# Del operator apply directly to orthogonal curvilinear coordinate does not match

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 ..

• – user10354138 Dec 9 '18 at 14:00
• 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... – lau Dec 9 '18 at 15:38
• 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 ? – lau Dec 9 '18 at 16:20
• 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 $\mathbf{a}_{u_1}\frac{\partial A_1}{\partial u_1}$. – user10354138 Dec 9 '18 at 16:55