How to proof equality of del dot a cross b

Question

I am trying to prove that $\nabla \dot{}(A\times B) = B\dot{}(\nabla \times A) - A\dot{}(\nabla\times B)$

I tried expanding the RHS but the $x$ component of the vector I am getting is $B_x(\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z}) - A_x(\frac{\partial B_z}{\partial y} - \frac{\partial B_y}{\partial z})$ and this does not seem to be the $x$ component of the LHS.

Could anyone please help?

Answer 1

Following your notation let $A = \langle A_x,A_y,A_z \rangle$ and $B = \langle B_x,B_y,B_z \rangle$ (i.e. subscripts indicate components of vector fields not partial derivatives).

Then $A \times B = \langle A_yB_z-A_zB_y,-(A_xB_z-A_zB_x), A_xB_y-A_yB_x \rangle$ so that $$\nabla \bullet (A \times B) = \dfrac{\partial}{\partial x}\left( A_yB_z-A_zB_y \right) + \dfrac{\partial}{\partial y}\left( A_zB_x-A_xB_z \right) + \dfrac{\partial}{\partial z}\left( A_xB_y-A_yB_x \right)$$

Also, $$B \bullet (\nabla \times A) = \langle B_x,B_y,B_z \rangle \bullet \left\langle \dfrac{\partial A_z}{\partial y} - \dfrac{\partial A_y}{\partial z}, -\left(\dfrac{\partial A_z}{\partial x} - \dfrac{\partial A_x}{\partial z}\right), \dfrac{\partial A_y}{\partial x} - \dfrac{\partial A_x}{\partial y} \right\rangle$$ $$=B_x\left(\dfrac{\partial A_z}{\partial y} - \dfrac{\partial A_y}{\partial z}\right)+B_y\left(\dfrac{\partial A_x}{\partial z} - \dfrac{\partial A_z}{\partial x}\right)+B_z\left(\dfrac{\partial A_y}{\partial x} - \dfrac{\partial A_x}{\partial y}\right)$$ likewise, you can write down $A \bullet (\nabla \times B)$.

Write down and completely expand: $B \bullet (\nabla \times A)-A \bullet (\nabla \times B)$. Then use the product rule to finish calculating $\nabla \bullet (A \times B)$ and see that the left and right hand sides match.

Edit: How to expand the LHS?

$$\dfrac{\partial}{\partial x}\left( A_yB_z-A_zB_y \right) = \dfrac{\partial}{\partial x}\left(A_yB_z \right) - \dfrac{\partial}{\partial x}\left(A_zB_y \right) = \dfrac{\partial A_y}{\partial x}B_z + A_y\dfrac{\partial B_z}{\partial x} - \dfrac{\partial A_z}{\partial x}B_y - A_z\dfrac{\partial B_y}{\partial x}$$

Answer 2

We know (in a orthonormal basis {$\hat{i},\hat{j},\hat{k}$}) $$A\times B =(a_y b_z-a_z b_y)\hat{i}+(a_z b_x-a_x b_z)\hat{j}+(a_x b_y -a_y b_x)\hat{k} $$

Furthermore we have for $$\nabla .(A\times B)=\frac{\partial}{\partial x}(a_y b_z-a_z b_y)+\frac{\partial}{\partial y}(a_z b_x-a_x b_z)+\frac{\partial}{\partial z}(a_x b_y -a_y b_x) $$

Now lets solve for the right hand side and see if we obtain e.g. the same argument for the partial derivative of $\frac{\partial}{\partial x}$:

For the first term we have: $$B.(\nabla \times A)=b_x(\frac{\partial a_z}{\partial y}-\frac{\partial a_y}{\partial z})+b_y(\frac{\partial a_x}{\partial z}-\frac{\partial a_z}{\partial x})+b_z(\frac{\partial a_y}{\partial x}-\frac{\partial a_x}{\partial y})$$

For the second term: $$-A.(\nabla \times B)=-a_x(\frac{\partial b_z}{\partial y}-\frac{\partial b_y}{\partial z})-a_y(\frac{\partial b_x}{\partial z}-\frac{\partial b_z}{\partial x})-a_z(\frac{\partial b_y}{\partial x}-\frac{\partial b_x}{\partial y})$$

Now add the two terms together, and assemble them so that you indentify the derivative product rule, e.g.: $$b_z\frac{\partial a_y}{\partial x}+a_y\frac{\partial b_z}{\partial x}-b_y\frac{\partial a_z}{\partial x}-a_z\frac{\partial b_y}{\partial x}$$

which is indeed equal to $$\frac{\partial}{\partial x}(a_y b_z-a_z b_y),$$ proceed in the same way for the rest of the terms and you will have proved your identity.

Answer 3

Why don't you simply do this:

$$\nabla .( A \times B) = \partial_i (A \times B)_i $$

$$= \partial_i (\epsilon_{ijk} A_jB_k) = \epsilon_{ijk} A_j (\partial_i B_k) + \epsilon_{ijk} B_k(\partial_i A_j) $$ $$= (\epsilon_{kij} \partial_i A_j) B_k - (\epsilon_{jik}\partial_i B_k)A_j = (\nabla \times A)_k \ B_k \ - \ (\nabla \times B)_j \ A_j $$ $$= (\nabla \times A) \ . \ B \ - \ (\nabla \times B) \ . \ A$$

Note: usual properties of levi-civita symbol $\epsilon_{ijk}$ have been used.