= Solution
Use <metric signature> $(+---)$ throughout this solution. <Stress-energy conservation> for the <perfect fluid> is $\nabla_bT^{ab}=0$. Expanding it and using <metric compatibility> gives
$$
0=U^a\nabla_b[(\rho+p)U^b]+(\rho+p)U^b\nabla_bU^a-\nabla^ap.
$$
The normalization of the <four-velocity> implies $U_aU^b\nabla_bU^a=0$. Contracting with $U_a$ therefore gives the <relativistic perfect-fluid energy equation>,
$$
U^b\nabla_b\rho+(\rho+p)\nabla_bU^b=0,
$$
while subtracting the component parallel to $U$ gives the <relativistic Euler equation>,
$$
\boxed{(\rho+p)U^b\nabla_bU_a=\nabla_ap-U_aU^b\nabla_bp.}
$$
Equivalently, apply the <orthogonal projection> $\delta_a{}^c-U_aU^c$ to the conservation equation. This form makes clear that a <pressure> gradient orthogonal to the <four-velocity> supplies the fluid's <four-acceleration>.
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