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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?

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5 Answers 5

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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.)

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  • $\begingroup$ 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? $\endgroup$ Commented May 11, 2011 at 15:25
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    $\begingroup$ I'm saying (a) that "pure energy" is a lousy synonym for "photons" and (b) photons aren't always what you get anyway! $\endgroup$
    – Ted Bunn
    Commented May 11, 2011 at 15:44
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    $\begingroup$ 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". $\endgroup$ Commented May 11, 2011 at 17:29
  • $\begingroup$ Does one photon, which carries angular momentum, count as "pure" energy for this context? The pair of photons might, in a sense, "annihilate" to return the angular momentum to zero, just as a proton and antiproton could annihilate to return other quantum numbers to zero. $\endgroup$ Commented Oct 10, 2022 at 22:52
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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).

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    $\begingroup$ 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. $\endgroup$ Commented May 11, 2011 at 15:46
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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.

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Energy is the ability of a system to perform work on another system. In other words, it's a property of a physical system. The annihilation of an electron and a positron performs work on the electromagnetic field by creating two photons. We usually do not talk about these processes in the same language that we use in classical mechanics, but there is absolutely nothing wrong with thinking about it in exactly the same way. We merely have to replace the names of the systems. In classical mechanics we may have been elevating a mass "above the floor" with a pulley by performing mechanical work on it. In the case of fields we are creating field excitations that are "above" the ground state.

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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

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