# Significance of Compton scattering being second order?

In the Compton scattering equation for changes in wavelength, for small angles the equation is second order in the angle. Is there any significance to this? To me it seems to say that,if a photon takes a smooth path with no abrupt turns, then it shouldn't have much of a change in its wavelength.

edit: I guess my big question is, why is this change second order and not first

The expected probability density is

$$\begin{equation*} \langle|\mathcal{M}|^2\rangle= 2e^4\left( \frac{\omega}{\omega'}+\frac{\omega'}{\omega} +\left(\frac{m}{\omega}-\frac{m}{\omega'}+1\right)^2-1 \right) \end{equation*}$$

The Compton formula is $$\begin{equation*} \frac{1}{\omega'}-\frac{1}{\omega}=\frac{1-\cos\theta}{m} \end{equation*}$$

It follows that $$\begin{equation*} \cos\theta=\frac{m}{\omega}-\frac{m}{\omega'}+1 \end{equation*}$$

Then by substitution $$\begin{equation*} \langle|\mathcal{M}|^2\rangle= 2e^4\left( \frac{\omega}{\omega'}+\frac{\omega'}{\omega}+\cos^2\theta-1 \right) \end{equation*}$$

The differential cross section for Compton scattering is $$\begin{equation*} \frac{d\sigma}{d\Omega} = \frac{\alpha^2}{2m^2} \left(\frac{\omega'}{\omega}\right)^2 \left( \frac{\omega}{\omega'}+\frac{\omega'}{\omega}+\cos^2\theta-1 \right) \end{equation*}$$

A complete derivation of $$\langle|\mathcal{M}|^2\rangle$$ for Compton scattering is here:

http://www.eigenmath.org/compton-scattering.pdf