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In classical field theory we can get, that adding gradient of some scalar field to magnetic vector potential does not change the physics at all. So, we have such a symmetry:

$\boldsymbol{A}\rightarrow\boldsymbol{A}+\nabla f$

Then there is such a thing written almost in every book on elecrodynamics:

"For example, we can use Coulomb gauge $\nabla \cdot \boldsymbol{A}=0$."

I can't understand this implication. Why this symmetry allow us to say, that divergence is zero? What is $f$ in this case?

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    $\begingroup$ "What is f in this case?" - Have you tried writing down $\nabla\dot(A+\nabla f) = 0$ and solving for $f$ in terms of $A$? $\endgroup$
    – ACuriousMind
    Dec 20, 2018 at 17:51
  • $\begingroup$ The symmetry allows one to constrain a degree of freedom. There are many ways to do that. $\endgroup$
    – user196418
    Dec 20, 2018 at 17:55
  • $\begingroup$ $∇ \cdot (A+∇f)=0 $ gives a poisson equation which can be solved for $f$. $\endgroup$ Dec 20, 2018 at 18:26
  • $\begingroup$ Related: physics.stackexchange.com/q/129819/2451 and links therein. $\endgroup$
    – Qmechanic
    Dec 20, 2018 at 19:57

1 Answer 1

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Why this symmetry allow us to say, that divergence is zero?

Stipulate a vector potential with non-zero divergence

$$\nabla\cdot\mathbf{A}\ne 0$$

'Gauge away' the divergence

$$\nabla\cdot\mathbf{A}'=\nabla\cdot\left(\mathbf{A}+\nabla f \right)=0$$

and it follows that

$$\nabla\cdot\mathbf{A}+\nabla^2f=0\Rightarrow\nabla^2f=-\nabla\cdot\mathbf{A}$$

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