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

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.

• Could you please define what is meant by a "Solenoidal field"? Jul 10, 2021 at 15:48
• @Enforce "Solenoidal" is a somewhat common alternative term for "divergence free". Jul 10, 2021 at 15:52
• What are $\nabla^k$ when $k>2$? Jul 10, 2021 at 15:52
• @arthur would you like to give any hint? Jul 10, 2021 at 16:56
• If $\nabla \cdot \vec F=0$ and $\vec F$ is sufficiently smooth, then $\nabla \times \nabla \times \vec F=-\nabla^2 \vec F$ and $$\nabla \times \nabla \times \nabla \times \vec F =-\nabla \times \nabla^2 \vec F= -\nabla^2 \nabla \times \vec F$$What are the definitions of $\nabla^3$ and $\nabla^4$?? Jul 10, 2021 at 17:34

## 1 Answer

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.

• $\nabla^3,\nabla^4...\nabla^k$ has their usual meaning,....would you please elaborate more? Jul 10, 2021 at 16:20
• For example,when I say $\nabla^3\vec F$, I simply means $(\frac{\partial^3}{\partial x^3}+\frac{\partial^3}{\partial y^3}+\frac{\partial^3}{\partial z^3})(F_1i+F_2j+F_3k)$ ,where $F_1,F_2,F_3$ are components of field $\vec F$ Jul 10, 2021 at 16:30
• @lbs One has to be a bit careful as "usual meaning" here can be ambiguous: For example, I would intuitively consider $\nabla^k \vec F$ to be the set (or tensor) containing all $k$-th partial derivatives of $\vec F$ :) Jul 10, 2021 at 16:59
• @maxmillion janisch I hope it's quite clear after my comment! Jul 10, 2021 at 17:05
• @MarkViola because the choices are laid out for you, you do not need to prove a general result. If it fails for my choice of $F$, that is enough. And for planar $F$ it is clear that the curl only has $z$ component without calculation, giving the result Jul 11, 2021 at 1:42