Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 4 / 18D (Special Relativity)

The Lorentz factor is $\gamma =(1-0.8^2{)}^{-1/2}=5/3$. Length contraction makes the moving javelin's length in the shed frameThis measures its two endpoints at the same shed-frame time.

The shed moves relative to the javelin frame at the same speed, so its simultaneous length there isThe javelin retains its proper length $2\mathrm{m}$. The two statements use different simultaneity slices, as prescribed by relativity of simultaneity.

Choose event $A$, when the trailing end reaches the entrance, as $(ct,x)=(0,0)$ and as the primed origin. Write $T=ct$, $T^{\prime }=c{t}^{\prime }$. Before impact, the shed-frame worldlines are entrance $x=0$, back $x=1.5$, trailing end $x=0.8T$, and leading end $x=0.8T+1.2$. The Lorentz transformation isThus in the javelin frame the ends have $x^{\prime }=0$ and $x^{\prime }=2$, while the entrance and back have $x^{\prime }=-0.8T^{\prime }$ and $x^{\prime }=0.9-0.8T^{\prime }$. These four lines in each frame produce the requested spacetime diagrams. The diagrams show initial motion; dashed portions extrapolate that motion beyond first impact and are not a model of collision dynamics.

In the shed frame, the leading end reaches the back when $0.8T+1.2=1.5$, giving $T_B=3/8\mathrm{m}$. Transforming both events yieldswith every coordinate in metres. The invariant interval isThus the events are spacelike separated. Event $A$ occurs first in the shed frame, but $B$ occurs first in the javelin frame; this is allowed for spacelike separation.

In the shed frame there is a pre-impact interval $0\le T<3/8$ during which both ends lie inside the shed, since its simultaneous moving length is only $1.2\mathrm{m}$. Those simultaneous endpoint events do not remain simultaneous after a boost. In the javelin frame the front hits the back before the rear enters, and there is no simultaneous pre-impact containment of the full $2\mathrm{m}$ rod in a $0.9\mathrm{m}$ shed. This is rod containment and simultaneity, not inconsistent length measurements.

After impact, the front's motion changes while the rear cannot respond instantaneously: stress signals propagate no faster than light. A claim that the entire javelin stops rigidly at event $B$ is inadmissible. Actual post-impact compression or fracture needs material data and is not represented by the straight-line extrapolations in the diagrams.