Let $A^{\mu}$ be a 4-potential that satisfies the Lorenz condition $$\partial_\mu A^\mu =0$$ We can make a gauge transformation $$A_\mu \to A'_\mu=A_\mu + \partial_\mu \Lambda$$ such that $$ \partial_\mu A'^\mu =\partial_{\mu}\left(A^{\mu}+\partial^{\mu}\Lambda\right)=0 $$ and still stay in the Lorenz gauge. Since $\partial_{\mu}A^{\mu}=0$, this condition for $\Lambda$ is \begin{equation} \partial_{\mu}\partial^{\mu}\Lambda=0\tag{1} \end{equation} This possibility to do some gauge transformations while having fixed the Lorenz condition is called residual gauge freedom.


We want to remove this residual gauge freedom by fixing another condition on $A_{\mu}$. My book claims that \begin{align} \begin{aligned}A_{3}(0,\vec{x}) & =0\\ \partial_{t}A_{3}(0,\vec{x}) & =0 \end{aligned} \tag{2} \end{align} is such a condition.


The solution to (1) is uniquely determined by its initial conditions \begin{align*} \Lambda(0,\vec{x}) & =\Phi_{1}(\vec{x})\\ \partial_{t}\Lambda(0,\vec{x}) & =\Phi_{2}(\vec{x}) \end{align*} so it is sufficient to prove that the conditions (2) imposed on the transformed 4-potential uniquely determine $\Phi_{1}$ and $\Phi_{2}$. Since $$ A'_{3}=A_{3}+\partial_{3}\Lambda $$ they read \begin{alignat*}{2} A_{3}(0,\vec{x})+\partial_{3}\Lambda(0,\vec{x}) & = & A_{3}(0,\vec{x})+\partial_{3}\Phi_{1}(\vec{x}) & =0\\ \partial_{t}A_{3}(0,\vec{x})+\partial_{t}\partial_{3}\Lambda(0,\vec{x}) & = & \partial_{t}A_{3}(0,\vec{x})+\partial_{t}\partial_{3}\Phi_{2}(\vec{x}) & =0 \end{alignat*} So we can choose an antiderivative wrt $x^{3}$ of the given functions $A_{3}(0,\vec{x})$ and $\partial_{t}A_{3}(0,\vec{x})$ to find the $\Lambda$ we're searching for.


It seems to me that this antiderivative is not uniquely determined, as you can add to it an arbitrary function of $x^{1}$ and $x^{2}$. So $\Lambda$ is not uniquely determined either. Is then the residual gauge not removed? Or am I missing something?

  • 1
    $\begingroup$ Which book? Which page? Related: physics.stackexchange.com/q/190001/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented May 26, 2020 at 15:30
  • $\begingroup$ It doesn't seem to me that this question is answered there. That question is answered by the fact that the Lorenz gauge gives the possibility of a residual gauge. In this case, this fact is the background for the question, which is about how to actually fix the residual gauge. $\endgroup$
    – renyhp
    Commented May 27, 2020 at 16:40
  • $\begingroup$ Book is Elettrodinamica Classica by Kurt Lechner and is only available in italian AFAIK. $\endgroup$
    – renyhp
    Commented May 29, 2020 at 9:30
  • $\begingroup$ Link to abstract page? Which page? $\endgroup$
    – Qmechanic
    Commented May 29, 2020 at 10:15
  • $\begingroup$ I'm not sure I understand which link you want me to post, but this is a link to the book: springer.com/gp/book/9788847052109 and the page is 137. $\endgroup$
    – renyhp
    Commented May 29, 2020 at 11:26

1 Answer 1


You are right that you can add a function of $\lambda(x,y)$ to your gauge parameter. But note that it should satisfy the Lorenz gauge which implies that $$(\partial_x^2+\partial_y^2)\lambda=0$$ whose general solutions consists of arbitrary holomorphic and antiholomorphic functions on the plane $$\lambda=f(z)+g(\bar z), \qquad z=x+i y,\quad \bar z =x-iy$$ The point is that this solution diverges somewhere. Expanding $f=\sum_n c_n (z-z_0)^n$ we see that the function diverges either at $z=z_0$ or at $z\to \infty$ except for the constant function.

Imposing some boundary conditions, say $A_\mu\sim\cal{O}(1/r)$ on the gauge field at $r=\sqrt{x^+y^2+z^2} \to \infty$ and requiring smoothness of the gauge field in the bulk of spacetime excludes the additional freedom you mentioned.

There is however a nice story about asymptotic symmetries surviving gauge fixing and boundary conditions which is not necessary for your question here.


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