# The runner on the train paradox.

1. Sep 24, 2007

### MathematicalPhysicist

the setting:
A train travels at high speed. A runner on the train sprints towards the back of the train with the same speed (wrt to the train).
as the train moves forward (wrt the earth). therefore the runner doesnt move wrt to earh.
The paradox: the runner's clock runs slower than compared to the train clocks.
therefore the runner's clock should run doubly slow wrt to earth's clock cause the trains clocks run slow compared to the earth's clocks.
but the runner is not mving wrt to the earth therefore the runner's clock must be run the smae as the earth's clocks.
how to resolve it?

well im given a hnit to resolve it by relativity of simulatentity, now if we were to send a light beam from the opposite direction of the runner towards him as soon as he starts running in the train, then obviously in the train it would we simultaneous while in the earth it won't, we can measure the time it takes the beam to reach the runner, is this approach correct have i resolved the paradox, i feel that i hvent really.

2. Sep 24, 2007

### ganstaman

I could be wrong, but is this not an issue of frames of reference? That is, the runner's clock runs slower than the train's clock from the point of view of a passenger on the train (?). The train's clock runs slower than one on the earth from the point of view of someone on earth (?). The runner's and earth's clock run at the same rate from the point of view of the runner and the earth, but not from the train.

As for your hint at the bottom, what is simultaneous with what? I see that you are sending a beam of light at the runner, but what event occurs simultaneously with that?

3. Sep 24, 2007

### MathematicalPhysicist

the hint from the book say that the solution of the paradox has to do with relativity of simultanity.
well, from the perspective of an observer from outside the train the beam and the runner start the running the same time but in the rest frame of the train they don't.

p.s this is not what written in the book, the only hint is that the solution uses relativity of simulatentity.

4. Sep 24, 2007

### Janus

Staff Emeritus
According to whom?
From the Earth frame, the runner has no relative velocity and thus the runner's clock runs at the same rate as the Earth clock.
From the train frame, both the runner and the Earth have the same relative velocity wrt to the train and thus both their clocks run slo by the same amount, and again their clocks run at the same rate.

For a problem that needs the Relativity of Simultaneity for resolution, try this:
The runner runs in the same direction as the train is moving at the same velocity relative to the train as the train has relative to the Earth. Then, in the Train's frame, both the runner and the Earth have the same relative speed, and the same time dilation factor, and thus their clocks run at the same rate. In the Earth frame, however, the runner has a relative velocity greater than that of the train and thus the runner's clock runs even slower than the train's.

5. Sep 24, 2007

### MathematicalPhysicist

ok, thanks janus i got it.

6. Sep 24, 2007

### JesseM

Try putting clocks at the front and back of the train which are synchronized in the train's frame, and figure out the time on the clock at the front when the runner starts at the front, and the time on the clock at the back when the runner reaches the back. The difference between these times will be greater than the time the runner's own watch says it took for him to get from front to back, showing that the runner's watch is running slow in the train's frame. But in the frame of the tracks, the runner's watch is running normally while the clocks at the front and back are running slow--however, the clocks at the front and back are also out-of-sync in this frame, with the time of the back clock ahead of the time on the front clock by vx/c^2 (where x is the distance between the clocks in the train's own rest frame, and v is the velocity of the train in the track frame). So, if you calculate the details you will find that the observer on the tracks predicts the same thing about the times on the two clocks as the runner passes them, even though in this frame the two clocks are running slower than the runner's watch.

7. Sep 25, 2007

### yogi

The problem is described in Wheeler and Tailor 2nd edition, Spacetime Physics. I posted the problem on these forums about a year ago.

8. Sep 25, 2007

### MathematicalPhysicist

yes, correct, yogi.