1. Cross Sections One of the most important quantities we measure in nuclear physics is the cross section. Cross sections always have units of area and in addition may also be characterized by additional observables, such as solid angle, energy, momentum, etc.. We start with a simple picture first. We want to calculate the scattering rate of incident projectiles from targets of area  distributed randomly, with a density  throughout a slab of volume of area S and thickness dx. The scattering rate here refers to any process which removes an incident projectile by actually scattering it or by changing its kinematic properties or character in any way. In this case we are actually calculating the total cross section. We assume that if the projectile hits any part of the target area  it is lost from the incident beam. We also assume that the incident beam is uniformly distributed over the target area S.

2. S dx   Suppose there is a target of area  in a volume dx X S. An incident flux, , of projectiles strikes the volume. The density of targets of area  Is 

4. The quantity  is called the absorption coefficient. Usually the total cross section is understood to be used in this expression and it is simply written as  Target thickness are often quoted in grams/cm 2. This is because the important quantity for rate calculation is  x. What we really use in rate calculations is (number of target nuclei)/ cm 2. We use Avogadro’s number, A 0 to convert mass density to particle density. A 0 = 6.02x10 23 atoms/mole

5. Differential Cross Sections We frequently measure particular reaction outcomes rather than the total cross section. These partial cross sections can be differential in the sense that they depend upon rather specific final state properties. Consider the case of elastic scattering.  dd N(  ) incident particles are scattered into the solid angle d  =sin(  )d  d  n a real experiment we must use a finite solid angle . If N 0 incident particles produce N(  scatters for a target of thickness  x, then the differential cross section is, 

6. Classical Cross Section and Potential Scattering Particles scatter because they change their momenta. This necessarily implies a force has acted. It can be shown from Newton’s equations of motion that if two particles interact with each other the problem can be separated into two pieces. The momentum of the center of mass is constant if there are no external forces. The coordinate that separates two particles has an equation of motion given by The reduced mass is  and the separation between particles 1 and 2 is r. Thus, to solve a two body problem in classical mechanics it is sufficient to solve two one body problems. We can then focus on how to solve the scattering problem for a particle of mass m from a potential V.

7. Potential Scattering This section utilizes the discussion in chapter 1 of the book by Das and Ferbel 1. The goal of this calculation is to determine the scattering angle  for a particle given an initial momentum mv 0 and impact parameter b for a central potential V(r). b  r rr0r0  mv 0  The particle follows the trajectory ( in blue ). At a given moment it is a distance r from the center of the force and at an angle . The distance of closest approach to the scattering center is r 0.  0 is the angle between the initial direction and the point of closest approach. The asymptotic angle of scattering is . We assume the potential goes to zero at infinity so that the total energy of the particle is its kinetic energy at infinity. For a central potential the torque is zero, hence the angular momentum is constant. 

8. For the position of the particle shown in the figure the negative square root solution is appropriate because the radial distance is decreasing. The positive square root is the solution for the particle leaving the scattering center. We could solve this differential equation by numerical methods for any central potential. Note that r = r(t). If we know r(t) we can then solve for  (t).

10. We solve the implicit equation for the distance of closest approach. From the diagram we see that the asymptotic scattering angle is  =    Substituting for L we obtain for  0,

11. From these equations for  we see that different values for the impact parameter correspond to different angles of scattering. Thus we can make a one-to-one correspondence between the scattering angle and b. In particular, we see that if particles approaching the target pass through an annulus of radius b and width db these will all emerge at an angle  (b). This annulus area is the correct differential cross section for the angle of observation. We can invert the relationship to get b = b(  ), then db = |(db/d  )d . Consider the area of the section of the annulus shown. b db d  (  ) =bd  db dd d  sin(  )d  d 

12. Historically this analysis was very important for the development of nuclear physics. E. Rutherford and his students discovered the very tiny size of the nucleus by studying Coulomb scattering. Rutherford did the classical analysis, but a quantum mechanical analysis of Coulomb scattering yields the same result in the first order approximation. Since the Coulomb potential goes as 1/r, any central potential that has this radial behavior will have the same form for its differential cross section. The gravitational potential is of the form 1/r. 1) “Introduction to Nuclear and Particle Physics”, Ashok Das and Thomas Ferbel, John Wiley and Sons, 1994