Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 37E (General Relativity)

In the static limit, the supplied linearized Riemann tensor givesThe trace of the Einstein field equations is $R=-\kappa T$. Here $T\simeq -\rho$, so $R=\kappa \rho$. The $00$ field equation is thereforewhich yields $R_{00}=\kappa \rho /2$. Combining the two expressions for $R_{00}$ gives

For a slowly moving particle in a static weak field, the spatial part of the geodesic equation reduces toTo first order in the perturbation,and henceNewtonian acceleration is $-{\delta }^{ij}{\partial }_j\Phi$, so $h_{00}=-2\Phi$. Substitution into part (a), followed by the Poisson equation ${\nabla }^2\Phi =4\pi G\rho$, givesThus the Newtonian limit of general relativity fixes

For a point mass, $\Phi =-GM/r$, and part (b) givesThe linearized Ricci tensor and scalar obeyFor a static perturbation with $h_{0i}=0$, this becomesAway from the source, ${\nabla }^2{h}_{00}=0$. For $h_{ij}=f(r)x_i{x}_j$, one has $h_{ii}=r^{2}f$, so the two supplied identities giveThe vacuum trace equation $R=0$ therefore reduces towhose general solution for $r>0$ isThe scalar equation alone leaves $C$ arbitrary. Imposing the remaining vacuum components $R_{ij}=0$ and matching the mass fixes $C=2GM$, as in the linearized point-mass metric in radial Cartesian form.