Transforming the potentials that satisfy Lorenz & Coulomb gauge to potentials that satisfy only Lorenz gauge

Question

If $\vec E(\vec r,t)=\vec E_0sin(\vec k \vec r- \omega t)$ and also that $\rho(\vec r,t)=0$ and $\vec j(\vec r,t)=0$

I was asked to find $\vec A(\vec r,t)$ and $\phi (\vec r,t)$ which satisfy both the Lorenz and Coulomb gauge. So the first thing I do is that I transform the potentials:

$$\vec A(\vec r,t)'= \vec A(\vec r,t) + \nabla f(\vec r,t)$$

$$\phi(\vec r,t)'=\phi(\vec r,t) - \frac{\partial f(\vec r,t)}{\partial t}.$$

In lorenz gauge:

$$\nabla \vec A(\vec r,t)'+\frac{1}{c^2}\frac{\partial \phi(\vec r,t)'}{\partial t}=0$$

And because I also assumed that simultaneously the Coulomb gauge is satisfied ($\nabla \vec A(\vec r,t)'=0$) then I get that $\frac{1}{c^2}\frac{\partial \phi(\vec r,t)'}{\partial t}=0$ .

Then in order to find an expression for $\vec A(\vec r,t)'$:

$$\vec E(\vec r,t)=-\frac{\partial \vec A(\vec r,t)'}{\partial t}- \nabla \phi(\vec r,t)'$$ $$\frac{\partial}{\partial t}\vec E(\vec r,t)=-\frac{\partial^2 \vec A(\vec r,t)'}{\partial t^2}- \nabla \frac{\partial}{\partial t}\ \phi(\vec r,t)'$$

Since $\frac{1}{c^2}\frac{\partial \phi(\vec r,t)'}{\partial t}=0$ then:

$$\frac{\partial}{\partial t}\vec E(\vec r,t)=-\frac{\partial^2 \vec A(\vec r,t)'}{\partial t^2}$$

From here, plugging in the expression for the electric field and integrating we get:

$$\vec A(\vec r,t)'=-\frac{\vec E_0}{\omega}\cos(\vec k \vec r- \omega t) + C_1(\vec r)t + C_2(\vec r)$$

Because $$\vec B(\vec r,t)=\nabla \times \vec A(\vec r,t)'=\nabla \times \vec A(\vec r,t)$$

We find out that $\nabla \times (C_1(\vec r)t + C_2(\vec r))=0$ and that is possible when we do the curl of the gradient hence we can say that $C_1(\vec r)t= \nabla c_1(\vec r)t$ and $C_2(\vec r)= \nabla c_2(\vec r)t$. From here we can say that $f(\vec r,t)=c_1(\vec r)t + c_2(\vec r)$

I also found that $\phi(\vec r,t)'=-\frac{\partial f(\vec r,t)}{\partial t}$.

Then I rename my potentials in the following manner:

$$\vec A(\vec r,t)=-\frac{\vec E_0}{\omega}\cos(\vec k \vec r- \omega t) + \nabla(c_1(\vec r)t + c_2(\vec r))$$

$$\phi(\vec r,t)=-\frac{\partial f(\vec r,t)}{\partial t}$$.

What I am unable to solve is, starting from these transformed potentials, I need to find the potentials which now solve the lorenz gauge but not the Coulomb gauge. Can someone help me, or give me a hint as to how I should proceed?

Answer 1

$\vec{E}=\vec{E_0}\sin{(\vec{k}\cdot\vec{r}-\omega t)}$. $\vec{J}=0,\rho=0$.

$\nabla \cdot (f\vec{c})=\nabla f \cdot \vec{c}$ if $\vec{c}$ is constant.

$\nabla \sin{(\vec{k}\cdot \vec{r}-\omega t)}=\vec{k}\cos{(\vec{k}\cdot \vec{r}-\omega t)}$

$\nabla \cdot \vec{E}=\rho/\epsilon _0=\vec{k}\cdot\vec{E_0}\cos{(\vec{k}\cdot \vec{r}-\omega t)}=0$

So $\vec{k}\cdot\vec{E_0}=0$ since the equation is time independent.

$\nabla \times (f\vec{c})=\nabla f \times \vec{c}$.

$\nabla \times \vec{E}= \vec{k}\times\vec{E_0}\cos{(\vec{k}\cdot \vec{r}-\omega t)}=-\partial \vec{B}/\partial t$

$\vec{B}=\frac{\vec{k}}{\omega}\times \vec{E_0}\sin{(\vec{k}\cdot \vec{r}-\omega t)}$

Reversing the identity, we can Let $\vec{A}=\frac{-1}{\omega}\vec{E_0}\cos{(\vec{k}\cdot \vec{r}-\omega t)}$ to get the vector potential for $\vec{B}$. Since its negative time derivative gives us $\vec{E}$, then $V=V_0$, a constant.

Since the wave vector and the electric field or orthogonal, $\vec{A}$ satisfies the Coulomb Gauge. Since V is constant, the combination solves the Lorentz Gauge as well.

$\vec{A}'=\vec{A}+\nabla f$

$V'=V-\partial f/ \partial t$

The Coulomb Gauge remains satisfied if $\nabla ^2 f=0$, but f will fail the Lorentz gauge unless $\nabla ^2 f -\frac{1}{c^2}\frac{\partial^2 f}{\partial t^2}=0$.

Let $f=g(x,y,z)e^{i(\vec{k}\cdot \vec{r}-\omega t)}$, a general solution to the equation.

$\nabla(mn)=n\nabla m + m \nabla n$

So $\nabla f= \nabla g e^{i(\vec{k}\cdot \vec{r}-\omega t)}+ ig \vec{k}e^{i(\vec{k}\cdot \vec{r}-\omega t)} $

$\nabla^2 f = \nabla^2g e^{i(\vec{k}\cdot \vec{r}-\omega t)} + 2i\vec{k}\cdot \nabla ge^{i(\vec{k}\cdot \vec{r}-\omega t)}-gk^2e^{i(\vec{k}\cdot \vec{r}-\omega t)}$

$\frac{1}{c^2}\frac{\partial^2 f}{\partial t^2}=-k^2ge^{i(\vec{k}\cdot \vec{r}-\omega t)}$

So we satisfy Lorentz iff:

$\nabla^2g e^{i(\vec{k}\cdot \vec{r}-\omega t)} + 2i\vec{k}\cdot \nabla ge^{i(\vec{k}\cdot \vec{r}-\omega t)}-gk^2e^{i(\vec{k}\cdot \vec{r}-\omega t)}=-k^2ge^{i(\vec{k}\cdot \vec{r}-\omega t)}$

or: $\nabla^2 g +2i\vec{k}\cdot \nabla g=0$. It fails the Coulomb gauge if $g$ is non-zero.

From there it depends on your coordinate system and $\vec{k}$.

If $\vec{k}$ is parallel to the $z$ axis and we use cylindrical coordinates, then the equation is solved if

$\frac{1}{\rho}\frac{d}{d \rho}(\rho \frac{d g}{d \rho})=0$, where $\rho$ is the point's distance from the z axis.

A possible solution $g(x,y,z)=c_1\ln{(x^2+y^2)}$.