The Wave Equation in Vacuum
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You should be comfortable with vector calculus identities, specifically the curl of a curl, and recall the four Maxwell equations in vacuum from Electromagnetism I.
Why this matters
Maxwell’s equations predict that light is an electromagnetic wave.
For centuries, electricity and magnetism were treated as separate phenomena. The unification achieved by Maxwell revealed that changing electric fields generate magnetic fields and vice versa. This interplay allows disturbances to propagate through empty space at a finite speed. Recognising this speed as the speed of light was the pivotal moment that established the electromagnetic nature of radiation, bridging the gap between static field theory and wave dynamics.
The Idea in Plain Language
Imagine a ripple on the surface of a pond. The water moves up and down, but the disturbance travels outward. In electromagnetism, there is no water, but there is a field. If you wiggle an electric field, it creates a magnetic field. That changing magnetic field, in turn, creates a new electric field further away. This chain reaction continues, carrying energy through the vacuum. The key insight is that the "stiffness" of the electric field and the "inertia" of the magnetic field determine how fast this ripple travels. Just as the tension and density of a guitar string set its wave speed, the constants \(\epsilon_0\) and \(\mu_0\) set the speed of light.
Deriving the Wave Equation
To see the wave nature explicitly, we must eliminate one field from Maxwell's equations to obtain a single equation for the other. We start with the source-free vacuum conditions: \(\rho = 0\) and \(\mathbf{J} = 0\). The relevant equations are Faraday’s law and the Ampère-Maxwell law:
We take the curl of Faraday’s law. Using the vector identity \(\nabla \times (\nabla \times \mathbf{A}) = \nabla(\nabla \cdot \mathbf{A}) - \nabla^2 \mathbf{A}\), and noting that \(\nabla \cdot \mathbf{E} = 0\) in vacuum, we get:
Now we substitute the Ampère-Maxwell law for \(\nabla \times \mathbf{B}\). Since the fields are smooth, we can interchange the spatial and temporal derivatives:
Rearranging gives the standard wave equation for the electric field. A parallel derivation starting with the curl of the Ampère-Maxwell law yields the identical form for the magnetic field.
The Speed of Propagation
The general form of a second-order wave equation in three dimensions is \(\nabla^2 \psi = \frac{1}{v^2} \frac{\partial^2 \psi}{\partial t^2}\), where \(v\) is the phase velocity. Comparing this with our derived equation:
We identify the propagation speed \(c\) as:
Using the SI values \(\epsilon_0 \approx 8.854 \times 10^{-12} \ \mathrm{F/m}\) and \(\mu_0 = 4\pi \times 10^{-7} \ \mathrm{H/m}\), we calculate:
This value matches the experimentally measured speed of light. This was not a coincidence; it was the theoretical prediction that confirmed light is an electromagnetic wave. The speed is fixed by the vacuum permittivity and permeability, meaning it is a fundamental property of spacetime itself, independent of the source or observer (in the context of special relativity).
Structure of the Solution
The wave equation is linear, meaning any linear combination of solutions is also a solution. The most fundamental solutions are plane waves. For a wave traveling in the \(z\)-direction, the spatial dependence is \(e^{i(kz - \omega t)}\). Substituting this into the wave equation reveals the dispersion relation:
Here, \(\omega\) is the angular frequency and \(k\) is the wavenumber. This linear relationship between \(\omega\) and \(k\) indicates that vacuum is a non-dispersive medium; all frequencies travel at the same speed \(c\). The electric and magnetic fields are coupled such that \(\mathbf{B} = \frac{1}{c} \hat{\mathbf{k}} \times \mathbf{E}\). This ensures that the fields are perpendicular to each other and to the direction of propagation, a property we will explore in detail in the next lecture.
The derivation
Electromagnetic Wave Equation in Vacuum
In a source-free vacuum (no free charge, ρ = 0; no free current, J = 0), Maxwell's four equations decouple into two identical second-order wave equations, one for each field: ∇2E = μ0ε0 ∂2E/∂t2 and ∇2B = μ0ε0 ∂2B/∂t2. Both describe disturbances propagating at a single speed c = 1/√(μ0ε0), fixed entirely by the electric and magnetic constants of the vacuum.
A common misconception is that the wave equation implies the fields are independent. In reality, the derivation requires the coupling via Maxwell’s equations. If you solve for \(\mathbf{E}\) alone, you lose the constraint that \(\mathbf{B}\) must be consistent with \(\mathbf{E}\). The wave equation describes the propagation of the disturbance, but the relationship between \(\mathbf{E}\) and \(\mathbf{B}\) is still governed by the first-order Maxwell equations. You cannot have a valid electromagnetic wave where \(\mathbf{E}\) and \(\mathbf{B}\) are arbitrary solutions to their respective wave equations; they must satisfy \(\nabla \cdot \mathbf{E} = 0\), \(\nabla \cdot \mathbf{B} = 0\), and the curl relations simultaneously. C
Worked example
Problem. A monochromatic plane electromagnetic wave propagates in the \(+z\) direction in vacuum. The electric field is given by \(\mathbf{E}(z,t) = E_0 \cos(kz - \omega t) \hat{\mathbf{x}}\), with \(E_0 = 10 \ \mathrm{V/m}\) and frequency \(f = 5.0 \times 10^9 \ \mathrm{Hz}\). Determine the wavenumber \(k\) and the amplitude of the magnetic field \(B_0\).
- Calculate the angular frequency \(\omega = 2\pi f\).
- Use the dispersion relation \(\omega = ck\) to find \(k\).
- Use the relation \(B_0 = E_0 / c\) to find the magnetic field amplitude.
Units check. \(k\) has units of \(\mathrm{rad/m}\) (radians are dimensionless). \(B_0\) has units of \(\mathrm{V/m} / (\mathrm{m/s}) = \mathrm{V \cdot s / m^2}\). Since \(1 \ \mathrm{V \cdot s} = 1 \ \mathrm{Wb}\) and \(1 \ \mathrm{Wb/m^2} = 1 \ \mathrm{T}\), the units are correct.
Three quick tests
- What is the ratio of the electric field amplitude to the magnetic field amplitude in a vacuum plane wave?
Answer
The ratio is \(c\), the speed of light. \(E_0 = c B_0\). - If the frequency of a light wave doubles, what happens to its wavelength in vacuum?
Answer
The wavelength halves. Since \(c = f\lambda\) is constant, \(\lambda\) is inversely proportional to \(f\). - Why does the wave equation for \(\mathbf{E}\) in vacuum not contain a source term?
Answer
Because we assumed \(\rho = 0\) and \(\mathbf{J} = 0\). The wave is a free disturbance propagating through empty space, not driven by local charges or currents.
Problem set for PU-204 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes checked line by line by the editor; every number in the worked example was recomputed.