Question

\frac{\partial m}{\partial t} = -\nabla \cdot (vm + \pi) + \rho g. \quad (5.32) Exercise 5.8 \quad Symmetry of the Pressure (Stress) Tensor Form the vector product of $r$ with the momentum balance (5.32), to obtain$^{15}$ $\frac{\partial}{\partial t}(r \times m) = -\nabla \cdot v(r \times m) - \nabla \cdot (r \times \pi^T) - \epsilon : \pi + r \times \rho g,$ which is the balance equation for angular momentum density $r \times m$. Show that the source associated with the pressure tensor $\epsilon : \pi$ vanishes if the pressure tensor is symmetric, $^{16} \pi^T = \pi.$

          \frac{\partial m}{\partial t} = -\nabla \cdot (vm + \pi) + \rho g. \quad (5.32)
Exercise 5.8 \quad Symmetry of the Pressure (Stress) Tensor
Form the vector product of $r$ with the momentum balance (5.32), to obtain$^{15}$
$\frac{\partial}{\partial t}(r \times m) = -\nabla \cdot v(r \times m) - \nabla \cdot (r \times \pi^T) - \epsilon : \pi + r \times \rho g,$ 
which is the balance equation for angular momentum density $r \times m$. Show that the
source associated with the pressure tensor $\epsilon : \pi$ vanishes if the pressure tensor is
symmetric, $^{16} \pi^T = \pi.$

(∂m)/(∂t) = -∇·(vm + π) + ρg. (5.32)
Exercise 5.8 Symmetry of the Pressure (Stress) Tensor
Form the vector product of r with the momentum balance (5.32), to obtain^15
(∂)/(∂ t)(r × m) = -∇· v(r × m) - ∇· (r ×π^T) - ϵ : π + r ×ρ g,
which is the balance equation for angular momentum density r × m. Show that the
source associated with the pressure tensor ϵ : π vanishes if the pressure tensor is
symmetric, ^16π^T = π.

Added by William R.

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