# Divergence of (A x B)

1. Apr 16, 2006

### r4nd0m

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

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

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

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

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

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

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

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?

Last edited: Apr 16, 2006
2. Apr 16, 2006

### 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.

3. Apr 16, 2006

### r4nd0m

Thanks a lot, I didn't realize that.

4. Apr 16, 2006

### 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.

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