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Did the Goudsmit-Uhlenbeck analysis of spin consider relativity?

  1. Nov 18, 2013 #1
    It's frequently mentioned in introductory quantum mechanics texts that Goudsmit and Uhlenbeck conjectured that the magnetic moment of an electron was due to angular momentum arising from the electron rotating around its own axis. But then when they tried to calculate how fast it would have to be spinning, assuming that the electron is a rigid sphere with radius equal to the classical electron radius, they found that a point on the equator would be moving with a speed greater than the speed of light, so they were embarrassed for publishing their work.

    My question is, did they do this calculation using Newtonian mechanics or special relativity? If we do take relativity into account, and consider a (Born-) rigid sphere with radius equal to the classical electron radius, and then we tried to find out what speed the sphere would need to rotate at in order to have an angular momentum that produces the magnetic moment of an electron, would we still get a speed faster than light? Momentum goes to infinity as speed approaches c, but what happens to angular momentum? I'm aware that angular momentum becomes really complicated in special relativity, with tensors and bivectors and the like, but is there a simple (or even approximate) expression that can give us some idea of what would happen in this case?

    This is of course just a curiosity, because there are other problems with the classical theory of spin, like the fact that a rotation of 720 degrees is required (for an electron) rather than a rotation of 360 to get you back to your initial state, due to the double cover property of SU(2).

    Any help would be greatly appreciated.

    Thank You in Advance.
     
    Last edited: Nov 18, 2013
  2. jcsd
  3. Nov 19, 2013 #2

    Bill_K

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    It's easy enough to reproduce the calculation. And the first thing to note is that the "classical electron radius" is gigantic! You get it from setting e2/r = mc2.

    r = e2/mc2 = (e2/ħc) (ħc/mc2) = (1/137)(200 MeV-f/0.5 MeV) = about 3 fermis.
    In many atoms this is larger than the nucleus!

    To get the right spin angular momentum, ħ/2 = mvr, which implies v = ħ/2mr = (ħc/e2)c ≈ 137 c, bigger than c, like you said.

    It's true that in special relativity, orbital angular momentum becomes a bivector, Lμν = xμ pν - xν pμ, but in the rest frame it's still just L = r x p. The difference is that p is now the relativistic momentum, γmv. So you now get γv ≈ 137 c, or γ ≈ 137, which is Ok.

    No, actually that's not true. It's amazing how many people believe this statement, which is totally counterintuitive and totally wrong. :smile: The world is (and must be!) invariant under a 360 degree rotation. The correct statement is that spinor wavefunctions are double valued. Under a 360 degree rotation they change sign. This is the same wavefunction, and it represents the same state.
     
    Last edited: Nov 19, 2013
  4. Nov 19, 2013 #3
    That works out to about .99997c. So Goudsmit and Uhlenbeck were wrong to be embarrassed about their publication (given the information they had in 1925)?
     
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