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I'm currently reading Alonso and Finn's Electromagnetism book.

It explains that the spin contributes to the magnetic moment and is somewhat comparable to a rotation of the particle around its own axis. It says that the spin of a particle is caused by a certain internal structure, which makes sense in the aforementioned analogy.

Right underneath the paragraph with the explanation of spin, it says "The electron has no known internal structure", but since it does have a spin, does that mean that we know the electron has an internal structure but we just don't know what it is?

Qmechanic
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Joshua
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    No, it means that as far as we know, the electron has no internal structure. – Javier Aug 12 '14 at 12:56
  • But then were does the spin come from? – Joshua Aug 12 '14 at 12:57
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    @Joshua Spin cannot be understood classically! This is an important, but difficult fact that everyone has to simply live with. – Danu Aug 12 '14 at 13:39
  • Possible duplicates: http://physics.stackexchange.com/q/1/2451 , http://physics.stackexchange.com/q/822/2451 and links therein. – Qmechanic Aug 12 '14 at 21:37
  • Aside from the fact that we have no positive empirical evidence for internal structure in the electron, there is an additional issue, which is the "confinement problem." If the hypothetical sub-parts of an electron (called preons) are confined to a space of size x, then the uncertainty principle says the mass-energy is at least about h/x. This would make the preons more massive than the electron they supposedly make up. The confinement problem can be worked around in some cases, but it's a reason not to expect to see substructure inside electrons. –  Aug 12 '14 at 22:30
  • From whatever scattering experiments we have done, the electron seems to exhibit no further break down. Spin is a property that doesn't require "spinning" particles. It can just exist as a property of the particle. – Manishearth Aug 13 '14 at 05:11
  • @Ben Crowell- This argument does not hold if the preons have zero rest masses, because in that case the potential energy for, say, a quark with color has such a low potential energy that it compensates for the high energy which the uncertainty implies. – Deschele Schilder May 23 '17 at 16:08
  • Either something is solid or it has an internal structure. I don’t believe anything in the universe is solid nor is there a way to describe that. – Bill Alsept Jun 19 '22 at 16:57

4 Answers4

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Spin is not about stuff spinning. (Confusing, I know, but physicists have never been great at naming things. Exhibit A: Quarks.)

Spin is a purely quantum mechanical phenomenon, it cannot be understood with classical physics alone, and every analogy will break down. It has also, intrinsically, nothing to do with any kind of internal structure.

(Non-relativistic) spin arises simply because quantum things must transform in some representation of the rotation group $\mathrm{SO}(3)$ in order for the operators of angular momentum to act upon them (and because we need to explain the degree of freedom observed in, e.g., the Stern-Gerlach experiment. Since the states in the QM space of states are only determined up to rays, we seek a projective representation upon the space, and this means that we actually represent the covering group $\mathrm{SU}(2)$. The $\mathrm{SU}(2)$ representations are labeled by a number $s \in \mathbb{N} \vee s \in \mathbb{N} + \frac{1}{2}$, which we call spin. Whether the thing we are looking at is "composite" or "fundamental" has no impact on the general form of this argument.

ACuriousMind
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  • Everything I read about it just feels grossly unintuitive. I find myself unable to grasp any of the concepts. I think too classically. I've read this page (http://en.wikipedia.org/wiki/Bell's_theorem) several times, and still don't understand a word of it. Maybe I just don't know where to start? Is there a question on here already about a good place to start? – Cruncher Aug 12 '14 at 15:44
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    @Cruncher You can't understand Bell's theorem without having at least an introduction to QM like Auletta's book. – user5402 Aug 12 '14 at 15:46
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    @Cruncher Or, if you don't like much mathematics, try Feynman's lecture (the last volume is QM). – user5402 Aug 12 '14 at 15:48
  • @metacompactness I know math up until calculus of several variables, and linear algebra 2(granted wasn't particularly strong in those anyway) – Cruncher Aug 12 '14 at 15:52
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    I think you're maybe a little too strongly worded here. Spin is called spin because it is the property that causes fundamental particles to have intrinsic angular momentum, and the $\mathbf{J} = \mathbf{L} + \mathbf{S}$ relation, plus the fact that it is $\mathbf{J}$ and only $\mathbf{J}$ that is the conserved quantity, demonstrate that intrinsic angular momentum is qualitatively the same thing as other angular momentum. (It's quantized, but so is $\mathbf{L}$ at the QM scale!) – zwol Aug 12 '14 at 16:31
  • (cont'd) ... and in fact you can apply a torque to a macroscopic object by shining circularly polarized light on it (further research along the same lines). – zwol Aug 12 '14 at 16:35
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    @Zack: You're correct that spin is, essentially, angular momentum - but that does not mean that anything is actually spinning, which is the strange thing about spin. – ACuriousMind Aug 12 '14 at 16:55
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    Yes, but my opinion is that at the intro-QM level, banging on that point may in fact be counterproductive, by confusing students into thinking that spin isn't a form of angular momentum. If they want to think of electrons as spinning balls of radius $O(10^{-15},\mathrm{m})$ for a while, until they get comfortable with delocalization, that's not that wrong by comparison. – zwol Aug 12 '14 at 17:08
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    The first two paragraphs ("Spin is not about stuff spinning" and "Spin is a purely quantum mechanical phenomenon" are wrong. Maxwell's equations are a spin-1 field theory. General relativity is a spin-2 field theory. The scalar Klein-Gordon equation is a spin-0 field theory. The Dirac equation is a spin-½ field theory. They are all classical, despite the different contexts of their discovery. Classically, the spin is a ratio between the linear and angular momentum of a circularly polarized plane wave (at least in the massless case): $L/p = s\lambda/2\pi$, where $s$ is the half-integer spin. – benrg Aug 12 '14 at 18:41
  • @Cruncher Try the Feynman lectures. – user5402 Aug 12 '14 at 19:48
  • @benrg: And how would, classically, the $\mathrm{SU}(2)$ reps, corresponding to half-integer fermionic spin, arise? Sure, you can write down classical fields transforming in such reps, but the need for it only arises through QM. – ACuriousMind Aug 12 '14 at 20:17
  • @ACuriousMind, I'm not sure I understand your first question, but solutions of the Dirac equation (which, again, is classical—no entanglement) can be written as linear combinations of plane-wave solutions with spins parallel to the z axis. The eight real components of a Dirac spinor can be identified with the eight dimensions of the real even Clifford algebra of spacetime rotations, which makes it easier to see what's happening geometrically. The details are interesting but comments are probably too short. – benrg Aug 12 '14 at 22:07
  • Re "the need for it"—historically Dirac thought he needed a first-order relativistic wave equation because of QM, but he was wrong: the Dirac field is not analogous to the Schrödinger equation, but instead shows up in the Lagrangian alongside other quasiclassical fields with second-order time derivatives. Dirac had to introduce spin to get a first-order equation, and thought that meant that relativistic QM predicts spin, but that was wrong too: spin-0 elementary quantum fields exist. The whole thing is historical accident. Mathematicians could have discovered spin ½ at any time in principle. – benrg Aug 12 '14 at 22:09
  • @benrg: Nothing you say is wrong, but representing the Clifford algebra is only always representing the Lorentz group (which is what we really want) if we seek projective reps. And the necessity of projective reps arises solely by the fact that QM/QFT take place on the rays in a Hilbert space. Classically, (I think) you would have only thought of bosons. But, you're right, this comment margin is too small to fit these arguments into. I am now considering asking a series of questions on the individual steps of these arguments, but I'll have to read up some things first for that. – ACuriousMind Aug 12 '14 at 22:34
  • Considering that the universe is a quantum mechanical system, even saying that "color" or "heat" are quantum mechanical phenomenon are not technically "wrong". All classical systems are quantum mechanical. Only the maths that describe them sometimes aren't. Not because the systems aren't but the maths break down at the quantum level. When the maths don't break down then that does not mean it's not quantum mechanical. Separating "classical" and "modern" is only useful as a reminder that classical physics is a very rough approximation. – slebetman Aug 13 '14 at 04:46
  • @slebetman: When I say "inherently QM phenomenon", then I mean that there is no fully satisfying description of it within the classical formalism. For mere angular momentum, there would be. The half-integer spins need QM though, as I understand them. Separating classical from quantum is very useful for understanding what the procedure of quantization (be it geometric, deformation, or sth. else) actually does, and is formally well-defined in most situations. – ACuriousMind Aug 13 '14 at 12:56
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"The electron has no known internal structure", but since it does have a spin, does that mean that we know the electron has an internal structure but we just don't know what it is?

An electron has no known internal structure simply means that nobody knows if the electron has an internal structure. So far they know none and therefore they suppose it has none

Spin is not related to an internal structure. If you consider the electron as the classical ball, the ball can spin both with or without an internal structure.

But spin is now considered an intrinsic property of the electron, which means that the effects are those of a classical spin, but the particle must not necessarily spin.

bobie
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    Similarly, a classical ball with an unremarkable surface will appear identical to a similar ball which has some angular momentum, though it defies visual observation. – KidElephant Aug 12 '14 at 15:07
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Spin is a wave property. It exists in classical relativistic wave theories as well. A circularly polarized wave carries an angular momentum that's related to the spin of the field. A gravitational wave (spin-2) can carry twice the angular momentum of a classical electromagnetic wave (spin-1).

Being "pointlike" is a particle property. You can think of the field value at a point as being related to the presence of a particle there. If the field's associated particle is an extended object (like a pion) then it doesn't just occupy the point where the field is nonzero, but also nearby points, which means that the interaction with a pointlike test particle depends not only on the field value at the test particle's location, but also on nearby field values. If the field's particle is pointlike (like the photon) then the force depends only on the field value at that point. Even classically, you could say that the electromagnetic field is "pointlike" since the Lorentz force only depends on the field at a point, though the terminology makes less sense without wave-particle duality.

So while spin is certainly a property of the particle (inasmuch as the particle and the wave are the same thing), it's not a property that depends on any internal structure of the particle.

benrg
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No, it does not imply that there is an internal structure. There may be one however, and it may be complex, since an internal structure does not have to reveal itself with external electric or magnetic moments. (If it is purely electromagnetic, these would be the only fields to consider, maybe, gravitational moments.)

Mathematically, the fields at any point external to a charge/current distribution are calculated in terms of a fourier-like expansion, whch ends up being the moment values when using sphercal harmoncs (J.D. Jackson). A charge/current distribution can have non-zero values inside the distribution and the moment value calculated outside the distribution can still be zero since the moment value is an integral over the whole distribution (Bleistein, reference stealth technology).

peterh
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Jack Swann
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    Norman Bleistein and Jack Cohen, "Nonuniqueness in the Inverse Source Problem in Acoustics and Electromagnetics", Journal of Mathematical Physics, Vol. 18, #2, Feb. 1977, pp. 194 - 201. Describes the possible internal structure to a charge/current distribution where externally, the fields due to the structure do not appear. – Jack Swann Dec 04 '16 at 20:37