Recent content by A.Brown

Excellent, thank you so much!

Oh, the clocks in S' would read 2 minutes to get to the mid-point + 0.75 minutes for B, and 2 + 3 minutes for the signal to reach A?

Ok, so then solving for Δt' for both expressions, I'm getting 0.75 minutes for the signal to reach B and 3 minutes for the signal to reach A according to the observers in S' who see the tunnel as 2.4 c*min long (making L'/2 = 1.2 c*min)?

Oh, ok, so then for A does that make it L'/2 + 0.6cΔt' = cΔt' making L'/2 = cΔt'-0.6cΔt' because in reaching A, the signal has to travel extra distance as the end of the tunnel rushes away from the emission point?

If the signal is defined as moving in the positive direction, to the right towards B, then isn't the tunnel approaching in a negative x direction? If not, then L'/2 = cΔt'+0.6cΔt' but then I'm not sure how to do the calculation for the signal approaching A, which is receding from the point...

Sorry, I'm still confused. When the signal is emitted: A( []--> )B |--L'/2--|--L'/2--| B appears to be moving towards the signal at 0.6c in S'. When the signal arrives: A( cΔt'--><--0.6cΔt')B Which seems like L'/2 = cΔt'-0.6cΔt' ? What am I missing?

Is L' = cΔt' - 0.6cΔt' for the end B the train is moving towards?

I was afraid of that :) How do I account for that? Can I use γt + (vγt)/c for A and γt - (vγt)/c for B?

Homework Statement A very fast train (system S') travels on a straight track (system S) with speed 0.6c. When it enters a tunnel (at end A), which is 3 c*min long relative to S, observers in S and S' set their clocks to zero. The train will emerge from the tunnel at end B. At the midpoint of...