9.2.3 Angular momentum and symmetry of the stress tensor

As mentioned previously, the symmetric deformation or strain tensor vanishes for motion consisting of rigid rotation plus uniform translation. In such cases, the generalized Hooke's law (9.15) says that the stress tensor T^ij reduces to $-p \, \delta^{ij}$ since $\tau^{ij}$ vanishes. The purpose of this section is to provide some further details regarding these ideas and their connection to conservation of angular momentum.

The continuum form of Newton's law is given by

$$\rho \, \frac{D u^{i}}{Dt} \rho \, f^{i} + T^{ij}_{\; ,j}$$ = (9.16)

where $\rho$ is the mass density, and fⁱ are components to external or body forces such as those arising from gravity and the Coriolis force. $T^{ij}_{\; ,j}$ is the divergence of the stress tensor, where T^ij is written in the form (9.14) which incorporates the pressure. A component of the angular momentum for a fluid parcel is given by

$$\epsilon_{ijk} \, x^{j} \, u^{k} \, \rho \, dV,$$ L_i = (9.17)

where $\rho \, dV$ is the mass of the infinitesimal parcel. The material time derivative of this angular momentum is given by

$$\frac{D L_{i}}{Dt} \epsilon_{ijk} \, x^{j} \, \frac{D u^{k}}{Dt} \, \rho \, dV,$$ = (9.18)

where $D(\rho \, dV)/Dt = 0$ through conservation of mass. For a Boussinesq fluid, $\rho$ appears as the constant $\rho_{o}$, and D(dV)/Dt = 0 then follows from volume conservation. Substituting Newton's law into this expression leads to

$$\frac{D L_{i}}{Dt} \epsilon_{ijk} \, (x^{j} \, \rho \, f^{k} + x^{j} \, T^{km}_{\; ,m} ) \, dV.$$ = (9.19)

The first term accounts for torques placed on the parcel from external forces. The second term arises from torques on the fluid from internal stresses. To further interpret the second term, consider the budget for total angular momentum of the fluid, which is obtained by integrating over the fluid volume

$$\frac{D L^{T}_{i}}{Dt} \int \epsilon_{ijk} \, (x^{j} \, \rho \, f^{k} + x^{j} \, T^{km}_{\; ,m} ) \, dV.$$ = (9.20)

Now integrate by parts on the stress tensor term to find

 

$$\int \epsilon_{ijk} \, x^{j} \, T^{km}_{\; ,m} \, dV \int \epsilon_{ijk} \, [ \partial_{m} \, (x^{j} \, T^{km}) - T^{kj}] \, dV.$$ = (9.21)

The first term integrates to a boundary contribution, which is non-vanishing for cases in which there are torques arising from boundary stresses. The second term is a volume contribution and it picks out the term $\epsilon_{ijk} \, \tau^{kj}$, since $\epsilon_{ijk} \, \delta^{kj} = 0$. For most fluids, such as ocean water, the internal torques are balanced and so there will be no net contribution to angular momentum from internal stresses. This case can be ensured if the frictional stress tensor is symmetric

$$\tau^{mn} \tau^{nm},$$ = (9.22)

which renders $\epsilon_{ijk} \, \tau^{kj} = 0$.