Divergence of (A x B): Proving the Equation for Vector Calculus

[r4nd0m]

Hi, I'm having trouble proving that:
\nabla \cdot (\textbf{A} \times \textbf{B}) = \textbf{B}\cdot<br /> (\nabla \times \textbf{A}) - \textbf{A}\cdot (\nabla \times<br /> \textbf{B})

This is how I proceeded:
\textbf{A} \times \textbf{B} = \overrightarrow{i}(A_y B_z - A_z<br /> B_y) + \overrightarrow{j} (A_z B_x - A_x B_z) + \overrightarrow{k}<br /> (A_x B_y - A_y B_x)

\nabla \cdot (\textbf{A} \times \textbf{B}) = \frac{\partial A_y B_z - A_z<br /> B_y}{\partial x} + \frac{\partial A_z B_x - A_x B_z}{\partial y} +<br /> \frac {\partial A_x B_y - A_y B_x}{\partial z}

\nabla \times \textbf{A} = \overrightarrow{i}(\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z})<br /> + \overrightarrow{j}(\frac {\partial A_x}{\partial z} - \frac<br /> {\partial A_z}{\partial x}) + \overrightarrow{k}(\frac {\partial<br /> A_y}{\partial x} - \frac {\partial A_x}{\partial y})

\textbf{B} \cdot (\nabla \times \textbf{A}) = (\frac{\partial A_z B_x}{\partial y} - \frac{\partial A_y B_x}{\partial z})<br /> + (\frac {\partial A_x B_y}{\partial z} - \frac {\partial A_z<br /> B_y}{\partial x}) + (\frac {\partial A_y B_z}{\partial x} - \frac<br /> {\partial A_x B_z}{\partial y}) =

= \frac{\partial A_y B_z - A_z<br /> B_y}{\partial x} + \frac{\partial A_z B_x - A_x B_z}{\partial y} +<br /> \frac {\partial A_x B_y - A_y B_x}{\partial z} = \nabla \cdot<br /> (\textbf{A} \times \textbf{B})

\textbf{A} \cdot (\nabla \times \textbf{B}) = (\frac{\partial B_z A_x}{\partial y} - \frac{\partial B_y A_x}{\partial z})<br /> + (\frac {\partial B_x A_y}{\partial z} - \frac {\partial B_z<br /> A_y}{\partial x}) + (\frac {\partial B_y A_z}{\partial x} - \frac<br /> {\partial B_x A_z}{\partial y}) = -\nabla \cdot (\textbf{A} \times<br /> \textbf{B})

Which finally yields:
\textbf{B}\cdot (\nabla \times \textbf{A}) - \textbf{A}\cdot (\nabla \times \textbf{B}) = 2\nabla \cdot (\textbf{A} \times \textbf{B})

Where did I make a mistake?

 

[matt grime]

The obvious problem is that you've written B.(\/xA) and somehow got the components of the B's past the differential symbols, which is worrying

B_x\partial_yA_z

is not the same as

\partial_yA_zB_x

though it is the same as

(\partial_yA_z)B_x

that might be one problem.

 

[r4nd0m]

Thanks a lot, I didn't realize that.

 

[matt grime]

It's just the product rule. if you differentiate f(x)g(x) the answer is not f'(x)g'(x), which I'm sure you do know.