3
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ 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. $\endgroup$
    – knzhou
    Oct 25, 2017 at 8:42
  • $\begingroup$ 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. $\endgroup$ Oct 25, 2017 at 19:09

1 Answer 1

5
$\begingroup$

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: $$ \sigma_{1+2\to \text{all}} \lesssim \ln^{2}(s) $$

But I would think that particles scatter more when they are given more energy?

The more energy of the particle is, the less corresponding de Broglie wave-length is. Therefore qualitatively it's rather expected that the colliding particles with very large energies will overlook each other.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.