Consider now radial geodesics, for which $\ell=0$, so that \begin{equation} \dot{r}^2 = e^2 - \left(1-\frac{2m}{r}\right) \end{equation} A radial geodesic represents a freely falling object with no angular momentum. The energy $e$ determines the radius $r_0$ at which the object is at rest, since if $\dot{r}=0$ at $r=r_0$ then \begin{equation} r_0 = \frac{2m}{1-e^2} \end{equation} Since $r$ must be nonnegative, we must have \begin{equation} e^2 \le 1 \end{equation} and we will assume that $e$ is positive. If $e^2>1$, then $\dot{r}$ is nowhere $0$.

We consider here the special case $e=1$, which corresponds to a freely falling object that starts at rest at $r=\infty$, and for which \begin{equation} \dot{r}^2 = \frac{2m}{r} \end{equation} so that 1) \begin{equation} \dot{r} = -\sqrt{\frac{2m}{r}} \end{equation} Since $e=1$, we also have \begin{equation} \dot{t} = \frac{1}{1-\frac{2m}{r}} \end{equation} Thus, a far-away observer ($r\gg m$) sees the object falling with speed \begin{equation} \frac{dr}{dt} = \frac{\dot{r}}{\dot{t}} = -\left(1-\frac{2m}{r}\right) \sqrt{\frac{2m}{r}} \end{equation} which approaches $0$ as $r$ approaches $2m$. What does a shell observer ($r=\hbox{constant}$) see?

A shell observer uses the line element to measure both distance $ds$ (with $t$ constant) and time $d\tau$ (with $r$ constant). For radial motion in the equatorial plane, both $d\theta$ and $d\phi$ are $0$, so we have \begin{align} ds &= \frac{dr}{\sqrt{1-\frac{2m}{r}}} \\ d\tau &= \sqrt{1-\frac{2m}{r}}\,dt \end{align} More generally, we have written the infinitesimal vector displacement $d\rr$ in terms of an orthonormal basis of vectors, so its components represent physical distances and times. For this reason, we often give these components names, and write \begin{equation} d\rr = \sigma^t\,\That + \sigma^r\,\rhat + \sigma^\theta\,\that + \sigma^\phi\,\phat \end{equation} and refer to $\{\sigma^i\}$ as an orthonormal basis of 1-forms. 2) The shell observer therefore sees the object go past with speed \begin{equation} \frac{\sigma^r}{\sigma^t} = \frac{dr\bigg/\sqrt{1-\frac{2m}{r}}}{\sqrt{1-\frac{2m}{r}}\,dt} = \frac{1}{1-\frac{2m}{r}} \frac{dr}{dt} = -\sqrt{\frac{2m}{r}} \end{equation} this expression approaches the speed of light ($-1$) as $r$ approaches $2m$. Thus, the observer far away never sees the object cross the horizon ($r=2m$), whereas the shell observer sees the object approach the speed of light as it approaches the horizon! What is going on here?