# Massless Kerr black hole

Kerr metric has the following form:

$$ds^2 = -\left(1 - \frac{2GMr}{r^2+a^2\cos^2(\theta)}\right) dt^2 + \left(\frac{r^2+a^2\cos^2(\theta)}{r^2-2GMr+a^2}\right) dr^2 + \left(r^2+a^2\cos(\theta)\right) d\theta^2 + \left(r^2+a^2+\frac{2GMra^2}{r^2+a^2\cos^2(\theta)}\right)\sin^2(\theta) d\phi^2 - \left(\frac{4GMra\sin^2(\theta)}{r^2+a^2\cos^2(\theta)}\right) d\phi\, dt$$

This metric describes a rotating black hole.

If one considers $$M=0$$:

$$ds^2 = - dt^2 + \left(\frac{r^2+a^2\cos^2(\theta)}{r^2+a^2}\right) dr^2 + \left(r^2+a^2\cos(\theta)\right) d\theta^2 + \left(r^2+a^2\right)\sin^2(\theta) d\phi^2$$

This metric is a solution of the Einstein equations in vacuum.

What is the physical interpretation of such a solution?

• See page 15 of this review. Nov 13, 2020 at 23:51

It's simply flat space in Boyer-Lindquist coordinates. By writing

$$\begin{cases} x=\sqrt{r^2+a^2}\sin\theta\cos\phi\\ y=\sqrt{r^2+a^2}\sin\theta\sin\phi\\ z=r\cos\theta \end{cases}$$

you'll get good ol' $$\mathbb{M}^4$$.

• Note also that these coordinates are related to oblate spheroidal coordinates by the simple substitution $r = a \sinh \mu$ and $\theta = \pi/2 - \nu$. Nov 15, 2020 at 18:41

This is presumably a flat spacetime described in funny coordinates. You can check this by calculating the Riemann tensor to see if it's zero. If I was going to do this, I would code it in the open-source computer algebra system Maxima, using the ctensor package.

• I disagree with the “not an answer” flags and comments. Partial answers are still answers. This is essentially the same as the accepted answer, except with a nudge towards an analysis technique rather than the name of the solution.
– rob
Nov 16, 2020 at 22:31

A reference which answers this is Visser (2008). It discusses the limits of vanishing mass $$M \rightarrow 0$$, and rotation parameter $$a \rightarrow 0$$. Your example is in $$\S5$$. Visser comments "This is flat Minkowski space in so-called “oblate spheroidal” coordinates...", as described in a different answer here.