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Given a static spherically symmetric spacetime with metric being solution for interior of perfect fluid sphere $$ds^2=e^{2\nu}c^2dt^2-e^{2\lambda} dr^2-r^2 (d\theta^2+\sin{\theta}~d\phi^2) \tag{1}$$ is the hyperspace defined by equation $${e}^{2\nu}(r_0)=0\tag{2}$$ a trapped surface, apparent horizon or event horizon?

I presume that for $r_0=r_S$ it is the event horizon but not for the lower values.

The coordinate $r_0$ above means the curvature radius of a two-sphere and ranges from $0$ to $r_S$ (Schwarzschild radius).

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  • $\begingroup$ Event horizon depends on global properties of spacetime while $g_{00}=0$ is a local condition only $\endgroup$
    – paul230_x
    Commented Apr 11, 2023 at 11:15
  • $\begingroup$ You are right, I have forgotten it. I should clarify my question. Let think about spacetime described by Schwarzschild interior solution for compactness parameter $\alpha$ ($r_S/R)$ from 8/9 to $1$ ($r_0=0$ to $r_0=r_S$). Is that hyperspace a trapped surface or apparent horizon or something else? $\endgroup$
    – JanG
    Commented Apr 11, 2023 at 14:12
  • $\begingroup$ In the case of static spherically symmetric spacetimes, I believe the event horizon coincides with the apparent horizon. You can find the AH by looking at where null congruences change focusing properties, i.e., where the expansion scalar becomes zero. $\endgroup$
    – jboy
    Commented Apr 11, 2023 at 14:45
  • $\begingroup$ OK, but what is expansion scalar? $\endgroup$
    – JanG
    Commented Apr 11, 2023 at 14:50
  • $\begingroup$ I believe the condition of the metric, and thus geometry, being static and spherically symmetric puts constraints on the form of the metric you could have. And by assuming the sphere is a perfect fluid, you are adding extra constraints which ultimately would make the metric have the property: $g_{tt}=-1/g_{rr}$. Therefore $g_{00}=0$ does imply that there is something wrong in our space when this condition is satisfied. We then go on examine in detail, checking for example if we could complete a geodesic along a light ray trajectory. If we could extend our spacetime past the singularity. $\endgroup$
    – Rescy_
    Commented Apr 24, 2023 at 15:11

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These things depend on timelike foliations of the spacetime.

Consider some timelike normal $n_{a}$ such that $n_{a}n^{a} = -1$ and $n_{a} \propto \nabla_{a}\tau$ for some function $\tau$ that gives the foliation.

You wish to ask whether some closed surface is a trapped surface. In that case, within the three slice, the intersection of the trapped surface and your leaf of constant $\tau$ will have a spacelike normal $s_{a}$ within the slice and associated 2-metric${}^{1}$ $q_{ab} = g_{ab} + n_{a}n_{b} - s_{a}s_{b}$, and the 2-surface will have the outgoing null normal $\ell^{a} = c\left(n^{a} + s^{a}\right)$ for any constant $c$. Then, any surface that satisfies:

$$\ell^{a}\nabla_{a}\left(q^{ab}\nabla_{a}\ell_{b}\right) = 0$$

will be a trapped surface, which captures the idea that the stack of surfaces on different leaves of $\tau$ will have $\ell$ as a Killing vector, which, in turn, captures the idea that that is the surface where "outgoing rays are frozen on the horizon"

Then, knowing all of this, it's just a matter of computing it yourself. But the definition does require that you either define a timelike foliation, or work out the somewhat dark art of null intrinsic geometry.

${}^{1}$technically, this is the pullback of the intrinsic 2-metric onto the 4-space

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  • $\begingroup$ That is a very useful explanation. However, I have to think about it first to comprehend it in my terms. Thanks! $\endgroup$
    – JanG
    Commented Apr 11, 2023 at 15:19
  • $\begingroup$ I think I understand it. In simply words, the condition $g_{00}(r_0)=0$ in static spherically symmetric case, for example in Schwarzschild (static) interior solution, constitutes a trapped surface. Am I right, or it is more complicated? $\endgroup$
    – JanG
    Commented Apr 11, 2023 at 15:43
  • $\begingroup$ @JanG in what context are you setting $g_{00} = 0$? are you starting from a general spherically symmetric vacuum solution? Are you requiring asymptotic flatness? How are you introducing the black hole mass? $\endgroup$ Commented Apr 11, 2023 at 17:14
  • $\begingroup$ No. I am interested in static spherically symmetric perfect fluid solutions as for example Schwarzschild interior solution. I assume asymptotic flatness ($p(R)=0$). As long $r_0<r_S$ there is no (eternal) black hole mass. Please see Appendix in my question physics.stackexchange.com/q/757634/281096 and/or workout in physics.stackexchange.com/a/679431/281096. $\endgroup$
    – JanG
    Commented Apr 11, 2023 at 18:00
  • $\begingroup$ I consider the black hole forming process in quasi-static steps. By growing parameter $\alpha$ the perfect fluid sphere is continuously in equilibrium (negative pressure gradient). $\endgroup$
    – JanG
    Commented Apr 11, 2023 at 18:06

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