\noindent {\bf Problem (\ref{ex_TE11mode}): Cylindrical waveguide, $\rm TE_{11}$ mode power transmission.} \smallskip \noindent The cutoff frequency follow 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 Poynting vector one can show show (see \cite{Jackson} chapter 8.5) that the time-averaged power flow through the cylindrical waveguide in the TE mode is (Gaussian units) $$ P = \frac{c\,\mu}{8\pi\,\sqrt{\mu\epsilon}}\,\left(\frac{\omega} {\omega_{11}}\right)^2\,\left(1-\frac{\omega_{11}^2}{\omega^2} \right)^{1/2} \int_A da\,\overline{\psi}\,\psi $$ 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 Problem (\ref{ex_rad_quad}): Electric quadrupole approximation for the magnetic field of the radiation zone.} \smallskip \noindent We define the integrals $$ \vec{I}_B = \hat{r}\times\int \vec{x}^{\,'}\, (\hat{r}\cdot\vec{x}^{\,'})\,\rho(\vec{x}^{\,'})\,d^3x' $$ and show $$ \vec{I}_B = \frac{1}{3}\,\hat{r}\times\vec{Q}(\hat{r})\,, ~~{\rm where}~~ \hat{r} = \frac{\vec{r}}{r} = \left( \begin{matrix} \hat{r}^1\cr \hat{r}^2\cr \hat{r}^3 \end{matrix} \right)\,. $$ First, note that $$ Q^i(\hat{r}) = \sum_{j=1}^3 Q^{ij}\,\hat{r}^j = \int \left[ 3\,x'^i\,(\hat{r}\cdot\vec{x}^{\,'})-r'^2\,\hat{r}^i\right]\, \rho(\vec{x}^{\,'})\,d^3x'$$ holds. Now $$ I_B^1 = \hat{r}^2 \int x'^3\,(\hat{r}\cdot\vec{x}^{\,'})\, \rho(\vec{x}^{\,'})\,d^3x' - \hat{r}^3 \int x'^2\,(\hat{r}\cdot\vec{x}^{\,'})\, \rho(\vec{x}^{\,'})\,d^3x' $$ which agrees with $$ \frac{1}{3}\,\left[\hat{r}\times\vec{Q}(\hat{r})\right]^1 = \frac{1}{3}\,\left[ \hat{r}^2\,Q^3(\hat{r}) - \hat{r}^3\,Q^2(\hat{r}) \right] $$ $$ = \hat{r}^2 \int x'^3\,(\hat{r}\cdot\vec{x}^{\,'})\, \rho(\vec{x}^{\,'})\,d^3x' - \hat{r}^3 \int x'^2\,(\hat{r}\cdot\vec{x}^{\,'})\, \rho(\vec{x}^{\,'})\,d^3x' $$ $$ - \frac{1}{3}\,\hat{r}^2\,\hat{r}^3 \int r'^2\,\rho(\vec{x}^{\,'}) \,d^3x' + \frac{1}{3}\,\hat{r}^3\,\hat{r}^2 \int r'^2\, \rho(\vec{x}^{\,'})\,d^3x' $$ because the last line adds to zero. The other components follow in the same way. \bigskip \noindent {\bf Problem (\ref{ex_Hertz}):} Hertz vector. \smallskip \noindent (1) We have $$ F^{\alpha\beta} = \partial^{\alpha} A^{\beta} - \partial^{\beta} A^{\alpha}\,,\qquad \left(A^{\alpha}\right) =\ \left( \begin{matrix} \Phi & \\ \vec{A} & \end{matrix} \right)\ .$$ 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 $$ \nabla^2\Phi - \frac{1}{c^2}\frac{\partial^2\Phi}{\partial t^2} = -4\pi\,\rho\,,\qquad \nabla^2\vec{A} - \frac{1}{c^2}\frac{\partial^2\vec{A}}{\partial t^2} = -\frac{4\pi}{c}\,\vec{J}\ . $$ (2) Continuity equation: $$ \frac{\partial\rho}{\partial t} + \nabla\cdot\vec{J} = - \frac{\partial~}{\partial t}\,\left(\nabla\cdot\vec{p}\,\right) + \nabla\cdot\left(\frac{\partial\vec{p}}{\partial t}\right) = 0\ .$$ (3) 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} \,=\, 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} \,=\, - \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: $$ \frac{1}{c}\frac{\partial\Phi}{\partial\,t} + \nabla\cdot\vec{A} = - \frac{1}{c}\frac{\partial~}{\partial t} \left(\nabla\cdot\vec{\Pi}\right) + \frac{1}{c}\,\nabla\cdot \frac{\partial\vec{\Pi}}{\partial\,t} = 0\ . $$ (4) For 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. \hfil\break % \vskip 10pt