All Problems Dispersion and Absorption of Light
Problem 5.204
An electron experiences a quasi-elastic force \(k x\) and a "friction force" \(\gamma x\) in the field of electromagnetic radiation. The \(E\)-component of the field varies as \(E=E_{0} \cos \omega t .\) Neglecting the action of the magnetic component of the field, find: (a) the motion equation of the electron; (b) the mean power absorbed by the electron; the frequency at which that power is maximum and the expression for the maximum mean power.
\(5.204 .\) (a) \(x=a \cos (\omega t+\varphi),\) where \(a\) and \(\varphi\) are defined by the formulas \[ a=\frac{e E_{0} / m}{\sqrt{\left(\omega_{0}^{2}-\omega^{2}\right)^{2}+4 \beta^{2} \omega^{2}}}, \quad \tan \varphi=\frac{2 \beta \omega}{\omega^{2}-\omega_{0}^{2}} . \] Here \(\beta=\gamma / 2 m, \omega_{0}^{2}=k / m, m\) is the mass of an electron. (b) \(\langle P\rangle=\) \[ =\frac{m \beta\left(e E_{0} / m\right)^{2} \omega^{2}}{\left(\omega_{0}^{2}-\omega^{2}\right)^{2}+4 \beta^{2} \omega^{2}}, \quad\langle P\rangle_{\max }=\frac{m}{4 \beta}\left(\frac{e E_{0}}{m}\right)^{2} \text { for } \omega=\omega_{0} \]