Classical Electrodynamics · Problem 10.20

Radiating Systems, Multipole Fields and Radiation — John David Jackson

Physics — E&M · scattering-dielectric-cylinder-suspension

Problem

A suspension of transparent fibers in a clear liquid is modeled as a collection of scatterers, each being a right circular cylinder of radius aa and length LL of uniform dielectric material whose electric susceptibility differs from the surrounding medium by a small fractional amount δϵ/ϵ\delta\epsilon/\epsilon. (a) Show that to first order in δϵ/ϵ\delta\epsilon/\epsilon the scattering cross section per scatterer for unpolarized radiation of wave number kk is dσdΩ=∣δϵϵ∣2k4a4L232(1+cos⁡2θ)∣2J1(q⊥a)q⊥a⋅sin⁡(q∥L/2)q∥L/2∣2\frac{d\sigma}{d\Omega} = \left|\frac{\delta\epsilon}{\epsilon}\right|^2 \frac{k^4 a^4 L^2}{32}(1+\cos^2\theta)\left|\frac{2J_1(q_\perp a)}{q_\perp a}\cdot\frac{\sin(q_\parallel L/2)}{q_\parallel L/2}\right|^2 where J1(z)J_1(z) is the Bessel function of order unity and q∥q_\parallel (q⊥q_\perp) is the component of the wave number transfer parallel (perpendicular) to the cylinder axis. (b) In the limit of very slender cylinders (ka≪1ka \ll 1), show that the scattering cross section, averaged over the orientation of the cylinder (appropriate for an ensemble of randomly oriented fibers), is ⟨dσdΩ⟩=∣δϵϵ∣2k4a4L232(1+cos⁡2θ)[2qL Si(qL)−(sin⁡(qL/2)qL/2)2]\left\langle\frac{d\sigma}{d\Omega}\right\rangle = \left|\frac{\delta\epsilon}{\epsilon}\right|^2\frac{k^4 a^4 L^2}{32}(1+\cos^2\theta)\left[\frac{2}{qL}\,\text{Si}(qL) - \left(\frac{\sin(qL/2)}{qL/2}\right)^2\right] where Si(x)=∫0x[(sin⁡x)/x] dx\text{Si}(x) = \int_0^x [(\sin x)/x]\,dx is the sine integral (Abramowitz and Stegun, p. 231) and q2=2k2(1−cos⁡θ)q^2 = 2k^2(1-\cos\theta). (c) Plot the square-bracketed quantity in part b as a function of q2L2q^2L^2 on the range (0,100)(0, 100). Verify that the cross section is the expected one when kL≪1kL \ll 1 and show that when kL≫1kL \gg 1 (but ka≪1ka \ll 1) the total scattering cross section is approximately σscatt≈11π260∣δϵϵ∣2k3a4L{1+O[1kL,ln⁡(kL)kL]}\sigma_{\text{scatt}} \approx \frac{11\pi^2}{60}\left|\frac{\delta\epsilon}{\epsilon}\right|^2 k^3 a^4 L\left\{1 + O\left[\frac{1}{kL}, \frac{\ln(kL)}{kL}\right]\right\} Comment on the frequency dependence.

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