広告
(∇・テンソル)
微分作用素とテンソルの内積は次式となります。
\[
\begin{aligned}
[\nabla\cdot\vec{\tau}]
&=\left[\sum_{i=0}^{3}\vec{\delta}_{i}\frac{\partial}{\partial x_i}\cdot\sum_{j=0}^{3}\sum_{k=0}^{3}\vec{\delta}_{j}\vec{\delta}_{k}\tau_{jk}\right]\\[6pt]
&=\sum_{i=0}^{3}\sum_{j=0}^{3}\sum_{k=0}^{3}\left[\vec{\delta}_{i}\frac{\partial}{\partial x_i}\cdot\vec{\delta}_{j}\vec{\delta}_{k}\tau_{jk}\right]\\[6pt]
&=\sum_{i=0}^{3}\sum_{j=0}^{3}\sum_{k=0}^{3}\left[(\vec{\delta}_{i}\cdot\vec{\delta}_{j})\vec{\delta}_{k}\right]\frac{\partial}{\partial x_i}\tau_{jk}\\[6pt]
&=\sum_{i=0}^{3}\sum_{j=0}^{3}\sum_{k=0}^{3}\delta_{ij}\vec{\delta}_{k}\frac{\partial\tau_{jk}}{\partial x_i}\\[6pt]
&=\sum_{j=0}^{3}\sum_{k=0}^{3}\vec{\delta}_{k}\frac{\partial\tau_{jk}}{\partial x_i}\qquad(i=j)\\[6pt]
&=\sum_{k=0}^{3}\vec{\delta}_{k}\sum_{j=0}^{3}\frac{\partial\tau_{jk}}{\partial x_i}\\[6pt]
&=\sum_{k=0}^{3}\vec{\delta}_{k}\left(\frac{\partial\tau_{xk}}{\partial x}+\frac{\partial\tau_{yk}}{\partial y}+\frac{\partial\tau_{zk}}{\partial z}\right)\\[6pt]
&=\vec{\delta}_{x}\left(\frac{\partial\tau_{xx}}{\partial x}+\frac{\partial\tau_{yx}}{\partial y}+\frac{\partial\tau_{zx}}{\partial z}\right)\\[6pt]
&\quad+\vec{\delta}_{y}\left(\frac{\partial\tau_{xy}}{\partial x}+\frac{\partial\tau_{yy}}{\partial y}+\frac{\partial\tau_{zy}}{\partial z}\right)\\[6pt]
&\quad+\vec{\delta}_{z}\left(\frac{\partial\tau_{xz}}{\partial x}+\frac{\partial\tau_{yz}}{\partial y}+\frac{\partial\tau_{zz}}{\partial z}\right)
\end{aligned}
\]
上記の項のオーダーは、
(微分作用素・テンソル) = (1 - 2 + 2) = 1
と計算されるので、ベクトル量になります。また、下記が成り立ちます。
\[(\nabla\cdot\vec{\tau})\neq(\vec{\tau}\cdot\nabla)\]
| prev | | | up | | | next |

