Gauge fields and strings: Loop equations

Question

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?

Answer 1

1. 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 $$\begin{align} \frac{1}{2}F_{\mu\nu}\mathrm{d}x^{\mu}\wedge\mathrm{d}x^{\nu} ~=~&F~=~D \wedge D\cr ~=~&\frac{1}{2}[D\stackrel{\wedge}{,}D]\cr ~=~&[\mathrm{d},A] + \frac{1}{2}[A\stackrel{\wedge}{,}A]\cr ~=~&\mathrm{d}A + A \wedge A, \end{align} \tag{6.35}$$ $$ F_{\mu\nu}~=~\partial_{[\mu}A_{\nu]} + [A_{\mu},A_{\nu}]. \tag{6.36}$$

2. 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 $$\begin{align} \frac{dU(t_2,t_1)}{dt_2} ~=~&-\dot{x}^{\mu}(t_2) A_{\mu}(t_2) U(t_2,t_1), \cr \frac{dU(t_2,t_1)}{dt_1} ~=~&U(t_2,t_1)\dot{x}^{\mu}(t_1) A_{\mu}(t_1),\cr U(t_1,t_1)~=~&{\bf 1}.\end{align}\tag{B}$$

3. 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.$^2$

4. The infinitesimal (passive) change in holonomy is $$\begin{align} \delta U(C)~=~~~~&U(C^{\prime})-U(C)\cr ~\stackrel{(7.1')}{=}~~&-T\left[\exp\left(-\int_C\! A\right)\int_C\! \delta A\right]\cr ~\stackrel{(7.1')}{=}~~&-\int_{t_1}^{t_2}\! dt~U(t_2,t)\delta[\dot{x}^{\mu}(t)A_{\mu}(t)]U(t,t_1)\cr ~=~~~~&-\int_{t_1}^{t_2}\! dt~U(t_2,t)\left[\frac{d\delta x^{\mu}(t)}{dt}A_{\mu}(t)+\dot{x}^{\mu}(t)\delta A_{\mu}(t)\right]U(t,t_1)\cr ~\stackrel{\text{IBP}}{=}~~~& \text{bulk terms} ~+~ \text{boundary terms},\end{align}\tag{F}$$ where $$\begin{align} \text{bulk}&\text{ terms}\cr ~=~&\int_{t_1}^{t_2}\! dt~U(t_2,t)\left[ \frac{\stackrel{\leftarrow}{d}}{dt}\delta x^{\mu}(t)A_{\mu}(t) +\delta x^{\mu}(t)\dot{A}_{\mu}(t)\right.\cr &\left. -\dot{x}^{\mu}(t)\delta A_{\mu}(t) +\delta x^{\mu}(t)A_{\mu}(t)\frac{\stackrel{\rightarrow}{d}}{dt} \right]U(t,t_1)\cr ~\stackrel{(B)}{=}~&\int_{t_1}^{t_2}\! dt~U(t_2,t)\left[ \dot{x}^{\nu}(t) A_{\nu}(t)\delta x^{\mu}(t)A_{\mu}(t) +\delta x^{\mu}(t)\dot{x}^{\nu}(t)\partial_{\nu}A_{\mu}(t)\right.\cr &\left. -\dot{x}^{\mu}(t)\delta x^{\nu}(t)\partial_{\nu} A_{\mu}(t) -\delta x^{\mu}(t)A_{\mu}(t)\dot{x}^{\nu}(t) A_{\nu}(t) \right]U(t,t_1)\cr ~\stackrel{(6.36)}{=}&\int_{t_1}^{t_2}\! dt~U(t_2,t) \dot{x}^{\mu}(t) F_{\mu\nu}(t)\delta x^{\nu}(t) U(t,t_1)\cr ~=~&\int_{t_1}^{t_2}\! dt~ \dot{x}^{\mu}(t) \underbrace{U(t_2,t)F_{\mu\nu}(t)U(t,t_1)}_{=:{\cal F}_{\mu\nu}(t)}\delta x^{\nu}(t) \cr ~=~&T\left[\exp\left(-\int_C\! A\right) \int_C\! F_{\mu\nu}\mathrm{d}x^{\mu} \delta x^{\nu}\right] \cr ~\stackrel{(E)}{=}~&-T\left[\exp\left(-\int_C\! A\right) \iint_{\Sigma}\! F\right] ,\end{align}\tag{7.25'}$$
   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, such as choice of orientation.

References:

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

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$^1$ 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

$$ Te^{-\oint_{\partial\Sigma} \! A}~=~ P_2\prod_{x\in\Sigma} U(\ast,x) e^{-\iint_{\delta\Sigma_x}\!F}U(x,\ast) ~=:~P_2e^{-\iint_{\Sigma}\!{\cal F}}. \tag{I}$$ Here all infinitesimal plaquettes $\delta\Sigma_x$ at the point $x$ are parallel-transported to the same fiducial base point $\ast$. The construction depends on a choice of surface ordering $P_2$. Recall that for an infinitesimal plaquette $\Sigma$, $$Te^{-\oint_{\partial\Sigma} \! A} ~=~e^{-\iint_{\Sigma}\!F},\tag{J}$$ cf .e.g. my Phys.SE answer here.

Answer 2

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.