Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2015 / S16 / 2 / 4 / Solution

For a sinusoidal laser field of peak amplitude $E_0$, an electron's peak nonrelativistic quiver speed is $v_q=e{E}_0/(m_e\omega )$. Choose the one-component electron thermal speed $v_T=\sqrt{k_{B}T/m_e}=1.33\times 10^7\mathrm{m}{\mathrm{s}}^{-1}$. The vacuum-equivalent laser intensity is $I={ϵ}_{0}c{E}_0^2/2$, so the onset $v_q\sim v_T$ occurs atUsing $\sqrt{2k_{B}T/m_e}$ as the thermal-speed convention doubles this estimate; its order of magnitude is the physical content. Once the quiver is comparable to the thermal speed, a collision rate computed for an undriven thermal distribution is no longer adequate.

Relativistic motion begins when the normalized field $a_0=e{E}_0/(m_{e}c\omega )$ is of order one. ThusThe ratio $I_{\mathrm{rel}}/I_{\mathrm{th}}=m_e{c}^2/(k_{B}T)\simeq 511$ checks the separation of the two regimes. These estimates use peak field and cycle-averaged vacuum intensity. The small refractive-index and interface-transmission corrections in this underdense plasma change only the few-percent prefactor, below the accuracy intended here.