How specifically does an accelerated uniform rod Lorentz contract?

[PeroK]

I mean that an observer at each rod end has a clock and is recording his perceived velocity wrt the "stationary" frame at every time on his clock. The recorded times and velocities at the two rod ends are identical.

Is "the velocity of the stationary frame" well defined for an accelerating observer? Or, are different points or objects at rest in the stationary frame moving with different velocities relative to the accelerating observer?

If the accelerating observer stops accelerating and cruises at constant velocity, then there will be length contraction in the stationary frame. Which cannot be accounted for if all points in the stationary frame moved identically in the accelerating frame.

 

[Ibix]

I mean that an observer at each rod end has a clock and is recording his perceived velocity wrt the "stationary" frame at every time on his clock. The recorded times and velocities at the two rod ends are identical.

The problem is the relativity of simultaneity. If you mean that the acceleration profile of each clock, recorded against its own proper time, is identical then you get Bell's spaceships, as I said. If you require that the front and rear clocks accelerate at the same rate at the same time according to some synchronisation convention then you get other profiles, as @PeterDonis noted. You need to specify your synchronisation convention.

Never mind. You specified the first option. I'm going to bed...

 

[PeterDonis]

I mean that an observer at each rod end has a clock and is recording his perceived velocity wrt the "stationary" frame at every time on his clock.

Ok.

The recorded times and velocities at the two rod ends are identical.

This is only true if the proper accelerations of the two rod ends are identical, which they are in the Bell spaceship paradox scenario, but are not, as I understand it, in the scenario you are describing.

I strongly suggest that you refrain from making any further claims until you have done the math.

 

[pervect]

It is not possible to have Born rigid acceleration at all times with a force applied only at the center of the rod. The force takes time to propagate from the center to other parts of the rod and equilibrate throughout the rod, and during that time the other parts of the rod do not undergo constant proper acceleration.

The only way to get Born rigid acceleration for all times is to apply a precisely calculated force to each individual point of the rod separately.

That's a great way of putting it.

 

[pervect]

Okay everyone, I think I was able to come up with a solution myself. (I haven't taken a look at any papers other than the one I posted about.) Here's my approach:
Assume that force accelerating the Born rigid rod (that is what I meant when I initially said "uniform") is constant in time and applied to the center of the rod, we find that the center follows a hyperbolic trajectory with proper acceleration $a=F/m$.

It's really much simpler if you leave the forces out of it, and focus on the accelerations.

If you do that, you can say simply that the proper acceleration of any point on the rod is c^2/d, where d is the distance of the point of the rod from the Rindler horizon. You can use this equation to find where the Rindler horizon is, if you know the acceleration of the midpoint of the rod. If the rod is accelerating at 1 light year/ year^2, the rindler horizon is 1 light year behind the midpoint of the rod.

If you study the implications of this, it's only approximately true that the acceleration at the midpoint is the average of the acceleration at the front end and the back end, because the acceleration isn't linear. But for short rods, the non-linearity isn't important.

The other wrinkle, which is mentioned in other threads, is knowing how to relate the proper acceleration to the coordinate acceleration. As long as you are in a co-moving inertial frame, there isn't any difference in the numbers, but you _will_ confuse yourself if you assume that the coordinate acceleration is the same as the proper acceleration in any other frame than a co-moving one.

If you do want to learn about forces in special relativity, my advice is to learn about 4-forces first. To understand what a 4-force is, you'll need to study 4-vectors, and (ideally) a bit about geometric objects, also called tensors. The point of geometric objects like 4-forces is that they transform in a known manner, and can abstractly regarded as being independent of any frame of reference.

3-forces don't have any meaning unless you carefully express exactly what frame of reference you are measuring the force in.

For the case of a rod such as you describe, 4-forces won't actually be enough to do it right. To really describe the tension in the rod, you'll need to familiarize yourself with the stress-energy tensor $T^{ab}$, which is not a force and transforms differently. 4-forces transform as a rank 1 tensor, the stress energy tensor is a rank 2 tensor and transforms that way.

I don't know of any way to gracefully avoid talking about the forces without tensor language. But one can talk about the proper accelerations without getting so technical.

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[FactChecker]

This is only true if the proper accelerations of the two rod ends are identical, which they are in the Bell spaceship paradox scenario, but are not, as I understand it, in the scenario you are describing.

Thanks. This was not a claim of proven fact, but rather a supposition on my part. So I think that no mathematics is needed at that point. So I guess what you are saying is that my supposition is consistent with the Bell spaceship paradox, but not with a Born rigid rod. I will think about that.

 

[PeterDonis]

I guess what you are saying is that my supposition is consistent with the Bell spaceship paradox, but not with a Born rigid rod.

It could be consistent with either one, depending on what exactly you meant. Or it could be consistent with neither. But it can't be consistent with both, since the two are mutually exclusive.

You said the proper accelerations at the two ends of the rod were not equal. In the Bell spaceship paradox, they are. So it doesn't seem like your supposition is consistent with the Bell spaceship paradox.

However, you also said that observers at both ends of the rod think they are going at the same speed relative to some fixed inertial frame, at the same instant of time by each of their clocks. But if the rod is Born rigid, they don't. So it doesn't seem like your supposition is consistent with a Born rigid rod either.

 

[Orodruin]

However, you also said that observers at both ends of the rod think they are going at the same speed relative to some fixed inertial frame, at the same instant of time by each of their clocks. But if the rod is Born rigid, they don't. So it doesn't seem like your supposition is consistent with a Born rigid rod either.

This depends on what is meant by the original statement. The other interpretation is that they are not looking at their clocks but at their own clocks but the clocks of the instantaneous rest frame.

 

[PeterDonis]

The other interpretation is that they are not looking at their clocks but at their own vlocks but the clocks of the instantaneous rest frame.

Which instantaneous rest frame?

 

[Orodruin]

Which instantaneous rest frame?

The instantaneous rest frame of the rod. For Born rigid motion, this is the same for all parts of the rod.

 

[PeterDonis]

The instantaneous rest frame of the rod.

Ok. Then the only caution would be that the coordinate time in this frame is not the same as the proper time of observers in different parts of the rod; at most it will match the proper time of just one.

 

[Orodruin]

Ok. Then the only caution would be that the coordinate time in this frame is not the same as the proper time of observers in different parts of the rod; at most it will match the proper time of just one.

The coordinate time is always zero if you go by the standard setup, but I think it is still a valid interpretation of the claim.

 

[FactChecker]

I am afraid that I may have hijacked this thread. Thanks for the clarification. I wanted to describe the Born rigidity situation, but I did not know enough about it and seem to have gotten it wrong. In particular, if both ends maintain constant proper acceleration then the leading end must have a lower constant proper acceleration than the trailing end. This, in addition to the relativity of simultaneity, would give the reduced length when measured in the "stationary" reference frame.

 

[PeterDonis]

The coordinate time is always zero if you go by the standard setup

In a particular MCIF, yes; but you are really just using that frame's time to define "at the same time", instead of using the proper times along each worldline.

 

[PeterDonis]

I wanted to describe the Born rigidity situation

This...

if both ends maintain constant proper acceleration then the leading end must have a lower constant proper acceleration than the trailing end

...is correct for Born rigid acceleration, provided that by "proper acceleration" you mean "proper acceleration at the same time" and that by "at the same time" you mean what @Orodruin said.

However, because terms like "at the same time" are ambiguous in relativity if not qualified, it's always better to explicitly specify exactly what "time" is being used.

This, in addition to the relativity of simultaneity, would give the reduced length when measured in the "stationary" reference frame.

Yes, but this description is likely to be confusing, because you are using "at the same time" to mean time in the momentarily comoving inertial frame of the rod, but you are defining "length" in the stationary frame. That means you're mixing quantities from two different frames, which is generally a bad idea.

You can show that the length of the rod contracts in the stationary frame without having to go through any of this: just note that, at some instant, the stationary frame will be the rest frame of the rod, and at that instant, the front end of the rod has a smaller proper acceleration than the rear. That, combined with the fact that proper acceleration is constant along both worldlines, is enough to show that the rod's length contracts as it accelerates, and it makes no use of any frame-dependent quantities in any frame except the stationary frame.

 

[Orodruin]

In a particular MCIF, yes; but you are really just using that frame's time to define "at the same time", instead of using the proper times along each worldline.

Yes, it is just a particular simultaneity convention set up in such a way that "simultaneous" events on the rod have the same velocity.

 

[vanhees71]

The Born rigid body is fictitious and not so easy to treat as it might seem. It's a fascinating mathematical subject and a now classical result of mathematical physics to establish which motions such a fictitious body can make at all (Herglotz, Fritz Noether ~1910), but physicswise it's ficititious. I found the above cited treatment of an elastic body which comes closest to a kind of rigidity in assuming an ideal fluid with the speed of sound equal the speed of light. It's of course also fictitious but makes the issues much clearer since it's closer to a real "quasirigid" body.

 

[Orodruin]

The Born rigid body is fictitious

I think this is true for most (if not all) models of physical phenomena. It is then a question only if the model is close enough to reality to actually make useful predictions.

 

[DrGreg]

The Born rigid body is fictitious

I think this is true for most (if not all) models of physical phenomena. It is then a question only if the model is close enough to reality to actually make useful predictions.

I suppose this depends on how you apply the term "Born rigidity". If you're thinking of it as an intrinsic material property of an object, then, yes it is fictitious. But if you think of it as a type of motion, then there are circumstances in which an object can achieve Born rigid motion, if not exactly but to a very good approximation.

If you apply a force to an object that previously was subject to no force, then initially the object will deform and pressure waves will travel through the object. But if you continue to apply a constant proper force and wait for a long enough time, eventually the object will reach an equilibrium state (to a good approximation) and pressure waves will have reduced to a negligible level. The object will now differ from a truly Born rigid object by a negligible amount.

 

[vanhees71]

I think this is true for most (if not all) models of physical phenomena. It is then a question only if the model is close enough to reality to actually make useful predictions.

Well, as I said the model of an ideal fluid with $c$ as speed of sound is also fictitious, but it's not as fictitious as a Born rigid body, contradicting the fact how spinning tops behave, which are well described in non-relativistic mechanics within the model of a rigid body, but within relativity you have to use elastic bodies.

 

[vanhees71]

If you apply a force to an object that previously was subject to no force, then initially the object will deform and pressure waves will travel through the object. But if you continue to apply a constant proper force and wait for a long enough time, eventually the object will reach an equilibrium state (to a good approximation) and pressure waves will have reduced to a negligible level. The object will now differ from a truly Born rigid object by a negligible amount.

Sure, that's one of the motions that are allowed for a Born rigid body. Nevertheless to set it in motion from rest you usually must assume some elasticity in the beginning.