The term divergence-free sounds more general and appears particularly in wavelet-related approaches to the Navier-Stokes equations. However I have yet to find a discussion focusing on the distinction, either mathematical or intuition-based, between curl fields and divergence-free fields. Can anyone clarify this distinction?


@JannisTeunissen gave an answer for infinite domains. However, for finite domains, the Helmholtz decomposition is not unique, and the scalar potential $\phi$ can take many forms.

First, it is true that the curl part $\nabla\times\vec A$ is of course divergence free. But if $\vec F$ is divergence free, all you know is that $\phi$ has to satisfy $-\Delta \phi=0$, but you don't know anything about its boundary values. In other words, any choice of boundary values for $\phi$, together with $-\Delta \phi=0$ yields a divergence free $\vec F$; in particular, $\phi$ is not necessarily zero. As a consequence, on finite domains, the set of divergence free vector fields is not just the set of curls of vector potentials, but substantially larger.

Mathematically, on infinite domains you have that $$ \{ \vec F \in H_\text{div}: \nabla\cdot\vec F=0 \} = \{ \nabla\times\vec A: \vec A \in H_\text{curl} \}. $$ On the other hand, on bounded domains, you have that $$ \{ \vec F \in H_\text{div}: \nabla\cdot\vec F=0 \} = \{ \nabla\times\vec A: \vec A \in H_\text{curl} \} \cup \{ \nabla\phi: \phi \in H^1, -\Delta\phi=0 \}. $$


For a vector field $\vec{F}$ that satisfies a couple of conditions (smooth, decaying rapidly enough) Helmholtz's theorem states that you can write $\vec{F}$ as

$$\vec{F} = \nabla \times \vec{A} - \nabla \phi.$$

The 'curl field' $\nabla \times \vec{A}$ is equal to the divergence-free part of $\vec{F}$, since we have: $$\nabla \cdot (\nabla \times \vec{A}) = 0,$$ and conversely, if $\nabla \cdot \vec{F} = -\nabla^2 \phi = 0$, we have $\nabla \phi = 0$ (when $\phi \to 0$ at infinity), so that $$\vec{F} = \nabla \times \vec{A}$$

  • $\begingroup$ So why is the term divergence-free sometimes used? Does it mean the same thing as curl field? $\endgroup$ – barnhillec Dec 6 '16 at 10:34
  • $\begingroup$ Because the curl-field has divergence zero $\endgroup$ – Jannis Teunissen Dec 6 '16 at 10:43
  • $\begingroup$ Sorry but I don't feel you are answering the question. I know the curl has zero divergence. This is not the same thing as saying that a divergence free field is identical to a curl field. Curl could be a subset of divergence free fields for example. $\endgroup$ – barnhillec Dec 6 '16 at 10:45
  • $\begingroup$ Maybe read my answer again: if a field is diverfence-free (and satisfies some basic constraints) then you can write it as the curl of some other field. $\endgroup$ – Jannis Teunissen Dec 6 '16 at 12:20

For physical intuition you can consider the following:

1)For a fluid flow divergence of velocity calculates the the total flux leaving and entering the control volume. Hence in absence of a mass source-the velocity has to be divergence free, that is mass leaving a fixed control volume should be equal to that entering it.

2) Curl calculates the 'magnitude' of rotation of the field. So a non-zero curl for velocity implies a rotational motion corresponding to a vortex. Then, curl free fields can be described by a potential function


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