Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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...

share|cite|improve this question
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

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: $$ \text{R.H.S} = \int_{\mathbb{R}^3} (xa_y-ya_x) \,dxdydz + \int_{\mathbb{R}^3} (y,x,0)\cdot \mathbf{a}\,dxdydz \\ =\int_{\mathbb{R}^3} (xa_y-ya_x) \,dxdydz + \int_{\mathbb{R}^3} (ya_x+xa_y)\,dxdydz \\ = 2\int_{\mathbb{R}^3} x a_y \,dxdydz = \text{L.H.S}. $$

share|cite|improve this answer

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.