In polar coordinates, we have \begin{equation} d\rr = dr\,\rhat + r\,d\phi\,\phat \end{equation} and the line element is \begin{equation} ds^2 = dr^2 + r^2\,d\phi^2 \end{equation} Both of these expressions depend explicitly on $r$, but not $\phi$. Thus, the line element does not change in the $\phi$ direction, and we expect “$\Partial{}{\phi}$” to be a Killing vector. But what vector is this?

It is easy to see that \begin{equation} df = \grad f\cdot d\rr \label{df} \end{equation} by computing both sides of the equation in rectangular coordinates. Since coordinates do not appear explicitly in (\ref{df}), it is a geometric statement, and must hold in any coordinate system. In fact, (\ref{df}) can be taken as the definition of the gradient in curvilinear coordinates. In polar coordinates, it now follows that \begin{equation} \grad f = \Partial{f}{r}\,\rhat + \frac{1}{r}\,\Partial{f}{\phi}\,\phat \end{equation} so that \begin{equation} \Partial{f}{\phi} = \grad f\cdot r\,\phat. \end{equation} Equivalently, to take the derivative in the $\phi$ direction, set $dr=0$ and divide both sides of ($\ref{df}$) by $d\phi$, then switch from ordinary derivatives to partial derivatives since we held $r$ constant. Thus, the vector representation of $\Partial{}{\phi}$ is the vector \begin{equation} \Pvec = r\,\phat \end{equation}

Is $\Pvec$ really a Killing vector? After working out the derivative of the basis vector $\phat$, it is straightforward to compute \begin{equation} d\Pvec = d(r\,\phat) = dr\,\phat + r\,d\phat = dr\,\phat - r\,d\phi\,\rhat \end{equation} which is indeed orthogonal to $d\rr$ as expected, in agreement with the general result in § Coordinate Symmetries.

Recall now that if $\XX$ is a Killing vector and $\vv$ is (the velocity vector of) a geodesic, then $\XX\cdot\vv$ must be constant along the geodesic. Since \begin{equation} \vv = \dot\rr = \dot{r}\,\rhat + r\,\dot\phi\,\phat \end{equation} we therefore have \begin{equation} \Pvec\cdot\vv = r^2\,\dot\phi = \ell \end{equation} with $\ell$ constant; $\ell$ is the constant of the motion associated with $\Pvec$, and turns out to represent angular momentum about the origin (per unit mass). We can obtain a further condition by using arclength as the parameter along the geodesic. Dividing the line element by $ds^2$ yields \begin{equation} 1 = \dot{r}^2 + r^2\,\dot\phi^2 = \dot{r}^2 + \frac{\ell^2}{r^2} \end{equation}

We have therefore replaced the system of second-order geodesic equations with the first-order system \begin{align} \dot\phi &= \frac{\ell}{r^2} \\ \dot{r} &= \pm\sqrt{1-\frac{\ell^2}{r^2}} \end{align} Compare this derivation with the ad hoc derivation given in § Solving the Geodesic Equation, then see § Geodesics in Polar Coordinates for the explicit solution of this system of first-order equations. What do these solutions represent? Straight lines in the plane, expressed in terms of polar coordinates.