# What does the Rayleigh Phase Function tell us

I am working on some radiative transfer equations, and struggling as I'm fairly new to this field. I have read about the Rayleigh Phase Function which is:

$P(\theta) = \frac{3}{4}(1 + cos^2 \theta)$

I can plot this function, and generate values from this function in my computer program (which is what I need to do) - but what is it actually telling me? If I know I have a ray of incident light at a certain angle, will this function tell me what angle it will scatter at? Or is it more of a statistical generalisation where the areas where the function is higher are the angles at which the scattering is more likely to happen?

Furthermore, when I see the function plotted it is always from 0-180, rather than 0-360 as I would expect for the scattering. This makes me think that the function may be giving the difference between the incoming angle and the outgoing angle rather than the absolute angle it is scattered at. Is that correct?

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The function is symmetrical, which would explain why it is typically only plotted from 0-180 degrees, it'd just be mirrored over 0 degrees. – PearsonArtPhoto Aug 25 '11 at 1:21
Thanks - that makes sense. – robintw Aug 25 '11 at 8:23

## 1 Answer

Not sure why the word "phase" is in the name of this function. The intensity of unpolarized light scattered by Rayleigh particles as a function of detector angle $\theta$ is proportional to this function. Colloidal scientists call this measurement "Static Light Scattering," but atmospheric scientists and others may have different names for it.

You can get some physical intuition for this function by picturing an EM wave interacting with a point scatterer. Imagine sweeping a detector in a circle of a radius $R$ centered on the scatterer. Then place an analyzer (polarizer) in front of the detector so you are detecting only either vertically or horizontally polarized light. If the incident light is vertically polarized, i.e. polarization normal to the plane of the circle, the amplitude (and intensity) of scattered light is isotropic. If the light is horizontally polarized, i.e. (polarization in the plane of the circle), the intensity detected depends on detection angle.

Convention is to call $\theta=0$ degrees the angle of forward scattered (also unscattered) light. That way, $\theta=180$ is the angle where we detect backscattered light, and $\theta=90$ and $\theta=270$ are each right in between those two "poles". Imagine horizontally polarized radiation interacting with our point scatterer. This polarization leads to maximum scattered intensity at $\theta=0$ and $\theta=180$, and minimum of zero scattered intensity at $\theta=90$ (and $\theta=270$). You can use the right-hand rule and dot products to convince yourself of this.

Let your fingers point in the direction of the polarization, and your thumb point in $\hat{k}$, the direction in which the scattered photon is propagating. You can let your thumb point in any direction around the circle; imagine that your thumb points toward the detector, so it determines which $\theta$ you are analyzing. If your fingers are in the plane of the circle you are looking at horizontally polarized light. Try to put your fingers in the plane and point your thumb at $\theta=90$. You should find that your fingers are pointing in a direction that is orthogonal to the horizontal radiation of the incident light, which means none of such light will scatter in this direction. The $cos^2(\theta)$ term characterizes this angular dependence of horizontal polarization, and the $1$ term states the isotropy of the vertically polarized radiation. We have to add these for unpolarized incident light.

This general dependence is true for dilute solutions of Rayleigh scatterers, or gases.

The intensity measurement comes from counting a large number of photons at each angle. The function comes from classical optics/electromagnetism. I suspect it could also be derived from a quantum mechanical (QM) treatment of photons. QM may be better suited for determining the "angles at which the scattering is more likely to happen" as you put it, since we think of the classical picture of as purely deterministic.

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