# For the free electromagnetic field, is it possible make single gauge transformation to achieve $\phi={\bf \nabla}\cdot{\bf A}=0$?

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.

• Gauge transformations form a group, the two gauge transformations performed one after the other can be viewed as a single gauge transformation. – jacob1729 Apr 15 '20 at 14:57
• 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 – SRS Apr 15 '20 at 15:02
• 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. – ACuriousMind Apr 15 '20 at 15:09
• @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. – SRS Apr 15 '20 at 15:14
• Possible duplicates: physics.stackexchange.com/q/1250/2451 and links therein. – Qmechanic Apr 15 '20 at 16:31

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.
• 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? – SRS Apr 15 '20 at 15:11