7

In higher dimensions black holes can take different forms. Besides spheroids, there are also “black rings” which have a toroidal event horizon. However, in our 3+1 dimension no black ring solution has been found. The only valid solution of a black hole with an angular momentum is the Kerr metric. The no hair theorem dictates that black holes are only described by their masses, spins and charges (which is assumed to be neutral for simplicity). However, no hair theorem is not actually proved mathematically despite its name.

Here is my thought experiment. Supposing that there is a fast spinning planet made of ideal fluids. The angular momentum it carries exceeds the theoretical maximum of a rotating black hole of the same mass. Because the planet is spinning too fast, it’s stretched into a torus. Although theoretical calculations revealed that such hydrostatic equilibrium is unstable against perturbations, it’s still possible under idealized conditions.

Now assuming that the mass $M$ and angular momentum $J$ are unchanged, but the density $ρ$ is fine tuned to be higher and higher. As a result, the torus will be thinner and thinner and the surface gravity will be stronger and stronger. The question is whether the surface gravity will reach infinity even with a finite ρ? If the answer is yes, will the object become a black ring, or collapse into a Kerr black hole (which sounds unlikely if $J>\frac{GM^2}{c}$)?

Qmechanic
  • 220,844
哲煜黄
  • 1,877

1 Answers1

6

Here is a naive intuitive argument. Heuristically we should be able to construct a black ring solution $(M^d,g^d)$ on a $d$-dimensional spacetime as follows. First take a black hole solution $(M^{d-1},g^{d-1})$ in 1 spatial dimension lower, and consider the product manifold $M^{d-1}\times \mathbb{R}$. Next compactify the spatial dimension $\mathbb{R}$ to a (large at first) circle $\mathbb{S}^1$.

Now the issue is that black holes only$^1$ exist in $\geq 4$ spacetime dimensions, so the above argument suggests that black rings only exist in $\geq 5$ spacetime dimensions.

--

$^1$ Unless we consider the BTZ black hole in 2+1D with a negative cosmological constant $\Lambda<0$, but it has no Newtonian limit.

Qmechanic
  • 220,844