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Faster than light subatomic spin

  1. Jan 1, 2013 #1
    I’ve heard in several places that we must view the spin of subatomic particles not as we would that of a classical object since, if we did so, we would find that the velocity of the spin would be so great as to violate the light speed limit to achieve their magnetic moments or whatnot. Can someone re-paraphrase that argument for me one, and two, please address this follow up question.

    I’m guessing that the violation of the c limit problem comes from the tangential velocity of the spinning particle? I’ve done some calculations and it seems as though the particle would have to be spinning at an absurd rate to exceed the speed of light in meters per second tangential velocity. The reason is the extremely short radius of subatomic particles. Tangential velocity = angular velocity times the radius. Thus, for a proton spinning at 1 billion revolutions per second, its tangential velocity would only be 4.5 x 10^-8 m/s. Thus, the proton would have to be spinning at 16 orders of magnitude faster than that in order to approach the speed of light and therefore violate it.

    What’s going on here? Is my reasoning way out in left field here, or am I missing something?
     
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  3. Jan 1, 2013 #2

    K^2

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    Lets pretend that particle is a solid sphere. The angular momentum is given by ##\small L = I\omega## and ##\small I = \frac{2}{5}MR^2##. Angular momentum of an electron is ##\small L = \hbar \sqrt{3/4}##. Mass ##\small M = 9.109\times 10^{-31}kg##. And lets take classical radius for size ##\small R = 2.818\times 10^{-15}m##. This gives us ##\small \omega = 3.156\times 10^{25}s^{-1}##. So the surface velocity is ##\small v = \omega R = 8.89\times 10^{10} m/s##.

    This really isn't so hard to compute. Just don't make up numbers. If you make up the angular velocity of a particle, your answer doesn't have any significance either. Try to figure out what it is that is known and work from there. Estimates are fine, but they need to be based on something. When you are guessing something on order of 109 and the answer is on the order of 1025 (1023 for proton), you are clearly doing something wrong.
     
  4. Jan 1, 2013 #3

    Bill_K

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    You're having a bad day, K^2. Whether the surface velocity of an elementary particle is 109 m/s or 1023 hardly matters, they're both nonsense.
     
  5. Jan 1, 2013 #4

    K^2

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    Bill, I recommend reading the original post. Then reply. Then thinking about it. DiracPool specifically asked why spin of a subatomic particle cannot be vied as physical rotation.
     
  6. Jan 1, 2013 #5

    dextercioby

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    K2, don't take it personally, but your calculation assumed the elementary particle as it was described as a rigid solid ball. Which is nonsense to start with. Quantum spin is not a classical notion, it's a notion of Quantum Physics which, last time I checked, didn't account for the kinematics and dynamics of a rigid solid ball, as, for example, Euler's equations would.
     
    Last edited: Jan 1, 2013
  7. Jan 1, 2013 #6
    The original poster asked for someone to describe why it is wrong to assume elementary particles are classical objects. K^2 described why it is wrong using a "the sphere's surface would move faster than light which is not possible" argument. I'm not sure I understand why he is criticized.
     
  8. Jan 1, 2013 #7

    dextercioby

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    Not quite true. <why it is wrong to assume elementary particles are classical objects> would be explained simply as <assume quantum mechanics (ok, not even Bohr-Sommerfeld models) didn't exist. According to classical physics, the H atom would be described by the Rutherford model according to which no matter (atoms, molecules, ions) would exist. But matter exists. Hence H atom and matter in general cannot be accounted for by classical theories using classical objects>.

    Classical physics uses the concept of point particle extensively. Quantum spin is an observable without classical counterpart (i.e. there's no Hamiltonian obsevable which would become the spin operator(s) upon application of Dirac's quantization receipt: Hamiltonian observables go to (...) operators). Without a classical counterpart, it's meaningless to make up a classical model (such as a rigid solid ball/sphere) which would make an attempt to describe a notion which classically doesn't exist.
     
    Last edited: Jan 1, 2013
  9. Jan 1, 2013 #8
    I'm not sure why having an even stronger argument that elementary particles cannot be classical objects makes K^2's argument (that elementary particles cannot be classical objects) invalid. There is more than one reason why they can't be classical objects:

    - The measured angular momentum (along with other measured properties) cannot be accounted for by a classical object's rotation.
    - A classical orbit (or any other classical model) of the electron's "orbit" would be unstable.
     
  10. Jan 1, 2013 #9

    K^2

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    dexter, OP asked to clarify a particular argument. I did. I don't think it's the best argument either, but it is how it goes. Electron cannot be a tiny ball of "solid stuff," because that would require it to spin impossibly fast.

    A better complaint would be that I'm treating angular momentum itself classically when relativistic treatment is appropriate. But this is the kind of treatment that is assumed with OP's question.
     
  11. Jan 1, 2013 #10

    samalkhaiat

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    From the uncertainty relation
    [tex]m v R \sim \hbar \ \Rightarrow \ v \sim \frac{ \hbar }{ m R }[/tex]
    If you take R to be the so-called “classical radius of electron” you end up with non-sense [itex]v > c[/itex]. Relativity says that [itex]v < c[/itex]. Therefore the notion of extended object can only make sense when it is a lot larger than the Compton wavelength of that object
    [tex]R \gg \frac{ \hbar }{ m c }[/tex]
    In the world of massive elementary particles, the uncertainty relation should be understood as
    [tex]v \sim \frac{ \hbar }{ m r } < c[/tex]

    Sam
     
  12. Jan 1, 2013 #11

    K^2

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    Sam, where do you get that uncertainty relation from?
     
  13. Jan 1, 2013 #12

    dextercioby

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    Yes, but the argument (OP claim + your clarification) would be: Quantum spin of an electron is no physical rotation of the particle (assumed a finite sized rigid ball), because this would contradict the principles of classical special relativity, which has nothing to do with the quantum theory whose concept quantum spin is :bugeye: And quantum spin can be explained within QM in the absence of special relativity as space-time 'background'...

    If you did, perhaps the 'relativistic tangential speed at the equator' might be even smaller than c!

    And how did you come up with the "classical radius" ? I suspect something fishy here as well.

    @Sam, hand-waving arguments are not part of your PF style.
     
  14. Jan 1, 2013 #13

    K^2

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    Classical Radius.

    In a nutshell, if electron was a physical ball of "stuff", it would have to be that big to agree with scattering experiments.
     
  15. Jan 1, 2013 #14

    samalkhaiat

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    K^2, I was generous, my uncertainty relation is
    [tex]r \sim \frac{ 1 }{ m }.[/tex]

    Dex

    I was doing "order of magnitude". It is ok since it is the first day of the new year.
    By the way, Happy new year everybody.

    Sam
     
  16. Jan 1, 2013 #15

    K^2

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    Sam, which operators?
     
  17. Jan 1, 2013 #16

    samalkhaiat

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    Operators? there are no operators in the uncertainty relations.
     
  18. Jan 1, 2013 #17

    K^2

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    0_o

    Uncertainty principle for two observables corresponding to operators A and B is given by:

    [tex](\langle A^2\rangle - \langle B \rangle^2)(\langle B^2\rangle - \langle B \rangle^2) \leq \frac{1}{4}|\langle [A, B]\rangle|^2[/tex]

    Now, I'm sure you are thinking of some special case, but I can't tell which observables you have in mind.
     
  19. Jan 2, 2013 #18
    May be the whole problem is with the word spin.Quantum physicist must have used some other word from greek to call it.It really becomes confusing at some point.But they used it because of it's classical limit.
     
  20. Jan 2, 2013 #19
    K^2 basically addressed what I asking. Couple questions. 1) Your value for the angular momentum is the total momentum, I was referring specifically to the spin about a particles central axis, not an orbital component, so wouldn’t we just be talking about the intrinsic spin component? Even, so, at 1/2h-bar, ω comes out to the exact same value, to three digits at least (3.156), so that point may be moot anyway, at least for the electron.

    My second point/question relates to a more general interpretation. My initial example was the spin of a proton. I used that as an example because we classify the proton as an object with an actual size to it, whereas the electron is a bit tenuous. In any case, is there a value for the gross or total momentum or especially the rotational spin momentum of the proton? This example would serve as a better model, especially because it may address a recent hot topic in particle physics, the proton spin CRISIS!!

    http://en.wikipedia.org/wiki/Proton_spin_crisis
     
  21. Jan 2, 2013 #20

    samalkhaiat

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    Recent? This makes me feel younger, thank you! I have been working on the Nucleon Spin Problem for 20 years. I even “SEE” it in my dreams. But I can not see the connection between your original question and the confirmed experimental fact that the Nucleon constituents contribute only 1/3 to the spin of the Nucleon. This is the “spin crisis”. If you are interested in this problem, start by looking at (my teacher then research supervisor) text

    E. Leader, “Spin in Particle Physics”, Cambridge University Press (2001)

    The source of the problem can be found in:

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    The whole problem rest on the fact that there is no satisfactory guage invariant decomposition (spin + orbital) of the angular momentum of quarks and gluons in nucleon state.

    Sam
     
    Last edited: Jan 2, 2013
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