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\begin{document} \Large

\begin{center}
{Mathematical Physics --- PHZ 3113}\\\vspace {2mm} 
{Solutions Midterm~2 (March 18, 2013)}  % \\
\end{center}
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\centerline{------------------------------------------------------}

\bigskip

\begin{enumerate}

\item Use the definition $\nabla\times\vec{A}=\epsilon_{ijk}\hat{x}_i
\partial_jA_k$ (with Einstein convention) and properties of the 
Levi-Civita tensor $\epsilon_{ijk}$ to transform
\noindent 
\begin{eqnarray} \nonumber
  \nabla\times\nabla\times\vec{A}
\end{eqnarray} 
into applications of the $\nabla$ operator, which do no longer
involve the curl. 
\bigskip

\noindent Solution:
\begin{eqnarray} \nonumber
&~& \nabla\times\nabla\times\vec{A}\ =\ \epsilon_{ijk}\hat{x}_i\partial_j
    \epsilon_{klm} \partial_l A_m \\ \nonumber
&=& \left(\delta_{il} \delta_{jm}-\delta_{im} \delta_{jl}\right)\,
    \hat{x}_i\partial_j\partial_l A_m \\ \nonumber
&=& \hat{x}_i\partial_j\partial_i A_j-\hat{x}_i\partial_j\partial_j A_i 
    \\ \nonumber &=& (\hat{x}_i\partial_i)\,(\partial_jA_j)
                  -  (\partial_j\partial_j)\,(\hat{x}_iA_i)\\ \nonumber 
&=& \nabla\,\left(\nabla\cdot\vec{A}\,\right) - \nabla^2\,\vec{A}\,.
\end{eqnarray} 
\bigskip

\item Calculate the volume of an ellipsoid
$$ V\ =\ \int_{\left(\frac{x_1}{a}\right)^2+\left(\frac{x_2}{b}\right)^2
              +\left(\frac{x_3}{c}\right)^2 \le 1} d^3x\,.            $$
Hint: With suitable substitutions the integral can be
mapped on the integral for the volume of a sphere.
\bigskip

\noindent Solution: Let $x_1=a\,x_1'$, $x_2=b\,x_2'$ and $x_3=c\,x_3'$. 
With these substitution we find
\begin{eqnarray} \nonumber
  V &=& a\,b\,c \int_{x_1'^{\,2}+x_2'^{\,2}+x_3'{\,^2} \le 1} d^3x'\\ 
    \nonumber &=& a\,b\,c \int_0^1 r'^{\,2}\,dr' \int_{-1}^{+1} 
                  d\cos\theta' \int_0^{2\pi} d\phi'\\ 
    \nonumber &=& \frac{4\pi}{3}\,a\,b\,c\,.
\end{eqnarray} 
\bigskip

\item Calculate $$\vec{v}=\vec{\omega}\times\vec{r}$$ and express the
result in spherical coordinates. Here $\vec{\omega}=\dot{\phi}\,\hat{z}$
is the angular velocity and $\vec{r}=r\,\hat{r}$ the position vector.
% Use then cylindrical coordinates to calculate $\nabla\times\vec{v}$.
\bigskip

\noindent Solution: $$\vec{v}=
\dot{\phi}\,r\,\sin(\theta)\,\hat{\phi}=\dot{\phi}\,\rho\,\hat{\phi}\,.$$
% As $\vec{v}$ is of the form $v_{\phi}\,\hat{\phi}$ we have
% \begin{eqnarray} \nonumber
%  \nabla\times\vec{v} 
%  &=& \frac{1}{\rho}\frac{\partial~}{\partial\rho}\,\left(\rho\,v_{\phi}
%  \right)\,\hat{z} - \frac{\partial\,v_{\phi}}{\partial\,z}\,\hat{\rho} 
%  \\ \nonumber &=&\frac{1}{\rho}\frac{\partial~}{\partial\rho}\,
%  \left(\rho^2\,\dot{\phi}\right)\,\hat{z} - 
%  \frac{\partial~}{\partial z}\,\left(\rho\,\dot{\phi}\right)\,\hat{\rho} 
%  \\ \nonumber &~&\\ \nonumber &=&2\,\dot{\phi}\,\hat{z}=2\,\vec{\omega}\,. 
% \end{eqnarray} 
\bigskip

\item Calculate the time derivative
$$ \dot{\hat{r}}\ =\ \frac{d\,\hat{r}}{d\,t}$$
of the unit vector
$$ \hat{r}\ =\ \frac{\vec{r}}{\left|\vec{r}\right|} $$
and express the result in spherical coordinates. 
\bigskip

\noindent Solution: From
\begin{eqnarray} \nonumber
  \hat{r}&=&\sin(\theta)\,\cos(\phi)\,\hat{x} +
            \sin(\theta)\,\sin(\phi)\,\hat{y} +
            \cos(\theta)\,\hat{z}\,, \\ \nonumber
  \hat{\theta}&=&\cos(\theta)\,\cos(\phi)\,\hat{x} +
                 \cos(\theta)\,\sin(\phi)\,\hat{y} -
                 \sin(\theta)\,\hat{z}\,, \\ \nonumber
  \hat{\phi}&=& -\sin(\phi)\,\hat{x}+\cos(\phi)\,\hat{y}
\end{eqnarray} 
we get
\begin{eqnarray} \nonumber
  \frac{\partial\,\hat{r}}{\partial\,r} &=& 0\,, \\ \nonumber
  \frac{\partial\,\hat{r}}{\partial\,\theta}
   &=&\cos(\theta) \cos(\phi)\,\hat{x} +
      \cos(\theta) \sin(\phi)\,\hat{y} \\ \nonumber
   &-&\sin(\theta)\,\hat{z}\ =\ \hat{\theta}\,,\\ \nonumber
  \frac{\partial\,\hat{r}}{\partial\,\phi} 
  &=&-\sin(\theta)\,\sin(\phi)\,\hat{x} +
      \sin(\theta)\,\cos(\phi)\,\hat{y} \\ \nonumber
  &=& \sin(\theta)\,\hat{\phi}\,.
\end{eqnarray} 
Therefore, 
\begin{eqnarray} \nonumber
  \dot{\hat{r}}\ =\ 
  \frac{\partial\,\hat{r}}{\partial\,\theta}\,\dot{\theta} 
  + \frac{\partial\,\hat{r}}{\partial\,\phi}\,\dot{\phi} \ =\ 
  \dot{\theta}\,\hat{\theta} + \sin(\theta)\,\dot{\phi}\ \hat{\phi}\,.
\end{eqnarray} % \vfil\eject 
\bigskip

\item The orbit of a planet is assumed to be an ellipse, which in 
cylindrical coordinates is given by the equation
\begin{eqnarray} \nonumber
  \rho\ =\ \frac{p}{1+\epsilon\,\cos\phi}
\end{eqnarray} 
where $0\le\epsilon<1$ is the eccentricity and $p$ the latus rectum.
From the analytical solution of Kepler's problem it follows that
the magnitude of the angular momentum is (up to a correction for 
the finite sun mass) given by
$$ L={\rm const}\,m\,\sqrt{p}\,, $$
where $m$ is the mass of the planet. Use this information and angular 
momentum conservation to derive Kepler's 3rd law.
\bigskip

\noindent Solution: For $\phi=0$ we have ($a$ major half-axis and
$\epsilon$ eccentricity)
\begin{eqnarray} \nonumber
  a-\epsilon\,a =\ \frac{p}{1+\epsilon}\ \Rightarrow\
  p\ =\ a\,(1-\epsilon^2)\,.
\end{eqnarray} 
Angular momentum conservation implies 
\begin{eqnarray} \nonumber
  \frac{L}{m}\ =\ \rho^2\,\dot{\phi} &=& {\rm const}\,
  \sqrt{a\,(1-\epsilon^2)} \\ \nonumber 
  \int_0^{2\pi} \rho^2\,d\phi &=& {\rm const}\,
  \sqrt{a\,(1-\epsilon^2)}\,\int_0^T dt
\end{eqnarray} 
where $T$ is the orbital period. As 
$$ df=\frac{1}{2}\,\rho^2\,d\phi $$ 
is the infinitesimal sectorial area of an ellipse, the integral on the 
left side is two times the area of an ellipse, $2\pi\,a\,b$, where $b=a
\,\sqrt{1-\epsilon^2}$ is the minor axis of the ellipse. Collecting 
the terms we have
\begin{eqnarray} \nonumber
  2\pi\,a^2\ =\ {\rm const}\,\sqrt{a}\,T\,.
\end{eqnarray} 
Squaring both sides, we arrive at Kepler's 3rd law:
\begin{eqnarray} \nonumber
  a^3\ =\ \ {\rm const'}\,T^2\,,
\end{eqnarray} 
where $\rm const'$ is a new constant. %  \bigskip
\vfil

\end{enumerate}

\end{document}
