Independent fields' components in Maxwell's equations

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In a source-free, isotropic, linear medium, Maxwell's equations can be reformulated to express the relationship between electric and magnetic fields. By assuming a wave solution traveling in the z direction, the equations simplify to show that the transverse field components can be expressed as functions of the longitudinal components. The discussion highlights the challenge of proving that all six field components depend solely on two independent scalar functions, E_z and H_z. A proposed method involves manipulating the equations to eliminate certain components, leading to expressions for E_x and E_y in terms of E_z and H_z. This approach aims to establish a direct relationship among the field components without relying on external references.
EmilyRuck
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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, traveling 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 http://my.ece.ucsb.edu/York/Bobsclass/201B/W01/potentials.pdf: 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?
 
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You chose z as the propagation axis, so the fields along that axis won't behave like those in x and y.
 
DuckAmuck said:
You chose z as the propagation axis, so the fields along that axis won't behave like those in x and y.

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.
 
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