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Dispersion and Absorption of Light

Problem 5.204

An electron experiences a quasi-elastic force kxk x and a "friction force" γx\gamma x in the field of electromagnetic radiation. The EE-component of the field varies as E=E0cos⁡ωt.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.

Reveal Answer
5.204.5.204 . (a) x=acos⁡(ωt+φ),x=a \cos (\omega t+\varphi), where aa and φ\varphi are defined by the formulas a=eE0/m(ω02−ω2)2+4β2ω2,tan⁡φ=2βωω2−ω02. 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 β=γ/2m,ω02=k/m,m\beta=\gamma / 2 m, \omega_{0}^{2}=k / m, m is the mass of an electron. (b) ⟨P⟩=\langle P\rangle= =mβ(eE0/m)2ω2(ω02−ω2)2+4β2ω2,⟨P⟩max⁡=m4β(eE0m)2 for ω=ω0 =\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}
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