http://www.academia.edu/1912047/The_Ehrenfest_Adiabatic_Hypothesis_and_the_Old_Quantum_Theory_before_Bohr I did aske someone professional to look at this as well and he obtained the same answer. Therefore I am poting this:
I believe they approach g=2am looking at page 138the theory for the classical relation between the magnetic momentum $\mu$ and the spin $S$. It is said that the $g$-141factor is $g=2$ for the equation: $\mu=g\frac{e}{2m_e}S$ if you look at an electron. Here I am trying to prove it with classical reasoning:
$$\begin{align} S&=I\omega=\omega\int \rho_m r^2 dV\\ \mu&=\frac{\omega}{2}\int \rho_e r^2 dV \end{align}$$
The next two formulas are based on this page: https://en.wikipedia.org/wiki/Electron_magnetic_moment#The_classical_theory_of_the_g-factor
$$\begin{align} \rho_e&=eN_ee^{-\frac{r^{2}}{r_e^{2}}}\\ \rho_m&=m_eN_me^{-\frac{r^{2}}{r_e^{2}}} \end{align}$$
Hence,
$$\begin{align} \mu&=4\pi\frac{\omega}{2}\int_0^\infty eN_ee^{-\frac{r^{2}}{r_e^{2}}}r^2 r^2dr\\ S&=4\pi\omega\int_0^\infty m_e N_me^{-\frac{r^{2}}{r_m^{2}}} r^2 r^2dr \end{align}$$
I have to normalize these two
$$\begin{align} \int_0^\infty N_ee^{-\frac{r^{2}}{r_e^{2}}}dr&\\ \int_0^\infty N_me^{-\frac{r^{2}}{r_e^{2}}}dr& \end{align}$$
It is of any useobtained that:
$$\begin{align}
N_e&=\frac{1}{\sqrt{\pi}r_e}\\
N_m&=\frac{1}{\sqrt{\pi}r_m}
\end{align}$$
We get:
$$\begin{align} \mu&=\frac{e}{\sqrt{\pi}r_e}4\pi\frac{\omega}{2}\int_0^\infty e^{-\frac{r^{2}}{r_e^{2}}}r^4dr\\ S&=\frac{m_e}{\sqrt{\pi}r_m}4\pi\omega\int_0^\infty e^{-\frac{r^{2}}{r_m^{2}}}r^4dr \end{align}$$
I obtained from an online integral calculator that: $\int_0^\infty e^{\frac{-x^2}{a}}x^4=\frac{3\sqrt{\pi} a^{\frac{5}{2}}}{8}$
So
$$\begin{align} \mu&=\frac{e}{\sqrt{\pi}r_e}4\pi\frac{\omega}{2}\frac{3\sqrt{\pi} r_e^5}{8}\\ S&=\frac{m_e}{\sqrt{\pi}r_m}4\pi\omega \frac{3\sqrt{\pi} r_m^5}{8} \end{align}$$
We want to solve
$$\mu=DS$$
$$\frac{e}{\sqrt{\pi}r_e}4\pi\frac{\omega}{2}\frac{3\sqrt{\pi} r_e^5}{8}=D\frac{m_e}{\sqrt{\pi}r_m}4\pi\omega \frac{3\sqrt{\pi} r_m^5}{8}$$ We obtain:
$$D=\frac{e}{2m_e}\frac{r_e^4}{r_m^4}$$
But is $\frac{r_e^4}{r_m^4}=2$?
from the wikipedia article above it says that one needs
$\frac{r_e^8}{r_m^8}$. They mention Abraham But my calculations fail to get to the same result. Any input are most welcome. I guess that it would have taken me a step closer had it been the same result as wellthe wikipedia page. It might be worthThe wikipedia page also informs that $\frac{r_e}{r_m}\approx 1.09051$ and that would lead to read until 145 were another approximation of the gyromagnetic constant is introduced$\frac{r_e^8}{r_m^8}\approx 2$.