Ehrenfest / rod thought experiment.

In summary: The original claim was that if you had some sort of accelerator device in a lab of radius r meters, and some rods with rest lengths of Pi meters each, and could comfortably fit more than 2*r rods around the perimeter, then the rods would expand when they were accelerated. This is not a valid thought experiment, because the rods do not expand because they are not linked to each other. In the Ehrenfest paradox, the train carriages are more rigid than the elastic links between the carriages and so the links have to expand to take up the extra space between the carriages at higher velocities. In the rod thought experiment, it is the gaps between the rods that expands as the rods contract and so you can fit
  • #1
yuiop
3,962
20
This is new thread on an issue that was that getting slightly off topic in the original thread.

yuiop said:
What if we had some sort of accelerator device in a lab of radius r meters, and some rods with rest lengths of Pi meters each and what if we could comfortably fit more than 2*r rods around the perimeter if they were moving at high enough velocity

starthaus said:
Once you start accelerating the rods around the perimeter they expand since they are not Born rigid, so you could not fit in more than 2*r rods. This is not a valid thought experiment for demonstrating length contraction. The above is a variant of the Ehrenfest paradox and the resolution is the same.

yuiop said:
This is just nonsense. The rods do not expand because they are not linked to each other. In the Ehrenfest paradox the train carriages are more rigid than the elastic links between the carriages and so the links have to expand to take up the extra space between the carriages at higher velocities. In the rod thought experiment, it is the gaps between the rods that expands as the rods contract and so you can fit additional rods in the gaps. I don't think Born rigidity is even relevant here. Born rigidity is a method of acceleration, rather than a property of a material. It does not matter how the rods are accelerated as long as they get to a final velocity and are allowed to stabilize to their length contracted length. It is only when we are considering methods of spinning up solid discs to relativistic speeds that applying Born rigid acceleration becomes a problem. For rods that are not connected to each other, it is not a problem.

If you wanted to make the Ehrenfest paradox a bit more like the rod thought experiment, you could remove one link so that the train is not connected all the way around, but occupies all the track and is is just touching at the point where the link has been removed. When the train gets to high enough velocity it will have length contracted sufficiently that you could (in principle but with considerable technical difficulty) slip another (high speed) carriage onto the track into the expanded gap.

Passionflower said:
I think this topic is not the place to discuss the Erhenfest paradox, but one comment: how can a rod remain Born rigid if it rotates? I thought only rods that accelerate in one direction can remain Born rigid? It seems to me that because the rod is spatially separated there will be some form of Thomas precession which will make Born rigidity impossible.

Am I wrong?

DaleSpam said:
You are correct. Linear acceleration can be done without material strain (Born rigid), but there is no way to have angular acceleration without mechanical strain. Something must stretch.


Let's consider a slight variation of the Ehrenfest experiment. The fairly rigid carriages are all linked together by elastic couplings and are on a suitably highly banked track. As the velocity of the train increases, the elastic couplings get progressively more stretched putting a measurable strain on the carriages. At high enough velocity, the strain on the couplings get so high that they all snap. Once the couplings have snapped the stretching strain on the carriages vanishes and we end up with essentially the rod thought experiment I first proposed with no longitudinal strain parallel to the track, but there will of course be transverse strain as the carriages/ rods will of course be compressed down on to the track by the reaction force to the centripetal force exerted by the track. The transverse strain is not an issue here because I am only considering longitudinal length contraction.

The Thomas rotation is also not an issue here, because the orientation of the carriages/ rods is maintained by the banked track.

The key issue here is, can you in principle fit more carriages on the track when they are moving at relativistic velocities, than the number of carriages that will fit on the same track when they are at rest wrt the track?
 
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  • #2
yuiop said:
The key issue here is, can you in principle fit more carriages on the track when they are moving at relativistic velocities, than the number of carriages that will fit on the same track when they are at rest wrt the track?
Yes. Although I am not sure why this is "key".
 
  • #3
yuiop said:
The key issue here is, can you in principle fit more carriages on the track when they are moving at relativistic velocities, than the number of carriages that will fit on the same track when they are at rest wrt the track?
DaleSpam said:
Yes. Although I am not sure why this is "key".
My original claim was:
yuiop said:
... we could comfortably fit more than 2*r rods around the perimeter if they were moving at high enough velocity
and Starthaus's counterargument was:
starthaus said:
Once you start accelerating the rods around the perimeter they expand since they are not Born rigid, so you could not fit in more than 2*r rods. This is not a valid thought experiment for demonstrating length contraction. The above is a variant of the Ehrenfest paradox and the resolution is the same.
Obviously we are diametrically opposed on the issue of how many moving rods/ carriages that will comfortably fit on the track / perimeter and I saw this as the key issue that needs resolving. You seem to be agreeing with my original claim. I also think it is a demonstration of length contraction, but whether it a demonstration of the "reality" of length contraction is a mute point. It certainly, to me provides a nice way to visualize length contraction, when the carriages or rods are considered as rulers, observers at wrt the track and observers at rest wrt the train all agree that the proper length of the train (with additional carriages) is longer than the proper length of the track it is riding on.
 
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  • #4
There doesn't seem to be a contradiction between your claim and starthaus's. If your train cars are linked together all the way around the circumference of the circular track, they will be forcefully stretched. If they're not linked together, and accelerate individually using their own engines for a while, you will get some extra space between them.
 
  • #5
DaleSpam said:
You are correct. Linear acceleration can be done without material strain (Born rigid), but there is no way to have angular acceleration without mechanical strain. Something must stretch.
How are you defining "material strain"? I would have thought that Born rigid acceleration would involve nonzero stress throughout the object since some parts of the object are experiencing greater G-forces than others, it's just that the internal forces have reached a sort of equilibrium where the stress doesn't change with time. It's sort of like how if you have an upright spring at rest on the surface of the Earth and eventually the spring should stop oscillating and reach an equilibrium where the coils nearer the bottom are more compressed than the coils nearer the top. And couldn't the same sort of equilibrium in stresses be reached by an object rotating around a central point with constant angular velocity?
 
  • #6
Fredrik said:
There doesn't seem to be a contradiction between your claim and starthaus's. If your train cars are linked together all the way around the circumference of the circular track, they will be forcefully stretched. If they're not linked together, and accelerate individually using their own engines for a while, you will get some extra space between them.

My original claim was for rods that are not linked together and although I did not make it clear that the rods are not linked, I probably would have said they are linked or chained together if that was what I intended. Anyway, if with the clarification that the rods are not all inked together, I wonder if Starthaus would now agree that my original claim that there would be extra space between the rods, that you you could fit more similar moving rods into?
 
  • #7
JesseM said:
How are you defining "material strain"? I would have thought that Born rigid acceleration would involve nonzero stress throughout the object since some parts of the object are experiencing greater G-forces than others, it's just that the internal forces have reached a sort of equilibrium where the stress doesn't change with time. It's sort of like how if you have an upright spring at rest on the surface of the Earth and eventually the spring should stop oscillating and reach an equilibrium where the coils nearer the bottom are more compressed than the coils nearer the top. And couldn't the same sort of equilibrium in stresses be reached by an object rotating around a central point with constant angular velocity?

I tend to agree with your thoughts here about the stresses. There are non-zero material stresses in the linear acceleration of a rocket and the angular acceleration of a cylinder around its long axis of symmetry. If strain is measured by mechanical devices such as rulers that are subject to the same acceleration profile, then the strain appears to be zero in both cases. On the other hand, while the length of the linearly accelerating rocket as measured by rulers is in agreement with the radar length of the rocket, the perimeter length of the rotating cylinder as measured by the sum of a series of radar measurements is longer than perimeter length as measured by the sum of a series of rulers. So if we define material strain as radar length versus ruler length, then Dalespam is correct that the rotating cylinder perimeter has measurable strain, while the linearly accelerating rocket does not. It is as you say, dependent on how you define or measure strain.
 
  • #8
JesseM said:
How are you defining "material strain"?
A change in the proper distance between different particles in the material.

JesseM said:
I would have thought that Born rigid acceleration would involve nonzero stress throughout the object since some parts of the object are experiencing greater G-forces than others, it's just that the internal forces have reached a sort of equilibrium where the stress doesn't change with time.
No, using the above definition of strain, Born rigid acceleration is strain-free by definition.

JesseM said:
It's sort of like how if you have an upright spring at rest on the surface of the Earth and eventually the spring should stop oscillating and reach an equilibrium where the coils nearer the bottom are more compressed than the coils nearer the top. And couldn't the same sort of equilibrium in stresses be reached by an object rotating around a central point with constant angular velocity?
Sure, those are all in equilibrium but all those equilibria are strained states. Equilibrium and strain are separate and independent concepts.
 
  • #9
yuiop said:
while the length of the linearly accelerating rocket as measured by rulers is in agreement with the radar length of the rocket.
Are you claiming that radar distance and ruler distance between the front and back of a Born rigid rocket undergoing constant proper acceleration is equal ?

Furthermore, do you agree or disagree that rods around the perimeter cannot be Born rigid because they are rotating?
 
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  • #10
I think it actually is possible to impart an angular acceleration to a Born-rigid one-dimensional object, just not to a Born-rigid two-dimensional object. The argument that it's impossible to have an angular acceleration is given in Ø. Grøn, Relativistic description of a rotating disk, Am. J. Phys. 43 869 (1975), and I think it depends on the assumption that the object encloses some area.

I think the notion of a Born-rigid ruler is exactly equivalent to the notion of measuring distances by radar; anything that can be done with one technique can be done with the other. The difference is that radar (a) actually exists, and (b) doesn't trap unwary people into incorrect arguments.
 
  • #11
DaleSpam said:
A change in the proper distance between different particles in the material.
Why would this necessarily be true for an object rotating at a constant angular speed?
JesseM said:
It's sort of like how if you have an upright spring at rest on the surface of the Earth and eventually the spring should stop oscillating and reach an equilibrium where the coils nearer the bottom are more compressed than the coils nearer the top. And couldn't the same sort of equilibrium in stresses be reached by an object rotating around a central point with constant angular velocity?
DaleSpam said:
Sure, those are all in equilibrium but all those equilibria are strained states. Equilibrium and strain are separate and independent concepts.
Well, for the spring in the gravitational field it would also be an equilibrium in your sense, right? There'd be no change in the proper distance between particles over time? If so, why couldn't a similar equilibrium be reached by a rotating object? For example, if we have a wheel-shapes space station rotating at a constant rate, why couldn't the proper distance between any given pair of particles be constant?
 
  • #12
Passionflower said:
Are you claiming that radar distance and ruler distance between the front and back of a Born rigid rocket undergoing constant proper acceleration is equal ?

Yes, I am. They could calibrate a short rulers using a radar measurement and when they lay these rulers end to end to measure the length of the rocket, the ruler length would agree with radar length of the rocket. If you are asking if the ruler length of the accelerating rocket is the same as the ruler length and radar length of the rocket when it was not accelerating, that is a more technical question, because the transfer from not accelerating to accelerating is not constant acceleration and Born rigid motion requires constant acceleration. A rocket with constant Born rigid acceleration was never not accelerating by definition.

If the ruler length and radar length length of a rocket is measured while it is initially at rest and not accelerating, it will find that its radar length is shorter after it starts accelerating, even with Born rigid acceleration. What it finds its ruler length to be depends on how and where the rulers are mounted on the rocket and how the rulers are accelerated. For Born rigid motion, every atom of the rocket has to accelerated by the right amount, so each atom technically needs its own rocket motor. If the rulers are accelerated in the same manner they will obviously not notice any change in the length of the rocket. However, if they check the calibration of the Born rigid accelerating rulers with radar, they will find they are shorter than when the rocket was not accelerating. Consider one ruler mounted only at the nose of the rocket and one ruler mounted only at the tail of the rocket. They overlap at the middle.When the rocket is not accelerating, they put a aligned marks on the two rulers where they overlap and another mark on the wall aligned with the first two on the rulers. If the rockets takes off and goes to Born rigid acceleration, and if no no special acceleration is applied to the rulers, the nose mounted one will stretch under the acceleration forces and tail mounted ruler will compress. Now while the two aligned marks on the rulers may still be aligned they will not still be aligned with the mark on the wall. This means strain and stress can be detected in a rocket that goes from a state of not being accelerated to a state of being accelerated even if the acceleration phase is Born rigid.
Passionflower said:
Furthermore, do you agree or disagree that rods around the perimeter cannot be Born rigid because they are rotating?
I never claimed that the rods around the perimeter are Born rigid and never claimed Born rigidity is an important part of the experiment. Quite the opposite. I claimed you could fit more moving rods around the perimeter than you can fit stationary rods around the perimeter. The moving rods are not squashed together because they are not touching. If you slowed down the moving rods they would expand and eventually when they came to a stop they would be squashed up together, with no gaps and under serious longitudinal compression.
 
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  • #13
OK, here's a more formal argument that it is possible to give an angular acceleration to a one-dimensional Born-rigid ruler.

Let a one-dimensional ruler be initially rotating at a certain angular velocity with one of its end-points fixed at the origin. At times judged to be simultaneous in the nonrotating lab frame, apply impulses at evenly spaced points P1, ... Pn along the ruler, in such a way that the lab observer sees the result as an impulsive angular acceleration about the origin. To prove that this is consistent with Born rigidity, we have to prove that the result of this is to preserve the distance between Pk and Pk+1 as measured using radar by an observer comoving with Pk and Pk+1. (If n is made large enough, it makes sense to have an observer who is approximately comoving with both of these points.) But this is certainly true, because the effect on Pk and Pk+1 is simply a Lorentz boost in the direction perpendicular to the line connecting them. The Lorentz boost does not affect the simultaneity of the two impulses, because the two events are separated in space along a line perpendicular to the boost. Nor does the boost affect the radar distance between the two points, because Lorentz contraction doesn't apply along a line perpendicular to the boost.
 
  • #14
JesseM said:
Why would this necessarily be true for an object rotating at a constant angular speed?
Sorry if I caused any confusion. You can have Born rigid constant angular velocity. You cannot have Born rigid angular acceleration.
 
  • #15
DaleSpam said:
You cannot have Born rigid angular acceleration.

...of an object that encloses a finite area.
 
  • #16
bcrowell said:
...of an object that encloses a finite area.
I don't know about that. The derivations I have seen have been for cylindrically symmetric objects, but I didn't find the above very convincing. I don't want to make a claim either way on a rod.
 
  • #17
DaleSpam said:
I didn't find the above very convincing.

Which part of it did you find unconvincing?
 
  • #18
yuiop said:
My original claim was for rods that are not linked together and although I did not make it clear that the rods are not linked, I probably would have said they are linked or chained together if that was what I intended. Anyway, if with the clarification that the rods are not all inked together, I wonder if Starthaus would now agree that my original claim that there would be extra space between the rods, that you you could fit more similar moving rods into?
Whether they are linked together or not, to keep them in place around the perimeter would require restraint of some kind yes??
They would be subjected to the same outward inertial forces in effect on the disk of the Ehrenfest conditions wouldn't they?.
SO would those inertial forces cancel or mitigate the contraction or not??
starthoaus seems to be saying they would negate the contraction entirely.
From what I have read of the resolutions to the original Ehrenfest scenario it appears they would be a significant factor in all cases. Is this wrong?
 
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  • #19
Originally Posted by JesseM
I would have thought that Born rigid acceleration would involve nonzero stress throughout the object since some parts of the object are experiencing greater G-forces than others, it's just that the internal forces have reached a sort of equilibrium where the stress doesn't change with time.
Originally Posted by JesseM
It's sort of like how if you have an upright spring at rest on the surface of the Earth and eventually the spring should stop oscillating and reach an equilibrium where the coils nearer the bottom are more compressed than the coils nearer the top. And couldn't the same sort of equilibrium in stresses be reached by an object rotating around a central point with constant angular velocity?

DaleSpam said:
A change in the proper distance between different particles in the material.

No, using the above definition of strain, Born rigid acceleration is strain-free by definition..
Hi
I have questions
As I understand the Born hypotheses it proposes to eliminate the strain of contraction by controlled distributed acceleration, at the same time countering an assumed force of disruptive expansion.
If this is the case then, even given an ideal [impossible] perfect control of force, wouldn't the best result acheivable be an equilibrium??
But can a balance of counter forces ever be strain free??
What is the meaning of proper distance in this context?
In the context of inertial frames the meaning is clear but here it seems to be without a reference of any kind.
DaleSpam said:
Sure, those are all in equilibrium but all those equilibria are strained states. Equilibrium and strain are separate and independent concepts.
But isn't an accelerated state equialent to the exact conditions JesseM is referring to [in gravity]??
 
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  • #20
yuiop said:
I tend to agree with your thoughts here about the stresses. There are non-zero material stresses in the linear acceleration of a rocket and the angular acceleration of a cylinder around its long axis of symmetry. If strain is measured by mechanical devices such as rulers that are subject to the same acceleration profile, then the strain appears to be zero in both cases. On the other hand, while the length of the linearly accelerating rocket as measured by rulers is in agreement with the radar length of the rocket, the perimeter length of the rotating cylinder as measured by the sum of a series of radar measurements is longer than perimeter length as measured by the sum of a series of rulers. So if we define material strain as radar length versus ruler length, then Dalespam is correct that the rotating cylinder perimeter has measurable strain, while the linearly accelerating rocket does not. It is as you say, dependent on how you define or measure strain.
WHy would the radar length with rotation be longer than the ruler length?
WHy would the radar length agree with rulers with linear acceleration?
Aren't the clock rates different at the front and the back in a Born rigid rocket?
Radar distance longer from the Front to Back to Front , than from B->F->B ??
 
  • #21
Austin0 said:
Whether they are linked together or not, to keep them in place around the perimeter would require restraint of some kind yes??

Yep, that is why I said:
yuiop said:
The ... carriages ... are on a suitably highly banked track.
The "suitably highly banked track" is the restraint. [/quote]

Austin0 said:
They would be subjected to the same outward inertial forces in effect on the disk of the Ehrenfest conditions wouldn't they?.

They would be subject to outward inertial forces. That is why I said:
yuiop said:
... there will of course be transverse strain as the carriages/ rods will of course be compressed down on to the track by the reaction force to the centripetal force exerted by the track. The transverse strain is not an issue here because I am only considering longitudinal length contraction.

By the clock hypothesis, the ticking rates of the the clocks on the perimeter of the disk is purely a function of instantaneous tangential velocity. The same is true for length contraction. Those that say there is an additional effect due to acceleration over and above the velocity effects that requires GR to analyse are confused. Wikipedia is wrong on this point. It is a purely SR experiment. A real experiment was carried out that stored muons in a cyclotron at relativistic speeds. The time dilation (extended half lives) was exactly what SR predicts due to velocity alone, despite the fact the muons were subjected to centripetal forces of the millions of g. You can analyse the time dilation and effects as being due to velocity or due to pseudo-gravitational effects and get the same answers, but you cannot do both at the same time and add them together or get them cancel out. On top of that, the Schwarzschild metric does not predict length contraction for stationary rods that are horizontal to the gravitational field.

Austin0 said:
SO would those inertial forces cancel or mitigate the contraction or not??
No.

Austin0 said:
starthoaus seems to be saying they would negate the contraction entirely.
He would, but he has not shown any convincing arguments or calculations of how that would come about.

Austin0 said:
From what I have read of the resolutions to the original Ehrenfest scenario it appears they would be a significant factor in all cases. Is this wrong?

From http://en.wikipedia.org/wiki/Ehrenfest_paradox:
1916: While writing up his new general theory of relativity, Albert Einstein notices that disk-riding observers measure a longer circumference, C′ = 2π r /√(1−v2). That is, because rulers moving parallel to their length axis appear shorter as measured by static observers, the disk-riding observers can fit more small rulers of a given length around the circumference than stationary observers could.
I am backing Einstein on this one. Are you saying he is wrong about his own theory?
 
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  • #22
yuiop said:
Passionflower said:
Are you claiming that radar distance and ruler distance between the front and back of a Born rigid rocket undergoing constant proper acceleration is equal ?
Yes, I am. They could calibrate a short rulers using a radar measurement and when they lay these rulers end to end to measure the length of the rocket, the ruler length would agree with radar length of the rocket. If you are asking if the ruler length of the accelerating rocket is the same as the ruler length and radar length of the rocket when it was not accelerating, that is a more technical question, because the transfer from not accelerating to accelerating is not constant acceleration and Born rigid motion requires constant acceleration. A rocket with constant Born rigid acceleration was never not accelerating by definition.

If the ruler length and radar length length of a rocket is measured while it is initially at rest and not accelerating, it will find that its radar length is shorter after it starts accelerating, even with Born rigid acceleration. What it finds its ruler length to be depends on how and where the rulers are mounted on the rocket and how the rulers are accelerated. For Born rigid motion, every atom of the rocket has to accelerated by the right amount, so each atom technically needs its own rocket motor. If the rulers are accelerated in the same manner they will obviously not notice any change in the length of the rocket. However, if they check the calibration of the Born rigid accelerating rulers with radar, they will find they are shorter than when the rocket was not accelerating. Consider one ruler mounted only at the nose of the rocket and one ruler mounted only at the tail of the rocket. They overlap at the middle.When the rocket is not accelerating, they put a aligned marks on the two rulers where they overlap and another mark on the wall aligned with the first two on the rulers. If the rockets takes off and goes to Born rigid acceleration, and if no no special acceleration is applied to the rulers, the nose mounted one will stretch under the acceleration forces and tail mounted ruler will compress. Now while the two aligned marks on the rulers may still be aligned they will not still be aligned with the mark on the wall. This means strain and stress can be detected in a rocket that goes from a state of not being accelerated to a state of being accelerated even if the acceleration phase is Born rigid.
Sorry yuiop, it has been a long time since I have seen such a demonstration of trying to maintain an untenable position by obfuscation and redefining.

Facts:
  • A Born rigid rocket with length X,remains X during constant proper acceleration by definition!
  • The radar distance measured by an observer at the front and at the back is different, the observer at the front would measure the distance longer and the observer at the back would measure a distance shorter compared to the rocket distance in an inertial frame.
yuiop said:
I never claimed that the rods around the perimeter are Born rigid
I went back to the original thread to discover what prompted me to ask you this question, I did notice you edited your posting recently. Since I do not remember the original posting I am going to assume it was I who was mistaken and supposedly completely misread what you wrote.
 
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  • #23
Austin0 said:
Whether they are linked together or not, to keep them in place around the perimeter would require restraint of some kind yes??

yuiop said:
Yep, that is why I said:
yuiop said:
The ... carriages ... are on a suitably highly banked track.

The "suitably highly banked track" is the restraint.
I was talking about rods or rulers that would need restraint that was itself rotating and subject to the same inertial forces. As per the original conditions.
You are now talking about a situation where the restrainst is external to the moving system , a radically different set of conditions. Here the outward expansion of the rods is drastically constrained.



Austin0 said:
They would be subjected to the same outward inertial forces in effect on the disk of the Ehrenfest conditions wouldn't they?.

yuiop said:
They would be subject to outward inertial forces. That is why I said:
yuiop said:
... there will of course be transverse strain as the carriages/ rods will of course be compressed down on to the track by the reaction force to the centripetal force exerted by the track. The transverse strain is not an issue here because I am only considering longitudinal length contraction.

Austin0 said:
WHy would the radar length with rotation be longer than the ruler length?
WHy would the radar length agree with rulers with linear acceleration?
Aren't the clock rates different at the front and the back in a Born rigid rocket?
Radar distance longer from the Front to Back to Front , than from B->F->B ??

yuiop said:
By the clock hypothesis, the ticking rates of the the clocks on the perimeter of the disk is purely a function of instantaneous tangential velocity. The same is true for length contraction. Those that say there is an additional effect due to acceleration over and above the velocity effects that requires GR to analyse are confused. Wikipedia is wrong on this point. It is a purely SR experiment. A real experiment was carried out that stored muons in a cyclotron at relativistic speeds. The time dilation (extended half lives) was exactly what SR predicts due to velocity alone, despite the fact the muons were subjected to centripetal forces of the millions of g. You can analyse the time dilation and effects as being due to velocity or due to pseudo-gravitational effects and get the same answers, but you cannot do both at the same time and add them together or get them cancel out. On top of that, the Schwarzschild metric does not predict length contraction for stationary rods that are horizontal to the gravitational field.
""WHy would the radar length with rotation be longer than the ruler length?""
You seem to have misunderstood my question , since all of the above was exactly my point.
I.e. If the clocks used to measure radar length are dilated then this would mean that they would measure any length as shorter , no?
So if contraction and dilation are effected by the same factor then the radar measured length of a contracted ruler should be even shorter than the metric length indicated on that ruler , is this not so?


Austin0 said:
SO would those inertial forces cancel or mitigate the contraction or not??

yuiop said:
No.

Would you still say this with regard to rods and disk without external restraint??

Austin0 said:
From what I have read of the resolutions to the original Ehrenfest scenario it appears they would be a significant factor in all cases. Is this wrong?

yuiop said:
From http://en.wikipedia.org/wiki/Ehrenfest_paradox:

1916: While writing up his new general theory of relativity, Albert Einstein notices that disk-riding observers measure a longer circumference, C′ = 2π r √(1−v2)−1. That is, because rulers moving parallel to their length axis appear shorter as measured by static observers, the disk-riding observers can fit more small rulers of a given length around the circumference than stationary observers could.

I am backing Einstein on this one. Are you saying he is wrong about his own theory?
It is not me questioning Einstein's analysis in this question. I have been so far , unable to find enough actual material about the various resolutions to even have an idea. But having found some general references to the history of the question it is clear that many qualified people found the 1916 conclusion lacking and it has been a subject of controversy for many years after that. Among the various different perspectives listed was mentioned taking more physical parameters into consideration than was originally factored. I never found the final and definitve resolution which seemed to involve Fermi observers and other complex coordinates.
So I am just trying to get the consensus picture here. I will say that in the 1916 quote above there seems to be no accomodation or regard for how the rulers were to be kept in place for measurements or any consideration of the forces acting on them.
So I am just curious as to what the definitve analysis concludes. ANd whose it is?
 
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  • #24
Passionflower said:
Facts:
  • A Born rigid rocket with length X,remains X during constant proper acceleration by definition!
  • The measured radar distance during acceleration between the front and the back of the rocket is no longer X/c.
  • The radar distance measured by an observer at the front and at the back is different.
Length X remains constant by definition but what does that actually mean?
It still is contracted relative to other frames correct?
So even if possible, Born rotating rods would be the same as any other accelerated rods , no?
The radar distance is no longer X/c so by what reference is it determined that X remains the same?
In fact it appears that to some extent rulers also have a different length at the front and the back even if the difference is negligable so proper distance seems to be a problematic concept in all ways or am I missing something here, Again?
 
  • #25
Austin0 said:
am I missing something here, Again?
The relative velocity between the front and the back of the rocket is zero at all times, hence the length of the rocket remains the same. This is fairly standard material.

I suggest you look up Born rigid motion or Bell's spaceship paradox (where the relative velocity is not equal to zero) to get more info if you need to.
 
  • #26
Passionflower said:
The relative velocity between the front and the back of the rocket is zero at all times, hence the length of the rocket remains the same. This is fairly standard material.

I suggest you look up Born rigid motion or Bell's spaceship paradox (where the relative velocity is not equal to zero) to get more info if you need to.
I have researched Born rigid acceleration and I don't understand your statement that the relative velocity between the front and the back is zero when there is both unequal acceleration and unequal distance traveled for the front and the back.
Besides which I never mentioned relative velocity so why did you bring it up?

Also unequal dilation and contraction at front and back. It has been stated by many others [not me] that, in fact, a Born rigid system cannot be represented by a single CMIRF for this reason,,,are you saying this is a concept totally without merit?
Here again you say the length remains the same but do not specify the same relative to what?
 
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  • #27
Austin0 said:
I have researched Born rigid acceleration and I don't understand your statement that the relative velocity between the front and the back is zero when there is both unequal acceleration and unequal distance traveled for the front and the back.
Besides which I never mentioned relative velocity so why did you bring it up?
Well I can try one more time, but it appears I am not very successful in explaining things to you.

The relative velocity between the front and the back is indeed zero and the proper acceleration of the front and the back is not equal.That is standard special relativity. If you wish to learn why I can try to help but if you wish to argue against it than I am sorry but I have no desire to debunk those arguments.

Austin0 said:
Besides which I never mentioned relative velocity so why did you bring it up?
In my, apparently unsuccessful, attempt to actually help you.
 
  • #28
FAQ: How is Ehrenfest's paradox resolved?

As described in [Einstein 1916], the relativistic rotating disk was an example that was influential in leading Einstein to describe gravity in terms of curved spacetime. Einstein writes:

"In a space which is free of gravitational fields we introduce a Galilean system of reference K (x,y,z,t), and also a system of coordinates K' (x',y',z',t') in uniform rotation relative to K. Let the origins of both systems, as well as their axes of Z, permanently coincide. We shall show that for a space-time measurement in the system K' the above definition of the physical meaning of lengths and times cannot be maintained. For reasons of symmetry it is clear that a circle around the origin in the X, Y plane of K may at the same time be regarded as a circle in the X', Y' plane of K'. We suppose that the circumference and diameter of this circle have been measured with a unit measure infinitely small compared with the radius, and that we have the quotient of the two results. If this experiment were performed with a measuring-rod at rest relative to the Galilean system K, the quotient would be π. With a measuring-rod at rest relative to K', the quotient would be greater than π. This is readily understood if we envisage the whole process of measuring from the stationary'' system K, and take into consideration that the measuring-rod applied to the periphery undergoes a Lorentzian contraction, while the one applied along the radius does not."

Einstein's friend Paul Ehrenfest posed the following paradox [Ehrenfest 1909]. Suppose that observer L, in the lab frame, measures the radius of the disk to be r when the disk is at rest, and r' when the disk is spinning. L can also measure the corresponding circumferences C and C'. Because L is in an inertial frame, the spatial geometry does not appear non-Euclidean according to measurements carried out with his meter sticks, and therefore the Euclidean relations C=2πr and C'=2πr' both hold. The radial lines are perpendicular to their own motion, and they therefore have no length contraction, r=r', implying C=C'. The outer edge of the disk, however, is everywhere tangent to its own direction of motion, so it is Lorentz contracted, and therefore C' is less than C.

The resolution of the paradox is that it rests on the incorrect assumption that a rigid disk can be made to rotate. If a perfectly rigid disk was initially not rotating, one would have to distort it in order to set it into rotation, because once it was rotating its outer edge would no longer have a length equal to 2π times its radius. Therefore if the disk is perfectly rigid, it can never be set into rotation.

Thorough modern analyses are available,[Grøn 1975,Dieks 2009] and in particular it is not controversial that, as claimed in [Einstein 1916], C/r is measured to be *greater* than 2π by an observer in the rotating frame.

A common source of confusion in discussions of Ehrenfest's paradox is the role of the rigid meter-sticks, since it is not clear whether sufficiently rigid meter-sticks can exist, or how to verify that they have remained rigid. This confusion can be avoided simply by replacing the meter-stick measurements with radar measurements.

In connection with these discussions, one often hears about the concept of a Born-rigid object, meaning an object that is subject to prearranged external forces in such a way that observers moving with the object find radar distances between points on the object to remain constant. It is kinematically impossible to impart an angular acceleration to a Born-rigid disk,[Grøn 1975] and therefore it is also impossible to do so for any plane figure that encloses a finite area, since it would enclose a disk. The reason for this is that in order to maintain Born-rigidity, the torques would have to be applied simultaneously at all points on the perimeter of the area, but Einstein synchronization (i.e., synchronization by radar) is not transitive in a rotating frame; that is, if A is synchronized with B, and B with C, then C will not be synchronized with A if the triangle ABC encloses a nonzero area and is rotating. (This does not make it impossible to manipulate the rigid meter-sticks as described in [Einstein 1916], since they can be one-dimensional, and therefore need not enclose any area.)

A. Einstein, "The foundation of the general theory of relativity," Annalen der Physik, 49 (1916) 769; translation by Perret and Jeffery available in an appendix to the book at http://www.lightandmatter.com/genrel/ (PDF version)

P. Ehrenfest, Gleichförmige Rotation starrer Körper und Relativitätstheorie, Z. Phys. 10 (1909) 918

Ø. Grøn, Relativistic description of a rotating disk, Am. J. Phys. 43 (1975) 869

Dieks, "Space, Time, and Coordinates in a Rotating World," in Rizzi and Ruggiero, ed., Relativity in Rotating Frames: Relativistic Physics in Rotating Reference Frames, 2009, http://www.phys.uu.nl/igg/dieks/rotation.pdf
 
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  • #29
bcrowell said:
Which part of it did you find unconvincing?
That you didn't calculate the proper distances between neighboring particles before and after.
 
  • #30
Austin0 said:
What is the meaning of proper distance in this context?
Take a particle's worldline, pick some event on that worldline, and find the tangent vector at that event. Then form the hyperplane which is orthogonal to the tangent vector. Find the event that forms the intersection of a neighboring particle's worldline with that hyperplane. Then calculate the spacetime interval between the two events. That is the proper distance between two neighboring particles. I should mention that it needs to be a differential distance, not a large distance.

Born rigidity is a kinematic condition, not a dynamic condition. I.e. it is a statement about the motion of each particle, where it is at any given instant in time relative to its neighbors. It is not a statement in any way about forces acting on the particles. I have hopefully consistently used the word "strain" and avoided the words "stress" or "force".
 
  • #31
Austin0 said:
WHy would the radar length with rotation be longer than the ruler length?
WHy would the radar length agree with rulers with linear acceleration?
Aren't the clock rates different at the front and the back in a Born rigid rocket?
Radar distance longer from the Front to Back to Front , than from B->F->B ??

Passionflower said:
Sorry yuiop, it has been a long time since I have seen such a demonstration of trying to maintain an untenable position by obfuscation and redefining.

Facts:
  • A Born rigid rocket with length X,remains X during constant proper acceleration by definition!
  • The radar distance measured by an observer at the front and at the back is different, the observer at the front would measure the distance longer and the observer at the back would measure a distance shorter compared to the rocket distance in an inertial frame.

Actually Austin;s probing questions made me realize I have made a complete hash with claims about radar length versus ruler length in posts #7 and #11. (I still stand by my claims in the OP, but made a mess of the red herrings that have thrown in since). So yes my claims about radar length versus ruler length are untenable and I realized that while I was away from my PC :bugeye: Basically I had been shooting from the hip and missed my target completely. My foot is in bandages right now. I should have used a trick I have learned on this forum that when someone who usually knows what they are talking about asks for your exact position on a subject, the best defense is not to state it clearly! :-p

So basically ignore my comments in #7 and #11 that are way off the mark, except for one (I think valid observation) that by definition Born rigid acceleration can not take any object from of not being accelerated to a state of being accelerated. I made that observation when reflecting on Jesse's observation that an linearly accelerating is not necessarily stress free even with Born rigid acceleration and is in a state of equilibrium stress. However, the motion can be described as "strain free" in the sense that its length is not changing over time as long as the rocket always has constant acceleration. This is clearly not true for the the disc with angular acceleration where the radar length of a short segment of the perimeter is always increasing according to an observer on the disc at one end of the segment.

To Passionflower, I agree with your statement of *facts* in the post quoted above. Sorry for the confusion I caused in posts #7 and #11.
 
  • #32
DaleSpam said:
Take a particle's worldline, pick some event on that worldline, and find the tangent vector at that event. Then form the hyperplane which is orthogonal to the tangent vector. Find the event that forms the intersection of a neighboring particle's worldline with that hyperplane. Then calculate the spacetime interval between the two events. That is the proper distance between two neighboring particles. I should mention that it needs to be a differential distance, not a large distance.

Dalespam, you are probably just the person to ask (although anyone should feel free to answer) about this statement in mathpages that I think might be incorrect:

This is shown clearly in the figure above. The line of simultaneity for the accelerating particles simply rotates about the pivot event

Image5456.gif


from http://www.mathpages.com/home/kmath422/kmath422.htm

As far as I can tell, the line of simultaneity pivoting about the origin is only true, if the clocks of the accelerating rocket nearer the origin are artificially sped up relative to the clocks nearer the nose. If the clocks are left to do their own thing so that they agree with local natural processes, the natural line of simultaneity would be tilted in the opposite direction and would not pass through the origin, no? With this artificial speeding up of the clocks nearer the tail, the radar distance measured at the nose would agree with radar distance measured at the tail. Agree?

This is similar to applying a correction factor to the frequency of clocks on GPS satellites so that they agree with clocks in land based receiving equipment.
 
  • #33
yuiop said:
As far as I can tell, the line of simultaneity pivoting about the origin is only true, if the clocks of the accelerating rocket nearer the origin are artificially sped up relative to the clocks nearer the nose.
True.
yuiop said:
If the clocks are left to do their own thing so that they agree with local natural processes, the natural line of simultaneity would be tilted in the opposite direction and would not pass through the origin, no?
Yes, although now "simultaneity" would bear no relationship to Einstein synchronisation.
yuiop said:
With this artificial speeding up of the clocks nearer the tail, the radar distance measured at the nose would agree with radar distance measured at the tail. Agree?
No. Radar measurements are made using "proper clocks", not "coordinate clocks" and the front-back-front measurement will differ from from the back-front-back measurement. The attached diagram illustrates why. The time taken by the red signal (measured by the red observer) is much longer than the time taken by the blue signal (measured by the blue observer).
 

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  • #34
DaleSpam said:
That you didn't calculate the proper distances between neighboring particles before and after.

I see. It's an instantaneous impulse, so the positions of the particles are the same before and after, in the lab frame. Therefore the lab-frame distances between them are also the same before and after. What does change discontinuously is the frames of the comoving observers, so it is necessary to check that the comoving-frame distances also stay the same after the Lorentz boost -- but that's what I did explicitly.

What I probably didn't make very clear was that the impulses I was describing were only the impulses that would provide the angular acceleration. There is a separate set of steady radial forces required as well, even when the ruler doesn't have any angular acceleration. I wasn't describing those.

If you read the Gron paper, it should be pretty clear why the issues he describes go away completely when you have a one-dimensional object.
 
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  • #35
bcrowell said:
FAQ: How is Ehrenfest's paradox resolved?

As described in [Einstein 1916], the relativistic rotating disk was an example that was influential in leading Einstein to describe gravity in terms of curved spacetime. Einstein writes:

"In a space which is free of gravitational fields we introduce a Galilean system of reference K (x,y,z,t), and also a system of coordinates K' (x',y',z',t') in uniform rotation relative to K. Let the origins of both systems, as well as their axes of Z, permanently coincide. We shall show that for a space-time measurement in the system K' the above definition of the physical meaning of lengths and times cannot be maintained. For reasons of symmetry it is clear that a circle around the origin in the X, Y plane of K may at the same time be regarded as a circle in the X', Y' plane of K'. We suppose that the circumference and diameter of this circle have been measured with a unit measure infinitely small compared with the radius, and that we have the quotient of the two results. If this experiment were performed with a measuring-rod at rest relative to the Galilean system K, the quotient would be π. With a measuring-rod at rest relative to K', the quotient would be greater than π. This is readily understood if we envisage the whole process of measuring from the stationary'' system K, and take into consideration that the measuring-rod applied to the periphery undergoes a Lorentzian contraction, while the one applied along the radius does not."

Einstein's friend Paul Ehrenfest posed the following paradox [Ehrenfest 1909]. Suppose that observer L, in the lab frame, measures the radius of the disk to be r when the disk is at rest, and r' when the disk is spinning. L can also measure the corresponding circumferences C and C'. Because L is in an inertial frame, the spatial geometry does not appear non-Euclidean according to measurements carried out with his meter sticks, and therefore the Euclidean relations C=2πr and C'=2πr' both hold. The radial lines are perpendicular to their own motion, and they therefore have no length contraction, r=r', implying C=C'. The outer edge of the disk, however, is everywhere tangent to its own direction of motion, so it is Lorentz contracted, and therefore C' is less than C.

Call the above the problem statement of the Ehrenfest paradox. The proposed resolution is:
bcrowell said:
The resolution of the paradox is that it rests on the incorrect assumption that a rigid disk can be made to rotate. If a perfectly rigid disk was initially not rotating, one would have to distort it in order to set it into rotation, because once it was rotating its outer edge would no longer have a length equal to 2π times its radius. Therefore if the disk is perfectly rigid, it can never be set into rotation.
This resolution fails, because it introduces the straw man of the "rigid disk". Nowhere in the problem statements does it claim a rigid disk. We know that in practice in everyday life, non rigid disks such as the wheels of a car can set into rotation.

bcrowell said:
Thorough modern analyses are available,[Grøn 1975,Dieks 2009] and in particular it is not controversial that, as claimed in [Einstein 1916], C/r is measured to be *greater* than 2π by an observer in the rotating frame.
If this is non controversial then it is clear that Einstein did not intend the spinning disk to be perfectly rigid and the "accepted resolution" is not valid.

We *can* get a non rigid disk to rotate. If observers on the spinning disk measure the circumference they will find it to be greater than 2*pi*R, where R is the radius of the spinning disk as measured by observers in a non rotating frame at rest with the rotation axis of the disk.

A maybe interesting question to ask, is what the observers on a disk with constant angular velocity, measure the radius R' of the spinning disk to be. It is often glibly stated that the radial measurement is not parallel to the motion and therefore not subject to length contraction and so R=R'.

This assumes a ruler measurement of the radius by the observers on the disk, but a ruler with non zero mass will itself be subjected to stresses and strains by the angular motion and is not a reliable measurement. If observers on the perimeter of the disk send a signal to the center which is reflected back, i.e a radar measurement of R', then they find that due to the time dilation of their own clocks, that R' = [itex]\gamma[/itex]R. Therefore when they compare their radar measurements of the circumference C' of the spinning disk to the their radar measurement's of the spinning disk's radius R', they conclude that C' = 2*pi*R' and so by radar measurements alone, the circumference/radius relationship agrees with the Euclidean equation, (but not by ruler measurements). However, the geometry of the disk as measured by the observer's on the disk is not Euclidean in general, because light paths that are not parallel to the radius will follow highly curved paths in some cases when compared to a ruler map of the disk.
 
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