Past exam of the relativity course of the University of Cambridge / 2021 / Section B / 6

Take $\alpha >0$ and work in a connected patch with $x\ne 0$. Setting the line element to zero givesThese are the two families of Rindler null geodesics. They remain on the same side of $x=0$ and approach it only at infinite coordinate time. $x=0$ itself is not an additional geodesic within this chart, since the metric is degenerate there.

The paths can also be written $t=\pm {\alpha }^{-1}\log |x/C|$. In two dimensions a null direction has only one independent orthogonal null direction, namely itself, so each smooth null curve is a geodesic after reparametrization. More explicitly, the transformation in part (d) makes these paths straight null lines. The coordinate $t$ is generally not an affine parameter.

Put $r_s=2\mu$ and $f(r)=1-r_s/r$. Radial null geodesics of the Schwarzschild metric obeyNear the horizon, $\delta =r-r_s$ satisfies$C>0$ describes the exterior and $C<0$ the interior. The exponential approach in Schwarzschild coordinate time resembles part (a), but $\delta$ is not a proper radial distance.

Outside, set $\rho =2\sqrt{r_s\delta }$ and retain the leading near-horizon terms. The radial metric becomes the Rindler limit of a Schwarzschild horizon:It agrees with part (a) for $x=\rho$, $\alpha =c/(2r_s)$. Accordingly $\rho \propto e^{\pm ct/(2r_s)}$, while $\delta \propto {\rho }^2$ explains the factor two in its exponent.

Inside, write $r=r_s-{\rho }^2/(4r_s)$. The leading radial metric is insteadHere the radial coordinate is timelike and the Schwarzschild $t$ coordinate is spacelike. Thus $x<0$ in the Rindler metric does not represent the black-hole interior: $g_{tt}=(\alpha x{)}^2$ stays positive on both Rindler wedges. In the black-hole interior all future-directed radial light rays move towards decreasing $r$; the two signs of $dr/dt$ do not mean opposite future radial directions because $t$ is no longer time.

The horizon is a coordinate singularity in Schwarzschild coordinates and in the accelerated chart, not a curvature singularity. The full Schwarzschild geometry nevertheless has genuine tidal forces; the flat Rindler metric captures its leading local radial near-horizon structure.

For coordinates $(t,x)$ the inverse metric is $g^{tt}=1/({\alpha }^2{x}^2)$, $g^{xx}=-1$. Only ${\partial }_x{g}_{tt}=2{\alpha }^{2}x$ is nonzero, giving the Christoffel symbolswith all remaining components zero.

On the specified constant-$x_0$ worldline, assume $x_0>0$. The four-velocity is $u^t=c/(\alpha x_0)$, $u^x=0$, and its norm is $c^2$, so $\tau$ is indeed proper time. The coordinate components are constant, but the covariant acceleration is nonzero:The proper acceleration of a static Rindler observer isThe acceleration has constant components in this accelerated chart and constant invariant magnitude. It is not a covariantly constant vector: its inertial-frame components change along the hyperbolic trajectory. For $x_0<0$, the magnitude is $c^2/|x_0|$, with the appropriate choice of future time orientation.

Differentiating the supplied coordinates givesSubtract the squares. The cross terms cancel and ${\cosh }^2-{\sinh }^2=1$, yieldingThe spacetime is flat Minkowski spacetime in Rindler coordinates. $x>0$ covers $X>|cT|$, the right Rindler wedge, while $x<0$ covers the left wedge $X<-|cT|$; neither is the whole spacetime.

The null combinations areFor $x=C{e}^{-\alpha t}$ the first is constant; for $x=C{e}^{\alpha t}$ the second is constant. Thus part (a)'s exponentials are simply the inertial light rays $X\pm cT=\text{constant}$. Their coordinate-time asymptotes reflect the chart's Rindler horizon, not a failure of light to move at $c$ in an inertial frame.

Let $a>0$ be the constant proper acceleration away from the emitter and set $R=c^2/a$. Take the signal emission to be simultaneous, in the initial inertial frame, with the ship being at rest and its clock reading $\tau =0$. Place that initial ship event at $T=0$, $X=R$. Its hyperbolic motion isA photon emitted from $X=R-d$ at $T=0$ towards the departing ship follows $X_{\gamma }=R-d+cT$. At reception,Therefore light reception by a uniformly accelerated observer occurs afterThe same result follows in Rindler coordinates: the ship stays at $x=R$, the emitter starts at $x=R-d$, and the right-moving null ray has $x=(R-d)e^{\alpha t}$. The onboard lapse $d\tau =\alpha Rdt/c$ converts its coordinate catch-up time to the expression above.

If $d=c^2/a$, the ray lies on the limiting null horizon and never intersects it at finite proper time; if $d>c^2/a$, it never reaches the ship. The reception time diverges as the emitter approaches this Rindler horizon. For $ad/c^2\ll 1$, ${\tau }_{\mathrm{rec}}=d/c+a{d}^2/(2c^3)+\cdots$, recovering the inertial light-travel time. A different emission time or initial relative velocity would require specifying that extra datum; the formula uses the initial-rest simultaneous-emission convention stated above.