# Expansion tensor on rotating disk

1. May 27, 2014

### Staff: Mentor

Hi Everyone,

Suppose that we have cylindrical coordinates on flat spacetime (in units where c=1): $ds^2 = -dt^2 + dr^2 + r^2 d\theta^2 + dz^2$

I would like to explicitly calculate the expansion tensor for a disk of constant radius R<1 and non-constant angular velocity $\omega(t)<1$. I don't know how to do this. All I know that it starts with defining a congruence that represents the material of the disk.

Each point on the disk would have a four-velocity of
$$v=\left( \frac{1}{\sqrt{1-r^2 \omega^2}},0, \frac{r\omega}{\sqrt{1-r^2 \omega^2}},0\right) = \gamma \partial_t + r\omega\gamma\partial_{\theta}$$ where $\gamma=(1-r^2 \omega^2)^{-1/2}$

I think that is the congruence, but where do I go from there?

Last edited: May 27, 2014
2. May 27, 2014

### Staff: Mentor

Wikipedia has a brief definition of the expansion tensor (as well as the other elements of the kinematic decomposition):

http://en.wikipedia.org/wiki/Congru...atical_decomposition_of_a_timelike_congruence

The computations are tedious (at least, I find them so ) but straightforward.

For this particular case, it looks to me like there should be zero shear, so all the information about expansion (if any) should be in the expansion scalar, which is easier to calculate; it's just $\nabla_a u^a$, i.e., the (covariant) divergence of the 4-velocity.

3. May 27, 2014

### WannabeNewton

Hi Dale, see http://arxiv.org/pdf/gr-qc/0312087v2.pdf

The authors do not explicitly write out the details of the calculations, rather they simply explicitly write down the expansion and rotation tensors from the definitions, but this does let you check the result of your own calculations.

As Peter remarked, the shear tensor vanishes identically as we would expect given the rotational symmetry of the entire problem. The expansion tensor is non-vanishing of course due to the fact that the 4-velocity field of the time-like congruence is not parallel to a time-like Killing field of flat space-time, which by the Herglotz-Noether theorem implies the non-Born rigidity of the congruence since the 4-velocity field has non-vanishing vorticity.

Last edited: May 27, 2014
4. May 28, 2014

### Staff: Mentor

Thanks Peter and WBN, I will get back to you if I get stuck.