Evaluating $\nabla\times f\vec u$ where $f$ is a scalar function

Question

Evaluating $\nabla\times f\vec u$ where $f$ is a scalar function

Let $\vec u=(a,b,c)$. Then $fu$ is $(fa,fb,fc)$

When I write the determinant form of the cross product, I can take the $f$ outside the determinant.

Hence $\nabla\times f\vec u=f\nabla\times u$.

But my book specifies an extra term of $\nabla f\times u$. Where does this come from?

Answer 1

I see what you mean that in the determinant formula $$\nabla\times (f\vec u) = \begin{vmatrix}\hat i & \hat j & \hat k \\ \partial_x & \partial_y & \partial_z \\ fa & fb & fc\end{vmatrix}$$ it looks like we can just factor out the $f$. But this formula isn't exactly correct. It is in fact just a heuristic for remembering the actual formula which is

$$\nabla\times (f\vec u) =\big[\partial_y (fc)-\partial_z(fb)\big]\hat i + \big[\partial_z(fa)-\partial_x(fc)\big]\hat j + \big[\partial_x(fb)-\partial_y(fa)\big]\hat k$$ In this form it's clear that we can't just factor out the $f$, because it is being differentiated. Starting from here (and remembering to use the product rule), you should be able to derive the correct identity.

Answer 2

Easy writing with Einstein notation: $$\nabla \times (fu) = \epsilon_{ijk} \partial_j (fu)_k \\ = \epsilon_{ijk} \partial_jfu_k+\epsilon_{ijk}f\partial_j u_k \\ = \epsilon_{ijk}\nabla f u_k +f\epsilon_{ijk}\partial_j u_k \\ = \nabla f \times u + f\nabla \times u$$