Why is $(p’-p)^2 = -|\mathbf {p’} - \mathbf p|^2 + \mathcal O(\mathbf p^4)$? On page 121 of Peskin and Schroeder, they claim that
$$(p’-p)^2 = -|\mathbf {p’} - \mathbf p|^2 + \mathcal O(\mathbf p^4),$$
where $p = (m,\mathbf p)$ is the incoming 4-momentum of a particle in a Yukawa interaction, and $p’ = (m, \mathbf {p’})$ its outgoing momentum. My question is where the terms depending on higher powers of the momentum come from. After all, it seems that
$$(p’-p)^2 = (m-m,\mathbf {p’} - \mathbf p)^2 = 0^2 -|\mathbf {p’} - \mathbf p|^2$$
exactly.
 A: The confusion comes because they took the low momentum approximation early. The four-vector should look like,
$p = (E,\textbf{p}) = (\sqrt{m^2 + \textbf{p}^2},\textbf{p})$
When the momentum is low you can approximate the energy as $E \approx m$, which is why they write their 4-vector as $p=(m,\textbf{p})$. Using the full four-vector you should get,
\begin{align}
(p'-p)^2 &= p'^2 + p^2 -2p'_\mu p^\mu \\
&= m^2 + m^2 - 2(E'E - \textbf{p}'\cdot\textbf{p})\\
&=2m^2 -2\sqrt{(m^2+\textbf{p}'^2)(m^2+\textbf{p}^2)} + 2\textbf{p}'\cdot\textbf{p}\\
&=2m^2 -2m^2\sqrt{1+\frac{\textbf{p}'}{m^2}^2+\frac{\textbf{p}^2}{m^2}+\frac{\textbf{p}'^2\textbf{p}^2}{m^4}} + 2\textbf{p}'\cdot\textbf{p}\\
\end{align}
Then if $\textbf{p}^2\ll m^2$ and $\textbf{p}'^2 \ll m^2$ then we can simplify,
\begin{align}
(p'-p)^2 &\approx 2m^2 - 2m^2 -2m^2\bigg(\frac{\textbf{p}^2}{2m^2}+\frac{\textbf{p}'^2}{2m^2}\bigg) +\mathcal O(\textbf{p}^4) + 2\textbf{p}'\cdot \textbf{p}\\
&\approx -\textbf{p}'^2 +2\textbf{p}'\cdot\textbf{p}-\textbf{p}^2 = -|\textbf{p}'-\textbf{p}|^2
\end{align}
I was admittedly a little sloppy with my notation for the Taylor expansion in treating $\textbf{p}$ and $\textbf{p}'$ like the same variable.
