Relativity paradox

[Tach]

if x is the distance of some point of the rod form the end in the train frame, then the time it takes for the impact to propagate form the end to this point is

$$\frac{x}{\gamma (c-v)}$$

The time delay between the end and the same point is:

$$\frac{xv}{\gamma c^2}$$

For any value of x or v, the RoS time delay will be shorter than the time needed for the impact to propagate to the same point.

OK,

Let's continue (hopefully without the RJBeery and Neddy Bate drones buzzing around).

I have already pointed out that I disagree with the first formula, you are trying to construct a relativistic theory of elasticity on very shaky grounds. The next thing you are trying to do is to say: "look, the reaction force of the ground takes much longer to propagate than the delay $$\frac{xv}{\gamma c^2}$$ necessary for point x of the rod to hit the ground ". This doesn't mean anything, the rod points hit the ground sequentially in the platform frame, as opposed to the train frame where all points are subjected to the reaction force simultaneously. Take two points separated by the distance $$\Delta x$$, there is going to be a NON_NULL interval $$\frac{v \Delta x}{\gamma c^2}$$ during which one endpoint will be subjected to the reaction force of the ground while the other point isn't subjected to any such force. This should result into a deformation of the rod in the platform frame while there is no deformation in the car frame. I really hate getting into all this elasticity theory, I do not believe using SR is the right formalism in addressing the problem. You really can't use SR in the presence of gravitational fields, this is why , when using SR, the results come out wrong. Not because SR is "wrong" but because SR is the wrong formalism for the problem.

 

[RJBeery]

You make up strawmen and you beat them up. You can keep your strawmen.

This is no strawman. This is Tach tap dancing all over the place. In your own words:

Nope, that end (the far end) starts falling first in the embankment frame. Yet, the Thomas precession gradually counter-rotates the rod in the opposite direction.

Yet you also claim

Perhaps if you knew the math behind it, you knew how that was possible. Before you deny it, you must be able to write down the equations of motion. Good luck with that. Hint: could it be that the solution is independent of "the distance fallen"? eh?

If Thomas precession perfectly counter-balances things such that both ends of the rod hit the ground simultaneously in all frames from any arbitrary height, which is what you are clearly contending, then the body must remain parallel to the ground at all times and in all frames. Your words.

 

[Tach]

This is no strawman.

Sure it, is , that is what you always do since you are totally incompetent in terms of formulating any mathematical argument.
For the time being, I am going to restrict the conversation to the one with Janus, who, unlike you, seems to be able to formalize his arguments. Shut up, stop droning, you may learn something.

If Thomas precession perfectly counter-balances things such that both ends of the rod hit the ground simultaneously in all frames from any arbitrary height, which is what you are clearly contending, then the body must remain parallel to the ground at all times and in all frames. Your words.

Yes, I understand that this is difficult for you to get through your thick skull but it is pretty much the case. I would point you to a couple of documents on the internet that support this point but knowing that you are severely math challenged, I doubt that you would understand them. Some of them require 9-th grade algebra, way above your level of understanding.

 

[RJBeery]

Sure it, is , that is what you always do since you are totally incompetent in terms of formulating any mathematical argument.
For the time being, I am going to restrict the conversation to the one with Janus, who, unlike you, seems to be able to formalize his arguments. Shut up, stop droning, you may learn something.

I accept your concession.

 

[Tach]

I accept your concession.

You are not only a drone, you are totally delusional.

 

[Pete]

I really hate getting into all this elasticity theory, I do not believe using SR is the right formalism in addressing the problem. You really can't use SR in the presence of gravitational fields, this is why , when using SR, the results come out wrong. Not because SR is "wrong" but because SR is the wrong formalism for the problem.

In this problem, the gravitational field is uniform, so the spacetime is flat, so SR should be sufficient.
But it doesn't matter, because the problem remains in the absence of gravity:

Consider the rod, train, and platform all moving inertially in flat space, such that all elements of the rod collide with the floor of the train simultaneously in the train rest frame.
In the platform frame, the collisions between the rod elements and the floor of the train are not simultaneous.

Take two points separated by the distance $$\Delta x$$, there is going to be a NON_NULL interval $$\frac{v\Delta x}{\gamma c^2}$$ during which one endpoint will be subjected to the reaction force of the ground while the other point isn't subjected to any such force.

This is true.

This should result into a deformation of the rod in the platform frame while there is no deformation in the car frame.

This is true to some extent.
A simultaneous slice of the rod's worldlines in the train rest frame will always be straight.
A simultaneous slice of the rod's worldlines in the platform rest frame will be bent during the collision.

But is this a problem? Does it lead to any paradoxes?

 

[Tach]

But it doesn't matter, because the problem remains in the absence of gravity:

Consider the rod, train, and platform all moving inertially in flat space, such that all elements of the rod collide with the floor of the train simultaneously in the train rest frame.
In the platform frame, the collisions between the rod elements and the floor of the train are not simultaneous.


This is true.

This is true to some extent.
A simultaneous slice of the rod's worldlines in the train rest frame will always be straight.
A simultaneous slice of the rod's worldlines in the platform rest frame will be bent during the collision.

But is this a problem? Does it lead to any paradoxes?

Yes, it does. The rod gets bent in the platform frame as a result of the point-sequential application of reaction force. The rod is straight in the train car frame because the force is applied simultaneously to all points. Same experiment, two contradicting outcomes.

 

[Pete]

Yes, it does. The rod gets bent in the platform frame as a result of the point-sequential application of reaction force. The rod is straight in the train car frame because the force is applied simultaneously to all points. Same experiment, two contradicting outcomes.

At the end of the experiment, the rod is straight in both reference frames.
I don't see the contradiction.

 

[Tach]

At the end of the experiment, the rod is straight in both reference frames.
I don't see the contradiction.

It is obviously straight in the train car frame.
Can you prove it is straight in the platform frame?

 

[Pete]

It is obviously straight in the train car frame.
Can you prove it is straight in the platform frame?

Sorry, I didn't think that was a point of contention.

All rod elements are in contact with the floor of the train at the end of the experiment. The floor is straight, so the rod is straight.

 

[Tach]

Sorry, I didn't think that was a point of contention.

All rod elements are in contact with the floor of the train at the end of the experiment. The floor is straight, so the rod is straight.

Not a very good answer, the observer in the platform frame disagrees.

 

[Markus Hanke]

In this problem, the gravitational field is uniform, so the spacetime is flat

Uniformity of the gravitational field does not mean a flat space-time. That is a contradiction in itself, since a flat space-time implies the complete absence of gravity.

 

[Markus Hanke]

I like Janus's idea that in the embankment frame the rod might not hit the ground flat, yet still produce a physical outcome indistinguishable from the one in the train frame ( i.e. no bending etc ). I have no problem conceding that this may be a better, more meaningful solution than my own which involves Thomas precession.

 

[Pete]

the observer in the platform frame disagrees.

What do you mean?

 

[Tach]

What do you mean?

Sigh, I thought that you understood: the platform observer measures the reaction force of the platform being applied sequentially, piece by piece, point by point, thus bending the rod. I thought you already agreed to that in post 86 when you replied "This is true".
Alternatively, the platform observer sees the rod hitting one endpoint first, thus making it bounce differently than the way observed by the train car observer.

 

[Tach]

I like Janus's idea that in the embankment frame the rod might not hit the ground flat, yet still produce a physical outcome indistinguishable from the one in the train frame ( i.e. no bending etc ). I have no problem conceding that this may be a better, more meaningful solution than my own which involves Thomas precession.

While the approach is interesting, it has quite a few errors of its own, as pointed out. This puzzle is a very tough nut to crack.

 

[Pete]

Uniformity of the gravitational field does not mean a flat space-time. That is a contradiction in itself, since a flat space-time implies the complete absence of gravity.

No, flat space-time implies the absence of spacetime curvature.
A uniform gravitational field also implies the absence of curvature.

Note that the equivalence principle says that a constantly accelerating reference frame in zero gravity is indistinguishable from an inertial reference frame permeated by a uniform gravitational field.

 

[Tach]

No, flat space-time implies the absence of spacetime curvature.
A uniform gravitational field also implies the absence of curvature.

I didn't get involved in this part because I thought it is a sideshow but I need to point out to you two facts: both the platform and the car are accelerated frames (upwards, by 1g), so they aren't inertial, so, you can't use the Lorentz transforms, therefore, the machinery of SR does not apply. Secondly, the rod falls accelerated (obviously) , so even in a naive attempt of using SR, you would need to use Rindler coordinates, a messy proposition.

Note that the equivalence principle says that a constantly accelerating reference frame in zero gravity is indistinguishable from an inertial reference frame permeated by a uniform gravitational field.

That would be the frame of the rod. Not a very useful frame for solving the puzzle. You are being asked to compare the results between the platform and the car frames, both frames being accelerated (see above).

 

[Pete]

Sigh, I thought that you understood: the platform observer measures the reaction force of the platform being applied sequentially, piece by piece, point by point, thus bending the rod. I thought you already agreed to that in post 86 when you replied "This is true".
Alternatively, the platform observer sees the rod hitting one endpoint first, thus making it bounce differently than the way observed by the train car observer.

Ah, I was considering an inelastic collision that ended with the rod flat on the floor of the train.
So, let the rod bounce elastically.

After the collision in the train reference frame, the rod is straight and moving inertially with the reverse of its previous velocity.
It seems obvious that this straight inertial rod will be straight in all reference frames.

Why would you suspect the rod to be bent in the platform reference frame after the collision?

 

[Pete]

I didn't get involved in this part because I thought it is a sideshow but I need to point out to you two facts: both the platform and the car are accelerated frames (upwards, by 1g), so they aren't inertial, so, you can't use the Lorentz transforms, therefore, the machinery of SR does not apply. Secondly, the rod falls accelerated (obviously) , so even in a naive attempt of using SR, you would need to use Rindler coordinates, a messy proposition.

Let's remove gravity from the problem.
Consider the rod, train, and platform all moving inertially in flat space, such that all elements of the rod collide with the floor of the train simultaneously in the train rest frame.
In the platform frame, the collisions between the rod elements and the floor of the train are not simultaneous.