0
$\begingroup$

I want to find a wave equation propagating in a coupled medium.

Normally, to find a wave equation in an isotropic and sourceless medium, derivations shown below can be done. $$\nabla \times \vec E = -j\omega \vec B$$ $$\nabla \times \nabla \times \vec E = -j\omega \nabla \times \vec B$$ If the constitutive relation can be expressed as shown below, $$ \vec B = \mu \vec H$$ Then we have, $$\nabla^2 \vec E +\omega^2 \mu \varepsilon\vec E=0$$ However, in a coupled medium, fields can be coupled to each other as shown below which represents the constitutive relations of this medium. $$ \vec D = \varepsilon \vec E - j\vec B$$ and $$\vec B = \mu \vec H + j\vec E$$

So, if we try to take curl operator both sides of the Maxwell's equation shown below, we get, $$\nabla \times \nabla \times \vec E = -j\omega \nabla \times \vec B$$ $$\nabla \times \nabla \times \vec E = -j\omega \nabla \times (\mu \vec H + j\vec E)$$ $$\nabla \times \nabla \times \vec E = -j\omega (\mu \nabla \times \vec H + j \nabla \times \vec E)$$ Which yields, $$\nabla \times \nabla \times \vec E = -j\omega (\mu \nabla \times \vec H + j \nabla \times \vec E)$$ As we stated above, the medium is sourceless. Therefore, the related Maxwell's are $$\nabla \times \vec E = -j\omega \vec B$$ and $$\nabla \times \vec H = j\omega \vec D$$ If we reorganize the equation according to Maxwell's, $$\nabla \times \nabla \times \vec E = -j\omega (\mu j\omega \vec D + j (-j\omega \vec B ))$$ If we plug the constitutive relations into the equation, we have, $$\nabla \times \nabla \times \vec E = -j\omega (\mu j\omega (\varepsilon \vec E - j\vec B) + j (-j\omega (\mu \vec H + j\vec E) ))$$ Finally, $$\nabla \times \nabla \times \vec E = \omega^2 \mu \varepsilon \vec E + \omega^2 \mu \vec B + \omega \mu \vec H + j\omega \vec E $$ So, you can see from the equation above, the magnetic field still remains in the wave equation. To obtain a closed form wave equation of the electrical field, the equation derived should only contain the electrical field components. For example, form of the wave equation should be like this $$\nabla \times \nabla \times \vec E +p\nabla \times \vec E +q \vec E= 0 $$

How can I emit the magnetic fields and obtain a wave equation formed like shown below?

$\endgroup$
3
  • $\begingroup$ In an anisotropic medium the quantities $\mu$ and $\epsilon$ become symmetric matrices. You don't seem to have taken that into account. Also where does ${\bf D}= \epsilon {\bf E}-i{\bf B}$ come from? $\endgroup$ – mike stone Dec 26 '20 at 21:55
  • $\begingroup$ Yes, in an anisotropic medium, quantities should be like as you said. Actually, I miswrote the definition of the medium, I do not know where anisotropy came from. Anyway, I edited it, the correct definition of this kind of medium should be "chiral medium". @mikestone $\endgroup$ – Aldrich Taylor Dec 27 '20 at 7:52
  • $\begingroup$ Also, $$\vec D = \varepsilon \vec E - j \vec B$$ is a constitutive relation which gives a relation between D field and E field. en.wikipedia.org/wiki/Constitutive_equation @mikestone $\endgroup$ – Aldrich Taylor Dec 27 '20 at 7:57
0
$\begingroup$

I think the problem becomes easier upon a little re-organization.

The relevant Maxwell equations may are \begin{gather} \nabla\times\vec E=-j\omega\vec B, \\ \nabla\times\vec H = j\omega\vec D, \end{gather} where we have assumed time harmonic fields and are using the $e^{j\omega t}$ time convention. The constitutive equations may be written in terms of $\vec E$ and $\vec B$ (trying to reduce the number of fields to solve for) as \begin{gather} \vec D = \varepsilon \vec E - j\alpha\vec B, \\ \vec H = \frac{1}{\mu}\vec B - \frac{j}{\mu}\beta\vec E, \end{gather} where $\alpha$ and $\beta$ are set equal to one in the original problem. (I have generalized the problem in case you want a more general solution.) Substituting $\vec D$ and $\vec H$ into the second of the Maxwell's equations given above and assuming all material properties are uniform in space yields \begin{gather} \nabla\times\vec B - \frac{j}{\beta}\nabla\times\vec E = j\omega\mu\varepsilon\vec E + \mu j\alpha(-j\omega\vec B). \end{gather} Multiplying by $-j\omega$ then yields \begin{gather} \nabla\times(-j\omega\vec B) - \frac{\omega}{\beta}\nabla\times\vec E = \omega^2\mu\varepsilon\vec E + \omega\mu\alpha(-j\omega\vec B). \end{gather} Substituting the first of the Maxwell equations above then yields an equation entirely in terms of the electric field: \begin{gather} \nabla\times\nabla\times\vec E - \frac{\omega}{\beta}\nabla\times\vec E = \omega^2\mu\varepsilon\vec E + \omega\mu\alpha\nabla\times\vec E. \end{gather} Finally, cleaning up a bit yields \begin{equation} \nabla\times\nabla\times\vec E = \omega^2\mu\varepsilon\vec E + \left(\omega\mu\alpha + \frac{\omega}{\beta}\right)\nabla\times\vec E. \end{equation}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.