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Thomas Precession

We must begin our discussion by noting that the magnetic moment of an electron is (according to the ``Uhlenbeck-Goudsmit hypothesis'')

\begin{displaymath}
\mbox{\boldmath$\mu$} = \frac{ge}{2mc} {\bf s}
\end{displaymath} (16.103)

where $\mbox{\boldmath$s$}$ is the (half integer) spin of the electron in units of $\hbar$ and where $g$ is the ``g-factor'' introduced to accomodate two distinct results. The splitting of the observed spectra in an applied magnetic field B via the anomalous Zeeman interaction:
\begin{displaymath}
U_{\rm AZ} = - \frac{ge}{2mc} {\bf s} \cdot {\bf B}
\end{displaymath} (16.104)

was correctly predicted only if $g = 2$. On the other hand (as we shall see), the simple classical argument that led to this result also led to a spin orbit interaction
\begin{displaymath}
U_{\rm SO} = \frac{g}{2m^2c^2} ({\bf s} \cdot {\bf L}) \frac{1}{r}
\frac{dV}{dr}
\end{displaymath} (16.105)

(where ${\bf L} = m({\bf r} \times {\bf v})$ is the orbital angular momentum of the electron) that was a factor of, curiously enough, 2 too large. That is, the fine-structure intervals observed in nature were only half the theoretically predicted values. If $g =1 $ was chosen instead, the splittings were correct but the Zeeman effect was then normal (instead of anomalous, as observed).

I don't have the time to go into more detail on, for example, what the Zeeman effect (splitting of energy levels in an applied magnetic field) is. In any event, it is strictly a quantum effect, and you should study it soon in elementary quantum theory, if you haven't already.

Thomas (who taught for years over at NC State) showed in 1927 that the discrepancy is due to a relativistic kinematic correction like that we previously considered. In a nutshell, the rest frame of the electron rotates as well as translates (boosts) and we must therefore take into account both kinematical effects. This results in an additional (Thomas) ``precession'' of the frames. When Thomas precession is taken into account, not only are both the fine structure and anomalous Zeeman effect in atoms accomodated, but a deeper understanding of the spin-orbit interaction in nuclear physics (and rotating frames in general) also results.

Let us begin by (naívely) deriving the spin-interaction energy. Suppose the electron moves with velocity ${\bf v}$ in external fields E and B. Then the torque on the electron in its rest frame is just

\begin{displaymath}
\left( \frac{d{\bf s}}{dt} \right)_{\rm rest~frame} = \mbox{\boldmath$\mu$} \times
{\bf B}'
\end{displaymath} (16.106)

where ${\bf B}'$ is the magnetic field in that frame.

As we will show very soon, the magnetic field transforms like

\begin{displaymath}
{\bf B}' = \left( {\bf B} - \frac{{\bf v}}{c} \times {\bf E} \right)
\end{displaymath} (16.107)

to order $v^2/c^2$. Then
\begin{displaymath}
\left( \frac{d{\bf s}}{dt} \right)_{\rm rest~frame} = \mbox...
...
\left( {\bf B} - \frac{{\bf v}}{c} \times {\bf E} \right) .
\end{displaymath} (16.108)

Associated with this torque there is an interaction energy
\begin{displaymath}
U' = - \mbox{\boldmath$\mu$} \cdot \left( {\bf B} - \frac{{\bf v}}{c} \times {\bf E}
\right).
\end{displaymath} (16.109)

The electric force $e{\bf E}$ is very nearly the negative gradient of a spherically averaged potential energy $V(r)$. For one electron atoms this is exact; it is a good approximation for all the others. Thus we will try using

\begin{displaymath}
e{\bf E} = - \frac{{\bf r}}{r} \frac{dV}{dr}
\end{displaymath} (16.110)

in the equation for the spin interaction energy:
\begin{displaymath}
U' = -\frac{ge}{2mc} {\bf s} \cdot {\bf B} + \frac{g}{2 m^2 c^2}({\bf s}
\cdot {\bf L}) \frac{1}{r} \frac{dV}{dr}
\end{displaymath} (16.111)

(where ${\bf L} = m({\bf r} \times {\bf v})$ for the orbiting electron). This gives the anomalous Zeeman effect correctly (from the first term) but the spin orbit (fine structure) splitting is a factor of two too large. Too bad!

The error is, in a nutshell, that we have assumed the electron to be in a ``rest'' frame (that is, a frame travelling in a straight line) when that frame is, in fact, rotating. There is an additional correction to vector quantities that arises from the rotation of the frame. This correction, in macroscopic systems, gives rise to things like coriolis force.

Let us recall (from classical mechanics) that if a coordinate system rotates at some angular velocity $\mbox{\boldmath$\omega$}$, the total rate of change of any vector quantity is given by

\begin{displaymath}
\left( \frac{d{\bf G}}{dt} \right)_{\rm non-rot} = \left(
...
...{\rm rest~frame} + \mbox{\boldmath$\omega$} \times {\bf G} .
\end{displaymath} (16.112)

This is a geometric relation that says that a vector in a non-rotating frame is related to the same vector expressed in a (rotating) ``rest'' frame by adding its time rate of change in direction resulting from the rotation of the frame. A moment of quiet reflection should convince you that this should have the magnitude

\begin{displaymath}
G \frac{d \theta}{dt}
\end{displaymath}

and should be perpendicular to $\mbox{\boldmath$\omega$}$ and G. This just adds the rotation of the frame to the vector in the frame to get the vector in a non-rotated frame.

Well, as I noted above, the expression we have given above for the time rate of change of the spin was correct for the field and moment expressed in the rest frame of the electron. In the lab (non-rotating) frame, which is where we measure its energy, we therefore should have:

\begin{displaymath}
\left( \frac{d {\bf s}}{dt} \right)_{\rm non-rot} = {\bf s}...
... \frac{ge{\bf B'}}{2mc} - \mbox{\boldmath$\omega$}_T \right)
\end{displaymath} (16.113)

where $\omega_T$ is the angular velocity of the precession of the frames. This adds a $({\bf s} \cdot \mbox{\boldmath$\omega$}_T)$ correction to the interaction energy:
\begin{displaymath}
U = -\frac{ge}{2mc} {\bf s} \cdot {\bf B} + \frac{g}{2 m^2 ...
... \frac{dV}{dr} + ({\bf s} \cdot \mbox{\boldmath$\omega$}_T) .
\end{displaymath} (16.114)

$U$ is thus the laboratory potential energy of interaction. What, then, is the correct value of $\mbox{\boldmath$\omega$}_T$?

To answer that we must consider carefully what defines the ``rest'' frame of the accelerating electron. We will do so by chopping the motion of the electron into infinitesimal segments. If the electron is moving at velocity ${\bf v}(t) = c \mbox{\boldmath$\beta$}$ at any instant of time $t$, then at $t + \delta
t$ the electron is moving at ${\bf v}(t) = c(\mbox{\boldmath$\beta$} + \delta
\mbox{\boldmath$\beta$})$. To get from the lab frame ($x$) to the instantaneous rest frame of the electron ($x'$) we must therefore boost:

\begin{displaymath}
x' = A(\mbox{\boldmath$\beta$}) x
\end{displaymath} (16.115)

(at $t$) or
\begin{displaymath}
x'' = A(\mbox{\boldmath$\beta$} + \delta \mbox{\boldmath$\beta$}) x
\end{displaymath} (16.116)

(at $t + \delta
t$). Note that for each of these transformations, there is no rotation, just the boost.

The coordinate frame precession is going to be determined by the Lorentz transformation between these two (infinitesimally separated) results:

\begin{displaymath}
x'' = A_T x'
\end{displaymath} (16.117)

where (as I hope is obvious)
\begin{displaymath}
A_T = A(\mbox{\boldmath$\beta$} + \delta \mbox{\boldmath$\b...
...delta \mbox{\boldmath$\beta$}) A(- \mbox{\boldmath$\beta$}) .
\end{displaymath} (16.118)

To evaluate this (in the limit of vanishing $\delta t$) we will pick an initial $\mbox{\boldmath$\beta$}$ along the 1 direction and add to it $\delta \mbox{\boldmath$\beta$}$ in the 1-2 plane. Clearly this is general, for a suitable initial orientation of the coordinate system.

Then

\begin{displaymath}
A(- \mbox{\boldmath$\beta$}) =
\left( \begin{array}{cccc}...
... 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{array} \right)
\end{displaymath} (16.119)

and (keeping only first order terms in $\delta \mbox{\boldmath$\beta$}$)
\begin{displaymath}
A(\mbox{\boldmath$\beta$} + \delta \mbox{\boldmath$\beta$})...
...a \beta_2 & 1 & 0
\\
0 & 0 & 0 & 1
\end{array} \right) .
\end{displaymath} (16.120)

We multiply these matrices together to obtain:
\begin{displaymath}
A(\mbox{\boldmath$\beta$} + \delta \mbox{\boldmath$\beta$})...
...a \beta_2 & 1 & 0
\\
0 & 0 & 0 & 1
\end{array} \right) .
\end{displaymath} (16.121)

(Note that the action of $A(-\mbox{\boldmath$\beta$})$ is only in the upper left corner). Finally, if we decompose this in terms of the $S$ and $K$ matrices, we get:
\begin{displaymath}
A_T = I - \left( \frac{\gamma - 1}{\beta^2} \right) (\mbox{...
...+
\gamma \delta \mbox{\boldmath$\beta$}_\perp) \cdot {\bf K}
\end{displaymath} (16.122)

where $\delta \mbox{\boldmath$\beta$}_\parallel$ and $\delta \mbox{\boldmath$\beta$}_\perp$ are the components of $\delta \mbox{\boldmath$\beta$}$ parallel to and perpendicular to $\mbox{\boldmath$\beta$}$, respectively.

To first order in $\delta \mbox{\boldmath$\beta$}$, we see that the total transformation $A_T$ is equivalent to a boost and a rotation:

\begin{displaymath}
A_T = A(\Delta \mbox{\boldmath$\beta$}) R(\Delta \mbox{\boldmath$\Omega$})
\end{displaymath} (16.123)

which can be performed in either order (because they are ``infinitesimal'' and hence commute to first order. In this expression,
\begin{displaymath}
A(\Delta \mbox{\boldmath$\beta$}) = I - \Delta \mbox{\boldmath$\beta$} \cdot {\bf K}
\end{displaymath} (16.124)

and
\begin{displaymath}
R(\Delta \mbox{\boldmath$\Omega$}) = I - \Delta \mbox{\boldmath$\Omega$} \cdot {\bf S} .
\end{displaymath} (16.125)

Obviously,
\begin{displaymath}
\Delta \mbox{\boldmath$\beta$} = \gamma^2 \delta \mbox{\bol...
...ta$}_\parallel +
\gamma \delta \mbox{\boldmath$\beta$}_\perp
\end{displaymath} (16.126)

and
\begin{displaymath}
\Delta \mbox{\boldmath$\Omega$} = \left( \frac{\gamma - 1}{...
...ox{\boldmath$\beta$} \times \delta
\mbox{\boldmath$\beta$} .
\end{displaymath} (16.127)

Finally we see explicitly that at least for infinitesimal transformations, a pure Lorentz boost $A(\mbox{\boldmath$\beta$} + \delta \mbox{\boldmath$\beta$})$ is equivalent to a boost to an infinitesimally differing frame $A(\mbox{\boldmath$\beta$})$ followed by a simultaneous infinitesimal boost and rotation.

Now comes the tricky part. The equation of motion for the spin that we began with (in the ``rest frame'') can be expected to hold provided that the evolution of the rest frame is described by a series of infinitesimal boosts alone (without rotations). In other words, we have to add the relativistic equivalent of counterrotating the frames (like we did above with the $\mbox{\boldmath$\omega$}_T \times {\bf G}$ term). These ``relativistically nonrotating coordinates'' are related to the instantaneous rest frame coordinates of the electron by the infinitesimal boost

\begin{displaymath}
x''' = A(\Delta \mbox{\boldmath$\beta$}) \left \{ A(\mbox{\boldmath$\beta$}) x = x' \right \}
\end{displaymath} (16.128)

alone. In terms of the lab coordinates,
\begin{displaymath}
x''' = R(-\Delta \mbox{\boldmath$\Omega$}) A(\mbox{\boldmath$\beta$} + \delta \mbox{\boldmath$\beta$}) x .
\end{displaymath} (16.129)

Thus the ``rest'' system of coordinates of the electron are defined by $x'''$. They are rotated by $- \Delta \mbox{\boldmath$\Omega$}$ relative to the boosted laboratory axes $x''$. If a physical vector G has a (proper) time rate of change of $d{\bf G}/d\tau$ in the rest frame, the precession of the rest frame axes with respect to the laboratory makes the total time rate of change

\begin{displaymath}
\left( \frac{d{\bf G}}{dt} \right)_{\rm non-rot} = \left(
...
...)_{\rm rest~frame} + \mbox{\boldmath$\omega$} \times {\bf G}
\end{displaymath} (16.130)

as before with
\begin{displaymath}
\mbox{\boldmath$\omega$}_T = \lim_{\delta t \rightarrow 0} ...
...ac{\gamma^2}{\gamma + 1} \frac{{\bf a} \times
{\bf v}}{c^2}
\end{displaymath} (16.131)

(Recall that the connection to laboratory time is $d{\bf G}/dt = \gamma^{-1}
d{\bf G}/d \tau$ in the rest frame itself).

The acceleration perpendicular to the instantaneous velocity appears in this expression because it is this quantity that produces the ``rotation'' in the infinitesimal transformation between frames that occured in the infinitesimal time interval. Note that this is a purely kinematical effect, and has nothing to do with the laws of nature, just like the non-relativistic ``coriolis force'' and ``centrifugal force''. If one wishes to relate the laws of nature as measured in some accelerating frame to those measured in a non-accelerating frame, then it is necessary to insert a fictitious ``force'' (or more properly interaction ``energy'') that is kinematic in origin.

In this case, curiously enough, the laws of nature are known in the accelerating frame, and the fictitious force appears in the lab frame, where it is not properly speaking fictitious. However, it is still kinematic. That is, there is no actual energy associated with the fictitious interaction (whatever that means); however, this interaction is necessary nonetheless if we wish to obtain the equation of motion from the energy equation alone without explicit consideration of the transformations of frames.

To conclude, for electrons the acceleration is caused by the (screened) Coulomb force on the electron that keeps it bound. Thus

\begin{displaymath}
\mbox{\boldmath$\omega$}_T = \frac{-1}{2c^3} \frac{{\bf r} ...
...dr} = \frac{-1}{2m^2c^2} {\bf L} \frac{1}{r}
\frac{dV}{dr} .
\end{displaymath} (16.132)

This has exactly the same form as the ``rest frame'' spin orbit interaction with half the magnitude and the opposite sign. It beautifully cancels the extra factor of 2. The final result is:
\begin{displaymath}
U' = -\frac{ge}{2mc} {\bf s} \cdot {\bf B} + \frac{(g-1)}{2 m^2 c^2}({\bf
s} \cdot {\bf L}) \frac{1}{r} \frac{dV}{dr} .
\end{displaymath} (16.133)

With $g = 2$, both the spin-orbit interaction and the anomalous Zeeman effect are correctly predicted in accord with what is experimentally observed. Relativistic effects, which are generally thought of as being ``subtle'', are not subtle at all when it comes to kinematics. The relativistic kinematic correction is as large as the other quantities naturally present independent of the particular orbit or speed of the electron.

This effect is even more pronounced in atomic nuclei. There the electromagnetic forces are much weaker than the binding nuclear forces, and can be neglected to lowest order. However, even uncharged neutrons experience a spin-orbit interaction

\begin{displaymath}
U_{rm N} = - \frac{1}{2M^2c^2} {\bf s} \cdot {\bf L} \frac{1}{r}
\frac{dV_{\rm N}}{dr}
\end{displaymath} (16.134)

that is now purely kinematic and has nothing whatsoever to do with the electromagnetic force! There will be a small electromagnetic correction to this for protons. This simple prediction is in reasonable agreement with what is observed in many nucleii for simple models for $V_{\rm N}$. Unfortunately, the system is actually so complicated that this simple minded, single particle description itself is not really valid.

This is just a drop in the proverbial bucket of accelerated systems. Clearly, accelerated, relativistic systems have a much more involved structure than that described by the Lorentz transformations alone. This becomes even more so when Einstein's revered equivalence principal is invoked, so that gravitational force and ``real'' acceleration are not (locally) distinguishable. But that is general relativity and far beyond the scope of this course.


next up previous contents
Next: Covariant Formulation of Electrodynamics Up: The Lorentz Group Previous: Infinitesimal Transformations   Contents
Robert G. Brown 2013-01-04