# Antimatter and negative energy

1. Aug 14, 2010

In his popular-physics book "The Universe in a Nutshell", Hawking presents the creation/annihilation of virtual particle pairs in two ways:
(1) one particle has positive energy, the other negative energy, and
(2) a matter/antimatter pair.
This leads to the obvious conclusion that he is identifying antimatter as particles with negative energy, out of a sort of Dirac sea. However, this identification makes me wonder:
(a) on one side, negative energy results in repulsive gravity, but
(b) on the other side, antimatter is supposed to have the same gravitation as matter.
Something is not right. What?

While I am at it, another question on such pairs, whatever they are. One explanation for Hawking radiation is that one of the particles, either the one with negative energy and/or the antimatter one, falls into the black hole, while the other particle, matter with positive energy, escapes as radiation. However, these explanations never mention why it is the antimatter and/or particle with negative energy that falls into the hole, and the other one escapes, instead of the other way around. Why?

2. Aug 14, 2010

### bcrowell

Staff Emeritus
In electromagnetism, the reason the Dirac sea is undetectable is that the field equations are perfectly linear. For example, we need to be able to assume that the (zero) field of the Dirac sea is the same as the field of an electron plus the field of the Dirac sea with an electron missing from the corresponding position. I think the whole thing is spoiled by the slightest deviation from nonlinearity, and this may be the underlying issue. I don't think you can have a Dirac sea applied to gravity, because GR is nonlinear. This is the same kind of issue you get into with white holes, where you can convince yourself of all kinds of paradoxes by assuming linearity. (E.g., "Let's superimpose a black-hole solution and a white-hole solution to get a flat spacetime!")

3. Aug 15, 2010

Thank you, bcrowell. The fact that the Dirac sea is no longer used, being replaced by creation and annihilation operators, is the reason I wrote "a sort of Dirac sea" merely to give the flavour of the way Hawking presents negative energy. This was not meant to be taken literally. Leaving aside the Dirac sea imagery, then, my questions still remain. In short, in Hawking radiation, are we dealing with particles with negative energy, or with antimatter, or are they the same thing? If the latter, how are my objections to identifying them answered? Finally, why does the black hole consistently select the unusual half of the pair to swallow? (For better expressions, see my original post, and delete the "out of a sort of Dirac sea", which was a failed image.)

4. Aug 15, 2010

### bcrowell

Staff Emeritus
The processes that create Hawking radiation are completely agnostic about the types of particles they create. You can have Hawking radiation of neutrinos or sterile particles, for example. The customary generic way to describe this is that Hawking radiation consists of particle-antiparticle pairs that normally would have reannihilated, but are prevented from doing that because one falls past the event horizon. But photons are the dominant contribution to, say, the rate at which a black hole loses mass. Photons are their own antiparticle, so the discussion of particles and antiparticles becomes irrelevant in the photon case. Also, clearly an "antiphoton" isn't going to have repulsive gravity.

Regardless of whether we use the Dirac sea interpretation, a negative-energy particle in field theory is always interpreted as an antiparticle. Antiparticles do not have negative mass, they have positive mass. I don't think there's anything wrong with your use of the Dirac sea. The real issue is the difficulty in connecting the QFT idea of energy with the GR idea of energy. This is one of the fundamental things that makes it difficult to construct a theory of quantum gravity.

When you say that negative energy results in repulsive gravity, this is not really true in general -- or it depends on what you mean by "negative energy." In field theory, states of negative energy are just interpreted as antimatter. If you say "negative energy" to a relativist, they're more likely to take it as referring to exotic matter that violates an energy condition, http://en.wikipedia.org/wiki/Weak_energy_condition . These are two completely different ideas. A positron, for example, does not violate the WEC.

This is a good question, and I'm not certain of the answer, although I may be able to point you in the right direction. When you have a pair of photons being created, conservation of momentum says that they come off back to back with equal energies. Since the photon is its own antiparticle, there is no physical way for a local observer to look at them and decide which is the positive-energy one. The situation is completely symmetrical in local terms, and the distinction could only become relevant in a global sense, where you can see that one is headed inward and one is headed outward. I think the description in terms of positive and negative energies comes from the Bogoliubov transformations, which I'm only familiar with in a nuclear-physics context. John Baez has a discussion here http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html of why he's not even sure the Bogoliubov picture matches up with the usual description in terms of particle-antiparticle pairs. The standard nuclear-physics way of thinking about the Bogoliubov transformations is exactly analogous to the Dirac sea, by the way.

5. Aug 15, 2010

### Haelfix

It doesn't matter which particle falls into the blackhole, in either case the particle at infinity takes energy away from the gravitational field while its antiparticle enters the blackhole horizon on a timelike path of negative energy relative to infinity.

In fact its a bit of a misnomer to even talk about 'particle/antiparticle' here, since the concept is ridiculously ill defined near the horizon (the particle has wavelength on the order of the mass of the blackhole!! which should indicate that we aren't really thinking about positron-electron pairs or anything like that). Its really just a useful 'analogy' of the complicated mathematics viewed from a nice tranquil place off at some asymptote somewhere.

6. Aug 15, 2010

### bcrowell

Staff Emeritus
I don't follow you here. What frame of reference gives the wavelength you describe, and why does that wavelength lead to the conclusion that they're not a particular type of particle?

7. Aug 15, 2010

### Haelfix

Yea, this is going to be difficult to be precise so keep in mind this is horribly handwavey and the interested reader will need to consult a textbook.

Anyway, you most likely know that it is difficult to talk about particles in curved space, and this is an example.

There are two seperate analogies (by that I mean words to describe a set of rather complicated and ambigous looking mathematical statements), neither of them quite right, and neither altogether wrong about what is taking place in semiclassical gravity and the Hawking process.

The first is the usual one, often talked about in books, where you have vacuum pair production, where one 'particle' gets displaced relative to the other by a 'distance' x, and of course promptly passes a horizon. A calculation yields a value for this 'wavelength' to be roughly on the order of ~ M the mass of the blackhole, working in natural units.

Things to keep in mind about this analogy: This is general relativity, and distances are ambiguous notions that you need to handle with care (choice of coordinates, observers etc). Secondly, these particles are in no sense localized, either in space or in time. They are not defined in any sort of flat reference frame. Instead its really off at infinity somewhere where we measure a Hawking flux and its really only the integral of all quantum field modes that we actually measure. Now of course the incoming particle flux can and usually is just about anything in the world (Planckian particles, positrons, electrons, whatever) all in a nice thermalized black body spectrum. What we can't actually talk about, is whether those particles were really 'there' at the horizon or not.

The second analogy, less talked about, but in some sense better. You can think of the blackhole as a sort of classical barrier, that is violated by quantum tunneling. This is known as the Hawking instanton. Anyway, after a calculation you will discover the quanta that has tunneled has characteristic wavelength ~ M, and appears in a thermal distribution. There you don't really have to talk about pair production or anything like that.

In any event, I would say that most people agree that indeed a terrestrial measurement would in fact pick up a Hawking flux, even though it is tiny in practise however the exact details of this whole story are still not entirely settled and likely will need a full theory of quantum gravity.

8. Aug 16, 2010