Coulomb gauge fixing and "normalizability"

Question

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!

Answer 1

A quick answer, if I may.

You need $\theta$ to be smooth since you want to derive it. So mathematics imposes you to choose $\theta$ smooth.

Now the trick: choosing $\theta$ to be smooth means you can always impose $\mathbf{A}$ to be smooth, and use several patches related to each other by a gauge transform. Then you should always discuss smooth vector potential... do you ? Well, you should, if you want to make proper math. But physicists usually don't care about that, and choose a singular vector potential to prove that the field configuration hosts a monopole. The prototype example is the vortex associated with the U(1) Lie group / algebra. See for instance the paper by Dirac:

Dirac, P. A. M. Quantised Singularities in the Electromagnetic Field. Proc. R. Soc. London. Ser. A 133, 60–72 (1931)

where the vector potential is singular at the north or south pole. Note that the theory of connection on fiber bundle was still to be discovered at that time ! The correct mathematical picture came late in physics, as far as I know at least. Here a beautiful reading

Wu, T. T. & Yang, C. N. Concept of nonintegrable phase factors and global formulation of gauge fields. Phys. Rev. D 12, 3845–3857 (1975)

where they choose two parameterisation of the circle: one for the south and one for the north pole, these two parameterisations of the vector potential being related to the other one by a gauge transformation.

How about normalisable then? Well, I never heard about that, and mainly one defines everything in compact space(s), where it scarcely makes sense to impose norm.

Answer 2

You don't actually need smoothness in general, you just need a $C^2$ transition function so that the Laplacian is well-defined - without that, you don't have a well-defined source field, which clearly kind of goes against the whole point of introducing gauge fields in the first place.

In a classical context, the normalizability is there to make Coulomb gauge a unique gauge fixing in the physically realistic situation where the charges are spatially bounded. There are at least three reasons to motivate this particular choice of complete gauge fixing:

1. It satisfies the physical intuition from locality that if the sources are localized, then the gauge fields should naturally fall off to zero at spatial infinity.
2. It often allows us to integrate by parts and drop the surface terms, which is a trick that's constantly used in E&M.
3. It gives by far the simplest Green's function for the gauge field, leading to the simple Coulomb-like equation $\varphi({\bf x}) = \int d^3x' \frac{\rho({\bf x'})}{|{\bf x} - {\bf x}'|}$, and similarly for the vector potential and the electric current.

I'm less familiar with the quantum situation, but I know that there are all kinds of subtleties with large gauge transformations in nonabelian gauge theory, instantons, etc. Here, the non-uniqueness of Coulomb gauge requires a more careful consideration of the boundary conditions.

Answer 3

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.