= Solution
Use signature $(+---)$, coordinates $x^0=ct$ and <four-velocity> $u^\mu=dx^\mu/d\tau$ normalized by $u_\mu u^\mu=c^2$. The derivative is the metric-compatible <Levi-Civita connection>. Fix the curvature sign by $[\nabla_\alpha,\nabla_\beta]V^\mu=R^\mu{}_{\nu\alpha\beta}V^\nu$. Then the required contractions are
$$
\boxed{R_{\nu\beta}=R^\mu{}_{\nu\mu\beta},\qquad R=g^{\nu\beta}R_{\nu\beta}.}
$$
With $G^{\mu\nu}=R^{\mu\nu}-g^{\mu\nu}R/2$, the <contracted Bianchi identity> gives $\nabla_\mu G^{\mu\nu}=0$. The <Einstein field equations> $G^{\mu\nu}+\Lambda g^{\mu\nu}=(8\pi G/c^4)T^{\mu\nu}$, with constant $\Lambda$, therefore require <covariant conservation of stress-energy>. This is a consistency relation between geometry and the total matter/field source. It does not mean $\partial_\mu T^{\mu\nu}=0$ in arbitrary coordinates, nor guarantee a global conserved <energy> in an arbitrary curved <spacetime>. The geometric identity alone, without the field equations or matter dynamics, is not an equation of motion for a freely specified material source.
For <relativistic dust>, $\rho$ is proper rest-mass <density>, measured in the fluid <rest frame>, and $u^\mu$ is its timelike <four-velocity> with the $c^2$ normalization above. In a local Cartesian <inertial frame> $u^\mu=\gamma(c,\mathbf v)$, so
$$
T^{00}=\rho\gamma^2c^2,\qquad \boxed{T^{0i}=\rho\gamma^2cv^i.}
$$
Accordingly $T^{0i}/c$ is <momentum density> and $cT^{0i}$ is <energy flux> in direction $i$. The equality of momentum-density and energy-flux components after the appropriate factors of $c$ reflects the symmetry of the <stress-energy tensor>. In the comoving frame the spatial stress is zero, as required for pressureless matter.
Expand the dust conservation equation by the <product rule>:
$$
0=u^\nu\nabla_\mu(\rho u^\mu)+\rho u^\mu\nabla_\mu u^\nu.
$$
Contract with $u_\nu$. <metric compatibility> and constant normalization give $u_\nu\nabla_\mu u^\nu=\nabla_\mu(u_\nu u^\nu)/2=0$, leaving $c^2\nabla_\mu(\rho u^\mu)=0$. Substitute this rest-mass continuity law back and divide by nonzero $\rho$:
$$
\boxed{u^\mu\nabla_\mu u^\nu=0,\qquad \frac{d^2x^\nu}{d\tau^2}+\Gamma^\nu_{\alpha\beta}\frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau}=0.}
$$
This proves <geodesic motion of uncharged dust> along each fluid flow line. <pressure> <forces> or another exchange of momentum would change the argument; at zero <density> there is no dust <worldline> to constrain.
The <electromagnetic field tensor> is antisymmetric and has six independent components. In the stated local inertial coordinates choose $F^{0i}=-E_i/c$ and $F^{ij}=-\epsilon_{ijk}B_k$, giving
$$
F^{\mu\nu}=\begin{pmatrix}
0&-E_x/c&-E_y/c&-E_z/c\\
E_x/c&0&-B_z&B_y\\
E_y/c&B_z&0&-B_x\\
E_z/c&-B_y&B_x&0
\end{pmatrix}.
$$
The fields are those measured by that local observer; changing observer mixes them through the <Lorentz transformation> of $F$. With $\epsilon^{0123}=+1$ and $\widetilde F^{\mu\nu}=\epsilon^{\mu\nu\alpha\beta}F_{\alpha\beta}/2$, two independent quadratic contractions are
$$
\boxed{F_{\mu\nu}F^{\mu\nu}=2(B^2-E^2/c^2),\qquad F_{\mu\nu}\widetilde F^{\mu\nu}=-\frac4c\mathbf E\cdot\mathbf B.}
$$
The second is a <pseudoscalar>: it is invariant under proper <Lorentz transformations> and changes sign under an <orientation> reversal. Opposite choices for the <orientation> <tensor> change its written sign. A nonzero <null electromagnetic field> can have both contractions zero, so the two numbers do not determine whether the <tensor> itself vanishes.
For <charged relativistic dust>, insert $j^\alpha=(q/m)\rho u^\alpha$ into the force-density equation:
$$
u^\nu\nabla_\mu(\rho u^\mu)+\rho u^\mu\nabla_\mu u^\nu=\frac qm\rho F^{\nu\alpha}u_\alpha.
$$
Contraction with $u_\nu$ again gives the continuity law, because $u_\nu F^{\nu\alpha}u_\alpha=0$ by antisymmetry. After removing the continuity term,
$$
\boxed{u^\mu\nabla_\mu u^\nu=\frac qm F^{\nu\alpha}u_\alpha.}
$$
In a <local inertial frame> the spatial components reduce to $d(\gamma m\mathbf v)/dt=q(\mathbf E+\mathbf v\times\mathbf B)$ and the temporal component gives $d(\gamma mc^2)/dt=q\mathbf E\cdot\mathbf v$. This is the relativistic <Lorentz force>, with gravity represented by the connection. The <force> is orthogonal to $u$ and preserves the rest-mass normalization. Matter alone now exchanges <energy> and momentum with the electromagnetic field; adding the field <tensor>'s stress-energy restores zero divergence of the combined source.
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