合成関数の偏微分の応用として，（デカルト座標で定義された）3次元のラプラシアン $\nabla^2 = \dfrac{\partial^2}{\partial x^2} + \dfrac{\partial^2}{\partial y^2}+ \dfrac{\partial^2}{\partial z^2}$ を極座標 $r, \theta, \phi$ を使って表してみる。練習問題にしようかと思ったが単に計算量が多くなるだけなので，備忘録として。
3次元のラプラシアン
3次元のラプラシアンはデカルト座標 $x, y, z$ の2階偏微分で以下のように定義される：
$$\nabla^2 \equiv \dfrac{\partial^2}{\partial x^2} + \dfrac{\partial^2}{\partial y^2}+ \dfrac{\partial^2}{\partial z^2}$$
3次元極座標
3次元デカルト座標 \(x, y, z\) から極座標 \(r, \theta, \phi\) への座標変換（つまり元の座標 \(x, y, z\) を使って新しい座標 \(r, \theta, \phi\) を表す式）は
\begin{eqnarray}
r &=& \sqrt{x^2 + y^2 + z^2} \\
\theta &=& \tan^{-1} \frac{\sqrt{x^2 + y^2}}{z}\\
\phi &=& \tan^{-1} \frac{y}{x}
\end{eqnarray}
その逆変換（つまり新しい座標 \(r, \theta, \phi\) を使って元の座標 \(x, y, z\) を表す式）は，
\begin{eqnarray}
x &=& r \sin\theta \cos\phi\\
y &=& r \sin\theta \sin\phi\\
z &=& r \cos\theta
\end{eqnarray}
1階偏微分を極座標で書き直す
極座標 \(r, \theta, \phi\) を \(x, y, z\) で1階偏微分した結果を極座標で表す。
\(r(x, y, z) \) の偏微分は，$u \equiv x^2 + y^2 + z^2$ とおいて合成関数の偏微分の要領で…
\begin{eqnarray}
\frac{\partial r}{\partial x} &=& \frac{d}{du} \sqrt{u} \cdot \frac{\partial u}{\partial x} \\
&=& \frac{1}{2} \frac{1}{\sqrt{u}} \cdot 2 x \\
&=&\frac{x}{r} = \sin\theta \cos\phi \\
\frac{\partial r}{\partial y} &=& \frac{y}{r} = \sin\theta \sin\phi \\
\frac{\partial r}{\partial z} &=& \frac{z}{r} = \cos\theta
\end{eqnarray}
\(\theta(x, y, z) \) の偏微分は，$u \equiv \dfrac{\sqrt{x^2+y^2}}{z}$ とおいて合成関数の偏微分の要領で…
\begin{eqnarray}
\frac{\partial \theta}{\partial x} &=& \frac{d}{du} \tan^{-1} {u} \cdot \frac{\partial u}{\partial x} \\
&=& \frac{1}{1+u^2} \frac{1}{z}\cdot \frac{x}{\sqrt{x^2+y^2}} \\
&=&\frac{z}{r^2} \cdot \frac{r \sin\theta \cos\phi}{r \sin\theta}\\
&=& \frac{\cos\theta \cos\phi}{r} \\
\frac{\partial \theta}{\partial y} &=& \frac{\cos\theta \sin\phi}{r} \\
\frac{\partial \theta}{\partial z} &=& -\frac{\sin\theta}{r}
\end{eqnarray}
\(\phi(x, y, z) \) の偏微分は，$u \equiv \dfrac{y}{x}$ とおいて合成関数の偏微分の要領で…
\begin{eqnarray}
\frac{\partial \phi}{\partial x} &=& \frac{d}{du} \tan^{-1} {u} \cdot \frac{\partial u}{\partial x} \\
&=& \frac{1}{1+u^2} \cdot \left( -\frac{y}{x^2} \right)\\
&=&-\frac{y}{x^2 + y^2}\\
&=& -\frac{\sin\phi}{r \sin\theta} \\
\frac{\partial \phi}{\partial y} &=& \frac{\cos\phi}{r \sin\theta} \\
\frac{\partial \phi}{\partial z} &=& 0
\end{eqnarray}
従って，
\begin{eqnarray}
\frac{\partial }{\partial x} &=&
\frac{\partial r}{\partial x}\frac{\partial}{\partial r}
+\frac{\partial\theta}{\partial x} \frac{\partial}{\partial \theta}
+\frac{\partial \phi}{\partial x}\frac{\partial}{\partial \phi} \\
&=& \sin\theta \cos\phi \frac{\partial}{\partial r}
+\frac{\cos\theta \cos\phi}{r} \frac{\partial}{\partial \theta}
-\frac{\sin\phi}{r \sin\theta} \frac{\partial}{\partial \phi} \\
\frac{\partial }{\partial y} &=&
\sin\theta \sin\phi \frac{\partial}{\partial r}
+\frac{\cos\theta \sin\phi}{r} \frac{\partial}{\partial \theta}
+\frac{\cos\phi}{r \sin\theta} \frac{\partial}{\partial \phi} \\
\frac{\partial }{\partial z} &=&
\cos\theta \frac{\partial}{\partial r}
-\frac{\sin\theta}{r} \frac{\partial}{\partial \theta}
\end{eqnarray}
2階偏微分を極座標で書き直す
\begin{eqnarray}
\frac{\partial^2}{\partial x^2} &=& \left(\sin\theta \cos\phi \frac{\partial}{\partial r}
+\frac{\cos\theta \cos\phi}{r} \frac{\partial}{\partial \theta}
-\frac{\sin\phi}{r \sin\theta} \frac{\partial}{\partial \phi} \right)^2 \\
&=& \left(\sin\theta \cos\phi \frac{\partial}{\partial r} \right)
\left(\sin\theta \cos\phi \frac{\partial}{\partial r} \right) \\
&& +\left(\sin\theta \cos\phi \frac{\partial}{\partial r} \right)
\left(\frac{\cos\theta \cos\phi}{r} \frac{\partial}{\partial \theta} \right) \\
&& +\left(\sin\theta \cos\phi \frac{\partial}{\partial r} \right)
\left( -\frac{\sin\phi}{r \sin\theta} \frac{\partial}{\partial \phi}\right) \\
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&& +\left(\frac{\cos\theta \cos\phi}{r} \frac{\partial}{\partial \theta} \right)
\left(\sin\theta \cos\phi \frac{\partial}{\partial r} \right) \\
&& + \left(\frac{\cos\theta \cos\phi}{r} \frac{\partial}{\partial \theta} \right)
\left(\frac{\cos\theta \cos\phi}{r} \frac{\partial}{\partial \theta} \right) \\
&& + \left(\frac{\cos\theta \cos\phi}{r} \frac{\partial}{\partial \theta} \right)
\left( -\frac{\sin\phi}{r \sin\theta} \frac{\partial}{\partial \phi}\right) \\
%
&& +\left( -\frac{\sin\phi}{r \sin\theta} \frac{\partial}{\partial \phi}\right)
\left(\sin\theta \cos\phi \frac{\partial}{\partial r} \right) \\
&& + \left( -\frac{\sin\phi}{r \sin\theta} \frac{\partial}{\partial \phi}\right)
\left(\frac{\cos\theta \cos\phi}{r} \frac{\partial}{\partial \theta} \right) \\
&& + \left( -\frac{\sin\phi}{r \sin\theta} \frac{\partial}{\partial \phi}\right)
\left( -\frac{\sin\phi}{r \sin\theta} \frac{\partial}{\partial \phi}\right) \\
&=& \sin^2\theta \cos^2\phi \frac{\partial^2}{\partial r^2}
+ \frac{\cos^2\theta \cos^2\phi}{r^2} \frac{\partial^2}{\partial \theta^2}
+ \frac{\sin^2\phi}{r^2 \sin^2\theta} \frac{\partial^2}{\partial \phi^2}\\
&& + \frac{2}{r} \sin\theta\cos\theta \cos^2\phi \frac{\partial^2}{\partial r \partial\theta} \\
&& -\frac{2}{r} \sin\phi \cos\phi \frac{\partial^2}{\partial r \partial\phi} \\
&& -\frac{2}{r^2} \frac{\cos\theta}{\sin\theta} \sin\phi \cos\phi \frac{\partial^2}{\partial \theta \partial\phi} \\
&& +\frac{1}{r} \left(\cos^2\theta \cos^2\phi + \sin^2\phi \right) \frac{\partial}{\partial r} \\
&& + \frac{1}{r^2} \left(\frac{\cos\theta}{\sin\theta} \sin^2\phi -2 \sin\theta \cos\theta \cos^2\phi \right)\frac{\partial}{\partial \theta} \\
&& +\frac{2 \sin\phi \cos\phi}{r^2 \sin^2\theta} \frac{\partial}{\partial \phi} \\
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\frac{\partial^2}{\partial y^2} &=&\sin^2\theta \sin^2\phi \frac{\partial^2}{\partial r^2}
+ \frac{\cos^2\theta \sin^2\phi}{r^2} \frac{\partial^2}{\partial \theta^2}
+ \frac{\cos^2\phi}{r^2 \sin^2\theta} \frac{\partial^2}{\partial \phi^2}\\
&& + \frac{2}{r} \sin\theta\cos\theta \sin^2\phi \frac{\partial^2}{\partial r \partial\theta} \\
&& +\frac{2}{r} \sin\phi \cos\phi \frac{\partial^2}{\partial r \partial\phi} \\
&& +\frac{2}{r^2} \frac{\cos\theta}{\sin\theta} \sin\phi \cos\phi \frac{\partial^2}{\partial \theta \partial\phi} \\
&& +\frac{1}{r} \left(\cos^2\theta \sin^2\phi + \cos^2\phi \right) \frac{\partial}{\partial r} \\
&& + \frac{1}{r^2} \left(\frac{\cos\theta}{\sin\theta} \cos^2\phi -2 \sin\theta \cos\theta \sin^2\phi \right)\frac{\partial}{\partial \theta} \\
&& -\frac{2 \sin\phi \cos\phi}{r^2 \sin^2\theta} \frac{\partial}{\partial \phi} \\
%%
\frac{\partial^2}{\partial z^2}
&=&
\cos^2\theta \frac{\partial^2}{\partial r^2}
+ \frac{1}{r^2} \sin^2\theta \frac{\partial^2}{\partial \theta^2} \\
&& -\frac{2}{r} \sin\theta \cos\theta \frac{\partial^2}{\partial r \partial\theta} \\
&& + \frac{1}{r} \sin^2\theta \frac{\partial}{\partial r} + \frac{2}{r^2} \sin\theta \cos\theta \frac{\partial}{\partial\theta}
\end{eqnarray}
3次元のラプラシアンを極座標で
最終的に，
\begin{eqnarray}
\nabla^2 &=& \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2} \\
&=& \frac{\partial^2}{\partial r^2} +\frac{2}{r} \frac{\partial}{\partial r}
+\frac{1}{r^2} \frac{\partial^2}{\partial\theta^2} + \frac{1}{r^2} \frac{\cos\theta}{\sin\theta}\frac{\partial}{\partial \theta}
+ \frac{1}{r^2 \sin^2\theta} \frac{\partial^2}{\partial \phi^2} \\
&=& \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial}{\partial r}\right)
+\frac{1}{r^2\sin\theta} \frac{\partial}{\partial\theta} \left(\sin\theta\frac{\partial}{\partial\theta} \right)
+ \frac{1}{r^2 \sin^2\theta} \frac{\partial^2}{\partial \phi^2}
\end{eqnarray}
参考：計量テンソルを使って表す
将来，「ベクトルの発散やラプラシアンを共変微分で理解する」のページで計量テンソルや共変微分を学んだあとでこのページを見かえしてみると，3次元極座標における計量テンソル $g_{ij}$ の成分が
$$
g_{ij} = \left(\begin{array}{ccc}
g_{rr} & g_{r\theta} & g_{r\phi} \\
g_{\theta r} & g_{\theta\theta} & g_{\theta\phi} \\
g_{\phi r} & g_{\phi\theta} & g_{\phi\phi}
\end{array}\right)
= \left(\begin{array}{ccc}
1 & 0& 0 \\
0 & r^2& 0 \\
0 & 0 & r^2 \sin^2\theta
\end{array}\right)
$$
であり，計量テンソルを行列とみなしたときの行列式 $g$ は
$$g \equiv \det (g_{ij}) = r^4 \sin^2\theta, \ \therefore\ \sqrt{g} = r^2 \sin\theta$$
また，計量テンソルの逆行列 $g^{ij}$ は
$$
g^{ij} = \left(\begin{array}{ccc}
g^{rr} & g^{r\theta} & g^{r\phi} \\
g^{\theta r} & g^{\theta\theta} & g^{\theta\phi} \\
g^{\phi r} & g^{\phi\theta} & g^{\phi\phi}
\end{array}\right)
= \left(\begin{array}{ccc}
1 & 0& 0 \\
0 & \dfrac{1}{r}& 0 \\
0 & 0 & \dfrac{1}{r^{2} \sin^{2}\theta}
\end{array}\right)$$
これらを使うと3次元のラプラシアンが
\begin{eqnarray}
\nabla^2 &=& \frac{1}{\sqrt{g}} \partial_r \left( \sqrt{g} \ r^{rr}\ \partial_r\right)
+ \frac{1}{\sqrt{g}} \partial_{\theta} \left( \sqrt{g} \ r^{{\theta}{\theta}}\ \partial_{\theta}\right)
+ \frac{1}{\sqrt{g}} \partial_{\phi} \left( \sqrt{g} \ r^{{\phi} {\phi} } \ \partial_{\phi} \right)\\
&=& \frac{1}{r^2 \sin\theta} \partial_r \left( r^2 \sin\theta \ \partial_r\right)
+ \frac{1}{r^2 \sin\theta} \partial_{\theta} \left( \frac{r^2 \sin\theta}{ \ r^{2}}\ \partial_{\theta}\right)
+ \frac{1}{r^2 \sin\theta}\partial_{\phi} \left( \frac{r^2 \sin\theta}{r^{2 } \sin^{2}\theta}\ \partial_{\phi} \right)\\
&=& \frac{1}{r^2 } \partial_r \left( r^2 \ \partial_r\right)
+ \frac{1}{r^2 \sin\theta} \partial_{\theta} \left( \sin\theta \ \partial_{\theta}\right)
+ \frac{1}{r^2 \sin^2\theta}\partial_{\phi} \left( \partial_{\phi} \right)
\end{eqnarray}
と書けているのだなぁということがわかるようになると思う。