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How to calculate the area / volume of a black hole? Is there a corresponding mathematical function such as rotating $1/x$ around the $x$-axis or likewise to find the volume?

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Presumably the area and volumes are that of and inside the event horizon. –  namehere Dec 5 '12 at 3:55
How much do you know about integrating over forms? Also, the volume of a black hole is coordinate dependent. –  Jerry Schirmer Dec 5 '12 at 4:11

2 Answers 2

up vote 10 down vote accepted

The event horizon is a lightlike surface, and so its area is coordinate-invariant. For a Schwarzschild black hole, $$ds^2 = -\left(1-\frac{2m}{r}\right)dt^2 + \left(1-\frac{2m}{r}\right)^{-1}dr^2 + r^2(d\theta^2+\sin^2\theta\,d\phi^2)$$ The horizon suface is at $r = 2m$ of Schwarzschild radial coordinate, and so at any particular Schwarzschild time ($dt = 0$) has the metric $$dS^2 = (2m)^2(d\theta^2 + \sin^2\theta\,d\phi^2),$$ which is just the metric on a standard 2-sphere of radius $2m$. You can find the square area element explicitly as the determinant of the metric, $dA^2 = (2m)^4\sin^2\theta\,d\theta^2d\phi^2$, and integrating: $$A = 16\pi m^2 = \frac{16\pi G^2}{c^4}m^2.$$

The volume of the black hole is not invariant, as Jerry Schirmer says. If you try to apply anything the above Schwarzschild coordinates above, then since the coefficients of $dt^2$ and $dr^2$ switch signs across the horizon, $t$ is spacelike and $r$ is timelike. Therefore, since the black hole is eternal, it could be said to have infinite volume (classically, but a real astrophysical black hole would have a finite but still extraordinarily high lifetime), as you'll be integrating $dt$ across its lifetime.

Technically, the above argument is a bit flawed, because the Schwarzschild coordinate chart is not defined across the event horizon, so one should be more careful how they're continued across the horizon (e.g., with Kruskal-Szekeres coordinates). But this can be made more rigorous.

In another coordinate chart, e.g., the Gullstrand-Painlevé coordinates adapted to a family of freely falling observers, $$ds^2 = -\left(1-\frac{2m}{r}\right)dt^2-2\sqrt{\frac{2m}{r}}\,dt\,dr + \underbrace{dr^2 + r^2(d\theta^2 + \sin^2\theta\,d\phi^2)}_{\text{Euclidean in spherical coord.}},$$ at any instant of time ($dt = 0$), space is precisely Euclidean; since the horizon is still $r = 2m$ in these coordinates, "the" volume is $$V_{\text{GP}} = \frac{4}{3}\pi(2m)^3.$$

If you pick yet another coordinate chart, you may get a yet different answer.

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A better answer than mine! +1 –  John Rennie Dec 5 '12 at 9:11

If you are a Schwarzschild observer the radial co-ordinate $r$ is defined as the circumference of a circle around the black hole divided by $2\pi$. Also the event horizon is at a Schwarzschild radius of $2M$ (in geometrised units), so for a Schwarzschild observer the area of the event horizon is simply $16\pi M^2$.

But this is a somewhat trivial answer as in effect I'm just saying "well that's how we define $r$". I get the impression you're looking for a deeper answer than this, but you'll need to expand on your question a bit.


I've just noticed Is a black hole's surface area invariant for distant intertial observers? that states the area of the event horizon is an invarient. This means all observers will measure the area to be $16\pi M^2$.

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