vec lap the other way B

Vector Laplacian "the other way" PhL 3.27.12 Although I still don't know how to show the two tensorized forms of the vector Laplacian are the same, I have shown that they are the same for spherical and cylindrical coordinates in my new Appendix I of tensor doc. 3/30/12. Comments: I would like to be a able to add this section to the end of my tensor doc Section 15 on the vector Laplacian, but I have been unable to make it fly. There are a few fleeting references to this approach on the web, but I was unable to find anyone who carried it through. A few comments: (1) It seems to give the result in a form I had been looking for which is [ yes, this is correct ] (B)n = 2Bn + other terms (2) It seems to involve Γ Γ products which are the same products which appear in the Riemann curvature tensor Rabcd which I don't know anything about yet. [ and I still don't know how to show this ] (3) I would like to claim that there is a unique "tensorization" of a Cartesian equation and this would have been a good place to verify that the claim is in fact true. [ I think it is unique ] (4) Here is one of those fleeting web references from Wolfram , search for vec Lap See also http://mathworld.wolfram.com/TensorLaplacian.html on the "tensor Laplacian" (5) M&S wrote a paper that might shed light on this subject, but I cannot find it online. Moon, P. and Spencer, D. E. "The Meaning of the Vector Laplacian." J. Franklin Inst. 256, 551-558, 1953. (6) The book by Arken might have something on this subject Arfken, G. Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, 1985. (7) I store my failed efforts below, and will return to this subject perhaps after I have cleaned up other messes in vector doc that still persist. Note added 3.28.12. Consider item (4) above (gλκAμ;λ);κ = gλκ;κAμ;λ + gλκ Aμ;λ;κ = gλκ Aμ;λ;κ I am still mystified by the third line in the quote above. I think I am obtaining the Arfken book right now, maybe it will have something to say. // It does not. (8) Hodge Theory. Interesting comment on uniqueness and the two forms for vec Lap. (9) Pondering a Maple computation . This long section contains the original work which led to what is now Appendix I. Nothing in this section is missing from App I except a quote of spherical results. __________________________________________________________________________ An alternate derivation of this result would be to start with the Cartesian form, (B)n = ∂j∂jBn = Bn,j,j and simple add semicolons to get an appropriate tensorized form, (B)n = Bn;j;j = gjk Bn;j;k . Note inserted: I agree that you can at this point slide the gjk inside the ;k derivative because I do know that gjk;k = 0 since it is generally true that gjk;i = 0 for each term, so (A)n = gjk (An;j);k = (gjk An;j);k and this at least replicates the second line of the Wolfram quote above. But I don't understand the meaning of the third line! I don't know what the objects gλ and K are. This is a true tensor equation, so in x'-space it can be written (B)'n = B'n;j;j = g'jk B'n;j;k Somehow it must be possible to show that Bn;j;j = (Bj;j);n – g-1/2εnab(g-1/2gbcεcdeBe;d);a since the correctly tensorized form must be unique. One must then show that (Bj;j);n – (Bn;j);j = g-1/2εnabεbde (g1/2Be;d);a = g-1/2 εbnaεbde (g1/2Be;d);a = g-1/2 [δndδae – δneδad] (g1/2Be;d);a = g-1/2 [(g1/2Ba;n);a – (g1/2Bn;a);a] = g-1/2 [(g1/2Ba;n);a – (g1/2Bn;a);a] How about (Bj;j);n – (Bn;j);j = g-1/2 [(g1/2Ba;n);a – (g1/2Bn;a);a] (Bj;j);n – (Bn;j);j = g-1/2 [(g1/2Ba;n);a – (g1/2Bn;a);a] (Bj;j);n – (Bn;j);j = g-1/2 [(g1/2[Ba;n –Bn;a] ];a (Bj;j);n – (Bn;j);j = g-1/2 [(g1/2[Ba,n –Bn,a] ];a Maybe write (g1/2Ba;n);a = (g1/2);a Ba;n + g1/2 Ba;n;a (g1/2Bn;a);a = (g1/2);a Bn;a + g1/2 Bn;a;a Now (g1/2);a = (g1/2),a = ∂a(g1/2) = (1/2)g-1/2 (∂ag) Then RHS = g-1/2{ (g1/2);a [Ba;n - Bn;a] + g1/2[Ba;n;a – Bn;a;a]} = (1/2)g-1(∂ag) [Ba;n - Bn;a] + [Ba;n;a – Bn;a;a] Our equation is then (Bj;j);n – (Bn;j);j = (g1/2);a [Ba;n - Bn;a] + [Ba;n;a – Bn;a;a] The 2nd term on the LHS cancels the last term on the RHS by doing a tilt, so have to show (Bj;j);n = Bj;n;j + (g1/2);a [Ba;n - Bn;a] or (Bj;j);n – (Bj;n;j) = (g1/2);a [Ba;n - Bn;a] which is perhaps simpler than what we started with! Try to get indices down: (Bj;j);n – (Bj;n;j) = (g1/2);a [Ba;n - Bn;a] [Bj;j;n – Bj;n;j] = (g1/2);a [Ba;n - Bn;a] gji[Bi;j;n – Bi;n;j] = (g1/2);a [Ba;n - Bn;a] = (g1/2);a [Ba,n - Bn,a] On each side we have the difference of covariant derivatives in reverse order. Weinberg says that Bi;j;n – Bi;n;j = –BσRσijn But this causes a contradiction. It says that the LHS does not involve derivatives of B, but the RHS clearly involve only derivatives of B, so how can the two sides be equal? OK, I gave it the old college try and failed, and the web seems to have nothing, so let's let it go and move on to cleaning up the mess I already have. ________________________________________________________________________ I have at this point found that I made the same error in Section 13 and Section 15 and that my old form called V(1) was not valid, but that it could be corrected to say V(1) = (1/) [∂'c{B'c;n} – ∂'c{B'n;c} ] en which I think is valid, though computationally unattractive due to the Γ terms. Thus, I claim that I have shown there that ( B)'n = ∂'n{ (1/) ∂'i ( B'i)} – (1/) [∂'c{B'c;n} – ∂'c{B'n;c} ] which could be written as ( B)'n = { (1/) ( B'i)i},n – (1/) [{B'c;n},c – {B'n;c},c ] I know that the first term here is really (B'i;i),n so we have ( B)'n = (B'j;j),n – (1/) [{B'c;n},c – {B'n;c},c ] Question: does this last result have any bearing on the subject of this document? What I have way above is this claim: (B)'n = B'n;j;j = g'jk B'n;j;k so I would have to show in x'-space that ( I have changed ,n to ;n below which is OK on a scalar) B'n;j;j = (B'j;j);n – (1/) [{B'c;n},c – {B'n;c},c ] so I would then have to show that - B'n;j;j = –(B'j;j);n + (1/) [{B'c;n},c – {B'n;c},c ] or (B'j;j);n – (B'n;j);j = (1/) [{B'c;n},c – {B'n;c},c ] which appears to be something "fairly symmetrical in overall shape". This is similar to what I had last time in x-space, but last time I had ;c in place of ,c in both places on the RHS. Maybe I could show that they could be ;c here as well and then it is a tensor equation. Suppose they really are semicolons just for the sake of proceeding a tiny amount. Then what we would like to show would be (B'j;j);n – (B'n;j);j = (1/) [{B'j;n};j – {B'n;j};j ] Now throw in some g factors g'jk(B'k;j);n – g'jk (B'n;j);k = (1/) [{g'jkB'k;n};j – { g'jk B'n;k};j ] and finally lower n g'jk(B'k;j);n – g'jk (B'n;j);k = (1/) [{g'jkB'k;n};j – { g'jk B'n;k};j ] g'jk[ (B'k;j);n – (B'n;j);k] = (1/) [{g'jkB'k;n};j – { g'jk B'n;k};j ] Now suppose on the RHS we do this {g'jkB'k;n};j = (g'jk);j B'k;n + g'jk (B'k;n);j { g'jk B'n;k};j = (g'jk);j B'n;k + g'jk (B'n;k);j Then what we want to show is this g'jk [ (B'k;j);n – (B'n;j);k] = (g'jk);j B'k;n + g'jk (B'k;n);j – (g'jk);j B'n;k - g'jk (B'n;k);j which rewrite as g'jk [ (B'k;j);n – (B'n;j);k] = (g'jk);j B'k;n – (g'jk);j B'n;k + g'jk (B'k;n);j - g'jk (B'n;k);j or g'jk [ (B'k;j);n – (B'n;j);k] = (g'jk);j [B'k;n –B'n;k] + g'jk[(B'k;n);j - (B'n;k);j] or g'jk [(B'k;j);n –(B'k;n);j – (B'n;j);k + (B'n;k);j ] = (g'jk);j [B'k;n –B'n;k] Now if I use Weinberg page 140 this becomes g'jk [ – B'σRσkjn + B'σRσnjk ] = (g'jk);j [B'k;n –B'n;k] or g'jk B'σ [ –Rσkjn + Rσnjk ] = (g'jk);j [B'k,n –B'n,k] and then I end up with the same contradiction as before: the LHS involves components of B', while the RHS involves derivatives of B'. Since B' is arbitrary, this cannot possibly be a valid equation. ______________________________ more fiddling 3.27.12 _________________________ Suppose instead I start with this Section 15 equation (B)'n = (B'j;j);n – g'-1/2ε'nab(g'-1/2g'bcε'cdeB'e;d);a in which at least everything is a tensor index etc etc. Then I want to show that B'n;j;j = (B'j;j);n – g'-1/2ε'nab(g'-1/2g'bcε'cdeB'e;d);a and now we are at least all tensors. Rewrite as B'n;j;j = (B'j;j);n – g'-1/2ε'nab(g'-1/2ε'bdeB'e;d);a You cannot extract the ε'bde as we know without generating lots of extra terms. It just looks very hard to show still! What do you do with all those ε indices! You have to special case it. How many terms on the RHS? For fixed n, ε'nab has only two possible non-zero values. For each of those, b has some value. For each b value, ε'bde has only two non-zero values. So maybe only 4 terms on the RHS? First write as B'n;j;j = (B'j;j);n – g'-1/2ε'nab(g'+1/2ε'bdeB'e;d);a Try n = 1 and ignore – g'-1/2 and look at RHS RHS terms are ε'1ab(g'+1/2ε'bdeB'e;d);a ε'123(g'+1/2ε'312B'2;1);2 (g'+1/2B'2;1);2 ε'1ab(g'+1/2ε'bdeB'e;d);a ε'123(g'+1/2ε'321B'1;2);2 – (g'+1/2B'1;2);2 ε'1ab(g'+1/2ε'bdeB'e;d);a ε'132(g'+1/2ε'213B'3;1);3 (g'+1/2B'3;1);3 ε'1ab(g'+1/2ε'bdeB'e;d);a ε'132(g'+1/2ε'231B'1;3);3 –(g'+1/2B'1;3);3 So here is what I would have to show ( do tilts on the LHS) B'1;j;j = (B'j;j);1 – g'-1/2{ (g'+1/2B'2;1);2 – (g'+1/2B'1;2);2 + (g'+1/2B'3;1);3–(g'+1/2B'1;3);3} Write out all terms now B'1;1;1 + B'1;2;2 + B'1;3;3 – (B'1;1);1 –(B'2;2);1– (B'3;3);1 = – g'-1/2{ (g'+1/2B'2;1);2 – (g'+1/2B'1;2);2 + (g'+1/2B'3;1);3–(g'+1/2B'1;3);3} or B'1;2;2 + B'1;3;3 –B'2;2;1 – B'3;3;1 = – g'-1/2{ (g'+1/2B'2;1);2 – (g'+1/2B'1;2);2 + (g'+1/2B'3;1);3–(g'+1/2B'1;3);3} Wow, it is still very hard to see anything here! We could expand the RHS to get RHS = - g'-1/2{ (g'+1/2);2B'2;1 – (g'+1/2);2B'1;2 + (g'+1/2);3B'3;1–(g'+1/2);3B'1;3} - g'-1/2{ g'+1/2(B'2;1);2 –g'+1/2(B'1;2);2 + g'+1/2(B'3;1);3–g'+1/2(B'1;3);3} RHS = - g'-1/2{ (g'+1/2);2B'2;1 – (g'+1/2);2B'1;2 + (g'+1/2);3B'3;1–(g'+1/2);3B'1;3} - { (B'2;1);2 – (B'1;2);2 + (B'3;1);3– (B'1;3);3} Then we have to show that B'1;2;2 + B'1;3;3 –B'2;2;1 – B'3;3;1 = -(B'2;1);2 + (B'1;2);2 - (B'3;1);3+ (B'1;3);3 - g'-1/2{ (g'+1/2);2B'2;1 – (g'+1/2);2B'1;2 + (g'+1/2);3B'3;1–(g'+1/2);3B'1;3} or –B'2;2;1 – B'3;3;1 = -(B'2;1);2 - (B'3;1);3 - g'-1/2{ (g'+1/2);2B'2;1 – (g'+1/2);2B'1;2 + (g'+1/2);3B'3;1–(g'+1/2);3B'1;3} or (B'2;1);2 + (B'3;1);3–B'2;2;1 – B'3;3;1 = - g'-1/2{ (g'+1/2);2B'2;1 – (g'+1/2);2B'1;2 + (g'+1/2);3B'3;1–(g'+1/2);3B'1;3} At least I can see that the "second derivative" pieces on the LHS cancel each other so none are left, and this matches the fact that there are no second derivatives on the RHS. Suppose we look now at just the first derivative terms. ( Γ means just the Γ part) (B'2,1);2Γ + (B'3,1);3Γ –B'2,2;1Γ – B'3,3;1Γ + (B'2;1Γ),2 + (B'3;1Γ),3–B'2;2Γ,1 – B'3;3Γ,1 = - g'-1/2{ (g'+1/2);2B'2,1 – (g'+1/2);2B'1,2 + (g'+1/2);3B'3,1–(g'+1/2);3B'1,3} Let's look at the ∂1 B'2 stuff only: (B'2,1);2Γ – B'2;2Γ,1 = - g'-1/2{ (g'+1/2);2B'2,1 } ? Examples says that (B'2,1);2Γ = + Γ22n Bn1 + Γ12n B2n B'2;2Γ,1 = ∂1 (B'2;2Γ) = ∂1(g2βΓ2βn Bn) So this would have to be true: Γ22n Bn1 + Γ12n B2n – ∂1(g2βΓ2βn Bn) = - g'-1/2{ (g'+1/2);2B'2,1 } ? The derivative terms would have to cancel, which would require – (g2βΓ2βn ∂1Bn) = - g'-1/2{ (g'+1/2);2∂1B'2 } But for this to be true, the STOP. The ∂1 will involve all three lower derivatives. OK enough! Even confirming a mechanical single case is close to impossible! Maple could perhaps do this, but what I need are the right tools to show it in general. And I don't have them yet, so put this back to bed! _________________________________ 8. Hodge Theory An interesting hit. Notice uniqueness comment. 9. Maple computation? You could do it it you wanted. Here is the math, What about doing things the other way? (B)n = Bn;j;j = (Bn;j);j Recall that Bab;α ≡ ∂α Bab + Γaαn Bnb + Γbαn Ban so that Ba;b;α ≡ ∂α Ba;b + Γaαk Bk;b + Γbαk Ba;k so that Bn;j;j ≡ ∂j Bn;j + Γnjk Bk;j + Γjjk Bn;k Now use Ba;α = ∂α Ba + gαbΓabs Bs Bn;j = ∂j Bn + gjbΓnbs Bs check? OK. Bk;j = ∂j Bk + gjbΓkbs Bs Bn;k = ∂k Bn + gkbΓnbs Bs Then Bn;j;j = ∂j[∂j Bn + gjbΓnbs Bs] + Γnjk[∂j Bk + gjbΓkbs Bs] + Γjjk[∂k Bn + gkbΓnbs Bs] But for an orthogonal system we have ( we know that Γjjk is a simple thing) = ∂j∂j(Bn) + ∂b (Γnbs Bs) + Γnjk[∂j Bk + hj-2Γkjs Bs] + Γjjk[∂k Bn + hk-2Γnks Bs] Now, all the time doing this we are in x-space keeping all the Γ terms even though they are 0 there. No terms have been dropped. At any time then we can go to x'-space by adding primes to everything. B'n;j;j = ∂'j∂'j(B'n) + ∂'b (Γ'nbs B's) + Γ'njk[∂'j B'k + h'j-2Γ'kjs B's] + Γ'jjk[∂'k B'n + h'k-2Γ'nks B's] Here is my Big Question of the moment: where in this mess do we find this object ? [1/] ∂'j [ (∂'j B'n)] = ∂'j∂'j B'n + (1/) ∂'j() ∂'j B'n Well, recall that Γ'jjk = (1/) ∂k() . So look at the term in red which is then (1/) ∂k()∂'k B'n = (1/) ∂j()∂'j B'n and there it is!!! So I have now shown B'n;j;j = [1/] ∂'j [ (∂'j B'n)] + ∂'b (Γ'nbs B's) + Γ'njk[∂'j B'k + h'j-2Γ'kjs B's] + (1/) ∂k()h'k-2Γ'nks B's] and finally I have identified the famous "extra terms": extra terms = (∂'bΓ'nbs) B's + Γ'njk h'j-2Γ'kjs B's + (1/) ∂k()h'k-2Γ'nks B's + Γ'njk(∂'j B'k) = [(∂'bΓ'nbs) + Γ'njk h'j-2Γ'kjs + (1/) ∂k()h'k-2Γ'nks] B's + Γ'njk(∂'j B'k) and we see that there is a big term in B, and then a small term in ∂B. Now consider for the orthogonal case, Γdab = (1/2) gdc [ ∂agbc + ∂bgca – ∂cgab] = (1/2) hd-2 δdc [ ∂a(hb2δbc) + ∂b(hc2δca) – ∂c(ha2δab)] = (1/2) hd-2 δdc [δbc ∂a(hb2) + δca ∂b(hc2) – δab ∂c(ha2)] = (1/2) hd-2 [δdc δbc ∂a(hb2) + δdc δca ∂b(hc2) – δdc δab ∂c(ha2)] = (1/2) hd-2 [Σcδdc δbc ∂a(hb2) + Σcδdc δca ∂b(hc2) – Σcδdc δab ∂c(ha2)] = (1/2) hd-2 [δdb ∂a(hb2) + δda∂b(ha2) – δab ∂d(ha2)] Each term is a K delta in two of the three indices. So, when all three Γ indices are different, Γdab vanishes. There are 3x2x1 = 6 such terms out of the 27. But in sphericals, many others will vanish as well due to the simple nature of the scale factors. Yes, I could throw this into Maple, but I learn nothing when it works right. The Γ object in spherical coordinates. There is something else going on here I know. I think the Γ object is pretty sparse in sphericals. Here from the CRC thing http://books.google.com/books?ei=oMtzT9D6MuGXiQLXy5GXCw&id=aFDWuZZslUUC&dq=encyclopedia+mathematics&q=spherical+coordinates#v=onepage&q&f=false page 2794 // wrong! So out of 33 = 27 possible entries, only 6 entries are non-zero in sphericals. Probably that is why there are so few terms to add to the scalar Laplacian term. But it would still be a pain to implement in Maple. Today 3.29.12 for the first time ever I have a little Maple code to compute Γ according to the fully general formula. I find that Γ122 = -r, in disagreement with the result -1/r quoted above! It is easy also to do by hand, Γ'dab = (1/2) h'd-2 [δdb ∂'a(h'b2) + δda∂'b(h'a2) – δab ∂'d(h'a2)] Γ122 = (1/2) h'1-2 [δ12 ∂'2(h'22) + δ12∂'2(h'22) – δ22∂'1(h'22)] = (1/2) h'1-2 [ – δ22∂'1(h'22)] = (1/2) h'r-2 [ –∂r(hθ2)] = (1/2) 1 [ –∂r(r2)] = - r I just found another source in a PDF which I think agrees with me! So here is what this table says for all non-vanishing: Γ122 = -r Γ212 = Γ221 = 1/r Γ133 = -r sin2θ Γ313 = Γ331 = 1/r Γ233 = -cosθsinθ Γ323 = Γ332 = cotθ I will now verify these claims and make each red as I do, DONE and back to black. So only 9 or 27 are non-zero. I wonder what Eric is thinking here. I see that on the web, Wolfram math has a new section, to wit This is very screwy because the angles φ,θ have the reverse meaning from my usual convention. Also, the ordering of the rows is unclear So I could understand the above quoted results this way: (1) φ is the polar angle, θ is the azimuth (2) coordinate order is r,θ,φ with such angles. Then to translate the above to my notation , I take φ↔θ. But then the rows are in order r,φ,θ which I don't want. Here would be my matrices, where r,θ,φ = 1,2,3 ( θ = polar, φ = az) is the ordering of rows and columns Γr = Γθ = Γφ = Γ122 = -r Γ212 = Γ221 = 1/r Γ133 = -r sin2θ Γ313 = Γ331 = 1/r Γ233 = -cosθsinθ Γ323 = Γ332 = cotθ OK, we have figured out the Γ object. The world can do it however it wants. Maple Program for XXX Go back to our starting point B = Bn;j;j un B = B'n;j;j en Bn;j;j = ∂j[∂j Bn + gjbΓnbs Bs] + Γnjk[∂j Bk + gjbΓkbs Bs] + Γjjk[∂k Bn + gkbΓnbs Bs] Rewrite in this order Bn;j;j = [∂j ∂j Bn + ∂j (gjbΓnbs Bs)] + Γnjk[∂j Bk + gjbΓkbs Bs] + Γjjk[∂k Bn + gkbΓnbs Bs] and again Bn;j;j = [∂j ∂j Bn + Γjjk(∂kBn )] + ∂j(gjbΓnbs Bs) + Γnjk[∂j Bk + gjbΓkbs Bs] + Γjjk[gkbΓnbs Bs] Since no terms have been dropped, we are still "covariant" and we could have done this all in x'-space B'n;j;j = [∂'j ∂'j B'n + Γ'jjk(∂'kB'n )] + ∂'j(gjbΓ'nbs B's) + Γ'njk[∂'j B'k + gjbΓ'kbs B's] + Γ'jjk[gkbΓ'nbs B's] Now use Γ'jjk = (1/) ∂'k() in the first term, and then realize that [∂'j ∂'j B'n + (1/) ∂'k() (∂'kB'n )] = lap (Bn) in the sense that we earlier wrote ∂'j∂'jf ' + (1/) ∂'j() ∂'jf = lap (f) This is simply a gathering together of certain terms and giving that group of terms the name lap (Bn) . Then we have B'n;j;j = lap (Bn) + extra terms extra terms = ∂'j(gjbΓ'nbs B's) + Γ'njk[∂'j B'k + gjbΓ'kbs B's] + Γ'jjk[gkbΓ'nbs B's] extra terms = ∂'j(gjbΓ'nbs B's) + Γ'njk[g'js ∂'s B'k + gjbΓ'kbs B's] + Γ'jjk[gkbΓ'nbs B's] ET1 ET2 ET3 ET4 I now want to have Maple compute these extra terms for spherical coordinates! Call it ET. Now one more little fiddling. We have B = B'n;j;j en = h'n B'n;j;j n = h'n (B'n);j;j n = h'n (B'n/h'n);j;j n so the thing we really want is this (B)'n = h'n (B'n/h'n);j;j Unfortunately, this means I don't have the right "lap term" above. I had gathered this lap (Bn) = [∂'j ∂'j B'n + (1/) ∂'k() (∂'kB'n )] But now I have to write this as lap (Bn) = lap (Bn/hn) = [∂'j ∂'j (Bn/hn) + (1/) ∂'k() (∂'k (Bn/hn) )] and this makes a bunch of annoying new terms. ∂'j ∂'j (Bn/hn) = ∂'j ∂'j (Bnh-1n) = ∂'j [(∂'jBn) h-1n + Bn(∂'j h-1n) ] = (∂'j∂'jBn)h-1n + (∂'jBn) (∂'j h-1n) + (∂'j Bn)(∂'j h-1n) + Bn (∂'j ∂'j h-1n) = (∂'j∂'jBn) h-1n + 2(∂'jBn) (∂'j h-1n) + Bn (∂'j ∂'j h-1n) ∂'k (Bn/hn) = [(∂'kBn) h-1n + Bn(∂'k h-1n) ] lap (Bn) = (∂'j∂'jBn) h-1n + 2(∂'jBn) (∂'j h-1n) + Bn (∂'j ∂'j h-1n) + (1/) ∂'k()[(∂'kBn) h-1n + Bn(∂'k h-1n) ] = h-1n {(∂'j∂'jBn) + (1/) ∂'k()(∂'kBn) } + 2(∂'jBn) (∂'j h-1n) + Bn (∂'j ∂'j h-1n) + (1/) ∂'k()Bn(∂'k h-1n) = h-1n lap (Bn) + 2(∂'jBn) (∂'j h-1n) + Bn (∂'j ∂'j h-1n) + (1/) ∂'k()Bn(∂'k h-1n) = h-1n lap (Bn) + other terms other terms = 2(∂'jBn) (∂'jh-1n) + Bn (∂'j ∂'j h-1n) + (1/) ∂'j()Bn(∂'j h-1n) Now let's look at these "other terms". other terms = 2(∂'jh'-1n) (∂'jBn) + [(∂'j ∂'j h'-1n) + (1/) ∂'j()(∂'j h'-1n)] Bn I don't see any real way to simplify these, they have to be added into Maple! So let's back up and restate what we have so far: B'n;j;j = lap (Bn) + extra terms extra terms = ∂'j(gjbΓ'nbs B's) + Γ'njk[∂'j B'k + gjbΓ'kbs B's] + Γ'jjk[gkbΓ'nbs B's] extra terms = ∂'j(gjbΓ'nbs B's) + Γ'njk[g'js (∂'s B'k) + gjbΓ'kbs B's] + Γ'jjk[gkbΓ'nbs B's] ET1 ET2 ET3 ET4 lap (Bn) = h-1n lap (Bn) + other terms other terms = 2(∂'jh'-1n) (∂'jBn) + (∂'j ∂'j h'-1n) Bn + (1/) ∂'j()(∂'j h'-1n) Bn = 2(∂'jh'-1n)g'js (∂'sBn) + g'js (∂'s ∂'j h'-1n) Bn + (1/) ∂'j()g'js (∂'s h'-1n) Bn OT1 OT2 OT3 Then we can write (B)'n = h'n (B'n/h'n);j;j = h'n B'n;j;j = h'n [lap (Bn) + extra terms ] = h'n [h-1n lap (Bn) + other terms + extra terms ] = lap (Bn) + h'n[other terms + extra terms] For the first time now I see the isolated lap (Bn) terms!! This is good. But now I have a lot more stuff to put into Maple. Status 2:30 PM. Mark called, I had to decline with all these plates spinning. For the first time, the r-term came our right: I have not gotten all three to come out right. Make a file and do it all in cylindricals! What does Γ look like now? Γdab = (1/2) hd-2 [δdb ∂a(hb2) + δda∂b(ha2) – δab ∂d(ha2)] In cylindricals we have for r,θ,z that hr=1 and hθ = r and hz= 1. I let Maple compute all 27 and I now note that that are not 0: Γ122 = -r Γ212 = 1/r Γ221 = 1/r // that's all folks! Γr = Γθ = Γz = and this agrees with mathworld. Looking at the above, the only non-zero terms will come from first term when b = 2 (hence d=2) and a = 1: Γ212 = (1/2)r-2 ∂r(r2) = +1/r second term Γ122 = -r Γ212 = Γ221 = 1/r Γ133 = -r sin2θ Γ313 = Γ331 = 1/r Γ233 = -cosθsinθ Γ323 = Γ332 = cotθ