13
$\begingroup$

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?

$\endgroup$
1
29
$\begingroup$

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' $\mathbb{M}^4$.

$\endgroup$
1
  • 3
    $\begingroup$ Note also that these coordinates are related to oblate spheroidal coordinates by the simple substitution $r = a \sinh \mu$ and $\theta = \pi/2 - \nu$. $\endgroup$ Nov 15 '20 at 18:41
15
$\begingroup$

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.

$\endgroup$
1
  • 5
    $\begingroup$ 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. $\endgroup$
    – rob
    Nov 16 '20 at 22:31
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.