Chapter 4: Differentation

### Div, Grad, Curl in Orthogonal Coordinates

What is the gradient in an orthogonal coordinate system? From $$df = \Partial{f}{u}\,du + \Partial{f}{v}\,dv + \Partial{f}{w}\,dw = \grad f\cdot d\rr$$ and $$d\rr = h_u\,du \,\Hat u + h_v\,dv \,\Hat v + h_w\,dw \,\Hat w$$ we see that $$\grad f = \frac{1}{h_u} \Partial{f}{u} \Hat u + \frac{1}{h_v} \Partial{f}{v} \Hat v + \frac{1}{h_w} \Partial{f}{w} \Hat w$$

What about divergence and curl? These formulas can of course be computed using $d$ and $*$ in the usual way, but there is also a shortcut, which we present in the language of vector calculus.

Since $$\grad u = \frac{1}{h_u} \Hat u$$ we see that $$\grad\times \left( \frac{1}{h_u} \Hat u \right) = \grad\times\grad u = 0$$ This means that $\{\frac{1}{h_u}\Hat{u},\frac{1}{h_v}\Hat{v},\frac{1}{h_w}\Hat{w}\}$ is a curl-free basis, and computing the curl is easiest if we use this basis. We have \begin{align} \grad\times\FF &= \grad\times\left( F^u\,\Hat{u}+F^v\,\Hat{v}+F^w\,\Hat{w} \right) \nonumber\\ &= \grad\times\left( (h_u F^u)\frac{\Hat{u}}{h_u} + (h_v F^v)\frac{\Hat{v}}{h_v} + (h_w F^w)\frac{\Hat{w}}{h_w} \right) \nonumber\\ &= \grad(h_u F^u)\times\frac{\Hat{u}}{h_u} + \grad(h_v F^v)\times\frac{\Hat{v}}{h_v} + \grad(h_w F^w)\times\frac{\Hat{w}}{h_w} \end{align} Considering just the first term, we have \begin{align} \grad\times\left(F^u\,\Hat{u}\right) &= \left|\begin{matrix} \Hat u & \Hat v & \Hat w \\ \frac{1}{h_u} \Partial{}{u}\left(h_u F^u\right) & \frac{1}{h_v} \Partial{}{v}\left(h_u F^u\right) & \frac{1}{h_w} \Partial{}{w}\left(h_u F^u\right) \\ \frac{1}{h_u} & 0 & 0 \\ \end{matrix}\right| \nonumber\\ &= \frac{1}{h_u h_v h_w} \left|\begin{matrix} h_u\,\Hat u & h_v\,\Hat v & h_w\,\Hat w \\ \Partial{}{u}\left(h_u F^u\right) & \Partial{}{v}\left(h_u F^u\right) & \Partial{}{w}\left(h_u F^u\right) \\ 1 & 0 & 0 \\ \end{matrix}\right| \nonumber\\ &= \frac{1}{h_u h_v h_w} \left|\begin{matrix} h_u\,\Hat u & h_v\,\Hat v & h_w\,\Hat w \\ \Partial{}{u} & \Partial{}{v} & \Partial{}{w} \\ h_u F^u & 0 & 0 \\ \end{matrix}\right| \end{align} Putting back the missing terms, we obtain the following remarkable formula for curl in an arbitrary orthogonal coordinate system: $$\grad\times\FF = \frac{1}{h_u h_v h_w} \left|\begin{matrix} h_u\,\Hat u & h_v\,\Hat v & h_w\,\Hat w \\ \Partial{}{u} & \Partial{}{v} & \Partial{}{w} \\ h_u F^u & h_v F^v & h_w F^w \\ \end{matrix}\right| \label{orthocurl}$$

We can proceed similarly for the divergence. Since $$\grad\times \left( v \, \grad w \right) = \grad v \times \grad w = \frac{\Hat{v}}{h_v} \times\frac{\Hat{w}}{h_w} = \frac{\Hat{u}}{h_v h_w}$$ we have $$\grad\cdot \left( \frac{\Hat{u}}{h_v h_w} \right) = \grad\cdot\grad\times \left( v \, \grad w \right) = 0$$ Thus, $\{\frac{\Hat{u}}{h_v h_w},\frac{\Hat{v}}{h_w h_u},\frac{\Hat{w}}{h_u h_v}\}$ is a divergence-free basis, which we can use to simplify the computation of the divergence. We have \begin{align} \grad\cdot\FF &= \grad\cdot\left( F^u\,\Hat{u}+F^v\,\Hat{v}+F^w\,\Hat{w} \right) \nonumber\\ &= \grad\cdot\left( (h_v h_w F^u)\frac{\Hat{u}}{h_v h_w} + (h_w h_u F^v)\frac{\Hat{v}}{h_w h_u} + (h_u h_v F^w)\frac{\Hat{w}}{h_u h_v} \right) \nonumber\\ &= \grad(h_v h_w F^u)\cdot\frac{\Hat{u}}{h_v h_w} + \grad(h_w h_u F^v)\cdot\frac{\Hat{v}}{h_w h_u} + \grad(h_u h_v F^w)\cdot\frac{\Hat{w}}{h_u h_v} \nonumber\\ &= \frac{1}{h_u h_v h_w} \left( \Partial{}{u}\left(h_v h_w F^u\right) + \Partial{}{v} \left(h_w h_u F^v\right) + \Partial{}{w} \left(h_u h_v F^w\right) \right) \end{align} which we can rewrite in the form \begin{align} \grad\cdot\FF &= \frac{1}{h_u h_v h_w} \left( \Hat{u}\,\Partial{}{u} + \Hat{v}\,\Partial{}{v} + \Hat{w}\,\Partial{}{w} \right) \nonumber\\ &\qquad \cdot \left( h_v h_w F^u \Hat{u} + h_w h_u F^v \Hat{v} + h_u h_v F^w \Hat{w} \right) \label{orthodiv} \end{align} The Laplacian can now be computed as $\Delta f=\grad\cdot\grad f$.

It is instructive to verify formulas ($\ref{orthocurl}$) and ($\ref{orthodiv}$) using differential forms.