# Relativistic Angular Momentum

1. May 14, 2013

### michael879

Ok so I have two very different questions, both closely tied to the concept of relativistic angular momentum.

Question #1: Why do people claim there is a minimum radius for a classical model of electrons with spin? The typical argument is that if the radius were to fall below some minimum value, the surface velocity of the electron would be moving faster than light.

For simplicity, assume an electron is modeled by a rotating ring of charge (the problem is nearly identical for any other configuration, as any cylindrically symmetric object can be composed of rings). Say that at rest, this ring has mass m, mass density λ, radius r, and charge e. Rotating, the tangential momentum is given as $\vec{p} = \gamma\lambda\vec{v}$, where v is the tangential velocity of the matter. It is easy to show that the energy density along the ring is $e=\gamma\lambda$. Therefore since the total momentum of the system is 0, and the total energy is $\gamma{m}$, the total mass M of the system is just $\gamma{m}$.

The angular momentum is given as $\int{\vec{r}\times\vec{p}}$, so $L=\gamma m r v$. Leaving L constant, and rearranging the terms we get:
$r = \dfrac{L}{mv}\sqrt{1-v^2}$
And it becomes clear that as r→0, v→1! This shows that there is no lower bound on the radius of a spinning particle! Note however that as r→0, M→∞. While it is clear that the mass of an electron is not infinite, this is a well known problem of point particles and shows up everywhere from classical physics to QFT. Also, there is no reason to believe the electron is actually a point particle. All we know is that its radius is below some experimental bound that has been pushed past the "limit" classical physics would set.

Another interesting result is if you consider the "material" making up the electron as massless. Much of the above treatment is the same, except that you need to rephrase everything in terms of E and p. What you find is that the total energy of the system M is $2\pi p$, and that L=Mr. Again you find that there is no minimum radius, but as r→0, M still diverges.

Question #2: Are the typical dismissals of GR predictions of the interior of a black hole valid? The typical argument is that a singularity is inconsistent with pauli exclusion principle (there are plenty of other arguments of similar nature).

Although the above results are derived in flat space, a very similar argument can be made. We find that the singularity of a rotating black hole is "made" of a massless "material" and the overall mass comes entirely from the rotational energy! What I find interesting about this result is that there isn't aren't any massless fermions!

Now I realize that you need to be careful when interpreting combined GR and QM results. However what I get out of this is that there must be some mechanism to convert any incoming fermions into massless bosons inside of a black hole. Hawking radiation could easily provide such a mechanism! So if the singularity of blackholes is purely bosonic, what is the contradiction between QM and GR?? As I understand it this would be a kind of Bose-Einstein condensate

2. May 15, 2013

### Agerhell

In quantum mechanics the electron (as well as all other particle) have a "de Broglie wavelength". A whole number of wavelengths must fit along the circumference of the electron's orbit. The least possible radius is the radius where the orbit is "one wavelength long".

Your argument is "semi-classical", in that you are using relativistic momentum. If you had used classical momentum instead, there would be a radius where the speed of the electron would have to be faster than light.

3. May 15, 2013

### michael879

I believe you are talking about a bound electron in a hydrogen atom. This really has nothing to do with what I'm talking about. Also, I am looking at the problem from a classical picture, ignoring all the "small" effects of quantum mechanics (hawking radiation could be called a "large" effect as it occurs macroscopically, and it too was originally derived in a semi-classical picture).

This is just semantics, but yes you could call my argument "semi-classical". However, since special relativity has been shown to be valid in ALL branches of physics, I would argue it is a fundamental theory of nature that can never be ignored, especially when you're talking about relativistic velocities. Using classical mechanics to show that an electron has a minimum radius is just a ridiculous argument, as classical mechanics completely breaks down at those scales!

When I speak of "classical physics", I always mean E&M, special relativistic mechanics, and depending on the context Newtonian or Einsteinian gravity. Classical mechanics and E&M can just not both be true, so your "classical physics" isn't even self-consistent! Using special relativity and either GR or the low-field approximation of GR (i.e. making newtonian gravity lorentz invariant) provides a self-consistent theory of our world.

4. May 15, 2013

### Staff: Mentor

Strictly speaking, this is only true below the electroweak symmetry breaking energy, at which the Higgs boson acquires a vacuum expectation value and gives mass to the fermions in the Standard Model. Above that energy the fermions are massless.

I'm not sure this is necessary, since the temperature of any collapsing body would have to increase as it collapsed, and there would be some point before the singularity forms at which the temperature exceeds the electroweak energy, so that fermions inside the collapsing body became massless.

5. May 15, 2013

### Bill_K

Even assuming it makes any sense to talk about a classical model of an electron, the obvious thing wrong with this one is that a charged ring has large multipole moments (quadrupole, octupole, etc) which an electron does not have. That's why when Lorentz did it, he used a charged sphere.

6. May 15, 2013

### michael879

Ok thats a good point, didn't think of that. I guess that answers the second question :P

7. May 15, 2013

### michael879

I think I made it clear that this discussion WAS about a classical model of an electron, as non-physical as it may be. I was asking about statements commonly made to show that the classical model of the electron can't work.

I used a charged ring for simplicity, not as a model for the electron, and for the purposes of this problem it is identical. MAGNETICALLY speaking there is no fundamental difference between a rotating sphere and a rotating ring (you can construct a sphere from rings trivially). Also, as far as the surface velocity goes there is no difference here..

8. May 15, 2013

### jartsa

http://www7b.biglobe.ne.jp/~kcy05t/spin.html

It says "One electron is very light and small".

I agree that small mass and small size are the important properties of electrons here. I don't think so much about the calculation on that page. I agree with the OP: relativity should not be ignored like that.

So therefore I'll do a better calculation, I will also ignore relativity, but when it's done this way there's no error:

A flywheel that is light and small and has a decent amount of angular momentum, has a very large rotatational energy.

An electron has maximum rotational energy 511 KeV.

An electron has angular momentum 1/2 hbar.

If electron is a ring with the classical radius, spinning so that a point on the ring travels at the speed of light, then the ring turns (3*10^8 m/s) / ( 2.8*10^-15m) = 10^23 radians/second.

If we apply a constant torque on this ring in order to stop the ring in one second, then the torque is
1/2 hbar / 1 second = 5.6*10^-35 Nm

The energy we get when stopping the ring is torque * angle = 5.6*10^-35 Nm * 10^23 = 5.6*10^-12 J

= 34 MeV

Edit: I just noticed that I assumed without any reason that a point on the ring has velocity c.

But anyway, at velocity c we had 66 times too much spinning energy, so the velocity must be at most 1/66 of c, so that the spinning energy is at most 511 KeV.

But then the angular momentum becomes 1/66 of what it should be. So now we must give the ring a rest mass of 66 *34 MeV.

So my possibly final conclusion is that a ring shaped electron with classical radius and angular momentum 1/2 hbar requires electron to have about 4500 times the measured rest mass of electron.

Last edited: May 16, 2013
9. May 16, 2013

### Staff: Mentor

I agree that his argument is semi-classical, but yours is semi-relativistic. You use the Newtonian definition for momentum, but c is not a limiting speed in Newtonian physics.

Not that it matters, electrons are quantum mechanical.

10. May 16, 2013

### Agerhell

Just for the record, I do nout use the Newtonian definition for momentum ($\gamma=1$ instead of $\gamma=$ Lorentz factor). Neither do I believe in superluminal velocites. The thread starter mentioned "classical model" but still used relativistic momentum. Later on he specified that he meant classical as in classical Maxvellian electrodynamics and not classical as in classical mechanics.