Eddington-Finkelstein coordinates and radial null geodesics 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?
 A: 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.
A: Take this equation:
$$\underbrace{\left(1-\frac{2\,G\,M}{r}\right)}_{a}\,dv^2=a\,dv^2=2\,dr\,dv$$
divide by $dr^2$ you get:
$$\frac{1}{dr^2}\left(a\,dv^2-2\,dr\,dv\right)=0$$
$$\frac{dv}{dr}\,\left(a\,\frac{dv}{dr}-2\right)=0$$
you get then two solutions for $\frac{dv}{dr}$
$$\frac{dv}{dr}=0$$ and
$$\frac{dv}{dr}=\frac{2}{a}=2\,\left(1-\frac{2\,G\,M}{r}\right)^{-1}$$
