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## Homework Statement

Confirm that the retarded potentials satisfy the Lorentz gauge condition.

## Homework Equations

[itex]\vec{A}(\vec{r}, t) = \frac{\mu_{0}}{4\pi}\int\frac{\vec{J}(\vec{r'},t_{r})}{R}d\tau'[/itex]

[itex]V(\vec{r}, t) = \frac{1}{4\pi\epsilon_{0}}\int\frac{\rho(\vec{r'},t_{r})}{R}d\tau'[/itex]

Where:

[itex] \vec{r} [/itex]: position of measurement

[itex] \vec{r'} [/itex]: integration variable (running on charge and current densities)

[itex] R = |\vec{R}| = |\vec{r} - \vec{r'}| [/itex]

[itex] t_{r} [/itex] is the retarded time: [itex] t_{r} = t - \frac{R}{c} [/itex]

[itex] \vec{J}(\vec{r'},t_{r}) [/itex] is the current density (estimated at the point [itex] \vec{r'}[/itex] at the retarted time), and [itex] \rho(\vec{r'},t_{r}) [/itex] is the charge density.

Lorentz gauge means:

[itex] \nabla \cdot \vec{A} = -\epsilon_{0}\mu_{0} \frac{\partial V}{\partial t} [/itex]

## The Attempt at a Solution

Now, Griffiths gives a rather lengthy hint containing different formulas that one could apply in order to prove this. But I found his method cumbersome and could not understand why it's necessary (he uses relations between primed and unprimed operators and eventually gets there). For me, my "straight-forward" method is simpler, but for some reason it doesn't work. I'd really appreciate your insights!

[itex] \nabla \cdot \vec{A} = \frac{\mu_{0}}{4\pi}\int\nabla \cdot (\frac{\vec{J}(\vec{r'},t_{r})}{R}) d\tau' = \frac{\mu_{0}}{4\pi}\int [\frac{1}{R}\nabla \cdot \vec{J} - \frac{1}{R^{2}} \hat{R} \cdot \vec{J}]d\tau'[/itex]

Meanwhile:

[itex] -\epsilon_{0}\mu_{0} \frac{\partial V}{\partial t} = - \frac{\mu_{0}}{4\pi}\int \frac{\partial}{\partial t} (\frac{\rho(\vec{r'},t_{r})}{R})d\tau' = - \frac{\mu_{0}}{4\pi}\int \frac{1}{R}\frac{\partial \rho}{\partial t} d\tau' = \frac{\mu_{0}}{4\pi}\int \frac{1}{R}\nabla \cdot \vec{J} d\tau' [/itex]

Where in the last inequality I've used the continuity equation: [itex]\frac{\partial \rho}{\partial t} = -\nabla \cdot \vec{J}[/itex]

So somehow the next equation should hold:

[itex] \frac{\mu_{0}}{4\pi}\int \frac{1}{R}\nabla \cdot \vec{J} d\tau' = \frac{\mu_{0}}{4\pi}\int [\frac{1}{R}\nabla \cdot \vec{J} - \frac{1}{R^{2}} \hat{R} \cdot \vec{J}]d\tau'[/itex]

However, unless the integral containing [itex]\frac{1}{R^{2}} \hat{R} \cdot \vec{J} [/itex] vanishes, this is obviously wrong. And I can't seem to prove it vanishes.

Am I doing something wrong? Is any step here false? THANK YOU!

Tomer