1
$\begingroup$

My understanding of Hawking Radiation is this:

All the time, small amounts of matter and antimatter pop into existence, find each other, and annihilate. When this happens near/on the edge of a black hole, the matter particle slips out into space, while the antimatter particle falls into the black hole and reduces it's mass.

That's fine. My question is; How is the thing with negative mass being pulled in by gravity, while the thing with positive mass is escaping? shouldn't it be opposite of that?

$\endgroup$

2 Answers 2

2
$\begingroup$

You have some basic misconceptions. The big one is that antimatter particles don't have negative mass: they have positive mass just like matter particles.

Also, Hawking radiation can be antimatter as well as matter. Then the matter is the one that "falls into" the black hole.

$\endgroup$
1
$\begingroup$

As far as we know, antimatter does not have a negative mass, it just has an opposite electric and quantum charge (although scientists at CERN’s alpha institute are looking into the idea of antimatter having negative mass). When these two virtual particles are created, either can fall into the black hole, its not always the antimatter. Stephen Hawking noticed is that if these two particles are created at the event horizon of the black hole, one could fall in. This would defy one of the important concepts of quantum mechanics - the conservation of information - which states that these two virtual particles should be able to exchange information with each other if they were created at the same time. If this was the case, than information would have to leave that black hole which is impossible according to relativity as the information would have to break the speed of light. Hawking proposed that black holes can emit radiation, Hawking radiation, that would allow the particles to still share information without breaking the speed of light. This unfortunately brings up a few paradoxes such as the fact that black holes should have a shell of radiation that we haven’t directly observed.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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