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 p1for \(p_i\), pffor \(p_f\),
kifor \(k_i\) and kffor \(k_f\)) 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 p1instead of piin 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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