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!

 

[Sagittarius A-Star]

$E = \sum_{i=1}^N \frac{2\pi\lambda r_1i}{\sqrt{1-\omega^2r_1^2i^2}}$
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.

For a rotating disk with only 3 rings (because of Wolfram Alpha calculation limitations):
$\lim_{v \rightarrow 1} {E/M} =$
$\lim_{v \rightarrow 1} {\sum_{i=1}^3 \frac{2i}{12\sqrt{1-v^2i^23^{-2} }}} = \infty$
Calculation

 

[Demystifier]

For a rotating disk with only 3 rings (because of Wolfram Alpha calculation limitations):
$\lim_{v \rightarrow 1} {E/M} =$
$\lim_{v \rightarrow 1} {\sum_{i=1}^3 \frac{2i}{12\sqrt{1-v^2i^23^{-2} }}} = \infty$
Calculation

That's infinite for any $N$, not only for $N=3$, due to the infinite last term $i=N$ in the sum.

 

[PAllen]

I want to add a few more observations about this topic. I think the key point is that there is nothing wrong or paradoxical about the OP result when properly understood. Comparing the continuum case to "what about a particle on the rim" neglects that with a continuum model, the total mass all around the rim is zero. Given this, it is normal in SR for limiting cases of mass approaching zero and speed approaching c to end up with finite energy; done properly, this leads to the kinematics of massless particles.

The other key aspect is linear versus nonlinear motion. If a continuum shape is moving in a straight line, all its points are, so the total mass moving at that one velocity is finite. In the case of circular motion, each ring of the continuum is moving at a different speed, so the density profile matters.

Finally, to quantify this last point, I want to show how circular versus linear motion is fundamental to understanding this problem with a straightforward computation. We start from the formula for the total energy of an annulus that I gave in post #59:
$$E=\frac{2\pi\rho}{\omega^2}\left(\sqrt{1-\omega^2 r_1^{~2}}-\sqrt{1-\omega^2 r_2^{~2}}\right)$$
Now, consider holding the mass and width of the annulus constant as we increase the outer radius ($\rho=\frac{m}{\pi(r_2^{~2}-r_1^{~2})}$, $r_1=r_2-h$). Then let the rim velocity be c: ($\omega=1/r_2$). Finally, dropping the subscript on $r_2$ letting it be simply r, we get (after a little messy algebra):
$$E=2m\sqrt{\frac{r}{2h-(h^2/r)}}$$
Note, that $h=r$ gives 2m, as expected. Now, let $r=kh$ so k is a measure of linearity of an annulus of fixed mass and width:
$$E=2m\sqrt{\frac k{2-1/k}}$$
Again, note k=1 gives 2m, but as k increases (increasing linearity), the energy grows without bound.

 

[PeroK]

I want to add a few more observations about this topic. I think the key point is that there is nothing wrong or paradoxical about the OP result when properly understood. Comparing the continuum case to "what about a particle on the rim" neglects that with a continuum model, the total mass all around the rim is zero. Given this, it is normal in SR for limiting cases of mass approaching zero and speed approaching c to end up with finite energy; done properly, this leads to the kinematics of massless particles.

Following this principle, you could apply the Navier-Stokes equations and replace the molecular physics of water with continuous fluid flow. You can't do this because the continuum model breaks down at some appropriately small scale.

Given, therefore, that the continuum model is an approximation to whatever there is at the smallest scales of matter, it's somewhat irrational to insist on applying it beyond the scales where it could possibly be a valid physical model.

 

[PAllen]

Following this principle, you could apply the Navier-Stokes equations and replace the molecular physics of water with continuous fluid flow. You can't do this because the continuum model breaks down at some appropriately small scale.

Given, therefore, that the continuum model is an approximation to whatever there is at the smallest scales of matter, it's somewhat irrational to insist on applying it beyond the scales where it could possibly be a valid physical model.

On the other hand, a particle model is, itself, an approximation. If you believe QFT, there are only continuous fields at the root. And if you consider particles as blobs of continuum, then the issues of the continuum model arise in the same way.

I think it is perfectly ok to consider the continuum model as such, contrasting it with particle models, with context indicating which reasonable for a given real world situation. Obviously, no model of a disc with rim velocity near c matches any achievable physical set up, even in principal.

 

[pervect]

I've come to realizie that when we spin up the disk, we have a choice to determine what stresses hold it together. If we make all the stresses radial, and make the circumferential stresses zero (or negligible), the problem becomes a lot more intuititve, and better behaved to boot.

This was somewhat inspired by thinking about the "rocket" analogy, and making a related one which was a closed system which had a stress energy tensor with a vanishing divergence.

My last, longer attempt at a post started having preview problems, so I'll make this shorter.

We start with a non-rotating disk with a uniform stress energy distribution of $T_0 = \rho_0 u_0 \otimes u_0$, using index free notation. If we use cylindrical coordinates $(t,r,\theta,z)$ $u_0$ would have components of (1,0,0,0) in the lab frame.

Then we spin up the disk. We'll call the stress energy tensor of the spun-up disk $T_1$. It's conceptually a different stress energy tensor after the spin-up. We have alternatives of how to spin it up, as I mentioned. We choose to make the circumfential stresses negligible, and have all the stresses holding it together to be radial. Furthermore, we can imagine that the disk doesn't change it's radius as we spin it up. So the disk is "rigid" in the radial direction, but "stretches" circumfentially.

To do the spin up, we define $u_1$ as the 4-velocity of a point on the spun up disk.

After spin-up, we can write the new stress energy tensor, T_1, as
$$ T_1 = \frac{\rho_0} {\gamma(r)} u_1(r) \otimes u_1(r) + P_1(r) r \otimes r $$

$u_1$ is now a function of r in the lab frame, and $\gamma(r)$ is the associated gamma factor. We're sort of overloading the notation a bit here, as gamma is being used to represent both the velocity (coordinate dependent) and stretching (coordinate independent).



The density of the disk in a co-moving frame is dropping when we spin it up because the disk stretches. But it rises in the lab frame as expected, by a factor of $\gamma$. We have one factor of gamma for the time component of u_1 in the lab frame, which gets squared, and a factor of 1/gamma for the density. So we wind up with a single factor of gamma.

And, as a bonus, we don't care about $P_1(r)$ if we are interested in the energy density. In the case where the stress is all radial, none of the components transform to affect the energy density, unlike in the case where there is significant circumfrential stress.

If we did care about the radial tension, we could compute it as always by$\nabla_a T^{ab}$ = 0. But since it doesn't have any terms to contribute to the lab frame energy density, we son't need to, unless we want to.

This analysis makes the assumption the disk doesn't radially stretch as we spin it up, which isn't really physically realistic for any sort of material, but it's easier to analyze, and shouldn't directly violate any physical laws other than possibly the weak energy condition. In a more general case, we'd need to consider the total stretching in all directions which reduces the energy density, and the amount of work we did in the process of doing said stretching which increases the energy density. In this case, there is only stretching in a direction with no tension, so there's no work being done, though the stretching itself affects the density.

I'm not sure exactly what people have been analyzing, having been a bit distracted, and also not reading what thread was being referred to as the "previous thread" (I'm not even sure what it was called). I think this scenario is a bit closer to the problem that people have been working on, though probalby not the same exact assumptions. But I think these assumptions give a better idea of what happens when we spin up a disk that had a uniform energy density when it was at rest, and consider a case that is simpler to analyze than some of the previous cases.

I do hope it is useful in its own right, though ideally it'd best be combined with consideration of other assumptions about the internal stresses, and why this would matter for the analysis even if we think we only care about energy. The meta point is that energy, momentum, and stresses are all related, part of a larger covariant entity, the stress-energy tensor.

 

[PAllen]

@pervect, thanks! I was having a similar thought that to make a spun up disc be valid in the sense of local conservation of momentum (divergence of SET 0), that it might be simplest to imagine it as a system spokes, with no circumferential connectedness, thus only radial stress. But I didn’t pursue it, so glad to see you did. Almost the whole thread, including all of my posts, have chosen to ignore the need for SET to have a valid closed system. In effect, this analyzes the total energy due to the distribution of rest mass, ignoring, for simplicity, any other contributions to the system.

 

[PAllen]

Actually, I think the radial tension must be related to a potential energy contribution to the T00 term, forced by zero divergence requirement. Imagine any spring constant, however large, acting in the radial direction. During spin up, you accumulate both tension and potential energy in the springs. Then, the total energy of the system reflects both the kinetic energy and the potential energy related to tension (assuming, COM frame, where other components of SET can e ignored for computing total energy). In other words, I am disputing the claim that you can assume zero radial expansion - I think you could not the satisfy the zero divergence requirement. So, in a consistent model of a spinning array of spokes (as the simplest representation of a spinning disc), you must include potential energy in the T00 term.

By the way, I don’t think this changes any of main issues under discussion in this thread. It would just add some multiplier of the same order of magnutude of as the KE, at most. So, the conceptual issues around which cases have unbounded energy versus bounded energy are unchanged.