Question on special relativity

[kknull]

understood. thank you! So my mistake was to consider my question as a paradox, while it was simply an empirical demonstration that simultaneity is relative.

 

[EL]

Exactly!..

 

[stevmg]

Again, from the Einstein train example, the staionary observer sees te light flashed at the same time. The one in motion catches the lead flash before the rear flash.

Question is: what is the mathematics behind this?

Assume the train is 150,000 km in lngth which places the train observer 17000 km in the middle. It will take the light from the front flash 0.5 sec to reach the stationary observer. If the train is moving at 0.5 c, how long will it take the front falsh light to reach the moving observer on the train - DO THE MATH! Not just words - DO THE MATH

I have a feeling it is still 0.5 seconds but it shouldn't be

 

[Doc Al]

Assume the train is 150,000 km in lngth which places the train observer 17000 km in the middle. It will take the light from the front flash 0.5 sec to reach the stationary observer. If the train is moving at 0.5 c, how long will it take the front falsh light to reach the moving observer on the train - DO THE MATH! Not just words - DO THE MATH

How long will it take according to whom? The observer on the train or the observer on the platform? What's the proper length of the train? It the train is 150,000 km long, the middle would be 75,000 km from each end.

Since you use a time of 0.5 seconds, perhaps you meant the train to be 300,000 km long? If so, then the light from each flash would take 0.5 sec to reach the middle of the train according to train observers.

According to the platform observer, it would take less time since the length is shorter and the train moves towards the incoming light. The length of the train would be shorter by a factor of γ = 1.155. So if the proper distance between the middle and ends of the train is L = 0.5 light-seconds, that distance would be L/γ according to the platform observer. To find the time for the front flash to reach the middle of the train, use:
L/γ - vt = ct

Solving for time, you get:
t = L/[γ(c+v)] ≈ 0.29 sec