# Particle falling towards a mass: two types of velocity

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 10; Problem 10.4.

The radial equation of motion in the Schwarzschild metric is

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

We can use this to derive an equation for ${dr/dt}$, the rate of change of ${r}$ with respect to the Schwarzschild time coordinate. The coordinate ${t}$ isn’t the time as measured by any particular object (that time is the proper time ${\tau}$ in the reference frame of the object) so we wouldn’t expect it to be the same as ${dr/d\tau}$.

To get the equation, we can use

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

where the last line uses the definition of ${e}$. Plugging this into the top equation we get

$\displaystyle \frac{dr}{dt}=\frac{1}{e}\left(1-\frac{2GM}{r}\right)\left[e^{2}-1+2GM\left(\frac{1}{r}+\frac{l^{2}}{r^{3}}\right)-\frac{l^{2}}{r^{2}}\right]^{1/2} \ \ \ \ \ (4)$

As ${r}$ approaches ${2GM}$, ${dr/dt\rightarrow0}$.

In the special case where we drop an object from rest at ${r=r_{0}}$, we can work out both ${dr/dt}$ and ${dr/d\tau}$. In this case, motion is radially inward so ${l=0}$. To find ${e}$, we use the fact that for an object at rest at ${r=r_{0}}$:

 $\displaystyle e$ $\displaystyle =$ $\displaystyle \left(1-\frac{2GM}{r_{0}}\right)\frac{dt}{d\tau}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1-\frac{2GM}{r_{0}}\right)u^{t}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1-\frac{2GM}{r_{0}}\right)\left(1-\frac{2GM}{r_{0}}\right)^{-1/2}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1-\frac{2GM}{r_{0}}\right)^{1/2} \ \ \ \ \ (8)$

We have therefore

 $\displaystyle \frac{dr}{dt}$ $\displaystyle =$ $\displaystyle \frac{1}{e}\left(1-\frac{2GM}{r}\right)\left(e^{2}-1+\frac{2GM}{r}\right)^{1/2}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1-\frac{2GM}{r}}{\sqrt{1-\frac{2GM}{r_{0}}}}\sqrt{2GM\left(\frac{1}{r}-\frac{1}{r_{0}}\right)}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1-\frac{2GM}{r}\right)\sqrt{\frac{2GM}{r}}\sqrt{\frac{r_{0}-r}{r_{0}-2GM}} \ \ \ \ \ (11)$

From 1 we get

 $\displaystyle \frac{dr}{d\tau}$ $\displaystyle =$ $\displaystyle \sqrt{e^{2}-1+\frac{2GM}{r}}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{2GM\left(\frac{1}{r}-\frac{1}{r_{0}}\right)} \ \ \ \ \ (13)$

Comparing the two, we see that

$\displaystyle \frac{dr}{dt}=\frac{1-\frac{2GM}{r}}{\sqrt{1-\frac{2GM}{r_{0}}}}\frac{dr}{d\tau} \ \ \ \ \ (14)$

For the case where the object is released from rest at ${r_{0}=\infty}$, the speed at ${r=6GM}$ is

$\displaystyle \frac{dr}{d\tau}=\frac{1}{\sqrt{3}} \ \ \ \ \ (15)$

which agrees with the earlier calculation done by a different method.