# Stress energy tensor of a rotating hoop

1. Jul 17, 2008

### jdstokes

This problem comes from the GR problem book, question 5.1 (b).

The setup is a ring of N identical masses rotating in the x-y plane at a distance $a$ from the origin. Assume that N is large enough so the ring may be treated as continuous. Ignore the stress energies keeping the particles in orbit.

According to the solution, the energy density of the ring is

$T^{00} = \frac{Nm\delta(r - a)\delta(z)}{2\pi a\gamma}$.

I'm having trouble understanding why this is the case. It is certainly true that for a stationary ring the rest-energy density is

$\rho_0 = \frac{Nm\delta(r-a)\delta(z)}{2\pi a}$

which can be easily seen by integrating over all of space in cylindrical coordinates. Alternatively it can be shown using the fact that $\rho_0 2\pi r dr dz = Nm$ and then dividing through by the infinitesimal $2\pi r dr dz$.

Why would it be that the energy density decreases as the angular velocity is increased?

Last edited: Jul 17, 2008
2. Jul 17, 2008

### jdstokes

It looks like I must have miss-read the solution. I think the idea is to realize that since there is no change to the volume of the ring, the rest-energy density picks up a factor of $\gamma$ as opposed to $\gamma^2$ which would be the case for a contracted rod e.g.

3. Jul 17, 2008

### Antenna Guy

It appears to me that the contracted circumference is $\frac{2\pi a}{\gamma}$.

Regards,

Bill

4. Jul 17, 2008

### jdstokes

I thought perpendicular motion is unaffected by Lorentz contraction? Since the radius is everywhere perpendicular to the direction of motion it should remain constant and consequently so should the circumference?

5. Jul 17, 2008

### Antenna Guy

Is it not true that a differential element along the circumference is parallel to velocity at that point?

Regards,

Bill

6. Jul 17, 2008

### jdstokes

True, but I can't seem to reconcile this with the radius problem. Clearly one quantity can't change without affecting the other.

In any case, if what you purport is true, then the factor of $\gamma$ on the denominator would cancel the $\gamma$ coming from the energy implying that the energy density is independent of angular velocity (false).

7. Jul 17, 2008

### jdstokes

For reference, the expression given for $T^{00}$ in the book is

$T^{00}= \frac{\gamma Nm\delta(r-a)\delta(z)}{2\pi a}$.

What I can't understand is why they claim that

$\rho_0 = \frac{Nm\delta(r-a)\delta(z)}{2\pi a\gamma}$

(note the gamma on the denominator). Surely the rest-mass density should be independent of gamma as I indicated in the OP.

8. Jul 17, 2008

### jdstokes

After reading a bit more about this, it turns out that the circumference does indeed contract and the radius remains constant. This can happen because of the fact that the measurement of the radius requires the addition of distances made over a continuum of different frames of reference while the radius is measured in one, comoving frame.

This implies that the reasoning of my earlier post is incorrect.

Here is the justification given by Lightman et al.

Suppose we are in the frame where the disc rotates with angular speed $\omega$. The total energy of the disc is $N m \gamma$. Thus

$T^{00} 2\pi r dr dz = N m \gamma \implies T^{00} = \frac{Nm\gamma\delta(r-a)\delta(z)}{2\pi a}$.

The confusing thing is why they use $2\pi r$ even though the disc is rotating so its circumference is contracted to $2\pi r/\gamma$. It looks like the authors have made a mistake here.

A further piece of evidence suggesting that the authors are in error is their bizarre expression for $\rho_0$ which they conclude depends on the inverse of gamma. If one assumes that the circumference contracts, one will find that $\rho_0$ is indeed independent of gamma as it should be.

Interesting.

9. Jul 17, 2008

### Mentz114

If the radius decreases and the circumfernece does not, this implies that the space-time is curved. Cosidering we have accelerating frames this is not a surprise.

The best paper I've seen on rotating frames is here ( reprinted this year)

arXiv:gr-qc/9805089

but they don't look at the SET so maybe not relevant.

10. Jul 17, 2008

### Antenna Guy

The problem at hand appears to be the opposite - the circumference contracts, but the radius does not.

The energy density of the rotating hoop has been given in strange units that look [to me] like:

kg*differential_area/(contracted_circumference*dilated_second) (or, kg*m/s).

I'm out of my league here, so don't assume "dilated_second" is where the extra gamma comes from.

Regards,

Bill

11. Jul 17, 2008

### Mentz114

Hi Bill,
Well, same conclusion, different sign of curvature. Actually I calculated the Riemann tensor for the 2-d rotating disc and of course it is all zero so curvature is irrelevant it seems.

This topic was covered about a year ago by Pervect and others but I can't find the post.

12. Jul 17, 2008

### Antenna Guy

I don't know if it matters, but the differential element in z seems to indicate 3-d (i.e. a differential segment along the length of a cylinder).

Regards,

Bill

13. Jul 17, 2008

### Antenna Guy

Last edited by a moderator: Apr 23, 2017
14. Jul 17, 2008

### Mentz114

Bill,
that's the one. I'm sorry the paper I cited isn't relevant, being more about the relativity of rotating frames. Worth finding that thread though, it's got some big guns.

Yes, that might make a difference. No time to play with it now, unfortunately.

M

15. Jul 18, 2008

### jdstokes

Getting back to the OT, the solution is now to clear to me.

Since the hoop is rotating at a radius $a$ from the origin in the lab frame, the circumference is dilated in the frame of the hoop to $2\pi a \gamma$, but the radius remains fixed at $a$. Thus the rest-mass density must be

$\rho_0 = \frac{Nm \delta(r-a) \delta(z)}{2\pi a \gamma}$.

At each point along the hoop, the the constituents of the hoop may be regarded as dust traveling with velocity tangential to the hoop. The energy density of the hoop is thus given by 00 component of $\rho_0 u\otimes u$, ie

$T^{00} = \gamma^2 \rho_0 = \frac{\gamma Nm \delta(r-a) \delta(z)}{2\pi a}$