Verify the following relationship: $\nabla \cdot (a \times b) = b \cdot \nabla \times a - a \cdot \nabla \times b $

Question

Verify the vector identity:

$\nabla \cdot (a \times b) = b \cdot \nabla \times a - a \cdot \nabla \times b $

Given that:

$a = (R_a, S_a, T_a)$, $b = (R_b, S_b, T_b)$ and $\nabla = (\frac{\partial} {\partial x},\frac{\partial} {\partial y},\frac{\partial} {\partial z} )$.

Where $R_i, S_i, T_i$ have continuous partial derivatives.

I attempted it on paper but cannot get it to work. Can anyone show me how this is done?

My working:

LHS:

$\nabla \cdot (a \times b) = \frac{\partial}{\partial x}(S_a T_b - S_b T_a) + \frac{\partial}{\partial y}(T_a R_b - R_a T_b) + \frac{\partial}{\partial z}(R_a S_b - R_b S_a) $

RHS:

$\nabla \times a = (\frac{\partial}{\partial y} T_a - \frac{\partial}{\partial z} S_a)\hat i - (\frac{\partial}{\partial x} T_a - \frac{\partial}{\partial z} R_a)\hat j + (\frac{\partial}{\partial x} S_a - \frac{\partial}{\partial y} R_a)\hat k$

$b \cdot (\nabla \times a) = R_b(\frac{\partial}{\partial y} T_a - \frac{\partial}{\partial z} S_a) - S_b(\frac{\partial}{\partial x} T_a - \frac{\partial}{\partial z} R_a) + T_b(\frac{\partial}{\partial x} S_a - \frac{\partial}{\partial y} R_a)$

Now I computed the other part of the RHS as well however I can't see how I would be able to maninipulate it into the LHS.

Anyone have any idea?

Answer 1

Use the Leibniz rule in the firts term:

$$ \frac{\partial}{\partial x}(S_aT_b - S_bT_a) = T_b\frac{\partial S_a}{\partial x} + S_a\frac{\partial T_b}{\partial x} - T_a\frac{\partial S_b}{\partial x} - S_b \frac{\partial Ta}{\partial x}. $$

Expanding the other two you should get the desired equality.

Another approach can be using index notation.

Answer 2

There's a much more convenient derivation using the chain rule and the properties of the mixed product. In particular it holds that cyclic shift doesn't change the result of a mixed product: $$ a \cdot (b \times c) = b \cdot (c \times a) = c \cdot (a \times b) \: . $$ Now applying the chain rule to the expression in question, we get: $$ \nabla \cdot (a \times b) = \nabla_a \cdot (a \times b) + \nabla_b \cdot (a \times b) \: , $$ where $\nabla_a$ takes a derivative of $a$ and sees $b$ as a constant, and $\nabla_b$ vice versa. (This works exactly the same as the dot notation in this answer.) Now, shifting the factors in the first term to the right and the factors in the second term to the left, we get: $$ \nabla \cdot (a \times b) = b \cdot (\nabla_a \times a) + a \cdot (b \times \nabla_b) \: . $$ Finally, we change the order of multiplication in the second cross product. Since $a\times b = -b \times a$, we get: $$ \nabla \cdot (a \times b) = b \cdot (\nabla_a \times a) - a \cdot (\nabla_b \times b) \: . $$ Since both nablas now only act on the factor that is directly to the right, we can drop the subscript and arive at the equation in question.