Proper Time for Free Fall into a Schwarschild Black Hole [closed]

In the solution to the question here, they define the proper time for free fall into a Schwarschild black hole as $$\Delta \tau=(2 / 3) r_{s}\left(r / r_{s}\right)^{3 / 2}$$ but they don't provide a derivation. Instead the link to a University of Tennessee graduate course which has since removed the course notes from the Internet. There's several different coordinate systems to write down the Schwarschild metric in though, so I don't know where to start applying conserved Killing vectors to. Any help for how to derive this?

• The $r$ in your formula is the usual Schwarzschild radial coordinate, as here. Pick your favorite way of finding the equation for a radial geodesic in this metric. – G. Smith Nov 14 '19 at 4:14
• Since this is a common homework problem in any GR course, you probably won’t get a complete derivation. And if someone provides one, it may get deleted by a moderator. – G. Smith Nov 14 '19 at 4:27
• they define the proper time for free fall That formula doesn’t define the proper time. As you say, it is the result of a calculation. The metric is what defines proper time (and geodesics). – G. Smith Nov 14 '19 at 4:37
• In case. like me, you are not on a course with access to lecturers and fellow students, there is a pretty exhaustive treatment here: mathpages.com/rr/s6-04/6-04.htm – m4r35n357 Nov 14 '19 at 9:42

The derivation I would consider can be found from General Relativity: An Introduction for Physicists by Hobson, Efstathiou, and Lansenby in Chapters 9.6 and 9.7.

The energy equation equation for the radial coordinate satisfying the Schwarschild metric is given by $$\dot{r}^{2}+\frac{h^{2}}{r^{2}}\left(1-\frac{2 G M}{c^{2} r}\right)-\frac{2 G M}{r}=c^{2}\left(k^{2}-1\right)$$ The Schwarschild solution must also obey the constraint $$\left(1-\frac{2 \mu}{r}\right) \dot{t}=k$$ In free fall, the angular coordinate is constant, so the constant $$h$$ is zero. This simplifies to $$\dot{r}^{2}=c^{2}\left(k^{2}-1\right)+\frac{2 G M}{r}$$ Considering a particle dropped at rest from infinity as in our situation gives $$k = 1$$ and therefore

$$\frac{d r}{d \tau}=-\left(\frac{2 \mu c^{2}}{r}\right)^{1 / 2}$$ integrating this from a point $$r_0$$ to $$r$$ $$\tau=\frac{2}{3} \sqrt{\frac{r_{0}^{3}}{2 \mu c^{2}}}-\frac{2}{3} \sqrt{\frac{r^{3}}{2 \mu c^{2}}}$$

Now, $$2\mu = r_s$$ and I can believe they might have set $$c=1$$, but if I plugged in the event horizon, $$r = r_s$$, I still don't see what they're doing, so I am confused.

If this was a homework question, it wouldn't have a derivation in a textbook. I'm asking specifically about the equation posted on stack exchange

• This answer does not explain what $h$ is. Or what $k$ is. Or what $\mu$ is. Or where the “energy equation” comes from. Or where “the constraint” comes from. – G. Smith Nov 15 '19 at 18:58
• The $\Delta\tau$ in the equation you are trying to understand is the proper time that elapses falling from $r$ to $r=0$ along a null radial geodesic with $k=1$. So just take $r_0$ to be $r$ and $r$ to be 0. – G. Smith Nov 15 '19 at 19:02
• If this was a homework question, it wouldn't have a derivation in a textbook. This site’s policy on what is a homework-like problem has nothing to do with whether some textbook has the derivation. – G. Smith Nov 15 '19 at 19:06