# 2.6: Impedance

We need to remind ourselves of one other thing from electromagnetic theory before we can proceed, namely the meaning of *impedance* in the context of electromagnetic wave propagation. The impedance \(Z\) is merely the ratio \(E/H\) of the electric to the magnetic field. The SI units of \(E\) and \(H\) are V/m and A/m respectively, so the SI units of \(Z\) are V/A, or ohms, \(\Omega\). We are now going to see if we can express the impedance in terms of the permittivity and permeability of the medium in which an electromagnetic wave is travelling.

\[\begin{align} \boldsymbol{\nabla} \cdot \mathbf{\text{D}} &= \rho \\ \boldsymbol{\nabla} \cdot \mathbf{\text{B}} &= 0. \\ \boldsymbol{\nabla} \times \mathbf{\text{H}} &= \mathbf{\dot{\text{D}}}+\mathbf{\text{J}}. \\ \boldsymbol{\nabla} \times \mathbf{\text{E}} &= -\mathbf{\dot{\text{B}}}. \end{align}\]

In an isotropic, homogeneous, nonconducting, uncharged medium (such as glass, for example), the equations become:

\[\begin{align} \boldsymbol{\nabla} \cdot \mathbf{\text{E}} &= 0 \\ \boldsymbol{\nabla} \cdot \mathbf{\text{H}} &= 0 \\ \boldsymbol{\nabla} \times \mathbf{\text{H}} &= \epsilon \mathbf{\dot{\text{E}}}. \\ \boldsymbol{\nabla} \times \mathbf{\text{H}} &= -\mu \mathbf{\dot{\text{H}}}. \end{align}\]

If you eliminate \(\mathbf{H}\) from these equations, you get

\[ \nabla^2 \mathbf{\text{E}} = \epsilon\mu \mathbf{\ddot{\text{E}}}, \label{NewEq1}\]

which describes an electric wave of speed

\[\dfrac{ 1}{ \sqrt{\epsilon\mu}}. \label{Speed1}\]

In free space, this becomes

\[\dfrac{1}{\sqrt{\epsilon_0 \mu_0}}\]

which is \(2.998 \times 10^8 \,m \,s^{−1}\).

The ratio of the speeds in two media is

\[\dfrac{v_1}{v_2} = \dfrac{n_2}{n_1} = \sqrt{\dfrac{\epsilon_2\mu_2}{\epsilon_1\mu_1}},\]

and if, as is often the case, the two permeabilities are equal (to \(\mu_0\)), then

\[\dfrac{v_1}{v_2} = \dfrac{n_2}{n_1} = \sqrt{\dfrac{\epsilon_2}{\epsilon_1}}.\]

In particular, if you compare one medium with a vacuum, you get: \(n = \sqrt{\dfrac{\epsilon}{\epsilon_0}}\).

Light is a high-frequency electromagnetic wave. When a dielectric medium is subject to a high frequency field, the polarization (and hence \(D\)) cannot keep up with the electric field \(E\). \(D\) lags behind \(E\). This can be described mathematically by ascribing a complex value to the permittivity. The amount of lag depends, unsurprisingly, on the frequency - i.e. on the color - and so the permittivity and hence the refractive index depends on the wavelength of the light. This is *dispersion*.

If instead you eliminate **E** from Maxwell’s equations, you get

\[ \nabla^2 \mathbf{\text{H}} = \epsilon \mu \mathbf{\ddot{\text{H}}}. \label{NewEq2}\]

This is a magnetic wave of the same speed.

If you eliminate the time between \ref{NewEq1} and \ref{NewEq2}, you find that \(\dfrac{E}{H} = \sqrt{\dfrac{\mu}{\epsilon}}\), which, in free space, has the value \(\sqrt{\dfrac{\mu_0}{\epsilon_0}} = 377 \Omega\), which is the impedance of free space. In an appropriate context I may use the symbol \(Z_0\) to denote the impedance of free space, and the symbol \(Z\) to denote the impedance of some other medium.

The ratio of the impedances in two media is

\[\dfrac{Z_1}{Z_2} = \sqrt{\dfrac{\epsilon_2\mu_1}{\epsilon_1 \mu_2}},\]

and if, as is often the case, the two permeabilities are equal (to \(\mu_0\)), then

\[\dfrac{Z_1}{Z_2} = \sqrt{\dfrac{\epsilon_2}{\epsilon_1}}= \dfrac{n_2}{n_1}=\dfrac{v_1}{v_2}.\]

We shall be using this result in what follows.