Electromagnetic Waves in Matter

Step 1: Maxwell's equations in linear media

In a linear medium, $\mathbf{D} = \epsilon\mathbf{E}$ and $\mathbf{H} = \mathbf{B}/\mu$. With no free charges/currents:

$$\nabla \cdot \mathbf{E} = 0, \quad \nabla \cdot \mathbf{B} = 0$$

$$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}, \quad \nabla \times \mathbf{B} = \mu\epsilon\frac{\partial \mathbf{E}}{\partial t}$$

Step 2: Take the curl of Faraday's law

$$\nabla \times (\nabla \times \mathbf{E}) = -\frac{\partial}{\partial t}(\nabla \times \mathbf{B}) = -\mu\epsilon\frac{\partial^2 \mathbf{E}}{\partial t^2}$$

Step 3: Apply the vector identity

$$\nabla(\nabla \cdot \mathbf{E}) - \nabla^2\mathbf{E} = -\mu\epsilon\frac{\partial^2 \mathbf{E}}{\partial t^2}$$

Since $\nabla \cdot \mathbf{E} = 0$:

$$\nabla^2\mathbf{E} = \mu\epsilon\frac{\partial^2 \mathbf{E}}{\partial t^2}$$

Step 4: Identify the wave speed

$$v = \frac{1}{\sqrt{\mu\epsilon}} = \frac{1}{\sqrt{\mu_r\mu_0\epsilon_r\epsilon_0}} = \frac{c}{\sqrt{\mu_r\epsilon_r}} = \frac{c}{n}$$

Step 5: Derive the index of refraction

$$n = \frac{c}{v} = \sqrt{\mu_r\epsilon_r}$$

For non-magnetic materials ($\mu_r \approx 1$): $n \approx \sqrt{\epsilon_r}$. For example, water has $\epsilon_r \approx 1.77$ at optical frequencies, giving $n \approx 1.33$.

Step 6: Modified amplitude relation

From Faraday's law for a plane wave $e^{i(kz - \omega t)}$ in the medium:

$$ikE_0 = i\omega B_0 \implies B_0 = \frac{k}{\omega}E_0 = \frac{1}{v}E_0 = \frac{n}{c}E_0$$

Step 1: Equation of motion for a bound electron

Model each electron as a damped harmonic oscillator driven by the wave's electric field:

$$m\ddot{x} + m\gamma\dot{x} + m\omega_0^2 x = qE_0 e^{-i\omega t}$$

where $\omega_0$ is the natural frequency, $\gamma$ is the damping rate, and $q = -e$.

Step 2: Solve for the steady-state displacement

Try $x = \tilde{x}_0 e^{-i\omega t}$:

$$m(-\omega^2 - i\gamma\omega + \omega_0^2)\tilde{x}_0 = qE_0$$

$$\tilde{x}_0 = \frac{q/m}{\omega_0^2 - \omega^2 - i\gamma\omega}E_0$$

Step 3: Compute the polarization

The dipole moment per electron is $p = qx$. For $N$ molecules per unit volume with $f_j$ electrons having natural frequency $\omega_j$:

$$\mathbf{P} = Nq\mathbf{x} = \frac{Nq^2}{m}\sum_j \frac{f_j}{\omega_j^2 - \omega^2 - i\gamma_j\omega}\mathbf{E}$$

Step 4: Relate to electric susceptibility

Since $\mathbf{P} = \epsilon_0\chi_e\mathbf{E}$:

$$\chi_e(\omega) = \frac{Nq^2}{m\epsilon_0}\sum_j \frac{f_j}{\omega_j^2 - \omega^2 - i\gamma_j\omega}$$

Step 5: Obtain the complex dielectric function

$$\tilde{\epsilon}_r(\omega) = 1 + \chi_e = 1 + \frac{Nq^2}{m\epsilon_0}\sum_j \frac{f_j}{\omega_j^2 - \omega^2 - i\gamma_j\omega}$$

Step 6: Separate real and imaginary parts

$$\tilde{\epsilon}_r = \epsilon_r' + i\epsilon_r''$$

$$\epsilon_r' = 1 + \frac{Nq^2}{m\epsilon_0}\sum_j \frac{f_j(\omega_j^2 - \omega^2)}{(\omega_j^2 - \omega^2)^2 + \gamma_j^2\omega^2}$$

$$\epsilon_r'' = \frac{Nq^2}{m\epsilon_0}\sum_j \frac{f_j\gamma_j\omega}{(\omega_j^2 - \omega^2)^2 + \gamma_j^2\omega^2}$$

The real part governs dispersion; the imaginary part governs absorption.

Step 1: Complex refractive index

Since $\tilde{\epsilon}_r$ is complex, define the complex refractive index:

$$\tilde{n} = \sqrt{\tilde{\epsilon}_r} = n + i\kappa$$

where $n$ is the real refractive index and $\kappa$ is the extinction coefficient.

Step 2: Complex wave number

The wave number in the medium is:

$$\tilde{k} = \frac{\tilde{n}\,\omega}{c} = \frac{(n + i\kappa)\omega}{c} = k + i\alpha/2$$

where $k = n\omega/c$ (propagation) and $\alpha = 2\kappa\omega/c$ (absorption coefficient).

Step 3: Substitute into the plane wave

$$\tilde{\mathbf{E}} = E_0\,e^{i(\tilde{k}z - \omega t)}\,\hat{x} = E_0\,e^{-\kappa\omega z/c}\,e^{i(kz - \omega t)}\,\hat{x}$$

The wave propagates with phase velocity $v_p = c/n$ but decays exponentially with penetration depth.

Step 4: Derive Beer's law

The intensity is proportional to $|E|^2$:

$$I(z) = I_0\,e^{-2\kappa\omega z/c} = I_0\,e^{-\alpha z}$$

This is Beer's law (Beer-Lambert law), where $\alpha = 2\kappa\omega/c$ is the absorption coefficient in units of $\text{m}^{-1}$.

Step 5: Connect n and kappa to the dielectric function

From $(n + i\kappa)^2 = \epsilon_r' + i\epsilon_r''$:

$$n^2 - \kappa^2 = \epsilon_r', \qquad 2n\kappa = \epsilon_r''$$

Solving simultaneously:

$$n = \sqrt{\frac{|\tilde{\epsilon}_r| + \epsilon_r'}{2}}, \qquad \kappa = \sqrt{\frac{|\tilde{\epsilon}_r| - \epsilon_r'}{2}}$$

Step 1: Construct a wave packet from two nearby frequencies

$$E(z,t) = E_0\cos(k_1 z - \omega_1 t) + E_0\cos(k_2 z - \omega_2 t)$$

where $\omega_{1,2} = \omega_0 \pm \Delta\omega$ and $k_{1,2} = k_0 \pm \Delta k$.

Step 2: Apply the trig identity for sum of cosines

$$E(z,t) = 2E_0\cos(\Delta k\cdot z - \Delta\omega\cdot t)\cos(k_0 z - \omega_0 t)$$

The first factor is the slowly-varying envelope; the second is the fast carrier wave.

Step 3: Identify the two velocities

The carrier wave (phase fronts) moves at the phase velocity:

$$v_p = \frac{\omega_0}{k_0} = \frac{c}{n(\omega_0)}$$

The envelope (signal) moves at the group velocity:

$$v_g = \frac{\Delta\omega}{\Delta k} \to \frac{d\omega}{dk}$$

Step 4: Express group velocity in terms of refractive index

From $k = n\omega/c$, differentiate: $dk = (n + \omega\,dn/d\omega)\,d\omega/c$. Therefore:

$$v_g = \frac{d\omega}{dk} = \frac{c}{n + \omega\,dn/d\omega}$$

Step 5: Normal vs anomalous dispersion

Normal dispersion ($dn/d\omega > 0$): $v_g < v_p$. Higher frequencies travel slower. This is the common case away from resonances.

Anomalous dispersion ($dn/d\omega < 0$): $v_g > v_p$. Occurs near absorption resonances where $n$ decreases with frequency.

Step 6: Express in terms of wavelength

Using $\omega = 2\pi c/\lambda$ and the chain rule:

$$v_g = v_p\left(1 + \frac{\lambda}{n}\frac{dn}{d\lambda}\right)^{-1}$$

The group index is $n_g = c/v_g = n - \lambda\,dn/d\lambda$.

Step 1: Add the conduction current to Ampere's law

$$\nabla \times \mathbf{B} = \mu\sigma\mathbf{E} + \mu\epsilon\frac{\partial \mathbf{E}}{\partial t}$$

The free current density is $\mathbf{J}_f = \sigma\mathbf{E}$ (Ohm's law).

Step 2: Derive the wave equation in a conductor

Taking the curl of Faraday's law and using the same vector identity as before:

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

Step 3: Substitute a plane wave ansatz

For $\mathbf{E} = E_0\,e^{i(\tilde{k}z - \omega t)}\,\hat{x}$:

$$-\tilde{k}^2 = -\mu\epsilon\omega^2 + i\mu\sigma\omega$$

$$\tilde{k}^2 = \mu\epsilon\omega^2 + i\mu\sigma\omega = \mu\omega^2\left(\epsilon + \frac{i\sigma}{\omega}\right)$$

Step 4: Good conductor approximation

For a good conductor, $\sigma \gg \epsilon\omega$, so we drop the $\epsilon$ term:

$$\tilde{k}^2 \approx i\mu\sigma\omega$$

$$\tilde{k} = \sqrt{i\mu\sigma\omega} = \sqrt{\frac{\mu\sigma\omega}{2}}(1 + i)$$

using $\sqrt{i} = (1+i)/\sqrt{2}$.

Step 5: Identify the skin depth

Writing $\tilde{k} = k + i/\delta$ where $k = 1/\delta = \sqrt{\mu\sigma\omega/2}$:

$$\delta = \sqrt{\frac{2}{\mu\sigma\omega}}$$

Step 6: The wave in a good conductor

$$\mathbf{E} = E_0\,e^{-z/\delta}\cos\left(\frac{z}{\delta} - \omega t\right)\hat{x}$$

The wave decays to $1/e$ of its surface amplitude after one skin depth. The wavelength inside the conductor is $\lambda = 2\pi\delta$, much shorter than in free space.

Step 7: Intensity attenuation (Beer's law form)

$$I(z) = I_0\,e^{-2z/\delta} = I_0\,e^{-\alpha z}, \quad \alpha = \frac{2}{\delta} = \sqrt{2\mu\sigma\omega}$$

At $z = 5\delta$, the intensity is reduced by a factor of $e^{-10} \approx 4.5 \times 10^{-5}$, which is 99.995% absorption.