The Wave Equation | Physics - Wyatt's Notes

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. $\blacksquare$

Worked Example: Verifying a wave solution

Problem. A string under tension has the wave equation $\partial^2 y/\partial t^2 = 100\,\partial^2 y/\partial x^2$ . Verify that $y(x,t) = 0.03\sin(5x - 50t)$ is a solution. Find the wave speed, wavelength, frequency, and propagation direction.

Solution. The wave speed is $v = \sqrt{100} = 10$ m/s. Comparing with the standard form $y = A\sin(kx - \omega t)$ : $A = 0.03$ m, $k = 5$ rad/m, $\omega = 50$ rad/s.

Verification: $\partial^2 y/\partial x^2 = -25 \times 0.03\sin(5x - 50t)$ and $\partial^2 y/\partial t^2 = -2500 \times 0.03\sin(5x - 50t)$ So $\partial^2 y/\partial t^2 = 100 \times \partial^2 y/\partial x^2$ . $\checkmark$

Wavelength: $\lambda = 2\pi/k = 2\pi/5 \approx 1.26$ m. Frequency: $f = \omega/(2\pi) = 50/(2\pi) \approx 7.96$ Hz. Direction: the argument $5x - 50t$ indicates propagation in the $+x$ direction.

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 $\tilde{E}$ :

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

Worked Example: Complex amplitude and intensity

Problem. A plane wave propagating in the $+z$ direction has complex amplitude $\tilde{\mathbf{E}} = (10 + 5i)\hat{\mathbf{x}}$ V/m in vacuum. Find the real electric field, the Intensity, and the phase of the wave relative to $\cos(kz - \omega t)$ .

Solution. $|\tilde{E}| = \sqrt{10^2 + 5^2} = \sqrt{125} \approx 11.18$ V/m. Phase: $\phi = \arctan(5/10) = 26.57°$ .

Real field: $\mathbf{E}(z,t) = 11.18\cos(kz - \omega t + 26.57°)\,\hat{\mathbf{x}}$ V/m.

Intensity: $I = \frac{1}{2}c\varepsilon_0 |\tilde{E}|^2 = \frac{1}{2}(3 \times 10^8)(8.854 \times 10^{-12})(125) = 0.166$ W/m $^2$ .

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.

Worked Example: Guitar string harmonics

Problem. A guitar string of length $L = 0.65$ m is fixed at both ends. The fundamental frequency Is $f_1 = 330$ Hz (the note E4). Find the wave speed, the frequency of the third harmonic, and the Positions of the nodes for the third harmonic.

Solution. Wave speed: $v = 2Lf_1 = 2 \times 0.65 \times 330 = 429$ m/s.

Third harmonic: $f_3 = 3f_1 = 990$ Hz. Wavelength: $\lambda_3 = 2L/3 = 0.433$ m.

Nodes at $x = m\lambda_3/2 = m(0.433/2) = 0.217m$ for $m = 0, 1, 2, 3$ . Positions: $x = 0, 0.217, 0.433, 0.650$ m (the endpoints and two interior nodes).

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.

Worked Example: Wave energy and power on a string

Problem. A steel wire of diameter 1.0 mm and density $7800$ kg/m $^3$ is under 500 N of tension. A sinusoidal wave of amplitude 5.0 mm and frequency 200 Hz propagates along it. Find (a) the wave Speed, (b) the average power, and (c) the intensity (power per unit cross-sectional area).

Solution. Cross-sectional area: $A_{\mathrm{wire} = \pi(0.5 \times 10^{-3})^2 = 7.85 \times 10^{-7}}$ m $^2$ . Linear density: $\mu = \rho A_{\mathrm{wire} = 7800 \times 7.85 \times 10^{-7} = 6.12 \times 10^{-3}}$ kg/m.

(a) $v = \sqrt{T/\mu} = \sqrt{500/(6.12 \times 10^{-3})} = \sqrt{8.17 \times 10^4} = 286$ m/s.

(b) $\omega = 2\pi \times 200 = 1257$ rad/s. $\langle P \rangle = \frac{1}{2}\mu v \omega^2 A^2 = \frac{1}{2}(6.12 \times 10^{-3})(286)(1257)^2(5.0 \times 10^{-3})^2$ $= \frac{1}{2}(6.12 \times 10^{-3})(286)(1.58 \times 10^6)(2.5 \times 10^{-5}) = 34.7$ W.

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.

Worked Example: Group velocity in a dispersive medium

Problem. A dispersive medium has the dispersion relation $\omega = \alpha k^2$ with $\alpha = 5.0 \times 10^3$ m $^2$ /s. For a wave packet centred at $k_0 = 200$ rad/m, Find (a) the phase velocity, (b) the group velocity, and (c) the time for the packet width to Double, given an initial width $\sigma_0 = 0.01$ m.

Solution.

(a) Phase velocity: $v_p = \omega/k = \alpha k_0 = 5.0 \times 10^3 \times 200 = 1.0 \times 10^6$ m/s.

(b) Group velocity: $v_g = d\omega/dk = 2\alpha k_0 = 2 \times 5.0 \times 10^3 \times 200 = 2.0 \times 10^6$ m/s.

(c) GVD: $\alpha_{\mathrm{GVD} = d^2\omega/dk^2 = 2\alpha = 1.0 \times 10^4}$ m $^2$ /s. The packet width doubles when $1 + (\alpha_{\mathrm{GVD}\, t / 2\sigma_0^2)^2 = 4}$ : $\alpha_{\mathrm{GVD}\, t / 2\sigma_0^2 = \sqrt{3}}$ $t = 2\sigma_0^2\sqrt{3}/\alpha_{\mathrm{GVD} = 2(10^{-4})(1.732)/(10^4) = 3.46 \times 10^{-8}}$ s $= 34.6$ ns.

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 $c$ .

flowchart TD
    A[1_The Wave Equation] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Intuition

Waves are disturbances that travel without the medium itself moving. Throw a stone into a pond and the ripples spread outward, but the water molecules mostly bob up and down. The wave equation captures this essential behavior: the second time derivative equals a constant times the second spatial derivative, meaning the acceleration of any point is proportional to how different its neighbors are from it. Standing waves are like a dance where two identical waves moving in opposite directions create a pattern of fixed nodes and oscillating antinodes. A guitar string vibrates this way: the ends cannot move, so only certain frequencies fit. Group velocity is how fast information or energy travels, while phase velocity is how fast individual crests move. In dispersive media, these differ, which is why ocean waves can have crests moving faster than the overall wave front.

1.7 Common Mistakes

Mistake 1: Confusing phase velocity with group velocity Phase velocity $v_p = \omega/k$ is the speed of individual wave crests, while group velocity $v_g = d\omega/dk$ is the speed of the wave packet envelope (and hence of information). In non-dispersive media they are equal, but in dispersive media they differ. A common error is assuming energy travels at the phase velocity, which is incorrect when dispersion is present.

Mistake 2: Forgetting that standing waves do not transport energy A standing wave is a superposition of two counter-propagating waves of equal amplitude. The net energy flux is zero because the energy flowing in one direction is exactly cancelled by energy flowing in the other. Students sometimes assume standing waves carry energy because individual travelling waves do. The energy in a standing wave oscillates between kinetic and potential forms but does not propagate.

Mistake 3: Misapplying superposition to non-linear media The superposition principle applies only to linear media. In non-linear optics (e.g., high-intensity lasers in certain crystals), waves interact and the total field is not directly the sum of individual fields. The wave equation itself is derived from Maxwell’s equations, which are linear in vacuum, but material responses can introduce non-linearity.

Cross-References

Maxwell’s Equations: The electromagnetic wave equation is derived from Maxwell’s equations in free space by taking curls of Faraday’s and Ampere-Maxwell’s laws.
Electrodynamics: Electrodynamics provides the foundation for understanding how changing electric and magnetic fields propagate as waves.
Interference: The superposition of wave solutions to the wave equation produces interference patterns used in Young’s double slit and Fabry-Perot interferometers.
Calculus
Linear Algebra
Vector Calculus
Quantum Computing

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