Paper 1, Section II, 38D

General Relativity | Part II, 2020

Let (M,g)(\mathcal{M}, \boldsymbol{g}) be a four-dimensional manifold with metric gαβg_{\alpha \beta} of Lorentzian signature.

The Riemann tensor R\boldsymbol{R} is defined through its action on three vector fields X,V,W\boldsymbol{X}, \boldsymbol{V}, \boldsymbol{W} by

R(X,V)W=XVWVXW[X,V]W\boldsymbol{R}(\boldsymbol{X}, \boldsymbol{V}) \boldsymbol{W}=\nabla_{\boldsymbol{X}} \nabla_{\boldsymbol{V}} \boldsymbol{W}-\nabla_{\boldsymbol{V}} \nabla_{\boldsymbol{X}} \boldsymbol{W}-\nabla_{[\boldsymbol{X}, \boldsymbol{V}]} \boldsymbol{W}

and the Ricci identity is given by

αβVγβαVγ=RραβγVρ.\nabla_{\alpha} \nabla_{\beta} V^{\gamma}-\nabla_{\beta} \nabla_{\alpha} V^{\gamma}=R_{\rho \alpha \beta}^{\gamma} V^{\rho} .

(i) Show that for two arbitrary vector fields V,W\boldsymbol{V}, \boldsymbol{W}, the commutator obeys

[V,W]α=VμμWαWμμVα[\boldsymbol{V}, \boldsymbol{W}]^{\alpha}=V^{\mu} \nabla_{\mu} W^{\alpha}-W^{\mu} \nabla_{\mu} V^{\alpha}

(ii) Let γ:I×IM,I,IR,(s,t)γ(s,t)\gamma: I \times I^{\prime} \rightarrow \mathcal{M}, \quad I, I^{\prime} \subset \mathbb{R},(s, t) \mapsto \gamma(s, t) be a one-parameter family of affinely parametrized geodesics. Let T\boldsymbol{T} be the tangent vector to the geodesic γ(s=\gamma(s= const, t)t) and S\boldsymbol{S} be the tangent vector to the curves γ(s,t=\gamma(s, t= const )). Derive the equation for geodesic deviation,

TTS=R(T,S)T\nabla_{T} \nabla_{T} \boldsymbol{S}=\boldsymbol{R}(\boldsymbol{T}, \boldsymbol{S}) \boldsymbol{T}

(iii) Let XαX^{\alpha} be a unit timelike vector field (XμXμ=1)\left(X^{\mu} X_{\mu}=-1\right) that satisfies the geodesic equation XX=0\nabla_{\boldsymbol{X}} \boldsymbol{X}=0 at every point of M\mathcal{M}. Define

Bαβ:=βXα,hαβ:=gαβ+XαXβ,Θ:=Bαβhαβ,σαβ:=B(αβ)13Θhαβ,ωαβ:=B[αβ].\begin{array}{ll} B_{\alpha \beta}:=\nabla_{\beta} X_{\alpha}, & h_{\alpha \beta}:=g_{\alpha \beta}+X_{\alpha} X_{\beta}, \\ \Theta:=B^{\alpha \beta} h_{\alpha \beta}, \quad \sigma_{\alpha \beta}:=B_{(\alpha \beta)}-\frac{1}{3} \Theta h_{\alpha \beta}, & \omega_{\alpha \beta}:=B_{[\alpha \beta]} . \end{array}

Show that

BαβXα=BαβXβ=hαβXα=hαβXβ=0Bαβ=13Θhαβ+σαβ+ωαβ,gαβσαβ=0\begin{gathered} B_{\alpha \beta} X^{\alpha}=B_{\alpha \beta} X^{\beta}=h_{\alpha \beta} X^{\alpha}=h_{\alpha \beta} X^{\beta}=0 \\ B_{\alpha \beta}=\frac{1}{3} \Theta h_{\alpha \beta}+\sigma_{\alpha \beta}+\omega_{\alpha \beta}, \quad g^{\alpha \beta} \sigma_{\alpha \beta}=0 \end{gathered}

(iv) Let S\boldsymbol{S} denote the geodesic deviation vector, as defined in (ii), of the family of geodesics defined by the vector field XαX^{\alpha}. Show that S\boldsymbol{S} satisfies

XμμSα=BμαSμX^{\mu} \nabla_{\mu} S^{\alpha}=B_{\mu}^{\alpha} S^{\mu}

(v) Show that

XμμBαβ=BβμBαμ+RμβανXμXνX^{\mu} \nabla_{\mu} B_{\alpha \beta}=-B_{\beta}^{\mu} B_{\alpha \mu}+R_{\mu \beta \alpha}{ }^{\nu} X^{\mu} X_{\nu}

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