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I am interested in finding event horizon of static space times of the following forms : \begin{equation} ds^2=-f(r)dt^2+\frac{1}{g(r)}dr^2+r^2d\Omega^2, \end{equation} where we have $f(r)\neq g(r)$. According to the Carroll's book :

1-determinig the point at which r=constant hypersurfaces become null is easy; $\partial_{\mu}r$ is a one form normal to such hypersurfaces, with norm \begin{equation} g^{\mu\nu}\partial_{\mu}r\partial_{\nu}r=g^{rr}, \end{equation} we are looking for the place where the norm of our one vanishes \begin{equation} g^{rr}(r_H)=0, \end{equation}

so from this definition, $g(r_H)=0$ is the location of event horizon.

2-But $K^{\mu}=\delta^{\mu0}$ is a time like Killing vector which becomes null $g_{\mu\nu}K^{\mu}K^{\nu}=g_{tt}(r_{H'})=0$ for some hypersurface which is located at $r=r_{H'}$.

It seems to me that 1&2 are contradictory according to Carroll's book which claims

A-Every event horizon $\Sigma$ in a stationary, asymptotically flat space time is a Killing horizon for some Killing vector field $\xi^{\mu}$.

B-If the space time is static, $\xi^{\mu}$ will be the Killing vector field $\partial_t^{\mu}$ representing time translations at infinity.

if $f(r_{H'})\neq g(r_H)$, the above conditions are not satisfied.

3-By looking at the null geodesics \begin{equation} \frac{dr}{dt}=\sqrt{f(r)g(r)}, \end{equation} it seems to me that the event horizon should be the outer radios for which one of the functions $f(r)$ or $g(r)$ is equal to zero.

which one is the definition of event horizon? I am interested in an explicit calculation and not the general explanations as event horizon is the hypersurface for which spacetime is divided to two separately causally disconnected regions.

I found some answers to the similar questions without explicit calculations :

How to derive the Schwarzschild radius?

What is the radius of the event horizon?

Does condition $g_{00}(r_0)=0$ define the event horizon on $r_0$?

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  • $\begingroup$ In any case i have encountered $g(r)$ will eventually end up being something like $f(r)*h(r)$ with $h(r)$ being an analytic non zero function everywhere. $\endgroup$
    – Noone
    Aug 23, 2023 at 8:09

1 Answer 1

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If you use arbitrary metrics and coordinates the event horizon is at the contravariant $g^{\rm rr}=0$. In static Schwarzschild spacetime this gives the same result as the covariant $g_{\rm tt}=0$, otherwise the latter gives the static limit which can also happen at the ergosphere if there is one.

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