Disclaimer: this a [cross post from MSE](https://math.stackexchange.com/questions/3421905/is-the-wikipedia-depiction-of-the-ergosphere-of-a-kerr-black-hole-a-cassini-oval), 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](https://en.wikipedia.org/w/index.php?title=Rotating_black_hole&oldid=930769965)) 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](https://en.wikipedia.org/wiki/Cassini_oval). Is it actually one? If yes, how can we prove it?



Is the Wikipedia depiction of the ergosphere of a Kerr black hole a Cassini oval?
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](https://en.wikipedia.org/wiki/Cassini_oval), 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.


