6

In searching for images of spinning black holes, one finds a discrepancy. In many cases, the outer event horizon is spherical, with a simple formula for the radius. For example, Forbes and many other sites describe the outer event horizon as spherical with radius: $r = m + \sqrt{m^2 - a^2}$ with m being $GM/c^2$ and $a$ being the angular momentum.

Yet wikipedia (and many other sites) draw the event horizon as asymmetric and bulging at the equator. That particular link also shows the ergoshere as extending out beyond $2GM/c^2$.

If the difference is due to coordinate systems, is there any big picture way to describe the differing co-ordinate system? For example of "big picture" language that would be perfect for me, is one co-ordinate system selected so that all points that are equidistant from the center have the same time speed, while the other is drawn so all points have the same defined distance (to a local observer) to the center?

Ralph Berger
  • 480
  • 1
    Where does Forbes describe the event horizon as spherical? Their diagram clearly is not spherical, and seems to be clearly giving the radii as being that of the equator. And dimensional analysis easily shows that $a$ is not the angular momentum. Rather, it's the angular momentum divided by $Mc$. – Acccumulation Feb 15 '22 at 07:33
  • Typo - ergoshere/ergosphere – AJM Feb 15 '22 at 12:15
  • 1
    For the spherical radii, I was eye-balling this figure from Forbes: https://imageio.forbes.com/blogs-images/startswithabang/files/2019/04/visser.jpg?fit=bounds&format=jpg&width=960 . But there are lots of places where the event horizon is draw as a circle, like http://www.astronomical.org/astbook/images/fig3.gif. The answer may be as simple as "don't trust poplar science articles to be exact." – Ralph Berger Feb 18 '22 at 18:14

1 Answers1

12

$r$ isn't a physical radius; it's just a coordinate. If you plug $r = m + \sqrt{m^2 - a^2}$ and $t=\text{const}$ into the Boyer-Lindquist metric, you get

$$ds^2 = ρ^2 dθ^2 + \frac{r_s^2}{ρ^2} r^2 \sin^2 θ\,d\phi^2$$

which has a dependence on $θ$. Of course, $θ$ is also just a coordinate, so that doesn't prove that the surface isn't geometrically spherical. But if you calculate the Gaussian curvature (which is half the Ricci scalar), you'll find it's not constant, which does prove it isn't a sphere.

benrg
  • 26,103
  • Do any of those terms, s, $\rho$, and $\phi$ have any physcial descriptions? I believe $\theta$ is the equivalent of a lattitude. Also - is there any meaning behind Forbe's choice to present the event horizon as a sphere and Wikipedia's decision to present it as a non-sphere? – Ralph Berger Feb 14 '22 at 22:39
  • @RalphBerger $s$ is metric distance and $\phi$ is longitude. I don't know whether $ρ$ and $θ$ have a straightforward physical interpretation. (They may.) You can certainly think of them informally as a radius and latitude of some sort. The forbes.com post has an image that shows the event horizon as non-spherical, although it also has one that says "no matter the initial matter distribution, it collapses to a spherical black hole." Forbes.com is in any event just a blogging site, and posts there aren't vetted by anyone. – benrg Feb 15 '22 at 00:48
  • Thank you very much for answering, but I am afraid I am confused by the equation including a metric distance, a thing we can consider a radius, and a coordinate "r." I am also wondering, what is the equation useful for? – Ralph Berger Feb 15 '22 at 06:06
  • @RalphBerger The equation is the metric of a 2D surface with Euclidean signature and coordinates $θ,\phi$. It's useful for answering the question "what is the shape of the event horizon?" $r$ isn't a coordinate, but a constant ($m+\sqrt{m^2-a^2}$) in this metric, which is confusing, and possibly I should have picked a different name for it. – benrg Feb 15 '22 at 23:58
  • Again, thank you. But there are lots of places online that give me that equation for a constant value of "r" while implying that it is the radius (by drawing circles with the event horizon shown as a circle and the ergosphere shown as either a flattened oval or a dimpled sphere). That is, it is certainly not your fault! – Ralph Berger Feb 18 '22 at 19:54