0$. Calling $x_1$ as $x$ and $a_1$ as $a$ for notational simplicity, the geodesic equations for~(\ref{suone}) are given by \begin{eqnarray} x^{\prime\prime} + 2 \frac{a^\prime}{a} x^\prime - a a^\prime {x^\prime}^3 &=& 0,\nonumber\\ z z^{\prime\prime} + {z^\prime}^2 + {x^\prime}^2 a^2 - a a^\prime z z^\prime {x^\prime}^2 -1 &=& 0.\phantom{\frac{1}{1}} \label{suthree} \end{eqnarray} Here, we have taken time as a parameter and derivatives are with respect to time. General solutions of these equations can be written as \begin{equation} x (t) = \pm \int \frac{a(t^*) dt}{a(t) {\sqrt{a^2(t^*) -a^2(t)}}} \label{sufour} \end{equation} and \begin{equation} z = + {\sqrt{ -2 \int dt \bigg[\frac{a(t)}{\sqrt{a^2(t^*) - a^2(t)}} \bigg(\int^{t} dt' \frac{a(t')}{\sqrt{a^2(t^{\prime *}) - a^2(t')}}\bigg)\bigg] }}. \label{sufive} \end{equation} To write~(\ref{sufour}) in this form, we have used the fact that there is a turning point of the geodesic in the bulk and at that point $dx/dt$ diverges. For the solution above, we have taken the point to be $t = t^*$. Given a functional form for $a(t)$, one would then try to integrate the left hand side of the above equations. In this process, three integration constants would appear. However, all of these can be fixed by boundary conditions. The constant appearing from~(\ref{sufour}) can be set to zero by using $x \rightarrow x + {\rm constant}$ symmetry of the metric. Other two constants arise from the two integrations in~(\ref{sufive}). Both of them can be fixed - (1) by demanding $dz/dx =0$ at the turning point of the geodesic in $z-x$ plane and (2) by requiring $z =0$ for $t = t_0$. We now argue that if the metric has the scaling symmetry~(\ref{suscal}), then the constant $a(t^*)$ can be scaled away. Taking $a(t) = t^p$ and defining new coordinates $\bar z, \bar t$ and $\bar x$ as \begin{equation} z = t^* \bar z,~~t = t^* \bar t, ~~x = {t^*}^{1-p} \bar x \label{susix} \end{equation} we can re-write~(\ref{sufour}) and~(\ref{sufive}) as \begin{equation} \bar x (\bar t) = \pm \int \bigg[\frac{d\bar t}{ {\bar t}^p {\sqrt{1 - \bar t^{2p}}}}\bigg], \label{suseven} \end{equation} and \begin{equation} \bar z(\bar t) = + {\sqrt{ -2 \int d\bar t \bigg[ \frac{\bar t^p} {\sqrt{1 - \bar t^{2p}}} \bigg( \int^{\bar t} d\bar t' \frac{{\bar t}'^p} {\sqrt{1 - {\bar t}'^{2p}}}\bigg)\bigg] }}. \label{sueight} \end{equation} With this,~(\ref{suseven}) and~(\ref{sueight}) can be easily integrated. This gives, for \emph{generic} $p$ \begin{eqnarray} \bar x(\bar t) = \frac{{\bar t}^{1-p}}{1 - p} ~{}_2F_1 \bigg(\frac{1}{2}, \frac{1-p}{2p}, \frac{1+p}{2p}, {\bar t}^{2p}\bigg) - \frac{\sqrt{\pi} \Gamma(\frac{1-p}{2p})}{(1-2p) \Gamma(\frac{1-2p}{2p})}, \label{sunine} \end{eqnarray} and \begin{eqnarray} \bar z(\bar t) = && \bigg[{\bar t}^2\bigg[ 1 - {}_3F_2\bigg( \bigg\{1, \frac{1}{2p}, \frac{1}{p}\bigg\}, \bigg\{\frac{1}{2} + \frac{1}{2p}, 1 + \frac{1}{p}\bigg\}, {\bar t}^{2p}\bigg)\bigg] \nonumber\\ &&\qquad+\frac{4{\sqrt \pi} p \Gamma(\frac{1+p}{2p}) {\bar t}^{1-p} } { (1-2p)\Gamma(\frac{1-2p}{2p})} \bigg[ {}_2F_1\bigg(\frac{1}{2}, \frac{1-p}{2p}, \frac{1+p}{2p}, {\bar t}^{2p}\bigg) - {\sqrt {1 - {\bar t}^{2p}}}\bigg] + c\bigg]^{\frac{1}{2}}. \label{suten} \end{eqnarray} Here, $c$ is a constant which can fixed using $\bar z = 0$ for $\bar t = \bar t_0$. ${}_2F_1$ and ${}_3F_2$ are the hypergeometric function and the generalized hypergeometric function respectively. For some specific values of $p$, the solutions however simplify. In appendix B, we provide a way to solve~(\ref{suthree}) and get to these results. Having reached this far, we proceed to find the geodesic length. For the correlator, $\langle {\cal O}(x, t_0) {\cal O}(-x, t_0)\rangle$, we first need to calculate the integral~(\ref{sureg}) \begin{equation} {\cal L} = \int \frac{2 dt}{z} \Bigg[{\sqrt{ -1 + \bigg(\frac{dz}{dt}\bigg)^2 + t^{2p} \bigg(\frac{dx}{dt}\bigg)^2}}\ \Bigg], \end{equation} with appropriate limits. Now, as for the lower limit, the turning point of the geodesic is at $t = t^*$. In terms of scaled time, it is at $\bar t = 1$. For the upper limit, we note that the correlator is being calculated at a constant $t = t_0$ slice. This, in terms of scaled variable, is $ \bar t = \bar t_0$. We further need to UV-regulate the integral by introducing a cut-off, $\bar \delta$. The geodesic length is therefore \begin{equation} {\cal L} = \int_{\bar t =1}^{\bar t = \bar t_0 - \bar \delta} \frac{2 d\bar t}{\bar z} \Bigg[{\sqrt{ -1 + \bigg(\frac{d\bar z}{d\bar t}\bigg)^2 + {\bar t}^{2p} \bigg(\frac{d\bar x}{d\bar t}\bigg)^2}}\ \Bigg]. \end{equation} In general, $\cal L$ is infinite. In order to render it finite, we need to subtract, from $\cal L$, the equivalent AdS part. This removes the $\bar \delta \rightarrow 0$ singularity in the geodesic length. Consequently, the regulated $\cal L$ will only depend on $\bar t_0$. Above observation, in turn, means that the gauge theory correlator has the form \begin{equation} \langle {\cal O}(x, t_0) {\cal O}(- x, t_0) \rangle = e^{-m {\cal{L}}_{\rm reg}} = f(\bar t_0), \label{sueleven} \end{equation} for some function $f$. For general $p$, we are unable to evaluate this function analytically. Nevertheless, numerically it can be calculated. We provide our results later. However, for some values of $p$, expressions simplify and analytic computations can be done. As an illustrative example, we do it for $p = 1/3$. Results are given below. \begin{eqnarray} && \bar z(\bar t) = {\sqrt{ 3 ({\bar t}^{~\frac{4}{3}} - {\bar t}_0^{~\frac{4}{3}}) + ({\bar t}^2 - {\bar t}_0^2)}}\nonumber\\ &&\bar x(\bar t)= \pm 3 {\sqrt{1 - {\bar t}^{~\frac{2}{3}}}}. \end{eqnarray} The geodesic length turn out to be \begin{eqnarray} {\cal L} &&= \int_{\bar t =1}^{\bar t_0 - \bar \delta} d\bar{t} \frac{ 2 \bar t^{\frac{1}{3}} {\sqrt{1 - {\bar t_0}^{~\frac{2}{3}}}} (2 + {\bar t_0}^{~\frac{2}{3}}) } { {\sqrt{1 - {\bar t}^{~\frac{2}{3}}}} ( 3 {\bar t}^{~\frac{4}{3}} + {\bar t}^{~2} - 3 {\bar t_0}^{~\frac{4}{3}} - {\bar t_0}^{~2})}\nonumber\\ && = 2 ~{\rm tanh}^{-1}~\left[ \frac{ {\sqrt{ 1 - ({\bar t_0} - {\bar\delta})^{\frac{2}{3}} }} \,\left(2 + (\bar t_0 -\bar \delta)^{\frac{2}{3}}\right) } { {\sqrt{1 - {\bar t}_0^{\frac{2}{3}} }} \left(2 + {\bar t}_0^{~\frac{2}{3}}\right) }\right]. \end{eqnarray} Finally, to obtain the regularized length, we need to subtract appropriate AdS contribution. Therefore, \begin{eqnarray} {\cal L}_{\rm reg} &&= {\cal L} - 2 ~{\rm log}~\left[\frac{ {\bar t_0}^{~\frac{1}{3}}} { \bar z(\bar t_0 - \bar \delta)}\right]\nonumber\\ &&= {\rm log}~\left[ \frac{ 4 ( 4 -{\bar t}_0^2 - 3 {\bar t}_0^{~\frac{4}{3}})} {{\bar t}_0^{~\frac{2}{3}}}\right]. \end{eqnarray} Therefore, we find that $f(\bar t_0)$ goes to zero as we take $\bar t_0 \rightarrow 0$. As we mentioned previously, for arbitrary $p$, it is not possible to evaluate the correlator analytically. However, it is straightforward to carry out a numerical computation. The result is shown in figure 1. Indeed, we find the correlators do not pick up singularities as we take ${\bar t}_0 \rightarrow 0$. \begin{figure} \centering \includegraphics[width=.55\textwidth]{p-KAdSLength.pdf} ]]>

0$, we proceed to make some general remarks about the correlator. First, let us notice that we can re-write~(\ref{sueleven}) as \begin{equation} \langle {\cal O}(t^{* (1-p)} \bar x, t^*{\bar t}_0) {\cal O}(-t^{* (1-p)} {\bar x}, t^*{\bar t}_0) \rangle = f(\bar t_0). \label{sutwelvea} \end{equation} However, since $t^*$ is a free parameter, we are free to choose it. Let us take $t^* = {\bar t_0}^{-1}$. Then the correlator takes the form \begin{equation} \langle {\cal O}({\bar t}_0^{(p-1)} \bar x, 1) {\cal O}(-{\bar t}_0^{(p-1)} \bar x, 1) \rangle = f(\bar t_0). \end{equation} Calling ${\cal O}({\bar t}_0^{(p-1)} \bar x, 1) = {\cal {\tilde O}} ({\bar t}_0^{(p-1)} \bar x)$ and so on we get, \begin{equation} \langle {\cal {\tilde O}} ({\bar t}_0^{(p-1)} \bar x) {\cal {\tilde O}} (-{\bar t}_0^{(p-1)} \bar x)\rangle = f(\bar t_0). \label{sutwelve} \end{equation} Dependence on the arguments of correlator in this fashion is indeed expected in a scale invariant theory. Note that for $p >0$, as we push the space-like surface close to $\bar t_0 =0$, the separation between the two points in the correlator increases. So we capture the large separation behaviour of the correlator. The second scaling solution of the previous section has similar metric but there is a non-trivial dilaton. Though this scalar goes to zero at $t =0$, it diverges at a later time --- leading to the divergence in Yang-Mills coupling. This however is not a concern for the third scaling solution. This is the Kasner-AdS soliton and we now compute the gauge theory correlator at the boundary of this background. ]]>