# Is the Thomas precession a correction to a Lorentz transformation or an interaction?

In the heuristic derivations I have seen for the fine structure of hydrogen includes three terms spin-orbit coupling, Darwin term and corrections to the kinetic energy. For the spin-orbit coupling the construction goes as follows, write the relative magnetic field that an electron magnetic moment sees from an orbiting proton, change back to the frame of the proton using relativity, and multiply by a factor 1/2 due to Thomas precession (the fact that the Lorentz transformation used was in an specific non inertial frame).

All three of these corrections can be obtained from Dirac equation in the weak relativistic limit. However, today I stumbled on the spin orbit article of Wikipedia that introduces a fourth energy correction known as the Thomas interaction, which is minus half of spin-orbit and reproduces the 1/2 in the short derivation.

Is this just another way to introduce the 1/2 factor? Does it matter? Or is it a true interaction that has a different origin? and if so are there cases where for example when the "Thomas interaction" is larger or smaller than 1/2 the "bare" spin-orbit term?

• Thomas precession is a semiclassical idea to justify the electron magnetic moment. It accounts for the fact that in the Bohr model the electron is accelerated. The "electron rest frame", where the magnetic energy is initially evaluated, is related to the lab frame by means of instantaneous non-commuting boosts that cause the $\frac{1}{2}$ when the problem is studied in the lab frame, which is what we're interested in. Mar 16, 2023 at 20:57
• @Mr.Feynman that I get, but it is there any specific system in which seeing it as an interaction is more useful/appropriate? See also the first answer to this physics.stackexchange.com/questions/719879/… Mar 16, 2023 at 20:59

The calculation method using a non-relativistic Schrodinger equation and then adding corrections is a bit questionable, but for what it is worth the argument goes as follows.

First we adopt the instantaneous rest frame of the electron. Note, this is a sequence of inertial frames. Each frame being inertial, we can apply ordinary inertial-frame reasoning quite correctly and we get the interaction energy $$-\mu \cdot {\bf B}$$ where $$\mu = -g \mu_B \hat{\bf s}$$ and $${\bf B} = - {\bf v} \times {\bf E}/c^2$$. Consequently we find there will be a precession of the spin at the rate $$\omega_0 = -\mu \cdot {\bf B}/\hbar$$.

Now this precession is relative to the instantaneous rest frame. But the sequence of such frames has to be thought about carefully. We make the axes of each frame parallel to those of the one just before, so there is just a boost not a rotation between them. The sequence of boosts however results in an overall precession of axes: the Thomas precession. This precession has some rate $$\omega_{\rm T}$$. Consequently in the rest frame of the nucleus we find the electron's spin is precessing at the rate $$\omega_0 + \omega_T$$ But according to ordinary quantum theory a precession at that rate has to be associated with an energy level gap of $$E = \hbar(\omega_0 + \omega_T)$$ (a standard argument using ordinary quantum treatment of a spin-half system in an inertial frame shows this).

When one does the calculation one finds that $$\omega_T$$ has the opposite sign to $$\omega_0$$ and it does not have the $$g$$ factor, so the result overall is $$E = \hbar \omega_0 \frac{g-1}{g}.$$ Note by the way that since $$g$$ is not exactly 2, this correction is not exactly a factor $$1/2$$. Indeed the Thomas precession factor is not a multiplicative factor at all: it is an additive one. (The difference between $$g/2$$ and $$(g-1)$$ is readily seen in modern high-precision spectroscopy).

The question asks whether the Thomas precession can be seen as a kind of interaction. This is like asking whether Lorentz contraction can be seen as a kind of interaction. The answer for contraction is that the effect of differing lengths in different frames is a kinematic one, but if a body is being accelerated then all length changes it undergoes in some given fixed frame are owing to the forces acting on it. Similarly, for Thomas precession the rotation of a spin vector in any given fixed inertial frame is owing to the torques acting on that spin vector, but if you want to track the spin from a sequence of inertial frames then the Thomas precession is part of the story and it is a kinematic effect (i.e. a result of a choice of how to slice up spacetime into a sequence of spatial hypersurfaces).

• Thanks! Maybe it is too much to ask, but could you tell me where the Thomas term appears when deriving it from Dirac's equation (from a Foldy-Wouthyusen transformation)? Mar 16, 2023 at 23:47
• Sorry I've never heard of Foldy-Wouthyusen. But I would add that I think there cannot be any fixed recipe for deciding, in the Dirac equation, which part of the term in $\alpha^4$ to apportion to which part of the reasoning one would make when starting from a non-relativistic theory and adding corrections. Mar 17, 2023 at 13:33

I hesitate to improve the answer of my much more distinguished colleague @Andrew Steane, but this answer adds several things to his.

The ' general argument '. This derives from the idea that expectation values satisfy classical equations of motion and that energy level differences generate frequencies of oscillation in expectation values. Thus the precessing spin in an expectation value taken in a suitable superposition state must correlate with a energy level splitting. This is why this standard textbook approach arrives at an energy level splitting without first finding the individual level shifts.

However the argument cannot find the Hamiltonian from which the level shifts can be calculated. It is therefore in this sense incomplete.

There are two issues to be addressed. The Thomas precession is a vector but the Hamiltonian is a scalar. Thus we need an argument to deliver that. Secondly we need dimensions of energy/frequency or equivalently angular momentum. An obvious candidate is the spin angular momentum.

A very classical argument does both: the kinetic energy of a spinning body is $$(1/2)\ \omega^T \bf I \omega$$ so an extra precession $$\omega_T$$ adds a kinetic energy (to first order) of $$\omega^T\bf I\omega_T= \bf S.\omega_T$$ and this is indeed the right answer for the Hamiltonian, but surely not a satisfying way of arriving at it! No textbook known to me gives this argument for obvious reasons!

So there is a specific Thomas Hamiltonian. Whether this is equivalent to an interaction is a moot point. And it is given by the dot product of the spin and the Thomas precession. And this is not generally present in the textbook literature. But we need a better argument for deriving it

• I tried a FW construction but there is no natural term that appears that relates to Thomas. Anyway here is another question about this: physics.stackexchange.com/questions/755696/… Apr 11, 2023 at 8:59