The Hadrons — Question 6

Question 6

A low-momentum form factor estimates the hadron charge radius

For a normalized spherical charge distribution, use $F(q)\simeq1-q^2\langle r^2\rangle/6$ at small q, where q is in inverse fm. A measurement gives F=.980 at $q=.400\,\mathrm{fm^{-1}}$. Cross section relative to a point charge is $|F|^2$.

1. Infer the rms charge radius.

2. Find the cross-section suppression and check the expansion parameter.

Question 6 - Solution

A low-momentum form factor estimates the hadron charge radius

(a) Method and calculation

$$\boxed{\langle r^2\rangle=6(1-F)/q^2=6(.020)/.160=.750\,\mathrm{fm^2}}.$$$$\boxed{r_{\mathrm{rms}}=\sqrt{.750}=.866025\,\mathrm{fm}}.$$

(b) Interpretation and further calculation

$$\boxed{\sigma/\sigma_{\mathrm{point}}=.980^2=.9604}\quad\text{a 3.96 percent suppression}.$$$$\boxed{q^2\langle r^2\rangle/6=.020\ll1}.$$

Physical check

A finite radius is evidence of extended structure. Higher moments and larger momentum transfer require more than this leading expansion.