# Oppenheimer Snyder Collapse - finite time?

1. Oct 15, 2011

### Passionflower

The Oppenheimer Snyder Collapse into a black hole does it happen in finite time for local and remote observers?
Does a trapped surface really form in finite time?

Last edited: Oct 15, 2011
2. Oct 17, 2011

### George Jones

Staff Emeritus
The last sentence of the abstract from the original 1939 paper by Oppenheimer and Snyder:

3. Oct 17, 2011

### Passionflower

Right we can see that if we consider the integrals:

For an observer in a Schwarzschild solutions with rs being the Schwarzschild radius riding on the surface of the dust ball starting at rstart we get a proper time of:
$$\Large \int _{{\it ri}}^{{\it ro}} -{\frac {1}{\sqrt {{\frac {{\it rs}}{r}}-{\frac {{\it rs}}{{\it r_{start}}}}} }} {dr}$$

And the same for an stationary observer at an arbitrary robserver coordinate we get, if I am not mistaken, a proper time of:
$$\Large \int _{{\it ri}}^{{\it ro}} - \sqrt{{\frac {r_{{{\it start}}}}{{\it rs}}}-1} \sqrt{1-{\frac {{\it rs}}{r_{{{\it observer}}}}}} \left( 1-{\frac {{\it rs}}{r}} \right) ^{ -1} \left( \sqrt{{\frac {r_{{{\it start}}}}{r}}-1} \right) ^{-1} {dr}$$

The observer riding on the dust ball will always see a finite proper time to reach the event horizon, however outside stationary observers always see an infinite proper time for the surface to reach the event horizon, even if such an observer is for instance stationary at rs+epsilon.

It would be interesting to have an observer robserver starting to free fall from zero how his notion of proper time will be, any takers for expressing that integral?

4. Oct 17, 2011

### DrStupid

I know that this would be valid for a black hole with constant mass, but does it also apply for a growing black hole? The matter would need infinite time to reach a stationary event horizon, but it does not need to reach it to enter the black hole. As soon as the distance falls below the Schwarzschild radius of the incoming mass the event horizon will expand by this value and the matter is gone. This should happen in finite time even for an external observer.

5. Oct 17, 2011

### DrStupid

According to this equation the half-life of the distance to the event horizon is

$t_{{\textstyle{1 \over 2}}} = \frac{{\gamma \cdot M}}{{c^3 }}$

for an observer at infinite distance. For the Sun this is 5 microseconds. That means even small objects would enter the event horizon very fast.

6. Oct 17, 2011

### Passionflower

I am not sure what you are trying to say here can you show your calculations?

Last edited: Oct 17, 2011
7. Oct 18, 2011

### DrStupid

Your equation is valid for natural units only. To get the above mentioned formula I start with the following equation which is independent from the system of measurement:

$\frac{{dt}}{{dr}} = - \frac{{\sqrt {\frac{{r_{start} }}{{r_s }} - 1} \sqrt {1 - \frac{{r_s }}{{r_{observer} }}} }}{{c \cdot \left( {1 - \frac{{r_s }}{r}} \right)\sqrt {\frac{{r_{start} }}{r} - 1} }}$

For a far distant observer (robserver>>rs), a starting point high above the event horizon (rstart>>rs) and positions far below the starting point (r<<rstart) this can be simplified to

$dt = - \frac{{dr}}{{c \cdot \left( {r - r_s } \right)}}\sqrt {\frac{{r^3 }}{{r_s }}}$

As I am interested in the distance h from the event horizon I substitute with

$h: = r - r_s$

and get

$dt = - \frac{{dh}}{{c \cdot h}}\sqrt {\frac{{\left( {r_s + h} \right)^3 }}{{r_s }}}$

For small steps (and near the event horizon the are small) the differential coefficient can be replaced by the difference quotient and with

$\Delta h: = - {\textstyle{1 \over 2}}h$

I get the time for a bisection of h:

$t_{{\textstyle{1 \over 2}}} = \frac{1}{{2 \cdot c}}\sqrt {\frac{{\left( {r_s + h} \right)^3 }}{{r_s }}}$

$\mathop {\lim }\limits_{x \to 0} t_{{\textstyle{1 \over 2}}} = \frac{{r_s }}{{2 \cdot c}} = \frac{{\gamma \cdot M}}{{c^3 }}$

This is what I called the half-life of the distance to the event horizon.

8. Oct 19, 2011

### DrStupid

I just realized that the use of the difference quotient wasn't a good idea because x/2 isn't really small compared to x. Better to start with the substitution $r: = r_s + x$

$\frac{{dt}}{{dr}} = - \frac{{\sqrt {r_s + x} ^3 \sqrt {\frac{{r_{start} }}{{r_s }} - 1} \sqrt {1 - \frac{{r_s }}{{r_{observer} }}} }}{{c \cdot x \cdot \sqrt {r_{start} - r_s - x} }}$

than to approximate for $r_{start} > > r_s$, $r_{observer} > > r_s$ and $x < < r_s$ and finally to integrate the simplified equation:

$\Delta t \approx - \frac{{r_s }}{c}\int\limits_{x_0 }^x {\frac{{dx}}{x}} = \frac{{r_s }}{c} \cdot \ln \left( {\frac{{x_0 }}{x}} \right)$

The result differs from my first approximation by the factor ln(2)/2 but it doesn't change the result in general. The distance from the event horizon decreases exponentially and will run below every reasonable limit in a rather short period of time. In case of a black hole with the mass of the Sun it would be less than a millisecond for the last meter down to Planck length or to the Schwarzschild radius of the incoming matter. That's negligible compared to the 1770 seconds for a free fall from the original surface near to the final event horizon. If a wizard transforms the Sun into very cold dark matter without rotation, the gravitational collapse would take half an hour from the view of a distant observer.

9. Oct 19, 2011

### Passionflower

I agree with you that the 'last step' is very very small in case of the Sun.

In the graph below contrasting the proer time riding on the surface of the dust ball and the proper time far removed the two lines simply overlap as the discrepancy wil only occur extremely close to the EH.
[PLAIN]http://img829.imageshack.us/img829/3846/dustball.png [Broken]

But I find your conclusion wrong, as it will take forever for the external observer.
Who are you to decide that the last small part is negligible?
How do you support that attitude is scientific?

Last edited by a moderator: May 5, 2017
10. Oct 19, 2011

### DrStupid

As I already mentioned in my first answer the last step is not necessary. If the incoming mass is closer to the event horizon as its own Schwarzschild radius it is already inside the common Schwarzschild radius. When this happens the event horizon of the black hole expands to the common Schwarzschild. For a single particle this is a very complex process because it can not be described correctly in Schwarschild metric but this should be possible for a spherical symmetric collapse.

Who are you to decide that the ART is valid below Planck length? How do you support that attitude is scientific?

11. Oct 19, 2011

### Passionflower

I am talking about the Schwarzschild solution nothing more nothing less.

If you teach people that dust balls in a Schwarzschild solution become black holes in finite proper time for far away observers you are simply not telling the truth.

12. Oct 19, 2011

### DrStupid

Quite apart from the fact that below Planck length this is no longer physics but a mathematical exercise only (because ART is known to be not valid in these dimensions) we do not need to discuss the Schwarzschild solution anymore because the initial question was already answered by George Jones. It is well known that a static spherical metric leads to infinite time for the collapse and I wrote in my first sentence, that I am aware about it. I wonder if this is also valid for a non static metric and I gave some reasons for my doubt. I'm quite sure that a real spherical symmetric collapse will take finite time even for a distant observer.