General Relativity Fall 2019 Lecture 8: covariant derivatives Yacine Ali-Ha¨ımoud September 26th 2019 METRIC IN NON-COORDINATE BASES Last lecture we defined the metric tensor field g as a \special" tensor field, used to convey notions of infinitesimal 2 µ ν (µ) (ν) spacetime \lengths". In a coordinate basis, we write ds = gµν dx dx to mean g = gµν dx ⊗ dx . While we will mostly use coordinate bases, we don't always have to. In a non-coordinate basis, we would write explicitly ∗(µ) ∗(ν) g = gµν e ⊗ e : Let us consider for example flat 3-D space, in which the line element is d`2 = dx2 + dy2 + dz2 = dr2 + r2dθ2 + r2 sin2 θd'2 in cartesian and spherical polar coordinates, respectively. In homework 3, we assumed there existed some scalar product h::i on the space of vectors; well, this scalar product is just the metric: hV; W i ≡ g(V; W ), and jjV jj2 ≡ g(V; V ). We can read off the norms of the coordinate basis vectors from the line element: 2 2 2 2 2 2 jj@(r)jj = 1; jj@(θ)jj = r ; jj@(')jj = r sin θ: Thus the unit-norm vectors along the coordinate basis vectors are 1 1 e ≡ @ ; e ≡ @ ; e ≡ @ : (r) (r) (θ) r (θ) (') r sin θ (') The dual basis vectors are then e∗(r) = dr; e∗(θ) = r dθ; e∗(') = r sin θ d': ∗(µ) µ Indeed, you can check explictly that these vectors satisfy e · e(ν) = δν { again, the \·" operation represents the action of dual vectors on vectors, not the scalar product. In this non-coordinate basis, the line element is then d`2 ≡ g = e∗(r) ⊗ e∗(r) + e∗(θ) ⊗ e∗(θ) + e∗(') ⊗ e∗('): RAISING AND LOWERING INDICES WITH THE METRIC AND ITS INVERSE αβ αβ α We define the rank-(2, 0) inverse metric tensor field g such that g gβγ = δγ . ∗ We saw last week that there isn't any basis-independent mapping from Vp to Vp (in contrast to the basis-independent ∗∗ mapping between Vp and Vp ). Now that we have a special tensor gαβ, we can use it and its inverse to go from Vp ∗ α to Vp , without having to define a basis. Specifically, given a vector X , we define the dual vector Xα (using the same letter!) β Xα ≡ gαβX : α Similarly, given a dual vector Yα, we may define the vector Y α αβ Y ≡ g Yβ : You can check that these two definitions are self-consistent. More generally, we can use the metric to define a rank (k − 1; l + 1) tensor field from a rank-(k; l) tensor field, and the inverse metric to do the opposite. For instance, we have α β αρβλ T γ δ ≡ gγρgδλT : You see that the position of the up and down indices matters. For instance, αβ α β αβρλ αρβλ αβρλ αρβλ T γδ − T γ δ = gγρgδλT − gγρgδλT = gγρgδλ T − T 6= 0 in general; unless the rank-(4,0) tensor T αβγδ is symmetric in its middle two indices. 2 COVARIANT DERIVATIVES Given a scalar field f, i.e. a smooth function f { which is a tensor of rank (0, 0), we have already defined the dual vector rαf. We saw that, in a coordinate basis, @f V αr f = V µ ≡ r f α @xµ V gives the directional derivative of f along V . We now want to generalize this idea of directional derivative to tensor fields of arbitrary rank, and we want to do so in a geometric, basis-independent way. We cannot just recklessly take derivatives of a tensor's components: partial derivatives of components do not transform as tensors under coordinate transformations. Indeed, given a vector field V α, under a coordinate transformation, the partial derivatives of its components transform as 0 " 0 # 0 0 @V µ @xν @ @xµ @xµ @xν @V µ @xν @2xµ = V µ = + V µ: @xν0 @xν0 @xν @xµ @xµ @xν0 @xν @xν0 @xν @xµ The presence of the second piece means that the partial derivatives of the components do not transform as a tensor. µ We thus need to correct correct for this: we have to find some non-tensor coefficients Γνσ such that µ µ µ µ σ rν V ≡ V ;ν ≡ @ν V + ΓνσV as a whole transforms as a tensor. These coefficients will thus transform as µ0 ν σ µ0 2 µ µ0 @x @x @x µ @x @ x Γ 0 0 = Γ + : (1) ν σ @xµ @xν0 @xσ0 νσ @xµ @xν0 @xσ0 µ µ σ The proof that the combination @ν V + ΓνσV does indeed transform as a tensor is identical to Exercise 1 (iii) in homework 2. • Axiomatic definition. We will define a covariant derivative in an axiomatic way { we will see that this definition does not uniquely specify the covariant derivative yet. Given a tensor T of rank (k; l), we define the tensor rT of rank (k; l + 1), denoted as follows (rT )α1...αk ≡ r T α1...αk ≡ T α1...αk ; β1...βlγ γ β1...βl β1...βl;γ satisfying the following properties: (i) the operator r is linear, i.e. for two tensors T ; S of equal rank (and the same index structure), and two real numbers a; b, r(aT + bS) = arT + brS. (ii) for a scalar field f, rf is just the usual gradient. (iii) it satisfies Leibniz's rule: given two tensors T ; S of arbitrary ranks (not necessarily equal), r(T ⊗ S) = rT ⊗ S + T ⊗ rS; α (iv) it commutes with contractions: given a rank-(1, 1) tensor T β - or more generally, a tensor of rank (k ≥ 1; l ≥ 1), α α β α α α rδ(T α) = rδ(δβ T α) = δβ (rT ) βδ = (rT ) αδ: In words, we are requiring that the gradient of a scalar field (which is itself a contraction) is equal to the contraction α of the rank-(1,2) tensor (rT ) βγ in its first two indices, which is a non-trivial requirement. • Connection coefficients. Let us now apply our axiomatic definition to the covariant derivative of a vector field. Suppose that we are given a coordinate basis f@(µ)g that is smoothly defined around a neighborhood, so that each one of the basis vectors is a smooth vector field. The dual basis fdx(µ)g will then also be defined around some µ neighborhood, and consist of smooth dual vector fields. Now consider a vector field V = V @(µ). The components of 3 µ (µ) µ V are V = V · dx , and are thus n smooth scalar fields. We can thus formally see the expression V = V @(µ) µ as a sum of tensor products of rank-(0,0) tensors (the components) with rank (1, 0) tensors: V = V ⊗ @(µ). Let us now compute a covariant derivative of V , which is a rank (1, 1) tensor. Using Leibniz's rule, we get µ µ µ rV = r(V ⊗ @(µ)) = (rV ) ⊗ @(µ) + V ⊗ (r@(µ)): By construction, any covariant derivative must just give the usual gradient when applied to scalar fields, thus rV µ = µ (ν) (ν) @ν V dx . Since r@(µ) is a rank (1, 1) tensor, we may find its components on the basis fdx ⊗ @(σ)g: σ (ν) r@(µ) = Γνµ dx ⊗ @(σ) : σ This defines the connection coeefficients Γνµ. We thus found µ (ν) σ µ (ν) µ µ σ (ν) rV = @ν V dx ⊗ @(µ) + ΓνµV dx ⊗ @(σ) = (@ν V + ΓνσV ) dx ⊗ @(µ); after renaming dummy indices. Thus we have µ µ µ µ σ rν V ≡ V ;ν = @ν V + ΓνσV : You see that the connection coefficients \connect" the covariant derivative to the partial derivative. α • Covariant derivative of a dual vector field. Consider a dual vector field Wα. For any vector field V , the α contraction V Wα is a scalar field. Thus, in a coordinate basis, µ µ µ µ rν (V Wµ) = @ν (V Wµ) = (@ν V )Wµ + V (@ν Wµ); per property (ii) of a covariant derivative, followed by Leibniz's rule for the usual partial derivative. On the other hand, from property (iii) combined with (iv), we also have µ µ µ µ µ σ µ rν (V Wµ) = (rν V )Wµ + V (rν Wµ) = (@ν V )Wµ + ΓνσV Wµ + V (rν Wµ): Equating the two, and renaming dummy indices, we get µ σ µ V rν Wµ + ΓνµWσ = V @ν Wµ: This equality must hold for any vector field V α. Thus we found the components of the rank-(0, 2) tensor rW : σ rν Wµ ≡ Wµ;ν = @ν Wµ − ΓνµWσ : Note the minus sign! • Covariant derivative of a general tensor field. It is straightforward to generalize this to arbitrary tensor fields { you'll do this explicitly in the homework for rank-(0,2) and rank(1,1): µ ···µ λ r T µ1···µk = @ T µ1···µk + Γµ1 T λµ2···µk + ··· Γµk T 1 k−1 σ ν1···νl σ ν1···νl σλ ν1···νl σλ ν1···νl − Γλ T µ1···µk − · · · Γλ T µ1···µk σν1 λν2···νl σνl ν1ν2···νl−1λ In words, the covariant derivative is the partial derivative plus k + l \corrections" proportional to a connection coefficient and the tensor itself, with a plus sign for all upper indices, and a minus sign for all lower indices. THE TORSION-FREE, METRIC-COMPATIBLE COVARIANT DERIVATIVE The properties that we have imposed on the covariant derivative so far are not enough to fully determine it. In fact, there is an infinite number of covariant derivatives: pick some coordinate basis, chose the 43 = 64 connection coefficients in this basis as you wis.