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Save previous Appendix F sections that got replaced. (g) The covariant derivative of a vector is a rank-2 tensor The rest of this Appendix is in the Picture A context 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 The claim is that Vβ;α transforms as a covariant rank-2 tensor. One must then show that V'b;a = Raα Rbβ Vβ;α or [∂'aV'b – Γ' cab V'c] = Raα Rbβ [∂αVβ – Γ cαβ Vc] (*) 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 (f) 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 first identity of section (c) states that gan Γ bcn + gbn Γacn = –(∂cgab) gcb Γ dab + gdb Γcab = – (∂agcd) // c↔a then n→b 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 (h) 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 (f). 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 – ........... – ΓnxαBacc..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 energetic reader.