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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 ...
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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 primitiv...
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/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 a...
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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 le...
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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) ...
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��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...
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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 isom...
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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...
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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 ⊂ G...
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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, th...
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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...
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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.
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-- 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 ma...
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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, su...
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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 ...
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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-' ...
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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 ...
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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 ...
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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 ch...
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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 n...
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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 b...
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. 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 u...
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(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, ...
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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 formul...
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= (-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...
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. 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...
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(-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 mtr...
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~ 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 err...
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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 ...
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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 multip...
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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...
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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. ...
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+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 com...
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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 colum...
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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 th...
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, 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 th...
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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...
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. 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. T...
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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....
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(-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 an...
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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....
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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 de...
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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 (**), ...
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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)....
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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...
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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 produ...
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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...
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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 (D...
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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’...
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) 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) ...
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) 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 ∥...
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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, an...
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)((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 ...
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=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∇ ...
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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(...
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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...
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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 w...
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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 ener...
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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...
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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...
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(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 ...
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(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 grant...
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(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 t...
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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 . . . . . . . . . . . . . . . . . . . . . . ....
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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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ...
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ows 10 Marangoni Flows II 10.1 Tears of Wine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Surfactants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 Surfactant-in...
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. . . . . . . . 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 . . . . . . . . . . . . ....
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aylor Instability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2 Kelvin-Helmholtz Instability 15 Contact angle hysteresis, Wetting of textured solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ...
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. . . . . . . . 17 Coating: Dynamic Contact Lines 17.1 Contact Line Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Spreading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.1 Spreading of small drops on solids . . . . . . . . . . . . . . . . . . 18.2 Immis...
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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 con...
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• 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...
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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 wor...
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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 ar...
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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 ... ...
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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 betwe...
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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 energ...
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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 equation...
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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 ...
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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...
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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 ...
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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 prescrib...
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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 a...
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) 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 st...
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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 cy...
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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 necess...
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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. ...
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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 < ...
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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 e...
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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 Develop...
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, 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 m...
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= 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 (Newto...
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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 characterize...
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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 : ...
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∇ ∂ ∂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 integra...
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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 s...
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[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 ...
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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 ...
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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 satis...
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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...
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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 c...
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