Creation of an Electron-Positron Pair by a Photon

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SUMMARY

The discussion centers on the impossibility of creating an electron-positron pair from a single photon in isolation, as established by conservation laws and the principles of special relativity. At threshold energy, the minimum photon energy required is given by the equation hf(min) = 2m(e)c^2, where m(e) is the mass of the electron. The analysis demonstrates that a Lorentz frame cannot be established where the electron and positron have equal and opposite momenta, violating the laws of physics. The conclusion is that additional mass or radiation must be present for such a pair production to occur.

PREREQUISITES
  • Understanding of special relativity principles
  • Familiarity with particle physics concepts, specifically electron-positron pairs
  • Knowledge of conservation laws in physics
  • Basic algebra and manipulation of equations in physics
NEXT STEPS
  • Study the implications of conservation of momentum in particle interactions
  • Learn about threshold energy calculations in particle physics
  • Explore the concept of pair production in high-energy photon interactions
  • Investigate the role of Lorentz transformations in special relativity
USEFUL FOR

Students of physics, particularly those focusing on particle physics and special relativity, as well as educators seeking to explain the principles of particle-antiparticle creation.

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Homework Statement



Show that the creation of an electron-positron pair (or any particle-antiparticle pair, for that matter) by a single photon is not possible in isolation, ie, that additional mass (or radiation) must be present. (Hint: Consider the reaction at threshold, then apply conservation laws.)

Homework Equations



hc/lambda - pc = 2m(e)c^2
at threshold, hf(min) = 2m(e) c^2

The Attempt at a Solution



I know this can't work b/c you can't find a Lorentz frame in which the e+ and e- have equal and opposite momenta, b/c then the photon would be at rest which would disobey the law of special relativity.

At threshold, all the energy of the photon becomes the mass of the electron and positron so the electron and positron have KE=0.

h/lambda = p- + p+

2m(e) c = m(e) u- + m(e) u+

2c= u- + u+
 
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I prefer doing it by showing that W^2=(E1+E2)^2-(p1+p2)^2 cannot equal zero, as it must if the two particles came from a photon.
Algebra gives W^2=2m^2+2(E1E2+p1p2 cos\theta), which can't equal zero.
 

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