Two-Body Coulomb Collisions

Consider a two-body Coulomb collision between a particle of species $1$ , with mass $m_1$ and charge $e_1$ , and a particle of species $2$ , with mass $m_2$ and charge $e_2$ . The equations of motion of the two particles take the form

$$m_1\,\ddot{\bf r}_1 = k\,\frac{{\bf r}}{\vert{\bf r}\vert^{\,3}},$$ (3.56)

$$m_2\,\ddot{\bf r}_2 = -k\,\frac{{\bf r}}{\vert{\bf r}\vert^{\,3}},$$ (3.57)

where

$$k = \frac{e_1\,e_2}{4\pi\,\epsilon_0},$$ (3.58)

Here, ${\bf r}_1$ and ${\bf r}_2$ are the respective position vectors, and ${\bf r}={\bf r}_1-{\bf r}_2$ is the relative position vector. It is easily demonstrated that

$${\bf r}_1 = {\bf R} + \frac{\mu_{12}}{m_1}\,{\bf r},$$ (3.59)

$${\bf r}_2 = {\bf R}-\frac{\mu_{12}}{m_2}\,{\bf r},$$ (3.60)

where

$${\bf R} = \frac{m_1\,{\bf r}_1+m_2\,{\bf r}_2}{m_1+m_2}$$ (3.61)

is the vector position of the center of mass (which does not accelerate), and $\mu_{12} = m_1\,m_2/(m_1+m_2)$ the reduced mass. Equations (3.56) and (3.57) can be combined to give a single equation of relative motion,

$$\mu_{12}\,\ddot{\bf r} = k\,\frac{{\bf r}}{\vert{\bf r}\vert^{\,3}}.$$ (3.62)

Two relations that immediately follow from the previous equation are

$$\frac{d {\bf h}}{dt} ={\bf0},$$ (3.63)

$$\frac{d E}{dt} =0,$$ (3.64)

where

$${\bf h} = {\bf r}\times\dot{\bf r}$$ (3.65)

is the conserved angular momentum per unit mass, and

$$E = \frac{1}{2}\,\mu_{12}\,\vert\dot{\bf r}\vert^{\,2} + \frac{k}{\vert{\bf r}\vert}$$ (3.66)

the conserved energy.

Equation (3.65) implies that ${\bf r}\cdot{\bf h}=0$ . This is the equation of a plane that passes through the origin, and whose normal is parallel to the constant vector ${\bf h}$ . We, therefore, conclude that the relative position vector ${\bf r}$ is constrained to lie in this plane, which implies that the trajectories of both colliding particles are coplanar. Let the plane ${\bf r}\cdot{\bf h}=0$ coincide with the $x$ -$y$ plane, so that we can write ${\bf r}=(x,\,y)$ . It is convenient to define the standard plane polar coordinates $r=(x^2+y^2)^{1/2}$ and $\theta=\tan^{-1}(y/x)$ . When expressed in terms of these coordinates, the conserved angular momentum per unit mass becomes

$${\bf h} = h\,{\bf e}_z,$$ (3.67)

where

$$h = r^2\,\skew{3}\dot{\theta}.$$ (3.68)

Furthermore, the conserved energy takes the form

$$E= \frac{1}{2}\,\mu_{12}\left(\skew{3}\dot{r}^{\,2} + r^2\,\skew{3}\dot{\theta}^{\,2}\right) + \frac{k}{r}.$$ (3.69)

Suppose that $r=z^{\,-1}$ , where $z=z(\theta)$ and $\theta=\theta(r)$ . It follows that

$$\skew{3}\dot{r} = -\frac{\dot{z}}{z^2} = -r^2\,\frac{dz}{d\theta}\,\frac{d\theta}{dt} = -h\,\frac{dz}{d\theta}.$$ (3.70)

Hence, Equation (3.69) transforms to give

$$E = \frac{1}{2}\,\mu_{12}\,h^2\left[\left(\frac{dz}{d\theta}\right)^2 + z^2\right] + k\,z.$$ (3.71)

It is convenient to define the relative velocity at large distances,

$$u = \left(\frac{2\,E}{\mu_{12}}\right)^{1/2},$$ (3.72)

as well as the impact parameter,

$$b = \frac{h}{u}.$$ (3.73)

The latter parameter is simply the distance of closest approach of the two particles in the situation in which there is no Coulomb force acting between them, and they, consequently, move in straight-lines. (See Figure 3.1.) The previous three equations can be combined to give

$$b^2\left(\frac{dz}{d\theta}\right)^2 = 1-b^2\,z^2- \left(\frac{k}{E}\right)z.$$ (3.74)

Figure 3.1: A two-body Coulomb collision.

Figure 3.1 shows the collision in a frame of reference in which particle $2$ remains stationary at the origin, $O$ , whereas particle $1$ traces out the path $ABC$ . Point $B$ corresponds to the closest approach of the two particles. It follows, by symmetry (because Coulomb collisions are reversible), that the angles $\alpha$ and $\beta$ shown in the figure are equal to one another. Hence, we deduce that

$$\chi = \pi-2\,{\mit\Theta}.$$ (3.75)

Here, $\chi$ is the angle through which the path of particle $1$ (or particle $2$ ) is deviated as a consequence of the collision, whereas ${\mit\Theta}$ is the angle through which the relative position vector, ${\bf r}$ , rotates as particle $1$ moves from point $A$ (which is assumed to be infinity far from point $O$ ) to point $B$ . Suppose that point $A$ corresponds to $\theta=0$ . It follows that

$${\mit\Theta} = \int_0^{z_{\rm max}} \frac{d\theta}{dz}\,dz.$$ (3.76)

Here, $z_{\rm max} =1/r_{\rm min}$ , where $r_{\rm min}$ is the distance of closest approach. Now, by symmetry, $(dz/d\theta)_{z_{\rm max}}=0$ , so Equation (3.74) implies that

$$1-b^2\,z_{\rm max}^{\,2}- \left(\frac{k}{E}\right)z_{\rm max}=0.$$ (3.77)

Combining Equations (3.74) and (3.76), we obtain

$${\mit\Theta} =\int_0^{z_{\rm max}} \frac{b\,dz}{\sqrt{1-b^2\,z^2-k\,z/E}}= \int_0^{\zeta_{\rm max}}\frac{d\zeta}{1-\zeta^{\,2}-\alpha\,\zeta},$$ (3.78)

where $\alpha=k/(E\,b)$ , and

$$1-\zeta_{\rm max}^{\,2} -\alpha\,\zeta_{\rm max} = 0.$$ (3.79)

Integration (Speigel, Liu, and Lipschutz 1999) yields

$${\mit\Theta} = \frac{\pi}{2}-\sin^{-1}\left(\frac{\alpha}{\sqrt{4+\alpha^{\,2}}}\right).$$ (3.80)

Hence, from Equation (3.75), we get

$$\chi = 2\,\sin^{-1}\left(\frac{\alpha}{\sqrt{4+\alpha^2}}\right),$$ (3.81)

which can be rearranged to give

$$\cot\left(\frac{\chi}{2}\right) = \frac{2\,E\,b}{k}= \frac{4\pi\,\epsilon_0\,\mu_{12}\,u^{2}\,b}{e_1\,e_2}.$$ (3.82)