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Chapter 15 ∇ in Other Coordinates

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Chapter 15 ∇ in Other Coordinates r in other coordinates 1 Chapter 15 r in other coordinates On a number of occasions we have noticed that del is geometrically determined | it does not depend on a choice of coordinates for Rn. This was shown to be true for rf, the gradient of a function from Rn to R (Section 2H). It was also veri¯ed for r ² F , the divergence of a function from Rn to Rn (Section 14B). And in the case n = 3, we saw in Section 13G that it is true for r £ F , the curl of a function from R3 to R3. These three instances beg the question of how we might express r in other coordinate systems for Rn. A recent example of this is found in Section 13G, where a formula is given for r £ F in terms of an arbitrary right-handed orthonormal frame for R3. We shall accomplish much more in this chapter. A very interesting book about r, by Harry Moritz Schey, has the interesting title Div, Grad, Curl, And All That. A. Biorthogonal systems We begin with some elementary linear algebra. Consider an arbitrary frame f'1;'2;:::;'ng for Rn. We know of course that the Gram matrix is of great interest: G = ('i ² 'j): This is a symmetric positive de¯nite matrix, and we shall denote its entries as gij = 'i ² 'j: Of course, G is the identity matrix () we have an orthonormal frame. We also form the matrix © whose columns are the vectors 'i. Symbolically we write © = ('1 '2 :::'n): We know that © is an orthogonal matrix () we have an orthonormal frame (Problem 4{20). Now denote by ª the transpose of the inverse of ©, and express this new matrix in terms of its columns as ª = (Ã1 Ã2 :::Ãn): We then have the matrix product 0 1 t Ã1 t B . C ª © = @ . A ('1 :::'n) t Ãn = (Ãi ² 'j): 2 Chapter 15 But since ªt equals the inverse of ©, we conclude that Ãi ² 'j = ±ij: DEFINITION. Two frames f'1;:::;'ng and fÃ1;:::;Ãng are called a biorthogonal system if Ãi ² 'j = ±ij: Clearly, f'1;:::;'ng and f'1;:::;'ng form a biorthogonal system () the frame f'1;:::;'ng is an orthonormal one. So the present de¯nition should be viewed as a generalization of the concept of orthonormal basis. Also it is clear that the relation between the 'i's and the Ãi's given here is completely symmetric. PROBLEM 15{1. Let a frame for R2 be f^{; a^{ + b|^g, where of course b 6= 0. Compute the corresponding fÃ1;Ã2g which produces a biorthogonal system. Sketch all four vectors on a copy of R2. 3 PROBLEM 15{2. Let f'1;'2;'3g be a frame for R . Prove that the biorthogonal frame is given by '1 £ '2 Ã3 = etc: ['1;'2;'3] PROBLEM 15{3. Suppose f'1;:::;'ng is an orthogonal frame. Show that the corresponding vectors Ã1;:::;Ãn are given as 'i Ãi = 2 : k'ik PROBLEM 15{4. Given a frame f'1;:::;'ng denote J = det ©: Prove that det G = J 2. r in other coordinates 3 Finally, a little more notation. It has become customary to denote the inverse of G = (gij) by G¡1 = (gij) (called \raising the indices"). That is, Xn Xn ik kj g gkj = gikg = ±ij: k=1 k=1 Inasmuch as G = ©t©; we conclude that ª = (©t)¡1 = ©G¡1: In terms of entries of these matrices this equation means that Xn kj (ª)ij = (©)ikg : k=1 Thus we ¯nd that the columns satisfy Xn kj Ãj = 'kg : k=1 Rewriting this equation gives Xn ij Ãi = g 'j: j=1 The corresponding inverse equation is of course Xn 'i = gijÃj: j=1 PROBLEM 15{5. Prove that for all v 2 Rn Xn v = (Ãi ² v)'i: i=1 4 Chapter 15 The result of this problem is that when you wish to express v as a linear combination of the basis vectors 'i, the corresponding coe±cients are given directly in terms of inner products with the Ãi's. You might think of the formal expression Xn (Ãi ² ( ))'i i=1 as representing the identity linear function on Rn. In other words, PROBLEM 15{6. Prove that Xn t I = 'iÃi : i=1 B. The gradient We continue with an arbitrary biorthogonal system on Rn, and maintain the notation of Section A. f Suppose that Rn ¡! R is di®erentiable at a point x 2 Rn. Our task is to express the n vector rf(x) 2 R as a linear combination of the vectors 'i in the given frame. This is quite an easy task. In fact, Problem 15{5 gives immediately Xn rf(x) = rf(x) ² Ãi'i: i=1 The inner products in this formula are of course directional derivatives of f in the directions Ãi. Thus, in the notation of Section 2C Xn rf(x) = Df(x; Ãi)'i: i=1 PROBLEM 15{7. Show that everything in this formula can be expressed entirely in terms of the frame f'1;:::;'ng by a double sum Xn ij rf(x) = g Df(x; 'j)'i: i;j=1 r in other coordinates 5 C. The divergence We want to discuss a vector ¯eld f de¯ned on an open subset of Rn. We can thus regard f as a function from Rn to Rn, and as such it has a derivative. At a point x in its domain, the derivative Df(x) is a linear transformation of Rn to Rn, represented in terms of the standard coordinate basise ^1;:::; e^n, by the n £ n Jacobian matrix µ ¶ @f i : @xj It is the trace of this matrix which is r ² f, the divergence of f. This observation is the key to our representation of r ² f, and we need a simple fact from linear algebra: THEOREM. Suppose A is a real n £ n matrix, and regard the 'i's and Ãi's as column vectors. Then Xn traceA = AÃi ² 'i: i=1 PROOF. We use the matrices © and ª from Section A, so that ¡t AÃi ² 'j = ji entry of the matrix © Aª: Thus Xn t AÃi ² 'i = trace(© Aª) i=1 = trace(Aª©t) (Problem 4{2) = trace(A); since ª and ©t are inverses. QED As a result we now have Xn @f r ² f(x) = i @x i=1 i Xn = Df(x)Ãi ² 'i i=1 Xn = Df(x; Ãi) ² 'i: i=1 6 Chapter 15 D. The curl Before going into the representation of curl, we summarize what we have obtained to this point. In terms of a ¯xed biorthogonal system we have found Xn rf(x) = rf(x) = Df(x; Ãi)'i i=1 and Xn divF (x) = r ² F (x) = DF (x; Ãi) ² 'i: i=1 These two formulas can be incorporated as a single expression Xn r = 'i times D(; Ãi) i=1 where r and the right side are both regarded as operators. They operate on scalar-valued functions to produce the gradient and on vector ¯elds to produce the divergence. In the gradient situation \times" is scalar multiplication, whereas in the divergence situation it is dot product. A good way to remember this formula is to replace the directional derivative Df(x; Ãi) by the symbol @f @Ãi Then we have Xn @ r = ' times : i @à i=1 i This leads us to an educated guess for curl, where we just use the same formula. Thus we 3 suppose that f'1;'2;'3g and fÃ1;Ã2;Ã3g form a biorthogonal system for R . Then we have THEOREM. For a vector ¯eld F on R3, X3 curlF (x) = r £ F (x) = 'i £ DF (x; Ãi): i=1 r in other coordinates 7 PROOF. Every vector ¯eld F can be expressed in the given frame in the form X3 F (x) = fj(x)'j: i=1 Thus it su±ces to prove the formula for the special case of vector ¯elds of the form fj(x)'j, for j = 1, 2, 3. Thus we assume F (x) = f(x)'j: We have from the product rule of Problem 13{14, r £ F (x) = rf(x) £ 'j (since 'j is constant) and our formula for gradient gives X3 r £ F (x) = Df(x; Ãi)'i £ 'j i=1 X3 = 'i £ Df(x; Ãi)'j i=1 X3 = 'i £ D(f'j)(x; Ãi) i=1 X3 = 'i £ DF (x; Ãi): i=1 QED PROBLEM 15{8. Are you surprised that the formula for curl does not require the frame f'1;'2;'3g to be right-handed? Explain what happens if '3 is replaced with ¡'3. PROBLEM 15{9. Prove that '1 £ '2 = JÃ3; '2 £ '3 = JÃ1; '3 £ '1 = JÃ2: 8 Chapter 15 PROBLEM 15{10. Show how the formulas of the preceding problem behave when each 'i is replaced with the vector ai'i, where the ai's are nonzero scalars. E. Curvilinear coordinates All that we have done up to now is represent r in terms of a ¯xed frame for Rn. In practice, however, the coordinates themselves are changed in a perhaps nonlinear fashion. We still need convenient expressions for r in terms of the new coordinates. We shall call our coordinate system T . What we mean is that T is a C1 function from Rn to Rn, and has a C1 inverse as well. (Usually the domains of T and T ¡1 will not be all of Rn.) The notation we shall use is x = T (u); representing the formula which determines x = (x1; : : : ; xn) in terms of the curvilinear coor- dinates u = (u1; : : : ; un).
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