Vector Potential

• The vector potential $\vec{A}$ for a magnetic field $\vec{B}$ is defined with the following equation:

$$\vec{B} = \mathrm{rot}\,\vec{A}$$

• In order to make this equation unique, the Coulomb gauge has to be applied which is defined as

$$\mathrm{div}\,\vec{A} = 0$$ In electrostatics, the electric field $\vec{E}(\vec{r})$ can be written as the gradient of the electric potential $\phi(\vec{r})$: $$\vec{E}(\vec{r}) = \mathrm{grad}\,\phi(\vec{r})$$ Now the question is whether we can define a potential for the magnetic field in a similar way. Ampere's law implies that the integral $$\oint \vec{B}\cdot\,\mathrm{d}\vec{r} \neq 0$$ can never be 0 as long as the integration path surrounds a flowing current of electric charges, because magnetic field lines cannot start or end anywhere. This is equal to the statement $$\mathrm{rot}\, \vec{B} \neq 0$$ It is, therefore, not possible to determine a unique magnetic potential $\phi_m$ in the following way $$\vec{B}(\vec{r}) = -\mu_0\,\mathrm{grad}\,\phi_m(\vec{r})$$ because in this case the rotation of $\vec{B}$ would lead to $$\mathrm{rot}\,\vec{B} = \vec{\nabla}\times\vec{\nabla}\phi_m(\vec{r}) = 0$$ We can, however, define a vector potential in the following form: $$\boxed{\vec{B} = \mathrm{rot}\,\vec{A}}$$ This is possible because the divergence of a rotation is always 0: $$\mathrm{div}\,\vec{B} = \vec{\nabla}\cdot \left(\vec{\nabla}\times \vec{A}\right) = \mathrm{div}\,\mathrm{rot}\,\vec{A}$$ However, this equation alone is not sufficient, because we can always add an arbitrary scalar field in the form $$\vec{A'} = \vec{A} + \mathrm{grad}\,f$$ since $\mathrm{rot}\,\mathrm{grad}\,f$ is always identical 0. It was therefore decided to use the following important gauge condition in addition to the definition of $\vec{A}$: $$\boxed{\mathrm{div}\,\vec{A} = 0}$$ This condition is called Coulomb gauge.