回転 (vector解析)の各種演算
tensor積$ \otimesは略す
$ \pmb{\nabla}\times\pmb{a}={\Large\pmb{\epsilon}}:(\pmb{\nabla}\pmb{a})
$ \pmb{\nabla}\times\pmb{\nabla}\phi={\Large\pmb{\epsilon}}:(\pmb{\nabla}\pmb{\nabla}\phi)=\pmb{0}
$ \because
偏微分演算子を交換可能な場合、$ \pmb{\nabla}\pmb{\nabla}\phiは対称tensorとなる。
ざっくりいうと、cross積の性質より$ \pmb{\nabla}\times\pmb{\nabla}=\pmb{0}だから0になった、とも説明できる
$ \pmb{\nabla}\times(\pmb{\nabla}\cdot\pmb{T})={\Large\pmb{\epsilon}}:(\pmb{\nabla}\pmb{\nabla}\cdot\pmb{T})
$ \epsilon_{ijk}\frac{\partial^2T_{lk}}{\partial e_j\partial e_l}\pmb{e}_i
$ \pmb{\nabla}\cdot(\pmb{\nabla}\times\pmb{a})={\Large\pmb{\epsilon}}\vdots(\pmb{\nabla}\pmb{\nabla}\pmb{a})=0
$ \because\pmb{\nabla}\pmb{\nabla}\pmb{a}は1階目と2階目が対称なので、$ \epsilon_{ijk}+\epsilon_{jik}=0より打ち消される
(偏微分演算子が交換可能な場合)
$ \pmb{\nabla}\cdot(\pmb{a}\times\pmb{b})={\Large\pmb{\epsilon}}\vdots(\pmb{\nabla}\pmb{a}\pmb{b})
これは$ \pmb{\nabla},\pmb{a},\pmb{b}の各成分を並べた3次元行列式と同じ値になる
積の微分を、行列式の交換法則で再現する
$ ={\Large\pmb{\epsilon}}\vdots(\pmb{b}\pmb{\nabla}\pmb{a})-{\Large\pmb{\epsilon}}\vdots(\pmb{a}\pmb{\nabla}\pmb{b})
$ =\pmb{b}\cdot(\pmb{\nabla}\times\pmb{a})-\pmb{a}\cdot(\pmb{\nabla}\times\pmb{b})
$ \bm\nabla\cdot(\bm a\times\bm T)=\partial_i\epsilon_{ijk}a_jT_{kl}
$ = \epsilon_{ijk}(\partial_ia_j)T_{kl}+\epsilon_{ijk}a_j\partial_iT_{kl}
$ = (\epsilon_{kij}\partial_ia_j)T_{kl}-a_j\epsilon_{jik}\partial_iT_{kl}
$ = (\bm\nabla\times\bm a)\cdot\bm T-\bm a\cdot(\bm\nabla\times\bm T)
$ \bm\nabla\cdot(\bm T\times\bm a)=\partial_k\epsilon_{ilj}T_{kl}a_j
$ = \epsilon_{ilj}(\partial_kT_{kl})a_j+\epsilon_{ilj}T_{kl}\partial_ka_j
$ =(\bm\nabla\cdot\bm T)\times\bm a+{\Large\pmb\epsilon}:(\bm T^\top\cdot\bm\nabla\bm a)
$ (\bm a\times\bm T)\cdot\overleftarrow{\bm\nabla}=\partial_l\epsilon_{ijk}a_jT_{kl}
$ = \epsilon_{ijk}\partial_la_jT_{kl}+\epsilon_{ijk}a_j\partial_lT_{kl}
$ = \bm a\times(\bm T\cdot\overleftarrow{\bm\nabla})+{\Large\pmb\epsilon}:(\bm a\overleftarrow{\bm\nabla}\cdot\bm T^\top)
$ \bm\nabla\cdot(\bm T\times\bm U)=\partial_iT_{il}\epsilon_{jlm}U_{mk}
$ \bm a=\bm r($ \bm rは位置)のとき
$ \bm\nabla\cdot(\bm r\times\bm T)=-\bm r\cdot(\bm\nabla\times\bm T)
$ \bm\nabla\cdot(\bm T\times\bm a)=(\bm\nabla\cdot\bm T)\times\bm r+{\Large\pmb\epsilon}:\bm T^\top
$ \bm a\times\bm T=(\bm T^\top\times\bm a)^\topという関係にある
$ \bm a\times\bm T=\epsilon_{ijk}a_jT_{kl}\bm e_i\bm e_l
$ \bm T\times\bm a=\epsilon_{ilj}T_{kl}a_j\bm e_k\bm e_i
$ \pmb{\nabla}\times(\pmb{\nabla}\times\pmb{a})={\Large\pmb{\epsilon}}:(\pmb{\nabla}{\Large\pmb{\epsilon}}:(\pmb{\nabla}\pmb{a}))
$ =2{\cal\pmb{W}}\vdots(\pmb{\nabla}\pmb{\nabla}\pmb{a})
$ =\pmb{\nabla}\cdot(\pmb{\nabla}\wedge\pmb{a})^\top
$ =\pmb{\nabla}(\pmb{\nabla}\cdot\pmb{a})-\Delta\pmb{a}
実は、$ \bm a\times(\bm a\times\bm b)=\bm a\bm a\cdot\bm b-|\bm a|^2\bm bであり、それと対応している
$ \pmb{\nabla}\times(\pmb{a}\times\pmb{b})=2{\cal\pmb{W}}\vdots(\pmb{\nabla}(\pmb{a}\pmb{b}))
$ =\pmb{\nabla}\cdot(\pmb{a}\wedge\pmb{b})^\top
$ =\pmb{\nabla}\cdot(\pmb{b}\pmb{a})-\pmb{\nabla}\cdot(\pmb{a}\pmb{b})
$ =(\pmb{\nabla}\cdot\pmb{b})\pmb{a}+(\pmb{b}\cdot\pmb{\nabla})\pmb{a}-(\pmb{\nabla}\cdot\pmb{a})\pmb{b}-(\pmb{a}\cdot\pmb{\nabla})\pmb{b}
$ \pmb{a}\times(\pmb{\nabla}\times\pmb{b})=2{\cal\pmb{W}}\vdots(\pmb{a}\pmb{\nabla}\pmb{b})
$ =(\pmb{\nabla}\wedge\pmb{b})\pmb{a}
$ =(\pmb{\nabla}\pmb{b})\pmb{a}-(\pmb{a}\cdot\pmb{\nabla})\pmb{b}
ここで、$ \pmb{\nabla}(\pmb{a}\cdot\pmb{b})=(\pmb{\nabla}\pmb{a})\pmb{b}+(\pmb{\nabla}\pmb{b})\pmb{a}を用いると、
$ \pmb{a}\times(\pmb{\nabla}\times\pmb{b})+\pmb{b}\times(\pmb{\nabla}\times\pmb{a})=\pmb{\nabla}(\pmb{a}\cdot\pmb{b})-(\pmb{a}\cdot\pmb{\nabla})\pmb{b}-(\pmb{b}\cdot\pmb{\nabla})\pmb{a}
になる
さらに$ \pmb{a}=\pmb{b}なら↓のように簡略化される
$ \pmb{a}\times(\pmb{\nabla}\times\pmb{a})=\pmb{\nabla}\left(\frac12|\pmb{a}|^2\right)-(\pmb{a}\cdot\pmb{\nabla})\pmb{a}
$ \pmb{\nabla}(\pmb{a}\times\pmb{b})=\pmb{\nabla}{\Large\pmb{\epsilon}}:(\pmb{a}\pmb{b})
$ ={\Large\pmb{\epsilon}}:(\pmb{a}(\pmb{\nabla}\pmb{b})^\top)-{\Large\pmb{\epsilon}}:(\pmb{b}(\pmb{\nabla}\pmb{a})^\top)
$ ={\Large\pmb{\epsilon}}:(\pmb{a}(\pmb{\nabla}\pmb{b})^\top-\pmb{b}(\pmb{\nabla}\pmb{a})^\top)
$ \pmb{a}\times\pmb{T}:={\Large\pmb{\epsilon}}:\pmb{a}\pmb{T}とすると
$ \pmb{\nabla}(\pmb{a}\times\pmb{b})=\pmb{a}\times(\pmb{\nabla}\pmb{b})^\top-\pmb{b}\times(\pmb{\nabla}\pmb{a})^\top
$ \int_{\partial V}(\bm a\times\bm\nabla\times\bm b)\cdot\mathrm d\bm S=\int_V((\bm\nabla\times a)\cdot(\bm\nabla\times b)-\bm a\cdot(\bm\nabla\times\bm\nabla\times\bm b))\mathrm dV