Particle momenta and cross section

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
10 replies · 2K views · last reply
Schwarzschild90
Messages
111
Reaction score
1
Will someone explain this step to me?
 
Attachments
  • Three lines of particle physics algebra: |M|^2 = 8g^4/(p1 - p3)^4 times [(p1 . p2)(p3 . p4) + (p1 . p4)(p2 . p3)]; then (p1 - p3)^2 = p1^2 + p3^2 - 2p1 . p3 = -2p1 . p3; then |M|^2 = 2g^4/(p1 . p3)^2 times [(p1 . p2)(p3 . p4) + (p1 . p4)(p2 . p3)].
    Calculation.PNG
    3.3 KB · Views: 513
Physics news on Phys.org
Some more background would be helpful. I suppose this is for massless particles where p^2=0?
 
But neither electrons nor myons are massless. Or is this some approximation?
 
it has nothing to do with CM frame ([itex]p^2= m^2[/itex] in any frame -frame independent minkowski product).
Probably they consider electron/muon with high enough momenta/Energy so that they can be considered massless (i.e. the last [itex]=[/itex] in 2nd line should be [itex]\approx[/itex] instead)
 
I was in a rush. CM has nothing to do with it.

Indeed, the particles considered are massless.

I figured it out.
 
Last edited:
[itex]E = \sqrt{p^2 + m^2} \Rightarrow E = p + \mathcal{O}\big(\frac{m^2}{2p}\big)[/itex]
(taylor expanding the square root for m<<p)