KE of rotating disc

[PeterDonis]

The proposed integral in the OP impies the following physical process: a disc of dust with uniform density in some inertial frame is spun up (totally non rigidly) under the constraints that density of dust per unit area in the inertial frame remains constant.

I'm not sure this is true. There is no spin-up process described in the OP or in the integral there--the angular velocity $\omega$ is treated as a constant, not a function of time (and there is no time in the integral anyway). As far as I can see, the constraint of uniform mass distribution only has to imply that the density is uniform in the inertial rest frame of the center of rotation, when the disk is rotating with constant angular velocity $\omega$.

 

[PAllen]

I'm not sure this is true. There is no spin-up process described in the OP or in the integral there--the angular velocity $\omega$ is treated as a constant, not a function of time (and there is no time in the integral anyway). As far as I can see, the constraint of uniform mass distribution only has to imply that the density is uniform in the inertial rest frame of the center of rotation, when the disk is rotating with constant angular velocity $\omega$.

I am simply describing a procedure to arrive at the final state starting from a rest state.

 

[PeterDonis]

I am simply describing a procedure to arrive at the final state starting from a rest state.

I understand that, I just don't see how it's relevant to what's in the OP. The OP is just describing the final state, not how it's achieved. And the final state can be analyzed on its own merits, without having to consider how it's achieved. As far as I can tell, what I said in post #31 is sufficient to apply the "uniform mass distribution" condition to analyze the final state. How things would have had to change while spinning up the disc doesn't seem to me to affect that.

 

[Thor Jackson]

TL;DR: Off the back of yesterday's thread about relativistic discs I tried to work out the KE of a rotating disc and got an odd result.

Consider a disc of radius $R$ with a uniform mass distribution and total mass $M$ rotating in its own plane. In its COM's inertial rest frame it has angular velocity $\omega$ about its center. In the obvious polar coordinates in that frame the kinetic energy of a small part of the disc at radius $r$ is $$dK=\frac{M}{\pi R^2}\left(\frac 1{\sqrt{1-\omega^2r^2}}-1\right)r\,dr\,d\phi$$in units where $c=1$. I just integrate to get the total KE of the disc. The $\phi$ integral is trivial and the $r$ one is not much harder:$$\begin{eqnarray*}K&=&\frac{2M}{R^2}\int_0^R\left(\frac 1{\sqrt{1-\omega^2r^2}}-1\right)r\,dr\\
&=&\frac{2M}{\omega^2R^2}\left(1-\sqrt{1-\omega^2R^2}-\frac 12\omega^2R^2\right)
\end{eqnarray*}$$This formula has the correct limit for $|\omega R|\ll 1$ - plugging in the Taylor expansion $\sqrt{1-x^2}\approx 1-\frac 12x^2-\frac 18x^4$ gives $K\approx\frac 14MR^2\omega^2=\frac 12I\omega^2$, as expected. But it is finite in the ultra-relativistic limit where $\omega R\rightarrow 1$ - in fact it tends to the absurdly low $M$. Clearly this is wrong, since you can chip off a flake of matter near the rim and it can have arbitrarily high kinetic energy.

What's going on? The integrand looks sensible and diverges at $|\omega r|=1$, but the integral behaves in an unexpected manner. Have I overlooked something? Just screwed up the maths? Can I not study a massive rotating disc in SR at these speeds because spacetime will be significantly curved by the energy?

I had a look for the result online. I found a lot of discussion of Ehrenfest and a lot of derivations of the KE of a spinning disc in Newtonian physics, but I couldn't find the result for the total kinetic energy in the relativistic case.

The maths of your integral is fine — what’s breaking is the physical model you’re feeding into it.

You’ve assumed:

• a rigid disc
• uniform mass density in the lab/COM frame
• a single angular velocity ω all the way out to the rim, even as v→c

All three of those are incompatible with special relativity at ultra‑relativistic rim speeds.

In SR there is no such thing as a perfectly rigid rotating disc. Born rigidity already tells you that you can’t spin up a finite‑radius disc from rest to arbitrarily high angular velocity without introducing stresses that themselves gravitate and, in the extreme, require a GR treatment.

More concretely:

• If the disc is “rigid” in its own local rest frame, then in the COM frame the tangential lengths are Lorentz‑contracted, so the mass density cannot stay uniform with radius.
• If you insist on uniform density in the COM frame, then the material is not rigid in its own rest frame and the stress–energy tensor is no longer that of “dust”; you need to include stresses, not just kinetic energy.

Your integral is effectively integrating the kinetic energy of a dust distribution with a fixed ω and fixed lab‑frame density all the way out to r=c/ω. That’s a perfectly well‑posed mathematical problem, so it’s not surprising the integral converges. What it isn’t is a physically realizable rotating disc in SR.

If you instead model a thin ring and let its tangential speed approach c, the kinetic energy of that ring does diverge as expected. The paradox only appears when you try to extend that picture to a rigid, uniformly filled disc, which SR simply doesn’t allow at arbitrarily high rim speeds.

 

[PAllen]

I understand that, I just don't see how it's relevant to what's in the OP. The OP is just describing the final state, not how it's achieved. And the final state can be analyzed on its own merits, without having to consider how it's achieved. As far as I can tell, what I said in post #31 is sufficient to apply the "uniform mass distribution" condition to analyze the final state. How things would have had to change while spinning up the disc doesn't seem to me to affect that.

It’s possibly important to understand what M means. The construction of the integral means it summed from constituent rest masses.

 

[PeterDonis]

In SR there is no such thing as a perfectly rigid rotating disc.

Yes, there is. It is perfectly possible to have a disk whose motion is rigid at a constant angular velocity $\omega$. SR does not rule that out. That is the physical model that is being analyzed in this thread.

What is not possible is to have a rigid disk with a changing angular velocity--so you can't spin up a disk from rest to some nonzero $\omega$ with the motion being rigid the whole time. But, as I said in post #33, I don't see the spin-up process being included in the analysis being done in this thread, nor does it need to be.

• If the disc is “rigid” in its own local rest frame, then in the COM frame the tangential lengths are Lorentz‑contracted, so the mass density cannot stay uniform with radius.
• If you insist on uniform density in the COM frame, then the material is not rigid in its own rest frame and the stress–energy tensor is no longer that of “dust”; you need to include stresses, not just kinetic energy.

The question of density distribution, or more generally the stress-energy tensor of the disk, is separate from the question of Born rigidity; Born rigidity is a purely kinematic property, and it is simple to show that the congruence of worldlines describing a disk rotating at constant $\omega$ is Born rigid. (This congruence is usually called the "Langevin congruence" in the literature.) Born rigidity is also frame independent, since it only depends on the kinematic decomposition of the congruence, which is done entirely in terms of covariant objects. So as you state them, these points cannot be correct.

Your point about the stress-energy tensor components possibly having to become arbitrarily large as the rim of the disk is approached, in the limit $\omega R \to 1$, might be worth expanding on with some actual math, if you can. (The Greg Egan treatment that was referenced earlier was at least one attempt at this, and might be worth looking at.)

 

[PeroK]

I claim the integral in the OP, evaluated between specified r values below R corresponding to speed c, will always be finite, and no limiting process for the either inner or outer r value will be unbounded. This contradicts your claim.

Sure, if you do the integral first and then take the limit as the velocity tends to $c$. But, if you consider the construction of the integral for velocity tending to $c$ (i.e. swapping the order you take the limits), then the definite integral does not converge.

 

[PeterDonis]

It’s possibly important to understand what M means.

I think the properties of the final state are important for that. But I don't think the spin-up process is.

The construction of the integral means it summed from constituent rest masses.

I agree that, in the final state, exactly what "uniform mass distribution in the COM frame" means, and whether the integral in the OP properly captures that, is a good question (which I don't think has been fully answered yet). But again, I don't think analyzing the spin-up process is necessary for that.

 

[Thor Jackson]

Yes, there is. It is perfectly possible to have a disk whose motion is rigid at a constant angular velocity $\omega$. SR does not rule that out. That is the physical model that is being analyzed in this thread.

What is not possible is to have a rigid disk with a changing angular velocity--so you can't spin up a disk from rest to some nonzero $\omega$ with the motion being rigid the whole time. But, as I said in post #33, I don't see the spin-up process being included in the analysis being done in this thread, nor does it need to be.


The question of density distribution, or more generally the stress-energy tensor of the disk, is separate from the question of Born rigidity; Born rigidity is a purely kinematic property, and it is simple to show that the congruence of worldlines describing a disk rotating at constant $\omega$ is Born rigid. (This congruence is usually called the "Langevin congruence" in the literature.) Born rigidity is also frame independent, since it only depends on the kinematic decomposition of the congruence, which is done entirely in terms of covariant objects. So as you state them, these points cannot be correct.

Your point about the stress-energy tensor components possibly having to become arbitrarily large as the rim of the disk is approached, in the limit $\omega R \to 1$, might be worth expanding on with some actual math, if you can. (The Greg Egan treatment that was referenced earlier was at least one attempt at this, and might be worth looking at.)

Thanks — that’s a fair clarification. I agree that a disk in uniform rotation can satisfy Born rigidity; the Langevin congruence is the standard example, and nothing in my earlier comment was meant to dispute that kinematically.

The point I was trying to get at is that once you go beyond the kinematics and actually look at the stress–energy required to support that Born‑rigid motion at high rim speeds, the situation becomes much less benign.

For a Born‑rigid rotating disk, the tangential velocity field is fixed by the congruence, but the material stresses needed to maintain that motion grow with radius. In fact, if you take the usual expression for the required circumferential stress in the local rest frame, it scales roughly like

Tϕϕ∼γ2(r) r2ω2,
so as r→c/ω, the Lorentz factor diverges and the stress does as well. That means the stress–energy tensor cannot remain finite all the way out to the light‑speed radius.

So even though the kinematic description of a rigidly rotating disk is well‑defined, the physical realization of such a disk with finite material stresses is not. Any real material would fail long before the rim approached c, and the stress–energy blow‑up is exactly what shows up when you try to compute the total energy.

That’s why the integral in the OP behaves oddly: the model being integrated is kinematically consistent but not dynamically realizable once the rim speed becomes ultra‑relativistic. The stress–energy required to maintain rigidity is what actually diverges, not the kinetic‑energy density of dust.

 

[PeroK]

The maths of your integral is fine

There's an ambiguity over which limit to take first. Even if the physical model is not valid, there's a question over the validity of the mathematics.

That said, a more sophisticated physical model might lead to a different integral and no ambiguity in the result.

 

[PeterDonis]

one can imagine that the "disc" is a swarm of many little rockets (the illustration below is generated by ChatGPT), each with its own engine, and without the attractive forces between them, but with a communication system which makes their motions synchronized. In that case there are no internal stresses at all.

But there is additional stress-energy in the problem, because the little rockets have to have fuel and rocket exhaust, and the direction of their thrust has to vary with time. So you're just substituting that extra stress-energy (which is going to be much more intractable to analyze) for the material stresses in an ordinary disk where the inter-atomic forces are what constrain the motion of each element.

 

[PeterDonis]

it scales roughly like

Please use the PF LaTeX feature to post equations. There is a LaTeX Guide link at the bottom left of each post window.

 

[PeroK]

The proposed integral in the OP impies the following physical process: a disc of dust with uniform density in some inertial frame is spun up (totally non rigidly) under the constraints that density of dust per unit area in the inertial frame remains constant.

Dust is not a continuous distribution of matter.

 

[renormalize]

Dust is not a continuous distribution of matter.

Neither is an ideal fluid, since it's made of discrete molecules that interact in a way that prevents compression and friction, yet we can usefully model such a fluid as a continuum. Why shouldn't it be possible to similarly model a system of dust particles as continuous?

 

[PeterDonis]

Dust is not a continuous distribution of matter.

It is if we are using a typical continuum model, where "dust" is just another name for "a perfect fluid with zero pressure and stress". That seems to be basically what the OP of the thread is modeling.

 

[PeroK]

Neither is an ideal fluid, since it's made of discrete molecules that interact in a way that prevents compression and friction, yet we can usefully model such a fluid as a continuum. Why shouldn't it be possible to similarly model a system of dust particles as continuous?

You can, but perhaps you can't meaningfully have the boundary of the dust cloud moving at $c$. In one sense you can, because the boundary has zero mass and zero energy. This thread is trying to make sense of the consequences of that.

 

[PeterDonis]

perhaps you can't meaningfully have the boundary of the dust cloud moving at $c$.

There is such a thing as null dust in the literature. But it's not intended to cover this kind of case.

I think @Demystifier made the point earlier that we can actually remove the boundary and just consider the open region of the disc inside it, where all the worldlines are timelike. He did it in the context of his finite model, but you could do the same thing in the continuum model. The gamma factor in the COM frame that's associated with the worldlines as the rim is approached still will increase without bound.

 

[anuttarasammyak]

But of course we want K to diverge even without adding material

In the c limit two OP disks and this disk both have mass 2M. KE of the former is finite 2M, that of the latter diverges.

If we don’t provide material to the disk, it becomes a ring KE of which diverges.