Past exam of the relativity course of the University of Cambridge / 2021 / Section A

Take $\alpha >0$, $t\ne 0$, and the negative-norm convention for a timelike curve used by the printed metric. Writing $v=dx/dt$ and $d{s}^2=-c^{2}d{\tau }^2$ givesSpatial translation is a Killing vector field. Equivalently, the geodesic Lagrangian has the cyclic coordinate $x$, so its conserved momentum isTherefore the correct first integral for this metric isThe first-integral assertion in the PDF is false. Its dimensionless square root requires $v/c$, and even after setting $c=1$ it is missing the factor $|t|$. For example, the nonvertical geodesics below have $v/\sqrt{1-v^2/c^2}$ proportional to $1/|t|$.

Set $b=\sqrt{\alpha }P/c$. Solving for the velocity, with increasing $t$ as the time orientation, gives $v=cb/\sqrt{t^2+b^2}$. The complete unparametrized family of timelike geodesics of a quadratic-time conformal metric isEach curve is restricted to a connected positive- or negative-$t$ patch; $t=0$ is excluded because the metric becomes degenerate. For $t\ne 0$, $|v|<c$, and the two independent constants are $x_c$ and $b$. If an explicit affine parametrization is desired,while the vertical family has $d\tau =|t|dt/\sqrt{\alpha }$. A reversal of the time orientation reverses the parameter without changing these paths.

Put $\kappa =8\pi G/c^4$ and $T=T_{\nu }{}^{\nu }$. The contracted Bianchi identity and metric compatibility giveTaking the trace of the field equation in four dimensions gives $(1-4\alpha )R=-\kappa T$. For $\alpha \ne 1/4$, eliminate $R$:Ordinary covariantly coupled matter obeys stress-energy conservation, ${\nabla }_{\nu }{T}_{\mu }{}^{\nu }=0$. Its trace need not be constant, so the trace coefficient fixed by stress-energy conservation isThis restores the Einstein tensor and is the only coefficient compatible with arbitrary conventional matter without adding a new constraint or changing the matter conservation law. Other coefficients can work for specially restricted sources with constant trace, but do not give the same universal matter coupling.

The excluded case $\alpha =1/4$ must be treated before dividing: its trace equation requires $T=0$. If stress-energy is also conserved, its divergence equation requires ${\partial }_{\mu }R=0$. It can describe a restricted traceless-source sector with constant scalar curvature, not general matter such as ordinary dust. In particular it is not the general trace-free Einstein equation with source $T_{\mu \nu }-g_{\mu \nu }T/4$, which is a different equation.

Interpret the orbit in the Schwarzschild metric as an unforced circular timelike geodesic and choose its plane as $\theta =\pi /2$. Put $f=1-2\mu /r$. The radial geodesic equation at constant radius uses the Christoffel symbolsand gives the Angular velocity of a circular Schwarzschild geodesicThe proper-time normalization then yieldsOver one revolution, the Proper period of a circular Schwarzschild geodesic is thereforeBoth the gravitational and orbital kinematic contributions to time dilation are included. Circular timelike geodesics exist for $r>3\mu$; they are stable for $r>6\mu$ and unstable for $3\mu <r<6\mu$, with marginal stability at $6\mu$.

If “circular orbit” instead means a forced trajectory with an externally chosen angular velocity, the radius alone does not fix the answer. Its elapsed time is $\Delta \tau =\Delta t\sqrt{1-2\mu /r-r^2{\Omega }^2/c^2}$, or $2\pi \sqrt{1-2\mu /r-r^2{\Omega }^2/c^2}/|\Omega |$ per revolution. The geodesic interpretation supplies the missing angular velocity.