# Calculating coordinate increase of light ray escaping black hole

Consider a light ray near a black hole described by Eddington-Finkelstein coordinates $$(v,r,\theta,\varphi)$$. My aim is to calculate the increase of the coordinate $$v$$ along a radial path from the horizon $$r=2m$$ to some farther point $$r = R$$. Please tell me 1. if my approach is correct and 2. if the diverging result is a manifestation of the fact that a light ray cannot escape from beyond the black hole horizon.

The metric is regular at the horizon and of the form:

$$ds^2 \enspace = \enspace - \underbrace{\Big( 1 - \frac{2m}{r} \Big)}_{=: \, V(r)} \, dv^2 + 2 \, dv \, dr + r^2 \, d\Omega^2.$$

A light ray travels on a null geodesic, therefore one has

$$0 \enspace = \enspace -V(r) \, dv^2 + 2 \, dv \, dr + r^2 \, d\Omega^2$$

and by "dividing" by $$dv$$ one finds

$$dv \enspace = \frac{2}{V(r)} \, dr.$$

Integrating this yields

$$\Delta v \enspace = \enspace 2 \int_{2m}^R \frac{dr}{V(r)} \enspace = \enspace 2 \bigg( R - 2m + 2m \ln( r - 2m ) \,\Big|_{2m}^R \; \bigg).$$

The step of "dividing" by $$dv$$ is formally not correct, I guess. What would be the mathematical correct way of obtaining the result?

What you are doing in the "dividing by $$dv$$" step is, rigorously, a part of imposing that the tangent $$k^\mu$$ to the photon's worldline is a null vector. If you want to do it in a "cleaner" way, you'd first write \begin{align} 0 &= g_{\mu\nu}k^\mu k^\nu, \\ &= - V(r) \dot{v}^2 + 2 \dot{v} \dot{r} + \text{angular terms}. \end{align} In the step you mentioned, you neglected the angular terms, since the photon is moving radially (which means the angles are constant). Hence, they vanish by assumption. We get $$0 = - V(r) \dot{v}^2 + 2 \dot{v} \dot{r}.$$ Now there is nothing wrong with dividing by $$\dot{v}$$ to get $$V(r) \dot{v} = 2\dot{r}.$$ This can be simplified using the derivative of an inverse function and the chain rule (which, in Leibniz notation, means "we can cancel the $$d\lambda$$'s" where $$\lambda$$ is the parameter with respect to which we are differentiating) to get $$\frac{dv}{dr} = \frac{2}{V(r)}.$$
One can solve the differential equation by integrating with respect to $$r$$ on both sides and using the Fundamental Theorem of Calculus on the left hand-side. The solution is then exactly the expression you found.
The divergence is indeed because the photon can't leave the event horizon at $$r = 2m$$: your calculation is essentially asking how long (in terms of advanced time $$v$$) it takes for the photon to reach $$r=R$$, but the photon can't leave $$r=2m$$, so it takes infinite time. Notice that you get finite results if the lower limit of the integral is $$2m + \epsilon$$ for some $$\epsilon > 0$$. For $$M=1$$ and $$R=4$$, here's the plot of the result of the integral as a function of $$\epsilon$$: