Nuclear charge form factor ($F(\mathbf{q})$)

In the static charge density model, define the normalized Fourier transformHere $\mathbf{q}=(\mathbf{p}-{\mathbf{p}}^{\prime })/\hslash$ is the spatial momentum transfer divided by the reduced Planck constant: $q=|\mathbf{q}|$ has units of inverse length. Physical momentum transfer is $\hslash \mathbf{q}$. We keep $\hslash$ and the speed of light $c$ explicit; compute $q=|\mathbf{p}-{\mathbf{p}}^{\prime }|c/(\hslash c)$, expressing the numerator in $\mathrm{MeV}$ and $\hslash c$ 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,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.

Consider an ideal nuclear charge distribution that is constant inside a ball of radius $R>0$ and zero outside:This is a deliberately sharp-edged model. Substitution into the spherical nuclear charge form factor gives, for $q\ne 0$,At $q=0$ take the continuous limit $F(0)=1$. Direct radial integration gives $⟨r^2{⟩}_{\mathrm{ch}}=3R^2/5$, and the Taylor series provides an independent check:The first positive zero solves $\sin x-x\cos x=0$, equivalently $\tan x=x$, with $x=qR\simeq 4.49340946$. Consequently the first diffraction minimum moves toward smaller momentum transfer as $R$ increases.

For comparison, a smooth Gaussian function charge density with the same nuclear charge radius isThe Fourier transform of a Gaussian stays positive and has no zero. Both models have the same leading low-$q$ curvature, but they disagree at higher momentum transfer: at $qR=2$, the sphere gives $F\simeq 0.653097$ and the Gaussian function gives $F_G\simeq 0.670320$. Measuring only a nuclear charge radius cannot distinguish these two nuclear charge distributions.

As a chosen scale illustration, take $R=3.00\mathrm{fm}$. The predicted first zero is $q=1.49780{\mathrm{fm}}^{-1}$, or $\hslash cq\simeq 295.557\mathrm{MeV}$ using $\hslash c\simeq 197.327\mathrm{MeV}\mathrm{fm}$. With a $200\mathrm{MeV}$ ultrarelativistic Electron beam and negligible nuclear recoil, it occurs at $\theta \simeq 95.27^{\circ }$. These are consequences of chosen model parameters, not measured properties of carbon or a claim that all atomic nuclei are uniform balls.

For an ultrarelativistic Electron scattered elastically by a static, spherical, spin-zero atomic nucleus, a plane-wave Born approximation retains a single electromagnetic interaction. The Fourier transform of its Coulomb potential is proportional to $ZeF(q)/q^2$. Relative to the same point charge, the scattering amplitude is therefore multiplied by $F(q)$ and the differential scattering cross-section by $|F(q){|}^2$:Here $E\gg m_e{c}^2$ is the incident total Electron energy, $\theta$ the scattering angle, $d\Omega$ a solid angle element, and $\alpha$ the fine-structure constant. The cosine factor already accounts for the projectile's spin; a spinless projectile would give a different point-charge factor. The formula neglects nuclear recoil, electron rest-mass corrections, radiative corrections and Coulomb distortion in electron scattering. The latter can be substantial at large $Z\alpha$: electron waves then require treatment in the target's Coulomb potential, rather than undistorted plane waves.

Nonzero target spin can introduce magnetic and additional electric contributions, so division by this point-charge expression does not generally isolate one charge form factor for a proton or deuteron. Even for spin zero, finite energy resolution, recoil and Coulomb distortion in electron scattering can change a measured minimum. A zero of the uniform-sphere model is a clear illustration of amplitude cancellation, not a universal prediction of a zero experimental counting rate.