Faraday rotation

The electromagnetic force acting on an electron is given by

$${\bfm f} = -e\,({\bfm E} + {\bfm v}\wedge{\bfm B}).$$ (668)

If the ${\bfm E}$ and ${\bfm B}$ fields in question are due to an electromagnetic wave propagating through a dielectric medium then

$$\vert B\vert = \frac{n}{c}\, \vert E\vert.$$ (669)

It follows that the ratio of the magnetic to the electric forces acting on the electron is $n\, v/c$. In other words, the magnetic force is completely negligible unless the wave amplitude is sufficiently high that the electron moves relativistically in response to the wave. This state of affairs is rare, but can occur when intense laser beams are made to propagate through plasmas.

Suppose, however, that the dielectric medium contains an externally generated magnetic field ${\bfm B}$. This can easily be made much stronger than the optical magnetic field. In this case, it is possible for a magnetic field to affect the propagation of low amplitude electromagnetic waves. The electron equation of motion (4.11) generalizes to

$$m\,\ddot{\bfm s} +f\,{\bfm s} = - e({\bfm E} + \dot{\bfm s}\wedge{\bfm B}),$$ (670)

where any damping of the motion has been neglected. Suppose that the direction of ${\bfm B}$ is in the positive $z$-direction, and that the wave propagates in the same direction. With these assumptions the ${\bfm E}$ and ${\bfm s}$ vectors lie in the $x$-$y$ plane. The above equation reduces to

$$(\omega_0^{~2} -\omega^2)\,s_x -{\rm i}\, \omega{\mit\Omega}\, s_y = -\frac{e}{m}\, E_x,$$ (671)

$$(\omega_0^{~2} -\omega^2)\,s_y +{\rm i}\, \omega{\mit\Omega} \,s_x = -\frac{e}{m}\, E_y,$$ (672)

provided that all perturbed quantities have an $\exp(-{\rm i}\,\omega t)$ time dependence. Here,

$${\mit\Omega} = \frac{e B}{m}$$ (673)

is the electron cyclotron frequency. Let

$$E_\pm = E_x \pm {\rm i}\,E_y,$$ (674)

and

$$s_\pm = s_x \pm {\rm i} \,s_y.$$ (675)

Note that

$$E_x = \frac{1}{2} \,(E_+ + E_-),$$ (676)

$$E_y = \frac{1}{2\,{\rm i}}\, (E_+-E_-).$$ (677)

Equations (4.43) reduce to

$$(\omega_0^{~2} -\omega^2 -\omega\,{\mit \Omega})\,s_+ = -\frac{e}{m}\,E_+,$$ (678)

$$(\omega_0^{~2} -\omega^2 + \omega\,{\mit \Omega})\,s_- = -\frac{e}{m}\,E_-.$$ (679)

Defining $P_\pm = P_x\pm{\rm i}\,P_y$, it follows from Eq. (4.10) that

$$P_\pm = \frac{(N e^2/m)\,E_\pm}{\omega_0^{~2} -\omega^2 \mp \omega\, {\mit\Omega}}.$$ (680)

Finally, from Eq. (4.15), we can write

$$\epsilon_\pm\equiv n^{~2}_\pm= 1 + \frac{P_\pm}{\epsilon_0\,E_\pm},$$ (681)

giving

$$n^{~2}_\pm(\omega) = 1 + \frac{(Ne^2/\epsilon_0 m)} {\omega_0^{~2} -\omega^2 \mp \omega\, {\mit\Omega}}.$$ (682)

According to the dispersion relation (4.51), the refractive index of a magnetized dielectric medium can take one of two possible values, which presumably correspond to two different types of wave propagating along the $z$-axis. The first wave has the refractive index $n_+$ and an associated electric field [see Eqs. (4.45)]

$$E_x = E_0 \cos[(\omega/c)(n_+ z -c t)],$$ (683)

$$E_y = E_0 \sin[(\omega/c)(n_+ z-ct)].$$ (684)

This corresponds to a left-handed circularly polarized wave propagating in the $z$-direction with the phase velocity $c/n_+$. The second wave has the refractive index $n_-$ and an associated electric field

$$E_x = E_0 \cos[(\omega/c)(n_- z -c t)],$$ (685)

$$E_y = - E_0 \sin[(\omega/c)(n_- z-ct)].$$ (686)

This corresponds to a right-handed circularly polarized wave propagating in the $z$-direction with the phase velocity $c/n_-$. It is clear from Eq. (4.51) that $n_+ > n_-$. Thus, we conclude that in the presence of a $z$-directed magnetic field, a $z$-directed left-handed circularly polarized wave propagates with a phase velocity which is slightly less than that of the corresponding right-handed wave. It should be remarked that the refractive index is always real (in the absence of damping), so the magnetic field gives rise to no net absorption of electromagnetic radiation. This is not surprising since the magnetic field does no work on charged particles, and can therefore transfer no energy to the particles from any waves propagating through the medium.

We have seen that right-handed and left-handed circularly polarized waves propagate with different phase velocities through a magnetized dielectric medium. But, what does this imply for the propagation of a plane polarized wave? Let us superimpose the left-handed wave whose electric field is given by Eqs. (4.52) on the right-handed wave whose electric field is given by Eqs. (4.53). In the absence of a magnetic field $n_+=n_-=n$, and we obtain

$$E_x = 2 E_0 \cos[(\omega/c)(n z -c t)],$$ (687)

$$E_y = 0.$$ (688)

This, of course, is the field of a plane polarized wave (polarized along the $x$-direction) propagating along the $z$-axis with the phase velocity $c/n$. In the presence of a magnetic field we obtain

$$E_x = 2E_0 \cos[(\omega/c)(nz -ct)]\cos[(\omega/2c)(n_+-n_-)z],$$ (689)

$$E_y = 2E_0 \cos[(\omega/c)(nz-ct)]\sin [(\omega/2c)(n_+-n_-) z],$$ (690)

where

$$n = \frac{1}{2}\, (n_+ + n_-)$$ (691)

is the mean index of refraction. Equations (4.55) can be recognized as the field of a plane polarized wave whose angle of polarization with respect to the $x$-axis,

$$\chi = \tan^{-1}(E_y/E_x),$$ (692)

rotates as the wave propagates along the $z$-axis with the phase velocity $c/n$. In fact, the angle of polarization is given by

$$\chi = \frac{\omega}{2c}\, (n_+ - n_-)\, z,$$ (693)

which clearly increases linearly with the distance traveled by the wave along the direction of the magnetic field. This rotation of the plane of polarization of a linearly polarized wave propagating through a magnetized dielectric medium is known as Faraday rotation (since it was discovered by Michael Faraday in 1845).

Assuming that the cyclotron frequency $\mit \Omega$ is relatively small compared to the wave frequency $\omega$, and also that $\omega$ does not lie close to the resonant frequency $\omega_0$, it is easily demonstrated that

$$n \simeq 1 + \frac{(Ne^2/\epsilon_0 m)}{\omega_0^{~2}-\omega^2},$$ (694)

and

$$n_+-n_- \simeq \frac{Ne^2}{\epsilon_0 m\, n} \frac{\omega {\mit \Omega}}{(\omega_0^{~2} -\omega^2)^2}.$$ (695)

It follows that the rate at which the plane of polarization of an electromagnetic wave rotates with the distance traveled by the wave is given by

$$\frac{d\chi}{dl} = \frac{\kappa(\omega)\,N B_\parallel} {n(\omega)},$$ (696)

where $B_\parallel$ is the component of the magnetic field along the direction of propagation of the wave, and

$$\kappa(\omega) = \frac{e^3}{2\epsilon_0 m^2 c} \frac{\omega^2} {(\omega_0^2 -\omega^2)^2}.$$ (697)

If the medium in question is a tenuous plasma then $n\simeq 1$ and $\omega_0=0$. Thus,

$$\frac{d\chi}{dl} \simeq \frac{e^3}{2\epsilon_0 m^2 c} \frac{N B_\parallel} {\omega^2}$$ (698)

Clearly, the rate at which the plane of polarization rotates is proportional to the product of the electron number density and the parallel magnetic field strength. Moreover, the plane of rotation rotates faster for low frequency waves than for high frequency waves. The total angle by which the plane of polarization is twisted after passing through a magnetized plasma is given by

$$\Delta\chi \simeq \frac{e^3}{2\epsilon_0 m^2 c\,\omega^2} \int N(l) B_\parallel(l) \,dl,$$ (699)

provided that $N$ and $B_\parallel$ vary on length-scales which are large compared to the wavelength of the radiation. This formula is regularly employed in radio astronomy to infer the magnetic field-strength in interstellar space.