# Eddington-Finkelstein Coordinates: Radial Time-like Geodesics falling into a black hole

I am trying to get an expression for the radial timelike geodesics in EF coordinates: $$g_{\mu\nu}dx^\mu dx^\nu = -\left(1-\frac{2GM}{r}\right)dv^2 +2dvdr +r^2d\Omega^2$$ for an observer initially stationary at $$r_0 > 2GM$$ falling into a black hole with a 4-velocity of $$u^\mu = (\dot v(\lambda), \dot r(\lambda),0,0)$$.

I know that $$\xi^\mu \partial_\mu = \partial_v$$ is a killing vector with $$\xi^\mu = (1,0,0,0)$$ and thus $$\xi_\mu = \left(-\left(1- \frac{2GM}{r} \right),0,0,0 \right)$$. The conservation condition and normalization of the four-velocity gives a system of equations:

$$g_{\mu\nu}\xi^\mu u^\nu = -\left(1-\frac{2GM}{r}\right)\dot v = C_0$$

$$g_{\mu\nu}u^\mu u^\nu = -\left(1-\frac{2GM}{r}\right)\dot v^2 +2\dot v \dot r = -1$$

I solved for $$\dot r$$ and then used the intial condition $$\dot r (r=r_0)= 0$$ to get $$C_0 = \sqrt{1-\frac{2GM}{r_0}}$$. Then I plugged $$C_0$$ back into the first equation, yielding:

$$\dot r = \frac{-1}{2C_0}\left(\frac{2GM}{r}-\frac{2GM}{r_0}\right)$$

$$\dot v = - \frac{C_0}{1-\frac{2GM}{r}}$$

$$\frac{\dot v}{\dot r} = \frac{dv}{dr} = \frac{1}{2\left(1-\frac{2GM}{r}\right) \left(\frac{2GM}{r}-\frac{2GM}{r_0}\right)}$$

However, these equations don't seem to make sense to me. For one, since $$\frac{dr}{d\lambda}$$ is normalized, integrating it from $$r_0$$ to $$0$$ should yield the proper time it takes to reach the singularity but the integral seems to diverge at $$r = r_0$$. Furthermore, $$\frac{dv}{dr}$$ goes to infinity at $$r=2GM$$ just like with the schwarzchild metric, which is what I thought this metric was suppose to remove.

Any ideas as to where this calculation went wrong?

• How have you proven that the metric diverges at $r = 2GM$? en.wikipedia.org/wiki/… Nov 24, 2020 at 2:21
• I'm not saying I've proved anything, I believe by calculation is wrong because the dv/dr equation above seems to go to infinity at R= 2GM. From the link, I am not sure how they got $r(\tau)$, seems they redefined the coordinate system which I don't want to do here. Nov 24, 2020 at 2:30
• Who is they? If you're following a text please cite it in your question, include a link to a free version if possible. If you are asking how they compute $r(\tau)$ then you need to make that question clear. Also, you said that the equation seems to diverge. Have you attempted the integral? Type in integrate: (1)/((1 - 1/x)*(1/a - 1/x)), [a,infinity] into wolframalpha.com Nov 24, 2020 at 2:36
• I was referring to the wiki link yu have in your first comment. Also, the integral isn't to infinity. Since my observing is falling into the black hole, I would integrate from $r=r_0$ to $r =0$, which diverges when I integrate Nov 24, 2020 at 2:41
• "integrate (1)/((1 - 1/x)*(-1/a +1/x)) with respect to x from a to 0" - results in 'does not converge' Nov 24, 2020 at 2:49

You've gotten the components of your Killing vector field $$\xi_\mu$$ wrong. Since $$\xi^\mu=(1,0,0,0)$$, lowering that index with the metric $$g_{\mu\nu}=\begin{bmatrix}\frac{2GM}{r}-1&1&0&0\\1&0&0&0\\0&0&r^2&0\\ 0&0&0&r^2\sin^2\theta\end{bmatrix}$$ actually gives us $$\xi_\mu=g_{\mu\nu}\xi^\nu=(\frac{2GM}{r}-1,1,0,0).$$ Our 4-velocity $$u^\mu(\tau)$$ then obeys $$\xi_\mu u^\mu=\left(\frac{2GM}{r}-1\right)\dot v+\dot r=-C$$ for some constant $$C>0$$. Note that we're dotting two timelike vectors, so we should get something negative. Your solution to $$C$$ was of the wrong sign. The other condition is \begin{align} u_\mu u^\mu&=\left(\frac{2GM}{r}-1\right)\dot v^2+2\dot v\dot r=-1\\ &=\dot v(\dot r-C). \end{align}
When $$r=r_0,\dot r=0,$$ we get $$\dot v = \left(1-\frac{2GM}{r_0}\right)^{-\frac{1}{2}},$$ letting us pin down $$C=\sqrt{1-\frac{2GM}{r_0}}$$ again. I throw the equations into Mathematica and I solve \begin{align} \dot v&=\frac{r\sqrt{1-\frac{2GM}{r_0}}-\sqrt{2GMr\left(1-\frac{r}{r_0}\right)}}{r-2GM},\\ \dot r&=-\sqrt{2GM\left(\frac{1}{r}-\frac{1}{r_0}\right)},\\ \frac{\dot v}{\dot r}=\frac{dv}{dr}&=\frac{r\left(\sqrt{\frac{r\left(\frac{r_0}{2GM}-1\right)}{r_0-r}}-1\right)}{2GM-r}. \end{align}
Approaching the event horizon, \begin{align} \lim_{r\to2GM}\dot v&=\frac{1}{2\sqrt{1-\frac{2GM}{r_0}}}=\frac{1}{2C},\\ \lim_{r\to2GM}\dot r&=-\sqrt{1-\frac{2GM}{r_0}}=-C,\\ \lim_{r\to2GM}\frac{\dot v}{\dot r}=\lim_{r\to2GM}\frac{dv}{dr}&=\frac{r_0}{4GM-2r_0}=-\frac{1}{2C^2}, \end{align} so it does appear that there's no coordinate singularity here. Success! Let's plug in some numbers and make Mathematica calculate how long it would take to reach the singularity of a solar black hole ($$M=1\,\textrm{M}_\odot$$) if we let a probe fall from Earth's orbital radius ($$r_0\approx1\,\textrm{AU}$$): $$\int_{r_0}^0\frac{d\tau}{dr}dr=\int_{r_0}^0\frac{1}{\dot r}dr\approx64.6\,\mathrm{days}.$$
How should we verify this result? Let's redo this procedure with the metric in Schwarzschild coordinates $$(t,r,\theta,\phi)$$, which is related to EF by $$t=v-r-2GM\ln\left|\frac{r}{2GM}-1\right|$$. We know $$g_{\mu\nu}=\begin{bmatrix}\frac{2GM}{r}-1&0&0&0\\0&\frac{1}{1-\frac{2GM}{r}}&0&0\\0&0&r^2&0\\0&0&0&r^2\sin^2\theta\end{bmatrix},$$ and we also find a Killing vector field $$\xi^\mu=(1,0,0,0)$$, with $$\xi_\mu=(\frac{2GM}{r}-1,0,0,0).$$ If a geodesic is given by $$x^\mu(\tau)=(t(\tau),r(\tau),0,0)$$ with 4-velocity $$u^\mu=(\dot t,\dot r,0,0),$$ we get the following equations: \begin{align} \xi_\mu u^\mu&=\left(\frac{2GM}{r}-1\right)\dot t=-C,\\ u_\mu u^\mu&=\left(\frac{2GM}{r}-1\right)\dot t^2+\frac{\dot r^2}{1-\frac{2GM}{r}}=-1\\ &=\frac{\dot r^2}{1-\frac{2GM}{r}}-C\dot t. \end{align} At $$r=r_0,\dot r=0$$, we find the same $$C=\sqrt{1-\frac{2GM}{r_0}}.$$ This gives the same $$\dot r=-\sqrt{2GM\left(\frac{1}{r}-\frac{1}{r_0}\right)},$$ so it appears we've done something right.