# Weyssenhoff fluid and Frenkel condition

A Weyssenhoff fluid is a continuos fluid with spin. The spin is described by an antisimmetric tensor $s{_{ab}}=s{_{[ab]}}$ satisfying the Frenkel condition

$$s{_{ab}}u{^b}=0$$

where $u^b$ is the tangent vector of the family of curve $\lambda$. This is a timelike congruence (I use the 1+3 formalism).

Why do I take the Frenkel condition? Is there any physical interpretation? Or is it a fact of convenience for the following calculus?

• My guess is this make the spin purely spatial.
– MBN
Sep 21, 2017 at 10:37
• Why is this necessary? Sep 21, 2017 at 10:42
• I don't know, presumably because whatever spinning it models is spinning in space.
– MBN
Sep 21, 2017 at 11:03

In special relativity (flat space-time), we define the spin tensor of a body enclosed in a spatial volume $V$ as $$S^{\mu\nu} = \int_V T^{0 \mu} (x^\nu-x^\nu_c(t)) - T^{0 \nu} (x^\mu-x^\mu_c(t)) dV$$ where $x^\mu_c$ is some central referential point and the spatial volume is located at $x^0=t$. When we choose the center of our coordinates in the center of mass $x^i_c = \int T^{00} x^i dV/ \int T^{00} dV$, and $x^0_c = t$, we get $$S^{\mu0} = 0$$ I.e., a part of the spin tensor can be "gauged away" by an appropriate choice of a referential central point. This is a general pattern which also extends to general relativity: even though the spin tensor formally has 6 independent components, only 3 of them are physical and the rest correspond to some nonphysical wobble of the referential worldline of the object. Choosing $S^{\mu\nu}u_\mu = 0$ reduces to $S^{\mu0}$ in the rest frame and is one of the particular choices how to constrain this "gauge freedom".
(Sometimes, these instrinsic spin models are derived from some Grassmanian "supersymmetric" extension of coordinate space. However, when you do this even for a single particle you will find that the condition $S^{\mu\nu}u_\mu = 0$ cannot hold during the whole particle evolution up to trivial cases.)