Frame dragging slowly rotating shell

[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!

 

[PeterDonis]

Inside the shell the metric is just that of flat space-time

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

 

[WannabeNewton]

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

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.

 

[PeterDonis]

I should have mentioned that it is the above form if using $\psi$ i.e. $ds^2 = -\alpha_R^2 dt^2 + d\rho^2 + \rho^2(d\theta^2 + \sin^2\theta d\psi^2)$.

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 $\sin^2 \theta$?

 

[WannabeNewton]

I'm not sure I see how you get this from the Kerr metric.

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.

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 $\sin^2 \theta$?

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.

Thanks for the reply!

 

[PeterDonis]

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

I don't see how this can be. First, the metric within the shell is not the Kerr metric; that's a vacuum solution and the shell is not vacuum. Even if you model the shell as infinitely thin, there still has to be a discontinuity between the interior and exterior because of the presence of stress-energy in the shell.

Second, inside the shell, spacetime should not be flat if the shell is rotating, because of frame dragging. Put another way, for spacetime to be flat, there must be a congruence of timelike worldlines whose kinematic decomposition vanishes, i.e., the congruence has zero proper acceleration (i.e., it is geodesic), zero expansion, zero shear, and zero vorticity. Such a congruence should not exist in the region inside the shell if the shell is rotating.