Question about the Lorenz gauge in classical electrodynamics

Question

The Lorenz gauge is the gauge such that $$\nabla \cdot \mathbf{A} = -\mu_0\epsilon_0\frac{\partial\Phi}{\partial t}.$$ This condition dictates what $\lambda$ is in $$\mathbf{A}' = \mathbf{A} + \nabla \lambda.$$ However, $\Phi$ can be adjusted via the equation $$\Phi' = \Phi - \frac{\partial \lambda}{\partial t}.$$ This seems circular to me because $\lambda$ is found using $\Phi$ which itself depends on $\lambda$. I don't know if $\lambda$ can be solved for in this case. I'd appreciate it if someone would dispel my confusion.

Answer 1

The Maxwell equations are independent of the gauge function $\lambda$. Our gauge condition is given by

$$\nabla \cdot \mathbf{A} = -\frac{1}{c^2}\frac{\partial\Phi}{\partial t}$$

However, note that there is still a subspace of nontrivial gauge transformations that can be made when we take $\mathbf{A} \to \mathbf{A}'$ and $\varphi \to \varphi'$. Note that this still doesn't completely fix the gauge. Since we can can still make the transformations

$$ \mathbf{A}' = \mathbf{A} + \nabla \lambda$$

$$\Phi' = \Phi - \frac{\partial \lambda}{\partial t}.$$

for some special kinds (but not uniquely defined) of $\lambda$s and still have a valid solution to Maxwell's equations. That is, if we can fix $\lambda$ then we fix the gauge.

But what do I mean by special? To figure that out we apply the gauge transformation to the Lorentz condition

\begin{align*} \nabla \cdot \mathbf{A} = -\frac{1}{c^2}\frac{\partial\Phi}{\partial t} \to &\nabla \cdot \mathbf{A'} = -\frac{1}{c^2}\frac{\partial\Phi'}{\partial t}\\ &= \nabla\cdot \mathbf{A} + \nabla^2\lambda= - \frac{1}{c^2}\frac{\partial\Phi}{\partial t} + \frac{1}{c^2} \frac{\partial^2 \lambda}{\partial t^2} \end{align*}

So we see that we still have the freedom to choose any function $\lambda$ that satisfies

$$ \nabla^2\lambda = \frac{1}{c^2} \frac{\partial^2 \lambda}{\partial t^2} \tag{c}. $$

Note that any such $\lambda$ satisfying (c) is still a valid solution to Maxwell's equations (it should be clear that there is not just one $\lambda$ that does this). That is, there is still a redundancy in the theory. To completely fix the gauge you need to add boundary conditions to equation (c).