Past exam of the general relativity course of the University of Oxford / 2021 / 3

The intended orbit is a freely falling timelike circular Schwarzschild geodesic. Use the Schwarzschild metric with $t$ rather than $ct$ as the zeroth coordinate, and put the orbit at $\theta =\pi /2$. Thus $g_{tt}=-f{c}^2$, $g_{rr}=1/f$, $g_{\varphi \varphi }=r^2$, with $f=1-2GM/(r{c}^2)$. The radial geodesic equation at constant $r$ isConsequently ${\Omega }^2=(u^{\varphi }/u^t{)}^2=GM/r^3$. The four-velocity normalization givesand henceHere $t$ is Schwarzschild time, $\tau$ the orbiting clock's proper time, $G$ the gravitational constant, $M$ the central mass, $r$ the constant areal radius, and $c$ the speed of light. The dimensionless $u^0$ follows the question's time-valued zeroth-coordinate convention; with $x^0=ct$ the zeroth four-velocity component would be $cdt/d\tau$.

The condition is $r>3GM/c^2$ for a massive circular geodesic; radial stability further requires $r>6GM/c^2$, with equality marginal. The phrase “any circular orbit” is false without free fall: a powered circular trajectory has $dt/d\tau =[f-r^2{\Omega }^2/c^2{]}^{-1/2}$ with independently chosen $\Omega$. For example, at $r=10GM/c^2$ choose $\Omega =\frac{1}{2}\sqrt{GM/r^3}$; this valid accelerated circular trajectory has $dt/d\tau =\sqrt{40/31}$, instead of the displayed $\sqrt{10/7}$.

Let $\mu =GM/c^2$, $f=1-2\mu /r$, $q=\sqrt{1-3\mu /r}=1/u^0$, and choose $\Omega >0$. A torque-free test gyroscope moving on a geodesic has a spin four-vector obeying parallel transport,The orthogonality condition says the temporal spin component vanishes in the body's rest frame, not necessarily in Schwarzschild coordinates. Here it givesIn addition to the radial Christoffel symbols in part (a), use ${\Gamma }_{tr}^t=\mu /(r^{2}f)$ and ${\Gamma }_{r\varphi }^{\varphi }=1/r$. On the equator the parallel transport equations becomeThe radial equation follows by substituting $s^t$ into $d{s}^r/dt+{\Gamma }_{tt}^r{s}^t+\Omega {\Gamma }_{\varphi \varphi }^r{s}^{\varphi }=0$. Eliminating $s^{\varphi }$ gives $d^2{s}^r/d{t}^2+{\Omega }^2{q}^2{s}^r=0$.

Let $s_0$ be the invariant spin magnitude. Initially radial spin means $s^r(0)=s_0\sqrt{f}$ and $s^t(0)=s^{\varphi }(0)=s^{\theta }(0)=0$. The complete solution isTo interpret the orientation, convert coordinate components into the orbiting orthonormal frame. The Lorentz factor relative to a Schwarzschild static observer is $\gamma =\sqrt{f}/q$, so the body's measured radial and tangential spin components areTheir squared sum is $s_0^2$, and their angle relative to the outward radial direction decreases uniformly at rateThe differing coordinate amplitudes in $s^r,s^{\varphi }$ must not be interpreted as a nonuniform physical spin rotation. After one orbital period the radial basis has turned by $2\pi$ while the spin has turned by $-2\pi q$ relative to it. The residual geodetic precession is $\Delta \psi =2\pi (1-q)$ in the orbital direction.

For a timelike circular Schwarzschild geodesic, the coordinate orbital period and onboard proper time per orbit areUsing $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$ and $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$, the Schwarzschild radius and orbital areal radius are $R_s=2GM/c^2\simeq 1.21789\times 10^{10}\mathrm{m}$ and $R=200R_s\simeq 2.43578\times 10^{12}\mathrm{m}$. ThereforeThe crew accumulates about $3836$ fewer seconds, or $63.9$ fewer minutes, per orbit. Both gravitational time dilation and kinematic time dilation contribute: a static clock at $R$ has $d{T}_{\mathrm{static}}=\sqrt{1-R_s/R}dt$, while the orbiting clock additionally has $d\tau =d{T}_{\mathrm{static}}/\gamma$, where $\gamma =\sqrt{398/397}$. A distant observer measures the coordinate period; comparing arrivals emitted at the same orbital phase also removes the repeated light propagation delay.

The geodetic precession per orbit isThe orbital position and four-velocity repeat after a full orbit, so the direction to distant Earth in the instantaneous rest frame repeats. The freely transported gyroscope does not: its beam axis has turned through $\Delta \psi$ relative to the original aiming direction. This applies to any initial in-plane spin direction by linearity of parallel transport, not only the radial initial condition used in part (b). The beam's angular diameter is $1^{\circ }$, giving a half-angle of $0.5^{\circ }$. Thus Earth lies outside the beam after one orbit; comparing the offset with the diameter would already show a miss, but the correct acceptance angle is the half-angle.

The initial radial ray must be outward from the black hole for a direct path to Earth. A precisely inward radial null geodesic falls into the black hole and cannot provide the assumed initial communication. For outgoing near-radial Schwarzschild light deflection, the angular correction is especially small. If a static observer at $R$ measures a small emission angle $\psi$ from the outward radial direction, its impact parameter is $b=R\sin \psi /\sqrt{f(R)}$. The outgoing null geodesic givesRelative to flat propagation, curvature changes the small pointing angle and beam radius by the fraction $f(R{)}^{-1/2}-1\simeq 0.00251$. At an offset of $1.35^{\circ }$ this is only about $0.0034^{\circ }$; the beam's $0.5^{\circ }$ radius changes by about $0.0013^{\circ }$. The exactly radial central ray has zero angular deflection. This is not the geometry of the usual $4GM/(b{c}^2)$ scattering formula, which assumes a ray coming from infinity, passing the black hole, and returning to infinity.

Relativistic aberration also distinguishes a coordinate-radial light ray from a beam aimed radially in the moving crew's frame. Here $v/c=1/\sqrt{398}\simeq 0.0501$. Successful initial aiming includes that fixed correction; the same correction applies at the repeated orbital position. More precisely, an outgoing static-radial ray initially has crew-frame angle ${\psi }_0^{\prime }=-\arcsin (v/c)$ from the radial axis. The local small-angle aberration factor about that aim is $d\psi /d{\psi }^{\prime }=\gamma =\sqrt{398/397}$; including the outgoing propagation factor gives $\gamma /\sqrt{f}=1/q\simeq 1.00377$. Thus the asymptotic miss is about $1.358^{\circ }$ and the beam half-angle about $0.502^{\circ }$, with the same conclusion. The small distortion of the outward beam cannot correct the much larger accumulated geodetic precession.