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

{\bf Set 8} \bigskip

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\noindent
{\bf 20.\ 1D wave in a dispersive medium (E.80).}
\smallskip

\noindent (a) Due to the $e^{-iwt}$ dependence we get at fixed $\omega$
$$ \frac{d^2u}{d^2x}\ =\ - \frac{\omega^2}{v^2}\,u         $$
with the fundamental solutions $\exp[\pm i\,k(\omega)\,x]$,
$k(w)=\omega/v(\omega)$ so that the general solution is the
superposition
$$ u(x,t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} d\omega\,
   e^{-i\omega t}\,\left[ A(\omega)\,e^{ik(\omega)x} +
   B(\omega)\,e^{-ik(\omega)x} \right]\ .$$
With $k(\omega)=n(\omega)\,\omega/c$ the desired results follows.

\noindent (b) At $t=0$ we have
$$ u(0,t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} d\omega\,
   e^{-i\omega t}\, \left[ A(\omega) + B(\omega) \right]\,, $$
$$ \frac{\partial u(0,t)}{\partial\,x} = \frac{i}{\sqrt{2\pi}} 
   \int_{-\infty}^{+\infty} d\omega\, e^{-i\omega t}\, \left[ 
   \frac{n(\omega)\,\omega}{c}\,A(\omega) - 
   \frac{n(\omega)\,\omega}{c}\,B(\omega) \right]\,. $$
Inverse Fourier transformation:
$$ A(\omega)+B(\omega) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} 
   dt\, e^{i\omega t}\, u(0,t)\,,$$
$$ i\,\frac{n(\omega)\,\omega}{c}\,\left[A(\omega)-B(\omega)\right]
   = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} 
   dt\, e^{i\omega t}\, \frac{\partial u(0,t)}{\partial\,x}\,,$$
Addition and subtraction give the final results:
$$ A(\omega) = \frac{1}{2\,\sqrt{2\pi}} \int_{-\infty}^{+\infty} dt\, 
   e^{i\omega t}\, \left[ u(0,t) - \frac{i\,c}{n(\omega)\,\omega}\,
   \frac{\partial u(0,t)}{\partial\,x} \right]\,,  $$
$$ B(\omega) = \frac{1}{2\,\sqrt{2\pi}} \int_{-\infty}^{+\infty} dt\, 
   e^{i\omega t}\, \left[ u(0,t) + \frac{i\,c}{n(\omega)\,\omega}\,
   \frac{\partial u(0,t)}{\partial\,x} \right]\,,  $$
\bigskip \eject

\noindent
{\bf 21.\ Energy density and Poynting vector for plane waves (E.83).}
\smallskip

\noindent (a) Energy and Poynting vector are given by
$$ u = \frac{1}{8\pi}\left(\vec{E}^{\,2}+\vec{B}^{\,2}\right)\,,\qquad
   \vec{S}=\frac{1}{4\pi}\left(\vec{E}\times\vec{B}\right)\ . $$
We have to insert the physical fields
$$ \vec{E}=\vec{E}_0\,\cos(\vec{k}\,\vec{x}-\omega t)\,,\qquad
   \vec{B}=\vec{B}_0\,\cos(\vec{k}\,\vec{x}-\omega t) $$
$$ {\rm with}~~~\omega=c\,k\,,~~~k=|\vec{k}|\,,~~~
   \vec{B}_0=\frac{\vec{k}\times\vec{E}_0}{k}\,,~~~
   \vec{E}_0\perp\vec{k}\ .$$
Using
$$ \left(\vec{k}\times\vec{E}_0\right) \cdot
   \left(\vec{k}\times\vec{E}_0\right) = k^2\,\vec{E}^{\,2}_0
   - \left(\vec{k}\cdot\vec{E}_0\right)^2 = k^2\,\vec{E}^{\,2}_0 $$
The energy density becomes
$$ u = \frac{1}{8\pi}\left(\vec{E}_0^{\,2}+\vec{E}_0^{\,2}\right)\,
  \cos^2\left(\vec{k}\cdot\vec{x}-\omega t\right) = \frac{1}{4\pi}\,
  \vec{E}_0^{\,2}\,\cos^2\left(\vec{k}\cdot\vec{x}-\omega t\right) $$
The time average over a period $T=2\pi/\omega$ is
$$ \widehat{u} = \frac{1}{T} \int_0^T u(t)\,dt = 
   \frac{1}{8\pi}\,\vec{E}_0^{\,2}\ .$$
Using
$$ \vec{E}_0\times \left(\vec{k}\times\vec{E}_0\right) =
   - \left(\vec{k}\cdot\vec{E}_0\right)\,\vec{E}_0 +
   \vec{k}\,\left(\vec{E}_0\right)^2 =
   \vec{k}\,\left(\vec{E}_0\right)^2 $$
we find for the Poynting vector
$$ \vec{S} = \frac{c\,\vec{k}}{4\pi\,k}\,\left(\vec{E}_0\right)^2
   \cos^2\left(\vec{k}\cdot\vec{x}-\omega t\right)\ . $$
The time average over a period $T=2\pi/\omega$ is
$$ \widehat{\vec{S}} = \frac{1}{T} \int_0^T \vec{S}(t)\,dt =
   \frac{c\,\vec{k}}{8\pi\,k}\,\left(\vec{E}_0\right)^2\ . $$
(b) In complex notation we have
$$ \vec{E}=\vec{E}_0\,e^{i\vec{k}\cdot\vec{x}-i\omega t}
   ~~~{\rm and}~~~\vec{B}=\frac{\vec{k}\times\vec{E}_0}{k}\,
   e^{i\vec{k}\cdot\vec{x}-i\omega t} $$
and, therefore,
$$ \vec{S} = {c\over 8\pi}\,\vec{E}\times\overline{\vec{B}}\ 
    = {c\over 8\pi}\,\vec{E}_0\times \left(
   \frac{\vec{k}\times\vec{E}_0}{k}\right)
    = {c\,\vec{k}\over 8\pi\,k}\,\left(\vec{E}_0\right)^2 $$
which agrees with the time average.

\noindent (c) Assuming perfect absorption, we want to calculate the 
radiation pressure on a surface perpendicular to $\vec{k}$. The
average momentum density is
$$ \vec{p} = \frac{1}{c^2}\,\vec{S} = {\vec{k}\over 8\pi\,k\,c}\,
    \left(\vec{E}_0\right)^2\,, $$
which gives for the radiation pressure (total momentum transfer per 
time unit times unit area) 
$$ \left|\vec{p}\right|\,(c\,\triangle t)\,\triangle a /
   (\triangle t\,\triangle a) = \left|\vec{p}\right|\,c
   = \frac{1}{8\pi}\,\left|\vec{E}_0\right|^2\ .$$
\bigskip


\noindent
{\bf 22.\ Reflection and transmission of a circularly 
polarized wave (E.89).}
\smallskip

The incident, circularly polarized wave is
$$ \vec{E} = E_0\,\left(\hat{\epsilon}_1\pm i\,\hat{\epsilon}_2\right)\,
             e^{i\vec{k}\cdot\vec{x}-i\omega t} $$
where the directions $\hat{\epsilon}_1$ and $\hat{\epsilon}_2$ 
correspond to perpendicular and parallel polarized light, respectively.

For the transmitted wave we choose (instead of linearly polarized 
components) the convention 
$$ \vec{E}'=\left( E'_1\,\hat{\epsilon}_1\pm 
                i\,E'_2\,\hat{\epsilon}^{\,'}_2\right)\,
                e^{i\vec{k}'\cdot\vec{x}-i\omega t}\,, $$
so that $E_0$, $E'_1$ and $E'_2$ are in phase.
For the perpendicularly polarized amplitude we have
$$ \frac{E'_1}{E_0} = \frac{2n\,\cos(\theta_i)}{n\cos(\theta_i)+
                      n'\,\cos(\theta_t)} $$
and for the parallel polarized amplitude 
$$ \frac{E'_2}{E_0} = \frac{2n\,\cos(\theta_i)}{n'\cos(\theta_i)
                    + n\,\cos(\theta_t)} $$
For $\theta_i<\theta_0$, where $\theta_0$ is the angle of total
reflection, the phases stay unchanged. Therefore, the axes of the 
ellipse are along $\hat{\epsilon}_1$ and $\hat{\epsilon}^{\,'}_2$.
The ratio of the axes is 
$$ \frac{E'_2}{E'_1} = \frac{n\,\cos(\theta_i)+n'\,\cos(\theta_t)}
   {n'\,\cos(\theta_i)+n\,\cos(\theta_t)}~~~{\rm with}~~~
   \frac{\sin(\theta_t)}{\sin(\theta_i)}=\frac{n}{n'}\ . $$
Linear polarization,
$$ E'_1\ =\ 0~~~{\rm or}~~~E'_2\ =\ 0\,,$$
are both impossible as they imply $\cos(\theta_i)=0$.

\noindent For the reflected wave we choose the convention 
$$ \vec{E}'' = \left(E''_1\,\hat{\epsilon}_1\pm 
                  i\,E''_2\,\hat{\epsilon}^{\,''}_2\right)\,
                  e^{i\vec{k}''\cdot\vec{x}-i\omega t}\ . $$
For the perpendicular polarized amplitude we have
$$ \frac{E''_1}{E_0} = \frac{n\,\cos(\theta_i)-n'\,\cos(\theta_t)}
                            {n\,\cos(\theta_i)+n'\,\cos(\theta_t)} $$
and for the parallel polarized amplitude 
$$ \frac{E''_2}{E_0} = \frac{n'\,\cos(\theta_i)-n\,\cos(\theta_t)}
                            {n'\,\cos(\theta_i)+n\,\cos(\theta_t)} $$
As the phases are not changed, the axes of the ellipse are along 
$\hat{\epsilon}_1$ and $\hat{\epsilon}^{\,''}_2$. The ratio of the 
axes is 
$$ \frac{E''_2}{E''_1} = \frac{
   (n'\,\cos\theta_i-n\,\cos\theta_t)\,
   (n\,\cos\theta_i+n'\,\cos\theta_t) }
 { (n'\,\cos\theta_i+n\,\cos\theta_t)\,
   (n\,\cos\theta_i-n'\,\cos\theta_t) } $$
Linearly polarized light: For
$$ E''_2\ =\ 0 ~~~{\rm we\ have}~~~ 
   n'\,\cos\theta_i-n\,\cos\theta_t = 0 $$
with the non-trivial solution
$$ \tan (\theta_i) = \frac{n'}{n}~~~{\rm (Brewster's\ angle)}\ .$$
For
$$ E''_1\ =\ 0 ~~~{\rm we\ have}~~~ 
   n\,\cos\theta_i-n'\,\cos\theta_t = 0 $$
with only the trivial solution $n'=n$, $\theta_t=\theta_i$ for which 
the wave remains circularly polarized as there is no reflection and 
$E'_1=E'_2=E_0$.
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