Wave Equations and Plane Waves in Free Space

Free space is the cleanest laboratory Maxwell offers: no charges (\(\rho_v=0\)), no currents (\(\vec{J}=0\)), no loss (\(\sigma=0\)), and the constants of vacuum \(\varepsilon=\varepsilon_0\), \(\mu=\mu_0\). The phasor curl equations from Chapter 17 collapse to a tidy pair:

Take the curl of the first, substitute the second, and use the identity \(\nabla\times\nabla\times\vec{E}_s=\nabla(\nabla\cdot\vec{E}_s)-\nabla^2\vec{E}_s\). Because there is no charge, \(\nabla\cdot\vec{E}_s=0\), and the gradient term vanishes:

The Helmholtz equation is a partial differential equation in all three coordinates — still hard. We tame it with the single most productive assumption in the subject: a uniform plane wave. "Plane" means every surface of constant phase is a flat plane; "uniform" means the fields have the same value everywhere on that plane. Orient the propagation along \(z\); then \(\vec{E}\) varies with \(z\) alone, and all \(x\)- and \(y\)-derivatives die:

This ordinary differential equation has the familiar exponential solutions. Keeping only the wave that travels toward \(+z\):

The combination \((\omega t-\beta z)\) is the whole story: hold it constant and you ride along with a point of fixed phase. A second solution \(e^{+j\beta z}\) describes a wave returning toward \(-z\); we will need it the moment a boundary appears in Chapter 22, but in unbounded free space the outgoing wave stands alone.

Demanding constant phase, \(\omega t-\beta z=\text{const}\), and differentiating gives the speed at which a wavefront advances — the phase velocity. In vacuum it evaluates to a number you already know by another name:

That Maxwell's two electrostatic-and-magnetostatic constants combine into the measured speed of light was the thunderclap of nineteenth-century physics: light is an electromagnetic wave. The spatial period of the wave is the wavelength \(\lambda\), tied to the phase constant (or wavenumber) \(\beta\):

The phase constant \(\beta\) measures radians of phase accumulated per metre travelled; the wavelength is simply how far you go to rack up \(2\pi\) of it. In a perfect (lossless) dielectric the same formulas hold with \(\varepsilon=\varepsilon_r\varepsilon_0\), so \(u=c/\sqrt{\varepsilon_r}\) and the wave slows and shortens — the seed of refraction.

Does the plane wave have a field component along its own direction of travel? Gauss's law settles it. With \(\nabla\cdot\vec{E}_s=0\) and fields depending only on \(z\), the divergence reduces to \(\partial E_{zs}/\partial z=0\); the wave equation then forces any such constant to be zero. The field lies entirely in the plane of the wavefront:

Feed the solution \(\vec{E}_s=E_0e^{-j\beta z}\hat{a}_x\) into the phasor Faraday law \(\nabla\times\vec{E}_s=-j\omega\mu_0\vec{H}_s\). The only surviving curl term points along \(\hat{a}_y\), so the magnetic field is forced perpendicular to \(\vec{E}\):

The ratio that appeared above — electric amplitude over magnetic amplitude — is a property of the medium alone, with the units of resistance. It is the intrinsic impedance \(\eta\), the electromagnetic cousin of a transmission line's characteristic impedance:

Because \(\eta_0\) is real, \(\vec{E}\) and \(\vec{H}\) are in phase in free space — their peaks and zeros line up in time. (Chapter 19 will make \(\eta\) complex in a conductor, tilting \(\vec{H}\) behind \(\vec{E}\).) The impedance lets you convert between the two fields instantly: divide \(E\) by \(377\,\Omega\) to get \(H\), or multiply \(H\) by \(377\,\Omega\) to get \(E\).

Assemble the pieces. The electric field, the magnetic field, and the propagation direction \(\hat{a}_k\) form a right-handed orthogonal triad, with the two fields oscillating in step and in perpendicular planes as the whole pattern marches forward at \(c\):

A few habits to lock in. The two fields reach their maxima at the same places — this in-phase relationship is special to lossless media. Their amplitudes are not free: they are tied by \(\eta\). And the cross product \(\vec{E}\times\vec{H}\) always points the way the wave goes, which is the directional fingerprint of power flow that Chapter 20 will quantify with the Poynting vector. Memorise the triad \(\hat{a}_E\times\hat{a}_H=\hat{a}_k\) and you can reconstruct any one of the three from the other two.

For each item, identify whether you are extracting a propagation parameter, converting between \(\vec{E}\) and \(\vec{H}\), or fixing a direction — then apply the matching rule. Difficulty rises down the list.