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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?

infinitezero
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Nikita
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3 Answers3

30

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' $\mathbb{M}^4$.

Mauro Giliberti
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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.

user279733
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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.

Colin MacLaurin
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