What is the meaning of a cross section having "no angular dependence"?

This is in reference to page 112, equation 4.99 of Peskin and Schroeder.

We have just completed the first calculation (to first order) of a differential cross section. Specifically that of 2-particle to 2-particle scattering in $$\phi^4$$ theory:

$$\left(\frac{d\sigma}{d\Omega}\right)_{CM}=\frac{\lambda^2}{64\pi^2E_{CM}^2} \tag{4.99}.$$

Peskin and Schroeder state that this result is "rather dull", having "no angular dependence at all". The exact meaning of this statement is confusing to me. I hope I have understood correctly that a differential cross section tells us the probability of the final state particles being scattered into some small region of momentum space. We then integrate this quantity over a solid angle to find out the total probability of finding our final state particles within this region of momentum space (which I believe is referred to as phase space).

Does "no angular dependence" mean that in our phase space integral (which we do in spherical coordinates), the cross section depends only on the absolute momentum of the final state particles and not on the direction they ultimately end up travelling in? This seems like a reasonable interpretation, but I am not really sure. Peskin/Schroeder then state that this lack of angular dependence will change when fermions are considered, which I assume relates to their spin.

It means that $$d\sigma/d\Omega$$ doesn’t depend on $$\theta$$ or $$\phi$$, the angles in spherical polar coordinates that specify the direction of scattering. Think of $$(\theta,\phi)$$ as the direction to $$d\Omega$$.

When the differential cross section has no angular dependence in some frame, it means that the particles are scattering in all directions with equal probability.

Contrast your differential cross with this one for nonrelativistic electron-electron scattering:

$$\frac{d\sigma}{d\Omega}=\frac{m^2\alpha^2}{E_\text{CM}^2p^4}\frac{1+3\cos^2\theta}{\sin^4\theta},$$

which does depend on $$\theta$$.

You seem overly focused on thinking about scattering in terms of momentum space and phase space. You can just visualize these angles in position space. For example, the particles come together along the $$z$$-direction but go off in some other direction with polar angle $$\theta$$. Think of $$\theta$$ as specifying where the detector is positioned.

• The counterexample is helpful thank you. Would it be correct to interpret the quantity $d\sigma/d\Omega$ having no angular dependence as meaning that the particles scatter in all directions equally? Commented Dec 27, 2020 at 21:30
• Yes. But realize that that happens only in the CM frame. Commented Dec 27, 2020 at 21:31
• Ah, I did see this question while reading related questions that I think talks about the frame dependence. Thanks! Commented Dec 27, 2020 at 21:34
• I added a paragraph about scattering equally in all directions. Commented Dec 27, 2020 at 21:38