The n-Category Café

guest post by Greg Egan

If we happen to know the Kerr metric, it’s easy to show that the blueshift factor on the equatorial plane of the Kerr geometry is exactly the same as for the Schwarzschild metric. In other words, the angular momentum of a rotating hole has (on the equatorial plane) absolutely no effect on $g_{tt}$. Now, while this doesn’t surprise me in the least, and seems very intuitively reasonable, I haven’t been able to find any simple way to prove it — not without deriving the entire Kerr solution!

Along with $g_{tt}$, there are two other things that I know of that are unchanged by the angular momentum of a black hole when you’re in the equatorial plane. Working in an orthonormal tetrad whose time vector is parallel to the timelike Killing vector, the tidal squeezing in the direction that points equatorially around the hole is just $M/r^3$. And (working in the same frame) if you compute the covariant derivative of the radial vector with respect to time, the magnitude of that covariant derivative is unaffected by the hole’s angular momentum. (In the Schwarzschild case the derivative just measures the acceleration of the chosen frame; in the Kerr case, it’s a mixture of acceleration and frame dragging. But the magnitude of the derivative is independent of the angular momentum.)

One of these three facts can be taken as a choice of gauge, a physical definition of the $r$ coordinate on the equatorial plane. So I don’t expect to be able to prove one statement in isolation; you need to define $r$, in order for the other statements to have any physical content. But short of pulling the Kerr metric out of a hat, I haven’t been able to show why any one of these three follows from any other. The fact about the blueshift seems so “obvious” that it’s very hard to believe that you really need to know the complete Kerr metric just to prove it, but maybe my intuition about that is simply wrong.

One fun thing I did discover was that if you take these three facts as given (and already know the Schwarzschild geometry), you can derive the whole Riemann tensor for the Kerr geometry on the equatorial plane, with very little work.