A train of length 15 cs* moves at speed 3c/5. 39 How much time does
it take to pass a person standing on the ground (as measured by that
person)? Solve this by working in the frame of the person, and then
again by working in the frame of the train.
*cs stands for light-second. It is 3*10^8 m.
x = γ(x' + vt')
t = γ(t' + vx'/c^2)
and their inverses:
x' = γ(x - vt)
t' = γ(t - vx/c^2)
Length Contraction and Time Dilation:
L = L0/γ
T = γ*t0
where γ = 1/√(1 - v^2/c^2), L0 and t0 are the proper lengths and times, respectively
The Attempt at a Solution
So, for simplification later, I can find γ, which turns out to be 5/4.
I then found the length contraction of the train from the persons perspective, which is L = 12 cs
Then its seems to be simply applying t = d/v, where the train needs to travel 12 cs to pass the person, and its moving at 3c/5. t = 20/c.
I believe that's right, the problem comes when I change frames to the train. The train should see itself at its proper length (I think), so if I imagine the train as standing still, and the person moving past it at -3c/5. Assuming the person is infinitesimally thin, it won't undergo length contraction and so the time it takes for the train to pass the person in the trains reference is found with t = d/v again, but with the proper length this time, coming out to t = 25/c.
Somehow it feels like there should be more funky stuff going on with the trains frame of reference since this is special relativity.
I also would appreciate it if someone could explain how to do this with Lorentz Transformations.
Thanks for your time.