# Zeroth component of Spin 4-Vector

1. Jun 5, 2014

### thebiggerbang

The spin 4-vector is defined in the rest frame of the particle as $s^{\mu}= (0, \vec{s})$.

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

2. Jun 5, 2014

### stevendaryl

Staff Emeritus
3. Jun 5, 2014

### Staff: Mentor

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
4. Jun 5, 2014

### WannabeNewton

$S^{\mu}$ 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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