# Matter/Anti-matter annihilation question

In this thread I started a small hijack that I’d rather stood as its own OP.

Why do matter/antimatter particles pairs annihilate each other completely? When a proton/proton merge occurs you get a proton, a neutron, a positron and a neutrino. Aside from energy, what conservation mechanism is it that requires both particles to be converted completely into 2 photons?

Charge has to be conserved as well. So in a electron/positron collision, the electron has a charge of -e, and the positron a charge of +e, making the entire system electrically neutral. The “after” result must have the same charge as the “before” scenario, in terms of energy, electrical charge, momentum, and fundamental particles (quarks etc). Hence, the only thing that an electron/positron collision can produce is two photons, travelling in opposite directions (to conserve momentum).

Well, off the top of my head, the following quantites are conserved:[ul][li]mass/energy[]momentum[]charge[]spin[]lepton number (electrons and neutrinos)[/ul](More things are conserved, but I think this is everything relevant for electron-positron annihilation.)[/li]
We have an electron with mass m[sub]e[/sub], momentum p, charge -e, spin +1/2, and lepton number +1. The positron has mass m[sub]e[/sub], momentum -p (in an appropriate frame of reference), charge +e, spin -1/2 and lepton number -1. So our sums are mass 2 m[sub]e[/sub], momentum 0, charge 0, lepton number 0.

So, we could create a single photon with energy E = 2 m[sub]e[/sub] c[sup]2[/sup]. Except that a photon cannot have zero momentum. So two photons (each with energy m[sub]e[/sub] c[sup]2[/sup]) are created, with equal and opposite momentums, such that the total momentum is still zero.

Thanks for adding the details! Its been a long time since I did particle physics, and my memory’s a little hazy on the detail.

Minor correction, I left off the units on spin. The electron spin should be +(1/2) h[sub]bar[/sub]. In this inadequate notation h[sub]bar[/sub] is Planck’s constant divided by 2 pi.

As I mentioned in the other thread, I think an electron-positron pair may be more stable in a magnetic field. This is because if their spins are the same (both positive, or both negative) the total spin will not be zero. (The magnetic field is required to keep the spins aligned.) I believe photons must have zero spin, and so the pair would not be able to annihilate into photons.

Hey, it’s been years for me as well.

And I thought you were using proper particle physicists’ units, where you simply put i]h[sub]bar[/sub]*=c=1. Strange, strange people.

Yeah, but in the meantime, my brain’s been addled by astrophysics!

Actually, I believe there are conditions where the result is three photons. Besides, I wouldn’t say that having photons at the end means “annihilating completely”.

The reason that photons are the product so much more than anything else is that photons carry pretty much no conserved quantities. Antimatter doesn’t only flip charge, but all conserved quantities associated to internal symmetries of the system. Electric charge comes from U(1)-invariance of physics, so it gets flipped. Weak “charge” comes from an SU(2)-invariance, so it gets flipped. Now, when you run two particles whose conserved quantities exactly cancel out, you need to have a product with the same property. If you made anything but photons, it would be far more difficult (in a very handwavy argument. If you work out the probabilities that’s a lot more enlightening).

Oh, and I suppose it’s also strange that people use meters instead of feet and use a constant multiplier to convert measurements in feet to measurements in meters?

Planck’s constant and the speed of light are just conversion factors reflecting the underlying geometry.

:rolleyes: Chip on your shoulder? I know these things. I am a physicist for goodness sake! That was supposed to be a totally flippant, throwaway, joke, I don’t know if you’re aware of it, but over here, there tends to be a flippant, friendly rivalry between astrophysicists and particle physicists (probably because we have the same funding body). The particle physicists call us wierd and strange, and we call them wierd and strange. And no one means anything by it.

Sheesh. Don’t get so strung up. I use eVs, cm[sup]-2[/sup] and other non-SI units myself.

Sorry, coming off a long night out among non-techies. “Oh, you’re a mathematician? this friend of mine over here likes math and physics…” leading into an inescapable smile-and-nod session with someone who doesn’t know what the hell he’s ranting about.

Also, outside the “proper” physics fold I read physicist as physicist, making little distinction other than theoretical or experimental. I see someone saying physicists are strange for that sort of thing I jump in.

You poor thing. I can really sympathise with that. Its the annoying one when you tell people you’re an astrophysicist, and they ask you about aliens.

Can’t say I blame you, I do it myself, normally, but at the moment, I’m in a room with three other astrophysicists, and so, my humour goes decidedly into physicist mode. Didn’t mean to get quite so annoyed, but as a woman, I get pretty damned snarky when its implied that I know nothing about physics/physicists/andything technical.

It’s important to note that an electron & a positron don’t have to produce photons when they annihilate; they can produce any particle-antiparticle pair. The only limiting factor is that the total energy (rest + kinetic) of the incoming particles must equal the total energy of the outgoing particles; so if the total energy of the electron & positron are lower than the total rest mass of the desired resulting particles, that particular process won’t happen.

If you crank up the energy of your incoming electrons & positrons, though, all kinds of interesting stuff can happen. That’s what these folks (among others) are doing.

I hadn’t even really noticed a gender attatched, so…

Actually, I also end up acting as a campaigner for geometrizing physics (and pretty much anything else I can think of). There are a lot of amateur physics people out there who somehow haven’t seen how elegant the geometric theories are, so any chance I get, I make my pitch.

Proably just a matter of poor notation. A great many folks have never seen the Einstein summation convention for tensors, for instance, without which geometric theories are much less elegant.

As for the OP, another factor is that electron-positron interactions are predominantly mediated by the electromagnetic force. If you annihilate a hadron with its antiparticle, you generally won’t get photons as your decay products, at least not immediately. Your immediate decay products are likely to be various pions, which themselves will decay into other pions and/or leptons (charged or uncharged), and the charged leptons (such as electrons and positrons) may or may not eventually meet each other and also annihilate.

Actually, I’m not a fan of the summation convention (except in that it saves a lot of time writing things out). The summations are all over bases of one type or another which are derived from local coordinate systems which are (by the principle of relativity) highly noncanonical. Better to think of the vector itself than its components in any local coordinate frame.

Oops, left out tensor, differential form, connection (Levi-Civita and gauge), curvature…