5. Ch. 2, part 2

————————— Lecture 5 - Fourier expansions

Eigenfunctions come naturally in the treatment of vibrations in a confined space. Electromagnetic eigenmodes arise in the same way as their mechanical counterpart, such as sound waves in organ pipes. The basic equation for waves of speed c is

\frac{∂^{2}ψ}{∂x^{2}} = \frac{1}{c^{2}} \frac{∂^{2}ψ}{∂t^{2}}

in one dimension. An eigenmode is sinusoidal in time, so that the wave equation becomes

\frac{∂^{2}ψ}{∂x^{2}} + k^{2}ψ = 0

where c |k| is the angular frequency (usually called ω). The solutions are plane waves exp(ikx), with positive and negative k.

5.1. Convergence

If the waves are confined to a ring of length a, the eigenfunctions are simply plane waves with k taking the special values (eigenvalues)

k_{n} = 2πn/a, n integer

The expansion

f(x) = ∑_{n=−∞}^{∞} C_{n} exp(ik_{n}x)  (5.1)

with

C_{n} = ∫_{−a/2}^{a/2} exp(−ik_{n}x)f(x)dx/a  (5.2)

is the basic Fourier expansion. It converges uniformly for any continuous periodic function of period a. If f(x) is discontinuous or not periodic (i.e. if f(0) ≠ f(a)), or worse if it is singular, the convergence is slow, not uniform, and likely to cause problems in numerical evaluations. At a simple jump of f from 0 to 1, for instance, any partial sum S(x) of the Fourier series will “overshoot” up to 1.179 and show some wiggles. The overshoot and wiggles occur over a narrower range as more terms are included in S, but never die out. One should be wary when using Fourier expansions near jumps and even near cusps. On the other hand for any square-integrable f there is convergence in the mean, i.e. the mean square error

M = ∫_{−a/2}^{a/2} |f(x) − S(x)|^{2} dx

can be made arbitrarily small. In practice, this means that most integrals involving f can also be evaluated with arbitrarily small error.

5.2. Normalization

The eigenfunctions exp(ik_{n}x) are mutually orthogonal but are normalized to a:

∫_{−a/2}^{a/2} [exp(ik_{m}x)]^{∗} exp(ik_{n}x) dx = ∫_{−a/2}^{a/2} exp(2πi(n − m)x/a) dx = aδ_{nm}

This is not a problem, it simply entails that we must use dx/a in integrals such as (5.2). On the other hand, we may prefer to introduce normalized eigenfunctions

U_{n}(x) = \frac{1}{\sqrt{a}} exp(ik_{n}x)

and obtain, in place of (5.1) and (5.2), the more symmetric pair

f(x) = ∑_{n=−∞}^{∞} A_{n}U_{n}(x)  (5.3)

A_{n} = ∫_{−a/2}^{a/2} U_{n}^{∗}(x)f(x) dx  (5.4)

Often one deals with functions defined over all space, i.e. with the limit a → ∞. The limit can be tricky and sometimes it is better to keep a finite until the very end of the calculation. But usually one can simply make the replacements k_{n} → k and

∑_{n=−∞}^{∞} → \frac{a}{2π} ∫_{−∞}^{∞} dk

The most usual way to write the resulting Fourier-integral transform pair is

f(x) = \frac{1}{2π} ∫_{−∞}^{∞} f(k) e^{ikx}dk

f(k) = ∫_{−∞}^{∞} f(x) e^{−ikx}dx

where f(k) corresponds to aC_{n} of eq. (5.2). Mathematicians (and Jackson, at times) have the habit of using a factor 1/\sqrt{2π} in front of each integral, instead of leaving the entire 1/2π attached to the k integral.

5.3. Other boundary conditions

In place of the basis functions exp(ik_{n}x) one can use cos(k_{n}x) and sin(k_{n}x) with k_{n} ≥ 0. This serves only to produce messier formulas.

On the other hand, there are cases where the functions of interest are required to vanish at the boundaries (−a/2, a/2) or (0, a). The relevant normalized eigenfunctions in (0, a) are

U_{n}(x) = \sqrt{\frac{2}{a}} sin(q_{n}x)

with

q_{n} = nπ/a, n positive integer

The transform pair (5.3), (5.4) is still valid in this case and is in fact entirely general for expansions in orthonormal eigenfunctions.

5.4. Higher dimensions

In 3 dimensions, the wave equation leads to the Helmholtz equation

∇^{2}ψ = −k^{2}ψ

which has plane wave solutions exp(ik · x). Periodic boundary conditions on a box of sides a_{x}, a_{y}, a_{z} are not physically attainable in this case, but they are mathematically convenient. They lead to the set of orthogonal functions exp(ik_{n}·x), with the j Cartesian component of k_{n} given by

(k_{n})_{j} = 2πn_{j}/a_{j}, n_{j} integer

The expansion of a function f(x) is

f(x) = ∑_{n_{x}=−∞}^{∞} ∑_{n_{y}=−∞}^{∞} ∑_{n_{z}=−∞}^{∞} C_{n} exp(ik_{n} · x)  (5.5)

with

C_{n} = ∫_{−a_{x}/2}^{a_{x}/2} \frac{dx}{a_{x}} ∫_{−a_{y}/2}^{a_{y}/2} \frac{dy}{a_{y}} ∫_{−a_{z}/2}^{a_{z}/2} \frac{dz}{a_{z}} exp(−ik_{n} · x)f(x)  (5.6)

In the continuum limit we obtain the Fourier integral representation

f(x) = \frac{1}{(2π)^{3}} ∫ f(k) e^{ik·x}d^{3}k

f(k) = ∫ f(x) e^{−ik·x}d^{3}x

by the replacements

∑_{n_{x}=−∞}^{∞} ∑_{n_{y}=−∞}^{∞} ∑_{n_{z}=−∞}^{∞} → \frac{V}{(2π)^{3}} ∫ d^{3}k

C_{n}V → f(k)

where V = a_{x}a_{y}a_{z} is the volume of the “quantization box” and the integrals are over all space.

Similar formulae apply in two dimensions.

5.5. The Dirac delta function

This is a particularly important singular function. In one dimension, it has the general expression

δ(x − x′) = ∑_{all n} U_{n}^{∗}(x′)U_{n}(x)

for any complete set of eigenfunctions U_{n}. For the plane waves exp(ik_{n}x) this becomes

δ(x − x′) = \frac{1}{a} ∑_{n=−∞}^{∞} exp [ik_{n}(x − x′)]  (5.7)

and in the continuum limit

δ(x − x′) = \frac{1}{2π} ∫ e^{ik(x−x′)}dk

The three-dimensional generalization is the usual δ(x) =δ(x)δ(y)δ(z), so that, for instance

δ(x − x′) = \frac{1}{(2π)^{3}} ∫ e^{ik·(x−x′)}d^{3}k

It is worth noting that the partial sum S_{N} of (5.7) has a central peak that becomes higher and narrower for larger N, but also a lot of side wiggles that never go away:

S_{N}(x) = \frac{1}{a} \frac{e^{−i2πNx/a} − e^{i2πNx/a}}{1 − e^{i2πx/a}} = \frac{sin [(2N + 1)πx/a]}{a sin(πx/a)}

5.6. Application to electrostatics

How do we use all this math in electrostatics, where in the simplest case we need solutions of Laplace’s equation, corresponding to k = 0 in the Helmholtz equation, with appropriate boundary conditions? Let us learn from examples.

5.6.1. Rectangular box

Suppose the potential is assigned on the sides of an empty conducting rectangular box of sides a, b, c. It is enough to consider the case where Φ is not zero on one face only: the general case can be built by superposition. Suppose then that Φ = V (x, y) on the face z = c, and vanishes on the other faces (Fig. 2.9). We Fourier-expand in the dimensions x and y, cleverly picking eigenfunctions that vanish on the lateral faces:

Φ(x, y, z) = ∑_{mn} sin(mπx/a) sin (nπy/b) A_{mn}(z)

The expansion coefficients are now functions of z, to be determined by using Laplace’s equation and the remaining boundary conditions on the top and bottom faces.

• From ∇^{2}Φ = 0 for all x and y it follows that

[− (\frac{mπ}{a} )^{2} − (\frac{nπ}{b} )^{2} + \frac{∂^{2}}{∂z^{2}} ] A_{mn}(z) = 0

which has the solutions exp(±γ_{mn}z), with

γ_{mn} = \sqrt{(\frac{mπ}{a} )^{2} + (\frac{nπ}{b} )^{2}}

• To satisfy Φ = 0 on the z = 0 face, we pick the combination

A_{mn}(z) = A_{mn} sinh(γ_{mn}z)

• To satisfy the condition on the z = c face, which is

∑_{mn} sin(mπx/a) sin (nπy/b) A_{mn} sinh(γ_{mn}c) = V (x, y)

we use Fourier’s theorem and find

A_{mn} = \frac{4}{ab sinh(γ_{mn}c)} ∫_{0}^{a} dx ∫_{0}^{b} dy V (x, y) sin(mπx/a) sin (nπy/b)

5.6.2. Remarks

• The complete solution was easily obtained because:

– The starting equation (Laplace’s equation in this case) is completely separable in the chosen variables x, y, z. That means that there are solutions of the form X(x)Y (y) Z(z). In our case the differential operator is the sum of three separate pieces that depend only on x, y and z respectively. This is the simplest case of separability.

– The boundary conditions are also separable in the same variables x, y, z.

• Even partial separability may help. For instance, we may be able to expand in one dimension only, thus reducing the problem to two-dimensional.

• We are still left with an infinite double sum that must usually be evaluated numerically, with all the caveats about the behavior at discontinuities. It is worthwhile to look for approximations that permit the series to be evaluated analytically near singular points.

• The physical meaning of the solution is easiest to see if we consider the slightly different problem where Φ is assigned on the bottom face and the top face is taken to infinity. The solution is then

Φ(x, y, z) = ∑_{mn} A_{mn} sin(mπx/a) sin (nπy/b) exp(−γ_{mn}z)

with

A_{mn} = \frac{4}{ab} ∫_{0}^{a} dx ∫_{0}^{b} dyV (x, y) sin(mπx/a) sin (nπy/b)

Note that exp(ikx) represents a travelling wave, or a free photon, and sin(kx) represents a standing wave, or a confined photon. What does exp(−γz) represent? It is an evanescent wave, or a virtual photon that does not have enough energy to penetrate into a classically forbidden region. Photons are trying to squeeze into the box, but fail. If they adjust themselves to the x and y boundary conditions, they cannot make headway in the z direction. (If c is finite there is backreflection at the far end, giving sinh(γz) or the like).

• It would not be convenient in our case, but we could have expanded in y and z, for instance, and would then find evanescent waves in the x direction. This is another of many equivalent ways to describe the field: picture it as photons trying to come in from the x = 0 and x = a faces.

5.6.3. Rectangular trough

This is a two-dimensional problem: nothing depends on z. The boundary conditions are that Φ vanishes at x = 0 and x = a, as well as at y = ∞, and it has an assigned value V (x) at y = 0 (Fig. 2.10). The solution is

Φ(x, y) = ∑_{n=1}^{∞} A_{n} sin(nπx/a) exp(−nπy/a)

with

A_{n} = \frac{2}{a} ∫_{0}^{a} V (x) sin(nπx/a) dx

For instance, if V (x) is constant, we have

A_{n} = \frac{4V}{nπ} for odd n

and 0 otherwise. In simple cases like this the series can be summed. As in most twodimensional problems, it is convenient to work with the complex variable

z = x + iy

instead of x and y. In our case we note that

Φ = \frac{4V}{π} Im ∑_{\begin{matrix}n =1 \\ odd \end{matrix}}^{∞} \frac{1}{n} exp(inπz/a)

We put Z = e^{iπz/a} and recall that

ln(1 + Z) = ∑_{n=1}^{∞} \frac{1}{n}(−1)^{n+1}Z^{n}
ln(1 − Z) = ∑_{n=1}^{∞} \frac{1}{n}Z^{n}

so that

ln \frac{1 + Z}{1 − Z} = 2 ∑_{\begin{matrix}n =1 \\ odd \end{matrix}}^{∞} \frac{1}{n}Z^{n}

Then

Φ = \frac{2V}{π} Im (ln \frac{1 + Z}{1 − Z} ) = \frac{2V}{π} arctan (\frac{2 Im Z}{1 − |Z|^{2}} )

With Im Z = e^{−πy/a} sin(πx/a) and |Z|^{2} = e^{−2πy/a}, we obtain finally

Φ = \frac{2V}{π} arctan (\frac{sin(πx/a)}{sinh(πy/a)})

Plots of this potential are in Jackson’s Fig. 2.11. In the corner near the origin, Φ ≃ V (1 − 2φ/π), where φ = arctan(y/x) as usual. The field is azimuthal, E_{φ} = 2V/πρ. The energy density diverges as 1/ρ^{2} and the total energy for ρ > ρ_{0} also diverges as ρ_{0} → 0, although only logarithmically. More generally, it costs infinite energy to join two plates at any angle α when there is a potential difference V between them. In practice, a small gap must be left and the energy of the adjoining plates goes like (V/α)^{2} ln w per unit length of the common edge, where w the width of the gap. This logarithmic divergence also occurs for bent plates or shells with curved adjoining edges.

Jackson has many problems and examples of adjoining plates and shells. Such problems are not realistic, for the reason discussed above, and will be omitted in the following.