Circle tangential velocity approaching c

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When a merry go rounds tangential velocity approach c, and horses look closer than when at rest, what is reduced, angular velocity or radius?
 

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  • #2
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The merry go round breaks, the angular momentum remains constant as the radius increases while the pieces fly apart.
 
  • #3
Ibix
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This kind of question is tricky - it depends a lot on exactly what you mean.

Imagine you have a spinning disc and two beetles, one on the disc and one on the (non-rotating) floor. If you have both beetles walk the circumference of the disc then the one on the disc will take more steps to do it. If the beetles walk a radius they will agree on the distance.

However, if you start with a non-rotating disc and spin it up, you will find that both the disc's radius and circumference change compared to when it was stationary. As Dale says, this will eventually lead to the disintegration of the disc as you increase the spin rate. Infinitely strong infinitely rigid materials aren't consistent with relativity.

Note that the distance between the horses on your merry go round also depends what you mean. If there are N horses evenly spaced around the merry go round then this is true whether the thing is spinning or not, so their angular separation is not changed. What this means in terms of the measured distance depends on which frame is doing the measuring and what we are comparing to.
 
  • #4
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Obviously am not proposing this thread to discuss the resistance of the material spinning at higher speed. My point is if I am stationary and I measure that 2 points in the mass-less rotating circle get closer is it because angle I measure shrinks or because the radius of the sniping circle, measured from my stationary position is shorter?
 
  • #5
Ibix
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and I measure that 2 points in the mass-less rotating circle get closer
Massless doesn't work - that would move at the speed of light in a straight line.

I don't think any points are likely to get closer to each other. It's a bit difficult to answer specifically because it's not entirely clear what you are comparing. My guess is that you are assuming that the disc doesn't expand significantly even when rotating at relativistic speeds, and you are comparing measurements (made in the non-rotating floor frame) of the disc when not rotating to measurements when it is rotating. Under that model the distances between horses don't change - how could they? The circumference doesn't change and the number of horses doesn't change.

If you make measurements on the disc, you'll find that the horses have moved apart. Strain gauges in the floor will show that the floor is stretched, however, which one can interpret as length contraction.
 
  • #6
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Obviously am not proposing this thread to discuss the resistance of the material spinning at higher speed. My point is if I am stationary and I measure that 2 points in the mass-less rotating circle get closer is it because angle I measure shrinks or because the radius of the sniping circle, measured from my stationary position is shorter?
It doesn't work that way. The disk cannot possibly remain rigid. It must inevitably deform. As it does so the geometry of the disk changes, and the question can only be answered by analyzing the specific material involved and the resulting stresses and strains, which become infinite as the tangential velocities approach c.
 
  • #7
pervect
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When a merry go rounds tangential velocity approach c, and horses look closer than when at rest,
OK so far
what is reduced, angular velocity or radius?
You just lost me here. My best guess is that you are trying to imagine a rigid rotating disk, though you haven't specified such a thing. As other posters have pointed out, a realistic rotating disk would break apart under the strain at tangential velocities of a few km/sec. So you are imagining somethign other than a realistic disk - but it's not clear what. A rigid rotating disk is my best guess.

If that's what you're imagining, we can go into more details on the notion of rigidity in special relativity, but I'm not sure that that's actually your question at this point.
 
  • #8
Ibix
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OK so far
Are you acknowledging the question, or agreeing the premise that the horses will look closer together when the merry go round is moving? Because I don't think I agree with the premise under any interpretation of "look". As measured from the ground frame I think they'll be at least the same distance apart, and literally looked at from the ground frame they'll be uneven distances apart due to Doppler. Or am I missing something?
 
  • #9
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OK so far
I am not convinced even about that. I doubt that there is any material that would produce that result.
 
  • #10
jbriggs444
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and horses look closer than when at rest
The idea is that the horses are arranged on the moving platform so that an observer positioned on the rim and using the momentary tangent inertial rest frame will measure their separation as "normal" and so that an observer positioned nearby on he ground will measure their separation as "closer".

Is this what you intend?
 
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  • #11
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When a merry go round's tangential velocity approach c, and horses look closer than when at rest...
If you want to know what the horses look like then you will need to account for light travel times from the different points on the merry-go-round. But if you want to know what is happening that's an easier question, and speeds close to c are not needed to conduct the experiment.
 
  • #12
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Seems to be the merry go round was not a good example. Forget about it and the horses. Let us think what happens at he periphery of a neutron star with 2 marks at the Equator of the star and separated 1 km. In response to Dale, there are about 100 Million Neutron stars in the Milky Way that spin at high rate. One of them PSR J1748-2446ad spin at 716 RPS or 24 % c at the perihery (radius 16 km) and as far as I know has not yet disintegrated as Dale thinks might happen.
Then if a stationary observer were able to measure the distance between the spots while rotating at 0.24c, beside measuring the length contraction, my question is is there an angle contraction or neutron star radius contraction?
 
  • #13
jbriggs444
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Seems to be the merry go round was not a good example. Forget about it and the horses. Let us think what happens at he periphery of a neutron star with 2 marks at the Equator of the star and separated 1 km. In response to Dale, there are about 100 Million Neutron stars in the Milky Way that spin at high rate. One of them PSR J1748-2446ad spin at 716 RPS or 24 % c at the perihery (radius 16 km) and as far as I know has not yet disintegrated as Dale thinks might happen.
Then if a stationary observer were able to measure the distance between the spots while rotating at 0.24c, beside measuring the length contraction, my question is is there an angle contraction or neutron star radius contraction?
These spots are separated by 1 km in what frame?

Edit: In my limited understanding, it is important to realize that the "rest frame" of the neutron star is an accelerating frame in which it is not possible to use the Einstein synchronization convention in a self-consistent manner. Accordingly, a naive expectation that the coordinate system corresponding to the rest frame will match the coordinate system corresponding to a tangent inertial frame may not be well founded.
 
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  • #14
Ibix
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Seems to be the merry go round was not a good example. Forget about it and the horses. Let us think what happens at he periphery of a neutron star with 2 marks at the Equator of the star and separated 1 km. In response to Dale, there are about 100 Million Neutron stars in the Milky Way that spin at high rate. One of them PSR J1748-2446ad spin at 716 RPS or 24 % c at the perihery (radius 16 km) and as far as I know has not yet disintegrated as Dale thinks might happen.
Then if a stationary observer were able to measure the distance between the spots while rotating at 0.24c, beside measuring the length contraction, my question is is there an angle contraction or neutron star radius contraction?
There is no angle contraction. There is no radius contraction.

Neutron stars are problematic because they have very strong gravity and special relativity does not apply. Your merry-go-round, albeit made of a ridiculously strong material, is a better example.

If you use single nails to attach short rulers to the edge of the merry-go-round you will find the rulers to be length contracted with respect to rulers at rest. Thus, an observer on the merry-go-round would measure the circumference to be greater than ##2\pi r## (because they're using a curved coordinate system, not because spacetime is curved). Using this measurement method they will measure the distance between two horses to be greater than an observer at rest beside the merry-go-round. Note that this also applies to the distance between atoms and scales like the Lorentz ##\gamma##, which is one explanation for why the merry-go-round must disintegrate before the rim reaches lightspeed.

As @jbriggs444 noted, the merry-go-round is rotating so a frame attached to it is non-inertial. Many of the usual rules for inertial frames do not apply.
 
  • #15
ibix said:
the merry-go-round is rotating so a frame attached to it is non-inertial. Many of the usual rules for inertial frames do not apply.
This leads to a larger question of; does Einstein's equations provide for rotational accelerations in General Relativity?
 
  • #16
Ibix
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This leads to a larger question of; does Einstein's equations provide for rotational accelerations in General Relativity?
I'm not sure what you mean. This problem (excluding the neutron star) is entirely in flat spacetime, so would be classified as a special relativity problem. No GR is needed. You can handle rotating objects fine in SR.
 
  • #17
PeterDonis
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does Einstein's equations provide for rotational accelerations in General Relativity?
If you mean, does the EFE predict that, for example, the material in a rotating neutron star will have nonzero proper acceleration, of course it does.
 
  • #18
I'm not sure what you mean. This problem (excluding the neutron star) is entirely in flat spacetime, so would be classified as a special relativity problem. No GR is needed. You can handle rotating objects fine in SR.
What I'm saying is that rotations introduce a radial acceleration; since acceleration is the domain of general relativity, has this acceleration been addressed by the equations?
 
  • #19
PeterDonis
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What I'm saying is that rotations introduce a radial acceleration
If you mean coordinate acceleration, that depends on your choice of coordinates.

If you mean proper acceleration, that depends on the spacetime geometry; you can have spacetime geometries in which matter is rotating with zero proper acceleration.

since acceleration is the domain of general relativity
No, it isn't. You can handle accelerations (either kind) in flat spacetime just fine using SR.

has this acceleration been addressed by the equations?
In any solution of the EFE which is not flat spacetime, the solution tells you the acceleration of a piece of matter in any state of motion you like.
 
  • #20
Ibix
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What I'm saying is that rotations introduce a radial acceleration; since acceleration is the domain of general relativity, has this acceleration been addressed by the equations?
As Peter says, all scenarios in flat spacetime are in the domain of special relativity, accelerating or not. A lot of SR is taught using exclusively inertial reference frames, but there's no obligation to restrict yourself that way. If you choose a more general approach, however, you need to be aware that simple rules relating inertial frames don't work.

The OP is using a rotating frame, which is fine, but I suspect he has imported a few intuitions from inertial frames without examining them. I might be wrong.
 
  • #21
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One of them PSR J1748-2446ad spin at 716 RPS or 24 % c at the perihery (radius 16 km) and as far as I know has not yet disintegrated as Dale thinks might happen.
The Lorentz factor at 0.24 c is about 1.03, which is hardly what I had in mind by a “tangential velocity approaching c”.

In any case, to answer the question you would still need to consider the material properties of the neutron star. The point I have been making is that you cannot spin up something without distorting it. You will not simply have length contraction, but you will have an unavoidable stretching of the material. A spinning neutron star will be stretched compared to a non spinning one.

This is a fundamental difference between linear motion and rotational motion. For linear motion you can start from rest, and accelerate rigidly, meaning that strain gauges attached to the object will detect no strain. For rotational motion that is not possible. You cannot start from rest and undergo angular acceleration without having some strain somewhere. So what happens to the disk depends on the material properties of the disk
 
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  • #22
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since acceleration is the domain of general relativity,
As others stated, acceleration can be treated just fine in SR, but if we want to pursue the neutron star idea, that would be GR.
 
  • #23
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Let us think what happens at he periphery of a neutron star with 2 marks at the Equator of the star and separated 1 km.
Say forget about gravity of neutron star which might affect metric of nearby space and set a ring at rest in IFR almost touching periphery of neutron star. The ring would have number of 2*3.14*16km/1km marks. In the rotating neutron star frame of reference the marks are equidistant but more than 1km. When periphery speed is reaching c, the distance between marks goes to infinity. It is interpreted that pi, i.e. ratio of periphery to diameter, reaches infinity in geometry of the rotating reference frame. If you define angle be length of part of periphery divided by radius as we do for radian angle, angle between the marks increases. But of course sum of all the angles coming back to the origin is more than 2pi or 360 degree and the sum reaches infinity angle as periphery speed goes to c.

(We can set the same number of equidistant marks on the periphery with more than 1 km distance between in the rotating frame of reference. We need more number of marks for 1km distant set in the rotating frame of reference.)
 
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  • #24
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It is interpreted that pi, i.e. ratio of periphery to diameter, reaches infinity
No it isn’t. ##\pi## is a defined constant and it is always about 3.14 regardless of whether your geometry is curved or flat. The ratio of the circumference to the diameter is equal to ##\pi## for flat Euclidean geometry. For other geometries the ratio changes, but it is not ##\pi##
 
  • #25
PeterDonis said:
You can handle accelerations in flat spacetime just fine using SR.
Ibix said:
As Peter says, all scenarios in flat spacetime are in the domain of special relativity, accelerating or not.
I must admit, I'm confused - I thought that special relativity dealt exclusively with inertial reference frames, and that's why it's deemed "special"? The Equivalence Principle finds that acceleration due to gravity is equivalent to acceleration due to motion, doesn't it? So then the "warping" of spacetime by massive objects should also occur due to accelerations, which means that the formerly flat spacetime is no longer flat.
PeterDonis said:
In any solution of the EFE which is not flat spacetime, the solution tells you the acceleration of a piece of matter in any state of motion you like.
I know the Equivalence Principle holds true for linear accelerations, but I wasn't sure about accelerations that aren't along the path of motion. I thought by extension they might. I also thought that they'd be described by the off-diagonal components of the Riemann tensor. Is that correct? Could you specify which components that would be?
 

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