If an electron has spin and volume than the point on the surface is rotating at constant speed according to Planck constant defined angular momentum.If this electron accelerates then this point on the surface has to add its rotating velocity to the velocity of translation of the electron but this sum must not reach the speed of light to not to contradict with special relativity so it seems its spin has to slow down in the reference frame that is not moving with the electron.Is that right?

  • 1
    $\begingroup$ In the Standard Model of particle physics, an electron is a point particle, it has no volume or surface. $\endgroup$ – PM 2Ring Sep 7 at 11:50
  • 2
    $\begingroup$ Confusingly, ‘spin’ of an electron has nothing to do with, say, the spin of a top. $\endgroup$ – Jon Custer Sep 7 at 14:19
  • $\begingroup$ In that case the charge density should go infinite along with coulomb force wich implies a point with infinite acceleration and also implies that an electron cannot collide with an electron because the probability of colliding with a zero volume is also zero maybe just fields collide.Also at point zero where should the vector of infinite force be oriented to, it cannot be undefined or point in all directions? $\endgroup$ – Janko Bradvica Sep 7 at 16:48
  • 1
    $\begingroup$ @JohnCuster Your statement is too strong. There is no mechanical model of spin. In every other aspect spin is comparable to orbital angular momentum. $\endgroup$ – my2cts Sep 7 at 19:21
  • $\begingroup$ @Janko Sorry, I didn't see your comment because you didn't use the @UserName syntax. (You get notified automatically because it's your question). A quantum particle has different properties to a classical particle. You can either change this question to ask about that stuff in your comment, or just ask a fresh question. $\endgroup$ – PM 2Ring Sep 8 at 8:25

I think there are two questions that are mixed together here: a question about the nature of quantum-mechanical spin, and a question about how to deal with the relative motion of subsystems when you have a system moving very close to the relativistic limit of $c$.

First, spin. It's very tempting to think of an electron as a "little ball of stuff," because macroscopic bits of matter come in discrete shapes that have surfaces, and because the people who illustrate textbooks feel like they have to have some picture of an electron, and choose a little ball. But those appealing models are wrong. We don't have any evidence that an electron has a surface or an interior, the way that a water droplet or a dust mote or even a nucleus has. (The nucleus is an interesting case because nucleons participate in interactions that electrons ignore, but we won't go there for now.)

The modern picture of an electron is as a quantized disturbance in a "field," where a "field" is a continuous property of spacetime. When one applies conservation laws to interactions with the electron field, it becomes parsimonious to talk about these disturbances as being associated with an intrinsic mass, charge, and angular momentum --- the same mass, charge, and angular momentum that one ascribes to the electron in the little-ball model. But little balls have an intrinsic size parameter, which the electron doesn't appear to.

If that little paragraph doesn't satisfy you, I'm not sure I can do better. Usually we tell people that quantum-mechanical spin is like a spinning little ball, but different, and if people press the issue we sign them up for a graduate class in QFT.

But let's think about the relativity side of your question, too. Here's an example from accelerator physics. Suppose you inject a group of ultra-relativistic electrons into an accelerator. (I like to use the CEBAF accelerator, where the electrons are injected into the accelerator with $\gamma = (1-v^2/c^2)^{-1/2} \approx 100$ and exit with $\gamma \approx 20\,000$. These electrons are traveling at speeds upwards of $0.9999c$, in bunches that are about 0.3mm long. (Usually the bunch length is measured in how many picoseconds it takes for a bunch to pass a point on the accelerator.)

The part of this that's relevant to your question is what happens to those little bunches of electrons as they spend a microsecond or so traveling around the accelerator. In their rest frame, the electrons in each little bunch think they are at rest and surrounded by other electrons --- whom they hate, because they all have the same sign of electric charge and repel each other. So without special focusing magnets which accelerate the front and rear of each bunch differently, the electron bunches would spread out as the beam travels: some would go faster and slower than the average, just like the surfaces on your imaginary spinning ball.

How can you have velocity dispersion in a system that's traveling at a speed experimentally indistinguishable from $c$? It works because of relativistic velocity addition. If the bunch is moving at speed $u$ relative to me, and the fastest electron in the bunch is moving at speed $v$ relative to its friends, then my measurement of the speed of the fastest electron in the bunch is

$$ w = \frac{u+v}{1 + uv/c^2} $$

Part of any first course on relativity is playing with this formula to convince yourself that, if the speeds $u$ and $v$ are less than the limit $c$, then so is $w$. You can't constrain an object's motion in its rest frame just by looking at it from a reference frame that's moving too fast. I'm pretty sure that was the main concern in your question.

  • $\begingroup$ I'm up-voting this as the clearest answer consistent with the Standard Model, but, as composing things with spatial extent (like us) out of things lacking it seems a little too mystical, I'd (almost) fo' sho' vote for EC over GR, if I had the math to grasp either! $\endgroup$ – Edouard Oct 8 at 15:52
  • 1
    $\begingroup$ @Edouard Note that the "little-ball" model makes more sense for nucleons, nuclei, and atoms than it does for electrons. For nucleons and nuclei there is a very strong short-range interaction which vanishes suddenly beyond a distance of a few femtometers; this maps well to your macroscopic intuition for things that act differently when you "touch" them. Likewise, a neutral atom is electrically neutral but polarizable, which leads to "contact interactions" between nearby atoms as described first by van der Waals. The contrast with the electron is too complex for a comment, but very interesting. $\endgroup$ – rob Oct 8 at 18:34
  • $\begingroup$ As a fan of Poplawski's "Cosmology with torsion", dependent on ECSK gravity, I'd been about to write off rival cosmologies as persisting only due to a lag in public education, so, contrary to the policy against thanks in comments, I've got to say I appreciate your reply to mine, and would just like to ask for whatever title(s) or site(s) might come to mind that might elaborate on the electron's distinction from other fermions (-just pick the one using the largest proportion of English, that you haven't noticed to contain errors). (If none does, I can post a "resource request" question.) $\endgroup$ – Edouard Oct 9 at 3:16
  • 1
    $\begingroup$ @Edouard I'm not aware of a single book that describes quite the picture I have in mind. I'd be curious about the outcome of a resource-request question. But the heart of the difference is that the electron participates in a long-range interaction, electromagnetism, without any short-range stronger interactions that would localize the electron in the way that strong interactions localize nucleons. $\endgroup$ – rob Oct 9 at 16:07

Presently in the standard model of particle physics, spin is a necessary adjunct of keeping conservation of angular momentum at the framework of quantum mechanics. The historical survey given by Rob is the way it was discovered, interaction by interaction.

In all particle interactions there will be angular momentum missing where fermions are taking part (see table of particles), if a fixed spin is not assigned to the fermions (defined by having half integral spin). The way spin has been historically assigned conserves angular momentum in all quantum mechanical interactions. This has not been falsified in all data up to now.

Note that conservation laws hold in all inertial frames of special relativity, so the speed the particle is going makes no difference to the value of spin. There is no rotational velocity associated with the spin assignment, it is just a number needed to conserve angular momentum in quantum mechanical frameworks.

  • $\begingroup$ The Wikipedia article "Spin (physics)" mentions, in its "History" paragraph, that Pauli objected to spin as originally formulated because of the potential for particle rotations at a velocity exceeding the speed of light, and claims that the problem was circumvented by a "fiber bundle" or "tangent bundle" description of it, but I haven't been able to find a comprehensible (-for me, plain English) explanation of what those "descriptions" might be. Could you maybe explain how they'd apply to something point-like? (The 5-tag limit may've prevented use of the GR tag in this question.) $\endgroup$ – Edouard Oct 8 at 14:59
  • $\begingroup$ I've got to add that, since I realize that anything rotating has a point at its center, I understand that your reply to my comment might be, "It's what I just said", but I don't want to miss any point even more abstract than the one we're discussing, since Cartan's take seems a bit more naturalistic. $\endgroup$ – Edouard Oct 8 at 15:20
  • 1
    $\begingroup$ In our present model of elementary particles there exist point particles which have spin, so it makes no sense to talk about rotation of a point or fiber bundles outside. . en.wikipedia.org/wiki/Elementary_particle . It is a quantum "number" assigned to each particle. $\endgroup$ – anna v Oct 8 at 15:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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