in Peskin Schroeder after the derivation of the differential cross section there is a comment for the central mass system (CMS), which says:

In the special case, where all four particles have identical masses (...), this [the general formula] reduces to the formula[...] (p.107): $$ \left(\frac{d\sigma}{d\Omega}\right)_{CM}=\frac{\left|\mathcal{M}\right|^2}{64\pi^2E_{CM}^2}$$ However in Srednicki the general formula for the differential cross section in the CMS is (p.97): $$\left(\frac{d\sigma}{d\Omega}\right)_{CM}=\frac{\left|\mathcal{M}\right|^2}{64\pi^2E_{CM}^2}\frac{\left|\textbf{k}^\prime\right|}{\left|\textbf{k}\right|} $$ where $\left|\textbf{k}^\prime\right|$ is the outgoing three-momentum in the CMS and $\left|\textbf{k}\right|$ the incoming. Now to get from the formula from Srednicki to Peskin's formula I don't need the masses to be all the same but merely the incoming masses equal to the outgoing masses, so to say elastic scattering. I didn't see a further restriction in Srednicki's formula apart from being in the CMS.

If I take e.g. Compton scattering I get with Srednicki's formula: $$\left(\frac{d\sigma}{d\Omega}\right)_{CM}=\frac{\left|\mathcal{M}\right|^2}{64\pi^2E_{CM}^2} $$ but Peskin gets on page 164 with his formula for the differential cross section: $$\left(\frac{d\sigma}{d\Omega}\right)_{CM}=\frac{1}{32} \frac{1}{E_1E_2}\frac{\left|\textbf{k}^\prime\right|}{E_{CM}}$$ where $E_1+E_2=E_{CM}$.

I don't see, that this is equal? Is it actually equal? Where did I miss something?

  • $\begingroup$ Can you check the page number for Srednicki? I'm looking at a digital library copy that I can't flip through easily, and page 97 is in the section "Loop corrections to the propagator". EDIT: Nevermind, found it on pages 79-84. $\endgroup$
    – jwimberley
    Commented Oct 1, 2013 at 16:44

1 Answer 1


The formula given by Srednicki $$d \sigma_{\textrm{CM}} = \frac{1}{64 \pi^2 s } \frac{|\bf{p}'|}{|\bf{p}|}|\mathcal{M}|^2 d \Omega_{\textrm{CM}}$$ is the general result for the CM $2 \to 2$ scattering where we have

$\bf{p}_1 = - \bf{p}_2 = \bf{p}$, $\bf{p}_3 = - \bf{p}_4 = \bf{p}'$, $(E_1 + E_2) = (E_3 + E_4) =\sqrt{s} $ and

$$ |{\bf{p}}| = \tfrac{1}{2\sqrt{s}}\sqrt{\lambda(s,m_1^2,m_2^2)}, \quad \quad |{\bf{p}'}| = \tfrac{1}{2\sqrt{s}}\sqrt{\lambda(s^2,m_3^2,m_4^2)} $$ with $\lambda(x,y,z) = (x-y-z)^2 - 4y z$ being the Källen function. Furthermore, we have $$E_{1/2} = \frac{s \pm (m_1^2 - m_2^2)}{2\sqrt{s}} \quad \quad E_{3/4} = \frac{s \pm (m_3^2 - m_4^2)}{2\sqrt{s}} $$

As you have correctly observed, for $m_1=m_3$ and $m_2=m_4$ we indeed obtain $\frac{|\bf{p}'|}{|\bf{p}|} =1 $ and the formula reduces to

$$d \sigma_{\textrm{CM}} = \frac{1}{64 \pi^2 s } |\mathcal{M}|^2 d \Omega_{\textrm{CM}} $$

On page 164 Peskin is referring to Eq. 4.84 which reads $$ \frac{d \sigma_{\textrm{CM}} }{d \Omega_{\textrm{CM}}} = \frac{1}{2 E_1 2 E_2 v_{12}} \frac{|\bf{p}'|}{(2\pi)^2 4 \sqrt{s}}, $$ with $v_{12}$ being the absolute value of the relative velocity of incoming particles. This expression is actually the same as the Sredinicki formula, just written in a slightly different way. To see this, observe that in any frame $v_{12}$ can be rewritten as $$ v_{12} =|{\bf{v}}_1 - {\bf{v}}_2| = \left | \frac{{\bf{p}_1}}{E_1} - \frac{{\bf{p}_2}}{E_2} \right | = \frac{\sqrt{(E_1 {\bf{p}_2} - E_2 {\bf{p}_1})^2}}{E_1 E_2} $$ and when we go to the CM frame ${\bf{p}} \equiv \bf{p_1} = - \bf{p}_2$ we end up with $$ v_{12} = \frac{|{\bf{p}}|(E_1 + E_2)}{E_1 E_2} = \frac{|{\bf{p}}|\sqrt{s}}{E_1 E_2} $$ Plugging this into Peskin's formula you arrive to $$ \frac{d \sigma_{\textrm{CM}} }{d \Omega_{\textrm{CM}}} = \frac{1}{64 \pi^2 s} \frac{|\bf{p}'|}{|\bf{p}|} |\mathcal{M}|^2 $$ in perfect agreement with Srednicki.

So yes, it is actually equal.

By the way, notice that in any frame where the particle velocities are parallel or antiparallel, $v_{12}$ can be written in a covariant form since $$ (p_1 \cdot p_2)^2 - m_1^2 m_2^2 = E_1^2 E_2^2 {(1 - \bf{v}_1 \cdot \bf{v}_2)^2} - E_1^2 E_2^2 {(1- {\bf{v}}_1^2) (1- {\bf{v}}_2^2)} = E_1^2 E_2^2 ({\bf{v}_1} - {\bf{v}}_2)^2 $$ where we used $({\bf{v}_1} \cdot {\bf{v}_2})^2 = {\bf{v}_1^2} {\bf{v}_2^2}$, so that $$ v_{12} = \frac{\sqrt{ (p_1 \cdot p_2)^2 - m_1^2 m_2^2}}{E_1 E_2} $$

  • $\begingroup$ Hi vsht, I believe that in the expression for $(p_1\cdot p_2)^2$ we should have $|E_1\mathbf{p}_2-E_2\mathbf{p}_1|^2$ so that the cross term is $-2E_1 E_2\mathbf{p}_1\cdot\mathbf{p}_2$ when expanded out, not $-2E_1 E_2|\mathbf{p}_1||\mathbf{p}_2|$? Also, could you help me with how you derived the expression $v_{12}=\frac{\sqrt{(p_1\cdot p_2)^2-m_1^2 m_2^2}}{E_1 E_2}$? In which frame are $E_1$ and $E_2$ defined? Thank you very much!! $\endgroup$ Commented Nov 5, 2017 at 14:31
  • 1
    $\begingroup$ Many thanks for your comment. I agree with your correction, but in fact this whole formula is not needed, since one can use a much simpler one. Also I was too sloppy regarding $v_{12}$. We can write it in a covariant form, but only in the frames where the particle velocities are parallel or antiparallel. I updated my answer accordingly. $\endgroup$
    – vsht
    Commented Nov 8, 2017 at 7:51

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