= Rayleigh scattering by a small sphere
{c}
{title2=$\sigma=\frac{8\pi}{3}k^4a^6\left(\frac{m^2-1}{m^2+2}\right)^2$}
= Rayleigh scattering cross-section
{c}
{synonym}
Consider a fixed, homogeneous, nonabsorbing, nonmagnetic <sphere> of <radius> $a$ in <vacuum>. Let $m$ be its relative <refractive index>, a fixed <real number>; in <vacuum> this equals the <sphere>'s <refractive index>. Treat $m$ as independent of <wavelength> over the range being compared, neglecting <optical dispersion>. Air can be approximated by <vacuum> here because its <refractive index> is close to one. These assumptions describe a toy particle, rather than a literal model of an individual air <molecule>.
The <Rayleigh scattering cross-section> $\sigma$ is an effective <area>: the total scattered <power> is $P_{\rm sca}=I_{\rm inc}\sigma$, where $I_{\rm inc}$ is the incident <light intensity>. For $ka\ll1$ its leading approximation is
$$
\sigma=\frac{8\pi}{3}k^4a^6\left(\frac{m^2-1}{m^2+2}\right)^2,\qquad k=\frac{2\pi}{\lambda}.
$$
The units are <area>, since $k^4a^6$ has units $\mathrm{m}^{-4}\mathrm{m}^6=\mathrm{m}^2$. This <Rayleigh scattering cross-section> measures scattering, not conversion of the incident <electromagnetic radiation> into <heat>. The definition follows https://farside.ph.utexas.edu/teaching/jk1/lectures/node107.html[Fitzpatrick, Thomson Scattering: total and differential scattering cross-sections]{ref}.
To see the mechanism, the nearly uniform applied <electric field> produces an <electric dipole moment> with peak magnitude
$$
p_0=4\pi\epsilon_0a^3\frac{m^2-1}{m^2+2}E_0,
$$
where $E_0$ is the incident <electric field> peak <wave amplitude> and $\epsilon_0$ is <vacuum permittivity>. This is the <sphere>'s response in <electrostatics>, with its <permittivity> divided by <vacuum permittivity> equal to $m^2$ under the nonmagnetic assumption. https://farside.ph.utexas.edu/teaching/jk1/lectures/node42.html[Fitzpatrick, Boundary Value Problems with Dielectrics, equation 554: induced sphere dipole]{ref}. The averaged <electric dipole radiation> <power> is $P_{\rm sca}=\omega^4p_0^2/(12\pi\epsilon_0c^3)$, with <angular frequency> $\omega=ck$ and <speed of light> $c$. Dividing by $I_{\rm inc}=\epsilon_0cE_0^2/2$ gives the displayed <Rayleigh scattering cross-section>. Thus doubling <radius> multiplies the <electric dipole moment> by eight and the scattered <power> by $64$, provided the larger <sphere> still satisfies $ka\ll1$. https://farside.ph.utexas.edu/teaching/em/lectures/node95.html[Fitzpatrick, Electric dipole radiation: angular power and its integral]{ref}.
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