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拉普拉斯算子的球坐标形式
球坐标与笛卡尔坐标的转换关系为:
{ x = r sin θ cos φ y = r sin θ sin φ z = r cos θ \left\{ \begin {aligned} x &= r\sin\theta\cos\varphi \\ y & = r\sin\theta\sin\varphi \\ z &= r\cos\theta \end{aligned}\right . ⎩⎪⎨⎪⎧xyz=rsinθcosφ=rsinθsinφ=rcosθ
{ r = x 2 + y 2 + z 2 θ = arccos z x 2 + y 2 + z 2 φ = arctan y x \left\{ \begin {aligned} r &= \sqrt{x^2 + y^2 + z^2} \\ \theta &= \arccos\frac{z}{\sqrt{x^2 + y^2 + z^2}} \\ \varphi &= \arctan\frac{y}{x} \end{aligned}\right . ⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧rθφ=x2+y2+z2
=arccosx2+y2+z2
z=arctanxy
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∂ r \partial r ∂r
∂ θ \partial \theta ∂θ
∂ φ \partial \varphi ∂φ
∂ x \partial x ∂x
sin θ cos φ \sin\theta\cos\varphi sinθcosφ
cos θ cos φ r \frac{\cos\theta\cos\varphi}{r} rcosθcosφ
− sin φ r sin θ \frac{-\sin\varphi}{r\sin\theta} rsinθ−sinφ
∂ y \partial y ∂y
sin θ sin φ \sin\theta\sin\varphi sinθsinφ
cos θ sin φ r \frac{\cos\theta\sin\varphi}{r} rcosθsinφ
cos φ r sin θ \frac{\cos\varphi}{r\sin\theta} rsinθcosφ
∂ z \partial z ∂z
cos θ \cos\theta cosθ
− sin θ r \frac{-\sin\theta}{r} r−sinθ
0 0 0
根据微分形式不变性 ∂ ∂ x = ∂ ∂ r ∂ r ∂ x + ∂ ∂ θ ∂ θ ∂ x + ∂ ∂ φ ∂ φ ∂ x \frac{\partial}{\partial x} = \frac{\partial}{\partial r}\frac{\partial r}{\partial x} + \frac{\partial}{\partial \theta}\frac{\partial \theta}{\partial x} + \frac{\partial}{\partial \varphi}\frac{\partial \varphi}{\partial x} ∂x∂=∂r∂∂x∂r+∂θ∂∂x∂θ+∂φ∂∂x∂φ
依上表就有 { ∂ ∂ x = sin θ cos φ ∂ ∂ r + cos θ cos φ r ∂ ∂ θ + − sin φ r sin θ ∂ ∂ φ ∂ ∂ y = sin θ sin φ ∂ ∂ r + cos θ sin φ r ∂ ∂ θ + cos φ r sin θ ∂ ∂ φ ∂ ∂ z = cos θ ∂ ∂ r + − sin θ r ∂ ∂ θ \left\{ \begin {aligned} \frac{\partial}{\partial x} &= \sin\theta\cos\varphi\frac{\partial}{\partial r} + \frac{\cos\theta\cos\varphi}{r}\frac{\partial}{\partial\theta} + \frac{-\sin\varphi}{r\sin\theta}\frac{\partial}{\partial\varphi} \\ \frac{\partial}{\partial y} &= \sin\theta\sin\varphi\frac{\partial}{\partial r} + \frac{\cos\theta\sin\varphi}{r}\frac{\partial}{\partial\theta} + \frac{\cos\varphi}{r\sin\theta}\frac{\partial}{\partial\varphi} \\ \frac{\partial}{\partial z} &= \cos\theta\frac{\partial}{\partial r} + \frac{-\sin\theta}{r}\frac{\partial}{\partial\theta} \end{aligned}\right . ⎩⎪⎪⎪
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