2019/11/02 空間の運動の極座標表示2

本日のお題

空間の運動を極座標表示した場合に,速度と加速度がどのように表されるか考えましょう

本日のお題は,昨日の続きです
空間の運動の極座標表示 で次のことを示しました

\[\displaystyle v_r = \dot{r\mathstrut},\quad v_{\theta} = r\,\dot{\theta\mathstrut},\quad v_{\varphi} = r\sin\theta\,\dot{\varphi\mathstrut}\]

このことを含めて,速度・加速度の極座標成分を基底から求めましょう

最初に,極座標の基底 \(\boldsymbol{e}_r\)\(\boldsymbol{e}_{\theta}\)\(\boldsymbol{e}_{\varphi}\)\(xyz\) 座標系の基底 \(\boldsymbol{e}_x\)\(\boldsymbol{e}_y\)\(\boldsymbol{e}_z\) との関係を調べておきます

平面で座標軸を原点周りに \(\alpha\) だけ回転する一次変換は \(\begin{pmatrix}\cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha\end{pmatrix}\) で表されましたから,同様に考えると,空間において座標軸を \(y\) 軸及び \(z\) 軸の周りにそれぞれ \(\alpha\) だけ回転する一次変換は \[\begin{pmatrix}\cos\alpha & 0 & \sin\alpha \\ 0 & 1 & 0 \\ -\sin\alpha & 0 & \cos\alpha\end{pmatrix},\quad \begin{pmatrix}\cos\alpha & \sin\alpha & 0 \\ -\sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{pmatrix}\] で表されます

ここで,座標軸を \(z\) 軸周りに \(\varphi\) 回転したのちに,回転した後の新たな \(y\) 軸の周りに \(\displaystyle \frac{\pi}{2} - \theta\) 回転する一次変換を考えます

\(\theta\)

\(\varphi\)

\[\begin{eqnarray} && \begin{pmatrix}\cos\left(\frac{\pi}{2} - \theta\right) & \sin\left(\frac{\pi}{2} - \theta\right) & 0 \\ -\sin\left(\frac{\pi}{2} - \theta\right) & \cos\left(\frac{\pi}{2} - \theta\right) & 0 \\ 0 & 0 & 1\end{pmatrix} \begin{pmatrix}\cos\varphi & 0 & \sin\varphi \\ 0 & 1 & 0 \\ -\sin\varphi & 0 & \cos\varphi\end{pmatrix} \\ &=& \begin{pmatrix}\sin\theta & \cos\theta & 0 \\ -\cos\theta & \sin\theta & 0 \\ 0 & 0 & 1\end{pmatrix} \begin{pmatrix}\cos\varphi & 0 & \sin\varphi \\ 0 & 1 & 0 \\ -\sin\varphi & 0 & \cos\varphi\end{pmatrix} \\ &=& \begin{pmatrix}\sin\theta\cos\varphi & \sin\theta\sin\varphi & \cos\theta \\ -\sin\theta & \cos\varphi & 0 \\ -\cos\theta\cos\varphi & -\cos\theta\sin\varphi & \sin\theta\end{pmatrix} \end{eqnarray}\] この一次変換によって,\(\boldsymbol{e}_x\)\(\boldsymbol{e}_y\)\(\boldsymbol{e}_z\)\[\left\{\begin{array}{l} \displaystyle \sin\theta\cos\varphi\,\boldsymbol{e}_x + \sin\theta\sin\varphi\,\boldsymbol{e}_y + \cos\theta\,\boldsymbol{e_z} \\ -\sin\varphi\,\boldsymbol{e}_x + \cos\varphi\,\boldsymbol{e}_y \\ \displaystyle -\cos\theta\sin\varphi\,\boldsymbol{e}_x - \cos\theta\sin\varphi\,\boldsymbol{e}_y + \sin\theta\,\boldsymbol{e_z} \\ \end{array}\right.\] にそれぞれなり,これらは \(\boldsymbol{e}_r\)\(\boldsymbol{e}_{\varphi}\)\(-\boldsymbol{e}_{\theta}\) に一致します
したがって \[\left\{\begin{array}{l} \displaystyle \boldsymbol{e}_r = \sin\theta\cos\varphi\,\boldsymbol{e}_x + \sin\theta\sin\varphi\,\boldsymbol{e}_y + \cos\theta\,\boldsymbol{e_z} \\ \displaystyle \boldsymbol{e}_{\theta} = \cos\theta\sin\varphi\,\boldsymbol{e}_x + \cos\theta\sin\varphi\,\boldsymbol{e}_y - \sin\theta\,\boldsymbol{e_z} \\ \displaystyle \boldsymbol{e}_{\varphi} = -\sin\varphi\,\boldsymbol{e}_x + \cos\varphi\,\boldsymbol{e}_y \end{array}\right.\] これらの式を \(t\) で微分します \[\begin{eqnarray} \dot{\boldsymbol{e}_r\mathstrut} &=& \left(\dot{\theta\mathstrut}\cos\theta\cos\varphi - \dot{\varphi\mathstrut}\sin\theta\sin\varphi\right)\boldsymbol{e}_x \\ && \quad + \left(\dot{\theta\mathstrut}\cos\theta\sin\varphi + \dot{\varphi\mathstrut}\sin\theta\cos\varphi\right)\boldsymbol{e}_y - \dot{\theta\mathstrut}\sin\theta\,\boldsymbol{e}_{\varphi} \\ &=& \dot{\theta\mathstrut}\left(\cos\theta\sin\varphi\,\boldsymbol{e}_x + \cos\theta\sin\varphi\,\boldsymbol{e}_y - \sin\theta\,\boldsymbol{e_z}\right)\\ && \quad + \dot{\varphi\mathstrut}\sin\theta\left(-\sin\varphi\,\boldsymbol{e}_x + \cos\varphi\,\boldsymbol{e}_y\right) \\ &=& \dot{\theta\mathstrut}\,\boldsymbol{e\mathstrut}_{\theta} + \dot{\varphi\mathstrut}\sin\theta\,\boldsymbol{e}_{\varphi} \\[4px] \dot{\boldsymbol{e}_{\theta}\mathstrut} &=& \left(-\dot{\theta\mathstrut}\sin\theta\cos\varphi - \dot{\varphi\mathstrut}\cos\theta\sin\varphi\right)\boldsymbol{e}_x \\ && \quad + \left(-\dot{\theta\mathstrut}\sin\theta\sin\varphi + \dot{\varphi\mathstrut}\cos\theta\cos\varphi\right)\boldsymbol{e}_y - \dot{\theta\mathstrut}\,\boldsymbol{e}_{\varphi} \\ &=& -\dot{\theta\mathstrut}\left(\sin\theta\cos\varphi\,\boldsymbol{e}_x + \sin\theta\sin\varphi\,\boldsymbol{e}_y + \cos\theta\,\boldsymbol{e_z}\right) \\ && \quad \dot{\varphi}\cos\theta\left(-\sin\varphi\,\boldsymbol{e}_x + \cos\varphi\,\boldsymbol{e}_y\right) \\ &=& -\dot{\theta\mathstrut}\,\boldsymbol{e}_r + \dot{\varphi\mathstrut}\cos\theta\,\boldsymbol{e}_{\varphi} \\[4px] \dot{\boldsymbol{e}_{\varphi}\mathstrut}&=& -\dot{\varphi\mathstrut}\cos\theta\,\boldsymbol{e}_x - \dot{\varphi\mathstrut}\sin\varphi\,\boldsymbol{e}_y \\ &=& -\dot{\varphi\mathstrut}\sin\theta\,\boldsymbol{e}_r - \dot{\varphi}\cos\theta\,\boldsymbol{e}_{\theta} \end{eqnarray}\]

これで準備が整いましたので,動点の位置ベクトル \(\boldsymbol{p} = r\boldsymbol{e}_r\) を時刻 \(t\) で微分していきましょう \[\begin{eqnarray} \dot{\boldsymbol{p}\mathstrut} &=& \frac{d}{dt}r\boldsymbol{e}_r \\ &=& \dot{r\mathstrut}\,\boldsymbol{e}_r + r\,\dot{\boldsymbol{e}_r} \\ &=& \dot{r\mathstrut}\,\boldsymbol{e}_r + r\left(\dot{\theta\mathstrut}\,\boldsymbol{e\mathstrut}_{\theta} + \dot{\varphi\mathstrut}\sin\theta\,\boldsymbol{e}_{\varphi}\right) \\ &=& \dot{r\mathstrut}\,\boldsymbol{e}_r + r\dot{\theta\mathstrut}\,\boldsymbol{e\mathstrut}_{\theta} + r\dot{\varphi\mathstrut}\sin\theta\,\boldsymbol{e}_{\varphi} \\[4px] \ddot{\boldsymbol{p}\mathstrut} &=& \frac{d}{dt}\left(\dot{r\mathstrut}\,\boldsymbol{e}_r + r\dot{\theta\mathstrut}\,\boldsymbol{e\mathstrut}_{\theta} + r\dot{\varphi\mathstrut}\sin\theta\,\boldsymbol{e}_{\varphi}\right) \\ &=& \ddot{r\mathstrut}\,\boldsymbol{e}_r + \dot{r\mathstrut}\,\dot{\boldsymbol{e}_r\mathstrut} + \dot{r\mathstrut}\dot{\theta\mathstrut}\,\boldsymbol{e}_{\theta} + r\ddot{\theta\mathstrut}\,\boldsymbol{e\mathstrut}_{\theta} + r\dot{\theta\mathstrut}\,\dot{\boldsymbol{e\mathstrut}_{\theta}\mathstrut} \\ && \quad + \dot{r\mathstrut}\dot{\varphi\mathstrut}\sin\theta\,\boldsymbol{e}_{\varphi} + r\ddot{\varphi\mathstrut}\sin\theta\,\boldsymbol{e}_{\varphi} + r\dot{\theta\mathstrut}\dot{\varphi\mathstrut}\cos\theta\,\boldsymbol{e}_{\varphi} + r\dot{\varphi\mathstrut}\sin\theta\,\dot{\boldsymbol{e}_{\varphi}} \\ &=& \ddot{r\mathstrut}\,\boldsymbol{e}_r + \dot{r\mathstrut}\left(\dot{\theta\mathstrut}\,\boldsymbol{e\mathstrut}_{\theta} + \dot{\varphi\mathstrut}\sin\theta\boldsymbol{e}_{\varphi}\right) \\ && \quad + \dot{r\mathstrut}\dot{\theta\mathstrut}\,\boldsymbol{e}_{\theta} + r\ddot{\theta\mathstrut}\,\boldsymbol{e\mathstrut}_{\theta} + r\dot{\theta\mathstrut}\left(-\dot{\theta\mathstrut}\,\boldsymbol{e}_r + \dot{\varphi\mathstrut}\cos\theta\,\boldsymbol{e}_{\varphi}\right) \\ && \quad + \dot{r\mathstrut}\dot{\varphi\mathstrut}\sin\theta\,\boldsymbol{e}_{\varphi} + r\ddot{\varphi\mathstrut}\sin\theta\,\boldsymbol{e}_{\varphi} + r\dot{\theta\mathstrut}\dot{\varphi\mathstrut}\cos\theta\,\boldsymbol{e}_{\varphi} \\ && \qquad + r\dot{\varphi\mathstrut}\sin\theta\left(-\dot{\varphi\mathstrut}\sin\theta\,\boldsymbol{e}_r - \dot{\varphi}\cos\theta\,\boldsymbol{e}_{\theta}\right) \\ &=& \left(\ddot{r\mathstrut} - r\dot{\theta\mathstrut}\,^2 - r\dot{\varphi\mathstrut}\,^2\sin^2\theta\right)\boldsymbol{e}_r \\ && \quad + \left(r\ddot{\theta\mathstrut} + 2\dot{r\mathstrut}\dot{\theta\mathstrut} - r\dot{\varphi\mathstrut}\,^2\sin\theta\cos\theta\right)\boldsymbol{e}_{\theta} \\ && \quad + \left(r\ddot{\varphi\mathstrut}\sin\theta + 2\dot{r\mathstrut}\dot{\varphi\mathstrut}\sin\theta + 2r\dot{\theta\mathstrut}\dot{\varphi\mathstrut}\cos\theta\right)\boldsymbol{e}_{\varphi} \end{eqnarray}\] 以上から,速度の \(r\)\(\theta\)\(\varphi\) 成分 \(v_r\)\(v_{\theta}\)\(v_{\varphi}\) と加速度の \(r\)\(\theta\)\(\varphi\) 成分 \(a_r\)\(a_{\theta}\)\(a_{\varphi}\) はそれぞれ次のようになることが分かりました

\[\begin{eqnarray} v_r &=& \dot{r\mathstrut} \\ v_{\theta} &=& r\dot{\theta\mathstrut} \\ v_{\varphi} &=& r\dot{\varphi\mathstrut}\sin\theta \\[4px] a_r &=& \ddot{r\mathstrut} - r\dot{\theta\mathstrut}\,^2 - r\dot{\varphi\mathstrut}\,^2\sin^2\theta \\ a_{\theta} &=& r\ddot{\theta\mathstrut} + 2\dot{r\mathstrut}\dot{\theta\mathstrut} - r\dot{\varphi\mathstrut}\,^2\sin\theta\cos\theta \\ a_{\varphi} &=& r\ddot{\varphi\mathstrut}\sin\theta + 2\dot{r\mathstrut}\dot{\varphi\mathstrut}\sin\theta + 2r\dot{\theta\mathstrut}\dot{\varphi\mathstrut}\cos\theta \end{eqnarray}\]

Last modified: Friday, 5 March 2021, 5:26 PM