Chapter 4: Differentation

### Laplacian in Polar Coordinates

The Laplacian of a function $f$ is defined by $$\Delta f = \grad\cdot\grad f$$ Rewriting the Laplacian in terms of differential forms, we obtain $$\Delta f = {*}d{*}df$$

As an example, consider polar coordinates. We have \begin{align} \Delta f &= {*}d{*}df \nonumber\\ &= {*}d{*} \left(\Partial{f}{r}\,dr+\Partial{f}{\phi}\,d\phi\right) \nonumber\\ &= {*}d{*} \left(\Partial{f}{r}\,dr + \frac{1}{r}\Partial{f}{\phi}\,r\,d\phi\right) \nonumber\\ &= {*}d \left(\Partial{f}{r}\,r\,d\phi - \frac{1}{r}\Partial{f}{\phi}\,dr\right) \nonumber\\ &= {*} \left(d\left(\Partial{f}{r}\,r\right)\wedge d\phi - d\left(\frac{1}{r}\Partial{f}{\phi}\right)\wedge dr\right) \nonumber\\ &= {*} \left( \Partial{}{r}\left(\Partial{f}{r}\,r\right)\,dr\wedge d\phi - \Partial{}{\phi}\left(\frac{1}{r}\Partial{f}{\phi}\right) \,d\phi\wedge dr\right) \nonumber\\ &= {*} \left( \frac{1}{r}\left[ \Partial{}{r}\left(\Partial{f}{r}\,r\right) + \Partial{}{\phi}\left(\frac{1}{r}\Partial{f}{\phi}\right)\right] \,dr\wedge r\,d\phi\right) \nonumber\\ &= \frac{1}{r} \Partial{}{r}\left(r\,\Partial{f}{r}\right) + \frac{1}{r^2}\frac{\partial^2f}{\partial\phi^2} \end{align}

A similar computation can be used to determine the Laplacian in any dimension and signature.