極座標での成分表示メモ

nablaの値とか計量行列の値とかを調べる

定義

自然基底

$ \pmb e_r:=\frac{\partial\pmb r}{\partial\bar e_r}

$ \pmb e_\theta:=\frac{\partial\pmb r}{\partial\bar e_\theta}

$ \mathrm d\bm r=\bm e_r\mathrm d\bar e_r+\bm e_\theta\mathrm d\bar e_\theta

dot積

$ \pmb e_r\cdot\pmb e_\theta=0

$ |\pmb e_r|=1

$ |\pmb e_\theta|=\bar e_r

直交基底だが正規ではない

計量行列

$ [\pmb I]^{\sf PP}=\begin{pmatrix}1&0\\0&{\bar e_r}^2\end{pmatrix}

$ \therefore[\pmb I]^{\sf\bar P\bar P}=\left([\pmb I]^{\sf P P}\right)^{-1}=\begin{pmatrix}1&0\\0&\frac1{{\bar e_r}^2}\end{pmatrix}

行列式

$ \det[\pmb I]^{\sf\bar P\bar P}=\frac{1}{{\bar e_r}^2}

$ \det[\bm I]^{\sf PP}=\bar e_r^2

反変基底

$ \begin{pmatrix}\bar{\pmb e}_r\\\bar{\pmb e}_\theta\end{pmatrix}=\begin{pmatrix}1&0\\0&\frac1{{\bar e_r}^2}\end{pmatrix}\begin{pmatrix}\pmb e_r\\\pmb e_\theta\end{pmatrix}=\begin{pmatrix}\pmb e_r\\\frac1{{\bar e_r}^2}\pmb e_\theta\end{pmatrix}

共変パラメタ

$ \begin{pmatrix}\mathrm d e_r\\\mathrm de_\theta\end{pmatrix}=\begin{pmatrix}1&0\\0&\frac1{{\bar e_r}^2}\end{pmatrix}\begin{pmatrix}\mathrm d\bar e_r\\\mathrm d\bar e_\theta\end{pmatrix}=\begin{pmatrix}\mathrm d\bar e_r\\\frac1{{\bar e_r}^2}\mathrm d\bar e_\theta\end{pmatrix}

解けそうにない……

微分

$ \pmb\Gamma^{\sf P P}_{rr}=\frac{\partial\pmb e_r}{\partial\bar e_r}=\pmb 0

動径方向には曲がらない

$ \pmb\Gamma^{\sf PP}_{r\theta}=\frac{\partial\pmb e_r}{\partial\bar e_\theta}=\frac1{\bar e_r}\pmb e_\theta

偏角方向に微小回転する

https://kakeru.app/bb70d19beb4dd4c80d00c72678b3bbb3 https://i.kakeru.app/bb70d19beb4dd4c80d00c72678b3bbb3.svg

$ \pmb\Gamma^{\sf PP}_{\theta r}=\pmb\Gamma^{\sf PP}_{r\theta}=\frac1{\bar e_r}\pmb e_\theta

$ \pmb\Gamma^{\sf PP}_{\theta\theta}=\frac{\partial\pmb e_\theta}{\partial\bar e_\theta}=-\bar e_r\pmb e_r

$ \pmb\nabla=\bar{\pmb e}_r\frac{\partial}{\partial\bar e_r}+\bar{\pmb e}_\theta\frac{\partial}{\partial\bar e_\theta}

$ =\pmb e_r\frac{\partial}{\partial\bar e_r}+\frac1{{\bar e_r}^2}\pmb e_\theta\frac{\partial}{\partial\bar e_\theta}

勾配

$ \pmb\nabla\pmb v=\pmb e_k[\pmb I]^{\sf \bar P\bar P}_{ik}\frac{\partial [\pmb v]^{\sf\bar P}_j}{\partial\bar e_i}\pmb e_j+\pmb e_k[\pmb I]^{\sf \bar P\bar P}_{ik}[\pmb v]^{\sf\bar P}_j\pmb\Gamma_{ij}^{\sf PP}

$ = \begin{matrix}\frac{\partial[\pmb v]^{\sf\bar P}_r}{\partial\bar e_r}\pmb e_r\pmb e_r+\pmb 0&+\frac{\partial [\pmb v]^{\sf\bar P}_\theta}{\partial\bar e_r}\pmb e_r\pmb e_\theta+\frac1{\bar e_r}[\pmb v]^{\sf\bar P}_\theta\pmb e_r\pmb e_\theta\\+\frac1{{\bar e_r}^2}\frac{\partial[\pmb v]^{\sf\bar P}_r}{\partial\bar e_\theta}\pmb e_\theta\pmb e_r+\frac1{{\bar e_r}^3}[\pmb v]^{\sf\bar P}_r\pmb e_\theta\pmb e_\theta&+\frac1{{\bar e_r}^2}\frac{\partial [\pmb v]^{\sf\bar P}_\theta}{\partial\bar e_\theta}\pmb e_\theta\pmb e_\theta-\frac1{\bar e_r}[\pmb v]^{\sf\bar P}_\theta\pmb e_\theta\pmb e_r\end{matrix}

$ \therefore [\pmb\nabla\pmb v]^{\sf\bar P\bar P}_{ij}=\begin{pmatrix}\frac{\partial[\pmb v]^{\sf\bar P}_r}{\partial\bar e_r}&\frac{\partial [\pmb v]^{\sf\bar P}_\theta}{\partial\bar e_r}+\frac1{\bar e_r}[\pmb v]^{\sf\bar P}_\theta\\\frac1{{\bar e_r}^2}\frac{\partial[\pmb v]^{\sf\bar P}_r}{\partial\bar e_\theta}-\frac1{\bar e_r}[\pmb v]^{\sf\bar P}_\theta&\frac1{{\bar e_r}^3}[\pmb v]^{\sf\bar P}_r+\frac1{{\bar e_r}^2}\frac{\partial [\pmb v]^{\sf\bar P}_\theta}{\partial\bar e_\theta}\end{pmatrix}

$ v^j|^i=[\pmb\nabla\pmb v]^{\sf\bar P\bar P}_{ij}

検算：発散とも整合性が取れてる

$ \pmb\nabla\cdot\pmb v=[\pmb I]^{\sf PP}_{ii}v^i|^i

$ =\frac{\partial[\pmb v]^{\sf\bar P}_r}{\partial\bar e_r}+{\bar e_r}^2\left(\frac1{{\bar e_r}^3}[\pmb v]^{\sf\bar P}_r+\frac1{{\bar e_r}^2}\frac{\partial [\pmb v]^{\sf\bar P}_\theta}{\partial\bar e_\theta}\right)

$ =\frac{\partial[\pmb v]^{\sf\bar P}_r}{\partial\bar e_r}+\frac1{{\bar e_r}}[\pmb v]^{\sf\bar P}_r+\frac{\partial [\pmb v]^{\sf\bar P}_\theta}{\partial\bar e_\theta}

$ v_{\hat r}=\pmb v\cdot\hat{\pmb e}_r=\pmb v\cdot\bar{\pmb e}_r=v^r

$ v_{\hat\theta}=\pmb v\cdot\hat{\pmb e}_\theta=\pmb v\cdot r\bar{\pmb e}_\theta=rv^\theta

$ \sf\hat P:=(\hat{\pmb e}_r,\hat{\pmb e}_\theta)

$ [\pmb I]^{\sf \hat PP}=[\pmb I]^{\sf P\hat P}=\begin{pmatrix}1&0\\0&r\end{pmatrix}

$ \begin{pmatrix}1&0\\0&r\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}1&0\\0&r\end{pmatrix}=\begin{pmatrix}a&rb\\rc&r^2d\end{pmatrix}

$ \therefore [\pmb\nabla\pmb v]^{\sf\hat P\hat P}_{ij}=\begin{pmatrix}\frac{\partial v_{\hat r}}{\partial r}&r\cdot\frac{\partial v_{\hat \theta}/r}{\partial r}+r\cdot\frac1rv_{\hat \theta}/r\\r\cdot\frac1{r^2}\frac{\partial v_{\hat r}}{\partial\theta}-r\cdot\frac1rv_{\hat \theta}/r&r^2\cdot\frac1{r^3}v_{\hat r}+r^2\cdot\frac1{r^2}\frac{\partial v_{\hat\theta}/r}{\partial\theta}\end{pmatrix}

$ =\begin{pmatrix}\frac{\partial v_{\hat r}}{\partial r}&\frac{\partial v_{\hat \theta}}{\partial r}-\frac1rv_{\hat\theta}+\frac1rv_{\hat \theta}\\\frac1r\frac{\partial v_{\hat r}}{\partial\theta}-\frac1rv_{\hat \theta}&\frac1rv_{\hat r}+\frac1r\frac{\partial v_{\hat\theta}}{\partial\theta}\end{pmatrix}

$ =\begin{pmatrix}\frac{\partial v_{\hat r}}{\partial r}&\frac{\partial v_{\hat \theta}}{\partial r}\\\frac1r\frac{\partial v_{\hat r}}{\partial\theta}-\frac1rv_{\hat \theta}&\frac1rv_{\hat r}+\frac1r\frac{\partial v_{\hat\theta}}{\partial\theta}\end{pmatrix}

$ \bm\nabla\cdot\bm v=\frac{1}{\sqrt{\det[\bm I]^{\sf PP}}}\frac{\partial}{\partial\bar e_i}\left([\bm v]^{\sf\bar P}_i\sqrt{\det[\bm I]^{\sf PP}}\right)

$ =\frac{1}{|\bar e_r|}\frac{\partial}{\partial\bar e_i}\left([\bm v]^{\sf\bar P}_i\bar e_r\right)

$ \because\det[\bm I]^{\sf PP}=\bar e_r^2

$ = \frac1{\bar e_r}\left(\frac{\partial}{\partial\bar e_r}\bar e_r[\bm v]^{\sf\bar P}_r+\frac{\partial}{\partial\bar e_\theta}\bar e_r[\bm v]^{\sf\bar P}_\theta\right)

$ \because\bar e_r\ge0

$ = \frac1{\bar e_r}\bar v_r+\frac{\partial\bar v_r}{\partial\bar e_r}+\frac{\partial\bar v_\theta}{\partial\bar e_\theta}

$ \bm\nabla\cdot\bm v=\frac1rv_{\hat r}+\frac{\partial v_{\hat r}}{\partial r}+\frac1r\frac{\partial v_{\hat\theta}}{\partial\theta}

幾何学的に求める

https://kakeru.app/0a77bb8d98949a6396f360730c24f504 https://i.kakeru.app/0a77bb8d98949a6396f360730c24f504.svg

ざっと検索した限りだと、第1項と第2項はまとめて表示していることが多かった

$ \bm\nabla\cdot\bm v=\frac1r\frac{\partial rv_{\hat r}}{\partial r}+\frac1r\frac{\partial v_{\hat\theta}}{\partial\theta}

一般式を使わずに導出すると、

$ |\pmb\nabla|^2=\pmb e_r\frac{\partial}{\partial r}\cdot\pmb e_r\frac{\partial}{\partial r}+\pmb e_r\frac{\partial}{\partial r}\cdot\pmb e_\theta\frac{\partial}{\partial\theta}+\pmb e_\theta\frac{\partial}{\partial\theta}\cdot\pmb e_r\frac{\partial}{\partial r}+\pmb e_\theta\frac{\partial}{\partial\theta}\cdot\pmb e_\theta\frac{\partial}{\partial\theta}

$ =\pmb e_r\frac{\partial}{\partial r}\cdot\bar{\pmb e}_r\frac{\partial}{\partial r}+\pmb e_r\frac{\partial}{\partial r}\cdot\frac1{r^2}\bar{\pmb e}_\theta\frac{\partial}{\partial\theta}+\pmb e_\theta\frac{\partial}{\partial\theta}\cdot\bar{\pmb e}_r\frac{\partial}{\partial r}+\pmb e_\theta\frac{\partial}{\partial\theta}\cdot\frac1{r^2}\bar{\pmb e}_\theta\frac{\partial}{\partial\theta}

$ = \frac{\partial^2}{{\partial r}^2}+\frac1{r^2}\pmb e_r\cdot\pmb\Gamma^{\sf\bar P\bar P}_{r\theta}\frac{\partial}{\partial\theta}+\pmb e_\theta\cdot\pmb\Gamma^{\sf \bar P\bar P}_{\theta r}\frac{\partial}{\partial r}+\frac1{r^2}\pmb e_\theta\cdot\pmb\Gamma^{\sf\bar P\bar P}_{\theta\theta}\frac{\partial}{\partial\theta}+\frac1{r^2}\frac{\partial^2}{{\partial\theta}^2}

$ = \frac{\partial^2}{{\partial r}^2}+\frac1r\frac{\partial}{\partial r}+\frac1{r^2}\frac{\partial^2}{{\partial\theta}^2}

$ \bm\nabla^2\bullet= \frac1{\sqrt{\det[\bm I]^{\sf PP}}}\frac{\partial}{\partial\bar e_i}\left(\sqrt{\det[\bm I]^{\sf PP}}[\bm I]^{\sf \bar P\bar P}_{ij}\frac{\partial\bullet}{\partial\bar e_j}\right)

$ = \frac1{|\bar e_r|}\frac{\partial}{\partial\bar e_r}\left(|\bar e_r|\cdot1\cdot\frac{\partial\bullet}{\partial\bar e_r}\right)+\frac1{|\bar e_r|}\frac{\partial}{\partial\bar e_\theta}\left(|\bar e_r|\cdot\frac1{\bar e_r^2}\cdot\frac{\partial\bullet}{\partial\bar e_\theta}\right)

$ = \frac{\partial^2\bullet}{{\partial\bar e_r}^2}+\frac1{\bar e_r}\frac{\partial\bullet}{\partial\bar e_r}+\frac1{\bar e_r^2}\frac{\partial^2\bullet}{{\partial\theta}^2}

$ \because\bar e_r\ge0

$ \pmb\nabla\cdot\pmb A=\frac{1}{\sqrt{\det[\pmb I]^{\sf \bar P\bar P}}}\frac{\partial}{\partial e_i}\left([\pmb A]^{\sf PP}_{ij}\sqrt{\det[\pmb I]^{\sf \bar P\bar P}}\right)\bar{\pmb e}_j+[\pmb A]^{\sf PP}_{ij}\pmb\Gamma^{\sf \bar P\bar P}_{ij}

$ = \frac1r\left(\frac{\partial}{\partial r}r[\pmb A]^{\sf PP}_{rj}+\frac{\partial}{\partial \theta}r[\pmb A]^{\sf PP}_{\theta j}\right)\bar{\pmb e}_j+\frac1r\bar{\pmb e}_\theta([\pmb A]^{\sf PP}_{r\theta}+[\pmb A]^{\sf PP}_{\theta r})-\bar{\pmb e}_r[\pmb A]^{\sf PP}_{\theta\theta}

$ =\left(\frac1r\frac{\partial}{\partial r}r[\pmb A]^{\sf PP}_{rr}+\frac{\partial}{\partial \theta}[\pmb A]^{\sf PP}_{\theta r}-[\pmb A]^{\sf PP}_{\theta\theta}\right)\bar{\pmb e}_r+\left(\frac1r\frac{\partial}{\partial r}r[\pmb A]^{\sf PP}_{r\theta}+\frac{\partial}{\partial \theta}[\pmb A]^{\sf PP}_{\theta\theta}+\frac1r[\pmb A]^{\sf PP}_{r\theta}+\frac1r[\pmb A]^{\sf PP}_{\theta r}\right)\bar{\pmb e}_\theta

$ r:=\bar e_r

$ \theta:=\bar e_\theta

$ \bm e_{\hat r}:=\hat{\bm e_r}=\bm e_r

$ \bm e_{\hat\theta}:=\hat{\bm e_\theta}=\frac1r\bm e_\theta

$ [\bm I]^{\sf\hat PP}=\begin{pmatrix}1&0\\0&r\end{pmatrix}