In terms of the variable $r^{2}\equiv a\bar{a},$ the Nambu-Goto action takes the form

$$S={T\over 2\pi}\int d\sigma
\sqrt{(\partial_{\sigma}r)^{2}+{1\over f\tau_{2}}}
={T\over 2\pi}\int d\sigma
\sqrt{(\partial_{\sigma}r)^{2}+{r^{4}\over a}-{b\over a^{2}}}$$

$$a={4N\pi\over \tau_{2cl}},\quad
  b=\cos (4\theta)m_{1}m_{2}m_{3}m_{4}N\Big(
{\alpha\tau_{2cl}\over 2}-{6\over \tau_{2cl}}\Big)$$

where $T=\int d\tau$. We have dropped terms of $O(r^{-4})$ in the square root above. The solution to the Euler-Lagrange equations of motion following from this action is obtained in the usual way: The action does not depend explicitly on $\sigma$ so that the Hamiltonian in the $\sigma$ direction is a constant of the motion

$${{r^{4}\over a}-{b\over a^{2}}\over
\sqrt{(\partial_{\sigma}r)^{2}+{r^{4}\over a}-{b\over a^{2}}}}=
const=\sqrt{{r_{0}^{4}\over a}-{b\over a^{2}}}$$

where $r_{0}$ is the minimal value of $r$. By symmetry we have $r(\sigma=0)=r_{0}$. It is now straight forward to obtain

$$\sigma=\sqrt{{r_{0}^{4}\over a}-{b\over a^{2}}}
\int_{r_{0}}^{r} {dy\over
\sqrt{({y^{4}\over a}-{b\over a^{2}})
      ({y^{4}\over a}+{r_{0}^{4}\over a})} }.$$

The string endpoints are at the boundary of the asymptotically AdS$_{5}$ space ($r\to\infty$) so that we can trade the integration constant $r_{0}$ for the distance $L$ between the quark and anti-quark

$$L=2\sqrt{{r_{0}^{4}\over a}-{b\over a^{2}}}
\int_{r_{0}}^{\infty} {dy\over
\sqrt{({y^{4}\over a}-{b\over a^{2}})
      ({y^{4}\over a}+{r_{0}^{4}\over a})} }.$$

The energy is now computed by evaluating our action at this classical solution. After subtracting twice the self-energy of a quark, we obtain the following result for the quark-antiquark potential

$$E={1\over\pi\sqrt{({r_{0}^{4}\over 