Past exam of the relativity course of the University of Cambridge / 2015 / Paper 1 / 1

Use arc length $s$ as the affine parameter, and put a dot over derivatives with respect to $s$. The geodesic Lagrangian on the sphere isThe cyclic coordinate $\varphi$ gives $\partial L/\partial \dot{\varphi }=a^2{\sin }^2\theta \dot{\varphi }=\text{constant}$. With the constant normalized as in the paper,Here $h$ has units of inverse length, because $s$ is physical arc length. Unit-speed normalization givesFor $h\ne 0$, division by ${\dot{\varphi }}^2$ yields the spherical-geodesic first integralThe turning latitudes satisfy $\sin \theta =|ah|$. In the usual polar-coordinate range $0\le \theta \le \pi$,The maximum departure from the equator, often the geometric quantity intended by an angular amplitude, is insteadThus polar angle and signed latitude must not be interchanged. At $|ah|=1$ the geodesic is the equator. At $h=0$ it is a meridian through the poles; $\varphi$ cannot serve as a global parameter and the divided equation is inapplicable. The limiting polar range is then $[0,\pi ]$.

The original PDF prints an ambiguous derivative with the square on $\theta$ rather than a conventional derivative notation. If the intended equation was second order, differentiating the first integral, or eliminating $\dot{\theta }$ from the Euler-Lagrange equation, gives the equally useful formIt has the same normalized geodesics and the same turning angles. Giving both forms resolves the typographical ambiguity without pretending that a second derivative is the square of a first derivative. Geometrically these trajectories are great circles, with inclination $\arccos |ah|$ to the equator.

Assume an emitter at rest on the stellar surface and a static observer at infinity. Write $f(r)=1-2GM/(r{c}^2)$. The time-translation Killing energy of a photon is conserved, while the locally measured photon energy is proportional to $1/\sqrt{f(r)}$. ConsequentlyUsing $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$, $M=4\times 10^{30}\mathrm{kg}$ and $R=1.1\times 10^4\mathrm{m}$,The Schwarzschild radius is approximately $5.94\mathrm{km}$, below the given surface radius. The positive shift is therefore a gravitational redshift of an escaping photon, with received wavelength about $1.475$ times the emitted wavelength. A moving emitter would introduce an additional Doppler factor; that is outside this static-surface model. The weak-field estimate $z\simeq GM/(R{c}^2)=0.270$ is insufficient at this compactness. These stated constants are numerical conventions, not a claim about an unavailable historical handbook's last digits.

Use metric signature $(+---)$ and four-velocities $U_i^a={\gamma }_{i}c(1,{\mathbf{\beta }}_i)$, with ${\gamma }_i=(1-{\beta }_i^2{)}^{-1/2}$. Their Lorentz invariant inner product isIn the rest frame of either particle this same scalar is the relative Lorentz factor ${\gamma }_{\mathrm{rel}}$. HenceLet $d={\mathbf{\beta }}_1\cdot {\mathbf{\beta }}_2$. The numerator after combining terms isUse $|{\mathbf{\beta }}_1\times {\mathbf{\beta }}_2{|}^2={\beta }_1^2{\beta }_2^2-d^2$ to obtainThe two rest frames agree on the relative speed, although their velocity vectors are expressed in different spatial frames. Checks include equal velocities giving zero, collinear velocities giving $|{\beta }_1-{\beta }_2|/|1-{\beta }_1{\beta }_2|$, and low speeds reducing to the Galilean relative speed. For timelike particles the four-velocity inner product also ensures $0\le {\beta }_{\mathrm{rel}}<1$.