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?


1 Answer 1


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 $\epsilon$: Solution of said integral in terms of epsilon in the range 0.01 to 1. The curve is high and diverging at the left side (small epsilon), but small on the right side (large epsilon).

  • $\begingroup$ Thank you very much for your elaborate answer! I upvoted and accepted it. $\endgroup$
    – Octavius
    Commented Aug 2, 2022 at 8:41

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