% Bernd Berg Oct 27 2018.
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\centerline{\bf Special and General Relativity (PHZ 4601/5606) Fall 2018 
Solutions} \smallskip

\centerline{\bf Set 10} 

\begin{description} 
\item {\bf 24. Christoffel Symbols.}
\end{description}

\noindent (a) 
$$ ds^2 = dx^2 + dy^2 = dr^2 + r^2\,d\theta^2\ .$$
(b) \begin{eqnarray} \nonumber
  \left(g_{ij}\right) = \left(\,\begin{matrix} 1& 0\cr
                        0& r^2  \end{matrix}\,\right)\ .
\end{eqnarray}
(c) \begin{eqnarray} \nonumber
  \left(g^{ij}\right) = \left(\,\begin{matrix} 1& 0\cr
                        0& 1/r^2\end{matrix}\,\right)\ .
\end{eqnarray}
(d) We use 
\begin{eqnarray} \label{Gamma2}
  \Gamma^i_{jk} = \frac{1}{2}\,g^{im}\,\left( g_{mj,k} + 
  g_{mk,j} - g_{jk,m} \right) 
\end{eqnarray}
to calculate the Christoffel symbols. Note that $g_{ij}$ is constant
except for $g_{\theta\theta}$. Therefore, $g_{ij,k}=0$ except for
$$ g_{\theta\theta,r} = 2\,r\ .$$
In the following we use the summation convention for $i,\,j,\,k$ and 
$m$, but not for $\theta$ and $r$.

Inspecting Eq.~(\ref{Gamma2}) for $g_{\theta \theta,r}$ contributions, 
we find
\begin{eqnarray} \nonumber
  \frac{1}{2}\,g^{\theta\,m}\,g_{m\theta,r} &=& \frac{1}{2}\,
  g^{\theta\theta}\,g_{\theta\theta,r} = \frac{1}{r}\\ \nonumber
  \frac{1}{2}\,g^{r\,m}\,g_{\theta\theta,m} &=& \frac{1}{2}\,
  g^{r\,r}\,g_{\theta \theta,r} = r\ .
\end{eqnarray}
Therefore,
$$ \Gamma^\theta_{\theta\,r}=\Gamma^\theta_{r\,\theta}=\frac{1}{r}~~
   {\rm and}~~\Gamma^r_{\theta \theta}=-r\ . $$
All other $\Gamma^i_{jk}$ are zero.  \medskip

\noindent (e) So the equations of motion
$$ \ddot{x}^i + \Gamma^i_{jk}\,\dot{x}^j\,\dot{x}^k = 0 $$
become
$$ \ddot{r}+\Gamma^r_{\theta \theta}\,\dot{\theta}\,\dot{\theta}
   =  \ddot{r}-r\,\dot{\theta}\,\dot{\theta} = 0 $$
and
$$ \ddot{\theta}+\Gamma^\theta_{\theta\,r}\,\dot{\theta}\,\dot{r}
   +\Gamma^\theta_{r\,\theta}\,\dot{r}\,\dot{\theta} =
   \ddot{\theta}+\frac{2}{r}\,\dot{\theta}\,\dot{r} = 0\ .$$

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