1
$\begingroup$

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?

$\endgroup$
1

1 Answer 1

0
$\begingroup$

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

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

$\endgroup$
1
  • $\begingroup$ 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 $\Delta f=0$ ? $\endgroup$
    – imbAF
    Dec 31, 2021 at 0:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.