MadSci Network: Physics
Query:

Re: Why is the minimum energy for pair production 2mc^2(1+m/M)?

Date: Thu Jun 24 17:22:59 1999
Posted By: Dave Dixon, Assistant Research Physicist,University of California
Area of science: Physics
ID: 927904908.Ph
Message:

Actually, the formula you give is an approximation.
The short answer is that in order to conserve energy AND momentum,
the nucleus must recoil.  Consider the case where the initial photon
is traveling in the z-direction, the nucleus is initially at rest,
and the created electron and positron have zero momentum.  The
photon is destroyed in the reaction, the electron/positron are not
moving, so the final nucleus momentum must be equal to that of the
photon.  However, since the nucleus is moving, it also gained some
kinetic energy.  Thus the initial photon needs to have energy equal
to twice the electron rest mass PLUS a little extra for the nucleus.

Here is the mathematical derivation for the situation described above.
I use units such that c=1.

p_g == initial photon momentum == E_g == initial photon energy
p_N == final nucleus momentum
E_N == final nucleus momentum = sqrt(M^2+p_N^2)
M == nucleus rest energy, m==electron rest energy

Conservation of momentum in z-direction:
p_g = p_N, or E_g = p_N   (remember c=1)

Conservation of energy:
M + E_g = sqrt(M^2+p_N^2) + 2m

Substituting from the momentum eqn:

M + E_g = sqrt(M^2+E_g^2) + 2m

Solving for E_g:

E_g = 2m(1-m/M)
       -------
       (1-2m/M)

This is the exact answer.  Note that M>>m, so we can use the
binomial expansion:

1/(1-2m/M) = 1 + 2m/M + ...

E_g = 2m(1-m/M)(1+2m/M+...) = 2m(1 + 2m/M - m/m - 2m^2/M^2 + ...)

Dropping all terms of order m^2/M^2 and higher, we arrive at

E_g ~= 2m(1+m/M)

Putting c back in

E_g ~= 2mc^2(1+m/M)


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