# QFT Question: What is meant by dipole form ?

1. Nov 6, 2012

### spooky51

QFT Question: What is meant by "dipole form"?

Hello physics people! Probably a very basic question, but here goes.

I'm taking a course on QFT based on Ryder. I've heard my professor refer to propagators as having a "polar" or "dipole" form. Things like (k^2 - m^2 + ie)^(-1)

For anyone who has a copy of Ryder hand, the specific case I'm puzzling over at at the moment is equation 1.20 on page 16, where the text above says that the form factor G_M has a dipole form. ( G_M = (1+q^2/M_q^2)^(-1)).

I'm not sure what is meant by something having a polar/dipole form. I can't seem to find anything on google or wikipedia, but I might not be digging deep enough. Can someone shed some light on this?

2. Nov 6, 2012

### Bill_K

Re: QFT Question: What is meant by "dipole form"?

The form factor is the Fourier transform of the charge distribution,

F(q) = ∫ρ(x) eiq·x d3x

For small q, (long wavelength photons) expand the exponential,

F(q) = ∫ρ(x)(1 + iq·x -(q·x)2/2 + ...) d3x

Keeping only these leading terms is called the dipole approximation. Assuming the charge distribution is spherically symmetric, you get a form factor,

F(q) = 1 + <r2> q2 + ...

where <r2> is the mean square radius of the charge cloud.

Whether you write it this way or in the denominator, (1 - <r2> q2)-1, the idea is the same. Keeping only the leading term in q2 is the dipole approximation.

Last edited: Nov 6, 2012
3. Nov 7, 2012

### andrien

Re: QFT Question: What is meant by "dipole form"?

what you say is dipole approximation which is essentially the point that atomic dimensions are small compared with wavelength.Moreover,the form factor also arises in rosenbluth formula where it has to do with the structured form of proton where form factors really has it's connection to anomalous magnetic moment and also the structure of charge distribution(there are two).I don't know what it has to do with the propagators of a fermion.