vec lap debug

Vec Lap Debug PhL 3.25.12 This is one of those amazing canoe voyages with many wrong turns. Overview (3.27.12) 1. The Problem. Here, I note that Maple is failing to generate a certain vector Laplacian term in spherical coordinates that everyone says is there. I make a list of 3 possible error sources. [ I coded into Maple my V(1) 2nd term from Sec 13 (a) ] . The term in question is - 2(r2sinθ)-1 (∂φBφ) , by the way. 2. Experiment A. Here I show that the term is also missing if I do a direct manual calculation directly from my equations with V(1) . Therefore Maple has nothing to do with this error! 3. Experiment B. Being doubtful, I redo my manual calculation using the Section 13 (a) V(1) expression directly, and I find that the term of interest is still missing! So finally I am convinced that the V(1) form in tensor doc must in fact be wrong! I am horrified at this thought. 4. Comments: But how can V(1) be wrong if I verified it with my "covariant work" in Section 15 (g), which I thought I had done! I review this verification briefly looking just at the last steps of Sec 15 and all seems in order. 5. Digression: Here I go off and use the M&S expressions which I derived from my V(2) form to see what happens with the questioned term. The M&S expressions do give that term - 2(r2sinθ)-1 (∂φBφ). This lends some credence to my V(2) form and says nothing about my suspect V(1) form. 6. Tentative Conclusions. Even though I "verified" the V(1) in Section 15, I now think this term is wrong. Therefore I have made an error both in Section 13 deriving the V(1) term, and in Section 15 deriving it again by covariant methods. What are the chances of that happening I ask you, unless it is the same mistake made in both places perhaps. [ Yup, it was basically the same mistake in both places.] 7. Seeking the V(1) error in Section 13. Here I carefully went through my Section 13 (a) derivation of the V(1) expression and I could find nothing wrong! Comment: At this point I did a lot of scratch work relating to the Section 15 derivation. First, I redid this derivation with a new and careful threading, and I ended up with V(2) instead of V(1) !!! So my so-called Section 15 "verification" of the V(1) result "went away" ! 8. Find the mistake made in old Section 15 (g) : Then, after rewriting Section 15 (g), I once again reviewed the "old" Section 15 (g) and I found a critical error. The truth is that εbfe Be;f ≠ εbfe Be,f, and by assuming this was true, I got bad results. It is true that εbfe Be;f = εbfe Be,f . [ Also, in general there were several thread sequencing errors in my derivation. ] 9. Factoid: Here I show how it is possible to have εbfe Be;f = εbfe Be,f but not the other way. You are not allowed to reverse tilts unless you have true vector indices. At some point, I reinforced this last fact in tensor doc where I talk about the tilt-reversal rule. 10. Where is the error in Section 13 (a) regarding V(1) ??? OK, I once again went through my Section 13 (a) derivation of the V(1) form and I discovered that I made the same mistake I made in Section 15 as noted above which I state once again just to hammer it in: (Xb = εbfe Be,f) (Xb = εbfe Be;f) (Xb = εbfe Be;f) (Xb = εbfe Be,f) or (Xb = εbfe Be,f) (Xb = εbfe Be,f) 11. Action: At this point I expunged V(1) from tensor doc. __________________________________________________________________________________ 1. The problem: My Maple program fails to generate the following term in the r-component of the spherical coordinates vector Laplacian, -2/(r2sinθ)∂φBφ I am confident this term should be there from two sources : (1) Lai page 67 second equation from page bottom (2) my ancient TK binder. This is not the only error. Possible sources of the error: (a) my starting equations are wrong (b) my entry into Maple is wrong (c) Maple is making some mysterious internal error I don't understand. Here are my starting equations that I coded into Maple: (B)'n = hn-1 T1n – h'n (1/) (T2n - T3n) (B) = Σn(B)'n n T1n = ∂'n{ (1/) ∂'c (B ' c/h'c)} T2n = ∂'c{h'n-2∂'n(B 'c/h'c)} T3n = ∂'c{ h'c-2∂'c (B 'n/h'n)} 2. Experiment A. I will hand-compute the ∂φBφ term and see what I see with the above equations. The data I have is pretty simple, from tensor doc. h'r = 1 h'θ= r h'φ= rsinθ = r2sinθ 1 2 3 My problems exists in the n=1 component of vec Lap, so I will be setting n = 1 always in the above. In the T1 expression then I care only about the c=3=φ term since only that term creates Bφ . T11 = ∂r{ (1/) ∂φ (B ' φ/h'φ)} = ∂r{( r-2sinθ-1) ∂φ (r2sinθ B ' φ/ rsinθ)} = ∂r{( r-2sinθ-1) ∂φ (r Bφ)} (***) = ∂r{ (r-2sinθ-1) r (∂φBφ)} = sinθ-1∂r{ r-1 (∂φBφ)} = sinθ-1[ ∂r(r-1)(∂φBφ) + r-1∂r(∂φBφ) ] = sinθ-1[-r-2(∂φBφ) + r-1(∂r∂φBφ) ] = - 1/(r2sinθ) (∂φBφ) + (1/rsinθ) (∂r∂φBφ) = - 1/(r2sinθ) [ (∂φBφ) - r (∂r∂φBφ) ] Next, we have T21 = ∂'φ{h'r-2∂'r(B 'φ/h'φ)} = ∂φ{ r2sinθ ∂r(B 'φ/ rsinθ)} (*) = r2(sinθ /sinθ)∂φ{ ∂r(r-1B 'φ/)} = r2 { ∂r(r-1(∂φBφ)} = r2 [ ∂r(r-1)(∂φBφ) + r-1∂r(∂φBφ)] = r2 [ -r-2(∂φBφ) + r-1(∂r∂φBφ)] = – (∂φBφ) + r(∂r∂φBφ) There is no T3 contribution because n=r only in that term so it involves only Br. Now our first equation says (B)'n = hn-1 T1n – h'n (1/) (T2n - T3n) (B)'1 = hr-1 T11 – h'1 (1/) (T21 - T31) = T11 – (r2sinθ)-1 (T21) = {- 1/(r2sinθ) [ (∂φBφ) - r (∂r∂φBφ) ]} – (r2sinθ)-1 { – (∂φBφ) + r(∂r∂φBφ)} (**) = 1/(r2sinθ)* [- [ (∂φBφ) - r (∂r∂φBφ) ] – ] { – (∂φBφ) + r(∂r∂φBφ)} ] = 1/(r2sinθ)* [- (∂φBφ) + r (∂r∂φBφ) + (∂φBφ) - r(∂r∂φBφ)} ] = 0 So! My hand calculation agrees with Maple. THEREFORE, my equations must be wrong. I have eliminated items (b) and (c) as the cause of this particular problem. Maple and I agree on the algebra. Just the amount of spaced burned above reinforces the value of Maple! Now let's make another option list for this error. (d) I have something wrong inside tensor doc. (e) Tensor doc is correct, but I did something wrong transcribing out of it. 3. Experiment B. Let's look at (e) first since it seems most likely. Here is what tensor doc says, Section 13 (a) B = G – V V = either V(1) or V(2) G = hn ∂n{ (1/) ∂i (B i/hi)} n B = Bn n V(1) = hn [ (1/) εncd εdab ∂c{∂a(B b/hb)}] n V(2) = hn [ (1/) εncd εeab ∂c { (1/) gde(∂a[gbfBf/hf]) }] n where B b are my vector components of interest and of interest to Lai in everything else he has done, and in my TK in all its other results. I don't use the V(2) form in my calculation, so simplify the above B = G – V(1) G = hn ∂n{ (1/) ∂i (B i/hi)} n B = Bn n V(1) = hn [ (1/) εncd εdab ∂c{∂a(B b/hb)}] n Let's try my direct calculation from these equations with no further steps to see if the error exists in the equations above. I only care about n=1 and in the first term i=3 and the second term b=3. So simplify B = G – V(1) G = hr ∂r{ (1/) ∂φ (B φ/hφ)} r B = Bn n V(1) = hr [ (1/) ε1cd εda3 ∂c{∂a(B φ/hφ)}] r G = ∂r{ ([r2sinθ]-1) ∂φ ([r2sinθ]Bφ/[rsinθ])} r B = Bn n V(1) = [ ([r2sinθ]-1 ε1cd εda3 ∂c{[r2sinθ]∂a(Bφ/[rsinθ])}] r G = ∂r{ ([r2sinθ]-1) ∂φ (rBφ)} r B = Bn n V(1) =[r2sinθ]-1 (ε1cd εda3 ∂c{[r2sinθ]∂a(Bφ/[rsinθ])} r G'r = ∂r{ ([r2sinθ]-1) ∂φ (rBφ)} V(1)'r =[r2sinθ]-1 (ε1cd εda3 ∂c{[r2sinθ]∂a(Bφ/[rsinθ])} The problem is most likely to be in the second term so be careful. But do the first term first. We know ∂r = grr'∂r'= hr-2∂r = ∂r so G'r = ∂r{ ([r2sinθ]-1) ∂φ (rBφ)} = T11 = - 1/(r2sinθ) [ (∂φBφ) - r (∂r∂φBφ) ] (***) above so we have already computed this tem. Now for the more complicated second term. There is only one choice for index d and that is d = 2. It cannot be 1, and it cannot be 3, so we have ε1cd εda3 = ε1c2 ε2a3 = > c = 3 and a = 1 = ε132 ε213 δc,3δa,1 = + ε123 ε123 δc,3δa,1 = δc,3δa,1 Therefore, V(1)'r = [r2sinθ]-1 (ε1cd εda3 ∂c{[r2sinθ]∂a(Bφ/[rsinθ])} = [r2sinθ]-1 δc,3δa,1∂c{[r2sinθ]∂a(Bφ/[rsinθ])} = [r2sinθ]-1 ∂φ{[r2sinθ]∂r(Bφ/[rsinθ])} = [r2sinθ]-1 ∂φ{[r2sinθ]∂r(Bφ/[rsinθ])} = [r2sinθ]-1 T21 // see (*) above = [r2sinθ]-1 [– (∂φBφ) + r(∂r∂φBφ)] Therefore, putting the pieces together result = G'r – V(1)'r = - 1/(r2sinθ) [ (∂φBφ) - r (∂r∂φBφ) ] – [r2sinθ]-1 [– (∂φBφ) + r(∂r∂φBφ)] But according to (**), this is the same result I got last time, which is 0. I have now eliminated (e) and have reached this conclusion: 4. Comments. Something is wrong with my tensor doc results at the end of Section 13 (a). Note: I thought I got verification of my G – V(1) from section (15). Of course that was using B'i components. Wow. This means I must have made the same error in both places, ouch! That means it will be hard to find. Let's make sure the path from this "verified by Section 15" result to the result I used is correct. Here is the verified result quoted from Section 13 G = ∂n{ (1/) ∂i ( Bi)} en B = Bnen V(1) = [ (1/) εncd εdab ∂c{∂aBb}] en and here is my quote from Section 15 G = ∂'n{ (1/) ∂'i ( B'i)} en B = B'nen V(1) = [ (1/) ε'ncd ε'dab ∂'c{∂'aB'b}] en They match. Then all I did at this point was replace B'n = B 'n/h'n which I have used a zillion times B = B'n en = B'n h'n n = B 'nn => B'n = B 'n/h'n and then I get G = h'n ∂'n{ (1/) ∂'i (B 'i/h'i)} n B = B'n n V(1) = h'n [ (1/) ε'ncd ε'dab ∂'c{∂'a(B 'b/h'b)}] n and this agrees with what I used above to get the contradictory result! 5. Digression. Let's see what my V(2) Moon and Spencer result says about this error. [ 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] ) For my purposes, I will have T = (1/) ∂3 (B3/h3) Γ1 = (1/) h12 ∂2 [h3 B3] Γ2 = –( 1/) h22∂1 [h3 B3] ) Γ3 = 0 T = (r2sinθ)-1 ∂φ (r2sinθBφ/rsinθ) = (r2sinθ)-1∂φ(rBφ) = (rsinθ)-1(∂φBφ) Γ1 = (r2sinθ)-1 ∂θ [rsinθ Bφ] = (r2sinθ)-1 r∂θ [sinθ Bφ] = (rsinθ)-1 ∂θ [sinθ Bφ] Γ2 = –(r2sinθ)-1 r2 ∂r [rsinθ Bφ] ) = –(sinθ)-1 sinθ∂r [r Bφ] ) = – ∂r [r Bφ] Then keeping only the r term [ B](x) = [(1/h1) ∂1T + (h1/) (∂3 Γ2 – ∂2 Γ3) ] 1 [ B]1 = [(1/h1) ∂1T + (h1/) (∂3 Γ2 – ∂2 Γ3) ] = ∂rT + (r2sinθ)-1 (∂3 Γ2) = ∂r[(rsinθ)-1(∂φBφ)] + (r2sinθ)-1 ∂φ [– ∂r [r Bφ]] = (sinθ)-1∂r[r-1(∂φBφ)] + (r2sinθ)-1 [– ∂r [r (∂φBφ)] = (sinθ)-1∂r[r-1(∂φBφ)] – (r2sinθ)-1 ∂r [r (∂φBφ)] = (sinθ)-1{-r-2(∂φBφ) + r-1 (∂r∂φBφ)} – (r2sinθ)-1 {(∂φBφ) + r (∂r∂φBφ)} = {-(sinθ)-1r-2(∂φBφ) + (sinθ)-1r-1 (∂r∂φBφ)} – {(r2sinθ)-1 (∂φBφ) + (r2sinθ)-1r (∂r∂φBφ)} = {-(sinθ)-1r-2(∂φBφ) + (sinθ)-1r-1 (∂r∂φBφ) – (r2sinθ)-1 (∂φBφ) – (r2sinθ)-1r (∂r∂φBφ)} = {-(sinθ)-1r-2(∂φBφ) + (sinθ)-1r-1 (∂r∂φBφ) – (r2sinθ)-1 (∂φBφ) – (rsinθ)-1 (∂r∂φBφ)} = {-(sinθ)-1r-2(∂φBφ) – (r2sinθ)-1 (∂φBφ) } = - 2(r2sinθ)-1 (∂φBφ) and this finally is the correct result. 6. Tentative conclusion: I think my G term and my V(2) terms are correct because from them I obtained the correct M&S result which matches Lai p 67 on this trouble term. This suggests that the G term is OK, and that the problem really is with the V(1) term. I now have three sources confirming Lai p 67 on that last term: M&S, TK and Lai. 7. Finding the V(1) error in tensor doc. Let's start with Section 13. I will paste here and turn to red. 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) . If B is a vector, div B is a scalar, grad(div B) is a vector, so we expect 2B to be a vector. An annoying observation is that curl B is a vector density of weight -1 (see Appendix D (h)) and curl(curlB) a vector density of weight -2, and so apples are being subtracted from oranges to make 2B. This tensor conundrum is resolved in Section 15 (g) so we ignore it for now and proceed undaunted. 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 (a) gives this expression for G, // not worried about 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 = ∂'n{ (1/) ∂'i ( B'i)} en The second term V is little more complicated. First, from Section 12 (d), C = curl B = ε'nab [(1/) ∂'a{ B'b} ] en = J-1C'n en J = V = curl C = ε'ncd [(1/) ∂'c{ J-1C'd} ] en = J-1V'n en (*) // after sep doc !!!! where recall from Appendix D (d) that ε'abc... = εabc.. = the usual permutation tensor, but written up and primed so as to be in covariant form. The reason for the factor J-1 in ∂'c{ J-1C'd} was explained in Comment 2 at the end of Section 12 (c) (namely, C is a vector density of weight -1). ___________________________________ Comment 2: The equation curl B = C = [(1/) ε'nab ∂'aB'b ] en obtained above assumed that B was a vector and the expansion B = B'n en was used. If B were a vector density of weight -1, the expansion would be B = J-1B'nen and the result would be curl B = C = [(1/) ε'nab ∂'a(J-1B'b) ]. This situation will arise in consideration of the vector Laplacian in Section 13. Once again, J = . B = J-1B'jej where J-1B'j(x') = B(x) ej ______________________________________________________ The above lines can be rewritten as C'n = ε'nab ∂'a{ B'b} V'n = ε'ncd ∂'c{ J-1C'd} // J = We shall now derive two different but equivalent forms for V. First Form for V. As shown in Appendix D (h), one can replace ∂'aB'b by B'b;a which is a rank-2 tensor and cause no change in the first equation above. This reminds us that both sides of C'n = ε'nab ∂'aB'b are vector densities of weight -1 so it is a "valid tensor density equation". This means one is free to "reverse the tilt" of contracted pairs of indices, so the first equation can be written C'n = ε'nab ∂'a{ B'b} and this result can be inserted into the V'n equation to give V'n = ε'ncd ∂'c{(1/) ε'dab ∂'aB'b} Then, as shown in Appendix D (e) 2, ε'dab = g' ε'dab where ε'dab = permutation tensor = constant, so V'n = ε'ncd ε'dab ∂'c{∂'aB'b} and then V = curl C = J-1V'n en = [ (1/) ε'ncd ε'dab ∂'c{∂'aB'b}] en STOP. This agrees with a result I used above V(1) = hn [ (1/) εncd εdab ∂c{∂a(B b/hb)}] n (*) Since I have discredited, something must be wrong with my derivation in red above, even though I have checked every stop and then made it red. For now, let's just say I have reviewed my Section 13 derivation very carefully, and I found nothing wrong with it. ––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– 8. Find the mistake made in old Section 15 (g) : (g) vector Laplacian (Section 13) This differential operator is going to take a bit more work. In Section 13 the vector Laplacian is written (instead of 2) and in Cartesian coordinates is defined by B ≡ grad(div B) – curl (curl B) In components this reads (notice that ∂aεbde = εbde∂a ) (B)n = ∂n(∂jBj) - εnab∂a [curl B]b = ∂n(∂jBj) - εnab∂a (εbde∂dBe) ok In Cartesian space up and down index position does not matter and we are just jockeying the indices in search of a tensorized form with contracted indices where possible. Since simple semicolon equations want to have a lower semicolon index, we replace ∂d = gdf∂f to get (B)n = ∂n(∂jBj) - εnab∂a (εbde gdf∂fBe) ok Writing the above in comma notation gives (B)n = (Bj,j),n - εnab(εbde gdfBe,f),a ok (B)n = (Bj,j),n - εnab(εbfe Be,f),a STOP! In moving to the last line, what is allowed and what is not? The part εbde gdf = εbde is OK. The part saying then that εnab εbde = εnab εbde is also OK because since only a comma, you can think of both ε factors outside the ,a structure. You then have εnab(εbde Be,f),a = εnab(εbde ∂fBe),a. But now tilting the e index is illegal as now described in tensor doc Section on the tilt rule. However, if we assume that there is a semicolon all along here so it is Be;f, then this last e tilt is allowed. Assume that and continue: (B)n = (Bj,j),n - εnab(εbfe Be;f),a In the last step, Be,f is replaced by Be;f since Be;f = Be,f – ΓnefBn and then εbfeΓnef = 0 by symmetry. At this point, then, Be;f is a rank 2 tensor (weight 0) and εbfe has the usual weight -1, so we have to do a "repair" by adding a factor g-1/2 as follows: (B)n = (Bj,j),n - εnab(g-1/2εbfe Be;f),a which of course has no effect in Cartesian space. But now (g-1/2εbfe Be;f) is a vector of weight 0 since the factor g-1/2 is a scalar density of weight +1. If we then put in a semicolon before the a, (B)n = (Bj,j),n - εnab(g-1/2εbfe Be;f);a ok Note: I could at this point write εnab = gbc εnac and I could slide the gbc inside the ;a and this would give (B)n = (Bj,j),n - εnac(g-1/2 gcb εbfe Be;f);a and this result matches my current Section 15 (g) result apart from index names, so OK to here. (the outer g-1/2 has not yet been added). the object (g-1/2εbfe Be;f);a is a normal rank-2 tensor like Tba. We now lower the n index to get (B)n = (Bj,j),n - εnab(g-1/2εbfe Be;f);a = (Bj,j),n - gaa'εna'b(g-1/2εbfe Be;f);a where a gaa' gets the first ε to have all lower indices. Note: This was another illegal index lowering (n) really because the second term is not yet a vector because the outside g-1/2 is not added yet. But since I add it below, let it be. Next, semicolons can be installed on the first term just as it is to give = (Bj;j);n - gaa'εna'b(g-1/2εbfe Be;f);a ok except no g-1/2 where (Bj;j) is a scalar, and then (Bj;j);n is a covariant vector. One problem remains. Since gaa' and (g-1/2εbfe Be;f);a are tensors of zero weight, but εna'b has weight -1, we find that the first term (Bj;j);n is a vector, but the second term is a vector density of weight -1. This gets repaired as described earlier by adding a scalar density factor g-1/2, (B)n = (Bj;j);n - g-1/2gaa'εna'b(g-1/2εbfe Be;f);a ok Since the above is now a "valid tensor equation" and by covariance (Section 7 (u)) it can be written in x'-space as follows (everything gets primed) (in the first time, ' is a prime on B, not a comma before j) (B)'n = (B' j;j);n - g'-1/2g'aa'ε'na'b(g'-1/2ε'bfe B'e;f);a ok Now I am flailing around a bit, trying to arrive at my V(2) result. Let's raise the n's again (B)'n = (B' j;j);n - g'-1/2g'aa'ε'na'b(g'-1/2ε'bfe B'e;f);a ok Then let g' raise a' to get (B)'n = (B' j;j);n - g'-1/2ε'nab(g'-1/2ε'bfe B'e;f);a ok and for usual reason replace ;n by ,n (B)'n = (B' j;j),n - g'-1/2ε'nab(g'-1/2ε'bfe B'e;f);a ok The first term gives my desired G'n and now the battle is with this second term!!! I need get that index b raised up, so try (B)'n = (B' j;j),n - g'-1/2ε'nab(g'-1/2ε'bfe B'e;f);a ok Note: The above essentially agrees with one of my new Section 15 (g) equations. Now the object (g'-1/2ε'bfe B'e;f);a = T'b;a = T'b,a – ΓnbaT'n and symmetry kills second term so (B)'n = (B' j;j),n - g'-1/2ε'nab(g'-1/2ε'bfe B'e,f),a ok, my first arrow STOP!!! Between the above two equations I did ε'bfe B'e;f → ε'bfe B'e,f which is WRONG!!! I now know that this symmetry argument is wrong. I did not even comment on it in the text! I will fix up the next steps by re-inserting that semicolon. Now finally replace ε'bfe → g' ε'bfe [ seems ok ] and pull out to get (B)'n = (B' j;j),n - g'-1/2ε'nab ε'bfe (g'1/2B'e;f),a ok with red ; (B)'n = ∂n(B' j;j) - g'-1/2ε'nab ε'bfe ∂a(g'1/2∂fB'e + extra term) (B)'n = ∂n(B' j;j) - g'-1/2ε'nab ε'bfe ∂a(g'1/2gfs∂sB'e+ extra term) Now take a→c and b→d to get (B)'n = ∂n(B' j;j) - g'-1/2ε'ncd ε'dfe ∂'c(g'1/2g'fs∂sB'e+ extra term) The second term is close to but does not agree with V(2)which is this [ (1/) ε'ncd ε'eab ∂'c { (1/) g'de(∂'aB'b) }] Comment: OK, I made an error in the above sequence and when I fix the error, I get an extra term which contains a Γ object, and I know this cannot agree with any Section (15) result. I just "threaded incorrectly" and goofed up. The above sequence of steps is riddled with bad moves, and the new section does it right and gets agreement with the V(2) result of Section (15). _______________________________________________________________________ Factoid: Consider Vb = (εbdeBe;d) where B is a rank-2 tensor of weight 0. Then Vb is a vector equation of weight -1. We can then say Vb = (εbdeBe;d) = (εbdeBe,d) due to the Γ symmetry. After removing the semicolon, Be,d is no longer a tensor, so e and d are no longer tensor indices, so we can no longer "tilt things back" . In other words Vb =(εbdeBe,d) Vb = (εbdeBe,d) _______________________________________________________________________ 10. Where is the error in Section 13 (a) regarding V(1) ??? I will cut and paste that section here and try to find the error. I put red ok's to mark progress. Blue stuff is irrelevant to finding the error. 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) . If B is a vector, div B is a scalar, grad(div B) is a vector, so we expect 2B to be a vector. An annoying observation is that curl B is a vector density of weight -1 (see Appendix D (h)) and curl(curlB) a vector density of weight -2, and so apples are being subtracted from oranges to make 2B. This tensor conundrum is resolved in Section 15 (g) so we ignore it for now and proceed undaunted. 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 (a) 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 = ∂'n{ (1/) ∂'i ( B'i)} en The second term V is little more complicated. First, from Section 12 (d), C = curl B = ε'nab [(1/) ∂'a{ B'b} ] en = J-1C'n en J = V = curl C = ε'ncd [(1/) ∂'c{ J-1C'd} ] en = J-1V'n en (*) where recall from Appendix D (d) that ε'abc... = εabc.. = the usual permutation tensor, but written up and primed so as to be in covariant form. The reason for the factor J-1 in ∂'c{ J-1C'd} was explained in Comment 2 at the end of Section 12 (c) (namely, C is a vector density of weight -1). The above lines can be rewritten as C'n = ε'nab ∂'a{ B'b} V'n = ε'ncd ∂'c{ J-1C'd} // J = We shall now derive two different but equivalent forms for V. First Form for V. As shown in Appendix D (h), one can replace ∂'aB'b by B'b;a which is a rank-2 tensor and cause no change in the first equation above. [ok]. This reminds us that both sides of C'n = ε'nab ∂'aB'b are vector densities of weight -1 so it is a "valid tensor density equation". This means one is free to "reverse the tilt" of contracted pairs of indices, so the first equation can be written C'n = ε'nab ∂'a{ B'b} STOP!!! I am arguing that C'n = ε'nab ∂'aB'b => C'n = ε'nab ∂'aB'b. This is an illegal tilt reversal on index b since it is inside a derivative, as now noted in tensor doc. We can repair it by installing a semicolon (using symmetry) before doing the tilt. The first equation is then taken to be C'n = ε'nab B'b;a and then I can reverse tilt to get C'n = ε'nab B'b;a and this lets me continue. Looking at the STOP paragraph in item 8 above, you see that this is exactly the same error I made in Section 15 in effect, and so I made the same error in both Sections. and this result can be inserted into the V'n equation to give V'n = ε'ncd ∂'c{(1/) ε'dab B'b;a} // corrected Then, as shown in Appendix D (e) 2, ε'dab = g' ε'dab where ε'dab = permutation tensor = constant, so V'n = ε'ncd ε'dab ∂'c{ B'b;a } // corrected and then V = curl C = J-1V'n en = [ (1/) ε'ncd ε'dab ∂'c{B'b;a}] en // corrected But ε'ncd ε'dab = ε'dnc ε'dab = δn,aδc,b – δn,bδc,a so V = curl C = J-1V'n en = (1/) [∂'c{B'c;n} – ∂'c{B'n;c} ] en // corrected but we shall use the ε ε form instead since it is a single term. Second Form for V. Go back to C'n = ε'nab ∂'a{ B'b} V'n = ε'ncd ∂'c{ J-1C'd} // J = Then write C'd = g'deC'e = g'de ε'eab (∂'aB'b) and insert this into the V'n equation to get V'n = ε'ncd ∂'c{ (1/) g'de ε'eab (∂'aB'b) } = ε'ncd ε'eab ∂'c { (1/) g'de(∂'aB'b) } so V(2) = curl C = J-1V'n en = [ (1/)ε'ncd ε'eab ∂'c { (1/) g'de(∂'aB'b) }] en We can then summarize our results: B = G – V V = either V(1) or V(2) G = ∂'n{ (1/) ∂'i ( B'i)} en B = B'nen V(1) = [ (1/) ε'ncd ε'dab ∂'c{B'b;a}] en // corrected V(2) = [ (1/) ε'ncd ε'eab ∂'c { (1/) g'de(∂'aB'b) }] en B = B'nen = [ (1/) ε'ncd ε'eab ∂'c { (1/) g'de(∂'a[g'bfB'f]) }] en B = B'nen Comment: I now have a "corrected" V(1)form which is now pretty ugly because B'b;a = ∂'aB'b + g'aβΓ'bβn B'n so in fact I have V(1) = [ (1/) ε'ncd ε'dab ∂'c{(∂'aB'b + g'aβΓ'bβn B'n)}] en and then finally V(1) = [ (1/) ε'ncd ε'dab ∂'c{(∂'a[B 'b/h'b] + g'aβΓ'bβn [B 'n/h'n])}] en So I suspect that the "extra term" here is where my missing - 2(r2sinθ)-1 (∂φBφ) is sitting. Often one wants vectors expanded onto the unit vectors n B = B'n en = B'n h'n n = B 'nn => B'n = B 'n/h'n and the above set of equations becomes B = G – V G = h'n ∂'n{ (1/) ∂'i (B 'i/h'i)} n B = B'n n V(1) = h'n [ (1/) ε'ncd ε'dab ∂'c{(B 'b/h'b);a}] n // corrected V(2) = h'n [ (1/) ε'ncd ε'eab ∂'c {(1/) g'de(∂'a[g'bf B 'f/h'f]) }] n Conclusion: This V(1) is probably correct, but it seems to me not worth mentioning in tensor doc because it involves the Γ object and my whole effort has been to avoid Γ objects in my expressions for operators in curvilinear coordinates.