The Killing current of a static dust congruence in flat space

Let $M={\mathbb{R}}^4$ carry the identity chart, the flat metric $g=\mathrm{diag}(-1,1,1,1)$, the time orientation $N={\mathrm{\partial }}_0$ and the chart orientation, with metric-selected connection $\mathrm{\nabla }$ (Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model, Metric compatibility, torsion freedom and the metric-selected connection class). Let $\xi :={\mathrm{\partial }}_0$, let $\rho :{\mathbb{R}}^4\to (0,\mathrm{\infty })$ be smooth with ${\mathrm{\partial }}_0\rho =0$, and let $u:={\mathrm{\partial }}_0$, so that $(\rho ,u)$ is a dust model on $M$ with tensor $T^{ab}=\rho u^a{u}^b$ (The dust matter model: rest density, four-velocity, stress-energy and dust system). Then:

1. $\xi ={\mathrm{\partial }}_0$ is a Killing field.
2. The current $J^a:=T^{ab}{\xi }_b$ equals $J=-\rho u$, and ${\mathrm{\nabla }}_a{J}^a=0$.
3. ${\iota }_J{\mathrm{v}\mathrm{o}\mathrm{l}}_g=-\rho d{x}^1\wedge d{x}^2\wedge d{x}^3$: the current $3$-form is (minus) the rest density carried by the constant-$x^0$ slices, and its flux through corresponding regions of two slices is equal because ${\mathrm{\partial }}_0\rho =0$.

• Signs. The minus sign in clauses 2-3 is the sign of the orientation conventions: the adapted unit normal of a constant-$x^0$ slice pointing in the $+{\mathrm{\partial }}_0$ direction is timelike, so the flux formula of the density theorem carries a factor $\epsilon =-1$. Reversing the orientation of the slice or of $M$ flips the sign of the flux but not the conservation statement.

• Reading. The current here is an energy current: the dust coefficient $\rho$ is the rest-energy density, and $J^a=T^{ab}{\xi }_b=-\rho u^a$. Under the physical normalization $U=cu$ and ${\rho }_m=\rho \mathrm{/}{c}^2$, this is $J^a=-c{\rho }_m{U}^a$, not the rest-mass current ${\rho }_m{U}^a$. The example asserts no empirical fact about matter distributions.