Can Born Rigid Motions Occur in Curved Spacetime?

[Tahmeed]

Can a sphere or disk rotating with uniform speed follow born condition of rigidity?

 

[pervect]

Can a sphere or disk rotating with uniform speed follow born condition of rigidity?

Yes.

 

[stevendaryl]

Yes.

Are you sure about that? I thought that Ehrenfest paradox showed that a rotating disk can't actually be perfectly rigid.

 

[Tahmeed]

Are you sure about that? I thought that Ehrenfest paradox showed that a rotating disk can't actually be perfectly rigid.

As far as i know, Ehrenfest showed that it can't start rotating without breaking Born condition for rigidity. But my knowledge is very limited. And I would however request 'Mr.Pervect' to elaborate. I didn't find sufficient resources on this issue.

 

[PeterDonis]

I thought that Ehrenfest paradox showed that a rotating disk can't actually be perfectly rigid.

No, it just shows that you can't take a disk that isn't rotating and make it rotate while keeping it perfectly rigid. In other words, the motion of a disk that is rotating at a changing angular velocity is not a rigid motion. But the motion of a disk that is rotating at a constant angular velocity is a rigid motion.

 

[pervect]

A not particularly readable (IMO) reference on the topic is "The Rich Structure of Minkowskii Space" https://arxiv.org/abs/0802.4345. The notion of motion is described abstractly by a vector field $u^a$, the integral curves of this vector field are the worldlines of particles on the body whose rigidity is to be tested. "Rigid motion" is a specific mathematical requirement that the vector field representing the motion must meet.

There's a couple of different ways of describing the mathematical requirements for the motion to be Born rigid. One of the more convenient ways is to say that the expansion and shear of the vector field u is zero. There's also an approach using Lie derivatives, which may be simpler in some respects. I believe from context that Born's original expression of the condition was in terms of the Lie derivative, for what it's worth.

The definition in terms of the expansion and shear is proposition 17 in the above reference:

Proposition 17.
Let u be a normalised timelike vector field

The motion described by its flow is rigid iff u is of vanishing shear and expansion

Also of interest is the following:

Vector fields generating rigid motions are now classified according to whether or not they have a vanishing vorticity ω : if
ω = 0 the flow is called irrotational, otherwise rotational. The following theorem is due to Herglotz and Noether:

Theorem 18 (Noether & Herglotz, part 1)
.
A rotational rigid motion in Minkowski space must be a Killing motion.

Rotational Killing motioins do exist - they're just the motion of a rigid rotating disk. The paper even writes down the vector field of such a moton (using notation in which partial derivative operators are used to describe vectors) - it mentions that

This motion corresponds to a rigid rotation with constant angular velocity κ

Sorry if this is all too advanced, the complexities of a fuller explanation are why my first answer was just "Yes" and not more detailed.

 

[Tahmeed]

A not particularly readable (IMO) reference on the topic is "The Rich Structure of Minkowskii Space" https://arxiv.org/abs/0802.4345. The notion of motion is described abstractly by a vector field $u^a$, the integral curves of this vector field are the worldlines of particles on the body whose rigidity is to be tested. "Rigid motion" is a specific mathematical requirement that the vector field representing the motion must meet.

There's a couple of different ways of describing the mathematical requirements for the motion to be Born rigid. One of the more convenient ways is to say that the expansion and shear of the vector field u is zero. There's also an approach using Lie derivatives, which may be simpler in some respects. I believe from context that Born's original expression of the condition was in terms of the Lie derivative, for what it's worth.

The definition in terms of the expansion and shear is proposition 17 in the above reference:
Also of interest is the following:
Rotational Killing motioins do exist - they're just the motion of a rigid rotating disk. The paper even writes down the vector field of such a moton (using notation in which partial derivative operators are used to describe vectors) - it mentions that
Sorry if this is all too advanced, the complexities of a fuller explanation are why my first answer was just "Yes" and not more detailed.

It is actually too advanced for me anyway, haha. However, does this only apply to spheres or anybody rotating uniformly?

 

[pervect]

It is actually too advanced for me anyway, haha. However, does this only apply to spheres or anybody rotating uniformly?

Given that a sphere can rotate so as to satisfy Born's conditions any subset of the sphere will also satisfy Born's conditions. So the basic answer is yes.

Some necessary conditions for a sphere to satisfy Born's conditions are that the sphere rotate with a constant angular velocity (it won't satisfy Born's conditions if it's rate of rotation varies with time), and that the sphere be small enough so that all parts of it move at less than the speed of light.

An assumption I make in giving this answer is that the space-time is the flat space-time of special relativity, not the curved space-time of General Relativity in the presence of significant mass. Rather than say this is necessary, I'd say that this is assumed.

 

[PeterDonis]

An assumption I make in giving this answer is that the space-time is the flat space-time of special relativity, not the curved space-time of General Relativity in the presence of significant mass. Rather than say this is necessary, I'd say that this is assumed.

Born rigid motions are possible in at least some curved spacetimes as well, but they will in general not be the same in terms of the "shape" of the object as they would be in flat spacetime. Also, the kinds of trajectories that would allow Born rigid motions are more limited in curved spacetimes, because there will not be as many Killing vector fields (in fact most curved spacetimes have none).