Scattering in the Center of Mass Frame

As before, the first particle is of mass , and is located at position
vector , whereas the second particle is of mass ,
and is located at .
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 then the second must approach
with momentum . Likewise, after the collision, if the
first particle recedes from the collision point with momentum
then the second must recede with momentum --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, , is equal to the magnitude of the initial momentum vector, . Because of this, the collision event is completely specified
once the angle 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.

Suppose that the two particles interact via the potential , where 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 is scattered in the fixed potential , where is now the distance from the origin. The vector position of the particle in the equivalent problem corresponds to the relative position vector in the original problem. It follows that the angle through which the particle is scattered in the equivalent problem is the same as the scattering angle in the original problem.

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

Suppose that the polar coordinates of the particle in the equivalent problem
are
. Let the particle approach the origin from the direction
, and attain its closest distance to the origin when
. From symmetry, the angle in Figure 22 is equal to the angle . However, from simple geometry,
. Hence,

where , and is the angular momentum per unit mass in the equivalent problem. It is easily seen that

(342) |

The above equation can be rearranged to give

Integration yields

Here, , where is the distance of closest approach. Since, by symmetry, , it follows from Equation (344) that

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

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 to . These particles are scattered in such a manner that their scattering
angles lie in the range to
, where
is determined from inverting the function , and

(347) |

(348) |

(349) |

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

and measures the effective target area for scattering in

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

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

(353) |

(354) |

Hence, Equations (350) and (355) yield

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

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

(358) |

(359) |

(360) |

(361) |

(362) |

(363) |

(364) |

There are a number of things to note about the above formula. First, the scattering cross-section is proportional to . This means that

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

(366) |

This very famous formula is known as the

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
as
. 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.