# Average of product of two waves

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 9, Post 11.

A common calculation that is required when analyzing any system that varies with a sinusoidal period is a time average over one cycle. For example, a monochromatic plane wave with amplitude ${A}$, direction ${\mathbf{k}}$, frequency ${\omega}$ and phase ${\delta}$ can be written as

 $\displaystyle f$ $\displaystyle =$ $\displaystyle A\cos\left(\mathbf{k}\cdot\mathbf{r}-\omega t+\delta\right)=\Re\tilde{A}e^{i\left(\mathbf{k}\cdot\mathbf{r}-\omega t\right)}\ \ \ \ \ (1)$ $\displaystyle \tilde{A}$ $\displaystyle =$ $\displaystyle Ae^{i\delta} \ \ \ \ \ (2)$

Now suppose we have two waves with the same direction and frequency, but different amplitudes and phases. Then

 $\displaystyle f$ $\displaystyle =$ $\displaystyle A\cos\left(\mathbf{k}\cdot\mathbf{r}-\omega t+\delta_{a}\right)\ \ \ \ \ (3)$ $\displaystyle g$ $\displaystyle =$ $\displaystyle B\cos\left(\mathbf{k}\cdot\mathbf{r}-\omega t+\delta_{b}\right) \ \ \ \ \ (4)$

The average of the product of these waves over a single cycle is then

$\displaystyle \left\langle fg\right\rangle =\frac{\omega AB}{2\pi}\int_{0}^{2\pi/\omega}\cos\left(\mathbf{k}\cdot\mathbf{r}-\omega t+\delta_{a}\right)\cos\left(\mathbf{k}\cdot\mathbf{r}-\omega t+\delta_{b}\right)dt \ \ \ \ \ (5)$

We can transform this integral by defining

 $\displaystyle \theta$ $\displaystyle \equiv$ $\displaystyle \mathbf{k}\cdot\mathbf{r}-\omega t\ \ \ \ \ (6)$ $\displaystyle d\theta$ $\displaystyle =$ $\displaystyle -\omega dt\ \ \ \ \ (7)$ $\displaystyle \left\langle fg\right\rangle$ $\displaystyle =$ $\displaystyle \frac{AB}{2\pi}\int_{0}^{2\pi}\cos\left(\theta+\delta_{a}\right)\cos\left(\theta+\delta_{b}\right)d\theta \ \ \ \ \ (8)$

We’ve used the limits of 0 and ${2\pi}$ since any interval of ${2\pi}$ covers one complete cycle of ${\theta}$.

The two cosines have the same period and differ only in their phase, so we will get the same result from the integral if we replace them by

 $\displaystyle \cos\left(\theta+\delta_{a}\right)\cos\left(\theta+\delta_{b}\right)$ $\displaystyle \rightarrow$ $\displaystyle \cos\theta\cos\left(\theta+\delta_{a}-\delta_{b}\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \cos^{2}\theta\cos\left(\delta_{a}-\delta_{b}\right)-\cos\theta\sin\theta\sin\left(\delta_{a}-\delta_{b}\right) \ \ \ \ \ (10)$

We now have

 $\displaystyle \left\langle fg\right\rangle$ $\displaystyle =$ $\displaystyle \frac{AB}{2\pi}\cos\left(\delta_{a}-\delta_{b}\right)\int_{0}^{2\pi}\cos^{2}\theta d\theta-\frac{AB}{2\pi}\sin\left(\delta_{a}-\delta_{b}\right)\int_{0}^{2\pi}\cos\theta\sin\theta d\theta\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}AB\cos\left(\delta_{a}-\delta_{b}\right)-0\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}AB\cos\left(\delta_{a}-\delta_{b}\right)\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\Re\left(fg^*\right)=\frac{1}{2}\Re\left(f^*g\right) \ \ \ \ \ (14)$

Thus we can get the answer using complex notation without doing any integrals.

This applies to vector products as well, since the components of vector products are just products of scalar functions. For example, the time average of the Poynting vector becomes, when the electric and magnetic fields are written in complex notation:

$\displaystyle \left\langle \mathbf{S}\right\rangle =\frac{1}{2\mu_{0}}\Re\left(\tilde{\mathbf{E}}\times\tilde{\mathbf{B}}^*\right) \ \ \ \ \ (15)$

The electromagnetic energy density in the fields has a time average of

 $\displaystyle \left\langle u_{em}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{4}\Re\left(\epsilon_{0}\tilde{\mathbf{E}}\cdot\tilde{\mathbf{E}}^*+\frac{1}{\mu_{0}}\tilde{\mathbf{B}}\cdot\tilde{\mathbf{B}}^*\right)\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{4}\left(\epsilon_{0}\tilde{\mathbf{E}}\cdot\tilde{\mathbf{E}}^*+\frac{1}{\mu_{0}}\tilde{\mathbf{B}}\cdot\tilde{\mathbf{B}}^*\right)\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{4\mu_{0}}\left(\frac{1}{c^{2}}\tilde{\mathbf{E}}\cdot\tilde{\mathbf{E}}^*+\tilde{\mathbf{B}}\cdot\tilde{\mathbf{B}}^*\right) \ \ \ \ \ (18)$

We dropped the ${\Re}$ in line 2 since the quantity in parentheses is automatically real anyway, and in the last line we used

$\displaystyle \mu_{0}\epsilon_{0}=\frac{1}{c^{2}} \ \ \ \ \ (19)$

# Potentials for an electromagnetic wave

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 10, Post 4

We can define an electromagnetic wave in terms of electric and magnetic potentials as follows. Using rectangular coordinates, let

 $\displaystyle V$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (1)$ $\displaystyle \mathbf{A}$ $\displaystyle =$ $\displaystyle A_{0}\sin\left(kx-\omega t\right)\hat{\mathbf{y}} \ \ \ \ \ (2)$

These potentials give rise to the fields

 $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle \nabla\times\mathbf{A}=A_{0}k\cos\left(kx-\omega t\right)\hat{\mathbf{z}}\ \ \ \ \ (3)$ $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\nabla V-\frac{\partial\mathbf{A}}{\partial t}=A_{0}\omega\cos\left(kx-\omega t\right)\hat{\mathbf{y}} \ \ \ \ \ (4)$

We can check that these fields satisfy Maxwell’s equations in vacuum. First,

$\displaystyle \nabla\cdot\mathbf{E}=\nabla\cdot\mathbf{B}=0 \ \ \ \ \ (5)$

For the curls, we have

 $\displaystyle \nabla\times\mathbf{E}$ $\displaystyle =$ $\displaystyle -A_{0}k\omega\sin\left(kx-\omega t\right)\hat{\mathbf{z}}=-\frac{\partial\mathbf{B}}{\partial t}\ \ \ \ \ (6)$ $\displaystyle \nabla\times\mathbf{B}$ $\displaystyle =$ $\displaystyle A_{0}k^{2}\sin\left(kx-\omega t\right)\hat{\mathbf{y}} \ \ \ \ \ (7)$

The second equation should be equal to ${\mu_{0}\epsilon_{0}\frac{\partial\mathbf{E}}{\partial t}=\frac{1}{c^{2}}\frac{\partial\mathbf{E}}{\partial t}}$ so

 $\displaystyle A_{0}k^{2}\sin\left(kx-\omega t\right)\hat{\mathbf{y}}$ $\displaystyle =$ $\displaystyle \frac{1}{c^{2}}\frac{\partial\mathbf{E}}{\partial t}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{c^{2}}A_{0}\omega^{2}\sin\left(kx-\omega t\right)\hat{\mathbf{y}} \ \ \ \ \ (9)$

which can be true only if

$\displaystyle k^{2}=\frac{\omega^{2}}{c^{2}} \ \ \ \ \ (10)$

which is the usual relation between wave number ${k}$ and angular frequency ${\omega}$.

# Total internal reflection

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.37.

When we looked at the behaviour of waves passing from one medium to another at an angle, one of the consequences was Snell’s law of refraction which says

$\displaystyle \frac{\sin\theta_{I}}{\sin\theta_{T}}=\frac{n_{2}}{n_{1}} \ \ \ \ \ (1)$

where ${\theta_{I}}$ is the angle of incidence (angle between the wave vector and the normal to the surface) in medium 1 with index of refraction ${n_{1}}$, and ${\theta_{T}}$ is the angle of the refracted wave in medium 2. If ${n_{1}>n_{2}}$, that is, the wave is incident from a medium (such as water) with a higher index of refraction than medium 2 (such as air), then we can reach a critical incident angle ${\theta_{c}}$ where the refracted angle is ${\pi/2}$, so that the refracted wave moves parallel to the interface. This happens when

$\displaystyle \sin\theta_{c}=\frac{n_{2}}{n_{1}} \ \ \ \ \ (2)$

If ${\theta_{I}>\theta_{c}}$, no wave is transmitted through the interface and the entire wave is reflected back into medium 1. This is known as total internal reflection.

To see what happens in this case, we can follow through the same derivation as before, except we allow the wave vector ${\mathbf{k}_{T}}$ of the transmitted wave to be complex. That is, for a given ${\theta_{T}}$ we say that (assuming that the incident and transmitted plane is the ${xz}$ plane):

$\displaystyle \mathbf{k}_{T}=k_{T}\left(\sin\theta_{T}\hat{\mathbf{x}}+\cos\theta_{T}\hat{\mathbf{z}}\right) \ \ \ \ \ (3)$

The wave vector has, as usual, the magnitude of

$\displaystyle k_{T}=\frac{\omega}{v_{2}}=\frac{\omega n_{2}}{c} \ \ \ \ \ (4)$

Now suppose ${\theta_{I}>\theta_{c}}$. In that case,

 $\displaystyle \sin\theta_{T}$ $\displaystyle =$ $\displaystyle \frac{n_{1}}{n_{2}}\sin\theta_{I}>\frac{n_{1}}{n_{2}}\frac{n_{2}}{n_{1}}=1\ \ \ \ \ (5)$ $\displaystyle \cos\theta_{T}$ $\displaystyle =$ $\displaystyle \sqrt{1-\sin^{2}\theta_{T}}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\sqrt{\sin^{2}\theta_{T}-1}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\sqrt{\frac{n_{1}^{2}}{n_{2}^{2}}\sin^{2}\theta_{I}-1}\ \ \ \ \ (8)$ $\displaystyle \mathbf{k}_{T}$ $\displaystyle =$ $\displaystyle k_{T}\left(\frac{n_{1}}{n_{2}}\sin\theta_{I}\hat{\mathbf{x}}+i\sqrt{\frac{n_{1}^{2}}{n_{2}^{2}}\sin^{2}\theta_{I}-1}\hat{\mathbf{z}}\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\omega n_{2}}{c}\left(\frac{n_{1}}{n_{2}}\sin\theta_{I}\hat{\mathbf{x}}+i\sqrt{\frac{n_{1}^{2}}{n_{2}^{2}}\sin^{2}\theta_{I}-1}\hat{\mathbf{z}}\right)\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle k\hat{\mathbf{x}}+i\kappa\hat{\mathbf{z}} \ \ \ \ \ (11)$

where

 $\displaystyle k$ $\displaystyle \equiv$ $\displaystyle \frac{\omega n_{1}}{c}\sin\theta_{I}\ \ \ \ \ (12)$ $\displaystyle \kappa$ $\displaystyle \equiv$ $\displaystyle \frac{\omega}{c}\sqrt{n_{1}^{2}\sin^{2}\theta_{I}-n_{2}^{2}} \ \ \ \ \ (13)$

That is, we venture into the realm of complex variables, and ${\theta_{T}}$ can no longer be interpreted as a geometric angle. However, let’s proceed with the analysis and see what happens.

First, we look at the electric field, which has the general form

$\displaystyle \tilde{\mathbf{E}}_{T}\left(\mathbf{r},t\right)=\tilde{\mathbf{E}}_{0_{T}}e^{i\left(\mathbf{k}_{T}\cdot\mathbf{r}-\omega t\right)} \ \ \ \ \ (14)$

Plugging in 11, we get

$\displaystyle \tilde{\mathbf{E}}_{T}\left(\mathbf{r},t\right)=\tilde{\mathbf{E}}_{0_{T}}e^{-\kappa z}e^{i\left(kx-\omega t\right)} \ \ \ \ \ (15)$

That is, the transmitted wave propagates in the ${x}$ direction (parallel to the interface) and is attenuated in the ${z}$ direction.

How much of the wave is reflected in this case? For a wave with polarization parallel to the incident plane (that is, ${\mathbf{E}}$ has only an ${x}$ component), we found earlier That the reflected amplitude is

$\displaystyle E_{R}=\frac{\alpha-\beta}{\alpha+\beta}E_{I} \ \ \ \ \ (16)$

where

 $\displaystyle \alpha$ $\displaystyle \equiv$ $\displaystyle \frac{\cos\theta_{T}}{\cos\theta_{I}}\ \ \ \ \ (17)$ $\displaystyle \beta$ $\displaystyle \equiv$ $\displaystyle \frac{\mu_{1}v_{1}}{\mu_{2}v_{2}}=\frac{\mu_{1}n_{2}}{\mu_{2}n_{1}} \ \ \ \ \ (18)$

In this case, ${\alpha}$ is purely imaginary and ${\beta}$ is real, so the reflection coefficient is

$\displaystyle R=\left|\frac{E_{R}}{E_{I}}\right|^{2}=\left|\frac{\alpha-\beta}{\alpha+\beta}\right|=1 \ \ \ \ \ (19)$

since

$\displaystyle \left|\alpha-\beta\right|^{2}=\left|\alpha\right|^{2}+\beta^{2}=\left|\alpha+\beta\right|^{2} \ \ \ \ \ (20)$

For perpendicular polarization, we have

$\displaystyle E_{R}=\frac{1-\alpha\beta}{1+\alpha\beta}E_{I} \ \ \ \ \ (21)$

and again, since 1 is real and ${\alpha\beta}$ is purely imaginary

$\displaystyle R=\left|\frac{1-\alpha\beta}{1+\alpha\beta}\right|^{2}=1 \ \ \ \ \ (22)$

Thus the reflection is indeed total for both polarizations.

Still with perpendicular polarization, the electric field is entirely in the ${y}$ direction so

$\displaystyle \tilde{\mathbf{E}}_{T}=\tilde{E}_{0}e^{-\kappa z}e^{i\left(kx-\omega t\right)}\hat{\mathbf{y}} \ \ \ \ \ (23)$

The magnetic field is given by (using 11 and ${\mathbf{k}_{T}=\frac{\omega n_{2}}{c}\hat{\mathbf{k}}_{T}}$ and ${v_{2}=c/n_{2}}$)

 $\displaystyle \tilde{\mathbf{B}}_{T}$ $\displaystyle =$ $\displaystyle \frac{1}{v_{2}}\hat{\mathbf{k}}_{T}\times\tilde{\mathbf{E}}_{T}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{v_{2}}\frac{c}{\omega n_{2}}\tilde{E}_{0}e^{-\kappa z}e^{i\left(kx-\omega t\right)}\left(k\hat{\mathbf{z}}-i\kappa\hat{\mathbf{x}}\right)\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\tilde{E}_{0}}{\omega}e^{-\kappa z}e^{i\left(kx-\omega t\right)}\left(k\hat{\mathbf{z}}-i\kappa\hat{\mathbf{x}}\right) \ \ \ \ \ (26)$

Taking the real parts to get the actual fields, we have

 $\displaystyle \mathbf{E}_{T}$ $\displaystyle =$ $\displaystyle E_{0}e^{-\kappa z}\cos\left(kx-\omega t\right)\hat{\mathbf{y}}\ \ \ \ \ (27)$ $\displaystyle \mathbf{B}_{T}$ $\displaystyle =$ $\displaystyle \frac{E_{0}}{\omega}e^{-\kappa z}\left(k\cos\left(kx-\omega t\right)\hat{\mathbf{z}}+\kappa\sin\left(kx-\omega t\right)\hat{\mathbf{x}}\right) \ \ \ \ \ (28)$

To check that these fields satisfy Maxwell’s equations requires grinding away at some derivatives, which we can do in Maple. The divergences are fairly easy and we get

$\displaystyle \nabla\cdot\mathbf{E}_{T}=\nabla\cdot\mathbf{B}_{T}=0 \ \ \ \ \ (29)$

The curls give

 $\displaystyle \nabla\times\mathbf{E}_{T}$ $\displaystyle =$ $\displaystyle E_{0}e^{-\kappa z}\left(-k\sin\left(kx-\omega t\right)\hat{\mathbf{z}}+\kappa\cos\left(kx-\omega t\right)\hat{\mathbf{x}}\right)\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\partial\mathbf{B}_{T}}{\partial t}\ \ \ \ \ (31)$ $\displaystyle \nabla\times\mathbf{B}_{T}$ $\displaystyle =$ $\displaystyle \frac{E_{0}}{\omega}e^{-\kappa z}\sin\left(kx-\omega t\right)\hat{\mathbf{y}}\left(k^{2}-\kappa^{2}\right)\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{n_{2}^{2}}{c^{2}}\omega E_{0}e^{-\kappa z}\sin\left(kx-\omega t\right)\hat{\mathbf{y}}\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{v_{2}^{2}}\frac{\partial\mathbf{E}_{T}}{\partial t} \ \ \ \ \ (34)$

Thus all four of Maxwell’s equations are satisfied.

The Poynting vector is

 $\displaystyle \mathbf{S}$ $\displaystyle =$ $\displaystyle \frac{1}{\mu}\mathbf{E}_{T}\times\mathbf{B}_{T}\ \ \ \ \ (35)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{E_{0}^{2}e^{-2\kappa z}}{\mu\omega}\left(k\cos^{2}\left(kx-\omega t\right)\hat{\mathbf{x}}-\kappa\sin\left(kx-\omega t\right)\cos\left(kx-\omega t\right)\hat{\mathbf{z}}\right) \ \ \ \ \ (36)$

Integrating this over one cycle (${t=0}$ to ${2\pi/\omega}$) gives zero for the ${z}$ component, thus no energy is transmitted perpendicular to the interface, and all the energy flows in the ${x}$ direction, parallel to the interface.

# Microwave shielding for perfect transmission

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 9, Post 35.

Here’s a simple example of the transmission coefficient for waves passing from medium 1 through medium 2 into medium 3. Suppose we want to protect a microwave antenna from the weather by enclosing it in a plastic shield, where the dielectric constant of the plastic is 2.5. If the antenna radiates at 10 GHz, what is the best (that is, smallest) thickness of plastic?

The transmission coefficient is given by

$\displaystyle T^{-1}=\frac{1}{4n_{1}n_{3}}\left[\left(n_{1}+n_{3}\right)^{2}+\sin^{2}\left(\frac{n_{2}\omega d}{c}\right)\frac{\left(n_{1}^{2}-n_{2}^{2}\right)\left(n_{3}^{2}-n_{2}^{2}\right)}{n_{2}^{2}}\right] \ \ \ \ \ (1)$

where ${n_{j}}$ is the index of refraction of medium ${j}$, given by

$\displaystyle n=\sqrt{\frac{\mu\epsilon}{\mu_{0}\epsilon_{0}}} \ \ \ \ \ (2)$

If ${\mu=\mu_{0}}$ then

$\displaystyle n_{2}=\sqrt{\frac{\epsilon_{2}}{\epsilon_{0}}}=\sqrt{\epsilon_{r_{2}}}=\sqrt{2.5} \ \ \ \ \ (3)$

where ${\epsilon_{r_{2}}}$ is the dielectric constant of the plastic.

We can see that if the sine is zero and ${n_{1}=n_{3}}$ (which we’re assuming here, since both mediums 1 and 3 are air with ${n=1}$), then ${T=1}$ and we get perfect transmission. The smallest thickness ${d}$ is such that

 $\displaystyle \frac{n_{2}\omega d}{c}$ $\displaystyle =$ $\displaystyle \pi\ \ \ \ \ (4)$ $\displaystyle \frac{2\pi\nu d\sqrt{2.5}}{c}$ $\displaystyle =$ $\displaystyle \pi\ \ \ \ \ (5)$ $\displaystyle d$ $\displaystyle =$ $\displaystyle \frac{c}{2\sqrt{2.5}\nu}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 9.49\times10^{-3}\mbox{ m} \ \ \ \ \ (7)$

A plastic shield of about 9.5 mm thickness would allow perfect transmission at that frequency.

# Spherical electromagnetic wave

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 9, Post 33.

One form of spherical wave has an electric component given by

$\displaystyle \mathbf{E}\left(r,\theta\phi,t\right)=A\frac{\sin\theta}{r}\left[\cos\left(kr-\omega t\right)-\frac{\sin\left(kr-\omega t\right)}{kr}\right]\hat{\boldsymbol{\phi}} \ \ \ \ \ (1)$

where ${A}$ is a constant. We can derive the corresponding magnetic field from Maxwell’s equations in vacuum:

 $\displaystyle \nabla\cdot\mathbf{E}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (2)$ $\displaystyle \nabla\cdot\mathbf{B}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (3)$ $\displaystyle \nabla\times\mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\partial\mathbf{B}}{\partial t}\ \ \ \ \ (4)$ $\displaystyle \nabla\times\mathbf{B}$ $\displaystyle =$ $\displaystyle \mu_{0}\epsilon_{0}\frac{\partial\mathbf{E}}{\partial t} \ \ \ \ \ (5)$

The calculations are straightforward but can get messy, so we’ll use Maple to do the derivatives. We get

 $\displaystyle \nabla\times\mathbf{E}$ $\displaystyle =$ $\displaystyle \frac{1}{r\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta E_{\phi}\right)\hat{\mathbf{r}}-\frac{1}{r}\frac{\partial}{\partial r}\left(rE_{\phi}\right)\hat{\boldsymbol{\theta}}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2A\cos\theta}{r^{2}}\left[\cos\left(kr-\omega t\right)-\frac{\sin\left(kr-\omega t\right)}{kr}\right]\hat{\mathbf{r}}+\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{A\sin\theta}{r}\left[\left(k-\frac{1}{kr^{2}}\right)\sin\left(kr-\omega t\right)+\frac{\cos\left(kr-\omega t\right)}{r}\right]\hat{\boldsymbol{\theta}}\nonumber$

Integrating this with respect to ${t}$ we get

 $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle \frac{2A\cos\theta}{r^{2}\omega}\left[\sin\left(kr-\omega t\right)+\frac{\cos\left(kr-\omega t\right)}{kr}\right]\hat{\mathbf{r}}+\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{A\sin\theta}{r^{3}\omega}\left[\left(\frac{1}{k}-kr^{2}\right)\cos\left(kr-\omega t\right)+r\sin\left(kr-\omega t\right)\right]\hat{\boldsymbol{\theta}} \ \ \ \ \ (9)$

We can verify that ${\mathbf{E}}$ and ${\mathbf{B}}$ satisfy the other 3 Maxwell equations by direct calculation.

 $\displaystyle \nabla\cdot\mathbf{E}$ $\displaystyle =$ $\displaystyle \frac{1}{r\sin\theta}\frac{\partial}{\partial\phi}\left(E_{\phi}\right)=0\ \ \ \ \ (10)$ $\displaystyle \nabla\cdot\mathbf{B}$ $\displaystyle =$ $\displaystyle \frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^{2}B_{r}\right)+\frac{1}{r\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta B_{\theta}\right)\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2A\cos\theta}{r^{2}\omega}\left[\left(k-\frac{1}{kr^{2}}\right)\cos\left(kr-\omega t\right)-\frac{\sin\left(kr-\omega t\right)}{r}\right]+\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{2A\cos\theta}{r^{2}\omega}\left[\frac{1}{r^{2}}\left(\frac{1}{k}-kr^{2}\right)\cos\left(kr-\omega t\right)+\frac{\sin\left(kr-\omega t\right)}{r}\right]\nonumber$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (13)$ $\displaystyle \nabla\times\mathbf{B}$ $\displaystyle =$ $\displaystyle \frac{1}{r}\left[\frac{\partial}{\partial r}\left(rB_{\theta}\right)-\frac{\partial B_{r}}{\partial\theta}\right]\hat{\boldsymbol{\phi}}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{2A\sin\theta}{\omega r^{3}}\left(\left(\frac{1}{kr}-kr\right)\cos\left(kr-\omega t\right)+\sin\left(kr-\omega t\right)\right)\hat{\boldsymbol{\phi}}+\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{A\sin\theta}{\omega r^{3}}\left[-kr\cos\left(kr-\omega t\right)-\left(\frac{1}{k}-kr^{2}-1\right)\sin\left(kr-\omega t\right)\right]\hat{\boldsymbol{\phi}}+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{2A\sin\theta}{\omega r^{3}}\left[\sin\left(kr-\omega t\right)+\frac{\cos\left(kr-\omega t\right)}{kr}\right]\hat{\boldsymbol{\phi}}\nonumber$ $\displaystyle$ $\displaystyle =$ $\displaystyle A\frac{\sin\theta}{r^{2}c}\left[kr\sin\left(kr-\omega t\right)+\cos\left(kr-\omega t\right)\right]\hat{\boldsymbol{\phi}} \ \ \ \ \ (16)$

where in the last line we used ${\omega/k=c}$ and collected terms. From 1 we get

 $\displaystyle \frac{\partial\mathbf{E}}{\partial t}$ $\displaystyle =$ $\displaystyle A\frac{\sin\theta}{r}\left[\omega\sin\left(kr-\omega t\right)+\frac{\omega\cos\left(kr-\omega t\right)}{kr}\right]\hat{\boldsymbol{\phi}}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle A\frac{\sin\theta}{r^{2}}\left[krc\sin\left(kr-\omega t\right)+c\cos\left(kr-\omega t\right)\right]\hat{\boldsymbol{\phi}}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle c^{2}\nabla\times\mathbf{B} \ \ \ \ \ (19)$

so the final Maxwell equation is satisfied.

The Poynting vector is, after simplifying

 $\displaystyle \mathbf{S}$ $\displaystyle =$ $\displaystyle \frac{1}{\mu_{0}}\mathbf{E}\times\mathbf{B}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hat{\mathbf{r}}\frac{A^{2}\sin^{2}\theta}{\mu_{0}\omega kr^{5}}\left[\left(\frac{1}{2}-r^{2}k^{2}\right)\sin\left[2\left(kr-\omega t\right)\right]-kr\cos\left[2\left(kr-\omega t\right)\right]+k^{3}r^{3}\cos^{2}\left(kr-\omega t\right)\right]+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \hat{\boldsymbol{\theta}}\frac{A^{2}\sin2\theta}{\mu_{0}\omega kr^{5}}\left[\frac{1}{2}\left(k^{2}r^{2}-1\right)\sin\left[2\left(kr-\omega t\right)\right]+kr\cos\left[2\left(kr-\omega t\right)\right]\right] \ \ \ \ \ (21)$

The intensity, or time average of the Poynting vector is

 $\displaystyle \mathbf{I}=\left\langle \mathbf{S}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{\omega}{2\pi}\int_{0}^{2\pi/\omega}\mathbf{S}dt\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{A^{2}k\sin^{2}\theta}{2\mu_{0}\omega r^{2}}\hat{\mathbf{r}} \ \ \ \ \ (23)$

The energy flows radially outwards and falls off as ${r^{-2}}$.

The total power radiated is the integral of ${\mathbf{I}\cdot d\mathbf{a}}$ over a sphere:

 $\displaystyle P$ $\displaystyle =$ $\displaystyle \int\mathbf{I}\cdot d\mathbf{a}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \pi\frac{A^{2}k}{\mu_{0}\omega}\int_{0}^{\pi}\frac{r^{2}}{r^{2}}\sin^{3}\theta d\theta\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4\pi A^{2}}{3\mu_{0}}\frac{k}{\omega}\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4\pi A^{2}}{3\mu_{0}c} \ \ \ \ \ (27)$

# Transmission coefficient for a wave passing through 3 media

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 9, Post 34.

We can extend the analysis of reflection and transmission of waves at a boundary by considering the case of an electromagnetic wave starting out in medium 1 (with wave speed ${v_{1}}$, wave number ${k_{1}}$ and index of refraction ${n_{1}=c/v_{1}}$), then passing at normal incidence to medium 2 at ${z=-d}$ and then to medium 3 at ${z=0}$. (We’ve changed the origin from that stated in Griffiths’s problem to make the analysis a bit easier, as we’ll see). We’d like to find the transmission coefficient between mediums 1 and 3, that is, we’d like to see how much of the wave’s energy gets transmitted all the way through the middle medium. We’ll assume the mediums are all homogeneous and linear, and that ${\mu=\mu_{0}}$ in all of them.

The analysis is much the same as the earlier method, but a bit more complicated. We have a wave ${\tilde{\mathbf{E}}_{1R}}$ travelling in towards the right in medium 1. At the boundary with medium 2, it gives rise to a reflected wave ${\tilde{\mathbf{E}}_{1L}}$ travelling to the left in medium 1 and a trasmitted wave ${\tilde{\mathbf{E}}_{2R}}$ travelling to the right in medium 2. When this wave hits the boundary with medium 3, there is a reflected wave ${\tilde{\mathbf{E}}_{2L}}$ travelling to the left and a transmitted wave ${\tilde{\mathbf{E}}_{3R}}$ travelling to the right. There are, of course, corresponding magnetic waves ${\tilde{\mathbf{B}}_{1L}}$ and so on. We can then apply the boundary conditions to work out the amplitudes. [Actually, the wave reflected back to the left from the 2-3 boundary will hit the 1-2 boundary and be reflected and transmitted there too, so that there is, in principle, an infinite number of reflected and transmitted waves resulting from the wave bouncing back and forth between the two boundaries. However, we can subsume all the left-moving waves into ${\tilde{\mathbf{E}}_{1L}}$ and ${\tilde{\mathbf{E}}_{2L}}$ and all the right moving waves into ${\tilde{\mathbf{E}}_{1R}}$, ${\tilde{\mathbf{E}}_{2R}}$ and ${\tilde{\mathbf{E}}_{3R}}$. The important thing is that these waves must satisfy the boundary conditions.]

We can take the electric component to be polarized along the ${x}$ direction, so that the magnetic component is then along the ${y}$ direction. The waves are

 $\displaystyle \tilde{\mathbf{E}}_{1L}$ $\displaystyle =$ $\displaystyle E_{1L}e^{i\left(-k_{1}z-\omega t\right)}\hat{\mathbf{x}}\ \ \ \ \ (1)$ $\displaystyle \tilde{\mathbf{E}}_{1R}$ $\displaystyle =$ $\displaystyle E_{1R}e^{i\left(k_{1}z-\omega t\right)}\hat{\mathbf{x}}\ \ \ \ \ (2)$ $\displaystyle \tilde{\mathbf{E}}_{2L}$ $\displaystyle =$ $\displaystyle E_{2L}e^{i\left(-k_{2}z-\omega t\right)}\hat{\mathbf{x}}\ \ \ \ \ (3)$ $\displaystyle \tilde{\mathbf{E}}_{2R}$ $\displaystyle =$ $\displaystyle E_{2R}e^{i\left(-k_{2}z-\omega t\right)}\hat{\mathbf{x}}\ \ \ \ \ (4)$ $\displaystyle \tilde{\mathbf{E}}_{3R}$ $\displaystyle =$ $\displaystyle E_{3R}e^{i\left(k_{3}z-\omega t\right)}\hat{\mathbf{x}}\ \ \ \ \ (5)$ $\displaystyle \tilde{\mathbf{B}}_{1L}$ $\displaystyle =$ $\displaystyle -\frac{1}{v_{1}}E_{1L}e^{i\left(-k_{1}z-\omega t\right)}\hat{\mathbf{y}}\ \ \ \ \ (6)$ $\displaystyle \tilde{\mathbf{B}}_{1R}$ $\displaystyle =$ $\displaystyle \frac{1}{v_{1}}E_{1R}e^{i\left(k_{1}z-\omega t\right)}\hat{\mathbf{y}}\ \ \ \ \ (7)$ $\displaystyle \tilde{\mathbf{B}}_{2L}$ $\displaystyle =$ $\displaystyle -\frac{1}{v_{2}}E_{2L}e^{i\left(-k_{2}z-\omega t\right)}\hat{\mathbf{y}}\ \ \ \ \ (8)$ $\displaystyle \tilde{\mathbf{B}}_{2R}$ $\displaystyle =$ $\displaystyle \frac{1}{v_{2}}E_{2R}e^{i\left(k_{2}z-\omega t\right)}\hat{\mathbf{y}}\ \ \ \ \ (9)$ $\displaystyle \tilde{\mathbf{B}}_{3R}$ $\displaystyle =$ $\displaystyle \frac{1}{v_{3}}E_{3R}e^{i\left(k_{3}z-\omega t\right)}\hat{\mathbf{y}} \ \ \ \ \ (10)$

The negative signs for the left-moving magnetic waves are to keep the Poynting vector pointing to the left. The coefficients ${E_{1L}}$ and so on are actually complex numbers, but we’ve dropped the tilde (and the subscript 0 that Griffiths uses) to make the notation simpler.

Because the medium 2-3 boundary consists of an incident right-moving wave, a reflected left-moving wave and a transmitted wave, it is identical to the case we treated earlier, provided we take ${z=0}$ at this point (which we’ve done). We can therefore write down the results:

 $\displaystyle E_{2L}$ $\displaystyle =$ $\displaystyle \frac{v_{3}-v_{2}}{v_{2}+v_{3}}E_{2R}\ \ \ \ \ (11)$ $\displaystyle E_{3R}$ $\displaystyle =$ $\displaystyle \frac{2v_{3}}{v_{2}+v_{3}}E_{2R} \ \ \ \ \ (12)$

Now for the medium 1-2 boundary at ${z=-d}$. From the boundary condition ${\mathbf{E}_{1}^{\parallel}=\mathbf{E}_{2}^{\parallel}}$ and (since ${\mu=\mu_{0}}$
everywhere) ${\mathbf{B}_{1}^{\parallel}=\mathbf{B}_{2}^{\parallel}}$ we get

 $\displaystyle E_{2R}e^{-ik_{2}d}+E_{2L}e^{ik_{2}d}$ $\displaystyle =$ $\displaystyle E_{1R}e^{-ik_{1}d}+E_{1L}e^{ik_{1}d}\ \ \ \ \ (13)$ $\displaystyle \frac{1}{v_{2}}\left(E_{2R}e^{-ik_{2}d}-E_{2L}e^{ik_{2}d}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{v_{1}}\left(E_{1R}e^{-ik_{1}d}-E_{1L}e^{ik_{1}d}\right) \ \ \ \ \ (14)$

These 4 equations are linear in the ${E}$ coefficients so it’s straightforward (although tedious) to solve them. It’s easiest to let Maple handle this part, and we get (since we’re interested only in expressing ${E_{3R}}$ in terms of ${E_{1R}}$):

 $\displaystyle E_{1R}$ $\displaystyle =$ $\displaystyle E_{3R}\frac{e^{ik_{1}d}}{4v_{2}v_{3}}\left[e^{ik_{2}d}\left(v_{2}v_{3}+v_{1}v_{2}-v_{2}^{2}-v_{1}v_{3}\right)+e^{-ik_{2}d}\left(v_{2}v_{3}+v_{1}v_{2}+v_{2}^{2}+v_{1}v_{3}\right)\right]\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle E_{3R}\frac{e^{ik_{1}d}}{4v_{2}v_{3}}\left[2\cos\left(k_{2}d\right)\left(v_{2}v_{3}+v_{1}v_{2}\right)-2i\sin\left(k_{2}d\right)\left(v_{2}^{2}+v_{1}v_{3}\right)\right] \ \ \ \ \ (16)$

The intensity of a wave is

$\displaystyle I=\frac{\left|E\right|^{2}}{2\mu v} \ \ \ \ \ (17)$

and the transmission coefficient is

$\displaystyle T=\frac{I_{3R}}{I_{1R}} \ \ \ \ \ (18)$

so we get

 $\displaystyle T^{-1}$ $\displaystyle =$ $\displaystyle \frac{v_{3}}{v_{1}}\frac{4\cos^{2}\left(k_{2}d\right)\left(v_{2}v_{3}+v_{1}v_{2}\right)^{2}+4\sin^{2}\left(k_{2}d\right)\left(v_{2}^{2}+v_{1}v_{3}\right)^{2}}{16v_{2}^{2}v_{3}^{2}}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{4v_{1}v_{3}}\left[\frac{\left(v_{2}v_{3}+v_{1}v_{2}\right)^{2}}{v_{2}^{2}}\left(1-\sin^{2}\left(k_{2}d\right)\right)+\frac{\left(v_{2}^{2}+v_{1}v_{3}\right)^{2}}{v_{2}^{2}}\sin^{2}\left(k_{2}d\right)\right]\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{4v_{1}v_{3}}\left[\left(v_{3}+v_{1}\right)^{2}+\sin^{2}\left(k_{2}d\right)\frac{\left(v_{2}^{2}+v_{1}v_{3}\right)^{2}-\left(v_{2}v_{3}+v_{1}v_{2}\right)^{2}}{v_{2}^{2}}\right]\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{4v_{1}v_{3}}\left[\left(v_{3}+v_{1}\right)^{2}+\sin^{2}\left(k_{2}d\right)\frac{\left(v_{1}^{2}-v_{2}^{2}\right)\left(v_{3}^{2}-v_{2}^{2}\right)}{v_{2}^{2}}\right] \ \ \ \ \ (22)$

We can express this in terms of the indexes of refraction by noting that since ${n_{i}=c/v_{i}}$ and the power of the ${v}$s is the same in numerator and denominator so the factors of ${c}$ cancel out and we can replace ${v_{i}}$ with ${1/n_{i}}$. We can then multiply the first term top and bottom by ${n_{1}^{2}n_{3}^{2}}$ and the second term top and bottom by ${n_{1}^{2}n_{3}^{2}n_{2}^{4}}$ to get

$\displaystyle T^{-1}=\frac{1}{4n_{1}n_{3}}\left[\left(n_{1}+n_{3}\right)^{2}+\sin^{2}\left(k_{2}d\right)\frac{\left(n_{1}^{2}-n_{2}^{2}\right)\left(n_{3}^{2}-n_{2}^{2}\right)}{n_{2}^{2}}\right] \ \ \ \ \ (23)$

Finally we note that the wave speed in medium 2 is ${v_{2}=\omega/k_{2}=c/n_{2}}$ so ${k_{2}=n_{2}\omega/c}$ and we get

$\displaystyle T^{-1}=\frac{1}{4n_{1}n_{3}}\left[\left(n_{1}+n_{3}\right)^{2}+\sin^{2}\left(\frac{n_{2}\omega d}{c}\right)\frac{\left(n_{1}^{2}-n_{2}^{2}\right)\left(n_{3}^{2}-n_{2}^{2}\right)}{n_{2}^{2}}\right] \ \ \ \ \ (24)$

# 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:

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

# 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 frequency 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.

# 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

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