Undergrad Clarification on a Peskin Equation

[thatboi]

Hi all,
I am currently reading through Peskin's QFT book and have a question regarding an equation that seems very simple in nature:
On page 121, below equation (4.121), there is an equation
$(p'-p)^2 = -|\textbf{p}'-\textbf{p}|^2 + \mathcal{O}(\textbf{p}^4)$
I was just wondering where exactly the higher powers in $\textbf{p}$ came from?

Thanks.

 

[malawi_glenn]

You could start with say $p = (E,\vec p) = (\sqrt{m^2+|\vec p|^2} ,\vec p)$ and expand $E = m\sqrt{1+|\vec p|^2/m^2}$ around $|\vec p|^2/m^2 = 0$ which is $m+ |\vec p|^2/2m + \ldots$
Instead of just doing what the authors did in eq 4.121

Then you would get
$(p'-p)^2 = - |\vec p \, {}' - \vec p |^2 + {E'}^2 + E^2 - 2E'E$
$= - |\vec p \, {}' - \vec p |^2 + (m+ |\vec p \, {}'|^2/2m + \ldots)^2 + (m+ |\vec p |^2/2m + \ldots)^2 - 2(m+ |\vec p \, {}'|^2/2m + \ldots) (m+ |\vec p |^2/2m + \ldots)$
$= - |\vec p \, {}' - \vec p |^2 + \mathcal{O}(|\vec p |^4, |\vec p \, {}'|^4 ,|\vec p |^2|\vec p \, {}'|^2 )$

 

[thatboi]

You could start with say $p = (E,\vec p) = (\sqrt{m^2+|\vec p|^2} ,\vec p)$ and expand $E = m\sqrt{1+|\vec p|^2/m^2}$ around $|\vec p|^2/m^2 = 0$ which is $m+ |\vec p|^2/2m + \ldots$
Instead of just doing what the authors did in eq 4.121

Then you would get
$(p'-p)^2 = - |\vec p \, {}' - \vec p |^2 + {E'}^2 + E^2 - 2E'E$
$= - |\vec p \, {}' - \vec p |^2 + (m+ |\vec p \, {}'|^2/2m + \ldots)^2 + (m+ |\vec p |^2/2m + \ldots)^2 - 2(m+ |\vec p \, {}'|^2/2m + \ldots) (m+ |\vec p |^2/2m + \ldots)$
$= - |\vec p \, {}' - \vec p |^2 + \mathcal{O}(|\vec p |^4, |\vec p \, {}'|^4 ,|\vec p |^2|\vec p \, {}'|^2 )$

Great, thanks a lot!