# Wave equation in a coupled chiral medium

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?

• 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? – mike stone Dec 26 '20 at 21:55
• 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 – Aldrich Taylor Dec 27 '20 at 7:52
• 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 – Aldrich Taylor Dec 27 '20 at 7:57

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 $$$$\nabla\times\nabla\times\vec E = \omega^2\mu\varepsilon\vec E + \left(\omega\mu\alpha + \frac{\omega}{\beta}\right)\nabla\times\vec E.$$$$