# Homework Help: Divergence of a cross product help

1. Mar 13, 2013

### simba_lk

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

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:

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

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.

2. Mar 13, 2013

### vela

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

3. Mar 13, 2013

### 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?

4. Mar 13, 2013

### SteamKing

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

5. Mar 13, 2013

### simba_lk

Of course they are functions, consider for instance spherical coordinates

6. Mar 13, 2013

### vela

Staff Emeritus
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.

7. Mar 13, 2013

### simba_lk

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

8. Mar 13, 2013

### 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.

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

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

9. Mar 13, 2013

### simba_lk

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

10. Mar 13, 2013

### lurflurf

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

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

11. Mar 13, 2013

### vela

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

12. Mar 13, 2013

thanks a lot