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

\begin{equation} \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} \end{equation}

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

\begin{equation} \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}} \end{equation}

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 $F(q^2)$?


1 Answer 1


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.

I am not sure it's related to your question but the equivalence or not of the charge distribution and the form factor for defining the proton charge radius is addressed in this article: see G. Miller, Phys. Rev. C 99, 035202 (2019).

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