2
$\begingroup$

Context The origin of the question below stems from this lecture here by Raman Sundrum between $48.20$ to $51$ minutes.


Let at some initial instant $t_0$, the electric and magnetic fields (E and B) are such that they can be derived from an initial field configuration of the four-potential $A^{(1)}_\mu(t_0,{\bf x})$. Obviously, this initial configuration is not unique; it is one of the infinitely many possible choices. However, having chosen this configuration as the initial condition, is it possible to solve for the equation of motion $$\frac{\partial}{\partial t}(\nabla\cdot{\bf A})+\nabla^2\phi=\frac{\rho}{\epsilon_0},\\ \nabla\Big(\frac{1}{c^2}\frac{\partial\phi}{\partial t}+\nabla\cdot{\bf A}\Big)+\frac{1}{c^2}\frac{\partial^2{\bf A}}{\partial t^2}-\nabla^2{\bf A}=\mu_0{\bf J}\tag{1}$$ to unambiguously determine ${\bf E}(t,{\bf x})$ and ${\bf B}(t,{\bf x})$ fields without fixing a gauge? If not, why? If necessary, one may consider vacuum i.e. $\rho={\bf J}=0$ to answer my question.

If the answer to the above question is 'yes', then I have the following question. Suppose instead of choosing $A^{(1)}_\mu(t_0,{\bf x})$, we choose a gauge-transformed four-potential $$A_\mu^{(2)}(t_0,{\bf x})=A^{(1)}_\mu(t_0,{\bf x})+\partial_\mu\theta({\bf x})\tag{2}$$ as the initial condition. This is a valid initial condition, too. Now, we again solve $(1)$ but this time with the initial condition $A_\mu^{(2)}(t_0,{\bf x})$. Are we gurranteed to obtain the same ${\bf E}(t,{\bf x})$ and ${\bf B}(t,{\bf x})$ as obtained with the previous initial condition?

Question In a nutshell, my question can be summarized as follows.

Without fixing a gauge, and starting with two different initial conditions, if we can solve $\Box A_\mu=0$, are we guaranteed to obtain the same physical fields ${\bf E}(t,{\bf x})$ and ${\bf B}(t,{\bf x})$ at time $t$?

In other words, if $A_\mu^{(1)}(t_0,{\bf x})$ and $A^{(2)}_\mu(t_0,{\bf x})$ both give the same ${\bf E}, {\bf B}$ at $t_0$, then can we say that $A_\mu^{(1)}(t,{\bf x})$ and $A^{(2)}_\mu(t,{\bf x})$ also give same ${\bf E}, {\bf B}$ at a later time $t>t_0$?

$\endgroup$
  • $\begingroup$ $\partial^2 A_\mu = 0$ is only the equation of motion in Lorenz gauge. Are you implicitly maintaining that condition throughout? (You still have some gauge freedom if you do that.) $\endgroup$ – Javier Apr 27 at 15:30
  • $\begingroup$ Edited! I am interested in demonstrating that without fixing a gauge, it is not possible to determine E, B unambiguously. @Javier $\endgroup$ – SRS Apr 27 at 15:40
3
$\begingroup$

In terms of gauge invariant objects we have $$ \frac{\partial {\bf B}}{\partial t} = -{\rm curl}\,{\bf E}\\ \epsilon_0\frac{\partial {\bf E}}{\partial t} =-{\bf J}+\frac 1 {\mu_0} {\rm curl}{\bf B}. $$ These six equations determine the evolution of ${\bf E}({\bf x},t)$ and ${\bf B}({\bf x},t)$ from ${\bf E}({\bf x},0)$ and ${\bf B}({\bf x},0)$ uniquely without gauge fixing. Further, if $$ {\rm div}{\bf B}=0, \quad {\rm div} {\bf E}= \rho/\epsilon_0 $$ at $t=0$, and provided that $\partial_t \rho+ {\rm div} {\bf J}=0$ then these conditions are preserved at all times. There is no need to introduce the potential $A^\mu$.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

You seem to have multiple questions here.

Is it possible to solve the equations of motion to unambiguously determine $\mathbf E(\mathbf x,t)$ and $\mathbf B(\mathbf x,t)$ without choosing a gauge?

Yes, certainly. It's not difficult to rearrange Maxwell's equations to yield

$$\left(\nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right)\left.\cases{\mathbf E \\ \mathbf B}\right\} = \left.\cases{\frac{1}{\epsilon_0}\nabla \rho +\mu_0 \frac{\partial}{\partial t}\mathbf J\\-\mu_0\nabla\times\mathbf J}\right\}$$

Therefore the $\mathbf E$ and $\mathbf B$ terms are solutions to the inhomogeneous wave equation, with sources terms involving $\rho$ and $\mathbf J$. If the latter are prescribed and valid initial/boundary conditions are applied, then $\mathbf E$ and $\mathbf B$ can be immediately written down e.g. via Green's functions.

If $A_\mu^{(1)}(t_0,{\bf x})$ and $A^{(2)}_\mu(t_0,{\bf x})$ both give the same ${\bf E}, {\bf B}$ at $t_0$, then can we say that $A_\mu^{(1)}(t,{\bf x})$ and $A^{(2)}_\mu(t,{\bf x})$ also give same ${\bf E}, {\bf B}$ at a later time $t>t_0$?

Yes, this is also true (of course it must be - otherwise it would matter which gauge we chose at time $t=t_0$, and so there wouldn't actually be any gauge freedom at all).


As clarified by your comment, the question you're trying to ask is actually the following:

If $A_\mu^{(1)}(t_0,{\bf x})= A^{(2)}_\mu(t_0,{\bf x})$ and $\dot A_\mu^{(1)}(t_0,{\bf x})= \dot A^{(2)}_\mu(t_0,{\bf x})$ at $t_0$, then is $A^{(1)}_\mu(t,\mathbf x) = A^{(2)}_\mu(t,\mathbf x)$ for all $t$?

The answer to this question is an emphatic no. Specifying the 4-potential $A_\mu$ and its derivatives at some initial moment is not sufficient to determine it for all $t$, and so it does not correspond to a well-posed initial value problem.

That being said, we are rescued by the answer to your question above. While there is an entire family of $A_\mu$'s which have precisely the same initial conditions (making the IVP ill-defined), every member of that family yields precisely the same $\mathbf E$ and $\mathbf B$. In other words, the ambiguity in the time evolution of $A_\mu$ is the introduction of a (physically irrelevant) time-varying gauge transformation.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ The origin of this question stems from the lecture here by Raman Sundrum between $48.20$ to $50$ minutes. What is he trying to convey? @J.Murray $\endgroup$ – SRS Apr 28 at 14:51
  • $\begingroup$ @SRS I've updated my answer to address your question. $\endgroup$ – J. Murray Apr 28 at 16:05
1
$\begingroup$

So, the thing is that the equations (1) along with the initial condition $A^{(1)}_\mu(t_0,\vec{x})$ do not admit a unique solution. Namely, if you have a solution $A(t,\vec{x})$, then $A(t,\vec{x})+\partial_\mu\theta(t,\vec{x})$, for some $\theta$ whose support does not overlap with the time slice $t=t_0$. The key point is that the latter also satisfies the same initial condition.

For your second question, you have to be careful because Maxwell's equations are not $\square A_\mu=j_\mu$. They are $\square A_\mu-\partial_\mu(\partial\cdot A)=j_\mu$. Everything is however clearer in the language of differential forms. In it, the equations of motion are $d\star dA=J$. However, $F=dA$ and thus, these are really equations $d\star F=J$ for $F$. The later can be shown to have a unique solution given initial conditions.

Summary The equations $d\star dA=J$ with a given initial condition cannot be solved, in the sense that solutions are not unique. The equations $d\star F=J$ can however. In particular, two solutions $A^{(1)}$ and $A^{(2)}$ of the former leading to the same initial $F$ have to lead to the same $F$ at later times.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ The question is whether the solution of $A_\mu$ at time $t$ obtained with initial condition $A^{(1)}(t_0)$ and that obtained from $A^{(2)}(t_0)$ which are related by a gauge transformation at $t_0$ are also related by a gauge transformation at time $t$. $\endgroup$ – SRS Apr 27 at 16:22
  • $\begingroup$ The answer is yes since both solutions have to determine the same $F$ (from uniqueness of solutions to $d\star F=0$) and therefore their difference is exact $d(A^{(1)}-A^{(2)})=F-F=0$. From Poincaré's lemma, locally there is a function $\theta$ such that $A^{(1)}-A^{(2)}=d\theta$, or in other words, $A^{(1)}=A^{(2)}+d\theta$. Thus, the solutions are related by a gauge transform. $\endgroup$ – Iván Mauricio Burbano Apr 27 at 18:38
  • $\begingroup$ As a side comment, recall that it doesn't make sense to say the solution of $A$ obtained with a given initial condition since the solutions to Maxwell's equations for the potential, even with a given initial condition, are not unique. $\endgroup$ – Iván Mauricio Burbano Apr 27 at 18:41

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.