This was posted before, way down in another thread and went unanswered. So I thought I will start a new thread of the interesting question. Quote: Originally Posted by Farsight I release a test particle of mass mp a given distance away from a black hole of mass M and measure their initial closing acceleration as a. If I now substitute the test particle with a second black whole of mass M, is the initial closing acceleration of the two black holes a or 2a or something else?. I think the locally Lorentz acceleration of a test particle starting from rest outside a black hole with mass M at circumferential distance 2 Pi r is given by: (with c=G=1) a = -(1-2M/r)^(-0.5) M/r^2 What is the equivalent for two black holes, momentarily at rest relative to each other, like Farsight asked above? I am pretty sure we can't just add the calculated accelerations like in the Newton case. Isn't there a "commandment" in relativity that says: "Thou shalt not add thine own acceleration directly to that of thine fellow traveler"?
I'd suggest using the equivalent Newtonian situation as a rough guide. As far as departures from Newtonian behavior - as far as I know colliding black holes have only been handled numerically, not analytically. Once you've got the numerical simulation down :-), you still have to pin down how, exactly, you are going to measure the acceleration to answer the question. You'd have to specify all the gory details - the idea of an accelration presupposes some sort of coordinate system, and you'll have to specify either the coordinate system or the measurement technique to get an answer.
Thanks pervect, but yea, gory problem, especially the coordinate choice! The nice "locally Lorentz" spacetime surrounding a test particle is down the (BH) drain. Perhaps, since the stated problem is completely static, 'pseudo Schwarzschild' coordinates with the origin in-between the two BHs can be used in a numerical attempt. There will of course be two event horizons...
If the BH are actually free-falling, the problem isn't static :-(. If you have two charged BH maintaining a constant distance, you can probably find the force between them. The static metric allows you to think in terms of forces. Of course the electric fields are contributing to the metric in this case, so you can't really think of the solution as being just that of the two BH - it's the combined solution of the two black holes and their electromagnetic fields. I thought there was possibly an analytic solution known for this case, but I'm not sure. Google finds references to: "The double-Kerr solution of Kramer and Neugebauer" but so far I haven't dug up an actual metric.
OK, bad word choice. If the two BHs are momentarily stationary, we do not have to worry about relative velocity or movement in curved space. Then the problem should be simpler, but probably not easy! I’m also searching for a metric, but have not found anything yet. BTW, for simplicity, let's leave charge and rotation out for the moment...
Two black holes, momentarily stationary... The Newtonian equivalent suggests that if a single black hole with mass M causes a momentarily stationary particle to fall with a starting acceleration a at a radial distance r, then two momentarily stationary black holes with that same mass and at that same distance apart, will fall towards each other with a starting relative acceleration of 2a. In the Newtonian case, a = -M/r^2, geometrically. In the relativistic (Schwarzschild) case: a = -(M/r^2)/(1-2M/r)^(0.5), i.e., the Newtonian case divided by the gravitational redshift factor. Since we are not working with colliding black holes, but rather with momentarily stationary ones, will it be a reasonable assumption to just double this Schwarzschild acceleration, like in the Newtonian case? As in a previous threat https://www.physicsforums.com/showthread.php?t=125715&page=3), we have to ignore the effect of tidal deformation of the objects for simplicity.