ベクトル解析7 種々の公式
5つの公式とその証明を書いていきます。
公式
\(\phi\)と\(\psi\) はスカラー関数、\(\boldsymbol{A}\)と\(\boldsymbol{B}\)はベクトル。
① \(\mathrm{grad}(\phi\psi)=\phi\nabla\psi+\psi\nabla\phi\)
② \(\mathrm{div}(\phi\boldsymbol{A})=\nabla\phi\cdot\boldsymbol{A}+\phi\mathrm{div}\boldsymbol{A}\)
③ \(\mathrm{rot}(\phi\boldsymbol{A})=\nabla\phi\times\boldsymbol{A}+\phi\mathrm{rot}\boldsymbol{A}\)
④ \(\mathrm{div}(\boldsymbol{A}\times\boldsymbol{B})=\boldsymbol{B}\cdot\mathrm{rot}\boldsymbol{A}-\boldsymbol{A}\cdot\mathrm{rot}\boldsymbol{B}\)
⑤ \(\mathrm{rot}(\boldsymbol{A}\times\boldsymbol{B})=(\boldsymbol{B}\cdot\nabla)\boldsymbol{A}-(\boldsymbol{A}\cdot\nabla)\boldsymbol{B}+\boldsymbol{A}\mathrm{div}\boldsymbol{B}-\boldsymbol{B}\mathrm{div}\boldsymbol{A}\)
証明
1番
\(\mathrm{grad}(\phi\psi)=\phi\nabla\psi+\psi\nabla\phi\)
\(\biggl[\nabla(\phi\psi)\biggr]_{x}=\displaystyle\frac{\partial}{\partial x}(\phi\psi)=\phi\displaystyle\frac{\partial \psi}{\partial x}+\psi\displaystyle\frac{\partial \phi}{\partial x}\)
同様に
\(\biggl[\nabla(\phi\psi)\biggr]_{y}=\phi\displaystyle\frac{\partial \psi}{\partial y}+\psi\displaystyle\frac{\partial \phi}{\partial y}\)
\(\biggl[\nabla(\phi\psi)\biggr]_{z}=\phi\displaystyle\frac{\partial \psi}{\partial z}+\psi\displaystyle\frac{\partial \phi}{\partial z}\)
これら3つの式より、\(\mathrm{grad}(\phi\psi)=\phi\nabla\psi+\psi\nabla\phi\) が示される。
2番
\(\mathrm{div}(\phi\boldsymbol{A})=\nabla\phi\cdot\boldsymbol{A}+\phi\mathrm{div}\boldsymbol{A}\)
\(\mathrm{div}( \phi\boldsymbol{A})=\displaystyle\frac{\partial}{\partial x}( \phi A_{x})+\displaystyle\frac{\partial}{\partial y}( \phi A_{y})+\displaystyle\frac{\partial}{\partial z}( \phi A_{z})\) \(\cdots\) 定義による変形。
\(=\biggl(\displaystyle\frac{\partial \phi}{\partial x}(A_{x})+\phi\displaystyle\frac{\partial A_{x}}{\partial x}\biggr)+\biggl(\displaystyle\frac{\partial \phi}{\partial y}(A_{y})+\phi\displaystyle\frac{\partial A_{y}}{\partial y}\biggr)+\biggl(\displaystyle\frac{\partial \phi}{\partial z}(A_{z})+\phi\displaystyle\frac{\partial A_{z}}{\partial z}\biggr)\) \(\cdots\) それぞれ計算。
\(=\biggl(\displaystyle\frac{\partial \phi}{\partial x}(A_{x})+\displaystyle\frac{\partial \phi}{\partial y}(A_{y})+\displaystyle\frac{\partial \phi}{\partial z}(A_{z})\biggr)+\phi\biggl(\displaystyle\frac{\partial A_{x}}{\partial x}+\displaystyle\frac{\partial A_{y}}{\partial y}+\displaystyle\frac{\partial A_{z}}{\partial z}\biggr)\) \(\cdots\) 順番を入れ替えた。
よって、答えは
\(=\nabla\phi\cdot \boldsymbol{A}+\phi\nabla\cdot\boldsymbol{A}\)
3番
\(\mathrm{rot}(\phi\boldsymbol{A})=\nabla\phi\times\boldsymbol{A}+\phi\mathrm{rot}\boldsymbol{A}\)
まず、x成分に関して示す。
\(\biggl[\mathrm{rot}(\phi\boldsymbol{A})\biggr]_{x}=\displaystyle\frac{\partial}{\partial y}(\phi A_{z})-\displaystyle\frac{\partial}{\partial z}(\phi A_{y})\) \(\cdots\) 定義より。
\(=\displaystyle\frac{\partial \phi}{\partial y}(A_{z})+\phi\displaystyle\frac{\partial A_{z}}{\partial y}-\displaystyle\frac{\partial \phi}{\partial z}(A_{y})-\phi\displaystyle\frac{\partial A_{y}}{\partial z}\) \(\cdots\) それぞれ計算。
\(=\biggl(\displaystyle\frac{\partial \phi}{\partial y}(A_{z})-\displaystyle\frac{\partial \phi}{\partial z}(A_{y})\biggr)+\biggl(\phi\displaystyle\frac{\partial A_{z}}{\partial y}-\phi\displaystyle\frac{\partial A_{y}}{\partial z}\biggr)\) \(\cdots\) 順番を入れ替えた。
\(=\biggl[\nabla\phi\times \boldsymbol{A}+\phi\nabla\times\boldsymbol{A}\biggr]_{x}\)
ほかの成分についても同様なので、\(\mathrm{rot}(\phi\boldsymbol{A})=\nabla\phi\times\boldsymbol{A}+\phi\mathrm{rot}\boldsymbol{A}\) が成立する。
4番
\(\mathrm{div}(\boldsymbol{A}\times\boldsymbol{B})=\boldsymbol{B}\cdot\mathrm{rot}\boldsymbol{A}-\boldsymbol{A}\cdot\mathrm{rot}\boldsymbol{B}\)
レビチビタ記号を使用して示す。
\(\mathrm{div}(\boldsymbol{A}\times\boldsymbol{B})=\partial_{k}(\boldsymbol{A}\times\boldsymbol{B})_{k}\)
\(=\partial_{k}(\epsilon_{ijk}{A}_{i}{B}_{j})\)
片方ずつ微分。
\(={B}_{j}\epsilon_{ijk}\partial_{k}{A}_{i}+{A}_{i}\epsilon_{ijk}\partial_{k}B_{j}\)
\(={B}_{j}\epsilon_{kij}\partial_{k}{A}_{i}-{A}_{i}\epsilon_{kji}\partial_{k}B_{j}\)
\(=B_{j}(\nabla\times\boldsymbol{A})_{j}-A_{i}(\nabla\times\boldsymbol{B})_{i}\)
\(=\boldsymbol{B}\cdot(\nabla\times\boldsymbol{A})-\boldsymbol{A}\cdot(\nabla\times\boldsymbol{B})\)
が成立。
5番
\(\mathrm{rot}(\boldsymbol{A}\times\boldsymbol{B})=(\boldsymbol{B}\cdot\nabla)\boldsymbol{A}-(\boldsymbol{A}\cdot\nabla)\boldsymbol{B}+\boldsymbol{A}\mathrm{div}\boldsymbol{B}-\boldsymbol{B}\mathrm{div}\boldsymbol{A}\)
\(\mathrm{rot}(\boldsymbol{A}\times\boldsymbol{B})\)
\(=\epsilon_{ijk}\partial_{i}(\boldsymbol{A}\times\boldsymbol{B})_{j}=\epsilon_{ijk}\partial_{i}(\epsilon_{lmj}A_{l}B_{m})\)
\(=\epsilon_{ijk}\epsilon_{lmj}\partial_{i}(A_{l}B_{m})=\delta_{kl}\delta_{im}-\delta_{km}\delta_{il}(B_{m}\partial_{i}A_{l}+A_{l}\partial_{i}B_{m})\)
\(i=m\) かつ\(k=l\) のとき
\(B_{i}\partial_{i}A_{k}+A_{k}\partial_{i}B_{i}=(\boldsymbol{B}\cdot\nabla)\boldsymbol{A}+\boldsymbol{A}\mathrm{div}\boldsymbol{B}\)
\(i=l\) かつ\(k=m\) のとき
\(B_{k}\partial_{i}A_{i}+A_{i}\partial_{i}B_{k}=\boldsymbol{B}\mathrm{div}\boldsymbol{A}+(\boldsymbol{A}\cdot\nabla)\boldsymbol{B}\)
これらより最終的な答えは
\((\boldsymbol{B}\cdot\nabla)\boldsymbol{A}-(\boldsymbol{A}\cdot\nabla)\boldsymbol{B}+\boldsymbol{A}\mathrm{div}\boldsymbol{B}-\boldsymbol{B}\mathrm{div}\boldsymbol{A}\)