= Solution
Use signature $(+---)$ and local inertial coordinates $x^0=ct$. Write $u^\mu=\gamma(c,\mathbf v)$ and $w=\rho c^2+p$, where $\rho$ is rest <mass density>. The <perfect fluid in general relativity> has components
$$
T^{00}=w\gamma^2-p,\qquad T^{0i}=T^{i0}=\frac{w\gamma^2v_i}{c},\qquad
T^{ij}=\frac{w\gamma^2v_iv_j}{c^2}+p\delta_{ij}.
$$
Here $T^{00}$ is <energy density>, $cT^{i0}$ is <energy flux>, $T^{0i}/c$ is <momentum density>, and $T^{ij}$ is the flux of the $j$th <momentum> component through a surface normal to $i$. In the local rest frame these reduce to <energy density> $\rho c^2$, zero energy and momentum flow, and isotropic <pressure> $p$. The equation $\nabla_\mu T^{\mu\nu}=0$ expresses local <stress-energy conservation>. Its temporal and spatial components give energy and momentum balance; in flat coordinates, for example, $\partial_tT^{00}+\partial_i(cT^{i0})=0$.
For <relativistic dust>, $T^{\mu\nu}=\rho u^\mu u^\nu$. Contracting its divergence with $u_\nu$, and using <metric compatibility> together with $u_\nu u^\nu=c^2$, gives
$$
0=u_\nu\nabla_\mu(\rho u^\mu u^\nu)
=c^2\nabla_\mu(\rho u^\mu)+\frac\rho2u^\mu\nabla_\mu(u_\nu u^\nu)
=c^2\nabla_\mu(\rho u^\mu).
$$
Consequently $\nabla_\mu(\rho u^\mu)=0$. Substituting this continuity equation back into the divergence leaves $\rho u^\mu\nabla_\mu u^\nu=0$. Wherever matter is present,
$$
\boxed{u^\mu\nabla_\mu u^\nu=0,}
$$
so the individual particles follow <geodesics>.
The electromagnetic <four-force> law $f^\mu=qF^{\mu\nu}u_\nu$ must preserve the particle's <rest mass>. Thus $u_\mu f^\mu=0$, and requiring this for arbitrary timelike <four-velocities> makes the <electromagnetic field tensor> antisymmetric. Its six independent components comprise the <electric field> and <magnetic field>. With the stated signature and $x^0=ct$, one consistent convention is
$$
F^{i0}=E_i/c,\qquad F^{0i}=-E_i/c,\qquad F^{ij}=-\epsilon_{ijk}B_k.
$$
It gives the spatial force $d\mathbf p/dt=q(\mathbf E+\mathbf v\times\mathbf B)$ and energy transfer $dE/dt=q\mathbf E\cdot\mathbf v$. Under a <Lorentz transformation>, $F^{\mu\nu}$ transforms as a rank-two <contravariant tensor>; its electric and magnetic components consequently mix. A four-potential supplies $F_{\mu\nu}=\nabla_\mu A_\nu-\nabla_\nu A_\mu$, unchanged by $A_\mu\mapsto A_\mu+\nabla_\mu\chi$. The <Maxwell equations> are
$$
\nabla_\mu F^{\mu\nu}=\mu_0j^\nu,\qquad
\nabla_\alpha F_{\beta\gamma}+\nabla_\beta F_{\gamma\alpha}+\nabla_\gamma F_{\alpha\beta}=0.
$$
These combine the source equations and the homogeneous equations into covariant tensor statements; $j^\mu$ is the <four-current>.
For <charged relativistic dust>, the matter tensor exchanges <four-momentum> with the electromagnetic field: $\nabla_\mu T_{\rm m}^{\mu\nu}=F^{\nu\alpha}j_\alpha$, while their total <stress-energy tensor> is conserved. Identical particles of <rest mass> $m$ and <electric charge> $q$ have $j^\alpha=(q/m)\rho u^\alpha$. Contraction with $u_\nu$ again gives mass continuity because $u_\nu F^{\nu\alpha}u_\alpha=0$. Removing that term gives
$$
\boxed{u^\mu\nabla_\mu u^\nu=\frac qmF^{\nu\alpha}u_\alpha.}
$$
Using zero divergence for the matter tensor alone in this last step would omit the electromagnetic <Lorentz force>.
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