Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 1 / 35C / Solution

Stationarity gives the conserved Killing energy per unit rest mass $E=(1+x^2)\dot{t}$, where dots here denote proper-time derivatives. It measures energy relative to the time-translation Killing field; a static local observer measures $E/\sqrt{1+x^2}$. Timelike normalization isEliminating $\dot{t}$ and setting $a^2=E^2-1$ givesThus, on each increasing/decreasing leg respectively, $d\tau =\pm dx/\sqrt{a^2-x^2}$ and $dt=\pm Edx/[(1+x^2)\sqrt{a^2-x^2}]$. The printed positive differentials describe the increasing leg, not both legs at once. Differentiation along a leg gives $\ddot{x}=-x$, extending smoothly through the turning points. Hence $x=a\sin (\tau -{\tau }_0)$ and the proper-time period is $2\pi$.

The requested velocity sketches are the branches $\dot{x}=\pm \sqrt{a^2-x^2}$ and $dx/dt=\pm (1+x^2)\sqrt{a^2-x^2}/E$. They share the turning points and opposite signs on the two legs; their speed ratio is $(1+x^2)/E$, smaller than one near zero and larger than one near the endpoints. The supplied integral is the coordinate time from $-a$ to $a$, equal to $\pi$. The return leg takes the same time, so the coordinate-time period is also $2\pi$.

For the perturbed metric the same Killing conservation gives $\dot{t}=E/(1+x^2)$, and normalization becomes $(1+f(x){)}^2{\dot{x}}^2=a^2-x^2$. Since $1+f>0$,Both weights without $f$ are even in $x$. Oddness of $f$ makes its integral over the symmetric excursion $[-a,a]$ vanish. Each increasing leg therefore still takes $\pi$ in either clock; the decreasing leg takes the same integrals with reversed orientation. Both periods remain exactly $2\pi$, although the proper-time waveform is no longer generally sinusoidal. This is the odd radial perturbations preserve timelike-geodesic periods mechanism. Smoothness of the metric and positivity of $1+f$ ensure regular turning and monotone coordinate time.