I was reading Representing Electrons: A Biographical Approach to Theoretical Entities, by Theodore Arabatzis.

At a certain point, where he is explaining the history of the magnetic moment of the electron, he describes the process that led to $$ \boldsymbol \mu=g\frac{e}{2m}\boldsymbol S $$

The orbital magnetic moment satisfies the relation above, with $g=1$; somehow, the spin magnetic moment has $g=2$. On page 226, he states that (emphasis mine):

The electron, thus, acquired an intrinsic magnetic moment (one Bohr magneton) that was twice its magnetic moment due to its orbital motion. The question whether that property could be accommodated within the classical electromagnetic representation of the electron then arose. Indeed, on Ehrenfest's suggestion, Uhlenbeck managed to explain this property, by capitalizing on Abraham's analysis of the gyromagnetic ratio of a spherical (surface) distribution of charge. On the assumption that the electron was a rotating sphere whose charge was distributed on its surface, the required value of its magnetic moment followed.

If I'm getting this right, the author is saying that if we think of the electron as a sphere with a surface charge distribution, we should get the $g=2$ factor, using solely classical arguments. The thing is, I tried to check this, and my result is that $g=1$.

My analysis is as follows: suppose that the electron is a solid sphere with mass $m$ and radius $r_e$; then its moment of inertia is $$ I=\frac{2}{5}mr_e^2 $$

If we assume that the electron is spinning with angular frequency $\omega$, we find that the spin angular momentum is $$ S=I\omega=\frac{2}{5}mr_e^2\omega $$

On the other hand, the magnetic moment of a hollow charged sphere is $$ \mu=\frac{1}{5}er_e^2\omega $$

Finally, the ratio of $\mu$ to $S$ is $$ \frac{\mu}{S}=\frac{1}{5}er_e^2\omega\ \frac{5}{2}\frac{1}{mr_e^2\omega}=\frac{e}{2m} $$ which means that $g=1$.

My question is: where did my analysis fail?

As a matter of fact, the same claim is given on George Uhlenbeck and the discovery of electron spin, by Abraham Pais:

Following a hint from Ehrenfest, George found in an old article by Max Abraham that an electron considered as a rigid sphere with only surface charge does have $g=2$.

As A. Pais is a respected science historian, I am to believe the statement is accurate, but I'm still unable to prove this (rather) simple claim. Is there any chance the claim is false? Or is it possible to somehow prove that $g=2$ is true for a classical sphere?

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    $\begingroup$ Note: I already know that $g=2$ is very well explained by Quantum Mechanics; my question is: can it also be explained, as the author sais, by classical mechanics? I found it cannot be explained by a solid sphere, but I believe the author must be right, so at which point did my analysis break down? $\endgroup$ Dec 17 '15 at 23:00
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    $\begingroup$ Your first calculation uses non-relativistic formula for angular momentum $I\omega$. Since the second calculation shows that required sphere size $r_e$ and angular momentum $\hbar$ imply superluminal speed of the surface of the sphere, you have set of assumptions in violation of special relativity. You can recover either by increasing $r_e$ so that non-relativistic formula gets applicable, or redo the calculation with relativistic formula for angular momentum. You should be able to get arbitrarily high angular momentum while all parts of the sphere have subluminal speeds. $\endgroup$ Dec 22 '15 at 22:22
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    $\begingroup$ @JánLalinský thank you for your response. "redo the calculation with relativistic formula for angular momentum" is not possible (I believe), because the concept of a rigid body is not valid in SR (so there is no relativistic generalisation of $I\omega$). If we (as Uhlenbeck suposedly did) want to calculate the gyromagnetic ratio of the electron as if it were a solid sphere, we must settle for non-relativistic mechanics (the fact that $v>c$ probably means that the problem is ill-posed to begin with. Is perhaps the author's claim inexact?). $\endgroup$ Dec 22 '15 at 23:19
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    $\begingroup$ Do you use hollow sphere, or sphere filled with uniformly distributed mass & charge? From your description, it seems that you mix it, mass being distributed and charge being entirely on surface. $\endgroup$
    – Vladimir
    Dec 23 '15 at 0:08
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    $\begingroup$ "the concept of a rigid body is not valid in SR (so there is no relativistic generalisation of Iω)" it is true body cannot be rigid(non-deformable) in SR, but you only need assumption of rigid stationary rotation, which does not contradict SR. Or you can try to make and analyze non-rigid model of the particle, but that gets hard real quick. $\endgroup$ Dec 23 '15 at 0:38

I went through unanswered questions, and stumbled over this...
Did you find the original books?

The mistake should be in your formula for the $\mu$ of a hollow sphere; the value with $1/5$ you gave is that of a solid sphere...
The problem gets more simple I think, if you compare the two things directly:

You get both, the angular momentum and $\mu$, from highly analogous integrals over all points, in which there is a $r^2\mathrm dm$ or an $r^2\mathrm dq$:

$$ S = I\omega = \omega \int r^2\,\mathrm dm $$ and with the definition of $\mathrm d\mu$ as current times enclosed area: $$ \mu = \int\mathrm d\mu = \int A\,\mathrm dI = \int \pi r^2\cdot\frac{\mathrm dq}T = \int \pi r^2\cdot\frac{\mathrm dq}{2\pi}\omega = \frac \omega 2 \int r^2\,\mathrm dq $$

The g-factor is defined to be one if the charges coinside with the masses (the ratio of their densities is equal everywhere), i.e. the definition accounts for the $1/2$ in the second formula.

Thus, if you distribute the charge further from the axis that the mass, you get a g-factor greater than one. The integrals are always equivalent and depend on the geometry of the distribution.
For the same geometry you will always get a pre-factor for the intertia which is twice the factor for the magnetic moment -- and thus by definition a $g=1$.

Now comes the strange thing: the pre-factor in the moment of inertia of a full sphere is $1/5$ and for a hollow sphere $1/3$. The g-factor with the distribution of the mass in the sphere and of the charge on the shell thus gives a $g=5/3$.
This is obviously in contrast with the claim, that it equals two. It explains, that it is greater than one, though.
Maybe back then they could not measure $g$ so well and saw only, that it is considerably greater than one, and so could explain at least this ... ?

So the point seems to be, that the charges are further from the axis than the masses. The sphere is just a nice example, which explains the (measured) factor to be greater than one by a beautiful/plausible distribution.

...The argument with the relativistic velocities (from the comments) goes in another direction: since other measurments suggest a maximal radius for the electron, you can compute the neccessary velocities, which disproves the naive explanation of the spin (for both, the inertia and the magnetic aspect; this has nothing to do with their ratio) as a real motion.

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    $\begingroup$ But the moment of inertia of a hollow sphere is $2/3$, not $4/5$. $\endgroup$
    – knzhou
    Apr 4 '16 at 21:40
  • $\begingroup$ you are right... I did the computation and edited the post. But still I am convinced the main point which seemed to be not obvius in the comments is right: the magnetic moment is for example calculated as current times area, and is thus for a fixed $\omega$ proportional to $r^2q$, just analogous to the moment of inertia. $\endgroup$
    – Ilja
    Apr 4 '16 at 22:02
  • $\begingroup$ @Ilja (+1) Thank you very much for showing interest. I've got to say I'm pretty confident about my value for $\mu$ (I doubt it is twice as much) because I found the same result on many pages online (if I google magnetic moment of hollow sphere I find the same value $\mu=\frac{1}{5}er^2\omega$ everywhere...). $\endgroup$ Apr 6 '16 at 16:52
  • $\begingroup$ ... this is strange; I tried to google it too, and found the paper you linked in your original question, and in its title there is "full sphere"... I edited the post to make the reasoning more clear. I dont see where it can be wrong, and we'll have to read the original papers to see what they meant... $\endgroup$
    – Ilja
    Apr 8 '16 at 9:54

It seems that some people liked this question so I shall post my thoughts so far. I don't have a definitive answer, but I did get some interesting results.

Let $\rho_m(\boldsymbol r)$ and $\rho_e(\boldsymbol r)$ be the mass and charge densities of the electron. The $g$ factor is given by $$ g=\frac{m}{e}\frac{\int\mathrm d\boldsymbol r\ r^2\sin\theta\ \rho_m(\boldsymbol r)}{\int\mathrm d\boldsymbol r\ r^2\sin\theta\ \rho_e(\boldsymbol r)} \tag{1} $$

From this, it's easy to see that if $\rho_m\propto \rho_e$, we get $g=1$. This means that if we have a solid sphere with constant charge density and constant mass density, the $g$ factor is 1; al hollow sphere with surface charge has also $g=1$. If we want $g\neq 1$ we must take a charge density that is not proportional to the mass density.

The first model that comes to mind is to take a volume mass density and a surface charge density, that is, a filled sphere with its charge on the surface: \begin{align} \rho_m&=\frac{m}{V}\Theta(R-r)\\ \rho_e&=\frac{e}{S}\delta(r-R)\tag{2} \end{align} where $V=\frac{4}{3}\pi R^3$ and $S=4\pi R^2$. If we plug these functions into $(1)$ we get $g=5/3$ as already acknowledged by Ilja and Anubhav. This means that Arabatzis' and Pais' claims are inaccurate: this model does not predict $g=2$ but $g=1.67$ instead.

To go a step further, we may take the same model before, but with a different mass and charge radii, that is, \begin{align} \rho_m&=\frac{m}{V}\Theta(R_m-r)\\ \rho_e&=\frac{e}{S}\delta(R_e-r)\tag{3} \end{align} with $R_m\neq R_e$. In this case, we find $g=5R_e^2/3R_m^2$, which equals 2 if $R_e=1.095 R_m$. This model seems highly artificial though.

The next possible example could be to take exponential densities, which could be the result of some kind of screening at some fundamental level: \begin{align} \rho_m&\propto\exp\left[-\frac{r^2}{R_m^2}\right]\\ \rho_e&\propto\exp\left[-\frac{r}{R_e}\right]\tag{4} \end{align} from which we find $g=8R_e^2/R_m^2$; if we take $R_m=2R_e$ we get $g=2$. This is still very artificial but there might be some electrostatic model that is able to accommodate this.

Other possible models could consist of non-spherical densities, such as cylinders or string-like wires. I leave to the reader to explore this models. In any case, it is clear that the most natural models don't predict $g=2$, and it's not easy to find another one that fixes this while not getting too ad-hoc. But it is possible to write down exotic models with tunable parameters so as to get $g=2$, which means at least that $g=2$ is achievable at the classical level.



My search gives

$\mu = \frac{e\omega R^2}{3}$

This gives $g = 5/3 = 1.667$

Did not you provided link given below?


Which explains that non-uniform charge distribution can explain the value of g = 2 without any Dirac equation.

Following a hint from Ehrenfest, George found in an old article by Max Abraham that an electron considered as a rigid sphere with only surface charge does have .

It may be so by above statement he meant that the radius ratio $\frac{r_e}{r_m} ≈ 1.09051$ has been approximated to surface of charge.


I did aske someone professional to look at this as well and he obtained the same answer. Therefore I am poting this:

I am looking at the theory for the classical relation between the magnetic momentum $\mu$ and the spin $S$. It is said that the $g$-factor 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

$$\begin{align} \rho_e&=eN_ee^{-\frac{r^{2}}{r_e^{2}}}\\ \rho_m&=m_eN_me^{-\frac{r^{2}}{r_e^{2}}} \end{align}$$


$$\begin{align} \mu&=4\pi\frac{\omega}{2}\int_0^\infty eN_ee^{-\frac{r^{2}}{r_e^{2}}}r^2 r^2dr\\ S&=4\pi\omega\int_0^\infty m_e N_me^{-\frac{r^{2}}{r_m^{2}}} r^2 r^2dr \end{align}$$

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 obtained 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: $\int_0^\infty e^{\frac{-x^2}{a}}x^4=\frac{3\sqrt{\pi} a^{\frac{5}{2}}}{8}$


$$\begin{align} \mu&=\frac{e}{\sqrt{\pi}r_e}4\pi\frac{\omega}{2}\frac{3\sqrt{\pi} r_e^5}{8}\\ S&=\frac{m_e}{\sqrt{\pi}r_m}4\pi\omega \frac{3\sqrt{\pi} r_m^5}{8} \end{align}$$

We want to solve


$$\frac{e}{\sqrt{\pi}r_e}4\pi\frac{\omega}{2}\frac{3\sqrt{\pi} r_e^5}{8}=D\frac{m_e}{\sqrt{\pi}r_m}4\pi\omega \frac{3\sqrt{\pi} r_m^5}{8}$$ We obtain:


But is $\frac{r_e^4}{r_m^4}=2$?

from the wikipedia article above it says that one needs

$\frac{r_e^8}{r_m^8}$. 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 the wikipedia page. The wikipedia page also informs that $\frac{r_e}{r_m}\approx 1.09051$ and that would lead to $\frac{r_e^8}{r_m^8}\approx 2$.


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