Past exam of the relativity course of the University of Cambridge / 2021 / Section A / 1 / Solution

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.