Gauge fields and strings: Loop equations I am trying to derive Eq. (7.25) (p. 117) of Polyakov's book:
$$ \delta \Psi (C) ~=~ \int_{0}^{2\pi} {\rm P} \left(F_{\mu\nu}(x(s)) \exp \oint_C A_\mu dx^\mu \right)\dot{x}_\nu \delta x_\mu(x) \, {\rm d} s, \tag{7.25} $$
where the non-abelian phase factor around a closed loop $C$ is defined as
$$ \Psi(C) ~=~ {\rm P}\exp \left(\oint A_\mu dx^\mu \right) = {\rm P}\exp \left(\int_{0}^{2\pi} A_\mu \dot{x}_\mu\, {\rm d}s \right). \tag{7.1} $$
It seems that he is using the relation given on p. 116:
$$ \delta \, {\rm P} \exp \int_{0}^{2\pi} M(\tau) {\rm d}\tau ~=~ \int_{0}^{2\pi}{\rm d}t\,{\rm P} \left(\delta M(t) \exp \int_{0}^{2\pi}M(\tau){\rm d}\tau\right). \tag{7.24b} $$
Matching with (7.25) I find $\delta A_\nu = F_{\mu\nu} \delta x_\mu$. This relation seems to be saying that if I change the position of the loop at the parameter $s$ by $\delta x_\mu(s)$ then the vector potential changes by $\delta A_\nu(x(s)) = F_{\mu\nu}(x(s)) \delta x_\mu(s)$.  
I don't know how to derive this relation. Is it legitimate?
 A: *

*We start with a non-abelian gauge theory. The covariant derivative is
$$D~=~\mathrm{d}+A, \qquad A~=~\mathrm{d}x^{\mu} A_{\mu},\tag{A}$$
while the field strength is
$$\begin{align} \frac{1}{2}F_{\mu\nu}\mathrm{d}x^{\mu}\wedge\mathrm{d}x^{\nu} ~=~&F~=~D \wedge D\cr
~=~&\frac{1}{2}[D\stackrel{\wedge}{,}D]\cr
~=~&[\mathrm{d},A] + \frac{1}{2}[A\stackrel{\wedge}{,}A]\cr
~=~&\mathrm{d}A + A \wedge A, \end{align} \tag{6.35}$$
$$ F_{\mu\nu}~=~\partial_{[\mu}A_{\nu]} + [A_{\mu},A_{\nu}]. \tag{6.36}$$


*Next consider a non-abelian Wilson-line$^1$
$$ U(t_2,t_1)~=~ \left\{\begin{array}{rcl} T\exp\left(-\int_{t_1}^{t_2}\! A\right)&{\rm for}& t_1\leq t_2,\cr AT\exp\left(-\int_{t_1}^{t_2}\! A\right)&{\rm for}& t_2\leq t_1,\end{array}\right.  \tag{7.1'} $$
over a (possibly open) curve $C$. Here $(A)T$ denotes (anti)time-ordering. Let us for simplicity assume from now on that $t_1\leq t_2$. Then we may write
$$ U(C)~=~T\exp\left(-\int_C\! A\right) \tag{7.1'} $$
with a parametrized curve $C:[t_1,t_2]\to\mathbb{R}^4$.
The Wilson-line (7.1') is the solution to the following ODE
$$\begin{align} 
\frac{dU(t_2,t_1)}{dt_2}
~=~&-\dot{x}^{\mu}(t_2) A_{\mu}(t_2) U(t_2,t_1), \cr 
\frac{dU(t_2,t_1)}{dt_1}
~=~&U(t_2,t_1)\dot{x}^{\mu}(t_1) A_{\mu}(t_1),\cr  
U(t_1,t_1)~=~&{\bf 1}.\end{align}\tag{B}$$


*We now make an infinitesimal variation of the curve $C$ to a new curve $C^{\prime}$. The varied curve $C^{\prime}$ is assumed to have the same end points as $C$, and the same parametrization interval $[t_1,t_2]$. We may define an infinitesimally thin 2-surface $\Sigma$ with oriented boundary
$$ \partial \Sigma~=~ C^{\prime}-C \tag{C}$$
given by the two curves $C$ and $C^{\prime}$. This induces a (passive) change $\delta A$ of the gauge field $A$.
NB: Be aware that the 2 sides
$$   \int_{C}\! \delta A ~=~ \int_{C^{\prime}}\! A-\int_{C} \!A  ~=~  \oint_{\partial\Sigma} \!A ~=~ \iint_{\Sigma}\! \mathrm{d} A \tag{D} $$
and
$$
\iint_{\Sigma}\! F ~=~ \int_C\! \delta x^{\mu} F_{\mu\nu}   \mathrm{d} x^{\nu}  \tag{E} $$
of Stokes' circulation theorem are not necessarily equal for non-Abelian gauge-fields.$^2$


*The infinitesimal (passive) change in holonomy is
$$\begin{align}  \delta U(C)~=~~~~&U(C^{\prime})-U(C)\cr
~\stackrel{(7.1')}{=}~~&-T\left[\exp\left(-\int_C\! A\right)\int_C\! \delta A\right]\cr
~\stackrel{(7.1')}{=}~~&-\int_{t_1}^{t_2}\! dt~U(t_2,t)\delta[\dot{x}^{\mu}(t)A_{\mu}(t)]U(t,t_1)\cr
~=~~~~&-\int_{t_1}^{t_2}\! dt~U(t_2,t)\left[\frac{d\delta x^{\mu}(t)}{dt}A_{\mu}(t)+\dot{x}^{\mu}(t)\delta A_{\mu}(t)\right]U(t,t_1)\cr
~\stackrel{\text{int. by parts}}{=}&
\text{bulk terms} ~+~ \text{boundary terms},\end{align}\tag{F}$$
where
$$\begin{align}
\text{bulk}&\text{ terms}\cr
~=~&\int_{t_1}^{t_2}\! dt~U(t_2,t)\left[
\frac{\stackrel{\leftarrow}{d}}{dt}\delta x^{\mu}(t)A_{\mu}(t)
+\delta x^{\mu}(t)\dot{A}_{\mu}(t)\right.\cr
&\left. -\dot{x}^{\mu}(t)\delta A_{\mu}(t)
+\delta x^{\mu}(t)A_{\mu}(t)\frac{\stackrel{\rightarrow}{d}}{dt}
\right]U(t,t_1)\cr
~\stackrel{(B)}{=}~&\int_{t_1}^{t_2}\! dt~U(t_2,t)\left[
\dot{x}^{\nu}(t) A_{\nu}(t)\delta x^{\mu}(t)A_{\mu}(t)
+\delta x^{\mu}(t)\dot{x}^{\nu}(t)\partial_{\nu}A_{\mu}(t)\right.\cr
&\left. -\dot{x}^{\mu}(t)\delta x^{\nu}(t)\partial_{\nu} A_{\mu}(t)
-\delta x^{\mu}(t)A_{\mu}(t)\dot{x}^{\nu}(t) A_{\nu}(t)
\right]U(t,t_1)\cr
~\stackrel{(6.36)}{=}&\int_{t_1}^{t_2}\! dt~U(t_2,t)
\dot{x}^{\mu}(t) F_{\mu\nu}(t)\delta x^{\nu}(t)
U(t,t_1)\cr
~=~&T\left[\exp\left(-\int_C\! A\right)
\int_C\! F_{\mu\nu}\mathrm{d}x^{\mu} \delta x^{\nu}\right]
\cr
~\stackrel{(E)}{=}~&-T\left[\exp\left(-\int_C\! A\right)
\iint_{\Sigma}\! F\right]
,\end{align}\tag{7.25'}$$
and
$$\begin{align}
\text{boundary terms}~=~&-\left[U(t_2,t)\delta x^{\mu}(t)A_{\mu}(t)U(t,t_1)\right]_{t=t_1}^{t=t_2}\cr
~=~& U(t_2,t_1)\delta x^{\mu}(t_1)A_{\mu}(t_1)
-\delta x^{\mu}(t_2)A_{\mu}(t_2)U(t_2,t_1)\cr
~\stackrel{(H)}{=}~&0,\end{align}\tag{G}$$
since the endpoints are not varied
$$ \delta x^{\mu}(t_1)~=~0~=~\delta x^{\mu}(t_2).\tag{H}$$
Eq. (7.25') answers OP's main question about eq. (7.25). The minus signs are caused by different sign conventions.
References:

*

*A.M. Polyakov, Gauge Fields and Strings, 1987; Chapter 7.

--
$^1$ A Wilson line is physics jargon for holonomy. If the curve $C$ is closed, we speak of a Wilson loop rather than a Wilson line. We prefer to use time-ordering rather than path-ordering, since the latter is ambiguous. Ref. 1. uses a path-ordering $P$ from left to right,
$$ \Psi(C)~:=~ P e^{\int_{C} \!A},  \tag{7.1} $$
which induces an opposite sign in front of the gauge field $A$ as compared to eq. (7.1').
$^2$ Let us mention for completeness that there exists a
non-Abelian Stokes' Theorem, which takes an
exponentiated form
$$  P\exp\oint_{\partial\Sigma} \! A~=~ P_2\exp\iint_{\Sigma}\! F. \tag{I}$$
It  depends on a choice of surface ordering $P_2$.
A: Consider the non-abelian phase factor around a closed path $C$,
\begin{equation}
 \psi(C) = \mathrm{P} e^{\oint A_\mu dx^\mu} = \mathrm{P} e^{\int_0^{2\pi} dt \, A_\nu(x(t)) \dot{x}^\nu(t) }
\end{equation}
Let us take the functional derivative with respect to $x^\mu(s)$
\begin{align}
 \frac{\delta}{\delta x^\mu(s)} \psi(C)
  %& = \int_0^{2\pi} dt \, \mathrm{P} \left[ \partial_\mu A_\nu(x(t))\delta(s-t)\dot{x}^\nu(t) + A_\nu (x(t))\delta^\nu_\mu \dot{\delta}(s-t) \right] e^{\int_0^{2\pi} dt \, A_\nu(x(t)) \dot{x}^\nu(t) } \\
  & = \int_0^{2\pi} dt \,  \left\{ \mathrm{P}e^{\int_0^{t} dt' \, A_\nu\dot{x}^\nu} \left[ \partial_\mu A_\nu(x(t))\delta(s-t)\dot{x}^\nu(t) + A_\mu (x(t))\dot{\delta}(s-t)\right] \mathrm{P}e^{\int_t^{2\pi} dt' \, A_\nu \dot{x}^\nu} \right\} 
\end{align}
We now integrate by parts the $t$-derivative on the delta function,
\begin{align}
 \frac{\delta}{\delta x^\mu(s)} \psi(C)
  & = \int_0^{2\pi} dt \, \mathrm{P}e^{\int_0^{t} dt' \, A_\nu\dot{x}^\nu} \left(\partial_\mu A_\nu - \partial_\nu A_\mu + [A_\mu, A_\nu]\right)_{x(t)} \dot{x}^\nu(t)\delta(s-t)  \mathrm{P}e^{\int_t^{2\pi} dt' \, A_\nu\dot{x}^\nu} \notag \\
  & \quad + \mathrm{P} e^{\int_0^{2\pi} dt A_\nu \dot{x}^\nu}A_\mu(x(2\pi))\delta(s-2\pi) - A_\mu(x(0))\mathrm{P} e^{\int_0^{2\pi} dt A_\nu \dot{x}^\nu}\delta(s) \\
  & = \mathrm{P}e^{\int_0^{s} dt \, A_\nu\dot{x}^\nu} F_{\mu\nu}(x(s)) \dot{x}^\nu(s)\mathrm{P}e^{\int_s^{2\pi} dt \, A_\nu\dot{x}^\nu}
\end{align}
where we have discarded boundary terms by periodicity.
