Electromagnetic Waves
The wave equation from Maxwell's equations, plane waves, polarization, energy transport, and Fresnel equations for reflection and transmission.
1.1 The Electromagnetic Wave Equation
In free space ($\rho = 0$, $\mathbf{J} = 0$), Maxwell's equations combine to yield wave equations for both $\mathbf{E}$ and $\mathbf{B}$:
The wave speed is $c = 1/\sqrt{\mu_0\epsilon_0} = 2.998 \times 10^8$ m/s, identifying light as an electromagnetic wave.
Derivation 1: Wave Equation from Maxwell's Equations
We derive the wave equation by decoupling the curl equations.
Step 1: Start from the source-free Maxwell equations
$\nabla\times\mathbf{E} = -\partial\mathbf{B}/\partial t$ and $\nabla\times\mathbf{B} = \mu_0\epsilon_0\,\partial\mathbf{E}/\partial t$.
Step 2: Take the curl of Faraday's law
$$\nabla\times(\nabla\times\mathbf{E}) = -\frac{\partial}{\partial t}(\nabla\times\mathbf{B}) = -\mu_0\epsilon_0\frac{\partial^2\mathbf{E}}{\partial t^2}$$
Step 3: Apply the vector identity
$\nabla\times(\nabla\times\mathbf{E}) = \nabla(\nabla\cdot\mathbf{E}) - \nabla^2\mathbf{E}$. Since $\nabla\cdot\mathbf{E} = 0$ in free space:
$$-\nabla^2\mathbf{E} = -\mu_0\epsilon_0\frac{\partial^2\mathbf{E}}{\partial t^2}$$
Step 4: Identify the wave equation
$$\boxed{\nabla^2\mathbf{E} - \frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = 0, \qquad c = \frac{1}{\sqrt{\mu_0\epsilon_0}}}$$
Step 5: Maxwell's prediction
Substituting the known values of $\mu_0$ and $\epsilon_0$, Maxwell obtained $c \approx 3\times 10^8$ m/s, matching the measured speed of light. This was the first evidence that light is an electromagnetic phenomenon.
1.2 Monochromatic Plane Waves
The simplest solutions are monochromatic plane waves: $\mathbf{E} = E_0\hat{\epsilon}\,e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}$with $\omega = c|\mathbf{k}|$. Maxwell's equations impose the transversality conditions:
Derivation 2: Energy and Momentum of Plane Waves
We compute the Poynting vector and energy density for a monochromatic plane wave.
Step 1: Energy density
The electromagnetic energy density is $u = \frac{1}{2}(\epsilon_0|\mathbf{E}|^2 + |\mathbf{B}|^2/\mu_0)$. For a plane wave, $|\mathbf{B}| = |\mathbf{E}|/c$, so:
$$u = \epsilon_0|\mathbf{E}|^2 = \epsilon_0 E_0^2\cos^2(\mathbf{k}\cdot\mathbf{r}-\omega t)$$
Step 2: Poynting vector
$$\mathbf{S} = \frac{1}{\mu_0}\mathbf{E}\times\mathbf{B} = c\epsilon_0 E_0^2\cos^2(\mathbf{k}\cdot\mathbf{r}-\omega t)\,\hat{k}$$
Step 3: Time-averaged intensity
$$\langle\mathbf{S}\rangle = \frac{1}{2}c\epsilon_0 E_0^2\,\hat{k} = \frac{E_0^2}{2\mu_0 c}\,\hat{k}$$
The intensity is $I = |\langle\mathbf{S}\rangle| = E_0^2/(2Z_0)$ where $Z_0 = \sqrt{\mu_0/\epsilon_0} \approx 377$ $\Omega$.
Step 4: Radiation pressure
The momentum density is $\mathbf{g} = \mathbf{S}/c^2$. For a wave absorbed by a surface, the radiation pressure is:
$$P_{\text{rad}} = \frac{\langle S\rangle}{c} = \frac{\langle u\rangle}{1}$$
Step 5: Relation S = cu
For a plane wave, the energy flux equals the energy density times the wave speed: $|\mathbf{S}| = cu$. This confirms that energy propagates at speed $c$ in vacuum.
1.3 Polarization
The polarization of a plane wave describes the trajectory of the electric field vector in the plane perpendicular to $\hat{k}$. A general monochromatic wave traveling in the $z$-direction has:
When $\delta = 0$ or $\pi$: linear polarization. When $E_x = E_y$ and $\delta = \pm\pi/2$: circular polarization. In general: elliptical polarization. The Stokes parameters $(I, Q, U, V)$ provide a complete description including partial polarization.
Derivation 3: Fresnel Equations at a Dielectric Interface
We derive the reflection and transmission coefficients for s- and p-polarized waves at a planar interface.
Step 1: Boundary conditions
At $z=0$, the tangential components of $\mathbf{E}$ and $\mathbf{H}$ must be continuous. The incident, reflected, and transmitted waves have the form $e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}$.
Step 2: Snell's law from phase matching
Matching the phase at the interface for all $x$: $k_i\sin\theta_i = k_r\sin\theta_r = k_t\sin\theta_t$. This gives $\theta_r = \theta_i$ and $n_1\sin\theta_i = n_2\sin\theta_t$.
Step 3: s-polarization (TE mode)
E perpendicular to the plane of incidence. Matching $E_\parallel$ and $H_\parallel$:
$$r_s = \frac{n_1\cos\theta_i - n_2\cos\theta_t}{n_1\cos\theta_i + n_2\cos\theta_t}, \quad t_s = \frac{2n_1\cos\theta_i}{n_1\cos\theta_i + n_2\cos\theta_t}$$
Step 4: p-polarization (TM mode)
E in the plane of incidence:
$$r_p = \frac{n_2\cos\theta_i - n_1\cos\theta_t}{n_2\cos\theta_i + n_1\cos\theta_t}, \quad t_p = \frac{2n_1\cos\theta_i}{n_2\cos\theta_i + n_1\cos\theta_t}$$
Step 5: Brewster's angle
Setting $r_p = 0$: $n_2\cos\theta_i = n_1\cos\theta_t$. Combined with Snell's law: $$\boxed{\theta_B = \arctan(n_2/n_1)}$$
At Brewster's angle, only s-polarized light is reflected, which is why polarizing sunglasses work.
1.4 Waves in Linear Media
In a linear medium with permittivity $\epsilon$ and permeability $\mu$, the wave speed becomes $v = 1/\sqrt{\mu\epsilon}$and the index of refraction is $n = c/v = \sqrt{\mu_r\epsilon_r}$.
Derivation 4: Total Internal Reflection and Evanescent Waves
When light passes from a denser medium to a rarer one, total internal reflection occurs beyond a critical angle.
Step 1: Critical angle from Snell's law
When $\theta_t = \pi/2$: $n_1\sin\theta_c = n_2$, so $\theta_c = \arcsin(n_2/n_1)$. For $n_1 > n_2$, this has a solution.
Step 2: Beyond the critical angle
For $\theta_i > \theta_c$, $\sin\theta_t > 1$ and $\cos\theta_t = i\sqrt{\sin^2\theta_t - 1}$ becomes purely imaginary.
Step 3: Evanescent wave in medium 2
The transmitted wave becomes $e^{i(k_x x - \omega t)}e^{-\kappa z}$ with $\kappa = k_2\sqrt{(n_1\sin\theta_i/n_2)^2 - 1}$. The field decays exponentially but carries no net energy across the interface.
Step 4: Penetration depth
$$d = \frac{1}{\kappa} = \frac{\lambda}{2\pi\sqrt{n_1^2\sin^2\theta_i - n_2^2}}$$
Typically of order one wavelength.
Step 5: Phase shift and FTIR
The reflection coefficients acquire complex phase factors. If a second interface is placed within the evanescent region, frustrated total internal reflection (FTIR) allows tunneling transmission, analogous to quantum mechanical tunneling.
1.5 Dispersion and Group Velocity
In a dispersive medium, the index of refraction depends on frequency: $n(\omega)$. A wave packet travels at the group velocity $v_g = d\omega/dk$, which generally differs from the phase velocity $v_p = \omega/k$.
Derivation 5: Group Velocity of a Wave Packet
We derive the group velocity by analyzing the superposition of two nearby frequencies.
Step 1: Superpose two waves
Consider $E = E_0[\cos(k_1 z - \omega_1 t) + \cos(k_2 z - \omega_2 t)]$ with $k_2 = k_1 + \Delta k$, $\omega_2 = \omega_1 + \Delta\omega$.
Step 2: Apply the sum-to-product identity
$$E = 2E_0\cos\!\left(\frac{\Delta k}{2}z - \frac{\Delta\omega}{2}t\right)\cos(\bar{k}z - \bar{\omega}t)$$
where $\bar{k} = (k_1+k_2)/2$ and $\bar{\omega} = (\omega_1+\omega_2)/2$.
Step 3: Identify the envelope
The slowly varying envelope $\cos(\Delta k\,z/2 - \Delta\omega\,t/2)$ modulates the fast carrier wave.
Step 4: Envelope velocity
The envelope moves at $v_g = \Delta\omega/\Delta k \to d\omega/dk$ in the limit.
Step 5: Relation to phase velocity
$$\boxed{v_g = \frac{d\omega}{dk} = v_p + k\frac{dv_p}{dk} = \frac{c}{n + \omega\,dn/d\omega}}$$
In a plasma, $v_g v_p = c^2$, so the group velocity is always less than $c$ even though the phase velocity exceeds it. Information travels at $v_g$.
Historical Notes
Maxwell (1865): In his epochal paper "A Dynamical Theory of the Electromagnetic Field," Maxwell predicted electromagnetic waves traveling at the speed of light, unifying optics with electromagnetism.
Hertz (1887): Heinrich Hertz experimentally demonstrated the existence of electromagnetic waves, confirming Maxwell's prediction. He showed they exhibited reflection, refraction, and polarization, just like light.
Fresnel (1821): Augustin-Jean Fresnel developed the mathematical theory of light reflection and refraction based on wave theory, deriving the famous coefficients that bear his name, decades before Maxwell's electromagnetic theory.
Rayleigh (1881): Lord Rayleigh introduced the concept of group velocity, clarifying the distinction between the speed of individual wave crests and the speed at which a signal or energy propagates.
Applications
Anti-Reflection Coatings
Quarter-wave coatings exploit destructive interference between reflections at two interfaces. For a coating of thickness $\lambda/4$ and index $n_c = \sqrt{n_1 n_2}$, the reflectance vanishes at the design wavelength.
Fiber Optics
Total internal reflection guides light in optical fibers. Single-mode fibers exploit the evanescent field for coupling and sensing. Dispersion management enables Tbit/s data transmission over thousands of kilometers.
Polarimetry
Measuring the Stokes parameters of reflected or transmitted light reveals material properties, stress distributions, and surface features. Used in remote sensing, astronomy, and materials science.
Radar Technology
Electromagnetic wave propagation, reflection from targets, and Doppler shifts form the basis of radar. The Fresnel equations determine the radar cross section of dielectric objects.
Simulation: Electromagnetic Waves
This simulation visualizes plane wave E and B fields, Fresnel reflection coefficients with Brewster's angle, the plasma dispersion relation with group velocity, and polarization state trajectories.
EM Waves: Plane Waves, Fresnel Equations & Polarization
PythonClick Run to execute the Python code
Code will be executed with Python 3 on the server
Practice Problems
Problem 1:Compute the Fresnel reflection coefficients $r_s$ and $r_p$ for light going from glass ($n_1 = 1.5$) to air ($n_2 = 1.0$) at an angle of incidence $\theta_i = 30°$.
Solution:
1. Snell's law: $n_1\sin\theta_i = n_2\sin\theta_t$. $\sin\theta_t = \frac{1.5}{1.0}\sin 30° = 0.75$, so $\theta_t = 48.59°$.
2. s-polarization (TE): $r_s = \frac{n_1\cos\theta_i - n_2\cos\theta_t}{n_1\cos\theta_i + n_2\cos\theta_t}$.
3. $\cos\theta_i = 0.866$, $\cos\theta_t = 0.661$. $r_s = \frac{1.5(0.866) - 1.0(0.661)}{1.5(0.866) + 1.0(0.661)} = \frac{1.299 - 0.661}{1.299 + 0.661} = \frac{0.638}{1.960} = 0.326$.
4. p-polarization (TM): $r_p = \frac{n_2\cos\theta_i - n_1\cos\theta_t}{n_2\cos\theta_i + n_1\cos\theta_t} = \frac{0.866 - 1.5(0.661)}{0.866 + 1.5(0.661)} = \frac{0.866 - 0.991}{0.866 + 0.991} = \frac{-0.125}{1.857} = -0.067$.
5. The reflectances are $R_s = r_s^2 = 0.106$ (10.6%) and $R_p = r_p^2 = 0.0045$ (0.45%).
6. Note that $|r_p| < |r_s|$ at this angle because we are below the Brewster angle but above it would flip. The large difference means reflected light is predominantly s-polarized.
Problem 2:Find Brewster's angle for light traveling from air ($n_1 = 1.0$) into glass ($n_2 = 1.5$). What is special about the reflected light at this angle?
Solution:
1. Brewster's angle: $\tan\theta_B = \frac{n_2}{n_1} = \frac{1.5}{1.0} = 1.5$.
2. $\theta_B = \arctan(1.5) = 56.31°$.
3. At Brewster's angle, $r_p = 0$ exactly: no p-polarized light is reflected. All reflected light is s-polarized (electric field perpendicular to the plane of incidence).
4. Physical explanation: at $\theta_B$, $\theta_i + \theta_t = 90°$. The refracted and reflected rays are perpendicular. Since the p-polarized component would require the reflected wave to oscillate along its propagation direction (which EM waves cannot do), it must vanish.
5. Verify: $\theta_t = \arcsin(\sin 56.31°/1.5) = \arcsin(0.555) = 33.69°$. Indeed $56.31° + 33.69° = 90°$.
6. Applications: Brewster windows in lasers (loss-free for p-polarization), polarizing filters, glare reduction (reflected sunlight from roads/water is partially s-polarized, so polaroid sunglasses block it).
Problem 3:Find the critical angle for total internal reflection at a glass ($n_1 = 1.5$) to air ($n_2 = 1.0$) interface. What happens to the transmitted wave beyond this angle?
Solution:
1. Critical angle: $\sin\theta_c = \frac{n_2}{n_1} = \frac{1.0}{1.5} = 0.667$.
2. $\theta_c = \arcsin(0.667) = 41.81°$.
3. For $\theta_i > \theta_c$, Snell's law gives $\sin\theta_t > 1$, so $\cos\theta_t = i\sqrt{\sin^2\theta_t - 1}$ becomes purely imaginary.
4. The transmitted wave becomes evanescent: $\mathbf{E}_t \propto e^{-\kappa z}e^{i(k_x x - \omega t)}$, decaying exponentially into the air with $\kappa = \frac{\omega}{c}\sqrt{n_1^2\sin^2\theta_i - n_2^2}$.
5. The penetration depth is $d = 1/\kappa$. At $\theta_i = 45°$ with $\lambda = 500\,\text{nm}$: $d = \frac{500}{2\pi\sqrt{1.5^2 \times 0.5 - 1}} = \frac{500}{2\pi\sqrt{0.125}} \approx 225\,\text{nm}$.
6. Total internal reflection has $|r_s| = |r_p| = 1$ (no energy transmitted), but the Fresnel coefficients acquire complex phases, meaning the reflected wave has a phase shift (the Goos-Hanchen effect).
Problem 4:Compute the energy reflection coefficient $R$ and transmission coefficient $T$ for normal incidence ($\theta_i = 0$) from air to glass ($n = 1.5$). Verify that $R + T = 1$.
Solution:
1. At normal incidence, both polarizations give the same result: $r = \frac{n_1 - n_2}{n_1 + n_2} = \frac{1.0 - 1.5}{1.0 + 1.5} = \frac{-0.5}{2.5} = -0.2$.
2. The energy reflection coefficient: $R = |r|^2 = 0.04$ (4% reflected).
3. The amplitude transmission coefficient: $t = \frac{2n_1}{n_1 + n_2} = \frac{2.0}{2.5} = 0.8$.
4. The energy transmission coefficient: $T = \frac{n_2\cos\theta_t}{n_1\cos\theta_i}|t|^2 = \frac{1.5}{1.0}(0.8)^2 = 1.5 \times 0.64 = 0.96$.
5. Check: $R + T = 0.04 + 0.96 = 1.00$. Energy is conserved.
6. Note: $|t|^2 = 0.64 \neq T = 0.96$. The factor $n_2/n_1$ accounts for the different impedances (or equivalently, the different beam cross-sections and wave speeds in the two media).
Problem 5:Derive the reflectance of a single dielectric slab of thickness $d$ and index $n$ in air, and find the condition for zero reflection (anti-reflection coating).
Solution:
1. The total reflection results from multiple reflections at both interfaces. The net reflection coefficient (at normal incidence) using the transfer matrix or Fabry-Perot formula is:
2. $r_{\text{total}} = \frac{r_{12}(1 - e^{2i\delta})}{1 - r_{12}^2 e^{2i\delta}}$ where $r_{12} = \frac{1-n}{1+n}$ and $\delta = \frac{2\pi n d}{\lambda}$ is the phase accumulated in one pass.
3. The reflectance is $R = |r_{\text{total}}|^2 = \frac{4r_{12}^2\sin^2\delta}{(1 - r_{12}^2)^2 + 4r_{12}^2\sin^2\delta}$.
4. Zero reflection when $\sin\delta = 0$, i.e., $\delta = m\pi$, giving $d = m\lambda/(2n)$ (half-wave thickness). This is because the two reflected beams interfere destructively.
5. For a coating on glass ($n_{\text{glass}}$): anti-reflection requires $n_c = \sqrt{n_{\text{glass}}}$ and $d = \lambda/(4n_c)$ (quarter-wave thickness).
6. For $n_{\text{glass}} = 1.5$: $n_c = \sqrt{1.5} = 1.225$. MgF$_2$ ($n \approx 1.38$) is the closest practical material, reducing reflectance from 4% to about 1.3% per surface.