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

Question

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?

Answer 1

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.