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The production of electron-positron pairs occurs if a photon with high enough energy passes near a nucleus so that the energy of the photon is converted into particle mass, in accordance with the mass-energy equivalence,

$$\gamma \rightarrow e^- + e^+ $$

Is there a similar process for the production of proton-antiproton pairs? If not, how are they produced?

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  • $\begingroup$ Your two questions aren't really linked like you think they are. In short, such a pair production does exist (with caveats, see the answer I'm about to post), but the way we produce proton-antiproton pairs in the laboratory isn't really related to pair production from a photon. $\endgroup$ – probably_someone Jan 27 at 20:03
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Before we get to the actual answer, there's something that needs to be clarified about the reaction as you've written it down. The reaction $\gamma\to e^-+e^+$ is not a valid reaction, because it violates conservation of momentum. To see this, consider a photon with energy $2m_e$. Since $2m_e$ is precisely the energy of an electron-positron pair at rest, the momentum of the system is zero after pair production. But since the energy of the initial photon was nonzero, the momentum of the photon was also nonzero. The momentum of the photon is not equal to the total momentum of the electron-positron pair, so this reaction cannot happen as stated (this argument also works when the photon's energy is greater than $2m_e$, it's just more straightforward to see in this particular special case).

To see how pair production actually works, let's look at the reverse reaction: electron-positron annihilation, which is the reaction $e^-+e^+\to\gamma+\gamma$. You can see that, unlike the process you described, two photons are produced by an annihilating electron-positron pair. This means that there's no momentum-conservation issue even when the electron and positron annihilate at rest; the two photons have equal energy and travel away from the vertex back-to-back, so the total momentum before and after annihilation is zero.

Since quantum electrodynamics is time-reversal invariant, the reaction for pair production should look like the reverse of the reaction for electron-positron annihilation. Therefore, the actual process for pair production is $\gamma+\gamma\to e^-+e^+$. But this isn't the way that pair production is usually described (people usually talk about a single photon producing a particle-antiparticle pair). Why is this?

The answer is that the second photon is usually a virtual photon, a mathematical tool used to describe one of the produced charges' interaction with some other nearby charge. Pair production cannot occur spontaneously in a complete vacuum. There has to be some other charged object somewhere in the universe that can carry off the difference in momentum between the initial photon and the electron-positron pair. An example diagram of this process is below:

taken from https://physics.stackexchange.com/questions/326267/the-feynman-diagram-for-pair-production

Fortunately, even in intergalactic space, there's still at least some amount of charge floating around, so this typically isn't a concern, and as a result we tend to omit the other virtual photon when we discuss pair production colloquially.


With that said, there are two essentially unrelated questions being asked here:

Can you produce a proton-antiproton pair from a photon in a way analogous to electron-positron pair production?

The answer to this is yes, it's definitely possible. All you would need is a photon with energy greater than $2m_p$, which is roughly 1.876 GeV, and there would be some chance of it happening. However, a photon with energy that high can also produce a particle-antiparticle pair of anything with electrically-charged constituents that is lighter than a proton, which includes electrons and many types of mesons (e.g. pions). So you wouldn't be guaranteed to see proton-antiproton pair production for a single given photon of this energy or higher, but if you had enough of them, then you would eventually see the process you described happen (the same caveats apply - you can't do this in an empty universe, one of the produced particles has to exchange a virtual photon with some external charge). This process tends to be really rare simply because there aren't really any natural processes that generate photons that have such high energies.

How are proton-antiproton pairs produced?

Probably the most common mechanism, both in cosmic-ray antimatter generation and in the laboratory, is inelastic scattering of a proton off of a nucleus, $p+A\to p+\bar{p}+p+A$. For cosmic-ray protons, $A$ is typically a nucleus in the interstellar medium. For laboratory protons, $A$ is typically a nucleus in some sort of collision target in an accelerator beamline.

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  • $\begingroup$ Layman's observation here, but there also needs to be a tad bit of energy above and beyond the mass of the particles to be created, so that the proton and anti-proton have the momentum to overcome their mutual attraction. (Or is this a negligible amount?) $\endgroup$ – EvilSnack Jan 28 at 4:38
  • $\begingroup$ I suppose there's a perfectly good reason I can't slam gamma rays together. $\endgroup$ – Joshua Jan 28 at 5:29
  • $\begingroup$ @Joshua but they are planning a gamma gamma collider . slac.stanford.edu/pubs/beamline/26/1/26-1-kim.pdf $\endgroup$ – anna v Jan 28 at 5:37
  • $\begingroup$ @EvilSnack Typically we ignore the electrostatic attraction between the particle and antiparticle because it's a negligible correction to the total energy (for example, the lowest-energy, most-tightly-bound electron-positron state takes 6.8 eV to break, and the photon producing an electron-positron pair has at least 1,022,000 eV). That said, if you try and account for it, it's kind of the other way around. A bound system of an electron and a positron has slightly less mass than a free electron and a free positron. $\endgroup$ – probably_someone Jan 28 at 5:47
  • $\begingroup$ @Joshua The fact that photons are themselves electrically neutral means that they have to momentarily fluctuate into something else at the moment they cross paths in order to interact with each other, which, as you might imagine, is pretty rare, especially at low energies. That said, it gets easier as the photon energy goes up, and we've even directly observed light-by-light scattering in heavy-ion collisions as of last year: physicsworld.com/a/light-is-seen-to-scatter-off-light. Heavy ions have huge electric fields, which makes them a good source of high-energy virtual photons. $\endgroup$ – probably_someone Jan 28 at 5:55
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A basic rule about antiparticle production is that quantum numbers have to be conserved, that means baryon number, lepton number, strangeness..... This means that in addition to the methods described by probably_someone, collider experiments of e+e- , and the future ones discussed of gamma gamma collisions produce dominantly particle antiparticle pairs of all types, from hadrons to leptons as far as the energy conservation allows. Of course proton antiproton colliders also did that ( but antiprotons are already there!)

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In practice, antiprotons are produced in the antiproton decelerator at CERN by colliding a high-energy proton beam from the super synchrotron (PS) with a water-cooled iridium target, iridium being chosen for being the 2nd highest density metal but much less chemically reactive than the densest, osmium. This is because mass density is proportional to density of nuclei, and so the higher the density the more chance there is of a proton in the beam interacting with a nucleus to create an antiproton. The proton-nuclear scattering which produces antiprotons is described in another answer.

Some properties of that target are described in this arxiv paper, including a schematic, and a prototype of a future target is described in this open access paper.

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