Divergence-free dust is exactly the dust system, and divergence-free perfect fluid is exactly the fluid equations

(Dust divergence.) By [L1], ${\mathrm{\nabla }}^a{T}_{ab}={\mathrm{\nabla }}^a(\rho u_a{u}_b)=u_b{\mathrm{\nabla }}_a(\rho u^a)+\rho u^a{\mathrm{\nabla }}_a{u}_b$. Contracting with $u^b$ and using [L4], $u^b{\mathrm{\nabla }}^a{T}_{ab}=-{\mathrm{\nabla }}_a(\rho u^a)+\rho u^a{u}_b{\mathrm{\nabla }}_a{u}^b$, and $u_b{\mathrm{\nabla }}_a{u}^b=\frac{1}{2}{\mathrm{\nabla }}_a(u_b{u}^b)=0$ because $g(u,u)=-1$ is constant; so $u^b{\mathrm{\nabla }}^a{T}_{ab}=-{\mathrm{\nabla }}_a(\rho u^a)$.

(Fluid divergence.) By [L1], ${\mathrm{\nabla }}^a{T}_{ab}={\mathrm{\nabla }}^a((\epsilon +p)n_a{n}_b)+{\mathrm{\nabla }}_{b}p$, and ${\mathrm{\nabla }}^a(n_a{n}_b)=n_b{\mathrm{\nabla }}_a{n}^a+n^a{\mathrm{\nabla }}_a{n}_b=\vartheta n_b+a_b$. Hence ${\mathrm{\nabla }}^a{T}_{ab}=n_b(D\epsilon +Dp+(\epsilon +p)\vartheta )+(\epsilon +p)a_b+{\mathrm{\partial }}_{b}p,$ where $Df=n^a{\mathrm{\partial }}_{a}f=n^a{\mathrm{\nabla }}_{a}f$ and $a_b=g_{bc}{a}^c$ by [L2].

(Dust equivalence.) If ${\mathrm{\nabla }}^a{T}_{ab}=0$, then step 1.1 gives ${\mathrm{\nabla }}_a(\rho u^a)=0$; substituting back into step 1.1 gives $\rho u^a{\mathrm{\nabla }}_a{u}_b=0$, and $\rho >0$ (The dust matter model: rest density, four-velocity, stress-energy and dust system) gives $u^a{\mathrm{\nabla }}_a{u}^b=0$. Conversely, if both dust-system equations hold, step 1.1 gives ${\mathrm{\nabla }}^a{T}_{ab}=u_b\cdot 0+\rho \cdot 0=0$. This proves (a).

(Two projections.) Contracting step 1.2 with $n^b$ and using [L4], $n^b{n}_b=-1$ and $n^b{a}_b=0$: $n^b{\mathrm{\nabla }}^a{T}_{ab}=-(D\epsilon +Dp+(\epsilon +p)\vartheta )+n^b{\mathrm{\partial }}_{b}p=-(D\epsilon +(\epsilon +p)\vartheta ),$ since $n^b{\mathrm{\partial }}_{b}p=Dp$ and $Dp=n^b{\mathrm{\nabla }}_{b}p$ by [L2]. Applying the projection $h^{cb}=g^{cb}+n^c{n}^b$ to step 1.2 kills the $n_b$-term and gives $h^{cb}{\mathrm{\nabla }}^a{T}_{ab}=(\epsilon +p)h^{cb}{a}_b+h^{cb}{\mathrm{\partial }}_{b}p=(\epsilon +p)a^c+h^{cb}{\mathrm{\partial }}_{b}p$, where $h^{cb}{a}_b=a^c$ by the remark of The perfect-fluid matter model: rest energy density, pressure, four-velocity and stress-energy ($n^b{a}_b=0$).

(Fluid equivalence.) If ${\mathrm{\nabla }}^a{T}_{ab}=0$, then contracting step 1.2 with $n^b$ gives, by step 2.2, $D\epsilon +(\epsilon +p)\vartheta =0$, and projecting step 1.2 with $h^{cb}$ gives, again by step 2.2, $(\epsilon +p)a^c+h^{cb}{\mathrm{\partial }}_{b}p=0$. Conversely, suppose both fluid equations hold. By the projection identity $h_b{}^c{\mathrm{\partial }}_{c}p={\mathrm{\partial }}_{b}p+n_{b}Dp$, which follows from $h_b{}^c={\delta }_b{}^c+n_b{n}^c$, the second equation reads $(\epsilon +p)a_b=-{\mathrm{\partial }}_{b}p-n_{b}Dp$. Substituting this and the first equation into step 1.2, ${\mathrm{\nabla }}^a{T}_{ab}=n_b(D\epsilon +(\epsilon +p)\vartheta )+n_{b}Dp-({\mathrm{\partial }}_{b}p+n_{b}Dp)+{\mathrm{\partial }}_{b}p=0.$ This proves (b).