In orthogonal curvilinear coordinates $(\zeta_1, \zeta_2, \zeta_3)$ with scale factors $h_1, h_2, h_3$ the grad, curl and div can be written as: $$\nabla (\psi)=\frac{\hat e_1}{h_1} \frac{\partial \psi}{\partial \zeta_1}+\frac{\hat e_2}{h_2} \frac{\partial \psi}{\partial \zeta_2}+\frac{\hat e_3}{h_3} \frac{\partial \psi}{\partial \zeta_3} $$ $$\nabla \cdot \vec F=\frac{1}{h_1 h_2 h_3}(\frac{\partial (h_2h_3F_1)}{\partial \zeta_1} +\frac{\partial (h_1h_3F_2)}{\partial \zeta_2}+\frac{\partial (h_1h_2F_3)}{\partial \zeta_3})$$ $$\nabla\times \vec F=\frac{\vec e_1}{h_2h_3}(\frac{\partial h_3F_3}{\partial \zeta_2}-\frac{\partial h_2F_2}{\partial \zeta_3} )+\frac{\vec e_2}{h_1h_3}(\frac{\partial h_1F_1}{\partial \zeta_3}-\frac{\partial h_3F_3}{\partial \zeta_1} )+\frac{\vec e_3}{h_2h_1}(\frac{\partial h_2F_2}{\partial \zeta_1}-\frac{\partial h_1F_1}{\partial \zeta_2} )$$ Although not that hard to remember, I was wondering if there was a quick way to derive these equations? (I don't mind how complex the maths is as long as it is quick).

Note: The scale factor for the coordinate $\zeta_i$ is the square root of the coefficient of $d\zeta^2_i$ in the first fundamental such that: $$ds^2=\sum_i h_i^2 d \zeta_i^2$$

  • $\begingroup$ As fas as I'm aware, these formulas hold for orthogonal coordinates, not for general curvilinear coordinates. Also, I think you should define the "scale factors" $h_i$ for the benefit of those less familiar with the topic. $\endgroup$ – joriki Sep 20 '15 at 19:06
  • $\begingroup$ @joriki Thanks for your suggestion, I have made the changes. $\endgroup$ – Quantum spaghettification Sep 20 '15 at 19:15

I think it is most convenient to write each of these expressions in the following form: $$\nabla (\psi)=\begin{pmatrix}\frac{\partial_1}{h_2} \\ \frac{\partial_2}{h_2}\\ \frac{\partial_3}{h_3}\end{pmatrix} \psi$$ $$ \nabla \cdot \vec F=\frac{1}{h_1 h_2 h_3} \left[ \begin{pmatrix} \partial_1 \\ \partial_2 \\ \partial_3 \end{pmatrix} \cdot \begin{pmatrix} F_1 h_2 h_3\\ h_1 F_2 h_3\\ h_1 h_2 F_3 \end{pmatrix}\right]_{\{\hat e_x \hat e_y \hat e_z\}}$$ $$\nabla \times \vec F=\frac{1}{h_1 h_2 h_3} \left| \begin{matrix} h_1 \hat e_1 & h_2 \hat e_2 & h_3 \hat e_3 \\ \partial_1 & \partial_2 & \partial_3 \\ h_1 F_1 & h_2 F_2 & h_3 F_3\end{matrix}\right|$$ Note for the grad the vectors are written in the basis $\hat e_1, \hat e_2 \hat e_3$ but for the div they are written as the Cartesian basis, as indicated.


This answer is very different from that given by me around 2 years ago. As such I am posting it as a new answer rather then editing the old one.

In this answer we will find the grad, div and curl in general orthogonal curvilinear coordinates using differential forms.


We have that: $$df=\frac{\partial f}{\partial \zeta^1} d\zeta^1+\frac{\partial f}{\partial \zeta^2}d\zeta^2+\frac{\partial f}{\partial \zeta^3}d\zeta^3$$ To read the grad of from this we need to include scale factors with the differentials (we will see that this is a recuring theme): $$df=\left( \frac{1}{h_1} \frac{\partial f}{\partial \zeta^1}\right) (h_1 d\zeta^1)+\left( \frac{1}{h_2}\frac{\partial f}{\partial \zeta^2}\right) (h_2d\zeta^2)+\left(\frac{1}{h_3} \frac{\partial f}{\partial \zeta^3}\right) (h_3d\zeta^3)$$ From which we can read of that we have: $$\nabla f=\begin{pmatrix}\frac{1}{h_1} \frac{\partial f}{\partial \zeta^1} \\ \frac{1}{h_2} \frac{\partial f}{\partial \zeta^2} \\ \frac{1}{h_3} \frac{\partial f}{\partial \zeta^3} \end{pmatrix}$$


Here we start with: $$\Phi= F_1 h_1 d\zeta^1+ F_2 h_2 d\zeta^2+ F_3 h_3 d\zeta^3$$ (again note the scale factors with the differentials)

We then take the exterior derivative of this: $$d \Phi=\left( \frac{\partial (h_3 F_3)}{\partial \zeta^2} - \frac{\partial (h_2F_2 )}{\partial \zeta^3}\right) d\zeta^2 \wedge d\zeta^3+\left( \frac{\partial (h_1 F_1)}{\partial \zeta^3} - \frac{\partial (h_3F_3)}{\partial \zeta^1}\right) d\zeta^3 \wedge d\zeta^1+\left( \frac{\partial (h_2 F_2)}{\partial \zeta^1} - \frac{\partial (h_1F_1 )}{\partial \zeta^2}\right) d\zeta^1 \wedge d\zeta^2$$ where I have collected relevant terms. Including the scale factors:

$$d \Phi=\frac{1}{h_2h_3}\left( \frac{\partial (h_3 F_3)}{\partial \zeta^2} - \frac{\partial (h_2F_2 )}{\partial \zeta^3}\right) (h_2 h_3 d\zeta^2 \wedge d\zeta^3)+\frac{1}{h_3h_1}\left( \frac{\partial (h_1 F_1)}{\partial \zeta^3} - \frac{\partial (h_3F_3)}{\partial \zeta^1}\right)(h_3h_1 d\zeta^3 \wedge d\zeta^1)+\frac{1}{h_1h_2}\left( \frac{\partial (h_2 F_2)}{\partial \zeta^1} - \frac{\partial (h_1F_1 )}{\partial \zeta^2}\right)(h_1h_2 d\zeta^1 \wedge d\zeta^2)$$ from which we can read of the coefficients of the curl to give the expression in the question.


Here we start with: $$\Omega=F_1 h_2 h_3 d\zeta^2 \wedge d \zeta^3+ F_2 h_3 h_1 d \zeta^3 \wedge d \zeta^1+ F_3 h_1h_2 d\zeta^1 \wedge d \zeta^2$$ Take the exterior derivative to get: $$d\Omega=\left( \frac{\partial (F_1h_2 h_3)}{\partial \zeta^1}+\frac{\partial (F_2h_3 h_1)}{\partial \zeta^2}+\frac{\partial(F_3 h_1 h_2)}{\partial \zeta^3}\right) d\zeta^1 \wedge d \zeta^2 \wedge d\zeta^3$$ Rewriting with the correct scale factors on the differentials we get: $$d\Omega=\frac{1}{h_1 h_2 h_3} \left( \frac{\partial (F_1h_2 h_3)}{\partial \zeta^1}+\frac{\partial (F_2h_3 h_1)}{\partial \zeta^2}+\frac{\partial(F_3 h_1 h_2)}{\partial \zeta^3}\right) (h_1 h_2 h_3 d\zeta^1 \wedge d \zeta^2 \wedge d\zeta^3)$$ from which we can read of the divergence.


  1. The answer https://math.stackexchange.com/a/1302344/203397 by Liviu Nicolaescu.
  2. Section 1.9 of the pdf: https://www.cefns.nau.edu/~schulz/diff.pdf

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