old App F affine

Christoffel v2 PhL 3.26.05 This is an early draft version of Appendix F. It was replaced with something much better and much more complete. But wave. Appendix F: The Affine Connection Γcab and Covariant Differentiation 1 (a) Definition of Γ and Γ ' 1 (b) Reverse-Tilt-of-R Derivative theorems 2 (c) Upper-index Γ theorem 3 (d) Relations between Γ and Γ ' 4 (e) The covariant derivative of a vector is a rank-2 tensor 5 (f) Product Rule and Covariant Derivative of a rank-n tensor 7 (g) Theorem: Γdab = (1/2) gdc [ ∂agbc + ∂bgca – ∂cgab] 8 (h) An interpretation of Γ: (∂jen) = Σk Γkjn ek 10 Appendix F: The Affine Connection Γcab and Covariant Differentiation (a) Definition of Γ and Γ ' Here is our former Picture D but with symbol names changed to make things more compatible with the literature on this subject: where R = R' R-1 = R' S => R' = RR S = R R'-1 = R S' The tangent and reciprocal base vectors in ξ-space associated with the transformations F and F' are these: (en)i = Rni (en)i = Rni (e'n)i = R'ni (e'n)i = R'ni The two affine connection objects are defined as Γcab ≡ (∂xc/∂ξn) (∂2ξn/∂xa∂xb) = Rcn ∂a (∂ξn/∂xb) = Rcn(∂aRbn) = [ec]i (∂a[eb]i) = ec (∂aeb) Γ'cab ≡ (∂x'c/∂ξn) (∂2ξn/∂x'a∂x'b) = R'cn ∂'a (∂ξn/∂x'b) = R'cn(∂'aR'bn) = [e'c]i (∂'a[e'b]i) = e'c (∂'ae'b) The expression (∂aRbn) does not look symmetric under a↔b but it is, (∂aRbn) = ∂a (∂ξn/∂xb) = ∂a∂bξn = ∂b∂aξn = ∂b (∂ξn/∂xa) = (∂bRan) We are mainly interested in the transformation represented by the upper part of Picture D, which we usually refer to as Picture A However, we have to keep in mind that there is a ghost ξ-space behind the black curtain which has transformations to x-space and to x'-space. One can regard this ξ-space as able to "slide around" without affecting x-space and x'-space. For example, since R = R' S, one could double R' and halve S without affecting R. Or one could regard ξ-space as being fixed and then x-space and x'-space are what "slide around" if we double R' and halve S. This is the point of view usually taken. In a general relativity presentation of this subject, ξ space is identified with a local inertial frame of reference, but here we can just think of ξ-space as some generic quasi-Cartesian space (metric tensor G) as described in Section 1. (b) Reverse-Tilt-of-R Derivative theorems The theorems claim (∂'aRdn) = – Ren Rdm (∂'aRem) (∂'aRdn) = – Ren Rdm (∂'aRem) where on each line the R in one derivative has down-tilt indices and the other up-tilt indices. One line can be obtained from the other by reflecting each R index pair in a horizontal line through the indices. Proof: These theorems are a simple consequence of the fact that RS = 1 which in standard notation is written δcb = RcαRbα (one of the orthogonality rules). So, 0 = ∂'a(δde) = ∂'a(RdmRem) = Rdm (∂'aRem) + Rem (∂'aRdm) (*) Apply Σe Ren to both sides of (*) to get 0 = Ren Rdm (∂'aRem) + (Ren Rem) (∂'aRdm) = Ren Rdm (∂'aRem) + δnm (∂'aRdm) = Ren Rdm (∂'aRem) + (∂'aRdn) => (∂'aRdn) = – Ren Rdm (∂'aRem) QED 1 Alternatively, apply Σd Rdn to both sides of (*) to get 0 = (Rdn Rdm) (∂'aRem) + Rdn Rem (∂'aRdm) = δnm (∂'aRem) + Rdn Rem (∂'aRdm) = (∂'aRen) + Rdn Rem (∂'aRdm) => (∂'aRen) = – Rdn Rem (∂'aRdm) now swap d and e: => (∂'aRdn) = – Ren Rdm (∂'aRem) QED 2 (c) Upper-index Γ theorem The claim of this theorem is gcb Γ dab + gdb Γcab = – (∂agcd) where the two Γ objects have the same lower indices, but the upper index gets shuffled. Proof: This proof uses the context of the right side of Picture D1 shown above. First, the Γ objects can be replaced by their definitions, Γcab ≡ Rcn(∂aRbn) // from section (a) Γdab ≡ Rdn(∂aRbn) // c→d The LHS of the claimed theorem may then be written LHSacd = gcb Rdn(∂aRbn) + gdb Rcn(∂aRbn) The metric tensors can be replaced by gcb = Rci Rbj Gij = Rci RbiGii gdb = Rdi Rbj Gij = Rdi RbiGii so that LHSacd = Rci Rbi Rdn(∂aRbn) Gii + Rdi Rbi Rcn(∂aRbn) Gii Now process the right side of the claimed theorem, – RHSacd = (∂agcd) = (∂a[Rci RdiGii]) = (∂a[Rci Rdi]) Gii = Rci(∂aRdi) Gii + Rdi(∂aRci) Gii Since the RHS and LHS involve R matrices of reverse tilts, the second "Reverse-Tilt-of-R Derivative theorem" of section (b) is recruited and adjusted to the current Picture context, (∂'aRdn) = – Ren Rdm (∂'aRem) // direct quote from section (b) (∂aRdn) = – Ren Rdm (∂aRem) // R → R and x' → x (destination space) (∂aRdi) = – Rei Rdm (∂aRem) // n→ i (∂aRci) = – Rei Rcm (∂aRem) // d → c Installing these last two lines into the RHS gives RHSacd = Rci Rei Rdm (∂aRem) Gii + Rdi Rei Rcm (∂aRem) Gii Now rename summation indices m→ n and e→ b RHSacd = Rci Rbi Rdn (∂aRbn) Gii + Rdi Rbi Rcn (∂aRbn) Gii and visual inspection shows that this is the same as LHSacd computed above, QED. (d) Relations between Γ and Γ ' The claimed relations are the following in the context of Picture D1 above (first terms are the same): Γ'cab = Rcd Raα Rbβ Γdαβ + Rcα (∂'aRbα) // Weinberg (4.5.2) Γ'cab = Rcd Raα Rbβ Γdαβ – Rbβ(∂'aRcβ) Γ'cab = Rcd Raα Rbβ Γdαβ – Raα Rbβ (∂αRcβ) // Weinberg (4.5.8) Comments: (1) If the second term were not present, the relation would state that Γdαβ transforms as a mixed rank-3 tensor in the usual manner (Section 7 (j)). Since the second term is present, Γdαβ is not a tensor. (2) Unlike the definitions of Γ and Γ', the relations above makes no reference to ξ-space ! Proof of the first relation : Start with the definition given above, Γ'cab ≡ R'cn(∂'aR'bn) = (RR)cn ∂'a(RR)bn = RcdRdn ∂'a(RbβRβn) // R' = RR = RcdRdnRbβ(∂'aRβn) + Rcd(RdnRβn)( ∂'aRbβ) = RcdRdnRbβ([Raα∂α]Rβn) + Rcd(δdβ)( ∂'aRbβ) // ∂'a = Rak∂k = RcdRaαRbβRdn(∂αRβn) + Rcβ (∂'aRbβ) = RcdRaαRbβ Γdαβ + Rcα (∂'aRbα) // Γdαβ ≡ Rdn(∂αRβn) QED. The second term in the above relation can be written a different manner as follows. First, 0 = ∂'a(δcb) = ∂'a(RcαRbα) = Rcα (∂'aRbα) + Rbα (∂'aRcα) => Rcα (∂'aRbα) = – Rbα(∂'aRcα) = – Rbβ(∂'aRcβ) = – Rbβ Raα(∂αRcβ) and this gives the other two relations stated above. The actual Weinberg equations show Rij = (∂x'i/∂xj) and Rji = (∂xi/∂x'j), as in Section 7 (q). (e) The covariant derivative of a vector is a rank-2 tensor The "covariant derivative" of a vector is defined as Vβ;α ≡ [∂αVβ – Γ cαβ Vc] // in x-space V'b;a ≡ [∂'aV'b – Γ' cab V'c] // in x'-space So the claim of the theorem is that Vβ;α transforms as a covariant rank-2 tensor. One must then show that [∂'aV'b – Γ' cab V'c] = Raα Rbβ [∂αVβ – Γ cαβ Vc] // V'b;a = Raα Rbβ Vβ;α (*) Proof: The brute force method is to replace primed objects on the LHS with unprimed objects: (∂'aV'b) = (Raα∂α) (RbβVβ) = Raα Rbβ (∂αVβ) + Raα(∂α Rbβ)Vβ Γ'cab = Rcd Raα Rbβ Γdαβ + Rcα (∂'aRbα) // From section (b) above V'c = RceVe and ∂'a = Raα∂α The LHS of then becomes LHSb;a = [∂'aV'b – Γ' cab V'c] = Raα Rbβ (∂αVβ) + Raα(∂α Rbβ)Vβ – [Rcd Raα Rbβ Γdαβ + Rcα (∂'aRbα) ] RceVe = Raα Rbβ (∂αVβ) + Raα(∂α Rbβ)Vβ – (RceRcd) Raα Rbβ Γdαβ Ve – (RceRcα) (∂'aRbα) Ve] = Raα Rbβ (∂αVβ) + Raα(∂α Rbβ)Vβ – δedRaα Rbβ Γdαβ Ve – δeα (∂'aRbα) Ve] = Raα Rbβ (∂αVβ) + Raα(∂α Rbβ)Vβ – Raα Rbβ Γeαβ Ve – (∂'aRbe) Ve] // ∂'a = Raα∂α = Raα Rbβ (∂αVβ) + Raα(∂α Rbβ)Vβ – Raα Rbβ Γeαβ Ve – Raα(∂αRbe) Ve] The second and fourth terms cancel leaving = Raα Rbβ (∂αVβ) – Raα Rbβ Γeαβ Ve = Raα Rbβ [(∂αVβ) – Γcαβ Vc] = RHS QED Thus it has been shown that Vβ;α transforms as a covariant rank-2 tensor V'b;a = Raα Rbβ Vβ;α where Vβ;α ≡ [∂αVβ – Γ cαβ Vc] Summary. It immediately follows that all four of these transformations are valid (shown on the left) : V'b;a = Raα Rbβ Vβ;α Vβ;α ≡ [∂αVβ – Γ cαβ Vc] 1 V'b;a = Raα Rbβ Vβ;α Vβ;α = [∂αVβ + Γ βαc Vc] 2 V'b;a = Raα Rbβ Vβ;α Vβ;α = [∂αVβ – gαd Γ cdβ Vc] 3 V'b;a = Raα Rbβ Vβ;α Vβ;α = [∂αVβ + gαd Γ βdc Vc ] 4 We shall now derive the lower three expressions shown on the right. Start with Vβ;α ≡ [∂αVβ – Γ cαβ Vc] Apply Σβgbβ to both sides to get Vb;α = gbβVβ;α ≡ [gbβ(∂αVβ) – gbβΓ cαβ Vc] (*) The "Upper-Index Γ theorem" of section (***) states that gcb Γ dab + gdb Γcab = – (∂agcd) gcβ Γ bαβ + gbβ Γcαβ = – (∂αgcb) // => gbβ Γcαβ = – gcβ Γ bαβ – (∂αgcb) // Γcβα = Γcαβ Installing this into (*) gives Vb;α = [gbβ(∂αVβ) + gcβ Γ bαβ Vc + (∂αgcb) Vc] = [gbc(∂αVc) + Γ bαβ Vβ + (∂αgcb) Vc] The first and third terms are recognized as the RHS of this evaluation (∂αVb) = ∂α(gbcVc) = (∂αgbc) Vc + gbc(∂αVc) and therefore Vb;α = [(∂αVb) + Γ bαβ Vβ ] QED 2 To this result apply Σαgaα to both sides to get Vb;a = gaα Vb;α = [gaα (∂αVb) + gaα Γ bαβ Vβ ] = [ (∂aVb) + gaα Γ bαβ Vβ ] QED 4 Starting over with Vβ;α ≡ [∂αVβ – Γ cαβ Vc] apply Σα gaα to both sides to get Vβ;a = gaαVβ;α ≡ [gaα(∂αVβ) – gaαΓ cαβ Vc] = [(∂aVβ) – gaαΓ cαβ Vc] QED 3 (f) Product Rule and Covariant Derivative of a rank-n tensor Product Rule. Here are two examples of the product rule : (AaBb);n ≡ Aa;nBb + AaBb;n (AabcBde);n ≡ Aabc;n Bde + Aabc Bde;n More generally if A and B are arbitrary tensors each with an arbitrary set of up and down indices, then (A----B----);n ≡ A----;n B---- + A---- B----;n // Weinberg p 105 (4.6.14) where each tensor maintains all its indices and the ;n is simply distributed as shown. Once this fact is known, one can apply ;n to the product of more tensors, for example (A----B---- C----);n ≡ A----;n B---- C---- + A---- B----;n C---- + A---- B---- C----;n The implication here is that the object defined in this product rule transforms as a tensor in the obvious manner. For the cases shown above one would have (A'aB'b);n = Raa'Rbb' Rn'n(Aa'Bb');n' (A'abcB'de);n = Raa'Rbb'Rcc' Rdd'Ree' Rn'n (Aa'b'c'Bd'e');n' (A'----B'----);n = RRRR... RRRR... Rn'n (A-'-'-'-'B-'-'-'-');n' No proof of the product rule is provided here. The brute force method would be to write out the LHS and remove all primed expressions using known facts such as the relation between Γ' and Γ of section (d). Certainly there are more elegant methods. One might consider an induction proof if all else fails. Covariant derivative of a tensor of rank-n. Here are some examples of properly defined covariant derivatives of covariant tensors : Aab;α ≡ ∂α Aab – ΓnaαAnb – ΓnbαAan // Weinberg p 108 Babc;α ≡ ∂α Babc – ΓnaαBnbc – ΓnbαBanc – ΓncαBabn Notice that there is one Γ term for each index of the tensor being differentiated. The summation index n on the right "marches through" the indices of the tensor shown. The general case would be Babc..x;α ≡ ∂α Babc..x – ΓnaαBnbc..x – ΓnbαBanc..x – ........... – ΓnbαBacc..n (*) Again, the implication of this definition is that the resulting object transforms as a tensor: A'ab;α = Raa'Rbb'Rαα' Aa'b';α' B'abc;α = Raa'Rbb' Rcc'Rαα' Ba'b'c';α' B'abc..x;α = R R R R...... Rαα' Ba'b'c'..x';α' (**) The proof that (**) is indeed satisfied by (*) is left to the reader. (g) Theorem: Γdab = (1/2) gdc [ ∂agbc + ∂bgca – ∂cgab] Comments: (1) Once established, this theorem shows that Γ is a property of x-space with its metric tensor g. There is no need for ξ-space in order to think about Γdab. (2) Of course gab = gab(x) and Γdab = Γdab(x), so all objects are fields. (3) The proof in x-space below also works in x'-space so one will have both these results, Γdab = (1/2) gdc [ ∂agbc + ∂bgca – ∂cgab] Γ'dab = (1/2) g'dc [ ∂'ag'bc + ∂'bg'ca – ∂'cg'ab] , so the theorem represents an equation that is "covariant" (Section 7 (u)) even though most objects which appear in the equation are not tensors (gdc is a tensor). The following corollary is derived at the end of this section, and concerns contraction of the upper Γ index with either of the lower ones: Γaan = (1/2) gad ∂ngad = (1/2)(1/g)∂ng = (1/) ∂n() Proof: Here we ignore x'-space and focus on the right side of Picture D above which is replicated here The quasi-Cartesian metric tensor G (see Section 1) is Gij = Gij = δij Gii were Gii = ±1 independently for each i We know that gab = RaeRbfGef = RaeRbe Gee gab = RaeRbfGef = RaeRbe Gee From the first of these lines, gdc = RdiRci Gii . The first line below is computed from the second line in the pair above, then the next two lines below are obtained by doing forward cyclic permutations of the first line : ∂cgab = [Rbe (∂cRae) + Rae(∂cRbe) ]Gee ∂agbc = [Rce (∂aRbe) + Rbe(∂aRce) ]Gee ∂bgca = [Rae (∂bRce) + Rce(∂bRae) ]Gee . The last four lines can be inserted into the Right Hand Side of our desired theorem to obtain (RHS)dab = (1/2) gdc { ∂agbc + ∂bgca – ∂cgab} = (1/2) RdiRci Gii Gee * [Rce (∂aRbe) + Rbe(∂aRce) + Rae (∂bRce) + Rce(∂bRae) – Rbe (∂cRae) – Rae(∂cRbe) ] 1 2 3 4 5 6 Due to the symmetry noted above in section (a), (∂iRje) = (∂jRie), terms 2 and 5 cancel as do terms 3 and 6, while terms 1 and 4 are equal. Therefore, (RHS)dab = (1/2) RdiRci Gii Gee * 2 Rce (∂aRbe) = Rdi (Rce Rci) Gii Gee (∂aRbe) = Rdi δei Gii Gee (∂aRbe) = Rde Gee Gee (∂aRbe) = Rde(∂aRbe) = Γdab QED. The abovementioned corollary is this (g ≡ det(gab) ) Γaan = (1/2) gad( ∂agnd + ∂ngad – ∂dgan ) = (1/2) gad ∂ngad = (1/2) (1/g)∂ng = (1/) ∂n() where the first and third terms cancel due to symmetry gad∂agnd – gad∂dgan = gad∂agnd – gda∂agdn = gad∂agnd – gad∂agnd = 0 and the fact that (1/g) ∂ng = gab(∂ngab) is proved as follows: (1) gab = (g-1)ab = cof(gab)T/det(gab) = cof(gab)/g => cof(gab) = g gab (2) g = det(gab) = Σa gabcof(gab) => ∂g/∂gab = cof(gab) = g gab (3) ∂ng = ∂g/∂xn = (∂g/∂gab)( ∂gab/∂xn) = g gab (∂ngab) => (1/g) ∂ng = gab(∂ngab) The final form shown is just calculus : g-1/2 ∂n(g1/2) = g-1/2 (1/2) g-1/2 ∂n(g) = (1/2) (1/g) (∂ng) (h) An interpretation of Γ: (∂jen) = Σk Γkjn ek If in picture D1 we choose to have R=S=1, then ξ-space and x-space are the same, g = G, and R = R'. This gives Picture A above but with g = G, In this case, one can write (e'n)i = R'ni = Rni = (en)i (e'n)i = R'ni= Rni = (en)i where en and en are the the usual tangent and reciprocal base vectors in x-space. Then, Γ'cab = e'c (∂'ae'b) = ec (∂'aeb) (*) Γ'cab = R'cn(∂'aR'bn) = Rcn(∂'aRbn) Apply Rcd to both sides of the second line and sum on c, Γ'cabRcd = RcdRcn(∂'aRbn) = δdn(∂'aRbn) = (∂'aRbd) Shuffling indices c→k, b→n, d→i, a→j this gives ∂'jRni = ΣkΓ'kjnRki or ∂'j(en)i = ΣkΓ'kjn(ek)i or (∂'jen) = ΣkΓ'kjn ek and from (*) above Γ'kjn = ek (∂'jen) The derivative of a tangent base vector en with respect to a curvilinear coordinate x'j can of course be written as some linear combination of the en since the en form a complete set in x-space. What the above line says is that when this linear combination is formed, the coefficients are Γ'kjn ! If one then converts this result to Picture C, the results can be stated this way : (∂jen) = Σk Γkjn ek where Γkjn = ek (∂jen) = Rka(∂jRna) // from (*) above So one can interpret Γkjn as the set of coefficients which arises when the derivative of a tangent base vector is expanded on the tangent base vectors. Comment: One must be cognizant of "the Picture" one is talking about at any given time. This determines what objects get primes, no primes, or perhaps other markings.