Let’s say an electron and a positron attempt to occupy the same space. As I understand it, they annihilate each other and energy is released.
What kind of energy? Gamma, X-ray, 1920’s death ray?
Does it depend on the incident energy of the two particles? And, if so, do you add the two incident energies together (which makes more sense to me), or subtract the lesser from the greater–since that’s what’s “left over” after the annihilation?
Nah. That second idea doesn’t work. Nevermind that one.
Anyway, I got into this discussion at work with some rad tech students following a discussion of pair production–and what would happen subsequently.
The Large Hadron Collider is doing just that though with protons and anti-protons rather than electrons and positrons. Those could produce an electron anti-electron pair or gamma rays or a few other things even if they collided with no kinetic energy, but the LHC is doing this with a lot of kinetic energy trying to produce much more massive things – like the sought for Higgs particle.
It depends on what particles you’re dealing with. An electron and positron are nice and clean: You just get a couple of gamma rays. A proton and antiproton, or a neutron and antineutron, is more complicated: You’ll start by getting three pions (a 1/3 chance of all of them being neutral, or a 2/3 chance of getting a positive, negative, and a neutral), which will then decay. The neutral pions will each decay to a pair of gamma rays, while the charged ones will each decay to a muon (or antimuon) and antineutrino (or neutrino). And the muon (antimuon) will then decay to an electron (positron), another neutrino, and another antineutrino. And then, of course, if you produced a positron, there’s a good chance that it’ll sooner or later find some ambient electron and annihilate with it (probably not the one that was produced in the annihilation/decay event, though, since that’s probably headed off in a different direction). So in total, the proton-antiproton or neutron-antineutron pair will eventually end up as either photons, or a mixture of photons and neutrinos.
Some particle accelerators do use protons and antiprotons, but the LHC uses the same things in both beams: Either protons, or lead nuclei. It’s the latter that give it its name: “Large” is modifying the hadrons, not the collider itself.
The OPs title is interesting and raises a question that the text makes clear isn’t the question. So I will ask the question in the title:
What happens when two anti-matter particles collide? We know what happens with two matter particles (say a neutron and a proton) collide (I don’t, so that is another question) but does anything different happen when two anti-matter particles collide?
Assuming it was in a vacuum so there was no interaction with normal matter, they’d behave just about like their anti-particles (i.e., anti-anti-particles – the equivalent regular particles) with all results in anti-particles instead of regular particles. I say just about because there are slight differences due to CPT symmetry.
I know that pair production can occur when using high-energy X-rays. But then, if, say, the positron emitted from the pair production encountered a run-of-the-mill electron, you’re saying that gamma rays result?
That confuses me because it seems like you’re getting something more energetic from two less energetic particles since gamma rays are further along the EM spectrum.
It’s indeed the collider that is large in “Large Hadron Collider”.
It’s the violation of CP that is relevant here. As far as we know (and within the Standard Model), CPT is a good symmetry of nature.
For fundamental particles, say electrons/positrons, what you can get out is a function of the “center-of-mass” energy available. Any translational energy of the entire system isn’t usable in making new stuff. So, for an electron and positron colliding head-on with identical energy E, the center-of-mass energy is the sum of the two particles’ energies, or 2E. If the electron is at rest and the positron strikes it with energy E, the center-of-mass energy is less. Indeed, the final system must have a momentum equal to the momentum of the incoming positron, so some of the initial energy must go toward the kinetic energy of the products rather than into their raw production.
As for what you can make… for relatively low-energy situations like the ones your rad tech colleagues were probably discussing, the available energy is only a bit more than the sum of the electron and positron masses. This can only yield photons. If the available energy is higher, you can make heavier things: pions, muons, taus, Z bosons, …
Proton-antiproton collisions are similar, except these aren’t fundamental particles so the laundry list of possible final states is more complex. (The fact that these involve the strong interaction doesn’t help the complexity.)
“X-ray” and “gamma ray” aren’t strictly defined energy ranges, so you’re just hitting a language issue. However, you need at least 1.02 MeV of energy to produce an e[sup]+[/sup]/e[sup]-[/sup] pair, and 1.02 MeV is a bit beyond what is usually considered the X-ray range. Nonetheless, your intuition is correct that you cannot get energy for free. (Keep in mind, though, the energy you get from the mass of your run-of-the-mill electron when tallying the numbers.)
I disagree. The Large Hadron Collider (LHC) is built in place of (and uses the same tunnels as) the previous Large Electron Positron Collider (LEP).
The proposed upgrade (proposed for in 10 years or so, I believe) is the Super Large Hadron Collider, which is a change to the machine, not to the hadrons.
Large definitely refers to the machine. Not the hadrons. I’m not aware that there’s any such thing as a “Large Hadron”.
[I initially mistyped “Large Hadron”. Accidentally switching the D and the R makes for an entirely different thread…]
