A train with velocity v is moving forward. A light beam shoots from the bottom and bounces it off the ceiling, returning back to the original point inside the train. The distance between the bottom and the ceiling is S.
What is the light travel time to a man standing on the train?
t0 = 2S / C
What is the light travel time to a man standing outside the train?
t = 2S / ((C^2 – V^2)^(1/2))
Therefore,
t0/t = ((C^2 – V^2)^(1/2)) / C = (1 – (V/C)^2)^(1/2)
t = t0 / (1 – (V/C)^2)^(1/2)
if V = 0.8 C, then t = 5/3 t0
3 seconds inside the train = 5 seconds outside of the train
if V => C, then t => infinity
when you look at the train from outside, it looks as if the train has stopped moving
Therefore, people who flight often may be younger in age.
Benc Tan 