勾配・発散・回転



$ラプラシアンを求めることが目的ですので、詳しい説明や物理的な意味は別の機会にします。$
$ここで出てくる関数はすべて、少なくとも2回微分可能とします。$

$(1)\ \ 勾配$

$スカラー関数 \psi (x,y,z)に対し、ベクトル \big(\cfrac{\partial \psi}{\partial x},\ \cfrac{\partial \psi}{\partial y},\ \cfrac{\partial \psi}{\partial z}\big) を \psi(x,y,z)の勾配(gradient)といい、$

$\nabla \psi \ \ または \ \ \mathrm{grad} \ \ \psi \ \ とかきます。\nabla はナブラと読みます。$

$\hspace{4em} 勾配 \qquad \nabla \psi = \cfrac{\partial \psi}{\partial x}\vec i+ \cfrac{\partial \psi}{\partial y} \vec J + \cfrac{\partial \psi}{\partial z}\vec k$


　
$\psi(x,y,z)=c \ \ (cは定数)\ は曲面を表しますが、$
$この曲面上で点P(x,y,z)を通る$
$曲線\ \ \vec r=(x(t),\ y(t),\ z(t)) \ \ を考えると$

$\qquad \cfrac{d\psi}{dt}=\cfrac{\partial \psi}{\partial x}\cfrac{dx}{dt}+ \cfrac{\partial \psi}{\partial y}\cfrac{dy}{dt}+\cfrac{\partial \psi}{\partial z}\cfrac{dz}{dt}=0$

$\qquad \nabla \psi \cdot \cfrac{d\boldsymbol{r}}{dt}=0$
$したがって　\nabla \psi は点Pを通る曲面上の任意の曲線の接線ベクトルに直交するから、$
$Pにおけるこの曲面の接平面に垂直な法線ベクトルとなります。$


$(2)\ \ 発散$

$関数 \ \ A_x(x,y,z),\ A_y(x,y,z),\ A_z(x,y,z) \ を成分にもつ、ベクトル関数$

$\qquad \boldsymbol{A}(x,y,z)=A_x\vec{i}+A_y\vec{j}+A_z\vec{k} \quad を考えます。$

$\quad \cfrac{\partial A_x}{\partial x}+\cfrac{\partial A_y}{\partial y}+\cfrac{\partial A_z}{\partial z} \ \ を \boldsymbol{A}の発散(divergence)といい、\nabla \cdot \boldsymbol{A} \ \ または \ \ \mathrm{div} \ \boldsymbol{A} \ \ とかきます。$

$\hspace{4em} 発散 \qquad \nabla \cdot \boldsymbol{A}=\cfrac{\partial A_x}{\partial x}+\cfrac{\partial A_y}{\partial y}+\cfrac{\partial A_z}{\partial z} $

$とくに　\boldsymbol{A}=\nabla \psi では、A_x=\cfrac{\partial \psi}{\partial x},\quad A_y=\cfrac{\partial \psi}{\partial y},\quad A_z=\cfrac{\partial \psi}{\partial z} \ \ だから$

$\qquad \nabla \cdot \boldsymbol{A} =\cfrac{\partial}{\partial x}\big(\cfrac{\partial \psi}{\partial x}\big)+\cfrac{\partial}{\partial y}\big(\cfrac{\partial \psi}{\partial y}\big)+\cfrac{\partial}{\partial z}\big(\cfrac{\partial \psi}{\partial z}\big) =\cfrac{\partial ^2 \psi}{\partial x^2}+\cfrac{\partial^2 \psi}{\partial y^2}+\cfrac{\partial^2 \psi}{\partial z^2}$

$左辺は \ \ \nabla \cdot \boldsymbol{A}=\nabla \cdot (\nabla \psi) \ \ となり、これを \ \ \nabla ^2 \psi \ \ とかき、ラプラシアンといいます。$

$\qquad ラプラシアン　\nabla ^2 \psi =\nabla \cdot (\nabla \psi)=\cfrac{\partial ^2 \psi}{\partial x^2}+\cfrac{\partial^2 \psi}{\partial y^2}+\cfrac{\partial^2 \psi}{\partial z^2}$



$(3)\ \ 回転$

$ベクトル関数 \ \ \boldsymbol{A}(x,y,z)=A_x\vec{i}+A_y\vec{j}+A_z\vec{k}\ \ に対して$

$\qquad \cfrac{\partial A_z}{\partial y}-\cfrac{\partial A_y}{\partial z},\quad \cfrac{\partial A_x}{\partial z}-\cfrac{\partial A_z}{\partial x},\quad \cfrac{\partial A_y}{\partial x}-\cfrac{\partial A_x}{\partial y}$

$を成分とするベクトルを \ \ \boldsymbol{A}\ \ の回転(rotation)といい、\nabla \times \boldsymbol{A} \ \ または \ \ \mathrm{rot}\ \boldsymbol{A} \ \ とかきます。$

$\qquad 回転 \qquad \nabla \times \boldsymbol{A}=\big(\cfrac{\partial A_z}{\partial y}-\cfrac{\partial A_y}{\partial z}\big)\vec i+\ \big(\cfrac{\partial A_x}{\partial z}-\cfrac{\partial A_z}{\partial x}\big )\vec j+\ \big (\cfrac{\partial A_y}{\partial x}-\cfrac{\partial A_x}{\partial y}\big )\vec k$


$各成分が込みいっていますが、ベクトルの外積の行列式をつかって$
\[ \nabla \times \boldsymbol{A} = \left| \begin{array}{ccc} \vec i & \vec j & \vec k \\ \cfrac{\partial }{\partial x} & \cfrac{\partial }{\partial y} & \cfrac{\partial }{\partial z} \\ A_x & A_y & A_z \end{array} \right| \ \ とあらわせます。\]

$(4)\ \ 勾配・発散・回転の公式$

(i)$\ \ \nabla \cdot (\psi \ \boldsymbol{A})=(\nabla \psi)\cdot \boldsymbol{A} + \psi(\nabla \cdot \boldsymbol{A})$

$(証明)$
\begin{eqnarray*} \nabla \cdot (\psi \ \boldsymbol{A}) &=&\nabla \cdot (\psi A_x\vec i+\psi A_y\vec j+ \psi A_z \vec k)\\ \\ &=&\cfrac{\partial }{\partial x}(\psi A_x)+\cfrac{\partial }{\partial y}(\psi A_y)+\cfrac{\partial }{\partial z}(\psi A_z)\\ \\ &=&\cfrac{\partial \psi}{\partial x}A_x+\psi \cfrac{\partial A_x}{\partial x}+\cfrac{\partial \psi}{\partial y}A_y+\psi \cfrac{\partial A_y}{\partial y}+ \cfrac{\partial \psi}{\partial z}A_z+\psi \cfrac{\partial A_z}{\partial z}\\ \\ &=&\big(\cfrac{\partial \psi}{\partial x}A_x+\cfrac{\partial \psi}{\partial y}A_y+\cfrac{\partial \psi}{\partial z}A_z \big)+ \psi \big(\cfrac{\partial A_x}{\partial x}+ \cfrac{\partial A_y}{\partial y}+\cfrac{\partial A_z}{\partial z}\big)\\ \\ &=&(\nabla \psi) \cdot \boldsymbol{A}+\psi (\nabla \cdot \boldsymbol{A})\\ \end{eqnarray*}
(ii)$\ \ \nabla \cdot (\ \boldsymbol{A} \times \boldsymbol{B})=\boldsymbol{B} \cdot (\nabla \times \boldsymbol{A})-\boldsymbol{A} \cdot (\nabla \times \boldsymbol{B})$

$(証明)$
$\hspace{3em}\nabla \cdot (\ \boldsymbol{A} \times \boldsymbol{B})$
$\hspace{3em}=\cfrac{\partial }{\partial x}(A_yB_z-A_zB_y)+\cfrac{\partial }{\partial y}(A_zB_x-A_xB_z)+\cfrac{\partial }{\partial z}(A_xB_y-A_yB_x)$

$\hspace{3em}=\big(\cfrac{\partial A_y}{\partial x}B_z+A_y\cfrac{\partial B_z}{\partial x}-\cfrac{\partial A_z}{\partial x}B_y -A_z\cfrac{\partial B_y}{\partial x}\big) +\big(\cfrac{\partial A_z}{\partial y}B_x + A_z\cfrac{\partial B_x}{\partial y}-\cfrac{\partial A_x}{\partial y}B_z -A_x\cfrac{\partial B_z}{\partial y}\big)$

$\hspace{8em} +\big(\cfrac{\partial A_x}{\partial z}B_y + A_x\cfrac{\partial B_y}{\partial z}-\cfrac{\partial A_y}{\partial z}B_x -A_y\cfrac{\partial B_x}{\partial z}\big)$

$\hspace{3em}=B_x\big(\cfrac{\partial A_z}{\partial y}-\cfrac{\partial A_y}{\partial z}\big) +B_y\big(\cfrac{\partial A_x}{\partial z}-\cfrac{\partial A_z}{\partial x}\big) +B_z\big(\cfrac{\partial A_y}{\partial x}-\cfrac{\partial A_x}{\partial y}\big)$

$\hspace{8em}-A_x\big(\cfrac{\partial B_z}{\partial y}-\cfrac{\partial B_y}{\partial z}\big) -A_y\big(\cfrac{\partial B_x}{\partial z}-\cfrac{\partial B_z}{\partial x}\big) -A_z\big(\cfrac{\partial B_y}{\partial x}-\cfrac{\partial B_x}{\partial y}\big)$

$\hspace{3em}=\boldsymbol{B} \cdot (\nabla \times \boldsymbol{A})-\boldsymbol{A} \cdot (\nabla \times \boldsymbol{B})$


(iii)$\ \ \nabla \times (\nabla \ \psi)=\vec 0$

$(証明)$
\begin{eqnarray*} \nabla \times (\nabla \ \psi) &=&\nabla \times \big(\cfrac{\partial \psi}{\partial x}\vec i+ \cfrac{\partial \psi}{\partial y} \vec J + \cfrac{\partial \psi}{\partial z}\vec k \big)\\ \\ &=&\big(\cfrac{\partial }{\partial y}\cfrac{\partial \psi}{\partial z}-\cfrac{\partial }{\partial z}\cfrac{\partial \psi}{\partial y}\big)\vec i +\big(\cfrac{\partial }{\partial z}\cfrac{\partial \psi}{\partial x}-\cfrac{\partial }{\partial x}\cfrac{\partial \psi}{\partial z}\big)\vec j +\big(\cfrac{\partial }{\partial x}\cfrac{\partial \psi}{\partial y}-\cfrac{\partial }{\partial y}\cfrac{\partial \psi}{\partial x}\big)\vec k\\ \\ &=&\vec 0\\ \end{eqnarray*}


　

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