8—Multivariable Calculus 2008.13 Solid Angle
The extension of the concept of angle to three dimensions is called “solid angle.” To explain what this
is, I’ll first show a definition of ordinary angle that’s different from what you’re accustomed to. When
you see that, the extension to one more dimension is easy.
Place an object in the plane somewhere not at the origin. You are at the origin and look at it. I
want a definition that describes what fraction of the region around you is spanned by this object. For
this, draw a circle of radiusRcentered at the origin and draw all the lines from everywhere on the
object to the origin. These lines will intersect the circle on an arc (or even a set of arcs) of lengths.
Definethe angle subtended by the object to beθ=s/R.
s
R
A
R
Now step up to three dimensions and again place yourself at the origin. This time place a sphereof radiusRaround the origin and draw all the lines from the three dimensional object to the origin.
This time the lines intersect the sphere on an area of sizeA. Definethe solid angle subtended by the
object to beΩ =A/R^2. (If you want four or more dimensions, see problem8.52.)
For the circle, the circumference is 2 πR, so if you’re surrounded, the angle subtended is 2 πR/R=
2 πradians. For the sphere, the area is 4 πR^2 , so this time if you’re surrounded, the solid angle subtended
is 4 πR^2 /R^2 = 4πsterradians. That is the name for this unit.
All very pretty. Is it useful? Only if you want to describe radiative transfer, nuclear scattering,
illumination, the structure of the atom, or rainbows. Except for illumination, these subjects center
around one idea, that of a “cross section.”
Cross Section, Absorption
Before showing how to use solid angle to describe scattering, I’ll take a simpler example: absorption.
There is a hole in a wall and I propose to measure its area. Instead of taking a ruler to it I blindly fire
bullets at the wall and see how many go in. The bigger the area, the larger the fraction that will go
into the hole of course, but I have to make this quantitative to make it useful.
Define the flux of bullets: f=dN/(dtdA). That is, suppose that I’m firing all the bullets in
the same direction, but not starting from the same place. Pick an area∆Aperpendicular to the stream
of bullets and pick a time interval∆t. How many bullets pass through this area in this time? ∆N,
and that’s proportional to both∆Aand∆t. The limit of this quotient is the flux.
lim
∆t→ 0
∆A→ 0