Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 4 / 17B (Special Relativity)

The Lorentz factor for $\beta =v/c=12/13$ isThe javelin's length at simultaneous barn-frame times follows from length contraction:Its proper length remains four metres.

In the javelin's rest frame, the barn moves at speed $12c/13$. Applying length contraction to its proper length givesThe lengths in the two frames use different pairs of simultaneous endpoint events.

Choose the barn entrance as $x=0$, its exit as $x=3$, and the positive $x$ direction as the javelin's motion. Put $t=0$ at the trailing end entering the barn. Write $s=ct$ and $s^{\prime }=c{t}^{\prime }$ in metres. In the javelin frame let the trailing end have $x^{\prime }=0$, the leading end $x^{\prime }=4$, and use the same origin event. The Lorentz transformation isIn the barn frame, the four world lines areIn the javelin frame they areThe two spacetime diagrams below use vertical time coordinates and horizontal spatial coordinates. Event labels refer to the same physical events in both panels; dashed lines show light-ray directions.

Event $A$ is the chosen origin. At $B$, the leading-end world line reaches $x=3$, soThe Lorentz transformation givesFor example $s_B^{\prime }=(13/5)(19/12-36/13)=-37/12$. The squared spatial separation minus squared time separation isequally $16-1369/144$ in the other frame. Thus $A$ and $B$ have spacelike separation.

In the barn frame, for $0<s<19/12$ the trailing end has entered while the leading end has not yet exited, so the whole javelin is inside at each simultaneous time slice. In the javelin frame the barn is shorter than the javelin and there is no simultaneous slice on which it contains the whole javelin. Also $B$ precedes $A$ there. This is consistent because relativity of simultaneity changes which endpoint events are simultaneous; the ordering of events with spacelike separation is not invariant. The physical intersections of the four world lines are the same in both diagrams.