Understanding Special Relativity: Train Example and Time Dilation Explained

[Chuck37]

I'm sure this has been done a thousand times here, but I have a fast train example I'd like to use to help me understand. I may have a followup question depending on whether resolving my ignorance here suffices...

Say I have a train with two cars separated by distance D and it’s going really fast. A pulse of light is sent from one car to the other. A stopwatch on the receiving car starts when the pulse is sent and stops when it is received.

My understanding is that the speed of light is fixed in any inertial frame.

If I pick the reference frame of the train cars, the pulse of light will hit the receiving car at D/c and that’s it. In this frame the cars appear to be relatively still.

Now I pick the track frame, so the cars are going at v. c is still fixed, so it appears to take D/(c+v) or D/(c-v) for the pulse to reach the other car, depending on whether the lead or trailing car is the transmitter.

But shouldn’t we agree in both frames what the stopwatch will say when the light pulse arrives? Maybe time and distance dilation can resolve it. Time and distance are dilated at speed by sqrt(1-(v/c)^2). Call that value A. In the track frame the cars appear DA apart, so it really looks to take DA/(c+v) or DA/(c-v) for the light to make the trip. I also see their stopwatch as going slow, so it will only have counted tA relative to mine, so I get DA^2/(c+v) or DA^2/(c-v).

These two can convert to D/c+Dv/c^2 and D/c-Dv/c^2, which is obviously not equal to D/c. What did I do wrong? Or is it wrong to even expect the stopwatch/pulse interaction to be the same in all frames?

Thanks for any help.

 

[Doc Al]

Now I pick the track frame, so the cars are going at v. c is still fixed, so it appears to take D/(c+v) or D/(c-v) for the pulse to reach the other car, depending on whether the lead or trailing car is the transmitter.

Don't forget that the distance between cars is contracted according to the track frame.

But shouldn’t we agree in both frames what the stopwatch will say when the light pulse arrives?

Of course.

Maybe time and distance dilation can resolve it. Time and distance are dilated at speed by sqrt(1-(v/c)^2). Call that value A. In the track frame the cars appear DA apart, so it really looks to take DA/(c+v) or DA/(c-v) for the light to make the trip. I also see their stopwatch as going slow, so it will only have counted tA relative to mine, so I get DA^2/(c+v) or DA^2/(c-v).

Don't forget that according to the track frame, the stopwatch was not started at the same time that the pulse was emitted. This is the relativity of simultaneity. There are three effects to consider, not just time dilation and length contraction.

 

[Chuck37]

Don't forget that according to the track frame, the stopwatch was not started at the same time that the pulse was emitted. This is the relativity of simultaneity. There are three effects to consider, not just time dilation and length contraction.

Can you elaborate on this? Let me try a different way to make it easier to explain. Instead of a stopwatch, let's have the cars have perfectly synchronized clocks and the transmitting car transmits when his second hand crosses zero. In the car frame the pulse leaves at 0 and arrives at D/c (on both clocks). How does it look in the track frame?

Thanks.

 

[Erland]

Perhaps, this would help:
https://www.physicsforums.com/showthread.php?t=468826

 

[Doc Al]

Can you elaborate on this? Let me try a different way to make it easier to explain. Instead of a stopwatch, let's have the cars have perfectly synchronized clocks and the transmitting car transmits when his second hand crosses zero.

The clocks are synchronized in the train frame. According to the track frame, the clock in the front of the train lags behind the one in the rear by an amount equal to Dv/c^2.

In the car frame the pulse leaves at 0 and arrives at D/c (on both clocks).

OK.

How does it look in the track frame?

Let's assume that the light pulse is transmitted towards the front of the train. The track frame observers say that when the light was emitted, the front train clock read -Dv/c^2, not zero.

The distance between cars is contracted by a factor of gamma (γ), where γ = 1/√(1 - v^2/c^2). So the travel time would be Δt = (D/γ)/(c-v). Of course, due to time dilation, the train clocks would show an elapsed time (according to track observers) of Δt/γ = (D/γ^2)/(c-v) = D/c + Dv/c^2 during the flight of the pulse. (Work this out for yourself to confirm.) Since the front clock is out of synch by Dv/c^2, the track observers will predict that the front clock will show a time of D/c when the pulse arrives. Which exactly matches what the train observers see. (It better!)

And track observers will say that when the pulse arrives at the front clock, the rear clock reads D/c + Dv/c^2, not D/c.

 

[Chuck37]

I pulled up this page: http://en.wikipedia.org/wiki/Relativity_of_simultaneity

And I see that the simultaneous clock ticks in the car frame are offset by vx/c^2 in the track frame. That corresponds to the Dv/c^2 term I was off by. I haven't resolved the sign issue, but clearly this is what I was missing.So Part 2 of my question is how to think about this when the clocks on the cars are not actually free running but are slaved to GPS, which operates in the track frame (sort of, if the track is on Earth I think it's in an inertial frame where the track would also be moving according to Earth rotation, but ignore that for now). I think in this case, if the GPS systems are doing their jobs properly, the clocks will actually be synced in the track frame and the answer will revert to DA/(c +/- v). (?)

I need to think on this some more, but if anyone has thoughts I'd love to hear it. Thanks.

ADDED: Thanks Doc Al. I posted when you were posting.