In the course of learning electrodynamics, I was asked to solve the problem following:

  • $\vec A$ is a vector which satisfies $\vec A\cdot \vec{\textbf{n}}=0$, where $\vec{\textbf{n}}$ is the normal vector of the surface of the volume $V$. Besides, $\nabla\cdot\vec A=0$ within the volume $V$. Please prove $$\int_V\text{d}V\ \vec A=0.$$

  • A standard solution is $$\int_V\text{d}V \vec A=\int_V\text{d}V\,\nabla\cdot(\vec A\vec r )=\oint_S\text{d}\vec\sigma\cdot(\vec A\vec r )=\oint_S\text{d}\sigma\,\vec{\textbf{n}}\cdot(\vec A\vec r)=\oint_S\text{d}\sigma\ (0\times\vec r)=0.$$

However, does this solution mean that a person that has never learned tensor analysis can never solve the problem? I wonder whether there are any other solutions without using tensors.

Thank you for your reading the question. Waiting for your excellent answers.

  • $\begingroup$ @Frank Can you please explain the second step of your standard solution? Looks like you have integrated without having a differential dV, using the divergence theorem, that is. $\endgroup$ Mar 11, 2017 at 6:12
  • $\begingroup$ @Frank Also since A and $\hat n$ are both normal to the boundary, they should be parallel to each other. Hence $A \times \hat n=0$. $\endgroup$ Mar 11, 2017 at 6:14
  • $\begingroup$ You need to clarify the problem much more. $\endgroup$
    – Lelouch
    Mar 11, 2017 at 7:14
  • $\begingroup$ @SchrodingersCat I've made a mistake. I wanted to say that $\vec A\cdot\vec n=0$ originally. So sorry about that. $\endgroup$
    – Frank
    Mar 11, 2017 at 13:04
  • 1
    $\begingroup$ I don't understand the question. $\int_V\text{d}V\ \vec A=0$ is a tensor analysis statement. How would a proof not use those methods? $\endgroup$ Mar 11, 2017 at 15:11

1 Answer 1


"Tensor" analysis is just fancy language, and you can do this just fine without it. The essence of tensor analysis (or at least one way of looking at said essence; see this answer for more on that) is to do things component by component. Since $$ \int_V\text{d}V\ \mathbf A=\sum_j \hat{\mathbf e}_j\int_V\text{d}V\ (\hat{\mathbf e}_j\cdot\mathbf A), $$ it is sufficient to just consider the integral of the scalar quantity $A_j=\hat{\mathbf e}_j\cdot\mathbf A$. In this language, the proof is a reformulation of what you've given: because $$ \nabla \cdot(x_j\mathbf A) = (\nabla x_j)\cdot\mathbf A + x_j \nabla\cdot\mathbf A = \hat{\mathbf e}_j\cdot\mathbf A, $$ we can write \begin{align} \int_V\text{d}V\ (\hat{\mathbf e}_j\cdot\mathbf A) & = \int_V\text{d}V\ \nabla \cdot(x_j\mathbf A) \\ & = \oint_S\text{d}\mathbf S\cdot(x_j\mathbf A), \end{align} which vanishes because $\mathbf A$ and $\mathrm d\mathbf S$ are orthogonal, and you're done. See? Easy! The tensor-analysis layer is just some fancy language on top to make everything come together slightly more coherently, but it is fundamentally the same proof.


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