Matrix Operation for Stress Tensor

Suppose we define a 2-order stress tensor \(P_{ij}\), where \(i\) stands for scalar components constituting the vector and \(j\) represents the dimension. Giving a specific normalized direction \(n_j\), projection of that stress tensor on surface of this normal direction can be then described by \(P_{ij}\circ n_j\), where \(\circ\) denotes matrix multiplication. Therefore the rate of work can be expressed as

\[\begin{aligned} \dot{w} &= - \int_s \vec{(P_{ij}\circ n_j)}\cdot \vec{v_j} dA~~~~~~~~~~~~ = - \int_s [v_j]^\mathsf{T} \circ [P_{ij}]\circ [n_j] dA \\ &= - \int_s [[P_{ij}]^\mathsf{T} \circ [v_j]]^\mathsf{T} \circ [n_j] dA~~ = - \int_s \vec{([P_{ij}]^\mathsf{T} \circ [v_j])} \cdot \vec{n_j} dA \\ &= - \int_v \nabla \cdot (\bar{P} \vec{v}) dV, \end{aligned}\]

where \(\nabla \cdot (\bar{P} \circ \vec{v}) = \vec{v} \cdot (\nabla \cdot \bar{P}^\mathsf{T}) + \bar{P}^\mathsf{T} : \nabla \vec{v} ~~~~(1)\) .

Proof:

\[\nabla \cdot \left[ \begin{array}{cc} P_{11} & P_{12} \\ P_{21} & P_{22} \end{array} \right] \left[ \begin{array}{c} v_1 \\ v_2 \end{array} \right]= \nabla \cdot \left[ \begin{array}{c} P_{11}v_1+P_{12}v_2 \\ P_{22}v_1+P_{22}v_2 \end{array} \right]= \frac{\partial}{\partial x_1}(P_{11}v_1+P_{12}v_2)+\frac{\partial}{\partial x_2}(P_{21}v_1+P_{22}v_2).\] \[\vec{v} \cdot (\nabla \cdot \bar{P})=v_1(\frac{\partial P_{11}}{\partial x_1}+\frac{\partial P_{12}}{\partial x_2})+v_2(\frac{\partial P_{21}}{\partial x_1}+\frac{\partial P_{22}}{\partial x_2})\] \[\bar{P} : \nabla \vec{v} = P_{11} \frac{\partial v_1}{\partial x_1} + P_{12}\frac{\partial v_1}{\partial x_2} + P_{21}\frac{\partial v_2}{\partial x_1} +P_{22}\frac{\partial v_2}{\partial x_2}\]

To equate two sides of \((1)\), \(\bar{P}\) has to take its transpose at RHS. But there also convention that \(A:B = A_{ij}B_{ji}\), where no extra tranpose is needed.