On the sphere, we have \begin{equation} d\rr = r\,d\theta\,\that + r\sin\theta\,d\phi\,\phat \end{equation} and the line element is \begin{equation} ds^2 = r^2\,d\theta^2 + r^2\,\sin^2\theta\,d\phi^2 \end{equation} Both of these expressions depend explicitly on $\theta$, but not $\phi$ (or $r$, which is constant). Thus, the line element does not change in the $\phi$ direction, and we again expect “$\Partial{}{\phi}$” to be a Killing vector.

As before, we compute \begin{equation} \grad f = \frac{1}{r}\Partial{f}{\theta}\,\that + \frac{1}{r\sin\theta}\,\Partial{f}{\phi}\,\phat \end{equation} so that \begin{equation} \Partial{f}{\phi} = \grad f\cdot r\sin\theta\,\phat. \end{equation} Thus, the vector representation of $\Partial{}{\phi}$ is the vector \begin{equation} \Pvec = r\sin\theta\,\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{align} d\Pvec &= d(r\sin\theta\,\phat) \nonumber\\ &= r\cos\theta\,d\theta\,\phat + r\sin\theta\,d\phat \nonumber\\ &= r\cos\theta\,d\theta\,\phat - r\sin\theta\cos\theta\,d\phi\,\that \end{align} which is indeed orthogonal to $d\rr$ as expected.

As before, since \begin{equation} \vv = \dot\rr = r\,\dot\theta\,\that + r\sin\theta\,\dot\phi\,\phat \end{equation} we must have \begin{equation} \Pvec\cdot\vv = r^2\sin^2\theta\,\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 $z$-axis (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 = r^2\dot\theta^2 + r^2\sin^2\theta\,\dot\phi^2 = r^2\dot\theta^2 + \frac{\ell^2}{r^2\sin^2\theta} \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\sin^2\theta} \\ \dot\theta &= \pm\frac{1}{r}\sqrt{1-\frac{\ell^2}{r^2\sin^2\theta}} \end{align} Compare this derivation with the ad hoc derivation given in § Solving the Geodesic Equation, then see § Geodesics on the Sphere for the solution of this system of first-order equations.

What do these solutions represent? What are the “straight lines” on a sphere? Great circles, that is, diameters such as the equator, dividing the sphere into two equal halves. (Lines of latitude other than the equator are not great circles, and are not “straight”. Among other things, they do not represent the shortest distance between two points on the sphere, as anyone who has flown across a continent or an ocean is likely aware.)