vec lap the other way A

Another approach to vector Laplacian PhL 3.24.12 I was unable to make this fly, but will keep it around for some future day. Well, consider Qi ≡ 2(ui) = div grad ui. If ui is a vector, grad ui we know is not a rank-2 tensor, and then div [ grad ui] is also not a tensor. Think about the covariant approach. Start with Qi = 2(ui) = ∂j∂j ui = ∂j ui,j = ui,j,j This thing is not a tensor until we put in the semicolons Qi = ui;j;j Only now is Qi a true vector. Maybe this is a simpler approach compared to what I did in Section 15. How does this involve that curl I wonder? Let's use Appendix XXX to write ui;j;k = gkα ui;j;α = gkα [ ∂α ui;j + Γiαn un;j – Γnjα ui;n ] Then you go install this: ui;j = ∂j ui + Γimjum to get ui;j;k = gkα [ ∂α {∂j ui + Γinjun} + Γiαn {∂j un + Γnmjum } – Γnjα {∂n ui + Γimnum } ] = [ ∂k {∂j ui + Γinjun} + gkα Γiαn {∂j un + Γnmjum } – gkα Γnjα {∂n ui + Γimnum } ] Now contract k with j to get ui;j;j = [ ∂j {∂j ui + Γinjun} + gjα Γiαn {∂j un + Γnmjum } – gjα Γnjα {∂n ui + Γimnum } ] And now we wonder if this equals (u)i ??? Maybe consider at an earlier point: Qi = ui;j;j (u)i = (uj;j);i – g-1/2εiab(g-1/2εbdeue;d);a If these really were equal, that would imply that ( I do the two semicolon replacements on the right) uj;j;i – ui;j;j = g-1/2εiab(g-1/2εbdeue,d),a That is pretty non-obvious to me! Is it even true in Cartesians? Then it would say uj,j,i – ui,j,j = εiabεbde(ue,d,a) But in this case we know that εiabεbde = εbiaεbde = δidδae – δieδad so RHS = εiabεbde(ue,d,a) = [δidδae – δieδad](ue,d,a) = (ua,i,a) – (ui,a,a) = (uj,i,j) – (ui,j,j) = uj,j,i – ui,j,j and yes, it works.! So armed with this info, let's try to do it more generally. Start with ui;j;k = gkα [ ∂α {∂j ui + Γinjun} + Γiαn {∂j un + Γnmjum } – Γnjα {∂n ui + Γimnum }] First, replace i→I uI;j;k = gkα [ ∂α {∂j uI + ΓInjun} + ΓIαn {∂j un + Γnmjum } – Γnjα {∂n uI + ΓImnum }] Now set I = j and k = i to get uj;j;i = giα [ ∂α {∂j uj + Γjnjun} + Γjαn {∂j un + Γnmjum } – Γnjα {∂n uj + Γjmnum }] Now start again with uI;j;k = gkα [ ∂α {∂j uI + ΓInjun} + ΓIαn {∂j un + Γnmjum } – Γnjα {∂n uI + ΓImnum }] This time, replace I = i and k = j ui;j;j = gjα [ ∂α {∂j ui + Γinjun} + Γiαn {∂j un + Γnmjum } – Γnjα {∂n ui + Γimnum }] We can then write uj;j;i – ui;j;j = giα [ ∂α {∂j uj + Γjnjun} + Γjαn {∂j un + Γnmjum } – Γnjα {∂n uj + Γjmnum }] – gjα [ ∂α {∂j ui + Γinjun} + Γiαn {∂j un + Γnmjum } – Γnjα {∂n ui + Γimnum }] Somehow I expect the entire RHS to reduce to g-1/2εiab(g-1/2εbdeue,d),a There are 12 terms on the RHS now. Here are the 6 double Γ terms (only i is a free index) giα Γjαn Γnmjum – giα Γnjα Γjmnum – gjα Γiαn Γnmjum + gjα Γnjα Γimnum In the first term do j↔n, = giα Γnαj Γjmnum – giα Γnjα Γjmnum – gjα Γiαn Γnmjum + gjα Γnjα Γimnum The first and fourth term can then be combined like this: giα Γnαj Γjmnum + STOP. Back up to an earlier point, uI;j;k = gkα [ ∂α uI;j + ΓIαn un;j – Γnjα uI;n ] First, replace i→j then k→i : uj;j;i = giα [ ∂α uj;j + Γjαn un;j – Γnjα uj;n ] Now start again uI;j;k = gkα [ ∂α uI;j + ΓIαn un;j – Γnjα uI;n ] and this time take I → i and k→ j ui;j;j = gjα [ ∂α ui;j + Γiαn un;j – Γnjα ui;n ] Now we can say uj;j;i – ui;j;j = giα [ ∂α uj;j + Γjαn un;j – Γnjα uj;n ] – gjα [ ∂α ui;j + Γiαn un;j – Γnjα ui;n ] I would like now to show that RHS = g-1/2εiab(g-1/2εbdeue,d),a . Maybe start with this g-1/2εiab(g-1/2εbdeue,d),a = g-1/2εiab(g1/2εbdeue,d),a = g-1/2εiabεbde (g1/2ue,d),a = g-1/2εbiaεbde (g1/2ue,d),a = g-1/2[ δid δae - δie δad ] (g1/2ue,d),a = g-1/2 { (g1/2ua,i),a – (g1/2ui,a),a } = g-1/2 { ∂a(g1/2ua,i) – ∂a (g1/2ui,a) } = g-1/2 { g1/2∂a(ua,i) +∂a(g1/2) ua,i – g1/2∂a (ui,a) – ∂a (g1/2) ui,a)} = [∂a(ua,i) – ∂a (ui,a) ] + ∂a(g1/2)[ ua,i – ui,a] = [∂j(uj,i) – ∂j (ui,j) ] + ∂j(g1/2)[ uj,i – ui,j] = [ uj,i,j –ui,j,j ] + ∂j(g1/2)[ uj,i – ui,j] = [ uj,j,i –ui,j,j ] + ∂j(g1/2)[ giαuj,α – gjαui,α] So I am done if I can show this: uj;j;i – ui;j;j = [ uj,j,i –ui,j,j ] + ∂j(g1/2)[ giαuj,α – gjαui,α] (*) Earlier I showed that uj;j;i – ui;j;j = giα [ ∂α {∂j uj + Γjnjun} + Γjαn {∂j un + Γnmjum } – Γnjα {∂n uj + Γjmnum }] – gjα [ ∂α {∂j ui + Γinjun} + Γiαn {∂j un + Γnmjum } – Γnjα {∂n ui + Γimnum }] The first terms on each line may be combined to get giα ∂α {∂j uj) – gjα ∂α (∂j ui) = ∂i {∂j uj) –∂j (∂j ui) = uj,j,i – ui,j,j This matches the first term on the RHS of (*), so we would then have to show that = giα [ ∂α {Γjnjun} + Γjαn {∂j un + Γnmjum } – Γnjα {∂n uj + Γjmnum }] – gjα [ ∂α { Γinjun} + Γiαn {∂j un + Γnmjum } – Γnjα {∂n ui + Γimnum }] = ∂j(g1/2)[ giαuj,α – gjαui,α] = g1/2 Γaaj[ giαuj,α – gjαui,α] I do know from Appendix E (d) that ∂j(g1/2) = g1/2 Γaaj to give the second RHS shown above. How on earth could this complicated thing ever be true? What action can generate g1/2 ? The square root of a determinant of gij. The only helpful theorems I know of are these: (∂cgab) = – [gan Γ bcn + gbn Γacn] 1 (∂cgab) = + [gan Γncb + gbn Γnca] 2 Γdab = (1/2) gdc [ ∂agbc + ∂bgca – ∂cgab] Γaan = (1/2) gad ∂ngad = (1/2)(1/g)∂ng = (1/) ∂n() Go back now to what we want to show is true: = giα [ ∂α {Γjnjun} + Γjαn {∂j un + Γnmjum } – Γnjα {∂n uj + Γjmnum }] – gjα [ ∂α { Γinjun} + Γiαn {∂j un + Γnmjum } – Γnjα {∂n ui + Γimnum }] = ∂j(g1/2)[ giαuj,α – gjαui,α] This would have to be true for any vector ui. It seems then that all the non-derivative terms would then have to add up to 0: giα [ ∂α {Γjnj}un + Γjαn {Γnmjum } – Γnjα {Γjmnum }] – gjα [ ∂α { Γinj}un + Γiαn { Γnmjum } – Γnjα {Γimnum }] = 0 ? giα [ ∂α {Γjmj}um + Γjαn {Γnmjum } – Γnjα {Γjmnum }] – gjα [ ∂α { Γimj}um + Γiαn { Γnmjum } – Γnjα {Γimnum }] = 0 ? giα [ ∂α {Γjmj} + ΓjαnΓnmj – ΓnjαΓjmn ] – gjα [ ∂α { Γimj} + Γiαn Γnmj – ΓnjαΓimn] = 0 ? Looking at Weinberg p 133, the above structure looks a lot like the curvature tensor. He says Rλμνκ = ∂κΓλμν – ∂νΓλμκ + ΓnμνΓλκn – ΓnμκΓλνn Rjmνα = ∂αΓjmν – ∂νΓjmα + ΓnmνΓjαn – ΓnmαΓjνn Rjmjα = ∂αΓjmj – ∂jΓjmα + ΓnmjΓjαn – ΓnmαΓjjn