It is well known that the divergence of the curl is always 0. Mathematically I understand why this happens ($d^2=0$ where $d$ is the exterior derivative) but today I was wondering what is the physical meaning of this.

The divergence represents the volume density of the outward flux of a vector field from an infinitesimal volume around a given point (from Wikipedia) and the curl describes the infinitesimal rotation of a 3-dimensional vector field (also from Wikipedia).

Does this mean that the rotation of a vector field is always stable and doesn't go inwards or outwards? What is the physical meaning of the divergence of the curl equals 0?

  • 5
    $\begingroup$ Given that it is true for any vector field I don't think there is any physical significance. $\endgroup$
    – ProfRob
    Dec 24, 2015 at 11:06
  • $\begingroup$ You can check Feynman's Lectures on curl free & divergence free fields; there he presents a nice example as why this relation is true. $\endgroup$
    – user36790
    Dec 24, 2015 at 11:56
  • $\begingroup$ Not sure I'd use "stable" here, but seems to me that ${\rm div}({\rm curl}(\mathbf F))$ signifies that the flux through any infinitesimal volume of a field that is rotating is zero. Perhaps it could be better to say that it means that there are no sources or sinks with a field that is rotating? $\endgroup$
    – Kyle Kanos
    Dec 24, 2015 at 12:59
  • $\begingroup$ @KyleKanos: But you're not taking the divergence of a rotating field, you're taking the divergence of its curl, which may or may not rotate. $\endgroup$
    – Javier
    Dec 24, 2015 at 13:52
  • $\begingroup$ @Javier: Well if $\mathbf F$ is not rotating, then $\nabla\times\mathbf F=\mathbf 0$ and ${\rm anything}\cdot\mathbf 0=0$ identically & it's not terribly interesting. $\endgroup$
    – Kyle Kanos
    Dec 24, 2015 at 14:02

3 Answers 3


I don't think there is any physical meaning. There is geometric meaning, however, which is almost as good, in this case.

Consider the integral theorems, namely the Kelvin-Stokes-theorem: $$ \int_\Sigma \nabla\times \mathbf{V}\cdot d\mathbf{A}=\int_{\partial\Sigma}\mathbf{V}\cdot d\mathbf{l}, $$ and the Gauss-theorem: $$ \int_U \nabla\cdot\mathbf{V}\ d^3x=\int_{\partial U}\mathbf{V}\cdot d\mathbf{A},$$ where $\Sigma$ is some smooth surface, $U$ is a 3-dimensional domain of integration, and $\partial$ is the boundary operator, eg. the one that maps a set to its boundary.

Now let $U$ be once again a volume, and let $\mathbf{V}$ be a completely arbitrary vector field, and apply these theorems twice in succession: $$ \int_{\partial\partial U}\mathbf{V}\cdot d\mathbf{l}=\int_{\partial U}\nabla\times\mathbf{V}\cdot d\mathbf{A}=\int_U\nabla\cdot(\nabla\times\mathbf{V})\ d^3x=0. $$ These integrals are all zero, because $\nabla\cdot(\nabla\times\mathbf{V})$ is zero for any $\mathbf{V}$, however, if you look at the first expression, $$ \int_{\partial\partial U}\mathbf{V}\cdot d\mathbf{l}=0 $$, this must be true for all vector fields $\mathbf{V}$. The only way it is true for all such vector fields is that if the domain of integration $\partial\partial U$ is a set of zero measure.

Therefore, the identity $\nabla\cdot(\nabla\times\mathbf{V})=0$ essentially says that for a three-dimensional volume $U$, the boundary of its boundary is always zero. If you think about it, this is pretty obvious, but this line of thought actually proves this.

This is even more elegant for differential forms, where you can repeat the same procedure by applying the generalized stokes theorem twice, and you will essentially get that $\mathrm{dd}=0$ corresponds to $\partial\partial=0$.

  • $\begingroup$ The discussion of $d^2 = 0$ and that the boundaries of a boundary is empty in MTW is very good and has great illustrations. $\endgroup$ Dec 24, 2015 at 14:50

First I should say that it is a mathematical definition and does not carry any physical meaning necessarily. But as an example it relates the Maxwell equations: $$\nabla\times E=-\frac{\partial B}{\partial t}$$ and $$\nabla \cdot B=0$$ $$\Longrightarrow \nabla\cdot(\nabla\times E)=-\frac{\partial \nabla\cdot B}{\partial t}=0$$

and the continuity equation can be deduced from $$\nabla\times H=J+\frac{\partial D}{\partial t}$$ and $$\nabla \cdot D=\rho$$$$\Longrightarrow \nabla\cdot(\nabla\times H)=\nabla\cdot J+\frac{\partial \nabla\cdot D}{\partial t}=0$$ which is the continuity equation: $$\nabla\cdot J=-\frac{\partial \rho}{\partial t}$$

If one day in future a magnetic monopole is discovered in reality, as some people are interested in it for some reasons, then $\nabla \cdot B$ is not zero anymore, but, mathematically $\nabla\cdot(\nabla\times E)$ has to be zero.

This means that a new entity has to be defined instead of $B$ in order to hold the $\nabla\cdot(\nabla\times E)=0$. This is how math can help physics.

  • 3
    $\begingroup$ Although correct, I don't think the above addresses the question because the left hand sides always vanish; as such, we demand the rhs to vanish as well (namely those are the equation of motion for the fields and the current, as you pointed out). The question, though, is about the meaning of the lhs vanishing for the field that occurs in it (the $\textbf{E}$ field, in your example), rather than its consequences on the possible rhs of an equation of motion. $\endgroup$
    – gented
    Dec 24, 2015 at 11:55
  • 1
    $\begingroup$ You are right. Anyway I don't think that divergence of curl has any physical meaning as long as we attribute something to it. $\endgroup$
    – Amin
    Dec 24, 2015 at 12:14
  • $\begingroup$ May be we can look into conditions where $\nabla\cdot(\nabla \times F)=0$ fails, (when $F$ is not differentiable?) $\endgroup$
    – Courage
    Dec 24, 2015 at 12:18
  • $\begingroup$ I agree with the fact that there ought not necessarily be any particular physical meaning in $d^2=0$, especially because it is always true by definition. Regarding the point of $\textbf{F}$ not being differentiable: interesting, although all the calculus on manifolds (and hence Stoke's theorems and hence all the rest) are defined based on the differentiability assumption. $\endgroup$
    – gented
    Dec 24, 2015 at 12:28

It quite simply means that the radial direction is orthogonal to the tangential direction.

The universe in which we live behaves, as far as we can tell and measure, as a 3D volume (time aside for now:). A 3D volume is defined by 3 orthogonal directions. We can choose to look at a 2D cross-section of this volume, and we get a 2D plain, in which, every point is defined by 2 orthogonal direction. One can show these directions can be X\Y, but also R\Teata. R is the radial direction, and Theata is the tangential direction.

Because these directions are orthogonal, BY DEFINITION, it's not hard to understand that if you measure the radial component of a purely tangential scale (the deliverance of the curl), of the other way around, you get nothing - because radialiry is not tangential, and tangentiality is not radial :)

  • $\begingroup$ Could you expand on why you think this is the case? $\endgroup$
    – Kyle Kanos
    Dec 24, 2015 at 15:20
  • $\begingroup$ Expansion has occurred. $\endgroup$ Dec 24, 2015 at 15:47
  • $\begingroup$ I don't see how this is related to $\textrm{div}(\textrm{curl}\,\textbf{v})=0$. $\endgroup$
    – gented
    Dec 24, 2015 at 15:55
  • $\begingroup$ Because, as written, the divergence (which is a measure of radial component of a 2D vector) of the curl (which is a measure of the tangential component of a 2D vector) is by definition 0, because they are orthogonal. $\endgroup$ Dec 24, 2015 at 15:58

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