# Independent fields' components in Maxwell's equations

Tags:
1. Dec 3, 2015

### EmilyRuck

In a source-free, isotropic, linear medium, Maxwell's equations can be rewritten as follows:

$\nabla \cdot \mathbf{E} = 0$
$\nabla \cdot \mathbf{H} = 0$
$\nabla \times \mathbf{E} = -j \omega \mu \mathbf{H}$
$\nabla \times \mathbf{E} = j \omega \epsilon \mathbf{E}$

If we are looking for a wave solution, travelling along the $z$ direction, with $k = k_z = \beta$, that means

$\displaystyle \frac{\partial}{\partial z} = e^{-j \beta z}$

and the above equations (after some steps) become

$\displaystyle \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} = j \beta E_z$
$\displaystyle \frac{\partial H_x}{\partial x} + \frac{\partial H_y}{\partial y} = j \beta H_z$

$E_x = - j \displaystyle \frac{1}{\omega \epsilon} \frac{\partial H_z}{\partial y} + \displaystyle \frac{\beta}{\omega \epsilon} H_y$
$E_y = - \displaystyle \frac{\beta}{\omega \epsilon} H_x + j \displaystyle \frac{1}{\omega \epsilon} \frac{\partial H_z}{\partial x}$
$E_z = - j \displaystyle \frac{1}{\omega \epsilon} \frac{\partial H_y}{\partial x} + j \displaystyle \frac{1}{\omega \epsilon} \frac{\partial H_x}{\partial y}$

$H_x = j \displaystyle \frac{1}{\omega \mu} \frac{\partial E_z}{\partial y} - \displaystyle \frac{\beta}{\omega \mu} E_y$
$H_y = \displaystyle \frac{\beta}{\omega \mu} E_x - j \displaystyle \frac{1}{\omega \mu} \frac{\partial E_z}{\partial x}$
$H_z = j \displaystyle \frac{1}{\omega \mu} \frac{\partial E_y}{\partial x} - j \displaystyle \frac{1}{\omega \mu} \frac{\partial E_x}{\partial y}$

It should be possible to express the transverse field components as functions of the longitudinal field components:

$E_x = E_x (E_z, H_z)$
$E_y = E_y (E_z, H_z)$
$H_x = H_x (E_z, H_z)$
$H_y = H_y (E_z, H_z)$

Which is equivalent to state that just two scalar functions $E_z = f(x,y)e^{-j \beta z}$ and $H_z = g(x,y)e^{-j \beta z}$ are actually independent. But how could it be proved? It is not evident from the equations I wrote above: they show instead that $E_z, H_z$ appear to be functions of $E_x, E_y, H_x, H_y$.
The only link I could find is this: at the bottom of page 11, it shows that all the fields can be expressed in terms of two scalar functions. But it is not a direct approach, because it uses the Hertz Vector potentials.

Is there a direct approach to prove that all the 6 field components are function of just 2 of them?

2. Dec 3, 2015

### DuckAmuck

You chose z as the propagation axis, so the fields along that axis won't behave like those in x and y.

3. Dec 4, 2015

### EmilyRuck

This is reasonable. But why?
Anyway, there is a direct approach to express $E_x, E_y, H_x, H_y$ as functions of $E_z, H_z$ only.
For example, let's combine the first and the fifth equations in order to cancel $H_y$: we will obtain

$E_x = -j \displaystyle \frac{1}{k^2 - \beta^2} \left( \omega \mu \frac{\partial H_z}{\partial y} + \beta \frac{\partial E_z}{\partial x} \right)$

Where $k^2 = \omega^2 \mu \epsilon$. A similar result for $E_y$ can be obtained from equations 2 and 4. The same is for the magnetic field components.