Proton Radius Puzzle Solution and NFT Series

Solution to the Proton Radius Puzzle

Proton Radius: Modeling the proton as a quantized vortex with \( m = m_p \), \( v = c \), \( n = 4 \):

\( r_p = \frac{4 \hbar}{m_p c} \approx 0.8415 \text{ fm} \)

Matches the muonic hydrogen value of 0.84184 fm, vs. old value of 0.877 fm.

Magnetic Moment: As a thin shell of charge \( q = e \) rotating at \( v = c \):

\( \mu_p = \frac{1}{3} e c r_p = \frac{8}{3} \mu_N \approx 2.6667 \mu_N \)

Within 4.5% of actual 2.7928 \( \mu_N \).

Refinement: Using the fine-structure constant:

\( \alpha^2 = \pi \cdot r_p \cdot R_\infty \cdot \frac{m_p}{m_e} \)

Confirms \( r_p \approx 0.841 \text{ fm} \).

Highlights the discrepancy between old (0.877 fm) and new (0.841 fm) proton radius measurements.

\( r_p^{\text{old}} \approx 0.877 \text{ fm}, \quad r_p^{\text{new}} \approx 0.841 \text{ fm} \)

Proton as a circular quantized vortex with \( n = 4 \), yielding the new radius.

\( r_p = \frac{4 \hbar}{m_p c} \approx 0.841 \text{ fm} \)

Magnetic moment from a shell model, within 4.5% of actual value.

\( \mu_p = \frac{1}{3} e c r_p = \frac{8}{3} \mu_N \approx 2.6667 \mu_N \)

Refines the radius using fundamental constants.

\( \alpha^2 = \pi \cdot r_p \cdot R_\infty \cdot \frac{m_p}{m_e} \)

Model predicts new radius and near-match magnetic moment.

\( r_p \approx 0.841 \text{ fm}, \quad \mu_p \approx 2.6667 \mu_N \)