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Suppose an isolated system with a number of particles with parallel spins. Can the macroscopic angular momentum of the system increase at expense of the number of particles having parallel spins (that is by inverting the direction of the spins of some of them) or by converting all particles into spinless ones?

Conversely in a rotating system of spinless particles can the macroscopic algular momentum be decreased by converting some of the particles into those having parallel spins?

Can one choose a rotating frame of reference in such a way so in it a particle that has spin in inertial frame, does not have one?

Anixx
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2 Answers2

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First of all yes, you can convert angular momentum to spin orientations, as only the total angular momentum $\vec J=\vec L+\vec S$ is conserved. The setup you describe is very similar to a famous experimental effect dubbed the "Einstein-de-Haas" effect. You can read more about it on Wikipedia. The main idea is that you have a ferromagnet hung on a string and when you magnetize it (which really only means that you orient the spins of the electrons in a coordinated way) the magnet has to pick up angular (mechanical) momentum, since angular momentum is conserved. What you observe is that it starts to turn seemingly out of nothing once you magnetize it.

Now the part where you are mistaken is that you cannot change the spins of particles. The spin of a particle is a property of the particle itself (much like its mass. In fact those are the two properties particles have as a result of their being representations of the symmetry group of spacetime, c.f. Wigner's classification). You can only orient a component of the spin, but not the total spin magnitude. A Spin 1/2 particle (fermion) will always stay a fermion.

luksen
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If we model a particle as a uniform ball spinning around its own centre (which it probably isn't) then:

$$I = \frac{2mr^2}{5}$$

$$L = I\omega$$

The speed of a point at the edge of the ball is:

$$v = r\omega$$

We know the spin of fundamental particles so:

$$rL = \frac{2mr^2v}{5}$$

$$rv = \frac{5L}{2m}$$

We know that for the electron the spin angular momentum:

$$ L = \frac{h}{4\pi}\sqrt{3} $$ $$ L = 9.1327631527 \times 10^{-35} kg \, m^2/s $$

and

$$ rv = 2.50847711816 \times 10^{-4} m^2/s $$

The upper limit on the electron's radius is $10^{-22} m$.

So, that would predict a velocity of $2.50847711816 10^{18}m/s$ which is much higher than the speed of light.

As a naive picture of fundamental particles this is obviously flawed and I would guess that if there was any sensible speed to assign to the speed of the outside of a point particle it would be the speed of light or something but anyway I don't think it's possible to travel fast enough to obtain a reference frame where fundamental particles are not rotating or the notion of speed does not make sense for the rotation of point particles.

Similar to Frame of reference of the photon? I don't think it's possible to reach the reference frame of a point particle.

Note also that the gyromagnetic ratio of the electron suggests that it doesn't make sense to treat an electron as a uniform ball of matter.

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Ms. Molly Stewart-Gallus
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