Derivation

General Solution for a Planewave

To obtain a solution for EM planewaves within a homogeneous medium, let us begin with the following vector Helmholtz equations for $$\mathbf{E}$$ and $$\mathbf{H}$$:

(112)$\begin{split}\boldsymbol{\nabla}^2 \mathbf{E} + k^2 \mathbf{E} &= 0\\ \boldsymbol{\nabla}^2 \mathbf{H} + k^2 \mathbf{H} &= 0\end{split}$

where the complex wavenumber is given by:

(113)$k = \sqrt{\mu \epsilon \omega^2 - i \mu \sigma \omega}$

For simplicity, let us assume that the planewave propagates along the z-direction. According to Griffiths [Gri99] (pp. 378), the electric and magnetic fields supported by a planewave are transverse to the direction of propagation; thus the electric and magnetic fields lie in the xy-plane. In this case, the governing equation for the electric field simplifies to:

(114)$\frac{\partial^2 \mathbf{E}}{\partial z^2} + k^2 \mathbf{E} = 0$

where $$\mathbf{E} \equiv \mathbf{E}(z,\omega)$$; thus it does not depend on $$x$$ or $$y$$. The previous equation has a general solution of the form:

(115)$\mathbf{E} = \mathbf{E}_0^- \, e^{i(kz-\omega t)} + \mathbf{E}_0^+ \, e^{-i(kz + \omega t)}$

where $$\mathbf{E}_0^-$$ and $$\mathbf{E}_0^+$$ are the vector amplitudes of down-going and up-going waves, respectively. The $$e^{-i\omega t}$$ in both terms controls the temporal phase. The complex wavenumber has both real and imaginary components. Thus it can be expressed as:

(116)$k = \alpha - i\beta$

where $$\alpha \geq 0$$ and $$\beta \geq 0$$ depend on the frequency and physical properties of the media. Substituting Eq. (116) into Eq. (115), the solution to our wave equation for $$\mathbf{E}$$ becomes:

(117)$\mathbf{E} = \mathbf{E}_0^- \, e^{\beta z} e^{i(\alpha z -\omega t)} + \mathbf{E}_0^+ \, e^{-\beta z} e^{-i (\alpha z + \omega t)}$

For both the down-going and up-going waves, there are two important behaviours within the solution. The first term, which contains $$e^{\pm i \alpha z}$$, controls the oscillatory behaviour (i.e. wavelength) and velocity of each wave. The second term, which contains $$e^{\pm \beta z}$$, controls the decay behaviour (i.e. attenuation) of each wave. Notice that as $$z \rightarrow -\infty$$ for the down-going wave, its amplitude goes to zero. The same behaviour occurs for the up-going wave as $$z \rightarrow \infty$$.

Using the same approach on the Helmholtz equation for $$\mathbf{H}$$, the magnetic field has a general solution of the form:

(118)$\begin{split}\mathbf{H} &= \mathbf{H}_0^- \, e^{i(kz-\omega t)} + \mathbf{H}_0^+ \, e^{-i(kz+\omega t)}\\ &= \mathbf{H}_0^- \, e^{\beta z} e^{i(\alpha z-\omega t)} + \mathbf{H}_0^+ \, e^{-\beta z} e^{-i (\alpha z+\omega t)}\end{split}$

Note

Eq. (117) is still a general solution. To determine $$\mathbf{E}_0^-$$ and $$\mathbf{E}_0^+$$ explicitly, you must envoke a set of boundary conditions. For example, $$\mathbf{E}(z \rightarrow -\infty,\omega) = 0$$ and $$\mathbf{E}(z =0,\omega) = \mathbf{E}_0$$. This would give you a solution $$\mathbf{E}(z,\omega) = \mathbf{E}_0 \, e^{\beta z} e^{ i(\alpha z-\omega t)}$$ (i.e. just the down-going wave). From this solution, $$\mathbf{H}(z,\omega)$$ can be determined using Faraday’s law. You could also envoke boundary conditions to solve for $$\mathbf{H}$$ and use the Ampere-Maxwell law to obtain $$\mathbf{E}$$.

Supporting Derivation for the App

The app simulates the downward propagation of an EM planewave. As we can see in Fig. 44, the planewave is polarized such that the electric lies along the x-direction and the magnetic field lies along the y-direction. Physically, we can think of this wave as being caused by a horizontal sheet of harmonic current $$\mathbf{I}(\omega) = I_x \, \textrm{cos} (\omega t) \mathbf{u_x}$$, where $$\mathbf{u_x}$$ is the unit vector in the x-direction.

To solve for the electric field, we begin with the general solution from Eq. (117):

$\mathbf{E} (z,\omega) = \mathbf{E}_0^- e^{ikz} + \mathbf{E}_0^+ e^{-ikz}$

where $$\mathbf{E}_0^-$$ and $$\mathbf{E}_0^+$$ are the amplitudes of the down-going and up-going waves, respectively. Given that we are only modeling the downgoing wave and the corresponding electric field only has components in the x-direction, our solution takes the form:

$\mathbf{E} (z,\omega) = E_x (z,\omega) \, \mathbf{u_x} = E_{x,0}^{-} e^{ikz} \mathbf{u_x}$

where $$E_x$$ is a scalar function and $$E_{x,0}^{-}$$ is the scalar amplitude of the electric field. Using Faraday’s law, we can confirm that the corresponding magnetic field only has components in the y-direction, where:

$\frac{\partial E_x}{\partial z} + i \omega \mu H_y = 0$

Solving for the y-component of the magnetic field, we obtain:

$H_y (z,\omega ) = H_{y,0}^- e^{ikz} = -\frac{k}{\omega \mu} E_{x,0}^- \, e^{ikz}$

Thus:

$\mathbf{H}(z,\omega) = H_y (z,\omega) \, \mathbf{u_y} = - \frac{k}{\omega \mu} E_{x,0}^- \, e^{ikz} \, \mathbf{u_y}$

where $$\mathbf{u_y}$$ is the unit vector in the y-direction.