Is it always possible to do that decomposition? I'm asking it because Helmholtz theorem says a field on $\mathbb{R}^3$ that vanishes at infinity ($r\to \infty$) can be decomposed univocally into a gradient and a curl. But I also know, for example, that a constant field $\mathbf{E}$ on $\mathbb{R}^3$ is a gradient (not univocally definied): $\mathbf{E}(x+y+z+\mbox{constant})$. And the electric field is $-\nabla G+ d\mathbf{A}/dt$, where $\mathbf{A}$ can be (Coulomb Gauge) free-divergence.

So, is it always possible to do the decomposition of a (regular, of course) field on $\mathbb{R}^3$ into two fields, free-curl and free-divergence? And on a limited domain?

  • $\begingroup$ This wikipedia page summarizes standard Helmholtz decomposition: en.wikipedia.org/wiki/Helmholtz_decomposition $\endgroup$
    – Qmechanic
    Mar 22, 2011 at 20:34
  • $\begingroup$ From my understanding the general mathematical theory of these sorts of questions is Hodge theory: en.wikipedia.org/wiki/Hodge_theory, though I'm afraid the mathematical lingo on that page goes over my head! All I really know about it is you get different decompositions depending on the topology of the space. $\endgroup$
    – Michael
    Jan 1, 2013 at 3:39

2 Answers 2


As long as the field can be Fourier transformed, $$\tilde{\mathbf F}(\mathbf k) = \frac1{(2\pi)^{3/2}} \iiint e^{-i\mathbf k\cdot\mathbf r} \mathbf F(\mathbf r) d^3\mathbf r, $$ we can separate $\tilde{\mathbf F}$ into the longitudinal and traverse parts $$ \tilde{\mathbf F}(\mathbf k) = \tilde{\mathbf F}_\parallel(\mathbf k) + \tilde{\mathbf F}_\perp(\mathbf k)$$ where $$ \tilde{\mathbf F}_\parallel(\mathbf k) = \hat{\mathbf k} (\hat{\mathbf k}\cdot\tilde{\mathbf F}(\mathbf k)).$$ This makes $$ \mathbf k\cdot\tilde{\mathbf F}_\perp(\mathbf k) = 0, \quad \mathbf k\times\tilde{\mathbf F}_\parallel(\mathbf k) = 0,$$ which is equivalent to ($\mathbf k \mapsto -i\nabla$) $$ \nabla\cdot{\mathbf F}_\perp(\mathbf r) = 0, \quad \nabla\times{\mathbf F}_\parallel(\mathbf r) = 0,$$ when performing the inverse Fourier transform (again requiring it works). Thus shows $\mathbf F$ is split into a divergence-free part $\mathbf F_\perp$ and curl-free part $\mathbf F_\parallel$.

  • $\begingroup$ OK thank you for the answer! Now I ask (ultimate question :-D) : if you can't trasform F, sometimes you can do the decomposition (for example, the field $$e^{r} \mathbf u_r = \nabla (e^r+c)$$; it's a sum of gradient with free-divergence part =zero.So, the scomposition can be do with every field? PS why can you write ($\mathbf k \mapsto -i\nabla$) ? I didn't understand why, sorry :-( $\endgroup$
    – Boy S
    Nov 19, 2010 at 23:28
  • $\begingroup$ @Boy: The 1D Fourier transform of $df/dt$ is $i\omega\tilde f$. The $\nabla\mapsto i\mathbf k$ in 3D is equivalent to this. For fields that cannot be Fourier-transformed, I don't know, I can't find any articles on this, perhaps because an exponentially increasing field isn't physical :) $\endgroup$
    – kennytm
    Nov 20, 2010 at 9:14

The splitting of a vector field


into a curl-free part,


and a divergence-free part,


is given as

$$\begin{align}\tag{4}\vec{V}_{\parallel} &:=\vec{\nabla}(\Delta^{-1}(\vec{\nabla}\cdot\vec{V})),\\ \tag{5}\vec{V}_{\perp} &:=\vec{V}-\vec{V}_{\parallel}.\end{align}$$

Here $\Delta:=\vec{\nabla}\cdot\vec{\nabla}$ is the Laplacian, and $\Delta^{-1}$ is a right inverse $\Delta\circ\Delta^{-1} = {\rm id}$.

The operators $\Delta$ and $\Delta^{-1}$ take scalars $f:\mathbb{R}^3\to\mathbb{R}$ into scalars. Potential problems are related to zero-modes and whether the inverse $\Delta^{-1}$ is well-defined. For sufficiently well-behaved scalars $f$ (and implicit choice of boundary conditions), the following integral formula applies

$$\tag{6}(\Delta^{-1}f)(\vec{x}) ~=~ - \iiint_{\vec{y}\neq\vec{x}}\frac{d^3y}{4\pi} \frac{f(\vec{y})}{|\vec{y}-\vec{x}|},$$

cf. Poisson equation.


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