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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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Hans Castrop
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2 Answers2

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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.

Emilio Pisanty
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  • 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. – Hans Castrop Mar 26 '17 at 09:43
  • 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. – Emilio Pisanty Mar 26 '17 at 10:21
  • 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. $\nabla \phi=0$) ? – Hans Castrop Mar 26 '17 at 12:33
  • Given that I compute $\phi$ and $\mathbf A$ following the method explained in http://onlinelibrary.wiley.com/doi/10.1029/2005JA011382/full , I must admit I don't really know how to fix the boundary conditions to achieve $\nabla \phi = 0$. – Hans Castrop Mar 26 '17 at 12:39
  • 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. – Emilio Pisanty Mar 26 '17 at 12:49
  • Alright. Thank you again, Emilio. I suspect a change of reference system (rotation + translation?) might help achieve the desired condition. I need to figure out if this is true, and how to compute such a transformation automatically and efficiently. – Hans Castrop Mar 26 '17 at 13:16
  • No, a change in reference system won't help at all. You need to solve the differential equation $\nabla\times \mathbf A= \nabla \phi$ for $\mathbf A$, so you'll need a bit more machinery. – Emilio Pisanty Mar 26 '17 at 13:42
  • Hi Emilio, just a quick update, for your info: eventually I verified that applying a rotation of the axes has indeed the effet of "moving the field information" from $\phi$ to $\mathbf A$, and vice versa; and I could (almost completely) vanish the scalar potential as I desired. The reason why this happens becomes manifest if one writes the expressions of the scalar and of the vector potentials in the Fourier space. Of course, once the field is reconstructed from the rotated vector potential, one needs to counter-rotate the result to go back to the original XYZ set of coordinates. – Hans Castrop Mar 31 '17 at 13:25
  • @HansCastrop That looks pretty bogus to me but without more details it's impossible to tell what it is you're actually doing. – Emilio Pisanty Mar 31 '17 at 13:30
  • It is all based on the idea presented in the paper I cited few comments earlier. In particular formulae (1) and (4), which provide a recipe to extract the scalar and the vector potentials of a field map, based on the Fourier Transform. But some fields can feature both scalar and vector potentials simultaneously. As the FT commutes with any axes rotation, one can calculate the rotation that minimizes the dot product in (4), which in turn minimizes the FT of $\phi$ (and therefore $\phi$). Once $\phi$ eliminated, one can use $\mathbf A$ from the rotated field. – Hans Castrop Mar 31 '17 at 14:34
  • @HansCastrop That sounds like complete hogwash to me, but suit yourself. – Emilio Pisanty Mar 31 '17 at 16:38
  • "Hogwash" sounds a little dismissive to me. If you have concrete objections to the correctness of this method, could you please articulate them? – Hans Castrop Mar 31 '17 at 18:06
  • @HansCastrop As I said, there's not enough details to judge the method's merits. I would recommend that you post a separate answer with how you're resolving this (which is the ideal course of action, anyways, so that future visitors that find this through search engines can find an answer), I can probably comment further. – Emilio Pisanty Mar 31 '17 at 18:50
  • Try a gauge transformation $\mathbf A \rightarrow \mathbf A + \nabla \phi$? – Cinaed Simson Mar 17 '19 at 19:20
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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?

flippiefanus
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