Past exam of the general relativity course of the University of Oxford / 2011 / 2 / Solution

For this question use the Levi-Civita connection, its metric compatibility and the usual curvature symmetries. In the index ordering of Question 1, put $R_{\mu \nu \rho \sigma }=g_{\sigma \lambda }{R}_{\mu \nu \rho }{}^{\lambda }$ and $R_{\nu \rho }=g^{\mu \sigma }{R}_{\mu \nu \rho \sigma }$. Contract the second Bianchi identity in its derivative index $\mu$ and curvature index $\rho$. The curvature symmetries giveNow contract $\lambda ,\sigma$. The left-hand side becomes $-{\nabla }^{\rho }{R}_{\nu \rho }$, while the right-hand side becomes ${\nabla }^{\sigma }{R}_{\nu \sigma }-{\nabla }_{\nu }R$. ThusUsing $\nabla g=0$ yields the contracted Bianchi identityTaking the covariant divergence of the Einstein field equations, with constant $G$, then gives $8\pi G{\nabla }^{\mu }{T}_{\mu \nu }=0$. The matter equation is therefore consistent with the geometric identity rather than imposing an extra incompatible condition on the metric.

Stress-energy conservation expresses the local balance of energy and momentum, including their transport and the effects of the gravitational Levi-Civita connection. In a freely falling frame at a point it reduces there to the familiar ordinary divergence law. A global conserved energy requires additional geometric structure, such as a timelike Killing vector field; it does not follow in an arbitrary spacetime just by replacing a covariant derivative with an ordinary derivative.

For relativistic dust,With signature $(+,-,-,-)$, $u^{\nu }{u}_{\nu }=1$ implies $u_{\nu }{u}^{\mu }{\nabla }_{\mu }{u}^{\nu }=0$. Contracting gives ${\nabla }_{\mu }(\rho u^{\mu })=0$; where $\rho >0$, substitution then gives $u^{\mu }{\nabla }_{\mu }{u}^{\nu }=0$. Thus the continuity law combines rest-mass conservation and free geodesic motion for this matter model.

For the Newtonian limit, take $c=1$ and $\eta =\mathrm{diag}(1,-1,-1,-1)$, consistently with Question 3. Use the static weak field sourced by the time-independent, slowly moving matter, retaining first order in $ϵ$ and leading order in the spatial speed. Write spatial gradients with ${\partial }^i={\delta }^{ij}{\partial }_j$ here; these are Euclidean spatial indices in the Newtonian equation. The Levi-Civita connection formula givesThe exact change from proper time to $t$ in the spatial geodesic equation isTerms containing spatial velocities are negligible at leading order, leavingAn additive constant in $\Phi$ is immaterial to this acceleration and is normally fixed by the asymptotic time normalization. With units restored, $g_{00}=1+2{\Phi }_{\mathrm{physical}}/c^2$.

At the same order, staticity and the curvature definition giveThe trace of the four-dimensional Einstein equations is $R=-8\pi GT$, so their trace-reversed form isFor nonrelativistic dust, $u^0\simeq 1$, $u^i\simeq 0$, so $T_{00}\simeq \rho$ and $T\simeq \rho$. Keeping leading matter terms therefore gives $R_{00}=4\pi G\rho$. Combining this with the geometric expression yieldsThis trace reversal is essential: equating $R_{00}$ directly to $8\pi G\rho$ would miss the factor of two. The approximation is the quasistatic Newtonian sector; arbitrary independently imposed gravitational radiation is not part of that sector.