Editing Magnetic vector potential (section)

== Magnetostatics in the Coulomb Gauge ==
In [[magnetostatics]], if the Coulomb gauge <math>\ \nabla \cdot \mathbf{A} = 0 </math> is imposed, then there is an analogy between <math> \mathbf{A}, \mathbf{J} </math> and <math> V, \rho </math> in [[electrostatics]]:<ref name="Semon1996">{{cite journal
 | author = Mark D. Semon and John R. Taylor
 | title = Thoughts on the magnetic vector potential
 | journal = American Journal of Physics
 | volume = 64
 | issue = 11
 | pages = 1361–1369
 | year = 1996
 | doi = 10.1119/1.18400
 | bibcode = 1996AmJPh..64.1361S
 | url = https://doi.org/10.1119/1.18400
| url-access = subscription
 }}</ref>
<math display="block"> \nabla^2 \mathbf{A} = -\mu_0 \mathbf{J} </math>
just like the electrostatic equation
<math display="block"> \nabla^2 V = -\frac{\rho}{\epsilon_0} </math> 

Likewise one can integrate to obtain the potentials:
<math display="block"> \mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4 \pi} \int_R  \frac{\mathbf{J}(\mathbf{r}')}{\left | \mathbf{r} - \mathbf{r}' \right |} \mathrm{d}^3 r'</math>
just like the equation for the [[electric potential]]:
<math display="block"> V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \int_R \frac{\rho(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} \mathrm{d}^3 r' </math>