My error
<br />
\sqrt {r \left( r-1 \right) }+\ln \left( \sqrt {r}+\sqrt {r-1}<br />
\right) <br />
works just fine. And I think I see how to sketch out a solution that most of us will agree with.
(It appears that some other people wrote up similar ideas while I was writing this - and it appears that if yuiop agrees with me, strathaus will claim he's wrong, so I don't expect universal agreement ::smiley face::).
If we take r_s =1 or M = 1/2, and R = infinity, we can use the simpler equations MTW 25.38 to find the Schwarzschild t coordinate as a function of the r coordinate. It also gives us proper time, \tau, though we don't really need it.
<br />
\tau = -\frac{2}{3} \, r^{\frac{3}{2}}<br />
<br />
t = -\frac{2}{3} \, r^{\frac{3}{2}} - 2 \, r^{\frac{1}{2}} + ln\frac{r^{\frac{1}{2}}+1}{r^{\frac{1}{2}}-1}<br />
Now, all we need to do to make this a one-parameter group of curves is to use time-translation symmetry. We simply add some small number \epsilon to the schwarzschild time coordinate to represent the second particle, which works because of the time-translation symmetry of the metric, in fancier language \partial / \partial t is a "Killing vector". So we have
<br />
t\left(r, \epsilon\right) = -\frac{2}{3} \, r^{\frac{3}{2}} - 2 r^{\frac{1}{2}} + ln\frac{r^{\frac{1}{2}}+1}{r^{\frac{1}{2}}-1} + \epsilon<br />
Assume that the leading edge of the rod, with the parameter \epsilon =0, has some r-coordinate r_0 at some t-coordinate t_0.
Then we need to find the radial coordinate r_1 of the trailing rod at the same time coordinate t_0 by solving
<br />
t\left(r_1, \epsilon) = t_0 = t\left(r_0, 0\right)<br />
i.e.
<br />
-\frac{2}{3} \, r_{1}^{\frac{3}{2}} - 2 r_{1}^{\frac{1}{2}} + ln\frac{r_{1}^{\frac{1}{2}}+1}{r_{1}^{\frac{1}{2}}-1} + \epsilon = -\frac{2}{3} \, r_{0}^{\frac{3}{2}} - 2 r_{0}^{\frac{1}{2}} + ln\frac{r_{0}^{\frac{1}{2}}+1}{r_{0}^{\frac{1}{2}}-1} <br />
Then we use
<br />
\sqrt {r_{1} \left( r_{1}-1 \right) }+\ln \left( \sqrt {r_{1}}+\sqrt {r_{1}-1} \right) - \sqrt {r_{0} \left( r_{0}-1 \right) } - \ln \left( \sqrt {r_{0}}+\sqrt {r_{0}-1}<br />
\right) <br />
to integrate between r_0 and r_1 to get the length as seen by a static observer for the falling rod. To get the proper length, we multiply by gamma, based on the radial velocity which we know is 1 / \sqrt{r}.
The integrated distance between freefalling objects in GP coordinates is the same as the integrated distance between two stationary objects in Schwarzschild coordinates. Not quite sure what the significance of that is yet.
