$r = (x, y, z)$、$r = |r|$のとき、以下の(a), (b), (c)を求めよ。ただし、$r \neq 0$とする。 (a) $\nabla(\frac{1}{r^2})$ (b) $\nabla(\frac{r}{r^2})$ (c) $\nabla \times \frac{r}{r^2}$

1. 問題の内容

$r = (x, y, z)$、$r = |r|$のとき、以下の(a), (b), (c)を求めよ。ただし、$r \neq 0$とする。

(a) $\nabla(\frac{1}{r^2})$

(b) $\nabla(\frac{r}{r^2})$

(c) $\nabla \times \frac{r}{r^2}$

2. 解き方の手順

(a) $\nabla(\frac{1}{r^2})$について。

まず、$r^2 = x^2 + y^2 + z^2$なので、$\frac{\partial r}{\partial x} = \frac{x}{r}$となる。同様に、$\frac{\partial r}{\partial y} = \frac{y}{r}$、$\frac{\partial r}{\partial z} = \frac{z}{r}$。

$\frac{\partial}{\partial x}(\frac{1}{r^2}) = -\frac{2}{r^3} \frac{\partial r}{\partial x} = -\frac{2x}{r^4}$

$\frac{\partial}{\partial y}(\frac{1}{r^2}) = -\frac{2}{r^3} \frac{\partial r}{\partial y} = -\frac{2y}{r^4}$

$\frac{\partial}{\partial z}(\frac{1}{r^2}) = -\frac{2}{r^3} \frac{\partial r}{\partial z} = -\frac{2z}{r^4}$

したがって、

$\nabla(\frac{1}{r^2}) = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z})(\frac{1}{r^2}) = (-\frac{2x}{r^4}, -\frac{2y}{r^4}, -\frac{2z}{r^4}) = -\frac{2}{r^4}(x, y, z) = -\frac{2r}{r^4} = -\frac{2r}{r^4}$

(b) $\nabla(\frac{r}{r^2})$について。$\frac{r}{r^2} = \frac{(x, y, z)}{x^2 + y^2 + z^2}$

$\nabla(\frac{r}{r^2}) = \nabla(\frac{x}{r^2}, \frac{y}{r^2}, \frac{z}{r^2})$

$\nabla(\frac{r}{r^2}) = (\frac{\partial}{\partial x}(\frac{x}{r^2}) + \frac{\partial}{\partial y}(\frac{y}{r^2}) + \frac{\partial}{\partial z}(\frac{z}{r^2}))$

$\frac{\partial}{\partial x}(\frac{x}{r^2}) = \frac{1}{r^2} - \frac{2x^2}{r^4}$

$\frac{\partial}{\partial y}(\frac{y}{r^2}) = \frac{1}{r^2} - \frac{2y^2}{r^4}$

$\frac{\partial}{\partial z}(\frac{z}{r^2}) = \frac{1}{r^2} - \frac{2z^2}{r^4}$

$\nabla \cdot (\frac{r}{r^2}) = \frac{3}{r^2} - \frac{2(x^2 + y^2 + z^2)}{r^4} = \frac{3}{r^2} - \frac{2r^2}{r^4} = \frac{3}{r^2} - \frac{2}{r^2} = \frac{1}{r^2}$

(c) $\nabla \times \frac{r}{r^2}$について。

$\frac{r}{r^2} = (\frac{x}{r^2}, \frac{y}{r^2}, \frac{z}{r^2})$

$\nabla \times \frac{r}{r^2} = (\frac{\partial}{\partial y}(\frac{z}{r^2}) - \frac{\partial}{\partial z}(\frac{y}{r^2}), \frac{\partial}{\partial z}(\frac{x}{r^2}) - \frac{\partial}{\partial x}(\frac{z}{r^2}), \frac{\partial}{\partial x}(\frac{y}{r^2}) - \frac{\partial}{\partial y}(\frac{x}{r^2}))$

$\frac{\partial}{\partial y}(\frac{z}{r^2}) = z \cdot (-\frac{2}{r^3} \frac{y}{r}) = -\frac{2yz}{r^4}$

$\frac{\partial}{\partial z}(\frac{y}{r^2}) = y \cdot (-\frac{2}{r^3} \frac{z}{r}) = -\frac{2yz}{r^4}$

$\frac{\partial}{\partial z}(\frac{x}{r^2}) = x \cdot (-\frac{2}{r^3} \frac{z}{r}) = -\frac{2xz}{r^4}$

$\frac{\partial}{\partial x}(\frac{z}{r^2}) = z \cdot (-\frac{2}{r^3} \frac{x}{r}) = -\frac{2xz}{r^4}$

$\frac{\partial}{\partial x}(\frac{y}{r^2}) = y \cdot (-\frac{2}{r^3} \frac{x}{r}) = -\frac{2xy}{r^4}$

$\frac{\partial}{\partial y}(\frac{x}{r^2}) = x \cdot (-\frac{2}{r^3} \frac{y}{r}) = -\frac{2xy}{r^4}$

$\nabla \times \frac{r}{r^2} = (0, 0, 0) = 0$

3. 最終的な答え

(a) $\nabla(\frac{1}{r^2}) = -\frac{2r}{r^4}$

(b) $\nabla \cdot (\frac{r}{r^2}) = \frac{1}{r^2}$

(c) $\nabla \times \frac{r}{r^2} = 0$