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?
Best Answer
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.