Suppose that, in the center of mass frame, the first particle has velocity
before the collision, and velocity after the
collision. Likewise, the second particle has velocity before the
collision, and after the collision. We know that

(368) |

(369) | |||

(370) |

Let us transform
to a new inertial frame of reference--which we shall call the *laboratory frame*--which is moving with the uniform velocity with respect to
the center of mass frame. In the new reference frame, the first
particle has initial velocity
, and
final velocity
. Furthermore, the second particle
is initially at *rest*, and has the final velocity
. The relationship between scattering in the center of mass frame
and scattering in the laboratory frame is illustrated in Figure 23.

In the center of mass frame, both particles are scattered through the same angle . However, in the laboratory frame, the first and second particles are scattered by the (generally different) angles and , respectively.

Defining - and -axes, as indicated in Figure 23, it is easily
seen that the Cartesian components of the various velocity vectors in the
two frames of reference are:

In the center of mass frame, let be the total energy, let and be the kinetic energies of the first and second particles, respectively, before the collision, and let and be the kinetic energies of the first and second particles, respectively, after the collision. Of course, . In the laboratory frame, let be the total energy. This is, of course, equal to the kinetic energy of the first particle before the collision. Likewise, let and be the kinetic energies of the first and second particles, respectively, after the collision. Of course, .

The following results can easily be obtained from the above definitions and
Equations (371)-(377). First,

These equations specify how the total energy in the center of mass frame is distributed between the two particles. Note that this distribution is

These equations specify how the total energy in the laboratory frame is distributed between the two particles after the collision. Note that the energy distribution in the laboratory frame is

Equations (371)-(377), and some simple trigonometry, yield

(384) |

Differentiating Equation (383) with respect to , we obtain

(386) |

(387) |

(388) |

(389) |

Equations (378)-(385) enable us to relate the particle energies and scattering angles in the laboratory frame to those in the center of mass frame. In general, these relationships are fairly complicated. However, there are two special cases in which the relationships become much simpler.

The first special case is when . In this limit, it is easily seen
from Equations (378)-(385) that the second mass is *stationary*
both before and after the collision, and that the center of mass frame
*coincides* with the laboratory frame (since the energies and
scattering angles in the two frames are the same). Hence, the simple analysis
outlined in Section 6.4 is applicable in this case.

The second special case is when . In this case, Equation (383)
yields

(390) |

In other words, the scattering angle of the first particle in the laboratory frame is

(392) |

In other words, the total energy in the laboratory frame is

(394) |

(395) | |||

(396) |

Thus, in the laboratory frame, the unequal energy distribution between the two particles after the collision is simply related to the scattering angle .

What is the angular distribution of scattered particles when a beam
of particles of the first type scatter off stationary particles of the second type?
Well, we can define a differential scattering cross-section,
, in the
laboratory frame, where
is an element of solid angle in
this frame. Thus,
is the effective cross-sectional area in the laboratory frame
for scattering into the range of scattering angles to .
Likewise,
is
the effective cross-sectional area in the center of mass frame for
scattering into the range of scattering angles to
.
Note that
.
However, a cross-sectional area is not changed when we transform between
different inertial frames. Hence, we can write

(397) |

(398) |

The above equation allows us to relate the differential scattering cross-section in the laboratory frame to that in the center of mass frame. In general, this relationship is extremely complicated. However, for the special case where the masses of the two types of particles are

Let us now consider some specific examples. We saw earlier that, in the
center of mass frame, the
differential scattering cross-section for *impenetrable spheres* is [see Equation (356)]

Note that this cross-section is

(403) |

As we have seen, the Rutherford scattering cross-section takes
the form [see Equation (367)]

(404) |

(405) |