Faraday Rotation

Consider a high-frequency, circularly polarized, electromagnetic wave propagating, along the $z$-axis, through a plasma with a longitudinal equilibrium magnetic field of strength, ${\bf B} = B_0\,{\bf e}_z$. The equations of motion of an individual electron making up the plasma take the form (see Appendix C)

$$m_e\,\frac{d^{\,2}x}{dt^{\,2}} = -e\,E_x-e\,B_0\,\frac{dy}{dt},$$ (9.54)

$$m_e\,\frac{d^{\,2} y}{dt^{\,2}} = -e\,E_y+e\,B_0\,\frac{dx}{dt},$$ (9.55)

where $m_e$ is the electron mass, and $-e$ the electron charge. Here, $x$ and $y$ are the wave-induced displacements of the electron in the $x$- and $y$-directions, respectively. (As before, it is a good approximation to neglect the wave-induced displacements of the ions, because of their relatively large mass.) The former terms on the right-hand sides of the previous equations represent the $x$- and $y$-directed forces exerted on the electron by the wave electric field, ${\bf E}$, whereas the latter terms represent the forces exerted by the equilibrium magnetic field when the electron moves (Fitzpatrick 2008). (As before, we can neglect the forces due to the wave magnetic field, provided that the electron motion remains non-relativistic.)

Consider a right-hand circularly polarized wave, of angular frequency $\omega$, whose electric field takes the form

$$E_x(z,t) = E_R\,\cos(\omega\,t-k_R\,z),$$ (9.56)

$$E_y(z,t) =E_R\,\sin(\omega\,t-k_R\,z).$$ (9.57)

(See Section 7.7.) Let us search for solutions of Equations (9.54) and (9.55) of the form

$$x = x_0\,\cos(\omega\,t-k_R\,z),$$ (9.58)

$$y = y_0\,\sin(\omega\,t-k_R\,z).$$ (9.59)

It is readily demonstrated (see Exercise 6) that

$$x_0 = \frac{e\,E_R}{m_e\,\omega\,(\omega-{\mit\Omega}_e)},$$ (9.60)

$$y_0 = \frac{e\,E_R}{m_e\,\omega\,(\omega-{\mit\Omega}_e)},$$ (9.61)

where

$${\mit\Omega}_e = \frac{e\,B_0}{m_e}$$ (9.62)

is the so-called electron cyclotron frequency at which the electrons gyrate in the equilibrium magnetic field (Stix 1962). If $n_e$ is the number density of electrons in the plasma then it follows that the $x$- and $y$-components of the electric dipole moment per unit volume are

$$P_x(z,t) =-e\,n_e\,x= -\frac{n_e\,e^{\,2}\,E_R}{m_e\,\omega\,(\omega-{\mit\Omega}_e)}\,\cos(\omega\,t-k_R\,z),$$ (9.63)

$$P_y(z,t) =-e\,n_e\,y= -\frac{n_e\,e^{\,2}\,E_R}{m_e\,\omega\,(\omega-{\mit\Omega}_e)}\,\sin(\omega\,t-k_R\,z),$$ (9.64)

respectively. For the case of a circularly polarized wave, Equations (9.24) and (9.25) generalize to give (see Appendix C)

$$\frac{\partial E_x}{\partial t} = - \frac{1}{\epsilon_0}\left(\frac{\partial P_x}{\partial t}+\frac{\partial H_y}{\partial z}\right),$$ (9.65)

$$\frac{\partial E_y}{\partial t} = -\frac{1}{\epsilon_0}\left(\frac{\partial P_y}{\partial t} - \frac{\partial H_x}{\partial z}\right),$$ (9.66)

$$\frac{\partial H_x}{\partial t} =\frac{1}{\mu_0}\,\frac{\partial E_y}{\partial z},$$ (9.67)

$$\frac{\partial H_y}{\partial t} = -\frac{1}{\mu_0}\,\frac{\partial E_x}{\partial z}.$$ (9.68)

Thus, writing $E_x$ and $E_y$ in the form (9.56) and (9.57), respectively, the components of the magnetic intensity in the form

$$H_x(z,t) = -Z_R^{\,-1}\,E_R\,\sin(\omega\,t-k_R\,z),$$ (9.69)

$$H_y(z,t) = Z_R^{\,-1}\,E_R\,\cos(\omega\,t-k_R\,z),$$ (9.70)

and $P_x$ and $P_y$ in the form (9.63) and (9.64), respectively, Equations (9.65)–(9.68) yield the nonlinear dispersion relation

$$\omega^{\,2} = k_R^{\,2}\,c^{\,2} + \frac{\omega\,\omega_{p\,e}^{\,2}}{\omega-{\mit\Omega}_e},$$ (9.71)

where $\omega_{p\,e}$ is the (electron) plasma frequency. [See Equation (9.28).] It follows that the refractive index of the plasma for right-hand circularly polarized waves is

$$n_R = \frac{k_R\,c}{\omega} = \left[1-\frac{\omega_{p\,e}^{\,2}}{\omega\,(\omega-{\mit\Omega}_e)}\right]^{1/2},$$ (9.72)

whereas the effective impedance becomes

$$Z_R = \frac{Z_0}{n_R}.$$ (9.73)

Here, $Z_0$ is the impedance of free space.

Consider a left-hand circularly polarized wave whose electric field takes the form

$$E_x(z,t) = E_L\,\cos(\omega\,t-k_L\,z),$$ (9.74)

$$E_y(z,t) =-E_L\,\sin(\omega\,t-k_L\,z).$$ (9.75)

(See Section 7.7.) Repeating the previous analysis (with suitable modifications), we deduce that the dispersion relation for left-hand circularly polarized waves is

$$\omega^{\,2} = k_L^{\,2}\,c^{\,2} + \frac{\omega\,\omega_{p\,e}^{\,2}}{\omega+{\mit\Omega}_e}.$$ (9.76)

It follows that the refractive index of the plasma for such waves is

$$n_L = \frac{k_L\,c}{\omega} = \left[1-\frac{\omega_{p\,e}^{\,2}}{\omega\,(\omega+{\mit\Omega}_e)}\right]^{1/2},$$ (9.77)

whereas the effective impedance becomes

$$Z_L = \frac{Z_0}{n_L}.$$ (9.78)

According to the previous analysis, in the presence of a longitudinal equilibrium magnetic field, the refractive indices of right-hand and left-hand circularly polarized electromagnetic waves propagating through a plasma are slightly different. Consider what happens when a linearly polarized electromagnetic wave, whose electric field is initially of the form

$$E_x(z,t) = E_0\,\cos(\omega\,t-k_0\,z),$$ (9.79)

$$E_y(z,t) =0,$$ (9.80)

propagates through the plasma. We can represent a linearly polarized wave as a superposition of right- and left-hand circularly polarized waves of equal amplitudes. In other words,

$$E_x(z,t) = E_R\,\cos(\omega\,t-k_R\,z) + E_L\,\cos(\omega\,t-k_L\,z),$$ (9.81)

$$E_y(z,t) = E_R\,\sin(\omega\,t-k_R\,z) - E_L\,\sin(\omega\,t-k_L\,z),$$ (9.82)

where $E_R=E_L=E_0/2$. Furthermore, in the high-frequency limit $\omega\gg \omega_{p\,e}$, ${\mit\Omega}_e$, Equations (9.72) and (9.77) yield

$$k_R \simeq k_0-{\mit\Delta} k,$$ (9.83)

$$k_L \simeq k_0+{\mit\Delta} k,$$ (9.84)

where $k_0 = (\omega/c)[1-(1/2)\,(\omega_{p\,e}/\omega)^2]$, and

$${\mit\Delta} k = \frac{1}{2}\left(\frac{\omega_{p\,e}}{\omega}\right)^2\left(\frac{{\mit\Omega}_e}{c}\right).$$ (9.85)

Equations (9.81)–(9.84), in combination with some standard trigonometric identities (see Appendix B), give

$$E_x(z,r) = E_0\,\cos(\omega\,t-k_0\,z)\,\cos({\mit\Delta} k\,z),$$ (9.86)

$$E_y(z,r) = E_0\,\cos(\omega\,t-k_0\,z)\,\sin({\mit\Delta} k\,z).$$ (9.87)

It can be seen that the presence of the longitudinal magnetic field (whose strength is parameterized by ${\mit\Delta} k$) causes the plane of polarization of the wave to rotate as it propagates through the plasma. This effect is known as Faraday rotation. Defining the angle of polarization,

$$\varphi = \tan^{-1}\left(\frac{E_y}{E_x}\right) = {\mit\Delta} k\,\,z,$$ (9.88)

the rate at which this angle advances as the wave propagates is

$$\frac{d\varphi}{dz} = {\mit\Delta} k = \frac{\omega_{p\,e}^{\,2}\... ...\,2}\,c} = \frac{e^{\,3}}{2\,\epsilon_0\,m_e^{\,2}\,c\,\omega^{\,2}}\,n_e\,B_0.$$ (9.89)

Thus, a linearly polarized electromagnetic wave that propagates through a plasma with a slowly varying electron number density, $n_e(z)$, and longitudinal magnetic field, $B_0(z)$, has its plane of rotation rotated through a net angle

$${\mit\Delta}\varphi=\phi(z)-\phi(0)=\frac{e^{\,3}}{2\,\epsilon_0\,m_e^{\,2}\,c\,\omega^{\,2}}\int_0^z n_e(z')\,B_0(z')\,dz'.$$ (9.90)

Observe the very strong inverse depedence of ${\mit\Delta}\varphi$ with the wave frequency, $\omega$.

Pulsars are rapidly rotating neutron stars that emit regular blips of highly polarized radio waves (Longair 2011). Hundreds of such objects have been found in our galaxy since the first was discovered in 1967. By measuring the variation of the angle of polarization, $\varphi$, of radio emission from a pulsar with frequency, $\omega$, astronomers can effectively determine the line integral of $n_e\,B_0$ along the straight line joining the pulsar to the Earth using formula (9.90) (ibid.). Here, $n_e$ is the number density of free electrons in the interstellar medium, whereas $B_0$ is the parallel (to the line joining the pulsar to the Earth) component of the galactic magnetic field. In order to perform this calculation, astronomers must make the reasonable assumption that the radiation was emitted by the pulsar with a common angle of polarization, $\varphi_0$, over a wide range of different frequencies. By fitting Equation (9.90) to the data, and then extrapolating to large $\omega$, it is possible to determine $\varphi_0$, and, hence, the amount, ${{\mit\Delta}}\varphi(\omega)$, through which the polarization angle of the radiation has rotated, at a given frequency, during its passage to Earth.

Astronomers can also determine the line integral of $n_e$ by looking at the variation of the arrival time of the various components of a pulsar radio blip with frequency (Longair 2011). This calculation depends on the reasonable assumption that the components were emitted simultaneously, and then traveled through interstellar space at the frequency dependent group velocity $v_g = (1-\omega_{p\,e}^{\,2}/\omega^{\,2})^{1/2}\,c$. [See Equation (9.34).] It follows that the arrival time can be written

$$t \simeq t_0+ \frac{e^{\,2}}{2\,\epsilon_0\,m_e\,c\,\omega^{\,2}}\,\int_0^z n_e(z')\,dz'.$$ (9.91)

By fitting Equation (9.91) to the data, and then extrapolating to large $\omega$, it is possible to determine $t_0$, and, hence, $t-t_0$ at a given frequency. Finally, once the line integrals of $n_e\,B_0$ and $n_e$ have been independently determined, estimates can be made of the mean electron number density, and the mean galactic magnetic field, along the straight line joining the pulsar to the Earth.