Does a black hole have a surface area?

[Stephanus]

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

So it's not that bigger black holes emit less hawking radiation than smaller ones? It's just that the surface area becomes more spread out as the black hole becomes larger and is distributed less intensely to an object that remains a constant size? In other words, theoretically, if an object geometrically grew at the same rate as the black hole, would it experience the same amount of hawking radiation the whole time?

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

 

[PeterDonis]

does black hole have surface area?

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

Did you mean the sphere defined by Schwarzschild Radius?

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.

 

[Stephanus]

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

The event horizon.

Did you mean the sphere defined by Schwarzschild Radius?

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.

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

 

[PeterDonis]

The event horizon.

Yes.

The area defined by $4\pi {R_{schwarzshild}}^2$?

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.

 

[Stephanus]

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.

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 R_s itself is defined by https://en.wikipedia.org/wiki/Schwarzschild_radius#Formula
$R_s = \frac{2GM}{c^2}$, still Newton, right? Only instead of V_e², we replace it with the speed of light.

 

[PeterDonis]

So the surface area of a black hole is the surface area of a sphere which has the Schwarzschild radius

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.

Rs itself is defined by https://en.wikipedia.org/wiki/Schwarzschild_radius#Formula
$R_s = \frac{2GM}{c^2}$,

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.

still Newton, right? Only instead of Ve2, we replace it with the speed of light.

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