Trains, tunnels, and help needed with special relativity

[A.Brown]

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 the tunnel, a transmitter on the train emits a radio signal towards observers at both ends of the tunnel. At what times are the signals received at ends A & B in S and in S'?

Homework Equations

$x' = γ(x-vt)$
$t' = γ(t - xv/c^2)$
$u'= (u-v)/(1 - uv/c^2)$

The Attempt at a Solution

I'm sure I'm being stupid about this, but I have managed to confuse myself pretty well. For these conditions, the midpoint is 1.5 c*min into the tunnel to observers in S. Since the train is traveling at 0.6c, observers in S will observe the train reaching the midpoint 2.5 minutes after it enters. The train then releases a radio signal which travels at c to the observers in S, taking another 1.5 minutes to reach the end of the tunnel, for a total of 4 minutes elapsed. Is this correct? I feel like I'm missing something.

For observers in S', the tunnel only appears to be 2.4 c*min due to length contraction, so they observe reaching the midpoint 2 minutes after entering the tunnel. They then set off the signal, which they observe traveling at c a distance of 1.2 c*min for a total of 3.2 minutes to reach the observers at the ends of the tunnel.

Again, I'm not sure I'm going about this correctly, our lecturer has been taken ill, and we've had a disjointed string of guest lecturers, and I feel a bit turned around. Thanks for your help!

 

[TSny]

Your S-frame analysis looks very good . But in the S'-frame the ends of the tunnel are moving while the radio signal is traveling :yuck:

 

[A.Brown]

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?

 

[TSny]

Consider the end of the tunnel that is approaching the radio signal. Let Δt' be the time it takes from the emission of the signal until the signal meets the end of the tunnel. Consider the distance the light travels during this time and also the distance the end of the tunnel travels in this time. How are those distances related to L' (the length of the tunnel in S')?

 

[A.Brown]

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

 

[TSny]

Not quite. Draw a sketch showing the end of the tunnel at the time the signal was emitted and the location of the point of emission of the signal. Then draw another sketch showing the signal meeting the end of the tunnel and the distances traveled by the signal and the end of the tunnel.

 

[A.Brown]

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?

 

[TSny]

When the signal arrives:

A( cΔt'--><--0.6cΔt')B

Which seems like L'/2 = cΔt'-0.6cΔt' ? What am I missing?

I don't understand the minus sign. That would mean that L'/2 is less than cΔt'.

 

[A.Brown]

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 where the signal was released.

 

[TSny]

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?

Yes, B is heading toward negative x. But I still don't see why you want to put in a minus sign. You can see in your diagram that the distance L'/2 is the sum of the distance the signal travels and the distance B travels during the time Δt'. Each distance is a positive number: cΔt' for the signal and 0.6cΔt' for B.

L'/2 = cΔt'+0.6cΔt'

Right, that should give you Δt' for B.

but then I'm not sure how to do the calculation for the signal approaching A, which is receding from the point where the signal was released.

Try drawing another diagram for A and use it to relate the distance the signal travels to the distance A travels and L'/2.

 

[A.Brown]

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?

 

[TSny]

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?

Yes, good.

 

[A.Brown]

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)?

 

[TSny]

Once you get the times for the signals to reach A and B according to the S' frame, you can check your result by using the Lorentz transformation for t' to directly calculate t' from the times and locations of the events in the S frame.

 

[TSny]

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)?

Yes, that looks right.

 

[TSny]

So, what do the clocks in the S' frame read for the arrival of the signals at A and B?

 

[A.Brown]

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?

 

[TSny]

That looks correct. You should be able to check these times using

$t' = γ(t - xv/c^2)$

 

[A.Brown]

Excellent, thank you so much!