I read an article saying that the inner product between divergence-free current and a gradient field is zero.

A divergence-free surface current is $\nabla\cdot\vec{J}=0$, and $\vec{J}$ could be represented as $\vec{J}=\nabla\times(\psi\hat{n})$, where $\hat{n}$ is the normal vector of the surface. So the statement becomes: $\left( \nabla\times(\psi\hat{n}) \right) \cdot \nabla \varphi=0$.

I think according to the identity: $$\nabla\cdot(\vec{A}\times\vec{B})=\vec{B}\cdot(\nabla\times\vec{A})-\vec{A}\cdot(\nabla\times\vec{B})$$ we have $$\nabla\times(\psi\hat{n})\cdot \nabla \varphi=\nabla\cdot(\psi\hat{n}\times\nabla\varphi)+\psi\hat{n}\cdot\nabla\times\nabla\varphi=\nabla\cdot(\psi\hat{n}\times\nabla\varphi),$$ but what next?

Update Thank you Luboš Motl. I suppose I now understand why, but I don't have enough reputation to reply below, so just update here my answer.

The goal is to prove $\int_s \vec{J}\cdot\nabla\varphi ds=0$ The whole process is as follows:

First, $\vec{J}$cannot go across the surface edge, so $\vec{J}\cdot\hat{t}=0$, where $\hat{l}$ is the surface edge direction and $\hat{t}=\hat{l}\times\hat{n}$ is the edge out direction.

Second, according to the identity $$\nabla\cdot(\vec{A}\times\vec{B})=\vec{B}\cdot(\nabla\times\vec{A})-\vec{A}\cdot(\nabla\times\vec{B}) \, ,$$ we have \begin{align} \vec{J}\cdot\nabla\varphi &=\nabla\times(\psi\hat{n})\cdot \nabla \varphi \\ &=\nabla\cdot(\psi\hat{n}\times\nabla\varphi)+\psi\hat{n}\cdot\nabla\times\nabla\varphi \\ &=\nabla\cdot(\psi\hat{n}\times\nabla\varphi) \end{align} since $$\nabla\times(f\vec{A})=\nabla{f}\times\vec{A}+f(\nabla\times A)$$ $$\psi\hat{n}\times\nabla\varphi= -\nabla \times (\varphi\psi\hat{n}) + \varphi \nabla \times(\psi\hat{n}) \, .$$ Then \begin{align} \nabla\cdot(\psi\hat{n}\times\nabla\varphi) &=\nabla\cdot(-\nabla\times(\varphi\psi\hat{n})+\varphi\nabla\times(\psi\hat{n})) \\ &=\nabla\cdot(\varphi\nabla\times(\psi\hat{n})) \end{align} Finally, \begin{align} \int_s \vec{J}\cdot\nabla\varphi ds &=\int_s\nabla\times(\psi\hat{n})\cdot \nabla \varphi ds \\ &= \int_s \nabla\cdot(\varphi\nabla\times(\psi\hat{n}))ds \\ &=\oint_l \varphi\nabla\times(\psi\hat{n})\cdot\hat{t}dl \\ &=\oint_l \varphi\vec{J}\cdot\hat{t}dl \\ &=0 \, . \end{align}

I think here the important things are:

  1. Generally speaking, divergence-free current usually can be expressed as $\vec{J}=\nabla\times\vec{T}$, and $\vec{J}=\nabla\times(\psi\hat{n})$ is specially for surface current.

  2. the $\hat{n}$ is only valid on the surface(there is no meaning of $\hat{n}$ for point in side of a body). the integral is on the surface rather than on the body. According to the original article, it is just talking about PEC and surface current.

  • $\begingroup$ This question was extremely difficult to understand because all the TeX was mashed together and the punctuation and capitalization was essentially random. Please take the time to use proper English punctuation etc. so that people can understand what you are asking. I edited the math but I think the sentences are still screwed up. $\endgroup$
    – DanielSank
    Jul 9, 2015 at 6:45

3 Answers 3


A divergence-free current is still a pretty general vector field, so its inner product with another general field, a gradient, is surely not zero in general.

A trivial counterexample. $\psi n = (y/2,-x/2,0)$. Then $\nabla\times (\psi n) = (0,0,1)$. On the other hand, the gradient field may be $(0,0,1)=\nabla\cdot (0,0,z)$ and the inner product of the two unit $z$-direction vectors isn't zero anywhere.

What the statement that you encountered could have said was $$ \nabla \times (\nabla\cdot \phi) = 0$$ which is one of the basic identities that can be easily proven.


The OP has provided us with the source and it's clear that they made a different, true statement. The inner product wasn't meant to be just the simple product of two 3-vectors but the inner product in the Hilbert space sense $$ b(\vec u,\vec v) = \int d^3 x \, \vec u(x)^* \cdot \vec v(x) $$ integrated over the space. This vanishes if $\vec u$ is a multiple of a curl and $\vec v$ is a multiple of a gradient. This is trivially seen in the momentum space where it is $$ b(\vec u_k, \vec v_k) = \int d^3 k \, A(\vec k \times \vec B) \cdot (C\vec k \cdot D) $$ Here, $k\times$ arises from the curl and $\vec k\cdot$ arises from the gradient and the integral above vanishes (the integrand vanishes for each $\vec k$, in this representation) because $\vec k \cdot (\vec k \times \vec M) \equiv 0$. The analogous proof in the $x$-representation requires some integration by parts.

  • $\begingroup$ Thank you Luboš Motl. I agree with you. I must misunderstand the authors' idea. I read the statement from this article: docs.lib.purdue.edu/cgi/… On page.2, left column, the end of last but one paragraph, it is mentioned: '...The two problems are readily fixed in this work by removing the gradient-field component of the incident field and Green’s function when computing their inner products with a divergence-free current since the inner product between a divergence-free current and a gradient field is analytically known to be zero.' $\endgroup$
    – user50510
    Jul 1, 2014 at 11:22
  • $\begingroup$ It is again mentioned on page.5, around eqn.32 and eqn.33 '...Since $W_0$ represents a divergence-free current, which can be written as $\nabla\times\psi\hat{n}$ and hence $\hat{n}\times\nabla\psi$ with $\psi$ being a scalar, its inner product with a gradient field can be analytically proved to be zero...' $\endgroup$
    – user50510
    Jul 1, 2014 at 11:23
  • $\begingroup$ OK, that's a different statement, @user50510. The inner product is meant to be the quantum-mechanics-like integral of vector fields over the space, not just the pointwise scalar product function of space. See my updated answer for a fast proof. $\endgroup$ Jul 1, 2014 at 12:04
  • $\begingroup$ thank you Luboš Motl. I now understand why, but it looks to process is a bit complicated, I am wondering if there are easier process. $\endgroup$
    – user50510
    Jul 1, 2014 at 16:46
  • $\begingroup$ You can do an integration by parts to reduce it to either $\operatorname{curl} \operatorname{grad} = 0$ or $ \operatorname{div}\operatorname{curl} = 0$. $\endgroup$ Jul 1, 2014 at 16:52

I think write $\nabla\times(\psi\hat{n})$ as $\nabla\psi\times\hat{n}$ is to express the antisymmetry between $\varphi$ and $\psi$ and prepare the $\nabla\psi$ for $\vec{J}$ component in the final expression , would make it a little clearer to get the answer, $\vec{J}\cdot\nabla\varphi=\underline{\nabla\times(\psi\hat{n})\cdot \nabla \varphi}=(\nabla\psi\times\hat{n})\cdot\nabla\phi=\nabla\psi\cdot(\hat{n}\times\nabla\varphi)=\underline{-\nabla\psi\cdot\nabla\times(\varphi\hat{n})}$

and then use $\nabla\cdot(\vec{A}\times\vec{B})=\vec{B}\cdot(\nabla\times\vec{A})-\vec{A}\cdot(\nabla\times\vec{B})$ to include the expression into surface divergence operator $\nabla\cdot()$, so

-$\int_s\nabla\psi\cdot\nabla\times(\varphi\hat{n})ds=\int_s \nabla\cdot(\nabla\psi\times\varphi\hat{n})ds-\int_s\varphi\hat{n}\cdot\nabla\times\nabla\psi ds=\oint_l (\nabla\psi\times\varphi\hat{n})\cdot\hat{t}dl=\oint_l \varphi\vec{J}\cdot\hat{t}dl=0$


OK, another path is to present $\nabla\times(\psi\hat{n})\cdot\nabla\varphi=\nabla\psi\times\hat{n}\cdot\varphi=\hat{n}\cdot(\nabla\varphi\times\nabla\psi)=\hat{n}\cdot(\nabla\times(\varphi\nabla\psi)-\varphi\nabla\times\nabla\psi)=\hat{n}\cdot\nabla\times(\varphi\nabla\psi)$, where the identity $\nabla\times(f\vec{A})=\nabla f\times\vec{A}+f\nabla\times\vec{A}$ is used. then with Stokes' theorem $\int_s \nabla\times(\varphi\nabla\psi)\cdot\hat{n}ds=\oint_l \varphi\nabla\psi\cdot d\vec{l}=\oint_l \varphi\nabla\psi\cdot(\hat{n}\times\hat{t})dl=\oint_l \varphi\nabla\psi\times\hat{n}\cdot\hat{t}dl=\oint_l \varphi\vec{J}\cdot\hat{t}dl=0$


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