According to the text "On the Production of the Positive Electron" by Oppenheimer and Plesset, from 1933, the authors give a lower bound for the energy $E$ for the pair production near a nucleus of charge $eZ$ producing a Coulomb field. The bound is:

$$E > mc^2\left( 1 + \left(1 + \alpha^2 Z^2/\sqrt{1-\alpha^2 Z^2}\right)^{-1/2} \right)$$

where $\alpha$ should be, through the context, the fine structure constant. This should mean that near a nucleus of lead, for instance, the actual minimum value for $E$ should be $\approx1.8mc^2$.

I haven't been able to derive this formula, and this is where I need help. The only way I could think of working with $eZ$ here is by saying the nucleus has a radius of about $r_0 = \frac{Z^2e^2}{mc^2}$, (the classical electron radius for charge $Ze$), through this I tried assuming the electron and positron are created at $r_0$ (like a rutherford scattering problem), but of course it didn't work.

So any help would be appreciated, thanks in advance.

  • $\begingroup$ This is related to the binding energy of the ground state of an electron in a Coulomb potential, found by solving the Dirac equation. But I forget exactly how this relates to the pair production threshold. $\endgroup$ – G. Smith Nov 17 '19 at 22:58
  • $\begingroup$ If you consider, as Dirac did, a positron to be an unfilled negative energy state, then I think this pair production threshold is the sum of the mass-energy it takes to create an electron plus the energy to excite a positron from the least-negative state in the Dirac sea to zero energy. Or something like that. $\endgroup$ – G. Smith Nov 17 '19 at 23:57

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