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Disclaimer: this a cross post from MSE, where this question was asked on November 4th 2019 and has so far received no upvote, no comment and no answer whatsoever.

Glancing at https://en.wikipedia.org/wiki/Rotating_black_hole (current revision) I thought that the frontier of the ergosphere appearing in the picture at the beginning of the considered article looks very much like a Cassini oval. Is it actually one? If yes, how can we prove it?

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    $\begingroup$ An answer has been given on MSE after this cross posting. $\endgroup$ Commented Dec 17, 2019 at 12:05
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    $\begingroup$ The answer given on MSE is confusing and/or incomplete: it gives the value of $r$ as a function of $\theta$ for the static limit and horizon, but the function being used to polar-plot the diagram as given on Wikipedia is not $r$ but $\sqrt{r^2+a^2\,\sin^2(\theta)}$ as I explain here (and evidenced by the fact that the horizons are clearly not shown as circles). I will try to answer the question later. $\endgroup$
    – Gro-Tsen
    Commented Dec 17, 2019 at 15:09
  • $\begingroup$ Keep in mind that the geometry of this surface when mapped onto a certain coordinate chart has little to do with its intrinsic geometry. A similar question, suffering from a similar problem: mathoverflow.net/questions/342527/… $\endgroup$
    – user21349
    Commented Apr 25, 2020 at 13:27

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It's not a Cassini oval.

To see this, recall that we're talking about the outer static limit of the black hole, whose (Boyer-Lindquist) $r$ coordinate in function of $\theta$ is given by $r = M + \sqrt{M^2 - a^2\,\cos^2\theta}$ (where, as usual, $M$ is the black hole mass and $a$ its angular momentum per unit mass). However, this $r$ coordinate is not what is being plotted in a slice, because it does not reduce to the Euclidean distance-to-the-center coordinate when $M\to 0$, but rather the $\sqrt{r^2 + a^2\,\sin^2\theta}$ coordinate, let's call it $R$, which does (this $R$ is not to be confused with $\rho := \sqrt{r^2 + a^2\,\cos^2\theta}$ which features prominently in the expressions of the Kerr metric).

Now eliminating $r$ from the algebraic equations $(r-M)^2 = M^2 - a^2\,\cos^2\theta$ and $R^2 = r^2 + a^2\,\sin^2\theta$ gives the implicit polar equation $$ R^4 + 2(a^2\cos(2\theta) - 2M^2)R^2 + a^2(a^2\cos^2(2\theta) + 4M^2\sin^2(\theta)) = 0 $$

In contrast, a Cassini oval, if I believe Wikipedia, has the implicit polar equation $$ R^4 - 2u^2\cos(2\theta)\, R^2 + (u^4-v^4) = 0 $$ where $u,v$ (called $a,b$ in Wikipedia) are real parameters. Clearly the curve cannot be put into this form.

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  • $\begingroup$ I first thought that these two curves could be two instances of "generalized complex Cassini ovals" allowing $u$ and $v$ to be complex and setting $a=2M\in i\mathbb{R}$, but I missed the fact that we face a $\sin^{2}\theta$ instead of the required $\sin^{2} 2\theta$. $\endgroup$ Commented Dec 17, 2019 at 20:37
  • $\begingroup$ And also the fact that the term $-4M^2R^2$ depends on $R$. Anyway, thank you very much for your answer. $\endgroup$ Commented Dec 17, 2019 at 20:47

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