Divergence of a cross product help

[simba_lk]

I need to prove the identity: [itex]\nabla[/itex]([itex]\vec{A} \times \vec{B}[/itex])=[itex]\vec{B} \bullet[/itex]([itex]\nabla \times \vec{A}[/itex]) - [itex]\vec{A} \bullet[/itex]( [itex]\nabla \times \vec{B}[/itex])

I need to prove for an arbitrary coordinate system, meaning I have scaling factors.

The proof should be quite straight forward if you use the levi chevita symbol but so far this is what i have:

[itex]\nabla[/itex]([itex]\vec{A} \times \vec{B}[/itex])=[itex]\frac{1}{h_{1}h_{2}h_{3}} \frac{\partial}{\partial x_{i}}[/itex]( [itex]\frac{h_{1}h_{2}h_{3}}{h_{i}} \epsilon_{ijk} A_{j} B_{k}[/itex]) = [itex]\frac{1}{h_{1}h_{2}h_{3}} \frac{\partial}{\partial x_{i}}[/itex]( [itex]\frac{h_{1}h_{2}h_{3}}{h_{i}} \epsilon_{ijk} A_{j} B_{k}[/itex]) [itex]\ast \hat{e_{i}} \bullet \hat{e_{i}}[/itex]

But I can't see how to take it into the desired format
I think I should squeeze in a kroncker delta somehow, but I'm not sure.

 

[vela]

Do you know how to do the proof without the scaling factors?

 

[simba_lk]

Yes, I just tried and it works fine even without the addition of unit vectors and the delta.
My problem is that I cannot take the scaling factors in or out of the derivative.
Perhaps I missed somthing?

 

[SteamKing]

Would these scaling factors perhaps be constants? Or are they functions of one or more of the coordinates?

 

[simba_lk]

Of course they are functions, consider for instance spherical coordinates

 

[vela]

I'd try simplifying ##\frac{h_1 h_2 h_3}{h_i}## to ##h_j h_k##. Then I think you just have to apply the product rule and grind it out.

 

[simba_lk]

This is my idea too but how can take the scaling factor out of the derivative?

 

[lurflurf]

I don't know why you want to fool with h or moving it. Just use the chain rule to hold each vector fixed in turn.

$$\dfrac{\mathrm{d}}{\mathrm{dt}}u=\dfrac{\mathrm{dx}}{\mathrm{dt}}\dfrac{\partial}{\partial x}u+\dfrac{\mathrm{dy}}{\mathrm{dt}} \dfrac{\partial} {\partial y}u$$

in your example

$$\frac{1}{h_{1}h_{2}h_{3}} \frac{\partial}{\partial x_{i}}( \frac{h_{1}h_{2}h_{3}}{h_{i}} \epsilon_{ijk} A_{j} B_{k}) =\frac{1}{h_{1}h_{2}h_{3}} \frac{\partial}{\partial x_{i}}( \frac{h_{1}h_{2}h_{3}}{h_{i}} \epsilon_{ijk} A_{j} (B_{k})_\mathrm{fixed}) +\frac{1}{h_{1}h_{2}h_{3}} \frac{\partial}{\partial x_{i}}( \frac{h_{1}h_{2}h_{3}}{h_{i}} \epsilon_{ijk} (A_{j})_\mathrm{fixed} B_{k}) $$

 

[simba_lk]

Ok but this doesn't prove the identity, if you look at the definition of the curl in curvilinear coordinates

 

[lurflurf]

Once variables are fixed they can move through derivatives and the identity is easily proven

$$\frac{1}{h_{1}h_{2}h_{3}} \frac{\partial}{\partial x_{i}}( \frac{h_{1}h_{2}h_{3}}{h_{i}} \epsilon_{ijk} A_{j} B_{k}) =(B_{k})_\mathrm{fixed} \frac{1}{h_{1}h_{2}h_{3}} \frac{\partial}{\partial x_{i}}( \frac{h_{1}h_{2}h_{3}}{h_{i}} \epsilon_{ijk} A_{j} ) +(A_{j})_\mathrm{fixed} \frac{1}{h_{1}h_{2}h_{3}} \frac{\partial}{\partial x_{i}}( \frac{h_{1}h_{2}h_{3}}{h_{i}} \epsilon_{ijk} B_{k}) $$

 

[vela]

Hint:
$$\nabla\cdot(\vec{A} \times \vec{B})
= \frac{1}{h_1 h_2 h_3} \frac{\partial}{\partial x_i} [h_j h_k \epsilon_{ijk} A_{j} B_{k}]
= \frac{1}{h_1 h_2 h_3} \frac{\partial}{\partial x_i} [\epsilon_{ijk} (h_j A_j) (h_k B_k)]$$

 

[simba_lk]

thanks a lot