The Wave Equation

1.1 Derivation from Maxwell’s Equations

In vacuum, with no sources ($\rho = 0$, $\mathbf{J} = \mathbf{0}$), Maxwell’s equations give:

$\nabla \times (\nabla \times \mathbf{E}) = \nabla(\nabla \cdot \mathbf{E}) - \nabla^2 \mathbf{E} = -\nabla^2 \mathbf{E}$

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

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

Hence:

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

This is the electromagnetic wave equation with wave speed $c = 1/\sqrt{\mu_0 \varepsilon_0}$.

The same equation holds for $\mathbf{B}$:

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

1.2 General Solutions

The one-dimensional wave equation:

$\frac{\partial^2 u}{\partial x^2} = \frac{1}{v^2}\frac{\partial^2 u}{\partial t^2}$

Has the general solution (d’Alembert’s solution):

$u(x, t) = f(x - vt) + g(x + vt)$

Where $f$ is a wave travelling in the $+x$ direction and $g$ in the $-x$ direction.

Proof. Substitute $u = f(x - vt)$. Let $\xi = x - vt$. Then $\partial u/\partial x = f'(\xi)$ $\partial^2 u/\partial x^2 = f''(\xi)$, $\partial u/\partial t = -vf'(\xi)$ $\partial^2 u/\partial t^2 = v^2 f''(\xi)$. The wave equation gives $f'' = (v^2/v^2)f''$Which Is identically satisfied. The same holds for $g(x + vt)$. By linearity, the sum is also a solution. $\blacksquare$

1.3 Complex Representation

It is convenient to write monochromatic waves as:

$\mathbf{E}(\mathbf{r}, t) = \mathrm{Re}\left[\tilde{\mathbf{E}}\, e^{i(\mathbf{k}\cdot\mathbf{r} - \omega t)}\right]$

Where $\tilde{\mathbf{E}}$ is the complex amplitude, $\mathbf{k}$ is the wave vector, and $\omega$ is The angular frequency. The dispersion relation is $\omega = ck = c|\mathbf{k}|$.

The wave vector satisfies $|\mathbf{k}| = 2\pi/\lambda$ and $\omega = 2\pi\nu$.

When computing intensities, the complex representation simplifies the algebra. For a plane wave with Complex amplitude $\tilde{E}$:

$I = \frac{1}{2}c\varepsilon_0 |\tilde{E}|^2$

1.4 Standing Waves

When two waves of equal amplitude and frequency travel in opposite directions, their superposition Produces a standing wave. Consider:

$u_1 = A\sin(kx - \omega t), \quad u_2 = A\sin(kx + \omega t)$

Using the identity $\sin\alpha + \sin\beta = 2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$:

$u(x,t) = 2A\sin(kx)\cos(\omega t)$

This is a standing wave with the following properties:

• Nodes (points of zero displacement): $kx = m\pi$I.e., $x = m\lambda/2$ for $m = 0, 1, 2, \ldots$
• Antinodes (points of maximum displacement): $kx = (m + 1/2)\pi$I.e., $x = (2m+1)\lambda/4$.
• All points between two nodes oscillate in phase (or in antiphase with adjacent segments).
• The standing wave does not transport energy in either direction.

Standing waves on a string of length $L$ fixed at both ends. The boundary conditions $u(0,t) = 0$ And $u(L,t) = 0$ require $\sin(kL) = 0$So:

$k_n L = n\pi \implies \lambda_n = \frac{2L}{n}, \quad f_n = \frac{nv}{2L}, \quad n = 1, 2, 3, \ldots$

The allowed frequencies are integer multiples of the fundamental frequency $f_1 = v/(2L)$. The general solution is a superposition of all normal modes:

$u(x,t) = \sum_{n=1}^{\infty} \left(A_n \sin k_n x \cos \omega_n t + B_n \sin k_n x \sin \omega_n t\right)$

Standing waves on a string fixed at one end ($x = 0$) and free at the other ($x = L$). The free end requires $\partial u/\partial x|_{x=L} = 0$Giving $\cos(kL) = 0$So:

$k_n L = (n + 1/2)\pi \implies f_n = \frac{(2n+1)v}{4L}, \quad n = 0, 1, 2, \ldots$

Only odd harmonics are present. A pipe open at one end and closed at the other behaves analogously For sound waves.

1.5 Energy Transport by Waves

For a transverse wave on a string of linear mass density $\mu$ under tension $T$With $v = \sqrt{T/\mu}$:

Kinetic energy of an element $dx$:

$dK = \frac{1}{2}\mu\,dx\left(\frac{\partial u}{\partial t}\right)^2$

Potential energy (from stretching the string):

$dU = \frac{1}{2}T\,dx\left(\frac{\partial u}{\partial x}\right)^2$

Proof that kinetic and potential energies are equal. For a rightward-travelling wave $u = f(x - vt)$:

$\frac{\partial u}{\partial t} = -vf'(x - vt) = -v\frac{\partial u}{\partial x}$

Since $v^2 = T/\mu$ (i.e., $T = \mu v^2$):

$dU = \frac{1}{2}\mu v^2\left(\frac{\partial u}{\partial x}\right)^2 = \frac{1}{2}\mu\left(\frac{\partial u}{\partial t}\right)^2 = dK \quad \blacksquare$

The total energy density (energy per unit length) is:

$\frac{dE}{dx} = dK + dU = \mu\left(\frac{\partial u}{\partial t}\right)^2$

Energy flux (power): The rate of energy transport past a point is:

$P = -T\frac{\partial u}{\partial x}\frac{\partial u}{\partial t} = \mu v\left(\frac{\partial u}{\partial t}\right)^2$

For a sinusoidal wave $u = A\sin(kx - \omega t)$The time-averaged power is:

$\langle P \rangle = \frac{1}{2}\mu v \omega^2 A^2$

The intensity (power per unit area, generalising to 3D) is:

$I = \frac{1}{2}\rho v \omega^2 A^2$

Where $\rho$ is the mass density of the medium.

1.6 Wave Packets and Group Velocity

A real wave is never perfectly monochromatic. A wave packet is a superposition of plane waves With a narrow range of frequencies and wave vectors:

$\psi(x,t) = \int_{-\infty}^{\infty} A(k)\, e^{i(kx - \omega(k)t)}\,dk$

Where $A(k)$ is the spectral amplitude, peaked around $k_0$ with width $\Delta k$.

Expanding the dispersion relation around $k_0$:

$\omega(k) \approx \omega_0 + v_g(k - k_0) + \frac{1}{2}\alpha(k - k_0)^2$

Where $v_g = \left.d\omega/dk\right|_{k_0}$ is the group velocity and $\alpha = d^2\omega/dk^2$ is the group velocity dispersion (GVD).

Substituting and carrying out the Gaussian integral (for a Gaussian envelope $A(k)$):

$|\psi(x,t)|^2 \propto \exp\left(-\frac{(x - v_g t)^2}{2\sigma_x^2(t)}\right)$

Where $\sigma_x(t) = \sigma_x(0)\sqrt{1 + (\alpha t / 2\sigma_x^2(0))^2}$.

The envelope moves at $v_g$While individual wave crests move at the phase velocity $v_p = \omega/k$. The packet broadens over time due to GVD.

• In a non-dispersive medium ($\omega \propto k$So $v_g = v_p$): the packet propagates without distortion.
• In a normally dispersive medium ($d^2\omega/dk^2 \gt 0$): $v_g \lt v_p$ and the packet broadens.

Relation to the wave equation. The 1D wave equation $\partial^2 u/\partial t^2 = v^2\partial^2 u/\partial x^2$ Has dispersion relation $\omega = \pm vk$Giving $v_g = v_p = v$ — it is non-dispersive.

:::caution Common Pitfall The group velocity $v_g = d\omega/dk$ describes the motion of the wave packet envelope and equals The energy transport velocity in lossless media. However, it is only equal to the signal velocity In regions of weak, normal dispersion. Near absorption resonances, $v_g$ can exceed $c$ or become Negative — this does not violate causality, since the true signal velocity (front velocity) never Exceeds $c$.

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