text stringlengths 30 4k | source stringlengths 60 201 |
|---|---|
By definition of inf, there exists Mk → �f �∞ such that f (x) <
Mk a.e, or, equivalently, there exists Nk with µ(Nk ) = 0 such that
c . Let N = ∪∞ Nk . Then µ(N ) = 0. If
f (x)| ≤ Mk for all x ∈ Nk
|
x ∈ N c =
∩∞
. Thus,
|
�
f (x)| ≤ �f �∞ for all x ∈ N c .
|
k=1(Nk )c, then f (x)| ≤ Mk since Mk
→ �f �∞
k=1
|
|
Dat... | https://ocw.mit.edu/courses/18-125-measure-and-integration-fall-2003/0c85400465928cf42a19e5b87bdc349c_18125_lec15.pdf |
Lp(µ). Then f + g ∈ Lp(µ)
and �f + g�
.
p ≤ �f �
p + �g�p
Proof. If 1 < p < ∞, this is simply Minkowski’s inequality. If p = 1,
�
�
then
f +
|
|
f +
|
f + g| ≤
|
�
g
f �
f + g�∞ ≤ �
| | ⇒ �
�
|
g is true. If p = ∞, then f + g
|
�∞
�g
∞ +
| ≤ |
.
|
Normed space and Banach spaces. A normed space is a vector
... | https://ocw.mit.edu/courses/18-125-measure-and-integration-fall-2003/0c85400465928cf42a19e5b87bdc349c_18125_lec15.pdf |
metric d(f, g) = �f − g�.
Recall that xi → x ∈ M if limn→∞ d(xn, x) = 0. A sequence {xi} is
Cauchy if for every � > 0 there exists N (�) such that d(xj , xk ) ≤ � for
all j, k ≥ N (�).
Claim: if xn → x, then it is Cauchy. We know that limn→∞ d(xn, x) =
0, so given � > 0, there exists N such that d(xk , x) < �/2 fo... | https://ocw.mit.edu/courses/18-125-measure-and-integration-fall-2003/0c85400465928cf42a19e5b87bdc349c_18125_lec15.pdf |
nk+1
Theorem 0.5. For 1 ≤ p ≤ ∞ and for any measure space (X, M, µ),
the space Lp(µ) is a Banach space.
Proof. Let 1 ≤ p < ∞ and let {fn} ∈ Lp(µ) be a Cauchy sequence.
By the lemma, there exists
< · · ·
�
2
�
< 2−k
such that fnk+1
and g =
, fnk
�
�
∞
fni+1 − f
limk→∞ gk
�
i=1
< n
with
�
�
k
1
�
fni+1 − ... | https://ocw.mit.edu/courses/18-125-measure-and-integration-fall-2003/0c85400465928cf42a19e5b87bdc349c_18125_lec15.pdf |
AND INTEGRATION: LECTURE 15
The partial sum
k−1
�
fn1 (x) +
(fni+1 (x) − fni (x)) = fnk (x),
and so
i=1
lim fnk (x) = f (x) a.e.
k→∞
Thus we have shown that every Cauchy sequence has a convergent
nk →
subsequence, and we NTS that f
Given � > 0, there exists N such that �fn − fm�p < � for all n, m >
p
f in L ... | https://ocw.mit.edu/courses/18-125-measure-and-integration-fall-2003/0c85400465928cf42a19e5b87bdc349c_18125_lec15.pdf |
1
|
For x ∈ N c , fn is a Cauchy sequence of complex numbers. Thus,
fn → f by completeness of C uniformly. Since �f
is bounded,
c
c
|
fk (x) < M for all x ∈ N . Thus, f (x) < M for all x ∈ N . Letting
0 as n → ∞. �
f = 0 on N , we have �f �∞ < ∞ and �f
Theorem 0.6. Let 1 ≤ p ≤ ∞ and {fn} be a Cauchy sequence in Lp(... | https://ocw.mit.edu/courses/18-125-measure-and-integration-fall-2003/0c85400465928cf42a19e5b87bdc349c_18125_lec15.pdf |
HW
problem).
(3) A sequence can belong to Lp1 ∩ Lp2 and converge in Lp1 without
converging in Lp2 . Let fk = k−1χ(k,2k). Then fk → 0 pointwise
and �fk � = k−1k1/p = k1/p−1 . If p > 1, then �fk �
0 as
k → ∞, so fk → 0 in Lp norm. But �fk � = 1 so fk �→ 0 in L1 .
p →
p
1 | https://ocw.mit.edu/courses/18-125-measure-and-integration-fall-2003/0c85400465928cf42a19e5b87bdc349c_18125_lec15.pdf |
Lecture 5
8.251 Spring 2007
Lecture 5 - Topics
• Nonrelativistic strings
• Lagrangian mechanics
Reading: Zwiebach, Chapter 4
Non-Relativistic Strings
Study nonrelativistic strings first to develop intuition and math notation before
moving to the relativistic strings that we actually care about.
Non-relativistic... | https://ocw.mit.edu/courses/8-251-string-theory-for-undergraduates-spring-2007/0cca13a5a5fa624cb90d0b457692b8d6_lec5.pdf |
( 1 + (dy/dx)2 − 1)
= dx(dy/dx)2
1
2
((small))
General form of wave equation:
v: velocity of wave, v = T0/µ0
�
∂2f
∂x2
−
1 ∂2f
v2 ∂t2
= 0
General Solution:
y(x, t) = h+(x − v0t) + h (x + v0t)
−
Note: the h’s are function of 1 variable (x ± v0t) not 2 variables x and t inde
pendently.
Boundary Conditions: ... | https://ocw.mit.edu/courses/8-251-string-theory-for-undergraduates-spring-2007/0cca13a5a5fa624cb90d0b457692b8d6_lec5.pdf |
Let u = −a − v0t
h+(u + 2a) = h+(u)
Variational Principle
Consider point mass m doing 1D motion x(t).
Assume x(ti) = xi, x(tf ) = xf . Under the influence of potential V (x)
Know:
3
Lecture 5
8.251 Spring 2007
Possible motions:
Not possible:
Given a path:
4
Lecture 5
8.251 Spring 2007
Functional: S : x(t)... | https://ocw.mit.edu/courses/8-251-string-theory-for-undergraduates-spring-2007/0cca13a5a5fa624cb90d0b457692b8d6_lec5.pdf |
δ(x(t)))]dt must go away for S[x + δx] = S[x] + θ[(δx)2] to
Call this the variation δS
δS =
�
tf
ti
�
d
dt
dt
Integrate by parts
(m ˙
xδx) − m¨
xδx − V �(x(t))δ(x(t))
�
dS = mx˙ (tf )δx(tf ) − mx˙ (ti)δx(ti) +
�
tf
dtδx(t)[−mx¨ − V �(x(t))]
ti
δx(tf ) = δ(ti) = 0 from before.
The integral
�
tf dtδx(t... | https://ocw.mit.edu/courses/8-251-string-theory-for-undergraduates-spring-2007/0cca13a5a5fa624cb90d0b457692b8d6_lec5.pdf |
0, t) or δy(x = a, t)
δS =
�
tf
a
�
dt
ti
0
�
∂L
dx
∂y˙
δy˙ +
�
δy�
∂L
∂y�
6
Lecture 5
Let:
δS =
�
tf
�
dt
ti
0
8.251 Spring 2007
tP = ∂L/∂y˙
P x = ∂L/∂y�
�
tf
δS =
� �
a
P t
∂(δy)
∂t
�
+ P x
∂(δy)
∂x
ti
0
a
�
dx −δy(x, t)
�
∂P t
∂t
+
∂P x
∂x
�� �
+
a
0
dxP t[δy]tf ... | https://ocw.mit.edu/courses/8-251-string-theory-for-undergraduates-spring-2007/0cca13a5a5fa624cb90d0b457692b8d6_lec5.pdf |
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Physics Department
Physics 8.07: Electromagnetism II
Prof. Alan Guth
October 17, 2012
LECTURE NOTES 9
TRACELESS SYMMETRIC TENSOR APPROACH
TO LEGENDRE POLYNOMIALS
AND SPHERICAL HARMONICS
In these notes I will describe the separation of variable technique for solving Laplace’s
equati... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
Edition (John Wiley & Sons, 1999), Sections 3.1, 3.2, 3.5, and 3.6.
1. LAPLACE’S EQUATION IN SPHERICAL COORDINATES:
In spherical coordinates, Laplace’s equation can be written as
∇2
ϕ(r, θ, φ) =
(cid:2)
(cid:1)
r2
∂ϕ
∂r
+
1
r2
1 ∂
r2 ∂r
∇2
θ ϕ = 0 ,
where the angular part is given by
∇2
θ ϕ ≡
1
sin θ
∂
∂θ
(cid:2)
(cid:... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
way that the equation can be satisfied is if each term
is a constant. Thus we can write
∇2
1
F
(cid:1)
R
r2 d
dr
1 d
R d
r
θ F = Cθ ,
(cid:2)
= −C .
θ
2. THE EXPANSION OF F (θ, φ):
We now wish to find the most general solution to the equation
∇2
θ F = Cθ F ,
(9.6)
(9.7)
(9.8)
which is a rewriting of Eq. (9.6). If such a ... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
1i2...i n(cid:1) ˆi1 nˆi2 . . . nˆi(cid:1) + . . . ,
(9.10)
((cid:9))
where repeated indices are summed from 1 to 3 (as Cartesian coordinates). Note that
Ci1i2...i ni
(cid:1) ˆ 1 nˆi2 . . . nˆi(cid:1) represents the general term in the series, where the first three terms
correspond to (cid:16) = 0, (cid:16) = 1, and (ci... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
the sense that if any two indices are set equal
the result is equal to zero. Since the tensors are already
to each other and summed,
assumed to be symmetric, it does not matter which indices are summed, so we can
choose the last two:
(cid:1)
((cid:9))
Ci1i2...i
(cid:1) 2jj−
= 0
.
(9.12)
To explain why these restriction... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
C˜(2)
ij :
(cid:3)
(2)
Cij ninj
ˆ ˆ =
˜(2)
Cij + λδ
ij
(cid:4)
nˆinˆ
˜(2)
j = Cij nˆinˆj + λ ,
1
3
where we used the fact that δij nˆinˆj = 1, since nˆ is a unit vector. The extra term, 1 λ3 ,
can then be absorbed into a redefinition of C(0):
C˜(0) = C(0) + λ .
1
3
Finally, we can write
C(0) +
(1)
i
C nˆi +
(2)
Cij nˆ n... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
call them xi. So
(cid:21)r = xieˆi = x1eˆ1 + x2eˆ2 + x3eˆ3 .
(9.21)
8.07 LECTURE NOTES 9, FALL 2012
p. 5
Then, given any F(cid:9)(ˆn) of the form given in Eq. (9.19), we can define a function F˜(cid:9)((cid:21)r ) by
F˜
(cid:9)((cid:21)r ) = Ci1i2...i xi
(cid:1)
1 xi2 . . . xi(cid:1) = r F(cid:9)(ˆn) .
(cid:9)
((cid:9)... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
2C(0) = 0 ,
since the derivative of a constant vanishes. Similarly for (cid:16) = 1,
∇2F˜
1((cid:21)r ) = ∇ Ci xi = 0 ,
2
(1)
(9.23)
(9.24)
since the first derivative produces a constant, so the second derivative vanishes. The first
nontrivial case is (cid:16) = 2:
∇2F˜ ((cid:21)r ) = ∇2
2
(2)
Cij xixj
=
Cij
(2) ∂
∂
∂xm ... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
ECTURE NOTES 9, FALL 2012
p. 6
((cid:9))
Ci1i2...i(cid:1) that caused the term to vanish. Thinking
where again it is the tracelessness of
about the general term, one can see that after the derivatives are calculated, there are
(cid:16)−2 factors of xi that remain, but there are still (cid:16) indices on
Ci1i2...i(cid:1... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
F(cid:9)(ˆn)
,
and therefore
∇2
θ F(cid:9)(ˆn) = −(cid:16)((cid:16) + 1)F(cid:9)(ˆn) .
(9.28)
(cid:6)
and eigen-
Thus, we have found the eigenfunctions
(cid:5)
−(cid:16)((cid:16) + 1)
(cid:6)
values
of the differential operator ∇2
θ. This is a very useful resul
t!
(cid:5)
F(cid:9) n
(ˆ) =
((cid:9))
Ci1i2...i nˆi1 nˆi2 .... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
can be written as a linear sum of these two. Thus we can write
R(cid:9)(r) = A(cid:9)r(cid:9)
+
B
(cid:9)
r(cid:9)+1
,
(9.31)
allowing for different values of A(cid:9) and B(cid:9) for each (cid:16). The most general solution to
Laplace’s equation, in spherical coordinates, can then be written as
(cid:11) (cid:1)
∞
(cid... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
to parameterize the most general traceless symmetric tensor of rank
(cid:16)
?
We begin by calculating Nsym((cid:16)), the number of linearly independent symmetric
tensors of rank (cid:16). Note that we are postponing the consideration of tracelessness. For
grounding, we can start by saying that it takes one number to ... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
6) tensor. There is then a 1:1 correspondence between
independent tensor elements and the different ways that the balls can be put into the
hats. For example, if there are 9 balls, with 3 in the first hat, 2 in the second, and 4 in
the third, then this arrangement of balls corresponds to
S111223333. So now we just have
t... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
)!
= ((cid:16) + 1)((cid:16) + 2) .
1
2
(9.33)
It is easily checked that this gives Nsym(0) = 1, Nsym(1) = 3, and Nsym(2) = 6, consistent
with the examples we started with.
To impose tracelessness, we require that our traceless tensors also satisfy
((cid:9))
Si1...i(cid:1)−2jj = 0 .
(9.34)
How many conditions is this? ... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
general expansion (9.10) for a
function of (θ, φ) to the azimuthally symmetric case, we need to construct traceless sym-
metric tensors which are invariant under rotations about the z-axis. One straightforward
way to this is to build the traceless symmetric tensor from the vector zˆ, the unit vector in
the z direction.... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
curly brackets { . . . } to denote the traceless symmetric part of . . ..
Thus,
{ 1 } = 1 ,
{ zˆi } = ˆzi .
(9.36)
But for rank 2, the trace of zˆizˆj is equal to zˆizˆi = ˆz · zˆ = 1. But we can subtract a
constant times δij so that the result is traceless:
{ zˆizˆj } = ˆzizˆj − 1 δij .
3
The coefficient is 1/3, because... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
with either one or two Kronecker
δ-functions, and both are needed to give a traceless result. We can start with arbitrary
coefficients, and see what they have to be:
{ zˆizˆj zˆkzˆm } = ˆzizˆjzˆkzˆm + c1
(cid:12)
zˆizˆjδkm + ˆzizˆkδmj + ˆzizˆmδjk + ˆzj zˆkδim
(cid:13)
(cid:12)
(cid:13)
+ ˆzj zˆmδik + ˆzkzˆmδij
+ c2
δijδk... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
zˆmδik + ˆzkzˆmδij
zˆizˆjδkm + ˆzizˆkδmj + ˆzizˆmδjk + ˆzj zˆkδim
+ 1 δij δkm δikδjm δimδjk
+
+
(cid:13)
(cid:12)
35
(cid:13)
.
(9.43)
It can be shown that any traceless symmetric tensor of rank (cid:16) that is invariant
under rotations about the z-axis is proportional to { zˆi1 . . . zˆi(cid:1)
}. (To see this, note ... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
. nˆi(cid:1) + . . . ,
(9.44)
where the c(cid:9)’s are constants. This corresponds to what is standardly called an expansion
in Legendre polynomials. In Lecture Notes 8 I show exactly how to relate these terms to
the standard conventions for normalizing the Legendre polynomials, but we can see here
exactly what these f... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
must all vanish. The B(cid:9) factors can be absorbed into the definition of
Ci ...i , so we can
write the expansion as
((cid:9))
1
(cid:1)
Φ(
(cid:21)r ) =
∞
(cid:11)
(cid:9)=0
1
r(cid:9)+1
C
((cid:9))
i1...i n(cid:1) ˆi1 . . . nˆi(cid:1) .
(9.46)
8.07 LECTURE NOTES 9, FALL 2012
p. 12
Since each successive term comes ... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
:21)r − (cid:21)r (cid:4)| in a power series
in (cid:21)r (cid:4).
I’ll begin by doing it as Griffiths does, which gives the simplest — but not the most
useful — form of the multipole expansion. Griffiths rewrote the denominator as
1
|(cid:21)r − (cid:21)r (cid:4)|
(cid:16)
=
2
|
(cid:21)r
|
1
+ |2 − 2 ·
(cid:21) (cid:4)
(... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
1 +
1
(cid:13)2
(cid:12)
r(cid:2)
r
− 2 r(cid:2)
r
cos θ(cid:4)
=
1
r
(cid:9)=0
(cid:1)
∞
(cid:11)
(cid:2)
(cid:9)
(cid:4)
r
r
P (cos θ) .
(cid:9)
(9.51)
8.07 LECTURE NOTES 9, FALL 2012
p. 13
Inserting this relation into Eq. (9.47), we find
V ((cid:21)r ) =
1
π%
0
4
∞
(cid:11) 1
r +1
(cid:9)
(cid:9)=0
(cid:15)
(cid:4)(... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
harmonics, but
here I will derive the equivalent relations using the traceless symmetric tensor approach.
Instead of expanding 1/ |(cid:21)r − (cid:21)r (cid:4)| in powers of r(cid:4), we will think of it as a function of
i of (cid:21)r (cid:4), and we will expand it as a Taylor series in 3
three variables — the compon... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
)
(cid:17)
(cid:17)
(cid:17)
(cid:16)
(cid:16)r (cid:2)=0
(cid:17)
(cid:17)
(cid:17)
(cid:17)
(cid:17)
(cid:16)r (cid:2)=(cid:16)0
f is a function of (cid:21)r − (cid:21)r (cid:4), the
i can be replaced by derivatives with respect to xi with a
r(cid:4)2
∂2f
(cid:4)
(cid:4)
2! ∂x ∂x
j
i
ˆ(cid:4) ˆ(cid:4)
n n + . . . .
i... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
operation (cid:16) times:
(cid:17)
(cid:17)
∂(cid:9)f
(cid:17)
(cid:17)
(cid:4)
. . . ∂x
i(cid:1) (cid:16)r =(cid:16)
(cid:2) 0
∂x(cid:4)
i
1
= (−1)
(cid:9)
∂(cid:9)
∂xi1 . . . ∂xi(cid:1)
1
|(cid:21)r
|
.
Combining Eqs. (9.58) with (9.56),
we can write
(cid:1)
1
|(cid:21)r − (cid:21)r (cid:4)|
=
∞
(cid:11)
(
(cid:9)=0
... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
, we will work out the first several terms until we recognize
the pattern. We write
and adopt the abbreviation
(cid:21)r ≡ rn ,ˆ
∂i ≡
∂
xi
.
It is useful to start by evaluating the derivatives of the basic quantities r and nˆi:
∂ r = ∂
i
i(xjxj)1 2
/
(cid:5)
(cid:6)
1
= (
2
x
kxk)− /2∂i(xjxj) = 2x
1
jδij =
x
i
r
= ˆni ,... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
cid:9)(2(cid:16) − 1)!!
r +1
(cid:9)
∂x
{ nˆii . . . nˆi
(cid:1)
} ,
(9.65)
where
(2(cid:16) − 1)!! ≡ (2(cid:16) − 1)(2(cid:16) − 3)(2(cid:16) − 5) . . . 1 =
(2(cid:16))!
2(cid:9)(cid:16)!
, with ( 1)!!
−
≡ 1 .
(9.66)
Inserting this result into Eq. (9.59), we find
1
|(cid:21)r − (cid:21)r (cid:4)|
=
∞
(cid:11) (2(cid:16... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
1) by terms proportional to Kronecker δ-functions,
which vanish when summed with the traceless tensor { nˆi1 . . . nˆi(cid:1)
Starting with
Eq. (9.67), one can if one wishes drop the curly brackets around either factor (but not
both!).
}.
Inserting this expression for 1/|(cid:21)r − (cid:21)r (cid:4)| into Eq. (9.47), ... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
ρ((cid:21)r (cid:4)) d3x(cid:4) .
Vdip((cid:21)r ) =
· nˆ
1 p(cid:21)
4π%0 r2
(cid:15)
,
pi =
(1)
Ci =
ρ((cid:21)r (cid:4))(cid:21)ri d3x .
(9.71)
(9.72)
(9.73)
(9.74)
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8.07 Electromagnetism II
Fall 2012
For information about citing these materials or our Terms of Use, visit: http:... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/0ccc978da12536a8d95333ec7726c555_MIT8_07F12_ln9.pdf |
LECTURE 10
• Readings: Section 3.6
Lecture outline
• More on continuous r.v.s
• Derived distributions
Review
Conditioning “slices” the joint PDF
• Recall the stick-breaking example:
• Pictorially:
Buffon’s Needle (1)
• Parallel lines at distance
Needle of length (assume )
• Find P(needle intersects one of the lines)... | https://ocw.mit.edu/courses/6-041-probabilistic-systems-analysis-and-applied-probability-spring-2006/0cced51cb2e7e6062d04055e7e4c1684_lec10.pdf |
that if is normal,
then is also normal. | https://ocw.mit.edu/courses/6-041-probabilistic-systems-analysis-and-applied-probability-spring-2006/0cced51cb2e7e6062d04055e7e4c1684_lec10.pdf |
Letkvr
(0
4k '..
Uotrzh
(M&L DaviI
CAeA
--Axt~
'I\
C
^-
£
A LI) c t
'C-i
t
+ At,
_ Y
( t)
v\c\,
'Grc
a'.
a
:W w
Lk'
[
6oic~pLP
( 1e4- MI
-\
s
fS
t,') k,
t
JL L4
P =
I,-
a C , w J\L
4&
,
-�
� 1_^�-r-111·---4-�-
^1·-
.. <1
AlF
,
. .. ..
..
4ewvn Orq~or
\1~
.erS
~),-
s-e. ric lassic... | https://ocw.mit.edu/courses/8-322-quantum-theory-ii-spring-2003/0ce4f20acbdfa718cd95bd716de2c74b_83227Lecture6.pdf |
F\n,,,
4d
C
P EK (L\
z1
±L ,
t
--ie
K 1.
-3e
i
- ......
I .-.. -- 5ojv
os~e.~
t)_.oe-F.
_
>QLYK~.
..
I..
......
.-- . .-.- I
...
AP.
-9
-
.r
r
4
, '..AJ~'-E.....
J'."'-
.
4
C
'!V' ......
.~k..............
~~'c-
"' '' : '
-~._.~lwge~n
,~ut..a.
............
Tr~
AV
c~f~tJ-9-')1--
e \~~e! ... | https://ocw.mit.edu/courses/8-322-quantum-theory-ii-spring-2003/0ce4f20acbdfa718cd95bd716de2c74b_83227Lecture6.pdf |
2
R. STANLEY, HYPERPLANE ARRANGEMENTS
LECTURE 1
Basic definitions, the intersection poset and the
characteristic polynomial
1.1. Basic definitions
The following notation is used throughout for certain sets of numbers:
N nonnegative integers
P
positive integers
Z
integers
Q
rational numbers
R
real numbers
... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0d0d20fa9004b352a20c85b22d8d6a17_lec1.pdf |
(v1, . . . , vn) =
κivi.
V : κ v = a
·
where κ is a fixed nonzero vector in V and a
v
{
J =
K.
≤
,
}
·
≤
If the equations of the hyperplanes of A are given by L1(x) = a1, . . . , Lm(x) =
am, where x = (x1, . . . , xn) and each Li(x) is a homogeneous linear form, then we
call the polynomial
QA(x) = (L1(x)
a1)... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0d0d20fa9004b352a20c85b22d8d6a17_lec1.pdf |
v
V : v
y = 0
y
Y
.
(1)
codimW (H
W ) = 1
⊕
⊕
⊕
≤
A
}
≤
W : H
A. In other words, H
W is a hyperplane of W , so the set AW :=
for all H
is an essential arrangement in W . Moreover, the arrangements A
H
{
and AW are “essentially the same,” meaning in particular that they have the same
intersection po... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0d0d20fa9004b352a20c85b22d8d6a17_lec1.pdf |
r(A) = #R(A),
the number of regions. For instance, the arrangement A shown below has r(A) = 14.
R(A) is open and convex
It is a simple exercise to show that every region R
(continuing to assume K = R), and hence homeomorphic to the interior of an n-
dimensional ball Bn (Exercise 1). Note that if W is the subspace ... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0d0d20fa9004b352a20c85b22d8d6a17_lec1.pdf |
two (open) components
whose closures are half-spaces. It follows that the closure R of a region R of A is
a finite intersection of half-spaces, i.e., a (convex) polyhedron (of dimension n). A
bounded polyhedron is called a (convex) polytope. Thus if R (or R) is bounded,
then R is a polytope (of dimension n).
for s... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0d0d20fa9004b352a20c85b22d8d6a17_lec1.pdf |
of the hyperplanes
�
=
+ m + 1.
m
2
Bn : xi
xj = 0, 1
i < j
n.
−
→
→
n
2
⎜
�
Thus Bn has
hyperplanes. To count the number of regions when K = R, note
that specifying which side of the hyperplane xi
xj = 0 a point (a1, . . . , an) lies
on is equivalent to specifying whether ai < aj or ai > aj . Hence the nu... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0d0d20fa9004b352a20c85b22d8d6a17_lec1.pdf |
.
}
· · ·
The braid arrangement has a number of “deformations” of considerable interest.
We will just define some of them now and discuss them further later. All these
arrangements lie in K n, and in all of them we take 1
n. The reader who
i < j
→
→
⇔
LECTURE 1. BASIC DEFINITIONS
5
likes a challeng... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0d0d20fa9004b352a20c85b22d8d6a17_lec1.pdf |
a linear arrangement (an arrangement of linear hyperplanes, i.e., hyperplanes
passing through the origin). Many other writers call an arrangement central, rather
H⊆A H, then rank(A) =
than linear, if 0
codim(X). If A is central, then note also that b(A) = 0 [why?].
H⊆A H. If A is central with X =
H⊆A H =
≤
The... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0d0d20fa9004b352a20c85b22d8d6a17_lec1.pdf |
When n = 3, we can draw P 2 as a disk with
antipodal boundary points identified. The circumference of the disk represents the
hyperplane at infinity. This provides a good way to visualize three-dimensional real
linear arrangements. For instance, if A consists of the three coordinate hyperplanes
x1 = 0, x2 = 0, and x3... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0d0d20fa9004b352a20c85b22d8d6a17_lec1.pdf |
URE 1. BASIC DEFINITIONS
7
We now define an operation which is “inverse” to projectivization. Let A be
an (affine) arrangement in K n, given by the equations
L1(x) = a1,
. . . , Lm(x) = am.
Introduce a new coordinate y, and define a central arrangement cA (the cone over
A) in K n
K = K n+1 by the equations
×
L1(x... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0d0d20fa9004b352a20c85b22d8d6a17_lec1.pdf |
y will be
y and y
x, then x = y.
≤
→
[x, y] =
.
}
is not a closed interval. For basic information on posets
P : x
z
{
z
y
→
→
≤
Note that the empty set
�
not covered here, see [18].
Definition 1.1. Let A be an arrangement in V , and let L(A) be the set of all
nonempty intersections of hyperplanes in A,... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0d0d20fa9004b352a20c85b22d8d6a17_lec1.pdf |
its cover relations. The
(Hasse) diagram of a finite poset is obtained by drawing the elements of P as dots,
with x drawn lower than y if x < y, and with an edge between x and y if x � y.
Figure 2 illustrates four arrangements A in R2, with (the diagram of) L(A) drawn
below A.
≤
A chain of length k in a poset P is ... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0d0d20fa9004b352a20c85b22d8d6a17_lec1.pdf |
rank function of
L(A) is given by
= K n
where dim(x) is the dimension of x as an affine subspace of V .
rk(x) = codim(x) = n
dim(x),
−
0
−
−
⊕
Proof. Since L(A) has a unique minimal element ˆ = V , it suffices to show that
(a) if x�y in L(A) then dim(x)
dim(y) = 1, and (b) all maximal elements of L(A)
rank(A). By li... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0d0d20fa9004b352a20c85b22d8d6a17_lec1.pdf |
3. The characteristic polynomial
A poset P is locally finite if every interval [x, y] is finite. Let Int(P ) denote the
Z, write f (x, y) for
set of all closed intervals of P . For a function f : Int(P )
f ([x, y]). We now come to a fundamental invariant of locally finite posets.
∃
Definition 1.2. Let P be a locally ... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0d0d20fa9004b352a20c85b22d8d6a17_lec1.pdf |
I(P ) = I(P, K) denote
the vector space of all functions f : Int(P )
K. Write f (x, y) for f ([x, y]). For
I(P ) by
f, g
I(P ), define the product f g
∃
≤
≤
f g(x, y) =
f (x, z)g(z, y).
It is easy to see that this product makes I(P ) an associative Q-algebra, with mul
tiplicative identity ζ given by
x⊇z⊇y
... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0d0d20fa9004b352a20c85b22d8d6a17_lec1.pdf |
of linear transformations by
∃
K forms a vector space on which I(P )
where f
the statement
≤
K P and �
(�f )(x) =
�(x, y)f (y),
y∗x
�
I(P ). The M¨obius inversion formula is then nothing but
≤
αf = g
√
f = µg.
�
We now come to the main concept of this section.
Definition 1.3. The characteristic polynomi... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0d0d20fa9004b352a20c85b22d8d6a17_lec1.pdf |
a standard
result in enumerative combinatorics with many proofs. We will give here a naive
proof from first principles. Let y
L(A), r(y) = k. We claim that
(4)
≤
µ(y) = (
1)k .
−
The assertion is clearly true for rk(y) = 0, when y = ˆ
show that
0. Now let y > ˆ
0. We need to
(5)
1)rk(x) = 0.
−
(
x⊇y
�
... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0d0d20fa9004b352a20c85b22d8d6a17_lec1.pdf |
MIT OpenCourseWare
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8.512 Theory of Solids II
Spring 2009
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
Lecture 1: Linear Response Theory
Last semester in 8.511, we discussed linear response theory in the context of charge screening and
the fre... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/0d1bb83b0e6471874b6c3e02aff52ad5_MIT8_512s09_lec01_rev2.pdf |
spent most of 8.511 thinking about how to solve for the behavior of a system governed by ˆH.
As interesting as that behavior may be, we will now consider that to be a solved problem. That
is, we will assume the existence of a set of eigenkets {|n�} that diagonalize H with associated
eigenvalues (energies) En.
In ad... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/0d1bb83b0e6471874b6c3e02aff52ad5_MIT8_512s09_lec01_rev2.pdf |
electron density operator, ˆρ (q�), is related to ρˆ(�r)
through the Fourier transforms:
such that
ρˆ(�r) =
�
e q·rρˆ(�q)
i� �
Ψ (�r) =
q
�
�
i� �r
e k· c�
k
�
k
ρˆ(q�) =
�
q·
e−i� �r
r
�
�
c†
c�
�
k−� kq
=
�
k
(1.5)
(1.6)
(1.7)
(1.8)
Equation (1.7) is the first quantized form of ˆρ (�q), and equatio... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/0d1bb83b0e6471874b6c3e02aff52ad5_MIT8_512s09_lec01_rev2.pdf |
to H from the state ket φ (t)� to form an
interaction representation state ket φ˜ (t)�I by
|
|
ˆ
Ht
|φ (t)�I = e i ˆ φ (t)�
˜
|
|φ (t)�I = e−i ˆ |
Ht φ˜ (t)�
(1.10)
(1.11)
Note that in the absence of Vˆ , these interaction picture state kets are actually the Heisenberg
picture state kets of the system. Also, we ... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/0d1bb83b0e6471874b6c3e02aff52ad5_MIT8_512s09_lec01_rev2.pdf |
�) φ˜ (t�)�
ˆ
|
(1.15)
(1.16)
At first it seems like we have not done much to benefit ourselves, since all we have done is
to convert the ordinary Schrodinger equation, a PDE, into an integral equation. However, if VˆI
is small, then we can iterate equation (1.16):
|φ˜ (t)� ≈ |φ0� − i
� t
−∞
dt� VˆI (t�) |φ0� + ·... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/0d1bb83b0e6471874b6c3e02aff52ad5_MIT8_512s09_lec01_rev2.pdf |
our system. After
applying some probe via the external potential Vˆ , we want to measure the response of some
ˆ
observable of the system ˆ
A. We characterize this response through the expectation value of A,
� ˆA�:
ˆ |
� ˆ
A� = �φ (t) A φ (t)�
|
| ˜
= φ˜ (t) AI φ (t)�
ˆ
|
�
(1.18)
(1.19)
The key now is to substit... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/0d1bb83b0e6471874b6c3e02aff52ad5_MIT8_512s09_lec01_rev2.pdf |
t) , VˆI (t�)].
ˆ
These two terms simply come from the two possible terms linear in VˆI arising from the substitu
tions
and
|φ˜ (t)� ≈ |φ0� − i
� t
−∞
dt�VˆI (t�) φ0�
|
�φ˜ (t) | ≈ �φ0| + i
� t
−∞
ˆ
dt�VI (t�) �φ0|
Note that the integration is with respect to t�, since it comes from the expression for φ˜(t)�... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/0d1bb83b0e6471874b6c3e02aff52ad5_MIT8_512s09_lec01_rev2.pdf |
Hamiltonian, so the interaction picture representation of Vˆ is
given by
�
V
�
V
�
VˆI = e i ˆ
Ht
=
=
� � �
�
dr�� ρˆ r�� U r��, t e−i ˆ
Ht
�
�
r��, t
dr�� e i ˆ ρ(r��)e−i ˆ
Ht ˆ
HtU
�
dr�� ρˆI (r��) U r��, t
�
V
Substituting this expression for VˆI back into equation (1.22), we obtain:
�
� t
�δ ˆ
−i... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/0d1bb83b0e6471874b6c3e02aff52ad5_MIT8_512s09_lec01_rev2.pdf |
terms of some observable ˆA at the point �r and time t is modulated by the response
function χ(�r, �r�, t − t�).
Thus from comparing this definition with equation (1.26), we see that
χ(�r, �r�, t − t�) ≡
(1.28)
− i �φ0 [AI (�r, t) , ρˆI (�r�, t�)] φ0� e η(t� −t) θ(t − t�)
| ˆ
|
Note that in equation (1.27) we exten... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/0d1bb83b0e6471874b6c3e02aff52ad5_MIT8_512s09_lec01_rev2.pdf |
I (�r�, t�) by ˆmI (r��, t�) in defi nition (1.28).
1.3 Electron Density Response to an Applied Electric Potential
In this section, we will specialize further to the case where we observe the response of the electron
density to an applied potential that couples to the density. Thus we are picking ˆ = ρˆ.
We begin by ... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/0d1bb83b0e6471874b6c3e02aff52ad5_MIT8_512s09_lec01_rev2.pdf |
��n|
ρˆ(�r�) φ0� e
|
i(En −E0 )t�� −(iω−η)t��
(1.33)
�φ0|ρˆ(�r�)|n��n|ρˆ(�r) φ0� e−i(En −E0 )t�� −(iω−η)t��
|
(1.34)
All of the time dependence has now been brought up into the exponentials, so it is trivial to
perform the integration over time. This yield the spectral representation of χ(�r, r�, ω):
χ(�r, �r�, ω) ... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/0d1bb83b0e6471874b6c3e02aff52ad5_MIT8_512s09_lec01_rev2.pdf |
.36)
(1.37)
ρ(�
where ˆ q) ≡ ˆq is given by equations (1.7) and (1.8) in first quantized or second quantized
notation, respectively.
ρ�
Since the electron density ˆ r) is a real function, we have the important relation
ρ(�
ρˆ−� = ρˆ†
q
�
q
which is a simple consequence of the nature of the Fourier transform. This im... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/0d1bb83b0e6471874b6c3e02aff52ad5_MIT8_512s09_lec01_rev2.pdf |
of the peaks
tells us about the types of excitations being produced. As we will see shortly, it actually turns
out that knowledge of χ��(�
q, ω), can be
reconstructed from χ��(�q, ω) alone.
q, ω) is all we need; the real part of χ(�
q, ω), denoted χ�(�
Sanity Check: Free Fermions
7
1.4 Sanity Check: Free Fermion... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/0d1bb83b0e6471874b6c3e02aff52ad5_MIT8_512s09_lec01_rev2.pdf |
Fermi sea with wave vector �k + q� to jump
into. Thus
� c†
k � q k
k−�
�
�n� ρˆ† φ0� = (1 − f�
k| �|
q
q
k+�
)f�
k
(1.44)
where f� is 1 if the state with momentum �k is occupied in the ground state, and 0 if it is empty.
k
Substituting this in, we get
�
Letting
�
k
Π0(�
q, ω) =
�
(1 − f�
k+�
q
ω − (�� q − �... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/0d1bb83b0e6471874b6c3e02aff52ad5_MIT8_512s09_lec01_rev2.pdf |
define the correlation function
S(�r, t) = �φ0 ρˆH (�r, t)ˆρH (0, 0) φ0�
|
|
(1.49)
S(�r, t) described fluctuations of the electron density across the sample in space and time. Due
to the translational invariance of the sample, we arbitrarily set one of arguments to (r��, t�) = (0, 0)
and observe the density correla... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/0d1bb83b0e6471874b6c3e02aff52ad5_MIT8_512s09_lec01_rev2.pdf |
(�q, ω) is identical to the first (absorptive) term in the expression for the
imaginary part of the response function χ��(�q, ω). This can be restated as the ZeroTemperature
FluctuationDissipation Theorem:
χ��(�
1
q, ω) = −
2
(S(�
q, ω) − S(−�
q, −ω))
(1.54)
This shows that the energy absorbed in a probing exper... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/0d1bb83b0e6471874b6c3e02aff52ad5_MIT8_512s09_lec01_rev2.pdf |
ω)
q, −ω)
χ��(�
Equation (1.59) is simply a statement of the law of detailed balance.
(1.56)
(1.57)
(1.58)
(1.59)
(1.60)
1.6 Measuring S(�q, ω)
It is possible to measure S(�q, ω) directly through scattering experiments. Depending on the
particle density of interest, the scattering can be performed using electro... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/0d1bb83b0e6471874b6c3e02aff52ad5_MIT8_512s09_lec01_rev2.pdf |
regime of weak coupling. Thus we can
apply the first order Born Approximation and Fermi’s Golden Rule to obtain the scattering rate
v� =
r
2πb
Mn
δ(�r)
(1.65)
Wi
f =
→
π
2
¯
h
|�f | ˆ
Hint
|i�|
2δ(Ei − Ef )
(1.66)
We take the initial and final states of our scattering probe to be plane waves ki� and |�kf �,
|�... | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/0d1bb83b0e6471874b6c3e02aff52ad5_MIT8_512s09_lec01_rev2.pdf |
0
(1.70)
(1.71)
Thus the scattering rate for scattering with a momentum transfer �
hω is
related to the correlation function S( �Q, ω) very simply through a scaling by the square of the
�
Qth Fourier component of the interaction potential.
Q and energy loss ¯ | https://ocw.mit.edu/courses/8-512-theory-of-solids-ii-spring-2009/0d1bb83b0e6471874b6c3e02aff52ad5_MIT8_512s09_lec01_rev2.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
8.821 String Theory
Fall 2008
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
8.821 F2008 Lecture 10: CFTs in D > 2
Lecturer: McGreevy
October 11, 2008
In today’s lecture we’ll elaborate on what conformal invariance really... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/0d1bda106dfd2af510daf14b261c56fb_lecture10.pdf |
at different distances. Conformal
symmetry relates these kinds of observables!
To implement a conformal transformation, we need to change the metric via a Weyl rescaling
x → x
gµν → Ω(x)gµν
(2)
which changes physical distances ds2 → Ω(x)ds2 . Now follow this by a coordinate transformation
′ ). This takes us to a ... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/0d1bda106dfd2af510daf14b261c56fb_lecture10.pdf |
=
T µν δgµν
(3)
δgµν = 2λgµν →
δS =
µ2λ
Tµ
(4)
where λ is a constant. Therefore we see that if Tµ
µ = 0 then the theory is scale invariant.
�
In fact, since nothing we said above depended on the fact that λ is a constant, and since our theory
is a local theory, we can actually make the following transformation... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/0d1bda106dfd2af510daf14b261c56fb_lecture10.pdf |
�
(6)
(7)
1.2 Geometric Interpretation
Lastly we’d like to give an alternative geometric interpretation of the conformal group. The con
formal group in Rd,1 is isomorphic to SO(d + 1, 2), the Lorentz group of a space with two extra
dimensions. Suppose this space Rd+1,2 has metric
ηab = diag(− + +... + −)ab
(8)
... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/0d1bda106dfd2af510daf14b261c56fb_lecture10.pdf |
κ is
a redundant variable. Conformal invariants actually are cross ratios of invariants in Rd+1,2, for
example
ζ1 · ζ2ζ3 · ζ4 .
ζ1 · ζ3ζ2 · ζ4
(11)
This perspective was lifted from a paper by Callan. It may be useful if you are in the business of
relating CFTs to theories in higher dimensional spacetimes, but yo... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/0d1bda106dfd2af510daf14b261c56fb_lecture10.pdf |
∂)φ(x).
The Green’s function thus transforms under the action of Dˆ by:
n
δG =
µ
(xi
∂
µ
∂x
i
i=1
�
+ Δi)�
φi(xi)� + �0|Dˆ
φi(xi)|0� + �0|
φi(xi)Dˆ |0�.
(14)
�
�
�
Assuming that the vacuum is conformally invariant, the last two terms above vanish. We will
assume this in the rest of the discussion. N = 4 is... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/0d1bda106dfd2af510daf14b261c56fb_lecture10.pdf |
Assuming φ’s are scalars, let’s constrain the conformal invariants by
using the symmetries of the theory:
• Translation invariance implies I depends only on differences: xi − xj. Therefore there are
d(n − 1) numbers I can depend on.
• Rotational invariance implies I depends only on magnitudes of differences: rij = |x... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/0d1bda106dfd2af510daf14b261c56fb_lecture10.pdf |
scalar primaries. Its spatial dependence must be completely
determined by the logic above:
�φ1(x1)φ2(x2)� = c12f (|r12|) = c12 Δ1+Δ2
1
r12
(17)
1Ginsparg proves why there are n(n − 3)/2 if you are curious.
4
where the first equality follows from translation plus rotation invariance, and the second equality
f... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/0d1bda106dfd2af510daf14b261c56fb_lecture10.pdf |
still have to ‘do’
one. First we note that the two-point function �φ1(x1)φ2(x2)� depends only on |x1 − x2|, so we
can perform a 180 degree rotation to obtain �φ2(x1)φ1(x2)�, which must give the same answer.
Performing a conformal transformation to the two Green’s functions above gives
−Δ1/2 −Δ2/2
Ω2
Ω1
�φ1(x1)φ2... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/0d1bda106dfd2af510daf14b261c56fb_lecture10.pdf |
Using translation and rotation
invariance, we can write the three-point function as
G123 = �φ1(x1)φ2(x2)φ3(x3)� =
Cabc
b
a
r12r23r31
c .
a,b,c
�
(21)
Scale invariance then implies all terms on the RHS transform in the same way as the LHS, so
a + b + c = Δ1 + Δ2 + Δ3. You can then check that special conformal... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/0d1bda106dfd2af510daf14b261c56fb_lecture10.pdf |
etermined function of the two invariants.
�
i<j rij
�
(23)
�
The results in this section are true for D > 1, and actually for D = 2 there are extra symmetries
which allow us to learn something about these undetermined functions of invariants. They’re often
hypergeometric. Also, in principle, there exist similar ex... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/0d1bda106dfd2af510daf14b261c56fb_lecture10.pdf |
this talk makes Penrose
diagrams sound way more informative than they may seem.
Start with the simple example of Euclidean space Rp+1 with metric ds2 = dr2 + r2dΩ2
is the metric on the p-sphere. Define r = tan θ/2. Then the metric is
p where dΩ2
p
ds2(dθ2 + sin2 θdΩ2
p)
1
2 θ/2
4 cos
(24)
1
which is the metri... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/0d1bda106dfd2af510daf14b261c56fb_lecture10.pdf |
diagram of Minkowski space, with angular coordinates on Sp−1 suppressed. The
size of the Sp−1 shrinks at sin θ = 0. The analytical continuation of the finite τ coordinate is
indicated by the dashed lines, and gives the Einstein static universe.
which looks like R × Sp with a conformal factor which blows up at the con... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/0d1bda106dfd2af510daf14b261c56fb_lecture10.pdf |
��
�
SO(p, 1) × SO(1, 1)
����
(26)
• SO(p+1)×SO(2) is obvious in the conformal coordinates we discussed above. Here SO(p+1)
rotates Sp and SO(2) corresponds to τ translations. In Rp+1,2 , SO(p + 1) rotates the spatial
coordinates and SO(2) rotates the time-like coordinates.
SO(2)
(− +......+ −)
�
�
SO(p + 1)
�
... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/0d1bda106dfd2af510daf14b261c56fb_lecture10.pdf |
One lesson from this is that it’s difficult in a CFT to agree on what the Hamiltonian is, because we
have a bunch of charges to pick from. The physics looks different in different coordinates though
they are of course related by conformal transformations.
4 State-Operator Correspondence
Consider a QFT on R × Sp, which ... | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/0d1bda106dfd2af510daf14b261c56fb_lecture10.pdf |
and states. See Fig(2).
3OK, we cheated for the sign of dτ 2 in ds2 but it just means using a slightly different coordinate transformation.
8 | https://ocw.mit.edu/courses/8-821-string-theory-fall-2008/0d1bda106dfd2af510daf14b261c56fb_lecture10.pdf |
6.852: Distributed Algorithms
Fall, 2009
Class 5
Today’s plan
• Review EIG algorithm for Byzantine agreement.
• Number-of-processors lower bound for Byzantine
agreement.
• Connectivity bounds.
• Weak Byzantine agreement.
• Time lower bounds for stopping agreement and
Byzantine agreement.
• Reading: Sections 6.3-6.7,... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/0d1d84b9fd1e74bf8e2f8ad916cfca0d_MIT6_852JF09_lec05.pdf |
23
24
31
32 34
41
42
43
Lies
1
1
0
0
1
1
0
0
1
1
1
0
1
1
1
0
1101
0
1
0
00011
1101
1
1
0
00011
1111
1
1
0
10011
Process 1
Process 2
(Process 3)
Process 4
Example: n = 4, f = 1
• Now calculate newvals, bottom-up, choosing majority
values, v0 = 0 if no majority.
Corrected by taking majority
1
1
1
1
1
1
0
1
1
1
0
1
1
1
... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/0d1d84b9fd1e74bf8e2f8ad916cfca0d_MIT6_852JF09_lec05.pdf |
.
• Termination:
– Obvious.
• Agreement:
Agreement
λ
• Path covering: Subset
of nodes containing at
least one node on each
path from root to leaf:
1
2
3
4
12
13
14
21
23
24
31
32 34
41
42
43
• Common node: One for which all nonfaulty processes
have the same newval.
– All nodes whose labels end in nonfaulty process... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/0d1d84b9fd1e74bf8e2f8ad916cfca0d_MIT6_852JF09_lec05.pdf |
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