원통좌표계에서 2차 텐서의 발산

원통좌표계에서 2차 텐서(2nd rank tensor)의 발산을 유도해 본다. 원통좌표계의 델 연산자\((\nabla)\)를 적용하고 텐서를 단위벡터의 형태로 나타내면 다음과 같다.

\[\nabla\cdot\boldsymbol\sigma=\left(\hat{r}\frac{\partial}{\partial r}+\hat\theta\frac{\partial}{r\partial\theta}+\hat{z}\frac{\partial}{\partial z}\right)\cdot\left(\begin{matrix}\quad\sigma_{rr}\hat{r}\hat{r}+\sigma_{r\theta}\hat{r}\hat\theta+\sigma_{rz}\hat{r}\hat{z}\\+\sigma_{\theta r}\hat\theta\hat{r}+\sigma_{\theta\theta}\hat\theta\hat\theta+\sigma_{\theta z}\hat\theta\hat{z}\\+\sigma_{zr}\hat{z}\hat{r}+\sigma_{z\theta}\hat{z}\hat\theta+\sigma_{zz}\hat{z}\hat{z}\end{matrix}\right)\]

2차 텐서 각 항에 단위벡터의 좌표계 미분과 적(積)의 미분법을 이용하여 델 연산자의 성분을 적용하면

\[\begin{split}&\hat{r}\frac{\partial}{\partial r}\left(\sigma_{rr}\hat{r}\hat{r}\right)=\hat{r}\frac{\partial\sigma_{rr}}{\partial r}\\&\hat{r}\frac{\partial}{\partial r}\left(\sigma_{r\theta}\hat{r}\hat\theta\right)=\hat\theta\frac{\partial\sigma_{r\theta}}{\partial r}\\&\hat{r}\frac{\partial}{\partial r}\left(\sigma_{rz}\hat{r}\hat{z}\right)=\hat{z}\frac{\partial\sigma_{rz}}{\partial r}\\&\hat\theta\frac{\partial}{r\partial\theta}\left(\sigma_{rr}\hat{r}\hat{r}\right)=\hat{r}\frac{\partial\sigma_{rr}}{\partial r}\\&\hat\theta\frac{\partial}{r\partial\theta}\left(\sigma_{r\theta}\hat{r}\hat\theta\right)=\hat\theta\frac{\sigma_{r\theta}}{r}\\&\hat\theta\frac{\partial}{r\partial\theta}\left(\sigma_{rz}\hat{r}\hat{z}\right)=\hat{z}\frac{\sigma_{rz}}{r}\\&\hat\theta\frac{\partial}{r\partial\theta}\left(\sigma_{\theta r}\hat\theta\hat{r}\right)=\hat{r}\frac{\partial\sigma_{\theta r}}{r\partial\theta}+\hat\theta\frac{\sigma_{\theta r}}{r}\\&\hat\theta\frac{\partial}{r\partial\theta}\left(\sigma_{\theta\theta}\hat\theta\hat\theta\right)=\hat\theta\frac{\partial\sigma_{\theta\theta}}{r\partial\theta}-\hat{r}\frac{\sigma_{\theta\theta}}{r}\\&\hat\theta\frac{\partial}{r\partial\theta}\left(\sigma_{\theta z}\hat\theta\hat{z}\right)=\hat{z}\frac{\partial\sigma_{\theta z}}{r\partial\theta}\\&\hat{z}\frac{\partial}{\partial z}\left(\sigma_{zr}\hat{z}\hat{r}\right)=\hat{r}\frac{\partial\sigma_{zr}}{\partial z}\\&\hat{z}\frac{\partial}{\partial z}\left(\sigma_{z\theta}\hat{z}\hat\theta\right)=\hat\theta\frac{\partial\sigma_{z\theta}}{\partial z}\\&\hat{z}\frac{\partial}{\partial z}\left(\sigma_{zz}\hat{z}\hat{z}\right)=\hat{z}\frac{\partial\sigma_{zz}}{\partial z}\end{split}\]

이 된다(이 외의 항들은 \(0\) 이다). 따라서 정리하면 다음식이 유도된다.

\[\begin{align}\nabla\cdot\boldsymbol\sigma&=\hat{r}\left(\frac{\partial\sigma_{rr}}{\partial r}+\frac{\partial\sigma_{\theta r}}{r\partial\theta}+\frac{\partial\sigma_{zr}}{\partial z}+\frac{\sigma_{rr}-\sigma_{\theta\theta}}{r}\right)\\&+\hat\theta\left(\frac{\partial\sigma_{r\theta}}{\partial r}+\frac{\partial\sigma_{\theta\theta}}{r\partial\theta}+\frac{\partial\sigma_{z\theta}}{\partial z}+\frac{2\sigma_{r\theta}}{r}\right)\\&+\hat{z}\left(\frac{\partial\sigma_{rz}}{\partial r}+\frac{\partial\sigma_{\theta z}}{r\partial\theta}+\frac{\partial\sigma_{zz}}{\partial z}+\frac{\sigma_{rz}}{r}\right)\end{align}\]

[주의] 위의 7번째 식의 미분과정을 살펴보자. 텐서 성분항에 있는 단위벡터의 곱은 다이애딕 곱(diadic product)를 의미한다. 즉,

\[\hat\theta\hat{r}=\hat\theta\otimes\hat{r}=(0,\,1,\,0)\otimes(1,\,0,\,0)=\begin{Bmatrix}0\\1\\0\end{Bmatrix}\begin{Bmatrix}1&0&0\end{Bmatrix}=\begin{bmatrix}0&0&0\\1&0&0\\0&0&0\end{bmatrix}\]

우선 편미분을 취하기 위해 다이애딕 곱이 아닌 개별 단위벡터로 계산한다.

\[\begin{split}\hat\theta\frac{\partial}{r\partial\theta}(\sigma_{\theta r}\hat\theta\hat{r})&=\frac{\hat\theta}{r}\left(\frac{\partial\sigma_{\theta r}}{\partial\theta}\hat\theta\hat{r}+\sigma_{\theta r}\frac{\partial\hat\theta}{\partial\theta}\hat{r}+\sigma_{\theta r}\hat\theta\frac{\partial\hat{r}}{\partial\theta}\right)=\frac{\hat\theta}{r}\left(\frac{\partial\sigma_{\theta r}}{\partial\theta}\hat\theta\hat{r}-\sigma_{\theta r}\hat{r}\hat{r}+\sigma_{\theta r}\hat\theta\hat\theta\right)=\hat{r}\frac{\partial\sigma_{\theta r}}{r\partial\theta}+\hat\theta\frac{\sigma_{\theta r}}{r}\end{split}\]

여기서 단위벡터의 곱은 아래와 같이 계산되었다.

\[\begin{align}&\hat\theta\hat\theta\hat{r}=\begin{Bmatrix}0&1&0\end{Bmatrix}\begin{bmatrix}0&0&0\\1&0&0\\0&0&0\end{bmatrix}=\begin{Bmatrix}1&0&0\end{Bmatrix}=\hat{r}\\&\hat\theta\hat{r}\hat{r}=\begin{Bmatrix}0&1&0\end{Bmatrix}\begin{bmatrix}1&0&0\\0&0&0\\0&0&0\end{bmatrix}=\begin{Bmatrix}0&0&0\end{Bmatrix}=0\\&\hat\theta\hat\theta\hat\theta=\begin{Bmatrix}0&1&0\end{Bmatrix}\begin{bmatrix}0&0&0\\0&1&0\\0&0&0\end{bmatrix}=\begin{Bmatrix}0&1&0\end{Bmatrix}=\hat\theta\end{align}\]