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I am trying to show that $\nabla\cdot \vec{A}=-\mu_0 \epsilon_0 \frac{\partial V}{\partial t}$

$V=\frac{1}{4\pi\epsilon_0}\int \frac{\rho(\vec{r}',t_r)}{r}d\tau'$

$\vec{A}=\frac{\mu_0}{4\pi}\int \frac{J(\vec{r}',t_r)}{r}d\tau'$

Where $d\tau'=dxdydz$, $t_r=t-\frac{r}{c}$ is the retarded time, $\vec{r}$ is the position vector of the charge and $\vec{r}'$ is the position of the observer (us) and $r=|\vec{r}-\vec{r}'|$ the distance from us to the source.

Probably best to start with $\nabla\cdot \vec{A}$:

$\nabla\cdot \vec{A} =\frac{1}{4\pi \epsilon_0}\int \nabla \cdot \left(\frac{\vec{J}(\vec{r}',t_r)}{r}\right)d\tau'$

Apparently,

$\nabla\cdot ({\vec{J}\over r})=\frac{1}{r}(\nabla\cdot \vec{J})+\frac{1}{r}(\nabla'\cdot \vec{J})-\nabla'\cdot ({\vec{J}\over r})$

This is the first thing I need to do to get any further but I cannot do this. $\nabla$ is differentiation with respect to $r$ and $\nabla'$ is with respect to $r'$.

I thought that

$\nabla\cdot ({\vec{J}\over r})=\frac{1}{r}(\nabla\cdot\vec{J})+\vec{J}\cdot (\nabla(\frac{1}{r}))$

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It is called Lorenz gauge, after Ludvig Lorenz, not after Hendrik Lorentz. –  Luboš Motl Apr 2 '13 at 19:35
    
OK changed the title –  shilov Apr 2 '13 at 19:40

1 Answer 1

up vote 1 down vote accepted

The identity $$ \nabla\cdot \vec{A}=-\mu_0 \epsilon_0 \frac{\partial V}{\partial t} $$ holds because $V,\vec A$ are calculated from $\rho, \vec J$ by the same integral transform and because $$\nabla\cdot \vec J = - \frac{\partial\rho}{\partial t} $$ The integral transform may be written in the combined way $$ \{V \epsilon_0(\vec r, t),\vec A(\vec r,t)/\mu_0\} = \int d t'\,d^3 r' \{\rho(\vec r',t'),\vec J(\vec r',t')\} D(\vec r'-\vec r,t'-t)$$ where the braces represent a "list" (like in Mathematica) and where the kernel of the integral transform is $$ D(\vec y,t) = \frac{1}{4\pi |\vec y|} \delta(t-|\vec y|) \theta(t). $$ Note that in your formulae for $V,\vec A$, you forgot to integrate over the 3-space and divide by $1/|\vec r - \vec r'|$. The theta function guarantees that the potential is retarded, omitting any advanced contributions. The delta-function produces the right distance.

Now, the integral transform may be expressed by an operator ${\mathcal D}$ that is translationally symmetric. It therefore commutes with the operator of the four-divergence $$\nabla_4\cdot\{ V\epsilon_0,\vec A/\mu_0 \}\equiv \frac{\nabla\cdot \vec A}{\mu_0}+ \frac{\partial V\epsilon_0}{dt}=\dots$$ may be written, because $V\epsilon_0={\mathcal D}\rho$ and $\vec A/\mu_0={\mathcal D}\vec J$, as $$\dots = \nabla_4\cdot {\mathcal D} \{\rho,\vec J\} = {\mathcal D}\nabla_4\cdot \{\rho,\vec J \} = 0$$ but this vanishes because of the continuity equation for the sources (even before we act on it with ${\mathcal D}$). The proof may be rewritten without unconventional "operators" but I think that its idea is much more transparent in this form.

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1  
For those at home, the "list" of the vector and scalar potentials constructed here is, of course, the four-vector potential in Minkowski spacetime (which, unfortunately, also tends to be called $A$). –  Muphrid Apr 2 '13 at 21:03
    
I am going to examine this for a bit –  shilov Apr 3 '13 at 14:26

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