Question

Why is Lorenz gauge $\partial_\mu A^\mu = 0$ not attainable for 'permissible boundary conditions'?

Answer 1

I'm reading Paul Townsend's string theory lecture notes and I'm confused about a paragraph. Below, the φi are first class constraints and the χi are gauge fixing conditions.

Whenever {χi,φj} has zero determinant, two problems arise. One is that the gauge fixing conditions don’t completely fix the gauge, and the other is that you can’t always arrange for the gauge fixing conditions to be satisfied by making a gauge transformation.

Consider the Lorenz gauge ∂μAμ=0 in electrodynamics. A gauge transformation A→A+dα of the gauge condition gives ∂2α=0, which does not imply that α=0; the gauge has not been fixed completely. For the same reason, you can’t always make a gauge transformation to get to the Lorenz gauge if ∂μAμ is not zero, even if it is arbitrarily close to zero: the reason is that the operator ∂2 is not invertible because there are non-zero solutions of the wave equation that cannot be eliminated by imposing the b.c.s permissible for hyperbolic partial differential operators. The Coulomb gauge ∇⋅A=0 does not have this problem because ∇2 is invertible for appropriate boundary conditions.

I know that Lorenz gauge is not complete but I've never heard of it being unreachable. I think the notes are saying something tricky about physically permissible boundary conditions, but I'm not sure. Why do ∂2 and ∇2 have different types of boundary conditions, what are they, and why are they relevant here? Does this issue have physical consequences?

Answer 2

While reading the Paul Theory I went through many points that to be noted or remembered by the practical practices.