Past exam of the general relativity course of the University of Oxford / 2022 / 1

For the Levi-Civita connection, the all-lowered Riemann curvature tensor has the algebraic symmetriesThe first Bianchi identity is the cyclic identityThese statements concern the Riemann curvature tensor at each point and are unchanged by reversing its overall curvature sign convention.

A semicolon denotes the covariant derivative for the Levi-Civita connection. For example, the four lower indices of the Riemann curvature tensor contribute four connection terms:Here ${\partial }_{\eta }=\partial /\partial x^{\eta }$ is a partial derivative, ${\Gamma }_{b\eta }^a$ are Christoffel symbols, and repeated indices are summed from $0$ to $3$. Write $g_{\mu \nu }$ for the metric tensor, $g^{\mu \nu }$ for its inverse metric, $R_{\mu \nu }=g^{\lambda {\nu }^{\prime }}{R}_{\lambda \mu {\nu }^{\prime }\nu }$ for the Ricci tensor, and $R=g^{\mu \nu }{R}_{\mu \nu }$ for the Ricci scalar. This tensor contraction agrees with the paper's $R_{\mu \nu }=-R^a{}_{\mu \nu a}$.

By metric compatibility, the inverse metric may pass through each covariant derivative. Contracting the second Bianchi identity with $g^{\lambda \nu }$ giveswhere ${\nabla }_{\eta }$ is the same covariant derivative as the semicolon. Next contract with $g^{\mu \kappa }$. Antisymmetry of the first pair gives $g^{\mu \kappa }{R}_{\lambda \mu \kappa \eta }=-R_{\lambda \eta }$, soThus the contracted Bianchi identity is ${\nabla }^{\mu }{R}_{\mu \eta }=\frac{1}{2}{\partial }_{\eta }R$. Raising the remaining index and using metric compatibility provesThe Einstein tensor $G^{\mu \nu }$ therefore has identically zero covariant divergence, without imposing any field equation. This makes the Einstein field equations compatible with stress-energy conservation, ${\nabla }_{\mu }{T}^{\mu \nu }=0$, where $T^{\mu \nu }$ is the stress-energy tensor. The connection terms describe how energy and momentum are compared at neighboring points; ordinary componentwise partial derivatives would not give a coordinate-independent conservation law in curved spacetime.

Take the tensor contraction of the Einstein field equations with the inverse metric $g^{\mu \nu }$. In four-dimensional spacetime, $g^{\mu \nu }{g}_{\mu \nu }=4$, henceHere $T$ is the tensor contraction of the stress-energy tensor. Substituting this result back into the Einstein field equations, while keeping both free indices lowered, gives the trace-reversed stress-energy tensor:The factor $1/2$ uses the dimension four: in dimension $n>2$, the Trace-reversed Einstein field equations in D dimensions have $1/(n-2)$ in place of $1/2$.

Use $x^0=ct$ and metric signature $(-,+,+,+)$. With the source's curvature sign convention, the Ricci tensor isIts time-time component is thereforeFor a slowly moving, pressure-negligible source, the stress-energy tensor is approximated by relativistic dust: $T^{\mu \nu }=\rho U^{\mu }{U}^{\nu }$, where $\rho$ is mass density and $U^{\mu }$ is the four-velocity. In its approximate rest local inertial frame, $U^{\mu }\simeq (c,0,0,0)$ and $U_0\simeq -c$. Thusand the trace-reversed stress-energy tensor satisfies

Independently, introduce the metric perturbation $g_{\mu \nu }={\eta }_{\mu \nu }+h_{\mu \nu }$, with $|h_{\mu \nu }|\ll 1$ and constant Minkowski metric ${\eta }_{\mu \nu }$. The Christoffel symbols are first order in $h$, so their products are discarded in the weak-field approximation. With a comma denoting a partial derivative, the linearized Riemann curvature tensor isTaking the tensor contraction gives, to this order,In the static Newtonian limit, all time partial derivatives vanish. The constant ${\eta }_{00}$ also has zero partial derivatives, leavingHere ${\nabla }^2={\sum }_i{\partial }_i^2$ is the spatial Laplace operator in the approximately Cartesian background coordinates. This sign follows the paper's curvature sign convention.

For the Newtonian gravitational potential $\Phi$, the weak static metric tensor has $g_{00}\simeq -1-2\Phi /c^2$. The Newtonian limit derived above consequently givesMatching the Poisson equation for Newtonian gravity, ${\nabla }^2\Phi =4\pi G\rho$, fixesHere $G$ is the gravitational constant and $c$ is the speed of light. This negative coupling agrees with the source formula sheet. The familiar positive coupling uses the opposite curvature sign convention for the Ricci tensor; the physical Poisson equation for Newtonian gravity is the same.