How can a train appear shorter to an outside observer due to its speed?

[arkain]

If you mean to change the scenario as you described above, it seems you now want to compare two different time intervals:
(1) From the train frame, the time it takes point A on the Earth to travel the length of the train (which has proper length L).
(2') From the Earth frame, the time it takes the nose of the train to travel a distance L along the tracks.

In this case, both time intervals are the same: Te = Tt. (How could they not be? The situation is perfectly symmetric.) Note that (2') is not the same as (2); (2') requires the use of two clocks/timers in the Earth frame. Note that these time intervals do not correspond to the time between the same two events:

I think I can almost see where my misunderstanding might be. Hopefully this next set o questions will shake it out.

Supposing that L is something like 20ly. To move the train it's full length in the FoR of Earth would take ~23 years. During that interval, how much time has passed for the train's conductor? I.e. If the conductor was 30 before starting this run, has he experienced 53 years of life after disembarking, or has he only experienced around 42 years?

 

[Doc Al]

I think I can almost see where my misunderstanding might be. Hopefully this next set o questions will shake it out.

Good.

Supposing that L is something like 20ly. To move the train it's full length in the FoR of Earth would take ~23 years.

It would take about 11.55 years in the Earth frame. (Recall that the length is contracted to L/2 in the Earth frame.)

During that interval, how much time has passed for the train's conductor?

According to the train frame, when the tail of the train passes location A, about 23.1 years have passed.

I.e. If the conductor was 30 before starting this run, has he experienced 53 years of life after disembarking, or has he only experienced around 42 years?

It depends on when he disembarks. (I'll play along and assume that he can be 'disembarked' instantly at any time without killing him.)

Here's how it all works. Imagine there is a conductor at the nose of the train with a clock and a conductor at the rear of the train with another clock. In the train frame those clocks are synchronized. Further imagine that the nose clock reads zero when the nose passes point A. According to Earth observers, the clocks are not synchronized. The clock at the rear of the train is about 17.3 years ahead of the clock at the nose. That means that according to Earth observers, when the nose passes A the rear clock already reads 17.3 years. When the tail of the train passes A, the rear clock reads 23.1 years. So, according to Earth observers, only 5.8 years have elapsed on each clock during the time that the train passed point A. Of course, Earth observers will see that time as being dilated to 11.55 years.

So, if the conductor at the nose was 30 years old when the nose passed A, he'll be 35.8 years old when the tail passes A according to Earth observers. If he is magically disembarked at that instant, he'll show only 35.8 years of wear and tear. (Ignoring any damage from the violent acceleration!)

Hopefully I haven't botched the calculation. (I'm multi-tasking at the moment.) Maybe someone can check my numbers.

 

[arkain]

Good.

It would take about 11.55 years in the Earth frame. (Recall that the length is contracted to L/2 in the Earth frame.)

According to the train frame, when the tail of the train passes location A, about 23.1 years have passed.


It depends on when he disembarks. (I'll play along and assume that he can be 'disembarked' instantly at any time without killing him.)

Here's how it all works. Imagine there is a conductor at the nose of the train with a clock and a conductor at the rear of the train with another clock. In the train frame those clocks are synchronized. Further imagine that the nose clock reads zero when the nose passes point A. According to Earth observers, the clocks are not synchronized. The clock at the rear of the train is about 17.3 years ahead of the clock at the nose. That means that according to Earth observers, when the nose passes A the rear clock already reads 17.3 years. When the tail of the train passes A, the rear clock reads 23.1 years. So, according to Earth observers, only 5.8 years have elapsed on each clock during the time that the train passed point A. Of course, Earth observers will see that time as being dilated to 11.55 years.

So, if the conductor at the nose was 30 years old when the nose passed A, he'll be 35.8 years old when the tail passes A according to Earth observers. If he is magically disembarked at that instant, he'll show only 35.8 years of wear and tear. (Ignoring any damage from the violent acceleration!)

Hopefully I haven't botched the calculation. (I'm multi-tasking at the moment.) Maybe someone can check my numbers.

Don't worry about the numbers, I've got the idea. What of the case where the train is brought to a complete stop before the front conductor disembarks? How old is the conductor in that case?

 

[Mentz114]

What I've been trying to measure is the interval of differential time as seen from each frame. Put another way, I'm trying to measure, from train's perspective how long the train took to pass a fixed point on the Earth (Tt), and from the Earth's perspective how long a length of Earth equal to the rest length of the train took to pass a fixed point on the train (Te). My expectation is that Te = 2 * Tt for a relative velocity of 0.86666c.

You can answer these questions if you work with space-time diagrams. All the rules of relativity are in the Minkowski geometry and the LT. It might help you if you have a look at this pic and watch the animation (mpeg in the zip).

The diagram shows the world as it is at different times in the ground frame. Where the 'now' line intercepts a worldline is the position (on the horizontal axis) of the object at that time, measured on the vertical axis. The motion of the objects can be seen by sliding a horizontal line from the bottom of the diagram to the top. The times on clocks is given by the proper intervals 1->6, 2->5, 3->4 and for the ground clock 1->4. In this setup the intervals are

1->6 21
2->5 15
3->4 9.9

1->4 14

The relative velocity is ~0.4.