3
$\begingroup$

Spinors are famously like spinning tops, but not actually like spinning tops since they are point particles and thus cannot rotate around their axis. It is easy to show algebraically how spinors must interact with the B (magnetic) part of the electromagnetic field tensor, just like a scalar particle with orbital angular momentum would, but it eludes me how this is actually connected to spinning/rotating/circular motion in a geometrical sense.

This is how it's usually presented, I believe:

  • Start with Lorentz covariance.
    • Either work with preserving the quadratic form (Minkowski metric), in which case we can observe that Clifford Algebra helps take square roots of said form. This leads to the Dirac equation from the Klein-Gordon equation
    • Alternatively: Find other representations of the Lorentz group (same Lie algebra), which also leads to Clifford Algebra. Finding a Lagrangian for this new representation leads to the Dirac equation.
  • Replace the derivative with the covariant derivative to incorporate the electromagnetic field. Act on this Dirac equation with $(i\gamma^{\mu} D_{\mu} + m )$ (observe the plus sign), do the algebraic juggling and get a new equation with a $\sigma^{\mu \nu} F_{\mu \nu}$ in it.
  • Because of how Clifford algebra works we get $\sigma^3 (F_{12}-F_{21})$ instead of $\sigma^3 F_{33}$ which we would get if we worked with ordinary numbers.
  • $F_{12}-F_{21}$ and similar expressions are of course associated with the Curl/Cross product/Rotation. It's possible to calculate that presence of orbital angular gives the same expression, which is where we get the Einstein–de Haas effect.

To me, it seems almost coincidental that we get something resembling rotations. We need Clifford algebra to solve square roots, and Clifford algebra is "rotation-like".

Is there a better and more "geometric" way of understanding this? Is there a rotation hiding in the Minkowski metric, or perhaps in all square roots? I feel like all the algebra obscures what actually happens.

I'm aware of this Math.SE question, but I believe mine to be slightly different: Is there Geometric Interpretation of Spinors?

$\endgroup$
2
  • $\begingroup$ Maybe my answer here, on how spin was necessarily assigned to elementary particles will help physics.stackexchange.com/questions/403267/… $\endgroup$
    – anna v
    Jul 25, 2022 at 10:35
  • $\begingroup$ "Spinors are famously like spinning tops, but not actually like spinning tops " Well, actually, Andrew Steane has pointed out that Felix Klein "originally designed the spinor to simplify the treatment of the classical spinning top in 1897." referenced along with other non-quantum instances in physics.stackexchange.com/questions/459727/… $\endgroup$
    – iSeeker
    Aug 6, 2022 at 15:16

1 Answer 1

2
$\begingroup$

In short: the spinning fields correspond to representations of the Lorentz group that we see arising for an isolated rotating body.


Internal angular momentum as Lorentz group generator

As a demonstration, consider a cloud of non-interacting particles in Minkowski space-time that you are viewing from afar. Let me use 3+1 splitted Minkowski coordinates so that I can define a Hamiltonian formalism and denote the coordinates as $t,x^i$ with $i,j,k=1,2,3$. Now each particle with mass $m_a$, $a = 1...N$ will have an on-shell 4-momentum $$P^\alpha = \left(\sqrt{m_a - \sum (P^i_a)^2}, P^1_a,P^2_a,P^3_a\right)$$ I also define the formal 4-position as $x^\alpha_a(t) = (t,x^i_a(t))$, where $t$ is simply the value of the time coordinate used to parametrize the motion of all the particles.

This Hamiltonian system naturally has the Poisson algebra $\{x^i_a,P_{ja}\} = \delta^i_j$ for every $a$ and otherwise all the variables commute. (Note that $P_i=P^i$ in Minkowski coords.) Now define an angular momentum tensor $$M^{\alpha \beta} = \sum_a (x^\alpha_a - x^\alpha_{\rm c.})P_a^\beta - (x^\beta_a - x^\beta_{\rm c.})P_a^\alpha $$ where all the functions on the right-hand side are given as functions of $t$, and $x^\alpha_{\rm c.} = (0,x^i_{\rm c.})$ is some referential centroid position (which is treated as Poisson-commuting with all the other variables here, for simplicity).

Finally, we define the total momentum $P^\alpha_{\rm tot.} = \sum_a P_a^\alpha$. Now we see that $P^\alpha, M^{\beta \gamma}$ fulfill the commutation relations of the generators of the Poincaré group (showing only non-zero brackets): $$ \{M^{\alpha \beta},M^{\gamma \delta}\} = \eta^{\alpha \gamma} M^{\beta \delta} + \eta^{\beta \delta} M^{\alpha \gamma} - \eta^{\beta \gamma} M^{\alpha \delta} - \eta^{\alpha \delta} M^{\beta \gamma} \\ \{M^{\alpha \beta},P^\gamma\} = \eta^{\alpha \gamma} P^\beta - \eta^{\beta \gamma} P^\alpha $$ where in this case you have to use the equations of motion to prove some of the identities. With a little bit of extra work you can see that these brackets will be the same in every 3+1 split so they can be considered as fully covariant.


Casimir elements and their relation to internal angular momentum

Now we see that there are a couple of Casimir invariants of the algebra such as $P_\alpha P^\alpha \equiv - m_{\rm tot}^2$ and $\mathcal{S} \equiv \sqrt{-M^{\alpha \beta}M_{\alpha \beta}/2}, \, \mathcal{S}^* = \sqrt{\epsilon_{\mu\nu\kappa\lambda}M^{\mu\nu}M^{\kappa\lambda}/2} $. The interpretation of $m_{\rm tot}$ as the total mass of the ensemble of particles (including their kinetic energy by $E=mc^2$!) is quite obvious. However, the interpretation of $\mathcal{S},\mathcal{S}^*$ is less obvious. We notice that in the 3+1 split we are using we can reparametrize the angular momentum tensor by the two 3-vectors $$ J^i \equiv \frac{1}{2} \epsilon_{ijk} M^{jk} \\ D^i \equiv M^{0i} $$ We naturally identify these as the mass dipole moment and angular momentum 3-vector with respect to the centroid $x^\alpha_{\rm c.}$. In terms of these 3-vectors we see that the Casimir invariants actually are $$ \mathcal{S} = \sqrt{J^2 - D^2} \\ \mathcal{S}^* = \sqrt{2\vec{J}\cdot \vec{D}} $$


Universality

Now, you have to trust me that such constructions are general enough so that this is the picture we get for any isolated classical system. Hence, general classical isolated bodies rotating around a center of mass will have non-zero $\mathcal{S}$ and possibly even $\mathcal{S}^*$.

Conversely, if $\mathcal{S}$ is real and positive for a massive isolated body, we see that the body will appear as having a non-zero angular momentum with respect to the centroid in every frame. Even more, since $M^{\alpha \beta}$ is the generator of Lorentz transformations and these Casimirs are all the Casimirs the algebra has, $m_{\rm tot}, \mathcal{S}, \mathcal{S}^*$ are the only properties of the ensemble that are invariant with respect to Lorentz transforms. When writing down coarse-grained (but still Lorentz-invariant) interactions of external fields with such rotating bodies, the numbers $m_{\rm tot}, \mathcal{S}, \mathcal{S}^*$ provide a universal set of parameters for the coupling.


Correspondence to irreducible representations

If you then switch to the quantum picture, you will see that the $(m,n)$ labels of the representations of the Lorentz group actually correspond roughly to the values of an alternative base of Casimir invariants $$ m,n \sim \sqrt{\frac{\mathcal{S}^2 \pm i (\mathcal{S}^*)^2}{2 \hbar^2}} $$ or $$\mathcal{S} \sim \hbar \sqrt{m^2 + n^2} $$ Now it is easy to see that nonzero $m,n$, such as $m=n=1/2$ for vector fields, must, inevitably correspond to a complex analytical continuation of the notion of an isolated rotating body with nonzero "internal" angular momentum.

(In fact, the highest weight of $SO(3)$ appearing within the $(m,n)$ representation would actually be $m+n$, so one can actually state that $J$ can be understood as reaching up to $\sim \hbar (m+n)$, depending on the choice of centroid. This justifies calling the $(m,n)$ fields as "spin $m+n$".)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.