What is the radius of the event horizon?

Question

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 1

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.

Answer 2

Really, to answer this carefully, we have to really think through what a horizon is. And for a general spacetime, there are several different notions of horizon, and "event horizon" is probably the most difficult of them to work with.

The formal definition of "event horizon" says "Let's go to the distant future, take every freely-falling path that intersects with a singularity, and then trace out the past of those paths. The outer boundary of that set of paths is the event horizon." Therefore, if there are dynamics to the spacetime, it can be a complicated question about whether i"m inside the event horizon. You can even construct spacetimes where there are completely geometrically flat regions that are inside of event horizons!

Instead, let's relax this, and look at something that is both more physical and local than this. We'll call it the "apparent horizon". To define this, let's look for two tangent vectors of light paths through spacetime. We will call them $\ell^{a}$ and $k^{a}$, and we will say that $\ell_{a}k^{a} = -1$ (this can always be done by multiplying $k_{a}$ by some constant number). Then, these two vectors will define a two-dimensional surface that is perpendicular to both tangent directions, and will have submetric $q_{ab}$. Then, the apparent horizon is any closed surface which satisfies $q^{ab}\nabla_{a}\ell_{b} = 0$ and $q^{ab}\nabla_{a}k_{b} < 0$ (strictly, you need some conditions on the derivatives of these guys, but I feel that I"m already complicating this too much).

This looks really mathy, but what it tells us is that for this surface, the light ray $\ell^{a}$ just spatially hovers in place, neither falling into the horizon, nor escaping it, while the light ray $k^{a}$ points toward another surface with smaller area. The surface is focusing the ray. What this tells you is that the best you can do on this surface is to be a light ray not falling in. All other paths go toward the inside of the surface. Therefore, for a general spacetime, you just have to run your metric through this procedure, and find apparent horizons. You can test for yourself (if you know enough differential geometry) that the apparent horizon of Schwarzschild spacetime cooresponds exactly with the well-known event horizon.

Answer 3

There is already a good answer from John Rennie.

1. Here we will just mention that if the spherically symmetric metric (2) is supposed to be a vacuum solution to Einstein field equations with $\Lambda=0$, then Birkhoff's theorem shows that the metric (2) [after a possible reparametrization of the time coordinate $t$] is exactly the Schwarzschild metric. See e.g. this Phys.SE post. In particular $$B(r,t)~=~1-\frac{R_S}{r}$$ is automatically satisfied for some length parameter $R_S$. So it is still easy to identify the event horizon in the case (2). It is secretly just the Schwarzschild spacetime, whose causal structure we assume OP already know.

2. On the other hand, for general Lorentzian spacetime manifolds $(M,g)$, it is a non-trivial to determine event horizons, partly because they are global (as opposed to local) properties of the manifolds ("Classically, you don't feel anything special when crossing an event horizon").