Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 4 / 17C / c / Solution

In the station frame, intersecting the signal trajectories with the endpoint world lines gives

u t_{A}^{'} = v t_{A}^{'} + \frac{L}{2 γ}, - u t_{B}^{'} = v t_{B}^{'} - \frac{L}{2 γ} .

Since

u > v

, both receptions occur at positive times,

t_{A}^{'} = \frac{L}{2 γ ( u - v )}, t_{B}^{'} = \frac{L}{2 γ ( u + v )} .

The inverse Lorentz transformation then gives

t_{A} = γ t_{A}^{'} (1 - uv / c^{2}) = \frac{L ( 1 - uv / c ^{2} )}{2 ( u - v )}, t_{B} = γ t_{B}^{'} (1 + uv / c^{2}) = \frac{L ( 1 + uv / c ^{2} )}{2 ( u + v )} .

Subtracting and simplifying yields

t_{B} - t_{A} = - \frac{Lv ( 1 - u ^{2} / c ^{2} )}{u ^{2} - v ^{2}}, t_{B}^{'} - t_{A}^{'} = - \frac{Lv}{γ ( u ^{2} - v ^{2} )} .

For

u < c

, both differences are negative: the rear receives first in both frames, but by different time intervals.

 Ancestors (11)

 View article source

 Discussion (0) New discussion

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version