# Why don't back holes decrease in mass as you add things?

1. Sep 22, 2015

### Psip

In hawking radiation a black hole decreases in mass and is described by E= mc^2-GMm/r and as r gets closer to zero the energy of the particle that enters becomes negative and takes away from the net energy of the black hole. My question is why does this only apply to these particles that pop out of the vacuum and not every particle that falls into the black hole. I realize though that this would violate conservation of energy if everything that fell in made the energy go down.

2. Sep 22, 2015

### phinds

I don't have a solid answer for you and there are others here who will, but I would give you one piece of information that might help. The whole concept of Hawking Radiation being due to virtual particle-pairs is a pop-science type description which was put forth by Hawking himself because, as he said, he just wasn't able to think of any other way to translate the math into English.

3. Sep 23, 2015

### Demystifier

As you said, that would violate energy conservation, which is the answer to your question. In the case of Hawking radiation the energy is however conserved because the particles are created in pairs, one with positive energy going outside and another with negative energy going inside, so that their total energy is zero.

But negativity of energy is not directly related to the fact that mc^2-GMm/r is negative for small r. The origin of negative energies is different.

4. Sep 24, 2015

### Psip

Thanks for the replies. I got an answer from somewhere else and apparently total energy would be E=mc^2-GMm/r+1/2mv^2 so when particles come in from far away they will have positive energy. So things really need to be made on the edge of the event horizon. However, would the star junk near the center of mass right before collapse be able to initially be close enough to still have negative energy?