[mathjax]
目次
ベクトル解析公式
公式の前に補足です。以下は表記が違うだけです。
$\mathrm{grad}\boldsymbol{A}=\nabla\boldsymbol{A}$
$\mathrm{div}\boldsymbol{A}=\nabla\cdot\boldsymbol{A}$
$\mathrm{rot}\boldsymbol{A}=\nabla\times\boldsymbol{A}$
公式
※\(\phi\)と\(\psi\) はスカラー関数、\(\boldsymbol{A}\)と\(\boldsymbol{B}\)はベクトル。
① \(\mathrm{div} \mathrm{rot} \boldsymbol{A}=0\)
② \(\mathrm{rot} \mathrm{grad}\phi=\boldsymbol{0}\)
③ \(\mathrm{rot}\mathrm{rot}\boldsymbol{A}=\nabla(\nabla\cdot\boldsymbol{A})-\nabla^2 \boldsymbol{A}\)
④ \(\mathrm{grad}(\phi\psi)=\phi\nabla\psi+\psi\nabla\phi\)
⑤ \(\mathrm{div}(\phi\boldsymbol{A})=\nabla\phi\cdot\boldsymbol{A}+\phi\mathrm{div}\boldsymbol{A}\)
⑥ \(\mathrm{rot}(\phi\boldsymbol{A})=\nabla\phi\times\boldsymbol{A}+\phi\mathrm{rot}\boldsymbol{A}\)
⑦ \(\mathrm{div}(\boldsymbol{A}\times\boldsymbol{B})=\boldsymbol{B}\cdot\mathrm{rot}\boldsymbol{A}-\boldsymbol{A}\cdot\mathrm{rot}\boldsymbol{B}\)
⑧ \(\mathrm{rot}(\boldsymbol{A}\times\boldsymbol{B})=(\boldsymbol{B}\cdot\nabla)\boldsymbol{A}-(\boldsymbol{A}\cdot\nabla)\boldsymbol{B}+\boldsymbol{A}\mathrm{div}\boldsymbol{B}-\boldsymbol{B}\mathrm{div}\boldsymbol{A}\)
証明
以下、\(\displaystyle\frac{\partial}{\partial y}=\partial_{y}\)と表記します。
公式1
\(\mathrm{rot} \boldsymbol{A}=\)(\(\partial_{y}A_{z}-\partial_{z}A_{y}\) , \(\partial_{z}A_{x}-\partial_{x}A_{z}\) , \(\partial_{x}A_{y}-\partial_{y}A_{x}\)) であるので、
\(\mathrm{div} \mathrm{rot}\boldsymbol{A}\)\(=\partial_{x}\)\((\partial_{y}A_{z}-\partial_{z}A_{y})\)\(+\partial_{y}\)\((\partial_{z}A_{x}-\partial_{x}A_{z})\)\(+\partial_{z}\)\((\partial_{x}A_{y}-\partial_{y}A_{x})\)
\(=\partial_{x}\partial_{y}A_{z}-\partial_{x}\partial_{z}A_{y}+\partial_{y}\partial_{z}A_{x}-\partial_{x}\partial_{y}A_{z}+\partial_{x}\partial_{z}A_{y}-\partial_{y}\partial_{z}A_{x}\)
\(=0\)
$$\mathrm{div} \mathrm{rot} \boldsymbol{A}=0$$
公式2
\(\mathrm{rot} \mathrm{grad}\phi=\boldsymbol{0}\)
これの\(x\)成分について考えると
\([\mathrm{rot} \mathrm{grad}\phi]_{x}=\displaystyle\frac{\partial}{\partial y}\cdot \displaystyle\frac{\partial \phi}{\partial z}-\displaystyle\frac{\partial}{\partial z}\cdot \displaystyle\frac{\partial \phi}{\partial y}=\displaystyle\frac{\partial^2\phi}{\partial y\partial z}-\displaystyle\frac{\partial^2 \phi}{\partial y\partial z}=0\)
他の成分についても同様なので
$$\mathrm{rot} \mathrm{grad}\phi=\boldsymbol{0}$$
公式3
この証明で「レビチビタ記号」を使います。
\([\mathrm{rot}\mathrm{rot}\boldsymbol{A}]_{k}=[\nabla\times(\nabla\times\boldsymbol{A})]_{k}=\epsilon_{ijk}\partial_{i}(\nabla\times\boldsymbol{A})_{j}=\epsilon_{ijk}\partial_{i}\epsilon_{lmj}\partial_{l}\boldsymbol{A}_{m}\)
\(=\epsilon_{kij}\epsilon_{lmj}\partial_{i}\partial_{l}\boldsymbol{A}_{m}=(\delta_{kl}\delta_{im}-\delta_{km}\delta_{il})\partial_{i}\partial_{l}\boldsymbol{A}_{m}\)
等式、\(\epsilon_{ijk}\epsilon_{ilm}=\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}\)を使用。\((\delta_{kl}\delta_{im}-\delta_{km}\delta_{il})\) で
第一項が生き残るのは \(k=l\) かつ \(i=m\) のときで、この時 \(1\cdot \partial_{i}\partial_{k}\boldsymbol{A}_{i}\)
第二項が生き残るのは \(k=m\) かつ \(i=l\) のときで、この時 \(-1\cdot \partial_{i}\partial_{i}\boldsymbol{A}_{k}\)
\((\delta_{kl}\delta_{im}-\delta_{km}\delta_{il})\partial_{i}\partial_{l}\boldsymbol{A}_{m}=\partial_{i}\partial_{k}\boldsymbol{A}_{i}-\partial_{i}\partial_{i}\boldsymbol{A}_{k}\)\(=[\nabla(\nabla\cdot \boldsymbol{A})-\nabla^2 \boldsymbol{A}]_{k}\)
他の成分も同様なので最終的に
$$\mathrm{rot}\mathrm{rot}\boldsymbol{A}=\nabla(\nabla\cdot\boldsymbol{A})-\nabla^2 \boldsymbol{A}$$
公式4
\(\biggl[\nabla(\phi\psi)\biggr]_{x}=\displaystyle\frac{\partial}{\partial x}(\phi\psi)=\phi\displaystyle\frac{\partial \psi}{\partial x}+\psi\displaystyle\frac{\partial \phi}{\partial x}\)
\(\biggl[\nabla(\phi\psi)\biggr]_{y}=\phi\displaystyle\frac{\partial \psi}{\partial y}+\psi\displaystyle\frac{\partial \phi}{\partial y}\)
\(\biggl[\nabla(\phi\psi)\biggr]_{z}=\phi\displaystyle\frac{\partial \psi}{\partial z}+\psi\displaystyle\frac{\partial \phi}{\partial z}\)
$$\mathrm{grad}(\phi\psi)=\phi\nabla\psi+\psi\nabla\phi$$
公式5
\(\mathrm{div}( \phi\boldsymbol{A})=\displaystyle\frac{\partial}{\partial x}( \phi A_{x})+\displaystyle\frac{\partial}{\partial y}( \phi A_{y})+\displaystyle\frac{\partial}{\partial z}( \phi A_{z})\)
\(=\biggl(\displaystyle\frac{\partial \phi}{\partial x}(A_{x})+\phi\displaystyle\frac{\partial A_{x}}{\partial x}\biggr)+\biggl(\displaystyle\frac{\partial \phi}{\partial y}(A_{y})+\phi\displaystyle\frac{\partial A_{y}}{\partial y}\biggr)+\biggl(\displaystyle\frac{\partial \phi}{\partial z}(A_{z})+\phi\displaystyle\frac{\partial A_{z}}{\partial z}\biggr)\)
\(=\biggl(\displaystyle\frac{\partial \phi}{\partial x}(A_{x})+\displaystyle\frac{\partial \phi}{\partial y}(A_{y})+\displaystyle\frac{\partial \phi}{\partial z}(A_{z})\biggr)+\phi\biggl(\displaystyle\frac{\partial A_{x}}{\partial x}+\displaystyle\frac{\partial A_{y}}{\partial y}+\displaystyle\frac{\partial A_{z}}{\partial z}\biggr)\)
\(=\nabla\phi\cdot \boldsymbol{A}+\phi\nabla\cdot\boldsymbol{A}\)
$$\mathrm{div}(\phi\boldsymbol{A})=\nabla\phi\cdot\boldsymbol{A}+\phi\mathrm{div}\boldsymbol{A}$$
公式6
x成分に関して示す。
\(\biggl[\mathrm{rot}(\phi\boldsymbol{A})\biggr]_{x}=\displaystyle\frac{\partial}{\partial y}(\phi A_{z})-\displaystyle\frac{\partial}{\partial z}(\phi A_{y})\)
\(=\displaystyle\frac{\partial \phi}{\partial y}(A_{z})+\phi\displaystyle\frac{\partial A_{z}}{\partial y}-\displaystyle\frac{\partial \phi}{\partial z}(A_{y})-\phi\displaystyle\frac{\partial A_{y}}{\partial z}\)
\(=\biggl(\displaystyle\frac{\partial \phi}{\partial y}(A_{z})-\displaystyle\frac{\partial \phi}{\partial z}(A_{y})\biggr)+\biggl(\phi\displaystyle\frac{\partial A_{z}}{\partial y}-\phi\displaystyle\frac{\partial A_{y}}{\partial z}\biggr)\)
\(=\biggl[\nabla\phi\times \boldsymbol{A}+\phi\nabla\times\boldsymbol{A}\biggr]_{x}\)
他の成分についても同様なので
$$\mathrm{rot}(\phi\boldsymbol{A})=\nabla\phi\times\boldsymbol{A}+\phi\mathrm{rot}\boldsymbol{A}$$
公式7
\(\mathrm{div}(\boldsymbol{A}\times\boldsymbol{B})=\partial_{k}(\boldsymbol{A}\times\boldsymbol{B})_{k}=\partial_{k}(\epsilon_{ijk}{A}_{i}{B}_{j})\)
\(={B}_{j}\epsilon_{ijk}\partial_{k}{A}_{i}+{A}_{i}\epsilon_{ijk}\partial_{k}B_{j}={B}_{j}\epsilon_{kij}\partial_{k}{A}_{i}-{A}_{i}\epsilon_{kji}\partial_{k}B_{j}\)
\(=B_{j}(\nabla\times\boldsymbol{A})_{j}-A_{i}(\nabla\times\boldsymbol{B})_{i}=\boldsymbol{B}\cdot(\nabla\times\boldsymbol{A})-\boldsymbol{A}\cdot(\nabla\times\boldsymbol{B})\)
$$\mathrm{div}(\boldsymbol{A}\times\boldsymbol{B})=\boldsymbol{B}\cdot\mathrm{rot}\boldsymbol{A}-\boldsymbol{A}\cdot\mathrm{rot}\boldsymbol{B}$$
公式8
\([\mathrm{rot}(\boldsymbol{A}\times\boldsymbol{B})]_{k}=\epsilon_{ijk}\partial_{i}(\boldsymbol{A}\times\boldsymbol{B})_{j}=\epsilon_{ijk}\partial_{i}(\epsilon_{lmj}A_{l}B_{m})\)
\(=\epsilon_{ijk}\epsilon_{lmj}\partial_{i}(A_{l}B_{m})=(\delta_{kl}\delta_{im}-\delta_{km}\delta_{il})(B_{m}\partial_{i}A_{l}+A_{l}\partial_{i}B_{m})\)
\(i=m\) かつ\(k=l\) のとき
\(B_{i}\partial_{i}A_{k}+A_{k}\partial_{i}B_{i}=[(\boldsymbol{B}\cdot\nabla)\boldsymbol{A}+\boldsymbol{A}\mathrm{div}\boldsymbol{B}]_{k}\)
\(i=l\) かつ\(k=m\) のとき
\(B_{k}\partial_{i}A_{i}+A_{i}\partial_{i}B_{k}=[\boldsymbol{B}\mathrm{div}\boldsymbol{A}+(\boldsymbol{A}\cdot\nabla)\boldsymbol{B}]_{k}\)
これらより
$$\mathrm{rot}(\boldsymbol{A}\times\boldsymbol{B})=(\boldsymbol{B}\cdot\nabla)\boldsymbol{A}-(\boldsymbol{A}\cdot\nabla)\boldsymbol{B}+\boldsymbol{A}\mathrm{div}\boldsymbol{B}-\boldsymbol{B}\mathrm{div}\boldsymbol{A}$$
問題
次のdivとrotを求めよう。
１，$a\boldsymbol{r}$
２，$\boldsymbol{r}r^n$
３，$\boldsymbol{c}\times \boldsymbol{r}$
１番
$\mathrm{div}(a\boldsymbol{r})=a(\partial_{x} x+\partial_{y} y+\partial_{z} z)=3a$
$\mathrm{rot}(a\boldsymbol{r})=a(\partial_{y} z-\partial_{z} y)+\cdots)=0$
2番
公式５，６を使用する。
$\mathrm{div}(\boldsymbol{r}r^n)=r^n \nabla\cdot \boldsymbol{r}+\boldsymbol{r}\cdot \nabla r^n=3r^n+\boldsymbol{r}\cdot nr^{n-2}\boldsymbol{r}=(n+3)r^n$
$\mathrm{rot}(\boldsymbol{r}r^n)=r^n(\nabla\times \boldsymbol{r})-\boldsymbol{r}\times (\nabla r^n)=-\boldsymbol{r}\times nr^{n-2}\boldsymbol{r}=0$
3番
公式７，８をそれぞれ使用する。
$\mathrm{div}(\boldsymbol{c}\times \boldsymbol{r})=\boldsymbol{r}\cdot (\nabla\times \boldsymbol{c})-\boldsymbol{c}\cdot (\nabla\times \boldsymbol{r})=0$
$\mathrm{rot}(\boldsymbol{c}\times \boldsymbol{r})=(\boldsymbol{r}\cdot \nabla)\boldsymbol{c}-(\boldsymbol{c}\cdot\nabla)\boldsymbol{r}+\boldsymbol{c}(\nabla\cdot \boldsymbol{r})-\boldsymbol{r}(\nabla\cdot \boldsymbol{c})$
$=-(\boldsymbol{c}\cdot\nabla)\boldsymbol{r}+\boldsymbol{c}(\nabla\cdot \boldsymbol{r})$
$=-\boldsymbol{c}+3\boldsymbol{c}=2\boldsymbol{c}$
