Charge Renormalization in Peskin (Chapter 7.5)

In Chapter 7.5 in Peskin and Schroeder, the authors define the physical charge in eq. (7.76), $$\text{(physical charge)}=\sqrt{Z_3}\cdot\text{(bare charge)}$$ where $\dfrac{1}{1-\Pi(0)}\equiv Z_3$. Here, $$\Pi^{\mu\nu}(q)=(q^2g^{\mu\nu}-q^\mu q^\nu)\Pi(q^2)$$ $i\Pi^{\mu\nu}(q)$ being the sum of all 1PI insertions into the photon propagator.

Now, just below eq. (7.76), the authors say that in a scattering process with non-zero $q^2$, we get a quantity, $$\frac{-ig_{\mu\nu}}{q^2}\left(\frac{e_0^2}{1-\Pi(q^2)}\right){\underset{\mathrm{\mathcal{O}(\alpha)}}{=}}\frac{-ig_{\mu\nu}}{q^2}\left(\frac{e^2}{1-[\Pi_2(q^2)-\Pi_2(0)]}\right)$$ where $\Pi_2(q^2)$ is the $\mathcal{O}(\alpha)$ value of $\Pi(q^2)$.

My specific question is, why do we subtract $\Pi_2(0)$ in the denominator of the RHS of the above expression? Shouldn't we just replace $\Pi(q^2)$ with $\Pi_2(q^2)$?

Thanks in advance.

EDIT:
I think the answer lies a posteriori in the discussion leading up to eq. (7.90). They show that the $\mathcal{O}(\alpha)$ shift in the electric charge is given by, $$\Pi_2(0)\approx-\frac{2\alpha}{3\pi\epsilon}$$ which blows up as $\epsilon\rightarrow0$ ($\epsilon=4-d$, $d$ being the number of spacetime dimensions). The argument is:

What can be observed is the $q^2$ dependence of the effective electric charge (7.77).

[(7.77) basically refers to the denominator I was talking about in my original question.]

1 Answer

Firstly, replacing the e0 with e; secondly, using the Taylor series and dropping the higher order terms. Below is the specific derivation

$$\frac{e_0^2}{1-\Pi(q^2)}\approx\frac{e_0^2}{1-\Pi_2(q^2)}\approx\frac{e^2}{[1-\Pi_2(q^2)]\cdot[1-\Pi_2(0)]^{-1}}\approx\frac{e^2}{[1-\Pi_2(q^2)]\cdot[1+\Pi_2(0)]}\approx\frac{e^2}{1-[\Pi_2(q^2)-\Pi_2(0)]}$$