general relativity - What is the radius of the event horizon?

I know that the Schwarzschild radius is given by

$$r~=~\frac{2GM}{c^{2}}.\tag{1}$$

However, If we had the metric

$$ds^2~=~−A(r,t)dt^2+\frac{dr^2}{B(r,t)}+r^2(dθ^2+\sin^2{θ}dϕ^2),\tag{2}$$

where

$$A(r,t)~\neq~(1−\frac{2GM}{c^2r})\tag{3}$$

and

$$B(r,t)~\neq~(1−\frac{2GM}{c^{2}r}),\tag{4}$$

then what is the event horizon?

Answer

Let's start with what we mean by a horizon:

The event horizon of an asymptotically-flat spacetime is the boundary between those events from which a future-pointing null geodesic can reach future null infinity and those events from which no such geodesic exists.

A null geodesic is the path followed by a light ray, so the horizon marks the surface at which light just cannot escape to infinity. So what we need to do is look at the trajectories followed by light rays and find where they become trapped.

In this case the spherical symmetry makes the problem easy because radial light rays will be normal to the horizon. So we can just look for the radius at which the coordinate velocity of the radial light rays is zero. For a radial trajectory $d\theta = d\phi = 0$, and the metric becomes:

$$ ds^2=−A(r)dt^2+\frac{dr^2}{B(r)} $$

And we know that for light $ds = 0$ so we get the equation:

$$ 0 = −A(r)dt^2 + \frac{dr^2}{B(r)} $$

which gives us:

$$ \frac{dr}{dt} = \sqrt{A(t)B(t)} $$

The left side $dr/dt$ is the coordinate velocity of the light ray, so the location of the event horizon is where this is zero:

$$ \sqrt{A(t)B(t)} = 0 \tag{1} $$

For example, the Schwarzschild metric has $A(t)$ and $B(t)$ equal to:

$$ A(t) = B(t) = 1 - \frac{2GM}{c^2r} $$

and equation (1) becomes:

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

which has the solution we already know:

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

Anyone interested in this area may want to look at Anonymous' previous question on the subject as my answer to that provides more detail on what the metric tells us.

Footnote:

Michael Seifert points out that the analysis I've given above is only applicable when the metric is time independant i.e. when $A$ and $B$ are functions only of $r$ and not of $t$. This type of metric is called a static spacetime.

My analysis is based on finding the value of $r$ for which the coordinate velocity is zero, but in a non-static spacetime the coordinate velocity at a particular value of $r$ may be zero at some time but non-zero at other times, and vice versa, and my argument doesn't apply.

Finding the location of the event horizon in a time dependant spacetime is a complex problem, and I can't give a simple answer as some function of $A$ and $B$. If you're interested there is an article on finding horizons on the Living Reviews in Relativity web site. This is in the context of numerical solutions rather than analytical ones, but it still gives a good idea of what is required.