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I know that the Schwarzschild radius is given by


but I never seen a derivation for this equation.

1- Does anyone know how to derive this equation from general relativity?

2- If we had the metric $$ds^2=−A(r)dt^2+\frac{dr^2}{B(r)}+r^2(dθ^2+\sin^2{θ}dϕ^2)$$ , where $A(r)≠(1−\frac{2GM}{c^2r})$ and $B(r)≠(1−\frac{2GM}{c^{2}r})$, then what is the event horizon?


marked as duplicate by John Rennie general-relativity Jun 24 '15 at 15:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ This new edit seems to change the question a bit too much. It is probably worth a new question, rather than appending it to an existing question with a >7-hour old answer. $\endgroup$ – Kyle Kanos Jun 24 '15 at 0:32
  • $\begingroup$ I've closed this as a duplicate of the new question, since the answers to the new question cover this one as well. $\endgroup$ – John Rennie Jun 24 '15 at 15:04

The geometry of spacetime is described by an equation called the metric. This is analogous to Pythagoras' theorem but with some key differences.

Start with a 2d plane, where we identify positions of points by their $(x, y)$ coordinates. Suppose you move a distance $dx$ then a distance $dy$, then the distance from your starting point, $ds$, is given by Pythagoras' theorem:

$$ ds^2 = dx^2 + dy^2 $$

If you introduce a third spatial dimension, $z$, then Pythagoras' theorem generalises to:

$$ ds^2 = dx^2 + dy^2 + dz^2 $$

And if you now introduce a time dimension, $t$, you might be tempted to think the distance $ds$ is given by:

$$ ds^2 = dt^2 + dx^2 + dy^2 + dz^2 $$

But this is wrong. Relativity tells us that the distance $ds$ is actually given by:

$$ ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2 $$

Note that the $dt^2$ term gets a minus sign. We also multiply by the speed of light $c$, but that's just to convert the time to a distance so the equation is dimensionally consistent (the units of $ct$ are light-seconds i.e. a distance). This equation is called the Minkowski metric, and it is the basis of Special Relativity. Indeed, all of Special Relativity is described by this one equation i.e. time dilation, length contraction and all the other weird stuff.

Anyhow, the Minkowski metric tells us the distance $ds$ in flat spacetime. In curved spacetime the equation is more complicated, and for the spacetime around a black hole $ds$ is given by:

$$ ds^2 = -(1-\frac{2GM}{c^2r})c^2dt^2 + \frac{1}{1-\frac{2GM}{c^2r}}dr^2 + r^2 (d\theta^2 + \sin^2 \theta d\phi^2) $$

This is the notorious Schwarzschild metric. The equation is in polar coordinates, which makes it look more complicated, but if you compare it with the Minkowski metric you'll see it's not so different except that the $c^2dt^2$ term is now multiplied by $1 - 2GM/(c^2r)$, and the spatial term, $dr^2$, is divided by $1 - 2GM/(c^2r)$.

The Schwarzschild metric is obtained by solving Einstein's equation for a spherically symmetric mass. The details are long and complicated, so I'm afraid you're going to have to take it on trust that the Schwarzschild metric really does describe a black hole geometry.

Anyhow, if you take the distance $r$ to be:

$$ r = \frac{2GM}{c^2} $$

then something odd happens because that factor of $1 - 2GM/(c^2r)$ becomes:

$$ 1 - \frac{2GM}{c^2r} = 1 - \frac{2GM}{c^2}\frac{c^2}{2GM} = 1 - 1 = 0 $$

and the equation becomes (I've omitted the angular bit because it isn't relevant for this argument):

$$ ds^2 = -0c^2dt^2 + \frac{1}{0}dr^2 + ... $$

Do you see the problem? Our equation now contains a division by zero so the value of $ds^2$ is undefined. This is a coordinate singularity, and that's what defines the event horizon. That's why the position of the event horizon, aka the Schwarzschild radius, is given by:

$$ r_s = \frac{2GM}{c^2} $$

  • $\begingroup$ Are you saying that Schwarzschild radius is distance where there is a coordinate singularity? $\endgroup$ – MrDi Jun 23 '15 at 17:39
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    $\begingroup$ I would actually argue that the event horizon is defined by the surface such that one can never escape from within to infinity. As it turns out, this matches the coordinate singularity in question, but it didn't have to. For example, the polar axis $\theta = 0,\pi$ is also a coordinate singularity in Schwarzschild coordinates, but it has nothing to do with the horizon. $\endgroup$ – user10851 Jun 23 '15 at 18:10
  • $\begingroup$ @Anonymous: Chris White is quite correct to pick me up for being loose with my argument. Defining exactly what the event horizon is turns out to be a lot more complicated than I've made it sound. However, in this case the Schwarzschild radius is indeed the distance where the coordinate singularity occurs. $\endgroup$ – John Rennie Jun 23 '15 at 19:18
  • $\begingroup$ @JohnRennie, if we had the metric $$ ds^2 = -A(r)dt^2 + \frac{1}{B(r)}dr^2 + r^2 (d\theta^2 + \sin^2 \theta d\phi^2) $$, where $A(r)\neq(1-\frac{2GM}{c^2r})$ and $B(r)\neq(1-\frac{2GM}{c^2r})$ then what is the event horizon? $\endgroup$ – MrDi Jun 24 '15 at 0:14
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    $\begingroup$ @Anonymous: that's too big a question to answer in a comment. You might want to post it as a new question. $\endgroup$ – John Rennie Jun 24 '15 at 4:57

You are standing on surface of a planet of mass $M$ and radius $R$. With which velocity $v$ you need to throw away from the planet an object of mass $m$ that it will not return? The gravitational force is $F=G\frac{mM}{r^2}$. The work to be done to move the object in the gravitational field of the planet from distance R to infinity is $A = \int{_R}^{+\inf}G\frac{mM}{r^2}dr = \frac{GMm}{R}$. You give the energy to the object by throwing it with some velocity $v$, so this energy is provided by the kinetic energy of your object: $\frac{mv^2}{2}$. You obtain: $G\frac{mM}{R}=\frac{mv^2}{2}$, and finally $R=\frac{2GM}{v^2}$. For a velocity $v$ and the planet mass $M$ the formula tells you that if you are standing on a distance $x>R$, the object will return. You know the maximum object velocity is limited by the speed of light $c$, use it and obtain the Schwarzschild radius.

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    $\begingroup$ You seem to be doing Newtonian gravity, i.e. thinking of gravity as just another force field. Then you are claiming the maximum velocity is the speed of light. This is inconsistent because the speed of light is not the maximum velocity in the Newtonian world, and gravity is not a simple force field in the relativistic world. $\endgroup$ – ACuriousMind Jun 23 '15 at 22:23
  • $\begingroup$ Your answer neglects the fact that OP wanted an answer rooted in general relativity and not Newtonian mechanics. Otherwise, it is essentially the same argument lurscher provided. $\endgroup$ – Kyle Kanos Jun 24 '15 at 0:10

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