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

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?

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

$$\nabla \cdot (f\vec{c})=\nabla f \cdot \vec{c}$$ 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 $$g(x,y,z)=c_1\ln{(x^2+y^2)}$$.

• can't you make the argument that f is linearly dependent from t, and therfore a double time derivative gives you zero and that's why you can also assume $\Delta f=0$ ? Dec 31, 2021 at 0:41