Scattering in the Center of Mass Frame

Let us now consider scattering due to the collision of two particles. We shall restrict our discussion to particles which interact via conservative central forces. It turns out that scattering looks particularly simple when viewed in the center of mass frame. Let us, therefore, start our investigation by considering two-particle scattering in the center of mass frame.

As before, the first particle is of mass $m_1$, and is located at position vector ${\bf r}_1$, whereas the second particle is of mass $m_2$, and is located at ${\bf r}_2$. By definition, there is zero net linear momentum in the center of mass frame at all times. Hence, if the first particle approaches the collision point with momentum ${\bf p}$ then the second must approach with momentum ${\bf -p}$. Likewise, after the collision, if the first particle recedes from the collision point with momentum ${\bf p}'$ then the second must recede with momentum $-{\bf p}'$--see Figure 21. Furthermore, since the interaction force is conservative, the total kinetic energy before and after the collision must be the same. It follows that the magnitude of the final momentum vector, ${\bf p}'$, is equal to the magnitude of the initial momentum vector, ${\bf p}$. Because of this, the collision event is completely specified once the angle $\theta$ through which the first particle is scattered is given. Of course, in the center of mass frame, the second particle is scattered through the same angle--see Figure 21.

Figure 21: A collision viewed in the center of mass frame.

Suppose that the two particles interact via the potential $U(r)$, where $r$ is the distance separating the particles. As we have seen, the two-body problem sketched in Figure 21 can be converted into the equivalent one-body problem sketched in Figure 22. In this equivalent problem, a particle of mass $\mu= m_1\,m_2/(m_1+m_2)$ is scattered in the fixed potential $U(r)$, where $r$ is now the distance from the origin. The vector position ${\bf r}$ of the particle in the equivalent problem corresponds to the relative position vector ${\bf r}_2-{\bf r}_1$ in the original problem. It follows that the angle $\theta$ through which the particle is scattered in the equivalent problem is the same as the scattering angle $\theta$ in the original problem.

Figure 22: The one-body equivalent to the previous figure.

The scattering angle, $\theta$, is largely determined by the so-called impact parameter, $b$, which is the distance of closest approach of the two particles in the absence of an interaction potential. In the equivalent problem, $b$ is the distance of closest approach to the origin in the absence of an interaction potential--see Figure 22. If $b=0$ then we have a head-on collision. In this case, we expect the two particles to reverse direction after colliding: i.e., we expect $\theta=\pi$. Likewise, if $b$ is large then we expect the two particles to miss one another entirely, in which case $\theta=0$. It follows that the scattering angle, $\theta$, is a decreasing function of the impact parameter, $b$.

Suppose that the polar coordinates of the particle in the equivalent problem are $(r,\,\vartheta)$. Let the particle approach the origin from the direction $\vartheta=0$, and attain its closest distance to the origin when $\vartheta={\mit\Theta}$. From symmetry, the angle $\alpha$ in Figure 22 is equal to the angle $\beta$. However, from simple geometry, $\alpha={\mit\Theta}$. Hence,

$$\theta = \pi - 2\,{\mit\Theta}.$$ (340)

Now, by analogy with Equation (267), the conserved total energy $E$ in the equivalent problem, which can easily be shown to be the same as the total energy in the original problem, is given by

$$E = \frac{\mu\,h^2}{2}\left[\left(\frac{du}{d\vartheta}\right)^2 + u^2\right] + U(u),$$ (341)

where $u=r^{-1}$, and $h$ is the angular momentum per unit mass in the equivalent problem. It is easily seen that

$$h = b\,v_\infty = b\left(\frac{2\,E}{\mu}\right)^{1/2},$$ (342)

where $v_\infty$ is the approach velocity in the equivalent problem at large $r$. It follows that

$$E = E\,b^2\left[\left(\frac{du}{d\vartheta}\right)^2 + u^2\right] + U(u).$$ (343)

The above equation can be rearranged to give

$$\frac{d\vartheta}{d u } = \frac{b}{\sqrt{1 - b^2\,u^2-U(u)/E}}.$$ (344)

Integration yields

$${\mit\Theta} = \int_0^{u_{max}}\frac{b\,du}{\sqrt{1-b^2\,u^2- U(u)/E}}.$$ (345)

Here, $u_{max}=1/r_{min}$, where $r_{min}$ is the distance of closest approach. Since, by symmetry, $(du/d\vartheta)_{u_{max}}=0$, it follows from Equation (344) that

$$1 - b^2\,u_{max}^2 - U(u_{max})/E = 0.$$ (346)

Equations (340) and (345) enable us to calculate the function $b(\theta)$ for a given interaction potential, $U(r)$, and a given total energy, $E$, of the two particles in the center of mass frame. The function $b(\theta)$ tells us which impact parameter corresponds to which scattering angle, and vice versa.

Instead of two particles, suppose that we now have two counter-propagating beams of identical particles (with the same properties as the two particles described above) which scatter one another via binary collisions. What is the angular distribution of the scattered particles? Consider pairs of particles whose impact parameters lie in the range $b$ to $b+db$. These particles are scattered in such a manner that their scattering angles lie in the range $\theta$ to $\theta+d\theta$, where $\theta$ is determined from inverting the function $b(\theta)$, and

$$d\theta = \frac{db}{\vert db(\theta)/d\theta\vert}.$$ (347)

Incidentally, we must take the modulus of $db(\theta)/d\theta$ because $b(\theta)$ is a decreasing function of $\theta$. Assuming, as seems reasonable, that the scattering is azimuthally symmetric, the range of solid angle into which the particles are scattered is

$$d{\mit\Omega} = 2\pi\,\sin\theta\,d\theta = \frac{2\pi\,\sin\theta\,db}{\vert db/d\theta\vert}$$ (348)

Finally, the cross-sectional area of the annulus through which incoming particles must pass if they are to have impact parameters in the range $b$ to $b+db$ is

$$d\sigma = 2\pi\,b\,db.$$ (349)

The previous two equations allow us to define the differential scattering cross-section:

$$\frac{d\sigma}{d{\mit\Omega}} = \frac{b}{\sin\theta}\left\vert\frac{db}{d\theta}\right\vert$$ (350)

The differential scattering cross-section has units of area per steradian, and specifies the effective target area for scattering into a given range of solid angle. For two uniform beams scattering off one another, the differential scattering cross-section thus effectively specifies the probability of scattering into a given range of solid angle. The total scattering cross-section is the integral of the differential cross-section over all solid angles,

$$\sigma = \int \frac{d\sigma}{d{\mit\Omega}}\,d{\mit\Omega},$$ (351)

and measures the effective target area for scattering in any direction. Thus, if the flux of particles per unit area per unit time, otherwise known as the intensity, of the two beams is $I$, then the number of particles of a given type scattered per unit time is simply $I\,\sigma$.

Let us now calculate the scattering cross-section for the following very simple interaction potential:

$$U(r) = \left\{\begin{array}{lll} 0&\mbox{\hspace{1cm}} & r>a\\ [0.5ex] \infty && r\leq a \end{array}\right..$$ (352)

This is the interaction potential of impenetrable spheres which only exert a force on one another when they are in physical contact (e.g., billiard balls). If the particles in the first beam have radius $R_1$, and the particles in the second beam have radius $R_2$, then $a=R_1+R_2$. In other words, the centers of two particles, one from either beam, can never be less than a distance $a$ apart, where $a$ is the sum of their radii (since the particles are impenetrable spheres).

Equations (340), (345), and (352) yield

$$\theta = \pi - 2\int_0^{1/a}\frac{b\,du}{\sqrt{1-b^2\,u^2}} = \pi - 2\,\sin^{-1}(b/a).$$ (353)

The above formula can be rearranged to give

$$b(\theta) = a\,\cos(\theta/2).$$ (354)

Note that

$$b\left\vert\frac{db}{d\theta}\right\vert = \frac{1}{2}\left\... ...}\,\sin(\theta/2)\,\cos(\theta/2) = \frac{a^2}{4}\,\sin\theta.$$ (355)

Hence, Equations (350) and (355) yield

$$\frac{d\sigma}{d{\mit\Omega}} = \frac{a^2}{4}.$$ (356)

We thus conclude that when two beams of impenetrable spheres collide, in the center of mass frame, the particles in the two beams have an equal probability of being scattered in any direction. The total scattering cross-section is

$$\sigma = \int \frac{d\sigma}{d{\mit\Omega}}\,d{\mit\Omega} = \pi\,a^2.$$ (357)

Obviously, this result makes a lot of sense--the total scattering cross-section for two impenetrable spheres is simply the area of a circle whose radius is the sum of the radii of the two spheres.

Let us now consider scattering by an inverse-square interaction force whose potential takes the form

$$U(r) = \frac{k}{r}.$$ (358)

It follows from Equations (345) and (346) that

$${\mit\Theta} = \int_0^{u_{max}}\frac{b\,du}{\sqrt{1-b^2\,u^2 - k\,u/E}} = \int_0^{x_{max}}\frac{dx}{\sqrt{1-x^2- \alpha\,x}},$$ (359)

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

$$1 - x_{max}^{\,2} - \alpha\,x_{max}= 0.$$ (360)

Integration yields

$${\mit\Theta} = \frac{\pi}{2} - \sin^{-1}\left(\frac{\alpha}{\sqrt{4+\alpha^2}}\right).$$ (361)

Hence, from Equation (340), we obtain

$$\theta = 2\,\sin^{-1}\left(\frac{\alpha}{\sqrt{4+\alpha^2}}\right).$$ (362)

The above equation can be rearranged to give

$$b^2 = \frac{k^2}{4\,E^{\,2}}\,\cot^2(\theta/2).$$ (363)

Thus,

$$2\, b\left\vert\frac{db}{d\theta}\right\vert = \frac{k^2}{8\,E^{\,2}}\frac{\sin\theta}{\sin^4(\theta/2)}.$$ (364)

Finally, using Equation (350), we get

$$\frac{d\sigma}{d{\mit\Omega}} = \frac{k^2}{16\,E^{\,2}}\frac{1}{\sin^4(\theta/2)}.$$ (365)

There are a number of things to note about the above formula. First, the scattering cross-section is proportional to $k^2$. This means that repulsive ($k>0$) and attractive ($k<0$) inverse-square interaction forces of the same strength give rise to identical angular distributions of scattered particles. Second, the scattering cross-section is proportional to $E^{-2}$. This means that inverse-square interaction forces are much more effective at scattering low energy, rather than high energy, particles. Finally, the differential scattering cross-section is proportional to $\sin^{-4}(\theta/2)$. This means that, with an inverse-square interaction force, the overwhelming majority of ``collisions'' consist of small angle scattering events (i.e., $\theta\ll 1$).

Let us now consider a specific case. Suppose that we have particles of electric charge $q$ scattering off particles of the same charge. The interaction potential due to the Coulomb force between the particles is simply

$$U(r) = \frac{q^2}{4\pi\,\epsilon_0\,r}.$$ (366)

Thus, it follows from Equation (365) [with $k=q^2/(4\pi\,\epsilon_0)$] that the differential scattering cross-section takes the form

$$\frac{d\sigma}{d{\mit\Omega}} = \frac{q^4}{16\,(4\,\pi\,\epsilon_0)^2\,E^{\,2}}\frac{1}{\sin^4(\theta/2)}.$$ (367)

This very famous formula is known as the Rutherford scattering cross-section, since it was first derived by Earnst Rutherford for use in his celebrated $\alpha$-particle scattering experiment.

Note, finally, that if we try to integrate the Rutherford formula to obtain the total scattering cross-section then we find that the integral is divergent, due to the very strong increase in $d\sigma/d{\mit\Omega}$ as $\theta\rightarrow 0$. This implies that the Coulomb potential (or any other inverse-square-law potential) has an effectively infinite range. In practice, however, an electric charge is generally surrounded by charges of the opposite sign which shield the Coulomb potential of the charge beyond a certain distance. This shielding effect allows the charge to have a finite total scattering cross-section (for the scattering of other electric charges). However, the total scattering cross-section of the charge depends (albeit, logarithmically) on the shielding distance, and, hence, on the nature and distribution of the charges surrounding it.