# Frame dragging slowly rotating shell

1. May 18, 2015

### WannabeNewton

I was given a problem recently that I had to solve and I haven't been able to get a fully satisfactory conceptual grasp of a certain aspect of it. Consider the slowly rotating Kerr solution $ds^2 = -\alpha^2 dt^2 + \alpha^{-2}dr^2 + r^2d\Omega^2 - \frac{4Ma}{r}\sin^2\theta dt d\phi + O(a^2)$ where $\alpha^2 = 1 - \frac{2M}{r}$ for a slowly rotating thin spherical shell of radius $R$. Inside the shell the metric is just that of flat space-time i.e. $ds^2 = -\alpha_R^2 dt^2 + d\rho^2 + \rho^2 d\Omega^2$.

If one makes the transformation to a rotating coordinate system $\psi = \phi -\Omega t$ with an angular velocity $\Omega = \frac{2M a}{R^3}$ then the exterior Kerr metric becomes diagonalized when evaluated on the hypersurface swept out by the shell as is easy to show.

However it can be shown that in the $(t,\theta,\psi)$ coordinate system the shell still has an angular velocity $\omega = \frac{6Ma}{R^3}\frac{\alpha_R^2}{(1 - \alpha_R)(1 + 3\alpha_R)}$. It can be further shown that it is in fact the inertial observers inside the shell that have an angular velocity $\Omega$ with respect to the spatial infinity of the $(t,\theta,\phi)$ coordinate system and that the shell has an angular velocity $\Omega + \omega$ with respect to this system.

What I don't fully understand is why the Kerr metric on the hypersurface swept out by the shell is diagonalized by the corotating system of angular velocity $\Omega$ if this isn't the angular velocity of the shell but rather that of the inertial observers inside. Usually if we have of a rotating frame in which the metric on a rotating object is diagonalized then we would naively think of this frame as cortating with the object. Clearly that isn't the case here and I would like to understand why. In other words, why is it the angular velocity of the inertial observers inside the shell that diagonalizes the shell 3-metric and not that of the shell itself? Thanks for any insights!

2. May 18, 2015

### Staff: Mentor

Is it? Won't there be an extra term from frame dragging inside the shell?

3. May 18, 2015

### WannabeNewton

Sorry I should have mentioned that it is the above form if using $\psi$ i.e. $ds^2 = -\alpha_R^2 dt^2 + d\rho^2 + \rho^2(d\theta^2 + \sin^2\theta d\psi^2)$. This solves Einstein's equation inside the shell and satisfies the first Israel junction condition with the exterior slowly rotating Kerr 3-metric when evaluated on the shell so it should be fine.

4. May 19, 2015

### Staff: Mentor

I'm not sure I see how you get this from the Kerr metric. But first, I'm not sure I agree with the approximate Kerr metric you've written down in the usual coordinates; shouldn't the $dt d\phi$ term have just $\sin \theta$, not $\sin^2 \theta$?

5. May 19, 2015

### WannabeNewton

It's not from the Kerr metric, it's just the flat interior metric in the $(t,\theta,\psi)$ coordinates above; this solves Einstein's equations and also clearly agrees with the exterior slowly rotating Kerr metric given above in the exterior $(t,\theta,\psi)$ coordinates when evaluated on the shell, thus satisfying the 1st Israel junction condition which is necessary and, with the 2nd junction condition, sufficient in order for this to be a valid gluing of two space-time solutions with a hypersurface boundary separating the two space-times.

The $dtd\phi$ term in the exact metric is as given here: http://en.wikipedia.org/wiki/Kerr_metric#Mathematical_form and from this we have $\frac{4M r a\sin^2\theta}{r^2 + a^2\cos^2\theta} = \frac{4M a\sin^2\theta}{r}(1 + \frac{a^2}{r^2}\cos^2\theta)^{-1} = \frac{4M a\sin^2\theta}{r} + O(a^2)$ as desired.