Calculation of the black hole horizon's surface

[PhysiSmo]

Hi everyone!

I'm looking for a way to calculate the horizon's surface for an arbitrary black hole in more than 4 dimensions. For example, if one has a metric of the form

$$ds^2=A(r)dt^2+B(r)dr^2+C(r)d\Omega_d^2,$$

where $$A(r),B(r),C(r)$$ various functions and $$d$$ the spacetime dimension, how can one calculate the surface of the horizon, given that the horizon is at position $$r=r_0.$$

Such cases occur in supersymmetric generalizations of various black holes (extremal Reissner Nordstrom for example) in more dimensions, say, 5.

Thank you in advance!

 

[Jonathan Scott]

Hi everyone!

I'm looking for a way to calculate the horizon's surface for an arbitrary black hole in more than 4 dimensions. For example, if one has a metric of the form

$$ds^2=A(r)dt^2+B(r)dr^2+C(r)d\Omega_d^2,$$

where $$A(r),B(r),C(r)$$ various functions and $$d$$ the spacetime dimension, how can one calculate the surface of the horizon, given that the horizon is at position $$r=r_0.$$

Such cases occur in supersymmetric generalizations of various black holes (extremal Reissner Nordstrom for example) in more dimensions, say, 5.

Thank you in advance!

If by "surface" you mean the boundary of the sphere, that has a 2-D area for ordinary 3-D space, but a 3-D volume for a 4-D space and so on.

I'll assume that the tangential part should have been notated as $C(r) r^2 d\Omega_d^2$ as in that case all of the metric factors are dimensionless and typically start from 1 at a distance.

In the above metric, for the ordinary static 3-D space plus time case, I think that only the term $C(r_0) r^2$ matters. The square root of this is the effective radius multiplied by the tangential metric, so all we have to do to get the surface is substitute that square root expression into the standard formula for the surface area of a sphere.

I'd guess (somewhat optimistically, perhaps) that if you substitute $(\sqrt{C(r_0)} r_0)$ into the formula for the boundary of a higher dimensional sphere, that would give the corresponding result. For example, if the boundary of a sphere in 4-dimensional space is $2 \pi^2 r^3$ (which I vaguely remember it might well be) then the surface in this case would be $2 \pi^2 C(r_0)^{3/2} {r_0}^3}$.

However, I'm really out of my depth here, so I can only hope that the above vague ideas might be useful.

 

[PhysiSmo]

Thank you for your answer! The formula you provided is pretty correct, it passed many tests including 4d and 5d Schwarschild, Reissner Nordstrom, Supersymmetric generalizations of etc. So I guess that it should be correct.

Any ideas where does this formula come from? Why lim(r-->r_0) [sqrt{C(r)}r] is the effective radius?

 

[Jonathan Scott]

Thank you for your answer! The formula you provided is pretty correct, it passed many tests including 4d and 5d Schwarschild, Reissner Nordstrom, Supersymmetric generalizations of etc. So I guess that it should be correct.

Any ideas where does this formula come from? Why lim(r-->r_0) [sqrt{C(r)}r] is the effective radius?

That's just geometry. To measure the surface using local rulers, you take the surface measured in coordinate space then and adjust each r factor by the tangential component of the metric. I can visualise that for 3-D space, but for higher dimensions I just have to blindly trust the mathematics, so I hope that's right!