Looking at Carroll's chapter 5.6 he derives the Eddingtion-Finkelstein coordinates and writes the Schwarzschild metric out, resulting in ($v-r$ coordinates) $$\mathrm{d} s^{2}=-\left(1-\frac{2 G M}{r}\right) \mathrm{d} v^{2}+(\mathrm{d} v \mathrm{d} r+\mathrm{d} r \mathrm{d} v)+r^{2} \mathrm{d} \Omega^{2}.$$ He then goes on to say that "the condition for radial null curves is solved by" $$\frac{d v}{d r}=\left\{\begin{array}{ll} {0} & {\text { (infalling) }} \\ {2\left(1-\frac{2 G M}{r}\right)^{-1} .} & {\text { (outgoing) }} \end{array}\right.$$

I can't follow. "Radial null geodesics" implies that $d\Omega=0$ in the metric and that $ds^2=0$, so we have $$\left(1-\frac{2 G M}{r}\right) \mathrm{d} v^{2} = 2\mathrm{d} v \mathrm{d} r.$$ My problem is now that I don't know what the extra contraints on this equations are in the case of "infalling" and "outgoing" particles. For the "ingoing" case I have no idea what we are assuming... I think I've read somewhere that $dv=0$, but I don't see how this is supposed to work. And for the "outgoing" part we can just assume that $dv\neq 0$? Why is that?


2 Answers 2


Here is my reasoning, as for why we assign different labels to $dv=0$ and $dv\neq 0$. The $v$ coordinate is defined to be $v=t+r^{*}$, where $r^{*}=r+2GM\ln(\frac{r}{2GM}-1)$.

If we assume that $v=const$, then the coordinate one-form $dv$ induced on this hypersurface is $dv=0$. Let us inspect what the $v=const$ condition means in the $t,r^{*}$ coordinates

$$t+r^{*}=const,$$ $$1 + \frac{dr^{*}}{dt} = 0.$$ Then, just by using the definition of $r^{*}$, we arrive at the equation:

$$ \frac{dr}{dt}=\frac{2GM}{r}-1.$$ For any $r>2GM$ this means $\frac{dr}{dt}<0$ - that is, the value of the radial coordinate decreases with coordinate time $t$. Hence we deal with ingoing geodesics.

Now, for the case $dv\neq 0 $, this corresponds to $t+r^{*}\neq const = f(t)$, where I assumed that the function $f$ measures the variability of $v$. Differentiating the above equation with respect to time I will arrive at:

$$\frac{dr}{dt} = \big{(}1 -\frac{2GM}{r} \big{)}\big{(}\frac{df}{dt} -1 \big{)}.$$

The first term in the parentheses is bigger than zero for $r>2GM$, and so, for a suitably chosen function $f$ and coordinate time $t$ we have a chance of obtaining $\frac{dr}{dt} > 0 $ - an outgoing geodesic.


Take this equation:


divide by $dr^2$ you get:



you get then two solutions for $\frac{dv}{dr}$

$$\frac{dv}{dr}=0$$ and


  • $\begingroup$ Thank you for the answer. I don't see why $\frac{dv}{dr}=0$ and $\frac{dv}{dr}=\frac{2}{a}$ are sensible? Where do you get these expressions from? $\endgroup$
    – Sito
    Jan 7, 2020 at 8:49
  • $\begingroup$ @sito those are the solutions of the equation „take this equation „, so where is the problem? $\endgroup$
    – Eli
    Jan 7, 2020 at 12:33
  • $\begingroup$ From the you last equation it became clear what $a$ is, but it would be nice if you cloud define it before using it (or at least make it more explicit by replacing the corresponding $=$ with a $=:$). $\endgroup$
    – Sito
    May 25, 2020 at 12:25
  • 1
    $\begingroup$ @Sito see new document $\endgroup$
    – Eli
    May 25, 2020 at 16:53

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