KE of rotating disc

[PAllen]

Note that if we change the mass assumption to be constant density in each element's local comoving frame, we get for an arbitrary annulus:
$$E=\frac{\pi\rho}{\omega^2}\left(\ln(\omega^2 r_1^{~2}-1)-\ln(\omega^2 r_2^{~2}-1)\right)$$
This, of course, diverges if the outer edge speed approaches c. The difference is all in the mass distribution.

So if we spin up a uniform density dust disc to an angular velocity such that its rim approaches c, the total energy of the disc remains finite. While if we build a spinning disc by adding material as we add each annulus, with locally observed density constant, then the energy of the disc grows without bound as the outermost added annulus approaches speed c in the COM frame.

 

[pervect]

One interesting point from that post - I actually explicitly wrote the stress energy tensor as
$$\rho u \otimes u + P \omega \otimes \omega$$

I recall a bit more about that now. The vector u is just the 4-velocity of a point on the rotating hoop. w is a vector perpendicular to u, representing the direction of the stress.

Then $\rho u \otimes u$ is just a coordinate independent way of writing the stress-energy tensor contribution of the mass densityh of the disk, and $P w \otimes w$ is just a coordinate independent way of writing the contribution of the stress in the hoop to the total stress-energy tensor. In the rest frame, u has components (1,0) and w has components (0,1), when we plug those components we get the components of the stress energy tensor in the lab frame. When we want the components of the stress energy tensor in the lab frame, we just replace them with the components of u in the lab frame and the components of w in the lab frame.

It's a more subtle question to ask if $\rho$ really stays constant when we spin up the hoop.

 

[renormalize]

While the stress energy tensor is mostly used in GR, it originates in SR. That said, it's usually only brought up in the context of GR, the only author I've read that talks about it in a SR context is Rindler, and I don't particularly care for his treatmen.

Your experience is very different from mine. I learned about the stress-energy-momentum tensor from, among others:

• Jackson, Classical Electrodynamics, §12.10: Canonical and Symmetric Stress Tensors, Conservation Laws (GR never introduced.)
• Weinberg, Gravitation and Cosmology, Chap. 2: Special Relativity, §8, §10-§11 (GR not introduced until chap. 3).
• Landau & Lifshitz, The Classical Theory of Fields, Chap. 4: The Electromagnetic Field Equations, §32-§35 (GR not introduced until chap. 10).
• Wald, General Relativity, §4.2: Special Relativity, pp. 61-64 (GR introduced in §4.3).

 

[PeroK]

Note that we can get a similar paradox by considering a length of elastic material that stretches out until instantaneously it has a certain rest-mass density $\rho(x)$ and a velocity profile $v(x)$. Nominally, the instantaneous energy of the expanding material is:
$$E = \int_0^L \gamma(x)\rho(x) \ dx$$Now choose linear functions for $\rho$ and $v$:
$$\rho(x) = \rho(1 - \frac x L), \ \ v(x) = \frac x L$$For which the energy is:
$$E = \rho \int_0^L \frac{1 - \frac x L}{\sqrt{1 - \frac {x^2}{L^2}}} \ dx = \rho L(\frac \pi 2 - 1)$$And, apparently, we have a finite energy for a massive object, one end of which is moving at $c$.

Again, the physics might need to be more sophisticated to be physically valid. But, I think the lesson is that we have to be careful with improper integrals. (Although, it may not even be an improper integral, as the integrand can be rewritten as $\sqrt{(1 - x/L)/(1+x/L)}$.

 

[A.T.]

So if we spin up a uniform density dust disc to an angular velocity such that its rim approaches c, the total energy of the disc remains finite.

Isn't that just an artifact of the unphysical uniform density assumption?

A single massive particle going in circles approaching c will have energy not bound by any finite value. Adding more such particles on/within that circle, also moving in circles all at the same orbital $\omega$, will not counter/reduce that. It's only when you model them as a uniform mass density distribution, that the total energy is reduced to stay below a certain bound.

To me this seems like a inappropriate model then. Or a general limitation of continuum models for systems where some particles have diverging properties.

 

[PAllen]

Note that we can get a similar paradox by considering a length of elastic material that stretches out until instantaneously it has a certain rest-mass density $\rho(x)$ and a velocity profile $v(x)$. Nominally, the instantaneous energy of the expanding material is:
$$E = \int_0^L \gamma(x)\rho(x) \ dx$$Now choose linear functions for $\rho$ and $v$:
$$\rho(x) = \rho(1 - \frac x L), \ \ v(x) = \frac x L$$For which the energy is:
$$E = \rho \int_0^L \frac{1 - \frac x L}{\sqrt{1 - \frac {x^2}{L^2}}} \ dx = \rho L(\frac \pi 2 - 1)$$And, apparently, we have a finite energy for a massive object, one end of which is moving at $c$.

Again, the physics might need to be more sophisticated to be physically valid. But, I think the lesson is that we have to be careful with improper integrals. (Although, it may not even be an improper integral, as the integrand can be rewritten as $\sqrt{(1 - x/L)/(1+x/L)}$.

I don’t think the improper integral is the issue at all. In some sense, you never need to get there. Intuition suggests that any chosen energy should be exceeded before anything actually reaches c. The math shows otherwise.

 

[PAllen]

Isn't that just an artifact of the unphysical uniform density assumption?

A single massive particle going in circles approaching c will have energy not bound by any finite value. Adding more such particles on/within that circle, also moving in circles all at the same orbital $\omega$, will not counter/reduce that. It's only when you model them as a uniform mass density distribution, that the total energy is reduced to stay below a certain bound.

To me this seems like a inappropriate model then. Or a general limitation of continuum models for systems where some particles have diverging properties.

But the uniform COM frame density follows necessarily from any simple (i.e. not including stress energy of holding the disc together) model of spinning up a disc. Note, there is not attempt to maintain rigidity during this process. I think the core of the intuitive disconnect must be connected to manner of finite particle to continuum transition. Earlier, @Demystifier showed a finite particle model with the same behavior as the continuum model. I am going to review this in more detail now, to try to find what assumptions in such a transition lead to bounded vs unbounded energy

 

[PAllen]

I think I have the key notion for the finite to continuum transition that explains the counterintuitive continuum result. The uniform density model implies that the total rest mass in any annulus stays constant during spin up. An annulus can be modeled as a ring of particles. The spin up process says this ring of particles preserves particle number and radius as spun up. The key question is when do you stop. The continuum model has the outer edge of the smoothed annulus reaching c when the the center of the smoothed annulus has some specific speed less than c. The continuum model says no further speed up is allowed. But the particles are then moving at a specific maximum speed less than c. Put conversely, there is a specific particle speed above which the transition to a smooth annulus would put the rim having speed greater than c. This is the core of the intuitive disconnect.

 

[PAllen]

Also note that the constraint of constant density per a local, comoving observer, means adding particles to outer rings as they are sped up. And, for the outermost ring, the number of high speed particles you add, plus their KE, can exceed any bound before the smoothed rim velocity reaches c. But this is definitely not a model of spinning up a disc.

 

[PAllen]

One further way of looking at this is that transitioning from a ring of point particles to a smooth annulus adds width to the particles. The amount of width depends on the chosen particle mass and chosen density. For any such width, the maximum particle speed is capped at a specific value below c by the constraint of smoothed rim velocity being c.

 

[PAllen]

I think there is no nice way to allow a continuum model to capture the notion that there is no upper bound to particle energy before c is reached. To do this, you would need to have nonzero rest mass exactly at the rim. With standard functions, this is impossible. Thus, to get a continuum model showing unbounded energy as the disc is sped up, it appears you would need to involve a delta function for the rim.

 

[Sagittarius A-Star]

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 $N\to \infty$.

I include in the summation the outer ring N and set it's speed slightly below the speed of light.

$E = \sum_{i=1}^N \frac{2\pi\lambda r_1i}{\sqrt{1-\omega^2r_1^2i^2}}$
with $\lambda = \frac{M}{\pi r_1 N(N+1)}$, $v:=\omega R$ and $R=r_1N$
$\Rightarrow$
$E/M = \sum_{i=1}^N \frac{2i}{N(N+1)\sqrt{1-v^2i^2N^{-2} }}$

Numerical calculation for $N=R=1000$ and $v= 0.999999999$
$E/M \approx 46.61$
This is larger than the OP result.

 

[PAllen]

And one more thing… the issue of transitioning to a continuum is a problem for the radial direction, but not the tangential. So the following mixed approach captures the expected intuition. Consider a finite set of one dimensional rings, with mass for each chosen per constant linear density per COM frame. Then, consider increasing angular velocity. Now, any chosen total energy will be exceeded before the speed of the outer ring reaches c.

 

[Demystifier]

I include in the summation the outer ring N and set it's speed slightly below the speed of light.

$E = \sum_{i=1}^N \frac{2\pi\lambda r_1i}{\sqrt{1-\omega^2r_1^2i^2}}$
with $\lambda = \frac{M}{\pi r_1 N(N+1)}$, $v:=\omega R$ and $R=r_1N$
$\Rightarrow$
$E/M = \sum_{i=1}^N \frac{2i}{N(N+1)\sqrt{1-v^2i^2N^{-2} }}$

Numerical calculation for $N=R=1000$ and $v= 0.999999999$
$E/M \approx 46.61$
This is larger than the OP result.

Interesting! How much the last term $i=N$ contributes to the full sum? What happens if you take $v=1$ but compute the sum only up to $N-1$? What happens with larger $N$, say $N=10^4$?

 

[PAllen]

It seems obvious to me that the finite particle approach can be made to exceed any energy bound, based on extrapolation from my post #73. For me, my posts #70 and #73 answer all intuitive issues around the continuum model.

 

[Sagittarius A-Star]

What happens if you take $v=1$ but compute the sum only up to $N-1$?

I get from Wolfram Alpha 1.93277
Calculation 1

What happens with larger $N$, say $N=10^4$?

I get with else original values 4.23555
Calculation 2

 

[Sagittarius A-Star]

Interesting! How much the last term $i=N$ contributes to the full sum?

I get from Wolfram Alpha 44.676683503771391589009225229630072178116008914652013623324856498
Calculation

 

[Demystifier]

I get from Wolfram Alpha 1.93277
Calculation 1

That is in accordance with the result by OP, who obtained $KE=E-M=M$, which implies $E=2M$. This demonstrates that my removal of the $N$-th ring with the speed of light is justified. With bigger $N$ one should get a result even closer to 2. Indeed, $N=10^4$ gives 1.979,
https://www.wolframalpha.com/input?i2d=true&i=Sum[Divide[2j,10000\(40)10000+1\(41)Sqrt[1-Power[1,2]Power[j,2]Power[10000,-2]]],{j,1,9999}]
and $N=10^5$ gives 1.993
https://www.wolframalpha.com/input?i2d=true&i=Sum[Divide[2j,100000\(40)100000+1\(41)Sqrt[1-Power[1,2]Power[j,2]Power[100000,-2]]],{j,1,99999}]

Thank you for the cooperation on this mini project!