Allen B. answered • 02/03/23
Physics Degree + Lifelong Hobbyist + 10+ Years Tutoring Experience
The question here asks whether a proton-antiproton pair can be created, not how likely it is to happen. We don't have to worry about whether it's a common or vanishingly rare process; all we care about is if something will make the pair production impossible.
So let's turn the question around. What would prevent a proton-antiproton pair from being created? If you can eliminate the impossible, you're a long way towards the answer.
(By the way, let's use the abbreviations p and p-bar from now on; it'll make everything shorter.)
A lot of things are cancelled out when a particle meets its antiparticle. For example, p has electrical charge +1, while p-bar has -1. If it cancels out when they disappear, you don't need it to make them appear. A lot of quantum numbers (baryon number, color charge, spin, isospin....) can vanish this way.
But even if they cancel, all the conservation laws still apply. And two very important things that can't disappear -- or appear -- are the conserved quantities of energy and momentum. Even with pair production, you can't get something for nothing; and antiparticles have the same mass as their counterparts. So we'll need at least double the rest mass energy of a proton -- E=2mpc2 -- for pair production to take place.
Why did I say "at least"? Well, consider what would happen if there's no other energy involved: no extra potential energy from forces or fields, no kinetic energy from motion. The p+ and p-bar- appear next to one another, both with zero velocity. They have opposite charges, so they attract one another... and collide. Poof! Alas, we hardly knew ye, particles.
That's a special case of scenario C. Virtual particles are those that only exist as intermediates; they are never directly detected, but only can be deduced from their effects on other particles and fields. Whether due to zero kinetic energy or any other reason, virtual p+/p-bar- pairs by definition aren't "created"; they only exist an infinitesimal time, and we only know about them by their effects on surrounding fields. No particles are created in scenario C, so it can be ruled out.
Okay, so we know we need some momentum and other energy in the process, if only to keep the newly created particles apart. The other three scenarios all involve at least one photon, so we can't avoid having momentum. (After all, photons are light and must move at lightspeed, with momentum p=E/c.)
What about scenario B, with a black hole's gravity potential well providing energy? This is the mechanism of Hawking radiation, which extracts energy from within the black hole and transfers it to an escaping particle. However, for one half of the newly created p+/p-bar- pair to have momentum away from the black hole, the other must have momentum towards the hole... and fall in. Since the unlucky partner never interacts with anything else, it remains a virtual particle. Only one of the pair is created by scenario B, and therefore it is also ruled out.
Is there anything wrong with scenario D, where a photon adds its energy to an electron and pair production results? The scenario is written ambiguously, but the key point is that the pair production cannot be the only product of such a reaction. After all, the electron is not just a force boson of unattached energy; it has an electrical charge (-1), a lepton number (+1), and other properties that must be conserved in the overall reaction.
A p+/p-bar- pair has no lepton number at all -- not just together, but also as separate particles. So where does the lepton number go? There are many possible answers -- the simplest is that the electron scatters off and goes its merry way. Its kinetic energy and momentum are altered, but otherwise the electron's unaffected. But there must be +1 lepton number at the end of the process, no matter what. That demands at least one more particle exist to carry a lepton number, since neither the p+ nor the p-bar- can. If we read the question as implying that the pair is the only product, that would exclude scenario D.
The only scenario that clearly has nothing preventing pair production is thus scenario A. Depending on angle, the colliding photons together could have nonzero total momentum -- indeed, that's far more likely than a perfect head-on collision. And the scenario itself specifies there's enough energy. Finally, the photons have no additional quantum numbers dragged along that would invoke additional conservation laws.
Because of the ambiguity in scenario D, we can't be sure that it's an invalid answer. But the best answer, the most clearly correct one, is scenario A.
