Event horizon from a metric in Cartesian coordinates How can we derive the event horizon from a metric dependant on $(t,x,y,z)$? I've seen the Schwarzschild solution and the Kerr solution, but both of these are given in $(t,r,θ,φ)$. I tried converting them from $x,y,z$ to $r$ (keeping $θ$ and $φ$ constant along with $t$)
The problem is I can't use the same method as with Kerr and/or Schwarzschild because the $g_{xx}=g_{yy}=g_{zz}$ is still $(1+M/2r)^4$ which only goes to infinity at $r=0$ (there is no incoherence in the coordinates like in the Schwarzschild solution)... And I can't use the sign and say from that point on is timelike and before it, it's spacelike, so the light cone is tilted towards the interior when r<(event horizon).
There is no sign to change, the term it is in the fourth power. I just don't know where to begin I suppose, everywhere I've searched the only mathematical definitions of the event horizon I could find are those two and they don't seem to apply here.
I just need a point to begin.
 A: For the purposes of this post, I will mean "event horizon" when I say "horizon".
As you correctly observe, many properties of the horizon in the Schwarzschild coordinates that one might think of as the defining properties of the horizon go away in better coordinate systems. The key is that the horizon is the horizon not for local reasons but for global reasons. There is nothing special that happens at the horizon; someone passing through a horizon wouldn't feel special at all (well, there is a potential exception if you are a group of people whose initials read AMPS but I'd ignore it for current purposes). Thus, you cannot define the horizon through some local condition on the metric, e.g., that the first derivative of the metric should be singular or that the metric should change sign, etc.
Now, globally, of course, the event horizon is a very special and interesting surface, it is the surface that a physical particle can only cross once. Speaking at the leading order in complexity, you'd need to conformally transform the given metric to draw its Carter-Penrose diagram to see what is the global causal structure associated with the given metric and you'd be able to see in this process if there are horizons or not. If there are horizons, you can invert the conformal transforms to get the defining equation for the surface that is the horizon in the original coordinates in which the metric is given to you.
