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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?

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I) We start with a non-Abelian Wilson line $^1$

$$\tag{7.1} \Psi(C)~:=~ P e^{\int_{C} \!A} $$

over a parametrized (possibly open) curve $C$. Here $P$ denotes path-ordering.

II) We now make an infinitesimal variation of the curve $C$ to a new curve $C^{\prime}$. We may define an infinitesimally thin 2-surface $\Sigma$ with oriented boundary $^2$

$$\tag{A} \partial \Sigma~=~ C^{\prime}-C $$

given by the two curves $C$ and $C^{\prime}$.

III) One may show a non-Abelian version of Stokes' circulation theorem $^3$

$$\tag{B} \delta\!\int_{C}\! A ~:=~\int_{C^{\prime}}\! A-\int_{C} \!A~=~ \oint_{\partial\Sigma} \! A~=~ \iint_{\Sigma}\! F . $$

The sketched proof of the non-Abelian eq. (B) is in two steps as its Abelian cousin:

  1. Split the $2$-surface $\Sigma$ into infinitely many infinitesimal small polygons, where contributions from internal edges of the polygons cancel because of opposite orientations.

  2. Notice that for a sufficiently small polygon the holonomy around the polygon is approximately equal to the non-Abelian field strength $F$ times the area.

IV) Now we are ready to evaluate the (passive) change in holonomy because of the change $C\to C^{\prime}$ in curves:

$$\tag{7.25} \delta\Psi(C)~:=~\Psi(C^{\prime})-\Psi(C) ~=~ P\left( e^{\int_{C} \!A} ~\delta\!\int_{C} \!A \right) ~\stackrel{(B)}{=}~ P\left( e^{\int_{C} \!A} ~\iint_{\Sigma}\! F\right). $$

Because of the path-ordering $P$ in eq. (7.25), we are strictly speaking applying a "chopped up version" of eq. (B) consisting of infinitely many infinitesimally small polygons, which reflect the proof technique of eq. (B).

References:

  1. A.M. Polyakov, Gauge Fields and Strings, 1987.

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$^1$ A Wilson line $\Psi(C)$ is physics jargon for holonomy. If the curve $C$ is closed, we speak of a Wilson loop rather than a Wilson line.

$^2$ For an open curve $C$, the varied curve $C^{\prime}$ is assumed to have the same end points as $C$. We let both curves $C$ and $C^{\prime}$ be parametrized so that we have a common notion of path-ordering $P$.

$^3$ Let us mention for completeness that the exponentiated version

$$ \tag{C} P\exp\oint_{\partial\Sigma} \! A~=~ P_2\exp\iint_{\Sigma}\! F $$

of non-Abelian Stokes' Theorem depends on a choice of surface ordering $P_2$. Note that an infinitesimal thin 2-surface $\Sigma$ has a natural choice of surface ordering $P_2$.

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  • $\begingroup$ Thanks for your answer. I'm not quite sure I fully understand it yet. In the meantime, however, I think I have a more pedestrian explanation (see below). $\endgroup$ – user11881 Dec 15 '13 at 20:11
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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.

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