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

[arkain]

Careful here, as you are now introducing acceleration into the mix. If the train manages to slow down in such a way that clock synchronization is preserved in the train, the Earth observers will not agree. There's no way to avoid the issue of simultaneity.

That's irrelevant once the train has come to rest. Stick to time measurements made in the original frames!

Neither did I.


Because 'at the same instant' is different for different frames. In the train frame, the difference between the timer readings represents the time it takes for the train to pass the Earth timer; the Earth observers disagree.

Yes, as I explained in an earlier post.


As stated above, that just unnecessarily complicates things. All your claims about the time difference between the train timers were made in the frame of the moving train anyway.


You'll have to try, as I don't understand it. (I haven't looked at your post #25 yet. One scenario at a time.)


You have a basic misunderstanding of 'time dilation'. Time dilation is a relationship between frames in relative motion. Both frames see the other's clocks/timers as running slow. Time dilation works both ways!

(As Mentz114 says, you are mixing up time dilation with differential ageing.)

Ok. I think we've finally uncovered one of my largest mistakes: Improper use of terminology. Problem: for differential aging to occur, the FoR that experienced acceleration must have traveled less along the time axis than the "rest" frame. Worded differently, differential aging occurred because the train experienced less time than the Earth did between the initial and ending rest (WRT Earth) periods of the train. This "actual" relative decrease in the passage of time for the train is what I've been referring to as time-dilation. If this was an incorrect term usage, then I apologize for the confusion.

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.

Was this a better use of terminology? Did it help relieve the confusion about what I'm trying to do? Is my expectation incorrect? Is there anything improper about the experiment itself?

 

[arkain]

Thanks, that helps clarify considerably. In step 7 (Accelerate the train to 0.8666c) you need to specify the acceleration profiles for the front and the back timers, or at a minimum tell whether or not there is any change in the rest length of the train after the acceleration. In step 8 (Stop the train after all 4 timers have been started) you need to specify the deceleration profiles for the front and rear timers as this will affect the results. Please note that the acceleration/deceleration profiles will not in general be the same, particularly if you are considering Born rigid deceleration.

EDIT: I also just noticed that in step 9 you need to specify the motion of all 4 timers from the various locations where they stopped to the workstation since that will also affect the outcome.

I reiterate my advice to find a simpler scenario, the description is much clearer now but the proposed scenario is overly complicated. There is so much going on in this scenario that I doubt you will learn much from the answer. Particularly since it depends on the details of several specific acceleration profiles, so it is not terribly generalizable.

Yay! More progress! I was beginning to get worried that I'm irrecoverably off my gourd! :)

First thing to note is that in this thought experiment, regardless of the length L of the train, the train itself is to be considered perfectly rigid. There should be no time t at which the Born rigidity should be broken. I know that this is impossible and violates more rules than I'd care to count, but the only way around this is to use 2 separate, identical trains of unimportant but equal length, both with timers in the nose and with an initial separation between the trains of length L. Then you'd have to set up some kind of simultaneous launch of both trains. Both trains would need to have previously synchronized timers used to control the stopping of both trains so that they both shut down at the same time (in their FoR). While such a convoluted thing could be done and doesn't violate any rules, it's way simpler to simply consider 1 perfectly rigid train. It'll have the same effect on the experiment. Put simply, the timers in motion need to experience identical acceleration and deceleration profiles relative to each other. They need to remain in the same frame WRT each other at all points during this experiment.

The motion of the timers to the workstation would be non-relativistic (human walking speed WRT Earth) and should not significantly affect the timers.

 

[Dale]

First thing to note is that in this thought experiment, regardless of the length L of the train, the train itself is to be considered perfectly rigid. ...

Put simply, the timers in motion need to experience identical acceleration and deceleration profiles relative to each other.

These are contradictory requirements. If the ship is undergoing Born-rigid motion then the acceleration profiles will be different for the front and the back. In either case, it is necessary to completely specify the exact details of the acceleration profile because the answer depends on the result.

To determine the time accumulated on a clock in general you use the following approach. First, you write down the position of the clock as a function of time x(t). Then you evaluate the following integral:

\int_a^b \sqrt{1-\frac{1}{c^2}\left(\frac{dx}{dt}\right)^2} \, dt

Or you can generalize it even further by writing the position and time coordinate as a function of some arbitrary parameter. Then you would evaluate this integral.

\int_a^b \sqrt{\left(\frac{dt}{d\lambda}\right)^2-\frac{1}{c^2}\left(\frac{dx}{d\lambda}\right)^2} \, d\lambda

Which obviously reduces to the first integral if time is the arbitrary parameter.

So, unless you specify x(t) for each clock, the question cannot be answered other than by pointing to these general formulas. Once you specify x(t) for each clock then the question is simply a matter of computation.

The motion of the timers to the workstation would be non-relativistic (human walking speed WRT Earth) and should not significantly affect the timers.

You can specify this as "slow transport" or "ultra-slow transport".

 

[Doc Al]

What I've been trying to measure is the interval of differential time as seen from each frame.

Despite talk about accelerating and decelerating the train (which complicates things, as DaleSpam points out), you still seem to just want to measure time intervals from the two frames when they are in relative motion. Not really 'differential aging' of anything.

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).

Is this a new scenario? As I understood it, in your original scenario you wanted to compare:
(1) From the train frame, the time it takes for a point A on the Earth to travel the length of the train.
(2) From the Earth frame, the time it takes for the train to pass point A.

The two events defining this time span are: (E1) The nose of the train passes point A and (E2) The tail of the train passes point A. This is perfectly well-defined.

My expectation is that Te = 2 * Tt for a relative velocity of 0.86666c.

No, as I explained earlier it will be just the opposite. Tt = 2*Te. Time dilation applies from either frame, but only in the Earth frame are we recording the time elapsed on a single clock. Note that both events take place at location A, so the single Earth timer is able to measure that time interval. Not so in the train frame, where multiple clocks are required.

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:

For (1), the events defining the time span are exactly what they were before, (E1) and (E2).
For (2'), the events defining the time span are: (E1) The nose of the train passes point A and (E3) The nose of the train passes point B, where A and B are a distance L apart in the Earth frame.

Was this a better use of terminology? Did it help relieve the confusion about what I'm trying to do? Is my expectation incorrect? Is there anything improper about the experiment itself?

See my comments above.

Note: In my opinion it is essential that you understand everything described above. This is just basic relativity, so any additional complications (such as having the train accelerate and decelerate) will only be that much harder to unravel.

 

[arkain]

Again, forget about acceleration--you're just adding complications that only serve to hide what's going on. In your scenario the train is already moving at 0.866c. It's a lovely inertial frame. You used it yourself when you described how the difference between the train timers would show the travel time in the train frame. (That's the frame of the train while it's moving with respect to the earth.)

I'm not trying to hide anything at all, especially not with undue complexity. What I've been saying in a nutshell is this:

2 identical timers some distance L apart are accelerated to 0.86666c WRT Earth using identical acceleration profiles. The 2 timers maintain this velocity and eventually cross point P WRT Earth, at which point, the timer crossing P is started. At some time after both timers are started, both timers are decelerated to 0c WRT Earth using identical acceleration profiles.

Unless what I'm about to say is another of my mistakes, the following should be true:

1) the 2 timers were always in the same FoR.
2) the timer FoR was non-inertial during the acceleration and deceleration phases.
3) both timers were started while the timer FoR was in an inertial state.
4) before the deceleration phase starts, the difference between the two timers as seen from the timer's FoR represents the amount of time in that FoR between the starting of the 2 timers.
5) (and this seems to be part of the point of contention here) since 1 is true, 4 must also be true after the deceleration phase ends.
6) when this time difference is compared with the time taken for both timers to cross the fixed point P (given that the crossing of the first timer is T=0) in Earth's FoR, the intervals will not be identical, thus indicating differential aging.
7) the timer FoR will show the shorter time interval.

 

[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.