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Main Question or Discussion Point
When a merry go rounds tangential velocity approach c, and horses look closer than when at rest, what is reduced, angular velocity or radius?
Massless doesn't work - that would move at the speed of light in a straight line.and I measure that 2 points in the mass-less rotating circle get closer
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.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?
OK so farWhen a merry go rounds tangential velocity approach c, and horses look closer than when at rest,
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.what is reduced, angular velocity or radius?
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?OK so far
I am not convinced even about that. I doubt that there is any material that would produce that result.OK so far
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".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.When a merry go round's tangential velocity approach c, and horses look closer than when at rest...
These spots are separated by 1 km in what frame?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.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?
This leads to a larger question of; does Einstein's equations provide for rotational accelerations in General Relativity?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.
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.This leads to a larger question of; 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.does Einstein's equations provide for rotational accelerations in General Relativity?
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?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.
If you mean coordinate acceleration, that depends on your choice of coordinates.What I'm saying is that rotations introduce a radial acceleration
No, it isn't. You can handle accelerations (either kind) in flat spacetime just fine using SR.since acceleration is the domain of general relativity
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.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.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?
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”.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.
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.since acceleration is the domain of general relativity,
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.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.
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##It is interpreted that pi, i.e. ratio of periphery to diameter, reaches infinity
PeterDonis said:You can handle accelerations in flat spacetime just fine using SR.
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.Ibix said:As Peter says, all scenarios in flat spacetime are in the domain of special relativity, accelerating or not.
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?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.