# Particles falling towards a mass

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 10; Problem 10.3.

Another example of using the Schwarzschild metric to calculate some properties of a particle’s trajectory. This time we start off with two masses at infinity, with one at rest (so ${e=1}$) and another moving radially inwards towards the central mass ${M}$ with ${e=2}$. What are the speeds of these two particles when they pass the point ${r=6GM}$?

We can start by looking at the invariant equation

$\displaystyle E=-\mathbf{p}\cdot\mathbf{u}_{\mbox{obs}} \ \ \ \ \ (1)$

where ${\mathbf{p}}$ is the particle’s momentum as measured by an observer at rest at ${r=6GM}$, and ${\mathbf{u}_{\mbox{obs}}}$ is the four-velocity of the observer. In the Schwarzschild metric an observer at rest has four-velocity

$\displaystyle u^{t}=\left[\left(1-\frac{2GM}{R}\right)^{-1/2},0,0,0\right] \ \ \ \ \ (2)$

The relation above then becomes

 $\displaystyle E$ $\displaystyle =$ $\displaystyle -g_{tt}p^{t}u^{t}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1-\frac{2GM}{r}\right)\left(m\frac{dt}{d\tau}\right)\left(1-\frac{2GM}{r}\right)^{-1/2}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle m\left(1-\frac{2GM}{r}\right)^{1/2}\frac{dt}{d\tau} \ \ \ \ \ (5)$

where ${m}$ is the particle’s mass.

We know that the conserved quantity ${e}$ is given by

$\displaystyle \left(1-\frac{2GM}{r}\right)\frac{dt}{d\tau}=e \ \ \ \ \ (6)$

so we can substitute for ${dt/d\tau}$ above to get

 $\displaystyle E$ $\displaystyle =$ $\displaystyle m\left(1-\frac{2GM}{r}\right)^{1/2}\frac{e}{\left(1-\frac{2GM}{r}\right)}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle me\left(1-\frac{2GM}{r}\right)^{-1/2} \ \ \ \ \ (8)$

The energy of a particle moving at speed ${v}$ is (from special relativity)

$\displaystyle E=\frac{m}{\sqrt{1-v^{2}}} \ \ \ \ \ (9)$

so combining the two results we get

 $\displaystyle \frac{m}{\sqrt{1-v^{2}}}$ $\displaystyle =$ $\displaystyle me\left(1-\frac{2GM}{r}\right)^{-1/2}\ \ \ \ \ (10)$ $\displaystyle \frac{1}{e^{2}}\left(1-\frac{2GM}{r}\right)$ $\displaystyle =$ $\displaystyle 1-v^{2}\ \ \ \ \ (11)$ $\displaystyle v$ $\displaystyle =$ $\displaystyle \left[1-\frac{1}{e^{2}}\left(1-\frac{2GM}{r}\right)\right]^{1/2} \ \ \ \ \ (12)$

For ${e=1}$ and ${r=6GM}$:

$\displaystyle v=\frac{1}{\sqrt{3}} \ \ \ \ \ (13)$

For ${e=2}$:

$\displaystyle v=\sqrt{\frac{5}{6}} \ \ \ \ \ (14)$

## 3 thoughts on “Particles falling towards a mass”

1. Chris Kranenberg

Part d asks to express the energy in terms of l, GM and R. Therefore, replacing the time component of the four-momentum with Moore Eq. 10.35 provides the desired expression and satisfies the requirements for Part e.