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Two-Body Elastic Collisions

Before specializing to two-body Coulomb collisions, it is convenient to develop a general theory of two-body elastic collisions. Consider an elastic collision between a particle of type $ 1$ and a particle of type $ 2$ . Let the mass and instantaneous velocity of the former particle be $ m_1$ and $ {\bf v}_1$ , respectively. Likewise, let the mass and instantaneous velocity of the latter particle be $ m_2$ and $ {\bf v}_2$ , respectively. The velocity of the center of mass is given by

$\displaystyle {\bf U} = \frac{m_1\,{\bf v}_1+m_2\,{\bf v}_2}{m_1+m_2}.$ (3.10)

Moreover, conservation of momentum implies that $ {\bf U}$ is a constant of the motion. The relative velocity is defined

$\displaystyle {\bf u} = {\bf v}_1-{\bf v}_2.$ (3.11)

We can express $ {\bf v}_1$ and $ {\bf v}_2$ in terms of $ {\bf U}$ and $ {\bf u}$ as follows:

$\displaystyle {\bf v}_1$ $\displaystyle = {\bf U} + \frac{\mu_{12}}{m_1}\,{\bf u},$ (3.12)
$\displaystyle {\bf v}_2$ $\displaystyle = {\bf U} - \frac{\mu_{12}}{m_2}\,{\bf u}.$ (3.13)

Here,

$\displaystyle \mu_{12} = \frac{m_1\,m_2}{m_1+m_2}$ (3.14)

is the reduced mass. The total kinetic energy of the system is written

$\displaystyle K = \frac{1}{2}\,m_1\,v_1^{\,2} + \frac{1}{2}\,m_2\,v_2^{\,2} = \frac{1}{2}\,(m_1+m_2)\,U^{\,2} + \frac{1}{2}\,\mu_{12}\,u^2.$ (3.15)

Now, the kinetic energy is the same before and after an elastic collision. Hence, given that $ U$ is constant, we deduce that the magnitude of the relative velocity, $ u$ , is also the same before and after such a collision. Thus, it is only the direction of the relative velocity vector, rather than its length, that changes during an elastic collision.


next up previous
Next: Boltzmann Collision Operator Up: Collisions Previous: Collision Operator
Richard Fitzpatrick 2016-01-23