Train Stopping Time: Different Observers, Different Times?

[adelmakram]

Are you familiar with matrices and linear algebra?

yes,

 

[Dale]

Then it is not that difficult. Using units where c=1 you simply set up the spacetime coordinate 4-vector: $r=(t,x,y,z)$ and the Lorentz transform matrix
$$\Lambda = \left(
\begin{array}{cccc}
\gamma & - \gamma v & 0 & 0 \\
-\gamma v & \gamma & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right) $$

And the electromagnetic field tensor:
$$F=\left(
\begin{array}{cccc}
0 & E_x & E_y & E_z \\
-E_x & 0 & B_z & -B_y \\
-E_y & -B_z & 0 & B_x \\
-E_z & B_y & -B_x & 0 \\
\end{array}
\right)$$

Then $r'=\Lambda r$ and $F'=\Lambda F \Lambda^T$

 

[adelmakram]

Then it is not that difficult.

For the train observer, the force acting on the point charge after the A-end starts to move with velocity v toward B-end is a magnetic force described by Lorentz.
F`= y qv x B`(the primed symbols indicate the field in the train frame of reference)
The direction is upward by right hand rule.

For the platform observer, there is no relative motion between the 2 ends, however, there is an expected change in the magnetic field after the train stops.

If we consider that the train moves along x-axis, and B` is directed along z-axis, then the transformation equation is:

B`z = γ [Bz – (v/c2) Ey ]

So the inverse transformation is:

Bz = γ [B`z + (v/c2) E`y ]

But E`y = 0

So Bz before = ^ [B`z ]
Where γ is Lorentz factor.

After the train stop,

Bz after = B`z

So Bz before > Bz after

So the point charge will experience a net electric force that depends on the rate of change of the Bz. The direction is governed by Lenz law. As there is reduction of the magnetic field, then the direction of the induced force on the charge will aim to oppose the change of the magnetic field. So the charge should move upward similar to what is seen by the train observer.

First, I am wondering if B is really reduced after the train comes to a stop? Once this is cleared the problem will be solved.

One more question, what will be the direction of B relative to the platform observer while the train is moving. I assumed in my calculation that the direction of B is the same as B` along the z direction.

 

[adelmakram]

Will the point electric charge really intersect the magnetic lines as it starts to move from A toward B?
I think the magnetic lines themselves will start moving too because they are confined within the space of the train, so as the magnetic lines moves toward B in the same velocity as the point charges does leading to no Lorentz force on it!.
On the other hand, the platform observer should notice the upward motion of the charge because it is an electric force induced on it from a changing magnetic field after the train comes to a stop.

Will that be a paradox?

 

[Dale]

Will the point electric charge really intersect the magnetic lines as it starts to move from A toward B?
I think the magnetic lines themselves will start moving too because they are confined within the space of the train, so as the magnetic lines moves toward B in the same velocity as the point charges does leading to no Lorentz force on it!.
On the other hand, the platform observer should notice the upward motion of the charge because it is an electric force induced on it from a changing magnetic field after the train comes to a stop.

Will that be a paradox?

I don't know why you keep rambling on about paradoxes. Maxwell's equations are invariant under the Lorentz transform so there will obviously never be any paradox. Using the usual tensor notation with c=1 and a (-+++) signature:

In the train frame:
$$F^{\mu\nu}=\left(
\begin{array}{cccc}
0 & 0 & 0 & 0 \\
0 & 0 & B_z & 0 \\
0 & -B_z & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{array}
\right)$$
$$U^{\mu}=(\gamma,\gamma v, 0,0)$$
so the four-force is
$$q U_{\mu}F^{\mu\nu} = (0,0,q\gamma B_z v,0)$$

In the particle's frame:
$$F'^{\mu\nu}=\left(
\begin{array}{cccc}
0 & 0 & -\gamma B_z v & 0 \\
0 & 0 & \gamma B_z & 0 \\
\gamma B_z v & -\gamma B_z & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{array}
\right)$$
$$U'^{\mu}=(1,0, 0,0)$$
so the four-force is once again
$$q U'_{\mu}F'^{\mu\nu} = (0,0,q\gamma B_z v,0)$$

No paradox. Do the math.

 

[adelmakram]

No paradox. Do the math.

Thank you for your formulation. But that was irrelevant to my concern. I did not ask about the invariance of Lorentz force relative to the point charge frame.

I asked whether the point charge will still intersect the magnetic field lines as it starts moving from A to B. And I wondered whether the magnetic field lines themselves will move too as long as they are confined in the same space. For if that would be the case, there will be no force on the charge. Then I questioned whether there would be a paradox if the charge would not experience a Lorentz force because it does not intersect the B`-lines in the frame of the train but it experience that force in the platform frame because of the change of the amplitude of B due to sudden stop of the train.

 

[Dale]

I wondered whether the magnetic field lines themselves will move too as long as they are confined in the same space. For if that would be the case, there will be no force on the charge.

Where in the Lorentz force law is the velocity of the magnetic field? Write down the equation and ask about the meaning of any term that you don't recognize.

Thank you for your formulation. But that was irrelevant to my concern. I did not ask about the invariance of Lorentz force relative to the point charge frame.
... I questioned whether there would be a paradox if the charge would not experience a Lorentz force because it does not intersect the B`-lines in the frame of the train but it experience that force in the platform frame because of the change of the amplitude of B due to sudden stop of the train.

It is completely relevant. It disproves your suggestion of a paradox by explicitly calculating the supposedly paradoxical quantity. And even if the fields are not what you had in mind (your description is unclear as I mentioned earlier), the method is instructive and relevant.

Regarding the dynamic EM effects due to the stopping of the train. Whatever the field and forces are in one frame, regardless of how complicated, as long as they satisfy Maxwell's equations and the Lorentz force law in one frame they will satisfy them in all frames. The complicating details are not important, but you are welcome to calculate them following my example.

 

[adelmakram]

I don't know why you keep rambling on about paradoxes.

I can not promise that I will stop doing that, but I might call them potential paradoxes.

Consider the same scenario but this time there are 2 points charges, one of them is a positive charge put at A-end and the other is a negative charge put at B-end and I am interested to examine the electric force between them relative to different observers.

For the platform observer, the force value should not change after the train comes to a complete stop because the electric field in the direction of the motion does not change; Ex =E`x

For the train observer, the length of the train changes and so does the electric force between the 2 points charges according to the Coulomb law which creates a potential paradox.

 

[Dale]

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.

 

[adelmakram]

Since you agree that there must be a restraint, then you agree that there is a lateral force (along the direction of the rails) between the rails and the wheels to keep the train from expanding back to about 6250 feet.

But how do we know that the original length at the start of the motion was 6250 feet? Is that possible that the train would start moving with both ends accelerating simultaneously? If so, the train length would remain unaltered before the motion, during the motion and after the stop as long as both ends have the same states of acceleration all times. So the existence of the lateral force that hinders the train from getting back to its original length depends on the prior information at the start of the motion which makes it non-deterministic !

 

[ghwellsjr]

But how do we know that the original length at the start of the motion was 6250 feet? Is that possible that the train would start moving with both ends accelerating simultaneously? If so, the train length would remain unaltered before the motion, during the motion and after the stop as long as both ends have the same states of acceleration all times. So the existence of the lateral force that hinders the train from getting back to its original length depends on the prior information at the start of the motion which makes it non-deterministic !

My comments from post #8 apply:

In order for all parts of a train (or any object) to stop simultaneously according to a frame, there must be something like clamps set up all along the track which simultaneously stop the train (by preprogrammed timers) that bring all the parts to a halt (in that frame) and keep the different parts from expanding back to their natural length.

I can't help it if you specified an unrealistic scenario but that's what you did and Special Relativity doesn't address the realisticness of a scenario. That's why we can specify instantaneous accelerations (or decelerations) and see what happens in different IRF's even if they can't actually happen in reality.

If you want the unrestrained length of the train to be 6250 feet before and after it moves and then for both ends to accelerated simultaneously according to the original rest frame of the tracks, then there will be a strain on the train during the period of time between the acceleration and deceleration.

As long as you specify your scenarios precisely and completely, there will never be anything non-deterministic. But if you leave details like this out, then of course there will be different scenarios that all meet your imprecise specs but you can't blame that on the physics.

I would also like to ask you how a train can be moving down the track with a strain on it? How are those two locomotives going to maintain that separation merely with the wheels on the rails? Or are there more unspecified details that you have in mind that you haven't told us about?

 

[adelmakram]

If you want the unrestrained length of the train to be 6250 feet before and after it moves and then for both ends to accelerated simultaneously according to the original rest frame of the tracks, then there will be a strain on the train during the period of time between the acceleration and deceleration.

That stain is due to a force that prevents the train to go to a contracted length relative to the platform observer right? So it should be felt during the motion and an opposite force would be felt if such strain had to be relieved and the length would be allowed to go to a contracted length also during the motion. But if the train comes to a stop with its length unchanged, because all its parts follow the same acceleration, then no force because no stretch. For example, if you are holding a rubber band with fingers of your 2 hands. You will fell a force when you stretch it but suppose that the band suddenly looses its elasticity, then you will not feel the force any more.

 

[adelmakram]

Length contraction is not well-defined relative to a given observer if the two ends of the train are in relative motion, with respect to that observer. That's because the train's "length" itself is not well-defined, relative to that observer, if the two ends of the train are in relative motion.

So how length contraction occurs in the beginning of the motion? To my understanding from this thread, the rear end of the train should start moving before the near end relative to the platform observer which leads to a contracted length. So for that observer, 2 ends have different states of motion in order for the length contraction to start happening. This again makes me question: is the length contraction a visual illusion or a mechanical reality?

 

[Dale]

: is the length contraction a visual illusion or a mechanical reality?

There are many threads on this topic. It is not a visual illusion. Whether or not you call it a mechanical reality is largely a philosophical question which depends on your definition of "mechanical reality".

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. Alternatively, if you do not want to allow "mechanical reality" to be frame variant then length cannot be part of "mechanical reality".

 

[ghwellsjr]

That stain is due to a force that prevents the train to go to a contracted length relative to the platform observer right?

That strain is due to the two locomotives at both ends of the train applying forces in opposite directions on the rest of the train. If there had been only one locomotive on the front of the train, there would still be a strain but it would eventually go away with the length of the train changing in some unspecified way. This is not what we call Length Contraction. As I already explained in post #16:

Length Contraction is the ratio of the length of an object in a frame in which the object is moving compared (or divided by) the length of the object in a frame in which the object is not moving. It has nothing to do with the length of an object before and after it experiences an acceleration. As I said before, Special Relativity cannot address that issue, that is a materials or structural issue. It's no different than asking the question of how the length of a nail changes when you hit it with a hammer. Special Relativity cannot answer that question.

So it should be felt during the motion and an opposite force would be felt if such strain had to be relieved and the length would be allowed to go to a contracted length also during the motion.

While the two locomotives continue to apply a force in opposite directions, the strain would continue but I don't know what you mean by the bold part.

But if the train comes to a stop with its length unchanged, because all its parts follow the same acceleration, then no force because no stretch.

This again is not a question that Special Relativity can answer. There's no guarantee that when you subject a train to the strains that you specify that you will leave it with no forces at the end. That's like claiming that if I apply a force to the two ends of a rod so that I stretch it by 25% of its original length and then I apply a force in the opposite direction so that I'm forcing it back to its original length that there will be no force on the rod when I'm done. It seems highly likely to me that if you had just released the stretching force then the rod would relax to maybe 10% over its original length and that if you tried to get it back down to it original length, you would have to apply a constant compressive force.

For example, if you are holding a rubber band with fingers of your 2 hands. You will fell a force when you stretch it but suppose that the band suddenly looses its elasticity, then you will not feel the force any more.

Sure, but what has that got to do with anything else that we have been discussing in this thread?

 

[adelmakram]

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 Bell`s 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.

 

[adelmakram]

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.

 

[adelmakram]

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.

 

[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]

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 got to give, something's got to give, something's got to givvvvvvve!"

 

[adelmakram]

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]

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?

 

[Dale]

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]

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.

 

[adelmakram]

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.

 

[adelmakram]

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.

 

[Dale]

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.