If $\vec F$ is a solenoidal field, then curl curl curl $\vec F$=?

Question

Question:If $\vec F$ is a solenoidal field, then curl curl curl $\vec F$=

a)$\nabla^4\vec F$

b)$\nabla^3\vec F$

c)$\nabla^2\vec F$

d) none of these.

My approach:I first calculate $\nabla×\nabla×\vec F$. We know that $$\nabla\times\left(\nabla\times\textbf{F}\right)=\nabla\left(\nabla\cdot\textbf{F}\right)-\nabla^2\textbf{F}$$ and since $\vec F$ is solenoidal,$\nabla\cdot\textbf{F}=0$,there fore we have $$\nabla\times\left(\nabla\times\textbf{F}\right)=-\nabla^2\textbf{F}$$ Now for $\nabla×\nabla×\nabla×\vec F$ I am unable to proceed,and totally stuck how to proceed further!I guess the answer is (b) but I have no idea.

Please guide me with correct answer!

Thanks in advance.

Answer 1

It suffices to check an example. $F=(\cos y, \sin x,0)$ is divergence free. As you have shown, the second curl is a laplacian, and $-\nabla^2F=F$. So the third curl only has $z$ component.

Now it is time to Interpret Notation. If $\nabla^3$ means gradient of laplacian then it gives a matrix so definitely wrong. If $\nabla^4$ is the bilaplacian, then it has no third component. So it’s none of them.

If $\nabla^k$ means the tensor of all $k$th derivatives (Max’s definition) then they all have the wrong shape ($ (\nabla\times)^3F$ is a vector).

And if you really mean $\nabla^k=\sum_i\partial_i^k$ (Notation that I have never seen before...) then again it’s all wrong because the third component of $\nabla^kF$ is always zero.