Relativity paradox

[Pete]

Please correct my naive understanding of your disagreeing positions. I am confused. If the train observer sees the rod and floor of the train straight and in full contact along their touching lengths, but the platform observer sees a crooked rod, then the platform observer must also see a crooked train too. Is that not the logical thing? The floor of the train must also appear crooked to the platform observer because the train observer testifies that the floor and rod touch all along their lengths while platform observer says the rod is bent. If rod appears bent to platform observer, and if train observer say that rod and floor touch everywhere along rod, then floor must follow rod shape for platform observer, making both rod and train appear bent to platform observer? Is something wrong with this logical conclusion from combining the meanings of your respective posts above? If correct, then both observers are correct that the rod is in contact with train floor perfectly all the contact lengths. The train observer sees the contact line as "straight". The platform sees the contact line as "bent". Both correct because full contact between rod and floor "seen" in both frames?

Hi Undefined,
My understanding is that:
- as measured in the train frame, the rod is parallel to the floor, and hits the floor all at the same time.
- as measured in the platform frame, the rod is at an angle to the floor, and hits the floor with one end first.

 

[Undefined]

Hi Undefined,
My understanding is that:
- as measured in the train frame, the rod is parallel to the floor, and hits the floor all at the same time.
- as measured in the platform frame, the rod is at an angle to the floor, and hits the floor with one end first.

Thankyou Pete. But what I asked Tach is to explain what he means by "measurement" when he talks about different frame perspectives and information used for measurement. The process of measurement inside proper frame are not measurement process from relative different remote frame. Please read my post again for what is confusing me when I read Tach's arguments using the word "measurement" as if the process was equivalently based in both frames. Tach will explain to me soon I hope. I will wait for that. Thankyou just same Pete.

 

[Pete]

Tach, I think we're making this much more complicated than it needs to be.
The essence of the problem comes down to two simple questions:

Do the ends of the rod strike the floor simultaneously in the train frame? Yes.
Do the ends of the rod strike the floor simultaneously in the platform frame? No.

You said this yourself back in post 81:

...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.

 

[eram]

Don't forget, Tach, the test of the rod falling in the train frame needs to match the top 20! In the train frame there is a specific elapsed time that it takes for that rod to travel, and a specific distance that it travels. I hate to break it to you, but that is gonna leave a mark!

Haha.

Not sure how this is gonna help lol

 

[Tach]

Tach, I think we're making this much more complicated than it needs to be.
The essence of the problem comes down to two simple questions:

Do the ends of the rod strike the floor simultaneously in the train frame? Yes.
Do the ends of the rod strike the floor simultaneously in the platform frame? No.

You need to take in the FULL context, I pointed out that the rod striking gradually the platform results into the deformation of the rod in the platform frame while there is no deformation in the car frame, a contradiction that you keep trying to sweep under the rug.

You said this yourself back in post 81:

A classical rookie error in solving relativity problems is the naive attempt at applying:

1. length contraction
2. time dilation
3. relativity of simultaneity

You have been given the correct solution, the problem is solved properly using Thomas precession. Feel free to continue using hacks.

 

[Tach]

I looked through those notes, and I actually couldn't find the formula you cited. Where in the notes is it? The closest formula I could find was

$$\phi=\gamma\tan^{-1}(\gamma \tan \theta)$$

$$y=tan(\theta)x$$ in frame S
$$y'=\gamma tan(\theta)x'$$ in frame S'

What does this tell you?

 

[Pete]

Tach,
Do the ends of the rod strike the floor simultaneously in the train frame?
Do the ends of the rod strike the floor simultaneously in the platform frame?

Also, you need to read (edit) [post=3062610]post 157[/post], because your solution has an obvious flaw.

 

[Undefined]

Tach,
Do the ends of the rod strike the floor simultaneously in the train frame?
Do the ends of the rod strike the floor simultaneously in the platform frame?

Also, you need to read post 160, because your solution has an obvious flaw.

To settle actual data before applying theory, can we use a train carriage with one sidewall missing and open, and roof overhanging platform such that normal but open side train floor moves past platform side like sliding contact? Then using two identical rods released identically from the roof fixings, such that one rod falls on train-side (train floor) of the sliding contact line, and the other rod falls on platform side (platform surface). When we compare rods we will see if either rod is actually bent or not and go from there in analysis? If neither rod bent then both rods fall level with both train floor level and platform surface level. If either bent then analysis should explain why the different outcome using physical reasons in practice and in theory?

 

[Fednis48]

$$y=tan(\theta)x$$ in frame S
$$y'=\gamma tan(\theta)x'$$ in frame S'

What does this tell you?

Ah, I see where you get that now. Thank you. The equation you're referencing, $$y'=\tan(\theta_1)\gamma x'$$, comes from the polygon model of an orbiting vector, which is later turned into a limit to produce a circular orbit. More specifically, it describes the bottom face of the polygon, where the vector is moving parallel to x. Along this surface, it makes sense that $$\theta=0$$ only if $$\theta'=0$$, because a vector Lorentz-boosted along its own length should simply contract. But in our situation, the vertical motion of the bar is a critical part of the situation. It would probably be more accurately described by the equation for y' immediately following the one you cited, which uses a face on the polygon that is not parallel to the x or y axes:

$$y'=\tan(\theta_2)(\frac{1-v^2\cos(\alpha)}{\gamma})x'+\frac{v^2x'}{\gamma} \sin(\alpha)$$

And surprise - that equation does not imply that $$\theta$$ can be zero in both frames at once.

On a different note, it's a little annoying to have you pose question after question to me without you ever responding to my own queries. I've now slogged through two sets of notes on Thomas precession which, while informative, are pretty dense. On the other hand, in post 151 I put forward a really simple argument for why the scenario you're arguing for leads to absurd results. Your only response to this so far was to complain that I didn't specify my reference frames, which wasn't even true. So before you throw any more notes at me, what, if anything, was wrong with my post 151?

 

[Fednis48]

To settle actual data before applying theory, can we use a train carriage with one sidewall missing and open, and roof overhanging platform such that normal but open side train floor moves past platform side like sliding contact? Then using two identical rods released identically from the roof fixings, such that one rod falls on train-side (train floor) of the sliding contact line, and the other rod falls on platform side (platform surface). When we compare rods we will see if either rod is actually bent or not and go from there in analysis? If neither rod bent then both rods fall level with both train floor level and platform surface level. If either bent then analysis should explain why the different outcome using physical reasons in practice and in theory?

I'm not sure I understand you exactly, but it sounds like you want to drop two bars from a stationary roof: one onto a stationary platform, and one onto a moving train. I can see where you're coming from, but this won't solve anything, because part of what makes the problem interesting is that the train-frame bar is dropped from a moving roof.

 

[Undefined]

I'm not sure I understand you exactly, but it sounds like you want to drop two bars from a stationary roof: one onto a stationary platform, and one onto a moving train. I can see where you're coming from, but this won't solve anything, because part of what makes the problem interesting is that the train-frame bar is dropped from a moving roof.

No, not from a stationary roof, from the roof of the moving train with one sidewall cut away like I described and its roof extended over the platform a little so as it passes the platform both rods are released from the moving roof and one falls train side and one falls platform side of the sliding contact line constituted by the floor of train and platform sliding past each other as train moves past.

 

[Pete]

To settle actual data before applying theory, can we use a train carriage with one sidewall missing and open, and roof overhanging platform such that normal but open side train floor moves past platform side like sliding contact? Then using two identical rods released identically from the roof fixings, such that one rod falls on train-side (train floor) of the sliding contact line, and the other rod falls on platform side (platform surface). When we compare rods we will see if either rod is actually bent or not and go from there in analysis? If neither rod bent then both rods fall level with both train floor level and platform surface level. If either bent then analysis should explain why the different outcome using physical reasons in practice and in theory?

After the rod lands, all observers agree that it is straight.
The only time it is bent is during the impact, as measured in the platform rest frame.
I might have to make some animations to illustrate this.

 

[Undefined]

After the rod lands, all observers agree that it is straight.
The only time it is bent is during the impact, as measured in the platform rest frame.
I might have to make some animations to illustrate this.

To make possible a naive understanding for myself based on real data, I would use two rods. One falls on "moving" train floor and one falls on "stationary" platform. We have to look at both rods. Not just the one rod you mention so far. Then we go from there like I said in my post about the two rod exercise? Then if both platform and train observer sees there respective rod as straight, then nothing really happens to bend either rod to prevent flat contact by rod on train or rod on platform. Is that naive inference incorrect?

 

[Fednis48]

No, not from a stationary roof, from the roof of the moving train with one sidewall cut away like I described and its roof extended over the platform a little so as it passes the platform both rods are released from the moving roof and one falls train side and one falls platform side of the sliding contact line constituted by the floor of train and platform sliding past each other as train moves past.

Oh, ok. That might be an interesting experiment. I still don't think it would shed light on the current debate, though, for the following reason. In the question at hand, we are comparing a rod dropped from a stationary roof onto a stationary floor (train frame) to one dropped from a moving roof onto a moving floor (platform frame). According to general relativity, the results of any experiments must agree in both frames, period; any explanation that predicts differing results is incorrectly derived. In your proposed experiment, then, any differences between the two rods' behaviors would have to come from the materials science of a moving object hitting a stationary floor rather than from general relativity.

Edit: Relevance to this thread aside, I would love to read a paper entitled "Deformation of a metal rod hitting a stationary train platform at relativistic speeds".

 

[Undefined]

Oh, ok. That might be an interesting experiment. I still don't think it would shed light on the current debate, though, for the following reason. In the question at hand, we are comparing a rod dropped from a stationary roof onto a stationary floor (train frame) to one dropped from a moving roof onto a moving floor (platform frame). According to general relativity, the results of any experiments must agree in both frames, period; any explanation that predicts differing results is incorrectly derived. In your proposed experiment, then, any differences between the two rods' behaviors would have to come from the materials science of a moving object hitting a stationary floor rather than from general relativity.

Edit: Relevance to this thread aside, I would love to read a paper entitled "Deformation of a metal rod hitting a stationary train platform at relativistic speeds".

You have something there. We should do both. The rods dropped from stationary platform overhanging an open sided train floor like you said before, and my moving train drop. The results from both dropping setups comparing two rods in either case would be interesting. Yes! Very strange to find relativistically moving trains on the local rail network! If only we could also find a relativistically moving platform we could do it all ways. Yessiree! But just thought experiment analysing reasons why both rods seem straight to respective observers where two rods are involved is enough for the moment I think to give us the clue where the essentials lay.

 

[Tach]

After the rod lands, all observers agree that it is straight.
The only time it is bent is during the impact, as measured in the platform rest frame.
I might have to make some animations to illustrate this.

No, there is no reason whatsoever why the rod would not stay bent AFTER impact.
To make matters worse, there is no reason why the rod would not pivot around the endpoint cut first (in the platform frame) and spiral down "head over heels", rather than fall parallel to itself (in the platform frame).
The reason that you get all these contradictions is that you (and others) are trying to FORCE the application of SR OUTSIDE its domain of applicability: the platform frame and the train frame are both accelerated frames so you cannot blindly apply RoS. It is not that RoS (or SR) are wrong, you are trying to apply them outside the domain of applicability , resulting into contradictory answers: RoS says that both ends cannot hit simultaneously (in the paltform frame), Thomas precession says that the rod is parallel to the platform, one observer measures the rod to be un-deformed, the other other one measures it to be deformed. And the list will continue for as long as you try to replace the actual problem (ACCELERATED fall in a GRAVITATIONAL field) with its dumbed down version.

 

[Tach]

Ah, I see where you get that now. Thank you. The equation you're referencing, $$y'=\tan(\theta_1)\gamma x'$$, comes from the polygon model of an orbiting vector, which is later turned into a limit to produce a circular orbit. More specifically, it describes the bottom face of the polygon, where the vector is moving parallel to x. Along this surface, it makes sense that $$\theta=0$$ only if $$\theta'=0$$, because a vector Lorentz-boosted along its own length should simply contract. But in our situation, the vertical motion of the bar is a critical part of the situation. It would probably be more accurately described by the equation for y' immediately following the one you cited, which uses a face on the polygon that is not parallel to the x or y axes:

$$y'=\tan(\theta_2)(\frac{1-v^2\cos(\alpha)}{\gamma})x'+\frac{v^2x'}{\gamma} \sin(\alpha)$$

And surprise - that equation does not imply that $$\theta$$ can be zero in both frames at once.

Err, we aren't talking about a polygon, we are talking about THE SPECIFIC SIDE $$y=$$constant. (i.e. motion perpendicular to the y axis), no?

On a different note, it's a little annoying to have you pose question after question to me without you ever responding to my own queries. I've now slogged through two sets of notes on Thomas precession which, while informative, are pretty dense. On the other hand, in post 151 I put forward a really simple argument for why the scenario you're arguing for leads to absurd results. Your only response to this so far was to complain that I didn't specify my reference frames, which wasn't even true. So before you throw any more notes at me, what, if anything, was wrong with my post 151?

See my answer to Pete. If you two guys want to continue trying to solve a problem DIFFERENT from the OP, i.e. a DUMBED down version, don't be surprised why you get contradictions.

 

[Tach]

duplicate

 

[Tach]

This is absurd; a metal bar pinned to an inertial reference frame at only one end will rotate under gravity, whether we use GR or SR. The puzzle, then, is not whether the two ends of the bar hit at different times, but how the two ends can hit at different times without deforming the final shape of the bar. I'm inclined to buy Janus' explanation for this.

...which is exactly my argument as well, except for two things:

1. You cannot and should not treat this problem as SR
2. I have pointed out to you, in response, twice already that I have pointed out errors in Janus' argument. Feel free to continue "buying it" blindly.

Now, for chuckles and grins a direct counterpoint to your angle-based counterpoint: in the (non-applicable) framework of SR, the rod is parallel to the car floor at all times in the car frame. The car floor is parallel to the platform in both the car and the platform frame. Therefore, the rod is parallel, at all times, to the platform. This answer coincides with the one based on Thomas precession and it is contradicted by the one based on RoS. You put garbage in, you get garbage out. You try to dumb down a problem, don't be surprised at the contradictory answers.

 

[Fednis48]

Err, we aren't talking about a polygon, we are talking about THE SPECIFIC SIDE $$y=$$constant. (i.e. motion perpendicular to the y axis), no?

Well, the equation you quoted was describing the $$y=$$constant side of a polygon, so the answer to the first part of your question might be yes or no depending on how you meant it. But what's crucial is that whether or not it's on a polygon, a $$y=$$constant surface does not describe the interesting physics of the problem at hand. In order to ever reach the floor, the rod must be moving non-parallel to the floor, which means that the equations after the one you cited come into play.

See my answer to Pete. If you two guys want to continue trying to solve a problem DIFFERENT from the OP, i.e. a DUMBED down version, don't be surprised why you get contradictions.

You can certainly argue that the problem with a rod falling under gravity is qualitatively different from the case of a rod moving toward the ground at constant velocity with no gravity. But both cases are certainly physical. Solving the constant-velocity case with SR might give a different result than the GR in gravity case, but if you get a contradiction when you solve the SR constant-velocity case, you're doing something wrong.

...which is exactly my argument as well, except for two things:

1. You cannot and should not treat this problem as SR
2. I have pointed out to you, in response, twice already that I have pointed out errors in Janus' argument. Feel free to continue "buying it" blindly.

1. My argument from post 151 doesn't assume SR. All it assumes is that the height of the rod is a monotonically decreasing function of time, and that the wires are cut at different times in the platform frame. These by themselves, which are entirely consistent with GR, lead to a contradiction if you assume the rod hits the ground flat in the platform frame.
2. I'm trying to take one point at a time, demonstrating that Thomas precession is an inadequate description before moving on to other possibilities. But if you really want me to defend Janus' argument, here goes:

You say that Janus is wrong about which side of the rod that hits the ground first. You say that c is a limiting overestimate of the propagation speed of a momentum shock, and that $$c-v$$ is a vastly oversimilified expression for propagation speed in a moving frame. The thing is, none of these points really get in the way of Janus' central conclusion: if the momentum shock propagation speed is at most $$c-v$$ (which it must be, to preserve causality), consecutive segments of the rod hit the ground before the momentum shock gets to them. Now, you're quite right that proving this rigorously would be hard. It would require a theory of relativistic materials science, which you yourself said probably doesn't exist yet. But it still makes gut-level sense to me, and though you've pointed out plenty of reasons it's not rigorous, you have yet to show that its basic thesis is wrong. On the other hand, I'm still convinced that I've shown Thomas precession is definitely not the whole story, because assuming it is leads to absurd scenarios.

Now, for chuckles and grins a direct counterpoint to your angle-based counterpoint: in the (non-applicable) framework of SR, the rod is parallel to the car floor at all times in the car frame. The car floor is parallel to the platform in both the car and the platform frame. Therefore, the rod is parallel, at all times, to the platform. This answer coincides with the one based on Thomas precession and it is contradicted by the one based on RoS. You put garbage in, you get garbage out. You try to dumb down a problem, don't be surprised at the contradictory answers.

I don't think parallel-ness is transitive in relativity. In the car frame, the rod, the car floor, and the platform are all parallel. In the platform frame, the car floor and the platform are parallel, while the rod is at an angle to both. There's no contradiction there.