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

[r4nd0m]

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

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

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

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

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

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

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

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

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

[tex]B_x\partial_yA_z[/tex]

is not the same as

[tex]\partial_yA_zB_x[/tex]

though it is the same as

[tex] (\partial_yA_z)B_x[/tex]

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.