A fundamental principle of general relativity is that freely falling objects travel along timelike geodesics. So what are those geodesics?

Taking the square root of the line element, the infinitesimal vector displacement in the Schwarzschild geometry is \begin{equation} d\rr = \sqrt{1-\frac{2m}{r}}\,dt\,\That + \frac{dr\,\rhat}{\sqrt{1-\frac{2m}{r}}} + r\,d\theta\,\that + r\sin\theta\,d\phi\,\phat \label{drschw} \end{equation} where \begin{align} \That\cdot\That &= -1 \\ \rhat\cdot\rhat &= 1 = \that\cdot\that = \phat\cdot\phat \end{align} with all other dot products between these basis vectors vanishing due to orthogonality. Spherical symmetry allows us to assume without loss of generality that any geodesic lies in the equatorial plane, so we can set $\sin\theta=1$ and $d\theta=0$, resulting in \begin{equation} d\rr = \sqrt{1-\frac{2m}{r}}\,dt\,\That + \frac{dr\,\rhat}{\sqrt{1-\frac{2m}{r}}} + r\,d\phi\,\phat \end{equation} or equivalently \begin{equation} \vv = \dot{\rr} = \sqrt{1-\frac{2m}{r}}\,\dot{t}\,\That + \frac{\dot{r}\,\rhat}{\sqrt{1-\frac{2m}{r}}} + r\,\dot\phi\,\phat \end{equation} where a dot denotes differentiation with respect to proper time along the geodesic.

What other symmetries does the Schwarzschild geometry have? The line element depends on $r$ (and $\theta$), but not on $t$ or $\phi$. Thus, we have two Killing vectors, namely 1) \begin{align} \Tvec &= \sqrt{1-\frac{2m}{r}}\,\That \\ \Pvec &= r\,\phat \end{align} since \begin{align} \Tvec\cdot\grad f &= \Partial{f}{t} \\ \Pvec\cdot\grad f &= \Partial{f}{\phi} \\ \end{align} where we have used the fact that $\theta=\pi/2$ (and that $\grad f\cdot d\rr=df$). Thus, we obtain two constants of the motion, namely \begin{align} \ell &= \Pvec\cdot\vv = r^2 \dot\phi \\ e &= -\Tvec\cdot\vv = \left(1-\frac{2m}{r}\right)\dot{t} \end{align} where $\vv=\dot\rr$, and where we have introduced a conventional minus sign so that $e>0$ (for future-pointing timelike geodesics). We interpret $\ell$ as the angular momentum and $e$ as the energy of the freely falling object. 2)

Our final equation is obtained from the line element itself. Since a freely falling object moves along a timelike geodesic, we have chosen proper time $\tau$ as our parameter. Since \begin{equation} d\tau^2 = -ds^2 = \left(1-\frac{2m}{r}\right)\,dt^2 - \frac{dr^2}{1-\frac{2m}{r}} - r^2\,d\phi^2 \end{equation} we can divide both sides by $d\tau^2$ to obtain \begin{equation} 1 = \left(1-\frac{2m}{r}\right)\,\dot{t}^2 - \frac{\dot{r}^2}{1-\frac{2m}{r}} - r^2\,\dot\phi^2 \end{equation} where we have again used the fact that $\theta=\pi/2$.

Putting these expressions together, the geodesic equation in the Schwarzschild geometry reduces to the system of equations \begin{align} \dot\phi &= \frac{\ell}{r^2} \\ \dot{t} &= \frac{e}{1-\frac{2m}{r}} \\ \dot{r}^2 &= e^2 - \left(1+\frac{\ell^2}{r^2}\right)\left(1-\frac{2m}{r}\right) \label{rdote0} \end{align} Before considering solutions of these equations, we first discuss the Newtonian case.