Past exam of the general relativity and cosmology course of the University of Oxford / 2013 / 1

Use metric signature $(-,+,+,+)$, $x^0=ct$, and retain terms linear in $\Phi /c^2$ or $\Psi /c^2$. The inverse metric is $g^{00}=-(1-2\Phi /c^2)$ and $g^{ij}=(1+2\Psi /c^2){\delta }^{ij}$. In the Christoffel symbol formulathe differentiated metric perturbation is already first order, so the prefactor can be replaced by the flat inverse metric. For example ${\Gamma }^0{}_{ij}=\frac{1}{2}{\partial }_0{g}_{ij}=-{\delta }_{ij}{\partial }_0\Psi /c^2$ and ${\Gamma }^k{}_{00}=-\frac{1}{2}{\partial }_k{g}_{00}={\partial }_k\Phi /c^2$. Evaluating the remaining components givesThe lower-index symmetry follows from the torsion-free connection. Products of these first-order Christoffel symbols are beyond the retained weak-field approximation.

A pressureless relativistic fluid has stress-energy tensor $T^{\mu \nu }=\rho u^{\mu }{u}^{\nu }$, with four-velocity normalized by $u^{\mu }{u}_{\mu }=-c^2$. In its rest frame, $u^i=0$, so $\boxed{T^0{}_0=-\rho c^2,T^i{}_j=0}$. Here $\rho$ is rest-frame mass density, or energy density divided by $c^2$.

For the linearized Ricci tensor and scalar, discard the quadratic connection terms. The contracted connection is ${\Gamma }^{\mu }{}_{\mu \nu }={\partial }_{\nu }(\Phi -3\Psi )/c^2$. Differentiation then givesThe Ricci scalar uses the zeroth-order inverse metric, $R=-R_{00}+{\delta }^{ij}{R}_{ij}$, because multiplying first-order curvature by another metric perturbation would again be second order. These are the linearized curvature of a two-potential weak-field metric in the paper's Riemann curvature tensor sign convention.

The Einstein tensor is $G^{\mu }{}_{\nu }=R^{\mu }{}_{\nu }-\frac{1}{2}{\delta }_{\nu }^{\mu }R$. Thus $G^0{}_0=-R_{00}-R/2$, where the time raising introduces the minus sign. Substitution cancels the $\Phi$ and time-derivative terms in that component and givesThe right sides follow from the Einstein field equations $G^{\mu }{}_{\nu }=(8\pi G/c^4)T^{\mu }{}_{\nu }$ with zero cosmological constant. The full equations also contain the momentum constraint $G^0{}_i=-2{\partial }_0{\partial }_i\Psi /c^2=0$ for dust at rest to this order. Specifying arbitrary time-dependent potentials without that constraint would not give a solution of all the Einstein field equations.

For a massive particle parametrized by proper time, the geodesic equation isThe weak-field approximation alone does not imply Newtonian motion. Also take a nonrelativistic limit, $|v|/c\ll 1$, in the usual quasistatic regime so the potentials' time variation does not compensate for the small velocity factors. Reparametrizing by coordinate time and retaining the leading velocity-independent spatial term givesIn empty space the first field equation gives ${\nabla }^2\Psi =0$. For static potentials, the spatial equations imply ${\nabla }^2(\Phi -\Psi )=0$ and then ${\partial }_i{\partial }_j(\Phi -\Psi )=0$. The difference is at most affine; isolated-system boundary conditions fixing both potentials to zero at infinity remove it. Thus $\Phi =\Psi$ andthe empty-space Newton-Poisson equation. With the same quasistatic assumptions and dust present it becomes ${\nabla }^2\Phi =4\pi G\rho$. Keeping rapid time dependence instead leaves ${\nabla }^2\Phi =-3{\partial }_0^2\Psi$ in vacuum, so the static qualification matters.