Recognize Black Hole: Metric Conditions & Features

[sadegh4137]

consider have a metric
$ds^2=f(r,t)dt^2+g(r,t)dr^2+k(r,t)dΩ^2$
if g(r,t) =0, we have black hole?
any metric that has this condition, are black hole? or not this is first clue to determine black hole and we should check some other features!

solution of g(r,t)=0, may have some roots, for example BTZ black hole has two horizon, what we can interpret two horizon? for example,BTZ has inner and outer horizon!
what's this mean? what happen between two horizon?

 

[Nugatory]

consider have a metric
$ds^2=f(r,t)dt^2+g(r,t)dr^2+k(r,t)dΩ^2$
if g(r,t) =0, we have black hole?
any metric that has this condition, are black hole? or not this is first clue to determine black hole and we should check some other features!

solution of g(r,t)=0, may have some roots, for example BTZ black hole has two horizon, what we can interpret two horizon? for example,BTZ has inner and outer horizon!
what's this mean? what happen between two horizon?

This question might be clearer if you could rephrase without using the term "black hole".

Generally if you examine a metric written in a particular coordinate system, you can find singularities; then you get to amuse yourself seeing if you can find a coordinate transformation in which the singularities disappear. Any that can't be made to disappear are bona-fide singularities in the space-time described by that metric. Whether that space-time contains event horizons that we'd describe as "black holes" is a different question.

 

[GRGuy]

consider have a metric
$ds^2=f(r,t)dt^2+g(r,t)dr^2+k(r,t)dΩ^2$
if g(r,t) =0, we have black hole?
any metric that has this condition, are black hole? or not this is first clue to determine black hole and we should check some other features!

solution of g(r,t)=0, may have some roots, for example BTZ black hole has two horizon, what we can interpret two horizon? for example,BTZ has inner and outer horizon!
what's this mean? what happen between two horizon?

Singularities in a metric contravariant or covariant can correspond to event horizons or they can correspond to physical singularities depending on whether the singularity can be transformed away. In the Schwarzschild solution your g term isn't zero at the event horizon. Its singular there. For that solution it is your f that is zero there. Having an event horizon for your coordinates doesn't necessarily mean that you have a black hole. It could just mean you are using odd coordinates like those appropriate for someone's standards of spacetime who is just accelerating in a rocket. The line element you wrote down only gives you a black hole solution if there is no time dependence in the functions. The Vaidya solution is a known time dependent solution that's what you'd want to look at for what you're trying to do. The main difference is that in Kerr-Schild coordinates where its expressed something like that, it carries a dtdr cross term.
As for the last bit, personally I consider making theories and assertions about 2+1 dimensional black holes in negative cosmological constant spacetime as being something like working out how superman shaves.