B1 - potentials

the solenoidal condition:

\vec{\nabla} \cdot \vec{B} = 0

vector law:

\vec{\nabla} \cdot (\vec{\nabla} \times \vec{A}) = 0

so, $\vec{A}$ is the vector potential of $\vec{B}$

\vec{B} = \vec{\nabla} \times \vec{A}

eg: consider a uniform field, $\vec{B} = B_{0} (0, 0, 1)$

\begin{matrix} \vec{B} = (\frac{\partial A_{z}}{\partial y} - \frac{\partial A_{y}}{\partial z}, \frac{\partial A_{x}}{\partial z} - \frac{\partial A_{z}}{\partial x}, \frac{\partial A_{y}}{\partial x} - \frac{\partial A_{x}}{\partial y}) \\ {\vec{A}}_{1} = B_{0} (0, x, 0) \to B_{0} \hat{z} \\ {\vec{A}}_{2} = B_{0} (- y, 0, 0) \to B_{0} \hat{z} \\ {\vec{A}}_{3} = \frac{1}{2} B_{0} (- y, x, 0) \to B_{0} \hat{z} \end{matrix}

these are the different vector potentials $\vec{A}$ that all produce the same magnetic field $\vec{B}$
switching the polar coordinates:

\begin{matrix} | \vec{A} | = \frac{B_{0}}{2} \sqrt{x^{2} + y^{2}} \\ \vec{A} = \frac{B_{0}}{2} r \hat{θ} \end{matrix}

choosing a particular $\vec{A}$ is choosing the gauge to work on
Faraday's law:

\begin{matrix} \vec{\nabla} \times \vec{E} + \frac{\partial \vec{B}}{\partial t} = 0 \\ \vec{\nabla} \times (\vec{E} + \frac{\partial \vec{A}}{\partial t}) = 0 \\ ⟹ \vec{E} + \frac{\partial \vec{A}}{\partial t} = - \vec{\nabla} ϕ \end{matrix}

so it is the gradient of a scalar potential $ϕ$
so, the recipe for creating fields from potentials:

\begin{matrix} \vec{E} = - \vec{\nabla} ϕ - \frac{\partial \vec{A}}{\partial t} \\ \vec{B} = \vec{\nabla} \times \vec{A} \end{matrix}

note that on the LHS, there are $6$ components (3 each from $\vec{E}$ and $\vec{B}$ ), but on the RHS, there are $3 + 1 = 4$ components ( $3$ from $\vec{A}$ and $1$ from $ϕ$ ), so not all are independent
so, $\vec{A}$ is not unique, but its choice affects $ϕ$ as well
consider:

{\vec{A}}^{'} = \vec{A} + \vec{\nabla} ψ

check:

{\vec{B}}^{'} = \vec{\nabla} \times {\vec{A}}^{'} = \vec{\nabla} \times \vec{A} + \vec{\nabla} \times (\vec{\nabla} ψ) = \vec{B}

{\vec{E}}^{'} = - \vec{\nabla} ϕ^{'} - \frac{\partial {\vec{A}}^{'}}{\partial t} = - \vec{\nabla} (ϕ^{'} + \frac{\partial ψ}{\partial t}) - \frac{\partial \vec{A}}{\partial t}

to get $\vec{E} = {\vec{E}}^{'} :$

ϕ = ϕ^{'} + \frac{\partial ψ}{\partial t}

the selection of $ψ$ is called 'choosing the gauge', usually done by choosing $\vec{\nabla} \cdot \vec{A}$
coulomb gauge: $\vec{\nabla} \cdot \vec{A} = 0$ , useful for statics or plane waves
lorenz gauge: $\vec{\nabla} \cdot \vec{A} = - \frac{1}{c^{2}} \frac{\partial ϕ}{\partial t}$ , useful for realtivity