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. Pingback: Particle falling towards a mass: two types of velocity | Physics tutorials

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  3. 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.

    Reply

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