# Vector potential

### Vector potential in magnetostatics

Since $\nabla\cdot \mathbf{B}=0$ is always true, one can think of $\mathbf{B}$ as a curl of another vector field. (recall $\nabla\cdot(\nabla\times A)=0$)

${\color{Blue}{B(r)=\nabla\times A(r) }}$

Using vector identities

$\nabla\left({1\over R}\right)=-{\nabla}'\left({1\over R}\right)=-{{\hat{\mathbf{R}}}\over{R^2}}=-{\mathbf{R}\over{R^3}}$
$\nabla\times(uA)=(\nabla u)\times A+u(\nabla\times A)$

Biot-Savart law can be rearranged as follows:

${{ds'\times\hat{\mathbf{R}}}\over{R^2}}=-ds'\times\nabla\left({1\over R}\right)=\nabla\times\left({ds' \over R}\right)-{{\nabla\times{ds'}}\over R}=\nabla\times\left({ds' \over R}\right)$
$\mathbf{B(r)}={\mu_0 \over{4\pi}}{\oint_{C'} {{I' ds' \times \hat{\mathbf{R}}}\over R^2}}={\mu_0 \over{4\pi}}\oint_{C} {{I ds \times\mathbf{R(r)}}\over{R^3}}$
$\mathbf{B}={{\mu_0 I'}\over{4\pi}}\oint_{C'} \nabla\times\left({ds'\over R}\right)=\nabla\times\left({{\mu_0 I'}\over{4\pi}}{\oint_{C'} {ds'\over R}}\right)$

Therefore, the vector potential due to current I can be written as:

$\mathbf{A(r)}={\mu_0 \over{4\pi}}\oint_{C'} {{I' ds'}\over R}$
$\mathbf{A(r)}=\Sigma{{\mu_0 \over{4\pi}}\oint_{C'} {{I' ds'}\over R}}$

### Potentials in electrodynamics

When $\mathbf{B}=\nabla\times\mathbf{A}$ is plugged into Faraday's law, we have $\nabla\times\mathbf{E}+\frac{\partial}{\partial t}(\nabla\times\mathbf{A})=0$ which leads to $\nabla\times\left(\mathbf{E}+\frac{\partial\mathbf{A}}{\partial t}\right)=0$, so

$\mathbf{E}=-\mathbf{\nabla} V-{{\partial \mathbf{A}}\over{\partial t}}$

Gauss's law becomes:

${\nabla}^2 V+{\partial\over{\partial t}}(\nabla\cdot \mathbf{A})=-{\rho\over{{\epsilon}_0}}$

Ampere-Maxwell's law becomes:

${\nabla}^2 \mathbf{A}-\mu_0 {\epsilon}_0 {{{\partial}^2 \mathbf{A}}\over{\partial t^2}}-\nabla\left(\nabla\cdot \mathbf{A}+\mu_0 \epsilon_0 {{\partial V}\over{\partial t}}\right)=-\mu_0 \mathbf{J}$

By defining D'Alembertian operator as

$\square^2 \equiv \nabla^2 - \mu_0 \epsilon_0 {{\partial}^2\over{\partial t^2}}$

we obtain

$\square^2 \mathbf{A} - \nabla L = -\mu_0 \mathbf{J}$
$\square^2 V + {\partial L \over \partial t}= - {\rho\over \epsilon_0}$

where we have defined a quantity $L\equiv\nabla\cdot \mathbf{A}+\mu_0 \epsilon_0 {{\partial V}\over{\partial t}}$.

### Ambiguity of vector potential

In magnetostatics,

$\nabla\times\mathbf{B} = \nabla\times(\nabla\times\mathbf{A}) = \nabla(\nabla\cdot\mathbf{A})-\nabla^2\mathbf{A}=\mu_0\mathbf{J}$

So, if we choose $\nabla\times\mathbf{A}=0$, we have

$\nabla^2\mathbf{A} = -\mu_0\mathbf{J}$

which is a Poisson's equation, whose solution can be read off of it.

$\mathbf{A}(\mathbf{r})=\frac{\mu_0}{4\pi}\int\frac{\mathbf{J}(\mathbf{r}')}{R}\, d\tau'$

### Gauge transformation

$\mathbf{E}$ and $\mathbf{B}$ do not change under the following transformation:

$\mathbf{A}' = \mathbf{A} + \nabla\lambda$
$V' = V - {\partial\lambda\over\partial t}$

These set of transformation is called gauge transformation.

#### Coulomb gauge

Most often used in magnetostatics.

$\nabla\cdot\mathbf{A} = 0$

This choice of gauge leads us to:

${\nabla}^2 V = -{\rho\over{{\epsilon}_0}}$
${\nabla}^2 \mathbf{A}-\mu_0 {\epsilon}_0 {{{\partial}^2 \mathbf{A}}\over{\partial t^2}}=-\mu_0 \mathbf{J} + \mu_0 \epsilon_0 \nabla\left({{\partial V}\over{\partial t}}\right)$

#### Lorentz gauge

It treats $\mathbf{E}$ and $\mathbf{B}$ field on equal footing, so the whole equation becomes very neat using d'Alembertian.

$\nabla\cdot\mathbf{A} = -\mu_0\epsilon_0 {\partial V \over \partial t}$

This choice of gauge leads us to:

$\square^2 V = - {\rho\over \epsilon_0}$
$\square^2 \mathbf{A} = -\mu_0 \mathbf{J}$