= Nuclear charge form factor
{title2=$F(\mathbf q)$}
In the static <charge density> model, define the normalized <Fourier transform>
$$
F(\mathbf q)=\frac1{Ze}\int\rho_{\mathrm{ch}}(\mathbf r)e^{i\mathbf q\cdot\mathbf r}\,d^3r,
\qquad F(\mathbf 0)=1.
$$
Here $\mathbf q=(\mathbf p-\mathbf p')/\hbar$ is the spatial <momentum transfer> divided by the reduced <Planck constant>: $q=|\mathbf q|$ has units of inverse length. Physical <momentum transfer> is $\hbar\mathbf q$. We keep $\hbar$ and the <speed of light> $c$ explicit; compute $q=|\mathbf p-\mathbf p'|c/(\hbar c)$, expressing the numerator in $\mathrm{MeV}$ and $\hbar c$ in $\mathrm{MeV}\,\mathrm{fm}$, to obtain $q$ in inverse <femtometres>. The spatial $q$ is distinct from the invariant <four-momentum> transfer used in <relativistic scattering cross-sections>.
For <spherical symmetry>, take the polar axis along $\mathbf q$ and use <spherical coordinates>. With $\mu$ the cosine of the polar angle,
$$
\int_{-1}^{1}e^{iqr\mu}\,d\mu=2\frac{\sin(qr)}{qr},
\qquad
F(q)=\frac{4\pi}{Ze}\int_0^\infty r^2\rho_{\mathrm{ch}}(r)\frac{\sin(qr)}{qr}\,dr.
$$
Every radial shell contributes an amplitude. At $q=0$ all shells add with the same sign; at larger $q$ their contributions can cancel. This cancellation is the origin of a <diffraction> minimum in the <nuclear charge form factor>, rather than an absence of <electric charge> in the <nucleus>.
The static <Fourier transform> interpretation is appropriate for the nuclear model and kinematic regime specified here. A fully relativistic composite system requires frame and current conventions; its electric form factor is not universally the transform of a frame-independent three-dimensional <charge density>.
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