"Which of these identities hold for all vector fields $\mathbf{F}$ and functions $f$ in $\mathbf{R}^3$ with continuous first and second derivatives?"
- $\nabla\times(\nabla f)=0$ (Ans.: True)
- $\nabla.(\nabla\times\mathbf{F})=0$ (Ans.: True)
- $\nabla(\nabla.\mathbf{F})=0$ (Ans.: False)
- $\nabla\times(\nabla.\mathbf{F})=0$ (Ans.: False)
Can someone please explain how you would prove/disprove these identities?
Substitute $\nabla = \frac{\partial}{\partial x}\hat{\mathbf{i}}+\frac{\partial}{\partial y}\hat{\mathbf{j}}+\frac{\partial}{\partial z}\hat{\mathbf{k}}$ and $\mathbf{F} = F_x\hat{\mathbf{i}}+F_y\hat{\mathbf{j}}+F_z\hat{\mathbf{k}}$ into the identities. E.g.: \begin{align} \nabla\times(\nabla f)&=\left\vert\array{ \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} }\right\vert = \left[\begin{array}\\ \frac{\partial^2 f}{\partial y \partial z}-\frac{\partial^2 f}{\partial y \partial z} \\ \frac{\partial^2 f}{\partial x \partial z}-\frac{\partial^2 f}{\partial x \partial z} \\ \frac{\partial^2 f}{\partial x \partial y}-\frac{\partial^2 f}{\partial x \partial y} \end{array}\right] =0 \\ \nabla.(\nabla\times\mathbf{F})&= \left[\begin{array}\\ \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{array}\right] .\left[\begin{array}\\ \frac{\partial F_z}{\partial y}-\frac{\partial F_y}{\partial z} \\ \frac{\partial F_x}{\partial z}-\frac{\partial F_z}{\partial x} \\ \frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y} \end{array}\right] = \frac{\partial^2 F_z}{\partial x\partial y}-\frac{\partial^2 F_y}{\partial x\partial z}+\frac{\partial^2 F_x}{\partial y\partial z}-\frac{\partial^2 F_z}{\partial x\partial y}+\frac{\partial^2 F_y}{\partial x\partial z}-\frac{\partial^2 F_x}{\partial y\partial z}=0 \end{align}