Deriving the Electromagnetic Wave Equation

Fig 1 — A uniform plane wave travelling along z: 𝐄 along x (blue), 𝐇 along y (orange), in phase and perpendicular to each other and to the direction of travel

Introduction to Electromagnetic Waves

In Part 7: The Four Maxwell's Equations Explained we saw that the two curl equations couple the electric and magnetic fields: a changing 𝐁 creates 𝐄 (Faraday) and a changing 𝐃 creates 𝐇 (Ampère–Maxwell). In this article we follow that coupling to its conclusion. By combining the two equations we eliminate one of the fields and obtain a wave equation — the same type of equation that governs waves on a string and sound in air. Its solutions are electromagnetic waves, and the equation itself tells us how fast they travel.

This derivation is one of the great results of physics: from constants measured with capacitors and coils, Maxwell calculated the speed of light. For the RF engineer it supplies the basic quantities of every propagation problem — wavelength, phase velocity, wave impedance, attenuation and skin depth.

What Is an Electromagnetic Wave?

An electromagnetic wave is a self-sustaining oscillation of electric and magnetic fields that carries energy through space, at a finite speed, without needing any medium.

Electric Field Component

The electric field of a wave oscillates in time at the source frequency and varies periodically in space along the direction of travel. For a wave travelling in +z with its electric field along x (a linearly polarised wave), E_x(z, t) = E₀ cos(ωt − βz). The direction of 𝐄 defines the wave's polarisation — vertical, horizontal, or rotating for circular polarisation.

Magnetic Field Component

The magnetic field oscillates with the same frequency and phase, but perpendicular to 𝐄: for the wave above, H_y(z, t) = (E₀/η) cos(ωt − βz). Neither field can exist alone in a propagating wave; each continuously regenerates the other.

Relationship Between Electric and Magnetic Fields

Perpendicular: 𝐄 ⟂ 𝐇.
In phase (in a lossless medium): both reach their maxima at the same place and time.
Fixed ratio: |𝐄|/|𝐇| = η, the intrinsic impedance of the medium (≈ 377 Ω in free space).
Power flow: the direction of travel is the direction of the Poynting vector 𝐒 = 𝐄 × 𝐇.

Transverse Electromagnetic Wave Propagation

Because both fields lie in the plane perpendicular to the direction of propagation, the wave is transverse electromagnetic (TEM). A uniform plane wave — fields constant over infinite planes perpendicular to the direction of travel — is the simplest TEM wave and an excellent model for the far field of any antenna, where the spherical wavefronts are locally almost flat. Coaxial cables and stripline also carry TEM waves; waveguides cannot, and support TE and TM modes instead (Part 10).

How Maxwell's Equations Predict Electromagnetic Waves

Consider a region of space far from any sources: no free charge (ρ_v = 0) and no conduction current (𝐉 = 0), filled with a linear, homogeneous, isotropic, lossless medium with permittivity ε and permeability μ. Maxwell's equations become:

Maxwell's equations in a source-free, lossless, homogeneous medium

\begin{array}{l} (1) \nabla \cdot 𝐄 = 0 & (3) \nabla \times 𝐄 = - μ \frac{\partial 𝐇}{\partial t} \\ (2) \nabla \cdot 𝐇 = 0 & (4) \nabla \times 𝐇 = ε \frac{\partial 𝐄}{\partial t} \end{array}

We have used 𝐃 = ε𝐄 and 𝐁 = μ𝐇 with constant ε and μ. Equations (3) and (4) are two coupled first-order equations in two unknown fields.

The strategy is the same as with any pair of coupled equations: differentiate one, substitute the other, and obtain a single second-order equation in one unknown. The time derivative of 𝐇 in equation (3) can be replaced using equation (4) — but only after taking a curl, because equation (4) gives the curl of 𝐇, not 𝐇 itself.

Derivation of the Electromagnetic Wave Equation

Starting With Faraday's Law

Take the curl of both sides of Faraday's law (3). Because the spatial curl and the time derivative act on different variables, they can be swapped on the right-hand side:

Step 1 — curl of Faraday's law

\nabla \times (\nabla \times 𝐄) = - μ \frac{\partial}{\partial t} (\nabla \times 𝐇)

Applying Ampere-Maxwell Law

Now substitute the Ampère–Maxwell law (4), ∇ × 𝐇 = ε ∂𝐄/∂t, into the right-hand side:

Step 2 — substitute ∇ × 𝐇

\nabla \times (\nabla \times 𝐄) = - μ ε \frac{\partial^{2} 𝐄}{\partial t^{2}}

This is the step that needs the displacement current. Without the ε ∂𝐄/∂t term in Ampère's law, the right-hand side would be zero and there would be no wave.

Using Vector Calculus Identities

The "curl of a curl" is simplified with a standard vector identity, valid for any vector field in Cartesian coordinates:

Step 3 — the curl–curl identity

\nabla \times (\nabla \times 𝐄) = \nabla (\nabla \cdot 𝐄) - \nabla^{2} 𝐄 \nabla^{2} = \frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} + \frac{\partial^{2}}{\partial z^{2}}

∇² is the Laplacian, applied to each Cartesian component of 𝐄. In a source-free region, Gauss's law (1) gives ∇·𝐄 = 0, so the first term on the right vanishes.

Deriving the Wave Equation for the Electric Field

With ∇·𝐄 = 0, the left-hand side becomes −∇²𝐄, and Step 2 gives:

Wave equation for 𝐄

\nabla^{2} 𝐄 = μ ε \frac{\partial^{2} 𝐄}{\partial t^{2}}

Compare the general wave equation ∇²ψ = (1/v²) ∂²ψ/∂t². The electric field obeys a wave equation with 1/v² = με. All four Maxwell equations were used: (3) and (4) for the coupling, (1) to simplify, and (2) in deriving the matching equation for 𝐇.

Deriving the Wave Equation for the Magnetic Field

Starting instead from the curl of the Ampère–Maxwell law, substituting Faraday's law, and using ∇·𝐇 = 0 gives an identical equation for 𝐇:

Wave equation for 𝐇

\nabla \times (\nabla \times 𝐇) = ε \frac{\partial}{\partial t} (\nabla \times 𝐄) = - μ ε \frac{\partial^{2} 𝐇}{\partial t^{2}} ⟹ \nabla^{2} 𝐇 = μ ε \frac{\partial^{2} 𝐇}{\partial t^{2}}

For sinusoidal fields, ∂²/∂t² → −ω², and the wave equations become the Helmholtz equations, the form solved in frequency-domain RF analysis:

Helmholtz (time-harmonic) wave equations

\nabla^{2} 𝐄 + β^{2} 𝐄 = 0 \nabla^{2} 𝐇 + β^{2} 𝐇 = 0 β = ω \sqrt{μ ε}

β (rad/m), also written k, is the phase constant or wavenumber. A plane-wave solution is E_x = E₀ cos(ωt − βz): substituting gives −β²E = −ω²με E, so β = ω√(με) is required.

Electromagnetic Wave Velocity

Speed of Light in Free Space

Wave velocity in a medium and in vacuum

v = \frac{1}{\sqrt{μ ε}} c = \frac{1}{\sqrt{μ_{0} ε_{0}}} = \frac{1}{\sqrt{1.2566 \times 10^{- 6} \times 8.854 \times 10^{- 12}}} = 2.998 \times 10^{8} m/s

Since 2019 the SI fixes c = 299,792,458 m/s exactly; μ₀ and ε₀ are measured values tied by μ₀ε₀ = 1/c². In Maxwell's day the agreement between this calculation and the measured speed of light was the decisive evidence that light is electromagnetic.

Wave Propagation in Dielectric Materials

In a non-magnetic dielectric (μ_r = 1) with relative permittivity ε_r, the wave slows down by √ε_r. This ratio is the refractive index n in optics and the inverse of the velocity factor in cable datasheets:

Velocity in a dielectric

v = \frac{c}{\sqrt{μ_{r} ε_{r}}} VF = \frac{v}{c} = \frac{1}{\sqrt{ε_{r}}} n = \sqrt{ε_{r}}

Solid-polyethylene coax (ε_r = 2.25): VF = 0.667. Foam-PE coax: VF ≈ 0.80–0.85. Air-dielectric line: VF ≈ 1.

Effect of Permittivity and Permeability on Wave Velocity

Higher ε or μ slows the wave and shortens the wavelength in proportion to 1/√(μ_rε_r).
Miniaturisation: resonant structures scale with λ, so high-ε_r ceramics (ε_r = 20–90) make compact dielectric resonators, ceramic chip antennas and GPS patches.
Phase length: a cable or trace of physical length ℓ has electrical length βℓ = 2πℓ√ε_r/λ₀ — the quantity that matters for matching stubs, delay lines and phased-array feeds.
Dispersion: when ε_r varies with frequency, different frequencies travel at different speeds, distorting wideband signals.

Wavelength, Frequency, and Phase Velocity

For a sinusoidal wave cos(ωt − βz), a point of constant phase moves according to ωt − βz = constant, so dz/dt = ω/β. That speed is the phase velocity. The distance over which the phase advances by 2π is the wavelength:

Phase velocity, wavelength and wavenumber

v_{p} = \frac{ω}{β} = \frac{1}{\sqrt{μ ε}} λ = \frac{2 π}{β} = \frac{v_{p}}{f} λ_{g} = \frac{λ_{0}}{\sqrt{ε_{r}}}

The frequency is fixed by the source and never changes between media; the wavelength and phase velocity change with the medium. λ₀ is the free-space wavelength.

Frequency	Band / use	λ₀ (free space)	λ in FR-4 (ε_r = 4.4)
13.56 MHz	NFC / RFID	22.1 m	10.5 m
433 MHz	ISM telemetry	69.2 cm	33.0 cm
2.45 GHz	Wi-Fi, Bluetooth	12.2 cm	5.83 cm
5.8 GHz	Wi-Fi 5/6	5.17 cm	2.46 cm
28 GHz	5G FR2	10.7 mm	5.1 mm
77 GHz	Automotive radar	3.89 mm	1.86 mm

Phase velocity is not always the speed at which information travels. In dispersive structures such as waveguides the phase velocity can exceed c, while energy and modulation travel at the group velocity v_g = dω/dβ, which in a waveguide is always less than c (v_p·v_g = c² for an air-filled guide). In a non-dispersive medium the two are equal.

Intrinsic Impedance of Free Space

Substitute the plane wave E_x = E₀ cos(ωt − βz) into Faraday's law. The only non-zero component of ∇ × 𝐄 is ∂E_x/∂z = βE₀ sin(ωt − βz) in the y-direction, and integrating −μ∂H_y/∂t = βE₀ sin(ωt − βz) gives H_y = (β/ωμ) E₀ cos(ωt − βz). The ratio E_x/H_y = ωμ/β = √(μ/ε) is the intrinsic impedance of the medium:

Intrinsic (wave) impedance

η = \frac{E_{x}}{H_{y}} = \sqrt{\frac{μ}{ε}} η_{0} = \sqrt{\frac{μ_{0}}{ε_{0}}} = 376.73 Ω \approx 120 π Ω η = \frac{η_{0}}{\sqrt{ε_{r}}} (non-magnetic)

η is a property of the medium, not of any circuit. It sets the power density of a wave: S = E_rms²/η = η·H_rms². Mismatches in η between media cause reflections (Part 9), just as mismatches in Z₀ do on transmission lines.

Time-average power density of a plane wave (Poynting vector)

S_{avg} = \frac{E_{0}^{2}}{2 η} = \frac{E_{rms}^{2}}{η} (W/m²)

Electromagnetic Wave Propagation in Different Media

Free Space

μ = μ₀, ε = ε₀, σ = 0. The wave travels at c with no attenuation (power density still falls as 1/r² from a point source because the energy spreads over a growing sphere — that is the "spreading loss" in the Friis equation, not absorption). β₀ = 2π/λ₀ and η₀ = 376.73 Ω.

Lossless Dielectric Media

σ = 0, ε = ε₀ε_r, μ = μ₀. The wave is slower (v = c/√ε_r), the wavelength shorter, and the impedance lower (η = η₀/√ε_r), but there is still no attenuation and 𝐄 and 𝐇 remain in phase. Low-loss laminates, PTFE and polyethylene approximate this case.

Lossy Media

With conductivity σ, the Ampère–Maxwell law becomes ∇ × 𝐇 = (σ + jωε)𝐄. Repeating the derivation gives a complex propagation constant γ:

Propagation in a lossy medium

γ = α + j β = \sqrt{j ω μ (σ + j ω ε)} E_{x} (z) = E_{0} e^{- α z} \cos (ω t - β z) η = \sqrt{\frac{j ω μ}{σ + j ω ε}}

α (Np/m) is the attenuation constant; 1 Np = 8.686 dB. The impedance becomes complex, so 𝐇 lags 𝐄. For a low-loss dielectric, α ≈ (σ/2)√(μ/ε) = (β tan δ)/2.

In a good conductor (σ ≫ ωε) the expressions simplify, and the field decays within a skin depth δ of the surface:

Good conductor: skin depth and surface resistance

α = β \approx \sqrt{π f μ σ} δ = \frac{1}{α} = \frac{1}{\sqrt{π f μ σ}} R_{s} = \frac{1}{σ δ} = \sqrt{\frac{π f μ}{σ}}

After one skin depth the field has fallen to 1/e (−8.7 dB); after five skin depths, to under 1%. R_s (Ω per square) sets conductor loss in transmission lines, waveguides and antennas.

Applications in RF and Microwave Engineering

Line lengths and stubsQuarter-wave transformers, stubs and baluns are sized from λ_g = λ₀/√ε_eff in the actual medium.

Antenna sizeResonant antennas scale with wavelength; dielectric loading shrinks patches and chip antennas by roughly √ε_r.

Conductor lossSkin depth sets RF resistance: plating thickness, surface roughness and copper finish all matter once they are comparable to δ.

ShieldingA shield several skin depths thick absorbs the fields that penetrate it, while the impedance mismatch reflects most of the incident wave.

Propagation and exposurePower density S = E²/η links field-strength measurements (V/m) to link budgets and RF exposure limits (W/m²).

Penetration of materialsAttenuation in walls, foliage, the body and water decides indoor coverage, SAR, and why submarines use very low frequencies.

Numerical Examples

Wavelength and Phase Constant at 2.4 GHz

Problem: Find the free-space wavelength, phase constant and period of a 2.4 GHz wave.

λ = c/f; β = 2π/λ; T = 1/f

λ_{0} = \frac{2.998 \times 10^{8}}{2.4 \times 10^{9}} = 12.49 cm β_{0} = \frac{2 π}{0.1249} = 50.3 rad/m T = 417 ps

Answer: λ₀ = 12.49 cm | β₀ = 50.3 rad/m (≈ 2.88°/mm) | T = 417 ps

A 1 mm length error in a 2.4 GHz air line shifts phase by about 2.9°; in FR-4 (√4.4 ≈ 2.1× slower) by about 6°.

Wave in a PTFE Dielectric at 10 GHz

Problem: A 10 GHz wave travels in PTFE (ε_r = 2.1, μ_r = 1, lossless). Find the phase velocity, wavelength and intrinsic impedance.

√2.1 = 1.449

v_{p} = \frac{2.998 \times 10^{8}}{1.449} = 2.069 \times 10^{8} m/s λ = \frac{2.069 \times 10^{8}}{10^{10}} = 20.7 mm η = \frac{376.7}{1.449} = 260 Ω

Answer: v_p = 2.07 × 10⁸ m/s (VF = 0.69) | λ = 20.7 mm (vs 30.0 mm in air) | η = 260 Ω

This is why PTFE-dielectric coax has a velocity factor of about 0.69–0.70.

Magnetic Field and Power Density of a Plane Wave

Problem: A plane wave in free space has a peak electric field of 10 V/m. Find the peak magnetic field and the time-average power density. What power would a receiving antenna with effective aperture 50 cm² collect?

H₀ = E₀/η₀; S_avg = E₀²/(2η₀); P = S·A_e

H_{0} = \frac{10}{376.7} = 26.5 mA/m S_{avg} = \frac{10^{2}}{2 \times 376.7} = 0.133 W/m² P = 0.133 \times 0.005 = 0.66 mW

Answer: H₀ = 26.5 mA/m | S_avg = 0.133 W/m² (13.3 μW/cm²) | P ≈ 0.66 mW (−1.8 dBm)

Field-strength meters measure E (V/m) and infer power density through η₀, which is only valid in the far field where the wave is plane. Close to an antenna, E/H differs from 377 Ω and both fields must be measured.

Skin Depth and Surface Resistance of Copper

Problem: Find the skin depth and surface resistance of copper (σ = 5.8 × 10⁷ S/m) at 1 GHz. Is a 35 μm (1 oz) copper trace thick enough?

δ = 1/√(πfμ₀σ)

δ = \frac{1}{\sqrt{π \times 10^{9} \times 4 π \times 10^{- 7} \times 5.8 \times 10^{7}}} = \frac{1}{4.785 \times 10^{5}} = 2.09 μm R_{s} = \frac{1}{5.8 \times 10^{7} \times 2.09 \times 10^{- 6}} = 8.25 mΩ/□

Answer: δ = 2.09 μm | R_s = 8.25 mΩ per square | 35 μm ≈ 17 skin depths — more than enough

Skin depth scales as 1/√f: 66 μm at 1 MHz, 0.66 μm at 10 GHz. At microwave frequencies, the copper surface roughness (often 1–3 μm) becomes comparable to δ and noticeably raises conductor loss — the reason for "low-profile" copper foils on RF laminates.

Attenuation of Radio Waves in Seawater

Problem: Seawater (σ ≈ 4 S/m, μ_r = 1) is a good conductor at low frequency (Part 6). Find the attenuation constant, skin depth and loss per metre at 1 MHz and at 100 Hz.

Good conductor: α = √(πfμσ); loss = 8.686 α dB/m

α_{1 MHz} = \sqrt{π \times 10^{6} \times 4 π \times 10^{- 7} \times 4} = 3.97 Np/m δ = 0.25 m 34.5 dB/m

At 100 Hz the attenuation is √(10⁴) = 100× smaller

α_{100 Hz} = 0.0397 Np/m δ = 25 m 0.345 dB/m

Answer: 1 MHz — δ = 0.25 m, 34.5 dB/m | 100 Hz — δ = 25 m, 0.35 dB/m

At 1 MHz a signal loses 100 dB in under 3 m of seawater. This is why submarine communication uses VLF (3–30 kHz) and ELF frequencies, accepting tiny bandwidths in exchange for penetration of tens of metres.

Summary

Quantity	General	Free space	Lossless dielectric (μ_r = 1)
Wave equation	∇²𝐄 = με ∂²𝐄/∂t²	με = 1/c²	με = ε_r/c²
Phase velocity	1/√(με)	c = 2.998 × 10⁸ m/s	c/√ε_r
Phase constant	β = ω√(με)	2π/λ₀	β₀√ε_r
Wavelength	2π/β	λ₀ = c/f	λ₀/√ε_r
Intrinsic impedance	√(μ/ε)	376.7 Ω	376.7/√ε_r Ω
Attenuation	Re{√(jωμ(σ + jωε))}	0	0
Skin depth (good conductor)	1/√(πfμσ)	—	—

Taking the curl of Faraday's law and substituting the Ampère–Maxwell law yields the wave equation ∇²𝐄 = με ∂²𝐄/∂t²; the same holds for 𝐇.
The waves travel at v = 1/√(με), which in vacuum is the speed of light — light is an electromagnetic wave.
In a plane wave 𝐄 and 𝐇 are perpendicular, in phase (if lossless), in the ratio η = √(μ/ε), and carry power density S = 𝐄 × 𝐇.
Dielectrics slow waves and lower impedance; lossy media attenuate them; good conductors confine fields to a skin depth.

Waves eventually meet boundaries — a substrate, a metal wall, a radome. Part 9: Maxwell's Equations and Electromagnetic Boundary Conditions derives what happens to the fields at an interface and calculates how much of a wave is reflected and transmitted.

Put It Into Practice

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