 Deviatoric eddy-viscosity closure

For an incompressible mean flow, an eddy viscosity can model the non-isotropic part of the Reynolds stress through the strain-rate tensor $S_{ij}=(U_{i,j}+U_{j,i})/2$. The isotropic part is fixed separately by the turbulent kinetic energy $e_t=\overline{u_k^{\prime }{u}_k^{\prime }}/2$ in $d$ fluctuating dimensions. Taking the trace proves that the isotropic term is necessary: $S_{ii}=0$, whereas $-\rho \overline{u_i^{\prime }{u}_i^{\prime }}=-2\rho e_t$. In the mean momentum equation the isotropic term can be absorbed into $P_{\mathrm{eff}}=P+2\rho e_t/d$.