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For any electromagnetic field, it is easy to impose the Coulomb gauge condition ${\bf\nabla}\cdot{\bf A}=0$. To start with, if ${\bf \nabla}\cdot{\bf A}_{\rm old}\neq 0$, the trick is to make a gauge transformation $${\bf A}_{\rm old}\to {\bf A}_{\rm new}={\bf A}_{\rm old}+{\bf \nabla}\theta({\bf x},t)\tag{1a}$$ such that the new vector potential ${\bf A}_{\rm new}$ satisfies, ${\bf \nabla}\cdot{\bf A}_{\rm new}=0$. This is achieved by choosing the gauge function $\theta({\bf x},t)$ to be a solution of the Poisson's equation $$\nabla^2\theta({\bf x},t)=-{\bf \nabla}\cdot{\bf A}_{\rm old}\tag{1b}$$ which can almost always be solved.

Now let us consider a slightly more nontrivial job. For a free electromagnetic field, it is always possible to choose a gauge such that both $\phi_{\rm new}=0$ and ${\bf \nabla}\cdot{\bf A}_{\rm new}=0$. I wonder if this can be achieved by a single gauge transformation. To start with, again if $\phi_{\rm old}\neq 0$ and ${\bf \nabla}\cdot{\bf A}_{\rm old}\neq 0$, can we make a single gauge transformation $$\phi_{\rm old}\to\phi_{\rm new}=\phi_{\rm old}-\frac{\partial\theta}{\partial t}, {\bf A}_{\rm old}\to {\bf A}_{\rm new}={\bf A}_{\rm old}+\nabla\theta\tag{2}$$ to achieve $$\phi_{\rm new}=0, ~{\bf \nabla}\cdot{\bf A}_{\rm new}=0\tag{3}$$ in one shot i.e. by choosing a single gauge function $\theta({\bf x},t)$? Unlike the previous case, $\theta$ has to satisfy two equations now: $$\frac{\partial}{\partial t}\theta({\bf x},t)=\phi_{\rm old}, ~~ \nabla^2\theta({\bf x},t)=-{\bf \nabla}\cdot{\bf A}_{\rm old}.\tag{4}$$ What is the gurrantee that both the equations in $(4)$ can be simulaneusly solved? Or should it really be done in two succesive steps?

In a two-step process, if we do a first gauge transformation to make $\phi=0$, what is the guarantee that by doing a second gauge transformation to satisfy $\nabla\cdot{\bf A}=0$, I would not regenerate $\phi$? It isn't very transparent.

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    $\begingroup$ Gauge transformations form a group, the two gauge transformations performed one after the other can be viewed as a single gauge transformation. $\endgroup$ – jacob1729 Apr 15 '20 at 14:57
  • $\begingroup$ Sure. But I was wondering if we do a first gauge transformation to make $\phi=0$, what is the guarantee that by doing a second gauge transformation to satisfy $\nabla\cdot{\bf A}=0$, I would not regenerate $\phi$. It isn't very transparent. @jacob1729 $\endgroup$ – SRS Apr 15 '20 at 15:02
  • $\begingroup$ I believe jacob's point (or at least the one I would make) is that the phrase "single gauge transformation" or the idea of "successive steps" is meaningless. You're just asking whether $\phi = 0$ and $\nabla\cdot A = 0$ is an admissible gauge, i.e. whether this state of the field can be achieved with gauge transformations at all. $\endgroup$ – ACuriousMind Apr 15 '20 at 15:09
  • $\begingroup$ @ACuriousMind That's fair! But for the first case, I have sketched the intermediate step i.e. how to go to the desired gauge. In the second case, how to go to the desired gauge i.e., $\phi=\nabla\cdot{\bf A}=0$ is not very clear to me. $\endgroup$ – SRS Apr 15 '20 at 15:14
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/1250/2451 and links therein. $\endgroup$ – Qmechanic Apr 15 '20 at 16:31
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EDIT: This answer has been substantially edited since first posted.

For the two-step process. This gauge transformation is only possible if we are in the absence of charges. Then $\nabla \cdot \mathbf E=0$, which means for a general gauge $$\nabla^2 \phi +\frac{\partial}{\partial t} \nabla \cdot \mathbf A =0$$ Now do a gauge transformation that removes $\phi$. This gauge transformation can be both space and time dependent. Now you are in a gauge where $\phi=0$ and $\nabla \cdot \mathbf A \neq 0$. The above implies

$$\frac{\partial}{\partial t} \nabla \cdot \mathbf A =0$$

Therefore, you can now do a time-independent gauge transformation to set $\nabla \cdot \mathbf A =0$. Since the transformation is time-independent, it won't affect $\phi$.

For the single-step process, you want to solve $$\frac{\partial}{\partial t}\theta({\bf x},t)=\phi_{\rm old}, ~~ \nabla^2\theta({\bf x},t)=-{\bf \nabla}\cdot{\bf A}_{\rm old}$$ Roughly speaking, the first equation is solved by $$\theta({\bf x},t) = \int_{t_0}^t \phi({\bf x},\tau) d\tau+f(\mathbf x)$$ where $f(\mathbf x)$ is arbitrary. Then $$\nabla^2 \theta = \int_{t_0}^t \nabla^2\phi({\bf x},\tau) d\tau+\nabla^2 f(\mathbf x)$$ Now we can use the condition above, valid in absence of charges, to get, $$\nabla^2 \theta = -\int_{t_0}^t \frac{\partial}{\partial \tau} \nabla \cdot \mathbf A({\bf x},\tau) d\tau+\nabla^2 f(\mathbf x)=$$ $$=-\nabla \cdot \mathbf A({\bf x},t)+\nabla \cdot \mathbf A({\bf x},t_0)+\nabla^2 f(\mathbf x)$$ Now you can pick $f$ so that $$\nabla \cdot \mathbf A({\bf x},t_0)+\nabla^2 f(\mathbf x)=0$$ and then the condition $$\nabla^2 \theta =-\nabla \cdot \mathbf A$$ is satisfied.

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  • $\begingroup$ How is an equation of the form $\frac{\partial}{\partial t}\theta_1(t)=\phi_{\rm old}(t,{\bf x})$ make sense where RHS is a function of ${\bf x}$ (in general) but not the LHS? Am I missing something? $\endgroup$ – SRS Apr 15 '20 at 15:11
  • $\begingroup$ Apologies, that was incorrect. I have updated my answer. $\endgroup$ – John Donne Apr 15 '20 at 17:28

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