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Pair production and photon

1. Homework Statement

To show that the minimum energy a photon must have to create an electron-positron pair in the presence of a stationary nucleus of mass M is 2mc^2[1 + (m/M)], m is the electron rest mass.

2. Homework Equations

Conservation of energy and momentum.Also the minimum energy a photon should have is 1.02MeV to form an electron and a positron ( wont be needing the figures here though)


3. The Attempt at a Solution

I tried doing it in the center of mass frame ( coz the CM frame happens to give the minimum energy) but I'm getting nowhere. Should I be using the lorentz invariance E^2 - p^2c^2 in CM and lab frame?
 

Answers and Replies

979
1
Your thinking is along the correct lines. The CofM frame should include the nucleus. Thus the question is really asking what the minimum energy in the lab frame is. You should be able to convince yourself that the minimum energy case correspond to three stationary particles in the CofM frame: electron, positron and nucleus. From this you should be able to deduce the necessary initial energy and momentum for the photon in the CofM frame. Lorentz transform and it should give the correct answer.
 
Okay is the energy of initial photon energy in Cm frame 2mc^2 + Mc^2?(I havent really many solved problems in center of mass frames so i'm finding it a lil difficult )but then how do I get momentum of photon? the momentum of all three particles are zero in CM frame.
 
pam
455
1
Should I be using the lorentz invariance E^2 - p^2c^2 in CM and lab frame?
Yes. The threshold cm energy is given (2m+M)^2=(k+M)^2-k^2, where k is the photon lab energy. (c=1).
 
I'm getting the photon momentum in CM frame to be zero( cos Center of mass is nucleus and its at rest in CM frame?)hows that possible!?
 
979
1
(In natural units) the 4-momentum of the nucleus is (sqrt(M^2 + p^2), p). The CofM frame is possibly a bad name, because really we want the zero momentum frame (ZMF) --- which is the appropriate relativistic concept; in non-relativistic cases the two are the same. So the momentum of the photon must be -p, and so its 4-momentum is (p, -p). So for pair creation to *just about* happen, you want it to be the case that (sqrt(M^2+p^2)+p)=(M+2m). Now, the motion of the ZMF relative to the lab is given by p=(gamma)Mv, so you just need to Lorentz transform the 4-momentum of the photon to the lab frame.
 

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