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Old Oct 12th 2018, 02:48 AM   #1
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Join Date: Sep 2018
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Problem with different reference frames - special relativity

I have the following conditions: A plane with length L in its own rest frame moves with constant velocity with respect
to the inertial system S. The inertial system of the plane is called S'. Two flight attendants A and B start simultaneously from the middle of the plane and move with constant velocity
toward the front end and back end of the plane, respectively, such that the magnitude of their velocities, u, is the same, as seen in S'. In reference frame S' it takes a time τ before they reach the ends of the plane.
The questions goes:

Seen from S, A reaches the back end of the train a time T before B reaches the front end.

a) Find the velocity v of the plane with respect to S in terms of L,T and c.

b) Seen from S, find how far the plane moves in the time it takes for A to reach the back end of the plane.

In A, I used that the spacetime interval in the two reference frames are invariant. in S' $\Delta x=L$ and $\Delta ct=0$ since the attendants reach the ends simultaneously. for S I am given that $\Delta t=T$ and the length of the plane can be found by a Lorentz transform. By setting the two spacetime intervals equal I get:
$(cT)^2-\frac{L^2}{(1-\frac{v^2}{c^2})}=-L^2\Longrightarrow v=\pm\sqrt{c^2+\frac{c^2L^2}{-L^2-(cT)^2}}$

In question B, my idea was that since I have found (I hope) the velocity of the plane relative to S, I could make a Lorentz transform of $\tau$ to get the time it takes the attendant to reach the back seen from S and then multiply by the velocity. I tried this out, but the calculations get very nasty and I think the result is supposed to be a bit prettier than what I got. Is it possible to calculate the distance the plane travels in the time A reach the back end simply by making a Lorentz-transform of $\tau$ and multiplying by the velocity of reference frame S' relative to S?
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