Calc Rotational E of 6km Bar: Einstein's Method

[Jens]

[Moderator's note: spun off from another thread.]

How would Einstein calculate the rotational energy of a 6 kilometer long rigid bar with rest mass M rotating about its center with 15000 revolutions per second in gravity free space?

 

[Ibix]

An interesting feature of relativity is that it denies the possibility of something perfectly rigid, since you would be able to communicate faster than light by tapping one end and seeing the other end moving instantaneously. All materials will flex, compress, stretch or break. It isn't possible to discuss a rigid rod in the relativistic regime, even as an idealisation. So your rod would bend (assuming it had the utterly unbelievable tensile strength not to simply disintegrate). The first question, then, is what are the material properties of this rod? When you've defined that you can begin to address the question of how it would behave when rotated at relativistic speeds.

Edit: Actually, it might just stretch. But we still need a material model and to decide whether the 6km is a rest length or a length when it's in motion.

 

[Jens]

An interesting feature of relativity is that it denies the possibility of something perfectly rigid, since you would be able to communicate faster than light by tapping one end and seeing the other end moving instantaneously. All materials will flex, compress, stretch or break. It isn't possible to discuss a rigid rod in the relativistic regime, even as an idealisation. So your rod would bend (assuming it had the utterly unbelievable tensile strength not to simply disintegrate). The first question, then, is what are the material properties of this rod? When you've defined that you can begin to address the question of how it would behave when rotated at relativistic speeds.

Ok but let's assume the bar has a length of 1 meter at rest, has rest mass M = 1 kilo and is made of steel with a circular cross section. Now assuming the angular velocity is so small and the bar so rigid that any bending and tension effects are such that the deformation remains constant after the final small angular velocity has been reached.
The resulting S-shape and elongation could be counteracted by prebending and shortening the bar appropriately before spinning it. Thus one can, in principle, end up with a straight bar of length L and mass M rotating at angular speed w. So if relativistic Lorenzt contraction and mass increase with speed is not neglected (however small) then one should be able to derive a formula for the stationary rotational energy. Thus the question still stands.

 

[pervect]

Ok but let's assume the bar has a length of 1 meter at rest, has rest mass M = 1 kilo and is made of steel with a circular cross section. Now assuming the angular velocity is so small and the bar so rigid that any bending and tension effects are such that the deformation remains constant after the final small angular velocity has been reached.
The resulting S-shape and elongation could be counteracted by prebending and shortening the bar appropriately before spinning it. Thus one can, in principle, end up with a straight bar of length L and mass M rotating at angular speed w. So if relativistic Lorenzt contraction and mass increase with speed is not neglected (however small) then one should be able to derive a formula for the stationary rotational energy. Thus the question still stands.

The closest thing I've ever seen is Greg Egan's analysis using a particular material model (dubbed hyperelasticity, an idealization of Hooke's law) as applied not to rotating rods, but rotating rings, disks, and hoops. I'm not aware of anything at all written about rotating rods. See http://www.gregegan.net/SCIENCE/Rings/Rings.html

I did a bit of work on the topic back when Egan first posted about it to PF, but I've long since forgotten all the details. I did assist with a Lagrangian formulation of the model. Egan's article is still online at the link I gave earlier. Note that one of the conclusions drawn is that the hyperelastic model used eventually becomes ill-behaved, this happens because the speed of sound predicted by the simple model becomes greater than light if the hoop stretches too much.

Also note that the result hasn't been peer reviewed. It's a genuine attempt to address the problem, but it shouldn't be regarded as an authoritative answer.

 

[Jens]

Actually, since the final shape and elongation of the bar is stationary, one may assume the center of a small mass element dm at a distance r from the center can be expressed by r and its polar angle u. Then a given r(u) will represent the effective shape of the bar and as a function of u.
Hence it seems that regardless of how the final effective shape of the bar has evolved, an assumed knowledge of r(u) is sufficient to calculate the rotational energy no matter the physical viability.
In fact my question really centers on whether the
application of dm = gamma dm0 and r = r0/gamma is correct or not (the index 0 indicate variables at rest and gamma is the usual factor).

 

[pervect]

In Newtonian physics, tension and density are two totally separate things. In relativistic physics, they're both part of the stress-energy tensor. See https://en.wikipedia.org/wiki/Stress–energy_tensor. So a Newtonian analysis isn't going to give valid relativistic answer.