optics - Rayleigh scattering in three dimensions

How does the Rayleigh scattering intensity depend on the polarization angle of the incident, linearly polarized light, and the observation angle in three dimensions?

Answer

In short, Rayleigh scattering is like ideal dipole radiation: the radiation pattern for Rayleigh scattering is exactly the same as dipole radiation when the light is perfectly and linearly polarised, so that the intensity has a $(\cos(\theta))^2$ dependence. For natural, randomly polarised light (e.g. from the Sun), the intensity varies as $1 + (\cos(\theta))^2$. Here $\theta$ is the altitude angle - the angle the scattering makes with the wavevector (Poynting vector) of the incoming plane field. See the Wikipedia page on Rayleigh Scattering for details with the formula for unpolarised light. For other scenarios, you'll have to dig into the theory as follows.

At the level of individual particles, you can get all you need from Section 13.5 "Diffraction of a Conducting Sphere: Theory of Mie" in Born and Wolf "Principles of Optics" (Mine is the sixth edition). in Section 13.5.2 under the subheading "limiting cases" we have:

"[begin quotation] It is easy to see that equations (88) [first order approximation in general Mie theory] are identical with those for the radiation zone ($r\gg\lambda$) of an electric dipole at O [the origin], oscillating parallel to the x-axis, i.e. parallel to the electric vector of the primary incident field, and of moment $p=p_0 e^{-i\,\omega\,t}$ where:

$p_0 = a^3 \left|\frac{\hat{n}^2-1}{\hat{n}^2+2}\right|$

[end quotation]" The emphasis (italics) is Born and Wolf's, not mine.

Here $a$ is the Mie scatterer's radius ($a \ll \lambda$ for Rayleigh scattering), and $\hat{n}$ is the ratio of the particle's refractive index to that of the medium the particle is steeped in (in general complex, because the discussion is mainly about metal particles, but the theory is perfectly general). The $p_0$ formula assumes that the incident field has an electric field vector of unit amplitude, and Born and Wolf use Gaussian units. So beware to make the right conversion factor! The moment in general of course scales linearly with the incident electric field.

So you just look up the Maxwell equation solutions for the small dipole and use the recipe above to set the dipole moment in your formulas.

For Rayleigh scattering from a general shape rather than a sphere, the above will hold just as well but with an effective radius $a = \sqrt[3]{\frac{3 V}{4 \pi}}$, where $V$ is the scatterer's volume. This is because the particle just "looks like a dipole" to the electromagnetic field, whose wavelength is too big to "see" the details of the particle's shape: it simply sees that there is a "volume of stuff" at the origin. So, the scattered field from $N$ tiny spheres in a small volume at the origin will give $N$ times the scattered field (to first order, where the scattered field from one sphere is too small to influence the behaviour of another), by linearity of Maxwell's equations. Therefore, we'll get the right answer if we replace $a$ by a calibrated scaling of the cube root of the particle's volume.

Thanks for asking the question. I'm ashamed to say I've never really thought about it, but now that you mention it, I think you can see this effect when you are riding in an aeroplane: the blue cast of the sky looks layered and the blueness looking horizontally out of the window seems less bright than it does if you cast your gaze almost horizontally but slightly towards the ground or sky (i.e. you get less scattering from scatterers at your selfsame height above ground).