text stringlengths 16 3.88k | source stringlengths 60 201 |
|---|---|
n
j=0
ε.
�
χm(a)s
Thus for any s ∈ k, we can define φs : A
=
where χ0
by φs(a) =
m≥0
pairwise distinct homomorphisms.
finite dimensional algebra. We are done.
k((t))
mtm/m!, and we find that φs is a family of
This is a contradiction, as A is a
�
Corollary 1.27.5. If a finite ring category C over a field of charac
teristic zero has a unique simple object 1, then C is equivalent to the
category Vec.
→
59
Corollary 1.27.6. A finite dimensional bialgebra H over a field of
characteristic zero cannot contain nonzero primitive elements.
Proof. Apply Theorem 1.27.4 to the category H − comod and use
�
Proposition 1.27.1.
Remark 1.27.7. Here is a “linear algebra” proof of this corollary. Let
x be a nonzero primitive element of H. Then we have a family of
grouplike elements estx ∈ H((t)), s ∈ k, i.e., an infinite collection of
characters of H ∗((t)), which is impossible, as H is finite dimensional.
Corollary 1.27.8. If H is a finite dimensional commutative Hopf
algebra over an algebraically closed field k of characteristic 0, then
H = Fun(G, k) for a unique finite group G.
Proof. Let | https://ocw.mit.edu/courses/18-769-topics-in-lie-theory-tensor-categories-spring-2009/041cee4c86b5e5af04423e900bcb36fc_MIT18_769S09_lec06.pdf |
0, then
H = Fun(G, k) for a unique finite group G.
Proof. Let G = Spec(H) (a finite group scheme), and x ∈ T1G =
(m/m2)∗ where m is the kernel of the counit. Then x is a linear function
on m. Extend it to H by setting x(1) = 0. Then x s a derivation:
x(f g) = x(f )g(1) + f (1)x(g).
This implies that x is a primitive element in H ∗. So by Corollary
1.27.6, x = 0. this implies that G is reduced at the point 1. By using
translations, we see that G is reduced at all other points. So G is a
�
finite group, and we are done.
Remark 1.27.9. Theorem 1.27.4 and all its corollaries fail in char
acteristic p > 0. A counterexample is provided by the Hopf algebra
k[x]/(xp), where x is a primitive element.
1.28. Pointed tensor categories and pointed Hopf algebras.
Definition 1.28.1. A coalgebra C is pointed if its category of right co
modules is pointed, i.e., if any simple right C-comodule is 1-dimensional.
Remark 1.28.2. A finite dimensional coalgebra C is pointed if and
only if the algebra C ∗ is basic, i.e., the quotient C ∗/Rad(C ∗) of C ∗ by
its radical is commutative. In this case, simple C-comodules are points
of Specm(H ∗/Rad(H ∗)), which justifies the term “pointed”.
Definition 1.28.3. | https://ocw.mit.edu/courses/18-769-topics-in-lie-theory-tensor-categories-spring-2009/041cee4c86b5e5af04423e900bcb36fc_MIT18_769S09_lec06.pdf |
/Rad(H ∗)), which justifies the term “pointed”.
Definition 1.28.3. A tensor category C is pointed if every simple
object of C is invertible.
Thus, the category of right comodules over a Hopf algebra H is
pointed if and only if H is pointed.
Example 1.28.4. The category Vecω is a pointed category. If G is
a p-group and k has characteristic p, then Repk(G) is pointed. Any
cocommutative Hopf algebra, the Taft and Nichols Hopf algebras, as
well as the quantum groups Uq (g) are pointed Hopf algebras.
G
60
1.29. The coradical filtration. Let C be a locally finite abelian cat
egory.
Any object X ∈ C has a canonical filtration
(1.29.1)
0 = X0 ⊂ X1 ⊂ X2 ⊂ · · · ⊂ Xn = X
such that Xi+1/Xi is the socle (i.e., the maximal semisimple subobject)
of X/Xi (in other words, Xi+1/Xi is the sum of all simple subobjects
of X/Xi). This filtration is called the socle filtration, or the coradical
filtration of X. It is easy to show by induction that the coradical
filtration is a filtration of X of the smallest possible length, such that
the successive quotients are semisimple. The length of the coradical
filtration of X is called the Loewy length of X, and denoted Lw(X). | https://ocw.mit.edu/courses/18-769-topics-in-lie-theory-tensor-categories-spring-2009/041cee4c86b5e5af04423e900bcb36fc_MIT18_769S09_lec06.pdf |
ation of X is called the Loewy length of X, and denoted Lw(X).
Then we have a filtration of the category C by Loewy length of objects:
C0 ⊂ C1 ⊂ ..., where Ci denotes the full subcategory of objects of C of
Loewy length ≤ i + 1. Clearly, the Loewy length of any subquotient of
an object X does not exceed the Loewy length of X, so the categories
Ci are closed under taking subquotients.
Definition 1.29.1. The filtration of C by Ci is called the coradical
filtration of C.
If C is endowed with an exact faithful functor F : C → Vec then we
can define the coalgebra C = Coend(F ) and its subcoalgebras Ci =
Coend(F |Ci ), and we have Ci ⊂ Ci+1 and C = ∪iCi (alternatively, we
can say that Ci is spanned by matrix elements of C-comodules F (X),
X ∈ Ci). Thus we have defined an increasing filtration by subcoalgebras
of any coalgebra C. This filtration is called the coradical filtration, and
the term C0 is called the coradical of C.
The “linear algebra” definition of the coradical filtration is as fol
lows. One says that a coalgebra is simple if it does not have nontrivial
subcoalgebras, i.e. if it is finite dimensional, and its dual is a simple
(i.e., matrix) algebra. Then C0 is the sum of all simple subcoalgebras
of C. The coalgebras Cn+1 for n ≥ 1 are then defi | https://ocw.mit.edu/courses/18-769-topics-in-lie-theory-tensor-categories-spring-2009/041cee4c86b5e5af04423e900bcb36fc_MIT18_769S09_lec06.pdf |
of C. The coalgebras Cn+1 for n ≥ 1 are then defined inductively to
be the spaces of those x ∈ C for which
Δ(x) ∈ Cn ⊗ C + C ⊗ C0.
Exercise 1.29.2. (i) Suppose that C is a finite dimensional coalgebra,
and I is the Jacobson radical of C ∗. Show that Cn
⊥ = I n+1 . This
justifies the term “coradical filtration”.
(ii) Show that the coproduct respects the coradical filtration, i.e.
Δ(Cn) ⊂
n
=0 Ci ⊗ Cn−i.
i
�
(iii) Show that C0 is the direct sum of simple subcoalgebras of C.
In particular, grouplike elements of any coalgebra C are linearly inde
pendent.
61
Hint. Simple subcoalgebras of C correspond to finite dimensional
irrreducible representations of C ∗.
Denote by gr(C) the associated graded coalgebra of a coalgebra C
with respect to the coradical filtration. Then gr(C) is a Z+-graded
coalgebra. It is easy to see from Exercise 1.29.2(i) that the coradical
¯
filtration of gr(C) is induced by its grading. A graded coalgebra C
with this property is said to be coradically graded, and a coalgebra C
such that gr(C) = C is called a lifting of C.
¯
Definition 1.29.3. A coalgebra C is said to be cosemisimple if C is a
direct sum | https://ocw.mit.edu/courses/18-769-topics-in-lie-theory-tensor-categories-spring-2009/041cee4c86b5e5af04423e900bcb36fc_MIT18_769S09_lec06.pdf |
��nition 1.29.3. A coalgebra C is said to be cosemisimple if C is a
direct sum of simple subcoalgebras.
Clearly, C is cosemisimple if and only if C − comod is a semisimple
category.
Proposition 1.29.4. (i) A category C is semisimple if and only if
C0 = C1.
(ii) A coalgebra C is cosemisimple if and only if C0 = C1.
Proof. (ii) is a special case of (i), and (i) is clear, since the condition
means that Ext1(X, Y ) = 0 for any simple X, Y , which implies (by
the long exact sequence of cohomology) that Ext1(X, Y ) = 0 for all
�
X, Y ∈ C.
Corollary 1.29.5. (The Taft-Wilson theorem) If C is a pointed coal
gebra, then C0 is spanned by (linearly independent) grouplike elements
g, and C1/C0 = ⊕h,gPrimh,g(C)/k(h − g). In particular, any non
cosemisimple pointed coalgebra contains nontrivial skew-primitive ele
ments.
Proof. The first statement is clear (the linear independence follows from
Exercise 1.29.2(iii). Also, it is clear that any skew-primitive element
is contained in C1. Now, if x ∈ C1, then x is a matrix element of a
C-comodule of Loewy length ≤ 2, so it is a sum of matrix elements 2
dimensional comodules, i.e. of grouplike and skew-primitive elements.
Primh,g(C)/k(h − g) | https://ocw.mit.edu/courses/18-769-topics-in-lie-theory-tensor-categories-spring-2009/041cee4c86b5e5af04423e900bcb36fc_MIT18_769S09_lec06.pdf |
dimensional comodules, i.e. of grouplike and skew-primitive elements.
Primh,g(C)/k(h − g) ⊂ C/C0
is direct. For this, it suffices to consider the case when C is finite
dimensional. Passing to the dual algebra A = C ∗, we see that the
statement is equivalent to the claim that I/I 2 (where I is the radical
of A) is isomorphic (in a natural way) to ⊕g,hExt1(g, h)∗.
It remains to show that the sum
�
h,g
Let pg be a complete system of orthogonal idempotents in A/I 2, such
that h(pg) = δhg. Define a pairing I/I 2 × Ext1(g, h) → k which sends
a ⊗ α to the upper right entry of the 2-by-2 matrix by which a acts
in the extension of g by h defined by α. It is easy to see that this
62
pairing descends to a pairing B : ph(I/I 2)pg × Ext1(g, h) → k. If the
extension α is nontrivial, the upper right entry cannot be zero, so B is
right-nondegenerate. Similarly, if a belongs to the left kernel of B, then
a acts by zero in any A-module of Loewy length 2, so a = 0. Thus, B
�
is left-nondegenerate. This implies the required isomorphism.
Proposition 1.29.6. If C, D are coalgebras, and f : C
D is a
coalgebra homomorphism such that f |C1 is injective, then f is injective.
Proof. One may assume that C and D are finite | https://ocw.mit.edu/courses/18-769-topics-in-lie-theory-tensor-categories-spring-2009/041cee4c86b5e5af04423e900bcb36fc_MIT18_769S09_lec06.pdf |
ive, then f is injective.
Proof. One may assume that C and D are finite dimensional. Then
the statement can be translated into the following statement about
finite dimensional algebras: if A, B are finite dimensional algebras and
B is an algebra homomorphism which descends to a surjective
f : A
homomorphism A B/Rad(B)2, then f is surjective.
→
→
→
To prove this statement, let b ∈ B. Let I = Rad(B). We prove by
induction in n that there exists a ∈ A such that b − f (a) ∈ I n . The
base of induction is clear, so we only need to do the step of induction.
So assume b ∈ I n . We may assume that b = b1...bn, bi ∈ I, and let
ai ∈ A be such that f (ai) = bi modulo I 2 . Let a = a1...an. Then
�
b − f (a) ∈ I n+1, as desired.
Corollary 1.29.7. If H is a Hopf algebra over a field of characteristic
zero, then the natural map ξ : U (Prim(H)) H is injective.
→
Proof. By Proposition 1.29.6, it suffices to check the injectivity of ξ in
degree 1 of the coradical filtration. Thus, it is enough to check that
ξ is injective on primitive elements of U (Prim(H)). But by Exercise
�
1.24.4, all of them lie in Prim(H), as desired.
1.30. Chevalley’s theorem.
Theorem 1.30.1. (Chevalley) Let | https://ocw.mit.edu/courses/18-769-topics-in-lie-theory-tensor-categories-spring-2009/041cee4c86b5e5af04423e900bcb36fc_MIT18_769S09_lec06.pdf |
1.30. Chevalley’s theorem.
Theorem 1.30.1. (Chevalley) Let k be a field of characteristic zero.
Then the tensor product of two simple finite dimensional representa
tions of any group or Lie algebra over k is semisimple.
Proof. Let V be a finite dimensional vector space over a field k (of any
characteristic), and G ⊂ GL(V ) be a Zariski closed subgroup.
Lemma 1.30.2. If V is a completely reducible representation of G,
then G is reductive.
Proof. Let V be a nonzero rational representation of an affine algebraic
group G. Let U be the unipotent radical of G. Let V U ⊂ V be the space
of invariants. Since U is unipotent, V U = 0. So if
V is irreducible, then
V U = V , i.e., U acts trivially. Thus, U acts trivially on any completely
reducible representation of G. So if V is completely reducible and
�
G ⊂ GL(V ), then G is reductive.
�
63
Now let G be any group, and V, W be two finite dimensional irre
ducible representations of G. Let GV , GW be the Zariski closures of
the images of G in GL(V ) and GL(W ), respectively. Then by Lemma
1.30.2 GV , GW are reductive. Let GV W be the Zariski closure of the
image of G in GV × GW . Let U be the unipotent radical of GV W . Let
pV : GV W
GW be the projections. Since pV is
surjective, pV (U ) is a normal unipotent subgroup of GV , so pV | https://ocw.mit.edu/courses/18-769-topics-in-lie-theory-tensor-categories-spring-2009/041cee4c86b5e5af04423e900bcb36fc_MIT18_769S09_lec06.pdf |
projections. Since pV is
surjective, pV (U ) is a normal unipotent subgroup of GV , so pV (U ) = 1.
Similarly, pW (U ) = 1. So U = 1, and GV W is reductive.
GV , pW : GV W
→
→
Let G�
V W be the closure of the image of G in GL(V ⊗ W ). Then
G�
V W is a quotient of GV W , so it is also reductive. Since chark = 0,
this implies that the representation V ⊗ W is completely reducible as
a representation of G�
V W , hence of G.
This proves Chevalley’s theorem for groups. The proof for Lie alge
�
bras is similar.
1.31. Chevalley property.
Definition 1.31.1. A tensor category C is said to have Chevalley prop
erty if the category C0 of semisimple objects of C is a tensor subcategory.
Thus, Chevalley theorem says that the category of finite dimensional
representations of any group or Lie algebra over a field of characteristic
zero has Chevalley property.
Proposition 1.31.2. A pointed tensor category has Chevalley prop
erty.
Proof. Obvious.
�
Proposition 1.31.3. In a tensor category with Chevalley property,
(1.31.1)
Lw(X ⊗ Y ) ≤ Lw(X) + Lw(Y ) − 1.
Thus Ci ⊗ Cj ⊂ Ci+j .
Proof. Let X(i), 0 ≤ i ≤ m, Y (j), 0 ≤ j ≤ n, be the successive
quotients of the coradical filtrations of X, Y . Then Z := X ⊗ Y has
a filtration | https://ocw.mit.edu/courses/18-769-topics-in-lie-theory-tensor-categories-spring-2009/041cee4c86b5e5af04423e900bcb36fc_MIT18_769S09_lec06.pdf |
filtrations of X, Y . Then Z := X ⊗ Y has
a filtration with successive quotients Z(r) = ⊕i+j=rX(i) ⊗ Y (j), 0 ≤
m + n. Because of the Chevalley property, these quotients are
r
�
semisimple. This implies the statement.
≤
Remark 1.31.4. It is clear that the converse to Proposition 1.31.3
holds as well: equation (1.31.3) (for simple X and Y ) implies the
Chevalley property.
Corollary 1.31.5. In a pointed Hopf algebra H, the coradical filtration
is a Hopf algebra filtration, i.e. HiHj ⊂ Hi+j and S(Hi) = Hi, so gr(H)
is a Hopf algebra.
64
In this situation, the Hopf algebra H is said to be a lifting of the
coradically graded Hopf algebra gr(H).
Example 1.31.6. The Taft algebra and the Nichols algebras are corad
ically graded. The associated graded Hopf algebra of Uq (g) is the Hopf
algebra defined by the same relations as Uq (g), except that the commu
tation relation between Ei and Fj is replaced with the condition that Ei
and Fj commute (for all i, j). The same applies to the small quantum
group uq (sl2).
Exercise 1.31.7. Let k be a field of characteristic p, and G a finite
group. Show that the category Repk(G) has Chevalley property if and
only if G has a normal p-Sylow subgroup.
MIT OpenCourseWare
http://ocw.mit.edu
18.769 Topics in Lie Theory: Tensor Categories
Spring | https://ocw.mit.edu/courses/18-769-topics-in-lie-theory-tensor-categories-spring-2009/041cee4c86b5e5af04423e900bcb36fc_MIT18_769S09_lec06.pdf |
p-Sylow subgroup.
MIT OpenCourseWare
http://ocw.mit.edu
18.769 Topics in Lie Theory: Tensor Categories
Spring 2009
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/18-769-topics-in-lie-theory-tensor-categories-spring-2009/041cee4c86b5e5af04423e900bcb36fc_MIT18_769S09_lec06.pdf |
-- The inverse of a matrix
We now consider the pro5lern of the existence of multiplicatiave
inverses for matrices. A t this point, we must take the non-commutativity
of matrix.multiplication into account.Fc;ritis perfectly possible, given
a matrix A, that there exists a matrix B such that A-B equals an
identity matrix, without it following that B - A equals an identity matrix.
Consider the following example:
Example 6. Let A and B be the matrices
r
1-1
0 ] 0
check.
=
-Then A - B = I2 , but B-A # I3 , as you can
I
Definition. Let A be a k by n matrix. A matrix B of size n
by k is called an inverse for A if both of the following equations hold:
A-B = Ik
and
B-A = In .
W e shall prove that if k # n, then it is impossible for both these
equations to hold. Thus only square matrices can have inverses.
We also show that if the matrices aro, square and one of these equations
holds, then the other equation holds as v e l l !
Theorem 13. Let A be a matrix of size k by n. Then A has
an inverse if and only if k = n = rank A. If A has an inverse, that
inverse is unique.
,Proof. Step 1. If B is an n by k matrix, we say B is a
right inverse for A if A-B = Ik . We say B is a leEt inverse for A if
B-A = In .
Q A & a . k
be the rank
that if A h a s a right inverse,
then r = k; ar.d if A has a left inverse, then r = n.
I
of the theorem follows.
Fjrst, suppose B is a right inverse for | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
left inverse, then r = n.
I
of the theorem follows.
Fjrst, suppose B is a right inverse for A . Then A - B = I
follows that the system of equations A.X = C has a solution for arbitrary
k *
It
C, for the vector X = B.C is one such solution, as you can check.
Theorem 6 then implies that r must equal k.
Second, suppose B is a left inverse for A. Then B - A = In . It
follows that the system of equations A - X = -0 has only the trivial solution,
for the equation A - X = g implies that B-(A*X) = 2 , whence X = d .
Nc;w the dimension of the solution space of the system A.X = Q is n - r ;
it follows that n - r = 0.
*
Step 2. Now let A be an n by n matrix of rank n. We show there
is a matrix B such that A.B = In .
i
i
Because the rows of A are independent, the system of equations
A.X = C has a solution for arbitrary C. In particular, it has a solution
when C is one of the unit coordinate vectors Ei in Vn. Let us choose
Bi SG that
for i = l...,n. Then if B is the n by n matrix whose successive
columns are B1,... ,B ' the product A.B equals the matrix whose successive
n
columns are El,...,E
that is, A.B = I,
.
n i
1
Step 3. We show that if A and B are n by n matrices and .
A-B = I, , then BmA = In.
The " i E n part of the theorem follows.
\
Let us note that if we | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
then BmA = In.
The " i E n part of the theorem follows.
\
Let us note that if we apply Step 1 to the case of a square matrix
of size n by n , it says that if such a matrix has either a right
inverse or a left inverse, then its rank must be n.
Now the equation A.B = In says that A has a right inverse and that
B has a left inverse. Hence both A and B must have rank n. Applying
Step 2 to the matrix B, we see that there is a matrix C such that
B.C = I n . Now we compute
The equation B * C = In now becomes B - A = In , as desired.
Step 4. The computation we just made shows that if a matrix has
an inverse, that inverse is unique. Indeed, we just showed that if
B has an left inverse A and a ri~ht inverse C, then A = C.
k:tus state the result proved in Step 3 as a separate theorem:
Theorem 14. If A and B are n by n matrices such that
A - B = I n ' then B . A = I n . El
':
W e now have a t h e o r e t i c a l criterion for t h e e x i s t e n c e
of A .
But how can one f i n d A-Ii n p r a c t i c e ? For
I.
i n s t a n c e , how does one' compute B = d l 'if A is a given
nonsingular 3 by 3 matrix? By Theorem 14,
it will suffice
to find a matrix
such t h a t A . B = 13. But t h i s problem is j u s t t h e sroblem of
s o l v i n g | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
But t h i s problem is j u s t t h e sroblem of
s o l v i n g three systems of l i n e a r equations
Thus the Gauss-Jordan algorithm applies. An efficient way
to apply this aqlgorithm to the computation af A-I is out-
lined on p . 612 of Apostol, which you should read now.
There is a l s o a f o n u l a f o r A-'
t h a t involves
determinants. It is given in the next s e c t i o n .
R~.mark. It remains to consider the question whether the existence
of the inverse of a matrix has any practical significance, or whether it is
of theoretical interest only. In fact, the problem of finding the inverse
of a matrix in an efficient and accurate lay is of great importance in
engineering. One way to explain this is t o note that often in a real-life
situation, one has a fixed matrix A, and one wishes to solve the system
A.X = C repeatedly, for many different values of C. Rather than solving
each one of these systems separately, it is much more efficient to find
the inverse of A, for then the solution X = A".C
can be computed by
sirple matrix multiplication.
Exercises
1. Give conditions on a,b,c,d,e,E such that the matrix
is a right inverse to the matrix A of Example 6. Find two right inverses for A.
2. Let A be a k by n matrix with k < n. Show that A has
no left inverse. *.ow that if A has a right inverse, then that right inverse
is not unique.
3. Let B be an n by k matrix with k 4 n. Show that B has
no right inverse. Show that if B has a left inverse, then that left
inverse is not unique.
-Determinants
The determinant is a function that assigns, to each square matrix
.-
A, | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
that left
inverse is not unique.
-Determinants
The determinant is a function that assigns, to each square matrix
.-
A, a real number. It has certain properties that are expressed in the
following theorem:
Theorem 15. There exists a function that assigns, to each n by
n rtlatrix A , a real number that we denote by det A. It has the following
properties:
(1) If B is the matrix obtained from A by exchanging rows
i and j of A, then det B = - det A.
( 2 ) If B is the matrix obtained form A by replacing row i of A
hy itself plus a scalar multiple of row j (where i # j), then det B = det A .
(3) If B is the matrix obtained from A by multiplying row i
i
of A by the scalar c, then det B = c-det A .
4
If In is the identity matrix, then det In = 1 .
We are going to assume this theorem for the time being, and explore
some of its consequences. We will show, among other things, that these
four properties characterize the determinant function completely. kter
we shall construct a function satisfying these properties.
First we shall explore some consequences of the first three of these
properties. We shall call properties (1)-(3) listed in Theorem 15 the
elementary row properties of the detsrminant function.
Theorem 16. t
f 5e a function that assigns, to each n by n
matri;: A, a real number. Scppose f satisfies the elementary row
properties of the determinant function. Then for every n by n matrix A,
( *)
f(A) = f(In).det A .
This theorem says that any function f that satisfies properties
(I), ( 2 ) , and (3) of Theorem 15 is a scalar multiple of the determinant
function. It also says that if f | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
2 ) , and (3) of Theorem 15 is a scalar multiple of the determinant
function. It also says that if f satisfies property (4)as well, then
E must equal the determinant function. Said differently, there is at
--
most one function that satisfies all four conditions.
-Proof. -St= 1. First we show that if the rows of A are dependent,
then f ( A ) = 0 and det A = 0 . Equation ( * ) then holds trivially in this case.
Let us apply elementary row operations to A to brin~ it to echelon
form B. We need only the first two elementary row operations to do this,
and they change the ~ l u e s of f and of the determinant function by at
most a sign. Therefore it suffices to prove that f ( B ) = 0 and det B = 0.
The last row of B is the zero row, since A has rank less than n. If
we multiply this row by the scalar c, we leave the matrix unchanged, and
hence we leave the values of f and det urlchanged. On the other hand,
this operation multiplies these values by c. Since c is arbitrary, we
conclude that f ( B ) = 0 acd d ~ tB = 0.
Step 2. Now let us consider the case where the rows of A are
independent. Again, we apply elementary row operations to A. Hcwever,
we will do it very carefully, so that the values of f and det do not
change.
A s usual, we begin w i t h t h e first column.
I f a l l
e n t r i e s are zero, nothing remains t o be done with t h i s column.
We move on to consider columns 2,...,n and begin the process again.
Otherwise, w e f i n d a non-zero e n | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
2,...,n and begin the process again.
Otherwise, w e f i n d a non-zero e n t r y i n t h e f i r s t column.
I f necessary, we exchange rows t o bring t h i s entry up t o the
upper left-hand corner: this changes the sign of both the func-
t i o n s f and d e t ,
s o we then m u l t i p l y this r o w by -1 to
change the s i g n s back. Then we add m u l t i p l e s of t h e f i r s t row
t o each of t h e remaining rows s o as t o make a l l t h e remaining
e n t r i e s i n t h e f i r s t column i n t o zeros. By t h e preceding theorem
and i t s c o r o l l a r y , this does n o t change t h e values of e i t h e r f
o r det.
Then w e repeat the process, working w i t h t h e second
column and w i t h rows 2 . n .
The o p e r a t i o n s we a p p l y w i l l
n o t a f f e c t t h e z e r o s w e already have i n column 1.
Ssnce the rows of the original matrix were independent, then we do
not have a zero row at the bottom when we finish, and the "stairsteps"
of the echelon form go wer just one step at a time.
In this case, w e have brought t h e matrix t o a form where a l l of
the e n t r i e s below the main | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
h e matrix t o a form where a l l of
the e n t r i e s below the main diagonal a r e zero.
c a l l e d upper t r i a n g u l a r -form.) Furthermore, all the diagonal
e n t r i e s are non-zero.
Since the values of
f and d e t remain
(This i s what i s
the same i f we r e p l a c e A by this new matrix B ,
it now s u f -
fices t o prove our formula f o r a m a t r i x of the form
where t h e diagonal e n t r i e s a r e non-zero.
St.ep 3. We show that our formula holds for the matrix B. To do
this we continue the Gauss-Jordan elimination process. By adding a multiple
of the last row to the rows above it, then adding multiples of the next-
to-last row to the rows lying above it, and so on, we can bring the matrix to
the form where all the non-diagonal entries vanish. This form is called
diaqonal form. The values of both f and det remain the same if we replace
B by t h i s new matrix C. So now it suffices to prove our
formula for a m a t r i x of the form
0
0
0
.
0
0
..' bnn
. .
(Note that the diagonal entries of B remain unchaaged when
we apply the Gauss-Jordan process to eliminate a11 t h e
non-zero entries above the diagonal. Thus the diagonal
entries of C are the same a s t h o s e o f B.)
WE?multiply. the first row of C by
l/bl
This action m l t iplies the
values of both f and det by a Eactor of l/bll. Then we multiply the ,
second row by l/b22, the third by | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
det by a Eactor of l/bll. Then we multiply the ,
second row by l/b22, the third by l/b33, and so on. By this process,
we transform the matrix C into the identity matrix In. We conclude that
det In
(l/bll)...( l/bNI) det C.
and
Since det In = 1 by hypothesis, it follows from the second equation that
det C = bll b22 ... bnn '
Then it follows from the first equation that
E(C) = f(In)- det C,
as desired. a
Besides proving the determinant function unique, this theorem also
tells us one way to compute determinants. O r d applies this version
of the Gauss-Jordan algorithm to reduce the matrix to
echelon form. If the matrix that results has a zero row, then the
determinant is zero. Otherwise, the matrix that results is in upper triangular
form with non-zero diagonal entries, and the determinant is the product
of the diagonal entries.
.-
, :- -lz .
, .:
.....
<,
'.,;;
:,-::= , - .
.. ..... !,
c::.=
'-'2. 2 cr:. , ,l
-
.). . - : * . . .
.. . <,,, ,
The proof of this theorem tells us something else: If the rows of
A are not independent, then det A = 0, while if they are independent,
then det A # 0. We state this result as a theorem:
Theorem 16. Let A be an n by n matrix. Then A ha.s rank n
if and only i f det A # 0 .
An n by n matrix A fur which det A # 0 is said to be non-sinqular .
This theorem tells us that A has rank n if and only if A is non-singular.
Now we prove a totally unexpected result:
Theorem 17. L e t A m d B k n by | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
.
Now we prove a totally unexpected result:
Theorem 17. L e t A m d B k n by n matrices. Then
det (A-B) = (det A). (det B) .
Proof. This theorem is almost impossible to prove by direct computation.
Try the case n = 2 if you doubt me ! Instead, we proceed in another direction:
Let B be a fixed n by n matrix. Let us define a function f of
n by n matrices by the formula
f(A) = det(A9~).
We shall prove that f satisfies the elementary row properties of the
determinant function. From this it follows that
which means that
f(A) = f(In)- det A ,
det(A*B) = det(In.B)* det A
= det B . det A ,
and the theorem is proved.
First, let us note that if A1, ...,A
n
are the rows of A, considered
multiplication) the row matrices A .B,...,A;B
as row matrices, then the rows of A-B are (by the definition of matrix
. Now exchanging rows
has the effect of exchanging rows
1
i and j of A, namely Aj arid A
j'
i and j of A.B. Thus this operation changes the value of f by a
factor of -1. Similarly, replacing the ithrow Ai of A by Ai + FAI
has the effect on A-B of replacing its ith row Ai.B by
( A ~t CA.).B = A ~ . Bt c A , - B
3
3
Hence it leaves the value of f unchanged. Finally, replacing the ith row
Ai of A by cAi h s the effect on A.B of replacing the ith row Ai.B
= (row i of A . B ) + c(row j of A - B ) .
by
(cAi)- | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
(row i of A . B ) + c(row j of A - B ) .
by
(cAi)-B = c (Ai-B) = c (row i of A-B).
Hence it multiplies the value of f by c.
The determinant function has many further properties, which we shall
not explore here. (One reference book on determinants runs to four volumes!)
We shall derive just one additional result, concerning the inverse matrix.
-Exercises
-
1. Suppose that f satisfies Lhe elementary row properties of
the determinant function.
suppose also t h a t x, y, z are numberssuch t h a t
mrnpute the value o f f for each o f the following matrices:
2. L e t f be the Function of Esercise 1. Calculate f(In). Express
E i n terms of the determinant function.
7 . Compute the determinant of the following matrix, using Gauss-
4
Jordan elimination.
4 . Determine whether the following sets o f vectors are l i n e a r l y
independent, using determinants .,
(a) Al = ( l , - l , O ) r % = ( O f l f - l l r
( b ) q = ( 1 , - 1 . 2 , 1 ) , A.2 = ( - l t 2 , - 1 , O ) f
A3. = ( 2 t 3 f - 1 ) 9
A 3 = ( 3 t - l t l t Q ) t
A4 = . ( 1 , 0 , 0 , 1 )
(c) P, = ( ~ . O , O ~ , O , ~ ) ,
42 = ( I ~ ~ ~ o ~ o ~ o )
A3 = ( I ~ O ~ ' ~ ~ O , ~ )
,
A4 = ( l . l f O , l t l ) )
(dl | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
I ~ O ~ ' ~ ~ O , ~ )
,
A4 = ( l . l f O , l t l ) )
(dl q = (1,-11,
A5 = (1,010,010) .
A2 = ( O t l ) ,
A3 = ( l f l ) .
4
i ' j
-A formula for A-
1
We know that an n by n matrix A has an inverse if and only if
it has rank n, and we b o w that A has rank n if and only if
det A # 0. Now we derive a formula for the inverse that involves determinants
directly.
We begin with a lema about the evaluation of determinants.
Lemma -- 18. Given the row matrix [ a
... an] , let us define a
function f of (n-1) by (n-1) mtrtrices B by- the formula
f(B) = det
,I
where B1 consists of the first j-1 columns of B, arid B
2
cc~nsists
of the remainder of B. Then
Proof. You can readily check that f satisfies properties ( 1)-( 3)
of the determinant function. Hence- f(B)= f(I ) -det B.
n- 1
W compute
f(In) = det
n-j
where the large zeros stand for zero matrices of the appropriate size.
A sequence of .j-1 interchanges of adjacent rows gives us the equation
One can apply elementary operations to this matrix, without changing the
value of the determinant, to replace all of the entries al,...,aj-l,aj+l,...,a
n
by zeros. Then the resulting matrix is in diagonal form. We conclude that
Corollary
Consider an n by n mtrix of the form
where Bl,...,B4 are matrices of appropriate size. Then
det A = (-1)
Proof. A sequence of i-1 interchanges of adjacent rows wilt king
the matrix A to the | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
= (-1)
Proof. A sequence of i-1 interchanges of adjacent rows wilt king
the matrix A to the form given in the preceding l m a . a
Definition. In general, if A is an n by n matrix, then the
matrix of size (n-1) by (n-1) obtained by deleting the ith row and
the j th column of A is called the (i,j)-minor of A, and is denoted Ai j.
I
me preceding corollary can then be restated as follows:
[
Corollary 20. If all the entries in the jth column of A are zero except
for the entry a i j in row i, then det A = (-1)
aij-detAij.
i+j
~ t ~ e
n w r ( - 1) i + j
det Aij
that appears in this corollary is also
given a special name. It is called the (i,j)-cofactor of A. Note that
the signs (-l)itj follows the pattern
Nc;w we derive our formula for A - ~ .
Tl~eorern21. k t A be an n by n matrix with det A # 0.
1
If A * B = I
n'
then
bi j
= (-l)jci det A . ./det A .
3 1
(That is, the entry of B in row i and column j equals the ( j,i)-
cofactor of A, divided by det A. This theorem says that you can compute
B by computing det A and the determinants of n2 different (n-1) by
(n-1) matrices. This is certainly not a practical procedure except in
low dimensions !)
Proof. k t X denote the jth col-
of B. Then xi -
- 'ij.
the column matrix X satisfies the equation
Because A-B = I
n'
e | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
. Then xi -
- 'ij.
the column matrix X satisfies the equation
Because A-B = I
n'
e ere E is the .column matrix consisting of zeros except for an entry
Furthermore, if Ri denote the ith mlumn of A, then ,
j
of 1 in row j.)
I
because A - In = A , we1 have the equation
~ - ( i ~ ~
column of In) = A.Ei = A , .
1
Now we introduce a couple of weird matrices for reasons that will become
clear. Using the two preceding equations, we put them together to get
the following matrix equation:
It turns out that when we take determinants of both sides of this equation,
we get exactly the equation of our theorem! First, we show that
det [El ... Ei-l X Ei+l ... En] = xi .
Written out in full, this equation states that
det
If x. = 0, this equation holds because the matrix has a zero row. If
1
xi # 0, we can by elementary operations replace all the ectries above
and beneath xi in its column by zeros. The resulting matrix will k
in diagonal form, and its determinant will be xi.
TFus the determinant of the left side of sqxation ( * ) equals (det A) .xi,
which equals (det A)*bij. We now compute the determinant of the right
side of equation ( * ) . Corollary 20
applies, because the ith column of this matrix consistsof zeros except for
an entry of 1 in row j . Thus the right side of ( * ) equals (-l)jti times
the determinant of the matrix obtained by deleting raw j and column i.
This is exactly the same matrix as we would obtain by deleting rcw j and
column i of A. Hence the right side of ( * ) equals (-l)jti det R . . ,
J 1
and our theorem is proved. a
-Rc;mark 1 | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
(-l)jti det R . . ,
J 1
and our theorem is proved. a
-Rc;mark 1. If A is a matrix with general entry a i
in
row i and colrmn j , then the transpose of A (denoted
is the matrix
.
whose entry In row i and column j is a , ,
J 1
Thus i f A has size K by n, then A'=
has sire n by k; it
can be pictured as the mtrix obtained by flipping A around the line
y = -x. For example,
Of murse,i£ A is square, then the transpose of A has the same dimensions
as A.
Using this terminology, the theorem just proved says that the inverse of
A can be computed by the following four-step process:
(1) Fcrm the matrix whose entry in row i and column j is the
nr*r
det *ij.
(This is called the matrix zf minor determinants.)
( 2 ) Prefix the sign (-1)
each entry of the matrix. (This is called the matrix -of cofactors.)
to the entry in row i and column j, for
(5) Transpose the resulting matrix.
( 4 ) Divide each entry of the matrix by det A.
In short, this theorem says that
det A
This formula for A ' ~ is used for rornputational purposes only for 2 by 2
A-1 = -(cof A ) ~ ~ .
or 3 by 3 matrices; the work simply gets too great otherwise. But it is
important for theoretical purposes. For instance, if the entries of A
are contin~ous functions of a parameter t, this theorem tells us that
the e ~ t r i e sof A-'
are also continuous functions of t, provided det A
is never zero.
R ~ m r k2. This formula does have one practical consequence of great
importance. It tells us that i f deb A | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
~ m r k2. This formula does have one practical consequence of great
importance. It tells us that i f deb A is small as cunpared with the entries
of A, then a small change in the.entries of A is likely to result in a
large change in the ccmputed entries of A-I.This means, in an engineering problem ,
that a small error in calculating A (even round-off error) may result in a
gross error in the calculated value of A-l.A matrix for which d e t A is
relatively small is saidtobeill-conditioned. If such a'mtrix arises in practice,
one usually tries to refomlate the problem to avoid dealing with such a matrix.
(
Exercises
\
I
use t h e formula f o r A
to f i n d the inverses of the follow-
ing matrices , assuming the usual definition of the deter-,inant in LOW
dirnens ions.
a L 3
(b) c 0 c : d n , a s s m F n g ace f O .
2 . L e t A ba a square matrix all of whose entries are integers.
show that i f dat A = 2 1 ,
then all the entries of A-'
are
integers.
3 . Consider the matrices A,B,C,D,E
of p. A.23. Which of these
matrices have inverses?
4. Consider the following matrix function!
For what values of t does A-'
exist? Give a formula for A-Iin t e r m
5 . Show that the conclusion of Theorem20 holds if A has an entry
in row i and calm j, =d all the other entries in row i equal 0.
Or ail
'b.
Theorem L e t A, B I C be matricas of .size lc by k, and
m
by I(, and m b | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
C be matricas of .size lc by k, and
m
by I(, and m b l mr respectively.
Then
\ rB
z]
d
= (det A ) (det (2).
(Here 0 is the zero matrix of appropriate size.)
Prwf. k t B and C ke fixed. Fcr each k by k matrix
A, define
( a ) Show
f satisfies the elementary row properties of the determinant
function.
(b) U s e Exercise 5 to show that f( I ~ )= det C.
( c ) Cctrrplete the proof.
\
Ccnstruction of the determinant when n 5 3.
, T'i~e actual definition of the determinant function is t're least interesting
part of this entire discussion. The situation is similar to the situation
with respect to the functions sin x, cos x, and ex. You trill recall that
their actual definitions (as limits of Pbrseries) were not nearly as interesting
0
as the properties we derived from simple basic assumptions about them.
We first consider the case where n < 3, which isdoubtless familiar
to you. This case is in fact all we shall need for our applications to calculus.
We begin with a lema:
Lemma 21. Let f ( A ) be a real-valued function of n by n matrices.
Suppose that:
( i ) Exchanging any two rows of A changes the mlue of f by a factor
of -1.
( i i ) For each i , £ is linear as a function of t h e ith row.
Then f satisfies the elementary row properties of the determinant function.
Proof.
By hypothesis, f satisfies the first elementary row property.
We check the other tu'0.
Let A,, ...,A,
be t h e rows of A. To say that E is linear
as a function of row f alone is to say t h a t | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
of A. To say that E is linear
as a function of row f alone is to say t h a t (when f is written as a,function
of the rows of A):
( * ) f(A ,,..., cx + dY, ... ,An)
=
cf(A1,...,X,. .., n + dE(A1,...,Y,...,A,,).
A )
where cX + dY and X and Y appear in the ith component.
The special case d = 0 tells us that multiplying the ith row
of A by c has the effect of multiplying the mlue of f by c.
We now consider the third type of elementary operation.
Suppose that B is the matrix obtained by replacing row i of A 5 y
itself plus c times row j. We then compute (assuming j > i for
convenience in notation),
The second term vanishes, since two rows are the same. (Exchanging them does
not change the matrix, but by Step 1 it changes the value of f by a factor
of - 1 . ) ( J
Definition. We define
det [a]
= a.
det I 1
bl b2 1 =
L
a l b ~- a2bl.
q,eorem 22. The preceding definitions satisfy the four conditions
of the determinant f uncti on.
-Proof. The fact that the determinant of the identity matrix is 1
follows by direct computation. . It then suffices to check that (i) and (ii)
of the preceding theorem hold .
Irl the 2 by 2 case, exchanging rows leads to the determinant bla2- b2al ,
which is the negative of what is given.
In the 3 by 3 case, the fact that exchanging the last two rows changes the
sign of the determinant follows from the.2 by 2 case. The fact t h a t exchanging
the first two rows also changes the sign follows similarly if we rewrite the
formula defining. t | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
case. The fact t h a t exchanging
the first two rows also changes the sign follows similarly if we rewrite the
formula defining. t h e determinant in the form
Finally, exchanging rows 1 and 3 can be accomplished by three exchanges of
adjacent rows [ namely, (A,B,C)
--> (A,C,B)-9 (C,A,B) -3 (C,B,A) 1, so it changes
the sign of the determinant.
To check (ii) is easy. Consider the 3 by 3 case, for example. We
h o w that any function o f the form
E(X) =
[ a b c ] - X = axl + b w + a
3
2
is linear, where X is a vector in V
3
The function
f ( X ) = d e t
has this form, where the coefficients a, b, and c involve the constants
bi and c
Hence f is linear as a function of the first row.
j *
The "row-exchange propertyw then implies that E is linear as a function
of each of the other rows. 0
Exercise
*l. Let us define
det
(a) Show that det Iq = 1.
(b) Show that excha 7 'ng any two of the last three rows changes the sign of the
determinant.
(c) Shwthat exchanging the first two rows changes the sign. [Hint: Write the
expression as a sum of terms involving det pi
I
Pi 'j7.
bjJ
(d)
Show that exchanging any two rows changes the sign.
(e) Show that det is linear as a function of the first row.
(f) Conclude that det is linear as a function of the ith row.
( g ) Conclude that this formula satisfies all the properties of the determinant
function.
Construction of fhe Determinant unction^^ Suppose we take the posi-
tive integers 1, 2, . . . , k and write them down in some arbitrary order,
say jl | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
ive integers 1, 2, . . . , k and write them down in some arbitrary order,
say jl, j z , . . . , j h . This new ordering is called a permutation of these
integers. For each integer ji in this ordering, let us count how many
integers follow it in this ordering, but precede it in the natural ordering
1, 2, . . . , k. This number is called the number of inaersions caused by the
integer j;. If we determine this ilumber for each integer ji in the ordering
and add the results together, the number we get is called the total number
of inversions which occur in this ordering. If the number is odd, we say
the permutation is an odd permutation; if the number is even, we say it is
an even permutalion.
For example, consider the following reordering of the integers between
1 and 6:
2, 5 , 1, 3, 6, 4.
If me count up the inversions, we see that the integer 2 causes one inver-
sion, 5 causes three inversions, 1 and 3 cause no inversions, 6 causes one
inversion, and 4 causes none. The sum is five, so the permutation is odd.
Xf a permutation is odd, me say the sign of that permutation is - ; if
it is even, we say its sign is +. A useful fact about the sign of a permuta-
tion is the following:
Theorem 22.If we interchange two adjacent elements of a per-
mutation, we %hange the sign of the permutation.
ProoJ. Let us suppose the elements ji and ji+1 of the permutation
jl, . . .' , ji, ji+l, . . . , j k are the two we interchange, obtaining the permu-
tation
j ~ ,. . . ) | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
+l, . . . , j k are the two we interchange, obtaining the permu-
tation
j ~ ,. . . ) j+l,j ~ ,. . . , jk.
The number of inversions caused by the integen j l , . . . , ji-1 clearly is
the same in the new permutation as in the old one, and so is the number
of inversions caused by ji+t, . . . , js. I t remains to compate the number of
inversions caused by ji+land by ji in the two permutations.
Case I: j r precedes j,-+lin the natural ordering 1, . . . , k. I n this case,
the number of inversions caused by ji is the same in both permutations,
but the number of inversions caused by ji+lia one larger in the second
permutation than in the first, for ji follows j4+*in the second permutation,
but not in the first. Hence the total number of inversions is increased by
one.
Case 11: j i follows js+l in the natural ordering 1, . . . , k. I n this case,
the number of inversion caused by jiclis the same in both permutations,
but the number of inversions caused by ji is one less in the second permu-
tation than in the first.
I n either case the total number of inversions changes by one, 80 t h a t the
sign of the permutation changes. U
EXAMPLE.If we interchange the second and third elements of the
permutation considered in the previous example, we obtain 2, 1, 5, 3, 6, 4,
in which the total number of inversions is four, so the permutation is even.
Definition. Consider a k by k matrix
Pick out one entry from each row of A ; do this in such a way that these
entries all lie in different columns of A . Take the product of these entries,
and prefix a + sign according | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
that these
entries all lie in different columns of A . Take the product of these entries,
and prefix a + sign according as the permutation jl,. . . , j k is even or
odd. (Note that we arrange the entries in the order of the rows they come
from, and then"we compute the sign of the resulting permutation of the
column indices.)
If we write down all possible such expressions and add them together,
the number we get is defined to be the determinant of A .
REMARK.We apply this definition to the general 2 by 2 matrix, and
obtain the formula
If we apply i t to a 3 by 3 matrix, we find that
The formula for the determinant of a 4 by 4 matrix involves 24 terms,
and for a 5 by 5 matrix it involves 120 terms; we will not write down these
formulas. The reader will readily believe that the definition we have
given is not very useful for computational purposes!
The definition is, however, very convenient for theoretical purposes.
Theorem 24. The determinant of the identity matrix is 1.
. ,,:!,
-"
Proof. Every term in the expansion of det In has a factor .
of zero in it except for the term alla22...alck, and this term equals 1.
Theorem $5. If A ' is obtained from A by interchanging rows
/
i and i+l, then det A ' = - det A.
Proof. Note that each term
in the expansion of det A' also appears in the expansion of det A, because
we make all possible choices of one eritry from each row and column when
we write down this expansion. The only thing we have to do is to compare
this term has when i t appears in the two expansions.
what signs
-
Let a~r, - . n i j r a i f l , j , + ,. . . nkjk be a term | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
Let a~r, - . n i j r a i f l , j , + ,. . . nkjk be a term in the expansion of det A .
If we look a t the correspondi~ig term in the expansion of det A', we see
that we have the same factors, but they are arranged diflerenlly. For to
compute the sign of this term, we agreed to arrange the entries in the
order of the row8 they came from, and then to take the sign of the cor-
respondiilg permutation of the column indices. Thus in the expansion of
det A', this term mill appear as
l'lie permutation of the columr~ indices here is the same as above except
that eleme~lts ji and j i + l have been interchanged. By Theorem 8.4, this
means that this term appears in the expansion of det A' with the sign
opposite to its sign ill the expansion of det A .
Since this result holds for each term in the expansion of det A', we have
det A' = - det A . -
. 0
Theorem 26'. The function det is linear as a function of the ith row.
--
Proof. Suppose we take the constant matrix A, and replace its i
. When we take the determinant of this
row by the row vector [xl ... A\]
new matrix, each term in the expression equals a constant times x , for
th
j
some j. (This happens because in forming this term, we picked out exactly one
entry from each row of A . ) Thus this function is a linear combination
of th2 components xi; that is, it has the form
LC, .... c l . x
k
, for some constants c
i ' a
Exercises
1, Use Theorem 2s.' to show that exchanging | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
, for some constants c
i ' a
Exercises
1, Use Theorem 2s.' to show that exchanging
two rows of A
changes the sign of the determinant.
2. Consider the term
all; a2j2 "' %jk
in the definition of
the determinant.
(The integers j l , j2, . . . , j k are distinct.) Suppose
we arrange the factors in this term in the order of their column indices,
obtaining an expression of the form
Show that the sign of the perm-tion
il,i2,...,ik equals the sign of the
permutation jl,j2,...,jk
Ccnclude that det
= det A in general.
3.' k t A be an n by n matrix, with general entry aij in
row i and column j.
Let m be a fixed index. Show that
Here A
mj
denotes , as usual, the (m,j )-minor of A. This formula is
called theMformula for expanding det A according to the cofactors o f
the nth row.
[Hint: Write the mth row as the sum of n vectors, each
of which has a single non-zero component. Then use the fact that the
determinant function is linear as a function of the mth row. 1
-Tl~ecross-product 2 V
3
If A = (alI a2' a 3 ) and B = (blI bZI b ) are vectors in
3
v3'
we define their cross product r2 be the vector
a.] , - det 2) det[bl bd)
A X B = (det
al a2
2 3
We shall describe the geometric significance of this product shortly.
But first, we prove some properties of the cross product:
Theorem 27. Fc'r all vectors A, B in V3, we have
( a ) B % A =
- A x B .
(11) Ar(B + C) = A X B + | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
a ) B % A =
- A x B .
(11) Ar(B + C) = A X B + A X C ,
( E i + C ) X A = B k A + C % A .
(CA)X B = C ( A X B ) = AX(CB) .
(d) A X B is orthogonal to both A and B.
( c )
Proof. (a) follows becauseexhanging two rows of a determinant
ck-anges the sign; ar:d (b) and (c) follows because the determinant is linear
as a function of each row separately. To prove ( d ) , Lie note that if
C = (cII c2, c3) , then
fcl
'31
by definition of the determinant. It; follows that A-(AxB) = B - ( A X B ) = 0
because the determinant vanishes if two rows are equal. The only proof
that requires some work is (e). For this, we recall that
(a + bl2 = a2 + b2 + 2ab, and (a + b + c12 = a2 + b2 + c2 c 2ab + 2ac + 2bc .
Equatiori ( e ) can be written in the form
We first take the squared terms on the left side and show they equal
the right side. Then we take the "mixed" terms on the left side and show
they equal zero. The squared terms on the left side are
which equals the right side,
x3
(a,b.) .
2
i,j = 1 1 3
The mixed terms on the left side are
In the process of proving the previous theorem, we proved also
the following:
Theorem 2.8:. Given A, B, C , we have A- (Bx C) = ( A xB).C.
--Proof. This fbllows from the fact that
&finition. The ordered 3-tuple of independent vectors (A,B,C) | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
. This fbllows from the fact that
&finition. The ordered 3-tuple of independent vectors (A,B,C)
of vectors of V3 is called a positive triple if
A-(B% C) > 0. Otherwise, it is called a neqative triple. A positive
triple is sometimes said to be a riqht-handed triple, and a negative one
is said to be left-handed.
The reason for this.terminology is the following: (1) the triple
(i,i, &) is a positive triple, since i - ( d x k ) = det I3 = 1 , and
(2) if we draw the vectors i,i, and & in V3 in the usual way,
and if one curls the fingers of one's right hand in the direction from the
first to the second, then one's thumb points in the direction of the
third.
n
&" 1
j-
>
Furthermore, if one now moves the vectors around in V
3'
perhaps changing their
lengths and the angles between them, but never lettinq them become dependent,
=d if one moves one's right hand around correspondingly, then the
fingers still correspond. to the new triple (A,B,C) in the same way, and
this new triple is still a positive triple, since the determinant cannot
have changed sign while the vectors moved around.(Since they did not become
dependent, the determinant did not vanish.)
Theorem .29. Let A and B be vectors in V
3'
If A: and: B
are dependent, then A X B = -0. Otherwise, AXB is the unique vector
orthogonal to both A and B having length llA ll IlBll sin €3
(where 8
is the angle between A and B), such that the triple (A,B,A%B)
forms a positive (i.e.,right-handed) triple.
ProoF. We know that A X B is orthogonal to both A and B. \v'e
also have
II~xaU2 = ~ I A | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
is orthogonal to both A and B. \v'e
also have
II~xaU2 = ~ I A C - I I B ~ ?-
~ ( 1 1-
\ ( A [ \ ~ .~
=
(A.B) 2
2
€I~
ros 1
1 =
1 ~
1
~ ( B A ~ ,
1
s i n 2 e .
~
Finally, i f C = A X B , then ( A , B , C ) is a positive t r i p l e , since
Polar coordinates
Let A = (a,b) be a point of V2 different from Q. We wish to define what we mean
by a "polar angle" for A. The idea is that it should be the angle between the vector A
and the unit vector i = (1,O). But we also wish to choose it so its value reflects whether
A lies in the upper or lower hdf-plane. So we make the following definition:
Definition. Given A = (a,b) # Q. We define the number
8 = * arcos (Aei/llAII)
(*I
t o be a polar annle for A, where the sign in this equation is specified to be + if b > 0,
and to be -if b < 0. Any number of the form 2mn + 0 is also defined to be a polar angle
for A.
If b = 0, the sign in this equation is not determined, but that does not matter. For
if A = (a,O) where a > 0, then arccos (A-i/llAII) = arccos 1 = 0, SO the sign does not
matter. And if A = (-a,O) where a > 0, then arccos (A.L/IIAII) = arccos (-1) = T . Since
the two numbers + T and - T differ by a multiple of 27, the | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
(-1) = T . Since
the two numbers + T and - T differ by a multiple of 27, the sign does not matter, for
since one is a polar angle for A, so is the other.
Note: The polar angle B for A is uniauely determined if we require -n < f? < T.
But that is a rather artificial restriction.
Theorem.
A = (a,b) # Q
-a polar annle for A. Then
2
a ~ o i n t of V2. Lgt r = (a +b ) = IIAll; l a 8
2
' I 2
A = (r cos 8, r sin 8).
Proof. If A = (a,O) with a > 0, then r = a and 0 = 0 + 2ms; hence
r cos 8 = a and r sin 0 = 0.
If A = (-a,O) with a > 0, then r = a and 0 = a +2m7r, so that
r cos 0 = -a and r sin B = 0.
Finally, suppose A = (a,b) with b f 0. Then A.C/I(AII = a/r, so that
0 = 2ma e arccos(a/r).
Then
Furthermore,
so
a/r = cos(*(&2rn~))= cos 0, or a = r cos 0.
b2 = r2 a2 = r (l-cos 8) = r sin 8,
-
2
2
2
2
b = *r sin 8.
We show that in fact b = r sin 8. For if b > 0,then 0 = 2ma + arccos(a/r), so that
and sin 19 is positive. Because b, r, and sin 0 are all positive, we must have b = r sin B | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
positive. Because b, r, and sin 0 are all positive, we must have b = r sin B
rather than b = -r sin 8.
On the other hand, if b < 0, then 0 = 2m7r - arccos(a/r), so that
2 m n - a < B < 2 m a
and sin 0 is negative. Since r is positive, and b and sin 8 are negative, we must have
b = r sin 0 rather than b = -r sin 8. o
( ~ l a n e t a r vMotion (
In the text, Apostol shows how Kepler's three (empirical) laws of planetary motion
can be deduced from the following two laws:
(1) Newton's second law of motion: F = ma.
(2) Newton's law of universal gravitation:
Here m, M are the masses of the two objects, r is the distance between them, and G is a
universal constant.
Here we show (essentially) the reverse-how
Newton's laws can be deduced from
Kepler's.
" . , . k m L M
More precisely, suppose a planet
xy plane with the
origin. Newton's laws tell us that the acceleration of P is given by the equation
That is, Newton's laws tell us that there is a number A such that
X
! ! = - 7 " r ,
r
and that X is the same for all pIanets in the solar system. (One needs to consider other
systems to see that A involves the mass of the sun.)
This is what we shall prove. We use the formula for acceleration in polar
coordinates (Apostol, p. 542):
We also use some facts about area that we shall not actually prove until Units VI and
VII of this course.
Theorem. S u ~ ~ o s | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
shall not actually prove until Units VI and
VII of this course.
Theorem. S u ~ ~ o s e
a la net P moves in the xy plane with the sun at the orign.
(a) Ke~ler's second law im~lies that the acceleration is radial.
(b) Ke~ler's first and second laws imply that
-a,=--
A~
2 4 ,
I
\I
where Xp ~ a number that mav depend on the particular planet P.
(c) Keder's three laws i m ~ l v that Xp is the same for d la nets.
Proof. (a) We use the following formula for the area swept out by the radial
vector as the planet moves from polar
angIe Q1 to polar angle 02:
Here it is assumed the curve is specified by giving r as a function of 0.
Now in our present case both 0 and r are functions of time t. Hence the area swept
out as time goes from to to t is (by the substitution rule) given by
Differentiating, we have = I
dA 1 2 dB
,which is constant by Kepler's second law. That
is,
(*I
for some K.
Differentiating, we have
The left side of this equation is just the transverse component (the po component) of a!
Hence a is radial.
(b) To apply Kepler's first law, we need the equation of an ellipse with focus at
the origin.
We put the other focus at (a,O), and use
the fact that an ellipse is the locus of all
, - ,
points (x,y) the sum of whose distances
from (0,O) and (a,O) is a constant b > a.
The algebra is routine:
or
r + Jr2 - 2a(r cos 0) + a'= b,
r2 - 2a(r cos 8) +
a2 = (b-r)2 = | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
a'= b,
r2 - 2a(r cos 8) +
a2 = (b-r)2 = b - 2br + 2 r ,
2
2br -
2ar cos 0 = b - a ,
2
2
r = -
1
C
e cos 0
, where c = -26-. and e
b2 - a2
= a/b.
i
+
e
(The numberhis called the eccentricity of the ellipse, by the way.) Now we compute the
radial component of acceleration, which is
Differentiating (**), we compute
Simplifying,
-1
(1-e cos 0)
dr = a1(-1)r 2(e sin e)=dB
Then using (*) from p. B60, we have
dr = $e1 sin 6')K.
Differentiating again, we have
, d2r
d0
= - -(e cos O)zK,
1
C
dt
d2r
1
-Z = -&e cos 8)
d t
K , using (*) to get rid of de/dt.
Similarly,
[q
- r
= - r[K
2
] 7 using (*) again to get rid of dO/dt.
Hence the radial component of acceleration is (adding these equations)
K~ e cos e 1
+ TI
K~ K~
- 3 e cos B ) = - - = - ~ [
3
r
r
K' e cos 8
r
-5[
C
=
C
+
81
cOs
C
Thus, as desired,
(c) To apply Keplerys third law, we need a formula for the area of an ellipse,
which will be proved later, in Unit VII. It is
Area = T(
ma'or
a axis) (minor axis)
2
The minor axis is easily determined to be
given by:
minor axis = 4-
It is also easy | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
The minor axis is easily determined to be
given by:
minor axis = 4-
It is also easy to see that
major axis = b.
=dm.
Now we can apply Kepler's third law. Since area is being swept out at the constant
rate ;K, we know that (since the period is the time it takes to sweep out the entire
area),
Area = (2K)(Period).
1
Kepler's third law states that the following number is the same for all planets:
(major axis)"(major
axi s)3
--- 2
4 r (major axia)l(minor A a w i ~ ) ~ / ~ ~
16
r2lT i ni*
(major axi s)'
1
= b major axis ;;Z
Thus the constant Xp is the same for all planets.
(1) L e t L b e a l i n e i n V, with d i r e c t i o n vector A ; l e t P b e a p o i n t n o t on L. S h o w t h a t t h e
p o i n t X o n t h e l i n e L closest to P s a t i s f i e s the c o n d i t i o n t h a t X-P
i s p e r p e n d i c u l a r to A.
(2) F i n d p a r a m e t r i c e q u a t i o n s f o r t h e c u r v e C c o n s i s t i n g of a l l p o i n t s of V % e q u i d i s t a n t
f r o m t h e p o i n t P = ( 0 , l ) a n d t h e l i n e y =-I. | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
n t P = ( 0 , l ) a n d t h e l i n e y =-I. If X i s a n y p o i n t of C, s h o w t h a t t h e t a n g e n t
v e c t o r to C' a t X m a k e s e q u a l a n g l e s w i t h t h e v e c t o r X d P a n d t h e v e c t o r j. ( T h i s i s t h e
3
r e f l e c t i o n p r o p e r t y of t h e p a r a b o l a . )
(3) C o n s i d e r t h e c u r v e f ( t ) =(t,t cos ( d t ) ) f o r 0 c t I 1,
T h e n f is c o n t i n u o u s . L e t P be t h e p a r t i t i o n
= (0,O)
f o r t = 0.
P = { O , l / n , l / ( n - 1 ),...,1/3,1/2,1].
D r a w a p i c t u r e of t h e i n s c r i b e d polygon x ( P ) in t h e c a s e n = 5. Show t h a t i n g e n e r a l , x(P)
h a s l e n g t h
I x(P) I 2 1 + 2(1/2 + 113 + ... + l / h ).
C o n c l u d e t h a t f is n o t rectifiable.
($)
Let q be a fixed unit vector. A particle moves in Vn | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
is n o t rectifiable.
($)
Let q be a fixed unit vector. A particle moves in Vn in such a way that its position
satisfies the equation
vector ~ ( t )
constant angle 0 with u, where 0 < 9 < 7r/2.
r(t)-u
3
= 5t for all t, and its velocity vector makes a
(a) Show that ((I[(= 15t /cos 8.
2
(b) Compute the dot product . a ( t ) - ~ ( t )in terms o f t and 0.
(9 A particle moves in i-space so as to trace out a curve of constant curvature K = 3.
Its speed at time t is e2t. Find Ila(t)ll, and find the angle between q and 3 at time t.
(6) Consiaer the curve given in polar coordinates by the equaticn
r = e-
is a positive integer.
Find the length of this curve. h3at happens as M becomes
f o r 0 5 Q 52UM , where N
a r b i t r a r i l y large?
( a ) Derive the following formula, vhich can be used t o compute the
cctrvature of a curve i n R":
(t) Find the c u n a t u r e of the curve ~ ( t )= ( l c t , 3 t , Z t t 2 , 2 t 2 ) .
MIT OpenCourseWare
http://ocw.mit.edu
18.024 Multivariable Calculus with Theory
Spring 2011
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/18-024-multivariable-calculus-with-theory-spring-2011/043cde74c9d86eb3405f99edbc1a4c2e_MIT18_024s11_ChB2notes.pdf |
Lecture Notes for LG’s Diff. Analysis
trans. Paul Gallagher
DiGeorgi-Nash-Moser Theorem
1 Classical Approach
Our goal in these notes will be to prove the following theorem:
Theorem 1.1 (DiGeorgi-Nash-Moser). Let
∑
Lu :=
@i(aij@ju) and 0 < (cid:21) (cid:20) aij (cid:20) (cid:3)
(DGH)
Then there exists (cid:11)(n; (cid:21); (cid:3)) > 0 and C(n; (cid:21); (cid:3)) such that if Lu = 0, then
∥u∥C(cid:11)(B1=2) (cid:20) C((cid:21); (cid:3); n)∥u∥C0(B1)
Note that this estimate does not in any way involve derivatives of the aij.
We start by reminding of the Dirichlet energy of a function:
De(cid:12)nition 1.1 (Dirichlet Energy). If u : Ω ! R, then E(u) =
∫
Ω
j∇uj2.
With this, we have the following easy proposition.
Proposition 1.1. If u; w 2
E(w).
(cid:22)C 2(Ω), u = w on @Ω, and ∆u = 0, then E(u) (cid:20)
1
Proof. : Let w = u + v, so vj@Ω = 0. Then
∫
∫
∫
E(w) =
⟨∇w; ∇w⟩ =
j∇uj2 + j∇v
j2 + 2
∇ (cid:1) ∇v
u
∫
Ω
Ω
Ω
(cid:20)
j∇uj2 = E(u)
Ω
where we got from the (cid:12)rst line to the second by integration by parts.
In a similar way, we can de(cid:12)ne
De(cid:12)nition 1.2 (Gen. Dirichlet Energy). If L; a satis(cid:12)es (DGH), then
∫
∑ | https://ocw.mit.edu/courses/18-156-differential-analysis-ii-partial-differential-equations-and-fourier-analysis-spring-2016/045cf240688351f14d53c0b03b5a7b8a_MIT18_156S16_lec10-12.pdf |
2 (Gen. Dirichlet Energy). If L; a satis(cid:12)es (DGH), then
∫
∑
Ea(u) =
aij(@iu)(@ju)
Ω
and get a similar proposition with identical proof:
Proposition 1.2. If u; w 2
Ea(w) (cid:21) Ea(u).
(cid:22)C 2(Ω), and u = w on @Ω, and Lu = 0, then
We now prove an L2 estimate relating ∇u to u.
Proposition 1.3. If L follows (DGH) and Lu = 0 on B1 then
∫
∫
j∇ j
u 2 .
j
j
u 2
B1=2
B1
Proof. We will use integration by parts and localization. Let (cid:17) = 1 on B1=2
and be 0 outside of B1.
∫
∫
∫
j∇uj2 (cid:20)
(cid:17)2j∇uj2 (cid:25)
(cid:17)2
B1=2
∫
∫
∑
aij@iu@ju
(cid:17)2(Lu)u +
j∇(cid:17)j(cid:17)j∇ujjuj
)
) (∫
1=2
1=2
(cid:20)
(cid:20)
(∫
(cid:17)2j∇uj2
j∇(cid:17)j2u2
2
A classical approach would be to then prove the following:
Proposition 1.4. If (DGH), Lu = 0 and ∥aij∥C1 (cid:20) B then
∫
B
1=2
∫
jD2u
j2
(cid:20) C(B
;
n;
(cid:21); (cid:3))
j∇ j
u 2
B3=4
Proof. We have that 0 = @kLu = L(@ku) + (@kaij)@i@ju. Then,
∫
B1=2
jD2uj2 .
.
.
∫
∫
∫
∑ | https://ocw.mit.edu/courses/18-156-differential-analysis-ii-partial-differential-equations-and-fourier-analysis-spring-2016/045cf240688351f14d53c0b03b5a7b8a_MIT18_156S16_lec10-12.pdf |
ju. Then,
∫
B1=2
jD2uj2 .
.
.
∫
∫
∫
∑
(cid:17)2
aij@i@ku@j@ku
∫
j∇(cid:17)j(cid:17)jD2ujj∇uj +
(cid:17)2L(@ku)@ku
∫
j∇(cid:17)j(cid:17)jD2ujj∇uj +
(cid:17)2BjD2ujj∇uj
The result comes from applying Cauchy-Schwartz to this last pair of terms.
However, this won’t get us closer to proving DiGeorgi-Nash-Moser be-
cause we’re using an estimate on the derivatives of a in our inequality. Looks
like we’ll have to be clever!
2 L1 Bound
Theorem 2.1 (DGNM L1 bound). Let L satisfy (DGH), Lu (cid:21) 0, u > 0.
Then
∥u∥L1(B1=2) (cid:20) ∥u∥L2(B1)
Proof. We start with a lemma:
Lemma 2.1. Under the hypotheses, and if 1=2 (cid:20) r < r + w (cid:20) 1 then
∥∇u∥
L2(Br) . ∥u∥L2(Br+w)w(cid:0)
1
3
Proof. Let (cid:17) = 1 on Br and 0 on Bc
r+w. Note that (cid:17) can be constructed so
that j∇(cid:17)j < 2w(cid:0)1. Then the proof proceeds in exactly the same fashion as
Proposition 1.3.
Lemma 2.2. Under hypotheses, and 1=2 (cid:20) r < r + 2 (cid:20) 1, we have
∥u∥
L2n=(n(cid:0)2)(Br)
. w(cid:0) ∥u∥L2(Br+w)
1
Proof. Consider (cid:17)u with (cid:17) = 1 on Br, and 0 outside of Br+w= | https://ocw.mit.edu/courses/18-156-differential-analysis-ii-partial-differential-equations-and-fourier-analysis-spring-2016/045cf240688351f14d53c0b03b5a7b8a_MIT18_156S16_lec10-12.pdf |
)
1
Proof. Consider (cid:17)u with (cid:17) = 1 on Br, and 0 outside of Br+w=2. Then by the
Sobolev inequality, we have
∥(cid:17)u∥
L2n=(n 2) . ∥∇((cid:17)u)
(cid:0)
∥L2
(cid:20) ∥(∇(cid:17))u∥L2 + ∥(cid:17)(∇u)∥L2
Also, we have that
∥(∇(cid:17))u∥
∥(cid:17)(∇u)∥L2 (cid:20) ∥∇u∥
L2 (cid:20) ∥∇(cid:17)∥ ∥u1 ∥L2(Br+w=2) . w(cid:0) ∥u∥L2(Br+w)
1
L2(B
r+w=2) . w(cid:0) ∥u∥L (Br+w)
1
2
Lemma 2.3. If (cid:12) > 1, Lu (cid:21) 0 and u > 0, then Lu(cid:12) (cid:21) 0.
Proof. Compute:
∑
Lu(cid:12) =
@i(aij@j(u(cid:12))) =
∑
= (Lu)((cid:12)u(cid:12)(cid:0)1) +
∑
@i(aij(cid:12)u(cid:12)(cid:0)1@ju)
aij@iu@ju(cid:12)((cid:12) (cid:0) 1)u(cid:12)(cid:0)2 (cid:21) 0
where the last inequality comes from ellipticity of aij.
Now, apply Lemma 2.2 to u(cid:12) to get
∥u(cid:12)∥
L2n=(n(cid:0)2)(B
r)
. w(cid:0) ∥u ∥L (Br+w)
(cid:12)
1
2
Rewriting this with s = n we get
n(cid:0)2
4
Lemma 2.4. If | https://ocw.mit.edu/courses/18-156-differential-analysis-ii-partial-differential-equations-and-fourier-analysis-spring-2016/045cf240688351f14d53c0b03b5a7b8a_MIT18_156S16_lec10-12.pdf |
)
1
2
Rewriting this with s = n we get
n(cid:0)2
4
Lemma 2.4. If 1=2 (cid:20) r < r + w (cid:20) 1 and p (cid:21) 2, then
∥u∥
Lsp(Br) (cid:20) (Cw(cid:0) )
1 2=p
∥u∥Lp(Br+w)
For the next step, we iterate this lemma. If we have 1 = r0 > r1 > (cid:1) (cid:1) (cid:1) >
rk > 1=2, then we get the sequence of inequalities
∥u∥L2(B1) (cid:21) A0∥u∥L2s(Br ) (cid:21) (cid:1) (cid:1) (cid:1) (cid:21) A0
1
(cid:1) (cid:1) (cid:1)
A
k(cid:0)1∥u∥
kL2s (Br )k
where the A
r (cid:0) r (cid:25) j(cid:0)2. Thus, A = (C(r (cid:0) r
j
j+1
j
j
j are given by Lemma 2.4. Let’s pick rj = +
1
2
. Therefore,
)(cid:0)1)s(cid:0)j
j(cid:0)1
1
j+2
, so that
∏
log(
Aj) (cid:20)
∑
log(Aj) (cid:20)
∑
1
j=0
s(cid:0)j(C + C log(rj
(cid:0) rj+1))
(cid:20)
∑
1
j=0
s(cid:0)j(C + C log j) <
1
3 Finishing the Proof
Recall the Harnack inequality:
Theorem 3.1 (Harnack). If ∆u = 0 on B1 and u > 0 then minB1=2 u (cid:21)
(cid:13)(n) maxB1 u.
We will show a Harnack inequality for our L | https://ocw.mit.edu/courses/18-156-differential-analysis-ii-partial-differential-equations-and-fourier-analysis-spring-2016/045cf240688351f14d53c0b03b5a7b8a_MIT18_156S16_lec10-12.pdf |
B1=2 u (cid:21)
(cid:13)(n) maxB1 u.
We will show a Harnack inequality for our L which satis(cid:12)es (DGH).
Theorem 3.2 (DGNM Harnack). If L satis(cid:12)es (DGH), Lu = 0, 1 > u > 0
on B1, and
jfx 2 B1=2ju(x) > 1=10gj (cid:21)
then minB1=2 u (cid:21) (cid:13)(n).
1
10
jB1=2j
(P)
For now, let’s assume this theorem, and see how it implies the DiGeorgi-
Nash-Moser estimate.
5
De(cid:12)nition 3.1. oscΩu := supΩ u (cid:0) inf Ω u.
Corollary 3.1. If Lu = 0 on Ω, Br(x) (cid:26) Ω, then
oscBr=2(x)u (cid:20) (1 (cid:0) (cid:13))oscBr(x)u
(O)
Proof. We start with some simple reductions via scaling. Without loss of
generality, we can take:
inf u = 0; sup u = 1; r = 1
Br(x)
Br(x)
jfx 2 B1=2ju(x) (cid:21) 1=2gj (cid:21) B1=2=2
Thus by DGNM Harnack, minB1=2 u (cid:21) (cid:13), and thus oscB1=2u (cid:20) 1 (cid:0) (cid:13) =
(1 (cid:0) (cid:13))oscB1u
Now we can complete the proof with the following:
Proposition 3.1. Let u : B1 ! R satisfy (O). Then ∥u∥C(cid:11)(B1=2) . ∥u∥C0(B1)
for some (cid:11) = (cid:11)((cid:13)) > 0.
Proof. Let x; y 2 B1=2, jx (cid:0) yj = | https://ocw.mit.edu/courses/18-156-differential-analysis-ii-partial-differential-equations-and-fourier-analysis-spring-2016/045cf240688351f14d53c0b03b5a7b8a_MIT18_156S16_lec10-12.pdf |
)((cid:13)) > 0.
Proof. Let x; y 2 B1=2, jx (cid:0) yj = d and a = (x + y)=2. Then
ju(x) (cid:0) u(y)j (cid:20) (oscBd(a)u)(1 (cid:0) (cid:13)) (cid:20) (cid:1) (cid:1) (cid:1) (cid:20) (1 (cid:0) (cid:13))koscB k (a)u
2 d
Choose k such that 1=4 < 2kd (cid:20) 1=2. Then k = log2(1=d) + O(1), and so
ju(x) (cid:0) u(y)j (cid:20) (1 (cid:0) (cid:13))koscB1u (cid:20) 2(1 (cid:0) (cid:13))kju∥C0(B1):
Also,
(1 (cid:0) (cid:13))k (cid:20) 4(1 (cid:0) (cid:13))log2(1=d) = 4d(cid:0) log2(1(cid:0)(cid:13)):
Therefore, setting (cid:11)((cid:13)) = (cid:0) log2(1 (cid:0) (cid:13)) (cid:25) (cid:13) + O((cid:13)2), we get our proposition.
Now let’s prove the Harnack inequality. Before we do the DGNM Har-
nack, we’ll remember how the normal ∆ Harnack inequality works:
Lemma 3.1. If ∆u = 0 and u > 0 then ∥∇ log u∥L1(B1=2) . 1.
6
Note that the lemma implies the Harnack inequality by integrating.
Proof. We have ∇ log u = ∇u . Also, by elliptic regularity, we have that
u
j∇uj(x) . ∥u∥L1(B1=2(x)) =
so that j∇uj=u . 1.
∫
B1=2(x)
u = jB1=2 | https://ocw.mit.edu/courses/18-156-differential-analysis-ii-partial-differential-equations-and-fourier-analysis-spring-2016/045cf240688351f14d53c0b03b5a7b8a_MIT18_156S16_lec10-12.pdf |
=2(x)) =
so that j∇uj=u . 1.
∫
B1=2(x)
u = jB1=2(x)ju(x)
With this method in mind, let’s prove the DGNM Harnack.
DGNM Harnack.
Lemma 3.2. If L satis(cid:12)es (DGH), Lu = 0, u > 0 on B1 then ∥∇ log u∥L2(B1=2) .
1.
Proof. Pick a nice cutoff function (cid:17) as usual.
∫
∫
∫
∫
B1=2
j∇ log uj2 =
=
.
(cid:20)
∫
(cid:17)2j∇ log uj2 .
(cid:17)2
ij
a
(cid:17)2
∑ @iu @ju
u u
(cid:17)j∇(cid:17)jj∇uju(cid:0)1 =
)
1
(∫
∑
∫
= (cid:0)
∫
aij@i log u@j log u
∑
2
(cid:17)
aij@iu@ju(cid:0)1
(cid:17)j∇(cid:17)jj∇ log uj
(∫
)
1=2
=2
(cid:17)2j∇ log uj2
j∇(cid:17)j2
Letting w = (cid:0) log u, we have that ∥∇w∥L2(B9=10) . 1. We want an L1
bound on w. By (P), we have that
jfx 2 B1=2jw (cid:20) log 10gj (cid:21)
1
10
j
B
1=2j
Now we use the Poincare Inequality:
7
Theorem 3.3 (Poincare). If (P) then
∫
jwj2 .
B8=10
∫
j∇wj2 + 1
B9=10
Therefore, we have an L2 bound on w instead of ∇w. Now we have
Lemma 3.3. Lw (cid | https://ocw.mit.edu/courses/18-156-differential-analysis-ii-partial-differential-equations-and-fourier-analysis-spring-2016/045cf240688351f14d53c0b03b5a7b8a_MIT18_156S16_lec10-12.pdf |
10
Therefore, we have an L2 bound on w instead of ∇w. Now we have
Lemma 3.3. Lw (cid:21) 0
Proof. Compute:
∑
(cid:0)
@i(aij@j log u) = (cid:0)
∑
@i(aij(@ju)u(cid:0)1)
∑
= Lu (cid:1) u(cid:0)1 +
aij(@iu)(@ju)u(cid:0)2 (cid:21) 0
Finally, w = (cid:0) log u > 0 because u < 1, and so we can apply Theorem
2.1 and get
∥w∥L1(B1=2) . ∥w∥L2(B8=10) . 1
thus completing the proof of the Harnack inequality.
8
MIT OpenCourseWare
http://ocw.mit.edu
18.156 Differential Analysis II: Partial Differential Equations and Fourier Analysis
Spring(cid:3) 2016
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/18-156-differential-analysis-ii-partial-differential-equations-and-fourier-analysis-spring-2016/045cf240688351f14d53c0b03b5a7b8a_MIT18_156S16_lec10-12.pdf |
II.C Spontaneous Symmetry Breaking and Goldstone Modes
For zero field, ~h = 0, although the microscopic Hamiltonian has full rotational sym
metry, the low–temperature phase does not. As a specific direction in n–space is selected
for the net magnetization M~ , there is a spontaneously broken symmetry, and a correspond
ing long–range order is established in the system. The original symmetry is still present
globally, in the sense that if all local magnetizations ~m(x), are rotated together (i.e.
m(x)
~
if
m(x)), there is no change in energy. Such a rotation transforms one ordered
~
state into an equivalent one. If a uniform rotation costs no energy, by continuity we expect
(x) only has
a rotation that is slowly varying in space (e.g. ~
m(x), where
m(x)
(x) ~
7→ ℜ
ℜ
long wavelength variations) to cost very little energy. Such low energy excitations are called
Goldstone modes. They are present in any system with a broken continuous symmetry.
7→ ℜ
There are no Goldstone modes when a discrete symmetry is broken, since it is impossible
to produce slowly varying rotations from one state to an equivalent one. Phonons are an
example of Goldstone modes, corresponding to the breaking of translation and rotation
symmetries by a crystal structure.
Let us explore the origin and consequences of Goldstone modes in the context of
superfluidity. In analogy to Bose condensation, the superfluid | https://ocw.mit.edu/courses/8-334-statistical-mechanics-ii-statistical-physics-of-fields-spring-2014/0467b954164431b59100b5725779e6e4_MIT8_334S14_Lec3.pdf |
context of
superfluidity. In analogy to Bose condensation, the superfluid phase has a macroscopic
occupation of a single quantum ground state. The order parameter,
ψ(x)
≡
ψℜ + iψℑ
ψ(x)
|
≡ |
e iθ(x),
(II.11)
is the ground state component (overlap) of the actual wavefunction in the vicinity of x.
The phase of the wavefunction is not an observable quantity and should not appear in any
physically measurable probability. For example, the effective coarse grained Hamiltonian
can be obtained as an expansion,
=
β
H
dd x
(cid:20)
Z
K
2 |∇
ψ
|
2
+
t
2 |
ψ
2
|
+ u
ψ
|
|
4
+
.
· · ·
(cid:21)
(II.12)
Clearly, eq.(II.12) is equivalent to the Landau–Ginzburg Hamiltonian with n = 2 ( ~m
≡
(ψℜ, ψℑ)). The superfluid transition is signaled by the onset of a finite value of ψ for t < 0.
Minimizing the Hamiltonian fixes the magnitude of ψ, but not its phase θ. Now consider
a state with a slowly varying phase, i.e. with ψ(x) = ψeiθ(x). Inserting this form in the
Hamiltonian yields an energy
¯
β
H
= β
H0 +
dd x(
θ)2 ,
∇
¯
K
2
19
Z
(II.13)
where K¯ = Kψ¯2 . Taking advantage | https://ocw.mit.edu/courses/8-334-statistical-mechanics-ii-statistical-physics-of-fields-spring-2014/0467b954164431b59100b5725779e6e4_MIT8_334S14_Lec3.pdf |
K
2
19
Z
(II.13)
where K¯ = Kψ¯2 . Taking advantage of translational symmetry, Eq.(II.13) can be decom
q eiq·xθq/√V ,
posed into independent modes (in a region of volume V ) by setting θ(x) =
as
β
H
= β
H0 +
K¯
2
q
X
2
q
|
θ(q)
2
.
|
P
(II.14)
Clearly the long wavelength Goldstone modes cost little energy and are easily excited by
thermal fluctuations.
Assuming that the amplitude of the order parameter is indeed uniform, the probability
of a particular configuration is given by,
[θ(x)]
exp
∝
P
¯
K
2
Z
−
(cid:20)
dd x(
.
θ)2
(cid:21)
∇
(II.15)
Alternatively, in terms of the Fourier components,
[θ(q)]
exp
P
∝
¯
K
− 2
"
2
q
θ(q)
|
2
| ∝
#
p(θq).
(II.16)
q
X
Each mode θq, is an independent random variable with a Gaussian distribution of zero
mean, and with †
q
Y
θqθq ′
i
h
=
δq,−q ′
Kq2 .
¯
(II.17)
† Note that the Fourier transform of a real field θ(x), is complex θq = θq,ℜ + iθq,ℑ.
However, the number of fields is not doubled, due to the constraint of θ−q = θq
iθq,ℑ. A | https://ocw.mit.edu/courses/8-334-statistical-mechanics-ii-statistical-physics-of-fields-spring-2014/0467b954164431b59100b5725779e6e4_MIT8_334S14_Lec3.pdf |
due to the constraint of θ−q = θq
iθq,ℑ. A Gaussian translational invariant weight has the generic form
∗ = θq,ℜ
−
[
θq
{
}
]
P
∝
exp
−
(cid:20)
q
Y
K
(q)
2
θqθ−q =
exp
(cid:21)
q>0
Y
(q)
2K
2
−
(cid:20)
(cid:0)
2
2
θq
,ℜ + θq
,ℑ
.
(cid:21)
(cid:1)
While the first product is over all q, the second is restricted to half of the space. There
are clearly no cross correlations for differing q, and the Gaussian variances are
2
2
,ℑ =
,ℜ = θq
θq
1
2K(q)
,
(cid:10)
from which we can immediately construct
(cid:10)
(cid:11)
(cid:11)
θqθ∓q
h
i
2
= θq
,ℜ ±
(cid:11)
(cid:10)
(cid:10)
2
,ℑ =
θq
1
1
±
)
(q
K
2
.
(cid:11)
20
From eq.(II.17) we can calculate the correlations in the phase θ(x) in real space. Clearly
θ(x)
= 0 by symmetry, while
h
i
θ(x)θ(x ′ )
h
1
V
=
i
′
q,q
X
iq·x+iq
e
′
′ ·x
θqθq =
′
i
h
1
V
′
eiq·(x−x )
.
¯
Kq2
q
X
( | https://ocw.mit.edu/courses/8-334-statistical-mechanics-ii-statistical-physics-of-fields-spring-2014/0467b954164431b59100b5725779e6e4_MIT8_334S14_Lec3.pdf |
h
1
V
′
eiq·(x−x )
.
¯
Kq2
q
X
(II.18)
In the continuum limit, the sum can be replaced by an integral (
V
dd q/(2π)d),
q 7→
and
The function,
θ(x)θ(x ′ )
h
i
=
d
(2
d q
)d
π
Z
(x−x )
′
eiq·
¯
Kq2
=
Cd(x
¯−
K
−
P
x ′ )
.
R
Cd(x) =
−
dd q eiq·x
(2π)d q2 ,
Z
is the Coulomb potential due to a unit charge at the origin in a d–dimensional space, since
it is the solution to
2Cd(x) =
∇
Z
2
dd q q
(2π)d q2
e iq·x = δd(x).
(II.21)
We can easily find a solution by using Gauss’ theorem,
dd x
∇
2Cd =
I
Z
dS
· ∇
Cd
.
Cd = (dCd/dx)ˆx, and the above equation simplifies
∇
For a spherically symmetric solution,
to
where
1 = Sdx
d−1 dCd
dx
,
Sd =
2πd/2
(d/2
1)!
−
,
is the total solid angle (area of unit sphere) in d dimensions. Hence
dCd
dx
=
1
Sdxd−1 , =
⇒
Cd(x) =
x2−d
(2
−
d)Sd
+ c0,
where c0 is a constant of integration.
The long distance behavior of Cd(x) changes dramatically at d = 2, as
(II.19)
(II.20)
(II.22 | https://ocw.mit.edu/courses/8-334-statistical-mechanics-ii-statistical-physics-of-fields-spring-2014/0467b954164431b59100b5725779e6e4_MIT8_334S14_Lec3.pdf |
(x) changes dramatically at d = 2, as
(II.19)
(II.20)
(II.22)
(II.23)
(II.24)
lim Cd(x) = (2
x→∞
c0
x
2−d
d)Sd
−
ln(x)
2π
21
d > 2
d < 2
d = 2
.
(II.25)
The constant of integration can obtained by looking at
h
which goes to zero as x
′
x . Hence,
→
[θ(x)
θ(x ′ )]2
i
−
= 2
h
θ(x)2
i −
2
h
θ(x)θ(x ′ )
,
i
2 x
|
(cid:0)
where a is of the order of the lattice spacing.
θ(x ′ )]2
i
[θ(x)
−
=
h
a2−d
−
¯
K(2
x ′ 2−d
|
d)
−
−
Sd
,
(cid:1)
(II.26)
(II.27)
For d > 2, the phase fluctuations are finite, while they become asymptotically large
for d
2. Since the phase is bounded by 2π, this implies that long range order in the
phase is destroyed. This result becomes more apparent by examining the effect of phase
fluctuations on the two point correlation function
≤
ψ(x)ψ ∗ (0)
= ψ¯2
h
i
h
e i[θ(x)−θ(0)]
.
i
( | https://ocw.mit.edu/courses/8-334-statistical-mechanics-ii-statistical-physics-of-fields-spring-2014/0467b954164431b59100b5725779e6e4_MIT8_334S14_Lec3.pdf |
(0)
= ψ¯2
h
i
h
e i[θ(x)−θ(0)]
.
i
(II.28)
(Since amplitude fluctuations are ignored, we are in fact looking at a transverse correlation
function.) We shall prove later on that for any collection of Gaussian distributed variables,
exp(αθ)
= exp
i
h
α2
2 h
.
θ2
i
(cid:19)
(cid:18)
Taking this result for granted, we obtain
ψ(x)ψ ∗ (0)
i
h
= ψ¯2 exp
1
− 2
(cid:20)
[θ(x)
h
θ(0)]2
i
−
(cid:21)
= ψ¯2 exp
−
(cid:20)
and asymptotically
lim
x→∞h
ψ(x)ψ ∗ (0)
i
= ψ¯′2
0
(cid:26)
for d > 2
for d
2
≤
2−
d
2
−
d
x
¯
K
(2
−
−
a
)
d
,
S
d
(cid:21)
(II.29)
.
(II.30)
The saddle point approximation to the order parameter ψ¯, was obtained by ignoring fluctu
ations. The above result indicates that inclusion of phase fluctuations leads to a reduction
of order in d > 2, and its complete destruction in d
2.
The above example typifies a more general result known as the Mermin–Wagner theo
rem. The theorem states that there is no spontaneous breaking of a continuous symmetry
in systems with short range interactions in dimensions d
2. Some corollaries to this
theorem are:
(1) The borderline dimensionality of two, known as the lower critical dimension, has | https://ocw.mit.edu/courses/8-334-statistical-mechanics-ii-statistical-physics-of-fields-spring-2014/0467b954164431b59100b5725779e6e4_MIT8_334S14_Lec3.pdf |
(1) The borderline dimensionality of two, known as the lower critical dimension, has to
be treated carefully. As we shall demonstrate later on in the course, there is in fact
a phase transition for the two dimensional superfluid, although there is no true long
range order.
≤
≤
(2) There are no Goldstone modes when the broken symmetry is discrete (e.g. for n = 1).
In such cases, long range order is possible down to the lower critical dimension of
dℓ = 1.
22
MIT OpenCourseWare
http://ocw.mit.edu
8.334 Statistical Mechanics II: Statistical Physics of Fields
Spring 2014
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/8-334-statistical-mechanics-ii-statistical-physics-of-fields-spring-2014/0467b954164431b59100b5725779e6e4_MIT8_334S14_Lec3.pdf |
18.357 Interfacial Phenomena, Fall 2010
taught by Prof. John W. M. Bush
June 3, 2013
Contents
1
Introduction, Notation
1.1 Suggested References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 Definition and Scaling of Surface Tension
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1 History: Surface tension in antiquity
2.2 Motivation: Who cares about surface tension? . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Surface tension: a working definition
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4 Governing Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5 The scaling of surface tension
2.6 A few simple examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 Wetting
4 Young’s Law with Applications
4.1 Formal Development of Interfacial Flow Problems . . . . . . . . . . . . . . . . . . . . . . .
5 Stress Boundary Conditions
5.1 Stress conditions at a fluid-fluid interface
5.2 Appendix A : Useful identity
5.3 | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
Stress conditions at a fluid-fluid interface
5.2 Appendix A : Useful identity
5.3 Fluid Statics
5.4 Appendix B : Computing curvatures
5.5 Appendix C : Frenet-Serret Equations
. . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6 More on Fluid statics
6.1 Capillary forces on floating bodies
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7 Spinning, tumbling and rolling drops
7.1 Rotating Drops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2 Rolling drops
8 Capillary Rise
8.1 Dynamics
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9 Marangoni Flows
10 Marangoni Flows II
10.1 Tears of Wine . . . . . . . . . . . . . | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
ows
10 Marangoni Flows II
10.1 Tears of Wine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.2 Surfactants
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.3 Surfactant-induced Marangoni flows
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.4 Bubble motion
3
3
4
4
5
6
7
7
9
11
12
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14
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21
21
22
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1
Contents
Contents
11 Fluid Jets
11.1 The shape of a falling fluid jet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.2 The Plateau-Rayleigh Instability
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.3 Fluid Pipes
12 Instability Dynamics
. . . . . . . . . . . . . . . . . . . . . .
12.1 Capillary Instability of a Fluid Coating | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
. . . . . . . .
12.1 Capillary Instability of a Fluid Coating on a Fiber
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
12.2 Dynamics of Instability (Rayleigh 1879)
12.3 Rupture of a Soap Film (Culick 1960, Taylor 1960) . . . . . . . . . . . . . . . . . . . . . .
13 Fluid Sheets
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13.1 Fluid Sheets: shape and stability
13.2 Circular Sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13.3 Lenticular sheets with unstable rims (Taylor 1960)
. . . . . . . . . . . . . . . . . . . . . .
13.4 Lenticular sheets with stable rims . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13.5 Water Bells
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13.6 Swirling Water Bell . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14 Instability of Superposed Fluids
14.1 Rayleigh-Taylor Instability . . . . . . . . . . . . . . . . . . . . . . . . . . . . | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
aylor Instability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.2 Kelvin-Helmholtz Instability
15 Contact angle hysteresis, Wetting of textured solids
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15.1 Contact Angle Hysteresis
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15.2 Wetting of a Rough Surface
15.3 Wenzel State (1936)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15.4 Cassie-Baxter State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16 More forced wetting
16.1 Hydrophobic Case: θe > π/2, cos θe < 0
16.2 Hydrophilic Case: θe < π/2
16.3 Forced Wetting: the Landau-Levich-Derjaguin Problem
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . .
17 Coating: Dynamic Contact Lines
17.1 | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
. . . . . . . .
17 Coating: Dynamic Contact Lines
17.1 Contact Line Dynamics
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18 Spreading
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18.1 Spreading of small drops on solids
. . . . . . . . . . . . . . . . . .
18.2 Immiscible Drops at an Interface Pujado & Scriven 1972
18.3 Oil Spill
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18.4 Oil on water: A brief review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18.5 The Beating Heart Stocker & Bush (JFM 2007) . . . . . . . . . . . . . . . . . . . . . . . .
19 Water waves
40
40
41
43
45
45
46
47
49
49
50
51
51
53
54
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57
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59
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MIT OCW: 18.357 Interfacial Phenomena
2
Prof. John W. M. Bush
1. Introduction, Notation
We consider fluid systems dominated by the influence of interfacial tension. The roles of curvature pressure
and Marangoni stress are elucidated in a variety | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
of interfacial tension. The roles of curvature pressure
and Marangoni stress are elucidated in a variety of situtations. Particular attention will be given to the
dynamics of drops and bubbles, soap films and minimal surfaces, wetting phenomena, water-repellency,
surfactants, Marangoni flows, capillary origami and contact line dynamics. Theoretical developments will
be accompanied by classroom demonstrations. The role of surface tension in biology will be highlighted.
Notation
Nomenclature: σ denotes surface tension (at fluid-gas interface)
γ denotes interfacial tension (at fluid-fluid or fluid-solid interface).
Note on units: we will use predominantly cgs system.
Unit of force: 1 dyne = 1 g cm s = 10−5N as the cgs unit of force, roughly the weight of 1 mosquito.
Pressure: 1 atm ≈ 100kPa = 105N/m2=106 dynes/cm2 .
Units: [σ]=dynes/cm=mN/m.
What is an interface?: roughness scale δ, from equality of surface and thermal energy get σδ ∼kT⇒
δ ∼ (kT/σ)1/2 . If δ ≪ scales of experiment, can speak of a smooth interface.
−2
2
1.1 Suggested References
While this list of relevant textbooks is far from complete, we include it as a source of additional reading
for the interested student.
• Capillarity and Wetting Phenomena: Drops, Bubbles, Pearls, Waves
by | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
• Capillarity and Wetting Phenomena: Drops, Bubbles, Pearls, Waves
by P.G. de Gennes, F. Brochard-Wyart and D. Qu´er´e. Springer Publishing.
A readable and accessible treatment of a wide range of capillary phenomena.
• Molecular theory of capillarity
by J.S. Rowlinson and B. Widom. Dover 1982.
• Intermolecular and surface forces
by J. Israelachvili. Academic Press, 2nd edition 1995.
• Multimedia Fluid Mechanics
Cambridge University Press, Ed. Bud Homsy.
A DVD with an extensive section devoted to capillary effects. Relevant videos will be used throughout
the course.
3
2. Definition and Scaling of Surface
Tension
These lecture notes have been drawn from many sources, including textbooks, journal articles, and lecture
notes from courses taken by the author as a student. These notes are not intended as a complete discussion
of the subject, or as a scholarly work in which all relevant references are cited. Rather, they are intended as
an introduction that will hopefully motivate the interested student to learn more about the subject. Topics
have been chosen according to their perceived value in developing the physical insight of the students.
2.1 History: Surface tension in antiquity
Hero of Alexandria (10 AD - 70 AD) Greek mathematician and engineer, “the greatest
experimentalist of antiquity”. Exploited capillarity in a number of inventions described in his
book Pneumatics, including the water clock.
Pliny the Elder (23 AD - 79 AD) Author, natural philosopher, | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
neumatics, including the water clock.
Pliny the Elder (23 AD - 79 AD) Author, natural philosopher, army and naval commander
of the early Roman Empire. Described the glassy wakes of ships. “True glory comes in doing
what deserves to be written; in writing what deserves to be read; and in so living as to make
the world happier.” “Truth comes out in wine”.
Leonardo da Vinci (1452-1519) Reported capillary rise in his notebooks, hypothesized that
mountain streams are fed by capillary networks.
Francis Hauksbee (1666-1713) Conducted systematic investigation of capillary rise, his
work was described in Newton’s Opticks, but no mention was made of him.
Benjamin Franklin (1706-1790) Polymath: scientist, inventor, politician; examined the
ability of oil to suppress waves.
Pierre-Simon Laplace (1749-1827) French mathematician and astronomer, elucidated the
concept and theoretical description of the meniscus, hence the term Laplace pressure.
Thomas Young (1773-1829) Polymath, solid mechanician, scientist, linguist. Demonstrated
the wave nature of light with ripple tank experiments, described wetting of a solid by a fluid.
Joseph Plateau (1801-1883) Belgian physicist, continued his experiments after losing his
sight. Extensive study of capillary phenomena, soap films, minimal surfaces, drops and bubbles.
4
2.2. Motivation: Who cares about surface tension? Chapter 2. Definition and Scaling of Surface T | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
Who cares about surface tension? Chapter 2. Definition and Scaling of Surface Tension
2.2 Motivation: Who cares about surface tension?
As we shall soon see, surface tension dominates gravity on a scale less than the capillary length (roughly
2mm). It thus plays a critical role in a variety of small-scale processes arising in biology, environmental
science and technology.
Biology
• early life: early vessicle formation, confine
• all small creatures live in a world dominated
by surface tension
ment to an interface
• oil recovery, carbon sequestration, groundwa
• surface tension important for insects for many
ter flows
basic functions
• desi gn of insecticides intended to coat insects,
• weight support and propulsion at the water
leave plant unharmed
surface
• chemical leaching and the water-repellency of
• adhesi on and deadhesion via surface tension
soils
• the pistol shrimp: hunting with bubbles
• oil spi ll dynamics and mitigation
• underwater breathing and diving via surface
tension
• natu ral
strategies
plants and animals
for water-repellency
in
• disease transmission via droplet exhalation
• dynamics of magma chambers and volcanoes
• the exploding lakes of Cameroon
• the dynamics of lungs and the role of surfac
tants and impurities
Technology
This image has been removed due to copyright restrictions.
Please see the image on page http://news.sciencemag.
org/sciencenow/2011/06/spiders.html#panel-2.
Figure 2.1: The | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
encemag.
org/sciencenow/2011/06/spiders.html#panel-2.
Figure 2.1: The diving bell spider
Geophysics and environmental science
• capillary effects dominant in microgravity set
tings: NASA
• ca vitation-induced damage on propellers and
submarines
• ca vitation in medicine: used to damage kidney
stones, tumours ...
• design of superhydrophobic surfaces e.g. self-
cleaning windows, drag-reducing or erosion-
resistant surfaces
• lab-on-a-chip technology: medical diagnostics,
drug delivery
• microfluidics:
continuous and discrete fluid
• the dynamics of raindrops and their role in the
transport and mixing
biosphere
• tracking submarines with their surface signa
• most biomaterial is surface active, sticks to the
ture
surface of drops / bubbles
• chemical, thermal and biological transport in
the surf zone
• ink jet printing
• the bubble computer
MIT OCW: 18.357 Interfacial Phenomena
5
Prof. John W. M. Bush
2.3. Surface tension: a working definition
Chapter 2. Definition and Scaling of Surface Tension
Figure 2.2: a) The free surface between air and water at a molecular scale. b) Surface tension is analogous
to a negative surface pressure.
2.3 Surface tension: a working definition
Discussions of the molecular origins of surface or interfacial tension may be found elsewhere (e.g. Is
raelachvili 1995, Rowlinson & Widom 1982 ). Our discussion follows that of de | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
ili 1995, Rowlinson & Widom 1982 ). Our discussion follows that of de Gennes, Brochard-Wyart
& Qu´er´e 2003.
Molecules in a fluid feel a mutual attraction. When this attractive force is overcome by thermal
agitation, the molecules pass into a gaseous phase. Let us first consider a free surface, for example
that between air and water (Fig. 2.2a). A water molecule in the fluid bulk is surrounded by attractive
neighbours, while a molecule at the surface has a reduced number of such neighbours and so in an
energetically unfavourable state. The creation of new surface is thus energetically costly, and a fluid
system will act to minimize surface areas. It is thus that small fluid bodies tend to evolve into spheres;
for example, a thin fluid jet emerging from your kitchen sink will generally pinch off into spherical drops
in order to minimize the total surface area (see Lecture 5).
If U is the total cohesive energy per molecule, then a molecule at a free surface will lose U/2 relative to
molecules in the bulk. Surface tension is a direct measure of this energy loss per unit area of surface. If the
characteristic molecular dimension is R and its area thus R2, then the surface tension is σ ∼ U/(2R)2 . Note
that surface tension increases as the intermolecular attraction increases and the molecular size decreases.
For most oils, σ ∼ 20 dynes/cm, while for water, σ ∼ 70 | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
oils, σ ∼ 20 dynes/cm, while for water, σ ∼ 70 dynes/cm. The highest surface tensions are
for liquid metals; for example, liquid mercury has σ ∼ 500 dynes/cm. The origins of interfacial tension
are analogous.
Interfacial tension is a material property of a fluid-fluid interface whose origins lie in
the different energy per area that acts to resist the creation of new interface. Fluids between which no
interfacial tension arises are said to be miscible. For example, salt molecules will diffuse freely across a
boundary between fresh and salt water; consequently, these fluids are miscible, and there is no interfacial
tension between them. Our discussion will be confined to immiscible fluid-fluid interfaces (or fluid-gas
surfaces), at which an effective interfacial (or surface) tension acts.
Surface tension σ has the units of force/length or equivalently energy/area, and so may be thought
of as a negative surface pressure, or, equivalently, as a line tension acting in all directions parallel to the
surface. Pressure is generally an isotropic force per area that acts throughout the bulk of a fluid: small
surface element dS will feel a total force p(x)dS owing to the local pressure field p(x). If the surface S is
closed, | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
owing to the local pressure field p(x). If the surface S is
closed, and the pressure uniform, the net pressure force acting on S is zero and the fluid remains static.
Pressure gradients correspond to body forces (with units of force per unit volume) within a fluid, and so
appear explicitly in the Navier-Stokes equations. Surface tension has the units of force per length, and
its action is confined to the free surface. Consider for the sake of simplicity a perfectly flat interface. A
surface line element dℓ will feel a total force σdℓ owing to the local surface tension σ(x). If the surface
line element is a closed loop C, and the surface tension uniform, the net surface tension force acting on
C is zero, and the fluid remains static. If surface tension gradients arise, there may be a net force on the
surface element that acts to distort it through driving flow.
MIT OCW: 18.357 Interfacial Phenomena
6
Prof. John W. M. Bush
2.4. Governing Equations
Chapter 2. Definition and Scaling of Surface Tension
2.4 Governing Equations
The motion of a fluid of uniform density ρ and dynamic viscosity µ is governed by the Navier-Stokes
equations, which represent a continuum statement of Newton’s laws.
∂u
∂t
ρ
(
+ u · ∇u = −∇ | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
of Newton’s laws.
∂u
∂t
ρ
(
+ u · ∇u = −∇p + F + µ∇2 u
)
∇ · u = 0
(2.1)
(2.2)
This represents a system of 4 equations in 4 unknowns (the fluid pressure p and the three components of
the velocity field u). Here F represents any body force acting on a fluid; for example, in the presence of
a gravitational field, F = ρg where g is the acceleration due to gravity.
Surface tension acts only at the free surface; consequently, it does not appear in the Navier-Stokes
equations, but rather enters through the boundary conditions. The boundary conditions appropriate at a
fluid-fluid interface are formally developed in Lecture 3. We here simply state them for the simple case of
a free surface (such as air-water, in which one of the fluids is not dynamically significant) in order to get
a feeling for the scaling of surface tension. The normal stress balance at a free surface must be balanced
by the curvature pressure associated with the surface tension:
n · T · n = σ(∇ · n)
(2.3)
where T = −pI + µ ∇u + (∇u)T
is the
deviatoric stress tensor, and n is the unit normal to the surface. The tangential stress at a free surface
must balance the local surface tension gradient:
= −pI + 2µE | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
a free surface
must balance the local surface tension gradient:
= −pI + 2µE is the stress tensor, E = ∇u + (∇u)T
]
[
]
[
1
2
n · T · t = ∇σ · t
(2.4)
where t is the unit tangent to the interface.
2.5 The scaling of surface tension
Fundamental Concept The laws of Nature cannot depend on arbitrarily chosen system of units. Any
physical system is most succinctly described in terms of dimensionless variables.
Buckingham’s Theorem For a system with M physical variables (e.g. density, speed, length, viscosity)
describable in terms of N fundamental units (e.g. mass, length, time, temperature), there are M − N
dimensionless groups that govern the system.
E.g. Translation of a rigid sphere through a viscous fluid:
Physical variables: sphere speed U and radius a, fluid viscosity ν and density ρ and sphere drag D; M = 5.
Fundamental units: mass M , length L and time T ; N = 3.
Buckingham’s Theorem: there are M − N = 2 dimensionless groups: Cd = D/ρU 2 and Re = U a/ν.
System is uniquely determined by a single relation between the two: Cd = F (Re).
We consider a fluid of density ρ and viscosity µ = ρν with a free surface characterized by a surface tension
σ. The flow is marked | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
with a free surface characterized by a surface tension
σ. The flow is marked by characteristic length- and velocity- scales of, respectively, a and U , and evolves
in the presence of a gravitational field g = −gzˆ. We thus have a physical system defined in terms of six
physical variables (ρ, ν, σ, a, U, g) that may be expressed in terms of three fundamental units: mass, length
and time. Buckingham’s Theorem thus indicates that the system may be uniquely described in terms of
three dimensionless groups. We choose
Re =
Fr =
Bo =
=
U a
ν
U 2
ga
ρga2
σ
=
Inertia
Viscosity
Inertia
Gravity
= Reynolds number
= Froude number
=
Gravity
Curvature
= Bond number
(2.5)
(2.6)
(2.7)
MIT OCW: 18.357 Interfacial Phenomena
7
Prof. John W. M. Bush
2.5. The scaling of surface tension
Chapter 2. Definition and Scaling of Surface Tension
The Reynolds number prescribes the relative magnitudes of inertial and viscous forces in the system,
while the Froude number those of inertial and gravity forces. The Bond number indicates the relative
importance of forces induced by gravity and surface tension. Note that these two forces are comparable
when Bo = 1, which arises at a lengthscale corresponding to the capillary length: ℓc = (σ/(ρg))
. For
an air-water surface, | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
to the capillary length: ℓc = (σ/(ρg))
. For
an air-water surface, for example, σ ≈ 70 dynes/cm, ρ = 1g/cm3 and g = 980 cm/s2, so that ℓc ≈ 2mm.
Bodies of water in air are dominated by the influence of surface tension provided they are smaller than the
capillary length. Roughly speaking, the capillary length prescribes the maximum size of pendant drops
that may hang inverted from a ceiling, water-walking insects, and raindrops. Note that as a fluid system
becomes progressively smaller, the relative importance of surface tension over gravity increases; it is thus
that surface tension effects are critical in many in microscale engineering processes and in the lives of
bugs.
1/2
Finally, we note that other frequently arising dimensionless groups may be formed from the products
of Bo, Re and Fr:
We =
Ca =
ρU 2a
σ
ρνU
σ
=
=
Inertia
Curvature
Viscous
Curvature
= Weber number
= Capillary number
(2.8)
(2.9)
The Weber number indicates the relative magnitudes of inertial and curvature forces within a fluid, and
the capillary number those of viscous and curvature forces. Finally, we note that if the flow is marked by
a Marangoni stress of characteristic magnitude Δσ/L, then an additional dimensionless group arises that
characterizes the relative magnitude of Marangoni and | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
then an additional dimensionless group arises that
characterizes the relative magnitude of Marangoni and curvature stresses:
Ma =
aΔσ Marangoni
Curvature
Lσ
=
= Marangoni number
(2.10)
We now demonstrate how these dimensionless groups arise naturally from the nondimensionalization of
Navier-Stokes equations and the surface boundary conditions. We first introduce a dynamic pressure:
pd = p − ρg · x, so that gravity appears only in the boundary conditions. We consider the special case of
high Reynolds number flow, for which the characteristic dynamic pressure is ρU 2 . We define dimensionless
primed variables according to:
′
u = U u ,
pd = ρU 2 pd ,
′
x = ax ,
′
t =
′
t
a
U
,
(2.11)
where a and U are characteristic lenfth and velocity scales. Nondimensionalizing the Navier-Stokes equa
tions and appropriate boundary conditions yields the following system:
∂u ′
∂t′
(
+ u ·
′ ∇ ′
′
u = −∇pd +
′
)
∇ ′2
u , ∇ ′ · u = 0
′
′
1
Re
The normal stress condition assumes the dimensionless form:
′
−pd +
1
Fr
′
z +
2
Re
n ·
E ′
· n =
1
We
∇ ′
· n
(2.12)
(2.13)
The relative importance of surface tension to gravity is prescribed by the Bond number Bo, | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
)
The relative importance of surface tension to gravity is prescribed by the Bond number Bo, while that
of surface tension to viscous stresses by the capillary number Ca. In the high Re limit of interest, the
normal force balance requires that the dynamic pressure be balanced by either gravitational or curvature
stresses, the relative magnitudes of which are prescribed by the Bond number.
The nondimensionalization scheme will depend on the physical system of interest. Our purpose here
was simply to illustrate the manner in which the dimensionless groups arise in the theoretical formulation
of the problem. Moreover, we see that those involving surface tension enter exclusively through the
boundary conditions.
MIT OCW: 18.357 Interfacial Phenomena
8
Prof. John W. M. Bush
2.6. A few simple examples
Chapter 2. Definition and Scaling of Surface Tension
Figure 2.3: Surface tension may be measured by drawing a thin plate from a liquid bath.
2.6 A few simple examples
Measuring surface tension. Since σ is a tensile force per unit length, it is possible to infer its value by
slowly drawing a thin plate out of a liquid bath and measure the resistive force (Fig. 2.3). The maximum
measured force yields the surface tension σ.
Curvature/ Laplace pressure: consider an oil drop in water (Fig. 2.4a). Work is required to increase
the radius from R to R + dR:
dW = −podVo − pwdVw + γowd | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
from R to R + dR:
dW = −podVo − pwdVw + γowdA
(2.14)
where dVo = 4πR2dR = −dVw and dA = 8πRdR.
For mechanical equilibrium, we require
dW = −(p0 − pw)4πR2dR + γow8πRdR = 0 ⇒
ΔP = (po − pw) = 2γow/R.
mech. E
v
surf ace E
' v '
'
'
Figure 2.4: a) An oil drop in water b) When a soap bubble is penetrated by a cylindrical tube, air is
expelled from the bubble by the Laplace pressure.
MIT OCW: 18.357 Interfacial Phenomena
9
Prof. John W. M. Bush
2.6. A few simple examples
Chapter 2. Definition and Scaling of Surface Tension
Note:
1. Pressure inside a drop / bubble is higher than that outside ΔP ∼ 2γ/R ⇒ smaller bubbles have
higher Laplace pressure ⇒ champagne is louder than beer.
Champagne bubbles R ∼ 0.1mm, σ ∼ 50 dynes/cm, ΔP ∼ 10−2 atm.
2. For a soap bubble (2 interfaces) ΔP =
4σ
R
, so for R ∼ 5 cm, σ ∼ 35dynes/cm have ΔP ∼ 3×10−5atm.
More generally, we shall see that there is a pressure jump across any curved interface:
Laplace pressure Δp = σ∇ · n.
Examples:
1.
Soap bubble jet - Exit | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
Laplace pressure Δp = σ∇ · n.
Examples:
1.
Soap bubble jet - Exit speed (Fig. 2.4b)
Force balance: Δp = 4σ/R
4σ
∼ ρairU 2 ⇒ U ∼ ρair R
(
1/2
∼
)
(
4×70dynes/cm
0.001g/cm3·3cm
)
∼ 300cm/s
2.
Ostwald Ripening: The coarsening of foams (or emulsions) owing to diffusion of gas across inter
faces, which is necessarily from small to large bubbles, from high to low Laplace pressure.
3.
Falling drops: Force balance M g ∼ ρairU 2a gives
fall speed U ∼ ρga/ρair.
drop integrity requires ρairU 2 ∼ ρga < σ/a
raindrop size a < ℓc =
If a drop is small relative to the capillary length, σ maintains it against the destabilizing influence
v
of aerodynamic stresses.
σ/ρg ≈ 2mm.
v
2
MIT OCW: 18.357 Interfacial Phenomena
10
Prof. John W. M. Bush
3. Wetting
Puddles. What sets their size?
Knowing nothing of surface chemistry, one anticipates that Laplace pressure balances hydrostatic pressure
if σ/H ≥ ρgH ⇒ H < ℓc =
σ/ρg = capillary length.
Note:
J
1. Drops with R < ℓc remain heavily spherical
2. Large drops spread to depth H ∼ ℓc so that
Laplace + hydrostatic pressures balance at the | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
drops spread to depth H ∼ ℓc so that
Laplace + hydrostatic pressures balance at the
drop’s edge. A volume V will thus spread
to a radius R s.t. πR2ℓc = V , from which
R = (V /πℓc)
1/2
.
3. This is the case for H2O on most surfaces,
where a contact line exists.
Figure 3.1: Spreading of drops of increasing size.
Note: In general, surface chemistry can dominate and one need not have a contact line.
More generally, wetting occurs at fluid-solid contact. Two possibilities exist: partial wetting or total
wetting, depending on the surface energies of the 3 interfaces (γLV , γSV , γSL).
Now, just as σ = γLV is a surface energy per area or tensile force per length at a liquid-vapour surface,
γSL and γSV are analogous quantities at solid-liquid and solid-vapour interfaces.
The degree of wetting determined by spreading parameters:
S = [Esubstrate]dry − [Esubstrate]wet = γSV − (γSL + γLV )
(3.1)
where only γLV can be easily measured.
Total Wetting: S > 0 , θe = 0 liquid spreads completely in order to minimize its surface energy. e.g.
silicon on glass, water on clean glass.
Note: Silicon oil is more likely to spread than H2O since σw ∼ 70 dyn/cm > σs.o. ∼ 20 dyn/cm. Final
result | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
since σw ∼ 70 dyn/cm > σs.o. ∼ 20 dyn/cm. Final
result: a film of nanoscopic thickness resulting from competition between molecular and capillary forces.
Partial wetting: S < 0, θe > 0.
In absence of
g, forms a spherical cap meeting solid at a con
tact angle θe. A liquid is “wetting” on a particular
solid when θe < π/2, non-wetting or weakly wetting
when θe > π/2. For H2O, a surface is hydrophilic
if θe < π/2, hydrophobic if θe > π/2 and superhy
drophobic if θe > 5π/6.
Note: if g = 0, drops always take the form of a spherical cap ⇒ flattening indicates the effects of gravity.
Figure 3.2: The same water drop on hydrophobic
and hydrophilic surfaces.
11
4. Young’s Law with Applications
Young’s Law: what is the equilibrium contact angle θe ? Horizontal force balance at contact line:
γLV cos θe = γSV − γSL
γSV − γSL
γLV
S
γLV
= 1 +
(Y oung 1805)
(4.1)
Note:
cos θe =
1. When S ≥ 0, cos θe ≥ 1 ⇒ θe undefined and
spreading results.
2. Vertical force balance not s | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
1 ⇒ θe undefined and
spreading results.
2. Vertical force balance not satisfied at contact
line ⇒ dimpling of soft surfaces.
E.g. bubbles in paint leave a circular rim.
3. The static contact angle need not take its equi
librium value ⇒ there is a finite range of pos
sible static contact angles.
Back to Puddles: Total energy:
Figure 4.1: Three interfaces meet at the contact line.
E = (γSL − γSV )A + γLV A +
ρgh2A
1
2
= −S + ρgV h
V
h
1
2
(4.2)
Minimize energy w.r.t. h:
dE
dh
surf ace energy
v
1
2 + ρgV = 0 when −S/h2 = ρg ⇒
= SV h
grav. pot. energy
v
1
2
1
2
h0 =
J
−2S
ρg = 2ℓc sin θ
e
2 gives puddle depth, where ℓc =
σ/ρg.
J
Capillary Adhesion: Two wetted surfaces can
stick together with great strength if θe < π/2, e.g.
Fig. 4.2.
Laplace Pressure:
ΔP = σ
)
i.e. low P inside film provided θe < π/2.
If H ≪ R, F = πR2 2σ
between the plates.
is the attractive force
1 − cos θe
H/2
R
≈ − 2σ
cos θe
H
cos θe
H
(
Figure 4.2 | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
2
R
≈ − 2σ
cos θe
H
cos θe
H
(
Figure 4.2: An oil drop forms a capillary bridge
between two glass plates.
E.g. for H2O, with R = 1 cm, H = 5 µm and θe = 0, one finds ΔP ∼ 1/3 atm and an adhesive force
F ∼ 10N , the weight of 1l of H2O.
Note: Such capillary adhesion is used by beetles in nature.
4.1 Formal Development of Interfacial Flow Problems
Governing Equations: Navier-Stokes. An incompressible, homogeneous fluid of density ρ and viscosity
µ = ρν (µ is dynamic and ν kinematic viscosity) acted upon by an external force per unit volume f evolves
according to
∇ · u = 0
∂u
∂t
ρ
(
+ u · ∇u = −∇p + f + µ∇2 u
)
(continuity)
(4.3)
(Linear momentum conservation)
(4.4)
12
4.1. Formal Development of Interfacial Flow Problems
Chapter 4. Young’s Law with Applications
This is a system of 4 equations in 4 unknowns (u1, u2, u3, p). These N-S equations must be solved subject
to appropriate BCs.
Fluid-Solid BCs: “No-slip”: u = Usolid.
E.g.1 Falling sphere: u = U on sphere surface, where U is the sphere velocity.
E.g.2 Convection in a box: u = | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
, where U is the sphere velocity.
E.g.2 Convection in a box: u = 0 on the box surface.
But we are interested in flows dominated by interfacial effects. Here, in general, one must solve N-S
equations in 2 domains, and match solutions together at the interface with appropriate BCs. Difficulty:
These interfaces are free to move ⇒ Free boundary problems.
Figure 4.3: E.g.3 Drop motion within a fluid.
Figure 4.4: E.g.4 Water waves at an air-water in
terface.
Continuity of Velocity at an interface requires that u = uˆ.
And what about p ? We’ve seen Δp ∼ σ/R for a static bubble/drop, but to answer this question in
general, we must develop stress conditions at a fluid-fluid interface.
Recall: Stress Tensor. The state of stress within an incompressible Newtonian fluid is described by
the stress tensor: T = −pI + 2µE where E =
is the deviatoric stress tensor. The
associated hydrodynamic force per unit volume within the fluid is ∇ · T .
One may thus write N-S eqns in the form: ρ Du = ∇ · T + f = −∇p + µ∇2u + f.
Now: Tij = force / area acting in the ej direction on a surface with a normal ei.
(� | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
= force / area acting in the ej direction on a surface with a normal ei.
(∇u) + (∇u)T
1
2
Dt
[
]
Note:
1.
normal stresses (diagonals) T11, T22, T33 in
volve both p and ui
2. tangential
stresses (off-diagonals) T12, T13,
etc., involve only velocity gradients, i.e. vis
cous stresses
3.
Tij is symmetric (Newtonian fluids)
4.
t(n) = n·T = stress vector acting on a surface
with normal n
∂u
x
∂y
E.g. Shear flow. Stress in lower boundary is tan
gential. Force / area on lower boundary:
Tyx = µ
y-surface in x-direction.
Note: the form of T in arbitrary curvilinear coordi- Figure 4.5: Shear flow above a rigid lower bound-
nates is given in the Appendix of Batchelor.
|y=0 = µk is the force/area that acts on
ary.
MIT OCW: 18.357 Interfacial Phenomena
13
Prof. John W. M. Bush
5. Stress Boundary Conditions
Today:
1. Derive stress conditions at a fluid-fluid inter
face. Requires knowledge of T = −pI + 2µE
2. Consider several examples of fluid statics
Recall: the curvature of a string under tension may
support a normal force. (see right) | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
the curvature of a string under tension may
support a normal force. (see right)
Figure 5.1: String under tension and the influence
of gravity.
5.1 Stress conditions at a fluid-fluid interface
We proceed by deriving the normal and tangential stress boundary conditions appropriate at a fluid-fluid
interface characterized by an interfacial tension σ.
Figure 5.2: A surface S and bounding contour C on an interface between two fluids. Local unit vectors
are n, m and s.
Consider an interfacial surface S bounded by a closed contour C. One may think of there being a force
per unit length of magnitude σ in the s-direction at every point along C that acts to flatten the surface S.
Perform a force balance on a volume element V enclosing the interfacial surface S defined by the contour
C:
Du
Dt
ρ
V
dV =
f dV +
V
S∗
[t(n) + tˆ(nˆ)] dS +
σs dℓ
C
(5.1)
Here ℓ indicates arc-length and so dℓ a length increment along the curve C.
t(n) = n · T is the stress vector, the force/area exerted by the upper (+) fluid on the interface.
The stress tensor is defined in terms of the local fluid pressure and velocity field as T = −pI+µ ∇u + (∇u)T
The stress exerted on the interface by the lower (-) fluid is ˆt(nˆ) = | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
The stress exerted on the interface by the lower (-) fluid is ˆt(nˆ) = nˆ · Tˆ = −n · T
where Tˆ = −pˆI + ˆµ ∇uˆ + (∇uˆ)T
[
.
.
]
[
]
Physical interpretation of terms
ρ Du dV
f dV
body forces acting within V
:
:
V Dt
I
inertial force associated with acceleration of fluid in V
S
S
I
I
I
t(n) dS :
ˆt(nˆ) dS :
σs dℓ :
C
I
hydrodynamic force exerted by upper fluid
hydrodynamic force exerted by lower fluid
surface tension force exerted on perimeter.
14
5.1. Stress conditions at a fluid-fluid interface
Chapter 5. Stress Boundary Conditions
Now if ǫ is the characteristic height of our volume V and R its characteristic radius, then the accel-
eration and body forces will scale as R2ǫ, while the surface forces will scale as R2. Thus, in the limit of
ǫ → 0, the latter must balance.
Now we have that
t(n) + tˆ(nˆ) dS +
σs dℓ = 0
Z
C
Z
S
t(n) = n · T ,
ˆ
ˆt(nˆ) = nˆ · T = −n · T
Moreover, the application of Stokes Theorem (see below) allows us to write
where the tangential (surface) gradient operator, defined
σs dℓ =
Z
S
Z
C
∇Sσ − σn (∇S · n) dS
∇
S = [I − nn] · = − n
∇ ∇
∂
∂n
(5.2)
(5.3)
(5.4)
(5.5)
appears | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
∇
∂
∂n
(5.2)
(5.3)
(5.4)
(5.5)
appears because σ and n are only defined on the surface S. We proceed by dropping the subscript s on
∇, with this understanding. The surface force balance thus becomes
ˆ
n · T − n · T
(cid:17)
dS =
Z
S
Z (cid:16)
S
n (∇ · n) − ∇σ dS
σ
(5.6)
Now since the surface S is arbitrary, the integrand must vanish identically. One thus obtains the interfacial
stress balance equation, which is valid at every point on the interface:
Stress Balance Equation
n · T − n · T = σn (∇ · n) − ∇σ
ˆ
(5.7)
Interpretation of terms:
n · T stress (force/area) exerted by + on - (will generally have both ⊥ and k components)
ˆn · T stress (force/area) exerted by - on + (will generally have both ⊥ and k components)
σn (∇ · n)
∇σ
normal curvature force per unit area associated with local curvature of interface, ∇ · n
tangential stress associated with gradients in σ
Normal stress balance Taking n·(5.7) yields the normal stress balance
ˆ
n · T · n − n · T · n = σ(∇ · n)
(5.8)
The jump in the normal stress across the interface is balanced by the curvature pressure.
Note: If ∇ · n = 0, there must be a normal stress jump there, which generally involves both pressure and
viscous terms.
MIT OCW: 18.357 Interfacial Phenomena
15
Prof. John W. M. Bush
6
5.2. Appendix A : Useful identity
Chapter 5. Stress Boundary Conditions
Tangential stress balance Taking d·(5.7), where d is any linear combination of s and m (any tangent
to S), yields the tangential stress balance at the interface:
ˆ
n · T · d − n · T · d = ∇σ · d
(5.9)
Physical Interpretation
• LHS represents the jump in tangential components of the hydro | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
T · d = ∇σ · d
(5.9)
Physical Interpretation
• LHS represents the jump in tangential components of the hydrodynamic stress at the interface
• RHS represents the tangential stress (Marangoni stress) associated with gradients in σ, as may
result from gradients in temperature θ or chemical composition c at the interface since in general
σ = σ(θ, c)
• LHS contains only the non-diagonal terms of T - only the velocity gradients, not pressure; therefore
any non-zero ∇σ at a fluid interface must always drive motion.
5.2 Appendix A : Useful identity
Recall Stokes Theorem:
Z
Along the contour C, dℓ = m dℓ, so that we have
Z
C
F · dℓ =
F · m dℓ =
Z
C
n · (∇ ∧ F ) dS
S
Z
S
n · (∇ ∧ F ) dS
Now let F = f ∧ b, where b is an arbitrary constant vector. We thus have
(f ∧ b) · m dℓ =
Z
S
Z
C
n · (∇ ∧ (f ∧ b)) dS
Now use standard vector identities to see (f ∧ b) · m = −b · (f ∧ m) and
(5.10)
(5.11)
(5.12)
∇ ∧ (f ∧ b) = f (∇ · b) − b (∇ · f ) + b · ∇f − f · ∇b = −b (∇ · f ) + b · ∇f
(5.13)
since b is a constant vector. We thus have
b ·
Z
C
(f ∧ m) dℓ = b ·
Z
S
[n (∇ · f ) − (∇f ) · n] dS
Since b is arbitrary, we thus have
(f ∧ m) dℓ =
Z
S
Z
C
[n (∇ · f ) − (∇f ) · n] dS
(5.14)
(5.15)
We now choose f = σn, and recall that n ∧ m = −s. One thus obtains
−
[n∇ · (σn) − ∇ (σn) · n] dS = [n∇σ · n + σn | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
[n∇ · (σn) − ∇ (σn) · n] dS = [n∇σ · n + σn (∇ · n) − ∇σ − σ (∇n) · n] dS.
σsdℓ =
S
C
We note that ∇σ · n = 0 since ∇σ must bRe tangent to the surface S and (
(1) = 0, and so obtain the desired result:
R
S
R
1 ∇2
∇n) · n = ∇2
1
(n · n) =
(5.16)
σs dℓ =
Z
S
Z
C
[∇σ − σn (∇ · n)] dS
MIT OCW: 18.357 Interfacial Phenomena
16
Prof. John W. M. Bush
5.3. Fluid Statics
Chapter 5. Stress Boundary Conditions
5.3 Fluid Statics
We begin by considering static fluid configurations, for which the stress tensor reduces to the form T = −pI,
so that n · T · n = −p, and the normal stress balance equation (5.8) assumes the simple form:
pˆ − p = σ∇ · n
(5.17)
The pressure jump across a static interface is balanced by the curvature force at the interface. Now since
n · T · d = 0 for a static system, the tangential stress balance indicates that ∇σ = 0. This leads to
the following important conclusion: There cannot be a static system in the presence of surface tension
gradients. While pressure jumps can sustain normal stress jumps across a fluid interface, they do not
contribute to the tangential stress jump. Consequently, tangential surface (Marangoni) stresses can only
be balanced by viscous stresses associated with fluid motion. We proceed by applying equation (5.17 | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
only
be balanced by viscous stresses associated with fluid motion. We proceed by applying equation (5.17) to
describe a number of static situations.
1. Stationary Bubble : We consider a spherical air bubble of radius R submerged in a static fluid.
What is the pressure drop across the bubble surface?
The divergence in spherical coordinates of F = (Fr, Fθ, Fφ) is given by
∇ · F =
Hence ∇ · n|S =
so the normal stress jump (5.17) indicates that
1
∂
r sin θ ∂θ
r2|r=R =
∂
r sin φ ∂φ
(sin θFθ) +
2
R
r2Fr +
1 ∂
)
r2 ∂r
1 ∂
r2 ∂r
Fφ.
1
(
ΔP = pˆ − p =
2σ
R
(5.18)
The pressure within the bubble is higher than that outside by an amount proportional to the surface
tension, and inversely proportional to the bubble size. As noted in Lec. 2, it is thus that small bubbles
are louder than large ones when they burst at a free surface: champagne is louder than beer. We note
that soap bubbles in air have two surfaces that define the inner and outer surfaces of the soap film;
consequently, the pressure differential is twice that across a single interface.
2. The static meniscus (θe < π/2)
Consider a situation where the pressure within a
static fl | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
iscus (θe < π/2)
Consider a situation where the pressure within a
static fluid varies owing to the presence of a gravi- meniscus (below).
tational field, p = p0 + ρgz, where p0 is the constant
ambient pressure, and g = −gzˆ is the grav. acceler
ation. The normal stress balance thus requires that
the interface satisfy the Young-Laplace Equation:
early with z. Such a situation arises in the static
ρgz = σ∇ · n
(5.19)
The vertical gradient in fluid pressure must be bal
anced by the curvature pressure; as the gradient is
constant, the curvature must likewise increase lin-
Figure 5.3: Static meniscus near a wall.
The shape of the meniscus is prescribed by two factors: the contact angle between the air-water
interface and the wall, and the balance between hydrostatic pressure and curvature pressure. We treat
the contact angle θe as given; noting that it depends in general on the surface energy. The normal
force balance is expressed by the Young-Laplace equation, where now ρ = ρw − ρair ≈ ρw is the density
difference between water and air. We define the free surface by z = η(x); equivalently, we define a
functional f (x, z) = z − η(x) that vanishes on the surface. The normal to the surface z = η | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
η(x) that vanishes on the surface. The normal to the surface z = η(x) is thus
n =
∇f
|∇f |
=
zˆ − η′(x)xˆ
[1 + η′(x)2]
1/2
(5.20)
MIT OCW: 18.357 Interfacial Phenomena
17
Prof. John W. M. Bush
5.3. Fluid Statics
Chapter 5. Stress Boundary Conditions
As deduced in Appendix B, the curvature of the free surface ∇ · nˆ , may be expressed as
∇ · nˆ =
−ηxx
(1 + η2)3/2
x
≈ −ηxx
(5.21)
Assuming that the slope of the meniscus remains sufficiently small, η2
x
(5.21), so that (5.19) assumes the form
≪ 1, allows one to linearize equation
ρgη = σηxx
(5.22)
Applying the boundary condition η(∞) = 0 and the contact condition ηx(0) = − cot θ, and solving (5.22)
thus yields
η(x) = ℓc cot θee
−x/ℓc
(5.23)
σ/ρg is the capillary length. The meniscus formed by an object floating in water is exponen-
where ℓc =
tial, decaying over a length scale ℓc. Note that this behaviour may be rationalized as follows: the system
arranges itself so that its total energy (grav. potential + surface) is minimized.
p
3. Floating Bodies
Floating bodies must be supported by some combination of buoyancy and curvature forces. Specifically,
since the fluid pressure beneath the interface is related to the atmospheric pressure p0 above the interface
by
p = p0 + ρgz + σ∇ · n ,
one may express the vertical force balance as
The buoyancy force
M g = z ·
Z
C
−pndℓ = Fb + | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
may express the vertical force balance as
The buoyancy force
M g = z ·
Z
C
−pndℓ = Fb + F
c
.
buo
yancy
|{z}
c
urvature
|{z}
Fb = z ·
Z
C
ρgzn dℓ = ρgVb
is thus simply the weight of the fluid displaced above the object and inside the line of tangency (see figure
below). We note that it may be deduced by integrating the curvature pressure over the contact area C
using the first of the Frenet-Serret equations (see Appendix C).
F
c = z ·
Z
C
∇ ·
σ (
n) n dℓ = σz ·
dt
C dℓ
Z
dℓ = σz · (t1 − t2) = 2σ sin θ
(5.27)
At the interface, the buoyancy and curvature forces must balance precisely, so the Young-Laplace relation
is satisfied:
0 = ρgz + σ∇ · n
(5.28)
Integrating this equation over the meniscus and taking the vertical component yields the vertical force
balance:
where
b + F m
F m
c = 0
F m
b = z ·
Z
Cm
ρgzn dℓ = ρgVm
(5.29)
(5.30)
F m
c = z ·
Z
Cm
σ (∇ · n) n dℓ = σz ·
dt
Cm dℓ
Z
dℓ = σz · (t1 − t
2) = −2σ sin θ
(5.31)
where we have again used the Frenet-Serret equations to evaluate the curvature force.
MIT OCW: 18.357 Interfacial Phenomena
18
Prof. John W. M. Bush
(5.24)
(5.25)
(5.26)
5.3. Fluid Statics
Chapter 5. Stress Boundary Conditions
Figure 5.4: A floating non-wetting body is supported by a combination of buoyancy and curvature forces, | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
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