0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – user4552 Dec 10 '18 at 2:42
1
$\begingroup$

The horizon forms first where the absolute value of the gravitational potential is the highest. In case of a collapsing star, it is in the center. However, this is not necessarily true in other cases.

Consider a thin massive shell collapsing in a perfect vacuum. The time dilation inside the shell is the same everywhere. Once the shell reaches the gravitational radius, the event horizon forms at this radius. There are no arguments for why the horizon should form in the center first in this case, because the gravitational potential is the same everywhere inside until the horizon is formed.

Now imagine that a perfect vacuum is disturbed by a spec of dust at some random location inside the shell. As the shell collapses, the gravitational potential is no longer the same everywhere inside, but has a higher absolute value at the location of the dust particle, because the mass of the particle contributes to the potential at its location. Therefore, in this case, the horizon will form first at the location of this particle (which is not necessarily in the center) and then expand out to the shell.

The last example, of course, is not spherically symmetric, but just an illustration that even a small disturbance can potentially significantly offset the location of where the horizon initially forms.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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