Gauges

Equations 1.1,1.2 are not unique. Suppose $A'=A + a$ and $V'=V+b$, we do not violate Maxwell's equations if

$$\nabla \times a = 0 \ \Rightarrow a \ = \nabla \lambda$$

and

$$\nabla b + \frac{\partial a}{\partial t} = 0 \ \ni \nabla \left( ... ...\ \Rightarrow \ b + \frac{\partial \lambda}{\partial t} = f(t), \ \nabla f = 0.$$

Typically one redefines $\lambda \rightarrow \lambda + \int_0^t f(t')dt'$, returning the general gauge transform,

$$A' = A+\nabla \lambda$$ (2.1)

$$V' = V - \frac{\partial \lambda}{\partial t}$$ (2.2)

Two common gauges are the Coulomb and Lorenz. The Coulomb gauge has us take $\lambda$ so that $\nabla \cdot A = 0$. Thus Equation 1.1 simplifies to a harmonic equation:

$$\nabla^2V = -\frac{\rho}{\epsilon_0} \ \ni \ V(r,t)=\frac{1}{4\pi \epsilon_0}\int \frac{\rho(r',t)}{\vert r-r'\vert}d\tau'$$

Unfortunately this does not do much to simplify the defining equation for the vector potential,

$$\nabla^2A -\mu_0\epsilon_0 \frac{\partial^2 A}{\partial t^2}=-\mu_0 J + \mu_0 \epsilon_0 \nabla\left(\frac{\partial V}{\partial t}\right)$$

The Lorenz gauge has us chose $\nabla \cdot A + \mu_0\epsilon_0 \frac{\partial V}{\partial t} = 0$. Define the d'Alembertian as $\nabla^2 - \mu_0 \epsilon_0 \frac{\partial^2}{\partial t^2} \equiv \Box$. Then the Lorenz gauge reduces the potential equations to

$$\Box^2 V = -\frac{\rho}{\epsilon_0}$$ (2.3)

$$\Box^2 A = -\mu_0 J$$ (2.4)