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As stated here, Uniqueness of Helmholtz decomposition? , the solution of the Helmholtz decomposition is not unique.

Suppose that, for given vector field $\mathbf F$ with $\nabla \cdot \mathbf F =0$, I have a solution of its Helmholtz decomposition: the pair $\phi$ and $\mathbf A$. They are such that: $$\mathbf F = -\nabla \phi + \nabla\times\mathbf A.$$

What kind of transformation can I apply to $\phi$ and $\mathbf A$ to find another pair $\phi_1$ and $\mathbf A_1$ such that $$\nabla \phi_1 = 0$$ and $$\mathbf F=\nabla \times \mathbf A_1 $$?

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The transformation you want is, in general, impossible. The reason is that curls have zero divergence, but if the gradient is nontrivial then your vector field will have nonzero divergence, so it can't be modelled as only a curl.

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  • $\begingroup$ Hi Emilio, thanks a lot for your answer. I should have specified (which I'll do now, editing my question) that my vector field $\mathbf F$ has zero divergence. If this is the case, I suppose a transformation should exist. $\endgroup$ – Hans Castrop Mar 26 '17 at 9:43
  • $\begingroup$ If it has zero divergence, and you choose boundary conditions that are strong enough for uniqueness, then phi will naturally be zero. If you have good reasons why you don't want to impose such conditions then I guess it becomes a bit more complex of a question but I'd say you really need to say what those reasons are. $\endgroup$ – Emilio Pisanty Mar 26 '17 at 10:21
  • $\begingroup$ Thank you for the hint, it is probably indicating what direction I should take to solve my problem. My field F is actually a magnetic field, therefore must have div(F) = 0. In particular it's a magnetic field given in form of field map (three 3d Cartesian meshes: one for Bx, one for By, and one for Bz). Unfortunately this field map represents a truncated field, which doesn't go to zero at the boundaries. Do you think that, if I had a field map which zeroes at the boundaries, my computation of $\phi$ and $\mathbf A$ should be as I desire (i.e. $\nabla \phi=0$) ? $\endgroup$ – Hans Castrop Mar 26 '17 at 12:33
  • $\begingroup$ Given that I compute $\phi$ and $\mathbf A$ following the method explained in onlinelibrary.wiley.com/doi/10.1029/2005JA011382/full , I must admit I don't really know how to fix the boundary conditions to achieve $\nabla \phi = 0$. $\endgroup$ – Hans Castrop Mar 26 '17 at 12:39
  • $\begingroup$ I guess there is some legitimate cause for wanting what you're asking for, but I don't have time at the moment - I'll look into it if I find some time. $\endgroup$ – Emilio Pisanty Mar 26 '17 at 12:49
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Seems to me one can perform any transformation $$ \phi\rightarrow\phi+\phi_0 ~~~~~~ {\bf A}\rightarrow{\bf A}+\nabla f , $$ provided that $\nabla\phi_0=0$ (which means $\phi_0$ is a constants). The result after the transformation should give the same ${\bf F}$.

Am I missing something?

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