How does the retarded scalar potential satisfy the Lorentz gauge condition?

[kent davidge]

As homework, I shall show that the retarded scalar potential satisfîes the Lorentz gauge condition as well as the inhomogenous wave equation. We saw in class how to do it. But I was thinking about this, and it seems to me that it's redundant to prove both of those things. For, if the scalar potential satisfies the Lorentz gauge condition, it will automatically lead to the inhomogenous wave equation. So if I show that the retarded scalar potential satisfies the inhomogenous wave equation, that automatically implies that it satisfies the Lorentz gauge condition. So why would I have to prove both assertions?

 

[vanhees71]

First of all to do historical justice one should call the gauge Lorenz gauge (after the Danish physicist Ludvik Lorenz) rather than Lorentz gauge (after the Dutch physicist Hendrik Antoon Lorentz).

I don't understand what you mean by "the retarded scalar potential satisfies the Loren(t)z gauge condition". A scalar potential alone cannot fulfill a gauge condition (except in the temporal gauge condition, $\Phi=0$, but than in general you cannot in addition fulfill the Lorenz gauge condition if charges and currents are present).

If you impose the Lorenz gauge condition the four-potential separate component wise to wave equations,
$$\Delta A^{\mu}=\frac{1}{c} j^{\mu},$$
and the usually needed solution is the retarded one, using the corresponding Green's function of the D'Alembert operator. The solution then reads
$$A^{\mu}(x)=\int_{\mathbb{R}^4} \mathrm{d}^4 x' G_{\text{ret}}(x-x') \frac{1}{c} j^{\mu}(x').$$
It's important to check that this solution fulfills the gauge condition,
$$\partial_{\mu} A^{\mu}=0,$$
since otherwise these solutions do not lead to a valid solution of Maxwell's equations.

For details, see my SRT FAQ:

https://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf