How Does Gauge Symmetry Allow Solutions to the Lorentz Gauge Condition?

[QuasarBoy543298]

hi, I'm currently taking a classical field theory class (electromagnetism in the language of tensors and actions and etc) and we have just encountered the gauge symmetry, that is for the 4 vector potential we can add a gradient of some smooth function and get the same physics (if we take A_μ → A_μ + ∂_μf the actions stays the same, except some irrelevant constant).
now after that long preview, my question is - how can we conclude from that freedom that we can find a solution for the field A, with any boundary conditions, that satisfies the Lorentz gauge ( ∂_μA^μ = 0)?

 

[Meir Achuz]

That is done on most textbooks. It is the Lorenz gauge.

 

[vanhees71]

Why it's Lorenz gauge rather than Lorentz gauge, see

https://arxiv.org/abs/hep-ph/0012061

NB: Don't be appalled by the use of MS Word, which usually is a bad sign in theoretical physics. This paper is a really good one. It appeared in readable form in Rev. Mod. Phys.

https://doi.org/10.1103/RevModPhys.73.663

 

[Orodruin]

how can we conclude from that freedom that we can find a solution for the field A, with any boundary conditions, that satisfies the Lorentz gauge ( ∂μAμ = 0)?

Assume that you have a 4-potential $A_\mu$ that does not satisfy the Lorenz gauge condition and write down a new 4-potential $A'_\mu = A_\mu + \partial_\mu f$. What is the condition on $f$ in order for $A'_\mu$ to satisfy the Lorenz gauge condition? Can you find such a function?