What is the volume inside the event horizon of a black hole?

As I understand it, we measure things like distance (and volume) by the speed of light. A meter is the distance light travels in 1/299792458 seconds. In "normal space" this is (barely) straightforward, but around black holes it's confusing.

I suppose we reason about black holes by doing external measurements. We have theory that tells us where the event horizon is relative to a center of mass, right? But "inside" that event horizon, the means by which we measure distance and volume itself is nonsense. Is knowing the volume of a black hole using external measurements actually sufficient to know the amount of space inside the event horizon of a black hole?

Or I guess, if a meter is the distance light travels in some time and light takes an undefined or infinite amount of time to travel a meter inside a black hole, then is a black hole "bigger on the inside" (even infinite) than it is on the outside?

Edit: Clarify the question to be about the volume inside the event horizon, not about the volume of the singularity itself.

The event horizon of a Black Hole is a lightlike surface, and so its area is coordinate-invariant.

Take as an example the Schwarzschild black hole

\begin{align} 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) \end{align}

The horizon suface is at $$r = 2m$$ of Schwarzschild radial coordinate, and so at any particular Schwarzschild time has the metric $$\begin{equation} dS^2 = (2m)^2(d\theta^2 + \sin^2\theta\,d\phi^2) \end{equation}$$ this is clearly just the metric on a standard 2-sphere of radius $$2m$$. 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. 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, $$ds^2 = -\left(1-\frac{2m}{r}\right)dt^2-2\sqrt{\frac{2m}{r}}\,dt\,dr + dr^2 + r^2(d\theta^2 + \sin^2\theta\,d\phi^2),$$ 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 a different set of coordinates, you may get a yet different answer. For further reading see the paper by Di Nunno and Matzner

• The part about the area is interesting in and of itself, but there are some points that I'm not quite clear on. You describe the area as invariant, but it's not clear to me in what sense it would be invariant. Wouldn't it be more correct to say that we have a preferred set of observers, the observers who are at rest (i.e., the their world-lines are along the timelike Killing vector)? The area at $r>r_s$ is the area measured by a set of such observers who are all at that $r$. – Ben Crowell Apr 10 at 14:56
• The event horizon of a Black Hole is a lightlike surface, and so its area is coordinate-invariant. Maybe you could fill in the logic leading from the first statement to the second? I don't see the connection, or how you're using the fact that it's lightlike. It seems to me that there is a preferred definition of area $A$ for any $r>r_s$, which is the one measured by stationary observers, and when we talk about the area of the event horizon, we mean $\lim_{r\rightarrow r_s} A$. – Ben Crowell Apr 10 at 14:58
• The accepted answer to the earlier PSE question "The arrow of time and the cosmology of a black hole interior" elaborates helpfully on the points made by Kevin about the artificiality of coordinates. Somehow it's not on the list of related questions. – Edouard Jun 1 at 3:13

A black hole is a point mass, or a singularity. It has an 'infinitely' large density and 'infinitely' small volume. The mass can vary which to me means that the volume and/or density must vary somewhat from black hole to black hole, but that is how they are defined. Although the actual volume of the black hole is infinitely small, it's field effects are much wider. There is a brief article related to this here: https://www.skyandtelescope.com/astronomy-resources/how-big-is-a-black-hole/

• Right, my question was poorly phrased -- I was more asking what is the "volume inside the event horizon" I have edited the question to be clearer. – kojiro Apr 10 at 14:33
• It's not true that the singularity of a (Schwarzschild) black hole is a point, or that it has zero volume. See physics.stackexchange.com/questions/144447/… – Ben Crowell Apr 10 at 14:39