(adsbygoogle = window.adsbygoogle || []).push({}); [SOLVED] How to deduce rot and div formulas under spherical or cylinderical coordinat

It is said that in general, these formulas are:

if [tex]

dl_1=h_1du_1, dl_2=h_2du_2, dl_3=h_3du_3

[/tex]

then

[tex]\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

[/tex]

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

[tex]\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}]

[/tex]

[tex]\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

[/tex]

==========================

some one says that

[tex]

h_1=\sqrt{(\frac{\partial x_1}{u_1})^2+(\frac{\partial x_2}{u_1})^2+(\frac{\partial x_3}{u_1})^2}

[/tex]

and etc.

But I don't know how to deduce this too.

thanks a lot!

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

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