# Minimum energy for pair production near charge $eZ$

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.

• 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. – G. Smith Nov 17 '19 at 22:58
• 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. – G. Smith Nov 17 '19 at 23:57