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

Community
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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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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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