# In any collapse does the horizon “emerges” from $r =0$?

Consider the Vaidya geometry describing the collapse of a massless null shockwave: $$ds^2=-\left(1-\dfrac{2M\theta(v)}{r}\right)dv^2+2dvdr+r^2(d\theta^2+\sin^2\theta d\phi^2).$$

It is claimed in several papers (this one for example - see discussion after equation 3.1 on page 8) that the horizon of the black hole formed by this collapse originates at $$r=0, v =-4M$$ continues along $$r = \frac{v}{2}+2M$$ from $$v = -4M$$ to $$v = 0$$ and finally remains at $$r = 2M$$.

So consider the (spherical) surfaces $$\Sigma(v,r) = \{(v,r,\theta,\phi) : (\theta,\phi)\in S^2\}$$

What this tells is that there is one non-decreasing function $$R_h(v)$$ such that the horizon "at time $$v$$" is $$\Sigma(v,R_h(v)).$$

For this particular case $$R_h(v)=\begin{cases}0,& v\in (-\infty,-4M),\\ \frac{v}{2}+2M,&v\in (-4M,0),\\ 2M, &v\in (0,+\infty)\end{cases}$$

Physically this means that for this black hole geometry the horizon is a spherical surface that emanates from $$r =0$$ and grows in radius as time passes until it reaches the Schwarzschild radius.

Is that true for any (spherically symmetric) gravitational collapse? Does the horizon of the black hole always appears at $$r =0$$ and grows with time until it reaches the Schwarzschild radius with mass equal to the total mass of the collapsing body?

If this is true, is there some general proof of this, and in particular some differential equation that the function $$R_h(v)$$ should satisfy?

• It isn't well established that gravitational collapse to a singularity always leads to the formation of a horizon at all. See, e.g., arxiv.org/abs/1201.3660 – Ben Crowell Dec 10 '18 at 2:42