0
$\begingroup$

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.

Improve this question
$\endgroup$
3
  • $\begingroup$ Could you perhaps show a little bit more of your work? In particular, if you're only making a change of spatial coordinates then $g_{tt}$ remains unchanged and therefore vanishing at the event horizon, rendering the metric singular ... $\endgroup$
    – J. Murray
    Commented May 25, 2021 at 18:32
  • $\begingroup$ that is exactly what I'm doing in order to give ds^2 in relation of only dr^2, just like we analyse the Schwarzschild metric $\endgroup$
    – Agatha Harkness
    Commented May 25, 2021 at 20:08
  • $\begingroup$ Perhaps I was not clear. Assuming that you're defining $r(x,y,z)=\sqrt{x^2+y^2+z^2}$, $\theta(x,y,z) =\cos^{-1}(z/\sqrt{x^2+y^2+z^2})$, and $\phi(x,y,z)=\tan^{-1}(y/x)$ (where, I should add, $x,y,$ and $z$ have a somewhat different physical significance than the cartesian coordinates in flat space), the metric components $g_{xx},g_{yy},g_{zz}$ are still singular at $r=2M$. $\endgroup$
    – J. Murray
    Commented May 25, 2021 at 20:53

1 Answer 1

Reset to default
4
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • $\begingroup$ thank you! I will read a little more on that and try again. $\endgroup$
    – Agatha Harkness
    Commented May 25, 2021 at 20:10
  • $\begingroup$ Strictly, isn't the event horizon the * outmost * surface which a particle can only cross once? Because even after crossing it, a notional observer would be unable to see what's ahead of them, it will still look and behave like an event horizon, because anything else closer to the singularity/centre will only have inward facing futures too, so the surfaces between will also be "surfaces that can only be crossed once". $\endgroup$
    – Stilez
    Commented May 26, 2021 at 7:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.