# Vertical particle motion

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 10; Problem 10.5.

And now for a general relativistic look at the standard physics problem of throwing an object up in a gravitational field. Suppose we start at radius ${r=r_{0}}$ and throw up the object so that it comes to rest momentarily at ${r=r_{1}}$ before turning around and falling back to ${r_{0}}$. What is the total proper time (as measured by the object) in this trip?

We begin by working out the energy ${e}$. For an object at rest at ${r=r_{1}}$:

 $\displaystyle e$ $\displaystyle =$ $\displaystyle \left(1-\frac{2GM}{r_{1}}\right)\frac{dt}{d\tau}\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1-\frac{2GM}{r_{1}}\right)u^{t}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1-\frac{2GM}{r_{1}}\right)\left(1-\frac{2GM}{r_{1}}\right)^{-1/2}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1-\frac{2GM}{r_{1}}\right)^{1/2} \ \ \ \ \ (4)$

The radial equation of motion in the Schwarzschild metric is

$\displaystyle \frac{1}{2}\left(\frac{dr}{d\tau}\right)^{2}+\frac{1}{2}\frac{l^{2}}{r^{2}}-GM\left(\frac{1}{r}+\frac{l^{2}}{r^{3}}\right)=\frac{1}{2}\left(e^{2}-1\right) \ \ \ \ \ (5)$

For radial motion, ${l=0}$ so we get

 $\displaystyle \frac{1}{2}\left(\frac{dr}{d\tau}\right)^{2}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(e^{2}-1\right)+\frac{GM}{r}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle GM\left(\frac{1}{r}-\frac{1}{r_{1}}\right)\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle GM\frac{r_{1}-r}{rr_{1}} \ \ \ \ \ (8)$

To find the time that elapses on the upward leg of the journey, we must evaluate

$\displaystyle \Delta\tau=\sqrt{\frac{r_{1}}{2GM}}\int_{r_{0}}^{r_{1}}\sqrt{\frac{r}{r_{1}-r}}dr \ \ \ \ \ (9)$

This is, as Moore says, a bit of a nasty integral. Using software produces a bit of a jumble, so I resorted to the old-fashioned method of looking the integral up in a table, which produced a somewhat nicer result. We get

 $\displaystyle \Delta\tau$ $\displaystyle =$ $\displaystyle \sqrt{\frac{r_{1}}{2GM}}\left[-\sqrt{r\left(r_{1}-r\right)}-r_{1}\arctan\left(-\sqrt{\frac{r}{r_{1}-r}}\right)\right]_{r_{0}}^{r_{1}}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{r_{1}}{2GM}}\left[-\sqrt{r\left(r_{1}-r\right)}+r_{1}\arctan\sqrt{\frac{r}{r_{1}-r}}\right]_{r_{0}}^{r_{1}} \ \ \ \ \ (11)$

At the upper limit, the first term is zero and the second term is

$\displaystyle r_{1}\arctan\left(\infty\right)=\frac{\pi}{2}r_{1} \ \ \ \ \ (12)$

so the result is

$\displaystyle \Delta\tau=\sqrt{\frac{r_{1}}{2GM}}\left[\sqrt{r_{0}\left(r_{1}-r_{0}\right)}-r_{1}\arctan\sqrt{\frac{r_{0}}{r_{1}-r_{0}}}+\frac{\pi}{2}r_{1}\right] \ \ \ \ \ (13)$

Here we’re assuming that the arctan lies in the range ${\left[-\frac{\pi}{2},\frac{\pi}{2}\right]}$. The total time is twice this, so

$\displaystyle \tau=\sqrt{\frac{2r_{1}}{GM}}\left[\sqrt{r_{0}\left(r_{1}-r_{0}\right)}-r_{1}\arctan\sqrt{\frac{r_{0}}{r_{1}-r_{0}}}+\frac{\pi}{2}r_{1}\right] \ \ \ \ \ (14)$