old Section 13 vec lap

The tensor doc section 13 below was replaced on 3/8/12 and the old section is stored here. But first, here is the old entry from the Section 14 summary: vector Laplacian general: // N=3 only [ B](x) = en [gkn ∂k{ (1/) ∂i (Bi)} – (1/) εncd εeab ∂c{ (1/) gde (∂a [gbfBf]) } ] B = Bnen [ B](x) = n hn [gkn ∂k{ (1/) ∂i (Bi/hi)} – (1/) εncd εeab ∂c{ (1/) gde (∂a [gbf Bf/hf]) } ] B = Bnn [ B](x) = 2(B(x)) // Cartesian, meaning [ B]i(x) = 2[Bi(x) ] B = Bn 13. The Vector Laplacian in curvilinear coordinates This operator is defined in terms of the vector curl which is only defined for N=3. The context is Picture B. (a) Derivation of the Vector Laplacian in general curvilinear coordinates The definition of the vector Laplacian of a vector field B(x) is 2B ≡ grad(div B) – curl (curl B) , so, as expected, the vector Laplacian is a vector field. In Cartesian coordinates, one finds that [2B]i = 2(Bi) ≡ Σn ∂n2Bi but expressed in general curvilinear coordinates the form gets modified. To avoid confusion, some authors use different symbols for the vector Laplacian operator. For example, M&S use in place of 2 and we will honor these authors by using that symbol here, so B ≡ grad(div B) – curl (curl B) In order to make use of the results of earlier sections, define G ≡ grad(f) where f = div B V ≡ curl C where C ≡ curl B so that B = G – V Section 10 (c) gives this expression for G, G = grad(f) = (∂'kf ') ek in which expression Section 9 (b) allows replacement of f ' as follows, f ' = f '(x') = f(x) = div B = [1/] ∂'i [ B'i] so that => G = ∂'k{ (1/) ∂'i ( B'i)} ek // 2 implied sums, i and k The second term V is little more complicated. First, from Section 12 (d), C = curl B = ε'nab [(1/) ∂'a{B'b} ] en = C'n en V = curl C = ε'ncd [(1/) ∂'c{C'd} ] en = V'n en where recall from Appendix D (d) that ε'abc... = εabc.. = εabc.. = the usual permutation tensor, but written up and primed so as to be in covariant form. Appendix D (h) shows that C and V are both true contravariant vectors, with components C'n and V'n in x'-space. The first line above says C'e = ε'eab [(1/) ∂'aB'b ] so that C'd = g'de C'e = g'de ε'eab [(1/) ∂'aB'b ], which can then be inserted then into the second line to get V = curl C = ε'ncd [ (1/) ∂'c{ g'de ε'eab [(1/) ∂'aB'b ]} ] en = V'n en = (1/) ε'ncd ε'eab ∂'c{ (1/) g'de (∂'aB'b) } en // note g'dc = g'dc(x') , etc which has 6 implied sums, a,b,c,d,e, and n. For each value of n, there are not really 35 terms because most terms vanish due to the ε factors. Looking at ε'ncd ε'eab = εncd εeab, one sees that for each n, the c and d sums generate only 2 terms, and for each of these εeab generates 3*2*1 = 6 terms, so there are 12 terms total for each n. Later when g'de is assumed diagonal, the effective factor is εncd εeab implying 2 * (2*1) = 4 terms, which shall be written out in that case. Combining these terms, the vector Laplacian is now B = G – V = ∂'k{ (1/) ∂'i ( B'i)} ek – (1/) ε'ncd ε'eab ∂'c{ (1/) g'de (∂'aB'b) } en Setting ek = g'kn en in the first term gives B = ∂'k{ (1/) ∂'i ( B'i)} g'kn en – (1/) ε'ncd ε'eab ∂'c{ (1/) g'de (∂'aB'b) } en = en [g'kn ∂'k{ (1/) ∂'i ( B'i)} – (1/) ε'ncd ε'eab ∂'c{ (1/) g'de (∂'aB'b) } ] so at least now both terms use the same expansion base vector en . As a next step, en = h'n n so B = h'n n [g'kn ∂'k{ (1/) ∂'i ( B'i)} – (1/) ε'ncd ε'eab ∂'c{ (1/) g'de (∂'aB'b) } ] The component B'b in the second term can be made contravariant using B'b = g'bfB'f to get B = h'n n [g'kn ∂'k{ (1/) ∂'i ( B'i)} – (1/) ε'ncd ε'eab ∂'c{ (1/) g'de (∂'a [g'bfB'f]) } ] and then, as was done in earlier sections, replace B'n = b'n/h'n B = B'n n to get this final form in "practical units", B = h'n n [g'kn ∂'k{ (1/) ∂'i (B'i/h'i)} – (1/) ε'ncd ε'eab ∂'c{ (1/) g'de (∂'a [g'bf B'f/h'f]) } ] There are so many options here it is difficult to summarize, but here are two forms from above: [ B](x) = en [g'kn ∂'k{ (1/) ∂'i ( B'i(x'))} – (1/) ε'ncd ε'eab ∂'c{ (1/) g'de (∂'a [g'bfB'f(x')]) } ] B = B'n en [ B](x) = n h'n [g'kn ∂'k{ (1/) ∂'i (B'i(x')/h'i)} – (1/) ε'ncd ε'eab ∂'c{ (1/) g'de (∂'a [g'bf B'f(x')/h'f]) } ] B = B'n n [ B](x) = 2(B(x)) // Cartesian, meaning [ B]i(x) = 2Bi(x) B = Bn Converting from Picture B to Picture MS gives (see Section 9 (c)) [ B](x) = en [gkn ∂k{ (1/) ∂i (Bi)} – (1/) εncd εeab ∂c{ (1/) gde (∂a [gbfBf]) } ] B = Bn en [ B](x) = n hn [gkn ∂k{ (1/) ∂i (Bi/hi)} – (1/) εncd εeab ∂c{ (1/) gde (∂a [gbf Bf/hf]) } ] B = Bn n [ B](x) = 2(B(x)) // Cartesian, meaning [ B]i(x) = 2[Bi(x) ] B = Bn In the first two equations above, all the ∂i mean ∂/∂ui and the argument u of all functions is suppressed. The Cartesian form will be verified below. (b) The Vector Laplacian in orthogonal curvilinear coordinates We continue in Picture M&A and process only the second equation of the above block, since it is the one with the practical components B = Bn n . Setting gij = hi2δi,j and gij = (1/hi)2δi,j things simplify somewhat B = hn n [gkn ∂k{ (1/) ∂i (Bi(u)/hi)} – (1/) εncd εeab ∂c{ (1/) gde (∂a [gbf Bf(u)/hf]) } ] = hn n [δk,n ∂k{ (1/) ∂i (Bi(u)/hi)} (1/hn)2 – (1/) εncd εeab ∂c{ (1/) hd2δd,e (∂a [hb2δb,f Bf(u)/hf]) } ] = hn n [ ∂n{ (1/) ∂i (Bi(u)/hi)} (1/hn)2 – (1/) εncd εeab ∂c{ (1/) hd2 (∂a [hb Bb(u)]) } ] = n [ (1/hn) ∂n{ (1/) ∂i (Bi/hi)} – (hn/) εncd εeab ∂c{ (1/) hd2 (∂a [hb Bb]) } ] The first term can be written as n (1/hn) ∂nT where T = (1/) ∂i (Bi/hi) To expand the second term, set n = 1 and then write things out explicitly. For the moment, we suppress the leading factor – (h1/) and write ε1cd εdab ∂c{ (1/) hd2 (∂a [hb Bb]) } = ε3ab ∂2{ (1/) h32 (∂a [hb Bb]) } – ε2ab ∂3{ (1/) h22 (∂a [hb Bb]) } c=2 d=3 c=3 d=2 = [ ∂2{ (1/) h32 (∂1 [h2 B2]) } – ∂2{ (1/) h32 (∂2 [h1 B1]) } ] a = 1 b = 2 a = 2 b = 1 – [ ∂3{ (1/) h22 (∂3 [h1 B1]) } – ∂3{ (1/) h22 (∂1 [h3 B3]) } ] a = 3 b = 1 a = 1 b = 3 = ∂2{ (1/) h32 ( ∂1 [h2 B2] – ∂2 [h1 B1] ) } – ∂3{ (1/) h22 (∂3 [h1 B1] – ∂1 [h3 B3] ) } Define now Γn in cyclic fashion. Γ1 ≡ (1/) h12 ( ∂2 [h3 B3] – ∂3 [h2 B2] ) Γ2 ≡ (1/) h22 ( ∂3 [h1 B1] – ∂1 [h3 B3] ) Γ3 ≡ (1/) h32 ( ∂1 [h2 B2] – ∂2 [h1 B1] ) and then we have shown that 2nd term (n=1) = – (h1/) ε1cd εdab ∂c{ (1/) hd2 (∂a [hb Bb]) } 1 = – (h1/) (∂2 Γ3 – ∂3 Γ2) 1 = + (h1/) (∂3 Γ2 – ∂2 Γ3) 1 Therefore the entire first term (n=1) of B is given by B (first term) = [(1/h1) ∂1T + (h1/) (∂3 Γ2 – ∂2 Γ3) ] 1 The other two terms are obtained by cyclic permutation so the final result is then [ B](x) = [(1/h1) ∂1T + (h1/) (∂3 Γ2 – ∂2 Γ3) ] 1 + cyclic = [(1/h1) ∂1T + (h1/) (∂3 Γ2 – ∂2 Γ3) ] 1 + [(1/h2) ∂2T + (h2/) (∂1 Γ3 – ∂3 Γ1) ] 2 + [(1/h3) ∂3T + (h3/) (∂2 Γ1 – ∂1 Γ2) ] 3 // M&S 1.11 where T = (1/) ∂i (Bi/hi) Γ1 = (1/) h12 ( ∂2 [h3 B3] – ∂3 [h2 B2] ) Γ2 = (1/) h22 ( ∂3 [h1 B1] – ∂1 [h3 B3] ) Γ3 = (1/) h32 ( ∂1 [h2 B2] – ∂2 [h1 B1] ) With the replacements B → E Bn→ En n→ an hi → T → ϒ the result agrees with M&S p 3 (1.11). A more compact summary is this: B = [ (1/hn) ∂nT – (hn/) εnab∂a Γb ] n T = (1/) ∂i (Bi(u)/hi) Γb = (1/) hb2 εbcd ( ∂c [hd Bd(u)]) (c) The Vector Laplacian in Cartesian coordinates First, one can verify that the last result of section (b) gives the starting point formula for B if g = 1 (in u-space). One then has, hi = 1 g = 1 u = F(x) = x (n)i = Sni = δni => n = , B = Bn n = Bn => Bn = Bn so the above 3-line equation block becomes B = [ ∂nT – εnab∂a Γb ] T = ∂i (Bi(u)) Γb = εbcd ( ∂c Bd(u)) or B = [ ∂n{∂iBi} – εnab∂a { εbcd ( ∂c Bd)} ] (*) = [ ∂n{div B} – εnab∂a { (curl B)b } ] = [ ∂n{div B} – [curl (curl B)]n } ] = {div B} – [curl (curl B)] QED Second, one can verify the claim made earlier that in Cartesian coordinates [ B]n = 2 Bn . To show this, it is necessary to show that (left side from (*) above) ∂n ∂i Bi – εnab∂a εbcd(∂c Bd) = ∂i2Bn ? εnab εbcd ∂a (∂c Bd) = ∂n ∂i Bi – ∂i2Bn ? εbna εbcd ∂a (∂c Bd) = ∂n ∂i Bi – ∂i2Bn ? But since index b now appears only in the ε's, use ( up and down indices same in Cartesian x-space) εbna εbcd = δncδad – δndδac // Appendix D (j) item 4 so (δncδad – δndδac) ∂a (∂c Bd) = ∂n ∂i Bi – ∂i2Bn ? δncδad∂a (∂c Bd) – δndδac∂a (∂c Bd) = ∂n ∂i Bi – ∂i2Bn ? ∂a (∂n Ba) – ∂a (∂a Bn) = ∂n ∂i Bi – ∂i2Bn ? ∂n (∂a Ba) – ∂a2Bn = ∂n (∂i Bi) – ∂i2Bn ? Since this last equation is true on inspection, QED.