Plane Electromagnetic Waves in Anisotropic Non-conducting Medium

An anisotropic medium is one in which electromagnetic field properties depend on direction. Let us consider a non-magnetic homogeneous isotropic medium in which

$$ J = 0,\: \rho = 0\: \text {and}\: \mu = \mu_0 \: \dots (1) $$

Permittivity is tensor not scalar so, that the components of electric displacement D are in general related to components of E by the equations

\begin{align*}\left.\begin{matrix} D_x &= \varepsilon_{xx}E_x+ \varepsilon_{xy}E_y + \varepsilon_{xz}E_z \\ D_y &= \varepsilon_{yx}E_x+ \varepsilon_{yy}E_y + \varepsilon_{yz}E_z \\ D_z &= \varepsilon_{zx}E_x+ \varepsilon_{zy}E_y + \varepsilon_{zz}E_z \\ \end{matrix}\right )\dots (2) \end{align*}

where the coefficients are scalar constants for a homogeneous medium. If we choose the co-ordinate axes as the principle axes of Freshnel ellipsoid for the medium, then D and E are related rather simple equations

\begin{align*}\left.\begin{matrix} D_x &= \varepsilon_{x}E_x = K_x\varepsilon_0 E_x \\ D_y &= \varepsilon_{y}E_y= K_y\varepsilon_0 E_y \\ D_z &= \varepsilon_{z}E_z = K_z\varepsilon_0 E_z\end{matrix}\right )\dots (3) \end{align*}

So Maxwell’s equations in an anisotropic dielectric medium take the form

\begin{align*}\left.\begin{matrix} \text {Div}\: D &= 0\:\dots (a)\\ \text {Div}\: H & = 0\:\dots (b)\\ \text {Curl}\: E &= -\mu_0 \frac {\partial H}{\partial t}\:\dots (c)\\ \text {Curl}\: E &=\frac {\partial D}{\partial t}\:\dots (d)\end{matrix}\right )\dots (4) \end{align*}

So, Maxwell’s equations in an isotropic dielectric medium take form shown in equation (4). D and E do not possess the same direction so; div E is not the symmetrical as of equation (4a).

Now consider a plane wave advancing with angular frequency \(\omega \) and phase velocity \(\nu\) along the direction of propagation vector K. Then

\begin{align*} E = E_0\: e^{ik.r – \omega t} \dots (5)\end{align*}

where r is the radius vector from origin and

\begin{align*}k &= kn = \frac {2\pi}{\lambda}n =\frac{\omega}{\nu} n \\\text{Here n is unit vector along k.} \\ \text {from equation} (4c)\\ \frac {\partial H}{\partial t} &= \frac{1}{\mu_0}\: \text {curl}\: E \\ &= \frac{1}{\mu_0}\: \text {curl}\: (E_0\: e^{ik.r – \omega t})\dots (6)\end{align*}

\begin{align*}\text {Using vector identity}\\ \text {curl}\:(\phi A) = \phi \: \text {curl} A - A \times \text {grad}\: \phi\dots (7)\end{align*}

Here \(\text {Curl}\: E_0 = 0\) as \(E_0 \) is spatially constant. So, equation (6) gives

\begin{align*}\frac {\partial H}{\partial t} &= \frac {E_0}{\mu_0}\times \text {grad}(E_0\:e^{ik.r - i\omega t})\\ \text {grad}(E_0\:e^{ik.r - i\omega t})&= ike^{ik.r - i\omega t} \\ \frac {\partial H}{\partial t} &= \frac {E_0}{\mu_0}\times (ike^{ik.r - i\omega t})=\frac {ie^{ik.r - i\omega t}}{\mu_0}E_0 \times K. \end{align*}

\begin{align*}\text {Integrating,}\\ H&=-\frac {i}{\mu_0}\frac {e^{ik.r - i\omega t}}{i\omega}E_0 \times k \\ &=-\frac {1}{\mu_0\omega}e^{ik.r - i\omega t}E_0 \times k\\ &=-\frac {1}{\mu_0\omega}k\times E\: [\text {using equation}\: (5)]\\ H &=-\frac {1}{\mu_0\omega}E\times k\end{align*}

This shows that H is normal to the plane of E and k. Now from equation (4d)

\begin{align*}\frac {\partial D}{\partial t} &= \text {curl}\: H \\&=\text {curl}\: (H_0e^{(ik.r -i \omega t)})\:\:\:[\because H =H_0e^{(ik.r -i \omega t)}] \end{align*}

Using vector identity in equation (7) and nothing that \(\text {curl}\: H = 0\), we obtain

\begin{align*}\frac {\partial D}{\partial t} &= -H_0 \times \text {grad} (e^{(ik.r -i \omega t)})=-H_0\times ike^{(ik.r -i \omega t)}\\ D &= - H_0 \times ik\frac {e^{(ik.r -i \omega t)}}{(-i\omega)}\\ &= \frac {1}{\omega} H \times k \:\dots (10) \end{align*}

This equation shows that D is normal to k and H and both H and D are normal to the direction if propagation vector k. Therefore the electromagnetic wave in anisotropic non conducting medium is transverse with respect to H and D.

Now substituting values of H from (9) to(10), we get

\begin{align*}D &= \frac {1}{\omega}\left (\frac{1}{\mu_0 \omega}k\times E \right )\times k \\ &= \frac {1}{\mu_0\omega ^2}k \times (k\times E) = \frac {1}{\mu_0\omega ^2}[(k.k)E - (E.k)E]\\ &= - \frac {1}{\mu_0\omega ^2}[k^2E - (k.E)k]\:\dots (11)\end{align*}

This equation shows D, E and k all lie in same plane.

Now pointing vectors \(S = E \times H\) is normal to the plane containing E and H. This implies that vector D,E, k and S are coplanar. Also since k is normal to D; and S is normal to E; therefore the angle between S and k is equal to the angle between E and D. In other word S is in the direction of k. As a result an anisotropic medium energy is not propagated in general, in the direction of wave propagation: since poynting vectors k represents the direction of energy flow.

Fresnel’s law for phase velocity \(\nu\).

Equation (11) can be written as

\begin{align*}D &= \frac {K^2}{\mu_0\omega ^2}[(E - n.E)n]\: [\text {since}\: K = kn] \\ &= \frac {1}{\mu_0\nu ^2}[E - (n.E)E]\: [\text {since}\: k = \frac {\omega}{\nu}] \end{align*}

If \(\cos \alpha,\: \cos \beta \: \text {and}\: \cos \gamma\) are direction cosines of unit wave vector n; then

$$ n = \hat i \:\cos \alpha + \hat j \: \cos \beta + \hat k \: \cos \gamma \dots (12)$$

Now the components of D are

\begin{align*}\left.\begin{matrix}D_x &= \frac {1}{\mu_0\nu ^2}[E_x - (n.E)\cos \alpha]\\D_y &= \frac {1}{\mu_0\nu ^2}[E_y - (n.E)\cos \beta] \\ D_z &= \frac {1}{\mu_0\nu ^2}[E_z - (n.E)\cos \gamma]\end{matrix}\right )\dots (12) \end{align*}

Now from equation (3)

\begin{align*}E_x = \frac {D_x}{K_x\varepsilon_0},\:E_y = \frac {D_y}{K_y\varepsilon_0}\: \text {and}\: E_z &= \frac {D_z}{K_z\varepsilon_0}\end{align*}

From definition of refractive index

\begin{align*}n_x &= \sqrt {K_x} = \frac {c}{\nu_x},\:n_y = \sqrt {K_y} = \frac {c}{\nu_y}\:\text {and}\:n_z = \sqrt {K_z} = \frac {c}{\nu_z}\\ c &= \frac {1}{\sqrt {(\mu_0\varepsilon_0)}}\end{align*}

Using these equations, we get

\begin{align*}E_x = \mu_0v_x^2Dx\:;E_y = \mu_0v_y^2Dy\: \text {and}\:E_z = \mu_0v_z^2Dz\end{align*}

So that equation (12), (13) and (14) becomes

begin{align*}\left.\begin{matrix} D_x &= \frac {1}{\mu_0}\frac {(n.E)\cos\alpha}{v_x^2 - v^2}\\ D_y &= \frac {1}{\mu_0}\frac {(n.E)\cos\beta}{v_y^2 - v^2}\\ D_z &= \frac {1}{\mu_0}\frac {(n.E)\cos\gamma}{v_z^2 - v^2}\end{matrix}\right )\dots (13) \end{align*}

Since vector D is normal to n, we get

\begin{align*}D.n = D_x\cos \alpha + D_y\cos \beta + D_z\cos \gamma = 0 \\ \text {using equation}\: (13), \text {we obtain}\\ \frac {\cos ^2\alpha}{v_x^2 - v^2}+ \frac {\cos^2 \beta}{v_y^2 - v^2}+ \frac {\cos ^2\gamma}{v_z^2 -v^2} = 0\:\dots (14)\end{align*}

This is well known Freshnel’s law for phase velocity. This equation indicates that the phase velocity v, in general, can have two values in any given direction. However for most anisotropic media, it is found that there are two directions for which equation (14) has only one solution i.e. it gives only one velocity phase. These directions are called optic axes and the medium is said to be biaxial.

Bibliography

P.B. Adhikari, Bhoj Raj Gautam, Lekha Nath Adhikari. A Textbook of Physics. kathmandu: Sukunda Pustak Bhawan, 2011.

Jha, V. K.; 'Lecture title'; Maxwell's Equation; St. Xavier's College, Kathmandu; 2016.