## Rectangular wave guides: TE modes and the cutoff frequency

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.28.

The possible TE modes in a rectangular wave guide are given by

$\displaystyle B_{z}\left(x,y\right)=B_{0}\cos\frac{m\pi x}{a}\cos\frac{n\pi y}{b} \ \ \ \ \ (1)$

where ${m}$ and ${n}$ are integers and ${a}$ and ${b}$ are the dimensions of the rectangle, with ${a\ge b}$. These values are related to the wave number by

$\displaystyle k=\sqrt{\frac{\omega^{2}}{c^{2}}-\pi^{2}\left(\frac{m^{2}}{a^{2}}+\frac{n^{2}}{b^{2}}\right)} \ \ \ \ \ (2)$

so any modes where

$\displaystyle \omega^{2}

are not allowed, as they would give an imaginary value of ${k}$. The frequency

$\displaystyle \omega_{mn}=\pi c\sqrt{\frac{m^{2}}{a^{2}}+\frac{n^{2}}{b^{2}}} \ \ \ \ \ (4)$

is therefore the cutoff frequency for the mode ${\mbox{TE}_{mn}}$.

As an example, suppose we had a wave guide with ${a=2.28\mbox{ cm}}$, ${b=1.01\mbox{ cm}}$ and a frequency of ${\omega=1.70\times10^{10}\mbox{ Hz}}$. Then the condition for a real ${k}$ is

 $\displaystyle \frac{\omega^{2}}{\pi^{2}c^{2}}$ $\displaystyle \ge$ $\displaystyle \frac{m^{2}}{a^{2}}+\frac{n^{2}}{b^{2}}\ \ \ \ \ (5)$ $\displaystyle \frac{4\nu^{2}}{c^{2}}$ $\displaystyle \ge$ $\displaystyle \frac{m^{2}}{a^{2}}+\frac{n^{2}}{b^{2}}\ \ \ \ \ (6)$ $\displaystyle 1.284\times10^{4}$ $\displaystyle \ge$ $\displaystyle 10^{4}\left(\frac{m^{2}}{5.2}+\frac{n^{2}}{1.02}\right)\ \ \ \ \ (7)$ $\displaystyle 1.02m^{2}+5.2n^{2}$ $\displaystyle \le$ $\displaystyle 6.813 \ \ \ \ \ (8)$

Since at least one of ${m}$ and ${n}$ must be non-zero, the allowed modes are ${\mbox{TE}_{10}}$, ${\mbox{TE}_{20}}$, ${\mbox{TE}_{01}}$ and ${\mbox{TE}_{11}}$ (where the first subscript is ${m}$ and the second is ${n}$).

If we wish to allow only one mode, this would have to be ${\mbox{TE}_{10}}$ (since that gives the lowest value on the LHS). The angular frequency range that would do this is between ${\omega_{10}}$ and ${\omega_{20}}$, that is

$\displaystyle \omega_{10}=\frac{\pi c}{a}\le\omega<\frac{2\pi c}{a}=\omega_{20} \ \ \ \ \ (9)$

For the values above

 $\displaystyle \omega_{10}$ $\displaystyle =$ $\displaystyle 4.134\times10^{10}\mbox{ s}^{-1}\ \ \ \ \ (10)$ $\displaystyle \omega_{20}$ $\displaystyle =$ $\displaystyle 8.268\times10^{10}\mbox{ s}^{-1} \ \ \ \ \ (11)$

The corresponding frequencies and wavelengths are

 $\displaystyle \nu_{10}$ $\displaystyle =$ $\displaystyle \frac{\omega_{10}}{2\pi}=6.58\times10^{9}\mbox{ Hz}\ \ \ \ \ (12)$ $\displaystyle v_{20}$ $\displaystyle =$ $\displaystyle 1.32\times10^{10}\mbox{ Hz}\ \ \ \ \ (13)$ $\displaystyle \lambda_{10}$ $\displaystyle =$ $\displaystyle \frac{c}{\nu_{10}}=0.0456\mbox{ m}\ \ \ \ \ (14)$ $\displaystyle \lambda_{20}$ $\displaystyle =$ $\displaystyle 0.0228\mbox{ m} \ \ \ \ \ (15)$

## Rectangular wave guides: transverse electric waves

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.27.

The wave equations for an electromagnetic wave in a wave guide are

 $\displaystyle \left[\partial_{xx}+\partial_{yy}+\omega^{2}/c^{2}-k^{2}\right]E_{z}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (1)$ $\displaystyle \left[\partial_{xx}+\partial_{yy}+\omega^{2}/c^{2}-k^{2}\right]B_{z}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (2)$

If ${E_{z}=0}$ we have a transverse electric or TE wave, so we need to solve only the ${B_{z}}$ equation. For a rectangular wave guide with dimensions of ${a}$ in the ${x}$ direction and ${b}$ in the ${y}$ direction, we can use separation of variables to get a solution (the technique is mathematically the same as that used in solving the infinite square well in quantum mechanics). That is, we assume that

$\displaystyle B_{z}\left(x,y\right)=X\left(x\right)Y\left(y\right) \ \ \ \ \ (3)$

and substitute this into 2, then divide through by ${XY}$:

$\displaystyle \frac{X"}{X}+\frac{Y"}{Y}+\frac{\omega^{2}}{c^{2}}-k^{2}=0 \ \ \ \ \ (4)$

Because this equation must hold for all values of ${x}$ and ${y}$, the two derivative terms must separately be constant, so we have

 $\displaystyle X"$ $\displaystyle =$ $\displaystyle -k_{x}^{2}X\ \ \ \ \ (5)$ $\displaystyle Y"$ $\displaystyle =$ $\displaystyle -k_{y}^{2}Y \ \ \ \ \ (6)$

for some constants ${k_{x}}$ and ${k_{y}}$, which satisfy

$\displaystyle -k_{x}^{2}-k_{y}^{2}+\frac{\omega^{2}}{c^{2}}-k^{2}=0 \ \ \ \ \ (7)$

The general solution to these ODEs can be written as either trigonometric functions or complex exponentials. If we choose trig functions, we have

 $\displaystyle X\left(x\right)$ $\displaystyle =$ $\displaystyle A\sin k_{x}x+B\cos k_{x}x\ \ \ \ \ (8)$ $\displaystyle Y\left(y\right)$ $\displaystyle =$ $\displaystyle C\sin k_{y}y+D\cos k_{y}y \ \ \ \ \ (9)$

The boundary conditions require that at ${x=0}$ and ${x=a}$

 $\displaystyle B_{1}^{\perp}-B_{2}^{\perp}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (10)$

The component of ${\mathbf{B}}$ perpendicular to the walls of the guide in the ${x}$ direction is ${B_{x}}$, and since ${\mathbf{B}=0}$ inside the conducting wall, this condition requires ${B_{x}=0}$. In our derivation of the wave equation for a wave guide, we found that

$\displaystyle B_{x}=\frac{i}{\omega^{2}/c^{2}-k^{2}}\left(k\partial_{x}B_{z}-\frac{\omega}{c^{2}}\partial_{y}E_{z}\right) \ \ \ \ \ (11)$

and since ${E_{z}=0}$, we must also have ${\partial_{x}B_{z}=0}$, which in turn requires that

$\displaystyle X'\left(0\right)=X'\left(a\right)=0 \ \ \ \ \ (12)$

From 8, this means that ${A=0}$ and

$\displaystyle -k_{x}B\sin k_{x}a=0 \ \ \ \ \ (13)$

so

$\displaystyle k_{x}=\frac{m\pi}{a} \ \ \ \ \ (14)$

for ${m}$ equal to a non-negative integer.

Exactly the same analysis on ${Y}$ gives us

$\displaystyle k_{y}=\frac{n\pi}{b} \ \ \ \ \ (15)$

so the separation of variables solution gives us

$\displaystyle B_{z}\left(x,y\right)=B_{0}\cos\frac{m\pi x}{a}\cos\frac{n\pi y}{b} \ \ \ \ \ (16)$

This is known as the ${\mbox{TE}_{mn}}$ mode. From 7 we get the wave number

$\displaystyle k=\sqrt{\frac{\omega^{2}}{c^{2}}-\pi^{2}\left(\frac{m^{2}}{a^{2}}+\frac{n^{2}}{b^{2}}\right)} \ \ \ \ \ (17)$

In fact, at least one of ${m}$ or ${n}$ must be non-zero, as we can see by the following argument. Suppose ${m=n=0}$. Then ${k=\frac{\omega}{c}}$. We need the results we got in the previous post:

 $\displaystyle \partial_{y}E_{z}-ikE_{y}$ $\displaystyle =$ $\displaystyle i\omega B_{x}\ \ \ \ \ (18)$ $\displaystyle -\partial_{x}E_{z}+ikE_{x}$ $\displaystyle =$ $\displaystyle i\omega B_{y}\ \ \ \ \ (19)$ $\displaystyle \partial_{x}E_{y}-\partial_{y}E_{x}$ $\displaystyle =$ $\displaystyle i\omega B_{z} \ \ \ \ \ (20)$

 $\displaystyle \partial_{y}B_{z}-ikB_{y}$ $\displaystyle =$ $\displaystyle -i\frac{\omega}{c^{2}}E_{x}\ \ \ \ \ (21)$ $\displaystyle -\partial_{x}B_{z}+ikB_{x}$ $\displaystyle =$ $\displaystyle -i\frac{\omega}{c^{2}}E_{y}\ \ \ \ \ (22)$ $\displaystyle \partial_{x}B_{y}-\partial_{y}B_{x}$ $\displaystyle =$ $\displaystyle -i\frac{\omega}{c^{2}}E_{z} \ \ \ \ \ (23)$

With ${k=\frac{\omega}{c}}$ and ${E_{z}=0}$ we get from 21 and 19

 $\displaystyle \partial_{y}B_{z}-i\frac{\omega}{c}B_{y}$ $\displaystyle =$ $\displaystyle -i\frac{\omega}{c^{2}}E_{x}\ \ \ \ \ (24)$ $\displaystyle i\frac{\omega}{c^{2}}E_{x}$ $\displaystyle =$ $\displaystyle i\frac{\omega}{c}B_{y} \ \ \ \ \ (25)$

$\displaystyle \partial_{y}B_{z}=0 \ \ \ \ \ (26)$

Similarly, from 22 and 18 we get

$\displaystyle \partial_{x}B_{z}=0 \ \ \ \ \ (27)$

Therefore, ${B_{z}}$ is constant in both the ${x}$ and ${y}$ directions, so it is a constant overall. To find what constant this is, we can use Faraday’s law in integral form:

$\displaystyle \oint\mathbf{E}\cdot d\boldsymbol{\ell}=-\frac{d\Phi}{dt}=-\int\frac{\partial\mathbf{B}}{\partial t}\cdot d\mathbf{a} \ \ \ \ \ (28)$

The area of integration on the RHS that we’ll choose is a cross-section of the wave guide, so the path of integration on the LHS is around the rectangular boundary of the guide. We know that

$\displaystyle \mathbf{B}=\mathbf{B}_{0}e^{i\left(kz-\omega t\right)} \ \ \ \ \ (29)$

so

$\displaystyle \frac{\partial\mathbf{B}}{\partial t}=-i\omega\mathbf{B}_{0}e^{i\left(kz-\omega t\right)} \ \ \ \ \ (30)$

In this case ${d\mathbf{a}}$ points in the ${z}$ direction, so we get

 $\displaystyle \oint\mathbf{E}\cdot d\boldsymbol{\ell}$ $\displaystyle =$ $\displaystyle i\omega e^{i\left(kz-\omega t\right)}\int B_{z}da\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\omega abe^{i\left(kz-\omega t\right)}B_{z} \ \ \ \ \ (32)$

since ${B_{z}}$ is constant over the area of integration.

Since we’ve chosen the path of integration to be the boundary of the wave guide, we can use the boundary condition

$\displaystyle \mathbf{E}_{1}^{\parallel}-\mathbf{E}_{2}^{\parallel}=0 \ \ \ \ \ (33)$

The field inside the conducting wall of the wave guide is zero, so the parallel field at the boundary must also be zero and, since ${\mathbf{E}\cdot d\boldsymbol{\ell}=E^{\parallel}d\ell}$, the line integral must also be zero, so we must have

$\displaystyle B_{z}=0 \ \ \ \ \ (34)$

That is, if ${m=n=0}$, both the electric and magnetic fields must be transverse (known as TEM mode). Griffiths shows in section 9.5 of his book that in this case, for a hollow wave guide of any cross section, the electric field must actually be zero everywhere, which means no wave can propagate down the guide. Thus the ${TE_{00}}$ mode cannot exist in a hollow wave guide.

## Wave guides: derivation of the wave equation

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.26.

A wave guide is a hollow tube which allows electromagnetic waves to travel down it. Wave guides are usually made of conductors, so we’ll assume that they are made of a perfect conductor so that ${\mathbf{E}=0}$ everywhere inside the conducting boundary.

The boundary conditions implied by Maxwell’s equations are

 $\displaystyle \epsilon_{1}E_{1}^{\perp}-\epsilon_{2}E_{2}^{\perp}$ $\displaystyle =$ $\displaystyle \sigma_{f}\ \ \ \ \ (1)$ $\displaystyle B_{1}^{\perp}-B_{2}^{\perp}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (2)$ $\displaystyle \mathbf{E}_{1}^{\parallel}-\mathbf{E}_{2}^{\parallel}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (3)$ $\displaystyle \frac{1}{\mu_{1}}\mathbf{B}_{1}^{\parallel}-\frac{1}{\mu_{2}}\mathbf{B}_{2}^{\parallel}$ $\displaystyle =$ $\displaystyle \mathbf{K}_{f}\times\hat{\mathbf{n}} \ \ \ \ \ (4)$

In particular, 3 tells us that the parallel component of ${\mathbf{E}}$ is zero at the boundary of the wave guide. As usual, we’ll take the ${z}$ axis to be parallel to wave guide’s axis, so the waves have the form

 $\displaystyle \tilde{\mathbf{E}}$ $\displaystyle =$ $\displaystyle \tilde{\mathbf{E}}_{0}\left(x,y\right)e^{i\left(kz-\omega t\right)}\ \ \ \ \ (5)$ $\displaystyle \tilde{\mathbf{B}}$ $\displaystyle =$ $\displaystyle \tilde{\mathbf{B}}_{0}\left(x,y\right)e^{i\left(kz-\omega t\right)} \ \ \ \ \ (6)$

Maxwell’s equations inside the guide (assumed to be a vacuum) are

 $\displaystyle \nabla\cdot\mathbf{E}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (7)$ $\displaystyle \nabla\cdot\mathbf{B}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (8)$ $\displaystyle \nabla\times\mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\partial\mathbf{B}}{\partial t}\ \ \ \ \ (9)$ $\displaystyle \nabla\times\mathbf{B}$ $\displaystyle =$ $\displaystyle \frac{1}{c^{2}}\frac{\partial\mathbf{E}}{\partial t} \ \ \ \ \ (10)$

Previously, we applied the divergence equations to show that the waves were transverse (no ${z}$ component), but that relied on the waves being unbounded plane waves, which is not the case here. It turns out that waves in a wave guide are not transverse in general, in that at least one of ${\mathbf{E}}$ and ${\mathbf{B}}$ must have a longitudinal component. We therefore write

 $\displaystyle \tilde{\mathbf{E}}_{0}\left(x,y\right)$ $\displaystyle =$ $\displaystyle E_{x}\hat{\mathbf{x}}+E_{y}\hat{\mathbf{y}}+E_{z}\hat{\mathbf{z}}\ \ \ \ \ (11)$ $\displaystyle \tilde{\mathbf{B}}_{0}\left(x,y\right)$ $\displaystyle =$ $\displaystyle B_{x}\hat{\mathbf{x}}+B_{y}\hat{\mathbf{y}}+B_{z}\hat{\mathbf{z}} \ \ \ \ \ (12)$

where all the components on the RHS depend on ${x}$ and ${y,}$ and may be complex functions. Putting these together with 5 and 6 into 9, we get (remembering that the components do not depend on ${z}$):

 $\displaystyle \nabla\times\mathbf{E}$ $\displaystyle =$ $\displaystyle \left(\partial_{y}E_{z}-ikE_{y}\right)\hat{\mathbf{x}}+\left(-\partial_{x}E_{z}+ikE_{x}\right)\hat{\mathbf{y}}+\left(\partial_{x}E_{y}-\partial_{y}E_{x}\right)\hat{\mathbf{z}}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\partial\mathbf{B}}{\partial t}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\omega B_{x}\hat{\mathbf{x}}+i\omega B_{y}\hat{\mathbf{y}}+i\omega B_{z}\hat{\mathbf{z}} \ \ \ \ \ (15)$

Equating components, we get

 $\displaystyle \partial_{y}E_{z}-ikE_{y}$ $\displaystyle =$ $\displaystyle i\omega B_{x}\ \ \ \ \ (16)$ $\displaystyle -\partial_{x}E_{z}+ikE_{x}$ $\displaystyle =$ $\displaystyle i\omega B_{y}\ \ \ \ \ (17)$ $\displaystyle \partial_{x}E_{y}-\partial_{y}E_{x}$ $\displaystyle =$ $\displaystyle i\omega B_{z} \ \ \ \ \ (18)$

We can apply exactly the same procedure to 10 to get the analogous equations

 $\displaystyle \partial_{y}B_{z}-ikB_{y}$ $\displaystyle =$ $\displaystyle -i\frac{\omega}{c^{2}}E_{x}\ \ \ \ \ (19)$ $\displaystyle -\partial_{x}B_{z}+ikB_{x}$ $\displaystyle =$ $\displaystyle -i\frac{\omega}{c^{2}}E_{y}\ \ \ \ \ (20)$ $\displaystyle \partial_{x}B_{y}-\partial_{y}B_{x}$ $\displaystyle =$ $\displaystyle -i\frac{\omega}{c^{2}}E_{z} \ \ \ \ \ (21)$

We can solve these 6 equations to get the ${x}$ and ${y}$ components in terms of the ${z}$ components. For example, multiplying 17 through by ${k}$ and 19 through by ${\omega}$ and adding, we get

 $\displaystyle -k\partial_{x}E_{z}+ik^{2}E_{x}-i\frac{\omega^{2}}{c^{2}}E_{x}$ $\displaystyle =$ $\displaystyle \omega\partial_{y}B_{z}\ \ \ \ \ (22)$ $\displaystyle E_{x}$ $\displaystyle =$ $\displaystyle \frac{1}{i\left(k^{2}-\omega^{2}/c^{2}\right)}\left(k\partial_{x}E_{z}+\omega\partial_{y}B_{z}\right)\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{i}{\omega^{2}/c^{2}-k^{2}}\left(k\partial_{x}E_{z}+\omega\partial_{y}B_{z}\right) \ \ \ \ \ (24)$

Similarly, we can get the other 3 equations. Multiply 16 by ${k}$ and 20 by ${\omega}$ and subtract to get

$\displaystyle E_{y}=\frac{i}{\omega^{2}/c^{2}-k^{2}}\left(k\partial_{y}E_{z}-\omega\partial_{x}B_{z}\right) \ \ \ \ \ (25)$

Multiply 16 by ${\omega/c^{2}}$ and 20 by ${k}$ and subtract to get

$\displaystyle B_{x}=\frac{i}{\omega^{2}/c^{2}-k^{2}}\left(k\partial_{x}B_{z}-\frac{\omega}{c^{2}}\partial_{y}E_{z}\right) \ \ \ \ \ (26)$

Multiply 17 by ${\omega/c^{2}}$ and 19 by ${k}$ and add to get

$\displaystyle B_{y}=\frac{i}{\omega^{2}/c^{2}-k^{2}}\left(k\partial_{y}B_{z}+\frac{\omega}{c^{2}}\partial_{x}E_{z}\right) \ \ \ \ \ (27)$

To get the wave equations we can apply 7 and 8:

 $\displaystyle \nabla\cdot\mathbf{E}$ $\displaystyle =$ $\displaystyle \frac{i}{\omega^{2}/c^{2}-k^{2}}\left[\left(k\partial_{xx}E_{z}+\omega\partial_{yx}B_{z}\right)+\left(k\partial_{yy}E_{z}-\omega\partial_{xy}B_{z}\right)\right]+ikE_{z}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{i}{\omega^{2}/c^{2}-k^{2}}\left[k\partial_{xx}E_{z}+k\partial_{yy}E_{z}\right]+ikE_{z}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (30)$

The wave equation for ${E_{z}}$ is thus

$\displaystyle \left[\partial_{xx}+\partial_{yy}+\omega^{2}/c^{2}-k^{2}\right]E_{z}=0 \ \ \ \ \ (31)$

Exactly the same procedure applied to ${\nabla\cdot\mathbf{B}=0}$ gives

$\displaystyle \left[\partial_{xx}+\partial_{yy}+\omega^{2}/c^{2}-k^{2}\right]B_{z}=0 \ \ \ \ \ (32)$

Solving these two equations subject to the boundary conditions will give us all 3 components of each field.

## Group velocity of electromagnetic waves in a dispersive medium

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.25.

In a dispersive medium, the permittivity is a complex quantity given by

$\displaystyle \tilde{\epsilon}=\epsilon_{0}+\frac{Nq^{2}}{m}\sum_{j}\frac{f_{j}}{\omega_{j}^{2}-\omega^{2}-i\omega\gamma_{j}} \ \ \ \ \ (1)$

and the wave vector is also complex:

$\displaystyle \tilde{k}=\sqrt{\tilde{\epsilon}\mu}\omega \ \ \ \ \ (2)$

If the sum term in 1 is small compared to ${\epsilon_{0}}$, we can approximate it in the square root using ${\sqrt{1+x}\approx1+\frac{1}{2}x}$.

 $\displaystyle \tilde{k}$ $\displaystyle =$ $\displaystyle \omega\sqrt{\mu\epsilon_{0}}\left[1+\frac{Nq^{2}}{m\epsilon_{0}}\sum_{j}\frac{f_{j}}{\omega_{j}^{2}-\omega^{2}-i\omega\gamma_{j}}\right]^{1/2}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle \frac{\omega}{c}\left(1+\frac{Nq^{2}}{2m\epsilon_{0}}\sum_{j}\frac{f_{j}}{\omega_{j}^{2}-\omega^{2}-i\omega\gamma_{j}}\right) \ \ \ \ \ (4)$

where we’ve taken ${\mu\approx\mu_{0}}$ and used ${\sqrt{\mu_{0}\epsilon_{0}}=\frac{1}{c}}$.

The solution to the wave equation is

$\displaystyle \tilde{\mathbf{E}}\left(z,t\right)=\tilde{\mathbf{E}}_{0}e^{i\left(\tilde{k}z-\omega t\right)} \ \ \ \ \ (5)$

If the damping factors ${\gamma_{j}}$ are all small and we’re away from any resonance frequencies (${\omega_{j}}$), then we can ignore the imaginary term in the denominator and get a real wave vector:

$\displaystyle k=\frac{\omega}{c}\left(1+\frac{Nq^{2}}{2m\epsilon_{0}}\sum_{j}\frac{f_{j}}{\omega_{j}^{2}-\omega^{2}}\right) \ \ \ \ \ (6)$

We can get the group velocity ${\frac{d\omega}{dk}}$ of a wave packet by implicit differentiation:

 $\displaystyle 1$ $\displaystyle =$ $\displaystyle \frac{d\omega}{dk}\left[\frac{1}{c}+\frac{Nq^{2}}{2m\epsilon_{0}c}\left(\sum_{j}\frac{f_{j}}{\omega_{j}^{2}-\omega^{2}}+\omega\sum_{j}\frac{2f_{j}\omega}{\left(\omega_{j}^{2}-\omega^{2}\right)^{2}}\right)\right]\ \ \ \ \ (7)$ $\displaystyle \frac{d\omega}{dk}$ $\displaystyle =$ $\displaystyle c\left[1+\frac{Nq^{2}}{2m\epsilon_{0}}\sum_{j}\frac{f_{j}}{\omega_{j}^{2}-\omega^{2}}\left(1+\frac{2\omega^{2}}{\omega_{j}^{2}-\omega^{2}}\right)\right]^{-1}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle c\left[1+\frac{Nq^{2}}{2m\epsilon_{0}}\sum_{j}\frac{f_{j}}{\omega_{j}^{2}-\omega^{2}}\left(\frac{\omega_{j}^{2}-\omega^{2}}{\omega_{j}^{2}-\omega^{2}}+\frac{2\omega^{2}}{\omega_{j}^{2}-\omega^{2}}\right)\right]^{-1}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle c\left[1+\frac{Nq^{2}}{2m\epsilon_{0}}\sum_{j}\frac{f_{j}\left(\omega_{j}^{2}+\omega^{2}\right)}{\left(\omega_{j}^{2}-\omega^{2}\right)^{2}}\right]^{-1} \ \ \ \ \ (10)$

Since everything inside the square brackets is positive, we see that ${\frac{d\omega}{dk} for all frequencies. On the other hand, from 6 we see that the phase velocity is

$\displaystyle \frac{\omega}{k}=c\left(1+\frac{Nq^{2}}{2m\epsilon_{0}}\sum_{j}\frac{f_{j}}{\omega_{j}^{2}-\omega^{2}}\right)^{-1} \ \ \ \ \ (11)$

which can actually exceed ${c}$ for some frequencies near the resonant frequencies, since the denominator in the sum can be negative there. Griffiths states that ordinarily, energy (and thus information) carried by the wave travels at the group velocity which is always less that ${c}$. However, it is possible for the group velocity to exceed ${c}$ in some cases.

## Resonances in a dispersive medium

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.24.

In a dispersive medium, the index of refraction is

 $\displaystyle n$ $\displaystyle \approx$ $\displaystyle 1+\frac{Nq^{2}}{2\epsilon_{0}m}\sum_{j}\frac{f_{j}\left(\omega_{j}^{2}-\omega^{2}\right)}{\left(\omega_{j}^{2}-\omega^{2}\right)^{2}+\left(\omega\gamma_{j}\right)^{2}} \ \ \ \ \ (1)$

and the absorption coefficient is

 $\displaystyle \alpha$ $\displaystyle \approx$ $\displaystyle \frac{Nq^{2}\omega^{2}}{c\epsilon_{0}m}\sum_{j}\frac{f_{j}\gamma_{j}}{\left(\omega_{j}^{2}-\omega^{2}\right)^{2}+\left(\omega\gamma_{j}\right)^{2}} \ \ \ \ \ (2)$

We can look at the behaviour of these coefficients near one of the resonances, that is, when ${\omega\approx\omega_{j}}$ for some ${j}$. To simplify things, we’ll assume that there is only one term in the sum (that is, only one resonance). In practice, when we’re near one resonance, the other resonances don’t affect things much unless they are very close together.

In this case, we’ll take the one natural frequency to be ${\omega_{0}}$ and the associated damping coefficient to be ${\gamma_{0}}$, and define

$\displaystyle x\equiv\frac{\omega}{\omega_{0}} \ \ \ \ \ (3)$

Then we get, after dividing top and bottom by ${\omega_{0}^{2}}$:

 $\displaystyle n$ $\displaystyle \approx$ $\displaystyle 1+\frac{Nq^{2}f_{0}}{2\epsilon_{0}m}\frac{1-x^{2}}{\omega_{0}^{2}\left(1-x^{2}\right)^{2}+\gamma_{0}^{2}x^{2}}\ \ \ \ \ (4)$ $\displaystyle \alpha$ $\displaystyle \approx$ $\displaystyle \frac{Nq^{2}f_{0}\gamma_{0}}{m\epsilon_{0}c}\frac{x^{2}}{\omega_{0}^{2}\left(1-x^{2}\right)^{2}+\gamma_{0}^{2}x^{2}} \ \ \ \ \ (5)$

The index of refraction ${n}$ rises to a peak when ${\omega}$ is just before ${\omega_{0}}$, then dips sharply, reaching a minimum just after ${\omega_{0}}$, after which it rises slowly again. We can find the maximum and minimum by setting the derivative to zero and solving for ${x}$. Using Maple to simplify the result, we get

$\displaystyle \frac{dn}{dx}=\frac{Nq^{2}f_{0}}{\epsilon_{0}m}\frac{x\left(x^{4}-2x^{2}+1-\frac{\gamma^{2}}{\omega_{0}^{2}}\right)}{\omega_{0}^{2}\left(x^{4}\left(\frac{\gamma^{2}}{\omega_{0}^{2}}-2\right)x^{2}+1\right)}=0 \ \ \ \ \ (6)$

The roots are

$\displaystyle x=0,\pm\sqrt{1+\frac{\gamma}{\omega_{0}}},\pm\sqrt{1-\frac{\gamma}{\omega_{0}}} \ \ \ \ \ (7)$

The negative and zero roots aren’t of interest, so we see that ${n}$ reaches its maximum at ${x=\sqrt{1-\frac{\gamma}{\omega_{0}}}}$ and minimum at ${x=\sqrt{1+\frac{\gamma}{\omega_{0}}}}$ . If ${\gamma\ll\omega_{0}}$, these approximate to

 $\displaystyle \omega_{n\; max}$ $\displaystyle \approx$ $\displaystyle \omega_{0}-\frac{1}{2}\gamma\ \ \ \ \ (8)$ $\displaystyle \omega_{n\; min}$ $\displaystyle \approx$ $\displaystyle \omega_{0}+\frac{1}{2}\gamma \ \ \ \ \ (9)$

so the width of the anomalous region is ${\gamma}$.

From 5, we see that the absorption reaches a maximum when ${x=1}$ (this can be checked by calculating the derivative) and has a value of

$\displaystyle \alpha_{max}=\frac{Nq^{2}f_{0}}{m\epsilon_{0}c\gamma_{0}} \ \ \ \ \ (10)$

Substituting the positive roots from 7 into 5 and dividing by ${\alpha_{max}}$ we find

 $\displaystyle \frac{\alpha_{n\; max}}{\alpha_{max}}$ $\displaystyle =$ $\displaystyle \frac{\omega_{0}-\gamma}{2\omega_{0}-\gamma}\approx\frac{1}{2}\ \ \ \ \ (11)$ $\displaystyle \frac{\alpha_{n\; min}}{\alpha_{max}}$ $\displaystyle =$ $\displaystyle \frac{\omega_{0}+\gamma}{2\omega_{0}+\gamma}\approx\frac{1}{2} \ \ \ \ \ (12)$

Thus for small ${\gamma}$, the index of refraction reaches its maximum and minimum values roughly where the absorption is half its maximum.

## Refraction and dispersion coefficients

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.23b.

In a dispersive medium, the permittivity depends on the frequency of electromagnetic radiation.

$\displaystyle \tilde{\epsilon}=\epsilon_{0}+\frac{Nq^{2}}{m}\sum_{j}\frac{f_{j}}{\omega_{j}^{2}-\omega^{2}-i\omega\gamma_{j}} \ \ \ \ \ (1)$

where there are ${f_{j}}$ electrons per atom with natural frequency ${\omega_{j}}$ and damping factor ${\gamma_{j}}$, and there are ${N}$ atoms per unit volume. Because ${\tilde{\epsilon}}$ is complex, the medium isn’t linear in the sense that the polarization is directly proportional to the applied field, but if we take both the polarization ${\tilde{\mathbf{P}}}$ and field ${\tilde{\mathbf{E}}}$ to be complex, then the medium is linear in the sense that

$\displaystyle \tilde{\mathbf{P}}=\epsilon_{0}\tilde{\chi}_{e}\tilde{\mathbf{E}} \ \ \ \ \ (2)$

With this assumption, we can substitute the complex permittivity ${\tilde{\epsilon}}$ for the ordinary real permittivity ${\epsilon}$ in Maxwell’s equations and follow through the same steps to get the wave equation, which now becomes

$\displaystyle \nabla^{2}\tilde{\mathbf{E}}=\mu\tilde{\epsilon}\frac{\partial^{2}\tilde{\mathbf{E}}}{\partial t^{2}} \ \ \ \ \ (3)$

Just as before, we can get plane wave solutions of the form

$\displaystyle \tilde{\mathbf{E}}\left(z,t\right)=\tilde{\mathbf{E}}_{0}e^{i\left(\tilde{k}z-\omega t\right)} \ \ \ \ \ (4)$

where ${\tilde{k}}$ is a complex wave vector

$\displaystyle \tilde{k}=\sqrt{\tilde{\epsilon}\mu}\omega \ \ \ \ \ (5)$

The actual real and imaginary parts of ${\tilde{k}}$ are complicated expressions since ${\tilde{\epsilon}}$ is a sum of complex numbers, but we can use the shortcut notation

$\displaystyle \tilde{k}=k+i\kappa \ \ \ \ \ (6)$

giving (assuming ${\mathbf{E}}$ is polarized in the ${x}$ direction):

$\displaystyle \tilde{\mathbf{E}}\left(z,t\right)=\tilde{E}_{0}e^{-\kappa z}e^{i\left(kz-\omega t\right)}\hat{\mathbf{x}} \ \ \ \ \ (7)$

The intensity of the radiation is proportional to ${\left|\tilde{\mathbf{E}}\right|^{2}}$ so the intensity falls off according to ${e^{-2\kappa z}}$ as we penetrate the medium. The absorption coefficient is defined as

$\displaystyle \alpha\equiv2\kappa \ \ \ \ \ (8)$

and gives a measure of the reciprocal of the distance at which the intensity is attenuated.

We can write the complex permittivity in 1 as

 $\displaystyle \tilde{\epsilon}$ $\displaystyle =$ $\displaystyle \epsilon_{0}+\frac{Nq^{2}}{m}\sum_{j}\frac{f_{j}}{\omega_{j}^{2}-\omega^{2}-i\omega\gamma_{j}}\left[\frac{\omega_{j}^{2}-\omega^{2}+i\omega\gamma_{j}}{\omega_{j}^{2}-\omega^{2}+i\omega\gamma_{j}}\right]\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \epsilon_{0}+\frac{Nq^{2}}{m}\sum_{j}\frac{f_{j}\left(\omega_{j}^{2}-\omega^{2}\right)}{\left(\omega_{j}^{2}-\omega^{2}\right)^{2}+\left(\omega\gamma_{j}\right)^{2}}+i\frac{Nq^{2}}{m}\sum_{j}\frac{f_{j}\omega\gamma_{j}}{\left(\omega_{j}^{2}-\omega^{2}\right)^{2}+\left(\omega\gamma_{j}\right)^{2}} \ \ \ \ \ (10)$

If we stay away from the resonant frequencies, where ${\omega\approx\omega_{j}}$, the sum terms are quite small so we can approximate them in 5 by the first order term in a Taylor expansion. If we also take ${\mu\approx\mu_{0}}$ as is true of most materials, and use ${c=1/\sqrt{\mu_{0}\epsilon_{0}}}$, we get

$\displaystyle \tilde{k}=\sqrt{\frac{\tilde{\epsilon}}{\epsilon_{0}c^{2}}}\omega \ \ \ \ \ (11)$

Using ${\sqrt{1+x}\approx1+\frac{1}{2}x}$ for small ${x}$, we get

$\displaystyle \tilde{k}\approx\frac{\omega}{c}\left[1+\frac{Nq^{2}}{2\epsilon_{0}m}\sum_{j}\frac{f_{j}\left(\omega_{j}^{2}-\omega^{2}\right)}{\left(\omega_{j}^{2}-\omega^{2}\right)^{2}+\left(\omega\gamma_{j}\right)^{2}}\right]+i\frac{\omega}{c}\left[\frac{Nq^{2}}{2\epsilon_{0}m}\sum_{j}\frac{f_{j}\omega\gamma_{j}}{\left(\omega_{j}^{2}-\omega^{2}\right)^{2}+\left(\omega\gamma_{j}\right)^{2}}\right] \ \ \ \ \ (12)$

From 7 the speed of the wave is

$\displaystyle v=\frac{\omega}{k} \ \ \ \ \ (13)$

so the index of refraction is

 $\displaystyle n$ $\displaystyle =$ $\displaystyle \frac{c}{v}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle 1+\frac{Nq^{2}}{2\epsilon_{0}m}\sum_{j}\frac{f_{j}\left(\omega_{j}^{2}-\omega^{2}\right)}{\left(\omega_{j}^{2}-\omega^{2}\right)^{2}+\left(\omega\gamma_{j}\right)^{2}} \ \ \ \ \ (15)$

and the absorption coefficient is

 $\displaystyle \alpha$ $\displaystyle =$ $\displaystyle 2\kappa\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle \frac{Nq^{2}\omega^{2}}{c\epsilon_{0}m}\sum_{j}\frac{f_{j}\gamma_{j}}{\left(\omega_{j}^{2}-\omega^{2}\right)^{2}+\left(\omega\gamma_{j}\right)^{2}} \ \ \ \ \ (17)$

If we stay away from resonances, the damping term becomes insignificant so the index of refraction is approximately

$\displaystyle n\approx1+\frac{Nq^{2}}{2\epsilon_{0}m}\sum_{j}\frac{f_{j}}{\omega_{j}^{2}-\omega^{2}} \ \ \ \ \ (18)$

If the frequency ${\omega}$ of the wave is significantly less than all the resonant frequencies ${\omega_{j}}$ we can further approximate this using ${\frac{1}{1-x}\approx1+x}$ for small ${x}$:

 $\displaystyle n$ $\displaystyle \approx$ $\displaystyle 1+\frac{Nq^{2}}{2\epsilon_{0}m}\sum_{j}\frac{f_{j}}{\omega_{j}^{2}\left(1-\omega^{2}/\omega_{j}^{2}\right)}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle 1+\frac{Nq^{2}}{2\epsilon_{0}m}\sum_{j}\frac{f_{j}}{\omega_{j}^{2}}+\omega^{2}\frac{Nq^{2}}{2\epsilon_{0}m}\sum_{j}\frac{f_{j}}{\omega_{j}^{4}} \ \ \ \ \ (20)$

In a vacuum, ${c=\lambda\nu=\lambda\omega/2\pi}$ so

 $\displaystyle n$ $\displaystyle \approx$ $\displaystyle 1+\frac{Nq^{2}}{2\epsilon_{0}m}\sum_{j}\frac{f_{j}}{\omega_{j}^{2}}+\frac{4\pi^{2}c^{2}}{\lambda^{2}}\frac{Nq^{2}}{2\epsilon_{0}m}\sum_{j}\frac{f_{j}}{\omega_{j}^{4}}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1+A\left(1+\frac{B}{\lambda^{2}}\right) \ \ \ \ \ (22)$

where

 $\displaystyle A$ $\displaystyle =$ $\displaystyle \frac{Nq^{2}}{2\epsilon_{0}m}\sum_{j}\frac{f_{j}}{\omega_{j}^{2}}\ \ \ \ \ (23)$ $\displaystyle B$ $\displaystyle =$ $\displaystyle \frac{4\pi^{2}c^{2}}{A}\frac{Nq^{2}}{2\epsilon_{0}m}\sum_{j}\frac{f_{j}}{\omega_{j}^{4}} \ \ \ \ \ (24)$

Eqn 22 is known as the Cauchy formula, although Cauchy had many equations named after him (particularly in the area of complex variable theory), so the name is easily confused with other formulas. The parameter ${A}$ is the coefficient of refraction and ${B}$ is the coefficient of dispersion. The more usual form of Cauchy’s equation seems to be ${n=1+A+\frac{B}{\lambda^{2}}}$ (I couldn’t find any sources that gave the formula in the form used by Griffiths).

Example Applying this model to hydrogen at 0 C and atmospheric pressure (that is, standard temperature and pressure, or STP), the number of electrons per molecule of ${\mbox{H}_{2}}$ is ${f_{j}=2}$. In the previous post, we found that the resonant frequency is ${\omega_{0}=4.13\times10^{16}\mbox{ s}^{-1}}$. At STP, an ideal gas occupies ${22.414\mbox{ m}^{3}\mbox{kmol}^{-1}}$, so the number density is

$\displaystyle N=\frac{\left(6.02\times10^{23}\right)\left(1000\right)}{22.414}=2.69\times10^{25} \ \ \ \ \ (25)$

The parameters are

 $\displaystyle A$ $\displaystyle =$ $\displaystyle \frac{Nq^{2}}{2\epsilon_{0}m}\frac{2}{\omega_{0}^{2}}=5\times10^{-5}\ \ \ \ \ (26)$ $\displaystyle B$ $\displaystyle =$ $\displaystyle \frac{4\pi^{2}c^{2}}{A}\frac{Nq^{2}}{2\epsilon_{0}m}\frac{2}{\omega_{0}^{4}}=\frac{4\pi^{2}c^{2}}{\omega_{0}^{2}}=2.08\times10^{-15}\mbox{ m}^{2} \ \ \ \ \ (27)$

The experimental values quoted by Griffiths are

 $\displaystyle A$ $\displaystyle =$ $\displaystyle 1.36\times10^{-4}\ \ \ \ \ (28)$ $\displaystyle B$ $\displaystyle =$ $\displaystyle 7.7\times10^{-15} \ \ \ \ \ (29)$

so the values from the model are at least around the right order of magnitude.

## Frequency dependence of electric permittivity

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.23a.

Experimentally, it is known that the permittivity of a material when an electromagnetic wave passes through it depends on the frequency of the wave. To develop a (relatively crude) theory of how this comes about, it’s worth recalling the definition of permittivity ${\epsilon}$, which arises from the ability of an external electric field ${\mathbf{E}}$ to polarize a dielectric, producing a polarization density ${\mathbf{P}}$:

$\displaystyle \mathbf{P}=\epsilon_{0}\chi_{e}\mathbf{E} \ \ \ \ \ (1)$

${\chi_{e}}$ is the electric susceptibility, and the permittivity is defined in terms of it by

$\displaystyle \epsilon=\epsilon_{0}\left(1+\chi_{e}\right) \ \ \ \ \ (2)$

Therefore, if we want to discover the dependence of ${\epsilon}$ on frequency, we might start by trying to find a relation between ${\mathbf{P}}$ and ${\mathbf{E}}$, where ${\mathbf{E}}$ arises from the electromagnetic wave passing through the material. The idea is to look at a typical electron bound to one of the atoms in the dielectric and work out the dipole moment of this atom in terms of the applied field in the wave.

The electron (with charge ${q}$) is subject to several forces. First, there is the force from the wave. The wave’s electric component has the form (assuming it’s polarized in the ${x}$ direction):

$\displaystyle \tilde{\mathbf{E}}=\tilde{E}_{0}e^{i\left(kz-\omega t\right)}\hat{\mathbf{x}} \ \ \ \ \ (3)$

For a fixed point, say ${z=0}$, the field oscillates in place so the force on the electron is the real part of the field times the charge:

$\displaystyle \mathbf{F}_{E}=qE_{0}\cos\omega t\hat{\mathbf{x}} \ \ \ \ \ (4)$

Second, the electron experiences a binding force with the nucleus. A simple model that we used earlier took the electron to be a sphere of uniform charge density, of radius ${a}$ (the Bohr radius in hydrogen, which is ${5.29\times10^{-11}\mbox{ m}}$) centred on the nucleus. In this case, when the electron is displaced a distance ${x}$ from equilibrium, the binding force is

$\displaystyle \mathbf{F}_{b}=-\frac{Zq^{2}}{4\pi\epsilon_{0}a^{3}}x\hat{\mathbf{x}} \ \ \ \ \ (5)$

where ${Z}$ is the atomic number (number of protons in the nucleus) and the minus sign is because ${\mathbf{F}_{b}}$ pulls the electron back towards equilibrium. This force is a harmonic oscillator force, since

 $\displaystyle \mathbf{F}_{b}$ $\displaystyle =$ $\displaystyle -k_{b}x\hat{\mathbf{x}}\ \ \ \ \ (6)$ $\displaystyle k_{b}$ $\displaystyle \equiv$ $\displaystyle \frac{Zq^{2}}{4\pi\epsilon_{0}a^{3}} \ \ \ \ \ (7)$

The harmonic oscillator force has a natural frequency of

$\displaystyle \omega_{0}=\sqrt{\frac{k_{b}}{m}} \ \ \ \ \ (8)$

so we can write the binding force as

$\displaystyle \mathbf{F}_{b}=-m\omega_{0}^{2}x\hat{\mathbf{x}} \ \ \ \ \ (9)$

Example 1 We can work out this natural frequency for a hydrogen-like atom with ${Z=1}$. We get

 $\displaystyle k_{b}$ $\displaystyle =$ $\displaystyle \frac{\left(1.6\times10^{-19}\right)^{2}}{4\pi\left(8.85\times10^{-12}\right)\left(5.29\times10^{-11}\right)^{3}}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.55\times10^{3}\mbox{ kg s}^{-2}\ \ \ \ \ (11)$ $\displaystyle \omega_{0}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{k_{b}}{m}}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{1.55\times10^{3}}{9.1\times10^{-31}}}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 4.13\times10^{16}\mbox{ s}^{-1}\ \ \ \ \ (14)$ $\displaystyle \nu_{0}$ $\displaystyle =$ $\displaystyle \frac{\omega_{0}}{2\pi}=6.58\times10^{15}\mbox{ s}^{-1} \ \ \ \ \ (15)$

This frequency is in the near ultraviolet, just beyond the violet end of the visible spectrum.

Finally, there will be a damping force because, once the wave is turned off, we expect the electron to eventually return to its equilibrium position. This can happen by radiating away energy or from interactions with other fields in the material. The simplest damping force is proportional to, and opposite in direction to, the velocity, so we can let

$\displaystyle \mathbf{F}_{d}=-\gamma m\dot{x}\hat{\mathbf{x}} \ \ \ \ \ (16)$

where ${\gamma}$ is the damping constant.

Since all the forces act in the ${x}$ direction, we can drop the vector notation and apply ${F_{total}=m\ddot{x}}$ to get

$\displaystyle m\ddot{x}=-\gamma m\dot{x}-m\omega_{0}^{2}x+qE_{0}\cos\omega t \ \ \ \ \ (17)$

At this point, we can do the usual trick of introducing complex numbers by defining ${x}$ to be the real part of a complex variable ${\tilde{x}}$. This equation then becomes

$\displaystyle \ddot{\tilde{x}}+\gamma\dot{\tilde{x}}+\omega_{0}^{2}\tilde{x}=\frac{q}{m}E_{0}e^{-i\omega t} \ \ \ \ \ (18)$

The solution is

$\displaystyle \tilde{x}\left(t\right)=\tilde{x}_{0}e^{-i\omega t} \ \ \ \ \ (19)$

since if we substitute this into the ODE we get

 $\displaystyle -\omega^{2}\tilde{x}_{0}e^{-i\omega t}-i\omega\gamma\tilde{x}_{0}e^{-i\omega t}+\omega_{0}^{2}\tilde{x}_{0}e^{-i\omega t}$ $\displaystyle =$ $\displaystyle \frac{q}{m}E_{0}e^{-i\omega t}\ \ \ \ \ (20)$ $\displaystyle \tilde{x}_{0}$ $\displaystyle =$ $\displaystyle \frac{qE_{0}}{m\left(\omega_{0}^{2}-\omega^{2}-i\omega\gamma\right)} \ \ \ \ \ (21)$

The dipole moment ${p}$ of the atom is the charge times the separation, which is given by ${x}$, so it is the real part of

 $\displaystyle \tilde{p}$ $\displaystyle =$ $\displaystyle q\tilde{x}_{0}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{q^{2}E_{0}}{m\left(\omega_{0}^{2}-\omega^{2}-i\omega\gamma\right)}e^{-i\omega t}\ \ \ \ \ (23)$ $\displaystyle \tilde{\mathbf{p}}$ $\displaystyle =$ $\displaystyle \frac{q^{2}}{m\left(\omega_{0}^{2}-\omega^{2}-i\omega\gamma\right)}\tilde{\mathbf{E}} \ \ \ \ \ (24)$

We have now achieved our objective (for a single electron) since we have the dipole moment in terms of the applied electric field. However, because of the damping term, the real part of ${\tilde{p}}$ is not directly proportional to ${E=E_{0}\cos\omega t}$ so the medium isn’t linear. We still need to calculate the polarization density ${\mathbf{P}}$ to get the permittivity. If there are ${f_{j}}$ electrons with natural frequency ${\omega_{0}=\omega_{j}}$ and damping constant ${\gamma=\gamma_{j}}$ per atom (or molecule) and ${N}$ atoms per unit volume, then

 $\displaystyle \tilde{\mathbf{P}}$ $\displaystyle =$ $\displaystyle N\sum_{j}f_{j}\tilde{\mathbf{p}}_{j}\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{Nq^{2}}{m}\tilde{\mathbf{E}}\sum_{j}\frac{f_{j}}{\left(\omega_{j}^{2}-\omega^{2}-i\omega\gamma_{j}\right)} \ \ \ \ \ (26)$

Generalizing 1 so that the susceptibility is complex, we have

 $\displaystyle \tilde{\mathbf{P}}$ $\displaystyle =$ $\displaystyle \epsilon_{0}\tilde{\chi}_{e}\tilde{\mathbf{E}}$ $\displaystyle \tilde{\chi}_{e}$ $\displaystyle =$ $\displaystyle \frac{Nq^{2}}{\epsilon_{0}m}\sum_{j}\frac{f_{j}}{\left(\omega_{j}^{2}-\omega^{2}-i\omega\gamma_{j}\right)}$

This gives a complex dielectric constant and permittivity:

 $\displaystyle \tilde{\epsilon}_{r}$ $\displaystyle =$ $\displaystyle 1+\tilde{\chi}_{e}\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1+\frac{Nq^{2}}{\epsilon_{0}m}\sum_{j}\frac{f_{j}}{\left(\omega_{j}^{2}-\omega^{2}-i\omega\gamma_{j}\right)}\ \ \ \ \ (28)$ $\displaystyle \tilde{\epsilon}$ $\displaystyle =$ $\displaystyle \epsilon_{0}\tilde{\epsilon}_{r}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \epsilon_{0}+\frac{Nq^{2}}{m}\sum_{j}\frac{f_{j}}{\left(\omega_{j}^{2}-\omega^{2}-i\omega\gamma_{j}\right)} \ \ \ \ \ (30)$

## Phase velocity and group velocity in a wave packet

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.22.

A monochromatic wave has velocity called the phase velocity given by

$\displaystyle v_{p}=\frac{\omega}{k} \ \ \ \ \ (1)$

where ${\omega}$ is the frequency and ${k=2\pi/\lambda}$ is the wave number. However, if we have a compound wave that is composed of individual waves with a range of frequencies, each individual wave has a velocity given by 1, but the amplitudes of the waves add up to produce a wave packet which has a velocity all its own. This velocity is called the group velocity and is usually different from the individual phase velocities of the waves that make up the packet.

This effect arises from the fact that typically the frequency is a function of the wave number: ${\omega=\omega\left(k\right)}$. Suppose we have a wave packet made up of a range of individual waves. We can write this as

$\displaystyle \psi\left(x,t\right)=\int A\left(k\right)e^{i\left(kx-\omega t\right)}dk \ \ \ \ \ (2)$

where ${A\left(k\right)}$ is a function giving the contribution to the packet of waves with wave number ${k}$. Now suppose that most of these waves have values of ${k}$ that are close to some value ${k_{0}}$. In that case, we can expand ${\omega\left(k\right)}$ and keep only the first order term:

$\displaystyle \omega\left(k_{0}+\Delta k\right)=\omega_{0}+\Delta k\omega_{0}' \ \ \ \ \ (3)$

where

$\displaystyle \omega_{0}'\equiv\left.\frac{d\omega}{dk}\right|_{k_{0}} \ \ \ \ \ (4)$

Plugging this into 2 we get

$\displaystyle \psi\left(x,t\right)=e^{i\left(k_{0}x-\omega_{0}t\right)}\int A\left(k\right)e^{i\left(\Delta k\cdot x-\Delta k\cdot\omega_{0}'t\right)}d\left(\Delta k\right) \ \ \ \ \ (5)$

We can see that the wave packet is composed of a monochromatic wave represented by the exponential outside the integral modulated by the integral factor. The speed of the monochromatic wave is just the phase velocity of the main wave:

$\displaystyle v_{p}=\frac{\omega_{0}}{k_{0}} \ \ \ \ \ (6)$

The velocity of the modulation is now a constant given by

$\displaystyle v_{g}=\frac{\Delta k\cdot\omega_{0}'}{\Delta k}=\left.\frac{d\omega}{dk}\right|_{k_{0}} \ \ \ \ \ (7)$

It’s this latter velocity that is the group velocity. This derivation relies on the waves making up the packet all having wave numbers (and hence wavelengths) lying close to each other.

Example 1 A wave travelling on the surface of water has a phase velocity that is proportional to the square root of its wavelength (provided ${\lambda}$ is less than the depth of the water). That is,

 $\displaystyle v_{p}$ $\displaystyle =$ $\displaystyle \frac{\omega}{k}=A\sqrt{\lambda}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle A\sqrt{\frac{2\pi}{k}}\ \ \ \ \ (9)$ $\displaystyle \omega\left(k\right)$ $\displaystyle =$ $\displaystyle A\sqrt{2\pi k}\ \ \ \ \ (10)$ $\displaystyle v_{g}$ $\displaystyle =$ $\displaystyle \frac{d\omega}{dk}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}A\sqrt{\frac{2\pi}{k}}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}v_{p} \ \ \ \ \ (13)$

Thus the phase velocity of deep water waves is twice the group velocity.

Example 2 We’ve seen that a free particle can be represented in quantum mechanics by a superposition of waves, each of which has the form

$\displaystyle \Psi\left(x,t\right)=Ae^{i\left(px-Et\right)/\hbar} \ \ \ \ \ (14)$

where ${p}$ is the momentum and ${E}$ is the energy, given by

$\displaystyle E=\frac{p^{2}}{2m} \ \ \ \ \ (15)$

In terms of frequency and wave number, we have

 $\displaystyle k$ $\displaystyle =$ $\displaystyle \frac{p}{\hbar}\ \ \ \ \ (16)$ $\displaystyle \omega$ $\displaystyle =$ $\displaystyle \frac{p^{2}}{2\hbar m}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar k^{2}}{2m}\ \ \ \ \ (18)$ $\displaystyle v_{g}=\frac{d\omega}{dk}$ $\displaystyle =$ $\displaystyle \frac{\hbar k}{m}=\frac{p}{m}\ \ \ \ \ (19)$ $\displaystyle v_{p}$ $\displaystyle =\frac{\omega}{k}=$ $\displaystyle \frac{p}{2m}=\frac{v_{g}}{2} \ \ \ \ \ (20)$

So in this case, the phase velocity is half the group velocity. Classically a particle’s velocity is given by ${v=p/m}$ so it is the quantum group velocity that corresponds to classical velocity.

## Reflection at a conducting surface: the physics of mirrors

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.21.

We can analyze reflection of an electromagnetic wave at a nonconductor-conductor interface in a similar way to that used for a nonconductor-nonconductor interface. We’ll look only at the case of normal incidence here.

As before, we start with the boundary conditions in linear media derived from Maxwell’s equations:

 $\displaystyle \epsilon_{1}E_{1}^{\perp}-\epsilon_{2}E_{2}^{\perp}$ $\displaystyle =$ $\displaystyle \sigma_{f}\ \ \ \ \ (1)$ $\displaystyle B_{1}^{\perp}-B_{2}^{\perp}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (2)$ $\displaystyle \mathbf{E}_{1}^{\parallel}-\mathbf{E}_{2}^{\parallel}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (3)$ $\displaystyle \frac{1}{\mu_{1}}\mathbf{B}_{1}^{\parallel}-\frac{1}{\mu_{2}}\mathbf{B}_{2}^{\parallel}$ $\displaystyle =$ $\displaystyle \mathbf{K}_{f}\times\hat{\mathbf{n}} \ \ \ \ \ (4)$

We’ll take medium 1 as the nonconductor (air, say) and medium 2 as the conductor. We’re allowing for the presence of free surface charge density ${\sigma_{f}}$ and free current density ${\mathbf{K}_{f}}$ at the boundary.

If we’re dealing with a conductor that obey’s Ohm’s law, the volume current density is proportional to the electric field

$\displaystyle \mathbf{J}_{f}=\sigma\mathbf{E} \ \ \ \ \ (5)$

where here ${\sigma}$ is the conductivity, not a charge density. Recall that ${\mathbf{J}_{f}}$ is the amount of current flowing through a unit area in the conductor. If we had a surface current density ${\mathbf{K}_{f}}$, this current flows along the boundary as a sheet of moving charge with infinitesimal thickness, so that the cross-sectional area occupied by ${\mathbf{K}_{f}}$ is essentially zero, making the volume charge density infinite. For a finite conductivity ${\sigma}$ it would take an infinite electric field to produce this surface current, so we can safely assume that ${\mathbf{K}_{f}=0}$ in what follows.

The incident and reflected waves are both in medium 1, so if we polarize the wave in the ${x}$ direction, we have for the incident wave:

 $\displaystyle \tilde{\mathbf{E}}_{I}$ $\displaystyle =$ $\displaystyle \tilde{E}_{0_{I}}e^{i\left(k_{1}z-\omega t\right)}\hat{\mathbf{x}}\ \ \ \ \ (6)$ $\displaystyle \tilde{\mathbf{B}}_{I}$ $\displaystyle =$ $\displaystyle \frac{1}{v_{1}}\tilde{E}_{0_{I}}e^{i\left(k_{1}z-\omega t\right)}\hat{\mathbf{y}} \ \ \ \ \ (7)$

where ${v_{1}}$ is the speed of the wave in medium 1.

The reflected wave is travelling in the ${-z}$ direction and has equations

 $\displaystyle \tilde{\mathbf{E}}_{R}$ $\displaystyle =$ $\displaystyle \tilde{E}_{0_{R}}e^{i\left(-k_{1}z-\omega t\right)}\hat{\mathbf{x}}\ \ \ \ \ (8)$ $\displaystyle \tilde{\mathbf{B}}_{R}$ $\displaystyle =$ $\displaystyle -\frac{1}{v_{1}}\tilde{E}_{0_{R}}e^{i\left(-k_{1}z-\omega t\right)}\hat{\mathbf{y}} \ \ \ \ \ (9)$

The transmitted wave is inside the conductor, so its equations can be written as

 $\displaystyle \tilde{\mathbf{E}}_{T}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \tilde{E}_{0_{T}}e^{i\left(\tilde{k}_{2}z-\omega t\right)}\hat{\mathbf{x}}\ \ \ \ \ (10)$ $\displaystyle \tilde{\mathbf{B}}_{T}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \frac{\tilde{k}_{2}}{\omega}\tilde{E}_{0_{T}}e^{i\left(\tilde{k}_{2}z-\omega t\right)}\hat{\mathbf{y}} \ \ \ \ \ (11)$

where the wave vector is complex:

$\displaystyle \tilde{k}=\frac{\omega\sqrt{\mu\epsilon}}{\sqrt{2}}\sqrt{\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}+1}+i\frac{\omega\sqrt{\mu\epsilon}}{\sqrt{2}}\sqrt{\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}-1}\equiv k+i\kappa \ \ \ \ \ (12)$

We can now apply the boundary conditions. Equation 1 tells us that ${\sigma_{f}=0}$ since there is no perpendicular component of ${\mathbf{E}}$ (remember the wave is transverse). Equation 2 tells us nothing (${0=0}$). From 3, assuming that the boundary is at ${z=0}$, we get, since all components of ${\mathbf{E}}$ are in the ${x}$ direction:

$\displaystyle \tilde{E}_{0_{I}}+\tilde{E}_{0_{R}}=\tilde{E}_{0_{T}} \ \ \ \ \ (13)$

Finally, from 4 we get, since all components of ${\mathbf{B}}$ are in the ${y}$ direction and ${\mathbf{K}_{f}=0}$:

$\displaystyle \frac{1}{\mu_{1}v_{1}}\left(\tilde{E}_{0_{I}}-\tilde{E}_{0_{R}}\right)=\frac{\tilde{k}_{2}}{\mu_{2}\omega}\tilde{E}_{0_{T}} \ \ \ \ \ (14)$

which we can rewrite as

 $\displaystyle \tilde{E}_{0_{I}}-\tilde{E}_{0_{R}}$ $\displaystyle =$ $\displaystyle \frac{\mu_{1}v_{1}}{\mu_{2}\omega}\tilde{k}_{2}\tilde{E}_{0_{T}}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle \tilde{\beta}\tilde{E}_{0_{T}} \ \ \ \ \ (16)$

Adding 13 and 16 we get

 $\displaystyle \tilde{E}_{0_{I}}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(1+\tilde{\beta}\right)\tilde{E}_{0_{T}}\ \ \ \ \ (17)$ $\displaystyle \tilde{E}_{0_{T}}$ $\displaystyle =$ $\displaystyle \frac{2}{1+\tilde{\beta}}\tilde{E}_{0_{I}} \ \ \ \ \ (18)$

Subtracting 13 and 16 we get

 $\displaystyle \tilde{E}_{0_{R}}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(1-\tilde{\beta}\right)\tilde{E}_{0_{T}}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1-\tilde{\beta}}{1+\tilde{\beta}}\tilde{E}_{0_{I}} \ \ \ \ \ (20)$

These are deceptively simple equations, since everything with a tilde on it is a complex number. To get the actual amplitudes and phases we need to extract the real and imaginary parts.

Example To put some numbers into these equations, let’s consider an air-silver interface. For a good conductor such as silver, ${\sigma\gg\epsilon\omega}$ and in 12

$\displaystyle k\approx\kappa\approx\sqrt{\frac{\mu\sigma\omega}{2}} \ \ \ \ \ (21)$

In air, ${v_{1}\approx c}$ and we can take ${\mu_{1}=\mu_{2}=\mu_{0}}$ so from 16

$\displaystyle \tilde{\beta}\approx c\sqrt{\frac{\mu_{0}\sigma}{2\omega}}\left(1+i\right) \ \ \ \ \ (22)$

For silver ${\sigma=6\times10^{7}\mbox{ S m}^{-1}}$ and at an optical wavelength of ${\omega=4\times10^{15}\mbox{ s}^{-1}}$ we get

 $\displaystyle \tilde{\beta}$ $\displaystyle =$ $\displaystyle 29.1\left(1+i\right)\ \ \ \ \ (23)$ $\displaystyle \frac{1-\tilde{\beta}}{1+\tilde{\beta}}$ $\displaystyle =$ $\displaystyle \frac{-28.1-29.1i}{30.1+29.1i}\times\frac{30.1-29.1i}{30.1-29.1i}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{1753}\left(-1693-58.2i\right) \ \ \ \ \ (25)$

To get the reflection coefficient we can write the complex amplitudes in modulus-phase form as

 $\displaystyle \tilde{E}_{0_{R}}$ $\displaystyle =$ $\displaystyle E_{0_{R}}e^{i\delta_{R}}\ \ \ \ \ (26)$ $\displaystyle \tilde{E}_{0_{I}}$ $\displaystyle =$ $\displaystyle E_{0_{I}}e^{i\delta_{I}} \ \ \ \ \ (27)$

The intensity of a wave is the average over one cycle of the magnitude of the Poynting vector, so the fact that the incident and reflected waves may have different phases doesn’t matter (since they have the same frequency). This means that

 $\displaystyle R$ $\displaystyle =$ $\displaystyle \frac{\frac{1}{2}c\epsilon_{0}E_{0_{R}}^{2}}{\frac{1}{2}c\epsilon_{0}E_{0_{I}}^{2}}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{\left|E_{0_{R}}\right|^{2}}{\left|E_{0_{I}}\right|^{2}}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1693^{2}+58.2^{2}}{1753^{2}}\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0.93 \ \ \ \ \ (31)$

Silver reflects 93% of the incident light, so it makes a good mirror.

## Fermat’s principle of least time and Snell’s law

References: Tom Lancaster and Stephen J. Blundell, Quantum Field Theory for the Gifted Amateur, (Oxford University Press, 2014) – Problem 1.1

One of the guiding principles of quantum field theory is that a particle travelling between two points actually traverses all possible paths between these two points, although with varying probabilities for different paths. Although this idea is expressed mathematically using the calculus of variations, a simpler example of the same idea is that of Fermat’s principle of least time applied to the derivation of Snell’s law of refraction in optics.

The idea is that given that the speed of light in a medium with index of refraction ${n}$ is ${c/n}$, if a light beam starts at a point ${A}$ in medium 1 and hits the interface between mediums 1 and 2 at an angle ${\theta_{1}}$ to the normal, and continues through into medium 2 at an angle ${\theta_{2}}$ to the normal eventually arriving at point ${B}$, then these angles are such that the travel time from ${A}$ to ${B}$ is a minimum. There isn’t any particular reason why this assumption is made (apart from the the fact that it gives the right answer!).

To see how it works, suppose we orient the interface so that it lies in the ${xy}$ plane, so that the normal to the interface is the ${z}$ axis. We’ll take the incident beam of light starting at point ${A}$ to lie in the ${yz}$ plane, as does the refracted beam which travels from the interface to point ${B}$. We’ll let ${y_{AB}}$ be the difference in ${y}$ coordinate of the points ${A}$ and ${B}$, and let ${y_{AO}}$ be the difference in ${y}$ coordinate of the point ${O}$ where the beam hits the interface. Thus the difference in ${y}$ coordinate between ${O}$ and ${B}$ is ${y_{OB}=y_{AB}-y_{AO}}$. Similarly, let ${z_{AO}}$ and ${z_{OB}}$ be the differences in ${z}$ coordinates between the corresponding points. Finally, let ${a}$ be the distance from ${A}$ to ${O}$, and ${b}$ the distance from ${O}$ to ${B}$.

Then by Pythagoras

 $\displaystyle a$ $\displaystyle =$ $\displaystyle \sqrt{z_{AO}^{2}+y_{AO}^{2}}\ \ \ \ \ (1)$ $\displaystyle b$ $\displaystyle =$ $\displaystyle \sqrt{z_{OB}^{2}+\left(y_{AB}-y_{AO}\right)^{2}} \ \ \ \ \ (2)$

The total travel time of the light beam is

 $\displaystyle t$ $\displaystyle =$ $\displaystyle \frac{a}{v_{1}}+\frac{b}{v_{2}}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{a}{c}n_{1}+\frac{b}{c}n_{2}\ \ \ \ \ (4)$ $\displaystyle ct$ $\displaystyle =$ $\displaystyle n_{1}\sqrt{z_{AO}^{2}+y_{AO}^{2}}+n_{2}\sqrt{z_{OB}^{2}+\left(y_{AB}-y_{AO}\right)^{2}} \ \ \ \ \ (5)$

Since the points ${A}$ and ${B}$ are fixed, as is the location of the interface, the only thing we can vary is ${y}$ coordinate of the point where the light beam hits the interface, that is, ${y_{AO}}$. We can therefore minimize ${ct}$ with respect to ${y_{AO}}$:

 $\displaystyle \frac{d\left(ct\right)}{dy_{AO}}$ $\displaystyle =$ $\displaystyle \frac{y_{AO}n_{1}}{\sqrt{z_{AO}^{2}+y_{AO}^{2}}}-\frac{n_{2}\left(y_{AB}-y_{AO}\right)}{\sqrt{z_{OB}^{2}+\left(y_{AB}-y_{AO}\right)^{2}}}=0\ \ \ \ \ (6)$ $\displaystyle \frac{y_{AO}}{\sqrt{z_{AO}^{2}+y_{AO}^{2}}}n_{1}$ $\displaystyle =$ $\displaystyle \frac{\left(y_{AB}-y_{AO}\right)}{\sqrt{z_{OB}^{2}+\left(y_{AB}-y_{AO}\right)^{2}}}n_{2}\ \ \ \ \ (7)$ $\displaystyle n_{1}\sin\theta_{1}$ $\displaystyle =$ $\displaystyle n_{2}\sin\theta_{2} \ \ \ \ \ (8)$

where the last line uses the trigonometric definition of the sine from the sides of the triangles.