Help with derivation of two-body relativistic cross section

[charlesmartin14]

Homework Statement

I am trying to derive the 2 body cross section, as given in
https://web2.ph.utexas.edu/~schwitte/PHY362L/QMnote.pdf

dσ/dΩ

Homework Equations

I am stuck on

d√s/dp=v

where

√s=(Ea+Eb)=E

The Attempt at a Solution

p=Ev (relativisitic energy-momentum relationship)
√s=E

E=Ec+Ed=p/vc + p/vd

pc = pd = pf

d√s/dp = d(Ec+Ed)/dp = d(p/vc + p/vc)/dp=(1/vc + 1/vd)=(vc+vd)/(vc*vd)=vf/vc*vd

.

 

[vela]

Did you have a specific question?

 

[charlesmartin14]

I am trying to derive the 2+2 body relativisitic cross section, as given in
https://web2.ph.utexas.edu/~schwitte/PHY362L/QMnote.pdf

a+b -> c+d

I want the form of the cross section

dσ/dΩ ~ 1/((v_i)(v_f))

I am stuck on step (13)-(14)
and I don't understand why d√s/dp_f=v_f

Here, s is the mandelstam variable √s=(Ec+Ed)=(Ea+Eb)
and p_f is the (amplitude) of the final (I assume 3 vector ) momentum

 

[vela]

$\vec{p}_\text{f}$ is the three-momentum of particle C.

Try differentiating $-m_c^2 = p_c^2-E_c^2$ to calculate $dE_c/dp_c$. I think the problem with your attempt is that you're assuming $dv_c/dp_c = 0$ and $dv_d/dp_c = 0$.

 

[charlesmartin14]

$\vec{p}_\text{f}$ is the three-momentum of particle C.

Try differentiating $-m_c^2 = p_c^2-E_c^2$ to calculate $dE_c/dp_c$. I think the problem with your attempt is that you're assuming $dv_c/dp_c = 0$ and $dv_d/dp_c = 0$.

 

[charlesmartin14]

Thanks ! That makes perfect sense. this is a great forum