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(This must have been asked somewhere before but for some reason I could not find an answer)

In physics one often meets the criterion that a vector field has zero divergence, e.g.

$$\nabla\cdot\mathbf{v}= 0 \quad (1)$$

Based on this one often claim that the vector field can be written as the curl of another field,

$$\mathbf{v}=\nabla\times\mathbf{A} \quad (2)$$

since $\nabla\cdot (\nabla \times \mathbf{f})= 0$ for any (smooth) $\mathbf{f}$. But the implication here is only obvious from (2) -> (1) and not the other way around.

I tried the Helmholtz decomposition theorem where the vector field can be decomposed into curl-free and divergence-free components

$$\mathbf{v} = \nabla \phi + \nabla\times \mathbf{A}$$

Taking the divergence of this and combining with (1) gives

$$\nabla\cdot\mathbf{v} = \nabla \cdot\nabla \phi = 0$$

But I can't see how this would imply $\nabla \phi = 0$ which I guess would thus imply (2).

Am I missing something or completely on the wrong track?

DiggingDeep
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1 Answers1

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I am giving an answer which is differential form notation and Poincaré Lemma, so perhaps not very useful for OP. It is definitely possible to do this without using differential forms language so hopefully someone will provide another answer doing that.

It is a basic result in de Rham cohomology that closed differential forms, that is satisfying $d \alpha=0$ are locally exact, that is there is if $d \alpha = 0 $ at some point $p$ then there is a neighbourhood of $p$ where $\alpha = d \beta$. If the topology of the space is trivial (let us say contractible, $R^n$ is contractible) then $\alpha = d \beta $ is true globally.

An inner product gives a canonical isomorphism between a vector space $V$ and its dual, I will denote the map $V \to V^*$ by $\flat$ and its inverse by $\sharp$. In this language a vector being divergence-free translates to \begin{equation} d * v^\flat = 0 \end{equation} hence we can, globally on $R^3$, write \begin{equation} * v^\flat = d A^\flat \end{equation} for some vector $A$. The map $*$ is the Hodge star, an isomorphism between $k$ forms and $(n-k)$-forms. In this setting it squares to the identity, so, also applying $^\sharp$, we get \begin{equation} v = (* d A^\flat)^\sharp \end{equation} and $(* d A^\flat)^\sharp = \mathrm{curl} A$ so we get the result we wanted.

GFR
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