Proof that we can always find a gauge transformation such that $A_0=0$?

Question

I'm trying to follow Coleman's proof from his lectures "Aspects of Symmetry" on page 200-201. He proofs it is always possible to work in the temporal gauge for a general Yang-Mills-Higgs theory. I shall quickly repeat his argument. Consider some Higgs field, $\phi$, for which the directional covariant derivative vanished on some path $P$:

\begin{equation} \frac{dx^\mu}{ds} D_\mu \phi=0 \Rightarrow \frac{d \phi}{ds}=-\frac{dx^\mu}{ds} A_\mu \phi \end{equation}

where $s$ is the parameter of the path, such that the path starts at the point $x_0$ and ends at $x_1$ for $s \in [0,s_f]$. The solution of this equation is given by:

\begin{equation} g(P) = \mathcal{P} \exp \left( -\int\limits_{P(0)}^{P(s_f)} A_\mu(P(s)) \; \mathrm{d} x^\mu \right) \end{equation}

where $\mathcal{P}$ denotes the path ordering symbol. Furthermore, we can show that transformation properties are given by:

\begin{equation} g(P)' = g(x_1) g(P) g(x_0)^{-1} \end{equation}

Now the proof: ``For any space-time point $x$, define $P_x$ to be the straight-line path from $(\mathbf{x},0)$ to $x$. The desired gauge transformation is defined by:

\begin{equation} g(x)=g(P_x)^{-1} \end{equation}

for, under this transformation:

\begin{equation} g(P_x)' = g(P_x)^{-1} g(P_x) g(P_0)=1 \end{equation}

from which $A_0=0$ follows by differentiation.''

I understand the mathematics before the actual proof, but I find his proof quite confusing (maybe because English is not my first language). From what I understand, he is defining a path $P_x$ at every point $x$ in space-time. Furthermore, $P_x$ is a straight line evolving in time only, i.e. $P_x$ stays at the same point $\mathbf{x}$ in space but evolves with $t$. Is that correct? If so, then $g(P_x)$ is given by:

\begin{equation} g(P_x) = \mathcal{P} \exp \left( -\int\limits_{P(0)}^{P(s_f)} A_0(P_x(s)) \; \mathrm{d} x^0 \right) \end{equation}

and indeed this implies:

\begin{equation} \partial_0 g(P_x)' = A_0' = 0 \end{equation}

If my interpretation is correct until now, then I have the following (perhaps stupid) question:

How do we know that $\phi$ at each space-time point $x$ always obey the first equation I wrote? In other words, the whole proof is based on the idea the $\phi$ satisfied that equation for the path $P_x$, but how do we know that is true?

Answer 1

This assumption he mades there in setting up the Wilson (path) dependent line is a trick, to argue that if $\phi$, respective $g\left(\mathcal{P}\right)$ is a true solution of the first order differential equation $\frac{dx^{\mu}}{ds}D_{\mu}\phi=0$, it is unique and indeed has the transformation law $g(P)^{\prime}=g(x_1)g(P)g(x_0)^{−1}$, which he then uses to complete the proof. I can recommend you also to follow the similar discussion about the Wilson Loop in Chapter 15.3, p.491 in the textbook by Peskin & Schroeder, where the argumentation sequence is basically the same.