8—Multivariable Calculus 236The cross section for being sent in a direction between these two angles is the area of the ring:dσ= 2πbdb.
Anything that hits in there will scatter into the outgoing angles shown. How much solid angle is this? Put the
z-axis of spherical coordinates to the right, so thatθis the usual spherical coordinate angle fromz. The element
of area on the surface of a sphere isdA=r^2 sinθdθdφ, so the integral over all the azimuthal anglesφaround
the ring just gives a factor 2 π. The element of solid angle is then
dΩ =dA
r^2= 2πsinθdθAs a check on this, do the integral over all theta to get the total solid angle around a point.
Divide the effective area for this scattering by the solid angle, and the result is the differential scattering
cross section.
dσ
dΩ
=
2 πbdb
2 πsinθ dθ=
b
sinθdb
dθIf I knowθas a function ofb, I can compute this. There are a couple of very minor modifications that you need
in order to finish this. The first is that the derivativedb/dθ can easily be negative, but both the area and the
solid angle are positive. That means that you need an absolute value here. One other complication is that one
value ofθcan come from several values ofb. It may sound unlikely, but it happens routinely. It even happens in
the example that comes up in the next section.
dσ
dΩ=
∑
ibi
sinθ∣
∣
∣
∣
dbi
dθ∣
∣
∣
∣ (29)
8.14 Rainbow
An interesting, if slightly complicated example is the rainbow. Sunlight scatters from small drops of water in the
air and the detector is your eye. The water drops are small enough that I’ll assume them to be spheres, where
surface tension is enough to hold them in this shape for the ordinary small sizes of water droplets in the air.
The first and simplest model uses geometric optics and Snell’s law to figure out where the scattered light goes.
This model ignores the wave nature of light and it does not take into account the fraction of the light that is