= Spin-orbit g-factor ratio from two P states
{title2=$g_s/g_L=(4R-2)/(R+1)$}
An original reduced model makes the cancellation explicit. Assume <LS coupling>, neglect nuclear <spin>, and write the electronic <magnetic dipole moment> as $-\mu_B(g_L\mathbf L+g_s\mathbf S)/\hbar$. For $L=1$, $S=1/2$, the projection used to derive the <Landé g-factor> gives
$$
g_{3/2}=\frac{2g_L+g_s}{3},\qquad
g_{1/2}=\frac{4g_L-g_s}{3}.
$$
These coefficients follow by substituting $J=3/2$ or $1/2$ into the angular-momentum projection formula; neither level measures pure <spin> alone. In a weak uniform <magnetic field>, adjacent electronic projections have spacing $h f_J=g_J\mu_B B$ when $g_J>0$. Therefore
$$
R=\frac{f_{3/2}}{f_{1/2}}=\frac{g_{3/2}}{g_{1/2}}
=\frac{2+x}{4-x},\qquad x=\frac{g_s}{g_L},\qquad
\boxed{x=\frac{4R-2}{R+1}.}
$$
The common <magnetic field> and <Bohr magneton> cancel. The ratio determines $g_s/g_L$, not $g_s$ without further knowledge of $g_L$. For $g_L=1$, it gives $a_e=(R-2)/(R+1)$. Small departures from $R=2$ are magnified differently from small departures of $g_s$ from $2$; the two fractional changes should not be identified.
With the stipulated inputs $g_L=1$, $g_s=2.0024$ and $\mu_BB/h=100\,\mathrm{MHz}$, the two electronic frequencies are $66.5867$ and $133.4133\,\mathrm{MHz}$. Their ratio is approximately $2.0036043$, and inversion recovers $a_e=0.0012$. These are ideal electronic lines, not historical <gallium> data. Nuclear <spin>, mixing between levels and other deviations from <LS coupling> require a more complete model in an actual atom.
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