Why Coulomb gauge is a possible gauge choice?

Question

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?

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

score 3 · Answer 1 · answered Dec 20 '18 at 18:49

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

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

Why Coulomb gauge is a possible gauge choice?

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