Baby Astronaut Posted October 15, 2009 Share Posted October 15, 2009 It seems the meeting of particle/antiparticle keeps the violation of conservation from occurring. But if Hawking Radiation is a vast group of particles composed entirely of the one particle that didn't fall into the black hole, then can't we safely deduce the escaping particles won't be annihilated in the near future? Yet if that's the case, has the violation occurred? Another question on Hawking Radiation: it follows that about half of the particles falling into the black hole by that process will be "anti". If so, the black hole itself surely is made of regular particles. So as soon the anti- enters the black hole, shouldn't it annihilate with a random particle? Link to comment Share on other sites More sharing options...
ajb Posted October 15, 2009 Share Posted October 15, 2009 I would be careful trying to use the heuristic interpretation of particle/anti-particle pairs falling into a black hole to make any concrete statements. What I will tell you is that the radiation is thermal. That is it is a black-body and no information about what fell into the black-hole can be deduced. Can you just clear up what you mean by "violation of conservation "? Conservation of what? Link to comment Share on other sites More sharing options...
Spyman Posted October 15, 2009 Share Posted October 15, 2009 The particle/anti-particle pairs could very simplified be viewed as to be virtual particles being "boosted" by the black hole's strong gravitation, close to the event horizon, into becoming real particles. So you could say that the BH pays the energy depts for the particles by loosing the mass equivalent to the pair, so there is no violation of conservation laws here. It doesn't matter if the particle that falls back into the BH is "anti" or not, it will still be a part of the BH. If the particle meets an opposite inside the BH there would be energy radiating equivalent to the mass of the pair from the annihilation and according to Relativity that energy causes the same amount of gravity that the particles would have otherwise. Link to comment Share on other sites More sharing options...
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