22.6 Example

We give here as an example of a simple calculation in high energy physics the computation of the Compton scattering cross-section as given in Bjorken and Drell Eqs. (7.72) through (7.74). We wish to compute the trace of \[ \frac {\alpha ^2}{2} \left (\frac {k^\prime }{k}\right )^2 \left (\frac {\gamma \cdot p_f+m}{2m}\right )\left (\frac {\gamma \cdot e^\prime \gamma \cdot e \gamma \cdot k_i}{2k.p_i} + \frac {\gamma \cdot e\gamma \cdot e^\prime \gamma \cdot k_f}{2k^\prime \cdot p_i}\right ) \] \[ \left (\frac {\gamma \cdot p_i+m}{2m}\right ) \left (\frac {\gamma \cdot k_i\gamma \cdot e\gamma \cdot e^\prime }{2k.p_i} + \frac {\gamma \cdot k_f\gamma \cdot e^\prime \gamma \cdot e}{2k^\prime \cdot p_i} \right ) \] where \(k_i\) and \(k_f\) are the four-momenta of incoming and outgoing photons (with polarization vectors \(e\) and \(e^\prime \) and laboratory energies \(k\) and \(k^\prime \) respectively) and \(p_i\), \(p_f\) are incident and final electron four-momenta.

Omitting therefore an overall factor \(\displaystyle \frac {\alpha ^2}{2m^2}\left (\frac {k^\prime }{k}\right )^2\) we need to find one quarter of the trace of \[ \left ( \gamma \cdot p_f + m\right ) \left (\frac {\gamma \cdot e^\prime \gamma \cdot e\gamma \cdot k_i}{2k.p_i} + \frac {\gamma \cdot e\gamma \cdot e^\prime \gamma \cdot k_f}{2k^\prime .p_i}\right )\times \] \[ \qquad \left ( \gamma \cdot p_i + m\right ) \left (\frac {\gamma \cdot k_i\gamma \cdot e\gamma \cdot e^\prime }{2k.p_i} + \frac {\gamma \cdot k_f\gamma \cdot e^\prime \gamma \cdot e}{2k^\prime .p_i}\right ) \] A straightforward REDUCE program for this, with appropriate substitutions (using p1 for \(p_i\), pf for \(p_f\), ki for \(k_i\) and kf for \(k_f\)) is

 on div; % this gives output in same form
         % as Bjorken and Drell.
 mass ki= 0, kf= 0, p1= m, pf= m; vector e,ep;
 % if e is used as a vector, it loses its scalar
 %      identity as the base of natural logarithms.
 mshell ki,kf,p1,pf;
 let p1.e= 0, p1.ep= 0, p1.pf= m^2+ki.kf, p1.ki= m*k,
     p1.kf= m*kp, pf.e= -kf.e, pf.ep= ki.ep,
     pf.ki= m*kp, pf.kf= m*k, ki.e= 0, ki.kf= m*(k-kp),
     kf.ep= 0, e.e= -1, ep.ep=-1;
 operator gp;
 for all p let gp(p)= g(l,p)+m;
 comment this is just to save us a lot of writing;
 gp(pf)*(g(l,ep,e,ki)/(2*ki.p1) + g(l,e,ep,kf)/
     (2*kf.p1))
   * gp(p1)*(g(l,ki,e,ep)/(2*ki.p1) + g(l,kf,ep,e)/
     (2*kf.p1))$
 write "The Compton cxn is ",ws;

(We use p1 instead of pi in the above to avoid confusion with the reserved variable pi).

This program will print the following result

                         2    1      -1    1   -1
The Compton cxn is 2*e.ep  + ---*k*kp   + ---*k  *kp - 1
                              2            2

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