## 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.

## Electromagnetic waves in conductors: energy density and intensity

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.20.

We can write the electromagnetic wave inside a conductor as (if we orient the axes so that ${\mathbf{E}}$ is polarized in the ${x}$ direction)

 $\displaystyle \tilde{\mathbf{E}}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \tilde{E}_{0}e^{-\kappa z}e^{i\left(kz-\omega t\right)}\hat{\mathbf{x}}\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle E_{0}e^{-\kappa z}e^{i\left(kz-\omega t+\delta_{E}\right)}\hat{\mathbf{x}}\ \ \ \ \ (2)$ $\displaystyle \tilde{\mathbf{B}}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \frac{\tilde{k}}{\omega}\tilde{E}_{0}e^{-\kappa z}e^{i\left(kz-\omega t\right)}\hat{\mathbf{y}}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\epsilon\mu\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}}E_{0}e^{-\kappa z}e^{i\left(kz-\omega t+\delta_{E}+\phi\right)}\hat{\mathbf{y}} \ \ \ \ \ (4)$

where

$\displaystyle \tilde{k}=\frac{\omega\sqrt{\mu\epsilon}}{\sqrt{2}}\sqrt{\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}+1}+i\frac{\omega\sqrt{\mu\epsilon}}{\sqrt{2}}\sqrt{\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}-1}\equiv k+i\kappa\equiv Ke^{i\phi} \ \ \ \ \ (5)$

The actual fields are the real parts of these equations, so

 $\displaystyle \mathbf{E}\left(z,t\right)$ $\displaystyle =$ $\displaystyle E_{0}e^{-\kappa z}\cos\left(kz-\omega t+\delta_{E}\right)\hat{\mathbf{x}}\ \ \ \ \ (6)$ $\displaystyle \mathbf{B}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \sqrt{\epsilon\mu\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}}E_{0}e^{-\kappa z}\cos\left(kz-\omega t+\delta_{E}+\phi\right)\hat{\mathbf{y}} \ \ \ \ \ (7)$

The energy density in the wave is

$\displaystyle u=\frac{1}{2}\left(\epsilon E^{2}+\frac{1}{\mu}B^{2}\right) \ \ \ \ \ (8)$

Taking the time average (over one cycle) of this we have (since the average of ${\cos^{2}\omega t}$ over one cycle ${\tau=2\pi/\omega}$ is ${\frac{1}{2}}$):

$\displaystyle u=\frac{E_{0}^{2}e^{-2\kappa z}}{4}\left(\epsilon+\epsilon\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}\right) \ \ \ \ \ (9)$

For a good conductor, ${\sigma\gg\epsilon\omega}$ so

 $\displaystyle u$ $\displaystyle \approx$ $\displaystyle \frac{E_{0}^{2}e^{-2\kappa z}}{4}\left(\epsilon+\frac{\sigma}{\omega}\right)\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle \frac{E_{0}^{2}e^{-2\kappa z}}{4}\frac{\sigma}{\omega} \ \ \ \ \ (11)$

From 10, we see that the magnetic contribution (${\sigma/\omega}$) is much larger than the electric contribution (${\epsilon}$) for a good conductor.

We can express this in terms of the wave vector ${k}$ by using 5 for a good conductor.

 $\displaystyle k$ $\displaystyle \approx$ $\displaystyle \frac{\omega\sqrt{\mu\epsilon}}{\sqrt{2}}\sqrt{\frac{\sigma}{\epsilon\omega}}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\omega\mu\sigma}{2}}\ \ \ \ \ (13)$ $\displaystyle \sigma$ $\displaystyle =$ $\displaystyle \frac{2k^{2}}{\omega\mu}\ \ \ \ \ (14)$ $\displaystyle u$ $\displaystyle \approx$ $\displaystyle \frac{E_{0}^{2}e^{-2\kappa z}}{2}\frac{k^{2}}{\mu\omega^{2}} \ \ \ \ \ (15)$

The intensity is the energy crossing a unit area in unit time, which is the energy density times the volume crossing a unit area per unit time, which is

$\displaystyle I=uv \ \ \ \ \ (16)$

where ${v}$ is the speed of the wave, which is ${\omega/k}$ so

$\displaystyle I=\frac{E_{0}^{2}e^{-2\kappa z}}{2}\frac{k}{\mu\omega} \ \ \ \ \ (17)$

## Electromagnetic waves in conductors: phases and amplitudes

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.19c.

Electromagnetic waves in a conductor (where there is free current but no free charge) can be written as

 $\displaystyle \tilde{\mathbf{E}}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \tilde{\mathbf{E}}_{0}e^{i\left(\tilde{k}z-\omega t\right)}\ \ \ \ \ (1)$ $\displaystyle \tilde{\mathbf{B}}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \tilde{\mathbf{B}}_{0}e^{i\left(\tilde{k}z-\omega t\right)} \ \ \ \ \ (2)$

where the wave vector is complex:

$\displaystyle \tilde{k}=\frac{\omega\sqrt{\mu\epsilon}}{\sqrt{2}}\sqrt{\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}+1}+i\frac{\omega\sqrt{\mu\epsilon}}{\sqrt{2}}\sqrt{\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}-1}\equiv k+i\kappa \ \ \ \ \ (3)$

so

 $\displaystyle \tilde{\mathbf{E}}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \tilde{\mathbf{E}}_{0}e^{-\kappa z}e^{i\left(kz-\omega t\right)}\ \ \ \ \ (4)$ $\displaystyle \tilde{\mathbf{B}}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \tilde{\mathbf{B}}_{0}e^{-\kappa z}e^{i\left(kz-\omega t\right)} \ \ \ \ \ (5)$

By applying Maxwell’s equations in a conductor we can get a few more properties of these waves. The equations are

 $\displaystyle \nabla\cdot\mathbf{E}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (6)$ $\displaystyle \nabla\cdot\mathbf{B}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (7)$ $\displaystyle \nabla\times\mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\partial\mathbf{B}}{\partial t}\ \ \ \ \ (8)$ $\displaystyle \nabla\times\mathbf{B}$ $\displaystyle =$ $\displaystyle \mu\sigma\mathbf{E}+\mu\epsilon\frac{\partial\mathbf{E}}{\partial t} \ \ \ \ \ (9)$

Using the same techniques as in analyzing waves in vacuum. Both ${\nabla\cdot\mathbf{E}=0}$ and ${\nabla\cdot\mathbf{B}=0}$ from which we get

 $\displaystyle \nabla\cdot\tilde{\mathbf{E}}$ $\displaystyle =$ $\displaystyle \left(ik-\kappa\right)\tilde{E}_{0z}e^{i\left(kz-\omega t\right)}=0\ \ \ \ \ (10)$ $\displaystyle \nabla\cdot\tilde{\mathbf{B}}$ $\displaystyle =$ $\displaystyle \left(ik-\kappa\right)\tilde{B}_{0z}e^{i\left(kz-\omega t\right)}=0 \ \ \ \ \ (11)$

Since this must be true for all ${z}$, we must have

$\displaystyle \tilde{E}_{0z}=\tilde{B}_{0z}=0 \ \ \ \ \ (12)$

That is, the wave has only ${x}$ and ${y}$ components, so it must be a transverse wave: a wave that oscillates in a plane perpendicular to the direction of propagation. If we orient the axes so that ${\mathbf{E}}$ is polarized in the ${x}$ direction then

$\displaystyle \tilde{\mathbf{E}}\left(z,t\right)=\tilde{E}_{0}e^{-\kappa z}e^{i\left(kz-\omega t\right)}\hat{\mathbf{x}} \ \ \ \ \ (13)$

Applying 8 to this gives

 $\displaystyle \nabla\times\tilde{\mathbf{E}}$ $\displaystyle =$ $\displaystyle i\left(k+i\kappa\right)\tilde{E}_{0}e^{-\kappa z}e^{i\left(kz-\omega t\right)}\hat{\mathbf{y}}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\tilde{k}\tilde{E}_{0}e^{-\kappa z}e^{i\left(kz-\omega t\right)}\hat{\mathbf{y}}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\partial\mathbf{B}}{\partial t}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\omega\tilde{\mathbf{B}}_{0}e^{-\kappa z}e^{i\left(kz-\omega t\right)}\ \ \ \ \ (17)$ $\displaystyle \tilde{\mathbf{B}}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \frac{\tilde{k}}{\omega}\tilde{E}_{0}e^{-\kappa z}e^{i\left(kz-\omega t\right)}\hat{\mathbf{y}} \ \ \ \ \ (18)$

As in vacuum, ${\mathbf{E}}$ and ${\mathbf{B}}$ are perpendicular and transverse to the direction of propagation. Unlike in the vacuum, however, the two components of the wave may not be in phase, due to the presence of the complex variable ${\tilde{k}}$ in the equation for ${\tilde{\mathbf{B}}}$. If we write ${\tilde{k}}$ in modulus-phase form we have

$\displaystyle \tilde{k}=Ke^{i\phi} \ \ \ \ \ (19)$

where

 $\displaystyle K$ $\displaystyle =$ $\displaystyle \sqrt{k^{2}+\kappa^{2}}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \omega\sqrt{\epsilon\mu\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}}\ \ \ \ \ (21)$ $\displaystyle \phi$ $\displaystyle =$ $\displaystyle \arctan\frac{\kappa}{k} \ \ \ \ \ (22)$

Then the complex amplitudes of the two components can be written as

 $\displaystyle \tilde{E}_{0}$ $\displaystyle =$ $\displaystyle E_{0}e^{i\delta_{E}}\ \ \ \ \ (23)$ $\displaystyle \tilde{B}_{0}$ $\displaystyle =$ $\displaystyle \frac{K}{\omega}E_{0}e^{i\left(\delta_{E}+\phi\right)} \ \ \ \ \ (24)$

and the ratio of the real amplitudes is

$\displaystyle \frac{B_{0}}{E_{0}}=\frac{K}{\omega}=\sqrt{\epsilon\mu\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}} \ \ \ \ \ (25)$

Example For a good conductor, ${\sigma\gg\epsilon\omega}$ so from 3 ${k\approx\kappa}$ so from 22 the phase difference between ${\mathbf{B}}$ and ${\mathbf{E}}$ is ${\pi/4}$. The ratio of amplitudes is

$\displaystyle \frac{B_{0}}{E_{0}}=\sqrt{\frac{\sigma\mu}{\omega}} \ \ \ \ \ (26)$

For a typical good conductor ${\sigma\approx10^{7}\mbox{S m}^{-1}}$ and ${\mu\approx\mu_{0}}$ and at visible frequencies ${\omega\approx10^{15}\mbox{ s}^{-1}}$ so

$\displaystyle \frac{B_{0}}{E_{0}}=1.12\times10^{-7}\mbox{ s m}^{-1} \ \ \ \ \ (27)$

## Skin depth of water and metals

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.19a-b.

Electromagnetic waves in a conductor (where there is free current but no free charge) can be written as

 $\displaystyle \tilde{\mathbf{E}}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \tilde{\mathbf{E}}_{0}e^{i\left(\tilde{k}z-\omega t\right)}\ \ \ \ \ (1)$ $\displaystyle \tilde{\mathbf{B}}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \tilde{\mathbf{B}}_{0}e^{i\left(\tilde{k}z-\omega t\right)} \ \ \ \ \ (2)$

where the wave vector is complex:

$\displaystyle \tilde{k}=\frac{\omega\sqrt{\mu\epsilon}}{\sqrt{2}}\sqrt{\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}+1}+i\frac{\omega\sqrt{\mu\epsilon}}{\sqrt{2}}\sqrt{\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}-1}\equiv k+i\kappa \ \ \ \ \ (3)$

For a poor conductor, the conductivity ${\sigma}$ is small, so for large enough frequencies ${\sigma\ll\epsilon\omega}$ and we can approximate ${\kappa}$ by

 $\displaystyle \kappa$ $\displaystyle \approx$ $\displaystyle \frac{\omega\sqrt{\mu\epsilon}}{\sqrt{2}}\sqrt{1+\frac{1}{2}\left(\frac{\sigma}{\epsilon\omega}\right)^{2}-1}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\sigma}{2}\sqrt{\frac{\mu}{\epsilon}} \ \ \ \ \ (5)$

Since the imaginary part of ${\tilde{k}}$ governs the attenuation of the wave as it penetrates the material, the skin depth for a poor conductor is

$\displaystyle d=\frac{1}{\kappa}=\frac{2}{\sigma}\sqrt{\frac{\epsilon}{\mu}} \ \ \ \ \ (6)$

For pure (deionized) water ${\sigma=5.5\times10^{-6}\mbox{S m}^{-1}}$ and ${\epsilon=80.1\epsilon_{0}}$ (at ${20^{\circ}\mbox{C}}$) (we can take ${\mu\approx\mu_{0}}$) so the skin depth of water is

$\displaystyle d=8635\mbox{ m} \ \ \ \ \ (7)$

Because the skin depth is so large, water is transparent.

For a good conductor, ${\sigma\gg\epsilon\omega}$ and we can approximate

$\displaystyle \kappa\approx\frac{\omega\sqrt{\mu\epsilon}}{\sqrt{2}}\sqrt{\frac{\sigma}{\epsilon\omega}}=\sqrt{\frac{\mu\sigma\omega}{2}}\approx k \ \ \ \ \ (8)$

so the skin depth is

$\displaystyle d=\sqrt{\frac{2}{\mu\sigma\omega}}\approx\frac{1}{k}=\frac{2\pi}{\lambda} \ \ \ \ \ (9)$

where ${\lambda}$ is the wavelength within the material. For a typical metal, ${\sigma\approx10^{7}\mbox{S m}^{-1}}$ and ${\mu\approx\mu_{0}}$ so the skin depth at visible frequencies ${\omega\approx10^{15}\mbox{s}^{-1}}$ is

$\displaystyle d\approx1.26\times10^{-8}\mbox{m} \ \ \ \ \ (10)$

With a skin depth this small, even a thin film of metal is effectively impervious to any penetration by visible light.

## Skin depth of electromagnetic waves in conductors

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.18b-c.

We’ve seen that any free charge within a conductor migrates to the surface with a characteristic time that depends on the conductor’s conductance and permittivity. Once that transient effect subsides, we can take the free charge density to be zero: ${\rho_{f}=0}$. This doesn’t mean that the free current density ${\mathbf{J}_{f}}$ is zero, though. We can still have a current in an electrically neutral conductor caused by electrons moving relative to stationary atomic nuclei.

In linear media, Maxwell’s equations are

 $\displaystyle \nabla\cdot\mathbf{E}$ $\displaystyle =$ $\displaystyle \frac{1}{\epsilon}\rho_{f}\ \ \ \ \ (1)$ $\displaystyle \nabla\cdot\mathbf{B}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (2)$ $\displaystyle \nabla\times\mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\partial\mathbf{B}}{\partial t}\ \ \ \ \ (3)$ $\displaystyle \nabla\times\mathbf{B}$ $\displaystyle =$ $\displaystyle \mu\mathbf{J}_{f}+\mu\epsilon\frac{\partial\mathbf{E}}{\partial t} \ \ \ \ \ (4)$

Within a conductor with conductivity ${\sigma}$ (not surface charge density!) Ohm’s law can be written as

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

so with ${\rho_{f}=0}$, Maxwell’s equations become

 $\displaystyle \nabla\cdot\mathbf{E}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (6)$ $\displaystyle \nabla\cdot\mathbf{B}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (7)$ $\displaystyle \nabla\times\mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\partial\mathbf{B}}{\partial t}\ \ \ \ \ (8)$ $\displaystyle \nabla\times\mathbf{B}$ $\displaystyle =$ $\displaystyle \mu\sigma\mathbf{E}+\mu\epsilon\frac{\partial\mathbf{E}}{\partial t} \ \ \ \ \ (9)$

We can take the curl of the last two equations in the same way as when we derived the wave equation in a vacuum.

 $\displaystyle \nabla\times\left(\nabla\times\mathbf{E}\right)$ $\displaystyle =$ $\displaystyle \nabla\left(\nabla\cdot\mathbf{E}\right)-\nabla^{2}\mathbf{E}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\nabla\times\frac{\partial\mathbf{B}}{\partial t}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\mu\epsilon\frac{\partial^{2}\mathbf{E}}{\partial t^{2}}-\mu\sigma\frac{\partial\mathbf{E}}{\partial t} \ \ \ \ \ (12)$

Since ${\nabla\cdot\mathbf{E}=0}$ we get

$\displaystyle \nabla^{2}\mathbf{E}=\mu\epsilon\frac{\partial^{2}\mathbf{E}}{\partial t^{2}}+\mu\sigma\frac{\partial\mathbf{E}}{\partial t} \ \ \ \ \ (13)$

A similar calculation for ${\mathbf{B}}$ gives

$\displaystyle \nabla^{2}\mathbf{B}=\mu\epsilon\frac{\partial^{2}\mathbf{B}}{\partial t^{2}}+\mu\sigma\frac{\partial\mathbf{B}}{\partial t} \ \ \ \ \ (14)$

We thus get the wave equation modified by the addition of an extra first-derivative term. Conveniently, these equations have a similar solution to the ordinary wave equation. For a plane wave travelling in the ${z}$ direction

 $\displaystyle \tilde{\mathbf{E}}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \tilde{\mathbf{E}}_{0}e^{i\left(\tilde{k}z-\omega t\right)}\ \ \ \ \ (15)$ $\displaystyle \tilde{\mathbf{B}}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \tilde{\mathbf{B}}_{0}e^{i\left(\tilde{k}z-\omega t\right)} \ \ \ \ \ (16)$

Substituting 15 into 13 we get, after cancelling terms

$\displaystyle \tilde{k}^{2}=\omega^{2}\mu\epsilon+i\omega\mu\sigma \ \ \ \ \ (17)$

The fact that the wave vector ${\tilde{k}}$ is complex means that the resulting wave has both an oscillatory and an exponentially decaying factor. Finding the square root of this using Maple gives

 $\displaystyle \tilde{k}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\,\sqrt{2\,\sqrt{{\omega}^{4}{\mu}^{2}{\epsilon}^{2}+{\omega}^{2}{\mu}^{2}{\sigma}^{2}}+2\,{\omega}^{2}\mu\,\epsilon}+\frac{1}{2}\, i\sqrt{2\,\sqrt{{\omega}^{4}{\mu}^{2}{\epsilon}^{2}+{\omega}^{2}{\mu}^{2}{\sigma}^{2}}-2\,{\omega}^{2}\mu\,\epsilon}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\omega\sqrt{\mu\epsilon}}{\sqrt{2}}\sqrt{\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}+1}+i\frac{\omega\sqrt{\mu\epsilon}}{\sqrt{2}}\sqrt{\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}-1}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle k+i\kappa \ \ \ \ \ (20)$

The wave is then given by

 $\displaystyle \tilde{\mathbf{E}}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \tilde{\mathbf{E}}_{0}e^{-\kappa z}e^{i\left(kz-\omega t\right)}\ \ \ \ \ (21)$ $\displaystyle \tilde{\mathbf{B}}\left(z,t\right)$ $\displaystyle =$ $\displaystyle \tilde{\mathbf{B}}_{0}e^{-\kappa z}e^{i\left(kz-\omega t\right)} \ \ \ \ \ (22)$

If an electromagnetic wave hits a conductor (starting in air or vacuum, say), the wave will attenuate as it penetrates the conductor with a characteristic distance, called the skin depth ${d}$ of

$\displaystyle d=\frac{1}{\kappa}=\frac{1}{\omega}\left[\frac{\mu\epsilon}{2}\left(\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^{2}}-1\right)\right]^{-1/2} \ \ \ \ \ (23)$

The skin depth depends not only on the conductivity and permittivity, but also on the frequency of the incident radiation.

Example 1 The skin depth for good conductors such as metals is very small for a wide range of frequencies. For silver, the conductivity is ${\sigma=6.30\times10^{7}\mbox{ S m}^{-1}}$. To get the permittivity of silver, we can return to its definition:

$\displaystyle \epsilon=\epsilon_{0}\left(1+\chi_{e}\right) \ \ \ \ \ (24)$

where the electric susceptibility ${\chi_{e}}$ is defined by the amount of polarization in the material that is produced by an applied electric field:

$\displaystyle \mathbf{P}=\epsilon_{0}\chi_{e}\mathbf{E} \ \ \ \ \ (25)$

For a perfect conductor, no polarization is produced by any amount of electric field, so ${\chi_{e}=0}$ and ${\epsilon=\epsilon_{0}}$. Thus it’s a reasonable approximation for a good conductor like silver to take ${\epsilon\approx\epsilon_{0}}$. Most materials also have a permeability ${\mu\approx\mu_{0}}$. At a microwave frequency of ${10^{10}\mbox{ Hz}=2\pi\times10^{10}\mbox{ s}^{-1}}$

$\displaystyle \frac{\sigma}{\epsilon_{0}\omega}=1.13\times10^{8} \ \ \ \ \ (26)$

so to a good approximation

$\displaystyle d=\frac{1}{\omega}\sqrt{\frac{2\omega}{\mu_{0}\sigma}}=\sqrt{\frac{2}{\omega\mu_{0}\sigma}}=6.34\times10^{-7}\mbox{m} \ \ \ \ \ (27)$

The real part of ${\tilde{k}}$ gives the oscillatory part of the wave, so the wavelength is

$\displaystyle \lambda=\frac{2\pi}{k} \ \ \ \ \ (28)$

and the wave speed is

$\displaystyle v=\nu\lambda=\frac{\omega}{2\pi}\lambda=\frac{\omega}{k} \ \ \ \ \ (29)$

The index of refraction is the ratio of ${c}$ to ${v}$ as usual:

$\displaystyle n=\frac{c}{v}=\frac{ck}{\omega} \ \ \ \ \ (30)$

Example 2 For copper, ${\sigma=5.96\times10^{7}\mbox{ S m}^{-1}}$. For radio waves with a frequency of ${1\mbox{ MHz}=2\pi\times10^{6}\mbox{ s}^{-1}}$ we have

 $\displaystyle \frac{\sigma}{\epsilon_{0}\omega}$ $\displaystyle =$ $\displaystyle 1.07\times10^{12}\ \ \ \ \ (31)$ $\displaystyle k$ $\displaystyle \approx$ $\displaystyle \sqrt{\frac{\omega\mu_{0}\sigma}{2}}\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.533\times10^{4}\mbox{ m}^{-1}\ \ \ \ \ (33)$ $\displaystyle \lambda$ $\displaystyle =$ $\displaystyle \frac{2\pi}{k}=4.1\times10^{-4}\mbox{m} \ \ \ \ \ (34)$

The wave speed is

$\displaystyle v=\frac{\omega}{k}=410\mbox{ m s}^{-1} \ \ \ \ \ (35)$

This is obviously very slow for electromagnetic radiation.

In air or vacuum, ${v=c=3\times10^{8}\mbox{m s}^{-1}}$ and ${\lambda=v/\nu=300\mbox{ m}}$ which is closer to what we’d expect for radio waves.

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.18a.

When we looked at conductors in electrostatics, we saw that any free charge in a conductor must reside on the surface. In electrodynamics, however, we can place some free charge inside a conductor and then watch it (figuratively) move to the surface. How long does this migration take?

 $\displaystyle \nabla\cdot\mathbf{D}$ $\displaystyle =$ $\displaystyle \rho_{f}\ \ \ \ \ (1)$ $\displaystyle \nabla\cdot\mathbf{B}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (2)$ $\displaystyle \nabla\times\mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\partial\mathbf{B}}{\partial t}\ \ \ \ \ (3)$ $\displaystyle \nabla\times\mathbf{H}$ $\displaystyle =$ $\displaystyle \mathbf{J}_{f}+\frac{\partial\mathbf{D}}{\partial t} \ \ \ \ \ (4)$

For linear media, ${\mathbf{D}=\epsilon\mathbf{E}}$ and ${\mathbf{H}=\mathbf{B}/\mu}$, so these equations become

 $\displaystyle \nabla\cdot\mathbf{E}$ $\displaystyle =$ $\displaystyle \frac{1}{\epsilon}\rho_{f}\ \ \ \ \ (5)$ $\displaystyle \nabla\cdot\mathbf{B}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (6)$ $\displaystyle \nabla\times\mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\partial\mathbf{B}}{\partial t}\ \ \ \ \ (7)$ $\displaystyle \nabla\times\mathbf{B}$ $\displaystyle =$ $\displaystyle \mu\mathbf{J}_{f}+\mu\epsilon\frac{\partial\mathbf{E}}{\partial t} \ \ \ \ \ (8)$

Within a conductor with conductivity ${\sigma}$ (not surface charge density!) Ohm’s law can be written as

$\displaystyle \mathbf{J}_{f}=\sigma\mathbf{E} \ \ \ \ \ (9)$

so Maxwell’s equations become

 $\displaystyle \nabla\cdot\mathbf{E}$ $\displaystyle =$ $\displaystyle \frac{1}{\epsilon}\rho_{f}\ \ \ \ \ (10)$ $\displaystyle \nabla\cdot\mathbf{B}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (11)$ $\displaystyle \nabla\times\mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\partial\mathbf{B}}{\partial t}\ \ \ \ \ (12)$ $\displaystyle \nabla\times\mathbf{B}$ $\displaystyle =$ $\displaystyle \mu\sigma\mathbf{E}+\mu\epsilon\frac{\partial\mathbf{E}}{\partial t} \ \ \ \ \ (13)$

The conservation of charge (continuity condition) says

$\displaystyle \nabla\cdot\mathbf{J}_{f}=-\frac{\partial\rho_{f}}{\partial t} \ \ \ \ \ (14)$

so from 9 and 10 we get

$\displaystyle \frac{\partial\rho_{f}}{\partial t}=-\sigma\nabla\cdot\mathbf{E}=-\frac{\sigma}{\epsilon}\rho_{f} \ \ \ \ \ (15)$

For a fixed location within the conductor we can integrate this with respect to time to get

$\displaystyle \rho_{f}\left(t\right)=\rho_{f}\left(0\right)e^{-\sigma t/\epsilon} \ \ \ \ \ (16)$

That is, the free charge density decays exponentially with a characteristic time ${\tau=\epsilon/\sigma}$.

Example For glass, the conductivity is around ${10^{-11}\mbox{ S m}^{-1}}$ (S stands for Siemens, which is the SI unit of conductance, where ${1\mbox{ S}=1\mbox{ kg}^{-1}\mbox{m}^{-2}\mbox{s}^{3}\mbox{A}^{2}}$) and the permittivity is ${4.7\epsilon_{0}=4.16\times10^{-11}\mbox{m}^{-3}\mbox{kg}^{-1}\mbox{s}^{4}\mbox{A}^{2}}$ so the charge would move to the surface in a time of about

$\displaystyle \tau=4.16\mbox{ s} \ \ \ \ \ (17)$

Some forms of glass have much smaller conductivities, so the migration time would be correspondingly larger.