Mathematical Tools for Physics

8—Multivariable Calculus 238

The derivativedb/dθ= 1

/

[dθ/db]. Compute this.

dθ

db

=

2

√

R^2 −b^2

−

4

√

n^2 R^2 −b^2

(33)

In the parametrized form this is

db

dθ

=

db/dβ

dθ/dβ

=

Rcosβ

2 −4 cosβ/

√

n^2 −sin^2 β

In analyzing this, it’s convenient to have both forms, as you never know which one will be easier to interpret.
(Have you checked to see if they agree with each other in any special cases?)

0 90 180

0

R

q

b

n = 1 to 1.5, left to right

0 90 180

q

ds/dW

These graphs are generated from Eq. ( 31 ) for eleven values of the index of refraction equally spaced from
1 to 1.5. The key factor that enters the cross-section calculation, Eq. ( 29 ), isdb/dθ, because it goes to infinity
when the curve has a vertical tangent. For water, withn= 1. 33 , theb-θcurve has a vertical slope that occurs
atθa little less than 140 ◦.Thatis the rainbow.
To complete this I should finish withdσ/dΩ. The interesting part of the problem is near the vertical part
of the curve. To see what happens near such a point I can use the power series expansion near there. Notb(θ)
butθ(b). This has zero derivative here, so near the vertical point

θ(b) =θ 0 +γ(b−b 0 )^2