Electromagnetic Waves

In free space ($\rho = 0$, $\mathbf{J} = \mathbf{0}$), take the curl of Faraday’s law:

$\nabla \times (\nabla \times \mathbf{E}) = -\frac{\partial}{\partial t}(\nabla \times \mathbf{B}) = -\mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2}$

Using the identity $\nabla \times (\nabla \times \mathbf{E}) = \nabla(\nabla \cdot \mathbf{E}) - \nabla^2 \mathbf{E}$ And $\nabla \cdot \mathbf{E} = 0$:

$\nabla^2 \mathbf{E} = \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2}$

Similarly: $\nabla^2 \mathbf{B} = \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{B}}{\partial t^2}$.

These are wave equations with wave speed $c = 1/\sqrt{\mu_0 \varepsilon_0} \approx 3 \times 10^8$ m/s.

Theorem 5.1. Electromagnetic waves in free space are:

1. Transverse: $\mathbf{E}$ and $\mathbf{B}$ are perpendicular to the direction of propagation.
2. Mutually perpendicular: $\mathbf{E} \perp \mathbf{B}$.
3. In phase: $E = cB$ at every point.
4. Linearly polarised (; other polarisations are superpositions).

Energy. The energy density of an EM wave is $u = \frac{1}{2}(\varepsilon_0 E^2 + B^2/\mu_0)$.

The Poynting vector $\mathbf{S} = \frac{1}{\mu_0}\mathbf{E} \times \mathbf{B}$ represents the energy Flux (power per unit area).

Problem. Show that $\mathbf{E} = E_0 \cos(kz - \omega t)\,\hat{\mathbf{x}}$ satisfies the wave Equation and find the associated $\mathbf{B}$ field.

Solution. $\nabla^2 \mathbf{E} = \frac{\partial^2 E_x}{\partial z^2}\hat{\mathbf{x}} = -k^2 E_0 \cos(kz - \omega t)\,\hat{\mathbf{x}}$.

$\frac{\partial^2 \mathbf{E}}{\partial t^2} = -\omega^2 E_0 \cos(kz - \omega t)\,\hat{\mathbf{x}}$.

The wave equation requires $k^2 = \mu_0 \varepsilon_0 \omega^2$I.e., $\omega/k = c$.

From Faraday’s law: $\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$.

$(\nabla \times \mathbf{E})_y = -\frac{\partial E_x}{\partial z} = k E_0 \sin(kz - \omega t)$

$\frac{\partial B_y}{\partial t} = -k E_0 \sin(kz - \omega t) \implies B_y = \frac{k}{\omega} E_0 \cos(kz - \omega t) = \frac{E_0}{c}\cos(kz - \omega t)$

So $\mathbf{B} = \frac{E_0}{c}\cos(kz - \omega t)\,\hat{\mathbf{y}}$. $\blacksquare$

Poynting’s theorem is the statement of energy conservation for electromagnetic fields.

Derivation. Start with the two Maxwell equations containing time derivatives:

$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}, \quad \nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}$

Compute $\mathbf{B} \cdot (\nabla \times \mathbf{E}) - \mathbf{E} \cdot (\nabla \times \mathbf{B})$:

$\mathbf{B} \cdot (\nabla \times \mathbf{E}) = -\mathbf{B} \cdot \frac{\partial \mathbf{B}}{\partial t} = -\frac{\partial}{\partial t}\left(\frac{B^2}{2}\right)$

$-\mathbf{E} \cdot (\nabla \times \mathbf{B}) = -\mu_0 \mathbf{E} \cdot \mathbf{J} - \mu_0 \varepsilon_0 \mathbf{E} \cdot \frac{\partial \mathbf{E}}{\partial t} = -\mu_0 \mathbf{E} \cdot \mathbf{J} - \frac{\partial}{\partial t}\left(\frac{\varepsilon_0 E^2}{2}\right)$

Using the vector identity $\nabla \cdot (\mathbf{E} \times \mathbf{B}) = \mathbf{B} \cdot (\nabla \times \mathbf{E}) - \mathbf{E} \cdot (\nabla \times \mathbf{B})$:

$\nabla \cdot (\mathbf{E} \times \mathbf{B}) = -\mu_0 \mathbf{J} \cdot \mathbf{E} - \mu_0 \varepsilon_0 \frac{\partial}{\partial t}\left(\frac{E^2}{2}\right) - \frac{\partial}{\partial t}\left(\frac{B^2}{2}\right)$

Dividing by $\mu_0$ and rearranging:

$\boxed{-\nabla \cdot \mathbf{S} = \mathbf{J} \cdot \mathbf{E} + \frac{\partial u}{\partial t}}$

Where $\mathbf{S} = \frac{1}{\mu_0}\mathbf{E} \times \mathbf{B}$ is the Poynting vector and $u = \frac{1}{2}\left(\varepsilon_0 E^2 + \frac{B^2}{\mu_0}\right)$ is the energy density.

Interpretation: The rate of energy leaving a volume equals the work done on charges plus The rate of increase of field energy. In integral form:

$-\oint_S \mathbf{S} \cdot d\mathbf{A} = \frac{d}{dt}\int_V u\,dV + \int_V \mathbf{J} \cdot \mathbf{E}\,dV$

Intensity. For a plane wave, the time-averaged Poynting vector is:

$\langle\mathbf{S}\rangle = \frac{E_0^2}{2\mu_0 c}\,\hat{\mathbf{k}} = \frac{1}{2}\varepsilon_0 c E_0^2\,\hat{\mathbf{k}}$

In a conductor with conductivity $\sigma$Ohm’s law gives $\mathbf{J} = \sigma\mathbf{E}$. Substituting into the Ampere-Maxwell law:

$\nabla \times \mathbf{B} = \mu_0\sigma\mathbf{E} + \mu_0\varepsilon_0\frac{\partial \mathbf{E}}{\partial t}$

For a monochromatic wave $\mathbf{E} = \mathbf{E}_0\,e^{-i\omega t}$This leads to a complex Wave number:

$\tilde{k}^2 = \mu_0\varepsilon_0\omega^2 + i\mu_0\sigma\omega$

Writing $\tilde{k} = k + i\kappa$ where $k$ is the real part (wave number) and $\kappa$ is the Imaginary part (attenuation constant):

$\mathbf{E}(z,t) = \mathbf{E}_0\,e^{-\kappa z}\cos(kz - \omega t)$

The field decays exponentially. The skin depth is the distance over which the amplitude Falls by a factor of $1/e$:

$\delta = \frac{1}{\kappa}$

For a good conductor ($\sigma \gg \varepsilon_0\omega$):

$\delta = \sqrt{\frac{2}{\mu_0\sigma\omega}}$

Electromagnetic waves can be guided by hollow conducting pipes (waveguides). Consider a Rectangular waveguide with dimensions $a$ (width) and $b$ (height).

TE modes (transverse electric, $E_z = 0$, $B_z \neq 0$). The lowest-order mode is \mathrm{TE_}{10}With fields:

$E_y = E_0 \sin\!\left(\frac{\pi x}{a}\right)\cos(k_g z - \omega t)$

$B_x = -\frac{k_g}{\omega}E_0 \sin\!\left(\frac{\pi x}{a}\right)\cos(k_g z - \omega t)$

$B_z = \frac{\pi}{\omega a}E_0 \cos\!\left(\frac{\pi x}{a}\right)\sin(k_g z - \omega t)$

Where the guide wave number is $k_g = \sqrt{(\omega/c)^2 - (\pi/a)^2}$.

Cutoff frequency. Waves propagate only when $\omega \gt \omega_c$ where:

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

For the \mathrm{TE_}{10} mode: $f_c = \frac{c}{2a}$.

Phase and group velocities. In a waveguide, the phase velocity exceeds $c$:

$v_p = \frac{\omega}{k_g} = \frac{c}{\sqrt{1 - (\omega_c/\omega)^2}} \gt c$

The group velocity (signal velocity) is less than $c$:

$v_g = \frac{d\omega}{dk_g} = c\sqrt{1 - (\omega_c/\omega)^2} \lt c$

They satisfy $v_p\,v_g = c^2$.

:::caution Common Pitfall The phase velocity in a waveguide exceeds $c$But this does not violate special relativity. No information or energy travels faster than $c$; the signal velocity is the group velocity $v_g \lt c$. The phase velocity is the speed of the wave crests, which is a purely Kinematic quantity.

An oscillating electric dipole is the simplest source of electromagnetic radiation.

Consider a dipole $\mathbf{p}(t) = p_0\cos(\omega t)\,\hat{\mathbf{z}}$. In the radiation zone ($r \gg \lambda$), the fields are:

$\mathbf{E} = -\frac{\mu_0 p_0 \omega^2}{4\pi}\frac{\sin\theta}{r}\cos[\omega(t - r/c)]\,\hat{\boldsymbol{\theta}}$

$\mathbf{B} = -\frac{\mu_0 p_0 \omega^2}{4\pi c}\frac{\sin\theta}{r}\cos[\omega(t - r/c)]\,\hat{\boldsymbol{\phi}}$

The fields fall off as $1/r$ (not $1/r^2$ as for static fields), which is characteristic of Radiation.

Radiation pattern. The intensity varies as $\sin^2\theta$With maximum radiation in the Equatorial plane ($\theta = \pi/2$) and zero along the dipole axis ($\theta = 0, \pi$).

Total radiated power. Integrating the Poynting vector over a sphere:

$P = \frac{\mu_0 p_0^2 \omega^4}{12\pi c}$

Larmor formula. For a point charge $q$ undergoing acceleration $a$:

$P = \frac{q^2 a^2}{6\pi\varepsilon_0 c^3}$

This is the non-relativistic limit and is valid whenever $v \ll c$.

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