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)

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