ベクトル解析公式証明 - kimamanikakuの日記

$\nabla \times \left( \nabla \times A\right)$

$=\nabla(\nabla \cdot A)-\Delta A$

まず、

　 $\nabla A=\left(\dfrac{\partial}{\partial x}\boldsymbol{i}+\dfrac{\partial}{\partial y}\boldsymbol{j} +\dfrac{\partial}{\partial z}\boldsymbol{k}\right)\times A$

$=\boldsymbol{i}\times \dfrac{\partial A}{\partial x}+\boldsymbol{j}\times \dfrac{\partial A}{\partial y}+\boldsymbol{k}\times \dfrac{\partial A}{\partial z}$

したがって、

$\boldsymbol{A} \times \left(\boldsymbol{B}\times \boldsymbol{C}\right)$

$=\left(\boldsymbol{A}\cdot\boldsymbol{B}\right)\boldsymbol{C}-\left(\boldsymbol{A}\cdot\boldsymbol{C}\right)\boldsymbol{B}$

を用いると、

$\nabla\times \left(\nabla \times \boldsymbol{A}\right)$

$=\nabla\times\left(\boldsymbol{i}\times \dfrac{\partial A}{\partial x}+\boldsymbol{j}\times \dfrac{\partial \boldsymbol{A}}{\partial y}+\boldsymbol{k}\times\dfrac{\partial\boldsymbol{A} }{\partial z}\right)$

$=\left(\nabla\times\boldsymbol{i}\times \dfrac{\partial \boldsymbol{A}}{\partial x}\right)+\left(\nabla\times \boldsymbol{j}\times \dfrac{\partial \boldsymbol{A}}{\partial y}\right)+\left(\nabla\times\boldsymbol{k}\times\dfrac{\partial \boldsymbol{A}}{\partial z}\right)$

$=\left(\nabla \cdot \dfrac{\partial \boldsymbol{A}}{\partial x}\right)\boldsymbol{i}-\left(\nabla\cdot\boldsymbol{i}\right)\dfrac{\partial\boldsymbol{A}}{\partial x}+\left(\nabla \cdot \dfrac{\partial \boldsymbol{A}}{\partial y}\right)\boldsymbol{j}-\left(\nabla\cdot\boldsymbol{j}\right)\dfrac{\partial \boldsymbol{A}}{\partial y}+\left(\nabla \cdot \dfrac{\partial \boldsymbol{A}}{\partial z}\right)\boldsymbol{k}-\left(\nabla\cdot\boldsymbol{k}\right)\dfrac{\partial \boldsymbol{A}}{\partial z}$

$=\dfrac{\partial}{\partial x}\left(\nabla\cdot\boldsymbol{A}\right)\boldsymbol{i}+\dfrac{\partial}{\partial y}\left(\nabla\cdot\boldsymbol{A}\right)\boldsymbol{j}+\dfrac{\partial}{\partial z}\left(\nabla\cdot\boldsymbol{A}\right)\boldsymbol{k}-\dfrac{\partial^2\boldsymbol{A}}{\partial x^2}-\dfrac{\partial^2\boldsymbol{A}}{\partial y^2}-\dfrac{\partial^2\boldsymbol{A}}{\partial z^2}$

$=\nabla\left(\nabla\cdot\boldsymbol{A}\right)-\nabla^2\boldsymbol{A}$

より簡単に書いてみる。基底ベクトルを $\boldsymbol{e}_i (i=1,2,3)$ と表せば

$=\nabla \times \left( \nabla \times A\right)$

$=\nabla\times \left(\boldsymbol{e}_i \times \dfrac{\partial \boldsymbol{A}}{\partial x^i}\right)$

$\left(\nabla\cdot\dfrac{\partial \boldsymbol{A}}{\partial x^i}\right)\boldsymbol{e}_i -\left(\nabla\cdot\boldsymbol{e}_i \right)\dfrac{\partial \boldsymbol{A}}{\partial x^i}$

$=\nabla\left(\nabla\cdot\boldsymbol{A}\right)-\nabla^2\boldsymbol{A}$

つぎ。

$(2)\nabla\times\left(\boldsymbol{A}\times\boldsymbol{B}\right)$

$=\dfrac{\partial}{\partial x^i}\Bigl\{\boldsymbol{e}_i \times\left(\boldsymbol{A}\times\boldsymbol{B}\right)\Bigr\}$

$\dfrac{\partial}{\partial x^i}\Bigl\{ \left(\boldsymbol{e}_i \cdot \boldsymbol{B}\right)\boldsymbol{A}-\left(\boldsymbol{e}_i \cdot \boldsymbol{A}\right)\boldsymbol{B}\Bigr\}$

$=\dfrac{\partial}{\partial x^i}\left(B_i \boldsymbol{A}-A_i \boldsymbol{B}\right)$

$=\dfrac{\partial B_i}{\partial x^i}\boldsymbol{A}+B_i \dfrac{\partial \boldsymbol{A}}{\partial x^i}-\dfrac{\partial A_i}{\partial x^i}\boldsymbol{B}-A_i \dfrac{\partial \boldsymbol{B}}{\partial x^i}$

$=\left(\nabla\cdot\boldsymbol{B}\right)\boldsymbol{A}+\left(\boldsymbol{B}\cdot\nabla\right)\boldsymbol{A}-\left(\nabla\cdot\boldsymbol{A}\right)\boldsymbol{B}-\left(\boldsymbol{A}\cdot\nabla\right)\boldsymbol{B}$

3つ目

$\nabla\times\left(\nabla Φ\right)=\boldsymbol{0}$

$\nabla\times\left(\nabla Φ\right)$

$=\left(\dfrac{\partial}{\partial x^i}\boldsymbol{e}_i \right)\times\left(\dfrac{\partial Φ}{\partial x^j}\boldsymbol{e}_j \right)$

$=\dfrac{\partial^2 Φ}{\partial x^i \partial x^j}\boldsymbol{e}_i \times \boldsymbol{e}_j$

$= \dfrac{1}{2}\left(\dfrac{\partial^2 Φ}{\partial x^{i}\partial x^{j}}\boldsymbol{e}_{i} \times \boldsymbol{e}_{j} +\dfrac{\partial^2 Φ}{\partial x^{j}\partial x^{i}} \boldsymbol{e}_{j} \times \boldsymbol{e}_{i} \right)$

$\dfrac{1}{2}\dfrac{\partial^2 Φ}{\partial x^{i}\partial x^{j}}(\boldsymbol{e}_i\times \boldsymbol{e}_j -\boldsymbol{e}_i\times \boldsymbol{e}_j)$

$=\boldsymbol{0}$

4つ目

$\nabla\cdot\left(\boldsymbol{A}\times\boldsymbol{B}\right)$

$=\dfrac{\partial}{\partial x^{i}}\Bigl\{\boldsymbol{e}_i \cdot\left(\boldsymbol{A}\times \boldsymbol{B}\right)\Bigr\}$

$\boldsymbol{e}_i \cdot \left(\dfrac{\partial \boldsymbol{A}}{\partial x^i}\times\boldsymbol{B} \right) +\boldsymbol{e}_i \cdot \left(\dfrac{\partial \boldsymbol{B}}{\partial x^i}\times\boldsymbol{A} \right)$

$=\left(\nabla\times \boldsymbol{A}\right)\cdot\boldsymbol{B}-\left(\nabla\times \boldsymbol{B}\right)\cdot\boldsymbol{A}$