KE of rotating disc

[Ibix]

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.

 

[Sagittarius A-Star]

The rest-mass depends on $\omega$ because of internal stress.

 

[PAllen]

I think the cause is that the mass distribution is specified in the inertial frame. Then, in the local frame of a rotating piece, as the angular speed increases to a rim limit of c, the local density decreases to zero. I think I came up with a similar result some years ago.

 

[anuttarasammyak]

In your formula K=M for $\omega R=1$. I think that even the rim part speed reaches light speed, its element contribution is infinitesimal or at least finite in the volume integral. Supplemental minus stress energy would take place but I do not think it matters essentially in this discussion.

 

[Sagittarius A-Star]

The proper area in the non-Euclidean geometry doubles close to the speed of light:
$A(R)= 2\pi \int_0^R \gamma r \, dr= {2 \pi c^2 \over \omega^2}(1-\sqrt{1-\omega^2R^2/c^2})$
Source

 

[PAllen]

Qualitatively, what I am getting at is that as the rim speed approaches c, the number of small area elements of given measure (per a comoving inertial frame) along the rim, grows without bound. So the mass per such element goes to zero (because mass is specified in the COM inertial frame). This balances the growth of gamma.

But wait, this doesn’t fully explain it. It gets finite KE for each area element, in the limit, but the number of area elements grows without bound by gamma as well, so the KE of the rim should grow without bound.

 

[PAllen]

So, if you do just treat the rim as one dimensional, with a linear mass distribution per the COM frame, you do get KE unbounded in the limit. So it must be as @anuttarasammyak suggests, the falloff from the rim must be so fast as to give a finite result. Note that the total KE of M, in the natural units, is actually very large compared to normal speeds. It is also 4 times the Newtonian formula for the case of rim speed of c.

 

[Demystifier]

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.

What's going on?

Compute the integral in the interval $[R-\epsilon R,R]$ for small $\epsilon$, it should explain what's going on.

 

[Demystifier]

The proper area in the non-Euclidean geometry doubles close to the speed of light:
$A(R)= 2\pi \int_0^R \gamma r \, dr= {2 \pi c^2 \over \omega^2}(1-\sqrt{1-\omega^2R^2/c^2})$
Source

That's irrelevant, because the KE is computed in the laboratory frame (not in the disk frame), where the spatial geometry is Euclidean.

 

[Demystifier]

In your formula K=M for $\omega R=1$. I think that even the rim part speed reaches light speed, its element contribution is infinitesimal or at least finite in the volume integral. Supplemental minus stress energy would take place but I do not think it matters essentially in this discussion.

Exactly. To understand it quantitatively, one has to focus on the contribution of the integral near the $r=R$ region. For that purpose, one can introduce a new integration variable $y=R-r$ and consider the region of small $y$. Taking also $\omega R=1$, I find
$$\frac{1}{\sqrt{1-\omega^2 r^2}} \approx \frac{1}{\sqrt{2\omega y}} \propto y^{-1/2}$$
Even though it diverges for $y\to 0$, it diverges rather slowly. In that limit we also have
$$rdr \approx -Rdy \propto dy$$
so close to $y=0$ we have the integral proportional to
$$\int y^{-1/2} dy \propto y^{1/2}$$
which actually has zero contribution from the region $y\to 0$. This explains why the contribution from the region near the rim of the disc has a negligible contribution, despite the fact that it moves with the speed of light.

 

[pervect]

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.

I remember looking at relativistic hoops and disks a long time ago, but I no longer recall most of the details. I do recall Greg Egan had a treatment at https://www.gregegan.net/SCIENCE/Rings/Rings.html using a "hyper-elastic" material model. This may be significantly harder than what you want to do.

Veary basic suggestions from what I do recall:

Analyze a hoop, it's easier. I believe Sagittarius A-Star is on the correct track when he mentions that the rest mass of the disk depends on w because of the internal stresses.

Compute the stress energy tensor, including the tension terms. It's probably best to come up with a stress energy in the lab frame.

You'll need the stress energy tensor to formulate an analysis, because that's the only covariant entity. Without that concept, you won't be able to formulate the relativistic and covariant equations.

In lay language - pressure (or in this case stress) affects inertia.

I think the only nonzero terms in the stress-energy tensor for the thin hoop will be $T^{tt}, T^{t\theta},T^{\theta\theta}$, which correspond to energy density, momentum and/or angular/momentum density, and tangential stress. Radial stress $T^{rr}$ needs to vanish at the boundary for a thin hoop, and if the hoop is thin, it's constant through the hoop, so it can be set to zero.

What's the principle that allows you to solve for the tension terms? It's the fact that $\nabla_a T^{ab}=0$. That's the covariant replacement for what you might do with F=ma in a Newtonian analysis.

I already mentioned the boundary conditions, I think.

You'll probably want to use both coordinate basis and an orthonormal basis. The difference is that the vector pointing in the $\theta$ direction in the coordinate basis, $\partial / \partial_{\theta}$, does not have a unit length. I'd write $T^{\theta\theta}$ for the coordinate basis and $T^{\hat{\theta}\hat{\theta}}$ for the orthonormal basis.

If you insist on analyzing a disk, you'll have to some how figure how the stress that holds the disk together splits between radial and tangential stress, rather than must making the radial stress vanish.

Once you have the stress energy tensor in the lab frame, you can convert it to the "rest" frame of the disk.

Beware of the weak energy condition. If the stress exceeds the density, you'll get a negative energy density in the rest frame - this is probably unphysical.

Anyway, even in the simplest case, it's quite involved.

I recall weird things did happen with the rotating ring, there was an upper bound on the angular momentum in Egan's hyperelastic version, and when the derivative of angular momentum with rotation speed vanished, the model basically failed.

 

[Demystifier]

Analyze a hoop, it's easier.

Do you mean a hoop with zero or non-zero extension in the radial direction? In the case of a realistic non-zero extension, it's not easier at all. In the case of zero extension the KE will be infinite for $\omega R=1$, which does not help to understand why the result is finite for the disc.

I believe Sagittarius A-Star is on the correct track when he mentions that the rest mass of the disk depends on w because of the internal stresses.

I think it's a red herring which makes the analysis unnecessarily complicated. First, it depends on elastic properties of the material, so it misses the essential purely kinematic effect that does not depend on the material. Second, 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.

 

[PeroK]

We have two limits here. The first is the limit involved in the definite integral of a continuous mass distribution across a disk. We also have the limit as the velocity of the rim of the disk tends to $c$. The answer, purely mathematically, depends on the order we take these limits.

If we take a disk composed of a finite number of rings of finite mass, then there is an outer ring at a radius of $R$, with a finite mass. As we speed up the disk, the KE of this outer ring increases without bound.

If we do the integral first and then take the limit as speed of the rim of the disk tends to $c$, then we get a finite answer.

If we combine the two limits and allow the speed to increase in direct relation to the rate that we decrease the width of each ring ($\Delta r$) (and hence the mass of each ring), then we can get any finite between the two extremes.

I'm not convinced the answer lies in a more elaborate physical analysis based on stress-energy. In any case, we have two mathematical limits in the equation; and there is no unambiguous physical description that demands the order in which these limits are evaluated.

Note that we could, as a thought experiment, assume that there are a large number of particles coincidentally all moving instantaneously approximately in the form of a rotating disk of uniform density. There would be no additional internal energy (except gravitation, which would be insufficient to hold the disk together). The "disk" would exist only instantaneously. But, a calculation of the instantaneous KE would yield the result in the OP, as we take the number of particles to increase without bound and their mass to decrease proportionally to zero to maintain a constant mass of approximately uniform density.

The same paradox would apply. If we stop at any finite number of particles and let the speed of the outer particles increase towards $c$, then the KE increases without bound, as expected.

But, if we take the limit involved in the integral first and model the disk as a continuous distribution of matter, and then take the limit as the rim increases to $c$, then we get a finite KE.

It's similar to taking a sequence of particles of mass $m_n$ and speed $v_n$ and calculating the KE as the speed $v_n \to c$ and the mass deceases to zero. The answer depends on whether we let $m_n \to 0$ first, in which case the KE of a massless particle moving at $c$ appears to be zero. Or, if we let $v_n \to c$ first, then the KE of a massless particle moving at $c$ appears to be infinite.

And, we can get any non-zero answer by taking the appropriate sequences $m_n, v_n$, so that the KE is constant for each $n$. The disk case is more complicated, but it's the same mathematical principle of the ambiguity in the order that two limits may be evaluated.