# Classical Electromagnetism

## Chapter 8: Laplace's equation in Cylindrical Coordinates

### By Mark Lawrence

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• Support our advertisers. Thanks, Mark ## Three Dimensional Boundary Value Problems in Spherical Coordinates

In this chapter we'll be working in spherical coordinates. We'll look for solutions to Laplace's equation. Later we'll apply boundary conditions to find specific solutions. s lies in the x-y plane, an angle φ up from the x axis. The relevant equations are:

 x = r sin θ cos φ y = r sin θ sin φ z = r cos θ r = √ (x2 + y2 + z2) θ = arctan( √ (x2 + y2) / z ) φ = arctan( y / x )

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

Ñ2Φ = 0

Ñ2 in spherical coordinates is:

 1 ∂ ∂ 1 ∂ ∂ 1 ∂ Ñ2 = r2 + sin θ + r2 ∂r ∂r r2sin θ ∂θ ∂θ r2sin2θ ∂φ2

We'll presume that our solution is separable, that is Φ( r, θ, φ ) = R(r)P(θ)Q(φ). 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,

Ñ2Φ(r,θ,φ) = Ñ2R(r)P(θ)Q(φ).

P is independent of r and φ, so it can be pulled through the r and φ partial derivatives. Similarly R is independent of θ and z, so it can be pulled through the θ and φ partial derivatives. Q is independent of r and θ, so it can be pulled through the r and θ partial derivatives.

 PQ ∂ ∂R RQ ∂ ∂P RP ∂Q Ñ2RPQ = r2 + sin θ + r2 ∂r ∂r r2sin θ ∂θ ∂θ r2sin2θ ∂φ2

Now multiply the entire equation by r2sin2θ / RPQ:

 sin2θ ∂ ∂R sin θ ∂ ∂P 1 ∂Q Ñ2RPQ = r2 + sin θ + R ∂r ∂r P ∂θ ∂θ Q ∂φ2

Now there's a portion which depends only on φ. We'll set that portion equal to a separation variable, -m2.

 ∂2Q = -m2Q ∂φ2
 sin2θ ∂ ∂R sin θ ∂ ∂P r2 + sin θ =  m2 R ∂r ∂r P ∂θ ∂θ

We can solve the Q equation immediately - this is just sines and cosines. We'll choose to represent the solutions as Q = Amcos mφ + Bm sin mφ. The remaining equation still needs a bit of work. If m = 0, the solution is Q = A0 + B0 φ. Normally B0 will be zero, as the linear solution rarely makes any physical sense. So m = 0 normally means we're dealing with a problem which has symmetry about the z axis, often referred to as azimuthal symmetry. These problems come up with some frequency, so we'll look into them in more detail in a moment.

We'll divide the remaining equation through by sin2θ,  resulting in:

 1 ∂ ∂R 1 ∂ ∂P m2 r2 + sin θ - = 0 R ∂r ∂r P sin θ ∂θ ∂θ sin2θ

Now there's a portion which depends only on r. We'll set that portion equal to a separation variable, l(l+1). This seems to come out of thin air a bit - it's because this equation has been solved for over a hundred years, and it's well known now what to do at each stage. While we're at it, we'll substitute U(r) = rR(r), another trick known for a hundred years. In the 1880's, when all physics problems were known to be solved and all that was left for the graduate students was the next few decimal places, this is all graduate students did: mess with Laplace's equation in various coordinate systems.

 ∂ ∂U/r ∂ ∂U ∂U ∂2U r r2 = l(l+1) U = r r - r = r2 ∂r ∂r ∂r ∂r ∂r ∂r2
 1 ∂ ∂P m2P sin θ - + l(l+1) P = 0 sin θ ∂θ ∂θ sin2θ

We'll try a polynomial, rk, in the equation for U. Immediately we see that k(k-1) = l(l+1). This is a simple quadratic, so U = Alrl+1 + Blr-l. More to the point, U/r, which is the r dependence of our final solution, is U/r = Alrl + Blr-l-1.

The equation for P is usually recast in terms of x = cos(θ), dx = -sin(θ)dθ, sin2θ = 1 - x2. Now,

 ∂ ∂P m2P (1 - x2) - + l(l+1) P = 0 ∂x ∂x 1 - x2

If m is zero, we are looking at Legendre's equation. We'll consider those solutions first. Remember, m = 0 means we're looking at problems with azimuthal symmetry, that is we're looking at problems that are symmetric about the z axis.

The solution to Legendre's equation is the Legendre polynomials. Remember, we're working in x = cos(θ), so -1 ≤ x ≤ 1, as -π ≤ θ ≤ π. We require that our solutions converge and are finite over this interval. It can be shown that this requirement implies that l is zero or a positive integer. So, we need only consider integer values of l. The first several Legendre polynomials are:

 P0(cos θ) = 1 P1(cos θ) = cos θ P2(cos θ) = 1/2 ( -1 + 3 cos2θ ) P3(cos θ) = 1/2 ( -3 cos θ + 5 cos3θ ) P4(cos θ) = 1/8 ( 3 - 30 cos2θ + 35 cos4θ ) P5(cos θ) = 1/8 ( 15 cos θ - 70 cos3θ + 63 cos5θ ) P6(cos θ) = 1/16 ( -5 + 105 cos2θ - 315 cos4θ + 231 cos6θ ) Important Properties of Legendre Polynomials
-1 ≤ Pn(x) ≤ 1 Pn(x) complete & orthogonal over [-1, 1]
Pn(0) = 0,  n odd
 1*3*5*...*(n-1) Pn(0) = -1n/2 ,  n even 2*4*6*...*n
Pn(-1) = -1n Pn(1) = 1
Pn(-x) = -1n Pn(x) Pn'(-x) = -1n+1  Pn'(x)

 P00( x ) = 1 = 1 P10( x ) = x = cos θ P11( x ) = -(1 - x2)1/2 = -sin θ P20( x ) = 1/2 (3x2 - 1 ) = 1/2(3cos2θ -1) P21( x ) = -3x (1 - x2)1/2 = -3sin θ cos θ P22( x ) = 3 (1 - x2) = 3sin2θ P30( x ) = 1/2 x (5x2 - 3) = 1/2 cos θ (5cos2θ -3) P31( x ) = 3/2 (1 - 5x2) (1 - x2)1/2 = -3/2(5cos2θ -1)sin θ P32( x ) = 15 x (1 - x2) = 15cos θ sin2θ P33( x ) = -15 (1 - x2)3/2 = -15sin3θ P40( x ) = 1/8 (35x4 - 30x2 + 3) = 1/8 (35cos4θ -30cos2θ +3) P41( x ) = 5/2 x (3 - 7x2) (1 - x2)1/2 = 5/2 cos θ (3 - 7cos2θ) sin θ P42( x ) = 15/2 (7x2 - 1) (1 - x2) = 15/2 (7cos2θ -1) sin2θ P43( x ) = 105 x (1 - x2)3/2 = 105cos θ sin3θ P44( x ) = 105 (1 - x2)2 = 105sin4θ P50( x ) = 1/8 x (63x4 - 70x2 + 15) = 1/8 cos θ (63cos4θ -70cos2θ + 15)

 The Spherical Harmonics √ 2l + 1 (l-m)! Ylm(θ,φ) = Plm( cos θ ) eimφ 4π (l+m)!
Y00(θ,φ) =
 1 √4π
Y10(θ,φ) =
 √3 cos θ √4π
Y11(θ,φ) =
 √3 - sin θeiφ √8π
Y20(θ,φ) =
 √5 (3cos2θ -1) 4√π
Y21(θ,φ) =
 √15 - sin θ cos θ eiφ √8π
Y22(θ,φ) =
 √15 sin2θ ei2φ √32π
Y30(θ,φ) =
 √7 (5cos3θ-3cos θ) 4√π
Y31(θ,φ) =
 √21 - (5cos2θ -1)sin θ eiφ 8√π
Y32(θ,φ) =
 √105 cos θ sin2θ ei2φ √8π
Y33(θ,φ) =
 √35 - sin3θ ei3φ 8√π
Y40(θ,φ) =
 3 (35cos4θ -30cos2θ +3) 16√π
Y41(θ,φ) =
 15 - cos θ (3 - 7cos2θ) sin θ eiφ 4√10π
Y42(θ,φ) =
 15 (7cos2θ -1) sin2θ ei2φ 8√10π
Y43(θ,φ) =
 105 - cos θ sin3θ ei3φ 8√35π
Y44(θ,φ) =
 105 sin4θ ei4φ 16√70π
Y50(θ,φ) =
 √11 cos θ (63cos4θ -70cos2θ + 15) 16√π