Special Relativity: Multiple FoR

In summary, the train passes the person in time 20 seconds and in space-time coordinates (t, x) = (25, -3).
  • #1

Homework Statement

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.

Homework Equations

Lorentz Transformations:

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.
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  • #2
So, I forgot the light-second units on my answers, which would make them go down to 20 seconds and 25 seconds, but I can't seem edit my post again.

Anyways, I think its all actually correct, but my second answer is in fact the time as measured by the train in the trains FoR. So, the train sees the person's watch move slower by a factor of γ, making the time from the person's watch in the trains FoR:

t = (25 seconds)/(5/4) = 100/5 = 20s which matches with the first answers. Success! (I think/hope)

I would still appreciate it if someone would explain how to do this through Lorentz Transformations, because I'm still not sure how to go about that.
  • #3
The two events are:

event A = the front of the train passes the person
event B = the back of the train passes the person

Find the space-time coordinates (t, x) for each event in one frame. The Lorentz transformations allow you to calculate the coordinate for the events in the other frame.

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