Can an observer sitting at rest at infinity, simultaneously measure the proper time and proper distance of a particle travelling in any geodesic?

Question

In the book General Relativity by M.P Hoson, the author tries to calculate the velocity of a particle radially infalling in Schwarzschild spacetime. The velocity is measured by the observer sitting at rest at infinity. For the same, he takes the ratio of proper distance and proper time of the particle, both measured by the observer. To calculate the proper distance, the path traveled by the particle should be spacelike, whereas, to calculate the proper time, the path traveled should be timelike. Since the particle can travel with only one of the paths, how does the observer simultaneously calculate the proper time and proper distance along the particle geodesic

Answer 1

You probably need to say exactly how the "velocity" is defined by Hoson. This would be the rate of change of some spatial coordinate with respect to the proper time measured by the observer.

In most texts, for an observer at infinity, and a particle falling from inifinity, that would be $$\frac{dr}{dt} = \left(1 - \frac{r_s}{r}\right)\left(\frac{r_s}{r}\right)^{1/2}$$ in the usual Schwarzschild coordinates. Whilst $dt$ is also an increment in proper time for the distant observer, $dr$ is not a proper distance. The proper distance increment between two radial coordinates could be written as $$ ds = \frac{dr}{(1 - r_s/r)^{1/2}}$$ and you could then say $$\frac{ds}{dt} = \left(1-\frac{r_s}{r}\right)^{1/2}\left(\frac{r_s}{r}\right)^{1/2}\ .$$ Is that what Hoson means? This would be the equivalent of estimating a velocity by the observer receiving successive pictures of the falling particle passing "milestones" in $r$, applying a correction to the distance travelled between the milepstones to account for the curvature of space, and then dividing this distance by the time between successive pictures.

Answer 2

Your question is lacking a lot of details and it is thus not clear what exactly you are asking for.

However, you seem to be having a common misunderstanding of what is happening. An observer at infinity will get to send and receive a lot of photon information about the particle, so that, from all the information extracted from all the observations, the observer can work out the details of the trajectory of the falling particle.

Then the observer can represent the data as a parametrised trajectory (geodesic) in the spacetime plot that the observer can make. The observer will label the trajectory with coördinate time and position, and if needed, can calculate the proper time as measured by the particle. The particle will not have any thing that corresponds to a proper distance.

The velocity of the particle is, by definition, zero from the particle's point of view. If you want to know what it is in the observer's view, there is some parallel transportation to do, and it is very complicated, and also choice-dependent. This is why it is not presented to students.