- #1

andyl01

- 6

- 0

$$

ds^2 = \left(1 - \frac{R_S}{r}\right)dt^2 - \left(1 - \frac{R_S}{r}\right)^{-1}dr^2 - r^2 [d\theta^2 + (\sin{\theta})^2d\phi^2],

$$

where ##R_S## is the Schwarzschild radius and ##t## is the time given by a clock located at infinite distance from the source of the field and stationary with respect to it. Using this coordinate system we find that a massive radially infalling particle (coming from ##r > R_S##) takes an infinite amount of coordinate time to reach the Schwarzschild radius. An observer located far away from the source "sees" the particle approaching ##R_S## more and more slowly.

Adopting Eddington-Finkelstein coordinates ##(t',r,\theta,\phi)##, we write the line element in the form:

$$

ds^2 = \left( 1- \frac{R_S}{r}\right)dt'^2 - \frac{2R_S}{r}dt'dr - \left( 1 +\frac{R_S}{r}\right)dr^2 - r^2 [d\theta^2 + (\sin{\theta})^2d\phi^2].

$$

When an observer is located far away from the source (##r \to \infty##) and is stationary with respect to it, ##t'## is the time given by the observer's clock, as well as ##t## is.

If I calculate how ##t'## and ##r## are related for a radially infalling massive particle, I find out that the particle takes a finite time to cross the Schwarzschild radius from an initial position ##r_0 > R_S## and to reach the singularity ##r=0##. So, after this calculation, I don't understand what is the situation which actually happens: my problem arises because for the "far away observer" ##t'## and ##t## have the same meaning, since by a suitable time translation they both represent the time given by his clock. Since the observer must see only one situation happening, does he see the particle cross the Schwarzschild radius or does he see the particle gradually "freeze" at the Schwarzschild radius?