Electron-Positron production by photon

[Silviu]

Hello! I am a bit confused about the decay of a photon into a electron-positron pair. In the center of mass of the photon, isn't this decay violating the energy conservation?

 

[vanhees71]

What do you mean by "center of mass of the photon". First of all there's no center of mass, because the photon mass is 0. There's also no center-momentum-frame, because the photon mass is 0, i.e., you can never find a reference frame, where the photon is at rest. As you correctly noted a photon cannot decay to an electron-positron pair, because you cannot fulfill energy-momentum conservation and the on-shell conditions for the photon and the particles simultaneously. So there is no photon decay.

 

[Nugatory]

Hello! I am a bit confused about the decay of a photon into a electron-positron pair. In the center of mass of the photon, isn't this decay violating the energy conservation?

As vanhees points out, there is no center of mass frame for a photon - that would be a frame in which the momentum of the photon is zero, and of course there is no such thing.

However, you are on to something here. An isolated photon cannot decay into an electron/positron pair, because there's no way that interaction can conserve both energy and momentum. (An easy way to see this is to think about how the interaction looks in the center of mass frame of the electron and positron after the collision).

Instead, pair production requires the involvement of some other massive charged particle, typically some nearby atomic nucleus. The reaction is properly written as $\gamma+Z\rightarrow{Z}+e^++e^-$ where Z is the other particle; its energy and momentum change in the interaction.

 

[ZapperZ]

Hello! I am a bit confused about the decay of a photon into a electron-positron pair. In the center of mass of the photon, isn't this decay violating the energy conservation?

To add to what have been mentioned already in the replies you received, we make e-p pair production by shooting high-energy photons into a material with high atomic number, such as Be. There is a reason for that, and that has been given in the two responses above.

Zz.

 

[mfb]

by shooting high-energy photons into a material with high atomic number, such as Be

Beryllium has atomic number 4. The only solid target with an even lower number would be lithium, but that is too reactive (chemically) to be practical: Be is as low as you can get.
High atomic numbers would be lead (82) or tungsten (74). You need more material with lighter elements, but the produced electrons/positrons pass through the material easier as well, so low atomic numbers can be favorable.

 

[stoomart]

Positrons and electrons (also neutrinos/antineutrinos) are released from the decay of W bosons after photon collisions.

http://cms.web.cern.ch/news/lhc-photon-collider

 

[ZapperZ]

Beryllium has atomic number 4. The only solid target with an even lower number would be lithium, but that is too reactive (chemically) to be practical: Be is as low as you can get.
High atomic numbers would be lead (82) or tungsten (74). You need more material with lighter elements, but the produced electrons/positrons pass through the material easier as well, so low atomic numbers can be favorable.

Sorry, I should have said tungsten. I had Be in my head because I've been shopping around for a Be window for one of our viewports.

Zz.

 

[mfb]

Positrons and electrons (also neutrinos/antineutrinos) are released from the decay of W bosons after photon collisions.

Technically true, but that is an incredibly rare process. Direct electron and pair production from photon collisions is a very rare process already, but that is much more frequent than the indirect process via W boson production. Note the "collision" part, it doesn't happen with single photons as OP asked about.

 

[stoomart]

Technically true, but that is an incredibly rare process. Direct electron and pair production from photon collisions is a very rare process already, but that is much more frequent than the indirect process via W boson production. Note the "collision" part, it doesn't happen with single photons as OP asked about.

Ahh thanks, didnt notice the OP was about a single photon.

 

[snorkack]

However, you are on to something here. An isolated photon cannot decay into an electron/positron pair, because there's no way that interaction can conserve both energy and momentum. (An easy way to see this is to think about how the interaction looks in the center of mass frame of the electron and positron after the collision).

And neither can two or more photons traveling in exact same direction. Nor a photon and another massless particle traveling in exact same direction.

Instead, pair production requires the involvement of some other massive charged particle, typically some nearby atomic nucleus. The reaction is properly written as $\gamma+Z\rightarrow{Z}+e^++e^-$ where Z is the other particle; its energy and momentum change in the interaction.

Obviously production of a pair from a photon and a massless chargeless photon traveling in a different direction must be possible because that´s time reversal of annihilation.
Is it possible to produce a pair from a photon and a massive neutral particle, such as neutron?

 

[mfb]

Is it possible to produce a pair from a photon and a massive neutral particle, such as neutron?

Neutrons have charged quarks inside. It is very unlikely in the MeV range, but becomes similar (within a factor of 2) to pair production at a proton at high energies. of a few hundred MeV.