\documentclass[12pt,a4paper]{article} \usepackage{graphics} \usepackage{cite,a4wide} % % Here are some macros % that I like to use % \newcommand{\beq}{\begin{equation}} \newcommand{\eeq}{\end{equation}} \newcommand{\bea}{\begin{eqnarray}} \newcommand{\eea}{\end{eqnarray}} \renewcommand{\ni}{\noindent} \newcommand{\rf}[1]{(\ref{#1})} \newcommand{\ra}{\rightarrow} \newcommand{\pa}{\partial} % % \begin{document} % % \noindent Physics 375 \hfill {\bf \large Project 37} \hfill Feb. 31, 1999 \begin{center} \vspace{24pt} {\bf \large Thermal Wilson Loops from Supergravity } \vspace{24pt} {\sl Gil Bates} \end{center} \vspace{24pt} \noindent \emph{(Comments:~~ You should describe the problem you want to solve, up to and including the actual equations to be treated numerically. Its not necessary to actually derive those equations, unless that is part of the assignment.)} \bigskip Maldacena has recently suggested a method for calculating Wilson loops in ${\cal N}=4$ super Yang-Mills theory, using a proposed correspondence to supergravity in anti-deSitter space ($AdS_5$). These loops can be used to compute the potential energy between a quark and an antiquark. In this project I will use his method to compute spacelike Wilson loops at finite temperature. Briefly, the method boils down to computing the action of a classical Nambu-Goto string in a Wick-rotated $AdS_5 \times S_5$ black-hole metric \beq ds^2 = \alpha' \left\{ {U^2 \over R^2}[f(U)dt^2 + dx_i^2] + {R^2\over U^2} f^{-1}(U) dU^2 + R^2 d\Omega_5^2 \right\} \eeq where the boundary $C$ of the string worldsheet lies $U=\infty$, and \beq f(U) = 1 - {U_T^4 \over U^4}, ~~~~~~ R^2 = \sqrt{4\pi gN} \eeq The proposal is that the expectation value of the Wilson loop at large N is given by \beq W(C) \sim \exp(-S) \eeq where the exponent is the action of the classical worldsheet bounding loop $C$. We consider rectangular $L\times T$ contours $C$, which lie in the $x-x'$ plane for spacelike loops, where $x,x' \in \{x_i\}$. Worldsheet coordinates $\sigma, \tau$ are identified (in static gauge) with $\sigma=x,~ \tau=t$ for timelike loops, and $\sigma=x,~\tau=x'$ for spacelike loops. It is assumed that $T\gg L$, and the string is static in the sense that the transverse position $U(\sigma,\tau)$ of the worldsheet depends only on $x$. One then finds, for timelike loops, that the Nambu action becomes \beq S = {T\over 2\pi} \int_{-L/2}^{L/2} {dx \over \sqrt{f(U)} } ~\sqrt{ (\pa_x U)^2 +(U^4 - U_T^4)/R^4 } \label{Ss} \eeq It was explained in class\footnote{No it wasn't. But we'll pretend it was.} that, since the Lagrangian \rf{Ss} does not depend on $x$ explicitly, there is a constant of motion \beq {U^4 - U_T^4 \over \sqrt{f(U)}\sqrt{ (\pa_x U)^2 +(U^4 - U_T^4)/R^4 } } = {R^2\over \sqrt{f(U_0)}} \sqrt{U_0^4-U_T^4} \label{com} \eeq where the right-hand side is the constants of motion in each case evaluated at the minimum $U=U_0$, where $\pa_x U=0$. Solving for $\pa_x U$, and writing \beq y=U/U_0, ~~~~ \epsilon = f(U_0) \eeq one finds \beq x = {R^2 \over U_0} \int_1^{U/U_0} {dy \over \sqrt{(y^4-1)(y^4-1+\epsilon)} } \label{xy} \eeq which determines $U(x)$ implicitly, given $U_0$. But $U_0=U(0)$, while the ends of the string are at $x=\pm L/2$, where $U=\infty$, so that \beq L = 2{R^2 \over U_0} \int_1^{\infty} {dy \over \sqrt{(y^4-1)(y^4-1+\epsilon)} } \label{L} \eeq gives $U_0$ as a function of quark separation $L$. The only difference in the spacelike and timelike expressions for $x$ and $L$ is an extra factor of $\sqrt{\epsilon}$ in the timelike case. The energy of the static worldsheet $E=S/T$ is calculated using the relationships \rf{com}, changing integration variable from $x$ to $y$ using \rf{xy}, and making an infinite subtraction corresponding to the mass of the W-boson according to Maldacena's prescription \beq E(L) = {U_0\over \pi} \int_1^\infty dy ~ \left( {y^4 \over \sqrt{(y^4 -1)(y^4-1+\epsilon)} } - 1 \right) - {U_0-U_T \over \pi} \label{energy} \eeq In principle, this gives use the potential energy between static quarks $E(L)$ as a function of quark separation $L$. \bigskip \emph{(Comment:~~ Now say something about your approach to solving equations \rf{L} and \rf{energy} numerically.)} blah blah blah numerics \bigskip \noindent \emph{(Comments:~~ Display and discuss your results.)} \begin{figure}[ht] \centerline{\scalebox{.5}{\rotatebox{270}{\includegraphics{figads1.ps}}}} \caption{String contours $U(x)$ for various $\epsilon = 1-(U_T^4/U_0^4)$, with $U_T=1,R^2=1$. The asymptotes of each curve lie at $x=\pm L/2$. Note the approach to the horizon (here at $U=1$), as $\epsilon \ra 0$.} \label{fig1} \end{figure} There are no subtleties of this sort for the spacelike loop, due the absence of the multiplicative factor $\sqrt{\epsilon}$ in \rf{L} for the spacelike expression. For any finite $L$, there corresponds a finite $U_0>U_T$, with $U_0 \ra U_T$ in the $L\ra \infty$ limit, and with energies given by the expression in \rf{energy}. Some typical solutions for $U(x)$ at various $U_0/U_T$, are shown in Fig.\ 1. Note that, as $U_0 \ra U_T$ (and $\epsilon \ra 0$) the string worldsheet approaches $U(x)=U_T$ in the finite interval $|x|