Birefringence

Up until now, our treatment of electromagnetic wave propagation through transparent dielectrics has been restricted to isotropic media in which the refractive index is independent of either the direction of propagation or the polarization of the wave. However, there exists a certain class of optically anisotropic materials (e.g., crystals with non-cubic lattices, and plastics under mechanical stress) which are such that the refractive index varies with both the direction of propagation and the polarization. Such materials are said to be birefringent. Let us investigate the propagation of electromagnetic waves through birefringent media.

In a birefringent medium, the constitutive relation (7.53) generalizes to give

$${\bf D} = \epsilon_0\,\bm{\epsilon}\cdot {\bf E}.$$ (7.136)

Here, ${\bf E}$ is the electric field-strength, ${\bf D}= \epsilon_0\,{\bf E}+{\bf P}$ the electric displacement, ${\bf P}$ the electric dipole moment per unit volume, and $\bm{\epsilon}$ is termed the dielectric tensor. In a general Cartesian coordinate system, the dielectric tensor takes the form of a real 3 by 3 matrix. (However, the components of this matrix transform under rotation of the coordinate axes in an analogous manner to that in which the components of a vector transform.) Thus, in component form, the previous equation becomes

$$D_x = \epsilon_0\left(\epsilon_{xx}\,E_x+\epsilon_{xy}\,E_y+\epsilon_{xz}\,E_z\right),$$ (7.137)

$$D_y = \epsilon_0\left(\epsilon_{yx}\,E_x+\epsilon_{yy}\,E_y+\epsilon_{yz}\,E_z\right),$$ (7.138)

$$D_z = \epsilon_0\left(\epsilon_{zx}\,E_x+\epsilon_{zy}\,E_y+\epsilon_{zz}\,E_z\right).$$ (7.139)

Note that $\epsilon_{yx}=\epsilon_{xy}$, $\epsilon_{zx}=\epsilon_{xz}$, and $\epsilon_{zy}=\epsilon_{yz}$ (i.e., the dielectric tensor is symmetric) in a lossless birefringent medium (i.e., one that does not absorb wave energy). Furthermore, the ${\bf D}$ and ${\bf E}$ vectors are not necessarily parallel to one another in such a medium.

It is a well-known fact that it is always possible to find a particular orientation of the Cartesian coordinate axes that diagonalizes a symmetric tensor (Riley 1974). For the case of the dielectric tensor, these special axes are called the principal axes of the dielectric medium in question. When the Cartesian axes are aligned along the principal axes, the dielectric tensor takes the form

$$\bm{\epsilon} = \left( \begin{array}{ccc} \epsilon_{xx},&0,&0... ...psilon_{yy},&0\\ [0.5ex] 0,&0,&\epsilon_{zz}\end{array}\right).$$ (7.140)

Here, $\epsilon_{xx}$, $\epsilon_{yy}$, and $\epsilon_{zz}$ are known as the principal values of the dielectric tensor. If the principal values are all equal to one another then the medium is termed isotropic. If one of the principal values is different from the other two then the medium is termed monaxial. This nomenclature arises because the medium possesses a single optic axis (i.e., a direction of wave propagation in which the phase velocity is independent of the wave polarization). Finally, if all of the principal values are different from one another then the medium is termed biaxial (because it possesses two optic axes). (Obviously, an isotropic medium possesses an infinite number of optic axes corresponding to all the possible directions of wave propagation.)

The equations that govern electromagnetic wave propagation through dielectric media can be written (see Appendix C)

$$\frac{\partial D_x}{\partial t} = \frac{\partial H_z}{\partial y}-\frac{\partial H_y}{\partial z},$$ (7.141)

$$\frac{\partial D_y}{\partial t} = \frac{\partial H_x}{\partial z}-\frac{\partial H_z}{\partial x},$$ (7.142)

$$\frac{\partial D_z}{\partial t} = \frac{\partial H_y}{\partial x}-\frac{\partial H_x}{\partial y},$$ (7.143)

$$\frac{\partial B_x}{\partial t} = \frac{\partial E_y}{\partial z}-\frac{\partial E_z}{\partial y},$$ (7.144)

$$\frac{\partial B_y}{\partial t} = \frac{\partial E_z}{\partial x}-\frac{\partial E_x}{\partial z},$$ (7.145)

$$\frac{\partial B_z}{\partial t} = \frac{\partial E_x}{\partial y}-\frac{\partial E_y}{\partial x},$$ (7.146)

where ${\bf B}=\mu_0\,{\bf H}$. Here, ${\bf B}$ is the magnetic field-strength, and ${\bf H}$ the magnetic intensity. Taking $E_x$ times Equation (7.142) plus $E_y$ times Equation (7.143) plus $E_z$ times Equation (7.144) plus $H_x$ times Equation (7.145) plus $H_y$ times Equation (7.146) plus $H_z$ times Equation (7.147), and rearranging, we obtain

$$= E_x\,\frac{\partial D_x}{\partial t} + E_y\,\frac{\partial D_y}... ... t} + H_y\,\frac{\partial B_y}{\partial t}+H_z\,\frac{\partial B_z}{\partial t}$$ 0

$$\phantom{=}+\frac{\partial}{\partial x}\,(E_y\,H_z-E_z\,H_y)+\fra... ...rtial y}\,(E_z\,H_x-E_x\,H_z)+\frac{\partial}{\partial z}\,(E_x\,H_y-E_y\,H_z).$$ (7.147)

However, assuming that the coordinate axes are aligned with the principal axes,

$$E_x\,\frac{\partial D_x}{\partial t} + E_y\,\frac{\partial D_y}{\partial t}+E_z\,\frac{\partial D_z}{\partial t} =\epsilon_0\left(\epsilon_{xx}\,E_x\,\frac{\partial E_x}{\partial... ...l E_y}{\partial t} + \epsilon_{zz}\,E_z\,\frac{\partial E_z}{\partial t}\right)$$

$$=\frac{1}{2}\,\epsilon_0\,\frac{\partial}{\partial t}\left(\epsilon_{xx}\,E_x^{\,2} +\epsilon_{yy}\,E_y^{\,2}+\epsilon_{zz}\,E_z^{\,2}\right)$$

$$=\frac{1}{2}\,\frac{\partial}{\partial t}\left(E_x\,D_x+E_y\,D_y+E_z\,D_z\right).$$ (7.148)

Here, we are also assuming that $\epsilon_{xx}$, $\epsilon_{yy}$, and $\epsilon_{zz}$ are time independent. Likewise,

$$H_x\,\frac{\partial B_x}{\partial t} + H_y\,\frac{\partial B_y}{\... ...frac{1}{2}\,\frac{\partial}{\partial t}\left(B_x\,H_x+B_y\,H_y+B_z\,H_z\right).$$ (7.149)

Hence, Equation (7.148) reduces to

$$\frac{\partial{\cal E}}{\partial t}+\frac{\partial{\cal I}_x}{\pa... ...frac{\partial {\cal I}_y}{\partial y}+\frac{\partial {\cal I}_z}{\partial z}=0,$$ (7.150)

where

$${\cal E} = \frac{1}{2}\,{\bf E}\cdot{\bf D} + \frac{1}{2}\,{\bf B}\cdot{\bf H},$$ (7.151)

$$\bm{\mathcal{I} } = {\bf E}\times {\bf H}.$$ (7.152)

We can recognize Equation (7.151) as a three-dimensional generalization of the energy conservation equation (6.125). It follows that ${\cal E}$ is the electromagnetic energy density (i.e., the electromagnetic energy per unit volume), whereas $\bm{\mathcal{I}}$ is the electromagnetic energy flux (i.e., electromagnetic energy flows at the rate $\vert\bm{\mathcal{I}}\vert$ joules per unit area per unit time in the direction of the vector $\bm{\mathcal{I}}$).

Let us search for wave-like solutions of Equations (7.142)–(7.147) of the form

$${\bf E}(x,y,z,t) = \hat{\bf E}\,\cos\left[k\,(v\,t-{\bf n}\cdot{\bf r})\right],$$ (7.153)

$${\bf D}(x,y,z,t) = \hat{\bf D}\,\cos\left[k\,(v\,t-{\bf n}\cdot{\bf r})\right],$$ (7.154)

$${\bf H}(x,y,z,t) = \hat{\bf H}\,\cos\left[k\,(v\,t-{\bf n}\cdot{\bf r})\right],$$ (7.155)

with ${\bf B}=\mu_0\,{\bf H}$. Here, the wavevector is ${\bf k}=k\,{\bf n}$, and the phase velocity is ${\bf v}_p = v\,{\bf n}$, where ${\bf n}$ is a unit vector. Equations (7.142)–(7.144) yield

$$v\,\hat{\bf D} = \hat{{\bf H}}\times {\bf n},$$ (7.156)

whereas Equations (7.145)–(7.147) give

$$v\,\hat{\bf H}=\epsilon_0\,c^{\,2}\,{\bf n}\times \hat{\bf E}.$$ (7.157)

The previous two equations immediately imply that

$${\bf n}\cdot\hat{\bf D}= {\bf n}\cdot\hat{\bf H} =0,$$ (7.158)

and

$$\hat{\bf H}\cdot\hat{\bf D}=\hat{\bf H}\cdot\hat{\bf E} = 0.$$ (7.159)

In other words, the electric displacement and the magnetic intensity are both perpendicular to the wavevector. Furthermore, the magnetic intensity is perpendicular to both the electric displacement and the electric field-strength. Equations (7.157) and (7.158) can also be combined to give

$$v^{\,2}\,\hat{\bf D} = \epsilon_0\,c^{\,2}\left[\hat{\bf E} - ({\bf n}\cdot\hat{\bf E})\,{\bf n}\right],$$ (7.160)

where use has been made of a standard vector identity. The quantities $v_x=c/\sqrt{\epsilon_{xx}}$, $v_y=c/\sqrt{\epsilon_{yy}}$, and $v_z=c/\sqrt{\epsilon_{zz}}$, where $\epsilon_{xx}$, $\epsilon_{yy}$, $\epsilon_{zz}$ are the principal components of the dielectric tensor, are termed the principal velocities of the dielectric medium in question. Now, given that $\epsilon_0\,c^{\,2}\,\hat{E}_x = c^{\,2}\,\hat{D}_x/\epsilon_{xx}= v_x^{\,2}\,\hat{D}_x$, et cetera, when the Cartesian axes are aligned along the principal axes, the three Cartesian components of the previous equation can be written

$$\left(v^{\,2}-v_x^{\,2}\right)\hat{D}_x = -\epsilon_0\,c^{\,2}\,({\bf n}\cdot\hat{\bf E})\,n_x,$$ (7.161)

$$\left(v^{\,2}-v_y^{\,2}\right)\hat{D}_x = -\epsilon_0\,c^{\,2}\,({\bf n}\cdot\hat{\bf E})\,n_y,$$ (7.162)

$$\left(v^{\,2}-v_z^{\,2}\right)\hat{D}_x = -\epsilon_0\,c^{\,2}\,({\bf n}\cdot\hat{\bf E})\,n_z.$$ (7.163)

Finally, because [see Equation (7.159)]

$${\bf n} \cdot\hat{\bf D}= n_x\,\hat{D}_x+n_y\,\hat{D}_y+n_z\,\hat{D}_z=0,$$ (7.164)

we obtain

$$\left(\frac{n_x^{\,2}}{v^{\,2}-v_x^{\,2}}+ \frac{n_y^{\,2}}{v^{\,... ...}{v^{\,2}-v_z^{\,2}}\right) \epsilon_0\,c^{\,2}\,({\bf n}\cdot\hat{\bf E}) = 0.$$ (7.165)

Assuming that ${\bf n}\cdot\hat{\bf E}\neq 0$, the previous equation yields

$$\frac{n_x^{\,2}}{v^{\,2}-v_x^{\,2}}+ \frac{n_y^{\,2}}{v^{\,2}-v_y^{\,2}}+\frac{n_z^{\,2}}{v^{\,2}-v_z^{\,2}}=0,$$ (7.166)

which is known as the Fresnel equation.

The Fresnel equation is a quadratic equation for $v^{\,2}$ that specifies the phase speeds, $v$, of the two independent electromagnetic wave polarizations that can propagate through a birefringent medium in a particular direction, ${\bf n}$. In general, these two speeds are different. Let $v_a$ be the first speed, $\hat{\bf D}_a$ the associated electric displacement, and $\hat{\bf E}_a$ the associated electric field-strength. Likewise, let $v_b$ be the second speed, $\hat{\bf D}_b$ the associated electric displacement, and $\hat{\bf E}_b$ the associated electric field-strength It follows from Equations (7.162)–(7.164) that

$$\hat{D}_{a\,x} = -\frac{\epsilon_0\,c^{\,2}\,({\bf n}\cdot\hat{\bf E}_a)\,n_x}{v_a^{\,2}-v_x^{\,2}},$$ (7.167)

$$\hat{D}_{a\,y} = -\frac{\epsilon_0\,c^{\,2}\,({\bf n}\cdot\hat{\bf E}_a)\,n_y}{v_a^{\,2}-v_y^{\,2}},$$ (7.168)

$$\hat{D}_{a\,z} = -\frac{\epsilon_0\,c^{\,2}\,({\bf n}\cdot\hat{\bf E}_a)\,n_z}{v_a^{\,2}-v_z^{\,2}},$$ (7.169)

$$\hat{D}_{b\,x} = -\frac{\epsilon_0\,c^{\,2}\,({\bf n}\cdot\hat{\bf E}_b)\,n_x}{v_b^{\,2}-v_x^{\,2}},$$ (7.170)

$$\hat{D}_{b\,y} = -\frac{\epsilon_0\,c^{\,2}\,({\bf n}\cdot\hat{\bf E}_b)\,n_y}{v_b^{\,2}-v_y^{\,2}},$$ (7.171)

$$\hat{D}_{b\,z} = -\frac{\epsilon_0\,c^{\,2}\,({\bf n}\cdot\hat{\bf E}_b)\,n_z}{v_b^{\,2}-v_z^{\,2}}.$$ (7.172)

Hence,

$$\hat{\bf D}_a\cdot\hat{\bf D}_b = D_{a\,x}\,D_{b\,x}+D_{a\,y}\,D_{b\,y}+D_{a\,z}\,D_{b\,z}$$ (7.173)

$$= \left[\frac{n_x^{\,2}}{(v_a^{\,2}-v_x^{\,2})\,(v_b^{\,2}-v_x^{\... ...lon_0\,c^{\,2})^{\,2}\,({\bf n}\cdot\hat{\bf E}_a)\,({\bf n}\cdot\hat{\bf E}_b)$$

$$= \left[\frac{n_x^{\,2}}{v_a^{\,2}-v_x^{\,2}}+ \frac{n_y^{\,2}}{v... ...({\bf n}\cdot\hat{\bf E}_a)\,({\bf n}\cdot\hat{\bf E}_b)}{v_b^{\,2}-v_a^{\,2}}.$$

However, according to the Fresnel equation, (7.167),

$$\frac{n_x^{\,2}}{v_a^{\,2}-v_x^{\,2}}+ \frac{n_y^{\,2}}{v_a^{\,2}... ... \frac{n_y^{\,2}}{v_b^{\,2}-v_y^{\,2}}+\frac{n_z^{\,2}}{v_b^{\,2}-v_z^{\,2}}=0.$$ (7.174)

Hence, we deduce that

$$\hat{\bf D}_a\cdot\hat{\bf D}_b=0.$$ (7.175)

In other words, the two independent wave polarizations have mutually orthogonal electric displacements.

As an example, consider the propagation of an electromagnetic wave through a monaxial material whose principal velocities are $v_x=v_y=v_\perp$ and $v_z=v_\parallel$, where $v_\perp\neq v_\parallel$. The corresponding principal components of the dielectric tensor are $\epsilon_{xx}=\epsilon_{yy}=\epsilon_\perp$ and $\epsilon_{zz}=\epsilon_{\parallel}$. Of course, $v_\perp=c/\sqrt{\epsilon_\perp}$ and $v_\parallel=c/\sqrt{\epsilon_\parallel}.$ In this case, the optic axis corresponds to the $z$-axis. It is convenient to specify the direction of wave propagation in terms of standard spherical angles,

$${\bf n} = (\sin\theta\,\cos\phi,\,\sin\theta\,\sin\phi,\,\cos\theta).$$ (7.176)

In particular, $\theta$ is the angle subtended between the direction of wave propagation and the optic axis.

The two independent electromagnetic wave polarizations that can propagate through a monaxial material are termed the ordinary wave and the extraordinary wave. The ordinary wave is such that ${\bf n}\cdot\hat{\bf E}=0$, which is one way of satisfying Equation (7.166). Assuming that the Cartesian axes correspond to the principal axes, it is easily demonstrated that

$$\hat{\bf D} = \hat{D}\,(\sin\phi,\,-\cos\phi,\,0),$$ (7.177)

$$\hat{\bf E} = \frac{\hat{D}}{\epsilon_0\,\epsilon_\perp}(\sin\phi,\,-\cos\phi,\,0)=\frac{\hat{\bf D}}{\epsilon_0\, \epsilon_\perp},$$ (7.178)

$$\hat{\bf H} = v_\perp\,\hat{D}\,(\cos\theta\,\cos\phi,\,\cos\theta\,\sin\phi,\,-\sin\theta),$$ (7.179)

$$\langle\bm{\mathcal{I}}\rangle = \frac{v_\perp\,\hat{D}^{\,2}}{2\,\epsilon_0\,\epsilon_\perp} (\... ...theta) = \frac{v_\perp\,\hat{D}^{\,2}}{2\,\epsilon_0\,\epsilon_\perp}\,{\bf n}.$$ (7.180)

Here, $\langle\cdots\rangle$ denotes an average over a wave period. Substitution of Equations (7.178) and (7.179) into Equation (7.161) reveals that

$$v=v_\perp.$$ (7.181)

In other words, the ordinary wave propagates at the fixed phase speed $v_\perp$, irrespective of its direction of propagation. Furthermore, the electric field-strength is parallel to the electric displacement, and the electromagnetic energy flux is parallel to the wavevector.

The phase speed of the extraordinary wave is obtained directly from the Fresnel equation, (7.167), which yields

$$\frac{\sin^2\theta\,\cos^2\phi}{v^{\,2}-v_\perp^{\,2}}+\frac{\sin... ...\phi}{v^{\,2}-v_\perp^{\,2}} +\frac{\cos^2\theta}{v^{\,2}-v_\parallel^{\,2}}=0,$$ (7.182)

or

$$v^{\,2} = \cos^2\theta\,\,v_\perp^{\,2} + \sin^2\theta\,\,v_\parallel^{\,2}.$$ (7.183)

It follows that the phase speed of the extraordinary wave varies with its direction of propagation. The phase speed matches that of the ordinary wave when the wave propagates along the optic axis (i.e., when $\theta =0$); otherwise, it is different. It is easily demonstrated that

$$\hat{\bf D} = \hat{D}\,(\cos\theta\,\cos\phi,\,\cos\theta\,\sin\phi,\,-\sin\theta),$$ (7.184)

$$\hat{\bf E} = \frac{\hat{D}}{\epsilon_0\,\epsilon}\left(\frac{\epsilon}{\epsi... ...\cos\theta\,\sin\phi,\,-\frac{\epsilon}{\epsilon_\parallel}\,\sin\theta\right),$$ (7.185)

$$\hat{\bf H} = v\,\hat{D}\,(-\sin\phi,\,\cos\phi,\,0),$$ (7.186)

$$\langle\bm{\mathcal{I}}\rangle = \frac{v\,\hat{D}^{\,2}}{2\,\epsilon_0\,\epsilon} \left(\frac{\e... ...el}\,\sin\theta\,\sin\phi,\,\frac{\epsilon}{\epsilon_\perp}\,\cos\theta\right).$$ (7.187)

Here,

$$\frac{1}{\epsilon^{\,2}} = \frac{\cos^2\theta}{\epsilon_\perp^{\,2}}+ \frac{\sin^2\theta}{\epsilon_\parallel^{\,2}}.$$ (7.188)

It can be seen that the $\hat{\bf D}$ and $\hat{\bf E}$ vectors are not parallel to one another. Moreover, the electromagnetic energy flux is not parallel to the wavevector. If $\alpha$ is the angle subtended between the directions of the $\hat{\bf D}$ and $\hat{\bf E}$ vectors then

$$\tan\alpha = \frac{\vert\hat{\bf D}\times \hat{\bf E}\vert}{\hat{... ...silon_\perp\vert}{\cos^2\theta/\epsilon_\perp+\sin^2\theta/\epsilon_\parallel}.$$ (7.189)

Likewise, if $\beta$ is the angle subtended between the directions of the electromagnetic energy flux and the wavevector then

$$\tan\beta = \frac{\vert\langle\bm{\mathcal{I}}\rangle\times{\bf n... ...silon_\perp\vert}{\cos^2\theta/\epsilon_\perp+\sin^2\theta/\epsilon_\parallel}.$$ (7.190)

It follows that $\alpha=\beta$. Moreover, these two angles are only zero when the extraordinary wave propagates parallel (i.e., $\theta=0, \pi$) or perpendicular (i.e., $\theta=\pi/2$) to the optic axis.

Figure 7.13: Birefringence.

Figure 7.13 illustrates what happens when unpolarized light is normally incident on a slab of birefringent material in such a manner that the incident light is neither parallel nor perpendicular to the optic axis. Because of the normal incidence, the wavevectors of both the ordinary and the extraordinary rays are not refracted, and remain parallel to the direction of incidence. However, the ray path is actually coincident with the associated electromagnetic energy flux. For the ordinary ray, the electromagnetic energy flux is parallel to the wavevector. However, for the extraordinary ray, the electromagnetic energy flux subtends a finite angle with the wavevector. Consequently, although the ordinary ray passes through the slab without changing direction, the direction of the extraordinary ray suffers a sideways deviation. This type of double refraction is known as birefringence.