试证明: 在物质描述下, 动量矩守恒定律等价于第二 Piola 应力张量的对称性.
证明: 由 $$\beex \bea \int_{G_t}\rho\sex{{\bf y}\times\cfrac{\rd {\bf v}}{\rd t}}\rd y &=\int_{G_0} \rho_0\sex{{\bf y}\times\cfrac{\p {\bf v}}{\p t}}\rd x,\\ \int_{S_t} ({\bf y}\times{\bf \sigma})\rd S_t&=\int_{S_t} ({\bf y}\times {\bf T}{\bf \nu})\rd S_t\\ &=\int_{S_0} ({\bf y}\times {\bf P}{\bf n})\rd S_0,\\ \int_{G_t}\rho({\bf y}\times{\bf b})\rd y&=\int_{G_0} \rho_0({\bf y}\times{\bf b})\rd x \eea \eeex$$ 知 $$\bex \int_{G_0}\rho_0{\bf y}\times\sex{\cfrac{\p{\bf v}}{\p t}-{\bf b}}\rd x =\int_{S_0}({\bf y}\times{\bf P}{\bf n})\rd S_0. \eex$$ 由动量矩守恒定律 (3. 43) 知 $$\bex {\bf I}\equiv \int_{G_0} {\bf y}\times \Div_x{\bf P}\rd x =\int_{S_0} ({\bf y}\times{\bf P}{\bf n})\rd S_0\equiv {\bf J}. \eex$$ 写成分量形式为 $$\beex \bea I_i&=\int_{G_0} \sum_{j,k,l}\ve_{ijk} y_j\cfrac{\p P_{kl}}{\p x_l}\rd x,\\ J_i&=\int_{S_0}\sum_{j,k}\ve_{ijk}y_j({\bf P}{\bf n})_k\rd S_0\\ &=\int_{S_0}\sum_{j,k,l} \ve_{ijk} y_jP_{kl}n_l\rd S_0\\ &=\sum_{j,k,l}\ve_{ijk}\int_{G_0} \cfrac{\p}{\p x_l}(y_jP_{kl})\rd x\\ &=\sum_{j,k,l}\ve_{ijk}\int_{G_0} f_{jl}P_{kl}+y_j\cfrac{\p p_{kl}}{\p x_l}\rd x. \eea \eeex$$ 于是 $$\bex \sum_{j,k,l}\ve_{ijk}f_{jl}P_{kl}=0. \eex$$ 分别取 $i=1,2,3$ 有 $$\beex \bea \sum_l (f_{2l}P_{3l}-f_{3l}P_{2l})&=0,\\ \sum_l (f_{3l}P_{1l}-f_{1l}P_{3l})&=0,\\ \sum_l (f_{1l}P_{2l}-f_{2l}P_{1l})&=0. \eea \eeex$$ 此即 $$\bex ({\bf F}{\bf P}^T)_{23}=({\bf F}{\bf P}^T)_{32},\quad ({\bf F}{\bf P}^T)_{31}=({\bf F}{\bf P}^T)_{13},\quad ({\bf F}{\bf P}^T)_{12}=({\bf F}{\bf P}^T)_{21}; \eex$$ 或等价地, $$\beex \bea ({\bf F}{\bf P}^T)^T&={\bf F}{\bf P}^T,\\ {\bf P}{\bf F}^T&={\bf F}{\bf P}^T,\\ {\bf P}&={\bf F}{\bf P}^T{\bf F}^{-T},\\ {\bf F}^{-1}{\bf P}&=({\bf F}^{-1}{\bf P})^T,\\ {\bf \Sigma}&={\bf \Sigma}^T. \eea \eeex$$
