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Given an arbitrary constant 3-D rotation matrix $\mathbf{R}$ and a 3-D vector field $\mathbf{A}$, how can I find the vector field $\mathbf{B}$ such that $$\nabla\times\mathbf{B}=\mathbf{R}\left(\nabla\times\mathbf{A}\right)$$ ?

I followed two ways, without really finding a way out

  1. I wrote the LHS and RHS explicitly. If I write $\mathbf{R}$ as the matrix $$\mathbf{R} = \begin{pmatrix}a & b & c\\ d & e & f\\ g & h & i \end{pmatrix}$$ (which is obviously an overkill, since the actual degrees of freedom of an arbitrary rotation are just three), then my equation looks like the nasty \begin{align*} \partial_{y}B_{z}-\partial_{z}B_{y} & =a\left(\partial_{y}A_{z}-\partial_{z}A_{y}\right)+b\left(\partial_{z}A_{x}-\partial_{x}A_{z}\right)+c\left(\partial_{x}A_{y}-\partial_{y}A_{x}\right)\\ \partial_{z}B_{x}-\partial_{x}B_{z} & =d\left(\partial_{y}A_{z}-\partial_{z}A_{y}\right)+e\left(\partial_{z}A_{x}-\partial_{x}A_{z}\right)+f\left(\partial_{x}A_{y}-\partial_{y}A_{x}\right)\\ \partial_{x}B_{y}-\partial_{y}B_{x} & =g\left(\partial_{y}A_{z}-\partial_{z}A_{y}\right)+h\left(\partial_{z}A_{x}-\partial_{x}A_{z}\right)+i\left(\partial_{x}A_{y}-\partial_{y}A_{x}\right) \end{align*} which intimidates me quite a bit... Rearranging it acquires an interesting structure, \begin{align*} \partial_{y}B_{z}-\partial_{z}B_{y} & =\partial_{x}\left(cA_{y}-bA_{z}\right)+\partial_{y}\left(aA_{z}-cA_{x}\right)+\partial_{z}\left(bA_{x}-aA_{y}\right)\\ \partial_{z}B_{x}-\partial_{x}B_{z} & =\partial_{x}\left(fA_{y}-eA_{z}\right)+\partial_{y}\left(dA_{z}-fA_{x}\right)+\partial_{z}\left(eA_{x}-dA_{y}\right)\\ \partial_{x}B_{y}-\partial_{y}B_{x} & =\partial_{x}\left(iA_{y}-hA_{z}\right)+\partial_{y}\left(gA_{z}-iA_{x}\right)+\partial_{z}\left(hA_{x}-gA_{y}\right) \end{align*} but I don't really know how to continue.. I don't think that replacing the actual rotation matrix elements would even help..

  2. Then I tried a different, let's say geometric, way. I Fourier transformed both sides of my equation and, as (1) the Fourier transform of the $\nabla$ operator is $i\mathbf{k}$ and (2) the Fourier transform commutes with a rotation matrix, I obtained $$i\mathbf{k}\times\tilde{\mathbf{B}}=\mathbf{R}\left(i\mathbf{k}\times\tilde{\mathbf{A}}\right)$$ which is a nice expression involving just vectors and not nasty partial derivatives. Now, I can rewrite this as $$\mathbf{k}\times\tilde{\mathbf{B}}=\mathbf{R}\left(\mathbf{k}\right)\times\mathbf{R}\left(\tilde{\mathbf{A}}\right)$$ but this expression, although simple and nice, does not really help me find $\mathbf{B}$...

Perhaps I am just rusted and I am missing something simple....

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  • $\begingroup$ Is $\mathbf{R}$ constant? $\endgroup$
    – N74
    May 3, 2018 at 23:11
  • $\begingroup$ It might be helpful to explain what you've tried and where you're getting stuck. For example, have you tried writing out the LHS and RHS in coordinates? $\endgroup$
    – Sophie M
    May 3, 2018 at 23:13
  • $\begingroup$ @N74 Yes, $\mathbf{R}$ is a constant. $\endgroup$ May 4, 2018 at 7:55
  • $\begingroup$ @Kyle MacDonand, yep I expanded my question with some details... $\endgroup$ May 4, 2018 at 7:55

2 Answers 2

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I think the reason you're having trouble is that there may be no such vector field $B$. Let's take an example:

$$A(x, y, z) = \pmatrix{z\sin x + y \cos x\\ 0\\ 0} $$ Then $$ curl ~ A(x, y, z) = \pmatrix{0 \\ \sin x \\ -\cos x} $$ if I've done the computation right. And notice that the divergence of that is $0$, because the divergence of the curl of any vector field on a contractible set like $\Bbb R^3$ is always zero (at least if everything is $C^2$, which it certainly is in this case).

Now let $$ R = \pmatrix{0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1}. $$ Then let's apply $R$ to the curl of $A$ and give it a name: $$ Q = R(\nabla \times A) = \pmatrix{0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1}\pmatrix{0 \\ \sin x \\ -\cos x} = \pmatrix{\sin x \\ 0 \\ -\cos x}. $$ Now the divergence of $Q$ is $$ \nabla \cdot (R(\nabla \times A)) = \cos x + 0 + 0 = \cos x, $$ which is not zero. That means that $Q$ cannot be the curl of any vector field. And that's why you couldn't find one. :)

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  • $\begingroup$ Hello John, thank you for your inputs. Yes, that's what I am starting to realize. Several crossed tests point all in that direction (the Fourier way, which I like a lot, confirms it also from an intuitive, geometric, viewpoint: the cross product is a projection and, when something gets projected on something else, some information gets lost). This might bring me to my next question.... $\endgroup$ May 6, 2018 at 9:54
  • $\begingroup$ I would say instead that a cross-product produces a co-vector (an element of the dual space) (where you define $(A\times B)(v)$ to be the determinant of the matrix whose rows are $A, B, v$), and that there's a hidden isomorphism from the dual to the primal in there and this messes things up. Either way...glad to have been of some help. $\endgroup$ May 6, 2018 at 11:41
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Let me add an interesting fact to the already existing answer: even though $x \mapsto R \: \text{rot}(F)(x)$ need not be a rotation field again (as John's answer demonstrates), the vector field $x \mapsto R \: \text{rot}(F)(R^{-1}x)$ is again a rotation field. In fact one has \begin{equation} R \: \text{rot}(F)(R^{-1}x) = \text{rot}(R F(R^{-1}x)) \end{equation} I don't know of a nice way to show this though, except by direct computation.

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