Show that: $\nabla \times (\phi F) = \nabla \phi \times F + \phi \nabla \times F$

Question

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?

Answer 1

The first term of RHS is not zero, it is

\begin{equation} (\nabla \phi) \times F=\det \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\[0.3em] \frac{\partial \phi}{\partial x} & \frac{\partial \phi}{\partial y} & \frac{\partial \phi}{\partial z} \\[0.3em] F_x & F_y & F_z \end{bmatrix}, \end{equation}

and the LHS is,

\begin{equation} \nabla \times (\phi F)=\det \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\[0.3em] \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\[0.3em] \phi F_x & \phi F_y & \phi F_z \end{bmatrix}, \end{equation}

where you have to keep in mind the product rule for differentiation, for example, \begin{equation} \frac{\partial (\phi F_y)}{\partial x} = \frac{\partial \phi }{\partial x}F_y + \phi \frac{\partial F_y}{\partial x}, \end{equation} if you keep all this in mind I think you will be able to figure it out.

Answer 2

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}