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ポテンシャル勾配

作成: 2017-12-19
更新: 2017-12-23

$S$

$f$

$S$

$xyz$

$f$

$f^{'} : {\bf x} \longmapsto \left( \frac{\partial f({\bf x})}{\partial x}, \frac{\partial f({\bf x})}{\partial y}, \frac{\partial f({\bf x})}{\partial z} \right) \ \ \ \ ({\bf x} \in S) \\$

$f^{'}$

^(註)

$S$

$f^{'} : {\bf x} \longmapsto \frac{df({\bf x})}{d{\bf x}} \ \ \ \ ({\bf x} \in S)$

$\Delta {\bf x} \rightarrow 0$

　註 :

基底変換

$( {{\bf e}^{'}}_1,\, {{\bf e}^{'}}_2,\, {{\bf e}^{'}}_3 ) = ( {\bf e}_1,\, {\bf e}_2,\, {\bf e}_3 ) \, A \\ \ \ \ \ A = ({a^i}_j)$ に対し，

$( x^1,\, x^2,\, x^3 ) = A\, ( {x^{'}}^1,\, {x^{'}}^2,\, {x^{'}}^3 ) \\ x^i = \sum_{k}{a^i}_k\, {x^{'}}^k \\ \ \ \ \ \ \Longrightarrow \frac{dx^i}{{dx^{'}}^j} = {a^i}_j\\$ よって，

$\left( \frac{df}{d{x^{'}}^1}, \frac{df}{d{x^{'}}^2}, \frac{df}{d{x^{'}}^3} \right) \\ = \left( \frac{df}{dx^1}\frac{dx^1}{d{x^{'}}^1} + \frac{df}{dx^2}\frac{dx^2}{d{x^{'}}^1} + \frac{df}{dx^3}\frac{dx^3}{d{x^{'}}^1}, \,\\ \ \ \ \ \ \ \frac{df}{dx^1}\frac{dx^1}{d{x^{'}}^2} + \frac{df}{dx^2}\frac{dx^2}{d{x^{'}}^2} + \frac{df}{dx^3}\frac{dx^3}{d{x^{'}}^2}, \,\\ \ \ \ \ \ \ \frac{df}{dx^1}\frac{dx^1}{d{x^{'}}^3} + \frac{df}{dx^2}\frac{dx^2}{d{x^{'}}^3} + \frac{df}{dx^3}\frac{dx^3}{d{x^{'}}^3} \right) \\ = \left( \frac{df}{dx^1} {a^1}_1 + \frac{df}{dx^2}\ {a^2}_1 + \frac{df}{dx^3} {a^3}_1, \,\\ \ \ \ \ \ \ \frac{df}{dx^1} {a^1}_2 + \frac{df}{dx^2} {a^2}_2 + \frac{df}{dx^3} {a^3}_2, \,\\ \ \ \ \ \ \ \frac{df}{dx^1} {a^1}_3 + \frac{df}{dx^2} {a^2}_3 + \frac{df}{dx^3} {a^3}_3 \right) \\ = \left( \frac{df}{dx^1}, \frac{df}{dx^2}, \frac{df}{dx^3} \right)\, A \\$ 即ち，ポテンシャル勾配は，共変。