The area of a black hole is an important parameter in the thermodynamic description of a black hole. In particular, reading popular literature, everyone knows that the entropy of a black hole is proportional to its area as discovered by Stephen Hawking. Can someone explain with a diagram which is really the area of a black hole? I know what is event horizon and Schwarzchild radius but I have real difficulty visualizing the area of a black hole.
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asked May 28, 2020 at 15:11
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2 Answers
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The area of the event horizon is simply $4\pi r_s{}^2$ where $r_s$ is the Schwarzschild radius. However this is because that's how the radial coordinate $r$ is defined.
$r$ is not the distance to the centre of the black hole (in fact the radial distance to the singularity is undefined). For any point $r$ is defined as the circumference of the circle passing through that point, and centred on the singularity, divided by $2\pi$. And that automatically makes the area of the sphere passing through the point $4\pi r^2$.
answered May 28, 2020 at 15:35
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3$\begingroup$ Notice that this area is just the spatial part of the event horizon itself, which is an hypersurface in spacetime. So it's not really the "event horizon's area" ! Saying that $4 \pi r_s^2$ is the horizon's area is a common abuse of language. $\endgroup$– ChamCommented May 28, 2020 at 16:03
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$\begingroup$ @JohnRennie When one speaks about the area of a black hole, does he/she really means the area of the event horizon? $\endgroup$ Commented Jun 1, 2020 at 5:52
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$\begingroup$ But am still confused how $r_s$ is defined, if the origin of the radial coordinate is not the centre of the black hole. $\endgroup$ Commented Jun 1, 2020 at 5:55
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You can calculate the area of the event horizon by taking the limit of the area of a sphere surrounding the event horizon as the radial coordinate tends to the Schwarzschild radius. This gives a coordinate independent result. E.g. you could calculate the area after defining a radial coordinate $$R=r-r_s,$$ and take the limit $R\rightarrow 0$. This would give you the same result, but if you were to think of the black hole using the $R$ coordinate, then you would think of the event horizon as a point which, being singular, happens to have an area. This is pretty much how Schwarzschild himself regarded it.
The point is that coordinates are not important. Neither the Schwarzschild radial coordinate $r$, nor $R$, has direct physical meaning. But you can calculate the area of a surface as an integral over small areas by using the metric.
In practice, both approaches are completely equivalent outside of the event horizon, and nothing can be observed at or inside the event horizon. Which you prefer is strictly a matter of opinion, outside the realm of science (for what it is worth, I prefer Schwarzschild's opinion).
answered May 28, 2020 at 17:00
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1$\begingroup$ In the original solution by Karl Schwarzschild, the gravitational radius (known today as the “Schwarzschild radius”) is zero indeed: $\alpha=r\equiv\sqrt{x^2+y^2+z^2}=0$. - See: On the gravitational field of a mass point according to Einstein's theory by Karl Schwarzschild, 1916 - This answer is correct +1. $\endgroup$ Commented Jun 4, 2020 at 16:17