= Charge density and four-current
Let an object be at rest in $S_0$, with proper <charge density> $\rho_0$ and no rest-frame current. In a frame moving at $\mathbf v$ relative to it, the object moves with <velocity> $-\mathbf v$. A simultaneous volume element is Lorentz-contracted along the motion, $dV=dV_0/\gamma$, while its charge is invariant. Therefore
$$
\boxed{\rho=\gamma\rho_0,\qquad\mathbf J=-\gamma\rho_0\mathbf v,\qquad
\gamma=(1-v^2/c^2)^{-1/2}.}
$$
The object's <four-velocity> in this frame is $U^\mu=\gamma(c,-\mathbf v)$, so
$$
\boxed{J^\mu=(c\rho,\mathbf J)=\rho_0U^\mu.}
$$
A scalar proper density times a <Lorentz four-vector> transforms as a <Lorentz four-vector>, demonstrating the claim. With the alternative convention that $\mathbf v$ denotes the object's <velocity> in the observer frame, the spatial component is $+\gamma\rho_0\mathbf v$ instead. Summing the four-currents of different carrier species preserves this transformation law even when the combined current has no rest frame, as for a lab-neutral current-carrying wire.
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