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Zeroth component of Spin 4-Vector

  1. Jun 5, 2014 #1
    The spin 4-vector is defined in the rest frame of the particle as [itex] s^{\mu}= (0, \vec{s})[/itex].

    Why is the zeroth component of the same zero in this frame?
     
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  3. Jun 5, 2014 #2

    stevendaryl

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  4. Jun 5, 2014 #3

    PeterDonis

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    Because the whole point of defining the spin 4-vector is to represent the spatial direction of the particle's spin axis. That spatial direction, like any spatial direction relative to the particle, must be orthogonal to the particle's 4-velocity ##u^{\nu}##, i.e., ##\eta_{\mu \nu} s^{\mu} u^{\nu} = 0##. Since the components of ##u^{\nu}## in the particle's rest frame are ##(1, 0, 0, 0)##, that forces the zeroth component of ##s^{\mu}## to be zero in that frame.

    Note that all of this is assuming that the concept of a "spin 4-vector" makes sense; the thread stevendaryl linked to discusses a number of significant limitations of that concept.
     
    Last edited: Jun 5, 2014
  5. Jun 5, 2014 #4

    WannabeNewton

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    ##S^{\mu}## is the spin 4-vector along the world-line of a given observer or particle with 4-velocity ##u^{\mu}##. It can be the expectation value of the quantum mechanical spin of a particle but it's easier conceptually to think of it as the spin of a gyroscope or of some extended body whose characteristic size is much smaller than the characteristic length scale of curvature. In this case ##u^{\mu}## belongs to the world-line of the center of mass of the extended body. The body itself defines a small world-tube around the center of mass world-line. Using axial Killing field symmetry considerations of the world-tube and the energy-momentum tensor of the body with compact support on the world-tube, we can define a 2-form ##S_{\mu\nu}## that quantifies the angular momentum or spin of the body about an instantaneous axis, as measured in the center of mass frame i.e. in the instantaneous rest frame of ##u^{\mu}##. This is also the canonical frame one would pick in Newtonian mechanics when calculating angular momentum (recall those ever-fun non-slipping rolling sphere problems).

    The details of the construction of ##S_{\mu\nu}## are rather involved. An extremely detailed exposition can be found in chapter 7, and specifically section 7.3, of "Relativity on Curved Manifolds"-de Felice and Clarke. If you can't access the book then see here: http://arxiv.org/pdf/1103.0543v4.pdf

    Once we have ##S_{\mu\nu}## it is natural to define the axial 4-vector ##S^{\mu} = \epsilon^{\mu\nu\alpha\beta}u_{\nu}S_{\alpha\beta}##. Clearly then ##S^{\mu}u_{\mu} = 0## which in the center of mass frame, or in the instantaneous rest frame of a comoving gyroscope in the limit as the size of the extended body goes to zero, we have ##S = (0, \vec{S})##.
     
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