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

{\bf Set 12} \bigskip

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


\noindent
{\bf 33.\ Cylindrical waveguide, $\rm TE_{11}$ mode power transmission.}
\smallskip

\noindent The cutoff frequency follows as the $d\to\infty$ limit from
our discussion of the cylindrical cavity,
$$ \omega_{11} = \frac{1.841\,c}{\sqrt{\mu\epsilon}\,R} = 
   \frac{\gamma_{11}^{\,'}}{R}\ . $$
Using the time-averaged Poynting vector 
$$ \vec{S}\ =\ \frac{c}{8\pi} {\rm Re}\left(\vec{E}\times
   \overline{\vec{H}} \right) $$
and the TE mode
$$ H^z = \psi\,e^{ikz}\,,~~\left.\frac{\partial\psi}{\partial n}
   \right|_S=0\,,~~~\vec{H}_t = \frac{ik}{\gamma^2}\,\nabla_t\psi\,,
   ~~~\vec{E}_t = -Z\,\hat{z}\times\vec{H}_t\,, $$
where $Z=\mu\omega/(ck)$ is the wwave impedance for the TE mode,
one finds
$$ \vec{S}\ =\ \frac{\omega\,k\,\mu}{8\pi\,\gamma^4}\,\left[\hat{z}
   \,\left|\nabla_t\psi\right|^2 - i\,\frac{\gamma^2}{k}\,
   \overline{\psi}\,\nabla_t\psi\right]\,. $$
We are only interested in the longitudinal energy $\sim\hat{z}$ and the 
power flow is obtained by integrating over the cross-sectional area:
$$ P\ =\ \int_A \hat{z}\cdot\vec{S}\ =\ \frac{\omega\,k\,\mu}{8\pi\,
   \gamma^4}\int_A(\overline{\nabla_t\psi})\cdot(\nabla_t\psi)\,da\,.$$
By means of Green's first identity applied to two dimensions, $P$ becomes
$$ P\ =\ \frac{\omega\,k\,\mu}{8\pi\,\gamma^4}\left[\oint_C\overline{
   \psi}\,\frac{\partial\psi}{\partial n}\,dl - \int_A
   \overline{\psi}\,\nabla_t^2\psi\,da\right]\,,$$
where the first integral is zero due to the BC. Using the wave 
equation for the second integral, the transmitted power for mode 
$\gamma^2_{\lambda}$ is
$$ P\ =\ \frac{\omega\,k_{\lambda}\,\mu}{8\pi\,
   \gamma^4_{\lambda}}\int_A \overline{\psi}\,\gamma^2_{\lambda}\,
   \psi\,da\ =\ \frac{c\,\mu}{8\pi\,\sqrt{\mu\epsilon}}\,
   \left(\frac{\omega}{\omega_{\lambda}}\right)^2\,
   \left(1-\frac{\omega^2_{\lambda}}{\omega^2}\right)^{1/2}
   \int_A \overline{\psi}\,\psi\,da\,,$$
where in the last step $\gamma^2_{\lambda}$ has been eliminated in
favor of the cutoff frequency, $\gamma_{\lambda}=\sqrt{\mu\epsilon}\,
\omega_{\lambda}/c$ and $k_{\lambda}=\sqrt{\mu\epsilon}\sqrt{\omega^2
-\omega^2_{\lambda}}/c$. For the $\rm TE_{11}$ mode
$$ P = \frac{c\,\mu}{8\pi\,\sqrt{\mu\epsilon}}\,\left(\frac{\omega}
   {\omega_{11}}\right)^2\,\left(1-\frac{\omega_{11}^2}{\omega^2}
   \right)^{1/2} \int_A da\,\overline{\psi}\,\psi   $$
and
$$ \psi = H_0\,J_1(\gamma'_{11}\,\rho)\,e^{i\,\phi} $$
for the mode in question. Thus
$$ P = \frac{c\,\mu}{8\pi\,\sqrt{\mu\epsilon}}\,\left(\frac{\omega}
   {\omega_{11}}\right)^2\,\left(1-\frac{\omega_{11}^2}{\omega^2}
   \right)^{1/2}\,2\pi\,|H_0|^2 \int_0^R \rho\,d\rho\,
   \left[J_1(\gamma'_{11}\,\rho)\right]^2\ . $$
Using the integral as given in the problem the final result is
$$ P = \frac{c\,\mu}{8\,\sqrt{\mu\epsilon}}\,\left(\frac{\omega}
   {\omega_{11}}\right)^2\,\left(1-\frac{\omega_{11}^2}{\omega^2}
   \right)^{1/2}\,|H_0|^2\,R^2\,\left(1-\frac{1}{x'^{\,2}_{11}}
   \right)\, \left[J_1(\gamma'_{11}\,R)\right]^2\ . $$
\bigskip


\noindent
{\bf 34.\ Hertz vector.} \smallskip

$$ F^{\alpha\beta} = \partial^{\alpha} A^{\beta} -
   \partial^{\beta} A^{\alpha}\,,\qquad \left(A^{\alpha}\right) =\
   \left( \begin{matrix} \Phi  & \\ \vec{A} & \end{matrix} \right)\ .$$
In the Lorentz gauge $\partial_{\alpha} A^{\alpha}=0$ holds and, 
therefore,
$$ \partial_{\alpha} F^{\alpha\beta} = \Box A^{\beta} -
   \partial^{\beta} \partial_{\alpha} A^{\alpha} =
   \Box A^{\beta} = \frac{4\pi}{c}\,J^{\alpha}\ . $$
In components this reads
$$ \nabla^2\Phi - \frac{1}{c^2}\frac{\partial^2\Phi}{\partial t^2} =
   -4\pi\,\rho\,,\qquad \nabla^2\vec{A} - 
   \frac{1}{c^2}\frac{\partial^2\vec{A}}{\partial t^2} =
   -\frac{4\pi}{c}\,\vec{J}\ .  $$
(1) Continuity equation:
$$ \frac{\partial\rho}{\partial t} + \nabla\cdot\vec{J} = 
  - \frac{\partial~}{\partial t}\,\left(\nabla\cdot\vec{p}\,\right) + 
    \nabla\cdot\left(\frac{\partial\vec{p}}{\partial t}\right) = 0\ .$$
(2) Let $\Phi = a\,\left(\nabla\cdot\vec{\Pi}\right)$ and $\vec{A}=b\,
    \partial\vec{\Pi}/\partial t$. Then,
\begin{eqnarray} \nonumber
  a\,\left[ \nabla^2 - \frac{1}{c^2}\frac{\partial^2~}{\partial t^2}
  \right]\,\left(\nabla\cdot\vec{\Pi}\right) \,=\, -
  a\,4\pi\,\nabla\cdot\vec{p} \,=\, a\,4\pi\,\rho \,=\, 
  -\,4\pi\,\rho\ &\Rightarrow&\ a\,=\,-1\,, \\ \nonumber
  b\,\left[ \nabla^2 - \frac{1}{c^2}\frac{\partial^2~}{\partial t^2}
  \right]\,\left(\frac{\partial\vec{\Pi}}{\partial t}\right) \,=\, -
  b\,4\pi\,\frac{\partial\vec{p}}{\partial t} \,=\, 
  - b\,4\pi\,\vec{J} \,=\, -\frac{4\pi}{c}\,\vec{J}\
  &\Rightarrow&\ b\,=\,\frac{1}{c}\,, 
  \\ \nonumber \Phi = - \left(\nabla\cdot\vec{\Pi}\right)\,,\qquad
  \vec{A} = \frac{1}{c}\, \frac{\partial\vec{\Pi}}{\partial t}\ .
\end{eqnarray} 
The Lorentz condition holds:
$$ \frac{1}{c}\frac{\partial\Phi}{\partial\,t} + \nabla\cdot\vec{A}
   = - \frac{1}{c}\frac{\partial~}{\partial t}
       \left(\nabla\cdot\vec{\Pi}\right) + \frac{1}{c}\,\nabla\cdot 
       \frac{\partial\vec{\Pi}}{\partial\,t} = 0\ . $$
(3) From the given Hertz vector we find
$$ \Phi = -\nabla\cdot\vec{\Pi} = - p_0\,\cos(\theta)\,
   \frac{\partial~}{\partial r} \left( r^{-1}\,e^{-i\,\omega\,t+i\,k\,r}
   \right) \approx - \frac{i\,k\,p_0}{r}\, \cos(\theta)\, 
   e^{-i\,\omega\,t+i\,k\,r}\,, $$ 
$$ \vec{A} = \frac{1}{c}\,\frac{\partial\vec{\Pi}}{\partial t} = -
   \frac{i\,\omega\,p_0}{c\,r}\,\hat{z}\,e^{-i\,\omega\,t+i\,k\,r}\,.$$ 
(5) The electric and magnetic fields are given by
$$ \vec{E} = -\frac{1}{c}\,\frac{\partial\vec{A}}{\partial t} -
   \nabla\Phi\,,\qquad \vec{B} = \nabla\times\vec{A}\ .$$
Therefore, we have in the far field approximation (use $\nabla$ in
spherical coordinates)
$$ \vec{E} \approx \frac{\omega^2\,p_0}{c^2\,r}\,\left( - \hat{z} +
   \cos(\theta)\,\hat{r}\right)\, e^{-i\,\omega\,t+i\,k\,r}\ .    $$ 
Using $\hat{z}=\cos(\theta)\,\hat{r}-\sin(\theta)\,\hat{\theta}$ this 
is
$$ \vec{E} \approx \frac{\omega^2\,p_0}{c^2\,r}\,\sin(\theta)\,
   \hat{\theta}\, e^{-i\,\omega\,t+i\,k\,r}\ .                    $$ 
For the magnetic field we have
$$ \vec{B} = - \frac{i\,\omega\,p_0}{c}\, \hat{r}\times\hat{z}\,
   \frac{\partial~}{\partial r}\,\frac{e^{-i\,\omega\,t+i\,k\,r}}{r}
   \approx - \frac{\omega\,k\,p_0}{c\,r}\,\sin(\theta)\,
   \hat{\phi}\, e^{-i\,\omega\,t+i\,k\,r}\ .                      $$ 
The $\vec{E}$ and $\vec{B}$ fields describe electric dipole radiation.
\bigskip
\eject

\noindent
{\bf 35.\ Vector potential of two current loops.}
\smallskip

\noindent We define
$$ x_0 = \cos(\theta_0) = \frac{d}{r_0}~~{\rm with}~~
   r_0 = \sqrt{R^2+d^2}~~~{\rm and}~~~x'=\cos(\theta')\ .$$
The currents flow in $\hat{\phi}$ direction and their densities are 
given by
$$ J^{(1,2)}_{\phi} = \pm I\,\sin(\theta')\,
   \delta(\cos\theta_0\mp\cos\theta')\,r_0^{-1}\,\delta(r'-r_0)\ . $$
Note the (correct) normalization
$$ \int_0^{\infty} r'dr' \int_0^{\pi} d\theta'\, J^{(1,2)}_{\phi} 
   = \pm I $$
for the currents. Using Cartesian coordinates we have
$$ \vec{J}^{(1,2)} = - J_{\phi}^{1,2}\,\sin(\phi')\,\hat{x}
                     + J_{\phi}^{1,2}\,\cos(\phi')\,\hat{y}\ . $$
(a) For the calculation of the vector potential $\vec{A}$ we exploit
cylindrical symmetry and choose the observation point at $\phi=0$, so
that the $\sin(\phi')\,\hat{x}$ component does not contribute. It 
remains to calculate the $\hat{y}$ component, $\hat{\phi}=\hat{y}$,
$r\gg r'$ now:
$$ {A}_{\phi} = \frac{1}{c} \int_0^{2\pi}d\phi'\,\cos(\phi')
   \int_{-1}^{+1} d\cos\theta' \int_0^{\infty} r'^2dr'\,J_{\phi}\, 
   \frac{e^{i\,k\,|\vec{x}-\vec{x}^{\,'}|}}{|\vec{x}-\vec{x}^{\,'}|}
   \,, $$
$$ \frac{e^{i\,k\,|\vec{x}-\vec{x}^{\,'}|}}{|\vec{x}-\vec{x}^{\,'}|}
   = 4\pi\,i\,k \sum_{l=0}^{\infty} j_l(kr')\,h_l^+(kr) 
   \sum_{m=-l}^{+l} Y_l^{\ m}(\theta,\phi)\, 
         \overline{Y}_l^{\ m}(\theta',\phi')\,. $$
After the $r'$-integration we have
\begin{eqnarray} \nonumber
  {A}_{\phi}&=&4\pi\,i\,k \frac{I\,r_0}{c} \sum_{l=0}^{\infty} 
  j_l(kr_0)\,h_l^+(kr) \sum_{m=-l}^{+l} Y_l^{\ m}(\theta,\phi)\, 
  \\ \nonumber &~& \int_{-1}^{+1} d\cos\theta' 
  \int_0^{2\pi}d\phi'\,\cos(\phi')\,
  \overline{Y}_l^{\ m}(\theta',\phi')\,.
\end{eqnarray}
As 
$$ \int_0^{2\pi}d\phi'\, \cos(\phi') = 0 $$
holds, it follows that $Y_0^{\ 0}$, $Y_1^{\ 0}$ and $Y_2^{\ 0}$ do not 
contribute. For $Y_1^{\ 1}=-\sqrt{3}\,\sin\theta\,e^{i\phi}/\sqrt{8\pi}
$, and similarly $Y_1^{-1}$, we have with $x'=\cos\theta'$
$$ \int_{-1}^{+1} dx'\,\sqrt{1-x'^2}
   \,\left[\delta(x_0-x')-\delta(x_0+x')\right]\ =\ 0\,, $$
and for $Y_2^{\ 2}$
$$  \int_0^{2\pi}d\phi'\,\cos(\phi')\,e^{i\,2,\phi'} = 0\,. $$
So, 
$$ Y_2^{\ 1} = - \sqrt{\frac{15}{8\pi}}\, \sin\theta\,
   \cos\theta\, e^{i\,\phi}, $$
and similarly $Y_2^{-1}$, give identical leading contributions, which 
are electric quadrupoles. We have:
$$ \int_0^{2\pi}d\phi'\,\cos(\phi')\,e^{-i\,\phi'} = \pi\,, $$
\begin{eqnarray} \nonumber
  &~& \int_{-1}^{+1} dx'\,\sqrt{1-x'^2}\,x\,
  \left[\delta(x_0-x')-\delta(x_0+x')\right]\\ \nonumber
  &=& 2\,\sqrt{1-x_0^2}\,x_0\ =\ 2\,\sin(\theta_0)\,\cos(\theta_0)\,,
\end{eqnarray}
and, adding $Y_2^{\ 1}$ and $Y_2^{-1}$ contributions, in this 
approximation the potential is
$$ A_{\phi} = -\frac{I\,r_0}{c}\,\sqrt{30\pi}\, 
   \sin(\theta_0)\,\cos(\theta_0)\,i\,k\,4\pi\,j_2(kr_0)\,
   h_2^+(kr)\,Y_2^{\ 1}(\theta,\phi=0)\,.  $$
In the $r\to\infty$ far field approximation we have
$$ h_2^+(kr) = i\,\frac{e^{i\,k\,r}}{r} $$
and, consequently,
$$ A_{\phi}(r,\theta) = A^0_{\phi}\,\sin(\theta)\,\cos(\theta)\,
                        \frac{e^{i\,k\,r}}{r}\  $$
with 
$$ A^0_{\phi}\ =\ - 30\pi\,k\,\frac{I\,r_0}{c}\,j_2(kr_0)\,
   \sin(\theta_0)\,\cos(\theta_0) \,. $$
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