Deriving the Navier-Stokes equation from energy-momentum tensor

1. Dec 5, 2011

PhyPsy

The energy-momentum tensor for a perfect fluid is $T^{ab}=(\rho_0+p)u^au^b-pg^{ab}$ (using the +--- Minkowski metric).

Using the conservation law $\partial_bT^{ab}=0$, I'm coming up with $(\rho+\gamma^2p) [\frac{\partial\mathbb{u}}{{\partial}t}+ (\mathbb{u}\cdot\mathbb{\nabla})\mathbb{u}]= -{\nabla}p$ instead of $\rho[\frac{\partial\mathbb{u}}{{\partial}t}+ (\mathbb{u}\cdot\mathbb{\nabla})\mathbb{u}]= -{\nabla}p$ (disregarding the body force part of the equation). Why is there a term for the part in brackets multiplied by $\rho$, but not for $\gamma^2p$?

2. Dec 5, 2011

dextercioby

Where did the gamma come from ? Can you post your derivation/scaned copy of it ?

3. Dec 6, 2011

PhyPsy

$u^a=\gamma(1,\mathbf{u})$
Keeping b constant and cycling a from x to z, I get $\partial_tT^{at}= \partial_t[\gamma^2(\rho_0+p)\mathbf{u}]$
$\partial_xT^{ax}= \partial_x[\gamma^2(\rho_0+p)u_x\mathbf{u}]+\partial_xp$
$\partial_yT^{ay}= \partial_y[\gamma^2(\rho_0+p)u_y\mathbf{u}]+\partial_yp$
$\partial_zT^{az}= \partial_z[\gamma^2(\rho_0+p)u_z\mathbf{u}]+\partial_zp$
$\rho=\gamma^2\rho_0$, so that is why you see a $\gamma^2$ coefficient for p but not for $\rho$. When I solve for $\partial_bT^{tb}=0$, I get $\partial_t(\rho+\gamma^2p)+ \partial_x[(\rho+\gamma^2p)u_x]+ \partial_y[(\rho+\gamma^2p)u_y]+ \partial_z[(\rho+\gamma^2p)u_z]=0$, so summing the four equations for $T^{ab}$ above and simplifying using $\partial_bT^{tb}=0$, I get $$(\rho+\gamma^2p)\partial_t\mathbf{u}+ (\rho+\gamma^2p)u_x\partial_x\mathbf{u}+ (\rho+\gamma^2p)u_y\partial_y\mathbf{u}+ (\rho+\gamma^2p)u_z\partial_z\mathbf{u}+ {\nabla}p=\mathbf{0}$$
This simplifies to what I put in the first post: $(\rho+\gamma^2p)[\partial_t\mathbf{u}+ (\mathbf{u}\cdot\mathbf{\nabla})\mathbf{u}]= -\mathbf{\nabla}p$

4. Dec 6, 2011

dextercioby

There must be a specially relativistic correction to the Galilei invariant Euler equations. You may wish to check L&L <Fluid Mechanics>, pp. 505 to 508.

5. Dec 6, 2011

PhyPsy

It looks like the reason the $\gamma^2p$ term is not in the Navier-Stokes equation is because a nonrelativistic limit is applied where $p<<\rho$, so $\rho+\gamma^2p\approx\rho$.