The Plane

In rectangular coordinates, it is clear that $\dot{v}$ will be zero if and only if $\ddot{x}=0=\ddot{y}$; the geodesics are straight lines. Consider this same problem in polar coordinates, where \begin{equation} d\rr = dr\,\rhat + r\,d\phi\,\phat \end{equation} Then \begin{equation} \vv = \dot{r}\,\rhat + r\,\dot\phi\,\phat \end{equation} so that \begin{align} d\vv &= d\dot{r}\,\rhat + \dot{r}\,d\rhat + d(r\dot\phi)\,\phat + r\dot\phi\,d\phat \nonumber\\ &= d\dot{r}\,\rhat + \dot{r}\,(d\phi\,\phat) + d(r\dot\phi)\,\phat - r\dot\phi\,(d\phi\,\rhat) \end{align} Dividing by $dt$ now yields \begin{align} \dot\vv &= (\ddot{r}-r\dot\phi^2)\,\rhat + \bigl((r\dot\phi)\dot{\hphantom{\phi}}+\dot{r}\dot\phi\bigr)\,\phat \nonumber\\ &= (\ddot{r}-r\dot\phi^2)\,\rhat + (r\ddot\phi+2\dot{r}\dot\phi)\,\phat \end{align} so that the geodesic equation in polar coordinates reduces to \begin{equation} \ddot{r}-r\dot\phi^2 = 0 = r\ddot\phi+2\dot{r}\dot\phi \label{Geopolar} \end{equation}

We will discuss solution strategies for this system of ordinary differential equations later on, and content ourselves here with checking special cases. A radial line through the origin has $\phi=\hbox{constant}$, so that ($\ref{Geopolar}$) is satisfied if $\ddot{r}=0$. But the speed in this case is just $v=|\vv|=|\dot{r}|$, so this condition says that radial lines are geodesics if they are traversed at constant speed.

A less obvious example is a line of the form $x=\hbox{constant}$, so that \begin{equation} r\cos\phi = \hbox{constant} \end{equation} which implies that \begin{equation} \dot{r}\cos\phi - r\sin\phi\,\dot{\phi} = 0 \end{equation} and further differentiation yields \begin{equation} (\ddot{r}-r\dot\phi^2)\cos\phi - (r\ddot\phi+2\dot{r}\dot\phi)\sin\phi = 0 \label{gp1} \end{equation} If we also assume that the line is traversed at constant speed, we must have \begin{equation} 0 = \ddot{y} = (\ddot{r}-r\dot\phi^2)\sin\phi + (r\ddot\phi+2\dot{r}\dot\phi)\cos\phi \label{gp2} \end{equation} Comparing ($\ref{gp1}$) and ($\ref{gp2}$), we see that ($\ref{Geopolar}$) is indeed satisfied.

The Sphere

On the sphere, we have \begin{equation} d\rr = r\,d\theta\,\that + r\,\sin\theta\,d\phi\,\phat \end{equation} so that \begin{equation} \vv = r\,\dot\theta\,\that + r\,\sin\theta\,\dot\phi\,\phat \end{equation} which implies that \begin{align} d\vv &= r\,d\dot\theta\,\that + r\,d(\sin\theta\,\dot\phi)\,\phat + r\,\dot\theta\,d\that + r\,\sin\theta\,\dot\phi\,d\phat \nonumber\\ &= r\,d\dot\theta\,\that + r\,d(\sin\theta\,\dot\phi)\,\phat + r\,\dot\theta\,\cos\theta\,d\phi\,\phat - r\,\sin\theta\,\dot\phi\,\cos\theta\,d\phi\,\that \end{align} Dividing by $dt$ yields \begin{align} d\vv &= r (\ddot\theta - \sin\theta\cos\theta\,\dot\phi^2) \,\that + r \bigl((\sin\theta\,\dot\phi)\dot{\hphantom{\phi}} + \cos\theta\,\dot\theta\,\dot\phi\bigr) \,\phat \nonumber\\ &= r\, (\ddot\theta - \sin\theta\cos\theta\,\dot\phi^2) \,\that + r\sin\theta\, (\ddot\phi + 2\cot\theta\,\dot\theta\,\dot\phi) \,\phat \end{align} so that the geodesic equation on the sphere reduces to \begin{equation} \ddot\theta - \sin\theta\cos\theta\,\dot\phi^2 = 0 = \ddot\phi + 2\cot\theta\,\dot\theta\,\dot\phi \label{Geosphere} \end{equation}

We again defer a discussion of how to solve ($\ref{Geosphere}$), and content ourselves with a simple example. Along the equator, we have $\theta=\pi/2$ and $\ddot\phi=0$, which shows that the equator is a geodesic, and hence that all great circles are geodesics.