I am trying to derive the inelastic cross section (the mass changes). I looked up the elastic derivation on Peskin's book (are there any alternatives?):

$$d \sigma = \frac{1}{2 E_a 2 E_b \left| v_a-v_b \right| } \left( \prod_f \frac{d^3 p_f}{(2 \pi)^3} \frac{1}{2 E_f} \right) (2 \pi)^4 \delta^{(4)} \left( p^\mu_a + p^\mu_b -\sum_f p^\mu_f \right) \left| M(p_A,p_B \rightarrow \left\{p_f\right\}) \right|^2$$

So it seems like the only difference for the inelastic version is the $\delta^{(4)}$. From this point the author defines:

$$\int d \Pi_n = \left( \prod_f \int \frac{d^3 p_f}{(2 \pi)^3} \frac{1}{2 E_f} \right) (2 \pi)^4 \delta^{(4)} \left( P -\sum_f p_f \right)$$

In the case of two particles in the final state this becomes:

$$\int d \Pi_2 = \left( \int \frac{d p_1 p^2_1 d\Omega}{(2 \pi)^3 2 E_2 2 E_1} d^3 p_2 \right) (2 \pi) \delta^{(4)} \left( P^\mu_{init} -(P^\mu_1+P^\mu_2) \right)$$

Now the phase space integrals are evaluated in the center of mass frame. Integrating over the 3 delta functions:

$$\int d^3 p_2 \delta^{(3)} \left(p_1+p_2\right) = \theta^{(3)} \left( p_1 + p_2 \right)$$

Here the result is a $\theta$ function which disappears and I am not sure how. The author simply assumes $\vec{p_1}=-\vec{p_2}$ for the momentum of the particles after collision, although it is not forced on the $p_1$ integral in any way. I guess that $p_1$ and $p_2$ are always integrated from 0 so the $\theta^{(3)} \left( p_1 + p_2 \right)=1$. The main issue is when doing the integration over the final delta function:

$$\int d p_1 p^2_1 \delta \left( E_{initial} - (E_1 + E_2) \right)$$

I am not sure how the author did this integration and if it would be different in the inelastic case (where the center of momentum frame is not the same after collision). I am guessing it was:

$$\int d p_1 p^2_1 \delta \left( \sqrt{p^2_1+m^2_1}+\sqrt{p^2_1+m^2_2} - (E_1 + E_2) \right)$$

But Mathematica can't solve that.

  • $\begingroup$ "The author simply assumes $\vec{p}_1=−\vec{p}_2$ for the momentum of the particles after collision" This still for a two-particle final state and evaluated in the intial-state CoM frame? If so, think a little on the matter. It should be pretty clear, why. $\endgroup$ – dmckee Jul 27 '15 at 4:15
  • $\begingroup$ It is clear why this is true in the CoM frame. What is not clear is how this follows from the $\theta$ function. Bad wording. $\endgroup$ – user1581390 Jul 27 '15 at 4:38

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