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.