\documentclass[12pt]{article} % BB Jan 18 2013.
\textheight 9.0in
\textwidth  6.0in
\topmargin -1.0in  
\headsep    0.4in  
\oddsidemargin  0.25in  
\evensidemargin 0.25in  

\def\sqr#1#2{{\vcenter{\hrule height.#2pt
   \hbox{\vrule width.#2pt height#1pt \kern#1pt
      \vrule width.#2pt}
   \hrule height.#2pt}}}
\def\square{{\mathchoice\sqr64\sqr64\sqr{4.2}3\sqr{3.0}3}}
\def\undertext#1{$\underline{\smash{\hbox{#1}}}$}
% \pagestyle{empty}

\begin{document} \huge

\begin{center}
{Mathematical Physics --- PHZ 3113}\\\vspace {2mm} 
Curl Homework (January 30, 2013) \\
\end{center}
\setcounter{equation}{0}

\noindent Use the identity (with Einstein convention)
\begin{eqnarray} \label{epsdel}
  \epsilon_{ijk}\epsilon_{klm} =
  \delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}
\end{eqnarray} 
to solve the following assignments.

\begin{enumerate}
\item Show (Book (1.89) p.51)
\begin{eqnarray} \nonumber
  \vec{B}\times\left(\nabla\times\vec{A}\right) +
  \vec{A}\times\left(\nabla\times\vec{B}\right)&=&\\ 
  \nabla\,\left(\vec{A}\cdot\vec{B}\right) -
  \left(\vec{B}\cdot\nabla\right)\,\vec{A} -
  \left(\vec{A}\cdot\nabla\right)\,\vec{B}&.&~~~
\end{eqnarray} 
Solution ($\partial_i = \frac{\partial~~}{\partial x_i}$ in
the following):
\begin{eqnarray} \nonumber
  &~&\vec{B}\times\left(\nabla\times\vec{A}\right) +
  \vec{A}\times\left(\nabla\times\vec{B}\right) \\ \nonumber
  &=&\epsilon_{ijk}\epsilon_{klm}\hat{x}_iB_j\partial_lA_m +
    \epsilon_{ijk}\epsilon_{klm}\hat{x}_iA_j\partial_lB_m
  \\ \nonumber &=&
  \left(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}\right)\,
  \hat{x}_iB_j\partial_lA_m\ + \\ \nonumber &~&
  \left(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}\right)\,
  \hat{x}_iA_j\partial_lB_m \\ \nonumber &=&
  \hat{x}_iB_j\partial_iA_j - \hat{x}_iB_j\partial_jA_i\ + 
  \\ \nonumber &~&
  \hat{x}_iA_j\partial_iB_j - \hat{x}_iA_j\partial_jB_i\\ \nonumber 
  &=& \hat{x}_i\partial_i\,\left(B_jA_j\right) - 
      \left(\vec{B}\cdot\nabla\right)\,\vec{A} -
      \left(\vec{a}\cdot\nabla\right)\,\vec{B} \\ \nonumber
  &=& \nabla\,\left(\vec{A}\cdot\vec{B}\right) -
  \left(\vec{B}\cdot\nabla\right)\,\vec{A} -
  \left(\vec{A}\cdot\nabla\right)\,\vec{B}\ .
\end{eqnarray} 

\item The vector potential $\vec{A}$ of a magnetic dipole moment
$\vec{m}$ is given by (Book 1.7.11 p.52)
\begin{eqnarray} \nonumber
  \vec{A} = \left(\frac{\mu_0}{4\pi}\right)\,
 \left(\frac{\vec{m}\times\vec{r}}{r^3}\right)\,.
\end{eqnarray} 
Calculate the magnetic field $\vec{B}=\nabla\times\vec{A}$.

Solution ($\partial_i = \frac{\partial~~}{\partial x_i}$ in
the following):
\begin{eqnarray} \nonumber
  &~& \left(\frac{4\pi}{\mu_0}\right)\nabla\times\vec{A} =
  \nabla\times\left(\frac{\vec{m}\times\vec{r}}{r^3}\right)\\ \nonumber
  &=&\epsilon_{ijk}\epsilon_{klm}\hat{x}_i\partial_j \frac{m_lx_m}{r^3}
     \\ \nonumber
  &=&\epsilon_{ijk}\epsilon_{klm}\hat{x}_i\frac{m_l\delta_{jm}}{r^3}
     - 3\,\epsilon_{ijk}\epsilon_{klm}\hat{x}_i\frac{m_lx_jx_m}{r^5}
     \\ \nonumber
  &=& \left(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}\right)\,
      \hat{x}_i\frac{m_l\delta_{jm}}{r^3}\ - \\ \nonumber
  &3& \left(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}\right)\,
      \hat{x}_i\frac{m_lx_jx_m}{r^5} \\ \nonumber
  &=& \frac{3\,\vec{m}}{r^3}- \frac{\vec{m}}{r^3}-\frac{3\,\vec{m}
      \,r^2}{r^5}+\frac{3\,\vec{r}\,\left(\vec{m}\cdot\vec{r}\right)}
      {r^5} \\ \nonumber
  &=& \frac{3\,\hat{r}\,\left(\vec{m}\cdot\hat{r}\right)-\vec{m}}{r^3}\,.
  \\ &~& \vec{B}\ =\ \left(\frac{\mu_0}{4\pi}\right)\,
     \frac{3\,\hat{r}\,\left(\vec{m}\cdot\hat{r}\right)-\vec{m}}{r^3}\,.
\end{eqnarray} 

Second solution:
\begin{eqnarray} \nonumber
  \nabla\times\left(\frac{\vec{m}\times\vec{r}}{r^3}\right) 
  &=& \frac{\vec{r}}{r^3}\cdot\nabla\,\vec{m}\ +\
  \vec{m} \left(\nabla\cdot\frac{\vec{r}}{r^3}\right) 
  \\ \nonumber &-& \vec{m}\cdot\nabla\,\frac{\vec{r}}{r^3}\ -\
     \frac{\vec{r}}{r^3}\,\nabla\cdot\vec{m} \\ \nonumber
  &=& \vec{m} \left(\nabla\cdot\frac{\vec{r}}{r^3}\right)\ -\
      \vec{m}\cdot\nabla\,\frac{\vec{r}}{r^3} 
\end{eqnarray} 
because the derivatives acting on the constant vector $\vec{m}$ are 
zero. Now,
\begin{eqnarray} \nonumber
  \left(\nabla\cdot\frac{\vec{r}}{r^3}\right) &=& \frac{3}{r^3}
  - \vec{r}\cdot\hat{r}\frac{3}{r^4}\ =\ 0\,,\qquad
  \\ \nonumber -\ \vec{m}\cdot\nabla\,\frac{\vec{r}}{r^3} &=&
  -\ \frac{1}{r^3} \sum_{i=1}^3 m_i\partial_i\sum_{j=1}^3\hat{x}_jx_j
  \\ \nonumber &~& -\ \vec{r}\,\sum_{i=1}^3 m_i\partial_i\frac{1}{r^3}
  \ =\\ \nonumber 
  -\ \frac{1}{r^3} \sum_{i=1}^3 \sum_{j=1}^3 m_i\hat{x}_j\delta_{ij}
  &+& \vec{r}\,\sum_{i=1}^3 m_i x_i \frac{3}{r^4}\ =
  \\ \nonumber -\ \frac{\vec{m}}{r^3}\ +\ 
  \vec{r}\,\frac{3\,\vec{m}\cdot\vec{r}}{r^4} &=& 
  \frac{3\,\hat{r}\,\left(\vec{m}\cdot\hat{r}\right)-\vec{m}}{r^3}
\end{eqnarray} 
and $\vec{B}$ as before.


\end{enumerate}

\end{document}
