Up | Next | Prev | PrevTail | Tail |
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
Up | Next | Prev | PrevTail | Front |