Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / Section B / 7

The collinear relativistic velocity addition formula isWith both velocities directed toward Earth,The denominator keeps the result below $c$. Adding the two speeds by the Galilean rule would give an inadmissible $1.4c$.

The separation simultaneous in the mothership frame at launch is the length contraction of the Earth-frame simultaneous separation:In that frame Earth approaches the launch position at $0.6c$ while the lander travels toward Earth at $0.8c$. Their coordinate separation closes at $1.4c$, soA closing rate between two objects can exceed $c$; it is not either object's speed in the other's rest frame. Simultaneity matters: simply transforming the Earth event at Earth-frame launch time would not give the simultaneous mothership-frame separation.

For the lander, $u/c=35/37$, soThe relativistic kinetic energy isThe rest energy $m{c}^2$ is excluded from kinetic energy.

Set the launch event to $(t,x)=(0,0)$ and measure $x$ toward Earth in light-years and $t$ in years, so light has coordinate slope magnitude one. Earth's worldline is $x=2$, and the lander's is $x=(35/37)t$. The outgoing signal is $x=t$ and reaches Earth at $B=(2,2)$. Earth's reply is sent at $C=(25/12,2)$.

The return signal has $x=2-(t-25/12)$. Its intersection with the lander gives $D=(1813/864,1715/864)$. Both signal lines are null; the lander's worldline lies inside the light cone. The drawing exaggerates the small gap near Earth in an inset so that B, C and D remain distinguishable.

Equating the lander and return-signal positions givesThis is measured from launch, so about 25.18 months pass on Earth-frame synchronized clocks. D occurs slightly before the lander reaches Earth at $t=74/35\mathrm{yr}$.

The two events A and D lie on the inertial lander's own worldline, and their proper time difference isEquivalently, applying the Lorentz transformation with $\Delta x=u\Delta t$ gives $\Delta t^{\prime }={\gamma }_u\Delta t(1-u^2/c^2)=\Delta t/{\gamma }_u$.