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?

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    $\begingroup$ No, it means that as far as we know, the electron has no internal structure. $\endgroup$
    – Javier
    Aug 12, 2014 at 12:56
  • $\begingroup$ But then were does the spin come from? $\endgroup$
    – Joshua
    Aug 12, 2014 at 12:57
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    $\begingroup$ @Joshua Spin cannot be understood classically! This is an important, but difficult fact that everyone has to simply live with. $\endgroup$
    – Danu
    Aug 12, 2014 at 13:39
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/1/2451 , physics.stackexchange.com/q/822/2451 and links therein. $\endgroup$
    – Qmechanic
    Aug 12, 2014 at 21:37
  • $\begingroup$ 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. $\endgroup$
    – user4552
    Aug 12, 2014 at 22:30

4 Answers 4


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.

  • 1
    $\begingroup$ @Cruncher You can't understand Bell's theorem without having at least an introduction to QM like Auletta's book. $\endgroup$
    – user5402
    Aug 12, 2014 at 15:46
  • 1
    $\begingroup$ @Cruncher Or, if you don't like much mathematics, try Feynman's lecture (the last volume is QM). $\endgroup$
    – user5402
    Aug 12, 2014 at 15:48
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    $\begingroup$ @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. $\endgroup$
    – ACuriousMind
    Aug 12, 2014 at 16:55
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    $\begingroup$ 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. $\endgroup$
    – zwol
    Aug 12, 2014 at 17:08
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    $\begingroup$ 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. $\endgroup$
    – benrg
    Aug 12, 2014 at 18:41

"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.

  • 4
    $\begingroup$ 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. $\endgroup$ Aug 12, 2014 at 15:07

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.


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).

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    $\begingroup$ 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. $\endgroup$
    – Jack Swann
    Dec 4, 2016 at 20:37

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