Proton Electric Form Factor GE(Q²)
Comparison of the j₀ standing wave model (Geometric Wave Theory) vs
the standard dipole parametrization against electron-proton elastic scattering data.
The proton is modeled as ψ(r) = A·sin(πr/Rc)/r — the fundamental j₀ mode in a spherical cavity.
j₀ Wave Model (GWT)
rrms = 0.841 fm
Rc = 1.581 fm (true boundary)
Dipole (Standard fit)
rp = 0.855 fm
GE = (1 + Q²/Λ²)⁻² Λ² = 0.71 GeV²
Proton Radius Measurements
0.831–0.877 fm
PRad: 0.831 μH: 0.841 e-p high-Q: 0.877 fm
Log–log scale. Drag sliders above to update fits in real time.
j₀ form factor (analytic, exact):
GE(Q) = (1/x) × [Si(x) − ½(Si(x+2π) + Si(x−2π))] x = Q·Rc
where Si(x) = ∫₀ˣ sin(t)/t dt (sine integral)
Derivation: GE = Fourier transform of ρ(r) = C·sin²(πr/Rc)/r² (charge density ∝ |ψ|²)
Dipole parametrization (phenomenological):
GEdip(Q²) = (1 + Q²/Λ²)⁻² Λ² = 0.71 GeV² rp = √(12/Λ²)/5.068 fm
RMS radius from j₀:
rrms² = Rc² × (1/3 − 1/(2π²)) → rrms/Rc = 0.5319
Why the j₀ model? The proton is a j₀ standing wave:
ψ(r) = A·sin(πr/Rc)/r with ψ(Rc) = 0 (cavity boundary condition).
The charge density ρ(r) ∝ |ψ(r)|² = sin²(πr/Rc)/r² gives an
exact, parameter-free form factor from the single value Rc = 1.581 fm.
Why the dipole fails at high Q²?
The dipole corresponds to an exponential charge distribution with no edge —
ρdip(r) ∝ r·e−r/r₀. The j₀ wave has a hard boundary at Rc:
no charge exists beyond 1.581 fm. At Q² ≫ 1/(Rc²·(ħc)²) the
wave model predicts a faster fall-off and characteristic oscillations.
At intermediate Q² (0.2–2 GeV²) the j₀ falls more steeply than the dipole —
testable with precision e-p scattering in this range.
Proton radius puzzle resolved:
The measured rp depends on Q² because both particles are waves.
High-Q² electrons carry more lattice pressure → inflate apparent boundary → larger rp.
Low-Q² electrons sample the geometric rrms = 0.841 fm (muonic H value).
The j₀ model predicts rrms = 0.841 fm with zero free parameters.
Experimental Data Used
χ² Statistics
Radial Charge Distribution
Note: Dipole ρdip(r) ∝ r·e−r/r₀ extends to ∞; j₀ ρ(r) ∝ sin²(πr/Rc)/r² cuts off at Rc.