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3.10.1. Cross Sections

Consider a uniform beam of particles, all of the same mass m, energyEand

momentummv

0, incident upon a fixed center of central force. The force is assumed tofall off sufficiently fast with distance so that the particles can be taken as free beyond

certain distance from the center of force. The incident beam is further characterized

by itsintensity(flux density)Idefined as

I=number of particles crossing a unit area normal to the beam in unit time

The (differential) cross section ( ) of scattering into the direction is defined

as

( ) d = number of particles scattered into per unit timed

I

(!."")

#ote that some author write ( ) asd

d

$ their cross section thus corresponds to

our total cross section T. %et the coordinate origin be at the force center with the

zaxis parallel to v0. &n terms of spherical coordinates, we ha'e ( ) ,

=

= , where

and are the polar and azimuthal angles, respecti'ely. The angle is called the

scattering angle. Thus,

sind d d = (!.")

The characteristics (or constants) of the orbit of a gi'en particle, and hence the

amount of its scattering, are determined by its energyEand angular momentum l.

urthermore, since the force center is immo'able,Eand lare themsel'es constants of

the motion. or a particle far away from the force center, its perpendicular distances

to thezaxis is called its impact parameter. &ts (constant) angular momentum is

therefore

= L r p 0 z

m

= r v

&n cylindrical coordinates, we ha'e ( ), ,s z=r and ( )0 00,0,v=v so that

$ $

0

cos sin

0 0

m s s z

v

=

x y z

L

$

$ $( )0 sin cosmsv = x y 0msv = e$

l=L 0msv= *s mE= (!.0)

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+nceEands(hence l ) are fixed, the scattering angle is also fixed (conser'ation of

Lmeans is a constant). To proceed, we stassume eachsleads to a distinct 'alue of

. -ssuming no particle is captured into a bounded orbit around the force center,

e'ery incident particle will be reco'ered as a scattered particle far away from the

force center. &ncident particles lying in the plane element sdsd will all be scattered

into some solid angle element sind d d = . Thus,

( ) sinIsdsd I d d = (!.)

where the and sign are chosen so that ( ) is always positi'e. #ow, sin 0

since 0 . Therefore, the and sign applies to attracti'e and replusi'e

forces, respecti'ely. or central forces, ( ) is independent of so that we can

write

( )sin

s ds

d =

(!.!)