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Alta
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Homework Statement
A train of proper length L moves at speed v1 with respect to the ground. A passenger runs from the back of the train to the front at speed v2 with respect to the train. How much time does this take as viewed by some one on the ground?
a) Find the relative speed of the passenger and the train (as viewed by someone on the ground), and then find the time it takes for the passenger to erase the initial "head start" that the front of the train had.
b) Find the time elapsed on the passenger's clock (by working in whatever frame you want), and then use time dilation to get the time elapsed on a ground clock.
Homework Equations
Vector addition:
u=(v1+v2)/(1+v1*v2/c^2)
time dilation:
t=gamma*t'
length contraction:
x=x'/gamma
gamma=1/sqrt(1-v^2/c^2)
The Attempt at a Solution
a)
For an observer on the ground the passenger is moving at speed, u=(v1+v2)/(1+v1*v2/c^2) and the train is moving at speed v1. The length of the train is L/sqrt(1-(v1/c)^2).
The time is just the length divided by the difference in the velocities:
t1=L'/(v1-u)=[L/sqrt(1-(v1/c)^2)]/[v1-(v1+v2)/(1+v1*v2/c^2)]
b)
According to the passenger, the length of the train is L/sqrt(1-(v2/c)^2) and he is run at velocity v2. So he thinks he reachs the end of the train in time (L/sqrt(1-(v2/c)^2))/v2.
According to someone standing still on the train, the time is
[(L/sqrt(1-(v2/c)^2))/v2]*(sqrt(1-(v2/c)^2)).
So according to someone on the ground t2=L/v2*(sqrt(1-(v1/c)^2))
...
My problem is that t1=/=t2 but it should. I am pretty sure that I did part a) correct. I am not sure if I am transforming the times correctly in part b). I am probably doing something really stupid but I just do not understand enough to know what is going on.
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