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Is there any way to annihilate matter without the use of anti-matter? And vice versa? I mean, for example is it possible to totally convert the mass of a proton into "pure energy" without use an anti-proton?

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It's probably "cheating", but there probably exist virtual states where this happens. It's just not a finished process, i.e. while this contributes to some probabilities it won't be actually observable... Then again you could claim $p + e^- + \gamma \to n + \nu$ made the proton vanish without any anti-matter involved... –  Tobias Kienzler Feb 27 at 12:17
Does radioactivity count? If so, yes. The answer is yes. –  Mr Lister Feb 27 at 14:05

6 Answers 6

I am assuming that by "energy" you mean photons. So you want to transform protons into photons.

It is not possible. It would violate several conservation laws - mainly the charge conservation (protons are positively charged), but also baryon number conservation.

The antiparticle is necessary to cancel these quantum charges to make the transition possible.

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Couldn't you throw the proton into a black hole and wait for Hawking radiation? –  jinawee Feb 26 at 11:24
@jinawee Even with a black hole you cannot violate the charge conservation law. If you create a black hole only from protons, the black hole will have positive charge. It is then impossible that the Hawking radiation will be only photons. Some of the radiation must be positively charged particles to carry away all the positive charge that was put into the black hole upon its creation. If the black hole is neutral, then there were some particles thrown into it previously that cancelled the positive charge of the proton. That is the same scenario as the usage of antiparticles. –  mpv Feb 26 at 12:53
@AricTenEyck If you throw an electron into the black hole, you cancel the electric charge of the proton. The baryon number does not seem to be conserved in this case: physics.stackexchange.com/questions/7290/… But you still need an extra particle carrying an anti-charge. You cannot just transform proton into a photon. Another problem is that in practice the black hole will eventually emit also baryons, even if it is neutral. The final explosion is high-energetic with many heavy particles flying everywhere. It won't be just photons. –  mpv Feb 26 at 16:10
Not to mention, when you have proton-antiproton annihilation you get a lot more than just photons. if i recall, a majority of the energy is actually carried off by neutrinos. Not a very efficient power plant (and the source of my pet peeve with Star Trek using anti-deuterium as fuel.) –  Michael Feb 26 at 20:27
@Michael proton antiproton annihilation still releases a lot of energy, about 1/3 into pi0s is about 600M3V which compared to e+e- is a lot and also to fusion or fission. The rest mostly charged mesons which can give up energy in ionization before decaying. average is maybe 4 charged pions the energy lost by each is half of tis mass so 50Mev to neutrinos*4 ( the electrons and positrons will end up in annihilation the e+ and ionization at rest , another half Mev lost) so most of the energy ends up in photons one way or the other. Only about 10% loss to neutrinos, fairly efficient! –  anna v Feb 27 at 15:16

The definition of an antiparticle is dependent on having the opposite quantum numbers of the particle so that they can annihilate, i.e. the sum of the conserved quantum numbers are zero. Thus the answer by @mpv is adequate.

The implication of your question is then: is baryon number conservation a strict law or an emergent law that may be violated at some small probability.

There exist models where protons can decay with very long lifetime, dependent on the model. for example, of a model

$$\mathrm p^+ \to \mathrm e^+ + π_0 \quad\text{and then}\quad π_0 → 2γ.$$

proton decay

so one would get two photons and an electron from this decay.

Now from the Feynman graph it is evident that it is a quark that disappears . The question though involves a proton disappearing. One can see from the diagram reading it from top down, right to left that if one scatters an $\mathrm e^-$ on a proton there exists a probability that the proton will disappear and a $\pi_0$ will manifest and decay into two photons.( a third particle should be involved in order to get a $\pi_0$ due to momentum conservation otherwise one would get two quark jets, maybe a second $\pi_0$)

So this is may be an annihilation of a proton into photons with the appearance of two pions minimum. It does conserve charge ( or B-L quantum numbers).

The limits of proton decay are pushed further and further with each experiment and thus this inverse reaction will have such a tiny probability that one cannot perform it in the lab and wait for results.

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Ok very exhaustive answer..so maybe a proton decay could be an example of matter annilihation without anti-matter but for what i know there's no experimental data yet or im wrong? –  user27494 Feb 26 at 15:28
You are correct. The limits get pushed every time I hear of a new experiment. Proton decay appears in higher than standard model theories but it has to be within the experimental limits. –  anna v Feb 26 at 15:38
The experimental lower limit on this decay mode of the proton (see pdg8.lbl.gov/rpp2013v2/pdgLive/… ) is $8.2 \cdot 10^{33}$ years (the corresponding publication is here: inspirehep.net/record/814697 ) . This is $5.9\cdot 10^{23}$ larger than the estimated age of the universe (13.8 billion years). –  Andre Holzner Feb 27 at 12:21

I just started here so I don't have the rep. to comment and I don't have the time for a full answer, but the black hole idea mentioned in the comments above is a fine answer. See, for example, http://arxiv.org/abs/0908.1803v1 and How would a black hole power plant work?

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Even black holes do obey charge conservation. Positively charged black hole (made from protons only) will emit positively charged particles. It cannot radiate completely away as just neutral photons. –  mpv Feb 26 at 12:58
I thought about the same thing and i have no doubt about the charge conservation laws,but (and maybe im going too far in details) what kind of radiation a black hole can emit if not photons ? –  user27494 Feb 26 at 15:30
You couldn't make a black hole out of protons - the electrostatic repulsion would be way too strong. A star is, overall, (nearly) neutral. As it collapses the protons and electrons would form neutrons (neutron star), then, with enough mass, a (nearly neutral) black hole. So charge conservation wouldn't be a problem. So, I suppose it would not directly convert protons to energy - they would become neutrons first - but it would convert matter to energy without anti-matter. Note: there are neutrinos released in the conversion to neutrons - it is a weak interaction. –  Wally Feb 27 at 3:28

A proton has a positive charge so by charge conservation it is not possible to reduce a proton to uncharged radiation particles such as photons (assuming that is what you mean by "pure-energy") Because of gauge invariance charge conservation is likely to hold good in all future physics, but we can't be totally sure of that.

It is possible that some charged massless particle will be discovered but that seems unlikely. If such a particle existed a proton might decay to that and you might consider that as "pure-energy".

If we ignore the "example" of a proton and consider the original question, the answer is that it may be possible to reduce an atom to photons but this is very difficult since it violates baryon number conservation. This has never been observed but there is theory that tells us that baryon number non-conservation is possible in the standard model using non-perturbative effects. It may also be possible to violate baryon number using beyond standard model physics or by throwing matter into a blackhole and recovering Hawking radiation. It is not possible to violate charge conservation this way (according to our best theories) but it should be possible to violate baryon number conservation (unless there is a hidden unknown reason why it is not possible) The same applies to lepton number for electrons.

So according to our current state of knowledge reducing an uncharged atom to photons is probably possible in principle but we have no experimental evidence to support this claim and it is unlikely to be something we can ever do in practice due to the low rate of baryon and lepton number violations in all known theories.

Some theories about the final state of the universe (such as Penrose's conformal cosmology) assume that on very very long timescales all matter will be reduced to photons in this way and these will lose their energy as the universe expands so that only dark energy (balanced by an opposing quantity of negative gravitational energy) remains.

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Well, the question is about annihilation not proton decay. For annihilation one needs two particles, –  anna v Mar 5 at 9:12
If the proton in an atom decays it can leave a positron which annihilates with the electrons in the atom. I could have explained this at more length perhaps. –  Philip Gibbs Mar 5 at 14:00

The simple answer to the main question is, yes. There are two ways to annihilate matter without using anti-matter. One is called fission, and the other is called fusion. Although only some of the matter is converted into energy in either of these processes, the efficiency of the "annihilation" is not in the main question. If 100% annihilation is required, then only anti-matter will do it.

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Note that this is possible even if we restrict ourselves to stay within the rigorous realms of the Standard Model. E.g. the deuteron is known to be unstable, it will decay via instanton tunneling to a positron and an anti-muon neutrino (or an anti-muon and anti-electron neutrino). The deuteron lifetime would be about 10^(218) years if this Standard Model process were the only process contributing to its decay.

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