The ergosphere is created because of the twisting of spacetime by a fast-rotating black hole. Then what if we rotate a non-singular object (I mean just a normal object that is not a black hole) really fast? Can we create an ergosphere without a black hole?

And also in the ergosphere, it is known that negative energy is possible. If we can create an ergosphere without a black hole (let me refer to this as artificial ergosphere; AE), is the negative energy possible in AE?


For stationary spherical mass distributions the Buchdahl bound tells us that $M<(4/9)(Rc^2/G)$, or $R>(9/8)R_s $ where $R_s$ is the Schwarzschild radius for the mass. This is because the pressure at the centre diverges as you approach the bound. For the Kerr metric the ergoregion is within $r<R_s$. So for low rotation rates there is a gap and there is no outside ergoregion (probably no inside one either, since the interior solution is pretty well behaved).

This also fits with numerical modeling of relativistic stars, that find that at least neutron stars cannot achieve any ergoregion: you need something denser, but the Buchdahl bound doesn't give much leeway. That paper also notes that stars with ergoregions would likely slow down by accelerating particles and fields, losing angular momentum: so these conditions are likely not lasting even if they happen. Still, see edit at the end: internal ergoregions may be possible

For rotating stationary axisymmetrical distributions there is a version of the Buchdahl inequality, if the energy density scales as $\epsilon(r)=\epsilon_c (1-(r/L)^q)$ then $$R\leq \frac{\Phi(q)}{\sqrt{16\pi \epsilon_c}}$$ where $\Phi(q)$ is a constant that can be calculated by solving a differential equation numerically and finding the first zero. $\lim_{q\rightarrow 0} \Phi(q)=\infty$ (dense core, fluffy exterior), and $\lim_{q\rightarrow \infty}\Phi(q)=\sqrt{8/3}\pi\approx 5.13$ (constant density). This is comparable to the static Buchdahl bound for neutron star densities.

So if I understand the paper right, this means there will always be a gap between the equatorial radius of stable mass distributions and the equatorial ergoregion radius. Some caveats here: clearly rapidly rotating objects will tend to be flat, producing a non-Kerr metric, but I think that just helps "bury" the polar part of the ergoregion inside.

Still, there might be tricky other cases. We can consider a rotating massive torus: it looks like there would be an ergoregion in the hole if it rotated fast enough (see this paper, although take it with a big pinch of salt: it is breaching the topological censorship theorem). Some serious issues of how stable such a situation could be: it might be that it is not possible to avoid it collapsing or flying apart - there could be a toroidal Buchdahl bound.

EDIT: It looks like differential rotation can allow "ergostars" to exist, with a toroidal ergoregion inside the volume. Whether the equations of state and the velocity fields (not to mention the numeric models) are any plausible, I do not know. However, it does look like this is a non-singular solution.

  • 1
    $\begingroup$ Thanks a lot! This really helped. $\endgroup$ – littlegiant Sep 28 '20 at 6:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.