# The relation between charge distribution and form factor

In Peskin and Schroeder Eq.(6.33), they introduced the concept of form factor

$$$$\Gamma^\mu\left(p^{\prime}, p\right)=\gamma^\mu F_1\left(q^2\right)+\frac{i \sigma^{\mu \nu} q_\nu}{2 m} F_2\left(q^2\right)\tag{6.33}$$$$

where $$q^2$$ is four momentum.

And I also know the charge distribution can be linked to the form factor by a three-dimensional Fourier transform

$$$$\rho(\mathbf{r})=\frac{1}{(2 \pi)^3} \int \mathrm{d}^3 \mathbf{q} F(\mathbf{q}) \mathrm{e}^{-\mathrm{i} \mathbf{q} \cdot \mathbf{r}}$$$$

where $$\mathbf{q}$$ is three momentum.

According to my naive understanding, in the Breit frame, we should have

$$q^2=-\mathbf{q}^2$$, $$F(q^2)=F(-\mathbf{q}^2)$$.

So, the charge distribution can be obtained from $$F(q^2)$$.

However, I have read some paper that say this is wrong for some systems, whose intrinsic size is comparable with the Compton wavelength. For example, https://inspirehep.net/literature/2005580, https://inspirehep.net/literature/2161738. If I understand correctly, they think that relativistic effects are not taken into account here.

My questions are:

1. For a photon, the on-shell condition should be $$q^2=0$$, does $$F(q^2)$$ mean the photon is off-shell?

2. Why is this naive understanding wrong for some systems, where is the approximation made and why does this approximation fail?

3. If I just want to get the static charge distribution, can I directly take the 3D Fourier transform of $$F(q^2)$$?

I think your understanding of the Breit frame is not correct. From https://en.wikipedia.org/wiki/Breit_frame, the Breit frame corresponds to the frame in which the momentum of a scattered particle is the opposite of its momentum before scattering, it doesn't impose $$q^2 = -\mathbf{q}^2$$. $$q^2$$ is equal to $$-\mathbf{q}^2$$ if the mass of the nucleus is infinite and we actually have $$q^2 = \omega^2-\mathbf{q}^2$$ where $$\omega$$ is the energy of the recoiled nucleus.
1. A photon is an elementary particle so there is no form factor. Moreover, on-shell doesn't mean that $$q^2=(p-p^\prime)^2=0$$ but $$p^2 = 0$$.
2. & 3. If you take the Fourier transform, you assume that the nucleus doesn't move during scattering (the energy of the recoiled nucleus is negligible). This is correct at high Z (for example $$^{208}$$Pb) but it is incorrect for light nuclei, especially for proton.