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in my revision guide the section on pair production only mentions it happening with gamma photons, so the question arose whether this is the only way it can happen?

This is what the book says: "Pair production is when a particle-antiparticle pair is produced from a single gamma photon. The gamma photon must have enough energy to produce that much mass. Pair-production usually happens near a nucleus, which helps to conserve momentum."

So does pair production only happen with gamma photons?

Thanks

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4 Answers 4

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"Pair production is when a particle-antiparticle pair is produced from a single gamma photon. The gamma photon must have enough energy to produce that much mass. Pair-production usually happens near a nuclear, which helps to conserve momentum."

It is a hand waving way of describing pair production. Gammas are photons of high energy by definition. Gammas always have to interact with a field in order to produce a particle antiparticle pair , otherwise momentum will not be conserved. It is not a matter of help. The energy of the gamma must be a bit over the sum of the masses of the particle antiparticle to be created.

e+e-

[Feynman Diagram of electron-positron pair production]. One can calculate multiple diagrams to get the cross section

Pair production can happen with other particles but not as easily studied .

Pair production is the creation of an elementary particle and its antiparticle, for example creating an electron and positron, a muon and antimuon, or a proton and antiproton. Pair production often refers specifically to a photon creating an electron-positron pair near a nucleus but can more generally refer to any neutral boson creating a particle-antiparticle pair

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  • $\begingroup$ Thanks! So it's always a photon with high energy, i.e. a gamma photon? Also why is momentum not conserved if it doesn't happen near a nucleus? $\endgroup$
    – user45220
    May 7, 2015 at 17:35
  • $\begingroup$ Take the electron/positron pair created. at the center of mass system it will always have an invariant mass, equal by minimum twice the mass of electron. The gamma has zero mass, it can never go to a center of mass system too. So it is a process that cannot happen by a single photon. the other answer explains the momentum business $\endgroup$
    – anna v
    May 7, 2015 at 17:53
  • $\begingroup$ "The gamma has zero mass" - why should that prohibit entering a mass-system? That would be absorption, not pair production? So, pair production only happens near nucleus, there is no or no easy explanation for that rule? "Mass system" sounds rather unusual. Is it another way to say "critical mass" - there must be some "field of mass" for pairs to be produced? $\endgroup$ Oct 21, 2022 at 15:40
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Pair producing an electron and a positron requires an energy at least equal to their masses, $2m_ec^2$. This would just create them stationary, and you'd need an even greater energy to give them some momentum.

Since $m_e = 0.511$ MeV/c$^2$, the minimum energy required for pair production to occur is $\sim 1$ MeV.

What photons have an energy close to that value? Gamma rays.

Now, if you tried to do the maths of energy & momentum conservation, a single photon pair converting to an electron and a positron would never conserve both.

You can see it by noting that, for the outgoing electron+positron system, there exists a centre of mass frame where the total momentum is 0. But for the photon, there is no such thing since zero momentum would mean 0 energy ($E =pc$). The presence of a nucleus makes you have more 'elements' in the energy-momentum conservation calculations, which allows it to give/take some energy/momentum thus enabling the pair production process to occur.

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  • $\begingroup$ Thanks! I'm still a bit confused by the "photon" stuff. Is it always a gamma photon that is needed in pair-production, or are there other types of photons that work? $\endgroup$
    – user45220
    May 7, 2015 at 17:38
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    $\begingroup$ By "gamma photon" I assume you mean in the "gamma ray" spectrum. You need a photon with enough energy to actually generate the masses of the particles, and photons with such energies are in the gamma spectrum. Technically lower energetic photons could create an electron-positron pair, violating energy conservation, as allowed by quantum mechanics. But they would have to be very short lived and therefore would not live on to be long lived states (they would annihilate again pretty soon). This is allowed by the energy-time uncertainty principle $\endgroup$
    – SuperCiocia
    May 7, 2015 at 17:42
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    $\begingroup$ As far as the other question in concerned. Try it. Free photon as initial state + electron & positron as final state. Energy conservation requires $E_{\gamma} = E_{e^+} + E_{e^-}$, while momentum conservation requires $p_{\gamma} = p_{e^+} + p_{e^-}$. But you can express $E_{\gamma}=p_{\gamma}c$ which leads to $E_{e^+} + E_{e^-} = p_{e^+}c + p_{e^-}c$. But $E = \beta p/c$, so the earlier equation is wrong for all massive particles ($\beta < 1$). $\endgroup$
    – SuperCiocia
    May 7, 2015 at 17:47
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    $\begingroup$ With a nucleus, you'd add $E_n$ and $p_n$ to the conservation equations, which gives you more freedom. The nucleus could provide the "missing" momentum, therefore making the process happen. $\endgroup$
    – SuperCiocia
    May 7, 2015 at 17:48
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    $\begingroup$ Same logic as before. The mass required is $2m_pc^2 \sim 2$ GeV. That's a very, very energetic gamma photon. $\endgroup$
    – SuperCiocia
    May 7, 2015 at 17:51
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Gamma radiation is the highest in energy on the electromagnetic spectrum. Any photons with energy over $100keV$ are classified as gamma radiation.

As stated in other answers, for a photon to produce a particle-antiparticle pair, the energy of the photon must be at least twice the mass-energy of the type of particle produced. There are 2 particles of equal mass produced, so it makes sense that you need at least this much energy in the photon that produces them.

The lightest fermion (because we don't usually mean bosons when we talk about pair production) is the electron neutrino; however, neutrinos don't interact with the EM force, so photons won't be directly producing them (indirectly, anything is possible). The lightest non-neutrino fermion is the electron. It's mass-energy is about $0.5MeV$, which means to produce an electron-positron pair, the initial photon has to have an energy of at least about $1MeV$.

Since we already established that anything over $100keV$ is a gamma photon, that means you need a gamma photon to produce an electron-positron pair. Furthermore, since the electron is the lightest particle that photon can produce, any other particle would require the photon to have even more energy. So no non-gamma photon would have enough energy.

This is why we say that pair production only occurs with gamma radiation. Any photon with enough energy to produce a pair of particles must automatically be in the gamma range of the EM spectrum.

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Pair production can also happen in extremely strong electric fields via the so-called "Schwinger mechanism". Unlike most other results in quantum field theory, in this case one can exactly compute the probability of creating pairs per unit volume and time, see e.g. here.

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  • $\begingroup$ Thanks. I just searched your name and found a TV series. Just wanted to mention that Iblis is also the word for devil in Arabic, but I'm sure you knew that. $\endgroup$
    – user45220
    May 7, 2015 at 18:43

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