Boussinesq eddy viscosity hypothesis

Remarks

The isotropic term $\frac23 k \delta_{ij}$ makes the traces of both sides agree, since $\bar S_{ii} = 0$ by incompressibility; it can be absorbed into a modified mean pressure. With this hypothesis the Reynolds stress term of the averaged equations becomes a viscous-type term with variable viscosity $\nu + \nu_t$, and the closure problem reduces to finding the single field $\nu_t$. Mixing-length models give $\nu_t$ by an algebraic formula, and the $k$–$\varepsilon$ model gives it from two transport equations.

Example: in a mean shear flow $\bar u_1(x_2)$, the hypothesis gives $-\overline{u_1' u_2'} = \nu_t \, d\bar u_1 / dx_2$. So momentum flows down the mean velocity gradient, as observed in channel and boundary layer flows away from the wall. Non-example: take a homogeneous flow with $\bar S_{ij} = 0$ but unequal normal stresses, for instance decaying anisotropic turbulence. The hypothesis would force $\overline{u_i' u_j'} = \frac23 k \delta_{ij}$, so it fails there. The anisotropy of the Reynolds stresses in such a flow is not linked to the mean strain.