Can a vector field $\mathbf{A}: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ always be expressed as the sum of a solenoidal (divergenceless) part $\mathbf{A}_S$, where $\nabla \cdot \mathbf{A}_S = 0$, and an irrotational part $\mathbf{A}_R$, where $\nabla\times\mathbf{A}_R = 0$?

$$ “\,\forall{\mathbf{A}}\; \exists{\mathbf{A}_S}\; \exists{\mathbf{A}_R} : \mathbf{A} = \mathbf{A}_S + \mathbf{A}_R \;\wedge\; \nabla \cdot \mathbf{A}_S = 0 \;\wedge\; \nabla \times \mathbf{A}_R = 0 \,”\,? $$

Since irrotational vector fields are gradients of some scalar field ($\mathbf{A}_R = \nabla R$), this is equivalent to asking:

$$ “\, \forall\mathbf{A}\; \exists\mathbf{A}_S\; \exists R \in \left(\mathbb{R}^3\rightarrow\mathbb{R}\right) : \mathbf{A} = \mathbf{A}_S + \nabla R \;\wedge\; \nabla\cdot\mathbf{A}_S = 0 \,”\,? $$

What’s a proof of the result? If true, are $\mathbf{A}_R$ and $\mathbf{A}_S$ unique? If it’s not always true, then what kind of fields can/cannot be decomposed in this way?

If true, this would be a very nice result: any vector field could then be thought of in terms of its unique solenoidal and irrotational components.


1 Answer 1


This is called Helmholtz decomposition, a.k.a., the fundamental theorem of vector calculus. Helmholtz’s theorem states that any vector field $\mathbf{F}$ on $\mathbb{R}^3$ can be written as $$ \mathbf{F} = \underbrace{-\nabla\Phi}_\text{irrotational} + \underbrace{\nabla\times\mathbf{A}}_\text{solenoidal} $$ provided 1) that $\mathbf{F}$ is twice continuously differentiable and 2) that $\mathbf{F}$ vanishes faster than $1/r$ as $r \rightarrow \infty$. If $\mathbf{F}$ is on a bounded domain $V \subset \mathbb{R}^3$, then the condition 2) that $\mathbf{F}$ vanishes can be relaxed.

Helmholtz’s theorem even gives $\Phi$ and $\mathbf{A}$ in terms of $\mathbf{F}$. More details can be found here: https://en.wikipedia.org/wiki/Helmholtz_decomposition


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