# Lorenz gauge fixing

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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Gauge conditions are a constraint we choose to impose. That is, we don't solve gauge conditions for unknowns, we impose gauge conditions. – DJBunk Mar 28 '13 at 15:19
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 – user8817 Mar 28 '13 at 15:25
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. – DJBunk Mar 28 '13 at 18:45
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. – FraSchelle Mar 28 '13 at 19:20
The question (v2) is basically: Does a gauge orbit intersect the Lorenz gauge condition at least once? – Qmechanic Apr 27 '13 at 17:05

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.
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 $dt$.
"...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? – user8817 Mar 28 '13 at 15:29