Conditions for Spherically Symmetric Black Hole Solution

[PeteSampras]

What is the condition for a spherically symmetric solution represents a black hole?

$ds^2=\exp(\nu(r))dt^2-\mu(r)^{-1}dr^2-r^2 d\Omega^2$

it is enough that it is fulfilled that $\nu$ and $\mu$ are nulled in the same value of r??.

There are other conditions?

 

[PeterDonis]

The coordinate-invariant definition of a black hole spacetime is that there is a region of spacetime that cannot send light signals to future null infinity. The boundary of this region is the event horizon.

$ds^2=\exp(\nu(r))dt^2-\mu(r)^{-1}dr^2-r^2 d\Omega^2$

First, you have implicitly assumed that this spacetime is not only spherically symmetric, but static (because all of the metric coefficients are functions of $r$ only). I don't know if that was your intent. The most general spherically symmetric spacetime would have $\nu$ and $\mu$ be functions of both $r$ and $t$.

If a spherically symmetric spacetime is vacuum (zero stress-energy), then it will also be static (more precisely, none of the metric coefficients will be functions of $t$ if you write the metric in the form you have written it). But if it's not vacuum, then you cannot assume that it will be static and the most general line element would have $\nu$ and $\mu$ as functions of both $r$ and $t$.

I would also recommend taking the exponential out of the $dt^2$ term. If it's there, technically the coefficient of that term can never go to zero. It's better just to write it as a simple function:

$$
ds^2 = \nu(t, r) dt^2 - \mu (t, r) dr^2 - r^2 d\Omega^2
$$

All that said, I'm not sure if there is a simple general condition on $\nu$ and $\mu$ that you can use to tell whether this metric describes a black hole. If you already know that both $\nu$ and $\mu$ are functions of $r$ only, then I think the metric will describe a black hole if there is a finite value of $r$ for which $\nu = 0$; that value of $r$ will correspond to the event horizon.