Why can't a static spacetime have an ergosphere?An ergoregion is

[latentcorpse]

Why can't a static spacetime have an ergosphere?

An ergoregion is just a region outside the event horizon where k can become negative (k being a Killing vector field that is timelike near infinity). Both stationary and static spacetimes have such a KVF so surely static spacetimes could also have regions where k can become spacelike.

I know that the answer is they can't - but I don't understand why? I'm assuming it has something to do with rotation of event horizon?

 

[grey_earl]

In a stationary spacetime you also have a Killing vector field ξ which is timelike near spatial infinity. So what is the exact difference between static and stationary?

 

[latentcorpse]

In a stationary spacetime you also have a Killing vector field ξ which is timelike near spatial infinity. So what is the exact difference between static and stationary?

static spacetimes are stationary spacetimes that are invariant under time reversal. but the killing vector field still exists in the static case! does that mean they do admit an ergoregion?

 

[grey_earl]

Invariant under time reversal, I see. You then have to show that this implies your Killing vector field is irrotational, $$\xi_{[a} \nabla_b \xi_{c]} = 0$$, i.e. that it is hypersurface orthogonal. (shouldn't be too hard, in fact I think it's a one-liner)

Then there are two classical articles by Carter "Killing Horizons and Orthogonally Transitive Groups in Space‐Time" and Vishveshwara "Generalization of the Schwarzschild Surface to Arbitrary Static and Stationary Metrics" showing this implies there cannot be an ergosurface. If you want to read them (they are quite technical), check if your university has access to the Journal of Mathematical Physics (where they we published), or drop me a private mail.

Basically, their argument shows that an ergosurface is a timelike surface where the Killing vector becomes null, but that in a static spacetime the length of the normal vector of a "would-be" ergosurface is proportional to the length of the Killing vector there, so this must be a null surface. Ergosurface and event horizon therefore coincide in a static spacetime, so there is no ergoregion between them.

 

[latentcorpse]

Invariant under time reversal, I see. You then have to show that this implies your Killing vector field is irrotational, $$\xi_{[a} \nabla_b \xi_{c]} = 0$$, i.e. that it is hypersurface orthogonal. (shouldn't be too hard, in fact I think it's a one-liner)

Then there are two classical articles by Carter "Killing Horizons and Orthogonally Transitive Groups in Space‐Time" and Vishveshwara "Generalization of the Schwarzschild Surface to Arbitrary Static and Stationary Metrics" showing this implies there cannot be an ergosurface. If you want to read them (they are quite technical), check if your university has access to the Journal of Mathematical Physics (where they we published), or drop me a private mail.

Basically, their argument shows that an ergosurface is a timelike surface where the Killing vector becomes null, but that in a static spacetime the length of the normal vector of a "would-be" ergosurface is proportional to the length of the Killing vector there, so this must be a null surface. Ergosurface and event horizon therefore coincide in a static spacetime, so there is no ergoregion between them.

Thanks for your reply. Interesting!

I have a related question concerning the Penrose process. Consider equation (4.50) in these notes
http://arxiv.org/PS_cache/gr-qc/pdf/9707/9707012v1.pdf
He claims that it needn't be positive in the ergoregion since p is timelike or null and k is spacelike meaning -p.k can be negative.

But timelike times spacelike is negative e.g. consider (1,0,0,0) and (1,1,1,1) in the minkowski metric. they have scalar product -1 so -p.k would be positive again!

I see no way in which -p.k will ever be negative?

Cheers.

 

[grey_earl]

And what do you think of (1,1/2,0,0) and (0,1,1,1) ?

 

[latentcorpse]

And what do you think of (1,1/2,0,0) and (0,1,1,1) ?

I see. The point being that it CAN be negative BUT won't always, right?

 

[grey_earl]

Yes, exactly. You have to shoot in a particle at the right angle (namely, against the rotation, if I remember correctly) for this to work.