Tension in slowly rotating cylinder in gravity

[wasia]

Hello,

I am sure this must have been considered before, but I haven't been able to find it. I would be most grateful to get a reference or the right keywords to search for.

Consider a slowly rotating cylinder, which is brought into a gravitational field (say from space to the surface of Earth). Is it true that due to time dilatation one side of the cylinder will spin slightly slower than the other and this effect is cumulative?

Thanks.

 

[bcrowell]

If you're talking about an ordinary-sized cylinder, then this obviously isn't possible because the cylinder would have to deform more and more. If you're talking about a very long cylinder, then Newtonian tidal effects will dominate, and relativity will be negligible.

 

[wasia]

Well, this is exactly what I mean by cumulative - the different becomes more pronounced with time. But what do you mean "obviously isn't possible"?

 

[bcrowell]

Well, this is exactly what I mean by cumulative - the different becomes more pronounced with time. But what do you mean "obviously isn't possible"?

(a) We don't observe it to occur. E.g., the moon isn't tied up in knots, nor were the gyroscopes aboard Gravity Probe B. (b) Internal stresses in the material are much stronger than Newtonian tidal effects, which in turn are much stronger than relativistic effects.

 

[yuiop]

Hello,

I am sure this must have been considered before, but I haven't been able to find it. I would be most grateful to get a reference or the right keywords to search for.

Consider a slowly rotating cylinder, which is brought into a gravitational field (say from space to the surface of Earth). Is it true that due to time dilatation one side of the cylinder will spin slightly slower than the other and this effect is cumulative?

Thanks.

You have not defined the orientation of the cylinder in the gravitational field so it is difficult to answer your question. If the circular cross section of the cylinder is parallel to the surface of the gravitational body and the axis of rotational symmetry is vertical, then yes, I imagine there could be cumulative effect in the number of rotations of one circular face relative to the other which would eventually tie the cylinder in knots.

For example if the circular face near the event horizon was rotating once per second as measured by a local clock and the other circular face much higher up was also forced to rotate once per second by a mechanical device (also as measured by a local clock) then twisting stress will be detectable in the cylinder which will eventually break the cylinder.

If the cylinder was rotating naturally due to its angular momentum and in a state of equilibrium, then the circular face nearest the gravitational body would appear to rotating faster than the higher end again as measured by local clocks, but in this case there would be no cumulative twisting stress on the cylinder.

This sort of cylinder could be used to directly observe gravitational time dilation as a real physical effect.

 

[bcrowell]

You have not defined the orientation of the cylinder in the gravitational field so it is difficult to answer your question. If the circular cross section of the cylinder is parallel to the surface of the gravitational body and the axis of rotational symmetry is vertical, then yes, I imagine there could be cumulative effect in the number of rotations of one circular face relative to the other which would eventually tie the cylinder in knots.

For example if the circular face near the event horizon was rotating once per second as measured by a local clock and the other circular face much higher up was also forced to rotate once per second by a mechanical device (also as measured by a local clock) then twisting stress will be detectable in the cylinder which will eventually break the cylinder.

If the cylinder was rotating naturally due to its angular momentum and in a state of equilibrium, then the circular face nearest the gravitational body would appear to rotating faster than the higher end again as measured by local clocks, but in this case there would be no cumulative twisting stress on the cylinder.

This sort of cylinder could be used to directly observe gravitational time dilation as a real physical effect.

This is incorrect, for the reasons given in #4.

Even if you make the cylinder so weak that its internal stresses become negligible, you'll still observe nothing but Newtonian gravity. For example, you can just put a chain of separate test particles in initially adjacent orbits, and call that a cylinder. Then there is no internal stress to keep the particles together. But what you will measure will simply be the cumulative effect of Kepler's laws, not anything about relativity.

 

[yuiop]

This is incorrect, for the reasons given in #4.

Even if you make the cylinder so weak that its internal stresses become negligible, you'll still observe nothing but Newtonian gravity. For example, you can just put a chain of separate test particles in initially adjacent orbits, and call that a cylinder. Then there is no internal stress to keep the particles together. But what you will measure will simply be the cumulative effect of Kepler's laws, not anything about relativity.

I disagree, but I suspect we are talking past each and considering different scenarios. To be more specific I am talking about a long thin cylinder with its axis of rotational symmetry orientated vertically and rotating about its vertical axis. I claim that if the rotation rate in rpm of the circular face is measured at the top of the cylinder far from the gravitational source, it s rpm will measured to be faster at the circular face nearest the gravitational source, if local clocks are used. Do you still disagree? If so, can you try and clearly describe an example scenario you are considering?

 

[bcrowell]

I disagree, but I suspect we are talking past each and considering different scenarios. To be more specific I am talking about a long thin cylinder with its axis of rotational symmetry orientated vertically and rotating about its vertical axis. I claim that if the rotation rate in rpm of the circular face is measured at the top of the cylinder far from the gravitational source, it s rpm will measured to be faster at the circular face nearest the gravitational source, if local clocks are used. Do you still disagree? If so, can you try and clearly describe an example scenario you are considering?

I took wasia's OP to be discussing a scenario in which the cylinder would undergo a deformation that would increase cumulatively. Maybe wasia could explain more about what was intended in the OP.