Reading off grad, div, curl, and the Laplacian in curvilinear coordinates
Every curvilinear vector-operator formula follows the same "divide or multiply by scale factors" pattern.
Don't memorize four separate formulas -- memorize the ONE pattern (divide a partial derivative by the scale factor for the gradient; the divergence-of-a-gradient construction for the Laplacian) and look up the scale factors for whatever system the problem actually uses.
The scale factors themselves
For a coordinate map \mathbf{r}(u_1,u_2,u_3), the scale factor for u_i is h_i = \left\|\dfrac{\partial\mathbf{r}}{\partial u_i}\right\| — the physical length of a unit step in u_i. The line element (squared distance for a small step (du_1,du_2,du_3)) in ORTHOGONAL curvilinear coordinates (where the \partial\mathbf{r}/\partial u_i are mutually perpendicular) is ds^2 = h_1^2\,du_1^2 + h_2^2\,du_2^2 + h_3^2\,du_3^2.
Show proof
d\mathbf{r} = \sum_i \dfrac{\partial\mathbf{r}}{\partial u_i}\,du_i by the ordinary multivariable chain rule (chapter 00). Then ds^2 = d\mathbf{r}\cdot d\mathbf{r} = \sum_{i,j} \dfrac{\partial\mathbf{r}}{\partial u_i}\cdot\dfrac{\partial\mathbf{r}} {\partial u_j}\,du_i\,du_j. Orthogonality means every cross term (i\neq j) has \dfrac{\partial\mathbf{r}}{\partial u_i}\cdot\dfrac{\partial\mathbf{r}}{\partial u_j}=0, leaving only the diagonal terms \left\|\dfrac{\partial\mathbf{r}}{\partial u_i}\right\|^2 du_i^2 = h_i^2\,du_i^2.
The gradient's pattern
For orthogonal curvilinear coordinates with scale factors h_1,h_2,h_3 and unit vectors \hat{\mathbf e}_i along each coordinate direction, \nabla f = \frac{1}{h_1}\frac{\partial f}{\partial u_1}\hat{\mathbf e}_1 + \frac{1}{h_2}\frac{\partial f}{\partial u_2}\hat{\mathbf e}_2 + \frac{1}{h_3}\frac{\partial f}{\partial u_3}\hat{\mathbf e}_3.
Example
Show proof
\nabla f is defined by df = \nabla f\cdot d\mathbf{r} for every displacement d\mathbf{r} (the total-derivative characterization, chapter 02). Writing d\mathbf{r} = \sum_i h_i\,du_i\,\hat{\mathbf e}_i (each coordinate step scaled to physical length by its own h_i, the previous theorem) and df = \sum_i \dfrac{\partial f}{\partial u_i}\,du_i (ordinary chain rule), matching the \hat{\mathbf e}_i component on both sides of df=\nabla f\cdot d\mathbf{r} forces the \hat{\mathbf e}_i-component of \nabla f to be \dfrac{1}{h_i} \dfrac{\partial f}{\partial u_i} — the h_i in the denominator exactly compensates for d\mathbf{r}'s own h_i factor.
The Laplacian's pattern
Divergence of the gradient -- the same "other two scale factors, over this one" pattern in each of the three terms.
Show proof
\nabla^2 f = \nabla\cdot(\nabla f) — apply the divergence formula directly to the vector field \nabla f, whose curvilinear components are \dfrac{1}{h_i}\dfrac{\partial f}{\partial u_i} (the gradient formula). Substituting F_i = \dfrac{1}{h_i}\dfrac{\partial f}{\partial u_i} into \dfrac{1}{h_1h_2h_3}\dfrac{\partial}{\partial u_1}(h_2h_3F_1)+\cdots turns h_2h_3F_1 into h_2h_3\cdot \dfrac{1}{h_1}\dfrac{\partial f}{\partial u_1} = \dfrac{h_2h_3}{h_1}\dfrac{\partial f}{\partial u_1}, and likewise for the other two terms — exactly the stated formula.
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