Get relation from definition of stress-energy tensor and the conservation of energy Starting from the following definition of stress-energy tensor for a perfect fluid in special relativity :
$${\displaystyle T^{\mu \nu }=\left(\rho+{\frac {p}{c^{2}}}\right)\,v^{\mu }v^{\nu }-p\,\eta ^{\mu \nu }\,}\quad(1)$$
with $$v^{\nu}=\dfrac{\text{d}x^{\nu}}{\text{d}\tau}$$ and
$$V^{\nu}=\dfrac{\text{d}x^{\nu}}{\text{d}t}$$ (we have $v^{\nu}=\gamma\,V^{\nu}$)
So, finally, I have to get the following relation :
$$\dfrac{\partial \vec{V}}{\partial t} + (\vec{V}.\vec{grad})\vec{V} = -\dfrac{1}{\gamma^2(\rho+\dfrac{p}{c^2})} \bigg(\vec{grad}\,p+\dfrac{\vec{V}}{c^2}\dfrac{\partial p}{\partial t}\bigg)\quad(2)$$
To get this relation, I must use the conservation of energy for $\nu=i$ and $\nu=0$ with :
$$\partial_{\mu}T^{\mu\nu}=0\quad(3)$$
If someone could help me to find the equation $(2)$ from $(1)$ and $(3)$, this would be nice to indicate the tricks to apply.
EDIT 1 :
For the moment, below where I am :
I recognize in the left member of wanted relation $(2)$ the Lagrangian derivative :
$$\dfrac{\text{D}\,\vec{V}}{\text{d}t}=\dfrac{\partial \vec{V}}{\partial t} + (\vec{V}.\vec{\nabla})\vec{V}\quad(4)$$
and I can rewrite $(1)$ with the $V^{\mu}$ components like :
$$T^{\mu\nu}=\left(\rho+\dfrac{p}{c^{2}}\right)\,\gamma^2\,V^{\mu}V^{\nu }-p\,\eta^{\mu\nu}\,\quad(5)$$
But from this point, I don't know how to make the link between $(4)$, $(5)$, $(3)$ (the divergence of stress-energy equal to zero),  and $(1)$ ...
Any help is welcome
 A: Here are the main steps.  I use $c = 1$ as usual in relativity, and $a, b = 0, 1, 2, 3$ are flat spacetime indices :
\begin{gather}
T^{ab} = (\rho + p) \, u^a \, u^b - \eta^{ab} p,\tag{1} \\[12pt]
\partial_a T^{ab} = u^b u^a \, \partial_a \, \rho + u^b u^a \, \partial_a \, p + (\rho + p)\big( (\partial_a \, u^a) \, u^b + u^a \, \partial_a \, u^b \,\big) - \partial^b p = 0. \tag{2}
\end{gather}
Contract (2) on $u_b$ and use properties $u_b \, u^b = 1$ and $u_b \, \partial_a \, u^b \equiv 0$.  You should get the continuity equation :
\begin{equation}\tag{3}
(\rho + p) \, \partial_a \, u^a = -\, u^a \, \partial_a \, \rho.
\end{equation}
Subsitute this constraint into equation (2).  You should get this :
\begin{equation}\tag{4}
(\rho + p) \, u^a \, \partial_a \, u^b = -\, u^b u^a \, \partial_a \, p + \partial^b \, p.
\end{equation}
Write these for simplicity (total proper-time derivative) :
\begin{align}\tag{5}
u^a \, \partial_a \, u^b &\equiv \frac{d u^b}{d\tau},
& u^a \, \partial_a \, p &\equiv \frac{d p}{d\tau}.
\end{align}
Then you get this, for index $b = i = 1, 2, 3$ :
\begin{align}\tag{6}
(\rho + p) \frac{d u^i}{d\tau} = -\, u^i \, \frac{dp}{d\tau} + \partial^i \, p.
\end{align}
Use $u^i = \gamma \, v^i$ and $\partial^i p = -\, \partial_i \, p$ and $d\tau = dt / \gamma$, so (6) becomes a vectorial equation :
\begin{align}\tag{7}
\gamma \, (\rho + p) \frac{d \gamma \, \vec{v}}{dt} = -\, \gamma^2 \, \vec{v} \, \frac{dp}{dt} - \vec{\nabla} \, p.
\end{align}
Now scalar-contract this with vector $\vec{v}$ and use $\dot{\gamma} = \gamma^3 \, \vec{v} \cdot \dot{\vec{v}}$ and the identity $1 + \gamma^2 \, v^2 \equiv \gamma^2$ :
\begin{gather}
\gamma^2 (\rho + p) \frac{d \vec{v}}{dt} + \gamma \, (\rho + p) \, \gamma^3 (\vec{v} \cdot \dot{\vec{v}}) \, \vec{v} = -\, \vec{\nabla} \, p - \gamma^2 \, \vec{v} \, \frac{dp}{dt} \tag{8} \\[12pt]
\gamma^4 (\rho + p)(\vec{v} \cdot \dot{\vec{v}}) = -\, \vec{v} \cdot \vec{\nabla} \, p - \gamma^2 \, v^2 \, \frac{d p}{dt}. \tag{9}
\end{gather}
Subsitute (9) into the second term of left part of equation (8).  After some simplification algebra and using
\begin{equation}\tag{10}
\frac{d p}{dt} = \frac{\partial p}{\partial t} + \vec{v} \cdot \vec{\nabla} \, p,
\end{equation}
you should get your equation :
\begin{equation}\tag{11}
\gamma^2 (\rho + p) \frac{d \vec{v}}{dt} = -\, \vec{\nabla} \, p - \vec{v} \, \frac{\partial p}{\partial t}.
\end{equation}
That's the relativistic Euler equation for a perfect fluid.  Usually : $p \ll \rho$ and $\gamma \approx 1$ for a slowly moving fluid.  The last term should be negligible.
