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 $LG$.

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    $\begingroup$ Mathematically ---> math.stackexchange.com $\endgroup$ – kennytm Nov 23 '10 at 20:01
  • $\begingroup$ ah sorry i didn't know it :-( $\endgroup$ – Boy Simone Nov 23 '10 at 20:09

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)

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  • $\begingroup$ 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 Simone Nov 24 '10 at 17:50
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    $\begingroup$ 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 $c^2\,\nabla^2\lambda_w = \partial_t^2\,\lambda_w$. $\endgroup$ – Selene Routley Mar 8 '17 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 '19 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...

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  • $\begingroup$ hmmm,...I haven't yet studied Em lagrangian, :-( but thank you for the answer :) $\endgroup$ – Boy Simone Nov 24 '10 at 17:51

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