9.3.3 Dissipation of total kinetic energy

The budget for kinetic energy of a fluid parcel is given by

$$\frac{\rho}{2} \, \frac{D ( \delta_{ij} \, u^{i} \, u^{j} )}{Dt} \rho \, \delta_{ij} \, u^{i} \, f^{j} + \delta_{ij} \, u^{i} \, T^{jk}_{\; ,k}$$ =

$$\rho \, u_{i} \, f^{i} + \partial_{k} \, (u_{j} \, T^{jk}) - u_{j,k} \, T^{jk}$$ =

$$\rho \, u_{i} \, f^{i} + \partial_{k} \, (u_{j} \, T^{jk}) - e_{jk} \, T^{jk}$$ =

$$\rho \, u_{i} \, f^{i} + \partial_{k} \, (u_{j} \, T^{jk}) + p \, u^{j}_{\; ,j} - e_{jk} \, \tau^{jk}.$$ = (9.25)

The first term on the right hand side arises from work done by external forces. The second term, when integrated over the fluid domain, accounts for work done at boundaries by the stresses. The third term arises from pressure work against changes in the parcel's volume. This term vanishes for a volume conserving fluid. The fourth term is present throughout the fluid domain, and it can be written

$$e_{ij} \, \tau^{ij} = \rho \, e_{ij} \, C^{ijmn} \, e_{mn}.$$ (9.26)

In general, this term is sign-indefinite. However, for a frictional stress tensor which manifests dissipative friction at each point in the fluid, one requires

$$e_{ij} \, C^{ijmn} \, e_{mn} \ge 0.$$ (9.27)

Since the strain tensor e_ij is symmetric, this constraint can be satisfied if

C^ijmn = C^mnij. (9.28)

This constraint brings the number of degrees of freedom in the viscosity tensor down to 21 = 6 + 5 + 4 + 3 + 2 + 1, which is the number of degrees of freedom in a symmetric $6 \times 6$ matrix.