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When mass in Kerr metric is put to zero we have $$ds^{2}=-dt^{2}+\frac{r^{2}+a^{2}\cos^{2}\theta}{r^{2}+a^{2}}dr^{2}+\left(r^{2}+a^{2}\cos^{2}\theta\right)d\theta^{2}+\left(r^{2}+a^{2}\right)\sin^{2}\theta d\phi^{2},$$ where $a$ is a constant. This is a flat metric. What exactly is the coordinate transformation that changes this into the usual Minkowski spacetime metric form $$ds^{2}=-dt^{2}+dx^{2}+dy^{2}+dz^{2}?$$

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2 Answers 2

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The transformation is given in page 15 of this paper: The Kerr spacetime: A brief introduction

BTW there is a 2017 paper that claims that the mass zero Kerr metric is not actually equivalent to Minkowski metric but is a wormhole instead: Zero mass limit of Kerr spacetime is a wormhole

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  • $\begingroup$ In the first paper you cited, the x, y, and z coordinates are not the typical x, y, and z coordinates. This still describes some curved surface. I tend to believe the second paper gets it right $\endgroup$
    – Habouz
    Aug 14, 2022 at 10:39
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As mentioned in @Umaxo's comment, according to Boyer-Lindquist coordinates - Line element:

The coordinate transformation from Boyer–Lindquist coordinates $r,\theta,\phi$ to Cartesian coordinates $x,y,z$ is given (for $m\to 0$} by: $$\begin{align} x &= \sqrt{r^2+a^2}\sin\theta\cos\phi \\ y &= \sqrt{r^2+a^2}\sin\theta\sin\phi \\ z &= r\cos\theta \end{align}$$

Proving that this is the desired transformation is a straight-forward but very tedious task.
First calculate the differentials of the above: $$\begin{align} dx &= \frac{r}{\sqrt{r^2+a^2}}\sin\theta\cos\phi\ dr \\ &+ \sqrt{r^2+a^2}\cos\theta\cos\phi\ d\theta \\ &- \sqrt{r^2+a^2}\sin\theta\sin\phi\ d\phi \\ dy &= \frac{r}{\sqrt{r^2+a^2}}\sin\theta\sin\phi\ dr \\ &+ \sqrt{r^2+a^2}\cos\theta\sin\phi\ d\theta \\ &+ \sqrt{r^2+a^2}\sin\theta\cos\phi\ d\phi \\ dz &= \cos\theta\ dr - r\sin\theta\ d\theta \end{align}$$

Then insert these differentials into the Minkowski metric: $$\begin{align} ds^2 &= -dt^2+dx^2+dy^2+dz^2 \\ &\text{... omitting the lengthy algebra here, exploiting $\cos^2\alpha+\sin^2\alpha=1$ several times} \\ &= -dt^2 +\frac{r^2+a^2\cos^2\theta}{r^2+a^2}dr^2 +\left(r^2+a^2\cos^2\theta\right)d\theta^2 +\left(r^2+a^2\right)\sin^2\theta\ d\phi^2 \end{align}$$ which is the Kerr-metric for zero mass $M$, angular momentum $J$, and charge $Q$.

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