Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 4 / 17C (Special Relativity)

If the primed frame moves at speed $v$ in the positive $x$ direction relative to the unprimed frame, the one-dimensional Lorentz transformation isFor a trajectory with velocity $q=dx/dt$, differentiating these expressions gives $q^{\prime }=(q-v)/(1-vq/c^2)$. Solving for $q$, when the particle has primed-frame speed $u$, proves the relativistic velocity-addition formulaFor the train problem the sign of the frame velocity reverses: the train frame $S$ moves at $+v$ relative to the station frame $S^{\prime }$. Put $\gamma =(1-v^2/c^2{)}^{-1/2}$. The transformation appropriate to the subsequent parts is therefore

The world lines in the train frame are $x_A=L/2$, $x_B=-L/2$ and $x_C=-vt$. In the station frame, length contraction givesThe emission event is $O=(0,0)$ in both frames. The signal world lines in $S^{\prime }$ are $x^{\prime }=u{t}^{\prime }$ and $x^{\prime }=-u{t}^{\prime }$; their intersections with the train endpoints define the reception events $R_A,R_B$. Their velocities in $S$, from the relativistic velocity-addition formula, are $(u-v)/(1-uv/c^2)$ and $-(u+v)/(1+uv/c^2)$.

The two platform-alignment events occur at $(x^{\prime },t^{\prime })=(\pm L/(2\gamma ),0)$ in $S^{\prime }$. The Lorentz transformation puts them at $(x,t)=(\pm L/2,\mp vL/(2c^2))$ in $S$, on the tilted simultaneity line $t=-vx/c^2$. The spacetime diagrams show all three observers, both signals, their receptions, and these distinct endpoint-alignment events:

Let $P$ be the platform's proper length, measured in its rest frame $S^{\prime }$. At the single station-frame time $t^{\prime }=0$, its ends coincide with train endpoints whose simultaneous station-frame separation is $L/\gamma$. ThusThe platform endpoint world lines are vertical at $x^{\prime }=\pm P/2$ in the station spacetime diagram. In the train frame they are $x=-vt\pm P/(2\gamma )$, so their separation at a common train-frame time is $\boxed{P/\gamma =L/{\gamma }^2}$. The two alignment events lie on the tilted line $t^{\prime }=0$ in the train diagram and are not simultaneous there; their separation $L$ cannot be used as the platform length measured at one train-frame time. This is the distinction enforced by length contraction.

In the station frame, intersecting the signal trajectories with the endpoint world lines givesSince $u>v$, both receptions occur at positive times,The inverse Lorentz transformation then givesSubtracting and simplifying yieldsFor $u<c$, both differences are negative: the rear receives first in both frames, but by different time intervals.

With light signals, $u=c$, the train-frame relativistic velocity-addition formula gives signal velocities $+c,-c$. The endpoints are equally distant from the source in that frame, andThe two reception events are simultaneous in $S$ but not in $S^{\prime }$: in the station frame the rear moves toward its light signal while the front moves away. Directly, $\Delta x=x_B-x_A=-L$ and $\Delta t=0$, so the Lorentz transformation gives $\Delta t^{\prime }=\gamma (\Delta t+v\Delta x/c^2)=-\gamma Lv/c^2$. This is relativity of simultaneity, concerning spatially separated events; it does not contradict the invariant light speed or the causal ordering of events on a single observer's world line.