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.

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