Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Two related questions:

  1. Small object of mass $m$ is falling into the supermassive black hole of mass $M$. What is the maximal kinetic energy can be acquired by the small object, from the point of view of an observer whose frame of reference is at rest near event horizon? Can it exceed $(M+m)c^2$?

  2. Two small black holes of equal mass $M$ are attracted each other. What is the maximal kinetic energy can be acquired by each black hole due to gravitational attraction, from the point of view of an observer between them?

share|improve this question
    
what do you mean by "rest near the event horizon" - Is the observer influenced by gravitational field..? –  Waffle's Crazy Peanut Aug 29 '12 at 5:06
    
@Crazy Buddy, it should be an opposite equal force against the gravitational force to support an observer at rest. –  voix Aug 29 '12 at 18:43
    
Since gravitational potential energy is dependent on distance, to figure out the maximum kinetic energy, one would need to know the initial distance between the small object and the black hole. –  abhishek Jun 11 '13 at 23:01

3 Answers 3

For case 1) The energy is well-defined, and it can be shown that (for the case of a Schwarzschild hole, but this can be generalized to a spinning charged hole if you wish):

$$-1 = \frac{-E^{2} + {\dot r}^{2}}{1-\frac{2M}{r}} + \frac{L^{2}}{r^{2}}$$ holds for the entire motion of a test particle, where, for a test particle with mass $m$, $Em$ is the total energy of the orbit, and $Lm$ is the angular momentum of the orbit. Answering your question is just a matter of applying it to this equation, but note that, as $L,r,{\dot r}\rightarrow 0$, you have $Em\rightarrow m$, which implies that the total energy of a stationary object far from the black hole is approximately its rest mass.

For 2), the case is very considerably more complicated. You can get a first order guess by considering the Area increase theorem, which tells you that the two black holes will coalesce to a black hole whose area is no less than the sum of the previous two black holes' area, which gives you the inequality $M_{1}^{2} + M_{2}^{2} \leq M^{2}$, up to some factors involving the spin (if you want to factor spin), this then gives you a cap on how much net energy you can transfer into gravitational radiation and the final translational kinetic energy of the final black hole (which is possible, because the gravitational radiation can be asymmetric, and typically, uneven mass black holes will end up with a translational velocity that is something like $10^{-3}c$, which can be above escape velocity from the enclosing galaxy.

Note that there are still bounds on this, because, since the spacetime will be asymptotically flat, the ADM Mass of the spacetime will be conserved.

share|improve this answer

The two objects have very low chance of moving directly onto each other. The nearest orbit would be the sum of the two Schwartzschildradei of the objects.

1 r=2G(M+m)/c^2 if you put this in E=GMm/r you get Mmc^2/2(M+m)<=mc^2/2 For (m<

2 r=2G(M+M)/c^2 and you get E=Mc^2/4

From these distances they will become accretion objects.

share|improve this answer
    
Any stable orbit between the two objects would break down before you got to the point where the horizons touched. Even for a point mass orbiting a Schwarzschild hole, the last stable orbit is at $r=6M$ –  Jerry Schirmer Oct 27 '13 at 22:52

technically, it would be infinite, as the energy of an object is determined by -GMm/(2r)

share|improve this answer
    
Note you are not answering the question. Your infinity would appear only if all the mass was concentrated at a point, which is an oxymoron in classical physics and also GR. By the type of small dimensions for masses quantum mechanics is dominant and there are no infinities. –  anna v Jul 24 '13 at 6:04

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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