# What is the shape of a Kerr black Hole, like in Schwarzschild BH we know it is just a spherical distribution of mass $M$?

We know that the Schwarzschild solution pictures the outside region of a spherical distribution of mass $$M$$, which is also non-rotating, charge-free and spherically symmetric. We can begin to describe this way objects such like stars or black holes. On the other hand, the Kerr solution is an axially symmetric spacetime for black holes only, which written in Boyer-Lindquist coordinates resembles a line element whose coordinate basis when $$M\to 0$$ is similar to oblate spheroidal coordinates:

$$x=\sqrt{r^2+a^2}\sin\theta\cos\phi,$$ $$y=\sqrt{r^2+a^2}\sin\theta\sin\phi,$$ $$z=r\cos\theta;$$

for $$r=a\sinh\mu$$ and $$\theta=\pi/2-\nu$$.

Therefore, for a "sub-special" case, a Kerr BH looks like an oblate spheroid (seed shaped, like an M&M). Is there a general form to look at a Kerr BH as a 3D surface? I think there should be different cases, depending on the value of the spin parameter $$a=J/Mc$$ and the mass $$M$$, i.e. on how the 3D sub-line element looks:

$$d\ell^2=\frac{\Sigma}{\Delta}\,dr^2 +\Sigma\, d\theta^2+\left(r^2+a^2+\frac{2GMra^2}{\Sigma\,c^2}\sin^2\theta\right)\sin^2\theta\,d\varphi^2.$$

However, I cannot imagine this the way I imagine a sphere like in the Schwarzschild solution. What am I missing?

• I am not sure what you want to imagine here exactly. Apr 7 at 18:49
• May I suggest rephrasing the title as "What is the shape of a Kerr black hole?"? I believe it conveys your question in a clearer way (the current title makes it seem that you're asking for which parameters one need to characterize a Kerr black hole (namely: mass, charge, spin) and comparing with how Schwarzschild is completely determined by the mass) Apr 7 at 20:35
• @Prahar A surface in 3D which I can associate (or at least which has an homeomorphism) to the spatial line element $d\ell^2$ of a Kerr black hole, of mass M and spin parameter $a$. In Schwarzschild I can imagine a sphere for this, despite I know the $r$ coordinate is not the physical radius length as in spherical coordinates, but $dl = dr/(1-Rs/r)$. , Apr 8 at 15:42
• At org.yukterez.net/… you have the ergosurfaces, horizons and curvature scalars for different spin parameters Apr 9 at 0:54

Strictly speaking, it is not true that the Schwarzschild solution is describing a spherical mass. The Schwarzschild solution is a vacuum solution of the Einstein Equation's, with a singularly at $$r=0$$. You can see this from $$R_{\mu\nu}=0$$ for all spacetime points. If we were to solve for a spacetime with a sphere with mass $$M$$, we would have a non-vanishing energy-momentum tensor and therefore also non-vanishing Ricci tensor. Of course, we know that if the scale of the object we consider is $$R_{\rm M}$$, the spacetime far away from the mass $$r\gg R_{\rm M}$$ is well described by the Schwarzschild metric.