Wave equation in a coupled chiral medium

Question

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.
$$nabla times vec E = -jomega vec B$$
$$nabla times nabla times vec E = -jomega nabla times vec B$$
If the constitutive relation can be expressed as shown below, $$ vec B = mu vec H$$
Then we have,
$$nabla^2 vec E +omega^2 mu varepsilonvec E=0$$
However, in a coupled medium, fields can be coupled to each other as shown below which represents the constitutive relations of this medium.
$$ vec D = varepsilon vec E - jvec B$$ and
$$vec B = mu vec H + jvec E$$
So, if we try to take curl operator both sides of the Maxwell's equation shown below, we get,
$$nabla times nabla times vec E = -jomega nabla times vec B$$
$$nabla times nabla times vec E = -jomega nabla times (mu vec H + jvec E)$$
$$nabla times nabla times vec E = -jomega (mu nabla times vec H + j nabla times vec E)$$
Which yields,
$$nabla times nabla times vec E = -jomega (mu nabla times vec H + j nabla times vec E)$$
As we stated above, the medium is sourceless. Therefore, the related Maxwell's are
$$nabla times vec E = -jomega vec B$$
and
$$nabla times vec H = jomega vec D$$
If we reorganize the equation according to Maxwell's,
$$nabla times nabla times vec E = -jomega (mu jomega vec D + j (-jomega vec B ))$$
If we plug the constitutive relations into the equation, we have,
$$nabla times nabla times vec E = -jomega (mu jomega (varepsilon vec E - jvec B) + j (-jomega (mu vec H + jvec E) ))$$
Finally,
$$nabla times nabla times vec E = omega^2 mu varepsilon vec E + omega^2 mu vec B + omega mu vec H + jomega vec E $$
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
$$nabla times nabla times vec E +pnabla times vec E +q vec E= 0 $$
How can I emit the magnetic fields and obtain a wave equation formed like shown below?

Michael M · Accepted Answer

I think the problem becomes easier upon a little re-organization.
The relevant Maxwell equations may are
begin{gather}
  nablatimesvec E=-jomegavec B, 
  nablatimesvec H = jomegavec D,
end{gather}
where we have assumed time harmonic fields and are using the $e^{jomega t}$ 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 - jalphavec B, 
  vec H = frac{1}{mu}vec B - frac{j}{mu}betavec 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}
  nablatimesvec B - frac{j}{beta}nablatimesvec E = jomegamuvarepsilonvec E + mu jalpha(-jomegavec B).
end{gather}
Multiplying by $-jomega$ then yields
begin{gather}
  nablatimes(-jomegavec B) - frac{omega}{beta}nablatimesvec E = omega^2muvarepsilonvec E + omegamualpha(-jomegavec B).
end{gather}
Substituting the first of the Maxwell equations above then yields an equation entirely in terms of the electric field:
begin{gather}
  nablatimesnablatimesvec E - frac{omega}{beta}nablatimesvec E = omega^2muvarepsilonvec E + omegamualphanablatimesvec E.
end{gather}
Finally, cleaning up a bit yields
begin{equation}
  nablatimesnablatimesvec E = omega^2muvarepsilonvec E + left(omegamualpha + frac{omega}{beta}right)nablatimesvec E.
end{equation}

Wave equation in a coupled chiral medium

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