How to derive the Schwarzschild radius? I know that the Schwarzschild radius is given by
$$r=\frac{2GM}{c^{2}}.$$
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?
 A: 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:
$$ r_s = \frac{2GM}{c^2} $$
A: 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.
