##### EM Waves in Conductors
emdm.emwm.ad.c

We now consider EM waves inside a bulk conductor and will consider nonvanishing free charge \(\rho_f\) and current \({\boldsymbol J}_f\) densities. By Ohm's law, we have \[ {\boldsymbol J}_f = \sigma {\boldsymbol E} \] which means that the Maxwell equations reduce to

\begin{align*} {\boldsymbol \nabla} \cdot {\boldsymbol E} &= \frac{\rho_f}{\varepsilon}, \hspace{10mm} & {\boldsymbol \nabla} \cdot {\boldsymbol B} &= 0, \nonumber\\ {\boldsymbol \nabla} \times {\boldsymbol E} &= -\frac{\partial {\boldsymbol B}}{\partial t} & {\boldsymbol \nabla} \times {\boldsymbol B} &= \mu \sigma {\boldsymbol E} + \mu \varepsilon \frac{\partial {\boldsymbol E}}{\partial t}. \end{align*}

Putting together the continuity equation for free charge \[ \frac{\partial \rho_f}{\partial t} + {\boldsymbol \nabla} \cdot {\boldsymbol J}_f = 0 \] with Ohm's law and Gauss's law yields \[ \frac{\partial \rho_f}{\partial t} = -\sigma ({\boldsymbol \nabla} \cdot {\boldsymbol E}) = -\frac{\sigma}{\varepsilon} \rho_f \longrightarrow \rho_f (t) = \rho_f (0) e^{-\frac{\sigma}{\varepsilon} t} \] so any initial free charge dissipates with characteristic time \(\tau \equiv \frac{\varepsilon}{\sigma}\).

After the free charge has dissipated, we have

\begin{align} {\boldsymbol \nabla} \cdot {\boldsymbol E} &= 0, \hspace{10mm} & {\boldsymbol \nabla} \cdot {\boldsymbol B} &= 0, \nonumber\\ {\boldsymbol \nabla} \times {\boldsymbol E} &= -\frac{\partial {\boldsymbol B}}{\partial t} & {\boldsymbol \nabla} \times {\boldsymbol B} &= \mu \sigma {\boldsymbol E} + \mu \varepsilon \frac{\partial {\boldsymbol E}}{\partial t} \end{align}

Applying \({\boldsymbol \nabla} \times\) to the curl equations gives the modified wave equations \[ {\boldsymbol \nabla}^2 {\boldsymbol E} = \mu \varepsilon \frac{\partial^2 {\boldsymbol E}}{\partial t^2} + \mu \sigma \frac{\partial {\boldsymbol E}}{\partial t}, \hspace{10mm} {\boldsymbol \nabla}^2 {\boldsymbol B} = \mu \varepsilon \frac{\partial^2 {\boldsymbol B}}{\partial t^2} + \mu \sigma \frac{\partial {\boldsymbol B}}{\partial t} \] with plane-wave solutions \[ {\boldsymbol E} ({\boldsymbol r}, t) = {\boldsymbol E}_0 e^{i (\tilde{\boldsymbol k} \cdot {\boldsymbol r} - \omega t)} \] with complex-valued wavevector \[ \tilde{k}^2 \equiv \tilde{\boldsymbol k} \cdot \tilde{\boldsymbol k} = \mu \varepsilon \omega^2 + i\mu \sigma \omega. \] We can thus write \[ \tilde{k} \equiv k + i\kappa, \hspace{5mm} k = k_+, \hspace{3mm}\kappa = k_-, \hspace{5mm} k_{\pm} \equiv \omega \sqrt{\frac{\mu\varepsilon}{2}} \left[\sqrt{1 + \left(\frac{\sigma}{\varepsilon \omega}\right)^2} \pm 1 \right]^{1/2}. \] The wave can thus be written (letting \(\hat{\boldsymbol k}\) represent the direction of propagation and using the notations \({\boldsymbol k} \equiv k \hat{\boldsymbol k}\) and \({\boldsymbol \kappa} \equiv \kappa \hat{\boldsymbol k}\)) \[ {\boldsymbol E} ({\boldsymbol r}, t) = {\boldsymbol E}_0 e^{-{\boldsymbol \kappa} \cdot {\boldsymbol r}} e^{i ({\boldsymbol k} \cdot {\boldsymbol r} - \omega t)} \] with a similar solution for \({\boldsymbol B}\). The quantity \(d = \frac{1}{\kappa}\) is known as the skin depth.

For simplicity, we will put the propagation direction along \(z\) and the polarization along \(x\): \[ {\boldsymbol E} (z,t) = E_0 e^{-\kappa z} e^{i(kz - \omega t)} \hat{\boldsymbol x}. \] From Maxwell (iii) we get \[ {\boldsymbol B} (z,t) = \frac{\tilde{k}}{\omega} E_0 e^{-\kappa z} e^{i(kz - \omega t)} \hat{\boldsymbol y}. \]

To simplify things further, we write the complex wavevector in polar form: \[ \tilde{k} = K e^{i\phi}, \hspace{5mm} K = \sqrt{k^2 + \kappa^2} = \omega \sqrt{\mu\varepsilon \sqrt{1 + \left( \frac{\sigma}{\varepsilon \omega}\right)^2}}, \hspace{5mm} \phi = \mbox{atan}~\frac{\kappa}{k}. \] The electric and magnetic fields thus develop a relative phase: if we write \(E_0 = E e^{i \delta_E}\) and \(B_0 = B e^{i\delta_B}\) then \[ \delta_B - \delta_E = \phi \] so the magnetic field lags behind the electric field. The real amplitudes are related by \[ \frac{B}{E} = \frac{K}{\omega} = \sqrt{\mu\varepsilon \sqrt{1 + \left( \frac{\sigma}{\varepsilon \omega}\right)^2}} \] so the final form of the fields is \[ {\boldsymbol E} (z,t) = E e^{-\kappa z} \cos (kz - \omega t + \delta_E) \hat{\boldsymbol x}, \hspace{5mm} {\boldsymbol B} (z,t) = B e^{-\kappa z} \cos (kz - \omega t + \delta_E + \phi) \hat{\boldsymbol y}. \]

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42