Ergoregions and Energy Extraction

[PeterDonis]

The paper linked to calls it in the title of Section 4.7 "ergoregions without superradiance" suggesting there is still an ergoregion but it just doesn't allow for superradiance/Penrose process.

That's because, as I said, "ergoregion" is not a standard term, and is not even used by all sources. This paper is using it in a way that, IMO, is going to cause confusion, but since there is no single standard definition of the term, their usage is not wrong, exactly, just confusing.

In any case, the important point is not words, but physics. The physics is clear: there are no timelike worldlines with negative energy at infinity in the boosted black string spacetime, whereas there are in Kerr spacetime. That is why supperradiance and the Penrose process are possible in the latter but not the former.

 

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That's because, as I said, "ergoregion" is not a standard term, and is not even used by all sources. This paper is using it in a way that, IMO, is going to cause confusion, but since there is no single standard definition of the term, their usage is not wrong, exactly, just confusing.

In any case, the important point is not words, but physics. The physics is clear: there are no timelike worldlines with negative energy at infinity in the boosted black string spacetime, whereas there are in Kerr spacetime. That is why supperradiance and the Penrose process are possible in the latter but not the former.

Earlier we said that there was no frame in Kerr in whcih the black hole was not rotating. But we know that an observer at infinity sees no rotation since $g_{t \phi} \rightarrow 0$. Is the point then that this is only true locally and as soon as we move away from infinity, we would observe some rotation whereas for the boosted black string, the comoving frame will see a static string regardless of the radial coordinate? I guess this is tied to the hypersurface orthogonality but I'm not clear what the connection is...

 

[PeterDonis]

we know that an observer at infinity sees no rotation since $g_{t \phi} \rightarrow 0$.

That doesn't mean an observer at infinity sees no rotation. It means an observer at infinity sees no frame dragging locally. But "locally" does not include the black hole; that is, if we consider a local "frame" for the observer at infinity, that frame does not contain the hole; it only contains a small local region of spacetime around the observer. When we said there was no frame in Kerr spacetime in which the hole is not rotating, we meant a frame that includes the hole.

 

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That doesn't mean an observer at infinity sees no rotation. It means an observer at infinity sees no frame dragging locally. But "locally" does not include the black hole; that is, if we consider a local "frame" for the observer at infinity, that frame does not contain the hole; it only contains a small local region of spacetime around the observer. When we said there was no frame in Kerr spacetime in which the hole is not rotating, we meant a frame that includes the hole.

To include the hole would require what? A global coordinate system in which $g_{t \phi}=0 \forall r$?

 

[PeterDonis]

To include the hole would require what? A global coordinate system

Yes--or at least one that covers enough of the spacetime to include the horizon and the region inside it.

in which $g_{t \phi}=0 \forall r$?

That's what is not possible in Kerr spacetime (whereas its analogue is in the black string spacetime).

 

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Yes--or at least one that covers enough of the spacetime to include the horizon and the region inside it.
That's what is not possible in Kerr spacetime (whereas its analogue is in the black string spacetime).

Right and we can show it's possible in the black string case since $\partial_T$ is hypersurface orthogonal for all r.

Still though, for Kerr, suppose I take the metric in Boyer-Lindquist coordinates and show that $\partial_t$ doesn't satisfy the Frobenius equation - what does that tell me? Surely it just says that $\partial_t$ isn't hypersurface orthogonal in B-L coordinates? Wouldn't I need to check this for the infinite number of possible coordinate systems in order to establish no such hypersurface orthogonal timelike KVF exists?

 

[PeterDonis]

Wouldn't I need to check this for the infinite number of possible coordinate systems in order to establish no such hypersurface orthogonal timelike KVF exists?

No. $\partial_t$ in Boyer-Lindquist coordinates is the only timelike KVF in Kerr spacetime*. Proving that it is not hypersurface orthogonal can be done in any chart, and since hypersurface orthogonality is a coordinate-free geometric property, if it holds in one chart, it holds in any chart.

* - strictly speaking, this isn't true; any linear combination of $\partial_t$ and $\partial_\phi$ with constant coefficients that is timelike is also a timelike KVF. But proving that $\partial_t$ is not hypersurface orthogonal is sufficient to prove that none of those other timelike KVFs are hypersurface orthogonal either.