# How to deduce rot and div formulas under spherical or cylinderical coordinat

1. Oct 4, 2005

### Dale12

[SOLVED] How to deduce rot and div formulas under spherical or cylinderical coordinat

It is said that in general, these formulas are:

if $$dl_1=h_1du_1, dl_2=h_2du_2, dl_3=h_3du_3$$

then
$$\nabla \Psi=\frac{1}{h_1}\frac{\partial\Psi}{\partial u_1}\vec{e}_1 + \frac{1}{h_2}\frac{\partial\Psi}{\partial u_2}\vec{e}_2 + \frac{1}{h_3}\frac{\partial\Psi}{\partial u_3}\vec{e}_3$$

with the first sentence, which above is easy to be understood, but how to deduce the two formulas below?

$$\nabla\cdot\vec{f}=\frac{1}{h_2h_3}[\frac{\partial(h_2h_3f_1)}{\partial u_1} + \frac{\partial(h_3h_1f_2)}{\partial u_2} + \frac{\partial(h_1h_2f_3)}{\partial u_3}]$$

$$\nabla\times\vec{f}=\frac{1}{h_2h_3}[\frac{\partial(h_3f_3)}{\partial u_2} - \frac{\partial(h_2f_2)}{\partial u_3}]\vec{e}_1 + \frac{1}{h_3h_1}[\frac{\partial(h_1f_1)}{\partial u_3} - \frac{\partial(h_3f_3)}{\partial u_1}]\vec{e}_2 + \frac{1}{h_1h_2}[\frac{\partial(h_2f_2)}{\partial u_1} - \frac{\partial(h_1f_1)}{\partial u_2}]\vec{e}_3$$
==========================
some one says that
$$h_1=\sqrt{(\frac{\partial x_1}{u_1})^2+(\frac{\partial x_2}{u_1})^2+(\frac{\partial x_3}{u_1})^2}$$
and etc.
But I don't know how to deduce this too.

thanks a lot!

Last edited: Oct 4, 2005
2. Oct 4, 2005

### robphy

Last edited by a moderator: Apr 21, 2017
3. Oct 5, 2005

### Dale12

I know that $$h_1=\sqrt{(\frac{\partial x_1}{u_1})^2+(\frac{\partial x_2}{u_1})^2+(\frac{\partial x_3}{u_1})^2}$$
is just the formula(6) in http://mathworld.wolfram.com/CurvilinearCoordinates.html now.

and by using the limit definitions of the divergence, I can deduce the formula like below:
$$\nabla\cdot\vec{f} = \lim_{\delta V\rightarrow 0} \frac{1}{\delta V}\oint\vec{f}\cdot\vec{n}dS$$
$$= \frac{1}{h_1h_2h_3du_1du_2du_3}[(h_2h_3f_1)^{'}-(h_2h_3f_1)]du_2du_3 + [(h_1h_3f_2)^{'}-(h_1h_3f_2)]du_2du_3 +[(h_1h_2f_3)^{'}-(h_1h_2f_3)]du_2du_3$$
$$= \frac{1}{h_1h_2h_3du_1du_2du_3}\cdot[\frac{\partial(h_2h_3f_1)}{\partial u_1} + \frac{\partial(h_1h_3f_2)}{\partial u_2} +\frac{\partial(h_1h_2f_3)}{\partial u_3}]du_1du_2du_3$$
$$= \frac{1}{h_1h_2h_3}[\frac{\partial(h_2h_3f_1)}{\partial u_1} + \frac{\partial(h_1h_3f_2)}{\partial u_2} +\frac{\partial(h_1h_2f_3)}{\partial u_3}]$$

but the formula of rot seems to be hard because it's different definition form, I am puzzled...

Last edited by a moderator: Apr 21, 2017
4. Oct 5, 2005

### robphy

$$\left(\nabla\times\vec{f}\right) \cdot \hat n = \lim_{\delta A\rightarrow 0} \frac{1}{\delta A}\oint\vec{f}\cdot d\vec{s}$$
finds the component of the curl (rot), parallel to $$\hat n$$.

The calculation follows a similar strategy to the one for the divergence.
Consider the case when $$\hat n=\hat e_1$$.
Consider a small oriented rectangle in the $$e_2e_3$$-plane, with unit normal $$\hat e_1$$ and area $$(h_2du_2)(h_3du_3)$$. Now do the line-integral... analogous to the surface integral you did for the divergence.

5. Oct 10, 2005

### Dale12

IC, thanks to robphy!