# Oppenheimer-Snyder collapse

1. May 12, 2014

### tom.stoer

I am looking for a mathematical solution of the following "paradox" regarding the Oppenheimer-Snyder collapse of a finite sphere of dust.

Suppose the sphere of mass M has just collapsed to its Schwarzschild radius

$$R_S(M) = \frac{2GM}{c^2}$$

1) Suppose there is another thin shell of dust just outside the Schwarzschild radius at $[R, R + dr], \, R > R_S$ which is falling into the black hole. Using the Schwarzschild time coordinate t it is clear that this shell never reaches the horizon at finite Schwarzschild time and therefore the mass flow through the event horizon is zero.

2) Suppose there is a stationary observer just outside the Schwarzschild radius at constant $R > R_S$ who sees the free falling dust. This shell has some mass dm. The shell will not fall through the Schwarzschild radius, but it will certainly fall through the radius R of the stationary observer and will reach $[R^\prime, R^\prime + dr^\prime]$ with $R_S < R^\prime < R^\prime + dr^\prime < R$ in finite Schwarzschild time.

(b/c the observer is stationary we can simply rescale Schwarzschild time to his proper time using a finite constant)

From the mass m one can derive the new Schwarzschild radius

$$R_S(M+dm) = \frac{2G(M+dm)}{c^2} = R_S(M) + R_S(dm) = R_S + dR_S$$

Of course for large enough mass dm we find for this new Schwarzschild radius $R_S < R^\prime < R^\prime + dr^\prime < R_S + dR_S$. Therefore the mass flow through the event horizon at $R_S^\prime$ is not zero - which contradicts the conclusion drawn from (1) that the mass flow is zero.

My question is whether there is a realistic solution for infalling matter of finite mass from which the growth of the event horizon can be derived analytically as a function of Schwarzschild time as some function $R_S(t)$

Last edited: May 12, 2014
2. May 12, 2014

### Staff: Mentor

This argument is not valid. If it were, it would prove too much: it would prove that the original shell could not have collapsed inside its Schwarzschild radius either. The correct resolution to this is to realize that Schwarzschild coordinates are singular at the horizon, so you can't use them to analyze this problem (or at least, you can't use them "naively" the way you are implicitly using them here). You either have to pick better coordinates, or you have to carefully distinguish in your analysis between coordinate quantities and actual physical quantities, like the proper time it takes a dust particle to fall to the horizon. The original Oppenheimer-Snyder paper used the latter method; more modern analyses use the former.

No, there isn't. There can't be, because, as I noted above, Schwarzschild coordinates are singular at the horizon. The key aspect of that for your question here is that Schwarzschild coordinates assign the same time coordinate (plus infinity) to *any* event on the horizon; i.e., Schwarzschild coordinates can't distinguish between distinct events on the horizon, so of course you can't construct a function $R_S(t)$ where $t$ is Schwarzschild time. You need to pick better coordinates.

The simplest solution I know of that describes mass-energy falling into a "black hole" is the ingoing Vaidya null dust, which is described here:

http://en.wikipedia.org/wiki/Vaidya_metric#Ingoing_Vaidya_with_pure_absorbing_field

This gives a solution $M(v)$ for the mass of the black hole as a function of "advanced time" $v$, which is basically the same as the ingoing Eddington-Finkelstein null coordinate $v$. The key point here, again, is that $v$ *can* be used as a valid coordinate on the horizon, i.e., each distinct event on the horizon is labeled by a distinct, finite value of $v$. So the function $M(v)$ naturally yields a function $R_S(v) = 2M(v)$ for the horizon radius as a function of $v$.

The infalling mass-energy in the Vaidya solution is null dust, not timelike dust, i.e., it's ingoing null radiation like photons. I have not seen an analogous solution using continuously infalling timelike dust, but it seems like there ought to be one, perhaps using ingoing Painleve coordinate time $T$ to label events on the horizon (which basically works the same as $v$ does, each distinct event on the horizon has a distinct, finite value of $T$).

3. May 12, 2014

### George Jones

Staff Emeritus
There might me exact solutions, but I am not aware of any, e.g., I don't see this type of exact solution in "Exact Space-Times in Einstein's General Relativity" by Griffiths and Podolsky. All the realistic work of which I am aware involve numerical codes, e.g. Chapter 8 "Spherically symmetric spacetimes" from the book "Numerical Relativity: Solving Einstein's Equations on the Computer" by Baumgarte and Shapiro.

4. May 12, 2014

### tom.stoer

Thanks for you hints.

That means that - unfortunately - it is hardly possible to describe this from the perspective of a Schwarzschild observer (or a static observer hovering at constant r).

I am aware of the Vaidya metric; anyway, thanks for your answers.