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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).

JanG
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1 Answers1

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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

Zo the Relativist
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