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In the Lorenz Gauge, we can write Maxwell's equations as: $$\tag1 \Box A^\beta=\mu_0j^\beta.$$

We then go on to solve this by treating each component $A^\beta$ as an independent solution of the scalar wave equation with source, and summing Green functions. That is: $$\tag2 A^\beta=\mu_0\int \frac{1}{4\pi |r-r'|}j^\beta(ct-|r-r'|, r')\ \mathrm d^{3}r'.$$

My question is, how have we ensured that the Lorenz gauge condition is still met?

Sure to write $(1)$, we needed $\partial_\mu A^\mu = 0$, but how have we ensured our solution $(2)$ meets this condition?

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  • $\begingroup$ Hint: what is $\partial_\mu j^\mu$? [also, its easier if you write $(2)$ in terms of $\mathrm dt'\mathrm d^3r'$] $\endgroup$ May 16, 2017 at 18:36
  • $\begingroup$ I thought that you were still struggling so I decided to post a more detailed hint. I'm glad you could solve it on your own though :-) $\endgroup$ May 18, 2017 at 21:31

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Sketch of the argument:

  1. Write your equation in the form $$ A^\mu(x)=\int_{\mathbb R^4} G(x-x')j^\mu(x')\ \mathrm dx' $$ where $G$ is one of the propagators of the wave equation. For example, $G_\mathrm{ret}(x)\propto\delta(x^2)\Theta(x^0)$ which, after integration over $\mathrm dx^0$, leads to your second formula.

  2. Show that $$ \partial_\mu A^\mu(x)=\int_{\mathbb R^4} G(x-x')\partial'_\mu j^\mu(x')\,\mathrm dx' $$ where we have use the fact that $G$ depends only on the difference $x-x'$, and we have integrated by parts. Here one should argue that boundary terms vanish because of the kinematics of $G$.

  3. Argue by current conservation that the Lorenz gauge condition holds. One should note that, independently of the chosen gauge, $$ \partial_\mu F^{\mu\nu}=j^\nu $$ and therefore, due to the skew-symmetry of $F$, the current must be conserved, as a consistency condition for the above PDE to be well-posed.

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