Physics Asked on June 10, 2021
Kerr metric has the following form:
$$
ds^2 = -left(1 – frac{2GMr}{r^2+a^2cos^2(theta)}right) dt^2 +
left(frac{r^2+a^2cos^2(theta)}{r^2-2GMr+a^2}right) dr^2 +
left(r^2+a^2cos(theta)right) dtheta^2
+ left(r^2+a^2+frac{2GMra^2}{r^2+a^2cos^2(theta)}right)sin^2(theta) dphi^2 –
left(frac{4GMrasin^2(theta)}{r^2+a^2cos^2(theta)}right) dphi, dt
$$
This metric describes a rotating black hole.
If one considers $M=0$:
$$
ds^2 = – dt^2 +
left(frac{r^2+a^2cos^2(theta)}{r^2+a^2}right) dr^2 +
left(r^2+a^2cos(theta)right) dtheta^2
+
left(r^2+a^2right)sin^2(theta) dphi^2
$$
This metric is a solution of the Einstein equations in vacuum.
What is the physical interpretation of such a solution?
It's simply flat space in Boyer-Lindquist coordinates. By writing
$begin{cases} x=sqrt{r^2+a^2}sinthetacosphi y=sqrt{r^2+a^2}sinthetasinphi z=rcostheta end{cases}$
you'll get good ol' $mathbb{M}^4$.
Correct answer by Mauro Giliberti on June 10, 2021
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.
Answered by user279733 on June 10, 2021
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.
Answered by Colin MacLaurin on June 10, 2021
Get help from others!
Recent Answers
Recent Questions
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP