method the heat capacity is measured as a function of temperature for the superconductor and for the normal conductor, which means the superconductor in a magnetic field larger than H,. From the differ- ence of the heat capacities we can compute the free energy difference, which is the stabilization free energy of the superconducting state. In the magnetic method the stabilization free energy is found from the value of the applied magnetic field that will destroy the superconducting state, at constant temperature. The argument follows. Consider the work done (Fig. 11) on a superconductor when it is brought reversibly at constant tem- perature from a position at infinity (where the applied field is zero) to a posi- tion r in the field of a permanent magnet: M-dBo , (3) =aJ Superconductor phase jm = H, -\~ormal phase (coexisting in equilibrium) Figure 11 (a) A superconductor in which the Meissner effect is complete has B = 0, as if the magnetization were M = -B,14~, in CGS units. (h) When the applied field reaches the value B .
, , the normal state can coexist in equilibrium with the superconducting state. In coexistence the free energy densities are equal: F,(T, B,,) = F,(T B,). per unit volume of specimen. This work appears in the energy of the magnetic field. The thermodynamic identity for the process is dF = -M .dB, , ( 4 ) as in TP, Chapter 8. For a superconductor with M related to B, by ( 1 ) we have The increase in the free energy density of the superconductor is on being brought from a position where the applied field is zero to a position where the applied field is B,. Now consider a normal nonmagnetic metal. If we neglect the small susceptibility4 of a metal in the normal state, then M = 0 and the energy of the normal metal is independent of field. At the critical field we have FN(B,) = FN(O) . (7) The results (6) and (7) are all we need to determine the stabilization energy of the superconducting state at absolute zero. At the critical value B,, of the applied magnetic field the energies are equal in the normal and super- conducting states: In SI units H, = B,,/p,, whereas in CGS units H, = B,,. The specimen is stable in either state when the applied field is equal to the critical field. Now by (7) it follows that 4This is an adequate assumption for type I superconductors. In type I1 superconductors in high fields the change in spin paramagnetism of the conduction electrons lowers the energy of the normal phase significantly In some, hut not all, type I1 superconductors the upper critical field is limited by this effect. Clogston has suggested that H,,(max) = 18,400 T,, where H,, is in gauss and T, in K. 10 Superconductivity b FN .- 2 -8 Figure 12 Tlrc frce energy density F, of a nonmag- p netic normal metal is approximately independent of tlrc c intensity of the applied magnetic field B,. At a temper- ature T : T , the nlctal is a superconductor in zero mag- a , e, netic field, so that FJT, 0) is Iowver than F,(T, 0). AII Lr, applied magnetic field increases Ii, by A ~ X T , in CGS ~mits, so that Fs(T, B,) = FF(T, 0) + B;/Xv. If B, is larger than the critical field R,, the free energy density is lowcr in the normal state than in the superconducting I state, and now the rror~~ral state is the stable state. The B , origin of the vertical scale in the drawing is at Fs(T, 0). Applied magnetic field B, -+ The figurc cqually applies to Us and U, at ?' = 0. where AF is the stabilization free e n e r a density of the superconducting state. For alu~ninum, B,, at absolute zcro is 105 gauss, so that at absolute zero AF = (105)~/8n = 439 erg cm 3, in excellent agreement with the result of thermal measurcments, 430 erg ~ m - ~ . At a finite temperature the normal and superconducting phases are in equilibrium when the magnetic field is such that their free encrgies F = U - TS are equal. The free energies of the two phases are sketched in Fig. 12 as a furiction of the magnetic field. Experimental curves of the free energes of the two phases for aluminiim are shown in Fig. 7. Because the slopes dF/dT are equal at the transition temperature, there is no latent heat at T,. London Equation We saw that the Meissner effect implies a magnetic susceptibility X = - 1 / 4 ~ in CGS in the superconducting state or, in SI, X = - 1. Can we modify a consti- tutive equation of electrodyna~nics (such as Ohm's law) in some way to obtain the Meissr~er effect? We do not want to modify the Maxwell equations them- selves. Electrical conduction in the normal state of a metal is described by Ohm's law j = uE. We need to modify this drastically to describe conduction and thc hleissner effect in the superconducting state. Let us make a postulate and see what happens. \Ve postulate that in the superconducting state the current density is di- rectly proportiorial to the vector potential A of the local magnetic field, where B = curl A. The gauge of A will bc specified. In CGS units we write the constant of proportionality as -c/4d; for reasons that will become clear. Here c is the speed of light and A, is a constant with the dimensions of length. In SI units we write -Up,+!. Thus C (CGS) j = --A ; 4 T r A ; This is the London equation. We express it another way by taking the curl of both sides to obtain C (CGS) curl j = - -B ; 4 5 ~ ~ : -a - 1 (SI) curl j = -- B (11) g,+: The London equation (10) is understood to be written with the vector po- tential in the London gauge in which div A = 0, and A, = 0 on any external surface through which no external current is fed. The subscript n denotes the component normal to the surface. Thus div j = 0 and j, = 0, the actual physi- cal boundary conditions. The form (10) applies to a simply connected super- conductor; additional terms may be present in a ring or cylinder, but (11) holds true independent of geometry. First we show that the London equation leads to the Meissner effect. By a Maxwell equation we know that 4 5 T (CGS) curl B = j ; under static conditions. We take the curl of both sides to obtain (CGS) curl curl B = -V'B = curl j ; air1 curl B = - V'B = k, curl j ; which may be combined with the London equation (11) to give for a super- conductor This equation is seen to account for the Meissner effect because it does not allow a solution uniform in space, so that a uniform magnetic field cannot exist in a superconductor. That is, B(r) = Bo = constant is not a solution of (13) unless the constant field Bo is identically zero. The result follows because V2B, is always zero, but B,/A; is not zero unless Bo is zero. Note further that (12) ensures that j = 0 in a region where B = 0. In the pure superconducting state the only field allowed is exponentially damped as we go in from an external surface. Let a semi-infinite superconductor 10 Superconductivity 275 Figure 13 Penetration of an applied magnetic field into a semi-infinite superconductor. The penetration depth A is defined as the distance in which the field decreases by the factor eCL. Typi- cally, A - 500 A in a pure superconductor. occupy the space on the positive side of the x axis, as in Fig. 13. If B(0) is the field at the plane boundary, then the field inside is for this is a solution of (13). In this example the magnetic field is assumed to be parallel to the boundary. Thus we see hL measures the depth of penetration of the magnetic field; it is known as the London penetration depth. Actual penetration depths are not described precisely by hL alone, for the London equation is now known to be somewhat oversimplified. It is shown by compari- son of (22) with (11) that (CGS) hL = (m2/4mq2)u2 ; for particles of charge q and mass m in concentration n. Values are p e n in Table 5. An applied magnetic field B, will penetrate a thin film fairly uniformly if the thickness is much less than hL; thus in a thin film the Meissner effect is not complete. In a thin film the induced field is much less than B,, and there is little effect of B, on the energy density of the superconducting state, so that (6) does not apply. It follows that the critical field H, of thin films in parallel - - . magnetic fields will be very high Table 5 Calculated intrinsic coherence length and London penetration depth, at absolute zero Intlins~c Pippard London coherence penetration length 50, depth A,, Metal in cm in 10-%m A ~ / 5 0 Sn 23. 3.4 0.16 A1 160. 1.6 0.010 Pb 8.3 3.7 0.45 Cd 76. 11.0 0.14 Nb 3.8 3.9 1.02 After R. Meservey and B. B. Schwartz. Coherence Length The London penetration depth A, is a fundamental lerlgth that character- izes a superconductor. An indcpcndent length is the coherence length 5. The coherence length is a measure of the distance within which the supercouduct- ir~g electron concentration cannot change drastically in a spatially-varying magnetic field. The London equation is a locul equation: it relates the current density at a point r to the vector potential at the same point. So long as j(r) is given as a constant time A(r), thc current is required to follow exactly any variation in the vector potential. Bi~t the cohcrcncc lcngth 5 is a measure of the range over which we should average A to ohtain j. It is also a mcasure of the minimum spa- tial extent of a transition layer between normal and superconductor. Thc coher- ence lerlgtll is best introduced into the theory through the Landau-Ginzhllrg equations, Appendix 1. Now we give a plausibility argument for the energy re- q~iircd to modulate the superconducting alectron concentration. Any spatial variation in the state or an electronic syste111 requires extra kinetic energy. A modillation of an cigcnfunction increases the kinetic energy because the modulation will increase the integral of d'p/dx2. It is reasonable to restrict the spatial variation of j(r) in such a way that thc cxtra energy is less than the stabilizatiori energy of the superconducting state. We compare the plane wave $(x) = dkX with the strongly modulated wavcfunction: ,(,) = 2-112 (e:(kiq)r + &) , (15a) The probability dcnsity associated with the plane wave is u~iiforrrl in space: $$ = e-"' eik = 1 , whereas qq is modulated with the wavevector q: pq = ;(,-Nk+q)x + ,-ik~)(~iik+ql' + & I ) = i(2 + eiQX + e-qX) = 1 + cos qx . (15b) The kinetic energy of the wave $(x) is 6 = fi2k2/2m; thc kinctic energy of the rrlodulated density distribution is higher, for where we neglcct q ' for q < k. The increase of energy requircd to modulate is R?kq/2nx. If this increase exceeds the energy gap E,, s~spercondnctivity will be destroyed. The critical value q, of the modulation wavevector is given by 10 Superconductivity We define an intrinsic coherence length 5, related to the critical modu- lation by 5, = lly,. W7e frave where cF is the electron vclocity at the Fermi surface. On the BCS theory a similar result is found: to = zliod~E, . u (17) Calculated values of 5, from (17) are p e n in Table 5. The intrinsic coherence length to is characteristic of a pure superconductor. In impure materials and in alloys the coherence length 5 is shorter than 6". This may be understood qualitatively: in impure material the electron eigen- functions already have wiggles in them: wc can construct a given localized variation of current density with less cncrgy from wavefunctions with wiggles than from s~nooth wavefunctions. The cohcrcnce length first appeared in the Landau-Ginzburg equations: these equations also follow from the BCS theory. They describe the structure of the transition layer between normal and superconducting phases in contact. The coherence length and the actual penetration depth A depend on the mean free path C of the electrons measured in the normal state; the relationships are indicated in Fig. 14. When the superconductor is very impure, with a very srr~all C, then 5 = (Eoe)lr%nd A = A, (t0/t)"" so that A 1 5 = A,/[. This is the "dirty superconductor" limit. The ratio A/< is denoted by K . BCS Theory of Superconductivity The basis of a quantum theory of superconductivity was laid by the classic 1957 papers of Bardeen, Cooper, and Schriefler. There is a "BCS theory of superconductivity" with a very wide range of applicability, from He3 atoms in their conderlsed $lase, to type I and type I1 metallic superconductors, and to high-temperature supercondnctors hased on planes of cuprate ions. Further, Figure 14 Penetration depth A and the coherence length 6 a s fi~nctinns of the m e a n free p a t h 8 of the 0.2 conduction electrons i n the normal state. All I 0.1 lenbehma in units of $, the intrinsic coherence length. The mNpS are sketched for $ = 10AP For 0 1 I short mean free ~ a t h s the coherence length be- 0 1 2 comles sl~orter and thc penetration depth becomes - [ - . longer The increase in the ratio dl[ favors type I1 50 superconductivity. there is a "BCS wavefunction" cornposed of particle pairs k l ' and -k&, which, when treated hy the BCS theory, gives the lamiliar electronic superconducti\r- ity observed in metals and exhihits the cncrgy gaps of Table 3. This pairing is known as s-wave pairing. There are other forms of particlc pairing possible with the BCS theory, but we do not have to consider other than the RCS wavc- function here. In this chapter we treat the specific accomplishments of BCS theory with a BCS wavefunction, which include: 1. An attractive interaction between electrons can lead to a ground state separated from excited states by an energy gap. The critical field, the thermal properties, and most of the electrorriagnetic properties are consequences of the energy gap. 2. The electron-lattice-electron interaction leads to an energy gap of the observed magnitude. The indirect interaction proceeds wlien one electron in- teracts with the lattice and deforms it; a second clcctron sees the deformed lattice and adjusts itself to take advantage of the deformation to lower its cn- ergy. Thus the second electron interacts with the first electron via the lattice deformation. 3. The penetration depth and the coherence length emerge as natural consequences of thc BCS theory. The London equation is obtained for mag- netic fields that vary slowly in space. Thus the central phenomenon in super- conductivity, the Meissner effect, is ohtained in a natural way. 4. The criterion for the transition temperatilre of an elcmcnt or alloy in- volves the electron density of orbitals D(eli) of one spin at the Fermi level and the electron-lattice interaction U , which can be estimated from the electrical resistivity because the resistivity at roorn temperature is a measure of the electron-phonon interaction. For UD(c,) 4 1 the BCS theory predicts where 19 is the Debye temperature and U is an attractive interaction. Thc rc- sult for T, is satisfied at least qualitatively by the experimental data. There is an interesting apparent paradox: the higher the resistivity at room temperature thc higher is U, and thus the ntore likely it is that the metal will be a super- condnctor when cooled. 5. Magnetic flnx through a superconducting ring is quantized and the ef- fective unit of charge is 2e rather than e. The RCS ground state involves pairs of electrons; thus flux quantization in terms of the pair charge 2e is a conse- quence of the theory. BCS Ground State The filled Fermi sea is the ground state of a Ferrr~i gas of noninteract- ing electrons. This state allows arbitrarily small excitations-we can forrr~ an 10 Superconductivity 219 Figure 15 (a) Probability P that an or- bital of kinetic energy E is occupied in the ground state of the noninteracting Fermi gas; (b) the BCS ground state differs from the Fermi state in a region of width of the order of the energy gap Ep. Both curves are for absolute zero. excited state by taking an electron from the Fermi surface and raising it just above the Fermi surface. The BCS theory shows that with an appropriate at- tractive interaction between electrons the new ground state is superconduct- ing and is separated by a finite energy Eg from its lowest excited state. The formation of the BCS ground state is suggested by Fig. 15. The BCS state in (b) contains admixtures of one-electron orbitals from above the Fermi energy E ~ . At first sight the BCS state appears to have a higher energy than the Fermi state: the comparison of (b) with (a) shows that the kinetic energy of the BCS state is higher than that of the Fermi state. But the attractive potential energy of the BCS state, although not represented in the figure, acts to lower the total energy of the BCS state with respect to the Fermi state. When the BCS ground state of a many-electron system is described in terms of the occupancy of one-particle orbitals, those near eF are filled some- what like a Fermi-Dirac distribution for some finite temperature. The central feature of the BCS state is that the one-particle orbitals are occupied in pairs: if an orbital with wavevector k and spin up is occupied, then the orbital with wavevector -k and spin down is also occupied. 1f k'f' is vacant, then -kJ is also vacant. The pairs are called Cooper pairs, treated in Appendix H. They have spin zero and have many attributes of bosons. Flux Quantization in a Superconducting Ring We prove that the total magnetic flux that passes through a superconduct- ing ring may assume only quantized values, integral multiples of the flux quan- tum 2 d c / q , where by experiment q = 2e, the charge of an electron pair. Flux quantization is a beautiful example of a long-range quantum effect in which the coherence of the superconducting state extends over a ring or solenoid. Let us first consider the electromagnetic field as an example of a similar boson field. The electric field intensity E(r) acts qualitatively as a probability field amplitude. When the total number of photons is large, the energy density may be written as E (r)E(r)/4~ n(r)h.w , where n(r) is the number density of photons of frequency o. Then we Inay write the electric field in a semiclassical approximation as where O(r) is the phase of thc ficld. A similar probability amplitude describes Cooper pairs. The arguments that follow apply to a hoson gas with a large number of boso~is i11 tlie same orbital. We then can treat the boson probability amplitude as a classical quantity, just as the electromagnetic field is used for photons. Both amplitude and phase are then rneani~igful and observable. The arguments do not apply to a metal in the normal state because an alectrorl in the normal state acts as a single nnpaircd fcrmion that cannot be treated classically. We first show that a charged boson gas obeys the London equation. Let $(r) be the particle probability ampliti~de. Wc suppose that the pair conceritratior~ n = $$ = constant. At absolute zero n is one-half of thc con- centration of electrons in tlie co~~duction band, for n refers to pairs. Then we may write The phase B(r) is important for what follows. I 1 1 SI units, set c = 1 irl the equa- tions that follo\i7. The velocity of a particle is, from the Hamilton cquations of mechanics, The particlc flux is given by so that the electric current density is \t'e nay take the curl of both sides to obtain the London equation: nq2 curl j = --B , rrx with use of the fact that the curl of thc gradirnt of a scalar is identically zero. The constant that multiplies B agrees with (14a). Wc rccall that the Meissner effect is a consequence of the London equation, which \i~e have hcrc derived. Quantization of the magnetic flux through a ring is a dramatic conse- quencc of Eq. (21). Let us take a closed path C througl~ tlle interior of the 10 Superconductivity 281 Flux lines Figure 16 Path of integration C through the interior of a superconducting ring. The flux through the ring is the sum of the flux @ .
, from external sources and the flux a , , from the superconducting currents which flow in the surface of the ring; @ - a,,, + @ , , . The flux a is quantized. There is normally no quantization condition on the flux from exter- nal sources, so that a , , must adjust itself appropriately in order that @ assume a quantized value. superconducting material, well away from the surface (Fig. 16). The Meissner effect tells us that B and j are zero in the interior. Now (21) is zero if ficV8 = qA . (23) We form for the change of phase on going once around the ring. The probability amplitude cl, is measurable in the classical approximation, so that cl, must be single-valued and Oz - 81 = 23rs , (24) where s is an integer. By the Stokes theorem, where du is an element of area on a surface bounded by the curve C, and @ is the magnetic flux through C. From (23), (24), and (25) we have 2&s = q@, or @ = (&&/q)s . (26) Thus the flux through the ring is quantized in integral multiples of 2&/q. By experiment q = -2e as appropriate for electron pairs, so that the quan- tum of flux in a superconductor is (CGS) @ , = 2dc/% -- 2.0678 X lo-' gauss cm2 ; This flux quantum is called a fluxoid or fluxon. The flux through the ring is the sum of the flux @ , , , from external sources and the flux @ , , from the persistent superconducting currents which flow in the surface of the ring: @ = a , , + Q,,. The flux cP is quantized. There is nor- mally no quantization condition on the flux from external sources, so that @ , , must adjust itself appropriately in order that @ assume a qiiantizcd value. Duration o f Persistent Currents Consider a persistcnt current that flows in a ring of a type I superconduc- tor of wire of length L and cross-sectional area A. The persistent current main- tains a flux through the ring of some integral number offluxoids (27). A fluxoid ca~lnot leak out of the ring and thereby reduce the persistcnt current unless by a thermal fluctuatio~~ a minimum volume of the supercondi~cting ring is mo- mentarily in the normal state. The probability per unit time that a fluxoid will leak out is the product P = (attempt frequency)(activation harrier factor) . (28) The activation harrier factor is exp(-AF/kBT), where t l ~ e free energy of the barrier is AF = (minimum volume)(excess free energy density of normal state) . The minimu111 volume of the ring that must turn normal to allow a fluxoid to escape is of the order of RS2, where 6 is the coherence length of the snpcr- cond~ictor and R the wire thickness. The excess free energy density of the nor- mal state is H ; / ~ P , whence the barrier free energy is Let the wire thickness he em, thc coherence length = lo-'' cm, and H, = lo3 6; then AF - erg. As we approach the transition temperature from below, AF will decrease toward zero, but the vali~e given is a fair cstimate between absolute zero and 0.8 T,. Thus the activation barrier factor is Thc characteristic frequency with which tl~e minimum volume can attempt to change its state must be of the order of Eg/fi. If Eg = erg, the attempt frequency is =10-15/10-27 = 1012 s-'. The leakage probability (28) beco~rles The reciprocal of this is a measure of the time required for a fluxoid to leak out, T = 1/P = 1 0 ~ . ~ ~ ~ ~ ~ ~ s . The age of the universe is only loL8 s, so that a fluxoid will not leak out in the age of the universe, under our assumed conditions. Accordingly, the cur- rent is maintained. There are two circumstances in which the activation energy is much lower and a fluxoid can be observed to lcak out of a ring-either very close to tlle critical temperature, where H, is ver)i small, or whcn the material ol the ring is a type I1 superconductor and already has fluxoids embedded in it. These spe- cial situations are discussed in the literature under the subject of fluctuatio~is in s~iperconductors. Type I1 Superconductors There is no difference in the mechanism of superconductivity in type I and type I1 superconductors. Both types have similar thermal properties at the superconductor-normd tra~isition in zero magnetic field. But the Meissner effect is eritiraly different (Fig. 5). A good type I superconductor cxclildes a magnetic field until super- conductivity is destroycd siiddenly, and then the field penetrates completely. A good type I1 supercond~ictor excludes the field completely up to a field H,,. Above HC1 the field is partially excluded, but the speci~rieri re~riains electrically snperconducting. At a much higher field, H,,, the flux penetrates completely and superconductivity vanishes. (An outer surface layer of the spccimcn may remain supercor~ductirlg up to a still higher field Hc3.) A I ~ i~riportant difference in a type I and a type I1 s~iperconductor is in the rnean free path of the conduction electrons in the normal state. If the coher- ence length 5 is longer than the penetration depth A, the superconductor will be typc I. Most pilre metals are type I, with A/[ < 1 (see Table 5 on p. 275). Biit, when the mean free path is short, the coherence length is short and the penetration depth is great (Fig. 14). This is the situation when A/( > 1, and the superconductor will be type 11. We can clia~ige some ~netals from type I to type I1 by a modest addition of an alloyi~ig element. In Figure 5 the addition of 2 u7t. percent of indium changes lead from typc I to type IT, although the transition temperature is scarcely changcd at all. Nothing fundamental has been done to the electronic struetiire of lead by this amount of alloying, hut the magnetic behavior as a s~iperconductor has changed drastically. The theory of type I1 superconductors was developed by Ginzbiirg, Landau, Abrikosov, and Gorkov. Later Kunzler and co-workers observed that NbnSrl wires can carry large supercurrcnts in fields approaching 100 kc; this led to the commercial development of strong-field superconducting magnets. Consider thr interface between a region in the superconducting state and a region in the normal state. The interface has a surface eriergy that nray be positive or negative and that decreases as the applied magnetic field is in- creased. A superconductor is type 1 if the surface energy is always positive as the niagrietic field is increased, and type I1 if the surface cncrgy becomes negative as the niagnetic field is increased. Thc sign of'the snrface energy has no importance lor the transition temperature. The frcc energy of a hrilk superconductor is increased when the magnetic field is expelled. IIowever, a parallel field can penetrate a very thin film nearly uniformly (Fig. 171, only a part of the flux is expelled, and the energy of the (a) (b) Figure 17 (a) Magnetic field penetration into a thin film of thickness equal to the penetration depth A. The arrows indicate the intensity of the magnetic field. (b) Magnetic field penetration in a homogeneous hulk structure in the mixed or vortex state, with alternate layers in normal and su- perconducting states. The superconducting layers are thin in comparison with A. The laminar structure is shown for convenience; the actual structure consists of rods of the normal state sur- rounded by the supercondncting state. (The N regions in the vortex state are not exactly normal, hut are described by low values of the stabilization energy density.) superconducting film will increase only slowly as the external magnetic field is increased. This causes a large increase in the field intensity required for the destruction of superconductivity. The film has the usual energy gap and will be resistanceless. A thin film is not a type I1 superconductor, but the film results show that under suitable conditions superconductivity can exist in high mag- netic fields. Vortex State. The results for thin films suggest the question: Are there sta- ble configurations of a superconductor in a magnetic field with regions (in the form of thin rods or plates) in the normal state, each normal region sur- rounded by a superconducting region? In such a mixed state, called the vortex state, the external magnetic field will penetrate the thin normal regions uni- formly, and the field will also penetrate somewhat into the surrounhng super- conducting material, as in Fig. 18. The term vortex state describes the circulation of superconducting currents in vortices throughout the bulk specimen, as in Fig. 19. There is no chemical or crystallographic difference between the normal and the supercon- ducting regions in the vortex state. The vortex state is stable when the penetra- tion of the applied field into the superconducting material causes the surface energy to become negative. A type 1 1 superconductor is characterized by a vortex state stable over a certain range of magnetic field strength; namely, between HC1 and H,,. Estimation of H,, and H,,. What is the condition for the onset of the vortex state as the applied magnetic field is increased? We estimate H,, from the penetration depth A . The field in the normal core of the fluxoid will be H,, when the applied field is H,,. 10 Superconductivity 285 Type I1 superconductor 0 Figure 18 Variation of the magnetic field and en- ergy gap parameter A(x) at the interface of super- conducting and normal regions, for type I and type I1 superconductors. The energy gap parameter is a measure of the stabilization energy density of the soperconducting state. The field will extend out from the normal core a distance h into the super- conducting environment. The flux thus associated with a single core is d2 HC1, and this must be equal to the flux quantum a,, defined by (27). Thus HC1 = @&rh2 . (30) This is the field for nucleation of a single fluxoid. At H,, the fluxoids are packed together as tightly as possible, consistent with the preservation of the superconducting state. This means as densely as the coherence length 5 will allow. The external field penetrates the specimen almost uniformly, with small ripples on the scale of the fluxoid lattice. Each core is responsible for carrying a flux of the order of ?rt2 H,, which also is quantized to @ , . Thus gives the upper critical field. The larger the ratio All, the larger is the ratio of H,, to Hcl. Figure 19 Flux lattice in NbSe, at 1,000 gauss at 02K, as ~iewed wit11 a scanning tunneling microscope. The photo shows the density of states at the Fermi level, as in Figure 23. The vortex cores have a high density of states and are shaded white; the superconducting regions are dark, with no states at the Fermi level. The amplitude and spatial extent of these states is determined by a potential well formed by A(x) as in Fig. 18 for a Type I1 superconductor. The potential well confines the core state wavefunctions in the image here. The star shape is a finer feature, a result special to NbSe, of the sixfold disturbance of the charge density at the Fermi surface. Photo cour- tesy of H. F. Hess. It remains to find a relation between these critical fields and the thermo- dynamic critical field H, that measures the stabilization energy density of the superconducting state, which is known by (9) to be H;/8n-. In a type I1 super- conductor we can determine H, only indirectly by calorimetric measurement of the stabilization energy. To estimate H,, in terms of H,, we consider the stability of the vortex state at absolute zero in the impure limit 6 < A; here K > 1 and the coherence length is short in comparison with the penetration depth. We estimate in the vortex state the stabilization energy of a fluxoid core viewed as a normal metal cylinder which carries an average magnetic field B,. The radius is of the order of the coherence length, the thickness of the bound- ary between N and S phases. The energy of the normal core referred to the energy of a pure superconductor is given by the product of the stabilization energy times the area of the core: 10 Superconductioity 287 per unit length. But there is also a decrease in magnetic energy because of the penetration of the applied field B, into the superconducting material around the core: For a single fluxoid we add these two contributions to obtain (CGS) f =fmre + fm,=f (HY - BY) . (34) The core is stable iff < 0. The threshold field for a stable fluxoid is at f = 0, or, with H,, written for B,, H,,IH, = .$/A . (35) The threshold field divides the region of positive surface energy from the re- gion of negative surface energy. We can combine (30) and (35) to obtain a relation for H,: We can combine (30), (31), and (35) to obtain ( H , ~ H , ~ ) ~ = H, , and H,, = (Al()H, = KH, . Single Particle Tunneling Consider two metals separated by an insulator, as in Fig. 20. The insulator normally acts as a barrier to the flow of conduction electrons from one metal to the other. If the barrier is sufficiently thin (less than 10 or 20 A) there is a significant probability that an electron which impinges on the barrier will pass from one metal to the other: this is called tunneling. In many experiments the insulating layer is simply a thin oxide layer formed on one of two evaporated metal films, as in Fig. 21. When both metals are normal conductors, the current-voltage relation of the sandwich or tunneling junction is ohmic at low voltages, with the current directly proportional to the applied voltage. Giaever (1960) discovered that if one of the metals becomes superconducting the current-voltage characteristic changes from the straight line of Fig. 22a to the curve shown in Fig. 22b. Figure 20 Two metals, A and B, separated by a thin layer of an insulator C. (a) ib) (c) id) Figure 21 Preparation of an AVAl,OdSn sandwich. (a) Glass slide with indium contacts. (b) An aluminum strip 1 mm wide and 1000 to 3000 A thick has been deposited across the contacts. (c) The aluminum strip has been oxidized to form an A1,0, layer 10 to 20 A in thickness. (d) A tin film has been deposited across the aluminum film, forming an AlIA1,OdSn sandwich. The external leads are connected to the indium contacts; two contacts are used for the current measurement and two for the voltage measurement. (After Giaever and Megerle.) Figure 22 (a) Linear current-voltage relation for junction of normal metals separated by oxide layer; (b) current- voltage relation with one metal normal and the other metal superconducting. Voltage ia) l Current (a) (b) Figure 23 The density of orbitals and the current-voltage characteristic for a tunneling junction. In (a) the energy is plotted on the vertical scale and the density of orbitals on the horizontal scale. One metal is in the normal state and one in the superconducting state. (b) 1 versus V; the dashes indicate the expected break at T = 0. (After Giaever and Megerle.) Figure 23a contrasts the electron density of orbitals in the superconductor with that in the normal metal. In the superconductor there is an energy gap centered at the Fermi level. At absolute zero no current can flow until the applied voltage is V = Eg/2e = A/e. The gap Eg corresponds to the break-up of a pair of electrons in the superconducting state, with the formation of two electrons, or an electron and 10 Superconductivity 289 a holc, in the normal state. The current starts whell eV = A. At finite temperatures there is a small current flow even at low voltages, because of electrons in the superconductor that are thermally excited across the energy gap. Josephson Superconductor Tunneling Under suitable conditions we observe remarkable effects associated with the tunneling of supercollducting electron pairs from a superconductor througll a layer of an insulator into another siiperconductor. Such a junction is called a weak link. The effects of pair tiinneling include: Dc Josephson effect. A dc current flows across the junction in the ab- sence of any clcctric or magnetic field. Ac Josephson effect. A dc voltage applied across the junction causes rf current oscillations across the junction. This effect has been utilized in a precision determination of the value of file. Further, an rf voltage applied with the dc voltage can then cause a dc current across thc junction. Macroscopic long-range quantum intcrference. A dc magnetic field applied through a superconducting circiiit containing two junctions causes the maximum supcrcurrcnt to show interference effects as a function of magnetic field intcnsity. This effect can be utilized in sensitive magnetometers. Dc Josephson Effect. Our discussion of Josephson ju~lction phenomena follows the discussion of flux quantization. Let +, be the probability amplitude of electron pairs on one side of a ju~lction, and let + , be the amplitudr on the other side. For sin~plicit~, let both superconductors bc identical. For the pres- erit we suppose that they are both at zero potential. The time-dependent Schrodinger eqnation ifia+/at = X+ applied to the two amplitudes gives Here fiT represents the effect of the electron-pair coupling or transfer interac- tion across the insulator; T has the dimensions of a rate or frequency. It is a measure or the leakagc of into the region 2, and of + , into the region 1. If the insulator is very- thick, T is zero and there is no pair tunneling. ~~t + , = n;/2e'h +2 = nl" , e ' 0 '.Then a, 1 - -- an, 80, at -,n,me'O1-+i+l-= at at -iT& ; We multiply (39) by nye-'6 to obtain, with S = 0, - el, We multiply (40) by rtin e-'82 to obtain Now equate the real and imagi~~ary parts of (41) and similarly of (42): an, -- - 2 ~ ( n ~ TL,)'!~ sill S ; - an' - - - 2 ~ ( n , n ~ ) ~ s i n 8 ; at at (43) 1/2 -- T a s . at (44) If n, n, as for identical superconductors 1 and 2. we haw from (44) that a8, - ae, a - (0, - 0,) = 0 at at . at From (43) we sec that The current flow fiom (1) to (2) is proportional to an,lat or, the saIrie thing, -an,lat. We therefore conclude from (43) that the current] of super- conductor pairs across the ji~nction depends on thc phase diffcrencr 6 as where 1 , is proportional to the transfer iriteractior~ T. The current J, is the maximum zero-voltage current that can be passed by the ju~iction. With 11o applied voltage a dc current will flow across the junction (Fig. 24), with a value between J, and -J, according to thc value of thc phase difference 0, - 8,. This is the dc Josephson effect. Ac Josephson Effect. Let a dc voltage V be applied across the jiinction. We can do this because the junction is an insulator. An electron pair experiences a potential energy difference yV on passing across the junction, where q = -2c. We can say that a pair on one side is at pote~itial energy -eV and a pair on the other side is at eV. The equations olnlotion that replace (38) are We proceed as above to find in place of (41) the equation 10 Superconductivity 291 This equation breaks up into the real part an,/at = 2T(n1 n2)ln sin 6 , (50) / / / / / / / / exactly as without the voltage V , and the imaginary part ae,/at = (eV/fi) - ~ ( n ~ h , ) ~ ' ' cos 6 , (51) which differs from (44) by the term eV/h. Further, by extension of (42), / / / / / / / / X7oItage v c Figure 24 Current-voltage characteristic of a Josephson whence an2/& = -2~(n,n,)"~ sin S ; (53) ae2/at = -(eV/h) - T(n,/nJm cos 6 . (54) junction. Dc currents flow under zero applied voltage op to a critical current i , : this is the dc Josephson effect. At voltages above V, the junction has a finite resistance, but the current has an oscillatory component of frequency w = 2eVlh: this is the ac Josepllson cffcct. From (51) and (54) with n, n2, we have a(e, - e,)iat = auat = -2etr1h We see by integration of (55) that with a dc voltage across the junction the relative phase of the probability amplitudes varies as 6(t) = S(0) - (2eVtlh) . (56) ] =lo sin [6(0) - (2eVtIfi)l . The superconducting current is given by (47) with (56) for the phase: (57) The current oscillates with frequency This is the ac Josephson effect. A dc voltage of 1 PV produces a frequency of 483.6 MHz. The relation (58) says that a photon of energy fiw = 2eV is emitted or absorbed when an electron pair crosses the barrier. By measuring the voltage and the frequency it is possible to obtain a very precise value of e/fi. Macroscopic Quantum Interference. We saw in (24) and (26) that the phase difference 0, - 0, around a closed circuit which encompasses a total magnetic flux CJ is given by The flux is the sum of that due to external fields and that due to currents in the circuit itself. We consider two Josephson junctions in parallel, as in Fig. 25. No voltage is applied. Let the phase difference between points 1 and 2 taken on a path through junction a be 6,. When taken on a path through junction b, the phase difference is ab. In the absence of a magnetic field these two phases must be equal. Now let the flux CJ pass through the interior of the circuit. We do this with a straight solenoid normal to the plane of the paper and lying inside the circuit. By (59), ab - 6, = (Ze/fic)CJ, or e e Sb=i30+-CJ ; 6,=60--CJ fic fic The total current is the sum of Ja and Jb. The current through each junc- tion is of the form (47), so that e@ = 2(Jo sin 6,) cos - fic Insulator a Figure 25 The arrangement for experiment on macroscopic quantum interference. A magnetic flux passes through the interior of the loop. 10 Superconductivity 293 - - l I I I I / I I I I I I l I I I I I I I I / I I I -500 4 0 0 3 0 0 2 0 0 1 0 0 0 100 ZOO 300 400 S O 0 Mabnetic field (mikgauss) Figure 26 Experimental trace of J , , versus magnetic field showing interference and diffraction effccts for hvo junctions A and B. The field periodicity is 39.5 and 16 mG h r A and B, respcc- titsely. Approximate ~rlaxi~rruln cnrrcnts are 1 mA (A) and 0.5 mA (B). The junction separation is 3 mm and junction width 0.5 mm for hoth cases. The zero offset of A is due to a background mag- netic ficld. (hftcr R. C. Jaklevic, J. Lambe, J. E. Mercereau, and A. H. Silver.) The current varies with and has maxima when eQl/fic = ST , s = integer . (61) The periodicity of the current is shown in Fig. 26. The short period varia- tion is produced by interference from the two junctions, as predicted by (61). The longer period variation is a diffraction effect and arises from the finite dimensions of each junction-this causes to depend on the particular path of integration (Problem 6). HIGH-TEMPERATURE SUPERCONDUCTORS High T, or IITS denotes superconductivity in materials, chiefly copper oxides, with high transition temperatures, accompanied by high critical cnr- rents and magnetic fields. By 1988 the long-standmg 23 K ceiling of T, in intermetallic comipounds had been elevated to 125 K in bulk superconducting oxides; these passed the standard tests for snperconductivity-the Meissner effect, ac Josephson effcct, persistent currents of long duration, and substan- tially zero dc resistivity. Memorable steps in the advance include: BaPb, 75Bi, T, = 12 K [BPBO] La, ,5B% 15CuO4 T, = 36 K [LBCO] YBaZCu307 T, = 90 K [YBCO] T1,Ba2Ca2Cu10,, T, = 120 K [TBCO] Hgo sT1, 2Ba2Ca,Cu,0,, T, = 138 K SUMMARY (In CGS Units) A superconductor exhibits infinite conductivity. A bulk specimen of metal in the superconducting state exhibits pcrfect dia- magnetism, with the rnagnetic induction B = 0. This is the Meissner effcct. The external magnetic field will penetrate the surface of the specimen over a distancc determined by the penetration depth A. There arc hvo types of superconductors, I and 11. In a bulk specimen of type I superconductor the superconducting state is destroyed and the ~ronr~lil state is restored by application of an external magnetic field in excess of a critical value H,. A type I1 superconductor has two critical fields, HC1 < H, < H,,; a vortex state exists in the range between I&, and H,,. The stabilization cncrgy density of the pure superconducting state is H?/S.rr in both type I and I1 snpercond~~ctors. In the superconducting state an energy gap, E, = 4kBTc, separates supcrcon- ducting electrons below from normal electrons above the gap. The gap is de- tected in experiments on heat capacity, infrared absorption, and tunneling. Three important lengths enter the theory of superconductivity; the London penetration depth A,: the intrinsic coherence length 5 , ; and the normal electron mean free path t. The London equation leads to the Meissner effect through the penetration equation V% ==/At, where AL = ( m ~ ~ / 4 ~ r n e ~ ) ~ ' ~ is the London penetration depth. In thc London equation A or B should be a weighted werage over the co- herence length t. The intrinsic coherence length 5, = ~ K C ~ J T ~ E ~ The BCS theory accounts for a supercouducting state forrnad frorrl pairs of electrons k ' ? and -kJ. Thcse pairs act as bosons. Tyye I1 snperconductors have t < A. The critical fields are related by H,, = (&/A)H, and H,, = (A/E)H,. The Ginzburg-Landau parameter K is de- fined as A/& Problems 1. Magnetic field penetration in a plate. The penetration equation rnay be written as h2V% = B, where A is thc penetratinn depth. (a) Slrow that B(x) inside a super- conducting plate perpendicular to the x axis and of thickness 6 is given by cosh (x/A) B(x' = " ' cosh (6/W) 10 Superconductivity where B, is the field outside the plate and parallel to it; here x = 0 is at the center of the plate. (b) The effective magnetization M(x) in the plate is defined by B(x) - B, = 47rM(x). Show that, in CGS, 47rM(x) = -B,(1/8A2)(S2 - 4x2), for S 4 A. In SI we replace the 47r by p,. 2. Criticalfield of thinfilms. (a) Using the result of Problem lb, show that the free energy density at T = 0 K within a superconducting film of thickness S in an exter- nal magnetic field B, is given by, for 6 4 A, (CGS) F,(x, B,) = Us(0) + (S2 - 4 ~ ~ ) ~ 2 6 4 7 r ~ ~ In SI the factor 7r is replaced by p , . We neglect a kinetic energy contribution to the problem. (b) Show that the magnetic contribution to Fs when averaged over the thickness of the film is ~:(6/~)~/967r. (c) Show that the critical field of the thin film is proportional to (A/S)H,, where H, is the bulk critical field, if we consider only the magnetic contribution to Us. Two-fluid model of a superconductor. On the two-fluid model of a supercon- ductor we assume that at temperatures 0 < T < T, the current density may be written as the sum of the contributions of normal and superconducting electrons: j = j , + j,, where j , = unE and js is given by the London equation. Here uo is an ordinary normal conductivity, decreased by the reduction in the number of normal electrons at temperature T as compared to the normal state. Neglect inertial ef- fects on bothy, andj,. (a) Show from the Maxwell equations that the dispersion re- lation connecting wavevector k and frequency w for electromagnetic waves in the superconductor is where A; is given by (148) with n replaced by ns. Recall that curl curl B = -V2B. (b) If T is the relaxation time of the normal electrons and n , i s their concentration, show by use of the expression a, = nNe2r/m that at frequencies w 4 1 1 7 the disper- sion relation does not involve the normal electrons in an important way, so that the motion of the electrons is described by the London equation alone. The super- current short-circuits the normal electrons. The London equation itself only holds true if h w is small in comparison with the energy gap. Note: The frequencies of intorest are such that w 4 wp, where wp is the plasma frequency. '4. Structure of a vortex. (a) Find a solution to the London equation that has cylin- drical symmetry and applies outside a line core. In cylindrical polar coordinates, we want a solution of 'This problem is somewhat difficult. that is singular at thc origin and for wlriclr the total flux is the flux quantum: The equation is in fact valid only outside the mornla1 core of radius 6. (b) Sho\i~ that the solution has the limits 5. London penetration depth. (a) Take the time derivative of the London equation (10) to show that aj/dt = (c2/4vA;)E. (b) I f mdvldt = qE, as for free carriers of charge y and mass rn, show that A 2 = d / 4 m q 2 . 6. Diffraction effect of Josephson junction. Consider a junction of rectangular cross sectiorr with a magnetic field B applied in the plane of the junction, normal to an edge of width w . Let the thickness of the junction be T. Assume for convenicncc that the phase difference of the two superconductors is d 2 when B = 0. Show that the dc current in the presence of the magnetic field is 7. Meissner effect in sphere. Consider a sphere of a typc 1 s~i~zrcond~~ctor with crit- ical field H,.. (a) Show that in the Meissner rcgimc the effective magiretieatiur~ M within the sphere is given by -8vMl3 = B,, the 1111iform applied lr~agrietic field. (h) Show that the magnetic field at the surfacc n T the sphere in the equatorial plane is 3B,/2. (It follows that the applicd ficld at which the hleissner affect starts to break down is 2HJ3.) Rernindcr: The demagnetization field of a ur~iformly magnetized sphere is -4vhfI3. Reference An excellent superconductor re.iew is the website supcrconductors.org. Diamagnetism and Paramagnetism LANGEVIN DIAMAGNETISM EQUATION 299 QUANTUM THEORY OF DIAMAGNETISM OF MONONUCLEAR SYSTEMS 301 PARAMAGNETISM 302 QUANTUM THEORY OF PARAMAGNETISM Rare earth ions Hund rules Iron group ions Crystal field splitting Quenching of the orbital angular momentum Spectroscopic splitting factor Van Vleck temperature-independent paramagnetism COOLING BY ISENTROPIC DEMAGNETIZATION 312 Nuclear demagnetization 314 PARAMAGNETIC SUSCEPTIBILITY OF CONDUCTION ELECTRONS 315 SUMMARY 317 PROBLEMS 318 1. Diamagnetic susceptibility of atomic hydrogen 2. Hund rules 3. Triplet excited states 4. Heat capacity from internal degrees of freedom 5. Pauli spin susceptibility 6. Conduction electron ferromagnetism 7. Two-level system 8. Paramagnetism of S = l system NOTATION: In the problems treated in this chapter the magnetic field B is always closely equal to the applied field B,, so that we write B for B, in most instances. fi ' Pauli paramagnetism (metals) ~ e ~ ~ l ~ e r a t u i e 2. P 6 Z e- 9 - w .P + 9 Diamagnetism \'an Meek paramagnetism t---------------------- Figure 1 Charactelistic magnetic susceptibilities of diamagnetic and paramagnetic substtances. CHAPTER 11: DIAMAGNETISM AND PARAMAGNETISM Magnetism is inseparable from quantum mechanics, for a strictly classical system in thermal equilibrium can display no magnetic moment, even in a magnetic field. The magnetic moment of a free atom has three principal sources: the spin with which electrons are endowed; their orbital angular mo- mentum about the nucleus; and the change in the orbital moment induced by an applied magnetic field. The first two effects give paramagnetic contributions to the magnetiza- tion, and the third gives a diamagnetic contribution. In the ground Is state of the hydrogen atom the orbital moment is zero, and the magnetic moment is that of the electron spin along with a small induced diamagnetic moment. In the ls2 state of helium the spin and orbital moments are both zero, and there is only an induced moment. Atoms with all filled electron shells have zero spin and zero orbital moment: finite moments are associated with unfilled shells. The magnetization M is defined as the magnetic moment per unit volume. The magnetic susceptibility per unit volume is defined as M (CGS) , y = - ; B where B is the macroscopic magnetic field intensity. In both systems of units ,y is dimensionless. We shall sometimes for convenience refer to M/B as the sus- ceptibility without specifying the system of units. Quite frequently a susceptibility is defined referred to unit mass or to a mole of the substance. The molar susceptibility is written as ,yM; the magnetic moment per gram is sometimes written as u. Substances with a negative mag- netic susceptibility are called diamagnetic. Substances with a positive suscep- tibility are called paramagnetic, as in Fig. 1. Ordered arrays of magnetic moments are discussed in Chapter 12; the ar- rays may be ferromagnetic, ferrimagnetic, antiferromagnetic, helical, or more complex in form. Nuclear magnetic moments give rise to nuclear paramag- netism. Magnetic moments of nuclei are of the order of times smaller than the magnetic moment of the electron. LANGEVIN DIAMAGNETISM EQUATION Diamagnetism is associated with the tendency of electrical charges par- tially to shield the interior of a body from an applied magnetic field. In electromagnetism we are familiar with Lenz's law: when the flux through an electrical circuit is changed, an induced (diamagnetic) current is set up in such a direction as to oppose the flux change. In a superconductor or in an electron orbit within an atom, the induced current persists as long as the field is present. The magnetic field of the in- duced current is opposite to the applied field, and the magnetic moment asso- ciated with the current is a diamagnetic moment. Even in a normal metal there is a diamagnetic contribution from the conduction electrons, and this diamagnetism is not destroyed by collisions of the electrons. The usual treatment of the diamagnetism of atoms and ions employs the Larmor theorem: In a magnetic field the motion of the electrons around a central nucleus is, to the first order in B, the same as a possible motion in the absence of B except for the superposition of a precession of the electrons with angular frequency (CGS) w = eB/2mc ; If the field is applied slowly, the motion in the rotating reference system will be the same as the original motion in the rest system before the application of the field. If the average electron current around the nucleus is zero initially, the application of the magnetic field will cause a finite current around the nucleus. The current is equivalent to a magnetic moment opposite to the applied field. It is assumed that the Larmor frequency (2) is much lower than the frequency of the original motion in the central field. This condition is not satisfied in free carrier cyclotron resonance, and the cyclotron frequency of the carriers is twice the frequency (2). The Larmor precession of Z electrons is equivalent to an electric current . . ) ! , . v . , , ..- ,. "? ... v . i . . . , ; . ... '.. . .
. - (SI) 1 = (charge)(revolutio~~s per unit time! - (-71.) The magnetic moment p of a current loop is given by the product (cur- rent) X (area of the loop). The area of the loop of radius p is rP2. We have Z e 2 ~ (CGS) p = --(p2) . (4) 4mc2 Here (p2) = (x2) + (y2) is the mean square of the perpendicular distance of the electron from the field axis through the nucleus. The mean square distance of the electrons from the nucleus is (2) = (x2) + (y2) + (2). For a spherically symmetrical distribution of charge we have (x2) = (y2) = (z2), SO that (2) = 3 p 2 ) . 11 Diamagnetism and Paramagnetism 301 From (4) the diamagnetic susceptibility per unit volume is, if N is the number of atoms per unit volume, This is the classical Langevin result. The problem of calculating the diamagnetic susceptibility of an isolated atom is reduced to the calculation of (2) for the electron distribution within the atom. The distribution can be calculated by quantum mechanics. Experimental values for neutral atoms are most easily obtained for the inert gases. Typical experimental values of the molar susceptibilities are the following: He Ne Ar Kr Xe ,yM in CGS in crn3/mole: -1.9 -7.2 -19.4 -28.0 -43.0 In dielectric solids the diamagnetic contribution of the ion cores is de- scribed roughly by the Langevin result. The contribution of conduction elec- trons in metals is more complicated, as is evident from the de Haas-van Alphen effect discussed in Chapter 9. QUANTUM THEORY OF DIAMAGNETISM OF MONONUCLEAR SYSTEMS We give the quantum treatment of the classical Langevin result. From Appendix (G.18) the effect of a magnetic field is to add to the Hamiltonian the terms iefi e2 x = - ( v - A + A . v ) + - A ~ ; 2mc 2mc2 for an atomic electron these terms may usually be treated as a small perturba- tion. If the magnetic field is uniform and in the z direction, we may write A =-' B z~ . A , = & B , A , = o , (7) and (6) becomes The first term on the right is proportional to the orbital angular momen- tum component L, if r is measured from the nucleus. In mononuclear systems this term gives rise only to paramagnetism. The second term gives for a spheri- cally symmetric system a contribution by first-order pertnrbation thcory. The associated magnetic moment is diamagnetic: aE1 e2V) p = - = = - s > (10) in agreement with the classical result (5). PARAMAGNETISM Electronic paramagnetism (positive contribution to X) is found in: 1. Atoms, molecules, and lattice defects possessing an odd number of electrons, as here the total spin of the system cannot bc zcro. Examples: free sodium atoms; gaseous nitric oxide (NO); organic free radicals si~ch as tri- phenylmethyl, C(C,H,),; F centers in alkali halides. 2. Free ato~ns and ions with a partly filled inner shell: transition ele- ments; ions isoelectronic with transition elements; rare earth and actinide ele- ments. Examples: Mn2', Gd3+, U4+. Paramagnetism is exhibited by many of these ions even when incorporated into solids, but not invariably. 3. A few compo~~nds with an even number of electrons, including molec- ular oxygen and organic biradicals. 4. Metals. QUANTUM THEORY OF PARAMAGNETISM The magnetic moment of an atom or ion in free space is given by where the total angular rriorrleriturn h.J is the surn of the orbital h.L and spin fiS angular momenta. The constant y is the ratio of the magnetic moment to the angular mo- mentum; y is called the gyromagnetic ratio or magnetogyric ratio. For electronic systems a quantity g called the g factor or the spectroscopic splitting factor is defined by ~ P B = - $ . (12) For an clcctron spin g = 2.0023, us~~ally taken as 2.00. For a Gee atom the g factor is given by the Land6 equation 2J(J + 1) (13) 11 Diamagnetism and Paramagnetism 303 Figure 2 Energy level splitti~ig for one elec- tron in a magnetic field B directed along the positivc z axis. For an electron the magnetic moment p is oppo~ite irr sign to thc spin S, so that p = -gp,S. In the low-energy state the rnagnetic moment is parallel to the magnetic field. 7 n 75 \upper state Figure 3 Fracho~ral pupulat~ons of a hyo-level system in thermal equilibrium at temperature T in a magnetic field B. The magnetic moment is proportional to the differe~~cc bctwcen the two curves The Bohr magneton p, is defined as efi/2m in CGS and efiflm in SI. It is closely equal to the spin magnetic moment of a free electron. The energy levels of the system in a magnetic field are where ml is the azimuthal quantum number and has the values J , J - 1, . . . , -1. For a single spin with no orbital moment we have VL] = 2; and g = 2, whence U = ?pBB. This splitting is shown in Fig. 2. If a system has only two levels the equilibrium populations are, with 7 ' kgT, here N,, Nz are the populations of the lower and upper levels, and N = N, + NP is the total number of atoms. The fractional populations are plotted in Fig. 3. The projcction of the magnetic moment of the upper state along the field direction is - p and of the lower state is p. The resultant magnetization for N atoms per unit volume is, with x = pB/k,T, F r - e - r - Np tanhx . M = (N, - N2)p = Np . - - eX + eKX For x < 1, tanh x = x, and we have In a magnetic field an atom wit11 angular momentum quantum niimher] has 2J + 1 equally spaced energy levels. The magnetization (Fig. 4) is given by Figure 4 Plot of magnetic moment versus B/T for spherical samples of (I) potassiulll chro~niurn alum, (11) ferric arnmoniunl alum, and (111) gadolirliuni sulfate octahydrate. Over 99.5% magnetic saturation is achieved at 1.3 K and about 50,000 gauss (5T). After W E. Henry. where the Brillouin function B, is defined by 2]+ 1 BJ(x) = - (2j + l?x 1 21 ctnh ( 21 ) - %ctnh ($) Equation (17) is a special case of (20) for] = i. For x = pB/kBT < 1, we have and the susceptibility is Here p is the effective number of Bohr magnetons, defined as p = g[]iJ + 1)11" . (23) 11 Diamagnetism and Paramagnetism 305 Figure 5 Plot of 1/x vs T for a gadoli~liu~n salt, Gd(C,H,SO,), . YH,O. The straight line is the Curie law (After L. C. Jackson and H. Kamrrlingh Onnes.) The constant C is know1 as the Curie constant. The form (19) is known as the Curie-Brillouin law, and (22) is known as the Curie law. Results for the paramagnetic ions in a gadolinium salt are shown in Fig. 5. Rare Earth Ions The ions of the rare earth elements (Table 1) have closcly similar chemical properties, and their chemical separation in tolerably pure form was accom- plished only long after their discovev. Their magnetic properties are fascinating: The ions exhibit a systematic variety and intelligible complexity The chemical properties of thc trivalent ions are similar because the outerrnost electron shells are identically in the 5~%~%onfiguration, like neutral xenon. In lan- thanum, just before the rare eartli group begins, the 4 f shell is empty; at cerium there is one 4 f electron, and the number of 4 f electrons increases steadily through the group until we have 4f13 at yttcrbium and the filled shell 4fL4 at lutetium. The radii of the trivalent ions contract fairly smoothly as we go through the group from 1.11 A at cerium to 0.94 A at ytterbium. This is known as the "lanthanide contraction." What distinguishes t l ~ e magnetic be- havior of one ion species from another is the riurnber of 4 f electrons com- pacted in the inner shell with a rahus of perhaps 0.3 A. Even in thc metals the Tablc 1 Effective magneton numbers p for trivalent lanthanide group ions (Near room tcmucrature) p(ca1c) = pkxp), Ion Cunfiguratiun Basic level d l ( l + 1)11'2 approximate 4 f core retains its integrity and its atomic properties: no other group of clc- lnents in the periodic table is as interesting. The preceding discussion of paramagnetism applies to atoms that have a (21 + 1)-fold degenerate ground state, the degeneracy being lifted by a mag- nctic field. The influence of all higher energy states of the system is neglected. These assllmptions appcar to be satisfied by a number of rare-earth ions, Table 1. The calc~llated magneton numbers are obtained with g values from the Land6 result (13) and the ground-state lcvcl assignment predicted below by the Hund theory of spectral terms. The discrepancy hetween the experi- nlental rnagrieton numbers and those calculated on these assumptions is quite marked for ~ u " aud Sm" ions. For these ions it is necessary to consider the influence of the high states of the L - S multiplet, as the intervals between successive states of the multiplet are not large conlpared to k,T at room tem- peratilre. A multiplet is the set of levels or different J values arising out of a given L and S. The levels of a mnltiplet arc split by the spin-orbit interaction. Hund Rules The Hiind rules as applied to electrons in a given shell of an atom affirm that electrons will occupy orbitals in such a way that the ground state is char- acterized by the following: 1. The maxi~nurn value of the total spin S allowed by the exclusion principle; 2. The maximum value of the orbital angular momentum L consistent with this value of S; I1 Dianragnetism and Paramugnetism 307 3. The value of the total angular ~rlo~nentum] is equal to I L - S I when the shell is less than half full and to L + S when the shell is more than half full. When the shell is just half full: the application of thc first rille gives L = 0, so that! = S. The first Hund rule has its origin in the exclusion principle and the coulomb repulsio~l between electrons. The exclusion principle prevents two electrons of the same spin from being at the same place at the same time. Thus electrons of the same spin are kept apart, further apart t h a electrons of opposite spin. Be- cause of the coulomb interaction the energy of electrons of the same spin is lower-the average potential energy is less positive for parallel spin than for antiparallel spin. A good example is the ion Mn2+. This ion has five electrons in the 3d shell, which is therefore half-filled. The spins can all be parallel if each electron enters a different orbital, and there are exactly five different orbitals available, characterized by the orbital quantum numbers r n ~ = 2, 1, 0, -1, -2. These will br occnpied by one electron each. We expect S = 9, and because Em, = 0 the only possible value of L is 0, as observed. The second Hund rule is best approached by model calc~llations. Pauling and Wilson,' for example, give a calculation of thc spectral terms that arise from the configuration p! The third Hund rule is a consequence of the sign of the spin-orbit interaction: For a singlc clectron the energy is lowest when the spin is antiparallel to the orbital angular momentum. But the low-energy pairs TTLL, mS are progressively used up as we add electrons to the shell; by the exclu- sion principlc when the shell is more than half full the state of lowest energy ncccssarily has the spin parallel to the orbit. Consider two examples of the Hund rules: The ion cc3+ has a single f electron; an f electron has E = 3 and s = i. Becausc the f shell is less than half 1 - full, theJ value by the preceding rule is IL - S I = L - , = i. The ion Pr3+ has two f electrons; one of the rules tells us that the spins add to give S = 1. Both f electrons cannot have = 3 without violating the Pauli exclusion principle, so that the maximum L consistent with the Pauli is not 6, hut 5. The J valuc is IL - S = 5 - 1 = 4. Iron Group Ions Table 2 shows that the experimental magneton numbers for salts of the iron transition group of tbe periodic tahle are in poor agreement with (23). The values often agrcc quite well with magneton numbers p = 2[S(S + l)]'" calculatcd as if the orbital moment were not there at all. Crystal Field Splitting The difference in behavior of the rare earth and thc iron group salts is that the 4 f shell respo~lsible fur paramagnetism in the rare earth ions lies deep 'L. Pauli~~g and E. B. \T7ilron; Introduction to quantum mechanics, McGraw-Hill, 1935, pp. 239-248. See also Dover Reptint. Table 2 Effective magneton numbers for iron group ions Basic p[calcj = p(m1c) = Ion Cnnfiguration level e[/(/ + 1)1112 2[S(S + l)]lA p(cxp)X "Representative values. inside the ions, within the 5s and 5p shells, whereas in the iron group ions the 3d shcll responsible lor paramagnetisni is the outermost shell. The 3d shell er- periences the intense inhomogeneous electric field produced. by neighboring ions. This inhomogeneous electric field is called the crystal field. The inter- action of the paramagnetic ions with the crystal field has two major effects: The coupling of L and S vectors is largely hroken up, so that the states are no longer specified by their J values; further, the 2L + 1 s~~hlevels belonging to a given L which are degenerate in the free ion may now be split by the crystal field, as in Fig. 6. This splitting dirni~iishes the contribution of the orbital mo- tion to the magnetic moment. Quenching ofthe Orbital Angular Momentum In an electric field directed toward a fixed nucleus, the plane of a classical orbit is fixed in space, so that all the orbital angular momentum components L,, Ly, Lz are constant. In quantum theory one angular momentum compo- nent, usually taken as L,, and the square of the total orbital angular momen- tum L ' are constant in a central field. In a noncentral field the plane of the orbit will move about; the angular momentum co~riponents are no longer con- stant and may average to zero. In a crystal L; will no longer be a constant of the motion, although to a good approximation nay continue to be constant. When L, averages to zero, the orbital angnlar momentum is said to he cluenched. The magnetic moment of a state is given hy the avrragc value of the magnetic moment operator pB(L + 25). In a magnetic field along the z direc- tion the orbital contribution to the magnetic moment is proportional to the quantum expectation value of L,; the orbital magnetic moment is quenched if the mechanical moment L, is quenched. 11 Diamagnetism and Paramagnetism 309 Figure 6 Consider an atom with orbital angular momentum L = 1 placed in the uniaxial crys- tallme electric field of the two positive ions along the z axis. In the free atom the states m , = f 1, 0 have identical energies-they are degenerate. In the crystal the atom has a lower energy when the electron cloud is close to positive ions as in (a) than when it is oriented midway between them, as in (b) and (c). The wavefunctions that give rise to these charge densities are of the form zfjr), x f ( r ) and yfjr) and are called the p,, p , , p, orbitals, respectively In an axially symmetric field, as shown, the p, andp, orbitals are degenerate. The energy levels referred to the free atom (dotted line) are shown in (d). If the electric field does not have axial symmetry, all three states will have different energies. As an example, consider a single electron with orbital quantum number L = 1 moving about a nucleus, the whole being placed in an inhomogeneous crystalline electric field. We omit electron spin. In a crystal of orthorhombic symmetry the charges on neighboring ions will produce an electrostatic potential cp about the nucleus of the form where A and B are constants. This expression is the lowest degree polynomial in x, y, z which is a solution of the Laplace equation V2cp = 0 and compatible with the symmetry of the crystal. In free space the ground state is three-fold degenerate, with magnetic quantum numbers m, = 1, 0, -1. In a magnetic field these levels are split by energies proportional to the field B, and it is this field-proportional splitting which is responsible for the normal paramagnetic susceptibility of the ion. In the crystal the picture may be different. We take as the three wavefunctions associated with the unperturbed ground state of the ion These wavefunctions are orthogonal, and we assume that they are normalized. Each of the U's can he shown to have the property where Y2 is the operator for the square of the orbital angular momentum, in units of fi. The result (26) confirms that the selected wavefunctions are in fact p functions, having L = 1. We observe now that the US are diagonal with respect to the perturbation, as by symmetry the nondiagonal elements vanish: (L'z!,lecplUy) = (U,lecplR) = (uy!,lecplUz) = 0 . (27) Consider for example, (U,lecplU,) = J xyl f(r)I2(Ax2 + By2 - (A + B)z2] dx dy dz ; (28) the integrand is an odd function of x (and also of y) and therefore the integral must be zero. The energy levels are then given by the diagonal matrix elements: (U,lecplU,) = J Ifir) I2{kx4 + By2x2 - (A + B)Z%] dx dy dz z A({, - 12) , (29) where I1 = J lfir) I2x4 dx dy dz ; I2 = J 1 f(r) I2x2zj2 dx dy c2z . In addition, (UylecplUy) = B(It - 12) ; (LrZlecpluz) = -(A + B)(I, - 1%) . The three cigcnstates in the crystal field are p functions with their angular lobes directed along each of the x, y, z axes, respectively. The orbital moment of each of the levels is zero, hecause The level still has a definite total angular momentum, since 2 ' is diagonal and gives L = 1, but the spatial components of the angular momentum are not constants of the motion and their time average is zero in the first approxima- tion. Therefore the components of the orbital magnetic moment also vanish in the same approximation. The role of the crystal field in the quenching process is to split the originally degenerate levels into nonmagnetic levels separated by energies % pH, SO that the magnetic field is a small perturbation in compari- son with the crystal field. At a lattice site of cubic symmetry there is no term in the potential of the form (24), that is, quadratic in the electron coordinates. Now the ground state of an ion with one p electron (or with one hole in a p shell) will be triply 11 Diamagnetism and Paramagnetism 311 degenerate. However, thc energy of the ion will be lowered if the ion displaces itself with respect to the surroundings, thereby creating a noncubic ~otential such as (24). Silch a spontaneous displace~rle~~t is k~lown as a Jahn-Teller effect and is often large and important, particularly with the ~ n ~ + and Cu2+ ions and with holes in alkali and silver halides. Spectroscopic Splitting Factor We suppose for convenicnce that the crystal field constants, A, B are such that U, = xfjr) is the orbital wave function of the ground state ol the atom in the crystal. For a spin S = : there are twv possible spin states S, = -ti repre- sented by the spin functions a, P, which in the absence of a magnetic field are degenerate in the zeroth approximation. The problem is to take into account the spin-orbit interactior~ energy AL . S. If the ground state function is $ , = UXa = ~f(r)a in the zeroth approxima- tion, the11 in the first approximation, considering the AL . S interaction by standard perturbation theory, we have where A1 is thc cnergy difference between the U, and U, states, and Az is the difference between the U, and Uz states. The term in U,P actually has only a second-order effect on the result and may be discardcd. The expectation value of the orbital angular momentum to the first order is given directly by and the magnetic moment of the statc as measured in the z direction is pB($ILZ + 2SZI+) = [-(A/Al) + l I p c L , . As the separation between the levels S : = ?$in a field H is AE = gpBH = 2[1- (A/Al)jpBH , the g value or spectroscopic splitting factor (12) in the z dircction is g = 2[1 - (AlA,)] . (31) Van Vleck Temperature-Independent Paramagnetism We consider an atomic or molccl~lar system which has no magnetic Ino- ment in the ground statc, by which we mean that the diagonal matrix element of the magnetic moment operator p, is zero. Suppose that there is a nondiagonal matrix element (sI,azI0) of thr mag- netic moment operator, connecting the ground state 0 with the cxcited state s of energy A = E, - E, above the ground state. Then by standard perturbation theory the wavefunction of the ground statc in a weak field (pZB 4) becon~es and the wavefunction of the excited state becomes t//l = GS - (~/A)(OIP~IS)$~ The perturbed ground state now has a moment (O'lpIO1) = ~ B ~ ( S ~ ~ ~ I O ) ~ ~ / A , (34) and the upper state has a inoinent There are two interesting cases to consider: Case (a). A < k,T. The surplus population in the ground state over the excited state is approximately equal to NA/W,T, so that the resultant magneti- zation is which gives for the siisceptihility x = NI (sI Pz 10) I2/kn~ Here N is the number of molecules per unit volume. This contrib~ition is of the usual Curie form, although the mechanism of magnetization here is hypo- larization of the states of the system, whereas with free spins the mechanism of magnetization is the redistribution of ions ariiong the spin states. We note that the splitting A does not enter in (37). Case (b). A B k,T. Here the population is nearly all in the ground state, so that The susceptibility is ~ N I ( ~ ~ P ~ I O ) ~ ~ x = A independent of temperature. This type of contribution is kriow~~ as Van \'leek pararnagnctism. COOLING BY ISENTROPIC DEMAGNETIZATION The first method for attaining temperatures much below 1 K was that of isentropic, or adiabatic, demagnetization of a paramagnetic salt. By its use, temperatures of lo-" and lower have been reached. The method rests on the fact that at a fixed temperature the entropy of a system of magnetic moments is lowcrcd by the application of a magnetic field. 11 Dbmagnetism and Paramagnetism ~~atticr I Time -+ Before t New cqlulihrium Time at which magnetic field is removed I Time -+ I Before 1 New eqoilihrii~m Time at \vhich magnetic field is removed Figure 7 During isentropic demagnetization the total entropy of the speci~ncn is constant. For effective cooling the ir~itial arrtropy of the lattice should he small in comparison \vith the entropy of the spin system. The entropy is a measure of the disorder of a system: the greater the dis- order, the higher is the entropy. In the magnetic field the moments will be partly lined up (partly ordered), so that the entropy is lowered by the field. The entropy is also lowered if the temperati~re is lowered, as more of the mo- rrlents line up. I l the lnagnctic field can then be removed without changing the entropy of thc spin system, the order of the spin syste~ri will look like a lower tempera- ture than the same degree of order in the presence of the field. When the specimen is demagnetized at constant entropy, entropy can flow into the spin system only from the system of lattice vibrations, as in Fig. 7. At the tempera- tures of interest the entropy of the lattice vibrations is usually negligible, thus the entropy ol the spin system will be essentially constant during isentropic demagnetization of the specimen. Magnetic cooling is a one-shot operation, not cyclic. We first find an expression for the spin entropy ola system of N ions, each of spin S, at a temperature sufficiently high that the spin system is entirely dis- ordered. That is, T is supposed to be much higher than some temperature A wl~ich characterizes the energy of the interactions (E,, - kBA) tending to orient the spins prcfcrcntially. Some of these interactions are discussed in Chapter 12. Thc definition of the entropy u of a system of G accessible states is u = kg In 6. At a temperature so high that all of the 2s + 1 states of each ion are nearly equally populated, G is the number orways of arranging N spins in 25 + 1 states. Thus G = (2s + I ) ~ , whence thc spin entropy us is: Figure 8 Entropy for a spin system as a function of temperature, assuming an internal random magnetic field BA of 100 gauss. The specimen is magnetized isothermally along ab, and is then insulated thermally The external magnetic field is turned off along bc. In order to keep the figure on a reasonable scale the initial temperature TI is lower than would be used in practice, and so is the external magnetic field. This spin entropy is reduced by a magnetic field if the lower levels gain in population when the field separates the 2s + 1 states in energy. The steps carried out in the cooling process are shown in Fig. 8. The field is applied at temperature T, with the specimen in good thermal contact with the surroundings, giving the isothermal path ab. The specimen is then insu- lated (Au = 0) and the field removed, the specimen follows the constant en- tropy path bc, ending up at temperature T2. The thermal contact at T , is pro- vided by helium gas, and the thermal contact is broken by removing the gas with a pump. The population of a magnetic sublevel is a function only of pB/kBT, hence of B/T. The spin-system entropy is a function only of the population distribu- tion; hence the spin entropy is a function only of BIT. If B, is the effective field that corresponds to the local interactions, the final temperature T2 reached in an isentropic demagnetization experiment is where B is the initial field and TI the initial temperature. Nuclear Demagnetization Because nuclear magnetic moments are weak, nuclear magnetic interac- tions are much weaker than similar electronic interactions. We expect to reach a temperature 100 times lower with a nuclear paramagnet than with an elec- tron paramagnet. The initial temperature T , of the nuclear stage in a nuclear spin-cooling experiment must be lower than in an electron spin-cooling 11 Diamagnetism and Paramagnetism 315 Initial magnetic field in kG I ~ t i a l BIT in 10%/~ Figure 9 Nuclear dcrnaglletizations of copper nuclei in the metal, starting frottr 0.012 K and various fields. (After M. V. Hobden and N. Kurti.) experiment. If we start at B = 50 kG and T, = 0.01 K, then pB/kBT1 = 0.5, and the entropy decrease on magnetization is over 10 percent of the maximum spin entropy. This is sufficient to overwhelln the lattice and from (41) we esti- mate a final temperature T2 = K. The first nuclear cooling experiment was carried out on Cu nuclei in the metal, starting from a first stage at about 0.02 K as attained by electronic cooling. The lowest temperature reached was 1.2 X lo-". The results in Fig. 9 fit a line of the form of (41): TZ = T1(3.1/B) with B in gauss, so that BA = 3.1 gauss. This is the effective interaction field of the magnetic moments of the Cti nuclei. The motivation for using nuclei in a metal is that conduction electrons help ensure rapid thermal contact of lattice and nuclei at the temperature of the first stage. PARAMAGNETIC SUSCEPTIBILITY OF CONDUCTION ELECTRONS \Ve are going to try to show how on the hasia of these statistics the fact that many metals arc diamagnetic, or only weakly paramagnetic, can be brought into agreement with tllr existence of a magnetic moment of the electrons. W. Pauli. 1927 Classical free clcctron theory gives an unsatisfactory account of the para- magnetic ~usceptihilit~ of the conduction electrons. An electron has aqsociated with it a magnetic moment of one Bohr magneton, pB. One might expect that the conduction electrons would make a Curie-type paramagnetic contribution (22) to the magnetization of the metal: M = N&B/kB~. Instead it is observed that the magnetization of most normal nonferromagnetic metals is indepen- dent of temperature. Pauli showed that the application of the Fermi-Dirac distribution (Chapter 6) would correct the theory as required. We first give a qualitative explanation. The result (18) tells us that the probability an atom will be lined up parallel to the field B exceeds the probability of the antiparallel orientation by roughly pBIkBT. For N atoms per unit volume, this gives a net magnetization =Np2BIkBT, the standard result. Most conduction electrons in a metal, however, have no possibility of turning over when a field is applied, because most orbitals in the Fermi sea with parallel spin are already occupied. Only the electrons within a range kBT of the top of the Fermi distribution have a chance to turn over in the field; thus only the fraction TITF of the total number of electrons contribute to the susceptibility. Hence which is independent of temperature and of the observed order of magnitude. We now calculate the expression for the paramagnetic susceptibility of a free electron gas at T < T,. We follow the method of calculation suggested by Fig. 10. An alternate derivation is the subject of Problem 5. Total energy, Idnetic + magnetic, of electrons I - Fermi level - Density of orbitals Figure 10 Pauli paramagnetism at absolute zero; the orbitals in the shaded regions in (a) are occupied. The numbers of electrons in the "up" and "down" band will adjust to make the energies equal at the Fermi level. The chemical potential (Fermi level) of the moment up electrons is equal to that of the moment down electrons. In (b) we show the excess of moment up electrons in the magnetic field. 11 Diamagnetism and Paramagnetism 317 The concentration of electrons with magnetic nlonrents parallel to the magnetic ficld is written for absolute zero. Here ~ D ( E + pB) is the density of orbitals of one spin orientation, with allowance for the downward shift of energy by -pB. The approximation is written for kgT 4 eF. The concentration of electrons with magnetic moments antiparallel to the magnetic field is The magnetization is given by 12.1 = p(N+ - I"), SO that with D(eF) = 3 N / 2 ~ ~ = 3N/2ksTF from Chapter 6. The result (45) gives the Pauli spin magnetization or the conduction electrons, for k,T 4 e,. In deriving the paramagnetic snsceptihility, we have supposed that the spatial motion of thc electrons is not affected by the magnetic field. But the wavefunctions are modified by the magnetic field; Landau has shown that for free electrons this causes a diamagnetic 1rlome11t equal to -$ of the para- magnetic moment. Thus the total magrletization of a free electron gas is Before comparing (46) with the experiment we must take account of the diamagnetism of the ionic cores, of band effects, and of electron-electron in- teractions. In sodiiim the interaction effects increase the spin susceptibility by perhaps 75 percent. The magnetic susceptibility is considerably higher for most transition metals (with unfilled inner electron shells) than for the alkali metals. The high values suggest that the density of orbitals is linusually high for transition met- als, in agreement with measurements of the electronic heat capacity. We saw in Chapter 9 how this arises from hand theory. SUMMARY (In CGS Units) The diamagnetic s~isceptihility of N atoms of atornic number Z is , y = -Ze2N(?)/6rnc" where (r2) is the mean square atomic radius. (Lange?n) Atoms with a permanent magnetic moment p have a paramagnetic susccpti- bility x = N / ~ , ~ l 3 k ~ T , for pB < . kRT. (Curie-Langevin) For a system of spins S = $, the exact rnag~letizatiorl is M = N p tanh(pA/kBT), where p = agpB. (Brillouin) The grolind state of clectrons in the same shell have the maximum value of S allowed by the Pauli principle and the maximum I, consistent with this S. The J value is L + S if the shell is more than half fill1 and I L - SI if thc shcll is less than half full. A cooling process operates by demagnetization of a paramagnetic salt at constant entropy. The final temperature reached is of the order of (B,/B)T,,,,,,,. where B, is the effective local field and B is the initial applied magnetic field. The para~nagnetic susceptibility of a Fer~ni gas of corlductiorl electrons is , y = 3Np2/2eF, independent of temperature for kgT 4 eF. (Pauli) Problems 1. Diamagnetic susceptibility of atomic hydrogen. The wave function of the hydrogen atom in its ground state (Is) is $= (.rra~)-'/'exp(-r/a,,), where a, = fi2/m" 0.529 X 10-'cm. The charge density is p(x, y, z ) = -el$12, accordmg to the statistical interpretation of the wave function. Show that for this state (?) = 3ai, and calculate the molar diamagnetic susceptibility of atomic hydrogen (-2.36 X 10 cm3/mole). 2. Hund rules. Apply the Hund rules to find the ground state (the basic level in the notation of Table 1) of (a) Eu++, in the configuration 4f7 5sZp6; (b) Yb3+; (c) Tb3+. The results for (b) and (c) are in Table I, but you should give the separate stcps in applying the rules. 3. Triplet excited states. Solnc organic lnolecules have a triplet (S = 1) excited state at an energy k,A above a singlct (S = 0) ground state. (a) Find an expression fnr the magnetic momcnt (p) in a ficld B. (b) Show that the susceptibility fnr T % A is ap- proximately independent of A. (c) With thc hclp of a diagram of energy levels versus ficld and a rongh sketch of entropy verslls field, explain how this system might be cooled by isentropic magnetixation (not den~agneti~ation). 4. Heat capacity from internal degrees of freedom. (a) Corlsider a two-level systerrl with an energy splitting kgA between upper and lower states; the splittirrg rnay arise h r n a magnetic field or in other ways. Show that the heat capacity per systern is The function is plotted in Fig. 11. Peaks of this type in the heat capacity are often known as Schottky anomalies. The maximum heat capacity is quite high, but for T < A and for T % A the heat capacity is low. (b) Show that for T A we have C - kB(A/2T)' + . . . . The hyperfine interaction between nuclear and electronic mag- netic moments in paramagnetic salts (and in systems hating electron spin order) causes 11 Diamagnetism and Paramagnetism 0 1 2 3 4 5 6 x = Tlh Figure 11 Heat capacity of a two-level system as a function of TlA, where A is the level splitti~~g. The Schottky anorllaly in the heat capacity is a very useful tool for determining energ level split- tings of ions in rare-earth and transition-grnnp metals, com~pounds, and alloys. Figure 12 The normal-statc heat capacity of gallium at 1 ' < 0.21 K. The nnclear q ~ ~ a d r n ~ o l e (C c c T -' ) and conduction electron (C T) contributions dominate the heat capacity at very low te~nperatures. (After N. E. Phillips.) splittings mlth A = 1 to 100 mK. These splittings are often detected experimentally by the presence of a ten11 in 1 / P in the heat capacity in the region T A. Nuclear electric quadrupole interactions wit11 crystal fields also cause splittings, as in Fig. 12. 5. Pauli spin susceptibility. Tlre spin s~wceptibility of a conduction electron gas at alsolute zero may be approacl~ed hy another mcthod. Let be the corrcentrations of spin-up and spin-down electrons. (a) Show that in a mag- netic field B the total energy of the spin-up band in a frcc clcctron gas is where E, = $A's,, in terms of the Ferini energy eF in zero magnetic field. Find a similar cxprcssion for E-. (b) hlinimize = E+ + E- with respect to [ and solve for the equilibrium valuc of 5 in thc approximation 5 < 1. Go on to show that the magnetization is M = 3NpEB12sF, in agreement with Eq. (45). 6. Conduction electron ferromagnetism. We approximate the effect of exchange interactions among the condnction clcctrons if we assurnc that electrons with paral- lel spins interact with each other with energy -V, and V is positive, while electrons wit11 aritiparallel spins do not interact with each other. (a) Show with the help of Problerrr 5 that the total energy of the spin-up hand is find a similar expression for E-. (b) Minimize the total energy and solve for 5 in the limit [ + 1. Show that the magnetization is so that the exchange interaction enhances the susccptihility. (c) Show that with B = 0 the total energy is unstalde at [ = 0 when V > 4sF/3N. If this is satisfied, a ferrorrragrretic state (i # 0) will have a lower energy than the paramagnctic statc. Because of the assn~rrptio~i 5 < 1, this is a s~~fficient condition for ferromagnctism, but it may not be a necessary co~~dition. It is known as tlie Stoner condition. 7. Two-level system. The result of Problerrl4 is often seen in another form. (a) If the two energy levels are at A and A , show that the energy arid heat ca~~acity are (b) If the systerrl has a random composition sr~ch that all values of A arc cqually likely up to some limit A,, show tlvat the heat capacity is linearly proportional to thc temperature, provided k,T 4 A,. This result was applied to the heat capacity of di- lute magnetic alloys by JV. Marshall, Phys. Rev. 118, 1519 (1960). It is also used in the theory of glasses. 8. Paramagnetism of S = 1 system. (a) Find the ~nagneti~,atiorr as a function of magnetic field and temperature for a system of spins with S - 1, rnoment fi2 and concentration n. (b) Show that in the limit pB 4 kT the result is M - (2r~p"~3kT)R. Ferromagnetism and Antiferrornagnetism FERROMAGNETIC ORDER Curie point and the exchange integral Temperature dependence of the saturation magnetization Saturation magnetization at absolute zero MAGNONS Quantization of spin waves Thermal excitation of magnons NEUTRON MAGNETIC SCATTERING FERRIMAGNETIC ORDER Curie temperature and susceptibility of ferrimagnets Iron garnets ANTIFERROMAGNETIC ORDER Susceptibility below the NBel temperature Antiferromagnetic magnons FERROMAGNETIC DOMAINS Anisotropy energy Transition region between domains Origin of domains Coercivity and hysteresis SINGLE-DOMAIN PARTICLES Geomagnetism and biomagnetism Magnetic force microscopy NOTATION: (CGS) B = H + 4 v M ; (SI) B = "(H + M). We call B, the applied magnetic field i 1 1 both systems of units: in CGS we have B, = Ha and in SI we have B, = pa,. The susceptibility is , .
y = MIB, in CGS and , y = MIH, = p&flB, in SI. One tesla = lo4 gauss. SUMMARY PROBLEMS 1. Magnon dispersion relation 2. Heat capacity of magnons 3. N6el temperature 4. Magnetoelastic coupling 5. Coercive force of a small particle 6. Saturation magnetization near T . 7. NBel wall 8. Giant magnetoresistance t t t t t t t l t l t t + t + t + Simple ferromagnet Simple anlifenomagnet Ferrimagnet Canted antiferromagnet Helical spin array Ferromagnetic energy band Figure 1 Ordered arrangements of electron spins. CHAPTER 12: FERROMAGNETISM AND ANTIFERROMACNETISM FERROMAGNETIC ORDER A ferromagnet has a spontaneous rr~agr~etic moment-a magnetic momcnt even in zero applied magnetic field. The existence of a spontaneous moment suggests that electron spins and magnetic moments are arranged in a regular manner. The order need not be simple: all of the spin arrangements sketched in Fig. 1 except the simple antiferromagnet have a spontaneous magnetic moment, called the saturation moment. Curie Point and the Exchange lntegral Consider a paramagnet with a concentration of M ions of spin S. Givcn an internal interaction tending to line up the magnetic momcnts parallel to each other, we shall have a ferromagnet. Let us postuIate such an interaction and call it the exchange field.' The orienting effect of the exchange field is opposed by thermal agitation, and at elcvatcd temperatures the spin order is destroyed. We treat the exchange field as equivalent to a magnetic field BE. The mag- nitude of the exchange field may be as high as 10' gauss (lo3 tesla). \Ve assume that BE is proportional to the magnetization M. The magnetization is defined as the magnetic moment per unit volume; unless otherwise specified it is understood to be the value in thermal cquilih- riuni at the temperature T. If domains (regions magnrtizcd in different direc- tions) are present, the magnetization refers to the value withn a domain. In the mean-field approximation we assume each magnetic atom expe- riences a field proportional to the magnetization: where A is a constant, independent of temperature. According to (I), each spin sees the average magnetization of all the other spins. In truth, it may scc only near neighbors, but our sirnplification is good for a first look at the problem. The Curie temperature T, is the tempcraturc above which the sponta- neous magnetization vanishes; it separates the disordered paramagnetic phase at T > T, from the ordcrcd ferromagnetic phase at T < T,. \Ve can find T,. in terms of the constant A in (1). 'Also called the molecular field or the Weiss field, after Pierre \'i'ciss who was the first to imagine such a field. The exchangc ficld BE silnulates a real magnetic field in the expressions for the energy -p . RE and the torque p X BE on a rrlag~etic IIIUIDBI~~ p. But BE is not really a magnetic field and therefore does not enter into the Maxwell eqnatinns; for example, there is no current density j related to BE by curl H = 4wjIc. The magnitude of HE is typically lo4 larger than the average magnetic field of the magnetic dipole> of the ferro~nagnct. Consider the paramagnetic phase: an applied field B, will cause a finite magnetization and this in turn will cause a finite exchange field BE. If xP is the paramagnetic susceptibility, The magnetization is equal to a constant susceptibility times a field only if the fractional alignment is small: this is where the assumption enters that the spec- imen is in the paramagnetic phase. The paramagnetic susceptibility (Chapt. 11) is given by the Curie law xp = CIT, where C is the Curie constant. Substitute (1) in (2); we find MT = C(B, + AM) and M C x = B , = (T-CA) The susceptibility (3) has a singularity at T = CA. At this temperature (and below) there exists a spontaneous magnetization, because if x is infinite we can have a finite M for zero B,. From (3) we have the Curie-Weiss law This expression describes fairly well the observed susceptibility variation in the paramagnetic region above the Curie point. The reciprocal susceptibility of nickel is plotted in Fig. 2. 350 400 450 500 Temperature in O C Figure 2 Reciprocal of the susceptibility per gram of nickel in the neighborhood of the Curie temperature (358°C). The density is p. The dashed line is a linear extrapolation from high temperatures. (After P. Weiss and R. Forrer.) 12 Ferromagnetiem and Antifelromagnetism From (4) and the definition (11.22) of the Curie constant C we may deter- mine the value of the mean field constant h in (1): For iron T, = 1000 K, g = 2, and S = 1; from (5) wc have h = 5000. With hf, = 1700 we have BE = AM = (5000)(1700) = lo7 CJ = lo3 T. The exchange field in iron is very much stronger than the real magnetic field due to the other magnetic ions in thc crystal: a magnetic ion produces a field = ~ , / a ~ or about lo3 G = 0.1 T at a neighboring lattice point. The exchange field gives an approxirrlate representation of the quantum- mechanical exchange interaction. On certain assumptions it is shown in texts on quantum theory that the energy of interaction of atoms i, j bcaring electron spins Si, S , coritai~ls a term U = -2JS,. S 1 , (6) where] is the exchange integral and is rclated to the overlap of the charge dis- tributions of the atolns i, j . Equation (6) is called the Heisenberg model. The charge distribution of a system of two spins depends on whether the spins are parallel or antiparallel2 for the Pauli principle excludes two electrons of the same spin from being at the same place at the same time. It does not ex- clude two electrons of opposite spin. Thus the electrostatic energy of a system will depend on the relative orientation of the spins: the difference in energy defines the exchange energy. The exchange energy of two electrons may bc writtcn in the form -2Jsl . s, as in (6), just as if there were a direct coiipling hehveen the directions of the two spins. For many purposes in ferromagnetism it is a good approximation to - - trcat the spins as classical angular momentum vectors. We can establish an approximate connection between the exchange inte- gral J and the Curie temperature T,. Suppose that the atom under considera- tion has z nearest neighbors, each connected with the central atom by the interaction 1. For more distant neighbors we take J as zero. The mean field theorv result is Better statistical approximations give somewhat different results. For thc sc. hcc, and fcc structures with S = i, Rushbrooke arid Wood give '1itwo spins arc antiparallel, the wavefunctions of the tu-o electrons must be symmetric, as in the combination u(r,)o(r,j + u(r,)o(r,). If the two spills arc parallcl, the Pauli principle requires that the orbital part of the wavefunction be antisymmetric, as in u(r,)o(r,j - u(r,)v(r,), for here if we i ~ ~ t e r c h a ~ ~ g c thc coordinales r,, r, the wavefunction changes s i p . If we set the positions equal so that r, = r, then the antisymmetric fu11ctio11 vanishes: for parallel spins there is zero probability of finding the two electrons at the same position. kBT,/z. = 0.28; 0.325; and 0.346, respectively, as compared with 0.500 from (7) for all three structures. II iron is represcntcd by the Heisenherg model with S = 1, then the obsenrcd Ci~rie temperature corresponds toJ = 11.9 meV Temperature Dependence ofthe Saturation Magnetization We can also use the mean field approxirrratio~l below the Curie tempera- ture to find the m~agnetizatiori as a function of temperature. We proceed as before, but instead of the Curie law we use the complete Brillouin expression for the magnetization. For spin $ this is A4 = Np tanh(pB/kBT). If we omit thc applied magnetic field and replace B by the molecular field BE = AM, then M = Np. tanh(pAA4/kBT) . (8) We shall see that solutions of this equation with nonzero M exist in the tem- perature range between 0 and T.. To solve (8) we write it in terms of the reduced magnetization m = M/Np and the reduced temperature t = k B ~ / ~ p 2 A , whence We then plot the right and left sides of this equation separately as functions of m, as in Fig. 3. The intercept of the two curves gives the value of 7 r ~ at the tem- perature of interest. The critical temperature is t = 1, or 2 : : = ~'p.~Alk,. tanhidt) fort = 0.5 furt = 1 fort = 2 0 0.2 0.4 0.6 0.8 1.0 1.2 Figure 3 Graphical solution of Eq. (9) for the reduced magnetization m as a function of tempera- ture. The rednced magnetization is defined as m = MlA'p. Thc Icft-hand side of Eq. (9) is plotted as a straight line m with unit slope. The right-hand side i s tanh(m/t) and is plotted vs, rn for three dffercnt values of the reduced temperature t = k,T/Np2h= 1771. The three curves correspond to the temperatnres 2T,, T,, and 0.5T,. The curve for t = 2 i~~tersccts thc straight line m only at m = 0, as appropriate for the paramagnetic region (there is no external applied magnetic field). The curve fort = 1 (or T = T,) is tangent to the straight line m at the origin; this temperature marks the onset of ferromagnetism. The curve fort = 0.5 is in the ferromagnetic region and inter- sects the straight line m at about m = 0.94%. As t + 0 the intercept moves I I ~ to m = 1, so that all magnetic rnolnerrts arc lincd up at absolute zero. 12 Fewornagnetism and Antiferrornagnetism Figure 4 Saturation magnetization of nickel a 7 a function of temperature, together with the theoretical curvc for S = on the mean field theory Experimental values hy P. Weiss and R. Forrer. The curves of 1 M versus T obtained in this way reproduce roughly the fea- tures of the experimental results, as shown in Fig. 4 for nickel. As T increases, the magnetization decreases smoothly to zero at T = T,. This behavior classifies the usual ferromagneti~/~aramagnetic transition as a second-order transition. The mean-field theory does not give a good description of the variation of M at low temperatures. For T + T, the argument of tanh in (9) is large, and tanh 5 - 1 - %-g. To lowest order the magnetization deviation AM = M(0) - M(T) is The argument of the exponential is equal to -2TJT. For T = 0.1TC we have AMINp 4 x The experimental results show a much more rapid dependence of AM on temperature at low temperatures. At T = 0.1TC we have AM/A4 2 X loF3 from the data of Fig. 5. The leading term in AM is observed from experiment to have the form where the constant A has the experimental value (7.5 t 0.2) X deg3I2 for Ni and (3.4 ? 0.2) X 10-%eg-"~ for Fe. The result (11) finds a natural expla- nation in terms of spin wave theory. ITpprr Middle h r 1 kG .
. Figure 5 Decrease in magnetization of nickel with trmperatnre. after Arple, Charap. In the plot AM = 0 at 4.2 K . Saturation Magnetization at Absolute Zero Table 1 gives representative vali~es of the saturation magnetization M,, the ferromagnetic Curie temperature, and the effective magneton number de- fined by M,(O) = n,Np,, where N is the number of formula units per unit volume. Do not confuse n, with the paramagnetic effective magneton r~urrlber p defined by (11.23). Tahle 1 Ferromagnetic cryslals Magnetization M,, in gauss Curie Room temperatllre, Fe 1707 1740 2.22 1043 Co 1400 1446 1.72 1388 Ni 385 510 0.606 627 Gd - 2060 7.fi3 292 Dy - 2920 10.2 88 MnAs 670 870 3.4 318 MnBi 620 680 3.52 630 MnSh 710 - 3.5 587 CrO, 515 - 2.03 386 MnOFe,O, 410 - 5.0 573 FeOFe,O,, 480 - 4.1 858 NiOFe,O, 270 - 2.4 (858) CuOFc,O, 135 - 1.3 728 MgOFr,O, 110 - 1.1 713 EuO - 1920 6.8 69 Y:3FesOls 130 200 5.0 560 12 Ferromagnstism and Antiferrornagnetism 329 Observed values of n , are often nonintegral. There are several possible causes. One is the spin-orbit interaction which adds or subtracts some orbital magnetic moment. Another cause in ferromagnetic metals is the conduction electron magnetization induced locally about a paramagnetic ion core. A third cause is suggested by the drawing in Fig. 1 of the spin arrangement in a ferri- magnet: if there is one atom of spin projection -S for every two atoms +S, the average spin is $9. Are there in fact any simple ferromagnetic insulators, with all ionic spins parallel in the ground state? The few simple ferromagnets known at present include CrBr,, EuO, and EuS. A band or itinerant electron model accounts for the ferromagnetism of the transition metals Fe, Co, Ni. The approach is indicated in Figs. 6 and 7. The relationship of 4s and 3d bands is shown in Fig. 6 for copper, which is not ferromagnetic. If we remove one electron from copper, we obtain nickel which has the possibility of a hole in the 3d band. In the band structure of nickel shown in Fig. 7a for T > T, we have taken 2 X 0.27 = 0.54 of an electron away from the 3d band and 0.46 away from the 4s band, as compared with copper. The band structure of nickel at absolute zero is shown in Fig. 7b. Nickel is ferromagnetic, and at absolute zero nB = 0.60 Bohr magnetons per atom. After allowance for the magnetic moment contribution of orbital electronic motion, D nickel has an excess of 0.54 electron per atom having spin preferentially ori- ented in one direction. The exchange enhancement of the susceptibility of metals was the subject of Problem 11.6. 4s 3d Filled-10 electrons (4 3dt 3 d .
1 5 electrons 5 electrons (b) Figure 6a Schematic relationship of 4s and 3d hands in metallic copper. The 3d band holds 10 electrons per atom and is filled. The 4s hand can hold two electrons per atom; it is shown half- filled, as copper has one valence electron outside the filled 3d shell. Figure 6b The filled 3d band of copper shown as two separate sub-hands of opposite electron spin orientation, each band holding five electrons. With both sub-hands filled as shown, the net spin (and hence the net magnetization) of the d hand is zero. Fermi Fermi - - surface surface 4.73 electrons Figure 7a Band relationships in nickel above the Curie temperature. The net magnetic moment is zero, as there are equal numbers of holes in the 3d and 3d f bands. Figure 7b Schematic relationship of bands in nickel at absolute zero. The energies of the 3d f and 3d L sub-bands are separated by an exchange interaction. The 3d T band is filled; the 3d .
1 band contains 4.46 electrons and 0.54 hole. The 4s band is usually thought to contain approxi- mately equal numbers of electrons in both spin directions, and so we have not troubled to divide it into sub-bands. The net magnetic moment of 0.54 pB per atom arises from the excess population of the 3d f band over the 3d .
1 band. It is often convenient to speak of the magnetization as arising from the 0.54 hole in the 3d .
1 band. MAGNONS A magnon is a quantized spin wave. We use a classical argument, just as we did for phonons, to find the magnon dispersion relation for w versus k. We then quantize the magnon energy and interpret the quantization in terms of spin reversal. The ground state of a simple ferromagnet has all spins parallel, as in Fig. 8a. Consider N spins each of magnitude S on a line or a ring, with nearest-neighbor spins coupled by the Heisenberg interaction: t t t t t t t t l t t t 8 8 V 8 8 V + o r + a t + + a r Figure 8 (a) Classical picture of the ground state of a simple ferromagnet: all spins are parallel. (b) A possible excitation: one spin is reversed. (c) The low-lying elementary excitations are spin waves. The ends of the spin vectors precess on the surfaces of cones, with successive spins ad- vanced in phase by a constant angle. 12 Ferromagnetism and Antijerromagnetism Figurc 9 A spin wave on a line of spins. (a) The spins viewed in perspective. (b) Spins viewed from ahow, showing U I I ~ wavclcngth. The wave is drawn through the ends of the spin vectors. Here J is the exchange integral and hSp is the angular ~no~nentum of the spin at sitc p. If we treat the spins Sp as classical vectors, then in the ground statc Sp . Spt = SP and the exchange energy of the system is Uo = -2NJS2. What is the energy of the first excited state? Consider an excited state with one particular spin reversed, as in Fig. 8b. We see from ( 1 2 ) that this increases the energy by ~JS', so that U1 = Uo + 8]s2. Rut we can form an excitation of mudl lower energy- i1 we let all the spins share the reversal, as in Fig. 8c. The elementary excitations of a spin system have a wavelike form and are called maglions (Fig. 9). These are analogous to lattice vibrations or phonons. Spin waves are oscillations in the relative orientations of spins on a lattice; lattice vi- brations are oscillations in the relative positions of atoms on a lattice. We now give a classical derivation of the magnon dispersion relation. The terms in (12) which involve the pth spin are 117e write magnetic mornent at site p as /+ = -gpBSp Then (13) becomes which is of the form -pp . Bp, where the effective magnetic field or exchange field that acts on the pth spin is Fro111 ~llechanics the rate of change of the angular momerltu~n fiSp is equal to the torquc pp X Bp which acts on the spin: fi dS,ldt = ~ c , X B,,, or In Cartesian components arid si~~lilarly for dSzMt and dS;ldt. These equations involve products of spin components and are nonlinear. If the amplitude of the cxcitation is small (if S;, S; S), we may obtain an approximate set of linear equations by taking all S; = S and by neglecting terms in the product of ST and S Y which appear in the equation for dSz/dt. The linearized equatio~is are By analogy with pho~ion yroble~~is we look for traveling wave solutions of (18) of the form where u, a are constants, p is an integer, and a is the lattice constant. On sub- stitution into (18) we have -iwu = (2JS/fi)(2 - e-lk" - e"") z. = (4]S/fi)(l - cos kak ; -i m = -(2]SIfi)(2 - e-ik" - eika)u = -(4JS/&)(l - cos ka)u . These equations have a solution for 1 1 and u if the determinant of the corf- ficients is equal to zero: (ajs/fi)(l- cos ku) ~ s f i ) ( l - cos ,a) io = 0 , (21) whence fro = 4jS(1 - cos ka) . (22) This result is plotted in Fig. 10. With this solution we find that o = -iu, corre- sponding to circular precession of each spin about the z axis. \Ve see this on taking real parts of (20), with v set equal to -iu. Then S; = u cos(pka - wt) ; S : = u sin(pka - ot) . Equation (22) is the dispersion relation for spin waves in one dimension with nearest-neighbor i~iteractions. Precisely the sarrle result is obtained froni the quantum-mechanical solution; see QTS, Clrapter 4. At long wavelengtlis ka < 1, so that (1 - cos ka) = : ( k ~ ) ~ and The frequency is proportional to k2; in the same limit the frequency of a phonon is dlrectly proportional to k. 12 Ferromagnetism and Ant$erromagnetism 333 2 - 3 ? 1 - 3 +c Figure 10 Dispersion relation for magnons in 0- a ferromagnet in one dimension with nearest- 0 ' 7 7 - neighbor interactions. k - The dispersion relation for a ferromagnetic cubic lattice with nearest- neighbor interactions where the summation is over the z vectors denoted by 8 which join the central atom to its nearest neighbors. For ka < 1, for all three cubic lattices, where u is the lattice constant. The coefficient of k2 often may be determined accurately by neutron scat- tering or by spin wave resonance in thin films, Chapter 13. By neutron scatter- ing G. Shirane and coworkers find, in the equation ?
i o = Dk2, the values 281, 500, and 364 meV AVor D at 295 K in Fe, Co, and Ni, respectively. Quantization o f Spin Waues. The quantization of spin waves proceeds as for photons and phonons. The energy of a mode of frequency w k with nk magnons is given by The excitation of a magrlon corresponds to the reversal of one spin i. Thermal Excitation of Magnons In thermal equilibrium the werage value of the number of magnons ex- cited in the niode k is giver1 by the Planck distribution" The total nnmhrr of magnons cxcitcd at a tcmpcraturr T is where D(w) is the number of magnon modes per unit frequency range. The integral is taken over the allowed range of k, which is the first Brillouin zone. At sufficiently low temperatures we niay evaluate the integral between 0 arld because (n(w)) + 0 exponentially as w + cc. Magnons havc a singlc polarization for each value of k. In three dimen- sions the niimher of modes of wavevector less than !
i is ( 1 / 2 ~ ) ~ ( 4 ? r k ~ / 3 ) per unit volume, whence the nunlber of magnons D(w)dw with frequency in dw at w is (1/25~)~(4?rk~)(dk/clw) dw. In the approximation (25), Thus the density of modes for magnons is so that the total number of magnons is, from (28), The definite integral is found in tables and has the value (0.0587)(44). The ~lurnber N of atorrls per unit volume is Q/u" where Q = 1, 2,4 for sc, bcc, fcc lattices, respectively. Now (C.nk)/hTS is equal to the fractional change of magnctixation AiM/A4(0), whrnce 3The argument is exactly as lor phonons or photons. The Planck dislribution lollows for any problem where the enerby levels are identical mith those of a harmonic oscillator or collection of harmonic oscillators. 12 Ferromagnetism and Antiferrornagnetism 335 This result is the Bloch T3/' law and has been confirmed experimentally. In neutron scattering experiments spin waves have been observed up to tempera- tures near the Curie temperature and even above the Curie temperature. NEUTRON MAGNETIC SCATTERING An x-ray photon sees the spatial distribution of electronic charge, whether the charge density is magnetized or unmagnetized. A neutron sees two aspects of a crystal: the distribution of nuclei and the distribution of electronic magne- tization. The neutron diffraction pattern for iron is sliown in Fig. 11. The magnetic moment of the neutron interacts with the magnetic moment of the electron. The cross section for the neutron-electron interaction is of the same order of magnitude as for the neutron-nuclear interaction. Diffraction of neutrons by a magnetic crystal allows the determination of the distribution, direction, and order of thc magnetic moments. A neutron can he inelastically scattered by the ~nagnetic structure, with the creation or annihilation of a magnon (Fig. 12); such events make possible thc experimental determination of magnon spectra. If the incident neutron has wavevector k , , and is scattered to k ; with the creation of a magnon of wavevector k, then by corlservation of crystal momentum k , = k k + k + G, where G is a reciprocal lattice vector. By conservation of energy Scdttering auglc Figure 11 Neutron diffraction pattern for iron. (After C. C.. Shidl, E. 0. WuUa~l, and W . C. Koehler.) Figure 12 Scattering of a neutron by an ordered magnetic structure, with creation of a magnon. Figure 13 Magnon energy as a function of the square of the wavevector, for the ferromagnet MnPt,. (After R. Antonini and V. J. Minkiewicz.) where huk is the energy of the magnon created in the process. The observed magnon spectriim for MnPt3 is shown in Fig. 13. FERRIMAGNETIC ORDER In many ferromagnetic cxystals the saturation magnetization at T = 0 K does not correspond to parallel alignment of the magnetic monie~lts of the constituent paramagnetic ions, even in crystals for which the individual para- magnetic ions have their normal magnetic moments. The most familiar example is magnetite, Fe,O, or FeO . Fe,03. From Table 11.2 we see that ferric (Fe3+) ions are in a state with spin S = and 12 Ferromugnetism and Antiferromugnetiam 8Fe3+ Figure 14 Spin arrallgc~llcnts in magnetite, FeO . Ye20,, showing how the moments uf the Fc3+ ions cancel out, leaving only the moments of the Fez+ ions. zero orbital moment. Thus each ion should contribute 5pCLB to the saturation moment. The ferrous (Fez+) ions have a spin of 2 and should contribute 4p,, apart from any residual orbital ~nornent contribution. Thus the effective num- ber of Bohr m a p e t o ~ ~ s per Fe,O, fornlula unit should be about 2 x 5 + 4 = 14 if all spins were parallel. The observed value (Table 1) is 4.1. The discrepancy is accounted for if the ~nolnents of the ~ e ~ + ions are antiparallel to each other: then the observed moment arises only from the Fez+ ion, as in Fig. 14. Neutron diffraction results agree with this model. A systematic discussion of the consequences of this type of spin order was given by L. NBel with reference to an important class of magnetic oxides known as ferrites. The usual chemical formnla of a ferrite is MO . Fe,O,, where M is a divalent cation, often Zn, Cd, Fe, Xi, Cu, Co, or Mg. The term ferrimagnetic was coined originally to describe the ferrite-type ferromag- netic spin order sllch as Fig. 14, and by extension the term covers almost any compound in which some ions have a rnornent antiparallel to other ions. Many fcrrimagnets are poor conductors of electricity, a quality exploited in applica- tions such as rf transformer cores. The cubic ferrites have the spinel crystal structure shown in Fig. 15. There are eight occupied tetrahedral (or A) sites and 16 occupied octahedral (or B) sites in a unit cuhc. The lattice constant is about 8 k. A remarkable fea- ture of the spinels is that all exchange integrals JM, J.41.4R, and JBR are negative and favor antiparallel alignment of the spins connected by the interaction. Rut the AB interaction is the strongest, so that the A spins are parallel to each other and the B spins are parallel to each other, just in order that the A spins may he antiparallel to the B spins. If] in U = -2JSi . Sj is positive, we say that the exchange integral is ferromagnetic; if J is negative, the exchange integral is antiferromagnetic. Figure 15 C~ystal structure of the mineral spinel MgAlZ0,; the Mg2+ ions occupy tetrahedral sites, each surrounded by four oxygen ions; the A13+ occupy octahedral sites, each surrounded by six oxygen ions. This is a normal spinel arrangement: the divalent metal ions occupy the tetrahe- dral sites. In the inverse spinel arrangment the tetrahedral sites are occupied by trivalent metal ions, while the octahedral sites are occupied half by divalent and half by trivalent metal ions. We now prove that three antiferromagnetic interactions can result in ferri- magnetism. The mean exchange fields acting on the A and B spin lattices may be written taking all mean field constants A, p, v to be positive. The minus sign then cor- responds to an antiparallel interaction. The interaction energy density is this is lower when MA is antiparallel to MB than when M A is parallel to MB. The energy when antiparallel should be compared with zero, because a possi- ble solution is MA = MB = 0 . Thus when the ground state will have MA directed oppositely to M,. (Under certain condi- tions there may be noncollinear spin arrays of still lower energy.) Curie Temperature and Susceptibility o f Ferrimagnets We define separate Curie constants C A and C B for the ions on the A and B sites. For simplicity, let all interactions be zero except for an antiparallel interac- tion between the A and B sites: BA = - p M B ; BB = -pMA, where p is positive. The same constant p is involved in both expressions because of the form of (33). 12 Ferronurgnetism and Antiferromagnetian We have in the mean field approximation where B, is the applied field. These equations have a nonzero solution for MA and M, in zero applied field if so that the ferrimagnetic Curie temperature is given by T, = ~ . ( c ~ C ~ ) ' " . We solve (35) for MA and M, to obtain the susceptibility at T > T,: a result more complicated than (4). Experimental values for Fe,O, are plotted in Fig. 16. The curvature of the plot of 1/x versus T is a characteristic feature of a ferrimagnet. We consider below the antiferromagnetic limit CA = CB. Iron Garnets. The iron garnets are cubic ferrimagnetic insulators with the general formula MJFe,0,2, where M is a trivalent metal ion and the Fe is the trivalent ferric ion (S = %, L = 0). An example is yttrium iron garnet Y3Fe,01,, hown as YIG. Here Y'+ is diamagnetic. The net magnetization of YIG is due to the resultant of two oppositely magnetized lattices of Fe3+ ions. At absolute zero each ferric ion contributes 25pB to the magnetization, hut in each formula unit the three ~ e ~ + ions on sites denoted as d sites are magnetized in one sense and the two Fe3 ions on a sites are magnetized in the opposite sense, giving a resultant of 5pB per formula unit in good agreement with the measurements of Geller et al. Temperature PC) Figure 16 Reciprocal s~isceptihility of magnetite, FeO . Fe,O, The mean field at an a site due to the ions on thc d sitcs is R, = -(1.5 x 104)M,,. The observed Curie temperature 5.59 K of YIG is due to the a-d interaction. The o11ly rrlagnetic ions in YIG are the ferric ions. Because these are in an L = 0 state with a spherical charge distribution, their interac- tion with latticc deformations and phonons is weak. As a result YIG is charac- terized by very narrow linewidths in fcrromapetic resonance experinlents. ANTIFERROMAGNETIC ORDER A classical example of magnetic structure deter~nination by rleutroris is shown in Fig. 17 for MnO, which has the NaCl structure. At S O K there are extra neutron reflections not present at 293 K. Thc rcflcctions at 80 K may be classified in terms of a cubic unit cell of lattice constant 8.85 A. 4t 293 K thc reflections correspond to an fcc unit cell of lattice constant 4.43 A. But the lattice cor~stant determined by x-ray reflection is 4.43 A at both temperatures, 80 K and 293 K. We conclude t2vat the chemical unit cell has the 4.43 latticc parameter, but that at 80 K the electronic magnetic r~~o~rlerits of Figure 17 Neutron diffraction patterns for MnO below, and above the spin-ordering temperature of 120 K, after C. 6. Shnll, \'. A. Strauser. and E. 0. U'ollan. Thc rcflcctio~r indicas arc based on an 8.85 A cell at 80 K and on a 4.43 A cell at 293 K. .kt the higher t~mperat~~re the bIn2' ions are still magnetic, but they are no longer ordered. 12 Ferromagnetisrn and Antiferrornagnetisrn Figure 18 determined Ordered arrangements of splns of the MnZ+ Ions m manganese by neutron d~ffraction The 02+ Ions are not shown oxide. MnO, as Figure 19 Spin ordering in ferromagnets (J > 0) and antiferromagnets ( J < 0). the ~ n ' + ions are ordered in some nonferromagnetic arrangement. If the ordering were ferromagnetic, the chemical and magnetic cells would give the same reflections. The spin arrangement shown in Fig. 18 is consistent with the neutron dif- fraction results and with magnetic measurements. The spins in a single [ I l l ] plane are parallel, but spins in adjacent [ I l l ] planes are antiparallel. Thus MnO is an antiferromagnet, as in Fig. 19. In an antiferromagnet the spins are ordered in an antiparallel arrangement with zero net moment at temperatures below the ordering or NBel temperature (Table 2). The susceptibility of an antiferromagnet is not infinite at T = T,, but has a weak cusp, as in Fig. 20. An antiferromagnet is a special case of a ferrimagnet for which both sub- lattices A and B have equal saturation magnetizations. Thus CA = CB in (37), and the NBel temperature in the mean field approximation is given by Table 2 Antiferromagnetic crystals Transition Paramagnetic temperature, Curie-Weiss - e ~ ( 0 ) Substance ion lattice T , , in K 0, in K TN x(TN) MnO MnS MnTe MnF, FeF, FeC1, FeO CoC1, c o o NiC1, NiO Cr fcc fcc hex. layer bc tetr. bc tetr. hex. layer fcc hex. layer fcc hex. layer fcc bcc Paramagnetism Ferromagnetism Antiferrornagnetism x = C x = - C T T-T, Curie law Curie-Weiss law (T > T,) Figure 20 Temperature dependence of the magnetic susceptibility in paramagnets, ferromag- nets, and antiferromagnets. Below the NBel temperature of an autiferromagnet the spins have an- tiparallel orientations; the susceptibility attains its maximum value at T, where there is a weU- defined kink in the curve of x versus T. The transition is also marked by peaks in the heat capacity and the thermal expansion coefficient. where C refers to a single sublattice. The susceptibility in the paramagnetic region T > T, is obtained from (37): The experimental results at T > T, are of the form 12 Ferromagnetbm and Antiferrornugneiiam 343 Experimental values of BIT, listed in Tablc 2 often differ substantially from the value unity expected from (39). Values of BIT,,, of the observed magnitude may be obtained when next-nearest-neighbor interactions are provided for, and when possible sublatticc arrangements are considered. If a mean field constant - E is i~ltroduccd to describe interactions within a sublattice, the11 BIT, = ( p + €)/(p - 6). Susceptibility Below the Nhel Temperature Thcre are two situations: with the applied magnetic field perpendicular to thc axis of the spins; and with the field parallel to the axis of the spins. At and ahove the N6el temperature the s~isceptibility is nearly independent of the di- rection of the field relativr to the spin axis. For B, perpendicular to the axis of the spins we can calculate the suscepti- bility by elementaly considerations. The energy density in the presence of the field is, with A 4 = = (&I, where 2rp is thc angle the spins make with each other (Fig. 21a). The energyis a minimum when so that (CGS) In thc parallel orientation (Fig. 21b) the magnetic energy is not changed if the spin systenls A and B make equal angles with the field. Thus the suscepti- bility at T = 0 K is zero: (b) Figure 21 Calcl~lation of (a) peycndicular and (b) parallel snsceptibilities at 0 K, in the mean field approximation. XI T, i n K Figure 22 Magnetic susceptibility of manganese flnoride. MnF,, parallel and perpendicular to the tetragonal axis. (After S. Foner.) The parallel susceptibility increases smoothly with temperature up to T N . Measurements on MnF, are shown in Fig. 22. In very strong fields the spin systems will turn discontinuously fro~r~ the parallel orientation to the perpen- dicular orientation where the energy is lower. Antiferromagnetic Magnons We obtain the dispersion relation of magnons in a one-dimensional anti- ferromagnet by making the appropriate substitutions in the treatment (16)-(22) of the ferromagnetic line. Let spins with even indices 2p composc sublattice A, that with spins up (Sz = S ) ; and let spins with odd indices 21, + 1 compose sublattice B, that with spins down (Sz = -S). We consider only nearest-neighbor interactions, with ] negative. Then (18) written for A becomes, with a careful look at (I:), The corresponding equations for a spin on B are We form S+ = S ' + iSY: then clSlj/clt = (2i]S/fi)(ZS& + S&, + Sip+,) ; dS~p+,/dt = -(2iJS/6)(2Sip+1 + S?, + S?,+9) 12 Ferromagnetirrm and Antifewomagnetism Figure 23 Magnon dispersion relatlon in the simple cubic antiferromagnet RhMnF, as deter- mined at 4.2 K by inelastic neutron scattering. (After C. G. Windsor and R. W7. H. Stevenson.) We look for solutions of the form S2:, = u exp[i2pka - iwt] ; Sl,+l = u exp[i(2p + 1) ka - iurt] , (49) so that (47) and (48) become, with we, - - 4JSIfi = 411 IS/fi, Equations (50) have a solution if w , cos ka = o ; we, cos ka w , + w thus w ~ = w ~ ~ ( ~ - c o s ~ ~ ) ; w=w,,lsinkaI. (52) The dispersion relation for magnons in an antiferromagnet is quite differ- ent from (22) for magnons in a ferromagnet. For ka < 1 we see that (52) is lin- ear in k: w w,,lkal. The magnon spectrum of RhMnF, is shown in Fig. 23, as determined by inelastic neutron scattering experiments. There is a large re- gion in which the magnou frequency is linear in the wavevector. Well-resolved magnons have been observed in MnF, at specimen temper- atures up to 0.93 of the NBel temperature. Thus even at high temperatures the magnon approximation is useful. FERROMAGNETIC DOMAINS At temperatures well below the Curie point the electronic magnetic mo- ments of a ferromagnet are essentially parallel when regarded on a micro- scopic scale. Yet, loolang at a specimen as a whole, the magnetic moment may be very much less than the saturation moment, and the application of an exter- nal magnetic field may be required to saturate the specimen. The behavior ob- served in polycrystalline specimens is similar to that in single crystals. Actual specimens are composed of small regions called domains, within each of which the local magnetization is saturated. The directions of magneti- zation of different domains need not be parallel. An arrangement of domains with approximately zero resultant magnetic moment is shown in Fig. 24. Do- mains form also in antiferromagnetics, ferroelectrics, antiferroelectrics, ferro- elastics, superconductors, and sometimes in metals under conditions of a strong de Haas-van Alphen effect. The increase in the gross magnetic moment of a ferromagnetic specimen in an applied magnetic field takes place by two independent processes: In weak applied fields the volume of domains (Fig. 25) favorably oriented with respect to the field increases at the expense of unfavorably oriented domains; Figure 24 Ferromagnetic dolna~n pattern on a single crystal platelet of nickel. The domain boundaries are made visible by the Bitter magnetic powder pattern technique. The direction of magnetization within a domain is determined by observing growth or contraction of the domain in a magnetic field. (After R. W. De Blois.) 12 Ferromagnetism and Antiferromugnetism 347 Applied ficld - Figure 25 Representative r~~ag~~etization cunre, showing the dominant magnetization processes in the different regions of the curve. Magnetic induction (CGS) B = H + 4 v B f Figure 26 The technical magnetization curve (or hysteresis loop). The coercivity H, is the re- verse field that reduces B to zero; a related coercivity HCj reduces M or B - H to zero. The remanence B, is the value of B at H = 0. The saturation induction B, is the limit of B - H at large H, and tllr saturation magnetization M, = Bi4~471. In SI the vertical axis is B = p,(H + iM). In strong applied fields the domain magnetization rotates toward the direction of thc field. Technical terms defined by the hysteresis loop are shown in Fig. 26. The cocrcivity is usually defined as the reverse field H, that reduces the induction %::::m l ; q q i j J E .
c - 800 2 Basal plane 400 0 Co 0 200 400 600 0 100 200 300 0 2000 4000 6000 8000 B, (gauss) Figure 27 Magnetization curves for single crystals of iron, nickel, and cobalt From the curves for iron we see that the [loo] directions are easy directions of magnetization and the [Ill] direc- tions are hard directions. The applied field is B,. (After Honda and Kaya.) (b) Figure 28 Asymmetry of the overlap of electron distributions on neighboring ions provides one mechanism of magnetocrystalline anisotropy. Because of spin-orbit interaction the charge distrib- ution is spheroidal and not spherical. The asymmetly is tied to the direction of the spin, so that a rotation of the spin directions relative to the crystal axes changes the exchange energy and also changes the electrostatic interaction energy of the charge distributions on pairs of atoms. Both ef- fects give rise to an anisotropy energy. The energy of (a) is not the same as the energy of (b). B to zero, starting from saturation. In high coercivity materials the coercivity H,, is defined as the reverse field that reduces the magnetization M to zero. Anisotropy Energy There is an energy in a ferromagnetic crystal which directs the magnetization along certain crystallographic axes called directions of easy magnetization. This energy is called the magnetocrystalline or anisotropy energy. It does not come about from the pure isotropic exchange interaction considered thus far. Cobalt is a hexagonal crystal. The hexagonal axis is the direction of easy magnetization at room temperature, as shown in Fig. 27. One origin of the anisotropy energy is illustrated by Fig. 28. The magnetization of the crystal sees the crystal lattice through orbital overlap of the electrons: the spin interacts 12 Ferromugnetisrn and Antiferromaptism 349 with the orbital motion by means of thc spin-orbit coupling. In cobalt the anisotropy energy density is givcn by where 6 is the anglc the magnetization makes with the hexagonal axis. At room temperature K; = 4.1 X 10%rg/cm3; = 1.0 X 10%rg/cm3. Iron is a cubic crystal, and the cube edges are thc directions of easy mag- nctization. To represent the anisotropy energy of iron magnetized in an arbi- trav direction with direction cosines a,, a, ar3 referred to the cube edges, we are guided by cubic symmetry. Thc expression for the anisotropy energy must be an even power of each mi, provided opposite ends of a crystal axis are equiv- alent ~nagnetically, and it must be invariant under interchanges of the ai arrlong themselvcs. The lowest order combination satisfying the symmetry re- quirements is a: + ar: + ar:, but this is identically equal to unity and does not describc anisotropy effects. The next conlhination is of the fourth degree: ar:cri + aryai + a&;, and then of the sixth degrce: a:a&$. Thus At room temperature in iron K, = 4.2 X lo5 crg/cm3 and K, = 1.5 x 10" erg/cm3. Transition Region Between Domains A Bloch wall in a crystal is the transition layer that separates adjacent regions (dorriai~~s) magnetized in different directions. The entire change in spin direction between domains does not occur in unc discontinuous jump across a single atomic plane, hut takes place in a gradual way over many atomic planes (Fig. 29). The exchange energy is lower when the change is distributed over many spins. This behavior may be understood by interpreting the Heisen- berg eq~lation (6) classically. We replace cos by 1 - i(p2; then w, = J S ~ ~ ' is the exchange energy between two spins making a small angle with each other. Here J is the exchange intcgral and S is the spin quantum number; we, is referred to the energy for parallel spins. If a total change of .
n occurs in N equal steps, the angle between ncighhor- ing spins is .n/AT1 and the exchange energy per pair of neighboring atoms is we, = ~ ~ ~ ( ? r / h ' ) ~ . The total exchange energy of a line of N + 1 atoms is The wall would thicken without limit were it not for the anisotropy energy, which acts to limit the width of the transition layer. The spins contained within the wall are largely directed away from the axes of easy rnagnetization. so thcrc is an anisotropy cncrgy associated with the wall, roughly proportional to the wall thickness. Figure 29 The structure of the Bloch wall separating domains. In iron the thickness of the transi- tion region is about 300 lattice constants. Consider a wall parallel to the cube face of a simple cubic lattice and sepa- rating domains magnetized in opposite directions. We wish to determine the number N of atomic planes contained within the wall. The energy per unit area of wall is the sum of contributions from exchange and anisotropy energies: o w = u e a + canis. The exchange energy is given approximately by (55) for each line of atoms normal to the plane of the wall. There are l/a2 such lines per unit area, where a is the lattice constant. Thus a , , = d]s2/Na2 per unit area of wall. The anisotropy energy is of the order of the anisotropy constant times the thickness Nu, or o,, - KNa; therefore ow - ( d ] ~ ~ / N a ~ ) + KNa . (56) This is a minimum with respect to N when For order of magnitude, N .
= 300 in iron. The total wall energy per unit area on our model is ow = z ~ ( K ] s ~ / a ) ~ ; (59) 12 Ferromagnetism and Antiferrornagnetism 35 1 (a) (b) (c) (dl (4 Figure 30 The origin of domains. in iron u , , , = 1 erg/cm2. Accurate calculation for a 180" wall in a (100) plane gives a, = ~(2K,]S~la)~'~. Origin o f Domains Landau and Lifshitz showed that domain structure is a natural conse- quence of the various contributions to the energy-exchange, anisotropy, and magnetic-of a ferromagnetic body. Direct evidence of domain structure is furnished by photomicrographs of domain boundaries obtained by the technique of magnetic powder patterns and by optical studies using Faraday rotation. The powder pattern method developed by F. Bitter consists in placing a drop of a colloidal suspension of finely divided ferromagnetic material, such as magnetite, on the surface of the ferromagnetic crystal. The colloid particles in the suspension concentrate strongly about the boundaries between domains where strong local magnetic fields exist which attract the magnetic particles. The discovery of transparent ferromagnetic compounds has encouraged the use also of optical rotation for domain studies. We may understand the origin of domains by considering the structures shown in Fig. 30, each representing a cross section through a ferromagnetic single crystal. In (a) we have a single domain; as a consequence of the mag- netic "poles" formed on the surfaces of the crystal this configuration will have a high value of the magnetic energy (lI8~r) J B2 dV. The magnetic energy den- sity for the configuration shown will be of the order of M: = lo6 erg/cm3; here M, denotes the saturation magnetization, and the units are CGS. In (b) the magnetic energy is reduced by roughly one-half by dividing the crystal into two domains magnetized in opposite directions. In (c) with N do- mains the magnetic energy is reduced to approximately 1IN of the magnetic energy of (a), because of the reduced spatial extension of the field. Figure 31 Domain of closure at the end of a single crystal iron whisker. The face is a (100) plane; the whisker axis is [0011. (Courtesy of R. V. Coleman, C. G. Scott, and A. Isin.) In domain arrangements such as (d) and (e) the magnetic energy is zero. Here the boundaries of the triangular prism domains near the end faces of the crystal make equal angles (45") with the magnetization in the rectangular do- mains and with the magnetization in the domains of closure. The component of magnetization normal to the boundary is continuous across the boundary and there is no magnetic field associated with the magnetization. The flux cir- cuit is completed within the crystal-thus giving rise to the term domains of closure for surface domains that complete the flux circuit, as in Fig. 31. Domain structures are often more complicated than our simple examples, but domain structure always has its origin in the possibility of lowering the energy of a system by going from a saturated configuration with high magnetic energy to a domain configuration with a lower energy. Coercivity and Hysteresis The coercivity is the magnetic field H, required to reduce the magnetiza- tion or the induction B to zero (Fig. 26). The value of the coercivity ranges over seven orders of magnitude; it is the most sensitive property of ferromag- netic materials which is subject to control. The coercivity may vary from 600 G in a loudspeaker permanent magnet (Alnico V) and 10,000 G in a special high stability magnet (SmCo,) to 0.5 G in a commercial power transformer (Fe-Si 4 wt. pet.) and 0.002 Gin a pulse transformer (Supermalloy). Low coercivity is desired in a transformer, for this means low hysteresis loss per cycle of opera- tion. Materials with low coercivity are called soft; those with high coercivity are called hard, although there is not necessarily a 1 : 1 relationship of mag- netic hardness with mechanical hardness. The coercivity decreases as the impurity content decreases and also as in- ternal strains are removed by annealing (slow cooling). Amorphous ferromag- netic alloys may have low coercivity, low hysteresis losses, and high permeabil- ity. Alloys that contain a precipitated phase may have a high coercivity, as in Alnico V (Fig. 32). 12 Ferromagnetism and Antiferrornagnetism Figure 32 Microstructure of Alnico \' in its optimum state as a permanent magnet. The composi- tion of Alnico \'is, by weight percent, 8 Al, 14 Ni, 24 Co, 3 Cu, 51 Fe. As aperinanent magnet it is a twn-phase system, with fine particles of one phase embedded in the other phase. The precipita- tion is carried out in a magnetic field, and the particles are oriented with their long axis parallel to the field direction. The width shown is 1.1 pm. (Courtesy of F. E. Luborsky.) Soft magnetic materials are used to concentrate and shape magnetic flux, as in motors, generators, transformers, and sensors. Useful soft materials in- clude electrical steels (usually alloyed with several percent of silicon to in- crease electrical resistivity and to decrease anisotropy); various alloys of Fe-Co-Mn, starting with permalloys of composition near NiT8Fezz, which have near-zero anisotropy energy and near-zero magnetostriction; NiZn and MnZn ferrites; and metallic glasses produced by rapid solidification. A commercial metallic glass (METGLAS 2605s-2) with composition Fe,,B1,Si9 has a hys- teresis loss per cycle much lower than the best grain-oriented silicon steel. The high coercivity of materials composed of very small grains or fine powders is well understood. A sufficiently small particle, with diameter less than 1 0 F or cm, is always magnetized to saturation as a single domain be- cause the formation of a flux-closure configuration is energetically unfavorable. In a single domain particle it is not possible for magnetization reversal to take place by means of the process of boundary displacement, which usually re- quires relatively weak fields. Instead the magnetization of the particle must rotate as a whole, a process that may require large fields depending on the anisotropy energy of the material and the anisotropy of the shape of the particle. The coercivity of fine iron particles is expected theoretically to be about 500 gauss on the basis of rotation opposed by the crystalline anisotropy energy, and this is of the order of the observed value. Higher coercivities have been reported for elongated iron particles, the rotation here being opposed by the shape anisotropy of the demagnetization energy. Rare earth metals in alloys with Mn, Fe, Co, and Ni have very large crystal anisotropies K and correspondingly large coercivitics, of the order of 2WM. These alloys are exceptionally good permanent magnets. For example, the hexagonal co~npourid SmCoS has an anisotropy energy 1.1 x 10serg ~ r n - ~ , equivalent to a coercivity 2WM of 290 kG (29 T). Magnets of Nd,Fel,B have energy products as high as 50 MGOe, exceeding all other commercially avail- able magnets. SINGLE-DOMAIN PARTICLES The dominant industrial and commercial applications of ferromagrietism are in magnetic recording devices, where the magnetic material is in the form of single-domain particles or regons. The total value of the production of magnetic devices for recording may be comparable with the total valuc of semiconductor device production and greatly exceeds the value of supercon- ducting device production, the latter being held back by low critical tempera- tures, as compared with magnetic Curie temperatures. The magnetic record- ing devices or memories typically are in the form of hard disks in coulputers and tape in video and audio recorders. An ideal single-domain particle is a fine particlc, usually elongated, that has its magnetic moment directed toward one end or the other of thc particle. The alternative orientations may be labeled as N or S; + or -; in digital recording, as 0 or 1. To have digital properties a ferromagnetic particle should be fine enough, typically 10-100 nm, so that only one domain is within the particlc. If the fine particle is elongated (acicular) or has uniaxial crystal sym- metry, only two valucs of the magnetic moment of the single domain are per- mitted, which is what one wants for digital properties. The first successful recording material was acicular T-Fe,O, with length-to-width ratio of about 5 : 1, coercivity near 200 Oe and a length <I pm; chrominm dioxide CrOz is the basis of a better material, in a form highly acicular (20 : 1) with coercivity near 500 Oe. Effective elongation can be attained with spheres by making a chain, like a string of beads. An ensemble of such chains or of elongated single do- main particles is said to exhibit superparamagnetism if the magnetic mo- ment of a unit is constant. If p is the magnetic moment in a magnetic field B, then the net magnetization of the ensemble will follow the Curie-Rrillouin- Langevin law of Chapter 11 if the articles are embedded in a liquid so that they are each free to rotate as a whole. If the particles are frozen in a solid, thcre will be a remanent magnetization (Fig. 26) after removal of an applied field. 12 Ferromagnetism und Ar8tiferromagnetism 355 Geomagnetism and Biomagnetism Singlc domain ferromagnetic properties are of special geological interest in scdimentary rocks because the rocks through their remanent magnetizatiori carry a memory of the direction of the earth's magnetic field at the time that they were laid down, and thus of the geographical location of the rocks at that epoch. The rr~agnetic record is perhaps the most important basis of the theory of the drift of continents. Annually, layers of sedin~e~lt are deposited in stream beds, layers that may hear some magnetic particles in single domain form. This record persists over at least 500 millior~ years of geological time and can tell us wherc on the surface of the earth the deposit was laid down at a given time. Lava flows also record rriagrletic field directions. The change in ~nagnetization from layer to layer gives a superb historical record of the drift of the continental plates on the earth's surface. The paleo- magnetic record is one basis of the branch of geology called plate tectonics. The original interpretation of the record was made Inore difficult, or more ex- citing, by the associated discovery (Brunhes, 1906) that the magnetic field of the earth itsclf can show reversals in direction, an effect contained within the standard dynamo theory of the earth's magnetism. Reversals have taken place once every 1 X lo4 to 25 X 10"ears. When a reversal occurs, it is relatively sudden. Fine single Jolnain particles, oftcn of magnetite Fe304, are even of impor- tance in biology. A direction-sceking effect known as magnetotaxis often con- trols, possibly sometimcs along with an astronomical guide system, the motion of bacteria, the migration of birds, and the ~novements of homing pigeons and bees. The cff'ect is due to the interaction of a single domain particle (or cluster of such particles, Fig. 33) in the orga~lism with the external magnetic field of the earth. Magnetic Force Mi~roscopy The success of the scanning tulineling microscope (STM) stimulated the development of related sca~lning probe dcvices, of which the scanning mag- netic force microscope is one of the most effective. A sharp tip of a magnetic material, such as nickel, is mounted on a cantilever lever (Fig. 34). Ideally, hut not yet, the tip is a singlc domain particle. Forces from the magnetic sample act on the tip and cailse a change, such as a deflection, in the cantilever status, and an image is formed by scanning the sample relative to the tip. The mag- netic force microscope (MFM) is the only magnetic imaging technique that can provide high resolution (10-100 nm) with little surface preparation. One can, for example, observe and image the magnetic flux that exits from the surface at the intersection of a Rloch wall with the surface (Fig. 29). An impor- tant application is to thc study of magnetic recording media-Figure 35 shows the magnetic signal from a test pattern of 2 p1n bits magnetized in the plane of Figure 33 Thin section of a cell of a magnetotactic bac- terium showing a chain of 50 nm particles of Fe,O,. Draw- ing by Marta Puebla from a photograph by R. B. Frankel and others. Deflection sensor Magnetic Flexible field, cantilever Figure 35 Test strip magnetization in the plane of a Co-alloy disk in 2 pm hits, as detected by MFM close above the plane of the disk. (After Rugar et al.) 't---. ,--- i \, , ,---., ,.--. _ ,---. Y v v v ! Sample - Figure 34 Basic concept of magnetic force microscopy. A magnetic tip attached to a flexible can- tilever is used to detect the magnetic field produced by the regions of alternating magnetization in the plane of the sample. (After Gmetter, Mamin, and Rugar, 1992.) - - - - 12 Ferromagnetism a d Antifemmagnetism 357 a Co-alloy disk; the parallel co~nporlent of the ficld seen by the sensor tip is what the photo shows. SUMMARY (In CGS Units) The susceptihility of a ferromagnet above the Curie temperature has the form x = C/(T - T,) in the mean field approximation. In thc mean field approximation the effective magnetic field seen by a mag- netic moment in a ferrornagnet is B , + AM, when A = T,/C and B , is the applied magnetic field. The elementary excitations in a ferromagnet are magnons. Their dispersion relation for ku < 1 has the form ho =Jk2az in zero external magnetic field. The thermal excitation of magnons leads at low temperatures to a heat ca- pacity and to a fractional magnetization change both proportional to T3". In an antiferrornagnet two spin lattices are equal, but antiparallel. In a ferri- rnagnet two lattices are antiparallel, but the magnetic moment of one is larger than the magnetic moment of the other. In an antifemornagnet the susceptibility above the Nee1 temperature has the form , y = 2C/(T + 8). The magnon dispersion relation in an antiferrornagnet has the form hw = jkn. The thermal excitation of rnagnons leads at low temperatures to a term in T 3 in the heat capacity, in addition to the phonon term in T3. A Rloch wall separates domains magnetized in different directions. The thickness of a wall is =(J/&?)'/' lattice constants, and the energy per unit area is =(KJ/u)~/', where K is the anisotropy energy density. Problems 1. Magnon dispersion rebtion. Derive the magnon dispersion relation (24) for a spin S on a simple cubic lattice, z = 6. Hint: Show first that (18a) is replaced by where the central atom is at p and the six nearest neighhors arc connected to it by six vectors 8. Look for solutions of the equations for r1S;ldt and dSEldt of the form exp(ik . p - iwt). 2. Heat capacity of magnons. Use the approximate lnagnon dispersion relation w = Akqo find the leading term in the heat capacity of a three-dimensional ferro- rrlagnet at low temperatures k,T 4 J. The result is 0.113 kB(kBTlh~)3/2, per unit volume. The zeta function that enters thc rcsuli may he rsti~nated numerically; it is tabulated in Jahnke-Emde. 3. Ne'el temperature. Taklng the effective ficld? on t l ~ r two-subldthce model ot an ant~ferromagnebc as show that 4. Magnetoelastic coupling. ITI a cubic crystal the elastic energy density in terms of the usual strain componer~ts uU is (Chapter 3) and the leading term in the magnetic anlsotropy e n e r a density is, frnm (54), Coupling hetweer~ elastic strain and magnetization direction may bc takcn formally into accnnnt by including in the total energy density a term arising from the strain dependence of L7,; here B, and B2 arc called magrretoelastic coupling constants. Show that the total energy is a minimum bvhe~r This explains the ongin of magnr~ostrirtin~i, t l ~ r cl~ange of length on magnet~zatlon. 5. Coercive force of a small particle. (a) Consider a small spherical single-domain particle of a uniaxial lcrrornagnrt. Sliow that the reverse field along the axis re- quired to reversc thc magnetization is B, = 2WM,, in CGS units. The coercive force of single-domain particles is observed to be of this magnitude. Take U, = K sin" as the anisotropy cnergy density and UM = B , M cos 0 as the interaction energy den- sity with the external field; here 6 is the angle between B, and M. IIint: Expand the energics for small angles about 0 = n , and find the value of B, for which UK + L7>, does not have a minirnunr r~ear 0 = n. (b) Show that the magnetic energy of a satu- rated sphere of diameter d is =M:d3. An arrangement with appreciably less mag- nctic energy has a single wall in an equatorial plane. The domain wall energy will bc nuud2/4, where cw is the wall energy per unit area. Estimate for cobalt the critical radius below wl~ich the particles are stable as single domains, taking thc valuc ol ]S2/n as for iron. 6. Saturation magnetization near T,. Show that in the mean field approximation the saturation rnagnetization just below the Curie temperature has thc dominant 12 Ferromognetisrn and Antiferrornagnetism temperature dependence (T, - T)". A S S U ~ C the spin is :. The result is the sarrrr as that for a second-order trarisitior~ in a rcrroclectric crystal, as discussed in Clia~pter 16. The experimental data for ferromagncts (Table 1) suggest that the exporlrrtt is closer to 0.33. 7 . Nkel wall. The direction of n~agnetizatinn change in a domain wall goes fro111 that of the Bloch wall to that of a YGel wall (Fig. 36) in thin films of material of negligi- blc crystalline anisotropy energy, suclr as Permallo)~ The intercept of the Bloch wall with the surface of the film creates a surface rcgion of high demagnetization energy. The Nee1 wall avoids this intercept contrih~~lion, but at the expense of'a demagneti- zation contribution throughout the volulrre n T thc wall. The NBel wall becomes en- ergctically favorable when the film lrecurrres s~lficicntly thin. Consider, however, the cncrgetics of the NBel wall in bulk rr~aterial or negligible crystalline anisotropy energy. There is now a demagnetization corrtrih~~tion to the wall energy density By a rl~~alitative argument similar to (56), show that ,~,=(dJ~'lRia') + (2~iZrlPh'a). Find A : fnr which uw is a minimum. Estimate the order of magnitude of u, for typical val- ues of J, Ms. and a. 8. Gbnt magnetoresistance. In a ferromagnetic metal, the conductivitj- up for elec- trons whose rnagnctic moments are oriented parallel to the magnetization is typi- cally larger than ua for those antiparallel to the magnetizatinn. Consider a ferromag- netic conductor consisting of two separate regions of ider~tical climcnsions in series \,hose rriagnetinations can be independently controlleci. Elet:trons ol a given spin flow first tllrorlglr one rcgion and then through the other. It is observed that the re- sistance wlvllen hoth rnagnctizations point upwards, RTT, is lower than the resistance when they poirtt opposite, KT&. This resistance change can be large for mp/cr, > 1, and the phenomenon is callcd giant magnetoresistance (GMR). A srrlall external magnetic field car) switch thc rcsistance from Rri to RrT by reorienting t l ~ e magneti- zation of the second layer. This cffcct is increasingly used in magnetic storagr appli- cations such as the ~rragnetic bit readout in hard drives. The giant rrtagnetnresis- tance ratio is defined as: Uloch wall NCel d l Figure 36 A Bloch wall and a Nee1 wall in a thin fi1111. The n~aglietization in the Uloch wall is nor- mal to the plane of thc Glnl and adds to the wall energy n demagnetizatio~~ energy -M%d per unit length of wall, where S is the \vdl tl~ickl~css and d the film thickness In the NPeI wall the magneti- zation is parallel to the surface: the addition to the wall encrgy is negligible \%.hen d < 8. The addi- tion to the Ntel wall cnerg). when d % S is the subject of Prohlclr~ 7. (After S . Middelhock.! (a) If there is no spin-flip scattering for the conduction electrons, show that GMRR = (u/ua + ua/up - 2)/4 (Hint: Treat the spin-up and spin-down conduction electrons as independent con- ducting channels in parallel.) (b) If ua + 0, explain physically wlry the resistance in the 7.1 magnetization configuration is infinite. Magnetic Resonance NUCLEAR MAGNETIC RESONANCE Equations of motion LINE WIDTH Motional narrowing HYPERFINE SPLITTING Examples: paramagnctic point defects F centers in alkali halides Donor atoms in silicon Knight shift NUCLEAR QUADRUPOLE RESONANCE FERROMAGNETIC RESONANCE Shape effects in FMR Spin wave resonance ANTIFERROMAGNETIC RESONLVCE ELECTRON PARAMAGNETIC RESONANCE Exchange narrowing Zero-field splitting PRINCIPLE OF MASER ACTION Three-level maser Lasers SUMMARY NOTATION: In this chapter the symbols B, and Bo refer to the applied field, and B, is the applied Geld plus thc demagnetizing field. In particular we write B, = Bog. For CGS rcaders it may be simpler to read H for B whenever it occurs in this chapter. PROBLEMS 391 1. Equivalent electrical circuit 391 2. Rotating coordinate system 391 3. Hyperfine effects on ESR in metals 391 4. FMR in the anisotropy field 391 5. Exchange frequency resonance 39 1 6. Rf saturation 391 Figure 1 CHAPTER 13: MAGNETIC RESONANCE In this chapter we discuss dpamical magnetic effects associated with the spin angular momentum of nuclei and of electrons. The principal phenomena are often identified in the literature by their initial letters, such as NMK: nuclear magnetic resonance NQR: nuclear quadnipole resonance EPR or ESR: electron paramagnetic or spin resonance (Fig. 1) FMR: ferromagnetic resonance SWR: spin wave resonance (ferromagnetic films) AFMR: antiferrorriagnetic resonance CESR: conduction electron spin resonance The information that can bc ohtained about solids by resonance studies may be categorized: Electronic structure of single defects, as revealed by the fine structure of the absorption. Motion of the spin or of the surroundings, as revealed by changes in the line width. Internal magnetic fields sampled by the spin, as revealed by the position of the resonance line (chemical shift; Knight shift). Collective spin excitations. It is best to discuss NMR as a basis for a brief account of the other reso- nance experimcnts. A great impact of NMR has been in organic chemistry and biochcrnistry, where NMR provides a powerful tool for the identification and the stnicture determination of cornplex molecules. This success is due to the extremely high resolutiori attainable in diamagnetic liquids. A major medical application of NMK is magnetic resonance imaging (MRI), which allows the resolution in 3D of abnormal growths, configurations, and reactions in the whole body. NUCLEAR MAGNETIC RESONANCE L i r consider a nucleus that possesses a magnetic moment p and an angu- lar momentum 7 5 1 . The two quantities are parallel, and we may writc p = y f i I ; (1) the magnetogyric ratio y is constant. By convention I denotes the nuclear angular mo~nentum measured in units of fi. Figure 2 Energy level splitting of a nucleus of spin I = i in a static magnetic field B,. The energy of interaction with the applied magnetic field is . . U = - p B a if B , = Bog, then The allowed values of I, are m, = I, I - 1, . . . , -I, and U = -mIyABo. In a magnetic field a nucleus with I = has two energy levels correspond- ing to m, = ?+, as in Fig. 2. If Awo denotes the energy difference between the two levels, then fiw, = $iB, or This is the fundamental condition for magnetic resonance absorption. For the proton1 y = 2.675 X lo4 s-I gauss-' = 2.675 X 10's-I tesla-l, so that where v is the frequency. One tesla is precisely lo4 gauss. Magnetic data for selected nuclei are even in Table 1. For the electron spin, 'The magnetic moment p, of the proton is 1.4106 X erg G-' or 1.4106 X J T-', and y = 2pdfi. The nuclear magneton pn is defined as e&WM,c and is equal to 5.0509 X lo-" erg G-' or 5.0509 X lO-"J T-'; thus p, = 2,793 nuclear magnetons. Table 1 Nuclear magnetic resonance data For every element the most abundant magnetic isotope is shown. After Varian Associates NMR Table. ~iiF"9":~Wi~~:iiB19~LB#~~I~rW~~!~~Mfl#ji~~Ilbilbii~N~6:~~W4i~?~~~El~~ilil~S!tiUtiuIE~"S?:"S?:in~ECirr;"4.~S~i~ZIES~M~ Equations o f Motion The rate of change of angular momentum of a system is equal to the torque that acts on the system. The torque on a magnetic moment p in a mag- netic field B is p X B, so that we have the gyroscopic equation M V d t = p X B , ; (3) or d p f d t = y p x B , . (6) The nuclear magnetization M is the sum Zp, over all the nuclei in a unit vol- ume. If only a single isotope is important, we consider only a single value of y, so that dMldt = yM X B, . (7) We place the nuclei in a static field B, = Bog. In thermal equilibrium at temperature T the magnetization will be along 4: where the susceptibility is , y o and the Curie constant C = Np2/3kB, as in Chapter 11. The magnetization of a system of spins with I = 5 is related to the population difference ATl - N, of the lowcr and upper levels in Fig. 2: M, = (N, - N,)p, where the NS refer to a unit volume. The population ratio in thermal equilibrium is just given by the Boltzmann factor for the energy difference 2pBn: The equilibrium magnetization is M, = N p tanh(~B/k,T). When the magnetization component M, is not in thermal equilibrium, we suppose that it approaches equilibrium at a ratc proportional to the departure from the equilibrium value M,: 1 1 1 the standard notation T, is called the longitudinal relaxation time or the spin-lattice relaxation time. If at t = 0 an unmagnetized specirnen is placed in a magnetic field B,4, the magnetization will increase from the initial value LM, = 0 to a final value MZ = M,. Before and just after the specimen is placed in the field, the popula- tion N, will be equal to N,, as appropriate to thermal equilibrium in zero mag- netic field. It is necessary to reverse some spins to establish the new equilibrium distribution in the field B,. On integrating (10): 13 Magnetic Resonance 367 Figure 3 At time t = 0 an unmagnetized specimen M,(O) = 0 is placed in a static magnetic field B,. The magnetization increases with time and approaches the new equilibrium value Mo = x$,. This experiment defines the longitudinal relaxation time T,. The magnetic energy density -M . B decreases as part of the spin population moves into the lower level. The asymptotic value at t 9 TI is -MOB,. The energy flows from the spin system to the system of lattice vibrations; thus TI is also called the spin-lattice relaxation time. as in Fig. 3. The magnetic energy -M . B, decreases as M, approaches its new equilibrium value. Typical processes whereby the magnetization approaches equilibrium are indicated in Fig. 4. The dominant spin-lattice interaction of paramagnetic ions in crystals is by the phonon modulation of the crystalline electric field. Relaxation proceeds by three processes (Fig. 4b): direct (emission or absorption of a phonon); Raman (scattering of a phonon); and Orbach (inter- vention of a third state). Taking account of (lo), the z component of the equation of motion (7) becomes where (Ma - Mz)/TL is an extra term in the equation of motion, arising from the spin-lattice interactions not included in (7). That is, besides precessing about the magnetic field, M will relax to the equilibrium value Ma. If in a static field BOP the transverse magnetization component M, is not zero, then M, will decay to zero, and similarly for My. The decay occurs Insulator Metal Figure 4a Some important processes that contribute to longitudinal magnetization relaxation in an insulator and in a metal. For the insulator we show a phonon scattered inelastically by the spin system. The spin system moves to a lower energy state, and the emitted phonon has higher energy by hw, than the absorbed phonon. For the metal we show a similar inelastic scattering process in which a conduction electron is scattered. Direct 1; a T Raman Orbach l/T1 T~ or T ' l/T1 a e ~ p - ~ ~ T ) Figure 4b Spin relaxation from 2 + 1 by phonon emission, phonon scattering, and a two-stage phonon process. The temperature dependence of the longitudinal relaxation time TI is shown for the several processes. because in thermal equilibrium the transverse components are zero. We can provide for transverse relaxation: dM,ldt = y(M X B,), - M,/T2 ; (13b) dM,ldt = y(M X B,), - M,/T, , (134 where T2 is called the transverse relaxation time. The magnetic energy -M . B , does not change as M, or My changes, pro- vided that B, is along %. No energy need flow out of the spin system during relaxation of M, or My, SO that the conditions that determine T2 may be less strict than for T,. Sometimes the two times are nearly equal, and sometimes T , % T,, depending on local conditions. The time T2 is a measure of the time during which the individual moments that contribute to M,, M, remain in phase with each other. Different local magnetic fields at the different spins will cause them to precess at different frequencies. If initially the spins have a common phase, the phases will be- come random in the course of time and the values of M,, My will become zero. We can think of T2 as a dephasing time. I 3 Magnetic Resonance 369 To rf supply and circuit for measuring inductanc IB,, (static) and losses. rf coil Figure 5 Schematic arrangement for magnetic resonance experiments The set of equations (13) are called the Bloch equations. They are not symmetrical in x, y, and z because we have biased the system with a static mag- netic field along 9. In experiments an rf magnetic field is usually applied along the f or j. axis. Our main interest is in the behavior of the magnetization in the combined rf and static fields, as in Fig. 5. The Bloch equations are plausible, but not exact; they do not describe all spin phenomena, particularly not those in solids. We determine the frequency of free precession of the spin system in a static field B, = B,9 and with M, = Ma. The Bloch equations reduce to We look for damped oscillatory solutions of the form M, = m exp(-tlT') cos wt ; My = -m exp(- t/T1) sin wt . (15) On substitution in (14) we have for the left-hand equation 1 1 -w sin wt - -7 cos wt = - yBo sin wt - - cos wt , T T 2 (16) so that the free precession is characterized by w , = yB, ; T' = T , . The motion (15) is similar to that of a damped harmonic oscillator in two dimensions. The analogy suggests correctly that the spin system will show res- onance absorption of energy from a driving field near the frequency w, = yB,, and the frequency width of the response of the system to the driving field will be A w .
= 1/T2. Figure 6 shows the resonance of protons in water. The Bloch equations may be solved to give the power absorption from a rotating magnetic field of amplitude B,: B, = B, cos wt ; B, = -B, sin wt . (18) Figure 6 Proton resonance absorption in water. (E. L. IIahn.) After a routine calculation one finds that the power ahsorption is (CGS) The half-width of the resonance at half-maximum power is (Aw), = 1/T2 . (20) LINE WIDTH The magnetic dipolar intcraction is usually the most i~rlportar~t cause of line broadening in a rigid lattice of magnctic dipoles. The lnaguetic field AB seer1 by a magnetic dipole p, due to a magnetic dipolc p, at a point r,, from the first dipole is (CGS) by a fundamental resillt of magnetostatics. The order of magnitude of the intcraction is, with B, written for AB, The strong dependence on r suggests that close-neighbor interactions will be dominant, so that (CGS) B, - )(Lla3 , (23) 13 Magnetic Resonance where a is the separation of' nearest neighbors. This result gives us a rrleasure ol the width of the spin resonance line, assuming randorr~ orientation of the neighbors. For protons at 2A separation, To exprrss (21 ), (22), and (23) in SI, multiply the right-lland sides by gd4~r Motional Narrowing The linc width decreases for nuclei in rapid relative motiur~. Tlie effect in solids is illlistrated by Fig. 7: diffusion resembles a random walk as atoms jump from one crystal site to another. An atorr~ ranlains in one site for an avcragc time T that decreases markedly as the teiliperature increases. The motional effects 0 1 1 the line width arc cvcn more spectacular in nor- mal liquids, because the n~olecules arc highly mobile. The width of the proton resonance line in water is only 1 K 5 of the width expected for water molecules frozen in position. Thc effect of nuclear motion on T, and on the line width is subtle, but call he iinderstood by an elementary argument. I'e know from the Bloch equa- tions that T, is a measure of the time in which an individual spin becomcs dephased by one radiar~ because of a local perturbation in thc magnetic field 6.0 , 5.0 - i % F 4.0 .- C = 3.0 2 ' 7 - 2.0 1.0 0 150 200 250 300 350 Te~nperaturc, K Figure 7 Effcct of dirrusion of nuclei on the ~ i ' NMK line width in metallic lithium. At low tem- peratures the width agrees with the theoretical value for a rigid lattice. .4s the temperature in- creases, the diffusion rate increases and the line width decreases. The abrupt decrease ill liue width above T = 230 K occurs d l e n the diffusion hopping time .
r becomes shorter than IIyR,. Thus the experiment gives a direct measure of tl~c Impping ti~ne lor an atom to change lattice sites. (hfter H. S . Gutowsky and B. R. blcGalu-ey) Phase q(t) .' Phase in constant , ' local field Bi = +I / Phase with randnm , / local field + 1 / / , / / / / / , A t- Figure 8 Phase of a spin ~ I I a constant local field, as compared with dephasing of a spin which after fixed time intervals T hops at random among sites having local fields il. intensity. Let (Aw), = yB, denote the local frequency deviation due to a perturbation Bi. The local field may be caused by dipolar interactions with other spins. If the atoms are in rapid relative motion, the local field Bi seen by a given spin will fluctuate rapidly in time. W e suppose that the local field has a value +B, for a m average time T and then changes to -B,, as in Fig. 8a. Such a ran- dom change could be caused by a change of the angle between / .
L and r in (21). In the time T the spin will precess by an extra phase angle 6 9 = -+ YB$T relative to thc phase angle of the steady precession in the applied field Bn. The motional narrowing effect arises for short T such that 69 < 1. After n intervals of duration T the mean square dephasing angle in the ficld B, will be by analogy with a random walk process: the mean square hsplacement from the initial position after n steps of length e in random directions is (P) = ne2. The average number of steps necessary to dephase a spin by one rachan is n = llyZ~:?. (Spins dephased by much more than one radian do not con- tribute to the absorption signal.) This numbcr of steps takes place in a time 13 Magnetic Resonance quite different frorr~ the rigid lattice result T2 3 llyB,. From (26) we obtain as the line width for rapid motion with a characteristic time T: where (Aw), is the line width in the rigid lattice. The argument assumes that (Aw),T < 1, as othenvisc 6p will not be < 1. Thus A w < (Aw),. The shorter is 7, the narrower is the resonance line! This remarkable effect is known as motional narrowing.' The rotational relax- ation time of water moleculcs at room temperature is known from dielectric constant measurerncnts to he of the order of 10-"' s; if (Aw), ; = lo%-', then ( A W ) ~ T = 10 and A w = (Aw)~,T - 1 S-I, Thus the motion narrows the proton resonance line to about lo-' of the static width. HYPERFINE SPLITTING The hypcrf'ine interaction is the magnetic interaction between the mag- nctic moment of a nucleus and the magnetic moment of an electron. To an observer stationed on the nucleus, the interaction is caused by the magnetic field produced by the magnetic ~rio~rlent of the electron and by the motion of the electron about the nucleus. There is an electron current about the mlcleus if the electron is in a state with orbital angular momentum about the nucleus. But even if the electron is in a statc of zero orbital angular momentum, there is an electron spin current ahout the nucleus, and this current gives rise to the contact hyperfine interaction, of particular importarlce in solids. We can understand the origin of the contact interaction by a qualitative physical argu- ment, given in CGS. The results of the Uirac theory or the electron suggest that the magnetic mo~rler~t of pg = eh/2~r~c or the electron arises from the circulation of an elec- tron with velocity c in a currcnt loop of radius approximately the electron Compton wavclcngth, X, = film - lo-" cm. The electric current associated with thc circ~llation is I - e X (turns per unit time) - ec/X, , (29) "The physicaI ideas are drre to N. BIoemhergen, E. M. PurceLI, and R. V Pound, Phys. Rev. 73, 679 (1948). The result differs from the theoly of optical line width cansed by strong collisiuns hrtween atoms (as irr a gas disclrargc), where a short T gives a broad line. In the nuclear spin prob- lem the collisions are weak. In most optical pn~bblns the collisions of atoms are strong enough to intcrruyi the phase of the oscillation. In nuclear resonance the phase may v a r y snrwtl~ly in a collision, although the frequency may valy suddenly from one value to another nearby value. Figure 9 Magnetic field B prodncrd hy a charge moving in a circular loop. The contact part of the hyperfine interaction with a nuclear magnetic moment arises Cro~n the rcgiorr within or near to the current loop. The field avcragcd over a spl~erical shell that encloses the loop is 7.ero. Thus for an s electron (L = 0) only the contact part contributes to the interaction. and the magnetic field (Fig. 9) produced by the currcnt is The observer on the nucleus has the probability of finding himself insidc thc electron, that is, within a sphere of volume X, about the electron. Here @(O) is the value of the electron wavefunction at the nucleus. Thus the average value of the magnetic field seen by thc nuclcus is where pB = eti/2rnc = :eke is the Bohr magneton. The contact part of the hyperfine interaction energy is where I is the nuclear spin in units of fi. The contact interaction in an atom has the form 13 Magnetic Resonance Figure 10 Energy levels in a magnetic field of a system with S = i, I = i. The diagram is drawn for the strong field approximation p , B P a, where a is the hyperfine coupling constant, taken to be positive. The four levels are labeled by the magnetic quantum numbers m,, m,. The strong electronic transitions have Am, = 0, Ams = 21. Values of the hyperfine constant a for the ground states of several free atoms are: nucleus H ' ~ i ' NaZ3 K 3 ' K4' I 1 3 3 - - 3 3 - I 2 2 a in gauss 507 144 310 83 85 a in MHz 1420 402 886 231 127 In a strong magnetic field the energy level scheme of a free atom or ion is dominated by the Zeeman energy splitting of the electron levels; the hyperfine interaction gives an additional splitting that in strong fields is U' amsm,, where ms, m, are the magnetic quantum numbers. For the energy level diagram of Fig. 10 the two electronic transitions have the selection rules Am, = 5 1, AmI = 0, the frequencies are w = yHo + a/2fi. The nuclear transitions are not marked; they have Ams = 0, so that w,,, = a/2h. The frequency of the nuclear transition 1 + 2 is equal to that of 3 -+ 4. The hyperfine interaction in a magnetic atom may split the ground energy level. The splitting in hydrogen is 1420 MHz, thls is the radio frequency line of interstellar atomic hydrogen. Examples: Paramagnetic Point Defects The hyperfine splitting of the electron spin resonance furnishes valuable structural information about paramagnetic point defects, such as the F centers in alkali halide crystals and the donor impurity atoms in semiconductor crystals. Figure 1 1 An F center is a negative ion vacancy with one excess electron bound at the vacancy. The distribution of the excess electron is largely on the positive metal ions adjacent to the vacant lattice site. F Centers in Alkali Halides. An F center is a negative ion vacancy with one excess electron bound at the vacancy (Fig. 11). The wavefunction of the trapped electron is shared chiefly among the six alkali ions adjacent to the vacant lattice site, with smaller amplitudes on the 12 halide ions that form the shell of second nearest neighbors. The counting applies to crystals with the NaCl structure. If ~ ( r ) is the wavefunction of the valence electron on a single alkali ion, then in the first (or LCAO) approximation where in the NaCl structure the six values of rp mark the alkali ion sites that bound the lattice vacancy. The width of the electron spin resonance line of an F center is determined essentially by the hyperfine interaction of the trapped electron with the nu- clear magnetic moments of the alkali ions adjacent to the vacant lattice site. The observed line width is evidence for the simple picture of the wavefunction of the electron. By line width we mean the width of the envelope of the possi- ble hyperfine structure components. As an example, consider an F center in KCl. Natural potassium is 93 per- cent K3' with nuclear spin I = ; . The total spin of the six potassium nuclei at the F center is I,, = 6 x = 9, so that the number of hyperfine components is 21,- + 1 = 19; this is the number of possible values of the quantum number m,. There are (21 + = 46 = 4096 independent arrangements of the six spins distributed into the 19 components, as in Fig. 12. Often we observe only the envelope of the absorption line of an F center. Donor Atoms in Silicon. Phosphorus is a donor when present in silicon. Each donor atom has five outer electrons, of which four enter diamagnetically into the covalent bond network of the crystal, and the fifth acts as a paramag- netic center of spin S = i. The experimental hyperfine splitting in the strong field limit is shown in Fig. 13. When the concentration exceeds about 1 x 10" donors ~ m - ~ , the split line is replaced by a single narrow line. This is a motional narrowing effect ( E ~ . 28) of the rapid hopping of the donor electrons among many donor atoms. The 13 Magnetic Resonance Qnantum nu~r~ber 7n1 Figure 12 The 4096 arrangements of the six nuclear spins of K~~ as distrihnted into I9 hyperfine components. Each component uill he split further into a very large number of components by virtue of the residual hyperfine interaction with the 12 neighboring C1 ~mclei, which may be CP5 (75 percent) or C13' (25 percent). The envelope of the pattern is approximately gaussian in form. 4 Phosphorus (6 x 10" cm3) - magnetic field Plrosplrorus (1 x 10" cm3); 1 = 1/2 Figure 13 Electron spin resonance lines of P donor atoms in silicon. At the higher donor con- celrtration near the metal-insulator transition, a donor electron C ~ I liop from site to site SO rapidly that the hyperfine structure is suppressed. (After R. C, Fletcher, W. A. Yager, G. L. Pearson, and F R. Merritt.) rapid hopping averages out the hyperfine splitting. The hopping rate increases at the higher concentratior~s as the overlap of the donor electron wavefunctions L, is increased, a view supported by conductivity measnrements (Chapter 14). Knight Shvt At a fixed frequency the resonance of a nuclear spin is observed at a slightly different magnetic field in a metal than in a diamagnetic solid. The effect is known as the Knight shift or metallic shift and is vahiable as a tool for the study of conduction electrons. The interaction energy of a nucleus of spin I and magnetopic ratio Yr is where the first term is the interaction with the applied magnetic field B, and the second is the average hyperfine interaction of thc nucleus with the con- duction electrons. The average conduction electron spin (Sz) is related to the Pauli spin susceptibility X, of the conduction electrons: Mz = gNp,(S,) = x$,, whence the interaction may be written as The Knight shift is defined as and simulates a fractional change in the magnetogyic ratio. By the definitio~l (34) of the hyperfine contact energy, the Knight shift is givcn approximately by K ; = xSl~(0)l2/~; that is, by the Pauli spin susceptibility increased in the ratio of the conduction electron concentration at the nucleus to the average conduc- tion electron concentration. Experimental values are given in Table 2. The value of the hyperfine cou- pling constant a is somewhat different in the ~netal than in the free at0111 be- cause the wave functions at the nucleus are different. From the Knight shift of metallic Li it is deduced that the value of 1$(0)12 in the metal is 0.44 or the value in the free atom; a calculated valiie of the ratio i~sing thcorctical wavc functions is 0.49. Table 2 Knight shifts in NMR in metallic elements (At room temperature) Knight shift in percent Nucleus Knight shift in percent 13 Magnetic Resonance 379 0 0 (a) (b) ( 4 Figure 14 (a) Lowest-energy orientation of a nuclear electric quadmpole moment (Q > 0) in the local electric field of the four ions shown. The electrons of the ion itself are not shown. (b) High- est energy orientation. (c) The energy level splitting for I = 1. NUCLEAR QUADRUPOLE RESONANCE Nuclei of spin I 2 1 have an electric quadrupole moment. The quadrupole moment Q is a measure of the ellipticity of the distribution of charge in the nucleus. The quantity of interest is defined classically by eQ = $ $ (3z2 - ?)P(r)d3~ , (39) where p(r) is the charge density. An egg-shaped nucleus has Q positive; a saucer-shaped nucleus has Q negative. The nucleus when placed in a crystal will see the electrostatic field of its environment, as in Fig. 14. If the symmetry of this field is lower than cubic, then the nuclear quadmpole moment will lead to a set of energy levels split by the interaction of the quadrupole moment with the local electric field. The states that are split are the 21 + 1 states of a spin I. The quadrupole splittings can often be observed directly because an rf magnetic field of the appropriate frequency can cause transitions between the levels. The term nuclear quadrupole resonance refers to observations of nuclear quadmpole splittings in the absence of a static magnetic field. The quadrupole splittings are particularly large in covalently bonded molecules such as Clz, Br2, and 12; the splittings are of the order lo7 or 10' Hz. FERROMAGNETIC RESONANCE Spin resonance at microwave frequencies in ferromagnets is similar in principle to nuclear spin resonance. The total electron magnetic moment of the specimen precesses about the direction of the static magnetic field, and energy is absorbed strongly from the rf transverse field when its frequency is equal to the precessional frequency. We may think of the macroscopic vector S representing the total spin of the ferromagnet as quantized in the static mag- netic field, with energy levels separated by the usual Zeeman frequencies; the magnetic selection rule Ams = i l allows transitions only between adjacent levels. The unusual features of ferromagnetic resonance include: The transverse silsceptihility components and are very large because the magnetization of a ferromagnet in a given static field is very milch lar- ger than the magnetization of electronic or nuclear paramagnets in the same field. The shape of the speci~ne~i plays an irriportar~t role. Because the magnetiza- tion is large, the demagnetization Geld is large. The strong exchange coupling between the ferromagnetic electrons tends to suppress the dipolar contribution to the linc width, so that thc fcrromag- netic resonance lines can be quite sharp (<1 6) under favorable conditions. Saturation effects occur at low rf power levels. It is not possible, as it is with nuclear spin systems, to drive a ferromagnetic spin system so hard that the magnetization M, is reduced to zero or reversed. The ferromagnetic reso- nance excitation breaks down into spin wave modes before the magnetiza- tion vector can bc rotated appreciably from its initial direction. Shape Effects in FMR We treat the effects of specimen shape on the resonance frequency. Con- sider a specinieri of a cubic ferromagnetic insulator in the form of an ellipsoid with principal axes parallel to x, y, z axes of a cartesiari coordinate system. The demagnetization factors N,, N y , ATz are identical with the depolarization fac- tors to be defined in Chapter 16. The components of the internal magnetic field Bi in the ellipsoid arc rclatcd to thc applicd field by The Lorentz field (4d3)M and the exchange field AM do not contribute to the torque because their vector product with M vanishes identically. In SI we re- place the components of M by poM, with the appropriate redefinition of the Ws. The components of the spin equation of motion M = y(M X B') become, for an applied static field B04, To first order we may set dMz/dt = 0 and M; = M. Solutions of (40) with time dependence cxp(-iot) cxist if 13 Magnetic Resonance 381 Applied magnetic field (in gauss) Figure 15 FMR in a polished sphere of the ferromagnet yttrium iron garnet at 3.33 GHz and 300 K for B, 11 [lll]. The total line width at half-power is only 0.2 G. (After R. C. LeCraw and E. Spencer.) so that the ferromagnetic resonance frequency in the applied field Bo is (CGS) 0 2 0 = ?[Bo + (Ny - N,)MI[Bo + (Nx - NJMI ; (41) The frequency wo is called the frequency of the uniform mode, in distinction to the frequencies of magnon and other nonuniform modes. In the uniform mode all the moments precess together in phase with the same amplitude. For a sphere N, = N , , = N,, SO that wo = yBo. A very sharp resonance line in this geometry is shown in Fig. 15. For a flat plate with Bo perpendicular to the plate N, = Ny = 0 ; N, = 47r, whence the ferromagnetic resonance frequency is (CGS) w, = Y(B, - 47rM) ; If Bo is parallel to the plane of the plate, the xz plane, then N, = N, = 0 ; Ny = 47r. and ' 5 " ," ?I+KE.'..'-.' (CGSJ w,, = y[B,,(Bo t 4 ~ 3 l ) l ' : !SI) q , y[Bo(B, + &Im . (43) Figure 16 Spin wave resonance in a thin film. The plane of the film is normal to the applied mag- netic field B,. A cross section of the film is shown here. The internal magnetic field is B,, - 4mM. The spins on the surfaces of the film are assumed to be held fixed in direction by surface anisotropy forces. A uniform rf field will excite spin wave modes having an odd number of half-wavelengths. The wave shown is for n = 3 half-wavelengths. The experiments determine y, which is related to the spectroscopic split- ting factor g by - y = gpB/h. Values of g for metallic Fe, Co, Ni at room tem- perature are 2.10,2.18, and 2.21, respectively. Spin Wave Resonance Uniform rf magnetic fields can excite long-wavelength spin waves in thin ferromagnetic films if the electron spins on the surfaces of the film see differ- ent anisotropy fields than the spins within the films. In effect, the surface spins may be pinned by surface anisotropy interactions, as shown in Fig. 16. If the rf field is uniform, it can excite waves with an odd number of half- wavelengths within the thickness of the film. Waves with an even number of half-wavelengths have no net interaction energy with the field. The condition for spin wave resonance (SWR) with the applied mag- netic field normal to the film is obtained from (42) by adding to the right-hand side the exchange contribution to the frequency. The exchange contribution may be written as Dk2, where D is the spin wave exchange constant. The as- sumption ka & 1 is valid for the SWR experiments. Thus in an applied field B, the spin wave resonance frequencies are: where the wavevector for a mode of n half-wavelengths in a film of thickness L is k = n?r/L. An experimental spectrum is shown in Fig. 17. 13 Magnetic Resonance Spin wave order number Figure 17 Spin wave resonance spectrum in a Pennalloy (80Ni20Fe) film at 9 GIIz. The order n~~mber is tlie number of half-wavelengths in the thickness of the film. (After R. Weber.) ANTIFERROMAGNETIC RESONANCE \ve consider a uniaxial antiferromagnet with spins on two sublattices, 1 and 2. \ Z k suppose that the magnetization M, on sublattice 1 is directed along the +z direction by an anisotropy field B,i; the anisotropy field (Chapter 12) results from an anisotropy energy density C1,(OI) = K sin2 O1. Here 8 , is the angle be- tween M, and the z axis, whence BA = 2K/M, with M = lMll = IM,I. The mag- netization M, is hrected along the -s direction by an anisotropy Field -BA4. If +z is an easy direction of magnetization, so is -z. If one sublattice is directed along +z, the other will bc directed along -z. The exchange interaction between M, and Mg is treated in the mean field approximation. The exchange fields are whcrc A is positive. Here B, is the field that acts on the spins of sublattice 1, and B , acts on sublattice 2. In the absence of an external magnetic field the total field acting on MI is B, = -AM, + B,?; the total field on Mp is Be = -AMI - B.42, as in Fig. 18. I B,(ex) = -AMz \ \ \ \ -BA \ \ B,(ex) S -AM1 Figure 18 Effective fields in antiferromagnetic resonance. \ The magnetization MI of sublattice 1 sees a field -AM, + f B,P; the magnetization M, sees -AM, - B,?. Both ends of the crystal axis are "easy axes" of magnetization. In what follows we set = M ; Mz = -M. The linearized equations of motion are dMi/dt = y[MY(AM + B,) - M(-AM$)] ; dMY/dt = y[M(-AM;) - M;(AM + B,)] ; (46) d M y d t = y[M$(-AM - B,) - (-M)(-AMY)] ; = Y[(-MI(-AM;) - M",(-AM - B,)] . (47) We define M: = M; + iM;; M: = M i + iM$. Then (46) and (47) bccomc, for time dependence exp(-iot), -iuM: = -iy[M:(BA + AM) + M l ( A M ) ] ; -ioMl = iy[Ml(BA + A M ) + M:(AM)] . These equations have a solution if, with the exchange field BE = AM, Thus the antiferromagnetic resonance frequency is given by 0; = $B(B, + 2BE) . (48) MnF, is an extensively studied antiferrornagnet. The structure is shown in Fig. 19. The observed variation of o, with temperature is shown in Fig. 20. Careful estimates were made by Keffer of B, and BE for MnF,. He estimated 13 Magnetic Resonance 385 @ Mn2+ Figure 19 Chem~cal and magnetic structure of MnF2 F- The arrows indicate the hrectlon and arrangement of the magnetic moments asslgned to the manganese atoms. Temperature in K I Figure 20 Antiferromagnetic resonance frequency for MnF, versus temperature. and Nethercot.) (After Johnson BE = 540 k G and BA = 8.8 k G at 0 K, whence (2BABE)lI2 = 100 kG. The observed value is 93 kG. Richards has made a compilation of AFMR frequencies as extrapolated to 0 K: Crystal CoF, NiF, MnF, FeF, MnO NiO Frequency in 10" Hz 85.5 93.3 26.0 158. 82.8 109 ELECTRON PARAMAGNETIC RESONANCE Exchange Narrowing We consider a pararr~agmet with an exchange interaction J among nearest- neighbor electron spins. The temperature is assumed to be well above any spin-ordering temperature T,. Under these conditions the width of the spin resonance line is usually much narrowcr than cxpcctcd for thc dipolc-dipolc interaction. The effect is called exchange narrowing; there is a close analogy with motional narrowing. We interpret the exchange frequency w , , =]/li as a hopping frequency 1 / ~ . Then by generalization of the motional-narrowing result (28) we have for tlie width of tlie exdrange-narrowed line: where = Y(B?) is the square 01 the static dipolar width in the absence of exchange. A usefill and striking example of exchange narrowing is the paramagnetic organic crystal known as the g marker or DPPH, diphenyl picryl hydrazyl, often used for magnetic field calibration. This free radical has a 1.35 G half-width of the resonance line at half-power, only a few percent of the pure dipole width. Zero-Field Splitting A number of paramagnetic ions have cvstal ficld splittings of thcir magnetic ground state encrgy levels in the range of 10'' - lo1' Hz, conveniently accessi- ble by microwave techniques. The Mn" ion has been studied in many crystals as an additive impurity. A ground state splitting in the range lo7 - lO9Hz is observed, according to the environment. PRINCIPLE OF MASER ACTION Crystals can be used as microwave and light amplifiers and as sourccs of coherent radiation. A maser amplifies microwaves by the stimulated emission of radiation; a laser amplifies light hy the same method. The principle, due to Townes, may be understood from the two-level magnetic system of Fig. 21 rel- evant for masers. There are nu atoms in the upper state and ne atoms in the lower state. U7e immerse the system in radiation at frequency w; tlie amplitude of the magnetic component of the radiation field is Brf The probability per atom per unit time of a transition between the upper and lower states is here p is the magnetic moment, and A o is the combined width of the two levels. The result (50) is from a standard result of quantum mechanics, called Fermi's golden rule. 13 Magnetic Resonance Upper state n" Figure 21 A two-level system, to explain maser op- eration. The populations of the upper and lower states are nu and n , , respectively. The frequency of 1 the emitted radiation is w; the combined width of the 7 states is AW = hw, + PW,. The net energy emitted from atoms in both upper and lower states is per unit time. Here 9 denotes the power out; hw is the energy per photon; and nu - nl is the excess of the number of atoms n, initially able to emit a pho- ton over the number of atoms nl able to absorb a photon. In thermal equilibrium nu < nl, so there is no net emission of radiation, but in a nonequilibrium condition with nu > nl there will be emission. If we start with nu > nl and reflect the emitted radiation back onto the system, we increase Brf and thereby stimulate a higher rate of emission. The enhanced stimulation continues until the population in the upper state decreases and becomes equal to the population in the lower state. We can build up the intensity of the radiation field by placing the crystal in an electromagnetic cavity. This is like multiple reflection from the walls of the cavity. There will be some power loss in the walls of the cavity: the rate of power loss is B P w (CGS) 9, = -- . - ; 8~ Q (52) where V is the volume and Q is the Q factor o nd B$ to be a volume average. The condition for maser action is that the emitted power 9 exceed the power loss 9 , . Both quantities involve B:. The maser condition can now be ex- pressed in terms of the population excess in the upper state: VAB (CGS) nu - nl > - 8 v Q ' where /L is the magnetic moment. The line width AB is defined in terms of the combined line width Aw of the upper and lower states as pAB = fiAw. The Figure 22 Three-level maser system. Two possible modes of operation are shown, starting from rf saturation of the states 3 and 1 to obtain n, = n,. central problem of the maser or laser is to obtain a suitable excess population in the upper state. This is accomplished in various ways in various devices. Three-Level Maser The three-level maser system (Fig. 22) is a clever solution to the excess population problem. Such a system may derive its energy levels from magnetic ions in a crystal, as Bloembergen showed. Rf power is applied at the pump frequency Lop = Eg - El in sufficient intensity to maintain the population of level 3 substantially equal to the population of level 1. This is called saturation-see Problem 6. Now consider the rate of change of the population n2 of level 2 owing to normal thermal relaxation processes. In terms of the indicated transition rates P, In the steady state dn2/dt = 0, and by virtue of the saturation rf power we have n3 = n,, whence The transition rates are affected by many details of the paramagnetic ion and its environment, but one can hardly fail with this system, for either n2 > nl and we get maser action between levels 2 and 1, or n2 < n, = n, and we get maser action between levels 3 and 2. The energy levels of the Er3+ ion are used in communication fiber optics amplifiers. The ion is optically pumped from level 1 to level 3; there is fast nonradiative decay from level 3 to level 2. The signal at a wavelength of 1.55 pm is amplified by stimulated emission from level 2 to level 1. The wavelength is favorable for long-distance propaga- tion in the optical fiber. The bandwidth is of the order of 4 X 1012 Hz. 13 Magnetic Resonance 389 30, Broad bands Intermediate levels , Figure 23 Energy level diagram of Cr3+ in rnhy, as used in laser operation. The initial excitation takes place to the broad hands; they decay to the intermediate levels by the emission of phonons, and the intermediate levels radiate photons as the ion makes the transition to the ground level. Lasers The same crystal, ruby, used in the microwave maser was also the first crystal to exhibit optical maser action, but a different set of energy levels of Cr3+ are involved (Fig. 23). About 15,000 cm-' above the ground state there lie a pair of states labeled 'E, spaced 29 cm-' apart. Above 'E lie two broad bands of states, labeled 4 ~ 1 and 4F2. Because the bands are broad they can be populated efficiently by optical absorption from broadband light sources such as xenon flash lamps. In operation of a ruby laser both of the broad 4F bands are populated by broadband light. Atoms thus excited will decay in sec by radiationless processes with the emission of phonons to the states 'E. Photon emission from the lower of the states 'E to the ground state occurs slowly, in about 5 x sec, so that a large excited population can pile up in 'E. For laser action this popula- tion must exceed that in the ground state. ions cm-3 are in an The stored energy in ruby is lo8 erg cm-3 if 10'' cr3+ . excited state. The ruby laser can emit at a very high power level if all this stored energy comes out in a short burst. The overall efficiency of conversion of a ruby laser from input electrical energy to output laser light is about one percent. Another popular solid state laser is the neodymium glass laser, made of calcium tungstate glass doped with Nd3+ ions. This operates as a four level system (Fig. 24). Here it is not necessary to empty out the ground state before laser action can occur. Figure 24 Four-level laser system, as in the neodymium glass laser. SUMMARY (In CGS Units) The resonance frequency of a free spin is wo = yB,, where y = p/hI is the magnetogyric ratio. The Bloch equations are dM,/dt = y(M X B), - MJT, ; dM,ldt = y(M X B), - My/T2 ; dMB/dt = y(M X B), + (M, - M,)/T, The half-width of the resonance at half-power is (Am),,, = 1/T2. The dipolar line width in a rigid lattice is (AB), = d a 3 . If the magnetic moments are ambulatory, with a characteristic time T < l/(Aw),, the line width is reduced by the factor ( A W ) , ~ . In this limit 1/T, = 1/T2 .
= (Aw)~,T. With exchange coupling in a paramagnet the line width becomes ==(Am)2,w,,. The ferromagnetic resonance frequency in an ellipsoid of demagnetization factors N,, Ny, Nz is 0; = Y [ B ~ + (N, - Nz)M][BO + (N, - N,)M]. The antiferromagnetic resonance frequency is 0 2 , = 4 B A ( B A + 2B,), in a spherical specimen with zero applied field. Here BA is the anisotropy field and BE is the exchange field. The condition for maser action is that nu - nl > VAB/BwpQ. 13 Magnetic Resonance 391 Problems 1. Equivalent electrical circuit. Consider an clnpty coil of inductance L, in a series with a resistance R,; show if the coil is co~nplctcly filled with a spin system charac- terized by the susceptibility components ~ ' ( w ) and x"(o) that the inductance at frequency w becomes L = [l + 4vx'(o)]Lo, in series with an effective resistance R = 4 ~ w ~ " ( o ) L , + A,. In this problem , y = X' + i,y" is defined for a linearly polar- ized rf field. Hint: Consider the impedance of the circuit. (CGS units.) 2. Rotating coordinate system. We define the vector F(t) = F&)f + Fy(t)f + F,(t)i. Let the coordinate system of the unit vectors % , y, 4 rotate with an instantaneous angular velocity a, so that d3dt = - 4 3 , etc. (a) Show that dF/& = (dF/dt), + R x F, where (dF/dt), is the time derivative of F as viewed in the rotating frame R. (h) Shou. that (7) may be written (dMldt), = yM X (R, + a&). This is the equation of motion of M in a rotating coordinate system. Tlre transformation to a rotating systcm is extraordinarily useful; it is exploited widely in the litcrature. (c) Let R = - yB,i; thus in the rotating fiame there is no static magnetic field. Still in the ro- tating frame, we now apply a dc pulse B , i for a time t. If the magnetization is initially along 9, find an expression for the pulse length t such that the magnetization will be directed along -4 at the end of the pulse. (Neglect relaxation effects.) (d) Decribe this I ~ ~ l s e a c viewed from the laboratory frame of reference. 3. Hyperfine effects on ESR in metals. We suppose that the electron spin of a con- duction electron in a metal sees an effective magnetic field from the hyperfine interaction of the electron spin with the nuclear spin. Let the z component of the firld seer, by the conduction clcctron be written where I,'is equally likely to be 2;. (a) Show that (B:) = (aI2N)'N. (b) Show that (B$ = 3(aI2N)'N" for N 1. 4. FMR in the anisotropy jeld. Consider a sphcrical specimen of a uniaxial ferro- magnetic cvstal with an anisotropy energy density of the form U, = K sin2 0, where 0 is the angle betu~een the rnagrletization and thc z axis. We assume that K is posi- tive. Show that the ferromagnetic resonance frequency in an external magnetic field B,9 is w,, = y(B, + B,), where B, = 2K/M,. 3. Exchange frequency resonance. Consider a ferrimagnct with two sublattices A and B of magnetizations M, and M,, where M, is opposite to Ma when the spin system is at rest. The gyromagnetic ratios are YA, y, and thc molecular fields are B, = -AMHr BE = -AMA. Show that there is a resonance at This is called the exchange frequency resonance. 6. Rf saturation. Gwen, at equilibriurr~ for temperaturc 7: a two-level spin system in a magnetic field H,i, with populations N1, N , and transition rates !
A ' , , , W,,. We apply an rf signal that giws a transition ratc Wa (a) Derive the equation for dMz/dt and show that ~ I I the steady state where l/Tl = WIP + WP1. It will he helpful to write i V = N, + N2; n = Nl - h;,; and no = N(W21 - W,,)/(FVPl + WII,,). We see that as long as 2W,+T1 < 1 the absorption of energy from the rf field does rrot snhstantially alter thc population distribution from its thermal equilibrium value. (b) Using the expression for n, write down the rate at which energy is absorbed frurrr the r f field. W7hat happcns as W , approaches 1/2Tl? This effect is called saturatiori, and i t .
7 onset may he uscd to measure TI. Plasmons, Polaritons, and Polarons DIELECTRIC FUNCTION OF THE ELECTRON GAS Definitions of the dielectric function Plasma optics Dispersion relation for electromagnetic waves Transverse optical modes in a plasma Transparency of metals in the ultraviolet Longitudinal plasma oscillations PLASMONS ELECTROSTATIC SCREENING Screened coulomb potential Pseudopotential component U(0) Mott metal-insulator transition Screening and phonons in metals POLARITONS LST relation ELECTRON-ELECTRON INTERACTION Fermi liquid Electron-electron collisions ELECTRON-PHONON INTERACTION. POLARONS PEIERLS INSTABILITY OF LINEAR METALS SUMMARY PROBLEMS 1. Surface plasmons 2. Interface plasmons 3. Alfvdn waves NOTE: The text and problems of this chapter assume facility in the use of electromagnetic theory at the level of a good senior course. 4. Helicon waves 425 5. Plasmon mode of a sphere 425 6. Magnetoplasma frequency 425 7. Photon branch at low wavevector 426 8. Plasma frequency and electrical conductivity 426 9. Bulk modulus of the Fermi gas 426 10. Response of electron gas 426 11. Gap plasmons and the van der Waals interaction 426 , Region of attenuation I Region of propagation Figure 1 Dielectric function E ( W ) or E ( W , 0) of a free-electron gas versus frequencyin units of the plasma frequency u p . Electromagnetic waves propagate without damping only when E is positive and real. Electromagnetic waves are totally reflected from the medium when E is negative. DIELECTRIC FUNCTION OF THE ELECTRON GAS The dielectric function E(w,K) of the electron gas, with its strong depen- dence on frequency and wavevector, has significant consequences for the physical properties of solids. In one limit, E(w,O) describes the collective exci- tations of the Fermi sea-the volume and surface plasmons. In another limit, E(O,K) describes the electrostatic screening of the electron-electron, electron- lattice, and electron-impurity interactions in crystals. We will also use the dielectric function of an ionic crystal to derive the po- lariton spectrum. Later we discuss the properties of polarons. But first we are concerned with the electron gas in metals. Definitions of the Dielectric Function. The dielectric constant E of elec- trostatics is defined in terms of the electric field E and the polarization P, the dipole moment density: (CGS) D = E + 47rP = EE ; Thus defined, E is also known as the relative permittivity. The introduction of the displacement D is motivated by the usefulness of this vector related to the external applied charge density p,, in the same way as E is related to the total charge density p = p,, + pind, where pind is the charge density induced in the system by p,,,. Thus the divergence relation of the electric field is (CGS) div D = div EE = 4npe, (2) div E = 47rp = 4n(p, + pied) (3) Parts of this chapter will be written in CGS; to obtain results in S I , write for 4n. .
. -- We need relations between the Fourier components of D, E, p, and the electrostatic potential cp. For brevity we do not exhibit here the frequency de- pendence. Define E(K) such that then (3) becomes div E = div C E(K) exp(iK - r) = 47r C p(K) exp(iK. r) , (3b) and (2) becomes div D = div Z e(K)E(K) exp(iK.r) = 4n-Z p,,(K) exp(iK-r) . (3c) Each of the equations must be satisfied term by term; we divide one by the other to obtain The electrostatic potential Q,,, defined by -VQ,, = D satisfies the Poisson equation V2~,,, = -47rp,,; and the electrostatic potential 9 defined by -Vq = E satisfies VZ9 = -4~rp. The Fourier components of the potentials must therefore satisfy by (3d). We use this relation in the treatment of the screened coulomb potential. Plasma Optics The long wavelength dielectric response e(w,O) or E(W) of an electron gas is obtained from the equation of motion of a free electron in an electric field: If x and E have the time dependence e-'"t, then - w2mx = -eE ; x = eElmw2 . (5) The dipole moment of one electron is -ex = -e2E/mw2, and the polarization, defined as the dipole moment per unit volume, is where n is the electron concentration. The dielectric function at frequency w is 14 Plasmons, Polaritons, and Polarons 397 The dielectric function of the free electron gas follows from (6) and (7): 4 m e 2 (CGS) E ( W ) = 1 - - ; m2 The plasma frequency wp is defined by the relation (CGS) w; = 4me2/m ; A plasma is a medium with equal concentration of positive and negative charges, of which at least one charge type is mobile. In a solid the negative charges of the conduction electrons are balanced by an equal concentration of positive charge of the ion cores. We write the dielectric function (8) as plotted in Fig. 1. If the positive ion core background has a dielectric constant labeled ~ ( m ) essentially constant up to frequencies well above wp, then (8) becomes E(W) = ~ ( m ) - 4mezlmw2 = ~ ( m ) [ l - ?$/w2] , (11) where Gp is defined as 0; = 4 m e 2 / ~ ( m ) m . Notice that E = 0 at w = Zp. Dispersion Relation for Electromagnetic Waves In a nonmagnetic isotropic medium the electromagnetic wave equation is . . "?""...Fr.-.u;i. .
, .
< , (SI) poaP~/.3? = v%'". : (13) We look for a solution with E cc exp(-iwt) exp(iK . r) and D = e(w,K)E; then we have the dispersion relation for electromagnetic waves: (CGS) e(w,K)w2 = cZI<2 ; This relation tells us a great deal. Consider E real and > 0. For w real, K is real and a transverse electromagnetic wave propagates with the phase velocity C/E'". E real and < 0. For w real, K is imaginary and the wave is damped with a characteristic length IIIKI. E complex. For w real, K is complex and the waves are damped in space. . E = m. This means the system has a finite response in the absence of an ap- plied force; thus thc poles of E(o,K) define the frequencies of the free oscillations of the medlum. E = 0. We shall see that longitudinally polarized waves are possible only at the zeros of E . Transuene Optical Modes in a Plasma The dispersion relation (14) hccomes, with (11) for ~ ( w ) , For o < ij, we have K2 < 0, so that K is imaginary. The solutions of the wave equation are of the form exp(-IKlx) in the frequency region O < w 5 G,. Waves incident on the medium in this frequency region do not propagate, but will be totally reflected. An electron gas is transparent when o > Gp, for here the dielectric func- tion is positive real. The dispersion relation in this region may be written as this describes transverse electromagnetic waves in a plasma (Fig. 2). Values of the plasma frequency wp and of thc free space wavelength A, = 2.rrc/wp for electron concentrations of interest are given below. A wave will prop- agate if its free space wavelength is less than A,; otherwise the wave is rcflected. Transparency of Metals i n the Ultraviolet. From the preceding discussion of the dielectric function we conclude that simple metals should reflect light in the visible region and be transparent to light at high frequencies. A comparison of calculated and observed cutoff wavelengths is given in Table 1. The reflection of light from a metal is entirely similar to the reflection of radio waves from the ionosphere, for the free electrons in thc ionosphere make the dielectric con- stant negative at low frequencies. Experimental restilts for InSb with n = 4 X 10'' c m - h e shown in Fig. 3, where the plasma frequency is near 0.09 eV. Longitudinal Plasma Oscillations The zeros or the dielectric function determine the frequencies of the longitudinal modcs of oscillation. That is, the condition determines the longitudinal frequency w~ near K = 0. 14 Plasmons, Pokaritons, and Pokarons 399 Figure 2 Dispersion relation for transverse electromagnetic waves in a plasma. The group veloc- ity o, = do/dK is the slope of the dispersion curve. Although the dielectric function is between zero and one, the group velocity is less than the velocity of light in vacuum. Table 1 Ultraviolet transmission limits of alkali metals, in A A,, calculated 1550 2090 2870 3220 3620 Ap, observed 1550 2100 3150 3400 - 0.05 0.10 0.15 0.20 Photon energy, eV Figure 3 Reflectance of indium antimonide with n = 4 X 10" CII-~. (After J. N. Hodgson.) By the geometry of a longitudinal polarization wave there is a depolarization field E = -47rP, mscussed below. Thus D = E + 4~rP = 0 for a longitudinal wave in a plasma or more generally in a crystal. In S I units, D = eOE + P = 0. For an electron gas, at the zero (17) of the dielectric function (10) whence wL = wp. Thus there is a free longitudinal oscillation mode (Fig. 4) of an electron gas at the plasma frequency described by (15) as the low- frequency cutoff of transverse electromagnetic waves. A longitudinal plasma oscillation with K = 0 is shown in Fig. 5 as a uni- form displacement of an electron gas in a thin metallic slab. The electron gas is moved as a whole with respect to the positive ion background. The displace- ment u of the electron gas creates an electric field E = 47rneu that acts as a restoring force on the gas. The equation of motion of a unit volume of the electron gas of concentra- tion n is (CGS) d2u nm- = -neE = -4m2e2u , dt2 (19) This is the equation of motion of a simple harmonic oscillator of frequency wp, the plasma frequency The expression for op is identical with (9), which arose in a different connection. In SI, the displacement u creates the electric field E = neul~,,, whence up = (ne2/eom)112. A plasma oscillation of small wavevector has approximately the frequency w,. The wavevector dependence of the dispersion relation for longitudinal oscillations in a Fermi gas is given by where u, is the electron velocity at the Fermi energy. Figure 4 A plasma oscillation. The arrows indicate the direction of displacement of the electrons. 14 Plasmons, Polaritons, and Polarons 401 Surface charge density u = -neu g = +neu Figure 5 In (a) is shown a thin slab or film of a metal. A cross section is shown in (b), with the positive ion cores indicated by + signs and the electron sea indicated by the gray back- ground. The slab is electrically neutral. In (c) the negative charge has been displaced upward uniformly by a small distance u, shown exaggerated in the figure. As in (d), this displacement establishes a surface charge density -neu on the upper surface of the slah and +neu on the lower surface, where n is the electron concentration. An electric field E = 4meu is produced inside the slah. This field tends to restore the electron sea to its equilibrium position (b). In SI units, E = neul~,. PLASMONS A plasma oscillation in a metal is a collective longitudinal excitation of the conduction electron gas. A plasmon is a quantum of a plasma oscillation; we may excite a plasmon by passing an electron through a thin metallic film (Figs. 6 and 7) or by reflecting an electron or a photon from a film. The charge of the electron couples with the electrostatic field fluctuations of the plasma oscillations. The reflected or transmitted electron wilI show an energy loss equal to integral multiples of the plasmon energy. Experimental excitation spectra for A1 and Mg are shown in Fig. 8. A comparison of observed and calculated values of plasmon energies is given in Table 2; further data are given in the reviews by Raether and by Daniels. Recall that Lip as defined by (12) includes the ion core effects by use of ~ ( m ) . I Scattered elecbon Figure 6 Creation of a plasmon in a metal film by inelastic scattering of an electron. The incident electron typically has an energy 1 to 10 keV; the plasmon energy may be of the order of 10 eV. An event is also shown in which two plasmons are created. L 1 Cathode -- -!
-A n o d e Retarding Figure 7 A spectrometer with electrostatic analyzer for the study of plasmon excitation by electrons. (After J. Daniels eta].) Spherical condensor It is equally possible to excite collective plasma oscillations in dielec- tric films; results for several dielectrics are included. The calculated plasma energies of Si, Ge, and InSb are based on four valence electrons per atom. In a dielectric the plasma oscillation is physically the same as in a metal; the entire valence electron gas oscillates back and forth with respect to the ion cores. 14 Plasmom, Polaritons, and Polarons 403 g 2 ::Fi s 6 .- e! $:; m .% + .2 9 4 -1 4 2 2 2 0 0 0 20 40 GU 80 100 120 0 10 20 30 40 50 60 70 Ele~trun energy loss (eV) Electron energy 10% (a) (a) (b) Figure 8 Energy loss spectra for electrons reflected from films of (a) aluminum and (b) nragne- sium, for primaty electron energies of 2020 cV The 12 loss peaks observed in A1 are made up of combinations of 10.3 and 15.3 eV losses, where the 10.3 eV loss is due to surface plas~nons and the 15.3 cV loss is due to voll~me plasmo~is. The ten loss peak? obsenrcd in Mg are made up of con~bi- nations of 7.1 cV surface plasmons and 10.6 cV volume plasmons. Surface plasmons are the suh- ject of Problem 1. (After C . J. Powell and J. R. Swan.) Table 2 Volume plasmon energies, in eV Calculated hlaterid Observed Metals Li Na K M g A1 Dielectrics Si Ge InSb ELECTROSTATIC SCREENING The electric field of a positive charge embedded in an electron gas falls off with increasing r faster than l/r, becanse the electron gas tends to gather around and thus to screen the positive charge. The static screening can be de- scribed hy the wavevector dependence of the static dielectric function e(0,K). We consider the response of the electror~s to an applied external electrostatic field. We start with a nniform gas of electrons of charge concentration -a$ superirriposed on a hackground of positive charge of concentration nfi Let the positive charge hackground be deformed mechanically to produce a sinusoidal variation of positive charge density in the x direction: P f ( x ) = noe + p,,(K) sin Kx . ( 2 2 ) The term P,,,(K) sin Kx gives rise to an electrostatic field that we call the exter- nal field applied to the electron gas. The electrostatic potential cp of a charge distribution is found from the Poisson cquation V Z q = -4.rrp, by ( 3 ) with E = - V q . For the positive charge we have q = qFa(K) sin Kx ; p = p,,(K) sin Kx . ( 2 3 ) The Poisson equation gives the relation The electron gas will be deformed by the combined influences of the elec- trostatic potential q,,,(K) of the positive charge distrihntion and of the as yet unknown induced clcctrostatic potential qind(K) sin Kx of the deformation of the electron gas itself. The electron charge density is P-(x) = -n$ + pbd(K) sin ~x , ( 2 5 ) where is the amplitude of the charge density variation induced in the electron gas. We want to find p J K ) in terms ofp,,,(K). The amplitude of the total electrostatic potential q ( K ) = cp,,(K) + qind(K) of the positive and negative charge distributions is related to the total charge density variation p(K) = p,,(K) + pind(K) by the Poisson equation. Then, as in Eq. (24), To go further we need another equation that relates the electron conceii- tration to the electrostatic potential. We develop this connection in what is called the Thomas-Fermi approximation. The approximation consists in assum- ing that a local internal chemical potential can be defined as a function of the electron concentration at that point. Now the total chemical potential of the electron gas is constant in equilibrium, independent of position. In a region where there is no electrostatic contrihlition to the chemical potential we have at absolute zero, accordi~~g to (6.17). In a region where the electrostatic poten- tial is ~ ( x ) , the total chemical potential (Fig. 9) is constant and equal to where cF(x) is the local value of the Fermi energy. The expression (28) is valid for static electrostatic potentials that vary slowly compared with the wavelength of an electron at the Fernli level; 14 Plasrnons, Polaritons, and Polarons 405 Che~r~ical potential / P 6 0 electron concentratinn Figure 9 1 1 1 thcr~nal and diffusive equilibrium the chemical potential is constant; to maintain it constant we increme the electron concentration in regions of space where the potential energy is low, and we decrease the concentration where the potential is high. specifically, the approximation is q kF. By a Taylor series expansion of E,, Eq. (28) may he written as de, -[n(x) - no] = ep(x) . (29) dno From (27) we have dcF/dno = 2eF/3n,, whence The left-hand side is the induced part of the electron concentration; thus the Fourier components of this equation are p,&) = (3noeZ/2e,jp(K) . (31) By (26) this becomes phd(K) = - (6m,e2/~FK2)p(K) . (32) By (3dj we have here, after some rearrangement, where a. is the Bohr radius and D(cF) is the density of states for a free electron gas. The approximation (33) for c(0,K) is called the Thomas-Fermi dielectric function, and l/k, is the Thomas-Fermi screening length, as in (40) below. For copper with no = 8.5 X loz2 ~ m - ~ , the screening length is 0.55 A. 1 % have derived two limiting expressions for the dielectric function of an electron gas: We notice that e(0,K) as K + 0 does not approach the same limit as e(w,O) as w 4 0. This means that great care must be taken with the dielectric function near the origin of the w-K plane. The full theory for the general function E(w,K) is due to ~i~~dharrl.' Screened Coulomb Potential. We consider a point charge q placed in a sea of conduction electrons. The Poisson equation for the unscreened coulomb potential is v2q, = -47rqa(r) : (36) and we know that cp, = q/n Let us write We use in (36) the Fourier representation of the delta function: whence K ' ~ , ( K ) = 4 ~ 4 . By (3e), cp,(K)lcp(K) = E(K) , where cp(K) is the total or screened potential. We nse E(K) in the Thomas- Fermi form (33) to find The screened coulomb potential is the transform of cp(K): K sin Kr 4 d K - = , exp(-kd) P + kf as in Fig. 10a. The screening parameter k, is defined by (34). The exponential factor reduces the range of the coulomb potential. The bare potential qlr is obtained on letting the chargc concentration no 4 0, for then k, -+ 0. In the vacuum limit q ( K ) = 47rq/K2. One application of the screened interaction is to the resistivity of certain alloys. The atoms of the series Cu, Zn, Ga, Ge, As have valences 1, 2, 3, 4, 5. An atom of Zn, Ga, Ge, or As added substitutionally to metallic Cu has an ex- cess charge, referred to Cu, of 1,2, 3, or 4 if all the valence electrons join the conduction band of the host metal. The foreign atom scatters the conduction electrons, with an interaction given by the screened coulomb potential. This scattering contributes to the rcsidual clcctrical resistivity, and calcnlations by Mott of the resistivity increase are in fair agreement with experiment. 'A good discussion of the Lindhard dielectric function is given by I. Ziman, Principles o f the thvuy ofsulirlu, 211d ed., Canbridge, 1072, Chapter 5. The algebraic stcps in the evaluation of Ziman's equation (5.16) are given in detail by C. Kittel, Solid state physics 22, 1 (196R), Section 6. 14 Plasmons, Polaritons, and Polarom 407 Figure 10a Comparison of screened and unscreened coulomb potentials of a static unit positive charge. The screening length Ilk, is set equal to unity. The static screened interaction is taken in the Thomas-Fermi ap- proximation, which holds for low wavevectors q < k , ; More complete calculations with all wavevectors in- cluded exhibit spatial oscillations, called Friedel oscillations, in 2kFr and are plotted in QTS, p. 114. Screened ~otent~al a 2 energy potential nscreened : 4 . 2 PI -0.3 Pseudopotential Component U(0). In the legend to Fig. 9.22b we stated a result that is important in pseudopotential theory: "For very small k the poten- tial approaches -$ times the Fermi energy." The result, which is known as the screened ion Iimit for metals, can be derived from Eq. (39). When converted to the potential energy of an electron of charge e in a metal of valency z with no ions per unit volume, the potential energy component at k = 0 becomes U(0) = -eznocp(0) = -4mnoe2/k,2 . The result (34) for kt in this situation reads whence U(0) = - ; E ~ . (43) Mott Metal-Insulator Transition A crystal composed of one hydrogen atom per primitive cell should always be a metal, according to the independent-electron model, because there will al- ways be a half-filled energy band within which charge transport can take place. A crystal with one hydrogen molecule per primitive cell is a different matter, because the two electrons can fill a band. Under extreme high pressure, as in the planet Jupiter, it is possible that hydrogen occurs in a metallic form. But let us imagine a lattice of hydrogen atoms at absolute zero: will this be a metal or an insulator? The answer depends on the lattice constant, with small values of a giving a metal and large values giving an insulator. Mott made an early estimate of the critical value a, of the lattice constant that separates the metallic state from the insulating state: a, = 4.5a0, where a, = ti2/me2 is the ra- dius of the first Bohr orbit of a hydrogen atom. On one approach to the problem, we start in the metallic state where a conduction electron sees a screened coulomb interaction from each proton: Figure lob Sclnilog plot of observed "zero tempera- ture" condiictivity m(D) versus donor concentration n for phosphorous donors in silicon. (After T F . Roscnbamii, ct al.) where k,2 = 3.939nhi3/ao, as in (34), where no is thc electron concentration. At high concentrations k, is large and the potential has no hound statc, so that we must have a metal. The potential is known to have a bound state when k, is smaller than 1.19/ao. With a bound state possible the electrons may condense about the protons to form an insulator. The inequality rnay be written in terms of no as With n, = l/a3 for a simple cnhic lattice, wc may have an insulator when a, > 2.78a0, which is not far from the Mott result 4.5ao found in a diffcrcnt way. The terrn metul-insulator transition has come to denote situations where the electrical conductivity of a rriaterial changes from metal to insulator as a function of some external parameter, which may be composition, pressure, strain, or magnetic field. The metallic phase may usually be pictured in terms of an independent-clcctron model; the insulator phase may suggest important electron-electron interactions. Sites randomly occupied introduce new and in- teresting aspects to the problem, aspects that lie within percolation theory. The percolation transition is beyond the scope of our book. When a semiconductor is doped with increasing concentrations of donor (or acceptor) atoms, a transition will occur to a conducting metallic phase. I 4 Plasmons, Pobritons, and Polarma Experimental rcsi~lts for P atoms in silicon are shown in Fig. lob. Here the i~isulator-mctal transition takes place when thc concentration is so high that the ground state wavefunctions of electrons on neighboring impurity atoms overlap significantly. The observed value of the critical concentration in the Si : P alloy system is n, = 3.74 x lo1' cm 3, as in the figure. If we take 32 X 10-%m as the radius of the ground state of a donor in Si in the spherical approximation, then by the Mott critcrion a, = 1.44 X 10-"m. The P atoms are believed to oc- cupy latticc sites at random, but if instead their lattice were simple cubic, the critical Mott concentratior~ would be appreciably less than the observed value. It is usual in the semiconductor liter- ature to refer to a heavily-doped semiconductor in the metallic range as a degenerate semiconductor. Screening and Phonons in Metals An interesting applicatio~l of our two limiting forms of the dielectric func- tion is to longitudirral acoustic phonons in metals. For lorigitudinal modes the total dielectric function, ions ph19 electrons, must be zero, by (17). Provided the sound velocity is less than the Fermi velocity of the eIectrons, we may use for the electrons the Thomas-Fermi dielectric function Provided also that thc ions are well-spaced axid move independently, we may use for them the plasmon E(w,O) limit with the approximate mass M. The total dielectric function, lattice plus electrons, but without the elec- tronic polarizability of the ion cores, is At low K and w wc neglect the term 1. At a zero of E(o,K) we have, with eF - ;mu$, This describes long wavelength longitudinal acoustic phonons. In thr alkali metals the result is in quitc good agreement with the observed longitudinal wave velocity. For potassium we calculate u = 1.8 X lo5 cm s-'; the observed longitudinal sound velocity at 4 K in the [loo] direc- tion is 2.2 X 1@ crrl s-'. There is another zero of E(w,K) for positive ions imbedded in an electron sea. For high frequencies we ~ise the dielectric contribiltion -w:/02 of the electron gas: and this function has a zero when This is the electron plasma frequency (20), but with the reduced mass correc- tion for the motion of the positive ions. POLARITONS Longitudinal optical phonons and transverse optical phonons were dis- cussed in Chapter 4, but we deferred treatment of the interaction of trans- verse optical phonons with transverse clectromagnctic waves. When the hvo waves are at resonance the phonon-photon colipling entirely changes the char- acter of the propagation, and a forbidden band is established for reasons that have nothing to do with the periodicity of the lattice. By resonance we mean a conditior~ in which the frequencies and wavevectors of both waves are approximately equal. The region of the crossover of the two dashed curves in Fig. 11 is the resonance region; the two dashed curves are the dispersion relations for photons and transverse optical phonons in the absence of any coupling bet-wccn thcm. In reality however, there always is coupling implicit - ~ in Maxwell's eqiiations and expressed by the dielectric function. The quantum of the coupled phonon-photon transverse wave field is called a polariton. In this section we see how the coupling is responsible for the dispersion relations shown as solid curves in the figure. All takes place at very low values of the wavevector in comparison with a zone boundary, because at crossover w(photon) = ck(photon) = o(phonon) = 1013 s-'; thus k = 300 cm-'. An early warning: although the symbol o, will necessarily arise in the the- ory, the effects do not conccrn longitudinal optical phonons. Longitudinal phonons do not coiiple to transverse photons in the bulk of a crystal. The coupling of the electric field E of the photon with the dielectric polar- ization P of the TO phonon is described by the electro~rlagnetic wave equation: (CGS) C'PE = W~(E + ~ V P ) . (53) At low wavevectors the TO phonon frequency w, is independent of K. The po- larization is proportional to the displacement of the positive ions relative to the negative ions, so that the equation of motion of the polarization is like that of an oscillator and may be written as, with P = Nqu, 14 Plusmorn, Polaritom, and Polarona 0 0.1 0 .
2 0.3 0.4 0.5 0 .
6 0.7 0.8 hcK, in eV Figure 1 1 A plot of the obsemed energies and wavevectors of the polantons and of the LO phonons in Gap The theoretical dispersion curves are shorn by the solid lines. The dispersion curves for the uncoupled phonons and photons are bllown by the short, dashed lines. (After C. H. Henyand J. J. Hopfield.) where there are N ion pairs of effective charge q and reduced mass M, per unit volnme. For simplicity we neglect the electronic contribution to the polarization. The equations (53) and (54) have a soliltion when This gives the polariton dispersion relation, similar to that plotted in Figs. 11 and 12. At K = 0 there are two roots, o = 0 for the photon and for the polariton. Here o , is the TO phonon frequency in the absence of coupling with photons. The dielectric function obtained from (54) is: If there is an optical electronic contribution to the polarization from the ion cores, this should be included. In the frequency range from zero up through the infrared, we write 3 - ---------____ Phononlike I I Figure 12 Couplcd lnodcs of plioto~~s and transverse optical phonon in an ionic crystal. The fine horizontal line represents oscillators of frequen o, in the absence of coupling to the electromag- netic field, and the fine line labeled w = cK/ F e(m) corresponds to clcctromag~~etic waves in the crystal, hut uncoupled to the lattice oscillators w,. The heavy lines are the dispersion relations in the presence of conpling hetween the lattice oscillators and the electromagnetic wave. One effect of the coupling is to create the frequency gap between w , and w,: within this gap the wavevector is pure imaginary of magnitude given by the hrnken line in the fignre. In the gap the wave attenuates as expi-IKlx), and we see from the plot that the attenuation is much stronger near w, than ncar w,. The character of the branches varies with K; tlrcre is a regiun of mixed electric-mechanical as- pects near the nominal crossover. Note, finally, it is intuitively obvious that the group velocity of light in the medium is always <c, because the slope a d a K for the actual dispersion relations (hemy lines) is everywhere lcss thau thc slope c fur the u~~cuupled photon in free space. in accord with the definition of ~ ( m ) as the optical dielectric constant, ob- tained as thc square of the optical refractive index. M 7 e set w = 0 to ohtain the static dielectric function: ~ ( 0 ) = e(m) + ~ T ~ N ~ ~ / M W ; , (59) which is combined with (58) to obtain ~ ( w ) in terms of accessible parameters: The zero of ~ ( w ) defines the longitudinal optical phonon frcqucncy wL, as the pole of e(w) defines w,. The zero gives ~ ( m ) w ; = E(O)O+ . (61) 14 Plasmons, Polaritons, and Polarons 413 (a) Figure 13a Plot of ~ ( w ) from (60) for ~ ( m ) = 2 and ~ ( 0 ) = 3. The dielectric constant is negative between w = wT and w, = (3/2)"2wT; that is, between the pole (infinity) of E(W) and the zero of ~ ( w ) . Incident electromagnetic waves of frequencies in the shaded regions w , < w < w, will not propagate in the medium, but will be reflected at the boundary. 28 24 g 20 " 16 X 3 12 8 0 4 - i a F4 -12 -16 -20 10' lo2 lo3 lo4 lo5 lo6 lo7 lo8 lo9 Frequency in Hz (h) Figure 13b Dielectric function (real art) of SrF, measured over a wide frequency range, exhibiting the decrease of the ionic polarizability at high frequencies. (A. von Hippel.) TO phonon LO phonon Figure 14 Relative displacements of the positive and negative ions at one instant of time for a wave in an optical mode traveling along the z axis. The planes of nodes (zero displacement) are shown; for long wavelength phonons the nodal planes are separated by many planes of atoms. In the transverse optical phonon mode the particle displacement is perpendicular to the wavevector K; the macroscopic electric field in an infinite medium will lie only in the ? x direction for the mode shown, and by the symmetry of the problem dE,/ax = 0. It follows that div E = 0 for a TO phonon. In the longitudinal optical phonon mode the particle displacements and hence the dielec- tric polarization P are parallel to the wavevector. The macroscopic electric field E satisfies D = E + 4 f l = 0 in CGS or eOE + P = 0 in SI: by symmetry E and P are parallel to the z axis, and aE Jaz # 0. Thus div E # 0 for an LO phonon, and ~ ( w ) div E is zero only if ~ ( o ) = 0. Waves do not propagate in the frequency region for which e(w) is nega- tive, between its pole at w = w , and its zero at w = o , , as in Fig. 13. For nega- tive E , waves do not propagate because then K is imaginary for real w, and exp(i&) + exp(- IKlx), damped in space. The zero of e(w), by our earlier argu- ment, is the LO frequency at low K, Fig. 14. Just as with the plasma frequency wp, the frequency w , has two meanings, one as the LO frequency at low K and the other as the upper cutoff frequency of the forbidden band for propagation of an electromagnetic wave. The value of w, is identical at both frequencies. LST Relation We write (61) as where ~ ( 0 ) is the static dielectric constant and ~ ( m ) is the high-frequency limit of the dielectric function, defined to include the core electron contribution. This result is the Lyddane-Sachs-Teller relation. The derivation assumed a cubic crystal with two atoms per primitive cell. For soft modes with WT+ 0 we see that ~ ( 0 ) + 00, a characteristic of ferroelectricity. Undamped electromagnetic waves with frequencies within the gap cannot propagate in a thick crystal. The reflectivity of a crystal surface is expected to be high in this frequency region, as in Fig. 15. 14 Planmons, Pobritons, and Polarons 415 I - - - - I 30 40 60 80 100 Photon wavelength in lo4 cm Figure 15 Keflectance of a crystal of NaCl at reveral temperatures, versus wavelength. The nom- inal values of w , and w, at room temperature correspond tn wavelengths uf 38 and 61 X lo-' cm, respectively. (After A. Mitsuishi et al.) Wave number in clllil Wavelength in lo4 cm Figure 16 Reflectance versus wavelength of a LiF film hacked by silver, for radiation incident near 30". The longitudinal optical ~ h o n o n absorbs strongly the radiation polarized (p) in the plane ~ ~ u r ~ n a l to the Cilm, hut absorbs hardly at all the radiation polarized (s) parallel to the film. (After D. W Rerreman.) For films of thickness less than a wavelength the situation i s changed. Be- cause for frequencics in the gap the wave attenuates a s exp(- IKlx), it is possi- ble for the radiation to be transmitted through a film for the small values of IKI near o , , but for the large values of IKI near o , the wave will be reflected. By reflection at nonnormal incidence the frequency w, of longitudinal optical phonons can be observed, as in Fig. 16. Experimental values of e(O), ~ ( m ) , and or are given in Table 3, with values of o, calculated using thc LST relation, Eq. (62). We compare values of Table 3 Lattice parameters, chiefly at 300 K Static Optical dielectric dielectric constant cu~~starit WT, in 10'5-' o,,, in 1013 sf' Crystal 4m) experimental LST relation LiH 12.9 3.6 11. 21. LiF 8.9 1.9 5.8 12. LiCl 12.0 2.7 3.6 7.5 LiBr 13.2 3.2 3.0 6.1 NaF 3.1 1.7 4.5 7.8 NaCl 5.9 2.25 3.1 5.0 NaBr 6.4 2.6 2.5 3.9 KF 5.5 1.5 3.6 6.1 KC1 4.85 2.1 2.7 4.0 KI 5.1 2.7 1.9 2.6 RbF 6.5 1.9 2.9 5.4 RbI 5.5 2.6 1.4 1.9 CsCl 7.2 2.6 1.9 3.1 CsI 5.65 3.0 1.2 1.6 TlCl 31.9 5.1 1.2 3.0 TlBr 29.8 5.4 0.81 1.9 AgCl 12.3 4.0 1.9 3.4 AgBr 13.1 4.6 1.5 2.5 Mg" 9.8 2.95 7.5 14. Gap 10.7 8.5 6.9 7.6 GaAs 13.13 10.9 5.1 5.5 GaSb 15.69 14.4 4.3 4.fi InP 12.37 9.6 5.7 6.5 InAs 14.55 12.3 4.1 4.5 InSb 17.88 15.6 3.5 3.7 Sic 9.6 6.7 14.9 17.9 C 5.5 5.5 25.1 25.1 Si 11.7 11.7 9.9 9.9 Ge 15.8 15.8 5.7 5.7 14 Plasmons, Pohritons, and Polarons w,/w, obtained by inelastic neutron scattering with experimental values of [E(o)/E(~)]'" obtained by dielectric measurements: NaI KRr GaAs wL/wT 1.44 ? 0.05 1.39 t- 0.02 1.07 ? 0.02 [E(O)/E(~)]"~ 1.45 ? 0.03 1.38 t- 0.03 1.08 The agreement with the LST relation is excellent ELECTRON-ELECTRON INTERACTION Fermi Liquid Because of the interaction of the conduction electrons with each other through their electrostatic interaction, the electrons suffer collisions. Further, a moving electron canses an inertial reaction in the surrounding electron gas, thereby increasing the effective mass of the electron. The effects of electron- electron interactions are usually described within the framework of the Landau theory of a Fermi liquid. The object of the theory is to give a unified account of the effect of interactions. A Fermi gas is a system of noninteracting fermions; the same system with interactions is a Fermi liquid. Landau's theory gives a good account of the low-lying single particle exci- tations of the system of interacting electrons. These single particle excitations are called quasiparticles; they have a one-to-one correspondence with the single particle excitations of the free-electron gas. A quasiparticle may be thought of as a single particle accompanied by a distortion cloud in the elec- tron gas. One effect of the coulomb interactions bctween electrons is to change the effective mass of the electron; in the alkali metals the increase is roughly of the order of 25 perccnt. Electron-Electron Collisions. It is an astonishing property of metals that conduction electrons, although crowded together only 2 A apart, travel long distances between collisions with each other. The mean free paths for electron-electron collisions are longer than lo4 A at room temperature and longer than 10 cm at 1 K. Two factors are responsible for these long mean free paths, without which the free-electron model of metals woi~ld have little value. The most powerful factor is the exclusion principle (Fig. 17), and the second factor is the screen- ing of the coulomb interaction between two electrons. We show how the exclusion pinciple reduces the collision frequency of an electron that has a low excitation energy E , outside a filled Fermi sphere (Fig. 18). We estimate the effect of the exclusion ~rinciple on the two-body collision 1 + 2 + 3 + 4 between an electron in the excited orbital 1 and an electron in the filled orbital 2 in the Fermi sea. It is convenient to refer all energies to the Fermi level p taken as the zero of energy; thus E , will be Figure 17 A collision between two electrons of wavevectors k , and k , . After the collision the particles have wavevectors k , and k , . The Pauli exclusion principle allows collisions only to final states k , , k , which were vacant before the collision. (4 Figure 18 In (a) the electrons in initial orbitals 1 and 2 collide. If the orbitals 3 and 4 are initially vacant, the electrons 1 and 2 can occupy orbitals 3 and 4 after the collision. Energy and momentum are conserved. In (b) the electrons in initial orbitals 1 and 2 have no vacant final orbitals available that allow energy to be conserved in the collision. Orbitals such as 3 and 4 would conserve energy and momentum, but they are already filled with other electrons. In (c) we have denoted with X the wavevector of the center of mass of 1 and 2. All pairs of orbitals 3 and 4 conserve momentum and energy if they lie at opposite ends of a diameter of the small sphere. The small sphere was drawn from the center of mass to pass through 1 and 2. But not all pairs of points 3,4 are allowed by the exclusion principle, for both 3,4 must lie outside the Fermi sphere; the fraction allowed is =E,/E~ 14 Plasmons, Polaritons, and Polarons positive and ep will bc negative. Because of the exclusion principle the orbitals 3 and 4 of thc electrons after collision must lie outside the Fermi sphere, all orbitals within the sphere being already occupied; thus both cncrgics e,, e, must be positive referred to zero on the Fermi spherc. The consewatiorl of energy requires that lc,l < el, for otherwise E, + eq = el + e2 could not be positive. This mcans that collisions are possible only if the orbital 2 lies within a shell of thickness el within the Fermi surface, as in Fig. 18a. Thus the fraction =el/eF of the electrons in filled urbitals provides a suit- able target for electron 1. But even if the target electron 2 is in the suitable energy shell, only a small fraction of the final orbitals compatible with conser- vation of energy and momentum are allowed by thc cxclilsion principle. This gives a second factor of e,/tF. In Fig. 18c we show a small sphcrc on which all pairs of orbitals 3, 4 at opposite ends of a diameter satisfy the conservation laws, but collisions can occur only if both orbitals 3, 4 lie outside the Fermi sea. The product or the two fractions is ( E ~ / E ~ ) ! If el corresponds to 1 K and tF to 5 X lo%, wc have ( E ~ / E ~ ) ~ = 4 X 10-lo, the factor by wludi the exclusion principle reduces the collision rate. The argument is not changed for a thermal distribiltion of electrons at a low temperature such that kgT < eF S V c replace el by the thermal energy =k,T, and now the rate at which electron-electron collisions take place is re- duced below the classical vahle by (kBT/e,)', SO that the effective collisio~i cross section u is u = (kBT/cF)2uo , (63) where uo is the cross section for the electron-electron interaction. The interaction of one electron with another has a range of the order of the screening length llk, as in (34). Numerjcal calcillations give the effective cross section with screening for collisions between electrons as of the order of 10-'" cm' or 10 A' in typical metals. The effect of the electron-gas background in electron-electron collisions is to reduce on below the value expected from the Rutherford scattering equation for the unscreened coulomb potential. However, much the greater reduction i 1 1 the cross section is caused by the Pauli factor (~,T/E,)~. At room temperature in a typical mctal kBT/eF is --lo-', so that u - l ~ - ~ u ~ -10-l9 cm2. The mean free path for electron-electron collisions is e = l/na - cm at room temperature. This is longer than the mean free path due to electron-phonon collisions by at least a factor of 10, so that at room temperature collisions with phonons are likely to be dominant. At liquid helium temperatures a contribution proportional to T2 has been found in the resistiblty of a number of metals, consistent with the form of the electron- electron scattering cross section (63). The mean free path of electrons in in- dium at 2 K is of the ordcr of 30 cm, as expected from (63). Thus the Pauli principle explains one of the central questions of thr theory of metals: how do the electrons travel long distances without colliding with each other? ELECTRON-PHONON INTERACTION: POLARONS The most common effect of the electron-phonon interaction is seen in the temperature dependence of the electrical resistivity, which for pure copper is 1.55 microhm-cm at 0°C: and 2.28 microhm-cm at 100°C. The electrons are scattered by the phonons, and the higher the temperature, the more phonons there are and hence more scattering. Above the Debye temperature the num- ber of thermal phonons is roughly proportional to the absolute temperature, and we find that the resistivity increases approximately as the absolute tem- perature in any reasonably pure metal in this temperature region. A more subtle effect of the electron-phonon intcraction is the apparent in- crease in electron mass that occurs because the electron drags the heavy ion cores along with it. In an insulator the combination of the electron and its strain field is known as a polaron, Fig. 19. The effect is large in ionic crystals because of the strong coulomb interaction between ions and electrons. I 1 1 co- valent crystals the effect is weak because neutral atoms have only a weak inter- action with electrons. Thc strength of thc clcctron-latticc intcraction is mcasured by the dimen- sionless coupling constant CY given hy 1 deformation energy za = 5% (64) where uL is the longiti~dinal optical phonon frequency near zero wavevector. We view ;a as "the number of phonons which surround a slow-moving elec- tron in a crystal." Values of a deduced frorri diverse experi~nerlts and theory are given in Table 4, after F. C. Brown. The values of a are high in ionic crystals and low in covalent crystals. The values of the effective mass miol of the polaron are fiom cyclotron resonance experiments. The values given for the band cffcctive mass rn werc calculatcd from mi,,. Thc last row in the tahle gives the factor m',,,lrn' by which the band mass is increased by the deformation of the lattice. Theory relates the effective mass of the polaron miul to the effective band mass rn of the electron in the undefornled lattice by t11e relation for CY < 1 this is approximately m(l + ;a). Because the coupling constant a is always positive, the polaron mass is greater than the bare mass, as we expect from the inertia of the ions. 14 Plosmons, Polaritons, and Polarons Figure 19 The formation of a polaron. (a) A conduction electron is shown in a rigid lattice of an ionic crystal, KCI. The forces on the ions adjacent to the electron are shown. (b) The electron is shown in an elastic or deformable lattice. The electron plus the associated strain field is called a polaron. The displacement of the ions increases the effective inertia and, hence, the effective mass of the electron; in KC1 the mass is increased by a factor of 2.5 with respect to the band theory mass in a rigid lattice. In extreme situations, often with holes, the particle can become self- trapped (localized) in the lattice. In covalent crystals the forces on the atoms from the electron are weaker than in ionic crystals, so that polaron deformations are small in covalent crystals. It is common to speak of large and small polarons. The electron associated with a large polaron moves in a band, but the mass is slightly enhanced; these are the polarons we have discussed above. The electron associated with a small polaron spends most of its time trapped on a single ion. At high temperatures the electron moves from site to site by thermally activated hopping; at low temperatures the electron tunnels slowly through the crystal, as if in a band of large effective mass. Table 4 Polaron coupling constants a, masses mio1, and band masses m for electrons in the conduction band Crvstal KC1 KRr AeCl AaBr ZIIO PbS InSb G d s Holes or electrons can become self-trapped by inducing an asymmetric local deformation of the lattice. This is most likely to occur when the band edge is degenerate and the crystal is polar (such as an alkali halide or silver halide), with strong coupling of the particle to the lattice. The valence band edge is more often degencratc than the conduction band edge, so that holes are more likely to be self-trapped than are electrons. Holes appear to be self- trapped in all the alkali and silver halides. Ionic solids at room temperature gerlerally have very low conductivities for the motion of ions through the crystal, less than (ohm-cm)-', but a family of compounds has been reported with conductivities of 0.2 (ohm-cm)-' at 20°C. The compounds have the composition MAGI,, where M denotes K, Rb, or NH,. The Ag' ions occupy only a fraction of the equivalent lattice sites available, and the ionic condnctivity proceeds by the hopping of a silver ion from one sitc to a nearby vacant site. The crystal structures also have parallel open channels. PEIERLS INSTABILITY OF LINEAR METALS Consider a one-dimcnsional metal with an electron gas filling all conduc- tion band orbitals out to the wavevector k , .
, at absolute zero of temperature. Peierls suggested that such a linear metal is unstable with respect to a static lattice deformation of wavevector G = 2kF. Such a deformation crcatcs an en- ergy gap at the Fermi surface, thereby lowering the encrgy of electrons below the energy gap, Fig. 20. The deformation proceeds until limited by the in- crease of elastic energy; the equilibrium deformation A is given by the root of Consider the elastic strain A cos 2kpx. The spatial-average elastic energy per unit length is Eei,,,, = $~A~(cos~2k,x) = $ 2 ~ " where C is the force constant of the linear metal. We next calculate Eel,,,,,,,,. Suppose that the ion 14 Plosrnons, Polaritons, and Polamns 423 Energy gap introduce Free electron energy C !
? 8 W after introduction Figure 20 Peierls instability. Electrons with wave- 0 vectors near the Fermi surface have their energy Wavevector lowered by a lattice deformation. contribution to the lattice potential seen by a conduction electron is propor- tional to the deformation: U(x) = 2AA cos 2kFx. From (7.51) we have eK = (fi2/2m)(kg + P) ? [4(fi2k22m)(fi2P/2m) + A2A2]ln . (67) It is convenient to defme x , = fi2@/m ; x, = fi2kg/m ; x - fi2~kF/m We retain the - sign in (67) and form whence, with dKlr as the number of orbitals per unit length, We put it all together. The equilibrium deformation is the root of $A - (2A2mA/d2kF) sinh-l(fi2k2mAA) = 0 . The root A that corresponds to the minimum energy is given by fi2k%lmAA = sinh(-fi2kFrC/4mA2) , whence IA IA = (2fi2k;/m) eq(-)i2kFrC/4mA2) , if the argument of the sinh in (68) is >> 1. We assume k, G k , , . The result is of the form of the energy gap equation in the BCS theory of superconductivity, Cl~apter 10. The deformation A is a collective effect of all the electrons. If we set JY = h2kj/2m = conduction hand width; N(0) = 2 m l ~ h ~ k , = density of orbitals at Fermi level; V = 2 A 2 !
C = effective electron- electron interaction energy, then we can write (69) as I A IA - 4W exp-l/N(O)V] , (70) which is analogous to the BCS energy gap equation. An example of a Peierls insulator is TaS,. SUMMARY (In CGS units) The dielectric function may he defined as in terms of the applied and induced charge density components at w,K. The plasma frequency Gp = [4mne?e(~)rn]"~ is the frequency of the uniform collective longitudinal oscillation of the electron gas against a background of fixed positive ions. It is also the low frequency cutoff for propagation of transverse electromagnetic waves in the plasma. The poles of the dielectric fur~ction define w~ and the zeroes define wL. In a plasma the coulomb interaction is screened; it becomes (qlr) exp(-k,r), where the screening length llk, = ( ~ F J 6 . r m ~ ~ ) ' ~ . A metal-insulator transition may occur when the nearest-neighbor scpara- tion a becomes of the order of 4ao, where no is the radius of the first Bohr orhit in the insulator. The metallic phasc cxists at smaller values of a. A polariton is a quantum of the coupled TO phonon-photon fields. The cou- pling is assured by the Maxwell equations. The spectral region w,- < w < w , is forbidden to electromagnetic wave propagation. The Lyddane-Sachs-Teller relation is w i !
w 2 , = e(O)/~(m). Problems 1. Surface plasmons. Consider a semi-infinite plasma on the positive side of the plane z = 0. A solution of Laplace's equation V " = 0 in the plasma is qi(x,z) = A cos kr KkZ, whence E, = kA cos kr e-k"; EXj = kA sin k w c-". (a)Shoa. that in the vac~~nrn qo(x,z) = A cos kx e " for z < 0 satisfies the boundary condition that the tangential component of E be continuous at the boundary; that is, find Ex0. (b) Note that D, = e(w)Ei; Do = E,. Show that the boundary condition that the 14 Plasmons, Polaritons, and Polaronn norn~al component of D be continuous at the hn~mdaly requires that c(w) = 1 , whence from (10) we have the Stern-Ferrell rewlt: for the frequency w , of a surface plasma oscillation 2. Interface plasmons. We consider the plane interface z = 0 between metal 1 at z > 0 and metal 2 at s < 0. Metal 1 has bl~lk plasmon frequency up,; metal 2 has op,. The dielectric constants in both ~netals are those or frcc-electron gases. Show that surface plasmons associated mith the interface havc thc frequency 3. Alfve'n waves. Consider a solid with an cqual concentration n of electrons of mass m, and holes of mass m,,. This situation may arisc in a semimetal or in a compen- sated semiconductor. Place the solid in a lrniform magnetic field B = Be. Intro- duce the coordinate 5 = x + i y appropriate for circularly polarized motion, with 6 having time dependence a-I"'. Let we = rBlm,c and wh = eB/mhc. (a) In CGS units, show that [ < : = eEe/m,wiw + w,); [,, = -rE+/rnhw(o - wh) are the displacements of the electrons and holes in the electric field E- ePot = (E, + iEy) e~'Ot. (b) Show that the dielectric polarization Pi = r~ejt,, - &) in thc regime w & w,, o h may be written as P+ = nc2(ml, + m,)E+/BZ, and the dielectric function c(w) = EL + 4vPiP+/E+ = E, + 4nc2p/B2, where c, is the dielectric constant of thc host lat- tice and p = n(m, + ml,) is the mass density of the carriers. If cl may bc ncglected, the dispersion relation wec(w) = c2K%ecomes, fur electroniagnetic waves propa- gating in the z direction, w2 = ( B ~ / ~ T ~ ) K ~ . Such waves are krro\z,n m Alfvkn waves; thcy propagate with the constant velocity B/(4.rrp)"'.. If B = 10 kG; n = 10" c111-~; m = g, the velocity is -10' cm s-'. Alfien waves have been observed in seini- metals and in electron-hole drops in germanium (Chapter 15). 4. Helicon wanes. (a) Employ the method of Problem 3 to treat a specimen with only onc carrier type, say holes in concentration p, and in t l ~ e limit o & oh = sBln~~c. Show that c(w) = -ITe2/nzhww,,, where Di(o) = t(w)E+(w). The term E, in c has hccn neglected. (b) Show further that the dispersion relation hecomes o = ( B C / ~ ~ I ~ ) K ' , the helicon dispersion relation; in CGS. For K = 1 cn-I and B = 1000 ( : , estimatc the helicon frequency in sodium metal. (The frequency is nega- tive; with circular-polarized modes the sign of the frequency refers to the sense o C the rotation.) 5. Plasmon mode o f a sphere. The frequency of the uniform plasmon mode of a sphere is dctcrmincd by the depolarization field E = -47rP/3 of a sphere, where the polarization P= -nrr, with r as the average displacement of the electrons of concentration n. S h o ~ from F = n u that the resonance frequency of the electron gas is w i = 4meV3m. Bccausc all electrons participate in the oscillation, such an excitatior~ is called a collective excitation or collective mode of the electron gas. 6. M~gnstoplanmafrequenc~. Use the method of Problem 5 to find the frequency of the urrifnrln plasulon ulode of a sphere placed in a constant uniform rnagnetic field B. Let B he along the z axis. The solution should go to the cyclotron frequency w,. = eB/mc in one limit and to wo = (4me2/3m)u' in another limit. Take the motion in the xy plane. 7. Photon branch at low wavevector. (a) Find what (56) becomes when E(W) is taken into account. (b) Show that there is a solution of (55) which at low wavevector is w = C W ~ , as expected for a photon in a crystal of refractive index n2 = E . 8. Plasma frequency and electrical conductivity. An organic conductor has been found by optical studies to have wp - 1.80 X 10'"~' for the plasma frequency, and T = 2.83 X 10-15 s for the electron relaxation time at room temperature. (a) Calculate the electrical conducti\lty from these data. The carrier mass is not knowu and is not needed here. Take ~ ( m ) = 1. Convert the result to units (n cm)-'. (b) From the crystal and chemical structure, the conduction electron couccntration is 4.7 X lo2' ~ m - ~ . Calculate the electron effective mass m'. 9. Bulk modulus of the Femi gas. Show that the contribution of the kinetic en- ergy to the bulk modulus of the electron gas at absolt~tr zero is B = inmt$. It is convenient to use (6.60). We can use our result for B to find the vclocity of sound. which in a compressible fluid is u = (B!p)l/Z, where u = (ni/3M)l"cF, in agreement with (46). These estimates neglect attrachve interactions. 10. Response of electron gas. It is sometimes stated error~eously in books on electro- magnetism that the static conductivity a, which in gaussian units has the dimensions of a frequency, measures the response frequency of a metal to an electric ficld sud- denly applied. Criticize this statement as it might apply to copper at room tempcra- ture. The resistivity is -1pohm-cm; the electron concentration is 8 X 10" 2 6 ' ; the mean free path is -400 A; the Fermi velocityis 1.6 X lo8 crn s-'. You will not neccs- sarily need all these data. Give the order of magnitude of the three frequencies u, w,, and 1 1 7 that might bc relevant in the problem. Set up and solve the prohlem of the response xjt) of the system to an electric field E(t < 0) = 0; E(t > 0) = 1. The system is a sheet of mpper; the field is applied normal to the sheet. Inclnde the damping. Solve the differential cquation by elementary methods. '11. Gap plasmons and the van der Waals interaction. Consider two sen~i-infinite media with plane surfaces z = 0, d. The dielectric function of the identical rriedia is ~ ( w ) . Show that for surface plasmons symmetrical with respect to the gap the frequency must satisfy ~ ( w ) = -tanh (Kd!2), where K2 = k: + k;. The electric po- tential will have the form p -,f(z) exp(ik& + ikYy - iwt) Look for nonretarded so1utio11-that is, solutions of the Laplace equation rather than of the wave equation. The sum of the zero-point energy of all gap modes is the nonretarded part of the van der Waals attraction between the two specimens-see N. G. van Karrlperr, R. R. A. Nijboer, and K. Schram, Physics Letters 26A, 307 (1968). h his prohlerr~ is so~ncwhat difficult. Optical Processes and Excitons OPTICAL REFLECTANCE 429 Kramers-Kronig relations 430 Mathematical note 432 Example: conductivity of collisionless electron gas 433 Electronic interband transitions 434 EXCITONS 435 Frenkel excitons 437 Alkali halides 440 Molecular crystals 440 Weakly bound (Mott-Wannier) excitons 441 Exciton condensation into electron-hole drops (EHD) 441 RAMAN EFFECT IN CRYSTALS Electron spectroscopy with x-rays ENERGY LOSS OF FAST PARTICLES IN A SOLID 448 SUMMARY 449 PROBLEMS 450 1. Causality and the response function 450 2. Dissipation sum rule 450 3. Reflection at normal incidence 450 4. Conductivity sum rule and superconductivity 450 5. Dielectric constant and the semiconductor energy gap 451 6. Hagen-Rubens relation for infrared reflectivity of metals 451 7. Davydov splitting of exciton lines 452 w', k ' Raman scattering (generic term): Brillouin scattering when acoustic phonon is involved; polariton scattering when optical phonon is involved. w = w ' f C l + for phonon emission (Stokes process) k = k ' K sials { _ for phonon absorption (anti-Stokes) Two phonon creation. Electron spectroscopy with x-rays (XPS): incident x-ray photon ejects valence or core electron from solid. Figure 1 There are many types of experiments in which light interacts with wavelike excitations in a crystal. Scvcral absorptio~i proccsscs arc illustrated Irere. CHAPTER 15: OPTICAL PROCESSES AND EXCITONS The dielectric function r(o,K) was introduced in the preceding chapter to describe the response of a crystal to an electromagnetic field (Fig. 1). The di- electric function depends sensitively on the electronic hand structure of a crystal, and studies of the dielectric function by optical spectroscopy are very useful in the determination of the ovcrall hand structure of a crystal. Indeed, optical spectroscopy has developed into the most important experimental tool for band structure determination. In the infrared, visible, and ultraviolet spectral regions the wavevcctor of the radiation is very small compared wit11 the shortest reciprocal lattice vector, and therefore it may usually be taken as zero. We are conccmed then with the real r' and imaginary e" parts of the dielectric fi~nction at zero wavevector; E(W) = el(w) + ~E"(w), also written as q(w) + ir,(w). However, the dielectric function is not directly accessible experimentally from optical measurcmcnts: the dlrectly accessible functions are the reflect- ance R(w), thc rcfractive index n(w), and the extinction coefficient K(w). Our first objective is to relate the experimentally observable quantities to the real and imaginary parts of the dielectric function. OPTICAL REFLECTANCE The optical measurements that give the fullest information on the elec- tronic system are mcasurcments of the reflectivity of light at norrrial incidence on single crystals. The reflectivity coefficient r(w) is a complex function de- fined at the crystal surface as the ratio of the reflected electric field E(rcf1) to the incident electric field E(inc): where we have separated the an~~litude p(w) and phase O(w) components of the reflectivity coefficient. The refractive index n(o) and the extinction coefficient K(o) in the crystal are related to the reflectivity at normal incidence by as derived in Problern 3 by elementary consideration of the continuity of the components of E and B parallel to the crystal surface. By definition, n(o) and K(w) are related to the dielectric functior~ r(w) by - n(w) + iK(w) = N(w) , (3) where N(w) is the complex refractive index. Do not confuse K(w) as used here with the symbol for a wavevector. If the incident traveling wave has the wavevector k, thcn the y component of a wave traveling in the x direction is The transmitted wave in the ~riedium is attenuated because, by the dispersion relation for electromagnetic waves, the wavevector in the medium is related to the incident k in vacuum by (n + iK)k: Ey(trans) exp ([i[(n + iK)kx - w t ] ) = exp(-Kkx) exp[i(nkx - wt)] . (5) One quantity measured in experiments is the reflectance R, defined as the ratio of the reflected intensity to the incident intensity: R = E(refl)E(refl)/Ea(inc)E(inc) = rir = P2 . (6) It is difficult to measure the phase O(w) of the reflected wave, but we show below that it can be calculated from the measured reflectance R(o) if this is known at all frequencies. Once we know hoth R(o) and O(w), we can proceed by (2) to obtain n(w) and K(w). We use these in (3) to obtain ~ ( w ) = E'(o) + iel'(w), where ~ ' ( w ) and ~ " ( w ) are the real and imagnary parts of the &electric fnnction. The inversion of (3) gives E'(w) = nL & ? ; ;"(w) = 2nK . (7) We now show how to find the phase O ( o ) as an integral over the re- flectance R(w); by a similar method we relate the real and imaginary parts of the dielectric function. In this way we can find everything from the experi- mental R(w). Kramers-Kronig Relations The Kramers-Kronig relations enable us to find the real part of the re- sponse of a linear passive system if we know the imaginary part of the response at all frequencies, and vice versa. They are central to the analysis of optical experiments on solids. The response of any linear passive system can be represented as the su- perposition of the responses of a collection of damped harmonic oscillators with masses M , . Let the response function a(w) = a'(w) + iuu(w) of the col- lection of oscillators be defined by where the applied force field is the real part of F, exp(-iwt) and the total dis- placement x = x , is the real part of x , exp(-iwt). From the equation of motion, I 15 Optical Processes and Excitons 431 we have the coniplex response function of the oscillator system: where the constaritsJ; = 1/M, and relaxation frcqllencies p, are all positive for a passive systern. If a(w) is the dielectric polarizahility of atoms in concentration n, then f has the form of an oscillator strength times ne'vm; s u d ~ a dielectric response function is said to be of the Kramers-Heisenberg form. The relations wc de- velop also apply to the electrical conductivity u(w) in Ohm's law, j, = u(o)E,. \C7e need not assume the specific form (9), but wc make ilse of three prop- erties of the response function viewed as a f~inction of the complex variable o . Any function wit11 the following propcrtics will satisfy the Kramers-Kronig relations (11): (a) The poles of a(w) are all helow the rcal axis. (b) The integral of a(w)/w vanishes when taken around an infinite semi- circle in the upper half of the complcx w-plane. It suffices that a(w) -+ 0 uni- for~nly as Iwl -+ m. (c) The function a'(@) is even and a"(@) is odd with respect to real w. Consider the Cauchy integral in thc form where P denotes the principal part of the integral, as discussed in the mathe- matical note that follows. The right-hand side is to he colnpleted by an integral over the semicircle at infinity in the upper half-plane, but urc have seen in (b) that this integral vanishes. We equate the real parts of (10) to ohtain In the last integral \VP s~ihstitute s for -p and use property (c) that a"(-s) = -aU(s); this integral then becomes and we have, with 1 1 - 2 s --+-------- s - w s + w s Z w 2 ' the result This is one of the Kramers-Kronig relations. The other relation follows on equating the imaginary parts of Eq. (10): 1 " a"" ' + = -kP fff(s) - J ~ E U ~ ~ ] a"(o) = -- p --- I - 2 - w s - w , s + o whence These relations are applied below to the analysis of optical reflectance data; this is their most important application. Let us apply the Kramers-Kronig relations to r(w) klewed as a response func- tion between the incident and reflected waves in (1) and (6). We apply (11) to to obtain the phase in terms of the reflectance: We integrate by parts to obtain a form that gives insight into the contribu- tions to the phase angle: Spectral regions in which the reflectance is constant do not contribute to the integral; further, spectral regions s 9 w and s w do not contribute milch be- cause the furictior~ 1 1 1 I(s + w)/(s - w)l is small in these regions. Mathematical Note. To obtain the Cauclly integral (10) we take the inte- gral Ja(s)(s - w)-'rls over the contour in Fig. 2. The function a ( s ) is analytic in the tipper half-plane, so that thc value of the integral is zero. The contribution of segment (4) to the integral vanishes if the integrand a(s)ls + 0 is faster than 1st-I as I s 1 + m. For the response function (9) the integrand + 0 as Is13; and for the conductivity a ( s ) the integrand + 0 as ~ S I - ~ . The segment (2) con- tributes, in the liriiit as u + 0, IS Optical Procexsea and Ezcitons 433 to the integral, where s = w + u el0. The segments (1) and (3) are by definition the principal part of the integral between -w and m. Because the integral over (1) + (2) + (3) + (4) must vanish, EXAMPLE: Conductivity of Collisionless Electron Gas. Consider a gas of free electrons in the lirr~it as the collision frequency goes to zero. From (9) the response function is, with f = llrrb, by the Dirac identity. We confir111 that the delta function in (16) satisfies the Kramers- Kronig relation (lla), by which in agreement ~r~ith (16). M 7 e obtain the electrical conductivity u(w) from the dielectric function where a ( ~ ) = xJ(-e)E, is thc response function. We use the equivalence for the Maxwell equation can be written either as c curl H = 4nu(o)E - iwE or as c curl H = -iue(o)E. We combine (16), (18), anrl (19) to find the conductivity of a collisionless electron gas: For collisionless electrons the real part of the conductivity has a delta function at o = 0. Electronic Interband Transitions It came as a sl~rprisc that optical spectroscopy developed as an important experimental tool for the determination of hand stmctl~re. First, the absorp- tion and reflection bands of crystals are broad and apparently featureless func- tions of the photon energy when this is greater than the band gap. Second, direct interband absorption of a photon fiw will occur at all points in the Bril- louin zone for which eneru is consewed: where c is an empty band and v is a filled hand. The total ahsorption at givcn o is an integral over all transitions in the zone that satisfy (21). Three factors unraveled the spectra: The broad bands are not like a spectral line greatly broadened by damping, but the bands convey much intelligence which emerges when derivatives are taken of the rcflcctance (Fig. 3); derivatives with respect to wavelength, electric field, temperature, prcssurc, or uniaxial stress, for example. The spectroscopy of derivatives is called modulation spectroscopy. I / f i o . in e \ ' Figure 3 Comparison of (a) reflectance, (h) wavelength derivative reflectance (first derivative), and (c) electroreflectance (third derivative), of the spectral region in germanium between 3.0 and 3.6 eV. iACLcr data by D. D. Scll, E. 0. Ka~ie, and D. E. .lspr~es.) 15 Optical Processes and Exeitons 435 The relation (21) does not exclude spectral structure in a crystal, because transitions accumulate at frequencies for which the bands c, v are parallel- that is, at frequencies where At these critical points in k space the joint density of states D,(E, + ho)D,(~,) is singular, according to the same argument we used in (5.37) to show that the density of phonon modes D(o) is singular when V k o is zero. The pseudopotential method for calculating energy bands helps identify the positions in the Brillouin zone of the critical points found in modulation spectra. Band-band energy differences can be calculated with an accuracy as good as 0.1 eV. The experimental results can then be fed back to give im- provements in the pseudopotential calculations. EXCITONS Reflectance and absorption spectra often show structure for photon energies just below the energy gap, where we might expect the crystal to be transparent. This structure is caused by the absorption of a photon with the creation of a bound electron-hole pair. An electron and a hole may be bound together by their attractive coulomb interaction, just as an electron is bound to a proton to form a neutral hydrogen atom. The bound electron-hole pair is called an exciton, Fig. 4. An exciton can move through the crystal and transport energy; it does not transport charge Figure 4a An exciton is a bound electron-hole pair, ' usually free to move together through the crystal. In . . . . some respects it is similar to an atom of positronium, formed from a positron and an electron. The exciton . . . shown is a Mott-Wannier exciton: it is weakly bound, with an average electron-hole distance large in com- e . parison with the lattice constant. Figure 4b A tightly-bound or Frenkel exciton shown local- ized on one atom in an alkali halide crystal. An ideal Frenkel exciton will travel as a wave throughout the crystal, but the electron is always close to the hole. because it is electrically neutral. It is similar to positronillm, which is formed from an electron and a positron. Excitoris can be forrned in every insulating crystal. When the band gap is indirect, excitons near the direct gap may be unstable with respect to decay into a free electron and free hole. All excitons are unstabIe with respect to the illtimate recombination process in which the electron drops into the hole. Ex- citons can also form complexes, snch as a hiexciton from two excitons. We have seen that a free electron and free hole are created whenever a photon of energy greater than the energy gap is absorbed in a crystal. The threshold for this process is fcw > Eg in a direct process. In the indirect phonon-assisted process of Chapter 8 the threshold is lower by the phonon en- ergy h0. But in the formation of excitons the energy is lowered with respect to these thresholds by the binding energy of the exciton, which may be in the range 1 meV to 1 eV, as in Table 1. Excitons can be formed by photon absorption at any critical point (22), for if Vke, = Vkec the group velocities of electron and hole are equal and the parti- cles may be bound by their coulomb attraction. Transitions leading to the formation of excitons below the energy gap are indicated in Figs. 5 and 6. The binding cncrgy of the exciton can be measured in three ways: In optical transitions from the valcncc band, by the difference between the energy required to create an exciton and the energy to create a free electron and free hole, Fig. 7. In recombination luminescence, by comparison of the energy of the free electron-hole recombination line with the energy of the exciton recombina- tion line. By photo-ionization of excito~ls, to form free carriers. This experinrent re- quires a high concentration of excitons. \Ye discuss excitons in two different limiting approximations, one by Frenkel in which the exciton is small and tightly hoiind, and the other by Mott and LYannier in which the exciton is weakly bound, wlth an electron-hole separation large in comparison with a lattice constant. Intermediate examples are known. Table 1 Binding energy of excitons, in meV Si 14.7 BaO 56. RbCl 440. Ge 4.15 InP 4.0 LiF (1000) GaAs 4.2 InSb (0.4) AgBr 20. Gap 3.5 KT 480. Ag Cl 30. CdS 29. KC1 400. TIC1 11. CdSe 15. KBr 400. TlBr 6. Data asscmblcd by Frcdcrick C. Brown and Arnold Schmidt. 15 Optical Processes and Excitons 437 Figure 5 Exciton levels in relation to the conduction band edge, for a simple band structure with both conduction and valence band edges at k = 0. An exciton can have translational kinetic en- ergy. Excitons are unstable with respect to radiative recombination in which the electron drops into the hole in the valence band, accompanied by the emission of a photon or phonons. Figure 6 Energy levels of an exciton created in a direct process. Optical transitions from the top of the valence band are shown by the arrows: the longest arrow corresponds to the energy gap. The binding energy of the exciton is E,, referred to a free electron and fiee hole. The lowest fre- quency absorption line of the crystal at absolute zero is not E,, but is Eg - E , . Frenkel Excitons In a tightly bound exciton (Fig. 4b) the excitation is localized on or near a single atom: the hole is usually on the same atom as the electron although the pair may be anywhere in the crystal. A Frenkel exciton is essentially an excited state of a single atom, but the excitation can hop from one atom to another by virtue of the coupling between neighbors. The excitation wave travels through the crystal much as the reversed spin of a magnon travels through the crystal. The crystalline inert gases have excitons which in their ground states cor- respond somewhat to the Frenkel model. Atomic krypton has its lowest strong ~ E x c i t o n absorption 4 I . . . . . . I I I oL I I I 1 I I I I 1.50 1.51 1.52 1.53 1.54 1.55 1.56 Photon energy in eV Figure 7 Effect of an exciton level on the optical ahsnrptinn of a semiconductor for photons of energy near the band gap Ep in gallium arsenide at 21 K. The vertical scale is the intensity absorp- tion coefficient a , as in I(r) = I, exp(-ax). Thc cnergy gap and exciton binding energy are dcduced fro111 the shape of the ahsorptinn cnrvo: the gap E, is 1.521 eV and the exciton binding energy i s 0.0034 eV (After M. D. Sturge,) atomic transition at 9.99 eV. Thc corresponding transition in the crystal is closely equal and is at 10.17 e\7, Fig. 8. The encrgygap in the crystal is 11.7 eV, so the exciton ground state energy is 11.7 - 10.17 = 1.5 eV, referred to a free electron and free hole separated and at rest in the crystal. The trar~slational states of Frenkel excitons have the form of propagating waves, like all other excitations i ~ i a structure. Consider a crystal of N atoms on a line or ring. If uj is the ground state of atomj, the ground state of the crystal is if interactions between the atoms arc ncglected. If a single atom j is in an excited state u,, the system is described by This function has the same energy as the function cpi with any other atom I ex- cited. However, the functions cp that describe a single excited atom and N - 1 atoms in their ground state are not the stationary quantum states of the prohlem. If there is any interaction between an excited atom and a nearby atom in its ground state, the excitation energy will be passed from atom to atom. The eigenstates will have a wavelike form, as we now show. 15 Optical Processes and Excitons Figure 8 Absorption spectrum of solid krypton at 20 K. (Atter G. Balhnl.) \he11 the ha~niltonian of the systcm operates on the function (oJ with the jtf~ atom excited, we obtain where E is the free atom excitation energy-; the interaction T measures the rate of transfer of the excitation from j to its nearest neighbors, j - 1 and j + 1. The solutions 01 (25) are waves of the Bloch form: $k = 2 exp(qka) q j To see this we let X operate on qhk: from (25). We rearrange the right-hand sidc to ohtain XJI, = 2 eqk"[c + ~(e"' + e-'b)]q = ( E + 2T cos ka)$k , J (28) so that the energy cigenvalues of the problem are Ek=e+2Tcoskn , Energy I 0 Wavevector k + Figure 9 Energy versus wavevector for a Frenkel exciton, calculated with pos~t~vc nearest- neighbor tra~lsfcr illteractio~l ' I : as in Fig. 9. The application of periodic boundary conditions determines the allowed values of the wavevector k: Alkali Halides. In alkali halide crystals the lowest-enera excitons are local- ized on the negative halogen ions, as in Fig. 4b. The negative ions have lower electronic excitation levels than do the positive ions. Pure alkali halide crystals are transparent in the visible spectral region, which means that the exciton en- ergies do not lie in the visible, but the crystals show considerable excitonic absorption structure in the vacuum ultraviolet. A doublet structure is particiilarly e.ident in sodium bromide, a structure similar to that of the lowest excited state of the krypton atom-which is iso- electronic with the Br- ion of KBr. The splitting is caused by the spin-orbit in- teraction. These excitons are Frenkel excitons. Molecular Crystals. In lnolecular crystals the covalent binding within a molecule is strong in comparison with the van der Waals binding between mol- ecules, so that thc excitons are Frenkel excitons. Electronic excitation lines of an inchvidual molecule appear in the crystalline solid as an cxciton, oftcn with little shift in frequency. At low temperatures the lines in the solid are qnite sharp, although there may be more structure to the lines in the solid than in the molecule because of the Davydov splitting, as discussed in Problem 7. 15 Optical Processes and Ezcitons 441 Weakly Bound (Mott-Wannier) Excitons Consider an electron in the conduction band and a hole in the valence band. The electron and hole attract each other by the coulomb potential (CGS) U(r) = -e%r , (31) where r is the distance between the particles and E is the appropriate dielec- tric constant. (The lattice polarization contribution to the dielectric constant should not be included if the frequency of motion of the exciton is higher than the optical phonon frcqucncies.) There will be bound states of the exciton sys- tem having total energies lower than the bottom of the conduction band. The problem is the hydrogen atom problem if the energy surfaces for the electron and hole are spherical and nondegenerate. The energy levels referred to the top of the valence band are given by a modified Rydberg equation (CGS) Here n is the principal quantu~n number and p is the reduced mass: formed from the effective Inasses m,, inh of the electron and hole. The excitor~ ground state energy is obtained on setting n = 1 in (32); this is the ionization energy of the exciton. Studies of the optical absorption lines in cuprous oxide, Cu20, at low tcmpcratures give results for the exciton level spacing in good agrcement with the Rydberg equation (32) except for transi- tions to the state n = 1. An empirical fit to the lines of Fig. 10 is obtained with the relation v(cm-') = 17,508 - (8001n". Taking c = 10, we find p -0.7 m from the coefficient of lln2. The constant term 17,508 cm-' corresponds to an energy gap E,, = 2.17 eV. Ezciton Condensation into Electron-Hole Drops (EHD) A condensed phase of an electron-hole plasma forrris in Ge and Si when maintained at a low temperature and irradiated by light. The following sequence of events takes place when an electron-hole drop (EHD) is formed in Ge: The absorption of a photon of energy hw > Eg produces a free electron and free hole, with hug11 efficiency These combine rapidly, pcrhaps in 1 ns, to form an exciton. The exciton inay decay with annihilation of the e-h pair with a lifetime of 8 ps. If the exciton conccntration is s~ifficiently high-over 1013 cm-3 at 2 K- most of thc cxcitons will condense into a drop. The drop lifetime is 40 ps, but in strained Ge may be as long as 600 ps. \'ithi11 the drop the excitons dissolve into a degenerate Fermi gas of electrons and holes, with metallic properties: this state was predicted by L. V. Keldysh. Figure 10 Logarithm of the optical transmission versus photon energy in cuprous oxide at 77 K, showing a series o T exciton lincs. Notc that 0 1 1 tl~e vertical axis the logarithm is plotted decreasing upward; thus a peak cnrresponds to absorption. The hand gap Ep is 2.17 el'. (After l ' . W. Uaumeister.) FE EHD I I I I I Figure 11 Recombination radiation of free electrons uith holes and of electron-hole drops in Ge at 3.04 K. The Fermi energy in the drop is E, and the cohesive energy of the drop with respect to a frcc cxciton is 9,. (After T. K. Lo.) Figure 11 sliows the reconibination radiation in Ge from free excitons (714 meV) and from the EHD phase (709 meV). The of the 714 meV line is accounted for by Doppler broadening, and the width of the 709 meV line is compatible with the kinetic cncrgy distribution of electrons and holes in a Fermi gas of concentration 2 x 10" Figure 12 is a photograph of a large EHD. 15 Optical Processes and Ezcitons 443 Figure 12 Photograph of an electron-hole drop in a 4 mm disk of pure germanium. The drop is the intense spot adjacent to the set screw on the left of the disk. The photograph is the image of the drop obtained by focusing its electron-hole recombination luminescence onto the surface of an infrared-sensitive camera. (After J. P. Wolfe et al.) '4verage concentration, all phases, p;~irs/cm3 Figure 13 Phase diagram for photoexcited electrons and holes in unstressed silicon. The diagram shows, for example, that with an average concentration near 1017 cm-3 at 15 K, a free-exciton gas with saturated-gas concentration of 1 0 1 6 cm-3 coexists with a (variable) volume of liquid droplets, each with a density of 3 X 10'' cm-! The liquid critical temperature is about 23 K. Theoretical and experimental values for the metal-insulator transition for excitons are also shown. (From J. P. Wolfe.) Table 2 Electron-hole liquid parameters Crystal Binding energy relative Concentration, Critical (unstressed) to free exciton, in meV n orp, in cm-3 temperature in K Conrtesy of D. Bimberg. The exciton phase dlagram for silicon is plotted in the temperature- concentration plane in Fig. 13. The exciton gas is insulating at low pressures. At high pressures (at the right of the diagram) the exciton gas breaks up into a conducting plasma of unpaired electrons and holes. The transition from exci- tons to the plasma is an example of the Mott transition, Chapter 14. Further data arc givcn in Tahle 2. RAMAN EFFECT IN CRYSTALS Raman scattering involves two photons-one in, one out-and is one step more complex tlran the one hoto on processes treated earlier in this chapter. In the Haman effect a is scattered inelastically by a crystal, with creation or annihilation of a phonon or magnon (Fig. 14). The process is identical to the inelastic scattering of x-rays and it is similar to the inelastic scattering of neutrons by a crystal. The selection rules for the first-order Raman effect are where w, k refer to the inciderit photon; w', k' refer to the scattered photon; and a, K refer to the phonon created or destroyed in the scattering event. In the second-order Raman effect, hvo phonons are involved in the inelastic scattering of the photon. The Raman effect is made possible by the strain-dependence of the elec- tronic polarizahility. To show this, we suppose that the ~olarizability cr associ- ated with a phonon mode may he written as a power series in the phonon amplitude u: If u(t) = u, cos Ot and the incident clcctric field is E ( t ) = E, cos wt, then the induced elcctric dipole moment has a component aiEouo cos wt cos Ot = ~a,~,u~[cos(w + O)t + cos(w - O)t] . (36) 16 Optical Processerr and Ercitons 445 Stokes Figure 14 Raman scattering of a photon with emission or absorp- tion of a phonon. Similar processes occur with Inagnons (spin waves). Thus photons at freq~iencies o + 0 and o - 0 can be emitted, accompanied by absorption or emission of a phonon of frequency 0. Thc photon at o - 0 is called the Stokes line and that at w + 0 is the anti-Stokes line. The intensity of the Stokes line involves the matrix element for phonon creation, which is just the matrix element for the harmonic oscilla- tor, as in Appendix C: where nK is the initial l~opi~lation of phonon niode K. The anti-Stokes line involves phonon annihilation, with a photon intensity proportional to ~ ( o + a) a J(nK - 1 Ju InK) l2 YLK (38) If the phonon population is initially in thermal equilibrium at temperature T, the intensity ratio of the two lines is with ( 7 1 ~ ) given by the Planck distribution function ll[exp(fiO/kBT) - I]. We see that the relativc intensity of the anti-Stokes lines vanishes as T + 0, be- cause here therc are no thermal phonorls available to be annihilated. Observations on the K = 0 optical phonon in silicon are shown in Figs. 15 and 16. Silicon has two identical atoms in the primitive cell, and there is no electric dipole momerlt associated with the primitive cell in the absence of deformation by phonons. But afl~ does not vanish for silicon at K = 0, so that we can observe the mode by first-order Raman scattering of light. The second-ordcr Raman effect arises fro~n the term a2u2 in the polariz- ability. Inelastic scattering of light in this order is accompanied by the creation or two phonons, or the absorption of two phonons, or the creation of one and tbc absorption of another phonon. The phonons may have different frequen- cies. The intensity distribution in thc scattered photon spectrum may be quite complicated if there are sevcral atoms in the primitive cell because of the Raman shift, cm-' Figure 15 First-order Raman spectra of the K = 0 optical mode of a silicon crystal observed at three temperatures. The incident photon has a wavelength of 5145 A. The optical phonon frequency is equal to the frequency shift; it depends slightly on the temper- ature. (After T. R. Hart, R. L. Aggarwal, and B. Lax.) 1 L F 200 400 600 800 Temperature, in K Figure 16 Intensity ratio of anti-Stokes to Stokes lines as a function of temperature, for the obser- vations of Fig. 15 on the optical m ~ d e of silicon. The observed temperature dependence is in good agreement with the prediction of Eq. (39): the solid curve is a plot of the function exp(-Afl/kBT). P .- B + H 4 0 0 -700 -600 500 4 0 0 3 0 0 -200 -100 0 Raman shift, cm" Figure 17 Raman spectrum of Gap at 20 K. The two highest peaks are the first-order Raman lines associated with the excitation of an LO phonon at 404 cm-I and a TO phonon at 366 cm-'. All the other peaks involve two phonons. (After M. V . Hohden and J. P. Russell.) 15 Optical Procennes and Excitons 447 corresponding number of optical phonon modes. Second-ordcr Raman spectra have been observed and analyzed in numerous crystals. Measurements on gallinm phosphide (Gap) are shown in Fig. 17. Electron Spectroscopy with X-Rays The next degree of complexity in optical processes involves a photon in and an electron out of the solid, as in Fig. 1. Two important techniques are x-ray photoemission from solids (XPS) and ultraviolet photoemission (UPS). In solid state physics they are used in band structure studies and surface physics, including catalysis and adsorption. XPS and UPS spectra can be compared directly with valence band densi- ties of states D(E). The specimen is irradiated with highly monochromatic x-rays or ultraviolet photons. The photon is absorbed, with the emission of a photoelectron whose kinetic energy is equal to the energy of the incident pho- ton minus the binding energy of the electron in dle solid. The electrons come from a thin layer near the surface, typically 50 A in depth. The resolution of thc best XPS spectrometer systems is less than 10 me\', which permits refined studies of band structure. The valence band structure of silver is shown by Fig. 18, with the zero of energy set at the Fermi lel~el. Electrons in the first 3 eV below the Fermi level come from the 5s conduction band. The strong peak with structure below 3 eV is from the 4d valencc electrons. Excitations are also seen from deeper levels, often accompanied by excita- tion of plasmons. For example, in silicon the 2p electron with a binding energy close to 99.2 eV is observed in replica at 117 eV with single plasmon excitation and at 134.7 eV with two plas~non excitation. The plasmon energy is 18 eV. 9 5 0 Figure 18 Valence-band elcctron emission from Binding energy, eV silver, after S~egbahn and co-workers ENERGY LOSS OF FAST PARTICLES IN A SOLID So far we have nsed photons as probes of the electronic structure of solids. We can also use electron beams for the same pilrpose. Thc rcsults also involve the dielectric function, now through the imaginary part of 1/e(w). The dielcc- tric function enters as Im{~(w)] into the energy loss by an electromagnetic wave in a solid, but as -In~{l/e(w)] into the energy loss by a charged particle that penetrates a solid. Consider this diffcrence. The general result from electromagnetic theory for the power dissipation density by diclcctric losses is per unit volume. With a transverse electromagnetic wave Ee-'"' in the crystal, we have dD/dt = -iw ~(w)Ee-"', whence the time-average power is 1 1 = - WE~((E"COS wt - e'sin wt)cos ot) = - we"(w)E2 , 471 87r (41) proportional to e"(w). The tangential component of E is continuous across the boundary of the solid. If a particle of charge e and velocity 1 ; enters a crystal, the dielectric dis- place~nerit is, by standard texts, because by the Poisson equation it is D, and not E, that is related to the free charge. In an isotropic medium the Fourier component E(w,k) is related to the Fourier conlponent D(w,k) of D(r,t) by E(w,k) = D(w,k)/e(w,k). The time-average power dissipatiorl associated with this Fourier compo- nent is -- - 47r I w D2(w,k)([(t)' cos wt + (t) sin wt],-sin wll) , whence 15 Optical Procenses and Excitons 449 Figure 19 t"(o) for Cu and Au; the bold lines are from energy loss r~~easurcrncnts by J. Daniels, and the other lines were daulated from optical measurements by D. Beaglehole, and L. R. Cm~firld ct d. The result is the motivation for the introduction of the energy loss function -I~n(l/c(w,k)} and it i s also a motivation for experinlents on energy losscs hy fast electrons in thin films. If thc dielectric function is indepelldellt of k : the power loss is where fiko is the maximum possible lllolnentum transfcr from the primary par- ticle to an electron of the crystal. Figure 19 shows the excellent experimental agreement between values of E"(w) deduced from optical reflectivit. measure- ments with values deduced from electron energy loss measure~nents. SUMMARY The Kramers-Kronig relations connect the real and imagnay parts of a responsc fi~nction: The complex refractive index N(w) = n(o) + iK(w), where n is the refrac- tive index and K is the extinction coefficient; further, ~ ( w ) = N2(w), whence ef(u) = 7 1 ' - f ? and ~ " ( 0 ) = 2nK. The reflectance at normal incidcncc is The energy loss function -In~{l/e(w)) gives the energy loss by a charged particlc moving in a solid. Problems 1. Causality and the response function. The Kramers-Kronig relations are consis- tent with the principle that an effect not precede its cause. Consider a delta- function force applied at time t = 0: whence F, = 112.x. (a) Show by direct irrtegration or by use of the KK relations that the oscillator response fur~ctiorr gives zero displacernent, r(t) = 0, for t < 0 under the above force. For t < 0 the contour integral rnay be completed by a semicircle in the upper half-plane. (h) Evahrate x(t) for t > 0. Note that a(w) has poles at - $p2)"2 - iiP, both in the lower half-plane. 2. Dissipation sum rule. By comparison of cr'(w) from (9) and from (114 in the limit w + m , shou~ that the following sum rule for the oscillator strengths must hold: 3. Rejlection at normul incidence. Consider an electromagnetic wave in vacuum, with field components of the form Let the wave be incident upon a medium of dielectric constant E and permeability r = 1 that fills the half-space x > 0. Show that the reflecthltp coefficient r(w) as defined by E(refl) - r(o)E(inc) is given by where n + i K = E"', with n and K rcal. Show further that the reflectance is '4. Conductioity sum rule and superconductioity. We write the electrical con- ductivity as u(w) = u'(w) + iu"(w), where a', u" are real. (a) Show by a Kramers- Kronig relation that 'This problem is somewhat difficult. 15 Optical Processes and Excitons 451 This result is used m the theory of ~~~pcrconductivity. If at very high frequencies (such as x-ray frequencies) u " ( w ) is identical for the superconducting and normal states, then we must have But at frequencies 0 < w < w, within the supcrconducting energy gap the real part of the conductivity of a superconductor vanishes, so that in this region the integral on the left-hand side is lower by = rrbwr. Thcrc must be an additional contribution to a j to balance this deficiency. (b) Show that if ul(w < wg) < uh(w < mg), as is ob- served experimentally, then c:(w) car1 lrave a dclta function contribution at w = 0, and from the delta function there is a contribution uC(o) = u b o$w. The delta fnnction corresponds to infinite conductivity at zero frequency. ( c ) By elementary consideration of the classical motion of corrdl~ction clcctrons at very high frequen- cies, show that (CGS) I , : uf(w) tiw = me2/zm , a rrs~llt round by Ferrell and Glover. 5. Dielectric constant and the semiconductor energy gap. Thc effect on ~ " ( w ) of an energy gap wg in a semiconductor may be approximated very roughly by sobstitnting $S(w - wg) for B(w) in the response function (Ifi); that is, we take E"(w) = (2me2/mw)nS(w - wg). This is crude because it puts all the absorption at tlir gall freqnency. The factor 1W enters as soon as we move the delta function away fro111 the origin, bccause the integral in the sum rule of Problem 2 starts at the origin. Show that thc rcal part of the dielectric constant on this irrodel is It follows that the ~tatic dielectric constant is ~ ' ( 0 ) = 1 + wk/w;, widcly used as a rule of tliurrrh. 6. Hagen-Rubenn relation for infrared reflectivity of metols. The complex refrac- tive index n + i K of a rnctal for w~ < 1 is given by where ug is the conductivity for static fields. .Vc assume here that intraband currents are dominant; interband transitions are neglected. Using the result of Problem 3 for the reflection coefficient at norn~al incidence, show that ~~rovirled that uo S w. This is the Hagen-Rubens relation. For sodium at room temperai~irc, uo = 2.1 X 10" sC1 in CGS and T = 3.1 X 10-14 s, as deduced from T = uOrn/neZ. Radiation of 10 pm has w = 1.88 X 1014 s-', so that the Hagen- Rubens result should apply: R = 0.976. The result calculated from experimental val- ues of n and K is 0.987. Hint: If uo w , then n2 = Kt. This simplifies the algebra. '7. Daoydoo splitting of exciton lines. The Frenkel exciton band of Fig. 9 is doubled when there are two atoms A, B in a primitive cell. Extend the theory of Eqs. (25) to (29) to a linear crystal AB.AB.AB.AB. with transfer integrals T, between AB and T, between B.A. Find an equation for the two bands as functions of the wavevector. The splitting between the bands at k = 0 is called the Daydov splitting. 'This problem is somewhat difficult. Dielectrics and Ferroelectrics Maxwell equations Polarization MACROSCOPIC ELECTRIC FIELD Depolarization field, E, LOCAL ELECTRIC: FIELD AT A N ATOM Lorentz field, E, Field of dipoles inside cavity, E, DIELECTRIC CONSTANT AND POLARIZABILITY Electronic polarizability Classical theory Examples STRUCTURAL PHASE TRANSITIONS FERROELECTRIC CRYSTALS Classification of ferroelectric crystals DISPLACIVE TRANSITIONS Soft optical phonons Landau theory of the phase transition Second-order transition First-order transition Antiferroelectricity Ferroelectric domains Piezoelectricity SUMMARY PROBLEMS 1. Polarizability of atomic hydrogen 2. Polarizability of conducting sphere 3. Effect of air gap 4. Interfacial polarization 5. Polarization of sphere 6. Ferroelectric criterion for atoms 7. Saturation polarization at Curie point 8. Dielectric constant below transition temperature 9. Soft modes and lattice transformations 10. Ferroelectric linear array NOTATION: E, = 10~/47rc~ ; (CGS) D = E + 4n-P = EE = ( 1 + 4 v ) E ; = 1.9 x 10-18 esu-cm Figure 1 The permanent dipole moment of a molecule of water has the magnitude 1.9 X lo-'' esu-cm and is directed from the 0'- ion toward the midpoint of the line connecting the H+ ions. (To convert to SI units, multiply p by f X lo".) Figure 2 Electrostatic potential and field components in CGS at position r, 0 for a dipole p directed along the z axis. For 0 = 0, we have E, = Ey = 0 and E, = 2p/r3; for 0 = ~ / 2 we have Ex = Ey = 0 and E, = TO convert to SI, replace p by pl47rq,. (After E. M . Purcell.) First we relate the applied electric field to the internal electric field in a dielectric crystal. The study of the electric field within dielectric matter arises when we ask: What is the relation in the material between the dielectric polarization P and the macroscopic electric field E in the Maxwell equations? What is the relation between the dielectric polarization and the local electric field which acts at the site of an atom in the lattice? The local field determines the dipole moment of the atom. Maxwell Equations Polarization The polarization P is defined as the dipole moment per unit volume, averaged over the volume of a cell. The total dipole moment is defined as where r, is the position vector of the charge q,. The value of the sum will be independent of the origin chosen for the position vectors, provided that the system is neutral: Let ri = r, + R; then p = Xq,rL = RXq, + Zq,r, = Zqnr,. The dipole moment of a water molecule is shown in Fig. 1. The electric field at a point r from a dipole moment p is given by a stan- dard result of elementary electrostatics: 3 ( p . r)r - pp (CGS) E(r) = 15 The lines of force of a dipole pointing along the z axis are shown in Fig. 2 MACROSCOPIC ELECTRIC FIELD One contribution to the electric field inside a body is that of the applied electric field, defined as where e(r) is the microscopic electric field at the point r. The field E is a much smoother quantity than the microscopic field e. .Ve could well have E, = field produced by fxed charges external to the body . written the dipole field (2) a. e(r) because it is a microscopic urisrnoothed field. \Ve call E the macroscopic electric field. It is adequate for all problems in the electrodylla~nics of crystals provided that we know thc connection be- tween E, the polarization P, and the current density j, and provided that thc wavelengths of interest are long in comparison with the lattice spacing.' To find the contribution of the polarization to the macroscopic field, we (3) can simplify the sum over all the dipoles in the specimen. By a famous theo- rem of electrostaticsZ the macroscopic electric field caused by a uniform polar- The other contribution to the electric field is the sum of the fields of all charges that constitute the body. If the body is neutral, the contribution to the average field may be expressed in terms of the sum of the fields of atomic dipoles. We define the average electric field E(r,) as the average field over the volume of the crystal cell that contains the lattice point r,: ization is equal to the electric field in vacuum of a fictitious surface charge 'A detailed derivation o C thc Maxwell equations fur the macroscopic fields E and B, starting from the Maxlvell equations in terms of the microscopic fields e and h, is given by E. M. Purcell, Elsrtrlcit!y and magnetism, 2nd ed., McGrawHill, 1985. 'The electrostatic potential in CGS units of a dipole p is ~ ( r ) = p . grad(1ir). For a volume distribution of polarization P we have which by a vcctor identity becomes If P is constant, then div P = 0 and by the Gaurs theorem we have where udS is an element of charge on the surface of the body. This completes the proof. 16 Dielectrics and Ferroelectrics 457 Figure 3 (a) A uniformly polarized dielectric slab, with the polarization vector P normal to the plane of the slab. (b) A pair of uniformly charged parallel plates which give rise to the identical electric field E, as in (a) The upper plate has the surface charge density u = +P, and the lower plate has u = -P. density u = fi . P on the surface of the body. Here fi is the unit normal to the surface, drawn outward from the polarized matter. We apply the result to a thin dielectric slab (Fig. 3a) with a uniform vol- ume polarization P. The electric field El(r) produced by the polarization is equal to the field poduced by the fictitious surface charge density u = fi . P on the surface of the slab. On the upper boundary the unit vector i i is directed upward and on the lower boundary fi is directed downward. The upper bound- ary bears the fictitious charge u = fi . P = P per unit area, and the lower boundary bears -P per unit area. The electric field El due to these charges has a simple form at any point between the plates, but comfortably removed from their edges. By Gauss's law (CGS) E, = - 4 ~ 1 ~ 1 = - 4 ~ p ; ( 4 4 We add E, to the applied field E, to obtain the total macroscopic field inside the slab, with i the unit vector normal to the plane of the slab: We define E, = field of the surface charge denisty fi . P on the boundary . This field is smoothly varying in space inside and outside the body and satisfies the Maxwell equations as written for the macroscopic field E. The reason El is a smooth function when viewed on an atomic scale is that we have replaced the discrete lattice of dipoles pi with the smoothed polarization P. Depolariaation Field, El If the polarization is uniform within the body, the only contributions to the macroscopic field are from E , and El: Here Eo is the applied field and El is the field due to the uniform polarization. The field E, is called the depolarization field, for within the body it tends to oppose the applied field E , as in Fig. 4. Specimens in the shape of ellipsoids, a class that includes spheres, cylinders, and discs as limiting forms, have an advantageous property: a uniform polarization produces a uniform de- polarization field inside the body. This is a famous mathematical result demon- strated in classic texts on electricity and magneti~m.~ If P , , Py, P, are the components of the polarization P referred to the principal axes of an ellipsoid, then the components of the depolarization field are written Here N,, N,, Nz are the depolarization factors; their values depend on the ratios of the principal axes of the ellipsoid. The N's are positive and satisfy the s u m r n l e N , + N y + N z = 4 ~ i n C G S , a n d N , + N y + N z = 1 i n S I . Values of N parallel to the figure axis of ellipsoids of revolution are plotted in Fig. 5. Additional cases have been calculated by Oshorn4 and by Stoner. In limiting cases N has the values: Shape Axis Sphere 4 ~ / 3 113 Thin slab normal 4 ~ 1 Thin slab in plane 0 0 Long circular cylinder longitudinal 0 0 Long circular cylinder transverse 271 1/2 We can reduce the depolarization field to zero in two ways, either by working parallel to the axis of a long fine specimen or by making an electrical connection between electrodes deposited on the opposite surfaces of a thin slab. 3R. Becker, Electromagneticfields and interactions, Blaisdell, 1964, pp. 102-107. 'J. A. Osborn, Phys. Rev. 67,351 (1945); E. C. Stoner, Philosophical Magazine 36,803 (1945). 16 Dielectrics and Ferroelectrics 459 CGS SI Figure 4 The depolarization field El is op- posite to P . The fictitious surface charges are indicated: the field of these charges is El within the ellipsoid. Figure 5 Depolarizatiou factor N parallel to the figure axis of ellip- soids of revolution, as a function of the axial ratio c/a. A uniform applied field E, will induce uniform polarization in an ellipsoid. We introduce the dielectric susceptibility x such that the relations (CGS) P = XE : connect the macroscopic field E inside the ellipsoid with the polarization P. Here ,ysl = 4?rxccs If E, is uniform and parallel to a principal axis of the ellipsoid, then (CGS) E = E , + E , = E , - N P ; by (8), whence (CGS) X P = x(E, - NP) ; P = - 1 + N ~ ~ ~ ' The value of the polarization depends on the depoIarization factor N. LOCAL ELECTRIC FIELD AT AN ATOM The value of the local electric field that acts at the site of an atom is signif- icantly different from the value of the macroscopic electric field. We can con- vince ourselves of this by consideration of the local field at a site with a cubic arrangement of neighbors5 in a crystal of spherical shape. The macroscopic electric field in a sphere is . . . . , . .
. , . - :.- . , . 1 E = E,, + El = E,, - 3E, P ' by (10). But consider the field that acts on the atom at the center of the sphere (this atom is not unrepresentative). If all dipoles are parallel to the z axis and have magnitude p, the z component of the field at the center due to all other dipoles is, from (2), In SI we replace p by p/4rreO. The x, y, z directions are equivalent because of the symmetry of the lattice and of the sphere; thus whence = 0. The correct local field is just equal to the applied field, El,,,, = E,, for an atom site with a cubic environment in a spherical specimen. Thus the local field is not the same as the macroscopic average field E . We now develop an expression for the local field at a general lattice site, not necessarily of cubic symmetry. The local field at an atom is the sum of the electric field E, from external sources and of the field from the dipoles within the specimen. It is convenient to decompose the dipole field so that part of the summation over dipoles may be replaced by integration. We write 'Atom sites in a cubic crystal do not necessarily have cubic syrnmetly. thus the 0'- sites in the barium titanate structure of Fig. 10 do not have a cubic environment. However, the Nat and C1- sites in the NaCl structure and the Cs+ and C1- sites in the CsCl structure have cubic symmetry 16 Dielectrics and Fewoeleetrics 461 L E3 from dipoles inside sphere Figure 6 The internal electric field on an atom in a clystal is the sum of the external applied field E, and of the field due to the other atoms in the crystal. The standard method of summing the di- pole fields of the other atoms is first to sum individually over a moderate number of neighboring atoms inside an imaginary sphere concentric with the reference atom: this defines the field E,, which vanishes at a reference site with cubic symmetry The atoms outside the sphere can be treated as a uniformly polarized dielectric. Their contribution to the field at the reference point is El + E,, where El is the depolarization field associated with the outer boundary and E, is the field associated with the surface of the spherical cavity. Here E, = field produced by fixed charges external to the body; E, = depolarization field, from a surface charge density i i . P on the outer surface of the specimen; E, = Lorentz cavity field: field from polarization charges on inside of a spherical cavity cut (as a mathematical fiction) out of the specimen with the reference atom as center, as in Fig. 6; El + E, is the field due to uniform po- larization of the body in which a hole has been created; E, = field of atoms inside cavity. The contribution El + E, + E, to the local field is the total field at one atom caused by the dipole moments of all the other atoms in the specimen: and in SI we replace p, by p , / 4 ~ € ~ . Dipoles at distances greater than perhaps ten lattice constants from the reference site make a smoothly varying contribution to this sum, a contribu- tion which may be replaced by two surface integrals. One surface integral is taken over the outer surface of the ellipsoidal specimen and defines E,, as in Eq. (6). The second surface integral defines E2 and may be taken over any interior surface that is a suitable distance (say 50 A) from the reference site. We count in E, any dipoles not included in the volume bounded by the inner and outer surfaces. It is convenient to let the interior surface be spherical. Figure 7 Calculation of the field in a spherical cavity in Charge on ring = a uniformly polarized medium. 2.irasinB.adB.PcosB Lorentz Field, E, The field E, due to the polarization charges on the surface of the fictitious cavity was calculated by Lorentz. If 0 is the polar angle (Fig. 7 ) referred to the polarization direction, the surface charge density on the surface of the cavity is -P cos 0 . The electric field at the center of the spherical cavity of radius a is 4%- (cGS) E, = /or(a-~)(2%-o sin @)(a ~ o ) ( P cos B)( cos 0 ) = -P ; 3 (16) This is the negative of the depolarization field E, in a polarized sphere, so that E , + E, = 0 for a sphere. Field of Dipoles Inside Cavity, E, The field E, due to the dipoles within the spherical cavity is the only term that depends on the crystal structure. We showed for a reference site with cubic surroundings in a sphere that E, = 0 if all the atoms may be replaced by point dipoles to each other. The total local field at a cubic site is, from (14) and (16), This is the Lorentz relation: the field acting at an atom in a cubic site is the macroscopic field E of Eq. (7) plus 4?rP/3 or P / ~ E , from the polarization of the other atoms in the specimen. Experimental data for cubic ionic crystals sup- port the Lorentz relation. 16 Dielectrics and Ferroekctrics 463 DIELECTRIC CONSTANT AND POLARIZABILITY The dielectric constant of an isotropic or cubic medium relative to vac- uum is defined in terms of the macroscopic field E: Remember that x,, = 43-r,yCGs, by definition, but esI = cCGS. The susceptibility (9) is related to the dielectric constant by P E - 1 (CGS) , y = ~ = - . 4 a ' In a noncubic crystal the dielectric response is described by the components of the susceptibility tensor or of the dielectric constant tensor: The polarizability a of an atom is defined in terms of the local electric field at the atom: p = aE~ocd . (21) where p is the dipole moment. This definition applies in CGS and in SI, but as = The polarizability is an atomic property, but the dielectric constant will depend on the manner in which the atoms are assembled to form a crystal. For a nonspherical atom a will be a tensor. The polarization of a crystal may be expressed approximately as the prod- uct of the polarizabilities of the atoms times the local electric field: where N, is the concentration and aj the polarizability of atomsj, and El,,(j) is the local field at atom sites j. We want to relate the dielectric constant to the polarizabilities; the result will depend on the relation that holds between the macroscopic electric field and the local electric field. We give the derivation in CGS units and state the result in both systems of units. If the local field is given by the Lorentz relation (17), then and we solve for P to find the susceptibility By definition = 1 + 4 7 r x in CGS; we may rearrange (23) to obtain (CGS) - - - € + 2 the Clausius-Mossotti relation. This relates the dielectric constant to the electronic polarizability, but only for crystal structures for which the Lorentz local field (17) obtains. Electronic Polaritability The total polarizability may usually be separated into three parts: elec- tronic, ionic, and dipolar, as in Fig. 8. The electronic contribution arises from the displacement of the electron shell relative to a nucleus. The ionic contri- bution comes from the displacement of a charged ion with respect to other ions. The dipolar polarizability arises from molecules with a permanent elec- tric dipole moment that can change orientation in an applied electric field. Total polarizability (real part) I UHF to Ultra- ImicroWaveS 1 i m r a d 1 1 violet 1 Figure 8 Frequency dependence of the several contributions to the polarizability. 16 Dielectrics and Perroelectrics In heterogeneons materials there is usually also an interfacial polarization arising from the accumulation of charge at structural interfaces. This is of little fundamental interest, but it is of considerable practical interest because com- mercial insulating materials are usually heterogeneo~is.~ The dielectric constant at optical frequencies arises almost entirely from the electronic polarizability. The dipolar and ionic contributions are small at high frequencies because of the inertia of the molecules and ions. In the opti- cal range (24) reduces to here we have used the relation n2 = e, where n is the refractive index. By applyng (25) to large numbers of crystals we determine in Table 1 em- pirical values of the electronic polarizabilities that are reasonably consistent with the observed values of the refractive index. The scheme is not entirely self-consistent, because the electronic polarizability of an ion depends somewliat on the environment in which it is placed. The negative ions are highly polarizable because they are large. Table 1 Electronic polarizabilities of atoms and ions, in cm3 Pauling 0.201 0.029 0.008 0.003 0.0013 0'- F - Ne NaC ~ g ~ + AP+ Si4+ Pauli~rg 3.88 1.04 0.390 0.179 0.094 0.052 0.0165 JS-(TKS) (2.4) 0.858 0.290 S2- CI- Ar Ki Ca2+ Se3+ Ti4+ Pauling 10.2 3.66 1.62 0.83 0.47 0.286 0.185 JS-(TKS) (5.5) 2.947 1.133 (1.1) (0.19) se2- Br- Kr Rbt S?' y 3 + Zr4+ Pauling 10.5 4.77 2.46 1.40 0.86 0.55 0.37 JS-(TKS) (7.) 4.091 1.679 (1.6) Te2- I- Xe Csf Ba2+ ~ . 3 + Ce4' Pauling 14.0 7.10 3.99 2.42 1.55 1.04 0.73 JS-(TKS) (9.) 6.116 2.743 (2.5) Values from L. Pauling, Proc. K. Soc. London A114, 181 (1927); S. S. Jaswal and T. P. Sharma, J. Phys. Chern. Solids 34, 509 (1973): and J. Tessman, A. Kahn, and \V. Shhackley, Phys. Rev 92, 890 (19.53). The TKS polarizabilities are at the frequcncy of the D lines of sodium. The valucs arc in CGS; to convert to SI, multiply by 9 X 'For rcfcrc~~ces see D. E. Aspnes, Am. J. Phys. 50, 704 (1982) Clossicol Theory of Electronic Polarizability. AII electron bound har- monically to an atom will show resonance absorption at a frequency wo = (Plm)'", where p is the force constant. The displacemcnt x of the electron occasioned by the application of a field El,, is given by so that the static electronic polarizability is The electronic polarizability will depend on frequency, and it is shown in the following example that for frequency w but in the visible region the frequency dependence (dispersion) is not usually very important in most tramparent materials. EXAMPLE: Frequency dependence. Find the frerlue~rt:y dependence of the elec- tronic polarizalrility of an electror~ having tlre resorrarlce frequency wn, treating the sys- tem as a simple harrnonic oscillator. The equation of motion in the local electric field El,, sin wt is d2x n t - + moix = -eEl,, sin wt , dt2 so that, for x = x, sin wt, m(-w2 + wi)x, = -eEl,,, The dipole mnment has the amplit~~de from which (28) follows In quantum theory the expression corresponding to (28) is (CGS) whercf;, is called the oscillator strength of the electric dipole transition be- tween the atomic states i and j. Near a transition the polarizability changes sign (Fig. 8). 16 Dielectrics and Ferroelectrics 467 STRUCTURAL PHASE TRANSITIONS It is not uncommon for crystals to transfor111 fro111 one crystal structure to another as the temperature or pressure is varied. The stablc str~~cture A at absolute zero generally has the lowest accessible internal energy of all the pos- sible structures. Even this selection of a structure A can he varied with appli- cation of pressure, because a low atomic volllme will favor closest-packed or even metallic structures. Hydrogen and xenon, for example, becor~ie rnetallic under extreme pressure. Some other structiire B may have a softer or lower frequency phonon spectrum than A. As the temperature is increased the phonons in B will be more highly excited (higher thermal average occupancies) than the phonons in A. Recanse the entropy increases with the occupancy, the entropy of B will be- come higher than the entropy of A as the temperature is increased. It is thereby possible for the stable structure to transfonn fro111 A to B as the temperature is increased. The stable structure at a temperature T is deter- n~ined by the minimum of the free energy F = U - TS. There will be a transi- tion from A to B if a temperature T, exists (below the nlelting point) such that FA(T,) = FLAT,). Often several structures have nearly the same intcrnal energy at absolute zero. The phonon dispersion relations for the structures may, however, be rather different. The phonorl energies are sensitive to the mrmber and arrangement of nearby atoms; these are the quantities that change as the structure is changed. Sorne structural phase transitions have only small effects on the macro- scopic physical properties of the material. However, if the transition is influ- enced by an applied stress, the crystal may yield nlechanically quite easily near the transition temperature because the relative proportions in the two phases will change under stress. Some other structural phase transitions may have spectacular effects on the macroscopic electrical properties. Ferroelectric transitions are a subgroup of structural phase transitions, a subgroup marked by the appearance of a spontaneous dielectric polarization in the crystal. Ferroelcctrics are of theoretical and technical interest bccause they often have nni~siiall~ high and unusually temperature-dependent values of the dielectric constant, the piezoelectric effect, the pyroclcctric effect, and clectro-optical effects, including optical frequency doubling. FERROELECTRIC CRYSTALS A ferroelectric crystal exhibits an electric dipolc moment even in the ab- sence of an external electric field. In the ferroclectric state the center of posi- tive charge of the crystal does not coincide with the center of negative charge. The plot of polarization versus electric field for the ferroelectric state sltows a hysteresis loop. A crystal in a normal dielectric state usually does not Temperature ("C) Figure 9 The temperature variation of (a) thc dielectric constant E, (b) the pyroelectlic coeffi- cient dPldT, and (c) the specific heat c,, of PhTiO,. (After Remeika and Class.) show significant hysteresis when the electric field is increased and then re- versed, both slowly. Ferroelectricity usually disappcars ahove a certain temperature called the transition temperature. Above the transition the crystal is said to be in a para- electric state. The term paraelectric suggests an analogy with paramagnetism: 16 Dielectrics and Ferroelectrics there is usually a rapid drop in the dielectric constant as the tenlperature increases. In some crystals the ferroelectric dipole nlonlent is not changed by an electric field of the maximum intensity which it is possible to apply before causing electrical breakdown. In these crystals we are often able to observe a change in thr spontaneoi~s moment when the temperature is changed (Fig. 9). Such cvstals are called pyroelectric. Lithium niobate, LiNbO,, is pyroelec- tric at room temperature. It has a high transition terr~perature (2; = 1480 K) and a high saturation polarizatio~~ (50 pC/c~n2). It can be "poled," which means giver1 a rerriarlerlt polarization, by an electric field applied over 1400 K. Classi$cation of Ferroelectric Crystals We list in Tablc 2 somc of the crystals commonly considered to be ferroelec- tric, along with the transition temperature or Curie point T, at which the crystal changes from the low-temperature ~olarized state to the h i g h -t t t l n p e r a t u r e l -t e r r l ~ e r a t u r e unpolarized state. Thermal motion tends to destroy the ferroelectric order. Some ferroelectric crystals lrave no Curie point because they melt before leaving the ferroelectric phase. The table also includes values of the spontaneous polar- ization P,?. Ferroelectric crystals may be classified into two main groups, order- disorder or displacive. One may dcfinc thc character of the transition in terms of the dynamics of the lowest frequency ("soft") optical phonon modes. If a soft mode can propa- gate in the crystal at the transition, then the transition is displacive. If the soft mode is only diffusive (non-propagating) tl~ere is really not a phonon at all, Table 2 Ferroelectric crystal6 To obtain the spontaneous polarization P, in the CGS unit of esu cm-" multiply the valuc givcn in PC. cm-2 by 3 X lo3. 'I-. in K P,. in fiC: cm ', at T K KDP type KH,PO, 123 4.75 1961 KD,PO, 213 4.83 [I801 RbH,PO, 147 5.6 [go] KH2hs0, 97 5.0 - [781 GeTe 670 TGS t p e Tri-glycinr sulfate 322 2.8 [291 Tri-glycine selenate 295 3.2 12831 Perovskites BaTiO,, 408 26.0 12961 KNbO? 708 30.0 15231 PbTiO, 765 >50 T.iTa0, 938 50 LiNbO, 1480 71 [296l but is only a large amplitude hopping motion between the wells of the order- disorder system. Many ferroelectrics have soft modes that fall between these two extremes. The order-disorder class of ferroelectrics includes crystals with hydrogen bonds in which the motion of the protons is related to the ferroelectric properties, as in potassium dihydrogen phosphate (KH,PO,) and isomorphous salts. The substitution of deuterons for protons nearly doubles T,, although the fractional change in the molecular weight of the compound is less than 2 percent: KH,P04 KD,PO, KH,AsO, KD,As04 Curie temperature 123 K 213 K 97 K 162 K This extraordinarily large isotope shift is believed to be a quantum effect in- volving the mass-dependence of the de Broglie wavelength. Neutron diffraction data show that above the Curie temperature the proton distribution along the hydrogen bond is symmetrically elongated. Below the Curie temperature the lstribution is more concentrated and asymmetric with respect to neighboring ions, so that one end of the hydrogen bond is preferred by the proton over the other end, giving a polarization. The displacive class of ferroelectrics includes ionic crystal structures closely related to the perovskite and ilmenite structures. The simplest ferro- electric crystal is GeTe with the sodium chloride structure. We shall devote ourselves primarily to crystals with the perovskite structure, Fig. 10. Consider the order of magnitude of the ferroelectric effects in barium titanate: the observed saturation polarization P, at room temperature (Fig. 11) Figure 10 (a) The crystal structure of barium titanate. The prototype crystal is calcium titanate (perovskite). The structure is cubic, with ~ a ' + ions at the cube corners, 02- ions at the face cen- ters, and a Ti4+ ion at the body center. (b) Below the Curie temperature the structure is slightly deformed, with Ba2+ and Ti4+ ions displaced relative to the 02- ions, thereby developing a dipole moment. The upper and lower oxygen ions may move downward slightly. 16 Dielectrics and Ferroelectrics Temperature ('C) Figure 11 Spontaneous polarization projected on cube edge of barium titanate, as a function of temperature. (After W J. Merz.) is 8 X lo4 esu cm-'. The volume of a cell is (4 X = 64 X cm3, SO that the dipole moment of a cell is (CGS) p (8 X lo4 esu cm-')(64 X lo-" ~ m - ~ ) 5 X lo-'' esu cm ; If the positive ions Ba2+ and Ti4+ were moved by 8 = 0.1 k with respect to the negative 0'- ions, the dipole moment of a cell would be 6e8 ^-. 3 X lo-'' esu cm. In LiNbO, the displacements are considerably larger, being 0.9 k and 0.5 k for the lithium and niobum ions respectively, giving the larger P,. DISPLACNE TRANSITIONS Two viewpoints contribute to an understanding of a ferroelectric displacive transition and by extension to displacive transitions in general. We may speak of a polarization catastrophe in which for some critical condition the polarization or some Fourier component of the polarization becomes very large. Equally, we may <peak of the condensation of a transverse optical phonon. Here the word condensation is to be understood in the Bose-Einstein sense (TP, p. 199) of a time-independent displacement of finite amplitude. This can occur when the corresponding TO phonon frequency vanishes at some point in the Brillouin zone. LO phonons always have higher frequencies than the TO phonons of the same wavevector, so we are not concerned with LO phonon condensation. In a polarization catastrophe the local electric field caused by the ionic displacement is larger than the elastic restoring force, thereby giving an asym- metrical shift in the positions of the ions. Higher order restoring forces will limit the shift to a finite displacement. The occurrence of ferroelectricity (and antiferroelectricity) in many perovskite-structure crystals suggests that this structure is favorably disposed to a displacive transition. Local field calculations make clear the reason for the favored position of this structure: the 0'- ions do not have cubic surround- ings, and the local field factors turn out to be unusually large. We give first the simple form of thc catashophc thcoly, supposing that the local field at all atoms is eqnal to E + 47rP/3 in CGS or E + P/3<, in SI. The theoty given now leads to a second-order transition; the physical ideas can be car- ried over to a first-order transition. In a second-order transition there is no latent heat; the order parameter (in this instance, the polarization) is not discorltiriuous at the transition temperature. In a first-order transition there is a latent heat; the order parameter changes discontinuously at the transition temperature. We rewrite (24) for the diclcctric constant in thc form where a, is the electronic plus ionic p~larizabilit~ of an ion of t,ype i and N, is the number of ions i per unit volume. The dielectric constant becomes infinite and permits a finite polarization in zero applied field when This is the condition for a polarization catastrophe. The value o f t in (30) is sensitive to small departures of Z N,a, from the critical value 3/4z-. If we write (CGS) (4~/3)8N,(u, = 1 - 3s , (32) where s < 1, the dielectric constant in (30) becorries E = ~ / s . (33) Slippose near the critical temperatures varies linearly with temperature: s = (T - T,)/< , (34) where (is a constant. Such a variation of s or X N,a might come from normal thermal expansion of thc lattice. The dielectric constant has the form close to the ohserved temperature variation in the paraelectric state, Fig. 12. 16 Dielectrics and Ferroelect~.ics 473 loo' in (K)-1 T - T , Figure 12 Dielectric constant versus 1/(T - TJ in the paraelectric state (T > T,) of perovskites, after G. R~lpprecht and R. 0. Bell. Soft Optical Phonons The Lyddane-Sachs-Teller relation (Chaptcr 14) is The static dielectric constant increases when the transverse optical phonon fre- quency decreases. When the static dielectric constant ~ ( 0 ) has a high value, such as 100 to 10,000, we find that w, has a low value. When w, = 0 the crystal is unstable and ~ ( 0 ) is infinite because there is no effective restorirlg force. The ferroelectric BaTi03 at 24OC has a TO mode at 12 cm-', a low frequency for an optical mode. If the transition to a ferroelectric state is first order, we do not find w,. = 0 or ~ ( 0 ) = w at the transition. The LST relation suggests only that e(0) extrapo- lates to a singularity at a temperature To below T,. The association of a high static dielectric constant with a low-frequency optical mode is supported by experiments on strontium titanate, SrTiO,. According to the LST relation, if the reciprocal of the static dielectric constant has a temperature dependence l/<(O) (T - To), then the square of the optical mode frequency will have a similar temperature dependence: 0 2 , a (T - To), if w, is independent of temperature. The result for w$ is very well confirmed by Fig. 13. Measurements of w, versus 1 ' for another ferroelectric crystal, SbSI, are shown in Fig. 14. Temperahlre, K Figure 13 Plot of the square of the frequency of the zero wavevector transverse optical mode against temperature, fur SrTiO,, as ubserved in ncutron diffraction cxpcriolents by Cowley The bro- ken line is the reciprocal of the dielectric constant from the mea~nrem~nts of Mitsni and Westphal. 0 I I I I I I I 160 140 120 100 80 60 40 20 0 IT- T C , in K Figure 14 Decrease of a transverse phonon frequency as the Curie temperature is appruachcd from below, in the ferroelectric clystal antimony sulphoiodide, SbSI. (After Raman scattering experiments by C. H. Peny and D. K. Agrawal.) Landau Theory o f the Phase Transition A ferroelectric with a first-order phase transition between the ferroelec- tric and the paraelectric state is distinguished by a discontirluous change of the saturation polarization at the transition temperature. The transition hctween the normal and superconducting states is a second-ordcr transition, as is the 1G Dielectrics and Ferroelectrics transition hetween the ferromagnetic and paramagnetic states. In these transi- tions the degree of order goes to zero without a discor~tir~uous change as the temperature is increased. \'e can obtain a consistent formal thermodylamic theory of thc behavior of a ferroelectric crystal by considering thc form of the expansion of the en- ergy as a lunction of the polarization P. \Ve assnme that the ~andau' free en- ergy density fi in one dimension may he expanded formally as where the coefficients g, depend on the temperature. The series does not contain terms in odd powers of P if the unpolarized crystal has a center of inversion syrrlnletry, but crystals are known in which odd powers are i~nportant. Power series expansio~ls of the Gee energy do not al- ways exist, for nonanalytic terms are known to occur. especially when very near a transition. For example, the transition in KH,P04 appears to have a logarith- mic singularity in the heat capacity at the transition, which is not classifiable as either first or second order. The value of P in thermal equilibrium is given by the minimum of k as a function of P; the value of k at this minirnurn defines the Helniholtz free en- ergy F(T,E). The equilibriu~r~ polarization in an applied electric field E satis- fies the extrernurrl condition In this section uTe assume that the specimen is a long rod with the external ap- plied field E parallel to the long axis. To obtain a ferroelectric state we must suppose that the coefficient of the term in P%n (37) passes through zero at some temperature To: g, = y(T - To) > (39) whcrc y is taken as a positive constant and T, may be equal to or lower than the transition temperature. h small positive value of g, means that the lattice is "soft" and is close to instability. A negative value of g, means that the unpolar- ized lattice is unstable. The variation of g, with temperature is accounted for by thermal expar~sior~ and other effects of anharillonic lattice interactions. Second-Order Transition If g4 in (37) is positive, nothing new is added by the term in g , , and this may then he neglected. The polarization for zero applied electric field is found from (38): 'In TP. see pp. 69 and 298 for a discussion of the Landau function. TITc - Figure 15 Spontaneous polarization versus temperature, for a second-order phase transition. Figure 16 Teniperaturc variation 01 the polar-axis static dielectric constant of LiTaO, (After Glass.) so that either P, = 0 or P : = (y/g4)(T0 - T). For T 2 To the only real root of (40) is at P, = 0, because y and g4 are positive. Thus To is the Curie temperature. For T < To the minim~im of the Landau free energy in zero applied field is at I P S 1 = ( Y ~ ~ . J ' ~ ( T o - T)'I2 , (41) as plotted in Fig. 15. The phase transition is a second-order transition because the polarization goes cor~tir~uousl~ to zero at the transition temperature. The transition in LiTaO? is an exanlple (Fig. IF) of a second-ordcr transition. 16 Dielectrics und Ferroelectrics 477 First-Order Transition The transition is first order if g4 i11 (37) is negative. We must now retain g6 and take it positive in order to restrain k from going to minus infinity (Fig. 17). The equilibrium condition for E = 0 is given hy (38): so that either P, = 0 or At the transition temperature T, the free energies of the paraelectric and ferroelectric phases will be equal. That is, the value of k for P, = 0 will be equal to the value of k at the nli~li~nu~rl given by (43). In Fig. 18 we show the charac- teristic variation with temperature of P, for a first-order phase transition; Figure 17 Landau free energy function versus (polariaation)2 in a fxst-order transition, at reprc- sentative ternperaturas. At T, thc Landau function has equal minima at P = 0 and at a finite P as shown. For T below T, the ahsolnte minimum is at larger valucs of P; as T passes through T, there is a discontinuous change in the position of the absolute minimum. The arrows mark the minima. Figure 18 Calculated values of the spontaneous polar- ization as a function of tem- perature, with parameters as for barium titanate. (After W. Cochran.) contrast this with the variation shown in Fig. 15 for a second-order phase tran- sition. The transition in BaTiO, is first order. The dielectric constant is calculated from the equilibrium polarization in an applied electric field E and 1 s found from (38). In equilibrium at tcmpera- tures over the transition, the terms in P4 and P%ay he neglected; thus E = y(T - T O E or (CGS) E(T > T,) = 1 + 4 d / E = 1 + 4.rr/y(T - To) , (44) of the for111 of (36). The result applies whether the transition is of the first or second order, but if second order we have To = T , ; if first order, then To < ?
; . Equation (39) defines To, but T, is the transition temperature. Applied field Ferrodistnrtive Antidistortive yloelectric Ferroelectric @ @ Charged atoms or poups 0 IJncharged atnrns or groups Figure 19 Schematic representation of fundamental types nf stroctr~ral phase transitinna from a centrospmetric prototype. (After Lines and Glass.) 16 Dielectrics and Ferroelectrics 479 A ferroelectric displacement is not the only type of instability that may develop in a &electric crystal. Other deformations occur, as in Fig. 19. These deformations, even if they do not give a spontaneous polarization, may be ac- companied by changes in the dielectric constant. One type of deformation is called antiferroelectric and has neighboring lines of ions displaced in oppo- site senses. The perovskite structure appears to be susceptible to many types of deformation, often with little difference in energy between them. The phase diagrams of mixed perovshte systems, such as the PbZr0,-PbTiO, system, show transitions between para-, ferro-, and antiferroelectric states (Fig. 20). Several crystals believed to have an ordered nonpolar state are listed in Table 3. Ferroelectric Domains Considcr a ferroelectric crystal (such as barium titanate in the tetragonal phase) in which the spontaneous polarization may be either up or down the c axis of the crystal. A ferroelectric crystal generally consists or regions called domains within each of which the polarization is in the same direction, but in adjacent domains the polarization is in different directions. In Fig. 21 the po- larization is in opposite directions. The net polarization depends on the differ- ence in the volumes of the upward- and downward-directed domains. The PbZrO, Mole percent PbTiO, PhTiO, Figure 20 Ferroelectric F, antiferroelectric A, and paraelectric P phascs of the lead zirconate-lead titanate solid solution system. The subscript T denotes a tetragonal phase; C a cubic phase; R a rho~nboliedrd phasc, of u41ich there are high-temperature (HT) and low-temperature (LT) forms. Near the rhomhhedral-tetragon ~ h a s e boundaries one finds \,cry high piezoelectric coupling coefficients. (After Jaffe.) Table 3 Antiferroelectric crystals Crystal Trans~t~on temperature to ant~ferroelectnc state, in K From a compilation by Walter J. Merz (a) (b) Figure 21 (a) Schematic drawing of atomic displacements on either side of a boundary between domains polarized in opposite directions in a ferroelectric crystal: (b) view of a domain structure, showing 180" boundalies between domains polarized in opposite directions. crystal as a whole will appear to be unpolarized, as measured by the charge on electrodes covering the ends, when the volumes of domains in opposite senses are equal. The total dipole moment of the crystal may be changed by the movement of the walls between domains or by the nucleation of new domains. Figure 22 is a series of photomicrographs of a single crystal of barium titanate in an electric field normal to the plane of the photographs and parallel to the tetragonal axis. The closed curves are boundaries between domains polarized into and out of the plane of the photographs. The domain bound- aries change size and shape when the intensity of the electric field is altered. 16 Dielectrics and Felroelectrics Figure 22 Ferroelectric domains on the face of a single crystal of barium titanate. The face is normal to the tetragona! or c axis. The net polarization of the crystal as judged by domain volumes is increased markedly as the electric field intensity parallel to the axis is increased from 550 volts/cm to 980 V/cm. The domain boundaries are made visible by etching the crystal in a weak acid solution. (R. C. Miller.) Piezoelectricity All crystals in a ferroelectric state are also piezoelectric: a stress Z applied to the crystal will change the electric polarization (Fig. 23). Similarly, an elec- tric field E applied to the crystal will cause the crystal to become strained. In schematic one-dimensional notation, the piezoelectric equations are (CGS) P=Zd+Ex; e = Z s + E d , (45) where P is the polarization, Z the stress, d the piezoelectric strain constant, E the electric field, ,y the dielectric susceptibility, e the elastic strain, and s the elastic compliance constant. To obtain (45) in SI, replace x by cox. These rela- tions exhibit the development of polarization by an applied stress and the de- velopment of elastic strain by an applied electric field. A crystal may be piezoelectric without being ferroelectric: a schematic ex- ample of such a structure is given in Fig. 24. Quartz is piezoelectric, but not ferroelectric; barium titanate is both. For order of magnitude, in quartz d = cm/statvolt and in barium titanate d = cm/statvolt. The general defi- nition of the piezoelectric strain constants is where i - x, y, z and k = xx, yy, zz, yz, zx, xy. To convert to cm/stat-V from values of dtk given in mN, multiply by 3 X lo4. The lead zirconate-lead titanate system (called the PZT system), Fig. 20, is widely used in polycrystalline (ceramic) form with compositions of very high piezoelectric coupling. The synthetic polymer polyvinylidenfluoride (PVF2) is I Stress I 1 Stress I (b) Figure 23 (a) Unstressed ferroelectric crystal and (b) stressed ferroelectric crystal. The stress changes the polarization by AP, the induced piezoelectric polarization. Figure 24 (a) The unstressed crystal has a threefold symmetry axls The arrows represent dipole moments; each set of three arrows represents a planar group of ions denoted by A $ - , with a B3- ion at each vertex. The sum of the three dipole moments at each vertex 1 s zero. (b) The crystal when stressed develops a polarization in the direction indicated. The sum of the dipole moments about each vertex is no longer zero. five times more strongly piezoelectric than crystalline quartz. Thin stretched films of PVF, are flexible and as ultrasonic transducers are applied in medicine to monitor blood pressure and respiration. SUMMARY (In CGS Units) The electric field averaged over the volume of the specimen defines the macroscopic electric field E of the Maxwell equations. The electric field that acts at the site rJ of an atom j is the local electric field, El,,. It is a sum over all charges, grouped in terms as El0,(q) = E, + E, + E, + E3(rJ), where only E3 varies rapidly within a cell. Here: E, = external electric field; El = depolarization field associated with the boundary of the specimen; E, = field from polarization outside a sphere centered about rJ; E,(q) = field at r, due to all atoms inside the sphere. 16 Dielectrics and Ferroelectrics 483 The macroscopic field E of the Maxwell equations is equal to E, + El, which, in general, is not equal to Eloc(ri). The depolarization field in an ellipsoid is E,, = -N,$,, where N,,, is the depolarization tensor; the polarization P is the dipole moment per unit vol- ume. In a sphere N = 4 ~ 1 3 . The Lorentz field is E, = 4 ~ P l 3 . The polarizability a of an atom is defined in terms of the local electric field as p = aEl,,. The dielectric susceptibility x and dielectric constant E are defined in terms of the macroscopic electric field E as D = E + 4wP = EE = ( 1 + 47rx)E, or x = PIE. In SI, we have x = PIE&. An atom at a site with cubic symmetry has El,, = E + (4wI3)P and satisfies the Clausius-Mossotti relation (24). Problems 1. Polaritability of atomic hydrogen. Consider a semiclassical model of the ground state o f the hydrogen atom in an electric field normal to the plane of the orbit (Fig. 25), and show that for this model a = a;, where a, is the radius of the un- perturbed orbit. Note: If the applied field is in the x direction, then the x compo- nent of the field of the nucleus at the displaced position of the electron orbit must be equal to the applied field. The correct quantum-mechanical result is larger than this by the factor g. (We are speaking of a, in the expansion a = a, + alE + . . ..) We assume x 4 a , . One can also calculate al on this model. Figure 25 An electron in a circular orbit of radius an is displaced a distance x on application of an electric field E in the -x direction. The force on the electron due to the nucleus is e2/ai in CGS or e 2 / 4 ~ c a i in SI. The problem assumes x an. 2. Polariaability of conducting sphere. Show that the polarizability of a conduct- ing metallic sphere of radius a is a = a3. This result is most easily obtained by not- ing that E = 0 inside the sphere and then using the depolarization factor 4 ~ 1 3 for a sphere (Fig. 26). The result gives values of a of the order of magnitude of the ob- served polarizabilities of atoms. A lattice of N conducting spheres per unit volume has dielectric constant E = 1 + 47rNa3, for Nu3 < 1. The suggested proportionality of a to the cube of the ionic radius is satisfied quite well for alkali and halogen ions. To do the problem in SI, use 5 as the depolarization factor. 3. Effect of air gap. Discuss the effect of an air gap (Fig. 27) between capacitor plates and dielectric on the measurement of high dielectric constants. What is the highest apparent dielectric constant possible if the air gap thickness is of the total thickness? The presence of air gaps can seriously distort the measurement of high dielectric constants. 4. Zntelfacial polarization. Show that aparallel-plate capacitor made up of two par- allel layers of material-one layer with dielectric constant E , zero conductivity, and thickness d, and the other layer with E = 0 for convenience, finite conductivity u, Figure 26 The total field inside a conduchng sphere is zero. If a field E , is applied externally, then the field E , due to surface charges on the sphere must just cancel E , , so that E , + El = 0 within the sphere. But E , can be simulated by the depolariza- tion field -4?rP/3 of a uniformly polarized sphere of polariza- tion P. Relate P to E , and calculate the dipole moment p of the sphere. In SI the depolarization field is -P/~E,. v Figure 27 A n air gap of thickness qd is in series Air in a capacitor with a dielectric slab of thickness d. qd f 0 16 Dielectrics and Ferroelectrice 485 and thickness qd-behaves as if the space between the condenser platcs were filled with a homogeneous dielectric with dielectric constant where w is the a~~gular freqnency. Valucs of ceir as high as lo4 or lo5 caused largely by this Maxwell-Wagner riiechanism are sometimes found, but the high values are always accompa~~iecl by largr ac losses. 5. Polarization of sphere. A sphere of dielectric constant E is placed in a uniform ex- ternal electric field E,. (a) What is the vnlrirne avcrage electric field E in the sphere? (b) Show that the polarization in the sphere is P = ,yEo/[l + ( 4 ~ x / 3 ) ] , where , y = ( E - 1 ) / 4 ~ . Hint: You do not need to calculate El,, in this problem; in fact it is con- fusing to do so, because s and , y are defined so that P = xE. We require E, to be un- changed by insertion of the sphere. We can produce a fxcd Eo by placing positive charges on one thin plate of an insulator arid negative charges on an opposite plate. If the plates are always far from the sphere, the field or the plates will remain un- changed when the sphere is inserted between them. The results abovc are in CGS. 6. Ferroelectric criterion for atoms. Consider a systenl of hvo neritral atoms sepa- rated by a fured distance a, each atom having a polarizability a. Find the rclation between a and a for such a system to be ferroelectric. Hint: The dipolar field is strongest along the axis of the dipole. 7. Saturation polarization at Curie point. In a first-order transition the equilibrium condition (43) with T set equal to T , gives one equation for the polarization P,(T,). A further condition at the Curie point is that F(P,, T,) = $(o, T,). (a) Corrlbirli~lg these two conditions, show that P:(T,) = 3Ig4l/4g,. (b) Using this result, show that T, = To + 3&16yg6. 8. Dielectric constant below transition temperature. In terms of the parameters in the Landau frcc energy expansion, show that for a second-order phase transition tlir dielectric constant below the transition temperature is This result may he cnmparcd with (44) above the transition. 9. Soft modes and lattice transformations. Sketch a monatomic linear lattice of lattice constant a. (a) Add to cach of six atoms a vector to indicate the direction of the displacement at a givcn timc caused by a longitudinal phonon with wavevector at the zone bonndary. (b) Skctch the c~ystal structure that results if this zone boundary phonon becomes nnstablc (o + 0) as the crystal is cooled through T,. (c) Sketch on orie graph the essential aspccts of the longitudinal phonon dispersion relation for the rnorratornic lattice at T wcll above T, and at T = T,. Add to the graph the same irrforrrratinn for phonons in the new structure at T well below T,. 10. Ferroelectric linear array. Consider a linc of atoms of polarizability a and sepa- ration a. Show that the array can polarizc spontaneously if a 2 a3/4Zn 3 , where the sum is over all positive integers and is given in tables as 1.202. . . . Surface and Interface Physics Reconstruction and relaxation SURFACE CRYSTALLOGRAPHY ~eflectiion high-energy electron diffraction SURFACE ELECTRONIC STRUCTURE Work function Thermionic emission Surface states Tangential surface transport MAGNETORESISTANCE IN A TWO-DIMENSIONAL CHANNEL Integral quantized Hall effect (IQHE) IQHE in real systems Fractional quantized Hall effect (FQHE) p-n JUNCTIONS Rectification Solar cells and photovoltaic detectors Schottky barrier HETEROSTRUCTURES n-iV heterojunction SEMICONDUCTOR LASERS LIGHT-EMITTING DIODES PROBLEMS 1. Diffraction from a linear array and a square array 2. Surface subbands in electric quantum limit 3. Properties of he two-dimensional electron gas Figure 1 Dangling bonds from the (111) surface of a co~~alently honded diamnnd c ~ ~ h i c structure. (Aftcr &I. Pruiton, Surfocephysics, Clarendon, 1975.) Reconstruction and Relaxation The surface of a crystalline solid in vacuum is generally defined as the few, approximately three, outermost atomic layers of the solid that differ sig- nificantly from the bulk. The surface may be entirely clean or it may have for- cign atoms deposited on it or incorporated in it. Thc bulk of the crystal is called the substrate. If the surface is clean the top layer may be either reconstructed or, sometimes, unreconstructed. In unrcconstn~cted surfaces the atomic arrange- ment is in registry with that of the hidk except for an interlayer spacing change (called multilayer rclaxation) at the top surface. The shrinking of the interlayer distance between the first and second layer of atoms with respect to subsequent layers in the bulk is a rather dominant phenomenon. The surface may be thought of as an intermediate between a di- atomic molecule and the bulk structure. Because the interatomic distances in diatomic molecules are much smaller than in the hulk, there is a rationale for the surface relaxation. This may be contrasted with reconstruction where the relaxation of atoms ylelds new surface primitive cells. In relaxation the atoms maintain their structure in the surface plane as it was (according to the projec- tion of the bulk cell on the surface); only their distance from the bulk changes. Sometimes in metals, but most often in nonmetals, the atoms in the snrface layer form superstructures in which the atoms in the layer are not in registry with the atorns in correspo~iding layers in the substrate. This is surface reconstruction; it car1 be a consequence of a rearrangement of broken covalent or ionic bonds at the surface. Under sllch conditions the atoms at the surface burich into rows with alternately larger and smaller spacings than in the bulk. That is, for some crystals held together by valence bonds, creation of a surface would leave unsaturated bonds dangling into space (Fig. 1). The energy may then he lowered if neighboring atoms approach each other and form bonds with their otherwise unused valence electrons. Atomic displacements can be as large as 0.5 A. Reconstruction does not necessarily require formation of a superstruc- ture. For example, on GaAs (110) surfaces a rotation of the Ga-As bond leaves the point group intact. The driving force is electron transfer from Ga to As, which fills the dangling bonds on As and depletes them on Ga. Surfaces of planes nominally of high indices may be built up of low index planes separated by steps one (or two) atoms in height. Such terrace-step arritngeme~~ts are iniportant in evaporation and desorption because the attach- ment energy of atoms is often low at the steps and at kinks in the steps. The chemical activity of such sites may be high. The presence of periodic arrays of steps may he detected by double and triple beams of diffraction in LEED (see below) experiments. SURFACE CRYSTALLOGRAPHY The surface structure is in general periodic only in two dimensions. The surfacc structure can be the structure of foreign rriaterial deposited on the substrate or it can bc the selvage of the pure substrate. I11 Chapter 1 we used the term Bravais lattice for the array of equivalent points in two or in three di- mensions, that is, for diperiodic or triperiodic striicturrs. In thc physics of sur- faces it is common to speak of a two-dimensional lattice. Flirther, the area unit niay be called a mesh. We showed in Fig. 1.7 four of the five nets possible for a diperiodic struc- ture; the fifth net is the general oblique net, with no special syrn~netry relation between thc mcsh basis vectors a,, a,. Thus the five distinct nets are the oblique, square, hexagonal, rectangular, and ccntcrcd rectangular. The substrate net parallel to the surface is ilsed as thc rcfcrcnce net for the description of the surface. For example, if the surface of a ciihic snhstratc crystal is the (111) surface, the substrate net is hexagonal (Fig. 1.7b), and the surface net is referred to these axes. Thc vectors cl, c2 that define the mesh of the surface structure may be ex- pressed in terms of the reference net al, a2 by a matrix operation P: Provided that the included a~lgles of the two meshes are equal, the short- hand notation due to E. A. Wood may be used. In this notation, which is widcly used, the relation 01 the mesh cl, c2 to the reference mesh a,, a2 is ex- pressed as in terms of the lengths of the mesh basis vectors and the angle a of relative ro- tation R of the two meshes. If a = 0, the angle is omitted. Examples of the Wood notation are given in Fig. 2. The reciprocal net vectors of the surface mcsh may be written as ci, c;, defined by Here the 27r (or 1) i~ldicates that two corlver~tior~s are in use. The definitions (3) used in Fig. 3 may be compared with the definitions in Chapter 2 of the reciprocal lattice vcctors of a triperiodic lattice. 17 Surface and Interface Physics 491 fcc(lll), hcp(0001) (a! Figure 2 Surface nets of adsorhed atoms. The circles reprcsent atoms in the top layer of the sub- strate. In (a) the designation fcc(ll1) means the (111) face of an fcc stmcturc. This Lace deter- mines a ref~rmce net. The liues rcprcsent ordered overlayers, with adatoms at the intersections of hvo lines. The intersection points represent diperiodic ncts (lattices in hvo dimensions). The des- igration p(l X 1) in (a) is a primitive mesh unit for which the basis is identical with the basis of the reference net. In (b) the c(2 X 2) mesh unit is a centered mesh with basis vectors twice as long as those of the reference net. Atomic adsorption on nwtals takes place most often into those sur- face sites (hollow sites) that maximize the number of nearest-neighhor aton~s on thc substrate. (After Van Hove.) The reciprocal net points of a diperiodic net may be thought of-when we are in three dimensions-as rods. The rods are infinite in extent and normal to the surface plane, where they pass through the reciprocal net points. It may be helpful to think of the rods as gcncrated by a triperiodic lattice which is ex- panded without limit along one of its axes. Then the reciprocal lattice points along this axis are moved closer together and in the limit form a rod. The ~isefulness of the rod concept comes out with the Ewald sphere construction explained in Fig. 2.8. Diffraction occurs everywhere the Ewald t + b2 -I,/' I X X I ' Surface-structure mesh (a) Figure 3 A (3 X 1) surface structure, E. A. Wood.) 00 01 03 0 Surface Q Q Q net points 0 - il 0 Substrate net points (b) (a) red-space; and (b) reciprocal-space diagrams. (After Figure 4 Ewald sphere construction for diffraction of incident wave k by a square net, when k is parallel to one axis of the mesh. The back scattered beams in the plane of the paper are k ; , kb, kb, k ' ; . Diffracted beams out of the plane of the paper will also occur. The vertical lines are the rods of the reciprocal net. sphere intercepts a reciprocal net rod. Each diffracted beam is labelled with the indices hk of the reciprocal net vector g=hc; + kc; (4) forming the beam. I 7 Surface and Interface Physics 493 Figure 5 LEED patterns frolri a Pt(ll1i crystal surface for incident electron energies of 51 and 63.5 eV The diffraction angle is greater at the lower energy (After G. A. Somorjai, Chemistq in two dimensions: sulface.~, Cornell, 1981.) Reciprocal netrods Screen Nearly flat ~ w a l d sphere (a) . - Crystal (b) Figure 6 The RHEED method. In (a) the high-energy incident electron beam at a glancing angle to the crystal surface is associated with an Ewald sphere of large radius, so large that the snr- face is nearly flat in relation to the separation between adjacent rods of the reciprocal net. The formation of diffraction lines on a plane screen is shown in (b). (After Prutton.) Low energy electron diffraction (LEED) is illustrated by Fig. 4. The electron energy is typically in the range 10-1000 eV. With this arrangement Davisson and Germer in 1927 discovered the wave nature of the electron. An experimental pattern is shown in Fig. 5. Reflection High-Energy Electron Diffraction. In the RHEED method a beam of high-energy electrons is directed upon a crystal surface at grazing incidence. By adjustment of the angle of incidence one can arrange the normal component of the incoming wavevector to be very small, which will minimize the penetration of the electron beam and enhance the role of the crystal surface. The radius k of the Ewald sphere for 100 keV electrons will be -lo3 k', which is much longer than thc shortest reciprocal lattice vector 2 d a = 1 kl. It follows that the Ewald sphere will be nearly a flat surface in the central scat- tering region. The intercept of the rods of the reciprocal net with the nearly flat sphere will be nearly a line when the beam is directed at grazing inci- dence. The experimental arrangement is shown in Fig. 6. SURFACE ELECTRONIC STRUCTURE Work Function The work function W of the uniform surface of a metal is defined as the diffcrcncc in potential energy of an electron between the vacuum level and the Fermi level. The vacniim level is the energy of an clcctron at rest at a point sufficiently far outside the surface so that the electrostatic image force on the electron may be neglected-more than 100 A from the surface. The Fermi level is the electrochemical potential of the electrons in the metal. Typical values of electron work functions are given in Table 1. The orien- tation of the exposed crystal face affects the value of the work function Table 1 Electron work functionsa (Values obtained by photoemission, except tungsten obtained by field emission.) Element Surfacc ulane Work function, in eV "After H. D. IIagstrum. 17 Surface and Interface Physics 495 because the strength of the electric double layer at the surface depends on the concentration of surface positivc ion cores. The double layer exists because the surface ions are in an asymmetrical environment, with vacuum (or an adsorbed foreign atom layer) on one side and the substrate on the other side. The work fnnction is equal to the threshold energy for photoelectric emis- sion at absolute zero. If liw is the energy ol an incident photon, then the Einstein equation is fiw = W + T, where T is thc kinetic energy of the emitted electron and W is the work [unction. Themionic Emission The ratc of emission of thermionic electrons depends exponentially on the work fiinction. The derivation follows. We first find the electron concentration in vacuum in equilibrium with electrons in a metal at temperature ?(=k,T) and chemical potential p. We treat the electrons in the vacuum as an ideal gas, so that their chemical potential is by TP, Chapter 5. Here ny = 2 ( r n ~ / 2 d i ' ) ~ ~ , for particles of spin 1/2. Now fiat - p = \ q by tlie definition of the work function W. Thus, from (5), n = ny exp( -W/T) . (7) The flux of electrons that leaves the metal surface when all electrons are drawn off is equal to the flux incident on the sinface from outside: by TP(14.95) and (14.121). Here E is the mean speed of the electrons in the vacuum. The electric charge flux is el,, or Je = (r2m/2n2fi3)exP(W/r) . (9) This is called the Richardson-Dushman equation for thermionic emission. Surface States At the free surface of a semiconductor there often exist surface-bound electronic states with energies in the forbidden gap between the valence and conduction bands of the bulk semiconductor. We can obtain a good impression of the nature of tlie surface states hy considering the wave functions in the weak binding or two-component approximation of Chapter 7, in one dimen- sion. (The wave functions in three dimensions will have extra factors exp[i(k,y + k,z)] in the y, z plane of the surface.) If the vacuum lies in the region x > 0, the potential energy of an electron in this region can be set equal to zero: In the crystal the potential cnrrgy has the usual periodic form: In onc dimension G = n d a , where n is any integer, including zero. In the vac~l~~rn the wavc fi~nction of a bound surface state must fall off exponentially: By the wave equation thc cncrgy of the state referred to the vacuum level is Within the crystal the two-component wave function of a bound surface state will have the form, for x < 0, by analogy with (7.49), but with the addition of the factor exp(yx) wl~icl~ senres to bind the electron to the surface. M7c now come to an important consideration that restricts the allowed val- ues of the wavevector k. If thc statc is bound, there can be no current flow in the x direction; normal to the snrface. This condition is assnred in q11ant11m mechanics if the wave function can be written as a real function of x, a condi- tion already satisfied by the exterior wave function (12). But (14) can be a real function or~ly if k = $G, so that This is real providcd c(~G) = c(-;G). Thus k, for a surlace state does not have a contin~l~~rn of values, hilt is limitcd to discrctc states associated with Brillouin zone boundaries. The state (15) is damped exponentially in the crystal. The constants s, q are related by the condition that $ and d$/dx are continuous at x = 0. The binding energy E is deternrined by solving the two-corrlponent secular equa- tion analogous to (7.46). The plot of Fig. 7.12 is helpful in this connection. 17 Surface and Interjace Phg~icn 497 Tangential Surface Transport We have seen that there may exist surface-hound electronic states with en- ergies in the forbidden gap between the valence and conduction bands of the substrate clystal. Thcsc states may be occupied or vacant; their existencc mrist affect the statistical mechanics of the problem. This means that thc states mod- ify the local eqtiilihrium concentration of electrons and holes, as expressed as a shift of the chemical potential relative to the band edgcs. Recaiise the chemical potential is independent of position in an equilihriiim system, the energy bands must be displaced or bent, as in Fig. 7. The thickness and carricr concentration in the surface layer may be changed by applying an electric field normal to the surface. The effect of an external field is utilized in the metal-oxide-semico~iductor field-effect transis- tor (MOSFET). This has a metal electrode just outside the scrniconductor surface and insulated from it by a layer of oxide. A voltagc, the gate voltage Vg, is applied between the nietal and semiconductor that modulates the n,, the surface charge density per unit area: where C, is the capacitance pcr unit area between the metallic gate and the se~niconductor. This surface charge layer forms the corlducting pathway of the MOSFET. The conductance of a surface layer of length L and width W between two electrical contacts is: where p is the carrier mobility. The carrier density n,, and hence the conduc- tance, is controlled by the gate voltage. This three-terminal electronic valve is Figure 7 Baud bending near a semiconductor surface that can give a highly conducti~lg aurfacc region. (a) Inversion layer on an r~-t)ye semiconductor. For the bending as shonn, the hole con- ccntration at the surface is far larger than the electnm conce~~tration in the interior. (b) Accumula- tion layer on an 7'-typc sclniconductor, with an electron concentration at the aurhce that is Car higher than in the interior. Vacuum Vacuum pntential potential cc LC E~ E,, ------------- EF €0 a principal component in microelectronic systems. The clcctronic states occu- pied by the carriers at the surface are quantized along the direction normal to the interface, as treated in Problem 2. MAGNETORESISTANCE IN A TWO-DIMENSIONAL CHANNEL The static magnetocondi~ctivity tcnsor in 3D was found in Problerr~ 6.9. Here we translate that result to a 2D surface conductance channel in the n-y plane, with the static magnetic field in the z direction, normal to thc MOS layer. Llie assurne the surface density of electrons is n, = AIIL! The snrface conducta~lce is defined as the volurr~e conductivity times the layer thickness. The surface current density is defined as the current crossing a line of unit length in thc surface. Thus, with (6.43) and (6.65) thc surface tensor conducta~lce components become where u,, = n,e2r/m and a , = eB/mc in CGS and eB/m in SI. The following dis- cussion is written in CGS only, cxccpt where oh~ns are used. These results apply specifically in the relaxation time approximation used in Cl~apter 6. When O,T S 1, as for strong magnetic ficld and low tempera- tures, the surface conductivity components approach the limits The limit for uly is a ger~eral property of free electrons in crossed electric E, and magnetic fields B;. We establish the result that such electrons drift in thc x direction with velocity c, = cEy/B,. Consider the electrons from a Lorentz framc that movcs in the x direction with this velocity. By electrornag- netic theory- there is in this frame an clcctric ficld E; = -v,B,lc that will can- cel the applied field E, for the above choice of r ; , . k'iewcd in the laboraton. frame, all electrons drift in the x direction with velocity v, in addition to any velocity co~npor~erlts they had before Ey was applied. Thus j, = uryEy = r',eun = (n,ec/B)E,, so that as in (17). The cxpcriments measure the voltage V in the y directior~ and the current I in the x direction (Fig. 8). Here I, = j,LY = (ri,ec/B)(EyLy) = (n,cc/B)Vy. The IIall resistance is We see thatj, can flow with zero E,, so that the effective conductance j,/E, can be infinite. Yaradoxically, this limit occurs only when a,, and a , , are zero. Consider the tensor relations 17 Surface and lnterface Physics 499 In the Hall effect geometry j, = 0, so that E, = (ux,lu,,)E,, with uzy = -uyz. Thus and in the limit a , , = u,, = 0 the effective conductance is infinite. Integral Quantized Hall Effect (IQHE) The results of the original measurements' under quantum conditions of temperature and magnetic field are shown in Fig. 9. The results are remarkable: at certain values of the gate voltage the voltage drop in the direction of current flow goes essentially to zero, as if the effective conductance were infinite. Further, there are plateaus of the Hall voltage near these same values of gate voltage, and the values of the Hall resistivity V& at these plateaus are accu- rately equal to (25,813linteger) ohms, where 25,813 is the value of hle2 expressed in ohms. The IQHE voltage minima VPp may be explained on a model that is, how- ever, oversimplified. Later we give a general theory. Apply a strong magnetic field such that the separation fiw, % - k,T It is meaningful to speak of Landau levels that are completely filled or completely empty. Let the electron surface concen- tration (proportional to the gate voltage) be adjusted to any of the set of values that cause the Fermi level to fall at a Landau level: from (9 33) and (9.34), where s is any integer and n , is the electron surface concentration. When the above conditions are satisfied, the electron collision time is greatly enhanced. No elastic collisions are possible from one state to another state in the same Landau level because all possible final states of equal energy are occupied. The Pauli principle prohibits an elastic collision. Inelastic colli- sions to a vacant Landau level are possible with the absorption of the necessary energy from a phonon, but there are very few thermal phonons of energy greater than the interlevel spacing by virtue of the assumption hw, 8 - k , T . 'K. von Klitzing, 6. Dorda, and M. Pepper, Phys. Rev. Lett. 45,494 (1980). 3.5 p-substrate V,, in volts - (b) Figure 9 In the original IQHE measurements a magnetic field of 180 kG (18 T) points out of the paper. The temperature is 1.5 K. A constant current of 1 pA is made to flow between the source and the drain. Voltages Vp, and V , are plotted versus the gate voltage V,, which is proportional to the Fermi level. (After K. von Klitzing, G. Dorda, and M. Pepper.) The quantization of the Hall resistance follows on combining (18a) and (21): pH = h/se2 = 2?r/sccu , (22) where a is the fine structure constant e2/?ic 1/137, and s is an integer IQHE in Real Systems The measurements (Fig. 9) suggest that the above theory of the IQHE is too good. The Hall resistivity is accurately quantized at 25,813/s ohms, whether or not the semiconductor is of very high purity and perfection. The sharp Landau levels (Fig. 10a) are broadened in the real crystal (Fig. lob), but this does not affect the Hall resistivity. The occurrence of plateaus in the Hall 17 Surface and Interface Physics 501 Figure 11 Geometry for Laughlin's thought-experiment. The 2D electron sys- tem is wrapped around to form a cylinder. A strong magnetic field B pierces the cylinder everywhere normal to its surface. A current I circles the loop, giving rise to the Hall voltage VH and a small magnetic flux Q through the loop. n B , a C 2. .- B n resistance, evident in the UH curve of Fig. 9, is not expected in ideal systems because partially filled Landau levels will exist for all gate voltages except those for which the Fermi level exactly coincides with a Landau level. Yet the experiments show that a range of Vg values gives the exact Hall resistance. Laughlin2 interpreted the results for real systems as the expression of the general principle of gauge invariance. The argument is subtle and somewhat reminiscent of the flux quantization in a superconductor in Chapter 10. In Laughlin's thought-experiment the 2D electron system is bent to form a cylinder (Fig. 11) whose surface is pierced everywhere by a strong magnetic field B normal to the surface. The current I (former I,) circles the loop. The magnetic field B acts on the charge carriers to produce a Hall voltage VH (for- mer V,,) perpendicular to the current and to B; that is, V , is developed be- tween one edge of the cylinder and the other. The circulating current I is accompanied by a small magnetic flux rp that threads the current loop. The aim of the thought-experiment is to find the Fermi A A A A Mobility level +- +- Lr 2. .- B n 'R. B. Laughlin, Phys. Rev. B 23, 5632 (1981); see also his article in the McGraw-Hill year-book of science and technology, 1984, pp. 209-214. A review is given by H. L. Stormer and D. C. Tsui, Science 220,1241 (1983). 3 ; h o , ; ho, ; fro, 3 h 4 Localized states (a) (b) Figure 10 Density of states in a 2D electron gas in a strong magnetic field. (a) Ideal 2D crystal. (h) Real 2D crystal, with impurities and other imperfections. relation between I and V,. \lie start with the electromagnetic relation that re- lates I to the total energy U of a resistanceless system: The value of I can now be found froin the variation SU of the electronic energy that accompanies a small variation 6 p of the flux. The carricr statcs divide into hvo classes: Localized states, which are not contin~lorls around thc loop. Extended states, continuous around the loop. Localized and extended states cannot coexist at the same energy, according to our present understanding of localization. The two classes of states respond differently to the application of the flux cp. The localized states are unaffected to first order because they do 11ot en- close any significant part of cp. To a localized state a change in cp looks like a gauge transformation, which cannot affect thc energy of the state. The extended states enclose 9, and their energy may be changed. How- ever, if the magnetic flux is varied by a flux quantum, 6 p = hcle, all extended orbits are identical to those before the flux quantum was added. The argument here is identical to that for the flux quantization in the superconducting ring trcatcd in Chapter 10, but with the 2e of the Cooper pair replaced bye. If the Fcrmi level falls within the localized states of Fig. lob, all extended states (Landau levels) below the Fermi level will be filled with electrons both before and after the flux change Sp. However, during the change an integral number of states, generally one per Landau level, enter the cylinder at onc edge and leave it at the opposite edge. The number must be integral because the system is physically identical before and after the flux change. If the transferred state is transferred while occupied by one electron, it contributes an euergy change eV,; if N occupied states are transfcrrcd, the energy change is NeVH. This electron transfcr is the only way the degenerate 2D electron system can change its energy. \tle can l~nderstand thc effect by looking at a model sys- tem without disorder in the Landau gauge for the vector potential: A = -By% . (24) -40 it~crease SA that corresponds to the flux increase S q is equivalent to a dis- placement of an extended state by SAIB in the y direction. By the Stokes theo- rem and the definition of the vector potential we have Sp = L,SA. Thus 6 9 causes a motion of the entire electron gas in the y direction. By SU = NeV, and 6 9 = hde, we havc 17 Surface and Interface Physics 503 so that the Hall resistance is Fractional Quantized Hall Effect (FQHE). A quantized Hall effect has been reported for similar systems at fractional values of the index s, by working at lowcr temperatures and higher magnetic fields. In the extreme quantum limit the lowest Landau level is only partially occlipied, and the inte- gral QHE treated above should not occur. It has heen ~bserved,~ however, that the IIall resistance PH is quantized in units of 3h/e2 when the occupation of the lowest Landau level is 113 and 213, and p,, vanishes for these occupations. Sim- ilar breaks have been reported for occupations of 2/5, 315, 4/(5, and 217. p-n JUNCTIONS A p-n junction is made from a single crystal ~r~odified in two scparate re- gions. Acceptor impurity atoms are incorporated into one part to produce the p region in which the majority carriers are holes. Donor impurity atoms in the other part produce the n mgon in which the majority carriers are electrons. The interface region may be less than cm thick. Away from the junction region on the p side there are (-) ionized acceptor impurity atoms arid an equal concentration of lree holcs. On the n side there are (+) ionized donor atorris and an equal concentration of free electrons. Thus the majority carricrs are holes on thc p side and electrons on the rr, side, Fig. 12. Holes concentrated on the p side would like to diffusc to fill the crystal uniformly. Electrons would like to diffuse from the n side. But diffusion will upset the local electrical neutrality of the system. A small charge transfer by diffusion leaves behind on the p side an excess of (-) ionized acceptors and on the n side an excess of (+) ionized donors. This charge double laycr creates an electric field directed from n top that in- hibits diffusion and therehy maintains the separation of the two carrier types. Because of this double layer the electrostatic potential in the crystal takes a jump in passing through the region of the junction. In thermal equilibrium the chemical potential of each carrier type is everywhere constarit in the crystal, even across the junction. For holes k,T h i p ( r ) + eq(r) = constant (27a) 3 ~ . C . TSII~, H. L. Sturrner, and A. C. Gossard, Phys. Hev Lett. 48. 1.562 (1982); .
4 . 41. Clrallg et al., Phys. Rev. Lett. 53, 997 (1984). For a &scussion of the theory see R. Laughlin in 6. Rar~rr et d., eds., Tco-dinrensional systems, heterostructures, and wpsrluitices, Springer, 1984. Concentration of holes Concentration I Figure 12 (a) Variation of the hole and elec- tron concentrations across an unbiased (zero (a) applied voltage) junction. The carriers are in thermal equilibrium with the acceptor and donor impurity atoms, so that the product pn of the hole and electron concentrations is constant 3 throughout the crystal in conformity with the B law of mass action. (b) Electrostatic potential $ from acceptor (-) and donor (+) ions near the 2 junction. The potential gradient inhibits diffu- sion of holes from the p side to then side, and it 5 inhibits diffusion of electrons from the n side to the p side. The electric field in the junction re- gion is called the built-in electric field. (b) across the crystal, where p is the hole concentration and cp the electrostatic potential. Thus p is low where cp is high. For electrons kBT In n(r) - ecp(r) = constant , (27b) and n will be low where cp is low. The total chemical potential is constant across the crystal. The effect of the concentration gradient exactly cancels the electrostatic potential, and the net particle flow of each carrier type is zero. However, even in thermal equi- librium there is a small flow of electrons from n to p where the electrons end their lives by combination with holes. The recombination current J,,, is bal- anced by a current of electrons which are generated thermally in the p re- gion and which are pushed by the built-in field to the n region. Thus in zero external applied electric field for otherwise electrons would accumulate indefinitely on one side of the barrier. A p-n junction can act as a rectifier. A large current will flow if we apply a voltage across the junction in one direction, but if the voltage is in the opposite direction only a very small current will flow. If an alternating voltage is applied across the junction the current will flow chiefly in one direction-the junction has rectified the current (Fig. 13). 17 Sul-Juce and Interface Physics 50.5 001 0.1 1.0 10 100 Current in ~nillia~rps/cm" Figure 13 Rectiticdtiul~ charactcris- tic of a p-71 junction in germanium, after Shockley. For back voltage bias a negative voltage is applied to the p region and a positive voltage to the r2 region, thereby incrcasing the potential difference be- tween the two regions. Now practically no electrons can climb the poteritial energy hill from the low sidc of the barrier to the high side. The rrcombina- tion current is rcduccd hy the Boltzmann factor: I,,,(\' back) = J,,(O) exp (-P[v~/~,T) . (29) Thr Roltzmann factor controls the nulnber of electrons with cnongh energy to get over the barrier. The thermal generation current ol" electrons is not particularly affected by the back voltage because the gencration electrons flow downhill (fro111 p to n ) anjway: J,,,(V back) = J,,,(O) . (30) W7e saw in (28) that J,,(O) = -J,,(O); thus the ger~eratioli current dominates the rccomhination current for a back bias. When a forward voltage is applied, the reco~nbination current increases because the potential energy barrier is lowcrcd, thereby enabling more elec- trons to flow fro111 the n side to the p sidc: Again thc grneration current is unchanged: The hole current flowing across the junction behaves similarly to the elec- tron current. The applied voltage which lowers the height of the barrier for electrons also lowers it for holes, so that large numbers of electrons flow from the n region under the same voltage conditions that produce large hole cur- rents in the opposite direction. The electric currents of holes and electrons are additive, so that the total forward electric current is where I, is the sum of the two generation currents. This equation is well satis- fied for p-n junctions in germanium (Fig. 13), but not quite as well in other semiconductors. Solar Cells and Photovoltaic Detectors Let us shine light on a p-n junction, one without an external bias voltage. Each absorbed photon creates an electron and a hole. When these carriers dif- fuse to the junction, the built-in electric field of the junction separates them at the energy barrier. The separation of the carriers produces a forward voltage across the barrier: forward, because the electric field of the photoexcited carri- ers is opposite to the built-in field of the junction. The appearance of a forward voltage across an illuminated junction is called the photovoltaic effect. An illuminated junction can deliver power to an external circuit. Large area p-n junctions of silicon are used as solar panels to convert solar photons to electrical energy Schottky Barrier When a semiconductor is brought into contact with a metal, there is formed in the semiconductor a barrier layer from which charge carriers are severely de- pleted. The barrier layer is also called a depletion layer or exhaustion layer. In Fig. 14 an n-type semiconductor is brought into contact with a metal. The Fermi levels are coincident after the transfer of electrons to the conduc- tion band of the metal. Positively charged donor ions are left behind in this re- gion that is practically stripped of electrons. Here the Poisson equation is (CGS) div D = 4 m e '< %,.>. , < + - (SI) div D = ne/~,, , (34) where n is the donor concentration. The electrostatic potential is determined by (CGS) d2cp/dx2 = -4melc which has a solution of the form (CGS) cp = -(2me/e)x2 17 Surface and Interface Physics Vacuum Bottom of level conduction A band Metal Semiconductor -= u z .
2 (a) Before contact (b) Just after (c) Equilibrium contact established Figure 14 Rectifying barrier between a metal and an n-type semiconductor. The Fermi level is shown as a broken line. The origin of x has been taken for convenience at the right-hand edge of the barrier. The contact is at -xb, and here the potential energy relative to the right-hand side is -ecpo, whence the thickness of the barrier is With E = 16; ecpo = 0.5 eV; n = 1 0 1 6 ~ m - ~ , we find xb = 0.3 pm. This is a some- what simplified view of the metal-semiconductor contact. HETEROSTRUCTURES Semiconductor heterostructures are layers of two or more different semi- conductors grown coherently with one common crystal structure. Heterostruc- tures offer extra degrees of freedom in the design of semiconductor junction devices, because both the impurity doping and the conduction and valence band offsets at the junction can be controlled. Because of this freedom many devices that utilize compound semiconductors incorporate heterostructures. Examples include semiconductor lasers in CD players and high-speed devices for cell-phone systems. A heterostructure may be viewed as a single crystal in which the occu- pancy of the atomic sites changes at the interface. As an example, one side of the interface can be Ge and the other side GaAs: both lattice constants are 5.65 A. One side has the diamond structure and the other side the cubic zinc sulfide structure. Both structures are built up from tetrahedral covalent bonds and fit together coherently as if they were a single crystal. There are a few edge dislocations (Chapter 21) to relieve the strain energy near the interface. The band gaps, however, are different, and this difference is the source of the real interest in the heterostructure, apart from the technical virtuosity in Normal Staggered Broken gap Figure 15 Three types of band edge offsets at hetero-interfaces. The forbidden gaps are shown shaded. The offset called normal occurs, for example, in GaAsI(A1,Ga)As. The "broken-gap" offset occurs in the GaSbIInAs heterojunction. forming the structure. The band gaps are 0.67 eV for Ge and 1.43 eV for GaAs, at 300 K. The relative alignment of the conduction and valence band edges of- fers several possibilities, as shown in Fig. 15. Calculations suggest that the top of the valence band E, in Ge should lie about 0.42 eV higher than in GaAs. The bottom of the conduction band E, in Ge should lie about 0.35 eV lower than in GaAs, so that the offsets are classified as normal in the scheme of Fig. 15. Band edge offsets act as potential barriers in opposite senses on electrons and holes. Recall that electrons lower their energy by "sinking" on an energy band diagram, whereas holes lower their energy by "floating" on the same dia- gram. For the normal alignment both electrons and holes are pushed by the barrier from the wide-gap to the narrow-gap side of the heterostructure. Other important semiconductor pairs used in heterostructure are AlAs/GaAs, InAs/GaSb, GaP/Si, and ZnSe/GaAs. Good lattice matching in the range 0.1-1.0 percent is often accomplished by use of alloys of different ele- ments, which may also adjust energy gaps to meet specific device needs. n-N Heterojunction As a practical example, consider two n-type semiconductors with a large offset of the two conduction bands, as sketched in Fig. 16a for a semiconduc- tor pair with a normal band line-up. The n-type material with the higher con- duction band edge is labeled with a capital letter as N-type, and the junction shown is called an n-N junction. The electron transport properties across the - - junction are similar to those across a Schottky barrier. Far from the interface the two semiconductors must be electrically neutral in composition. However, the two Fermi levels, each determined by the doping, must coincide if there is to be zero net electron transport in the absence of an external bias voltage. These two considerations fix the "far-off" conduction band edge energes relative to the Fermi level, as in Fig. 16b. The combination of a specified band 17 Surface and Interface Physics 509 Ec F.L. Figure 16 (a) Two semiconductors not in contact; the absolute band edge energies are labeled E, for the conduction band edge and E, for the valence band edge. An "absolute energy" means re- ferred to infinite distance. The Fermi levels in the two materials are determined by the donor con- centrations, as well as by the band structure. (b) The same semiconductors as a heterojunction, so that the two parts are in diffusive equilibrium. This requires that the Fermi level (F.L.) be inde- pendent of position, which is accomplished by transfer of electrons from the N-side to the n-side of the interface. A depletion layer of positively ionized donors is left behind on the N-side. offset (determined by the host material composition) at the interface and the distant band energes (determined by the Fermi level) can be reconciled only if the bands bend near the interface, as in the figure. The necessary band bending is created by space charges consequent to the transfer of electrons from the N-side to the lower n-side. This transfer leaves behind on the N-side a positive donor space charge layer, which through the Poisson equation of electrostatics is the source of the positive second derivative (upward curva- ture) in the conduction band edge energy on that side. On the n-side there is now a negative space charge because of the excess of electrons on that side. The layer of negative space charge gves a negative second derivative (downward curvature) in the conduction band edge energy. On the n-side the band as a whole bends down toward the junction. This dif- fers from the usual p-n junction. The downward bending and the potential step form a potential well for electrons. The well is the basis for the new physi- cal phenomena characteristic of heterostructure physics. If the doping on the n-side (low E,) is reduced to a negligible value, there will be very few ionized donors on that side in the electron-rich layer. The mo- bility of these electrons is largely limited only by lattice scattering, which falls off sharply as the temperature is lowered. Low-temperature mobilities as high as lo7 cm2 v-ls-' have been observed in GaAs/(Al, Ga)As. If now the thickness of the N-side semiconductor is reduced below the de- pletion layer thickness on that side, the N material will be entirely depleted of its low-mobility electrons. All of the electrical conduction parallel to the inter- face will be carried by the high-mobility electrons on the n-side, equal in num- ber to the number of ionized N-side donors, but spatially separated from them by the potential step. Such high-mobility structures play a large role in solid state studies of 2D electrorl gases and also in new classes of high-speed ficld effect transistors for computer applications at low temperatures. SEMICONDUCTOR LASERS Stimulated enlission of radiation can occur in direct-gap semiconductors from the radiation emitted when electrons recombine with holes. The electron and hole concentrations created by illumi~lation are larger than their equilib- rium concentrations. The recombination times for the excess carriers are I I I U ~ I longer than the times for the condrlction electrons to reach thermal equilibriurr~ with each other in the conduction hand, and for the holes to reach ther~nal equilibrium with each other in the valence hand. This steady-state condition for the electror~ and hole populations is described hy separate Fcnni levels pc and p, for the two bands, called quasi-Fermi levels. With p, and po rrfcrred to their band edges, the condition for population inversion is that P ~ > c L , + ~ ~ . (38) For laser action the quasi-Fermi levels must he separated by Inore than the hand gap. Population inversion and laser action can be achieved by forward bias of an ordinary GaAs or InP junction, but almost all practical injection lasers em- ploy the donblc hcterostructure proposed by H. Kroemer (Fig. 17). IIere the lasing semicondiictor is cmbcdded between two wider-gap semiconductor re- gions of opposite doping, creating a quantum well that confiries bot11 electrons and I~oles. An example is GaAs embedded in (A1,Ga)iis. In such a structure there is a potential barrier that prevents the oottlow of electrons to the p-type region, and an opposite potential barrier that prevents thc outflow of holes to the n-type region. Thc value of pc in the optically active layer lines up with g, in the n con- tact; similarly, pU lincs up with pP in the p contact. Iri\~ersion can be achieved if we apply a bias voltage largcr than the voltage equi.alent of the active layer energy gap. The diode wafer providcs its own electromagnetic cavity, for the reflectivity at the crystal-air interface is high. Crystals are usually polished to provide two flat parallel surfaces; the radiation is emitted in the plane of the heterojunctions. Crystals with direct band gaps are required normally for junction lasers. Indirect gaps involve phonons as well as photons; carriers recombine less effi- ciently heca~~sc. of competing processes, and no laser action has been observed in indirect gap semicondi~ctors. Gallium arsenide has been widely stildied as thc optically active layer. It emits in the near infrared at 8383 or 1.48 eV; the exact wavclcngth dcpends 17 Surface and Interface Physics 511 Figure 17 Double heterostmcture injection laser Electrons flow from the right into the optically- active layer, where they form a degenerate electron gas. The potential harrier provided by the wide energy gap on the p side prevents the electrons from escaping to the left. Holes flow from the left into the active layer, hut cannot escape to the right. on temperature. The gap is direct (Chapter 8). In a heterojunction the system is very efficient: the ratio of light energy output to dc electrical energy input is near 50 percent, and the differential efficiency for small changes is up to 90 percent. The wavelength can be adjusted over a wide range in the alloy system GaJnl-,PyAsl_y, so that we can match the laser wavelength to the absorption minimum of optical fibers used as a transmission medium. The combination of double heterostructure lasers with glass fibers forms the basis of the new light- wave communication technology that is rapidly replacing transmission of sig- nals over copper lines. LIGHT-EMITTING DIODES The efficiency of light-emitting diodes is now at the point of exceeding in- candescent lamps. Consider a p-n junction with a voltage source V splitting the two chemical potentials pn and pCLp by eV, as in Figure 18. Electrons from the n side are injected into the p side, and holes from the p side are injected into the n side. These injected carriers annlhilate each other across the junc- tion, thus generating photons if the quantum efficiency is unity. Distance 6) Figure 18 Electron-hole recombination into photons, across a p-n junction The generation or recombination process will be much stronger in a direct-gap semiconductor (Fig. 8.5a) than in an indirect gap semiconductor (Fig. 8.5b). In a direct-gap semiconductor such as GaAs, the band-to-band photons are absorbed in-;distance -1 pm, which is strong absorption. The direct-gap ternary semiconductor GaAsl-, P, gives light tuned to shorter wavelengths as the composition variable x is increased. This composition was made by Holonyak into one of the first p-n diode lasers and into the first visi- ble-spectrum (red) LED. Blue-emitting heterostructures have now been made, such as In,Gal-,N - Al,Gal-,N. The performance of LEDs has increased markedly over the years, from about 0.1 lumens/watt in 1962 to about 40 lumens/&tt in 2004; compared with 15 lumens/watt for a standard white unfiltered incandescent lamp. To quote Craford and Holonyak, "We are entering an entirely new era in lighting (illumination) with an ultimate form of lamp-a direct-gap 111-V alloy p-n heterostructure." 17 Surface and Interface Physics 513 a cos 0 = path differences = nA for constntctive interference Incident beam Fiber specimen (a) (h) Figure 19 The diffraction pattern from a single line of lattice constant a in a monochromatic x-ray beam perpendicular to the line. (a) The condition for constructive interference is a cos 0 = nA, where n is an integer. (h) For given n the diffracted rays of constant A lie on the surface of a cone. Problems 1. Diffraction from a linear array and a square array. The diffraction pattern of a linear structure of lattice constant a is explaineda in Fig. 19. Somewhat similar structures are important in molecular biology: DNA and many proteins are linear helices. (a) A cylindrical film is exposed to the diffraction pattern of Fig. 19b; the axis of the cylinder is coincident with the axis of the linear structure or fiber. De- scribe the appearance of the diffraction pattern on the film. (b) A flat photographic plate is placed behind the fiber and normal to the incident beam. Sketch roughly the appearance of the diffraction pattern on the plate. (c) A single plane of atoms forms a square lattice of lattice constant a. The plane is normal to the incident x-ray beam. Sketch roughly the appearance of the diffraction pattern on the photographic plate. Hint: The diffraction from a plane of atoms can be inferred from the patterns for two perpendicular lines of atoms. (d) Figure 20 shows the electron diffraction pattern in the backward direction from the nickel atoms on the (110) surface of a nickel crystal. Explain the orientation of the diffraction pattern in relation to the atomic positions of the surface atoms shown in the model. Assume that only the sur- face atoms are effective in the reflection of low-energy electrons 'Another viewpoint is useful: for a linear lattice the diffraction pattern is described by the single Laue equation a . Ak = 2mq, where q is an integer. The lattice sums which led to the other Laue equations do not occur for a linear lattice. Now a . Ak = constant is the equation of a plane; thus the reciprocal lattice becomes a set of parallel planes normal to the line of atoms. (4 ibl Figure 20 (a) Backward scattering pattern of 76 e V electrons incident normally on the (110) face of a nickel crystal; a model of the surface is shown in (b). (Courtesy of A. U. MacRae.) 2. Surface subbands in electric quantum limit. Consider the contact plane be- tween an insulator and a semiconductor, as in a metal-oxide-semiconductor transis- tor or MOSFET. With a strong electric field applied across the Si0,-Si interface, the potential energy of a conduction electron may be approximated by V(x) = eEx for x positive and by V(x) = m for x negative, where the origin of x is at the inter- face. The wavefunction is 0 for x negative and may be separated as +(x,y,z) = u(x) exp[i(k,,y + k,z)], where u(x) satisfies the differential equation With the model potential for V(x) the exact eigenfunctions are Airy functions, but we can find a fairly good ground state energy from the variational trial function x exp(-ax). (a) Show that ( E ) = (h2/2m)a2 + 3eEIZa. (b) Show that the energy is a minimum when a = (3eEm/2h2)1'3. (c) Show that = 1.89(fi2/2m)113 (3eE12)213. In the exact solution for the ground state energy the factor 1.89 is replaced by 1.78. As E is increased the extent of the wavefunction in the x direction is decreased. The function u(x) defines a surface conduction channel on the semi- conductor side of the interface. The various eigenvalues of u(x) define what are called electric subbands. Because the eigenfunctions are real functions of x the states do not carry current in the x direction, but they do carry a surface channel current in the yz plane. The dependence of the channel on the electric field E in the x direction makes the device a field effect transistor. 3. Properties of the two-dimensional electron gas. Consider a two-dimensional elec- tron gas (2DEG) with twofold spin degeneracy but no valley degeneracy. (a) Show that the number of orbitals per unit energy is given by: D(E) = m l ~ h ~ . (b) Show that the sheet density is related to the Fermi wavevector by: n , = kgI27r. (c) Show that, in the Drude model, the sheet resistance, i.e., the resistance of a square segment of the ZDEG, can be written as: R, = (h/e?/(k,e) where t? = vp7 is the mean free path. Nanostructures Written by Professor Paul McEuen of Cornell University IMAGING TECHNIQUES FOR NANOSTRUCTURES Electron microscopy Optical microscopy Scanning tunneling microscopy Atomic force microscopy ELECTRONIC STRUCTURE OF 1D SYSTEMS One-dimensional (lD) subbands Spectroscopy of Van Hove singularities 1D metals-Coulomb interactions and lattice couplings ELECTRICAL TRANSPORT IN 1D Conductance quantization and he Landauer formula Two barriers in series-resonant tunneling Incoherent addition and Ohm's law Localization Voltage probes and the Buttiker-Landauer Formalism ELECTRONIC STRUCTURE OF O D SYSTEMS Quantized energy levels Semiconductor nanocrystals Metallic dots Discrete charge states ELECTRICAL TRANSPORT IN O D Coulomb oscillations Spin, Mott Insulators, and the Kondo Effect Cooper pairing in superconducting dots VIBRATIONAL AND THERMAL PROPERTIES Quantized vibrational modes Transverse vibrations Heat capacity and thermal transport SUMMARY 562 PROBLEMS 562 1. Carbon nanotube band structure 562 2. Filling subbands 563 3. Breit-Wigner form of a transmission resonance 563 4. Barriers in series and Ohm's law 563 5. Energies of a spherical quantum dot 564 6. Thermal properties in one dimension 564 Figure 1 Schematic and scanning electron microscope (SEM) image of a gate electrode pattern on a GaAsIAIGaAs heterostructure used to create a quantum dot of complex shape in the underly- ing two-dimensional (2D) electron gas. (Courtesy of C. Marcus.) The previous chapter addressed solids with spatial confinement at the nanometer scale along one direction: surfaces, interfaces, and quantum wells. These systems were effectively two-dimensional, which we define as extended in two directions but of nanometer scale in the third. Only a small number of quantized states-often only one-are occupied in the confined direction. In this chapter we discuss solids confined in either two or three orthogonal direc- tions, creating effectively one-dimensional (ID) or zero-dimensional (OD) nanostructures. Important 1D examples are carbon nanotubes, quantum wires, and conducting polymers. Examples of O D systems include semiconduc- tor nanocrystals, metal nanoparticles, and lithographically patterned quantum dots. Some examples are shown in Figs. 1 to 3. We will almost exclusively focus on nanostructures that are created from confined periodic solids. Nonperiodic nanostructures are of great interest in other fields, such as molecular assem- blies in chemistry and organic macromolecules in biology. The techniques for the creation of nanostructures can be divided into two broad categories. Top-down approaches use lithographic patterning to struc- ture macroscopic materials at the nanoscale, such as the metallic electrodes on top of a semiconductor heterostructure shown in Fig. 1. Bottom-up ap- proaches utilize growth and self-assembly to build nanostructures from atomic or molecular precursors. A CdSe nanocrystal grown in solution is shown in Fig. 2. It is typically difficult to create structures smaller than 50 nm with Figure 2 Model and transmission electron lnicroscope (TEM) image of a CdSe nanocrystal. Individual rows of atoms are clearly resolved in the TEM image. (Courtesy of A. P. Alivisatos.) Figure 3 Atomic force microscope (AFM) ili~age of.1 pair of crossed carbon nanotuhes contacted by Au electrodes patterned by electron beam lithography. (Image courtesy of M. S. Fuhrer.) Also shown is a model of the nanotube cross region, showing the honeycomb lattice of the graphene sheets that form the nanotube walls. (Courtesy of P. Avouris.) top-down techniques, while it is often difficult to create structures larger than 50 nm by bottom-up techniques. A major challenge of nanoscience and tech- nology is to combine these approaches and develop strategies to reliably create complex systems over all length scales, from the molecular to the macroscopic. Figure 3 shows one example, where 100-nm-wide lithographic electrodes make contact to 2-nm-wide carbon nanotubes grown by chemical vapor deposition. When the extent of a solid is reduced in one or more dimensions, the physical, magnetic, electrical, and optical properties can be dramatically al- tered. This makes nanostructures a subject of both fundamental and practical interest; their properties can be tailored by controlling their size and shape on the nanometer scale. One class of effects is related to the large ratio of number of surface atoms to bulk atoms in a r~anostructure. For a spherical nanoparticle of radius R composed of atorrls with an average spacing a, thc ratio is given by For R = G a - 1 nm, half of the atoms are on the surface. The large surface area of nanoparticles is advantageous for applications in gas storage, where molcculcs are adsorbed on the surfaces, or in catalysis, wl~ere reactions occur on the surface of the catalyst. It also has dra~rlatic effects on the stability of thc nanoparticle. The cohesive energy is dra~natically lowercd because atoms on the surcace are inco~npletely bonded. Yanoparticlcs therefore melt at temper- atures far below the melting temperaturc of the corresponding bulk solid. The fundamental electronic and vibrational excitations of a nanostructure also become quantized, and these excitations determine many of the most irnpor- tant properties of the nanostructured material. These quantization phenomena will be the primary subject of this chapter. Typicdl5 they are important in the 1-100-nanometer size range. IMAGING TECHNIQUES FOR NANOSTRUCTURES The development of new techniques to image and probe nanostructures has been essential to the evolutior~ of the field. For periodic 3D structures, the diffraction of electrons or X-rays can be used to determine structure in rccip- rocal space, which can then be inverted to find the rcal-space atomic arrange- rnents, as discussed in Chapter 2. For individual nanoscale solids, diffraction is only of limited utility for both fundamental and practical reasons. The solid's small size intcrn~pts the periodicity of the lattice, blurring rllffraction peaks, and also produces a very small scattered signal. Real-space probes that can directly determine the properties of the nano- structure are therefore very valuable. These probes use thc interaction of a typically an electron or photon, with the object under study, to create an image. The techniques fall into two major classes, which we will refer to as focal and scanned probc. In focal microscopy, the probe particle is focused by a series of lenses onto the sample. Figure 4 shows a schematic. The ultimate resolution of the system is limited by the wavelike nature of the particle through the Heisenberg Un- certainty Principle, or, equivalently, diffraction. This smallest fcature spacing d that can be resolved is given by where h is the wavelength of the probe and P = sin 8 is the numerical aperture defined in Fig. 4. Achieving nanoscale resolution rcqnires using par- ticles with srnall wavelengths and maximizing the nnmerical aperture. 1 Source Figure 4 Schematic diagram of a focal mi- croscope. A beam emitted from a source is focused onto the sample by a series of lenses. An equivalent focal system can be used to focus particlelwaves emitted from the sample onto a detector. In scanned probe microscopy, by contrast, a tiny probe is brought close to the sample and scanned over its surface. The resolution of the microscope is determined by the effective range of the interaction between the probe and the structure under study, rather than by the wavelength of the probe particle. In addition to imaging, scanned and focal probes provide information about the electrical, vibrational, optical, and magnetic properties of individual nanostructures. Of particular importance is the electronic structure, expressed in the density of states. For a finite-sized system, the density of states is a se- ries of delta functions D(E) = Z ~ ( E - E,) , I (3) where the sum is taken over all the energy eigenstates of the system. For extended solids, the density of states can be represented by a continuous func- tion, but for a nanostructure the discrete sum form is necessary along the con- fined directions. This quantized density of states determines many of the most important properties of nanostructures, and it can be directly measured using the techniques described below. Electron Microscopy A very powerful focal tool is the electron microscope. A collimated beam of electrons is accelerated by high voltages and focused through a series of electrostatic or magnetic lenses onto the sample under study. In transmission electron microscopy, or TEM, the electron beam trav- els through the sample and is focused on a detector plate in much the same way as the image is focused onto the eyepiece of an optical microscope. The ul- timate resolving power d is set by the wavelength of the accelerated electrons where V is the accelerating voltage (measured in volts). For typical accelerat- ing voltages (100 kV), the theoretical resolving power is therefore subatomic. Other effects, such as imperfections in the lenses, keep the TEM resolution well above this limit, hut d- 0.1 nm has been acllieved. Figure 2 shows a TEM irnage of a semiconductor nanocrystal, where rows of atoms are clearly resolved. , 4 major limitation of TEM is that the electron beam must penetrate the sample, making it impossible to examinc stn~ctnres on solid substrates. This problem is overcome in the scanning electron microscope (SEM). In an SEM, a high-energy (100 V to 100 kV), tightly focused electron beam is scanned over the sample. The numhcr of hackscattered electrons and/or the secondary electrons generated by the beam that emerge from the sample depends on the local composition and topography of the sample. These electrons are collected by an electron detector, and an image is fornled by plotting this detector signal as a function of the beam location. This powerful technique can be used on most kinds of samples, but it typically has a lower resolution (>1 nm) than the TEM. Figure 1 is an SEM irnage of metallic electrodes on a GaAsIAlGaAs substrate. In addition to imaging, the SEM beam can be used to expose an electron- sensitive material and draw small features in a technique known as electron beam lithography. The ultimate resolution (<10 urn) is very high, but it is a slow process because the patterns must be drawn pixel by pixel. It is therefore used primarily in research, prototyping, and optical mask fahrication. Optical Microscopy The optical microscope is the prototypical focal instrument. Using visi- ble light and a high numerical aperture (P = I), the highest obtainable resolu- tion is 200400 nm. For direct imaging, optical microscopy thcrcfore only reaches the edge of the nanoscale realm. However, inany of the optical spectro- scopies discussed in Chapter 15 have been successfully adapted to study individual nanostructures. These include elastic light scattering, absorption, luminescence, and Raman scattering. Measurements of a single nanostructure, or even a single molecule, are possible if only one is in the field of view of the microscope. Here we briefly review the emission and absorption of electromagnetic radiation by matter in a manner suitable for applications to nanostnictures. Within the electric dipole approximation, Fermi's golden rule gives the transition rate between an initial state i and a higher energy statcj due to absorption: Transitions therefore occur between states that have a nonzero dipole matrix element and whose energies differ Ly the absorbed photon energy hw. Simi- larly, the emission rate from statej to i is given by 2 I+,= (~n-/fi)l(jle~.rli)l - E; + hw) + (4rrw;/c2)l{jlrli)12 , (6) where wji = (8, - ei)/fi and a is the fine striicturc constant. The first and sec- ond terms represent stimulated and spontaneous emission, respectivcly. By summing over all possible states, these relations can be used to calcll- late the total power u'E2 absorbed frorr~ tlle electromagnetic field and hence the real part of the conductivity: where i i is a unit vector pointing in the direction of the electric field. The absorption is proportional to the joint density of all initial and final states sepa- rated by an energy hw, weighted by the dipole matrix elemcnt and thc occupa- tion factors of the states. The Fermi functions inmcate that absorption only occurs when the initial state i is filled and the final statej is empty. The above relations show that absorptio~l and emission can be used to probe the rlectronic energy level spectra of nanostructures. Measurerr~erlts can readily he performed on macroscopic collections of no~ninall~ identical narios- tructures, hrlt the effects of inhomogcnous broadening due to the variation in the properties of the individual nanostnlct~lres are significant. Furthermore, sometimes only a few or even a single nanostructure is a-ailahle for measnre- rnent. Optical measurements that probe single nanostructures have therefore proven to be particularly valuable. Figure 5 shows an example of spor~taneous emission, or fluorescence, from individual optically excited semiconductor quantu~n dots. The ernission occurs from the lowcst cner,g state in the conduction band to the highest energy state in the valence band. The linewidths of the emission lines from single nanocrys- tals are very narrow, bnt they are distributed over a rangc of energies due to variations in the nanocrystal size, shape, and local environment. Meas~lremmts of an ensemble therefore show a broad peak that does not accurately reflect the properties of a single nanocrystal. In addition to their use in probing nanostructures, optical focal systerms arc also widely used for microfabrication. In projection photolithography, a pattern on a mask is projected onto a photosensitive resist using optical elements. Following exposure and development of the resist, the pattern is transferred into the material of interest by etching or deposition through the resist stencil. Optical lithography is the basis for the mass-fabrication of microelectronic and microrrlechanical systems. By using wavelengths into the 18 Nonostructures 523 5 35 65 Intensity (countslsec) (a) Figure 5 Left: Image of the fluorescence from individual CdSe nanocrystals dilutely distributed on a surface at T = 10 K. Right: Spectra of the fluorescence of a number of different individual nanocrystals. In each spectrum, the high-energy peak is the primary transition between the lowest electronic state in the conduction band and the highest energy state in the valence hand. The lower energy peaks are associated transitions involving the emission of an LO phonon. Variations in the nanocrystal size and local electronic environment shift the positions of the peaks. The broad peak is the spectrum obtained for an ensemble of nominally identical nanoclystals. (After S. Empedocles et al.) deep-UV, devices with features of 100 nm are in commercial production. Fur- ther improvements using Extreme UV light or even X-rays are possible, but the masks and focusing elements become more and more challenging to fabri- cate and control. Scanning Tunneling Microscopy The most famous scanned probe instrument is the scanning tunneling microscope (STM), schematically shown in Fig. 6. Its invention was a break- through in the field of nanoscience. In an STM, a sharp metal tip, preferably one with a single atom protruding from the end, is brought to within a nanometer of the conducting sample to be studied. The position of the tip is controlled with picometer precision using piezoelectric materials that expand or contract in response to electrical signals from a control system. A voltage bias V is applied to the sample, and a tunneling current I flowing between the tip and the sample is measured. The current is proportional to 3, the Figure 6 Schematic of a scanning tunneling microscope (STM). Ilihen operated in feedback mode, the piezos scan the tip over the sample and maintain a constant tunneling current between the tip and the sample. (Courtesy of D. LePage.) Lower: STM image of a carbon nanotube. (Cour- tesy of C. Dekker.) tunneling probability through the gap between the tip and sample. The tunneling probability is exponentially sensitive to the tunneling distance. In the WKB approximation, where z is the distance between the tip and sample and 4 is the effective bar- rier height for tunneling. For typical parameters, a 0.1-nm change in the tip position leads to an order of magnitude change in 3. When the STM operates in feedback mode, I is maintained at a con- stant value by changing the tip height z. The STM thus tracks the surface topography, and very small changes in the height of the surface can be detected Figure 7 A "quantum corral" of mean radius 7.1 nm was formed by moving 48 Fe atoms on a Cu (111) surface. The Fe atoms scatter the surface state electrons, confining them to the interior of the corral. The rings in the corral are the density distribution of the electrons in the three quan- tum states of the corral that lie close to the Fermi energy. The atoms were imaged and moved into position by a low-temperature, ultra-high vacuum scanning tunneling microscope. (Image cour- tesy of D. M. Eigler, IBM Research Division.) (<1 pm). This is illustrated in Fig. 6, where an STM image of a carbon nanotube is shown. The STM can also be used to manipulate individual atoms on a surface. An example is shown in Fig. 7, where the STM tip is used to con- struct a "quantum corral" by pushing Fe atoms on a Cu (111) surface into a ring. The STM tunneling current I as a function of bias V can give spatial and spectroscopic information about the quantum states of a nanostructure. At zero temperature, the derivative of the current with respect to voltage is It is proportional to the density of states at the tunneling electron energy eF + eV, weighted by the electron probability density of those states at the STM tip position r,. For the quantum corral, the electrons in the 2D surface state of Cu are re- flected by the Fe atoms, creating a discrete set of states in the interior of the corral. The observed ripples in the image in Fig. 7 are due to the modulations of the probability density l+,(rt)l2 of these localized states near the tunneling electron energy. Images at different bias voltages yield the spatial structure of quantized states at different energies. Atomic Force Microscopy The atomic force microscope (AFM) was developed soon after the STM. It is a much more flexible technique than STM and can be used on both con- ducting and insulating samples. However, it typically has poorer resolution. An AFM measures the force between the tip and the sample, rather than the tun- neling current. A sharp tip is mounted on the end of a millimeter-sized can- tilever, as shown in Fig. 8. A force F exerted on the tip by the sample deflects the cantilever by Az: where C is the force constant of the cantilever. The displacement of the can- tilever is measured as a function of tip position, often by using the back of the cantilever as a reflector for a laser beam (Fig. 8). Motion of the reflector changes the path of the laser beam, which is detected using a photodiode array; picometer-scale displacements can easily be measured. Since a typical value of the force constant is C = 1N/m, pN-scale forces can be transduced. Forces well below 1 fN have been measured under special circumstances. The simplest mode of operation is contact mode, where the tip is dragged along in contact with the surface and the cantilever deflection is measured. This gives a measure of the sample topography, but it can damage the sample. Noncontact or intermittent-contact imaging modes are less invasive, and they also can give information about the long-range forces between the sample and the tip. In these techniques, the cantilever oscillates just above the sample due to an applied driving force of amplitude F, near the cantilever resonance Figure 8 (a) Schematic of an atomic force microscope (AFM). Deflections of the cantilever are measured by a photodetector registering the position of a laser beam that reflects off the top o f the cantilever. (Courtesy of Joost Frenken.) Inset: SEM image of an AFM tip. The effective radius of curvature of the tip can he less than 10 nm. I8 Nunostructures frequency o,. Modeling the cantilever as a driven simple harmonic oscillator, the magnitude of the cantilever response at a frequency o is given by where Q, the quality factor of the oscillator, is the ratio of the energy stored in the cantilever to the enerm dissipated per cycle. Note that on-resonance, w = o,, the response is Q tirnes larger than at low frequencies, making the dctcction of small forces possible. The parameters charactcrixing the oscillating cantilever are sensitive to any forces that occur between the tip and the sample. These forces can he van der Waals, electrostatic, magnetic, or many others. Thc interaction shifts the resonance frequency w, andlor modifies Q. This change is recorded and uscd to construct an image. For example, in tapping mode imaging, the tip 'taps' the surface during the closest approach of the oscillation cycle, causing both a fre- qucncy shift and additional dissipation. The nanotube device shown in Fig. 3 is imagcd in tapping mode. Another important technique is Magnetic Force Microscopy (MFM), briefly discussed in Chapter 12. The tip is coated with a magnetic material so that it has a magnetic moment p normal to the surface of the sample. It then feels a force due to variations in local magnetic fields produced by the sample where 2 , is the tip's equilibrium position and A z is the displacement during the oscillation. The term p(aB/az) produces a static deflection of the cantilever, but does mot alter the oscillation frequency or the damping. The term p(a2~laz2)Az, OII the other hand, has the form of a force constant change 6C, since it is linear in the displacenlent Az of the cantilever. It therefore shifts the resonance frcqucncy of the cantilever. Monitoring this frequency shift produces an image. Gradicnts of other local force fields can be similarly measured. There are many other scanned probe techniqnes. Ncar-ficld scanning optical ~nicroscopy (NSOM) creates optical images with a resolution below the diffraction limit by using a scanned subwavelength aperture through which photons 'tunnel'. Scanning capacitance microscopy (SCM) measures capacitancc variations between the tip and the sample as a function of posi- tion. This ever-growing family of techniques is increasingly used to character- ize objects ranging from individual molecllles to Si transistors in integrated circuits. ELECTRONIC STRUCTURE OF 1D SYSTEMS The quantized electro~lic states of nanostructures deterrr~ir~e their electri- cal and optical properties, and they influence the physical and chemical prop- erties as well. To descrihe these states, we take as our starting point the band structure of the bulk material. An effective mass approximation is used tbr thc electronic dispersion of a given hand, and the associated wavefiinctions are treated as plane waves. These are simplifications; the bands are not always par- abolic, and the true eigenstates are Bloch states, not plane waves. However, these assumptions greatly simplify the ~nathenlatics and are qualitatively (and often quantitatively) correct. We will also often neglect the Coulomb interac- tions hctwccn clcctrons. However, thcre are many cases in the physics of nanostructi~res where electron-electron interactions cannot bc ignored, as dis- cussed later in this chapter. One-dimensional (ID) Subbands Consider a nanoscale solid in the geometry of a wire. Its dirnensions along the r and y are nanoscale, but it is continuous in z. The energies and eigen- states of such a wire arc: given by E = eiZj + fi2k2 / 2 m ; +(x,y,z) = +i,j(x,y)eikz , (13) where i a n d j are the quantum numbers labeling the eigenstates in the r,y plane and k is the wavevector in the z direction. For the rectangular wire shown i 1 1 Fig. 9, E , ~ and +,J ((~,y) are just particle-in-a-box energies and eigen- states discussed in Chapter 6. The dispersion relation consists of a series or 1D subbands, each corre- sponding to a different transverse energy state E,,,. The total density of elec- tronic states D(E) is the sum of thc dmsity of statcs of thc individual subhands: where Did(&) is given by k - iz,rz,~[ rn ]lb - Di,,(s) = -- - - -- dk dE 4L for 6 > cSi 2n 2fi2(& - E , ~ ) hvif (15) = 0 for E < E , ~ The first factor of two in the middle expression is due to spin degeneracy and the second from incltiding both positivc and ncgativc values of k. I 1 1 thc right expression, v,,~ is the velocity of the electron in the i,j suhhand with kinetic energy E - E ~ ~ . Note that the density of states diverges as (E - E,,~)-'~' at each subband threshold. These are called van Hove singularities. This behavior stands in contrast to three dimensions, where D(E) goes to zero at low energies 18 Nanostructures 529 Figure 9 Schematic of a rectangular quasi-one-dimensional wire, along with the dispersion rela- tions and the dens~ty of states of the 1D subbands. The peaks in the density of states at the sub- band thresholds are called Van Hove singularities. The probability density for the i = 2, j = 1 state is shown as a gray scale on the cross section of the wire. (Chapter 6), and two dimensions, where D(E) steps up a constant value at the bottom of each 2D subband (Prob. 17.3) Spectroscopy of Van Hove Singularities The Van Hove singularities described by (15) affect the electrical and opti- cal properties of 1D systems. Here, we discuss the case of a semiconducting carbon nanotube, whose band structure is calculated in Prob. 1 and shown in Fig. 10a. Van Hove singularities are seen in scanning tunneling spectroscopy, as shown in Fig. lob. Peaks in the differential conductance, which is propor- tional to the density of states by (9), are observed at bias voltages correspond- ing to the energies of these singularities. The optical absorption and emission of semiconducting nanotubes are also dominated by these singularities, since they depend on the initial and final density of states by (5)-(7). Figure 10c shows the photoluminescence intensity of a collection of carbon nanotubes as a function of the wavelength of the ex- citing and emitted light. The absorption of the incident light is enhanced when the energy matches E,, - E,,, the energy difference between the 2nd Van Hove singularities in the conductance and valence bands. The electrons and holes relax quickly to the bottom of the first subband, where they recombine, 0 2 4 6 8 1 0 Density of electronic states (a) Figure 10 (a) The density of states for a semiconducting carbon nanotube as a function of energy. The Van Hove singularities are seen in the STM tunneling spectra of a nanotube shown in (b). In (c), the emission intensity is plotted as a function of the emission wavelength and the exci- tation wavelength. Peaks in the intensity are observed when the absorption and emission energies correspond to those shown in the diagram (a). Different peaks correspond to nanotnbes with different radii and chirality. [After Bachilo et al. (a and c) and C. Dekker (b).] producing luminescence with energy - E , ~ . Peaks are therefore observed when the emitted and absorbed light simultaneously match the energies be- tween the 1st and 2nd van Hove singularities, respectively. Different peaks in the emission intensity in the plot correspond to nanotubes of varying diame- ters and chiralities. 1D Metals-Coulomb Interactions and Lattice Couplings In a quasi-one-dimensional metal, electrons fill up individual 1D sub- bands, with the Fermi energy and the total number of subbands occupied de- termined by the electron density. For a strictly 1D metal, there is only one (spin-degenerate) subband occupied. In this case, if there are nlD carriers per unit length: The Fermi surface of a 1D metal consists of just two points, at +k, and -kF, as shown in Fig. 1 1 . This is quite different from the Fermi surfaces in 3D and 2D free-electron metals, which consist of a sphere and a circle, respectively. Two consequences of this unusual Fermi surface are discussed below. Coulomb interactions cause scattering among electrons near the Fermi en- ergy. For 3D metals, scattering is strongly suppressed near & , by the restrictions of energylmomentum conservation combined with the Pauli exclusion principle. At an energy E measured relative to E,, IT,, = ( 1 1 ~ ~ ) (E/&,)~, where 1 1 7 , is the Figure 11 Electronic structure of a 1D metal near the Ferlni energy. The Fermi surface consists of two points at + k , . The scattering of electrons from filled states 1 and 2 to empty states 3 and 4 conserves energy as long as the energy difference is the same between 1 and 3 and between 2 and 4. Momentum is simultaneously consewed because the energy is locally linear ink. classical scattering rate. By the uncertainty principle, this prod~ices an iincer- tainty in thc energy of the electron: 8 ~ ( 3 ~ ) = IT,, - ( ~ L / T ~ ) ( E / E ~ ) ~ . (17) As the energy becomes small (measured relative to E ~ ) , the uncertai~ity ~ I I the energy goes to zero as the second power of E . The uncertainty 6.3 is therefore guaranteed to be small in conlparison to .
s sufficiently close to 8,. This ensures that the quasiparticles near the Fermi surfacc arc wcll dcfincd. Thc case of one dimension is shown in Fig. 11. Energy and momentum conservation are eqriivalent in this case since for small s the energy is locally linear in the momentum change Ak = k-k,: Referring to Fig. 11, energy conservation requires that for an electron in state 1 at energy E to scatter to state 3, a si~nultaneous scattering of an electron from state 2 to 4 must occur. The only restriction is that the final state encrgy 3 bc positive and less than E. This gives a rcduction factor I/%, - (l/r0) (sIsFJ, and, from the uncertainty principle, Since the uncertainty is linear in s, there is no guarantee that 8 s will be smaller than E as E + 0. The fundamental assu~r~ption behind Fermi liquid theory, that weakly interacting quasiparticles exist as E + 0, is therefore not guaranteed in ID. In fact, the ground state of the interacting I D clcctron gas is believed not to be a Fermi liquid, but rather a Luttingcr liquid whose low- energy excitations are collective in nature. Thc cxcitations are more analogous to phonons or plasmons-a collective motion of many objects-than isolated electrons moving independently of their neighbors. This collective nature has a number of effects. For example, tunneling into a 1D rr~etal is suppressed at low energies because the tunneling electron must excite the collective modes. In spite of this issue, the independent electron picture rerrlains a useful ap- proxiniatiori for the 1D electron gas. It has been successful in describing most, but not all, experiments on real 1D systems, and will bc adoptcd below. A second unusual propcrty of 1D metals is that they are unstable to per- turhations at a wavevector 2kp For example, a &stortion of the lattice at this wavevector will open up a bandgap in the electronic spectrum, converting the metal to an insulator. This is the Peierls instability treated in detail in Chapter 14. This effect is particularly important in 1U conducting polylners such as polyacetylene (Fig. 12). It has one conduction electron per carbon atom spac- ing a, and therefore from (16) kF = 7ri2n. Without any distortions, polyacety- lene would have a half-filled band and be a metal. However, a lattice distortion H H H H H H I I I C , c \ , l \ C c /C\~/L~/C\~/~ I H I H I H I H I 11 Figure 12 Structure of polyacetylene. Doe to the Peierls distortion, the lattice is dimerized, with carbon atoms joined by douhlc bonds in the diagram closer together than those linked by single bonds. The Peierls distortion opens a semiconducting gap ol approximately 1.5 e17. at 2k, = nla, corresponding to a wavelength 2a, opens up a gap at the Fermi cnergy. This corresponds to a dimerization of the lattice. This dimerization produces alternating single and double bonds along the chain and turns poly- acetylene into a se~r~iconductor with a bandgap of 1.5 e! Polyacetylene and related semiconducting polymers can be made into field-effect transistors, light-emitting diodes, and other semiconducting de- vices. They can also he doped chemically, producing metallic behavior with conductivities comparable to traditional metals. However, they retain the me- chanical flexibility and ease of processing characteristic of polymers. Their dis- covew has led to a revolution in flexible plastic elcctronics. A The Peierls distortion is large in polymers because their backbone consists of a single atomic chain, \vhich can casily distort. Other 1D systems such as nanotubes and nanowires arc much stiffer, and the Peierls transition is not ob- served at experimentally relevant temperatures. ELECTRICAL TRANSPORT IN 1D Conductance Quantization and the Landauer Formula A 1D channel has a finite current-carrying capacity for a given voltage ap- plied across its ends. It therefore llas a finite conductance even if there is no scattering in the wire. Cor~sider a wire with one siihband occupied connecting two larger reservvirs with a voltage difference V between them, as shown in Fig. 13. The right-going states will be populated up to an electrochemical po- tential pi and left-going states will he populated up to fi2, where p, - p2 = qV and 9 = -e for clcctrons and +e for holes. The net current flowing throngh the channel due to the excess right-moving carrier density An is then D,(s)qV I = Anqv = - 243% 2 vq2\r = -v L 4O=hz; h ' where DR(&), the density of states of right-moving carriers, is '/2 the total den- sity of states given in (15). Figure 13 (a) The net current propagating between two reservoirs for an applied bias voltage dif- ference V, - V,. (b) Schematic representation of transmission probability 3 and reflection proba- bility !I? from a barrier in the channel, where 3 + !I? = 1. Remarkably, in 1D the velocity exactly cancels with the density of states to create a current that depends only on the voltages and fundamental constants. The two-terminal conductance I/V and resistance V/I are then A perfectly transmitting one-dimensional channel has a finite conductance whose value is the ratio of fundamental constants. This is called the conduc- tance quantum GQ; its inverse is called the resistance quantum RQ. While derived here in the effective mass approximation, it is true for a 1D band of arbitrary dispersion. The quantization of conductance is dramatically illustrated in the data in Fig. 14. A short quasi-1D channel is formed between two regions of a 2D elec- tron gas in a GaAs/AlGaAs heterostructure. As the carrier density of the chan- nel is increased, the conductance increases in discrete steps of height 2e2/h. Each step corresponds to the occupancy of an additional 1D subband in the wire. Conductance quantization is also observed in atomic-scale bridges be- tween macroscopic metals. If the channel is not perfectly conducting, the overall conductance is the quantum of conductance times the probability 3(+) for electron transmission through the channel (Fig. 13): 18 Nunostructures 535 Gate voltage (volts) Figure 14 Conductance quantization in a short channel electrostatically defined in a GaAsI AlGaAs heterostmcture at different temperatures. A negative gate voltage Vg applied to the metal- lic gates on the s~lrface of the sample depletes the carriers in the underlying two-dimensional electron gas, creating a narrow channel. The channel is fully depleted of carriers at Vp = -2.1V Individual 1D subbands become occupied with increasing VR, with each new subband adding a conductance of 2e'llz. (Courtesy of H. Van Houten and C. Beenakker.) This equation is often called the Landauer Formula. For a quasi-1D system with multiple channels, we sum over the contributions of each channel, since conductances in parallel add: where i, j label the transverse eigenstates. For example, for N perfectly transmitted channels in parallel, 9 = N, as for the data in Fig. 14. At finite temperatures or biases, the Fermi-Dirac energ); distributions f of thc clcctrons in thr left and right lcads must hc takcn into account: m I(s,V,n = (%/hj 1 blip - rb7j - fn(sj]3(r) . (24) -a The net current is simply the difference behveen the left- and right-moving currents, intcgratcd ovrr all mcrgics. The Landauer formula (22) directly relates the resistance of a system to the transmission properties of the channel. Let us rewrite the resistance for the one-channel case in the following way: where A = 1 - 3 is thc rcflcction corfficicnt. Thc rcsistancc of the device is the sum of the first term, the quantized contact resistance, and the second term, the resistance due to scattering from barriers in the channel. The latter term is zero for a perfect conductor. Below we consider an application of the Landauer formula to the proble~ri of two barriers in series. We treat this ill both the colier- ent and incoherent limits ol electron propagation between the barriers. Two Barriers in Series-Resonant Tunneling Consider two barriers in series separated by a distance L, with transmis- sion/reflection amplitudes t,, r, and t , , r , ; as shown in Fig. 15. These ampli- tudes are complex: To calculate the transmission probability 3 tl~rougl~ t l ~ e entire double barrier structure, we need the corresponding tra~lsrrlissio~l amplitude. For an incident wave from the left whose amplitude is 1, the amplitudes defined in Fig. 15 arc givcn by where p = 2kL is the phase that an clcctron with kinctic energy h2k2/2m accll- mulates propagating the distance 2L on a round trip between the barriers. Combining these to solve for the transmitted amplitude yields: The tra~is~nission tl~rougl~ the double barrier is then 3 = I C i Y = It, l"t2 IZ l+ lrl 1 2 1 r 2 l2 - 21r1 I Ir2 lcos(va) where p = 2 7 , + c p , , + c p , 18 Nanostructures 537 Figure 15 Resonant tunneling through two identical barriers in series separated by a length L. The upper diagram shows the transmission amplitudes between and outside the barriers for a unity amplitude incident wave. The transmission resonances at the energies of the quasibound states between the barriers are shorn. This is plotted in Fig. 15. Note the round-trip phase accumulation cp includes the phase shifts associated with reflections from the barriers. The transmission probability (29) is greatly enhanced when cos(q~) ap- proaches unity, because the denominator becomes small. This occurs for the resonance codtion where n is an integer. This is a general property of waves, and is due to the constructive interference of many pathways through the sample. This can be m easily seen by rewriting (28) using the series expansion 1/(1 - x) = 2 xm: m=O The mth order in the expansion corresponds to a path with m round trips be- tween the barriers. On resonance, these paths add in phase to yield a strongly enhanced transmission. Consider the special case where the barriers are the same: tl = t,. We then have The transmission on resonance through a symmetric double-barrier structure is 1, even if the transmission through each of the individual barriers is small. This is called resonant tunneling. Ollresonance, the denominator of (29) is of order unity for opaque barriers, and the transmission is ronghly the product of the transmission coefficients of each of the two barriers in series: 3 - ltll"t212. The resonance condition 9 = 2 m corresponds to the energies of the quasibound electronic states confined between the two barriers. For very opaque walls, this is just the particle-i~i-a-box quantizatiorl condition: kL = m l . We derived the resonant tunneling condition for a one-dimensional case, but it is a general result. The transmission through a confincd clcctron systcm is strongly enhanced at energies corresponding to the hound-state energy levels of the confined electrons. This is also evident from the STM tunneling expres- sion (9); quasibound states produce peaks in the differential conductance. For the case of opaque barriers, lt,l~It,l2 91, the cosine term in the de- nominator of (29) can be expanded, as shown in Prub. 3, yielding the familiar Breit-W7igner form for a resonance: 4rJ-2 A< '(&) -- ( r L +r2)' + 4(& & , , ) 2 where r. = - It ' . J 2 T r J (33) The resonances are thus Lorentzian peaks with a width in energy of r = r, + r2 determined by the energy level spacing A& and the tra~~s~rlission prob- abilities through the two barriers. This is just the uncertainty principle broad- ening of the level due to the finite lifetime of the double-barrier bound state. Incoherent Addition and Ohm's Law If we instead treat the electron classically, we add probabilities rather than amplitudes. This is valid if the electron effectively loses track of its phase be- tween the barriers due to, for example, inelastic scattering from phonons. This corresponds to replacing (27) by I n 1 2 = l t 1 1 2 + j r l / Z / h 1 2 ; h 2 = a 2 r 2 ; I ~ i ~ = l a 1 ~ 1 t ~ ~ . (34) This gives Some elementary manipulations (Proh. 4) yield The resistance is just the sum of the quantized contact resistance and the intrinsic resistances of the indwidual barriers (see Eq. 25). This is Ohm's law- resistors in series add. It is valid if interference effects can be neglected. Equation (36) allows us to connect to the Drude Iormula. Consider a process that gives a backscattering rate l/rh This backscattering could rcsult from either an elastic scattering process such as impilrity scattering or from an inelastic process such as phonon scattering. For propagation over a small distance dL, the reflection probability d% (@ 1) is This gives a contribution to the resistance, yielding a resistivity: = t l ~ l d ~ = (h/2e2)lCb . (38) This is equal to the 1D Drude resistance a;; = (nlDe2r1m)-', as shown in Prob. 4. Ignoring interference effects, the resistances of individual segments add ohmically, giving R = R ~ + ( ~ / % " ) L / ~ ~ ) . (39) Localization Now consider when two barriers are connected in series, hut coherence is not neglected. However, we average over all possihle phases, corresponding to an average over different energies. From Eq. (29), the average resistance is Notably, the phase-averaged resistance (40) is larger than the resistance in the incoherent limit (36). To understand the scaling with length associatcd with (40), consider a long conductor of length L consisting of a series of only elastic (phase-prese~ng) scatterers cl~aracterized by an elastic backscattering length C,. Assume that the conductor has a large rcsistance (R), so that 3 1 = 1 and 3 4 1. For a small additional length dL, there will be an additional reflection and transmission d31 = dLlC,, as in (37), and d9 = 1 - d31. Combining these according to the prescription of (40), and assuming that d31 4 1, gives or equivalently, ( d ~ ) = ( ~ ) ( 2 d ~ I C,) Separating variables and integrating both sides of the eqnation yields (R) = (h / 2e2) exp(2L 1 tz) . (43) Remarkably, the resistance grows exponentially with the length of the sample, rather than linearly as in an ollrnic conductor. This bchavior is a result of localiza- tion. Due to quar~tu~rl interference among the states scattered by disorder, the states become localized on a size scale 8 - C,, where 6 is called the localization length. There are no extended states that traverse the cntirc lcngth of thc con- ductor, so the resistance is exponentially large. A similar result holds for quasi-1D systems, but with a localization length 5 - Ne,, where N is the number of 1D subbands occupied. At very low temperatures, only coherent scattering processes occur arid thc rcsistance is exponentially large by (43). At finite temperatures, electrons retain their phase memory only over thc phase coherence length P, due to their interaction with other degrees of freedom such as phonons or electrons. This length typically is a power-law function of temperature, Y , = AT-", since the number of electronic and vibrational excitations present is a power law in T The resistance of each phase coherent segment can be approximated by (43) with t , replacing L. The resistance of each phase coherent segrrlerlt decreases rapidly with increasing temperature (as the exponential of a power law in T). This dramatically decreases the overall rcsistancc, which is the (incoherent) series combination of L/t, such phase coherent sections. At a siifficiently high tcm- perature where t!, 5 e,, all phase coherence is lost between scattering events and the ohmic expression (36) is applicable. A related issue is the nature of the electronic states in 2D and 3D systems in the presence of disorder. In 2D, it is believed that, for riorliriteractiiig elec- trons, all states are also localized by disorder. I 1 1 3D, on the other hand, a criti- cal amount of disorder is required to localize the states. The subject of local- ization continlies to he of great fundamental intrrest and controversy, particularly when the effects of coulomb interactions between the electrons are included. Voltage Probes and the Buttiker-Lundauer Formalism In many measurements, Inore than two probes are connected to a conduc- tor. Some are used as voltage probes (which draw no nct cl~rrcnt from the sample) and others as current probes, as shown in Fig. 16. Biittiker extended the Landauer formalism to deal with this multiprobe case. Define 31nm' as the total transmission probability for an electron leaving contact in to arrive at contact n, including the contributions from all the 1D channels. For a current probe n with N,, channels, the electrochemical potential of the contact is fixed by an applied voltage, and the net current that flows through the con- tact is This is just the current flowing out of the contact rrlirius the currents flowing in that originated from each of the other contacts. Note that N , = z3'"2'"). which , , ! can be easily obtained from (44) by considering the equilibrium case where all the voltages are equal arid all the currents are zero. Figure 16 Schematic representation of a multiterminal conductor. Contacts 1 and 2 are current probes; contact 3 is a voltage probe. The transmission probability from contact 1 to 2 and from 1 to 3 is schematically indicated. For a voltage probe, the potential V, adjusts itself so that no net current flows (I" = 0): 2 3(nm)vm Vn = n i f n 2 CJ(n,m) (45) m f n The electrochemical potential measured by the probe is the weighted average - of the electrochemical potentials of the different contacts, where the weight- ing coefficients are the transmission probabilities. - Equations (44-45) have a number of surprising consequences. Since the measured currents and voltages depend on 3("."), the details of the path that an electron takes in traversing the sample influences the resistance. A voltage probe can disturb the paths, and the measured voltage can in turn be affected by transmission through all parts of the sample. Below we present three exam- ples that illustrate these properties. Consider a voltage probe connected to the center of an otherwise ballistic 1D conductor, as shown in Fig. 16. Assume that electrons leaving from probe 1 either arrive at probe 2 or 3, but none are directly backscattered. The voltage read by probe 3 is then where for the last step we assumed that the voltage probe couples symmetri- cally to the left and right moving channels, 3(3.1) = 3(3,2). The voltage mea- sured in the channel is just the average of the voltage of the two contacts. The current flowing out of contact 1 is given by: where (46) has been employed in the second step. Note that the presence of the voltage probe decreases the transmission below the unity value of a perfect Figure 17 Four-terminal Hall resistance measurements of submicron junctions of different shapes. In the junction shown schematically in the upper left, the Hall resistance is negative at small B and positive at large B. The reason is indicated in the diagram; at small B, the electrons bounce off the wall into the "wrong" probe. (After C. Ford et al.) channel. Some of the electrons scatter into the voltage probe, are re-emitted, and then return to contact 1. This shows that voltage probes are in general in- vasive; they influence what they measure unless they only couple very weakly to the system. Figure 17 shows a measurement of the Hall resistance of two nanoscale crosses patterned in a high mobility 2D electron gas whose geometries are shown in the insets. The junction region is ballistic, meaning that there is no scattering from disorder, only from the sample walls. The measured Hall resis- tance is not of the form B/n,e, where n , is the sheet carrier concentration, as expected for a macroscopic 2D electron gas, but has instead a number of notable features. Most surprisingly, the Hall voltage is of the opposite sign at low B compared to high B for the sample shown in the upper left inset. This can be easily understood from the shape of the classical electron paths sketched on the figure. At high B, the Lorentz force preferentially deflects the electron into the upper electrode, giving the expected sign of the Hall voltage. At low B, however, the electron bounces off the boundary of the conductor and arrives at the lower electrode, reversing the sign of the measured Hall voltage. For a small multiprobe conductor, the resistance is a measure of the electron trajectories through the sample and not simply related to intrinsic material properties like the electron density. Equations (44-45) can be used to treat arbitrarily complex microscopic (or even macroscopic) conductors. It has been widely used to describe measure- ments on small disordered metal samples at low temperatures as a function of magnetic field B. These samples have many transverse channels and contain impurities. The elastic scattering length 4, is less than the sample dimensions, but the phase coherence length 4 , is greater. Electrons therefore propagate diffusively, but phase-coherently, through the sample. This is called the meso- scopic regime. In a semiclassical picture, the transmission amplitude between two probes n and m corresponds to the sum of many different classical paths through the sample: Note that the phase associated with each path amplitude a] contains a contri- bution from the magnetic vector potential A, as described in Appendix G. Since 3 ( n 3 m ) = It(n,m)12, quantum interference among different conducting path- ways through the sample modulates the transmission. An interesting example is shown in Fig. 18. On the left, the four-terminal resistance of a nanoscale metallic wire is shown. Aperiodic fluctuations are seen in the conductance versus magnetic field B. These fluctuations are due to modulations of the interference between the many diffusive paths linking the contacts. Since there are many paths, the result is an essentially random varia- tion. These modulations are referred to as conductance fluctuations. When an additional loop is added that is outside the region between the con- tacts, as shown on the right of Fig. 18, the conductivity G qualitatively changes. A periodic modulation with magnetic field is seen. Ths is due to the Abaronov- Bohm effect. The vector potential modulates the quantum interference between those electron paths that encircle the ring and those that do not. For simplicity, consider the interference between just two such paths with transmission ampli- tudes al and a2 (Fig. 18) in the absence of a magnetic field. At finite B, In the last step we have employed Stokes's theorem, where Q, is the magnetic flux going through the loop and hc/e is the magnetic flux quantum (h/e in SI). With increasing flux, the transmission though the wire oscillates with the pe- riod of one flux quantum. This effect is closely related to the superconducting flux quantization discussed in Chapter 10, except that the charge appearing in the flux quantum here is e, not 2e, since the carriers here are electrons, not Cooper pairs. 507 428 2 - m ' x a ' - - % a> 8 506 + u 8 427 Y j i u 6 U 505 426 -1 .
6 -1 .
4 -1.2 -1 .6 1 . 4 -1.2 hlagnetic field (T) Figure 18 Upper: SEM micrographs of two vertical Au wires with current and voltage probes at- tached. In the device on the right, an extra loop has been added outsidc the regio~i betwee11 the probes. Thc diagram to the right shows two paths, one that encircles the ring and one that does not. Lower left: Conductance versus magnetic field for the left sample. Aperiodic conductance fluctuations are seen due to quantum interfercncc bctwee~i tlie conducting paths through the sample. Lower right: Periodic oscillations are ohserved associated with the Aharonov-Bohm effect for paths enclosing the loop nominally outside the region between the contacts, showing the non- local nature of diffusive coherent transport in mcsoscopic systerris. (After R. Webb.) It is remarkable that the addition of a loop outside of the region between the voltage contacts changes the measured properties. Resistances in the mesoscopic regime are nonlocal. Electrons coherently diffuse throughout the entire sample while journeying between contacts, arid their phase remembers the joiirney. 18 Nanostructuren 545 ELECTRONIC STRUCTURE OF O D SYSTEMS Quantized Energy Levels A system of electrons fully confined in all three dimensions will have dis- crete charge and electronic states, as do atonis and molecules. They are often called artificial atoms or quantum dots to reflect the importance of quantiza- tion phenomcna on their properties. As a simple example, consider an electron in a spherical potential well. Due to the spherical sym~netry, the Hamitonian separates into angular and ra- dial parts, giving eigenstates and eigenenergies: where Yf,,(O,+) are the spherical harrnouics and R,,,Jr) are the radial wave- functions. The energy levels and radial wavefunctions depend on the details of the particular confining potential. For an infinite spherical well, where V = 0 for r < R and is infinite othenvise, The function jl(x) is the lth spherical BesseI function and the coefficient P,:l is the nth zero ofjfix). For example, Po." = T (IS), = 4.5 (IP), = 5.8 (ID), Pl,o = 2~ (2S), and = 7.7 (2P). The labels in parentheses arc the atornic notations for the states, which have the usual degeneracies associated with spin and angular momentum orientation. Semiconductor Nanocrystals . A semiconductor r~anocrystal such as the one shown in Fig. 2 can, to a good approximation, be described by the spherical model given above. Both the electron states in the condiiction band and the liole states in the valence band are quantized. For a CdSe nanoparticle, the conduction hand effective rnass m: = 0.13 m, and the electron energy levels arc c,,, = (2.9 eV/R2) (~n,i/Po,o)', where R, the radius of the nanoparticle, is expressed in nanometers. For R = 2 nm, the spacing between t h e lowest two energy levels is E ~ , ~ - E ~ , ~ = 0.76 eV. The 1s electron state increases in energy with decreasiug K, while the IS hole state decreases. The handgap therefore grows and can be tuned over a wide range by changing R. This is shown in Fig. 19, where thc ahsorption spectra of CdSc nanocrystals of different sizes are presented. For the smallest radii, the threshold for absorption shifts by nearly 1 eV from its bulk value. A similar shift is seen in the err~ission spectrum. The optical spectra of nanocrys- tals can be tuned continuously across the visible spectrum, malong them use- ful in applications from fluorescent labeling to light-emitting diodes. Bulk band gap 1.i 1.9 2.1 2.3 2.5 2.7 2.9 3 1 Energy (eV) Figure 19 Optical absorption spectra for a series of CdSe nanocrystal samples of different aver- age radii. The lowest transition energy in the smallest nanoclystal samplc is shifted by rrearly 1 eV from the bulk bandgap. Two dominart tra~~sitions are labeled. (Co~lrtesy of A. P. Ali\lsatos.) The absorption intensity in nanocrystals becomes concentrated at thc spc- cific frequencies corresponding to the transitions between discrete states, as described by (7). An important result for the integrated absorption can be ob- tained from the Kramers-Kronig relations discussed in Chapter 15. From Eq. (15.11b), we have: At very high frequencies, o + GO, thc clectron's response is identical to that of a free electron. By (15.20), In addition, as w + m, the frequency s in the denominator of (52) can be ne- glected. Combining (52) and (53) then gives A bulk semiconductor and a nanoclystal therefore have the same overall absorption per unit voh~mc when integrated over all frequencies. It is distrib- uted very differently, however. The absorption spectrum of macroscopic semiconductor is continuous, but in a nanocrystal it consists of a series of dis- crete transitions with very high absorption intensity at the transition frequen- cies. These strong transitions at particular frequencies have motivated re- searchers to create lasers that work on the quantized electronic transitions of quantum dots. Metallic Dots For small spherical metallic dots, such as alkali-metal clusters created in an atomic beam, the electrons in the conduction band f a up the quantized energy levels described by (50), as shown in Fig. 20a. These quantized levels af- fect the electrical and optical properties, and can even influence the stability of the dot. Small clusters can be analyzed by mass spectroscopy to determine the number of atoms in the cluster (Fig. 20b). Since there is one conduction elec- tron per atom in an alkali metal, this is also the number of electrons in the con- duction band. Large abundances are seen at certain "magic numbers" of atoms in the cluster. These result from the enhanced stability for clusters with filled electronic shells. For example, the 8-atom cluster peak corresponds to the fill- ing of the 1S (n = 1, e = 0) and the 1P (n = 1, [ = 1) shells. These filled-shell clusters are analogous to chemically stable filled-shell atoms (the noble gases). For larger or irregularly shaped metallic dots, the shell structure is de- stroyed. The level spacing becomes small in comparison to the shifts in the levels due to shape imperfections, faceting of the crystals, or disorder. While the details of the level spectrum are difficult to predict, the average level spac- ing at the Fermi energy can be estimated using (6.21) as AE = l/D(eF) = 2eF13N . (55) For a spherical Au nanoparticle with R = 2 nm, the average level spacing is A& - 2 meV. This is much smaller than the spacing between the lowest states in the CdSe nanocrystal conduction band (0:76%V?calculated earlier. Energy- level quantization effects are much more important in semiconductor dots than they are in metallic ones. This is because the energy-level spacing is larger for low-lying states in a 3D potential well and also because the electron effective mass for semiconductors is typically smaller than for metals. The optical properties of a small metallic dot are typically dominated by its surface plasmon resonance. The polarizability of a sphere is, from (16.11), (CGS) P = xE0 , 1 + 457x13 ' where Eo is the external electric field and x is the electronic susceptibility. This relation was presented in Chap. 16 for the static case, but it applies at high frequencies as long as the dot is small enough for retardation effects to be Radius (nm) Figure 20 (a) Energy level diagram for the states in a small spherical alkali metallic cluster. The numbers at right of the diagram show the number of electrons required to fill successive elec- tronic shells. (b) Abundance spectrum of Na clusters, showing high intensities for clusters with completely filled electronic shells. (After W. A. de Heer et al.) absent. Modeling the carriers in the dot as a lossless free electron gas, the sus- ceptibility is, from (14.6), (CGS) ~ ( o ) = -ne2/mw2 ; (57) Combining (56) and (57) gives I8 Nanostructurex 549 where wl, is the plasma frequency of the bulk metal. The polarization diverges at a frequency: Ws = Or /v5 (59) This is the surface plasrriorl resonance frcqnency for a sphere. It shifts the bulk plasma resonance for metals like An and Ag from the UV into the visiblc por- tion of the spectrum. The rcst~lt (59) is independent of particle sizc. rn reality, however, the optical properties do depend so~riewliat on size dne to retar- dation effects at larger radii and losses and intraband transitions at smaller radii. Liquids or glasses contair~ing metallic nanoparticles are often brilliantly colored due to absorption by the surface plasmon resonance. They have been used for hundreds of years in stained glasses. Other optical applications of metallic nanoparticles make use of the large electric field just outside the nanoparticle near resonance. In techniques like surface enhanced Raman scat- tering (SERS) or sccond harmonic generation (SHG), wcak optical processes i 1 1 nanostructurcs near the surface of the nanoparticlc become measurable due to the locally high electric fields. Discrete Charge States If a quantum dot is relatively isolated electrically from its environment, it has a set of well-defined charge states, like an atom or niolecule. Each succes- sive charge state corresponds to the addition of one more electron to the dot. Because of the coulomb repulsion between electrons, the energy difference between successive charge states can be very large. It'ithin the Thomas-Fermi approximation (28). the electrochemical potential for adding the (N + 1)th electron to a dot containing A' electrons is given by: I.%+, = F ~ + ~ - ecp = F,+, + h7U - a~'17~ , (60) where U is the coulomb interaction energy between any two electrons on the dot, often called the charging energy. The dimensionless number a is the rate at which a voltage Vg applied to a nearby mctal, typically referred to as the gate (see Fig. 21), shifts the electrostatic potential cp of the dot. In general, C ' will vary for different electronic states in the dot, but we as- sume here it is a constant, as in a classical metal. In this case, we can describe the electrostatics and interactions in terms of capacitances: U = e2/c and a = Cg/C , (61) where C is the total electrostatic capacitance of the dot and Cg is the capaci- tance bctween the dot and the gate. The quantity e/C is the electrostatic po- tential shift of the dot when one electron is added. If the dot is in weak electrical contact with a metallic reservoir, electrons will tunnel or~to the dot until the electrochemical potential for adding another electron exceeds the elcctrochemical potential p of the rcsenroir (Fig. 21). Figure 21 (a) Schematic illustration of a quantum dot in tunnel contact with two metallic reser- voirs and capacitatively coupled to a gate. Main: Energy level diagrams illustrating the coulomb blockade. In (b) the gate voltage is such that the dot is stable with N electrons, so no current flows. In (c) the blockade is lifted when the electrochemical potential is lowered into the window be- tween the potentials in the leads, allowing successive charging and discharging of the dot and a net current flow. This sets the equilibrium occupancy N of the dot. The charge state can be changed using the gate voltage Vg. The additional gate voltage AVg required to add one more electron from a reservoir of fixed p is, from (60), Adding an extra electron to the dot requires enough energy to fill up the next single-particle state and also enough energy to overcome the charging energy. The charging energy U depends on both the size of the dot and the local electrostatic environment. Nearby metals or dielectrics will screen the coulomb interaction and reduce the charging energy. In general, U must be calculated for the specific geometly. As a simple model, consider a spherical dot of radius R surrounded by a spherical metal shell of radius R + d. This shell screens the coulomb interaction between electrons on the dot. An elementary application of Gauss's law (Problem 5) gives the capacitance and therefore the charging energy: e2 d (CGS) U = z m ; (63) For R = 2 nm, d = 1 nm, and E = 1, the charging energy is e2/C = 0.24 eV. This far exceeds kgT = 0.026 eV at room temperature, indicating that thermal fluctuations in the charge of the dot will be strongly suppressed. It is compara- ble to the energy level spacing (0.76 eV) between the lowest two states in a 2-nm- radius CdSe dot. In contrast, it is much larger than the level spacing (2 meV) for a 2-nm-radius metallic dot. The addition energy of a metallic dot is therefore dominated by the charging energy, but in a semiconductor dot the clrarging energy and the level spacing are of comparable importance. Charging effects are destroyed if the tunneling rate between the dot and the electrodes is too rapid. The charge resides on the dot for a time scale of order 6t = RC, where R is thc resistance for tunneling to the elrctrodes. By the uncertainty principle, the energy level wiIl be broadened by The uncertainty in the energy of the electron beco~nes comparable to the chargng energy when R - h/e2. For resistances below this value, quantum fluc- tuations due to the uncertainty principle smear out the colllomb charging effects. The conditions for well-defined charge states of a quantum dot are then n h/e2 and e2/C /C kk,T . (65) ELECTRICAL TRANSPORT IN O D Coulomb Oscillations At temperatures 2' < (Ll + A&)lkg, the charging energy CT and the level spacing A s coritrol the flow of electrons through a quantum dot, as shown in Fig. 21. Transport through the dot is suppressed where the Fermi levels of the leads lie between the electrochemical potential for thc N and N + 1 charge states (Fig. 21h). This is called the Coulomb blockade. Current can only flow when pe(N + 1) is lowered to lie between the Fermi levels of the left and right leads. Then an electron can hop on the dot from the left electrode and off the dot to the right electrode, resulting in current flow (Fig. 21c). This process re- peats with increasing V, for each new charge state. This leads to so-called Coulomb oscillations in the conductance as a function of shown in Fig. 22. If CT % -- ksT, these peaks can be very sharp. The spacing between the Coulo~nl peaks is determined by (62). Coulomb oscillations are first and foremost a result of charge quantiza- tion. They will occur if CT /C ksT even if the single particle level spacing is very small, A s + k , T This is often thc case in metallic quantum dots. A device showing Coulomb oscillations is called a single electron transistor (SET), since it turns on and off periodically as the occupancy of thc dot is changed by e. This effect is qnite remarkable, and can be used as an ultrasensitive electrometer. It detects electric fields much as a SQUID (Chap. 10) detects magnetic fields. One is based on the quantization of charge, the other on the quantization of flux. Figure 22 Conductance oscillations versns gate voltage Vp measured in a quantum dot formed in a gated GaAsIAIGaAs heterostructure at T = 0.1 K. The data are plotted on a log scalc. As the gate voltage increases, the bamers become lnorc transparent and the peaks get hroader. The lineshape of thc peak in (b) is determined hy thermal broadening alone, while the one in (c) also reflects the intrinsic Rreit-Wigner lineshape. (Adapted from Foxman et al.) SETS can also be used to make single electron turnstiles and pumps. Oscil- lating voltages at a frequency f applied to the gates of a properly designed quantum dot system can shuttle a single electron through dot per cycle of the oscillation. This results in quantized current flowing through the dot: Such devices are under investigation as current standards in metrology. For the quanturn dot in Fig. 22, the level spacing A& B kBT The Nth Coulomb oscillation then corresponds to resonant tunneling through a single Figure 23 (a) Schematic of a 2D circular quantum dot formed in a GaAsIAIGaAs heterostmc- ture. (b) The differential conductance dlldV as a function of both gate voltage and source drain bias, plotted as a gray scale. The white diamond regions correspond to different charge states of the dot. A larger charging energy is observed for N = 2 and 6 electrons on the dot, corresponding to filled electronic shells. The additional lines on the diagram correspond to excited energy levels of the dot. (Courtesy of L. Kouwenhoven.) quantized energy level EN. The coulomb oscillations in this case are analogous to the theoretical resonant tunneling peaks described by (29) and shown in Fig. 15. A crucial difference from (29) is that the positions of the coulomb peaks are determined by both the level spacing and the coulomb charging energy (62). The lower right panel of Fig. 22 shows a fit of one coulomb peak to the Breit-Wigner form for resonant tunneling (33). The I-V characteristics of a quantum dot are in general complex, reflect- ing the interplay of the charging energy, the excited state level spacing, and the source-drain bias voltage. In Fig. 23, measurements of the first few elec- trons added to a small, 2D circular dot are shown. The differential conduc- tance dI/dV is represented using a gray scale as both the gate voltage and the source drain bias are varied. Each of the lines seen corresponds to tunneling through an individual quantum state of the dot. The white diamonds along the ITg axis, indicating dIldV = 0, correspond to the Coulomb blockade. Each successive chamond corresponds to anothcr clcctron on the dot. The point at which different diamonds touch along the axis are thr Coulomb oscillations where the charge state of the dot changes. The height of thc diamond corre- sponds to eV,,, = e2/c + As, t11e maximum voltage that can he applied with- out current flowing in a given charge state. The diamonds corresponding to N = 2 and N = ' arc noticeably larger than neighboririg diamonds, indicating a larger adchtion energy for adding the third and seventh electron to the dot. This dot can be effectively modeled hy a 2D harmonic oscillator confining potential U(x, y) = ~rnwz(xz + yZ), with energy levels: where i and j are nonnegative integers. This levcl spectrum can be used in conjunction with (62) to find the addition energies that determinc the sizes of the diamonds in the figure. The first electron fills the spin-degenerate gronnd statc cnergy level .coo. The second electror~ fills the same quantum state, but with opposite spin, at a gate voltage AVg, = L1/ac: after the first electron is added. The third electron fills one of the degenerate states sol or s10 after a AVg, = (U + fiw)/ae. The next thrcc clcctrons fill the rest of these states, each spaced in gate voltage by U/ae. The seventh electron fills one of the degener- ate states sll, szO, or cO2, after a gate voltage AVg, = (U + Rw)lae. This simple model correctly predicts the larger addition energies for the 3rd and 7th electrons seen in the experiment. The addition energy is larger by the level spacing when the extra electron is added to a new energy level above a filled electronic shell (N = 2 and N = 6). Spin, Mott Insulators, and the Kondo Effect Consider a quantum dot that is occupied hy an odd number of electrons in the blockaded region, as shown in Fig. 24. The highest single particle cnergy level of the dot is doubly degenerate and an electron can either reside in a spin-up or spin-down state. The addition of a second electron of opposite spin is allowed by the Pauli exclusion pri~lciple but is energetically prohibited by the coulomb interaction between the electrons. This is analogous to the Mott Insulator where a half-filled band is insulating because coulomb interactions prohibit double-occupancy of the lattice sites hy clcctrons. The dot therefore has a spin-'/, magnetic moment with two dcgcnerate configurations, spin up and spin down, in the absence of coupling to the leads. If coupling to the leads is included, however, this degeneracy is lifted at low - ~ temperatures. The ground state is a quantu~n superposition of the two spin config~irations, with transitions between them accomplished by a virtual 18 Nanostructures 555 Final state (c) Density of states (dl Figure 24 The Kondo effect in a quantum dot. For an unpaired spin on the dot, a virtual process (h) can occur that converts the spin up (a) to the spin down (c) state and transfers an electron from one side of the dot to the other. The ground state of the system is a coherent superposition on the initial and final states shown, creating a spin singlet between the spin on the dot and the spins in the leads. This is called the Kondo effect, and produces a narrow peak of width -k,T, in the den- sity of states at EF in addition to the original broadened level of width T, as shown in (d). intermediate state involving an exchange of electrons with the leads, as is illus- trated in Fig. 24. This is known as the Kondo effect. The local moment pairs with electrons in the metallic electrodes to create a spin singlet. This occurs below a temperature known as the Kondo temperature TK: where r is the level width defined in (33) and E, is indicated in Fig. 24 (E,, < 0). A peak in the density of states of the dot of width kBTK appears at the Fermi energy due to the admixture of states in the electrodes at an energy EF. The Kondo temperature is very small unless the coupling r to the leads is large, since the process involves a virtual intermediate state. Because the Kondo effect involves exchange of electrons with the leads, it causes transmission through the dot even in the blockaded region, as illus- trated in Fig. 24a+. For symmetric harriers and T < T,, the transmission coefficient through the Kondo resonance can be unity, just as for resonant tun- neling. This effect has been seen in transport through qilantum dots and in STM measurements of magnetic impurities on metal surfaces. The Kondo ef- fect was first observed in metals containing magnetic impurities. The forma- tion of a spin singlet hetwecn the magnetic impurities and the conduction electrons enhances the scattering of the electrons. This will be discussed fur- ther in Chap. 22. Cooper Pairing in Superconducting Dots In a small metallic dot made of a superconductor, there is an interest- ing competition between single electron charging and the Cooper pairing of electrons. m7ith an odd number of elcctrons residing on the dot, there is Figure 25 Measnremnnt of coulomb oscillations in a superconducting metallic dot with dccreas- ing temperature. A crossover from e-periodic oscillations to 2s-periodic oscillations is seen as the temperature is lowered due to the Cooper pairing of electrons on the dot. (After M. Tinkham, J. M. Hergmrnther, and J. 6. Lu.) necessarily an unpaired electron. If the Cooper pair binding energy 2A is larger than the charging energy U, it is energetically favorable to add an elec- tron to the dot, paying the energy U in order to gain the pairing energy 2A. The odd-charge states are thus energetically unfavorable. Electrons will be added to the dot in Cooper pairs, and the Coulomb oscillations will be 2e- periodic. This is shown in Fig. 25. This is a re~narkable rnanifcstation of Cooper pairing. VIBRATIONAL AiiD THERMAL PROPERTIES To treat the vibrational propcrties of nanostructures, we will begin from a continuunl description of the elastic properties. This is analogous to employ- ing the band structure as the starting point to describe the electronic proper- ties. It is a good approximation for all but the smallest of nanostr~ictiires. In general, the components of stress and strain in a solid are related by a matrix. A stress along one axis will produce a strain along that axis, but it will aIso produce strains along other axes. For example, a cube stretched along one axis will typically contract somewhat along the orthogonal axes. To si~rlplify the discussion below, we will ignore the off-diagonal elements and treat the stress- strain matrix as diagonal and isotropic. In other words, strains will only occnr along the direction of the stress and the magnitude will be independcnt of the axis direction. For a more complete treatrrlent, we refer the rcader to advanced texts on mechanics. Quantized Vibrational Modes Just as the electronic degrees of freedom are quantized, the vibrational frequencies become discrete in a 1D or O D solid. The continuous low- frequency modes associated with the acoustic modes, w = o,K, becomc instead a series of discrete frequencies o , . The exact frequencies and wavevectors depend in detail on the shape and boundary conditions of'the solid. An illustrative example is the vibrations aronnd the circumference of a thin cylinder of radus K and thickness h < A, as shown in Fig. 26. In Fig. 26a, (a) (b) (c) Figure 26 Funda~ne~ital vibrational modes of a thin-walled cylinder. Image (a) is a longitudinal compressional mode, (h) is the radial breatlring mode (RBM), and (c! is a transverse mode. a quantized longitudinal acoustic mode is schematically illustrated. The al- lowed frequencies can be found by applying periodic b o n n d a ~ conditions around the circumference of the cylinder: Another mode, called the radial breathing mode (RBM), is shown in Fig. 26b. The radius of the cylinder uniformly expands and contracts, producing circumferential tension and compression. From elasticity theory, the elastic energy associated with a strain e for an isotropic medium is given by where Y is the elastic (Young's) modulus. The strain in the cylindcr for a radius change dr is e = dr/K, yielding where V is the volume of the cylinder. Equation (71) has the form of a Hooke's law spring energy, wherc the spring constant is give11 by YV/R2. The vibrational frequency is then where in the last step we have defined a longitudinal sound velocity uL = m. The final class of quantized modes around the circumference, correspond- ing to transverse acoustic modes, are shown in Fig. 26c. Their wavevectors and frequencies are given by Note the frequency scales like K;. The origin of this behavior will he discussed further below. Qnantizcd vibrational modes can be measured in a variety of ways. One technique widely used to probe the vibrational structure of individual nanoscale objects is Raman spectroscopy (see Chap. 15). Raman spectroscopy of single nanotubes is shown in Fig. 27. For a nanotube, tiL = 21 km/s, and the energy of the radial breathing mode is, from (721, The measured values are in good agreement with this expression. As a result, measurements of the RBM can be used as a diagnostic to infer the radius oia nanotube. 18 Nanostructures 559 100 150 200 250 300 350 Frequency (cm-') Figure 27 Raman spectra of individual carbon nanotubes. The radial breathing mode frequencies are labeled in the main panel, along with the struc- tural assignments (n, m). Note: 160 cm-' - 20 meV (After A. Jorio et al.) Transverse Vibrations We now address the phonons propagating in the direction of the axis of a long, thin object, such as the cylinder discussed in the previous section or a thin, solid beam (Fig. 28). The longitudinal phonons are similar to the 3D case, with a dispersion o = uLK where K is the continuous wavevector. How- ever, there is a fundamental modification of the transverse phonons at wave- lengths longer than the thickness h of the beam. Instead of shearing, as in a bulk transverse phonon, the solid bends, as shown in Fig. 28a. This is the classical problem of transverse flexural waves on a beam. The energy of bend- ing comes from the longitudinal compression/stretching of the solid along the innerlouter arcs of the bend. The linear wavevector dependence w , = uTK characteristic of a bulk solid is changed to hspersion quadratic in K, as we show below. Consider a transverse standing wave on a solid rectangular beam of thickness h, width w, and length L whose displacement is given by y(z,t) = yo cos(Kz - ot). The strain at a given point inside the beam is given by the local curvature and the distance t from the center line of the beam (Fig. 28a): e = -(a2ylaz2)t = P y t . (74) The total energy associated with this strain is again given by (70): where (y2) is averaged over one period of the oscillation. This again gives an effective spring constant and, from steps analogous to (70-72), an oscillation frequency: wr = u L h ~ 2 ~ ~ . (76) Figure 28 (a) Stresses in a bent beam, showing the inner portion under compression while the outer portion is under tension. (b) SEM micrograph of a series of suspended Si beams of varying lengths L and the measured resonance frequency as a function of L. The line is a £it to the func- tional form f = B/L2, where B is a constant. (After D. W. Carr et al.) Note that the frequency depends on the longitudinal sound velocity and not the transverse sound velocity, since the mode is now essentially compressional in nature. It is no longer linear in K since the effective restoring force grows stronger with increasing curvature, i.e., increasing K. In contrast, the torsional mode, correspondng to a twist of the beam along its length, retains its shear character and, o,,, Transverse vibrational modes described by (76) are frequently observed in microscale and nanoscale beams. A set of nanoscale beams constructed in Si using electron beam lithography and etching is shown in Fig. 28b. The frequencies associated with fundamental resonance K , = 2vlL of these beams scale as 1/L2 (Fig. 28c), as expected from (76). Note that the modification of the dispersion relation (76) for long wave- - length transverse modes is not restricted to nanoscale systems. The only re- quirement is that the system be in the geometry of a thin beam or slab with transverse dimension h smaller than the wavelength, i.e., Kh < 1. For exam- ple, AFM cantilevers operated in the noncontact mode, as discussed above, are well described by this relation. The dispersion relation (76) is also related to the w - f ? dependence seen in (73) for the class of modes shown in Fig. 26c. Both describe the transverse flexural vibrations, one of a beam and the other of a thin shell. A revolution is underway in the fabrication of small, complex mechanical structures using techniques adapted from microelectronic processing. The beams shown in Fig. 28b are a simple example. These structures can be inte- grated with electronic devices, creating rr~icroelectromechanical systems (MEMs) and nanoelectromechanical systems (NEMs). They are being ex- plored for a variety of applications, including scnqing, data storage, arid signal processing. Heat Capacity and Thermal Transport The above relations indicate that the quantized vibrational mode energies are typically less than k,T at room temperature except in the very smallest structures. Modes along the confined directions will thus he thermally excited at room temperature. As a result, the lattice thermal properties of nariostruc- tures will he similar to their bulk counterparts. In particular, the lattice heat capacity and thermal co~lductivit~ will be proportional to T ~ , as for a 3D solid (Chap. 5). At low temperatures, however, vibrational excitations of frequency o in the corlfi~ied directions xyill freeze out when T < hwlk,. In the case of a long, thin structure, the system will behave as 1D tlier~nal system at siifficiently low temperatures, with a set of 1D phono~l subhands analogoils to the 1D elec- tronic suhhands shown in Fig. 9. A calculation analogoils to the one perfomled in Chap. 5 for a 3D solid yields the heat capacity per 1D acoustic phorlon sub- band with a dispersio~l w = uk (Prob. 6): The thermal conductance Gth of the wire is defined as thc ratio of the net en- ergy flow through the wire divided by the temperatrlre difference AT between its ends. The thermal conductance per 1D phonon subband CjiD' is This result is derived in Prob. 6 using an approach analogous to the one em- ployed to derive the Landauer formula for conductance of a 1D channel. Note 3 is now the transmission probability for phonons through the structure. For a transverse flexural node, where o p, the result (77) is modified, but (78) is the same. Both (77) and (78) are linear in temperature, in contrast to the T3 result for 3D. The difference reflects the nurrllrer of modes with energies ko < k,T, or equivalently, with wavevectors K < kBT/hu. The number of modes scales like KD, where D is the dimensionality, producing T3 in 3D and Tin ID. Note that in the case of perfectly transmitted phonons through the chan- nel, the thermal conductance (78) is determined only by fundamental con- stants and the absolute temperature. This result is analogous to the quantized electronic conductance (21) of a 1D channel, which was independent of the electron velocity in the channel. Both the ballistic thermal conductance (78) and the 1D for~n of the heat capacity (77) have been observed in experiments on narrow wires at very low temperatures. SUMMARY Real spacc probcs can give atomic-scale images of nanostructures. The density of states of a ID subband, D(E) = 4L/ho, diverges at the subband thresholds. These arc called van Hove singularities. The electrical conductancc of a 1D system is given by the Landauer formula, G = (2e2/h)3, where 3 is the transmission coefficient through thc sample. The conductance of a quasi-1D system can he strongly influenced by quan- tum interference among the electron paths traversing the sample, leading to resonant tunneling, localization, and the Aharonov-Bohm effect. The optical properties of a quantum dot can be tuned by changing its size and hence its quantized energy levels. Adding an extra charge e to a quantum dot requires an additional electro- chemical potential given by U + A&, where U is the charging energy and A s is the level spacing. The vibrational modes of a nano~neter-scale object are quantized. Problems 1. Carbon nanotube band structure. Figure 29 shows the graphene lattice with the primitive lattice translation vectors of length a = 0.246 nm, along with the first Brillouin zone. (a) Find the set of reciprocal lattice vectors G associated with the lattice. (b) Find the length of the vectors K and K' shown in Fig. 29 in terms of a. For energies near the Fermi energy and wavevectors near the K point, thc 2D band structure can be approximated as where v, = 8 X lo5 d s . A similar approximation holds near the K' point. Consider a tube rolled up along the x-axis with a circumference nu. By applying periodic boundary conditions along the rollcd up direction, the dispersions nf the 1D sub- bands near the K point can bc found. (c) Show that, if n is divisible by 3, there exists a "massless" subband whosc cncrgy is linear in Aky. Sketch this sr~hband. These nanotubes arc 1 1 ) metals. (d) If n is not divisihle by 3, the s~~hl~and structure is that 18 Nanostructures 563 Figure 29 (a) The graphene lattice and (h) the first Brillouin zone of the graphene lattice showing the conical dis- persion of the energies near the K and K' points. as shown in Fig. 10. For the case of n = 10, find the magnitude of the semiconduct- ing bandgap E,, in eV and show that E ~ ~ / E ~ ~ = 2. (e) Again for the n = 10 case, show that the dispersion relation of the lowest electron subband is of the form of a rela- tivistic particle, E ' = (mc2)' + (p~)2, where v, plays the role of the speed of light, and find the ratio of effective mass rn" to the free electron mass m . 2. Filling subbands. For electrons in a square GaAs wire of width 20 nm, find the linear electron density at which the n, = 2, n, = 2 subband is first populated in equilibrium at T = 0. Assume an infinite confining potential at the wire boundary. 3. Breit-Wigner form of a transmission resonance. The purpose of this problem is to derive (33) from (29). (a) By expanding the cosine for small phase differences away from resonance, 69 = 9 ' - 2m, find a simplified form of (29) involving only lt,I2, lt2I2, and 69. (b) Show that, for states in a 1D box, the following relation holds between small changes in the phases and small changes in the energy: S ~ l h s = Sp/2~r, where AE is the level spacing. (c) Combine (a) and (b) to obtain (33). 4. Barriers in series and Ohm's Law. (a) Derive (36) from (35). (b) Show that the 1D Drude conductivity a , , = n,,e2r/m can be written as ulD = (2e2/h)tB. (Note: The momentum relaxation rate and the backscattering rates are related as 117 = 217, because the former corresponds to the relaxation from p to 0 while the latter corre- sponds to relaxation from p to -p.) 5. Energies of a spherical quantum dot. (a) Derive the formula (63) for the charg- ing energy. (b) Show that, for d 4 R, the result is the same as that obtained using the parallel plate capacitor result, C = ecOA/d. (c) For the case of an isolated dot, d + m, find the ratio of the charging energy to lowest quantized energy level. Express your answer in terms of the radius R of the dot and the effective Bohr radius a;. 6. Thermal properties in ID. (a) Derive the formula (77) for the low temperature heat capacity ofa single 1D phonon mode within the Debye approximation. (b) Derive the relation for the thermal conductance (78) of a 1D phonon mode between two reser- voirs by calculating the energy flow out of one reservoir at a temperature T, and subtracting the energy flow from the other reservoir at a temperature T,. Use an ap- proach analogous to that used to obtain (20) and (24) for the electrical conductance. Noncrystalline Solids DIFFRACTION PATIERN Monatomic amorphous materials Radial distribution function Structure of vitreous silica, SiO, GLASSES Viscosity and the hopping rate AMORPHOUS FERROMAGNETS AMORPHOUS SEMICONDUCTORS LOW ENERGY EXCITATIONS IN AMORPHOUS SOLIDS IIeat capacity calculation Thermal conductivity FIBER OPTICS Rayleigh attenuation PROBLEM 1. Metallic optic fibres? The terms amorphous solid, noncrystalline solid, disordered solid, glass, or liquid have no precise struct~iral meaning beyond the description that the structure is "not crystalline on any significant scale." Tlie principal structural order prcscnt is imposed by the approximately constant separation of nearest- neighbor atoms or molecules. We exclude from the present discussion disor- dered crystalline alloys (Chapter 22) where different atoms randomly occupy the sites of a regular crystal lattice. DIFFRACTION PATTERN The x-ray or neutron diffraction pattern of an amorphous material sucli as a liquid or a glass consists of one or more broad diffuse rings, when viewed on the planc normal to the incident x-ray beam. The pattern is different from the diffraction pattern of powdered crystalline material which shows a large num- ber of fairly sharp rings as in Fig. 2.17 of Chap. 2. The result tells us that a liq- uid does not have a unit of structure that repeats itself identically at periodic intervals in three dimensions. In a simple monatomic liquid the positions of the atoms show only a short range structure referred to an origin on any one atom. We never find the cen- ter of anothcr atom closer than a distance equal to the atomic diameter, but at roughly this distance we expect to find about the riuniber of nearest-neighbor atoms that we find in a crystalline forrn of the material. Although the x-ray pattern of a typical amorphous material is distinctly different from that of a typical crystalline material, there is no sharp division between them. For crystalline powder samples of smaller and smaller particle size, the powder pattern lines broaden continuously, and for small enough crystalline particles the pattern becomes similar to the amorphous pattern of a liquid or a glass. From a typical liquid or glass diffraction pattern, containing three or four diffuse rings, the only quantity which can be determined directly is the radial distribution function. This is obtained from a Fourier analysis of the experi- mental x-ray scattering curve, and gives directly the average number of atoms to be found at any distance from a given atom. The method of Fourier analysis is equally applicable to a liquid, a glass, or a powdered crystalline material. It is convenient to begm the analysis of the diffraction pattern with Eq. (2.43). Instead of writing it for the structure factor of the basis, we write the sum for all the atoms in the specimen. Further, instead of specializing the scat- tering to the reciprocal lattice vectors G characteristic of a crystal, we consider arbitray scattering vectors Ak = k' - k, as in Fig. 2.6. We do this because scattering from amorphous materials is not limited to the reciprocal lattice vectors, which in any event cannot here be defined. Therefore the scattered amplitude from an amorphous material is de- scribed by withf, the atomic form factor of the atom, as in Eq. (2.50). The sum runs over all atoms in the specimen. The scattered intensity at scattering vector Ak is given by in units referred to the scattering froni a single electron. If n denotes the anglc between Ak arid r,,, - r,,, then where K is the magnitude of Ak and r , , is the magnitude of r,,, - r,,. In an amorphous specimen the vector r,,, - r,, rnay take on all orientations, so we werage the phase factor over a sphere: 1 (exp(iKr cos a)) = tl(cus a) exp(iKrn, cos cu) sin Kr,, (4) - -- h i , , , ' Thus we have the Debye result lor the scattered intensity at Ak: Monatomic Amorphous Materials For atoms of ur~ly one type, we letf,,, = f , = f and separatr out from the surr~rnatiun (5) the terms with n = m For a spccimrn of hT atom?, (sin Krm)/Krm . 1 The sum runs over all atoms rn except the origin atom m = n. If p(r) is the concentration of atoms at distance r from a reference atom, we can write (6) as 19 Noncrystalline Solids where R is the (very largc) radius of the specimen. Let p, denote the average concentration; then (7) may be written as R I(K) = Nf{i + ioR &- 4w2p(r) - pn/& + (pdiX) dr 4 m sin Kr 0 The second integral in (8) gives the scattering from a uniform concentration and may be neglected except in the forward region of very small angles; it re- duces to a delta function at K = 0 as R + W . Radial Distribution Function It is convenient to introduce thc liquid structure factor defined by S(K) = I/Np . (9) Note that this is not at all the same as S(Ak) in (1). From (8) we have, after dropping the delta function contribution, S(K) = I + lom dr l ~ ' ~ ( r ) - &/Kr . (10) We define the radial distribution function g(r) such that p(r) = g(r)pn Then (10) becomes because (sin Kr)/Kr is the spherically syrnrnetric or s term in the expansion of exp(iK . r). By the Fourier integral theorem in three dimensions, This resi~lt allows us to calc~ilate the radial distribution function g(r) (also called the two-atom correlation function) from the measured structure factor S(K). One of the simplest liquids well suited to x-ray diffraction study is liquid sodium. The plot of the radial distribution 4rPp(r) vs. r is given in Fig. 1, to- gether with the distribution of neighbors in crystalline sodium. Figure 1 (a) Radial distribution cuwe 4nPp(r) for liquid sodium, (h) verage density 4?iPp,. (c) Distribution of neighbors in crystalline sodium. (After Tarasov and Warren.) Structure of Vitreous' Silica, SiO, Vitreous silica (fi~sed quartz) is a simple glass. The x-ray scattering curve is given in Fig. 2. The radial distribution curve 4n-rZp(r) vs. r is gven in Fig. 3. Because there are two kinds of atoms, p(r) is actually the superposition of two electron concentration curves, one about a silicon atorr~ as origin and one about an oxygen atom as origin. The first peak is at 1.62 A, close to the average Si-0 distance found in crystalline silicates. The x-ray workers conclude from thc intensity of the first peak that each silicon atom is tetrahedrally surrollnded hy four oxygen atoms. The relativc proportions of Si and 0 tell us that each 0 atom is bonded to two Si atoms. From the geometry of a tetrahedron, the 0-0 distance should be 2.65 A, compatible with the distance suggested by the shoulder in Fig. 3. The x-ray results are consistent with the standard model of an oxide glass, due to Zachariasen. Figure 4 illustrates in two dimensions the irregular struc- ture of a glass and the regularly repeating structure of a crystal of identical chemical composition. The x-ray results are completely explained by picturing glassy silica as a random network in which each silicon is tetrahedrally sur- rounded by four oxygens, each oxygen bonded to two silicons, the two bonds to an oxygen being roughly diametrically opposite. The orientation of one tetrahedral group with respect to a neighboring group about the connecting 19 Noncrystalline Solids Figure 2 Scattered x-ray ilitcl~sity vs, scattering anglc 0, lor vitreous SiO,. (After B. E. Warren.) Si-0-Si bond can be practically random. There is a definite structural scheme involved: each atom has a definite number of nearest neighbors at a definite distance, but no unit of structure repeats itself identically at regular intervals in three dimmsions, and hence the material is not crystalline. It is not possible to explain the x-ray results by assuming that vitreous silica consists of very small crystals of some crystalline form of quartz, such as cristoballite. Small angle x-ray scattering is not observed, but would be ex- pected f r o ~ r ~ discrete particles with breaks and voids between them. The scheme of bonding in glass must be essentially continuous, at least for the major part of the material, although the schcme of coordination about each atom is the same in vitreous silica and in crystalline cristoballite. The low ther- mal conductivity of glasses at room temperature, as discussed below, also is consistent with the continuous random network model. A comparison of experimental and calculated x-ray intensity results for anlorphous gernlaniurn is shown in Fig. 5. The calculations are for a random network model and for a microcrystallite model. The latter model gives a very 0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 r i n k Figure 3 Radial distribution curve for vitreous SiO,, as the Fourier transform of Fig. 2. The posi- tions of the peaks give the distances of atoms from a silicon or an oxygen. From the areas under the peaks it is possible to calculate the number of neighbors at that distance. The vertical lines in- dicate the first few average interatomic distances; the heights of the lines are proportional to the peak areas. (After B. E. Warren.) (a) (b) Figure 4 Schematic two-dimensional analogs illustrating the differences between: (a) the regularly repeating structure of a crystal and (b) continuous random network of a glass. (After Zachariasen.) 19 Noncrystalline Solids 573 Scattering vector 4~ sin O/A in .
k l Figure 5 Comparison of experimental (dashed curve) and calculated (solid curve) reduced inten- sity function for amorphous germanium. (a) Amorphous germanium compared with microciystal- lite model. (b) Amorphous germanium compared with random network model. (Results by J. Graczyk and P. Chaudhari.) poor agreement. The random network model is supported for amorphous sili- con by studies of the band gap and spectroscopic work on the 2p shell. GLASSES A glass has the random structure of the liquid from which it is derived by cooling below the freezing point, without crystallization. Also, a glass has the elastic properties of an isotropic solid. By general agreement, we say that a liquid on being cooled becomes a glass when the viscosity equals 1013 poise, where a poise is the CGS unit of viscosity.' This defines the glass transition temperature Tg. At temperatures above Tg we have a liquid; below Tg we have a glass. The transition is not a thermodynamic phase transition, only a transition for "practical purposes." 'The SI unit of viscosity is 1 Nsm-', so that 1 poise = 0.1 Nsm-'. It is quite common to find viscosities given in cp or centipoise, being lo-' poise. The value 1013 poise used to define T , is arbitrary, but not unreasonable. If we bond a slab of glass 1 cm thick to two plane parallel vertical surfaces, thc glass will flow perceptibly in one year under its own weight when the viscosity drops below 1013 poise. (For comparison, the viscosity of the mantle of the earth is of the order of loz2 poise.) Relatively few liquids can be cooled fast enough in the bulk to form a glass before crystallization intervenes. Molecules of most substances have high enough mmobility in the liquid so that on cooling a liquid-solid melting transi- tion occurs a long time before the viscosity increases to 1013 poise or 1 0 1 5 cp. Liquid water has a viscosity 1.8 cp at the freezing point; the viscosity increases enormously on frccxing. We can often make a glass by depositing a jet of atoms of a substrate cooled to a low temperature, a process which will sometimes produce an amorphous layer with glasslike properties. Amorphous ribbons of some metal alloys may be produced in this way in industrial quantities. Viscosity and the Hopping Rate Thc viscosity- of a liquid is related to the rate at which molecules undergo thermal rearrangement on a local scale, as by hopping into a vacant neighbor site or by interchange of two neighbor molecules. The physics of the transport process is somewhat different from that of viscosity in the gas phase, but the gas phase result gives a qualitative lower limit to the viscosity of the liquid phase, a limit that applies to nearest-neighbor hopping of atoms. The gas result (TP 14.34) is where 1 ) is the viscosity, p the density, i ? the mean thermal velocity, and e the mean free path. In the liquid I is of the order of magnitude of the intermole- cular separation a. With "tpical" values p = 2 g cm-" r = lo5 cm s-'; a ; -. 5 X cm, we have as a 1 1 estimate of the lower limit ofthe viscosity of a liquid. (Tahles in chemical handbooks only rarely list valucs below this.) \Ve givc now a very simple model of the viscosity of a liquid. In order to hop siiccessfillly, a molecule must surmount the potential energy barrier pre- sented by its neighbors in the liquid. The preceding esti~nate of the minimum viscosity applies when this barrier may be neglected. If the barrier is of height E, the molecule will have sufficient thermal energy to pass over thc barrier only a fraction 19 Noncrystalline Solids of the time. Here E is an appropriate free energy and is called the activation energy for the process that determines the rate of hopping. It is related to the activation energy for self-diffusion. The viscosity will be increased as the probability of successful hopping is decreased. Thus If 77 = 10'' poise at the glass transition, the order of magnitude off must be at the transition, using (15). The corresponding activation energy is If Tg = 2000 K , then k,Tg = 2.7 X 10-'"rg and E = 9.6 X 10-l2 erg - 6 eV. This is a high potential energy barrier. Glasses with lower values of T, will have correspo~ldingly lower values of E. (Activation energies obtained in this way are often labeled as Ens,.) Materi- als that are glass-lormers are characterized by activation energies of the order of 1 eV or more; non-glass-formers may have activation encrgies of the order of 0.01 cv. When being pressed into molds or drawn into tubes, glass is used in a range of temperatures at which its viscosity is 1090 10"oises. The working range for vitreous silica begins over 2000°C, so high that the practical useful- ness of the rriaterial is severely limited. In corrlrnon glass, about 25 percent of NazO is added as a network modifier to SiOe in order to reduce below 1000°C the temperature needed to make the glass fluid enough for the fbrming opera- tions necdcd to make electric lamp bnlhs, window glass, and bottles. AMORPHOUS FERROMAGNETS Amorphous metallic alloys are formed by very rapid quenching (cooling) of a liquid alloy, commonly by directing a molten stream of the alloy onto the surface of a rapidly rotating dnim. This process produces a continuous "melt- spun" ribbon of amorphous alloy in industrial quantities. Ferromagnetic amorphous alloys were developed because amorphous ma- terials have nearly isotropic properties, and isotropic materials should have es- sentially zero magnetocrystalline anisotropy energy. The absence of directions of hard and easy magnetization should result in low coercivities, low hysteresis , - losses, and high permeahilities. Because amorphoiis alloys are also random al- loys, their electrical resistivity is high. All these properties have technological value for application as soft magnetic materials. The trade name Metglas is at- tached to several of these. The transition metal-metalloid (TM-M) alloys are an important class of magnetic amorphous alloys. The transition metal component is usually about 80 percent of Fc, Co, or Ni, with the metalloid component B, C, Si, P, or Al. The presence of the metalloids lowers the melting point, making it possible to quench the alloy through the glass transition temperature rapidly enough to stabilize the amorphous phase. For example, the co~uposition FesnBzo (known as Metglas 2605) has T, = 441°C, as compared with the melting temperaturc 1538°C of pure iron. The Curie temperature of this composition in the amorphous phase is 647 K, and the value of the magnetization M, at 300 K is 1257, compared with T, = 1043 K and M, = 1707 for pure iron. The coercivity is 0.04 (3, and the maximum value of the permeability is 3 x 10'. Coercivities as low as 0.006 G have been reported for another composition. High coercivity materials can be produced by the same melt-spin process if the spin rate or quench rate is decreased to produce a fine-grained clys- talline phase, which may be of metastable composition. If the grain size is arranged to match the optimnm size for single domains, the coercivity can be Figure 6 Coercivity at room temperature vs, melt-spin vclocity u,~ for Sm,,,Fe,,. The madmum coercivity is 24 kC: and occurs at 1.65 m s-', which is helieved to correspond to si~~gle du~rrairr bc- havinr in each crystallite. At higher spin rates the coerci~ty decreases because the deposited ma- terial hecomes amorphous (more isutrupic). At lower spin ratcs the cr).stallites anneal to sizes above the single domain regime; domain boundaries give a Inwer coercivity. (After J. L. Croat.) 19 Noncrystalline Solids quite high (Fig. 6). J. L. Croat has reported H, = 7.5 k G for the metastable alloy Nd,,,Fe,, at the optimum melt-spin velocity 5 m sC1. AMORPHOUS SEMICONDUCTORS Amorphous semiconductors can be prepared as thin films by evaporation or sputtering, or in some materials as bulk glasses by supercooling the melt. What happens to the electron energy band model in a solid without regu- lar crystalline order? The Bloch theorem is not applicable when the structure is not periodic, so that the electron states cannot be described by well-defined k values. Thus, the momentum selection rule for optical transitions is relaxed; hence all infrared and Raman modes contribute to the absorption spectra. The optical absorption edge is rather featureless. Allowed bands and energy gaps still occur because the form of the density of states vs. energy is determined most strongly by local electron bonding configurations. Both electrons and holes can carry current in an amorphous semiconduc- tor. The carriers may be scattered strongly by the disordered structure, so that the mean free path may sometimes be of the order of the scale of the disorder. Anderson proposed that the states near band edges may be localized and do not extend through the solid (Fig. 7). Conduction in these states may take place by a thermally-assisted hopping process, for which the Hall effect is anomalous and cannot be used to determine the carrier concentration. Two distinct classes of amorphous semiconductors are widely studied: tetrahedrally-bonded amorphous solids such as silicon and germanium, and the chalcogenide glasses. The latter are multicomponent solids of which one major constituent is a "chalcogen" element-sulfur, selenium, or tellurium. The tetrahedrally-bonded materials have properties similar to those of their crystalline forms, provided the dangling-bond defects are compensated with hydrogen. They can be doped with small amounts of chemical impurities, and their conductivity can be sharply modified by injection of free carriers from a metallic contact. By contrast, the chalcogenide glasses are largely in- sensitive to chemical impurities and to free carrier injection. Amorphous hydrogenated silicon is a candidate material for solar cells. Amorphous silicon is a much less expensive material than single crystal silicon. Figure 7 Density of electron states as believed to occur in amorphous solids, when states are non-localized in the center of the band. Localized states are shown shaded. The mobil- ity band edges E,, E: separate the ranges of energy where states are localized and non-localized. (After N. Mott and E. A. Davis.) Attempts at using pure amorphous silicon, however, failed because of struc- tural defects (dangling bonds) which were impossible to eliminate. hltroduc- tion of hydrogen into amorphous silicon appears to remove the undesirable structural defects. Relatively large proportions of hydrogen are incorporatcd, of the order of 10 percent or more. LOW ENERGY EXCITATIONS IN AMORPHOUS SOLIDS The low temperature heat capacity of pure dielectric crystalline solids is lolown (Chapter 5) to follow the Debye de law, precisely as expected from the ex- citation of long wavelength phonons. The same behavior was expected in glasses and other amorphous solids. However, many insulating glasses show an nnex- petted linear term in the heat capacity below 1 K. Iudccd, at 25 mK the observed heat capacity of vitreous silica exceeds the Debye phonon contribution by a fac- tor of 1000. Anomalous linear terms of comparable magnitude are found in all, or nearly all, amorphous solids. Their presence is believed to be an intrinsic conse- quence of the amorphous states of matter, but the details of why this is so remain unclear. There is strong e d e n c e tlrat the a~lonlalous properties arise from two- level systems an3 not from multi-level oscillator systems; in brief, the evidence is that the systems can be saturated by intense phonon fields, just as a two-level spin system can be saturated by an intcnsc rf magnetic field. Heat Capacity Calculation Consider an amorphous solid with a concentration N of two-level systems at low energies; that is, with a level splitting A much less than the phonon Debye cutoff k , 8 . The partition function of one system is, with 7 = k8T, The thermal average energy is U = -:A tanh(M2.r) , and the heat capacity of the single system is Cv = kB(au/a7) = kB(&27)2 sech2(N2~) . (22) These results are given in detail in TP, pp. 62-63. Now suppose that A is distributed with uniforrrl probability in the range A = 0 to A = A,. The average value of Cy is The integral cannot be evaluated in closed form 18 Noncrystalline Solids Tho limits are of special interest. For T < A,, the sechZx term is roughly 1 from x = 0 to x = 1, and roughly zero for x > 1. The value of the integral is roughly 1/3, whence C, = 2kgT/3h, , (24) for T < A,,lkJk,. For T a Ao, the value of the integral is roughly $ ( A , , l 2 k , ~ ) ~ , so that in this limit which approaches zero as T increases. Thus the interesting region is at low temperatures, for here by (24) the two-level system contributes to the heat capacity a term linear in the tempera- ture. This term, originally introduced for dilute magnetic impurities in metals, has no connection with the usual conduction electron heat capacity which is also proportional to I : The empirical result appears to be that all disordered solids have N - 10" cm-3 "new type" low energy excitations uniformly distributed in the energy intend from 0 to 1 K. The anomalous specific heat can now be ob- tained from (24). For T = 0.1 K and AdkB = 1 K, Cr: = g ~ k ~ ( 0 . 1 ) .
= 1 erg c K 3 K-I . (26) For cornparison, the phonon contribution at 0.1 K is, from (5.35), -; 2.8 X lo-' erg cm-3 K-' , much smaller than (26). The experimental results (Fig. 8) for vitreous SiOz are represented by where c, = 12 erg g-I K-2 and c3 = 18 erg g-' K-4. Thermal Conductivity The thermal conductivity of glasses is very low. It is limited at room tem- perature and above by the scale of the disorder of the structure, for this scale determines the mean free path of the dominant thermal phonons. At low tem- peratures, below 1 K, the conductivity is carried by long wavelength phouons and is limited by phonon scattering from the mysterious two-level systems or tunneling states discussed earlier for their contribution to the heat capacity of amorphous solids. As in Chapter 5, the expression for the thermal conductivity K has the form K = $cut , (29) where c is the heat capacity per unit volume, v is an average phonon ve1ocit)i and e is the phonon mean free path. For vitreous silica at room temperature, K -- 1.4 X lo-? J cm-Is-' K-' ; c--1.6Jcm 3~ ; (u) -- 4.2 X 10; cm s-' . Thus the Inean free path € - 6 X lo-' cm; by reference to Fig. 3 we see that this is of the order of magnitude of the disordcr of thc structure. Figure 8 Heat capacity of vitreous silica and soda silica glass as a function of temperature. The heat capacity is roughly lincar in T bclow 1 K. The dashed line represents the calculated Debye heat capacity of vitreous silica. L f t + R Figure 9 Short phonon mean free path in a disordered structure. A short wavelength phonon that dirplaces aton1 L, as diown, will displace atom R by a much smaller distance, because of the phase cancellation of the upper and lower paths from L to R. The displacement of R is T + - 0, so that the wave incident from L is reflected at 1A. 19 Noncrystalline Solids 581 This value of the phonon mean free path is remarkably small. At room temperature and ahove (that is, above the Debye temperature), most of the phonons have half-wavelengths of the order of the interatomic spacing. It is through phase cancellation processes, as in Fig. 9, that the mean free path is limited to several interatomic spacings. No other structure for fused qllartz will give a 6 A mean free path. The normal modes of vibration of the glass structure are utterly unlike plane waves. Rut the modes, as distorted as they are, still have quantized amplitudes and therefore may be called phonons. FIBER OPTICS Fibers of silica-based lightguides carry a high proportion of the data and information transmitted on the surface of the earth and under the seas. The optical fibers consist of a thin core (=I0 pm) of high-refractive index glass surrounded by a cladding. The digital data are carried by light, with a mini- mum attenuation near 0.20 db km-' at wavelengths near 1.55 pm, which is in the infrared (Fig. 10). A range of 100 km corresponds to a loss of 20 db, power readily supplied by an Eu3+ laser amplifier. The optic window of high-p~~rity glasses near this wavelength is limited on the low frcq~~cncy side hy phonon absorption bands and on the high frequency side hy Rayleigh scattering, and, ultimately, by electronic absorption. In the Figure 10 The transmission characteristic of communication-quality optical fibers, showing the attenuation in units of decibels per k111 as a function of the wavelength of light, in pm. The Kayleigh scattering regime is dominant on the left of the curve, except for a strong in~purity ah- sorption line associated with OH ions that accompany SiO,: the line is the second harmonic of a line at 2.7 pm known as the "water line." Tlle wavelcrrgth nlarked at 1.31 pm is used in 1994 trans- mission lines: it was replaced by the wavelength 1.55 pm available from Eu" i i o ~ amplifiers, which arc uscd cvc~y 100 km in t).pical long distance applications. The power needed to pump the amplifiers is snpplied hy copper wires. (Courtesy of Tingyc Li, AT&T Bell Laboratories.) optic window the losses are determined by the Rayleigll scattering intrinsic to static fluctuations in the local dielectric constant of an inhomogeneous medium, and the attenuation varies as the fourth power of the freqitency. It is fortunate that an excellent source is available for radiation at 1.55 pm. As shown in Fig. 13.24, excited (pumped) erbium Er3+ ions can amplify in an erbium-doped section of fiber. Ray legh Attenuation The attenuation of light waves in glass is dominated at wavelengths in the infrared by the same scattering process, called Rayleigh scattering, that is re- sponsible for the blue light of the sky. The extinction coefficient, or attenua- tion coefficient, h, has the dimension of reciprocal length and for light scat- tered in a gas is given, after Rayleigh, by where n is the local refractive index and N is the number of scattering centers per unit volume. The energy flux as a function of distance has the form exp( -hx). Derivations of (30) are found in good texts on electrodynamics; the struc- ture of the result may be understood by a general argument: The radiant en- ergy scattered from a dipole element p is proportional to (dp2/dt")", and this accounts for the factor u4. The local polarizability a enters as a2; if there are N random scattering centers per unit volume, the scattered energy averaged over these random sources will go as N((A(~)~), or ((An)')lN. Thus we have the es- sential factors that appear in (30). As applied to a glass, An should refer to the variations in polarization around each group of Si-0 honds, and satisfactory numerical estimates of the attenuation may he made in this way. Problem 1. Metallic optic fibres? It has been specnlated that metallic wires can act as optic fibres, transmitting light ulth a lorrg delay appropriate to the high refractive index characteristic of metals. Unfortunately the refractive index of a t,ypical metal is dominated by a free-electron term in i ' " , so that the propagatiorr of a light wave is in fact highly damped in a metal. Show that in sodium at roo111 te~~~perature a wave of vacuum wavelength 10 p m will have a damping length of 0.1 y n . This may he con- trasted with the 100 km damping length found for light in high-quality glass fibres. Point Defects LAnICE VACANCIES DIFFUSION Metals COLOR CENTERS F centers Other centers in alkali halides PROBLEMS 1. FrenkeI defects 2. Schottky vacancies 3. F center P8S .u8!s aj!soddojo sans ~ u e 3 e ~ j o ~!ed paldnoa e pne 'saJ7s no: a~!$!sod lueaen. o w Buwoys '~ejsho ap!pq ![eyp a ~ n d e jo aneld v 1 a~nZ!a CHAPTER 20: POINT DEFECTS The common point imperfections in crystals are chemical impurities, vacant lattice sites, and extra atoms not in regular lattice positions. Linear im- perfections are treated under dislocations, Chapter 21. The crystal surface is a planar imperfection, with electron, phonon, and magnon surface states. Some important properties of crystals are controlled as much by imperfec- tions as by the composition of the host crystal, which may act only as a solvent or matrix or vehicle for the imperfections. The conductivity of some semicon- ductors is duc entirely to trace amounts of chemical impurities. The color and luminescence of many crystals arise from impurities or imperfections. Atomic diffusion may be accelerated enormously by impurities or imperfections. Mechanical and plastic properties are usually controlled by imperfections. LATTICE VACANCIES The simplest imperfection is a lattice vacancy, which is a missing atom or ion, also known as a Schottky defect. (Fig. 1). We create a Schottky defect in a perfect crystal by transferring an atom from a lattice site in the interior to a lattice site on the surface of the crystal. In thermal equilibrium a certain num- ber of latticc vacancies are always present in an otherwise perfect crystal, be- cause the entropy is increased by the presence of disorder in the structure. In metals with close-packed structures the proportion of lattice sites va- cant at temperatures just below the melting point is of the order of to 1 0 -! But in sorne alloys, in particular the very hard transition metal carbides such as Tic, the proportion of vacant sites of one component can be as high as 50 percent. The probability that a gven site is vacant is proportional to the Boltzmann factor for thermal equilibrium: P = exp(-EIT/kBT), where Ev is the energy re- quired to take an atom from a lattice site inside the crystal to a lattice site on the surface. If there are N atoms, the equilibrium number n of vacancies is given by the Boltzrnann factor If n < N, then nlN -- exp(-E,/k,T) . If E , = 1 eV and T = 1000 K, then n/N = e-" .
; ; The equilibrium concentration of vacancies decreases as the tempera- ture decreases. The actual concentration of vacancies will be higher than the Figure 2 Schottky and Frenkel defects in an ionic crystal The arrows indicate the displacement of the ions. In a Schottky defect the ion ends up on the surface of the crystal; in a Frenkel defect the ion is removed to an interstitial position. equilibrium value if the crystal is grown at an elevated temperature and then cooled suddenly, thereby freezing in the vacancies (see the discussion of diffu- sion below). In ionic crystals it is usually energetically favorable to form roughly equal numbers of positive and negative ion vacancies. The formation of pairs of va- cancies keeps the crystal electrostatically neutral on a local scale. From a sta- tistical calculation we obtain for the number of pairs, where Ep is the energy of formation of a pair. Another vacancy defect is the Frenkel defect (Fig. 2) in which an atom is transferred from a lattice site to an interstitial position, a position not nor- mally occupied by an atom. In pure alkali halides the most common lattice va- cancies are Schottky defects; in pure silver halides the most common vacancies are Frenkel defects. The calculation of the equilibrium number of Frenkel defects proceeds along the lines of Problem 1. If the number n of Frenkel defects is much smaller than the number of lattice sites N and the number of interstitial sites N', the result is where E, is the energy necessary to remove an atom from a lattice site to an interstitial position. Lattice vacancies are present in alkali halides when these contain addi- tions of divalent elements. If a crystal of KC1 is grown with controlled amounts of CaCl,, the density varies as if a K+ lattice vacancy were formed for each Ca2+ ion in the crystal. The Ca2+ enters the lattice in a normal K+ site and the two C1- ions enter two C1- sites in the KC1 crystal (Fig. 3). Demands of charge neutrality result in a vacant metal ion site. The experimental results show that the addition of CaCl, to KC1 lowers the density of the crystal. The density 20 Point Defects 587 Figure 3 Production of a lattice vacancy by the solution of CaC1, in KC1: to ensure electrical neutrality, a positive ion vacancy is introduced Into the lattice with each divalent cation Ca++. The two C1- ions of CaC1, enter normal negative ion sites. Figure 4 Three basic mechanisms of diffusion: (a) Interchange by rotation about a midpoint. More than two atoms may rotate together. (h) Migration through interstitial sites. (c) Atoms exchange position with vacant lattice sites. (After Seitz.) would increase if no vacancies were produced, because ca2+ is a heavier and smaller ion than Kt. The mechanism of electrical conductivity in alkali and silver halide crys- tals is usually by the motion of ions and not by the motion of electrons. This has been established by comparing the transport of charge with the transport of mass as measured by the material plated out on electrodes in contact with the crystal. The study of ionic conductivity is an important tool in the investigation of lattice defects. Work on alkali and silver halides containing known additions of divalent metal ions shows that at not too high temperatures the ionic conduc- tivity is directly proportional to the amount of divalent addition. This is not be- cause the divalent ions are intrinsically highly mobile, for it is predominantly the monovalent metal ion which deposits at the cathode. The lattice vacancies introduced with the divalent ions are responsible for the enhanced diffusion (Fig. 4c). The diffusion of a vacancy in one hrection is equivalent to the diffu- sion of an atom in the opposite direction. When lattice defects are generated thermally, their energy of formation gives an extra contribution to the heat capacity of the crystal, as shown in Fig. 5. Temperature, K Figure 5 Heat capacity of silver bromide exhibiting the excess heat capacity from thc Cormation of lattice defects. (After R. \V. Christy and A. \V. Lawson.) An associated pair of vacancies of opposite sign exhibits an electric dipole moment, with contributions to the dielectric constant and dielectric loss due to the motion of pairs of vacancies. The dielectric relaxation time is a measure of the time required for one of the vacant sites to jump by one atomic position with respect to the other. The dipole moment can change at low frequencies, but riot at high. In sodium chloride the relaxation frequency is 1000 s-' at 85'C. DIFFUSION When there is a concentration gradient of impurity atoms or vacancies in a solid, there will bc a flux of these through the solid. In equilibrium the impuri- ties or vacancies will he distributed uniformly. The net fluxJN of atoms of one species in a solid is related to the gradient of the concentration N of this species by a phenomenological relation called Fick's law: Here j , is the number of atoms crossing unit area in unit time; the constant D is the diffusion constant or diffusivity and has the units cm2/s or m2/s. The minus sign means that diffusion occurs away from regions of high concentra- tion. The form (5) of the law of diffusion is often adcquatr, but rigorously thc gradient of the chemical potential is the driving force for diffiision and not the concentration gradlent alone (TP, p. 406). The diffusion constant is often found to vary with temperature as here E is the activation energy for the process. Experiniental results on the diffusion of carbon in alpha iron are shown in Fig. 6. The data are represented 20 Point Defects 589 Temperature ("C! 0 % Z I- 2 8 ZzZgS I N 10P lo-" < 10-10 2 .- a 10-14 lo-'" 20 4.2 3.S 3.4 3.0 2.6 2.2 1.8 1.4 1.0 0.6 0 103rr(~) Figurc 6 Diffusion coefficient of carbon in iron, after Wert. The lugarithrn o C D is directly proportional to 1 1 T . Tablc 1 Diffusion constants and activation energies Host Du li ~ o s t D o E crystal Atom cm2 sC1 eV crystal Atom cm's-' eV by E = 0.87 cV, Do = 0.020 cmys. Representative values of Do and E are givcn in Tahle 1. To diffuse, an atom must overcome the potential encrgy harrier presented by its nearest neighbors. We treat the diffusion of imp~irity atoms between interstitial sites. The same argument will apply to the diffusion of vacant lattice sites. If the barrier is of height E, the atom will have sufficierit thennal energy to pass ovcr the barrier a fraction exp(-E/k,T) of the time. Quantum tunneling through the barrier is another possible process, but is usually impor- tant only for the lightest nuclei, particularly hydrogen. If v is a characteristic atomic vibrational frequency, then the probability p that during unit time the atom will have enough thermal energy to pass over the harrier is In unit time thc atom makes v passes at the barrier, with a probability exp(-E/kBT) of s~~rmoiinting the barricr on cach try. The quantity p is called the jump frequency. We consider two parallel planes of impurity atoms in interstitial sites. The planes are separated by lattice constant a. There are S impurity atoms on one plane and ( S + a dSldx) on the otl~er. The net nurnber of atoms crossing be- tween the planes in unit time is = -purlS/dx. If N is the total concentration of impurity atoms, then S = aW per unit area ora plane. The diffusion flux may now hc writtcn as On comparison with (5) wc havc the result of the fonn (6) wit11 Do = vu2. If the impurities are charged, we may find the ionic mobility p and the conductivity u from the diffusivity by using the Einstein relation kBTP = qD from TP, p. 406: where N is the concentration of impurity ions of charge q. The proportion of vacancies is independent of temperature in the range in which the niimher of vacancies is dctcrmincd by thc number of divalent metal ions. Then the slope of a plot of In u versus l/kBT gives E,, the harrier activa- tion energy for the jumping of positive ion vacancies (Table 2). Diffusion is very slow at low temperatures. At room temperature the jump frequency is of the order of Is-', and at 100 K it is of the order of lo-'' s-'. The proportion of vacancies in the temperature range in which the cull- centration of defects is determined by thermal generation is given by where Ef is the energy of formation of a vacancy pair, according to the theory of Schottky or Frcnkcl dcfccts. Here the slope of a plot of In u versus l/kBT will be E,+ + E ~ , according to (10) and (12). From mcastircmcnts in diffcrcnt 20 Point Defects 591 Table 2 Activation energy E+ for motion of a positive ion vacancy Values of the cncra of formation of a vacancy pair, E , , are also given. The numbers given in parenlhcscs for the silver salts refer to interstitial silver ions. NaCl LiF 1,iCI LiBr LiI KC1 AgCl AgRr Etzcl and hlaurer Haven Haven Haven Haven Wagner; Kelting and Witt Teltow Cornptori T o r Frenkel defect. temperature ranges we determine the energy of formation of a vacancy pair Ef and the jurrlp activation energy E,. The diffusion constant can bc mcasnred by radioactive tracer techniques. The diffusion of a hown initial distribution of radioactive ions is followed as a function of time or distance. \7alues of the diffusion co~lstant thus deter- mined may he compared with values from ionic conductivities. The two sets of vah~es do not usually agree within the experimental accuracy, suggesting the presence of a diffusion rneclianism that does not involvc the transport of charge. For example, the diffusion of pairs of positive and negative ion vacan- cies does not irlvolve the transport of' charge. Metals Self-diffusion in monatomic metals most commonly proceeds by lattice va- cancies. Self-diffusion means the diffusion of atoms of the 111etal itself, and not of impurities. The activation energy for self-diffusion in copper is expected to he in the range 2.4 to 2.7 eV for diffusior~ through vacancies and 5.1 to 6.4 eV for diffusion through interstitial sites. Observed valucs of the activation energy are 1.7 to 2.1 e\< Activation energies for diflusion in Li and Na can be determined from rrleasurements of the temperaturc dependence of the nuclear resonance line width. As discussed undcr magnetic resonance, the resonance line width narrows when the jump frequency of an atom between sites becomes rapid in comparison with the frequency c~rres~or~lli~lg to the static line width. The val- ues 0.57 eV and 0.45 eV were determined by NMR for Li and Na. Self-diffu- sion rneasurenie~~ts for sodium also give 0.4 cV. COLOR CENTERS Pure alkali halide crystals are transparent throughout the visible region of the spectrum. A color center is a lattice defect that absorbs visible light. An ordinary lattice vacancy does not color alkali halide crystals, although it af- fects the absorption in the ultraviolet. The crystals may be colored in a number of ways: by the introduction of chemical impurities; by the introduction of an excess of the metal ion (we may heat the crystal in the vapor of the alkali metal and then cool it quickly-an NaCl crystal heated in the presence of sodium vapor becomes yellow; a KC1 crystal heated in potassium vapor becomes magenta); by x-ray, y-ray, neutron, and electron bombardment; and by electrolysis. F Centers The name F center comes from the German word for color, Farbe. We usually produce F centers by heating the crystal in excess alkali vapor or by x-irradiation. The central absorption band (F band) associated with F centers in several alkali halides are shown in Fig. 7, and the quantum energies are listed in Table 3. Experimental properties of F centers have been investigated in detail, originally by Pohl. The F center has been identified by electron spin resonance as an electron bound at a negative ion vacancy (Fig. 8 ) , in agreement with a model suggested by de Boer. When excess alkali atoms are added to an alkali halide crystal, a corresponding number of negative ion vacancies are created. The valence electron of the alkali atom is not bound to the atom; the electron migrates in the crystal and becomes bound to a vacant negative ion site. A negative ion va- cancy in a perfect periodic lattice has the effect of an isolated positive charge: it attracts and binds an electron. We can simulate the electrostatic effect of Wavelength in A 4000 6000 4000 6000 4000 6000 4000 6000 4000 6000 D e d 3 8 4 3 2 4 3 2 4 3 2 4 3 2 4 3 2 Energy m eV Figure 7 The F bands for several alkali halides: optical absorption versus wavelength for crystals that contain F centers. 20 Point Defects 593 Table 3 Experimental li center absorption energies, in eV LlCl 3 1 NaBr 2 3 NaCl 2 7 KBr 2.0 KC1 2 2 RbBr 1.8 RbCl 2 0 LiF 5 0 CsCl 2 0 NaF 3 6 LiBr 2 7 KF 2 7 , - Figure 8 An F center 1 s a negahve Ion vacancy wth one excess electron bound at n the vacancy The dlstnbutlon of the excess Figure 9 An FA center in KCI: one of the six Kt ions which bind an F center is replaced by another alkali ion, here Na'. a negative ion vacancy by adding a positive charge q to the normal charge -q of an occupied negative ion site. The F center is the simplest trapped-electron center in alkali halide crys- tals. The optical absorption of an F center arises from an electric dipole transi- tion to a bound excited state of the center. Other Centers in Alkali Halides In the FA center one of the six nearest neighbors of an F center has been replaced by a different alkali ion, Fig. 9. More complex trapped-electron cen- ters are formed by groups of F centers, Fig. 10 and 11. Thus two adjacent F centers form an M center. Three adjacent F centers form an R center. Dif- ferent centers are distinguished by their optical absorption frequencies. Figure 10 An M center consists of two adjacent F centers. Figure 11 A n R center consists of three adjacent F centers; that is, a group of three negative ion vacancies in a [Ill] plane of the NaCl struc- ture, with three associated electrons. . .
. V , center Figure 12 A V, center formed when a hole is trapped by a pair of negative ions resembles a neg- ative halogen molecule ion, which is C1; in KC1. No lattice vacancies or extra atoms are involved in a VK center. The center at the left of the figure probably is not stable: the hexagon represents a hole trapped near a positive ion vacancy; such a center would be the antimorph to an F center. Holes have a lower energy trapped in a V, center than in an anti-F center. 20 Point Defects 595 Holes may be trapped to form color centers, but hole centers are not usu- ally as simple as electron centers. For example, a hole in the filled p%hell of a halogen ion leaves the ion in a p5 configuration, whereas an electron added to the filled p6 shell of an alkali ion leaves the ion in a p6s configuration. The chemistry of the two centers is different: p6s acts as a spherically sym- metric ion, but p5 acts as an asymmetric ion and, by virtue of the Jahn-Teller effect, will distort its immediate surroundings in the crystal. The antimorph to the F center is a hole trapped at a positive ion vacancy, but no such center has been identified experimentally in alkali halides; in insulating oxides the 0- (called V-) defect is known. The best-known trapped- hole center is the V , center, Fig. 12. The V , center is formed when a hole is trapped by a halogen ion in an alkali halide crystal. Electron spin resonance shows that the center is like a negative halogen molecular ion, such as C1, in KC1. The Jahn-Teller trapping of free holes is the most effective form of self- trapping of charge carriers in perfect crystals. Problems 1. Frenkel defects. Show that the number n of interstitial atoms in equilibrium with n lattice vacancies in a crystal having N lattice points and N' possible interstitial posi- tions is given by the equation E, = k,T ln [(N - n)(N' - n)ln2] , whence, for n < N , N', we have n = (NN')'12 exp(-EIl2kBT). Here E, is the energy necessary to remove an atom from a lattice site to an interstitial position. 2. Schottky vacancies. Suppose that the energy required to remove a sodium atom from the inside of a sochum crystal to the boundary is 1 eV. Calculate the concentra- tion of Schottky vacancies at 300 K. 3. F center. (a) Treat an F center as a free electron of mass m moving in the field of a point charge e in a mechum of dielectlic constant E = n2; what is the 1s-2p energy difference of F centers in NaCl? (b) Compare from Table 3 the F center excitation energy in NaCl with the 3s-3p energy difference of the free sodium atom. Dislocations SHEAR STRENGTH OF SINGLE CRYSTALS Slip DISLOCATIONS Burgers vectors Stress fields of dislocations Low-angle grain boundaries Dislocation densities Dislocation mul~iplicatiou and slip STRENGTH OF ALLOYS DISLOCATIONS AND CRYSTAL GROWTH Whiskers HARDNESS OF MATERIALS PROBLEMS 1. Lines of closest packing 2. Dislocation pairs 3. Force on dislocation Figure 1 (a) Relative shear of two planes of atoms (shown in section) in a uniformly strained clys- tal: (h) shear stress as a function of the relative displacement of the planes from their equilibrium position. The heavy broken hne drawn at the initial slope defines the shear modulus G. CHAPTER 21: DISLOCATIONS This chaptcr is concerned with the interpretation of the plastic mechani- cal properties of crystalline solids in terrrls of the thcory of dislocations. Plastic properties are irreversible defor~~iations; elastic properties are reversible. The easc with which pure single crystals deform plastically is striking. This i~itrinsic weakness of crystals is exhibited in various ways. Pure silver chloride melts at 455C, yet at room temperafirre it has a cheeselike consistency and can he rolled into sheets. Pure aluminum crystals are elastic (follow Hookc's law) only to a strain of about lo-', after which they deform plastically. Theoretical estimates of the strain at the elastic limit of perfect crystals may gyve val~ies 10%r lo4 higher than the lowest ohsenred values, although a factor 102 is more usual. There are few excrptions to the rule that pure crystals are plastic and are not strong: crystals of germanium and silicon are not plastic at room temperature and fail or yeld only by fracture. Glass at room tempcra- ture fails o~ily by fracture, but it is not crystalline. The fracture of glass is caused by stress concentration at minute cracks. SHEAR STRENGTH OF SINGLE CRYSTALS Frenkel gave a simple method of estiniating the theoretical shear strength of a perfect crystal. We consider in Fig. 1 the force needed to make a shear displacement of two planes of atoms past each other. For small elastic strains, the stress u is related to the displacement x by Here d is the interplanar spacing, and G denotes the appropriate shear modulus. When the displacement is large and has proceeded to the point that atom A is directly over atom B in the figure, the two planes of atoms arc in a configuratio~~ of unstable equilibriiim and the stress is zero. As a first approxi- mation we represent tlir stress-displacement relation by u = (Gd25-d) sin (2mlu) , (2) where a is the interatomic spacing in the direction of shear. This relation is con- structed to reduce to (1) for small values of x/u. The critical shear strcss uc at which the lattice becomes unstable is given by the niaximum valuc of u, or If a = d, then u, = (;/25-: the ideal critical shear stress is of the order or of the shear modulus. Table 1 Comparison of shear modulus G and observed elastic limit rrCA Shear modulus 6, Elastic limit a,., in dydr/crn2 in d>n/crn2 Sn, single crystal 1.9 X 10" 1.3 X 10' 15,000 Ag, single crystal A], single c~ystal Al, 1311re. I~olycrystt"l Al, corr~rr~ercial drawn Duralumin Fe, soft, polycrystal Heat-treated carbon stccl Nickel-chrome steel "After Mott The observatior~s in Table 1 show the experimental values of the elastic limit are much smaller tlrm (3) \i~ou!d suggest. The theoretical estimate may he improvcd by consideration of the actual fonr~ of the intermolecular forces and by consideration of other configurations of mecl~anical stability available to the lattice as it is sheared. Mackcnzie has shown that these two effects may reduce the theoretical ideal shear strength to about G/,30, corresponding to a critical shear strain angle of about 2 degrees. The ohscrvcd low values ol the shear strength can be explained only by the presence of impcrfkctions that can act as sources of rrrechariical weakness in real crystals. The movement of crystal imperlections called dislocations is responsible for slip at very low applied strcsses. Slip Plastic deformation in crystals occurs by slip, an example of which is shown in Fig. 2. In slip, one part of the crystal slides as a unit across an adja- cent pad. The surface on which slip takes place is known as the slip plane. The direction of motion is know11 as the slip direction. The great importance of lat- ticr properties lor plastic strain is indicated by the highly anisotropic nature of slip. Displacement takes place along crystallograpllic planes with a set of small Miller indices, such as thc {lll) planes in fcc metals and the (110), [llZ}, a ~ ~ d {I231 planes in bcc metals. The slip hrection is in the line of closest atomic packing, (110) in fcc 111etals and (111) in bcc metals (Problem 1). To maintain the crystal striicturc after slip, the displacement or slip vector must equal a lattice translation vec- tor. The shortest lattice trans la ti or^ vector expressed in terms of the lattice con- stant n in a fcc structure is of the form (al2)(? + y); in a bcc structure it is (a/2)(% + y + 2). Bnt in fcc crystals one also observes partial displacarr~ents which upset the regular seqllence ABCABC . . . of closest-packed planes, to Figure 2 Translational slip in zinc single clystals. (E. R. Parker.) produce a stacking fault such as ABCABABC. . . . The result is then a mix- ture of fcc and hcp stacking. Deformation by slip is inhomogeneous: large shear displacements oocur on a few widely separated slip planes, while parts of the crystal lying between slip planes remain essentially undeformed. A property of slip is the Schmid law of the critical shear stress: slip takes place along a given slip plane and direction when the corresponding component of shear stress reaches the critical value. Slip is one mode of plastic deformation. Another mode, twinning, is ob- served particularly in hcp and bcc structures. During slip a considerable dis- placement occurs on a few widely separated slip planes. During twinning, a partial displacement occurs successively on each of many neighboring crystal- lographic planes. After twinning, the deformed part of the crystal is a mirror image of the undeformed part. Although both slip and twinning are caused by the motion of dislocations, we shall be concerned primarily with slip. DISLOCATIONS The low observed values of the critical shear stress are explained in terms of the motion through the lattice of a line imperfection known as a dislocation. The idea that slip propagates by the motion of dislocations was published in 1934 independently by Taylor, Orowan, and Polanyi; the concept of dislocations was introduced somewhat earlier by Prandtl and Dehlinger. There are several basic types of dislocations. We first describe an edge dislocation. Figure 3 shows a simple cubic crystal in which slip of one atom distance has occurred over the left half of the slip plane but not over the right half. The boundary be- tween the slipped and unslipped regions is called the dislocation. Its position is marked by the termination of an extra vertical half-plane of atoms crowded into Figure 3 An edge dislocation EF in the glide plane ABCD. The figure shows the slipped region ABEF in which the atoms have been displaced by more than half a lattice constant and the un- slipped region FECD with displacement less than half a lattice constant. Figure 4 Structure of an edge disloca- tion. The deformation may be thought of as caused by inserhng an extra plane of atoms on the upper half of they axis. Atoms in the upper half-crystal are compressed by the insertion; those in the lower half are extended. the upper half of the crystal as shown in Fig. 4. Near the dislocation the crystal is highly strained. The simple edge dislocation extends indefinitely in the slip plane in a direction normal to the slip direction. In Fig. 5 we show a photo- graph of a dislocation in a two-dimensional soap bubble raft obtained by the method of Bragg and Nye. The mechanism responsible for the mobility of a dislocation is shown in Fig. 6. The motion of an edge dislocation through a crystal is analogous to the passage of a ruck or wrinkle across a rug: the ruck moves more easily than the whole rug. If atoms on one side of the slip plane are moved with respect to 21 Dislocations Figure 5 A dislocation in a hvo-dimensional bubble raft. The dislocation is ]nost eas~ly seen by turning the page by 30" in its plane and sighting at a low angle. (W. M. Lomer, after Bragg and Nye.) - -t -t Figure 6 Motion of a dislocation under a shear tending to move the upper surface of the s~ecimen to the - - - dght. (D. Hull.) those on the other side, atoms at the slip plane will experience repulsive forces from some neighbors and attractive forces from others across the slip plane. These forces cancel to a first approximation. The external stress required to move a dislocation has been calculated and is quite small, below lo5 dyn/cm2 when the bonding forces in the crystal are not highly directional. Thus dis- locations may make a crystal very plastic. Passage of a dislocation through a crystal is equivalent to a slip displacement of one part of the crystal. The second simple type of dislocation is the screw disIocation, sketched in Figs. 7 and 8. A screw dislocation marks the boundary between slipped and unslipped parts of the crystal. The boundary parallels the slip direction, in- stead of lying perpendicular to it as for the edge dislocation. The screw dis- location may be thought of as produced by cutting the crystal partway through with a knife and shearing it parallel to the edge of the cut by one atom spacing. A screw dislocation transforms successive atom planes into the surface of a helix: this accounts for the name of the dislocation. Figure 7 A screw dislocation. A part ABEF of the slip plane has slipped in the direction parallel to the dislocation line EF A screw dislocation may be visualized as a helical arrangement of lattice planes, such that we change planes on going completely around the dislocation line. (After Cottrell.) Figure 8 Another view of a screw dislocation. The bro- ken vertical line that marks the dislocation is surrounded by strained material. Burgers Vectors Other dislocation forms may be constructed from segments of edge and screw dislocations. Burgers has shown that the most general form of a linear dislocation pattern in a crystal can be described as shown in Fig. 9. We con- sider any closed curve within a crystal, or an open curve terminating oil the surface at both ends: (a) Make a cut along any simple surface bounded by the line. (b) Displace the material on one side of this surface by a vector b relative to the other side; here b is called the Burgers vector. (c) In regions where b is not parallel to the cut surface, this relative displacement will either ~roduce a gap or cause the two halves to overlap. In these cases we imagine that we either add material to fill the gap or subtract material to prevent overlap. (d) Rejoin the material on both sides. We leave the strain dsplacement intact at the time of rewelding, but afterwards we allow the medium to come to internal equilibrium. The resulting strain pattern is that of the dislocation character- ized jointly by the boundary curve and the Burgers vector. The Burgers vector must be equal to a lattice vector in order that the rewelding process will main- tain the crystallinity of the material. The Burgers vector of a screw dislocation (Figs. 7 and 8) is parallel to the dislocation line; that of an edge dislocation (Figs. 3 and 4) is perpendicular to the dislocation line and lies in the slip plane. 21 Dislocations 605 Figure 9 General method of forming a dislocation ring in a medium. The medium is represented by the rectangular block. The ring is represented by the closed curve in the interior in the block. A cut is made along the surface bounded by the curve and indicated by the contoured area. The ma- terial on one side of the cut is displaced relative to that on the other by vector distance b, which may be oriented arbitrarily relative to the surface. Forces will be required to effect the displace- ment. The medium is filled in or cut away so as to be continuous after the displacement. It is then joined in the displaced state and the applied forces are relaxed. Here b is the Burgers vector of the dislocation. (After Seitz.) 1 line Figure 10 Shell of elastically distorted cvstal surrounding screw dislocation with Burgers vector b; see also Fig. 16. Stress Fields of Dislocations The stress field of a screw dislocation is particularly simple. Figure 10 shows a shell of material surrounding an axial screw dislocation. The shell of circumference 2nr has been sheared by an amount b to give a shear strain e = bI2~r. The corresponding shear stress in an elastic continuum is This cxprcssion does not hold in the region immediately around the disloca- tion line, as the strains here are too large for continuum or linear elasticity theory to apply. The elastic energy of the shell is dE, = ~GC' dV = (GhP/4v) drlr per unit length. The total elastic energy per unit length of a screw dislocation is found on integration to be wherc R and r, are appropriate upper and lower limits for the variable r. A rea- sonable valile of r, is comparable to the magnitudc b of the Burgers vector or to the lattice constant; the value of R cannot exceed thc dimensions of thc crystal. The value of the ratio R/r, is not very important hecalise it enters in a logarithm term. \17e now show the form of the energy of an edge dislocation. Let u , and u , , denote the tensile stresses in the radial and circulnferential hrections, and let u , denote the shear stress. In an isotropic elastic continuum, u7, and u , , are proportional to (sin 8)lr: wc nerd a function that falls off as llr and that changes sign when y is replaced by -y. The shear stress mro is proportional to (COS 0)lr; considering the plane y = 0, we see from Fig. 4 that the shear stress is an odd fur~ction of x. The constants of proportionality in the stress are propor- tional to the shear modulus G and to the Burgers vector b of the displacement. The final result is where the Poisson ratio v = 0.3 for most cry-stals. The strain energy of a unit length of edge dislocation is N7e want an expression for the shear stress component u , , on planes paral- lel to the slip plane in Fig. 4. From the stress components a , , a , , , and ud evaluated on the plane a &stance y above the slip plane, we find It is shown in Problem 3 that the force caused by a resolved unifornl shear strcss u is F = bu per unit length of dislocation. The force that an edge dislo- cation at thc origin cxcrts upon a similar onc at the location (y, 8 ) is per unit length. Here F is the component of force in the slip direction. 21 Dislocations Low-angle Grain Boundaries Burgers suggested that low-angle boundaries between adjoining crystal- lites or crystal grains consist of arrays of dislocations. A simple example of the Burgers model of a grain boundary is shown in Fig. 11. The boundary occupies a (010) plane in a simple cubic Iattice and divides two parts of the crystal that have a axis in common. Such a boundary is called a pure tilt boundary: the misorientation can be described by a small rotation 0 about the common axis of one part of the crystal relative to the other. The tilt boundary is represented as an array of edge dislocations of spacing D = bl0, where b is the Burgers vector of the dislocations. Experiments have substantiated this model. Figure 12 shows the distribution of dislocations along small-angle grain boundaries, as observed with an eIectron microscope. Further, Read and Shockley derived a theory of the interfacial energy as a function of the angle of tilt, with results in excellent agreement with measurements. Direct verification of the Burgers model is provided by the quantitative x-ray and optical studies of low-angle boundaries in germanium crystals by Vogel Figure 11 (a) Low-angle grain boundary, after Burgers. (b) Electron micrograph of a low-angle grain boundaly in molybdenum. The three dislocations in the image each have the same Burgers vector as in the drawing in Fig. lla. The white circles mark the positions of atomic columns normal to the plane of the paper. Each array of circles defines the position of a dislocation, with four circles on the top of each array and three circles below. Closure failure is indicated by the arrows which define the Burgers vectors. (Courtesy of R. Gronsky.) Figure 12 Electron micrograph of dislocation structures in low-angle grain boundaries in an A1-7 percent Mg solid solution. Notice the lines of small dots on the right. Mag. X17,OOO. (R. Goodrich and 6. Thomas.) Figure 13 Dislocation etch pits in low-angle bounda~y on (100) face of germanium; the angle of the boundary is 27.5". The b o u n d v lies in a (011) plane; the line of the dislocations is . The Burgers vector is the shortest lattice translation vec- tor, or Ibl = a l l h = 4.0 A. (F. L. Vogel, Jr.) and co-workers. By counting etch pits along the intersection of a low-angle grain boundary with an etched germanium surface (Fig. 13), they determined the dislocation spacing D. They assumed that each etch pit marked the end of a dislocation. The angle of tilt calculated from the relation 0 = b/D agrees well with the angle measured directly by means of x-rays. The interpretation of low-angle boundaries as arrays of dislocations is fur- ther supported by the fact that pure tilt boundaries move normal to them- selves on application of a suitable stress. The motion has been demonstrated in 21 Dislocations 609 Figure 14 Motion of a low-angle grain boundan, under stress. The boundary is the straight vertical line, and it is photographed under vertical illumination, thereby making evident the 2" angular change in the cleavage surface of the zinc clystal at the boundaly. The irregular horizontal line is a small step in the cleavage surface which serves as a reference mark. The crystal is clamped at the left; at the right it is subject to a force normal to the plane of the page. Top, oliginal position of boundary; bottom, moved back 0.4 mm. (J. Washburn and E. R. Parker.) a beautiful experiment, Fig. 14. The specimen is a bicrystal of zinc containing a 2" tilt boundary with dislocations about 30 atomic planes apart. One side of the crystal was clamped, and a force was applied at a point on the opposite side of the boundary. Motion of the boundary took place by cooperative motion of the dislocations in the array, each dislocation moving an equal distance in its own slip plane. The motion was produced by stresses of the order of magni- tude of the yield stress for zinc crystals, strong evidence that ordinay defor- mation results from the motion of dislocations. Grain boundaries and dislocations offer relatively little resistance to diffu- sion of atoms in comparison with diffusion in perfect crystals. A dislocation is an open passage for hffusion. Diffusion is greater in plastically deformed ma- terial than in annealed crystals. Diffusion along grain boundaries controls the rates of some precipitation reactions in solids: the precipitation of tin from lead-tin solutions at room temperature proceeds about lo8 times faster than expected from diffusion in an ideal lattice. Dislocation Densities The density of dislocations is the number of dislocation lines that inter- sect a unit area in the crystal. The density ranges from well below 10' &slocations/cm2 in the best germanium and silicon crystals to 1011 or 1 0 1 2 dis- locations/cm2 in heavily deformed metal crystals. The methods available for estimating dislocation densities are compared in Table 2. The actual dislocation configurations in cast or annealed (slowly cooled) crystals correspond either to a group of low-angle grain boundaries or to a three-dimensional network of dislocations arranged in cells, as shown in Fig. 15. Lattice vacancies that precipitate along an existing edge dislocation will eat away a portion of the extra half-plane of atoms and cause the dislocation to climb, which means to move at right angles to the slip direction. If no dislocations are Table 2 Methods for estimating dislocation densitiesa W~dth of Maximum practical Techmque Speclmen thickness image density, per cm2 Electron microscopy >lo00 -100 A 10"- 10l2 X-ray transmission 0.1- 1.0 m m 5 ~m lo4-lo5 X-ray reflection <2 pm (min.) - 50 pm (max.) 2 pm l0~-10' Decoration -10 pm (depth of focus) 0.5 pm 2 x lo7 Etch pits no limit 0.5 p,mb 4 X lo8 "W. 6. Johnston. '~imit of resolution of etch pits. Figure 15 Cell structure of three-dimensional tangles of dislocat~ons in deformed aluminum. (P. R. Swann.) 21 Dislocations 611 Figure 16 Electron micrograph of dislocation loops formed by aggregation and collapse of vacancies in A15 percent Mg quenched from 550°C. The helical dislocations are formed by vacancy condensation on a screw dislocation. Mag. X43.000. (A. Eikum and 6. Thomas.) present, the crystal will become supersaturated with lattice vacancies; their pre- cipitation in cylindrical vacancy plates may be followed by collapse of the plates and formation of dislocation rings that grow with further vacancy precipitation, as in Fig. 16. Dislocation Multiplication and Slip Plastic deformation causes a very great increase in dislocation density, typically from lo8 to about 10" dislocations/cm2 during deformation. If a dis- location moves completely across its slip plane, an offset of one atom spacing is produced, but offsets up to 100 to 1000 atom spacings are observed. This means that dislocations multiply during deformation. Consider a closed circular dislocation loop of radius r surrounding a slipped area having the radius of the loop. Such a loop will be partly edge, partly screw, and mostly of intermediate character. The strain energy of the loop increases as its circumference, so that the loop will tend to shrink in size. However, the loop will tend to expand if a shear stress is acting that favors slip. Figure 17 Frank-Read mechanism for multiplication of dislocations, showing successive stages in the gen- eration of a dislocation loop by the segment BC of a dislocation line. The process can be repeated indefinitely Figure 18 A Frank-Read dislocation source in silicon, decorated with copper precipitates and viewed with infrared illumination. Two complete dislocation loops are visible, and the third, inner- most loop is near completion. (After W. C. Dash.) A common feature of all dislocation sources is the bowing of dislocations. A dislocation segment pinned at each end is called a Frank-Read source, and it can lead (Fig. 17) to the generation of a large number of concentric disloca- tions on a single slip plane (Fig. 18). Related types of dislocation multiplica- tion mechanisms account for slip and for the increased density of dislocations during plastic deformation. Double cross-slip is the most common source. 21 Dislocations 613 STRENGTH OF ALLOYS Pure crystals are very plastic and yield at very low stresses. There appear to be four important ways of increasing the yield strength of an alloy so that it will withstand shear stresses as high as lo-' 6. They are mechanical blocking of dislocation motion, pinning of dislocations by solute atoms, impeding dis- location motion by short-range order, and increasing the dislocation density so that tangling of dislocations results. All four strengthening mechanisms de- pend for their success upon impeding dislocation motion. A fifth mechanism, that of removing all dislocations from the crystal, may operate for certain fine hairlike crystals (whiskers) that are discussed in the section on crystal growth. Mechanical blocking of dislocation motion can be produced most directly by introducing tiny particles of a second phase into a crystal lattice. This process is followed in the hardening of steel, where particles of iron car- bide are precipitated into iron, and in hardening aluminum, where particles of A1,Cu are precipitated. The pinning of a dislocation by particles is shown in Fig. 19. In strengthening by the addition of small particles there are two cases to be considered: either the particle can be deformed with the matrix, which re- quires that the particle can be traversed by the dslocation, or the particle cannot be traversed by the dislocation. If the particle cannot be cut, the stress Figure 19 Dislocations pinned hy particles in magnesium oxide. (Electron micrograph by 6. Thomas and J. Washburn.) necessary to force a dislocation between particles spaced L apart on a slip plane should be approximately u/G = b/L . (10) The smaller the spacing L, the higher is the yield stress cr. Before particles pre- cipitate, L is large and the strength is low. Immediately after precipitation is complete and many small particles are present, L is a minimum and the strength is a maxi~nurn. If the alloy is then held at a high tempcraturc, some particles grow at the expense of others, so that L increa~es and the strength drops. Hard intermetallic phases, such as refractory oxides, cannot be cut by dislocations. The strength of dilute solid solutions is believed to result fro111 the pi~ining of dislocations by solute atoms. The solubility of a foreign atom will be greater in the neighborhood of a dislocation than elsewhere in a crystal. An atom that tends to expand the crystal will dissolve peferentially in the expandcd region near an edge dislocation. A small atom will tend to dissolve preferentially in the contracted region near the dislocation-a dislocation offers both expanded and contracted regions. As a resnlt of the affinity of solute atoms for dislocations, each dislocation will collect a cloud of associated solute atoms during cooling, at a time when the mobility of solute atoms is high. At still lower temperatures, diffusion of solute atorns effectively ceases, and the solute atom cloud becomcs fixed in the crystal. When a dislocation moves, leaving its solutc clond behind, the energy of the crystal must increase. The incrcasc in energy can only he provided by an increased stress acting on the dislocation as it pulls away from the solute atom cloud, and so the presence of the cloud strengthens the crystal. The passage of a dislocation across a slip plane in pure crystals does not alter the binding energy across the plane after the dislocation is gone. The in- ternal energy of the crystal remains unaffected. The same is true for random solid solutions, because the solution is equally random across a slip plane after slip. Most solid solutions, however, have short-rangc order. Atoms of different species are not arranged at random on the lattice sites, but tend to have an ex- cess or a deficiency of pairs of unlike atoms. Thus in ordered alloys disloca- tions tend to move in pairs: the second dislocation reorders the local disorder left by the first dislocation. The strength of a crystalline material increases with plastic delormation. The phenomenon is called work-hardening or strain-hardening. The strengtt~ is believed to increase because of the incrcascd density of disloca- tions and the greater difficulty of moving a given dislocation across a slip plane that is threaded by many dislocations. Strain-hardening frequently is em- ployed in the strengthening of materials, but its usefulness is limited to low enough temperatures so that annealing does not occur. An important factor in strain-hardening is the total density of dislocations. In most metals dislocations tend to form cells (Fig. 15) of dislocation-free areas of dinlensions orthe order of 1 pm. But unlcss we can get a uniform high density of dislocations wc cannot strain-harden a metal to its theoretical strength, because of slip in the dislocation-free areas. A high total density is accomplished by explosive deformation or by special therrnal-mechanical treatments, as of martensite in steel. Each of the rneclianisms of strengthening crystals can raise the yield strength to the range of G to lo-' G. All mechanisms begin to brcak down at temperatures where diffusion can occur at an appreciable rate. When diffusion is rapid, precipitated particles dissolve; solute clouds drift along with dislocations as they glide; short-range order repairs itself behind slowly mov- ing dislocations; and dislocation climb and annealing tend to decrease the dis- location density. The resulting time-dependent deformation is called creep. This irreversible motion precedes the elastic limit. The search for alloys for use at very high temperatures is a search Tor reduced diffusion rates, so that the four strengthening mechanisms will survivc to high temperatnres. But the central problcm of strong alloys is not strength, but ductility, for failure is oftcn by fracture. DISLOCATIONS AND CRYSTAL GROWTH In some cases the prcscncc of dislocations may he the controlling factor in crystal growth. \tihcn cry-stals are grown in conditions of low supersaturation, of the order of 1 percent, it has been observed that the growth rate is enor- mously faster than that calculated for an ideal crystal. The actual growth rate is explained in terms of the effect of dislocations on growth. The theory of growth of ideal crystals predicts that in crystal growth Gom vapor, a supersaturation (pressure/equilibrium vapor pressure) of the order of 10 is required to nucleate new crystals, of thc ordcr of 5 to form liqnid drops, and of 1.5 to form a two-dimensional monolayer of molecules on the face of a perfect crystal. \Jolmer and Schultze observed growth of iodine crystals at vapor supersaturations down to less than 1 percent, where the growth rate should have been down by the factor exp(-3000) frorn the rate defined as the rninirriurn observable growth. The large disagreement expresses the difficulty of nucleating a new mono- layer on a completed surface of an ideal clystal. Rut if a screw dislocation is present (Fig. 20), it is never necessary to nucleate a new layer: the crystal will grow in spiral fashion at the edge of the discontinuity shown. An atom can be bound to a step more strongly than to a plane. The calculated growth rates for this mechanism are in good agreement with observation. \%'e expect that nearly all crystals in irature grown at low supersaturation will contail1 dislocations, as otherwise the>. could not have grown. Spiral growth patterns havc bccn oh- sen~cd on a largc number of crystals. A heantifill example of the growth pat- tern from a single screw dislocation is given in Fig. 21. Figure 20 Development of a spiral step produced by intersection of a screw dislocation with the surface of a crystal as in Fig. 8. (F. C. Frank.) Figure 21 Phasecontrast micrograph of a hexagonal spiral growth pattern on a Sic crystal. The step height is 165 A. (A. R. Verma.) If the growth rate is independent of direction of the edge in the plane of the surface, the growth pattern is an Archimedes spiral, r = a0, where a is a constant. The limiting minimum radius of curvature near the dislocation is determined by the supersaturation. If the radius of curvature is too small, atoms on the curved edge evaporate until the equilibrium curvature is at- tained. Away from the origin each part of the step acquires new atoms at a con- stant rate, so that drldt = const. Whiskers Fine hairlike crystals, or whiskers, have been observed to grow under con- ditions of high supersaturation without the necessity for more than perhaps one dislocation. It may be that these crystals contain a single axial screw dislocation 21 Dislocations 617 Figure 22 .
4 nickel whisker of diameter 1000 bent in a loop. (R. W De Blois.) that aids their essentially one-dimensional growth. From the absence of dis- locations we would expect these crystal whiskers to have high yield strengths, of the order of the calculated value C/30 discussed earlier i11 this chapter. A single axial screw dislocatior~, if present, could not cause yielding, because in bending the crystal the dislocation is not subjected to a shear stress parallel to its Burgers vector. That is, the stress is not in a dircction that can cause slip. Hcrring and Galt obsen~cd whiskers of tin of radu~s -10-Qm with elastic properties near those expected from theoretically perfect crystals. They ob- served peld strains of the order of lo-', which correspond to shear stresses of order C, about 1000 times greater than in bulk tin, confirming the early estimates of the strength of perfect crystals. Theoretical or ideal elastic proper- ties have bee11 observed for a nu~nber of materials as lor carbon nanotubes. A single domain whisker of nickel is shown in Fig. 22. HARDNESS OF MATERIALS The hardness of materials is measured in several ways, the simplest test for nonmetals being the scratch test. Substance A is harder than substance B if A will scratch B but B will not scratch A. A standard scale is used for represen- tative minerals, wit11 diarr~or~d, the hardest, assigned the value 10 and talc, the softest, assigned the value 1: 10 diamond C $5 apatitc Ca,(PO,),F 9 cori~ndl~m AlzO, 4 fll~orite CaF, 8 topaz AlzS~O,Fz 3 calcite CaCO, 7 quartz SiO, 2 gjpsum CaSO, - 2H20 6 orthoclase K.41Si308 1 talc 3Mg0 . 4Si02 . H,O There is great current interest in the development of materials of great hardness, lor example as films for use as scratch-resistant coating? on Ien~es. It is widely felt that the scale between diamond and corundum is misleading, because diamond is much, much harder than corundum. It has been suggested that one might assign diamond the hardness 15, with the gap between 9 and 15 to be filled in eventually by synthetic materials, such as compounds of C and B. Modern scales of hardness, such as the VHN scale, are based on indcntcr tests in which an indenter is pressed into the surface of the material and the sizc of the impression is measured. The Vickers Hardness Numbers of se- lected materials are tabulated below, after conversion by E. R. m7eber to units of CPa [CN/m": Diamond 45.3 B e 0 7.01 SiC 20.0 Steel (quenched) 4.59 Si3N4 18.5 Cu (annealed) 0.25 A1203 14.0 A1 (annealed) 0.12 B 13.5 Pb 0.032 WC 11.3 The data are from J. C. Anderson and others. Problems 1. Lines of closest packing. Show that the lines of closest atomic pachng are (110) in fcc structures and (111) in bcc structures. 2. Dinlocation pairs. (a) Find a pair of dislocations equivalent to a row of lattice vacancies; (h) find a pair oidislocations cquivalcnt to a row of interstitial atoms. 3. Force on dislocation. Consider a crystal in thc form of a cube of side L containing an edge dislocation of Burgers vector h. If the crystal is subjcctcd to a shear stress a on the upper and lower faces in the directio~r of slip, sho\v, by considcring energy balance, that the force acting on the dislocatiorr is F = hrr per unit length. Alloys GENERAL CONSIDERATIONS SUBSTITUTIONAL SOLID SOLUTIONS- HUME-ROTHERY RULES ORDER-DISORDER TRANSFORMATION Elementary theory of order PHASE DIAGRAMS Eutectics TRANSITION METAL ALLOYS Electrical Conductiuity KONDO EFFECT PROBLEMS 1. Superlattice lines in Cu3Au 2. Configurational heat capacity Ordered Disordered (a) (b) Figure 1 Ordered (a) and disordered (b) arrangements of AB ions in the alloy AB. CHAPTER 22: ALLOYS GENERAL CONSIDERATIONS The theory of the band structure of solids assumes that the crystal has translational invariancc. Rut supposc that thc crystal is cornposcd of two elements A and B that occupy at random the regular lattice sites of the struc- ture, in proportions x and 1 - x for the composition A,B,-,. The translational symmetv is no longer perfect. Will we then lose the consequences of band theory, such as the existence of Ferrr~i surfaces and of energy gaps? Will insula- tors becorrie co~iductors because the eriergy gap is gone? We touched on these questions in the discussion or amorphous semiconductors in Chapter 19. Expcrimrnt and thcory agrcc that thc conscqncnccs of the destnlction of perfect translational symmetry are much less serious (nearly always) than we ex- pect at first sight. The viewpoint of the effective screened potential of Chapter 9 is helpful in these matters, first because the effective potentials are relatively weak in comparison with a free ion potential and, second and most important, tlie differerices betweeri the effective potentials of tlie host arid the additive atoms may he very weak in comparison with either alone. Alloys of Si and Ge or of Cu and Ag are classic examples of what we may call tlie relative ineffective- ncss of alloying. In any event, a low concentration of impurity atoms cannot have much effect on the Fourier components U , of the effective potential U(r) that is re- sponsible for the band gaps and for the form of the Fermi surface. (This state- ment irr~plies that the G ' s exist, which implies that a regular lattice exists. This is not an inlportalit assumptioii because we know that thermal phonoi~s do not have drastic erfects on the band structure, so that lattice distortions described as fiozcn-in phonons should not haw drastic cffrcts. If thc distortions arc more serious, as with amorphous solids, the electronic changes can be significant.) It is true that an impurity atom will introduce Fourier components of U(r) at wavevectors that are not reciprocal lattice vectors, but at low impurity con- centration such components are never large in comparison with the U,, arguing fro111 the statistics of random potentials. The Fourier cornpouents at the recip- rocal lattice vectors G will still be large and will give the band gaps, Fermi sur- faces, and sharp x-ray diffraction lines charactcristic of a rcgular lattice. The consequences of alloying will he particnlarly small when the impurity element belongs to the same column of the periodic table as the host element it replaces, because the atomic cores will make rather similar contributions to the effective pote~itids. One measure of the effect of alloying is the residual electrical resistivity, de- fined as the lo\v temperature limit of the resistivity. Here we must distinguish I I I I 0 27 .50 75 100 Cu Atomic percent Au Figure 2 Resistivity of a disordered binary alloy of copper and gold. The variation of the residual resistivity depends on t l ~ r cumpuritiun Cqi\u,_, as x ( l - x ) , which is known as Nordheim's Rule for a disordered alloy. IIere x(l - x) is a measure of the degree of maximum disorder possihle fix a givcn valuc of x. (Johansson and Linde.) between disordered and ordered alloys. An alloy is disordered if the A and B atoms are randonlly arranged, which occurs for a general value of x in the com- position A,B1-,. For special values of x, such as 114: 1/2, and 314 for a cubic structure, it is possible for ordered phases to form, phases in which the A and B atoms lorm an ordered array. Thc distinction between order and disorder is shown in Fig. 1. The effect of order on the electrical resistivity is s h o ~ n in Figs. 2 and 3. The residual resistivity increases with disorder, as discussed for amorphous materials in Chapter 19. The effect is shawl in Fig. 2 for the Cu-Au alloy system. When the specimen is cooled slowly fro~n a high te~nperature, or- dered structures are fornred at Cu,Au and CuAu; these structures have a lowcr residual resistivity by virtue or their order, as in Fig. 3. Thus we can usc thc residual electrical resistivity to measure the effect of alloying in a disordered structure. One atomic percent of copper dissolved in silver (which lies in the same column of the periodic table) increases the 22 Alloys 0 0 25 50 75 100 Cu Atomic percent .4u liigure 3 Effect of ordered pllares UII tl~c resistivity of a binar). alloy Cu,Au,-,. The alloys here have been annealed. whereas those in Fig. 2 have bee11 quc~lclrcd (cooled rapidly). The compositions of low rcsidual resistivity correspond to the ordered compositions Cul3Au and CuAu. (Johansson and Linde.) rcsidnal resistivity by 0.077 pohrr~-cm. This corresponds to a geometrical scattering cross section which is only 3 percent of thc naive "projected area" of the impurity atom, so that the scattering effect is very small. In insulators there is no experimental cbidence for a significant reduction of band gap caused by thc random potential components. For exa~nple, silicon and germanium form homogeneous solid solutior~s, known as substitutional alloys, ovcr the entire composition range, but t l ~ e band edge energies vary- con- tinlionsly with composition from the pure Si gap to the pure Gc gap. It is widely believed, however, that the density of states near the band edges in amorphous ~naterials is snlearecl by thc gross absence of translational syrrln~etry. So~ne of the newr statcs thns formed just inside the gap may not r~ecessaril~ he currcnt-carqing states because they may not extend throughout the crystal. SUBSTITUTIONAL SOLID SOLUTIONS-HUME-ROTHERY RULES We now discuss substitutional solid solutions of one metal A in another metal B of different valence, where A and B occup): at random; equivalent sites in the structure. Hurne-Rothery treated the empirical requirements for the stability of a solid solution of A and B as a single phase system. One requirement is that the atomic diameters be compatible, which means that they should not differ by more than 15 percent. For example, the diameters are favorable in the Cu (2.55 A) - Zn (2.65 A) alloy system: zinc dissolves in copper as an fcc solid solution up to 38 atomic percent zinc. The diameters are less favorable in the Cu (2.55 A) - Cd (2.97 A) system, where only 1.7 atomic percent cadmium is soluble in copper. The atomic diameters referred to copper are 1.04 for zinc and 1.165 for cadmium. Although the atomic diameters may be favorable, solid solutions will not form when there is a strong chemical tendency for A and B to form "intermetal- lic compoimds," which are compounds of definite chemical proportions. If A is strongly electronegative and B strongly electropositive, compounds such as AB and A,B may precipitate from the solid solution. (This is different fro111 the for- mation of an ordered alloy only by the greater chemical bonding strength of the intermetallic compounds.) Although the atomic diameter ratio is favor- able for As in Cu (1.02), only 6 atomic percent As is solublr. The diameter ratio is also favorable for Sb in Mg (1.06), yet the solllbility of Sb in Mg is very small. The electronic stnlctrlre of alloys can often be described by the average number of conduction electrons (or valence electrons) per atom, denoted by n. In the alloy CuZn the value of n is 1.50; in CuAl, n = 2.00. Changes in electron concentration determine stmctural changes in rnany alloy systems. The phase diagram of the copper-zinc system1 is shown in Fig. 4. The fcc structure of pure copper (n = 1) persists on the addition or zinc (n. = 2) until the electron concentration reaches 1.38. A bcc structure occurs at a minimum electron concentration of about 1.48. The y phase exists for the approximate range of n between 1.58 and 1.66, and the hcp phase E occurs near 1.75. The term electron compound denotes an intermediate phase (such as the p phase of CuZn) whose crystal structure is determined by a fairly well de- fined electron to atom ratio. For many alloys the ratio is close to the Hume- Rothery rules: 1.50 for the P phase, 1.62 for the y phase, and 1.75 for thc r phase. Representative experimental values are collcctcd in Table 1, based on the usual chemical valence of 1 for Cu and Ag; 2 for Zn and Cd; 3 for A 1 and Ga; 4 for Si, Ge, and Sn. The Hnme-Rothery rilles find a simple expression in terms of the band theory of nearly free electrons. The observed limit of the fcc ~11ase occurs 'The phases of interest are usually denoted by metallurgists by Creek characters: in the Cu-Zn system we havc a (fcc), p (bcc), y (complex cubic cell of 52 atoms), E (hcp) and 1) (hcp); E and q differ considerably in c/n ratio. The meaning of the characters depends UII the alloy systcm. 22 Alloys 625 Atomic percent zinc WcighL pcrcen t zinc Figure 4 Equilibrium diagram of phases in the cnpper-zinc alloy system. The a phase is fcc; p and p' are bcc; y is a complex structure; E and are both hcp, but E has a cia ratio near 1.56 and q (for pure ZII) has c/u = 1.86. Tlre B' phase is ordcrod boc, by which wc nlcau that most ol the Cu atorns occupy sites on our sr suhlattire and most of the Z n atoms occnpy s i t ~ s on a second sc sublattice that iuterpenetrates the first sublattice. The P phase is disordered bcc: any site is rq~rdlly likely to be occupied by a Cu or ZII aturn, al~rrort irrcspcctivc of what alorns arc in Lhe neighboring sites. Table 1 Electronlatom ratios of electron cnmpnunds Mirrirnurr~ fcc phase bcc phase y-phase hcp phase honndarv ho~rndarv hourrdarier bou~rdaries Cu-Zn 1.38 CII-A1 1.41 Cu-Ga 1.41 Cu-Si 1.42 Cu-Ce 1.36 Cu-Sn 1.27 ilg-Zn 1.38 Ag-Cd 1.42 Ag-A1 1.41 close to the electron concentration of 1.36 at which an inscribed Fermi sphere makes contact with the Brillouin zone boundary for the fcc lattice. The ob- sewed electron concentration of the bcc phase is close to thc concentration 1.48 at which an inscribed Fermi spherc makes contact with the zone bound- ary for the bcc lattice. Contact of the Fermi sphere with the zone boundary for the y phase is at thc concentration 1.54. Contact for the hcp phase is at the concentration 1.69 for the ideal cla ratio. Why is there a connection between the electron concentrations at u.hic11 a new phase appears and at which the Fermi surface makes contact with the boundary of the Brillouin zone? We recall that thc cncrgy hands split into hvo at the region of contact on the zone boundary (Chapter 9). If we add more electrons to the alloy at this stage, thcy ~411 have to be accommodated in the upper band or in states of high energy near the zone corners of the lower band. Both options are possible, and both involve an increase of energy. It may also he energetically favorable for the crystal structure to change to one which can contain a Fermi surface of larger volun~e (more electrons) before contact is made with the zone boundary. In this way H. Jones madc plausible the se- quence of structures fcc, bcc, y, hcp with increasing electron concentration. Measurements of the latticc parameter of Li-Mg alloys are shown in Fig. 5. In thc range shown the structure is bcc. The lattice contracts during the initial stages of the addition of Mg to Li. When the lithiurn content drops below 50 atomic percent, corresponding to an average electron concentration increasing above 1.5 per atom, the lattice starts to expand. We have sccn that Figure 5 Lattice parameter of body-centered cubic n1agnesrn111-lithiu111 alloys. (After D. W. Lewson.) G 3.520- + .- a E 2 3.510- + C c 5 2 3.500- I B g 3.490- 0, 2 2 3.480- 3.470 - > - I I I I I I I 20 30 40 50 GO 70 80 90 100 Atonlic percent Litllium 22 Alloys 627 Figure G Number of orbitals per unit energy range for the first Brilluui~l zone of the fcc and bcc lat- tices, as a function of enera. for a spherical Fermi surface, contact with the zone boundaly is established at 7~ = 1.48 electrons per atom, in a bcc lattice. It appears that the expansion of the lattice arises from the onset of overlap across the zone boundary. The transformation from fcc to bcc is illustrated hy Fig. 6; this shows the numher of orbitals per unit energy range as a fi~nction of enera, for the fcc and bcc structures. As the number of electrons is increased, a point is reached where it is easier to accornmodatc additional electrons in the Brillouin zone of the bcc lattice rather than in the Rrillouin zone of the fcc lattice. The figure is draw11 for copper. ORDER-DISORDER TRANSFORMATION The dashed horizontal line in the beta-phase (hcc) region of the phase dia- gram (Fig. 4) of the Cu-Zn system represents the transition temperature be- tween the ordered (low temperature) and disordered (high temperature) states of the alloy. In the common ordered arrange~rient of an AB alloy uith a bcc structure, all thc nearest-neighbor atoms of a B atom are A atoms, and vice versa. This arrangement results when the dominant interaction among the atoms is an attraction between A and B atoms. (If the AR interaction is weakly attractive or repulsive, a two-phase system is formrd in which some crystallites are largely A and other crystallites are largcly B.) The alloy is completely ordercd in equilibrium at absolute zero. It becomes less ordered as the temperafixre is increased, until a transition temperature is reached above which thr str~icture is disordered. The transition temperature niarks the disappearance of long-range order, which is order over many inter- atomic distances, but some short-range order or correlation among near ncighhors may persist above the transition. Thc long-range order in an AB alloy is shown in Fig. 7a. Long- and short-range order for an alloy of con~position AB, is given in Fig. 'ib. The degree of order is defined below. Short-range order 3 Temperature --t (b) Figure 7 (a) Jang-range order verslts temperature for an AB alloy. The tranrforn~atiun is seuu~~d order. (b) Long-range and short-range order for an An, alloy. The transformation for this composi- tion is first order. If an alloy is cooled rapidly lrom high temperatures to a temperature below the transition, a metastablc condition may he produced in which anon- cq~iilihrilim disorder is frozen in the structure. The reverse effect occurs when an ordered specimen is disordered at constant temperature by heavy irradia- tion with nuclear particles. The degree of order may be investigated experi- mentally by x-ray diffraction. The disordered structure in Fig. 8 has diffraction lines at the same positiorls as if t l ~ e lattice points were all occupied by only one type of atom, because the effective scattering power of each plane is eqlial to the average of the A and B scattering powers. The ordered str~tcture has extra diffraction lines not posscsscd by the disordered str~icture. The extra lines are called superstructure lines. The use of the terms order and disorder in this chapter always refers to regular lattice sites; it is the occupancy that is ra~ldonrly A or B. Do not con- fuse this usage with that of Chapter 19 on noncrystalline solids where therc are no regular lattice sites and the structure itself is random. Both possihilities occur in nature. 22 Alloys 629 (6 Figure 8 X-ray powder photographs in AuCu, alloy. (a) Disordered by quenching from T > T,; (b) ordered by annealing at T < T,. (Courtesy of 6. M. Gordon.) The structure of the ordered CuZn alloy is the cesium chloride structure of Chapter 1. The space lattice is simple cubic, and the basis has one Cu atom at 000 and one Zn atom at & ; . The diffraction structure factor This cannot vanish because fc, + fin; therefore all reflections of the simple cubic space lattice will occur. In the disordered structure the situation is different: the basis is equally likely to have either Zn or Cu at 000 and either Zn or Cu at & $ ; . Then the average structure factor is (~(hkZ)) = (f) + (f) e-zli(h+k+l) , (2) where (f) = $(fc, +fin). Equation (2) is exactly the form of the result for the bcc lattice; the reflections vanish when h + k + I is odd. We see that the or- dered lattice has reflections (the superstructure lines) not present in the disor- dered lattice (Fig. 8). Elementary Theory of Order We give a simple statistical treatment of the dependence of order on tem- perature for an AB alloy with a bcc structure. The case A,B differs from AB, the former having a first-order transition marked by a latent heat and the latter having a second-order transition marked by a discontinuity in the heat capacity (Fig. 9). We introduce a measure of the long-range order. We call one simple cubic lattice a and the other b: the bcc structure is composed of the two inter- penetrating sc lattices, and the nearest neighbors of an atom on one lattice lie on the other lattice. If there are N atoms A and N atoms B in the alloy, the Figure 9 Heat capacity versus tem- perature of CuZn alloy @-brass). Temperature in "C long-range order parameter P is defined so that the number of A's on the lattice a is equal to :(1 + P)N. The number of A's on lattice h is equal to :(I-P)N. When P = 2 1 , the order is perfect and each lattice contains only one type of atom. When P = 0, each lattice contains equal numbers of A and B atoms and there is no long-range order. We consider that part of the internal energy associated with the bond en- ergies ofAA, AB, and BB nearest-neighbor pairs. The total bond energy is where Nq is the number of nearest-neighhor i j bonds and U,i is the energy of an i j bond. The probability that an atom A on lattice a will have an AA bor~d is equal to the probability that an A occupies a particular nearest-neighbor site 0 1 1 b, times the number of nearest-neighbor sites, which is 8 Tor the bcc structure. We assume that the probabilities are independent. Thus, by the preceding ex- pressions for the number of A's on a and 6, 22 Alloys 631 The energy ( 3 ) becorrles E = E, + 2NP2u . where Eoz2N(C',,+li,B+2UAn); U=2UAu-UA4-CTuB. (6) \tie now calculate the entropy of this distribution of atoms. There are i ( 1 + P)N atoms A and i ( 1 - P)N atoms B on lattice a; there are i ( 1 - Y ) N atoms A and i ( 1 + P)N atoms B on lattice b. The number of arrangements G of these atoms is From thc dcfinition of the entropy as S = kB In G, we have, nsing Stirling's approximation, S = 2Nk, In 2 - hlkB[(l + P)ln(l + P) + ( 1 - P) In (1 - P)] . (8) This defines the entropy of mixing. For P = t-1: S = 0; for P = 0, S = 2NkB In 2. The equilibrium order is determined by the requirement that the free energy F = E - TS be a minimum with respect to the order parameter P. On differentiating F with respect to P, we have as the condition for the minimum 1 + P 4NPL' + NkBT In - = 0 . I - P The transcendental equation for P map be solved graphically; we find the sr~~oothly decreasirlg curve sl~owr~ in Fig. 7a. Near the transition we may expand ( 9 ) to find 4NPL' + 2Nk,?'Y = 0. At the transition tenlperature P = 0, so that For a transition to occur, the effective interaction U must be negative. The short-range order parameter r is a Irieasure of tlle fraction of the average number q of nearest-neighbor bonds that are A L 3 bo~lds. \Vlien com- pletely disordered, an AB alloy has an average of four AB bonds about each atom A. Thc total possiblc is eight. N7e may define so that r = 1 in complete order and r = 0 in complete disorder. Observe tlrat r is a measiire only of thc local ordcr about an atom, whereas the long-range order parameter P refers to the piirity of the entire popillation on a given si~h- lattice. Above the transition temperature T, the long-range order is rigorously zero, but the short-range order is not. PIUSE DIAGRAMS There is a large amount of information in a phase diagram even for a bi- nary system, as in Fig. 4. The areas enclosed by curves relate to the equilib- rium state in tltat region of conrposition and te~nperature. The curves miark the course of phase transitions as plotted in the T-x pla~ie, where x is the composi- tion parameter. The eq~iilibril~m statc is thc statc of minimum free energy of the binary system at given T, x. Thus the analysis of a phase diagram is the snbject of thermodynamics. Several extraordinary results come out of this analysis, in particular the existence of low-melting-point eutectic compositions. Because the analysis has been treated in Chapter 11 of TP, we only outline the principal results here. Two substances will dissolve in each other and form a homogeneous mix- ture if that is the configuration of lowcst frcc cncrgy accessible to thc compo- nents. The substances will form a heterogeneous mixtiire if the combined free energy of the two separate phases side by side is lower than the free e n e r g of the homogeneous mixture. Now we say that the mixture exhibits a solubility gap. In Fig. 4 we see that conipositions near Cuo6,Zn,,, are in a solubility gap and are mixtures of fcc and bcc phases of different structures and con~pusi- tions. The phase diagram represents the temperature dependence of the solu- bility gaps. When a small fraction of a homogcncous liquid frcczcs, thc composition of the solid that forms is almost always different from that of the liquid. Consider a horizontal section near the composition Cu, ,,Z%.,, in Fig. 4. Let x denote the weight percent of zinc. At a given temperature, there are three regions: x > x , , the equilibrium system is a l~omogeneous liquid. xs < x < x , , there is a solid phase of composition xs and a liquid phase of composition xL. x < x,, equilibrium system is a homogeneous solid. The point xL traces a curve called the liquidus curve, and the point xs traces the solidus curve. Eutectics. Mixtures with two liquidus branches in their phase diagram are called eutectics, as in Fig. 10 for the Au-Si system. The minimum solidification temperature is called the entectic temperature; here the composition is the eutectic composition. The solid at this composition consists of two separate phases, as in the microphotograph of Fig. 11. There are many binary systems in which the liquid phase persists to tem- peratures below the lower rrielting temperature of the constituents. Thus Auo69Sio,0, solidifies at 370°C as a two-phase heterogeneous mixture, although Au and Si solidify at 1063OC and 1404"C, respectively. One phase of thc eutec- tic is nearly pure gold; the other nearly purc silicon. 22 Alloys 633 Pure Au Atomic percent silicon Pure Si Figure 10 Eutectic phase diagram of gold-silicon alloys. The eutectic consists of two branches that come together at T, = 370 "C and xg = 0.31 atomic percent Si. (After Kittel and Kroemer, TP.) c ---.
l 10 fi111 Figure 11 Microphotograph of the Ph-Sn eutectic. (Courtesy of J. D. Hunt and K. A. Jackson.) The Au-Si eutectic is important in semiconductor technology because the entectic permits low temperature welding of gold contact wires to silicon de- vices. Lead-tin alloys have a similar eutectic of Pbo,,Sno 74 at 183'C. This or nearby compositions are used in solder: nearby if a range of melting tempera- tures is desired for ease in handling. TRANSITION METAT, ALLOYS When we add copper to nickel, the effective magneton number per atom decreases linearly and goes through zero near Cu, 60Ni0 4 0 , as shown in Fig. 12. At this composition the extra electron from the copper has filled the 3d band, or the spin-up and spin-down 3d sub-bands that were shown in Fig. 12.7b. The situation is shown schematically in Fig. 13. Figure 12 Bohr magneton numbers of nickel-copper alloys. 0.60 Electron Figure 13 Distribution of electrons in the alloy 60Cu40Ni. The extra 0.6 electron provided by the copper has filled the d band entirely and in- creased slightly the number of electrons in the s band with respect to Fig. 12.7b. 4s 0.2 0.1 0' 0 10 20 30 40 50 60 70 Percent copper in nickel \ \ \ \ \ . 22 Alloys 635 Energy relative to P'ermi energy (eV) Figure 14 Density of states in nickel. (V L. Moruzzi, J. F. Janak. and A. R. IVilliams.) For simplicity the block drawings represent (he density of statcs as uni- forrri in energy. The actual density is known to be far from nniform; the result of a 11iodern calculatio~l is shown in Fig. 14 for nickel. The width of the 3d band is about 5 eV. .4t thc top, where the magnetic effects are determined, the density of states is particularly high. The average density of states is an order of magnitude higher in the 3d band than in the 4s band. This enhanced density of states ratio gives a rough indication of the expected enhancement of the electronic heat capacity and of the paramagnetic susceptibility in the nonfcr- romagnetic trar~sition ~rletals as compared with the simple monovalent metals. Figure 15 shows the effect of thc addition of small amounts of other ele- ments to nickrl. On the hand model an alloymg metal with .
: valence electrons ontside a filled d shell is expected to decrease the magnetization of nickel by approximately z Bohr magnetons per solute atom. This sirriple relation holds well for Sn, Al, Zn, and Cu, with z = 4, 3, 2, and 1, respectively. For Co, Fe, and Mn the localized rrlorrler~t model of Friedel accounts lor effective z values of -1, -2, and -3, respectively. The average atomic magnctic moments of binaly alloys of the elements in thc iron gronp are plotted in Fig. 16 as a function of the concentratio11 of Added elenlents in atum percent Figure 15 Saturation nlagnctizatio~r of nickel alloys in Buhr nragnetons per atorra as a fu11ctiu11 of the atomic percent of salute element. electrons outside the 3p shell. This is called a Slater-Yauling plot. The main se- qucnce of alloys on the right-hand branch follows the rules discussed in con- nection with Fig. 15. As the electron concentration is decreased, a point is reached at which neither of the 3d sub-hands is entirely filled, and the mag- netic moment then decreases toward the left-hand side of the plot. Electrical Conductivity. It might be thought that in the transition metals the availability of the 3d band as a path for conduction in parallel with the 4s band would increase the conductivity, but this is not the way it works out. The resistiv- ity of the .s electron path is increased by collisions with the d electrons; this is a powerfill extra scattering mechanism not availablr when the d band is fillcd. 3.0 2.5 - e + . g 2.0 E 4 1.5 e, 3 1.0 2 4 0.5 0 Cr Mn Fe Co Ni Cu 6 7 8 9 1 0 11 Electmn concer~tratiori Figure 16 Averagc atomic moments of binav alloys of the elements in the iron group. (Bozorth.) llrc: compare the values of the electrical resistivities of Ni, Pd, and Pt in microhm-cm at 18'C with that of the noble metals Cu, Ag, and Au immedi- ately following them in the periodic table: The resistivities of the noble metals are lower than those of the transition metals by a factor of the order of 5. This shows the effectiveness of the s-d scattering mechanism. KONDO EFFECT In dilute solid solutions of a magnetic ion in a nonmagnetic metal crystal (such as Mn in Cu), the exchange coupling between the ion and the conduc- tion electrons has important consequences. The conduction electron gas is Figure 17 Magnetization of a free electro~~ Fer~rli gas at T = 0 in neighborhood of a point mag- netic moment at the origin r = 0, according to the RKKY theory The horizontal axis is 2k,r, where k , is the wavevector at the Fermi surface. (de Gennes.) magnetized in the Vicinity of the magnetic ion, with the spatial dcpendencc shown in Fig. 17. This magnetization causes an indirect exchange interaction" between two magnetic ions, because a second ion perceives the magnetization induced by the first ion. The interaction, known as the Friedel or RKKY inter- action, also plays a role in the magnetic spin order of the rare-earth metals, where the spins of the 4 f ion cores are coupled together by the rriag~ietizatiorl iri- duced in the conduction electron gas. A conseqlience of the magnetic ion-conduction clcctron interaction is the Kondo effect, discussed in a different context in Chapter 18. A minimnm in the electrical resistiVity-temperature curve of dilute magnetic alloys at low temperatures has been observed in alloys of Cu, Ag, Au, Mg, Zn with Cr, Mn, and Fe as impurities, among others. The occurrence of a resistance minimum is connected with the existence of localized magnetic moments on the impurity atoms. Where a resistance minimum is found, there is inevitably a local moment. Kondo showed that the anomalorisly high scattering probability of magnetic ions at low tcmperaturcs 'A r e v i c ~ of indirect excharige iriteractiu~is iri metals is giver1 by C. Kittel, Solid state pliysics 22, 1 (1968); a review of the Kondo effect is given by J. Kondo, "Theory of dihte magnetic alloys," Solid state physics 23, 184 (1969) and A. I. Heeger, "Localized moments and nonmoments in metals: the KUII~IJ effect," Solid state physics 23, 248 (1969). The notation RKKY stands for Ruderman, Kittel, Kasuya, and Yosida. Figure 18 A comparison of experimental and theoretical results for the increase of electrical re- sistivity at low temperatures in dilute alloys of iron in gold. The resistance minimum lies to the right of the figure, for the resisti.ity iircreases at high temperatures because of scattering of elec- trons by thermal phonons. The experiments are due to D. K. C. MacDonald, W. B. Pearson, and I. M. Templeton; the tlreo~y is by J. Kondo. 4 n exact sohitiu~~ was given by K. Wilson. is a consequence of the dynamic nature of the scattering by the exchange coupling and of the sharpness of the Fermi surface at low temperatures. The temperature region in which the Kondo effect is important is shown in Fig. 18. The central result is that the spin-dependent contribution to the resistivity is p,,,,, = cp, I + - ln T = cp, - cp, In T , [ : ! ] where J is the exchange energy; z the numher of nearest neighbors; c the con- centration; and p, is a measure of the strength of the exchange scattering. We see that the spin resistivity increases toward low temperatures ifJ is nega- tive. If the pho~lon contribution to the electrical resistivity goes as T5 in the region of interest and if the resistivities are additive, then the total resistivity has the form with a minimum at clp/dT = 5aT' - cp, l?' = 0 , (14) whence T,, = ( C ~ , / S ~ ) ~ ~ The ten~perature at which tlie resistivity is a minimum varies as the one-fifth power of the concentration of the magnetic impurity, in agreement with experi- ment, at least for Fe in Cu. Problems 1. Superlattice lines in C u d u . Cu,Au alloy (75% Cu, 25% Au) has an ordered state below 40OoC, in which the gold atoms occupy the 000 positiorrs and the copper 11 I I 1 1 atoms the ,,0, 505, and 05, positions in a face-centered cubic lattice. Give the in- dices of the new x-ray reflections that appear when the alloy goes fro111 the disor- dered to the ordered state. List all new reflections with indices s 2 . 2. Configurational heat capacity. Derive an expression in terms of P(T) for the heat capacity associated with ordeddisorder effects in an AB alloy [The entropy (8) is called the configurational entropy or entropy of mixing.] APPENDIX A: TEMPERATURE DEPENDENCE OF THE REFLECTION LINES . . . I came to the conclusion that the sharpness of the interference lines would not suffer but that their intensity should diminish with in- creasing angle of scattering, the more so the higher the temperature. P. Debye As the temperature of the crystal is increased, the intensity of the Bragg- reflected beams decreases, but the angular width of the reflected line does not change. Experimental intensities for aluminum are shown in Fig. 1. It is sur- prising that we can get a sharp x-ray reflection from atoms undergoing large amplitude random thermal motion, with instantaneous nearest-neighbor spac- ings differing by 10 perccnt at room temperature. Before the Laue experiment Figure 1 The dependence of intensity on temperature for the (hOO) x-ray reflections of aluminum. Reflections (hOO) with h odd are iorbidden for an Icc structure. (ACtcr R. M. Nicklow and R. A. Young.) was done, but when the proposal was discusscdl in a coffee house in Munich, the objection was made that the instantaneous positions of thc atoms in a crystal at room temperature are far from a regular periomc array, because of the large thermal fluctuation. Therefore, the argument went, one should not expect a well-defined diffracted beam. But such a beam is found. The reason was given by Debye. Consider the radiation amplitude scattered by a crystal: let the position of the atoni nomi- nally at r j contain a term u(t) fluctriating in time: r(t) = rj + u(t). We suppose each atom fluctuates independently about its own equilihri~~m position.2 Then the thermal average of the structure factor (2.43) contains terms f , exp(-iG. ?)(exp(-iG. u)) , (1) where ( . . . ) denotes thermal average. The series expansion of the exponential is (exp(-iG - u)) = 1 -i(G. u) - ; ( ( G . u ) ~ ) + . .. . (2) But (G . u) = 0, because u is a random thermal displacerrierit uncorrelated with the direction of 6. Further. The factor arises as the geometrical average of cos2% over a sphere. The function has the same series expansion as (2) for the first two terms shown here. For a harmonic oscillator all terms in the series (2) and (3) can be shown to be iden- tical. Then the scattered intensity, which is the square of the amplitude. is where I , is the scattered intensity from the rigid lattice. The exponential factor is the Debye-Waller factor. Here (u2) is the mean square displacement of an atom. The thermal aver- age potential energy (U) of a classical harmonic oscillator in three dimensions is ak,~, whence 'P P. Ewald, private comm~mication. 'This is the Einstein model of a solid; it is not a very good model at low temperatures, but it works well at high temperatures. Appendix 643 where C is the force constant, M is the mass of an atom, and w is the frequency of the oscillator. We have used the result w2 = C/M. Thus the scattered intensity is where hkl are the indices of the reciprocal lattice vector G. This classical result is a good approximation at high temperatures. For quantum oscillators (u2) does not vanish even at T = 0; there is zero- point motion. On the independent harmonic oscillator model the zero-point energy is ghw; this is the energy of a three-di~nensional qnantum harmonic oscillator in its ground state referred to the classical energy of the same oscilla- tor at rest. Half of the oscillator energy is potential energy, so that in the ground state whence, by (4), at absolute zero. If G = 10' cm-l, w = and M = g, the argu- ment of the exponential is approximately 0.1, so that 1/1, = 0.9. At absolute zero, 90 percent of the beam is elastically scattered and 10 percent is inelasti- calIy scattered. We see from (6) and from Fig. 1 that the intensity of the diffracted line decreases, but not catastrophically, as the temperature is increased. Reflec- tions of low G are alfectcd less than reflections of high G. The intensity we have calculated is that of the coherent diffraction (or the elastic scattering) in the well-defined Bragg directions. The intensity lost from these directions is the inelastic scattering and appears as a diffuse hackground. In inelastic scat- tering the x-ray photon causes tlie excitation or de-excitation of a lattice vibra- tion, and the photon changes direction and energ).. At a given temperature thc Debye-Waller factor of a diffraction line de- creases with an increase in the magnitude of the reciprocal lattice vector G as- sociated with the reflection. The larger IGI is, the weaker the reflection at high temperatures. Thc theory we have worked out here for x-ray reflection applies equally well to neutron diffraction and to the Mijssbauer effect, tlie recoil- less emission of gamma rays by nuclei bound in crystals. X-rays can be absorbed in a crystal also by the inelastic processes of photo- ionization of electrons and Compton scattering. In the photoeffect the x-ray photon is absorbed ancl an electron is ejected from an atom. In the Compton effect the photon is scattered inelastically by an electron: the photon loses energy ancl the clectron is ejected from an atom. The depth of penetration of the x-ray beam depends on the solid and on the photon energy, but 1 cm is typical. A diffracted beam in Rragg reflection may rerriove the energy in a much shorter distance, perhaps cm in an ideal crystal. APPENDIX B: EWALD CALCULATION OF LATTICE SUMS The problem is to calcnlate the electrostatic potential experienced by one ion in the presence of a U the other ions in the crystal. We consider a lattice made up of ions with positive or negative chargcs and shall assume that the ions are spherical. We compute the total potential cp = cp, + cp2 at an ion as the slim of tsvo dis- tinct but rdatcd potentials. The potential cp, is that of a structure with a Gaiissian distribution of charge situated at each ion site, with sips the same as those of the real ions. According to thc definition of the Madelung constant, the charge distribution on the reference point is not considered to contribute to the poten- tial or c p , (Fig. la). We therefore calculate the potential pl as the difference 'PI = (Pa - (Ph Figure 1 (a) Cliarge distlihution used for computing potential rp,; the potential rpn is cornputcd (it incllldes the dashed curve at the reference point), whilc cph is the potential of the dashed curve alone. (b) Charge distributio~~ for poterrtial qz. The reference point is denoted by an X. of two potentials, p , being the potential of a continuous series of Gaussian dis- tributions and a being the potential of the single Gaussian distribution on the reference point. The potential p2 is that of a lattice of point charges with an additional Gaussian distribution of opposite sign superposed upon the point charges (Fie. lb). \, The point of splitting the problem into the two parts 9, and q2 is that by a suitable choice of the parameter determining the width of each Gaussian peak we can get very good convergence of both parts at the same time. The Gaussian distrihiitions drop out completely on taking the surn of the separate charge distributions giving rise to cpl and p , , so that the value of the total potential is independent of the width parameter, but the rapidity of convergence depends on the value chosen for that parameter. \it: calculate first the potential q, of a continuoi~s Gaussim distribution. We expand pa and the charge density p in Fourier series: where G is 2 ~ r tinies a vector in the reciprocal lattice. The Poisson equation is so that c, = 4 7 7 p , / ~ ~ We suppose in finding pG that there is associated with each lattice point of the Bravais lattice a basis containing ions of charge q, at positions r, relative to the lattice point. Each ion point is therefore the center of a Gaussian charge distrihution of density where the factor in front of the exponential ensures that the total charge associ- ated with the ion is q,; the range parameter 7 is to he chosen judiciously to ensure rapid convergence of the final result (6), which is in value independent of T. L % ' e would normally evaluate pG by multiplying both sides of (2) by exp(-iG . r) and integrating over the volume A of one cell, so that the charge distrihution to be considered is that originating on the ion points within the cell and also that of the tails of the distributions originating in all other cells. It is easy to see, however, that the integral of the total charge density times exp[-(iG . r)] over a single cell is equal to the integral of the charge density originating in a singl~ cell times exp[-(iG . r)] over all space. We have therefore = I 2 q,in/~r"Yexp-~(r - r.)qexp(-i~ - r) ib 111 ' space This expression is readily evaluated: where S(G) = q,exp(-iG rt) is just the structure factor (Chapter 2) in appropriate units. Using (1) and (3), At the origin r = 0 we have 4 n cpa = S(G)G-2 eV(-G2/49) . C The potential cp,, at tht: reference ion point i due to thc central Gaussian distribution is and so The potential cp, is to be evaluated at the reference point, and it differs from zero because other ions have the tails of their Gaussian distributions overlapping the reference point. The potential is due to three contributions from each ion point: where the terms are from the point charge, frorrl the part of the Gaussian dis- tribution lying inside a sphere of radius r, about the lth ion point, and from that part lying outside the sphere, respectively. On substituting for p ( r ) and carrying out elementary manipulations, we have where Finally, is the desired total potential of the reference ion i in thc field of all the other ions in the crystal. In the application of the Ewald method the trick is to choose TJ such that both sums in (6) converge rapidly. Ewald-Kornfeld Method for Lattice Sums for Dipole Arrays Kornfeld extended the Ewald method to &polar and quadrupolar arrays. We discuss here the field of a dipolc array at a point which is not a lattice point. According to (4) and (5) the potential at a point r in a lattice of positive unit point charges is where rl is the distance from r to the lattice point 1. The first term on the right gives the potential of the charge distribution p = (TJ/T~)~!~ e ~ p ( - ~ ? ) about each lattice point. By a well-known relation in electrostatics we obtain the potential of an array of unit dipoles pointing in tlie s direction by taking -d/dz of the above potential. The term under discussion contributes and the s component of the electric field from this term is E, = @p/az2, or The second terrn on the right of (7) after one differentiation gves and the z component of this part of the field is 2 iz;[(3~(firl)/r;) f (6/r-f )(T/T)~' exp(-~r,? 1 The total Ez is given by the sum of (8) and (9). The effects of any number of lattices may be added. APPENDIX C: QUANTIZATION OF ELASTIC WAVES: PHONONS Phonor~s were introduced in Chapter 4 as quantized elastic waves. How do we quantize an elastic wave? As a simple model of phonons in a crystal, con- sider the vibrations of a linear lattice of particles connected by springs. We can quantize the particle motion exactly as for a harmonic oscillator or set of cou- pled harmonic oscillators. To do this we make a transformation from particle coordinates to phonon coordinates, also called wave coordinates because they represent a traveling wave. Let N particles of mass M be conncctcd by springs of force constant C and length a. To fix the boundary conditions, let the particles form a circular ring. We consider the tra~lsverse displacements of the particles out of the plane of the ring. The displacement of particle s is q, and its momentum is P,~. The Hamiltonian of the system is The Hamiltonian of a harmonic oscillator is and the energy eigcnvalues are, where n = 0,1,2,3, . . . , The eigenvalue problem is also exactly solvable for a chain with the diffrrent Hamiltonian (1). To solve (1) we make a Fourier transformation from the coordinates p,, q, to the coordinates Pk, Qk, which arc known as phonon coordinates. Appendix 649 Phonon Coordinates The transformation from the particle coordinates q, to the phonon coordi- nates Qn is used in all periodic lattice problems. We let consistent with the inversc transformation Qk = N-'/' z q, exp(-iksa) . (5) Here the N values of the wavcvector k allowed by the ~eriodic boundary con- dition q, = y , , , are given by: 1 % need the transformation from the particle momentum p, to the momen- turn Yk that is canonically conjugate to the coordmate Qk. The transformation is This is not quite what onc wonld obtain by the naive substitution of p for y and P for Q in (4) and (S), hecause k and -k have been interchanged between (4) and (7). L V e verify that our choice of PI and Qk satisfies the quantum commutation relation for canonical variables. We lorm the commutator Because the operators q, p are conjugate, they satisb the commutation relation [q,,p,l = ifiS(r, s) , (9) where S(r,s) is the Kronecker delta symbol. Thus (8) hecomes [Qk,Pk,] = N-I i'z exp[-i(k - kf)ra] = ih3(k, k') , (10) so that Q,, Pk also are conjugate variables. Here we have evaluated the summa- tion as z exp-i(k - k')ml = 2 exp[-i2v(n - nl)rlN] (11) = N6(n, n') = NS(k, k') , where we have used (6) and a standard result for the finite series in (11). We carry out the transformations (7) and (4) on the hamiltonian (I), and make use of the sum~nation (11): X exp(ikkss)[exp(ik'a) - 1 1 = 2 ~ Q ~ Q - ~ ( I - cos ka) . (13) k Thus the hamiltonian (1) becomes, in phonon coordinates, If we introduce the symbol wk defined by wk = ( 2 ~ / ~ ) ~ ' ~ ( 1 - cos ka)In , we have the phonon hamiltonian in the form 1 PkP-k + - M6.J: QkQ-k . k 2 1 (16) The equation or motion of the phonon coordinate operator Qk is found hy the standard prescription of quantum mechanics: ifii), = [Q,, H] = ifiP-klhl , (17) with H given by (14). Further, using the co~n~nutator (17), i h ~ ~ = [ Q ~ , H] = M-'[Y-~,H] = ihw:~, , (18) so that Q~ + wiQk=O . (19) This is the equation of motion of a harmonic oscillator with the frequency wk. The energy eigenvalues of a quantum harmonic oscillator are where the quantum number nk = 0, 1, 2, . . . . The energy of the entire system of all phonons is o=Z(nk+;)fiy k (21) This resnlt demonstrates the quantization of the energy of elastic waves on a line. Appendix Creation and Aaaihilation Operators It is helpful in advanced work to transform the phonon hamiltonian (16) into the form of a set of harmonic oscillators: Here a : , ak are harmonic oscillator operators, also called creation and destruc- tion operators or bosun operators. The transformation is derived below. The bosori creation operator a+ which "creates a phonon" is defined by the property when acting on a harmonic oscillator state of quantum number n, and the boson annihilation operator a which "destroys a phonon" is defined by the property ajn)=nmlr~-l) . (24) It follows that a+aln) = a+nl"ln - 1) = nln) , (25) so that In) is an eigenstate of the operator a+u with the integral eigenvalue n, called the quantum number or occupancy of the oscillator. When the phonon mode k is in the eigenstate labeled by nk, we may say that there are n k phonons in the mode. The eigenvalues of (22) are U = Z (nk + ;)nok, in agreement with (21). Because the commutator of the boson wave operators a: and ak satisfies the relation [a,aS]=aa+-ata= 1 . (27) We still have to prove that the hamiltonian (16) can be expressed as (19) in terms of the phonon operators a : , uk. This can be done by the transformation The inverse relations are Qk = (fi/2~@J~)'"(U~+ UZk) ; Pk = i(fiM0~/2)~"(u~- By (4), ( 5 ) , and (29) the particle position operator becomes This equation relates the particle dlsplacemcnt operator to the phonon cre- ation and annihilation operators. To obtain (29) horn (28), we use the properties QIk = Q, ; P k f = P-k (33) which follow from ( 5 ) and (7) by use of the quantum mechanical requirement that y, and p, be hermitian operators: % = y : ; p s = p : . (34) Then (28) follows from the transformations (4), ( 5 ) , and (7). We verify tlrat the commutation relation (33) is satisfied by the operators defined by (28) and (29): [ak, ail = ( 2 f L ) ~ - ' ( ~ ~ k [ ~ k , Q - k l - i[Qk,pkl + i[P-k,Q-kl + [P-k,PkI/MWk) . (35) By use of [QkrPk'] = ilid(k,k') from (10) we have [ak, a ; ] = 6(k, k') . (36) It remains to show that the versions of (16) and (22) of thc phonon hamil- tonian are identical. Sie note that wk = m k from (IS), and we form This e h b i t s thc cqt~ivalence of the two expressions (14) and (22) for H. l i e identify w k = (2C/M)'"(l - cos ka)"' in (15) with the classical frequency of the oscillator mode of wavevector k. The Fermi-Dirac distribution function' may bc derived in several steps by use of a modem approach to statistical mechanics. We outline the argument here. Thc notation is such that conventional entropy S is related to the funda- mental entropy cr by S = k , ~ , and the Kelvin temperature T is related to the fundamental temperature T by T = kBT, where kB is the Boltzmann constant with the value 1.38066 X 10 2 3 J K. The leading quantities are the entropy, the temperature, the Boltzmann fac- tor, the che~nical potential, the Gibbs factor, and thc distribution functions. The 'This appendix follows closely the introduction to C. Kitlel and H. Krue~r~er. T h r m l Physics, 2nd ed., Freeman, 1980. entropy measures the number of quantum states accessible to a system. A closed system might be in any of these quantum states and (we aqsume) with equal prob- ability. The fundamental assumption is that quantum states are either accessible or inaccessible to the system, and the system is equally likely to be in any onc ac- cessible state as in any other accessible state. Given g accessible statcs, the en- tropy is defined as a = log g. The entropy thus defined will be a fimction of the energy U, the number of particles N, and the volu~ne V of the system. When two systems, each of specified energy, are bronght into thermal contact, they may transfer energy; their total energy remains constant, but the constraints on their individual energies are lifted. A transfer of energy in one direction, or perhaps in the other, may increase the product g,g, that measures the number of accessible states of the combined systems. \ % ' h a t wc call the fundamental assumption biases the outcome in favor of that allocation of the total energy that maximizes the number of accessible states: more is better, and more likely. This statement is the kernel of the law of increase of entropy, which is the general expression of the second law of thermodynamics. We have brought two systems into thermal contact so that they may trans- fer energy. What is the most probable outcomc of the encounter? One system will gain energy at the expense of the other, and meanwhile the total entropy of the two systems will increase. Eventually the entropy will reach a maxim~lm for the given total energy. It is not difficult to show that the maximum is at- tained when the value of ( a ~ / a U ) ~ , , for one syste~n is equal to the value of the same quantity for the second system. This equality property for two systems in thermal contact is the property we expect of the temperature. Accordingly, the fundamental temperature T is defined by the relation The use of 1 1 7 assures that energy will flow from high T to low T; no more com- plicated relation is needed. Now consider a ver)r simple example of the Boltzmann factor. Let a small system with only two states, one at energy 0 and one at energy E , be placed in thermal contact with a large system that we call the reservoir. The total energy of the combined systems is U,; when the small system is in thc state of energy 0, the reservoir has energy U,, and will have g(U,) states accessible to it. When the srriall system is in the state of energy E , the reservoir d l have energy Un - E and will have g(UU - E) states accessible to it. By the funda~rlental as- sumption, the ratio of the probability of finding the small system with energy to the probability of finding it with energy 0 is The reservoir entropy a may be expanded in a Taylor series: u(uO - E ) = u ( u ~ ) - E ( ~ U / ~ U ~ ) = a ( u O ) - E/T , (3) by the definition ( 1 ) of the temperature. Higher order terms in the expansion may be dropped. Cancellation of the term e x p [ u ( l i , ) ] , which occurs in the nu- merator and denominator of (2) after the substitution of (3), leaves us with P(e)/P(O) = exp(-€17) . ( 4 ) This is Boltzmann's result. To show its use, we calculate the thermal aver- age energy ( E ) of the two-state system in thermal contact with a reservoir at temperature T: where we have imposed the normalization condition on the sum of the probabilities: p ( 0 ) + P(E) = 1 . (6) The argument can be generalized immediately to find the average energy of a harmonic oscillator at temperature T , as in the Planck law. The most important extension of the theory is to systems that can transfer particles as well as energy with the reservoir. For two systems in diffilsive and thermal contact, the entropy will be a mawi~rlum with respect to the transfer of particles as well as to thc transfer or energ).. Not only must ( a ~ / a U ) , ~ be equal for the two systems, hut (du/dN)U,v must also be equal, where N refers to the number of particles of a given species. The new equality condition is the occa- sion for the introduction2 of the chemical potential p: For two systems in thermal and diffusive contact, T , = T , and p1 = p2. The sign in (7) is chosen to ensure that the direction of particle flow is fro~n high chemical potential to low chemical potential as equilibrium is approached. The Gibbs factor is an extension of the Boltzmann factor ( 4 ) and allows us to treat systems that can transfer particles. The simplest example is a system with two states, one with 0 particles and 0 energy, and one with 1 particle and energy E. The system is in contact with a resemoir at temperature T and chem- ical potential p. \Ve extcnd (3) for the reservoir entropy: 'TP Chapter 5 has a careful treatment of the chemical potential. By analogy with (4), we have the Gibbs factor P(l,~)IPi0,0) = exp[(y -)/TI , for the ratio of the probability that the system is occupied by 1 particle at energy- E to the probability that the system is unoccupied, with energy 0. The rrsnlt (9) after normalization is readily expressed as This is the Ferrrri-Dirac distribution function. APPENDIX E: DERIVATION OF THE dk/dt EQUATION Thc simple and rigorous derivation that follows is due to Kroemer. In quantnm mechanics, for any operator A we have where H is the l~anriltonian. See also C. L. Cook, American J. Yhys. 55, 953 (1987). We let A be the lattice translation operator T defined by where a is a basis vector, here in one dimension. For a Bloch function Thic resnlt is usually written for one band, hnt it holds even if +hk is a linear combination of Bloch states from any nnmher of bands, but having the identi- cal wavevector k in the reduced zone scheme. The crystal hamiltonian No commutes with the lattice translation operator T, so that [Ho,T] = 0. If we add a uniform external force F, then and [H, TI = FaT . From (1) and ( 5 ) , d(T)/dt = (i/fi)(Fa)(T) . From (6) we form (T)d(T)/dt = (iFaIfi) l(T)I2 ; On addition, This is the equation of a circle in the complex plane. The coordinate axes in the plane arc the real and imaginary parts of the eigenvalue exp(ika). If (T) is initially on the unit circle, it will remain on the unit circle. For II/S that satisfy periodic houndaly conditions, (T) can lie on the unit circle only if (I,k is a single Bloch function or a s~lperposition of Bloch functions from different bands, but with the same reduced k. As (T) moves around the unit circle, the wavevector k changes exactly at the same rate for the co~nponents of & in all bands. With (T) = exp(ika), we have from (6) that an exact result. This does not mean that interhand mixing (such as Zener tunneling) does not occur under the influence of applied electric fields. It just means that k evolves at a constant rate for every component of a wave packet. The result is easily extended to three dimensions. APPENDIX F: BOLTZMANN TRANSPORT EQUATION The classical theory of transport processes is bascd on the Boltzmann trans- port equation. We work in the six-dimensional space of Cartesian coordinatcs r and velocity v. The classical distribution function f(r,v) is defined by the relation f(r,v)drdv = number of particles in drdv . (1) The Boltzlnann equation is derived by the following argument. L7e con- sider the effcct of a time displacement dt on the distribution function. The Liouville theorem of classical mechanics tells us that if we follow a volume element along a flowline the distrihntion is conscrvcd: f(t + dt,r + dr,v + dv) = f(t,r,v) , (2) in the absence of collisions. With collisions Thus dt(dflat) + dr - grad, f + d v grad, f = dt(?flflat)mu . (4) Let a denote the acceleration dv/dt; then This is the Boltzmann transport equation. In many problems the collision term (af/at),,,l may be treated by the intro- duction of a relaxation time r,(r,v), defined by the equation Here f, is the distribution function in thermal equilibrium. Do not confuse T, for relaxation time with T for temperature. Suppose that a nonequilihrium dis- tribution of velocities is set up hy external forces which are suddenly removed. The decay of the distribution towards eqiiilihrium is then obtained from (6) as a ( f - f o ) - f - f o at TC ' if we note that dfdat = 0 by definition of the eqnilihrinm distribution. This equation has the solution It is riot excluded that T, may be a function of r and v. We combine (I), ( S ) , and ( 6 ) to obtain the Boltzmann transport equation in the relaxation time approximation: In the steady state aflat = 0 by definition. Particle DifPusion Consider an isothermal system with a gradient of the particle concentra- tion. The steady-state Roltzmann transport equation in the relaxation time ap- proximation becomes where the nonequilibrium distribution function f varies along the x direction. Wc may write (10) to first order as fi = f o - uXrcdf2dx , (11) where we have replaced & ' a x by dfddx. We can iterate to obtain higher order solutions when dcsired. Thus the second order solution is fZ = fO - o,rJf1/dx = f o - u,~&f~/dx + vf<d?fO/dxZ . (12) The iteration may be used in the trcatment of nonlinear effects. Clarsical Distribution Let fO be the distribution function in the classical limit: We are at liberty to take whatever normalization for the distribution fi~nction is most convenient because the transport equation is linear in f and f,. We can take the normalization as in ( 1 3) rather than as in (1). Then and the first order solution (11) for the nonequilibrium distrihiltion becomes f = fa - (cx7zfo/~)(dddx) . (15) The particle flux density in the x direction is where D(6) is the density of electron states per unit volume per unit energy range: Thus The first integral vanishes because c, is an odd function and.fo is an even func- tion of 0,. This confirms that the net particle flux vanishes for the equilihriltm distribution f,. The second integral will not vanish. Before evaluating the second integral, we have an opportunity to make use of what we may know about the velocity dependence of the relaxation time 7,. Only for the sake of example we assume that T, is constant, independent of ve- locity; T, may then be taken out of the integral: Appendix 659 The integral rriay be written as because the integral is just thc kinetic energy density gnr of the particles. IIere j&D(~)de = n is the concentration. The particle flux density is J : , = - (nr,/M)(dgldx) = - (~,~/M)(dn/dx) , (21) because p = T log n + constant . (22) The result (21) is of the form of the diffusion equation with the diffusivity Another possible assumption about the relaxation time is that it is in- versely proportional to the velocity, as in T , = I/(;, where the mean free path 1 is constant. Instead of (19) we have and rlow the integral may he written as i lufuD(.)& = bii where c is the average speed. Thus J: = - $ ( ~ C V L / T ) ( ~ ~ / & ) = -$l~(dnldx) , (26) and the diffusivity is D, = i1c . (27) Fermi-Dirac Distribution The distribution function is To form dfn/dx as in (14) we need the derivative &Idp. We argue below that 4fn/dk = S ( E - p) , (29) at low temperatures T < p. Here 6 is the Dirac delta function, which has the property for a general function F ( E ) that Now consider the integral I , " F(e)(dfO/dp)de At low temperatures &/dp is very large for E - p and is sinall elsewhere. Unless the function F(E) is very rapidly varying near p, we may take F(E) outside the integral, with the value F(p): where we have used dfo/dp = -clfdcle. We have also used f, = 0 for E = m. At low temperatures f(0) = 1; thus the right-hand side of (31) is just F(p), consis- tent with the delta function approximation. Thus The particle flux density is, from (16), where T, is the relaxation time at thc surface E = p of the Fer~ni sphere. The integral has the value $$(3n/2~,) = n/rn , (34) by use of D(p) = 3n/2eF at absolute x r o , where E, = :mu; defines the velocity U F on the Fermi surface. Thus At absolute zero p(0) = (fi2/2m)(3?j.n)2'3, whence so that (33) bccomes 1; = - ( 2 ~ , / 3 m ) e ~ drddx = -$;T, dnldx . (37) The diffusivity is the coefficient of dn/dx: closely similar in form to the result (23) for the classical distribution of veloci- ties. In (38) the relaxation tirne is to be taken at the Fermi energy. Appsndir 661 We see we can solve transport problems where the Fermi-Dirac distribu- tion applies, as in metals, as easily as where the classical approximation applies. Electrical Conductivity The isothermal electrical conductivity u follows from the result for the particle diffusivity when we multiply the particle flux density by the particle charge q and replace the gradient d d d x of the chemical potential by the gra- dient qdpldx = -qE, of the external potential, where E, is the x component of the electric field intensity. The electric current density follows from (21): for a classical gas with relaxation time 7 , . For the Fermi-Dirac distribution, from (35), APPENDIX G: VECTOR POTENTIAL, FIELD MOMENTUM, AND GAUGE TRANSFORMATIONS This section is included because it is hard to find the magnetic vector po- tential A discusscd thoroughly in one place, and we need the vector potential in suprrcondnctivity. It may seem mysterious that the ha~niltonian of a particle in a magnetic field has the form derived in (18) below: where Q is the charge; hl is the mass; A is the vector potential; and 9 is the electrostatic potential. This expression is valid in classical mechanics and in quantu~n mechanics. Because the kinetic energy of a particle is not clranged by a static magnetic field, it is perhaps unexpected that the vector potential of the magnetic field enters the hamiltonian. As we shall see, the key is the observation that the momentum p is the sum of two parts, the kinetic momentum which is fa~rliliar to us, and the potential momentum or field momentum The total momentum is C and the kinetic energy is The vector is related to the magnetic field by B = curlA . (6) We assume that we work in nonmagnetic material so that H and B are trcated as identical. Lagrangian Equations o f Motion To find the Hamiltonian, the prescription of classical mechanics is clear: we must first find the Lagrangian. The Lagrangian in generalized coordinates is This is correct because it leads to the correct equation of motion of a charge in combined electric and magnetic fields, as we now show. In Cartesian coordinates the Lagrange equation of motion is and similarly for y and z. From (7) we form Thus (8) becomes 'For an elementary treatment of the vector potential see E M. Pnrcell. Electricity und mngdinm, 2nd ed., McGrdw-Hill, 1984. Appendix d2x Q M = QE, + - [v XB], , dr? c with B = curl A . (15) Equation (13) is the Lorentz force equation. This confirms that (7) is correct. We note in (14) that E has one contribution from the electrostatic potential p and another from the time derivative of the magnetic vector potential A. Derivation of the Hamiltonian The momentunl p is defined in terms of the Lagrangian as in agreemcnt with (4). The hamiltonian H(p,q) is defined by Field Momentum The momentum in the electromagnetic field that accompanies a particle moving in a magnetic field is given by the volume integral of the Poynting vec- tor, so that I ~ V E X B . Pfield = (19) We work in the nonrelativistic approximation with v 4 c, where v is the veloc- ity of the particle. At low values of vlc we consider B to arise from an external source alone, but E arises from the charge on the particle. For a charge Q at r', E = -Vp ; VZp = -4mQS(r - r') . (20) Thus dV V p X curl A . 4mc By a vector relation we have $ dV Vcp X curl A = -$ dV [ A X curl (Vq) - A div V q - (Vcp) &v A] . (22) But curl (Acp) = 0, and we can always choose the gauge such that div A = 0. This is the transverse gauge. Thus, we have This is the interpretation of the field contribution to the total momentlim p = Mv + QAlc. GAUGE TRANSFORMATION Suppose H = E, where Let us make a gauge transformation to A', where where x is a scalar. Now B = curl A = curl A', because curl (OX) = 0. The Schrodinger equation hecomes What +' satisfies with the same E as for +? Equation (27) is equivalent to We try Now so that Appendix 665 Thus +' = exp(iQx/fic)+ satisfies the Schrijdinger equation after the gauge transformation (25). The energy E is invariant under the transformation. The gauge transforniation on A merely changes the local phase of the wavefunction. N7e see that so that the charge density is invariant under a gauge transformation. Gauge in the London Equation Because of the equation of continuity in the flow of electric charge we require that in a superconductor divj = 0 , so that the vector potential in the London equation j = -cA/4~rhZ must satisfy divA = O . (32) Further, there is no current flow through a vacuum/superconductor interface. The normal component of the current across the interface must vanish: j, = 0, so that the vector potential in the London equation must satisfy A,, = 0 . (33) The gauge of the vector potential in the London equation of superconductivity is to be choscn so that (32) and (33) are satisfied. APPENDIX H: COOPER PAIRS For a co~nplete set of states of a two-electron system that satisfy periodic boundary conditions in a cube of unit volume, we take plane wave product functions We assurrie that the electrons are of opposite spin. We introduce center-of-mass and relative coordinates: K = k, + k, ; k = i(k1 - k,) , (3) so that k , - r l + k 2 - r , = K - R + k - r . (4) Thus (1) becomes p(K,k;R,r) = exp(iK . R) exp(ik . r) , (5) and the kinetic energy of the two-electron system is eg + Ek = (fi2/m)(;P + k ' ) . (6) We give special attention to the product functions for which the center-of- mass wavevector K = 0, so that k, = -k2. With an interaction H I between the two electrons, we set up the eigenvalue ~roblem in terms of the expansion ~ ( r ) = Xgk exp(ik . r) . (7) The Schrodinger equation is (H,, + H I - €)x(r) = 0 = zk, [(EL, - e)gkr + H1gk,]exp(ikl . r) , (8) where H , is the interaction energy of the two electrons. Here is the eigenvalue. We take the scalar product with exp(ik - r ) to obtain the secular equation of the problem. Now transform the sum to an integral: (E - e)g(E) + JdE' g(E1)HI(E,E')N(E') = 0 , (10) where N(E1) is the number of two electron states with total momentum K = 0 and with kinetic energy in dE' at E'. Now consider the matrix elements H1(E,E1) = (klHllkf). Studies of these by Bardeen suggest that they are important when the two electrons are confined to a thin energy shell near the Fermi surface-within a shell of thickness fiw, above E,, where w, is the Debye phonon cutoff frequency. We assume that for E,E' within the shell and zero otherwise. Here V is assumed to be positive. Thus (10) becomes with E, = eF + fiw,. Here C is a constant, independent of E Appendix 667 From (12) we have C g(E) = and \t7ith N(Ef) approximately constant and equal to N, over the small energy range between 2e, and 2cF, we take it out of the integral to obtain Let tlie eige~ivalue c of (15) be written as which defines the binding energy A of the electron pair, relative to two free electrons at the Fermi surface. Then (15) becomes 2c,, - 2~~ + A 2hwD + A 1 = NFV log A = AI;V log A ' (17) This result for the binding energy of a Cooper pair may be written as For V positive (attractive interaction) the energy of the system is lowered hy excitation of a pair of electrons above the Fermi level. Therefore the Fermi gas is unstable in an important way. The binding energy (19) is closely related to the superconducting energy gap Eg. The BCS calculations show that a high density of Cooper pairs may form in a metal. APPENDIX 1: GINZBURG-LANDAU EQUATION We owe to Ginzburg and Landau an elegant theory of the phenomenology ofthe snpercondlicting state and of the spatial variation of the order parameter in that state. An extension of the theory by Abrikosov describes the structure of the vortex state which is exploited technologically in superconducting mag- nets. The attractior~s of the GL theory are the natural introduction of the coherence length and of the wavefunction used in the theory of the Josephson effects in Chapter 12. We introduce the order parameter $(r) with the property that $(r)$(r) = ns(r) , (1) the local concentration of superconducting electrons. The mathematical for- mulation of the definition of the fiinction $(r) will come out of the BCS the- ory. We first set lip a form for the free energy density Fs(r) in a superconduc- tor as a function of the order parameter. We assume that in the general vicinity of the transition temperature with the phenomenological positive constants a, P, and m, of which more will be said. Here: 1. FN is the free energy density of the normal state. 2. -aI$I" $ $ P I $ I ~ is a typical Landau form for the expansion of the free energy in terms of an order parameter that vanishes at a second-order phase transition. This term may be viewed as -ans + ifin: and by itselc is a mini- mum with respect tons when ns(T) = alb 3. The term i11 lgrad $I2 represents an increase in encrgy caused by a spa- tial variation of the order parameter. It has the form of the kinetic energy in quantum mechanics.' The kinetic momentum -ifiV is accompanied by the field momentum -qNc to enslire the gauge invariance of the free energy, as in Appendix G. Here q = -2e for an electron pair. 4. The term -$M . dB,, with the fictitious magnetization M = (B - Ba)/4.rr, represents the increase in the superconductir~g free energy caused by the ex- pulsion of magnetic flux from the superconductor. The separate terms in (2) will be illustrated by examples as we progress further. First let us derive the GL equation (6). We minimize the total free en- ergy JdV Fs(r) with respect to variations in the function $(r). We have We integrate by parts to obtain if 89" vanishes on the boundaries. It follows that SJdVF, = JdVS$[-a$ + P1$12$ + (1/2m)(-ifiV - q ~ l c ) ~ $ ] + C.C. ( 5 ) 'A oo~~tribution of the form IVMI2, where M is the magnetization, was introduced hy Landau and Lifshitz to represent the exchange energy density in a fcrromagnet; see QTS, p. 65. Appendix 669 This integral is zero if the term in brackets is zero: This is the Ginzburg-Landau equation; it resembles a Schrvdinger equation for $. By minimizing (2) with respect to SA we obtain a gauge-invariant expres- sion for the supercurrent flux: At a frcr surface of the specimen we must choose the gauge to satisfy the hoi~ndary condition that no current flows out of the superconductor into the vacuum: ii . js = 0, where i i is the surface normal. Coherence Length. The intrinsic coherence length 5 may be defined from (6). Let A = 0 and suppose that /3l+l2 may be neglected in comparison with a. In one dimension the GL equation (6) reduces to This has a wavelike solution of the form exp(ix/(), where &is defined by 5 = (fi2/2ma)'" . (9) A more interesting special solution is obtained if we retain the nonlinear term p1$I2 in (6). Let us look for a solution with I + !
I = 0 at x = 0 and with I I , + I $ , as x + m. This situation represents a boundary between normal and supercon- ducting states. Such states can coexist if there is a magnetic field H , in the nor- mal regon. For the moment we neglect the penetration of the field into the si~perconducting region: we take the field penetration depth h < 5, which de- fines an extreme type I superconductor. The solution of subject to our boundary conditions, is $(x) = (a//3)lf2tanh(d~) . (11) This may be verified by dlrect substitution. Deep inside the superconductor we have Go = as follows from the minimization of the terms -a1+l2 + ipl$14 in the free energy. We see from (11) that 5 marks the extent of the co- herence of the superconducting wavefunction into the normal region. We have seen that deep inside the superconductor the free energy is a minimum when 1 $ , 1 " a/P, SO that by definition of the thermodynamic critical field H, as the stabilization free energy density of the superconducting state. It follows that the critical field is related to a and P by H, = (4m2/p)'" . (13) Consider the penetration depth of a weak magnetic field (B < H,) into a superconductor. We assume that 1+12 in the superconductor is equal to II)~~', the value in the absence of a field. Then the equation for the supercurrent flux reduces to js(r) = -(q2/~)lI)012A , (14) which is just the London equation js(r) = -(c/4.rrh2)A, with the penetration depth The dimensionless ratio K = A/[ of the two characteristic lengths is an important para~neter in the theory of superconductivity. From (9) and (15) we find We now show that the value K = 1 / f i divides type I superconductors ( K < 1 / f i ) from type I1 superconductors ( K > 1 / f i ) . Calculation of the Upper Critical Field. Superconducting regions nucle- ate spontaneously within a normal conductor when the applied magnetic field is decreased below a value denoted by H,,. At the onset of superconductivity I+I is small and we linearize thc GL equation (6) to obtain The magnetic field in a snperconducting region at the onset of superconduc- tivity is just the applied field, so that A = B(O,x,O) and (17) becomes This is of the same form as the Schrodinger equation of a free particle in a magnetic field. We look for a solution in the form exp[i(kyy + kzz)]p(x) and find Appendix 671 this is the equation for an harmonic oscillator, if we set E = a - (h2/2m) ( k t + k:) as thc eigenvalne of The term linear in x can be transformed away by a shift of the origin from 0 to x , = hkyqB/2mc, so that (20) hecomes, with X = x - x,, The largcst valiie of the magnetic field B for which solutions of (21) exist is given by the lowest eigenvalue, which is fhw = fiqB,,/2n~ = a - fi2k:/2m , (22) where w is the oscillator frequency yB/mc. With k, set equal to zero, B , , = H, = 2amclqh . (23) This result may be expressed by (13) and (16) in terms of the thermody- rianiic critical field H, and the GL pararnetcr K = A/(: When A/( > l / f i , a superconductor has H,, > H, and is said to be of type 11. It is helpful to write H,, in terms of the flux quantum @ , = 27rfic/q and E2 = h2/2ma: This tells us that at the upper critical field the flux density HC2 in the material is equal to one flux quantum per area 2n%, consistent with a fluxoid lattice spacing of the order of (. APPENDIX 1: ELECTRON-PHONON COLLISIONS Phonons distort the local crystal structure and hence distort the local band structure. This distortion is sensed by the conduction electrons. The important effects of the co~ipling of electrons with phonons are Electrons are scattered from one state k to another state k', leading to elec- trical resistivity. Phonons can be absorbed in the scattering event, leading to the attenuation of ultrasonic waves. An electron will carry with it a crystal distortion, and the effective mass of the electron is thereby increased. A crystal distortion associated with one electron can be sensed by a second electron, thereby causing the electron-electron interaction that enters the theory of superconductivity. The deformation potential approximation is that the electron energy ~ ( k ) is coupled to the crystal dilation A(r) or fractional volume change by ~ ( k , r ) = ~,,(k) + CA(r) , (1) where C is a constant. The approximation is useful for spherical band edges c0(k) at long phonon wavelengths and low electron concentrations. The dilation may be expressed in terms of the phonon operators uq, a : of Appendix C by A(r) =i C, (fi/2~o,)'~ Iq l[agexp(ig - r) - a;exp(-iq . r)] . (2) 9 as in QTS, p. 23. Here M is the mass of the crystal. The result (2) also follows from (C.32) on formingq, - q,?-, in the limit k 1. In the Born approximation for the scattering we are concerned with the matrix elements of CA(r) between the one-electron Bloch states Ik) and Ik'), with Ik) = exp(ik . r)uk(r) In the wave field representation the matrix ele- ment is = icC, cGck2 (~5/2Mw~)"~1qI(a,J d3x U ~ - U L ~ " ~ - ~ + ~ ) " k'k q (3) where where ct, ck are the fermion creation and annihilation operators. The product ui..(r)uk(r) involves the periodic parts of the Rloch filnctinns and is itself peri- odic in the lattice; thus the integral in (3) vanishes unless k - k ' q = vector in the reciprocal lattice. (" In semiconductors at low temperatures only the possibility zero (N proccsses) may be allowed energetically. Appendix 673 Let 1 1 s limit orirselves to N processes, and for convenience we approximate J d3x uk.uk by unity. Then the deformation potential perturbation is Relaxation Time. In the presence of the electron-phonon interaction the wavevector k is not a constant of the motion for the electron alone, but the sum of the wavevectors of the electron and virtual phonon is conserved. Suppose an electron is iliitially in the state I k ) ; how long will it stay in that state? We calculate first the probability w per unit time that the electron in k will emit a phonon q. If n, is the initial population of the phonon state, by time-dependent perturbation theory. Here The total collision rate W of an electron in the state I k ) with a phonon sys- tem at absolute zero is, with nq = 0, where p is t l ~ e Inass density. The argunient of the delta function is where y,, = 2hm' c , , with c, the velocity of sound. The minimum value of k for which the argument can be zero is k,,, = i(q + q,), which for q = 0 reduces to k . , , , =' ,q, = m'c,lh. For this value of k thc clcctron group velocityug = k , , / m ' is cqual to the velocity of soilnd. Thus the threshold for the emission of phonons by electrons in a crystal is that the electron group velocity should ex- ceed the acoustic velocity. This requirement resembles the Cerenkov thresh- old for the emission of photons in crystals by fast electrons. The electron energy at the threshold is im'c; - - 10" - 10-Ifi erg - 1 K. An electron of energ). below this threshold will not be slowed down in a perfect crystal at ahsolutc zero, even by higher order electron-phonon interactions, at least in the harmonic approximation for the phonons. Fork % q, we may neglect the qq, term in (9). The integrals in (8) become and the phonon emission rate is directly proportional to the electron energy ek. The loss of the component of wavevector parallel to the original direction of the electron when a phonon is emitted at an angle 8 to k is given by cj cos 8. The fractional rate of loss of k, is given by the transition rate integral with the extra factor (q/k) cos 0 in the inte- grand. Instead of (lo), we have so that the fractional rate of decrease of k, is W(k,) = 4C2m'k2/5~pcSfi2 . This quantity enters into the electrical resistivity. The above results apply to absolute zero. At a temperature k,T % Ac,k the integrated phonon emission rate is For electrons in thermal equilibrium at not too low temperatures the required inequality is easily satisfied for the rms value of k. If we take C = lo-'' erg; m = g; k = 10' cm-I; c, = 3 X 10; cm s-I; p = 5 g ~ m - ~ ; then W - 1012 s-'. At absolute zero (13) gives W = 5 x 10lOs-I with these same parameters. Absolute thermoelectric power, 215 Accrptor ionization energies, 212 Acceptor states, 211 Acoustical phonou branch, 95 Activatinn energ): 589 Adiabatic demagnetization, 312 Al~aru~~ov-Bohm cffcct, 543 Alfven waves, 425 Alkali halide crystals, table, 66 Alk~ys. 621 strength, 613 transitiou metal, 634 Alnico V, 353 Amorphous solids, 568 rerrornagnets, 575 semicnndllctnrs. 577 Anharmonic interactions, 119 Allisotropy cncrgy, 348 Annihilation operators, 651 Antiferroelectric crystals, 478, 479 Antiferrornagnetism. 340 rnagnons, 344 N6el temperature, 343 Anti-stokes line, 445 Atomic force microscope, 526 Atomic form factor, 41 Atomic radii, 70 tahlc, 71 Band bending, 507 Rand gap, 165,187 Band structure, germanium, 203 Barium titanate, 470 Basis, 4, 5 BCS tlwury, 270,277 Biomagnetism, 354 Bloch equations, 369 Bloc11 frequency, 217 Bloch function, 167 Bloch oscillator, 217 Bloch theorem, 173 Bloch T3,' law, 334 Bloch wall, 349 Bohr magneton, 303 Boltzmann transport equation. 656 Boson operators, 651 Boundary conditions, periodic, 110 Bragg law, 25 Bravais lattice, R Brillnuill function, 304 Brillo~~in scattering, 428 Brillouin zone, 33,223,252 first, 44, 93, 224 vol~~me, 44 Bulk modulus, 80 Bulk modulus, clcctron gas, 157 Bnrgers vector, 604 Biittiker-Landauer formalism, 540 Cauchy integral, 431 Causality, 450 Cell, primitive, 6 unit, 8 Wigner-Seitz, 34 Centipoise, 573 Ceutral eqwation, 174 Cesium chloride structure, 14 Chalcogenide glasses, 577 Charge density waves, 424 Charging energe 549 Chemical potential, 137, 157 Classical distribution, 658 Clausius-Mossotti relation, 464 Clogston relation, 272 Coercive force, small particle, 358 Cocrcivity, 347,352 Coherence length, 276,669 intrinsic, 277 Cohcsive cnergy, 49, 59, 73 Fermi gas, 159 Sodium metal, 237 Square well potential, 253 Cohesive energy, table, S O Collisionless electron gas, 433 Colur centers, 592 Compressibility 80 Concentration, table, 21 Conductance, quantum, 534 Conduction electron. ferromagnetism, 320 Conduction band edge, 190 Conduction electrons, 315 susceptibility, 315 Conductivity, electrical, 147, 209,420 impurity, 209 ionic, 420 thermal, 156 Condu~tivity suru rule, 450 Configurational heat capacity, 640 Contact hyperfine interaction, 374 Cooper p a i n , 279,556,665 Coulomb blockade, 551 Covalent bond, 67 Covalent crystals, 67 Critical points, 434 Contilluum wave equation, 103 Creation operators, 651 Creep, 615 Critical field, 262,295 thin films. 295 Critical shear stress, 599 Critical temperature, ferroelectric, 469 Crystallography, surface, 490 Crystal field, 309 splitting, 309 Crystal momentum, 100,173 Crystal struchlre, elements, 20 Cubic lattices, 10 Cubic zinc sulfide structure, 17 Curie constant, 305 law, 305 Curie-Weiss law, 324 Cyclotron frequency, 153 Cyclotron resonance, 200,219 spheroidal energy surface, 219 Dangling honds, 488 Uavydov splitting, 452 Dehye model, density of states, 112 Dehye temperature, 112 table, 116 Dehye ' P law, 114 Dehye-Waller factor, 642 Defects, paramagnetic, 375 Deficit semiconductors, 209 Degenerancy, 135 Degenerate semiconductor, 409 De Haas-van Alphen effect, 244 period, 253 De~nagnetization factors, 380 Demagnetization. isentropic, 312 Density of states, 108, 149 Dehye model, 112 Einstein model, 114 finite system, 520 general result, 117 one dimension, 108 singularity, 119 three dimensions, 11 1 Density, table, 21 Depolarization factors, 458 Depolarization field, 48.5 Diamagnetism, 299 Diamond stnlctnre, 16, 45, 182,187 Dielectric constant, 463 semiconductors, 211 susceptibility, 459 Dielectric function, 429 electron gas, 433 Lindhard, 406 Thomas-Fermi, 405 Diffraction cnnditions, wave, 25 Diffraction, Josephson junction, 282 Diffusion, 371. 588 Diffusivity, 588 Dilation, 75 Direct gap semiconductor, table, 201 Direct photon absorption, 189 Dislocation densities, 610 Dislocatio~~ rnultiplicatio~~. 611 Dislocations, 601 Dispersion relation, phonons, 92 electro~nagnetic waves, 92, 397 Displacive transition, 471 Dissipation sum rule, 450 Distribntion, classical, 130 Fermi-Dirac, 107 Planck, 107 dog:^ bone orbit, 250 Domains, origin of, 350 closure, 351 Donor ionizatinn energies, 211 Donor states, 209 Doping, 209 Dulong and Petit value. 117 Edge dislocation, 601 Effective mass, 197 negative, 199 semiconductors, table, 201 Einstein model, density of states, 114 thermal, 145 Elastic stiffness, 78, 84 Elastic strain, 73 Elastic \vatre quantization, 80 Electrical resisti~ity, 148 table, 149 Electric field, local, 460 macroscopic, 456 Electric conductivity, 147,661 table, 149 Electric quadruple moment, 379 Electronic polarizahilities table, 465 Electronic structure, surface, 494 Electron affinities, 62 Electron beam lithography, 521 Electrun co~npourrd, 624, 625 Electron-electron collisions, 417 Electron heat capacity, table, 148 Electron-hole drops, 441,443 Electron-electron interaction, 417 one dimension, 532 Electron-lattice interaction, 420 Electron orbits, 230 Electron-phonon collisions, 671 Electron spin resnnance, 362 Electron work functions, 494 table, 494 Electrostatic screening, 403 Elementary excitations, 90 Empirical pseudopotential method, 239 Empty core model, 240 Empty lattice approximation, 176 Energy band calculation. 232 Energy bands, 163,164 Energy gap, 165, 167 superconductors. 266,268 table, seniiconductors. 189 E n e r a levels, one dimension, 134 Energy loss, fast particles, 448 Entropy, superconductors. 265 Entropy of mixing, 631 Equations of motion, electron, 191 Equation of motion, hole, 196 ESCA, 447 Eutectics, 632 Ewald-Kornfeld method, 647 Ewald construction, 32 sphere, 493 sums. 644 Exchaigc cnergy, 326 field, 325 frequency resonance, 391 integral, 325 interaction, 325 narrowing, 371,386 Exclusion principle, Pauli, 56 Exotic snpercnnd~lctors. 147 Extremd orbits, 248 Excito~is, 435 Extended zone scheme, 226 Extinction coefficient, 429 Extre~nat orbits. 248 Factor, ato~nic tirrtn, 41 structure, 45 F centers, 376, 592, 595 Fermi-Dirac distribution, 136,652 Fermi energy, 135 Fermi gas, 134 Permi level, 137 Fcrmi liquid, 417 Fermi snrface, copper, 249 Fermi surface, gold Fermi surface yaramcters, table, 1.39 Fermi surfaces, 225 Fermiu~n, heavy, 147 Ferroelectric crystals, 467 domains, 479 linear array, 485 Ferromagnetism, crystals, 321,328 amorphnns, 5i5 conduction electron, 320 domains, 345 order, 338 resonance, 379 Fiber optics, 581 FickS law, 588 Fine structure constant, 499 First Brillouin zune, 36.93, 165 First-order transition, 477 Fluyoid, 281 Flux quantization, 279 Fourier analysis, 27, 39, 169 Fractional quantized Hall effect (FQHE), 503 Frank-Head source, 612 Free-cnerw, snpercnndnctors, 267 Frenkel defect, 586,595 Freukel cxciton, 437 Friedel oscillations, 407 Fullerenes, 262 Fused quartz, 570 Gap, direct, 190 Gap, indirect, 190 Gap plasmons, 426 Gauge transformation, 664 Geomagnetism, 354 Ginzburg-Landau equation, 667 Ginzburg-Landau parameter, 667 GLAG theow 283 Glass, 573 transition tcmpcraturc, 573 Grain honndary, small-angle, 607 Group velocity, 94 Grunciscn constant, 129 Gyromagnetic ratio, .?O2 Gyroscopic equation, 366 Hagen-Rubens relation, 451 Hall coefficient, 154 table, 155 Hall effect, 153 ballistic jnnction, 542 two-carrier types, 218 Hall resistance, 155 Hall resistivity, 498 Hardness, 617 Hcavy fcrmions, 147 Heat capacity, configurational, 640 electron gas, 141 glasses, 378 metals, 145 one-dimension, 561 phonon, 107 superconductors. 264 He3 liquid, 158 Heisenberg model, 325 IIelicon waves, 425 Heierojunction, 505 Heterostrnchlrrs. 507 Ilexagonal close-packed structure, 15 Hexagonal lattice, 44 High temperature super- conductors, 293 Hole, eqnation of motion, 104,194 Holc orbits, 230 Hole trapping. 422 Hooke's law, 73 HTS, 293 Hume-Rothery rules, 624 Hund roles, 306 Hydrogen bonds, 70 Hyperfine constant, 375 Hyperfine effects, ESR in metals, 391 Hypcrfinc splitting, 373 Hystrresis, 352 Ilmenite, 470 Impurity conducti~ty, 209 Impurity orbits, 218 Index system, 10 Indirect gap, 190 Indirect photon absorption, 189 Inelastic scattering, phonons, 100 Inert gas crystals, table, 53 Injection laser, 510 Insulators, 181 Interband transitions, 434 Interface plasmons, 425 Interfacial polarization, 484 Intrinsic carrier conce~ltratiun, 1 88,205 Intrinsic coherence length, 277 Intrinsic mobility, 208 Intrinsic semiconductors, 187 Inverse spinel, 338 Ionic bond, 60 Ionic character, table, 69 Ionic conductivity, 420, Ionic crystals, 6 Ionic radii, 72 table, 71 lnnization energies, acceptor, 209 donor, 209 table, 54 Iron garnets, 339 Iron group, 307 Isentropic demagnetization, 312 Isotope effect, 269 superconductors, 269 Isotope effect, thermal conduction, 127 Jahn-Teller effect, 209 trapping, 420 Josephsnn t~~nneling, 289 Jump frequency, 590 Junction, superconducting, 287 Jnnctinns, p-n, 503 k . p perturbation theory, 253 Kelvin relation, 215 Knight shift, 377, 378 Kohn anomaly, 103 Kramers-Heisenberg dispersion, 431 Kramers-Kmnig relations, 430,432 Kronig-Penney model, 174,182 reciprocal space, 174 LA modes, 95 Lagrangian equations, 662 Landaner forrnnla, 535 Landau gauge, 503 Landau-Ginzburg equations, 276, 667 Landau level, 245,254,493 Landau theory, Ferrni liquid, 417 Landau theory, phase transition, 474 Langevin diamagnetism, 299 Langevin result, 301 Lanthanide group, 306 Larmor frequency, 300 Larmor theorem, 300 Laser, 389 injection, 510 ruby, 389 semiconductor, 510 Lattice, Bravais, 8 Lattice constants, equilibrium, 58 Lattice frequencies, table, 41 6 Lattice momentum, 193 Lattice sums, dipole arrays, 647 Lattice types, R,10 Lattice vacancies, 585 Lattices, cubic, 10 Lane equation, 33,513,641 Law of mass action, 206 LCAO approxi~nation. 233 LEED, 511 Lennard-Jones potential, 58 Lcnz's law, 299 Lindhard dielectric function, 406 Line width, 370 Liquid He" 158 Liquidus, 632 Localization, 539 Local electric field, 460 London equation, 273,665 London gauge, 665 London penetration depth. 275,294 Longitudinal plas~rla oscillations, 398 Longitudinal relaxation time, 366 Long-range order, 630 LO modes, 95 &rent2 field, 462 Lorenz number, 157 Low-angle grain boundary, 607 LST relation, 414,416 Luminescence, 437 Luttinger liquid, 532 Lyddane-Sachs-Tcllcr relation, 414 Madclung constant, 64 Madelnng energy, 60 Magnetic breakdown, 251 Magnetic force niioroscupy, 355,527 Magnetic susceptibility, 298, 315,318 Magnetite, 337 Magnetoconductivity, tensor, 159,254 Magnetocrystalline energy, 347 Magnctoclastic coupling, 357 Magnetogyric ratio, 302,363 Magneton numbers, 308 Magnctoplas~na frcqucncy, 425 Magnetore~istance, giant, 359 Magnetoresistance, 408 two carrier types, 219 Magnetic resonance, 361 Magnetotaxis, 355 Magnetization. 303 saturation, 304 Magnons, 330 antiferromapetic, 344 dispersion relation, 357 Maser action, 386 Matthiassen's rule, 150 Maxwell equations, 455 Mean field appruximatiuo. 323 Meissner effect, 259 sphere, 296 Meltingpoints, table, 51 Melt-spin velocity, 576 Mesh, 490 Mesoscopic regime, 543 Metal-insulator transition, 407 Metals, 69 Metal spheres, 484 Metglas, 576 Mioroelectro~nechanical systems, 561 Miller indices (see index system), 11 Mobility edges, 501 Mobility gap, 501 Mobility, intrinsic, 208 Mobilities, table, 208 Molecular crystals, 440 Molecnlar hydrogen, 68, 86 Momentum crystal, 100,173 field, 661 lattice, 193 phonon, 100 MOSFET, 497 Motional narrowing, 371.373 Mott exciton, 441 Mott transition, 407 Mott-Wannier excitons. 441 np product, 206 Nanocrystals, 517 fluorescence, 522 e n e r a levels, 54.5 Nanostructures, 517 Nanotuhes, 518 density of states, 529 band structure, 562 N6el temperature, 341,342, 343 NBel wall, 358 Negative effective mass, 199 Neutron diffraction, 45 NMK tomography, 363 Noncrystalline solid, 519 Nonideal structures, 18 Normal mode enumeration, 108 Normal proccsscs, 124,125 Normal spinel, 338 Nuclear demagnetization, 314 Nuclear ~rlagrietic resonance, 363 table, 365 Nuclear magneton, 364 Nnclear paramagnetism, 314 Nuclear quadrupole resonance, 379 Ohm's law, 147, 538 Open urbits, 230 magnetoresistance, 254 Optical absorption, 190,521 Optical microscopy, 521 Optical phonon branch, 95 Optical phonons, soft, 473 Orbit, dog's bboe, 250 Order-disorder transformation, 627 Order, long-range. 630 short-range, 631 Order parameter, 668 Oscillations, Friedel, 407 Oscillator strength, 466 p-n junctions, 503 Paramagnetic defects, 375 Paramagnetism, 302 conduction electrons, 315 Van Vleck, 311 Particle diffusion, 657 Fauli exclusion principle, 56 Pauli spin magnetization, 316,319, 377 Peierls instability, 422,532 insulator, 424 Peltier coefficient; 215 Penetration depth, London, 296 Peliodic boundary conditions, 110 Periodic zone scheme, 225 Perovskite, 470 Persistent currents, 282 Phase diagrams, 625,632 Phase transitions, structural, 467 Phonons, 100,101,107 coordinates, 649 dispersion relations, 117 gas, ther~nal resistivity, 123 heat rapacity, 107 inelastic scattering, 100 rriean free path, 122 metals, 409 modes, soft, 103 morner~tu~n, 100 Photolithography, 522 Photovoltaic detectors, 506 Piezoelectricity, 481 Planck distribution, 107 Plasrnon frequency, 90, 397, 398 interface, 425 mode, sphere, 425 uptics, 396 oscillation, 398 surface, 403,424 Poise, 573 Poisson equation. 403 Poisson's ratio, 87 Polaritons, 410 Polarizability, 463 conducting sphere, 484 electronic, 464 Polarization, 455 interfacial, 484 saturation, 484 Polaron, coupling constant, 420, 422,426 Polytypism, 19 Power absorption, 370 Primitive cell, 5,6, 180 p-n junctions, 503 Pseudopotential, components, 239 metallic sodium, 240 method, 239 Pyroelectric, 469 PZT system, 479,481 Qi~anti~ation. elastic wave, 99,648 orbits in xr~agr~etic field, 242 spin waves, 382 Quantum corral, 525 Quautu~m dots, 517, 545 charge states, 549 Quantum interference, 282,292 Quar~turn solid, 85 Quantum theory, diamagnctism, 301 paramagnetism, 302 Quasi-lcermi levels, 510 Quasi particles, 417 Quenching, nrhital angular momentum, 308 QHE, 499 Radial breathing mude, 558 Radial distribution function. 569 Raman effect, 444 scatteling, 428 surface enhanced, 549 nanntuhcs, 558 Random network, 572 Random stacking, 19 Rarc carth ions, 305 Rayleigh attenuation. 582 I'teciprocal lattice points, 29 Rcciprooal lattice vectors, 29 Recombination radiation. 442 Reconstruction, 489 Rectification, 504 Reduced zone scheme. 223 Reflectance, 430 ReHection, normal incidence. 450 Reflectivity coefficient, 429 Refractivc indcx, 429 Relaxation, 366 direct, 368 Orbaclr, 368 Raman, 368 Relaxation time, longitudi~~al, 366 spin-lattice, 366 ltemanence, 347 Resistance per square, 159 Resistance quantum. 534 Ilesistance, surface, 159 Resistivity, electrical, 147, 159 Resistivity ratio, 150 Resonant tunneling, 538 Respo~~se, electron gas, 426 Response function, 430 Rf saturation, 391 RHEED, 493 Richardso~l-Dushman equation. 49.5 RKKY theory, 638 Rotatlng coordinate system, 391 Ruby laser, 389 Saturation magnetization, 326 Saturation polarization, 485 Saturatinn rf, 391 Scanned probe microscopy, 520 Scanning clcctro~i microscope, 521 Scanning tnnneling microscope, 523 Schottky barrier, 506 Schottky defect, 58.5 Schottb vacancies, 585,595 Screened Coulomb potential, 406 Screening, electrostatic, 403 Screw dislocation, 603 Second harmonic generation, 549 Second-order transition, 475 Self-diffusion, 591 Self-trapping, 209 Semiconductor crystals, 187 Scmico~lductor. degenerate, 409 Semiconduvtor lasers, 510 Semiconductors, deficit, 209 Se~~~i~netals; 162, 215 Shear constant, 87 Shear strength, silrglc crystals, 599 Shear stress, critical, 599 Short-range order, 631 Single-domain particles, 353, 358 Single-electron tranaistor, 551 Singularities, Van Hove, 119 Slatcr-Pauling plot, 636 Slip, 600 Sodium chloride structure, 13 Sodiu~n n~etal, 132 Soft modes. 474,485 Solar cells, 506 Solidus, 632 Solubility gap, 632 Spheres, metal, 484 Spectruscupic splitting factnr, 3 1 1 Spinel, 337 Spin-lattice interaction; 367 Spin-lattice relaxation time, 366 Spin wave (see also magnon), 330 quantization, 333 resonance, 382 Square lattice, 8, 182 Stability criteria, 88 Stabilization free energy, 272 Stacking fault, 601 STM, 523 Stokes line, 445 Strain component, 75 Strength of alloys, 613 Stress component, 75 Structural phase transitions, 467 Structure [actor, 39 bcc lattice, 40 diamond, 45 fcc lattice, 40 Substrate, 489 Surface plasmon resonance, 547 Superconducti\lty, table, 261 type I, 259 type 11,283 Superlattices, 628, 640 Superparamagnetism, 354 Surface crystallography, 490 surcace electronic stnlchlre, 494 surface nets, 490 surfacc plasmons, 279,302 surface resistance, 159 surface states, 495 surface transport, 497 Susceptibility, dielectric, 459 TA modes, 95 Temperaturc, Debye, 112 Temperature dependence. reflection lines, 643 Tetrahedral angles, 22 Thermal conductivity, 121,156 glasses, 534 isotope effect, 127 metals, 156 one-dimension, 561 tablc, 116 Thermal dilation, 128 Thermal effective mass, 145 Tl~rrxnal excitation, magnons, 334 Thermal expansion, 120 Thermal ionization, 213 Thermal resistiviv, phonon gas, 133 Thermionic emission, 495 Thermodynamics, superconducting transition, 270 Ther~noclcctric effects, 214 Thomas-Fermi approximation, 403 Thomas-Fermi dielechic function, 405 Three-level maser, 388 Tight-binding method, 232 TO modes, 6 1 5 To~nography, lnagnetic resonance, 363 Transistor, MOS, 507 Transition, displacive, 471 first-order, 477 metal insulator, 107 order-disorder, 471 second-order, 47.5 Transition metal alloys, 634 Transition temperature, glass, 573 Transistor, MOS, 497 Translation operation, 4 Translation vector, 4 Transmission electron microscopy, 520 Transmission probability, 534 Transparency, alkali metals, 388 Transport, surface, 497 Transverse optical modes, plasma, 398 Transverse relwatio~i time, 368 Triplet excited states, 318 Tunneling, 287 Josephson, 289 Tunneling probability, 524 Twinning, 601 nno-fluid model, 295 Two-level system, 320 Type I superco~~ductors. 264 Type I1 superconductors, 264,283 Ultraviolet transmission limits, 399 Umklapp processes, 125 Unit cell, 6 Upper critical field, 670 UPS, 447 Valence hand edge, 190 Van der \Vaals interaction, 53 Val1 Hove singularities, 119,528 Van Vleck paramagnetism. 311 Vector potential, 661 Vickers hardness nulnler, 618 Viscosity ,574 Vitreous Silica, 570 Vortex state, 264, 284, 295 Wanuier functions, 254 Wave eqwatio~i, co~itinuum, 103 periodic lattice, 169 Weiss field, 323 Whiskers, 616 Wiedemann-Franz law, 1.56 Wiper-Seitz boundaly condition, 237 Wigner-Seitz cell, 6, 8, 34, 238 Wiper-Seitz method, 236 Work hnction, 494 Work-hardening, 614 XPS, 447 Young$ modulus, 87 Yttriuni iron ganlct, 381 Zener breakdown, 217 Zener tnnneling, 217 Zero-field splitting, 386 Zero-puint motion, 56,85 Table of Values Quantity Symbol Value CCS SI \Jclocity of light c 2.997925 10'' crn s I 1 0 B rn s-' Proton charge e 1.60219 - 10- I= C 4.80:125 10-lo esu - Planck's constar~t h 6.62620 erg s J s R = h l 2 ~ 1.05459 lo-'' erg s J s Avogadro's number N 6.02217 X loz3 nrolk' - - Atomic mass unit amu 1.66053 lo-" g lo-z7 kg Electron rest mass rn 9.10956 10 g lo31 kg Proton rest mass MP 1.6726 1 10.~' kg Proton rnass/electron mass M,/m 1836.1 - - Rec~procal tine structure llu 137.036 - - constant hcle2 Electron radius e2/mc2 re. 2.81794 cm rn Electron Compton A? 3.86159 lo-" cm 10-l3 rn wavelrrrgth hlmc Bohr radius h2/rne" TO 5.29177 cm lo-" m Bohr rnagneton ehl2mc p, 9.27410 erg C-I lo-=' j T-I Rydberg constant me4/2h2 R, or Ry 2.17991 lo-'' erg J 13.6058 eV 1 electron volt eV 1.64219 lo-'' erg 10-l9 J eVIh 2.41797 x loL4 Hz - - cVlhc 8.06546 103 cm-' 1 W m-' eVlkB 1.16048 x 10' K - - -- Bol tzlnar~rl constant k~ 1.38062 10-16 erg K-I J K ~ I Permittivity of frep spacc e0 - 1 107/4~c2 Permeability of free bpare p,, - 1 4~ x S,,urcr. D. N. Taylor. W H Parker. and U N Langcnbrrg. Hev. Mud Phys.41, 375 (1969) See rlav E. R. Cohrn and 8. N. Taylor. Journal of Phyrical and Chcm~cal Reference Data 2(4l, 663 (1973). |
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Precalculus exercises
Precalculus: Imaginary and complex numbers
Mathematics III exercises
Classify complex numbers
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| Classify complex numbers |
| |
| Description |
| Exercise Name: | Classify complex numbers |
| Math Missions: | Algebra II Math Mission, Precalculus Math Mission, Mathematics III Math Mission |
| Types of Problems: | 1 |
The Classify complex numbers exercise appears under the Algebra II Math Mission, Precalculus Math Mission and Mathematics III Math Mission. This exercise practices classifying complex numbers.
Types of Problems[]
There is one type of problem in this exercise:
What type of number is __?
What type of number is __?: This problem has a number, and the student is asked to find which type of number it is. There are 3 choices: whole real, pure imaginary, or complex. There may be more than one answer.
Strategies[]
Basic knowledge of classifying complex numbers is needed to ensure success while doing this exercise.
A complex number z=a+b i{\displaystyle {z=a+bi}} has only two types of components: real (a{\displaystyle {a}}) and imaginary b{\displaystyle {b}}. Using this, students can classify z{\displaystyle {z}} as either:
Real, if b=0{\displaystyle {b=0}},
Pure imaginary, if a=0{\displaystyle {a=0}}, or
Complex, for all real values of a{\displaystyle {a}} and b{\displaystyle {b}}
Real-life Applications[]
Complex numbers have applications as models in non-Euclidean geometries.
Complex numbers are used to create fractal images.
Complex numbers are a natural extension of the real numbers.
The complex plane representation of complex numbers can be used to better understand polar coordinates.
Imaginary numbers are used in chaos theory for predicting chaotic situations, like earthquakes and stock markets.
Imaginary numbers can be used to model geometry on the Cartesian plane.
Categories
Categories:
Math exercises
Algebra II exercises
Algebra II: Introduction to complex numbers
Precalculus exercises
Precalculus: Imaginary and complex numbers
Mathematics III exercises
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Tiling with Triangles
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Tiling with Squares
A square or rectangle is said to be 'squared' into n squares if it is tiled into n squares of sizes s 1,s 2,s 3,..s n. A rectangle can be squared if its sides are commensurable (in rational proportion, both being integral mutiples of the same quantity) The sizes of the squaress i are shown as integers and the number of squaresn is called the order.
Techniques for making square tilings
Creating Squared Rectangles from Electrical Networks
It was shown by Brooks, Smith, Stone and Tutte that simple squared rectangles correspond to 3-connected planar electrical networks with one edge removed and an electromotive force applied. They reinterpreted the problem as one of electrical circuits. Squares in a rectangle dissection were replaced by edges of unit resistance in an electrical network, connected as nodes at the top and bottom edges of the squares and the rectangle.
69 x 61 Squared Rectangle and its Electrical Network
By making the correspondence between electrical networks and square dissections, it becomes possible by enumerating all electrical networks with a given number of edges, to produce all possible simple tilings of rectangles with a given number of squares.
3-connected planar graphs are equivalent to the '1-skeletons' (i.e. edges and vertices) of polyhedra, this is known as Steinitz's theorem. The problem of enumerating polyhedra is the same as enumerating 3-connected planar graphs. Tutte developed a theory for exhaustively creating 3-connected planar graphs (also called c-nets) using his Wheel Theorem and also developed enumeration formulae to estimate the numbers of c-nets by edge.
Duijvestijn and Bouwkamp created c-nets by computer using this theory. The method created many duplicate c-nets so a major computational effort went into eliminating them. Recently new techniques for producing c-nets have been developed, these include Brendan MacKay and Gunnar Brinkmann's plantri which generates exactly one member of each isomorphism class for many types of planar graph, and A Direct Decomposition of 3-Connected Planar Graphs by Bodirsky, Gropl, Johanssen, and Kang.
Once a supply of c-nets has been produced, each c-net with a selected edge is treated as an electrical network. The currents and voltages in the network branches correspond to the sizes of the squares and they can be calculated using Kirchhoff's Laws and the Matrix Tree Theorem. As the simple squared rectangles are produced, they can be searched for simple perfect squared squares.
Ian Gambini's Decomposition Approach
Ian Gambini in his thesis (French) has produced squared squares using 2 methods, his first method for a given integer n obtains every decomposition using n squares, the second method is a lot more efficient but incomplete and allows the calculation of decompositions of a square into distinct sized squares to generate squared squares, squared cylinders and tori. Gambini has claimed over 30,000 squared squares discovered with many in higher orders. Only a small selection of these are shown in his published work. It is not clear if all 30,000 are distinct tilings when rotations and reflections are eliminated.
Subdivision Tiling and Squared Rectangles
Cannon, Floyd and Parry use a coded tiling file and some simple finite subdivision rules with their program subdivide.c, to recursively subdivide a tiling into a complex subdivided tiling. Another program squarect.c, inputs the tiling file and outputs a postscript file for the associated squared rectangle. The order of these subdivided tilings increases exponentially with the number of sub-division levels. Subdivision creates imperfect and compound tilings. The squaring of the rectangle is done using the cyclic algorithm, as described in Squaring rectangles: the finite Riemann mapping theorem, by J.W. Cannon, W.J. Floyd, and W.R. Parry, Contemporary Mathematics, Volume 169, 1994.
A subdivision substitution tiling by S. Anderson |
8562 | https://www.slideserve.com/merton/3295971 | PPT - 第五章 可摘局部义齿 PowerPoint Presentation, free download - ID:3295971
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第五章 可摘局部义齿
Aug 16, 2014
• 1.71k likes • 2.25k Views
第五章 可摘局部义齿. 口腔修复学教研室. 概述. 定义 RPD 的优缺点 适应症和非适应症 分类. 相关定义. 牙列缺损 ： 上下颌牙列内的不同部位有不同 数目的牙齿缺失，但牙列中还有天然牙存在。 可摘局部义齿 （ removable partial denture RPD): 是牙列缺损修复最常用的方法，它利用天然牙和骨组织作支持，依靠固位体和基托的固位作用，患者能自行摘戴的一种修复体。. 优点： 磨除基牙组织较少 适用范围广 制作简单，容易修理 可修复组织缺损. 缺点： 体积大，异物感明显 稳定性较差 强度低，易磨耗 易变色.
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第五章 可摘局部义齿
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第五章 可摘局部义齿 口腔修复学教研室
概述 定义 RPD的优缺点 适应症和非适应症 分类
相关定义 牙列缺损：上下颌牙列内的不同部位有不同 数目的牙齿缺失，但牙列中还有天然牙存在。 可摘局部义齿（removable partial denture RPD):是牙列缺损修复最常用的方法，它利用天然牙和骨组织作支持，依靠固位体和基托的固位作用，患者能自行摘戴的一种修复体。
优点： 磨除基牙组织较少 适用范围广 制作简单，容易修理 可修复组织缺损 缺点： 体积大，异物感明显 稳定性较差 强度低，易磨耗 易变色 RPD的优缺点
适应症： 牙列游离缺失 暂时性修复、过渡性修复 伴有牙槽骨或软组织缺损 儿童的早期修复 不能耐受磨除牙体组织 腭裂缺牙患者 非适应症： 基牙固位不良 精神病患者 塑料过敏者 无法克服义齿异物感者 对修复要求较高者 RPD的适应症和非适应症
RPD的分类 按结构形式分类：可分为基托式和支架式两种。 按支持形式分类： 分类 运用 牙支持式 缺牙少，间隙小，两端基牙健康 粘膜支持式 多数牙缺失，余留牙不能作基牙 混合支持式 运用范围最广，临床最常采用
RPD的组成和作用 人工牙 基 托 RPD 固位体 连接体
固位体 人工牙 连接体 基 托
人工牙的作用和种类 • 人工牙具有代替缺失牙建立咬合关系，恢复 咀嚼功能和外形的作用 • 根据材料分： 塑料牙 瓷牙 金属合面牙 • 根据合面形态分： 解剖失牙 非解剖失 半牙解剖失牙
种类 特点 解剖式 牙尖斜度为30°或33°，牙尖锁结关 系好，咀嚼效率高，但侧向合力较大。 非解剖式 合面没牙尖和斜面，咀嚼效率低，但 侧向合力小，对牙槽嵴损伤小。 半解剖式 牙尖斜度为20°，牙尖有一定的锁结 关系咀嚼效率较高，侧向合力较小。
人工牙的选择 牙尖斜度的选择 人工牙的牙尖斜度应根据牙槽嵴的状况及缺牙数来确定 形态的选择 主要是前牙形态的选择,应综合患者脸形,性别,年龄,患者本人的 喜好等因素来决定 大小的选择 人工牙大小的选择应参考缺牙间隙的大小和余留牙或同名牙的 大小,后牙应适当减小颊舌径,以减轻牙槽嵴的负荷. 颜色的选择 人工牙的颜色应与患者的肤色、年龄相适应。
基托（denture base) 一 基托的功能： A、 连接义齿各部分为一整体。 B、 在基托上排列人工牙，承担、传递和分散合力。 C 、修复缺损的颌骨及软组织。 D、 加强义齿的稳定和固位。 E、支持义齿，防止义齿下沉 二 基托的种类： 有金属、塑料和金属塑料基托三种。
基托的种类 塑料基托 色泽美观，操作简单，质轻，易调磨修改；但受力易折裂，导热性差 金属基托 金属铸造而成，强度高，体积薄小，导热性好，感觉舒适；但制作复杂，不易修改 金属塑料联合基托 兼有塑料、塑料基托的优点
三 要求： 1、伸展范围：唇、颊侧止于粘膜转折处，后缘在上颌 应伸展至翼上颌切迹，颊侧盖过上颌结节，在下颌应 覆盖磨牙后垫1/3-1/2。 2、厚度 金属基托为0.5mm，塑料基托为2.0mm. 3、与天然牙的非倒凹区接触。 4、在一些骨性倒凹区应作缓冲。 5、磨光面的设计 颊舌面呈凹形并有牙根的外形。 6、修复前基托区的预备 主要是采用外科手术。
固位体（retainer) 一 定义： 固位体通常由金属制成，放在基牙上使RPD得以固 位，具有固位、稳定、支持的作用。 二 条件： 1、有足够的固位作用。 2、对基牙无侧向压力，不产生矫治力。 3、具有良好的生物学性能，暴露金属少，美观。
三 分类 主要分为直接固位体和间接固位体 （一）直接固位体（direct retainer) 定义：是防止义齿向合方向脱位，起主要固位作用的 部分，一般位于邻近的基牙上，可分为冠外固位体和 冠内固位体。 卡环（clasp) 卡环是最常用的直接固位体，直接卡抱在基牙上达到 固位作用，有些卡环还具有稳定和支持作用。
1卡环的结构和各组成部分的作用 以三臂卡环为例，卡环由卡环臂、卡环体和合支托三部分组成。 位置 作用 卡环臂 尖端位于倒凹区 固位 起始部位于非倒凹区 稳定、支持 卡环体 非倒凹区 稳定、支持 合支托 合面或舌隆突 稳定、支持、防食物嵌塞
合支托的要求 位置：应放置于基牙的近远中边缘嵴上，如咬 合过紧，也可放置于颊、舌沟或舌隆突上。 大小和形状：形状为薄而宽，呈匙形。要有一 定的强度，能起到足够的支持作用。 与基牙的关系：支托底应与基牙长轴垂直或成 20°角。 合支托不能影响就位和咬合
2卡环各组成部与基牙的关系 1)模型观测器：是用来确定义齿就位道的仪器， 可测量模型上基牙的倒凹，分析卡环放置的部位和 义齿的就位方向。 2)观测线：是将模型固定在观测台上，使基牙长 轴与水平面呈垂直关系，转动分析杆，分析杆在牙 冠轴面最突点所画的连线称为观测线，根据基牙倾 斜的方向和程度，观测线可分为三类。
一型观测线：基牙向缺隙相反方向倾斜时所 画出的观测线，基牙的近缺隙侧倒凹小，远缺 隙侧倒凹大。 二型观测线：基牙向缺隙方向倾斜时所画出 的观测线，基牙的近缺隙侧倒凹大，远缺隙侧 倒凹小。 三型观测线：基牙向颊侧或舌侧倾斜时画出 的观测线，基牙的近、远缺隙侧都有明显的倒 凹，倒凹区大，非倒凹区小。
3）观测线类型与卡环臂的选择 根据不同类型的观测线选择相应的固位卡环臂。 特点 Ⅰ型 固位、稳定和支持作用较好 Ⅱ型 固位作用较好，但稳定和支持作用稍差 Ⅲ型 固位和支持作用较好，但稳定作用较差
Ⅰ型 Ⅱ型 Ⅲ型
3卡环臂与倒凹深度的关系 不同类型和不同材料制 作的卡环固位臂弹性不同,需要进入倒凹的深度应不同 4卡环的共同就位道具有多个卡环的可摘局部义齿,其 卡环应具有共同的就位方向,就位方向的主要是通过观测 线来获得。 5卡环的对抗卡环在产生固位力的同时，也对基牙产 生作用力，所以需要有对抗臂以对抗固位臂对基牙产生 的作用力。
6卡环应具备的条件和材料要求 1）理想卡环应具备的条件 • 良好的支持、稳定和固位作用 • 适当的卡抱 • 有对抗作用 • 具有被动性 • 小的接触面积 • 垂直向力 2）卡环材料的要求 • 机械强度高 • 有一定的弹性 • 耐腐蚀，生物相容性好
卡环的种类 按制作方法分：铸造卡环和锻丝卡环 按卡环臂数目分：单臂卡环、双臂卡环和三臂卡环 圈形卡环：主要用于远中孤立的磨牙，有近、远中两个支托 回力卡环：主要用于前磨牙或尖牙作为基牙的末端游离缺失，远中合支托不与基托相连，可减轻基牙负荷，起应力中断作用。 杆形卡环：从基托伸出，经龈组织到达牙齿表面，由下向上呈推形固位。具有弹性好，美观的优点，但运用受软组织倒凹限制。
卡环的组合应用 RPI卡环：由近中合支托、邻面板和Ⅰ杆组成。具有对基牙的损伤小、固位作用好，舒适美观的优点。 RPA卡环：由近中合支托、邻面板和圆形卡环组成。它具有与RPI相同的优点，并且克服了RPI在口腔前庭深度不足或基牙下存在软组织倒凹时不能使用Ⅰ杆的不足之处。
（二）间接固位体（indirect retainer) 定义：是辅助直接固位体起固位作用，防止义齿翘起、 摆动、旋转、下沉而设计的一些装置，主要是加强 义齿的稳定性。 作用： Ａ、防止游离端基托合向脱位或翘动 Ｂ、防止游离端基托末端水平向摆动 Ｃ、防止义齿沿牙弓纵轴转动 Ｄ、分散合力，减轻基牙负担
间接固位体的位置和支点线的关系 a) 间接固位体到支点线的距离应等于基托游离 远端到支点线的距离。 b) 间接固位体应尽量位于支点线的垂直平分线 上。
连接体（connector) （一）大连接体 作用：a将义齿各部分连接为一整体。 b 传导和分散合力。 c 增加义齿强度，减少基托面积 要求：a要有一定的强度 b不妨碍软组织的运动 c横截面呈扁平形、板条形或半梨形 d不能进入软组织倒凹
种类： 上颌 腭杆（palatal bar)、腭板(palatal plate) 下颌 舌杆（lingual bar)、舌板(lingual plate)、颊杆(buccal bar)
（二）小连接体小连接体的作用是把金属支架上的各部分与大连接体相连接。（二）小连接体小连接体的作用是把金属支架上的各部分与大连接体相连接。 （三）弹性连接体也称为应力中断连接体，是一种用于游离缺失可摘局部义齿设计使用的有应力中断作用的连接体
牙列缺损与RPD的分类 • Kennedy牙列缺损分类法 • Applegate-Kennedy牙列缺损分类法 • 王征寿分类法
Kennedy 牙列缺损分类法 第一类 义齿鞍基在两侧基牙的远中，远中为游离端。 第二类 义齿鞍基在一侧基牙的远中，远中为游离端。 第三类 义齿鞍基在一侧或两侧，鞍基前后都有基牙。 第四类 义齿鞍基位于基牙的前面，基牙在缺隙远中。 优点 表达了缺隙所在的部位，体现了RPD鞍基与基牙的关系，简单易懂。 缺点不能反映缺牙的数目以及对功能的影响。
Kennedy第一类及其亚类
Kennedy第二类及其亚类
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8563 | https://math.stackexchange.com/questions/724294/if-a-ab-ba-is-a-nilpotent | linear algebra - If $A=AB-BA$, is $A$ nilpotent? - Mathematics Stack Exchange
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If A=A B−B A A=A B−B A, is A A nilpotent?
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Asked 11 years, 6 months ago
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Let matrix A n×n A n×n, be such that there exists a matrix B B for which
A B−B A=A A B−B A=A
Prove or disprove that there exists m∈N+m∈N+such
A m=0,A m=0,
I know
t r(A)=t r(A B)−t r(B A)=0 t r(A)=t r(A B)−t r(B A)=0
then I can't.Thank you
linear-algebra
matrices
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edited Nov 20, 2016 at 20:27
Martin Argerami
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asked Mar 24, 2014 at 4:42
math110math110
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1 I closed this question four years ago as a duplicate of math.stackexchange.com/q/227984 on the ground that the result here is a direct consequence of Jacobson’s lemma. On a second thought, as the assumption made here ([A,B]=A[A,B]=A) is stronger than the one ([[A,B],A]=0[[A,B],A]=0) in Jacobson’s lemma, simpler proofs are expected. So, I decided to reopen this question.user1551 –user1551 2024-09-08 20:04:53 +00:00 Commented Sep 8, 2024 at 20:04
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For any k k, you have A k+1=A A k=(A B−B A)A k=A B A k−B A k+1 A k+1=A A k=(A B−B A)A k=A B A k−B A k+1. So
tr(A k+1)=0,k=0,1,2,…tr(A k+1)=0,k=0,1,2,…
We deduce that tr(p(A))=0 tr(p(A))=0 for every polynomial p p with p(0)=0 p(0)=0. This implies that all eigenvalues of A A are zero (because otherwise we can get a polynomial p p that is 1 at all nonzero eigenvalues, and p(A)p(A) would not have zero trace). So A A is nilpotent, i.e. there exists m m with A m=0 A m=0.
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edited Nov 10, 2017 at 18:22
José Carlos Santos
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answered Mar 24, 2014 at 5:24
Martin ArgeramiMartin Argerami
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7 This argument requires characteristic 0 0 (which may be an unstated assumption in the problem). For example, in characteristic p p, every p×p p×p scalar matrix has trace zero. In particular, every polynomial of the identity matrix is trace zezro.Aaron –Aaron 2014-03-24 05:34:27 +00:00 Commented Mar 24, 2014 at 5:34
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Denote by p p the characteristic of the field. The statement does not necessarily hold if 0<p≤n 0<p≤n. For a counterexample, consider p=n=2 p=n=2 and A=(0 1 1 0),B=(1 1 1 0)A=(0 1 1 0),B=(1 1 1 0).
The statement is true, however, if p=0 p=0 or p>n p>n. From the condition A B−B A=A A B−B A=A, one can prove by mathematical induction that A k=A B A k−1−B A k A k=A B A k−1−B A k and in turn trace(A k)=0 trace(A k)=0 for k=1,2,…,n k=1,2,…,n. Now, it is know that this latter condition together with p=0 p=0 or p>n p>n imply that A A is nilpotent (for a proof, see achille hui's answer in this thread or my answer to another question).
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edited Apr 13, 2017 at 12:20
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answered Mar 24, 2014 at 6:51
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Answer ChangedSince the other answer (beat me by 10 minute) is already using eigenvalues, I've rewritten the answer to do it in an alternate way.
Let A,B∈M m×m(K)A,B∈M m×m(K) such that [A,B]=A B−B A=A[A,B]=A B−B A=A and K K is a field with characteristic 0 0 or greater than m m. For any k∈Z+k∈Z+, we have
[A k,B]=A k−1[A,B]+A k−2[A,B]A+⋯+A[A,B]A k−2+[A,B]A k−1=k A k[A k,B]=A k−1[A,B]+A k−2[A,B]A+⋯+A[A,B]A k−2+[A,B]A k−1=k A k
Let χ A(λ)χ A(λ) be the characteristic polynomial of A A, i.e.
χ A(λ)=det(λ I m−A)=λ m+α m−1 λ m−1+⋯+α 0=0 χ A(λ)=det(λ I m−A)=λ m+α m−1 λ m−1+⋯+α 0=0
By Cayley-Hamilton theorem, we have
χ A(A)=A m+α m−1 A m−1+⋯+α 0 I m.χ A(A)=A m+α m−1 A m−1+⋯+α 0 I m.
Repeat apply the commutator [⋅,B][⋅,B] to it, we find:
0 0 0 0===⋮=A m m A m m 2 A m m m A m++++α m−1 A m−1(m−1)α m−1 A m−1(m−1)2 α m−1 A m−1(m−1)m α m−1 A m−1+⋯++⋯++⋯++⋯+α 1 A α 1 A α 1 A α 1 A++++α 0 I m 0 0 0 0=A m+α m−1 A m−1+⋯+α 1 A+α 0 I m 0=m A m+(m−1)α m−1 A m−1+⋯+α 1 A+0 0=m 2 A m+(m−1)2 α m−1 A m−1+⋯+α 1 A+0⋮0=m m A m+(m−1)m α m−1 A m−1+⋯+α 1 A+0
This can be recasted in a matrix form:
⎛⎝⎜⎜⎜⎜⎜⎜⎜1 m m 2 m m 1 m−1(m−1)2⋮(m−1)m…………1 1 1 1 1 0 0 0⎞⎠⎟⎟⎟⎟⎟⎟⎟⎛⎝⎜⎜⎜⎜⎜⎜⎜A m α m−1 A m−1 α m−2 A m−2⋮α 0 I m⎞⎠⎟⎟⎟⎟⎟⎟⎟=⎛⎝⎜⎜⎜⎜⎜⎜⎜0 0 0⋮0⎞⎠⎟⎟⎟⎟⎟⎟⎟(1 1…1 1 m m−1…1 0 m 2(m−1)2…1 0⋮m m(m−1)m…1 0)(A m α m−1 A m−1 α m−2 A m−2⋮α 0 I m)=(0 0 0⋮0)
Since the characteristic of K K is 0 0 or greater than m m, the Vandermonde matrix appear in LHS above is invertible. This allow us to conclude A m=0 A m=0.
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edited Mar 24, 2014 at 7:30
answered Mar 24, 2014 at 5:35
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Here is another proof (using characteristic zero).
The matrix B B defines a linear transformation on the space of n×n n×n matrices by X↦φ(X)=X B−B X X↦φ(X)=X B−B X. The equation you gave just means that A A is an eigenvector for the eigenvalue 1. Now suppose X X is an eigenvector for eigenvalue λ λ. Then
φ(X 2)=X 2 B−B X 2=X(B X+λ X)−B X 2=X B X+λ X 2−B X 2=(B X+λ X)X+λ X 2−B X 2=B X 2+λ X 2+λ X 2−B X 2=2 λ X 2 φ(X 2)=X 2 B−B X 2=X(B X+λ X)−B X 2=X B X+λ X 2−B X 2=(B X+λ X)X+λ X 2−B X 2=B X 2+λ X 2+λ X 2−B X 2=2 λ X 2
So we get that X 2 X 2 is an eigenvector for an eigenvalue 2 λ 2 λ. The endomorphism φ φ has only finitely many distinct eigenvalues. Assuming the your field is of characteristic zero (so the 2 n λ 2 n λ is an infinite sequence), then you must have X n=0 X n=0 for some big enough n n.
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answered Mar 24, 2014 at 5:38
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2 More generally, if [B,X]=B X−X B[B,X]=B X−X B commutes with X X, then, because [B,−][B,−] is a (non-commutative) derivation, you have [B,X n]=n X n−1[B,X][B,X n]=n X n−1[B,X]. This is a useful computation to know (and avoid having to redo).Aaron –Aaron 2014-03-24 05:49:27 +00:00 Commented Mar 24, 2014 at 5:49
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With the risk of being repetitive, let me just record here the (original?) proof of N. Jacobson in this paper, where you only need to assume that [A,B][A,B] and A A commute to deduce [A,B][A,B] is nilpotent.
Let [A,B]=A′[A,B]=A′, and consider D(X)=[X,B]D(X)=[X,B]. Then for any polynomial F F, we have that D(F(A))=F′(A)A′D(F(A))=F′(A)A′. Now pick a polynomial such that F(A)=0 F(A)=0 (this can always be done, since the matrices {1,…,A N}{1,…,A N} are linearly dependent if N N is large).
Since A′A′ commutes with A A, iterating the above we see that if F F has degree d d, then we have that F(d)(A)A′(2 d−1)=d!A′(2 d−1)F(d)(A)A′(2 d−1)=d!A′(2 d−1). Hence, over a field of zero characteristic we get A′A′ nilpotent.
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answered Jul 28, 2020 at 8:36
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8A B−B A=A A B−B A=A implies A A is singular and A A is nilpotent.
0A B−B A=A A B−B A=A, Then A is singular?
4Proving an operator is nilpotent
2Proving nilpotency of a matrix
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9Problem from the shortlist of the Romanian Mathematical olympiad
9A linear algebra problem regarding A B−B A=A A B−B A=A
5A,B∈M 2(C)A,B∈M 2(C) be such that A B−B A=B 2 A B−B A=B 2 ; then is it true that A B=B A A B=B A?
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8564 | https://www.journalbonefragility.com/wp-content/uploads/journal/2021/1.2/59-66.pdf | 59 X-linked hypophosphatemic rickets.
What the orthopedic surgeon needs to know Introduction X-linked hypophosphatemic rickets (XLH) is a disease caused by a mutation of the PHEX gene located on the X chro-mosome. This gene encodes for a surface endopeptidase ex-pressed on several, cells including osteocytes and osteoblasts. PHEX gene mutation is associated with increased expression of fibroblast growth factor 23 (FGF23) that, through an unknown mechanism, leads to increased phosphaturia and a deficiency of the active form of vitamin D. XLH accounts for approximately 80% of all cases of hypophosphatemic rickets with a preva-lence of 1/20000 newborns. It shows complete penetrance and high clinical variability . The clinical signs generally appear during the first or second year of life and involve several systems including the neurolog-ical, dental and skeletal ones. The latter is generally affected by the most striking clinical manifestations of XLH, such as cupped and flared metaphyses (giving rise to what look like bracelets around the wrist and ankle), scoliosis and especially lower limb deformities . Therefore, orthopedic surgeons are often the first physicians to examine affected children, and their approach to these patients is important in ensuring that they are appropriately managed. However, the diagnosis can be diffi-cult. In fact, although in XLH the cortical bone is thickened, the trabeculae are larger without any signs of bone resorption, and genetic rickets is not distinguishable from the nutritional form radiographically [2,3]. Therefore, a comprehensive approach is Annalisa De Cicco 1, Giuseppe Toro 1,2, Anna Grandone 3, Adriano Braile 1, Giovanni Landi 1, Giovanni Iolascon 1, Emanuele Miraglia Del Giudice 3, Alfredo Schiavone Panni 1 1 Department of Medical and Surgical Specialties and Dentistry, University of Campania “Luigi Vanvitelli”, Naples, Italy 2 Department of Clinical Sciences and Translational Medicine, University of Rome Tor Vergata, Rome, Italy 3 Department of Woman, Child and of General and Specialized Surgery, University of Campania “Luigi Vanvitelli”, Naples, Italy ABSTRACT Purpose: X-linked hypophosphatemic rickets (XLH) is a rare genetic disease characterized by an increase in fibroblast growth factor 23 (FGF23) expression. The skeleton is one of the systems most affected and deformities of the lower limbs are one of the first reasons for consulting an orthopedic surgeon. The aim of the present study was to offer practical advice for a comprehensive orthopedic approach to XLH.
Materials: A literature search was conducted in PubMed, a freely available and cost-effective database. The articles in-cluded in the study were discussed by a research group with specific expertise in bone metabolism and pediatric deform-ities, in order to answer three fundamental questions and thus provide the orthopedic specialist with guidance on XLH: (1) How should the physician complete the diagnosis of XLH?; (2) When might a surgical procedure be recommended?; (3) What kind of surgical procedure should be performed?
Results: Sixty-three articles were included and discussed by the research group.
Conclusions: A correct and timely diagnosis of XLH is essential to appropriately manage affected patients. To complete this diagnosis a detailed medical history of the patient, a comprehensive clinical and radiographic evaluation, and specific biochemical tests are needed. Pharmacological treatment is based on supplementation of both phosphate and vitamin D, however, a monoclonal antibody that inactivates FGF23 (burosumab), has recently been introduced with promising results. Orthopedic surgery is needed in cases of moderate or severe deformities, to allow physiological growth and prevent early osteoarthritis and gait alterations. Surgical options are osteotomies and hemiepiphysiodesis, which is preferred when-ever possible. Three different devices for temporary hemiepiphysiodesis are available (staples, transphyseal screws and tension band plates). Obviously, surgical procedures need an appropriate medical therapy to be effective. In conclusion, the diagnosis, treatment and follow-up of XLH require a multidisciplinary approach and a comprehensive evaluation of anamnestic, clinical and radiographic data.
KEYWORDS Lower limb alignment, X-linked hypophosphatemic rickets, guided growth, burosumab, surgery, hemiepiphysiodesis.
Article history Received 15 Mar 2021 – 17 May 2021 Contact Annalisa De Cicco; annalisadecicco24@gmail.com University of Campania “Luigi Vanvitelli” Department of Medical and Surgical Specialties and Dentistry Via L. De Crecchio 4, 80138 Naples, Italy Int J Bone Frag. 2021; 1(2):59-66 https:doi.org/10.57582/IJBF.210102.059 Licens terms 60 needed in order to complete a correct diagnosis, which is essen-tial in order to appropriately treat these patients and avoid the development of severe deformities. The present study aims to summarize current knowledge around the orthopedic aspects of XLH, and give practical advice for such an approach.
Methods During a preliminary meeting a research group, previously identified among orthopedics and pediatrics with specific exper-tise in bone metabolism, endocrinological and genetic diseases, identified three fundamental questions that could guide the or-thopedic specialist dealing with XLH : How should the physi-cian complete the diagnosis of XLH? ; When might a surgical procedure be recommended?; and What kind of surgical pro-cedure should be performed? To answer these questions, a Pu-bMed search was conducted by three independent researchers using the following keywords: “X-linked hypophosphatemic rickets”, “skeletal angular deformities”, “hemiepiphysiodesis”, “guided growth” and “corrective osteotomies”. The literature search was conducted only in PubMed giv-en that 90% of high-quality studies can be retrieved from this database, as reported by Rollin et al. . Therefore, searching in PubMed can be considered cost-effective, and a practitioner should be able to efficiently retrieve most of the literature on a topic using it [4,5]. All articles in English, Spanish and Italian were eligible for inclusion. Articles involving subjects over 18 years of age were excluded. The references of the included articles were also reviewed. Relevant articles were identified by consensus be-tween at least 2 out of 3 researchers. Data were extracted from the included studies, and their relevant findings were discussed by the research group and accepted if consensus was reached between at least 66.6% of the researchers.
Results Eighty-one articles were identified after the first search. Of these, 20 were excluded because they were not considered use-ful for answering the proposed questions. Through evaluation of the references included in the articles, two further studies were identified. Therefore, data were extracted from a total of 63 articles and their relevant findings were discussed by the research group (Table I).
De Cicco A et al.
Table I List of the articles discussed by the research group.
1. Acar S, Demir K, Shi Y. Genetic causes of rickets. J Clin Res Pediatr Endocrinol. 2017;9(Suppl 2):88-105. 2. Al Juraibah F, Al Amiri E, Al Dubayee M, et al. Diagnosis and management of X-linked hypophosphatemia in children and adolescent in the Gulf Cooperation Council countries. Arch Osteoporos. 2021;16(1):52. 3. Berndt M, Ehrich JH, Lazovic D, et al. Clinical course of hypophosphatemic rickets in 23 adults. Clin Nephrol. 1996;45(1):33-41.
4. Cañete R, Caballero-Villarraso J, Aguilar-Quintero M, Vázquez-Rueda F. Beneficial effects of growth hormone therapy for ossification defects after bone distraction in X linked hypophosphataemic rickets. BMJ Case Rep. 2014;2014:bcr2013203069. 5. Carpenter TO, Imel EA, Holm IA, Jan de Beur SM, Insogna KL. A clinician’s guide to X-linked hypophosphatemia. J Bone Miner Res. 2011;26(7):1381-8. 6. Chesher D, Oddy M, Darbar U, et al. Outcome of adult patients with X-linked hypophosphatemia caused by PHEX gene mutations. J Inherit Metab Dis.
2018;41(5):865-76. 7. Cheung M, Roschger P, Klaushofer K, et al. Cortical and trabecular bone density in X-linked hypophosphatemic rickets. J Clin Endocrinol Metab. 2013;98(5):E954-61. 8. Evans GA, Arulanantham K, Gage JR. Primary hypophosphatemic rickets. Effect of oral phosphate and vitamin D on growth and surgical treatment. J Bone Joint Surg Am. 1980;62(7):1130-8.
9. Fuente R, Gil-Peña H, Claramunt-Taberner D, et al. X-linked hypophosphatemia and growth. Rev Endocr Metab Disord. 2017;18(1):107-15. 10. Gizard A, Rothenbuhler A, Pejin Z, et al. Outcomes of orthopedic surgery in a cohort of 49 patients with X-linked hypophosphatemic rickets (XLHR). Endocr Connect. 2017;6(8):566-73. 11. Greene WB, Kahler SG. Surgical aspects of limb deformity in hypophosphatemic rickets. South Med J. 1985;78(10):1185-9. 12. Hawley S, Shaw NJ, Delmestri A, et al. Prevalence and mortality of individuals with X-Linked hypophosphatemia: a United Kingdom real-world data analysis. J Clin Endocrinol Metab. 2020;105(3):e871-8. 13. Juárez Jiménez HG, Mier Cisneros R, Peralta Cruz S. [Mid-third femoral shaft fracture in a patient with hypophosphatemic rickets treated with a locking centromedullary nail]. Acta Ortop Mex. 2009;23(4):193-6.
14. Kanel JS, Price CT. Unilateral external fixation for corrective osteotomies in patients with hypophosphatemic rickets. J Pediatr Orthop. 1995;15(2):232-5.
15. Kinoshita Y, Fukumoto S. X-linked hypophosphatemia and FGF23-related hypophosphatemic diseases: prospect for new treatment. Endocr Rev. 2018;39(3):274-91. 16. Kocaoglu M, Bilen FE, Sen C, Eralp L, Balci HI. Combined technique for the correction of lower-limb deformities resulting from metabolic bone disease. J Bone Joint Surg Br. 2011;93(1):52-6. 17. Larson AN, Trousdale RT, Pagnano MW, Hanssen AD, Lewallen DG, Sanchez-Sotelo J. Hip and knee arthroplasty in hypophosphatemic rickets. J Arthroplasty. 2010;25(7):1099-103. 18. Laurent MR, De Schepper J, Trouet D, et al. Consensus recommendations for the diagnosis and management of X-linked hypophosphatemia in Belgium. Front Endocrinol (Lausanne). 2021;12:641543. 19. Lo SH, Lachmann R, Williams A, Piglowska N, Lloyd AJ. Exploring the burden of X-linked hypophosphatemia: a European multi-country qualitative study. Qual Life Res. 2020;29(7):1883-93. 20. Mao M, Carpenter TO, Whyte MP, et al. Growth curves for children with X-linked hypophosphatemia. J Clin Endocrinol Metab. 2020;105(10):3243-9. 21. Martel-Villagrán J, Arias-Medina A, García-Mardones G. Usefulness of X-rays in the differential diagnosis of hypophosphataemic rickets. Adv Ther. 2020;37(Suppl 2):89-94. 22. Nielsen LH, Rahbek ET, Beck-Nielsen SS, Christesen HT. Treatment of hypophosphataemic rickets in children remains a challenge. Dan Med J. 2014;61(7):A4874.
Int J Bone Frag. 2021; 1(2):59-66 61 23. Petje G, Meizer R, Radler C, Aigner N, Grill F. Deformity correction in children with hereditary hypophosphatemic rickets. Clin Orthop Relat Res. 2008;466(12):3078-85. 24. Rothenbuhler A, Schnabel D, Högler W, Linglart A. Diagnosis, treatment-monitoring and follow-up of children and adolescents with X-linked hypophosphatemia (XLH). Metabolism. 2020;103S:153892. 25. Santos Rodríguez F. X-linked hypophosphataemic rickets and growth. Adv Ther. 2020;37(Suppl 2):55-61. 26. Simon MH, Grünwald L, Schenke M, Dickschas J, Strecker W. Corrective osteotomies of femur and tibia: which factors influence bone healing? Arch Orthop Trauma Surg. 2020;140(3):303-11. 27. Song HR, Soma Raju VV, Kumar S, et al. Deformity correction by external fixation and/or intramedullary nailing in hypophosphatemic rickets. Acta Orthop. 2006;77(2):307-14. 28. Speirs JN, Nelson SC. The Rule of 57: orthopaedic trigonometry made easy. J Am Acad Orthop Surg. 2019;27(24):e1110-e1114. 29. Veilleux L-N, Cheung MS, Glorieux FH, Rauch F. The muscle-bone relationship in X-linked hypophosphatemic rickets. J Clin Endocrinol Metab. 2013;98(5):E990-5. 30. Emma F, Cappa M, Antoniazzi F, et al. X-linked hypophosphatemic rickets: an Italian experts’ opinion survey. Ital J Pediatr. 2019;45(1):67. 31. Haffner D, Emma F, Eastwood DM, et al. Clinical practice recommendations for the diagnosis and management of X-linked hypophosphataemia. Nat Rev Nephrol. 2019;15(7):435-55. 32. Popkov D, Lascombes P, Berte N, et al. The normal radiological anteroposterior alignment of the lower limb in children. Skeletal Radiol. 2015;44(2):197-206. 33. Lambert AS, Linglart A. Hypocalcaemic and hypophosphatemic rickets. Best Pract Res Clin Endocrinol Metab. 2018;32(4):455-76. 34. Espandar R, Mortazavi SM, Baghdadi T. Angular deformities of the lower limb in children. Asian J Sports Med. 2010;1(1):46-53. 35. Beck-Nielsen SS, Brixen K, Gram J, Mølgaard C. High bone mineral apparent density in children with X-linked hypophosphatemia. Osteoporos Int. 2013;24(8):2215-21. 36. Paley D, Tetsworth K. Mechanical axis deviation of the lower limbs. Preoperative planning of multiapical frontal plane angular and bowing deformities of the femur and tibia. Clin Orthop Relat Res. 1992;(280):65-71.
37. Thacher TD, Pettifor JM, Tebben PJ, et al. Rickets severity predicts clinical outcomes in children with X-linked hypophosphatemia: utility of the radiographic Rickets Severity Score. Bone. 2019;122:76-81.
38. Thacher TD, Fischer PR, Pettifor JM, Lawson JO, Manaster BJ, Reading JC. Radiographic scoring method for the assessment of the severity of nutritional rickets. J Trop Pediatr. 2000;46(3):132-9. 39. Saraff V, Nadar R, Högler W. New Developments in the treatment of X-linked hypophosphataemia: implications for clinical management. Paediatr Drugs. 2020;22(2):113-21. 40. Colares Neto GP, Pereira RM, Alvarenga JC, Takayama L, Funari MF, Martin RM. Evaluation of bone mineral density and microarchitectural parameters by DXA and HR-pQCT in 37 children and adults with X-linked hypophosphatemic rickets. Osteoporos Int. 2017;28(5):1685-92. 41. Bhambri R, Naik V, Malhotra N, et al. Changes in bone mineral density following treatment of osteomalacia. J Clin Densitom. 2006;9(1):120-7.
42. Bitzan M, Goodyer PR. Hypophosphatemic rickets. Pediatr Clin North Am. 2019;66(1):179-207. 43. Carpenter TO, Whyte MP, Imel EA, et al. Burosumab therapy in children with X-linked hypophosphatemia. N Engl J Med. 2018;378(21):1987-98.
44. Padidela R, Cheung MS, Saraff V, Dharmaraj P. Clinical guidelines for burosumab in the treatment of XLH in children and adolescents: British paediatric and adolescent bone group recommendations. Endocr Connect. 2020;9(10):1051-6. 45. Sharkey MS, Grunseich K, Carpenter TO. Contemporary medical and surgical management of X-linked hypophosphatemic rickets. J Am Acad Orthop Surg. 2015;23(7):433-42. 46. Rubinovitch M, Said SE, Glorieux FH, Cruess RL, Rogala E. Principles and results of corrective lower limb osteotomies for patients with vitamin D-resistant hypophosphatemic rickets. Clin Orthop Relat Res. 1988;(237):264-70.
47. Novais E, Stevens PM. Hypophosphatemic rickets: the role of hemiepiphysiodesis. J Pediatr Orthop. 2006;26(2):238-44. 48. Alsancak S, Guner S, Kinik H. Orthotic variations in the management of infantile tibia vara and the results of treatment. Prosthet Orthot Int. 2013;37(5):375-83. 49. Stamp WG, Whitesides TE, Field MH, Scheer GE. Treatment of vitamin-D resistant rickets. a long-term evaluation of its effectiveness. J Bone Joint Surg Am. 1964;46:965-77.
50. Gottliebsen M, Shiguetomi-Medina JM, Rahbek O, Møller-Madsen B. Guided growth: mechanism and reversibility of modulation. J Child Orthop. 2016;10(6):471-7. 51. Vaishya R, Shah M, Agarwal AK, Vijay V. Growth modulation by hemi epiphysiodesis using eight-plate in Genu valgum in Paediatric population. J Clin Orthop Trauma. 2018;9(4):327-33. 52. Phemister DB. [Epiphysiodesis for equalizing the length of the lower extremities and for correcting other deformities of the skeleton]. Mem Acad Chir (Paris). 1950;76(26-27):758-63.
53. Blount WP, Clarke GR. Control of bone growth by epiphyseal stapling; a preliminary report. J Bone Joint Surg Am. 1949;31A(3):464-78.
54. Mesa PA, Yamhure FH. Percutaneous hemi-epiphysiodesis using transphyseal cannulated screws for genu valgum in adolescents. J Child Orthop. 2009;3(5):397-403. 55. Stevens PM, Klatt JB. Guided growth for pathological physes: radiographic improvement during realignment. J Pediatr Orthop. 2008;28(6):632-9. 56. Shapiro G, Adato T, Paz S, et al. Hemiepiphysiodesis for coronal angular knee deformities: tension-band plate versus percutaneous transphyseal screw. Arch Orthop Trauma Surg. 2020 Sep 21. 57. Métaizeau J-P, Wong-Chung J, Bertrand H, Pasquier P. Percutaneous epiphysiodesis using transphyseal screws (PETS). J Pediatr Orthop. 1998;18(3):363-9. 58. Lee SW, Lee KJ, Cho CH, et al. Affecting factors and correction ratio in genu valgum or varum treated with percutaneous epiphysiodesis using transphyseal screws. J Clin Med. 2020;9(12):4093. 59. Campens C, Mousny M, Docquier PL. Comparison of three surgical epiphysiodesis techniques for the treatment of lower limb length discrepancy. Acta Orthop Belg. 2010;76(2):226-32.
60. Raimann A, Mindler GT, Kocijan R, et al. Multidisciplinary patient care in X-linked hypophosphatemic rickets: one challenge, many perspectives. Wien Med Wochenschr. 2020;170(5-6):116-23. 61. Al Kaissi A, Farr S, Ganger R, Klaushofer K, Grill F. Windswept lower limb deformities in patients with hypophosphataemic rickets. Swiss Med Wkly. 2013;143:w13904. 62. Özkul B, Çamurcu Y, Sokucu S, Yavuz U, Akman YE, Demir B. Simultaneous bilateral correction of genu varum with Smart frame. J Orthop Surg (Hong Kong). 2017;25(2):2309499017713915. 63. Horn J, Steen H, Huhnstock S, Hvid I, Gunderson RB. Limb lengthening and deformity correction of congenital and acquired deformities in children using the Taylor Spatial Frame. Acta Orthop. 2017;88(3):334-40. Int J Bone Frag. 2021; 1(2):59-66 X-linked hypophosphatemic rickets orthopedic approach 62 Discussion Rickets is a general term that refers to a heterogeneous group of diseases characterized by defective bone mineraliza-tion that leads to replacement of mineralized with non-mineral-ized bone matrix (osteoid tissue) in the early stages of growth . There are several causes of rickets, including nutritional ones (linked to vitamin D and/or phosphate deficiency), tumors secreting FGF23 and other phosphatonins, and other system-ic diseases such as hyperthyroidism, hyperparathyroidism and sarcoidosis . Genetic rickets is generally divided into “vita-min D-dependent” and “hypophosphatemic” types. The latter can be further divided into three forms: autosomal dominant, characterized by a mutation that makes FGF23 resistant to en-zyme cleavage and shows incomplete penetrance and clinical variability; autosomal recessive, characterized by increased production of FGF23 and defective maturation of osteocytes; and XLH, the most frequent, which shows an X-linked domi-nant pattern of inheritance.
How should the physician complete the diagnosis of XLH?
The diagnosis of XLH is based on medical history, physical and instrumental examinations, family history (even though 1/3 of cases are sporadic), and mutational analysis of the gene. Very often the child comes to the attention of the orthopedic specialist because of lower limb malalignment, which could affect 94.8% of patients with XLH .
The medical anamnesis is useful for understanding the on-set of the deformity, its history (whether it was preceded by a trauma, whether it is improving or worsening), and its associ-ation with other signs or symptoms. The lower limb malalign-ment might be angular (valgus or varus knee) or torsional (ret-ro-anteversion of the femur and extra-internal rotation of the tibia) . However, these conditions can also represent stages in normal growth. Therefore, it is important to be able to recog-nize pathological varus and valgus knees, and also to differenti-ate ones associated with rickets from ones associated with oth-er disorders . The physical examination should be conducted with the patient appropriately positioned (namely with the pa-tella of both limbs in the frontal plane) in order to reduce the in-fluence of femoral anteversion and adipose tissue which could lead to apparent malalignment (to reduce the femoral antever-sion and fat tissue contribution to the knee malalignment that could lead to an apparent malalignment ). When the patient’s position is considered appropriate, the orthopedic specialist should evaluate the presence and degree of the malalignment, calculating both the intercondylar and intermalleolar distanc-es. If pathological lower limb malalignment is observed, full-length antero-posterior and lateral X-rays are needed. Correct alignment of the lower limbs is demonstrated by the presence of a “physiological valgus angle” between the anatomical axis of the femur and the tibia (6° – 8°) . Howev-er, comprehensive evaluation of lower limb alignment needs a standardized analysis of the anatomical angles of both the femur and the tibia (see Fig. 1) [3,8]. The first radiological parameter to be evaluated is the me-chanical axis deviation (MAD), defined as the deviation of the knee center of rotation from the lower limb mechanical axis (see Fig. 1). Physiologically, the knee center of rotation should be located medially to the mechanical axis and the MAD should not be greater than 8 mm. Any deviation from this dis-tance will result in lower limb malalignment (varus knee in the presence of MAD values > 8mm; valgus knee when the MAD is < 8 mm). Obviously, the MAD tends to change as the child grows, going from positive deviation values (varus knee) dur-ing the first two years of age to negative ones (valgus knee) up to six years of age, when, in general, physiological alignment is reached . However, an orthopedic surgeon should always con-sider malalignment radiographically pathological if the MAD is greater than 8 mm, especially in children over 2 years of age.
When malalignment is observed, an appropriate descrip-tion of the deformity is required and this should be obtained by evaluating the joint orientation angles on both the full-length antero-posterior and the lateral lower limb radiographs (see Figure 1 and Table II for further details) [3,8,9]. Furthermore, a careful evaluation of the morphology of the physis is recommended to identify the characteristic changes De Cicco A et al.
Figure 1 Full-length lower limb X-ray of a 4-year-old patient with hypophosphatemic rickets.
Note the widening, cupping and indistinct margins of the distal femoral and proximal tibial physes. On the right side, the mechanical axis of the lower limb (straight line in red) and the evaluation of its distance from the knee center of rotation (MAD). On the left side, joint orientation angles of the femur (in blue) and the tibia (in green) calculated with respect to the mechanical axis of each bone.
Int J Bone Frag. 2021; 1(2):59-66 63 associated with rickets: growth plate widening and abnormal configuration of the metaphysis (indistinct margins, widening and cupping) [10,11]. If rickets is suspected, a radiograph of the distal radius is recommended in order to look for these metaph-ysis abnormalities in this area, too, and to calculate the Rickets score [10-12]. This score, which ranges from 0 to 10, is a useful tool for the follow-up of the patient with rickets, giving an idea of the severity of the disease . Computerized bone miner-alometry (CBM), a method of measuring bone mineral density (BMD), has been proposed as a diagnostic and monitoring tool for XLH because it is simple and quick to use, and painless . However, the interpretation of pediatric dual-energy X-ray absorptiometry scans is complex due to the continuous chang-es in the geometry of the growing skeleton. This can poten-tially make the values obtained misleading, especially in the presence of growth problems like those observed in rickets. In fact, children with XLH show an increase in bone miner-al apparent density (BMAD, a volumetric measure of bone density) of the spine compared with the femoral neck, prob-ably because the disease affects the growth of the upper and lower limbs more than that of the spine . CBM could be more useful for assessing, as part of a comprehensive eval-uation, the associated clinical manifestations, like short stat-ure, dental abscesses and head deformities, even though this test, also in adults with XLH osteomalacia, is used more to assess response to therapy than to make a diagnosis. Even in this case, however, doubtful values can be observed, as the BMD is often exceptionally high in the lumbar spine of these patients because of extra-skeletal calcifications and enthesopa-thies [13,14]. The diagnosis of XLH might be confirmed through evaluation of certain biochemical parameters that are typical-ly altered in this disease, such as hypophosphatemia (in this case, the evaluation must be age-adjusted), increased serum alkaline phosphatase and FGF23 levels, and phosphaturia. Although the serum levels of FGF23 may be up to five times the normal values, and a cut-off point for the diagnosis was set at 30 ng/ml, low values could also be observed in XLH . In any case, whenever XLH is suspected or biochemically con-firmed, a consultation with a pediatrician or endocrinologist with specific expertise is recommended. Figure 2 Comparative antero-posterior distal radius X-rays showing the physeal changes (indistinct margins, widening and cupping) observable in figure 1 (same patient).
Table II Joint orientation angles.
FRONTAL PLANE (ANTERO-POSTERIOR VIEW) ANGLE DEFINITION NORMAL VALUES Lateral proximal femoral angle (LPFA) The angle between a line connecting the hip center of rotation and the apex of the greater trochanter and the mechanical or anatomical axis of the femur 90° (85°–95°) Lateral distal femoral angle (LDFA) The lateral angle between the distal femoral joint line and either the mechanical or anatomical axis of the femur 88° (85°–90°) Medial proximal tibial angle (MPTA) The medial angle between the proximal tibial joint line and either the mechanical or anatomical axis of the tibia 87° (85°–90°) Lateral distal tibial angle (LDTA) The lateral angle between the distal tibial joint line and either the mechanical or the anatomical axis of the tibia 89° (86°–92°) SAGITTAL PLANE (LATERAL VIEW) ANGLE DEFINITION NORMAL VALUES Posterior proximal femoral angle (PPFA) The angle between a mid-diaphyseal line of the femoral neck and a line that crosses the proximal physis 90° Anterior neck shaft angle (ANSA) The angle between the mid-diaphyseal line of the femoral neck and the mid-diaphyseal line of the proximal part of the diaphysis 170° (165°–175°) Posterior distal femoral angle (PDFA) The angle between the distal femoral joint line and the mid-diaphyseal line of the distal femur 83° (79°–87°) Posterior proximal tibial angle (PPTA) The angle between the proximal joint line and the mechanical axis of the tibia 81° (77°–84°) Anterior distal tibial angle (ADTA) The angle between the distal articular joint line and mechanical axis of the tibia 80° (78°–82°) Int J Bone Frag. 2021; 1(2):59-66 X-linked hypophosphatemic rickets orthopedic approach 64 When might a surgical procedure be recommended?
Traditionally, medical management of XLH was based on early supplementation of oral phosphate and vitamin D. This kind of approach was demonstrated to be effective only on some of the manifestations of the disease, such as dental alterations and reduction of definitive height . However, up to two-thirds of children with XLH submitted to this kind of treatment required surgical intervention for lower limb deformities . The recent advent of burosumab has marked a pivotal change in the man-agement of XLH 16. Approved in 2018 by the European Med-icines Agency , burosumab is a monoclonal antibody that binds and neutralizes FGF23, reducing its effects on the bone . FGF23 has both direct and indirect actions on bone metab-olism and growth: it inhibits the proliferation of chondrocytes in the metaphysis, reduces the levels of phosphate and calcium (inorganic components of the bone matrix), and inhibits the conversion of vitamin D into its active form [2,9]. Therefore, bu-rosumab might reduce the severity or even prevent the devel-opment of skeletal deformities, while improving the effective-ness of the necessary surgical procedures. Non-responders to the supplementation have been seen to need up to nine surgical procedures to correct the deformities , with a recurrence rate of 27% after both closing- and opening-wedge osteotomies .
Correction of angular deformities of the lower limbs must be performed in order to improve the patient’s quality of life and limb function. In fact, a severe lower limb angular deform-ity could limit a child’s walking and participation in sports activities, and lead to abnormal joint loading with the occur-rence of early osteoarthritis in the compartment more affected by the deformity [18,20,21]. Moreover, valgus knee is frequently associated with patellofemoral syndrome . Use of traditional orthopedic footwear does not seem to be useful for the correc-tion of angular deformities of the lower limbs, and the only demonstrated effective conservative treatment is that based on orthoses with thrusts and counter-thrusts in the area of the de-formity .
In XLH, however, it should be noted that this kind of ap-proach would not be effective without appropriate medical management of the disease, and that patient compliance tends to become progressively worse from the age of 5 years .
What kind of surgical procedure should be performed?
The surgical procedures used for the correction of angular lower limb malalignments are hemiepiphysiodesis and osteot-omies. Hemiepiphysiodesis refers to procedures based on the concepts of “guided growth”, related to the mechanical ma-nipulation and modulation of bone growth . These concepts were first introduced in 1862 by Heuter, who stated that the growth of the epiphysis depends, among other things, on the pressure to which it is exposed (with increased pressure slow-ing down growth, and decreased pressure promoting it) . In 1933, Phemister et al. first described the surgical procedure of hemiepiphysiodesis as permanent growth arrest brought about by a partial physeal ablation . Nowadays, most authors prefer temporary hemiepiphysiodesis, which is based on the use of different fixation devices that straddle the physis on the convex side of the deformity with the ultimate aim of slowing down the growth of the more active side, while promoting recovery of the less active one . This surgical procedure was proposed by Blount and Clarke using staples , the aim being to arrest growth under the staples, thereby allowing gradual correction of angular deformities. Reported complications associated with the use of staples were loss of fixation, breakage or migration of the staples (especially if the epiphysis was too small), and re-bound effects [26,29]. Staple-related complications, including re-bound effects, were also observed by both Novais and Stevens and Stevens and Klatt [21,30] in their series of patients with XLH. This observation led Novais and Stevens to prefer the tension band plate (TBP) to treat this kind of patient . The TBP is a pre-shaped two- to four-hole plate known as an eight-plate. The plate, which has a long lever arm, acts as a focal hinge at the edge of the physis. Reported TBP-related complications are infections, screw breakage, partial correction and overcorrection [26,29,31]. In the series by Stevens and Klatt, the authors did not report any complications with the use of eight-plates in XLH . Metaizeau et al. described percutaneous epiphysiodesis using transphyseal screws (PETS) . The authors used cannu-lated screws across the physis. These screws are stronger than staples and provide a rigid temporary hemiepiphysiodesis that gives faster correction (albeit starting later) using a percuta-neous procedure. Under fluoroscopic guidance, the physis is crossed using a K-wire properly positioned in the center of the sagittal plane, i.e., in the external and the internal quarter of the coronal plane for genu varum and valgum, respectively. The positioning of the K-wire is a pivotal phase of the procedure because it serves as a guide for the cannulated screw; misposi-tioning could result in final malalignment [32,33]. One of the main complications of this surgery is the possibility of premature growth arrest . To the best of our knowledge, no studies have been published on the use of PETS in XLH.
The three guided-growth techniques reported (hemiepiphys-iodesis with staples , TBP and PETS) were directly compared for the treatment of lower limb discrepancy; less pain and faster operative time and recovery were observed with PETS, while no clear differences were reported in terms of complications .
Shapiro et al. compared TBD and PETS for the correction of angular deformities. They observed that the former was as-sociated with faster correction, while the complication rate was comparable .
Correction performed through guided-growth techniques can be considered satisfactory when the tibio-femoral angle is in the normal range and its correction is maintained after fixation removal until skeletal maturity. Of note, if appropriately per-formed, hemiepiphysiodesis may also aid in correcting diaphy-seal deformities . In order to improve outcomes, it is advisa-ble to avoid activities with excessive axial loading of the lower limbs, like jumping and running, during the first two weeks after the surgery and clinical and radiographic follow-up should be performed at regular intervals in a multidisciplinary setting [31,35]. No specific criteria have been proposed to guide the start of guided-growth treatment in XLH , but an important fac-tor that may have an impact on the final outcomes is the age at which the surgery is performed. Generally, the varus knee De Cicco A et al.
Int J Bone Frag. 2021; 1(2):59-66 65 should be corrected from 4 years of age, while the valgus knee from around 10 years of age, as the deformity is quite stable at this stage. However, Novais and Stevens suggested that a guided-growth procedure could be started before 10 years of age in XLH .
If “guided growth” led to unsatisfactory correction, the center of the deformity does not involve the growth plates, re-sidual growth is insufficient, or the deformity presents an angle > 20°, corrective osteotomies might be required . Corrective osteotomy should be performed at the center of the deformity — this is because the farther from the center it is performed, the wider the translation between the two fragments will be . However, in varus knee, osteotomy very often involves the proximal portion of the tibia, while in valgus knee it involves the distal portion of the femur [37,38]. There are two types of os-teotomy: opening-wedge (with the need of a bone graft) and closing-wedge (based on the subtraction of a piece of bone). Osteotomy is followed by deformity correction (acute or grad-ual) and fixation (internal and external, respectively). Gradual correction with external fixators reduces the rate of complica-tions associated with the acute type (i.e. neurovascular lesions, compartment syndrome, delayed union, etc.), and provides a better correction in cases with multiplanar deformities [37,38]. However, gradual correction with external fixators is not free of complications, and the main ones reported with its use are: fractures occurring during or after the correction, osteotomy nonunion and healing delays, osteomyelitis, recurrence, and residual deformity. Of note, the use of external fixation may be problematic before 8 years of age due to the lack of cooperation and motivation of patients . In these cases, too, patients must be followed up regularly in a multidisciplinary setting. Considering the unpredictability of bone metabolism in XLH and the need for medical therapy adjustments, close collaboration with an expert pediatric endocrinologist is man-datory in both types of surgery. However, in our opinion, the availability of burosumab will improve the outcomes of guid-ed-growth techniques, and thus reduce the need for corrective osteotomies. In conclusion, XLH is a disease that requires a multidis-ciplinary approach throughout the spectrum of its diagnosis, treatment and follow-up. The orthopedic surgeon is frequently the first physician involved in the identification of this disease because of the frequent occurrence of angular deformities of the lower limbs. However, to complete the diagnosis, detailed personal and family histories must be collected, and appropriate radiographic and laboratory tests must be performed and eval-uated, also involving an expert pediatric endocrinologist in the process. The orthopedic management is generally focused on the prevention and treatment of lower limb deformities through both conservative and surgical approaches. Surgery in XLH needs careful preoperative planning and close follow-up. In our opinion, when surgery is indicated, guided-growth techniques should be preferred whenever possible, and the availability of a targeted therapy would improve their outcomes, reducing the need for corrective osteotomies. However, starting burosumab early might reduce the severity or even prevent the develop-ment of limb deformities, potentially leading to a change in the orthopedic approach to XLH.
References 1. Emma F, Cappa M, Antoniazzi F, et al. X-linked hypophosphatemic rickets: an Italian experts’ opinion survey. Ital J Pediatr. 2019;45(1):67. 2. Haffner D, Emma F, Eastwood DM, et al. Clinical practice recommen-dations for the diagnosis and management of X-linked hypophospha-taemia. Nat Rev Nephrol. 2019;15(7):435-55. 3. Popkov D, Lascombes P, Berte N, et al. The normal radiological an-teroposterior alignment of the lower limb in children. Skeletal Radiol. 2015;44(2):197-206. 4. Rollin L, Darmoni S, Caillard J-F, Gehanno J-F. Searching for high-quality articles about intervention studies in occupational health--what is really missed when using only the Medline database? Scand J Work Environ Health. 2010;36(6):484-7. 5. Pillastrini P, Vanti C, Curti S, et al. Using PubMed search strings for efficient retrieval of manual therapy research literature J Manipulative Physiol Ther. 2015;38(2):159-66. 6. Lambert AS, Linglart A. Hypocalcaemic and hypophosphatemic rick-ets. Best Pract Res Clin Endocrinol Metab. 2018;32(4):455-76. 7. Espandar R, Mortazavi SM, Baghdadi T. Angular deformities of the lower limb in children. Asian J Sports Med. 2010;1(1):46-53.
8. Beck-Nielsen SS, Brixen K, Gram J, Mølgaard C. High bone mineral apparent density in children with X-linked hypophosphatemia. Osteo-poros Int. 2013;24(8):2215-21. 9. Paley D, Tetsworth K. Mechanical axis deviation of the lower limbs. Preoperative planning of multiapical frontal plane angular and bow-ing deformities of the femur and tibia. Clin Orthop Relat Res. 1992;(280):65-71.
10. Thacher TD, Pettifor JM, Tebben PJ, et al. Rickets severity predicts clinical outcomes in children with X-linked hypophosphatemia: utility of the radiographic Rickets Severity Score. Bone. 2019;122:76-81.
11. Thacher TD, Fischer PR, Pettifor JM, Lawson JO, Manaster BJ, Read-ing JC. Radiographic scoring method for the assessment of the severi-ty of nutritional rickets. J Trop Pediatr. 2000;46(3):132-9. 12. Saraff V, Nadar R, Högler W. New Developments in the treatment of X-linked hypophosphataemia: implications for clinical management. Paediatr Drugs. 2020;22(2):113-21. 13. Colares Neto GP, Pereira RM, Alvarenga JC, Takayama L, Funari MF, Martin RM. Evaluation of bone mineral density and microarchi-tectural parameters by DXA and HR-pQCT in 37 children and adults with X-linked hypophosphatemic rickets. Osteoporos Int. 2017;28 (5):1685-92. 14. Bhambri R, Naik V, Malhotra N, et al. Changes in bone mineral den-sity following treatment of osteomalacia. J Clin Densitom. 2006; 9(1):120-7.
15. Bitzan M, Goodyer PR. Hypophosphatemic rickets. Pediatr Clin North Am. 2019;66(1):179-207. 16. Carpenter TO, Whyte MP, Imel EA, et al. Burosumab therapy in children with X-linked hypophosphatemia. N Engl J Med. 2018; 378(21):1987-98. 17. European Medicines Agency (EMA). Crysvita. Published March 12, 2018.
18. Padidela R, Cheung MS, Saraff V, Dharmaraj P. Clinical guidelines for burosumab in the treatment of XLH in children and adolescents: Brit-ish paediatric and adolescent bone group recommendations. Endocr Connect. 2020;9(10):1051-6. 19. Sharkey MS, Grunseich K, Carpenter TO. Contemporary medical and surgical management of X-linked hypophosphatemic rickets. J Am Acad Orthop Surg. 2015;23(7):433-42. 20. Rubinovitch M, Said SE, Glorieux FH, Cruess RL, Rogala E. Princi-ples and results of corrective lower limb osteotomies for patients with vitamin D-resistant hypophosphatemic rickets. Clin Orthop Relat Res. 1988;(237):264-70.
21. Novais E, Stevens PM. Hypophosphatemic rickets: the role of hemie-piphysiodesis. J Pediatr Orthop. 2006;26(2):238-44. Int J Bone Frag. 2021; 1(2):59-66 X-linked hypophosphatemic rickets orthopedic approach 66 22. Panni AS, Toro G, de Cicco A, di Fino C, Braile A. Medial retinacu-lum reefing for patellar instability. In: Dejour D, Zaffagnini S, Arendt EA, Sillanpää P, Dirisamer F, eds. Patellofemoral pain, instability, and arthritis. Springer Berlin Heidelberg; 2020:219-224. 23. Alsancak S, Guner S, Kinik H. Orthotic variations in the management of infantile tibia vara and the results of treatment. Prosthet Orthot Int.
2013;37(5):375-83. 24. Stamp WG, Whitesides TE, Field MH, Scheer GE. Treatment of vita-min-D resistant rickets. a long-term evaluation of its effectiveness. J Bone Joint Surg Am. 1964;46:965-77.
25. Gottliebsen M, Shiguetomi-Medina JM, Rahbek O, Møller-Madsen B. Guided growth: mechanism and reversibility of modulation. J Child Orthop. 2016;10(6):471-7. 26. Vaishya R, Shah M, Agarwal AK, Vijay V. Growth modulation by hemi epiphysiodesis using eight-plate in Genu valgum in Paediatric population. J Clin Orthop Trauma. 2018;9(4):327-33. 27. Phemister DB. [Epiphysiodesis for equalizing the length of the lower extremities and for correcting other deformities of the skeleton]. Mem Acad Chir (Paris). 1950;76(26-27):758-63.
28. Blount WP, Clarke GR. Control of bone growth by epiphyseal stapling; a preliminary report. J Bone Joint Surg Am. 1949;31A(3):464-78.
29. Mesa PA, Yamhure FH. Percutaneous hemi-epiphysiodesis using transphyseal cannulated screws for genu valgum in adolescents. J Child Orthop. 2009;3(5):397-403. 30. Stevens PM, Klatt JB. Guided growth for pathological physes: radiograph-ic improvement during realignment. J Pediatr Orthop. 2008;28(6):632-9. 31. Shapiro G, Adato T, Paz S, et al. Hemiepiphysiodesis for coronal an-gular knee deformities: tension-band plate versus percutaneous trans-physeal screw. Arch Orthop Trauma Surg. 2020 Sep 21. 32. Métaizeau J-P, Wong-Chung J, Bertrand H, Pasquier P. Percutaneous epiphysiodesis using transphyseal screws (PETS). J Pediatr Orthop.
1998;18(3):363-9. 33. Lee SW, Lee KJ, Cho CH, et al. Affecting factors and correction ratio in genu valgum or varum treated with percutaneous epiphysiodesis us-ing transphyseal screws. J Clin Med. 2020;9(12):4093. 34. Campens C, Mousny M, Docquier PL. Comparison of three surgical epiphysiodesis techniques for the treatment of lower limb length dis-crepancy. Acta Orthop Belg. 2010;76(2):226-32.
35. Raimann A, Mindler GT, Kocijan R, et al. Multidisciplinary patient care in X-linked hypophosphatemic rickets: one challenge, many per-spectives. Wien Med Wochenschr. 2020;170(5-6):116-23. 36. Al Kaissi A, Farr S, Ganger R, Klaushofer K, Grill F. Windswept low-er limb deformities in patients with hypophosphataemic rickets. Swiss Med Wkly. 2013;143:w13904. 37. Okcu G, Ozkayin N, Okta C, Topcu I, Aktuglu K. Which implant is better for treating reverse obliquity fractures of the proximal femur: a standard or long nail? Clin Orthop Relat Res. 2013;471(9):2768-75. 38. Feldman DS, Madan SS, Ruchelsman DE, Sala DA, Lehman WB. Ac-curacy of correction of tibia vara: acute versus gradual correction. J Pediatr Orthop. 2006;26(6):794-8. 39. Horn J, Steen H, Huhnstock S, Hvid I, Gunderson RB. Limb length-ening and deformity correction of congenital and acquired deformi-ties in children using the Taylor Spatial Frame. Acta Orthop. 2017;88 (3):334-40.
Study limitations: Our study presents some limitations, mostly related to the quality of the included articles. Moreover, the use of a single scientific database could mean that other relevant articles are missed. However, the reliability, and cost-effectiveness of the PubMed database made us confident of the relevance of the discussed literature. Conflict of Interest: The authors declare that there is no conflict of interest De Cicco A et al.
Int J Bone Frag. 2021; 1(2):59-66 |
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Description Central binomial coefficient sports.svg The central binomial coefficients give the number of possible number of assignments of n-a-side sports teams from 2 n players, considering the playing area side by CMG Lee.
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8566 | https://artofproblemsolving.com/wiki/index.php/Spiral_similarity?srsltid=AfmBOopEMfZfWYflVlKnuFnISJufPzQfY5b8HFQDbTTl-1we3yZeNjny | Art of Problem Solving
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Spiral similarity
Page made by vladimir.shelomovskii@gmail.com, vvsss
Contents
1 Definition
2 Simple problems
2.1 Explicit spiral similarity
2.2 Hidden spiral symilarity
2.3 Linearity of the spiral symilarity
2.4 Construction of a similar triangle
2.5 Center of the spiral symilarity for similar triangles
2.6 Spiral similarity in rectangle
2.7 Common point for 6 circles
2.8 Three spiral similarities
2.9 Superposition of two spiral similarities
2.10 Spiral similarity for circles
2.11 Remarkable point for spiral similarity
2.12 Remarkable point for pair of similar triangles
2.13 Remarkable point’s problems
2.13.1 Problem 1
2.13.2 Problem 2
2.13.3 Problem 3
2.13.4 Problem 4
2.13.5 Solutions
2.14 Japan Mathematical Olympiad Finals 2018 Q2
Definition
A spiral similarity is a plane transformation composed of a rotation of the plane and a dilation of the plane having the common center. The order in which the composition is taken is not important.
Any two directly similar figures are related either by a translation or by a spiral similarity (directly similar figures are similar and have the same orientation).
The transformation is linear and transforms any given object into an object homothetic to given.
On the complex plane, any spiral similarity can be expressed in the form where is a complex number. The magnitude is the dilation factor of the spiral similarity, and the argument is the angle of rotation.
The spiral similarity is uniquely defined by the images of two distinct points. It is easy to show using the complex plane.
Let with corresponding complex numbers and so For any points and the center of the spiral similarity taking to point is also the center of a spiral similarity taking to This fact explain existance of Miquel point.
Case 1 Any line segment can be mapped into any other using the spiral similarity. Notation is shown on the diagram.
is circle is circle
is any point of is circle is the image under spiral symilarity centered at
is the dilation factor,
is the angle of rotation.
Case 2 Any line segment can be mapped into any other using the spiral similarity. Notation is shown on the diagram. is circle (so circle is tangent to is circle tangent to is any point of is circle is the image under spiral symilarity centered at is the dilation factor,
is the angle of rotation.
Simple problems
Explicit spiral similarity
Given two similar right triangles and Find and
Solution
The spiral similarity centered at with coefficient and the angle of rotation maps point to point and point to point
Therefore this similarity maps to
Hidden spiral symilarity
Let be an isosceles right triangle Let be a point on a circle with diameter The line is symmetrical to with respect to and intersects at Prove that
Proof
Denote Let cross perpendicular to in point at point
Then
Points and are simmetric with respect so
The spiral symilarity centered at with coefficient and the angle of rotation maps to and to point such that
Therefore
Linearity of the spiral symilarity
Points are outside
Prove that the centroids of triangles and are coinsite.
Proof
Let where be the spiral similarity with the rotation angle and
A vector has two parameters, modulo and direction. It is not tied to a center of the spiral similarity. Therefore We use the property of linearity and get Let be the centroid of so
is the centroid of the
Construction of a similar triangle
Let triangle and point on sideline be given. Construct where lies on sideline and lies on sideline
Solution
Let be the spiral symilarity centered at with the dilation factor and rotation angle
so image of any point lies on The spiral symilarity centered at with the dilation factor and rotation angle maps into and therefore the found triangle is the desired one.
Center of the spiral symilarity for similar triangles
Let triangle and point on sideline be given. where lies on sideline and lies on sideline The spiral symilarity maps into Prove
a)
b) Center of is the First Brocard point of triangles and
Proof
a) Let be the spiral symilarity centered at with the dilation factor and rotation angle
Denote Similarly
b) It is well known that the three circumcircles and have the common point (it is in the diagram).
Therefore is cyclic and
Similarly,
Similarly,
Therefore, is the First Brocard point of
is cyclic Similarly,
Therefore is the First Brocard point of and
Therefore the spiral symilarity maps into has the center the angle of the rotation
Spiral similarity in rectangle
Let rectangle be given. Let point
Let points and be the midpoints of segments and respectively.
Prove that
Proof
Let be the midpoint
is a parallelogram
and are corresponding medians of and
There is a spiral similarity centered at with rotation angle that maps to Therefore
Common point for 6 circles
Let and point on sideline be given. where lies on sideline and lies on sideline
Denote
Prove that circumcircles of triangles have the common point.
Proof
so there is the spiral symilarity taking to Denote the center of the center of is the secont crosspoint of circumcircles of and but this center is point so these circles contain point . Similarly for another circles.
Three spiral similarities
Let triangle be given. The triangle is constructed using a spiral similarity of with center , angle of rotation and coefficient
A point is centrally symmetrical to a point with respect to
Prove that the spiral similarity with center , angle of rotation and coefficient taking to
Proof
Corollary
Three spiral similarities centered on the images of the vertices of the given triangle and with rotation angles equal to the angles of take to centrally symmetric to with respect to
Superposition of two spiral similarities
Let be the spiral similarity centered at with angle and coefficient
Let be spiral similarity centered at with angle and coefficient
Let
Prove: a) is the crosspoint of bisectors and
b)
Algebraic proof
We use the complex plane Let Then Geometric proof
Denote
Then
Let be the midpoint be the point on bisector such that be the point on bisector such that Then
is the crosspoint of bisectors and
Corollary
There is another pair of the spiral similarities centered at and with angle coefficients and
In this case
Spiral similarity for circles
Let circle cross circle at points and Point lies on
Spiral similarity centered at maps into
Prove that points and are collinear.
Proof
Arcs
Corollary
Let points and be collinear.
Then exist the spiral similarity centered at such that
Let circle cross circle at points and
Points and lie on
Let be the tangent to be the tangent to
Prove that angle between tangents is equal angle between lines and
Proof
There is the spiral similarity centered at such that Therefore angles between these lines are the same.
Remarkable point for spiral similarity
Circles and centered at points and respectively intersect at points and Points and are collinear. Point is symmetrical to with respect to the midpoint point Prove:
a) b)
Proof
a) cross in midpoint
b) is parallelogram
Denote
Corollary
Let points and be collinear. Then
Therefore is the crosspoint of the bisectors and
Remarkable point for pair of similar triangles
Let
Let the points and be the circumcenters of and
Let point be the midpoint of
The point is symmetric to with respect point
Prove:
a) point be the crosspoint of the bisectors and
b)
Proof
is parallelogram Denote
Similarly,
The statement that was proved in the previous section.
Remarkable point’s problems
Problem 1
Let a convex quadrilateral be given,
Let and be the midpoints of and respectively.
Circumcircles and intersect a second time at point
Prove that points and are concyclic.
Problem 2
Let triangle be given.
Let point lies on sideline
Denote the circumcircle of the as , the circumcircle of the as .
Let be the circumcenter of
Let circle cross sideline at point
Let the circumcircle of the cross at point
Prove that
Problem 3
The circles and are crossed at points and points
Let be the tangent to be the tangent to
Point is symmetric with respect to
Prove that points and are concyclic.
Problem 4
The circles and are crossed at points and points
Let be the tangent to be the tangent to
Points and lye on bisector of the angle
Points and lye on external bisector of the angle
Prove that and bisector are tangent to the circle
bisector is tangent to the circle
Solutions
Solutions are clear from diagrams.
In each case we use remarcable point as the point of bisectors crossing.
Solution 1
We use bisectors and . The points and are concyclic.
Solution 2 We use bisectors of and
Solution 3
We use bisectors of and .
is the circumcenter of
Circle is symmetric with respect diameter
Point is symmetric to with respect diameter
Therefore
Solution 4
Let
Let be midpoint be midpoint .
We need prove that and
Denote
The angle between a chord and a tangent is half the arc belonging to the chord.
is tangent to
is diameter Similarly, is diameter is tangent to
Let be the spiral similarity centered at
Points and are collinear Points and are collinear
Therefore is tangent to
Similarly, is tangent to
Japan Mathematical Olympiad Finals 2018 Q2
Given a scalene let and be points on lines and respectively, so that
Let be the circumcircle of and the reflection of across
Lines and meet again at and respectively.
Prove that and intersect on
Proof
Let be the orthocenter of
Point is symmetrical to point with respect to height
Point is symmetrical to point with respect to height
is centered at
is symmetrical with respect to heightline
is symmetrical to point with respect to height
is symmetrical to point with respect to height
The isosceles triangles
a) are concyclic.
b) is the spiral center that maps to maps to
Therefore are concyclic and are concyclic.
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Nights at the Round Table
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Asked 3 years, 4 months ago
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And there you are, the famous Round Table, seeing it in all its glory makes you smile, even though you thought it was a lot bigger. Well, no matter, as soon as you arrive, you find a parchment on the wall, describing where everyone sat, though the instructions are a bit cryptic...
(The Round Table has six chairs, each chair is neighbor to two other chairs. For the sake of this riddle, consider the table a hexagon, and the chairs its vertices.)
The northmost chair is Arthur's, as it has always been, and will always be.
Merlin is closer to Lancelot than to Gawain
Gawain sits next to two other knights
Arthur can see his three most loyal knights without turning his head
Guinevere is closer to Lancelot than to Arthur
Percival is directly across from Merlin
Gawain is at Guinevere's right and Lancelot's left
And here are the members of the Round Table:
Arthur, King of Camelot
Guinevere, Queen of Camelot
Merlin, Camelot's trusted magician
Lancelot, Gawain and Percival are all three Arthur's knights
Now, can you figure out where everyone is?
logical-deduction
lateral-thinking
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asked May 4, 2022 at 8:45
AuribourosAuribouros
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can we assume they are all facing the same direction (say towards the centre)?Sid –Sid 2022-05-04 09:02:54 +00:00 Commented May 4, 2022 at 9:02
3 Every chair is facing the center yes Auribouros –Auribouros 2022-05-04 09:06:19 +00:00 Commented May 4, 2022 at 9:06
1 Is there a language component to this question? rot13(Pbhyq lbh ercynpr "vf" jvgu "fvgf" va nyy ohg gur svefg uvag?)xyldke –xyldke 2022-05-04 11:18:36 +00:00 Commented May 4, 2022 at 11:18
1 @xyldke Lbh pnaabg, abg sbe nyy bs gurz gung vf Auribouros –Auribouros 2022-05-04 11:26:00 +00:00 Commented May 4, 2022 at 11:26
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Here's what we can deduce:
Gawain sits next to two other knights (Lancelot and Percival) and is at Lancelot's left, so Gawain has Lancelot to his right and Percival to his left. (Somewhere in the counter-clockwise arrangement there is a consecutive string of Percival, Gawain, Lancelot)
That means, Merlin is on Lancelot's left, because Merlin sits opposite Percival, which means there are 2 people (Gawain and Lancelot) between Merlin and Percival. (Somewhere in the counter-clockwise arrangement there is a consecutive string of Percival, Gawain, Lancelot, Merlin). This also fulfills Merlin being closer to Lancelot than to Gawain. You'll note that that makes it impossible to fulfill the requirement that Guinevere sits closer to Lancelot than to Arthur, we'll get to that.
Arthur can see his knights without turning his head, so he sits opposite of the middle knight, which places him two seats from Gawain and thus to Merlin's right. (There is a consecutive string of Percival, Gawain, Lancelot, Merlin, Arthur). That leaves Guinevere to sit to Arthur's right. So in the end we have the following arrangement.
This is my solution
Starting in the north and moving in a counter-clockwise direction the seating arrangement would be: Arthur, Guinevere, Percival, Gawain, Lancelot and Merlin.
However, since "Guinevere is closer to Lancelot than to Arthur", there is something sneaky happening. In fact, it is well known that Guinevere and Lancelot were involved in a love affair that eventually tore the court of Camelot apart.
Guinevere is in fact under the table in front of Lancelot. Since she is facing Lancelot, Gawain is indeed at her right and at Lancelot's left simultaneously.
I think details are best left to the reader to whom I'll also leave the decision whether such an act of betrayal would remove Lancelot from the group of Arthur's most loyal knights.
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edited May 4, 2022 at 12:29
answered May 4, 2022 at 10:14
xyldkexyldke
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2 It makes sense because it has tag "lateral thinking" which definitely means something's out of box.I'm Nobody –I'm Nobody 2022-05-04 10:20:48 +00:00 Commented May 4, 2022 at 10:20
1 Well technically, I did use all statements. I just chose a creative interpretation for one of them ;)xyldke –xyldke 2022-05-04 11:42:30 +00:00 Commented May 4, 2022 at 11:42
1 Alright, I'll give you another small hint: Gur Ebhaq Gnoyr vf uvtu rabhtu sbe fbzrbar gb svg haqre. But given your answer, if you still can't find the "intended" answer, you still would deserve the checkmark Auribouros –Auribouros 2022-05-04 12:02:56 +00:00 Commented May 4, 2022 at 12:02
3 That last instruction is positively brilliant! Thanks for the hint. Without it, I don't think I would have ever considered fhpu hapbhegyl ebznapr.xyldke –xyldke 2022-05-04 12:37:52 +00:00 Commented May 4, 2022 at 12:37
1 It's still true. Picture her under the table with Lancelot is straight ahead of her... so yeah, Gawain is to her right.dcsohl –dcsohl 2022-05-05 15:27:50 +00:00 Commented May 5, 2022 at 15:27
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The table seating is represented as a clockwise sequence of chairs, starting at the north seat:
Arthur can see his three most loyal knights without turning his head gives:
Arthur | ? | Knight | Knight | Knight | ?
Gawain sits next to two other knights puts Gawain in the center of the three knights:
Arthur | ? | Knight | Gawain | Knight | ?
Gawain is at Guinevere's right and Lancelot's left. gives:
Arthur | ? | Lancelot | Gawain | Percival | ?
Percival is directly across from Merlin fixes a relation between Percival and Merlin:
Arthur | Merlin | Lancelot | Gawain | Percival | ?
The puzzle is a little vague about whether everybody is sitting on chair. If we assume that terms like the "position of X" and "the position of X's chair" mean the same, and that formulations like "X is", "X sits" and "X's chair is" all resolve to the position of the people, then Guinevere needs to take the last free seat, giving
Arthur | Merlin | Lancelot | Gawain | Percival | Guinevere
At first glance, this seating plan seem to be contradicting
Guinevere is closer to Lancelot than to Arthur
But this contradiction is based on the implicit premise that the hexagon was equilateral. This assumption is not backed by the question text. A solution is to put a big distance between the chairs of Arthur and Guinevere and pack Guinevere, Percival, Gawain and Lancelot quite close. If one interprets "closer" to refer to the physical distance -- which is not a stretch at all -- and not to the distance in the permutation, there is no contradiction left.
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answered May 4, 2022 at 22:25
Jonathan HerreraJonathan Herrera
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The chairs belong in clockwise order to:
Arthur | Merlin | Lancelot | Gawain | Percival | Guinevere
But:
everony is seated, except Arthur, who stands a bit behind his chair, so that he can see his knights but Guinevere is closer to Lancelot than to Arthur (e.g. his distance greater than the tables diameter).
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answered May 5, 2022 at 12:21
Matthias BäßlerMatthias Bäßler
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8568 | https://www.youtube.com/watch?v=yPwI0hw6d3Y | P5 Whole Numbers - Using Model To Solve Internal Transfer Word Problems
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Transcript:
okay hi everybody today we are doing a question like this now as you can see over here it says that John had 52 sorry 250 more stems than Harry so we are just going to show John and Harry we're going to show that John is everything from here to here okay you want to make sure that your model is drawn big all right you want to make sure that your model is drawn big so that it's clearer for you okay we know that this much over here is 250 okay so we are just going to do it this way and now sentence number two after John gave 57 the word gave means you minus right so I'm going to take away from John gave to who is it gave to somebody within the circle yes it is given to who who is this person Harry so I'm just going to add over here I'm going to add this and I'm going to say that this is 57 this will be transferred here when I see that this is 57 in the mod bar drawing you must always make sure you mirror image so that's the reason why there's a 57 over there now of course then over here we can find this question mark true right so you see if my motor is drawn big it's easier for me okay it's way easier for me all right so what I'm going to do over here I'm just going to do 250 minus away 57 minus away 57 when I do that I will get to an answer of 136 136 over here am I making sense here all right so we know that 36 sentence number three John still had three times three times basically means how many units three units who had three units John so John had three units so we are just going to go all the way you must understand that it stops here because this is already gone agree so far we say that this is three units can you see that if John is two units then Harry how many unit Harry has to be one can you see that so from here right I will just draw a line down I'll show I will draw a line down because I want to show you where the one unit ends can you see where the one unit ends the one unit is everything from the first yellow line to all the way to the last yellow line cor not so therefore what I have over here has to be how many units yes this is two units am I making sense so from here itself you continue you say that two units is actually equals to 136 so we always want to make sure we have Mr Greenman all right we will simply do 136 divided by two that gives me 68 the moment you get to one unit you want to read the question the question say how many stems do each of them have at first you must remember that this is internal transfer is that okay all right so let's see for John at first John will be how many units we will use three units because from here all the way to here and then later we will add 57 because I want John at first am I making sense all right so we say John at first will be simply just 68 3+ 57 all right 68 3 plus 57 all right we get to an answer of 261 for John okay now and then of course for this person called uh Harry okay we want to find what is Harry at first okay so if I know what is Harry at first what I want to do is obviously I want to write down Harry at first you can see Harry at first is from where to where it's one unit right so it's the whole thing so but I want at first only so I have to remove away the 57 okay why is it I'm removing away the 57 because I want to remember that we are finding the F first that means before receiving the 57 and this explains why Harry atas is 11 John atas is 2 six [Music] one |
8569 | https://math.stackexchange.com/questions/1412386/proof-of-conformal-mappings-onto-polygons-stein | complex analysis - Proof of conformal mappings onto polygons (Stein) - Mathematics Stack Exchange
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Proof of conformal mappings onto polygons (Stein)
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Recently I am reading Stein and Shakarchi's Complex Analysis and I find a great difficulty in understanding the proof of theorem 4.6 in Chapter 8 (p.242-244), which talks about conformal mappings onto polygons.
It first introduces the function h k h k and shows that it can be analytically continuous to an infinite strip. I understand that why h′k≠0 h k′≠0 but I don't know why that
F′′(z)F′(z)=−β k z−A k+E K(z)F″(z)F′(z)=−β k z−A k+E K(z)
I don't know why this expression is true and also don't understand the meaning of F′′(z)F′(z)F″(z)F′(z) for Im(z)<0 Im(z)<0, since only h k h k is defined for Im z<0 Im z<0 but not F F (if I define F=h α k k+a k F=h k α k+a k, will it be the same F F as the one I will discuss in the next paragraph?). Here I assume
h k(z)=e i θ h k(z¯¯¯)¯¯¯¯¯¯¯¯¯¯¯for Im(z)>0.h k(z)=e i θ h k(z¯)¯for Im(z)>0.
Secondly, in the next page (p.244), the writer is going to proof that F F is holomorphic outside a large circle so is at the infinity. I don't really understand what he means, only have an idea: since a polygon is bounded, any Im z<0 Im z<0 can correspond to a z¯z¯ which F(z)F(z) can be defined by reflecting F(z¯)F(z¯) along an edge of the polygon. I think I miss something because according to this, F F can be extend to become an entire function, which seems that I am wrong.
Can anyone please help me understand the proof of this theorem?
complex-analysis
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edited Aug 28, 2015 at 23:17
Y.H. ChanY.H. Chan
asked Aug 28, 2015 at 8:40
Y.H. ChanY.H. Chan
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I am also reading this book and here is my understanding about the first question:
The continuation of a homorphic function is unique in the sense that two holomorphic funtion corresponding on some non-empty open subset are the same function.
Here is what Stein said in Chapter 2:
Corollary 4.9 Suppose f f and g g are holomorphic in a region Ω Ω and f(z)=g(z)f(z)=g(z) for all z z in some non-empty open subset on Ω Ω (or more generally for z z in some sequence of distinct points with limit point in Ω Ω.) Then f(z)=g(z)f(z)=g(z) throughout Ω Ω.
Suppose we are given a pair of functions f f and F F analytic in regions Ω Ω and Ω′Ω′, respectively, with Ω⊂Ω′Ω⊂Ω′. If the two functions agree on the smaller set Ω Ω, we say that F F is an analytic continuation of f f into the region Ω′Ω′, The corollary then guarantees that there can be only one such analytic continuation, since F F is uniquely determined by f f.
We define F F for I m(z)<0 I m(z)<0 by F=h α k k+a k F=h k α k+a k and then we can get the unique continuation of F F on {z∣A k−1<R e(z)<A k+1}{z∣A k−1<R e(z)<A k+1}.
However I am also confused about the second assertion that
F F is continuable in the exterior of a disc |z|≤R|z|≤R, for large R R.
I have no idea what's the connection with reflection principle. Should we find the continuation of F F on the disc first? If so, we need to verify the continuation we made above is holomorphic on the line {z∣R e(z)=A k,1≤k≤n}{z∣R e(z)=A k,1≤k≤n}. If that's possible, we have already get an entire function and there is no need to make furthur continuation.
Hope more answers would come to help.
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answered May 8, 2023 at 11:21
Florian HuoFlorian Huo
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I'm not sure the extension you defined is valid. The image of the extension of h k h k has complex numbers with all possible arguments (i.e. "all possible directions are exhausted"). Thus, we would need to use different branches of the logarithm to define (h k(z))α k(h k(z))α k for the regions Im(z)≥0 Im(z)≥0 and Im(z)<0 Im(z)<0.Karthik Kannan –Karthik Kannan 2024-04-10 14:30:03 +00:00 Commented Apr 10, 2024 at 14:30
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After meditating on this for a long time, I believe that Stein and Shakarchi are applying the Schwarz reflection principle to the restriction of F F to the region H∖{z:|z|≤R}H∖{z:|z|≤R} where R R is chosen to be sufficiently large (say R>max 1≤k≤n|A k|R>max 1≤k≤n|A k|). In fact, F F can be extended to C∖{A 1,…,A n}C∖{A 1,…,A n}. This can be done by applying the reflection principle to (say) regions of the form
{z:π/4<arg(z−A k+1)≤π/2}∪{z:A k<Re(z)<A k+1}∪{z:π/2≤arg(z−A k)<3 π/4}.{z:π/4<arg(z−A k+1)≤π/2}∪{z:A k<Re(z)<A k+1}∪{z:π/2≤arg(z−A k)<3 π/4}.
As noted in the other answer the extensions of F F on the overlap of these regions should agree because of the nature of zeros of a holomorphic function.
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answered Apr 10, 2024 at 14:17
Karthik KannanKarthik Kannan
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8570 | https://en.wikipedia.org/wiki/Sinc_function | Jump to content
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Contents
(Top)
1 Definitions
2 Etymology
3 Properties
4 Relationship to the Dirac delta distribution
5 Summation
6 Series expansion
7 Higher dimensions
8 Sinhc
9 See also
10 References
11 Further reading
12 External links
Sinc function
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From Wikipedia, the free encyclopedia
Special mathematical function defined as sin(x)/x
"Sinc" redirects here; not to be confused with Site of Importance for Nature Conservation.
| Sinc |
| Part of the normalized sinc (blue) and unnormalized sinc function (red) shown on the same scale |
| General information |
| General definition | |
| Fields of application | Signal processing, spectroscopy |
| Domain, codomain and image |
| Domain | |
| Image | |
| Basic features |
| Parity | Even |
| Specific values |
| At zero | 1 |
| Value at +∞ | 0 |
| Value at −∞ | 0 |
| Maxima | 1 at |
| Minima | at |
| Specific features |
| Root | |
| Related functions |
| Reciprocal | |
| Derivative | |
| Antiderivative | |
| Series definition |
| Taylor series | |
In mathematics, physics and engineering, the sinc function (/ˈsɪŋk/ SINK), denoted by sinc(x), is defined as either or
the latter of which is sometimes referred to as the normalized sinc function. The only difference between the two definitions is in the scaling of the independent variable (the x axis) by a factor of π. In both cases, the value of the function at the removable singularity at zero is understood to be the limit value 1. The sinc function is then analytic everywhere and hence an entire function.
The normalized sinc function is the Fourier transform of the rectangular function with no scaling. It is used in the concept of reconstructing a continuous bandlimited signal from uniformly spaced samples of that signal. The sinc filter is used in signal processing.
The function itself was first mathematically derived in this form by Lord Rayleigh in his expression (Rayleigh's formula) for the zeroth-order spherical Bessel function of the first kind.
Definitions
[edit]
The sinc function has two forms, normalized and unnormalized.
In mathematics, the historical unnormalized sinc function is defined for x ≠ 0 by
Alternatively, the unnormalized sinc function is often called the sampling function, indicated as Sa(x).
In digital signal processing and information theory, the normalized sinc function is commonly defined for x ≠ 0 by
In either case, the value at x = 0 is defined to be the limiting value for all real a ≠ 0 (the limit can be proven using the squeeze theorem).
The normalization causes the definite integral of the function over the real numbers to equal 1 (whereas the same integral of the unnormalized sinc function has a value of π). As a further useful property, the zeros of the normalized sinc function are the nonzero integer values of x.
Etymology
[edit]
The function has also been called the cardinal sine or sine cardinal function. The term "sinc" is a contraction of the function's full Latin name, the sinus cardinalis and was introduced by Philip M. Woodward and I.L Davies in their 1952 article "Information theory and inverse probability in telecommunication", saying "This function occurs so often in Fourier analysis and its applications that it does seem to merit some notation of its own". It is also used in Woodward's 1953 book Probability and Information Theory, with Applications to Radar.
Properties
[edit]
The zero crossings of the unnormalized sinc are at non-zero integer multiples of π, while zero crossings of the normalized sinc occur at non-zero integers.
The local maxima and minima of the unnormalized sinc correspond to its intersections with the cosine function. That is, sin(ξ)/ξ = cos(ξ) for all points ξ where the derivative of sin(x)/x is zero and thus a local extremum is reached. This follows from the derivative of the sinc function:
The first few terms of the infinite series for the x coordinate of the n-th extremum with positive x coordinate are [citation needed] where and where odd n lead to a local minimum, and even n to a local maximum. Because of symmetry around the y axis, there exist extrema with x coordinates −xn. In addition, there is an absolute maximum at ξ0 = (0, 1).
The normalized sinc function has a simple representation as the infinite product:
and is related to the gamma function Γ(x) through Euler's reflection formula:
Euler discovered that and because of the product-to-sum identity
Euler's product can be recast as a sum
The continuous Fourier transform of the normalized sinc (to ordinary frequency) is rect(f): where the rectangular function is 1 for argument between −1/2 and 1/2, and zero otherwise. This corresponds to the fact that the sinc filter is the ideal (brick-wall, meaning rectangular frequency response) low-pass filter.
This Fourier integral, including the special case is an improper integral (see Dirichlet integral) and not a convergent Lebesgue integral, as
The normalized sinc function has properties that make it ideal in relationship to interpolation of sampled bandlimited functions:
It is an interpolating function, i.e., sinc(0) = 1, and sinc(k) = 0 for nonzero integer k.
The functions xk(t) = sinc(t − k) (k integer) form an orthonormal basis for bandlimited functions in the function space L2(R), with highest angular frequency ωH = π (that is, highest cycle frequency fH = 1/2).
Other properties of the two sinc functions include:
The unnormalized sinc is the zeroth-order spherical Bessel function of the first kind, j0(x). The normalized sinc is j0(πx).
where Si(x) is the sine integral,
λ sinc(λx) (not normalized) is one of two linearly independent solutions to the linear ordinary differential equation The other is cos(λx)/x, which is not bounded at x = 0, unlike its sinc function counterpart.
Using normalized sinc,
The following improper integral involves the (not normalized) sinc function:
Relationship to the Dirac delta distribution
[edit]
The normalized sinc function can be used as a nascent delta function, meaning that the following weak limit holds:
This is not an ordinary limit, since the left side does not converge. Rather, it means that
for every Schwartz function, as can be seen from the Fourier inversion theorem. In the above expression, as a → 0, the number of oscillations per unit length of the sinc function approaches infinity. Nevertheless, the expression always oscillates inside an envelope of ±1/πx, regardless of the value of a.
This complicates the informal picture of δ(x) as being zero for all x except at the point x = 0, and illustrates the problem of thinking of the delta function as a function rather than as a distribution. A similar situation is found in the Gibbs phenomenon.
We can also make an immediate connection with the standard Dirac representation of by writing and
which makes clear the recovery of the delta as an infinite bandwidth limit of the integral.
Summation
[edit]
All sums in this section refer to the unnormalized sinc function.
The sum of sinc(n) over integer n from 1 to ∞ equals π − 1/2:
The sum of the squares also equals π − 1/2:
When the signs of the addends alternate and begin with +, the sum equals 1/2:
The alternating sums of the squares and cubes also equal 1/2:
Series expansion
[edit]
The Taylor series of the unnormalized sinc function can be obtained from that of the sine (which also yields its value of 1 at x = 0):
The series converges for all x. The normalized version follows easily:
Euler famously compared this series to the expansion of the infinite product form to solve the Basel problem.
Higher dimensions
[edit]
The product of 1-D sinc functions readily provides a multivariate sinc function for the square Cartesian grid (lattice): sincC(x, y) = sinc(x) sinc(y), whose Fourier transform is the indicator function of a square in the frequency space (i.e., the brick wall defined in 2-D space). The sinc function for a non-Cartesian lattice (e.g., hexagonal lattice) is a function whose Fourier transform is the indicator function of the Brillouin zone of that lattice. For example, the sinc function for the hexagonal lattice is a function whose Fourier transform is the indicator function of the unit hexagon in the frequency space. For a non-Cartesian lattice this function can not be obtained by a simple tensor product. However, the explicit formula for the sinc function for the hexagonal, body-centered cubic, face-centered cubic and other higher-dimensional lattices can be explicitly derived using the geometric properties of Brillouin zones and their connection to zonotopes.
For example, a hexagonal lattice can be generated by the (integer) linear span of the vectors
Denoting one can derive the sinc function for this hexagonal lattice as
This construction can be used to design Lanczos window for general multidimensional lattices.
Sinhc
[edit]
Some authors, by analogy, define the hyperbolic sine cardinal function.
See also
[edit]
Anti-aliasing filter – Mathematical transformation reducing the damage caused by aliasing
Borwein integral – Type of mathematical integrals
Dirichlet integral – Integral of sin(x)/x from 0 to infinity
Lanczos resampling – Technique in signal processing
List of mathematical functions
Shannon wavelet
Sinc filter – Ideal low-pass filter or averaging filter
Sinc numerical methods
Trigonometric functions of matrices – Important functions in solving differential equations
Trigonometric integral – Special function defined by an integral
Whittaker–Shannon interpolation formula – Signal (re-)construction algorithm
Winkel tripel projection – Pseudoazimuthal compromise map projection (cartography)
References
[edit]
^ Olver, Frank W. J.; Lozier, Daniel M.; Boisvert, Ronald F.; Clark, Charles W., eds. (2010), "Numerical methods", NIST Handbook of Mathematical Functions, Cambridge University Press, ISBN 978-0-521-19225-5, MR 2723248..
^ Singh, R. P.; Sapre, S. D. (2008). Communication Systems, 2E (illustrated ed.). Tata McGraw-Hill Education. p. 15. ISBN 978-0-07-063454-1. Extract of page 15
^ Weisstein, Eric W. "Sinc Function". mathworld.wolfram.com. Retrieved 2023-06-07.
^ Merca, Mircea (2016-03-01). "The cardinal sine function and the Chebyshev–Stirling numbers". Journal of Number Theory. 160: 19–31. doi:10.1016/j.jnt.2015.08.018. ISSN 0022-314X. S2CID 124388262.
^ a b Poynton, Charles A. (2003). Digital video and HDTV. Morgan Kaufmann Publishers. p. 147. ISBN 978-1-55860-792-7.
^ Woodward, P. M.; Davies, I. L. (March 1952). "Information theory and inverse probability in telecommunication" (PDF). Proceedings of the IEE - Part III: Radio and Communication Engineering. 99 (58): 37–44. doi:10.1049/pi-3.1952.0011.
^ Woodward, Phillip M. (1953). Probability and information theory, with applications to radar. London: Pergamon Press. p. 29. ISBN 978-0-89006-103-9. OCLC 488749777. {{cite book}}: ISBN / Date incompatibility (help)
^ Euler, Leonhard (1735). "On the sums of series of reciprocals". arXiv:math/0506415.
^ Sanjar M. Abrarov; Brendan M. Quine (2015). "Sampling by incomplete cosine expansion of the sinc function: Application to the Voigt/complex error function". Appl. Math. Comput. 258: 425–435. arXiv:1407.0533. doi:10.1016/j.amc.2015.01.072.
^ "Advanced Problem 6241". American Mathematical Monthly. 87 (6). Washington, DC: Mathematical Association of America: 496–498. June–July 1980. doi:10.1080/00029890.1980.11995075.
^ Robert Baillie; David Borwein; Jonathan M. Borwein (December 2008). "Surprising Sinc Sums and Integrals". American Mathematical Monthly. 115 (10): 888–901. doi:10.1080/00029890.2008.11920606. hdl:1959.13/940062. JSTOR 27642636. S2CID 496934.
^ Baillie, Robert (2008). "Fun with Fourier series". arXiv:0806.0150v2 [math.CA].
^ a b c Ye, W.; Entezari, A. (June 2012). "A Geometric Construction of Multivariate Sinc Functions". IEEE Transactions on Image Processing. 21 (6): 2969–2979. Bibcode:2012ITIP...21.2969Y. doi:10.1109/TIP.2011.2162421. PMID 21775264. S2CID 15313688.
^ Ainslie, Michael (2010). Principles of Sonar Performance Modelling. Springer. p. 636. ISBN 9783540876625.
^ Günter, Peter (2012). Nonlinear Optical Effects and Materials. Springer. p. 258. ISBN 9783540497134.
^ Schächter, Levi (2013). Beam-Wave Interaction in Periodic and Quasi-Periodic Structures. Springer. p. 241. ISBN 9783662033982.
Further reading
[edit]
Stenger, Frank (1993). Numerical Methods Based on Sinc and Analytic Functions. Springer Series on Computational Mathematics. Vol. 20. Springer-Verlag New York, Inc. doi:10.1007/978-1-4612-2706-9. ISBN 9781461276371.
External links
[edit]
Weisstein, Eric W. "Sinc Function". MathWorld.
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Rencontres numbers
Known as:Problème des rencontres, Rencontres number, Partial derangement Expand
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Concept on two-force member
I have some homework question that i require some help understanding.
1) For the first question highlighted in red (referring to diagram viii), I know for sure that reactions at the supports is dependent on the length a because it determines the magnitude of moment it creates about the support and that it is not dependent on H because it is in the direction of the force(weight P) itself.
However, what I am uncertain about is the cable CE. Is this considered a two-force member in which tension at both ends will cancel each other out, hence varying the length L will not cause the reaction forces at A and B to change?
2) For internal loadings, I am guessing that it will depend on the length L, because for internal loading, we are viewing the forces at the particular point and hence varying the angle of elevation will change the Rx and Ry component of the forces that the point and that it does not cancel out.
And also, is it true that for a two-force member that is in equilibirum, the net translational forces and rotational moment will be 0 throughout the entire segment?
Thanks in advance.
1 Answer 1
I don't want to take away the opportunity to work out the exact details for yourself, but I'll talk about the thought process I would use to approach these questions, and hopefully that will help clarify things.
Firstly, while I understand that the problem statement says to complete (a) before (b), I think at the point that you're getting stuck, it makes sense to sketch a few free body diagrams. It's part of the learning process, and as you do more and more of them you'll start developing a sort of intuition. My tendency is to run a couple quick 'thought experiments' either in my head or on paper, investigating two extremes. That's some of what I'll be doing below. Bear in mind that there are often multiple valid approaches to arrive at a solution - this is just one possibility.
Idea 1: Dependency of external reactions on structure geometry
1.1 Do external reactions depend on length 'a'?
Perhaps the quickest mental check of this one is to think about...what if 'a' were suddenly set to '2a'...would it change the support reactions? Given that the applied load is unchanged, we certainly can't say the support reactions would also be multiplied by 2. This suggests the support reactions aren't dependent on the numerical value of 'a' but on something else.
Here's a thought experiment that will lead us toward what's really going on. Structure viii is a simply supported beam, so let's first imagine a more straightforward situation. What if we had a simply supported beam with single vertical point load at midspan.
Do we even need to know the quantitative beam length to determine the support reactions? No, we don't. We simply need to know the load is applied at midspan. That is, we need to know the ratio of the beam lengths to the left and right of the load. This is enough to tell us that half the applied load will go to each support. The same principle holds true for instances where the vertical load is offset from midspan. The length ratio (not the lengths themselves) is what determines the simply supported beam reactions. You'll see this when you complete part (b) of the assignment and go through the exercise of solving for the support reactions.
1.2 Do external reactions depend on the vertical position 'H' of the applied load?
You're on the right track with this one. Mathematically I guess it relates to how we can slide a force along its line of action. My first thought is usually the physical intuition that whether I'm holding, say, a can of paint on a 6" string or a 12" string, I've got to support the same vertical load from the can of paint.
1.3 Do external reactions depend on the height 'L' of the vertical member?
At first glance, this one seems like a harder question, but the important thing to remember is that with support reactions we're investigating external equilibrium. For this simply supported beam, we don't need to know what's happening inside the structure to find the external reactions. It's like how for a statically determinate truss, we can find the external reactions without investigating individual truss member forces or worrying about the specifics of truss geometry. So, for structure viii, we can solve for our support reactions using our equations of equilibrium without knowing 'L' at all. As with idea 1.1, this will probably become more apparent when you work through part (b) of the assignment.
Idea 2: Dependency of internal loading on height 'L' of vertical member
Now we're getting into internal equilibrium, so you're correct that the cable geometry matters. If we assume that the cable is continuous and that at the point where the cable connects to the beam, the cable is passing around some sort of frictionless pin - then we can say that the tension, T, at every point in the cable must be the same (and equal to the applied load). Changing height L of the vertical member change the angle of the cable. As you noted, when we hold T constant and vary the cable angle, the x- and y- components of T vary. This means the internal forces in beam segment CD and vertical member DE are also changing.
Idea 3: Two-force members and equilibrium
For any member in static equilibrium (two-force or otherwise) the equations of statics will be satisfied at every point in the structure (including sections cut through a member). That is, $\Sigma F_x = 0$, $\Sigma F_y = 0$, and $\Sigma M = 0$. Satisfying these equations is one way we can determine internal forces (shear, axial, moment) at points along a member.
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Determinant of 3×3 Matrix Exercises
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Determinant of 3×3 Matrix Practice Problems with Answers
Use the following ten (10) practice problems to get better at solving the Determinant of a 3×3 Matrix. With practice, you’ll get better at it until it becomes second nature.
You may use these problems for independent practice, homework, cooperative learning, and more!
Problem 1: Find the determinant of the 3×3 matrix below.
732−4122−5−5
Answer
Problem 2: Find the determinant of the 3×3 matrix below.
11−1−6−4−3−77−6
Answer
Problem 3: Find the determinant of the 3×3 matrix below.
−14−1−15−6−1−33
Answer
Problem 4: Calculate the determinant of the 3×3 matrix below.
7665−527−53
Answer
Problem 5: Calculate the determinant of the 3×3 matrix below.
4−42−6−774−53
Answer
Problem 6: Calculate the determinant of the 3×3 matrix below.
−1−1−3−32−7462
Answer
Problem 7: Determine the determinant of the 3×3 matrix below.
21072055−4
Answer
Problem 8: Determine the determinant of the 3×3 matrix below.
1−54−2−1−1250
Answer
Problem 9: Compute the determinant of the 3×3 matrix below.
−5−252−1−5−3−5−3
Answer
Problem 10: Compute the determinant of the 3×3 matrix below.
−41−41−34220
Answer
You might also like these tutorials:
Determinant of a 3×3 Matrix
Determinant of a 2×2 Matrix
Determinant of 2×2 Matrix Problems with Answers
Tags:
Advanced Algebra,
Lessons
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作文冲刺热点 …………………………………… 1
热点一 科技与人文的辩证关系 ……………… 6
热点二 传统文化的创新性传承 ……………… 16
热点三 个体与时代的命运共振 ……………… 22
热点四 生态文明与可持续发展 ……………… 30
热点五 社会公平与正义多元维度 …………… 39
保分小题 语言运用和名句默写 ……………… 52
增分大题 现代文阅读 ………………………… 77
信息类阅读 解题方法 ……………………… 102
文学作品阅读 答题 方法 …………………… 115
古诗文阅读答题技巧 ……………………… 122
语言文字运用技巧 ………………………… 130
考场作文高分技巧 ………………………… 140
高考获胜第一步：培养健康心理 ………… 146
冲刺复习备考指导 …………………………… 147
高考只是人生一站 …………………………… 149
聚焦可控因素：志愿填报 ………………… 150
送给考生的高考小贴士 …………………… 152
终极考问：遇上不会做的题怎么办？ …154
终极押题卷（一） ………………………… 155
终极押题卷（二） ………………………… 175
写给即将参加高考的学子 …………………… 196 学科 网（北 京）股 份有限 公司
1 -
【热点回顾 篇】
☞高考作文冲刺热点
近三年高考作文考查 方向和热 点
近三年高考作文考题呈现出鲜明的时代特征与教育导向，既注重对传统文化的传承，又强调对现实问
题的思辨，同时引导考生树立正确的价值观。总体来说，近五年的高考作文主要呈现以下趋势：
形式演变 ：情境任务型 （2020 ）→价值材料型 （2021 ）→多元关系型 （2022 ）→极简思辨型 （2023 —
2024 ）
能力进阶 ：应用写作（ 2020 ）→价值判断（ 2021 ）→系统思维（ 2022 ）→哲学思辨（ 2023 — 2024 ）
政策映射： 与《中国高考评价体系》提出的 “基础性 —综合性 —应用性 —创新性 ”考查要求完全同步
教育部《中国高考评价体系》明确提出 “服务选才、引导教学 ”的核心功能，这意味着作文命题将更
紧密地对接国家战略需求与社会现实关切。从近年 “AI 换脸技术争议 ”到“元宇宙文化重构 ”，从 “乡村
振兴中的青年返乡 ”到“生物科技的伦理边界 ”，这些鲜活的社会议题不仅是命题素材的重要来源，更是
检验考生批判性思维、逻辑建构能力与文化底蕴的试金石。高考作文紧扣社会热点， 迎合政策导向，凸显
时代脉搏的，体现时代精神 。近年的高考作文 命题来源于实际生活，涵盖政治、经济、民生、科技、文化
等众多社会热点话题。
2024 年作文 ：延续极简风 ，强化概念思辨 。教育部教育考试院命题专家表示 ，2024 年高考作文题不在
审题立意上设置门槛，让不同的学生结合生活体验或社会议题都能找到写作切入点，引导学生深入思考、
个性写作，助力创新人才培养。聚焦 2024 年高考作文题，我们不难发现其中几个突出的命题特点：
其一，继续强化 “立德树人 ”导向；
其二，呵护并引导青少年全面成长；
其三，注重思辨性和创新性思维品质的培养；
其四，激发学生崇尚科学、探索未知的兴趣；
其五，引导学生关注时代的脉动和社会热点；
其六，激励学生增强解决问题的意识和能力。
高考作文与整个高考命题思路一致 ，永远走在稳定中有变化 、传承中有创新的道路上 。2025 年的高考
作文一定会有 2024 年高考作文的 “影子 ”。
2025 年可能聚焦以下话题：气候变化与可持续发展 方面 如碳达峰、绿色生活方式等 ；人工智能与伦理
方面 如 AI 对就业 、隐私的影响 ；文化传承与创新 方面 如传统节日如何焕发新活力 ；青年责任与家国情怀 方学科 网（北 京）股 份有限 公司
2 -
面如“炫家乡 ”现象背后的文化自信。高考作文题目通常要求结合具体案例，体现思辨性。命题可能会继
续鼓励学生从多角度分析问题，提出创新解决方案。例如，科技与人文的关系、传统与现代的融合、个人
与集体的平衡等。 今年从 教育部的政策导向 看，“双减 ”政策之后教育评价体系的改革，在作文题中可能
会体现对学生综合素养的考察，强调实践能力和社会责任感。
（一）传统文化与现代精神的深度融合
1。特点：以经典文本或文化符号为载体，要求考生在古今对话中提炼普世价值。
案例：
2022 年全国甲卷：以《红楼梦》 “大观园试才题对额 ”为素材，探讨 “移用 ”“ 化用 ”“ 独创 ”的文
化传承逻辑。考生需从匾额题名的艺术选择，延伸至科技创新、文化自信等现代议题。
2024 年北京卷：议论文题目 “历久弥新 ”，要求考生结合经典常读、思想常新的现象，论证传统文化
在当代的生命力。如引用《论语》 “学而时习之 ”与现代终身学习理念的关联。
命题意图：通过文化符号的具象化，考查考生对 “创造性转化、创新性发展 ”的理解，避免空泛的
文化口号。
（二）科技与人文的思辨性对话
特点：聚焦技术发展对人类社会的双重影响，要求考生在利弊权衡中展现批判性思维。
案例：
2023 年全国甲卷 ：以“技术发展与时间掌控 ”为核心 ，探讨 “成为时间仆人 ”的悖论 。考生需分析 AI
工具如何提升效率，又可能导致注意力碎片化。
2024 年新课标 I 卷：针对 “人工智能能否减少人类问题 ”展开辩论，需辩证思考技术解决表层问题与
引发深层伦理困境的关系。
命题意图：呼应 “科技向善 ”的社会共识，引导考生超越技术乐观主义，关注人文关怀。
（三）全球化视野下的责任担当
特点：以国际关系或人类命运为切入点，强调个体与集体、国家与世界的关联。
案例：
2023 年全国乙卷：引用习近平总书记 “百花齐放春满园 ”的论述，要求考生以 “一花独放不是春 ”为
主题，论证合作共赢的必要性。可结合 “一带一路 ”倡议或全球气候治理案例。
2024 年天津卷：以 “被定义 ”与“自定义 ”为矛盾点，探讨个人、群体或国家如何在全球化中保持主
体性。如华为突破技术封锁的 “自定义 ”实践。
命题意图：培养 “人类命运共同体 ”意识，避免狭隘民族主义。 学科 网（北 京）股 份有限 公司
3 -
（四）成长叙事与价值引领
特点：以青少年成长为视角，通过具体情境引导考生反思自我与社会的关系。
案例：
2022 年全国新高考 Ⅱ卷：以共青团成立百年为背景，要求考生结合科学家、摄影家、建筑家的职业选
择，论述 “选择 ·创造 ·未来 ”的关系。可联系 “00 后”返乡创业、科研报国等现象。
2024 年全国甲卷：以 “坦诚交流 ”为核心，探讨人际冲突中的沟通策略。如校园合作项目中如何化解
意见分歧。
命题意图：通过贴近生活的议题，强化 “立德树人 ”的教育目标。
（五）材料类型的多元化与结构化
特点：材料形式从单一文本向多模态、跨学科延伸，考查信息整合能力。
案例：
2023 年新课标 II 卷：结合语言文字运用题中的 “安静空间 ”讨论，要求考生分析青少年的独处需求 。
需整合材料中的心理学观点与自身成长经验。
2024 年新课标 II 卷：关联现代文阅读中 “嫦娥四号 ”“天问一号 ”的航天探索 ，要求考生以 “抵达未
知之境 ”为主题，将科技突破与个人成长类比。
命题意图：模拟真实情境中的问题解决，打破学科壁垒。
（六）地方卷的地域特色与文化标识
特点：自主命题省份注重挖掘本土文化资源，体现地域文化自信。
案例：
2022 年浙江卷：以 “浙江青年 ”为素材，列举徐枫灿、杨杰等典型人物，要求考生论述新时代浙江精
神。需结合 “干在实处、走在前列 ”的地域文化。
2024 年上海卷：以 “认可度 ”为话题，探讨个人价值与社会评价的关系。可联系上海 “海纳百川 ”的
城市精神与职场竞争现实。
命题意图：强化地方文化认同，同时避免地域偏见。
聚焦时代：考前热点速记
主题一：贸易战热点
从“木受绳则直，金就砺则利 ”，到 “宝剑锋从磨砺出，梅花香自苦寒来 ”，再到 “世上无难事，只
要肯登攀 ”，中国人的基因里，从不惧怕任何风险挑战、困难矛盾，更能将各种外来压力视作推动自身进
步的动力。 学科 网（北 京）股 份有限 公司
4 -
从山重水复到柳暗花明，路是从荆棘丛生处开辟出来的。只要方向对了，就不怕山高路远。变压力为
动能，化挑战为机遇，制造业苦练内功，服务业提升质效，带动外贸结构实现蝶变，我们不仅能实现自身
的稳定发展，还将为全球经济发展注入更多稳定性。
对世界而言，中国 “杭州六小龙 ”现象无疑也是一个启示：人类真正需要的，并非目光短浅的投机与
刺激 ，而是志存高远的坚守与跋涉 ；人类的未来 ，不在闭门造车 、“小院高墙 ”的孤立中 ，而在敢为人先 、
开放共赢的格局中。
主题二：科技热点
进入 2025 年，中国创新成果井喷式涌现 ：神舟二十号载人飞船飞天在即 ，中国航天持续标注中国的新
高度 ；CR450 动车组样车试验时速 450 公里 ，蓄力刷新中国的新速度 ；DeepSeek “刷屏 ”全球 ，宇树机器人
火爆出圈 …… 新质生产力深刻改变着中国的经济发展。
“梦想 ”号大洋钻探船建成入列 ，探寻海洋深处的秘密 ；量子计算机信号闪烁 ，激荡算力觉醒的潮声 ；
人形机器人破圈迭代，孕育未来产业的雏形 …… 时间之河奔腾不息，创新之翼自由飞翔。中华大地上，处
处涌动着活跃跃的创造，书写着日新月异的进步。
嫦娥六号首次月背采样 ，“中国天眼 ”极目苍穹 ，“雪龙 2”号破冰前行 ，梦想号探秘大洋 ，全球最快
高铁列车 CR450 动车组横空出世，深中通道踏浪海天 …… 近日，外交部发布的一段名为 “未来已来 ”的高
燃视频，展示一系列中国创新成果，引发广泛共鸣。逐梦星辰大海，奋斗开创未来，新时代中国在高质量
发展之路上稳健前行。
主题三：科技与文化融合
《哪吒 2》以及之前爆火的游戏 《黑神话 ：悟空 》都是将中国传统神话故事以全新的叙事方式 、国际化
的美术风格呈现给观众。通过对经典形象的现代化塑造，使得中国文化受到全球观众和玩家的追捧。路透
社称，《哪吒 2》的爆火，印证了中国本土 IP 的强大号召力。人们看到，《哪吒 2》和《黑神话：悟空》
作为具有国际竞争力的作品，展现了中国文化的独特魅力，它们的成功让更多创作者意识到，中国文化不
仅可以满足国内市场需求，还可以被全球共享。
今年开春，生机勃勃。 DeepSeek （深度求索） “横空出世 ”，人形机器人春晚刷屏，科技创新的种子
破土成林 ；春节假期 ，国内出游 5.01 亿人次 ，入境游客同比增长 150% ，“流动的中国 ”活力四射 ；《哪吒
2》票房冲进世界影史前七，文化自信的花朵迎春怒放 ……
科技，始终稳居 “第一生产力 ”的王座。 DeepSeek 在世界一鸣惊人，机器人跳秧歌惊艳春晚，亚冬会
实现 100% 绿电供应，特效加持的电影《哪吒 2》票房屡创新高， “黑科技 ”赋能非遗更显中国范 …… 今年
开春以来，一个个 “现象级 ”的新科技、新场景、新气象纷纷涌现，这些生动案例彰显着我国科技创新持 学科 网（北 京）股 份有限 公司
5 -
续深化，更展现着中国经济社会发展向阳而兴的蓬勃生机。
主题四：开放与交流
一位是来自美国的 “甲亢哥 ”，他的环球直播之旅，被称作 “数字时代的马可 ·波罗游记 ”。透过他
的真实镜头，全球观众看见一个 “实况中国 ”，看到 “保镖要失业 ”的良好治安、功夫变脸等炫酷文化、
无人机送外卖的科技感，也在他与广场舞阿姨共舞、和夜市摊贩讨价还价、被路人用英文进行 “中国式文
明劝导 ”中，感受到中国的友好、活力与秩序 ……
从网上互晒合影的中美家庭在上海 “奔现 ”，到美国小伙保保熊 、加拿大博主瑞安成为中国 “自来水 ”，
再到为了证明自己没说谎 ，带邻居来中国旅游的澳大利亚 “老外 ”…… 中国免签过境政策不断放宽 ，“China
Travel ”成为潮流 ，越来越多外国人来到中国 、认识中国 、爱上中国 。摆脱了灰暗滤镜 ，穿透了偏见误读 ，
跨越了文化隔阂，一个真实的中国被重新发现，展现出动人心魄的魅力。
主题五：梦想与奋斗
梦想之翼，因奋斗而坚韧。 “尽管遭遇挑战，我们从没想过放弃，只会越战越勇 ”，这是民营企业家
李积回的决心； “不怕风高浪急，闯过去才有未来 ”，这是 “80 后”海归创业者袁玉宇的勇气； “生活给
了我多少积雪，我就能遇到多少个春天 ”，这是外卖员王计兵的心声； “命运永远掌握在我们自己手中，
科技强国步伐永不停止 ”，这是大学生姜未的志向； “别人觉得种田苦，但我看好农业的广阔前景 ”，这
是“90 后”新农人陈帅宇的选择 ……
是的 ，这种书写不只吕玉霞 ，也不只诗歌 。相信很多人都有同样的感受 ：近年来 ，素人写作渐成趋势 ，
也愈来愈受关注。快递小哥胡安焉《我在北京送快递》、 “外卖诗人 ”王计兵诗集《赶时间的人》、女儿
视角下的《我的母亲做保洁》等作品成畅销书，在吕玉霞之前，也有韩仕梅等分享 “田园诗人 ”的荣光。
而他们也只是在诸多巧合下被舞台追光灯照亮的极少数，在更多尚未被 “看见 ”的地方，相信存在一个更
为庞大的群体，他们默默写作，生活重压之下仍不放弃对精神世界的追求，绵密记录下真实的日常与复杂
的情愫。
主题六：巾帼榜样
航天员刘洋两度飞天，在苍穹之上镌刻下中国女性的时代坐标；科学家颜宁矢志不渝，在结构生物学
领域取得突破性成果 ；铁路 12306 技术领军人单杏花攻坚不懈 ，让“通宵达旦排长队购票 ”彻底成为历史 ……
在创新中自强不息、在创造中开拓进取，诠释了新时代女性的无限可能，也照亮了无数追梦的心灵。
主题七：沉淀与专注热点素材
选对适合自己的赛道是个人发展的前提，而核心竞争力的形成则需要时间的沉淀。如今许多中国企业
之所以能够在全球市场中追赶并超越对手，打造出难以复制的硬核实力，除选对了方向的原因外，更重要 学科 网（北 京）股 份有限 公司
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的是他们身处技术爆发的风口，且始终执着于对技术的精益求精，坚持对产品的迭代打磨。在面对困难时
不被吓倒，在面对周期波动时沉稳笃定、始终坚守，这是他们能够取得成功的另一个核心因素。试想如果
梁文锋的团队不专注于人工智能 ，他们如何能够成功研发出 DeepSeek 这一震惊全球的大模型？如果导演饺
子耐不住寂寞 ，没有长期专注于动画电影的创作 ，《哪吒 2》又怎能以破 100 亿元的超高票房 ，创造出中国
电影史的崭新纪录？
2025 高考作文五大核心主题预测
【热点一 科技与人文的辩证关系 】“Deepseek ”、“宇树科技 ” 、“《哪吒 2》饺
子”乙巳年的三把火
【热点 思维 】
受制于客观因素，如生理，天分、出身的限定，人类的局限性无可逃脱，换言之，局限性决定了人在 学科 网（北 京）股 份有限 公司
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某一特定时段条件下所能及的认识与行动上限。面对客观存在的局限，从接受局限性出发，向突破局限性
无限逼近，或许是对局限性的最好的尊重。积极的突破局限在于 “虽千万人，吾往矣 ”的克制的勇敢。
2025 年，国产 AI 大模型 DeepSeek 震动硅谷，走红全球。它源于开源研究和开源项目。开源是技术上
“开放 ”的结果，开源为创新提供了诞生的土壤，也让全球的开发者可以共同参与进来，探索人工智能的
下一个边界。或许，这既是 DeepSeek 成功的关键原因，也是它成功后给大众带来的重要启示。 DeepSee k
引发的不仅仅是一场前所未有的技术创新战，更是一场前所未有的认知战。
【事件 分析】
技术与人性的结合 ：工具与价值的平衡 。作为人工智能 ，DeepSeek 由数据 、算法和人类智慧共同构建 ，
本质是工具而非生命体。科技发展的终极目标是服务人类。学生可以思考如何利用技术解决现实问题（如
教育公平、信息获取），同时警惕技术对人的异化（如过度依赖、隐私风险）。
突破局限： 合作与创新的力量。 DeepSeek 的局限性：缺乏情感与创造力，但能通过海量知识库和逻辑
分析辅助人类决策。 人类与 AI 的关系是互补而非对立。学生可强调人的独特性（如情感、想象力）与 AI
的辅助性，呼吁通过合作实现突破。
伦理与责任 ：技术背后的选择 DeepSeek 的运行逻辑 ：依赖人类设定的规则与数据 ，可能隐含偏见或信
息偏差。技术发展需以伦理为基石。可反思技术背后的责任（如开发者初心、数据真实性），引申到个人
选择对社会的影响。
跨文化桥梁 ：全球化时代的包容性 。DeepSeek 的能力包括支持多语言交流 ，融合不同文化背景的知识 。
技术可以打破隔阂，但真正的理解仍需人类主动包容。学生可结合文明互鉴、文化自信展开讨论。
持续进化 ：终身学习的启示 。DeepSeek 的迭代方式 ：通过数据更新与算法优化不断进步 。 在快速变化
的时代，持续学习是个人与社会的必修课。学生可联系自身成长，强调适应力与终身学习的重要性。
【写作角度 】
角度一：文明进步发展
在这个瞬息万变的时代，科技的力量犹如破晓时分的璀璨曙光，为人类文明的航程指引方向。它不仅
有力地撬动了社会的进步车轮，更在不断塑造和刷新我们的生活方式。科技创新与社会发展之间的关系，
恰似种子与土壤共生共荣 、相互成就 。随着 DeepSeek 的发布 ，我们再次见证了科技创新对文明的深远影响 。
它不仅推动了技术飞跃，更预示了信息传播方式的革新、加速了文明的发展。然而，这也带来了平衡技术
与人性关系的挑战。因此，科技创新不仅是技术的突破，更是对文明进步和人性价值的双重考量。
角度二：社会文明进程
科技创新不仅深刻改变了人们的生活方式 ，还持续推动着社会文明向前迈进 。从工业革命到信息时代 ，学科 网（北 京）股 份有限 公司
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每一次科技的重大突破都伴随着社会翻天覆地的变化。历史学家阿诺德 ·汤因比指出，文明的进步依赖于
挑战和应对挑战的能力，而科技创新正是提供了应对挑战的工具和手段。
角度三：伦理道德考量
正如人工智能能够创造出惊艳的视频 ，我们也应创造出符合伦理的创新环境 。孔子曰 ：“工欲善其事 ，
必先利其器。 ”然而，利器亦需要明辨是非之心。科技发展必须以增进人类福祉为出发点，同时恪守道德
底线。中国历史上的四大发明促进了文明进步，但火药也用于战争，这说明了技术本身无善恶之分，关键
在于使用方式。这提醒我们，科技创新需要建立伦理导引规范以确保科技成果造福人类。
角度四：个性化服务与隐私保护
人工智能可以提供定制化的内容，但是对用户数据的搜集与应用不应逾越隐私保护的底线。 “不以规
矩，不能成方圆。 ”在提供个性化服务时，必须遵守数据保护的规矩，否则就可能失去用户的信任。
角度五：青年择业与人才兴国
DeepSeek 的发布为青年择业和人才兴国提供了新视角。它为青年提供了投身科技行业的新机会，鼓励
他们不断学习和更新知识。同时， DeepSeek 的成功研发也凸显了人才在国家发展中的关键作用，强调了培
养与吸引优秀人才的必要性。青年人才应将个人职业发展与国家需求相结合，实现个人梦想与国家梦想的
共融。这不仅有助于国家长远发展，也为青年创造了实现职业价值和人生目标的广阔舞台。
相关话题
创业精神逆境突围 2. 科技创新技术向善 3. 青年担当时代使命
坚持与信念破局与重生 5. 个人成长团队协作 6. 社会责任科技伦理
全球化视野本土化实践 8. 理想与现实平衡与抉择 9. 从代码到星辰，用 AI 照亮征途
抵押房子换未来，用算法重写人生 11. 至暗时刻点亮一盏技术的灯 12. 科技有温度，创新无边界 .
破局者在 AI 浪潮中锚定价值 14. 从零到 “深”15. 用算法解方程，用初心暖人间 .
技术向善 ，是底线更是天花板 17. 在数字丛林里种一朵人文之花 18. 与其预测未来 ，不如创造未来
科技与人文子话题 —— 深耕自己是最好的远见
学习的使命是什么？
思辨点： 现在机器人都在不断学习提升 —— 锲而不舍的学习，将目光投于未来，绝不能沉沦于现在的
无效娱乐！
（一）个人层面：学习是自我成长的基石
使命： 通过学习认识世界、发展潜能、实现人生价值。
突破认知局限 学科 网（北 京）股 份有限 公司
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导演饺子通过自学动画打破医学专业的限制，实现创作梦想。
启发： 学习让你看到更广阔的世界，拥有更多选择权。
培养解决问题的能力
（元旦晚会用等差数列数凳子）通过数学建模竞赛，学会用数据分析解决实际问题。
启发： 学习不仅是记忆知识，更是训练思维，应对未来挑战。
实现自我价值
学习决定你未来在哪个圈子，和哪些人，做什么事。
决定你在生活中有多少自由度、多少选择权、多少幸福感！
（二）家庭层面：学习是责任与爱的传递
使命： 通过知识改变家庭命运，传承精神力量。
改善家庭生活质量
云南贫困山区女孩张桂梅的学生通过高考走出大山，反哺家庭。
启发： 学习能为家庭带来经济改善和精神希望。
延续家庭文化基因
一代强于一代，而非下带啃老上代，让父母更加含辛茹苦！
启发：学习影响着你的婚姻、你自己未来的家庭、自己子女的成长环境和教育问题。
一个家庭的好榜样
一个家庭里大家的好榜样，别人眼中的好孩子，口口相传，人人称赞。
学校里的优秀校友、榜样学生，改变一个学校的地位！ （千亿总裁雷军与武汉大学 ）
（三）社会层面：学习是推动人类进步的动力
使命： 用知识服务社会，解决人类共同问题。
科技创新改变世界
学习电脑编程、 AI 模型、提高自己的数学能力，学会数学建模解决实际问题。
启发： 你今天学的物理公式，未来可能解决能源危机。
未来改变生活，强大国家的核心人才
民营企业家座谈会
学习的终极价值 —— 成为更好的自己，温暖他人，照亮世界。
“你读的每一本书，都是未来面对世界的底气。 ”
“学习不仅为 ‘我’，更为 ‘我们 ’—— 家人、同胞、人类。 ”学科 网（北 京）股 份有限 公司
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三、遇见更好的自己
1、打破现状 ，自我觉醒 ：与懒惰的自己斗争 ，改变现在的自己 ，每天坚持一点点 ，成为习惯 ，终会成
为优秀！
2、享受学习 ，主动成长 ：充满求知欲 ，以学为乐 ，培养钻研精神 ，养成 好习惯 ，锤炼 好品质 ，好习惯
成就好人生！
3、争做榜样 ，奋勇争先 ：不要习惯于默默无闻 ，而要被表扬 ，被看见 ，被认可 ，被鼓励 ，补充源源不
断的动力。
4、找准榜样 ，矢志不渝 ：未来的梁文峰 、未来的王兴兴 ，一定会在 我们当中诞生 ，要有足够的信心坚
持梦想！
【写作角度】
坚持兴趣与热爱： 发自内心的热爱胜过一切，热爱是起点，是动力，为了追求热爱，可以死磕答案 ，
不断探索，这样才会让自己的兴趣成为你真正的热爱！
勇于尝试和创新 ：成功往往来自于突破常规 ，不要害怕犯错 ，在学习和生活中要勇于尝试新的方法和
思路，培养创新思维，钻研精神，克服畏难情绪，愈挫愈勇！
不断学习和自我提升 ：无论处于什么阶段 ，学习都是关键 。应重视知识的积累 ，不仅是课本知识 ，还
包括各种技能和素养，像这些成功人士可能也在不断学习进步。
培养坚韧的毅力 ：在追求目标的道路上 ，难免会遇到挫折和困难 ，梁文峰 、王兴兴 、饺子等人的成功
也并非一帆风顺，要学会在困境中坚持，不轻易放弃。
善于规划和设定目标 ：一定学会给自己制定合理的短期和长期目标 ，并规划好实现路径 ，一步一个脚
印地朝着目标前进，会不断完成小目标的人，未来一定会成就大事业。
【时评文章】
不当 “空想家 ”，争做 “实干家 ”
鲁格言
来源：人民网
“伟大的事业都始于梦想、基于创新、成于实干。 ”新时代、新征程、新伟业，需要铆足实干劲头，
持续谋求新发展。
党的二十大提出了全面建设社会主义现代化国家、全面推进中华民族伟大复兴、实现第二个百年目标
的方向，中国开启了新伟业。如今站在新征程的起点，如何实现中国梦、抵达成功彼岸？唯一方法便是实
干。 学科 网（北 京）股 份有限 公司
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“空谈误国，实干兴邦。 ”当今社会有太多原地踏步、止步不前的 “空想家 ”，却缺乏脚踏实地、砥
砺前行的 “实干家 ”。这段时间网络上频繁出现 “弹幕人均专家 ”的讽刺，就是对 “空想家 ”们的质疑。
“空想家 ”们站在制高点，肆意地对问题 “评头论足 ”，发表 “高见 ”。在这个互联网高度发达，追求言
论自由的时代，这样的 “空想家 ”层出不穷。他们往往只是悲观、愤怒地提出问题，却因对未知的恐惧和
对困难的畏惧而选择逃避问题，最终问题还是问题。纵然梦中走了千万里，醒来还是在床上。我们想要谋
求发展，就必须铆足实干劲头，践踏实地，不断攻坚克难，才能让梦想之帆抵达成功彼岸。
那么如何成为 “实干家 ”？是坚持肯干，是埋头苦干，是灵活能干。
“实干家 ”是知难而进，坚持肯干。 “志行万里者，不中道而辍足。 ”我们要坚信功不唐捐，玉汝于
成。面对前方艰难曲折的开拓之路，应怀 “明知山有虎，偏向虎山行 ”的魄力，革故鼎新、攻坚克难，用
持之以恒的行动撬动一块块 “顽石 ”、创造一个个奇迹，在时代浪潮中掌握先机。
“实干家 ”是胸怀大局，埋头苦干。实干不仅是看得见头顶的月亮，也要能拥有脚下的六便士。以信
仰为帜，以目标为灯，以理论为证，以实干为先。以长远的眼光看事情，以切实的行动为根基，面对考验
无所畏惧、面对挑战一往无前。
“实干家 ”是初心不忘，灵活能干。实干不是蛮干，不是循规蹈矩、固守成规。面对不断变化的形式
和环境，实干也是巧干，我们要灵活变通，寻找规律与方法。巧干并非投机取巧。变通时我们要切记来时
路。电视剧 《狂飙 》既向我们展示了初心不改的警察安欣 ，也让我们看到了那些在长期工作中逐渐被腐蚀 、
被拉拢 ，一步步堕入深渊的贪污腐败者 。初心易得 ，始终难守 。若在复杂局势中做到坚持初心 、通时达变 ，
我们便掌握不断赢得胜利的法宝。
“大浪淘沙沉者为金，大道至简实干为先。 ”“ 空想家 ”的梦恰似象牙之塔、空中楼阁、镜花水月，
是无根之木 ，难成苍青 。生逢太平盛世 ，目视国富民强 ，更应守正出新 ，实干兴邦 。人人拒做 “空想家 ”，
争当 “实干家 ”，社会主义现代化强国、中华民族伟大复兴指日可待！
科技与人文子话题 —— 科技发展与人际理解
【真题演练】 阅读下面的材料，根据要求写一篇不少于 800 字的文章。
（2025 杭州市二模）当今社会，科技的发展为交流提供了多样的渠道和丰富的场景，人们的沟通更及
时，更便利。那么，人与人之间的相互理解是否会因此变得更为容易？以上材料引发了你怎样的联想与思
考？请写一篇文章。
审题：材料开篇点明 “当今社会，科技的发展为交流提供了多样的渠道和丰富的场景 ”，这是一个显
著的时代特征。从即时通讯软件如微信、 QQ ，实现随时随地的文字、语音、视频交流，打破时空限制，让
远在千里之外的人也能 “面对面 ”沟通 ；到社交媒体平台像微博 、抖音 ，人们可分享生活点滴 、观点见解 ，学科 网（北 京）股 份有限 公司
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极大拓展社交圈。丰富的场景体现在，无论是通勤路上、工作间隙，还是居家休闲，只要有网络，交流触
手可及。例如，线上会议使身处不同地区的工作团队能实时研讨；在线课堂让学生跨越地域向名师求学。
这些都表明科技让沟通在时间与空间维度上实现了前所未有的便捷 。“人们的沟通更及时 ，更便利 。那么 ，
人与人之间的相互理解是否会因此变得更为容易？ ”这一疑问是材料核心。其中， “相互理解 ”并非简单
知晓对方所言，而是能站在对方立场，体会其感受、洞察其想法与行为动机。科技虽赋予沟通便捷性，但
相互理解受多种因素影响，如交流深度、双方态度、文化背景差异等，并非仅由沟通便捷与否决定。这就
要求深入探究科技在人际相互理解中扮演的角色，是助力、阻碍，亦或兼具二者。
立意：
1 科技通讯千万里，理解未必近人心
科技是人际理解的催化剂还是绊脚石？
科技时代，理解的距离是近还是远？
穿越科技迷雾，找寻理解的微光
【范文 模板】
科技架桥梁，谨慎陷鸿沟
伴随着交通、互联网、虚拟现实、人工智能模型等科技的发展，人们的沟通更为及时便利，但人与人
的理解却陷入自足孤岛的因境。值此之际，我们应当以科技架好理解之桥梁。
我们的沟通借助网络媒介得到了及时的反馈和充分的互动，然而这一平台也汇集了 “乌合之众 ”的集
体非理性思潮，制造出回音室效应，剥夺或遏制了人们独立思考、理性判断的能力，使理解成为了一种单
向度的想象和集权式的引导。
我们的理解借由虚拟科技，再现了心灵的真实。视听的奇观和云端的连线使我们跨越想象的壁垒，抵
达彼此在现实无法实现的身份认识、职业认同。然而，这些虚拟的互动体验似乎取代了屏幕背后的情感交
流，使人们不愿意接受现实的琐屑，卑小，以及沟通所需要消耗的耐心和穷尽生命的探索。如果理解不需
要百转千回的路径，这样的理解真的不是一款热销版的商品吗？这样的理解真的是人类隐秘心灵的奇迹
吗？
当手机取代面对面的问候 ，当 AI 专家条分缕析人类的心理历程 ，给出最便捷的回复 、关心的模版 ，人
们的相互理解也就成为了机械的连线。需知，理解基于不同。只有独特的个性化认识才诞生出沟通交流的
必要。我们应谨慎只求便捷的沟通，在科技的辅助下，关注每一个情绪的生发，倾听那似乎不合时宜的声
音。
然反观当下，许多青少年沉溺于电子设备，在宏观的大世界中迷失了自己的坐标。许多吃瓜群众奔驰 学科 网（北 京）股 份有限 公司
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在“新闻 ”的第一线，发表出自己 “精辟 ”的互动言论，却唯独忘记了交流的目的是求同存异，是带去公
平、正直与关怀。人类学家项飙敏锐地观察当下人们的交流状态后提出 "附近的消失 "精神窘境。在科技反
使人们不易理解的当下，他尝试 "重建附近 "以阻止个体间连通性与互渗性的消亡；电影研究者戴锦华叩问
我们的虚拟化身，究竟是我们的想象，虚拟的自我，还是一种更真实的存在。如今手持头像、拥有线上身
份的我们应当深思，如何建立起真实的联系，如何为理解他人而交流，而非满足心理的镜像认识。科技使
交流更及时、便利，也给了我们充分自足的想象空间。值此之际，我们应当打破镜像，走出孤岛，真正地
生发人类理解的力量。
科技与人文子话题 ——“ 理想主义 ”“ 实干精神 ”与“融合创新 ”
阅读下面的材料，根据要求写作。
梁文峰在高校演讲时说： “这个世界需要两种人 —— 一种是抬头看星的理想主义者，另一种是低头修
路的实干家。而最幸运的是，我们正成为第三种人：把星光铺成道路的人。 ”
这句话引发了你怎样的联想与思考？请结合自身发展写一篇文章。
要求：选准角度，确定立意，明确文体，自拟标题；不少于 800 字。
【审题指导】
材料关键词为 “理想主义 ”“ 实干精神 ”与“融合创新 ”。需围绕三者关系展开，强调新时代青年既
要仰望星空（树立远大理想），又要脚踏实地（付诸实践行动），更需在创新中实现二者的统一。
以理想为帆，以实干为桨，驶向创新蓝海
将星光照进现实，让道路通向星辰
在理想与现实的交汇处播种未来
范文： 以理想为星，凭实干筑路
宇宙浩渺 ，理想如熠熠星辰 ，在无尽暗夜中闪耀 ，为我们指引前行方向 ；大地广袤 ，实干似坚实通衢 ，
承载着我们迈向目标的步伐。梁文峰说： “这个世界需要两种人 —— 抬头看星的理想主义者，低头修路
的实干家 。而最幸运的是 ，我们正成为第三种人 ：把星光铺成道路的人 。” 这句话振聋发聩 ，让我们明白 ，
唯有将理想与实干紧密结合，才能踏出独属于自己的康庄大道。
抬头仰望星空，是对理想的执着坚守。理想是人生的灯塔，驱散迷雾，给予我们前行的动力与勇气。
古往今来 ，无数仁人志士在理想感召下 ，为实现心中目标全力以赴 。“仰天大笑出门去 ，我辈岂是蓬蒿人 ”，
李白怀揣豪情 ，在诗海纵横驰骋 ，留下千古佳作 ，其文字照亮文学苍穹 ；周恩来总理年少立志 “为中华之
崛起而读书 ”，一生为国家和人民鞠躬尽瘁 ，用生命诠释理想的伟大 ，让生命绽放光芒 。正因有理想指引 ，
他们的人生才意义非凡，成为后人敬仰的丰碑。 学科 网（北 京）股 份有限 公司
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然而，仅有理想远远不够，还需秉持脚踏实地的实干精神。实干是跨越现实与理想鸿沟的桥梁，是将
理想变为现实的关键 。“杂交水稻之父 ” 袁隆平 ，几十年扎根田间 ，潜心钻研 ，解决数亿人温饱问题 ，名
字成为大地的传奇 ；“大国工匠 ” 徐立平 ，为导弹火药微整形 ，在极其危险岗位默默奉献 ，用精湛技艺为
航天事业奠基，助力中国航天翱翔天际。他们用行动诠释实干的力量，让我们明白，只有脚踏实地，才能
将理想照进现实。
身处机遇与挑战并存的时代，当代青年更应志存高远，努力成为 “把星光铺成道路的人 ”。我们像初
升的朝阳，既有广阔发展空间，也面临诸多问题。在实现民族复兴的征程中，我们肩负历史使命，要将个
人命运与国家 、民族命运相连 ，以国家富强 、民族振兴 、人民幸福为追求 ，让理想在时代沃土生根 ；同时 ，
发扬实干精神，努力学习知识，提升能力，用奋斗汗水浇灌理想之花，为实现理想全力以赴。
在学习和生活中，挫折与困难不可避免。当追求理想遭遇失败，切不可气馁放弃。要知道，每次失败
都是成长契机，我们应像海燕在暴风雨中飞翔，从失败中汲取教训，调整方向继续前行。正如鲁迅所说：
“愿中国青年都摆脱冷气，只是向上走 …… 有一分热，发一分光。 ” 我们要以积极心态拥抱生活，用行
动为社会贡献力量。
“路漫漫其修远兮 ，吾将上下而求索 。” 在时代浪潮中 ，让我们以理想为星 ，照亮征途 ；凭实干为路 ，
迈向成功彼岸。携手成为追梦者与实干家，为实现中国梦努力拼搏，奏响时代最强音！
科技与人文子话题 —— 时代的馈赠与未来的思辨
阅读下面的材料，根据要求写作：
国产 AI 大模型 DeepSeek 以 600 万美元的训练成本，实现了在多项测试中表现持平甚至超 越
OpenAI 模型的成绩，在性价比方面展现出惊人的优势。这一事件引发了广泛的关注和讨论。随着
人工智能技术的飞速发展， AI 已经渗透到我们生活的方方面面。从智能家居到无人驾驶汽车，从
智能语音助手到医疗诊断系统 ，人工智能为我们带来了前所未有的便捷 。然而 ，它也引发了诸多争
议，比如就业结构的调整 、隐私安全的威胁 、伦理道德的挑战等 。有人认为 ，人工智能是人类的助
手，能够帮助我们更好地生活和工作 ；也有人担心 ，人工智能可能会取代人类的许多工作 ，甚至对
人类的未来构成威胁。
要求 ：选准角度 ，确定立意 ，明确文体 ，自拟标题 ；不要套作 ，不得抄袭 ；不得泄露个人信息 ；
不少于 800 字。
【审题指导】
该题材聚焦于人工智能这一现代科技热点 ，它既呈现了技术进步为生活带来的革新 ，像智能家
居、智能语音助手等 ，也指出了随之而来的种种棘手问题 ，如就业结构变动 、隐私风险 、伦理困境 。学科 网（北 京）股 份有限 公司
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这一题材紧密贴合现实生活 ，极具时代性与思辨性 ，要求学生对人工智能的多面性有深入认知 ，能
从不同视角审视其影响。
审题立意
关键信息抓取 ：“人工智能 ” 是核心对象 ，“便捷 ”“争议 ”“助手 ”“威胁 ” 等词汇点明了对其
探讨的关键方向。
立意方向：
积极向立意：着重强调人工智能作为得力助手，助力提升生活与工作效率，如 “搭乘人工智能快车 ，
驶向便捷高效生活 ”。
消极向立意：关注人工智能带来的负面影响，如 “警惕人工智能阴霾，守护人类生存根基 ”。
辩证立意 ：全面考量人工智能 ，兼顾其优势与弊端 ，如 “以辩证之眼 ，看人工智能的机遇与挑战 ” 。
素材选用
正面素材 ：例如 ，电商平台利用人工智能算法精准推荐商品 ，提升用户购物体验 ；智能手术机器人辅
助医生进行复杂手术，提高手术成功率。
负面素材 ：如某些数据公司因人工智能算法漏洞导致大量用户信息泄露 ；一些传统制造业因引入自动
化设备，致使大量工人失业。
范文：
人工智能：时代的馈赠与未来的思辨
在时代的浪潮中 ，人工智能 （AI ）犹如一颗璀璨的星辰 ，悄然升起并照亮了人类生活的每一个
角落 。它以智能家居的温柔触角 ，轻抚我们的日常 ；以无人驾驶汽车的矫健身姿 ，驰骋于城市的脉
络；以智能语音助手的灵动之音 ，回应我们的需求 ；以医疗诊断系统的精准之眼 ，守护我们的健康 。
人工智能，无疑是这个时代赋予我们的珍贵馈赠。
然而 ，当我们在享受这份馈赠时 ，也难免陷入深深的思辨之中 。它如同一把双刃剑 ，既带来了
前所未有的便捷 ，也引发了诸多争议 。就业结构的调整 ，让无数劳动者面临转型的阵痛 ；隐私安全
的威胁 ，如同幽灵般潜伏在数据的海洋 ；伦理道德的挑战 ，更是让人类的价值观面临前所未有的考
验。面对这些挑战，我们不禁要问：人工智能，究竟是人类的助手，还是潜在的威胁？
人工智能 ，是时代的宠儿 ，也是人类智慧的结晶 。它以无与伦比的速度和精度 ，为我们的生活
带来了前所未有的便利 。智能家居系统 ，如同一位贴心的管家 ，默默守护着我们的生活 ；无人驾驶
汽车 ，如同一位经验丰富的老司机 ，安全地将我们送达目的地 ；智能语音助手 ，如同一位善解人意 学科 网（北 京）股 份有限 公司
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的朋友 ，随时回应我们的呼唤 ；医疗诊断系统 ，如同一位博学的医生 ，精准地为我们诊断疾病 。正
如古希腊神话中的普罗米修斯，人工智能为人类带来了火种，照亮了前行的道路。
然而 ，正如普罗米修斯的火种也带来了灾难 ，人工智能的发展也带来了诸多挑战 。就业结构的
调整 ，如同一场无声的风暴 ，席卷了无数传统岗位 。工厂中的流水线工人 、客服中心的接线员 、基
础的数据分析师 …… 他们的工作被机器取代 ，生活陷入迷茫 。隐私安全的威胁 ，如同一只无形的黑
手，时刻窥视着我们的数据。智能家居系统记录着我们的日常，智能语音助手聆听着我们的声音 ，
一旦这些数据被泄露 ，我们的隐私将无处遁形 。伦理道德的挑战 ，如同一座难以逾越的高山 ，横亘
在我们面前 。无人驾驶汽车在面临不可避免的碰撞时 ，应该如何做出决策？ AI 在医疗领域的应用 ，
又该如何平衡效率与伦理？
面对这些挑战 ，我们不能因噎废食 ，也不能盲目乐观 。我们需要在技术发展的过程中 ，积极寻
求解决方案 ，确保人工智能的发展能够造福人类 ，而不是成为潜在的威胁 。一方面 ，政府和社会应
加强对人工智能技术的监管，制定相关法律法规，保护用户的隐私和数据安全，规范 AI 的应用范
围和行为准则 。另一方面 ，企业和科研机构应承担起社会责任 ，在追求技术创新的同时 ，注重伦理
道德的考量，确保技术的发展符合人类的利益。
同时，我们还需要加强对人工智能相关知识的普及和教育，提高公众对 AI 的认知和理解。只
有当人们充分了解人工智能的优势和风险时 ，才能更好地参与到相关的讨论和决策中 ，共同推动人
工智能的健康发展。
“工欲善其事 ，必先利其器 。”人工智能 ，无疑是这个时代最锋利的 “器”。它既有可能成为
人类的得力助手 ，也有可能成为潜在的威胁 。我们应以理性和客观的态度看待人工智能的发展 ，在
充分发挥其优势的同时 ，积极应对和解决其带来的挑战 。只有这样 ，我们才能确保人工智能技术在
推动社会进步和人类发展的道路上发挥积极的作用，而不是成为人类未来的隐患。
让我们拥抱人工智能 ，但也要时刻警惕它的双刃剑特性 。在享受科技带来的便利时 ，更要坚守
人类的伦理道德底线 ，守护我们的隐私与尊严 。唯有如此 ，人工智能才能真正成为人类的助手 ，而
非潜在的威胁。
【热点二 、传统文化的创新性传承 】“英雄名字刻进城市街巷 ”、“中轴线申遗 ” 、
“人文景观研学 ”
传统文化创新性传承子话题 —— 把英雄的名字刻进城市的街巷
题目： 阅读下面的材料，根据要求写作。 学科 网（北 京）股 份有限 公司
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在许多城市中 ，街道的命名往往承载着深厚的文化内涵和历史记忆 。近年来 ，一些城市开始尝
试将英雄的名字刻进街巷的名称中 ，以此纪念英雄的事迹 ，传承英雄的精神 。例如 ，某市将一条街
道命名为 “卫国英雄街 ”，以纪念在边境冲突中英勇牺牲的戍边战士 。这种做法引发了广泛的关注
和讨论 。有人认为 ，这是对英雄的最好纪念 ，能够让英雄的精神在城市中延续 ；也有人担心 ，这种
做法可能会让英雄的名字被商业化利用，失去其应有的庄重感。
请结合材料 ，以“把英雄的名字刻进城市的街巷 ”为主题 ，写一篇文章 ，谈谈你对这一现象的
看法和思考 。要求 ：选准角度 ，确定立意 ，明确文体 ，自拟标题 ；不要套作 ，不得抄袭 ；不得泄露
个人信息；不少于 800 字。
题材 分析
这一题材聚焦于城市街道命名这一社会现象 ，核心在于将英雄名字用于街巷命名所引发的讨论。
它既关联到对英雄的纪念与传承 ，又涉及到商业开发可能带来的影响 ，是一个具有现实意义和文化
内涵的话题 。需要从文化传承 、社会价值 、商业伦理等多维度去思考 ，考量如何在现代社会中更好
地铭记和弘扬英雄精神。
审题 立意
关键信息抓取： “英雄名字刻进街巷 ”“ 纪念英雄事迹 ”“ 传承英雄精神 ”“ 商业化利用 ”
“庄重感 ” 等是关键要点。
立意方向：
支持命名：强调以英雄名字命名街巷对纪念英雄、传承精神的积极意义，如 “让英雄之名，
闪耀城市街巷之光 ”。
反对命名：突出对英雄名字被商业化利用的担忧，以及对其庄重感可能丧失的思考，如 “警
惕英雄之名，莫让商业侵蚀庄重 ”。
辩证看待：全面考虑命名的利弊，探讨如何在纪念英雄与避免商业化负面影响之间找到平衡 ，
如 “理性对待英雄命名，守护精神与庄重 ”。
素材 选用
支持命名素材 ：某城市以抗日英雄名字命名街道后 ，周边学校常组织学生参观学习 ，学生们
深受英雄精神鼓舞；某街道因英雄名字吸引游客，带动当地文化旅游发展，同时传播了英雄事迹 。
反对命名素材 ：一些地方将英雄名字用于商业促销活动 ，引发公众不满 ；个别以英雄名字命
名的街道周边出现低俗商业广告，破坏了英雄名字的庄重感。
范文 展示 学科 网（北 京）股 份有限 公司
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让英 雄之名 ，闪耀 城市街 巷之光
城市的街巷，宛如城市的脉络，静静诉说着岁月的故事。当英雄的名字被刻进这一条条街巷 ，
它们便如同璀璨的星辰 ，为城市的夜空增添了耀眼的光芒 。将英雄的名字融入街巷名称 ，无疑是一
种极具意义的纪念方式，它承载着对英雄的崇敬，更肩负着传承英雄精神的重任。
以英雄之名命名的街巷 ，仿若一座不朽的精神丰碑 。在那 “卫国英雄街 ” 上，每一块砖石都
仿佛在低语 ，讲述着戍边战士们无畏生死 、保家卫国的壮丽篇章 。过往的行人 ，无论是匆匆赶路的
上班族 ，还是嬉笑玩耍的孩童 ，都会在不经意间被这名字所触动 。英雄们的英勇事迹 ，如同春风化
雨，滋润着人们的心田 ，激励着大家在各自的岗位上 ，为了国家和社会 ，勇敢拼搏 ，无私奉献 。它
让人们在平凡的生活中 ，时刻铭记那些为了和平与安宁 ，付出宝贵生命的英雄们 ，成为推动城市前
进的强大精神动力。
英雄名字刻进街巷 ，又似一条坚韧的文化纽带 ，紧密连接着城市的过去 、现在与未来 。它让城
市的历史不再是尘封在书本中的冰冷文字 ，而是鲜活地展现在人们的日常生活里 。当孩子们在 “英
烈巷 ” 中穿梭嬉戏时，长辈们便会将英雄的故事娓娓道来，让那段波澜壮阔的历史，在一代又一
代的传承中焕发生机。这种传承，不仅丰富了城市的文化内涵，更让城市的精神得以延续和升华 ，
使城市在时代的浪潮中，始终坚守着那份对英雄的敬仰和对正义的执着。
把英雄名字赋予街巷 ，更如同一盏明灯 ，照亮了城市的精神角落 。在繁华喧嚣的都市中 ，商业
的浪潮时常汹涌澎湃 。然而 ，英雄名字所带来的庄重与神圣 ，如同一股清泉 ，流淌在城市的大街小
巷，时刻提醒着人们 ，在追求物质富足的同时 ，不能忘却精神的滋养 。它让城市的每一个角落都弥
漫着英雄的气息，提升了城市的文化品位，让城市在发展的道路上，始终保持着高尚的灵魂。
当然，我们也应警惕商业化可能带来的负面影响。但这绝不是否定英雄名字刻进街巷的理由 ，
而是需要我们加强管理和引导 ，让英雄的名字在城市中闪耀着最纯粹的光芒 。让我们珍视这一纪念
英雄的方式，让英雄之名，永远闪耀在城市的街巷之上，成为城市永恒的精神坐标。
传统文化创新性传承子话题 —— 北京中轴线申遗成功
题目： 阅读下面的材料，根据要求写作。
2024 年，北京中轴线申遗成功 ，这一消息引发了社会各界的广泛关注 。北京中轴线作为中国古
代城市规划和建筑设计的杰出代表 ，承载着深厚的历史文化内涵 。它不仅是北京城市空间的核心骨
架，更是中华文明的重要象征 。申遗成功不仅是对北京中轴线历史价值的肯定 ，也是对中华文化的
传承与弘扬 。然而 ，申遗成功也带来了新的挑战 ，如何在保护文化遗产的同时 ，实现文化的传承与
创新，成为了一个值得思考的问题。 学科 网（北 京）股 份有限 公司
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请结合材料 ，以“文化的传承与创新 ”为主题 ，写一篇文章 ，谈谈你的感悟与思考 。要求 ：选
准角度 ，确定立意 ，明确文体 ，自拟标题 ；不要套作 ，不得抄袭 ；不得泄露个人信息 ；不少于 80 0
字。
题材 分析
该题材以北京中轴线申遗成功为切入点 ，聚焦于文化遗产保护以及文化传承与创新这一核心议
题。它关联到历史文化 、城市规划 、现代发展等多个领域 ，需要从文化价值 、传承意义 、创新方式
等多维度去思考 ，探讨如何在现代社会背景下 ，让古老的文化遗产焕发生机 ，实现可持续发展 ，题
材具有很强的文化底蕴和现实指导意义。
审题 立意
关键信息抓取 ：“北京中轴线申遗成功 ”“历史文化内涵 ”“文化传承与创新 ”“保护文化
遗产 ” 等是关键要点。
立意方向：
强调文化传承的重要性 ：突出文化传承对于延续历史脉络 、增强民族认同感的重要意义 ，如 “传
承文化薪火，照亮民族前行之路 ”。
阐述文化创新的价值 ：着重说明文化创新在使传统文化适应现代社会 、吸引大众关注方面的积
极作用，如 “创新文化形式，赋予传统崭新活力 ”。
辩证看待传承与创新关系：全面分析文化传承与创新相互依存、相互促进的关系，如 “权衡
传承与创新，奏响文化发展和谐乐章 ”。
素材 选用
文化传承素材 ：故宫博物院通过举办传统文物展览 ，让公众近距离感受古代文化魅力 ；一些
传统手工艺人坚守技艺，传承古老制作工艺。
文化创新素材 ：故宫文创产品将故宫元素与现代设计结合 ，深受消费者喜爱 ；西安大唐不夜
城通过现代光影技术和表演，重现大唐文化辉煌。
范文 展示
权衡 传承与 创新， 奏响文 化发展 和谐乐 章
当北京中轴线成功申遗的消息如春风般传遍大街小巷 ，整个社会都沉浸在喜悦之中 。这条承载
着千年历史的中轴线 ，宛如一位饱经沧桑的智者 ，静静诉说着中华文明的辉煌过往 。它不仅是北京
城市空间的脊梁，更是中华文化的璀璨明珠。申遗成功，无疑是对其历史价值的崇高礼赞，然而 ，学科 网（北 京）股 份有限 公司
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这也为我们带来了新的思考 ：在现代社会的浪潮中 ，如何在精心守护这一文化遗产的同时 ，实现文
化的传承与创新，让古老的智慧在新时代焕发出新的生机？
文化传承 ，如一位忠诚的守护者 ，紧紧握住历史的接力棒 ，将北京中轴线的文化瑰宝代代相传 ，
延续民族的精神血脉 。北京中轴线凝聚了古代中国城市规划和建筑设计的精髓 ，从永定门到钟鼓楼 ，
一路串联起天坛 、故宫等众多文化地标 。每一块砖石 、每一处建筑 ，都承载着先辈们的智慧和情感 。
通过对中轴线的保护与传承 ，我们能够清晰地触摸到历史的脉络 ，深刻感受到中华民族的伟大创造
力。就像故宫博物院 ，多年来致力于文物保护和文化传播 ，通过举办各类展览 ，让古老的文物走出
深宫 ，与大众亲密接触 ，使人们在欣赏文物的同时 ，领略到中华文化的博大精深 ，让文化传承的火
种熊熊燃烧。
文化创新 ，似一位灵动的化妆师 ，为古老的北京中轴线文化精心装扮 ，使其在现代社会中光彩
照人 ，吸引众人目光 。在当今时代 ，文化创新是让传统文化融入现代生活的关键 。近年来 ，故宫文
创产品异军突起 ，将故宫的文化元素巧妙地融入到现代生活用品中 。从精美的书签 、文具到时尚的
服饰 、饰品 ，这些文创产品以新颖的形式展现了故宫文化的魅力 ，深受消费者喜爱 。同样 ，对于北
京中轴线文化 ，我们也可以通过创新手段 ，如利用虚拟现实技术让人们身临其境地感受中轴线的历
史风貌 ，或者举办与中轴线文化相关的创意活动 ，吸引更多年轻人关注 。创新 ，为古老文化注入了
新的活力，让其在现代社会中绽放出别样的光彩。
只有传承与创新携手共进 ，文化才能如同一棵生机勃勃的大树 ，以北京中轴线文化为根 ，在现
代社会的土壤中茁壮成长，绽放绚丽花朵。传承是创新的根基，没有传承，创新就成了无本之木 ；
创新是传承的动力 ，没有创新 ，传承就会失去活力 。我们要在传承中创新 ，在创新中传承 ，让北京
中轴线文化以及整个中华文化在历史的长河中源远流长，熠熠生辉。
在文化发展的征程中 ，让我们肩负起传承与创新的重任 ，以传承为笔 ，以创新为墨 ，书写中华
文化的壮丽篇章，奏响文化发展的和谐乐章，让古老的文化在新时代焕发出更加耀眼的光芒。
传统文化创新性传承子话题 —— 人文景观研学
题目 ： 阅读下面的材料，根据要求写作。
六尺巷位于安徽省桐城市 ，是清代大学士张英与邻居吴家因宅基地发生纠纷时 ，张英主动让出
三尺，吴家深受感动也让出三尺，从而形成了一条六尺宽的巷子。这条巷子不仅是一个物理空间 ，
更是一种文化象征 ，体现了 “和为贵 ”“邻里和睦 ”“宽容礼让 ”的传统美德 。如今 ，六尺巷已成
为著名的文化景点，吸引着众多游客前来参观。 学科 网（北 京）股 份有限 公司
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某学校组织学生前往六尺巷研学旅行 ，学生们在参观过程中 ，深刻感受到了传统文化的魅力和
智慧 。有的学生表示 ，六尺巷的故事让他学会了宽容和谦让 ；有的学生则认为 ，这种文化精神在现
代社会依然具有重要的意义。
请结合材料，以 “研学感悟 ”为主题，写一篇文章，谈谈你的思考和体会。要求：选准角度 ，
确定立意，明确文体，自拟标题；不要套作，不得抄袭；不得泄露个人信息；不少于 800 字。
题材 分析
该题材围绕六尺巷这一文化景点展开 ，结合学校组织的研学旅行活动 ，涉及传统文化与现代社
会的关联 。关键在于从六尺巷所蕴含的 “和为贵 ”“宽容礼让 ” 等传统美德出发 ，探讨其在现代
社会的价值与意义 ，可从个人品德修养 、社会和谐构建等多维度进行思考 ，题材兼具文化内涵与现
实意义。
审题 立意
关键信息抓取： “六尺巷 ”“ 和为贵 ”“ 宽容礼让 ”“ 研学旅行 ”“ 现代社会意义 ” 等是
关键要点。
立意方向：
强调传统美德传承：突出六尺巷所代表的传统美德在现代社会传承的重要性，如 “传承六尺
巷美德，点亮现代文明之光 ”。
阐述现代社会价值 ：着重说明 “和为贵 ” 等理念对解决现代社会矛盾 、促进和谐的积极作用 ，
如 “借六尺巷智慧，解现代社会纷争，筑和谐家园 ”。
结合研学感悟 ：从自身研学经历出发 ，讲述六尺巷对自己思想和行为的影响 ，以及对传统文化
的新认知，如 “研学六尺巷，感悟传统魅力，启迪成长之路 ”。
素材 选用
正面素材 ：现实生活中邻里之间因互相宽容礼让解决矛盾的案例 ；社区开展以六尺巷文化为
主题的活动，促进邻里和谐的事例。
反面素材 ：因一点小事邻里发生冲突 ，甚至对簿公堂的新闻报道 ；社会上因缺乏宽容导致人
际关系紧张的现象。
素材要求：素材要真实、典型，紧密围绕主题，有力支撑论点。
范文 展示：
研学 六尺巷 ，感悟 传统魅 力，启 迪成长 之路 学科 网（北 京）股 份有限 公司
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阳光洒满了古老的石板路，我们怀揣着对传统文化的好奇与期待，踏上了六尺巷的研学之旅 。
这条承载着历史记忆的小巷 ，宛如一位智慧的长者 ，静静地伫立在岁月的长河中 ，向我们诉说着 “和
为贵 ”“ 宽容礼让 ” 的动人故事。
六尺巷的宽容礼让 ，如同一颗璀璨的明珠 ，在现代社会的天空中闪耀 ，传承它能照亮我们的道
德之路 。漫步在六尺巷 ，我仿佛看到了当年张英大学士那豁达的身影 。面对宅基地纠纷 ，他没有选
择针锋相对，而是以一纸家书 “千里家书只为墙，让他三尺又何妨 ”，展现出了非凡的宽容与大
度。这份宽容 ，不仅化解了邻里间的矛盾 ，更成为了千古佳话 。在现代社会 ，我们也常常会遇到各
种摩擦与纷争 。在学校里 ，同学之间可能会因为一点小事而争吵不休 ；在生活中 ，邻里之间也可能
会因为一些琐事而闹得不愉快 。若我们能传承六尺巷的宽容精神 ，多一份理解 ，少一份计较 ，那么
这些矛盾或许就能迎刃而解，我们的生活也会更加和谐美好。
传承六尺巷 “和为贵 ” 的理念 ，似一条坚韧的纽带 ，将现代社会中人们的心紧密相连 ，促进
人际关系和谐 。六尺巷的故事 ，不仅仅是一个简单的邻里纠纷解决案例 ，更是一种 “和为贵 ” 文
化的生动体现 。在当今这个快节奏的社会中 ，人们似乎越来越忙碌 ，人与人之间的关系也变得越来
越冷漠 。然而 ，六尺巷所传达的 “和为贵 ” 理念 ，却如同一股温暖的春风 ，能吹散人与人之间的
隔阂 。当我们在公交车上主动为他人让座 ，当我们在与他人意见不合时能心平气和地沟通 ，当我们
在面对他人的错误时能给予宽容与谅解，我们就是在践行 “和为贵 ” 的理念。这种理念的传承 ，
能让我们的社会更加温暖，让人与人之间的关系更加紧密。
让六尺巷的故事在现代社会中传颂 ，像播撒温暖的种子 ，在每个人心中生根发芽 ，传承传统美
德的力量 。通过这次研学旅行 ，我深刻地认识到 ，六尺巷的故事不应仅仅停留在历史的书本中 ，更
应走进我们每个人的生活 。我们可以将这个故事讲给身边的人听 ，让更多的人了解六尺巷的文化内
涵；我们可以在日常生活中以实际行动践行宽容礼让的美德 ，为社会传递正能量 。当六尺巷的故事
在社会中广泛传颂，当宽容礼让的美德在每个人心中生根发芽，我们的社会必将变得更加美好。
研学六尺巷 ，让我收获了满满的感动与思考 。它不仅让我领略到了传统文化的魅力 ，更让我明
白了宽容礼让、 “和为贵 ” 等传统美德在现代社会的重要价值。我愿将这份感悟铭记于心，在未
来的生活中 ，以六尺巷的精神为指引 ，努力成为一个宽容 、友善 、懂得礼让的人 ，为构建和谐美好
的社会贡献自己的一份力量。
【热点 三、个体与时代的命运共振 】“危机应对与人类命运共同体 ”、“入局、破
局、新局 ” 、“中国探月开启新篇章 ”
个体与时代的命运共振子话题 —— 危机应对与人类命运共同体 学科 网（北 京）股 份有限 公司
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考点方向：全球公共卫生治理、气候变化应对、文明冲突化解。
解析： 培养国际视野，强调合作共赢思维与大国责任担当，这就需要青年好好理解我们的历史，
我们的过往，我们的前辈面对问题是如何解决问题的，我们又是如何一步步赶上来，担起来的，这里
是溯源，其实也是展望，给未来更多可能，给现实更多方法，考察考生思考问题，解决问题的能力，
这需要青年有更开阔是心胸，更大胆的设想。
原创押卷试题 ：阅读下面的材料，根据要求写作。
2024 年，国际舞台波谲云诡 ，复杂多元与动荡变革交织 。地缘冲突持续延宕升级 ，巴以冲突已超
400 天，加沙地带深陷战火，生灵涂炭；俄乌冲突超 1000 天，冲突范围不断扩大，外溢风险加剧。
与此同时 ，贸易保护主义 、经济民粹主义等思潮抬头 ，部分国家热衷于 “脱钩断链 ”，企图筑起经济
壁垒，阻碍全球经济的正常流通与合作。在全球南方，众多国家正加速崛起，它们秉持独立自主，积
极推动联合自强，在国际事务中发挥着越来越重要的作用。
面对这一复杂局势，中国始终坚定地站在和平与发展的一边。在巴以冲突中，中国推动安理会通
过首份加沙停火决议，积极促成巴勒斯坦各派别和解对话并签署《北京宣言》，持续为加沙提供人道
主义援助。在全球发展领域，中国积极推动与发展中国家的合作，深化 “一带一路 ” 倡议，为沿线
国家带去发展机遇。在二十国集团领导人里约峰会上，中国阐释全球治理观，围绕经济、金融、贸易
等维度提出完善全球治理的理念主张，为推动人类命运共同体建设注入动力。
身处这样的时代浪潮中，培养国际视野，树立合作共赢思维，担当大国责任，已成为时代赋予青
年一代的使命。
作为新时代的青年，你读了上述材料有什么感悟和思考？ 请结合材料写一篇文章。
要求：选准角度，确定立意，明确文体，自拟标题；不要套作，不得抄袭；不得泄露个人信息；
不少于 800 字。
文题分析
材料开篇描绘了 2024 年复杂严峻的国际形势 。地缘冲突方面 ，巴以冲突超 400 天，加沙地区民众
饱受战火摧残 ，俄乌冲突超 1000 天且范围扩大 、风险外溢 ，严重威胁地区与全球和平稳定 。经济领域 ，
贸易保护主义、经济民粹主义抬头，部分国家推行 “脱钩断链 ”，破坏全球经济合作秩序，阻碍正常
经济流通。与此同时，全球南方国家崛起，秉持独立自主原则，积极联合自强，在国际事务中影响力
渐增，为国际格局带来新变化。
面对复杂局势，中国坚定站在和平与发展一边。在巴以冲突中，积极推动安理会通过加沙停火决
议，促成巴勒斯坦派别和解并签署《北京宣言》，持续提供人道主义援助，展现大国担当与维护和平 学科 网（北 京）股 份有限 公司
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的决心。在全球发展领域，深化 “一带一路 ”倡议，与发展中国家合作，为沿线国家创造发展机遇，
促进共同繁荣。在二十国集团领导人里约峰会上，阐释全球治理观，从经济、金融、贸易等维度提出
完善全球治理主张，推动人类命运共同体建设。
题目要求考生以新时代青年身份，结合材料，谈谈对当下国际局势的感悟与思考。重点在于围绕
“培养国际视野，树立合作共赢思维，担当大国责任 ”，思考青年在这一时代背景下的使命与担当，
强调青年应如何积极应对国际挑战，为世界和平与发展贡献力量。
立意可从多个角度展开。可阐述青年培养国际视野的重要性，分析如何在复杂国际局势中拓宽视
野，了解世界动态；可探讨树立合作共赢思维的必要性，结合国际合作案例，说明青年如何在国际交
往中践行这一思维；可强调担当大国责任，突出青年在推动中国国际影响力提升、促进全球和平发展
中的责任与行动，如参与国际公益活动、传播中国理念等。
写成议论文较为合适。运用清晰的论点、丰富的论据（如具体事例、数据、名言等）和多样的论
证方法（如举例论证、对比论证、因果论证等），深入阐述青年在国际局势中的责任与担当，使文章
逻辑严谨、论证有力。
写作素材拓展
一、全球挑战相关素材
环境领域 ：全球气候变暖导致极端天气频发， 2024 年，多地遭遇破纪录的高温热浪，美国夏威夷
毛伊岛的野火肆虐，造成重大人员伤亡和财产损失，正如恩格斯所言： “我们不要过分陶醉于我们人
类对自然界的胜利。对于每一次这样的胜利，自然界都对我们进行报复。 ”这凸显了各国在应对气候
变化上加强合作的紧迫性，而部分国家却在环保政策上裹足不前，阻碍全球统一行动。
公共卫生领域： 全球仍面临新的传染病威胁，如猴痘病毒在一些国家零星爆发。疫苗分配不均问
题依旧存在，发展中国家在获取高质量医疗资源和疫苗研发合作上困难重重，单边主义导致的医疗资
源封锁，严重影响全球公共卫生安全网的构建。
经济领域： 贸易保护主义使得全球产业链供应链受阻，一些发达国家对新兴技术产品实施出口管
制，限制关键技术流向发展中国家，例如对芯片等高科技产品的限售，不仅破坏了全球市场的公平竞
争，也阻碍了全球科技进步和经济协同发展。
二、中国行动相关素材
外交斡旋： 在俄乌冲突中，中国一直秉持客观公正立场，积极劝和促谈，提出 “中国方案 ”，呼
吁尊重各国主权和领土完整，为推动和平解决冲突贡献智慧，展现大国担当。 学科 网（北 京）股 份有限 公司
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国际援助： 中国积极参与联合国人道主义救援行动，为非洲、中东等地受战乱和贫困影响的民众
提供粮食、医疗物资等援助，帮助当地改善民生，稳定社会秩序。
科技创新合作： 中国在 5G 通信、高铁等领域技术领先，通过 “一带一路 ”
与沿线国家开展技术合作，助力建设智能交通系统、数字经济产业园等，带动当地科技水平提升，促
进共同发展。
三、国际合作成功案例素材
国际空间站项目： 由美国、俄罗斯、欧洲、日本和加拿大等国家和地区合作建设，各国科学家在
太空探索、生命科学等领域共享资源、共同研究，取得了一系列重大科研成果，证明了国际合作在攻
克复杂科学难题上的巨大优势。
全球疫苗免疫联盟 ：旨在促进疫苗在全球范围内的公平分配，联合政府、企业、慈善组织等多方
力量，为发展中国家儿童提供疫苗接种，有效降低了儿童死亡率，提升全球公共卫生水平。
四、青年相关素材
青年国际志愿者行动： 许多中国青年积极参与海外志愿服务项目，如参与非洲的教育援助项目，
为当地儿童带去知识；投身东南亚的环保志愿活动，助力保护当地生态环境，在国际舞台上展现中国
青年的担当与风采。
青年国际学术交流： 越来越多的中国青年学者在国际学术会议上崭露头角，分享前沿研究成果，
与国际同行开展合作研究，推动不同学科领域的国际交流与发展，拓宽学术视野，为解决全球性问题
提供新思路。
范文示例：
怀国际视野，担时代重任
在时代的宏大叙事中 ，2024 年的国际舞台宛如一幅波澜壮阔却又充满纷争的画卷 。地缘冲突的烽
火持续燃烧 ，巴以冲突的硝烟已弥漫超 400 天，加沙大地沦为炼狱 ，无辜生灵惨遭涂炭 ；俄乌冲突历
经超 1000 天的漫长岁月，冲突范围不断扩张，外溢风险如高悬之剑，威胁着全球的和平与稳定。与
此同时，贸易保护主义与经济民粹主义等逆流涌动，部分国家妄图以 “脱钩断链 ” 筑起自我封闭的
堡垒，肆意破坏全球经济的正常流通与合作秩序，使世界经济的发展陷入阴霾。然而，在这重重困境
之中，全球南方国家如熠熠星辰，加速崛起，它们坚守独立自主的信念，携手推动联合自强，在国际
事务的舞台上绽放出愈发耀眼的光芒。
面对如此波谲云诡的复杂局势，中国宛如一座巍峨屹立的灯塔，始终坚定不移地锚定和平与发展
的航向。在巴以冲突的黑暗深渊中，中国挺身而出，凭借着坚定的外交努力与不懈的斡旋，推动安理 学科 网（北 京）股 份有限 公司
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会通过首份加沙停火决议，为这片饱经战火蹂躏的土地带来了一丝和平的曙光。积极促成巴勒斯坦各
派别和解对话并签署《北京宣言》，为巴以和平进程注入了强大的中国动力。同时，持续不断地为加
沙提供人道主义援助，将温暖与希望传递给每一个身处苦难的生命，彰显了大国的无疆大爱。在全球
发展的广阔领域，中国以深化 “一带一路 ” 倡议为有力抓手，与沿线国家紧密携手，共同绘制经济
繁荣的宏伟蓝图，为世界经济的复苏与发展注入源源不断的活力。在二十国集团领导人里约峰会上，
中国高瞻远瞩，阐释全球治理观，围绕经济、金融、贸易等关键维度提出完善全球治理的理念主张，
为推动人类命运共同体建设添砖加瓦，奏响了合作共赢的时代强音。
作为新时代的青年，我们置身于这风云变幻的时代浪潮之中，宛如航行在茫茫大海上的船只，培
养国际视野、树立合作共赢思维、担当大国责任，已成为我们义不容辞的使命。 “青年兴则国家兴，
青年强则国家强 。” 我们肩负着时代赋予的重任 ，应如展翅高飞的雄鹰 ，拥有广阔的国际视野 ，洞察
世界的风云变幻，不被狭隘的地域观念所束缚。以开放包容的心态，积极拥抱多元文化，汲取世界各
国的智慧与精华 ，为我所用 。在面对全球性挑战时 ，我们要深刻领悟 “孤举者难起 ，众行者易趋 ” 的
道理，摒弃狭隘的个人主义与单边思维，树立合作共赢的理念。如同全球疫苗免疫联盟（ Gavi ）汇聚
多方力量 ，共同为全球公共卫生事业贡献力量一般 ，我们应积极投身国际合作 ，与各国青年携手共进 ，
在合作中实现共同发展，在共赢中创造美好未来。
在国际舞台上，我们要以中国的大国担当为榜样，勇挑重担，展现新时代中国青年的风采。中国
青年志愿者在海外的教育援助、环保行动中，用汗水与爱心书写着中国青年的责任与担当；青年学者
在国际学术交流中，以扎实的学识与创新的思维，为全球知识创新贡献智慧。我们要以他们为楷模，
在各自的领域发光发热，为推动构建人类命运共同体贡献自己的青春力量。
“自信人生二百年 ，会当水击三千里 。” 新时代的青年们 ，让我们心怀国际视野 ，肩负大国责任 ，
以合作共赢为舟，以创新进取为桨，在时代的洪流中破浪前行，书写属于我们这一代的壮丽篇章，为
世界的和平与发展铸就新的辉煌！
个体与时代的命运共振子话题 —— 入局、破局、新局
题目 ： 阅读 下面的 材料， 根据要 求写作 。
在围棋比赛中 ，棋手们常说 “入局 、破局 、新局 ”。入局 ，是指进入比赛的开局阶段 ，布局谋
篇，奠定基础 ；破局 ，是指打破僵局 ，扭转局势 ，寻找突破 ；新局 ，则是在破局的基础上 ，开创全
新的局面 ，赢得胜利 。这不仅适用于围棋比赛 ，也适用于人生的各个阶段 。无论是学习 、工作还是
生活，我们都需要在 “入局 ”中找准方向，在 “破局 ”中突破困境，在 “新局 ”中创造未来。 学科 网（北 京）股 份有限 公司
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请结合材料 ，以“入局 、破局 、新局 ”为主题 ，写一篇文章 ，谈谈你的感悟与思考 。要求 ：选
准角度 ，确定立意 ，明确文体 ，自拟标题 ；不要套作 ，不得抄袭 ；不得泄露个人信息 ；不少于 80 0
字。
题材分析
该题材以围棋术语 “入局、破局、新局 ” 为切入点，延伸至人生的各个阶段，包括学习、工作和生
活等领域。这要求我们从围棋的策略理念拓展到对人生发展的思考，从个体成长、职业发展、社会进步等
多个维度去剖析在不同阶段如何运用 “入局 、破局 、新局 ” 的智慧 ，题材具有较强的启发性与现实意义 。
审题立意
关键信息抓取 ：“入局 ”“破局 ”“新局 ” 以及它们在人生不同阶段的对应表现 ，如找准方向 、突
破困境、创造未来等是关键要点。
立意方向：
强调阶段性发展 ：突出在人生不同阶段 ，依次实现 “入局 、破局 、新局 ” 的重要性 ，如 “依循入局
破局新局，奏响人生奋进乐章 ”。
聚焦破局关键 ：着重阐述在人生困境中 ，如何通过有效破局 ，实现从 “入局 ” 到 “新局 ” 的跨越 ，
如 “破局为翼，飞越人生困境，开启新局华章 ”。
整体辩证思考 ：全面分析 “入局 、破局 、新局 ” 之间的相互关系 ，以及它们对人生发展的综合影响 ，
如 “权衡入局破局新局，书写精彩人生篇章 ”。
素材选用
入局素材：一位大学生在选择专业时，通过自我评估和市场调研，精准选择了有发展前景且符合自
身兴趣的专业，为未来职业发展奠定基础。
破局素材：某企业在市场竞争中面临产品滞销困境，通过创新营销策略，如开展线上直播带货等方
式，成功打破销售僵局。
新局素材：科学家在攻克某一科研难题后，基于新的研究成果，开拓了全新的研究领域，取得一系
列重大突破。
范文示例：
弈人生棋局，书奋进华章
“博观而约取 ，厚积而薄发 。” 人生恰似一场扣人心弦的围棋对弈 ，“入局 、破局 、新局 ”，三局相
连，铺就我们逐梦前行的道路。每一局都暗藏玄机，每一步都关乎命运，在这黑白交错间，演绎着属于我
们的成长传奇。 学科 网（北 京）股 份有限 公司
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入局，是梦想启航的号角，是踏上征程的坚定步伐。它如同春日破土而出的嫩芽，带着对世界的好奇
与憧憬，勇敢地探出头来。恰似班超投笔从戎，不甘于平淡的书斋生活，毅然投身军旅，在保家卫国的棋
局中，落下了自己的第一子，开启了一段波澜壮阔的人生之旅；又仿若鉴真东渡，怀着传播佛法的宏愿，
不顾海上的惊涛骇浪 ，毅然踏上那充满未知的渡海之路 ，在文化传播的棋局中 ，布下了关键布局 ；而我们 ，
在青春的赛道上，怀揣着对未来的无限期许，走进校园，走进职场，在人生的棋盘上，精心谋划着属于自
己的开局。入局，是勇气的象征，是对未知的探索，它让我们在人生的舞台上，拥有了展示自我的机会。
但入局之后，并非一帆风顺，僵局常常不期而至，此时，破局便成为了关键。破局，是困境中的绝地
反击，是黑暗里的破晓之光。它像一只振翅高飞的雄鹰，冲破云层的束缚，展现出无畏的力量。韩信背水
一战 ，在绝境中激发士兵的斗志 ，以少胜多 ，打破了敌军的围困 ，书写了军事史上的传奇 ；海伦 ・凯勒 ，在
无声无光的世界里，凭借着顽强的毅力和对知识的渴望，学会了阅读和说话，打破了命运为她设置的重重
障碍，成为了无数人心中的励志楷模。当我们在学习中被难题困扰，在工作中陷入迷茫，在生活中遭遇挫
折时，我们要像这些破局者一样，不向困难低头，不被挫折打倒，用智慧和勇气，开辟出一条属于自己的
成功之路。
破局之后，便是新局的诞生。新局，是胜利的曙光，是成长的升华。它宛如夏夜璀璨的星空，充满了
无限的可能与希望。袁隆平一生致力于杂交水稻研究，突破传统农业技术的瓶颈，开创了高产水稻的新时
代，让无数人免受饥饿之苦；中国航天人，不断挑战极限，突破技术封锁，从 “神舟 ” 飞天到 “嫦娥 ”
揽月，从 “天问 ” 探火到 “羲和 ” 逐日，开创了中国航天事业的崭新局面，让中国在宇宙探索的道路
上迈出了坚实的步伐 。我们在人生的道路上 ，每一次成功破局后 ，都要抓住机遇 ，乘势而上 ，以创新为笔 ，
以奋斗为墨，书写属于自己的新辉煌。
“路漫漫其修远兮 ，吾将上下而求索 。” 人生的棋局没有终点 ，每一局都是新的起点 。让我们以无畏
的勇气入局，以坚韧的毅力破局，以创新的精神开启新局。弈人生棋局，书奋进华章，在人生的舞台上，
绽放出最耀眼的光芒。
个体与时代的命运共振子话题 —— 中国探月，开辟人类探索新篇章
题目： 阅读下面的材料，根据要求写作。
2024 年，中国航天领域取得了重大突破 ，嫦娥七号成功发射 ，开启了中国探月工程的新篇章 。中国探
月工程自启动以来，始终坚持自主创新、协同攻关的发展模式。从嫦娥一号到嫦娥七号，中国航天人攻克
了一个又一个技术难题，实现了从无人探测到载人登月的跨越。
请结合材料，以 “自主创新 ”为主题，写一篇文章，谈谈你对这一主题的理解和感悟。要求：选准角
度，确定立意，明确文体，自拟标题；不要套作，不得抄袭；不得泄露个人信息；不少于 800 字。 学科 网（北 京）股 份有限 公司
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题材分析
这一题材聚焦中国航天领域的重大成果，核心在于中国探月工程中 “自主创新、协同攻关 ” 的发展
模式。它涉及科技进步、国家实力提升以及团队协作等多个层面。从题材特性看，需深入探讨自主创新在
航天发展中的关键作用，并将其拓展至更广泛的社会、经济领域，思考其对国家和民族发展的深远意义。
此题材具有很强的时代性与现实意义，能激发对科技创新重要性的深入思考。
审题立意
关键信息抓取： “自主创新 ”“ 协同攻关 ”“ 中国探月工程 ”“ 技术难题攻克 ”“ 从无人到载人的
跨越 ” 等为关键要点。
立意方向：
强调自主创新核心地位 ：突出自主创新在国家科技发展 、综合国力增强方面的核心价值 ，如 “自主创
新，筑牢国家科技发展基石 ”。
阐述创新与协同关系：着重论述自主创新与协同攻关相互促进的关系，以及这种关系对重大科技项目
突破的推动作用，如 “自主创新为笔，协同攻关为墨，绘就科技华章 ”。
延伸创新价值意义 ：以航天自主创新为切入点 ，延伸到各行业及社会发展对自主创新的需求 ，如 “借
航天自主创新东风，推动社会全面创新发展 ”。
素材选用
航天领域素材 ：除嫦娥七号 ，可列举天问一号火星探测器在轨道设计 、着陆技术等方面的自主创新 ；
北斗卫星导航系统建设中自主研发的原子钟等关键技术。
其他领域素材：华为在 5G 通信技术上的自主创新，打破国外技术垄断；中国高铁通过自主创新，
实现高速、安全、舒适的技术领先，走向世界。
范文示例：
自主创新，逐梦九天
“大鹏一日同风起 ，扶摇直上九万里 。” 在中国探月工程的漫漫征途中 ，自主创新就是那强劲的东风 ，
助力中国航天大鹏展翅，翱翔于浩瀚宇宙。从嫦娥一号开启中国深空探测的新纪元，到嫦娥七号实现从无
人探测到载人登月的伟大跨越，中国航天人凭借自主创新，攻克一个又一个技术难关，让中华民族的飞天
梦想照进现实，也让我们深刻领悟到自主创新的磅礴力量。
自主创新，是中国探月工程的核心密码，宛如一颗璀璨的明珠，在历史的长河中熠熠生辉。面对太空
探索的重重挑战，中国航天人没有选择依赖他人，而是凭借着一股不服输的劲头，毅然踏上自主创新的征
程。在研制嫦娥号系列探测器时，他们就像一群无畏的勇士，向着未知领域发起冲锋。轨道设计、月球软 学科 网（北 京）股 份有限 公司
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着陆 、深空通信 …… 这些复杂的技术难题 ，在他们眼中不是阻碍 ，而是挑战自我的机遇 。他们日夜奋战在
科研一线，查阅海量资料，进行无数次试验，只为突破技术瓶颈。就如同古代的张衡，在没有任何现代科
技辅助的情况下，凭借自己的智慧和不懈努力，发明了地动仪，领先世界千余年。航天人以张衡为榜样，
在自主创新的道路上，用智慧和汗水铸就了中国探月的辉煌。
自主创新，是中国航天人不断突破自我的动力源泉，宛如一首激昂的战歌，激励着他们向更高的目标
迈进。在探月工程中，每一次技术突破都是一次自我超越。嫦娥三号实现了中国首次地外天体软着陆，嫦
娥四号成功着陆月球背面，嫦娥七号完成载人登月的壮举，每一次成功都离不开自主创新的支撑。航天人
在创新的道路上 ，不断挑战极限 ，探索未知 。他们就像勇敢的登山者 ，每攀登一步 ，都能看到更美的风景 ，
但他们从不满足，总是向着更高的山峰进发。正是这种不断突破自我的精神，让中国探月工程一次次创造
奇迹，在世界航天领域留下了浓墨重彩的一笔。
自主创新，也是中国航天事业引领世界的底气所在，宛如一面飘扬的旗帜，彰显着中国科技的实力。
中国探月工程的一系列成就，让世界看到了中国自主创新的能力。我们在航天领域的创新成果，不仅推动
了国内科技的进步 ，也为全球航天事业的发展贡献了中国智慧和中国方案 。就像古代的四大发明 ，造纸术 、
印刷术、火药、指南针，对世界文明的发展产生了深远影响。如今，中国航天的自主创新成果，也在为人
类探索宇宙开辟新的道路。我们凭借自主创新，在世界航天舞台上赢得了尊重和话语权，让中国成为航天
强国的有力竞争者。
“千磨万击还坚劲 ，任尔东西南北风 。” 在未来的航天征程中 ，中国航天人必将继续秉持自主创新的
精神，勇往直前。我们期待着更多的创新成果，期待着中国航天在宇宙探索中创造更多的辉煌。让我们以
航天人为榜样，在各自的领域中勇于创新，为实现中华民族伟大复兴的中国梦贡献自己的力量。让自主创
新的光芒，照亮我们前行的道路，引领我们走向更加美好的未来。
【热点 四、生态文明与可持续发展 】“生态智慧 ”、“和合理念 ” 、“劳动教育与
实践育人 ”
生态文明与可持续发展子话题 —— 生态智慧
考点方向：双碳战略实践、生物多样性保护、生态美学价值。
解析： 需结合中国古代生态智慧，构建人与自然生命共同体认知，这是近几年我们的优势发展，
也是我们领先的一点，这里需要我们注意更多，对于青年人来说，只有抬眼望世界，我们会发现更多
不一样，也激励我们更多的发展。
原创押卷试题 阅读下面的材料，根据要求写作。 学科 网（北 京）股 份有限 公司
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中国古代蕴含着丰富的生态智慧 。《论语 》中记载 “子钓而不纲 ，弋不射宿 ”，体现了孔子对自
然资源取用有度的理念 ；道家倡导 “道法自然 ”，认为人应顺应自然规律 ，与自然和谐共生 ；古代的
“月令 ” 制度，对不同时节的农事活动、资源利用等都有明确规定，以确保生态平衡。
如今，构建人与自然生命共同体已成为全球共识。从 “绿水青山就是金山银山 ” 的科学论断，
到大规模的国土绿化行动，再到对生物多样性的保护，我们正积极探索可持续发展之路。
结合上述材料，谈谈中国古代生态智慧对构建人与自然生命共同体的启示。请写一篇文章，阐述你的
观点和思考。
要求：选准角度，确定立意，明确文体，自拟标题；不要套作，不得抄袭；不得泄露个人信息；不少
于 800 字。
文题分析
这道作文题聚焦传统文化与现代生态理念的融合 。材料开篇列举了古代生态智慧 ：《论语 》中孔子 “子
钓而不纲，弋不射宿 ”，体现了对自然资源不过度索取，注重可持续利用。道家 “道法自然 ”理念强调人
类行为应契合自然规律，维持和谐共生关系。古代 “月令 ”制度则从制度层面规范不同时节农事与资源利
用，保障生态平衡，这些都是古人对自然深刻认知与尊重的体现。接着又阐述当下构建人与自然生命共同
体已成为全球共识，并列举了 “绿水青山就是金山银山 ”论断、国土绿化行动、生物多样性保护等具体实
践，展现现代社会在生态保护方面的积极探索与努力。
题目明确要求考生结合材料，探讨中国古代生态智慧对构建人与自然生命共同体的启示。这意味着写
作不能脱离材料，要在古今对比与联系中挖掘深层内涵。可从古代生态智慧在现代生态建设中的借鉴意义
入手 ，如古人对自然规律的遵循如何助力当下生态保护 ；也可分析古代理念在现代社会面临的挑战与发展 ，
思考如何在新时代更好地传承与创新古代生态智慧。
写成议论文较为合适，通过清晰的论点、充分的论据和严谨的论证，阐述古代生态智慧与现代生态建
设的关系。论据可选取古代生态保护案例，如古代对山林川泽的封禁制度；也可列举现代生态建设成功案
例，如塞罕坝植树造林、浙江 “千万工程 ”打造美丽乡村等。
写作素材拓展：
古代生态智慧素材
儒家经典论述
《孟子 ・梁惠王上》中提到
“不违农时 ，谷不可胜食也 ；数罟不入洿池 ，鱼鳖不可胜食也 ；斧斤以时入山林 ，材木不可胜用也 ”。学科 网（北 京）股 份有限 公司
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这清晰地表明儒家倡导在农业生产、渔业捕捞以及林业采伐等方面遵循自然规律，按照时节进行合理
活动，以实现资源的可持续利用，保障生态系统的平衡与稳定。
《礼记 ・王制》记载
“獭祭鱼 ，然后虞人入泽梁 ；豺祭兽 ，然后田猎 ；鸠化为鹰 ，然后设罻罗 ；草木零落 ，然后入山林 ”。
详细说明了古人依据动物的活动规律和季节变化来安排相应的生产活动，展现出对自然生态的敏锐观
察和尊重。
道家思想内涵
老子在《道德经》中说
“人法地，地法天，天法道，道法自然 ”。深刻阐述了人类应遵循自然法则，顺应自然规律，不可违
背自然而肆意妄为，强调人与自然的和谐统一是生存和发展的基础。
庄子主张
“天地与我并生，而万物与我为一 ”。这种思想体现了道家对万物平等的认知，认为人类与自然万物
同属一个整体，应摒弃人类中心主义，珍视每一个生命和自然元素。
古代生态保护制度与实践
西周时期就设立了专门负责山林川泽管理的官职 “山虞 ”和“泽虞 ”。山虞掌管山林之政令，
负责制定保护山林资源的政策和法规；泽虞负责管理川泽、湖沼等水域资源，监督人们合理利用水资
源，防止过度捕捞和污染。
秦朝的《田律》是中国历史上最早的一部环境保护法律，其中规定
“春二月，毋敢伐材木山林及雍（壅）堤水。不夏月，毋敢夜草为灰，取生荔、麛（卵）鷇，毋毒鱼
鳖，置穽罔（网），到七月而纵之 ”。对春季山林采伐、夏季用火、生物繁殖期的保护等方面都有明
确细致的规定，以维护生态平衡。
古代名人践行生态理念事例
苏轼在杭州任知州时，疏浚西湖，清理湖底淤泥，并用挖出的淤泥修筑了
“苏堤 ”。同时，他还在西湖中种植菱角，既防止了水草滋生，又增加了百姓收入，实现了生态治理
与民生改善的双赢。
白居易在担任苏州刺史期间 ，非常重视城市绿化 ，他亲自倡导并参与植树活动 ，在苏州城内外种
植了大量柳树、桃李等树木，不仅美化了环境，还起到了保持水土、调节气候的作用。
现代构建人与自然生命共同体素材
政策法规推动 学科 网（北 京）股 份有限 公司
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中国出台了 《中华人民共和国环境保护法 》《中华人民共和国野生动物保护法 》《中华人民共和
国森林法》等一系列法律法规，从法律层面为生态保护提供了坚实保障，严厉打击破坏生态环境的违
法行为。
各地政府积极推行河长制 、湖长制 ，明确各级河长 、湖长对河流 、湖泊生态保护的责任 ，加强对
水域生态环境的治理和监管，促进水资源的合理利用和保护。
生态建设工程实例
塞罕坝机械林场的建设堪称生态修复的奇迹。从 “黄沙遮天日，飞鸟无栖树 ”
的荒漠沙地到 “林海苍翠连绵、河水清澈甘甜 ”的生态宝地，经过三代人近 60 年的不懈努力，塞
罕坝人通过植树造林，成功修复了生态系统，创造了荒原变林海的人间奇迹，为全球生态治理提供了
宝贵经验。
浙江
“千万工程 ”以“千村示范、万村整治 ”为抓手，持续推进农村人居环境整治和生态建设。通过垃圾
处理、污水治理、村庄绿化等一系列措施，浙江的乡村面貌焕然一新，实现了从 “脏乱穷散 ”到“洁
净美富 ”的华丽转变，成为践行 “绿水青山就是金山银山 ”理念的生动实践。
科技创新助力生态保护
利用卫星遥感技术 、无人机监测等手段 ，对森林 、湿地 、野生动物栖息地等生态系统进行实时监
测，及时掌握生态环境变化情况，为生态保护决策提供科学依据。例如，通过卫星遥感可以准确监测
森林面积的增减、森林火灾的发生范围等。
研发和推广新能源技术 ，如太阳能 、风能 、水能 、生物能等 ，减少对传统化石能源的依赖 ，降低
碳排放，缓解气候变化对生态环境的影响。目前，我国在太阳能光伏产业和风力发电领域已取得显著
成就，成为全球新能源发展的重要力量。
民间环保行动与国际合作
众多环保志愿者组织积极开展各类环保活动 ，如垃圾分类宣传 、河流湖泊清理 、野生动物保护等 。
他们通过实际行动，向公众普及环保知识，提高人们的环保意识，带动更多人参与到生态保护中来。
在国际合作方面 ，中国积极参与全球气候治理 ，与世界各国共同应对气候变化挑战 。例如 ，中国
在《巴黎协定》的达成和实施过程中发挥了重要作用，承诺碳减排目标，并积极推动绿色
“一带一路 ”建设，与沿线国家分享生态保护经验和技术，共同促进全球生态环境的改善。
范文示例
承古智之脉，铸共生新章 学科 网（北 京）股 份有限 公司
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在历史的长河中，中国古代生态智慧如熠熠星辰，照亮了人类与自然相处的漫漫长路。从儒家对
自然规律的尊崇，到道家对万物平等的哲思；从古代生态保护制度的建立，到仁人志士的躬身践行，
皆为当下构建人与自然生命共同体提供了深厚滋养与无尽启示。
儒家经典中， “不违农时，谷不可胜食也；数罟不入洿池，鱼鳖不可胜食也；斧斤以时入山林，
材木不可胜用也 ”，寥寥数语 ，却道尽了遵循自然规律进行生产活动的要义 。其倡导的在农业 、渔业 、
林业等领域依时节而动，是对资源可持续利用的深刻洞察。《礼记 ・王制》所记 “獭祭鱼，然后虞人
入泽梁；豺祭兽，然后田猎 ”，更是古人对自然生态敏锐观察后的智慧抉择，依据动物活动规律与季
节变化安排生产，尽显对自然的敬畏之心。这种取之有时、用之有节的理念，宛如一盏明灯，为现代
社会合理开发利用自然资源指引方向。在当下，过度捕捞导致渔业资源枯竭、乱砍滥伐引发水土流失
等问题频发，重拾儒家这一生态智慧，有助于我们科学规划资源开采，实现经济发展与生态保护的良
性互动。
道家思想中 ，“人法地 ，地法天 ，天法道 ，道法自然 ” 的深刻哲理 ，将人与自然的关系提升到了
形而上的高度 。它警示人类不可肆意妄为 ，而应顺应自然法则 ，寻求与自然的和谐共生 。庄子 “天地
与我并生 ，而万物与我为一 ” 的主张 ，更是打破了人类中心主义的狭隘视角 ，让我们认识到人类与自
然万物同属一个整体，皆有其存在价值。在现代社会，当人类以主宰者姿态对待自然，引发生态危机
时，道家思想如黄钟大吕，振聋发聩。它提醒我们尊重每一个生命、每一处自然元素，以谦卑之心融
入自然，而非凌驾其上。
古代生态保护制度与实践同样值得我们借鉴。西周设立 “山虞 ”“ 泽虞 ” 管理山林川泽，秦朝
制定《田律》规范资源利用，从官职到律法，构建起一套较为完备的生态保护体系。这些制度虽历经
岁月变迁，但其中蕴含的对生态平衡的维护意识、对资源合理利用的管控思维，在现代依然具有强大
生命力。如今，我国制定一系列环境保护法律法规，推行河长制、湖长制，正是对古代生态保护制度
的传承与创新，以制度之力守护绿水青山。
苏轼疏浚西湖、白居易植树绿化，古代名人的生态实践为我们树立了榜样。苏轼以淤泥筑苏堤、
植菱角，既改善了西湖生态，又惠及民生；白居易亲自参与植树，美化苏州城环境。他们的行动诠释
了生态治理与民生改善并非矛盾对立，而是相辅相成。这启示我们在现代生态建设中，要注重生态效
益与社会效益的统一。塞罕坝机械林场的建设者们，历经三代人努力，将荒漠变为林海，不仅修复了
生态系统，还带动了当地经济发展，实现了生态与经济的双赢，与古代先贤的实践理念一脉相承。
在科技创新助力生态保护的当下，古代生态智慧依然具有重要价值。卫星遥感、无人机监测等技
术虽为现代科技产物，但我们可从古人对自然的细致观察中汲取灵感，优化监测手段，更好地掌握生 学科 网（北 京）股 份有限 公司
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态变化。研发新能源技术，减少对传统化石能源依赖，与道家顺应自然、减少过度开发的理念不谋而
合。民间环保行动蓬勃开展，国际合作积极推进，在这一过程中，古代生态智慧可成为凝聚共识的精
神纽带，让更多人认识到人与自然和谐共生的重要性。
从古代生态智慧到现代构建人与自然生命共同体的实践，是一场跨越时空的传承与发展。我们应
承古智之脉，汲取古人智慧精华，结合现代科技与理念，积极探索可持续发展之路，铸共生新章，让
人与自然和谐共生的美好愿景在新时代照进现实，为子孙后代留下一个美丽、宜居的地球家园。
生态文明与可持续发展子话题 —— “和合共生 ”理念
题目 ： 阅读 下面的 材料， 根据要 求写作 。
金砖国家领导人第十六次会晤正式提出了 “金砖伙伴国 ”的新概念 ，一国在获得金砖正式成员
资格之前，可先成为 “伙伴国 ”，参与金砖国家合作。这种合作模式体现了 “和合共生 ”的理念 ，
即通过包容 、合作 、共享的方式 ，实现共同发展 。金砖国家的合作不仅体现在经济领域 ，还涵盖了
文化、科技、教育等多个方面，为全球南方国家的发展提供了新的机遇和平台。
请结合材料 ，以“和合共生 ”为主题 ，写一篇文章 ，谈谈你对这种合作模式的理解和思考 。要
求：选准角度 ，确定立意 ，明确文体 ，自拟标题 ；不要套作 ，不得抄袭 ；不得泄露个人信息 ；不少
于 800 字。
题材分析
此题材聚焦于金砖国家合作模式中的 “金砖伙伴国 ” 新概念，核心在于 “和合共生 ” 理念在国际
合作中的体现。这一题材涉及国际关系、全球发展等宏观层面，既要求对金砖国家合作机制有一定了解，
又要深入理解 “和合共生 ” 这一理念的内涵，从政治、经济、文化等多维度去思考其在国际合作中的实
践与意义。
审题立意
关键信息抓取 ：“金砖伙伴国 ”“和合共生 ”“包容 、合作 、共享 ”“共同发展 ” 以及金砖国家在
多领域合作等是关键要点。
立意方向：
强调合作价值 ：突出 “和合共生 ”合作模式对金砖国家及全球南方国家发展的重要推动作用 ，如“‘和
合共生 ’，奏响国际合作发展强音 ”。
阐述理念内涵：深入剖析 “和合共生 ”理念在金砖合作中的具体体现，如包容不同发展水平国家、共
享发展成果等，立意可为 “以‘和合共生 ’之笔，绘国际合作多彩画卷 ”。学科 网（北 京）股 份有限 公司
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展望未来发展：基于当前合作模式，展望金砖国家及全球南方国家在 “和合共生 ”理念下的未来发展
前景，如 “秉持 ‘和合共生 ’，迈向全球合作新征程 ”。
素材选用
经济领域素材：例如金砖国家之间的贸易额持续增长，某国借助与其他金砖国家的合作，引进先进
技术设备，提升本国产业竞争力。
文化领域素材 ：金砖国家举办文化节 ，各国特色文化表演吸引大量观众 ，增进了民众间的相互了解 ；
各国互派留学生数量不断增加，促进文化交流融合。
科技领域素材：金砖国家联合开展科研项目，共同攻克某一技术难题，成果应用于多个国家的相关
产业，推动产业升级。
范文示例
以和合共生之姿，绘金砖合作宏图
在全球发展的宏大舞台上 ，金砖国家宛如一群耀眼的舞者 ，以独特的舞步 ，跳出了一曲激昂澎湃的 “和
合共生 ” 乐章 。金砖国家领导人第十六次会晤提出的 “金砖伙伴国 ” 新概念 ，恰似一颗璀璨的明珠 ，为
国际合作注入了全新活力，生动诠释了 “和合共生 ” 这一古老而又充满现代智慧的理念。
在经济的广袤海洋中 ，“和合共生 ” 宛如一座坚固的桥梁 ，连接着金砖国家及众多全球南方国家 。各
国凭借自身丰富的资源与独特的产业优势，在这座桥上相互往来。巴西的优质农产品，跨越重洋，走进中
国消费者的餐桌；中国的物美价廉的工业制成品，为南非的市场增添了丰富色彩。通过贸易合作，各国资
源得以优化配置，经济活力被充分激发。这不仅是商品的交换，更是发展机遇的共享，宛如一场经济的盛
宴，让各国在合作中收获满满。
于文化的缤纷花园里 ，“和合共生 ” 如同一位辛勤的园丁 ，精心培育着各国文化之花 。金砖国家的文
化节 ，便是这花园中最盛大的节日 。印度的神秘舞蹈 、俄罗斯的激昂交响乐 ，在这个舞台上尽情绽放光彩 。
各国的文化在这里相互交融，彼此欣赏。留学生们如同文化的使者，穿梭于各国之间，将不同的思想、观
念传播开来，促进了文化的包容与互鉴。文化的交流，让各国人民的心贴得更近，为更深层次的合作奠定
了坚实的情感基础。
在科技的前沿阵地 ，“和合共生 ” 化身为一把神奇的钥匙 ，开启了创新发展的大门 。金砖国家联合开
展的科研项目，如同一个个智慧的熔炉，将各国的科研力量汇聚在一起。在攻克难题的过程中，各国科学
家相互切磋，共享研究成果。某一先进的农业技术，从巴西的实验室走向印度的农田，助力粮食增产；某
一新型的信息技术，在中国研发后，迅速在俄罗斯得到应用，提升了产业效率。科技的合作，让金砖国家
在全球科技竞争中崭露头角，为各国的发展插上了腾飞的翅膀。 学科 网（北 京）股 份有限 公司
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“和合共生 ” 的金砖合作模式 ，是时代的选择 ，是发展的必然 。它如同一座灯塔 ，照亮了全球南方国
家前行的道路。让我们继续秉持这一理念，携手共进，在合作的道路上越走越远，共同绘制出一幅更加绚
丽多彩的国际合作宏图，迈向更加美好的未来。
生态文明与可持续发展子话题 —— 劳动教育与实践育人
考点方向：新时代劳动内涵、职业价值重构、工匠精神传承。
解析： 突破传统劳动认知，关注劳动创造与人的全面发展关系，这里面牵涉到了教育的根本，那就是
立德树人，什么德、什么人，这是值得考生深刻思考的问题，也牵涉青年未来的就业。
原创押卷试题 阅读下面的材料，根据要求写作。
在传统观念中，劳动往往被简单划分为体力劳动与脑力劳动，且体力劳动常被轻视，脑力劳动则备受
推崇 。然而 ，随着时代的发展 ，新兴劳动形态不断涌现 ，如数字化运营 、人工智能调试 、文化创意设计等 ，
这些劳动既融合了知识与技能，又强调创新与创造，模糊了传统劳动分类的界限。
教育的根本任务在于立德树人，培养全面发展的人。劳动教育作为教育体系的重要组成部分，对塑造
学生的价值观、培养创新精神和实践能力起着关键作用。但在现实中，部分学校和家庭对劳动教育重视不
足，学生缺乏对劳动的正确认知与体验。
新时代青年肩负着时代赋予的重任，面对这种情况，应如何突破传统劳动认知，理解劳动创造与人的
全面发展之间的紧密联系，在劳动中实现自我成长与价值？这是值得深入思考的问题。
请结合以上材料，写一篇文章，谈谈你的感悟与思考。
要求：选准角度，确定立意，明确文体，自拟标题；不要套作，不得抄袭；不得泄露个人信息；不少
于 800 字。
文题分析
这篇作文题聚焦于传统劳动认知的突破、劳动教育与青年发展，核心在于探讨劳动创造和人的全面发
展间的关系。材料开篇指出传统劳动认知中体力与脑力劳动的不合理区分，揭示其背后社会阶层分化与职
业发展不平衡的根源，强调这种认知对理解劳动本质的局限性。接着阐述新时代新兴劳动形态的涌现，它
们融合知识、技能与创新，打破传统劳动界限，体现劳动的多元魅力。随后提及劳动教育在培养全面发展
人才中的关键作用，以及当下学校和家庭对劳动教育的忽视现状。材料层层递进，从现象到本质，为考生
探讨劳动相关话题提供丰富视角。题目要求考生结合材料，谈谈感悟与思考。这意味着考生需深入分析传
统劳动认知弊端，阐述新时代劳动特点，探讨劳动教育的重要性，并明确新时代青年在其中的责任与行动
方向。 学科 网（北 京）股 份有限 公司
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立意可从多维度展开。如批判传统劳动认知偏见，树立正确劳动价值观；强调新兴劳动形态对个人和
社会发展的推动作用；论述劳动教育对塑造全面发展人才的不可或缺性；突出新时代青年突破传统、积极
参与劳动实践的使命担当等。从题目来看写成议论文较为合适。通过清晰的论点、充分的论据和严谨的论
证，深入阐述劳动相关话题。论据可选用历史故事、现实案例、名人名言、统计数据等，论证方法可采用
举例论证、对比论证、道理论证、引用论证等，使文章逻辑严密、内容丰富。
写作素材拓展
典型事例
大国工匠徐立平
徐立平为导弹固体燃料发动机的火药进行微整形，在极其危险的岗位上，凭借精湛的技艺和高度的专
注力，几十年如一日地进行着细致入微的体力与脑力结合的劳动。他的工作不仅需要强壮的体力支撑长时
间精细操作，更依赖深厚的专业知识与创新思维，不断优化操作方法，以确保发动机的精准性能，有力反
驳了传统劳动认知中体力与脑力劳动的简单划分。
一位普通的农村电商从业者 ，起初只是从事农产品搬运 、包装等看似简单的体力劳动 ，但随着业务发
展，他主动学习网络营销知识、数据分析技能，通过直播带货等新兴方式将家乡农产品推向全国市场。这
一过程中，体力劳动与脑力劳动深度融合，实现了个人职业发展的飞跃，也带动了当地经济发展，可作为
新时代新兴劳动形态打破传统认知的有力例证。
某中学开展 “校园农场 ”劳动教育项目 ，学生们亲自参与播种 、浇水 、施肥 、收获等农事活动 。在这
个过程中，学生们不仅学会了农业生产技能，还通过观察植物生长周期、研究土壤肥料等知识，将书本理
论与实践相结合，激发了对科学探索的兴趣，培养了团队协作精神与责任感，体现了劳动教育对学生全面
发展的积极作用。
范文示例：
破传统劳动认知之茧，绽全面发展之花
在人类社会的漫长演进中，劳动始终是推动历史车轮滚滚向前的强劲动力。然而，传统观念宛如一层
厚重的茧，将劳动简单区分为体力与脑力，且无端地为体力劳动贴上了被轻视的标签，而对脑力劳动过度
推崇。但时代的浪潮汹涌澎湃，正无情地冲击着这一陈旧认知的堤岸。
传统的劳动认知偏见，其根源在于社会阶层分化与职业发展的不平衡。在过去，从事体力劳动的群体
多为底层民众，工作环境恶劣、薪资待遇微薄，且往往缺乏上升通道，使得人们下意识地认为体力劳动是
低等的。与之相对，脑力劳动从业者多处于社会中上层，享有较高的社会地位与经济回报，这进一步加剧
了这种片面认知 。但这种二元对立的认知 ，极大地限制了人们对劳动本质的理解 。马克思曾深刻指出 ：“劳学科 网（北 京）股 份有限 公司
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动最先是人和自然之间的过程 ，是人以自身的活动来引起 、调整和控制人和自然之间的物质变换的过程 。”
无论是体力劳动还是脑力劳动，本质上都是人与自然、人与社会互动的方式，都具有同等的价值与尊严。
随着科技的迅猛发展与社会的深刻变革，新兴劳动形态如雨后春笋般破土而出。数字化运营者在虚拟
的网络世界中纵横捭阖，凭借对大数据的敏锐洞察与精准分析，为企业发展指引方向；人工智能调试员如
同技艺精湛的工匠，精心雕琢每一行代码，赋予机器以智慧与生命；文化创意设计师则以天马行空的想象
为画笔，在历史与现代、现实与虚幻之间勾勒出绚丽多彩的文化景观。这些新兴劳动打破了传统的体力与
脑力劳动的界限 ，它们既需要扎实的专业知识 ，又依赖于灵巧的实践技能 ，更离不开创新创造的思维火花 。
例如，在一款热门的虚拟现实游戏开发过程中，开发者不仅要精通编程算法等脑力知识，还要亲自动手调
试硬件设备，反复测试用户体验，将创意转化为可操作的产品。这种融合性的劳动形态，充分展示了劳动
的多元魅力与无限可能。
劳动教育，作为培育全面发展人才的关键一环，在这一时代背景下显得尤为重要。它不仅是简单的技
能传授，更是价值观塑造、创新精神培育和实践能力提升的综合性教育。通过劳动教育，学生能够亲身体
验劳动的艰辛与快乐，理解劳动成果的来之不易，从而树立正确的劳动价值观，培养尊重劳动、热爱劳动
的良好品质。在劳动实践中，学生不断尝试新方法、新思路，创新精神得以激发；在解决实际问题的过程
中，实践能力也得到锻炼与提升。然而，现实却令人遗憾，部分学校和家庭对劳动教育的忽视，使得学生
犹如温室中的花朵，缺乏对真实劳动世界的认知与体验。有的学校削减劳动课程，将更多时间用于所谓的
“主科 ” 学习 ；部分家庭过度溺爱孩子 ，不舍得让孩子参与家务劳动 ，导致孩子四体不勤 、五谷不分 。长
此以往，学生的全面发展将受到严重阻碍。
新时代青年，作为社会发展的中流砥柱，肩负着推动社会进步、实现民族复兴的伟大使命。面对传统
劳动认知的桎梏与劳动教育的缺失，必须勇敢地挺身而出，破茧而出。一方面，要主动更新观念，积极学
习新知识、新技能，勇于尝试新兴劳动领域，在实践中感受劳动的多元价值，摒弃对体力劳动的偏见，尊
重每一位劳动者的付出。另一方面，要积极投身劳动教育实践，无论是校园内的实验项目、社团活动，还
是家庭中的家务分担、社区里的志愿服务，都要热情参与，在劳动中磨砺意志、增长才干，实现自我成长
与价值。例如，参与社区组织的环保公益活动，通过垃圾分类宣传、环境清理等实践，不仅能增强环保意
识，还能提升沟通协作等综合能力，为社会贡献一份力量。
时代的号角已然吹响，新时代青年应勇做破茧的先锋，挣脱传统劳动认知的束缚，在劳动创造的广阔
天地中，实现个人的全面发展，为社会的繁荣进步注入源源不断的青春活力，共同描绘人类社会更加美好
的未来蓝图。 学科 网（北 京）股 份有限 公司
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【热点 五、社会公平与正义的多元维度 】“青春叙事的多维重构 ”、“大学生夜骑
开封，青春如何定义 ” 、“多元价值与理性思考 ”、“媒介素养与信息甄别 ”
社会公平与正义的多元维度子话题 —— 青春叙事的多维重构
考点方向：主体意识觉醒、数字原住民特征、代际认知差异。
解析： 关注青年主体意识觉醒，体现代际间的理解与精神传承，一代人有一点人的精神，一代人有一
代人的责任，每个时代的青年，都有自己的特色，但是，我们也不能否定其他，认知、包容是关键！
原创押卷试题 阅读下面的材料，根据要求写作。
在当今社会 ，“断舍离 ” 的生活理念在年轻人中颇为流行 ，他们崇尚简约 ，定期清理家中物品 ，只保
留真正需要的东西 ，追求一种清爽 、高效的生活方式 。然而 ，许多长辈对此却难以理解 ，在他们的观念里 ，
东西无论新旧，留着总有一天会派上用场，随意丢弃是浪费的行为。这种因年龄不同而产生的对生活物品
处理方式的认知差异，只是代际认知差异的一个缩影。
在价值观、消费观、职业选择、娱乐方式等诸多方面，代际之间都存在着明显的认知差异。这些差异
既可能引发家庭矛盾，也可能成为推动社会进步的动力。
在现实生活中，你也许正在经历或经历了这种认知差异带来的烦恼。读了上述材料让你有怎样的感悟
或思考 ?请结合材料写一篇文章。
要求：选准角度，确定立意，明确文体，自拟标题；不要套作，不得抄袭；不得泄露个人信息；不少
于 800 字。
文题分析：
这道作文题聚焦于 “代际认知差异 ” 这一具有现实意义的话题，从审题和立意角度来看，有如下要
点：
材料开篇以 “断舍离 ” 这一生活理念在年轻人和长辈间引发的不同看法为例，引出代际认知差异这
一核心主题。接着进一步拓展到价值观、消费观等多个领域都存在此类差异，最后指出其既可能导致家庭
矛盾 ，也能成为社会进步动力 。这表明材料旨在引导考生全面思考代际认知差异的表现 、影响及背后原因 。
题目要求考生结合自身经历，谈谈对材料的感悟或思考。这就需要考生从自身出发，将个人在现实生活中
经历或正在经历的因代际认知差异带来的烦恼融入文章，不能仅仅停留在对材料的理论分析上。
可以从以下几个方面思考，正视差异：可强调要正确认识代际认知差异的客观存在，如分析不同年代
成长环境、社会背景对两代人思维方式的塑造，从而导致在生活各方面认知不同，倡导以平和心态看待这
种差异。化解矛盾：基于材料中提到的家庭矛盾，探讨如何通过沟通、理解、相互尊重来减少因代际认知
差异引发的家庭冲突，比如讲述自己或他人如何在家庭中面对差异，通过有效方式化解矛盾，增进亲情。 学科 网（北 京）股 份有限 公司
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利用差异推动进步：着眼于代际认知差异能成为社会进步动力这一点，阐述年轻人的创新思维与长辈的经
验智慧相结合，在社会发展的各个层面（如文化传承、科技创新、经济发展等）如何碰撞出火花，推动社
会前进。
自我成长反思：从自身经历出发，讲述在面对代际认知差异烦恼时，如何通过反思实现自我成长，如
从最初的不理解到逐渐接纳长辈观点，或者如何在坚持自身正确理念的同时，引导长辈理解自己，实现代
际间的共同成长 。
写作素材拓展
事实论据
冬奥会志愿者数字孪生管理系统
北京冬奥会期间，一项创新的科技成果 —— 冬奥会志愿者数字孪生管理系统投入使用，为赛事的志愿
服务管理带来了全新变革。为实现对志愿者的精准调配、实时监控与科学管理，数字孪生管理系统应运而
生。
以数字孪生技术为核心 ，将现实中的志愿者团队 、服务流程 、场馆环境等信息 ，通过数据采集与建模 ，
在虚拟空间中构建出高度仿真的数字模型。例如，借助物联网设备收集志愿者的位置、工作状态等数据，
利用大数据分析技术对志愿者技能、服务时长等信息进行整合处理，再运用人工智能算法进行任务分配与
调度优化 。多团队协同合作 ，历经需求分析 、技术攻关 、系统测试等阶段 ，最终打造出这一智能管理系统 。
冬奥会结束后，该系统的价值仍在延续。一方面，其为后续大型体育赛事及活动的志愿者管理提供了
宝贵经验与技术模板，相关技术和理念被应用到其他赛事筹备中。另一方面，系统经改进后，可拓展至城
市公共服务领域，如大型展会、交通枢纽的志愿服务管理，进一步提升社会服务的智能化水平 。
大学生返乡创业的 “新知青运动 ”
“大学生返乡创业 ”这一现象，近年来逐渐形成一股引人注目的 “新知青运动。大学生返乡创业这一
“新知青运动 ”，为农村发展注入了新的活力与生机，虽面临诸多挑战，但在政府、社会和大学生自身的
共同努力下，正逐步发展壮大，成为推动乡村振兴的重要力量。
城市就业竞争日益激烈，对于刚毕业的大学生而言，在大城市站稳脚跟并非易事。高房价、高生活成
本以及巨大的工作压力，让许多年轻人开始重新审视自己的职业发展道路。与此同时，农村地区的发展迎
来了新契机。国家高度重视 “三农 ”问题，连续多年出台指导 “三农 ”工作的中央一号文件，一系列惠农
政策相继落地。农村的基础设施不断完善，交通愈发便利，网络覆盖逐渐普及，这为农村发展注入了新活
力，也为大学生返乡创业提供了基础条件。 学科 网（北 京）股 份有限 公司
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许多大学生利用所学专业知识，投身特色种植与养殖行业。比如，有的大学生引进先进的种植技术，
种植高附加值的经济作物 ，像有机蔬菜 、特色水果等 。他们注重农产品品质 ，通过绿色 、有机的种植方式 ，
提升产品竞争力。还有的专注于特色养殖，如养殖珍稀禽类、昆虫等，开拓新的市场需求。例如，江西农
业大学的刘瑶，返乡后成立丰采种养专业合作社，在专家团队指导下，合理规划果园布局，种植脐橙、鹰
嘴桃等多种水果，打造集采摘、农家乐、观光旅游于一体的现代化农业模式，带动了当地经济发展。
互联网的普及让农村电商成为热门创业方向。大学生们凭借对网络的熟悉，通过电商平台将农村的优
质农产品推向全国乃至全球市场。他们利用直播带货、短视频推广等新兴营销手段，打破地域限制，解决
农产品销售难题。不少大学生返乡后建立农村电商服务站，不仅帮助自己创业，还为当地农民提供销售渠
道，带动农民增收。
乡村拥有丰富的自然资源和独特的文化底蕴，这吸引了众多大学生投身乡村文旅产业。他们挖掘乡村
的历史文化、民俗风情，将其与旅游相结合，开发乡村旅游项目。有的对闲置农房进行改造，打造特色民
宿，为游客提供别样的住宿体验；有的组织乡村文化活动，如传统民俗表演、农事体验等，丰富游客的乡
村旅游经历。
3.B 站跨年晚会中的代际对话
在跨年晚会这一备受瞩目的舞台上 ，不同平台各展神通 ，而 B 站的跨年晚会却以独特的姿态脱颖而出 ，
其中蕴含的代际对话元素更是引发广泛关注 ，成为文化领域探讨代际认知差异与融合的典型案例 。B 站作为
以“二次元 ”文化为主打的视频平台 ，其用户群体主要为年轻人 。在筹备跨年晚会时 ，B 站团队敏锐捕捉到
不同代际在这一特殊节点都迎来新人生阶段的共性，决心打造一台专属年轻人的晚会，以标注这一特殊时
刻。
从节目内容来看， B 站跨年晚会巧妙融合了多元元素，为代际对话搭建了桥梁。一方面，晚会充分展
现了年轻人热衷的文化内容。动漫、游戏、二次元等元素贯穿始终。这些节目满足了年轻人对潮流文化、
虚拟世界的热爱，是他们日常兴趣爱好的集中呈现。
另一方面，晚会有意融入了诸多能引发长辈共鸣的 “长辈趣味 ”。琵琶大师方锦龙带来的国乐跨界串
烧表演，令人大开眼界。他在节目中弹奏了琵琶、高音琵琶、尺八、冲绳三味线等多种乐器，还将古今中
外的乐器与不同风格曲目巧妙融合 ，如把 《沧海一声笑 》《男儿当自强 》与中国古曲 《将军令 》“嫁接 ”，
并加入剧情表演，让观众在欣赏音乐的同时忍俊不禁。这一节目不仅展现了国乐的魅力，也让年轻观众领
略到传统文化的深厚底蕴，同时成功吸引了长辈们的目光，许多长辈被方锦龙的精湛技艺所折服，对国乐
产生了新的兴趣。 学科 网（北 京）股 份有限 公司
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B 站跨年晚会的成功，本质上是一次成功的代际对话实践。它打破了以往以代际划分消费者、文化圈
层的固有模式，让不同年龄层的观众在同一台晚会中找到属于自己的 “高光时刻 ”。年轻人在晚会中看到
自己热爱的文化被认可、被呈现，同时也通过这些节目了解到长辈们所经历的时代与文化；长辈们则借此
机会走进年轻人的世界 ，发现原来年轻人并非只沉迷于虚拟 、新潮事物 ，对经典文化同样怀有敬意与喜爱 。
这种跨越代际的交流与理解，不仅在晚会播出时引发热烈反响，也为社会各领域如何促进代际沟通提供了
宝贵经验，让人们看到通过文化这一载体，弥合代际认知差异、实现代际和谐共生的可能性。
生活中的代际差异
消费观念差异 ：
在购物节期间，年轻人热衷于抢购最新款的电子产品、时尚服装等，追求潮流和品质，愿意为品牌附
加值买单。而长辈们则更倾向于购买实用、性价比高的商品，对价格较为敏感，会花费大量时间在不同店
铺之间比较价格，对于年轻人动辄上千购买一双运动鞋的行为难以理解。
职业选择差异：
如今，许多年轻人选择投身新兴行业，如电竞选手、网络主播、自媒体运营等。他们看重这些行业的
发展潜力、灵活性和自我实现的机会。然而，长辈们大多认为医生、教师、公务员等传统职业才是 “铁饭
碗”，工作稳定，社会地位高，对年轻人从事新兴职业持怀疑态度，担心其未来发展不稳定。
娱乐方式差异：
年轻人闲暇时喜欢通过手机、电脑玩网络游戏、刷短视频、看直播等，享受虚拟世界带来的乐趣和便
捷社交。长辈们则更偏爱传统的娱乐方式，如看电视、下棋、散步、跳广场舞等，他们觉得这些活动更健
康、实在，对年轻人沉迷于电子设备的娱乐方式感到担忧，认为这是在浪费时间。
文化观念差异：
在对待传统文化上，年轻人可能更关注传统文化的创新表达形式，如故宫文创产品、国潮服饰等，将
传统文化与现代潮流相结合。而长辈们则更注重传统文化的原汁原味，对传统的节日习俗、礼仪规范等遵
循得更为严格，对于年轻人对传统文化的 “改造 ”可能不太认同。
范文示例
跨越代际沟壑，共筑和谐华章
在时代的洪流中，代际认知差异如同一座横亘在不同年龄群体之间的沟壑，清晰而深刻。从年轻人热
衷的 “断舍离 ” 生活理念与长辈传统惜物观念的碰撞，到价值观、消费观、职业选择、娱乐方式等多维
度的分歧，这些差异既可能成为家庭矛盾的导火索，也蕴含着推动社会进步的强大动力。如何正确看待并
跨越这道沟壑，是我们亟待思考与解决的问题。 学科 网（北 京）股 份有限 公司
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代际认知差异，是时代变迁留下的深刻烙印。每一代人都成长于特定的社会环境，其思维方式、价值
取向不可避免地受到所处时代的塑造。在物质匮乏的年代，长辈们经历过生活的艰辛，物资的短缺让他们
养成了节俭、惜物的习惯。
对他们而言，每一件物品都来之不易，即便破旧，也可能在未来某个时刻发挥作用，随意丢弃无疑是
一种浪费。而如今的年轻人，成长在物质丰富、信息爆炸的时代，他们追求简约高效的生活， “断舍离 ”
成为一种流行的生活哲学，在他们眼中，过多的物品不仅占据空间，更可能成为生活的负担。这种对生活
物品处理方式的差异，只是代际认知差异的冰山一角。
在价值观的领域，代际差异同样显著。年轻人更加注重自我价值的实现，追求个性化、多元化的发展
道路。他们勇于挑战传统，敢于突破常规，将个人兴趣与梦想置于重要位置。长辈们则深受传统观念的熏
陶，秉持着集体主义价值观，强调责任、奉献与传承，在他们看来，安稳的生活、良好的社会声誉以及对
家族和社会的贡献至关重要。
在消费观念上，年轻人倾向于超前消费，注重消费的品质与体验，愿意为新兴科技产品、时尚潮流买
单。长辈们则更为保守 ，量入为出是他们坚守的消费原则 ，追求性价比 ，对价格更为敏感 。职业选择方面 ，
新兴行业如电竞、直播、人工智能等吸引着年轻人的目光，他们看重行业的发展潜力与创新空间。长辈们
却依然对传统的 “铁饭碗 ” 职业情有独钟，认为这些职业稳定可靠，能为生活提供保障。娱乐方式上，
年轻人沉浸在虚拟世界，通过网络游戏、短视频等获得快乐与满足。长辈们则热衷于传统的休闲活动，如
散步、下棋、看戏等，享受现实生活中的人际互动。
不可否认，代际认知差异在一定程度上会引发家庭矛盾。当年轻人的选择与长辈的期望背道而驰时，
冲突便不可避免。孩子追求新兴职业，长辈却认为不务正业；年轻人崇尚 “断舍离 ”，长辈却觉得浪费败
家。这些矛盾如果处理不当，可能会破坏家庭的和谐氛围，影响亲情关系。但我们也应看到，代际认知差
异并非只有负面影响。
它为社会进步注入了源源不断的活力。年轻人的创新思维与长辈的丰富经验相结合，能够碰撞出智慧
的火花。在科技创新领域，年轻人凭借对新技术的敏锐感知和大胆探索，推动着行业的发展；长辈们则以
稳健的决策和深厚的行业积累，为创新提供坚实的支撑。在文化传承方面，年轻人以新颖的方式传承和弘
扬传统文化，如国潮文化的兴起、传统文化的创意表达，让古老的文化焕发出新的生机；长辈们则坚守文
化的内核，为传统文化的传承提供根基。
面对代际认知差异，我们应秉持理解、包容与沟通的态度。作为年轻人，要尊重长辈的经验与智慧，
理解他们观念形成的历史背景。在追求自己的理想与生活方式时，耐心倾听长辈的意见，以平和的方式表
达自己的想法，寻求双方的共识。长辈们也应与时俱进，以开放的心态接纳新事物，了解年轻人的世界， 学科 网（北 京）股 份有限 公司
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尊重他们的选择。家庭中，亲子之间应多开展深入的交流，分享彼此的生活经历与内心想法，增进相互之
间的理解与信任。社会层面，应营造尊重代际差异的氛围，通过教育、媒体等多种渠道，引导不同年龄群
体相互理解、相互学习。
代际认知差异是时代发展的必然产物，它既带来了挑战，也蕴含着机遇。让我们以包容之心跨越这道
沟壑，让不同代际的智慧与力量相互交融，共同构筑和谐美好的社会华章，在时代的舞台上书写属于我们
的精彩篇章。
社会公平与正义的多元维度子话题 —— 大学生夜骑开封：青春如何定义
题目：阅读下面的材料，根据要求写作。
2024 年，开封夜骑活动引发了广泛关注 。起初 ，当地政府为推动文旅产业发展 ，鼓励大学生参与夜骑
活动，吸引了大量年轻人涌入开封。然而，随着活动规模的迅速扩大，交通拥堵、安全隐患等问题逐渐凸
显，最终导致活动被紧急叫停 。这一事件引发了关于青春定义的广泛讨论 ：青春是追求自由与激情的冒险 ，
还是在规则与责任中的理性成长？有人认为，夜骑活动展现了大学生的活力与创造力；也有人指出，青春
的定义不应忽视规则与责任。
请结合材料，以 “大学生夜骑开封：青春如何定义 ”为主题，写一篇文章，谈谈你的思考和感悟。要
求：选准角度，确定立意，明确文体，自拟标题；不要套作，不得抄袭；不得泄露个人信息；不少于 80 0
字。
题材分析
该题材聚焦于开封夜骑活动这一社会现象，核心在于探讨青春的定义与此次活动的关联。它涉及到文
旅产业发展、大学生行为表现、规则与责任以及青春内涵等多个层面。需要从社会影响、个体成长、价值
观念等维度去思考 ，剖析在特定活动中青春的不同呈现方式 ，以及如何正确看待青春与规则 、责任的关系 ，
题材具有很强的现实讨论价值。
审题立意
关键信息抓取 ：“开封夜骑活动 ”“青春定义 ”“自由激情 ”“规则责任 ”“大学生活力创造力 ” 等
是关键要点。
立意方向：
强调青春的自由与激情 ：突出青春应勇敢追求自由体验 ，释放活力 ，如 “以自由激情为笔 ，绘就青春
绚丽画卷 ”。
注重青春的规则与责任 ：着重阐述青春在成长过程中需遵循规则 ，承担责任 ，如 “在规则责任中 ，雕
琢青春的理性模样 ”。学科 网（北 京）股 份有限 公司
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辩证看待青春 ：全面分析青春既要有自由激情 ，又不能忽视规则责任 ，如 “权衡自由激情与规则责任 ，
书写青春华章 ”。
素材选用
青春自由激情素材：大学生参加户外探险活动，挑战自我，突破极限；校园社团组织创意活动，展
现青春活力。
青春规则责任素材：大学生志愿者参与社区服务，承担社会责任；在校园活动中遵守组织规则，保
障活动顺利进行。
范文示例
在规则与责任中，书写青春华章
“青春须早为 ，岂能长少年 。” 青春 ，是人生中最璀璨的篇章 ，它充满了活力与激情 ，承载着我们的
梦想与追求 。然而 ，青春究竟该如何定义？ 2024 年开封夜骑活动引发的一系列现象 ，为我们提供了一个深
入思考的契机。
青春，是追求自由与激情的冒险，它如同一匹不羁的野马，在广袤的草原上肆意奔腾。大学生们响应
政府的号召 ，参与夜骑活动 ，穿梭在开封的大街小巷 。那飞扬的发丝 ，那爽朗的笑声 ，那勇往直前的身影 ，
无不彰显着青春的活力与创造力。他们在夜骑中感受着风的拥抱，探索着这座古老城市的夜晚魅力，仿佛
整个世界都在他们的脚下 。这是青春的激情 ，是对自由的向往 ，是对未知世界的勇敢探索 ，恰似李白 “仰
天大笑出门去，我辈岂是蓬蒿人 ” 的豪迈，展现着青春无畏的勇气。
但青春的自由并非毫无边界，它需要在规则的框架内绽放光彩。规则，是青春道路上的指示灯，引导
着我们前行的方向。开封夜骑活动初期，一切都充满着美好，然而随着规模的迅速扩大，交通拥堵、安全
隐患等问题接踵而至。原本有序的街道变得混乱不堪，车辆行人争道抢行，安全事故频发。这就如同脱缰
的野马 ，一旦失去了缰绳的约束 ，便会陷入危险的境地 。“不以规矩 ，不能成方圆 。” 规则不是对青春的
束缚，而是对青春的保护。它让我们在追求自由的同时，不致于迷失方向，不致于伤害他人，更不致于让
自己陷入危险的漩涡。
青春，更是在责任中的理性成长。当我们享受青春的自由与激情时，也不应忘记自己肩负的责任。大
学生作为社会的未来栋梁，他们的行为不仅关乎自身，更影响着社会的秩序与稳定。在夜骑活动中，大学
生们本应在享受自由骑行乐趣的同时，遵守交通规则，维护公共秩序，保障自身和他人的安全。这是对自
己负责 ，也是对社会负责 。正如林则徐 “苟利国家生死以 ，岂因祸福避趋之 ”，在青春的征程中 ，我们要
将个人的行为与社会责任紧密相连，以理性的态度对待青春的每一次选择。 学科 网（北 京）股 份有限 公司
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开封夜骑活动的紧急叫停，给我们敲响了警钟。它让我们明白，青春的定义是多元的，它既包含着自
由与激情的冒险，也包含着对规则的尊重和对责任的担当。我们不能只看到青春的绚烂，而忽视了背后的
规则与责任。只有在规则与责任的指引下，青春的冒险才会更加安全，青春的活力才能得到更好的释放。
“自信人生二百年 ，会当水击三千里 。” 作为新时代的青年 ，我们应在青春的道路上 ，勇敢地追求自
由与激情，同时也要时刻牢记规则与责任。让我们以理性的态度，在规则与责任的框架内，书写属于自己
的青春华章，让青春在时代的舞台上绽放出更加耀眼的光芒。
社会公平与正义的多元维度子话题 —— 多元价值与理性思考
考点方向 ：网络暴力 、过度追星 、盲目消费等具有争议性的话题或现象 ，考查考生的价值取向与思辨
能力。
解析： 身处信息爆炸的时代，各种价值观如潮水般涌来，相互碰撞、交融。如何引导学生树立正确的
价值观，学会在复杂的信息洪流中保持理性思考与准确判断，成为教育的重要使命。
原创押卷试题 阅读下面的材料，根据要求写作。
在当今信息爆炸的时代，互联网让信息传播的速度与广度达到了前所未有的程度。短视频平台上，有
人宣扬 “躺平 ” 的生活态度，认为不必努力奋斗，享受当下即为人生真谛；也有人推崇 “极致内卷 ”，
将高强度的工作与学习视为成功的唯一路径。社交媒体中，消费主义甚嚣尘上，鼓吹通过购买奢侈品来彰
显个人价值；与此同时，公益行动的报道也在传递着奉献、互助的正能量。各种价值观相互碰撞、交融，
如潮水般涌来，冲击着人们的认知。
青少年作为价值观形成的关键群体，正处于思想活跃、易受影响的阶段。面对复杂的信息洪流，如何
树立正确的价值观，学会理性思考与准确判断呢？
请结合以上材料，写一篇文章，谈谈你的看法与思考。
要求：选准角度，确定立意，明确文体，自拟标题；不要套作，不得抄袭；不得泄露个人信息；不少
于 800 字。
文题分析 ：
材料开篇点明信息时代互联网带来的信息传播变革 ，随后列举了多种价值观 。在短视频平台 ，“躺平 ”
观念主张摒弃努力，单纯享受当下； “极致内卷 ”则走向另一极端，把高强度工作学习当作成功的唯一通
途。社交媒体中，消费主义通过鼓吹奢侈品消费来定义个人价值，而公益行动报道传递的是奉献、互助等
正能量价值观。这些相互冲突的价值观，生动展现了信息时代价值观的多元与复杂。
材料着重指出青少年处于价值观形成关键期，思想活跃但也极易受外界影响。在信息洪流中，青少年
缺乏足够的辨别能力 ，很容易被错误价值观误导 ，这凸显了引导青少年树立正确价值观的紧迫性与重要性 。学科 网（北 京）股 份有限 公司
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要求考生结合材料，围绕青少年在信息时代如何树立正确价值观、学会理性思考与准确判断展开论述。
这意味着考生需要深入分析信息时代价值观多元的影响，提出切实可行的引导方法与培养策略。
立意可从多个层面展开。可以强调正确价值观对青少年成长的重要意义，如正确价值观如何助力青少
年在人生道路上做出正确选择。也可分析错误价值观的危害，如 “躺平 ” 可能导致青少年丧失奋斗动力，消
费主义易使青少年陷入物质虚荣陷阱。还能从引导主体出发，探讨家庭、学校、社会应如何协同合作，为
青少年营造良好的价值观培育环境。
文体选择：议论文是较为合适的文体。采用清晰的论点、丰富的论据（如具体事例、数据、专家观点
等）和多样的论证方法（如举例论证、对比论证、因果论证、引用论证等），深入阐述信息时代青少年价
值观引导的重要性与方法，使文章逻辑严谨、论证有力。记叙文也可尝试，通过讲述一个青少年在经历价
值观冲突后，最终在他人引导下树立正确价值观的故事，展现正确价值观的力量，但要注意在叙事中适当
穿插议论和抒情，点明主题，避免单纯叙事而缺乏深度。
写作素材拓展
典型案例
李子柒的价值传播：
网红李子柒通过短视频展示中国传统田园生活，将勤劳、质朴、热爱生活等价值观传递给全球观众。
青少年可以从她的成功中看到，通过自身努力创造有价值内容，传播积极价值观，能获得广泛认可，激励
自己在网络世界中也传播正能量。
衡水中学
“土猪拱白菜 ”演讲争议：
衡水中学学生在演讲中表达出一种带有强烈功利性、竞争性的价值观，引发社会广泛讨论。这一案例
可用于分析极端 “内卷 ” 价值观对青少年心理和认知的不良影响，以及在信息传播中如何引导青少年正确看
待竞争与成功。
“00 后”志愿者群体：
众多 “00 后”青少年主动投身志愿服务，他们不计个人得失，展现出奉献、担当的价值观。这一群
体的事迹可作为正面案例 ，阐述正确价值观对青少年成长和社会发展的积极作用 ，鼓励青少年向他们学习 ，
树立正确价值观。
某中学生沉迷奢侈品消费案例：
曾有报道，一名中学生为追求名牌奢侈品，在网络借贷平台借款，最终陷入高额债务困境。此案例可
直观体现消费主义对青少年的危害，用于论述错误价值观对青少年生活和未来发展的负面影响。 学科 网（北 京）股 份有限 公司
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研究数据
据某权威教育研究机构调查显示，超过 60% 的青少年表示在网络上看到过与自己价值观相悖的信息 ，
其中 25% 的青少年表示会受到一定程度影响。这一数据可用于强调信息时代价值观多元对青少年冲击的普
遍性和严重性。
一项针对青少年心理健康与价值观关系的研究表明，拥有积极健康价值观的青少年，其心理问题发生
率比价值观模糊或错误的青少年低 30% 。以此数据说明正确价值观对青少年身心健康的重要保障作用。
某网络平台统计，在青少年用户中，关注公益内容的用户数量逐年上升，增长率达到 15% ，但同时关
注消费主义相关内容的用户也占比高达 40% 。这组数据可用于分析青少年在价值观选择上的矛盾性，以及
引导青少年树立正确价值观的紧迫性。
范文示例：
于信息洪流中，锚定价值坐标
在这个信息如狂飙般席卷而来的时代，互联网宛如一张无形却强大的巨网，将世界紧密相连，让信息
的传播在速度与广度上实现了前所未有的飞跃 。短视频平台上 ，“躺平 ” 的喟叹与 “极致内卷 ” 的呐喊
相互交织，一方宣扬着及时行乐，不必在奋斗的荆棘中艰难攀爬，享受当下便是人生的全部真谛；另一方
则高举成功的大旗，将高强度的工作与学习奉为通往荣耀的唯一通途，仿佛只有在无尽的忙碌中才能找寻
到生命的意义 。社交媒体里 ，消费主义的鼓噪甚嚣尘上 ，它以物质的堆砌和奢侈品的闪耀来定义个人价值 ，
让人们在对名牌的追逐中逐渐迷失自我 ；而与此同时 ，公益行动的报道如同一束束温暖的光 ，传递着奉献 、
互助的正能量，展现着人性中至善至美的一面。各种价值观相互碰撞、交融，似汹涌潮水，一波又一波地
冲击着人们的认知防线，尤其是对正处于价值观形成关键期的青少年而言，这场价值观的风暴更为猛烈。
青少年，宛如初升的朝阳，思想活跃，充满着对世界的好奇与探索欲，然而，也正因如此，他们极易
受到外界信息的影响。在这复杂的信息洪流中，如何为青少年指引方向，帮助他们树立正确的价值观，学
会理性思考与准确判断，已然成为亟待解决的时代命题。
“非淡泊无以明志 ，非宁静无以致远 。” 诸葛亮的这句箴言 ，宛如一盏明灯 ，照亮了青少年在信息时
代的迷茫征途。在这喧嚣浮躁的环境中，青少年需摒弃外界的纷扰，保持内心的宁静与淡泊。当 “躺平 ”
的思潮如迷雾般弥漫，试图让他们放弃奋斗的热情时，唯有坚守内心的宁静，才能不被这股消极之风所裹
挟，从而明确自己的志向，坚定地踏上追求梦想的道路。就如同在繁华都市的喧嚣中，仍有一些青少年醉
心于知识的海洋，他们远离娱乐八卦的诱惑，在宁静中沉淀自我，为实现自己的理想默默努力，以淡泊之
心明高远之志。 学科 网（北 京）股 份有限 公司
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“千磨万击还坚劲 ，任尔东西南北风 。” 郑燮笔下的竹子 ，历经无数磨难却依然坚韧挺拔 ，这正是青
少年在面对多元价值观冲击时应有的姿态。在信息时代，各种价值观如同四面八方袭来的狂风，试图动摇
青少年的信念。从功利性的 “土猪拱白菜 ” 式竞争价值观，到消费主义的物质诱惑，都在考验着青少年
的意志。然而，只要青少年具备竹子般坚韧的品质，就能在这狂风骤雨中屹立不倒，坚守正确的价值观。
如那些在疫情期间挺身而出的 “00 后” 抗疫志愿者 ，他们面对危险与困难 ，毫不退缩 ，用行动诠释着奉
献与担当的价值 ，在外界的质疑与误解中 ，依然坚定地践行着自己的信念 ，恰似那 “咬定青山不放松 ” 的
翠竹。
“一个人的价值，应该看他贡献什么，而不应当看他取得什么。 ” 爱因斯坦的这句话，如洪钟大吕 ，
振聋发聩，有力地批判了消费主义等错误价值观。在消费主义盛行的当下，许多青少年被奢侈品的光环所
迷惑，认为拥有名牌就能彰显个人价值。但事实并非如此，真正的价值在于对社会的贡献。李子柒，这位
在网络世界绽放光芒的网红 ，她没有沉溺于物质的追求 ，而是通过短视频展示中国传统田园生活 ，将勤劳 、
质朴、热爱生活等价值观传递给全球观众。她用自己的努力和创意，为文化传播做出贡献，赢得了广泛的
赞誉与尊重。青少年应从她的经历中汲取力量，树立正确的价值衡量标准，明白奉献才是实现个人价值的
真谛。
“教育的本质意味着 ，一棵树摇动另一棵树 ，一朵云推动另一朵云 ，一个灵魂唤醒另一个灵魂 。” 雅
斯贝尔斯对教育的深刻阐释，提醒着我们，家庭与学校在青少年价值观引导中肩负着重要使命。家长和教
师作为青少年成长路上的引路人，要以身作则，用自己的言行传递正确的价值观。在家庭中，父母应营造
温馨和睦的氛围，注重培养孩子的品德与责任感；在学校里，教师要通过丰富多样的教学活动，引导学生
树立正确的世界观、人生观和价值观。例如，组织学生参与公益活动，让他们在实践中感受奉献的快乐，
唤醒他们内心深处的善良与担当。
在信息时代的浪潮中，青少年要以古圣先贤的智慧为指引，以时代楷模的事迹为榜样，在家庭与学校
的悉心引导下，于复杂的信息洪流中锚定正确的价值坐标。
让我们怀揣着坚定的信念，以理性为舟，以思考为桨，在价值观的海洋中破浪前行，驶向属于自己的光
明未来，同时也为社会的发展贡献属于自己的那一份力量，让这个世界因我们正确的价值观而变得更加美
好。
社会公平与正义的多元维度子话题 —— 媒介素养与信息甄别
考点方向：算法推荐困境、后真相时代批判、数字公民责任。 学科 网（北 京）股 份有限 公司
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解析： 培养信息时代的理性思维，构建清朗网络空间的责任意识、做一个负责的人，不是人云亦云，
我们不能去跟风，要有自己的判断，不要把自己困在小视频的算法里，无法自拔，我们更多需要的是独立
思考。
原创押卷试题
阅读下面的材料，根据要求写作。
在信息时代，互联网如同一把双刃剑，给我们的生活带来了翻天覆地的变化。一方面，海量信息如潮
水般涌来，为我们获取知识、拓展视野、交流互动提供了极大便利；另一方面，虚假信息、网络暴力、不
良文化等也充斥其中，侵蚀着网络空间的纯净，影响着人们的思想和行为。
青少年作为网络的主力军，在享受网络带来的便捷时，也面临着诸多挑战。如何培养理性思维，在纷
繁复杂的信息洪流中保持清醒，辨别是非真伪；如何树立责任意识，规范自身网络行为，为构建清朗网络
空间贡献力量，成为亟待思考的问题。
请结合以上材料，写一篇文章，谈谈你作为青少年的感悟与思考。
要求：选准角度，确定立意，明确文体，自拟标题；不要套作，不得抄袭；不得泄露个人信息；不少
于 800 字。
文题分析
材料开篇点明互联网在信息时代的双刃剑特性。海量信息带来便利，使我们能迅速获取知识，足不出
户了解世界 ，还能通过社交平台与世界各地的人交流互动 ，极大地拓展了视野 。然而 ，虚假信息肆意传播 ，
如一些谣言在网络上迅速扩散，误导公众认知；网络暴力现象频发，对当事人造成身心伤害；不良文化，
如低俗、暴力、色情等内容侵蚀着网络空间，污染网络环境，影响人们的价值观和行为方式。
明确指出青少年是网络主力军，他们在享受便捷的同时，必须面对诸多难题。在信息洪流中，青少年
易被虚假信息误导，因缺乏理性思维而难以辨别信息真伪。网络上的不良现象也可能影响他们的价值观，
使其在无意识中参与网络暴力或传播不良文化。
写作要求考生结合材料，从青少年自身角度出发，谈感悟与思考。重点围绕两个关键问题展开，一是
如何培养理性思维以应对复杂信息，二是怎样树立责任意识来规范网络行为，为构建清朗网络空间出力。
立意可从多个维度深入。可以强调培养理性思维的重要性，如阐述理性思维能帮助青少年在网络信息
海洋中不迷失方向，举例说明理性思维如何让青少年避免被虚假信息蛊惑。也可突出构建清朗网络空间责
任意识的意义，如分析责任意识对净化网络环境、营造良好网络文化氛围的积极作用。还能从具体行动层
面立意，探讨青少年培养理性思维、树立责任意识的具体方法和实践路径。 学科 网（北 京）股 份有限 公司
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议论文是较为合适的文体。采用清晰的论点、丰富的论据（如具体事例、数据、专家观点等）和多样
的论证方法（如举例论证、对比论证、因果论证、引用论证等），深入阐述青少年在信息时代培养理性思
维和树立责任意识的重要性及实践方法，使文章逻辑严谨、论证有力。记叙文也可尝试，通过讲述自己或
身边人的网络经历，展现理性思维和责任意识在网络行为中的作用，但要注意在叙事中适当穿插议论和抒
情，点明主题，避免单纯叙事而缺乏深度。
写作素材拓展
网络文明相关倡议话语
“网络不是法外之地 ，人人在网上知法守法 ，网络秩序才有规范 ，网络才能健康发展 。”—— 共青团
绵阳市委倡议内容可用于强调青少年树立责任意识，遵守网络法律法规，维护网络秩序的必要性 。
“要善于网上学习 ，不浏览不良信息 ；要诚实友好交流 ，不侮辱欺诈他人 ；要增强自护意识 ，不随意
约会网友；要维护网络安全，不破坏网络秩序；要有益身心健康，不沉溺虚拟时空。 ”—— 《全国青少年
网络文明公约》
这一公约全面涵盖了青少年在网络中的行为准则，可作为具体行动指南，融入文章阐述青少年如何在
网络中践行理性思维和责任意识。
“让我们携手践行 ，从自身做起 ，从每一次上网行为做起 。让文明之花在网络空间绽放 ，为青少年创
造一个绿色 、健康 、文明的网络环境 。”—— 澎湃新闻关于网络文明倡议话语可用于呼吁青少年积极行动 ，
以实际行动构建清朗网络空间，增强文章的感染力与号召力。
网络素养相关研究结论
“报告从 6 个维度对青少年网络素养进行分析评价 ，结果显示我国青少年上网注意力 、网络印象管理
能力、网络安全行为和隐私保护意识有所提高；网络信息保存和利用能力、网络规范认知能力仍需进一步
提升，网络信息的辨识和批判能力相对较差。 ”
。此研究结论可作为数据支撑，分析青少年网络素养现状，进而针对性地提出培养理性思维和责任意识的
方法与措施，增强文章的科学性与说服力。
“有研究通过观察 80 后、90 后、00 后三代青少年网民的网络素养形成历程发现 ，未成年人网络素养
的习得多发生在自身的网络实践活动与意义建构的过程中。 ”
可依据这一研究结论，强调青少年在网络实践中培养理性思维和责任意识的重要性，鼓励青少年积极参与
网络实践，并在实践中反思、成长。
范文示例：
持理性之灯，筑清朗网络 学科 网（北 京）股 份有限 公司
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在这个信息如洪流般奔涌的时代 ，互联网恰似一把双刃剑 ，在为我们的生活绘就斑斓色彩的同时 ，
也投下了不容忽视的暗影 。对于活跃在网络前沿的青少年而言 ，如何在这复杂的网络环境中披荆斩棘 ，
培养理性思维，扛起构建清朗网络空间的责任，已然成为亟待解答的时代命题。
朱熹曾言 ：“知之愈明 ，则行之愈笃 ；行之愈笃 ，则知之益明 。” 在网络世界里 ，对理性思维的
认知与实践，同样遵循着这般相辅相成的规律。当我们在铺天盖地的信息浪潮中穿梭时，虚假信息、
网络谣言常常如隐藏在暗处的礁石，稍不留意便会让我们的认知之舟触礁。就拿前阵子风靡一时的某
网红食品来说，网络上瞬间涌现出大量夸赞其 “神奇功效 ” 的信息，不少青少年因缺乏理性思考，
盲目跟风购买。然而，经权威机构检测，该食品不过是普通零食，所谓 “功效 ” 纯属无稽之谈。倘
若青少年能明晰理性思维辨别信息的重要性 ，在面对这类信息时 ，多一些思考 ，多渠道核实信息来源 ，
就能在实际网络行为中更好地避开陷阱。而在一次次成功辨别虚假信息的实践中，我们对理性思维的
理解与运用也会愈发深刻。
在网络这片虚拟天地中，善良的本性如同熠熠生辉的星辰，照亮每一个角落。罗素说： “在一切
道德品质之中 ，善良的本性在世界上是最需要的 。” 我们时常听闻网络暴力的种种乱象 ，一些青少年
在不明真相的情况下，受不良情绪裹挟，参与到对他人的恶意攻击中。可曾想过，那些伤人的话语，
如同利箭，会给当事人带来难以愈合的创伤。一个温暖的点赞、一句鼓励的评论，或许就能成为他人
在网络世界中的心灵慰藉；反之，一句恶语相向，可能就会成为压垮他人的最后一根稻草。我们应时
刻保持善良，用温暖的言行，为网络空间增添一抹亮色。
“谣言止于智者 。”《荀子 ・大略 》中的这句名言 ，在信息时代尤为振聋发聩 。如今 ，虚假信息如
病毒般肆意传播，从毫无根据的明星绯闻，到耸人听闻的社会谣言，让人防不胜防。青少年若能培养
起理性思维 ，练就一双 “火眼金睛 ”，就能成为阻挡谣言传播的坚固堤坝 。例如 ，在疫情期间 ，各种
关于病毒防治的谣言四处流传，一些青少年凭借所学知识，冷静分析，积极辟谣，有效遏制了谣言的
扩散，守护了网络空间的真实与纯净。
“勿以恶小而为之 ，勿以善小而不为 。” 刘备的这句告诫 ，在网络世界同样适用 。青少年的每一
次网络点击、每一条信息发布，看似微不足道，实则都在为网络环境添砖加瓦。也许只是随手转发一
条正能量的公益信息 ，就能让更多人关注到需要帮助的群体 ；也许只是在网络讨论中多使用文明用语 ，
就能营造出和谐的交流氛围 。反之 ，一条不经意的不当言论 ，可能就会引发一场骂战 ，破坏网络秩序 。
我们要从这些小事做起，树立起构建清朗网络空间的责任意识。
共青团绵阳市委倡议： “网络不是法外之地，人人在网上知法守法，网络秩序才有规范，网络才
能健康发展 。” 青少年作为网络的主力军 ，更应将此铭记于心 。遵守网络法律法规 ，不随意泄露他人 学科 网（北 京）股 份有限 公司
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隐私，不传播不良信息，是我们应尽的基本责任。同时，《全国青少年网络文明公约》也为我们的网
络行为提供了全面的指引，善于网上学习、诚实友好交流、增强自护意识等，都是我们在网络世界中
前行的准则。
相关研究表明，我国青少年在网络安全行为和隐私保护意识等方面有所进步，但在网络信息的辨
识和批判能力上仍有待提高。这就需要我们积极参与网络实践活动，在实践中不断反思、成长。比如
参与学校组织的网络信息甄别竞赛，在与同学的比拼中，学习如何快速准确地辨别虚假信息；参与网
络文明宣传活动，向他人传播网络文明知识的同时，也强化自身的责任意识。
在这信息时代的网络浪潮中，青少年应手持理性思维的明灯，心怀构建清朗网络空间的责任。让
我们从自身做起，从每一次上网行为做起，以理性辨别信息，以善良温暖他人，以责任规范言行，共
同营造一个绿色、健康、文明的网络家园，让网络真正成为助力我们成长的广阔天地。
☞保分小题题型热点
保分小题题型题组（语用 +默写 +文言文）
保分小题 题组 (一)
一、语言文字运用 (20 分)
(一)语言文字运用 Ⅰ(本题共 2 小题 ，10 分)
阅读下面的文字 ，完成 1、2 题。
素纱 衣作为湖南省博物馆的馆藏珍品之一 ，被誉为 “世界上最轻薄的衣服 ”，整件衣物薄如蝉翼 ，
仅重 49 克。 20 世纪 80 年代 ，湖南省博物馆委托南京云锦研究所复制素纱 衣。但该研究所复制出来的第
一件素纱 衣的重量超过 80 克。后来 ，专家共同研究才找到答案 ，原来现在的蚕宝宝比几千年前的要肥胖
许多 ， ① ，所以织成的衣物自然也就重多了。紧接着专家们着手研究一种特殊的食料喂养蚕 ，控制蚕
宝宝的个头 ，成功为蚕宝宝 “减肥 ”，得到了现代社会能够获得的最细的蚕丝 —— 只有 平时蚕丝的五分之
一，解决了材料问题 。他们又按照当时的门幅装造 ，定制了一台 48 厘米幅宽的机台 。纹样参考考古报告中
的记载 ，经电脑测绘 ，一比一还原。织工师傅花了三个月时间提前训练手感 ，方才上机织造。染色时 ，经
多次试验 ，最后选用化学染料和红茶浸泡相结合的方法进行面料染色做旧处理 。前后耗时 13 年之久 ，这件
薄如蝉翼的复制品方才大功告成。尽管如此精心 ，最后织成的复制品 __ ②__ ，达 49.5 克。
霓裳羽衣 ，锦绣华裳 ，素纱 衣能够走出历史 ，让我们有幸目睹其美丽与风 采，背后是一代代考古工
作者几十年如一 日的艰辛付出与坚韧追求。
1．请在文中画横线处补写恰当的语句 ，使整段文字语意完整连贯 ，内容贴切 ，逻辑严密 ，每处不超过 10学科 网（北 京）股 份有限 公司
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个字。 (4 分)
答： ①
②
2．请用 5 个动词性短语概括素纱 衣复制过程中的几个关键步骤。 (6 分)
答：
(二)语言文字运用 Ⅱ(本题共 3 小题 ，10 分)
阅读下面的文字 ，完成 3～5 题。
是不是夏天被钉子 钉．住了？每天都是二十四至三十二摄氏度 。不算太热 ，热得并不极端 ，但是没有喘息 ，
没有变化 ，没有哪怕是短暂的缓解。而居然有了转机：今晚有阵雨 ，转中到大雨。太好了 ，太好了 ，下场
痛痛快快的大雨吧！便抬头看西北方 ，有云吗？快来了吧？等了一个夜晚 ，又一个白天。十点钟的时候来
了一阵雨 ，轻描淡写 ，点点滴滴 ，来得麻利 ，去得轻巧。这样的阵雨好洒脱哟 ，它似乎代表着一种飘逸、
自由 、灵巧的风格 ，它简直是一个梦 。这样的阵雨好不负责任哟 ，它干脆只是走一走过场 ，它像一个骗局 。
此夜星光灿烂 ，莫非预报了又预报 ，等待了又等待的中雨大 雨又 “黄”了？便无奈地躺在床上 ，体味
汗与被 褥特别是与枕头结合起来的陈年芳馨。
嗒。
嗒嗒。
嗒—— 嗒—— 嗒。
什么？有一本书落到地上了么？
是雨！是雨点声清晰可辨的雨。
嗒嗒嗒嗒嗒 …… 听声音就是大雨点。雨点愈来愈密 ，雨点愈来愈混成一片一团 ，而且声音变得响亮和
尖厉起来。突然一道青绿色的强光 ，一声炸雷震响在屋顶上 ，大雨像敲击重物一样砸在地上 ，没有节奏 ，
没有间歇 ，没有轻重缓急 ，只有夹带着哗啦哗啦的乒乓叮咚。
睡意全无了 ，只觉得高兴 ，觉得有趣 ，觉得 老天爷还是有两下子 ，便光着脊梁去淋雨去了。
3．下列句子中加点字和文中加点字 “钉”，用法上相同的一项是 (3 分)( )
A．他总结失败的教训 ，把失败接起来 ，焊．上去 ，作为登山用的尼龙绳子和金属梯子。
B．她们看见不远的地方 ，那宽厚肥大的荷叶下面 ，有一个人的脸 ，下半截身子 长．在水里。
C．春、夏、秋、冬 ，都是在他们的竹枝扫帚下 ，一个一个地被 扫．走了 ，又被 扫．来了。
D．母亲手拿针线灯下久坐 ，为我熬夜缝制军衣；线儿 缝在军大衣上，情意全 缝．在我心里。 学科 网（北 京）股 份有限 公司
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4．“这样的阵雨好洒脱哟 ”“ 这样的阵雨好不负责任哟 ”两句中的 “哟”字看似随意 ，实则别有意趣 ，请
简要分析。 (3 分)
答：
5．文中画横线句子如合并为一段 ，改成 “嗒。嗒嗒。嗒 —— 嗒—— 嗒”语义基本相同 ，但原文独立成段 ，
表达效果更好 ，为什么？ (4 分)
答：
二、名篇名句默写 (本题共 1 小题 ，6 分)
6．补写出下列句子中的空缺部分。
(1) 《六国论 》中，苏洵认 为六国破亡的原因是 “__ ”，并规劝当朝统治者 “__ ”。
(2) 辛弃疾在《永遇乐 ·京口北固亭怀古》中 ，感叹刘义隆想要效仿古人 ，建立 “__ ”那样的大
功，但由于行事草率 ，仓促北伐 ，最终却落得 “__ ”的惨败 ，这对比透露出辛弃疾反对轻率用
兵的思想。
(3) “瑟”是我国传统拨弦乐器 ，也是古诗词中的常用意象 ，可用来祭祀 ，也可象征高洁的情操 ，或者寄托
丰富的情感。如 “__ ，__ ”。
1．答案 ①吐出的丝更粗 ②还是重了 0.5 克
2．答案 控制蚕丝细度 ，定制机台宽度 ，还原纹样设计 ，上机进行织造 ，染色做旧处理。
3．答案 B
解析 B 项中加点字 “焊”和文中加点字 “钉”都运用拟物修辞手法。其他三项中的加点字都是比喻。
4．答案 ①“ 哟”是一个语气词 ，用在感叹句句尾 ，可以强化感叹语气 。②“这样的阵雨好洒脱哟 ”表达
了作者对阵雨洒脱的赞叹， “哟”字让这种赞叹语气更加强烈。 ③“ 这样的阵雨好不负责任哟 ”表达了作
者对阵雨不负责任的无奈与不满， “哟”字让这种无奈与不满情绪表露无遗。
5．答案 ①原文独立成段能更清晰地表现出听雨时声音变化的层次性 ，让读者有更真切的感受 。②独立成
段能更好地表达出 “我”听雨时专注而又迫切的心情。 ③独立成段能够增强文章的节奏感和音韵美 ，让读
者在阅读时有更深的体会。
6．答案 (1) 弊在赂秦 为国者无使为积威之所劫哉
(2) 封狼居胥 赢得仓皇北顾 学科 网（北 京）股 份有限 公司
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(3) 示例：锦瑟无端五十弦 一弦一柱思华年
保分小题 题组 (二)
一、语言文字运用 (20 分)
(一)语言文字运用 Ⅰ(本题共 2 小题 ，10 分)
阅读下面的文字 ，完成 1、2 题。
跑步是牵动身体骨骼 、关节 、韧带和肌肉的组合运动 ，不同类型的跑步方式对膝盖健康有着不同程度
的影响 。跑步可以分 “快跑 ”和“慢跑 ”两种 ，其中 ， A ，这一点毋庸置疑 ，但慢跑时要注意强度是否
适当、跑姿是否正确 ，以及是否有膝部基础病。
长期以来 ，①“跑步运动 ”和“膝盖健康 ”的关系非常备受关注 。②最近 ，“越跑步越伤膝盖系谣言 ”
的话题引发了网友们的热议 。③记者搜索后发现 ，④有人 将“跑步 ”与“久坐 ”进行了两种行为习惯对比 ，
⑤认为 “久坐 ”更伤膝盖。 ⑥网友引用调查数据来证明该观点 ，在排除了肥胖、职业工作负荷或既往受伤
史等影响因素后， ⑦休闲跑人群 (非竞技跑步 )的关节炎发生率只在 3.5% 以上 ，⑧而久坐不动人群的关节炎
发生率则为 10.2% ，⑨参加竞技比赛跑人群的关节炎发生率为 13.3% 。
由此 ，专家特别提醒 ， B ，因为久坐时膝盖长时间弯曲 ，可能影响到下肢血液循环 ， C ，缺血
缺氧就容易使膝盖受损。另外 ，人体关节需要像泵一样有节律地缓慢震 荡，长时间坐着，会使膝关节的弹
性降低，影 响软骨、血液、滑液相互之间的营养交换 ，从而易出现损伤。
1．请在文中画横线处补写恰当的语句 ，使整段文字语意完整连贯 ，内容贴切 ，逻辑严密 ，每处不超过 10
个字。 (6 分)
答： A.
B.
C.
2．文中第二段有四处表述不当 ，请指出其序号并做修改 ，使语言表达准确流畅 ，逻辑严密 ，不得改变原意 。
(4 分)
答：
(二)语言文字运用 Ⅱ(本题共 3 小题 ，10 分)
阅读下面的文字 ，完成 3～5 题。
夜幕沉沉地拉下来。 ①要不是有雪光反射 ，什么东西也不会看到。 枯树在风中 挣扎着 ．．． ，发出像用力敲
打根根扯紧的细钢丝那样刺耳寒心的颤声。那狂风 无情地 ．．． 横扫着雪野 ，把高处的雪刮到凹处去 ，把屋顶上 学科 网（北 京）股 份有限 公司
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的白被子掀掉 ，茅草不结实的部分就被大把大把地撕下来 ，摔撒到空中去。 ②低狭的茅草屋 ，在寒风中难
以自控地战栗着。 家家户户的窗口 ，都射出昏黄的灯光。很寂静 ，没有了惯常的狗叫声 ，这是为着八路军
和游击队活动方便 ，人们早把狗打杀干净了。
母亲正在拾掇逃难用的干粮 。德强从外 面走进来 ，脚步是那样缓慢 ，就和腿上带着两百斤东西 似的 ，几
乎抬不动了。他一屁股坐在已经揭去锅的灶台上。母亲有些诧异儿子这种异常的举动。 ③仔细一看 —— 德
强沮丧着脸 ，眼泪快掉下来了。 母亲懵怔一下 ，又领会到什么似的笑笑 ，对他说：
“不去就算了吧。人家是要去打仗 ，也不是闹着玩的 ，掉了队怎么办？ 跟着我跑也是一样的。 帮我拿
拿东西也好啊。 ”
“你不知道 ，别说啦！ ”德强把身子一扭 ，几乎是向母亲发火了 ，④寻思了一刹那 ，又转过身软和下
来说：
“妈，打日本鬼子 ，不分男女老少 ，我又是儿童团长，怎么能和老百姓一起，叫 鬼子撵着跑 ，那太没
出息啦！ ”
3．文中加点的 “挣扎着 ”“ 无情地 ”不可以删掉 ，原因是什么？请简要说明。 (3 分)
答：
4．下列说法正确的一项是 (3 分)( )
A．①句中的 “也”可以换成 “都”。
B．②句运用了夸张修辞。
C．③句中破折号与 “我—— 我—— 我只要见见我的萍儿 ”用法相同。
D．④句中 “刹那 ”“ 软和 ”的读音分别是 sh à nà、ru ǎn huo 。
5．请将文中画横线的句子改成问句 ，强调母亲要表达的意思 ，并指出这样改的好处。 (4 分)
答：
二、名篇名句默写 (本题共 1 小题 ，6 分)
6．补写出下列句子中的空缺部分。
(1) 贾谊《过秦论》中极言秦始皇功业的两句是 “__ ，__ ”。
(2) 高适的《燕歌行并序》中 “__ ，__ ”两句 ，从将士和君王两个角度来写汉将
去战场时的 威武荣耀。
(3) 诗句 “无边落木萧萧下 ，不尽长江滚滚来 ”写景阔远清旷 ，自古推为名句 ，《登快阁》中也有类似的诗 学科 网（北 京）股 份有限 公司
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句“__ ，__ ”。
1．答案 A．慢跑对膝盖有益 B．“久坐不动 ”更不可取 C．导致膝盖缺血缺氧
2．答案 ①句，删去 “非常 ”或“备”。④句，把“进行了 ”移到 “对比 ”之前 。⑥句，把“或”改为 “和”。
⑦句，把“只在 3.5% 以上 ”改为 “仅为 3.5% ”。
解析 ①句的 “非常 ”与“备”语义重复 ，可删掉其中一个。 ④句语序不当 ，“进行了 ”修饰动词 ，所以
应移到 “对比 ”之前 。⑥句用词不当 ，“肥胖 ”“ 职业工作负荷 ”“ 既往受伤史 ”这三个因 素是并列关系 ，
所以 “或”应改为 “和”。⑦句不合逻辑 ，“只在 3.5% 以上 ”应改为 “仅为 3.5% ”。
3．答案 ①这两个词运用了拟人的手法 ，写出了枯树在风中的形态以及狂风的肆虐 ，形象生动 。②在对自
然景物的描摹中融入情感 ，渲染了战前逃难时的紧张氛围 ，暗示了社会环境的险恶。
4．答案 A
解析 A 项，正确 。“也”与“都”用在否定句里表示语气的加强 。B 项，“难以自控地战栗着 ”所用修辞
为拟人 。C 项，③句中破折号表示提示下文 ，“我—— 我—— 我只要见见我的萍儿 ”表示说话断断续续 。D
项，“刹那 ”的“刹”读音为 ch à。
5．答案 改为 ：跟着我跑还不是一样？好处 ：原文是陈述句 ，感情色彩较淡 ，而改为反问句则可以加强母
亲安慰儿子的语气。
6．答案 (1) 吞二周而亡诸侯 履至尊而制六合
(2) 男儿本自重横行 天子非常赐颜色
(3) 落木千山天远大 澄江 一道月分明
保分小题 题组 (三)
一、语言文字运用 (20 分)
(一)语言文字运用 Ⅰ(本题共 2 小题 ，10 分)
阅读下面的文字 ，完成 1、2 题。
①摄影术自传入中国后 ，②在照相馆的实践活动中 ，③背景布就成为照相馆师傅发挥与创造的重要阵
地。 ④方寸尺幅间 ，中西方文化鱼龙混杂 ，⑤轿车、洋楼、罗马柱常常与中 式亭台楼榭混搭出现。 ⑥这尺
幅不大的背景布 ，反映着社会政治经济发展状况， ⑦以及社会意识形态、大众审美观念的流变与萌生 ，⑧
也折射出社会大众对物质、精神文化元素的想象与向往。
在物资匮乏的时代，人们受生活空间所限不能信步天下，却可以通过照相馆里的一方背景布实现精神
朝圣。 20 世纪二三十年代 ， A ，成了衡量照相馆实力的标准 ，花样翻新的布景更能吸引顾客。 20 世纪学科 网（北 京）股 份有限 公司
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60 年代 ，背景布大多反映社会主义国家建设成就 ，展现祖国大好河山 。进入 80 年代 ，背景布领先大众旅游
步伐 ，逐渐出现国际化的都市和景点。 伴随时代发展 ，描绘了大众美好期待的照相馆背景布，在小县城以
及更广泛的乡村照相馆里，有着更长久 的活力与影响 ，直到 2000 年之后数字技术兴起 ， B 。
在这个 AI 生成图像建构世界的时代 ，回头看向那些承载着个体精神寄托 、集体生活记忆的一幕幕画布 ，
感受到的 ，是一种带着古典意味的浪漫。
1．请在文中画横线处补写恰当的语句 ，使整段文字语意完整连贯 ，内容贴切 ，逻辑严密 ，每处不超过 15
个字。 (4 分)
答： A.
B.
2．文中第一段有三处表述不当 ，请指出其序号并做修改 ，使语言表达准确流畅 ，逻辑严密 ，不得改变原意 。
(6 分)
答：
(二)语言文字运用 Ⅱ(本题共 3 小题 ，10 分)
阅读下面的文字 ，完成 3～5 题。
“宝宝 ”都“上山 ”了(指蚕爬上稻草秆子 ，准备吐丝结茧 )，老通宝他们还是捏着一把汗 。他们钱都花
光了 ，精力也绞尽了 ，能不能有收获呢？到此时还没有把握 。虽则如此 ，他们还是硬着头皮去干 。“山棚 ”
下爇了火， 老通宝和儿子阿四伛着腰慢慢地从这边蹲到那边 ，又从那边蹲到这边。他们听见山棚上有些 屑．
屑索索 ．．． 的细声音 ，就忍不住想笑 ，过一会儿听不见了 ，他们的心又开始重甸甸地往下沉了。心是这样焦灼
着，却不敢向山棚上望。偶或他们仰着的脸上淋到了一滴蚕尿了 ，虽然觉得有点难过 ，但心里却快活： 巴．
不得 ．． 多淋一些。
“上山 ”后三天 ，熄火了 。老通宝的儿媳四大娘再也忍不住了 ，偷偷地挑开芦帘角看了一眼 ，她的心像
要从嗓子里蹦出来。 那是一片雪白 ，几乎连 “缀头 ”都瞧不见 ，那是四大娘有生以来从没见过的 “好大蚕
花”呀！老通宝全家立刻充满了欢笑。
3．下列各句中的 引号，和文中 “上山 ”的引号作用相同的一项是 (3 分)( )
A．遇到难事就推 ，遇到好事就抢 ，这样的 “聪明人 ”还是少一些为好。
B．AI 可以学习任何 “投喂 ”给它的内容 ，以便训练出一个完整、可用的模型。
C．“祥林嫂 ，你放着罢！我来摆。 ”四婶慌忙的说。
D．“包身工 ”的身体 ，以一种奇妙的方式包给了带工的老板。 学科 网（北 京）股 份有限 公司
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4．下列选项中 ，和文中画波浪线的句子所用修辞手法相同的一项是 (3 分)( )
A．你望着她那洁净得仿佛一分钟前才诞生的面孔 …… 心中会升起一种美好的感情。 (《哦 ，香雪》 )
B．叶子出水很高 ，像亭亭的舞女的裙。 (《荷塘月色》 )
C．层层的叶子中间 ，零星地点缀着些白花 ，有袅娜地开着的 ，有羞涩地打着朵儿的。 (《荷塘月色》 )
D．祥子愿意早早的拉车跑一趟 ，凉风飕进他的袖口。 (《骆驼祥子》 )
5．请结合文本 ，分别说说文中加点词语 “屑屑索索 ”“ 巴不得 ”的含义及作用。 (4 分)
答：
二、名篇名句默写 (本题共 1 小题 ，6 分)
6．补写出下列句子中的空缺部分。
(1) 青年如初升红日 ，肩负着 伟大复兴的重责 ，应当以 《〈论语 〉十二章 》中曾子所说的 “__ ，
__ ”自勉。
(2) 《劝学》设喻取譬层出不穷 ，比如以 “__ ，__ ”之喻 ，引出广泛学习且省察
自己就会见识通明且行为无过错的论断 ，更有说服力。
(3) “中华民族母亲河 ”—— 黄河 ，历来为文人墨客所青睐 ，以“黄河 ”为吟诵对象的古诗文名句有很多 ，
比如 “__ ，__ ”。
1．答案 A．背景布数量与类型的多少 B．才逐渐淡出乡村照 相馆的舞台
2．答案 语句 ①可修改为：自摄影术传入中国后。语句 ④可修改为：方寸尺幅间 ，中西方文化交错杂糅 。
语句 ⑦可修改为：以及社会意识形态、大众审美观念的萌生与流变。
解析 语句 ①，中途易辙 。结合 “摄影术自传入中国后 ”可知 ，此处主语是 “摄影术 ”，“自传入中国后 ”
是状语 ，这个分句没有说完 ，语句 ③主语变成了 “背景布 ”，表意混杂 ，故可改为 “自摄影术传入中国后 ”。
语句 ④，成语使用不当 ，“鱼龙混杂 ”比喻坏人和好人混在一起 ，故可 改为 “方寸尺幅间 ，中西方文化交
错杂糅 ”。语句 ⑦，语序不当， “流变与萌生 ”语序不当，应是先 “萌生 ”再“流变 ”，故可改为 “以及
社会意识形态、大众审美观念的萌生与流变 ”。
3．答案 B
解析 文中 “上山 ”的引号表示特殊含义 。A 项，表示讽刺和否定 ；B 项，表示特殊含义 ；C 项，表示引用
人物语言； D 项，表示特定称谓。
4．答案 A学科 网（北 京）股 份有限 公司
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解析 画波浪线的句子主要运用了夸张的修辞手法。四大娘的 “心像要从嗓子里蹦出来 ”，是她挑开芦帘
角看到 “好大蚕花 ”时的主观感受 ，写出了四大娘看到这一景象时内心的激动和喜悦。 A 项，运用了夸张
的手法 ，“那洁净得仿佛一分钟前才诞生的面孔 ”属于夸张的手法； B 项，运用了比喻的手法 ，将荷叶比
作舞女的裙； C 项，运用了拟人的手法 ，“袅娜 ”“ 羞涩 ”将荷花人格化； D 项，没有运用修辞手法。
5．答案 ①“ 屑屑索索 ”是拟声词 ，写出了蚕吐丝的声音 。老通宝和儿子连这样细小的声 音都能听见 ，表
现出他们对蚕吐丝结茧的关注、重视。 ②“ 巴不得 ”表示对某事物迫切盼望。老通宝和儿子希望蚕生命力
旺盛，表达了他们对蚕茧丰收的迫切期待。
6．答案 (1) 士不可以不弘毅 任重而道远
(2) 故木受绳则直 金就砺则利
(3) 示例：君不见黄河之水天上来 奔流到海不复回
保分小题 题组 (四)
一、语言文字运用 (20 分)
(一)语言文字运用 Ⅰ(本题共 2 小题 ，7 分)
阅读下面的文字 ，完成 1、2 题。
1981 年的夏天 ，关中西府的古周原上 ，绿汪汪 ．．． 蓬勃了一个春天的小麦 ，被阵阵暖风催逼着 ，摇身一变 ，
就都 透透地 ．．． 黄熟了 ，风摇着金黄的身姿 ，招引来勤劳的农人 ，下地挥镰 ，弯腰收割。我是割麦农人中的一
个，在我的身后 ，是我割倒打成捆子 ，一簇簇复又立起在地里的麦子。 我甚至听得见麦捆子的絮语 ，一簇
说我可香可香 了呢 ，说着还问着另一簇，你闻得见我的麦香味儿吗？被问着的那一簇麦捆子，很是不屑地
摇摇散乱的麦穗儿 ，回答说谁又不香呢？我也香呀 ！…… 就在麦捆子相互争香的时候 ，御风而来的风先生 ，
热烘烘 ．．． 撵到我的身边 ，帮我把弯着的腰扳直 ，指示我往麦地边的大渠看。我看见送报送信的邮递员骑着他
的绿色单车 ，风驰电掣般骑行在大渠边的道路上 ，赶到我家的麦地边了。
邮递员一脚离开单车的脚蹬 ，撑在路上 ，从他单车的帆布兜里 ，抽出两本杂志 ，冲着地里割麦的我喊 ：
“吴木匠 ，你的 杂志来咧！ ”
1．文中三个加点的重叠形式的词语 “绿汪汪 ”“ 透透地 ”“ 热烘烘 ”有怎样的表达效果？请简要说明。 (3
分)
答：
2．请简要赏析文中画横线句子运用的艺术手法。 (4 分)
答： 学科 网（北 京）股 份有限 公司
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(二)语言文字运用 Ⅱ(本题 共 3 小题 ，13 分)
阅读下面的文字 ，完成 3～5 题。
杭州西湖就是一座典型的 潟湖。它的南 、西、北三面环山 ，东邻城区 ，因而有 “三面云山一面城 ”的
说法 。自古以来 ， A 。这里风景迷人 ，绿柳丛中 ，车水马龙 ；堤岸之上 ，游人如织 。置身其中 ：令人不
禁吟咏起千古名句 —— ( )
__ B__ 。有学者认为 ，西湖所在地曾经出现过火山喷发 ，因火山口坍塌成为洼地 ，后来海水汇入而成为一座
火山口湖。还有学者认为 ，西湖是地质构造运动形成的凹陷湖盆汇水而成的构造湖。目前被广 泛接受的一
种观点是：西湖所在地原本是一处海湾 ，由于 钱塘江携带大量泥沙不断淤积 ，地形不断增高，海湾与海洋
之间的通道被逐渐堵塞， __ C__ 。
①历史上 ，人们在西湖边上修建治水工程 ，也加速了海洋与西湖的隔绝。 ②据史料记载 ，东汉时期 ，
当地居民修建了 “防海大塘 ”，③目的就是阻止海水入侵。 ④我国近代著名气象学家、地理学家和教育家
竺可桢先生 ，⑤根据钱塘江三角洲每年在海中伸涨的速度 ，⑥推算出西湖形成的时间至少在 12 000 年左右 。
3．请在文中横线处补写恰当的语句 ，使整段文字语意完整连贯 ，内容贴切，逻辑严密，每处不超过 15 个
字。 (6 分)
答： A.
B.
C.
4．文中第一段括号内 ，填写 的句子最恰当的一项是 (3 分)( )
A．“接天莲叶无穷碧 ，映日荷花别样红。 ”
B．“湖上春来似画图 ，乱峰围绕水平铺。 ”
C．“欲把西湖比西子 ，淡妆浓抹总相宜。 ”
D．“处处回头尽堪恋 ，就中难别是湖边。 ”
5．文中第三段有两处表述不当 ，请指出其序号并做修改 ，使语言表达准确流畅 ，逻辑严密 ，不得改变原意 。
(4 分)
答：
二、名篇名句默写 (本题共 1 小题 ，6 分)
6．补写出下列句子中的空缺部分。
(1) 古人对于水面的雾气常有不同的表达 ，如《赤壁赋》 “__ ”中的 “露”和《桂枝香 ·金陵怀 学科 网（北 京）股 份有限 公司
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古》 “__ ”中的 “烟”都指水汽。
(2) 《 阿 房 宫 赋 》 在 运 用 六 组 排 比 尽 情 揭 露 秦 王 朝 的 奢 靡 给 人 民 带 来 的 深 重 灾 难 后 ， 笔 锋 一 转 ， 让
“__ ”的“天下之人 ”同“__ ”的“独夫 ”进行较量 ，结果阿房宫变成了一片焦土 。
(3) 习近平总书记在 2024 年新年致辞中说 “黄河九曲 、长江奔流 ，总让人心潮湃 、豪情满怀 ”，古诗文中不
乏借黄河或长江抒发情感的句子 ，如“__ ，__ 。”
1．答案 ①这三个词都起到强化描写效果的作用 ，“绿汪汪 ”“ 透透地 ”分别强化了小麦蓬勃生长和成熟
时的状态 ，“热烘烘 ”突出了夏日的风的特点。 ②三个重叠词读起来朗朗上口 ，增强了语言的韵律美和节
奏感。
2．答案 ①运用了想象和拟人的手法 ，作者想象麦捆子之间的对话 ，将麦捆子人格化 。②使叙述和描写更
加生动 ，突出了作者对麦捆子的喜爱之情。
3．答案 A．西湖就是富有诗情画意的地方 B．关于西湖的成因曾经存在争议 C．于是就形成了封闭的
西湖
4．答案 C
解析 A 项，描写夏季的西湖 ；B 项，描写春季的西湖 。D 项，是说与西湖离别 。C 项，赞美西湖的整体美 ，
前文并未提及西湖的具体季节 ，只是赞美西湖 ，所以 C 项最合适。
5．答案 句①，“海洋与西湖的隔绝 ”改为 “西湖与海洋的隔绝 ”；句 ⑥，“至少 ”删掉。
解析 句①“加速了海洋与西湖的隔绝 ”语序不当 ，应把 “海洋与西湖的隔绝 ”改为 “西湖与海洋的隔
绝”。句 ⑥“至少在 12 000 年左右 ”前后矛盾 ，把“至少 ”删掉。
6．答案 (1) 白露横江 但寒烟衰草凝绿
(2) 不敢言而敢怒 日益骄固
(3) 示例：君不见黄河之水天上来 奔流到海不复回
保分小题 题组 (五)
一、语言文字运用 (20 分)
(一)语言文字运用 Ⅰ(本题共 2 小题 ，10 分)
阅读下面的文字 ，完成 1、2 题。
雨，下了一夜 。一团团黑云像筋疲力尽的逃兵 ，蜷缩在天际的一隅 。花园南端 ，曙光照临柚子树波动的
新叶 ，惊动了树下的阴影。时值斯拉万月 ，喷薄的旭日像不速之客 ，簌簌的笑声在枝头流荡。于是 ，沐浴
阳光的情思 ，在邈远的心空飘游。 学科 网（北 京）股 份有限 公司
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时光仿佛凝固了。
下午 ，突然响 起的隆隆雷声 ，似在发出信号 。顷刻之间 ，云团离开倒卧的所在 ，膨胀着 ，呼啸着 ，飞驰
而来。堤坝里的水变得黑黝黝的，沉重的幽暗落在榕树底下。远 处的树叶奏起了下雨的前奏。转眼间大雨
滂沱 ，天空白茫茫的，地上一片汪洋。年老的林木甩动着蓬发似的枝梢像正在戏耍的顽童，硕大的棕榈叶
也失去了往常的恬静。在不多久的时候，风止雨停了，青空像被擦拭过一般。 一钩纤弱的月亮虽然刚离开
病榻 ，但脸上仍挂着慵倦的笑意在天宇漫步。
心儿对我说 ，我见到的一切细小的东西都不愿自行消亡 。无数鲜活的瞬间登上我七十岁的渡口 ，随即驶
向“无形 ”。只有几许懈怠的时日被我留住 ，留在了平庸的诗歌里；它们告诉后人一件不平常的事 —— 我
曾观赏过这些美妙的景象。
1．文中 “时光仿佛凝固了 ”七个字独立成段 ，在内容上有什么好处？ (4 分)
答：
2．简要分析文中画横线句子的表达效果。 (6 分)
答：
(二)语言文字运用 Ⅱ(本题共 3 小题 ，10 分)
阅读下面的 文字，完成 3～5 题。
“人一天到底睡多久最好？ ”这个问题可以说是睡眠话题里最高频的问题了。那么人一天到底睡多久
最好呢？ __ ①__ ，不同年龄每天所需的睡眠时间不同。总体来看 ，婴幼儿所需的睡眠时间最多 ，随着年龄
的增长 ，人类所需的睡眠时间有减少的趋势 ，一直到 18 岁以后 ，变化才慢慢趋于稳定。
这个世界上存在部分人 ，他们不需要睡那么多 ，睡 5～6 个小时就够了 。所谓的 “够了 ”，意思是不影
响白天状态 ，白天不需要打盹 ，也不需要咖啡 ，工作发挥和其他常人一样。这种人叫作 “短睡眠者 ”。和
短睡眠者相反 ，还有部分人 ，天生需要睡很久 ，比如 ：常常需要睡 9 小时甚至 10 小时以上才能保证白天的
精神状态 ， A 。
如何判断自己每天需要睡多久呢？简单说：看醒后的表现。如果你醒来后在多数时间都能保持足够的清醒
状态 ，不影响工作和生活 ，那么就可以认为你睡得可以； __ ②__ ，如果醒后容易嗜睡 ，注意力下降 ，那么
就意味着你睡得不够好 —— 或者是时间不够 ，或者是质量不佳 ，或者两者都有。
__ ③__ ，你需要多少睡眠 ，可以和别人不一样 。结合之前 说的 ，现在可以总结两句话 ：你的睡眠时间 ，
不要和过去比；你的睡眠时间， __ B__ 。学科 网（北 京）股 份有限 公司
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3．请在文中序号 ①②③ 三处 ，填入恰当的虚词。 (3 分)
答：
4．请在文中 A、B 两处补写恰当的语句 ，使整段文字语意完整连贯 ，内容贴切 ，逻辑严密 ，每处不超过 12
个字。 (4 分)
答： A.
B.
5．请具体说明材料第三段破折号的作用 ，并简要说明在这里使用破折号有什么好处。 (3 分)
答：
二、名篇名句默写 (本题共 1 小题 ，6 分)
6．补写出下列句子中的空缺部分。
(1) 孔子非常善于鼓励学生 ，在《子路、曾皙、冉有、公西华侍坐》中 ，当曾点说自己和别人想法不同时 ，
孔子循循善诱： “__ ？__ 。”
(2) 陶渊明在《归园田居 (其一 )》中 ，上下两句均使用叠词来表现乡居生活中祥和安宁的烟火气的句子是：
“__ ，__ 。”
(3) 岁寒雪至 ，老师带着同学们赏雪 ，大家纷纷吟诵起古人咏雪的诗词 ，老师问小光能否记起古人以雪之洁
白来做比喻的诗词句 ，小光回答道： “__ ，__ 。”
1．答案 七个字独立成段 ，突出时光的沉静和停滞 ，与后文突然响起的隆隆雷声和看到的滂沱大雨形成鲜
明对比 ，进一步凸显了天气的变化之快。
2．答案 运用拟人手法 ，把月亮比拟成刚离开病榻的虚弱病人 ，但仍笑着在天宇漫步 ，突出作者所见景物
的美丽和活力。
3．答案 ①其实 ②反之 ③所以
4．答案 A．这种人叫作 “长睡眠者 ” B．不要和别人比
5．答案 材料第三段的破折号表示解释说明 。破折号后面的内容进一步解释了睡得不够好的原因 ，有助于
读者更好地理解材料内容。
6．答案 (1) 何伤乎 亦各言其志也
(2) 暧暧远人村 依依墟里烟 学科 网（北 京）股 份有限 公司
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(3) 示例：惊涛拍岸 卷起千堆雪
保分小题 题组 (六)
一、语言文字运用 (20 分)
(一)语言文字运用 Ⅰ(本题共 2 小题 ，9 分)
阅读下面的文字 ，完成 1、2 题。
古树名木是记录自然生态变迁的 “活化石 ”，是承载民族历史记忆的绿色 “国宝 ”。从“万条垂下绿
丝绦 ”的春意盎然 ，到“晴川历历汉阳树 ”的壮阔悠远 ，从《诗经》里 “其叶 牂牂 ”的东门之杨 ，到《项
脊轩志》里 __ ①__ ，树木寄托的是人与人、人与故乡、人与自然的情与意 ，是中国历史传承至今的朴素生
态观和绿色发展理念。
古树名木是自然与文化的共同遗存 ， ② 。而古树名木的保护是一项系统性工程 ，今天做好古树名
木保护要坚持守正创新 ，保护古树名木 ，要讲好古树故事 ，挖掘古树名木背后的文化和精神价值。陕西黄
帝陵的轩辕柏 ，山西洪洞的大槐树 ，安徽黄山的迎客松 ，这些古树穿越时空 ，是中华文明的一部分；塞罕
坝的落叶松 ，大漠深处的胡杨林 ，种子 “飞”上太空的普陀鹅耳枥 ，这些名 木记录着国家和民族发展进步
的印记。应 __ ③__ ，讲好并传承好这些故事 ，让．它们真正成为有记忆的地标、可触摸的历史 ，成为坚定文
化自信的有力支撑。保护古树名木不仅是为子孙后代留下一棵棵树、一片片林 ，更是通过对古树名木的保
护，延续文化传统、传承发展理念 ，共建人与自然和谐共生的美丽中国。
1．下列句子中的 “让”与文中加点的 “让”，意义和用法相同的一项是 (3 分)( )
A．让我说 ，我们应该这样做。
B．她还真是巾帼不让须眉啊！
C．我的失误让整个团队陷入了 困境。
D．他总是先人后己 ，总是把方便让给别人。
2．请在文中横线处补写恰当的语句 ，使整段文字语意完整连贯 ，内容贴切 ，逻辑严密 ，每处不超过 15 个
字。 (6 分)
答： ①
②
③
(二)语言文字运用 Ⅱ(本题共 3 小题 ，11 分)
阅读下面的文字 ，完成 3～5 题。
医学理想的萌发 ，来自青春期的一场大病 。16 岁时 ，屠呦呦患肺结核 ，经过两年多的治疗才得以康复 。学科 网（北 京）股 份有限 公司
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“医药的作用很神奇 ，我当 时就想 ，如果我学会了，不仅可以让自己远离病痛，还可以救治更多人，何乐
而不为呢？ ”屠呦呦对记者回忆道。
1951 年，屠呦呦如愿考入北大医学院。在关于屠呦呦的原始材料里 ，有一张照片保存得很好 ，照片中 ，
屠呦呦佩戴着北京大学的校徽 ，照片旁是她的亲笔备注： “佩戴北京大学校徽的大学生。 ”在这里 ，她打
下了扎实的中西医知识功底。
20 世纪 60 年代 ，疟疾横行、生命临危。背负着党和国家的希望 ，屠呦呦挂帅 “523 ”小组 ，与死神的较
量拉开帷幕。 190 次失败的打击 ，因缺乏防护所致的中毒性肝炎的折磨 ，以身试药的风险 …… 终于 ，197 8
年，屠呦呦小组成功研制青蒿素 。2015 年，屠呦呦获得诺贝尔生理学或医学奖。 “在全球疟疾的综合治疗
中，青蒿素至少降低了 20% 的死亡率及 30% 的儿童死亡率。 ”诺奖评选委员会在评选资料上特别指出。
2016 年，屠呦呦踏上研究青蒿素治疗红斑狼疮的新征程 。2018 年初 ，双氢青蒿素对红斑狼疮的治疗效
果最终得以确认。屠呦呦告诉记者： “青蒿素对治疗红斑狼疮存在有效性趋势 ，我们对试验成功持谨慎的
乐观。 ”
3．新闻强调真实性 ，请结合材料简要概括 作者采集材料的不同渠道。 (4 分)
答：
4．如以 “以创造性劳动造福人类 ”为题写一 篇新闻评论 ，请结合材料 ，写出三个评论角度 ，每个角度内容
不超过 12 个字。 (3 分)
答：
5．文中画横线的部分表现了屠呦呦坚定的科研精神 ，这一表达效果是怎么取得的？ (4 分)
答：
二、名 篇名句默写 (本题共 1 小题 ，6 分)
6．补写出下列句子中的空缺部分。
(1) 读史以明鉴 ，前朝灭亡的教训往往给后世深刻的启示。诸如：杜牧《阿房宫赋》中的 “__ ”
一针见血地指出秦国的灭亡完全是咎由自取 ；苏洵在 《六国论 》中警告统治者 “__ ”，否则国
力日渐亏损终至灭亡。
(2) 在古典诗文中 ，有许多关于飞鸟的意象 ，作者借它传情达意 ，读来意趣横生。诸如：《蜀道难》
“__ ”一句中 ，它善于高飞却难 以逾越蜀山，侧面烘托蜀道高险难行；《梦游天姥吟留别》 学科 网（北 京）股 份有限 公司
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“__ ”一句中 ，它是只有在梦境才能看到的神奇之鸟；《登高》 “__ ”一句中 ，
它低飞盘旋 ，仿佛作者漂泊无助的身影 ；《蜀相 》“__ ”一句中 ，它叫声美好却无人赏识 ，暗
含作者怀才不遇的深深悲愤。
1．答案 C
解析 文中加点的 “让”，指使、致使 ﹑容许或听任。 A 项，跟“看、说 ”搭配使用 ，表示主观看法 ﹐相
当于 “依”或“照”。B 项，指亚于 ，不如 (用于否定式 )。C 项，指使 、致使 、容许或听任 。D 项，指把方
便或好处给别人。
2．答案 ①“ 亭亭如盖 ”的枇杷树 ②应受到重视与保护 ③深挖古树名木背后的故事
3．答案 ①直接采访被采访者 (屠呦呦 )本人 ，获取其真实回忆 、经历 。②参考关于被采访者 (屠呦呦 )的原始
材料 (实物 )，包括亲笔备注的照片等 。③查阅诺贝尔奖评选委员会的资料 ，了解青蒿素在全球疟疾治疗中的
作用和贡献。
4．答案 ①创造性劳动的体现。 ②创造性劳动的条件。 ③劳动成果对人类的作用。
5．答案 ①通过细数打击﹑折磨 、风险等经历﹐从不同的角度写出了科研路上的困难之大 ；省略号表示前
面的困难列举未尽 ，突出科研路上的挑战之多。用困难之大、挑战之多来反衬屠呦呦科研精神的坚定。 ②
用副词 “终于 ”与具体的年份 “1978 年”，强调经过了漫长的过程 ，最后研究获得了成功 ，从而表现了屠
呦呦坚定的科研精神。
6．答案 (1) 族秦者秦也 为国者无使为积威之所劫哉
(2) 黄鹤之飞尚不得过 虎鼓瑟兮鸾回车 渚清沙白鸟飞回 隔叶黄鹂空好音
保分小题题组 —— 文言文阅读
(一)文言文阅读 (本题共 5 小题 ，20 分)
(人物型 )阅读下面的文言文 ，完成 1～5 题。
材料一：
徐渭 ，字文长 ，山阴人 ，为诸生 ，有盛名 。总督胡宗宪招致幕府 ，宗宪得白鹿 ，将献诸朝 ，令渭草表 ，
并他客草寄所善学士 ，择其尤上之。学士以渭表进 ，世宗大悦 ，益宠异宗宪 ，宗宪以是益重渭。 督府势严
重，将吏莫敢仰视。渭角巾布衣，长揖纵谈。 幕中有急需 ，夜深开戟门以待。渭或醉不至 ，宗宪顾善之。
渭知兵 ，好奇计 ，宗宪擒徐海 ，诱王直 ，皆预其谋 。藉宗宪势 ，颇横 。及宗宪下狱 ，渭惧祸 ，遂发狂 ，
引巨锥 剚耳， 深数寸。已 ，又击杀继妻 ，论．死系狱 ，里人张元忭力救得免。乃游金陵 ，抵宣、辽 ，纵观诸
边厄塞 ，善李成梁诸子 。入京师 ，主元忭 。元忭导以礼法 ，渭不能从 ，久之怒而去 。后元忭卒 ，白衣往 吊．，学科 网（北 京）股 份有限 公司
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抚棺恸哭 ，不告姓名去。
渭天才超轶 ，诗文 绝出伦辈 。善草书 ，工写花草竹 石。当嘉靖时 ，王、李倡七子社 ，谢榛以布衣被摈 。
渭愤其以轩冕压韦布，誓不入二人党。后二十年， 公安袁宏道游越中得渭残帙以示祭酒陶望龄相与激赏刻
其集行世 。
(节选自《明史 ·徐渭传》 )
材料二：
文长既已不得志于有司 ，遂乃放浪曲 糵，恣情山水 ，走齐、鲁、燕、赵之地 ，穷览朔漠。其所见山奔
海立、沙起雷行、雨鸣树偃、幽谷大都、人物鱼鸟 ，一切可惊可愕之状 ，一一皆达之于诗。其胸中又有勃
然不可磨灭之气 ，英雄失路 、托足无门之悲 ，故其为诗 ，如嗔如笑 ，如水鸣峡 ，如种出土 ，如寡妇之夜哭 ，
羁人之寒起。虽其体格时有卑者，然匠心独出，有王者气，非彼巾帼而事人者所敢望也。文有卓识，气沉
而法严，不以摸拟损才，不以议论伤格，韩、曾之流亚也。 文长既雅不与时调合 ，当时所谓骚坛主盟者 ，
文长皆叱而奴之 ，故其名不出于越 ，悲夫！喜作书 ，笔意奔放如其诗 ，苍劲中姿媚跃出 ，欧阳公所谓 “妖
韶女老 ，自有余态 ”者也。间以其余 ，旁溢为花鸟 ，皆超逸有致。然文长竞以不得志于时 ，抱愤而卒。
石公曰： “先生数奇不已 ，遂为狂疾；狂疾不已 ，遂为囹圄。古今文人牢骚 困苦 ，未有若先生者也。
虽然 ，胡公 间世豪杰 ，永陵英主 ，幕中礼数异等 ，是．胡公知有先生矣 ；表上 ，人主悦 ，是人主知有先生矣 ，
独身未贵耳。先生诗文崛起 ，一扫近代芜秽之习 ，百世而下 ，自有定论 ，胡为不 遇．哉？ ”余谓文长无之而
不奇者也。无之而不奇 ，斯无之而不奇也。悲夫！
(节选自《古文观止 ·徐文长传》 )
1．材料一画波浪线的部分有三处需要断句 ，请用铅笔将相应位置的答案标号涂黑。 (3 分)
公安袁宏道游越 A 中 B 得渭残帙 C 以示 D 祭酒 E 陶望龄 F 相与激赏 G 刻其集 H 行世
2．下列对材料中加点的词语及相关内容的解说 ，不正确的一项是 (3 分)( )
A．“论死系 狱”和“单于使使晓武，会论虞常 ”(《苏武传》 )两句中的 “论”字含义相同。
B．“白衣往吊 ，抚棺恸哭 ”和“茕茕孑立 ，形影相吊 ”(《陈情表》 )两句中的 “吊”字含义相同。
C．“是胡公知有先生矣 ”和“觉今是而昨非 ”(《归去来兮辞并序》 )两句中的 “是”字含义不同。
D．“胡为不遇哉 ”和“臣以神遇而不以目视 ”(《庖丁解牛》 )两句中的 “遇”字含义不同。
3．下列对材料有关内容的概述 ，不正 确的一项是 (3 分)( )
A．徐渭是明代著名的文学家、书画家。他喜好书法 ，擅长草书 ，擅长画花草竹石 ，都超逸有情致。
B．徐渭诗作格调高雅 ，匠心独运 ，有王者之气。他的文章有卓越的见解 ，气度沉静而且章法谨严。
C．徐渭性格狂放不羁 ，命途多舛 ，仍有胡宗宪这样百年难遇的豪杰、世宗这样英明的君主赏识他。 学科 网（北 京）股 份有限 公司
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D．徐渭纵情山水 ，走遍齐 、鲁、燕、赵等地 ，饱览塞外大漠风光 ，一切令人惊讶的情状 ，他都一一化入诗
中。
4．把材料中画横线的句子翻译成现代汉语。 (8 分)
(1) 幕中有急需 ，夜深开戟门以待。渭或醉不至 ，宗宪顾善之。
译文：
(2) 文长既雅不与时调合 ，当时所谓骚坛主盟者 ，文长皆叱而奴之 ，故其名不出于越。
译文：
5．“无之而不奇 ，斯无之而不奇也 ”是对徐渭的高度推崇 ，请结合两则材料简要概括徐渭之 “奇”。(3 分)
答：
(二)文言文阅读 (本题共 5 小题 ，20 分)
(事理型 )阅读下面的文言文 ，完成 1～5 题。
材料一：
子路曰： “桓公杀公子纠 ，召忽死之 ，管仲不死。 ”曰： “未仁乎？ ”子曰： “桓公九合诸侯 ，不以
兵车 ，管仲之力也。如其仁。 ”
子贡曰： “管仲非仁者与？桓公杀公子纠 ，不能死 ，又相之。 ”子曰： “管仲相桓公 ，霸诸侯 ，一匡
天下 ，民到于今受其赐。 微．管仲 ，吾其被发左衽矣。岂若匹夫匹妇之为谅也 ，自经于沟渎而莫之知也？ ”
(节选自《论语 ·宪问》 )
材料二：
古之君子 ，其责己也重以周 ，其待人也轻以约。 重以周 ，故不怠；轻以约 ，故人乐为善。闻古之人有
舜者 ，其为人也 ，仁义人也。求其所以为舜者 ，责于己曰： “彼，人也；予 ，人也。彼能是 ，而我乃不能
是！ ”早夜以思 ，去其不如舜者 ，就．其如舜者。闻古之人有周公者 ，其为人也 ，多才与艺人也。求其所以
为周公者 ，责于己曰： “彼，人也；予 ，人也。彼能是 ，而我乃不能是！ ”早夜以思 ，去其不如周公者 ，
就其如周公者。舜、周公大圣人也 ，后世无及焉。是人也 ，乃曰： “不如舜，不如周公，吾之 病．也。 ”是
不亦责于身者重以周乎？其于人也 ，取其一 ，不责其二；即其新 ，不究其旧。恐恐然惟惧其人之不得为善
之利。不亦待于人者轻以约乎？
今之君子则不然。其责人也详 ，其待己也廉。详 ，故人难于为善；廉 ，故自取也少。己未有善 ，曰： 学科 网（北 京）股 份有限 公司
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“我善是 ，是亦足矣。 ”己未有能 ，曰： “我能是 ，是亦足矣。 ”外以欺于人 ，内以欺于心 ，未少有得而
止矣。不亦待其身者已廉乎？其于人也 ，曰： “彼虽能是 ，其人不足 称．也；彼虽善是 ，其用不足称也。 ”
举其一 ，不计其十；究其 旧，不图其新。恐恐然惟惧其人之有闻也。是不亦责于人者已详乎？夫是 之谓不
以众人待其身 ，而以圣人望于人，吾未见其尊己也。
虽然，为是者，有本有原，怠与忌之谓也。怠者不能修，而忌者畏人修。 是故事修而谤兴德高而毁来
士之处此世而望名誉之光道德之行难已 。
(节选自《原毁》 )
1．材料二画波浪线的部分有三处需要断句 ，请用铅笔将相应位置的答案标号涂黑。 (3 分)
是故事 A 修而谤 B 兴 C 德高 D 而毁来 E 士之处 F 此世 G 而望 H 名誉之光道德之行难已
2．下列对材料中加点的词语及相关内容的解说 ，不正确的一项是 (3 分)( )
A．微，是没有的意思 ，与《烛之武退秦师》中 “微夫人之力不及此 ”的“微”意思相同。
B．就，是效仿的意思 ，与《劝学 》中 “金就砺则利 ”的“就”意思不同。
C．病，是缺点的意思 ，与《种树郭橐驼传》中 “故病且怠 ”的“病”意思相同。
D．称，是称赞的意思 ，与《屈原列传》中 “其称文小而其指极大 ”的“称”意思不同。
3．下列对材料有关内容的概述 ，不正确的一项是 (3 分)( )
A．孔子认为 ，管仲帮助齐桓公召集诸侯会盟 ，息兵戈而解纷争 ，使天下由此而安 ，为维护和平作出了贡献 ，
这就是他的仁德。
B．舜和周公都是圣人仁人 ，古时的君子努力探究他们成为圣人的原因 ，日夜反思 ，改掉自己不如他们的地
方，努力向他们看齐。
C．无论古代的君子还是现代的君子 ，他们对待别人的标准与自己不同 ，原因在于 “怠”与“忌”，这也是
当时不良社会风气产生的根源。
D．材料二多处采用对比手法 ，全篇行文严肃而恳切 ，句式整齐中富有变化 ，语言生动而形象 ，刻画当时的
士风 ，可谓入木三分。
4．把材料中画横线的句子翻译成现代汉语。 (8 分)
(1) 管仲非仁者与？桓公杀公子纠 ，不能死 ，又相之。
译文：
(2) 古之君子 ，其责己也重以周 ，其待人也轻以约。
译文： 学科 网（北 京）股 份有限 公司
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5．在待人方面 ，两则材料有何相同之处？请简要分析。 (3 分)
答：
(三)文言文阅读 (本题共 5 小题 ，20 分)
(议论型 )阅读下面的文言文 ，完成 1～5 题。
材料一：
以全．举人固难 ，物之情也。人伤尧以不慈之名 ，舜以卑父之号 ，禹以贪位之意。由此观之 ，物岂可全
哉？故君 子责人则以人 ，自责则以义。责人以人则易足，易足则得人。不肖者则不然。责人则以义，自责
则以 人。责人以义则难赡 ，难赡则失亲。尺之木必有节目，寸之玉必有瑕 瓋。先王知务之不可全也，故择
务而贵取一也。宁戚欲干齐桓公，穷困无以自进，于是为商旅将 任车 ．． 以至齐。桓公郊迎客 ，夜开门 ，辟任
车，爝火甚盛 ，从者甚众。 宁戚饭牛居车下 ，望桓公而悲 ，击牛角疾歌。 桓公闻之 ，曰： “异哉！之歌者
非常人也！ ”命后车 ．． 载之。桓公赐之衣冠 ，将见之。宁戚见 ，说桓公以治境内。明日复见 ，说桓 公以为天
下。桓公大说 ，将任之。群臣争之曰： “客，卫人也。 卫之去齐不远君不若使人问之而固贤者也用之未晚
也。”桓公曰 ：“不然 。问之 ，患其有小恶 。以人之小恶 ，亡人之大美 ，此人主之所以失天下之士也已 。”
人固难全 ，权．而用其长者当举也。
(节选自《吕氏春秋 ·举难》 )
材料二：
或曰： “人材有能大而不能小 ，犹函牛之鼎不可以烹鸡。 ”愚以为此非名也。岂有能大而不能小乎？
凡所谓能大而不能小 ，其语出于性有宽急 。性有宽急 ，故宜有大小 。宽弘之人 ，宜为郡国 ，使下得施其功 ，
而总成其事。急小之人，宜理百里，使事办于己。然则郡之与县， 异体之大小者也。以实理宽急论辨之 ，
则当言大小异宜，不当言能大不能小也。若夫鸡之与牛，亦异体之小大也，故鼎亦宜有大小。 若以烹犊 ，
则岂不能烹鸡乎？ 故能 治．大郡 ，则亦能治小郡矣。推此论之 ，人材各有所宜 ，非独大小之谓也。
(节选自《人物志 ·材能》 )
1．材料一画波浪线的部分有三处需要断句 ，请用铅笔将相应位置的答案标号涂黑。 (3 分)
卫之去 A 齐 B 不远 C 君 D 不若使人问 E 之 F 而固贤者 G 也 H 用之未晚也
2．下列对材料中加点的词语及相关内容的解说 ，不正确的一项是 (3 分)( )
A．全，指完美 ，与成语 “求全责备 ”中的 “全”词义相同。 学科 网（北 京）股 份有限 公司
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B. 任车 ，指装载货物的车 ，与下文宁戚乘坐的 “后车 ”意思不同。
C．权，指权衡 ，与《过秦论》中 “比权量力 ”的“权”词义不同。
D．治，指治理 ，与成语 “励精图治 ”中的 “治”词义不同。
3．下列对材料有关内容的概述 ，不正确的一项是 (3 分)( )
A．一尺长的树木必定会有节疤 ，一寸大的玉石必定会有瑕疵 ，即使古代圣贤 —— 尧、舜、禹等 ，也有受人
诋毁之处。
B．君子得到人民拥护 ，不贤的人失去亲近的人 ，其根本在于要求不同 ，君子用义的标准要求自己 ，用一般
标准要求别人。
C．齐桓公没有接纳群臣的建议 ，不调查宁戚品行就准备对他加以重用 ，是因为担心会查出宁戚的小毛病从
而失去人才。
D．材料二将性情宽宏的人与急躁的人进行对比 ，认为前者意义价值更大 ，因为性情宽宏的人能统筹大事 ，
治理郡国。
4．把材料中画横线的句子翻译成现代汉语。 (8 分)
(1) 宁戚饭牛居车下 ，望桓公而悲 ，击牛角疾歌。
译文：
(2) 若以烹犊 ，则岂不能烹鸡乎？
译文：
5．(新考向 ·情境任务 )学校 将举行模拟招聘会 ，请从招聘者或应聘者中任选一个角度 ，谈谈材料对你的启发 。
(3 分)
答：
(一)文言文阅读
1．答案 BFG
解析 “越中 ”作“游”的宾语 ，不可断开 ，应在 B 处断开。 “祭酒陶望龄 ”，官职与姓名间不断开 ，应
在 F 处断开。 “刻其集行世 ”句意完整 ，中间不断开 ，应在 G 处断开。
2．答案 B
解析 B 项，“含义相同 ”错误 ，悼念 /安慰。
3．答案 B学科 网（北 京）股 份有限 公司
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解析 B 项，曲解文意 ，“徐渭诗作格调高雅 ”错误 ，由材料二原文 “虽其体格时有卑者 ”可知 ，其诗作
的体裁格调有时也有卑下的。
4．答案 (1) 幕府中若有紧急之需 ，会在夜深时打开大门来等待徐渭 。徐渭有时喝醉了不能来 ，胡宗宪却好
好对待他。
(2) 徐文长向来不能与时俗苟合 ，当时所谓诗坛的领袖 ，徐文长都加以抨击并把他们看作奴才 ，所以他的名
声不能传出越地。
5．答案 ①才能奇异； ②性情奇怪； ③遭际奇特。
【参考译文】
材料一：
徐渭 ，字文长 ，山阴人 ，做学生时 ，有盛大的名气。总督胡宗宪招他到幕府里 ，胡宗宪猎得白鹿 ，将
要献给朝廷 ，让徐渭起草奏章 ，与其他门客所起草的奏章一道寄给跟自己亲善的学士 (评判 )，挑选其中特别
好的上送朝廷 。学士把徐渭所写的奏章进献 (给皇上 )，世宗非常高兴 ，更加尊崇和宠爱胡宗宪 ，胡宗宪因此
更看重徐渭 。督府位高权重 ，将领和官吏们没有谁敢仰视 。而徐渭头戴角巾身着布衣 ，(只)行长揖之礼放纵
无拘地交谈。幕府中若有紧急之需 ，会在夜深时打开大门来 等待徐渭。徐渭有时喝醉了不能来 ，胡宗宪却
好好对待他。
徐渭懂军事，喜欢出奇妙计策，胡宗宪擒住徐海，诱捕王直， (徐渭 )都参与了其中的谋划。 (徐渭 )依
凭胡宗宪的权势，很骄横。等到胡宗宪被投入监狱，徐渭害怕灾祸 (及身 )，于是发狂， (竟)拿巨大的锥子
刺自己的耳朵，深入几寸。后来，又打死了续娶的妻子，被判死罪拘押在监狱里， (他的 )同乡张元忭竭力
挽救才使他得以免除死罪。于是游历金陵，到达宣、辽等地，纵观各个 边关险塞 ，与李成梁的几个儿子关
系良好 。(徐渭 )进入京师后 ，在张元忭家中居住 。张元忭用礼法来教导他 ，(但)徐渭不能听从 (张元忭的教
导)，时间长了就生气离开。后来张元忭去世， (徐渭 )身穿白衣前往悼念，扶着棺木大哭，也不告诉 (主人
家)姓名就离开了。
徐渭天资才能过人，诗文水平超出同辈。 (他)善于写草书，擅长画花草竹石。在嘉靖年间，王世贞、
李攀龙倡导 (成立 )七子社，谢榛因为是平民身份被排斥。徐渭愤恨他们用官位爵禄来欺压寒士，发誓不入
这二人的团体 。之后二十年 ，公安的袁宏道在越中游历 ，得到徐渭的残卷后把它拿给祭酒陶望龄 看，(两人 )
都非常欣赏， (于是就 )刻印了徐渭的文集使之流行于世。
材料二：
徐文长既然已经在官场上不得志 ，于是就放纵饮酒 ，纵情山水 ，游历了齐、鲁、燕、赵等地 ，(又)尽
览了塞外的大漠 (风光 )。他所见的山峦起伏 、海浪壁立 、黄沙满天 、云雾升腾 、雷雨交加 、树木倒伏 、幽谷 学科 网（北 京）股 份有限 公司
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闹市 、奇人异士 、珍稀鱼鸟 ，一切令人惊讶的情状 ，(他)都一一化入诗中 。他胸中又有旺盛高涨不能消逝的
壮志 ，英雄不得志 、无处寄身的悲痛 ，所以他写的诗 ，像是在生气 (又)像是在嬉笑 ，像水在峡谷奔流而出的
声音 ，像种子破土而出 ，像寡妇在深夜的哭声 ，(像)客居在外的人迎寒启程 。虽然他诗作的体裁格调有时也
有卑下的 ，但是匠心独运 ，有王者之气 ，不是那种像以色事人的女子一般的媚俗的诗作所能赶得上的。他
的文章有卓越的见解 ，气度沉静且章法谨严 ，(他)不会因为墨守成规而损伤自己的才能 ，也不会因 (无节制
的)议论伤害文章的风格 ，如同韩愈 、曾巩这类人的文章 。徐文长向来不能与时俗苟合 ，当时所谓诗坛的领
袖，徐文长都加以抨击并把他们看作奴才，所以他的名声不能传出越地，可悲啊！ (徐文长 )喜好写书法，
用笔奔放就像他的诗一样，在苍劲豪迈中又使妩媚的姿态跃然纸上，正 是欧阳公所称的 “妖娆的美女 (即使
已经 )年老 ，(仍)风韵犹存 ”。(徐文长 )偶尔也会把他剩余的精力，倾注到花鸟画上，也都超脱不俗极有情
致。但徐文长终究因为在当时不得志，怀着愤恨去世。
石公说： “先生的命运不好 ，诸事不顺 ，最终得了狂病；狂病没有治愈 ，又被抓入狱。古今文人的牢
骚和苦难 ，没有像先生 (这样 )的。尽管如此 ，仍有胡公这样间隔几世 (难遇 )的豪杰 ，世宗这样英明的君主 (赏
识他 )，在胡公幕府中受到与众不同的礼遇 ，这是胡公对先生的赏识；奏表呈上 ，皇帝欢心 ，这表明皇帝也
赏识他 ，唯独身份未能显贵罢了 。先生的诗文兴起 ，一扫近代 (文坛 )混乱不堪的风气 ，百世之后 ，自会有确
定的判断 ，怎么说 (他)没有际遇呢？ ”我认 为徐文长没有一处不奇特的。 (正因为 )徐文长没有一处不奇特 ，
这(也就注定 )无论到哪里都不会被看重。可悲啊！
(二)文言文阅读
1．答案 CEG
解析 “事修而谤兴 ”与“德高而毁来 ”句式一致 ，独立成句 ，故应在 C、E 处断开 。“士之处此世 ”结构
完整 ，故应在 G 处断开。
2．答案 C
解析 C 项，“意思相 同”错误。缺点 /困苦。
3．答案 C
解析 C 项，“无论古代的君子还是现代的君子 ”错误。材料二第三段 “虽然 ，为是者 ，有本有原 ，怠与
忌之谓也 ”中“为是者 ”是承接第二段而言的 ，指的是 “今之君子 ”。
4．答案 (1) 管仲不是仁人吧？齐桓公杀了公子纠 ，他不能为了公子纠而死 ，还去辅佐齐桓公。
(2) 古时候的君子 ，他要求自己严格而全面 ，他对待别人宽容又简约。
5．答案 ①孔子与古代的君子待人的态度都是宽容又简约的 ，他们能看到别人的长处 ，而不是只关注别人
的不足。 ②子路、子贡与现代的君子待人却是要求严苛的。看不到别人的优点 ，只看到别人的缺点。 学科 网（北 京）股 份有限 公司
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【参考译文】
材料一：
子路说 ：“齐桓公杀死了公子纠 ，召忽为了公子纠而死 ，管仲没有为公子纠而死 。”(接着又 )说：“(管
仲的行为 )不合乎仁吧？ ”孔子说 ：“齐桓公多次联合诸侯 ，却不动用武力 ，是管仲的功劳 。这就是他 的仁 。”
子贡说： “管仲不是仁人吧？齐桓公杀了公子纠 ，他不能为了公子纠而死 ，还去辅佐齐桓公。 ”孔子
说： “管仲辅助齐桓公 ，称霸诸侯 ，匡正天下 ，百姓到现在还承受着他的恩惠。没有管仲 ，我大概会披散
着头发穿着衣襟向左边开的衣服了。难道要像普通百姓那样守着小节小信 ，在沟渠里上吊自杀而没有人知
道吗？ ”
材料二：
古时候的君子 ，他要求自己严格而全面 ，他对待别人宽容又简约。 (对自己要求 )严格而全面 ，所以不懈
怠；(对待别人 )宽容又简约 ，所以别人都乐意做好事 。听说古代有 位叫舜的人 ，从他为人处世来看 ，是讲究
仁德道义的人。探究他成为舜的原因， (君子 )责问自己说： “他，是个人；我，也是个人。他能够这样，
我却不能够这样！ ”他早晨夜晚都在思考，改掉那些他比不上舜的方面，效仿那些他与舜相似的方面。听
说古代有位叫周公的人 ，从他为人处世来看 ，是(具有 )很多才能和技艺的人 。探究他成为周公的原因 ，(君
子)责问自己说： “他，是个人；我，也是个人。他能够这样，我却不能够这样！ ”他早晨夜晚都在思考 ，
改掉 那些他比不上周公的方面 ，效仿那些他与周公相似的方面。舜、周公都是道德非常高尚、智慧非常高
超的人，后代的人没有人能比得上他们。这位古代的君子，于是说： “(我)比不上舜，比不上周公，是我
的缺点。 ”这不也是要求自己严格而全面吗？他对待别人，择取那人一个方面，而不苛求那人别的方面；
肯定那人今天的表现，而不追究那人过去的行为。小心翼翼地只担心别人不能得到做好事的利处。这不也
是对待别人宽容又简约吗？
现在的君子却不是这样 。他要求别人全面 ，他要求自己简约 。(要求别人 )全面 ，所以别人难以做好事 ；
(要求自己 )简约，所以自己获 得的就少。自己没有什么擅长的 ，却说： “我擅长这个，这就足够了。 ”自
己没有才能，却说： “我胜任这个，这就足够了。 ”对外欺骗别人，对己欺骗良心，还没有 一点收获就停
止了 。这不也是对待自己要求太少了吗？他们对待别人 ，说：“他虽然胜任这个 ，但他的人品不值得称赞 ；
他虽然擅长这个 ，但他的才用不值得称赞。 ”举出他一方面的欠缺 ，不考虑他多方面的长处；只追究他以
往的行为 ，不考虑他今日的表现。惶惶不安地只担心别人有好的名声。这不也是要求别人太全面了吗？这
就叫不用一般人的标准要求自身 ，却用圣人的标准希望别人 ，我看不出他尊重自己啊。
虽然 ，这样做的人 ，有他的根源 ，那就是所说的怠惰和妒忌啊。怠惰的人不能 修养自身 ，而妒忌的人
害怕别人修养自身。因此事情处置好诽谤就产生了，品德高尚诋毁就来了。读书人生活在当今的世道中， 学科 网（北 京）股 份有限 公司
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而希求名望荣誉发扬光大、道行品德推广太困难了。
(三)文言文阅读
1．答案 CFH
解析 “卫之去齐不远 ”“ 君不若使人问之 ”均结构完整 ，故应在 C、F 处断开 。“而固贤者也 ”是判断句 ，
省略了宾语 “贤者 ”的逻辑主语 “客”，动词 “晚”的逻辑主语是 “用之 ”这一事实 ，故应在 H 处断开。
2．答案 D
解析 D 项，“词义不同 ”错误。词义相同 ，都是治理的意思。
3．答案 D
解析 D 项，“认为前者意义价值更大 ”错误。根据材料二 “然则郡之与县 ，异体之大小者也。以实理宽
急论辨之 ，则当言大小异宜 ，不当言能大不能小也 ”“ 推此论之 ，人材各有所宜 ，非独大小之谓也 ”可知 ，
材料二认为人才各有其适合担当的职位 ，而非单方面认为性情宽宏的 人意义价值更大。
4．答案 (1) 宁戚在车下喂牛 ，他看到齐桓公感到很悲伤 ，就拍击着牛角大声唱起歌来。
(2) 假如能用来烹煮牛 ，那么怎么会不能用来烹煮鸡呢？
5．答案 示例一：招聘者： ①要善于发现并善用人的长处。 ②不以才能的大小判别人才 ，要知人善任。
示例二：应聘者： ①要善于展示自己的优点。 ②明确自己的能力和定位 ，寻找适合自己的岗位。
【参考译文】
材料一：
用完美的标准举荐人一定很困难 ，这是事物的实情 。有人用不爱儿子的名声诋毁尧 ，用不孝顺父亲的称
号诋毁舜 ，用贪图帝位的想法诋毁禹。从这些事情来看 ，事物怎么能够完美呢？所以君子按照一般人的标
准要求别人 ，按照义的标准要求自己。按照一般人的标准要求别人就容易满足 ，容易满足就能受到别人的
拥护。不贤德的人就不是这样了。他们按照义的标准要求别人 ，按照一般人的标准要求自己。按照义的标
准要求别人就难以满足 ，难以满足就会失去最亲近的人。一尺长的树木一定有节疤 ，一寸大的玉石一定有
瑕疵。先王明白事物不能 够完美 ，因此选择事物只看重其长处。宁戚想向齐桓公谋求官职，但处境窘迫没
有自谋仕进的办法，于是就替商人赶着装载货物的车来到齐国。齐桓公到郊外迎接宾客，在夜里打开了城
门，让装载货物的车避开 ，火把非常明亮 ，跟随的人非常多 。宁戚在车下喂牛 ，他看到齐桓公感到很悲伤 ，
就拍击着牛角大声唱起歌来。齐桓公听到歌声，说： “奇怪啊！这个唱歌的不是一般人！ ”就命令侍从的
车载着他。齐桓公赏赐给他衣服帽子，要接见他。宁戚进见，用治理国家的主张劝说齐桓公。 第二天再次
进见 ，用如何治理天下的话劝说齐桓公。齐桓公非常高兴，准备任用他。众位大臣规劝说： “这位宾客，
是卫国人 。卫国距离齐国不远 ，您不如派人去询问一下 。如果 (他)确实是贤德之人 ，(您)再任用他也不晚 。”学科 网（北 京）股 份有限 公司
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齐桓公说 ：“不能这样 。去询问他的情况 ，我担心他有小毛病 。因为一个人的小毛病 ，而丢掉他的大优点 ，
这是君主失去天下贤士的原因。 ”人本来就难以完美，权衡后而用他的长处是恰当的举荐人才的做法。
材料二：
有人说 ：“人的才能有能做大事而不能做小事的 ，犹如容纳牛的鼎不能用来烹煮鸡一样 。”我认为这样
不合乎名分。怎么会有能做大事而不能做小事的道理呢？凡是说能做大事而 不能做小事的 ，这些话出于人
的性格有宽宏和急躁之分。性格有宽宏和急躁之分，因此适合做的事有大、小的分别。性情宽宏的人，适
宜治理郡国，让下属能够发挥自己的才能，自己统筹成就事业。性情急躁的人，适宜治理县邑，事情由自
己处理 。然而郡国与县邑 ，只是区域的大小不同 。从实际治理时区分宽宏和急躁的性格的角度来论证分辨 ，
则应当表明能力大小所适宜的职位也不相同，不应该说能做大事而不能做小事。至于说鸡与牛，只是物体
的大小不同，因此所用的鼎也就有大小之别 。假如能用来烹煮牛 ，那么怎么会不能用来烹煮鸡呢？因此能
治理大的州郡，那么也能治理小的郡县。推究这些情况判断，人才各有其适合担当的职位，不能仅仅用大
小去评论。
☞增分大题题型热点
增分大题 题型题组 —— 现代文阅读 Ⅰ(信息类 )
(一)现代文阅读 Ⅰ(本题共 5 小题 ，19 分)
阅读下面的文字 ，完成 1～5 题。
“传统戏曲 ”在文本结构上从未以 “情节整一性 ”为前提 ，传奇剧本更是如此 ，它恰如中国传统绘画
中的卷轴手卷 ，具有典型的手卷式结构。
戏曲中的传奇与绘画中的手卷最为直观的相似点 ，就是它们特殊的长度。正如手卷的长度超出了一般
意义上应该让欣赏者 “一目了然 ”地完整欣赏的绘画应有的形态 ，传奇的长度也和通 常为实际演出撰写的
剧本不同 ，明显超出了在一个单位时间内讲述一个相对完整的故事的需要。
明清传奇作为戏剧文本的长度，根本就不曾预设要在 一个单位时间内完成演出 ，其前身宋元南戏的文
本，就超出了这一要求，而从传奇之始祖 —— 元末时期高则诚的《琵琶记》开始，传奇的文体更有如一匹
脱缰骏马，戏剧的逻辑从未对其构成制约。值得特别提示的是，高则诚创作《琵琶记》之前，以四折为一
本的格律严整的杂剧早就风行天下。
明清传奇均以数十出为常例。按《六十种曲》所收录的传奇， 40 出左右的剧本在传奇中比比皆是 ，明
清传奇不仅没有如元杂剧那种只 有 4 折的作品 ，20 出以下的都极为少见。这是因为多数传奇作家创作剧本
的最初动机 ，主要是供案头阅读的文学读本 ，他们要通过篇幅与容量证明才情 ，所以竞相以长度争胜。传 学科 网（北 京）股 份有限 公司
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奇除了长度远离一般对剧本文学的认知 ，更重要的是传奇作家安置戏剧故事的铺排方式 ，与从开端 、冲突 、
高潮直至结局的 “整一性 ”的戏剧理念大相径庭 ，更缺乏作为 “对立双方 ”之矛盾的戏剧冲突。
从传奇的文本结构看 ，尽管一部传奇叙述的故事本身是可以具有其完整性的 ，但是优秀的传奇作家 ，
似乎都偏好于在故事发展过程的时间延展中 ，有意地点缀着几个间隙性的段落 ，就像长卷中的山 水起伏一
样，时有枝蔓逸出，形成有张有弛、错落有致的节奏感。这样的结构安排，古代戏曲理论家和昆曲表演者
称之为 “冷热相剂 ”，显然为传奇作家普 遍认可 ，它也强化了传奇的文本结构与绘画手卷相似和相通的特
征。
明代传奇的结构对昆曲的演出形态有着决定性的影响。传奇作品动辄数十出，假如按一出演出半小时
计，明代的多数传奇作品按全本演出超过 20 小时 ，如同一二十米的长卷一般 ，这样的长度 ，无论是对表演
者还是观赏者 ，显然都并不适宜。其实 ，恰因传奇的叙事逻辑与手卷构图方式相似 ，所以传奇这种戏剧文
本的篇幅才会如此之长 ，正如手卷最适宜于 创作长卷 ，传奇手卷式的构成更宜于较长篇幅的创作，这也是
传奇动辄长达数十出的内在合理性之所在。
明清传奇以折子戏为主的演出形态与传奇的文体结构特点 ，简直是天作之合。传奇剧本虽以生旦戏为
主，但一直有不成文的惯例 ，即每部传奇要为所有角色安排其独立表现的出目 ，更导致传统戏曲在文本结
构上 ，呈现出手卷式多中心的特点。一些次要角色主演的出目 ，有时会选出剧情的主线 ，就像手卷的闲笔
和留白 ，让作品增添了叙事中的弹性与变化。假如其中没有起伏变化 ，没有 “冷热相剂 ”，欣赏过程将会
显得非常乏味。
戏曲尤其是昆曲各行 当表演艺术水平提升 ，与传奇的体例有着内在的因果关系，这是因为传奇不只关
心作为主要角色的生旦，几乎是下意识地顾及所有的行当，在剧本中为他们设置 可以展现其能力的段落。
反过来也可以说 ，传奇作家虽然不无对作品整体的把握，但是对局部的注意实不下于整体。传奇和手卷本
质上都可以被看成一个流动的整体，其多中心的特点，赋予了演出和欣赏过程中的自由裁量空间，灵动又
不随机 。其实当我们说中国画的手卷多中心或者散点透视的特点时 ，还需要加以补充的是 ，在每幅手卷里 ，
对具有相对完整性的局部的截取方式，并无固定模式，恰如折子戏一样，实际上是可以由表演者定义的。
元杂剧相对篇幅较短，所叙述的故事和事件亦较集中，当然和动辄长达数十出的传奇没有可比性，不
可能像传奇一样追求典型的长卷式 的铺叙结构。元杂剧的文本相当具有严整性 ，一本四折，每折一套曲，
由同一宫调的多支曲牌构成，结构规范。但这是音乐上的严整性，从戏剧角度看是松散和漫延的，文本结
构仍有与手卷相似的特点，元杂剧末本的戏剧人物设置，更证明了这一点。元杂剧清晰地体现出讲唱的文
本向戏剧文本的转化，其标志就是其代言体的特征，但这种转化并不能瞬间完成，从元杂剧大量末本中变
换主唱者的现象，不难看到在这一转化的过程中，还保留了讲唱艺术的若干痕迹。 学科 网（北 京）股 份有限 公司
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相对于从全剧中撷取一个段落的折子戏，戏曲行业把那种在一个单位时间里叙述一个相对完整故事的
戏剧形态称为 “本戏 ”，既是 “本戏 ”，就会自然而然地考虑演出剧目的 “情节整一性 ”。应该承认，本
戏，尤其是剧场演出的本戏，对情节整一性的追求更符合一般的戏剧观赏规律。所以，当剧场演出成为戏
剧的主要表现形态时，观众的欣赏方式既与欣赏绘画中的长卷截然不同，戏曲文本结构亦必然随之改变，
“情节整一性 ”就是其最可能呈现的结果。
(摘编自傅谨《论传统戏曲与传奇的手卷 式结构》 )
1．下列对原文相关内容的理解与分析 ，不正确的一项是 (3 分)( )
A．一般为实际演出撰写的剧本 ，需要在一个单位时间内讲述一个相对完整的故事 ，而宋元南戏的文本 ，则
超出了这一要求。
B．多数传奇作家竞相以长度争胜 ，致使传奇作者对故事的铺排方式与 “整一性 ”的戏剧理念大相径庭 ，缺
乏戏剧冲突。
C．传奇结构的 “冷热相剂 ”，类似于长卷中的山水起伏 ，为传奇作家普遍认可 ，也使观赏者的欣赏过程不
至于太乏味。
D．传奇作家重视作品局部 ，顾及所有角色 ，这样的传奇本质上可以被看成一个流动的整体 ，演出 过程中保
有自由裁量空间。
答案 B
解析 B 项，“致使 ……” 错误。根据原文第四段 “这是因为多数传奇作家 …… 要通过篇幅与容量证明才
情，所以竞相以长度争胜。传奇除了长度远离一般对剧本文学的认知 ，更重要的是传奇作家安置戏剧故事
的铺排方式 ，与从开端、冲突、高潮直至结局的 ‘整一性 ’的戏剧理念大相径庭 ”可知 ，原文陈述的是多
数传奇作家的创作特点 ，并不包含选项所述因果关系。
2．根据原文内容 ，下列说法不正确的一项是 (3 分)( )
A．在《琵琶记》之前，一本四折、格律严整的杂剧早已风行，但《琵琶记》之后，传奇的文体更加自由 。
可见，戏剧的逻辑并未对其构成制约。
B．不只关心生旦 ，而且在剧本中为所有行当设置展现他们能力的段落 ，传奇的这一特点对提升戏曲各行当
的表演艺术水平有着一定的帮助。
C．以四折为一本的元杂剧和动辄数十出的传奇之所以没有可比性 ，是因为元杂剧相对篇幅较短 ，故事集中 ，
不能追求典型的长卷式的铺叙结构。
D．随着戏剧的主要表现形态向剧场演出转变 ，观众的欣赏方式也会发生变化 ，戏曲文本结构向更符合戏剧
欣赏规律的 “情节整一性 ”的改变成为 必然。 学科 网（北 京）股 份有限 公司
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答案 D
解析 D 项，“戏曲文本结构向更符合戏剧欣赏规律的 ‘情节整一性 ’的改变成为必然 ”错误。 “必然 ”
一词强调 “戏曲文本结构 ”的改变是必然的；原文最后一段 “当剧场演出成为戏剧的主要表现形态时 ，观
众的欣赏方式既与欣赏绘画中的长卷截然不同 ，戏曲文本结构亦必然随之改变 ，‘情节整一性 ’就是其最
可能呈现的结果 ”可知 ，原文表述是向 “情节整一性 ”的变化是 “最可能呈现的结果 ”，“最可能 ”强调
的是变化的可能性 ，而不是 “成为必然 ”。
3．下列选项，没有使用原文所讲的传奇剧故事的铺排方式的一项是 (3 分)( )
A．梁辰鱼的 《浣纱记 》中，伍员和吴国的恩怨这条线与越国献西施这条线并无实质性关联 ；范蠡和西施两
个最核心的人物在全剧大部分出目里也分居吴越两地。
B．李玉的《一捧雪》中 ，前一半的主人公是莫怀古 ，而从第 15 出开始 ，故事的主要人物就出现了明显变
化，之前位居极次要地位的莫诚成为这一折的主角。
C．关汉卿的 《窦娥冤 》，以全剧中心人物窦娥为代表的善良人民 ，和以流氓恶棍张驴儿 、楚州太守桃杌为
代表的黑暗势力 ，构成彼此矛盾尖锐的双方。
D．汤显 祖创作的 《牡丹亭 》，在《闺塾 》《惊梦 》中穿插 《劝农 》，在《骇变 》《如杭 》中穿插 《淮警 》，
这些穿插与有迹可循的伏笔不同，粗看全无头绪。
答案 C
解析 根据原文 “传奇除了长度远离一般对剧本文学的认知 ，更重要的是传奇作家安置戏剧故事的铺排方
式，与从开端、冲突、高潮直至结局的 ‘整一性 ’的戏剧理念大相径庭 ，更缺乏作为 ‘对立双方 ’之矛盾
的戏剧冲突 ”可知 ，这里强调传奇剧故事的铺排方式有两个特点 ：一是不符合 “情节整一性 ”的戏剧理念 ，
二是缺少矛盾的戏剧 冲突 。A 项，戏剧两条线说明情节不整一 。B 项，“故事的主要人物就出现了明显变化 ”
说明情节不整一。 C 项，没有使用传奇剧故事的铺排方式。 D 项，穿插的内容无头绪 ，说明情节不整一。
4．根据文本相关内容 ，补充完成下面的文本结构图。 (4 分)
答： 学科 网（北 京）股 份有限 公司
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答案 ①文本结构的特点 ②特殊的长度 ③传奇作家通过篇幅与容量证明才情 ④助推戏曲各行当表演
艺术水平提升
解析 本题考查分析文章结构。解答本题首先需逐段概括文本内容 ，然后思考段落间的逻辑关系 ，结合结
构图中的先后顺序、总分关系等进行整合、归纳。
读图，明逻辑关系 结合原文，提炼内容关键词
①处
根据第二列已提供的关键词 “文本
结构的原因 ”“ 文本结构的影响 ”
可知，此处所填格式应为 “文本结
构的 ……” ；根据第三列的 “冷热
相剂 ”“ 手卷式多中心 ”可知，此
处应是对二者的概括
由第五段 “这样的结构安排 ”以及第七段 “更
导致传统戏曲在文本结构上 ，呈现出手卷式
多中心的特点 ”可知 ，此处可填 “文本结构
的特点 ”
②处
根据第三列的 “冷热相剂 ”“ 手卷
式多中心 ”可知，此处应填与之并
列的特点；由此处所填内容在这两
个词之前可知，答案需要从原文的
第五段之前筛选
由第二段 “戏曲中的传奇与绘画中的手卷最
为直观的相似点 ，就是它们特殊的长度 ”可
知，此处可填 “特殊的长度 ”
③处
根据第二列 “文本结构的原因 ”可
知，此处应从原因角度思考答案，
且在涉及 “传奇作家认可 ‘冷热相
剂’的结构模式 ”之前的段落筛选
由第四段 “这是因为多数传奇作家 …… 他们
要通过篇幅与容量证明才情 ”可知 ，此处可
填“传奇作家通过篇幅与容量证明才情 ”
④处
根据第二列 “文本结构的影响 ”可
知，此处应从影响角度思考答案，
且在涉及 “增添了叙事中的弹性与
变化 ”之后的段落筛选
由第八段 “戏曲尤其是昆曲各行当表演艺术
水平提升 ，与传奇的体例有着内在的因果关
系”可知 ，此处可填 “助推戏曲各行当表演
艺术水平提升
结合文本 ，说说曹禺的代表作《雷雨》是如何体现戏剧的 “情节整一性 ”原则的。 (6 分)
答： 学科 网（北 京）股 份有限 公司
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答案 ①在一个单位时间内 (一天之内 )讲述了周 、鲁两家相对完整的故事 。②作者对周 、鲁两家故事的铺排
方式 ，体现了 “开端 —冲突 —高潮 —结局 ”的整一性特征。 ③作为对立双方的周、鲁两家两代人之间三十
年的恩怨情仇 ，成为整部《雷雨》集中的戏剧冲突。
解析 本题考查理解文中重要概念的含义 。解答本题首先要明确什么是戏剧的 “情节整一性 ”，再结合 《雷
雨》 中的具体内容进行对应 ，明确各情节之间的关联。
“情节整一性 ”原则 《雷雨》的对应性体现
第二段：在一个单位时间内讲述一个相对完
整的故事
作品写的是北方某个闷热的夏日，一天之内
发生的事情，情节以早晨到傍晚再到深夜的
时间变化向前发展
第四段 ：从开端 、冲突 、高潮直至结局的 “整
一性 ”的戏剧理念
作品以明、暗两条线索，刻画了周、鲁两家
两代人之间三十年的恩怨情仇
第四段：作为 “对立双方 ”之矛盾的戏剧冲
突
作品中蕴含复杂的人物关系和矛盾冲突，如
周朴园与鲁侍萍，周朴园与蘩漪，周朴园与
周萍，周朴园与鲁大海等
(二)现代文阅读 Ⅰ(本题共 5 小题 ，18 分)
阅读下面的文字 ，完成 1～5 题。
材料一：
对于如何实现城乡协调发展的问题 ，当年的费孝通曾经专门将其作为一个研究课题提出来进行研究 ，
但今天的城市 化似乎走到了一个单向度发展的快车道上去 ，而体现中国特色的乡村发展也出现了一种发展
的瓶颈。当越来越多的人挤到大都市寻找工作、生活以及未来之时，北上广这些大城市是否真正能够满足
越来越城市化生活的需要，乡村的去留问题，都将是对于未来中国发展的一种考验。乡村如果是这样一番
处境，文化的未来又将会走向哪里同样作为一个极为严肃的问题而吸引着晚年费孝通的注意力。或者说，
从 1948 年前后撰写 《乡土中国 》和《乡土重建 》这两本小册子的时候开始 ，费孝通就一直探究在这样一个
世界性的大变革的时代中 ，中国文化到底该何去何从 ，这种探究到 了他晚年 90 多岁时才真正变成是一种自
觉性的思考 ，并为此而提出一种 “文化自觉 ”的概念去试图予以彻底解决。
对许多阅读过 《乡土中国 》的人来说似乎并不陌生 ，在这本书中所提到的一个极为重要的概念就是 “差
序格局 ”，或许受到潘光旦先生反复讲述的 “五伦 ”观念在中国社会伦理结构中的影响 ，费孝通由此而提
出了 “差序格局 ”的概念。提出这一概念的实质便是试图与一种西方 “团体格局 ”相对比之下而对中国乡 学科 网（北 京）股 份有限 公司
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土社会、熟人社会给出一种更为贴近真实的理解 ，以克服一种西方社会学教科书般的对于中国的远观。它
为我们的社会理解寻找到 了一个可以依附的框架 ，与此 同时 ，他也借此暗示了在这样一个社会之中，可以
把分散开来的个人一个个地聚拢在一起的基础究竟在哪里 ，而这个基础就在于一种强烈意识的自我的存在 。
但这种自我并不是孤立存在的，它是要被自我放置到他人的位置上加以绑定起来而实现的，因此这种自我
在天然的意义上便是一种社会性的自我，是 “我”的一种他者化。因此在我们乡土中国的文化里不太惧怕
有他人的存在，只要 “他”是和 “我”有关系的， “他”可以转化成为 “我”的，这种惧怕感也就自然消
失了 ，这里的关键是要看 “他”和“我”的关系究竟是远还是近 。这种关系结构绝非 “拉关系 ”那样简单 ，
它自 身具备一种真正私人性与安全感的结合：自我的存在 ，并以自我为中心，在一定意义上保证了中国人
自我价值的发挥；同时在他需要各种帮助之时，便可以瞬时启动一种差序格局的社会网络，由此而使得一
种由人情关系的亲疏远近安排的社会格局得以有助于每个自我与社会成就的获得。这种作为基础的社会关
系结构的差序格局，实际上也在深度影响着中国社会里的道德、法律以及伦理形态。换言之，所有这些都
无法脱离我们的社会关系中以自我为中心而又不断延伸出来的社会联系，这种关系网络的实践真正塑造出
了乡土中国的一些最为基本的特征，其中就包括对于私人的而非 公共道德的强调。换言之 ，在人们作出一
种道德与否的判断之前，首先想到的就是彼此之间私人关系的远近，这种观察无疑是对中国社会自身文化
逻辑的最为深刻的一种理解。
新的中国的建立 ，无论如何都不可能抛开一种现实的存在或者国情而自行构建 ，而这个摆在新的中国
面前的现实基础恰恰是与现代西方已经建立很久的法权制度的契约精神不能够相融合的。换言之，这个国
家的基础的文化可谓是建立在一种差序格局的社会关系模式之上的，由此才会有难以处理的一种所谓 “人
治”的社会、在野乡绅的自治、无讼的理想以及一种特殊的礼法文化等困境的出现，这些困境不 是在于一
种传统的文化有了问题 ，而是在于我们引入一种新的制度，这种制度实际上忽略了一点，即这种传统的差
序格局的文化观念是要慢节奏地发生转变的，言外之意便是，二者之间并没有可能做到同步发展，进而会
引发诸多的困境产生。
(摘编自赵旭东《文化自觉与人的相互看 —— 由作品
去理解费孝通思想的一种途径》 )
材料二：
《费孝通论文化自觉》一书集中阐述了费先生文化自觉思想： “通过我这 60 多年的经历 ，我深深体
会到我们生活在悠久 历史的中国文化中 ，而对中国文化本身 至今还缺乏实事求是的系统知识。我们的社会
生活还处于 ‘由之 ’的状态而还没有进入 ‘知之 ’的境界。而同时我们的生活本身却已进入一个世界性的
文化转型期 ，难免将人们陷入困惑的境地 ，其实不仅我们中国人是这样 ，这是面临 21 世纪的世界人类共同 学科 网（北 京）股 份有限 公司
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的危机。在多元文化中生活的人们还未能寻找到一个和平共处的共同秩序。 ”因而费先生特别强调文化自
觉的重要性 ，他详细解释道： “文化自觉只是指生活在一定文化中的人对其文化的 ‘自知之明 ’，明白它
的来历、形成过程 ，在生活各方面所起的作用 ，也就是它的意义和所受其他 文化的影响及发展的方向 ，不
带有任何 ‘文化回归 ’的意思 ，不是要 ‘复旧 ’，但同时也不主张 ‘西化 ’或‘全面他化 ’。自知之明是
为了加强对文化发展的自主能力，取得决定适应新环境对文化选择的自主地位。 ”他不仅对文化自觉的概
念进行了详细定义，还强调指出： “文化自觉是一个艰巨的过程：首先要认识自己的文化，根据其对新环
境的适应力决定取舍 。其次要理解所接触的文化 ，取其精华 ，去其糟粕 ，加以吸收 。各种文化都自觉之后 ，
这个文化多元的世界才能在相互融合中出现一个具有共同认可的基本秩序和形成一套各种文化的和平共处、
各舒所长、联手发展的共同守则。 ”目的就是通过 “文化自觉 ”，掌握文化转型的主动权 ，重建民族文化
自信心，巩固国家和民族认同，建立 “和而不同 ”的美好社会，更好地应对 “全球化 ”的挑战，实现中华
民族的伟大复兴。
(摘编自徐平《费孝通文化思想演变
及其文化自觉实践》 )
1．下列对材料相关内容的理解和分析 ，正确的一项是 (3 分)( )
A. 费孝通撰写《乡土中国》的时候 ，就在探究中国文化何去何从 ，这一问题到他晚年提出 “文化自觉 ”才
彻底解决。
B. “差序格局 ”是相对于西方 “团体格局 ”而提出的概念 ，让中国人甚至西 方人，非常真实地理解了中国
社会的特征。
C. 乡土中国文化里的关系结构 ，既有发挥自我价值的私人性 ，又有帮助获得自我与社会成就的安全感 ，是
二者的结合。
D. 我们已进入世界性文化转型期 ，21 世纪的人类也面临着共同危机 ，文化自觉已经成为人类的基本秩序与
共同守则。
答案 C
解析 A 项错在 “这一问题到他晚年提出 ‘文化自觉 ’才彻底解决 ”，原文是 “这种探究到了他晚年 90 多
岁时才真正变成是一种自觉性的思考 ，并为此而提出一种 ‘文化自觉 ’的概念去试图予以彻 底解决 ”。B
项错在 “让中国人甚至西方人 ，非常真实地理解了中国社会的特征 ”，原文是 “提出这一概念的实质便是
试图与一种西方 ‘团体格局 ’相对比之下而对中国乡土社会、熟人社会给出一种更为贴近真实的理解 ，以
克服一种西方社会学教科书般的对于中国的远观 ”。D 项错在 “文化自觉已经成为人类的基本秩序与共同
守则 ”，原文是 “在多元文化中生活的人们还未能寻找到一个和平共处的共同秩序 ”。学科 网（北 京）股 份有限 公司
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2．根据材料内容 ，下列说法不正确的一项是 (3 分)( )
A. 尽管费孝通当年就提出了城乡协调发展的课题 ，然而今天的城市化发展还是 引发了一系列亟待解决的问
题。
B. 中国社会之所以能将分散开的个人聚拢在一起 ，原因就是有强烈意识的自我存在 ，即尊重人的个性发展 。
C. 乡土中国的基本特征之一就是私人关系影响人们对道德的判断 ，这与社会关系网络的实践有关。
D. 传统差序格局的文化观念转变比较慢 ，无法和新制度同步发展 ，进而会引发诸多困境。
答案 B
解析 B 项错在 “尊重人的个性发展 ”，原文是 “但这种自我并不是孤立存在的 ，它是要被自我放置到他
人的位置上加以绑定起来而实现的 ，因此这 种自我在天然的意义上便是一种社会性的自我 ，是 ‘我’的一
种他者化 ”。
3．下列选项 ，最能体现 “差序格局 ”基本特点的一项是 (3 分)( )
A. 论亲戚关系 ，宝姐姐比林妹妹疏远一些。 (贾宝玉 )
B. 苟利国家生死以 ，岂因祸福避趋之。 (林则徐 )
C. 无尽的远方 ，无数的人们 ，都与我有关。 (鲁迅 )
D. 稳定的国家是以法律面前人人平等为基础的。 (亚里士多德 )
答案 A
解析 A 项最能体现 “差序格局 ”的基本特点 ，原文的表述是 “一种由人情关 系的亲疏远近安排的社会格
局”。B 项是爱国主义。 C 项是普遍联系。 D 项是法律的作用。
4．下列对材料一和材料二相关内容的梳理 ，不正确的一项是 (3 分)( )学科 网（北 京）股 份有限 公司
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答案 B
解析 B 项说 “潘光旦讲述的 ‘五伦 ’观念 ，启发费孝通提出了 ‘差序格局 ’的概念 ”错误。原文是 “或
许受到潘光旦先生反复讲述的 ‘五伦 ’观念在中国社会伦理结构中的影响 ，费孝通由此而提出了 ‘差序格
局’的概念 ”。
5．我国正在经历重要的 “社会转型 ”期，由乡村向城市社会的转化 ，文化是社会的灵魂所系 ，因此 ，社会
转型的背后正是更为深刻的文化转型问题。请结合两则材料 ，简要分析传统差序格局给中国社会 转型带来
的阻碍，并思考我们应该如何跨越这个阻碍。 (6 分)
答：
答案 阻碍：传统的差序格局文化观念虽然有助于每个自我与社会成就的获得 ，但随着亲疏远近的变化 ，
它会影响我们对道德、法律与伦理形态的判断 ，与法制的契约精神不相融合。
如何跨越 ：①首先要认识自己的文化 ，根据其对新环境的适应力决定取舍 。深刻认识差序格局的文化观念 ，
在文化转型的关键期和困境中做出合理的选择。 ②其次是理解所接触的文化，取其精华，去其糟粕，加以
吸收。文化自觉不是全盘他化，而是要根据国情进行构建。
(三)现代文阅读 Ⅰ(本题共 5 小题 ，19 分)
阅读下面的文字 ，完成 1～5 题。
不管是要找出第二次世界大战的原因 ，还是查明天花板上水印的来由 ，我们通常都要考察可能的解释 。学科 网（北 京）股 份有限 公司
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比如说天花板上的水印 ，是屋顶漏水了 ，还是管子漏水了？我们可能会这样推理： “这个水印在厨房天花
板上 ，正好是在浴室的下面 ，所以很可能是管子漏水 。”现在 到楼上去检查一下 ，如果发现了漏水的管子 ，
那么就可以合理地得出 结论 ，对于水印的最佳解释是管子漏水 ，当然 ，也可能是因为屋顶和管子同时漏水 。
这个简单而实际的例子展示了科学研究的推理过程：提出各种假说，一个一个地排除，直到得出最佳
解释。地质学的历史为科学研究如何运用这样的推理过程提供了一个清楚的例子。地球已经有上亿年的历
史、大陆在漂移，这些都是非常惊人的发现。它们被接受的过程是漫长而复杂的，要求仔细的观察、改良
的技术、大量的集体努力以及在很多学科中共享知识。地质学最近的发展历史就展现了这样的过程。
1912 年，德国科学家阿尔弗雷德 ·魏格纳提出了板块漂移理论来解释这个明显的事实 —— 非洲大陆和
南美洲大陆看上去好像很吻合。但是在他之前的理论家 ，通过观察过去的地图，也推测这些大陆原本是连
在一起的 。魏格纳对这一理论的补充是 ，在两个大陆相对应的边缘 ，岩石的形成和动植物化石都非常相似 。
因为他不能提出一个解释或者模型来说明像板块这样巨大的东西是如何 “漂移 ”的，他的理论遭到了普遍
的拒绝，甚至被嘲笑。虽然他的理论解释了一些观察到的现象，但是并没有被采信，因为它与当时人们所
相信的关于大洋和大陆的物理结构方面的观点不一致。拥有可接受的解释模型是科学断言能被接受的重要
标准。
魏格纳的理论在 20 世纪 60 年代被美国地质学家哈雷 ·赫斯复兴。赫斯提出 ，最近发现洋中脊在延伸 ，
而大陆居于板块之上 ，因此板块应是由底层的地幔缓慢运动的 “环流 ”所推动的。赫斯也承认 ，这个理论
一开始只是猜想 ，并且与现有的理论相悖。但它确实是可以从现有知识中推断出来的最合理的理论。
赫斯的理论为魏格纳的观察提供了一种解释 ，也解释了不断出现的惊人的异常 —— 根据现有理论不能
得到满意解释的数据 —— 比如岩石的磁性定向。人们发现 ，玄武岩 (由火山爆发所形成的岩石 )包含磁铁矿 ，
而正如这个名字所显示的 ，其行为模式让人感到它像是由小指南针所构成的。 在玄武岩温度很高、尚未凝
固之前 ，它们总是指向北极。基于一些尚未被理解的原因，每隔几百万年，北极和南极就调转方向，先前
指向北极的指南针，将会在调转之后指向南极。因此，玄武岩中的磁铁矿可以揭示它凝固时极点的位置。
此外 ，磁铁矿的定向不仅是水平的 ，也是垂直的。当它们凝固的时候 ，玄武岩越接近极点 ，磁铁矿的
“指南针 ”的定向就越垂直 ，不管是接近北极还是南极 ，在赫斯发表其学说的那个年代 ，地磁数据存在异
常。大量的地磁数据表明 ，相较于它们现在的位置 ，接近于赤道的岩石是在更为接近极点的地方形成的。
因为大陆运动的理论尚未被接受 ，所以对于这些观察缺乏适当的解释。
不过，赫斯的理论还是展示了一个成功的新理论模型的重要特点： (1) 它应该能对现有信息提供更全面
的解释 ，包含对目前尚不能理解的现象的解释； (2) 它应该能够提出新的、可以被测试的预测。赫斯的理论
成功地实现了这两点。美国《地质服务》上的一篇文章对此总结道： “1962 年，赫斯明确地意识到仍然缺 学科 网（北 京）股 份有限 公司
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乏坚实的证据来证实他的假说 ，因此无法说服虽然倾向于接受但是仍然存有怀疑的人。然而一年之后 ，范
恩·马修斯对于海底磁条带 的解释 ，以及在接下来的几年中其他的海洋勘测，最终提供了证据来证实赫斯关
于海底扩张的模型。之后的年代测定研究表明 ，离洋中脊地壳越远的海底存在的时间就越长，赫斯的理论
得到了进一步的巩固。最后，更完善的地震数据证实了赫斯的假说。他关于海底扩张以及洋中脊的基本观
点经受住了时间的考验。 ”
还应该指出，赫斯的理论不仅符合新的数据，而且还展现了巨大的解释力。板块构造理论对地震、山
脉形成以及火山现象都提出了新的解释。这样的理论是富有成果的，即一个理论具有支持很多研究的潜在
能力，以及提供大范围内的解释性洞见的能力。
(摘编自莎伦 ·白琳等《权衡：批判性
思维的探究与应用》 ，有删改 )
1．下列对原文相关内容的理解和分析 ，不正确的一项是 (3 分)( )
A. 文章开头讲述了推测天花板上水印由来的过程 ，用有生活气息的例子引出对科学研究推理过程的探讨。
B. 赫斯基于洋中脊在延伸这一发现 ，完善了板块漂移理论 ，解释了魏格纳的观察和岩石磁性定向的问题。
C. 玄武岩的磁性定向既可以揭示它凝固时极点的位置 ，也可用于测定其现在所在地与极点之间的距离。
D. 赫斯通过推理提出了海底扩张模型 ，之后的海洋勘测、年代测定研究 和地震数据不断证实了这一理论。
答案 C
解析 C 项，“也可用于测定其现在所在地与极点之间的距离 ”于文无据。
2．根据原文内容 ，下列说法不正确的一项是 (3 分)( )
A. 即便没有明确的数据支持 ，仅根据当时已有知识所能做出的推断而言 ，赫斯的板块构造理论依旧是最合
理的理论。
B. 赫斯的理论改变了板块漂移学说长期不被接受的处境 ，可见科学发展主要得益于提出与当时的技术及观
念相符的解释模型。
C. 赫斯的海底扩张理论很快得到了一系 列科学测量数据的支持，说明越来越多的人愿意进一步探索。
D. 诸多科学事实支持了板块构造理论 ，而该理论又为地震、山脉形成等现象提供了新解释 ，可见假说与发
现的互动促进了科学发展。
答案 B
解析 B 项，“可见科学发展主要得益于提出与当时的技术及观念相符的解释模型 ”错误 ，根据原文 “拥
有可接受的解释模型是科学断言能被接受的重要标准 ”可知 ，科学理论的被接受得益于提出与当时的技术
及观念相符的解释模型 ，而不是 “科学发展 ”。学科 网（北 京）股 份有限 公司
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3．下列对相关理论 的描述，最符合文中成功的新理论模型特点的一项是 (3 分)( )
A. 量子力学改变了人们对微观世界中物质结构及其相互作用的认识 ，并预言了新的物理现象。
B. 哈雷运用牛顿定律准确预测了 1759 年的彗星再现 ，为牛顿定律的可信性提供巨大支持。
C. 经营者根据马斯洛需求层次理论 ，评估消费者心理需求的级别 ，据此开发和推销产品。
D. 费孝通先生提出了 “差序格局 ”这一理论模型 ，有力地解释了中国乡土社会的人伦关系。
答案 A
解析 成功的新理论模 型不但应该能够对现有现象进行解释 ，还应该能提出新的、可以被测试的预测， A
项最符合 ，其余三项没有提出新的、可以被测试的预测 ，所以不符合。
4．梳理文中板块漂移理论的发展过程 ，填写下图中空缺部分的内容。 (4 分)
答：
答案 ①观察地图进行 推测 ②不能解释板块如何 “漂移 ” ③板块应是由底层的地幔缓慢运动的 “环流 ”
所推动的
5．以下是对屠呦呦团队青蒿素研究历程的简述 。请结合上文 ，分析这一历程所体现的科学研究路径 。(6 分)
20 世纪 50 年代 ，疟疾重新肆虐 。屠呦呦团队受命从中草药中寻找并提取具有抗疟疗效的成分 。团队收
集、挑选方药 ，测试提取物效果 ，发现青蒿提取物有一定效果 ，然而实验结果很难重复 ，而且似乎与文献
记录相悖。查阅大量文献后 ，团队发现加热提取的方式破坏了青蒿的活性成分 ，于是改为低温提取 ，效果
得到提升。之后成功从青蒿 中分离提纯出青蒿素 ，并将其转化为药物，该药疗效明确。团队进一步研发双
氢青蒿素，使疗效 提高近十倍 ，且病人复发率低。双氢青蒿素也被发展成新的药物，尝试用于治疗其他疾
病。
答：
答案 ①基于假说 ，进行验证 ，设想中草药中可能具有抗疟疗效的成分 ，通过收集 、挑选方药 ，逐一验证 ，
锁定青蒿提取物。 ②完善假说，改变青蒿的提取方式，分离提纯得到青蒿素，并研发双氢青蒿素，提升疗
效。 ③发展假说，将双氢青蒿素发展成新的药物，尝试用于治疗其他疾病。 学科 网（北 京）股 份有限 公司
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增分大题 题型题组 —— 现代文阅读 Ⅱ(小说 )
(一)现代文阅读 Ⅱ(本题共 4 小题 ，16 分)
阅读下面的文字 ，完成 1～4 题。
制琴记
阿 占
话说那天下午 ，胡三背着琴 ，像侠客佩剑一样 ，行于当街 ，去琴行与韩五见面。胡三亮出了琴 —— 一
把手作小提琴。琴体的造型和构 造比照了欧洲制琴巨匠鼎盛时期的风格 ，整体弧度圆 润。雕工很有自信。
琴腰狭窄 ，便于演奏高把位和低音弦。面板与背板中间有音柱支撑，位置不偏不倚。琴表油漆均匀，不太
硬也不太软。琴箱内部处理得同样细致，没有留下任何工具的痕迹 …… 真是一把有样貌的手作琴 ，韩五心
中暗暗叫绝。
“爷们儿，你代理的那些机械琴不利于天才琴童形成个人风格，机械琴看上去就像一个模子里出来的
饰物，手作琴却是艺术品。我有匠人手艺，你有音乐资本，不如我们一起做琴吧。 ”
是年，胡三五十初叩天命，韩五三十恰逢而立。
胡三看上去像个糙人 ，肿眼泡 ，狮子鼻，头顶是秃的，常见油光，一张凡夫黑脸 。胡三木匠出身 ，十
六岁学徒，三十岁上练成了一等一的高手。四十九岁那年，首届国际小提琴节在家门口举办，胡三走了进
去，结果被国际琴展上的名琴镇住了。太美了！他魔怔了一路，回家就跟老婆说： “我要做琴！ ”那年春
节，他用两瓶茅台换回来两摞小提琴图纸，大年初一就拉开架势，图纸铺了满床满地，逐步分解，归纳笔
记。二月初二，开凌梭鱼上市的时候，胡三取料、晒料、刨料，继而打眼、锯榫头、组装，把自己放在半
成品、木屑和工具之间，一边琢磨一边敲打，不分昼夜。终于，樱花盛开的时候，他做出了人生中的第一
把小提琴。当然，第一把琴的音色不均 、不圆、不润 ，自然也就不美。胡三很不服气，他决心一把一把地
做下去，于是便有了第二把、第三把、第四把。到了第五把，胡三觉得自己可以有一个搭档了，于是想起
了韩五，也就有了开头的那段当街背琴疾行。
与胡三不同，韩五看上去像个文人，戴眼镜，不高，偏瘦，食草动物的眼神，一介书生的白面。大学
毕了业却没脱下满身的学生气，韩五跟父亲借钱，开起了琴行。韩五似乎知道每把琴的脾性，知道如何顺
着琴的性子捋。没几年，琴行就有了口碑。乐器行当里，都知道城西有个韩五，性格孤僻，音乐学养却是
极高的，侍弄乐器很有道道儿。知音难逢，大多数时间 里，韩五都是寂寞的。直到胡三的闯入，让他预感
到，一些期待已久的事情就要发生了。
就这样 ，在太阳下面 ，在月光里面 ，在德式老房子中 ，在木头的淡淡暗香里 ，胡三 、韩五这一老一少 ，
一动一静，一黑一白，一武一文，运用数学、物理学、造桥工艺、美学、声学甚至化学，开始做琴。做一 学科 网（北 京）股 份有限 公司
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把琴至少需要三十五天，而每做完一把琴，胡三都会给自己一个彻底的放松 —— 通常是休息整整二十天，
望天、听海、穿风，各种出神。
只有一次，胡三做完琴之后没有休息。韩五去上海参观国际名琴展了，琴作坊里一下子没有了敲打木
头的声音，胡三感到很寂寞，便把北墙上的老 琴取了下来。这是一把被虫蛀了的老琴 ，千疮百孔，声音已
经喑哑，一直被韩五当镇店之宝供着。
刚修了两天，韩五回来了。
“胡三，你疯了！你在干什么！给我住手！ …… 你不应该自作主张去修它 ，你应该先问问我。因为你
不了解这把琴对一个家族意味着什么，它是我祖父用命换来的。 1914 年秋天 ，日军占领青岛 ，祖父护送一
个叫希姆森的德国建筑师一家乘船返回德国。希姆森将小提琴托祖父保管好 ，说是家传之物 ，日后来取。
20 世纪中期以后 ，祖父的苦难日子就没有间断过。祖母自缢 ，父亲和两个 伯父因为家庭成分不能上大学也
不能参军。祖父把能烧的书都烧了 ，琴总是藏得很好。最后一次，他从抄家人手中夺过这把琴，跳下二楼
的阳台，摔成了残疾。 ”
“为了一把琴，去跳楼？ ”
“人人都说祖父傻，为了一把琴赔了后半生。 ”
胡三发誓一定要修好这把琴。前前后后修了一年，果然，重生后的琴音绮丽饱满，也沧桑沉郁。韩五
感激胡三，胡三倒不好意思起来。
琴作坊开业的第七年，十一岁的小满来了。买不起手作琴的小满经常偷偷跑到琴作坊，只为看一眼漂
亮的琴。一天，小满又来了，但他的左臂上戴着孝，黑色布纱像一个死寂的静止符。忽然，胡三说： “小
满，你想试琴吗？有好几把琴等着你试呢。 ”小满很难把泪水 一下子咽回去 ，可他的眼睛被点亮了。他拉
起了布鲁赫的《 G 小调小提琴协奏曲》 ，即便是难度最高的第三乐章 ，小满仍能从容地使用双音技巧 ，他
似乎已经懂得捕捉瞬间之美而不事铺张。
“小满 ，你进步太快了！ ”
“爸爸走了以后 ，我一直拉这个曲子。妈妈外表坚强 ，其实一直吃不下饭睡不着觉 ，瘦得很厉害。教
琴的老师说这首曲子经常被心理学家用来给病人缓解痛苦 ，我就不停地拉 ，希望对妈妈有用 ……”
“是的 ，是的 ，音乐可以救人。小满 ，你要拉得更好一些。 ”
琴作坊开到第八年 ，订单越来 越多了 ，胡三、韩五爱挑剔的毛病却越来越厉害 —— 挑剔订单的数量 和
时间 ，挑剔琴主的品性。
琴作坊开到第九年 ，人们说胡三 、韩五越来越矫情了 ，琴做完了当年不卖 ，放一放 ，为了声音更好听 。
秋天，月亮升了起来。城里的儒商林先生亲自来琴作坊，说： “十把手作琴的订单，加拿大的朋友拜 学科 网（北 京）股 份有限 公司
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托我把这件事办好。请两位老师配合一下。价格翻番儿，时间紧。 ”
胡三接了一句： “做不了，情绪上不来。 ”
这两个怪人，一个抬头看着月亮，痴痴地、傻傻地、呆呆地不动。另一个笑了，伸出手指，弹了一下
月光，那铮铮鸣响，不觉间，把人世的一切都水银般流散了。他们似乎同时想起了那个孩子 —— 小满，还
有北墙上的老琴。
“小满 应该可以参加维尼亚夫斯基世界青少年小提琴比赛了吧？ ”
“他需要一把好琴。 ”
(有删改 )
1．下列对文本相关内容的理解 ，正确的一项是 (3 分)( )
A．胡三的手作琴制艺精良 ，他主动找上韩五合作 ，是因为可以通过韩五的琴行卖琴。
B．胡三在一年不到的时间里 ，就制作出 “有样貌的手作琴 ”，可见天赋决定人生成败。
C．“为了一把琴 ，去跳楼？ ”胡三的疑问表明不理解韩五祖父的行为 ，这成为两人间的隔阂。
D．胡三、韩五面对 “价格翻番儿 ”也不为所动，这种 “挑剔 ”“ 矫情 ”，有如陶渊明般的气节。
答案 D
解析 A 项，“是因为可以通过韩五的琴行卖琴 ”错误。原文中胡三、韩五是知音关系 ，寻求合作主要是
因为志趣相投。 B 项，“可见天赋决定人生成败 ”错误。原文是 “胡三木匠出身 ，十六岁学徒 ，三十岁上
练成了一等一的高手。四十九岁那年 ……” 可见胡三制琴成功 ，并非完全依靠天赋异禀 ，十几年苦练匠艺
更为他制琴成功提供了坚实基础 。C 项，“胡三的疑问表明不理解韩五祖父的行为 ，这成为 两人间的隔阂 ”
错误。原文是 “胡三发誓一定要修好这把琴。前前后后修了一年 ，果然，重生后的琴音绮丽饱满，也沧桑
沉郁。韩五感激胡三，胡三倒不好意思起来 ”，可见胡三的疑问只是出自本能的震惊，他能理解并认同韩
五祖父恪守诺言的可贵品质，且两人间并没有 “隔阂 ”之说。
2．下列对文本艺术特色的分析鉴赏 ，不正确的一项是 (3 分)( )
A．小说以 “琴”为线索 ，内涵丰富 ，既体现出胡三 、韩五二人的知音之情 ，又体现出胡三 、韩五 、韩五祖
父、小满的人性之美。
B．小说人物描写生动细致 ，如第四段将胡三的外貌与连续动作形 成反差 ，突出他外表粗糙 、内心细腻 、手
艺高超的特点。
C．小说情节安排巧妙 ，与《林教头风雪山神庙 》一样 ，都善于运用人物对话和环境描写推动故事情节有序
向前发展。
D．小说叙述中融入关于音乐的精妙表达 ，如“那铮铮鸣响 ，不觉间 ，把人世的一切都水银般流散了 ”，构学科 网（北 京）股 份有限 公司
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成了小说散文化的审美维度。
答案 C
解析 C 项，“都善于运用 …… 环境描写推动故事情节有序向前发展 ”错误。本文中的环境描写并未推动
故事情节的发展。
3．小满收到胡三、韩五为他量身定 做的琴，并向他们写了一封信。请结合文章相关内容，从小满的角度 ，
将下面的信补充完整。 (4 分)
敬爱的胡三叔、韩五叔：
真心感谢你们一直以来的帮助 ，如果不是你们的善良 ，我无法站在琴声悠悠的舞台上 ，
怀揣着自己的梦想至今。
…………
除了无私的善良 ，你们身上的各种闪闪光亮 ，无时无刻不照耀着我 ：那一把老琴 ，是恪
守诺言的可贵品质； __ ……
我想 ，这些不仅是我这茁壮禾苗所需要的营养 ，更是当下 ，很多人都需要的立世良方。
致以最诚挚的敬意！
幸运的小满
答：
答案 那一身本领 ，是“十年磨一剑 ”的努力坚持；那日日夜夜的雕琢 ，是专注品质、精益求精的匠心；
那种种挑 剔与矫情，是对理想的热爱与坚守，是对抗物欲世俗的初心。
4．本文是如何传达出 “古典审美追求 ”的？请结合全文简要说明。 (6 分)
答：
答案 ①从故事内容上：小说以 “琴”这一传统意象为线索贯穿全文 ，其中对琴、制琴、琴声等的描述洋
溢着浓郁的古典文化气质。 ②从人物形象上：胡三、韩五对制琴精益求 精的匠心，诚挚帮助小满的善心，
甘愿摒弃名利，一心为热爱和理想偃仰啸歌的价值追求，都极具古典精神风范。 ③从语言风格上：行文多
用四字短语，文白间杂，语言凝练典雅，呈现出浓厚的古典笔致韵味。
(二)现代文阅读 Ⅱ(本题共 4 小题 ，16 分)
阅读下面的文字 ，完成 1～4 题。 学科 网（北 京）股 份有限 公司
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老人与河
孔捷生
黎国铧一眼选中河汉苇塘 ，那里有棵歪脖子树 ，一根粗枝探入水中 ，像活着的倒影 ，正是下钓的好地
方。和碳素钓竿不同 ，罗杰斯的老钓竿是木质的 ，黄铜榫接还镶嵌银饰 ，透着 20 世纪中叶的年代感 ，只有
塑料浮漂是 21 世纪的。黎国铧给钓竿装上铅坠 ，鱼钩挂好假饵 ，避开水边红蓼下竿。
罗杰斯端坐轮椅 ，阳光摩挲脸上纵横沟壑 ，他吐纳江风水汽 ，闭目冥想。嘴角笑意似有若无 ，就像荡
漾的浮漂儿。
苇丛深处水鸟争喧 ，绿萍底下不时翻上气泡 ，散发微腐气息。黎国铧的感觉没错 ，色泽鲜艳的浮漂儿
很快被扯动 ，他三番五次收竿 ，吞钩鱼儿都偏小 ，放生了。后又钓起一尾两磅多的鲇鱼。他打手势问罗杰
斯，坐在轮椅上的老人摇头 ，他便摘钩 ，滑溜溜鲇鱼扑通扎入水草 ，惊走红蓼花穗上的蜻蜓。
罗杰斯示意有话说 ，黎国铧趋前俯身 。老人喁喁指点 ，下游石滩才是他以前钓鱼的福地 。老人兴致高 ，
执意弃轮 椅拄拐杖让黎国铧搀扶着挪动 ，一寸寸踏勘记忆的方位。老人全身功能都在衰退，尤其是被阿尔
茨海默症侵蚀的记忆力，唯剩飞行员的视力未被衰老压倒。他目力炯炯，认穴般指戳，呼哧带喘道： “就
是这里。 ”
这段河岸峭拔，怪石嶙峋。黎国铧拂去树墩青苔，扶罗杰斯坐下。眼底波涛拍击 巉岩，浪沫飞腾，怎
么看也不像钓客吉位。老人又示意别用假饵，河边湿泥里蚯蚓多的是。这倒合黎国铧心意，他也不喜欢塑
料假饵 ，便动手抠泥 ，两三下便有蚯蚓 。在乱石间隙下钓 ，湍流簇拥橙色浮漂儿 ，如沸汤翻腾 ，鱼能咬钩？
老人像时光穿越，固执追寻回不去的昔日，然而生命就像 破蛹羽化的蝉 ，再也钻不回草间蜕壳。
黎国铧已无获鱼之念。他除了做义工上门照看老人，间或也给这位老兵 —— 最后一位在世的飞虎队飞
行员做抗日口述访谈。他看罗杰斯的状态不错，便拉话。老人难得精神健旺，嗓音也清朗起来。他问飞虎
队旧事上次谈到哪里？答：是日寇在华最后一次大型战役 —— 豫湘桂会战。当时，日军攻陷贵州独山，飞
虎队出动战机封锁公路，掩护中国军民撤退 ……
罗杰斯嘟囔道，他记得清楚，当时驾机低飞扫射封锁道路，日军对空还击，子弹穿过机身钻进小腿。
他一时还没感觉，机枪手却没了声息，一看战友已歪倒在机枪上 …… 黎国铧没带笔记本电脑 ，便用手机录
音。老兵嘶哑声线化为波长在手机屏幕起伏，就像黔南莽苍群峰切割出波浪形天际线。罗杰斯龇牙咧嘴驾
机飞回成都，降落时冲出跑道，昏死过去。那是他在飞虎队最后一次执行任务。
语罢，老兵满脸皱褶现出光晕，好不容易聚拢的散乱记忆，又拐入另一段光阴隧道。养伤的日子他拄
拐到锦江边 ，在青羊宫外小摊吃麻辣凉粉 ，那是他和蓉贞初遇时刻 。老人清晰记得 ，蓉贞鬓边插着芙蓉花 ，
两条羊角辫晃来晃去，让他心跳。 学科 网（北 京）股 份有限 公司
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罗杰斯眼瞳好像散焦了，从虚空看到生命中最长的那个时辰，思路从青羊宫翘起的飞檐飘然坠地，老
人表情神秘道： “黎，告诉你 一个秘密。 ”黎国铧侧耳聆听 ，原来蓉贞并非盐商之女。罗杰斯为访佳人多
次到青羊宫 ，羊角辫少女的倩影就像锦江边一绺杨柳在眼前摇曳 。他记不清吃过多少碗凉粉 ，并爱上麻辣 。
蓉贞就是卖凉粉的女孩，盐商门第是她初次去俄亥俄州拜见公婆时即兴编的。罗杰斯随她怎么说，其实他
的父母也都是穷人，一辈子没走出阿巴拉契亚山地。
罗杰斯精神矍铄，逻辑清晰，令黎国铧诧异。故乡遥远的面影在老人语言中缓缓展开，青羊宫香烟缭
绕的炉鼎，高大的银杏树，湿漉漉的青石板路，凉粉挑子的小灯笼 ……
老人语罢吐纳调息，呼吸显得有些短促，他累了。黎国铧关闭手机录 音，钓竿依然翘挺，没有动静。
天风放牧碎云擦拭晴空，对岸层叠林木摇出绿光，掩映其间的红蓝屋顶，像波浪间的帆翼。罗杰斯腿脚不
行，腰板仍挺得直，未衰退的还有眼眸，投向宽阔的波托马克河。一江来水泛满夏天墨绿，涌向切萨皮克
湾，在河口天际线变蓝。
罗杰斯老眼忽然精光四射，指向水中石堆。一直在浪窝颠簸翻腾的橙色浮漂儿已下沉不见，钓竿怒弯
成弓，钓绳绷紧如弦斜插入水。黎国铧急急抄起插在石间的钓竿，一股愤怒之力将他猛然拽向湍流，钓竿
几欲脱手。他双腿夹竿绞动线轮曲柄，才拉几转就纹丝不动。僵持之下，罗杰斯颤巍巍打手语，黎国铧领
会， 便放绳和挪动位置角度 ，竿头吃不住活物泼剌剌挣扎，直弯入水。传导过来的能量这般猛烈，莫不是
水獭误咬鱼钩？
收放之间 ，活物经不住绞线器物力量 ，一点点升出水面 ，竟是一尾肥大的海鲈鱼。它绝望扭动 ，壮实
黑脊和肚腹银鳞闪闪发亮 ，陌生世界激发出它愤怒最高值 ，一头又扎入浪沫。黎国铧耐心反复拉锯 ，终于
把大鱼拽起。这种游弋咸淡水域的海鲈鱼 ，他此前钓过 ，却未见过这么大的 ，足足五磅多！
黎国铧自认资深钓客 ，却不知水底石堆别有洞天。海 鲈鱼 “噗噗 ”掉打自己 ，鱼鳍沾满草屑，鱼鳃大
开大合 ，鲜红腮片像龙牙花怒放 。它奋力一蹦 ，扑到罗杰 斯脚边 ，打湿他的裤腿 。罗杰斯额上都堆满笑纹 ，
仿佛推开了记忆迷宫的某扇窗户。罗杰斯的故事不知有几多遗失于积尘，心智和他的时代一同剥蚀退化，
只有河流如故，载走滔滔光阴 ……
(有删改 )
1．下列对本文相关内容的理解 ，不正确的一项是 (3 分)( )
A．文中写罗杰斯的钓竿与现代钓竿有许多不同 ，透着 20 世纪中叶的年代感 ，暗示了罗杰斯的身份特殊。
B．罗杰斯坚定自己的想法 ，执意要去黎国铧认为并不是合 适钓鱼的地方，写出他曾经作为军人的坚毅。
C．即使记忆力衰退 ，罗杰斯仍能想起 “青羊宫香烟缭绕的炉鼎 ”等，体现他对这片土地深深的思念。
D．钓到海鲈鱼后 ，罗杰斯露出了笑容 ，这似乎触动了老人内心的记忆 ，说明眼前的情景老人似曾相识。 学科 网（北 京）股 份有限 公司
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答案 B
解析 B 项，“写出他曾经作为军人的坚毅 ”错误。原文 “老人全身功能都在衰退 ，尤其是被阿尔茨海默
症侵蚀的记忆力 ，唯剩飞行员的视力未被衰老压倒 ”，明确提到罗杰斯受到阿尔茨海默症的影响 ，记忆力
衰退 ，所以他执意要 去的地方并非出于 “军人的坚毅 ”，而是出于他过去记忆的驱使。
2．下列对小说艺术特色的分析鉴赏 ，正确的一项是 (3 分)( )
A．小说以选择钓鱼地点开篇 ，在钓鱼过程中慢慢展现出一个患有严重疾病的老人形象 ，揭示了要懂得关爱
老人的主题。
B. 文中多处运用比喻 ，“嘴角笑意似有若无 ，就像荡漾的浮漂儿 ”“ 天风放牧碎云擦拭晴空 ”等，使文章
语言生动 ，富有诗意。
C．文中多次写到罗杰斯虽然记忆力衰退但视力依然很好 ，这既符合他曾经是飞行员的身份 ，也为后文精准
钓鱼做了铺垫。
D．黎国铧认为 自己是资深钓客 ，却只有在罗杰斯的指导下才能钓到大鱼 ，运用对比的手法突出了罗杰斯钓
鱼经验的丰富。
答案 C
解析 A 项，“慢慢展现出一个患有严重疾病的老人形象 ，揭示了要懂得关爱老人的主题 ”错误。文章虽
然以选择钓鱼地点开篇 ，但主要展示的是罗杰斯老人的性格特点和精神面貌 ，而非 “一个患有严重疾病的
老人形象 ”，且文章的主题并非 “要懂得关爱老人 ”，而是对老兵坚韧精神的赞美和对抗战往事的缅怀 。B
项，“文中多处运用比喻 …… 天风放牧碎云擦拭晴空 ”错误。不 是比喻 ，以 “放牧 ”“ 擦拭 ”写“天风 ”
“碎云 ”，这是比拟。 D 项，“只有在罗杰斯的指导下才能钓到大鱼 ，运用对比的手法 ”错误。文章并未
直接提到黎国铧在罗杰斯的指导下才能钓到大鱼 ，而且即使有这一情节 ，也并非为了突出罗杰斯钓鱼经验
的丰富 ，而是为了表现老人对过去时光的怀念 ，且此处是衬托 ，并非对比。
3．小说标题 “老人与河 ”有多重意蕴 ，请简要分析。 (4 分)
答：
答案 ①表面上老人虽然受阿尔茨海默症的影响记忆力衰退 ，但他仍能找到自己曾经钓鱼的那条河； ②河
流成为罗杰斯老人的象征 ，流去的不仅是河水 ，还有老人的记忆 ；③时光就像流去的滔滔河水 ，一去不返 ，
流露出人生易老、世事沧桑的感伤。
4．本文在叙事上很有特色 ，请结合文本加以分析。 (6 分)
答： 学科 网（北 京）股 份有限 公司
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答案 ①以黎国铧的视角叙事 ，通过黎国铧的所见所感组织材料 ，表达他对一个援华老兵的敬仰和对其英
勇往事的缅怀 ，深化了作品主题； ②采用访谈的叙事方式 ，通过黎国铧对罗杰斯老人的访谈引出老兵的辉
煌往事 ，推动故事情节发展； ③回忆与现实交织 ，文中既写了老人现实的境况 ，又通过回忆的方式回到他
的过去 ，结构灵活 ，避免了平铺直叙。
增分大题 题型题组 —— 现代文阅读 Ⅱ(散文 )
(一)现代文阅读 Ⅱ(本题共 4 小题 ，16 分)
阅读下面的文字 ，完成 1～4 题。
蒹葭苍苍
徐则臣
从五斗渠到大渠之间 ，浩浩荡荡地生着一片芦苇。在村庄方圆几里内 ，那是最为高大茂盛的芦苇。没
有人知道什么时候长出了第一棵 ，什么时候又蔓延了这么一大片。父亲小时候到田里捡麦穗 ，遇上了几十
年都罕见的大冰雹 ，就是躲在那片芦苇荡里。父亲说 ，谁会想到中午出门 时还艳阳高照 ，傍晚就降下了满
天鸡蛋大的冰雹呢。他把柳条编的小篮子顶在头上，捡了一个下午的麦穗撒落一地，篮子被砸坏了，他只
好钻进芦苇丛，把自己裹在里面使劲地摇动芦苇，用枝叶扫荡出一块安宁的空间。几十年后父亲说，起风
了，大风把芦苇荡卷起来，像煮沸的开水，发出鬼哭一样的呜咽声。
若干年后，我七岁，一个人在黑夜里经过那片芦苇荡两次。
我记不起来父亲出去干什么了，只有我和姐姐在家。母亲一个人在田里，我和姐姐把晚饭做好后，等
母亲回来。天很晚了，通常这个时候我们已经吃晚饭了，周围的邻居也早早收工，无数条炊烟飘在低矮的
屋顶上。可是母亲还没回来。我和姐姐都急了。小时候我十分依赖母亲，一天见不到心里就发慌。每次母
亲出门回来迟了，我都要站在家门口张望，直到她走进巷口我才安心。如果出远门， 比如去外婆家 ，更不
得了，走之前我要问清楚什么时候回来，到了母亲回家的那个傍晚，我会一个人沿着母亲回家的路一直走
到村头，扶着那棵歪脖子小树向前望去。
姐姐让我再等等，她问过前面小四子的妈了，说母亲正在赶活儿，趁着有点月亮干完了就算了。母亲
还让带来话 ，叫我们先吃 ，不用等她 。可是我等不下去 ，我要去湖地里找 ，我坐在锅灶旁感到一阵阵干冷 ，
我想，母亲一定也很冷，我要把她找回来。其实那会儿刚入秋，可我的赤脚在鞋子里直冒冷汗。姐姐问我
害不害怕，我说不怕。姐姐说那你一个人去吧，我把猪给喂了。我拿了母亲的一件衣服出了门。
从家 到母亲干活的田地大约四里路 ，先过后河的一座桥 ，再穿过平旷的打麦场 ，上了五斗渠直往北走 ，
一直走下去，向右拐就到了。但是我从来没在晚上一个人去过。真正的一个人，过了后河桥连一个人影都 学科 网（北 京）股 份有限 公司
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没看到。出村的时候一点不害怕，满脑子里都是尽快找到母亲。我一路飞奔上了打麦场。村庄在我身后，
狗叫和小孩的啼哭也在身后，听起来极不真切，像是跑进了另一个世界，然后我听见了自己的脚步声，在
打麦场上产生了更巨大的回声，好像有许多人随我一起跑，我出哪只脚，他们也出哪只脚。最可怕的莫过
于只听见自己的声音 。我停下来 ，看清了七岁时的那个夜晚 。它比 我想象的要黑 ，月光是那个夜晚的同谋 ，
暧昧的光亮只能增添旷野的恐怖。周围的树木和草堆黑黢黢地排列成一个圈，我站的地方仿佛是世界的中
心。树木和草堆板着黑脸，在风中摇头晃脑，我意识到这个夜晚风很大。我继续往前走，抱紧了母亲的衣
服。
真正的恐惧和我相遇在五斗渠上，我终于面对了那片芦苇荡。我得说，真是像海，它是一片桀骜不驯
的翻腾的巨浪。风也许并不像我当时认为的那么猛烈，只是田野过于平坦辽阔，风可以像旗帜一样从远处
卷来，而那片芦苇又过于惹眼，它是野外唯一的一堵墙。风必须经过它。于是我看见从第一棵开始，风拉
弯了所有芦苇的 腰，大风水一般地漫过它们，使之起伏具有了水一样的柔韧的表情、浪一样痛苦的姿势。
弯下又挺起，涌过去又退回来。恐惧终于降临，我把自己送到了一头奔腾的巨兽跟前，听到它在黑夜里粗
重又狂乱的呼吸，像一片森林突然倒下，像整座山峰缓缓裂开。
因为一片芦苇，大风得以在我七岁的夜晚存活。我听见了风的声音，杂乱，深不可测。如果有人告诉
我，黑暗里藏着十万魔鬼，我信。芦苇的声音比芦苇本身更像魔鬼。我后悔出门过于轻率，因为恐惧而发
抖，刚刚满怀的焦急和寻母的使命感被大风一扫而空。我不能半途而废，我对姐姐说了，我不害怕，我一
定要把母亲找 回来。我把衣服缠在手上 ，贴着远离芦苇的路的那一边磕磕绊绊地跑。跑几步就停了下来，
也就是从那次起我知道，恐惧时不能跑，越跑越害怕。我努力放轻脚步走，坚持忍着不回头看。身旁的一
大片庄稼像缓缓起伏的海，我听到风声、芦苇声、庄稼声和我的心跳声，感觉自己是走在梦里，整个身心
失去控制似的摇摆不定。
那大概是我有记忆以来走的最长的一 段路 ，好像怎么也走不完。当时我什么都想到了，包括死。而且
有关死想得最多，没有排除任何一种我所知道的死法。我想，如果我死了，母亲该到哪里找我呀。我没有
死，走到大渠上的老柳树下我 停下来 ，我还活着，一屁股坐到地上，衣服被汗湿透，牙咬得两腮生疼。
我没找到母亲，她从另一条路回家了。而我又从原路返回，同样是一身冷汗。我想，走过了黑夜里的
芦苇荡，任何恐怖的东西对我来说都无所谓了。但是当我走进村庄，看到第一户人家门缝里透出来的灯光
时，还是忍不住哭了 ，一直哭到家里 。母亲站在门口远远地问是不是我 ，我一声不吭 ，进了门就爬上了床 。
那个晚上我始终没说一句话，晚饭也没吃就睡了。夜间我的梦里长满了无边无际的芦苇和风，浩浩荡荡的
黑夜之声贯穿了整个梦境。
(有删改 )学科 网（北 京）股 份有限 公司
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1．下列对文本相关内容和艺术特色的分析鉴赏 ，正确的一项是 (3 分)( )
A．“我”小时候很依赖、担心母亲 ，每次母亲出门回来晚了 ，“我”都会到家门口甚至村头去张望。
B．母亲在地里干活很晚没回来 ，姐弟俩都很着急 ，但姐姐要喂猪 ，便叫 “我”独自去找妈妈。
C．“我”七岁时的一个夜晚第一次真切感受到大风席卷芦苇荡的恐怖 ，后悔自己出门找妈妈。
D．以儿童视角 写儿时的寻母经历， “我”感悟到走过黑夜里的芦苇荡，不再害怕任何恐怖的东西。
答案 A
解析 B 项，“便叫 ‘我’独自去找妈妈 ”错误。由原文 “我要去湖地里找 ”“ 我要把她找回来 ”“ 姐姐
问我害不害怕 ，我说不怕。姐姐说那你一个人去吧 ”可知 ，是“我”主动要去找妈妈的。 C 项，“后悔自
己出门找妈妈 ”错误。原文说的是 “我后悔出门过于轻率 ”。D 项，“感悟到走过黑夜里的芦苇荡 ，不再
害怕任何恐怖的东西 ”错误。由原文 “我想 ，走过了黑夜里的芦苇荡 ，任何恐怖的东西对 我来说都无所谓
了。但是当我走进村庄 …… 还是忍不住哭了 ，一直哭到家里 ”可知， “我”不是 “不再害怕 ”而是敢于面
对。
2．关于文中第一段 ，下列说法不正确的一项是 (3 分)( )
A．交代生活环境 ，展现故事背景 ，为下文讲述 “我”儿时一个人在黑夜里两次经过那片芦苇荡做铺垫。
B．“中午 ”与“傍晚 ”对比 ，突出时间之短 ；“艳阳高照 ”与“鸡蛋大的冰雹 ”对比 ，突出天气变化之大 。
C．“像煮沸的开水 ，发出鬼哭一样的呜咽声 ”用比喻 、比拟等修辞 ，表现风中的芦苇荡给 “我”造成的恐
惧。
D．父亲小时候遇到几十年都罕见 的大冰雹时躲进芦苇荡，用篮子保护头，反衬 “我”小时候的胆小怕事 。
答案 D
解析 D 项，“反衬 ‘我’小时候的胆小怕事 ”错误。从全文来看 ，主要是渲染大风席卷芦苇荡的恐怖气
氛，为下文写 “我”七岁时一个人晚上两次经过那片芦苇荡的特殊经历做铺垫。
3．文章为何说 “那大概是我有记忆以来走的最长的一段路 ”？(4 分)
答：
答案 ①“ 最长的一段路 ”指的是 “我”在五斗渠上经过芦苇荡的那条乡间小路； ②其实这段路本身并不
长，只是七岁的 “我”夜晚走过如巨兽般的芦苇荡时一路担惊受怕 ，才感觉是 “最长的一段路 ”；③这句
话是对 “我”儿时一段独特经历的感叹。
4．本文标题 “蒹葭苍苍 ”也可改为 “寻母记 ”，请问哪个标题更好？请简要说明。 (6 分)
答： 学科 网（北 京）股 份有限 公司
102 -
答案 示例一 ：“蒹葭苍苍 ”更好 。①蒹葭 ，古代指的就是芦苇 ；苍苍 ，形容茂盛的样子 。以“蒹葭苍苍 ”
为题 ，符合文中描写的芦苇荡的实情。 ②“ 蒹葭苍苍 ”揭示了故事发生的背景 ，也交代了所描写的主要对
象。 ③“ 蒹葭苍苍 ”出自《诗经》 ，有着浓郁的文化意蕴 ，以此为题可以增强作品的诗情画意 ，提升作品
的品位。
示例二： “寻母记 ”更好。 ①本文是一篇以儿童视角写的回忆性散文 ，以“寻母记 ”为题 ，与“我”七岁
时的一个 晚上去田里找妈妈的故事内容相吻合。 ②作为记叙性散文，其主要情节都与 “寻母 ”有关，包括
起因、过程和结果等环节。 ③以“寻母记 ”为题，更有乡土气息和生活味道，更接地气，因而更受读者欢
迎。
(二)现代文阅读 Ⅱ(本题共 4 小题 ，16 分)
阅读下面的文字 ，完成 1～4 题。
寒风吹彻
刘亮程
雪落在那些年雪落过的地方 ，我已经不注意它们了。三十岁的我 ，似乎对这个冬天的来临漠不关心 ，
却又好像一直在倾听落雪的声音 ，期待着又一场雪悄无声 息地覆盖村庄和田野。许久以后我还记起我在这
样的一个雪天 ，围抱火炉 ，吃咸菜啃馍馍想着一些人和事情 ，想得深远而入神 。柴火在炉中啪啪地燃烧着 ，
炉火通红，我的手和脸都烤得发烫了，脊背却依旧凉飕飕的。 寒风正从我看不见的一道门缝吹进来。它比
我更熟悉墙上的每一道细微裂缝。
我紧围着火炉，努力想烤热自己。我的一根骨头，却露在屋外的寒风中，隐隐作痛。那是我多年前冻
坏的一根骨头，它永远地冻坏在那段天亮前的雪路上了。那个冬天我十四岁，赶着牛车去沙漠里拉柴火。
牛车一走出村子，寒冷便从四面八方拥围而来，把我从家里带出的那点温暖搜刮得一干二净，浑身上下只
剩下寒冷。
那个夜晚并不比其他夜晚更冷，只是我一个人赶着牛车进沙漠，似乎寒冷把其他一切都收拾掉了，现
在全部地对付我。我掖紧羊皮大衣，一动不动趴在牛车里，不敢大声吆喝牛，免 得让更多的寒冷发现我。
天亮后 ，牛车终于到达有柴火的地方 。我的一条腿却被冻僵了 ，失去了感觉 。我试探着用另一条腿跳下车 ，
拄着一根柴火棒活动了一阵，又点了一堆火烤了一会儿，勉强可以行走了，腿上的一块骨头却生疼起来，
是我从未体验过的一种疼，像一根根针刺在骨头上又狠命往骨髓里钻 —— 这种痛感一直延续到以后所有的
冬天以及夏季里阴冷的日子。
太阳落地时，我装着半车柴火回到家里，父亲一见就问我：怎么拉了这点柴，不够两天烧的。我没吭
声，也没向家里说腿冻坏的事。隔着多少个季节，今夜的我，围抱火炉，再也暖不热那个遥远冬天的我。 学科 网（北 京）股 份有限 公司
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我准 备了许多柴火 ，是准备给这个冬天的。我才三十岁，肯定能走过冬天。
但在我周围，肯定也有人不能像我一样度过冬天。他们被留住了。冬天总是一年一年地弄冷一个人，
先是一条腿、一块骨头、一副表情、一种心境 …… 尔后整个人生。
我曾在一个寒冷的早晨，把一个浑身结满冰霜的路人让进屋子，给他倒了一杯热茶。那是个上了年纪
的人，身上带着许多个冬天的寒冷，当他坐在我的火炉旁时，炉火须臾间变得苍白。我没有问他的名字，
在火炉的另一边，我感到迎面逼来的一个老人的透骨寒气。他一句话不说。我想他的话肯定全冻硬了，得
过一阵才能化开。大约坐了半个 时辰 ，他站起来，朝我点了一下头，开门走了。我以为他暖和过来了。
第二天下午，听人说村西边冻死了一个人。我跑过去，看见这个上了年纪的人躺在路边，半边脸埋在
雪中。
我第一次看到一个人被冻死。我不敢相信他已经死了。他的生命中肯定还深藏着一点温暖，只是我们
看不见。他的身上怎么能留住一点点温暖呢？靠什么去留住？他的烂了几个洞、棉花露在外面的旧棉衣？
底快磨通、一边帮已经脱落的那双鞋？还有他的比多少个冬天加起来还要寒冷的心境 ……
落在一个人一生中的雪 ，我们不能全部看见 。每个人都在自己的生命中 ，孤独地过冬 。我们帮不了谁 。
我的一小炉火 ，对这个贫寒一生的人来说 ，显然微不足道。他的寒冷太巨大。
我有一个姑妈 ，住在河那边的村庄里 ，许多年前的那些个冬天 ，我们兄弟几个常手牵手走过封冻的玛
河去看望她。每次临别前 ，姑妈总要说一句：天热了让你妈过来喧喧。
姑妈年老多病 ，她总担心自己过不了冬天。天一冷她便足不出户 ，偎在一间矮土屋里 ，抱着火炉，等
待春天来临。
一个人老的时候，是那么渴望春天来临。尽管春天来了她没有一片要抽芽的叶子，没有半瓣要开放的
花朵。春天只是来到大地上，来到别人的生命中。但她还是渴望春天，她害怕寒冷。
我一直没有忘记姑妈的这句话，也不止一次地把它转告给母亲。母亲只是望望我，又忙着做她的活。
姑妈死在几年后的一个冬天。我回家过年，记得是大年初四，我陪着母亲沿一条即将解冻的马路往回走。
母亲在那段路上告诉我姑妈去世的事。她说： “你姑妈死掉了。 ”
母亲说得那么平淡 ，像在说一件跟死亡无关的事情。 “怎么死的？ ”我似乎问 得更平淡。母亲没有直
接回答我 ，她只是说： “你大哥和你弟弟过去帮助料理了后事。 ”此后的好一阵，我们再没说话，只顾静
静地走路。
快到家门口时 ，母亲说了句 ：天热了 。对母亲来说 ，这个冬天已经过去了 。母亲拉扯大她的七个儿女 。
她老了。我们长高长大的七个儿女 ，或许能为母亲挡住一丝的寒冷。但母亲斑白的双鬓分明让我感到她一
个人的冬天已经来临 ，那些雪开始不退、冰霜开始不融化 —— 无论春天来了 ，还是儿女们的孝心和温暖备 学科 网（北 京）股 份有限 公司
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至。
随着三十年这样的人生距离 ，我感觉着母亲独自在冬天的透心寒冷。我无能为力。雪越下越大。天彻
底黑透了。
我围抱着火炉 ，烤热漫长一生的一个时刻 。我知道这一时刻之外 ，我其余的岁月 ，我的亲人们的岁月 ，
远在屋外的大雪中，被寒风吹彻。
(摘编自《一个人的村庄》 )
1．下列对文本相关内容的理解 ，不正确的一项是 (3 分)( )
A．十四岁的 “我”赶牛车去拉柴火 ，“一个人 ”“ 一动不动 ”“ 不敢大声吆喝牛 ”等词语写出 “我”的孤
独以及对寒冷的恐惧。
B. 路人进屋后 ，炉火的 “苍白 ”，与外面的大雪、路人身上的 “冰霜 ”相呼应 ，又与路人内心的孤独和生
命的苍凉相映衬。
C．姑妈邀请母亲去她家 ，“我”将话带给了母亲 ，但是母亲的反应却非常冷淡 ，暗示姑妈和母亲之间的感
情并不好。
D．文章描写了 “我”、路人 、姑妈和母亲遭遇的寒冷 ，多场寒冷带来的不仅仅是死亡和绝望 ，还有对待生
命严酷的坦然。
答案 C
解析 C 项，“暗示姑妈和母亲之间的感情并不好 ”错误 ，应是体现母亲多次经历 “风雪 ”后对待生命的
坦然。
2．下列对文本艺术特色的分析鉴赏 ，不正确的一项是 (3 分)( )
A．本文以寒冷的记忆为线索 ，向读者展现了一个历经艰辛、 被生活折磨的 “我”内心成长的过程。
B．本文通过 “路人 ”“ 上了年纪的人 ”“ 老人 ”等称呼的变化逐渐拉近 “我”与“路人 ”的距离 ，加深了
感情。
C．“寒风 ”这一意象不仅指自然中的寒风 ，也象征着人间的衰老与苦难 ，寄托着在命运归宿前人的脆弱和
无奈。
D．本文情景交融 ，意境恬淡闲适 ，苍凉宁静 ，作者对乡村生命自然状态的描摹 ，透着一股清新之风。
答案 D
解析 D 项，“恬淡闲适 ”“ 清新之风 ”错误 ，本文的意境是沉重而悲痛的 ，并非 “恬淡闲适 ”；作者对
乡村生 命自然状态的描摹透露着生活的贫寒、内心的孤独、心境的悲凉 ，甚至是衰老死亡，所以 “清新之
风”是错误的。 学科 网（北 京）股 份有限 公司
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3．马尔克斯 《百年孤独 》的开篇 “多年以后 ，面对行刑队 ，奥雷里亚诺 ·布恩迪亚上校将会回想起父亲带他
去见识冰块的那个遥远的下午。 ”运用跨时空叙事笔法连接现在、未来与过去。本文在叙事上也有时空的
跨越 ，请结合全文谈谈你的理解。 (4 分)
答：
答案 ①如今 ，“我”在雪天围抱火炉 ，回忆着在寒冷的雪天冻坏了腿的过去 ，想象着许久以后 ，“我”
还会记起这个雪天的情形。作者通过想象 ，运用跨时空叙事笔法连接了现在、过去与未来。 ②“ 我”在这
个雪天 ，回忆着不同时间和空间的人和事：那个曾经在雪中冻死的老人 ，害怕寒冷的姑妈 ，还有走到 “冬
天”的母亲。
4．有人评价这篇散文 “每个字都重得好像要沉下去 ，但又不悲痛到绝望 ，字里行间透着人生孤独之外的大
爱”。这种 “大爱 ”包含着多重温暖 ，请结合全文加以分析。 (6 分)
答：
答案 ①饱尝寒冷的 “我”愿意帮助同样遭受寒冷的陌生老人； ②老人或许早已有了死亡的预感 ，表示感
谢后 ，在生命的最后选择离开 ；③自然的 、生活的 “冬天 ”不断出现 ，但人们心中依然期待 “冬去春会来 ”，
依然对生活充满希望； ④母亲含辛茹苦养大儿女 ，孝顺的儿女尽己所能为母亲挡住一丝的寒冷。
【考前技巧篇】
☞信息类文本阅读 技巧
解答 思辨 类文本阅读题指津
（1）突破 “知识 ”关：论证结构、论述思路、论证方法
（2）突破 “读文关 ”。把握观点 ，梳理段与段之间的逻辑 ，段落内部句与句之间的逻辑 。两遍阅读后 学科 网（北 京）股 份有限 公司
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快速准确把握全文的思路、中心。
（3）突破 “转述和整合 ”关。细读比对选项表述，分析、判断选项文字在整合和转述过程中的正误 。
熟悉客观题的常见陷阱设置。打通各类错误点，能具有敏锐发现错误的能力。
【总原则】
解答此类题目，应 先梳理文章的内容 ，圈出 每段的中心句 ，把握文章的 观点（结论） 与论证结构 ，根
据文章的 行文思路（内容） ，梳理文章的 论证思路（论证手法、论据等） ，分析 中心论点 和分论点间 的关
系，分析 论点 和论据 之间的关系，明了 论证方法 的类型；
【思考】 论点 是什么（是否正确）、 论据证明 的是什么 观点 （得出怎样的结论 ）、用了什么 论证方法
证明观点 、如何安排思路 、怎样展开论证 、怎样结构文章 。
指津一：整体把 握，微观勾画
1．阅读原文后 ，可提出如下问题 ：本文说明或论证的对象是什么？有什么最新成果或最新观点？今后
的发展前景如何？作者对此新成果或新观点的态度和看法如何？
2．理清全文的脉络，把握主要内容，迅速提取每一节的主要信息。
3．微观勾画是指随时勾勒一些关键词语 ，以备答题时所用 。特别要关注指示代词 、关联词语 （如“一
旦”“ 如果 ”“ 因此 ”“ 但是 ”“ 然而 ”等）、副词（如 “凡时 ”“ 全”“ 将”“ 基本上 ”“ 已经 ”“ 也
许”“ 可能 ”等）以及一些修饰性的词语。
此外，由于论述类文章中有些内容表达起来比较抽象，为了说得具体，有时会运用比喻的修辞手法，
理解时要找出其 “本体 ”。要确切理解含有修辞的句子，要注意前后对照，特点对应。
指津二：紧扣语境，把握内涵 学科 网（北 京）股 份有限 公司
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要准确理解词、句在文中的意思，就要紧密联系语境，注意上下文的修饰、指代等暗示信息，从而把
握其内涵。论述类文章阅读考查的词语往往都具有极为重要的作用，这些词话要么是关键信息点（如指代
性词语 、概念性词语 ），要么就或承前或蒙后省略了相关内容 。这些词语往往已突破了其原来的意义限制 ，
与具体语境结合而有了新的意义。因此，阅读中要对这类词语慎重考虑。可采用如下方法：
1．瞻前顾后法。 联系上下文选择恰当的义项。
2．比照辨析法。 仔细比较辨析文中的一词多义现象和同义词、近义词在语言运用中的差异。
3．参考语境法。 根据语境揣摩 词语的语境义、比喻义、借代义等，分析词语派生或隐含的内容。
论述类文章中重要的句子有如下几种：一是结构比较复杂的句子，二是内涵较为丰富的句子，三是与
文章中心和结构密切相关的句子 （如文眼句 、中心句 、过渡句等 ）。对于第一种 ，可以用 “抽取主干法 ”，
抓住句子主干，理清那些修饰、限制等附加成分，进而理解其含意。对于第二种，应该按照 “句不离段 ”
的原则，结合上下文语境，仔细领会，整体解析。第三种句子体现了文章的思路，有的画龙点睛，有的承
上启下，有的阐明要旨。要理解这些句子的含意，既要注意它在文中的位置，还要看清来龙去脉。
指津三：抓住概念，识别混淆
论述类文章阅读题目，在设置选项上具有很大的迷惑性，综观近几年的高考试卷，其干扰项的设置方
法主要有：
1．程度深浅 、范围大小有意混淆 。选项在概念的外延上做文章 ，或是外延过小 ，以偏概全 ；或是外延
太大，判断过宽，以 “面”代“点”。阅读时要特别留心材料和选项中的 “凡”“ 一切 ”“ 全”“ 都“等
修饰词语。
2．偶然、必然有意混淆。 即把原文中的 “可能 ”有意说成 “必然 ”，把 “偶尔 ”说成 “往往 ”。
3．指代上的混淆 ，颠倒主客 ，偷换概念 。选项偷换概念 ，用形同义异词或形近义异词来迷惑考生 。解
题时，要注意选项是否混淆了概念的所指对象，是否颠倒了陈述主体与修饰语，是否省略了一些关键的修
饰词，犯了偷换概念的错误。
4．现实和设想的混淆 。选项在对概念的判断上时间超前或滞后 。把已经成功的现实和没有成为现实的
设想或可能性混为一谈。
5．肯定和否定 、主要和次要关系上的混淆 ，无中生有 ，牵强附会 。有些选项的设置 ，把肯定说成否定 ，
或把否定说成肯定，有些混淆主要和次要关系，有些属于无中生有，牵强附会，解答时，一定要在原文中
找到依据，切忌主观臆断，望文生义。
6．条件和结果 ：原因和结果的关系的混淆 、颠倒 。有些选项在设置时 ，将条件说成结果 ，或把结果说
成原因，或强加条件及因果关系。要重点辨析，找准答案。
指津四：筛选判断，逐步排除
对选择题要分析、比较、选择，首先排除明显错误的选项，然后分析剩下的选项及与之相关的语言环
境，瞻前顾后 ，寻找有效信息 ，并归纳信息的要点 ，进行筛选 ，再次排除干扰选顼 ，剩下的便是正确答案 。
这样做题，可以提高答题的准确率。
基本方法如下： 学科 网（北 京）股 份有限 公司
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1．复位验证法 。在理解 文中的重要概念时 ，如果对自己的选择没有十分的把握 ，可把选出的答案 “复
位”到原文中 “验证 ”一下，如果语意连贯、意思准确则该项即为正确答案。
2．事理分析法 。在论述类文章中 ，常会遇到事理之间的逻辑关系 ，如因果关系 、条件关系 、假设关系 、
选择关系等，要紧紧抓住表示事理之间逻辑关系的关键词语，作出正确的判断。
3．巧用选项法 。在考查理解文中重要句子的命题中 ，命题者常常在句中确定两个考查点 ，每个考查点
按两种理解列为四个选项 。遇到这种题目 ，可以巧妙地利用选项提供的 “方便 ”，根据自己对某一个考查点的
正确理解，排除错误的选项。
【题型解读 】
【题型 1】行文脉络（行文思路） ：实际上就是 概括文章各层大意 。
【答题模式 】回答先写 什么 ，接着写 什么 ，最后写 什么 —— 把全文划分好层次 ，概括层次内容 （段意 ），
依次回答即可。
【注意】 必须使用 “首先，然后，最后 ”等表示先后顺序的词语。
【2020 新高考 1 卷.山东卷】 5. 请简要梳理材料一的 行文脉络 。（6 分）
【策略】读懂大意，梳理层次，概括内容，分点作答。
【答案 】①材料一首先介绍了 《禹贡 》《汉书 •地理志 》《水经注 》中的内容 ，说明了历史地理学在中
国的起源； （首先 …… 说明 …概括第一层内容）
②然后介绍了沿革地理的概念，论述了其存在的意义以及和历史地理学的区别；（然后 …… 概括第二
层的内容 ）
③最后总结了沿革地理和历史地理学的关系，说明了历史地理学的发展过程。（最后 …… 概括最后一
层的内容 ）
【变式问】简要说明材料二是如何对 “中医药走进中小学 ”逐步展开论述 的。（ 4 分）
【审题】所谓 “逐步 ”，即是 有先后次序的 ，这就是 “思路 ”的问题。
【答案】
首先 从中医药的特点、意义与现实困境说起，引发对中医药出路的探讨（ 1 分），
然后 从中小学这个切入点引出 “中医药走进中小学 ”话题（ 1 分），
最后 有针对性地提出两条落实中医药进中小学的具体建议（ 2 分）。
【题型 2】论证思路 【注意 :与“行文思路（脉络） ”相似，但又略有不同】
不同 ：侧重把 全文的观点、论据、论证方法之间的关系是如何呈现 的说清楚。
【2021 新高考 1 卷】 4. 请简要分析材料一和材料二的 论证思路 。
①材料一围绕莱辛《拉奥孔》提出的 “诗画异质 ”观点 ，从缘由、推论到 结论 ，纵向展开 ，引述其观 学科 网（北 京）股 份有限 公司
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点，并结合作者个人的理解，以 举例、引证的方法 加以阐释 ;
②材料二点出莱辛 “诗画异质 ”的核心观点 后，以 札记形式 列举中国古人关于诗画关系的相关讨论，
与莱辛观点形成照应。
【题型 3】论证结构
先做判断： 指出 是什么论证结构 （并列式、层进式、对照式、总分式 ）；
然后 解释每部分是怎样体现这样的结构 的：每部分之间 如何体现层进 ，如何体现对照，如何体现总分
关系。
【2021 新高考 2 卷】 4. 请简要分析文章的 论证结构 。
【答案】
①文章采用 先总后分（总分式） 的论证结构 ，先提出网络行为尤其是青年的网络行为需要规范这一论
点， 然后 从网络行为的 “底线要求 ”和“基准意识 ”两个角度展开论证 ；
②“ 底线要求 ”和“基准意识 ”又构成逻辑上的 递进关系。
【题型 4】论证方法
答题方式与 “论证结构 ”同，即：
先做判断，是 什么论证方法 ：事实论证（举例） ，道理论证（引证） ，对比论证，比喻论证，假设论
证，因果论证，类比论证 ；
然后 说明每种论证方法在材料中是如何体现 的，比如举例论证，举了什么典型事例，证明（论证）了
什么观点。
【常见论证方法】
举例论证：列举确凿、充分、有代表性的事例证明论点 。
引用论证：引用名言等作为论据来分析问题、说明道理 。
对比论证：用正反两方面的论点或论据做对比，在对比中证明论点 。
比喻论证：用人们熟知的事物做比喻来证明论点。（注：异类比方）
因果论证：通过对事理原因或结果的分析，来证明论点的正确合理。
类比论证：通过对已知事物 (或事例 )与跟它有某些相同特点的事物 (或事例 )进行比较类推，从而证明
论点。 （注：同类相比）
【常见论证方法的论证效果】
事实论证（例证法） ：有力地证明中心论点（ “事实胜于雄辩 ”），增强文章说服力、趣味性、权威
性，让文章浅显易懂。 学科 网（北 京）股 份有限 公司
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道理论证（引证法） ：可以增强文章说服力或文采，使论证更有力或更有吸引力。
对比论证（对比法） ：正确错误分明，是非曲直明确，给人印象深刻，使论证更有力或更有吸引力。
比喻论证（喻证法） ：道理讲得通俗易懂，语言生动形象，容易被人接受。
材料二使用了 哪些论证手法 ？请简要说明。（ 4 分）
【答案】
①对比论证，将文学与一般娱乐、好作家与 “文匠 ”、人与电脑或机器人进行对比；
②举例论证，以生活中很多日常现象为例，论证人类具有的特殊能力；
③比喻论证，把人类的心领神会比喻为 “像对付一个趔趄或一个喷嚏那样自然 ”。
【题型 5】论证特点
论证特点 几乎包括了前面所有的内容，如 论证思路、结构特点、论证方法乃至语言特点 等。
作答时要依据材料内容实际，从以上几个方面全面思考，灵活作答，踩准得分点。
另外，表述要力求简练达意，比如 论证语言特点 ：严密，准确，逻辑性强 ；运用修辞 ，生动形象，通
俗易懂等。
【考试中心命制山东海南联考模考题】 4. 材料二在 论证上有哪些特点 ？请简要说明。（ 4 分）
【答案 】①以设问开篇 ，引发关注；
②采用辩驳的 论证结构 ，先立再驳 ；
③论证中综合运用了多种论证方法 ，如 例证法、引证法、对比法 等。
【考试中心命制海南模考题】 4. 材料三在 论证上有哪些特点 ?请简要说明。（ 4 分）
【答案】 ①论证思路清晰 。先指出共享经济存在的问题，再 提出观点 ，然后论证并指出其意义，形成
递进式的论证结构 。②论证手法多样 。在论证时采取了 举例论证、假设论证 等论证方法，如网约车的例子
论证了这些 “共享模式 ”是通过增量服务释放了潜在需求 ;用“倘若共享成色更浓一些 ，比如对顺风车 、拼
车 、合租等优化闲置社会资源的方式 ”进行假设论证等，从而推导出这样做的好处。（每点 2 分）
【论证结构 +论证手法 +结合文本分析手法】
例 4. 材料一在 论证上有何特点 ?请简要说明。 (4 分)
【答案】
(1) 总分式的 论证结构。 文章开篇引出话题，然后从三个方面进行论述，论点明晰，结构清晰严谨。
(2) 论证方法 多样。采用了例证法、引证法等，论证有力。
(3) 论证语言 准确严谨，如 “接近 ”“ 已经 ”“ 还不能算是 ”等词句。 (本题 4 分，每点 2 分，答出任
意两点即可 )学科 网（北 京）股 份有限 公司
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【变式问】 论证的严谨（严密）性
议论文的严谨周密，体现在以下四个角度 ：
（1）结构严谨有序 ，具有逻辑的力量；（ 2）语言准确严密 ，具有精确的力量；
（3）论证方法得当 ，具有说服的力量；（ 4）材料选择确凿 ，具有真实的力量。
【答题模板】
①论证结构严谨，加上文本分析。
②论证语言准确严密，有具体的词语、句子分析。
③围绕中心论点展开论述，透彻地阐明其内涵。
④论点与论据高度统一，加上事例分析。
例 9. 请结合材料内容 ，简要分析材料一第三段 论证的严密性 。（ 4 分）
【答案 ]】①论证结构严谨 。文段采用 总分结构 ，先以设问引出观点，然后按事理逻辑逐层分析；
②论点与论据高度统一 。所举 “夸父追日 ”等事例、所引孔孟等言论都能恰切论证观点；
③用词精炼准确 。运用 “事实上 ”等词语使表达恰切、有分寸。
(答出一点给 1 分，答出两点给 3 分，答出三点给 4 分。 )
其他与 论述文文体特点有关的题型 ，如：
材料二引用了王国维 “隔与不隔 ”的论据，请简要 分析其论证目的 。（ 4 分）
【思路】首先找出文中与试题有关的文字（第 3-4 段），然后 借助试题提示 （论据与论证 ）加以分析
即可。
论证 ：先 “破”后“立”， “破”“ 立”结合。
【答案】 ①引用 王国维 论据 在于以此为论敌并指出其不足； ②进而 论证 作者的 观点 ，即诗歌既必须寓
新颖的情趣于具体的意象，也有 “显”和“隐”的分别。
解答 实用类 文本阅读题指津 学科 网（北 京）股 份有限 公司
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Ⅰ.人物传记类文本阅读
1．常见题型及答题方法
（1）题型： 文章某段在文中的作用是 什么 ？
方法： 从内容上、结构上，表达效果上等多元分析。
示例： 1．文章第四段写到了梁漱溟父亲自杀身亡这段文字在文中有何作用 ?
【答案 】①表明梁漱溟关心国事是有家庭传统的 ;②是梁漱溟格外关注文化问题的原因 ;③使读者对人物了解
更全面。
【解析】 结合本段内容看，主要是谈传主在文化方面的 “本事 ”，因此答案中要有 “文化 ”一词。
【点睛】 这是一篇关于梁漱溟的评价，从写作学上讲，写梁父之死为了更好地表现梁漱溟，所以，我们的
答案大致应为 “更好地表现了梁漱溟 ……” 。
（2）题型 ：某个人物在文本中的特点如何，从哪些方面可以看出？
示例 2．作者在评述 《中国文化要义 》等著作特点的同时 ，也指出了梁漱溟的不足 。他的不足之处具体
表现在哪几个方面 ?
【点睛 】回答这类题 ，一是要注意分析好文章层次 ，看看文章哪里是说不足 ，分几个小层次说的 ;还要注
意抓住中心句和关键句，特别是别人的观点。
（3）题型： 文章这样写给你怎样的启示？联系实际谈谈你的看法 。
示例 3．作为一篇评传性文章，作者是从哪几个方面 “认识 ”梁漱溟的 ?这样写对你的写作有何启示 ?学科 网（北 京）股 份有限 公司
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【答案】 主要是从学问和人格两个方面去认识梁漱溟的。启示有四点。
（1）从文体来说，要评传结合。
（2）选取自己熟悉的材料。
（3）中心内容要体现人物的身份特征
（4）通过细节表现人物
【解析】 本文思路比较清晰，第一段最后一句 “对梁先生的学问与人格也有一些了解 ”可视为总起句，后
面几段均以此为中心展开评述。因此，第一问要点为 “学问和人格 ”。
【点睛】要想做好这类题可以从如下两点考虑：
（1）文章在细节、材料的选用上的特点 ;
（2）文章材料的使用与传主身份特点的关系 。同时题目问的是 “对写文章的启示 ”，这就启发我们还
应考虑文体特点，本文是一篇评传，所以还要想到与一般传记的不同。
Ⅱ．人物访谈类文本阅读
（一）、掌握访谈文体的相关知识。
“访谈 ”的概念 ：因为某个人 、事件 、特定问题去访问专家或知情者 ,请他们对提出的问题进行解答 ，运用
谈话纪实的方式进行报道的文章。这里要注意以下要点：访谈的目的性非常强：就某一特定人、事、问题 ;
访谈的对象有很强的限定：专家或知情者。通常以答记者问、采访纪实、访谈录、对话录等形式出现 。
（二）、访谈类 的题型 、方法 及作答技巧
1． 常见 题型 ：
（1）你如何评价被访谈人物的 什么 特点？
（2）你如何 看待 这个人物的成功或失败？
（3）你如何 评价 他在特殊阶段的努力对成功的帮助等等？
（4）他的这种人生反映了怎样的时代精神，在当前社会下有什么积极意义？
2．阅读方法 ：
阅读访谈关键是迅速把握访谈的话题 、理清访谈的线索 、归纳访谈的主要内容 、分析访谈的技巧 、评
价访谈的收效。
3．作答 技巧 ：依据 高考考纲要点作答 。
（1）“评价文本的主要观点和基本倾向 ”。“文本的主要观点和基本倾向 ”包括访谈者的观点和访谈
对象的观点，必须认真梳理，全面把握其要素和倾向，能找出他们支持自己相关观点的相应材料，分析其
逻辑推理的科学性，然后作出自己的价值判断。 学科 网（北 京）股 份有限 公司
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（2）“评价文本产生的社会价值和影响 ”。“文本产生的社会价值和影响 ”可以是正面的也可以是负面
的，高考考察一般会以正面的为主。访谈的题材通常都是名人、大事，与当前整个社会或某些阶层息息相
关，因此会产生较大的社会价值和影响。阅读时要将视野扩大，将该访谈放到当时社会的大背景中，联系
当时社会的政治经济状况 、主流思想文化 、各阶层的关注程度来进行对照 ，然后实事求是地进行评价阐释 。
（3）“对文本的某种特色作深度的思考和判断。 ”“文本的某种特色 ”主要落实于文本的思想情感、文
本风格、语言特色、设问技巧等方面。 “作深度的思考和判断 ”首先要能指出这种特色，然后联系文章说
一说文章为什么要营造这种特色，还要进一步指出这种特色的好处、效果或不足，如果能提出一些建设性
的意见就更好。
（4）“从不同的角度和层面发掘文本的深层意蕴 ”。“不同的角度 ”可以是正面的角度 、反面的角度 ，当
时人的角度、旁观者的角度等等。 “不同的层面 ”指的是文本可达到的层次，如深层、浅层，实用层面、
美学层面 、哲学层面 ；可能涉及到的范畴 ，如政治 、经济 、科学 、文化 、风俗 、娱乐 、艺术 、行文等 。“发
掘文本的深层意蕴 ”就是要尽量多从上述角度揭示更多的内容。
（5）“探讨文本反映的人生价值和时代精神 。”就是要分析归纳出提问者和访谈对象 （主要是后者 ）的人
生观，探讨他的人生价值，揭示他的这种人生反映了怎样的时代精神，在当前社会下有什么积极意义。
（6）“探究文本中的疑点和难点 ，提出自己的见解 ”。这个考点是考查考生的思维的全面性 、深刻性和质
疑精神。要善于对有疑问的地方提出怀疑、对难点尽量依据文章或借助工具书试图解决，要善于对文章内
容进行辩证分析、提出补充，并在此基础上提出更为科学、合理的观点。要注意的是，高考考查一般访谈
对象应该是该领域的专家，必有其独到之处，对其观点不要轻易完全否定。
Ⅲ．新闻纪实类
（一）概念：新闻属于纪实类作品。 它是借助各种传媒对新近发现的有价值的信息所作出的及时、客观、
准确、简洁的报道，它是报刊、广播、电视、网络等媒体广泛采用的一种文体。它具有真实性、时效性和
受众性的特点。内容一般包括 6 个要素 :时间、地点、人物、以及事件的起因、经过和结果。新闻中所用到
的材料有三类： 新闻事实，背景材料和作者的主观评价。
（二）阅读方法
（1）看标题，抓要素，整体把握新闻内容。
（2）理清脉络，把握文章结构。
（3）把握中心，把握主旨。
抓住关键句 、中心句 、过渡句去把握文本的中心内容 ，研究文本的主旨 ，揣摩作者的写作动机和感情态度 。
（4）研究写作技巧，分析表现手法。 学科 网（北 京）股 份有限 公司
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可以从叙述、描写、抒情、议论等表达方式，对比衬托、以小见大、烘托渲染、借景抒情等文学手段，锤
词炼句及比喻、拟人等修辞方法几个角度去考虑。
（三）常规题型 及答题方法
（1）如何分析新闻的真实性
①从再现采访过程的角度思考； ②从再现现场情形的角度思考； ③从叙述人称选择的角度思考； ④从细节
描写作用的角度思考； ⑤从新闻数据运用的角度思考。
（2）如何把握新闻语言的准确性
①从表现形象特点的角度思考； ②从表现形象变化的角度思考； ③从前后内容联系的角度思考； ④从新闻
传达感情的角度思考。
（3）如何筛选整合新闻的信息要点
①看清题目，明确筛选要求； ②理解层次，把握新闻要点； ③根据题目，筛选相关信息。
（4）如何分析新闻材料的详略性
①从材料角度是否不同的角度思考； ②从材料联系是否层进的角度思考； ③从能否开阔读者视野的角度思
考。
（5）如何分析新闻结构的合理性
①从能否深化主题的角度思考； ②从结构是否完整的角度思考； ③从群众能否想到的角度思考。
（6）如何鉴赏新闻构思的独特性
①抓对比点，分析相反内容的内在联系； ②抓相似点，分析物人之间的相似之处； ③抓相关点，分析事物
之间的内在联系。
（7）如何分析新闻标题的艺术性
①分析新闻标题的表达技巧及其作用； ②分析新闻标题对表达新闻主题的作用； ③分析新闻标题对吸引读
者的作用； ④分析新闻标题对表达记者观点和感情的作用； ⑤分析新闻标题对群众获取信息的作用。
（8）如何评价新闻技巧的巧妙性
①分析表达技巧对表现人物特点的作用； ②分析表达技巧对表现主题的作用； ③分析表达技巧对群众的影
响，如：能否让群众如临新闻现场 （生动形象 ），能否让群众体会出作者态度 ，能否让群众把握新闻主题 ，
能否突现新闻的社会效应等。
（9）如何探究文本反映的人生价值
①概述相关事实； ②分析人物精神（品质、素质）； ③阐发所受启发（相同、相反）。
（10 ）如何评价文本产生的社会功用 学科 网（北 京）股 份有限 公司
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①明观点，即明白无误地表明观点。这是第一个得分点。 ②引论据，即恰当引用论据，遵循 “内引外联 ”
的原则，既要充分利用原文信息，又要合理联系现实状况。 ③作论述，即就所引论据作必要的阐述以证明
自己的观点。 ④作小结，即归结回扣观点，简要作结。
解答非连续性文本阅读指津
非连续性文本 ，主要以新闻类为主 ，重点考查考生的筛选整合文中的信息 ，概况文本内容要点的能力 ，
包括读图。
非连续性文本是相对于以句子和段 落组成的 “连续性文本 ”而言的阅读材 料，以文字、统计图表、图
画等形式呈现 ，这些材料从不同角度呈现事物或主题 ，单独看是完整的 ，合在一起又能够综合地表达意义 。
其特点是直观、简明，概括性强，易于比较，实用性特征明显。学会从非连续文本中获取我们所需要的信
息，得出有意义的结论，是现代公民应具有的阅读能力。
指津一：关注文体特征
非连续性文本通常以图画 、数据表格 、符号 、图解文字等多种方式呈现 ，中间缺少明显的连续性线索 ，
需要经过思考提炼才能找到相关信息或内在联系的文本形式。
指津二：注意阅读方法
第一，寻找所需信息。 发现文本关键信息是核心，善于抓住文中负载关键信息的词句，剔除与阅读目
的无关的多余信息，从字面看出表达的言外之意，以快速锁定要寻找的目标。
第二，整合得出结论。 对于非连续性文本的阅读，在没有详细且明确的陈述言语的情况下，需要将
文本中有关联的信息通过比较 、归纳 、综合进行加工处理 ，判断出题人的真实意图 ，从而整合得出有效
的结论。 学科 网（北 京）股 份有限 公司
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第三，构建文本意义。 对于多种材料组合的较为复杂的非连续性文本，要注意识别文本材料的主题，
联系实际需求，找出文本写作的目的，为材料信息内容排序，联系文本中的不同资料，结合自己的知识、
想法和经验，提出独特见解，自主构建文本的意义。
指津三：注意解题方法
（1）比较材料 。在当今这个大数据时代里 ，阅读新闻 、报告类文本 ，要加强比较分析能力的培养 。要
从材料来源、主要内容、中心观点、数据图表等多角度比较分析，辨明异同，深入探究规律。
（2）筛选并整合文中信息 。一方面浏览全文 ，找到相应语句并在原文中标注出来 。另一方面 ，逐一把
选项和原文比对，不一致的即可判定错误。
（3）分析概括作者在文中的观点态度 。首先读懂文本 ，标记包含相关信息的重要语句 。可以从概括性
的句子、文中运用的材料和作者的评价三方面入手。其次，从作者的写作背景和写作意图出发，结合新闻
材料提炼观点。最后，观点必须于文有据，合情合理。要做到客观公正，不能任意拔高或者贬低。
指津四：培养理解能力
对于实用类文本阅读而言，要更加重视对文本的理解能力。
1．理解重要概念，要注意它在文中的含义 。
因为考点是 “理解文中重要概念的含义 ”强调的是 “在文中 ”，故在理解时，既要注意概念的 “词典
义”，又要注意它的 “语境义 ”。要把 “重要概念 ”还原到文本中，在具体的上下文语境中进行理解，做
到“词不离句、句不离段、段不离篇 ”。要在把握文章内容的基础上，正确理解词语的使用意义，特别是
临时意义。
对于关乎主旨、作者主张的重要概念（词语）的理解，首先要掌握基本要求： ①从词语的本义、引申
义、比喻义、借代义来思考，选准准确的词义，理解临时意义； ②必须结合具体的语言环境理解词语的隐
含意义，同时兼顾词语的本义； ③从词语的词性、语法结构分析、斟酌词义。
2．理解文中重要句子的 含意 ，可以运用以下方法 ：
（1）关键词语理解法
理解句子，从某种意义上说就是对句中关键词语的理解。关键词语，有的是有深刻含义的词语，有的
是指代性的词语，有的是多义词语，有的是概括性词语，有的是有隐含义的词语。抓住这些关键词语后，
再结合支撑该词语的相关词语、句子，把它们分析透彻即可。
（2）句子结构分析法
对那些结构复杂的句子 ，不论是单句还是复句 ，都可以使用 “抽取主干法 ”进行解读 ，抓住句子主干 ，
理解那些修饰、限制等附加成分，进而理解句子含意。如果是复句，可以划分内部层次，分层理解。
（3）根据位置确定句意法
句子在文章或语段中的位置对于语句的理解相当重要。总领句，要结合其领起的范围，结合 “首句标其
目”作答 ；总括句 ，则要总结上文 ，或者 “卒章显其志 ”；过渡句 ，则在结构上承上启下 ，内容上延伸扩展 。
（4）表达意图还原法
对于那些表达上有特色的句子 ，如用了比喻 、拟人 、双关 、反语等修辞手法的句子 ，要用还原法理解 ，学科 网（北 京）股 份有限 公司
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即把作者原来想表达的意思、意图还原出来。如用了比喻句，作者用它真正要表达的意思、意图是什么，
只要写出来即可。另外，要注意多角度答题，起码是正反、你我或者褒贬等两个角度的思考。对于个别表
意较为含蓄而未用特殊表达技巧的句子，也应在写出句子本身含意的同时写出作者的表达意图。
☞文学类文本阅读 技巧
（一） 小说阅读
方法点津
“文字 ·文学 ·文化 ”“ 三层次 ”阅读法
一、文字层次 —圈点勾画，明浅层义
第一个层次是读懂字面意思 ，聚焦词语 、句子所表达的浅层意义 。阅读过程本身就是获取信息的过程 ，
阅读质量的高低取决于获取信息的多少。在阅读的过程中，我们可以通过 “圈点勾画 ”的形式尽可能多地
捕捉关键信息，如圈出小说中的主要人物，便于理清人物关系；勾画出标志情节发展的语句，便于梳理情
节脉络；圈画出文末标示的写作时间，可能暗示社会环境。因此，考生首先要从文字层面读懂文本，通过
精读文本充分获取小说内容信息，并在此基础上理清小说各个要素之间的关系。
二、文学层次 —分析鉴赏，解深层义
第二个层次是读懂深层含义。这个层次的阅读除了关注字面的含义外，还需要考生积极补白，通过句
与句之间的起承转合，来想象、分析或者推论作者的创作意图以及文章的主旨大意，从关注文字表面进入
文学层面的深度解读 。例如 ，通过人物外在的动作 、神态描写能够深入人物内心世界 ，准确揣摩人物心理 ；
通过特定的时间 、地点等环境描写 ，能够准确把握文章的情感基调 ；通过文章的特定物象 ，例如 《百合花 》
中的 “百合花 ”、《哦，香雪》中的 “铅笔盒 ”，能够把握人物形象特点与文章主题内涵。从文学层面解
读文本 ，可引导考生先从分析小说人物 、情节 、环境等基本要素入手 ，从而推断作者创作意图和文本主旨 ，
建构文本意义。
三、文化层次 —— 反思感悟，知引申义
第三个层次是读懂本质精神。这个层次的阅读关注文本中的 “言外之意 ”。文学作品往往蕴含着深刻 学科 网（北 京）股 份有限 公司
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的哲学思想，作者通过故事情节、人物塑造、象征手法等表达自己对人生、人性、社会等方面的思考。例
如《红楼梦》通过对贾府兴衰的描绘，展现了封建社会的种种矛盾，传达了作者对人性的反思、对封建社
会制度的思考。通过阅读这些作品，考生可以提升自己的综合素养，能够对人生、对人性、对社会有更深
刻的认识。因此，考生要从文化层面解读文本，在深入理解小说文本主题意义之后，把从文本中所获取的
知识与感悟和生活实践相结合，建构文化意义，形成积极向上的世界观、人生观和价值观。
思维导图
指津一： 人物形象塑造的考查
误区：分析人物形象时，不从实际出发，过分拔高人物的思想品质；没有立足原文，无中生有， “概
括”出原文中没有的东西；以偏概全，不能全面分析评价人物。
解题思路应分四步走：
首先总体把握小说人物形象特点，确定作者的情感倾向是褒还是贬，是颂扬还是讽刺。然后画出小说
中关于这个人物言行的语句，以及作者的议论或者作者借作品中其他人物对他的评价语句。接着看用了什
么描写方法，在此基础上进行归类概括。最后选择恰当的词句表述出来。
指津二： 重要情节（细节）作用的分析
解题时要注意其思考的方向：
①是对表现主题的作用。其作用一般来说是点题或突出主题。
②是对塑造人物形象方面的作用。或是发展了人物性格，或是表现了人物性格。
③是对整个故事情节的构成上的作用。一般来说是推动了故事情节的发展。
指津三： 分析环境描写的作用 学科 网（北 京）股 份有限 公司
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首先找到环境描写的语句在文章中的位置。处于不同位置的环境描写其作用将会是不同的。然后再概
括所描写的环境。
分析环境描写作用时，可从以下几个方面考虑：
①环境描写交代故事发生的时间、地点、背景。
②添或烘托某种气氛。
③托或突出人物的性格、心理等。
④有时有推动故事情节向前发展的作用。
⑤揭示主旨。
指津四： 思想内容（主题）和写作技巧的鉴赏和评价
理解主题主要看重要情节和主要人物；而小说主要的写作技巧也表现在重要情节的安排和主要人物形
象的塑造上。
把握主题时，不从作品的客观实际出发，就不能避免认识上的偏见和情感、情绪上的偏激，没有认真
阅读原文，拿自己已知的道理去硬套，评价作品缺乏针对性，这是同学们做这类题最大的误区。鉴赏评价
小说思想内容和写作特点时的误区是，过分拔高小说所表现的主题和哲理；偏激地看待创作技法的成败，
认为发表的文章总是好的，不敢提出批评的意见；以自己头脑中固有的观念看待作品和作者，要求作品和
作者符合自己先入为主的思想观念 ；以自己的兴趣爱好作为评价小说的标准 ，合自己口味的则说好 ，否则 ，
就说不好。
[
把握文学类 作品主旨的诀窍 ：
a 从重点语段上找突破；（ 1）抓标题与文体；（ 2）抓 “文眼 ”；（ 3）抓对全篇有概括性的句、段；
（4）抓文中表现作者情感的议论、抒情性的句、段；
b.从写作材料（题材）上看作者用什么来表达感情，以此来把握文章的主旨；（ 1）借景抒情，通过对
所写景物的分析去体味作者的感情 ；（2）寓情于物 （托物言志 ），通过对所写之物的分析去体味作者的感
情；（ 3）借事抒情，通过对所写之事的分析去体味作者的感情；
c. 从写作背景及作者的思想发展上来了解作者的写作意图。 学科 网（北 京）股 份有限 公司
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读文能力： 要有文体意识，学会整体阅读，读通前后逻辑，明晰基本手法
读题能力： 摆脱思维定式，读清题干构成，梳理逻辑层次，创建答题框架
（二）散文阅读
找线索结构清
一篇好的散文往往需要清晰的结构和引人入胜的线索 。散文常见线索有人 、事、物、景、时、空、情、
理，寻找散文线索有如下六大方法：
体裁猜 “线”。写景散文多以游踪（空间变化）或某一景物为线，状物散文多以该物或对该物的情感
为线，写人散文多以人物交往为线，叙事散文多以时间为线，议论抒情散文多以情、理（哲思）为线。 学科 网（北 京）股 份有限 公司
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标题判 “线”。有的标题即线索，如以景、物命名。
时空缀 “线”。文中有一些表示时空转换的词语，阅读时只要把这些词语连缀起来看，就能把握文章
的线索。
以物求 “线”。不少托物言志散文或叙事类散文，常用一个具体事物或象征事物贯穿全文，作为行文
线索以突出主旨，这个事物常作为标题或在文中反复出现。
反复出 “线”。可以通过反复出现的具有丰富内涵的事物或抒情议论的语句去寻找和把握线索。
以情导 “线”。感情线索常常隐伏于记叙的内容之中 ，这就需要阅读时细心分析文章之间的内在联系 ，
理清感情发展变化的轨迹，以此导出文章的线索。
答题策略 学科 网（北 京）股 份有限 公司
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1．问什么答什么，切忌答非所问。语言准确到位，表述规范，简明扼要。
2．分条回答，条理清晰，字迹清楚，便于阅卷，更易得分。
3．主观题的高考阅卷是 “踩点 ”给分 ，因此 ，不怕你多答 ，就怕你不答 。遇到拿不准的问题 ，如不限
定字数可尽可能多陈述自己的感觉可能拿分的点，但不是多多益善，把不相干的都写上，错的多了反而淹
没了对的，阅卷人不易发现又反感。
4．文通字顺也是提高答题得分率的关键 ，许多时候甚至直接作为 “得分点 ”。考生先拟草稿 ，组织好
语言，再正式作答。
5．除非有明确要求 ，否则作答时不能用描写或比喻 、拟人的句子作为答案 。因为这种语言的表意是间
接的甚至是模糊的，不能直接准确地回答问题。
6．阐释 、概括具有一定文学色彩的词 、句的意思 ，要努力探求其本义 ：是比喻的 ，要透过喻体看到本
体；是象征的，要透过象征体看到象征意义，等等。只有探本、返本，才算从根本上理解、掌握了实质性
的东西，同时也可进一步看出文章的行文之妙。
思维导图
指津一： “答案在文中 ”
张建华先生说： “现代文阅读的答案就在原文中，不要凭空去想。 ”离开了原材料恐怕谁也答不准，
答不全。因此，准确解答高考阅读题最重要最有效的方法是在原文中找答案。当然，找出的语句不一定能
够直接使用，还必须根据题目要求进行加工，具体方法：
（1）中心主旨句移用法
命题人常常围绕文章主旨制题，中心主旨句移用法是既省时又准确的好方法。 学科 网（北 京）股 份有限 公司
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（2）关键语句、中心词组合法
高考现代文阅读能力的考查，既要求能够筛选并提取文中的信息，还要求能够归纳内容要点，概括中
心思想。归纳、概括能力是一种较高的能力，而许多文章自身，作者都会对要点和主题作必要的归纳和概
括。要答好这类高考题，如果运用关键语句、中心词组合法，就可事半功倍。
（3）综合分析，条分缕述法
散文阅读不但要考查考生的分析能力，更要考查其综合能力，而且还要在此基础上考查表述能力。因
此，我们在回答高考主观试题的时候，就要在综合分析之后，条分缕析地表述出来。从阅卷的角度看，分
条答题显得条理清晰，更能阅卷人的好感。
指津二： “主题在心中 ”
答题时也要有主题意识，心中时时想着文章的主题，想着能否从主题的角度解答本题。许多难题就会
迎刃而解。
同时，还要有 “文体意识 ”，从文体基本特征与表现手法的角度思考问题，避免答题不得要领。
指津三： “方法在胸中 ”
既然 “现代文阅读题的答案在原文之中，不要凭空去想 ”，我们要善于利用原文词句来组织答案。做
阅读主观题能抄原文答题是上策，事实上，许多高考主观题的评分参考答案，都是直接抄自原文；能概括
原文答题是中策，或摘取词语连缀，或压缩主干调整，或抽取要点组合；离开原文答题是下策，即使有些
题目找不出原词句组成答案，也要弄通语境，得其要旨，不能自由生发，离 “题”万里。 学科 网（北 京）股 份有限 公司
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☞古代诗文阅读 技巧
一、文言文阅读 学科 网（北 京）股 份有限 公司
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解答文言断句题的五个 “锦囊 ”
锦囊一：靠文段大意断句
拿到一道文言断句题 ，首先要讲文段整体阅读一遍 ，联系上下文粗略地揣摩出大意 。对于叙事性文段 ，
要大致清楚其叙的什么事，涉及到哪几个人；对于说理性文段，要大致明白其论点和主要论据；对于对话
性文段 ，应先清楚是哪几人的对话 ，各人的身份和观点如何 ；等等 。应该指出的是 ，这里的 “整体阅读 ”，
强调的是陶渊明式的 “不求甚解 ”的“读”，这对于某些文段来说，也许就相当于已经得出了答案。
锦囊二：以动词宾语断句
和现代汉语一样，在古汉语中，动词（有时是形容词）是构成一句话的重要成分。一般地说，只要找
准了动词（或形容词），弄清楚它们的宾语，就可以宾语之后断句。
锦囊三：凭特殊虚词断句
这里的所谓 “特殊虚词 ”，指的是一些多用于句首或句末的文言虚词和一些表示说话的词语，以及文
段中的诸如地名、人名、官职名等词语，它们往往可以成为给文言断句的重要参考标志。比如，常用于句
首的文言虚词有 “若夫、乃夫、至于、于是、虽然、是故、是以、已而、向使、夫、焉、盖、惟 ”等，见
了这些虚词 ，在它们前边就可断句 ；常用于句末的文言虚词有 “矣、焉、乎、也、耶（邪）、欤、哉、夫、
耳、者也 ”等，在它们后边也大多可断句；相当于 “说”的词语有 “曰、道、云、语、白 ”等，在它们之
后也大多应断句。对于文段中出现的地名、人名、官职名、谦称和敬称等词语，断句时要分析其是作主语
还是宾语。
锦囊四：据句式特点断句
文言中有一些固定句式 ，如果能比较熟悉地掌握 ，对于断句也很有帮助 。比如 ，了解了 “不亦 …… 乎？ ”学科 网（北 京）股 份有限 公司
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“得无 …… 耶？ ”“…… 之谓也。 ”“ 如（奈、若） …… 何？ ”“ 何…… 为？ ”“ 何…… 之有？ ”等固定
句式，可以给我们在给文言断句时以十分明显的指示。此外，还应该知道，古汉语比较讲究句式的对称，
给文言断句时可据此查看文段中有无对称句式。
锦囊五：参考他题断句
有时，我们还可以根据文本中的内容和试题中其他题所提供的信息，作为断句的参考。
总之，如果我们能在复习备考时，按照以上 “锦囊 ”所提到的方法，深入细致地思考，适当地多做练
习，那么，曾被视为畏途的文言断句题，就一定会比较容易地得出正确答案。
文言实词、虚词、 文化知识题解题方法
从近两年的高考题来看，它既考查了词语在语境中的意义 ，又涉及 文化知识 、文言现象 ，成语中的重
点词义理解 等。
一、紧扣教材，抓住重点，总结延伸。
二、强化训练、随题整理，建立资源库。
三、活学活用，善于联想，学会迁移。
四、树立语境意识，前后勾连，明确语境中的特定含义。
翻译文言语句的解题技巧
对文言文句子的理解是翻译文言文语句的第一步，理解文言语句必须遵循古文的一般规律并把握其特
殊性。对句意的理解要注意以下几点：
技巧一：要在全文中理解句子。 无论何种句子，都不能脱离全文或文段去理解，要注意作者的基本观
点和感情倾向。要做到 “字不离词，词不离句，句不离篇 ”。
技巧二：要注意省略句、倒装句、词类活用、通假字、互文见义和偏义复词的理解，对这些现象的理
解把握，往往是翻译文言语句的关键。
技巧三：对句子中难懂的地方，不能采取忽略或笼统翻译的办法去逃避，而应该通过前后推导，或选
项推敲分析等方法，进行认真理解。
技巧四：利用文言排比句、对偶句、 并列结构，把握句意。 也就是说，在理解文言句子时，要充分利
用对应词语的用法和意义去把握文句意思。
技巧五：对复杂的不好理解的句子，最好做成分分析，先抓主干，再看枝叶；或抓住关键词语，分析
句间关系。
技巧六 ：要善于调动已学知识 ，进行比较 ，辨析异同 。特别是对 “一词多义 ”、古代文化知识的积累 ，
有助于我们去把握文言词语在句中的具体用法和含义。
二、古代诗歌阅读
思维导图 学科 网（北 京）股 份有限 公司
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指津一： 鉴赏诗歌形象解题指津
1．鉴赏人物形象解题指津
（1）鉴赏依据 —— 人物（肖像、行动、语言、神态、心理）、环境（自然环境、社会环境）、情节。
（2）鉴赏角度 —— 所写人物身份性格、所写人物的思想感情、作者对所写人物的态度。
例如 ，阅读苏轼的 《念奴娇 ·赤壁怀古 》，通过作者所塑造的人物形象 ，体会作者所表达的思想感情 。
本词所塑的人物形象如下 ：少年英雄周瑜 （客观人物形象 ）—— 雄姿英发 、羽扇纶巾 、谈笑间 、强虏灰飞烟
灭；“我”（抒情主人公 ）—— 多情应笑我 、早生华发 、人生如梦 、一尊还酹江月 。通过周瑜形象和自我形
象的对比刻画，表达了作者怀才不遇、人生易老而壮志难酬所表现的抑郁难平的思想感情。
2．鉴赏意象解题指津
（1）鉴赏的依据 —— 景物形象的自然属性。
（2）鉴赏的角度 —— 景物形象的社会属性（作者借此表达的情感、抒发的寓意）。
例如：王之涣《登鹳雀楼》 “白日依山尽，黄河入海流。欲穷千里目，更上一层楼 ”。诗人在这首二
十字的千古绝唱中，粗线条地描绘了落日、晚山、黄河、大海等意象，显示出登高远眺的特点， “落日 ”
在西， “大海 ”在东，视野开阔。既 有落日近山的奇异风光，又有千里黄河归大海的壮丽景象，甚至还有
滔滔黄河水的声响，不仅可以使人们看到一幅登楼远眺的无限广阔的艺术图画，而且能够有力地激发人们
“更上一层楼 ”的思想愿望，提高人们的精神境界。
指津二： 鉴赏诗歌语言解题指津
（一）炼字、炼句题
1．设问方式 学科 网（北 京）股 份有限 公司
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（1）这一联中最生动传神的是什么字，为什么？
（2）本诗的某字，换成另外一字好不好？为什么？
（3）某句（联）来历为人称道，你认为它好在哪里？
（4）请对某字（句）进行赏析。
解答分析：古人作诗讲究锤炼字句，这种题型是要求品味这些经锤炼的字或句子的妙处。答题时不能
把该字或句子孤立起来谈，需要将字放在句中，或将句放在整首诗中，结合全诗的意境情感来分析。
2．答题步骤：
（1）解释该字在句中的含义。
（2）展开联想把该字放入原句中描述景象。
（3）点出该字烘托了怎样的意境，或表达了怎样的感情。
（二） 赏析诗歌的语言特色
1．设问方式
（1）这首诗在语言上有何特色？
（2）请分析这首诗的语言风格（语言艺术）。
（3）本诗语言的特点主要表现在哪里？请作简要分析。
（4）用有两个词准确点明语言特色。
2．答题步骤：
（1）点明语言特色。
（2）结合诗中有关语句具体分析这种特色。
（3）指出所表现的思想感情。
指津三：鉴赏诗歌表达技巧解题指津
1．提问方式：
这首诗歌采用了何种表现手法？（提问变体：这首诗歌运用了怎样的艺术手法（技巧）？诗人是怎样
来抒发自己的情感的？）
正确解答：这类提问，着重点是诗歌整体的艺术表现特色，主要应该从诗歌的整体构思、诗歌整体的
艺术技巧方面来解答。
常见错误：解答这类提问时，很多学生常犯的一个错误是对诗歌某个局部的修辞手法进行阐述。但阐
述具体某句诗时，是可以谈及修辞手法的运用的。
2．答题步骤：
简单地说就是 ：明手法 （用一两个词准确指出用了何种手法 ）+阐运用 （结合诗句阐释说明诗人运用了
这种手法的依据） +析效果（此手法表达了诗人怎样的感情，或刻画了什么形象，或表现了什么主旨）
具体步骤是：
【第一步】准确解读文本：在了解 “诗家语 ”多省略、多倒装特点的基础上，抓关键点。
（1）上看 ：看诗歌题目 ，圈出题眼 （某一词语 ），认真研究古诗的题目 ，有的题目实际上就概括了诗 学科 网（北 京）股 份有限 公司
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的主要内容，或者给你理解该诗提供了感情基调。再看作者，回忆作者所处的朝代和作品风格。注意时代
对作家的影响（如南宋的爱国思想）。
（2）下看 ：看注解提示 ，了解诗歌的背景 ，寻找诗歌内容和情感的线索 。（即我们平时讲的五读 ：读
题目、读作者、读内容、读注释、读命题）
【第二步】明了答案构成要点（即给分点）。
（1）采用的写作手法。
（2）手法揭示的内容 ：结合诗句 ，分析该手法写出了意象 （人，物，景）的什么特点 ，或抒发 （突出
了）什么思想感情（哲理）。
指津四：鉴赏诗歌的思想感情和情感态度题目解题指津
1．读题目
题目是诗歌的有机组成部分，有时我们可以从题目中得到写作的时间、地点以及相关内容。而这些有
时又是理解诗歌内容、思想感情必不可少的提示。
2．关注作者
（1）关注作者，可以从他的 风格特点 上获得提示，对鉴赏诗歌的语言和理解思想感情有一定的帮助 。
如王维 ，以创作山水田园诗为主， 语言清新自然，诗中有画 ，内容上主要写 山水隐逸生活 ，特别是清
静优美的景物， 渗透着佛家思想 ，艺术成就更高。
（2）诗人的生平 、思想和性格等等 ，无不深深打上当时社会风气的烙印 ；具体了解写作某一首诗时的
生活状况、思想情绪和创作意图等对把握情感有重要的帮助。
如：李白、杜甫、李清照等诗人词人的生平、人生经历，有 利于帮助我们了解时代背景以及诗人的思
想内容。
3．看注释
注释是鉴赏中最值得注意的内容。注释可以揭示背景，理解诗歌，了解典故内容等，这些都有助于把
握情感。
4．抓关键句
从五七言律诗的一般规律来看 ，艺术中心在中间两联 ，思想中心在首尾两联 。（注重绝句的三 、四句 ，
律诗的首尾联）
5．把握形象
注意意象的特征和寓意，注意形象的外在特征、内质品性，以及环境特点等，有助于情感的把握，特
别是写景抒情诗和咏物诗。另外柳、月、长亭、水、白云等常见意象的含义暗示了情感的流露。
6．抓诗眼
诗眼是诗中最精练传神、能巧妙表达主旨的词语，即能直接表达情感。
三、名句名篇默写
默写功夫在平时 学科 网（北 京）股 份有限 公司
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1．以大纲要求背诵的内容为主，必须逐一过关，不可放过。
2．印象不深的名句，要反复记忆。
3．一定要动手写写其中的易错字，坚决消灭错别字。
4．默写题要看准要求，必须用钢笔写正楷字，此题万不可连笔。如果是 “X 题中选填 Y 题”的选择模
式，一定要选准你最有把握的 Y 题。没有把握的千万不要填，这样既为了节省时间，还因为高考阅读卷的
时候按前 Y 题批阅，后面的即使正确也不算。
意象类开放式理解性默写仍然值得注意。
例如 ——
自古以来，秋天一直是文人墨客常常吟味不已的季节。借助秋景表达内心悲伤的诗句不胜枚举，如
“____ ，____ ”就写出了秋天的凄清。
（示例一）无边落木萧萧下 不尽长江滚滚来
（示例二）浔阳江头夜送客 枫叶荻花秋瑟瑟
（示例三）月落乌啼霜满天 江枫渔火对愁眠
峨眉山位于四川，是中国四大佛教名山之一，李白曾移居四川，很熟悉峨眉山，他的诗文中多有 “峨
眉”的身影，如 “ , ”等，不胜枚举。
示例一：西当太白有鸟道 可以横绝峨眉巅（李白《蜀道难》）
示例二：蜀国多仙山 峨眉邈难匹（李白《登峨眉山》）
示例三：峨眉山月半轮秋 影入平羌江水流（李白《峨眉山月歌》）
相比现代人 ，古人对生命易衰更为敏感 ，常从鬓发的细微变化生出深沉的感喟 ，这样的情感抒写在诗
词中比比皆是，如 “_ ，_ ”。
艰难苦恨繁霜鬓 潦倒新停浊酒杯
古代迁客骚人流连于山水之间 ，借以吟咏情性 ，舒展身心 ，发散出一种高度的生命意识 、民族意识和
责任感。这些借景抒情的诗篇常常借用 “松”这个意象，如 “___ ，______ ”。
连峰去天不盈尺 枯松倒挂倚绝壁（景翳翳以将入 /抚孤松而盘桓）
“鹿”音同 “禄”，世人喜好将其作为 “长寿安康 、仕途美好 ”的象征 ；“鹿”幽居山林 ，生活习性恬淡安静 ，
文人也常借其表达隐逸情怀，如 “_ ，___ ”。
呦呦鹿鸣 食野之苹
羌笛是出自古代西部的一种乐器，其音凄切，在诗歌中具有悲凉的象征意蕴，如 “_ ，___ ”。
中军置酒饮归客 胡琴琵琶与羌笛 /羌笛何须怨杨柳 春风不度玉门关
时空对举是指诗人从时间和空间两个角度描写景物来营造意境，让读者在时空交错中获得审美体验 。
“诗圣 ”杜甫即擅长在近体诗中运用时空对举的创作方法，如： “________ ，
________ 。”
【示例一】万里悲秋常作客 百年多病独登台
【示例二】窗含西岭千秋雪 门泊东吴万里船 学科 网（北 京）股 份有限 公司
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【示例三】锦江春色来天地 玉垒浮云变古今
【示例四】五更鼓角声悲壮 三峡星河影动摇
【示例五】乾坤万里眼 时序百年心
【示例六】竹深留客处 荷净纳凉时
樽”指酒杯，也写成 “尊”，在古代诗文中多指代饮酒，并以此表达作者饮酒时的心情，比如 “ ”或
“ ”。
金樽清酒斗十千 一尊还酹江月 举匏樽以相属
小明想从所学的诗文中选取两句话集成一副对联挂在书房 ，提醒自己学习 、做事贵在坚持 ，可选择诗
句“___ ，___ ”。
锲而不舍，金石可镂；千淘万漉虽辛苦，吹尽狂沙始到金；宝剑锋从磨砺出，梅花香自苦寒来 ……
“沧海 ”经常出现在唐宋诗文中，和其他意象组合成一个意境，或展现宽阔胸襟，或寄托忧思情怀，
或抒发离情别绪等，如 “__ ，____ ”。
沧海月明珠有泪 蓝田日暖玉生烟 （或“寄蜉蝣于天地 ，渺沧海之一粟 ”“ 曾经沧海难为水 ，除却巫
山不云 ”“ 东临碣石，以观沧海 ”）
龙是中国古代传说中的神异动物，是我国汉族的民族图腾，也是中国文化的突出符号。我国龙文化
源远流长， “龙”在古诗文中也是俯拾即是，如 “__ ，__ ”。
还似旧时游上苑 车如流水马如龙（但使龙城飞将在 不教胡马度阴山 /水不在深 有龙则灵）
在表现怀人主题的古代诗歌中，诗人经常采用 “对写 ”手法，从对方着笔，婉曲含蓄地表达情感，可
谓“此处思念，彼处着墨 ”，如 “_ ，_ ”。
示例一：还顾望旧乡 长路漫浩浩
示例二： ⑤遥知兄弟登高处 ⑥遍插茱萸少一人（王维《九月九日忆山东兄弟》）
示例三： ⑤今夜鄜州月 ⑥闺中只独看（杜甫《月夜》）
示例四： ⑤乡泪客中尽 ⑥归帆天际看（孟浩然《早寒有怀》）
示例五： ⑤想得家中夜深坐 ⑥还应说着远行人（白居易《邯郸冬至夜思家》）
“铁衣 ”是用铁甲编成的战衣，也借指战士。在古诗词中，透过 “铁衣 ”，我们仿佛可以看到边塞将士
不畏严寒、刻苦训练、奋勇杀敌的身影，如 “__ ，__ ”。
朔气传金柝 寒光照铁衣（《乐府诗集 ·木兰诗》）
或：将军角弓不得控 都护铁衣冷难着（唐 ·岑参《白雪歌送武判官归京》）
或：红颜岁岁老金微 砂碛年年卧铁衣（唐 ·王烈《塞上曲二首》）
或：铁衣远戍辛勤久 玉箸应啼别离后（唐 ·高适《燕歌行》）
“角”是古代军中的一种乐器，军营常用吹角来发号施令。 “角声 ”作为一个声音意象，常在唐诗中出
现，如 “ , ”就有 “角声 ”这一意象。
角声满天秋色里 塞上燕脂凝夜紫 学科 网（北 京）股 份有限 公司
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“杜鹃 ”，又叫子规，相传为周朝末年蜀王杜宇死后所化。 “杜鹃 ”的意象频繁在古诗词中出现，如
“ ， ”。
庄生晓梦迷蝴蝶，望帝春心托杜鹃 杜鹃啼血猿哀鸣
自然界中的一些常见事物常作为具有特定意蕴的意象存在于古诗中， “月”便是如此。诗人常借 “月”
来丰富深化某种特定的情绪，使其作品具有 “言外之意 ”的美学旨趣，如 “ ， ”。（本
小题请在高考规定篇目中选取作答）
由题干中 “月”“ 丰富深化某种特定的情绪 ”可知 ,所填句子应包含 “月”这一意象且表达出特定的情绪 ,据
此可根据所学篇目和日常积累 ,联想到苏轼《念奴娇 •赤壁怀古》中 “人生如梦，一尊还酹江月 ”或李煜《虞
美人 •春花秋月何时了 》 “小楼昨夜又东风 ,故国不堪回首月明中 ”或李白 《闻王昌龄左迁龙标遥有此寄 》“我
寄愁心与明月，随君直到夜郎西 ”。其余答案只要符合要求皆可。
古诗词中有很多以 “自”“ 空”营造意境氛围的，或写独守的孤寂，或写空寂苍凉的环境，或写英雄的
失意，如 “__ ，____ ”。
空山新雨后 天气晚来秋（杜甫《蜀相》 “映阶碧草自春色，隔叶黄鹂空好音 ”；白居易《琵琶行 》
“去来江口守空船，绕船月明江水寒 ”；陆游《书愤》 “塞上长城空自许，镜中衰鬓已先斑 ”；姜夔《扬州慢 》
“清角吹寒，都在空城 ”；李白《蜀道难》 “又闻子规啼夜月，愁空山 ”）
“鸟”是古诗词中的常见意象 ，古人常借鸟以抒发自己的情感 ，如王籍 《入若耶溪 》中的 “蝉噪林逾静 ，
鸟鸣山更幽 ”。其余的还有很多，如 “___ ，_____ ”。
杨花落尽子规啼 ，闻道龙标过五溪 /千山鸟飞绝 ，万径人踪灭 /感时花溅泪 ，恨别鸟惊心 /春眠不觉晓 ，
处处闻啼鸟 /荡胸生层云 ，决眦入归鸟 /山气日夕佳 ，飞鸟相与还 /蓬山此去无多路 ，青鸟殷勤为探看 /但见悲
鸟号古木，雄飞雌从绕林间 /鸷鸟之不群兮，自前世而固然
古代衣服因为原料质地，制成后较为硬挺，穿着前需置于石上舂捣，使之柔软，称为 “捣衣 ”。古典
诗词中常用 “捣衣 ”或“砧声 ”来表现妇人思念征夫 、游子思家怀乡 、征人对常年征战不满等情绪 ，比如唐人的
诗句 “ ， ”。
示例一：玉户帘中卷不去，捣衣砧上拂还来
示例二：长安一片月，万户捣衣声
示例三：寒衣处处催刀尺，白帝城高急暮砧
古代常以 “汗青 ”指书籍史册，在古诗中 “汗青 ”一词多次出现，如 “ ，______ ”。
人生自古谁无死，留取丹心照汗青 /芳名垂汗青 ,千载永不灭
☞语言文字运用 技巧
成语题解答有妙招
预测与备考建议 学科 网（北 京）股 份有限 公司
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1．从选文 来看 ，要加强对情景式的读题能力训练。
2．从考查能力来看 ，要最大限度地识记成语 ，成语作为语言的精华之一 ，是考查语文素养的方法之一 。
3．要养成良好的阅读习惯 ，语言运用能力与平时阅读密不可分 ，只有不断积累 ，不断运用 ，才会达到
熟练使用成语的能力 。学科 网（北 京）股 份有限 公司
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病句题解题四大技法
技法一：压缩法
一些结构复杂的长句，采用压缩法，删去枝节，留下主干，就会很快发现其中的问题。
例如： 他做事认真，待人诚挚，在生活和工作中，确实用自己的行动塑造了巨大的人格力量，感动和
引导着周围的人们。
分析：这个句子 “塑造了巨大的人格力量 ”中“塑造人格力量 ”动宾搭配不当，应作修改。
技法二：化解法
对一些语句内在关系复杂的句子，可采用分解成小步聚的形式加以化解。
例如： 加快西部地区发展的步伐，除了要尽力争取国外投资外，努力发展高新科技产业、节约用水，
也是能否发展西部经济的一条重要的路子。
分析：这个句子，我们可把 “高新科技产业、节约用水 ”拆开来，再分别搭配，即为 “努力发展高新
科技产业 ”和“努力发展节约用水 ”，就比较容易看出后半句的搭配问题。
技法三：替换法
对于一些吃不准的结构或句子，我们可根据其结构，用自己熟悉的语词替换陌生的语词，从而很快作
出判断。例如：教育部实施的学历证书电子注册即将推行，这将会给假文凭制造者以致命的一击。如果对
句中 “电子注册即将推行 ”的是否搭配吃不准，可以仿造一些句子，如， “课间做操即将推行 ”，“午间
自修即将推行 ”，等等，这样一对比，我们马上会得出正确的结论。
技法四：加强敏感
要对语病题中常出现的一些带有规律性的现象加强敏感 ，建立起类似于生物学中 “条件反射 ”般的
反应。
1．敏感点一：顿号或并列词句
反射之一：并列不当，指的是有从属、交叉关系的概念不能并列。
实例： 我们的报刊、杂志、电视和一切出版物，更有责任作出表率，杜绝用字不规范的现象，增强使
用语言文字的规范意识。
分析 ：“报刊 、杂志 、电视和一切出版物 ”不能并列在一起 ，因为 “一切出版物 ”显然包括了 “报刊 、
杂志 ”，可将 “一切出版物 ”改为 “一切新闻机构 ”。
反射之二：照应不当。
实例： 电子工业能否迅速发展，并广泛渗透到各行各业中去，关键在于要加速训练井造就一批专门技
术人才。
分析：句中的 “加速训练并造就一批专门技术人才 ”，是 “电子工业能迅速发展 ”的“关键 ”，如果
“不迅速发展 ”，也就无所谓这个 “关键 ”了。这犯了 “两面对一 面”的错误，前后缺少照应。
反射之三：搭配不当。
实例： 近年来，我国加快了高等教育事业发展的速度和规模，高校将进一步扩大招生，并重点建设一 学科 网（北 京）股 份有限 公司
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批高水平的大学和学科。
分析：说 “加快速度 ”可以，但不能说 “加快规模 ”，可说 “扩大规模 ”。
反射之四：逻辑不当。
实例： 许多海洋生物的药用价值正在被推广和发现，前途不可估量。
分析： “发现和推广 ”的关系就是先发现，后推广，这里前后应互换。
反射之五：有歧义。
实例： 近日新区法院审结了这起案件，违约经营的小张被判令赔偿原告好路缘商贸公司经济损失和诉
讼费三千多余元。
分析：这里是诉讼费三千余元，还是经济损失和诉讼费总共三千余元，表达不清。
2．敏感点二：否定词或反问句
反射之一：多重否定不当。
实例： 雷锋精神当然要赋予它新的内涵，但谁又能否认现在就不需要学习雷锋了呢？
分析：这里有双重否定，还有反问句式，表达的意思是 “每个人都认为现在就不需要学习雷锋了 ”，
这就犯了否定不当的毛病。
反射之二：有歧义。
实例： 这一桩发生在普通家庭中的杀人悲剧在亲戚当中也有着不解和议论，要说小莉的妈妈不爱她家
里人谁也不相信。
分析：句子停顿不同，可产生两种理解：一是 “要说小莉的妈妈不爱她，家里人谁也不相信 ”，二是
“要说小莉的妈妈不爱她家里人，谁也不相信 ”。
3．敏感点之三：数量词
反射之一：有歧义。
实例： 县里的通知说，让赵乡长本月 15 日前去汇报。
分析：这里有两种理解，一是 “15 日”去汇报，二是 “15 日以前 ”去汇报。
反射之二：数字用法有错。
实例： 小明的年龄刚好比他的哥哥小一倍。
分析 ：使用数字作倍数时 ，只能说张三比李四大几倍 ，而不能说张三比李四小几倍 ，故本句应改为 “小
明的年龄刚好是他哥哥的一半 ”。
4．敏感点之六：介词
反射之一：介词搭配不当。
实例： 随着通讯日渐发达，手机几乎成为大家不可缺少的必需品，但使用量增加之后，关于手机质量
的投诉也越来越多。
分析：这一句的后半句 “关于 ”用得不当，可改为 “对于手机质量的投诉 ……” 或“手机质量方面的
投诉 ……” 。
反射之二：缺少主语。 学科 网（北 京）股 份有限 公司
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实例： 观摩了这次关于农村经营承包合同法的庭审以后，对我们这些 “村官 ”的法律水平有了很大的提高。
分析：虚词 “对”的误用，淹没了主语，应去掉。
5．敏感点之五：关联词
反射：关联词误用。
实例： 我不但信任他，而且以前反对过他的人，现在也信任他了。
分析：句中前句主语是 “我”，后句主语是 “反对过他的人 ”，主语不一致，因此，关联词 “不但 ”
应放在 “我”之前。
6．敏感点之六：两面词
反射：前后照应不当。
实例： 文艺作品语言的好坏，不在于它用了一大堆华丽的词，用了某一行业的术语，而在于它的词语
用得是地方。
分析： “文艺作品语言的好坏 ”有两面， “而在于它的词语用得是地方 ”等只涉及一面，可在 “不在
于它 ”与“而在于它的词语用得 ”后面分别加上 “是否 ”与“是不是 ”。
表达题解题技法分类
综观近年高考语文试题，语言文字运用题呈现出稳中有变的特点。
“稳”表现在部分考点和题型的相对稳定上，如成语、病句和连贯这三个考点基本上年年都会考查，
虽然考查形式会有变化 ，如成语题由四选一变为近义成语辨析 、六选三 ，连贯题由句群关系变为句间关系 ，
三道选择题由单独考查变为综合考查，但考点总体保持稳定。
“变”主要表现在主观题上，逻辑推理、应用文改错、仿写语句、压缩语段等新考点 (也有多年以前考
过的旧考点 )轮番上场，让人真正感受到语言文字运用主观题在考场上的灵活性。这也反映出了高考语文试
题对课程标准的响应 —— 不是纠结于考点与题型，而是将命题重心放在考查语文核心素养上。这主要体现
在以下几个方面：
一是联系社会生活，突出实际应用。试题的灵活性与应用性继续加强，对能力的考查将更充分。
二是加大对思维的考查力度。语言文字运用的图文转换题通过描述图形内容，考查考生的思维建模能
力。
三是加强对中华文化的考查力度。重视汉语文化，加强对书面语准确表述、谦辞敬辞规范用法、文段
语体风格一致的考查，要求考生必须具备基本的理解语言文字文化内涵的能力。
总之，未来高考语文语言文字运用主观题的变化具有必然性与不确定性，为了应对这种不确定性，我
们将语言文字运用主观题可能出现的新变化分为六类进行讲解，让考生有较为全面的了解。
指津一：简明
1．去次留主法 —— 围绕中心，抓住要点
要做到语言简明 ，首先是每一句话都要围绕既定中心 ，不要节外生枝 。不过仅仅围绕中心还是不够的 ，学科 网（北 京）股 份有限 公司
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还应该抓住要点。俗话说 “简明扼要 ”，从表达上说，只有扼住 “要”，才能做到简明。
2．善于概括，巧用指代
无论是书面表达还是口头表达，都不能总是具体 叙述而不作必要的概括。只有把必要的叙述和概括结
合起来，表达才能简明。再者，运用必要的复指成分，也是表达中不可少的。不用复指成分，就会啰嗦。
3．删除繁冗法 —— 避免重复，删除多余
鲁迅在谈到自己的写作经验时曾说过：尽量删除可有可无的文字。这是确保表达简明的又一方法。
方法 ：（1）找主干 ，理枝叶 。这种方法就是通过句子成分分析 ，理清句子结构 ，以便发现和删除重复
的词语。（ 2）分析句间关系。这种方法可以有助于发现多余的短语和句子。
4．辨识歧义法 —— 防止误解，避免歧义
表意不明 ，令人误解或费解 ，是常见的语病 ，也是与 “简明 ”的要求相违背的 。要本着 “必须保留 ”
原则 ，可有可无的就无 。答题时可从句中有多义词 、指代不明 、重音不明 、切分不明 、关系不清等角度
辨识。
指津二： 连贯
1．排列句序
（1）抓中心
一个句群，虽然由若干句子组成，却表述一个中心。句序的安排必然围绕这一中心问题。因此抓住了
句群的中心，就抓住了要害，对句序的认识就会由暗到明。
分析句子的性质和作用（如总领句、总结句、过渡句、解说句、观点句、材料句等），是抓准中心的
重要手段，一个句群的中心，大多用一个关键句表达。这一关键句往往放在句首，也有放在句尾的。
（2）抓思路
从总体上看，句群小层次一般呈现出相并（并列、对照）、相承（顺接、层进）、相属（总分关系）
的关系。从局部看，句与句之间往往呈现出并列、承接、解说、对比、递进、转折、因果、总分等逻辑关
系。理顺句序，要尽可能多地确定出必然相连接的句子，找到 “句链 ”。
从文体来看，记叙文的句序常常以时间、空间为顺序，议论文的句序，常常把观点放在前面，把材料
句放在中间，把总结句放在后面，结构形式或总分、或并列、或对照、或层进；说明文同议论文一样，往
往把事理句放在前面把材料句放在后面，因为材料是用来说明事理的，材料的内部又遵循一定的顺序（时
间、空间、逻辑）。
（3）抓标志
语言标志常常表现为：
①关联词语的呼应。或并列、或转折、或条件、或假设、或递进、或因果 ……
②暗示性词语的使用。 “换句话说 ”表示等同关系，位在后； “同时 ”表示并列，位在后； “与此同
时”“与此相反 ”“反过来说 ”表示相反 、相对关系 ，中间不可插入别的词语 ；“首先 ”“其次 ”“再次 ”
表示主次轻重的顺序，不可倒置； “先前 ”与“后来 ”，“过去 ”“ 现在 ”与“将来 ”均表示时间先后；
“总之 ”“ 综上所述 ”“ 由此看来 ”表示要提出结论； “诸如此类 ”表示综合； “所谓 ”表示有所解释； 学科 网（北 京）股 份有限 公司
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“例如 ”表示举例 ……
③关键词语的重复出现，相同的句式重复出现。
④句子之间的对应关系（内容上、形式上），也往往体现语言顺序的一致性，肯定、否定的一致性。
⑤陈述对象前后一致。议论角度一致。
2．填充复位
（1）话题要统一
所谓话题要统一，是指组成段落的句子之间，或是组成复句的分句之间，有紧密联系，围绕着一个中
心，集中地表现一个事实、场景或思想观点，无关的话不掺杂在里面。
①陈述对象一致。
主句是句子的发端和陈述对象，它要求后面连带的一些句子必须兼顾上下文，与陈述对象保持一致，
防止出现暗换主语的现象。
②观点材料要统一
有的议论语段开头提出问题或论点，接着举事例，摆材料，进行分析。这里的 “观点 ”“ 问题 ”也可
以说是话题，话题必须统领后面的事例、材料；反过来说，事例、材料、语句表达要符合前面的话题，与
其相一致。
（2）表达要合乎事理、语境
①表达要合乎事理。 意思表达要符合客观事理，如果上下句在整理上出现 “裂痕 ”，就衔接不上。
②表达要合乎语境。 对于写景的复句或语段，要注意语境因素，要分析景物、情调、写法的特点。景
物，有远 、近、动、静的不同 ；色彩 ，有鲜明 、暗淡的区分 ；气氛 ，有热烈 、凄清之分 ；视角 ，有高 、低、
俯、仰之异；感情或悲或喜；态度或褒或贬。这一切，在同一文字里都应该保持和谐一致。
指津三： 得体
1．注意场合
特定的场合往往需要特定的话语形式来传递信息。俗话说 “到什么山上唱什么歌 ”。假如某盘山公路
边竖立的警示牌上面写着 “由于前面很陡，开车请减速！ ”，那么它绝对是一块催命牌。因为太长，司机
还没看完，死神已降临。其实有 “危险！慢！ ”几个字，司机就会立即减速。
2．注意谦敬
谦敬辞可归纳为 “家大舍小令外人 ”一句话 ，即：对别人称 比自己年龄 （或辈分 ）大的家人时冠以 “家”，
如家父（家严）、家母（家慈）、家叔、家兄等；对别人称比自己小的家人时则冠以 “舍”，如舍弟、舍
妹、舍侄等；称别人家中的人，则冠以 “令”，如令堂、令尊、令郎、令爱等。除了 “家”“ 舍”这两个
谦辞和 “令”这一敬辞外 ，“小”（如“小女 ”）、“拙”（如“拙见 、拙荆 ”）、“鄙”（如“鄙见 ”）、
“寒”（如“寒舍 ”）、“愚”（如“愚见 ”）等都指自己的 ，属于谦辞 ；“贵”（如“贵庚几何？ ”）、
“大”（如 “大作已拜读 ”）、 “高”（如 “愿闻高见 ”）、 “贤”（如 “贤弟 ”）“尊”（如 “尊姓大
名”）等都指对方的，属于敬辞。
3．注意对象 学科 网（北 京）股 份有限 公司
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即看对方的身份、年龄、性格、经历、心理状态、文化素养等而进行语言交流。向尊者、长者请示、
请教要用敬称 、敬辞 。一位年过半百的人丧妻后很悲伤 ，有人劝说 ：天涯何处无芳草？这是安慰还是调侃？
其实完全可以这样说：节哀吧，好好保重自己，才对得住您的妻子。
4．注意语体
同是书面表达，公文类要求庄重、严谨、准确、简明，科技类要求有术语性、客观性、逻辑性、符号性，
文艺类要求形象性和人物语言个性化，政论类要求具有逻辑性、鼓动性、综合性；口语要求通俗、明白。
指津四： “准确、鲜明、生动 ”
（一）语言表达准确
准确用词要注意几个 “符合 ”：
1．符合情境
选择词语要注意情境的制约，写作中使用的词语都处在全篇或上下文的具体语境中，只有根据特定语
境选择恰当的词语才能准确地表达意思，同样的词语用于不同的语境效果迥然不同。
2．符合语法、逻辑
组句应依照一定语言的语法、逻辑规则进行，否则会出现语病。应注意以下几点：句子结构要完整，
词语搭配要稳妥，正确使用关联词语。句义要有逻辑性，有些句子语法上没有问题，但概念不清，语意不
明，判断和推理不当，也是不通顺的。
3．符合对象
应仔细辨析同义词的基本意义和附加意义（包括风格、色彩和用法等）的同中之异，这有助于恰如其
分地叙事写景、表情达意、释物明理。
（二）语言表达鲜明
语言表达鲜明，应注意以下几点：
1．恰当选用词语
在选用动词、形容词、副词时，不要使用诸如 “可能 ”“ 大概 ”“ 也许 ”“ 左右 ”等不确定的词来表
明态度与观点。多使用 “坚决反对 ”“ 完全错误 ”“ 绝不能这样 ”等词语来表明自己所持的鲜明的态度。
选用富有感情色彩的词语时，要关注整个语境，根据在表达时的不同态度与感情，选择词义的褒贬。
感情色彩鲜明的褒义词贬义词可以增强语言表达的效果。选择那些感情色彩不鲜明的中性词时，只要结合
好语境，同样也可以找到具有强烈效果、表达鲜明的词语。
2．恰当运用修辞手法
比喻 、对偶 、对比等修辞能增强语言表达的鲜明性 ，独具特色 ，借助修辞格来增强语言表达的鲜明性 ，
是一条切实可行的途径。
（三）语言表达生动
生动的语言不但具体形象，活泼多变，有声有色，而且感情充沛，散发着生命的活力。具体来说，要
想语言生动，须做到以下几点：
1．要使用描绘性的词语和具体形象的写法 学科 网（北 京）股 份有限 公司
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描绘性词语绘声绘色，富有动感，用在合适的地方便显得生动活泼。具体形象的表达能给人一种身临
其境的感觉，使抽象的东西鲜活起来。
2．多用贴切的比喻和拟人等修辞格
要特别重视比喻的使用 ，因为它的主要功用就是把抽象的对象具体化 、形象化 ；而拟人的功用是使无 生
命的对象仿佛充满了生机和感情，而使它生动起来。
补写类解题技法
对应考点：语言的简明、连贯、得体、准确、鲜明、生动；压缩、概括能力。
命题规律
(1 ）“根据材料内容 ”要求补写句子应联系前后文语境
（2）空出的句子大多有特殊位置和性质，
（3）要求所补写的句子内容贴切、语意连贯、逻辑严密
（4）不能照抄材料，有字数限制。
（5）设题形式：以主观题为主，三空， 赋分五为 5 分或者 6 分。
“补写句子 ”是综合考点和能力的考查，这类题目一般要求 “根据材料内容 ”补写句子，要求所补写
的句子内容贴切、语意连贯、逻辑严密，并且不能照抄材料，有时另有字数限制。
所补写的句子的内容来源：文本。具体说，所补写的句子的内容、语言要从上下文的有关材料中去提
炼和概括，离开文本不可能补写正确。
所补写句子与上下文关系：或引领下文，或总结上文，或与上下文衔接连贯。
有效的答题技巧：
（1）阅读全文，了解文段性质和内容，确定中心。
（2）分清层次，判断句间上下文的逻辑关系。
（3）重点勾画：找到呼应，关联，暗示性语句（重视标点符号尤其是冒号、分号、问号）。
（4）结合文本 ，根据字数要求 ，概括答案 。写出的句子 ，文从字顺 ，不超字数 ，语义贯通 ，逻辑严密 。
突破中间嵌入式补写语句题
依据三个一致，结合语境嵌入
(1) 话题保持一致。叙述一件事，说明一个道理，要保持话题的一致性。话题往往是句子的主语，它要
求后面的句子必须兼顾上下文，与陈述对象一致，防止暗换主语。
(2) 行文照应一致。既要注意行文结构的前后照应，也要注意语意表达的前后勾连。
(3) 事理逻辑一致。或以认识事物的规律为序，由表及里、由浅入深、由感性认识到理性认识 ;或以空间
为序，从上到下、从左到右、从外到内等。
图文（表）转换解题技法
指津一： 图表题解题技法 学科 网（北 京）股 份有限 公司
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1．找主体 ，抓特征 。介绍漫画内容时 ，要抓住能反映画面寓意的特征进行详细说明 ，否则便不能清楚
地揭示漫画的内涵。
2．客观描述 。不可超越漫画所给图文信息进行添枝加叶 ，用主观想象代替画面中并不存在的东西 。注
意说明顺序，在整体上可按 “总─分─总”，即起笔一句点明介绍对象，然后依次介绍画面内容（先背景后
人物 ）；介绍人物时 ，可按照 “穿着 （从上到下 ）─动作 ─神态 ”这样的顺序进行说明 。还可用时间 、空间 、
逻辑顺序来说明。
3．勿略标题。 一定要写出来，内容要囊括所有画面，不要丢三落四。
指津二：漫画题解题技法
1．概括主题的方法
概括画面的主题 ，先要认真细致地观察分析画面的内容 ，找出其讽刺或颂扬的对象或行为 （有标题的 ，
一般多是标题），然后挖掘隐含信息，进一步提炼概括画面所揭示的主题。通常要采取联想的方法，由物
及人，彰显意义。
2．拟加标题的技巧
标题是漫画的眉目，它有暗示漫画寓意或揭示讽刺对象的作用。拟定标题先要弄清漫画讽刺的主体和
漫画的主题，然后围绕讽刺的主体或主题拟加标题。拟题可直接以讽刺或颂扬的主体命名，也可扣住漫画
的主题命名。
3．描述画面的原则
描述画面，具体说就是用描述性的语言介绍画面的内容。描述时要贯彻以下原则：注意对象，留意方
位，按照顺序，采用恰当的表达方式。
4．综合考查题的答题要领
此种题型多是将图文转换与编拟公益广告语结合起来，综合考查学生的语言表达简明、连贯、得体的
能力，其答案必须涉及画面的内容、主题和公益广告三个方面。做这种题型，要注意考查的角度和要求，
可在理解画意的基础上，紧扣画面内容作答。
指津三： 徽标类解题技法
1．仔细观察画面 ，抓住图标的特点联想 。（图画的构图 ，辨清画面上的背景与人 、物，人的服饰 、动
作、表情，物的地理位置、特征等，画的标题以及其他文字信息等）。
2．扣住行业特点准确理解内容和寓意。 理解应以画面为主，结合注释（文字、符号）。
3．抓住特征说明 。抓住能反映画面的特征进行详细说明 ，才能清楚地揭示画面的内涵与外延 。注意顺
序：总分、上下、左右。
对联类语言表达题解题技法
高考语文试题中，对联的考查主要有两种形式：一是根据所提供的语境要求写一副完整的对联；二是
根据所给上联对出下联。第二种考查形式是最常见的，具体说来有如下几种题型。
填空式。这种题型将对联的相关知识和教材内容、文学知识结合起来考查。 学科 网（北 京）股 份有限 公司
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补拟式。这种题型有两种情况：一是参照一联，补拟另一联；二是根据题目中提供的词语拟写对联 。
调整式。这种题型将对联的相关知识和语言表达连贯、正确使用修辞手法等考点结合起来考查。
情景式。这种题型为考生提供一个特定的情境，结合 “得体 ”这一考点要求拟写对联。
话题式 。这种题型只给考生提供一个话题 ，而其他方面都不加限制 ，这就为考生展现自己的才华提供
了自由而广阔的舞台，人人得以展个性，显才情，见灵气。
改写式 。这种题型一般是在一段文字中画出两个意义相对的句子 ，要求根据对联 ，语言简明 、连贯以
及句式的选用、仿用、变换等知识进行改写。
概括式 。这种形式是把拟写对联和压缩语段结合起来考查 。做这种对联题 ，首先要把语段中的主要信
息筛选出来并加以概括，然后用对联的形式加以组织。
不管是哪种考查方式，都离不开对对联基本知识的考查，因此，还要注意以下几个方面。
对联的内容都相关 。在解答对联题时 ，可以借助相关 、相似 、相反的联想 。先把一联拆成几个词 ，利
用联想的方法，给每个词分别做对，再把这些对出的词连缀成一联。
注意立意 。立意或写景 ，或言志 ，或抒情 ，或深思历史 ，或剖析现实 。但不管是歌颂什么还是批判什
么，都要有明确的感情，思想内容要积极向上。要做到这一点，就要注意选取符合这一立意要求的意象。
要善于化用 。所谓化用 ，指的是灵活地运用天文 、地理 、时令 、熟语 、诗文名句 、成语典故等知识来
拟写对联。如蒲松龄撰写的自勉联 “有志者、事竟成，破釜沉舟，百二秦关终属楚；苦心人、天不负，卧
薪尝胆，三千越甲可吞吴 ”，就巧妙地融入了 “有志者事竟成 ”“ 破釜沉舟 ”“ 皇天不负苦心人 ”“ 卧薪
尝胆 ”等成语典故。
修辞类解题技法
单独考查。着重考查修辞手法的运用，即根据规定的情境，运用一种或几种修辞手法表达思想感情 ，
或表达对事物的认识。所给材料、预设情境，多来自现实生活，与自然、社会、人生密切相关。
综合考查 。高考语文在考查修辞手法时注重对综合表达能力的考查 ，因此多与其他题型相结合 ，有与
“仿写 ”相结合的，有与 “连贯 ”相结合的，等等。
首先，要熟悉常见的修辞手法的特征，明辨易混淆的修辞手法，这是正确运用修辞手法的前提。
其次，要熟悉各种修辞手法的作用，明确其表达效果。掌握修辞手法的主要目的是提高鉴赏文学作品
的能力和更好地表达思想情感。
最后，要运用联想和想象等思维形式。语言是表达思想的工具，而思想是思维的结果。 学科 网（北 京）股 份有限 公司
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☞考场作文高分技巧
2025 年高考已经进入最后倒计时
第一阶段 ，熟悉高考 。高考命题的底层逻辑是什么？命题人的出题思路是什么？聚焦 2024 年高考作文
命题，结合考情精准分析，摸准 2025 年高考作文命题脉搏，洞悉命题新动向。
第二阶段，研究教材。教材永远是我们高考备考的第一出发点，在冲刺阶段也不例外。研究教材上每
个单元的单元研习任务 ，因为许多高考作文题目 ，其实都脱胎于此 ，如 2022 年北京卷的 “学习今说 ”、202 3
年新课标 Ⅰ卷的 “故事的力量 ”、2024 年新课标 Ⅰ卷的 “问题与答案 ”，都能从单元研习任务中找到 “蛛
丝马迹 ”。
从教材出发，对教材的导语、学习提示、单元研习任务进行细致入微的分析，把握其中隐藏的高考命
题倾向，把 “老课本 ”读出 “新味 ”来。
高考语文部编教材中出现的 19 个作文训练点梳理
学写诗歌 必上 30 页 情感、意象、韵律
写人要关注事例和 细节 必上 56 页 选取典型事例、感人细节细节
学写文学短评 必上 69 页 对作品准确把握、学会聚焦、学会叙议结合
议论要有针对性 必上 104 页 针对现实，针对读者 学科 网（北 京）股 份有限 公司
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如何做到情景交融 必上 124 页 触景生情，情因景生。因情写景，借景抒情
如何阐述自己的观 点 必下 19 页 想清、写清、充分论证、包容异见、明确思路、选用合适结构
如何清晰地说明事 理 必下 68 页 对事理有准确、深入的认识，说清关键要素，注意说明顺序
写演讲稿 必下 92 页 关注听众、主题鲜明、讲究语言技巧
叙事要引人入胜 必下 136 页 找好立足点，写出情节的曲折起伏，善于观察，善用技巧
学写综述 必下 142 页 全面、准确、客观、清晰
如何论证 必下 153 页 确定论点、选择合适论据、采用不同 论证方法、适当运用一些表达技巧
材料的积累与运 用 选上 41 页 材料来源于经典作品、社会生活。真 实可靠、恰当运用
审题与立意 选上 54 页 要我写什么，我要写什么
学写小小说 选上 92 页 明确立意、巧于构思、抓住传神之处
深化理性思考 选中 36 页 透过表象看实质，敢于追问、敢于质疑，理性辨析、提升思维品质
学写申论 选中 125 页 观察社会现象，分析解决社会问题， 有正确写作观
语言的锤炼 选下 70 页 推敲词语、活用句式、巧用修辞、充 满情趣和理趣
说真话、抒真情 选下 90 页 真诚的写作态度和情感
文章修改 选下 114 页 完善立意、增删材料、调整结构、推敲语言
教材母题归纳预测
教材母题预测一：探索与创新
教材母题预测二：青春与使命
教材母题预测三：理性与思辨
教材母题预测四：苦难与新生
教材母题预测五：德育与美育
第三阶段， 熟悉 哲理思辨类、关系辨析类、隐喻映射类、时事热点类、现实导向类五类作文题型，树
立题型意识， 做到心中有 “模”，进而能在考场上对常见题型快速上手。
第四阶段，理解把握国家政策和社会发展中的关键内容，也是我们在高考备考冲刺阶段的重点任务之
一。对 “科技与人文、文化传承与融合发展、大国自信与开放包容、自然情怀、个性与自我认知 ”等关键
词进行解读。对于社会热点的关注，不应该仅仅停留在显性的层面，还应该深入到隐性的层面；不应该仅
仅停留在对时代 “主流 ”的关注上，还应该努力把握住体现时代变化的 “暗流 ”和“潜流 ”。
万事俱备，在临考前，我们还需要借一次 “东风 ”，即 “临考必有的一次巅峰体验 ”，拿出自己写得
最好的一篇作文，一遍一遍地、不厌其烦地修改打磨，让自己在考前达到写作的巅峰状态。 学科 网（北 京）股 份有限 公司
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注意：行文前形成立意构思导图
作文出彩有技巧
高考作文分为基础等级和发展等级两个层次。简单地说，在基础等级中，考生要做到的是 “入格 ”；
在发展等级中，考生要做到的是 “出彩 ”。
下面结合高考考场写作的实际，介绍几种实用的方法技巧。
l．严格审题，深入发掘
所谓严格审题，深入发掘，就是要对试题及其要求认真推敲，准确领会命题意图，深入发掘提示或要
求中的含义，写出符合题意及要求，立意较高的作文来。立意上要尤其注意两点：一是准确性，二是全面
性。在此基础上，再谈思辨性、深刻性、创新性等等。
审题要完成两方面的任务，一是弄清作文类型（命题型？材料型 ？话题型 ？），确定文体（限制型自
主型？）；二是要分析材料，选准角度。
一定要注意材料中议论性语句，它往往给考生指明写作的方向，写作的方向是不能改动的，否则，就
会偏题 ！另外，看是否明确要求联系自身。不要抨击社会、发牢骚等，少用政治术语，想想阅卷老师的喜
好，说符合自己身份 的话。
2．广泛搜集，精心选材
材料要典型，要有意义，有意味，有文化性。尽量用自己最熟悉的材料，为了做到材料新鲜，建议多
使用近几年出现的新人新事。对一些过于 “新鲜 ”评卷老师可能不知道的，要适当地交代大体情况，以免
给人捏造之感，运用的材料尽量与提供的原材料保持一致，即使关系不够紧密，也要注意围绕题旨恰当勾 学科 网（北 京）股 份有限 公司
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连，表明不偏题。
不要去写科幻小说，写童话、寓言不要过于含蓄， “允许 ”编写故事，不等于提倡 “编写故事 ”，滥
编、胡编，绝对不会得高分。严禁照搬照抄别人的范文，杜绝 “无我文章 ”和“抄袭文 ”的出现。不要触
及敏感的政治事件，少谈宗教话题，不要单纯发牢骚，不要写早恋等敏感话题。
记叙文要有细节描写 ，推己及人 ，以情动人 。议论文要以事实和道理来写文章切忌大话 、套话 、废语 ，
要避免空发议论，乱提口号，乱发号召，空表决心等。
3．发挥所长，合理定体
文体不限不代表没有文体，要写什么像什么。记叙文要三分之二的篇幅落足于叙述，议论文反之。写
记叙文，最好将主人公设定为自己，用第一人称入文，不喊口号，情真意切。写议论文要注意事例贴切、
事理的分析、引用后的引申，做到首尾呼应。考生一定要发挥自己所长，写自己擅长的的文体，还要根据
自己占有的素材来确定文体。
4．结构完整，眉目清晰
标题是文章的眼睛，要有一个较好的标题（命题作文除外），标题能较好地引领文章的内容。
文章内容要精于分段 ，因为自然段分得越自然 ，越显得你成熟老练 ，最主要的还是能使评卷老师一目了然 ，
以五至八段为宜。
不管是写成记叙文还是议论文，一定要在标题（命题作文除外）、开头、结尾和每一段的开头反复用
材料中的关键词或标题中的词，材料作文如果拿不准，就找准材料陈述的对象和材料中的议论句，就事论
事，反复论述，多角度论述、正面说、反面说，假设推理（如果不这样会有什么样的后果）。
5．力求新颖，创造亮点
作文贵在创新。好的作文总有自己的独到之处，那种人云亦云，千篇一律的作文是不可取的，也难以
得到阅卷老师的首肯。求新还要从自己的实际出发，否则意欲求新求异，反而会因形害义，落了俗套。盲
目创新，认为这是一个省力的捷径的想法是危险的。
好的作文，总有自己的闪光之处，有自己的亮点。可在以下几方面出彩：
①标题要靓 。审美性 、哲理性强的标题 ，给人以耳目一新的感觉 ，使阅卷老师精神振奋 ，如名言为题 ，
比喻为题等。
②构思要巧。构思中比较容易被忽略的是写作的角度。好的切入角度容易出彩，容易表现自己独特的
感受、认识和写作者的个性。选取一个小的切入点，对材料进行深入挖掘，写具体，写深刻，往往能够以
小见大，写出好文章。
构思要耐心：不假思索就能想到的东西先写在草稿上；稍加思索想到的也写上。花上几分钟，列出提
纲，绝对严禁胡编乱造。即使编写故事也要做到一大胆想象，自圆其说。
③谋篇出彩：开头要引人（开门见山，直截了当；制造悬念，引人入胜；提出问题，引人注意；说明
情况，交待背景），结尾要有力 （画龙点睛，发人深思；总结全文，照应开头；叙述结束，自然收尾：抒
发情感，引起共鸣 ）。开头结尾忌讳冗长、拖沓，废话太多。
④情感要真 。高考作文中 ，屡见无病呻吟之作 。作文要抒发真实的感情 ，要求写作者要用心感悟生活 ，学科 网（北 京）股 份有限 公司
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要把自己的情感态度投 射或融入到所描写的对象中。人类有着许多共同的感情，但人的情感又往往有许多
差别，写出自己独到的内心感受和体验，也就做到了新颖。
⑤语言优美 。多用短句少用长句 。多引用贴切名言警句 ，可引进部分时代新词汇 ；引用流行的通俗
歌曲歌词 ；引用百姓口头民谣 （但取向一定要积极向上 ）。不用别人看不懂的方言 ；不使用别人看不懂
的词汇。
语句华丽一些不是坏事 ，或排比 ，或比喻 ，或整句 、或短句 、成名词化用为动词 、形容词化用为动词 、
或用动宾短语、或用省略号感叹号疑问号等。背诵的名篇名句要有意识地使用。
6．回头反观，弥补小足
主要有三方面的任务：一是看审题是否恰当，如有偏差可矫正，在结尾部分充分解说你写的内容与话
题或材料的关系，一旦跑题就不及格了；二是看标题、字数是否少，不写标题，按评分规定扣 2 分，字数
不足（ 50 字 1 分）；三是看是否有错字（ 1 个错别字 1 分）、病句。这三个方面可以说是作文中的硬伤，
扣分无形之中往往是双扣分，我们要尽可能地避免。
★考生在考试书写上要做到三清 —— 卷面清洁 ，字迹清楚 ，笔画清晰 。四不 —— 不写潦草字 ，不写异体
字、自造字，不规范的简化字，不添减笔画。
“八小 ”锁定高考一类作文
对高考作文的期望值要从追求满分作文降到追求 “一类作文 ”上来，因为满分作文要靠天赋和运气，
而“一类作文 ”则可靠训练。从操作层面来讲，就是要从追求满分作文的 “三大 ”（大手笔，大气象，大
智慧）降到追求 “一类作文 ”的“八小 ”上来。
一、有一点小生活
“小生活 ”就是要在文章中写属于学生的原汁原味的生活画面、酸甜苦辣、喜怒哀乐，去展现一个本
色、活蹦乱跳 、童心未泯 、当代风味 、可亲可爱的 “你”。它虽小 ，但香气扑真 ；带点青涩 ，但可信可近 。
二、有一点小感情
“小感情 ”抒发的是自然流露的感情 ，而不是做作的煽情 ：它是对 “大感情 ”的反拨 ，它抒发的是 “我
的切身的感情 ”，而不是 “我们的大而化之的感情 ”：它是对 “空感情 ”的反拨，是 “跳动着心灵颤音的
感情 ”，而不是 “口号式的感情 ”。
三、有一点小发现
高中生对生活的思考和对事物的看法一般不大可能高屋建瓴、石破天惊，但如果他们能有一点属于自
己的小发现，同样可以让阅卷者眼前一亮。大家看秃子看到的都是光头，而你却看到了头上的几缕青 丝，
这就是小发现，这就别有意味。
如话题作文 “借”的写作 ，许多考生都以 “中国制造 ”为骄傲 ，但有一位考生却提出了自己的一点 “小
发现 ”—— “中国创造 ”。他在文中说 ：“技术是可以借到的 ，但思想无法借得 。一个国家只有拥有了自己
的原发思想，才能真正强盛起来。否则，只能永远被别人勒着脖子。的确，无论是 “中国芯 ”，还是 “神
舟五号 ”“嫦娥一号 ”，靠的都是原发思想 ，靠的都是自主知识产权 。中国真正强大的标志不是有多少 “MAD E学科 网（北 京）股 份有限 公司
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IN CHINA ”，而是有多少 “中国创造 ”。从 “中国制造 ”到“中国创造 ”，虽是一字之变，但却别具慧眼 ，
星光闪烁。
四、有一点小哲理
那些带着泥土气息、带着情感余温，带着生活盛悟的小哲理，往往因为其真诚、真挚、真切，给人以
“于我心有戚戚焉 ”之感。它不大，就是日常所见；它不远，就是日常所及；它不深，就是日常所感。因
其切近、切身，故而沁人心脾。
“孩子 ，只要你有出息 ，天天都是妈妈的节日 。”母亲这句话虽然质朴 ，却蕴含着小哲理 —— “孩子的
出息 ”就是母亲的 —切，它胜过世上 —切物质的拥有 、形式的表示乃至精神的慰藉 。这句话中蕴含的哲理虽
然很小，但因为它带着母亲的体温，散发着乡村泥土的芬芳，因而使人怦然心动，回味绵长。
五、有一点小描写
记叙类文章要想打动阅卷者，依靠的不是叙述，而是描写。因为叙述搭建的仅仅是骨架，而描写凝结
的都是血肉。 “小描写 ”并不要求成篇成段，只要求 “时有风骚 ”，它可以是一句对话、一个动作、一丝
心动，甚至是一景一物，便可以突显人物神韵，传达文章旨趣。
如命题作文 “我的歌 ”的写作，有位考生写了一首 “用父亲的血为词、我的泪为曲 ”谱写而成的歌。
他在对 “背尸匠 ”父亲的刻画上就用足了 “小描写 ”：
在我一而再、再而三的要求下，父亲才从老家来到我的新房。上楼时，父亲显得十分吃力，我伸出手
准备拉他一把，谁知他倏地把手缩了回去，只笑着说了一句： “不必了。 ”到新房时，我高兴地打开门，
对父亲说： “瞧！ ”并随手拿了一双拖鞋让父亲换上，父亲刚准备脱鞋，随即又停了下来，用袖子揉了揉
眼说道： “我站在门口看看就心满意足了。 ”
“倏地把手缩了回去 ”是因为父亲怕把手上背尸体的晦气传给儿子。 “刚准备脱鞋 ”是父亲下意识的
行为，也是他欣喜之情的自然流露。 “随即又停了下来 ”是因为父亲怕把自己身上的 “晦气 ”带进儿子的
新房。 “用袖子揉了揉眼说道 ”是因为眼中有老泪溢出，这个 “泪”是欣慰之泪，儿子终于有出息了；这
个“泪”是骄傲的泪 ，自己一辈子被人瞧不起 ，儿子终于让自己 “在一村老小面前可以抬头做一回人了 ”；
这个 “泪”也是辛酸之泪 ，为了这一天 ，自己背了一辈子尸体 ，这一天等得实在太久了 。“小描写 ”虽小 ，
但传情，传神，让人物形象霎时丰满起来。
六、有一点小化用
引用可以增强文章的说服力，提升文章的理性层次，但有时也给人 “掉书袋 ”之嫌，而那些已被人引
用过千万遍的名人名言更是让人大倒胃口。为此，大可不必去 “嚼别人嚼过的甘蔗 ”，完全可以在这些名
人名言里加入一点你的 “情思 ”，而这些流淌着你的血液的 “凡人名言 ”就会顷刻间变成一道 “风味小吃 ”，
让阅卷者满口生津，这就是 “小化用 ”的魅力所在。
如材料作文 “经典与时尚 ”的写作，有位考生就在名句 “所有的果都曾经是花，但并非所有的花都能
成为果 ”中加入了自己的 “情思 ”，使之由一道 “大路菜 ”变成了一道 “时鲜小吃 ”：“所有的经典都曾
经是时尚，但并不是所有的时尚都能成为经典 ”。虽是 “小化用 ”，却将 “经典与时尚 ”的辩证关系演绎
得深入浅出。 学科 网（北 京）股 份有限 公司
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七、有一点小技巧
在文章主旨确定之后和写作素材选定之后，如何使既有的素材更好地表现主旨，有时需要使用一点小
技巧 。它可以是 ：行文上的小布局 —— 让文章有点眉目 ；情节上的小调整 —— 让文章有点想头 ；论证上的小
安排 —— 让文章有点层次，呈现上的小手法 —— 让文章有点味道。
如命题作文 “学会调整 ”的写作 ，有位考生在行文上就进行了一点 “小布局 ”，他根据 “水”有固体 、
液体、气体的 “三态 ”变化，为文章设计了三个小标题： “甸甸的冰 ”“ 流动的水 ”“ 升腾的气 ”，并在
三个小标题下分别写了一个创业者夯实基础、兼容壮大、促成飞跃的历程。其实，很多考生在文章中都写
了这个内容，但写得东扯西拉。而这位考生因为用了一点小技巧，使文章布局整齐匀称，眉目一清二楚，
故事简洁紧凑，主旨一看即明。
八、有一点小才气
何谓 “小才气 ”？就是在文章字里行间带点小聪明、小机智、小文气、小风格，它不一定很大，也不
一定很鲜明，只要有那么一点样子，有那么一点意思，有那么一点味道，有那么一点感觉，就会让阅卷者
顿首一笑，称道： “狡猾，狡猾，也亏这小家伙想得出来。 ”
上述八个方面，说其 “小”，是因为它们对你来说够得着，学得会，用得上。当然，它们也仅仅是示
例，旨在给你以启迪。最后，我们对周杰伦的金曲《蜗牛》 “小化用 ”一下： “小小的你有大大的梦想，
只要你带着这小小的启迪，一步一步往上爬，总有一天你会撑起你作文的一片天。 ”
【认知心态篇】
高考获胜第一步：培养健康心理
高考考的是知识，是能力，这一点应该没有谁不知道，没有谁不重视；但是，高考也是对考生心理的
考查，知道并且重视的恐怕就不多了。在这里，有必要提醒广大考生与他们的老师和家长：注意培养考生
的健康心理，过好 “心理关 ”，对成功参加高考，有着十分重大的意义。
我们常常看到，在高考前的复习过程中，特别是在每年的四、五月份，不少考生会产生一种焦虑、浮
躁的情绪，有人戏称为 “高考复习综合症 ”，其 “症状 ”是：心中莫名的烦躁，既害怕高考时间一天天的
临近，又巴不得明天就考了算了；他们听课时注意力难以集中，记忆力下降，好像什么都复习了，又好像
什么都没记住；失眠、多梦、常有体力不支的感觉；思维迟钝，有时甚至连面对简单题目，头脑都好像出
现“断路 ”现象，不会做了。这一切带来的后果是：信心下降，效率下降，导致各种测试的成绩下降，这
又使信心下降，产生严重的恶性循环，对高考十分不利。
为什么会产生这种现象？笔者认为，原因主要有四：
一是当下社会普遍存在浮躁心理给学校、学生和家长都产生不小影响，导致所有相关人员在考前都产
生一种十分严重的焦躁心理，大家都在自觉不自觉地制造考前紧迫情绪，这种情况越临近高考与明显。
二是家长和学生对高考的期望值过高，脱离了考生的实际水平。
三是目前我国 “一考定终生 ”的高考制度和严峻的就业形势对考生的压力太大 ，超过了考生的承受
能力。 学科 网（北 京）股 份有限 公司
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四是随着复习过程的一步步进展 ，要求考生尽快形成厚积薄发的态势 ，加快把知识转化为能力的步伐 ，
部分考生一时不能适应这种要求，因而产生焦虑和急躁的情绪。
怎样改变这种状况，过好 “心理关 ”，把 “高考复习综合症 ”治好？关键在于增强信心，培养健康心
理。办法主要有两点：
一、学会善于从纵向和横向两方面客观积极地、正确真实地认识和评价自己
所谓 “纵向 ”，就是自己和自己比，自己的现在和自己的以前比，在比较中发现自己的进步，在比较
中增强自己的信心。 特别是从考前两个来月开始，各科教师在复习过程中，一定要尽力用考试或作业、辅
导等各种方式向学生多提供一些有利于增强信心的信息，经常表扬学生，让学生心情愉快，受到鼓舞。所
谓“横向 ”，指的是学生对各科知识的掌握情况。教师在复习方法上，要更加注意科学性和有效性，要设
法让学生把学科中的薄弱环节转换成努力的目标，转换成一步步进步的台阶。即使是发现了学生知识的缺
陷、漏洞或欠缺，都既要给学生补正，更要注意方式方法。这样，学生的焦虑和急躁就可以逐步转化为愉
悦和信心。在这里， “形式 ”甚至比 “知识内容 ”更为重要。
二、明白知识从积累到释放的演变道理
从复习到考试的过程 ，是知识网络的形成过程 ，是将知识转化为能力的过程 ，也可以说是能量积
累到能量释放的过程 。在这个过程中 ，考生的心态便是一剂很重要的催化剂 。心态越好 ，知识与能力
的转化就越完全 ，越充分 ，因而也就越完美 。良好的心态来自于充足的信心 。因此 ，相信自己 、长自
己的志气便显得十分重要 。一些考生在考场上一碰上难题就左顾右盼 ，看见别人总在写 ，就总以为别
人比自己行 ，这在心态上就先输了一分 ，是考场大忌 。须知 ，你不会做的别人也很可能不会做 ，而你
会做的别人却不一定会做 。这种考试心态一定要建立 。另外 ，老师和家长也要鼓励学生注意以下规律 ：
“取法其上，得乎其中；取法其中，得乎其下。 ”在考前，如果只是鼓励学生 “别害怕，别怯场 ”，
这只是一种下策 ；鼓励学生 “正常发挥 ”，这也只能算是中策 ；只有鼓励学生 “超常发挥 ，尽情释放 ”，
才是最好的上策！
此外必须指出的是 ，克服 “高考复习综合症 ”绝不单单是考生一方面的事 ，需要教师和家长 （甚
至社会各方面 ）的共同努力和关注 。在这里 ，过分的关心和过多的指责 ，都是绝对不行的 ，决定地需
要的是科学的态度和方法 。我们不能只注重了知识而忽视了心理 ，一定要把对心理的重视提高到应有
的地步。
冲刺复习备考指导
同学们 ，高考的战鼓已经敲响 ，在这最后的冲刺阶段 ，如何高效复习语文 ，是大家最为关心的问
题。今天，我们就结合新高考卷的考情，来谈谈冲刺阶段的复习策略。
一、信息类文本阅读
新高考信息类文本阅读往往采用多文本组合的形式，考查大家对信息的筛选、整合与分析能力 。
从近年考情看，文本选材广泛，涉及社会科学、自然科学等多个领域，题型灵活多变。 学科 网（北 京）股 份有限 公司
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冲刺阶段 ，大家要以近三年高考真题为核心 ，进行复合文本材料关联性专项训练 。比如 ，针对同
一主题的多篇材料 ，分析它们之间的观点异同 、论证角度差异等 。强化信息筛选能力 ，学会精准定位
关键信息 ；提升逻辑推导能力 ，依据文本内容合理推断结论 ；培养批判性评价能力 ，对文本观点进行
客观评判 。重点突破材料异同比对题 ，关注选项与原文在表述上的细微差别 ；掌握图表数据解读技巧 ，
能从图表中准确提取有效信息并分析其与文本的关联 ；提高观点态度评析题的答题水平 ，明确作者观
点并阐述自己的理解 。此外 ，要增加时事热点文本阅读 ，像科技发展 、社会民生等热点话题 ，提升在
陌生语境中的阅读适应能力。
二、文学类文本阅读
文学类文本包括小说和散文 ，这部分大家常出现主旨把握偏差 、文体特征混淆 、关联阅读能力不
足等问题。
在接下来的复习中，构建 “教材 + 高考 + 经典 ” 三位一体训练体系。回归教材，梳理教材中
经典小说 、散文的文本特点 、写作手法等 ，强化教考衔接 。比如 ，学习 《祝福 》时，深入分析小说的
环境描写对人物塑造和主题表达的作用 ，将这种分析方法迁移到高考文本阅读中 。结合新教材各单元
涉及的重要概念 ，积累必备知识 。剖析 2020 - 2024 年新课标卷典型试题 ，建立小说场景建构 、散文意
象解码等思维模型 。例如 ，对于小说场景题 ，可从场景的构成 、作用 （渲染氛围 、推动情节 、塑造人
物等 ）方面思考答题思路 。进行分类教学 ，打破小说与散文的文体壁垒 ，注重关联阅读 ，同时加强历
史小说、散文化小说等新形态文本解读训练，拓宽阅读视野。
三、文言文阅读
文言文阅读一直是大家的难点 ，文言语感薄弱 、实词虚词理解不准确 、翻译存在 “以今释古 ” 等
问题较为突出。
我们采取 “真题精译 + 教材回扣 ” 双轨策略。精选 10 篇高考文言文进行深度翻译，逐字逐
句落实词义 、句式 ，同时加强文本诵读 ，培养文言语感 。开发 “以题解题 ” 技法 ，利用内容概述题
选项逆向推导文意 。创设文化常识情境题库 ，在具体语境中理解文化常识 ，避免机械记忆 。强化语法
断句训练，建立 “虚词定位 + 句式辨析 ” 断句模型。比如，看到 “之”“ 其”“ 而” 等虚词，
根据其常见用法初步断句 ，再结合句式结构进行调整 。收集各地模拟题中新视角 、新设问的简答题题
型，总结答题形式，做到举一反三。
四、古代诗歌阅读
古代诗歌阅读存在诗意理解不透彻、情感把握片面、答题逻辑混乱等问题。
构建 “三位一体 ” 解读体系 ，紧扣诗题导引 、注释暗示 、题干指向 ，提升诗意理解能力 。例如 ，
诗题若为 “怀人 ”，则可初步判断诗歌情感方向；注释中介绍诗人被贬经历，对理解诗歌情感有重
要提示 。精选试题 ，创设情境 ，让自己完成阅读 、鉴赏 、表达过程 ，之后借助参考答案进行反思 ，找
出认知缺漏并及时纠正 。在答题时 ，要做到言之有理 、言之有据 、言之有序 ，提高表达能力 ，清晰阐
述诗歌的意象、意境、情感及艺术手法等。
五、语言文字运用 学科 网（北 京）股 份有限 公司
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语言文字运用题型情境化趋势明显 ，对大家在真实情境中运用语言的能力要求较高 ，成语 、病句 、
补写等必考题型得分率有待提高，面对新题型时应对能力不足。
冲刺阶段 ，创设 “真实语境 +” 训练模式 ，将语言运用与国家经济 、社会发展 、生活实际 、科
技伦理等时代命题结合 。比如 ，以 “人工智能在生活中的应用 ” 为情境 ，进行成语填空 、语病修改
等训练。反复训练经典必考题型，巩固答题技巧。精选精做新题型，拓宽解题思路。加强读题指导 ，
明确答题思维方向 ，掌握各类常考题型解题策略 。例如 ，对于病句题 ，熟悉常见语病类型 （语序不当 、
搭配不当、成分残缺或赘余等），通过分析句子结构进行判断。
六、写作
写作部分，审题立意不准确、写作思维不清晰、语言表达平淡是常见问题。
强化审题立意训练 ，认真研读材料 ，抓住关键词句 ，准确把握材料内涵及命题意图 ，确定立意方
向，避免偏题跑题。解析 2022 - 2024 年新课标卷真题，学习优秀范文，了解高考作文命题特征与导
向。强化写作思维和框架结构训练 ，形成相对固定的议论文思维框架 ，如 “提出问题 - 分析问题 - 解
决问题 ” 结构，提升文章逻辑性和条理性。锤炼语言，多读优秀范文，进行仿写训练，提升语言表
现力。同时，关注社会热点，思考人生价值，积累写作素材，提升思想深度。比如，关注 “乡村振
兴”“ 文化传承 ” 等热点话题，积累相关事例和观点，以便在写作中灵活运用。
同学们 ，在这最后的冲刺阶段 ，保持良好的心态 ，合理规划时间 ，按照科学的复习策略进行备考 ，
相信大家一定能在高考中取得优异的成绩，为自己的高中生涯画上圆满的句号！
高考只是人生一站：建立 “可控思维 ”和“成长型心态 ”
“要是考砸了 ，我的人生就完了 。”每次听到这样的话 ，老师心里都很不是滋味 。今天我们就来
聊聊 —— 高考，到底意味着什么？ 一、它是重要的 “一站 ”，但不是唯一的 “终点 ”
有位叫小林的同学，考前两个月模拟考成绩突然下滑，每天失眠焦虑，觉得 “考不上 985 就没前
途”。后来我们一起分析他的情况 ：他擅长编程 ，曾在信息学竞赛中获奖 。我告诉他 ：“高考是展示
能力的平台，但你的编程才华不会因为一场考试被否定。 ”最终他调整心态，考上了省内普通本科 ，
但凭借大学期间开发的校园服务小程序 ，毕业时拿到了互联网大厂的 offer 。老师想告诉大家 ：高考像
高速公路上的一个服务区 ，你可以在这里休整 、补充能量 ，但前方还有无数风景等待你发现 。去年毕
业的小周同学 ，高考数学失误只考了 92 分，但她后来在考研时数学拿到 高分，逆袭进入 985 高校读研 。
人生不是百米冲刺，而是一场马拉松。
二、那些 “没考好 ”的人，后来都怎么样了？
给大家讲三个真实故事：
故事一： 2018 届毕业生阿远，高考因紧张作文跑题，最终去了专科。但他在校期间考取了导游
资格证，毕业后成为资深旅游博主，带网友走遍全国 56 个民族聚居地，现在全网粉丝超百万。
故事二： 小丁同学 ，当年高考差 3分与理想院校失之交臂 ，却在 普本院校遇到了改变她人生的导
师。导师带她做非遗保护项目，如今她已是省级非遗传承人，作品被博物馆收藏。 学科 网（北 京）股 份有限 公司
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故事三 ：科技圈知名企业家王兴 ，高考考上清华大学 ，但他的创业之路并非一帆风顺 。从校内网
到美团 ，他经历过多次失败 ，但每次都能重新出发 。他说 ：“高考教会我的不是 ‘必须成功 ’，而是
‘学会面对失败 ’。”
这些例子告诉我们 ：决定人生高度的 ，从来不是某一场考试 ，而是你面对生活的态度和持续成长
的能力。
三、考前三天，你可以这样做
每天给自己写三行 “成功日记 ”：
比如 “今天正确解析了一道复杂的文言文虚词题 ”“作文素材又记住了一个新案例 ”，用具体的
小成就积累信心。
模拟考场突发情况预案：
提前设想 “如果遇到没见过的题型怎么办 ”—— 可以在草纸上先列答题框架，或深呼吸 30 秒让
自己冷静。有备而来，才能临危不乱。
给未来的自己写封信：
用“亲爱的 _ （未来的自己 ）”开头 ，写下 “无论高考结果如何 ，我希望你依然保持 _ （比
如‘对知识的热爱 ’‘探索世界的勇气 ’）”。这封信会让你看清 ：比分数更重要的 ，是你不变的初
心。
同学们 ，走进考场时 ，请记住 ：你手里的笔不是 “决定命运的刀 ”，而是 “书写人生的笔 ”。无
论今天你是紧张还是从容 ，老师都为你骄傲 —— 因为这三年 ，你已经拼尽全力 ，这份坚持本身 ，就
值得点赞。
聚焦可控因素：志愿填报
有同学们问： “老师，万一我考砸了，志愿都没法填了怎么办？ ” 其实换个角度想：高考成绩是
我们目前无法改变的 “变量 ”，但志愿填报却是我们完全可以掌控的 “主动选择 ”。今天我们就用 “可控
思维 ”，提前为志愿填报做准备。
一、先搞懂：志愿填报的核心逻辑是什么？
举个真实案例 ：2022 届学生小夏 ，高考成绩 580 分（超本 科线13 5分），当时她的目标有两个 ：一
是去外省读综合类大学的冷门专业 ，二是留在省内读双非院校的王牌专业 。后来我们一起分析 ：她擅
长数据分析，未来想从事金融行业。省内双非院校的 “统计学 ”专业是国家级特色专业，且与本地 金
融机构 有校企合作项目。最终她选择后者，大四时通过校招进入银行实习，毕业即就业。
关键结论 ：志愿填报不是 “分数换学校 ” 的简单交易 ，而是 “用现有分数 ，为未来职业发展铺路 ”
的战略选择。
二、现在能做的三件事：把主动权握在手里
（一）制作 “专业 -兴趣匹配表 ”学科 网（北 京）股 份有限 公司
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我的兴趣点 可匹配专业 初步了解渠道
喜欢拆解
复杂问题 计算机科学与技术 观看 《编码 ：隐匿在计算机软硬件背后的
语言》纪录片
擅长与人沟通 市场营销、社会工作 关注 “中国市场营销学会 ”公众号
对历史感兴趣 考古学 、文物与博物馆学 参观省博物馆线上展览
操作建议 ：每天花 10 分钟 ，通过 “学习强国 ”职业教育板块 、各高校招生网 “专业介绍 ”栏目 ，积累 3
5个专业的具体信息 （如核心课程 、就业方向 ）。比如浙江大学 “机器人工程 ”专业 ，课程包括 《人工
智能导论》《机器人动力学与控制》，毕业生可进入大疆、 小米汽车 等企业。
（二）研究 “三位一体 ” 录取规则
以浙江省为例， 202 4年复旦大学 “三位一体 ”招生中，某考生高考成绩 645 分（裸分需 680 分），
但凭借竞赛获奖 +面试优秀 ，最终被录取 。如果你有学科竞赛 、艺术特长等 “加分项 ”，现在就可以梳
理：
哪些高校的 “三位一体 ”/“强基计划 ”接受我的特长？（如中国传媒大学对艺术类考生有特殊通道 ）
往年录取案例中 ，综合分计算方式是怎样的？ （比如 “高考分占 60%+ 校测占 30%+ 学业水平考试占
10% ”）
行动清单：本周内整理出 5所“冲刺型 ”（分数略高于你）、 5所“稳妥型 ”（分数匹配）、 5所“保底
型”（分数略低）院校，标注它们的特殊招生渠道。
（三）模拟填报：用 “假设分数 ” 练手
假设你高考成绩为 XX 分（可根据最近三次模拟考取平均值），尝试在 “省教育考试院志愿填报系
统”（可登录官网查看模拟版）中填报：
冲一冲：选择 202 4年录取最低分 ≈XX+5 分的院校专业组
例： XX=550 分，可看 202 4年录取线 553 分的 “XX 师范大学汉语言文学专业 ”
稳一稳：选择 202 4年录取最低分 ≈XX 分的院校专业组
例： “XX 财经大学财务管理专业 ” 202 4年录取线 550 分
保一保：选择 202 4年录取最低分 ≈XX-8 分的院校专业组
例： “XX 文理学院英语专业 ” 202 4年录取线 542 分
注意事项：每个院校专业组内，至少选择 3个“不排斥 ” 的专业，并勾选 “服从调剂 ”。通过模拟，
你会发现：即使分数不理想，依然有多种组合方式，主动权始终在你手中。
三、考前特别提醒：别让焦虑偷走你的 “可控力 ”学科 网（北 京）股 份有限 公司
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去年有位考生小吴，考前一周反复纠结 “考不好怎么办 ”，结果语文考试时大脑空白，作文只写了
一半。其实他后来的成绩达到了二本线，但因为考前过度焦虑，浪费了本可以用来准备志愿的时间 。
老师想对你说：现在每多花一分钟担心 “不可控的分数 ”，就少一分钟准备 “可控的选择 ”。不如把精
力集中在：
每天整理 1 个专业的详细资料（如武汉大学 “遥感科学与技术 ”专业，全球排名第一）
记录 3所院校的特殊招生要求（如哈尔滨工业大学 “强基计划 ”招物理竞赛生）
练习用 “分数 ±5 分” 做志愿组合方案
同学们 ，当你把注意力从 “不可控的恐惧 ”转移到 “可控的行动 ”上，就会发现 ：原来高考后的选择 ，
远比你想象的更丰富 。明天考语文 ，记得拿到试卷先深呼吸 ，把第一道现代文阅读题当作 “志愿填报
前的信息筛选训练 ”—— 你看，每一个当下，都在为未来铺路。
【考场注意篇】
送给考生的高考小贴士
同学们，我的亲，马上就要高考啦。你的身体、心理、 “装备 ”
都 准备好了嘛？！快来跟着我一起做考前最后的准备和检查吧！！
NO1 ．你需要准备的东西之 “装备篇 ”
有了装备才能 “打妖怪 ”，同学们 ，要提前准备好哟 ！别等到了
学 校，手忙脚乱痛哭流涕哦！哈哈！
一、准考证和身份证 。
考生可以找一条喜欢的丝带，把准考证和身份证挂在脖子前，以防丢失、遗忘。
二、文具 2B 铅笔至少两支、黑色签字笔至少两支、直尺、圆规等（不要带涂改液、胶带、修正带） 。
特别提醒：涂答题卡使用的铅笔削得太细，会延长涂卡时间，建议把铅笔削成扁扁的 “鸭嘴 ”形，也
有卖现成的楔形头的考试专用笔，请确保正规商场购买正品。进入考场的有效证件，必须妥善保存。
三、手表 。考试时最好戴一块手表合理安排时间。提前调好时间。
四、必要的生活用品，如清凉油、水、纸巾 。
五、雨具 。考试前两天考生或家长应注意天气预报，了解高考当天的天气情况，如果有雨，提前准备
好雨具。
六、着装 。如果高考当天温度较高，应准备舒适、宽松、透气性好的衣服，如棉、麻质地，避免考场
中暑。
七、眼镜 。戴眼镜的同学最好准备副备用眼镜，要提前试戴其是否舒服。戴隐形眼镜的同学要准备好
一副框架眼镜和明目眼药水。
八、出行 。如果骑自行车去考场 ，提前检查好 "座驾 "，车胎气是否足等 ；准备乘公共汽车或出租车的考
生，准备好零钱；家人驾车送，请提前看好线路。 学科 网（北 京）股 份有限 公司
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NO2 ．你需要准备的东西之 “心理篇 ”
马上高考 ，很多孩子出现焦躁不安 ，甚至是头痛 、失眠 、脾气暴躁等问题 ，你要知道现在再紧张再
压力再使劲熬夜补习，已木有用了，放松心态，深呼吸 ~~~ 吸气 ~~~ 吐气 ~~~ 吸气 ~~~ 吐气 ~~~ 告诉自己，
心态决定一切，遇事有个好心态才能事事顺利，不就是 “伸头一刀么 ”，高考来啊，我不怕你！！
一、给自己好的心理暗示，相信自己 。
“我一定能考好 ！一定能实现目标；之前学过的都记得都会用得着；题难难所有人，我不会不一定别
人的就会；我今天真精神一切都会顺利 ……”
二、要充满斗志，昂首挺胸进考场 。
天上写着五个字：这都不是事。见到监考老师，记得微笑以待哦！
三、为自己准备平时最喜欢穿的最舒服的衣服和鞋子，让自己保有安全感和舒适感 。
四、考前不要喝太多水 。免得进了考场突然想上厕所，造成心理上的慌张。
NO3 ．你需要准备的东西之 “身体篇 ”
俗话说身体是革命的本钱，身体棒精神好，做什么都好。考生也一样，有一个健康的身体迎接考试，
绝对是重中之重。
一、饮食上平时吃啥 考试也吃啥 。
考前大换食谱可是饮食的大忌，不要食用平时考生没有吃过的新食物。因为新食物平日没吃过，可能
存在过敏、不耐受、胃肠不适等问题。也要注意忌口，少吃荔枝、芒果等热性水果，防上火，冷饮也要少
吃噢 。不要临时大补 、迷信保健品 ，如果不吸收 、不适应的话 ，反而会导致腹泻 、过敏 、感冒上火等病症 ，
适得其反。
二、适当运动 。
早晚 ，天气不热的时候可以外出散散步 ，既可以缓解紧张情绪 ，平复内心紧张 ，也可以保持精神的
状态。
三、洗热水澡或热水泡脚泡 。
热水澡是最古老的镇静剂，要放松自己，最好浸泡在比自己的体温高一些些的热水里，时间不要超 过
15 分钟。热水泡脚同样能舒缓压力，放松身心的疲惫。
四、睡眠 。
保证充足的睡眠，不要临时改变自己的生物钟。考前不要晚上临时熬夜抱佛脚，这只会让你第二天考
试更疲惫。 学科 网（北 京）股 份有限 公司
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终极考问：遇上不会做的题怎么办？
高考不难 ，但是也一定会遇到不会的试题 ，这个时候 ，怎么办呢？讲究方法 ，就能将失分点降到最低 。
一问：要不要把全卷看一遍？
拿到卷子以后看一下，是看考卷一共几页，多少道题一定要先知道，千万不
能落题和落页。关于是否要把全卷的题目全看一遍，同学们按自己的习惯来做，
没有对错之分 。模拟考你们怎么做的 ，高考还是怎么做 ，不要改变你的习惯做法 。
对于第一场考试的语文试卷，作文题是要先看一看的，因为这是一道得分最高的
题目。看了作文，做到心里有数，等到真正开始作文的时候再认真考虑，不会出
现无暇细想的情况。
二问：如何提高一卷的得分率？
调查显示：一般试卷前几题的错误率比较高，因为一开始考生一般心情比较紧张，所以提醒大家，在
心情恢复正常时要着重检查一下前几题。比如语文第一题便是现代文阅读，虽说都是客观题，但是刚开始
做题便接触到阅读理解类比较大的题目，考生在紧张的情况下，这个题的失误率还是很高的，所以在平静
下来之后，要抽出时间重新检查下第一题。
三问：遇上不会做的题怎么办？
高考是选拔考试，碰到难题是非常正常的。碰到不会做的题不要紧张，要想到，我不会做，那好多人
也未必会做。我只要把能拿到的分拿到就行了，一定要稳定心态。
四问：有的题可以上手，但做半截又不会了，怎么办？
碰到这样的题不要慌 ，仔细审题 ，能做一步做一步 ，能做两步做两步 。高考试题题题设防 ，题题把关 ，
比如理科计算题按步计分做对一步便有一步的分，不要因为半截不会了，前面的就不算分数了。文科的主
观题，也是按点计分的，答对一点有一点的分。所以心态一定要放松，能做几步做几步，能得几分得分。
不要想着一道题会做，就一定能做到底。高考考题看重的是区分度。
五问：最后一题是最难的吗？
不一定。高考试卷有一个长度，指题量的答题时间的一个参数：中等程度以上的同学在规定的时间内
能答完试题。语文的最后一题是作文题，作为得分最高的题目，作文也不是最难的，所以遇到作文题不要
惊慌，合理安排作文时间，一个小时左右完全可以打造出一篇优秀的作文。所以对这最后一题一定要写，
并且最大努力把它写好，事实上，作文得高分、得满分的比比皆是。
六问：要不要最后检查一下全卷？
相当一部分同学在规定时间内答不完题 ，但一定要留下 15 分钟左右时间检查全卷 。往往检查一遍 ，能
检查出一个错误，从而多得几分，这也是高考成功的一个重要方法。
七问：有没有一个具体的答题要领？
基本的答题要领是： 慢做会的求全对，稳做中档题一分也不浪费，舍去全不会的。
会做的题慢慢做，保证全对。中档题可以上手，比如理科计算题按步计分，做一步给一步分。中档题 学科 网（北 京）股 份有限 公司
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能做一步就做一步 。舍去全不会指的是难题 ，不是说一看不会就舍去 。认真看认真思考 ，确实不会再舍去 。
终极押题篇
202 5 年高考终极押题卷 （一）【新高考通用卷 】
语 文
（考试时间： 150 分钟 试卷满分： 150 分）
注意事项：
1．答卷前，考生务必将自己的姓名、准考证号填写在答题卡上。
2．回答选择题时 ，选出每小题答案后 ，用铅笔把答题卡上对应题目的答案标号涂黑 。如需改动 ，用橡
皮擦干净后，再选涂其他答案标号。回答非选择题时，将答案写在答题卡上。写在本试卷上无效。
3．考试结束后，将本试卷和答题卡一并交回。
一、现代文阅读（ 35 分）
（一）现代文阅读 Ⅰ（本题共 5 小题， 19 分）
阅读下面的文字，完成下面小题。
材料一：
近年来 ，中国动漫娱乐产业佳作频出 ，《哪吒之魔童降世 》《黑神话 ：悟空 》等作品火爆出圈 ，在国
内外收获高人气与票房佳绩，掀起全球 “中国风 ”热潮。
为何中国动漫娱乐产业接连缔造市场 “神话 ”？
这背后 ，是中华优秀传统文化提供的深层动能 ，是改革推动的文化生产力的持续解放 ，是技术进步与
产业升级的水到渠成，是中国文化软实力崛起的具体体现。
①A 中华优秀传统文化的创新表达，炼成中国动画的 “乾坤圈 ”。
B 中国动画通过对中华优秀传统文化进行创新表达打造出核心竞争力。
《小蝌蚪找妈妈》《牧笛》《大闹天宫》 …… 从 20 世纪五六十年代起，中国动画曾享誉国际。但 到
20 世纪八九十年代，中国动画在一段时间的沉寂后，创作与生产模式已跟不上现代电影产业的发展步伐。
一段时间里 ，人们的视野被外国动画占据 ：“狮子王 ”“ 蜘蛛侠 ”“ 机器猫 ”…… 当花木兰 、功夫熊猫这些中国元
素被迪士尼、梦工厂制作成动画绽放于全球银幕，人们羡慕，人们沉思： 中国动画，何去何从？
随着经济社会的发展、文化产业的升级、文化市场的繁荣，以及科学技术的进步和电影人才的积累 ，
拥有深厚底蕴的中国动画，开始尝试用自己的语言、风格讲好中国故事。
2015 年，国产动画电影《西游记之大圣归来》上映，让中国动画找回了信心。至此，中国动画探索 学科 网（北 京）股 份有限 公司
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出吸引观众的创作路径：从中华优秀传统文化中汲取灵感，对接当代人的情感与价值观。不仅传承，更要
创新表达 。从撷取元素意象到提炼精神内涵 ，再到形成情感共鸣 ，中国动画在传承中创新 ，在探索中进步 ，
在发展中崛起。
②A 数字技术进步与工业化流程升级，驱动动漫文娱发展的 “风火轮 ”。
B 数字技术进步与工业化流程升级推动了动漫文娱产业的发展。
有好故事 ，并不一定就是好作品 。技术不能让一部坏电影变成好电影 ，但能让好电影变成伟大的电影 。
《哪吒之魔童闹海 》中，技术成就了令人叹为观止的视效 ：捕妖队在巨树上列阵 ，一帧画面装了两亿
多个人物，创造了从未有过的场景；在敦煌壁画、青绿山水间无缝切换镜头，重构了动画的蒙太奇语言；
敖丙的每片龙鳞都能自主调整折射率，应用了程序化生成算法 ……
“哪吒 ”的成功，是中国动画技术的突围，其背后是中国智造的托举。
《哪吒之魔童闹海 》拍摄之初 ，找过国外特效团队 ，但交出的样片不理想 。片方只得转向本土 ，最终
调动 138 家中国公司， 4000 多名中国动画人参与。这是中国动画和特效行业多年摸爬滚打积累的人才和技
术力量的集结。
实现创新的，不只是技术研发，以动画为代表的中国电影工业化流程也日益成熟、严谨。
③A 动漫故事激活了文旅路线，衍生品成了抢手货，织就新动能的 “混天绫 ”。
B 动漫故事激活了文旅路线，相关衍生品备受青睐，聚合起产业发展的新动能。
《哪吒之魔童闹海 》热映 ，火的不仅仅是一场电影 ，还赋能了整个文化产业 ，联动衍生品 、文旅 、品
牌等多方面资源，见证了中国经济社会发展的强劲活力。
近年来，动画已深度融入生活，《长安三万里》带动西安研学游，《黑神话：悟空》带火陕北说书 ，
“熊出没 ”IP 打造出主题乐园，文化产业 IP 生态链不断完善，建立健康的产业链，能支撑创作者精益求精 ，
催生更多精品。
从银幕到现实 ，从虚拟到实体 ，中国动画已成为撬动文化产业经济的新动能 ，赋能千行百业 ，构建起
更完整的文化生态体系，让文化自信在产业繁荣中扎根生长。
《哪吒之魔童闹海》， 5 年；《黑神话：悟空》， 7 年；《西游记之大圣归来》， 8 年…… 这些作品
都有漫长的制作周期，是发自内心的热情和精益求精的追求，激发出创作者心无旁骛的工匠精神，支撑着
创作者走到最后，促成了更多的精品问世。
动画形象有着天然的亲和力 ，是跨文化传播实现全球叙事的绝佳载体 ，能与不同文化背景下的观众同
频共振。中国哪吒的故事正在引发全球共鸣。
（摘编自《光明日报》李蕾、牛梦笛《中国动画何以创造了世界第一》） 学科 网（北 京）股 份有限 公司
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材料二：
鲁健（央视主持人）：你好，饺子老师！
饺子（《哪吒 2》导演）：你好。
鲁健：《哪吒》好久不见。太震撼了！这从另外一个赛道把中国的动画电影拉高到了一个新层次。
饺子：我们有好多地方做的还不足，还是要更专心地做好自己的事儿。
鲁健 ：《哪吒之魔童降世 》大家等了整整五年的时间 ，以为 《哪吒之魔童闹海 》可能会快一点 ，但是
没有想到又是一个五年。怎么用了这么长时间？是不是你们对自己太苛刻了，要求太高了？
饺子 ：观众这么期待 《哪吒 2》，机会是不能糟蹋的 ，必须倾尽所有去实现最好的效果 ，呈现给观众 。
鲁健：从这个角度讲，后边的《哪吒 3》也值得更多期待呀！
饺子 ：我跟团队一直在说 ，每一部作品都要当成自己的最后一部作品去创作 ，不要给自己留任何突破
的可能性。你跨过这座高山，下一部再去挑战自己新的极限！
鲁健 ：过去这十年 ，中国动画电影 ，可以用 “雨后春笋般 （出现 ）”形容 ，一批年轻的动画导演 ，一批
处女作，怎么一下子就集中在这十年，出现这种集体爆发？
饺子 ：中国经济发展在这十几年突飞猛进 ，有这环境 ，动画发展是自然而然的 。我们算是遇上一个好
时代，才有这样的机会去做自己喜欢的事儿，可以心无旁骛地投入创作，最终还能够得到回报。
鲁健 ：现在我们虽然有这么好的票房 ，作品又深受观众喜爱 ，但是好像我们在电影工业体系和流程方
面和国际团队还是有差距的，是吗？
饺子 ：是的 。像这一次 ，我们本来寄希望于一些国际制作团队 ，帮我们完成一些比较重点的镜头 。但
外包出去之后，发现效果不理想。哪怕是一些顶级的视效制作团队，但因为是中国项目，难免会对我们带
有傲慢和偏见，而且中国的文化产品让西方团队做，因为文化差异，他们理解上也有难度，所以外包的很
多镜头都不尽如人意 。最后还是收回来 ，交给国内的团队用心去打磨 、去兜底 ，最后才取得了更好的效果 。
由此也发现以往我们仰望的那些大山 ，其实慢慢地一步一步 “死磕 ”也能走出来的 。所有难做的事都是人做出
来的，就慢慢 “死磕 ”吧！
（摘编自《鲁健访谈》之《对话饺子》）
1．下列对材料相关内容的理解和分析，不正确的一项是（ ）（3 分）
A．《哪吒之魔童闹海》火爆出圈既是中国动漫娱乐产业兴盛发展的一个表征，又是中国文化软实力与
综合国力崛起的体现。
B．2015 年在中国动画发展史上是个有转折意义的年份，《西游记之大圣归来》的上映让中国动画找回
了信心，也探索出吸引观众的创作路径。 学科 网（北 京）股 份有限 公司
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C．中国动画技术的突围 ，离不开动画人心无旁骛的 “死磕 ”，虽然中国电影工业化体系和流程与国际团队
还有差距，但也在日益成熟。
D．《哪吒 2》用时 5 年才完成 ，主持人和饺子导演对此看法不一 ，主持人认为是创作团队对自己要求太
高，导演说是不想让观众的期待落空。
2．根据材料内容，下列说法不正确的一项是（ ）（3 分）
A．中国动画发展经历的波折起伏的过程 ，与中国经济发展的过程是同步的 ，二者一荣俱荣 ，一损俱损 ，
这符合社会发展的规律。
B．材料一中两处画波浪线的句子 ，第一处统领全文 ，引出对中国动漫娱乐产业接连缔造市场 “神话 ”原因
的探讨；第二处承上启下，启发读者思考中国动画发展的方向。
C．两则材料都提到国外特效团队制作效果不理想的例子，验证了毛泽东同志倡导过的 “独立自主 ”“ 自力
更生 ”的重要意义。
D．两则材料都谈到中国的经济发展对动画发展的影响，材料一从宏观角度来谈，材料二从创作者个人
体验来谈，结合来看很具说服力。
3．下列各项中，不能辅助诠释材料一中国动画创造世界第一的原因的一项是（ ）（3 分）
A．中国电影评论学会会长饶曙光说 ：“文化自信涵育了创作者 ，也涵育了观众 。人们血液里流淌着传统
文化基因 ，对历史典故和民间传说具有天然的亲近感 。近年来的动画爆款都既传统又现代 ，契合观众心
理及其背后的市场需求。 ”
B．国家广电总局发展研究中心主任祝燕南说 ：“动画是体现前沿科技应用的行业 。动画叙事靠技术呈现 ，
有了技术爆发力，好故事能让观众获得更强大的视听震撼。 ”
C．美国休斯敦的观众伊桑看完《哪吒之魔童闹海》后赞不绝口： “好久没这么激动了。电影让我联想到
父母无私的爱，也仿佛看到自己和朋友的影子。 ”
D．社会科学院城市文化研究所柳所长表示 ：“一批高质量动漫品牌与旅游业发展形成了共生共荣的状态 ，
作为新质生产力的重要来源之一，正在成为推动动画经济增长的新动能。 ”
4．请从材料一中任选一个角度 ，设计材料二主持人和嘉宾访谈的一组问答 ，使其能自然衔接到材料二的结
尾。 （4 分）
5．材料一中标序号的 ①②③ 三处横线上填写的句子有 A、B 两种表达形式，你认为哪种更好？为什么？
请结合材料加以分析。 （6 分）
（二）现代文阅读 Ⅱ（本题共 4 小题， 16 分） 学科 网（北 京）股 份有限 公司
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阅读下面的文字，完成下面小题。
隐秘之境
冷清秋
怎么办呢？
时间越久，她越爱上了这里，
置身于这样绿意重重的大山的怀抱，仿若无边绿境里的一朵小花。
无数的清晨和傍晚 ，她赤着脚静静地坐在湖边的草地上 ，让那柔软的凉意丝丝缕缕由下至上沁入心脾 ，
进入自己瘦弱的身体，重重叠叠的愉悦层层堆积，最终，在她的唇角开出一朵漾着快意的花儿。
现在 ，她每天沿着湖边漫无目的地游走 ，徜徉 ，无所事事的看绿色的湖水 ，看湖中倒着生长的树 ，看
树中间安然游弋的鱼儿，看船只远去，湖面留下无声的波。
尽管妈妈一再说：不要乱跑，不可以乱跑哦！尽管妈妈总是不由分说拉她回去。
大概是倦了，大概是看她并未惹出什么祸端，渐渐地，妈妈竟也由了她顺了她。
当初公司安排施工队进山装修民宿 ，妈妈是拒绝的 。她说 ，不行 ！亭亭怎么办？可对方说 ：有什么呀 ，
带着就是了！薪酬可是市里的一倍半哦！望着对方一脸的肯定，妈妈把后面的话咽了下去。对方仍然很知
心地讲：公司知道你情况特殊，每天你只要备完料，完成你的检验任务，余下的时间自由支配就是。
可妈妈还是一脸忧伤。妈妈问她：怎么办呢？亭亭，我们去不去？
可她并不理会妈妈的急躁，完全是不说去也不说不去的样子。
但妈妈还是决定去了 。当然要去了 ，一倍半的薪水 ，可以做很多事了 。更重要的是对方说 ：换个环境
啊，换个心境嘛！
可妈妈还是拽着衣袖告诫说：听说附近有野猪呢，不要乱走！
妈妈当然不会知道她一住下来就心生喜欢了 。她喜欢这里弯弯曲曲的山路 ，喜欢远处苹果果园里那一
片闪耀的红，喜欢羊群从山涧攀爬上来，拦住道路咩咩地叫，缓慢地走，一羊脸诧异地望她，喜欢圆滚滚
的大南瓜从墙里面生长出来，把自己悬挂在石头墙上，像一只黄澄澄的南瓜灯笼。尽管妈妈一再重申不让
她乱跑，尽管妈妈头疼她的乱跑，甚至一度生出想要把她锁在房间里的决心，可她怎么能放弃自己的喜欢
呢。
一天晚上望着她鞋子衣服上沾染的泥土和渗着血的膝盖 ，妈妈终是怒了 ，问她有什么好呢？不就是大
山里的一些树和水嘛！ 她原本想反驳来着，说很好啊，却终是没说 。她当然也没说在下午遇见的小松鼠，
草丛里的小刺猬，还想说还有土黄色的小兔子，一脸警惕地趴在草地上露半拉脑袋愣愣怔怔望她，然后纵
身一跃，蹿进草丛消失不见了。 学科 网（北 京）股 份有限 公司
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这些，她终是没有说出来 。她没说她遇见的树林里的管理员大爷。
管理员大爷脚边跟着一只嘎嘎叫的鹅，白色的大鹅，戴着橘红的鹅帽子，看见她扎着脖子就要啄她 。
而大爷，只是一伸手抓住鹅脖子就把大鹅带走了。她很想像大人那样对管理员道声感谢或者别的什么。可
没有。都没有。
她和妈妈睡觉的屋子进猫了 ，是一只小橘猫 ，蹑手蹑脚地叼走了她的火腿肠 。后来 ，那猫又来 ，居然
无所畏惧大摇大摆的，吃她安排好的酸奶和香肠，还满屋子巡逻。她则蜷在被卷中。
一声不响地看 ，一声不吭地听 ，看猫进来 ，看猫出去 。有次那猫吃喝完毕 ，远远站在月光下冲她轻轻
地喵 —— 喵—— 呜—— 激动不已的她捏着嗓子去回应，那猫收到信号， “蹭”就跳上了床。自此，那猫夜
夜来 ，天亮走 。吃饱喝足便卧在枕边给她打呼噜 ，那呼噜好听极了 ，所以即便是妈妈一再反对 ，坚决反对 ，
终是没拗过她。
她哪怕是自己不吃饭，也要操心猫的食物和水，并以更柔软更温和的态度待它。
自从父母分开后 ，她已经很久很久没有这么快乐这么幸福了 。她死死拽着自己沿着一条狭窄寒冷的路
朝更黑暗的地方走，不管不顾的，只切切实实厚铸所有的疼痛和难过，但现在，坚冰慢慢融化和模糊。
她和妈妈安安静静地坐在大大的落地窗前吃牛角包 ，简简单单的牛角包 ，搭配妈妈亲手制作的苏子酱
和姜片茶，而小猫逐来跳去研究一朵新摘的喇叭花。雨后初晴，苏铁的叶片在光照下闪闪发光，花桐木肥
大的叶片托着炸开的宽豆荚，两棵女贞一大一小挨在草地尽头的夕照下，她静静地望着一言不发，妈妈也
静静地望着没有说话。
暮色降临时 ，妈妈破天荒拽她到湖边踩水 。那时节 ，风遥遥从对岸过来了 ，一波一波凉意跟着水波涌
过来了，她悄悄瞥了一眼妈妈，打开手机，输入搜索，点开来，网上是这样介绍的：洛宁西子湖，美丽的
传说，真实的画卷，心灵的净土，梦中的天堂。
她忍不住笑了。
妈妈却落泪了 。但妈妈快速扬了下袖子赶走眼泪 ，带着湿漉漉的讨好凑过来说 ：话费早给你充了 ，电
源也很充足呢。说着，便伸手去捋她被风吹乱头发。她犹豫要不要躲开的瞬间，妈妈的温热就真实地落在
脸颊和耳上了。
清晰地颤抖中，她的眼泪没来由就滚落了。
她不喜欢妈妈也不喜欢爸爸 。他们大人们的事情 ，她一概不想插话 。可现在 ，不知怎的 ，她听见自己
用一个陌生的声音颤颤巍巍喊了一声：妈妈！
地动山摇电光石火的，妈妈就扑过来把她死死抱在了怀里了。
现在，一切都静止了。就像是过去了一万年。 学科 网（北 京）股 份有限 公司
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她一动也不想动，只是想 ——
好温暖啊。
（有删改）
6．下列对文本相关内容的理解，正确的一项是（ ）（3 分）
A．当接到新的工作任务时，妈妈的态度由开始的拒绝到后来的接受，是因为薪酬是市里的一倍半。
B．出于安全考虑，妈妈一再重申不让她到处乱跑，并且有时不由分说地拉她回去，将她锁在房间。
C．大山里的景、物、人，是妈妈刚开始时未察觉到的美好，也是她从黑暗的生活中挣脱出来的因素。
D．从 “话费早给你充了，电源也很充足 ”中可以看出她是一个沉迷于手机、和别人缺乏交流的人。
7．下列对文本艺术特色的分析鉴赏，不正确的一项是（ ）（3 分）
A．文中画线句子写出她自我封闭的特点 ，与后文喊出 “妈妈 ”的举动形成对比 ，这样使小说具有艺术张力 。
B．本文描写小橘猫时主要采用比拟的手法，通过生动传神的动作刻画，写出了小猫胆大又不失可爱的
特点。
C．文章后四段篇幅不长，却单独成段，这样安排使读者能够更深刻地感受到她们内心的变化和情感的
爆发。
D．和《哦，香雪》一样，本文也属于散文化小说，淡化了情节和主题，强化意境的营造，美化语言的
表达。
8．本文开头前 5 段写出她内心的愉悦和平和，这种心理是怎样表现出来的？请简要分析。 （4 分）
9．读书小组要为此文写一则文学短评 ，你是小组成员之一 ，请围绕关键词 “隐秘之境 ·成长 ”，写出你的短评
思路。 （6 分）
二、古诗文阅读（ 35 分）
（一）文言文阅读（本题共 5 小题， 20 分）
阅读下面的文言文，完成下面小题。
材料一：
世之为丘垄也 ，其高大若山 ，其树之若林 ，其设阙庭 、为宫室 、造宾阼也若都邑 。以此观世示富则可
矣，以此为死则不可也。
君之不 令．民，父之不孝子，兄之不悌弟，皆乡里之所釜䰛 ①．
者而逐之。 惮耕稼采薪之劳不肯官人事而
祈美衣侈食之乐智巧穷屈 ，无以为之，于是乎聚群多之徒，以深山广泽林薮，扑击遏夺，又视名丘大墓葬
之厚者 ，求舍便居 ，以微抇 ②之，日夜不休 ，必得所利 ，相与分之 。夫有所爱所重 ，而令奸邪 、盗贼 、寇乱
之人卒必辱之，此孝子、忠臣、亲父、交友之大事。 学科 网（北 京）股 份有限 公司
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尧葬于谷林 ，通树之 ；舜葬于纪市 ，不变其肆 ；禹葬于会稽 ，不变人徒 。是故先王以俭节葬死也 ，非
爱其费也 ，非恶其劳也 ，以为死者虑也 。先王之所恶 ，惟死者之辱也 。发则必辱 ，俭则不发 。故先王之葬 ，
必俭。此之谓爱人。故孝子、忠臣、亲父、交友不可不察于此也。
（选自《吕氏春秋 ·安死》）
材料二
吴王阖闾 ，违礼厚葬 ，十有余年 ，越人发之 。秦始皇帝葬于骊山之阿 ，下锢三泉 ，上崇山坟 ，其高五
十余丈，周回五里有余；石椁为游棺，人膏为灯烛，水银为江海，黄金为凫雁。珍宝之臧，机械之变，棺
椁之丽，宫馆之盛，不可胜原。又多杀宫人，生埋工匠，计以万数。天下苦其役而反之，骊山之作未成，
而周章 ③百万之师至其下矣 。项籍燔其宫室营宇 ，往者咸见发掘 。其后牧儿亡羊 ，羊入其凿 ，牧者持火照求
羊，失火烧其臧椁。自古至今，葬未有盛如始皇者也，数年之间，外被项籍之灾，内 离．牧竖之祸，岂不哀
哉！
是故德弥厚者葬弥薄 ，知愈深者葬愈微 。无德寡知 ，其葬愈厚 ，丘陇弥高 ，宫庙甚丽 ，发掘必速 。由
是观之，明暗之效，葬之吉凶，昭然可见矣。
（选自刘向《谏成帝营陵寝疏》）
材料三
朕闻 ：“死者终也 ，欲物之反于真也 ；葬者藏也 ，欲人之不得见也 。”上古垂风 ，未闻于封树 ；后世贻
范，始备于棺椁。 讥僭侈者非爱其厚费，美俭薄者实贵于无危 。
是以唐尧圣帝也 ，毂林有通树之说 ；秦穆明君也 ，橐泉无丘龙之处 。洎乎阖闾违礼 ，珠玉为凫雁 ；始
皇无度，水银如江海。因多藏以 速．祸。
朕居四海之尊，承百王之弊；未明求化，中宵载惕。虽送往之典，详诸仪制；失礼之禁，著在刑书 。
而勋戚之家，或流遁于习俗；闾阎之内，或侈靡而伤风。以厚葬为奉终，高坟为行孝。遂使衣衾棺椁，极
雕刻之华；刍灵明器，穷金玉之费。富者越法度以相高，贫者破资产以不 逮．。徒伤教义，无益泉壤。
为害既深，宜有惩革。其王公以下，爰及黎庶，送终之具有乖令式者，明加检察，随状科罪。
（选自唐太宗《戒厚葬诏》）
[注]①釜䰛：炊具。 ②抇：挖掘。 ③周章：秦末农民起义军将领，即周文。
10 ．材料一中画波浪线的部分有三处需要断句，请用铅笔将答题卡上相应位置的答案标号涂黑。 （3 分）
惮耕 A 稼采 B 薪之劳 C 不肯官 D 人事 E 而祈美衣 F 侈食之乐 G 智巧穷屈
11 ．下列对材料中加点的词语及相关内容的解说，不正确的一项是（ ）（3 分）
A．令，美好，与《孔雀东南飞》中 “年始十八九，便言多令才 ”的“令”意义相同。 学科 网（北 京）股 份有限 公司
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B．离，遭受，与《屈原列传》中 “‘ 离骚 ’者，犹离忧也 ”的“离”意义相同。
C．速，迅速，与《六国论》中 “至丹以荆卿为计，始速祸焉 ”的“速”意义不同。
D．逮，及，与《陈情表》中 “逮奉圣朝，沐浴清化 ”的“逮”意义相同。
12 ．下列对材料有关内容的概述，不正确的一项是（ ）（3 分）
A．材料一首先点明观点，建造高大的坟墓，向世人夸耀财富是可以的，这样安葬死者却是不行的。
B．材料二用铺排极力描述秦始皇营建陵墓之奢华，这种写法与《阿房宫赋》的手法一致。
C．材料二、三都提到秦始皇的例子，但是一详一略，这跟两篇文章写作目的不同有关。
D．三则材料皆告诉我们如果将坟茔埋葬得越丰厚，就越容易被挖掘，任何人都要俭葬。
13 ．把材料中画横线的句子翻译成现代汉语。 （8 分）
（1）是故德弥厚者葬弥薄，知愈深者葬愈微。
（2）讥僭侈者非爱其厚费，美俭薄者实贵于无危。
14 ．三则材料都提到厚葬的危害，侧重点各有不同，请结合文本进行分析。 （3 分）
（二）古代诗歌阅读（本题共 2 小题， 9 分）
阅读下面这首宋诗，完成下面小题。
湖上初春偶作
林逋 【注】
梅花开尽腊亦尽，春暖便如寒食天。
气色半归湖岸柳，人家多上郭门船。
文禽相并映短草，翠潋欲生浮嫩烟。
几处酒旗山影下，细风时已弄繁弦。
【注】林逋，字君复，北宋初年诗人，曾在杭州西湖隐居二十多年。
15 ．下列对这首诗的理解和赏析，不正确的一项是（ ）（3 分）
A．首联紧扣 “初春 ”二字，先以 “梅花开尽 ”起笔点出时令，后又以 “暖”字写了诗人感受。
B．第三句紧承第二句展开，年迈体衰的诗人在暖春看到岸柳飘拂，忽感自己气色半归。
C．颈联的首字，有人主张应为 “游”，有人主张应为 “文”，但依格律来看， “文”比“游”妙。
D．本诗是诗人即兴而作，四联都描摹了眼中之景，处处透露着闲情逸致，诗风清新自然。
16 ．诗中的第四句和第八句都描写了 “烟火气 ”，请分析它们在具体写法上的区别。 （6 分）学科 网（北 京）股 份有限 公司
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（三） 名篇名句默写（本题共 1 小题， 6 分）
17 ．补写出下列句子中的空缺部分。 （6 分）
（1）成都西郊的度假村开业了 ，在长满花草的道路深处 ，一家名为 “浣花溪 ”的民宿使用杜甫 《客至 》
中“ ， ”两句诗，悬挂门口，以示对游客的欢迎。
（2）小明最近在阅读《周易》，其中有一个观点强调宇宙万物处于不断变化之中，这不禁让他想到
苏轼在《赤壁赋》中的 “ ， ”两句话，也表达了同样的意思。
（3）唐人写诗用到的数字中， “三”是出现频率较高的，或描述确切数量，或表示数量多、程度深，
或表示序数，正如唐诗 “ ， ”两句。
三、语言文字运用（ 20 分）
阅读下面的文字，完成下面小题。
近来 ，一场由 “新中式 ”穿搭引领的时尚风潮悄然涌动 ，在街头巷尾 ，在旅游景点 ，总能见到其身影 。
“新中式 ”新在哪里？ “新中式 ”服装并没有全套传统服饰那么隆重华丽 ，主打既保留传统特色 ，（ 甲 ）。
比如一些 “新中式 ”服装中的花鸟、山水暗纹融入得十分巧妙，这些精致的图案在阳光下若隐若现，低调
典雅，既能穿上出游，也能穿着上班。从写意水墨，到梅兰竹菊；从棉麻丝绸到纤维毛料， “新中式 ”服
装设计元素 A ，将深厚的中华文化内涵融入 “日用而不觉 ”的日常穿搭中。
以马面裙这一极具辩识度的服装为例 ，其前后共有四个裙门 ，系带围合腰部 ，视觉上给人一种 “飞流
直下 ”的美感 。马面裙的色彩运用通常十分大胆鲜艳 ，以蓝色 、红色居多 ，从视觉上给人一种明艳的感觉 。
在马面裙的裙门和边缘配的各种纹饰图案更是它的重点。身着马面裙，婉若翩跹仙子。马面裙并不只是一
件新春战袍，也不只是一条漂亮的裙子，它从历史的衣橱里走出来，成为街上的一道靓丽风景，让传统与
时尚相互碰撞、相互交溶，焕发出了属于这个时代的光彩。
马面裙最初的基本裙式出现在宋代 ，成熟的马面裙形成于明代 ，是明代女服最具特色的服饰之一 ，马
面裙一直延续至 20 世纪初期 。可以说 ，①现在的马面裙经历了漫长的发展和深厚的积淀 ，②人们在保留马
面裙优雅挺拔造型的同时， ③对材质和设计进行了改良， ④更加适合现代人的日常穿着， ⑤同时也可机洗
降低洗护成本。
时下 ，（ 乙 ），犹如星火燎原 ，已经从服饰延伸到了家居 、美妆 、餐饮等多个领域 ，不断挖掘中华
优秀传统文化富矿 ，创新表达东方美学 。比如同仁堂结合传统养生与新消费需求 ， B ，推出枸杞拿铁 、罗
汉果美式咖啡、娇颜陈皮五红汤等饮品，不仅吸引了年轻消费者，也让老字号潮起来。 学科 网（北 京）股 份有限 公司
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18 ．请在文中画横线处填入恰当的成语。 （4 分）
19 ．文中标序号的部分有两处表述不当，请指出其序号并做修改，使语言准确流畅，逻辑严密，不得改变
原意。 （4 分）
20 ．请在文中括号内补写恰当的语句 ，使整段文字语意完整连贯 ，内容贴切 ，逻辑严密 ，每处不超过 10 个
字。 （4 分）
21 ．文中第二段有三处错别字，请指出并加以改正。 （3 分）
22 ．一位热爱马面裙的视频博主，想在自己的视频号上为马面裙配文，宣传马面裙之美，上句已给出，请
你结合材料替这位博主补写出下句，要求字数相等，内容相关，结构相同。 （5 分）
上句：时尚符号，惊艳中外，演绎东方韵味。
下句：
四、作文（ 60 分）
23 ．阅读下面的材料，根据要求写作。 （60 分）
古罗马帝国政治家、军事家、哲学家奥勒留在《沉思录》中说： “我们听到的一切都是一个观点，不
是事实；我们看见的一切都是一个视角，不是真相。 ”
以上材料引发了你怎样的联想和思考？请写一篇文章。
要求 ：选准角度 ，确定立意 ，明确文体 ，自拟标题 ；不要套作 ，不得抄袭 ；不得泄露个人信息 ；不少
于 800 字。
20 25 年高考终极押题卷 （一）（新高考通用 卷）
语文 ·全解全析
1 2 3 6 7 10 11 12 15
D A C C D CEG C D B
1．D 【解析】 本题考查学生理解文章内容，筛选并整合文中信息的能力。 D. “主持人和饺子导演对此看
法不一 ”概括有误，主持人的话只是提出的猜想，并不是确定的看法，导演是顺着主持人的话进一步解释
为什么团队对自己要求高，是因为不想让观众的期待落空。故选 D。
2．A【解析 】本题考查学生分析概括作者在文中的观点态度的能力 。A. “与中国经济发展的过程是同步的 ，
二者一荣俱荣，一损俱损 ”推断有误，材料中并未提及二者是兴衰同步关系的相关信息。故选 A。学科 网（北 京）股 份有限 公司
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3．C 【解析】本题考查学生分析论点、论据和论证方法的能力。
C. 该项是中国动画的国际影响，不能辅助诠释中国动画创造世界第一的原因。
A. 能辅助诠释 “中华优秀传统文化的创新表达，成就了中国动画 ”这一原因。
B. 能辅助诠释 “数字技术进步与工业化流程升级，促进动漫文娱发展 ”这一原因。
D. 能辅助诠释 “动漫故事激活了文旅路线 .衍生品成了抢手货，聚合新动能 ”这一原因。
故选 C。
4．（示例）鲁健：那么我们怎么才能让我们的电影更好地走向世界呢？
饺子：真正让中国电影走向世界的话，我们要继续以心无旁骛的工匠精神去精益求精地追求，利用好动画
这一绝佳载体，实现跨文化传播，实现全球叙事，吸引不同文化背景下的观众与我们同频共振，做出更多
能打动全世界观众的精品动画电影。 【解析】本题考查学生分析文章思路结构、情境补写的能力。
鲁健：材料一从中华优秀传统文化的创新表达、数字技术进步与工业化流程升级、动漫故事激活文旅及衍
生品等角度，阐述了中国动画取得成功的原因，最后强调动画是跨文化传播的绝佳载体，能实现全球叙事
引发全球共鸣。这组问答中主持人提出 “怎么才能让我们的电影更好地走向世界 ”，直接关联到材料一结
尾中国动画在跨文化传播方面的内容，实现了材料一与材料二的自然衔接 。
饺子：嘉宾饺子作为《哪吒》系列动画电影的导演，其回答围绕工匠精神、动画的跨文化传播优势以及制
作精品动画电影等方面展开，既呼应了材料一中提到的创作者工匠精神促成精品问世，以及动画适合跨文
化传播的观点 ，又符合饺子作为动画电影创作者的身份和他在材料二中所展现出的对作品精益求精的态度 ，
使回答内容合理且具有说服力。
5．示例一 ：A 组更好 。①三个句子都使用了比喻 ，生动形象 。将中华优秀传统文化的创新表达比作中国动
画的 “乾坤圈 ”，将数字技术进步与工业化流程升级比作动漫文娱发展的 “风火轮 ”，将动漫故事及衍生品形成
的新动能比作 “混天绫 ”，这些比喻意象鲜明，让抽象概念变得具体可感且与《哪吒》的动画内容相契合。
②句子富有文化底蕴 。“乾坤圈 ”“ 风火轮 ”“ 混天绫 ”这些神话道具本身承载着丰富的中国传统文化内涵 ，能唤
起读者对传统神话的情感记忆和共鸣，为语句增添了浓郁的文化气息，使表达更具感染力。 ③借助哪吒这
一人物的三件法宝串联文章结构，使文章整体性更为统一连贯，三方面内容层层递进，更有逻辑性。
示例二： B 组更好。 ①句子观点明确。直接阐述中华优秀传统文化的创新表达与中国动画核心竞争力、数
字技术进步及工业化流程升级与动漫文娱产业发展、动漫故事及衍生品与产业新动能之间的关系，观点清
晰直白，更利于读者快速抓住关键信息，明确认知动漫娱乐产业发展的实际推动因素。 ②语言简洁平实，
严谨准确 。以较为平实 、客观的语言进行表述 ，避免了因比喻可能带来的理解偏差 ，更利于精准传达信息 、
强调逻辑关系，更能清晰呈现各要素之间的因果关联。 【解析】本题考查学生分析语言特色，探究文本中 学科 网（北 京）股 份有限 公司
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的某些问题的能力。
示例一：认为 A 组更好。
①生动形象的表达： A 组表达运用了比喻的修辞手法，将中华优秀传统文化的创新表达等抽象概念分别比
作 “乾坤圈 ”“风火轮 ”“混天绫 ”，这些比喻与 《哪吒 》的动画内容紧密契合 ，使原本抽象的概念变得
具体可感，让读者能够更直观地理解其对于中国动画及动漫文娱发展的作用和意义。
②文化底蕴的赋予 ：“乾坤圈 ”“风火轮 ”“混天绫 ” 是哪吒的三件法宝 ，本身承载着丰富的中国传统文
化内涵。使用这些具有文化象征意义的比喻，能够唤起读者对传统神话的情感记忆和共鸣，为语句增添了
浓郁的文化气息，使表达更具感染力，也体现了中国动画与传统文化的紧密联系。
③结构上的优势 ：借助哪吒的这三件法宝来串联文章结构 ，使得文章在内容上更加统一连贯 。同时 ，从 “乾
坤圈 ” 到 “风火轮 ” 再到 “混天绫 ”，也体现了内容上的层层递进，具有更强的逻辑性，有助于读者
理解中国动画及动漫文娱产业发展的不同层面和阶段。
示例二：认为 B 组更好。
①观点明确直接： B 组表达直接阐述了中华优秀传统文化的创新表达等与中国动画核心竞争力、动漫文娱
产业发展、产业新动能之间的关系，没有使用比喻等修辞手法，而是以直白的方式呈现观点，让读者能够
快速抓住关键信息，明确了解动漫娱乐产业发展的实际推动因素，避免了读者在理解比喻含义上可能产生
的困惑。
②语言风格优势： B 组语言简洁平实、严谨准确，以较为客观的方式进行表述，更注重精准传达信息和强
调逻辑关系。这种语言风格能够清晰地呈现各要素之间的因果关联，使读者能够更理性地分析和理解其中
的逻辑，尤其适合那些更关注实际内容和逻辑关系，而不太注重语言文学性的读者。
6．C【解析】本题考查学生对文本相关内容的理解和分析能力。
A. “是因为薪酬是市里的一倍半 ”错。由原文 “但妈妈还是决定去了 。当然要去了 ，一倍半的薪水 ，可以做
很多事了。更重要的是对方说：换个环境啊，换个心境嘛！ ”可知，妈妈最后能够接受新的工作任务，除
了薪酬高之外，女儿可以带着以及 “换个环境，换种心境 ”也是重要的原因。
B. “将她锁在房间 ”错。原文为 “甚至一度生出想要把她锁在房间里的决心 ”，说妈妈曾一度有这种想法 ，
但并未实施。
D. “看出她是一个沉迷于手机 、和别人缺乏交流的人 ”错。文中提到 “话费早给你充了 ，电源也很充足 ”是
妈妈关心她的表现，并不能说明她沉迷于手机或缺乏交流。
故选 C。
7．D 【解析】本题考查学生对文本艺术特色的分析鉴赏能力。 学科 网（北 京）股 份有限 公司
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D. “淡化了情节和主题 ”错。本文虽然也有一定的意境描写 ，但情节线索清晰 ，主题明确 ，尤其是在母女情
感的变化方面，主题并未被淡化，而是通过细腻的描写和情感的铺垫，逐步深化了主题。
故选 D。
8．①直接抒写 （内心独白 ）：“爱上了这里 ”“ 重重叠叠的愉悦层层堆积 ”直接写出她的愉悦之情 ；②环境烘
托：通过 “绿意重重的大山 ”“ 无边绿境 ”等自然景象的描写 ，营造了宁静的意境 ，暗示了她内心的平和与满足 ；
③感官体验：通过触觉和视觉，细腻地刻画出她与自然的融合，这正是她内心惬意的表现。 ④动作描写 ：
通过 “赤着脚静静坐着 ”“ 游走 ”“ 徜徉 ”等写出她的悠闲。 【解析】本题考查学生分析描写人物心理的能力。
①本文开头通过直接抒写（内心独白），如 “爱上了这里 ”和“重重叠叠的愉悦层层堆积 ”，直接表达了
主人公对这片自然环境的深厚情感和内心的满足感。这种直接的表达方式让读者能够清晰地感受到主人公
内心的喜悦和宁静。
②作者通过环境烘托的手法，描绘了 “绿意重重的大山 ”和“无边绿境 ”等自然景象，营造出一种宁静而
美丽的意境。这种环境的描写不仅为故事提供了背景，也暗示了主人公内心的平和与满足，使读者能够通
过环境的描写感受到主人公的情感状态。
③作者还通过感官体验的描写，细腻地刻画了主人公与自然的融合。例如，主人公赤着脚静静地坐在湖边
的草地上，感受着 “柔软的凉意丝丝缕缕由下至上沁入心脾 ”，这种触觉的描写让读者能够身临其境地感
受到主人公的惬意和内心的宁静。同时，视觉上的描写，如 “绿色的湖水 ”“ 湖中倒着生长的树 ”等，也
进一步增强了这种情感的表达。
④作者通过动作描写，如 “赤着脚静静坐着 ”“ 游走 ”“ 徜徉 ”等，表现了主人公的悠闲与自在。这些动
作不仅展示了主人公的日常生活状态，也进一步传达了她内心的愉悦与平和。通过这些细腻的描写，作者
成功地塑造了一个在自然环境中找到内心宁静与满足的主人公形象，使读者能够深刻地感受到她的情感变
化和心理状态。
9．①“ 隐秘之境 ”：既指她和妈妈所居住的宁静 、美丽 ，远离城市喧嚣的深山 ；也指她封闭的 、不愿意和外
界交流的内心世界。 ②“ 成长 ”：她在深山中得到了治愈，获得了成长；妈妈也由最初的不理解女儿到落泪
拥抱，象征着两人的隔阂被打破，亲子关系得到修复。 ③环境的构成（景、物、人等）对心灵治愈和成长
具有重要意义。 【解析】本题考查学生拟写短评思路的能力。
①“ 隐秘之境 ”不仅指代了她和妈妈所居住的深山环境，这片远离城市喧嚣的宁静之地，也象征了她内心
封闭的世界。在经历了父母分离后，她将自己封闭在内心深处，拒绝与外界交流。而大山中的自然景致、
动物以及人，逐渐成为她与外界沟通的桥梁，帮助她打开心扉。
②“ 成长 ”是她在这片隐秘之境中获得的核心主题。她通过与自然的接触，逐渐治愈了内心的伤痛，重新 学科 网（北 京）股 份有限 公司
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找到了生活的快乐与意义。与此同时，妈妈的态度也发生了变化，从最初的不理解、担忧，到后来主动带
她到湖边踩水 ，并在情感上与她重新建立了联系 。母女之间的隔阂被打破 ，象征着亲子关系的修复与成长 。
③环境的构成对她的心灵治愈和成长起到了至关重要的作用 。大山中的绿意 、湖水 、动物以及淳朴的村民 ，
构成了一个充满生机的世界，帮助她从黑暗的生活中挣脱出来，重新感受到温暖与希望。这种环境不仅治
愈了她的内心，也促使她在情感上逐渐成熟，最终与妈妈达成了和解。
10 ．CEG 【解析】本题考查学生文言文断句的能力。
句意：他们害怕耕种、打柴之苦，不肯从事各种劳役，却追求享受锦衣玉食之乐；当智谋巧诈用尽。
“耕稼采薪之劳 ”是定语和中心语，整个作 “惮”的宾语， C 处断开；
“官人事 ”也是动宾结构， E 处断开；
“美衣侈食之乐 ”是定语和中心语，整个作 “祈”的宾语， G 处断开。
故选 CEG 。
11 ．C 【解析】本题考查学生了解并掌握常见的文言一词多义的能力。
A. 正确。句意：国君的不臣服的百姓。 /年仅十九岁，就已经能说会道且具备出众的才华。
B. 正确。句意：内部遭受了牧童失火的祸患。 /“离骚 ”就是遭遇忧愁。
C. 错误。都是 “招致 ”。句意：这些人都是因为在墓里埋藏了大量的财物而招致了灾祸。 /到燕国太子丹把
用荆轲刺杀秦王作为计策，才招致了祸患。
D. 正确。句意：贫穷之家倾家荡产而唯恐不及。 /等到了圣朝，沐浴着清明的教化。
故选 C。
12 ．D 【解析】本题考查学生理解文章内容的能力。
D. “任何人都要俭葬 ”说法过于绝对 ，三则材料主要是针对当时存在的厚葬现象进行论述 ，并非要求所有人
都必须俭葬。
故选 D。
13 ．（ 1）所以德行越笃厚的人埋葬越微薄，智慧越深的人埋葬越简约。
（2）谴责奢侈浪费的人 ，并不是舍不得钱财 ；赞美节俭薄葬的人 ，实际上看重这样做可以避免危险 。【解
析】本题考查学生理解并翻译文言文句子的能力。
（1）“是故 ”，所以； “弥”，越； “知”，通 “智”；“微”，简约。
（2）“讥僭侈 ”，谴责奢侈浪费； “爱”，吝惜，舍不得； “美”，赞美； “贵”，看重。
14 ．①材料一主要侧重于社会风气与道德层面的危害。厚葬不仅助长了奢侈之风，还可能成为奸邪、盗贼
的目标，导致死者受到侮辱，这是孝子、忠臣、亲父、交友所不愿见到的。 ②材料二则更侧重于对社会安 学科 网（北 京）股 份有限 公司
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定的负面影响。通过吴王阖闾和秦始皇的厚葬实例，展示了厚葬可能招致的灾祸，也指出厚葬可能加剧社
会矛盾，如秦始皇因厚葬而引发的民众反抗。 ③材料三则侧重于教化层面，社会风气的危害。当时社会存
在的厚葬风气及其危害，如耗费资财、伤风败俗等，并表明皇帝将对此进行整治的决心。 【解析】 本题考
查学生筛选并概括文中信息的能力。
①材料一 ：从“君之不令民 ，父之不孝子 ，兄之不悌弟 …… 于是乎聚群多之徒 …… 求舍便居 ，以微抇之 ，日夜
不休 ，必得所利 ，相与分之 ”以及 “夫有所爱所重 ，而令奸邪 、盗贼 、寇乱之人卒必辱之 ，此孝子 、忠臣 、亲
父、交友之大事 ”等内容可知 ，厚葬会引发奸邪 、盗贼挖掘坟墓 ，导致死者受辱 ，这是从社会风气和道德层
面来阐述厚葬的危害，强调厚葬不利于社会风气和道德规范。
②材料二 ：通过 “天下苦其役而反之 ，骊山之作未成 ，而周章百万之师至其下矣 ”“ 外被项籍之灾 ，内离牧竖
之祸 ”等内容 ，以吴王阖闾和秦始皇厚葬为例 ，指出厚葬不仅会招来灾祸 ，还因耗费大量人力物力导致民众
反抗，这是从社会安定和国家稳定的角度说明厚葬的负面影响，突出厚葬对社会秩序和国家统治的危害。
③材料三 ：从“而勋戚之家 ，或流遁于习俗 ；闾阎之内 ，或侈靡而伤风 。以厚葬为奉终 ，高坟为行孝 。遂使
衣衾棺椁，极雕刻之华；刍灵明器，穷金玉之费。富者越法度以相高，贫者破资产以不逮。徒伤教义，无
益泉壤 ”等内容可以看出 ，主要阐述了厚葬在社会中形成的不良风气 ，如耗费资财 、伤风败俗等 ，还提到皇
帝将对此进行整治，是从社会教化和风气引导的层面强调厚葬的危害。
参考译文：
材料一：
世人建造坟墓 ，高大如山 ，坟墓上种树 ，茂密如林 ，墓地修建墓阙 、庭院 ，建筑宫室 ，建造东西石阶 ，
像都邑一样。用这些向世人夸耀财富，那是可以的，但是用这些安葬死者却不行。
国君的不臣服的百姓 ，父亲的不孝之子 ，兄长的忤逆之弟 ，他们都是被乡里一致驱遂的人 。他们害怕
耕种、打柴之苦，不肯从事各种劳役，却追求享受锦衣玉食之乐；当智谋巧诈用尽，仍无法得到时，于是
就聚集起很多人，凭借深山、大泽、树林，拦路打劫，又探察葬器丰厚的大墓，想办法住到坟墓附近便于
盗墓的住所，暗中挖掘，日夜不止，一定要获得其中的器物，一起瓜分。如果所疼爱、所尊重的人，死后
却要遭到恶人、盗贼、匪寇的凌辱，这是孝子、忠臣、慈父、挚友当忧虑的大事。
尧葬在谷林 ，墓上处处种树 ；舜葬在纪市 ，市上的作坊 、店铺没有任何变动 ；禹葬在会稽 ，不烦扰众
人。由此看来，先王以节俭的原则安葬死者，不是吝啬钱财，也不是忧虑耗费人力，完全是为死者考虑。
先王所忧虑的 ，是担心死者受辱 。坟墓如果被盗取 ，死者肯定会受到凌辱 ；如果俭葬 ，墓穴就不会被盗掘 。
所以 ，先王安葬死者 ，一定要做到节俭 。这才叫做爱人 。所以 ，孝子 、忠臣 、慈父 、挚友对此不可不明察 。
材料二： 学科 网（北 京）股 份有限 公司
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吴王阖闾 ，违背周礼进行厚葬 。十多年后 ，越国人挖开了他的坟墓 。秦始皇帝埋葬在骊山的山顶 ，下
面用三泉禁锢，上面堆上高高的坟土，那高度达五十多丈，周边环绕有五里多；他用石椁作游棺，人膏作
灯烛，用水银做江海，用黄金做水面的大雁。奇珍异宝的储藏，精巧器械的变化多端，棺椁的华丽，宫殿
馆舍的盛大辉煌，简直无法完全探究清楚。（秦始皇帝）还大量杀戮宫女，将工匠们活生生地埋葬，估计
人数以万来计算。天下被他的劳役所困苦而反抗他，骊山的墓地修建还没有完成，而周文的百万军队已经
到了骊山脚下了。项羽焚烧了他的宫殿和新修的庙宇，前往的人都进行发掘。之后，放羊的小孩丢了羊，
羊进入到了秦始皇的墓穴，放牧的人手持火把照明找羊，失火烧了里面的葬椁。从古到今，厚葬没有像秦
始皇那么丰厚的了，然而几年之间，外部遭受了项羽的焚烧的灾难，内部遭受了牧童失火的祸患，这不是
太悲哀了吗！
所以德行越笃厚的人埋葬越微薄 ，智慧越深的人埋葬越简约 。只有那没有道德缺少知识的 ，他的埋葬
越丰厚，建立的坟墓越巍峨，修建的宫殿庙宇越高峻，被挖掘就一定最迅速。从这里看来，明白和蒙昧的
不同效果，埋葬的好与坏，非常明显地显现出来了。
材料三：
我听说 ：“死是人生的终结 ，它让人回归到自然 ；葬就是收藏 ，要让别人不能再看到自己 。”上古的
风俗，并没有堆坟树碑。只是到了后世遗留下葬礼仪式上下工夫，才开始准备棺椁。谴责奢侈浪费的人，
并不是舍不得钱财；赞美节俭薄葬的人，实际上看重这样做可以避免危险。
所以唐尧很圣明 ，死后葬在谷林 ，仅在坟边栽上树木作为标记 ；秦穆公是明君 ，去世后葬在橐泉 ，并
没修筑高大的陵墓。到吴王阖闾违背礼制，用珠玉做成野鸭大雁作为陪葬；秦始皇荒淫无度，坟墓里有水
银做的江河大海。这些人都是因为在墓里埋藏了大量的财物而招致了灾祸。
我位居四海之尊 ，承接百王之弊 ，如果不明白如何教化百姓 ，睡到半夜都会为之恐惧忧虑 。虽然现在
丧葬的法规，在仪制中已经有详细的记载；对违礼的处罚，也在刑书中写明了。但是皇亲贵族之中依然有
很多人还在沿袭着陈旧的习俗，民间很多百姓也在葬礼时奢侈靡费，伤风败俗。用厚葬来供奉死者，用高
坟来表示孝道，于是使衣衾棺椁，力求雕刻华丽；灵车冥器，也尽用金玉装饰。富贵人家破坏法度，相互
炫耀，贫穷之家倾家荡产而唯恐不及，这样做白白地损害了道德教化和礼义，对死者（所在的地下世界）
却没有任何益处。
造成的危害已经很深了，对此应予惩治革除。凡王公以下，直至百姓，葬礼如有不遵照律令格式的 ，
明确地加以考察检举，依据具体的情形判定罪行。
15 ．B【解析】本题考查学生对诗歌的综合理解和赏析能力。 学科 网（北 京）股 份有限 公司
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B. “年迈体衰的诗人在暖春看到岸柳飘拂 ，忽感自己气色半归 ”错误 。“气色半归湖岸柳 ”是说湖边的柳树
在春天里焕发出勃勃生机，呈现出葱茏的气色，强调初春的气色大半归于湖岸的柳树，并非指年迈体衰的
诗人看到岸柳飘拂，忽感自己气色半归。
故选 B。
16 ．①第四句 “人家多上郭门船 ”，通过描写郭门处很多人家上船出行的场景，从视觉角度直接呈现人们在
暖春时节乘船出行的热闹画面，是实写，展现出一种质朴的生活气息。
②第八句 “细风时已弄繁弦 ”，通过写微风中传来繁弦之音，从听觉角度侧面烘托出湖边酒肆的热闹，暗示
人们在酒肆中享受生活 、演奏音乐 ，给人以想象空间 ，让“烟火气 ”更具韵味 。【解析 】本题考查学生鉴赏诗
歌表达技巧的能力。
①第四句 “人家多上郭门船 ”是视觉呈现与实写：诗人以直白的笔触，描绘出亲眼所见的景象 —— 众多百
姓纷纷登上停靠在郭门旁的船只，属于实写。 “多上 ”二字，简洁而有力地勾勒出人群往来、熙熙攘攘的
动态感，生动地展现出暖春时节，人们趁着大好春光外出游玩、享受生活的热闹氛围，充满了浓郁的生活
气息，传递出初春时节人们对生活的热爱与快乐。
②第八句 “细风时已弄繁弦 ”是听觉描写与侧面烘托：本句诗人转换视角，从听觉入手，微风轻拂，诗人
仿佛听到了繁弦之音 。这里并没有直接描写演奏音乐的人群 ，而是以细腻的笔触 ，通过 “细风 ”与“繁弦 ”
的巧妙结合，侧面烘托出湖边酒肆的热闹场景。读者由此可以想象，在山影之下的酒旗旁，人们或开怀畅
饮，或谈笑风生，伴随着悠扬的音乐声，享受着惬意的时光，使诗歌中的 “烟火气 ”不再是简单的场景再
现，而是充满了韵味与诗意，余味悠长。
17 ．花径不曾缘客扫 蓬门今始为君开 盖将自其变者而观之 则天地曾不能以一瞬 三顾
频烦天下计 两朝开济老臣心（杀气三时作阵云，寒声一夜传刁斗 /烹羊宰牛且为乐，会须一饮三百杯 /
烽火连三月 ，家书抵万金 /飞流直下三千尺 ，疑是银河落九天 /白发三千丈 ，缘愁似个长 ）【解析 】本题考查
学生默写常见名篇名句的能力。
本题易错字： “缘”、“蓬”、“瞬”、“烦”、“刁”、“烹”。
18 ．A．异彩纷呈（丰富多彩） B．匠心独运 【解析】本题考查学生正确使用成语的能力。
A 处，前文提到 “从写意水墨 ，到梅兰竹菊 ；从棉麻丝绸到纤维毛料 ”，描述了 “新中式 ”服装设计元素涵
盖的范围广且种类多样，此处强调 “设计元素 ”的多样化，花色多，不受一种风格或格式限制，故填 “异
彩纷呈（丰富多彩） ”。异彩纷呈：奇异的光彩纷纷呈现，也形容新奇的事物纷纷涌现；丰富多彩：形容
内容丰富，花色繁多。
B 处，文中说同仁堂结合传统养生与新消费需求推出一系列饮品 ，此处强调同仁堂在产品研发上的独特构思 学科 网（北 京）股 份有限 公司
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和巧妙设计，故可填 “匠心独运 ”。匠心独运：独创性地运用精巧的心思。
19 ．（ 1）序号 ①修改为：现在的马面裙经历了漫长的发展，承载了深厚的积淀；
（2）序号 ④修改为 ：使其更加适合现代人的日常穿着 。 【解析 】本题考查学生辨析并修改病句的能力 。
（1）句 ①“ 经历了漫长的发展和深厚的积淀 ”搭配不当， “经历 ”不能和 “深厚的积淀 ”搭配，应改为 ：
现在的马面裙经历了漫长的发展，承载了深厚的积淀。
（2）句④“更加适合现代人的日常穿着 ”成分残缺 ，前句的主语是 “人们 ”，此处应在 “更加 ”前加 “使
其”。修改后，正确表述为：使其更加适合现代人的日常穿着。
20 ．甲：又时尚实用
乙： “新中式 ”流行 【解析】本题考查学生语言表达之情境补写的能力。
甲处，后文举例提到 “新中式 ”服装 “既能穿上出游，也能穿着上班 ”“ 将深厚的中华文化内涵融入 ‘日
用而不觉 ’的日常穿搭 ”，这表明 “新中式 ”服装除了保留传统特色外，还具有方便在日常生活场景中穿
着的特点，此处应填写与便于日常穿着相关的内容，故可填 “又时尚实用 ”。
乙处，前文围绕 “新中式 ”穿搭引发的时尚风潮展开，后文说这种风潮 “犹如星火燎原，已经从服饰延伸
到了家居 、美妆 、餐饮等多个领域 ”，此处应是对这种从服饰领域扩展到多领域的 “新中式 ”潮流的概括 ，
故可填 “‘ 新中式 ’流行 ”。
21 ．①“ 辩识 ”改为 “辨识 ”；
②“ 婉若 ”改为 “宛若 ”；
③“ 交溶 ”改为 “交融 ”。 【解析】 本题考查学生识记并正确书写现代常用规范汉字的能力。
①“ 辩”是辩论的意思， “辩识 ”应改为 “辨识 ”。“辨识 ”是辨认、识别的意思。
②“ 婉”是“婉转 ”的意思， “婉若 ”应改为 “宛若 ”。“宛若 ”是好像、仿佛的意思，
③“ 溶”是溶化、溶解的意思， “交溶 ”应改为 “交融 ”。“交融 ”是融合在一起的意思。
22 ．示例一：历史积淀，传承古今，焕发时代光彩。
示例二：靓丽风景，融汇古今，展现文化魅力。 【解析】本题考查学生概括要点、拟写对联的能力。
首先分析所给例句的结构和内容。 “时尚符号 ”为偏正结构，点明马面裙具有时尚属性这一身份。 “惊艳
中外 ”是动宾结构， “惊艳 ”为动词，描述马面裙给人的影响， “中外 ”表示范围。 “演绎东方韵味 ”是
动宾结构，阐述马面裙所起到的作用，其中 “东方韵味 ”为偏正结构。
然后进行具体的仿写示例一。结合材料中对马面裙历史的描述，从宋代出现到明代成熟并延续至今， “历
史积淀 ”既能体现马面裙的历史渊源，又符合偏正结构，且与上句中 “时尚符号 ”相对；马面裙在发展过
程中融合了传统与现代元素， “传承古今 ”为动宾结构，与上句中 “惊艳中外 ”结构一致，且能体现马面 学科 网（北 京）股 份有限 公司
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裙的特点，与上句相对；材料提到马面裙 “焕发出了属于这个时代的光彩 ”，“焕发时代光彩 ”为动宾结
构， “时代光彩 ”为偏正结构，与上句中 “演绎东方韵味 ”结构相同，且能体现马面裙在现代的魅力，与
上句呼应。
最终，拟写的下句为：历史积淀，传承古今，焕发时代光彩。
示例二的仿写如下： “靓丽风景 ”同样是对马面裙的一种形象表述，与上句中 “时尚符号 ”相对； “融汇
古今 ”说明马面裙融合了古代与现代的元素，和上句中 “惊艳中外 ”结构相似且在内容上呼应，体现马面
裙在时间维度上的特点； “展现文化魅力 ”与上句中 “演绎东方韵味 ”结构一致，都突出了马面裙所蕴含
的文化价值 ，从不同角度展现了马面裙之美 。故最终拟写的下句为 ：靓丽风景 ，融汇古今 ，展现文化魅力 。
23 ．例文：
穿透表象，趋近真相
在人类认知世界的漫长旅途中 ，古罗马帝国的奥勒留以其深邃的智慧留下了一句振聋发聩的箴言 ：“我
们听到的一切都是一个观点，不是事实；我们看见的一切都是一个视角，不是真相。 ” 这一论断犹如一道
凌厉的闪电，划破了人类思维的蒙昧夜空，直抵认知本质的核心地带，引发我们对主观认知与客观真相之
间复杂关系的深刻反思。
我们日常所听闻的信息 ，往往裹挟着表达者的立场 、情感与意图 。每一个观点都是个体基于自身阅历 、
价值观和利益诉求所构建的认知大厦。在信息传播的汹涌浪潮中，新闻报道看似客观，却可能因记者的观
察局限、媒体的议程设置而偏离事实全貌。以某明星税务风波为例，事件初期，部分媒体仅凭片面信息，
就抛出该明星 “恶意偷税漏税 ，毫无职业道德 ”的观点 ，舆论瞬间一边倒 ，网友们纷纷跟风指责 ，对该明星口
诛笔伐。但随着税务部门深入调查，真相逐渐浮出水面：该明星所属团队财务管理存在漏洞，在税务申报
流程上出现了一些非主观故意的失误，并非媒体最初渲染的那般恶劣。可在最初观点的影响下，大众早已
对这位明星形成了负面印象，即便后续真相澄清，也难以完全消除误解。
视觉同样难以逃脱主观视角的束缚 。我们眼中所见 ，不过是光线折射进视网膜后 ，经大脑加工处理的
产物。画家笔下的世界，是其独特审美视角的呈现，同一场景在不同画家的画布上展现出截然不同的风格
与意蕴。我们在生活中看待他人，也常因第一印象、刻板偏见等主观因素，对他人产生片面甚至错误的认
知。一位穿着朴素、不善言辞的人，可能被误解为平庸无奇，而其内在的丰富才华与高尚品格却被这单一
视角所遮蔽。
然而 ，承认认知的主观性并非意味着我们要陷入相对主义的泥沼 ，放弃对真相的追求 。相反 ，它是我
们踏上趋近真相征程的起点。我们应时刻保持对自身认知局限的警醒，以开放包容的心态接纳多元观点。
在面对信息洪流时，不盲目轻信，而是秉持批判性思维，多方求证，从不同视角审视同一事件，努力拼凑 学科 网（北 京）股 份有限 公司
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出事实的完整拼图。在人际交往中，我们要尝试打破固有偏见，深入了解他人，以同理心去感受他人的处
境，从而更全面、准确地认识他人。
在这个信息爆炸 、观点泛滥的时代 ，唯有保持清醒的头脑 ，穿透表象的迷雾 ，不懈追求真相 ，才能在
认知的道路上不断前行，避免被主观臆断所左右，构建起更加真实、理性的精神世界。
【解析】本题考查学生的写作能力。
审题：
这是一道引语类材料作文题。
这段话富有哲理性，引导我们对于所听和所见的事物持有更为审慎和深入的理解。首先，这段话强调了主
观性和客观性的区别。我们听到的一切和看见的一切都是经过主观解读的，而不是纯粹的客观事实。这就
意味着，我们在接收外界信息的同时，也在进行着自我理解和诠释，这是基于我们的经验、价值观和认知
背景的。因此，我们应当意识到，即使是最为真实、最确定的声音和图像，也可能只是一个人的看法或视
角，不一定代表全部的真相 。其次 ，这段话提醒我们要保持独立思考和质疑精神 。当我们接受外部信息时 ，
应当以自我思考为基础，对其进行独立的解读和判断。不能盲目接受别人的观点，而应通过自己的经验和
理解来形成自己的看法 。同时 ，我们也应该尊重他人的观点和看法 ，理解和包容不同的观点和意见 。最后 ，
这段话引导我们在认识和理解事物时，要超越表面现象，深入探究其本质。我们所听和所见的一切往往只
是事物的表面现象，而要真正理解事物的本质，需要我们超越表面，深入思考和研究。
行文构思上，先概括材料，由材料引出观点，我们生活在一个被观点和视角充斥的世界，唯有保持理性与
批判性思维，才能穿透迷雾，趋近事实与真相。接下来从为什么的角度展开论述，一、视角的局限性：以
“盲人摸象 ”故事为切入点，每个盲人基于自身触摸的部分得出对大象的片面认知，类比人们在生活中由
于所处位置、观察角度不同，对事物的认知也存在局限，难以看到全貌。二、追求真相的艰难：列举科学
史上哥白尼提出日心说推翻地心说的历程，说明突破既有视角，探寻真相需克服传统观念束缚、权威压制
等重重困难。
再从怎么办的角度展开论述，一、保持怀疑精神：鼓励对听到、看到的信息不要盲目接受，像科学家对新
理论进行反复验证一样，对观点和现象存疑，主动思考其合理性。二、分析信息来源：提醒读者关注信息
出处，权威、可靠的来源更有可能接近事实，而一些未经核实的小道消息往往可信度低。三、自我反思与
修正：以个人对某一问题观点的转变为例，说明要定期审视自己的认知，当有新证据出现时，勇于修正错
误观点，不断完善对世界的认知。最后总结全文，再次强调观点不等于事实，视角不等于真相，在信息爆
炸时代，理性与批判性思维是我们拨开认知迷雾的有力武器。
立意： 学科 网（北 京）股 份有限 公司
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多元性看待事物。
以批判性思维拂去蒙蔽的尘埃。
多方解和调查，才能接近事物的本质和真相。
202 5 年高考终极押题卷 （二）【新高考通用卷 】
语 文
（考试时间： 150 分钟 试卷满分： 150 分）
注意事项：
1．答卷前，考生务必将自己的姓名、准考证号填写在答题卡上。
2．回答选择题时 ，选出每小题答案后 ，用铅笔把答题卡上对应题目的答案标号涂黑 。如需改动 ，用橡
皮擦干净后，再选涂其他答案标号。回答非选择题时，将答案写在答题卡上。写在本试卷上无效。
3．考试结束后，将本试卷和答题卡一并交回。
一、现代文阅读（ 35 分）
（一）现代文阅读 Ⅰ（本题共 5 小题， 19 分）
阅读下面的文字，完成下面小题。
材料一：
传统校勘学通常是指对某部书籍用不同的版本和相关的资料加以比较核对 ，以考订其文字异同 ，辨别
正误。校勘学首先应该梳理异文，而异文的产生跟文本的流传过程密切相关。文本经历了口传时代、抄传
时代、印本时代，不同时代异文产生及表现形式往往有所区别。
口传时代 ，因为辗转的传讹 ，曲解本意的现象比较普遍 ，因此将其记录下来难免存在差异 ，这种现象
一直持续到写本时代 。写本抄传的时代 ，既有简册散乱导致的文本错简现象 ，又有字体演变造成的古今字 、
异体字、错别字的差异。在抄写过程中，囿于抄手的水平，又会产生俗字、减字、错别字等情况。抄本不
但数量繁众，传承难明，而且随意性强，极易产生异本异文。抄写者有抄写的脱漏，有无意的增添，也有
臆断的擅改，还有无知的妄改，过度的文字加工甚至改变了原创作品的面貌。故而写本文献的校勘有其独
特性，如口讹、形讹、笔讹、主观臆改等，在将其与印本文献一同校勘时要更加慎重。
宋代开始进入印本时代 ，学者 、编者 、校者继承的是丰富而混乱的写本遗产 ，不得不面对众多抄本异
文，并从中选出正确文本。较之于以往的正定字形，北宋时期可以说是有史以来第一次对文本的差异产生
了强烈的关怀，由此产生了对于善本的追求，校勘精良的善本也成为世人追逐的对象。另外，此时期校勘
之学上升到官学地位，以馆阁人员为主体，在国家藏书的基础之上对历代正史、诸子要籍、医药典籍进行 学科 网（北 京）股 份有限 公司
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了全面的校勘，并刊印颁行，成为影响广泛的版本。然而我们在肯定宋人校勘工作的同时，仍需对其加以
反思。宋代除了馆阁校勘之外，实际上对于异文的取舍存在比较大的裁量度，一些宋人的校勘编辑在后世
看来主观性较大 。比如说柳开校定韩愈的文集时 ，其改动多达五千多字 ；方崧卿校勘 《韩集举正 》时“尤尊
馆阁本 ，虽有谬误 ，往往曲从 ，它本虽善 ，亦弃不录 ”。印本经历了宋人对于异文的选择 、改造 ，在这一背
景下 ，我们就应考虑到写本的可贵 。朱德熙先生指出 “比起宋以后的刻本来 ，唐代类书用的本子以及敦煌唐
写本，跟竹简本或帛书本要接近得多。这说明印刷术的兴起一方面减少了古书失传的可能性，另一方面却
增加了比较剧烈地改变古书面貌的可能性 ”。
写本时代的校勘 ，文本校勘多是文字的纠误和字形的订正 ，其依据在文本内部或它书征引 ，这是由抄
传文献复杂性决定的。进入印本时代之后，才产生了真正意义上版本的概念，不同版本间可以进行版本源
流的梳理，逐步建立起今日基于版本谱系的校勘学。
（摘编自刘玉才《校勘何以为学》）
材料二：
校勘之学起于文件传写的不易避免的错误 。校勘学要改正这些传写的错误 ，恢复一个文件的本来面目 ，
或使他和原本相差最微 。校勘学的工作有三个主要的成分 ：一是发现错误 ，二是改正 ，三是证明所改不误 。
发现错误有主观的 ，有客观的 。读者读到不可解之处 ，或可疑之处 ，因此认为文字有错误 ，这是主观
的发现错误；因几种本子的异同，而发现某种本子有错误，这是客观的。
主观的疑难往往可以引起本子的搜索与比较 ，但读者去作者的时代既远 ，偶然的不解也许是由于不能
理解原意，未必由于传本的错误。况且错误之处未必都可以引起疑难，若必待疑难而后发现错误，而后搜
求善本 ，正误的机会就太少了 。况且传写的本子往往经 “通人 ”整理过 ，往往经人凭己意增删改削 ，成为文从
字顺的本子了 ，错误是不容易发现的 。试举一个例子为证 ，坊间石印 《聊斋文集 》附有 《柳泉蒲先生墓表 》，
其中记蒲松龄 “卒年八十六 ”，这是 “卒年七十六 ”之误 ，有《国朝山左诗钞 》所引墓表及原刻碑文可证 。若单
读“卒年八十六 ”之文而无善本可比较，决不能引起疑难，也决不能发现错误。又《山左诗钞》引这篇墓表 ，
字句多被删节，如云：
（先生）少与同邑李希梅及余从父历友结郢中诗社。
此处无可引起疑难，但扶轮社铅印本中此句乃作：
与同邑李希梅及余从伯父历视友，旋结为郢中诗社。（甲本）
依此文， “历视 ”为从父之名， “友”为动词， “旋”为“结”之副词，文理也可通。
石印本 《聊斋文集 》即从扶轮社本出来 ，但此本的编校者熟知 《聊斋志异 》的掌故 ，知道张历友是当
时诗人，故石印本墓表此句改成下式： 学科 网（北 京）股 份有限 公司
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与同邑李希梅及余从伯父历友亲，旋结为郢中诗社。（乙本）
最近我得墓表的拓本，此句原文是：
与同邑李希梅及余从伯父历友、视旋诸先生结为郢中诗社。（丙本）
视旋是张履庆，为张历友之弟，诗名不大，人多不知道 “视旋 ”是他的表字，且很少人用 “视旋 ”这样罕
见的表字。甲本校者不认得 “张历友 ”，就妄倒 “友视 ”二字而删 “诸先生 ”三字。乙本校者认得 “张历友 ”而不认
得“视旋 ”，所以他把 “视友 ”二字倒回来，而妄改 “视”为“亲”，用作动词。此两本文理都可通，虽少有疑难，
都可用主观的论断来解决。倘我们终不得见此碑拓本，我们终不能发现甲乙两本的真错误。这个小例子可
以说明校勘学的性质 。校勘的需要起于发现错误 ，而错误的发现必须倚靠不同本子的比较 ，古人称此学为 “校
仇”。刘向《别录》说： “一人读书，校其上下得谬误，为校；一人持本，一人读书，若怨家相对，为仇。 ”
其实单读一个本子 ，“校其上下 ”，所得谬误是很有限的 ，必须用不同的本子对勘 ，“若怨家相对 ”，一字不放
过，然后可以 “得谬误 ”。
（摘编自胡适《〈元典章校补释例〉序》）
1．下列对材料一相关内容的理解和分析，不正确的一项是（ ）(3 分)
A．口传时代，信息口口相传，缺乏相对固定的载体，辗转传讹的情况频繁发生，从而导致异文现象比
较普遍。
B．写本抄传时代，异文产生的原因较为复杂，既与抄写者水平有关，也与简册散乱引起的文本错简现
象等有关。
C．北宋时期，学者、编者以及校者面对众多的写本遗产，为了完全还原古书原貌，开始追求校勘精良
的善本。
D．印本时代，印刷术的广泛应用以及官方组织的校勘刊印颁行活动，在很大程度上减少了古书失传的
可能性。
2．根据材料二内容，下列说法正确的一项是（ ）(3 分)
A．从校勘学要改正传写错误的目的来看，其工作中 “改正错误 ”比“发现错误 ”和“证明所改不误 ”更重要 。
B．客观发现错误的方式之所以更准确、更可靠，是因为它不需要像主观发现错误那样依赖读者的个人
理解。
C．甲本改拓本原文为 “从伯父历视友 ，旋结 ”，乙本从甲本而来 ，但依然有所改动 ，可见校者主观的论断 。
D．发现错误必须倚靠不同版本的本子，只要广泛地搜求校本的不同版本，人们就可以发现并改正文本
的错误。
3．下列对两则材料的分析，不正确的一项是（ ）(3 分)学科 网（北 京）股 份有限 公司
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A．材料一以时间为序结构文章，指出校勘学者对不同时代文本应有的态度，并着重介绍宋代校勘学在
理念和影响力等方面的变化。
B．材料二中提到的两种发现错误的方式，主观发现表现为材料一中抄写者的主观臆改，客观发现表现
为依据版本谱系的校勘。
C．在材料二所举的例子中，若没有找到原碑拓本，仅凭甲本和乙本，很难确定该句的正确表述，这说
明了用不同本子对勘的重要性。
D．材料二中甲本和乙本对墓表的改动 ，可以证明材料一中所说的抄写者 “有臆断的擅改 ，还有无知的妄
改”的观点。
4．两则材料在介绍校勘过程中错误产生的原因时，语言风格迥异。请结合材料简要分析。 (4 分)
5．学习课文 《〈老子 〉四章 》中的第六十四章时 ，小刚查到该章有以下几个古籍版本 ，发现课文是在这些
版本的基础上经过校勘完成的。请根据材料分析课文经历了怎样的校勘过程。 (6 分)
版本一：战国中期郭店楚简（残缺文字用 “□”表示，异文部分加黑标示）
甲组（一）
为之者败之，执之者远之。是以圣人亡为故亡败，亡执故亡失。临事之纪，慎终如始，此亡败事矣 。
圣人欲不欲，不贵难得之货；教不教，复众之所过。是故圣人能辅万物之自然，而弗能为。
甲组（二）
其安也 ，易持也 。其未兆也 ，易谋也 。其脆也 ，易泮也 。其微也 ，易散也 。为之于其无有也 ，治之于
其未乱。合 □□□□□□ 末。九层之台，作 □□□□□□□□□ 足下。
版本二：西汉初期马王堆帛书（甲本）（残缺文字用 “□”表示）
其安也，易持也 □□□□□□□□□□□□□□□□□□□□□□□□□□□□□ 毫末。九成之台 ，
作于羸土。百仁之高，台于足 □□□□□□□□□□□□□□□□□ 也□无败 □无执也，故无失也。民之
从事也，恒于其成事而败之。故慎终若始，则 □□□□□□□□ 欲不欲，而不贵难得之 ；学不学，而复
众人之所过；能辅万物之自 □□ 弗敢为。
版本三：宋刻本老子道德经古本集注直解
其安易持 ，其未兆易谋 。其脆易泮 ，其微易散 。为之乎其未有 ，治之乎其未乱 。合抱之木 ，生于豪末 ；
九成之台，起于累土；千里之行，始于足下。为者败之，执者失之。是以圣人无为故无败，无执故无失。
民之从事，常于几成而败之。慎终如始，则无败事。是以圣人欲不欲，不贵难得之货；学不学，而复众人
之所过；以辅万物之自然，而不敢为也。
（二）现代文阅读 Ⅱ（本题共 4 小题， 16 分） 学科 网（北 京）股 份有限 公司
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阅读下面的文字，完成下面小题。
我把两粒莲子，放在叶先生身旁
高昌
11 月 30 日上午 10 时，在天津市第一殡仪馆滨河厅送别叶嘉莹先生。
哀乐声中 ，叶先生的灵车徐徐推入 ，安放在紫色雏菊和白色百合簇拥着的美丽莲形花台的中间 。灵台
的前面是由红玫瑰花瓣组成的巨大的心形图案。此时我胸中也如低回的哀乐，一样澎湃着，一样翻卷着，
泛起深深的情感涟漪。主持人在哀伤地诵读告别词，我的心也随着她悲凉的声音，一起追忆那颗芬芳而热
烈的诗心，一起追溯那纯净而淡泊的百年人生路 ……
我跟随主持人的声音给叶先生深深地三鞠躬 ，然后迈着沉重的脚步 ，随着吊唁人群慢慢向前走 。走得
近些，再近些，仔细凝视叶先生一眼。先生头戴棕色花帽，神情安然，静静沉睡。我也和大家一样，把紫
色纱网包裹着的两粒莲子，郑重地放在叶先生左侧的百合丛中。
这两粒莲子 ，像是晶莹的泪珠 ，凝结着洁白的哀思 ；像是沉静的琥珀 ，闪耀着沧桑的豁达 ；也像是一
串饱含深情的省略号，牵引着美丽的想象，蕴藏着悠长的情愫 …… 一百年起承转合的淡雅人生，如同一首隽
永通透的动人诗篇 。这诗篇结尾的标点不是决绝的句号 ，而是一粒粒莲子组成的意味深长的省略号 。“莲实
有心应不死 ，人生易老梦偏痴 。千春犹待发华滋 。”这是叶嘉莹先生的词句 。她说 ：“我的莲花总会凋落 ，我
要把莲子留下来。 ”小雪薄寒节气，南开大学马蹄湖畔的莲花想来已经凋落了。可是莲子在，芬芳就还在 ，
丰盈就还在 ，明年还会有新的莲花悠然盛开 。美丽的生命如一朵花开 ，又如一朵花谢 。叶先生飘然而去了 ，
留下的诗还在，光也还在。
叶嘉莹先生是中华诗词学会的发起人之一 ，是中华诗词学会的名誉会长 ，我们对叶先生有着深深的敬
仰之情。叶先生受业于著名学者、诗人、教育家顾随，而顾随先生的女儿顾之京、女婿许桂良也是我上大
学时候的老师，所以从个人感情来讲，我对叶嘉莹先生还有着一份特殊的敬重和亲切。 2017 年，叶嘉莹先
生主编、我赏析的《给孩子们的诗园 ·古诗卷》由人民文学出版社 ·天天出版社出版，也使我有机会感悟
叶先生传递的中华诗词之美 ，并留下一份关于叶先生的难忘回想与情思 。叶先生一路辛勤耕耘 ，弦歌不辍 ，
清荷卓立，芬芳万里，她是一位学者、诗人、教育家，也是一位令人尊敬的耕耘者、劳动者。诗词经典中
展现出来的精神品格、文化魅力，正如重新开花的古莲子一样，在她身上焕发出晶莹璀璨的生命光彩。
2021 年 5 月 16 日，97 岁的叶嘉莹先生在中华诗词学会残疾人诗词工作委员会成立时 ，曾经说过一段
话。她说诗词里 ，古代诗人的精神 、理想 、人格 、品质是普遍影响我们后世的 ，“每一个人 ，无论你身体上
有任何疾病 ，但是 ，这种精神是永远都存在 ，永远给我们鼓舞的 ，这就是我们中国诗词最了不起的地方 。”
她还勉励大家一同学习诗词 ，“在精神修养上都有进步 ”。虽然叶先生多次提到 “弱德之美 ”，但她本人对中华 学科 网（北 京）股 份有限 公司
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优秀传统文化的体悟和坚守，她内心的通透和自信，却又有着一份异乎寻常的坚韧和刚强。她的性灵境界
高洁纯净，别有一种尘俗之外的清凉和纯真。
叶先生既有深厚的诗词修养和传统积淀 ，又有坚实的新旧诗学底蕴和宽广的中西方文化视野 。她本人
是众所周知的古典诗词传承者，但并不狭隘地排斥新诗，而且还发表过谈新旧诗歌的比较文章，可见其学
术胸襟和理论气度。她讲课时从不囿于一家之言，每每能从一点常见的辞章，生发出许多不常见的奇思妙
想，而且还能把放出去的思绪之马再在课堂上重新拉回来。这种游刃有余的发散性教学，有助于培养学生
对传统文化的鉴别和分析能力 ，给学子们带来更多新鲜的营养和创新的思维理念 。叶先生从 “兴发感动 ”入手 ，
带领听众和读者一起进入瑰丽丰盈的诗词境界 ，也以其人格魅力让大家感受到那 “以生命为诗 、以生活实践
诗”的诗意人生。
11 月 30 日下午 3 点，我参加了在南开大学省身楼举行的叶嘉莹先生追思会 。大家对叶先生 “一生多艰 、
寸心如水 ”的赤子丹忱和家国情怀进行了更多追述和回望 。叶先生的女儿赵言慧含泪感谢大家对叶先生的爱
与关心，祝愿大家 “在未来，活出每个人的精彩 ”。
砚田溉甘澍，冰心曜晶莹。蕙风穆桃李，蹊径滋欣荣。大家赞赏作为教育家的叶嘉莹先生，但我想 ，
叶先生本质上首先是一位诗人，一位温暖、透明、真诚的本色诗人。她的诗词清雅深挚，她的诗学理论精
深宏阔 ，都有待我们进一步深入学习和认真研究 。正所谓 “高山仰止 ，景行行止 ”，中华诗词的温暖和光明代
代传承，叶先生也在我们的心中永生。
挽沧海起 ，看春潮来 。沐东风暖 ，醉晚梅香 。我们对叶先生最好的纪念 ，也正是薪火相传 ，传光传暖 ，
传真传美，继续为中华诗词事业一倾丹忱，尽献绵薄之力。诗讯播花讯，诗境化世境，岁岁东风绿，代代
荐赤诚。我们把莲子放在叶先生身旁，这莲子也象征着我们的誓言和责任。莲子在，花魂在，清荷的芬芳
在，中华诗词之美永流传 ……
（节选自《中国文化报》，有删改）
6．下列对文本相关内容和艺术特色的分析鉴赏，正确的一项是（ ）(3 分)
A．开篇时间精确，体现了真实性、严肃性，文章开头描写了人们自发送别叶嘉莹的场景。
B．将紫色纱网包裹的莲子放在百合丛中这一细节 ，表现了大家对叶先生充满了敬重 、怀念与哀思之情 。
C．作者运用借喻的手法，把莲子比作 “泪珠 ”“ 琥珀 ”“ 省略号 ”，表达了对逝者的深深怀念和敬意。
D．第四段句号的决绝与莲子省略号是意味深长的类比，句号代表着结束，省略号则表示意犹未尽、余
韵悠长。
7．关于最后三个段落的理解，下列说法不正确的一项是（ ）(3 分)
A．“含泪 ”展现出女儿对母亲深深的眷恋以及对众人关怀的感激，使读者更能共情于这种悲痛与不舍。 学科 网（北 京）股 份有限 公司
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B．“清雅 ”体现了叶先生诗词风格的清新高雅 ，“深挚 ”突出了其情感的深沉真挚 ，使读者感受到叶先生诗
词举足轻重的影响力。
C．“挽沧海起 ，看春潮来 。沐东风暖 ，醉晚梅香 ”，用字凝练 、音韵和谐 ，富有节奏感和韵律美 ，增强了
文章的艺术感染力。
D．莲子既代表叶先生的精神、留下的文化遗产，又体现了对诗词研究的代代传承。
8．有人认为本文的标题比 “诗歌界的典范 —— 叶嘉莹 ”这一标题更好。结合本文内容，简要分析这一看法的
理由。 (4 分)
9．读书小组要围绕此文进行一场文学探究 ，并进行组内交流展示 。经讨论 ，探究主题为 ：一个让人肃然起
敬的形象。作为小组成员，请写出你的发言要点。 (6 分)
二、古诗文阅读（ 35 分）
（一）文言文阅读（本题共 5 小题， 20 分）
阅读下面的文言文，完成小题。
材料一：
窦参阴狡而愎恃权而贪每迁除多与族子给事中申议之 。申招权受赂，时人谓之 “喜鹊 ”。上颇闻之 ，
谓参曰： “申必为卿累，宜出之以息物议。 ”参再三保其无他，申亦不悛。左金吾大将军虢王则之，李巨
之子也 ，与申 善．。左谏议大夫 、知制诰吴通玄与陆贽不叶 ，窦申恐贽进用 ，阴与通玄 、则之作谤书以倾贽 。
上皆察知其状。夏，四月，丁亥，贬则之昭州司马，通玄泉州司马，申道州司马。寻赐通玄死。乙未，贬
中书侍郎、同平章事窦参为郴州别驾，贬窦申锦州司户。
初，窦参恶左司郎中李巽 ，出为常州刺史 。及参贬郴州 ，巽为湖南观察使 。汴州节度使刘士宁遗参绢
五十匹，巽奏参交结藩镇。上大怒，欲杀参，陆贽以为参罪不至死，上乃止。既而复遣中使谓贽曰： “参
交结中外，其意难测，社稷事重，卿速进文书处分。 ”贽上言： “参朝廷大臣，诛之不可无名。昔刘晏之
死，罪不明白 ，至今众议为之愤邑 ，叛臣得以为辞 。窦参贪纵之罪 ，天下共知 ；至于潜怀异图 ，事迹暧昧 。
若．不推鞫，遽加重辟，骇动不细。窦参于臣无分，陛下所知，岂欲营救其人，盖惜典刑不滥。 ”三月，更
贬参驩州司马 ，男女皆配流 。上又命理其亲党 ，贽奏 ：“罪有首从 ，法有重轻 ，参既蒙宥 ，亲党亦应末减 。
况参得罪之初，私党并已连坐，人心久定，请更不问。 ”从之。上又欲籍其家赀，贽曰： “在法，反逆者
尽没其财，赃污者止征所犯，皆须结正施刑，然后收籍。今罪法未详，陛下已存惠贷，若簿录其家，恐以
财伤义。 ”
（节选自《资治通鉴》第二百三十四卷）
材料二： 学科 网（北 京）股 份有限 公司
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初，贽受张镒知 ，得居内职 ；及镒为卢杞所排 ，贽常忧惴 ；及杞贬黜 ，始．敢上书言事 。德宗好文 ，益
深顾遇。吴通微兄弟俱在翰林，亦承德宗宠遇， 文章才器不迨贽，而能交结权幸，共短贽于上前 。而贽为
朋党所挤，同职害其能，加以言事激切，动失上之欢心，故久之不为辅相。
时贽母韦氏在江东 ，上遣中使迎至京师 ，搢绅荣之 。俄丁母忧 ，东．归洛阳 ，寓居嵩山丰乐寺 。藩镇赙
赠及别陈饷遗，一无所取。初贽以受人主殊遇，不敢爱身，事有不可，极言无隐。朋友规之，以为太峻，
贽曰： “吾上不负天子，下不负吾所学，不恤其他。 ”精于吏事，斟酌决断，不失锱铢。
（节选自《旧唐书 ·陆贽传》）
10 ．材料中画波浪线的部分有三处需要断句，请用铅笔将答题卡上相应位置的答案标号涂黑。 (3 分)
窦参 A 阴狡 B 而愎 C 恃权 D 而贪 E 每迁除 F 多与族子 G 给事中 H 申议之。
11 ．下列对材料中加点的词语及相关内容的解说，不正确的一项是（ ）(3 分)
A．善，与 …… 交好，与《鸿门宴》 “不如因善遇之 ”的“善”意思不同。
B．若，如果，与《齐桓晋文之事》 “若民，则无恒产，因无恒心 ”的“若”意思相同。
C．始，才，与《愚公移山》 “寒暑易节，始一反焉 ”的“始”意思相同。
D．东，向东，名词作状语，与《赤壁赋》 “顺流而东也 ”的“东”用法不同。
12 ．下列对材料有关内容的概述，不正确的一项是（ ）(3 分)
A．窦申因收受贿赂而名声不好，却不知悔改。他还联合吴通玄、李则之一起诽谤陆贽，皇帝并没有相
信他们的话。
B．汴州节度使刘士宁送给窦参五十匹绢，李巽趁机上奏说窦参与藩镇结交，这是因为窦参在任常州刺
史期间与当时的左司郎中李巽交恶。
C．陆贽因受到张镒的赏识而担任朝中职务，张镒被卢杞排挤时，陆贽常常感到忧虑和不安，等到卢杞
被贬黜后，他才敢上书谈论政事。
D．陆贽因为受到皇帝的优待 ，不敢爱惜自己 ，遇到不可行的事 ，就极力直言不讳 。朋友们规劝他 ，他
也不想改变。
13 ．请把材料中画横线的句子翻译成现代汉语。 (8 分)
（1）上颇闻之，谓参曰： “申必为卿累，宜出之以息物议。 ”
（2）文章才器不迨贽，而能交结权幸，共短贽于上前。
14 ．读传观史 ，“互文印证 ”是阅读的绝佳方法 。材料二中陆贽的 “斟酌决断 ，不失锱铢 ”在材料一中是如何体
现的？ (3 分)
（二）古代诗歌阅读（本题共 2 小题， 9 分） 学科 网（北 京）股 份有限 公司
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阅读下面这首词，完成下面小题。
临江仙 ·怀辰州教授赵学舟 ①
【宋】张炎
一点白鸥何处去，半江潮落沙虚。淡黄柳上月痕初。遐观情悄悄，凝想步徐徐。
每一相思千里梦，十年有此相疏。休休寄雁 ②问何如。如何休寄雁，难写绝交书 ③。
【注 】①赵学舟 ：赵与仁 ，字元父 ，号学舟 ，南宋宗室 ，作者的文友 ，元朝时任辰州教授之职 。②《汉
书·苏武传》中传说苏武牧羊时以鸿雁传书。 ③嵇康《与山巨源绝交书》，是嵇康写给人生抉择相异后的
友人山涛（字巨源）绝交的一封信。
15 ．下列对这首词的理解和赏析，不正确的一项是（ ）(3 分)
A．词的起笔以白鸥象征友人。白鸥是自由飞翔于江海之上的水鸟，极像元初飘零江湖的赵学舟等人。
B．“凝想 ”句写出月上柳梢时，词人远眺不见友人踪影，暗中回忆其步履从容徐徐而来的情形。
C．“相思 ”句写出词人对友人的思念，词人与友人一直交往甚密，而今第一次因为两地距离遥远而疏远
了。
D．词的下阕词人综合运用用典设问等手法，委婉曲折地传达出自己欲语难言的复杂心绪。
16 ．后人评价这首词 ：“赵以宗人仕元 ，词是春秋笔法 。”春秋笔法是指寓褒贬于曲折的文笔之中的写法 。请
结合全词谈谈你对这一评价的理解。 (6 分)
（三） 名篇名句默写（本题共 1 小题， 6 分）
17 ．补写出下列句子中的空缺部分。 (6 分)
（1）司马迁在《屈原列传》中认为屈原作《离骚》的一个原因可能是 “自怨生 ”，而导致屈原怨愤的
直接原因是他 “ ， ”。
（2）小王参加世界电信日主题纪念活动时，感慨道： “昔日《春江花月夜》有云
‘ ， ’，古人即使以飞禽游鱼为信使，也往往难以传达思念；而今信息传输
瞬间千里，距离不再是情感的障碍了。 ”
（3）古代的边塞诗中有不少诗句出现了带有异域色彩的乐器名称，它们或者寄寓戍边将士的乡愁，
或者展示瑰丽的塞外风情，例如： “ ， 。”
三、语言文字运用（ 20 分）
阅读下面的文字，完成下面小题。
人脑中存在长期记忆和短期记忆 。（ 甲 ）叫作 “大脑皮质 ”，它相当于人脑的 “硬盘 ”，可以保存我们
已经记住的知识。但人脑不同于计算机，无法通过增加存储器来扩容。因此，为了灵活运用有限的存储空 学科 网（北 京）股 份有限 公司
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间，大脑会根据信息的价值 ，将其分成 “必要信息 ”和“非必要信息 ”，只有被大脑判定为 “必要 ”的信息才会顺
利通过 “关卡 ”，被运送到大脑皮质内长期保存。
具体判定信息是否必要的 “关卡检查员 ”就是人脑中的海马体。海马体的审查标准是 “该信息对生存而
言，是否不可或缺 ”。而那些大家在学校里必须记住的知识 ，很难通过海马体的审查 。人本身也是动物 ，有
着生存的本能。海马体以 “是否有利于生存 ”为尺度，（ 乙 ）。诸如在毫无生命危险的教室里学习之类的
行为，与人类的生存相比简直可以算是 _A___ 的事。
大脑将必要信息储存到长期记忆中会消费大量能量 ，如果将非必要的一些信息也储存到大脑里 ，那就
是对能量的浪费了 。如此一来，我们又可以把海马体看作是一个节能主义者，它也是为了节约能量而不允
许无用信息通过的 “财政大臣 ”。所以 ，从某种程度上来说 ，我们无法改变 “根本记不住 ”这种让人发愁的状况 ，
因为相对于 “记住 ”，（ 丙 ）。
要想让海马体将学校学到的知识划分为必要信息 ，方法只有一个 ，那就是 “欺骗 ”海马体 。我们要持续
不断地将信息传送过去，这样一来，海马体就会产生一种 “如此 _B___ 地传送来的信息一定是必要信
息”的错觉，进而允许信息通过 “关卡 ”，进入大脑皮质。俗话说： “学习就是要反复地训练。 ”从脑科学的角
度来看，事实的确如此。
18 ．请在文中横线处填入恰当的成语。 (4 分)
19 ．请在文中括号内补写恰当的语句 ，使整段文字语意完整连贯 ，内容贴切 ，逻辑严密 ，每处不超过 10 个
字。 (6 分)
20 ．文中画波浪线的句子有语病 ，请进行修改 ，使语言表达准确流畅 ，可少量增删词语 ，不得改变原意 。(3
分)
21 ．某少儿科普栏目计划举办一场 “认识大脑，助力学习 ”的科普活动，请根据以上材料解释大脑 “根本记不
住”学校学到的知识的原因 ，并提出学习建议 。要求 ：以“海马体 ”为开头 ，信息准确 ，语言流畅 ，不超过 80
个字。 (4 分)
22 ．下图是小明对引号常见用法的归纳，请你根据材料中使用引号的句子帮他补充完整。 (3 分)学科 网（北 京）股 份有限 公司
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四、作文（ 60 分）
23 ．阅读下面的材料，根据要求写作。
科技发展给传统文化带来了前所未有的冲击 。机器大生产正在逐渐取代传统手工艺 ，数字娱乐冲击着
戏曲市场，百年老店因无法适应现代经营模式而纷纷倒闭，传统手工艺因后继无人而面临失传 …… 有人忧
虑：科技发展正在恶化传统文化的生存土壤。
音乐话剧《敦煌》和水剧场版《青蛇》利用 AR 、VR 等技术，将戏剧艺术与数字科技融入文化和旅游
应用场景，突破了传统剧场的界限，打造了全新的演出空间；传统建筑创意表演秀《栋梁》利用虚拟现实
与数字孪生技术等，让观众不仅领略了建筑的外在之美，还深层次地理解了其中的文化精髓 …… 这些尝试
展现了传统文化与现代科技的完美融合 。人们欣喜地看到 ：瞬息万变的数字科技不仅没有让传统文化褪色 ，
反而为其注入了新的生命力。
以上材料引发了你怎样的联想与思考？请写一篇文章。
要求：选准角度，确定立意，明确文体，自拟标题；不要套作，不得抄袭；不得泄露个人信息；不少
于 800 字。
20 25 年高考终极押题卷 （二）（新高考通用 卷）
语文 ·全解全析
1 2 3 6 7 10 11 12 15
C C B B B CEF B B C
1．C【解析】本题考查学生对材料相关内容的理解和分析的能力。
C. “为了完全还原古书原貌 ，开始追求校勘精良的善本 ”错误 ，材料一指出 “较之于以往的正定字形 ，……
校勘精良的善本也成为世人追逐的对象 ”，但文中并没有提及北宋时期追求善本是为了 “完全还原古书原 学科 网（北 京）股 份有限 公司
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貌”，另外，原文说的是 “选出正确文本 ”，这和 “还原古书原貌 ”意思也不同。
故选 C。
2．C【解析】本题考查学生对材料内容的推断和辨析的能力。
A. “其工作中 ‘改正错误 ’比‘发现错误 ’和‘证明所改不误 ’更重要 ”错误 ，材料二提到 “校勘学的工作
有三个主要的成分：一是发现错误，二是改正，三是证明所改不误 ”，这三个成分是校勘学工作的主要方
面，它们相互关联、相辅相成的，并没有表明 “改正错误 ”比“发现错误 ”和“证明所改不误 ”更重要，
选项强加比较。
B. “客观发现错误的方式之所以更准确 、更可靠 ，是因为 …… ”错误 ，材料二只是说 “主观的疑难往往可以
引起本子的搜索与比较， …… 正误的机会就太少了 ”，强调了主观发现错误存在的一些局限性，但并没有
说客观发现错误更准确、更可靠是因为不依赖读者的个人理解，客观发现错误也需要读者对不同版本的本
子的差异进行判断，这同样依赖读者的理解能力，选项因果不当。
D. “只要广泛地搜求校本的不同版本 ，人们就可以发现并改正文本的错误 ”错误 ，材料二指出 “发现错误必
须倚靠不同本子的比较 ”，但 “只要广泛地搜求校本的不同版本，人们就可以发现并改正文本的错误 ”这
种说法过于绝对 ，发现和改正文本的错误不仅需要不同版本的比较 ，校勘过程中还需要严谨的比对和分析 。
故选 C。
3．B【解析】本题考查学生对材料相关内容的理解和分析的能力。
B. “主观发现表现为材料一中抄写者的主观臆改 ，客观发现表现为依据版本谱系的校勘 ”错误 ，材料一中抄
写者的主观臆改是指抄写过程中抄写者的行为，而材料二中主观发现错误是读者在阅读过程中因不可解或
可疑之处认为文字有错误，二者主体不同，不能简单地将主观发现表现为材料一中抄写者的主观臆改；材
料二中客观发现错误是因几种本子的异同而发现某种本子有错误，材料一中依据版本谱系的校勘是印本时
代才产生的，二者概念不同，不能简单对应。
故选 B。
4．①材料一语言风格较为专业严谨 ，多使用专业术语和概括性表述 。整体概述了口传时代 、抄传时代 、印
本时代的校勘过程中错误产生的原因，以专业、凝练的表述概述了多种复杂因素，体现了学术文章的严肃
性和规范性。
②材料二语言风格更为通俗易懂，术语较少。少用 “口讹、形讹、笔讹 ”这样的术语，而是用具体的例子来
解释校勘过程中错误产生的原因，使读者更易理解。 【解析】本题考查学生分析文章语言特色的能力。
①材料一：语言风格较为专业严谨，多使用专业术语和概括性表述。例如，在介绍口传时代异文产生原因
时用 “辗转的传讹，曲解本意 ”，写本抄传时代提到 “简册散乱导致的文本错简现象 ”“ 字体演变造成的 学科 网（北 京）股 份有限 公司
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古今字、异体字、错别字的差异 ”“ 口讹、形讹、笔讹、主观臆改 ”等专业术语，整体概述了不同时代校
勘过程中错误产生的原因 ，以专业 、凝练的表述概述了多种复杂因素 ，体现了学术文章的严肃性和规范性 。
②材料二：语言风格更为通俗易懂，术语较少。在解释校勘过程中错误产生的原因时，没有使用 “口讹、
形讹 、笔讹 ”这样的专业术语 ，而是通过具体的例子 ，如“坊间石印 《聊斋文集 》附有 《柳泉蒲先生墓表 》…… ”
来详细说明不同版本的差异以及错误是如何产生的，使读者更易理解。
5．①分析版本源流 ：郭店楚简本和马王堆帛书本都是抄传时代的写本 ，可能更接近古书原貌 ，但存在文字
残损现象 ，竹简本可能出现错简现象 ，与帛书本一样可能存在口讹 、主观臆改等抄写问题 ，需要慎重对待 ；
宋刻本是印本时代的版本，经过了前人的校勘和整理，文字表述相对清晰、完整，但可能存在主观性较大
的问题。
②比较异文：由于传抄、刊刻等行为会造成不同版本在文字上存在一定的差异，故而需要梳理异文现象。
③综合比较，改正错误：课文在最终校勘时可能综合比较了不同的版本，并修正了错误，根据版本特点和
文本内容等因素，选择更为经典和权威的版本并进行相关注释。 【解析】 本题考查学生分析运用文本信息
的能力。
①分析版本源流：郭店楚简本是战国中期的抄传写本，马王堆帛书本是西汉初期的抄传写本，这两个版本
都属于抄传时代的写本，根据材料一，写本抄传时代可能存在简册散乱导致的错简现象，以及抄写者水平
不同造成的俗字、错别字、主观臆改等问题，同时还存在文字残损现象（如郭店楚简有残缺文字）。但它
们可能更接近古书原貌 ，所以在校勘时需要慎重对待 。宋刻本老子道德经古本集注直解是印本时代的版本 ，
印本时代经过了前人的校勘和整理，文字表述相对清晰、完整，但根据材料一，印本时代对异文的取舍存
在较大裁量度，一些校勘编辑主观性较大，所以也不能完全依赖宋刻本。
②比较异文：由于不同版本是在不同时代、经过不同的传抄或刊刻过程形成的，传抄、刊刻等行为会造成
不同版本在文字上存在一定的差异，如郭店楚简本、马王堆帛书本与宋刻本在文字表述、字词使用等方面
可能存在不同，所以需要梳理异文现象，找出各版本之间的差异。
③综合比较，改正错误：课文在最终校勘时可能综合比较了郭店楚简本、马王堆帛书本和宋刻本等不同的
版本，根据各版本的特点（如写本的接近原貌但可能有错误，印本的相对完整但可能有主观改动）和文本
内容等因素，判断哪些文字是正确的，哪些是错误的，并进行修正。同时，可能还会参考其他相关资料，
选择更为经典和权威的版本并进行相关注释，以呈现出较为准确的《〈老子〉四章》第六十四章的内容。
6．B 【解析】本题考查学生对文本相关内容和艺术特色的分析鉴赏的能力。
A. “自发送别 ”错误 ，文章开头描写在天津市第一殡仪馆滨河厅送别叶嘉莹先生的场景 ，并未提及送别的人
群是否自发。 学科 网（北 京）股 份有限 公司
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C. “作者运用借喻的手法 ”错误，把莲子比作 “泪珠 ”“ 琥珀 ”“ 省略号 ”，运用的是明喻，而非借喻。
D. “意味深长的类比 ”错误 ，第四段中 “这诗篇结尾的标点不是决绝的句号 ，而是一粒粒莲子组成的意味深
长的省略号 ”，句号和省略号在这里是对比，而非类比。句号代表着结束，而省略号象征着叶先生所做出
的贡献以及她的精神品质，不会因为她的离去而画上一个戛然而止的句号，而是会像莲子一样，在人们心
中生根发芽，不断延续和传承下去，引发人们对她的事迹、思想和精神的长久回味与思索。
故选 B。
7．B 【解析】本题考查学生分析文章重要语段的作用的能力。
B. “使读者感受到叶先生诗词举足轻重的影响力 ”错误 ，“清雅 ”“深挚 ”这两个词主要是对叶先生诗词风
格和情感特点的描述，并没有直接体现出叶先生诗词举足轻重的影响力。
故选 B。
8．①情感上 ，文中作者在送别叶嘉莹先生时 ，将两粒莲子放在先生身旁 ，这一行为贯穿始终 ，莲子象征着
对叶先生的哀思 、敬重与追思等情感 ，标题直接体现了这一情感寄托 ，而“诗歌界的典范 —— 叶嘉莹 ”这一标
题无法体现这种细腻情感。
②手法上。莲子象征着叶先生的精神品格如莲子般坚韧，其对中华诗词文化的传承与贡献也如莲子生根发
芽，让诗词之美延续，原标题更好地呼应了叶先生的品格与精神。
③结构上，首尾呼应，强化了文章结构。 【解析】本题考查学生分析文章标题的含义和作用的能力。
①情感上：作者在送别叶嘉莹先生时，把莲子放在先生身旁这一行为是文章的重要情节，贯穿了整个送别
过程 。莲子在文中被赋予了丰富的情感内涵 ，象征着作者及众人对叶先生深深的哀思 、敬重与追思等情感 。
而“诗歌界的典范 —— 叶嘉莹 ”这一标题只是强调了叶先生在诗歌界的典范地位，没有体现出这种细腻、
深沉的情感寄托，原标题更能传达出作者对叶先生的情感。
②手法上：莲子具有深刻的象征意义。一方面，它象征着叶先生的精神品格，如文中所提到的 “莲实有心
应不死 ，人生易老梦偏痴 。千春犹待发华滋 ”，叶先生对中华诗词文化的热爱和坚守 ，如同莲子一般坚韧 ；
另一方面，莲子也象征着叶先生对中华诗词文化的传承，就像莲子生根发芽。原标题 “我把两粒莲子，放
在叶先生身旁 ”更好地呼应了叶先生的这种品格与精神，而 “诗歌界的典范 —— 叶嘉莹 ”则没有这种象征
意义和呼应效果。
③结构上：文章开头写送别时将莲子放在叶先生身旁，结尾又提到 “我们把莲子放在叶先生身旁，这莲子
也象征着我们的誓言和责任。莲子在，花魂在，清荷的芬芳在，中华诗词之美永流传 ”，以莲子为线索，
首尾呼应，使文章更加完整。 “诗歌界的典范 —— 叶嘉莹 ”则没有这种作用。
9．①性灵境界 ，高洁脱俗 。她的性灵境界高洁清净 ，别有一种尘俗之外的清凉和纯真 ，对中华优秀传统文 学科 网（北 京）股 份有限 公司
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化有着深刻的体悟和坚守，展现出了极高的精神境界和人格魅力。
②兼容并包，博大胸怀。叶嘉莹既有深厚的诗词修养和传统积淀，又有坚实的新旧诗学底蕴和宽广的中西
方文化视野，她不狭隘地排斥新诗，还发表过谈新旧诗歌的比较文章，体现了其广阔的学术胸襟和开放的
学术态度。 【解析】 本题考查学生分析人物形象的能力。
①性灵境界高洁脱俗 ：文中明确提到 “她的性灵境界高洁纯净 ，别有一种尘俗之外的清凉和纯真 。她对中华
优秀传统文化的体悟和坚守 ，她内心的通透和自信 ，却又有着一份异乎寻常的坚韧和刚强 ”，叶嘉莹先生一
生致力于中华诗词的研究与传承，她对诗词文化的热爱和执着体现了她高洁的精神境界，这种境界超越了
世俗的功利，给人一种尘俗之外的清凉和纯真之感，让人们对她充满敬意。
②兼容并包 ，博大胸怀 ：叶先生 “既有深厚的诗词修养和传统积淀 ，…… 可见其学术胸襟和理论气度 ”，在学
术研究方面，她能够接纳和研究新诗，并且从更广阔的中西方文化视野来审视诗词文化，这种兼容并包的
态度和广阔的学术胸襟，不仅体现了她的学识渊博，更展示了她作为学者的大气和风范，令人肃然起敬。
10 ．CEF 【解析】 本题考查学生文言文断句的能力。
句意：窦参阴险狡诈而且刚愎自用，凭借权势贪婪无度，每当任命官员时，大多和担任给事中的族侄窦申
商议。
“窦参 ”作主语， “阴狡而愎 ”作谓语，构成完整主谓结构，应在 C 处断开；
“阴狡而愎 ”与“恃权而贪 ”结构对称，其后 E 处断开；
“每迁除 ”作时间状语，说明后面行为发生的特定情境，后句 “多与族子给事中申议之 ”较长，应在 F 处
断开。故选 CEF 。
11 ．B 【解析】本题考查学生对文言词语中的一词多义现象和词类活用现象的理解能力。
A. 正确。与 …… 交好。 /好好地。句意：与窦申交好。 /不如趁机好好地对待他。
B. 错误 。如果 。/至于 。句意 ：如果不审讯 。/至于民众 ，如果没有固定不变的产业 ，就没有长久不变的善心 。
C. 正确。句意：才敢上书谈论政事。 /冬夏换季，才往返一次。
D. 正确 。名词作状语 ，向东 。/名词作动词 ，向东进军 。句意 ：向东回到洛阳 。/顺流而下 ，向东进军 。故选
B。
12 ．B 【解析】本题考查学生理解文章内容的能力。
B. “这是因为窦参在任常州刺史期间与当时的左司郎中李巽交恶 ”错误 。根据原文 “初，窦参恶左司郎中李
巽，出为常州刺史。及参贬郴州，巽为湖南观察使 ”，是窦参厌恶李巽，把李巽调出京城任常州刺史，并
非窦参任常州刺史时与李巽交恶。故选 B。
13 ．（ 1）皇上听到了很多有关他们的事，对窦参说： “窦申肯定会成为你的拖累，应该让他离开来平息众 学科 网（北 京）股 份有限 公司
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人的议论。 ”
（2）（吴通微兄弟）文章和才能赶不上陆贽，却能够结交权贵宠臣，一起在皇上面前诋毁陆贽。 【解析 】
本题考查学生理解并翻译文言文句子的能力。
（1）“累”，拖累； “出”，使 …… 离开； “息”，平息； “物议 ”，众人的议论。
（2）“迨”，赶上，比得上； “权幸 ”，权贵宠臣； “短”，诋毁。
14 ．陆贽在处理窦参之事上体现 “斟酌决断 ，不失锱铢 ”。窦参获罪 ，皇帝欲杀之 ，陆贽认为其罪不至死 ，应
依法处置，不可滥杀，体现他对刑罚尺度的精准把握；皇帝要惩治窦参亲党、抄没其家，陆贽依据法律和
情理，建议对亲党从轻处罚、不抄没其家，既维护法律尊严，又避免伤害道义，这些都体现他处理事务的
斟酌决断、精准得当。 【解析】 本题考查学生理解文章内容、筛选概括文中重要信息的能力。反对皇帝杀
窦参 ：当皇帝因窦参交结藩镇等事大怒欲杀之时 ，陆贽认为窦参虽有贪纵之罪 ，但潜怀异图之事证据不明 ，
不能不明不白诛杀朝廷大臣，否则会重蹈刘晏被冤杀的覆辙，引起众人非议。这体现陆贽对刑罚尺度的精
准判断，不轻易让皇帝加重刑罚，避免滥杀。
谏阻惩治窦参亲党：皇帝想命理窦参亲党，陆贽以罪有首从、法有重轻为由，指出窦参已蒙宥，其亲党应
从轻处罚，且之前窦参得罪时私党已连坐，人心已定，无需再问。这说明陆贽能依据法律和实际情况，合
理建议，避免过度牵连。
劝阻抄没窦参家产 ：皇帝想籍没窦参家赀 ，陆贽依据法律规定 ，指出反逆者才尽没其财 ，赃污者只征所犯 ，
且要结正施刑后才能收籍，如今罪法未详，存惠贷就不应簿录其家，以免伤义。这体现陆贽能准确依据法
律条文，从法与义的角度进行斟酌，提出合理建议。
参考译文：
材料一：
窦参阴险狡诈而且刚愎自用 ，凭借权势贪婪无度 ，每当任命官员时 ，大多和担任给事中的族侄窦申商
议。窦申揽权受贿 ，当时的人把他叫做 “喜鹊 ”。皇上听到了很多有关他们的事 ，对窦参说 ：“窦申肯定会成
为你的拖累 ，应该让他离开来平息众人的议论 。”窦参再三担保窦申没有其他问题 ，窦申也不思悔改 。左金
吾大将军虢王李则之，是李巨的儿子，与窦申交好。左谏议大夫、知制诰吴通玄与陆贽关系不和睦，窦申
担心陆贽被提拔任用，暗中与吴通玄、李则之一起写诽谤信来排挤陆贽。皇上全都察觉了解了这些情况。
夏季，四月，丁亥日，将李则之贬为昭州司马，吴通玄贬为泉州司马，窦申贬为道州司马。不久赐吴通玄
自杀。乙未日，将中书侍郎、同平章事窦参贬为郴州别驾，窦申贬为锦州司户。
当初 ，窦参厌恶左司郎中李巽 ，将他调出京城任常州刺史 。等到窦参被贬到郴州时 ，李巽担任湖南观
察使。汴州节度使刘士宁送给窦参五十匹绢，李巽上奏说窦参与藩镇结交。皇上大怒，想要杀掉窦参，陆 学科 网（北 京）股 份有限 公司
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贽认为窦参的罪行不至于处死，皇上这才作罢。不久皇上又派宦官对陆贽说： “窦参与朝廷内外官员结交 ，
他的意图难以猜测 ，国家的事情重要 ，你赶紧进呈文书来处理 。”陆贽上奏说 ：“窦参是朝廷大臣 ，诛杀他不
能没有罪名。从前刘晏被处死，罪名不明确，到现在众人的议论还为此愤恨不平，叛臣能够拿这件事作为
借口。窦参贪婪放纵的罪行，天下人都知道；至于他暗中怀有不轨意图，事情的迹象模糊不清。如果不审
讯，就仓促地加以重刑，造成的惊骇震动不小。窦参与我没有交情，这是陛下知道的，我哪里是想要营救
他这个人 ，只是珍惜法律不被滥用 。”三月 ，再次将窦参贬为驩州司马 ，他的儿女都被发配流放 。皇上又命
令惩治窦参的亲族党羽 ，陆贽上奏说 ：“罪行有主犯和从犯 ，法律有重有轻 ，窦参已经得到宽恕 ，亲族党羽
也应该从轻处罚 。况且窦参刚获罪时 ，他的私党已经一并受牵连治罪 ，人心早已安定 ，请不要再追究了 。”
皇上听从了他的建议 。皇上又想要没收窦参的家产 ，陆贽说 ：“按照法律 ，谋反叛逆的人全部没收他们的财
产，贪污受贿的人只追缴所贪赃物 ，都必须结案施刑之后 ，才能没收登记财产 。现在罪行和法律都不明确 ，
陛下已经心存宽宏，如果登记没收他的家产，恐怕会因为钱财而伤害道义。 ”
材料二：
当初 ，陆贽受到张镒的赏识 ，得以担任朝中官职 ；等到张镒被卢杞排挤 ，陆贽常常忧虑不安 ；等到卢
杞被贬黜，他才敢上书谈论政事。唐德宗喜好文章，对陆贽更加眷顾恩遇。吴通微兄弟都在翰林院，也受
到唐德宗的宠爱，他们的文章和才能赶不上陆贽，却能够结交权贵宠臣，一起在皇上面前诋毁陆贽。而陆
贽被朋党排挤，同僚嫉妒他的才能，再加上他谈论政事言辞激烈，常常失去皇上的欢心，所以很久没有担
任宰相。
当时陆贽的母亲韦氏在江东 ，皇上派宦官把她接到京城 ，官员们都以此为荣耀 。不久陆贽遭遇母亲去
世，向东回到洛阳，寄居在嵩山丰乐寺。藩镇赠送的丧葬财物以及另外进献的财物，他一概不接受。当初
陆贽因为受到君主特别的待遇，不敢爱惜自己，遇到不可行的事，就极力直言不讳。朋友们规劝他，认为
他言辞过于严厉 ，陆贽说 ：“我对上不辜负天子 ，对下不辜负自己的学识 ，不忧虑其他的事 。”他精通为官之
事，处理决断，丝毫不差。
15 ．C 【解析】本题考查学生鉴赏诗歌的意象、表现手法和情感的能力。
C．“而今第一次因为两地距离遥远而疏远了 ”错，这疏远不仅是两地之遥，更为重要的是价值观念相异 。
故选 C。
16 ．①词中是指词人不赞成友人仕元的选择 ，但又不直接指出 ，而是通过字里行间流露出的一种情感倾向 ；
②“ 十年 ”句写出词人内心的悲凉。词人与友人情谊深厚，而今第一次疏远。这疏远不仅是两地之遥，更为
重要的是价值观念相异，这一点词人没有直接说出但友人可以感知。
③“ 如何休寄雁，难写绝交书 ”作者自问自答写出了内心的矛盾。词人本拟寄书问候然又作罢。之所以不寄 学科 网（北 京）股 份有限 公司
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书信 ，是难写绝交书 。“绝交 ”即断绝交谊 ，词人很不赞成友人仕元的选择 ，也不想去信指摘 ，这深隐了一种
贬意。
【解析】本题考查学生鉴赏诗歌的内容和情感的能力。
由注释 ①可知，赵学舟，南宋的宗室，作者的文友，在元朝任职。词人不赞成友人仕元的选择，但又不直
接指出，而是通过字里行间流露出的一种情感倾向，体现了 “春秋笔法 ”。
“每一相思千里梦 ”句，写词人与友人情谊深厚 ，表达了诗人对友人的思念 。“十年有此相疏 ”句，写而今本诗
宋朝人的友人，却在遥远的元朝任职，词人与友人第一次产生了疏远感，其原因不只是空间距离远了，更
是二人价值观不同了。这一点词人没有直接说出，但友人可以感知，体现了 “春秋笔法 ”。
“如何休寄雁，难写绝交书 ”句，诗人先自问为何不托大雁带信问候友人，后自答说自己难以下笔写绝交信 ，
这表明了诗人内心的矛盾。 “绝交 ”即断绝交谊。诗人虽人不赞成友人仕元的选择，但也不想写信指责友人 ，
这也体现了 “春秋笔法 ”。
17 ．信而见疑 忠而被谤 鸿雁长飞光不度 鱼龙潜跃水成文 中军置酒饮归客 胡琴琵
琶与羌笛 （葡萄美酒夜光杯 欲饮琵琶马上催 /琵琶起舞换新声 总是关山离别情 /羌笛何须怨杨柳 春风不度
玉门关 /角声满天秋色里 塞上燕脂凝夜紫）
【详解】本题考查学生默写常见的名篇名句的能力。
易错字： “度”“ 潜”“ 文”“ 置”“ 饮”“ 声”“ 须”“ 燕”。
18 ．A. 无关痛痒 B. 锲而不舍 【解析】本题考查学生正确使用成语的能力。
A 处 ，前文提到海马体以 “是否有利于生存 ”为尺度来审查信息，而在教室里学习的知识对于人类的生存
本能来说，似乎没有直接的、至关重要的影响，属于相对次要、对生存 “无足轻重 ”的事情，所以用 “无
关痛痒 ”。“无关痛痒 ”：指与自身利害没有关系或无足轻重。
B 处，文中说要想让海马体将学校学到的知识划分为必要信息 ，需要持续不断地将信息传送过去 ，可用 “锲
而不舍 ”，与 “持续不断 ”的意思相呼应。 “锲而不舍 ”：不断地镂刻，比喻有恒心，有毅力。
19 ．甲：保存长期记忆的部位 乙：对信息进行判断和取舍 丙：人脑更擅长 “忘记 ” 【解析 】本题考查学
生语言表达之情境补写的能力。
甲处：前文提到 “人脑中存在长期记忆和短期记忆 ”，后文说 “叫作 ‘大脑皮质 ’，它相当于人脑的 ‘硬
盘’，可以保存我们已经记住的知识 ”，这里需要一个过渡句来引出 “大脑皮质 ”与长期记忆的关系。应
填“保存长期记忆的部位 ”。
乙处 ：前文指出 “具体判定信息是否必要的 ‘关卡检查员 ’就是人脑中的海马体 。海马体的审查标准是 ‘该
信息对生存而言，是否不可或缺 ’” ，说明海马体的作用是根据生存需求来判断信息的必要性。应填 “对学科 网（北 京）股 份有限 公司
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信息进行判断和取舍 ”。
丙处：前文先是阐述了海马体为了节约能量，会将非必要信息判定为不通过 “关卡 ”，不允许其进入长期
记忆，接着说 “我们无法改变 ‘根本记不住 ’这种让人发愁的状况 ”，这里强调了大脑在信息处理上更倾
向于让一些信息不被记住。应填 “人脑更擅长 ‘忘记 ’” 。
20 ．大脑将必要信息储存到长期记忆中会消耗大量能量 ，如果将一些非必要的信息也储存到大脑里 ，那就
是对能量的浪费。 【解析】本题考查学生辨析并修改病句的能力。画波浪线的句子有两处语病：一是搭配
不当， “消费大量能量 ”动宾不搭配，将 “消费 ”改为 “消耗 ”；
二是语序不当 ，“一些 ”是数量词 ，“非必要的 ”应直接修饰 “信息 ”，将“一些 ”提至 “非必要 ”之前 。
21 ．海马体为了节省大脑能量，只允许利于生存的信息进入大脑皮质内长期保存，而学校学到的知识对于
生存来说并非不可或缺 ，因此很难被记住 。我们在学习中一定要注重反复记忆 。 【解析 】本题考查学生概
括要点、提建议的能力。
解释原因：材料中提到 “具体判定信息是否必要的 ‘关卡检查员 ’就是人脑中的海马体 ”，明确了海马体
在信息筛选中的关键作用。文中指出 “海马体的审查标准是 ‘该信息对生存而言，是否不可或缺 ’” ，说
明海马体以生存必要性作为信息筛选的尺度，这是解释大脑记不住学校知识原因的核心依据。材料还提到
“大脑将必要信息储存到长期记忆中会消费大量能量，如果将非必要的一些信息也储存到大脑里，那就是
对能量的浪费了 ”，表明海马体为了节省大脑能量，会依据生存必要性对信息进行筛选，而学校学到的知
识通常不符合这一标准，所以难以被记住。
提出建议 ：材料中明确提到 “要想让海马体将学校学到的知识划分为必要信息 ，方法只有一个 ，那就是 ‘欺
骗’海马体。我们要持续不断地将信息传送过去 ”，由此得出答案 “在学习中一定要注重反复记忆 ”。
22 ．①俗话说： “学习就是要反复地训练。 ” ②突出强调 ③特殊含义
【解析】本题考查学生正确使用标点符号的能力。
①“ 俗话说： ‘学习就是要反复地训练。 ’” ，这里直接引用了俗语，引号内的内容是直接照搬的他人话语，
所以属于直接引用。
②“ 我们无法改变 ‘根本记不住 ’这种让人发愁的状况 ” ，句子中 “根本记不住 ”加上引号，是为了突出强调这
种记忆状态，引号起到着重突出的作用。
③“ 方法只有一个，那就是 ‘欺骗 ’海马体 ” ，这里的 “欺骗 ”并非其原本字面的欺诈之意，在该语境中是指通
过特殊方式让海马体产生错觉 ，从而允许信息通过进入大脑皮质 ，赋予了 “欺骗 ”新的 、特定的含义 ，所以属
于特殊含义。
23 ．例文： 学科 网（北 京）股 份有限 公司
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科技传统，互补相生
在瞬息万变的科技发展浪潮中 ，传统手工艺的梭子渐渐沉寂 ，戏曲舞台的聚光灯黯淡 ，百年老店的门
楣在数字化浪潮中摇摇欲坠。有人悲观地断言：科技正成为传统文化的掘墓人。然而，当我们将目光投向
数字艺术的前沿 ，会发现音乐话剧 《敦煌 》用 AR 技术打破剧场的物理边界 ，《栋梁 》借助数字孪生让古建
筑“活”在虚拟世界 —— 科技与传统并非水火不容，而是互补相生。
科技与传统的碰撞 ，本质上是两种文明形态的相遇 。工业革命时期 ，机器生产替代手工劳作 ，看似斩
断了匠人与作品的情感纽带，实则为传统工艺开辟了新可能。景德镇的陶瓷匠人不再执着于手工拉坯的单
一模式 ，转而利用 3D 打印技术实现复杂器型的精准塑造 ，传统纹样在数字建模中焕发新生 ；苏州的缂丝艺
人将 CAD 设计融入图案创作，让千年丝织技艺与现代审美完美契合。这些创新并非对传统的背叛，而是用
科技的 “钥匙 ”打开了传统文化更广阔的生存空间。
当传统文化主动拥抱科技，便能创造出震撼人心的艺术奇迹。沉浸式戏剧《不眠之夜》将 VR 技术与
莎翁经典结合，观众戴上头显便能 “走进 ”麦克白的城堡，触摸中世纪的迷雾；故宫博物院开发的 “数字文物
库”，让千里之外的观众能 360 度无死角欣赏《千里江山图》的矿物颜料之美。这些案例证明，科技不是
传统文化的终结者，而是其重生的催化剂。试想，如果昆曲《牡丹亭》依然困守传统戏台，没有全息投影
营造的 “良辰美景奈何天 ”，又怎能让 Z 世代在虚拟与现实交织的梦境中，读懂杜丽娘的春情？
更深刻的是 ，科技与传统的融合正在重塑人类文明的传承范式 。日本 “人间国宝 ”级匠人通过直播平台
展示漆器制作，让全球观众见证木胎上百层髹漆的匠心；法国卢浮宫利用区块链技术为文物建立数字身份
证，实现艺术品的溯源与保护 ；中国的 “数字非遗 ”项目通过 AI 技术分析传统戏曲的唱腔规律 ，为濒危剧种
建立声音基因库。这些实践表明，科技不仅延续了文化的生命，更赋予其突破时空限制的传播力。当古老
的故事在元宇宙中找到新的载体 ，当传统技艺在数字孪生中获得永生 ，我们是否还能用 “冲击 ”二字简单定义
这场变革？
站在文明演进的长河中 ，科技与传统的相遇恰似两条奔腾的江河 。表面的碰撞激荡起浪花 ，深层的交
融却孕育出新的生态。从甲骨文的刻刀到键盘的敲击，从竹简的墨痕到数字屏的光影，人类记录文明的工
具在变，但对美的追求、对精神家园的守护从未改变。或许，真正需要改变的不是科技或传统本身，而是
我们看待二者关系的视角 —— 当我们以开放的胸怀拥抱变革，科技终将成为传统文化飞向未来的翅膀。
【详解】本题考查学生的写作能力。
审题：
这是一道引语式材料作文题。
材料共两段。材料第一段阐述了科技发展给传统文化带来的冲击，直观展现出科技发展与传统文化生存之 学科 网（北 京）股 份有限 公司
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间的冲突，引发 “科技发展正在恶化传统文化生存土壤 ”的忧虑。第二段阐述了传统文化与现代科技完美融
合的例子 ，说明利用 AR 、VR 、虚拟现实与数字孪生等技术 ，传统文化能够与现代科技完美融合 ，突破传统
局限，获得新的生命力，呈现出科技与传统文化相互促进的一面。材料看上去是对传统文化与科技发展之
间关系唱衰还是看好的对立关系，但实则可以理解为一个问题的不同发展阶段。新事物的产生必然会给稳
定不变的生活带来巨大挑战，但如果大胆地拥抱新事物，打破认知局限，创新性地去运用，那么这种积极
转变的态度就会使新事物不但不成为阻力反而还会成为助推力。科技发展一开始确实让传统文化丧失了一
部分空间和活力，但是人们在经历冲击之后努力寻找突破与融合，让传统文化有了新的生命力。
所以，立意时若立足任意一层内涵单一立意，或表达对人们一味追逐新科技而忽视其对传统文化的伤害的
忧虑，或表达积极拥抱新科技进而促进科技与传统融合发展的乐观态度，都可视为符合题意。但最佳立意
还应是辩证思考科技发展与传承传统之间的关系：既看到科技之弊，也看到科技之利；既承认科技对传统
文化的冲击，如传统行业的式微，又阐述科技带来的机遇，如创新表达形式，引导思考如何在科技浪潮中
既保护传统文化内核，又借助科技实现创新发展，思考如何实现两者的融合转化，找到两者的平衡与共生
之道。
写作时，如果能有更深刻的思考，如以传统文化为根基，合理运用科技，突出传统文化深厚底蕴是科技应
用的基础，科技只是手段；或在文化创意产品设计中，要先深入挖掘传统文化内涵，再利用科技手段进行
呈现，避免过度依赖科技而忽视文化本身价值，都是比较有深度的立意。
具体写作时，考生可以开篇列举《敦煌》《青蛇》《栋梁》等借助科技实现文化创新的案例，点明科技与
传统文化并非对立，而是能为传统文化注入新生命力。接着从传播方式革新和创新发展两个维度论述科技
对传统文化的赋能作用。传播方式上，以故宫线上展览、河南卫视节目为例，说明科技突破时空限制，助
力传统文化广泛传播 ；创新发展方面 ，结合 3D 打印 、全息投影等技术在传统工艺 、戏曲表演中的应用 ，阐
述科技为传统文化赋予新内涵 ，增强观众认同感 。然后指出科技与传统文化融合中 ，需以传统文化为根基 ，
避免因追求科技新奇而忽视文化本质，让科技成为辅助传统文化发展的工具。最后，呼应开头，强调科技
与传统文化深度融合将书写传统新生篇章，展望传统文化在科技助力下未来将绽放光彩、传承民族精神智
慧。
立意：
警惕科技之刃，毋伤传统之根。
科技，传统文化复兴的强劲引擎。
立科技潮头，应挑战机遇。
洞察科技之弊，坚守传统文心。 学科 网（北 京）股 份有限 公司
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文化为根，科技为用，构建科传新生态。
【高考祝福篇】
给 亲爱的 2025 届高考生们：
当春日的柳絮第三次掠过教室窗棂 ，当倒计时牌上的数字即将折叠成纸飞机的模样，你们正站在时光
的渡口 —— 一边是叠成山脉的试卷，一边是翻涌着星辰的远方。此刻的世界或许像被按下了加速键，但请
相信，这段被油墨与晨光浸润的岁月 ，终成为你们青春诗篇里最激昂的章节。
少年逐梦：在时代浪潮中雕刻自我
你们成长的年代 ，恰逢百年未有之大变局 。当 AI 画笔在画布上绽放出人类想象力的边界 ，当量子计算
的光芒照亮暗物质的迷宫，当 “碳中和 ”的绿意在荒漠中萌芽 ，这个时代既像四通八达的立交桥，也似等
待拓荒的原野。你们或许曾在凌晨 五点的台灯下困惑：当知识可以被搜索引擎瞬间解码，当经验不再是稀
缺资源，属于 00 后的竞争力究竟在哪里？
请记住，真正的成长从不是知识的堆砌，而是灵魂的拔节。你们目睹过 “云端课堂 ”里闪烁的眼神，
见证过河南暴雨中 “救命文档 ”的生命接力，参与过 “清朗行动 ”中对网络文明的守护。这些经历早已在
你们血脉中注入独特的基因 —— 既有 “硬核 ”的科技素养，也有 “柔软 ”的人文情怀。就像敦煌壁画中
反弹琵琶的飞天，在传统与现代的交响中，你们正用多元视角重构属于自己的坐标系。
破茧之路：与焦虑和解的勇气
此刻的书桌前，或许摆着《五年高考三年模拟》，也藏着未写完的歌词本；抽屉里有提神的咖啡，也
有偷偷夹着的偶像明信片。你们在 “内卷 ”与“躺平 ”的博弈中寻找平衡，在 “别人家的孩子 ”和“真实
的自我 ”间反复校准。那些在深夜流过的眼泪，那些与父母争执后又悄悄和解的瞬间，那些考砸后在操场
狂奔的黄昏，都是成长最真实的注脚。
想起作家麦家说过： “人生海海，山山而川，不过尔尔。 ”真正的勇者不是从不疲惫，而是含着泪继
续奔跑 。你们见过凌晨五点的校园路灯 ，听过盛夏蝉鸣里的朗朗书声 ，丈量过每张试卷的经纬 —— 这些看
似重复的日子，其实都在为某一刻的质变蓄力。就像深海中的珍珠贝，唯有经历沙粒的刺痛，才能孕育出 学科 网（北 京）股 份有限 公司
202 -
温润的光华。当你们某天回望，会发现那些被焦虑浸泡的时光，早已沉淀成生命的钙质。
星辰在望：写给未来的三行诗
即将走进考场的你们，或许正把准考证小心夹在笔记本里，或许在草稿纸上画下最后一个笑脸。请收
下这份来自时光的礼物：
第一行，写给勇气
当笔尖触碰答题卡的瞬间，请相信这三年的积淀早已刻进肌肉记忆。就像敦煌莫高窟的画工，历经千
笔万笔的勾勒，最终让飞天的衣袂在壁上永恒飘动。不必害怕偶尔的卡顿，那些空白处恰恰是生命留白的
智慧，是你区别于标准答案的独特注脚。
第二行，写给从容
考试铃响时，请听听窗外的风声 —— 它可能是秦岭的松涛 ，是东海的潮音， 是扬州的檐铃叮当， 是塞
北的雪粒叩窗。这个世界从来不止一种成功的形状：有人在实验室里破译基因密码，有人在山区小学播撒
知识火种，有人在舞台上绽放艺术光芒。就像《诗经》里的植物各有其时，你们终将在属于自己的季节里
葳蕤生长。
第三行，写给热爱
走出考场的那天，请记得抱抱陪你奋斗的伙伴，谢谢书桌前的那盏台灯，亲吻一下校园里最爱的那棵
树。这场旅程的终点，亦是新的起点 —— 当你们带着油墨香的记忆走向广阔天地，请永远保持对世界的好
奇：去看量子纠缠如何编织时空的网，去听人工智能谱写的交响曲，去用代码勾勒乡村振兴的蓝图。请相
信，真正的成长永远在路上。
此刻，校园里的紫藤花正爬满长廊，就像你们三年的时光在记忆里悄然结网。当你们背着书包穿过校
门，不必频频回望 —— 那些熬夜刷题的夜晚，那些与老师争辩的课堂，那些和朋友分享的零食与秘密，早
已化作你们眼底的光。
愿你们在考场上如苏轼 “胸有成竹 ”，亦能如陶渊明 “心远地自偏 ”；愿你们既懂 “数风流人物，还
看今朝 ”的豪情 ，也知 “一屋不扫 ，何以扫天下 ”的沉淀 ；愿你们在未来某天回首时 ，能对着 2025 年的自
己说一声： “谢谢你曾如此认真地活过。 ”
天地为纸，青春作笔。你们只管向前奔跑吧，因为所有值得抵达的远方，都自有光芒为你们铺就。 学科 网（北 京）股 份有限 公司
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2025 年春日于梧桐树下
一一位永远为你们亮着灯的人 |
8576 | https://emcrit.org/ibcc/pericarditis/ | Internet Book of Critical Care (IBCC)
Online Medical Education on Emergency Department (ED) Critical Care, Trauma, and Resuscitation
You are here: Home / IBCC / Acute pericarditis
Acute pericarditis
by Josh Farkas
CONTENTS
Definitions
Epidemiology
Clinical presentation
Bedside examination
ECG
Laboratory studies
Pericardiocentesis
Cardiac MRI
Diagnostic criteria
Causes
Evaluation for the etiology of pericarditis
Treatment of acute pericarditis
Idiopathic pericarditis
Early post-MI pericarditis
Late post-MI pericarditis
Purulent pericarditis
Prognosis
Related topics:
Colchicine
definitions (back to contents)
Acute pericarditis: <4-6 weeks.
Incessant pericarditis: symptoms present >4-6 weeks but <3 months.
Chronic pericarditis: symptoms present >3 months.
Recurrent pericarditis: recurrence after a symptom-free period of >1 month.
Myopericarditis: pericarditis with elevated troponin and normal cardiac function. (Braunwald 12e)
Perimyocarditis: pericarditis and myocarditis with impaired cardiac function. (Braunwald 12e)
epidemiology (back to contents)
Pericarditis may occur at any age but is more common among younger people and men. (35595949)
Pericarditis accounts for ~5% of emergency department presentations for nonischemic chest pain. (Gaggin 2021)
One-third of patients will have a recurrence within 1.5 years. (Gaggin 2021)
15% of patients with acute pericarditis also have concomitant myocarditis. (38702128)
clinical presentation (back to contents)
chest pain due to pericarditis (~90% of patients)
Location:
Anywhere in the chest, but usually central.
It is often more localized than ischemic pain (parietal rather than visceral pain).
Radiation:
Classically, the pain radiates to the trapezius ridge (between the neck and shoulder) on one or both sides.
May radiate to the tip of the shoulder (due to diaphragmatic irritation).
Radiation down the left arm may occur (simulating ischemia). (Hurst 15e)
Quality:
Generally sharp and stabbing (due to parietal irritation).
However, can cause a steady ache that mimics MI.
Severity: often quite severe. (Braunwald 12e)
Chronicity:
Onset is often sudden.
May last hours to days.
Often occurs episodically.
Remitting/exacerbating factors:
Often alleviated by sitting up and leaning forward.
Generally pleuritic (exacerbated by deep breathing).
Exacerbated by swallowing (given the proximity of the heart to the esophagus).
Pain is more often absent in subacute etiologies (tuberculosis, malignancy, uremia, post-radiotherapy).
associated symptoms
Low-grade fever may occur (~15%).
Dyspnea (~33%) may result from respiratory splinting due to pain.
Less frequent symptoms may include hiccups, dysphagia, or cough. (Hurst 15e)
bedside examination (back to contents)
pericardial friction murmur
Sensitivity might be ~30%, with high specificity.
Rub may have 1-3 components (including ventricular systole, early ventricular filling, and atrial contraction).
It is often best heard at the left lower sternal border while the patient leans forward.
pericardial effusion on POCUS
Sensitivity might be ~60%. However, the presence of an effusion may support the diagnosis of pericarditis. (35595949)
Effusions are usually small (<10 mm, measuring the largest diameter in end-diastole). (Braunwald 12e)
If a large effusion (>20 mm) is present, this is important:
(1) There is an increased risk of pericardial tamponade.
(2) This suggests an underlying cause, such as malignancy or chronic inflammation (idiopathic or viral pericarditis generally doesn't cause a large effusion). (Sadhu 2023)
A complex/heterogeneous effusion suggests the possibility of malignancy or infection.
ECG (back to contents)
Sensitivity is reasonably high for patients presenting with chest pain (perhaps ~80%).
ECG findings in pericarditis are discussed here: 📖
laboratory studies (back to contents)
troponin
Troponin is frequently elevated (with sensitivity depending on the assay).
Troponin elevation is disproportionately low, as compared to MI.
Elevated troponin may imply a certain degree of myopericarditis (rather than purely pericarditis).
CRP
CRP elevation has a sensitivity of ~80%, but such elevations are often minimal (often utilizing a cutoff value of >3 mg/L). (Griffin 2022, 39235771, 38702128)
Causes of an initial normal CRP may include:
CRP levels rise over time, so CRP measured >12 hours after symptom onset may have a sensitivity closer to 95%. (35595949)
Status post-treatment with NSAIDs, colchicine, or steroids. (39235771)
Higher CRP levels may predict a greater risk of recurrence.
Serial measurement of CRP may help monitor disease activity over time. CRP generally normalizes within 1-2 weeks of therapy. (Braunwald 12e) Persistently elevated CRP predicts an elevated risk of recurrent pericarditis or pericardial constriction. (Hurst 15e)
leukocytosis
Modest leukocytosis is common.
WBC >13,000-14,000 may suggest a specific underlying etiology (rather than idiopathic pericarditis). (Braunwald 12e)
pericardiocentesis (back to contents)
Most patients will not require pericardiocentesis.
indications for pericardiocentesis may include:
Suspected bacterial, tuberculous, or neoplastic pericarditis (and effusion is large enough to access safely).
Tamponade.
Symptomatic moderate-large effusion plus the failure of anti-inflammatory therapy. (Gaggin 2021)
Laboratory studies to obtain: 📖
cardiac MRI (CRMI) (back to contents)
situations where CMRI may be useful
Diagnosis of pericarditis remains unclear despite the above investigations.
Definite pericarditis but concern for a clinically significant component of myocarditis (i.e., perimyocarditis).
Limited echocardiographic views.
key findings
LGE (late gadolinium enhancement):
This reflects the neovascularization of the pericardium.
LGE may be seen in acute, subacute, or chronic pericarditis (but generally resolves following resolution of pericarditis). LGE is ~95% sensitive for pericarditis. (38702128)
LGE isn't entirely specific for pericarditis (e.g., it may be seen in 44% of patients with prior cardiac surgery). (39235771)
T2-STIR (short-tau inversion recovery) signal intensity: Increased T2-STIR indicates edema, neovascularization, and/or granulation tissue. (usually edema; 39235771)
Enhancement of thickened pericardium on T1-weighted spin-echo sequence: This indicates active inflammation. (38702128)
CMRI allows for the staging of pericarditis
In acute pericarditis, there is pericardial edema on T2 sequences and late gadolinium enhancement (LGE, which indicates inflammation).
Subacute/chronic pericarditis is suggested by the resolution of pericardial edema but with the persistence of late gadolinium enhancement.
Resolution of both edema and LGE reveals resolution. (35595949)
diagnostic criteria (back to contents)
The diagnosis of pericarditis generally requires two of the following four criteria. (ESC 2023, 37622654) However, be careful because the precise performance of these criteria isn't clear.
Chest pain consistent with pericarditis (~90% sensitive).
ECG changes consistent with pericarditis, including isolated PR depression (~80% sensitive).
Pericardial effusion on echocardiography that is new or worsening (~60% sensitive).
Pericardial friction rub (~30% sensitive).
⚠️ These diagnostic criteria may be inadequate to differentiate pericarditis from perimyocarditis (which involves a combination of pericarditis -plus- myocarditis with impaired cardiac function). Importantly, perimyocarditis is a more worrisome disorder that should be treated similarly to myocarditis. (35595949)
causes (back to contents)
Idiopathic.
Infection:
Viral infection (in practice, an exhaustive evaluation for viruses isn't performed, so idiopathic acute pericarditis is likely due to an unidentified virus):
Enteroviruses (coxsackievirus, echovirus).
Adenoviruses.
Influenza.
COVID-19.
Herpesviruses (VZV, EBV, CMV).
HIV (either HIV itself or opportunistic infections).
HCV. (35595949)
Lyme.
Tuberculosis (generally causes a more subacute course).
Purulent pericarditis (1-2%) – following cardiothoracic surgery, esophageal rupture, or ruptured ring abscess in endocarditis.
Coxiella burnetii (Q fever).
Fungal (very rare)
Endemic fungal infections (histoplasmosis, coccidiomycosis, blastomycosis).
Candida (usually in the context of immunocompromise, IVDU, or after surgery).
Parasitic (amebiasis, toxoplasmosis, echinococcus, trichinosis)
Inflammatory:
Lupus, rheumatoid arthritis, scleroderma, Sjogren syndrome, dermatomyositis, mixed connective tissue disorder.
GPA (granulomatosis with polyangiitis), EGPA (eosinophilic granulomatosis with polyangiitis).
Sarcoidosis.
Adult-onset Still's disease.
Inflammatory bowel disease.
IgG-related pericarditis.
Familial Mediterranean fever (FMF).
Metabolic:
Uremia.
Myxedema.
Neoplastic (lung ~40%; breast ~25%; hematologic ~20%; also melanoma and thyroid). (39235771)
Post-cardiac injury:
Early post-MI pericarditis (discussed further below: ⚡️).
Post-cardiac injury syndrome: 📖
Medications-related:
Lupus-like syndrome (procainamide, hydralazine, isoniazid, phenytoin).
Antineoplastic drugs (often associated with cardiomyopathy; doxorubicin, daunorubicin, cytosine arabinoside, 5-fluorouracil, cyclophosphamide, immune checkpoint inhibitors).
Hypersensitivity pericarditis with eosinophilia: penicillins, amiodarone, mesalamine, clozapine, minoxidil, dantrolene, thiazides, sulfa drugs, cyclosporine, bromocriptine, GM-CSF, anti-TNF agents. (Hurst's 15e)
Radiation.
Traumatic (e.g., after surgery or blunt chest injury).
evaluation for the etiology of pericarditis (back to contents)
The extent of this evaluation will vary depending on the clinical context.
clinical features suggesting non-idiopathic disease:
Subacute-chronic presentation (developing over days to weeks).
Immunosuppression.
Related trauma; use of anticoagulant therapy.
Inadequate response to NSAID therapy or recurrent acute pericarditis.
Fever >38C.
Large pericardial effusion.
Evidence of myocarditis. (Hurst 15e)
focused historical questions
Evaluation for connective tissue diseases:
Small joint arthritis?
Skin changes (including rash/thickening/tightening/fissuring)?
Raynaud phenomenon? (Associated with scleroderma, Sjogren syndrome, MCTD).
Dry eyes /dry mouth?
Muscle weakness?
Evaluation for tuberculosis:
Geography:
Born in a country with high TB rates (strongest risk factor).
Lived in an endemic area.
Extensive travel to a highly endemic area.
Personal history of tuberculosis.
Close contact with active tuberculosis.
formal echocardiography
Echocardiography is generally recommended.
Ventricular dysfunction raises the possibility of perimyocarditis.
chest radiograph
Evaluate for evidence of pneumonia, pneumonitis, or tuberculosis.
laboratory studies may include:
Eosinophilia (may suggest drug reaction).
Renal function.
TSH (thyroid stimulating hormone).
ANA (antinuclear antibody) only if there are other features of systemic autoimmune disease (low-level titers are often seen in idiopathic pericarditis). (39235771)
RF (rheumatoid factor)
Lyme serology.
HIV serology.
HCV serology.
Evaluation for tuberculosis.
(Also consider obtaining a CRP level, discussed above: ⚡️)
treatment of acute pericarditis (back to contents)
idiopathic pericarditis
Front-line therapy is a combination of colchicine plus either aspirin or an NSAID.
aspirin or NSAIDs
Enormous doses should be utilized to treat underlying inflammation. Failure to use an adequate dose may lead to analgesia without adequate treatment of the underlying disease process (which may lead to disease recurrence).
Options include: (38702128)
Ibuprofen 600-800 mg TID.
Indomethacin 25-50 mg TID.
Aspirin 750-1,000 mg TID.
Gastric protection should be utilized to prevent ulceration (twice daily proton pump inhibitor).
Choice of agent:
There are no head-to-head trials to determine which agent is most effective. Consequently, the safest agent might be preferable.
Specific indications to use aspirin include:
Post-MI (NSAIDs can impair infarction healing).
Risk of acute kidney injury.
Coronary artery stents.
Older age. (Griffin 2022)
These therapies usually relieve chest pain within days. (Gaggin 2021) Failure to respond within a week suggests an alternative diagnosis (e.g., malignancy, post-cardiac injury syndromes, systemic inflammatory diseases, infection). (38702128)
Duration of therapy is usually 7-14 days at maximal dose. Depending on the persistence of symptoms and CRP level, a gradual taper may be instituted over 3-4 weeks. (Griffin 2022) Continuation for one month beyond the resolution of chest pain might reduce the risk of future constrictive pericarditis. (Hurst 15e)
colchicine
Discussed further below.
steroid
Acute pericarditis is generally exquisitely sensitive to steroid therapy. However, steroids should be avoided if possible because it frequently leads to steroid dependence. There is a high risk of recurrent pericarditis (especially when weaning off steroids).
Indications for steroid might include:
Pericarditis is unresponsive to a reasonable duration of therapy with maximal doses of aspirin/NSAID and colchicine (especially with persistently elevated CRP).
Patients with underlying connective tissue disease.
Patient with a contraindication to NSAIDs or aspirin.
Pregnancy. (35595949)
Constrictive/effusive pericarditis on echocardiography or CMRI. (Griffin 2022)
Low doses should be utilized, for example, 0.2-0.5 mg/kg (38702128) or even as low as 10-25 mg/day. Very gradual tapering of steroids is needed. Pericarditis will often flare during weaning off (at around 10-15 mg/day prednisone dose) (38702128) which ideally should be treated without an escalation in steroid dose (but rather utilizing colchicine and aspirin/NSAIDs).
analgesia
Additional analgesia may be required (beyond aspirin or NSAIDs as above).
Acetaminophen and even opioids may be helpful.
Chest pain that doesn't rapidly respond to the above treatments isn't an indication of steroids.
anticoagulants
It may be wise to avoid anticoagulants if possible, as this is a risk factor for complicated disease.
However, there doesn't seem to be solid evidence that anticoagulant use increases the risk of hemorrhage or tamponade, so they may be utilized if indicated.
early post-MI pericarditis
basics
It occurs within four days after MI (due to transmural necrosis that directly causes pericardial inflammation).
management
This is primarily transient and self-limiting.
Treatment may include aspirin 500 mg q8-12 hours.
Therapy beyond 5-7 days usually isn't required. (ESC 2023, 37622654)
post-cardiac injury syndrome
Discussed here: 📖
purulent pericarditis
Urgent pericardiocentesis with fluid analysis.
Systemic antibiotic therapy.
Consider additional pericardial interventions (e.g., intrapericardial thrombolysis, subxiphoid pericardiotomy, or even pericardiectomy to manage persistent infection).
prognosis (back to contents)
potential complications
Pericardial tamponade.
Atrial fibrillation.
Recurrent pericarditis.
Constrictive pericarditis.
predictors of complicated acute pericarditis
Major predictors:
Fever >38C.
Subacute onset.
Large pericardial effusion (>20 mm end-diastolic depth).
Cardiac tamponade.
Lack of response to aspirin/NSAIDs after at least one week of therapy.
Minor predictors:
Concomitant myocarditis (perimyocarditis).
Immunosuppression.
Trauma.
Oral anticoagulant. (26320112)
colchicine (back to contents)
contraindications, drug interactions, side effects 👎
contraindications
Renal failure (GFR <30 ml/min), especially if combined with hepatic dysfunction. This is a relative contraindication (reduced doses may be utilized for some indications).
Severe hepatic failure.
Interacting medications (see section below).
Preexisting cytopenia(s).
drug-drug interactions
P-glycoprotein inhibitors may markedly increase plasma levels.
CYP3A4 is the primary enzyme involved in colchicine metabolism.
Combined inhibitors of P-glycoprotein plus CYP3A4 are especially problematic with colchicine; these include the following:
Clarithromycin, erythromycin.
Antivirals (e.g., ritonavir).
Amiodarone, dronaderone.
Diltiazem, verapamil.
Ranolazine.
Azole antifungals.
Statins (and daptomycin):
Myotoxicity risk is increased by coadministration with statins.
Atorvastatin and simvastatin are also substrates for CYP3A4 and P-glycoprotein. These may increase colchicine exposure.
side effects
GI side effects (especially diarrhea; also nausea, emesis, anorexia, abdominal pain).
GI side effects necessitate dose reduction or discontinuation in 10% of patients. (39235771)
🚨 Gastrointestinal symptoms may represent the first sign of toxicity. If ignored, toxicity may progress. Progressive gastrointestinal toxicity may lead to hemorrhagic gastroenteritis, ileus, and bowel perforation.
Myotoxicity (muscle pain, weakness, elevated CK, myotonia, rhabdomyolysis).
Cytopenia(s) including neutropenia, thrombocytopenia, aplastic anemia, pancytopenia.
B12 deficiency (due to latering the ileal mucosa). (Wellington 3e)
Sensory motor neuropathy (more likely in chronic therapy).
Dermatological (alopecia, rash, purpura, maculopapular rash).
Hepatotoxicity.
indications, advantages 👍
Pericarditis.
Gout.
Familial Mediterranean fever.
Behcet's syndrome.
Sweet's syndrome.
dosing & monitoring
pericarditis (dosing & regimen)
Dose:
The most commonly utilized dose is weight-based: 0.6 mg daily (<70 kg) or 0.6 mg BID (>70 kg).
Alternatively, colchicine may be started at 0.6 mg daily and up-titrated to 0.6 BID in a few days if tolerated without diarrhea.
Continue for three months (well past the resolution of symptoms and inflammatory markers). (Griffin 2022) For patients at increased risk of toxicity, consider following the patient's renal function, creatine kinase, transaminases, and blood count. (Sadhu 2023)
acute gout flare
Start with 1.2 mg, then 0.6 mg one hour later. For ongoing symptoms, additional doses may be considered (but no more than 2.5 mg within 24 hours). Then stop and hold off on any additional colchicine for 3 days.
dose reductions
GFR <30 ml/min:
Relative contraindication for pericarditis.
0.3 mg daily could be used for prophylaxis against gout. (38019947)
Severe hepatic impairment: Reduce dose or use alternative therapy.
Elderly: Exercise caution (may have reduced hepatic and renal function).
Patients <50 kg: Consider lower doses.
monitoring
Follow renal function: may require dose adjustment.
Follow CBC to evaluate for cytopenias.
Follow CK.
pharmacology
Chemical properties:
Molecular weight: 399 g/mol.
LogP: 1.8
Absorption:
Well absorbed, but bioavailability is ~45% due to first-pass metabolism and efflux transport mechanisms (subject to efflux via P-glycoprotein).
Cmax occurs ~1.5 hours after ingestion.
Distribution:
Protein binding is relatively low (~40%), which facilitates tissue penetration.
Vd is ~5-8 L/kg (with a high affinity for microtubule proteins). Colchicine accumulates in leukocytes, the kidney, the liver, the spleen, and the brain. The drug becomes trapped in tissues, with prolonged pharmacological effects.
Metabolism:
Hepatic metabolism is primarily by CYP3A4.
Elimination:
40-65% is excreted unchanged in the urine.
Enterohepatic recirculation and biliary excretion also contribute to elimination.
Dialysis doesn't effectively remove colchicine (based on its large volume of distribution).
Half-life & duration of action:
The half-life after multiple doses is ~26-30 hours.
If the GFR is 10-30 ml/min, clearance is halved. (38019947)
questions & discussion (back to contents)
To keep this page small and fast, questions & discussion about this post can be found on another page here.
Guide to emoji hyperlinks
= Link to online calculator.
= Link to Medscape monograph about a drug.
= Link to IBCC section about a drug.
= Link to IBCC section covering that topic.
= Link to FOAMed site with related information.
📄 = Link to open-access journal article.
= Link to supplemental media.
References
31918837 Chiabrando JG, Bonaventura A, Vecchié A, Wohlford GF, Mauro AG, Jordan JH, Grizzard JD, Montecucco F, Berrocal DH, Brucato A, Imazio M, Abbate A. Management of Acute and Recurrent Pericarditis: JACC State-of-the-Art Review. J Am Coll Cardiol. 2020 Jan 7;75(1):76-92. doi: 10.1016/j.jacc.2019.11.021 [PubMed]
35595949 Lazarou E, Tsioufis P, Vlachopoulos C, Tsioufis C, Lazaros G. Acute Pericarditis: Update. Curr Cardiol Rep. 2022 Aug;24(8):905-913. doi: 10.1007/s11886-022-01710-8 [PubMed]
37622654 Byrne RA, Rossello X, Coughlan JJ, Barbato E, Berry C, Chieffo A, Claeys MJ, Dan GA, Dweck MR, Galbraith M, Gilard M, Hinterbuchner L, Jankowska EA, Jüni P, Kimura T, Kunadian V, Leosdottir M, Lorusso R, Pedretti RFE, Rigopoulos AG, Rubini Gimenez M, Thiele H, Vranckx P, Wassmann S, Wenger NK, Ibanez B; ESC Scientific Document Group. 2023 ESC Guidelines for the management of acute coronary syndromes. Eur Heart J. 2023 Oct 12;44(38):3720-3826. doi: 10.1093/eurheartj/ehad191 [PubMed]
38019947 Stamp LK, Horsley C, Te Karu L, Dalbeth N, Barclay M. Colchicine: the good, the bad, the ugly and how to minimize the risks. Rheumatology (Oxford). 2024 Apr 2;63(4):936-944. doi: 10.1093/rheumatology/kead625 [PubMed]
38702128 Malik AA, Lloyd JW, Anavekar NS, Luis SA. Acute and Complicated Inflammatory Pericarditis: A Guide to Contemporary Practice. Mayo Clin Proc. 2024 May;99(5):795-811. doi: 10.1016/j.mayocp.2024.01.012 [PubMed]
39235771 Cremer PC, Klein AL, Imazio M. Diagnosis, Risk Stratification, and Treatment of Pericarditis: A Review. JAMA. 2024 Sep 5. doi: 10.1001/jama.2024.12935 [PubMed]
Books:
Gaggin, H. K., & Januzzi, J. L., Jr. (2021). MGH Cardiology Board Review. Springer Science & Business Media.
Libby, P. (2021). Braunwald’s Heart Disease,2 Vol set: A Textbook of Cardiovascular Medicine. Elsevier.
Griffin BP, Kapadia SR, and Menon V. The Cleveland Clinic Cardiology Board Review. (2022). Lippincott Williams & Wilkins.
Sadhu, J., Husaini, M., & Williams, D. (2022). The Washington Manual Cardiology Subspecialty Consult. Lippincott Williams & Wilkins.
Fuster, V., & Narula, J. (2022). Fuster and Hurst’s The Heart, 15th edition. McGraw-Hill Education / Medical.
The Internet Book of Critical Care is an online textbook written by Josh Farkas (@PulmCrit), an associate professor of Pulmonary and Critical Care Medicine at the University of Vermont.
Who We Are
We are the EMCrit Project, a team of independent medical bloggers and podcasters joined together by our common love of cutting-edge care, iconoclastic ramblings, and FOAM.
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8577 | https://www.cs.toronto.edu/~nisarg/papers/error.aij.pdf | Artificial Intelligence 231 (2016) 1–16 Contents lists available at ScienceDirect Artificial Intelligence www.elsevier.com/locate/artint Voting rules as error-correcting codes ✩ Ariel D. Procaccia, Nisarg Shah ∗, Yair Zick Computer Science Department, Carnegie Mellon University, Pittsburgh, PA 15213, USA a r t i c l e i n f o a b s t r a c t Article history: Received 21 January 2015 Received in revised form 13 October 2015 Accepted 20 October 2015 Available online 27 October 2015 Keywords: Social choice Voting Ground truth Adversarial noise Error-correcting codes We present the first model of optimal voting under adversarial noise. From this viewpoint, voting rules are seen as error-correcting codes: their goal is to correct errors in the input rankings and recover a ranking that is close to the ground truth. We derive worst-case bounds on the relation between the average accuracy of the input votes, and the accuracy of the output ranking. Empirical results from real data show that our approach produces significantly more accurate rankings than alternative approaches.
© 2015 Elsevier B.V. All rights reserved.
1. Introduction Social choice theory develops and analyzes methods for aggregating the opinions of individuals into a collective decision. The prevalent approach is motivated by situations in which opinions are subjective, such as political elections, and focuses on the design of voting rules that satisfy normative properties .
An alternative approach, which was proposed by the marquis de Condorcet in the 18th Century, had confounded schol-ars for centuries (due to Condorcet’s ambiguous writing) until it was finally elucidated by Young . The underlying assumption is that the alternatives can be objectively compared according to their true quality. In particular, it is typically assumed that there is a ground truth ranking of the alternatives. Votes can be seen as noisy estimates of the ground truth, drawn from a specific noise model. For example, Condorcet proposed a noise model where — roughly speaking — each voter (hereinafter, agent) compares every pair of alternatives, and orders them correctly (according to the ground truth) with probability p > 1/2; today an equivalent model is attributed to Mallows . Here, it is natural to employ a voting rule that always returns a ranking that is most likely to coincide with the ground truth, that is, the voting rule should be a maximum likelihood estimator (MLE).
Although Condorcet could have hardly foreseen this, his MLE approach is eminently applicable to crowdsourcing and human computation systems, which often employ voting to aggregate noisy estimates; EteRNA is a wonderful example, as explained by Procaccia et al. . Consequently, the study of voting rules as MLEs has been gaining steam in the last decade [16,15,19,33,32,25,30,2–4,27,12,13].
Despite its conceptual appeal, a major shortcoming of the MLE approach is that the MLE voting rule is specific to a noise model, and that noise model — even if it exists for a given setting — may be difficult to pin down . Caragiannis et al. [12,13] have addressed this problem by relaxing the MLE constraint: they only ask that the probability of the voting ✩A preliminary version of the paper appears in the Proceedings of the 29th AAAI Conference on Artificial Intelligence (AAAI), 2015.
Corresponding author.
E-mail addresses: arielpro@cs.cmu.edu (A.D. Procaccia), nkshah@cs.cmu.edu (N. Shah), yairzick@cmu.edu (Y. Zick).
0004-3702/© 2015 Elsevier B.V. All rights reserved.
2 A.D. Procaccia et al. / Artificial Intelligence 231 (2016) 1–16 rule returning the ground truth go to one as the number of votes goes to infinity. This allows them to design voting rules that uncover the ground truth in a wide range of noise models; however, they may potentially require an infinite amount of information.
Our approach. In this paper, we propose a fundamentally different approach to aggregating noisy votes. Instead of assuming probabilistic noise, we assume a known upper bound on the “total noise” in the input votes, and allow the input votes to be adversarial subject to the upper bound. We emphasize that in potential application domains there is no adversary that actively inserts errors into the votes; we choose an adversarial error model to be able to correct errors even in the worst case. This style of worst-case analysis — where the worst case is assumed to be generated by an adversary — is prevalent in many branches of computer science, e.g., in the analysis of online algorithms , and in machine learning [23,9].
We wish to design voting rules that do well in this worst-case scenario. From this viewpoint, our approach is closely related to the extensive literature on error-correcting codes. One can think of the votes as a repetition code: each vote is a transmitted noisy version of a “message” (the ground truth). The task of the “decoder” is to correct adversarial noise and recover the ground truth, given an upper bound on the total error. The question is: how much total error can this “code” allow while still being able to recover the ground truth?
In more detail, let d be a distance metric on the space of rankings. As an example, the well-known Kendall tau (KT) distance between two rankings measures the number of pairs of alternatives on which the two rankings disagree. Suppose that we receive n votes over the set of alternatives {a, b, c, d}, for an even n, and we know that the average KT distance between the votes and the ground truth is at most 1/2. Can we always recover the ground truth? No: in the worst-case, exactly n/2 agents swap the two highest-ranked alternatives and the rest report the ground truth. In this case, we observe two distinct rankings (each n/2 times) that only disagree on the order of the top two alternatives. Both rankings have an average distance of 1/2 from the input votes, making it impossible to determine which of them is the ground truth.
Let us, therefore, cast a larger net. Inspired by list decoding of error-correcting codes (see, e.g., ), our main research question is: Fix a distance metric d. Suppose that we are given n noisy rankings, and that the average distance between these rankings and the ground truth is at most t. We wish to recover a ranking that is guaranteed to be at distance at most k from the ground truth. How small can k be, as a function of n and t?
Our results. We observe that for any metric d, one can always recover a ranking that is at distance at most 2t from the ground truth, i.e., k ≤2t. We also show that one can pick, in polynomial time, a ranking from the given noisy rankings that provides a weaker 3t upper bound. We complement the upper bounds by providing a lower bound of (roughly) k ≥t/2 that holds for every distance metric. We also show that an extremely mild assumption on the distance metric improves the lower bound to (roughly) k ≥t. In addition, we consider the four most popular distance metrics used in the social choice literature, and prove a tight lower bound of (roughly) k ≥2t for each metric. This lower bound is our main theoretical result; the construction makes unexpected use of Fermat’s Polygonal Number Theorem.
The worst-case optimal voting rule in our framework is defined with respect to a known upper bound t on the average distance between the given rankings and the ground truth. However, we show that the voting rule which returns the ranking minimizing the total distance from the given rankings — which has strong theoretical support in the literature — serves as an approximation to our worst-case optimal rule, irrespective of the value of t. We leverage this observation to provide theoretical performance guarantees for our rule in cases where the error bound t given to the rule is an underestimate or overestimate of the tightest upper bound.
Finally, we test our worst-case optimal voting rules against many well-known voting rules, on two real-world datasets , and show that the worst-case optimal rules exhibit superior performance as long as the given error bound t is a reasonable overestimate of the tightest upper bound.
Related work. Our work is related to the extensive literature on error-correcting codes that use permutations (see, e.g., , and the references therein), but differs in one crucial aspect. In designing error-correcting codes, the focus is on two choices: i) the codewords, a subset of rankings which represent the “possible ground truths”, and ii) the code, which converts every codeword into the message to be sent. These choices are optimized to achieve the best tradeoff between the number of errors corrected and the rate of the code (efficiency), while allowing unique identification of the ground truth. In contrast, our setting has fixed choices: i) every ranking is a possible ground truth, and ii) in coding theory terms, our setting con-strains us to the repetition code. Both restrictions (inevitable in our setting) lead to significant inefficiencies, as well as the impossibility of unique identification of the ground truth (as illustrated in the introduction). Our research question is reminiscent of coding theory settings where a bound on adversarial noise is given, and a code is chosen with the bound on the noise as an input to maximize efficiency (see, e.g., ).
List decoding (see, e.g., ) relaxes classic error correction by guaranteeing that the number of possible messages does not exceed a small quota; then, the decoder simply lists all possible messages. The motivation is that one can simply scan the list and find the correct message, as all other messages on the list are likely to be gibberish. In the voting context, one cannot simply disregard some potential ground truths as nonsensical; we therefore select a ranking that is close to every possible ground truth.
A.D. Procaccia et al. / Artificial Intelligence 231 (2016) 1–16 3 Our model is also reminiscent of the distance rationalizability framework from the social choice literature . In this framework, there is a fixed set of “consensus profiles” that admit an obvious output. Given a profile of votes, one finds the closest consensus profile (according to some metric), and returns the obvious output for that profile. Our model closely resembles the case where the consensus profiles are strongly unanimous, i.e., they consist of repetitions of a single ranking (which is also the ideal output). The key difference in our model is that instead of focusing solely on the closest ranking (strongly unanimous profile), we need to consider all rankings up to an average distance of t from the given profile — as they are all plausible ground truths — and return a single ranking that is at distance at most k from all such rankings.
A bit further afield, Procaccia et al. study a probabilistic noisy voting setting, and quantify the robustness of voting rules to random errors. Their results focus on the probability that the outcome would change, under a random transposition of two adjacent alternatives in a single vote from a submitted profile, in the worst-case over profiles. Their work is different from ours in many ways, but perhaps most importantly, they are interested in how frequently common voting rules make mistakes, whereas we are interested in the guarantees of optimal voting rules that avoid mistakes.
2. Preliminaries Let A be the set of alternatives, and |A| = m. Let L(A) be the set of rankings over A. A vote σ is a ranking in L(A), and a profile π ∈L(A)n is a collection of n rankings. A voting rule f : L(A)n →L(A) maps every profile to a ranking.1 We assume that there exists an underlying ground truth ranking σ ∗∈L(A) of the alternatives, and the votes are noisy estimates of σ ∗. We use a distance metric d over L(A) to measure errors; the error of a vote σ with respect to σ ∗is d(σ, σ ∗), and the average error of a profile π with respect to σ ∗is d(π, σ ∗) = (1/n) · σ∈π d(σ, σ ∗). We consider four popular distance metrics over rankings in this paper.
• The Kendall tau (KT) distance, denoted dKT, measures the number of pairs of alternatives over which two rankings disagree. Equivalently, it is also the minimum number of swaps of adjacent alternatives required to convert one ranking into another.
• The (Spearman’s) Footrule (FR) distance, denoted dFR, measures the total displacement of all alternatives between two rankings, i.e., the sum of the absolute differences between their positions in two rankings.
• The Maximum Displacement (MD) distance, denoted dMD, measures the maximum of the displacements of all alternatives between two rankings.
• The Cayley (CY) distance, denoted dCY, measures the minimum number of swaps (not necessarily of adjacent alternatives) required to convert one ranking into another.
All four metrics described above are neutral: A distance metric is called neutral if the distance between two rankings is independent of the labels of the alternatives; in other words, choosing a relabeling of the alternatives and applying it to two rankings keeps the distance between them invariant.
3. Worst-case optimal rules Suppose we are given a profile π of n noisy rankings that are estimates of an underlying true ranking σ ∗. In the absence of any additional information, any ranking could potentially be the true ranking. However, because essentially all crowdsourcing methods draw their power from the often-observed fact that individual opinions are accurate on average, we can plausibly assume that while some agents may make many mistakes, the average error is fairly small. An upper bound on the average error may be inferred by observing the collected votes, or from historical data (but see the next section for the case where this bound is inaccurate).
Formally, suppose we are guaranteed that the average distance between the votes in π and the ground truth σ ∗is at most t according to a metric d, i.e., d(π, σ ∗) ≤t. With this guarantee, the set of possible ground truths is given by the “ball” of radius t around π.
Bd t (π) = {σ ∈L(A) | d(π,σ) ≤t}.
Note that we have σ ∗∈Bd t (π) given our assumption; hence, Bd t (π) ̸= ∅. We wish to find a ranking that is as close to the ground truth as possible. Since our approach is worst case in nature, our goal is to find the ranking that minimizes the maximum distance from the possible ground truths in Bd t (π). For a set of rankings S ⊆L(A), let its minimax ranking, denoted MiniMaxd(S), be defined as follows.2 MiniMaxd(S) = arg min σ∈L(A) max σ ′∈S d(σ,σ ′).
1 They are known as social welfare functions, which differ from social choice functions that choose a single winning alternative.
2 We use MiniMaxd(S) to denote a single ranking. Ties among multiple minimizers can be broken arbitrarily; our results are independent of the tie-breaking scheme.
4 A.D. Procaccia et al. / Artificial Intelligence 231 (2016) 1–16 Table 1 Application of the optimal voting rules on π.
Voting rule Possible ground truths Bd t (π) Output ranking OPTdKT (1.5,π), OPTdCY (1,π) a ≻b ≻c, a ≻c ≻b, b ≻a ≻c a ≻b ≻c OPTdFR(2,π), OPTdMD(1,π) a ≻b ≻c, a ≻c ≻b a ≻b ≻c, a ≻c ≻b Let the minimax distance of S, denoted kd(S), be the maximum distance of MiniMaxd(S) from the rankings in S according to d. Thus, given a profile π and the guarantee that d(π, σ ∗) ≤t, the worst-case optimal voting rule OPTd returns the minimax ranking of the set of possible ground truths Bd t (π). That is, for all profiles π ∈L(A)n and t > 0, OPTd(t,π) = MiniMaxd Bd t (π) .
Furthermore, the output ranking is guaranteed to be at distance at most kd(Bd t (π)) from the ground truth. We overload notation, and denote kd(t, π) = kd(Bd t (π)), and kd(t) = max π∈L(A)n kd(t,π).
While kd is explicitly a function of t, it is also implicitly a function of n. Hereinafter, we omit the superscript d whenever the metric is clear from context. Let us illustrate our terminology with a simple example.
Example 1. Let A = {a, b, c}. We are given profile π consisting of 5 votes: π = {2 × (a ≻b ≻c), a ≻c ≻b, b ≻a ≻c, c ≻a ≻b}.
The maximum distances between rankings in L(A) allowed by dKT, dFR, dMD, and dCY are 3, 4, 2, and 2, respectively; let us assume that the average error limit is half the maximum distance for all four metrics.3 Consider the Kendall tau distance with t = 1.5. The average distances of all 6 rankings from π are given below.
dKT(π,a ≻b ≻c) = 0.8 dKT(π,a ≻c ≻b) = 1.0 dKT(π,b ≻a ≻c) = 1.4 dKT(π,b ≻c ≻a) = 2.0 dKT(π,c ≻a ≻b) = 1.6 dKT(π,c ≻b ≻a) = 2.2 Thus, the set of possible ground truths is BdKT 1.5 (π) = {a ≻b ≻c, a ≻c ≻b, b ≻a ≻c}. This set has a unique minimax ranking OPTdKT (1.5, π) = a ≻b ≻c, which gives kdKT(1.5, π) = 1. Table 1 lists the sets of possible ground truths and their minimax rankings4 under different distance metrics.
Note that even with identical (scaled) error bounds, different distance metrics lead to different sets of possible ground truths as well as different optimal rankings. This demonstrates that the choice of the distance metric is important.
3.1. Upper bound Given a distance metric d, a profile π, and that d(π, σ ∗) ≤t, we can bound k(t, π) using the diameter of the set of possible ground truths Bt(π). For a set of rankings S ⊆L(A), denote its diameter by D(S) = maxσ,σ ′∈S d(σ, σ ′).
Lemma 1. 1 2 · D(Bt(π)) ≤k(t, π) ≤D(Bt(π)) ≤2t.
Proof. Let σ = MiniMax(Bt(π)). For rankings σ, σ ′ ∈Bt(π), we have d(σ, σ), d(σ ′, σ) ≤k(t, π) by definition of σ . By the triangle inequality, d(σ, σ ′) ≤2k(t, π) for all σ, σ ′ ∈Bt(π). Thus, D(Bt(π)) ≤2k(t, π).
Next, the maximum distance of σ ∈Bt(π) from all rankings in Bt(π) is at most D(Bt(π)). Hence, the minimax distance k(t, π) = k(Bt(π)) cannot be greater than D(Bt(π)).
Finally, let π = {σ1, ..., σn}. For rankings σ, σ ′ ∈Bt(π), the triangle inequality implies d(σ, σ ′) ≤d(σ, σi) + d(σi, σ ′) for every i ∈{1, ..., n}. Averaging over these inequalities, we get d(σ, σ ′) ≤t + t = 2t, for all σ, σ ′ ∈Bt(π). Thus, we have D(Bt(π)) ≤2t, as required.
2 Lemma 1 implies that k(t) = maxπ∈L(A)n k(t, π) ≤2t for all distance metrics and t > 0. In words: 3 Scaling by the maximum distance is not a good way of comparing distance metrics; we do so for the sake of illustration only.
4 Multiple rankings indicate a tie that can be broken arbitrarily.
A.D. Procaccia et al. / Artificial Intelligence 231 (2016) 1–16 5 Theorem 1. Given n noisy rankings at an average distance of at most t from an unknown true ranking σ ∗according to a distance metric d, it is always possible to find a ranking at distance at most 2t from σ ∗according to d.
Importantly, the bound of Theorem 1 is independent of the number of votes n. Most statistical models of social choice restrict profiles in two ways: i) the average error should be low because the probability of generating high-error votes is typically low, and ii) the errors should be distributed almost evenly (in different directions from the ground truth), which is why aggregating the votes works well. These assumptions are mainly helpful when n is large, that is, performance may be poor for small n (see, e.g., ). In contrast, our model restricts profiles only by making the first assumption (explicitly), allowing voting rules to perform well as long as the votes are accurate on average, independently of the number of votes n.
We also remark that Theorem 1 admits a simple proof, but the bound is nontrivial: while the average error of the profile is at most t (hence, the profile contains a ranking with error at most t), it is generally impossible to pinpoint a single ranking within the profile that has error at most 2t with respect to the ground truth in the worst-case (i.e., with respect to every possible ground truth in Bt(π)). That said, it can be shown that there exists a ranking in the profile that always has distance at most 3t from the ground truth. Further, one can pick such a ranking in polynomial time, which stands in sharp contrast to the usual hardness of finding the optimal ranking (see the discussion on the computational complexity of our approach in Section 6).
Theorem 2. Given n noisy rankings at an average distance of at most t from an unknown true ranking σ ∗according to a distance metric d, it is always possible to pick, in polynomial time, one of the n given rankings that has distance at most 3t from σ ∗according to d.
Proof. Consider a profile π consisting of n rankings such that d(σ ∗, π) ≤t. Let x = minσ∈L(A) d(σ, π) be the minimum distance any ranking has from the profile. Then, x ≤d(σ ∗, π) ≤t. Let σ = arg minσ∈π d(σ, π) be the ranking in π which minimizes the distance from π among all rankings in π. An easy-to-verify folklore theorem says that d( σ , π) ≤2x. To see this, assume that ranking τ has the minimum distance from the profile (i.e., d(τ, π) = x). Now, the average distance of all rankings in π from π is 1 n σ∈π d(σ,π) = 1 n2 σ∈π σ ′∈π d(σ,σ ′) ≤1 n2 σ∈π σ ′∈π (d(τ,σ) + d(τ,σ ′)) = 2 n σ∈π d(τ,σ) = 2x ≤2t, where the second transition uses the triangle inequality. Now, σ has the smallest distance from π among all rankings in π, which cannot be greater than the average distance (1/n) σ∈π d(σ, π). Hence, d( σ, π) ≤2t. Finally, d( σ,σ ∗) ≤1 n σ∈π d( σ,σ) + d(σ,σ ∗) ≤2t + t = 3t, where the first transition uses the triangle inequality and the second transition uses the fact that d( σ, π) ≤2t and d(π, σ ∗) ≤t. It is easy to see that σ can be computed in O(n2) time.
2 3.2. Lower bounds The upper bound of 2t (Theorem 1) is intuitively loose — we cannot expect it to be tight for every distance metric. However, we can complement it with a lower bound of (roughly speaking) t/2 for all distance metrics. Formally, let d↓(r) denote the greatest feasible distance under distance metric d that is less than or equal to r. Next, we prove a lower bound of d↓(t)/2.
Theorem 3. For a distance metric d, k(t) ≥d↓(t)/2.
Proof. If d↓(t) = 0, then the result trivially holds. Assume d↓(t) > 0. Let σ and σ ′ be two rankings at distance d↓(t). Consider profile π consisting of only a single instance of ranking σ . Then, σ ′ ∈Bt(π). Hence, D(Bt(π)) ≥d↓(t). Now, it follows from Lemma 1 that k(t) ≥D(Bt(π))/2 ≥d↓(t)/2.
2 Recall that Theorem 1 shows that k(t) ≤2t. However, k(t) is the minimax distance under some profile, and hence must be a feasible distance under d. Thus, Theorem 1 actually implies a possibly better upper bound of d↓(2t). Together with Theorem 3, this implies d↓(t)/2 ≤k(t) ≤d↓(2t). Next, we show that imposing a mild assumption on the distance metric allows us to improve the lower bound by a factor of 2, thus reducing the gap between the lower and upper bounds.
Theorem 4. For a neutral distance metric d, k(t) ≥d↓(t).
6 A.D. Procaccia et al. / Artificial Intelligence 231 (2016) 1–16 Proof. For a ranking σ ∈L(A) and r ≥0, let Br(σ) denote the set of rankings at distance at most r from σ . Neutrality of the distance metric d implies |Br(σ)| = |Br(σ ′)| for all σ, σ ′ ∈L(A) and r ≥0. In particular, d↓(t) being a feasible distance under d implies that for every σ ∈L(A), there exists some ranking at distance exactly d↓(t) from σ .
Fix σ ∈L(A). Consider the profile π consisting of n instances of σ . It holds that Bt(π) = Bt(σ). We want to show that the minimax distance k(Bt(σ)) ≥d↓(t). Suppose for contradiction that there exists some σ ′ ∈L(A) such that all rankings in Bt(σ) are at distance at most t′ from σ ′, i.e., Bt(σ) ⊆Bt′(σ ′), with t′ < d↓(t). Since there exists some ranking at distance d↓(t) > t′ from σ ′, we have Bt(σ) ⊆Bt′(σ ′) ⊊Bt(σ ′), which is a contradiction because |Bt(σ)| = |Bt(σ ′)|. Therefore, k(t) ≥ k(t, π) ≥d↓(t).
2 The bound of Theorem 4 holds for all n, m > 0 and all t ∈[0, D], where D is the maximum possible distance under d. It can be checked easily that the bound is tight given the neutrality assumption, which is an extremely mild — and in fact, a highly desirable — assumption for distance metrics over rankings.
Theorem 4 improves the bounds on k(t) to d↓(t) ≤k(t) ≤d↓(2t) for a variety of distance metrics d. However, for the four special distance metrics considered in this paper, the next result, which is our main theoretical result, closes this gap by establishing a tight lower bound of d↓(2t), for a wide range of values of n and t.
Theorem 5. If d ∈{dKT, dFR, dMD, dCY}, and the maximum distance allowed by the metric is D ∈(mα), then there exists T ∈(mα) such that: 1. For all t ≤T and even n, we have k(t) ≥d↓(2t).
2. For all L ≥2, t ≤T with {2t} ∈(1/L, 1 −1/L), and odd n ≥(L · D), we have k(t) ≥d↓(2t). Here, {x} = x −⌊x⌋denotes the fractional part of x ∈R.
The impossibility result of Theorem 5 is weaker for odd values of n (in particular, covering more values of t requires larger n), which is reminiscent of the fact that repetition (error-correcting) codes achieve greater efficiency with an odd number of repetitions; this is not merely a coincidence. Indeed, an extra repetition allows differentiating between tied possibilities for the ground truth; likewise, an extra vote in the profile prevents us from constructing a symmetric profile that admits a diverse set of possible ground truths.
Proof of Theorem 5. We denote {1, ..., r} by [r] in this proof. We use σ(a) to denote the rank (position) of alternative a in ranking σ . First, we prove the case of even n for all four distance metrics. We later provide a generic argument to prove the case of large odd n. First, we need a simple observation.
Observation 1. If r 2 ≤⌊2t⌋and t ≥0.5, then r ≤4 √ t.
Proof. Note that (r −1)2 ≤r · (r −1) ≤2 · ⌊2t⌋≤4t. Hence, r ≤2 √ t + 1. We also have t ≥0.5, i.e., 1 ≤2t. This implies 1 ≤ √ 2t. Thus, we have r ≤2 √ t + √ 2t = (2 + √ 2) √ t ≤4 √ t.
2 The Kendall tau distance: Let d be the Kendall tau distance; thus, D = m 2 and α = 2. Let n be even. For a ranking τ ∈L(A), let τrev be its reverse. Assume t = (1/2) · m 2 , and fix a ranking σ ∈L(A). Every ranking must agree with exactly one of σ and σrev on a given pair of alternatives. Hence, every ρ ∈L(A) satisfies d(ρ, σ) + d(ρ, σrev) = m 2 . Consider the profile π consisting of n/2 instances of σ and n/2 instances of σrev. Then, the average distance of every ranking from rankings in π would be exactly t, i.e., Bt(π) = L(A). It is easy to check that k(L(A)) = m 2 = 2t = d↓(2t) because every ranking has its reverse ranking in L(A) at distance exactly 2t.
Now, let us extend the proof to t ≤(m/12)2. If t < 0.5, then d↓ KT(2t) = 0, which is a trivial lower bound. Hence, assume t ≥0.5. Thus, d↓(2t) = ⌊2t⌋. We use Fermat’s Polygonal Number Theorem (see, e.g., ). A special case of this remarkable theorem states that every natural number can be expressed as the sum of at most three “triangular” numbers, i.e., numbers of the form k 2 . Let ⌊2t⌋= 3 i=1 mi 2 . From Observation 1, it follows that 0 ≤mi ≤4 √ t for all i ∈{1, 2, 3}. Hence, 3 i=1 mi ≤ 12 √ t ≤m.
Partition the set of alternatives A into four disjoint groups A1, A2, A3, and A4 such that |Ai| = mi for i ∈{1, 2, 3}, and |A4| = m −3 i=1 mi. Let σ A4 be an arbitrary ranking of the alternatives in A4; consider the partial order PA = A1 ≻A2 ≻ A3 ≻σ A4 over alternatives in A. Note that a ranking ρ is an extension of PA iff it ranks all alternatives in Ai before any alternative in Ai+1 for i ∈{1, 2, 3}, and ranks alternatives in A4 according to σ A4. Choose arbitrary σ Ai ∈L(Ai) for i ∈{1, 2, 3} and define σ = σ A1 ≻σ A2 ≻σ A3 ≻σ A4, σ ′ = σ A1 rev ≻σ A2 rev ≻σ A3 rev ≻σ A4.
A.D. Procaccia et al. / Artificial Intelligence 231 (2016) 1–16 7 Note that both σ and σ ′ are extensions of PA. Once again, take the profile π consisting of n/2 instances of σ and n/2 instances of σ ′. It is easy to check that a ranking disagrees with exactly one of σ and σ ′ on every pair of alternatives that belong to the same group in {A1, A2, A3}. Hence, every ranking ρ ∈L(A) satisfies d(ρ,σ) + d(ρ,σ ′) ≥ 3 i=1 mi 2 = ⌊2t⌋.
(1) Clearly an equality is achieved in Equation (1) if and only if ρ is an extension of PA. Thus, every extension of PA has an average distance of ⌊2t⌋/2 ≤t from π. Every ranking ρ that is not an extension of PA achieves a strict inequality in Equation (1); thus, d(ρ, π) ≥(⌊2t⌋+ 1)/2 > t. Hence, Bt(π) is the set of extensions of PA.
Given a ranking ρ ∈L(A), consider the ranking in Bt(π) that reverses the partial orders over A1, A2, and A3 induced by ρ. The distance of this ranking from ρ would be at least 3 i=1 mi 2 = ⌊2t⌋, implying k(Bt(π)) ≥⌊2t⌋. (In fact, it can be checked that k(Bt(π)) = D(Bt(π)) = ⌊2t⌋.) We now proceed to prove the case of an even number of agents for the other three distance metrics. First, if M is the minimum distance between two distinct rankings under a distance metric d and t < M/2, then we have d↓(2t) = 0, which is a trivial lower bound. Hence, we assume t ≥M/2.
The footrule distance: Let dFR denote the footrule distance; recall that given σ, σ ′ ∈L(A), dFR(σ, σ ′) = a∈A |σ(a) −σ ′(a)|. The proof is along the same lines as the proof for the Kendall tau distance, but uses a few additional clever ideas. It is known that the maximum footrule distance between two rankings over m alternatives is D = m2/2 , and is achieved by two rankings that are reverse of each other . Hence, we have α = 2; thus, we wish to find T ∈(m2) for which the claim will hold. Formally writing the distance between a ranking and its reverse, we get dFR(σ,σrev) = m i=1 |m + 1 −2i| = m2 2 .
(2) Observation 2. The footrule distance between two rankings is always an even integer.
Proof. Take rankings σ, τ ∈L(A). Note that dFR(σ, τ) = a∈A |σ(a) −τ(a)|. Now, |σ(a) −τ(a)| is odd if and only if the positions of a in σ and τ have different parity. Since the number of odd (as well as even) positions is identical in σ and τ , the number of alternatives that leave an even position in σ to go to an odd position in τ equals the number of alternatives that leave an odd position in σ to go to an even position in τ . Thus, the number of alternatives for which the parity of the position changes is even. Equivalently, the number of odd terms in the sum defining the footrule distance is even. Hence, the footrule distance is an even integer.
2 Hence, Equation (2) implies that d↓ FR(2t) equals ⌊2t⌋if ⌊2t⌋is even, and equals ⌊2t⌋−1 otherwise. Let r = d↓ FR(2t). Hence, r is an even integer. We prove the result for t ≤(m/8)2. In this case, we invoke the 4-gonal special case of Fermat’s Polygonal Number Theorem (instead of the 3-gonal case invoked in the proof for the Kendall tau distance): Every positive integer can be written as the sum of at most four squares. Let r/2 = m2 1 + m2 2 + m2 3 + m2 4. Hence, r = (2m1)2 2 + (2m2)2 2 + (2m3)2 2 + (2m4)2 2 .
(3) It is easy to check that mi ≤√ r/2 for i ∈. Thus, 4 i=1 2mi ≤8√ r/2 ≤8 √ t ≤m. Let us partition the set of alternatives A into {Ai}i∈ such that |Ai| = 2mi for i ∈ and |A5| = m5 = m −4 i=1 2mi.
Fix σ A5 ∈L(A5) and consider the partial order PA = A1 ≻A2 ≻A3 ≻A4 ≻σ A5. Choose arbitrary σ Ai ∈L(Ai) for i ∈, and let σ = (σ A1 ≻σ A2 ≻σ A3 ≻σ A4 ≻σ A5), σ ′ = (σ A1 rev ≻σ A2 rev ≻σ A3 rev ≻σ A4 rev ≻σ A5).
Note that both σ and σ ′ are extensions of PA. Consider the profile π consisting of n/2 instances of σ and σ ′ each. Unlike the Kendall tau distance, Bt(π) is not the set of extensions of PA. Still, we show that it satisfies k(Bt(π)) = D(Bt(π)) = d↓ FR(2t) = r.
Denote by a j i the alternative ranked j in σ Ai . Take a ranking ρ ∈L(A). Consider dFR(ρ, σ) + dFR(ρ, σ ′). We have the following inequalities regarding the sum of displacement of different alternatives between ρ and σ , and between ρ and σ ′. For i ∈ and j ∈[2mi], ρ(a j i ) −σ(a j i ) + ρ(a j i ) −σ ′(a j i ) ≥ σ(a j i ) −σ ′(a j i ) = | j −(2mi + 1 −j)|.
(4) 8 A.D. Procaccia et al. / Artificial Intelligence 231 (2016) 1–16 Summing all the inequalities, we get dFR(ρ,σ) + dFR(ρ,σ ′) ≥ 4 i=1 2mi j=1 |2 j −2mi −1| = 4 i=1 (2mi)2 2 = r, (5) where the second transition follows from Equation (2), and the third transition follows from Equation (3).
First, we show that ρ ∈Bt(π) only if equality in Equation (5) holds. To see why, note that the footrule distance is always even and r = d↓ FR(2t) ≥⌊2t⌋−1. Hence, if equality is not achieved, then dFR(ρ, σ) + dFR(ρ, σ ′) ≥r + 2 ≥⌊2t⌋−1 + 1 > 2t. Hence, the average distance of ρ from votes in π would be greater than t.
On the contrary, if equality is indeed achieved in Equation (5), then the average distance of ρ from votes in π is r/2 ≤t. Hence, we have established that Bt(π) is the set of rankings ρ for which equality is achieved in Equation (5).
For ρ to achieve equality in Equation (5), it must achieve equality in Equation (4) for every i ∈ and j ∈[2mi], and it must agree with both σ and σ ′ on the positions of alternatives in A5 (i.e., σ A5 must be a suffix of ρ). For the former to hold, the position of a j i in ρ must be between σ(a j i ) and σ ′(a j i ) = σ(a2mi+1−j i ) (both inclusive), for every i ∈ and j ∈[2mi].
We claim that the set of rankings satisfying these conditions are characterized as follows.
Bt(π) = ρ ∈L(A) {ρ(a j i ),ρ(a2mi+1−j i )} = {σ(a j i ),σ(a2mi+1−j i )} for i ∈, j ∈[2mi], and ρ(a j 5) = σ(a j 5) = σ ′(a j 5) for j ∈[m5] .
(6) Note that instead of ρ(a j i ) and ρ(a2mi+1−j i ) both being in the interval [σ(a j i ), σ(a2mi+1−j i )], we are claiming that they must be the two endpoints. First, consider the middle alternatives in each Ai (i ∈), namely ami i and ami+1 i . Both must be placed between σ(ami i ) = σ ′(ami+1 i ) and σ(ami+1 i ) = σ ′(ami i ); but these two numbers differ by exactly 1. Hence, ρ(ami i ),ρ(ami+1 i ) = σ(ami i ),σ(ami+1 i ) .
Consider the two adjacent alternatives, namely ami−1 i and ami+2 i . Given that the middle alternatives ami i and ami+1 i oc-cupy their respective positions in σ or σ ′, the only positions available to ρ for placing the two adjacent alternatives are the endpoints of their common feasible interval [σ(ami−1 i ), σ(ami+2 i )]. Continuing this argument, each pair of alternatives (a j i , a2mi+1−j i ) must occupy the two positions {σ(a j i ), σ(a2mi+1−j i )} for every i ∈ and j ∈[mi].
That is, ρ can either keep the alternatives a j i and a2mi+1−j i as they are in σ , or place them according to σ ′ (equivalently, swapping them in σ ) for every i ∈ and j ∈[2mi]. Note that these choices are independent of each other. We established that a ranking ρ is in Bt(π) only if it is obtained in this manner and has σ A5 as its suffix.
Further, it can be seen that each of these choices (keeping or swapping the pair in σ ) maintain dFR(ρ, σ) + dFR(ρ, σ ′) invariant. Hence, all such rankings ρ satisfy dFR(ρ, σ) + dFR(ρ, σ ′) = r, and thus belong to Bt(π). This reaffirms our original claim that Bt(π) is given by Equation (6).
In summary, all rankings in Bt(π) can be obtained by taking σ , and arbitrarily choosing whether to swap the pair of alternatives a j i and a2mi+1−j i for each i ∈ and j ∈[2mi].
Note that σ, σ ′ ∈Bt(π) and dFR(σ, σ ′) = r (this distance is given by the summation in Equation (5)). Hence, D(Bt(π)) ≥r. Now, we prove that its minimax distance is at least r as well. Take a ranking ρ ∈L(A). We need to show that there exists some τ ∈Bt(π) such that dFR(ρ, τ) ≥r.
Consider alternatives a j i and a2mi+1−j i for i ∈ and j ∈[2mi]. We know that τ must satisfy {τ(a j i ), τ(a2mi+1−j i )} = {σ(a j i ), σ(a2mi+1−j i )} in order to belong to Bt(π). This allows two possible ways for placing the pair of alternatives. Let τ pick the optimal positions that maximize τi, j(ρ) = |τ(a j i ) −ρ(a j i )| + |τ(a2mi+1−j i ) −ρ(a2mi+1−j i )|.
That is, τi, j(ρ) should equal Mi, j(ρ), which we define as max |σ(a j i ) −ρ(a j i )| + |σ(a2mi+1−j i ) −ρ(a2mi+1−j i )|, |σ(a2mi+1−j i ) −ρ(a j i )| + |σ(a j i ) −ρ(a2mi+1−j i )| .
Note that the choice for each pair of alternatives (a j i , a2mi+1−j i ) can be made independently of every other pair. Further, making the optimal choice for each pair guarantees that dFR(ρ, τ) is at least A.D. Procaccia et al. / Artificial Intelligence 231 (2016) 1–16 9 4 i=1 2mi j=1 τi, j(ρ) = 4 i=1 2mi j=1 Mi, j(ρ), which we will now show to be at least r.
Algorithm 1 describes how to find the optimal ranking τ ∈Bt(π) mentioned above, which satisfies τi, j(ρ) = Mi, j(ρ) for every i ∈ and j ∈[2mi]. It starts with an arbitrary τ ∈Bt(π), and swaps every sub-optimally placed pair (a j i , a2mi+1−j i ) for i ∈ and j ∈[2mi]. In the algorithm, τa↔b denotes the ranking obtained by swapping alternatives a and b in τ .
ALGORITHM 1: Finds a ranking in Bt(π) at a footrule distance of at least ⌊2t⌋from any given ranking.
Data: Ranking ρ ∈L(A) Result: Ranking τ ∈Bt(π) such that dFR(τ, ρ) ≥⌊2t⌋ τ ←an arbitrary ranking from Bt(π); for i ∈ do for j ∈[2mi] do d j i ←|ρ(a j i ) −τ(a j i )|; d2mi+1−j i ←|ρ(a2mi+1−j i ) −τ(a2mi+1−j i )|; if d j i + d2mi+1−j i < Mi, j(ρ) then τ ←τa j i ↔a 2mi +1−j i ; end end end return τ ; Finally, we show that dFR(ρ, τ) ≥r. First, we establish the following lower bound on Mi, j(ρ).
Mi, j(ρ) ≥1 2 |σ(a j i ) −ρ(a j i )| + |σ(a2mi+1−j i ) −ρ(a2mi+1−j i )| + |σ(a2mi+1−j i ) −ρ(a j i )| + |σ(a j i ) −ρ(a2mi+1−j i )| ≥|σ(a2mi+1−j i ) −σ(a j i )| = |2mi + 1 −2 j|, where the first transition holds because the maximum of two terms is at least as much as their average, and the second transition uses the triangle inequality on appropriately paired terms. Now, we have dFR(τ,ρ) ≥ 4 i=1 2mi j=1 Mi, j(ρ) ≥ 4 i=1 2mi j=1 |2mi + 1 −2 j| = 4 i=1 (2mi)2 2 = r, where the third transition holds due to Equation (2), and the fourth transition holds due to Equation (3). Hence, the minimax distance of Bt(π) is at least r = d↓ FR(2t), as required.
The Cayley distance: Next, let dCY denote the Cayley distance. Recall that dCY(σ, τ) equals the minimal number of swaps (of possibly non-adjacent alternatives) required in order to transform σ to τ . It is easy to check that the maximum Cayley distance is D = m −1; hence, it has α = 1. We prove the result for t ≤m/4. Note that d↓ CY(2t) = ⌊2t⌋. Define rankings σ, σ ′ ∈L(A) as follows.
σ = (a1 ≻... ≻a2⌊2t⌋ ≻a2⌊2t⌋+1 ≻... ≻am), σ ′ = (a2⌊2t⌋≻... ≻a1 ≻a2⌊2t⌋+1 ≻... ≻am).
Let profile π consist of n/2 instances of σ and σ ′ each. We claim that Bt(π) has the following structure, which is very similar to the ball for the footrule distance.
Bt(π) = ρ ∈L(A) {ρ(ai),ρ(a2⌊2t⌋+1−i)} = {i, 2⌊2t⌋+ 1 −i} for i ∈[⌊2t⌋], and ρ(ai) = i for i > 2⌊2t⌋ .
(7) First, we observe the following simple fact: If rankings τ and ρ mismatch (i.e., place different alternatives) in r different positions, then dCY(τ, ρ) ≥r/2. Indeed, consider the number of swaps required to convert τ into ρ. Since each swap 10 A.D. Procaccia et al. / Artificial Intelligence 231 (2016) 1–16 can make τ and ρ consistent in at most two more positions, it would take at least r/2 swaps to convert τ into ρ, i.e., dCY(τ, ρ) ≥r/2.
Now, note that σ and σ ′ mismatch in each of first 2 ⌊2t⌋positions. Hence, every ranking ρ ∈L(A) must mismatch with at least one of σ and σ ′ in each of first 2 ⌊2t⌋positions. Together with the previous observation, this implies dCY(ρ,σ) + dCY(ρ,σ ′) ≥⌊2t⌋.
(8) Every ranking ρ that achieves equality in Equation (8) is clearly in Bt(π) because its average distance from the votes in π is ⌊2t⌋/2 ≤t. Further, every ranking ρ that achieves a strict inequality in Equation (8) is outside Bt(π) because its average distance from the votes in π is at least (⌊2t⌋+ 1)/2 > t. Hence, Bt(π) consists of rankings that satisfy dCY(ρ, σ) + dCY(ρ, σ ′) = ⌊2t⌋.
Now, any ranking ρ satisfying equality in Equation (8) must be consistent with exactly one of σ and σ ′ in each of first 2 ⌊2t⌋positions, and with both σ and σ ′ in the later positions. The former condition implies that for every i ∈⌊2t⌋, ρ must place the pair of alternatives (ai, a2⌊2t⌋+1−i) in positions i and 2 ⌊2t⌋+ 1 −i, either according to σ or according to σ ′. This confirms our claim that Bt(π) is given by Equation (7).
We now show that k(Bt(π)) ≥⌊2t⌋. Take a ranking ρ ∈L(A). We construct a ranking τ ∈Bt(π) such that τ mismatches with ρ in each of first 2 ⌊2t⌋positions. Together with our observation that the Cayley distance is at least half of the number of positional mismatches, this would imply that the minimax distance of Bt(π) is at least ⌊2t⌋, as required.
We construct τ by choosing the placement of the pair of alternatives (ai, a2⌊2t⌋+1−i), independently for each i ∈⌊2t⌋, in a way that τ mismatches with ρ in positions i and 2 ⌊2t⌋+ 1 −i both. Let I(X) denote the indicator variable that is 1 if statement X holds, and 0 otherwise. Let r = I (ρ(ai) = i) + I ρ(a2⌊2t⌋+1−i) = 2⌊2t⌋+ 1 −i . Consider the following three cases.
r = 0: Set τ(ai) = i and τ(a2⌊2t⌋+1−i) = 2 ⌊2t⌋+ 1 −i.
r = 1: Without loss of generality, assume ρ(ai) = i. Set τ(ai) = 2 ⌊2t⌋+ 1 −i and τ(a2⌊2t⌋+1−i) = i.
r = 2: Set τ(ai) = 2 ⌊2t⌋+ 1 −i and τ(a2⌊2t⌋+1−i) = i.
Finally, set τ(ai) = i for all i > 2 ⌊2t⌋. This yields a ranking τ that is in Bt(π), and mismatches ρ in each of first 2 ⌊2t⌋ positions; hence, dCY(ρ, τ) ≥⌊2t⌋, as required.
The maximum displacement distance: Finally, let dMD denote the maximum displacement distance. Note that it can be at most D = m −1; hence, it also has α = 1. However, this distance metric requires an entirely different technique than the ones used for previous distances. For example, taking any two rankings at maximum distance from each other does not work. We prove this result for t ≤m/4. Once again, note that d↓ MD(2t) = ⌊2t⌋.
Consider rankings σ and σ ′ defined as follows.
σ = (a1 ≻... ≻a⌊2t⌋ ≻a⌊2t⌋+1 ≻... ≻a2⌊2t⌋ ≻arest), σ ′ = (a⌊2t⌋+1 ≻... ≻a2⌊2t⌋ ≻a1 ≻... ≻a⌊2t⌋ ≻arest), where arest is shorthand for a2⌊2t⌋+1 ≻... ≻am. Note that the blocks of alternatives a1 through a⌊2t⌋and a⌊2t⌋+1 through a2⌊2t⌋are shifted to each other’s positions in the two rankings. Thus, each of a1 through a2⌊2t⌋have a displacement of exactly ⌊2t⌋between the two rankings. Thus, dMD(σ, σ ′) = ⌊2t⌋.
Consider the profile π consisting of n/2 instances of σ and σ ′ each. Clearly, σ and σ ′ have an average distance of ⌊2t⌋/2 ≤t from rankings in π. Hence, {σ, σ ′} ∈Bt(π). Surprisingly, in this case we can show that the minimax distance of Bt(π) without any additional information regarding the structure of Bt(π).
Take a ranking ρ ∈L(A). The alternative placed first in ρ must be ranked at a position ⌊2t⌋or below in at least one of σ and σ ′. Hence, max(dMD(ρ, σ), dMD(ρ, σ ′)) ≥⌊2t⌋. Thus, there exists a ranking in Bt(π) at distance at least ⌊2t⌋from ρ, i.e., the minimax distance of Bt(π) is at least ⌊2t⌋, as desired.
This completes the proof of the special case of even n for all four distance metrics. Now, consider the case of odd n.
Odd n: To extend the proof to odd values of n, we simply add one more instance of σ than σ ′. The key insight is that with large n, the distance from the additional vote would have little effect on the average distance of a ranking from the profile. Thus, Bt(π) would be preserved, and the proof would follow.
Formally, let L ≥2 and t ∈(1/L, 1 −1/L). For the case of even n, the proofs for all four distance metrics proceeded as follows: Given the feasible distance r = d↓(2t), we constructed two rankings σ and σ ′ at distance r from each other such that Bt(π) is the set of rankings at minimal total distance from the two rankings, i.e., Bt(π) = ρ ∈L(A) | d(ρ,σ) + d(ρ,σ ′) = r .
Let n ≥3 be odd. Consider the profile π that has (n −1)/2 instances of σ and σ ′ each, and an additional instance of an arbitrary ranking. In our generic proof for all four distance metrics, we obtain conditions under which Bt(π) = Bt(π′) A.D. Procaccia et al. / Artificial Intelligence 231 (2016) 1–16 11 where π′ is obtained by removing the arbitrary ranking from π (and hence has an even number of votes). We already proved that k(Bt(π′)) ≥d↓(2t). Hence, obtaining Bt(π) = Bt(π′) would also show the lower bound d↓(2t) for odd n.
In more detail, our objective is that every ranking ρ with d(ρ, σ) + d(ρ, σ ′) = r (which may have a worst-case distance of D from the additional arbitrary ranking) should be in Bt(π), and every ranking ρ with d(ρ, σ) + d(ρ, σ ′) > r should be outside Bt(π).
First, let d ∈{dKT, dCY, dMD}. If d(ρ, σ) + d(ρ, σ ′) > r, then d(ρ, σ) + d(ρ, σ ′) ≥r + 1. The total error incurred by rankings of distance r from π is n−1 2 · r, and a distance of D from the additional ranking. This means that t ≥ n−1 2 · r + D n .
For rankings with an error greater than r to be outside Bt(π), we must have t < n−1 2 · (r + 1) n .
Combining the inequalities, we obtain that n−1 2 · r + D n ≤t < n−1 2 · (r + 1) n ⇔n −1 2 · r + D ≤n · t < n −1 2 · (r + 1) ⇔r + 2D n −1 ≤ 2n n −1 · t < r + 1 ⇔r ≤2t −2D −2t n −1 < r + 1 − 2D n −1 .
(9) Choose n ≥2LD + 1. Then, 2D/(n −1) ≤1/L < {2t}. Note that 2t −2D −2t n −1 = ⌊2t⌋+ {2t} −2D −2t n −1 = ⌊2t⌋, where the last equality holds because we showed (2D −2t)/(n −1) < {2t}.
In all three distance metrics considered thus far, we had ⌊2t⌋= d↓(2t). Let r = ⌊2t⌋. We show that r = ⌊2t⌋satisfies Equation (9), thus yielding Bt(π) with minimax distance at least r = d↓(2t), as required. Note that r ≤2t −2D −2t n −1 is satisfied by definition from Equation (9). We also have 2t −2D −2t n −1 − r + 1 − 2D n −1 = 2t + 2t n −1 −⌊2t⌋−1 = {2t} + 2t n −1 −1 < 1 −1 L + 1 L −1 = 0.
Hence, we have k(t) ≥d↓(2t) for n ≥2LD + 1.
Next, consider the footrule distance. If ⌊2t⌋is even (i.e., if ⌊2t⌋= d↓(2t)), then the above proof works because r = ⌊2t⌋ is a feasible distance. If ⌊2t⌋is odd, then we must choose r = ⌊2t⌋−1. However, we have an advantage: since the footrule distance is always even, every ranking ρ with d(ρ, σ) + d(ρ, σ ′) > r must have d(ρ, σ) + d(ρ, σ ′) ≥r + 2. Hence, we only need n−1 2 · r + D n ≤t < n−1 2 · (r + 2) n ⇔r ≤2t −2D −2t n −1 < r + 2 − 2D n −1 .
(10) Note that r = ⌊2t⌋−1 clearly satisfies the first inequality in Equation (10). For the second inequality, note that r decreased by 1 compared to earlier but 1 −2D/(n −1) increased to 2 −2D/(n −1) instead. Hence, the second inequality is still satisfied, and we get Bt(π) with minimax distance at least r = ⌊2t⌋−1 = d↓(2t), as required.
2 12 A.D. Procaccia et al. / Artificial Intelligence 231 (2016) 1–16 Fig. 1. Positive correlation of t∗with the noise parameter.
4. Approximations for unknown average error In the previous sections we derived the optimal rules when the upper bound t on the average error is given to us. In practice, the given bound may be inaccurate. We know that using an estimate t that is still an upper bound ( t ≥t) yields a ranking at distance at most 2 t from the ground truth in the worst case. What happens if it turns out that t < t? We show that the output ranking is still at distance at most 4t from the ground truth in the worst case.
Theorem 6. For a distance metric d, a profile π consisting of n noisy rankings at an average distance of at most t from the true ranking σ ∗, and t < t, d(OPTd( t, π), σ ∗) ≤4t.
To prove the theorem, we make a detour through minisum rules. For a distance metric d, let MiniSumd, be the voting rule that always returns the ranking minimizing the sum of distances (equivalently, average distance) from the rankings in the given profile according to d. Two popular minisum rules are the Kemeny rule for the Kendall tau distance (MiniSumdKT ) and the minisum rule for the footrule distance (MiniSumdFR), which approximates the Kemeny rule .5 For a distance metric d (dropped from the superscripts), let d(π, σ ∗) ≤t. We claim that the minisum ranking MiniSum(π) is at distance at most min(2t, 2k(t, π)) from σ ∗. This is true because the minisum ranking and the true ranking are both in Bt(π), and Lemma 1 shows that its diameter is at most min(2t, 2k(t, π)).
Returning to the theorem, if we provide an underestimate t of the true worst-case average error t, then using Lemma 1, d MiniMax(B t(π)), MiniSum(π ) ≤2 t ≤2t, d MiniSum(π),σ ∗ ≤D(Bt(π)) ≤2t.
By the triangle inequality, d MiniMax(B t(π)),σ ∗ ≤4t.
5. Experimental results We compare our worst-case optimal voting rules OPTd against a plethora of voting rules used in the literature: plurality, Borda count, veto, the Kemeny rule, single transferable vote (STV), Copeland’s rule, Bucklin’s rule, the maximin rule, Slater’s rule, Tideman’s rule, and the modal ranking rule (for definitions see, e.g., ).
Our performance measure is the distance of the output ranking from the actual ground truth. In contrast, for a given d, OPTd is designed to optimize the worst-case distance to any possible ground truth. Hence, crucially, OPTd is not guaranteed to outperform other rules in our experiments.
We use two real-world datasets containing ranked preferences in domains where ground truth rankings exist. Mao, Procaccia, and Chen collected these datasets — dots and puzzle — via Amazon Mechanical Turk. For dataset dots (resp., puzzle), human workers were asked to rank four images that contain a different number of dots (resp., different states of an 8-Puzzle) according to the number of dots (resp., the distances of the states from the goal state). Each dataset has four different noise levels (i.e., levels of task difficulty), represented using a single noise parameter: for dots (resp., puzzle), higher noise corresponds to ranking images with a smaller difference between their number of dots (resp., ranking states that are all farther away from the goal state). Each dataset has 40 profiles with approximately 20 votes each, for each of the 4 noise levels. Points in our graphs are averaged over the 40 profiles in a single noise level of a dataset.
First, as a sanity check, we verified (Fig. 1) that the noise parameter in the datasets positively correlates with our notion of noise — the average error in the profile, denoted t∗(averaged over all profiles in a noise level). Strikingly, the results from the two datasets are almost identical!
Next, we compare OPTd and MiniSumd against the voting rules listed above, with distance d as the measure of error. We use the average error in a profile as the bound t given to OPTd, i.e., we compute OPTd(t∗, π) on profile π where t∗= d(π, σ ∗). While this is somewhat optimistic, note that t∗may not be the (optimal) value of t that achieves the lowest error. Also, the experiments below show that a reasonable estimate of t∗also suffices.
5 Minisum rules such as the Kemeny rule are also compelling because they often satisfy attractive social choice axioms. However, it is unclear whether such axioms contribute to the overall goal of effectively recovering the ground truth.
A.D. Procaccia et al. / Artificial Intelligence 231 (2016) 1–16 13 Fig. 2. Performance of different voting rules (Figs. 1(a) and 1(b)), and of OPT with varying t (Figs. 1(c) and 1(d)). (For interpretation of the references to color in this figure, the reader is referred to the web version of this article.) Figs. 2(a) and 2(b) show the results for the dots and puzzle datasets, respectively, under the Kendall tau distance. It can be seen that OPTdKT (solid red line) significantly outperforms all other voting rules. The three other distance metrics considered in this paper generate similar results; the corresponding graphs are presented in the appendix.
Finally, we test OPTd in the more demanding setting where only an estimate t of t∗is provided. To synchronize the results across different profiles, we use r = ( t −MAD)/(t∗−MAD), where MAD is the minimum average distance of any ranking from the votes in a profile, that is, the average distance of the ranking returned by MiniSumd from the input votes. For all profiles, r = 0 implies t = MAD (the smallest value that admits a possible ground truth) and r = 1 implies t = t∗(the true average error). In our experiments we use r ∈[0, 2]; here, t is an overestimate of t∗for r ∈(1, 2] (a valid upper bound on t∗), but an underestimate of t∗for r ∈[0, 1) (an invalid upper bound on t∗).
Figs. 2(c) and 2(d) show the results for the dots and puzzle datasets, respectively, for a representative noise level (level 3 in previous experiments) and the Kendall tau distance. We can see that OPTdKT (solid red line) outperforms all other voting rules as long as t is a reasonable overestimate of t∗(r ∈[1, 2]), but may or may not outperform them if t is an underestimate of t∗. Again, other distance metrics generate similar results (see the appendix for details).
Comments on the empirical results. It is genuinely surprising that on real-world datasets, OPTd (a rule designed to work well in the worst-case) provides a significantly superior average-case performance compared to most prominent voting rules by utilizing minimal additional information — an approximate upper bound on the average error in the input votes.
The inferior performance of methods based on probabilistic models of error is also thought provoking. After all, these models assume independent errors in the input votes, which is a plausible assumption in crowdsourcing settings. But such probabilistic models typically assume a specific structure on the distribution of the noise, e.g., the exponential distribution in Mallows’ model , and it is almost impossible that noise in practice would follow this exact structure. In contrast, our approach only requires a loose upper bound on the average error in the input votes. In crowdsourcing settings where the noise is highly unpredictable, it can be argued that the principal may not be able to judge the exact distribution of errors, but may be able to provide an approximate bound on the average error.
6. Discussion Uniformly accurate votes. Motivated by crowdsourcing settings, we considered the case where the average error in the input votes is guaranteed to be low. Instead, suppose we know that every vote in the input profile π is at distance at most t from the ground truth σ ∗, i.e., maxσ∈π d(σ, σ ∗) ≤t. If t is small, this is a stronger assumption because it means that there are no outliers, which is implausible in crowdsourcing settings but plausible if the input votes are expert opinions. In this setting, it is immediate that any vote in the given profile is at distance at most d↓(t) from the ground truth. Moreover, 14 A.D. Procaccia et al. / Artificial Intelligence 231 (2016) 1–16 Fig. A.3. Results for the footrule distance (dFR): Figs. A.3(a) and A.3(b) show that OPTdFR outperforms other rules given the true parameter, and Figs. A.3(c) and A.3(d) (for a representative noise level 3) show that it also outperforms the other rules with a reasonable estimate.
the proof of Theorem 4 goes through, so this bound is tight in the worst case; however, returning a ranking from the profile is not optimal for every profile.
Randomization. We did not consider randomized rules, which may return a distribution over rankings. If we take the error of a randomized rule to be the expected distance of the returned ranking from the ground truth, it is easy to obtain an upper bound of t. Again, the proof of Theorem 4 can be extended to yield an almost matching lower bound of d↓(t). While randomized rules provide better guarantees, they are often impractical: low error is only guaranteed when rankings are repeatedly selected from the output distribution of the randomized rule on the same profile; however, most social choice settings see only a single outcome realized.6 Complexity. A potential drawback of the proposed approach is computational complexity. For example, consider the Kendall tau distance. When t is small enough, only the Kemeny ranking would be a possible ground truth, and OPTdKT or any finite approximation thereof must return the Kemeny ranking, if it is unique. The NP-hardness of computing the Kemeny ranking therefore suggests that computing or approximating OPTdKT is NP-hard.
One way to circumvent this computational obstacle is picking a ranking from the given profile, which provides a weaker bound of 3t instead of 2t on the distance from the unknown ground truth (see Theorem 2). However, in practice the optimal ranking can also be computed using various fixed-parameter tractable algorithms, integer programming solutions, and other heuristics, which are known to provide good performance for these types of computational problems (see, e.g., [8,7]). More importantly, the crowdsourcing settings that motivate our work inherently restrict the number of alternatives to a relatively small constant: a human would find it difficult to effectively rank more than, say, 10 alternatives. With a constant number of alternatives, we can simply enumerate all possible rankings in polynomial time, making each and every computational problem considered in this paper tractable. In fact, this is what we did in our experiments. Therefore, we do not view computational complexity as an insurmountable obstacle.
Acknowledgements This work was partially supported by the NSF under grants IIS-1350598, CCF-1215883, and CCF-1525932, and by a Sloan Research Fellowship.
Appendix A. Additional experiments In the paper, we presented experiments (Fig. 2) that compare our proposed worst-case optimal rule against other voting rules when: i) it receives the true error of a profile t∗= d(π, σ ∗) as an argument (Figs. 2(a) and 2(b)), and ii) when it receives an estimate t of t∗(Figs. 2(c) and 2(d)). In these experiments, we used the Kendall tau distance as the measure of 6 Exceptions include cases where randomization is used for circumventing impossibilities [29,14,11].
A.D. Procaccia et al. / Artificial Intelligence 231 (2016) 1–16 15 Fig. A.4. Results for the Cayley distance (dCY ): Figs. A.4(a) and A.4(b) show that OPTdCY outperforms other rules given the true parameter, and Figs. A.4(c) and A.4(d) (for a representative noise level 3) show that it also outperforms the other rules with a reasonable estimate.
Fig. A.5. Results for the maximum displacement distance (dMD): Figs. A.5(a) and A.5(b) show that OPTdMD outperforms other rules given the true parameter, and Figs. A.5(c) and A.5(d) (for a representative noise level 3) show that it also outperforms the other rules with a reasonable estimate.
error. In this section we present additional experiments in an essentially identical setting but using the other three distance metrics considered in this paper as the measure of error. These experiments affirm that our proposed rules are superior to other voting rules independent of the error measure chosen. Figs. A.3, A.4, and A.5 show the experiments for the footrule distance, the Cayley distance, and the maximum displacement distance, respectively.
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Disjunctive syllogism vs. modus ponens [closed]
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Let us assume the material conditional. Is disjunctive syllogism the same rule as modus ponens? If no, why?
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asked Oct 25, 2019 at 17:24
MijitoMijito
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It would be better to put your Question into a more precise form. Presumably you have these terms ("material conditional", "disjunctive syllogism", "modus ponens") defined or illustrated in a text or other course materials. Unfortunately Readers will have to guess at the meaning of 'the same rule" without more context.hardmath –hardmath 2019-10-26 03:28:37 +00:00 Commented Oct 26, 2019 at 3:28
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This depends on whether the deductive system you're working in (1) takes →→ as primitive and uses it to define ∨∨, or (2) takes ∨∨ as primitive and uses it to define →→, or (3) takes both →→ and ∨∨ as primitive, or (4) does something else.
For systems of sort (1), disjunctive syllogism is, in view of the definition of ∨∨, the rule "from (¬p)→q(¬p)→q and ¬p¬p, infer q q." This is a special case of modus ponens.
For systems of sort (2), modus ponens is, in view of the definition of →→, the rule "from (¬p)∨q(¬p)∨q and p p, infer q q. This would be an instance of disjunctive syllogism if the second hypothesis were ¬¬p¬¬p instead of p p. So, if the system is set up (as some systems are) to make ¬¬p¬¬p the same formula as p p, then modus ponens is a special case of disjunctive syllogism. But in other systems, modus ponens would involve a combination of disjunctive syllogism and the rule "from p p infer ¬¬p¬¬p."
For systems of type (3), modus ponens and disjunctive syllogism would be different rules, simply because they involve different connectives.
Finally, for systems of type (4), I won't try to say anything because there are so many possibilities for doing "something else".
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answered Oct 25, 2019 at 18:31
Andreas BlassAndreas Blass
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8579 | http://dwest.web.illinois.edu/pubs/parity.pdf | Parity and Strong Parity Edge-Coloring of Graphs David P. Bunde∗ , Kevin Milans†, Douglas B. West‡, Hehui Wu§ Abstract A parity walk in an edge-coloring of a graph is a walk along which each color is used an even number of times. We introduce two parameters. Let p(G) be the least number of colors in a parity edge-coloring of G (a coloring having no parity path). Let b p(G) be the least number of colors in a strong parity edge-coloring of G (a coloring having no open parity walk). Note that b p(G) ≥p(G) ≥χ′(G).
We study examples and bounds; p(G) and b p(G) may be equal and or different, but equality is conjectured to hold for all bipartite graphs. If G is connected, then p(G) ≥⌈lg |V (G)|⌉, with equality for paths and even cycles (Cn needs one more color for odd n). The main result that b p(Kn) = 2⌈lg n⌉−1 for all n will appear in a subsequent paper; the conjecture that p(Kn) = b p(Kn) is proved for n ≤16.
Also, p(K2,n) = b p(K2,n) = 2 ⌈n/2⌉. In general, b p(Km,n) ≤m′ ⌈n/m′⌉, where m′ = 2⌈lg m⌉. Comparisons are given to other parameters, and many open questions are posed.
1 Introduction Our work began by studying which graphs embed in the hypercube Qk, the graph with vertex set {0, 1}k in which vertices are adjacent when they differ in exactly one position.
Coloring each edge with the position of the bit where its endpoints differ yields two necessary conditions for the edge-coloring on a subgraph G: 1) every cycle uses each color an even number of times, 2) every path uses some color an odd number of times.
In somewhat different language, Havel and Mov´ arek proved in 1972 that the existence of such an edge-coloring with k colors characterizes the connected graphs that are subgraphs of Qk. The problem was studied as early as 1953 by Shapiro .
∗Computer Science Department, Knox College, Galesburg, IL, dbunde@knox.edu. Partially supported by NSF grant CCR 0093348.
†Department of Mathematics, University of Illinois, Urbana IL, milans@uiuc.edu ‡Department of Mathematics, University of Illinois, Urbana, IL, west@math.uiuc.edu. Work supported in part by NSA grants MDA904-03-1-0037 and H98230-06-1-0065.
§Department of Mathematics, University of Illinois, Urbana, IL, hehuiwu2@math.uiuc.edu.
1 Studying the parity of the usage of colors along walks suggested two edge-coloring pa-rameters that have interesting properties and applications. Let the usage of a color on a walk be the parity of the number of times it appears along the walk. A parity walk is a walk in which every color has even usage.
Let a parity edge-coloring (pec) be an edge-coloring having no parity path. Using distinct colors on all edges produces a parity edge-coloring. Hence we introduce the parity edge-chromatic number p(G) to be the minimum number of colors in a pec of G. Paths of length 2 force p(G) ≥χ′(G), where χ′(G) is the edge-chromatic number.
A more restricted notion has better algebraic properties. A strong parity edge-coloring (spec) is an edge-coloring in which every parity walk is closed. Using distinct colors again works, so we let the strong parity edge-chromatic number b p(G) be the minimum number of colors in a spec. A spec has no parity path, so every spec is a pec, and always b p(G) ≥p(G).
Characterizing subgraphs of Qk using parity edge-coloring yields p(G) ≥⌈lg |V (G)|⌉ when G is connected, with equality for a path or even cycle (throughout, lg denotes log2).
When n is odd, p(Cn) = b p(Cn) = 1 + ⌈lg n⌉. Also p(K2,n) = b p(K2,n) = 2 ⌈n/2⌉. In these examples, p(G) = b p(G); we also give examples where equality fails.
In this paper, we primarily explore the elementary properties of these parameters. In a subsequent paper , we prove that b p(Kn) = 2⌈lg n⌉−1. This strengthens a special case of a theorem of Yuzvinsky about sums of binary vectors (see Section 3.2).
Among the questions we raise in Section 6 is whether also p(Kn) = 2⌈lg n⌉−1; we prove this for n ≤16. The complete bipartite graph Kn,n behaves like Kn in that p(Kn,n) = b p(Kn,n) = χ′(Kn,n) = n when n = 2k. Also, b p(Kn,n) ≤b p(Kn) + 1 for all n; we conjecture that equality holds. We show that b p(Km,n) ≤m′ ⌈n/m′⌉, where m ≤n and m′ = 2⌈lg m⌉.
As a possible tool for exploring conjectured equalities between p and b p, we introduce a generalization. A parity r-set edge-coloring assigns r colors to each edge so that every selection of one color from the set at each edge yields a parity edge-coloring. Let pr(G) be the minimum number of colors used. Always pr(G) ≤rp(G), and we prove equality for paths. Proving p2(Kn) = 2p(Kn) could be a step toward proving p(Kn) = 2⌈lg n⌉−1.
In Section 5, we distinguish parity edge-coloring from related edge-coloring problems.
Section 6 poses many open questions.
2 Elementary Properties and Examples First we formalize elementary observations from the Introduction.
Remark 2.1 For every graph G, b p(G) ≥p(G) ≥χ′(G), and the parameters b p and p are monotone under the subgraph relation.
Proof. We have p(G) ≥χ′(G) by considering paths of length 2, and b p(G) ≥p(G) since closed walks are not paths. For H ⊆G, a pec or spec of G restricts to such an edge-coloring on H, since every parity walk in the restriction to H is a parity walk in the coloring on G.
2 When G is a forest, every pec is also a spec, so p(G) = b p(G). Edge-coloring the hypercube by coordinates shows that p(Qk) ≤b p(Qk) ≤k. Hence p(G) ≤k if G ⊆Qk. For trees, we prove the converse.
Given a k-edge-coloring f and a walk W, we use π(W) to denote the parity vector of W, recording the usage of each color as 0 or 1. When walks W and W ′ are concatenated, the parity vector of the concatenation is the vector binary sum π(W) + π(W ′). The weight of a vector is the number of nonzero positions.
Theorem 2.2 A tree T embeds in the k-dimensional hypercube Qk if and only if p(T) ≤k.
Proof. We have observed necessity. Conversely, let f be a parity k-edge-coloring of T (there may be unused colors if p(T) < k). Fix a root vertex r in T. Define φ: V (T) →V (Qk) by setting φ(v) = π(W), where W is the r, v-path in T.
When uv ∈E(T), the r, u-path and r, v-path in T differ in one edge, so φ(u) and φ(v) are adjacent in Qk. It remains only to check that φ is injective. The parity vector for the u, v-path P in T is φ(u)+φ(v), since summing the r, u-path and r, v-path cancels the portion from r to P. Since f is a parity edge-coloring, φ(P) is nonzero, and hence φ(u) ̸= φ(v).
When k is part of the input, recognizing subgraphs of Qk is NP-complete , and this remains true when the input is restricted to trees . Therefore, computing p(G) or b p(G) is NP-hard even when G is a tree. Perhaps there is a polynomial-time algorithm for trees with bounded degree or bounded diameter.
The Havel–Mov´ arek characterization of subgraphs of Qk follows easily from Theorem 2.2 (they also proved statements equivalent to Theorem 2.2 and Corollary 2.5.) Their proof is essentially the same as ours, but our organization is different in the language of pecs.
Corollary 2.3 A graph G is a subgraph of Qk if and only if G has a parity k-edge-coloring in which every cycle is a parity walk.
Proof.
We have observed necessity.
For sufficiency, choose a spanning tree T.
Since p(T) ≤p(G) ≤k, Theorem 2.2 implies that T ⊆Qk. Map T into Qk using φ as defined in the proof of Theorem 2.2. For each xy ∈E(G) −E(T), the cycle formed by adding xy to T is given to be a parity walk. Hence the x, y-path in T has parity vector with weight 1. This makes φ(x) and φ(y) adjacent in Qk, as desired.
Mitas and Reuter later gave a much lengthier proof motivated by studying sub-diagrams of the subset lattice. They also characterized the graphs occurring as induced subgraphs of Qk as those having a k-edge-coloring satisfying properties (1) and (2) and (3), where property (3) essentially states that that if the parity vector of a walk W has weight 1, then the endpoints of W are adjacent.
Spanning trees yield a general lower bound on p(G), which holds with equality for paths, even cycles, and connected spanning subgraphs of Qk.
3 Corollary 2.4 If G is connected, then p(G) ≥⌈lg n(G)⌉.
Proof. If T is a spanning tree of G, then p(G) ≥p(T). Since T embeds in the hypercube of dimension p(T), we have n(G) = n(T) ≤2p(T) ≤2p(G).
Corollary 2.5 For all n, p(Pn) = b p(Pn) = ⌈lg n⌉. For even n, p(Cn) = b p(Cn) = ⌈lg n⌉.
Proof. The lower bounds follow from Corollary 2.4. The upper bounds hold because Qk contains cycles of all even lengths up to 2k.
A result equivalent to p(Pn) = b p(Pn) = ⌈lg n⌉appears in (without defining either parameter). When n is odd, Cn needs an extra color beyond ⌈lg n⌉. To prove this, we begin with simple observations about adding an edge.
Lemma 2.6 (a) If e is an edge in a graph G, then p(G) ≤p(G −e) + 1.
(b) If also G −e is connected, then b p(G) ≤b p(G −e) + 1.
Proof. (a) Put an optimal parity edge-coloring on G −e and add a new color on e. There is no parity path avoiding e, and any path through e uses the new color exactly once.
(b) Put an optimal spec on G −e and add a new color on e. Let P be a u, v-path in G −e, where u and v are the endpoints of e. Suppose that there is an open parity walk W.
Note that W traverses e an even number of times, since no other edge has the same color as e. Form W ′ by replacing each traversal of e by P or its reverse, depending on the direction of traversal of e. Every edge is used with the same parity in W ′ and W, and the endpoints are unchanged, so W ′ is an open parity walk in G −e. This is a contradiction.
Lemma 2.6(b) does not hold when G −e is disconnected (see Example 2.8).
Theorem 2.7 If n is odd, then p(Cn) = b p(Cn) = ⌈lg n⌉+ 1.
Proof. Lemma 2.6(b) yields the upper bound, since b p(Pn) = ⌈lg n⌉.
For the lower bound, we show first that b p(Cn) = p(Cn) (this and Lemma 2.6(a) yield an alternative proof of the upper bound). Let W be an open walk, and let W ′ be the subgraph formed by the edges with odd usage in W. The sum of the usage by W of edges incident to a vertex x is odd if and only if x is an endpoint of W. Hence W ′ has odd degree precisely at the endpoints of W. Within Cn, this requires W ′ to be a path P joining the endpoints of W. Under a parity edge-coloring f, some color has odd usage along P, and this color has odd usage in W. Hence f has no open parity walk, and every parity edge-coloring is a spec.
It now suffices to show that b p(Cn) ≥p(P2n). Given a spec f of Cn, we form a parity edge-coloring g of P2n with the same number of colors. Let v1, . . . , vn be the vertices of Cn in order, and let u1, . . . , un, w1, . . . , wn be the vertices of P2n in order. Define g by letting g(uiui+1) = g(wiwi+1) = f(vivi+1) for 1 ≤i ≤n −1 and letting g(unw1) = f(vnv1).
4 Each path in P2n corresponds to an open walk in Cn or to one trip around the cycle.
There is no parity path of the first type, since f is a spec. There is none of the second type, since Cn has odd length.
The “unrolling” technique of Theorem 2.7 leads to an example G with b p(G) > p(G), which easily extends to generate infinite families.
Example 2.8 Form a graph G by identifying a vertex of K3 with an endpoint of P8. Since p(K3) = p(P7) = 3, adding the connecting edge yields p(G) ≤4 (see Lemma 2.6(a)).
We claim that b p(G) ≥p(P18) = 5. We copy a spec f of G onto P18 with the path edges doubled. Beginning with the vertex of degree 1 in G, walk down the path, once around the triangle, and back up the path. This walk has length 17; copy the colors of its edges in order to the edges of P18 in order to form an edge-coloring g of P18.
Each path in P18 corresponds to an open walk in G or a closed walk that traverses the triangle once. There is no parity path of the first type, since f is a spec. There is none of the second type, since such a closed walk has odd length. This proves the claim.
Since b p(K3) = b p(P7) = 3, this graph G also shows that adding an edge can change b p by more than 1 when G is disconnected.
We know of no bipartite graph G with b p(G) > p(G). Nevertheless, it is not true that every optimal parity edge-coloring of a bipartite graph is a spec.
Example 2.9 Let G be the graph obtained from C6 by adding two pendant edges at one vertex. Let W be the spanning walk that starts at one pendant vertex, traverses the cycle, and ends at the other pendant vertex. Let f be the 4-edge-coloring that colors the edges of W in order as a, b, a, c, b, d, c, d. Although f is an optimal parity edge-coloring (∆(G) = 4), it uses each color twice on the open walk W, so it is not a spec. Changing the edge of color d on the cycle to color a yields a strong parity 4-edge-coloring.
3 Specs and Canonical Colorings The monotonicity of p and b p guarantees that Kn has the largest value of both parameters among all n-vertex graphs.
Thus determining p(Kn) or b p(Kn) solves the corresponding extremal problem for n-vertex graphs.
The optimal pecs for complete graphs and complete bipartite graphs seem to have a particularly nice form. Whether optimal or not, these constructions provide upper bounds.
Definition 3.1 A canonical coloring of a graph is an edge-coloring defined by assigning vertex labels that are binary vectors (of the same length) and giving each edge the color that is the binary vector sum of the labels of its endpoints.
5 Lemma 3.2 For any graph G, every canonical coloring generated using distinct vertex labels is a spec. If G is bipartite, then every canonical coloring generated from vertex labels such that each is used at most once in each partite set is a spec.
Proof. Suppose that W is an open walk whose endpoints have names differing in some bit i. The total usage of colors flipping bit i along W must then be odd, and hence some color has odd usage on W. In the bipartite case, we may also have open walks whose endpoints have the same names, but such walks have odd length, and odd length forces odd usage of some color.
Corollary 3.3 If n = 2k, then b p(Kn) = p(Kn) = χ′(Kn) = n −1, and b p(Kn,n) = p(Kn,n) = χ′(Kn,n) = n. In general, b p(Kn) ≤2⌈lg n⌉−1 and b p(Kn,n) ≤2⌈lg n⌉.
Proof. The canonical coloring of Kn using colors of length ⌈lg n⌉uses 2⌈lg n⌉−1 colors (the color 0 is not used), and this equals the trivial lower bound when n = 2k. The same is true for Kn,n, except that color 0 also is used.
In , we prove that always b p(Kn) = 2⌈lg n⌉−1. The main idea is to introduce an additional vertex without needing additional colors until a power of 2 is reached. At that point, the trivial lower bound implies that 2⌈lg n⌉−1 colors were in use all along. The proof involves studying the vector space Fk 2 of binary k-tuples under component-wise binary addition. A corollary of the proof is that every optimal spec of Kn, for every n, is a canonical coloring generated by vectors of length ⌈lg n⌉.
Theorem 3.4 () b p(Kn) = 2⌈lg n⌉−1.
To put this result in perspective and to motivate the conjectures that remain about Kn and Kn,n, we briefly describe Yuzvinsky’s Theorem. Yuzvinsky proved that for subsets A and B of Fk 2 with fixed sizes r and s, the number of vectors that can be obtained as the sum of a vector in A and a vector in B is at least a certain quantity r ◦s called the “Hopf–Stiefel function” of r and s. (In non-algebraic language, r ◦s has an equivalent definition as the least n such that n k is even for each k with n −s < k < r. The condition is vacuous if n ≥r + s −1, so trivially r ◦s ≤r + s −1.) Later, Plagne computed a nice formula for this function, and K´ arolyi gave a short proof of that result. We combine these results into a single statement relevant to our context.
Theorem 3.5 (Yuzvinsky , Plagne , K´ arolyi ) For A, B ⊆Fk 2, let C = {a + b : a ∈A, b ∈B}. If |A| = r and |B| = s, then |C| ≥r ◦s, where r ◦s = min k∈N n 2k l r 2k m + l s 2k m −1 o .
6 When A = B, with size r, the minimization yields r ◦r = 2⌈lg r⌉. Yuzvinsky’s Theorem for this case says that every canonical coloring of Kr uses at least 2⌈lg r⌉−1 colors. Our result strengthens this by showing that in the more general family of strong parity edge-colorings, it remains true that at least 2⌈lg r⌉−1 colors are needed.
The bound in Yuzvinsky’s Theorem is always tight (see ); that is, for r, s ≤2k there exist A, B ⊆Fk 2 with |A| = r, |B| = s, and |C| = r ◦s. By Lemma 3.2, b p(Kr,s) ≤r ◦s.
We conjecture that equality holds. A direct proof determining b p(Kr,s) in the graph-theoretic setting would strengthen all cases of Yuzvinsky’s Theorem.
Conjecture 3.6 b p(Kr,s) = r ◦s.
Yuzvinsky’s Theorem as stated describes a bipartite situation, with the application to complete graphs as a special case. This relationship extends to specs, which means that proving the special case of Conjecture 6.3 for r = s = n also implies the result of on Kn.
That implication uses the following proposition.
Proposition 3.7 b p(Kn) ≥b p(Kn,n) −1.
Proof. Let f be a spec of Kn with vertex set u1, . . . , un. Given Kn,n with partite sets v1, . . . , vn and w1, . . . , wn, let f ′(viwj) = f(uiuj) when i ̸= j, and give a single new color to all viwi with 1 ≤i ≤n. A parity walk W ′ under f ′ starts and ends in the same partite set.
Let W be the walk obtained by mapping it back to Kn, which collapses vi and wi into ui, for each i. The edges that had the new color disappear; this number of edges is even, since W ′ was a parity walk. Hence W is a parity walk under f.
Since f is a spec, W is a closed walk in Kn. Hence W ′ starts and ends at vertices in the same partite set that have the same index. Since Kn,n has only one vertex with each index in each partite set, W ′ is closed. Hence f ′ is a spec of Kn,n.
We have observed that canonical colorings yield b p(Kn,n) ≤2⌈lg n⌉for all n. Toward the conjecture that equality holds, we offer the following.
Proposition 3.8 If some optimal spec of Kn,n uses a color on at least n −1 edges, then b p(Kn,n) = b p(Kn) + 1 = 2⌈lg n⌉. If a color is used n −r times, then b p(Kn,n) ≥2⌈lg n⌉− r 2 .
Proof. We prove the general statement. Let f be such a spec, and let c be such a color. Let U be one partite set, with vertices u1, . . . , un. Whenever color class c is incident to at least one of distinct vertices ui, uj ∈U, let Pi,j be a ui, uj-path of length 2 in which one edge has color c under f. Choose these so that Pj,i is the reverse of Pi,j. When c appears at neither ui nor uj, leave Pi,j undefined.
Let G be the graph obtained from Kn with vertex set v1, . . . , vn by deleting the edges vivj such that Pi,j is undefined; there are r 2 such edges. Define a coloring f ′ on G by letting f(vivj) be the color other than c on Pi,j.
7 We claim that f ′ is a spec. Given a parity walk W ′ under f ′, define a walk W in Kn,n as follows. For each edge vivj in W ′, follow Pi,j. By construction, the usage in W of each color other than c is even. Hence also the usage of c is even. Hence W is a parity walk under f and therefore is closed. Since W starts and ends at the same vertex ui ∈U, also W ′ starts and ends at the same vertex vi.
We have proved that every parity walk under f ′ is closed, so f ′ is a spec. Hence f ′ has at least b p(G) colors, and f has at least one more. By Lemma 2.6(b) and Theorem 3.4, b p(G) ≥2⌈lg n⌉−1 − r 2 , which completes the proof of the lower bound.
For the upper bound, Corollary 3.3 shows that 2⌈lg n⌉colors suffice.
Corollary 3.9 b p(Kn,n) ≥maxr min{2⌈lg n⌉− r 2 , n2 n−r−1}.
Proof. If E(Kn,n) has a spec with s colors, where s < 2⌈lg n⌉− r 2 , then by Proposition 3.8 no color can be used at least n −r times, and hence n2/s ≤n −r −1. Thus b p(Kn,n) ≥ min{2⌈lg n⌉− r 2 , n2/(n −r −1)}.
With r = 1, we conclude that b p(Kn,n) ≥2k when n > 2k −3 −4/(n −2), since then n2/(n −2) > 2k −1. Thus b p(K5,5) = 8, and b p(Kn,n) = 16 for 13 ≤n ≤16. Using r = 2, we obtain 14 ≤b p(K9,9) ≤16.
Corollary 3.3 shows that when n is a power of 2, the lower bound of ∆(G) is optimal for strong parity edge-coloring of Kn,n. We next enlarge the class of complete bipartite graphs where this bound is optimal.
Theorem 3.10 If m = 2k and m divides n, then p(Km,n) = b p(Km,n) = ∆(Km,n) = n.
Proof. Let r = n/m and [r] = {1, . . . , r}. Label the vertices in the small part with Fk 2.
Label those in the large part with Fk 2 × [r]. Color the edges with color set Fk 2 × [r] by setting f(uv) = (u + v′, j), where v = (v′, j). In other words, we use r edge-disjoint copies of the bicanonical coloring on r edge-disjoint copies of Km,m.
We have used n colors, so it suffices to show that f is a spec. Let W be a parity walk under f. Erasing the second coordinate maps W onto a walk W ′ in Km,m. Furthermore, W ′ is a parity walk, because all edges in W whose color has the form (z, j) for any j are mapped onto edges with color z under the bicanonical coloring of Km,m, and there are an even number of these for each j. Hence W ′ is closed.
Hence W starts and ends at vertices labeled with the same element u of Fk 2, and they are in the same part since W has even length. If these vertices are different copies of u in the large partite set, then those copies of Km,m have contributed an odd number of edges to W, so for each of them some color confined to it has odd usage in W. This contradicts that W is a parity walk. Hence W is closed, and f is a spec.
8 Corollary 3.11 If m ≤n and m′ = 2⌈lg m⌉, then b p(Km,n) ≤m′ ⌈n/m′⌉.
Proof. Km,n ⊆Km′,m′⌈n/m′⌉.
Corollary 3.11 provides examples of complete bipartite graphs where the maximum degree bound is optimal even though the size of neither partite set is a power of 2. For example, b p(K3,12) = 12. We use Corollary 3.11 next to compute the exact values when m = 2. We will apply the result for K2,3 in Theorem 4.2.
Corollary 3.12 b p(K2,n) = p(K2,n) = 2 ⌈n/2⌉.
Proof. The upper bound is immediate from Corollary 3.11 with m′ = 2.
For the lower bound, since ∆(K2,n) = n for n ≥2, it suffices to show that n must be even when f is a parity edge-coloring of K2,n with n colors. Let {x, x′} be the partite set of size 2.
Each color appears at both x and x′. If color a appears on xy and x′y′, then f(xy′) = f(x′y), since otherwise the colors a and f(xy′) form a parity path of length 4.
Hence y and y′ have the same pair of incident colors. Making this argument for each color partitions the vertices in the partite set of size n into pairs. Hence n is even.
The upper bound in Corollary 3.12 can also be proved using an augmentation lemma.
If f is a spec of a connected graph G, and G′ is formed from G by adding new vertices x and y with common neighbors u and v in G (and no other new edges), then the coloring f ′ obtained from f by adding two new colors a and b alternating on the new 4-cycle is a spec of G′. This yields b p(G′) ≤b p(G) + 2. Like Lemma 2.6(b), this statement fails for disconnected graphs. Since we presently have no further applications for this lemma, we omit the proof.
4 Parity Edge-Coloring of Complete Graphs Theorem 3.4 states that b p(Kn) = 2⌈lg n⌉−1, and Conjecture 6.3 asserts that b p(Kn,n) = 2⌈lg n⌉.
When n is not a power of 2, these values for Kn and Kn,n exceed the maximum degree, which is the trivial lower bound. Hence it is conceivable that in the more general family of parity edge-colorings (not necessarily specs), there is an edge-coloring that uses fewer colors.
We conjecture that this is not the case, and that indeed p(Kn) = b p(Kn) = 2⌈lg n⌉−1, and similarly for complete bipartite graphs.
To prove that p(Kn) = 2⌈lg n⌉−1 for all n, it suffices to prove it when n has the form 2k+1.
Below we prove it for K5 and K9 by case analysis involving counting arguments. This proves the conjecture for Kn whenever n ≤16. Canonical colorings provide the constructions; we only need the lower bounds.
Proposition 4.1 p(K5) = 7.
9 Proof. Suppose that K5 has a parity edge-coloring f using at most six colors. Each color class is a matching and hence has size at most 2. Since K5 has 10 edges, using at most six colors requires at least four color classes of size 2. Since any two colors used twice must not form a parity path of length 4, each pair of colors used twice forms an alternating 4-cycle.
Hence the colors used twice are all restricted to the same four vertices. However, there are only three disjoint matchings of size 2 in K4. Thus f cannot exist.
Theorem 4.2 p(K9) = 15.
Proof. Let f be a parity edge-coloring using at most 14 colors; we obtain a contradiction.
Let Ci be the set of edges in the ith color class, and let Gi,j be the spanning subgraph with edge set Ci ∪Cj. By Lemma 2.4, a connected subgraph using any k colors has at most 2k vertices. Hence each Gi,j has at least three components. If |Ci ∪Cj| ≥7, then Gi,j has at most three components, since the only non-tree components are 4-cycles, allowing the edges to be ordered so that the first six edges reduce the number of components when added.
If each Gi,j has at least four components, then |Ci ∪Cj| ≤6. If any class has size 4, then the others have size at most two. Since K9 has 36 edges, and 4 + 2 · 13 = 30 < 36, always |Ci| ≤3. Furthermore, since 7 · 3 + 7 · 2 < 36, at least eight classes have size 3; let C1 be one of them. If also |Cj| = 3, then G1,j has a 4-cycle, since otherwise six edges reduce G1,j to three components. The three edges of C1 can form at most six 4-cycles with other colors, but seven other classes have size 3. The contradiction eliminates this case.
Hence we may assume that G1,2 has three components A1, A2, A3 with vertex sets V1, V2, V3 and |V1| ≤|V2| ≤|V3| ≤4. Note that |V2| ≥3. We show that for i < j, at least four colors join Vi to Vj. If |Vj| = 4, then the edges from Vj to a vertex of Vi have distinct colors. If |Vj| < 4, then |Vj| = 3 and |Vi| ≥2. The edges joining two vertices of Vi to Vj form K2,3.
By Corollary 3.12, p(K2,3) = 4.
No color class j outside {1, 2} connects one of {V1, V2, V3} to the other two, since that would yield a connected 9-vertex graph in the three colors {1, 2, j}, contradicting Corol-lary 2.4. With three disjoint sets of four colors joining the pairs of components of G1,2, we now have 14 colors in f. To avoid using another color, the remaining edges joining vertices within components of G1,2 must have colors used joining those components.
Since |V2| ≥3, we may choose u, v, w ∈V2 with uv ∈C1 and vw ∈C2 and uw ∈C3 and e being an edge of C3 that connects distinct sets Vi and Vj. Suppose first that e is incident to V2. If |V2| = 4, then wx ∈C1 or ux ∈C2, and appending e to one end of vu, uw, wx or vw, wu, ux yields a parity path. If |V2| = 3, then |V1| ≥2, and the end of e other than v is incident to an edge e′ in C1 or C2. Now e′, e, vu, uw or e′, e, vw, wu is a parity path.
Hence the endpoints of e are in V1 and V3. Let z be the endpoint in V3. If |V3| = 4, then each of the four colors joining V2 to V3 appears at each vertex of V2. Thus the color on uz appears also on some edge wy, and e, zu, uw, wy is a parity path.
10 Hence |V1| = |V2| = |V3| = 3. Since the nine edges joining V2 and V3 use only four colors, some color is used on three of the edges. Call it C4, with edges uu′, vv′, ww′ joining V2 and V3. Avoiding a parity path using C4 with C1 or C2 forces u′v′ ∈C1 and v′w′ ∈C2. If z ∈{u′, w′}, then e, zu, uw, ww′ or e, zw, wu, uu′ is a parity path. Hence z must be v′, and so C3 appears only once on the copy of K3,3 joining V2 and V3.
However, K3,3 has no parity 4-edge-coloring in which one color is used only once. The other three colors would have multiplicities 3, 3, 2. Two matchings of size 3 in K3,3 form a 6-cycle, and a 2-colored 6-cycle contains a parity path.
It may be possible to generalize these arguments, but the case analysis seems likely to grow. Instead, we suggest another approach that could help to prove p(Kn) ≥2⌈lg n⌉−1.
Definition 4.3 A parity r-set edge-coloring of a graph G is an assignment of an r-set of colors to each edge of G so that every selection of a color from the set on each edge yields a parity edge-coloring of G. Let pr(G) be the minimum size of the union of the color sets in a parity r-set edge-coloring of G.
Parity r-set edge-coloring is related to parity edge-coloring as r-set coloring is to ordinary proper coloring. An r-set coloring of a graph assigns r-sets to the vertices so that the sets on adjacent vertices are disjoint, with χr(G) being the least size of the union of the sets. The r-set edge-chromatic number χ′ r(G) is defined by χ′ r(G) = χr(L(G)). Thus pr(G) ≥χ′ r(G).
By using r copies of an optimal parity edge-coloring with disjoint color sets, it follows that pr(G) ≤rp(G). We have no examples yet where equality does not hold. Proving equality could help determine p(Kn) by using the following result.
Proposition 4.4 If Kn has an optimal parity edge-coloring in which some color class has size ⌊n/2⌋, then p(Kn) ≥1 + p2(K⌈n/2⌉).
Proof. Let f be an optimal parity edge-coloring in which c is used on ⌊n/2⌋edges. Let u1v1, . . . , u⌊n/2⌋v⌊n/2⌋be the edges with color c, and let u⌈n/2⌉be the vertex missed by c if n is odd. Contracting these edges yields K⌈n/2⌉, with uivi contracting to wi for i ≤⌊n/2⌋, and w⌈n/2⌉= u⌈n/2⌉when n is odd.
Form a 2-set edge-coloring f ′ of K⌈n/2⌉as follows. For i < j ≤⌈n/2⌉, let f ′(wiwj) = {f(uiuj), f(viuj)}. Since f ′ does not use c, to prove p2(K⌈n/2⌉) ≤p(Kn) −1 it suffices to show that f ′ is a parity 2-set edge-coloring.
If f ′ is not a parity 2-set edge-coloring, then there is a parity path P ′ under some selection of edge colors from f ′. Form a path P in Kn as follows. When P ′ follows the edge wiwj with chosen color a, P moves along the edge uivi of color c (if necessary) to reach an endpoint in {ui, vi} of an edge with color a under f whose other endpoint is in {uj, vj}, and then it follows that edge. This path has the same usage as P ′ for every color other than c. Since c misses only one vertex of Kn, at least one end of P ′ is a contracted vertex, and an edge of 11 color c can be added or deleted at that end of P to make the usage of c even if it had been odd. If P ′ is a wi, wj-path, then P starts in {ui, vi} and ends in {uj, vj} (one of the sets may degenerate to {u⌈n/2⌉}). Thus P is a parity path under f, which is a contradiction.
If n = 2k + 1, then ⌈n/2⌉= 2k−1 + 1. If there is always an optimal parity edge-coloring of Kn with a near-perfect matching, then proving p2(Kn) = 2p(Kn) would inductively prove that p(Kn) = 2⌈lg n⌉−1. Although we do not know whether p2(G) = 2p(G) in general, we provide support for the various conjectures by proving this when G is a path.
Theorem 4.5 pr(Pn) = rp(Pn).
Proof. We prove the stronger statement that for every parity r-set edge-coloring f of Pn, there is a set of p(Pn) edges whose color sets are pairwise disjoint.
Let e1, . . . , en−1 be the edge set of Pn in order. We say that a subset {ei1, . . . , eiq} of E(Pn) with i1 < · · · < iq is linked by f if f(eij) ∩f(eij+1) ̸= ∅for 1 ≤j ≤q −1.
We claim first that if E(Pn) decomposes into linked sets S1, . . . , St under f, then setting f ′(e) = i when e ∈Si yields a parity edge-coloring f ′ of Pn with t colors. If not, then there is a parity path Q under f ′. Since Q has an even number of edges with color i, we can pair successive edges in the list of edges having color i (first with second, third with fourth, etc.).
Since Si is linked, we can pick a common color from the two sets assigned to a pair. Doing this for each pair and each color under f ′ selects colors from the sets assigned to Q under f that form a parity path. This contradicts the choice of f as a parity r-set edge-coloring.
Thus every partition of E(Pn) into linked sets needs at least p(Pn) parts.
To obtain edges with disjoint color sets from such a partition, we first construct a bipartite graph H with partite sets v1, . . . , vn−1 and w1, . . . , wn−1 by letting viwj be an edge if and only if i < j and f(ei) ∩f(ej) ̸= ∅. If E(Pn) has a partition into t linked sets, then H has a matching of size n −1 −t, obtained by using the edge viwj when ei and ej are successive elements in one part of the partition.
The construction of a matching from a partition is reversible. As edges are added to the matching, starting from the empty matching and the partition into singletons, the structural property is maintained that for the edges in a part, only the first edge ej has wj unmatched, and only the last edge ei has vi unmatched.
Hence when an edge viwj is added to the matching, it links the end of one part to the beginning of another part, reduces the number of parts, and maintains the structural property.
Thus E(Pn) has a partition into t linked sets under f if and only if H has a matching of size n −1 −t. When t is minimized, the K¨ onig–Egerv´ ary Theorem yields a vertex cover of H with size n −t −1. Because the complement of a vertex cover is an independent set, H has an independent set T of size n + t −1. Since V (H) consists of n −1 pairs of the form {vi, wi}, at least t such pairs are contained in T. If {vi, wi}, {vj, wj} ⊆T, then f(ei) and f(ej) are disjoint. Therefore there is a set of t edges whose color sets are pairwise disjoint.
We conclude that pr(Pn) ≥rt ≥rp(Pn).
12 5 Other Related Edge-Coloring Parameters In this section we describe other parameters defined by looser or more restricted versions of parity edge-coloring, and we give examples to show that p(G) is a different parameter.
A nonrepetitive edge-coloring is an edge-coloring in which no pattern repeats immediately on a path. That is, no path may have colors c1, . . . , ck, c1, . . . , ck in order for any k. The notion was introduced for graphs in . Every parity edge-coloring is nonrepetitive, and every nonrepetitive edge-coloring is proper, so the minimum number of colors in a non-repetitive edge-coloring of G lies between p(G) and χ′(G).
The resulting parameter is called the Thue chromatic number in honor of the famous theorem of Thue constructing non-repetitive sequences (generalized to graphs in ). The concept is surveyed in .
More restricted versions of parity edge-colorings have also been studied.
A conflict-free coloring is an edge-coloring in which every path uses some color exactly once.
An edge-ranking is an edge-coloring in which on every path, the highest-indexed color appears exactly once. Letting c(G) and t(G) denote the minimum numbers of colors in a conflict-free coloring and an edge-ranking, respectively, we have t(G) ≥c(G) ≥p(G).
Conflict-free coloring has been studied primarily in geometric settings; see [7, 9, 24]. Edge-rankings were introduced in . It is known that t(Kn) ∈Ω(n2) ; since p(Kn) ≤2n −3, the gap here can be large. Equality can hold: t(Pn) = c(Pn) = p(Pn) = ⌈lg n⌉. Although computing p(G) or b p(G) is NP-hard when G is restricted to trees, there is a algorithm to compute t(G) that runs in linear time when G is a tree (at least four slower polynomial-time algorithms were published earlier). Computing t(G) is NP-hard on general graphs , as is finding a spanning tree T with minimal t(T) .
In this string of inequalities, c(G) and p(G) are neighboring parameters. In this section, we present examples to show that they may differ. In fact, in all these examples c(G) > b p(G).
Corollary 5.1 c(K2k) > b p(K2k) = p(K2k) when k ≥4.
Proof. We have observed that the proof of Theorem 3.4 in implies that every optimal spec f of K2k is canonical.
The spanning subgraph of K2k formed by the color classes whose names are vectors of weight 1 is isomorphic to the hypercube Qk, and the colors on it correspond to the coordinate directions. If there is a path in Qk that crosses each coordinate direction more than once, then f is not conflict-free and c(K2k) > b p(K2k). In fact, it is easy to find such paths when k ≥4.
Example 5.2 As noted in Corollary 2.5, b p(C8) = 3. Suppose that C8 has a conflict-free 3-edge-coloring. If a color is used only once, then the other two colors alternate on paths of length 4 avoiding it, thus forming parity paths of length 4. Hence the sizes of the three color classes must be (4, 2, 2) or (3, 3, 2). Now deleting a edge from a largest color class yields a spanning path on which no color appears only once.
13 By induction on the length, every path has an optimal parity edge-coloring that is conflict-free (use a color only on a middle edge and apply the induction hypothesis to each component obtained by deleting that edge). This statement does not hold for trees.
Definition 5.3 A broom is a tree formed by identifying an endpoint of a path with a vertex of a star. Let Tk be the broom formed using P2k−2k+2 and a leaf of a star with k edges. The parity of a vertex in Qk is the parity of the weight of the k-tuple naming it.
We prove that Tk embeds in Qk but needs more than k colors for a conflict-free edge-coloring (for k ≥4). W. Kinnersley (private communication) showed this initially for k = 5.
We must first show that Tk indeed embeds in Qk. This follows from the result of that “double-star-like” tree with 2k−1 vertices in each partite set embed as spanning trees of Qk, since adding k −2 leaf neighbors to the 2-valent neighbor of the k-valent vertex in Tk yields such a tree. Their proof is lengthy; we give a short direct proof for this special case.
Lemma 5.4 If x and y are distinct vertices of Qk having the same parity, then there is a path of length 2k −3 in Qk that starts at x and avoids y.
Proof. It is well known that Qk has a spanning cycle when k ≥2. Since Qk is edge-transitive, there is a spanning path from each vertex to any adjacent vertex (for k ≥1).
The desired path exists by inspection when k = 2. For larger k, we proceed inductively.
Vertices x and y differ in an even number of bits; by symmetry, we may assume that they differ in the first two bits. Let Q′ and Q′′ be the (k −1)-dimensional subcubes induced by the vertices with first bit 0 and first bit 1, respectively. We may assume that x ∈V (Q′).
There is a spanning x, u-path P ′ of Q′, where u is the neighbor of x obtained by changing the third bit. Note that P ′ has length 2k−1 −1.
Let v be the neighbor of u in Q′′. Since v has the same parity as y, and v ̸= y, the induction hypothesis yields a path P ′′ of length 2k−1 −3 in Q′′ that starts at v and avoids y. Together, P ′, uv, and P ′′ complete the desired path in Qk.
Lemma 5.5 For k ≥2, the broom Tk embeds in Qk, and hence b p(Tk) = p(Tk) = k.
Proof. Note that Tk = P4 ⊆Qk when k = 2; we proceed inductively. For k > 2, the tree Tk contains Tk−1, obtained by deleting one leaf incident to the vertex v of degree k and 2k−1 −2 vertices from the other end. With Q′ and Q′′ defined in Lemma 5.4, by the induction hypothesis Tk−1 embeds in Q′. The distance in Tk−1 from v to its leaf nonneighbor u is 2k−1 −2(k −1) + 2. This is even, so u and v have the same parity. Let x and y be the neighbors of u and v in Q′′, respectively; also x and y have the same parity. By Lemma 5.4, Q′′ contains a path P of length 2k−1 −3 starting from x and avoiding y. Now adding vy, ux, and P to the embedding of Tk−1 yields the desired embedding of Tk in Qk.
14 Theorem 5.6 If k ≥4, then c(Tk) = k + 1 = b p(Tk) + 1.
Proof. For k = 4, a somewhat lengthy case analysis is needed to show c(T4) > 4; we omit this. Let x be the vertex of degree k in Tk.
For k ≥5, we decompose Tk into several pieces. At one end is a star S with k −1 leaves and center x. Let P be the path of length 2k−2 beginning with x. Let R be the path of length 2k−1 beginning at the other end of P. Since k ≥5, we have 2k−2 +2k−1 ≤2k −2k +2, so P and R fit along the handle of the broom. Ignore the rest of Tk after the end of R.
Consider a conflict-free k-edge-coloring of S ∪P ∪R. Since R has 2k−1 + 1 vertices, at least k colors appear on E(R). Since P has 2k−2 + 1 vertices, at least k −1 colors appear on E(P). Hence on P ∪R there are k −1 colors that appear at least twice, and only one color c appears exactly once. Since x has degree k, all k colors appear incident to x, including c.
Hence c appears on some edge of S, and adding this edge to P ∪R yields a path on which every color appears at least twice.
For k ≥2, we obtain a conflict-free (k + 1)-edge-coloring using color k + 1 only on the edge e of P at x. Deleting e leaves the star S and a path P ′ with 2k −2k + 2 vertices. Since S has k −1 edges and the length of P ′ is less than 2k, each has a conflict-free edge-coloring using colors 1 through k. Paths from V (S) to V (P ′) use color k + 1 exactly once.
It remains unknown how large c(G) can be when b p(G) = k or p(G) = k, either in general or when G is restricted to be a tree.
6 Open Problems Many interesting questions remain about parity edge-coloring and strong parity edge-coloring.
We have already mentioned several and collect them here with additional questions.
In Section 4, we proved the first conjecture for n ≤16. In Section 3.2, we proved various special cases of the second conjecture, which yield further special cases of the first.
Conjecture 6.1 p(Kn) = 2⌈lg n⌉−1 for all n.
Conjecture 6.2 p(Kn,n) = b p(Kn,n) = 2⌈lg n⌉for all n.
For complete bipartite graphs in general, the full story would be given by proving Con-jecture 6.3, which we restate here for completeness.
Conjecture 6.3 b p(Kr,s) = r ◦s for all r and s.
We have exhibited families of graphs G such that b p(G) > p(G) (see Example 2.8), but the difference is only 1, and the graphs we obtained all contain odd cycles.
Question 6.4 What is the maximum of b p(G) when p(G) = k?
15 Conjecture 6.5 p(G) = b p(G) for every bipartite graph G.
If Conjecture 6.5 holds, than Conjecture 6.3 also determines p(Kr,s. If the conjectures are not both true, then it would still be interesting to know how p(Kk,n) and b p(Kk,n) grow with k for fixed n. In particular, when do they reach 2⌈lg n⌉? Theorem 3.10 may shed some light. Does equality hold in Corollary 3.11?
Several questions about parity edge-coloring of trees are related to the study of which tree with 2k vertices embed as spanning trees of Qk. Havel proposed studying that question, and many papers quickly followed; Havel presents a survey. Such a tree must have partite sets of equal size and must have maximum degree at most k, but these conditions are not sufficient. Later results on sufficient conditions include [5, 13, 17, 23].
Question 6.6 What is the maximum of p(T) among n-vertex trees T with ∆(T) = D?
We observed from Theorem 2.2 that testing p(T) ≤k is NP-hard. This suggests com-plexity questions for more restricted problems.
Question 6.7 Do polynomial-time algorithms exist for computing p(T) on trees with max-imum degree D or on trees with bounded diameter?
The algebraic arguments in yield that recognition of specs is in P. However, we do not know whether the same holds for parity edge-coloring. (It does hold for edge-colorings of trees using the labeling procedure of Theorem 2.2.) Question 6.8 What is the complexity of testing whether an edge-coloring is a pec?
Paths and complete graphs show that p(G) cannot be bounded by bounding the maximum degree or the diameter. However, bounding both parameters limits the number of vertices.
Hence the next question makes sense.
Question 6.9 What is the maximum of p(G) among graphs (or trees) with ∆(G) ≤k and diam (G) ≤d?
It is a classical question to determine the maximum number of edges in an n-vertex subgraph of Qk, where n ≤2k. Does the resulting graph have the maximum number of edges in an n-vertex graph with parity edge-chromatic number k? More generally, Question 6.10 What is the minimum of p(G) among all n-vertex graphs having m edges?
The lower bound in Corollary 2.4 naturally leads us to ask which graphs achieve equality.
Every spanning subgraph of a hypercube satisfies p(G) = lg n(G); is the converse true?
16 Question 6.11 Which connected graphs G satisfy p(G) = ⌈lg n(G)⌉? Which satisfy b p(G) = ⌈lg n(G)⌉?
Motivated by the uniqueness of the optimal spec of K2k, Dhruv Mubayi suggested study-ing the “stability” of the result.
Question 6.12 Does there exist an parity edge-coloring of K2k with (1+o(1))2k colors that is “far” from the canonical coloring?
In Section 5, we showed that paths satisfy all three properties below. Are there other such graphs?
Question 6.13 For which graphs G do the following (successively stronger) properties hold?
(a) p2(G) = 2p(G)?
(b) pr(G) = rp(G) for all r?
(c) every parity r-set edge-coloring of G contains a set of p(G) edges whose color sets are pairwise disjoint?
Lemma 2.6(a) states that deleting an edge reduces the parity edge-chromatic number by at most 1. Ordinary coloring has the same property. Thus we are motivated to call a graph G critical if p(G −e) < p(G) for all e ∈E(G). We say that G is doubly-critical if p(G −e −e′) = p(G) −2 for all e, e′ ∈E(G). Our results on paths and cycles imply that for all n ≥1, P2n+1 is critical and C2n+1 is doubly-critical. Naturally, any star is doubly-critical.
Question 6.14 Which graphs are critical? Which graphs are doubly-critical?
Since the factors can be treated independently in constructing a spec, b p is subadditive under Cartesian product. Note that b p(P2□P2) = 2 = b p(P2) + b p(P2).
Question 6.15 For what graphs G and H does equality hold in b p(G□H) ≤b p(G) + b p(H)?
What can be said about p(G□H) in terms of p(G) and p(H)?
It may be interesting to compare p(G) with related parameters such as conflict-free edge-chromatic number on special classes of graphs. We suggest two specific questions.
Question 6.16 What is the maximum of c(T) such that T is a tree with p(T) = k? What is the maximum among all graphs with parity edge-chromatic number k?
Finally, the definitions of parity edge-coloring and spec extend naturally to directed graphs: the parity condition is the same but is required only for directed paths or walks.
Hence p(D) ≤p(G) and b p(D) ≤b p(G) when D is an orientation of G.
17 For a directed path ⃗ Pm, the constraints are the same as for an undirected path. More generally, if D is an acyclic digraph, and m is the maximum number of vertices in a path in D, then p(D) = b p(D) = ⌈lg m⌉. The lower bound is from any longest path.
For the upper bound, give each vertex x a label l(x) that is the maximum number of vertices in a path ending at x (sources have label 0). Write each label as a binary ⌈lg m⌉-tuple. By construction, l(v) > l(u) whenever uv is an edge. To form a spec of D, use a color ci on edge uv if the ith bit is the first bit where l(u) and l(v) differ. All walks are paths.
Any x, y-path has odd usage of ci, where the ith is the first bit where l(x) and l(y) differ, since no edge along the path can change an earlier bit.
Thus the parameters equal ⌈lg n⌉for the n-vertex transitive tournament, which contains ⃗ Pn. This suggests our final question.
Question 6.17 What is the maximum of p(T) or b p(T) when T is an n-vertex tournament?
Acknowledgement We thank Dan Cranston, Will Kinnersley, Brighten Godfrey, Michael Barrus, and Mohit Kumbhat for helpful discussions.
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18 S. Har-Peled and S. Smorodinsky, Conflict-free coloring of points and simple regions in the plane. Discrete Comput. Geom. 34 (2005), 47–70.
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4 (15) (1988), 299–302.
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Discrete Appl. Math. 30 (1991), 43–52.
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8580 | https://www.youtube.com/watch?v=9k_UKs_D7Jk | Probabilities involving Equally Likely Outcomes | Probability Theory
Wrath of Math
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How do we assess probabilities of events in sample spaces that have equal outcomes? Like tossing a coin or a die? We'll be going over some of these basic probability concepts in today's probability theory lesson! We'll also touch on an example of the inclusion exclusion principle for probability.
The basic idea, when evaluating the probability of an event in a sample space of equally likely outcomes is that the probability of an event is the number of ways the event can occur divided by the total number of outcomes in the sample space.
SOLUTION TO PRACTICE PROBLEM:
We assume all numbers are equally likely to be chosen, so each number has a 1/100 chance of being chosen (since there are 100 possibilities: 1, 2, 3, ..., 100). Which numbers have two digits? That would be all numbers from 10 to 99. This is a total of 90 numbers. Thus, the probability that the chosen number has two digits is 90/100 = 90% = 0.9 = 9/10; however you choose to write it.
I hope you find this video helpful, and be sure to ask any questions down in the comments!
The outro music is by a favorite musician of mine named Vallow, who, upon my request, kindly gave me permission to use his music in my outros. I usually put my own music in the outros, but I love Vallow's music, and wanted to share it with those of you watching. Please check out all of his wonderful work.
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Transcript:
Introduction suppose we run a simple random experiment like rolling a fair six-sided die in this experiment we could call the sample space s the sample space just contains all the possible outcomes of the experiment in this case the possible outcomes are rolling a 1 2 3 4 5 or a 6 now what is meant by saying this is a fair six-sided die well typically what is meant by fair is that all of the outcomes in the sample space are equally likely in today's wrath of math lesson we'll talk a bit about probabilities involving experiments that have equally likely outcomes like rolling die or flipping a coin these are some of the first examples we'll typically study in Probability of Rolling a 3 probability theory a quick question we might ask about this experiment is what is the probability that we roll a 3 which we could denote as P of 3 well I think it's pretty intuitive to realize that in an experiment that has six possible outcomes that are all equally likely the probability of any one of those outcomes occurring like in this case the probability of rolling a 3 is 1 divided by the total number of possible outcomes which in this case is 6 so that would be a probability of 1 over 6 and of course since all the probabilities are equally likely that is also the probability of rolling a 1 or 2 or 4 or a 5 or a 6 I think that seems pretty reasonable but let's quickly justify for sure why that must be the case imagine we have some experiment with a sample space s then suppose we add up the probabilities of all the outcomes in that sample space so this summation notation just means add up all the probabilities P of X for every X in the sample space s in the case of our die example that would be P of 1 plus P of 2 plus P of 3 and so on what must this sum be equal to well certainly in experiment with the sample space s one of the outcomes in S has to happen so if we add up all the probabilities it must be equal to 1 because it's guaranteed that some outcome in the sample space needs to happen however if all of the probabilities are equal say that the probability of X is equal to some number P for every outcome X in the sample space then this sum is just adding the same probability P to itself over and over again for every possible outcome in the sample space then if the sample space has a total of say and possible outcomes in it so this is the cardinality of the sample space is equal to n then this sum of all the probabilities in the sample space is just equal to n times P that's the common probability shared by all of the outcomes P multiplied by the total number of outcomes n so the sum of all probabilities in the sample space with equally likely outcomes is n times P which must be equal to 1 and so the probability of each outcome is equal to 1 divided by n the total number of outcomes so that's how we know that the probability of rolling any one number on a fair six-sided die is 1 over 6 in this case n is equal to 6 that's the number of possible outcomes and P the probability of each outcome is 1 over 6 notice as well since we're dividing by n this rule only applies to experiments with more than zero possible outcomes which seems like a fine restriction because we're not very interested in experiments that have zero possible outcomes then knowing this rule is true Examples we can solve lots of simple problems involving experiments with equally likely outcomes for example if we roll a fair six-sided die what's the probability that we roll an even number well we have to ask how could this event happen how could we roll an even number well we could roll a 2 which has a probability of 1 over 6 or we could roll a 4 which also has a probability one over six or we could roll a six which again has a probability of one over six so since the probability of every individual outcome is the same the probability that this event occurs that we roll an even number is just equal to the number of ways we can roll an even number which is 3 1 plus 1 plus 1 divided by the total number of possible outcomes which is 6 notice that we could reduce this fraction if we wanted to to 1/2 but oftentimes we're not going to want to do that leaving this fraction as 3 over 6 is going to make it a lot easier to compare it to other probabilities in the same experiment that we might not be able to reduce for example what's the probability that we roll a multiple of 3 on our six-sided die well we just solve this problem the same way how many ways can we roll a multiple of 3 we could roll 3 or we could roll 6 so there are two ways that we can roll a multiple of three out of six possible outcomes so the probability is 2 over 6 now since we've left these two probabilities with the denominator of 6 we can very quickly see that the probability of rolling an even number is greater than the probability that we roll a multiple of 3 since 3 is greater than 2 we could even get a bit more complicated with a fun example like this what's the probability that we roll a number that's even or a multiple of 3 to find this probability we could use the same strategy we've been using just count the numbers that are even or multiples of 3 which would give us 2 3 4 and 6 for a total of 4 possibilities divided by these six possible outcomes so the probability would be 4 over 6 the number of ways the event were interested in can occur divided by the total number of possible outcomes of the experiment but initially you might have been tempted to solve this problem a different way by adding up the probabilities of the two pieces of this event so with the probability of rolling something that's even or a multiple of 3 maybe we could find that by just adding the probability of a multiple of 3 to the probability of an even number that would give us three over six plus two over six which is not equal to 4 over six the answer we got just a minute ago this type of strategy works and it's very effective and very useful there's just one thing we need to be careful about do you see what the problem is the problem is that 6 is both even and a multiple of 3 so it's being counted twice when we add these probabilities so to correct this answer since we counted the probability of 6 occurring twice we need to subtract it once that corrects our answer giving us what we expect a probability of 4 over 6 and what we're seeing in action here is a principle that you'll use a lot as you continue to study probability theory called the inclusion exclusion principle but I hope this has been a helpful introduction to Summary probabilities of experiments that have equally likely outcomes here's just a general statement of what we learned today let s be the sample space of an experiment with equally likely outcomes then for any event a which is a subset of s the probability that the event a occurs is equal to the number of outcomes in the event a divided by the number of outcomes total the number of possible outcomes in s to draw this back to one of the examples we looked at remember one of the events we asked about was the event that we roll an even number the number of outcomes in that event of rolling an even number was three because we could roll it to four or a six and then the probability was just that number of outcomes in our event divided by the total number of outcomes six we may also want to specify here that the cardinality of s is greater than zero as in there's at least one possible outcome just specifying that of course because if the number of outcomes is zero this formula doesn't hold because we can't divide by zero now with all that said here's a quick practice problem to try on your own a random number is chosen from the numbers 1 through 100 assume that all of the numbers are equally likely to be chosen so there are equally likely outcomes what's the probability that the chosen number has two digits let me know what you get down in the comments and I'll leave the solution in the description so I hope this video helped you understand how to calculate probabilities involving equally likely outcomes let me know in the comments if you have any questions need anything clarified or of any other video requests thank you very much for watching I'll see you next time and be sure to subscribe for the swankiest math lessons on the internet and a big thanks to valo who upon my request kindly gave me permission to use his music in my math lessons links to his music in the description [Music] |
8581 | https://chemistry.stackexchange.com/questions/140791/calcium-chloride-hydrolysis | Skip to main content
Calcium chloride hydrolysis
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I was wondering about CaClX2 hydrolysis.
The most primitive answer would have been:
Ca(OH)X2 does not dilute that great, therefore it is not strong enough to have the CaX2+s not hydrolise. Thus:
CaX2++HX2O↽−−⇀CaOHX++HX+
Yet, as we can see in this answer - pKIIB for Ca(OH)X2 is 1.4, which is not bad. Therefore, some reckon the diluted part of Ca(OH)X2 to be a pretty strong base, making hydrolysis impossible. To make things worse, the cation hydrolysis is known to be harder to trace rather than anion hydrolysis.
Still, I have come across some articles on CaClX2 hydrolysis (like this one - a top Google search line; yet, this one first shows that ΔG900oC is 90.951>0).
So, is it safe to say that at ≈25oC CaClX2 does not undergo hydrolysis?
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Zhiltsoff IgorZhiltsoff Igor
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You can easily construct the chart of Ca^2+/CaOH+/Ca(OH)2 distribution as the function of pH.
Poutnik
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Poutnik
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@Poutnik unfortunately, I do not know how. I would appreciate if you write an answer explaining how such charts are done - might be a good tip instead of the full solution.
Zhiltsoff Igor
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Take a look at this. Should give you a qualitative idea. researchgate.net/figure/…
Eashaan Godbole
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Eashaan Godbole
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Why not to use mhchem MathJax extension \ce{Ca(OH)2} Ca(OH)X2 ?
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Just calculating from the head by the simplifying formula pH=pKa+log([AX−][HA]), pH of 1 MCaClX2 is approximately 1/2( 12.6 - log 1)=6.3.
So concentration [CaOHX+] is about million times smaller then [CaX2+]. Hydrolysis happens, but is fully negligible. Note that the real pH depends on impurities and dissolved COX2.
About the calculation of respective fractions, we can use the formula for gradual dissociation of multiprotic acid:
If we formally suppose
HX2A↽−−⇀HAX−+HX+↽−−⇀AX2−+2HX+
then respective molar fractions for particular acid forms are
xHX2A=xCaX2+=[HX+]2[HX+]2+Ka1⋅[HX+]+Ka1⋅Ka2
xHAX−=xCa(OH)X+=Ka1⋅[HX+][HX+]2+Ka1⋅[HX+]+Ka1⋅Ka2
xAX2−=xCa(OH)X2=Ka1⋅Ka2[HX+]2+Ka1⋅[HX+]+Ka1⋅Ka2
where Ka1=Kw/Kb1,Ca(OH)X2 and Ka2=Kw/Kb2,Ca(OH)X2
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There is actually a somewhat unique and important aspect of anhydrous Calcium chloride hydrolysis, not yet addressed.
Namely, a very exothermic dissolution reaction that occurs with solid CaClX2 from replicating the reverse reaction of its formation from Hydrogen chloride acting on Calcium oxide, at least, to a limited extent:
CaO(s)+2HCl(g)↽−−⇀CaClX2(s)+HX2O(l)+Energy
So, where does the energy term arise to move the cited equilibrium reaction partially back to the left? The answer for solid calcium chloride apparently includes a lattice dissociation enthalpy of +2,258 kJ/mol.
Here is an example of a supporting report (also found in Concise Encyclopedia of Chemistry by DeGruyter):
Thus, when you add calcium chloride to water, the solution heats. When adding calcium chloride to water, hydrochloric acid and calcium oxide form. You must be careful when mixing the substances due to the heat of the reaction and the acid produced.
The chemical implications of a presence of some insoluble CaO also includes a possible basic salt formation (as in calcium oxychloride), or with air/CO2 exposure, CaCOX3. So, if ones leaves anhydrous Calcium chloride open in even a low humidity atmosphere, expect some deterioration over time to the exposed product (with possible further implications for concrete, for example).
A suggested experiment, to newly purchased CaClX2 take a small amount and completely dissolve in water. Note, the heating of the solution. Take a similar amount of CaClX2 and place it on a plate in a low humidity environment. In a week, try again to dissolve the salt completely.
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8583 | https://www.mathworksheetsland.com/algebra/13biexp.html | Math Worksheets Land
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Home > Grade Levels > High School Algebra >
Binomial Theorem for Expansion Worksheets
There will be times when you are looking at binomials that you must further simplify or evaluate, and they confuse you because they seem endless. This is especially true when you have a complex exponent to work with as part of the expression. This is where this theorem can be used as tool to reform the expression and help you speed up your day. This series of worksheets and lessons has students learn a quick way to expand binomial expressions with the help of the Binomial Theorem. Make special note of this tool because as you advance, you may forget that it is in your toolbox.
Aligned Standard: HSA-APR.C.5
Expand the Binomial Step-by-step Lesson- We introduce you to the Binomial Theorem.
Guided Lesson - Run wild with expanding these three binomials. See how they line up with the formula.
Guided Lesson Explanation - I start by explaining the basic procedure for solving such problems.
Practice Worksheet - A series of problems that will have you completely understanding the concept end to end.
Matching Worksheet - Match the binomials to their expand form. You will just need to expand all of them out.
Answer Keys - These are for all the unlocked materials above.
Homework Sheets
This is another skill that I didn't see coming when the Common Core was presented.
Homework 1 - Binomial expressions contain two terms. The first terms is seen as an and the last term is seen as bn. When binomial expressions are raised to a power, they can be expanded using the following expansion formulas.
Homework 2 - In this case, the binomial is raised to the third power, so we will use this formula raised to it as well.
Homework 3 - We complete these problems by simply inserting the values into the formula.
Practice Worksheets
In the past, the theorem was always given to you. You find the theorem in most engineering field manuals.
Practice 1 - Start to get some practice with this skill up on the next level.
Practice 2 - Why would you go any further with that one? They are pulled apart pretty well.
Practice 3 - Expand the expression using the Binomial Theorem.
Math Skill Quizzes
Expanding beyond the fourth power is much more difficult for kids at this level.
Quiz 1 - Take it a few steps further. This will help you use algebra to solve some pretty complex problems.
Quiz 2 - This is a very fair choice of problems that can be done quickly.
Quiz 3 - A nice quiz for review to help you gauge how well you know this stuff.
What is the Binomial Theorem for Expansion?
Ever come across very long algebra problems and got stuck? Well, this is where the binomial theorem comes in for the rescue. A lot of people are skeptical about using the theorem as it looks very complicated, but people who get used to it say that there is nothing simpler and easier than a binomial expansion to solve a horrendously long algebra problem. It helps us expand expressions that follow the form (a + b)n. It demonstrates the result of multiplying a binomial by itself repeatedly as many times as you want. For example, if we wanted to find out the end value of the expression (3x + y)5, the Binomial Theorem for Expansion would provide us a shortcut for getting this done.
There are going to be times in your life where you will come across very complex and overstated algebra problems. When you have a binomial that you would like to expand you can use Pascal’s Triangle to make your life a little easier for a few minutes. You will run into this whenever you need to repeatedly multiply a binomial by itself. This happens when you have an exponential binomial. Using the binomial theorem is intuitive. Once you get a hang of it, it is not too hard to use this method of expansion. At first look it truly is one of the most intimidating equations you will encounter. Take a look at the lesson below and you will see what I mean.
As far as using the theorem, it tells us that we expand an expression with power in form of (y + y)n into a sum involving the terms of the form axbyc. It should be stated that both of the exponents (b and c) in this form must non-negative with b + c = n. The coefficient a is a positive integer as well. If the exponent were to be zero, the theorem states that some terms will be omitted and you will be left with 3x< sup>2.
How Do You Process This Calculation?
Like we said before, this theorem immediately will overwhelm you. Just look at this scary thing:If we look to the left of the equals symbol this just says that when you have a binomial expression (x + y) that is being raised to a power (n). If the y value is a negative constant, it is important to remember that it is a negative value so that you can translate that into the expression on the right side of the equation. We then just substitute the values of the variables (x, y, and n) into their respective place on the right-hand portion of the equation. Looking at the right side of the equation after we have substituted the values leaves us with two things left unanswered. The sigma notation just tells us that we’ll be adding many (n) variables together. The n choose k combination helps you determine the number of different ways you could choose k elements. Once you complete a few of these problems you quickly learn that this is much easier than it looks.
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8584 | https://math.stackexchange.com/questions/4459036/mathematical-inclusion-and-exclusion-of-elements-from-a-given-set-a | Skip to main content
Mathematical "inclusion" and "exclusion" of elements from a given set A?
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If I have a set A, comprising of numbers from 1 to 10: A={1,2,3,4,5,6,7,8,9,10}. Let's say I want to make another set by "including" all even numbers:
{2,4,6,8,10}
Or I wanted to make a different set by "excluding" all odd numbers:
{2,4,6,8,10}
These sets are of course the same. So is it correct to say that inclusion/exclusion are synonymous when it comes to set theory, as they're just different ways of building a set?
This might sound trivial, but I have a reason for asking: I want to understand if inclusion and exclusion are "commutative" properties, i.e. it doesn't matter in which order you apply them.
For example, let's say I make an operation to "filter" my set, by including all even numbers as we did before, producing set B
B={2,4,6,8,10}
And then a separate operation to "exclude" any numbers less than 6 from set B, resulting in set C:
C={6,8,10}
What if I started with A and applied the operations the other way around? First remove all numbers less than 6:
B={6,7,8,9,10}
Then filter B to "include" only even numbers:
C={6,8,10}
It seems intuitively to me that the result will always be the same no matter which order you apply the operation. Is this true for all cases no matter the set, however? Is there a way to prove that applying "filters" to a set (I'm not sure of the proper term) will always be commutative?
So in summary, there are two questions here:
Are the notions of "inclusion" and "exclusion" really synonymous from the point of view of applying an operation to a set to produce a subset?
Will applying these operations to produce a subset of a set always be commutative, i.e. produce the same result?
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It looks lile you are taking repeated intersections of sets. Intersection is both associative and commutative.
– CrackedBauxite
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May 26, 2022 at 10:25
@CrackedBauxite I've had a think about this. Is it intersection, or is it a set difference / relative complement? If I've got the set {1,2,3,4,5} and I exclude all odd numbers, I've got a set of odd numbers {1,3,5}. It's my original set {1,2,3,4,5}−{1,3,5}={2,4}
– Lou
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May 26, 2022 at 16:45
But if I were to do "include all even numbers" from the set {1,2,3,4,5}, then I have a set of even numbers {2,4}. {1,2,3,4,5}∩{2,4}=2,4 I think - but please correct me if my logic is wrong
– Lou
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May 26, 2022 at 16:46
So I think "include" as an operation represents an intersection, but "exclude" represents a set difference?
– Lou
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Set differences are also intersections. A∖B=A∩Bc, where Bc is the complement of B.
– CrackedBauxite
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@KevinS offers an excellent answer from a logic point of view. Here's another that relies on the idea of a filter. That's a concept useful in programming (particularly in lisp). You pass the items in your set through a filter that lets some through and blocks others.
In this sense "inclusion" and "exclusion" are really different ways to describe the same result. You can specify what you keep or what you reject. "Keep (just) the odds" is the same as "reject (only) the evens".
If you have two filters each of which is described independently of the other and refers only to properties of the things you are filtering then you can filter in either order.
The independence matters. If you think that the wine must match the main course then your options will depend on whether you see the wine list or the menu first.
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You've hit upon my actual use case - I'm doing filtering in Python and trying to understand if "inclusion" and "exclusion" for my use case are order-sensitive, or commutative operations. You've all helped me understand that they are commutative and therefore order should not matter.
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This is perhaps easier to see in the math logic. When you apply conditions to a set, you get subsets that satisfy those conditions (possibly the emptyset if the conditions aren't met). In your example, define:
A:={n∈N | 1≤n≤10}.
Then B and C are had by adding conditionals in the set definition:
B:={n∈A | ∃k∈N:n=2∗k}
and
C:={n∈B | n≥6}.
⟹C={n∈N ∣∣∣ (1≤n≤10)∧(∃k∈N:(n=2∗k))∧(n≥6)}.
From this viewpoint, the reductions made were simply conjuctions (a logical operation). Conjunction is certainly commutative.
Also, the term "exclusion" is synonymous with set difference:
X−Y:=X∩Yc,
whereas "inclusion" usually refers to an injective map:
ι:S↪X
which can be used as an identifier of a subset. I wouldn't say the two notions are synonymous.
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I'm a beginner in set theory, would you mind clarifying what you mean about an "injective map"? I understand the term set difference.
– Lou
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May 26, 2022 at 14:21
It seems as though the terms used might have multiple meanings as they apply to math and programming. An injection is a 1-1 function. The math notion I gave generalizes to categories (using equivalence classes of monomorphisms to identify sub-objects). This is not inherently programmable and is best used for theory.
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To your first question, consider using the set-builder notation to build a subset. That is, define B⊆A such that B={a∈A:Φ(a)}, where Φ is the logical formula which dictates what how we pick the values of A.
Now say that A is some collection of numbers, and Φ is the rule "include all even numbers". Then some other rule Ψ which says "do not include all not even numbers" is completely logically equivalent to Φ.
And this generalises nicely by considering that, if we impose a condition like "even number", any element of a set will satisify that condition, or not satisfy that condition. We cannot have an element which does neither.
The second question follows as others have pointed out, by writing the conditions as operations which are commutative.
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I have open-ended survey responses that I have categorized and am trying to visualize. Some responses fall into multiple categories, so the counts of the categories could hypothetically total 115 responses when there were only 100 respondents. I want to visualize how many people out of the 100 respondents fell into each category.
What is the best practice for plotting proportions that total greater than 100%? Is a standard bar chart the way to go here? Is there any situation where a pie chart can be used? If I plot counts of each category using a pie chart, proportions are calculated using the total counts instead of the total number of respondents. Is there a better way that I have not thought of?
Some example data where there are 100 respondents (percent being calculated as Count / Total Respondents 100)
| Category | Count | Percent |
---
| |
| Category 1 | 80 | 80% |
| Category 2 | 21 | 21% |
| Category 3 | 10 | 10% |
Edit: I believe a lot of people are misunderstanding the question. If 10 people choose Category 1 and Category 2, I want to know that 100% of people mentioned Category 1. I don't need to know that Category 1 accounts for 50% of all the categories mentioned. The first scenario is what I want to visualize.
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Glomerular Filtration: Net Filtration Pressure
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Overview
Glomerular filtration, a key process in the kidneys, is regulated by three main pressures: Glomerular blood hydrostatic pressure (GBHP), Capsular hydrostatic pressure (CHP), and Blood colloid osmotic pressure (BCOP).
GBHP, with an average value of 55 mmHg, promotes filtration by pushing water and solutes through the filtration membrane. This is balanced by two opposing forces: CHP, a "back pressure" exerted against the filtration membrane by fluid already in the capsular space and renal tubule, averaging at 15 mmHg, and BCOP (averaging at 30 mmHg), which arises due to the presence of plasma proteins like albumin, globulins, and fibrinogen.
The Net Filtration Pressure (NFP) determines the total pressure that drives filtration, calculated as NFP = GBHP - CHP - BCOP. Under normal conditions, this results in an NFP of about 10 mmHg, allowing a standard amount of blood plasma (minus plasma proteins) to filter from the glomerulus into the capsular space.
Transcript
Glomerular filtration depends on three main pressures: glomerular blood hydrostatic pressure or GBHP, capsular hydrostatic pressure or CHP, and blood colloid osmotic pressure or BCOP.
GBHP, typically about 55 mmHg, is the blood pressure inside the glomerular capillaries. It promotes filtration by pushing water and solutes through the filtration membrane.
CHP, averaging around 15 mmHg, is a back pressure exerted against the filtration membrane by fluid already present in the capsular space and renal tubule, which opposes filtration.
Additionally, BCOP in glomerular capillaries also opposes filtration. It averages 30 mmHg.
The net filtration pressure or NFP is the total pressure that promotes filtration.
It's calculated as NFP = GBHP - CHP - BCOP. Normal NFP is usually about ten mmHg.
In certain kidney diseases, glomerular capillaries can be damaged, allowing plasma proteins to enter the filtrate. This increases the NFP, leading to more fluid being filtered.
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Glomerular FiltrationNet Filtration Pressure (NFP)Glomerular Blood Hydrostatic Pressure (GBHP)Capsular Hydrostatic Pressure (CHP)Blood Colloid Osmotic Pressure (BCOP)Filtration MembraneRenal TubulePlasma ProteinsAlbuminGlobulinsFibrinogen
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Guidelines for Mucormycosis Treatment
First-line treatment: High-dose liposomal amphotericin B is strongly recommended as the first-line treatment for mucormycosis, with a minimum dose of 5 mg/kg/day 1, 2.
Surgical debridement: Surgical debridement with clean margins is strongly recommended in addition to antifungal treatment, especially for rhino-orbito-cerebral, soft tissue, and localized pulmonary lesions 1, 2, 3.
Control of underlying conditions: Control of underlying conditions, such as diabetes, neutropenia, and immunosuppression, is strongly recommended to improve treatment outcomes 1, 2, 3.
Alternative treatments: Isavuconazole and posaconazole are recommended with moderate strength as alternative treatments, with specific dosing regimens outlined in the guidelines 1, 4, 5.
Amphotericin B deoxycholate: Amphotericin B deoxycholate is recommended against due to its substantial toxicity, but may be considered in resource-limited settings 1, 2.
Important Considerations
Dosing regimens: Specific dosing regimens for liposomal amphotericin B, isavuconazole, and posaconazole are outlined in the guidelines, with considerations for renal impairment and other factors 1, 4, 5.
Salvage treatment: Posaconazole and isavuconazole are strongly recommended as salvage treatments for patients who do not respond to first-line treatment 1, 2.
From the Research
Mucormycosis Guidelines
The Infectious Diseases Society of America (IDSA) 2019 guidelines for Mucormycosis are not directly mentioned in the provided studies. However, the studies provide information on the treatment and management of Mucormycosis:
The latest recommendations strongly recommend as first-line therapy the use of liposomal amphotericin B (≥5mg/kg) combined with surgery whenever possible 6.
Lipid formulations of amphotericin B (LFAB) are the mainstay of therapy, but the newer triazoles, posaconazole (POSA) and isavuconazole (ISAV), may be effective in patients refractory to or intolerant of LFAB 7.
Antifungal therapeutic options are few, with polyenes and their combinations being appropriate for empirical therapy, while posaconazole and isavuconazole are best reserved for de-escalation, refractory cases, or patients intolerant to amphotericin B 8.
Early recognition of infection is critical for treatment success, followed by prompt initiation of antifungal therapy with lipid formulation amphotericin B, and posaconazole and isavuconazole should be used for stepdown and salvage therapy 9.
Treatment Approaches
The treatment approaches for Mucormycosis include:
Antifungal therapy with liposomal amphotericin B as the first-line treatment 6, 7, 9.
Surgical debridement or excision as an important adjunctive role in the treatment of Mucormycosis 6, 7, 8, 9.
Use of posaconazole and isavuconazole as second-line therapy or for stepdown and salvage therapy 6, 7, 8, 9.
Important Considerations
Important considerations in the treatment of Mucormycosis include:
Early recognition of infection is critical for treatment success 9.
Prompt initiation of antifungal therapy is essential for improving outcomes 9.
Identifying underlying risk factors is important for proper treatment 8.
Surgical debridement should be done prior to the dissemination of infection to improve clinical outcomes 8.
References
1
Guideline
Guideline Directed Topic Overview
Dr.Oracle Medical Advisory Board & Editors, 2025
2
Guideline
Guideline Directed Topic Overview
Dr.Oracle Medical Advisory Board & Editors, 2025
3
Guideline
Guideline Directed Topic Overview
Dr.Oracle Medical Advisory Board & Editors, 2025
4
Guideline
Guideline Directed Topic Overview
Dr.Oracle Medical Advisory Board & Editors, 2025
5
Guideline
Guideline Directed Topic Overview
Dr.Oracle Medical Advisory Board & Editors, 2025
6
Research
Mucormycosis treatment: Recommendations, latest advances, and perspectives.
Journal de mycologie medicale, 2020
7
Research
Mucormycosis.
Seminars in respiratory and critical care medicine, 2020
8
Research
Mucormycosis: The hidden and forgotten disease.
Journal of applied microbiology, 2022
9
Research
Mucormycosis.
Infectious disease clinics of North America, 2021
Related Questions
What is the treatment for mucormycosis in a diabetic patient?What is the approach to testing for mucormycosis?What are the guidelines for the diagnosis and treatment of mucormycosis (Infectious Diseases Society of America) in pediatric patients?What is mucormycosis?What is the dosage of Amphotericin B (Amphotericin B) for the treatment of mucormycosis?What is Alfacalcidol?What does hyperglobulinemia indicate?What is the management of Transient Ischemic Attack (TIA) in a patient with recent Prosthetic Valve Replacement (PVR) surgery 1 week prior?What are the guidelines for the diagnosis and treatment of mucormycosis (Infectious Diseases Society of America) in pediatric patients?What is the diagnosis and management of fever?What is the diagnosis and management of Acute Fatty Liver of Pregnancy (AFLP)?
Professional Medical Disclaimer
This information is intended for healthcare professionals. Any medical decision-making should rely on clinical judgment and independently verified information. The content provided herein does not replace professional discretion and should be considered supplementary to established clinical guidelines. Healthcare providers should verify all information against primary literature and current practice standards before application in patient care. Dr.Oracle assumes no liability for clinical decisions based on this content.
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8589 | https://www.chemicalforums.com/index.php?topic=83080.0 | Chemical Forums: Effect of loss of heat on enthalpy of neutralization
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stevemont7
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Effect of loss of heat on enthalpy of neutralization
« on: November 10, 2015, 03:34:22 PM »
Hello,
On a test, the following information preceded a question:
_H+(aq) + OH-(aq) H 2 O(l)
A student is asked to determine the molar enthalpy of neutralization, ∆H neut , for the reaction
represented above. The student combines equal volumes of 1.0 M HCl and 1.0 M NaOH in an open polystyrene cup calorimeter. The heat released by the reaction is determined by using the equation q = mc∆T._
The last of the questions that followed asks:
Suppose that a significant amount of heat were lost to the air during the experiment. What effect would this have on the calculated value of the molar enthalpy of neutralization, ∆H neut ? Justify your answer.
I wrote, "the calculated value of ΔH neut would increase because the heat lost to the air will cause the value of ΔT to be smaller."
This is incorrect, apparently, and the correct answer is supposed to be that ΔH neut decreases. However, since the reaction is exothermic, would ΔH not be negative? So since the question does not ask for change in the magnitude of ΔH, why is saying it increases incorrect?
« Last Edit: November 10, 2015, 03:51:30 PM by stevemont7 »
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mikasaur
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Re: Effect of loss of heat on enthalpy of neutralization
« Reply #1 on: November 10, 2015, 06:35:27 PM »
You're correct (I think; if someone else can argue otherwise please do so).
The loss of heat to the air would decrease the magnitude of ΔT and subsequently decrease the magnitude of ΔH neut (or make it a less negative number which is an "increase" in the value). Perhaps you can bring it up to your teacher/professor and let him or her know that you understand the concepts behind the question but got tripped up on semantics. Hopefully once they see you understand the concept they might give you some/all of the points back.
Edited to change neut to neut
« Last Edit: November 10, 2015, 06:55:35 PM by mikasaur »
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stevemont7
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Re: Effect of loss of heat on enthalpy of neutralization
« Reply #2 on: November 10, 2015, 10:35:00 PM »
Thank you for your response, mikasaur. I will talk to my teacher about it. I did not lose terribly much credit because of this particular question; however, I made a few careless errors (something I very rarely do), which further reduced my score and damaged my 99 average, so I would like to get any credit back that I can.
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mjc123
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Re: Effect of loss of heat on enthalpy of neutralization
« Reply #3 on: November 11, 2015, 09:01:04 AM »
I agree that you are technically correct, but I would always in such a situation be quite explicit, to avoid any possibility of misunderstanding. So rather than saying
"the calculated value of ΔH neut would increase because..."
I would say
"the calculated value of ΔH neut would increase (i.e. become less negative) because..."
or maybe
"the calculated value of ΔH neut would decrease in magnitude because..."
Then the reader is clear that you have understood the situation correctly, even if you haven't used their preferred form of words.
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stevemont7
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Re: Effect of loss of heat on enthalpy of neutralization
« Reply #4 on: November 11, 2015, 03:24:43 PM »
I see what you're saying, mjc123. From now on I'll be sure to specify what I mean on any similar questions where confusion might occur.
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8590 | https://www.wyzant.com/resources/answers/946352/find-the-area-of-the-region-bounded-by-the-parabola-y-x2-the-tangent-line-t | Find the area of the region bounded by the parabola y = x2, the tangent line to this parabola at the point (5, 25), and the x-axis. | Wyzant Ask An Expert
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Calculus
Natalie N.
asked • 08/28/24
Find the area of the region bounded by the parabola y = x2, the tangent line to this parabola at the point (5, 25), and the x-axis.
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y' = 2x; at x = 5 y' = 10. Equation of tangent line: y = 25 + 10(x - 5); y = 10x - 25; x = y/10 + 2.5 and y = √x;
Area A = ∫0 25(y/10 + 2.5 - √y)dy = (y 2/20 + 2.5y - 2/3y 3/2)0 25 = 125/4 + 62.5 - 250/3 = 125/12
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8591 | https://www.geeksforgeeks.org/physics/variation-of-pressure-with-depth/ | Variation of Pressure With Depth - GeeksforGeeks
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Variation of Pressure With Depth
Last Updated : 21 Jul, 2021
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It's reasonable to assume that the deeper a person travels into a liquid or gas, the higher the pressure exerted on him by the surrounding fluid. The reason for the increasing pressure is that the deeper a person goes into a fluid, the more fluid he has over top of him, and therefore the more weight he has.
Pressure: The ratio of the force applied to the surface area over which the force is applied is known as the pressure. The pressure imposed by liquids is known as hydrostatic pressure. The SI unit of pressure isPascal.
Fluid Pressure
Solid things do not change form when pressure is applied, which is obviously not the case with fluids. In a closed container, fluid pressure can be generated by gravity, acceleration, or forces. The fluid acts equally in all directions since it has no set form. When you fill a bottle with water, the weight of the water acts evenly on both sides of the bottle.
The force is always exerted perpendicular to the container's surface. This may be seen in a balloon. As you fill the balloon with air, you'll note that it grows evenly, with no one side inflating more than the other. Liquids in a container also show this behaviour.
Hydrostatic Pressure
The hydrostatic pressure is the pressure exerted by a fluid in equilibrium owing to gravity at any given period. When a downward force is applied, hydrostatic pressure is proportional to the depth measured from the surface as the weight of the fluid increases.
Fluids exert equal pressure in all directions. Another intriguing event occurs as a result of this rule. When we examine the layer of water on top of a bottle, the pressure it exerts acts on the edges of the container, the air surface on top, and the layer of water at the bottom. The pressure imposed by the top layer on the bottom increases as we go down the bottle from top to bottom.
The fluid at bottom of the container receives higher pressure than the fluid above it as a result of this process.
Variation of Pressure with Depth
Take a look at the figure below for an example of a container. The weight of fluid inside it is supported by its bottom. Let's see how much pressure the weight of liquids exerts on the bottom. The weight of fluid 'mg' divided by the area 'A' equals the pressure.
Weight of the fluid, W = m g
Mass of the fluid is equal to product of volume and density of substance, i.e., m = ρ V
Volume of the fluid is equal to the dimension of the container, i.e., V = A h
Combine the last two equations for mass.
m = ρ A h
Therefore, weight of the fluid, W = ρ A h g
The pressure exerted on the bottom of the container is given as:
P = W ⁄ A
P = (ρ A h g) ⁄ A
P = ρ h g
where,
ρ is the density of fluid
A is the surface area of container
h is the height upto the fluid is filled in container
V is the volume of fluid
m is the mass of fluid
g is the acceleration due to gravity
W is the weight of fluid
P is the pressure exerted on the bottom of the container.
This is the pressure created by a fluid's weight. Beyond the specific conditions under which it is derived here, the equation has generic validity. The surrounding fluid would exert this pressure even if the container was not there, keeping the fluid static. This equation remains true to deep depths for liquids that are practically incompressible. This equation may be used for gases that are very compressible as long as the density variations are minimal across the depth covered.
Sample Problems
Problem 1: Define the relationship between the pressure and height of the liquid column.
Answer:
The pressure exerted by a liquid depends on the height of the liquid column.
Pressure can be written as P = ρ g h where h is height and ρ is density. The formula shows the direct relation between the pressure and height of the column.
Therefore, as the height increases, pressure will also increase .
Problem 2: Calculate the force exerted by water on the base of a tank of area 3 m 2 when filled with water up to a height of 2 m. (Density of water is 1000 kg m−3 and g = 10 m s−2).
Solution:
Given:
Area of base of tank, A = 3 m 2
Height of the tank filled with water, h = 2 m
Density of the fluid, ρ = 1000 kg m−3
The formula of the pressure is given as:
P = ρ g h
= (1000 × 10 × 2) N m−2
= 20000 N m−2
The force exerted by the water,
F = P A
= (20000 × 3) N
= 60000 N
Hence, the force exerted by the water is 60000 N.
Problem 3: Why the dam of the water reservoir is thick at the bottom?
Answer:
The dam of a water reservoir is thick at the bottom because the pressure of water is highest at the maximum depth, and the dam must be strong at the bottom to withstand this maximum pressure.
Problem 4: What is hydrostatic pressure?
Answer:
The hydrostatic pressure is the pressure exerted by a fluid in equilibrium owing to gravity at any given period. When a downward force is applied, hydrostatic pressure is proportional to the depth measured from the surface as the weight of the fluid increases.
Problem 5: What is the pressure acting on the water at a depth of 2 m at 4°C?
Solution:
Given:
The depth of water column, h = 2 m
The density of water at 4°C, ρ = 1000 kg ⁄ m 3
The formula of the pressure is given as:
P = ρgh
= (1000 × 9.81 × 1) Pa
= 9810 Pa.
Hence, the pressure acting on the water at a depth of 2 m is 9810 Pa.
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8592 | https://haematologica.org/article/view/6846 | Platelet interaction with von Willebrand factor is enhanced by shear-induced clustering of glycoprotein Ibα | Haematologica
Open access journal of the Ferrata-Storti Foundation, a non-profit organization
Open access journal of the Ferrata-Storti Foundation, a non-profit organization
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Open access journal of the Ferrata-Storti Foundation, a non-profit organization
Vol. 98 No. 11 (2013): November, 2013
Platelet interaction with von Willebrand factor is enhanced…
Articles
Platelet interaction with von Willebrand factor is enhanced by shear-induced clustering of glycoprotein Ibα
Eelo Gitz
Charlotte D. Koopman
Alèkos Giannas
Cornelis A Koekman
Dave J. van den Heuvel
Hans Deckmyn
Jan-Willem N. Akkerman
Hans C. Gerritsen
Rolf T. Urbanus
Thrombosis and Haemostasis Laboratory, Department of Clinical Chemistry and Haematology, University Medical Center Utrecht
Thrombosis and Haemostasis Laboratory, Department of Clinical Chemistry and Haematology, University Medical Center Utrecht
Thrombosis and Haemostasis Laboratory, Department of Clinical Chemistry and Haematology, University Medical Center Utrecht
Thrombosis and Haemostasis Laboratory, Department of Clinical Chemistry and Haematology, University Medical Center Utrecht
Department of Molecular Biophysics, Utrecht University, The Netherlands
Laboratory for Thrombosis Research, KU Leuven, Kortrijk, Belgium
Thrombosis and Haemostasis Laboratory, Department of Clinical Chemistry and Haematology, University Medical Center Utrecht
Department of Molecular Biophysics, Utrecht University, The Netherlands
Thrombosis and Haemostasis Laboratory, Department of Clinical Chemistry and Haematology, University Medical Center Utrecht
Vol. 98 No. 11 (2013): November, 2013
ARTICLEFIGURES AND DATAINFO AND METRICS
Abstract
Initial platelet arrest at the exposed arterial vessel wall is mediated through glycoprotein Ibα binding to the A1 domain of von Willebrand factor. This interaction occurs at sites of elevated shear force, and strengthens upon increasing hydrodynamic drag. The increased interaction requires shear-dependent exposure of the von Willebrand factor A1 domain, but the contribution of glycoprotein Ibα remains ill defined. We have previously found that glycoprotein Ibα forms clusters upon platelet cooling and hypothesized that such a property enhances the interaction with von Willebrand factor under physiological conditions. We analyzed the distribution of glycoprotein Ibα with Förster resonance energy transfer using time-gated fluorescence lifetime imaging microscopy. Perfusion at a shear rate of 1,600 s−1 induced glycoprotein Ibα clusters on platelets adhered to von Willebrand factor, while clustering did not require von Willebrand factor contact at 10,000 s−1. Shear-induced clustering was reversible, not accompanied by granule release or αIIbβ3 activation and improved glycoprotein Ibα-dependent platelet interaction with von Willebrand factor. Clustering required glycoprotein Ibα translocation to lipid rafts and critically depended on arachidonic acid-mediated binding of 14-3-3ζ to its cytoplasmic tail. This newly identified mechanism emphasizes the ability of platelets to respond to mechanical force and provides new insights into how changes in hemodynamics influence arterial thrombus formation.
Introduction
Platelet adhesion to subendothelial matrices in the damaged vessel wall is the prime event in the arrest of bleeding. Recruitment of platelets to sites of vascular injury is hampered by the rapid flow of blood in arteries and arterioles. In these vessels, the interaction between von Willebrand factor (VWF), a multimeric plasma glycoprotein, and the platelet glycoprotein (GP) Ib-IX-V receptor complex is critical for initial platelet adhesion.1 This interaction requires unfolding of the VWF A1 domain and allows platelets to decelerate until they attach firmly in a process assisted by platelet integrins.1,2 Defects in both the GPIb-IX-V complex (Bernard Soulier syndrome) and VWF (von Willebrand disease) result in a bleeding diathesis, which underscores the importance of this interaction in hemostasis.3,4
The GPIb-IX-V complex consists of four transmembrane subunits; GPIbα, GPIbβ, GPIX and GPV that are expressed in a 2:4:2:1 stoichiometry.5 Each platelet contains approximately 25,000 copies of GPIbα, the subunit that binds to the VWF A1 domain.6 The extracellular domain (residues 1–485) of GPIbα consists of an N-terminal flank, seven leucine-rich repeats, a C-terminal flank, a sulphated region and a highly glycosylated macroglycopeptide domain. Residues 486–514 form the transmembrane domain and the cytoplasmic tail consists of 96 amino acid residues (residues 515–610),7–9 which contain binding sites for multiple intracellular proteins, including filamin A10 and the adaptor protein 14-3-3ζ.11 The region that interacts with the VWF A1 domain resides within the concave face of the leucine-rich repeat domain of GPIbα.12,13 Despite the fundamental importance in initiating platelet adhesion, the molecular mechanism regulating the VWF-GPIbα interaction remains incompletely understood.
Binding of VWF to GPIbα requires the dynamic conditions of flowing blood. The unique biomechanical properties of VWF and GPIbα allow the interaction to strengthen upon increasing hemodynamic drag.14 An explanation for this counterintuitive finding is that VWF needs to change its conformation to allow GPIbα access to the A1 domain. Elevated shear force and immobilization on a surface trigger this conformational change in vivo, a process mimicked by the antibiotic ristocetin in vitro.15 The interaction between VWF and GPIbα is also regulated through changes in GPIbα. This molecule’s adhesive properties depend on translocation to cholesterol-rich membrane domains known as lipid rafts,16,17 which may increase the local density of GPIbα receptors and stimulate their signaling properties. Indeed, studies with Chinese hamster ovary cells in which GPIbα was artificially dimerized have suggested that receptor clustering increases the overall strength of the VWF-GPIbα interaction.18,19
During efforts to optimize the storage conditions of platelet concentrates used for transfusion, we recently demonstrated that GPIbα clusters in lipid rafts when platelets are kept at low temperature.20 Analysis of Förster resonance energy transfer (FRET) by fluorescence lifetime imaging microscopy (FLIM) revealed that cooling of platelets triggers [GPIbα-GPIbα] associations in lipid rafts within a range of 1–10 nm. In the present study, we assessed whether clustering of GPIbα occurs under physiological conditions, investigated its influence on VWF interaction and identified the responsible molecular mechanism.
Methods
The patient
Citrated blood (10.9 mM f.c.) was obtained from a patient with von Willebrand disease type 3. Permission was obtained from the local medical ethics committee. The patient had no detectable plasma VWF (<0.1%), 1% plasma factor VIII, <1% factor VIII activity, no ristocetin-induced platelet aggregation, and a normal platelet count and volume.21
Materials, antibodies, platelet preparation and incubations
A detailed description of the materials, antibodies, platelet preparation and incubations used in this study can be found in the Online Supplementary Methods.
Platelet adhesion and rolling under flow conditions
A parallel plate perfusion chamber22 was used to investigate platelet adhesion and rolling. Further details are available in the Online Supplementary Methods.
Exposure to shear force
Platelets were exposed to shear force by perfusion through a microcapillary (inner diameter 760 μm, blocked with 4% bovine serum albumin). Washed platelets were resuspended in HT buffer (2.5×10 cells/L, pH 7.3) supplemented with 4% human albumin. Platelet suspensions were prewarmed to 37°C for 5 min and perfused through the microcapillary at indicated shear rates for 5 sec. The length of the microcapillaries was matched with the shear rate, which means that the platelet suspensions had similar shear exposure times at different shear rates. Indicated shear rates are the maximal shear rates to which platelets were exposed near the wall of the microcapillary. The wall shear rate (γ w) inside a microcapillary is described as
Where Q is the volumetric flow rate and r is the inner radius of the microcapillary.
Agglutination
Platelet agglutination was measured in a Chrono-log Lumi-Aggregometer (model 700, Chrono-log Corporation, Haverton, PA, USA) with Aggrolink 8.0 software. Washed platelets in HT buffer were pre-incubated with the prostacyclin (PGI 2) analog iloprost and dRGDW (5 min, 37°C) and stimulated with VWF (10 μg/mL) and ristocetin (0.3 mg/mL) while stirring (900 rpm). Data are expressed as percentage of maximal agglutination, with light transmission through HT buffer set at 100%.
Flow cytometric analysis, immunoprecipitations and western blots
A detailed description of the flow cytometric analysis, immunoprecipitations and western blots can be found in the Online Supplementary Methods.
Analysis of GPIbα distribution by Förster resonance energy transfer and fluorescence lifetime imaging microscopy
GPIbα distribution was analyzed by FRET/FLIM as described elsewhere.20 In brief, 6B4-Fab fragments conjugated to either Alexa Fluor-488 or Alexa Fluor-594 (6B4-488 and 6B4-594, respectively) were incubated with fixed platelet samples under conditions in which each Fab labeled ~50% of the total number of receptors. GPIbα translocation to lipid rafts was determined by labeling GPIbα with 6B4-488 and monosialo-tetrahexosylganglioside (GM1) with Cholera toxin subunit B conjugated to Alexa Fluor-594 (CTB-594; 5 μg/mL). The fluorescence lifetimes of the donor fluorophore (6B4-488) were determined in the absence and presence of acceptor fluorophore (6B4-594 or CTB-594) and used to calculate the FRET efficiency, defined as
where τ is the donor fluorophore’s lifetime in nanoseconds in the absence (τD) and presence (τD/A) of the acceptor fluorophore. To determine variation in FRET efficiency, the lifetimes of three randomly chosen quadrants were quantified.
Statistical analysis
Data are means ± SEM. Statistical analysis was based on GraphPad Prism 5 (San Diego, CA, USA). Differences between control platelets and incubations were analyzed by the Mann-Whitney test. P-values less than 0.05 ( or ) and between incubations (|−−|)) were considered statistically significant.
Results
Platelet adhesion to von Willebrand factor under conditions of flow triggers GPIbα clustering
Platelet adhesion at shear rates above 1,000 s depends on the interaction between surface-bound VWF and GPIbα.1 We analyzed the effect of this interaction on the spatial distribution of GPIbα on the platelet plasma membrane with FRET/FLIM. Whereas GPIbα molecules were dispersed in resting platelets (Figure 1A,B), indicated by a FRET efficiency of 0.9±0.2%, GPIbα clustered upon adhesion to VWF (FRET efficiency 10.3±0.9% ). Clustering was not caused by close contact between adjacent platelets, as FRET efficiency did not differ between single platelets and platelets that adhered as small aggregates (Online Supplementary Figure S1A,B). As the platelet-VWF interaction is influenced by flow conditions, we analyzed the GPIbα distribution of platelets adhered to VWF at different shear rates. Adhesion to VWF at 300 s left GPIbα dispersed, but perfusion at 750 s and higher induced clustering (Figure 1C). The observed increase in clustering was not the result of more efficient adhesion, as the number of platelets binding to VWF was similar at each shear rate (Online Supplementary Figure S1C). To investigate whether changes in GPIbα distribution were specific for adhesion to VWF, platelets were perfused over collagen. Adhesion to collagen at a low shear rate (300 s) in the presence or absence of VWF resulted in FRET efficiencies similar to those observed in resting platelets (Figure 1D). Perfusion over collagen at 1,600 s in the absence of VWF had little effect on GPIbα distribution. In contrast, addition of VWF prior to perfusion at 1,600 s increased FRET efficiency to 8.3±0.6%, indicating that clustering of GPIbα requires the presence of VWF.
Exposure to high shear leads to reversible von Willebrand factor-independent GPIbα clustering
The change in GPIbα distribution measured on surface-attached platelets might be the result of shear, of rolling/attachment or both. To understand the contribution of shear, we perfused platelets in VWF-free buffer through a microcapillary tube at different shear rates in the absence of an adhesive surface. FRET/FLIM analysis showed that a shear rate of 300 s left GPIbα dispersed. Exposure to 1,600 s had a minor effect on GPIbα distribution, whereas a shear rate of 10,000 s increased FRET efficiency to 9.1±0.6% (Figure 2A). Addition of exogenous VWF prior to perfusion at a shear rate of 10,000 s did not further increase clustering. Platelet α-granules also contain VWF,23 which might be released during platelet isolation and thereby influence GPIbα clustering. To investigate this possibility, experiments were repeated using a nanobody against VWF which prevents its binding to GPIbα. Exposure of platelets to shear force in the presence of this nanobody led to the same increase in FRET efficiency as control platelets (Figure 2B). GPIbα clustering induced by a shear rate of 10,000 s was also similar in platelets from a patient with VWD type 3 (Online Supplementary Figure S2A), indicating that GPIbα clusters independently of the presence of VWF.
GPIbα clustering was reversible, as exposure to shear followed by incubations under static conditions resulted in a gradual decline to the range found in resting platelets (Figure 2C). Platelet exposure to a shear rate of 10,000 s did not result in P-selectin expression, αIIbβ3 activation, VWF binding (Figure 2D), cytoskeleton reorganization or altered whole protein tyrosine phosphorylation (Online Supplementary Figure S2B,C). Conversely, clustering was not induced by stimulation with cross-linked collagen-related peptide (CRP), thrombin receptor activating peptide (TRAP) or the thromboxane A 2 receptor (TPα) agonist U46619 under static conditions (Figure 2E). Ristocetin-induced VWF binding did induce clustering of GPIbα. Transient GPIbα clustering did not affect the ability of platelets to respond to agonists, because stimulation with TRAP or CRP before and after exposure to shear resulted in similar levels of P-selectin expression and αIIbβ3 activation (Figure 2F).
Figure 1.Platelet adhesion to VWF induces clustering of GPIbα. (A) Freshly isolated resting platelets (control; left panels) or platelets adhered to VWF after whole blood perfusion (1 min, 37°C) at a shear rate of 1,600 s−1 (right panels) were analyzed for GPIbα distribution by FRET/FLIM. Platelets were fixed with 2% paraformaldehyde and stained with 1 μg/mL 6B4-488 (donor) in the absence (top panels) and presence of 1 μg/mL 6B4-594 (acceptor; bottom panels). The fluorescence lifetimes in nanoseconds (ns) are shown in false color images. (B) Quantification of fluorescence lifetime values of donor probe in the absence and presence of acceptor probe of platelets treated under the conditions of (A). Corresponding FRET efficiencies are calculated as described in the Methods. (C–D) FRET/FLIM analysis of platelets adhered to VWF after whole blood perfusion (C) or to collagen after reconstituted blood perfusion (D) at indicated shear rates (1 min, 37°C). Reconstituted blood over collagen was performed in the absence (open bars) and presence (gray bars) of 10 μg/mL VWF (n=4).
Platelet interaction with von Willebrand factor is stimulated by clustered GPIbα
To clarify whether changes in GPIbα distribution contributed to platelet responsiveness to VWF, agglutination was measured in platelets with shear-induced clustered GPIbα. VWF and a suboptimal concentration of ristocetin were used and aggregation was prevented by pre-incubation with iloprost, a stable analog of prostacyclin, and dRGDW. Neither agent affected shear-induced GPIbα clustering (data not shown). Maximal agglutination of platelets with pre-clustered GPIbα was four-fold higher than with controls (Figure 3A,B).
VWF enables platelets to roll over the damaged vessel wall until they attach firmly in an integrin-dependent manner. The effect of GPIbα clustering was measured by platelet perfusion over a VWF-coated surface in the presence of dRGDW to block αIIbβ3-mediated attachment. Induction of GPIbα clustering prior to perfusion reduced the rolling velocity by 40% (Figure 3C,D). These data show that GPIbα clustering facilitates the platelet-VWF interaction.
GPIbα translocates to lipid rafts and forms clusters through 14-3-3ζ binding
Platelet binding to VWF depends on reallocation of GPIbα in membrane domains enriched in sphingomyelin and cholesterol, known as lipid rafts.16,17 To understand the role of raft allocation in GPIbα clustering, GPIbα was labeled with 6B4-488 (donor) and the raft marker GM1 with CTB conjugated to Alexa Fluor-594 (CTB-594; acceptor). The surface of resting platelets showed little co-localization but adhesion to VWF or exposure to high shear (10,000 s) induced GPIbα-GPIbα as well as GPIbα-GM1 associations, suggesting that clustering and raft translocation go hand in hand (Figure 4A). Disruption of lipid rafts by cholesterol depletion with methyl-β-cyclodextrin (mβCD) effectively abrogated GPIbα clustering. The time-dependent decay of shear-induced GPIbα clusters closely followed raft translocation, again suggesting a tight interrelationship (Figure 4B).
Binding of the adaptor protein 14-3-3ζ to the cytoplasmic tail of GPIbα is essential for platelet interaction with VWF.24,25 To assess the role of 14-3-3ζ in GPIbα clustering, we pre-incubated platelets with MPαC. This membrane permeable peptide represents the critical 14-3-3ζ binding site on GPIbα, which includes the constitutively phosphorylated Ser-609 residue on its cytoplasmic tail.24,26,27 As expected, the peptide prevented VWF-induced 14-3-3ζ binding to GPIbα (Online Supplementary Figure S2). Figure 4C shows that pre-incubation with MPαC had little effect on lipid raft translocation induced by platelet adhesion to VWF, but completely abrogated clustering of GPIbα. Control peptide MαC, which lacks phosphorylation at Ser-609, had no effect on GPIbα redistribution. FRET/FLIM analysis of platelets exposed to a shear rate of 10,000 s procuded similar results (Figure 4D), demonstrating that 14-3-3ζ binding to the cytoplasmic tail of GPIbα is essential for its clustering. To elucidate the importance of 14-3-3ζ-induced clustering, we determined the rolling velocity of platelets perfused over VWF in the presence of MPαC (Figure 4E). While control peptide MαC had no effect, the rolling velocity of platelets pre-incubated with MPαC increased more than two-fold. Moreover, inhibition of 14-3-3ζ-induced GPIbα clustering impaired stable adhesion to VWF during whole blood perfusion (Figure 4F).
Figure 2.High shear force induces reversible GPIbα clustering in the absence of VWF. (A) Platelets resuspended in HT buffer (pH 7.3) supplemented with 4% albumin were exposed to indicated shear rates in the absence or presence of VWF (10 μg/mL) by perfusion (37°C) through a microcapillary tube for 5 s and analyzed for GPIbα distribution by FRET/FLIM. (B) Shear-induced GPIbα clustering occurs in the absence of VWF. Platelets in the absence or presence of a nanobody against VWF (10 μg/mL) that prevents its interaction with GPIbα were exposed to indicated shear rates and analyzed by FRET/FLIM. (C) Shear-induced GPIbα clustering is reversible. Platelets exposed to 10,000 s−1 were monitored for clustering at indicated time intervals post-shear, which after 10 min decreased to control levels. (D) Platelets were analyzed by FACS for expression of P-selectin, αIIbβ3 activation and VWF binding after exposure to shear. (E) FRET/FLIM analysis of platelets stimulated by CRP (1 μg/mL), TRAP-6 (20 μM), U46619 (10 μM) or ristocetin (0.3 mg/mL) and VWF (10 μg/mL) for 5 min at 37°C in the absence of shear. (F) The effects of reversible GPIbα clustering on platelet responsiveness were analyzed by stimulating platelets with TRAP or CRP 0 and 30 min post-shear. FACS analysis revealed that agonist-induced surface expression of P-selectin or αIIbβ3 activation was similar to that of control platelets. Data are presented as the ratio of mean fluorescence intensity (MFI) of treated platelets over control platelets (n=4).
Arachidonic acid mediates 14-3-3ζ-induced GPIbα clustering
Platelet interaction with VWF or incubation at low temperature activates the stress kinase P38-mitogen-activated protein kinase (P38MAPK), which liberates arachidonic acid (AA) from membrane phospholipids through cytosolic phospholipase A 2.20,28,29 Incubation with inhibitors at 37°C indicated that P38MAPK-mediated AA release might support GPIbα-GPIbα interactions during exposure to shear. The P38MAPK inhibitor SB203580 and the cytosolic phospholipase A 2 inhibitor AACOCF 3 inhibited the rise in FRET efficiency induced by high shear. The low FRET efficiency observed under static conditions increased 12-fold upon addition of AA. The intracellular accumulation of free AA might therefore contribute to GPIbα clustering (Figure 5A). The FRET efficiency of shear-treated platelets decreased upon subsequent incubations under static conditions (Figure 2C).
In platelets, liberated AA is metabolized by cyclo-oxygenase-1 and lipo-oxygenase to thromboxane A 2 and other eicosanoids, which could account for the reversibility of shear-induced GPIbα clustering. Indeed, accumulation of AA by inhibition of these enzymes with indomethacin and 5,8,11-eicosatriynoic acid prevented GPIbα clusters from dispersing after exposure to high shear (Figure 5B). AA release triggered by platelet stimulation with CRP or TRAP in the presence of indomethacin and 5,8,11-eicosatriynoic acid also induced clustering of GPIbα (Figure 5C). Control studies confirmed that SB203580 blocked shear-induced P38MAPK phosphorylation/activation whereas the other treatments left the enzyme undisturbed (Figure 5D). Treatments that inhibited AA release prevented shear-induced 14-3-3ζ-GPIbα association and blockade of AA degradation and the separate addition of AA preserved the complex (Figure 5E). These data indicate that liberated AA binds 14-3-3ζ and facilitates its translocation to the cytoplasmic tail of GPIbα.20,28
The inhibitors of AA release and degradation affected platelet rolling velocity over a VWF surface. Inhibition of AA release induced by shear increased the rolling velocity, whereas treatments that preserved accumulation of AA resulted in reduced velocities (Figure 5F). These data indicate that the AA-mediated transfer of 14-3-3ζ to the GPIbα cytoplasmic tail induces GPIbα clustering which supports platelet interaction with VWF at high shear.
Discussion
Our results demonstrate that the exposure of platelets to high shear leads to clustering of GPIbα and enhances the interaction with VWF. Shear-induced clustering is reversible and not associated with granule release or activation of αIIbβ3. Clustering requires lipid raft translocation and critically depends on AA-mediated 14-3-3ζ binding to the cytoplasmic tail of GPIbα (Figure 6).
Previous studies found a role for receptor clustering in the GPIbα-VWF interaction.18,19 Experiments with Chinese hamster ovary cells showed that intracellular dimerization of a modified GPIbα construct increases the overall bond strength with VWF. We found that GPIbα formed clusters in platelets adhered to VWF after perfusion at a shear rate of 750 s or higher. At these shear rates, platelet adhesion strongly depends on the interaction between GPIbα and VWF, because only this interaction is sufficiently fast and strong to withstand the associated hemodynamic drag.1 Our results indicate that GPIbα clustering contributes to GPIbα-mediated platelet adhesion to VWF, as inhibition of clustering strongly attenuated both platelet rolling velocity and adhesion.
Figure 3.GPIbα clustering improves platelet interaction with VWF. Resting platelets (control) or platelets exposed to 10,000 s−1 were pre-incubated with dRGDW and iloprost for 5 min at 37°C. Platelet agglutination and rolling velocity experiments were performed immediately after shear exposure. (A–B) Platelets were stimulated with suboptimal concentrations of ristocetin (0.3 mg/mL) and VWF (10 μg/mL) and agglutination (A) was measured at 37°C while stirring. (B) Quantification of mean maximal agglutination (n=3). (C–D) Analysis of platelet rolling velocity over VWF at a shear rate of 1,600 s−1. (C) Single platelets (black circles) were tracked over distance in time from which the rolling velocity was determined. Black arrows indicate the location of a single platelet at 0, 15 and 30 s. (D) Quantification of rolling velocity in μm/s. Rolling velocity was significantly reduced after pre-exposure to a shear rate of 10,000 s−1 (n=5).
Initial adhesion to VWF increases the hemodynamic drag on platelets substantially and results in the formation of membrane tethers that are pulled from the cell surface.30,31 Upon adhesion to VWF, clustering was observed at a shear rate of 750 s or higher. In the absence of an adhesive surface, GPIbα clustering required exposure to 10,000 s, a shear rate found in stenotic arteries. By definition, only those platelets nearest to the vessel wall are subjected to this shear rate. Shear exposure is, therefore, probably limited to part of the platelet population perfused through a microcapillary. Nevertheless, this short exposure to shear stress induced clustering of GPIbα, illustrating the high sensitivity of platelets to mechanical stress. The combined force of shear exposure and tensile stress exerted on GPIbα when bound to VWF apparently cooperate in facilitating GPIbα clustering. Shear-induced clustering did not coincide with platelet activation, which is in line with earlier reports of VWF-dependent adhesion32 or aggregation of discoid platelets33 at high shear rates.
The effects of clustering on the interaction with VWF probably reflect avidity modulation, where an increased local density of GPIbα molecules increases the number of ligand-receptor bonds. Based on crystal structure studies, it is less likely that GPIbα clustering facilitates binding of two GPIbα molecules to a single A1 domain.12,34 Under the influence of elevated hemodynamic drag, VWF-bound GPIbα can subsequently undergo a conformational change that further strengthens the interaction.14
Figure 4.Clustering of GPIbα requires translocation to lipid rafts and 14-3-3ζ binding to its cytoplasmic tail. GPIbα translocation to lipid rafts was measured by labeling GPIbα with 6B4-488 (donor, 1 μg/mL) and raft-specific GM1 ganglioside with CTB-594 (acceptor; 5 μg/mL). (A–D) FRET/FLIM analysis of [GPIbα-GM1] and [GPIbα-GPIbα] associations. (A) GPIbα distribution of platelets adhered to VWF by whole blood perfusion at 1,600 s−1 (VWF) or washed platelets perfused through a microcapillary at 10,000 s−1. Disruption of lipid rafts by cholesterol depletion with mβCD (10 mM, 30 min at 37°C) prior to perfusion prevented changes in GPIbα distribution. (B) GPIbα translocation to lipid rafts induced by shear force is reversible after 10 min. (C–D) GPIbα clustering requires 14-3-3ζ binding to its cytoplasmic tail. Binding of 14-3-3ζ to GPIbα was prevented by pre-incubation with the cell-permeable peptide MPαC (100 μM, 5 min at 37°C). FRET/FLIM analysis of platelets bound to VWF by perfusion (C) or exposed to shear (D) in the presence of control peptide MαC or MPαC. (E–F) Platelet interaction with VWF is impaired by prevention of 14-3-3ζ binding to GPIbα. Analysis of platelet rolling velocity over VWF at 1,600 s−1 (E) and quantification of platelet surface coverage during whole blood perfusion (F) in the presence of MαC or MPαC. (F) Snapshots were taken after 5 min of perfusion (n=4).
Disruption of lipid rafts by cholesterol depletion strongly impairs platelet adhesion to VWF under conditions of flow, indicating that GPIbα localization to these regions is essential for its function.16 Lipid rafts are viewed as platforms that can physically concentrate receptors, adaptor proteins and effector enzymes, which lead to amplification of signaling events. We observed little GPIbα localization in rafts on the surface of resting platelets, which increased about three-fold upon ligation to immobilized VWF or exposure to high shear in solution. Disruption of rafts prevented GPIbα from clustering. Although translocation was essential for this process, clustering depended critically on the interaction between 14-3-3ζ and GPIbα. The importance of the [14-3-3ζ-GPIbα] association for the interaction of platelets with VWF is well established,24,25 but the exact mode of action remains poorly defined. It has been suggested that this association participates in αIIbβ3 integrin activation in GPIb-IX-expressing Chinese hamster ovary cells.35,36 We show that inhibition of 14-3-3ζ binding to GPIbα impaired adhesion to VWF and increased rolling velocity. The presence of iloprost and dRGDW excluded involvement of αIIbβ3 integrin in platelet rolling on VWF. Together, these data indicate that the 14-3-3ζ association with GPIbα directly improves platelet interaction with VWF by allowing receptor clustering. The dimeric nature of 14-3-3ζ supports this finding.37 Indeed, a similar mechanism has been described in muscle cells, in which clustering of the acetylcholine receptor depends on 14-3-3ζ.38
Figure 5.Binding of 14-3-3ζ to GPIbα is regulated by AA. (A) Clustering of GPIbα is prevented by inhibitors of upstream regulators of intracellular AA release. Pre-incubation (10 min at 37°C) with P38MAPK inhibitor SB203580 (SB; 10 μM) or cytosolic phospholipase A 2 inhibitor AACOCF 3 (ACF; 20 μM) abolished shear-induced clustering of GPIbα. Platelet incubation with 10 μM AA (5 min at 37°C) in the presence of TPα antagonist SQ30741 (25 μM) induced GPIbα clustering. (B) Shear-induced clustering of GPIbα is maintained in the presence of inhibitors of AA metabolism. Inhibitors were against cyclooxygenase-1 (indomethacin; 30 μM) and lipo-oxygenase (5, 8, 11-eicosatriynoic acid, ETI; 25 μM), abbreviated as I/E. (C) Platelet stimulation with CRP or TRAP in the presence of indomethacin and ETI induces clustering of GPIbα. (DE) Analysis of shear-induced phosphorylation of P38MAPK (D) and [14-3-3ζ-GPIbα] complex formation (E) under conditions described for (A) and (B). (F) Platelet rolling velocity increases in the presence of SB203580 and AACOCF3 and is reduced in the presence of AA metabolism inhibitors (I/E) or exogenous AA, under conditions described for (A) and (B) (n=4).
P38MAPK is a kinase that is responsive to stress stimuli, including alterations in thermal28 and shear conditions.39 We found that platelet exposure to shear in solution leads to P38MAPK phosphorylation. Phosphorylated P38MAPK subsequently activates cytosolic phospholipase A2 to release AA from membrane phospholipids.40 Our study reveals a central role for AA in GPIbα clustering. Inhibition of AA release during exposure to shear prevented the transfer of 14-3-3ζ to the cytoplasmic tail of GPIbα. Moreover, addition of AA enhanced clustering and inhibition of AA metabolism resulted in irreversible clustering. The findings that AA binding to 14-3-3ζ induces 14-3-3ζ multimerization41 and that AA-bound 14-3-3ζ directly associates with GPIbα,28 both support the concept that the adaptor protein provides a platform for GPIbα clustering. In addition, lipid rafts are enriched in AA,42 which may explain the dependence of GPIbα cluster formation on these membrane domains. Rolling experiments established the importance of AA-mediated GPIbα clustering, as inhibitors of AA release reduced platelet interaction with VWF, while its accumulation enhanced this initial step in adhesion.
Figure 6.Schematic representation of initial platelet adhesion to VWF at the damaged arterial vessel wall. Shear triggers GPIbα translocation to lipid rafts and activates P38MAPK. This stress kinase subsequently activates cytosolic phospholipase A 2 (cPLA 2), which liberates AA from membrane phospholipids. GPIbα clusters upon AA-mediated 14-3-3ζ binding to its cytoplasmic tail, leading to enhanced interaction with VWF.
Aspirin, a widely used antithombotic drug, also interferes with AA conversion by inhibiting COX-1 activity. Our studies suggest that the use of aspirin may prolong the presence of GPIbα clusters, which contradicts the antithrombotic effects of this drug. However, the inhibitory effects of aspirin are attributed to inhibition of thromboxane A 2-enhanced platelet activation,43 which is important for more advanced steps in thrombus formation. Interestingly, several studies have demonstrated that GPIbα-dependent platelet adhesion actually increases upon aspirin intake.44,45 These unexplained findings may be the result of aspirin-enhanced GPIbα clustering.
In conclusion, we have defined a central role for GPIbα clustering in platelet interaction with VWF under conditions of flow. Clustering of GPIbα requires translocation to lipid rafts and AA-mediated 14-3-3ζ binding to its cytoplasmic tail. These findings illustrate the mechanosensitive properties of platelets and give a new perspective on the molecular mechanism of arterial thrombus formation.
Acknowledgments
This study was supported by a grant from the Landsteiner Foundation of Blood Transfusion Research (LSBR grant n. 0807). Prof. Dr. JWN Akkerman is supported by the Netherlands Thrombosis Foundation. Dr. RT Urbanus is a research fellow of the Dutch Heart Foundation (grant n. 2010T068).
Footnotes
The online version of this article has a Supplementary Appendix.
Authorship and Disclosures Information on authorship, contributions, and financial & other disclosures was provided by the authors and is available with the online version of this article at www.haematologica.org.
Received March 1, 2013.
Accepted June 4, 2013.
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Data Supplements
2013.087221.GITZ_SUPPL.pdf
2013_087221-Contributions.pdf
2013_087221-Disclosures.pdf
Figures & Tables
Figure 1.Platelet adhesion to VWF induces clustering of GPIbα. (A) Freshly isolated resting platelets (control; left panels) or platelets adhered to VWF after whole blood perfusion (1 min, 37°C) at a shear rate of 1,600 s−1 (right panels) were analyzed for GPIbα distribution by FRET/FLIM. Platelets were fixed with 2% paraformaldehyde and stained with 1 μg/mL 6B4-488 (donor) in the absence (top panels) and presence of 1 μg/mL 6B4-594 (acceptor; bottom panels). The fluorescence lifetimes in nanoseconds (ns) are shown in false color images. (B) Quantification of fluorescence lifetime values of donor probe in the absence and presence of acceptor probe of platelets treated under the conditions of (A). Corresponding FRET efficiencies are calculated as described in the Methods. (C–D) FRET/FLIM analysis of platelets adhered to VWF after whole blood perfusion (C) or to collagen after reconstituted blood perfusion (D) at indicated shear rates (1 min, 37°C). Reconstituted blood over collagen was performed in the absence (open bars) and presence (gray bars) of 10 μg/mL VWF (n=4).
Figure 2.High shear force induces reversible GPIbα clustering in the absence of VWF. (A) Platelets resuspended in HT buffer (pH 7.3) supplemented with 4% albumin were exposed to indicated shear rates in the absence or presence of VWF (10 μg/mL) by perfusion (37°C) through a microcapillary tube for 5 s and analyzed for GPIbα distribution by FRET/FLIM. (B) Shear-induced GPIbα clustering occurs in the absence of VWF. Platelets in the absence or presence of a nanobody against VWF (10 μg/mL) that prevents its interaction with GPIbα were exposed to indicated shear rates and analyzed by FRET/FLIM. (C) Shear-induced GPIbα clustering is reversible. Platelets exposed to 10,000 s−1 were monitored for clustering at indicated time intervals post-shear, which after 10 min decreased to control levels. (D) Platelets were analyzed by FACS for expression of P-selectin, αIIbβ3 activation and VWF binding after exposure to shear. (E) FRET/FLIM analysis of platelets stimulated by CRP (1 μg/mL), TRAP-6 (20 μM), U46619 (10 μM) or ristocetin (0.3 mg/mL) and VWF (10 μg/mL) for 5 min at 37°C in the absence of shear. (F) The effects of reversible GPIbα clustering on platelet responsiveness were analyzed by stimulating platelets with TRAP or CRP 0 and 30 min post-shear. FACS analysis revealed that agonist-induced surface expression of P-selectin or αIIbβ3 activation was similar to that of control platelets. Data are presented as the ratio of mean fluorescence intensity (MFI) of treated platelets over control platelets (n=4).
Figure 3.GPIbα clustering improves platelet interaction with VWF. Resting platelets (control) or platelets exposed to 10,000 s−1 were pre-incubated with dRGDW and iloprost for 5 min at 37°C. Platelet agglutination and rolling velocity experiments were performed immediately after shear exposure. (A–B) Platelets were stimulated with suboptimal concentrations of ristocetin (0.3 mg/mL) and VWF (10 μg/mL) and agglutination (A) was measured at 37°C while stirring. (B) Quantification of mean maximal agglutination (n=3). (C–D) Analysis of platelet rolling velocity over VWF at a shear rate of 1,600 s−1. (C) Single platelets (black circles) were tracked over distance in time from which the rolling velocity was determined. Black arrows indicate the location of a single platelet at 0, 15 and 30 s. (D) Quantification of rolling velocity in μm/s. Rolling velocity was significantly reduced after pre-exposure to a shear rate of 10,000 s−1 (n=5).
Figure 4.Clustering of GPIbα requires translocation to lipid rafts and 14-3-3ζ binding to its cytoplasmic tail. GPIbα translocation to lipid rafts was measured by labeling GPIbα with 6B4-488 (donor, 1 μg/mL) and raft-specific GM1 ganglioside with CTB-594 (acceptor; 5 μg/mL). (A–D) FRET/FLIM analysis of [GPIbα-GM1] and [GPIbα-GPIbα] associations. (A) GPIbα distribution of platelets adhered to VWF by whole blood perfusion at 1,600 s−1 (VWF) or washed platelets perfused through a microcapillary at 10,000 s−1. Disruption of lipid rafts by cholesterol depletion with mβCD (10 mM, 30 min at 37°C) prior to perfusion prevented changes in GPIbα distribution. (B) GPIbα translocation to lipid rafts induced by shear force is reversible after 10 min. (C–D) GPIbα clustering requires 14-3-3ζ binding to its cytoplasmic tail. Binding of 14-3-3ζ to GPIbα was prevented by pre-incubation with the cell-permeable peptide MPαC (100 μM, 5 min at 37°C). FRET/FLIM analysis of platelets bound to VWF by perfusion (C) or exposed to shear (D) in the presence of control peptide MαC or MPαC. (E–F) Platelet interaction with VWF is impaired by prevention of 14-3-3ζ binding to GPIbα. Analysis of platelet rolling velocity over VWF at 1,600 s−1 (E) and quantification of platelet surface coverage during whole blood perfusion (F) in the presence of MαC or MPαC. (F) Snapshots were taken after 5 min of perfusion (n=4).
Figure 5.Binding of 14-3-3ζ to GPIbα is regulated by AA. (A) Clustering of GPIbα is prevented by inhibitors of upstream regulators of intracellular AA release. Pre-incubation (10 min at 37°C) with P38MAPK inhibitor SB203580 (SB; 10 μM) or cytosolic phospholipase A 2 inhibitor AACOCF 3 (ACF; 20 μM) abolished shear-induced clustering of GPIbα. Platelet incubation with 10 μM AA (5 min at 37°C) in the presence of TPα antagonist SQ30741 (25 μM) induced GPIbα clustering. (B) Shear-induced clustering of GPIbα is maintained in the presence of inhibitors of AA metabolism. Inhibitors were against cyclooxygenase-1 (indomethacin; 30 μM) and lipo-oxygenase (5, 8, 11-eicosatriynoic acid, ETI; 25 μM), abbreviated as I/E. (C) Platelet stimulation with CRP or TRAP in the presence of indomethacin and ETI induces clustering of GPIbα. (DE) Analysis of shear-induced phosphorylation of P38MAPK (D) and [14-3-3ζ-GPIbα] complex formation (E) under conditions described for (A) and (B). (F) Platelet rolling velocity increases in the presence of SB203580 and AACOCF3 and is reduced in the presence of AA metabolism inhibitors (I/E) or exogenous AA, under conditions described for (A) and (B) (n=4).
Figure 6.Schematic representation of initial platelet adhesion to VWF at the damaged arterial vessel wall. Shear triggers GPIbα translocation to lipid rafts and activates P38MAPK. This stress kinase subsequently activates cytosolic phospholipase A 2 (cPLA 2), which liberates AA from membrane phospholipids. GPIbα clusters upon AA-mediated 14-3-3ζ binding to its cytoplasmic tail, leading to enhanced interaction with VWF.
Article Information
Vol. 98 No. 11 (2013): November, 2013 : Articles
DOI
Pubmed
23753027
Pubmed Central
PMC3815184
Published
2013-11-01
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Ferrata Storti Foundation, Pavia, Italy
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0390-6078
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1592-8721
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Eelo Gitz, Charlotte D. Koopman, Alèkos Giannas, Cornelis A Koekman, Dave J. van den Heuvel, Hans Deckmyn, Jan-Willem N. Akkerman, Hans C. Gerritsen, Rolf T. Urbanus. Platelet interaction with von Willebrand factor is enhanced by shear-induced clustering of glycoprotein Ibα. Haematologica 2013;98(11):1810-1818;
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Finding the Measures of an Interior Angle and an Exterior Angle of a Regular Polygon
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00:04 Finding the measures…
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Nicole Koenig, Kathryn Boddie
Instructors Nicole Koenig
Nicole has taught for over 3 years in the U.S., Kazakhstan, and Czechia with K-12 and college students in both English and mathematics (e.g., from elementary math to Calculus II). She has a degree in mathematical sciences from the University of Illinois and a certification for teaching ESL learners.
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Kathryn Boddie
Kathryn has taught high school or university mathematics for over 10 years. She has a Ph.D. in Applied Mathematics from the University of Wisconsin-Milwaukee, an M.S. in Mathematics from Florida State University, and a B.S. in Mathematics from the University of Wisconsin-Madison.
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Example SolutionsPractice Questions
Finding the Measures of an Interior Angle and an Exterior Angle of a Regular Polygon:
Step 1: Note the information given (e.g., an interior angle, an exterior angle, the number of sides of the polygon, etc.). If needed, draw a graph or picture of the regular polygon.
Step 2: Substitute the information into an appropriate formula and then simplify and solve, or use your understanding of regular polygons to find the measure of the desired angle.
Finding the Measures of an Interior Angle and an Exterior Angle of a Regular Polygon: Equations and Vocabulary
Polygon: A polygon is a closed figure with at least three straight sides (e.g., triangle, quadrilateral, pentagon, etc. )
Regular Polygon: A regular polygon is a polygon where all of the sides have equal lengths and all of the angle's measures are equal. The term equilateral refers to a polygon that has all of its side lengths equal, and the term equiangular refers to a polygon that has all of its angle's measures equal. Thus, a regular polygon is both equilateral and equiangular.
Interior angle of a polygon: An interior angle of a polygon is an angle inside the border of the polygon.
The table shows the sums of the interior angles for regular polygons with different numbers of sides.
| Number of Sides | Calculation | Sum of Interior Angle measures |
---
| 3 | 1 x 180 | 180 degrees |
| 4 | 2 x 180 | 360 degrees |
| 5 | 3 x 180 | 540 degrees |
| 6 | 4 x 180 | 720 degrees |
| n | (n - 2) x 180 | sum of interior angles of an n-sided figure |
We see that the sum of the interior angles of a regular polygon with n sides is:
Sum of the Interior Angles of an n-sided Regular Polygon=(n−2)×180
To find the measure of one interior angle of a regular polygon with n sides, we divide the sum of the interior angles by the number of sides:
Measure of One Interior Angle=(n−2)⋅180∘n
Exterior Angle of a Polygon: An exterior angle of a polygon is an angle between the extended side of a polygon and the adjacent side (the next side) of the polygon.
Note: the sum of the exterior angles of any polygon is always 360 degrees.
Therefore, to find the measure of one exterior angle of a regular n-sided polygon, we divide 360 by n. That is:
Measure of One Exterior Angle=360∘n
The image below shows the location of an interior angle (green) and an exterior angle (red) of a regular 6-sided polygon.
From the image, we can see that adjacent interior and exterior angles form a straight line. Therefore, they will always sum to 180 degrees. This gives way to another formula relating interior and exterior angles of polygons. That is, if angle A is an interior angle of a regular polygon, and angle B is the exterior angle adjacent to angle A, then:
m∠A+m∠B=180∘
Now that we have a general understanding of regular polygons and their interior and exterior angles, let's look at some examples. The first two examples focus on finding an interior angle of a regular polygon in two different ways. In the third example, we will find both an exterior and interior angle measure when given a regular polygon.
Finding the Measures of an Interior Angle and an Exterior Angle of a Regular Polygon: Interior Angle Example
Find the measure of one interior angle of a regular polygon with 5 sides:
Step 1:Note the given information.
We have a 5-sided regular polygon. We need to find the measure of one interior angle.
Step 2:Substitute the given information into the formula for the measure of one interior angle of an n-sided regular polygon, and simplify:
Measure of one interior angle=(n−2)⋅180∘n=(5−2)⋅180∘5=108∘
We get that the measure of one interior angle of a regular 5-sided polygon is 108°.
Finding the Measures of an Interior Angle and an Exterior Angle of a Regular Polygon: Interior Angle Example
An exterior angle of a regular polygon measures 40 degrees. Find the measure of an interior angle of the polygon:
Step 1: Draw a picture or graph if necessary. And note the given information.
Note that the exterior and the interior angle of the regular polygon form a straight line, so when they are added together, the sum will be 180 degrees.
Step 2:Substitute the information into the formula and simplify.
The exterior angle is given as 40 degrees. If we let the interior angle be x degrees, then x ° + 40° will equal 180°. Thus, we have the following equation:
x∘+40∘=180∘
We can solve this for x by subtracting 40° from both sides of the equation.
x∘=180∘−40∘
Lastly, we simplify.
x∘=140∘
Thus, one interior angle of the regular polygon is 140°.
Finding the Measures of an Interior Angle and an Exterior Angle of a Regular Polygon:
Find the exterior and interior angle measurements of the regular polygon shown:
Step 1: Note the given information.
If we count the sides of the polygon, we see that it has 6 sides. Thus, this is a regular hexagon with n = 6 sides.
Step 2:First, we will find the sum of the angles of a regular polygon with 6 sides. That is, we will substitute n = 6 into the formula for finding the sum of the interior angles of a regular polygon with n sides, and then simplify.
Sum of interior angles=(n−2)×180∘=(6−2)×180∘=4×180∘=720∘
Thus, the sum of the interior angles of the given regular hexagon is 720°.
Now, we can find the measure of each interior angle of the hexagon. Since this is a regular polygon, all of the interior angles are the same. Thus, we can divide the sum of interior angles , 720 degrees, by the number of sides/angles of the hexagon, 6.
720∘6=120∘
So, each interior angle of the given regular hexagon is 120°.
We already know that the sum of the exterior angles of the given regular hexagon is 360°. To find each exterior angle of the regular hexagon, we can do one of two things. The first method we can use is to note that the interior and exterior angles of the polygon are supplementary, and use this fact to set up an equation. If we let the measure of one exterior angle of the regular hexagon be x, then we have:
Thus, we have the following equation:
x∘+120∘=180∘
Solving this for x, we have:
x∘+120∘=180∘x∘=180∘−120∘x∘=60∘
The other way to find one exterior angle of the regular hexagon is to divide the sum of the exterior angles, 360°, by the number of sides/angles of the hexagon, 6.
360∘6=60∘
In both cases, we see that each exterior angle of a regular hexagon is 60°.
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Table of Contents
Finding the Measures of an Interior Angle and an Exterior Angle of a Regular Polygon
Equations and Vocabulary
Interior Angle Example
Interior Angle Example
Finding the Measures of an Interior Angle and an Exterior Angle of a Regular Polygon
Test your current knowledge
Practice
Finding the Measures of an Interior Angle and an Exterior Angle of a Regular Polygon
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8594 | https://www.youtube.com/watch?v=pjVMZN-5kwc | Graph Theory Lecture #10 | Max Number of Edges in Triangle-Free Graphs | Turán’s Theorem
HAMEEDA MATHTUBER
41700 subscribers
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Posted: 13 Jul 2025
In this lecture, we explore an important result in graph theory:
✅ The maximum number of edges in a graph with p vertices and no triangles is \left\lfloor \frac{p^2}{4} \right\rfloor.
This theorem is a fundamental part of extremal graph theory, known as Turán’s Theorem for triangle-free graphs.
We’ll discuss:
• What the theorem means
• Why the bound is \frac{p^2}{4}
• Intuition and proof sketch
• Examples to illustrate the concept
Perfect for students of discrete mathematics, computer science, and anyone preparing for competitive exams or university courses!
Don’t forget to like, share, and subscribe for more lectures on graph theory!
GraphTheory #TuransTheorem #DiscreteMathematics #Lecture10 #MathLecture
Transcript:
Dear students, under the lecture series of graph theory, here we have lecture 10. This is a very significant theorem in the examination point of view. That is the theorem states that the maximum number of lines or edges among all P point graphs with no triangles is P² divided by 4. So the maximum number of lines among P point graph. So the number of vertices in the graph is P. Then the number of possible edges will be equal to P square the integral part of P square by 4 and with particularly no triangles that is in a graph if there are no triangles then the maximum number of lines possible for a point graph is integral part of P square by 4. Before proving this theorem, let us try to understand what is this integral part of P² by 4 and for example if you have a number that is if you have 2.36 then the integral part of this number is equal to 2 that is it will be the greatest integer not exceeding this number. What will be the integral part of 4.6? Obviously it will be equal to 4. So it is the greatest integer not exceeding 4.6 I mean not exceeding four. So it will be equal to four because this is the integer. So similarly when we have p points in a graph then the maximum number of edges possible edges will be equal to integral part of p² by 4. This theorem can be easily verified for values lesser than or equal to four. That is for all those values of P lesser than or equal to 4 easily we can verify but as the vertices keep on increasing the verification becomes uh difficult in which case we will be proving this theorem generally. So first we will see what happens when P is equal to 1. So if you consider a graph where the value of P is equal to 1. So if you see what is P? P is the number of points. So if there is only one point in the graph, how will the graph be? So it will be like this. So how many edges do this graph have? The number of edges is zero. Because if there is only one point, you cannot form any edge. So the number of edges is equal to zero. We are excluding the loop and the multiple edges. We are considering only graphs without loops and multiple edges. So there is no edge possible when p is equal to 1. So that is uh what is the integral part of this that is we will take p² p is 1. So 1 square is 1 divided by 4 which will be equal to some 0. So what is the integral part of this? This is equal to 0. So therefore the this is the value of Q that is the number of maximum number of lines with uh P point that is P is equal to 1 is 0 which is Q equal to0 which we saw uh by by taking an example of a graph. Now we will consider P is equal to two that is there are two points. So what are the possible number of edges? So if there are two points the maximum number of edges is one only. So Q is equal to 1. Let us verify from this that is we will consider the integral part of p² by 4. p² is 2² by 4 which is equal to integral part of 2² is 4 by 4 and that is equal to 1. So if you see q is equal to 1. We have arrived at it. So this result is true when p is equal to 2. Now if we take p is equal to 3 what will be q? So if for example if you you're taking three points. So now what are the maximum number of edges possible for this? You can have an edge like this and you can have an edge like this. Or if for example if the points are like this the edges can be like this also or you can also have so if it is like this you can also have the edges to be like this. So these are the possible number of edges. But if you see the maximum number of edges how many how how many edges you can have you can have the value of Q to be equal to 2. So we will check by the theorem also P² by 4 what is P3? So 3² is 9 by 4. And what is the integral part of this? This is equal to 2. So therefore uh if you see we have arrived at Q is equal to 2. So the theorem holds for P= 3 also. For example if you take P is equal to 4. And why why I considered these to be maximum number of edges. If I draw another edge like this, it'll become a triangle. But there should be no triangle. That is the condition that we have. So that's why we have drawn these many edges. Now for P equal to four, we will see what is Q. So you will have P equal to four points. So the maximum number of edges. So it can be like this. It can be like this and it can be like this. And if you see uh there is an edge here, here here three edges and it can be like this also with these points. But you cannot join like this because it'll form a triangle. If you join like this, it will form a triangle. So you cannot do that. So what are the maximum number of edges possible? 1 2 3 4. So the maximum number of a possible edges is 4. We will check with the theorem also. So p² by 4 p is 4. So 4² by 4 which is 16x 4 and that is equal to 4. Hence by the theorem also it is true. So we can easily prove the theorem for values lesser than or equal to four. For all values of p lesser than or equal to four the theorem holds. Now therefore we have to prove the theorem for all those values greater than four. So that we will write first. So therefore I have written that the theorem can be easily verified for P lesser than or equal to 4 which we have already done. Okay. So we are going to prove the result for P greater than four. So all those values which are going to be greater than four. Now what we are going to do this P we are going to segregate as odd vertices and even vertices. So odd P and then even P. So we are going to segregate it as odd and even and we are going to prove the result separately for odd and then separately for even. So let us write that. So I've written that we will prove the result separately for odd P and for even P. So let us proceed with the proof. So first the part one will be for odd P that is odd number of vertices. When P the value of P is odd. How will the odd numbers be? It will be like 2 n + 1 2 n + 3. These are the generalized form of odd numbers. Because if you give any value for n, it will be an odd number. For example, if you give n=0, it will be 1 which is an odd number. n= 1. So 2 + 1 3. So which is an odd number. n= 3. If you take 6 + 1 which will be an odd number because 3 2 s is 6 + 1 is 7 which is an odd number. Similarly 2 n + 3 is also an odd number. So these are the general forms of the odd number. So the idea of the proof is going to be by the method of induction. So what we are going to do? We are going to assume the result is true for odd number that is 2 n + 1 that is 2 n + 1 number of vertices and we are going to prove that the result is true for 2 n + 3. So till 2 n + 1 the result we are going to assume as true and then we are going to prove for 2 n + 3 which is known as by the method of induction. So we will be proving uh it is also true for 2 n + 3. So first let us assume that the result is true for 2 n + 1 number of vertices that is when p takes the value 2 n + 1 or lesser than that for p. So for P all p lesser than or equal to 2 n + 1 we going to the assume the result is true. So I've written that here. Let us assume that the result is true for all odd p lesser than or equal to 2 n + 1. So we are going to prove that the result is true for a particular value p which is equal to 2 n + 3. Now in order to prove that we are going to take a graph G and we are going to take take that it has P number of vertices and Q number of edges with the particular value P to be equal to 2 N + 3. So we want to prove the result for this value. So we're going to take it as 2 N + 3 and very importantly from the assumption of the theorem we have no triangles in the graph. So let us take consider all this. Now first if q is equal to zero that is if there is no number of edges then obviously q is already lesser than or equal to p ² by 4 which we already saw. So we are going to consider all those values of q to be greater than zero. So let us put it in words. Now after making this that is uh after uh pro uh taking q to be equal to zero we see that q is lesser than or equal to integral part of p² by 4. So therefore we are going to prove the result for all those values of Q greater than zero. So first what we are going to do is we are going to assume a pair of adjacent vertices U and V in this graph G. So in this graph G we are going to take two adjacent vertices. Adjacent vertices means what? If we take two vertices when are these two vertices set to be adjacent if they have an edge between them. For example if you take this as U and this as V. If there is an edge between them, these two vertices are set to be adjacent. So we are going to take two vertices which are going to be adjacent in G. So I have written that here. Now in this graph G we are going to form from this graph G we are going to form a subgraph. So subgraph that we are going to form newly we going to represent it as G dash. So the subgraph so the subgraph we are going to form newly now. And how are we going to format this G dash? We are going to remove the points U and V from G. And we are going to consider the remaining graph as G dash. So from G we are going to remove the points U and V. So the edge between them will also be removed and the remaining graph we are going to take as G dash. Now we know that the graph G has 2 N + 3 number of points. from those number of points if you're going to remove two points you have to subtract two from this value. So g dash will have so this has how many vertices now? So for g has 2 n + 3 vertices from that we are removing two vertices u and v. So minus2 points it will have and it will have no triangles also that also we should mention. So g dash finally how many vertices it will have 2 n + 3 - 2 that will be 2 n + 1. So that we will write. So G dash okay this new graph after removing the two points we get this. So G dash has 2 n + 1 because 2 n + 3 - 2 is 2 n + 1 points. So these many points it will have and it has no triangles. So that also we should mention. Now after removing these two points what is happening? We are getting 2 n + one points in g dash. But we have already assumed from our assumption. What do we assume? Assume let us assume that the result is true for all values of p lesser than or equal to 2n + 1. So we assume that the result is true for this which means it is satis it will satisfy the given result. So hence by induction hypothesis what we will have the result will hold according to induction hypothesis that is the number of edges of uh Q q of G dash will be either lesser or equal to so the maximum number of edges will be p² by 2 because the result will hold but what is p now 2 n + 1 we will substitute that so integral part of 2 n + 1 the whole square divided by it is actually four okay by mistake I wrote two it is four so divided by four so now this is this will be equal to that is it's going to be equal to for the see now a + b the whole square formula we can use and we can split up this so a² which will be 4 n² + 2 a b so 2 n in multiplied with 2 is 4 n so it will be 4 n + 1² square is 1 divided by 4. So the integral part of this now in this we can take this denominator 4 to each of the terms separately. So first term if you take 4 n² divided by 4 so 4 will get get cancelled so you will have n² plus the second term is 4n divided by 4 again 4 will get cancel so you will have n + 1 by 4. Now the what is the integer here this one. So we can remove this integer because we are taking only integral part. So after uh taking the integral part you get this to be equal to n² + n. But actually it is lesser than or equal to no. So that inequality we have to bring here. So lesser than or equal to n² + n. So therefore q of g dash is lesser than or equal to n² + and let us mark this as equation one. Okay. Now next. So after marking this as equation one, we are considering another one point that is since G has no triangles. We know that the graph G has no triangles and G dash is a subgraph of G. So no point of G dash can be adjacent to both U and V in G. So we know that U and V are adjacent to one another U and V. So now if there is another point in that graph. Now this point can be adjacent to any one of these two vertices only. It cannot be adjacent to both the vertices at the same time because now if for example if if these two vertices are adjacent if this is W adjacent this is possible only when there is no edge between U and W because if you form an edge it'll become a triangle but we know that there are no triangles in G. Therefore, if you take one point in G, it can be either adjacent to U or it can be adjacent to V. But it cannot be adjacent to both U and V. That only I have written. Since G has no triangles, no point of G dash can be adjacent to both U and V in G. So let us mark this as equation two because we are going to make use of this one and two. Now, now the lines of the graph G will be of three types. The first one is lines of G dash. So we know that from uh that is just now we saw that the lines of G dash is Q of G dash which is lesser than or equal to N² + N. So from 1 we can say that the lines of G dash that is Q of Gdash is lesser than or equal to N² + N. This is from one which we found just now. This one it can be one of this type. And the second type is lines between G dash and UV that is the lines between the subgraph Gdash and the removed vertices UV. So that is from two we have obtained that is so that is here. So from two since G has no triangles no point of G dash can be adjacent to both U and V in G. Which means that these are the lines which are lesser than or equal to 2 n + 1 which we assumed previously. So that only so lines of lines between gdash and uv is that which is lesser than or equal to 2 n + 1. This is from two and then the line uv. So how many there are the two vertices u and v has one edge between them. So this is one. So these are the three types of edges that we get. So now therefore how the edges of uh G will be so hence the edges of G will be Q because G has Q number of edges and that will be lesser than so that is lesser than or equal to the first one is n² + n. So n² + n. Okay, this one and then 2 n + 1. So we will add that. So + 2 n + 1 and then 1. So we will add this one. So now on simplifying this will be equal to n² n + 2 n is + 3 n and + 1 + 1 is 2. So therefore q is lesser than or equal to n ² + 3 n + 2. Now we want to bring it in the form of p² by 4. In this form we want to bring it. In order to do that we will multiply and divide this by 4. So Q will be lesser than or equal to if we multiply by 4 it will be 4 into N² + 3 N + 2 divided by 4 which is equal to 4 2 4 multiplied with n² is 4 n² 4 3 is 12 so + 12 n + 4 2 8 divided by 4. Now we will uh we want this in the form p square. We want to write it in this form. So in order to bring it in that form we can try to um do the bring it into a perfect square by adding and subract subtracting the square term. We know that how to make it a perfect square. So in order to make it a perfect square. So this this will be equal to this 4 n² can be written as 2 n the whole square + 2 a b. So a is 2 n. So 2 with 2 n and so the remaining will be from 12 n we have 4 n here 4 with 3 is 12. So multiplied with 3. Now this is like a squar + 2 a b. So b is what? 3. So b square we have to add. So 3 square if you are adding it you have to also subtract it. So minus 3 square and then plus this 8 the whole divided by 4 in order to make it a perfect square we are doing so and so that will be equal to 2 n the whole square + 2 and then uh 2 n 3 + 3 squar but what is this? This is 2 n + 3 the whole square and then - 9 because 3 square is 9 + 8 the whole divided by 4. Now combining these terms we can write it as 2 n + 3 the whole square and - 9 + 8. So it is -1 the whole divided by 4. Now we want uh we can divide the denominator. So this will be 2 n + 3 / 4 minus separating the denominator 1 by 4. But when we take the integral part of this this fraction will be negligible. So because we are going to take the integral part. So therefore q will be lesser than or equal to the integral part of 2 n + 3 the whole square. Okay the I missed the square here. The whole square divided by 4. So we have brought it in that form. So P ^2 by 4. But what is the P actually. So we have taken P here. P is equal to what? 2 N + 3. So this 2 N + 3 can be written as P. So therefore Q is from this we get Q is lesser than or equal to integral part of P² divided by 4 because 2 N + 3 is P. So we have proved the result. Also we will consider for a particular value of p that is uh so also for p = 2n + 3 we can rewrite this as see this 2n we can write it as uh n + 1 because uh n + 1 so what will be remaining n + 2 so I'm just splitting it so that I can get a complete bipartate graph that is I'm splitting this 2 n + 1 as n +1 number of vertices in one set and n +2 number of vertices in other set. So the what we will get we get a complete by bipartateed graph that is with k uh n +1 number of vertices in one set and n plus2 number of vertices in another set. So we know that in a bipartateed graph k y m y comma n the total number of edges will be y we have to multiply this yum and this n. So because it's a complete biparted graph so total number of edges is y n. So therefore multiplying these two we get the number of edges for this biraph and obviously a biparted graph will not have a triangle and then uh n + 1 multiplied with n +2 lines we have because we in for a bipartite graph mn is the number of lines. Therefore m here is n + 1 and n here is n +2. Multiplying these two we get the number of lines. So when we multiply n + 1 and into multiplied with n +2 what we will be getting? n² + 3 n + 2. So which is equal to what? P ² by 4 number of lines by this because just now we got it. We had n² + 3 n + 2 and we got it as uh q less than or equal to p ² by 4. Therefore in particular for one particular value of p = 2n + 3 we get it to be equal to p ² by 4. So we have proved the lesser than inequality and we have also proved the equality. So hence the maximum uh Q is attained. So I have written that hence the maximum Q is attained. Now remember we have completed one the proof for odd number of vertices when P is odd. In the same way we have to prove for when P is even. So I will show you that. So now part two is for even P. First in the same way we are going to assume that the result is true for all even p lesser than or equal to 2n because how the even numbers will be it will be 2 n and then 2 n + 2 the even numbers will be of this form. So because when you give value for uh n over here for example if you give n= 1 you get 2. n= 2 means 4 which is an even number. n= 3 it is six again an even number. Similarly here also if you give n= 2 4 + 2 is 6. So you get even number. So as you keep on change the changing the values of n you get even numbers. If you give n equal to 1 it is 2 + 2 which is four. So these are the generalized form of even numbers. So we are going to assume that the result is true for p lesser than or equal to 2n and we are going to prove that the result is true for pal 2n + 2. So by the method of induction we are going to prove this and obviously we consider that that there are no triangles. So the same how we proved for odd P in the same way we are going to pro prove for even P. So in the same way we are going to consider a graph G. Now here it is even no. So with P equal to 2 N + 2 number of vertices and no triangles. And as before we saw if Q is equal to zero then the result is true. Hence we are going to take all values of Q to be greater than zero. So now what we will do? We will consider U and V adjacent pair of vertices in G. So we will write that step. Then what we will do? We will consider a subgraph G dash by removing the vertices U and V from the graph G. So that step we will write. So as I told you we have we have taken U and V to be pair of adjacent vertices in G. Now we are forming a subgraph G dash after removing U and V from G. So we know that G has 2 N +2 number of vertices. If you remove two vertices from that so minus2 points you will have that is g dash will have 2 n + 2 -2 is what 2 and 2 will get cancelled and so remaining will be 2 n therefore the graph subgraph which we have we form will be g dash and it will have two end points and no triangles. Now by induction hypothesis we know that the result is true for two end points because of our assumption we assumed that the result is true for all those values of P lesser than or equal to two. So therefore by the induction hypothesis we can write that so by induction hypothesis we know that the number of edges of G dash will be lesser than or equal to P² by 4. But what we have obtained it to be 2 n points we are having the value of p for g dash is 2n. So when we substitute it will be 2 n the whole square. So which will be 4 n² divided by 4 and so 4 will get cancelled and uh the integral part of n² is square only. So therefore we obtain q of gdash to be uh to be lesser than or equal to n². Now let us mark this as equation three. Now already we have two and two. Now this has three. So therefore we will write the three types of lines of G that step we will write now. So as we saw in the previous odd case. Now in even case also the lines of G will be of three types. First is lines of G dash which is what here only we obtained Q of G dash to be lesser than or equal to N². So from three and lines between G dash and UV just previously we saw from two it will be lesser than or equal to here the assumed lines is 2 N. So it is 2N and the line UV is 1 because it is one edge. Now what we should do in order to find the lines of G we have to add all these three. So let us do that. So hence we have Q to be lesser than or equal to N² + 2 N + 1 because we have added all the three. First was n square then 2 n then 1 and this is what actually it is n + 1 the whole square so q is lesser than or equal to n + 1 the whole square now we want to bring it in the form p square by 4 so for that we should multiply and divide this by 4 so q will be lesser than or equal to 4 n + 1 the whole square divided by 4 so this four when taken inside the bracket it will become 2 multiplied with n + 1 the whole square. So it can be written like this okay and divided by 4 and so that will be equal to when you multiply this two inside it will be 2 n + 2 divided by 4 the whole I mean uh the integral part of this and this is square. Therefore we get q to be lesser than or equal to the integral part of 2 n + 2 the whole square divided by 4. But what is uh 2 n + 2? We have assumed it to be p. So that's equal to integral part of p ² by 4. So therefore we have obtained q to be lesser than or equal to the integral part of p² by 4. Hence the theorem holds for even p also. So hence the theorem holds for for even p also. Now for particular one value of p which is equal to 2 n +2 we should take and it can be partitioned as n + 1 n +1 vertices where uh it will be like ka n where the first set has n + 1 vertices and the second set also has n +1 vertices. So it's a complete by by graph bipartateed graph and it will have no triangle of course and what I told you how the vertices will be it is n into the number of edges will be n into so here n is n + 1 is n + 1. So when you multiply we get the number of lines but what is n + 1 multiplied with n + 1? It is n + 1 the whole square which is equal to n + 1 the whole square. And obviously we have already proved that it is equal to integral part of p² by 4. Just now we proved it here because when when it came it came as n + 1 the whole square we have proved that it is p² integral part of p² by 4. And so here also it holds. Hence this maximum q is attained. So hence we have proved this theorem. Hope you have understood this theorem and this result is a very significant result in the examination point of view. So kindly uh go through it very patiently surely you will understand and uh the proof of proof for odd p is uh same as for even p. So both are similar proof the only difference being there it is 2 n + 1 and 2 n + 3 since it is odd and here it is 2n and 2 n + 2 since it is even. So uh the it it will have likes of proof only. So kindly go through it patiently. Thank you. |
8595 | https://staticmy.zanichelli.it/catalogo/assets/9788808520272_04_CAP.pdf | 1 LA CHIMICA IN PRATICA: LA COSTANTE DI AVOGADRO Per onorare il lavoro del fisico e chimico italiano Amedeo Avogadro (1776-1856), all’inizio del secolo scorso la comunità scientifica internazionale decise di chiamare numero di Avogadro il numero di molecole contenute in una «grammo-molecola» di ossigeno (32 g): con questo termine si definiva, a quel tempo, una quantità in grammi pari alla massa molecolare, cioè quella che oggi chiamiamo mole.
Nel 1971, con l’introduzione della mole nel Sistema Internazionale delle unità di misura (abbreviato in SI), il nome è stato modificato in costante di Avogadro. Recentemente le definizioni di quelle grandezze sono state riviste basandole sui valori di costanti fisiche e non più sulle proprietà di particolari campioni fisici. Per quanto riguarda la mole, si rimanda appunto al valore della costante di Avogadro (6,02214076 ⋅ 1023 mol−1) determinato con misure di grande esattezza. Sono state utilizzate sfere di silicio purissimo (Figura 1.1) che, indagate per via cristallografica, hanno consentito di associare il numero di atomi di silicio in un dato volume alla corrispondente massa. Figura 1.1 Una sfera di silicio. [dpa picture alliance/Alamy Foto Stock] Moli, composizioni percentuali e formule 1 CAPITOLO [Martyn F. Chillmaid/SPL/AGF] 2 Capitolo 1 Moli, composizioni percentuali e formule 1 La mole Una grandezza che i chimici usano con molta frequenza è il numero delle par-ticelle che partecipano alle reazioni. Le combinazioni che producono le diverse sostanze sono fenomeni che coinvolgono individualmente le singole particelle. Per esempio, quando qualcosa di organico brucia, si produce diossido di carbo-nio (CO2) cioè un atomo di carbonio si lega a due atomi di ossigeno. Per «contare» queste particelle, la comunità scientifica ha adottato una spe-cifica unità di misura, la mole. Questa unità definisce la quantità di sostanza, espressione che indica il numero delle singole particelle di quella sostanza.
Nel Sistema Internazionale delle unità di misura (SI), la mole ( Video 1.1) è una delle unità fondamentali, cioè una di quelle che sono alla base dell’intero sistema e dalle quali si ricavano tutte le altre unità di misura.
La mole, il cui simbolo è mol, è l’unità di misura della quantità di sostanza. Una mole contiene esattamente 6,02214076 · 1023 entità elementari. Questo nume-ro è il valore numerico fisso della costante di Avogadro NA, quando è espressa in mol−1, ed è chiamato numero di Avogadro.
La quantità di sostanza, simbolo n, è una misura del numero di specificate en-tità elementari. Un’entità elementare può essere un atomo, una molecola, uno ione, un elettrone o ogni altra particella o specificato gruppo di particelle.
Per avere un’idea dell’entità della costante di Avogadro, proponiamo il proble-ma modello che segue.
Questa è la nuova definizione di mole nel Sistema Internazionale.
PROBLEMA MODELLO 1 Supponendo che lo spessore di un foglio di carta sia 0,05 mm, calcolare l’al-tezza di una pila di fogli pari al numero di Avogadro.
Risoluzione. Moltiplicando lo spessore di un foglio per NA si ottiene la risposta: 0,05 mm ⋅ 6,02214076 ⋅ 1023 = 3,01107038 ⋅ 1022 mm = 3,01107038 ⋅ 1016 km Cioè un’altezza pari a 78 miliardi di volte la distanza Terra-Luna!
2 L’unità di massa atomica Nell’attività pratica la misura delle quantità non può prescindere dall’uso di strumenti facilmente accessibili e di semplice utilizzo. Uno di questi è senz’altro la bilancia. Per questo motivo richiamiamo una definizione che riguarda l’uso della massa in campo atomico.
Si definisce unità di massa atomica unificata la massa di 1/12 della massa dell’isotopo 12 del carbonio, preso nel suo stato fondamentale nucleare ed elet-tronico. Il suo valore è mu = 1,6605402 · 10−27 kg.
Questa unità è indicata anche come dalton, in onore del chimico inglese John Dalton. Il simbolo che la rappresenta è u (oppure Da).
John Dalton (1766-1844) fu tra i primi a definire in modo completo l’ipotesi atomica.
Il dalton non è un’unità SI; è utilizzata spesso in biochimica per esprimere la massa di macromolecole.
GUARDA!
Video 1.1 Che cosa significa «mole» in chimica?
Tavola periodica interattiva 3 1.2 L’unità di massa atomica Per comprendere come questa unità di misura sia utile per collegare quantità (in moli) e massa (in grammi), mostriamo il seguente calcolo.
PROBLEMA MODELLO 2 Calcolare la massa di una mole di dalton.
Risoluzione. Si moltiplica la massa di un dalton per la costante di Avogadro: 1,6605402 · 10−27 kg · 6,02214076 · 1023 mol−1 = 1,00000068 · 10−3 kg · mol−1 Per numero di Avogadro si intende il solo valore numerico (Problema modello 1). La costante di Avogadro contiene anche la dimensione mol−1.
In definitiva, una mole di dalton corrisponde a un grammo. Quindi, se un elemento ha una massa atomica di x dalton, una mole dei suoi atomi corrisponde a una massa di x grammi. Per esempio, considerando il sodio (la cui massa atomica è 22,99 u), pesarne 22,99 g sulla bilancia equivale ad aver pesato una mole di atomi di sodio. La massa corrispondente a una mole dipende quindi dal valore della massa della particella specificata.
Le sostanze che osserviamo nel mondo macroscopico sono costituite da enti-tà elementari che, con poche eccezioni (per esempio, i gas nobili), sono formate dalla combinazione di più atomi. Le particelle così formate si indicano con il termine specie chimiche e si rappresentano con una formula.
Una specie chimica è una qualunque particella formata dai nuclei atomici di uno o più elementi, «legati» tra loro tramite elettroni. Pertanto sono specie chimiche le molecole monoatomiche e poliatomiche (come He, Ar, CO2, H2O, C6H12O6, HCl), gli ioni monoatomici e poliatomici (Na+, Br−, IO3 −, NH4 +) e i ra-dicali (H⋅, ⋅O⋅, ⋅OH, Cl⋅, ⋅CH3). È del tutto evidente che si può calcolare la massa di una specie chimica sem-plicemente sommando le masse dei singoli atomi che la costituiscono. Di norma ci si riferisce a essa come massa molecolare; tuttavia, poiché si può avere a che fare con specie non molecolari, come gli ioni, in questo testo useremo il termine più generale di massa formula, dato che a ogni specie corrisponde comunque una formula chimica.
La massa formula è la somma delle masse atomiche di tutti gli atomi che forma-no una specie chimica.
Ovviamente, se le masse atomiche sono espresse in dalton, anche la massa for-mula che si ottiene dalla loro somma è espressa in dalton.
Per la Tabella delle masse atomiche degli elementi si veda l’Appendice 2. PROBLEMA MODELLO 3 Determinare la massa formula in dalton di: O2 Ca(OH)2 Cr2O7 2− CuSO4⋅ 5 H2O Risoluzione. Queste formule si riferiscono a molecole, ioni o sostanze rap-presentate dalla loro formula minima. In base alla definizione di massa for-mula, indicando con MA la massa atomica, si calcola: 1. per O2: 2 ⋅ MAO = 2 ⋅ 16,00 u = 32,00 u 2. per Ca(OH)2: MACa+ 2 ⋅ MAO+ 2 ⋅ MAH = 40,08 u + 2 ⋅ 16,00 u + 2 ⋅ 1,008 u = 74,10 u 3. per lo ione Cr2O7 2−: 2 ⋅ MACr+ 7 ⋅ MAO = 2 ⋅ 52,00 u + 7 ⋅ 16,00 = 216,00 u 4. per CuSO4 ⋅ 5 H2O: MACu+ MAS+ 4 ⋅ MAO+ 10 ⋅ MAH+ 5 ⋅ MAO = 249,69 u 4 Capitolo 1 Moli, composizioni percentuali e formule QUESITO 1 Indicare quante sono le moli di atomi di ogni elemento contenute in 0,1 mol di CH3COOH (acido acetico).
QUESITO 2 Un composto è formato da tre elementi: C, H, O. In ogni molecola di tale com-posto ci sono 6 atomi di C. Sapendo che per ogni mole di atomi di C ci sono 2 mol di atomi di H e 1 mol di atomi di O, scrivere la formula del composto.
Numerose sostanze, come i metalli, la grafite, il quarzo e i composti ionici (i sa-li), sono in realtà aggregati di un numero indefinito di atomi o ioni. Tutte queste sostanze sono solide a temperatura ambiente, proprio perché sono costituite da un numero molto grande di particelle, tutte concatenate tra loro nel reticolo cristallino del solido. Non essendo possibile individuare in questi casi particelle elementari isolate l’una dall’altra, per rappresentare queste sostanze si usa la formula empirica o formula minima (di cui parleremo in dettaglio più avanti). Essa indica soltanto il rapporto di combinazione tra gli elementi che formano la sostanza.
Le formule di molti elementi, come per esempio Fe, Pb e C, sono formule empiriche; lo sono anche quelle di composti come SiO2 (quarzo) e NaCl (cloruro di sodio). Per tutte queste sostanze l’entità elementare a cui ci si riferisce quando si esprime la quantità in moli è proprio la formula empirica. Il termine massa formula resta comunque valido anche per queste sostanze.
3 La massa molare La massa atomica in dalton di una particella è tale che il suo valore in grammi corrisponde a una mole di quelle particelle. La tabella delle masse atomiche de-gli elementi è, in pratica, la tabella delle masse, espresse in grammi, di una mole di ciascun elemento. Passando alle formule, la stessa considerazione può essere fatta su massa formula e massa in grammi di una mole di formule.
Tutto questo ci porta alla definizione di un’importante grandezza, la massa molare, che è la massa di una mole di particelle (Figura 1.2).
La massa molare (MM) di una qualunque specie chimica è una massa in gram-mi pari alla sua massa in dalton. L’unità di misura della massa molare è g ⋅ mol−1.
Figura 1.2 Da sinistra a destra, una mole di saccarosio, cloruro di nichel(II), solfato di rame(II), permanganato di potassio, tornitura di rame, limatura di ferro. [SPL/AGF] NA = 6,022 · 1023 molecole di C6H12O6 5 1.3 La massa molare In conclusione, diciamo che la mole fa da «interprete» tra il mondo microscopi-co e quello macroscopico. In effetti, essa ci permette di convertire misure diret-tamente accessibili nel nostro mondo «macro» (come la massa) in grandezze del mondo «micro» (il numero di particelle), impossibili da misurare direttamente. La massa molare è il fattore di conversione per passare da uno all’altro dei due mondi (Figura 1.3).
In generale, se si vuole conoscere la massa (in grammi, g) di una certa quan-tità (in moli, mol), si calcola la massa molare corrispondente alla formula e si procede come segue: massa (g) = quantità (mol) ⋅ massa molare (g ⋅ mol−1) Viceversa, per calcolare la quantità di sostanza a partire dalla massa: 00/3 00/3 00/3 00/3 00/3 00/3 00/3 pp/00 180,16 g NA = 6,022 · 1023 molecole di C6H12O6 1 mol di fruttosio (C6H12O6) C6 H12 O6 Figura 1.3 Diversi modi di esprimere la stessa quantità. 1 massa (g) massa molare (g ⋅ mol ) − quantitˆ (mol) = PROBLEMA MODELLO 4 Calcolare la massa, in grammi, di 2,50 mol di acido nitrico (HNO3).
Risoluzione. Anzitutto si deve calcolare la massa formula dell’acido nitrico: si ottiene 63,02 u. Pertanto la massa molare di HNO3 è 63,02 g ⋅ mol−1. Poi si calcola la massa corrispondente alla quantità data: quantità (mol) ⋅ massa molare (g · mol−1) = massa (g) cioè: 2,50 mol ⋅ 63,02 g ⋅ mol−1 = 158 g.
PROBLEMA MODELLO 5 Determinare quante moli di acido acetico (CH3COOH) corrispondono a 10,50 g di tale sostanza.
Risoluzione. Il fattore di conversione tra massa e quantità è ancora la mas-sa molare, che si ottiene dalla massa formula. La massa formula dell’acido acetico è 60,05 u e quindi la massa molare di CH3COOH è 60,05 g ⋅ mol−1. Eseguendo i calcoli si ottiene: Come nel Problema modello 4, si tratta di eseguire una conversione tra massa e quantità, ma in questo caso si sono invertite le parti.
1 10,50 g 0,1749 mol 60,05 g ⋅ mol−= quantità di CH3COOH = 6 Capitolo 1 Moli, composizioni percentuali e formule SCHEDA 1.1 ALCUNE AVVERTENZE IMPORTANTI SUL CONCETTO DI MOLE Rileggendo la definizione di mole ci si accorge che essa non è solo un numero, ma un numero di entità elementari tutte uguali. Questo significa che si possono misurare a moli solo le sostanze pure.
Se per esempio abbiamo 100 g di NaCl puro al 90%, per calcolare quante sono in effetti le moli di NaCl occorre prima calcolare i grammi di NaCl puro (il 90% di 100 g, cioè 90 g) e poi dividere il risultato per la massa molare (58,44 g/mol): Una seconda avvertenza riguarda la necessità di specificare sempre il tipo di entità elementari. La frase «consideriamo una mole di idrogeno» è imprecisa, perché si possono considerare sia atomi singoli di idrogeno (e allora una mole di essi corrisponde a 1,008 g), sia molecole (che sono biatomiche, per cui la massa di una mole è 2,016 g). Nei due casi il numero di particelle è lo stesso (pari a NA), ma è diverso il tipo di particelle, che dev’essere dunque indicato esplicitamente. Nell’esempio proposto si dovrà specificare «una mole di atomi di idrogeno» oppure «una mole di molecole di idrogeno».
−= ⋅ 1 90 g 1 54 mol 58 44 g mol , , 4 Composizione percentuale e formula chimica Una sostanza pura ha proprietà chimiche e fisiche (come temperatura di fusio-ne e di ebollizione, densità, forma dei cristalli allo stato solido e così via) fisse e caratteristiche (invarianti).
Tra le proprietà chimiche è fondamentale la composizione elementare. Quando una sostanza pura viene analizzata per determinare gli elementi di cui è costituita, si trova che questi sono sempre gli stessi e sono presenti sempre nei medesimi rapporti quantitativi, quale che sia la provenienza del particolare cam-pione di quella sostanza.
L’analisi chimica elementare ha lo scopo di determinare quali elementi for-mano una determinata sostanza e in quale proporzione sono combinati tra loro. I risultati dell’analisi sono generalmente espressi come percentuale di ciascun elemento sulla massa totale.
La composizione percentuale e la formula chimica di un composto sono stret-tamente legate l’una all’altra: entrambe fissano la composizione qualitativa e quantitativa di una sostanza, sia pure in termini diversi.
QUESITO 3 Calcolare quante molecole di acido acetico sono presenti in una massa di acido uguale a quella del Problema modello 5.
QUESITO 4 Sono dati 20,00 g di cromato di potassio (K2CrO4). Calcolare quante moli di atomi di K, Cr e O sono presenti in tale quantità. La massa molare del cro-mato di potassio è 194,20 g ⋅ mol−1.
Nella Scheda 1.1 riportiamo alcune precisazioni relative all’uso della mole.
7 1.5 Dalla percentuale alla formula empirica La composizione percentuale fa riferimento alle masse degli elementi combi-nati in un composto. La formula chimica indica il numero di atomi di ciascun elemento combinati nel composto.
I due tipi di informazione sono equivalenti e occorre saperli convertire l’uno nell’altro. 5 Dalla percentuale alla formula empirica Consideriamo un generico composto fatto da tre elementi: A, B, C. Supponiamo che sia nota la massa percentuale di ciascun elemento nel composto e indichia-mo i tre valori rispettivamente con a%, b% e c%. Come possiamo passare da questi valori percentuali ai tre pedici (cioè i numeri scritti nelle formule in basso, sotto al simbolo di ciascun elemento) della formula empirica del composto? In altri termini, indicando con AxByCz la formula empirica del composto, voglia-mo calcolare i valori di x, y e z partendo da a%, b% e c%.
La formula empirica indica i rapporti tra gli atomi degli elementi, espressi con i numeri interi più piccoli possibile (per questo è detta anche formula minima). Per esempio, H2O è la formula empirica dell’acqua, oltre a essere la sua formula molecolare. Invece la formula empirica del perossido di idrogeno (la comune acqua ossigenata), che ha formula molecolare H2O2, è HO, perché il rapporto più semplice tra gli atomi degli elementi che la compongono è 1:1.
Ottenere la formula empirica di un composto significa, in pratica, rispondere alla domanda: in quale rapporto gli atomi di A, B e C sono presenti nel composto in esame?
Sappiamo che per collegare massa e particelle si ricorre alle moli: il fattore di conversione è la massa molare (MM). Ritornando all’esempio dell’acqua ossige-nata, se in ogni molecola di questa sostanza vi sono due atomi di idrogeno e due di ossigeno, dovrà esserci questo stesso rapporto 1:1 tra il numero totale di atomi di idrogeno e di ossigeno contenuti in 100, 1000, …, NA molecole. Il rapporto di combinazione tra due elementi in una qualunque formula, quindi, è uguale al rapporto molare tra le quantità di tali elementi.
Tornando allora al problema iniziale, si tratta di trasformare le masse degli elementi A, B e C, definite dalle rispettive percentuali, nelle corrispondenti moli di atomi, e stabilire in quale rapporto numerico stanno tra loro quelle moli. Considerando 100 g di sostanza e indicando con MMA, MMB, MMC le masse molari di A, B e C, si calcolano dapprima le moli di ciascun elemento in 100 g: Si ricordi che con il simbolo NA si indica la costante di Avogadro.
A mol A in 100 g B mol B in 100 g C mol C in 100 g MMA (g ⋅ mol−1) (g) a% n (g) b% n (g) c% n = = = MMB (g ⋅ mol−1) MMC (g ⋅ mol−1) 8 Capitolo 1 Moli, composizioni percentuali e formule A questo punto, per arrivare alla formula empirica, basta calcolare il rapporto tra le moli di atomi. Quali sono i più piccoli numeri interi che sono in rapporto tra loro come i numeri (che in genere non sono interi) nA, nB, nC?
Il metodo per determinarli è dividerli tutti e tre per il più piccolo di essi. Supponiamo che dei tre numeri il più piccolo sia nC. Con la divisione si ottengo-no tre nuovi numeri: Se x, y e z sono interi (o comunque possono essere approssimati a numeri interi nell’ambito dell’incertezza delle misure sperimentali), il lavoro è finito: la for-mula empirica cercata è AxByC (z, riferito a C, vale 1 e perciò viene sottinteso nella formula).
Se dopo un calcolo di questo tipo non si ottengono numeri interi, si conside-rano multipli successivi fino a ottenere tutti numeri interi. Per esempio, analiz-zando l’anidride fosforica (un composto fatto di ossigeno e fosforo), si ottiene un rapporto tra P e O di 1:2,5. È facile osservare che moltiplicando questi numeri per 2 si mantiene lo stesso rapporto, ma espresso con numeri interi. La formula empirica dell’anidride fosforica è P2O5. Se, anche dopo avere moltiplicato per 2, il rapporto non risultasse espresso da numeri interi, si riprova moltiplicando per 3, fino a che si ottiene un gruppo di numeri interi.
A B C C C C n n n x y z = 1 n n n = = = PROBLEMA MODELLO 6 Un dato composto organico risulta costituito da carbonio, idrogeno, ossige-no. L’analisi elementare ha fornito come risultato le seguenti composizioni percentuali: C 60,00% H 4,48% O 35,52% Determinare la formula empirica del composto.
Risoluzione. Si devono calcolare le moli di carbonio (nC), idrogeno (nH) e ossigeno (nO) combinate tra loro in quel composto. Esse si ricavano dalla composizione percentuale come segue: Dividendo il numero di moli di ciascun elemento per il valore più piccolo (2,220), si ottiene: Questo risultato indica che nel composto, per ogni atomo di ossigeno, vi so-no 2 atomi di idrogeno e 2,25 atomi di carbonio. Per rendere intero quest’ultimo numero si deve moltiplicare per 4: 2,250 ⋅ 4 = 9,000 (per C) 2,00 ⋅ 4 = 8,00 (per H) 1,000 ⋅ 4 = 4,000 (per O) 60,00 g C 4,996 mol C in 100 g di composto 12,01 g ⋅ mol−1 4,48 g H 4,44 mol H in 100 g di composto 35,52 g O 2,220 mol O in 100 g di composto = = = 1,008 g ⋅ mol−1 16,00 g ⋅ mol−1 4,996 2,250 (per C) 2,00 (per H) 1,000 (per O) = = = 4,44 2,220 2,220 2,220 2,220 (segue) 9 1.6 Dalla formula empirica alla composizione percentuale La formula empirica del composto è dunque C9H8O4 (Figura 1.4).
Figura 1.4 Massa percentuale degli elementi di un composto e formula empirica.
9 mol di C 8 mol di H 4 mol di O 12,01 g · mol−1 1,008 g · mol−1 16,00 g · mol−1 carbonio 60,00% idrogeno 4,48% ossigeno 35,52% QUESITO 5 Ricavare la formula empirica dell’etanolo (o alcol etilico, il comune alcol delle bevande alcoliche), sapendo che ha la seguente composizione percen-tuale: 52,1% carbonio, 34,7% ossigeno e il resto idrogeno.
QUESITO 6 In un composto molto comune, 3,93 g di sodio sono combinati con 6,07 g di cloro. Qual è la formula empirica del composto?
6 Dalla formula empirica alla composizione percentuale Supponiamo ora di conoscere la formula di un composto, che indichiamo con AaBbCc. Vogliamo rispondere alla domanda: quanti grammi di A, B e C sono contenuti in 100 g di composto (in altre parole, qual è la sua composizione per-centuale)?
Cominciamo discutendo, tramite un problema svolto, un uso molto frequen-te della relazione tra massa, percentuale e formula chimica: quello dell’acqua di cristallizzazione dei sali idrati.
Si tratta di composti solidi nei quali è «imprigionata» acqua trattenuta dal catione o, a volte, dall’anione del sale. La quantità di quest’acqua, detta appunto «di cristallizzazione», rispetta precisi rapporti quantitativi ed entra a far parte della formula del composto, come vedremo anche nel testo di Laboratorio.
C. Rubino, I. Venzaghi, R. Cozzi, Le basi della chimica analitica − Laboratorio, Zanichelli editore, 2022.
PROBLEMA MODELLO 7 Il solfato di rame pentaidrato (un tempo chiamato «vetriolo azzurro») ha formula CuSO4 · 5 H2O. Un campione di questo sale pesa 40,00 g. Calcolare: a. le moli di acqua di cristallizzazione; b. la massa dell’acqua di cristallizzazione; c. la percentuale di acqua di cristallizzazione.
La massa molare del sale idrato è pari a 249,69 g · mol−1, quella dell’acqua è 18,02 g · mol−1.
(segue) 10 Capitolo 1 Moli, composizioni percentuali e formule Risoluzione.
a. La formula CuSO4 · 5 H2O indica che ci sono cinque moli di acqua per ogni mole di sale. Per calcolare le moli di acqua nei 40,00 g di campione si devono quindi calcolare le moli di sale e moltiplicarle per 5: Per rispondere alle domande (b) e (c) si utilizza la massa molare come fatto-re di conversione dalla massa alle moli.
b. Massa di H2O in 40,00 g di sale: 0,8010 mol H2O · 18,02 g · mol−1 = 14,43 g H2O c. Ricordando che la percentuale di un componente in un campione è: si ricava: Esiste un altro modo, ancora più semplice, per rispondere a questa terza domanda. Infatti, per determinare la percentuale di acqua di cristalliz-zazione è sufficiente conoscere la massa di acqua presente in una certa massa di campione. Ovviamente, la percentuale di un componente in un composto resta la stessa riferendosi a qualunque quantità di esso. In par-ticolare, invece delle quantità presenti nei 40,00 g del campione di sale, ci si può riferire alle masse contenute in 1 mol di composto. Una mole del sale pentaidrato pesa 249,69 g e la massa di acqua in essa contenuta è quella corrispondente a 5 mol, ovvero: 5 mol H2O · 18,02 g · mol−1 = 90,10 g H2O La percentuale di acqua è: Come prevedibile, il risultato ottenuto con questo metodo alternativo è identico al precedente.
40,00 g 0,1602 mol CuSO4 ⋅ 5 H2O 249,69 g ⋅ mol−1 = 5 mol H2O 0,1602 mol sale ⋅ 1 mol sale = 0,8010 mol H2O massa del componente ⋅ 100 massa del campione % = Si può arrivare a questa espressione anche a partire dalla seguente proporzione (m = massa): mcomponente : mcampione = = % : 100 14,43 g H2O % H2O = ⋅ 100 = 36,08% 40,00 g sale 90,10 g H2O ⋅100 = 36,08% 249,69 g sale QUESITO 7 È dato un campione del sale K2CO3 · 7 H2O. Calcolare la massa di sale che contiene 25,00 g di acqua. La massa molare del sale è 264,32 g · mol−1, quella dell’acqua è 18,02 g · mol−1.
[Un aiuto: confrontare questo esercizio con il Problema modello 7. Qui la massa del sale è l’incognita, in quel caso, invece, era un dato, e l’incognita del Problema modello 7, la massa di acqua, è qui un dato. Invertendo il cal-colo, si possono calcolare le moli di acqua nei 25,00 g dati e poi, conoscendo il rapporto molare tra acqua e sale, …] 11 1.6 Dalla formula empirica alla composizione percentuale Possiamo ora riprendere il problema generale enunciato all’inizio del paragrafo: qual è la percentuale degli elementi A, B e C nel composto di formula AaBbCc? Partiamo dalla formula del composto e dalle definizioni di massa molare e di percentuale. La percentuale dell’elemento A nel composto AaBbCc, per defini-zione, è espressa da: Basta quindi riferirsi a una mole del composto AaBbCc, nella quale la massa di A è pari al prodotto della massa molare di A per il suo pedice e quella del com-posto è la sua massa molare, MMAaBbCc, per ottenere i dati che ci servono per il calcolo della percentuale ( Approfondimento 1.1).
Per esempio, le percentuali degli elementi nell’idrossido di sodio (NaOH, MM = 40,00 g · mol−1) si ottengono calcolando: massa A a% = ⋅ 100 massa AaBbCc cioè a ⋅ MMA (g) massa A % cioè MMAaBbCc (g) massa AaBbCc a = ⋅ 100 22,99 g % Na = 40,00 g 16,00 g % O = 40,00 g 1,008 g % H = 40,00 g ⋅ 100 = 57,48% ⋅ 100 = 40,00% ⋅ 100 = 2,52% PROBLEMA MODELLO 8 Determinare la composizione elementare del dicromato di potassio (K2Cr2O7).
Risoluzione. In base ai dati (formula chimica e masse atomiche) si calcola la massa molare di K2Cr2O7 = 294,20 g · mol−1 ( Approfondimento 1.2).
1. Si determina dapprima la massa di ciascun elemento contenuta nella mas-sa molare del composto, tenendo conto degli indici nella formula. Dalla formula si ricava che in 1 mol di K2Cr2O7 sono contenute 2 mol di K, 2 mol di Cr e 7 mol di O. Quindi in 294,20 g di composto (massa mo-lare) sono presenti: 2 mol · 39,10 g · mol−1 = 78,20 g K 2 mol · 52,00 g · mol−1 = 104,0 g Cr 7 mol · 16,00 g · mol−1 = 112,0 g O 2. La percentuale di ogni elemento si ottiene dai seguenti calcoli: 78,20 g K ⋅ 100 = 26,58% (K) 294,20 g composto 104,0 g Cr ⋅ 100 = 35,35% (Cr) 294,20 g composto 112,0 g O ⋅ 100 = 38,07% (O) 294,20 g composto GUARDA!
Approfondimenti 1.1 Un metodo alternativo per ricavare la percentuale degli elementi A, B e C nel composto di formula AaBbCc 1.2 Un metodo alternativo per determinare la composizione elementare del dicromato di potassio (K2Cr2O7) 12 Capitolo 1 Moli, composizioni percentuali e formule 7 Formula empirica e formula molecolare Non sempre la formula empirica di una sostanza coincide con la formula mo-lecolare. Affinché le due formule coincidano occorre, in primo luogo, che la so-stanza sia effettivamente costituita da particelle molecolari e, in secondo luogo, che il numero di atomi presente in ogni molecola sia uguale a quello presente nella formula empirica. Per esempio, acqua (H2O), acido solforico (H2SO4) e metano (CH4) sono composti per i quali valgono queste condizioni.
Se invece consideriamo un composto come il glucosio, il rapporto tra gli ele-menti che lo compongono, che sono C, H e O, è 1:2:1, cioè la sua formula empi-rica è CH2O. La formula molecolare del glucosio, invece, è C6H12O6. La particella di formula molecolare CH2O è il metanale (o formaldeide), le cui proprietà sono ovviamente molto diverse da quelle del glucosio.
Altri esempi del genere sono gli idrocarburi: alcheni e cicloalcani hanno la stessa formula empirica, CH2, ma si tratta di moltissime sostanze diverse (ognu-na con una propria temperatura di ebollizione, di fusione, densità, reattività e così via). Per distinguerle tra loro è necessaria la formula molecolare (e a volte anche questa non basta, come nel caso degli isomeri).
Esistono alcune sostanze la cui composizione chimica può essere rappresen-tata solo dalla formula empirica. È questo il caso dei composti ionici. Per esem pio, la formula chimica che rappresenta il cloruro di sodio è NaCl e corri-spon de alla formula empirica; i sali non esistono come molecole, tranne che in fase gassosa, perciò non possono avere una formula molecolare che li rap-presenti.
Per ottenere la formula molecolare di un composto non basta dunque co-noscere (o poter ricavare tramite calcoli) i rapporti molari tra gli elementi, ma bisogna disporre di una informazione aggiuntiva che consenta di sapere anche quanti atomi formano la molecola del composto. Un’informazione di questo tipo è, per esempio, la massa molare (MM) del composto.
PROBLEMA MODELLO 9 Un solvente organico ha formula empirica CH2 e MM = 84,20 g ⋅ mol−1. Qual è la sua formula molecolare?
Risoluzione. Si calcola la massa della formula empirica: MAC+ 2 · MAH = 12,01 + 2 · 1,008 = 14,03 La formula molecolare deve essere un multiplo della formula empirica, per-ché ne deve rispettare le proporzioni tra i diversi elementi (in questo caso 2 di H ogni 1 di C).
Quindi è sufficiente calcolare quante volte la formula empirica sta in quella molecolare per ottenere la formula del composto: La formula del composto è C6H12.
84,20 6 14,03 = 13 1.8 La percentuale formale di un composto (o di una formula) in una sostanza QUESITO 8 Determinare la formula empirica e la formula molecolare di un composto avente massa molecolare 90,04 u e la seguente composizione percentuale: C 26,68% H 2,24% O 71,08% 8 La percentuale formale di un composto (o di una formula) in una sostanza In chimica, la percentuale non si usa solo per esprimere la composizione ele-mentare di una sostanza. Il suo utilizzo più frequente è senz’altro quello relativo alla composizione dei miscugli (Capitolo 3). Altre situazioni nelle quali si usa la percentuale si hanno, per esempio, quando si tratta di valutare la qualità di un minerale, che può essere espressa tramite la percentuale della sostanza utile rispetto al resto. Questa percentuale può essere riferita sia a un singolo elemen-to, come potrebbe essere il ferro nel minerale pirite, sia a un composto (come la percentuale di magnetite, Fe3O4, in un minerale ferroso).
Il significato di percentuale comunque non cambia: si tratta sempre della massa di un componente (sia esso elemento o composto) presente in 100 unità di massa totali.
Un’estensione di tutto ciò si applica anche alle formule chimiche dei com-posti. In un certo senso potremmo considerare questi come delle miscele in proporzioni fisse. In queste miscele particolari si possono individuare, oltre ai singoli elementi chimici, anche delle aggregazioni di elementi che possono esse-re rappresentate con formule.
Un caso del genere l’abbiamo già incontrato nel Problema modello 7: i sali idra-ti sono composti nella cui formula empirica si può evidenziare quella dell’acqua. Tanto per chiarire ciò che intendiamo, la formula empirica del solfato di rame pentaidrato, di cui si è parlato nel Problema modello 7, può essere scritta come CuH10O9S. È ovvio che questa stessa formula, «scorporata» in CuSO4 ⋅ 5 H2O, è più significativa perché comunica più informazioni. Come abbiamo visto, per composti di questo tipo si usa esprimere la percentuale di acqua di cristallizza-zione.
Un caso simile è quello dei sali misti, come la dolomite, CaMgC2O6 (carbona-to doppio di calcio e magnesio, più chiaramente indicato come CaCO3 ⋅ MgCO3), e il sale di Mohr, Fe(NH4)2(SO4)2 ⋅ 6 H2O, e di gran parte dei minerali silicei. La composizione di queste sostanze può essere espressa tramite la percentuale di ciascun componente nella combinazione totale.
Chiameremo questo tipo di percentuale, riferita a una formula piuttosto che a un elemento, percentuale formale di quella formula nella sostanza che la contiene. Per completare questa carrellata di situazioni nelle quali si ha a che fare con formule nelle formule, consideriamo il caso di quelle sostanze (in genere sali di ossoacidi) la cui formula può essere considerata come la combinazione di altre formule più piccole. Un esempio tipico è quello del metasilicato di sodio, Na2SiO3, la cui formula corrisponde a Na2O ⋅ SiO2. 14 Capitolo 1 Moli, composizioni percentuali e formule In alcune applicazioni, come nella formulazione dei detersivi ma molto più spesso nei concimi (Figura 1.5), la presenza di sostanze di questo tipo è spesso indicata attraverso la percentuale formale di uno dei componenti più semplici, come Na2O o P2O5, piuttosto che riferendosi all’intero composto cui quella for-mula appartiene.
Figura 1.5 Percentuali formali di alcuni componenti di un concime per rose.
[Michela Tedeschi, Bologna] PROBLEMA MODELLO 10 Si consideri il sale del Problema modello 8, che ha formula K2Cr2O7. Esprimere il suo contenuto di cromo come percentuale formale dell’ossido CrO3.
Risoluzione. Si calcolano dapprima le masse molari: la massa molare di K2Cr2O7 è 294,20 g · mol−1, mentre quella di CrO3 vale 100,00 g · mol−1. Finora, trattando di singoli elementi, il rapporto molare era rappresentato direttamente dagli indici. In questo caso, invece, bisogna confrontare le formule dei due composti, l’ossido CrO3 e il sale K2Cr2O7: poiché 1 mol di K2Cr2O7 contiene 2 mol di Cr e a ogni Cr corrisponde una formula CrO3, il rapporto cercato è basato sulla relazione: 1 mol K2Cr2O7⇔ 2 mol CrO3 Applicando ora lo stesso metodo utilizzato nel Problema modello 8: 1. si determina la massa di CrO3 contenuta nella massa molare del sale K2Cr2O7: 2 mol ⋅ 100,00 g · mol−1 = 200,00 g CrO3 2. la percentuale formale dell’ossido si ottiene dal calcolo: 3 2 2 7 200 00 g CrO 100 67 98% 294,20 g K Cr O , , ⋅ = 15 1.8 La percentuale formale di un composto (o di una formula) in una sostanza PROBLEMA MODELLO 11 Ricavare la formula del minerale stilbite, che ha la seguente composizione percentuale: 57,41% SiO2; 16,43% Al2O3; 8,93% CaO; 17,23% H2O.
Risoluzione. Dal momento che «formula» significa rapporti molari, è possi-bile calcolare le moli di ciascuno dei componenti indicati in 100 g di minerale: La formula presente con il minore numero di moli è CaO, quindi si calcola il rapporto molare tra le altre formule e quella di CaO: La formula completa della stilbite, scritta come combinazione degli ossidi, è Al2O3 · CaO · 6 SiO2 · 6 H2O. Invece la formula empirica è Al2CaH12Si6O22.
= 0,9554 mol SiO2 in 100 g di minerale = 0,1611 mol Al2O3 in 100 g di minerale = 0,159 mol CaO in 100 g di minerale = 0,9562 mol H2O in 100 g di minerale 1 57,41 g 60,09 g ⋅ mol− 1 16,43 g 101,96 g ⋅ mol − 1 8,93 g 56,08 g ⋅ mol− 1 17,23 g 18,02 g ⋅ mol− mol SiO2 = 6,01 mol CaO = 0,9554 0,159 = 6,01 mol H2O mol CaO = 0,9562 0,159 mol CaO mol Al2O3 = 1,01 0,1611 0,159 = PROBLEMA MODELLO 12 Il metasilicato di sodio, Na2SiO3, molto usato per la produzione di de tersivi, è disponibile sia anidro, al costo di 0,54 € · kg−1, sia come pentaidra to (Na2SiO3 · 5 H2O), al costo di 0,48 € · kg−1. Valutando esclusivamente l’a-spetto del costo, quale dei due è più conveniente ( Approfondimento 1.3)?
Risoluzione. Si trasforma la domanda nella seguente: quanto costa 1 kg di Na2SiO3 nei due casi? Per il sale anidro la risposta è fornita dagli stessi dati: 0,54 €. Si tratta dunque di calcolare l’analogo valore per il sale idrato. Per farlo, si determina, in base alle masse molari, la percentuale di sale ani-dro in quello idrato: massa molare Na2SiO3 = 122,07 g · mol−1 massa molare Na2SiO3 ⋅ 5 H2O = 212,15 g · mol−1 Il risultato indica che, in 1 kg di sale idrato, solo 575,4 g sono di prodotto puro, il resto è acqua (pagata anch’essa, come il sale, a 0,48 € · kg−1).
Il costo effettivo del sale, in questo caso, è: 122,07 g ⋅ 100 = 57,54% Na2SiO3 212,15 g 0,48 € = 0,83 € ⋅ kg−1 (contro 0,54 € ⋅ kg−1 di quello anidro) 0,5754 kg GUARDA!
Approfondimento 1.3 Dalla parte del «consumatore chimico» 16 Capitolo 1 Moli, composizioni percentuali e formule 9 La percentuale sul secco di un composto umido Alcuni prodotti commerciali, soprattutto di genere alimentare, contengono di norma una certa dose di acqua, che tuttavia non è interessante per definire la composizione del prodotto stesso, dal momento che si tratta di una sorta di «eccipiente». In questi casi si preferisce esprimere la composizione percentuale in relazione al secco, cioè riferita al prodotto privo di umidità. Vediamo un esempio con il problema svolto che segue.
PROBLEMA MODELLO 13 Confrontare le seguenti percentuali di lipidi e protidi in una salsiccia fresca (contenente il 51,3% di acqua) e in una essiccata (con solo il 28,2% di ac-qua), per stabilire se hanno la stessa composizione: salsiccia fresca protidi 14,3% lipidi 30,8% salsiccia essiccata protidi 22,0% lipidi 47,3% Assumendo che la variazione di composizione percentuale tra il prodotto fresco e quello secco sia dovuta solo alla perdita d’acqua, quando la compo-sizione di entrambi viene riferita al secco si ottengono grandezze omogenee e confrontabili.
Risoluzione. Ci si riferisce a 100 g delle due salsicce. Il secco della salsiccia fresca è: 100 g − 51,3 g = 48,7 g, mentre per l’altra è 71,8 g.
Nella prima il contenuto di protidi è di 14,3 g che, riferiti al secco, sono: Analogamente, i lipidi sono: Per la salsiccia stagionata lo stesso calcolo fornisce: Data l’inevitabile variabilità di composizione di questi alimenti, la conclu-sione è che le due salsicce provengono dallo stesso tipo di materia prima.
14,3 g ⋅ 100 = 29,4% protidi sul secco 48,7 g 30,8 g ⋅ 100 = 63,2% lipidi sul secco 48,7 g 22,0 g ⋅ 100 = 30,6% protidi sul secco 71,8 g 47,3 g ⋅ 100 = 65,9% lipidi sul secco 71,8 g QUESITO 9 In un campione di fieno di erba medica, la percentuale di umidità è pari a 11,2% p/p. La determinazione delle ceneri nello stesso campione ha dato co-me risultato 8,20 g di ceneri in 100 g di fieno.
Calcolare la percentuale di sostanza organica riferita al secco.
[Un aiuto: considerare la massa di sostanza organica = massa campione − massa (umidità + ceneri)] 17 1.10 Definizioni e relazioni utili nei calcoli 10 Definizioni e relazioni utili nei calcoli Mole, massa atomica, massa formula, massa molare • Mole La mole, il cui simbolo è mol, è l’unità di misura della quantità di sostanza. Una mole contiene esattamente 6,02214076 · 1023 entità elementari. Questo numero è il valore numerico fisso della costante di Avogadro NA, quando è espressa in mol−1, ed è chiamato numero di Avogadro.
La quantità di sostanza, simbolo n, è una misura del numero di specificate entità elementari. Un’entità elementare può essere un atomo, una molecola, uno ione, un elettrone o ogni altra particella o specificato gruppo di particelle.
• Costante di Avogadro (NA) È il numero delle entità elementari contenute in una mole di sostanza. Il suo valore è: NA = 6,02214076 ⋅ 1023 mol−1 • Unità di massa atomica unificata Indicata con il simbolo u o Da, il suo valore è 1,6605402 · 10−27 kg. Per defini-zione, corrisponde a 1/12 della massa dell’isotopo 12C.
• Massa formula È la somma delle masse atomiche di tutti gli atomi che compaiono nella for-mula di una sostanza.
• Rapporti molari di combinazione tra gli elementi in una data formula In una mole di un composto generico di formula AxByCz sono contenute x moli di A, y moli di B e z moli di C. In un numero n di moli di composto, le moli di A, B e C si calcolano moltiplicando x, y e z per n.
• Massa molare Esprime il valore della massa di una mole di sostanza. Normalmente è espres-sa in g · mol−1.
Per qualunque sostanza, i valori numerici della massa formula (in dalton) e della massa molare (in grammi) sono gli stessi.
Relazione tra massa, quantità di sostanza e massa molare Se si vuole conoscere la massa (in grammi) di una certa quantità (in moli), si procede come segue: massa (g) = quantità (mol) ⋅ massa molare (g ⋅ mol−1) Viceversa, per calcolare la quantità di sostanza a partire dalla massa: Formula empirica di una sostanza In una formula empirica sono indicati i rapporti molari tra gli atomi degli ele-menti presenti in un composto, espressi con i numeri interi più piccoli possibile (perciò è detta anche formula minima).
massa (g) quantità (mol) = massa molare (g ⋅ mol−1) 18 Capitolo 1 Moli, composizioni percentuali e formule Come si ricava la formula empirica dalla composizione percentuale Si consideri un composto di formula incognita AxByCz.
1. Si trasformano le masse percentuali degli elementi nelle corrispondenti moli di atomi.
2. Si calcolano i rapporti (x, y, z) tra quelle moli di atomi, dividendo per il nu-mero più piccolo tra quelli ottenuti.
3. Se x, y e z sono interi (o comunque numeri che possono essere approssimati a interi), allora la formula empirica cercata è AxByCz.
4. Se non si ottengono subito numeri interi, si considerano multipli successivi del risultato, fino a ottenere tutti numeri interi.
Come si ricava la composizione percentuale dalla formula empirica 1. Anzitutto, in base alla formula, si determina per ciascun elemento la massa contenuta nella massa molare del composto.
2. Si calcolano le percentuali di tali masse di ciascun elemento sulla massa mo-lare del composto.
Formula molecolare Per ottenere la formula molecolare di un composto, a partire dalla sua formula empirica, bisogna conoscerne la massa molare.
1. Si calcola la massa della formula empirica.
2. La formula molecolare deve essere un multiplo di quella empirica, perciò si calcola quante volte la formula empirica è contenuta in quella molecolare. 3. I pedici della formula empirica si moltiplicano per il risultato.
Percentuale formale di un composto in una sostanza Talvolta una data formula può essere considerata come un «miscuglio» (in pro-porzioni definite e costanti) di formule più semplici. La massa percentuale di queste «sottoformule» nella formula completa si chiama percentuale formale e si calcola in modo analogo alla percentuale dei singoli elementi.
Se si deve calcolare la percentuale formale di un composto A in un altro compo-sto B, si procede come segue.
1. Confrontando le formule dei due composti si ricava il loro rapporto molare in relazione all’elemento che hanno in comune.
2. Si determina la massa di A contenuta nella massa molare di B.
3. Si calcola infine la percentuale della massa di A rispetto alla massa molare di B.
ESERCIZI 19 ESERCIZI Mole, massa molare, rapporti molari di combinazione tra elementi in una formula 1. Calcolare la massa di ferro contenente lo stesso numero di atomi che si hanno in: a. 19,70 g di oro; b. 508,4 mg di rame.
2. Indicare, motivando la risposta, in quali dei seguenti casi si ha lo stesso numero di molecole: a. 10 g di O2 e 10 g di O3; b. 2,8 mol di N2 e 2,8 mol di O2; c. 20,00 g di NaOH e 31,51 g di HNO3.
3. Quanti grammi di CaO contengono le stesse moli di 10,00 g di CrO3?
4. Si considerino masse uguali delle sostanze di ciascuna delle seguenti coppie. Senza effettuare calcoli, stabilire a quale sostanza di ogni coppia corrisponde una minore quantità in moli: a. □ Na □ K b. □ NaCl □ NaI 5. Si considerino masse uguali delle sostanze di ciascuna delle seguenti coppie. Senza effettuare calcoli, stabilire a quale sostanza di ogni coppia corrisponde una maggiore quantità in moli: a. □ SiO2 □ CO2 b. □ NO2 − □ NO3 − c. □ Al3+ □ Fe3+ 6. Indicare quante moli di atomi di ogni elemento sono contenute in 0,02 mol di Al2(SO4)3.
7. Indicare quante moli di atomi di ogni elemento sono contenute in 90,08 g di glucosio (C6H12O6).
8. Calcolare quante moli di acqua e quante moli di atomi di O sono presenti in 49,94 g di CuSO4 ⋅ 5 H2O.
9. Scrivere quante moli di atomi di ciascun elemento sono contenute in 97,1 g di K2CrO4.
10. Quanti grammi di K2CO3 contengono 782,0 mg di K?
11. Si mescolano 0,010 kg di NaCl con 5,0 g di NaBr. Quante sono le moli totali di atomi di Na?
12. Si mescolano 500 mg di NaI con 0,0010 kg di Na2SO4 e 1,50 g di Na3PO4. Calcolare le moli totali di atomi di Na.
GUARDA! Esercizi interattivi su Formule empiriche e molecolari a partire da composizioni percentuali (in massa) 13. Stabilire la formula empirica dei composti aventi le seguenti composizioni percentuali: a. 21,8% Mg; 27,9% P; 50,3% O; b. 40,2% K; 26,9% Cr; il rimanente è O; c. 52,8% Sn; 12,4% Fe; 16,0% C; il rimanente è N.
14. Determinare la formula minima di un composto organico che ha la seguente composizione percentuale: 4,38% H; 60,87% C; 34,75% O.
15. La massa molecolare di un composto è 194 u. All’analisi esso rivela la seguente composizione: 49,5% C; 5,15% H; 16,5% O; 28,9% N.
Stabilirne la formula empirica.
16. L’analisi di un composto organico ha dato la seguente composizione percentuale: 73,28% C; 10,68% N; 3,84% H; 12,20% O.
Determinarne la formula, sapendo che la massa molare è approssimativamente 260 g · mol−1.
17. L’analisi di un composto organico ha dato i seguenti risultati: 24% C; 71,0% Cl; 4,05% H.
Determinare la formula empirica e la formula molecolare del composto, sapendo che la sua massa molecolare è pari a 98,96 dalton.
18. Un composto organico, avente massa molare pari a circa 860 g · mol−1, ha la seguente composizione percentuale: 40,0% C; 6,72% H; 53,5% O.
Determinare la formula empirica e la formula molecolare di tale composto.
19. Scrivere la formula di un sale avente la seguente composizione: 38,76% Ca; 19,97% P; 41,27% O.
20. In 7,445 g di un composto organico puro sono presenti 2,554 g di Na, 1,335 g di C e ossigeno. Qual è la formula del composto, sapendo che la sua massa formula è 134 dalton?
21. In un campione puro di un sale sono presenti: 0,6517 g di K; 0,2672 g di S; 0,5333 g di O.
Stabilire qual è la formula del sale.
22. All’analisi, un campione di caffeina presenta la seguente composizione: 5,19% H; 28,85% N; 49,48% C; 16,48% O.
Sapendo che la massa molare del composto è circa 194 g · mol−1, ricavare la sua formula molecolare.
23. Un composto, avente massa pari a 25,49 g, si ottiene facendo reagire con ossigeno 13,49 g di un campione puro di alluminio. Qual è la formula empirica del composto?
20 Capitolo 1 Moli, composizioni percentuali e formule 24. Dalla combustione di 0,546 g di un composto organico (contenente C, H, O) si sono ottenuti 0,792 g di CO2 e 0,378 g di H2O. Stabilire qual è la formula empirica del composto.
25. L’analisi di un minerale ha dato i seguenti risultati: 38,07% Al2O3; 17,70% K2O; 10,46% CaO; 33,70% SiO2.
Determinare la formula empirica del minerale.
26. Un minerale ha la seguente composizione percentuale: 4,35% H2O; 27,18% CaO; 24,80% Al2O3; 43,70% SiO2.
Determinarne la formula empirica.
27. Le analisi di due campioni minerali di silicato hanno dato i seguenti risultati: campione A: 23,36% Al2O3; 21,53% K2O; 55,06% SiO2; campione B: 0,72 g Al; 1,49 g Si; 1,04 g K; 2,56 g O.
Verificare se i due campioni hanno la stessa formula minima.
28. In un minerale, elementi differenti possono occupare la stessa posizione nel reticolo cristallino. Tali elementi vengono detti «isomorfogeni» o «vicarianti» e sono scritti, nella formula, tra parentesi tonde. Per esempio, (Ca, Mn)SiO3 rappresenta la formula di un minerale con Ca e Mn come elementi isomorfogeni.
Scrivere la formula empirica del minerale avente la seguente composizione percentuale: 14,76% K2O; 1,51% Na2O; 18,45% Al2O3; 65,27% SiO2. Considerare Na e K come elementi isomorfogeni.
[Un aiuto: sommare le quantità (in moli) degli ossidi contenenti gli elementi isomorfogeni e considerare tale somma come se fosse riferita all’ossido di un solo elemento. Nel caso considerato, l’ossido va scritto come (K, Na)2O.] 29. Scrivere la formula empirica del minerale avente la seguente composizione percentuale: 27,71% CaO; 20,69% Al2O3; 7,03% Fe2O3; 44,55% SiO2.
Considerare Al e Fe come elementi isomorfogeni.
Calcoli delle percentuali (in massa) di elementi, composti o formule in una sostanza 30. Calcolare la percentuale di acqua di cristallizzazione nel sale (NH4)3PO4 · 3 H2O.
31. L’acqua di cristallizzazione in CuSO4 ⋅ 5 H2O è il 36,08%. In quanto sale sono contenuti 80,00 g di acqua?
32. Dato il sale Na2B4O7 · 10 H2O, calcolare: a. la percentuale di acqua di cristallizzazione; b. in quanto sale sono contenuti 50,00 g di acqua.
33. Calcolare la massa di acqua contenuta in 40,00 g di sale di Mohr la cui formula è la seguente: Fe(NH4)2(SO4)2 ⋅ 6 H2O Qual è la percentuale di acqua di cristallizzazione?
34. Determinare le percentuali degli elementi presenti nel minerale di formula KAlSi3O8.
35. Calcolare la composizione percentuale di C, H e O nell’eucaliptolo, composto estratto dalle foglie di eucalipto, la cui formula empirica è C10H18O.
36. Calcolare la composizione percentuale degli elementi presenti in un fertilizzante avente formula (NH4)H2PO4.
37. Calcolare la composizione percentuale degli elementi presenti nel sale NiSO4 · 6 H2O.
38. Calcolare le percentuali degli elementi presenti nel composto insetticida avente formula C6H6Cl6.
39. Esprimere il contenuto di argento come percentuale formale di ossido di argento Ag2O nel sale Ag2SO4.
40. Qual è il contenuto di magnesio, espresso come percentuale di MgO, in un campione puro di Mg2P2O7?
41. Un campione puro di Fe2O3 pesa 15,969 g.
Quanto ferro, espresso formalmente come FeO, è presente nel campione? Qual è la percentuale formale di FeO nel campione?
42. Un campione di un minerale presenta all’analisi 8,27% di umidità e 36,47% di Cu. Qual è la % Cu nello stesso minerale, privo di umidità? 43. La principale responsabile del colore giallo-arancio dello zafferano è la crocina (una molecola simile a quella del carotene contenuto nelle carote). Due bustine di zafferano, contenenti ciascuna 100 mg di prodotto, provengono da due diversi lotti A e B che hanno il seguente contenuto di crocina sul prodotto secco: A 8,3% e B 8,1%. Nelle bustine, il contenuto di umidità è differente: in quella che proviene dal lotto A è il 10%, mentre nell’altra è il 6%. Quale bustina contiene più crocina? |
8596 | https://www.wyzant.com/resources/answers/896046/help-asap-pleaseeeee | HELP ASAP PLEASEEEEE | Wyzant Ask An Expert
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Calculus
Linie T.
asked • 05/19/22
HELP ASAP PLEASEEEEE
Related Rates
Water is pouring into an inverted cone at the rate of 6 cubic meters per minute. If the height of the cone is 10 meters and the radius of its base is 6 meters, how fast is the water level rising when the water is 3-meter deep.
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Raymond B.answered • 05/19/22
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Water is pouring into an inverted cone at the rate of 6 m^3/min =the derivative of Volume = V' = dV/dt.
height of the cone = h = 10 m, radius of base = r = 6 m,
how fast is the water level rising when water is 3 meters deep? h=3
Volume of water in the cone = V = (1/3)Ah = pi(r)^2(h)/3
where A=Area of the base = pi(r)^2
and h= height of the water level
when r=6, h=10
r =6h/10 = .6h
V = (pi/3)(.6h)^2(h) = pi/3(.6)^2(h)^3
V = .12pi(h^3)
take the derivative with respect to time
V' = .36pi(h^2)h'
plug in 3 for h
V'(3) = .36pi(9)h' = 6
h' = 6/3.24pi = 1/.54pi = 100/54pi = 50/27pi = meters per minute
h'= about 0.59 meters per minute = rate of rising water level when height of water = 3 meters
h' = about 0.6 m/min
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William W.answered • 05/19/22
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The RATE of the incoming water is dV/dt (so dV/dt = 6 m 3/min)
Volume of a cone is calculated from V = 1/3πr 2 h
We are given information about the cone however that helps us relate radius (r) and height (h). Looking at this cross section:
We can say r/h = 6/10 and cross multiplying yields 10r = 6h or r = 3/5h
So we can substitute the expression "3/5h" into our volume equation in place of "r":
V = 1/3πr 2 h
V = 1/3π(3/5h)2(h)
V = 3/25πh 3
Taking the derivative with respect to time (using the chain rule):
Plugging in dV/dt = 6, h = 3 we get:
6 = 9/25π(3 2)(dh/dt)
6 = 81/25π(dh/dt)
6 = 10.178(dh/dt)
dh/dt = 6/10.178
dh/dt = 0.59 meters/minute
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Tom N.answered • 05/19/22
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Start with V= πr 2 h/3 use similar triangles to find a relationship between r and h such that h/r= 10/6 so r= 3h/5. Use this in the volume formula so that V= π(9h 2/25)h/3 so V=π3h 3/25 now dV/dt = 9πh 2 dh/25dt. using the information in the problem 6=9π9dh/25dt solving for dh/dt = 150/81π m/sec.
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8597 | https://courses.lumenlearning.com/uvu-combinedalgebra/chapter/4-4-zeros-of-a-quadratic-function-2/ | 4.4: Zeros of a Quadratic Function | Intermediate Algebra
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Intermediate Algebra
MAT 1015 Sections
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4.4: Zeros of a Quadratic Function
Learning Objectives
Determine the zeros of a given quadratic function by factoring.
Determine the zeros of a given quadratic function by completing the square
Determine the zeros of a given quadratic function by using the quadratic formula
Understand the relationship between zeros and x x-intercepts
Understand that quadratic functions with complex zeros have no x x-intercepts
Factoring
In chapter 3, we learned several factoring methods and used them to find the zeros of polynomial functions. A quadratic function is a special case of a polynomial function. It is a trinomial of the form f(x)=a x 2+b x+c f(x)=a x 2+b x+c, where a,b,c a,b,c are real numbers and a≠0 a≠0. Consequently, we have already learned how to determine the zeros of a quadratic function by factoring.
For example, to determine the zeros of f(x)=9 x 2−4 f(x)=9 x 2−4, we set f(x)=0 f(x)=0 and solve for x x by factoring and using the zero-product property:
9 x 2−4=0(3 x−2)(3 x+2)=0 3 x−2=0 or 3 x+2=0 x=2 3 or x=−2 3 9 x 2−4=0(3 x−2)(3 x+2)=0 3 x−2=0 or 3 x+2=0 x=2 3 or x=−2 3
The zeros of the function f(x)=9 x 2−4 f(x)=9 x 2−4 are x=−2 3,2 3 x=−2 3,2 3 or x=±2 3 x=±2 3, where the symbol ±± is read “plus or minus” or “positive or negative”.
We also learned that the zeros correspond with the x x-coordinates of the x x-intercepts on the graphs of the functions. So the x x-intercepts of f(x)=9 x 2−4 f(x)=9 x 2−4 are (−2 3,0)(−2 3,0) and (2 3,0)(2 3,0).
We will not repeat those factoring methods here. Factoring and using the zero-product property works well when the function factors. But in most cases, functions do not factor, so other methods need to be employed. In this section, we will focus on two methods that can be used to determine the zeros (and consequently the x x-intercepts) of any quadratic function that may or may not factor.
Completing the Square
We just learned in section 4.3 how to convert a quadratic function from its standard form f(x)=a x 2+b x+c f(x)=a x 2+b x+c into its vertex form f(x)=a(x−h)2+k f(x)=a(x−h)2+k to determine the vertex of the function. In this section, we will take advantage of that new found knowledge to determine the zeros of a quadratic function.
For example, to determine the zeros of the function f(x)=x 2−4 f(x)=x 2−4 we set the function equal to 0. To solve the equation, we may isolate the perfect square, so we have a perfect square equal to a constant:
x 2−4=0 x 2=4 x 2−4=0 x 2=4
To solve this equation we first need to learn about the square root property.
The Square Root Property
The square root property of equality tells us that if we have an equation A=B A=B, we can take the square root of each side of the equation and get an equivalent equation, √A=√B A=B.
The Square Root Property of Equality
If A=B A=B, then√A=√B A=B
So, since we have x 2=4 x 2=4, then √x 2=√4 x 2=4.
We already know that √4=2 4=2, but what is √x 2 x 2?
There are two cases to consider: x≥0 x≥0 and x<0 x<0. Suppose x=9 x=9, then x 2=(9)2=81 x 2=(9)2=81. Then √x 2=√81=9 x 2=81=9. And since 9=x 9=x, √x 2=x x 2=x when x=9 x=9. In fact, we could use any value of x≥0 x≥0 and get the same result. So, √x 2=x x 2=x when x≥0 x≥0. But what happens when x<0 x<0? Suppose x=−7 x=−7,then x 2=(−7)2=49 x 2=(−7)2=49. Then √x 2=√49=7 x 2=49=7. But x=−7 x=−7, not 7 7! So when x<0 x<0, √x 2=−x x 2=−x, which is a positive answer when x x is negative. Putting theses two results together, and not knowing whether x x is positive or negative, we have, √x 2=|x|x 2=|x|.
SQUARE ROOTS
√x 2=|x|x 2=|x|
Going back to our example,
x 2=4√x 2=√4|x|=2 x=±2 x 2=4 x 2=4|x|=2 x=±2
Therefore, the zeros are x=−2 x=−2 and x=2 x=2. These correspond with the two x x-intercepts on the graph of the function (2, 0) and (–2, 0).
Notice that we have an absolute value equal to a constant, (i.e., |x|=2|x|=2). This means that x x will equal either positive or negative of the constant, (i.e., x=±2 x=±2). This is because the absolute value of a constant and its opposite are the same, (i.e., |2|=|−2|=2|2|=|−2|=2.
Most often, we will not have a perfect square equal to a perfect square. For example, we may end up solving x 2=72 x 2=72, where 72 is not a perfect square. Let’s recall how to deal with simplifying a square root that is not a perfect square.
Simplifying a Square Root
A square root can be simplified by separating perfect squares from non-perfect squares inside a square root. A number may be factored such that one or several perfect squares are among the factors. We can then use the product property of radicals to take the square root of the factors.
Product property of radicals
√a b=√a√b a b=a b for all a,b≥0 a,b≥0
For example, the number 72 may be factored as 72=2 3×3 2 72=2 3×3 2. There are two perfect squares, 2 2 2 2 and 3 2 3 2, in the prime factorization of 72. According to the product property of radicals, we may express √72 72 as the product of square roots:
√72=√2 3×3 2=√2×2 2×3 2=√2×√2 2×√3 2 Product property of radicals=√2×2×3=6√2 72=2 3×3 2=2×2 2×3 2=2×2 2×3 2 Product property of radicals=2×2×3=6 2
Therefore, √72 72 may be simplified to 6√2 6 2.
So, if we are solving the equation x 2=72 x 2=72, we can take the square root of both sides and simplify the radical:
x 2=72√x 2=√72|x|=6√2 x=±6√2 x 2=72 x 2=72|x|=6 2 x=±6 2
Example 1
Simplify √1080 1080.
Solution
√1080=√2 3×3 3×5=√2×2 2×3×3 2×5=√2 2×3 2×2×3×5=√2 2×√3 2×√2×3×5 Product property of radicals=2×3×√30=6√30 1080=2 3×3 3×5=2×2 2×3×3 2×5=2 2×3 2×2×3×5=2 2×3 2×2×3×5 Product property of radicals=2×3×30=6 30
Try It 1
Simplify √220 220.
Show Answer
2√55 2 55
Example 2
Solve: (x−5)2=40(x−5)2=40
Solution
(x−5)2=40√(x−5)2=√40 Square root both sides|x−5|=√4⋅10 look for perfect square factors under the radical|x−5|=√4√10 Multiplication property of radicals|x−5|=2√10 Simplify the radicals x−5=±2√10 Remove the absolute value and add a±x=5±2√10 Add 5 to both sides(x−5)2=40(x−5)2=40 Square root both sides|x−5|=4⋅10 look for perfect square factors under the radical|x−5|=4 10 Multiplication property of radicals|x−5|=2 10 Simplify the radicals x−5=±2 10 Remove the absolute value and add a±x=5±2 10 Add 5 to both sides
Try It 2
Solve: (x+1)2=32(x+1)2=32
Show Answer
x=−1±4√2 x=−1±4 2
Let’s use our new found knowledge to determine zeros of quadratic functions.
Example 3
Determine the zeros of g(x)=4(x−3)2−32 g(x)=4(x−3)2−32, then state the x x-intercepts.
Solution
We start by setting g(x)=0 g(x)=0, then isolating the perfect square:
4(x−3)2−32=0 4(x−3)2=32(x−3)2=8 4(x−3)2−32=0 4(x−3)2=32(x−3)2=8
We can now take the square root of both sides of the equation:
√(x−3)2=√8|x−3|=√4⋅2 x−3=±2√2 x=3±2√2(x−3)2=8|x−3|=4⋅2 x−3=±2 2 x=3±2 2
The zeros of the function are x=3+2√2 x=3+2 2 and x=3−2√2 x=3−2 2.
The x x-coordinates are therefore (3+2√2,0)(3+2 2,0) and(3−2√2,0)(3−2 2,0).
Example 4
Determine the zeros of f(x)=2(x+4)2−50 f(x)=2(x+4)2−50, then state the x x-intercepts.
Solution
Start by setting f(x)=0 f(x)=0, the isolating the perfect square:
2(x+4)2−50=0 2(x+4)2=50(x+4)2=25 2(x+4)2−50=0 2(x+4)2=50(x+4)2=25
We can now take the square root of both sides of the equation:
√(x+4)2=√25|x+4|=5 x+4=±5 x=−4±5(x+4)2=25|x+4|=5 x+4=±5 x=−4±5
This results in two equations:
x=−4+5=1 x=−4+5=1 and x=−4−5=−9 x=−4−5=−9
The zeros of the function are x=1 x=1 and x=−9 x=−9.
The x x-intercepts are therefore (1,0)(1,0) and (−9,0)(−9,0).
Try It 3
Determine the zeros of f(x)=3(x−1)2−27 f(x)=3(x−1)2−27, then state the x x-intercepts.
Show Answer
Zeros are x=−2 x=−2 and x=4 x=4.
x x-intercepts are(–2, 0) and (4, 0).
Try It 4
Determine the zeros of f(x)=2(x+3)2−36 f(x)=2(x+3)2−36, then state the x x-intercepts.
Show Answer
Zeros are x=−3−3√2 x=−3−3 2 and x=−3+3√2 x=−3+3 2.
x x-intercepts are (−3−3√2,0)(−3−3 2,0) and(−3+3√2,0)(−3+3 2,0).
If the function is given in standard form, we first need to convert it to vertex form using completing the square.
Example 5
Determine the zeros of f(x)=2 x 2+12 x−20 f(x)=2 x 2+12 x−20, then state the x x-intercepts.
Solution
Start by using completing the square to convert the function into vertex form:
f(x)=2 x 2+12 x−20=2(x 2+6 x)−20=2[(x+3)2−9]−20=2(x+3)2−18−20=2(x+3)2−38 f(x)=2 x 2+12 x−20=2(x 2+6 x)−20=2[(x+3)2−9]−20=2(x+3)2−18−20=2(x+3)2−38
Now we can set the function equal to zero:
2(x+3)2−38=0 2(x+3)2=38(x+3)2=19√(x+3)2=√19|x+3|=√19 x+3=±√19 x=−3±√19 2(x+3)2−38=0 2(x+3)2=38(x+3)2=19(x+3)2=19|x+3|=19 x+3=±19 x=−3±19
The zeros of the function are x=−3−√19 x=−3−19 and x=−3+√19 x=−3+19.
The x x-intercepts are therefore (−3−√19,0)(−3−19,0) and (−3+√19,0)(−3+19,0).
Try It 5
Determine the zeros of f(x)=x 2+4 x−12 f(x)=x 2+4 x−12, then state the x x-intercepts.
Show Answer
The zeros are x=−6 x=−6 and x=2 x=2.
The x x-intercepts are (−6,0)(−6,0) and (2,0)(2,0).
Try It 6
Determine the zeros of f(x)=3 x 2+17 x+10 f(x)=3 x 2+17 x+10, then state the x x-intercepts.
Show Answer
The zeros are x=−2 3 x=−2 3 and x=−5 x=−5.
The x x-intercepts are (−2 3,0)(−2 3,0) and (−5,0)(−5,0).
Imaginary Numbers and Complex Zeros
While trying to determine the zeros of a quadratic function, it is possible that we end up with an equation where the perfect square is equal to a negative number. For example, determining the zeros of the the function f(x)=x 2+4 f(x)=x 2+4 leads to the equation x 2=−4.x 2=−4.Does this equation have a solution?
In the domain of real numbers, it is impossible to find a number whose square is a negative number. The square of every real number, whether positive or negative, is positive. Therefore, there are no real number solutions for this equation.
However, the square root of a negative number is defined in the set ofimaginary numbers. The square root of –1 is defined as the letter i i.
√−1=i−1=i
Consequently, all negative real numbers have a square root that is an imaginary number:
√−3=√−1⋅√3=i√3−3=−1⋅3=i 3 (we always write the i i in front of the radical)
√−100=√−1⋅√100=i√100=10 i−100=−1⋅100=i 100=10 i
When we combine through addition a real number a a and an imaginary number b i b i, where a,b a,b are real numbers, we form a complex number a+b i a+b i.
COMPLEX NUMBERS
The set of complex numbers C C is the union of the set of real numbers R R and the set of imaginary numbers {b i|i=√−1,b∈R}{b i|i=−1,b∈R}.
For any real numbers a a and b b, a+b i a+b i is a complex number.
Complex zeros will show up when we end up with a perfect square equal to a negative number.
Example 6
Determine the zeros of the function f(x)=(x+3)2+16 f(x)=(x+3)2+16.
Solution
To determine the zeros we set the function equal to zero and solve the equation:
(x+3)2+16=0(x+3)2=−16√(x+3)2=√−16|x+3|=4 i x+3=±4 i x=−3±4 i(x+3)2+16=0(x+3)2=−16(x+3)2=−16|x+3|=4 i x+3=±4 i x=−3±4 i
The result is two complex zeros x=−3−4 i x=−3−4 i and x=−3+4 i x=−3+4 i
Example 7
Determine the zeros of the function g(x)=x 2−4 x+24 g(x)=x 2−4 x+24.
Solution
We first convert the function to vertex form by completing the square:
x 2−4 x+24=(x−2)2−4+24=(x−2)2+20 x 2−4 x+24=(x−2)2−4+24=(x−2)2+20
To determine the zeros we set the function equal to zero and solve the equation:
(x−2)2+20=0(x−2)2=−20√(x−2)2=√−20|x−2|=√−4√5 x−2=±2 i√5 x=2±2 i√5(x−2)2+20=0(x−2)2=−20(x−2)2=−20|x−2|=−4 5 x−2=±2 i 5 x=2±2 i 5
The result is two complex zeros x=2−2 i√5 x=2−2 i 5 and x=2+2 i√5 x=2+2 i 5.
Try It 7
Determine the zeros of the function g(x)=x 2−6 x+30 g(x)=x 2−6 x+30.
Show Answer
The zeros are x=3−i√21 x=3−i 21 and x=3+i√21 x=3+i 21
A function having complex zeros is not unusual, but what does this mean for the x x-intercepts of the graph of such a function?
Let’s look at the graphs of the functions from examples 4 and 5.
g(x)=x 2+6 x+25 g(x)=x 2+6 x+25 with
complex zeros x=−3−4 i x=−3−4 i and x=−3+4 i x=−3+4 ig(x)=x 2−4 x+24 g(x)=x 2−4 x+24 with
complex zeros x=2−2 i√5 x=2−2 i 5 and x=2+2 i√5 x=2+2 i 5
Figure 1. Graphs of functions with complex zeros.
There are no x x-intercepts when a quadratic function has complex zeros. This is because the coordinate system, and in particular the x x-axis, graphs only real numbers.
graphs of quadratic functions with complex zeros
The graph of a quadratic function with complex zeros has no x x-intercepts; the parabola never intersects the x x-axis.
The Quadratic Formula
Another way to determine the zeros of a quadratic function f(x)=a x 2+b x+c f(x)=a x 2+b x+c is to use the quadratic formula.
The quadratic formula
For any quadratic function f(x)=a x 2+b x+c f(x)=a x 2+b x+c, the zeros of the function are given by
x=−b±√b 2−4 a c 2 a x=−b±b 2−4 a c 2 a
To use the quadratic equation, we need a quadratic function in the general form f(x)=a x 2+b x+c f(x)=a x 2+b x+c so that we can pick out the values of a,b,c a,b,c.
For example if we are asked to determine the zeros of the function h(x)=3 x 2−5 x+1 h(x)=3 x 2−5 x+1, we evaluate the quadratic equation when a=3,b=−5,c=1 a=3,b=−5,c=1:
x=−(−5)±√(−5)2−4(3)(1)2(3)=5±√25−12 6=5±√13 6 x=−(−5)±(−5)2−4(3)(1)2(3)=5±25−12 6=5±13 6
Therefore, the zeros are x=5−√13 6 x=5−13 6 and x=5+√13 6 x=5+13 6.
If we were asked to find the x x-intercepts, they are (5+√13 6,0)(5+13 6,0) and(5−√13 6,0)(5−13 6,0).
Example 8
Determine the zeros of the function g(x)=x 2−4 x+2 g(x)=x 2−4 x+2. Then state the x x-intercepts.
Solution
To use the quadratic formula we need a=1,b=−4,c=2 a=1,b=−4,c=2:
x=−b±√b 2−4 a c 2 a=−(−4)±√(−4)2−4(1)(2)2(1)=4±√16−8 2=4±√8 2=4±2√2 2=2±√2=2+√2,2−√2 x=−b±b 2−4 a c 2 a=−(−4)±(−4)2−4(1)(2)2(1)=4±16−8 2=4±8 2=4±2 2 2=2±2=2+2,2−2
The zeros of the function are x=2+√2 x=2+2 and x=2−√2 x=2−2.
The x x-intercepts are (2+√2,0)(2+2,0) and (2−√2,0)(2−2,0).
Example 9
Determine the zeros of the function g(x)=2 x 2−x+5 g(x)=2 x 2−x+5. Then state the x x-intercepts.
Solution
First pick out a=2,b=−1,c=5 a=2,b=−1,c=5, then evaluate the quadratic formula:
x=−b±√b 2−4 a c 2 a=−(−1)±√(−1)2−4(2)(5)2(2)=1±√1−40 4=1±√−39 4=1±i√39 4=1 4±i√39 4 x=−b±b 2−4 a c 2 a=−(−1)±(−1)2−4(2)(5)2(2)=1±1−40 4=1±−39 4=1±i 39 4=1 4±i 39 4
The zeros of the function are x=1 4−i√39 4 x=1 4−i 39 4 and 1 4+i√39 4 1 4+i 39 4.
There are no x x-intercepts since the zeros are complex.
Try It 8
Determine the zeros of the function g(x)=x 2−3 x+2 g(x)=x 2−3 x+2. Then state the x x-intercepts.
Show Answer
Zeros are x=1 x=1 and x=2 x=2.
x x-intercepts are(1,0)(1,0) and (2,0)(2,0).
Try It 9
Determine the zeros of the function g(x)=3 x 2+12 g(x)=3 x 2+12. Then state the x x-intercepts.
Show Answer
Zeros are x=−2 i x=−2 i and x=2 i x=2 i
Proof of the Quadratic Formula
This quadratic formula may be derived by converting the function f(x)=a x 2+b x+c f(x)=a x 2+b x+c into vertex form by completing the square, and then setting f(x)=0 f(x)=0 and solving for x x. The following proof shows how this formula is derived.
f(x)=a x 2+b x+c=a[x 2+b a x]+c=a[(x+(b 2 a))2−(b 2 a)2]+c=a(x+(b 2 a))2−a(b 2 a)2+c=a(x+(b 2 a))2−b 2 4 a+c f(x)=a x 2+b x+c=a[x 2+b a x]+c=a[(x+(b 2 a))2−(b 2 a)2]+c=a(x+(b 2 a))2−a(b 2 a)2+c=a(x+(b 2 a))2−b 2 4 a+c
To find the zeros, set the function equal to zero, and then solve for x x:
a(x+(d b 2 a))2−b 2 4 a+c=0 a(x+(b 2 a))2=b 2 4 a−c a(x+(b 2 a))2=b 2−4 a c 4 a x+(b 2 a)2=b 2−4 a c 4 a 2√(x+(b 2 a))2=√b 2−4 a c 4 a 2|x+b 2 a|=√b 2−4 a c√4 a 2 x+b 2 a=±√b 2−4 a c 2|a|x=−b 2 a±√b 2−4 a c±2 a x=−b 2 a±√b 2−4 a c 2 a x=−b±√b 2−4 a c 2 a a(x+(d b 2 a))2−b 2 4 a+c=0 a(x+(b 2 a))2=b 2 4 a−c a(x+(b 2 a))2=b 2−4 a c 4 a x+(b 2 a)2=b 2−4 a c 4 a 2(x+(b 2 a))2=b 2−4 a c 4 a 2|x+b 2 a|=b 2−4 a c 4 a 2 x+b 2 a=±b 2−4 a c 2|a|x=−b 2 a±b 2−4 a c±2 a x=−b 2 a±b 2−4 a c 2 a x=−b±b 2−4 a c 2 a
Q.E.D.
Candela Citations
CC licensed content, Original
Zeros of a Quadratic Function. Authored by: Hazel McKenna and Leo Chang. Provided by: Utah Valley University. License: CC BY: Attribution
All graphs created using desmos graphing calculator. Authored by: Hazel McKenna . Provided by: Utah Valley University. License: CC BY: Attribution
All examples. Authored by: Hazel McKenna and Leo Chang. Provided by: Utah Valley University. License: CC BY: Attribution
All Try Its. Authored by: Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution
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CC licensed content, Original
Zeros of a Quadratic Function. Authored by: Hazel McKenna and Leo Chang. Provided by: Utah Valley University. License: CC BY: Attribution
All graphs created using desmos graphing calculator. Authored by: Hazel McKenna . Provided by: Utah Valley University. License: CC BY: Attribution
All examples. Authored by: Hazel McKenna and Leo Chang. Provided by: Utah Valley University. License: CC BY: Attribution
All Try Its. Authored by: Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution
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•4 yr. ago
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Optimization modelling in Excel or Google Sheets
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I'm trying to build a dynamic optimization tool in Excel but I'm not sure where to start. It would be awesome if this could run in Google Sheets as well, but based on my rudimentary knowledge in this particular area, I don't think Sheets would be able to handle this problem. My data is set up as follows:
| . | Item A | Item B | Item C | Item D | Item E | Item F |
--- --- ---
| |
| Area 1 | 0 | 1 | 0 | 1 | 0 | 0 |
| Area 2 | 0 | 0 | 1 | 1 | 0 | 0 |
| Area 3 | 1 | 0 | 0 | 1 | 1 | 1 |
| Area 4 | 1 | 1 | 0 | 0 | 0 | 1 |
Where the 1 denotes an Item that can be used in a given Area. I want to minimize the number of Items used that gives complete Area coverage, however, I would also like to dynamically hold some Items as "necessary" (must be used in the optimized solution). The full data set has 30+ items and over 100 areas.
Any help would be much appreciated.
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8599 | https://launiusarcola.weebly.com/uploads/5/8/2/4/58246433/6.2_6.3_solving_systems_mixed_worksheet.pdf | ©9 k2p0N1r2C tK6u4t0aa 7S0oHfIt6whazr6eZ RLgLtC3.m I lAalVlN JrFi2gBhOtXsE ar8eKsFeArOveeadf.x 5 eM4a8dEeZ 1wZipt3hl TIlnCfbiVnmiHt5ec HAxlrglexbqrja7 F2W.N Worksheet by Kuta Software LLC Algebra 2 ID: 1 Name_____ Period_ Date_______ ©1 82M0v1p2H FKkultEay ISGoGfItbwHa2r2eM hL1LiC3.s B 7A3l3l6 pr9iPgShFtgsM 7r5eTsDevrKvSe5d6.x Solving Systems of Equation- Any Method Solve each system by elimination.
1) −7 x − 6 y = −5 −5 x − 7 y = −28 Solve each system by substitution.
2) − x − 8 y = −6 −2 x + y = −12 Solve each system by graphing.
3) y = 1 − 1 4 x x + 4 5 y = −12 5 −5 −4 −3 −2 −1 0 1 2 3 4 5 −5 −4 −3 −2 −1 1 2 3 4 5 -1-©t J2a091w27 oK1uJtlal FS4oYfetDwhaJrdej 9LuLVCQ.0 B uAGlglR 4rSikgNhztPsA yrkeisle2rovfe3dY.J C iM3aidmen xwCi1tBhY qI6nifyiynYi8theM 3AAlNgoeGbVryaH d2w.2 Worksheet by Kuta Software LLC Solve each system.
4) −10 x − 15 y = 25 x − 5 y = −9 5) y = −1 3 x + 3 y = −2 x − 2 6) −5 x − y = 12 7 x + y = −16 7) 3 x − 2 y = −4 x − 2 y = 4 8) y = −3 x + 20 y = 4 x − 15 9) −9 x + 10 y = −22 −7 x + 8 y = −18 10) −4 x + 4 y = −4 2 x − 8 y = 2 11) x − 3 y = 9 x + 3 y = −3 12) −7 x − 6 y = 10 y = x − 6 13) 2 x + 7 y = 19 5 x + 3 y = −25 -2-©T P2U081y29 wKPuft5ai aSRosfFt2wvaZrue6 AL1L3CS.6 a 7A4lVlC qrtiSgbh0t2sa Lrye0soe2rNv3e6dr.3 x WM6a5dMeB awsi9tvhE fIOn4fJi8nnist2e9 fABlVgle7bArgau G2g.p Worksheet by Kuta Software LLC -3-Answers to Solving Systems of Equation- Any Method (ID: 1) 1) (−7, 9) 2) (6, 0) 3) (−4, 2) 4) (−4, 1) 5) (−3, 4) 6) (−2, −2) 7) (−4, −4) 8) (5, 5) 9) (−2, −4) 10) (1, 0) 11) (3, −2) 12) (2, −4) 13) (−8, 5) |
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