We can think about a pulsar as a battery that drives currents along magnetic field lines. The two “terminals” of the battery are the pole and the edge of the polar cap. The radius of the polar cap is $ l \approx R\sqrt{R/cP} $, where $ R $ is the radius of the pulsar, $ c $ is the speed of light and $ P $ is the spin period. The electric field is given by $ E \approx B l / P c $ where $ B $ is the magnetic field on the surface of the pulsar. The voltage drop is $ \Delta V \approx lE \approx BR^3/P^2 c^2 $. This is enough to accelerate electrons to a Lorentz factor $ \gamma \approx q \Delta V/m c^2 $ where $ m $ is the mass of the electron and $ q $ is the elementary charge. These particles move along the field lines and emit curvature radiation in the process. The curvature radius of the magnetic field lines at the polar cap is given by $ r_c \approx \sqrt{R c P} $, and the frequency of the emitted radiation is $ f \approx \frac{c}{r_c} \gamma^3 $ (two factors of the Lorentz factor come from the fact that the radiation is beamed, and another one from the Lorentz boost).

We proceed to determine the condition under which the curvature radiation can interact with the magnetic field to produce pairs. We can think of the magnetic field as if it is made up of photons of energy $ \varepsilon $, wavelength $ \lambda \approx h c/\varepsilon $ (where $ h $ is Planck's constant) and occupying a volume $ \lambda^3 $, so the energy density is $ \varepsilon^4/c^3 h^3 \approx B^2 $ and hence the energy of a single virtual photon is $ \varepsilon \approx \left(c^3 h^3 B^2\right)^{1/4} $. Such a photon, together with a photon from curvature radiation, can create an electron - positron pair if $ \varepsilon h f > m^2 c^4 $.

The magnetic field cannot be measured directly, but it can be inferred from observations by measuring the change in period $ \dot{P} $. Assuming the radiated dipole energy comes at the expense of the spin rotation energy $ M \frac{R^2}{P^3} \dot{P} \approx \frac{B^2 R^4}{c^3} \left(\frac{R}{P^2}\right)^2 \Rightarrow B^2 \approx M \frac{c^3}{R^4} P \dot{P} $ where $ M $ is the mass of the pulsar. The condition for pair production becomes $ \dot{P} > 1.2 \cdot 10^{-16} \left(\frac{P}{1 \rm s}\right)^{19/7} \left(\frac{M}{M_{\odot}}\right)^{-1} \left(\frac{R}{10 \rm km}\right)^{6/7} $ In pulsar where this condition is violated radio emission is supposedly suppressed.