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Assume we are in a magnetostatic and electrostatic situation. In the multipole expansion of the magnetic vector potential, we end up with

$\vec{A}(\vec{r})=\sum_{n=0}^\infty \vec{A}_n(\vec{r})$ where $\vec{A}_n(\vec{r})=\frac{\mu_0}{4\pi r^{n+1}}\int_\text{space}(r')^nP_n(\hat{r}\cdot\hat{r'})\vec{J}(\vec{r}')\text{d}\tau'$

The monopole term is then when $n=0$ i.e.

$\vec{A}_0(\vec{r})=\frac{\mu_0}{4\pi r}\int_\text{space} \vec{J}(\vec{r}')\text{d}\tau'$

When the current distribution is simply a constant current loop, the above integral can have an $I$ factored out leaving just a loop integral which goes to zero. This shows that there is no monopole term. How can we in general show the monopole term vanishes though (or is it not true?) i.e. how can we show

$\int_\text{space} \vec{J}(\vec{r}')\text{d}\tau'=0$

assuming the static situation which is defined by the relations $-\frac{\partial \rho}{\partial t}=\nabla\cdot\vec{J}=0$ and $\frac{\partial \vec{J}}{\partial t}=0$. We also know we are working with the Coulomb gauge so $\nabla\cdot\vec{A}=0$.

I was thinking maybe that this is not in general true i.e. the monopole term doesn't in general vanish even for the static situation. Possibly you have to assume the current distribution is local i.e. doesn't extend to infinity or maybe it works only on loops, etc. Please let me know if you have any idea! Thanks in advance.

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  • $\begingroup$ It's a consequence of Gauss' Law for Magnetism, $\vec{\nabla} \cdot \vec{A} = 0$. $\endgroup$ – N. Steinle Dec 12 '18 at 3:35
  • $\begingroup$ That's not Gauss' Law and doesn't show that the monopole term is zero. I want to show specifically that $A_0=0$ $\endgroup$ – Aakash Lakshmanan Dec 12 '18 at 4:02
  • $\begingroup$ it would be nice to have definitions under your equations $\endgroup$ – Kosm Dec 12 '18 at 5:46
  • $\begingroup$ Sorry, i meant $\vec{\nabla} \cdot \vec{B}$ $\endgroup$ – N. Steinle Dec 12 '18 at 20:15
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Yes, you do have to assume that the current distribution is localized. To show that the first term is zero you may solve an almost different problem, that is, to evaluate the integral $$ \int g (\mathbf J\cdot \boldsymbol \nabla f )\, dv = \sum_i \int g J_i \frac{\partial f}{\partial x_i} \, dv. $$ where $f,g$ are functions of $\mathbf r$ (I'm dropping the prime), $i = 1,2,3$ and $dv$ is the cartesian volume element. Integrating by parts to transfer the derivative to $gJ_i$ and choosing the surface that encloses the integration volume to be where $J_i = 0$ (remember that $\mathbf J$ is localized!) you'll arrive at $$ \int g (\mathbf J\cdot \boldsymbol \nabla f )\, dv =-\sum_i \int f \frac{\partial g}{\partial x_i}J_i \, dv - \int fg (\boldsymbol \nabla \cdot \mathbf J) \, dv. $$ Now we use the continuity equation for magnetostatics: $\boldsymbol \nabla \cdot \mathbf J =- \partial \rho/ \partial t = 0$, so the last term vanishes and the final result is $$ \int g (\mathbf J\cdot \boldsymbol \nabla f )\, dv = -\int (\boldsymbol \nabla g \cdot \mathbf J)f \, dv. $$ Finally, if we choose $f = x_i$ and $g = 1$ you'll see that the integral over each component of $\mathbf J$ vanishes: $$ \int \mathbf J \cdot \mathbf e_i \, dv = \int J_i \, dv = - \int(0 \cdot \mathbf J)x_i \,dv = 0 $$

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  • $\begingroup$ Amazing, thank you! Where did you come across this? $\endgroup$ – Aakash Lakshmanan Dec 12 '18 at 4:16
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    $\begingroup$ My Electromagnetic Theory professor used this in class. I believe he extended the argument made in Jackson's Electrodynamics textbook since this relation is also useful to evaluate the next term in the series $\endgroup$ – ErickShock Dec 12 '18 at 10:57

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