How do gauge transformation imply gauge conditions?

Question

In classical EM I understand the electric and magnetic fields are invariant under the potential transformations $\varphi\to\varphi - \partial_t\chi$ and $\mathbf{A}\to\mathbf{A} + \nabla\chi$.

From here people often say this gives us a freedom to do something like choose $\nabla\cdot\mathbf{A} = 0$. I don't understand how we go from the above transformations to specifying properties that $\varphi$ and $\mathbf{A}$ satisfy.

Possible duplicate: http://physics.stackexchange.com/q/1250/2451 — Qmechanic, Mar 02 '17 at 15:40

score 2 · Accepted Answer · answered Mar 02 '17 at 14:55

2

If $\boldsymbol A$ does not satisfy $\nabla\cdot \boldsymbol A=0$, then redefine $$ \tilde{\boldsymbol A}\equiv\boldsymbol A+\nabla\chi $$ where $\chi$ is any solution of the PDE $$ -\nabla^2\chi=\nabla\cdot \boldsymbol A $$

The vector $\tilde{\boldsymbol A}$ satisfies, by construction, $\nabla\cdot \tilde{\boldsymbol A}=0$.

answered Mar 02 '17 at 14:55

AccidentalFourierTransform

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How do we know there are solutions to that PDE? – Nate Stemen Mar 02 '17 at 15:02
@NateStemen see https://en.wikipedia.org/wiki/Uniqueness_theorem_for_Poisson's_equation and http://farside.ph.utexas.edu/teaching/em/lectures/node31.html (last equation). – AccidentalFourierTransform Mar 02 '17 at 15:03
@AccidentalFourierTransform, I think that your references discuss the uniqueness of the solution and not its existence. – 0x90 Mar 02 '17 at 15:45
@0x90 the first one discusses uniqueness. The second one gives the explicit solution (for a certain boundary conditions). – AccidentalFourierTransform Mar 02 '17 at 15:46
@AccidentalFourierTransform who said the the integral converges? It depends on the boundary conditions too. – 0x90 Mar 02 '17 at 15:46
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@0x90 the fact that $\boldsymbol A$ is a well-defined vector potential (with trivial topology) means that it has to be fast decaying at infinity. The boundary conditions are important for uniqueness, but here we only need some solution of the PDE. Any solution works. – AccidentalFourierTransform Mar 02 '17 at 15:48
@AccidentalFourierTransform you are right I was talking in general. Thanks – 0x90 Mar 02 '17 at 15:49
@AccidentalFourierTransform what do you mean by a function having a trivial topology? – Nate Stemen Mar 02 '17 at 15:59
@NateStemen in some advanced gauge theories the behaviour of the gauge fields at infinity is important. It's not relevant for your question, so nevermind. – AccidentalFourierTransform Mar 02 '17 at 16:00

How do gauge transformation imply gauge conditions?

1 Answers1