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?

  • 1
    $\begingroup$ You misuderstood the first term of the rhs. It is the cross product of the gradient and the field $F$ and it's not zero in general. The lhs is not well computed as, e.g. $\partial_x(\phi P)=\phi_x P+\phi P_x$ $\endgroup$ – Rafa Budría Apr 30 '17 at 9:56

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.


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}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.