In special relativity, with metric tensor $\eta\_{\mu\nu}=\text{diag}(-c^2,1,1,1)$, take a perfect fluid stress-energy tensor : $T^{\mu\nu} = \left( \rho + \frac{p}{c^2} \right) \, U^\mu\otimes U^\nu + p \, \eta^{\mu\nu}$, where $U^\mu$ is the 4-speed, $\rho$ the volumic mass and $p$ the pressure of the fluid.

In the newtonian limit where $U^\mu \simeq (1,0,0,0)$, we find the newtonian fluid at rest $T^{\mu\nu}\simeq \text{diag}(\rho,p,p,p)$. However, if we want the more precise approximation $(U^t)^2=\gamma^2\simeq 1+\frac{v^2}{c^2}$, we get
$$ T^{tt} \simeq \left( \rho + \frac{p}{c^2} \right) \left( 1 + \frac{v^2}{c^2} \right) - \frac{p}{c^2} $$
Two things surprise me in this energy formula,

* The pressure term $\frac{pv^2}{c^4}$ remains. It is small but not zero.
* It figures $\rho +\rho\frac{v^2}{c^2}$ instead of the newtonian kinetic energy $\rho +\frac{1}{2}\rho\frac{v^2}{c^2}$.

Did I make a mistake in the approximation ?

**EDIT:**

On second thought, the newtonian limit would rather be $U^\mu \simeq (1,\vec{v})$ with $v\ll c$. In that case, the first line of the perfect fluid is $T^t=(\rho, (\rho+\frac{p}{c^2})\vec{v})$ and its zero divergence yields
$$ \frac{\partial\rho}{\partial t} +\text{div}(\rho\vec{v}) = -\text{div}\,\frac{p\vec{v}}{c^2} $$
On the right-hand we recognize the power received from the pressure forces (summed on the 6 faces of a small cube of mass). So it is an energy conservation equation, with energy approximately being $\rho c^2$, as usual. Still, I didn't get the newtonian kinetic energy, and now I have this strange $\frac{p}{c^2}\vec{v}$ in the momentum.