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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?

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  • $\begingroup$ 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. $\endgroup$ – G. Smith Nov 14 '19 at 4:14
  • $\begingroup$ 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. $\endgroup$ – G. Smith Nov 14 '19 at 4:27
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    $\begingroup$ 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). $\endgroup$ – G. Smith Nov 14 '19 at 4:37
  • $\begingroup$ 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 $\endgroup$ – m4r35n357 Nov 14 '19 at 9:42
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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

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  • $\begingroup$ 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. $\endgroup$ – G. Smith Nov 15 '19 at 18:58
  • $\begingroup$ 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. $\endgroup$ – G. Smith Nov 15 '19 at 19:02
  • $\begingroup$ 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. $\endgroup$ – G. Smith Nov 15 '19 at 19:06

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