Sunday, 14 October 2018

angular momentum - Classical proof of the gyromagnetic ratio $g=2$


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?



Answer



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.


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...