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Is it always possible to define function $\psi$ satisfying the Lorenz gauge equation $$ \partial_{\mu}\partial^{\mu} \psi + \partial_{\mu}A^{\mu} = 0? $$

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    $\begingroup$ Gauge conditions are a constraint we choose to impose. That is, we don't solve gauge conditions for unknowns, we impose gauge conditions. $\endgroup$
    – DJBunk
    Commented Mar 28, 2013 at 15:19
  • $\begingroup$ Can you explain your words? Gauge fixing is possible because the 4-potential is ambiguous, i.e. $$ A^{\mu} -> A^{\mu} + \partial_{\mu}\psi . $$ And I don't see evidence that the condition $\partial_{\mu}A^{\mu}$ is always satisfied $\endgroup$
    – user8817
    Commented Mar 28, 2013 at 15:25
  • $\begingroup$ The fact that we can write down equations like this at all is because of the gauge redundancy of fields like $A^\mu$. That is, there is extra freedom in these variables as opposed to just writing things in terms of the electric and magnetic fields $\mathbf{E}$ and $\mathbf{B}$. If we choose to eliminate some of this redundancy is our choice, or the `gauge' we choose. This is, for an equation like the one you wrote above you are choosing to restrict the extra degrees of freedom in some way. Choosing a particule $\psi$ is part of this choice. $\endgroup$
    – DJBunk
    Commented Mar 28, 2013 at 18:45
  • $\begingroup$ I do not understand this question. I guess that in your notation $A^{\mu}$ is the 4-potential. So the Lorenz gauge is $\partial_{\mu} A^{\mu}=0$. Are you wondering if a function $\partial _{\mu} \partial ^{\mu} \Psi =0$ exists ? In which case the answer is yes, and this function $\Psi$ describes a wave if $g_{\mu \nu} = diag \left( 1,-1,-1,-1 \right)$ among other. $\endgroup$
    – FraSchelle
    Commented Mar 28, 2013 at 19:20
  • $\begingroup$ The question (v2) is basically: Does a gauge orbit intersect the Lorenz gauge condition at least once? $\endgroup$
    – Qmechanic
    Commented Apr 27, 2013 at 17:05

2 Answers 2

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Yes. If you define $f=-\partial_\mu A^\mu$ then you can write the equation in the form $$ \partial_\mu\partial^\mu\psi = f$$ This is the Klein-Gordon equation with a nonzero source ($f$) and can be solved via Green's function methods. Once you have the Klein-Gordon propagator* $G(x)$ (this is derived in any e.g. quantum field theory textbook) appropriate to the boundary conditions the solution can be written as $$ \psi(x)=\int d^4 x' G(x-x') f(x')$$ since Green's functions by definition satisfy $$\partial_\mu\partial^\mu G(x-x')= \delta(x-x')$$ where we take all differentiations to be with respect to x.

*You need the propagator in the position space representation to write this down. It is usually more convenient to write it in momentum space; you can go back and forth using (inverse) Fourier transforms.

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Yes, sure, it is always possible to find $\psi$ so that your equation will be obeyed – i.e. that the new $A_\mu$ will obey the Lorenz gauge – assuming that $A_\mu$ obeys the appropriate continuity conditions etc.

There are many function like that (in flat infinite spacetime). For example, you may choose $\psi(x,y,z,t=0)$ arbitrarily and study the condition above at each point $(x,y,z)$ separately. Then the equation is just a very simple ordinary differential equation that depends on time which may be solved $dt$ after $dt$.

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  • $\begingroup$ "...For example, you may choose $\psi (x,y,z,t = 0)$ arbitrarily and study the condition above at each point $(x,y,z)$ separately..." Can you explain these words in details? $\endgroup$
    – user8817
    Commented Mar 28, 2013 at 15:29

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