There is something I would like to clarify with gauge fixing.

In E.M, we can introduce the potential vector.

As $div(\vec{B})=0$ we know that we can write $\vec{B}=\vec{curl}(\vec{A})$.

But as $\vec{curl}(\vec{grad})=0$, we have the fact that $A$ is defined at a gradient.

Thus : $\vec{A}=\vec{A_0}+\vec{grad}(\phi)$ describes the same $B$ field and thus the same physics.

To uniquely define $\vec{A}$ we can do gauge fixing. It means that we will add an equation on $A$ that will fix the potential $\phi$.

For example we have Coulomb gauge that is : $div(\vec{A})=0$.

But what I don't understand is : how do we know that fixing the divergence of $A$ will not change its curl ?

  • $\begingroup$ Are you familiar with Helmholtz decomposition? $\endgroup$ Sep 27, 2017 at 21:05
  • $\begingroup$ @Alfred Centauri. Not really but I'll take a look :) $\endgroup$
    – StarBucK
    Sep 27, 2017 at 21:44
  • $\begingroup$ Helmholtz decomposition is helpful, but I guess here his doubt is not really related, it is likely some misunderstanding of the definition. $\endgroup$
    – gamebm
    Sep 28, 2017 at 0:20
  • $\begingroup$ Then you might find this helpful: The Helmholtz Decomposition and the Coulomb Gauge $\endgroup$ Sep 28, 2017 at 0:47

1 Answer 1


In your example of the Coulomb gauge, the gauge condition $$\nabla\cdot({\bf A}+\nabla\phi)=0 $$ can be viewed as an equation for $\phi$ which picks out a specific $\phi$ field from an infinite number of possibilities before the choice of gauge. In other words, $\nabla\cdot A=0$ is an additional condition on top of the requirement physical observable (the curl of $A$) should be remain unchanged.

  • $\begingroup$ I have a little question in link with this remark. Lets assume i want $div(A)=0$. In my book they say it implicates that $\Delta \phi=0$ but for me if we choose to work with $A_2=A_1+grad(\phi)$ it would mean that $\Delta \phi=-A_1$ where $A_1$ is the "old" vector potential. Could you help me to clarify this ? I hope you see what I misunderstand $\endgroup$
    – StarBucK
    Sep 27, 2017 at 21:43
  • $\begingroup$ @StarBucK, if $\nabla \cdot \vec A_1 \ne 0$ but $\nabla \cdot \vec A_2 = 0$ then it follows that $\nabla \cdot \nabla\phi = - \nabla\cdot\vec A_1$. $\endgroup$ Sep 27, 2017 at 23:29
  • $\begingroup$ @StarBuck It is always difficult to guess with limited information :) My guess is the following. If you employ Coulomb gauge, $\nabla \cdot A=0$, and you want to introduce a scalar field $\phi$ so that $A\rightarrow A+\nabla\phi$ does not break the gauge, then you find $\Delta\phi=0$. In other words, you still have some extra freedom even after you adopt Coulomb gauge. $\endgroup$
    – gamebm
    Sep 28, 2017 at 0:16
  • $\begingroup$ @starbuck You're likely to be misreading your text - read it again, carefully. And, in any case, provide a full reference. $\endgroup$ Sep 28, 2017 at 4:00

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