# Time to the train stop

If you consider length to be part of "mechanical reality" then length contraction is also part of "mechanical reality" and consequently "mechanical reality" is frame variant.
Then in order for the length contraction to happen from the beginning, both ends should be in different states of motion relative to a platform observer. And in order for the train to regain its original length after coming to a stop, both ends should end in different states of motion too assuming in all cases that both ends have the same states of motion relative to the train observer or in other words, are space-like relative to the train.

But even this underestimates the effect of the motion on length because the length contraction should be associated with moving objects no matter when its 2 ends start to move relative to a platform observer, because its physical existence is related to the way the observer watches the light rays bouncing between 2 ends and not related to a mechanical translation of the rear end. And because in the first case, the length contraction is a result of physical constraint imposed by the invariance of speed of light while in the second the mechanical translation becomes a trigger to cause such invariance of c. So length contraction can not be the cause and the result at the same time.

Even the starting of the motion is not clear relative to all observers. For example, if the 2 ends of the platform start moving simultaneously relative to the train observer, the rear end of the train should start moving before the near one relative to the platform observer which makes the length contraction a mechanically wise for him. But if the 2 ends start moving simultaneously relative to the platform observer no length contraction appear, contrarily to many authors argued that there would be still length contraction and the train will go under strain,,, see Bells spaceship paradox.

So after those 2 arguments, the length contraction and inductively the length itself at the beginning and at the end of the motion is not only ambiguous but may be physically inconsistent too.

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Sure, but what has that got to do with anything else that we have been discussing in this thread?
The rubber band is an example of the train. The stretch on the band when holding between hands resembles the strain imposed on the train when the 2 ends maintain its rest length even during the motion and hindering it from reducing to a contracted length. The sudden release of the elasticity resembles the status when the train comes to a stop.

The correct expression for the fields from an arbitrarily moving point charge is called the Lienard Wiechert potential. Coulomb's law only applies to electrostatic situations.

You will need to do the math on your own. If you get a paradox then go back and check your work since you made an error.
I did a sketch of a solution. I compared the Lienard Wichert potential created by a moving charge on the one following it in the same direction of the motion. And I compared it with Coulomb potential after the 2 charges come to a rest.
The result is the potential during the motion relative to the platform observer is larger than after the stop contrarily to what is observed by the train observer.
I attached a pic file of the sketch.

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A.T.
The rubber band is an example of the train. The stretch on the band when holding between hands resembles the strain imposed on the train when the 2 ends maintain its rest length even during the motion and hindering it from reducing to a contracted length. The sudden release of the elasticity resembles the status when the train comes to a stop.
What is a "sudden release of the elasticity"? When the train/rubber-band come to a stop, the stress simply goes to zero again.

ghwellsjr
Gold Member
The rubber band is an example of the train. The stretch on the band when holding between hands resembles the strain imposed on the train when the 2 ends maintain its rest length even during the motion and hindering it from reducing to a contracted length. The sudden release of the elasticity resembles the status when the train comes to a stop.
If you had said that the strain in the rubber band was reduced when you eventually moved your hands back closer together, it would make sense, but you're implying that the strain simply disappears even when your hands remain the same distance apart, and that doesn't seem to relate to any of your train scenarios. Can you please try to explain?

A.T.
... but you're implying that the strain simply disappears even when your hands remain the same distance apart, and that doesn't seem to relate to any of your train scenarios.
He might mean "stress" not "strain".

Ibix
Perhaps consider ghwellsjr's two locomotives moving at the same relativistic velocity with respect to the track, joined by a spring that is not quite stretched. Cut to a commercial break and return. The locomotives are now at rest with respect to the track.

Before the break:
• An observer at rest with respect to the track claims that the locomotives are 5000 feet apart. The spring is 6250 feet long, but is length-contracted to a mere 5000 feet so is under no stress.
• An observer at rest with respect to the locomotives sees them as 6250 feet apart. The spring is 6250 feet long and is under no stress.
After the break:
• An observer at rest with respect to the track claims that the locomotives are 5000 feet apart. The 6250-foot spring is compressed to 5000 feet, so is in compression.
• An observer at rest with respect to the locomotives is also at rest with respect to the track, so sees the same thing.

The question is, where did the energy come from to compress the spring while we were busy looking at adverts? The answer is simple: something braked those trains to a stop. Whatever it was, it did enough "extra" work to compress the spring over and above simply stopping the locomotives. Think about it: the spring would have been forcing the back locomotive backwards and the front locomotive forwards, compared to what would have happened had they been unconnected. We chose to do some unnecessarily savage braking such that the spring ended up compressed.

The reasoning works like this:
• An observer at rest with respect to the track sees the spring change from a length-contracted-but-relaxed 5000 feet to a non-length-contracted-but-compressed 5000-feet. The brakes must have worked overtime.
• An observer at rest with respect to the locomotives sees the spring change from a non-length-contracted-but-relaxed 6250 feet to a non-length-contracted-but-compressed 5000-feet. The brakes must have worked overtime.

The same reasoning applies to a 6250 foot train trying to fit into a 5000 foot space. A real train isn't anywhere near so elastic as a spring, so it's going to break as it brakes. In the words of the great physicist, Nat King Cole, "Something's gotta give, something's gotta give, something's gotta givvvvvvve!"

1 person
If you had said that the strain in the rubber band was reduced when you eventually moved your hands back closer together, it would make sense, but you're implying that the strain simply disappears even when your hands remain the same distance apart, and that doesn't seem to relate to any of your train scenarios. Can you please try to explain?
While you are holding the rubber to a stretched length, you approach it near a heat source so as it chemically looses its elasticity even it is maintained at the same length.

ghwellsjr
Gold Member
While you are holding the rubber to a stretched length, you approach it near a heat source so as it chemically looses its elasticity even it is maintained at the same length.
And this is related to one of the train scenarios?

1 person
Dale
Mentor
Then in order for the length contraction to happen from the beginning, both ends should be in different states of motion relative to a platform observer.
No, this is not a prerequisite for length contraction. I have no idea how you would possibly go from what I said to such a conclusion.

the length contraction should be associated with moving objects no matter when its 2 ends start to move relative to a platform observer, because its physical existence is related to the way the observer watches the light rays bouncing between 2 ends and not related to a mechanical translation of the rear end. And because in the first case, the length contraction is a result of physical constraint imposed by the invariance of speed of light while in the second the mechanical translation becomes a trigger to cause such invariance of c. So length contraction can not be the cause and the result at the same time.
I have no idea what you are talking about here. Length contraction is a disagreement between two frames regarding the length of something.

So after those 2 arguments, the length contraction and inductively the length itself at the beginning and at the end of the motion is not only ambiguous but may be physically inconsistent too.
Can you provide a reference to back up this claim?

Dale
Mentor
I did a sketch of a solution. I compared the Lienard Wichert potential created by a moving charge on the one following it in the same direction of the motion. And I compared it with Coulomb potential after the 2 charges come to a rest.
The result is the potential during the motion relative to the platform observer is larger than after the stop contrarily to what is observed by the train observer.
I attached a pic file of the sketch.
Remember, if you get a contradiction then you made an error. So where do you think the error is?

My first suggestion is that Coulombs law doesn't apply, particularly not during and immediately following the acceleration. If you want to apply Coulombs law then you have to wait long enough for the changes to propagate through, but it seems like this is precisely the period you want to examine.

Second, you have claimed that the results in the platform frame contradict the results on the train frame without ever deriving the result in the train frame. Remember, the train frame is non inertial, so you will need to transform maxwells equations into that frame. You cannot have a one sided contradiction.

It would help if you made the acceleration concrete by specifying an exact equation of motion as well as the metric in the trains frame.

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The reasoning works like this:
• An observer at rest with respect to the track sees the spring change from a length-contracted-but-relaxed 5000 feet to a non-length-contracted-but-compressed 5000-feet. The brakes must have worked overtime.
• An observer at rest with respect to the locomotives sees the spring change from a non-length-contracted-but-relaxed 6250 feet to a non-length-contracted-but-compressed 5000-feet. The brakes must have worked overtime.
This is clear for me thanks. I was wondering in my previous query why a lateral force that ghwellsjr suggested for the platform observer is equivalent to the compression force relative to the train observer even if we have prior information that the initial length of the string is 5000 feet not 6250 feet.
But A possible answer is the force is only needed if the original length was 6250 feet because if it was 5000 feet that would imply the 2 ends of the train start moving simultaneously relative to the platform observer which has the following consequences:
* At the beginning of the motion:
1) For the platform observer, the length would contract to less than 5000 feet but it is maintained at 5000 feet which creates a tension.
2) For the train observer, the near end starts first which creates an expansion on the near end of the string causing a tension.

* At the end of the motion:
1) The platform observer sees the length maintained at 5000 feet and became relaxed with no length contraction which nulls the initial tension.
2) The train observer sees the near end moves back toward the rear end which nulls the tension force.

No, this is not a prerequisite for length contraction.
So would you please explain how the platform observer can see the length of the train that goes from 6250 feet before moving to 5000 feet after moving without seeing the rear end starts moving before the near end?

A.T.
So would you please explain how the platform observer can see the length of the train that goes from 6250 feet before moving to 5000 feet after moving without seeing the rear end starts moving before the near end?
The ends can accelerate a different rates.

1 person
Dale
Mentor
So would you please explain how the platform observer can see the length of the train that goes from 6250 feet before moving to 5000 feet after moving without seeing the rear end starts moving before the near end?
That is not what length contraction is. Before moving, in the platform's inertial frame the length of the train is 6250. Also before moving there are an infinite number of other inertial frames in which the length is less than 6250. This is length contraction.

Similar statements can be made after the acceleration.

During the acceleration there are relativity of simultaneity issues. Different frames still disagree on the length, but you cannot use the usual simplified length-contraction formula.

I hope this helps you understand what length contraction is and is not.

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The ends can accelerate a different rates.
Fine.

So how if the 2 ends are programmed to start moving at the same acceleration rate? Similarly to Bells spaceship paradox, there should be a tension on the string because its length now is larger than the contracted length. So How did this contraction come from?

So we are facing 2 situations: either the rear end start to accelerate before the near end which lead to a revealed length contraction or both accelerate at the same rate which create a non-revealed length contraction.

A.T.
So how if the 2 ends are programmed to start moving at the same acceleration rate? Similarly to Bells spaceship paradox, there should be a tension on the string because its length now is larger than the contracted length.
Yes there will be tension in Bells scenario.

So How did this contraction come from?
What?

So we are facing 2 situations: either the rear end start to accelerate before the near end which lead to a revealed length contraction or both accelerate at the same rate which create a non-revealed length contraction.
They can also start accelerating simultaneously, but at different rates, to shorten the string.

Dale
Mentor
So how if the 2 ends are programmed to start moving at the same acceleration rate? Similarly to Bells spaceship paradox, there should be a tension on the string because its length now is larger than the contracted length. So How did this contraction come from?
The expansion scalar is positive in all frames, despite the fact that the length is unchanged in the starting frame. In a relativistic version of Hooke's law the change in tension in the string/train is proportional to the expansion scalar, not the coordinate strain rate.

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Remember, if you get a contradiction then you made an error. So where do you think the error is?

My first suggestion is that Coulombs law doesn't apply, particularly not during and immediately following the acceleration. If you want to apply Coulombs law then you have to wait long enough for the changes to propagate through, but it seems like this is precisely the period you want to examine.

Second, you have claimed that the results in the platform frame contradict the results on the train frame without ever deriving the result in the train frame. Remember, the train frame is non inertial, so you will need to transform maxwells equations into that frame. You cannot have a one sided contradiction.

It would help if you made the acceleration concrete by specifying an exact equation of motion as well as the metric in the trains frame.
I did not apply Coulomb law immediately after the stop. I compared the electric fields on a charge before and after stop relative to 2 observers. I found that Lienard Wichert potential before the stop relative to the platform observer is larger than Coulomb potential after the stop. For the train observer, the Coulomb potential before A starts to move is smaller than after A comes close to B end. This is a potential paradox.

Dale
Mentor
It is not a potential paradox. Maxwells equations are invariant under the Lorentz transform. Therefore, it is not possible to set up a scenario which satisfies Maxwell's equations in one frame and not in another.

Your approach of actually working less than a quarter of the problem and then assuming you know the rest is wrong. That is all you have shown. To work this problem correctly requires the following steps:

1) write the expression of the motion of both charges in the platform frame
2) calculate the Lienard Wiechert potential from one charge
3) evaluate the potential at the other charge
4) write the transformation equations to the train frame
6) calculate the metric in the train frame
7) transform the motion of the charges to the train frame
8) transform the fields to the train frame
9) confirm that the transformed fields satisfy Maxwell's equations in the train frame
10) evaluate the fields at the other charge in the train frame
11) compare the results in each frame

You cannot claim even a possible paradox with less. In particular, you have never once written down the critical steps 1 and 4, and because of that if anyone else were to work it you would simply claim it is irrelevant, as you have done previously. This is a problem that you need to work through, and not just part way.

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