💼 Case: consider a $z$-directed monochromatic EM wave (being $z$-directed means that the wavevector is given by $\vec k=k(0,0,1)$)

• The electric field vector is given by

  $$ \vec E=\mathrm{Re}\left ( \vec E_0 e^{i(kz-\omega t)} \right ) $$

  where $\vec E_0$ is a complex amplitude which can be written as

  $$ \vec E_0=\frac{1}{\sqrt{2}}\left [ E_{0,x}e^{i\theta}\hat{\mathrm x}+E_{0,y}e^{i(\theta+\alpha)}\hat{\mathrm y} \right ] $$

  where $E_{0,x}$ and $E_{0,y}$ are real.

  The parameter $\theta$ is present in both terms and represents an overall phase that dictates whether the function is a cosine, sine or a combination of the two. The parameter $\alpha$ represents a relative phase. The amplitude term has two interesting cases $\alpha=0$ and $\alpha=\pi/2$

  • When $\alpha=0$ meaning that the two components are in phase so that the electric field vector becomes (for $\theta=0$)

    $$ \vec E=\frac{1}{\sqrt{2}}\left ( E_{0,x}\hat{\mathrm{x}}+E_{0,y}\hat{\mathrm{y}} \right )\cos(kz-\omega t) $$

    this is linear polarization along the axis $\frac{1}{\sqrt{2}}\left ( E_{0,x}\hat{\mathrm{x}}+E_{0,y}\hat{\mathrm{y}} \right )$

  • When $\alpha=\pm \pi/2$ and $E_{0,x}=E_{0,y}$ in which case the two components are out of phase by $+\pi/2$. The electric field vector is given by

    $$ \vec E=\frac{E_{0,x}}{\sqrt{2}}\left [ \hat{\mathrm{x}}\cos(kz-\omega t)\mp \hat{\mathrm{y}}\sin(kz-\omega t) \right ] $$

    at $\theta=0$.

    For the $+$ solution the electric field has right circular polarization and for the $-$ solution the electric field has left circular polarization.

    💃 Example: if we take $\alpha=-\pi/2$ then the components of the electric field vector are

    $$ E_x=E_r\cos(kz-\omega t) \qquad E_y=E_R\sin(kz-\omega t) $$

    where $E_R=\frac{1}{\sqrt{2}}E_{0,x}$

    The electric field will then trace out a circle of radius $E_R$ going anti-clockwise in the $E_x-E_y$ plane (shown in picture below)

    🗒️ Note: if we had taken $\alpha=\pi/2$ then it would have moved clockwise

    

  🗒️ Note: in general the electric field traces out an Ellipse in the $E_x-E_y$ plane and it is useful to characterise the polarization of a wave via the Stokes parameters $I,Q,U,V$

• We will operate in the complex representation such that

  $$ \cos(\vec k\cdot \vec r-\omega t)=\mathrm{Re}[e^{i(\vec k\cdot \vec r-\omega t)}] \quad \sin (\vec k \cdot \vec r-\omega t)=\mathrm{Re}[e^{i(\vec k \cdot \vec r-\omega t-\frac{\pi}{2})}] $$

• Defining two new bases ($\hat L,\hat R$) and $(\hat a,\hat b)$

  $$ \begin{aligned} \hat L&=\frac{1}{\sqrt{2}}(\hat{\mathrm{x}}+i \hat {\mathrm{y}}) \\ \hat a&= \frac{1}{\sqrt{2}}(\hat{\mathrm{x}}+\hat {\mathrm{y}}) \end{aligned} \qquad \begin{aligned} \hat R&=\frac{1}{\sqrt{2}}(\hat{\mathrm{x}}-i \hat {\mathrm{y}}) \\ \hat b&= \frac{1}{\sqrt{2}}(\hat{\mathrm{x}}-\hat {\mathrm{y}}) \end{aligned} $$

  🗒️ Note: $(\hat a,\hat b$) are at $45\degree$

• We define the Stokes parameters via

  $$ \begin{aligned} I&=|E_x|^2+|E_y|^2 \\ U&=|E_a|^2-|E_b|^2

  \end{aligned} \qquad \begin{aligned} Q&=|E_x|^2-|E_y|^2 \\ V&=|E_\ell|^2 -|E_r|^2 \end{aligned} $$

  where the components of the electric field vector in the various bases are given by

  $$ E_a=\hat a\cdot \vec E \quad E_b=\hat v\cdot \vec E \quad E_\ell =\hat L^* \cdot \vec E \quad E_r=\hat R^* \cdot \vec E $$

  The parameters $(Q,U)$ describe linear polarization and $V$ circular polarization. The parameter $I$ is the average intensity

  

• For monochromatic plane waves one can calculate that

  $$ I^2=Q^2+U^2+V^2 $$

• The fraction of a wave which is linearly polarized is

  $$ f_l=\frac{\sqrt{Q^2+U^2}}{I} $$

  and the fraction with circular polarization is

  $$ f_C=\frac{V}{I} $$

  with

  $$ f^2_L+f^2_C=1 $$

  🗒️ Note:

  $$ \begin{aligned} U&=E_yE^_x+E_xE^_y=2\mathrm{Re}(E_yE^_x) \\ V&=i(E^_yE_x-E_yE^_x)=2\mathrm{Im}(E_yE_x^) \end{aligned} $$

EM waves in dielectric and magnetic media

• For electrostatics in a dielectric medium we have

  $$ \rho_\text{bound}=-\vec \nabla\cdot \vec P \qquad \vec D=\epsilon_0 \vec E+\vec P $$

  and for magnetostatics in a magnetic medium

  $$ \vec j_\text{bound}=\vec \nabla \times \vec M \quad \vec H=\frac{1}{\mu_0} \vec B-\vec M $$

• In non-static situations there is an extra current due to the time dependence of the polarization $\vec P$:

  $$ \vec j_\text{bound}=\vec \nabla \times \vec M+\dot{\vec P} $$

  The last term is the displacement current created by the polarization

• Maxwell’s equations become

  $$ \begin{matrix} \vec \nabla \cdot \vec D=\rho_\text{free} & \vec \nabla \cdot\vec B=0 \\ \dot{\vec B}=-\vec \nabla \times \vec E & \dot{\vec D} =\vec \nabla \times \vec H-\vec j_\text{free}

  \end{matrix} $$

  where $\rho_\text{free}$ and $\vec j_\text{free}$ are the free charge and currents which are the quantities that can be externally controlled

• Let us set the free charge and current to zero $\rho_\text{free}=0 ,\vec j_\text{free}=0$ and consider linear isotropic materials so that we can take

  $$ \vec D=\epsilon_r \epsilon_0 \vec E \quad \vec H=\frac{1}{\mu_r \mu_0}\vec B $$

• Thus the Ampere-Maxwell equation becomes

  $$ \dot{\vec E}=\frac{c^2}{\epsilon_r \mu_r}\vec \nabla\times \vec B=v^2 \vec \nabla \times \vec B $$

  where

  $$ v=\frac{c}{n}=\frac{c}{\sqrt{\epsilon_r\mu_r}} $$

  and $n=\sqrt{\epsilon_r \mu_r}$ is the refractive index

  💎 Conclusion: EM waves move at a speed $v<c$ in dielectric/magnetic media

• The dispersion relation becomes

  $$ \omega=\pm vk=\pm \frac{ck}{n} $$

  🗒️ Note: the refractive index $n$ may be dependent upon the frequency

💼 Case: consider an EM wave in a medium with a refractive index $n_1$ hits a boundary with another medium with refractive index $n_T$. The EM wave then splits into reflected and transmitted components

• We write the field vectors as

  $$ \begin{aligned} \vec E^{(I)}&=\vec E^{(I)}_0e^{i(\vec k \cdot \vec r-\omega t)} \\ \vec E^{(R)}&=\vec E^{(R)}_0e^{i(\vec k' \cdot \vec r-\omega t)} \\ \vec E^{(T)}&=\vec E^{(T)}_0e^{i(\vec k'' \cdot \vec r-\omega t)} \end{aligned} $$

  where the wavevectors are

  $$ \begin{aligned} \vec k&=k(\sin \theta_I, \cos \theta_I, 0) \\ \vec k'&=k(\sin \theta_R, \cos \theta_R, 0) \\ \vec k''&=k(\sin \theta_T, \cos \theta_T, 0) \end{aligned} $$

  where

  $$ k=\frac{n_I\omega}{c} \qquad k''=\frac{n_T \omega}{c} $$

• As for reflective from a conductor we demand from the boundary that the phase is continuous

  $$ \vec k\cdot \vec r=\vec k'\cdot \vec r=\vec k'' \cdot \vec r \quad \text{at} \quad y=0 $$

  Hence

  $$ kx\sin\theta_I =kx\sin\theta_R \quad \Rightarrow \quad \theta_I =\theta_R $$

• From this we can find

  $$ kx\sin \theta_1=k'' x\sin\theta_T $$

  which if we insert the definitions of $k$ and $k''$ in terms of the refractive indices

  $$ n_t\sin\theta_T=n_I\sin\theta_I $$

  which is Snell’s law

  🗒️ Note: If we compute the electric fields $\vec E^{(R)}$ and $\vec E^{(T)}$ in terms of $\vec E^{(I)}$, which gives what are known as Fresnel’s equations, which give rise to a number of phenomenon such as the Brewster angle

• 💼 Case: consider an EM wave going from air into glass ($n_I=1$ and $n_T=1.5)$ so that

  $$ \sin\theta_T=0.66\sin\theta_I $$

  or in the opposite direction

  $$ \sin\theta_T =1.5\sin\theta_I $$

  In the second case $\sin\theta_I>1/1.5\sin\theta_T$ is greater than unity and there is no transmitted wave and everything is reflected back into the glass. This occurs at the critical angle (when $n_I>n_T$)

  $$ \theta_\text{crit}=\sin^{-1}(n_T/n_I) $$

  such that $\theta_I>\theta_\text{crit}$ only leads to a reflected wave: this is total internal reflection