# Coulomb gauge fixing and “normalizability”

The Setup

Let Greek indices be summed over $0,1,\dots, d$ and Latin indices over $1,2,\dots, d$. Consider a vector potential $A_\mu$ on $\mathbb R^{d,1}$ defined to gauge transform as $$A_\mu\to A_\mu'=A_\mu+\partial_\mu\theta$$ for some real-valued function $\theta$ on $\mathbb R^{d,1}$. The usual claim about Coulomb gauge fixing is that the condition $$\partial^i A_i = 0$$ serves to fix the gauge in the sense that $\partial^iA_i' = 0$ only if $\theta = 0$. The usual argument for this (as far as I am aware) is that $\partial^i A'_i =\partial^iA_i + \partial^i\partial_i\theta$, so the Coulomb gauge conditions on $A_\mu$ and $A_\mu'$ give $\partial^i\partial_i\theta=0$, but the only sufficiently smooth, normalizable (Lesbegue-integrable?) solution to this (Laplace's) equation on $\mathbb R^d$ is $\theta(t,\vec x)=0$ for all $\vec x\in\mathbb R^d$.

My Question

What, if any, is the physical justification for the smoothness and normalizability constraints on the gauge function $\theta$?

EDIT 01/26/2013 Motivated by some of the comments, I'd like to add the following question: are there physically interesting examples in which the gauge function $\theta$ fails to be smooth and/or normalizable? References with more details would be appreciated. Lubos mentioned that perhaps monopoles or solitons could be involved in such cases; I'd like to know more!

Cheers!

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Good question which is clearly more general than Coulomb gauge fixing! – Michael Brown Jan 26 at 9:03
@MichaelBrown Thanks! Yeah I'm learning some sugra at the moment, and this argument that exploits the vanishing of smooth, normalizable solutions to Laplace's equation seems to arise quite often, for instance when one wants to eliminate certain harmonic gauge field components. – joshphysics Jan 26 at 9:09
Smoothness of $\theta$ is needed because $A_\mu$ that is modified by derivatives of $\theta$ has to remain continuous - or smooth, but one level weaker requirement of smoothness than for $\theta$. The normalizability just means that $\theta$ never diverges in the bulk of the space and decreases sufficiently quickly at infinity. When it doesn't, one would have to discuss monopoles, instantons etc. but those things are a non-issue for U(1) gauge theory. – Luboš Motl Jan 26 at 9:51

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At any rate "sufficient smoothness" and "normalizability" are conditions that are often required in physics and physicists don't spend much time with such stuff - they're natural physical conditions for various reasons. Mathematicians are often obsessed with these mathematical details - they're needed for rigorous proofs - but physicists are not. In fact, physicists really think exactly in the opposite way than what you suggest. They would assume that functions in physics are sufficiently smooth and well behaved - and if some of them are not, they would get concerned or alert. – Luboš Motl Jan 26 at 9:53
@joshphysics: To focus the answers, perhaps you could point out a couple of references where you have encountered this use of the word normalizable? – Qmechanic♦ Jan 26 at 13:06
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## 1 Answer

It means that the gauge ambiguity is practically removed in the Coulomb gauge if you deal with a "nice" $\mathbf{A}$ (which is your purpose).

However, it does not mean you only deal with the radiation (propagating solutions). Transversal $\mathbf{A}$ is different from zero for a uniformly moving charge too.

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