Zeroth component of Spin 4-Vector

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Discussion Overview

The discussion revolves around the zeroth component of the spin 4-vector in the rest frame of a particle. Participants explore the definition and implications of the spin 4-vector, particularly why its zeroth component is zero in this frame. The conversation touches on theoretical aspects, conceptual clarifications, and references to previous discussions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants define the spin 4-vector as \( s^{\mu} = (0, \vec{s}) \) in the rest frame, questioning why the zeroth component is zero.
  • One participant suggests that the spin 4-vector represents the spatial direction of the particle's spin axis, which must be orthogonal to the particle's 4-velocity \( u^{\nu} \), leading to the conclusion that the zeroth component of \( s^{\mu} \) is zero in that frame.
  • Another participant elaborates on the concept of the spin 4-vector, relating it to the expectation value of quantum mechanical spin and comparing it to the spin of a gyroscope or extended body, emphasizing the role of the center of mass frame.
  • There is mention of significant limitations regarding the concept of the spin 4-vector, as discussed in a linked previous thread.
  • One participant references a detailed exposition on the construction of the spin 4-vector and its relation to angular momentum, indicating the complexity of the topic.

Areas of Agreement / Disagreement

Participants express various viewpoints regarding the definition and implications of the spin 4-vector, and while some explanations are provided, there is no consensus on the overall understanding or acceptance of the concept. The discussion remains unresolved with multiple competing views.

Contextual Notes

Participants note that the concept of a "spin 4-vector" may have significant limitations, and the discussion includes references to complex theoretical frameworks and assumptions that are not fully resolved.

thebiggerbang
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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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thebiggerbang said:
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?

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:
thebiggerbang said:
Why is the zeroth component of the same zero in this frame?

##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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