0
$\begingroup$

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: $$ds^2=\sum_i h_i^2 d \zeta_i^2$$

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

Grad

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: $$\nabla f=\begin{pmatrix}\frac{1}{h_1} \frac{\partial f}{\partial \zeta^1} \\ \frac{1}{h_2} \frac{\partial f}{\partial \zeta^2} \\ \frac{1}{h_3} \frac{\partial f}{\partial \zeta^3} \end{pmatrix}$$

Curl

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.

Div

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.

References

  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
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.