Relativity paradox

[RJBeery]

A rare admission of error from you, must mark the calendar.

So?

How is this different from what Markus is saying?

Because Markus said that both frames would agree that both ends of the body would hit the floor simultaneously due to Thomas precession. This is trivially false. Repeat the experiment but double the height of the box car. How would Thomas precession possibly "know" to induce negative torque such that the body always seems to land flat with the ground, regardless of distance fallen? No, the answer is that the body does not stay rigid when relativistic torque is involved.

 

[Tach]

Because Markus said that both frames would agree that both ends of the body would hit the floor simultaneously due to Thomas precession. This is trivially false.

Prove it. use math. Armwaving does not apply.

Repeat the experiment but double the height of the box car. How would Thomas precession possibly "know" to induce negative torque such that the body always seems to land flat with the ground, regardless of distance fallen?

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?

No, the answer is that the body does not stay rigid when relativistic torque is involved.

You are deeply confused, you are mixing material properties (rigidity) with dynamics, two very different fields. Time to stop waving your arms, RJ.

 

[Neddy Bate]

That's not what I am telling you, what I am telling you is that you are trying to use a sleigh of hand to re-transform the problem in the train frame. You are an observer in the embankment frame, you must judge the experiment based on what you observe in the embankment frame, consider the fact that you can't see what happens in the train frame.

I understand that each frame would like an explanation that does not rely upon any other frame. But the Lorentz transformations exist for a purpose, and if I choose to use them, I am allowed to do so.

No, it isn't. "Straight down" for Markus means parallel with the ground, un-rotated. The Thomas precession cancels out the effects of RoS.

I doubt that's what he's saying, because if it is, it is wrong.

You are considering RoS only (and trying to re-transform the problem back into the frame of the car through an inverse Lorentz transform, where the ends strike the floor simultaneously). Nowhere in your posts does the Thomas precession show up.

That is correct.

 

[Tach]

I understand that each frame would like an explanation that does not rely upon any other frame.

Good, because this is how physics works in the real world.

But the Lorentz transformations exist for a purpose, and if I choose to use them, I am allowed to do so.

No, you are not, imagine that you were enclosed in the embankment frame and all you had was the direct observation of what happens in your frame, you do NOT know what happens in the frame of the car. This is how physics really works, so get used to it.

 

[Neddy Bate]

Here is a Lorentz transformation for the event of the head hitting the floor at location x' in the train frame:

$$t_1 = \gamma(t' + \frac{vx'}{c^2})$$


For a body of proper length L, here is a Lorentz transformation for the event of the feet hitting the floor at location x'+L in the train frame:

$$t_2 = \gamma(t' + \frac{v(x'+L)}{c^2})$$


Now Tach, if you are claiming they hit at the same time in the embankment frame. then you are claiming this:

$$t_1 = t_2$$

which is the same as claiming this:

$$\gamma(t' + \frac{vx'}{c^2}) = \gamma(t' + \frac{v(x'+L)}{c^2})$$

which is the same as claiming this:

$$x' = x'+L$$

which is the same as claiming this:

$$L = 0$$

Where L is the length of the body.

 

[Tach]

Here is a Lorentz transformation for the event of the head hitting the floor at location x' in the train frame:

$$t_1 = \gamma(t + \frac{vx'}{c^2})$$


For a body of proper length L, here is a Lorentz transformation for the event of the feet hitting the floor at location x'+L in the train frame:

$$t_2 = \gamma(t + \frac{v(x'+L)}{c^2})$$


Now Tach, if you are claiming they hit at the same time in the embankment frame.

That is not what I am claiming, you need to engage comprehension. What I am claiming is that , contrary to your reductionist view, this is not a simple problem of kinematics, there are forces that you need to consider involved:

-the gravitational force
-the Thomas precession

Technically, this is not a problem that can be dealt with in SR, due to the presence of the gravitational field (and no, you cannot transform the field away). You need to cast the mathematical formalism correctly, not the caricature that you are attempting. You just built a strawman in order to divert attention from your inability to understand how experiments work, you can stop beating it now.

 

[Neddy Bate]

That is not what I am claiming, you need to engage comprehension. What I am claiming is that , contrary to your reductionist view, this is not a simple problem of kinematics, there are forces involved:

-the gravitational force
-the Thomas precession

You need to cast the mathematical formalism correctly, not the caricature that you are attempting.

Sorry but the impact events cannot be simultaneous in both frames, unless the length of the body is $$L=0$$ as I have proven. This is basic relativity. I'm surprised you didn't know that.

 

[Tach]

Sorry but the impact events cannot be simultaneous in both frames, unless the length of the body is $$L=0$$ as I have proven...

...only if you are stupid enough to miss all the other effects. Let me put it this way to you: you cannot even employ the Lorentz transforms because you aren't dealing with inertial frames, there is a gravitational field present. Take a hint.

 

[Neddy Bate]

Technically, this is not a problem that can be dealt with in SR, due to the presence of the gravitational field (and no, you cannot transform the field away). You need to cast the mathematical formalism correctly, not the caricature that you are attempting. You just built a strawman in order to divert attention from your inability to understand how experiments work, you can stop beating it now.

I admit that I was treating the thread as a basic SR problem. I did not think the gravity was supposed to be taken literally.

 

[Tach]

I admit that I was treating the thread as a basic SR problem. I did not think the gravity was supposed to be taken literally.

Good, this is progress.

 

[Neddy Bate]

Good, this is progress.

Yes, but I just went back and read the OP. It says, "Hence there is a paradox in special relativity."

 

[Tach]

Yes, but I just went back and read the OP. It says, "Hence there is a paradox in special relativity."

This is what the idiot added

 

[Neddy Bate]

This is what the idiot added

My point is that the gravity was not intended to be taken literally. Anyway, if sufficiently low, the gravity can be neglected.

 

[Tach]

My point is that the gravity was not intended to be taken literally.

How do you know? If the gravitation is absent, how does the rod fall? You are grasping at straws, unwilling to admit that you are wrong again.

Anyway, if sufficiently low, the gravity can be neglected.

Prove it. Stop waving your arms, use math. While you are at it, see if you can also neglect the Thomas precession. This is a much tougher problem than the mockery you are trying to make it to be.

 

[RJBeery]

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?

The only way the negative torque induced by Thomas precession could perfectly balance things independently of the distance fallen is if the body remained rigid and parallel to the floor at all times, from all frames, regardless of the appearance of when the respective wires were cut. In other words, one frame would see the wire cut on one end of the body...yet that body magically remains suspended in a horizontal position, apparently defying gravity. This is horse shit and you should be ashamed of yourself.

you need to engage comprehension

Bwahaha!

 

[Neddy Bate]

How do you know? If the gravitation is absent, how does the rod fall? You are grasping at straws, unwilling to admit that you are wrong again.

I know because the OP says "special relativity" in it. I doubt the guy was looking for a GR answer to such a simple problem. Maybe the body pushed himself off the ceiling with his hands and feet simultaneously.

Prove it. Stop waving your arms, use math. While you are at it, see if you can also neglect the Thomas precession. This is a much tougher problem than the mockery you are trying to make it to be.

In the absence of a gravitational field, GR reduces to SR. I gave you the SR proof already. Your claim of simultaneous impacts in both frames is only possible in SR when the length of the body $$L=0$$. If you think relativity of simultaneity goes away in the presence of a gravitational field, even in the limit as the strength of the field approaches zero, then the burden of proof is on you. Good luck, Tach.

 

[Tach]

I know because the OP says "special relativity" in it. I doubt the guy was looking for a GR answer to such a simple problem. Maybe the body pushed himself off the ceiling with his hands and feet simultaneously.

We've been over this already, the "guy" copied the description and added "special relativity". Besides, my question to you was more direct: "How does the body fall in the absence of the gravitational field"?

In the absence of a gravitational field, GR reduces to SR.

But the gravitational field is present.

I gave you the SR proof already.

You gave me a mockery of a "proof" , commensurable with your limited level of understanding.

Your claim of simultaneous impacts in both frames is only possible in SR when the length of the body $$L=0$$.

But this is not an SR problem. You keep trying to apply your reductionist view.

If you think relativity of simultaneity goes away in the presence of a gravitational field,

No one says that. Quite the opposite, RoS exists in GR but it takes a different form from SR. I don't expect you to know what that expression is.

even in the limit as the strength of the field approaches zero, then the burden of proof is on you. Good luck, Tach.

The "strength of the field" is NOT zero. The Thomas precession effect is also present. So, you can stop weaseling now.

 

[Neddy Bate]

Tach,

If you will first admit that my proof holds true under SR, then we can proceed. Otherwise, this is going nowhere.

 

[Tach]

Tach,

If you will first admit that my proof holds true under SR, then we can proceed.

You don't have a "proof" , what you have is a joke commensurable with your level of understanding of the problem statement.

 

[Tach]

The only way the negative torque induced by Thomas precession could perfectly balance things independently of the distance fallen is if the body remained rigid and parallel to the floor at all times, from all frames, regardless of the appearance of when the respective wires were cut.

RJBeery,

Maybe it is time you stopped making statements that you are unable to back up since you are unable to put together the mathematical formalism? You can't get away from your being a pretender forever.

In other words, one frame would see the wire cut on one end of the body...yet that body magically remains suspended in a horizontal position, apparently defying gravity.

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.

This is horse shit and you should be ashamed of yourself.

this is what comes regularly from your armwaving.