Can Gravitons Escape Black Holes?

[pervect]

I ran into a bit of a problem, maybe related. I tried to do the calculation in proper time and then I have to keep track of coordinate time in order. But I get dt/ds = (E + \sqrt{r/2}dr/ds)/(1 - 2/r) which goes to 0/0 at r=2. So before I can have particles pass the event horizon, I have to do the division.

I've finished updating the Painleve-Gullstrand thread, having resolved the sign issues and also having found a stable online reference that's likely not to disappear in a few weeks that gets the same results that I do.

I'm not sure what approach you are using, but you might want to check that some of the conseved quantities for orbital motion are actually being conserved by your integrator (unless you are already using the conservation of these quanties to get your solution).

You might find http://www.fourmilab.ch/gravitation/orbits/

helpful, in that it calculates GR orbits, and has some discussion of the formulas as well.

You can also download the source code for their applet at the bottom of the page, too.

Note that for the Schwarzschild metric, the conserved quantities are

E₀ = (dt/dtau) (1-2M/r)
L = r^2 (dphi/dtau)

As I mentioned, in the PG thread, for the PG metric, the expression for the conseved E₀ changes to

E₀ = (1-2*M/r)*(dt/dtau) - sqrt(2M/r) * (dr/dtau)

(The expression for L remains unchanged).

This is different by a sign convention from the conserved quantity I mentioned in the PG thread, however it uses the same sign convention as the Schwarzschild energy.

Using the above sign convention, when the velocity at infinity is zero, E₀ = 1. It can be regarded as the energy per unit rest mass.

The above conserved quantites along with the metric equation give enough information to determine the orbits, i.e. you can derive the effective potential for the Schwazschild equation

(dr/dtau)^2 + V^2(L,r) = E^2

from the above conseved quantites of the Schwarzschild metric and the Schwarzschild metric itself with a little algebra and the fact that ds^2 = -dtau^2.

You can derive a similar equation for (dr/dtau) in the PG metric by the same means.

 

[CarlB]

Good simulation link, except his java crashes my windows. It uses the usual methods of deriving the orbits that is in MTW (i.e. based on the constants of motion).

A method of deriving the orbits more similar to what one can do with Newtonian gravity (i.e. straightforward integration), which I think may be better for the computer, is here:
http://sb635.mystarband.net/relat.htm
http://sb635.mystarband.net/cip.htm

The above was originally published in "Computers in Physics", then by the American Institute of Physics, and put on the web by Steve Bell.

As it turns out, the above equations exhibit the same 0/0 behavior that mine do. This is inevitable when you have a division by (1-2/R). But I find mine suspicious, so I'm going to compare them with the above, when I get some time.

Carl

 

[pervect]

You can convert the first-order set of equations using conseved quantites into a second order set by just setting the derivatives of the conserved quantites to zero.

I think I recall this helping to get past some "turning points", but it's been too long for me to be sure it really helped. I'm not sure if it will help with your problem at all.

This set of 2nd order equations is equivalent to the geodesic equations you get from the Christoffel symbols (it may take some algebraic manipulation to see the equivalence).

i.e.

d^2 x^i / dtau^2 + \Gamma^i{}_{jk} (d x^j / dtau) (d x^k / dtau) = 0

 

[CarlB]

You can convert the first-order set of equations using conseved quantites into a second order set by just setting the derivatives of the conserved quantites to zero.

I think I recall this helping to get past some "turning points", but it's been too long for me to be sure it really helped. I'm not sure if it will help with your problem at all.

My intuition says that this is the way to do it. Break the 2nd order equations down to 1st order equations.

I've now got a first attempt at the simulation up here:
http://www.gaugegravity.com/testapplet/SweetGravity.html

It still has the division by zero problem, so I've set it for an initial condition that stays away from the event horizon. Also, in my equation dt/ds = (E + \sqrt{r/2}dr/ds)/(1 - 2/r) I had the wrong sign. The correct equation, of course, is dt/ds = (E - \sqrt{r/2}dr/ds)/(1 - 2/r). The choice of the sign amounts to a sort of arrow of time for the black hole.

Carl