old App F sections g,h,i,j

Old Appendix F sections g,h,i.,j PhL 3.26.12 (g) Theorem: The cov derivative of a cov rank-n tensor is a cov tensor of rank n+1. Proof: The rank-n covariant tensor of interest is Babc..x. We make the ansatz that the covariant form for the covariant derivative ;α of B is given as follows Babc..x;α ≡ ∂α Babc..x – ΓnaαBnbc..x – ΓnbαBanc..x – .... – ΓnxαBabc..n // x-space del a-term b-term x-term B'abc..x;α ≡ ∂'α B'abc..x – Γ 'naαB'nbc..x – Γ 'nbαB'anc..x – .... – Γ 'nxαB'abc..n // x'-space Pause for Examples: (raising indices comes in section(h) below). Let J be the number of indices on B. B;α ≡ ∂αB // here B is a scalar and ∂αB is a covariant vector J = 0 Ba;α ≡ ∂α Ba – ΓnaαBn J = 1 Bab;α ≡ ∂α Bab – ΓnaαBnb – ΓnbαBan J = 2 Babc;α ≡ ∂α Babc – ΓnaαBnbc – ΓnbαBanc – ΓncαBabn J = 3 Resume: We need to show that the object Babc..x;α transforms as a covariant tensor of rank n+1, and if this can be shown, then the ansatz is justified: B'abc..x;α = Raa'Rbb'..... Rxx' Rαα' Ba'b'c'..x';α' or B'ABC..X;α = Rαα'{RAa'RBb'..... RXx'} * Ba'b'c'..x';α' Therefore we need to show that ∂'αB'ABC..X – Γ 'nAαB'nBC..X – Γ 'nBαB'AnC..X – ........... – Γ 'nXαB'ABC..n // LHS del' a'-term b'-term x'-term = Rαα'{RAa'RBb'..... RXx'} * // RHS {∂α'Ba'bc'..x' – Γna'α'Bnb'c'..x' – Γnb'α'Ba'nc'..x' – .. – Γnx'α'Ba'b'c'..n } del a-term b-term x-term Expand the LHS del' term and show del'-del matches the RHS del. ∂'αB'ABC..X = (Rαβ∂β)( RAaRBb..... RXx Babc..x) del' = Rαβ (∂βRAa)RBb..... RXx Babc..x del'-a + Rαβ RAa(∂βRBb)..... RXx Babc..x del'-b ... + Rαβ RAaRBb. ... (∂βRXx) Babc..x del'-x + Rαβ RAaRBb.............RXx (∂β Babc..x) del'-del We first claim that this del'-del term on the LHS matches the del term on the RHS: del'-del LHS = Rαβ {RAaRBb...........RXx} (∂β Babc..x) del RHS = Rαα'{RAa'RBb'..... RXx'} }{∂α' Ba'bc'..x'} Unpriming all the Latin indices and setting α' = β shows that these two terms indeed match. Show that the a-related terms balance. The a'-term on the LHS is this – Γ 'nAαB'nBC..X The connection between Γ ' and Γ given in section (f) says Γ 'cab = Rcd Raκ Rbσ Γdκσ – Raκ Rbσ (∂κRcσ) or Γ 'nAα = Rnd RAκ Rασ Γdκσ – RAκ Rασ (∂κRnσ) We know how to expand the B' terms, so the LHS a'-term becomes – { Rnd RAκ Rασ Γdκσ – RAκ Rασ (∂κRnσ)} { Rnn'RBbRCc....Rxx Bn'bc..x } and to this we must add the contribution called del'-a above. Meanwhile, the RHS a-term is this Rαα'{RAa'RBb'..... RXx'}{– Γna'α'Bnb'c'..x'} // remove Latin primes, then n→n' = – { Rαα'{RAaRBb..... RXx}{Γn'aα'Bn'bc..x} So this then is what must be shown – { Rnd RAκ Rασ Γdκσ – RAκ Rασ (∂κRnσ)} { Rnn'RBbRCc....Rxx Bn'bc..x } + Rαβ (∂βRAn')RBb..... RXx Bn'bc..x // a→n' = – { Rαα'{RAaRBb..... RXx}{Γn'aα'Bn'bc..x} ? where now the del'-a term has been added in to the LHS, and so doing index a→ n'. We can see that the factors RBbRCc....Rxx Bn'bc..x are the same on both sides so they can be removed to give a simpler relation which we must show is valid: – { Rnd RAκ Rασ Γdκσ – RAκ Rασ (∂κRnσ)} { Rnn' } + Rαβ (∂βRAn') = – { Rαα'{RAa }{Γn'aα'} ? In the first term one sees Rnd Rnn' = δdn' which pins d to n' in that term only, so the above becomes – RAκ Rασ Γn'κσ + Rnn'RAκ Rασ (∂κRnσ) + Rαβ (∂βRAn') = – RασRAκ Γn'κσ ? The first term on the left cancels the term on the right, and do β→σ in the third term to get RAκ Rασ {Rnn'(∂κRnσ)} + Rασ (∂σRAn') = 0 . ? Then cancel the common Rασ and use the symmetry (∂κRnσ) = (∂σRnκ) to get (∂σRAn') = – Rnn'RAκ (∂σRnκ) ? Now in this order do σ→a, A→d, n→e, n'→n, κ→m to get (∂aRdn) = –Ren Rdm (∂aRem) and this is seen to be the third identity of section (b). Thus it has been shown that the three a-related terms in the above LHS = RHS balance: a'-term + del'-a = a-term. or – { Rnd RAκ Rασ Γdκσ – RAκ Rασ (∂κRnσ)} { Rnn'RBbRCc....Rxx Bn'bc..x } + Rαβ (∂βRAn')RBb..... RXx Bn'bc..x = – { Rαα'{RAaRBb..... RXx}{Γn'aα'Bn'bc..x} Show that the b-related terms balance. In the previous equation, which was shown true, do A,a↔B,b (indices) and Bn'bc..x → Ban'c..x to get the following known-valid equation : – { Rnd RBκ Rασ Γdκσ – RBκ Rασ (∂κRnσ)} { Rnn'RAaRCc....Rxx Ban'c..x } + Rαβ (∂βRBn')RAa..... RXx Ban'c..x = – { Rαα'{RBbRAa..... RXx}{Γn'bα' Ban'c..x } . The first line is in fact the b'-term – Γ 'nBαB'AnC..X, the second line is del'-b, and the RHS is the b-term. This shows that the three b-related terms match LHS = RHS. Similarly, the c,d.....x terms match, and this concludes the proof. QED How to apply to tensor densities. The above theorem is not valid for tensor densities with weight W ≠ 0. If one traces the above proof, one finds that the extra J-W factors do in fact cancel, but a whole extra term appears in the expansion of the ∂'αB'ABC..X Rαβ(∂βJ-W) RAaRBb..... RXx Babc..x , and this fouls things up. But an escape hatch is available! If Babc..x is a tensor density of weight W, then the quantity (gW/2Babc..x) is a true tensor, since g1/2 is a scalar density of weight -1 and since weights add as shown in Appendix D (b) item 3. Therefore the expansion of interest is this: (gW/2Babc..x);α = ∂α (gW/2Babc..x) – Γnaα(gW/2Bnbc..x) – Γnbα (gW/2Banc..x) – .... – Γnxα(gW/2Babc..n) and this new left hand side is a true tensor with the indices shown which transforms as (g'W/2B'abc..x);α = Raa'Rbb'..... Rxx' Rαα' (gW/2Ba'b'c'..x');α' Examples will be given in section (i). (h) Rule for raising any non-last index on a covariant derivative The Rule is stated at the end of this section before the examples. Consider the general form given in section (g) for a covariant derivative Babc..x;α ≡ ∂α Babc..x – ΓnaαBnbc..x – ΓnbαBanc..x – .... – ΓnxαBabc..n // x-space del a-term b-term x-term Notice that there are N indices on Babc..x and there are N corresponding terms on the RHS in addition to the del term. Each index of Babc..x thus has its own "correction term". What happens if one of the indices on Babc..x is raised? To find out, apply gβb to both sides. The effect of doing this is trivial for all terms except the del term and the b-term, since b is a regular tensor index on all such terms, so we get Baβc..x;α ≡ gβb ∂α Babc..x – ΓnaαBnβc..x – gβb ΓnbαBanc..x – .... – ΓnxαBaβc..n del a-term b-term x-term The del term can be written gβb (∂α Babc..x) = ∂α (gβb Babc..x) – (∂α gβb) Babc..x = ∂α Baβc..x – (∂α gβb) Babc..x The second term here can be combined with the b-term to give del-extra + b-term = – (∂α gβb) Babc..x – gβb ΓnbαBanc..x = – (∂α gβn) Banc..x – gβb ΓnbαBanc..x // b→n in first term only = [– (∂α gβn) – gβb Γnbα] Banc..x The first identity of section (c) reads (∂cgab) = – [gai Γ bci + gbi Γaci] or [– (∂cgab) – gai Γ bci] = gbi Γaci // now do c→α, b→n, a→β or [– (∂αgβn) – gβi Γ nαi] = gni Γβαi or [– (∂αgβn) – gβb Γ nαb] = gni Γβαi Therefore, del-extra + b-term = gni Γβαi Banc..x = Γβαi Baic..x = Γβαn Banc..x so it has been shown that Baβc..x;α ≡ ∂α Baβc..x – ΓnaαBnβc..x + Γβαn Banc..x – .... – ΓnxαBaβc..n Here then is a comparison Babc..x;α ≡ ∂α Babc..x – ΓnaαBnbc..x – ΓnbαBanc..x – .... – ΓnxαBabc..n Babc..x;α ≡ ∂α Babc..x – ΓnaαBnbc..x + Γbαn Banc..x – .... – ΓnxαBabc..n del a-term b-term x-term Rule for raising some non-last index q: (1) In all terms, raise the corresponding B index. (2) in the q correction term, make the replacement – Γnqα → + Γqnα ( = Γqαn ) Corollary: In order to construct the covariant derivative of any rank-N tensor, first write out the known expression for the all-covariant tensor, then use the above raising rule to raise indices as needed. What about tensor densities? This theorem is valid for tensor densities. The presence of gW/2 factors sitting in the terms does no harm to the derivation given above. One finds then that if one starts with the expansion given at the end of the previous section for a tensor density of weight W, (gW/2Babc..x);α = ∂α (gW/2Babc..x) – Γnaα(gW/2Bnbc..x) – Γnbα (gW/2Banc..x) – .... – Γnxα(gW/2Babc..n) then one can raise index b just as before to get (gW/2Babc..x);α = ∂α (gW/2Babc..x) – Γnaα(gW/2Bnbc..x) + Γbαn (gW/2Banc..x) – .... – Γnxα(gW/2Babc..n) (i) Examples of covariant derivative expressions It will be assumed that all B objects in the examples are true tensors unless otherwise specified. Example J = 0 (covariant derivative of a scalar B) B;α = ∂α B covariant vector // = B,α B;α = ∂α B contravariant vector // = B,α Example J=1: (covariant derivative of a vector B) Ba;α = ∂α Ba – ΓnaαBn covariant rank-2 tensor // second term is symmetric on a↔α Ba;α = ∂α Ba + Γaαn Bn mixed rank-2 tensor To obtain the other two possibilities, it is necessary to apply the metric tensor gαβ and gαβ∂β = ∂α , Ba;α = ∂α Ba – gαβΓnaβBn mixed rank-2 tensor Ba;α = ∂α Ba + gαβΓaβn Bn contravariant rank-2 tensor Example J=2: (covariant derivative of a rank-2 tensor) Bab;α ≡ ∂α Bab – ΓnaαBnb – ΓnbαBan covariant rank-3 tensor Bab;α ≡ ∂α Bab + Γaαn Bnb – ΓnbαBan etc. Bab;α ≡ ∂α Bab – ΓnaαBnb + Γbαn Ban Bab;α ≡ ∂α Bab + Γaαn Bnb + Γbαn Ban Again, application of gαβ would give expressions for the other four possibilities with :α being "up". These "other possibilities" are always present, but we shall no longer mention them in the following examples, Example J=3: (covariant derivative of a rank-3 tensor Babc;α ≡ ∂α Babc – ΓnaαBnbc – ΓnbαBanc – ΓncαBabn Babc;α ≡ ∂α Babc + Γaαn Bnbc – ΓnbαBanc – ΓncαBabn etc for the other 4 possibilities with ;α down. Special J=2 application to the metric tensor: gab;α ≡ ∂α gab – Γnaαgnb – Γnbαgan = 0 by section (c) identity 2 gab;α ≡ ∂α gab + Γaαn gnb – Γnbαgan = Γaαb – Γabα = 0 gab;α ≡ ∂α gab – Γnaαgnb + Γbαn gan = – Γbaα + Γbαa = 0 gab;α ≡ ∂α gab + Γaαn gnb + Γbαn gan = 0 by section (c) identity 1 The middle lines use the fact that gij = δij. Since gab;α is tensor, knowing that any one of the above vanishes implies that all four lines vanish! The net result is gab;α = gab;α = gab;α = gab;α = 0 // Weinberg p 105 (4.6.16,17,18) The covariant derivative of any form of the metric tensor vanishes. As Weinberg points one, one knows that in a quasi-Cartesian x-space gab;α = 0 since gab = Gaaδa,b and Γ = 0. Then in any x'-space g'ab;α = 0 as well since g'ab;α = Raa' Rbb' Rαα' ga'b';α' Example J=2: (double covariant derivatives) Consider again the J=1 examples from above Ba;α = ∂α Ba – ΓnaαBn Ba;α = ∂α Ba + Γaαn Bn This applies to any vector Ba. As the J=0 example shows, B;a and B;a are bona-fide vectors (covariant and contravariant components of the same vector ) and therefore B;a;α = ∂α B;a – ΓnaαB;n B;a;α = ∂α B;a + Γaαn B;n where we have simply inserted a semicolon in each term. Consider again the J=2 examples from above Bab;α ≡ ∂α Bab – ΓnaαBnb – ΓnbαBan Bab;α ≡ ∂α Bab + Γaαn Bnb – ΓnbαBan This applies to any rank-2 tensors Bab or Bab. According to sections (i) and (h) above, Ba;b and Ba;b are bona-fide rank-2 tensors, and therefore Ba;b;α ≡ ∂α Ba;b – ΓnaαBn;b – ΓnbαBa;n Ba;b;α ≡ ∂α Ba;b + Γaαn Bn;b – ΓnbαBa;n In a similar manner one can derive expressions for triple covariant derivatives and beyond. For example Ba;b;c;α ≡ ∂α Ba;b;c – ΓnaαBn;b;c – ΓnbαBa;n;c – ΓncαBa;b;n The next examples are for tensor densities: Example J = 0 (covariant derivative of a scalar density B of weight W) (gW/2B);α = ∂α(gW/2B) covariant vector (gW/2B);α = ∂α(gW/2B) contravariant vector Example J=1: (covariant derivative of a vector density B of weight W) (gW/2Ba);α = ∂α (gW/2Ba) – gW/2 Γnaα 2Bn covariant rank-2 tensor (gW/2Ba);α = ∂α (gW/2Ba) + gW/2 Γaαn Bn mixed rank-2 tensor Example J=2: (covariant derivative of a tensor density B of weight W) (gW/2Bab);α ≡ ∂α (gW/2Bab) – Γnaα(gW/2Bnb) – Γnbα(gW/2Ban) covariant rank-3 tensor (gW/2Bab);α ≡ ∂α (gW/2Bab) + Γaαn (gW/2Bnb) – Γnbα(gW/2Ban) etc. (gW/2Bab);α ≡ ∂α (gW/2Bab) – Γnaα(gW/2Bnb) + Γbαn (gW/2Ban) (gW/2Bab);α ≡ ∂α (gW/2Bab) + Γaαn (gW/2Bnb) + Γbαn (gW/2Ban) (j) The Leibnitz product rule for covariant derivatives If A and B are arbitrary tensors each with an arbitrary set of up and down indices, then the claim of the product rule is this: (A----B----);n ≡ A----;n B---- + A---- B----;n // Weinberg p 105 (4.6.14) Proof: Write out the LHS using the general form given in section (h), for example, Tabc..x;α ≡ Tabc..x,α – ΓnaαTnbc..x + Γbαn Tanc..x – .... – ΓnxαTabc..n ←-----------these are the correction terms---------------→ so that for the LHS one gets (A----B----);n = (A----B----),n + (correction terms for A indices) + (correction terms for B indices) Then write out the RHS using the same general form, A----;n B---- = A----,n B---- + (correction terms for A indices) A---- B----;n = A---- B----,n + (correction terms for B indices) The correction terms balance LHS = RHS. The remaining requirement is that (A----B----),n = A----,n B---- + A---- B----,n or ∂n(A----B----) = (∂n A----) B---- + A---- (∂n B----) but this is just the regular Leibnitz product rule. QED. What about tensor densities? If A has weight W and B has weight w, then gW/2A---- and gw/2B---- are both true tensors so the above theorem then applies as follows: (gW/2A---- gw/2B----);n ≡ (gW/2A----);n (gw/2B----) + (gW/2A----) (gw/2B----);n Examples with two vectors: (AaBb);n = Aa;nBb + AaBb;n (AaBb);n = Aa;nBb + AaBb;n (AaBb);n = Aa;nBb + AaBb;n (AaBb);n = Aa;nBb + AaBb;n Example with a scalar function A and a vector B: (ABb);n = A;nBb + ABb;n = A,nBb + ABb;n Example with a constant A and a vector B: (ABb);n =A(Bb;n) so a constant can always be extracted from (ABb);n to give A(Bb;n). A more general example: (AabcBde);n ≡ Aabc;n Bde + Aabc Bde;n An example with the metric tensor: (gabB----b----);α = gab;α B----b---- + gab B----b----;α But gab;α = 0 as shown at the end of section (h). Therefore (B----a----);α = gab B----b----;α which says that raising an index "commutes" with covariant differentiation -- you can raise an index ignoring the fact that :α is sitting there. But we already know this must be true because we know that the object B----b----;α is a true tensor, and gab can raise any index on a true tensor.