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\begin{document} \huge

\begin{center}
{Mathematical Physics --- PHZ 3113}\\\vspace {2mm} 
{Solutions Midterm~1 (February 18, 2013)}  % \\
\end{center}
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\centerline{------------------------------------------------------}

\bigskip

\begin{enumerate}

\item Calculate the gradient of the 3D potential (in arbitrary units) 
$\frac{1}{2}\,r^2$.

\noindent Solution (with Einstein convention):
\begin{eqnarray} \nonumber
  \nabla\,\frac{1}{2}\,r^2 &=& r\,\nabla\,r = r\,\hat{x}_i\,\partial_i\,r
  = r\,\hat{x}_i\,\frac{x_i}{r} \\ \nonumber &=&
    r\,\frac{\vec{r}}{r}= r\,\hat{r} = \vec{r}\,. 
\end{eqnarray} 

\item A force in 3D is (in arbitrary units) given by $\vec{F} =
-r\,\vec{r}$. Use (with Einstein convention)
\begin{eqnarray} \nonumber
  \nabla\times r\,\vec{r}\ =\ \epsilon_{ijk}\,\hat{x}_i\,\partial_j\,r\,x_k
\end{eqnarray} 
to calculate the curl of this force.

\noindent Solution:
\begin{eqnarray} \nonumber
  \epsilon_{ijk}\,\hat{x}_i\,\partial_j\,r\,x_k &=&    \\ \nonumber
      \epsilon_{ijk}\,\hat{x}_i\,x_k\,\partial_j\,r 
  &+& \epsilon_{ijk}\,\hat{x}_i\,r\,\partial_j\,x_k\ = \\ \nonumber
      \epsilon_{ijk}\,\hat{x}_i\,x_k\,\frac{x_j}{r}
  &+& \epsilon_{ijk}\,\hat{x}_i\,r\,\delta_{j\,k}\ = \\ \nonumber
      \frac{1}{r}\,\epsilon_{ijk}\,\hat{x}_i\,x_j\,x_k
  &+& \epsilon_{ijj}\,\hat{x}_i\,r\ =\ 0+0\ =\ 0\,. \\ \nonumber
\end{eqnarray} 
The first term is zero, because it is the contraction of an 
anti-symmetric with a symmetric tensor, the second because of
$\epsilon_{ijj}=0$.

\item Calculate the curl of the force
\begin{eqnarray} \nonumber
  \vec{F} = x_2\,\hat{x}_1+x_3\,\hat{x}_2+x_1\,\hat{x}_3 
          = \left(\matrix{x_2\cr x_3\cr x_1}\right)\,.
\end{eqnarray} 

\noindent Solution:
\begin{eqnarray} \nonumber
  \nabla\times\vec{F}&=&\hat{x}_1\,(\partial_2\,F_3-\partial_3\,F_2)
  \\ \nonumber       &+&\hat{x}_2\,(\partial_3\,F_1-\partial_1\,F_3)
  \\ \nonumber       &+&\hat{x}_3\,(\partial_1\,F_2-\partial_2\,F_1)
  \\ \nonumber &=&\hat{x}_1\,(0-1)+\hat{x}_2\,(0-1)+\hat{x}_3\,(0-1)
  \\ \nonumber &=&-\hat{x}_1-\hat{x}_2-\hat{x}_3\ = 
                  -\left(\matrix{1\cr 1\cr 1}\right)\,.
\end{eqnarray} 

\item A point mass on an inclined plane experiences under gravity
a force $\vec{F}=-F\,\hat{z}$ in downward direction (see the figure).
Find the magnitudes of the forces perpendicular and parallel to the
plane as function of the angle $\alpha$.

\noindent Solution:
\begin{eqnarray} \nonumber
  F_{\perp}\ =\ F\,\cos(\alpha)\,,\qquad
  F_{\|}\ =\ F\,\sin(\alpha)\,.
\end{eqnarray} 

\item A point mass is suspended as shown in the figure. As in the
previous problem it experiences a force $\vec{F}=-F\,\hat{z}$ in 
downward direction. Find the $z$ components of the tensions 
$\vec{T}^{\,1}$ and $\vec{T}^{\,2}$ as functions of the angles 
$\alpha_1$ and $\alpha_2$. In which limit becomes $T^1_x$ infinite?

\noindent Solution: There are four equations for the four unknowns
$T^1_x$, $T^1_z$, $T^2_x$ and $T^2_z$ as functions of three free
parameters $F$, $\alpha_1$, $\alpha_2$: 
\vfil\eject

\begin{eqnarray} \nonumber
  T^1_z + T^2_z &=& F\,, \\ \nonumber
  T^1_x + T^2_x &=& 0\,, \\ \label{T1x}
  \tan\alpha_1  &=& \frac{T^1_z}{T^1_x}\,,\\ \nonumber
  \tan\alpha_2  &=& \frac{T^2_z}{-T^2_x}\ =\ \frac{T^2_z}{T^1_x}\,.
\end{eqnarray} 
Now,
\begin{eqnarray} \nonumber
  \frac{\tan\alpha_1}{\tan\alpha_2} = 
               \frac{T^1_z}{T^2_z}\,,&~& T^2_z =  
  \frac{\tan\alpha_2}{\tan\alpha_1}\,T^1_z\,, \\ \nonumber
  T^1_z\,\left( 1 + \frac{\tan\alpha_2}{\tan\alpha_1}\right)
  &=& F\,,~~ T^1_z = 
  \frac{F\,\tan\alpha_1}{\tan\alpha_1+\tan\alpha_2}\,, \\ \nonumber
  T^2_z\,\left( 1 + \frac{\tan\alpha_1}{\tan\alpha_2}\right)
  &=& F\,,~~ T^2_z = 
  \frac{F\,\tan\alpha_2}{\tan\alpha_1+\tan\alpha_2}\,.
\end{eqnarray} 
It follows from (\ref{T1x}) that 
\begin{eqnarray} \nonumber
  T^1_x = T^1_z\,\tan\alpha_1 = \frac{F}{\tan\alpha_1+\tan\alpha_2}
\end{eqnarray} 
becomes infinite for $\alpha_1\to 0$ and $\alpha_2\to 0$.

\vfil\eject
Second solution:
\begin{eqnarray} \nonumber
  T^1\,\sin\alpha_1 + T^2\,\sin\alpha_2 &=& F\,, \\ \nonumber
  T^1\,\cos\alpha_1 - T^2\,\cos\alpha_2 &=& 0\,. 
\end{eqnarray} 
Solving for $T^1$:
\begin{eqnarray} \nonumber
  T^2 &=& T^1\ \frac{\cos\alpha_1}{\cos\alpha_2}\,, \\ \nonumber
  F   &=& T^1\,\left(\sin\alpha_1+\sin\alpha_2\,
  \frac{\cos\alpha_1}{\cos\alpha_2}\right)\,, \\ \nonumber
  T^1 &=& F\,\frac{\cos\alpha_2}{\sin\alpha_1\,\cos\alpha_2
          +\cos\alpha_1\,\sin\alpha_2}\,, 
\end{eqnarray} 
\begin{eqnarray} \label{T1}
  T^1 &=& F\,\frac{\cos\alpha_2}{\sin(\alpha_1+\alpha_2)}\,.
\end{eqnarray} 
Solving for $T^2$:
\begin{eqnarray} \nonumber
  T^1 &=& T^2\ \frac{\cos\alpha_2}{\cos\alpha_1}\,, \\ \nonumber
  F   &=& T^2\,\left(\sin\alpha_2+\sin\alpha_1\,
  \frac{\cos\alpha_2}{\cos\alpha_1}\right)\,, \\ \nonumber
  T^2 &=& F\,\frac{\cos\alpha_1}{\sin\alpha_2\,\cos\alpha_1
          +\cos\alpha_2\,\sin\alpha_1}\,, 
\end{eqnarray} 
\begin{eqnarray} \label{T2}
  T^2 &=& F\,\frac{\cos\alpha_1}{\sin(\alpha_1+\alpha_2)}\,.
\end{eqnarray} 
The components follow from
\begin{eqnarray} \nonumber
  T^1_x &=& T^1\,\cos\alpha_1\ =\ -T^2_x\,, \\ \nonumber
  T^1_z &=& T^1\,\sin\alpha_1\,, \\ \nonumber
  T^2_z &=& T^2\,\sin\alpha_2\,.
\end{eqnarray} 

\end{enumerate}

\end{document}
