Classical proof of the gyromagnetic ratio $g=2$

Question

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?

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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?

Answer 1

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.

Answer 2

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 $g=1$.

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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.

Answer 3

This confusion comes from misunderstanding the aim of Abraham's model, I think. The aim of Abraham's model is to explain the mechanical properties of the electron from electromagnetism only. In your calculation you are, in a way, thinking about the electron as a sphere or shell of charge carriers with mass. One then wonders what the structure of these charge carriers is... And sometimes it is said that the classical electron radius comes from setting the electrostatic energy equal to the rest mass, but Abraham's paper was published two years before Einstein discovered special relativity and $E = mc^2$. So we have to dig a bit deeper in the primary source.

What Abraham does is he starts from the action of the electromagnetic field $$ S = \frac{1}{2} \int E^2 - B^2 \, \mathrm{d}^3 x \, \mathrm{d} t \, , $$ determines the fields from a sphere (or shell), and performs the integral over $\mathrm{d}^3 x$. This leaves $S = \int L \, \mathrm{d} t$ where $L$ is the Lagrangian of the sphere (or shell). $L$ depends on the kinematic variables of the sphere (or shell), i.e., its position, velocity, and rotation. The equations of motion are then the Euler-Lagrange equations. For concreteness I'll talk about a shell below, but Abraham works things out for both the shell and the solid sphere.

Let's first do linear motion only. We need to determine the Lagrangian as a function of the velocity, to we need to solve for the fields from a moving charged shell. Actually, it's easier to Lorentz boost (the Lorentz transformation was invented a decade before special relativity!) to a frame where the shell is at rest. Since in this frame the problem is electrostatic we can discard the magnetic part of the action. Writing $\phi$ for the scalar potential and integrating by parts $$ S = \frac{1}{2} \int (\nabla \cdot \mathbf E) \phi \, \mathrm{d}^3 x' \, \mathrm{d} t' = \frac{1}{2} \int \rho \phi \, \mathrm{d}^3 x' \, \mathrm{d} t' $$ where $\rho$ is the charge density. (Abraham doesn't know it, but in modern notation he's calculating $S = \frac{1}{2} \int \mathrm d^4x \, j_\mu A^\mu$.) Now in the rest frame the charge sits on an ellipsoid squashed by a Lorentz factor $\gamma = 1/\sqrt{1 - v^2/c^2}$ and we need to remember that $\mathrm{d} t = \gamma \mathrm{d} t'$, so we get some non-trivial dependence on $v^2$. You can work it out by solving Poisson's equation in oblate spherical coordinates; the end result is $$ L_\text{shell} = \text{const} + \frac{2 e^2}{3 c^2 r} \frac{v^2}{2c^2} + \mathcal{O}(v^4/c^2) , $$ This is the Lagrangian for a particle of mass $m = \frac{2}{3} \frac{ e^2}{r c^2}$. (For a solid sphere $\frac{2}{3}$ should be replaced by $\frac{4}{5}$.)

It should be noted that this calculation is backwards to a modern view, where we would assume that in its rest frame the electron is spherical! But this is 1903 and the luminiferous aether is still a thing, so the natural thing for Abraham is to assume that in the frame of the aether the electron is spherical.

Abraham's goal in his model is to derive the mechanical mass from geometric and electromagnetic properties only. Conceptually the fundamental parameters of his model are the charge and the radius.

Now onto rotation. If we take an electron at rest, then we have to find the electric and magnetic fields from a rigid sphere or shell rotating with angular velocity $\omega$. Actually we only need the scalar and vector potentials, since $L = \frac{1}{2} \int \mathrm{d}^3 x \, j_\mu A^\mu$. This is electro- and magnetostatics so you can get these by solving Poisson's equation in cylindrical coordinates. The current density and the vector potential are both proportional to $\rho \omega$, so after integration we get $$ L_\text{shell} = \text{constant} + \frac{1}{9} \frac{e^2 r}{c^2} \omega^2 = \text{constant} + \frac{1}{3} mr^2\omega^2 \, . $$ This is the Lagrangian for a rotating body with moment of inertia $I = \frac{2}{9} \frac{e^2 r}{c^2}$. (For a solid sphere replace $\frac{1}{9}$ with $\frac{2}{35}$ and $\frac{1}{3}$ with $\frac{1}{7}$.) For both cases the moment of inertia is proportional to $m r^2$ as it must be, but as Abraham points out:

(Bei einer mit der Masse $M$ gleichförmig über Volumen oder Oberfläche belegten materiellen Kugel ist bekanntlich das Trägheitsmoment $$ P = \frac{2}{5} M a^2 \quad \text{bez.} \quad P = \frac{2}{3} M a^2 \, \text{.)} $$

(Trans: For a material sphere with mass $M$ uniformly distributed over its volume or surface the moments of inertia are, familiarly, ...)

Now we can calculate the magnetic moment of a spherical shell. At a polar angle $\theta$ the current is $ I = \rho \omega r^2 \sin \theta$, and the enclosed area is $\pi r^2 \sin^2 \theta$. Hence $$ \mu = \pi \rho \omega r^4 \int_0^{\pi} \sin^3 \theta \, \mathrm{d} \theta = \frac{4\pi}{3} \rho \omega r^4 = \frac{1}{3} e \omega r^2 \, . $$

Then with some simple algebra we have $$ \mu = \frac{e}{m} J \implies g = 2 \,. $$

Reference

• Abraham, M., Prinzipien der Dynamik des Elektrons, Ann. Physik 315, 105 (1903).

Answer 4

www.physicspages.com/2013/04/11/magnetic-dipole-moment-of-spinning-spherical-shell/

My search gives

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

This gives $g = 5/3 = 1.667$

Did not you provided link given below?

https://en.wikipedia.org/wiki/Electron_magnetic_moment#The_classical_theory_of_the_g-factor

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.

Answer 5

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

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

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

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

$\frac{r_e^8}{r_m^8}$. 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 $\frac{r_e^8}{r_m^8}\approx 2$.