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I have stumbled upon pair production recently and it intrigued me,but i need help understanding the subject further.

If i understood correctly you can produce a pair of a particle and an anti-particle by striking gamma rays at an object like a plate and depending the enegry of your gamma ray you can produce electrons/anti-electrons pairs,muons/anti-muons pairs and proton/anti-proton pairs.

Here is the point where the easy access to information at the internet stops.My questions are:

•How exaclty pair production works,why a material object is needed to make this phenomenon work?

•Are there specific objects or materials required to make pair production work(like platina from another forum i saw) and how much?

•The pair is turned into gamma rays again because they annihilate right after their creation.Can we capture them before they do so or the time frame we have to work with is so small to capture any particle or anti-particle (for example electrons or anti-electrons with magnets)?

I would really appreciate if someone helped me.I am not a physicist,i am just a guy that is interested in pair production and wants to learn a bit more.

Thanks for your time reading this.

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2 Answers 2

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In pair production from gamma rays, a material object is required to conserve energy and momentum. A pair of leptons has a center-of-momentum frame, in which:

$$ p_+^{\mu} = (E, \vec p) = (\sqrt{p^2+m^2}, \vec p) $$

$$ p_-^{\mu} = (E, \vec p) = (\sqrt{p^2+m^2}, -\vec p) $$

so that:

$$ q^{\mu} = p_+^{\mu} + p_-^{\mu} = (2\sqrt{p^2+m^2}, \vec 0) $$

which simply cannot work for a photon's 4-momentum (e.g., the wavelength is infinite and the frequency is finite). If you throw in an atomic nucleus, it can get the energy and momentum balance to work out.

Pair-production, like Bremsstrahlung, occurs in a strong electric field, and the best place to find that is near a nucleus of a high-$Z$ material. Lead (or leaded crystal/glass) is a popular choice.

Of course, CERN uses pair production to make and study anti-hydrogen. DESY studied deep inelastic scattering ($ep \rightarrow eX$), and for technical reasons it was better to use $e^+$, which were made from pair production before being put into the storage ring.

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You don’t actually need any material object. Two high-energy photons in vacuum can create pairs of any charged particle and antiparticle. This is called Breit-Wheeler pair production and was first calculated in 1934. (It’s just the time-reverse of the annihilation of a particle-antiparticle pair into two photons.) However, this two-photons-in-vacuum pair-production process has a lower probability of happening than the single-photon-near-a-nucleus pair-production process does.

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  • $\begingroup$ In Breit-Wheeler pair production, the "material object" is the other photon, so you still need some other object involved in order to avoid violating conservation laws. $\endgroup$ Commented Jun 16, 2020 at 4:14
  • $\begingroup$ @probably_someone Yes, you need a second object. But no physicists I have known have considered photons to be “material”. And I certainly did not get the impression that the OP was aware of the Breit-Wheeler process since they talked about “an object like a plate”. By the way, I was trying to complement your answer, not criticize it. $\endgroup$
    – G. Smith
    Commented Jun 16, 2020 at 4:29
  • $\begingroup$ It wasn't my answer anyway, and yes, it does appear that the OP thinks "material object" means something that it doesn't. I just wanted to emphasize that a single photon in a vacuum cannot by itself pair produce, as that's been a source of confusion on other questions. $\endgroup$ Commented Jun 16, 2020 at 4:38
  • $\begingroup$ @probably_someone Whoops, sorry about forgetting that the answer was JEB’s when I wrote that last comment. You make a very good point about the single-photon process being kinematically impossible. $\endgroup$
    – G. Smith
    Commented Jun 16, 2020 at 4:43

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