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Spin is the intrinsic angular momentum of a particle. The particle itself is elementary and is not spinning on its axis, and has this momentum even at rest. The absolute magnitude of this momentum cannot increase or decrease without changing the particle. The particle may have angular momentum in addition to its spin, increasing its total angular momentum linearly.

Rest mass is the intrinsic confined energy of a particle. The particle itself is elementary and does not physically contain internal energy (binding energy, elastic potential, vibrational, etc), and has this energy even at rest. The absolute magnitude of this energy cannot increase or decrease without changing the particle. The particle may have energy in addition to its rest mass, increasing its total energy linearly.

Would it therefore be correct to call spin "Rest angular momentum", in the same sense mass is "Rest energy". In what ways does this analogy break down?

Anthony Khodanian
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    possible duplicates: https://physics.stackexchange.com/q/1/84967, https://physics.stackexchange.com/q/67616/84967, https://physics.stackexchange.com/q/77231/84967 and links therein. – AccidentalFourierTransform Jun 17 '18 at 01:51
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    Possible duplicate of What is spin as it relates to subatomic particles? –  Jun 17 '18 at 02:24
  • My question is very different. I am neither asking for an explanation of what spin is, nor am I wondering why it is not an actual spinning of the particle. In the question I am beginning with the premise that spin is in fact intrinsic angular momentum and not at all a rotation. – Anthony Khodanian Jun 17 '18 at 05:45
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    "The particle... is not spinning on its axis", "The particle... does not physically contain internal energy". No experiment compels us to believe this. – my2cts Jun 17 '18 at 06:08
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    "spinning on its axis" tends to imply a classical composite body, where the components of that body all have orbital angular momentum around the axis. Of course, a fundamental quantum particle has no components, so there's nothing to undergo such orbital motion. But that doesn't imply that it's not spinning, it's just hard for our classical intuitions to comprehend spin that isn't orbital. – PM 2Ring Jun 17 '18 at 09:08
  • @PM2Ring has no known components – vol7ron Jun 18 '18 at 13:24
  • @vol7ron If A particle is fundamental, then by definition it has no components. ;) Of course, another theory (e.g. string theory) may ascribe structure to particles that have no structure in the Standard Model. – PM 2Ring Jun 18 '18 at 13:34
  • Yes good point. I was merely pointing out that our understanding of what is fundamental or base seems to change as our capabilities of detecting, analyzing, and observing our study improve. Known only provides a temporal recognition that we don’t know with 100% certainty. :) – vol7ron Jun 18 '18 at 14:09

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The analogy is pretty good, and when it applies it is pretty exact: in the framework of QFT, one can apply the angular momentum operator to a one-particle state at rest and find the spin in this way.

Where it fails (or rather can't be applied) is in the case of massless particles, which don't have a rest frame. The relevant quantity is then not spin but helicity: the projection of angular momentum in the direction of momentum. You can think of this procedure as the only way to eliminate orbital angular momentum: essentially, if $\mathbf{L} = \mathbf{r} \times \mathbf{p} + \mathbf{S}$, then $\mathbf{L}\cdot\mathbf{p} = \mathbf{S}\cdot\mathbf{p}$. Helicity is invariant under proper orthochronous Lorentz transformations, for the simple reason that if the particle is going at the speed of light you can't outrun it and make it go in the opposite direction, which would reverse its helicity. However, it switches sign under parity.

Since helicity is a projection of angular momentum and it is (proper orthochronous) Lorentz invariant, you can in principle have a type of particle which always has the same helicity $\sigma$. This contrasts with spin for massive particles, where the spin projection can take any value in $\{-s, -s+1, \dots, s\}$.

Javier
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    (+1) to me this argument actually strengthens the analogy between rest mass and "rest angular momentum" somewhat, since massless particles don't have a rest mass either. – N. Virgo Jun 17 '18 at 02:27
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Spin is just some statement on the representation a given state transforms under. The Lorentz group can be represented by $4\times4$ matrices (in the defining representation) or SL($2,\mathbb{C}$) for the so-called spin-$\frac{1}{2}$ states. But of course SL(2,$\mathbb{C}$) is complex, so one has $(1/2,0)$ and $(0,1/2)$ states, which are described by right- or left-chiral Weyl spinors. In 3D, you can work with SU(2) and you can get all representations from tensor products of SU(2) doublets. This shows that the analogy to rest mass is not really good. The mass of a particle is just a continuous parameter (or governed by a continuous parameter in the standard model) whereas spin is quantized as a consequence of representation theory and not continuous. Mass can also be an effective quantity (like an effective mass of a neutrino in a dense medium, say), but spin is not an effective quantity. So the bottom-line is that spin should not be thought as a "rest momentum".

ADDENDUM: And here is another reason why I do not think this analogy is good. If spin was some rest angular momentum, you'd (naively) expect it to contribute to the energy/mass of this particle. However, that's not what happens, e.g. in supersymmetry, superpartners differ in their spins but have the same mass etc. Of course, you could wiggle your way out by saying that this angular momentum does not contribute to the energy/mass, but then I am wondering what the analogy is good for.

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    Not sure I agree: mass can be discrete and spin continuous. Mass is defined as the eigenvalue of $P^2$, and there is nothing that prevents this operator from having a point spectrum; and massless particles admit the so-called continuous spin representations which, although somewhat pathological in some aspects, are well-defined at the level of representation theory. – AccidentalFourierTransform Jun 17 '18 at 01:58
  • @AccidentalFourierTransform I guess the question was about non-pathological particles which exist in Nature. And as for your statement about the eigenvalues of $P^2$: they are still controlled by the parameters of your setting or model. Of course, mathematically you can do what you want, but I believe that my answer is correct for all particles we know of. –  Jun 17 '18 at 02:04
  • Wouldn't spin fractionalization be a form of effective spin? – Kai Jun 17 '18 at 16:26
  • @Kai No. The question is about the ordinary spin in 4D Minkowski space, I think. If you have different setup, the symmetries and the representation theory change. –  Jun 17 '18 at 17:22
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I won’t tell you about what we can read from equations. I want to tell you what was observed before the equations were formulated.

Spin is the intrinsic angular momentum of a particle.

How the intrinsic spin of a subatomic particle was observed? It was observed by the deflection of the moving subatomic particle in an external magnetic field (with non-parallel direction of motion and direction of the the magnetic field). The notion “spin” was this way named because off the analogy to the gyroscopic effect of a spinning wheel under the influence of a force. And, which is the force in the case of the subatomic particle? The external force, which deflected the particle, is the external magnetic field!

The above statement is important. It is well known that the electric field of an electric charge does not interact with an magnetic field. Only electric fields interact with each over and only magnetic fields interact with each other. From this statement it could be concluded that the magnetic dipole of a subatomic particle interact with the external magnetic field. Indeed spin and magnetic dipole for subatomic particles are synonyms, they are uniquely linked. That is why an electron gets deflected in one direction and a positron in the opposite direction.

enter image description here

The particle itself is elementary and is not spinning on its axis, and has this momentum even at rest. The absolute magnitude of this momentum cannot increase or decrease without changing the particle. The particle may have angular momentum in addition to its spin, increasing its total angular momentum linearly.

If the particle is not spinning around it’s axis, then how it gets deflected under an external field? And why this field does not get exhausted? Discovered was that the particles kinetic energy get exhausted and the particle emits EM radiation during its deflection (which is an acceleration of course). In detail, the particle emits photons if it is under the influence of an external magnetic field and if the particle is in relative (and non-parallel] motion to this magnetic field. On the other side, the emission of photons is accompanied with a momentum, which decreases the kinetic energy of the subatomic particle. And deflects the particle in the case of the mentioned non-parallelity! Now we named the reason, the particle gets deflected, the intrinsic spin is nothing as the particles magnetic dipole moment, which gets aligned and disaligned periodicaly, emitting photons and exhausting kinetic energy.

Rest mass is the intrinsic confined energy of a particle. The particle itself is elementary and does not physically contain internal energy (binding energy, elastic potential, vibrational, etc), and has this energy even at rest. The absolute magnitude of this energy cannot increase or decrease without changing the particle. The particle may have energy in addition to its rest mass, increasing its total energy linearly.

Would it therefore be correct to call spin "Rest angular momentum", in the same sense mass is "Rest energy". In what ways does this analogy break down?

Following the said above the spin is the deflection of a moving particle in an external magnetic field. In annihilation processes the spin simply disappears and it is not known to me that for the amount of resulting energy the spin is taken in account. So the answer is no.

Remark 1 Now educated people wach argue that the arrangement of electrons around nuclei is based on the intrinsic spin. Of course the idea about revolving electrons around nucleus is obsolet for a long time and the electrons are dislocated in some volumes around th nucleus. But the spin as a angular momentum is alive with a strength that it overshadow the role of the magnetic dipole moments of the electrons. They are responsible for the self-alignment of the electrons, which is expressed in the Pauli’s exclusion principle and the octet rule in chemistry.

Remark 2 Javier wrote:

Where it fails (or rather can't be applied) is in the case of massless particles, which don't have a rest frame. The relevant quantity is then not spin but helicity: the projection of angular momentum in the direction of momentum ...

The spin of the photon comes from the simple fact that the photons electric and magnetic field components are fields with directions and the order they could follow one another could be only a lefthanded or a righthanded order:

N:S | +:- | S:N | -:+ or

S:N | +:- | N:S | -:+

For illustration see one of the two possibilities in this sketch from WP:

enter image description here

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HolgerFiedler
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    The statement "an electric charge does not interact with an magnetic field" is not correct, and neither are several of the statements which follow it in this answer. – rob Jun 17 '18 at 12:04
  • @rob It is the magnetic dipole of the charge which interacts with the external magnetic field. An electric Never interacts with a magnetic field. You are a good teacher. I’ve edited it, going more in detail – HolgerFiedler Jun 17 '18 at 12:21