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\centerline{\bf Electrodynamics B (PHY 5347) Winter/Spring 2017 
Solutions} \bigskip

{\bf Set 9} \bigskip

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% \end{figure} \centerline{Potential as function of $z$.}


\noindent
{\bf 23.\ Wave with a finite extension (E.84).}
\smallskip

\noindent 
We choose the wave vector in $z$-direction and make the ansatz
$$ \vec{E}(t,\vec{x}) = \left[ E_0(x,y)\,(\hat{x}\pm i\hat{y})
   + F(x,y)\,\hat{z}\right]\,e^{ikz-i\omega t} $$
with real functions $E_0$ and $F$. Then
\begin{eqnarray} \nonumber
  \nabla\cdot\vec{E} = 0 &\Rightarrow& \frac{\partial E_0}{\partial x}
  \pm i\,\frac{\partial E_0}{\partial y} + i\,k\,F(x,y) = 0
  \\ \nonumber &\Leftrightarrow& F(x,y) = \frac{i}{k}\,\left( 
  \frac{\partial E_0}{\partial x} \pm i\,\frac{\partial E_0}{\partial y}
  \right)\,. 
\end{eqnarray}
Next we use the Maxwell equation
$$\nabla\times\vec{E}=-\frac{1}{c}\frac{\partial\vec{B}}{\partial t}$$
together with the assumption that $F(x,y)$ is a slowly varying function 
of $x$ and $y$ (i.e., $\partial F/\partial x \approx 0$ and 
$\partial F/\partial y \approx 0$) to find
\begin{eqnarray} \nonumber
  \frac{i\omega}{c}\,B^x &=& \left( \frac{\partial F}{\partial y} 
  \pm k\,E_0\right)\,e^{ikz-i\omega t}\,\approx\,
  \pm k\,E_0\,e^{ikz-i\omega t}\,,\\ \nonumber
  \frac{i\omega}{c}\,B^y &=& \left( i\,k\,E_0 -
  \frac{\partial F}{\partial x} \right)\,e^{ikz-i\omega t}\,\approx\,
  i\,k\,E_0\,e^{ikz-i\omega t}\,,\\ \nonumber
  \frac{i\omega}{c}\,B^z &=& \left(\pm i\,
  \frac{\partial E_0}{\partial x} - \frac{\partial E_0}{\partial y}
  \right)\,e^{ikz-i\omega t}\,=\,\pm k\,F(x,y)\,e^{ikz-i\omega t}\,.
\end{eqnarray}
Using $\omega/c=k$ (for vacuum) gives
$$ \vec{B} \approx \mp i\,\vec{E}\ .$$
The other Maxwell equations are consistent with these results. Example:
$$ E_0(x,y) = \frac{1}{\cosh(\epsilon\rho)}\,,~~~\epsilon\ {\rm small}\,,
   ~~~\rho=\sqrt{x^2+y^2}\,. $$ % \newpage 
\bigskip 


\noindent
{\bf 24.\ Dispersion figure (E.90).}  \smallskip

The plot is that of Lecture Notes, Figure~5.2.
\bigskip \eject


\noindent
{\bf 25.\ Faraday effect (birefringent light) (E.91).}  \smallskip

\noindent We choose the magnetic field in $z$-direction, 
$\vec{B}_0=B_0\hat{z}$. Then
\begin{eqnarray} \nonumber
  m\,\ddot{x} + m\,\omega_0^2\,x &=& q_e\,E^x + \frac{q_e}{c}\, 
                                 \dot{y}\,B_0\,,\\ \nonumber
  m\,\ddot{y} + m\,\omega_0^2\,y &=& q_e\,E^y - \frac{q_e}{c}\, 
                                 \dot{x}\,B_0\,.
\end{eqnarray}
These equations separate in circular coordinates
$$ \hat{\epsilon}_{\pm} = \hat{x}_{\pm} = \frac{1}{\sqrt{2}}\,
   (\hat{x}\pm i\,\hat{y})\,,~~x_{\pm} = \frac{1}{\sqrt{2}}\,
   (x\mp i\,y)\,,~~E_{\pm} = \frac{1}{\sqrt{2}}\,(E^x\mp i\,E^y)\,.$$
With the ansatz $E_{\pm} = E^0_{\pm}\,\exp(-i\,\omega\,t)$,
$x_{\pm} = x^0_{\pm}\,\exp(-i\,\omega\,t)$ we get
$$ m\,\ddot{x}_{\pm} + m\,\omega_0^2\,x_{\pm} = 
   q_e\,E_{\pm} \pm i\,\frac{q_e}{c}\,\dot{x}_{\pm}\,B_0\,,         $$
$$ - m\,\omega^2\,{x}^0_{\pm} + m\,\omega_0^2\,x^0_{\pm} = 
   q_e\,E^0_{\pm} \mp \omega\,\frac{q_e}{c}\,x^0_{\pm}\,B_0\,.   $$
$$ x^0_{\pm} = \frac{q_e\,E^0_{\pm}}{m\,(\omega_0^2-\omega^2)
                 \pm \omega\,(q_e/c)\,B_0}                       $$
from which we find the polarizations $P_{\pm}=N\,q_e\,x^0_{\pm}$. 
This gives the indices of refraction for the two circular polarized 
components:
$$ \epsilon_{\pm} = 1 + \frac{4\pi\,N\,q_e^2}{
   m\,(\omega_0^2-\omega^2)\mp \omega\,(q_e/c)\,B_0}\,,~~~
   k_{\pm}=\frac{\omega}{c}\,\sqrt{\epsilon_{\pm}} =
   \frac{\omega}{c}\,n_{\pm}\ .  $$
The propagation of the two components of circular polarized light 
in $z$-direction is given by
$$ E_{\pm} = E_{\pm}^0\,e^{i\,k_{\pm}\,z-i\,\omega\,t}\ . $$ 
So we find
$$ \frac{E_-}{E_+} = \frac{E^0_-}{E^0_+}\,e^{i\,\alpha}~~~
   {\rm with}~~~ \alpha = (k_--k_+)\,z $$
which corresponds as shown after equation (5.50) of the lecture notes 
to a rotation of the axis by 
$$ \gamma = \frac{\alpha}{2} = (k_--k_+)\,\frac{l}{2} $$
when the light travels the distance $z=l$ along the $z$-axis. The wave
stays linearly polarized, while the direction of $\vec{E}$ changes when
the wave penetrates into the medium:
$$ \vec{E}\ =\ \left(E_1\,\hat{\epsilon}'_1+E_2\hat{\epsilon}'_2\right)
   \,e^{ikz-i\omega t} $$
with the $E_1$, $E_2$ coefficients and $k$ of the incoming wave and
$$ \hat{\epsilon}'_1\ = \hat{\epsilon}_1\,\cos(\gamma) +
                        \hat{\epsilon}_2\,\sin(\gamma)\,,$$
$$ \hat{\epsilon}'_2\ = -\hat{\epsilon}_1\,\sin(\gamma) +
                        \hat{\epsilon}_2\,\cos(\gamma)\,.$$
\bigskip


\noindent
{\bf 26.\ Principal value integral (E.93).}

Due to the fall-off behavior of the function $f(z)$ a half-circle
of radius $R$ about $0$ in the upper half-plane does not contribute
in the $R\to\infty$ limit:
\begin{eqnarray} \nonumber
  \lim_{R\to\infty} \left|\int_{C_{R+}} dz\,f(z)\right| &=&
  \lim_{R\to\infty} \left|\int_0^{2\pi} i\,R\,e^{i\phi}\,f(z)\,d\phi
  \right| \\ \nonumber \le \lim_{R\to\infty} \int_0^{2\pi}
  \left| i\,R\,e^{i\phi}\, f(z)\right|\,d\phi &\sim &
  \lim_{R\to\infty} \int_0^{2\pi} \frac{d\phi}{R} =
  \lim_{R\to\infty} \frac{2\pi}{R} = 0\,.
\end{eqnarray} 

\begin{enumerate}

\item For the contour of the first figure we obtain by the Cauchy 
integral theorem
$$ \frac{1}{2\pi i}\lim_{\eta\to 0} \int_{C_1} 
   \frac{f(z)\,dz}{z-\omega_R-i\eta}\ =\ \lim_{\eta\to 0} 
   f(\omega_R+i\eta)\ =\ f(\omega_R)\,.  $$

\item For the contour of the second figure we obtain by the Cauchy 
integral theorem
$$ \frac{1}{2\pi i}\lim_{\eta\to 0} \int_{C_2} 
   \frac{f(z)\,dz}{z-\omega_r-i\eta}\ =\ 0\,.            $$

\item We denote the half-circle of the first figure by
$C_+(\omega_R+i\eta,\eta)$. To evaluate the integral
\begin{eqnarray} \nonumber
  I_{C_+(\omega_R+i\eta,\eta)}\ =\ {1\over 2\pi\,i}\,\lim_{\eta\to 0} 
  \int_{C_+(\omega_R+i\eta,\eta)} \frac{f(z)\,dz}{z-\omega_R-i\eta} 
\end{eqnarray}
we use the Taylor expansion
$$ f(z)\ =\ f(\omega_r+i\eta) + f'(\omega_r+i\eta)\,
   (z-\omega_r-i\eta) + \dots                                   $$
and note that for $z$ on the half-circle $|z-\omega_r-i\eta|=\eta$
holds. Therefore, the integral becomes
\begin{eqnarray} \nonumber
  I_{C_+(\omega_R+i\eta,\eta)} &=& {1\over 2\pi\,i}\,\lim_{\eta\to 0} 
  f(\omega_r+i\eta)\int_{C_+(\omega_R+i\eta,\eta)}\frac{dz}{z-\omega_R
  -i\eta} \\ \nonumber &+&
  {1\over 2\pi\,i}\,\lim_{\eta\to 0} {\cal O}(\eta)
  \int_{C_+(\omega_R+i\eta,\eta)}\frac{dz}{z-\omega_R-i\eta} 
\end{eqnarray}
where by explicit integration (compare a previous exercise) the 
integrals for the half-circle $C_+(\omega_R+i\eta,\eta)$ are now 
$\pi\,i$, so that we find 
\begin{eqnarray} \nonumber
  {1\over 2\pi\,i}\,\lim_{\eta\to 0} \int_{C_+(\omega_R+i\eta,\eta)} 
  \frac{f(z)\,dz}{z-\omega_R-i\eta}\ =\ \frac{1}{2}\,f(\omega_R)\,.
\end{eqnarray}
Similarly, we denote the half-circle of the second figure by
$C_-(\omega_R+i\eta,\eta)$ and find
\begin{eqnarray} \nonumber
  {1\over 2\pi\,i}\,\lim_{\eta\to 0} \int_{C_-(\omega_R+i\eta,\eta)}
  \frac{f(z)\,dz}{z-\omega_R+i\eta} = -\frac{1}{2}\,f(\omega_R)\,,
\end{eqnarray}
because the integration is now in clockwise instead of anti-clockwise
direction.

For illustration, let us consider a half-circle about $z=0$. Then,
\begin{eqnarray} \nonumber
  I_{C_+(0,\eta)} &=& {1\over 2\pi\,i}\,\lim_{\eta\to 0} 
  \int_{C_+(0,\eta)} \frac{f(z)\,dz}{z}\\ \nonumber
  &=& {1\over 2\pi\,i}\,\lim_{\eta\to 0} \sum_{n=0}^{\infty}
  \frac{f^{n}(0)}{n!}\,\int_{C_+(0,\eta)}\frac{z^n\,dz}{z}
\end{eqnarray}
and for the last integral we find
\begin{eqnarray} \nonumber 
  \int_{C_+(0,\eta)}\frac{z^n\,dz}{z}\ =\ \int_0^{\pi} \eta\,i\,
  e^{i\phi}\,d\phi\,\eta^{n-1}\,e^{i\,(n-1)\,\phi}\ =\ 
  i \eta^n \int_0^{\pi} e^{i\,n\,\phi}\,d\phi\,.
\end{eqnarray}
For $n=0$ the result is $i\,\pi$, while for $n\ne 0$ we find
\begin{eqnarray} \nonumber 
  i \eta^n \int_0^{\pi} e^{i\,n\,\phi}\,d\phi\ =\ 
  \frac{i\,\eta^n}{i\,n}\,\left. e^{i\,n\,\phi}\right|_0^{\pi}
  \ =\ \begin{cases} 0~{\rm for}~n~{\rm even} \\
       2\eta^n/n~{\rm for}~n~{\rm odd}\,.
       \end{cases}
\end{eqnarray}
Therefore, only $n=0$ survives the $\eta\to 0$ limit and
\begin{eqnarray} \nonumber
  I_{C_+(0,\eta)} &=& {1\over 2\pi\,i}\,\lim_{\eta\to 0} 
  \int_{C_+(0,\eta)} \frac{f(z)\,dz}{z}\ =\ \frac{1}{2}\,f(0)\,.
\end{eqnarray}

\end{enumerate}

\end{document}
