Peskin/Schroeder 4.77 page (about cross-sections)

[Manu_]

Hello everyone,

I am trying to solve equation 4.77 (about cross-sections) in peskin/schroeder's book.
They state that:

$$\int d\bar{k}^{z}_{A} \delta \left( \sqrt{\bar{k}^{z}_{A}+m^{2}_{A}} + \sqrt{\bar{k}^{z}_{B}+ m^{2}_{B}} - \Sigma E_{f} \right) \Big{|}_{\bar{k}^{z}_{B}=\Sigma p^{z}_{f}-\bar{k}^{z}_{A}} = \frac{1}{\lvert \frac{\bar{k}^{z}_{A}}{ \bar{E}_{A}} - \frac{\bar{k}^{z}_{B}}{\bar{E}_{B}} \rvert}$$

With A and B the initial protons that collide, z the longitudinal direction, k the momentum, E the energy, and the subscript f denotes the final state (i.e., after collision).
I don't understand where does this come from. I tried to use some definitions of the delta function, but I'm not even getting close. Does someone have an idea?

Thank you.

 

[vanhees71]

Note that
$$\int_{\mathbb{R}} \mathrm{d} x \delta[f(x)] \phi(x)=\sum_{j} \frac{1}{|f'(x_j)|} \phi(x_j),$$
where $f$ is a function that has only 1st-degree zeros, $x_j$ (i.e., $f'(x_j) \neq 0$ for all $j$) and $\phi(x)$ is a appropriate test function. In other words as a distribution you have
$$\delta[f(x)]=\sum_{j} \frac{1}{|f'(x_j)|} \delta(x-x_j).$$

Now for your formula we use (I write $k_A^z=k_A$ for convenience)
$$f(k_A)=\sqrt{k_A^2+m_A^2}+\sqrt{k_B^2+m_B^2}-\sum E_f.$$
Now you have
$$f'(k_A)=\frac{k_A}{E_A}-\frac{k_B}{E_B},$$
where the above given square roots are $E_A$ and $E_B$. From this you get the formula, which is understood to fulfill all energy-momentum-conservation and on-shell constraints at place,
$$E_A+E_B-\sum E_f=0, \quad \vec{p}_A+\vec{p}_B-\sum_E \vec{p}_f=0$$.

 

[Manu_]

Thank you your answer, Vanhees, it got clearer for me now.