1
$\begingroup$

The electric field $\mathbf{E}$ and the magnetic induction $\mathbf{B}$ can be parameterized in terms of potentials $V$ and $\mathbf{A}$: $$ \mathbf{E}=-\nabla V-\frac{\partial \mathbf{A}}{\partial t},\quad \mathbf{B}=\nabla\times \mathbf{A}.$$ This parameterization is not unique, as we can find a scalar function $\theta$ and define a couple $(\tilde{V},\mathbf{\tilde{A}})$ via $\tilde{V} = V-\partial \theta/\partial t$ and $\mathbf{\tilde{A}}=A+\nabla \theta$ $(*)$. Then both $(V,\mathbf{A})$ and $(\tilde{V},\mathbf{\tilde{A}})$ will give rise to the same $(\mathbf{E},\mathbf{B})$.

Via Maxwell's equations we can find a coupled system of differential equations for $V$ and $\mathbf{A}$: $$ \begin{cases} \square \mathbf{A} = -\mu \mathbf{J}+\nabla\left( \nabla\cdot \mathbf{A}+\varepsilon\mu \frac{\partial V}{\partial t}\right) \\ \square V = -\frac{\rho}{\varepsilon} -\frac{\partial}{\partial t}\left( \nabla\cdot \mathbf{A}+\varepsilon\mu \frac{\partial V}{\partial t}\right)\end{cases},\quad \square=\nabla^2-\frac{\partial^2}{\partial t^2}.$$

These can be made independent by considering the Lorenz-gauge, in which we set $\nabla\cdot \mathbf{A}+\varepsilon\mu \frac{\partial V}{\partial t} = 0$. How can one explicitly show that for each $(V,\mathbf{A})$ there is $(\tilde{V},\mathbf{\tilde{A}})$ (i.e. give rise to the same fields) such that this couple will satisfy the Lorenz-gauge condition. Is it enough to consider the expressions $(*)$ and deduce that both couples of potentials need to satisfy the Lorenz-gauge condition, resulting in the condition $\square \theta = 0$, i.e. we can always choose a scalar function $\theta$ for which $\square \theta=0$ and consider a new potentials via $(*)$?

Thanks in advance.

$\endgroup$
1
$\begingroup$

Let's say that you have a pair of potentials $(V,\mathbf{A})$ that do not satisfy the Lorenz Gauge. i.e. $$\mathbf{\nabla \cdot A} + \frac{1}{c^2}\frac{\partial V}{\partial t} = f(x,t) \neq 0.$$

Let's now perform a gauge change to some new $(V',\mathbf{A'})$ using the function $\theta(x,t)$ as you mentioned.

\begin{equation*} \begin{aligned} \mathbf{A'} &= \mathbf{A} + \nabla\theta\\ V' &= V - \frac{\partial\theta}{\partial t} \end{aligned} \end{equation*}

Of course, these new potentials have been constructed such that they will also produce the same fields as $(V,\mathbf{A})$, by Gauge Invariance. Substituting these relations into the equation above, we get

$$\mathbf{\nabla \cdot \left(A' - \nabla\theta\right)} + \frac{1}{c^2}\frac{\partial}{\partial t}\left(V' + \frac{\partial\theta}{\partial t}\right) = f(x,t),$$

and rearranging, we have

$$\mathbf{\nabla \cdot A'} + \frac{1}{c^2}\frac{\partial V'}{\partial t} - \nabla^2\theta + \frac{1}{c^2}\frac{\partial^2\theta}{\partial t^2} = f(x,t).$$

From this it should be clear that if the function $\theta(x,t)$ that we chose satisfies the condition

$$\boxed{-\nabla^2\theta + \frac{1}{c^2}\frac{\partial^2\theta}{\partial t^2} = f(x,t),}$$

then

$$\mathbf{\nabla \cdot A'} + \frac{1}{c^2}\frac{\partial V'}{\partial t}=0.$$

The problem now reduces to finding a function $\theta(x,t)$ that solves the above wave equation with a source $f(x,y)$, and from the properties of the wave equation, we can always find such a $\theta(x,t)$, provided it's "source" $f(x,t)$ isn't too wacky.


EDIT: As @hyportnex points out in the comments and in this answer, in order to solve a wave equation like the one above, one needs to completely specify the boundary conditions on $\theta$ which could make solving the equation non-trivial, though I still feel that a solution should always exist. However, I'd appreciate it if anyone could correct me.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ This is all fine in vacuum but if matter is present you will also have to satisfy certain continuity/discontinuity boundary conditions. It is not obvious that those can be satisfied for some $\theta$ given $f$ with $\nabla^2\theta + \frac{1}{c^2}\frac{\partial^2\theta}{\partial t^2} = f(x,t)$. $\endgroup$ – hyportnex Jun 30 at 18:51
  • $\begingroup$ @hyportnex That's interesting, I didn't know that. I mean, clearly $\theta$ would need boundary conditions to be completely specified, but how does the presence of charge or current in space affect this? Are you saying that there may be charge or current configurations such that the Lorenz Gauge cannot be chosen? $\endgroup$ – Philip Jun 30 at 19:19
  • $\begingroup$ I am not saying that, all I am saying is that to the Helmholtz equation you quoted must be added some boundary conditions, and I do not see that as being a trivial amendment. see for example physics.stackexchange.com/questions/535577/… , but that has to be made relativistically invariant. $\endgroup$ – hyportnex Jun 30 at 20:06
  • $\begingroup$ Very interesting. I will edit my answer to include your point. $\endgroup$ – Philip Jun 30 at 20:26

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.