# Stress-Energy tensor of a rotating disk

1. Jul 4, 2006

### pervect

Staff Emeritus
I'm getting a rather crazy looking result, but I'm beginning to think it may be right.

Unfortunately, I haven't been able to find any specific references on the topic to check my sanity level.

Basically, I'm finding that in relativistic terms, there are no pressure (or tension) terms in the stress-energy tensor of a rotating disk. (Perhaps I should say - there are not necessarily any such terms).

This is different from the engineering result. But I believe that the difference is due to the fact that in engineering, the stress-energy tensor is taken to be comoving with the disk. i.e:

http://en.wikipedia.org/wiki/Stress-energy_tensor

If we adopt a cylindrical coordinate system (t,r,theta,z) the stress-energy tensor is just

$$\begin{array}{cccc} rho(r) & 0 & p(r) & 0\\ 0 & 0 & 0 & 0\\ p(r) & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{array}$$

rho(r) and p(r) are two arbitary functions, representing the energy density and the momentum density.

Rigidity of the disk will impose a relationship between p and rho, uniformity of the disk will give us another constraint.

If this is NOT correct, then my understanding of the continuity equation $\nabla_a T^{ab}[/tex] is wrong and needs to be fixed. The above stress-energy tensor satisfies the above equation - adding any radial tension terms would spoil this happy state of affairs. Comments? References? Brickbats? Last edited: Jul 4, 2006 2. Jul 4, 2006 ### coalquay404 This may or may not be useful to you, but if I recall correctly, the canonical reference for the (relativistic) rotating disk is a chapter in "General Relativity and Gravitation", Held, Volume 1. I believe that it's the first chapter but it's been a couple of years since I've read a copy of this book so I'm a bit iffy on the details. 3. Jul 5, 2006 ### pervect Staff Emeritus Thanks, it sounds like it might be worthwhile getting on interlibrary loan. (I think I found the book you mean in the Library of Congress catalog). I think I'm getting sensible-looking results, though. In non-rotating coordinates, the continuity equation I'm getting is: $${\frac { {\frac {d Prad}{dr}} r +{\it Prad} -{r}^{2}{\it P\theta} }{r}}$$ Prad being the radial pressure, P[itex]\theta$ being the circumfrential pressure.

This equation implies that if $P\theta$=0, Prad=k/r^2. For a finite pressure at r=0, Prad=0 if Ptheta=0.

Converting to co-rotating coordinates

(t1=t,r1=r,$\theta$1=$\theta-\omega$t,z1=z)

changes the metric

g_tt = -1 + r1*w^2
g_t$\theta = \omega r1^2$

The stress-energy tensor in this new metric should now have T^0j=0 because everything is stationary in the new coordinates.

In this corotating metric, the continuity eq becomes

$${\frac {-\rho {\omega}^{2}{{\it r1}}^{2}+ {\frac {d Prad}{d{\it r1}}} { \it r1} +Prad - {\it P\theta} {{\it r1}}^{2}}{{\it r1}}}=0$$

and we now see $\rho r \omega^2$ terms in the radial pressure which I initially expected.

Last edited: Jul 5, 2006