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\begin{document}

The laplacion is \( \nabla ^{i}\nabla ^{i} \).

This is simple in a rectangular basis. It can be found in arbitrary
basis by playing with partials. This is pretty messy.

Another method for orthogonal coordinate systems:

Let \( x^{i}(q_{j}) \) and \( q_{j}(x^{i}) \). Then the coordinates are orthogonal if \( e_{i}=\frac{\partial r^{j}}{\partial q_{i}} \) satisfy \( e_{i}e_{j}=\delta _{ij} \). 

The length element in the new coordinate system is \( ds^{2}=\frac{\partial x_{i}}{\partial q_{j}}\frac{\partial x_{i}}{\partial q_{k}}dq_{j}dq_{k}=g_{ij}dq_{i}dq_{j} \). So \( e^{k}_{i}e^{k}_{j}=C_{i}\delta _{ij}=g_{ij} \).

For spherical coordinates, 

\( g_{ij}=\begin{array}{ccc}
1 & 0 & 0\\
0 & r^{2} & 0\\
0 & 0 & r^{2}sin^{2}(\theta )
\end{array}
 \). Define \( h_{i}=\sqrt{g_{ii}} \).

For volume integrals, \( \int dxdydz=\int Jdq_{1}dq_{2}dq_{3} \) where \( J=\epsilon _{ijk}\frac{\partial x_{i}}{\partial q_{1}}\frac{\partial x_{j}}{\partial q_{2}}\frac{\partial x_{k}}{\partial q_{3}}=h_{1}h_{2}h_{3}\epsilon _{ijk}(orthonormalmatrix)=h_{1}h_{2}h_{3} \).

For the gradient: \( \nabla ^{i}\psi =\frac{\partial \psi }{\partial x_{i}}x_{i}= \) \( \frac{\partial \psi }{\partial q_{j}}\frac{\partial q_{j}}{\partial x_{i}}x_{i} \) The inverse matrix is needed so, \( \frac{\partial q_{i}}{\partial x_{k}}=\frac{1}{h_{i}}(e^{j}_{i}x_{k}^{j}) \) so gradient
= \( \frac{1}{h_{j}}\partial _{q_{j}}\psi e_{j} \)

For divergence: 

\end{document}
