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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.

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    $\begingroup$ Could you please define what is meant by a "Solenoidal field"? $\endgroup$
    – Enforce
    Jul 10, 2021 at 15:48
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    $\begingroup$ @Enforce "Solenoidal" is a somewhat common alternative term for "divergence free". $\endgroup$
    – Arthur
    Jul 10, 2021 at 15:52
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    $\begingroup$ What are $\nabla^k$ when $k>2$? $\endgroup$
    – Calvin Khor
    Jul 10, 2021 at 15:52
  • $\begingroup$ @arthur would you like to give any hint? $\endgroup$
    – Ibrahim Islam
    Jul 10, 2021 at 16:56
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    $\begingroup$ 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 $\nabla^4$?? $\endgroup$
    – Mark Viola
    Jul 10, 2021 at 17:34

1 Answer 1

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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.

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  • $\begingroup$ $\nabla^3,\nabla^4...\nabla^k$ has their usual meaning,....would you please elaborate more? $\endgroup$
    – Ibrahim Islam
    Jul 10, 2021 at 16:20
  • $\begingroup$ 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 $\vec F$ $\endgroup$
    – Ibrahim Islam
    Jul 10, 2021 at 16:30
  • $\begingroup$ @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 $\vec F$ :) $\endgroup$
    – Maximilian Janisch
    Jul 10, 2021 at 16:59
  • $\begingroup$ @maxmillion janisch I hope it's quite clear after my comment! $\endgroup$
    – Ibrahim Islam
    Jul 10, 2021 at 17:05
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    $\begingroup$ @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 $\endgroup$
    – Calvin Khor
    Jul 11, 2021 at 1:42

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