# Rotation of a curl?

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....

• Is $\mathbf{R}$ constant?
– N74
May 3, 2018 at 23:11
• 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? May 3, 2018 at 23:13
• @N74 Yes, $\mathbf{R}$ is a constant. May 4, 2018 at 7:55
• @Kyle MacDonand, yep I expanded my question with some details... May 4, 2018 at 7:55

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. :)
• 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. May 6, 2018 at 11:41
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 $$$$R \: \text{rot}(F)(R^{-1}x) = \text{rot}(R F(R^{-1}x))$$$$ I don't know of a nice way to show this though, except by direct computation.