Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I used to read the term "pure energy" in the context of matter-antimatter annihilation. Is the "pure energy" spoken of photons? Is it some form of heat? Some kind of particles with mass?

Basically, what does "pure energy" in the context of matter-antimatter annihilation refer to?

share|cite|improve this question
up vote 21 down vote accepted

If I ruled the world, I would ban the phrase "pure energy" in contexts like this. There's no such thing as pure energy!

When particles and antiparticles annihilate, the resulting energy can take many different forms -- one of the basic principles of quantum physics is that any process that's not forbidden (say, because of violation of some sort of conservation law) has some probability of happening. So when a proton and an antiproton annihilate, they can produce photons, or pairs of other particles and antiparticles, such as a neutrino-antineutrino pair, or a positron-electron pair.

Although all sorts of things are possible, by far the most common product of matter-antimatter annihilation is photons, especially if the collision occurs at low energy. One reason is that lower-mass particles are easier to create than high-mass particles, and nothing has less mass than a photon. (Other particles, particularly neutrinos, have so little mass that they are "effectively massless," but neutrinos are weakly interacting particles, which means that the probability of producing them is lower.)

share|cite|improve this answer
I would ban it too, thanks! – HDE May 11 '11 at 15:17
Doesn't the term "Pure energy" just refer to the fact that matter/anti-matter reactions result in nothing but photons (annihilating the matter)? Or, are you saying that this is not ALWAYS the case with matter/anti-matter reactions? – Polaris878 May 11 '11 at 15:25
I'm saying (a) that "pure energy" is a lousy synonym for "photons" and (b) photons aren't always what you get anyway! – Ted Bunn May 11 '11 at 15:44
awesome thanks! – Polaris878 May 11 '11 at 16:32
I think that "pure energy" is used to denote any intermediate state that carries no conserved charges. So a single virtual photon is "pure energy", too. A virtual graviton or their pair would also be "pure energy". "Pure energy" is meant to have the property that it's not hard for other objects to absorb it (or emit it) without changing their character qualitatively. It's like the energy in the sugars or chocolate that you may quickly transform to the energy to run. An extra proton in the state makes the energy "non pure". – Luboš Motl May 11 '11 at 17:29

At low energies of matter-antimatter collision, the end products can only be massless particles like photons -- hence annihilation is conventionally associated with the production of radiation/light.

At higher energies, the collision can have enough energy to generate pairs of massive particles like mesons and W bosons.

The LEP was an electron-positron collider and it has the record for producing the most massive particles from such an annihilation process (W+ and W- bosons).

share|cite|improve this answer
You can always just create the particle and antiparticle again, though--you always at least have the rest mass of the original particle-antiparticle pair. – Jerry Schirmer May 11 '11 at 15:46

In particle antiparticle annihilation the end products come from the table of elementary particles created so that quantum numbers are conserved.

photons, two to conserve momentum in the cmsystem, pions and other mesons etc.

share|cite|improve this answer

All the three above answers basically are correct and saying almost the same thing, i.e. the end product of matter-antimatter collision is photon at low energies and massive particles such as mesons... at high energies. However,by thinking of the reverse action namely production of matter from a form of energy ( radiation/light) and the source of that unimaginable huge energy,simply using E=mc*c,supports the above mentioned answers firmly. AA 7 June 20011 at10pm

share|cite|improve this answer

protected by Qmechanic Oct 19 '14 at 9:07

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

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