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Could someone provide a hint as to why $$\nabla \cdot \vec a(\vec x) = -i\,\,\,b\,\,\,c(\vec x)$$

where $b$ is a constant, $i$ is $\sqrt {-1}$, implies that $$2\int d^3x \,\,x_ia_j(\vec x)=\epsilon_{ijk} \Big[\int \,\,d^3x \,\, \vec x\times \vec a(\vec x)\Big]_k-ib \int\,\,d^3x\,\, x_ix_jc(\vec x)$$?

Context/Interpretation: $$\nabla \cdot \vec a(\vec x) = -i\,\,\,b\,\,\,c(\vec x)$$ is obtained from $$\nabla \cdot [\vec a(\vec x) \exp(-ibt)]= {\partial\over\partial t}[c(\vec x)\exp(-ibt)]$$ which can be interpreted as a conservation law. I don't have any more context information...

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Please give some context. Effective mathematics is the manipulation of ideas and not just symbol juggling. –  Fly by Night Dec 26 '12 at 23:03
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2 Answers 2

This is a community-wiki answer trying to remove this question from the unanswered queue.

I am not that comfortable with indices notation, so I will translate the integral into traditional multivariate calculus language. Your $\vec{x}$ will be bold face letter $\mathbf{r} = (x,y,z)$, $\mathbf{a} = (a_x,a_y,a_z)$, and $x_i x_j$ will be $xy$ in my answer if $i=1$, and $j=2$. The subscript $(\cdot)_z$ of a vector will just be its $z$-component.

$$2\int d^3x \,x_i a_j(\vec x)=\epsilon_{ijk} \Big[\int \,d^3x \,\, \vec x\times \vec a(\vec x)\Big]_k -i b \int\,d^3x\,\, x_i x_j c(\vec x)$$

Knowing $-ibc = \nabla \cdot \mathbf{a}$. This integral can be written as six integral identities if counting all the permutation and get rid of the Levi-Civita symbol: $$ 2\int_{\mathbb{R}^3} x a_y \,dxdydz = \int_{\mathbb{R}^3} (\mathbf{r}\times \mathbf{a})_z \,dxdydz - \int_{\mathbb{R}^3} xy\nabla \cdot \mathbf{a}\,dxdydz, \\ 2\int_{\mathbb{R}^3} y a_x \,dxdydz = -\int_{\mathbb{R}^3} (\mathbf{r}\times \mathbf{a})_z \,dxdydz - \int_{\mathbb{R}^3} yx\nabla \cdot \mathbf{a}\,dxdydz, \\ 2\int_{\mathbb{R}^3} y a_z \,dxdydz = \int_{\mathbb{R}^3} (\mathbf{r}\times \mathbf{a})_x \,dxdydz - \int_{\mathbb{R}^3} yz\nabla \cdot \mathbf{a}\,dxdydz, \\ 2\int_{\mathbb{R}^3} z a_y \,dxdydz = -\int_{\mathbb{R}^3} (\mathbf{r}\times \mathbf{a})_x \,dxdydz - \int_{\mathbb{R}^3} zy\nabla \cdot \mathbf{a}\,dxdydz, \\ 2\int_{\mathbb{R}^3} z a_x \,dxdydz = \int_{\mathbb{R}^3} (\mathbf{r}\times \mathbf{a})_x \,dxdydz - \int_{\mathbb{R}^3} zx\nabla \cdot \mathbf{a}\,dxdydz, \\ 2\int_{\mathbb{R}^3} x a_z \,dxdydz = -\int_{\mathbb{R}^3} (\mathbf{r}\times \mathbf{a})_x \,dxdydz - \int_{\mathbb{R}^3} xz\nabla \cdot \mathbf{a}\,dxdydz. $$ Here I will check if the first one is correct. Others can be checked in the exactly same fashion. Now $$\mathbf{r}\times \mathbf{a} = (ya_z - za_y, za_x - xa_z, xa_y-ya_x).$$ The right hand side of the first identity is: $$ \text{R.H.S}= \int_{\mathbb{R}^3} (\mathbf{r}\times \mathbf{a})_z \,dxdydz - \int_{\mathbb{R}^3} xy\nabla \cdot \mathbf{a}\,dxdydz \\ = \int_{\mathbb{R}^3} (xa_y-ya_x) \,dxdydz + \int_{\mathbb{R}^3} \nabla(xy)\cdot \mathbf{a}\,dxdydz, $$ in which we assume the decaying condition of $|\mathbf{a}| = o(r^{-4})$ where $r = |\mathbf{r}|$, this assumption just guarantees that if we are integrating on a really large ball in $\mathbb{R}^3$, the boundary integral term produced by Gauss-Divergence theorem vanishes when the radius of the ball goes to infinity. Expanding we have: $$ \text{R.H.S} = \int_{\mathbb{R}^3} (xa_y-ya_x) \,dxdydz + \int_{\mathbb{R}^3} (y,x,0)\cdot \mathbf{a}\,dxdydz \\ =\int_{\mathbb{R}^3} (xa_y-ya_x) \,dxdydz + \int_{\mathbb{R}^3} (ya_x+xa_y)\,dxdydz \\ = 2\int_{\mathbb{R}^3} x a_y \,dxdydz = \text{L.H.S}. $$

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Consider the form $\vec \nabla \cdot (\vec x \times \vec a)$ as it is a special case of the scalar triple product.

Or just integrate and use Stokes' theorem on the form $\int \vec \nabla \cdot \vec a$. I believe, depending on the choice of $\vec a$ that this could be interpreted as one of Maxwell's equations.

You also may want to consider clarifying the notation $\Big[\int \,\,d^3x \,\, \vec x\times \vec a(\vec x)\Big]_k$ although I am pretty sure the Levi-Civita symbol arises due to a change in coordinate basis here.

Depending on the meaning of the notation, Gauss' law may be applicable or even http://en.wikipedia.org/wiki/Divergence#Decomposition_theorem

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Thank you, Ricky, but I still don't see how I can get that result from the 2 suggestions... Would you mind elaborating a little bit more? Thanks –  Matthew 1629 Dec 27 '12 at 0:14
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