The problem posed here is this:
vaulter, holding a
fiberglass pole horizontally, approaches an open barn running at the
prodigous rate of 1 roddenberry (i.e. a proper-velocity w of 1 map lightyears
per traveler year, corresponding to a coordinate velocity of 0.707 c
map lightyears per map year). The pole (at rest) is 10 meters in length,
while the barn (at rest) is only 8 meters deep. Hence the pole vaulter has
no worry that the farmer, standing beside the barn, will be able to shut
the door and lock her in after she enters the barn. In fact, the barn
looks even shorter to the runner, since it is relativistically contracted
to 1/gamma=1/Sqrt[1+(w/c)2] of its resting length, or
8[m]/Sqrt = 5.65 meters. The farmer, on the other hand, sees the pole
as relativistically contracted
to 1/gamma of its resting length, or 10[m]/Sqrt = 7.07 meters, and hence
contends that he can easily slam the door on the runner after the runner
and pole enter the barn, and before the front of the pole impacts the
back wall. Who is right?
Consideration of this debate can be handled by considering 3 spacetime events, namely the first impact of the front of the pole on the back of the barn, and the position of the trailing end of the pole at that time in the frame of both runner and barn. A clue to resolving the debate lies in the simultaneity constraint, which of course means something different for travelers in different inertial frames. As you can see below, in context of their own frames, they are both right.
What happens to the pole after impact with the back wall of the barn?As the plot shows, the trailing end of the pole will not even see, let alone feel, the impact until it is significantly closer to the back wall than it's contracted length! I leave it as an exercise for the reader to determine the position of the trailing edge of the pole at "first light of impact", as noted on the plot.
What happens to the rod as "news of the collision" reaches it of course depends on it's structure as well as the back wall's immovability . If there is any resistance by the back wall at all, however, the rigidity (and original resting length) of the pole are gone. (If we are unsure about the actual magnitude of the back wall's immovability, we can symmetrize the problem by letting two poles collide in back to back barns. If the poles are identical down to the elementary particle scale, then symmetry allows us to expect no crossing of the back wall at all!)
As mentioned on our x-ct plotting page, the same information could be obtained for a plot which uses orthogonal axes for the rightward moving frame. Then of course the barn is moving to the left, and the pole-vaulter is stationary. As in other examples elsewhere on these pages, the picture looks quite different even though the information it provides is the same!