Wilson Loop in AdS/CFT

Question

In AdS/CFT correspondence one can compare results in $\mathcal{N}=4$ SYM with string theory type IIB in $AdS_5 \times S^5$. One of the observables that it's possible to get non-perturbative results is the Wilson loop.

Maldacena proposed in his paper that in the string side the expectation value of the Wilson loop is the area of the fundamental string that ends in the contour of the loop.

The metric of $AdS_5 \times S_5$ is given by:

$dS^2 = \alpha' [ \frac{U^2}{R^2}(dt^2 + dx_i dx_i) + \frac{R^2}{U^2} dU^2 + d\Omega_5^2]$

From which one can obtain the induced metric on the worldsheet: $h_{\alpha \beta} = G_{MN} \frac{\partial X^M}{\partial X^{\alpha}}\frac{\partial X^N}{\partial X^{\beta}}$ and in the large N limit one gets the area as the nambu goto action:

$S = \frac{1}{2\pi \alpha'} \int d\tau d\sigma \sqrt{det(h_{\alpha \beta})}$

Choosing a static gauge: $\tau = t$, $\sigma = x$, and fixing the coordinates in $S^5$ as constants $\theta^I_0$ one gets:

$ S = \frac{T}{2\pi} \int dx \sqrt{U^4/R^4 + (\partial_x U)^2}$

From which we need to solve the Euler-Lagrange equations, in the paper Maldacena says that as the actions does not depend on X the canonical momentum is conserved, and one gets the condition:

$\frac{U^4}{\sqrt{U^4/R^4 + (\partial_x U)^2)}} = constant$

But I can't exactly grasp what is he doing to get there, does anyone knows?

Answer 1

This just classical mechanics. We have a Lagrangian ${\cal L}(U,\partial_x U)$. Since the Lagrangian does not depend on time (here: $x$), the Hamiltonian $$ {\cal H} = \frac{\partial {\cal L}}{\partial (\partial_x U)} (\partial _x U) -{\cal L} $$ is conserved. Presto.