Larmor’s theorem
We can already conclude something from our results so far. First of all, in the classical theory the moment $\FLPmu$ was always proportional to $\FLPJ$, with a given constant of proportionality for a particular atom. There wasn’t any spin of the electrons, and the constant of proportionality was always $-q_e/2m$; that is to say, in Eq. (34.6) we should set $g=1$. The ratio of $\FLPmu$ to $\FLPJ$ was independent of the internal motion of the electrons. Thus, according to the classical theory, all systems of electrons would precess with the same angular velocity. (This is not true in quantum mechanics.) This result is related to a theorem in classical mechanics that we would now like to prove. Suppose we have a group of electrons which are all held together by attraction toward a central point—as the electrons are attracted by a nucleus. The electrons will also be interacting with each other, and can, in general, have complicated motions. Suppose you have solved for the motions with no magnetic field and then want to know what the motions would be with a weak magnetic field. The theorem says that the motion with a weak magnetic field is always one of the no-field solutions with an added rotation, about the axis of the field, with the angular velocity $\omega_L=q_eB/2m$. (This is the same as $\omega_p$, if $g=1$.) There are, of course, many possible motions. The point is that for every motion without the magnetic field there is a corresponding motion in the field, which is the original motion plus a uniform rotation. This is called Larmor’s theorem, and $\omega_L$ is called the Larmor frequency.  We would like to show how the theorem can be proved, but we will let you work out the details. Take, first, one electron in a central force field. The force on it is just $\FLPF(r)$, directed toward the center. If we now turn on a uniform magnetic field, there is an additional force, $q\FLPv\times\FLPB$; so the total force is \begin{equation} \label{Eq:II:34:18} \FLPF(r)+q\FLPv\times\FLPB. \end{equation} Now let’s look at the same system from a coordinate system rotating with angular velocity $\omega$ about an axis through the center of force and parallel to $\FLPB$. This is no longer an inertial system, so we have to put in the proper pseudo forces—the centrifugal and Coriolis forces we talked about in Chapter 19 of Volume I. We found there that in a frame rotating with angular velocity $\omega$, there is an apparent tangential force proportional to $v_r$, the radial component of velocity: \begin{equation} \label{Eq:II:34:19} F_t=-2m\omega v_r. \end{equation} And there is an apparent radial force which is given by \begin{equation} \label{Eq:II:34:20} F_r=m\omega^2r+2m\omega v_t, \end{equation} where $v_t$ is the tangential component of the velocity, measured in the rotating frame. (The radial component $v_r$ for rotating and inertial frames is the same.)  Now for small enough angular velocities (that is, if $\omega r\ll v_t$), we can neglect the first term (centrifugal) in Eq. (34.20) in comparison with the second (Coriolis). Then Eqs. (34.19) and (34.20) can be written together as \begin{equation} \label{Eq:II:34:21} \FLPF=-2m\FLPomega\times\FLPv. \end{equation} If we now combine a rotation and a magnetic field, we must add the force in Eq. (34.21) to that in Eq. (34.18). The total force is \begin{equation} \label{Eq:II:34:22} \FLPF(r)+q\FLPv\times\FLPB+2m\FLPv\times\FLPomega \end{equation} [we reverse the cross product and the sign of Eq. (34.21) to get the last term]. Looking at our result, we see that if \begin{equation*} 2m\FLPomega=-q\FLPB \end{equation*} the two terms on the right cancel, and in the moving frame the only force is $\FLPF(r)$. The motion of the electron is just the same as with no magnetic field—and, of course, no rotation. We have proved Larmor’s theorem for one electron. Since the proof assumes a small $\omega$, it also means that the theorem is true only for weak magnetic fields. The only thing we could ask you to improve on is to take the case of many electrons mutually interacting with each other, but all in the same central field, and prove the same theorem. So no matter how complex an atom is, if it has a central field the theorem is true. But that’s the end of the classical mechanics, because it isn’t true in fact that the motions precess in that way. The precession frequency $\omega_p$ of Eq. (34.11) is only equal to $\omega_L$ if $g$ happens to be equal to $1$. 