# How to impose $\phi=0$ to the solution of a Helmholtz decomposition of a divergence-free field?

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 $$\mathbf F=\nabla \times \mathbf A_1$$?

• 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 9:43
• 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 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
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 $${\bf F}$$.