A vaulter carrying a 4 meter pole runs toward a garage that is 2meters long, with a door at both ends.If the vaulter runs fast enough, his pole appears to be only 2 meters long, and it is possible (for an instant) to close both doors of the garage with him inside.But, from the pole vaulter’s point of view, his pole is still 4 meters long, and the garage has shrunk down to 1 meter long. So clearly it is impossible for both doors to be closed at the same time.
Who is right, and why?
Pole Vaulter Paradox 2
Let’s put the origin of the fixed frame at the first door, and the origin of the moving frame at the rear end of the pole, and t=0 is when the those points overlap. The second door is at x=2meters, and the front end of the pole is there at t=0.For the pole to shrink from 4 meters to 2, we need γ=2, which gives β=0.866. A pole end traverses the garage in time Now we transform the second door opening at x=2, t=0 into the moving frame.
x′ = γ (x − βct ) = 2(2 − 0.866 × 0) = 4 m
ct ′ = γ (ct − β x) = 2(0 − 0.866 × 2) = −3.464 m
Pole Vaulter Paradox 3
In the pole vaulter’s frame, the 4 meter pole is stationary and the 1 meter garage moves, backwards.The right door goes from closed to open at , which is the time that it reaches the right end of the pole. The left door overlaps the far end of the pole even earlier, but since the left door is open at all times ct’<0, there is no problem. The left door closes at ct’=0, which is when it passes the left end of the pole. The key point is that while the doors are simultaneously closed in the garage frame, this is not simultaneous in the pole frame.
from: http://www.phas.ubc.ca/~mattison/Courses/Phys330/p330-7.pdf
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