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(Note: $c =1$ throughout) The Schwarzschild metric is $$ds^2 = (1- \frac{2m}{r})dt^2 - \frac{1}{1-\frac{2m}{r}}dr^2 - r^2 d\Omega ^2,$$ with $d\Omega^2$ being the square of the solid angle element and $m = GM$, where $M$ is the mass of the object. Radial null geodesics in this geometry are given by $$t_\pm(r) = \pm(r+2m\log |r-2m|+C),$$ where $r = \pm k\lambda$, $\lambda$ an affine parameter, with the plus sign indicating that the particle is outgoing, and the minus sign indicating that it is ingoing. My question is: what does this physically represent? After thinking about it for some time, I have considered that it might be the time taken for a light particle to reach a distance $r$ from the singularity, but where is it falling from? If anyone could clarify my confusions, it would be greatly appreciated.

Edit: The photon would be falling from an initial position $r_0$, where $\mp (r_0 + \log|r_0 -2m|) = C$ as mentioned by Triatticus and myself.

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  • $\begingroup$ Wouldn't you determine where it is falling from by determining the constant C? Seems like the information on the starting point is there. $\endgroup$
    – Triatticus
    Oct 30, 2022 at 1:13
  • $\begingroup$ Yes I thought of that after posting the question. However, it seems that problems arise when trying to compute $t(C)$ (which should be $0$), so should the formula only apply for $r \not = C$? $\endgroup$
    – Daniel Waters
    Oct 30, 2022 at 1:36
  • $\begingroup$ I actually just found the answer to the question I posed in the last comment. The constant is not $r_0$, but rather $\mp (r_0 + \log|r_0 - 2m|)$. $\endgroup$
    – Daniel Waters
    Oct 30, 2022 at 1:39
  • $\begingroup$ Your metric has three problems: The second term on the right isn’t infinitesimal, and the second and third terms on the right are dimensionally incorrect. $\endgroup$
    – Ghoster
    Oct 30, 2022 at 2:15
  • $\begingroup$ The Schwarzschild metric in polar coordinates is: $$ {ds}^{2} = \left(1 - \frac{2m}{r} \right) \,dt^2 - \left(1-\frac{2m}{r}\right)^{-1} \,dr^2 - r^2 d\Omega^2 $$ Where $d\Omega^2$ denotes the spherical metric induced by the Euclidean on a $2$-sphere: $$ d\Omega^2 = d\theta^2 + \sin^2\theta \, d\varphi^2 $$ $\endgroup$
    – safesphere
    Oct 30, 2022 at 6:56

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The radial null geodesic $t_\pm(r)$ indeed represents the time taken for the light to reach a radial coordinate $r$ from the frame of reference of a distant observer (or an "observer at infinity"). The constant $C$ encodes the initial position and must be chosen in such a way that $t_\pm(r_0) = 0$. This implies that $C = \mp(r_0 +\log|r_0 - 2m|)$, in which case, all of the problems mentioned above (in both the question and the comments) are solved. For further reference see https://www.reed.edu/physics/courses/Physics411/html/411/page2/files/Lecture.31.pdf.

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