3次元極座標でのラプラシアン

math

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円柱座標を経由し，2次元極座標の場合を用いて計算すると，少しは楽．

  • 座標変換 (x,\,y,\,z)\to(r,\,\theta,\,\phi) を

x=r\sin\theta\cos\phi,\;y=r\sin\theta\sin\phi,\;z=r\cos\theta と定める．
\rho=r\sin\theta とおき，座標変換(x,\,y,\,z)\to(\rho,\,\phi,\,z)\to(r,\,\theta,\,\phi) を考える．

x=\rho\cos\phi,\;y=\rho\sin\phi より，
dx=\cos\phi d\rho-\rho\sin\phi d\phi,\; dy=\sin\phi d\rho+\rho\cos\phi d\phi,
\partial_x=\cos\phi\partial_\rho-\frac{\sin\phi}{\rho}\partial_\phi,\; \partial_y=\sin\phi\partial_\rho+\frac{\cos\phi}{\rho}\partial_\phi.
ゆえに，{\partial_x}^2+{\partial_y}^2 ={\partial_\rho}^2+\frac{1}{\rho}\partial_\rho +\frac{1}{\rho^2}{\partial_\phi}^2.
(x,\,y)\to(\rho,\,\phi) は z によらないので，\partial_z の意味は変わらない．
よって，\Delta={\partial_x}^2+{\partial_y}^2+{\partial_z}^2 ={\partial_\rho}^2+\frac{1}{\rho}\partial_\rho +\frac{1}{\rho^2}{\partial_\phi}^2+{\partial_z}^2.

z=r\cos\theta,\;\rho=r\sin\theta より，
\partial_\rho =\sin\theta\partial_r+\frac{\cos\theta}{r}\partial_\theta =\rho(\frac{1}{r}\partial_r+\frac{1}{r^2\tan\theta}\partial_\theta),
{\partial_z}^2+{\partial_\rho}^2 ={\partial_r}^2+\frac{1}{r}\partial_r +\frac{1}{r^2}{\partial_\theta}^2.
(z,\,\rho)\to(r,\,\theta) は \phi によらないので，\partial_\phi の意味は変わらない．
よって，\Delta={\partial_r}^2+\frac{2}{r}\partial_r +\frac{1}{r^2}{\partial_\theta}^2 +\frac{1}{r^2\tan\theta}\partial_\theta +\frac{1}{(r\sin\theta)^2}{\partial_\phi}^2.

  • Hodge star を用いれば次のように計算できる．

dr,\,rd\theta,\,r\sin\theta d\phi は余接空間の正規直交基底ゆえ，
\Delta=\star d\star d
=\frac{1}{r^2\sin\theta}(\partial_r(r^2\sin\theta\partial_r) +\partial_\theta (\frac{r\sin\theta}{r}\partial_\theta) +\partial_\phi (\frac{r}{r\sin\theta}\partial_\phi))
=\frac{1}{r^2}\partial_r(r^2\partial_r) +\frac{1}{r^2\sin\theta}\partial_\theta (\sin\theta\partial_\theta) +\frac{1}{(r\sin\theta)^2}{\partial_\phi}^2.