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Spin Direction for Massless Particles

  1. Aug 15, 2014 #1

    referframe

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    Mass-less particles travel at the speed of light. If such a particle has an intrinsic spin, the DIRECTION of the spin angular momentum vector must be in the same direction as the linear motion of the particle. Why exactly is that? Thanks in advance.
     
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  3. Aug 16, 2014 #2

    Simon Bridge

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  4. Aug 16, 2014 #3

    tom.stoer

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  5. Aug 16, 2014 #4
    Massive spin-1 particles are have 3 distinct nonzero eigenvalues and 3 respective eigenstates. These eigenstates correspond to 3 spin states in x, y, z spatial directions. On rotations they transform into each other as vectors. That's exactly why they are called vector particles. More formally, each of these states corresponds to a l=1 spherical harmonics.

    When the mass of a spin-1 particle reaches zero, one of its eigenvalues also goes to zero. So there are only 2 nonzero eigenvalue-eigenstate pairs. These 2 eigenstates correspond to 2 polarization states. They transform into each other when rotated 90-degrees over the axis of motion.

    As for spin-2 particles: when they are massive, they have 5 distinct nonzero eigenvalues and 5 eigenstates. The eigenstates may be arranged into a symmetric rank-2 tensor, that's why they are called tensor particles. Spin of such particle can described by an ellipsoid. Formally, they correspond to l=2 spherical harmonics.

    When the mass of a spin-2 particle reaches zero (as in the case of graviton), one of the eigenvalues goes to zero and 4 other go close to each other pairwise. In the end there are only 2 distinct nonzero eigenvalues. One eigenvalue is 0 and 2 other are degenerate (in particular, they are double). The 2 eigenstates transform into each other when rotated 45 degrees over the axis of motion.

    Now forgive me, but I never really understood spinors and I don't know how their eigenstates behave when going to 0 mass.
     
  6. Aug 16, 2014 #5

    tom.stoer

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    I don't agree with the formulation "when the mass of a particle reaches zero ..."; as far as I understood group theory this cannot be understood as a smooth limit.
     
    Last edited: Aug 16, 2014
  7. Aug 16, 2014 #6

    referframe

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    By "axis of motion", are you referring to the SPIN axis/direction or the LINEAR motion (or the opposite direction) of the particle?
     
  8. Aug 16, 2014 #7

    PeterDonis

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    No, this is not correct in general. There is a special state of a massless particle for which it is correct (under a suitable interpretation): the state called ##\vert R \rangle## in the thread Simon Bridge linked to, a state of right-handed circular polarization. This state is an eigenstate of the spin operator along the direction of the particle's linear motion, with eigenvalue ##\hbar##; so if the particle is in this state, measuring its spin along the direction of its linear motion will *always* give the value ##\hbar## (i.e., no chance of the opposite sign). Whereas, if you measure the spin along any other direction, you will have some chance of getting ##- \hbar## as the result. So the state ##\vert R \rangle## can be thought of as a state whose spin angular momentum does point in the same direction as the particle's linear motion.

    There is also a state called ##\vert L \rangle## (left-handed circular polarization), for which measuring the spin along the direction of the particle's linear motion always gives the result ##- \hbar##; this state can be thought of as a state whose spin angular momentum points in the opposite direction from the particle's linear motion. But a general state of the particle will be a complex linear combination of ##\vert R \rangle## and ##\vert L \rangle##, and for such a state, the "direction of the spin angular momentum" (to the extent that phrase has meaning) will *not* point along the direction of the particle's linear motion (either in the same or in the opposite direction).
     
    Last edited: Aug 16, 2014
  9. Aug 16, 2014 #8

    samalkhaiat

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    If you know group theory, then look at

    https://www.physicsforums.com/showpost.php?p=2223048&postcount=2

    Sam
     
  10. Aug 16, 2014 #9

    samalkhaiat

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    Who told you that? What is the connection between that particular "eigenvalue" and the "mass"?
     
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