Black holes and positive/negative-energy particles

There are two ways to approach your question. The first is to explain what Brian Greene means, and the second is to point out that the "particles being swallowed" explanation is a metaphor and isn't actually how the calculation is done. I'll attempt both, but I'm outside my comfort zone so if others can expand or correct what follows please jump in!

When a pair of virtual particles are produced there isn't a negative energy particle and a positive energy particle. Instead the pair form an entangled system where it's impossible to distinguish between them. This entangled system can interact with the black hole and split, and the interaction guarantees that the emerging particle will be the positive one. NB "positive" and "negative" doesn't mean "particle" and "anti-particle" (for what it does mean see below), and the black hole will radiate equal numbers of particles and anti-particles.

Now onto the second bit, and I approach this with trepidation. When you quantise a field you get positive frequency and negative frequency parts. You can sort of think of these as representing particles and anti-particles. How the positive and negative frequencies are defined depends on your choice of vacuum, and in quantum field theory the vacuum is unambiguously defined. The problem is that in a curved spacetime, like the region near a black hole, the vacuum changes. That means observers far from the black hole see the vacuum as different from observers near the black hole, and the two observers see different numbers of particles (and antiparticles). A vaccum near the event horizon looks like excess particles to observers far away, and this is the source of the radiation.

See the Wikipedia article on the Bogoliubov transformation for more information, though I must admit I found this article largely incomprehensible.

Exactly the same maths gives the Unruh effect, i.e. the production of particles in an accelerated frame. The fact that the Unruh effect also produces particles shows that a black hole is not necessary for the radiation, so it can't simply be virtual particles being swallowed.

Also, the book mentions that a negative-energy particle would appear to an observer inside the black hole as positive. Why?

Very roughly speaking and in as simple terms as possible, inside the black hole, gravity is so intense that the time coordinate and one of the spatial coordinates (the radial coordinate) swap "roles". That's one way to see why you can't get "back up" and exit the hole. "Back up" is now in the reverse time direction and you can't go back in time.

Anyhow, since energy is associated with the time coordinate and momentum with the spatial coordinates, the energy and radial momentum of a particle also swap "roles" when crossing the horizon. The negative energy of the particle becomes negative momentum and the positive momentum of the particle becomes positive energy.

I just watched PBS Space time on this, which also seems to answer your question.

https://www.youtube.com/watch?v=ztFovwCaOik

While I have no doubt the accepted answer is the most complete. I still want to try to explain it in the simple terms I understood it from the video.

Disclaimer: I am not a physicist and my answer is most likely wrong/incomplete.

a) Virtual particles are first and foremost a mathematical trick for us to be able to calculate the constantly wiggling space-time on a quantum scale. In reality it seems like, when we talk about virtual particles we actually talk about small field fluctuations canceling out each other.

b) On the event horizon however those fields are suddenly cut of, so the fluctuation can't cancel out and produce real, positive energy possessing particles instead. The Hawking radiation.

So this explains the radiation. Question is, why does the black hole loose energy.

The easy way out would be to call for the law of energy conservation. I have no clue, but I picture it like when you hold a sheet with others steady that wiggles due to wind. It costs you energy to hold it steady at the border as you absorb the fluctuations.