Given $4$-potential $A^\mu(x)=(\phi(x),\mathbf{A}(x))$, the vacuum Maxwell equations:
$$\nabla^2\phi+\frac{\partial}{\partial t}(\nabla\cdot \mathbf{A} )=0$$
$$\nabla^2 \mathbf{A} -\frac{\partial^2 \mathbf{A} }{\partial t^2} – \nabla(\nabla\cdot\mathbf{A} + \frac{\partial \phi}{\partial t})=0$$
There is redundant degree of freedom(d.o.f) in $A^\mu(x)$:
$$A^\mu(x)\rightarrow A^\mu(x) +\partial^\mu \lambda(x) $$
In Coulomb gauge:
$$\nabla\cdot \mathbf{A}(x)=0 \tag{1}$$
Vacuum Maxweel equation becomes:
$$\nabla^2\phi =0 \tag{2}$$
$$\nabla^2 \mathbf{A} -\frac{\partial^2 \mathbf{A} }{\partial t^2} = \nabla( \frac{\partial \phi}{\partial t}) \tag{3}$$
Then we can always choose $\phi(x)=0$. So question becomes that physical degree of freedom is $A^\mu(x)= (0,\mathbf{A}(x))$ with one constraint $\nabla\cdot \mathbf{A}(x)=0$. Every textbook then says the physical degree of freedom is $2$. But it seems there are still redundant d.o.f , we can always make
$$\mathbf{A}(\mathbf{x},t)\rightarrow \mathbf{A}(\mathbf{x},t)+\nabla \Lambda(\mathbf{x})$$
such that
$$\nabla^2 \Lambda(\mathbf{x}) =0\tag{4}$$
But above equation is Laplace equation that has nontrivial solutions, harmonic function.
For example, $\Lambda(\mathbf{x}) = xyz $.
My questions
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Using $\phi(x) =0 $ and $\nabla\cdot \mathbf{A}(\mathbf{x},t)=0$, I have substract two redundant d.o.f. , why fixing $\Lambda(\mathbf{x})$ further can not substract more redundant d.o.f. ?
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Many textbooks will argue that $A^\mu(x)$ should vanish at spacial infinity, so Laplace equation $(4)$ with zero boundary condition in infinity has only trivial solution. But why do we have to require $A^{\mu}$ vanish at spacial infinity? For example, a uniform magnetic field has $\mathbf{A}(x) = \mathbf{B}\times \mathbf{r}/2$ which does not vanish in infinity. If you require that $A^\mu$ vanish at spacial infinity, you even cannot get constant electrical or magnetic field solutions from Vacuum Maxwell equations. And even the elctromagnetic wave solution $e^{i k(t -x)}$ is also nonvanishing at spacial infinity. This question has some relation with Coulomb gauge fixing and “normalizability”
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Why textbook says "Lorentz gauge is Lorentz invariant but cannot fix all redundant d.o.f. Coulomb gauge can fix all redundant d.o.f but is not Lorentz invariant. "? But it's obvious that only Coulomb gauge $(1)$ also cannot fix all redundant d.o.f. We can see that the gauge fixing $\phi =0 $ is not a consequence of $(2)$. It's forced artificially and $\phi=0$ is independent from Coulomb gauge. For example, vacuum Maxwell equations $(2),(3)$ can have uniform electrical field solution $\phi(x)=-\mathbf{E}\cdot \mathbf{r}$, $\mathbf{A}=0$ satisfying only Coulomb gauge $(1)$ but $\phi(x)\neq 0$. If we requires $\phi=0$ and $\nabla \cdot \mathbf{A}=0$, the solution becomes $\phi=0$ and $\mathbf{A}= \mathbf{E} t$.
Best Answer
The main thing one has to be careful about here is the boundary conditions on the gauge field $A_\mu(x)$ and the gauge parameter $\Lambda(x)$. In particular, the set of gauge transformations that one is allowed to "gauge away" must vanish at "infinity". By infinity, here one often means both spatial infinity as well as null infinity. You cannot gauge fix "large gauge transformations" also known as global gauge transformations since those correspond to actual physical symmetries of the theory with physical consequences on the system (deduced via conservation laws). The simplest such example is when $\Lambda(x) = \lambda = $ constant. These gauge symmetries give rise to charge conservation and are very important. They cannot and should not be removed. Having said this, we go back to the equation in Coulomb gauge. The residual gauge transformations are generated by functions $\Lambda(x)$ satisfying $$ \nabla^2 \Lambda(x) = 0 \, , \qquad \Lambda(x) \to 0 ~~\text{at infinity.} $$ You can now convince yourself that the only solution to these equations is $$ \Lambda(x) = 0\, . $$ This answers your first question.
The answer to your second question about the "why" of these boundary conditions is the requirement of finite energy flux through the boundaries of the system. The idea is that we restrict to solutions which have the property that if finite energy is input into the system through a certain boundary, then finite energy should be released from the system. Here, I am not talking about the total energy which will always be fixed due to energy conservation. What we are talking about here is the local energy flux through each boundary. We are only interested in solutions to Maxwell's equations where even the local energy density is finite at all points and no singularities are created. All such solutions need to have the property that the "vanish" in a suitable way near a suitable boundary. This requirement, as you correctly point out, excludes constant electric or magnetic fields whose total energy is proportional to the volume of the system. The wave solutions do not die of at spatial infinity but that's OK since spatial infinity is not a suitable boundary for these solutions. Wave-like solutions travel outward and reach null infinity ${\mathscr I}^+$ as opposed to spatial infinity. In other words, even though wave-solutions are non-vanishing on spatial infinity, they do not contribute to the energy density there. Therefore, we need that the energy flux on ${\mathscr I}^+$ due to wave-solutions be finite and you can check that this is indeed the case.
$\phi(x) \neq 0$ does not mean that it is a degree of freedom. A degree of freedom is defined as a "part" or component of the field $A_\mu(x)$ that is not determined by the equations of motion, and it is therefore, completely free to choose. To state this more precisely, a degree of freedom is the data that must be prescribed (completely freely) on a Cauchy surface so as to ensure unique time evolution into the past and future. It is the piece of information about the field that completely determines every other aspect of it. Thus, $\phi$ is never a degree of freedom. In any gauge you choose, it is easy to see that it is determined entirely in terms of $A_i(x)$.