# Navier Stokes : what about angular momentum?

I play with CFD for a while, and suddenly, a transcendantal question raises: :-) Navier Stokes is basically Newton applied on a continuum in Eulerian. For solids, we would consider linear, but also angular momentum. Why don't we have to do that for fluids ?

Conversely, you can take the curl of Navier Stokes and have an equation expressed in vorticity, which looks like our angular momentum world. Does it mean that the equation with velocity somehow embed the both kind of momentum, and they are totally correlated for fluids ?

But how it's not the same for solids ? i.e., where is the intrinsic difference that makes it different degrees of freedom in one case and equivalent in the second ?

The basic answer was given here: In a fluid, why are the shear stresses $\tau_{xy}$ and $\tau_{yx}$ equal?. Angular momentum conservation follows from linear momentum conservation (expressed by the Euler/Navier-Stokes equation) combined with the symmetry of the stress tensor.
Momentum conservation is the equation $$\frac{\partial}{\partial t}\pi_i + \nabla_j\tau_{ij} = 0$$ where $\pi_i=\rho v_i$ is the momentum density. Using $$\tau_{ij} = P\delta_{ij}+\rho v_iv_j$$ this equation is equivalent to the Euler equation, and including dissipative stresses gives the Navier-Stokes equation.
The density of angular momentum (about the origin) is $l_i=\epsilon_{ijk}x_j\pi_k$ and $l_i$ is conserved if $\epsilon_{ijk}\tau_{jk}=0$. We get $$\frac{\partial}{\partial t}l_i + \nabla_j m_{ij} = 0$$ where $m_{ij}=\epsilon_{ikl}x_k\tau_{lj}$ is the angular momentum flux.