I understand that $\nabla$ in general orthogonal coordinate $(u_1,u_2,u_3)$ as follows:

$$ \nabla=\mathbf{a}_{u_1}\frac{\partial}{h_1\partial u_1}+\mathbf{a}_{u_2}\frac{\partial}{h_2\partial u_2}+\mathbf{a}_{u_3}\frac{\partial}{h_3\partial u_3}\tag{1} $$

It is also given that for general curvilinear coordinates $(u_1,u_2,u_3)$

$$ \nabla\cdot\mathbf{A}=\frac1{h_1h_2h_3}\left[\frac{\partial}{\partial u_1}(h_2h_3A_1)+\frac{\partial}{\partial u_2}(h_1h_3A_2)+\frac{\partial}{\partial u_3}(h_1h_2A_3)\right]\tag{2} $$

Assuming $\mathbf{A}$ is in curvilinear coordinates, apply directly equation (1) to $\mathbf{A}$ does not give me equation (2) .... or am I missing anything?

Clarification on what I am trying to do.... please point out any mistake..

I am taking that A= $\mathbf{a}_{u_1}$$A_1$ + $\mathbf{a}_{u_2}$$A_2$ + $\mathbf{a}_{u_3}$$A_3$

To make thing really simple, let assume A has only 1 term : A=$\mathbf{a}_{u_1}$$A_1$

so $\nabla\cdot\mathbf{A}$ = ($\mathbf{a}_{u_1}\frac{\partial}{h_1\partial u_1}$). ($\mathbf{a}_{u_1}$$A_1$)

which end up as $\nabla\cdot\mathbf{A}$ = ($\frac{\partial A_1}{h_1\partial u_1}$)

But the equation 2 first term shows $\frac1{h_1h_2h_3}[\frac{\partial}{\partial u_1}(h_2h_3A_1) ] $ .. somthing is wrong ..

  • $\begingroup$ See Maxim's comment here. $\endgroup$ – user10354138 Dec 9 '18 at 14:00
  • $\begingroup$ hi @user10354138 , i have seem Maxim comment , but not quite understand , i have make some edit on the post to make it clearer... kindly advise... $\endgroup$ – lau Dec 9 '18 at 15:38
  • $\begingroup$ hi @user10354138 , sorrry for my weak understanding , i thinking that i am dealing with orthogonal coordinates , $\mathbf{a}_{u_1}$ . $\mathbf{a}_{u_1}$ should give you 1 , given they are unit vectors. So you are saying that it is not true ? $\endgroup$ – lau Dec 9 '18 at 16:20
  • $\begingroup$ Your $\mathbf{a}_{u_i}$ is just the unit direction of $\nabla u_i$. However, they are not constant (e.g., polar coordinates where the $\mathbf{e}_r,\mathbf{e}_\theta$ varies with the angular coordinate $\theta$), so $\frac{\partial\mathbf{a}_{u_1}}{\partial u_i}$ are not necessarily zero and thus you cannot just consider the $\mathbf{a}_{u_1}$-component of $\nabla$, and also $\frac{\partial}{\partial u_1}(\mathbf{a}_{u_1}A_1)$ need not be $\mathbf{a}_{u_1}\frac{\partial A_1}{\partial u_1}$. $\endgroup$ – user10354138 Dec 9 '18 at 16:55

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