# Does a black hole have a surface area?

1. Aug 14, 2015

### Stephanus

[Moderator's note: Spin-off from another thread.]

I'm sorry, does black hole have surface area? Did you mean the sphere defined by Schwarzchild Radius?

Last edited by a moderator: Aug 15, 2015
2. Aug 14, 2015

### Staff: Mentor

Yes. More precisely, the hole's horizon does. See below.

Yes, but you're looking at it backwards. The "Schwarzschild radius" is really $\sqrt{A / 4 \pi}$, where $A$ is the area of the horizon (which is a 2-sphere). The area is the geometric quantity; the "radius" is just a coordinate label and doesn't correspond to a physical radius.

3. Aug 15, 2015

### Stephanus

The event horizon.
Perhaps I should state my question more clearly
The area defined by $4\pi {R_{schwarzshild}}^2$?

Last edited: Aug 15, 2015
4. Aug 15, 2015

### Staff: Mentor

Yes.

The answer is still the same: you are looking at it backwards. The area is not defined by $4 \pi R_s^2$. $R_s$ is defined as $\sqrt{A / 4 \pi}$. The area $A$, as the geometric quantity, is logically prior to $R_s$, which is just a coordinate label.

5. Aug 15, 2015

### Stephanus

Okay...
So the surface area of a black hole is the surface area of a sphere which has the Schwarzschild radius IS $4 \Pi R_s$
While Rs itself is defined by https://en.wikipedia.org/wiki/Schwarzschild_radius#Formula
$R_s = \frac{2GM}{c^2}$, still Newton, right? Only instead of Ve2, we replace it with the speed of light.

6. Aug 15, 2015

### Staff: Mentor

No. It is the surface area of a sphere which is labeled by the radial coordinate $R_s$. That number is not the actual geometric radius of that sphere; in fact, that sphere, the horizon, has no geometric radius in the ordinary sense. , because it has no geometric center. A black hole is not just a spherical object sitting in ordinary Euclidean space.

That formula tells you how to calculate $R_s$, yes. But it doesn't tell you how $R_s$ is defined, geometrically. It doesn't tell you that $R_s$ is the geometric radius of anything. It just tells you how to calculate a number.

To see what $R_s$ actually means, you have to look at the metric; you have to look at what determines which 2-sphere is the hole's horizon; and you have to look at what geometric properties that 2-sphere actually has. When you look at all those things, you find that, as I said before, $R_s$ is not the geometric radius of the horizon; it's just a coordinate label. The geometric property that actually characterizes the horizon is its area.

One way to see this is to find coordinates in which the horizon is labeled by a different coordinate; for example, isotropic coordinates. In these coordinates, the horizon is labeled by the radial coordinate $R = GM / 2 c^2$, i.e., 1/4 of the value $R_s$ that labels it in Schwarzschild coordinates. But the area of the horizon is the same, because that's an invariant geometric property.

No. A black hole is certainly not an object to which Newtonian gravity applies.