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 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.

 

[A.T.]

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.

Is this a general problem of continuum models, where a collection of discrete particles with some property X (mass, charge, etc) is represented mathematically as a smooth density distribution of X?

In physical situations, where X of individual particles tends to infinity, the integration over the continuum can become ambiguous, because the contribution of the locations with infinite X can tend to zero.

 

[PeroK]

Is this a general problem of continuum models, where a collection of discrete particles with some property X (mass, charge, etc) is represented mathematically as a smooth density distribution of X?

In physical situations, where X of individual particles tends to infinity, the integration over the continuum can become ambiguous, because the contribution of the locations with infinite X can tend to zero.

Mathematically, it's that the sequence or sum diverges. Swapping the order of limits is not always reliable. See the simple example for a massless particle as the limit of massive particles.

 

[PeroK]

Is this a general problem of continuum models, where a collection of discrete particles with some property X (mass, charge, etc) is represented mathematically as a smooth density distribution of X?

In physical situations, where X of individual particles tends to infinity, the integration over the continuum can become ambiguous, because the contribution of the locations with infinite X can tend to zero.

See also the recent thread about the electric field of an infinite, uniformly charged plate.

 

[A.T.]

Mathematically, it's that the sequence or sum diverges. Swapping the order of limits is not always reliable.

Yes, but here one of the limits is due to actual physics (KE of the rim particles of constant finite rest mass going to infinity), while to other limit is merely due to the (unphysical) continuum assumption (relative contribution of the outermost disc rim hoop going to zero).

It seems to be a limitation (no pun intended) of continuum modeling.

 

[anuttarasammyak]

From OP

Say the material of the disk is incompressible fluid the density of which is
$$ \rho = \frac{M}{\pi R^2 d}$$
where d is thickness of the disk. As $\omega r$ approaches c, space metric of the rotating system provides another $\gamma$ factor. Say we pour the material to fill the vacant space in the disk, the integrand ( ) is
$$\frac{1}{1-\omega^2 r^2}-\frac{1}{\sqrt{1-\omega^2 r^2}}$$
Mass of the disk approaches to 2M, K diverges, though I am afraid this case is different from OP.

 

[A.T.]

From OP
View attachment 370475

Say the material of the disk is incompressible fluid the density of which is
$$ \rho = \frac{M}{\pi R^2 d}$$
where d is thickness of the disk. As $\omega r$ approaches c, space metric of the rotating system has another $\gamma$ factor. Say we pour the material to fill the vacant space in the disk, the integrand ( ) is
$$\frac{1}{1-\omega^2 r^2}-\frac{1}{\sqrt{1-\omega^2 r^2}}$$
K diverges to infinity and the mass approaches 2M though I am afraid this case is different from OP.

This is what @Sagittarius A-Star alludes to in post #5. The tangential length contraction means that to keep proper mass density constant, you have to add mass and the total rest mass goes to infinity as well.

But the total KE should also go to infinity, even if we keep the total rest mass constant, like in the OP.

 

[Sagittarius A-Star]

This is what @Sagittarius A-Star alludes to in post #5. The tangential length contraction means that to keep proper mass density constant, you have to add mass and the total rest mass goes to infinity as well.

But the total KE should also go to infinity, even if we keep the total rest mass constant, like in the OP.

Close to the speed of light the proper area is $2\pi R^2$ instead of $\pi R^2$.
To keep proper mass density constant, you have at maximum to double the proper mass.

 

[A.T.]

To keep proper mass density constant, you have at maximum to double the proper mass.

Yes, as @anuttarasammyak also wrote. Sorry I missed that. So the reason why K diverges computationally in this case, is that the rest mass is not added uniformly, but mainly on the rim.

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

 

[Demystifier]

There is one more way to look at the problem, to avoid the continuum distribution of matter in the radial direction. Consider $N$ concentric rings, each with the radius $r_k=kr_1$, $k=1,2,\ldots, N$, all rotating with the frequency $\omega$. Suppose that the $N$-th ring rotates with the speed of light, namely $\omega r_N=1$. Since the $N$-th ring is in fact unphysical, we regularize the problem by peeling off the $N$-th ring, i.e. we suppose that the last ring is in fact the $(N-1)$-th ring. Now the question is well posed, does the energy of the last ring diverge in the limit $N\to \infty$?

The mass of the $k$-th ring is $m_k=\lambda 2\pi r_k$, where $\lambda$ is the linear mass density. The energy of the $k$-th ring is therefore
$$E_k = \frac{m_k}{\sqrt{1-\omega^2r_k^2}}$$
so
$$E_{N-1} = \frac{2\pi\lambda r_1(N-1)}{\sqrt{1-\omega^2r_1^2(N-1)^2}}$$
Using $\omega^2r_1^2N^2=1$ and taking $N\gg 1$, I obtain
$$E_{N-1} \approx \frac{\sqrt{2}\pi\lambda}{\omega} \sqrt{N}$$
If $\lambda$ was fixed, it would diverge for $N\to \infty$. But actually, $\lambda$ is not fixed, it depends on $N$ too. What is fixed is the total mass $M$ of all rings together
$$M = \sum_{k=1}^N m_k = \sum_{k=1}^N \lambda 2\pi r_k = 2\pi \lambda r_1 \sum_{k=1}^N k = 2\pi \lambda r_1 \frac{N(N+1)}{2}$$
so
$$\lambda = \frac{M}{\pi r_1 N(N+1)} \approx \frac{M}{\pi r_1 N^2}$$
Hence the energy of the last ring is
$$E_{N-1} \approx \frac{\sqrt{2}M}{\omega r_1} N^{-3/2}$$
Fixing also the radius $R$ of the $N$-th ring, $R=Nr_1$, we finally obtain
$$E_{N-1} \approx \frac{\sqrt{2}M}{\omega R} N^{-1/2} = \sqrt{2}MN^{-1/2} $$
which does not diverge for $N\to \infty$.

 

[PeroK]

Consider the outer portion of the disk. For each $n$, consider the annulus with inner radius $r_n = R(\frac {n-1}n)$. The mass of that annulus is:
$$m_n = M (1 - \frac{(n-1)^2}{n^2}) = M(\frac{2n -1}n)$$Now approximate the speed of that annulus by the speed at the inner radius (assuming the speed of the rim is approximately $c$):
$$v_n = \frac{n-1}{n}$$The gamma factor is:
$$\gamma_n = \frac 1 {\sqrt{ 1 - v_n^2}} = \frac n {\sqrt{2n - 1}}$$This gives us a lower bound on the integral of:
$$e_n = \gamma_n m_n = M(\frac{\sqrt{2n - 1}} n)$$This tends to zero as $n$ increases.

The upper bound has $V_n = 1$, for which the gamma factor and energy are undefined. If, instead, we consider some finite but arbitrarily large outer speed, then the upper bound for the energy of that annulus is arbitrarily large. This is why integration ultimately fails as the rim of the disk tends to $c$. The upper and lower bounds on the integral do not converge.

Whatever the physics, the naive application of the mathematics fails for this reason.

 

[PAllen]

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

I think the implication is a similar phenomenon - the circumference of the rim grow unbounded, but this leads only to limiting area of 2 times Euclidean. The KE result is mathematically similar, except the limiting KE is 4 times Newtonian.

 

[pervect]

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 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.

Greg Egan did the non-zero extension case with a hyperelastic material model which he describes in detail. It was definitely involved, especially for a disk rather than a hoop. I believe he's revised his analysis since I looked at it a long time ago. At the time, there was some discussion of it on PF - it might still be around. I eventually came up with a Lagrangian, as I recall, after a very large number of false starts.

The problem with the model appears at points where a factor in the denominatorof the angular acceleration:

(z2 β + r2 (K2 r2 + z1) ω2)

is zero, and the numerator is non-zero.The result is that the equations of motion here have no solution!This coincides with the derivative of L with respect to ω being zero.

I looked, and I did write up something at that time about a thin relativistic hoop that maintains a constant circumference. If one spins the hoop up slowly, one can make the radius change in such a manner that the distance between nearby points doesn't change appreciably in the limit of a sufficiently slow spin up. The thin hoop approaches Born rigidity as an approximation (for distances between points measured on the hoop) in the limit of a slow spinup process .

There was an argument over this point (approximate rigidity) as I read back over the thread.

That was at https://www.physicsforums.com/threa...t-circumference-solving-egans-results.976746/

The hoop had the same issue with L reaching a peak. (and also, E reaching a peak). I don't think I realized it at the time I wrote it, but I think that similarly dooms the model.

One interesting point from that post - I actually explicitly wrote the stress energy tensor as
$$\rho u \otimes u + P \omega \otimes \omega$$
I'm not sure this is really right, though. I don't recall anymore why I wrote that.

I'm also not so confident that $\rho$ was not a function of time if we spin up the hoop. It seems suspicious for older me to claim it was, as we are adding energy to the hoop as we spin it up.

 

[PAllen]

The upper bound has $V_n = 1$, for which the gamma factor and energy are undefined. If, instead, we consider some finite but arbitrarily large outer speed, then the upper bound for the energy of that annulus is arbitrarily large. This is why integration ultimately fails as the rim of the disk tends to $c$. The upper and lower bounds on the integral do not converge.

I don't believe this is correct, and you haven't made any attempt to demonstrate this part. The calculation for such an annulus with outer rim just less than c is similar to @Demystifier's computation in post #10. From which you do not get any possibility of unbounded KE for the annulus.

Comparing this to a one dimensional ring, which trivially had unbounded KE limit, the annulus limiting process required to arrive at this is to take outer annulus speed approaching c, thickness of annulus going to zero and mass of annulus staying the same. This last part is fundamentally at adds with the OP model.

 

[PeroK]

I don't believe this is correct, and you haven't made any attempt to demonstrate this part. The calculation for such an annulus with outer rim just less than c is similar to @Demystifier's computation in post #10. From which you do not get any possibility of unbounded KE for the annulus.

Comparing this to a one dimensional ring, which trivially had unbounded KE limit, the annulus limiting process required to arrive at this is to take outer annulus speed approaching c, thickness of annulus going to zero and mass of annulus staying the same. This last part is fundamentally at adds with the OP model.

A one-dimensional ring is fundamentally unphysical. Integration involves the limit of finite rings or annuluses, which is where the problem lies. If you deconstruct the integral as such a limit, then the problem is clear.

 

[SiennaTheGr8]

Consider a disc of radius $R$ with a uniform mass distribution and total mass $M$ rotating in its own plane.

To clarify, what do you (and presumably other commenters in this thread) mean by "total" mass here? Do you mean the mass of the disc, or the sum of the masses of its constituents?

 

[PAllen]

A one-dimensional ring is fundamentally unphysical. Integration involves the limit of finite rings or annuluses, which is where the problem lies. If you deconstruct the integral as such a limit, then the problem is clear.

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.

 

[PAllen]

To clarify, what do you (and presumably other commenters in this thread) mean by "total" mass here? Do you mean the mass of the disc, or the sum of the masses of its constituents?

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. This implies that the number density of dust in an area element of the rim approaching c goes to zero, for an element as measured by a dust comoving inertial frame. Note that the computation of KE here is tantamount to determining the increase in invariant mass of the spun up disc.