Reasoning behind the Lorenz gauge

Question

It is easy to show that Maxwell's equations are invariant under the following transformation of the four-potential $A^{\mu}$:

$$A^{\mu} \rightarrow A'^{\mu} = A^{\mu} + \partial^{\mu} \chi$$

for some scalar field $\chi$. The idea of gauge fixing is that we can choose different values of $\chi$.

In the Lorenz gauge we fix the gauge such that

$$\partial_{\mu} A'^{\mu} = 0$$

is satisfied. The way I understand it, the only freedom we have is to change the scalar field $\chi$ however we like in order to achieve the $\partial_{\mu} A'^{\mu} = 0$ condition. So it seems to me that we would have to transform $A^{\mu}$ to $A'^{\mu}$, such that

$$\partial_{\mu} A'^{\mu} = 0$$ $$\partial_{\mu} (A^{\mu} + \partial^{\mu}\chi) = 0$$ $$\partial_{\mu}\partial^{\mu}\chi = -\partial_{\mu} A^{\mu}$$

Therefore, according to my reasoning, we have to choose a scalar field $\chi$ which satisfies $\partial_{\mu}\partial^{\mu}\chi = -\partial_{\mu} A^{\mu}$, and then we work with $A'^{\mu} = A^{\mu} + \partial^{\mu} \chi$.

However, I'm wrong. All sources I've checked (including https://en.wikipedia.org/wiki/Lorenz_gauge_condition ) claim that we have to choose $\chi$ such that

$$\partial_{\mu}\partial^{\mu}\chi = 0.$$

The only way I can make sense of this (and to try and understand where I have gone wrong) is that the Lorenz gauge is a gauge of the type

$$A^{\mu} \rightarrow A'^{\mu} = A^{\mu} + \partial^{\mu} \chi$$

where we impose both

$$\partial_{\mu} A'^{\mu} = 0$$

and

$$\partial_{\mu} A^{\mu} = 0$$

such that $\partial_{\mu} A'^{\mu} = \partial_{\mu} (A^{\mu} + \partial^{\mu}\chi) = 0$ implies $\partial_{\mu}\partial^{\mu}\chi = 0$. If this is the correct interpretation of the reasoning behind the Lorenz gauge, then I do not understand it. How can we impose both $\partial_{\mu} A'^{\mu} = 0$ and $\partial_{\mu} A^{\mu} = 0$?

What am I getting wrong with my reasoning? Am I misunderstanding the idea of gauge fixing in the first place?

Answer 1

Your are confusing two things

1. The condition on $\chi$ such that the gauge transformed field $$A'_\mu := A_\mu +\partial_\mu \chi$$

is in Lorenz gauge. This is indeed $\Box \chi = -\partial^\mu A_\mu$.

2. Given that $A_\mu$ already satisfies the Lorenz gauge condition you can ask which $\chi$'s are allowed such that $A'_\mu$ still statisfies the Lorenz gauge condition. This is acutally a special cause of condition 1: $$\Box \chi = -\partial^\mu A_\mu = 0\,.$$

Answer 2

If you start out with a potential that does not obey the Lorenz condition you can fix that with a gauge transformation as you correctly state. Then you still have the remaining freedom of a gauge transformation for which $$\partial_\mu \partial^\mu \chi = 0 \,.$$ Of course you may wonder what kind of $\chi$ obeys this equation? A largely arbitrary neutral massless scalar field that obeys a free wave equation in the entire Minkowski space, that is never has and never will interact with anything anywhere. Except with any physicist's free will. There's no physics in there, it's only math.

Answer 3

Everything you said is correct. To set the potential $A_μ$ to satisfy the Lorenz condition, you choose a $χ$ such that $∂_μ∂^μχ = -∂_μA^μ$, and $A^\prime_μ = A_μ + ∂_μχ$ will satisfy the Lorenz condition. The only thing the source said is that the choice is not unique, because if $A^\prime_μ$ already satisfies the Lorenz condition, you can still add another gauge $χ^\prime$ such that $∂_μ∂^μχ^\prime = 0$ and the result $A^\prime_μ + ∂_μχ^\prime$ will also satisfy the Lorenz condition. So, for the original $A_μ$: both $A_μ + ∂_μχ$ and $A_μ + ∂_μχ + ∂_μχ^\prime$ satisfy the Lorenz condition.