Given the potential $$ \vec{A}=\frac{\mu_{0}}{4 \pi} \int_{V}^{\prime} \frac{\vec{J}\left(r^{\prime}\right)}{R} d v^{\prime} $$ where $\vec{J}$ is the stationary current density vector. I want to show that $\vec{\nabla} \cdot \vec{A}= 0$. I already did this integrating by parts, but the author gave the following solution which i can't follow:

$$ \require{cancel} \begin{aligned} \vec{\nabla} \cdot \vec{A} &=\vec{\nabla} \cdot\left[\frac{\mu_{0}}{4 \pi} \int_{V}^{\prime} \frac{\vec{J}\left(r^{\prime}\right)}{R} d v^{\prime}\right] \\ &=\frac{\mu_{0}}{4 \pi} \int_{V}^{\prime} \vec{\nabla} \cdot\left(\frac{\vec{J}\left(r^{\prime}\right)}{R}\right) d v^{\prime} \\ &=\frac{\mu_{0}}{4 \pi} \int_{v}^{\prime} \vec{J} \cdot\left[\vec{\nabla}\left(\frac{1}{R}\right)\right]+\frac{1}{R}\cancelto{0}{(\vec{\nabla} \cdot \vec{J})} d v \\ &=\frac{\mu_{0}}{4 \pi} \int_{V}^{\prime} \cancelto{0}{\vec{J}\left(r^{\prime}\right) \cdot \vec{r}} d v \\ \vec{\nabla}\cdot \vec{A} &=0 \end{aligned} $$

The 4th line is confusing me. Why do we have $\cancelto{0}{\vec{J}\left(r^{\prime}\right) \cdot \vec{r}}$? Its like the author is saying the density current vector is always ortoghonal to the position vector which is not necessarily true. I am not sure if that expression is zero because a mathematical reason or a physical reason.

Also, I think there is a missing term $\left ( \dfrac{-1}{R^2} \right) $ in the integrand of the 4th line.


  1. Wangsness, Electromagnetic Fields, Chap 16.
  • $\begingroup$ This is the question 16.1 from Wangsness, Electromagnetic Fields, and the answer is given in the page 77 of this pdf: freescribddownload.com/es/view-doc/… $\endgroup$ Dec 22, 2021 at 16:19
  • $\begingroup$ Because J(r') is not a function of r, which is what the divergence is referring to , can we use this rule for $\nabla \cdot (\phi \vec{V})$ ( the divergence of a scalar and vector) and if we can, I would also like to point out that the $ \nabla \cdot \vec{J}$ is not zero because of the fact that amperes law requires it. But it is zero because it is not a function of r $\endgroup$ Dec 22, 2021 at 16:31
  • $\begingroup$ can anyone clear the doubt about my question. And can anyone confirm my analysis is correct on the second half of my comment? $\endgroup$ Dec 22, 2021 at 16:33
  • $\begingroup$ Any useful related discussion comments given by this author in his book? Usually he would have explained something in previous pages you have missed out. $\endgroup$
    – Markoul11
    Dec 22, 2021 at 16:58
  • 1
    $\begingroup$ @Markoul11 not at all. Keep in mind the answer's book is independent from the main book. I definitely think there is something wrong. Either he is applying the divergence theorem and concluding with that without mentioning it, or he is using the fact that the integral vanishes because the current symmetry of a point when there are no net charges entering or leaving a location. I can't think in any relation between $\vec{J}\left(r^{\prime}\right)$ and $\vec{r}$ that vanishes the integral. $\endgroup$ Dec 22, 2021 at 17:08

1 Answer 1


I can't follow the author's solution either. I think they may be using an earlier result that showed that $J\cdot R$ is zero if J is divergence free.

I really dislike their notation, I think it'd be a lot clearer written as $$ \vec{A}(r) = \frac{\mu_0}{4\pi} \int d^3\vec{r'} \frac{\vec{J}(r')}{|\vec{r}-\vec{r}'|} $$ Then we end up with $$ \nabla \cdot \vec{A}(r) = \frac{\mu_0}{4\pi} \int d^3\vec{r'} \nabla_r \cdot \left(\frac{\vec{J}(r')}{|\vec{r}-\vec{r}'|}\right) $$ $$ = \frac{\mu_0}{4\pi} \int d^3\vec{r'} \vec{J}(r') \cdot \nabla_r\left(\frac{1}{|\vec{r}-\vec{r}'|}\right) $$ The last line followed from J being constant in $r$. Now the nontrivial point is that

$\nabla_r\left(\frac{1}{|\vec{r}-\vec{r}'|}\right) = -\nabla_{r'}\left(\frac{1}{|\vec{r}-\vec{r}'|}\right)$

This allows you to use integration by parts, along with $\nabla \cdot \vec{J} = 0$ (i.e. the magnetostatic condition, equivalent to the requirement that charge does not accumulate anywhere) to prove the result.

Note that this requirement can be relaxed, but the answer comes in terms of the electric potential $\phi$.

$$\nabla \cdot A = -\mu_0\epsilon_0 \frac{\partial \phi}{\partial t} \Leftrightarrow \frac{\partial (\phi/c)}{\partial (ct)} + \nabla \cdot A = 0$$

which is the Lorenz gauge condition.

  • $\begingroup$ Yes, they are different by a factor of minus 1 $\endgroup$ Dec 22, 2021 at 21:58
  • $\begingroup$ Fixed it in body, thanks for picking it up $\endgroup$ Dec 22, 2021 at 22:04
  • $\begingroup$ Would they be the same actually ?as one has an $\vec{r}'$ on top, and one has $\vec{r}$ on top? $\endgroup$ Dec 22, 2021 at 22:06
  • $\begingroup$ I think they are the same, both have $\pm\vec{r}\mp\vec{r'}$ on top: $\partial_j (\sum(x_i-x_i')^2 )^{-1/2} = -\frac{1}{2} (\sum(x_i-x_i')^2 )^{-3/2} 2(x_j-x'_j)$ $\endgroup$ Dec 22, 2021 at 22:09
  • $\begingroup$ Yes, you are correct, that's what I get for differentiating in my head. $\endgroup$ Dec 22, 2021 at 22:13

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