Cylindrical harmonics

From HandWiki

In mathematics, the cylindrical harmonics are a set of linearly independent functions that are solutions to Laplace's differential equation, 2V=0, expressed in cylindrical coordinates, ρ (radial coordinate), φ (polar angle), and z (height). Each function Vn(k) is the product of three terms, each depending on one coordinate alone. The ρ-dependent term is given by Bessel functions (which occasionally are also called cylindrical harmonics).

Definition

Each function Vn(k) of this basis consists of the product of three functions:

Vn(k;ρ,φ,z)=Pn(k,ρ)Φn(φ)Z(k,z)

where (ρ,φ,z) are the cylindrical coordinates, and n and k constants that differentiate the members of the set. As a result of the superposition principle applied to Laplace's equation, very general solutions to Laplace's equation can be obtained by linear combinations of these functions.

Since all surfaces with constant ρ, φ and z  are conicoid, Laplace's equation is separable in cylindrical coordinates. Using the technique of the separation of variables, a separated solution to Laplace's equation can be expressed as:

V=P(ρ)Φ(φ)Z(z)

and Laplace's equation, divided by V, is written:

P¨P+1ρP˙P+1ρ2Φ¨Φ+Z¨Z=0

The Z  part of the equation is a function of z alone, and must therefore be equal to a constant:

Z¨Z=k2

where k  is, in general, a complex number. For a particular k, the Z(z) function has two linearly independent solutions. If k is real they are:

Z(k,z)=cosh(kz)orsinh(kz)

or by their behavior at infinity:

Z(k,z)=ekzorekz

If k is imaginary:

Z(k,z)=cos(|k|z)orsin(|k|z)

or:

Z(k,z)=ei|k|zorei|k|z

It can be seen that the Z(k,z) functions are the kernels of the Fourier transform or Laplace transform of the Z(z) function and so k may be a discrete variable for periodic boundary conditions, or it may be a continuous variable for non-periodic boundary conditions.

Substituting k2 for Z¨/Z , Laplace's equation may now be written:

P¨P+1ρP˙P+1ρ2Φ¨Φ+k2=0

Multiplying by ρ2, we may now separate the P  and Φ functions and introduce another constant (n) to obtain:

Φ¨Φ=n2
ρ2P¨P+ρP˙P+k2ρ2=n2

Since φ is periodic, we may take n to be a non-negative integer and accordingly, the Φ(φ) the constants are subscripted. Real solutions for Φ(φ) are

Φn=cos(nφ)orsin(nφ)

or, equivalently:

Φn=einφoreinφ

The differential equation for ρ is a form of Bessel's equation.

If k is zero, but n is not, the solutions are:

Pn(0,ρ)=ρnorρn

If both k and n are zero, the solutions are:

P0(0,ρ)=lnρor1

If k is a real number we may write a real solution as:

Pn(k,ρ)=Jn(kρ)orYn(kρ)

where Jn(z) and Yn(z) are ordinary Bessel functions.

If k  is an imaginary number, we may write a real solution as:

Pn(k,ρ)=In(|k|ρ)orKn(|k|ρ)

where In(z) and Kn(z) are modified Bessel functions.

The cylindrical harmonics for (k,n) are now the product of these solutions and the general solution to Laplace's equation is given by a linear combination of these solutions:

V(ρ,φ,z)=nd|k|An(k)Pn(k,ρ)Φn(φ)Z(k,z)

where the An(k) are constants with respect to the cylindrical coordinates and the limits of the summation and integration are determined by the boundary conditions of the problem. Note that the integral may be replaced by a sum for appropriate boundary conditions. The orthogonality of the Jn(x) is often very useful when finding a solution to a particular problem. The Φn(φ) and Z(k,z) functions are essentially Fourier or Laplace expansions, and form a set of orthogonal functions. When Pn(kρ) is simply Jn(kρ) , the orthogonality of Jn, along with the orthogonality relationships of Φn(φ) and Z(k,z) allow the constants to be determined.[1]

If (x)k is the sequence of the positive zeros of Jn then:

01Jn(xkρ)Jn(xkρ)ρdρ=12Jn+1(xk)2δkk[2]

In solving problems, the space may be divided into any number of pieces, as long as the values of the potential and its derivative match across a boundary which contains no sources.

Example: Point source inside a conducting cylindrical tube

As an example, consider the problem of determining the potential of a unit source located at (ρ0,φ0,z0) inside a conducting cylindrical tube (e.g. an empty tin can) which is bounded above and below by the planes z=L and z=L and on the sides by the cylinder ρ=a.[3] (In MKS units, we will assume q/4πϵ0=1). Since the potential is bounded by the planes on the z axis, the Z(k,z) function can be taken to be periodic. Since the potential must be zero at the origin, we take the Pn(kρ) function to be the ordinary Bessel function Jn(kρ), and it must be chosen so that one of its zeroes lands on the bounding cylinder. For the measurement point below the source point on the z axis, the potential will be:

V(ρ,φ,z)=n=0r=0AnrJn(knrρ)cos(n(φφ0))sinh(knr(L+z))zz0

where knra is the r-th zero of Jn(z) and, from the orthogonality relationships for each of the functions:

Anr=4(2δn0)a2sinhknr(Lz0)sinh2knrLJn(knrρ0)knr[Jn+1(knra)]2

Above the source point:

V(ρ,φ,z)=n=0r=0AnrJn(knrρ)cos(n(φφ0))sinh(knr(Lz))zz0
Anr=4(2δn0)a2sinhknr(L+z0)sinh2knrLJn(knrρ0)knr[Jn+1(knra)]2.

It is clear that when ρ=a or |z|=L, the above function is zero. It can also be easily shown that the two functions match in value and in the value of their first derivatives at z=z0.

Point source inside cylinder

Removing the plane ends (i.e. taking the limit as L approaches infinity) gives the field of the point source inside a conducting cylinder:

V(ρ,φ,z)=n=0r=0AnrJn(knrρ)cos(n(φφ0))eknr|zz0|
Anr=2(2δn0)a2Jn(knrρ0)knr[Jn+1(knra)]2.

Point source in open space

As the radius of the cylinder (a) approaches infinity, the sum over the zeroes of Jn(z) becomes an integral, and we have the field of a point source in infinite space:

V(ρ,φ,z)=1R=n=00d|k|An(k)Jn(kρ)cos(n(φφ0))ek|zz0|
An(k)=(2δn0)Jn(kρ0)

and R is the distance from the point source to the measurement point:

R=(zz0)2+ρ2+ρ022ρρ0cos(φφ0).

Point source in open space at origin

Finally, when the point source is at the origin, ρ0=z0=0

V(ρ,φ,z)=1ρ2+z2=0J0(kρ)ek|z|dk.

See also

Notes

  1. Smythe 1968, p. 185.
  2. Guillopé 2010.
  3. Configuration and variables as in Smythe 1968

References