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This forum seems to agree that a billiard ball accellerated to ultra-relativistic speeds does not turn into a black hole. (See recent question "If a 1kg mass was accelerated close to the speed of light would it turn into a black hole?") On the other hand, LHC scientists take seriously the formation of black holes by colliding protons. Therefore I presume it will be agreed that two ultra-relativistic billiard balls colliding head on with the requisite energy will form a black hole. The question is, suppose the two relativistic billiard balls are on antiparallel tracks that just miss each other, but pass within the radius of the putative black hole formed by their combined mass-energy. Will the black hole still form? If not, why not?

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First of all, while there are superficially plausible models in which the LHC may produce microscopic black holes, it is extremely unlikely that this will take place. This possibility can only occur if there are large or warped extra dimensions of space that are relevant already at the LHC energy scale - a TeV. However, even if those dimensions exist, the black holes will be among the last things that will be created. The top researchers behind these models - such as Lisa Randall and Nima Arkani-Hamed (to choose the most famous people both from ADD and RS) - have written papers explaining the same fact.

Even if the black holes would be created, they would be indistinguishable - at least to the laymen - from other unstable elementary particles. Of course, all worries about the fate of the Earth are based on confusing the properties of tiny black holes and the big ones.

Second, if you collide two billiard balls and at some moment, both objects fit into a volume that is sufficiently smaller than a number comparable to the Schwarzschild radius of a black hole corresponding to the mass given by the center-of-mass energy of the two balls, then a black hole will form.

The previous sentence contains an order-of-magnitude estimate. It is very tough - and probably depending on the geometry of the balls and other things - to calculate the exact numerical coefficient. But if you're satisfied with the order-of-magnitude estimates, then it is true that if the balls even fit into the size of the black hole - from the center-of-mass inertial system's viewpoint - the black hole will form.

The formation of a black hole is a subtle thing. It will almost certainly not be the case that all the energy will be stored in the new black hole. An O(50%) part of the energy of the initial speedy balls will be radiated in the form of gravitational waves before the black hole is formed.

But morally speaking, what you say is true. The balls don't have to directly collide. When they get close to each other, regardless of the fact that they will change the direction a bit, the space around them just curves in the right way that leads to the emergence of the event horizon. After some short time, the initially anisostropic black hole radiates the energy (by "ringing modes") and settles to a cylindrically symmetric Kerr (rotating) black hole, and after that it relaxes more slowly via Hawking radiation towards the spherically symmetric Schwarzschild shape we know.

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The edit that suggests these balls will form a spinning black hole seems to assume linear momentum can be converted into angular momentum. Is the edit technically correct generically for two-ball collisions that just miss each other? – Brandon Enright Dec 4 '13 at 16:32
@Brandon: Yes, even ordinary, Newtonian billiard balls covered in sticky tape will form a spinning mass if they collide off-center. No angular momentum is created, angular momentum was always there even though nothing initially was spinning! – Kevin Kostlan Dec 4 '13 at 18:06
Re:'relax to spherically symmetric Schwarzschild' will actually look like quite the explosion. The black hole will last all of 1e-10 seconds or so as it evaporates. That's 1kg of energy released in a short time. Don't stand too close! – Tom Andersen Apr 6 '14 at 12:14

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