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Independent fields' components in Maxwell's equations

  1. Dec 3, 2015 #1
    In a source-free, isotropic, linear medium, Maxwell's equations can be rewritten as follows:

    [itex]\nabla \cdot \mathbf{E} = 0[/itex]
    [itex]\nabla \cdot \mathbf{H} = 0[/itex]
    [itex]\nabla \times \mathbf{E} = -j \omega \mu \mathbf{H}[/itex]
    [itex]\nabla \times \mathbf{E} = j \omega \epsilon \mathbf{E}[/itex]

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

    [itex]\displaystyle \frac{\partial}{\partial z} = e^{-j \beta z}[/itex]

    and the above equations (after some steps) become


    [itex]\displaystyle \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} = j \beta E_z[/itex]
    [itex]\displaystyle \frac{\partial H_x}{\partial x} + \frac{\partial H_y}{\partial y} = j \beta H_z[/itex]

    [itex]E_x = - j \displaystyle \frac{1}{\omega \epsilon} \frac{\partial H_z}{\partial y} + \displaystyle \frac{\beta}{\omega \epsilon} H_y[/itex]
    [itex]E_y = - \displaystyle \frac{\beta}{\omega \epsilon} H_x + j \displaystyle \frac{1}{\omega \epsilon} \frac{\partial H_z}{\partial x}[/itex]
    [itex]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}[/itex]

    [itex]H_x = j \displaystyle \frac{1}{\omega \mu} \frac{\partial E_z}{\partial y} - \displaystyle \frac{\beta}{\omega \mu} E_y[/itex]
    [itex]H_y = \displaystyle \frac{\beta}{\omega \mu} E_x - j \displaystyle \frac{1}{\omega \mu} \frac{\partial E_z}{\partial x}[/itex]
    [itex]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}[/itex]

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

    [itex]E_x = E_x (E_z, H_z)[/itex]
    [itex]E_y = E_y (E_z, H_z)[/itex]
    [itex]H_x = H_x (E_z, H_z)[/itex]
    [itex]H_y = H_y (E_z, H_z)[/itex]

    Which is equivalent to state that just two scalar functions [itex]E_z = f(x,y)e^{-j \beta z}[/itex] and [itex]H_z = g(x,y)e^{-j \beta z}[/itex] are actually independent. But how could it be proved? It is not evident from the equations I wrote above: they show instead that [itex]E_z, H_z[/itex] appear to be functions of [itex]E_x, E_y, H_x, H_y[/itex].
    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. jcsd
  3. Dec 3, 2015 #2
    You chose z as the propagation axis, so the fields along that axis won't behave like those in x and y.
     
  4. Dec 4, 2015 #3
    This is reasonable. But why?
    Anyway, there is a direct approach to express [itex]E_x, E_y, H_x, H_y[/itex] as functions of [itex]E_z, H_z[/itex] only.
    For example, let's combine the first and the fifth equations in order to cancel [itex]H_y[/itex]: we will obtain

    [itex]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)[/itex]

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