old Section 12 retired 3_2_12

old Section 12, retired on this day PhL 3.2.12 Change made are certain simple factors due to the fact that curl B is a tensor density (none of the final results are affected), and also I changed the very end since I now have general formulas for expansion of a tensor. **************************************** 12. The Curl in curvilinear coordinates The vector curl is defined only in N=3 dimensions (but see section (f) below). Picture B is used. In Cartesian coordinates one writes [curl B]i(x) = [ x B(x)]i = εijk∂jBk(x) but when expressed in terms of curvilinear coordinates and components, the form is different. (a) Definition of curl B Consider the x-space differential 3-piped shown on the right side of the figure in Section 8 (a), This 3-piped has three pairs of parallel faces. Within each pair, the "near" face touches the point x at which the tails of the spanning vectors meet, while the "far" face does not. As shown in Section 8 (c), the vector area associated with the face pair n is ( |J| = g'1/2 when g = 1) dAn = | det(Sij)| en ( Πi≠n dx'i) = |J| en ( Πi≠n dx'i) = g'1/2 en ( Πi≠n dx'i) n = 1,2,3 where J = det(Sij) is the Jacobian, as in Section 5 (k), and en is a reciprocal base vector, as in Section 6 or Appendix A (called En). Area dAn is the out-facing vector area for the far face of pair n, while - dAn is the out-facing vector for the near face. Consider now the line integral of a vector field B(x) around the boundary of near face n, where the circulation sense of the integral is determined from the right-hand-rule by the direction of dAn which is the same as the direction of en. For example, for the bottom face (near face 3) of the 3-piped shown above, this vector points "up", or toward the center of the 3-piped. Denote this line integral by ( Bdx)n Sometimes this line integral is referred to as "the circulation" or "the rotation" of B around near face n (and rot B is another notation used for curl B). In x-space the quantity C(x) ≡ curl B(x) is a vector field defined in the following manner in the limit that all the differentials dx'n → 0 : C dAn = ( Bdx)n C ≡ curl B Since dAn is given above in terms of en, C should be expanded on the ek and one writes C = Σk=13 C'k ek so that C dAn = (Σk C'k ek) (g'1/2 en ( Πi≠n dx'i) ) = C'n g'1/2 ( Πi≠n dx'i) and then C'n(x') ( Πi≠n dx'i) = ( Bdx)n Our task is to compute this line integral and thereby come up with an expression for C'n(x'), the components of curl B when B is expanded onto the ek in x-space. (b) Computation of the line integral This shall be done for the bottom face (n=3) of the x-space differential 3-piped. Since e3 is "up", the circulation is a counterclockwise line integral around the boundary of the bottom face. It is useful to have the above picture near at hand to allow visualization of the four contributions to the line integral: ( Bds)3 ≈ [B(xfront) - B(xback)] (e1 dx'1) + [B(xright) - B(xleft)] (e2 dx'2) where B(xfront) refers to the value of B at the center of the "front" edge of the parallelogram which is the bottom face, and similarly for the other three edges. In the limit that the dx'n → 0, this simple approximation of the line integral is "good enough" to produce the desired results. Motivated by ei ej = δij, expand B as follows B = B'jej where B'j(x') = B(x) ej where the B'j are the covariant components of B in x'-space. This gives ( Bdx)3 = [B'1(x'front) - B'1(x'back)] dx'1 + [B'2(x'right) - B'2(x'left)] dx'2 where x'front = F(xfront) and similarly for the other three points. In x-space one has dxBF ≡ xback - xfront = e2dx'2 dxRL ≡ xright - xleft = e1dx'1 Applying matrix R gives the corresponding x'-space equations (recall dx' = R dx and e'n = Ren) dx'BF ≡ x'back - x'front = e'2dx'2 e'2 = (0,1,0...) dx'RL ≡ x'right - x'left = e'1dx'1 e'1 = (1,0,0...) Using the fact that f(x'+dx') ≈ f(x') + Σn∂nf(x') dx'n one finds B'1(x'back) ≈ B'1(x'front) + (∂B'1/∂x'2) dx'2 dx' = e'2dx'2 B'2(x'right) ≈ B'2 (x'left)) + (∂B'2/∂x'1) dx'1 dx' = e'1dx'1 so the circulation integral is then ( Bdx)3 ≈ – (∂B'1/∂x'2) dx'2 dx'1 + (∂B'2/∂x'1) dx'1 dx'2 = [– (∂B'1/∂x'2) + (∂B'2/∂x'1)] dx'1 dx'2 = [– ∂'2B'1 + ∂'1B'2] dx'1 dx'2 = [∂'1B'2 – ∂'2B'1] dx'1 dx'2 = ε3ab ∂'aB'b ( Πi≠3 dx'i) Repeating this calculation for faces 1 and 2 produces cyclic results, and all three face line integrals can be summarized as (where equality holds in the limit dx'i → 0) ( Bdx)n = εnab ∂'aB'b ( Πi≠n dx'i) Appendix D discusses the tensor ε known as the Levi-Cevita ε tensor. In Cartesian space, the up and down position of the indices does not matter, as for any tensor. In non-Cartesian space up and down does matter, as with any tensor. The only fact needed here is that ε'abc... = εabc... where ε' is the tensor in x'-space, as shown in Appendix D (d). In Cartesian space one can regard εabc... = εabc... as a bookkeeping permutation tensor with the properties given in Section 7 (h). Installing the prime on ε, ( Bdx)n = ε'nab ∂'aB'b ( Πi≠n dx'i) and this integral is then given entirely in terms of x'-space coordinates and objects. (c) Solving for the curl The equation for the curl obtained at the end of section (a) was C'n(x') ( Πi≠n dx'i) = ( Bdx)n Insert the section (b) result for ( Bdx)n to get C'n(x') ( Πi≠n dx'i) = ε'nab ∂'aB'b ( Πi≠n dx'i) The differentials cancel out, so then take dx'i→ 0 and thus shrink the 3-piped around the point of interest x = F-1(x') so that C'n = [(1/) ε'nab ∂'aB'b ] C = C'n en B = B'n en curl B = C = [(1/) ε'nab ∂'aB'b ] en The comparison between the curvilinear and Cartesian expressed curls is this: [curl B](x) = [(1/) ε'nab ∂'aB'b(x') ] en = (1/) { [∂'1B'2 - ∂'2B'1] e3 + cyclic } [curl B](x) = εnab ∂aBb(x) = { [ ∂1B2 - ∂2B1 ] + cyclic } Appendix D (h) shows that the vector C = curl B really is a tensorial vector if B is a tensorial vector. Therefore the notation C = C'nen is justified. Comment: In the first line above one can replace [curl B](x) by [ x B](x) with the understanding that the LHS is the curl in Cartesian x-space and the RHS is expressing this LHS in terms of x'-space coordinates and objects. The RHS is certainly not equal to ' x B' = ε'nab ∂'aB'b(x') ' . It is to avoid this possible confusion that the curl is written out as the word curl, and the same comment applies to the other differential operators. (d) Various forms of the curl The first form is that just presented above, C'n = [(1/) ε'nab ∂'aB'b ] C = C'n en B = B'n en curl B = [(1/) ε'nab ∂'aB'b ] en If it is desired to have contravariant components of B, one gets C'n = [(1/) ε'nab ∂'a(g'bcB'c )] C = C'n en B = B'n en curl B = [(1/) ε'nab ∂'a(g'bcB'c )] en For practical applications, one usually wants both vectors expanded on the n unit vectors in this way C = C'n en = (C'n h'n) n ≡ C'n n C'n = h'n C'n B = B'n en = (B'n h'n) n ≡ B'n n B'n = h'nB'n so that C'n = [(1/) h'n ε'nab ∂'a(g'bcb'c/h'c )] C = C'n n B = B'n n curl B = [(1/) h'n ε'nab ∂'a(g'bcb'c/h'c )] n curl B = C To summarize: B'c = RcdBd g' = g'(x') etc. [curl B](x) = ε'nab [(1/) ∂'aB'b ] en B = B'nen [curl B](x) = ε'nab [(1/) ∂'a(g'bcB'c )] en B = B'nen [curl B](x) = ε'nab [(1/) h'n ∂'a(g'bc B'c/h'c )] n B = B'n n [curl B](x) = εnab ∂aBb(x) // Cartesian B = Bn Converting from Picture B to Picture MS (see Section 9 (c)) one gets : [curl B](x) = εnab [(1/) ∂aBb ] en B = Bnen [curl B](x) = εnab [(1/) ∂a(gbcBc )] en B = Bnen [curl B](x) = εnab[(1/) hn ∂a(gbc Bc/ hc )] n B = Bnn [curl B](x) = εnab ∂aBb(x) // Cartesian B = Bn Warning: The object εnab in the first three equations is now in u-space which is non-Cartesian, so up and down index positions do matter, but when indices are all up, it continues to be the normal permutation tensor. Each of the above results can be written as a determinant using the idea det(Q) ≡ Σi Q1i cof(Q1i) : [curl B](x) = (1/) B = Bn en [curl B](x) = (1/) B = Bn en [curl B](x) = (1/) B = Bn n // M&S 1.07 [curl B](x) = // here ∂n= ∂/∂xn and Bi = Bi(x) B = Bn (e) The curl in orthogonal coordinate systems For such systems gij = δi,jhi2 and det(gab) = h12h22h32 so = h1h2h3 . It is then a simple matter to convert all the above forms and the results are: Picture B: B'c(x') = RcdBd(x) hi' = hi'(x') etc. [curl B](x) = ε'nab [(1/(h1'h2'h3') ∂'aB'b ] en B = B'nen [curl B](x) = ε'nab [(1/(h1'h2'h3')) ∂'a(h'b2B'b )] en B = B'nen [curl B](x) = ε'nab [(1/(h1'h2'h3')) h'n ∂'a(h'b B'b) )] n B = B'nn [curl B](x) = εnab ∂aBb(x) // Cartesian B = Bn Picture M&S: Bc(u) = RcdBd(x) hi = hi(u) etc. [curl B](x) = εnab [(h1h2h3)-1 ∂aBb ] en B = Bnen [curl B](x) = εnab [(h1h2h3)-1 ∂a(hb2Bb )] en B = Bnen [curl B](x) = εnab[(h1h2h3)-1 hn ∂a(hb Bb)] n B = Bnn [curl B](x) = (h1h2h3)-1 B = Bn en [curl B](x) = (h1h2h3)-1 B = Bn en [curl B](x) = (h1h2h3)-1 B = Bn n // M&S 1.07a With the replacements B → E Bn→ En n→ an hi → (h1h2h3)-1 → (1/) the equations marked above agree with Moon & Spencer p 2 (1.07) and p 3 (1.07a). (f) The curl in N > 3 dimensions Looking at the basic form of the curl above [curl B]n(x) = εnab ∂aBb(x) // Cartesian it is hard to imagine a generalization to N>3 dimensions where the curl is still a vector. The only vectors available for construction purposes are ∂n and Bn . For N=4 one might try out various generalizing forms [curl B]n(x) = (1/) εnabc ∂a(∂b Bc) = (1/) εnabc ∂a∂b Bc ? [curl B]n(x) = (1/) εnabc ∂a (BbBc)) ? but these two forms vanish because antisymmetric ε is contracted against something symmetric. Thus the idea of using multiple cross products as used in Appendix A does not prove helpful. The rank-2 tensor Bb;a – Ba;b = ∂aBb – ∂bBa discussed in Appendix D (h) provides the logical extension of the curl to N > 3 dimensions. For N=3 it happens that the object can be associated with a vector, [curl B]n = εnab [Bb;a – Ba;b ]/2 = εnab Bb;a = εnab [∂aBb – ∂bBa ]/2 = εnab∂aBb . In relativity work, since N=4, there is no vector curl, and one sees Bb;a – Ba;b referred to as the covariant curl, and ∂aBb – ∂bBa as the ordinary curl ( Weinberg p 106). Writing the N-dimensional covariant curl in this manner [curl B]ij(x) = [Bj;i(x) – Bi;j(x)] one can ask how this same curl would be expressed in terms of x'-space coordinates and objects. In analogy with N=3 curl above, one might conjecture that [curl B]ij(x) = [B'b;a(x') – B'a;b(x')] (ea)i (eb)j where one could regard eabij ≡ (ea)i (eb)j as a basis vector of the direct product space x mentioned in Section 7 (j). The conjecture can be quickly verified. Start with the transformation rule for a rank-2 covariant tensor, which is what [Bj;i(x) – Bi;j(x)] is, [B'b;a(x') – B'a;b(x')] = RaiRbj[Bj;i(x) – Bi;j(x)] = RaiRbj[curl B]ij Apply Raa'Rbb' to the above line, Raa'Rbb'[B'b;a(x') – B'a;b(x')] = Raa'Rbb' RaiRbj[curl B]ij = (Raa'Rai) (Rbb'Rbj)[curl B]ij = δa'i δb'j [curl B]ij = [curl B]a'b' so that (See Section 7 (s)) [curl B]a'b' = Raa'Rbb'[B'b;a(x') – B'a;b(x')] = (ea)a' (eb)b'[B'b;a(x') – B'a;b(x')] and changing index names, [curl B]ij = (ea)i (eb)j[B'b;a(x') – B'a;b(x')] QED.