Show that: $\nabla \times (\phi F) = \nabla \phi \times F + \phi \nabla \times F$. Where F is any vector field, and \phi is any scalar field.
My attempt:
Let F = (P,Q,R).
Now by observation, the first term of the RHS of the identity is zero since the curl of a gradient field is 0.
Hence we are trying to prove that:
$$\nabla \times (\phi F) = \phi \nabla \times F$$
Now if I compute the LHS I get:
$$\nabla \times (\phi F) = (\phi _y R_y -\phi _z Q_z) \hat i – (\phi _x R_x – \phi _z R_z) \hat j + (\phi_x Q_x – \phi_y P_y) \hat k$$
and the RHS:
$$\phi \nabla \times F = \phi [(R_y -Q_z) \hat i – (R_x – R_z) \hat j + ( Q_x – P_y) \hat k]$$
But this is not equivalent??
Any know what I have done wrong?
Best Answer
Also \begin{align} \nabla\times(\phi\mathbf{F}) &=\sum_{i,j,k=1}^3\varepsilon_{ijk}\frac{\partial}{\partial x_j}(\phi F_k)\mathbf{e}_i=\\ &=\sum_{i,j,k=1}^3\varepsilon_{ijk}\left[\left(\frac{\partial}{\partial x_j}\phi\right)F_k+\phi\frac{\partial}{\partial x_j}F_k\right]\mathbf{e} _i=\\ &=\sum_{i,j,k=1}^3\varepsilon_{ijk}\left(\frac{\partial}{\partial x_j}\phi\right)F_k\mathbf{e}_i+\sum_{i,j,k=1}^3\varepsilon _{ijk}\phi\frac{\partial}{\partial x_j}F_k\mathbf{e}_i=\\ &=(\nabla\phi)\times\mathbf{F}+\phi\nabla\times\mathbf{F} \end{align}