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It has been a while since I have done any electromagnetism, and at one point I knew how to do this but for the life of me can't figure this out. The problem is as follows, if we have a time dependent charge at the origin given by:

$$q(t)=q_0\sin(\omega t)$$

How can we find the electric potential $\Phi$ in the Lorenz gauge? $\Phi$ must satisfy the following equation in the Lorenz gauge:

$$-\nabla^2\Phi +\mu_0\epsilon_0\frac{\partial^2\Phi}{\partial t^2}=\frac{\rho(\mathbf{x},t)}{\epsilon}$$

I'm pretty sure that we have:

$$\rho(\mathbf{x},t)=q_0\sin(\omega t)\delta^3(\mathbf{x})$$ but other than that I have no idea where to go from here. I can readily write the potential in the coulomb gauge, is there away I can make gauge transformation without knowing/finding $\mathbf{A}$?

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    $\begingroup$ Without a current there is no charge conservation and without charge conservation Maxwell's equations do not apply. Some clarification is needed. $\endgroup$
    – my2cts
    Commented Feb 15, 2022 at 23:15
  • $\begingroup$ The only context I was given was that there is a charge at the origin with time dependence. Can current not be given by $\frac{dq}{dt}$ $\endgroup$
    – Chris
    Commented Feb 15, 2022 at 23:31
  • $\begingroup$ You are correct. The current is not needed to find $\Phi$. $\endgroup$
    – my2cts
    Commented Feb 16, 2022 at 0:43
  • $\begingroup$ @my2cts ok so then what context is needed? $\endgroup$
    – Chris
    Commented Feb 16, 2022 at 0:53
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    $\begingroup$ In Lorenz Gauge the solution to the wave equation of the electrical potential is given by $\phi(\mathbf{x},t) = \frac{1}{4\pi} \int d^3y \frac{\rho(\mathbf{y},t-|\mathbf{x}-\mathbf{y}|/c)}{|\mathbf{x}-\mathbf{y}|}$. The delta-distribution in the charge density makes this integral trivial. The solution also makes sense intuitively. $\endgroup$
    – Samuel
    Commented Feb 16, 2022 at 0:55

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