Newtonian limit of a perfect fluid

Question

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.

Answer 1

Your approximations need to be consistent. In the non-relativistic limit, it's true that we should have $v/c<<1$, but that's not all. In particular, we should require that the mass energy of the fluid be much larger than the kinetic energy of the fluid.

Alternatively, one might demand that the sound speed in the fluid be much smaller than $c$, which would imply that $p/\rho c^2 << 1$.

To be consistent, we will assume that $\frac{\sqrt{p/\rho}}{c} \sim\frac{v}{c} \sim \delta$. To order $\delta$, we have that

$$U^\mu = (\gamma ,\gamma \vec v/c) \approx (1,\vec v/c)$$ $$T^{00} = (\rho + \frac{p}{c^2})U^0\otimes U^0 + p \eta^{00} \approx \rho $$

This is insufficient, so let's take it a step further and keep terms of order $\delta^2$:

$$U^\mu = (\gamma,\gamma \vec v/c) \approx (1+\frac{v^2}{2c^2},\vec v/c)$$ $$T^{00} = (\rho + \frac{p}{c^2})U^0\otimes U^0 + p \eta^{00} \approx (\rho +\frac{p}{c^2})(1+\frac{v^2}{2c^2})^2 - \frac{p}{c^2} $$ $$ = \rho + \rho\frac{v^2}{c^2} + \mathcal{O}(\delta^4)$$

The term proportional to $\frac{pv^2}{c^4}$ is too small; to keep it, but not incorporate terms of order $v^4/c^4$, is inconsistent, so we throw it away.

As you correctly state above, $T^{00}$ is the energy density in the rest frame of the fluid - $E = T^{00}d^3 x$. When we move to a frame where the fluid is moving with speed $v<<c$,

$$ E' = T'^{00} d^3 x' = T'^{00} \gamma^{-1} d^3x \approx T'^{00}(1-\frac{v^2}{2c^2})d^3x$$ when we keep terms of order $v^2/c^2$. Plugging in our calculated value for $T'^{00}$, we find that

$$ \frac{E'}{d^3x} = (\rho + \rho \frac{v^2}{c^2})(1 - \frac{v^2}{2c^2}) = \rho + \rho \frac{v^2}{2c^2} + \mathcal{O}(\delta^4)$$

Answer 2

Without answers for more than 3 weeks, I'll propose one. I'll only accept it if it is validated by the community by some upvotes.

$T^{tt}$ is the spatial density of energy, so the energy in a small volume $d_3x$ is $$ dE = T^{tt}d_3x \simeq \rho\gamma^2 d_3x $$ Now since the mass moves with 4-speed $U^\mu$, space contracts in the direction of $U^\mu$ by a factor $\gamma$. $dE$ is then explained in 2 steps :

• $\rho\gamma\simeq \rho + \frac{v^2}{2c^2}\rho$ is mass energy plus newtonian kinetic energy.
• $\gamma d_3x$ is an increased volume, which accounts for the Lorentz contraction of space : more mass is stored in $d_3x$ than if the mass was at rest.

If instead of a mass density $\rho$, a point particule of mass $m$ is considered, the Lorentz space contraction disappears and only the famous formula remains : $E=\gamma mc^2$.