If $B$ is magnetic field and $E$ electric Field, then

$$B=\nabla\times A,$$

$$E= -\nabla V+\frac{\partial A}{\partial t}.$$

There is Gauge invariance for the transformation

$$A'\rightarrow A+{\nabla L}$$

$$V'\rightarrow V-\frac{dL}{dt}.$$

Now, we can write:

  • Coulomb Gauge (CG): the choice of a $L$ that implies $\nabla\cdot A=0$.

  • Lorenz Gauge (LG): the choice of a $L$ that implies $\nabla \cdot A - \frac{1}{c^2} \frac{\partial V}{\partial t}=0$.

Now, I'm trying to mathematically prove that it's always possible to find such an $L$ satisfiying $CG$ or $LG$.

  • 1
    $\begingroup$ Mathematically ---> math.stackexchange.com $\endgroup$
    – kennytm
    Nov 23, 2010 at 20:01
  • $\begingroup$ ah sorry i didn't know it :-( $\endgroup$
    – Boy S
    Nov 23, 2010 at 20:09

4 Answers 4


The correct Gauge transformation formula should be $$\begin{aligned} \mathbf A &\mapsto \mathbf A + \nabla \lambda \\ \mathbf V &\mapsto V - \frac{\partial\lambda}{\partial t}, \end{aligned} $$ not something with "gradL/dt". The Coulomb gauge requires $\nabla\cdot\mathbf A=0$, not "rotA = 0". The Lorenz gauge requires $\nabla\cdot\mathbf A + \frac1{c^2}\frac{\partial V}{\partial t}=0$, not "gradA+1/c^2 dV/dt".

The Coulomb gauge can be chosen by solving the Poisson equation $$ \nabla^2 \lambda = -\nabla\cdot\mathbf A$$

The Lorenz gauge can be chosen by solving the inhomogeneous wave equation $$ \nabla^2 \lambda - \frac1{c^2}\frac{\partial^2\lambda}{\partial t^2} = -\nabla\cdot\mathbf{A} - \frac1{c^2}\frac{\partial{V}}{\partial{t}}$$

(Substitute the transformed potentials into the conditions to get the PDEs)

Existence of solutions of these PDEs are guaranteed as long as the source terms (stuff on the RHS) are "well-behaved" (e.g. $\nabla\cdot\mathbf A$ should grow slower than $1/r$ in the Poisson equation)

  • $\begingroup$ Hi!Sorry for the many mistakes that I wrote ;I don't know where my brain was,in fact if you could interpretate $rotA=0$ like a very heavy conceptual mistake, after you read $gradL/dt$ that hasn't got any sense, you could easily understand that what I wrote wasn't believable :-) Abaout the answer, I have understand all the things now;before I didn't know the theorems about boundary conditions in poisson and waves equations and solutions, I haven't yet studied them in "mathematical metods for physics" course, for that I asked the question!Now I know them! thank you very much;-) $\endgroup$
    – Boy S
    Nov 24, 2010 at 17:50
  • 2
    $\begingroup$ Perhaps also mention the not so widely appreciated fact that a choice of Lorenz gauge only forces a solution unique to within any free wave $\lambda_w$ fulfilling D'Alembert's equation $c^2\,\nabla^2\lambda_w = \partial_t^2\,\lambda_w$. $\endgroup$ Mar 8, 2017 at 23:51
  • $\begingroup$ If I take your example of Coulomb gauge, as it is an equation of order $2$, we have different sets of lambda possible for $A$ and $V$. Thus, in Coulomb gauge is there still indeed a degree of freedom for $A$ and $V$ ? Or for a reason that I don't see actually it completly fix the potentials ? $\endgroup$
    – StarBucK
    Apr 2, 2019 at 12:49

I'm not sure i understood your question. Look, all E&M Lagrangians have gauge freedom "built in", in the sense that you can re-write the $E$ and $B$ fields and the Lagrangian won't change. Therefore, it's always possible for you to make a choice of gauge, you always have that freedom.

On a slightly tangential note, remember what Helmholtz theorem has to say: you can always decompose a "well behaved" vector field into a sum of a curl and a grad part. And this is exactly what you're doing to the $E$ and $B$ fields in Maxwell's eqs. So, now your question becomes: what happens when you apply your gauge transformation in this context? That is, what is $\nabla\times\nabla$ and $\nabla\cdot\nabla$? What does this imply for $A$ (what equation you get for the vector-potential)?

This should get you going...

  • $\begingroup$ hmmm,...I haven't yet studied Em lagrangian, :-( but thank you for the answer :) $\endgroup$
    – Boy S
    Nov 24, 2010 at 17:51

Wikipedia on gauge fixing seems to imply that gauge fixing always works in abelian Yang-Mills theory, of which electrodynamics is the standard example. But that this does not always work in non-abelian Yang-Mills theory. There you have to restrict to submanifolds of the base spacetime to get the gauge fixing. Usually these go by the terms Gribov region and the like.

I'm not sure that I'm convinced by Wikipedia on the abelian theory since gauge fixing is merely a way of choosing a global section. But to choose a global section of a principal bundle simply shows that it is trivial, ie untwisted. Not all abelian principal bundles are trivial. For example the Dirac magnetic monopole is not trivial. Hence even here the technology of Gribov regions and horizons etc is useful.


Since, my similar question was closed, I will answer here.

The gauge transformation

$$ \mathbf {A} \rightarrow \mathbf {A} +\nabla \lambda, $$

$$ \varphi \rightarrow \varphi - \frac {\partial \lambda}{\partial t}, $$ (where $ \lambda=\lambda(\vec {r},t) $ is an arbitrary scalar function of coordinates $ \mathbf{r} $ and time $ t $) do not change the form of Maxwell's equations, and hence are admissible from a physical point of view.

In practice, no one chooses a special function $\lambda(\vec{r}, t)$ per se, although one is always implicitly assumed. But the described ambiguity of the potentials from mathematical point of view tells us that one can always be chosen to satisfy one arbitrary additional condition. One, since we can arbitrarily choose only one function $\lambda(\vec{r}, t)$.

For example, one can always choose the field potentials so that the scalar potential $\varphi = 0$ (is equal to zero). To make the vector potential equal to zero, is impossible, since the condition $\mathbf{A} = 0$ is three additional conditions (for the three components of $\mathbf{A}$).

Another possible way is to choose one arbitrary additional is

  1. Coulomb gauge: $\mathrm{div}\mathbf{A}' = 0 $. Where an arbitrary function can be chosen so that it satisfies the condition $\nabla^2 \lambda= -\nabla \cdot \mathbf{A}$. By solving the equation, one can get the following function.

In practice, they proceed as follows: imagine that we know that a given $\mathbf{A}$ solves the Maxwell's Equation. We can always find a gauge transformation that converts it to a new solution $\mathbf{A}'$ that satisfies the Lorentz Gauge condition $$\mathrm{div}\mathbf{A}' = 0. $$ Since we can always do this, rather than imposing this condition after solving Maxwell's Equations, we can require that we are looking for solutions of this type $$\mathrm{div}\mathbf{A} = 0. $$ before we solve that equation.

  1. For Lorenz gauge $\mathrm{div}\,\mathbf{A}' + {1 \over c^2}{\partial \mathbf{\varphi}' \over \partial t} = 0$, an arbitrary function can be chosen so that it satisfies the condition $\Box^2 \lambda= - \nabla \mathbf{A} - \frac{1}{c^2}\frac{\partial \varphi}{\partial t}$.

Just as a clever choice of coordinates can make solving a problem easier, so we will find that a clever choice of gauge can make finding solutions easier .

  • 1
    $\begingroup$ Would like to point out that the conditions you impose on $\lambda$, finds a new set of potentials that satisfy the chosen gauge, given that you are ALREADY in that gauge. $\endgroup$ Jun 28, 2022 at 18:40
  • $\begingroup$ @jensenpaull Yes, you are right. So I modified my answer. $\endgroup$
    – Sergio
    Jun 28, 2022 at 18:55

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