# Why does $2\to 2$ scattering cross-section have $E_{CM}^2$ in denominator?

For $$2\to 2$$ scattering with equal masses we have

$$\left(\frac{d\sigma}{d\Omega}\right)_{CM} = \frac{1}{64 \pi^2 E_{CM}^2} |\mathcal{M}|^2.$$

(Schwartz's QFT eq. 5.33) Can we make the general statement that total cross section increases as $$E_{CM}$$ decreases for scattering processes? I'm guessing not because $$\mathcal{M}$$ may have a complicated dependence on energy/momenta.

As an example, take the case of $$e^+e^-\to \mu^+\mu^-$$. With $$m_e = 0$$ we have

$$\frac{d\sigma}{d\Omega} = \frac{\alpha^2}{4 E_{CM}^2} \sqrt{1-m_{\mu}^2/E_{\mu}^2} \left(1+ m_{\mu}^2/E_{\mu}^2 + (1- m_{\mu}^2/E_{\mu}^2)\cos^2\theta\right)$$ in the c.o.m frame (Schwartz 13.78). Here it's pretty clear that total cross section increases as energy decreases.

But I would think that particles scatter more when they are given more energy? The limiting case of course being zero velocity $$\to$$ zero scattering. I would appreciate any intuition on the relation between $$E_{CM}$$ and total cross section.

• If the particles are going very fast they basically won’t be deflected significantly unless they hit almost head-on. Otherwise they’ll just glance past each other. I’m not sure how generally this reasoning works though. Oct 25, 2017 at 8:42
• Perhaps worth noting that this quantity is also know as the Mandelstam variable $s$, meaning that it is a Lorentz invariant and a good choice for writing expressions regardless. Oct 25, 2017 at 19:09

Can we make the general statement that total cross section increases as $$E_{CM}$$ decreases for scattering processes?
If the considered QFT is unitary (at least perturbatively), there is a constraint on the cross-section following from the unitarity, called Froissart bound. It is the direct consequence of the optical theorem. The latter states that $$\sigma_{1+2\to\text{all}} \simeq \frac{1}{s}\text{Im}(M_{\text{forward}}),$$ where $$\sigma_{1+2\to\text{all}}$$ is the $$1+2\to \text{all}$$ scattering cross-section, while $$s$$ is the invariant mass of the colliding particles. The amplitude $$M_{\text{forward}}$$ is the forward scattering amplitude bounded by $$M_{\text{forward}}\lesssim s\ln^{2}(s)$$ Therefore, for general $$1+2\to\text{all}$$ scattering one obtains the Frossairt bound: $$\sigma_{1+2\to \text{all}} \lesssim \ln^{2}(s)$$