Special relativity constrains massless electric dipoles, but not massless magnetic dipoles?

We expect a magnetic or electric dipole to make a field that has some universal transformation properties, and we expect these properties to be purely classical, and independent of the other characteristics of the source. So what does this tell us about massless dipoles?

Suppose you make an electric dipole by gluing charges $\pm q$ to the ends of a popsicle stick of length $L$. Then under a boost $v$ parallel to the stick, we have $qL\rightarrow 0$ as $v\rightarrow c$. This suggests that a massless particle has zero electric dipole moment parallel to its motion.

On the other hand, field theorists seem to expect that massless magnetic dipoles are OK, and can have dipole moments aligned with their spins and parallel to their motion. I imagine that neutrinos would have been considered to be examples, back when we thought they were massless. Mike Stone says in a comment: "A massless charged chiral particle has a magnetic moment of exactly μ=±e/(2E)×k/|k| where the ± is the helicity and E the energy."

But this all seems odd to me. Shouldn't duality hold between electric and magnetic fields, so that whatever is true for electric dipole fields is also true for magnetic ones? If our universe had magnetic monopoles, then we could recap the popsicle stick argument and convince ourselves that massless magnetic dipoles could not have a dipole moment in the direction of motion.

Can anyone clarify what is going on here?

related:

No magnetic dipole moment for photon

Electric dipole moment of electron: about what point is the moment taken?

• Are you sure that the formula ${\bf p} = q{\bf L}$ holds for a relativistically moving dipole? A surprising number of paradoxes in relativistic classical EM are resolved simply by noting that a lot of the textbook formulas actually only apply in the non-relativistic case - like in the Mansuripur's paradox debacle in PRL a few years ago. – tparker Jul 26 '18 at 17:08

I think that rather muddied some of the issues in my original question about electron electric dipole moments --- so here are some comments, and perhaps a further implied question, about about the Lorentz tranformation properties of such moments.

The $-\mu\cdot {\bf B}$ energy of a stationary magentic dipole interacting with a magnetic field can be written a Lorentz invariant contribution to a Lagrangian as $M_{\mu\nu}F^{\mu\nu}$ so this makes it clear that a dipole moment is a naturally a skew symmetric Lorentz $2$-tensor. In the particle rest frame a magnetic dipole will have $$\mu_x= M_{23},\quad\mu_y = M_{31}, \quad \mu_z= M_{12}$$ with $M_{01}=M_{02}= M_{03}=0$.

For an electric dipole a stationary particle will have and interaction term $D_{\mu\nu}F^{\mu\nu}$ with $$d_x= D_{01}, \quad d_y= D_{02}, \quad d_z= D_{03}$$ with the other three components vanishing

Whe the moment-possessing particle moves the $M_{0i}$ components of the $2$-tensor will become non-zero and so a moving magneitc dipole behaves as if it has an electric dipole moment. Indeed when a current loop moves, or is observed from a moving frame, it will appear to have plus and minus charges arranged so that it posseses an electric dipole moment that is perpendicular to its direction of motion.

Similarly a moving electric dipole will have something a magnetic dipole character.

All this assumes that the particle has a rest frame. A massless particle does not have a rest frame, so what happens to the moments? My statement about the moment of a massles charged spinning particle comes form thinking of such a particle in a circular cyclotron orbit. Its spin (and hence any magnetic moment) is forced to point in the direction of motion and so must precess at the cyclotron frequency $\Omega_{\rm cyclotron}=eB/E$ where $E$ is the energy. Now the Larmor precession rate is $\Omega_{\rm Larmor}= -B \mu/{\rm spin}$. So, using this as a definition of the effective moment and equating $\Omega_{\rm Larmor}$ with $\Omega_{\rm cyclotron }$ we have, for spin =1/2, $$\mu= e/2E.$$
This result can also be obtained from the Gordon decomposition of the current for a Weyl fermion, or by more sophisticated arguments [D.T Son, N. Yamamoto, arXiv:1210.8158]. Because the $\bar \psi \sigma_{\mu\nu}\psi$ Pauli-Weiskopf anomaous magnetic moment term is identically zero for Weyl particles there is no possibility of adding, by hand, an anomalous magnetic moment correction to the Weyl equation.

What is less clear to me, is how this phenomonological precession-rate definition of $\mu$ fits in with the 2-tensor moment $M_{\mu\nu}$. It's hidden in the machinary of the Weyl equations just as the $\mu =eg/2m\times 1/2$ with $g=2$ Dirac magnetic moment is hidden in the Dirac machinary.

Ben quite correctly says that the above does not answer his question. Below I try to explain why the issue of Lorentz transformations for massless dipoles (both magnetic electric) is not simple. I'd like to extract a physical picture (Ben's popsicle stick) from the rather complicated formalism but it is hard to see the wood for the trees...

Let's begin with the notion of relativistic "spin" for an extended body with a conserved energy-momentum tensor $T^{ab}$. The Lorentz tensor giving total angular momentum of the body about the origin is $$M^{ab}= x^a p^b-x^b p^a+S^{ab},$$ where $x^a=(x^0,x^1,x^2,x^3)$ is the body's spacetime position, $$p^a= \int_{x^0=const} T^{0a}\,d^3x$$ its four-momentum and the skew-symmetric tensor $S^{ab}$ is its "intrinsic angular momentum" --- the latter defined as the angular momentum about the point in the body labelled by the coordinates $x^a$. The problem is that for a relativistic spinning obect there is no natural choice for this point. The obvious choice, the "center of mass" $$X^a_{\rm cm} \equiv \frac 1 E \int_{x^0=const} x^a T^{00}\,d^3x, \quad E=\int_{x^0=const} T^{00}\,d^3x,$$ is a frame dependent. Changing the definition of the position'' of the body leads to a shuffling of angular momentum between the orbital part $x^a p^b-x^b p^a$ and the spin part $S^{ab}$. Different choices lead to different conditions on $S^{ab}$. It is shown in Misner, Thorn and Wheeler (MTW) that if we choose to define the "position" to be the body's center of mass in a frame moving with four-velocity $v^a$ then $v_aS^{ab}=0$. If we define the position to be the center of mass in the body's rest frame then we have the condition $p_aS^{ab}=0$; if we choose the center of mass in the lab frame then $S^{0b}_{\rm lab}=0$.
The totally antisymmetric Pauli-Lubanski tensor $$W^{abc}\stackrel{\rm def}{=} p^a S^{bc} + p^bS^{ca}+ p^c S^{ab}$$ has the useful property that it is unaffected by such reshuffling. In four spacetime dimensions $W^{abc}$ is usually repackaged as the Pauli-Lubanski psudovector $W^a={\textstyle\frac 16} \epsilon^{abcd}W_{bcd}$ but the 3-tensor is more general as it has the same properties in all space-time dimensions. Using $W^{abc}$ we find that the lab-frame center of mass spin $S^{ab}_{\rm lab}$ is related to the rest-frame center-of-mass spin $S^{ab}$ by
$$S^{ab}_{\rm lab} = \left(S^{ab} - \frac{p^a}{E} S^{0b} - S^{a0}\frac{p^b}{E}\right) = \frac 1 E W^{0ab}.$$ This latter quantity is a tensor only under rotations as its definition is tied to the frame where $v^a=(1,0, \ldots,0)$.

Now let's see how these ideas play out when applied to
positive energy solutions $u_\alpha({\bf k})$ of the Dirac equation. We use the rapidity $s$ in terms of which $$E=m\cosh s,\\ {\bf k}=\hat {\bf k }\, m \sinh s,\\ {\bf v}= \hat {\bf k}\,\tanh s,$$ and $\gamma \equiv (1-|{\bf v}|^2)^{-1/2}=\cosh s$. The 4-spinor part of the plane wave solution $$\psi_{\alpha, {\bf k}}(x) =u_\alpha({\bf k}) e^{i{\bf k}\cdot {\bf x} - iEt}$$ is then $$u_\alpha({\bf k})= \frac{1}{\sqrt{2m(E+m)}}\left[\matrix{(E+m)\chi_\alpha \cr ({\sigma}\cdot {\bf k} )\chi_\alpha}\right] =\left[\matrix{\phantom{({ \sigma}\cdot \hat {\bf k} )} \cosh (s/2)\chi_\alpha \cr \sinh(s/2) ({ \sigma}\cdot \hat {\bf k} )\chi_\alpha}\right].$$ I'm using covariant normalization $\bar u_\alpha u_\alpha=1$ in which the particle density in the plane-wave beam is $E/m$. The quantity $\chi_\alpha$ is a 2-spinor that determines the spin state in the particle's rest frame.

Now consider $$S^{ab}_{\alpha\beta} = \bar u_\alpha(k)\Sigma^{ab} u_\beta(k),$$ where the $$\Sigma^{ab}= \frac i 4 [\gamma^a,\gamma^b]$$ are the 4-spinor generators of Lorentz transformations. This expression
defines the $\alpha,\beta$ matrix elements a Lorentz tensor-valued opertor $\hat S^{ab}$ for the plane wave states. In the rest frame this tensor coincides with $\chi_\alpha^\dagger { \sigma} \chi_\beta$. Furthermore, the Dirac equation gives the condition $k_a \hat S^{ab}=0$ so it is natural to regard $\hat S^{ab}$ as the operator of intrinsic spin about the center of mass in the rest frame.

For Dirac-equation plane-wave solutions we have $$\frac 12 \bar u_\alpha\{\gamma_a, \Sigma_{bc}\} u_\beta = \frac 1m \bar u_\alpha(k_a \Sigma_{bc}+ k_b \Sigma_{ca}+k_c\Sigma_{ab})u_\beta=\frac{1}m (W_{abc})_{\alpha\beta},$$ so that $$\frac 1 \gamma u^\dagger_\alpha \Sigma_{ij} u_\beta = \bar u_\alpha\left(\Sigma_{ij}- \frac{k_i}{E} \Sigma_{0j}- \Sigma_{i0}\frac {k_j}{E}\right) u_\beta.$$ is the spin density in a beam with one particle per unit volume and is the angular momentum of a single particle about the lab frame center of mass.

Using the explicit solution $u_\alpha({\bf k})$ given above we find that $$\bar u_\alpha \Sigma_{ij} u_\beta= \frac 12 \epsilon_{ijk} \chi^\dagger_\alpha \left(\gamma \sigma_k -\frac{({\bf k}\cdot { \sigma}) k_k}{m^2(1+\gamma)}\right)\chi_\beta, \quad i,j=1,2,3, \\ \bar u_\alpha \Sigma_{0i} u_\beta= \frac1{2m} \chi^\dagger_\alpha (\epsilon_{ijk} k_j \sigma_k)\chi_\beta .$$ Both these quantities diverge as $\gamma^{-1}$ as $m\to 0$ at fixed $E$. This is natural as the rest frame is being pushed off to infinite momentum. Meanwhile $$\bar u_\alpha\left(\Sigma_{ij}- \frac{k_i}{E} \Sigma_{0j}- \Sigma_{i0}\frac {k_j}{E}\right) u_\beta= \epsilon_{ijk} \frac 1{\gamma}\left\{\frac 12 \left( \sigma_k + \frac{({\bf k}\cdot { \sigma}) k_k}{m^2(1+\gamma)} \right)_{\alpha\beta}\right\}$$ remains finite and tends to the $\alpha,\beta$ matrix elements of $\epsilon_{ijk}S_k,$ where ${\bf S}= (\hat {\bf k }\cdot { \sigma}) \hat {\bf k}/2$ is $\hat {\bf k}$ times the particle's helicity. It is this latter quantity multiplied by $e/E$ that gives the Weyl particle's Larmor-precession defined magnetic moment. Equations involving this moment will not be conventionally covariant as $\hat S^{ab}_{\rm lab}$ is not a Lorentz tensor. This is the source of unusual way that Lorentz invariance manifests itself (the side jump'' ) in the statistical mechanics of massless spinning particles.

The same issue has to be faced when we consider the electric dipole moment of a massless particle. For a charged particle the electric dipole moment depends on the chosen "position" of the particle. This position will change as we make a Lorentz transformation.

• This is helpful, thanks for writing it up. I don't think it's particularly relevant or physically important that a magnetic dipole energy is expressed as a contraction of F with and antisymmetric tensor while the electric dipole energy involves a symmetric tensor. We could just as easily have written Maxwell's equations using the Hodge dual *F, in which case the symmetry and antisymmetry would be expressed the other way around. The semiclassical argument is interesting, but it doesn't seem to me to clear up the underlying issues. – Ben Crowell Jul 26 '18 at 16:49
• @Ben Crowell Both my $M_{\mu\nu}$ and $D_{\mu\nu}$ are skew-symmetric. The violation of EM duality in the interaction with matter comes from all known particles having at most electric charge, and not magnetic charge. In the absence of matter E&M is exactly self dual. If we did have mag monoples then your argument would be exactly correct also. – mike stone Jul 26 '18 at 17:01
• Thanks for the correction re symmetry of M and D, but I still don't see the point of your discussion of their components. The point remains that your description of M would hold for D and vice versa, if we took the Hodge dual of F. I just don't see how this answer really gets at the point of the question. – Ben Crowell Jul 27 '18 at 1:05
• @Ben Crowell I want to answer your question properly -- it is an interesting one, The problem is that the proper answer depends more on the propeties of the spin rather than the charge distribution. Relativistic spin is complicated however. I am going to organise what I understand for the magnetic moment case. I will need to do this offline but once I have it sorted I will post it, and then we can try to relate this to your question about the electctric moment. – mike stone Jul 27 '18 at 14:29
• @Ben Crowel I've added a discusion of spin and how it affects this stuff. It's bit technical, but the issue is precisely how to relate the technical stuff to concrete pictures like your popsicle stick. – mike stone Jul 27 '18 at 19:18

If our universe had magnetic monopoles, then we could recap the popsicle stick argument and convince ourselves that massless magnetic dipoles could not have a dipole moment in the direction of motion.

But we don't appear to have magnetic monopoles in this universe, and assuming that we do requires a lot of very careful re-thinking of symmetry arguments. For example, since electric and magnetic fields behave differently under inversion of space, I think that magnetic charge would have to be a pseudoscalar quantity. I'm sure we could spend all day thinking of other problematic restrictions.

The way to extend your (very clever) argument about classical dipoles is not to posit the existence of charges that we don't observe, but instead to consider the behavior under boosts of a classical magnetic dipole: a current loop. Under boosts, any component of the normal vector to a current loop that's parallel to the velocity is unchanged, while any component of that normal vector perpendicular to the velocity is diluted by length contraction of one side of the loop. So a massless current loop may have a magnetic dipole moment which is parallel to its momentum, but not perpendicular to its momentum.

Note that your popsicle-stick dipole does not forbid an electric dipole moment which is perpendicular to the direction of the boost --- the length contraction only kills the component parallel to the boost.

Classical angular momentum undergoes the same alignment with the velocity direction under boosts as the magnetic moment does, for more or less the same reasons. My friends in the experimental electric-dipole-moment business, during this part of their talks, point out that the intrinsic angular momentum of a particle is its only preferred direction in space, and that any other (psuedo-)vector property of a particle must be parallel or antiparallel to the angular momentum. When pressed for an explanation, they'll either refer to the Wigner-Eckhart theorem, or else give a classical analogy where an electric dipole moment perpendicular to the spin axis which averages to zero.

I guess that's a couple of partial answers. Your classical argument is fine as long as you don't invent entities with different symmetries than the rest of classical electromagnetism. There may be a purely symmetry-based argument based on the way that $C,P,T$ transformations and the spin degree of freedom for fields arise from the Lorentz group and symmetry under boosts, but I feel a lot murkier in that territory.

• Interesting, thanks. I guess I'm still feeling unconvinced. I'm not sure I believe your argument about the current loop, for two reasons. (1) It seems to suggest that the transformation properties of a magnetic dipole field under a boost have to be different from the transformation properties of an electric dipole field under a boost, but I'm pretty sure that can't be true. The properties should be dual. This is just a statement about the fields, independent of what creates them. [...] – Ben Crowell Jul 26 '18 at 0:22
• [...] (2) If I imagine a current loop made out of a spinning, charged ring, shouldn't the current be affected by time dilation? I would imagine this could also be worked out in terms of the current four-vector. – Ben Crowell Jul 26 '18 at 0:23
• Oh, I forgot time dilation, silly me. But time dilation would reduce the magnetic moment parallel and perpendicular to the boost, while length contraction only affects the electric moment parallel to the boost. I'm not totally convinced myself (hence "partial answer"). – rob Jul 26 '18 at 11:17