The meaning of the "physical" electron charge in Peskin (Chap 7)

[niss]

Homework Statement

Why is the "physical" electron charge e_0 / (1-Pi(0)) and not e_0 / (1-Pi(q^2))

Relevant Equations

(physical charge) = e = sqrt(Z_3) e_0 = sqrt(Z_3) (bare charge)

On p. 246 in the Peskin QFT textbook, below is stated

the "physical" electron charge measured in experiments is $$\sqrt{Z_3} e$$

where Z₃ is defined as the residue of the q² = 0 pole, explicitly as
$$Z_3=\frac{1}{1-\Pi(0)}$$
and e is the bare charge.

In advance, the exact photon two point function is calculated as
$$\frac{-ig_{\mu\nu}}{q^2(1-\Pi(q^2))}$$

Though the photon propagator should be divided by
$$1-\Pi(q^2)$$
, why the defined "physical" charge is divided by
$$1-\Pi(0)$$
?
Why only the low-q² scattering is being considered here?

 

[nrqed]

Homework Statement:: Why is the "physical" electron charge e_0 / (1-Pi(0)) and not e_0 / (1-Pi(q^2))
Relevant Equations:: (physical charge) = e = sqrt(Z_3) e_0 = sqrt(Z_3) (bare charge)

On p. 246 in the Peskin QFT textbook, below is stated



where Z₃ is defined as the residue of the q² = 0 pole, explicitly as
$$Z_3=\frac{1}{1-\Pi(0)}$$
and e is the bare charge.

In advance, the exact photon two point function is calculated as
$$\frac{-ig_{\mu\nu}}{q^2(1-\Pi(q^2))}$$

Though the photon propagator should be divided by
$$1-\Pi(q^2)$$
, why the defined "physical" charge is divided by
$$1-\Pi(0)$$
?
Why only the low-q² scattering is being considered here?

One could define the electron charge at any momentum transfer. But historically, the value of the electric charge has been measured with precision in the limit $q^2 \rightarrow 0$. So the value quoted for the electric charge in the particle data booklet is the value measured in that limit.

 

[niss]

One could define the electron charge at any momentum transfer. But historically, the value of the electric charge has been measured with precision in the limit $q^2 \rightarrow 0$. So the value quoted for the electric charge in the particle data booklet is the value measured in that limit.

Thank you.