# 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?

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:

$$\underbrace{\left(1-\frac{2\,G\,M}{r}\right)}_{a}\,dv^2=a\,dv^2=2\,dr\,dv$$

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

$$\frac{dv}{dr}=\frac{2}{a}=2\,\left(1-\frac{2\,G\,M}{r}\right)^{-1}$$

• 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?
– Sito
Jan 7, 2020 at 8:49
• @sito those are the solutions of the equation „take this equation „, so where is the problem?
– Eli
Jan 7, 2020 at 12:33
• 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 $=:$).
– Sito
May 25, 2020 at 12:25
• @Sito see new document
– Eli
May 25, 2020 at 16:53