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Fermi Normal Coordinates General Relativity Fall 2019 Lecture 12: Curvature; Fermi normal coordinates Yacine Ali-Ha¨ımoud October 10th, 2019 RIEMANN = 0 , SPACETIME IS FLAT 2 µ ν If spacetime is flat, there exists a globally inertial coordinate system, such that ds = ηµν dx dx . In thees coordinates, the second derivatives of the metric vanish, thus the Riemann tensor vanishes identically. This proves the , implication. Let us now prove the ). We saw that the change of a vector upon parallel transport around a small closed loop is proportional to the Riemann tensor times the area of the loop. Equivalently, we could have shown that the difference in a vector parallel transported along two different paths is proportional to the Riemann tensor times the area of the loop enclosed by the two paths. Therefore, if the Riemann tensor vanishes everywhere, parallel transport is independent of the path (assuming the manifold is simply connected). This holds even for a finite path. To prove this, subdivide a finite loop of area A into N loops of area (A=N). If Riemann vanishes, then the difference in parallel transport per loop (for small enough loop) scales as (A=N)α, where the index α > 1. Hence the total error on a finite path scales as N(A=N)α / Aα=N α−1. This goes to zero as N ! 1. Suppose the Riemann vanishes everywhere. At a given point p 2 M, it is always possible to find an orthonormal basis of the tangent space, fe(µ)g. The fact that the basis is orthonormal means that g(e(µ); e(ν)) = ηµν at p. Now parallel transport the e(µ) everywhere in the manifold. This is a well-defined operation, as parallel transport is α α path-independent because the Riemann tensor vanishes. So, for any vector field V , we have V rαe(µ) = 0. This implies that re(µ) = 0. Since r is metric compatible, the n(n + 1)=2 scalar fields g(e(µ); e(ν)) have zero gradient everywhere, i.e. are constant and equal to ηµν , their value at the initial point p. But those are just the components of g. This means that the vector fields fe(µ)g form an orthonormal basis at any point of the manifold (not just the point p where they were initially defined). We denote by fe∗(µ)g the dual basis. Taking the gradient of ∗(µ) µ ∗(µ) ∗(µ) e ·e(ν) = δν , we find that, for any µ, ν,(re )·e(ν) = 0. Thus the dual vectors also have zero gradient, re = 0. µ 2 µ ν We are not done yet: we want to find globally inertial coordinates fx g such that ds = ηµν dx dx . Since 2 ∗(µ) ∗(ν) µ ∗(µ) we have already obtained ds = ηµν e ⊗ e , let us thus construct coordinates such that dx = e . Suppose we 0 0 start from some arbitrary coordinates fxµ g, with associated differential forms dxµ , which form a basis of the dual ∗ ∗(µ) (µ) µ0 (µ) 0 ∗(µ) vector space. We may decompose e (µ) on this basis: e = eµ0 dx , i.e. eµ0 is the µ -th component of e on the 0 0 0 basis fdxµ g. On the other hand, we have dxµ = (@xµ=@xµ )dxµ . Thus we want to find coordinates xµ that solve the following partial differential equations: µ @x (µ) = e 0 : (1) @xµ0 µ 0 Taking one more derivative with respect to xν , and invoking the fact that partial derivatives in the same coordinate (µ) system commute, we find that this implies that the components eµ0 must satisfy (µ) (µ) (µ) @ν0 eµ0 = @µ0 eν0 ) @[ν0 eµ0] = 0: (2) This is a necessary condition for the partial differential equations (1) to have a solution. It turns out that it is also ∗(µ) ∗(µ) a sufficient condition. Now, we know that rαeβ = 0, thus r[αeβ] = 0. By the symmetry of the Christoffel ∗(µ) µ0 symbols, this implies @[µ0 eν0] = 0 for any coordinates fx g. Thus the condition is indeed satisfied, and we can indeed solve for corrdinates fxµg such that dxµ = e∗(µ), which are therefore a globally inertial coordinate system. Note that µ0 µ µ0 (µ) µ even if the coordinates fx g only cover a subset of the manifold, we may solve @ξ =@x = eµ0 for x as a function 0 00 of xµ locally, then switch to other coordinates xµ , etc. µ µ ν To conclude, if Riemann = 0, we can find global coordinates fx g in which g = ηµν dx dx , i.e. the vanishing of Riemann implies that spacetime is flat. 2 FERMI NORMAL COORDINATES Consider a fiducial timelike geodesic G (i.e. G is a worldline on the manifold) in a curved manifold M. We will build a coordinate system (t; xi) defined in a neighborhood of G, such that the geodesic is at zero spatial coordinates i x jG = 0, the metric is Minkowski along the geodesic, gµν jG = ηµν , and the first derivatives of the metric µ vanish along the geodesic, i.e. @λgµν jG = 0, hence that ΓνσjG = 0. Note that this is more restrictive than a LICS: the metric is close to Minkowski not only at one event, but all along a curve! We will follow the proof of Manasse & Misner 1963. FIG. 1. Fermi normal coordinates are defined in the vicinity of a fiducial timelike geodesic G (left). The right figure illustrates i how they are constructed, by launching spacelike geodesics starting on G, with initial tangent vector T = x e(i). 0 On the geodesic, we define x = τ, the proper time, which we initialize at zero at some point p0 2 G. At that \initial" point, we put the metric in normal form, i.e. define three spacelike vectors e(1); e(2); e(3), in addition to e(0) = d/dτ ≡ U, such that g(e(µ); e(ν))jp0 = ηµν . We then define the four 4-vectors e(µ) all along G by parallel-transporting them. This is self-consistent for e(0) = U, which is already parallel-transported along G by definition of a geodesic. Since parallel transport preserves angles (by metric compatibility of r), g(e(µ); e(ν))jG = ηµν all along the geodesic. We now build coordinates fxµg in the vicinity of the geodesic, as follows. Given four small numbers fxµg, we define the point p(x0; xi) as follows. We build the spacelike geodesic starting on G at time τ = x0, with tangent vector i 0 µ T = x e(i) at p(x ; 0) 2 G. We parametrize the geodesic by λ, and define the point p(x ) to be the point reached 0 when λ = 1. Explicitly, in some arbitrary coordinate system fxµ g, we solve for the following differential equations 0 for the functions xµ (λ; x0; xi): 2 µ0 ν0 σ0 d x µ0 dx dx + Γ 0 0 = 0; (3) dλ2 ν σ dλ dλ µ0 µ0 0 0 x jλ=0 = x (p(x ; 0)); [p(x ; 0) 2 G]; (4) µ0 dx 0 j = xieµ ; (5) dλ λ=0 (i) 0 µ 0 µ0 0 i where e(i) is the µ -th component of e(i) on the coordinate basis f@(µ0)g. This defines the 4 functions x (λ; x ; x ), 0 0 i.e. xµ (λ) given the parameters x0; xi. We then define p(x0; xi) whose coordinates are xµ (λ = 1; x0; xi). This procedure does not necessarily work for arbitrary xi: the geodesics we build do not necessarily extend to λ = 1. However, for small enough xi, the geodesics are indeed well defined up to λ = 1, which is a point close to G. This defines the coordinates fxµg, i.e. the mapping from an open set of R4 to M. 3 Define λ~ = λ/s, where s is a constant, thus d=dλ~ = s d/dλ. Thus we find that the above equations take exactly the 0 0 same form for (λ, xi) and (λ,~ sxi). Hence, xµ (λ; x0; xi) = xµ (λ/s; x0; sxi). In particular, setting s = λ, we find 0 0 xµ (λ; x0; xi) = xµ (1; x0; λxi) : (6) 0 The right-hand-side are the fxµ g coordinates of the point p(x0; λxi), by construction. Let us differentiate this equation with respect to λ, and evaluate it at λ = 0: 0 0 dxµ @xµ j = xi : (7) dλ λ=0 @xi Note that @=@xi means \partial derivative with respect to the i-th variable". From Eq. (5), we find that, for any xi, µ0 @x 0 xi = xieµ : (8) @xi (i) This implies µ0 @x 0 j = eµ (9) @xi G (i) 0 µ0 @xµ On the other hand, @(i) = @xi . Thus we conclude that @(µ)jG = e(µ) ; (10) since we already known that it holds for µ = 0. Thus, the components of g on the basis fxµg are indeed gµν jG = gjG(@(µ);@(ν)) = ηµν . Let us now show that the Christoffel symbols vanish along G. Take one more derivative of Eq. (6) with respect to λ, and evaluate at λ = 0: 0 0 d2xµ @2xµ j = xixj : (11) dλ2 λ=0 @xi@xj From Eqs. (3) and (5), we thus find, for any xi; xj, 2 µ0 µ0 i ν0 j σ0 i j @ x Γ 0 0 x e x e = x x ; (12) ν σ (i) (j) @xi@xj hence, along the geodesic, 2 µ0 µ0 ν0 σ0 @ x Γ 0 0 e e = : (13) ν σ (i) (j) @xi@xj µ µ µ This expression holds for any coordinates, in particular, it holds in the coordinate system fx g, for which e(i) = @(i) = µ µ δi , and for which the right-hand-side is zero.
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