Why is the inner product between divergence-free current $\vec{J}$, and a gradient field$\nabla \varphi$ zero?

Question

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.

Answer 1

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.

Update

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.

Answer 2

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

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

Answer 3

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