Classical Electromagnetism

Three Dimensional Boundary Value Problems in Cylindrical Coordinates

In this chapter we'll be working in cylindrical coordinates. We'll look for solutions to Laplace's equation. Later we'll apply boundary conditions to find specific solutions. Traditionally, ρ is used for the radius variable in cylindrical coordinates, but in electrodynamics we use ρ for the charge density, so we'll use s for the radius. φ will be the angular dimension, and z the third dimension.

s lies in the x-y plane, an angle φ up from the x axis. The relevant equations are:

x = s cos φ y = s sin φ z = z

s = √ (x2 + y2) φ = arctan(y/x) z = z

Laplace's equation for the potential in the absence of sources is:

Ñ²Φ = 0

Ѳ in cylindrical coordinates is:

		1	∂		∂		1	∂2		∂2
Ñ2	=			s		+			+
		s	∂s		∂s		s2	∂φ2		∂z2

We'll presume that our solution is separable, that is Φ( s, φ, z ) = S(s)Q(φ)Z(z). We can check the separability assumption later. Laplace's equation is a second order differential equation, so if we find two unique solutions we've done it. Generally speaking, if the boundary conditions are separable, there's a good chance the solution is separable. If the boundary conditions are not separable, most likely we're hosed. Now,

ѲΦ(s,φ,z) = ѲS(s)Q(φ)Z(z).

Q is independent of s and z, so it can be pulled through the s and z partial derivatives. Similarly S is independent of φ and z, so it can be pulled through the φ and z partial derivatives. Z is independent of s and φ, so it can be pulled through the s and φ partial derivatives.

		ZQ	∂		∂		ZS	∂2		∂2Z
Ñ2SQZ	=			s		S +			Q + SQ		= 0
		s	∂s		∂s		s2	∂φ2		∂z2

Now multiply the entire equation by s² / SQZ:

		s	∂		∂		1	∂2		∂2Z
Ñ2SQZ	=			s		S +			Q + s2Z		= 0
		S	∂s		∂s		Q	∂φ2		∂z2

Now there's a portion which depends only on φ. We'll set that portion equal to a separation variable, -η².

∂2Q
	= -η2Q
∂φ2

s	∂		∂		∂2Z
		s		S + s2Z		= η2
S	∂s		∂s		∂z2

We can solve the Q equation immediately - this is just sines and cosines. We'll choose to represent the solutions as Q = A_ηcos ηφ + B_η sin ηφ. The remaining equation still needs a bit of work. We'll divide through by s²,  resulting in:
 
 

1	∂		∂		η2		∂2Z
		s		S -		+ Z		= 0
sS	∂s		∂s		s2		∂z2

Now, the z portion can be set to another seperation variable, k². This leaves us with two equations:

∂2Z
	= k2Z
∂z2

1	∂		∂		η2
		s		S -		S + k2S	= 0
s	∂s		∂s		s2

We can solve the Z portion immediately - this is Z = a_ke^kz + b_ke^-kz. Now we divide the s equation through by k², then change variable s -> ks, which eliminates k from the equation entirely.

1	∂		∂		η2
		s		S -		S + S	= 0
s	∂s		∂s		s2

This is Bessel's equation. The solutions are Bessel functions, Neumann functions, and Hankel functions, and we've officially entered Graduate Student Hell.

A Worked Problem