text:
\section{Classical string sources, iteration and holographic ratio}
In this section we look for a
simultaneous solution of the background equations of motion and
equations of motion for the string sources as $\gamma$ decreases from
zero to $-\frac{1}{3}$ with time proceeding from $-\infty$ to
$-t_{cl}$.
The nature of the evolution equations are such that we cannot get an
analytic exact solution for
$0>\gamma>-\frac{1}{3}$.
Therefore we start from the $\gamma=0$ end.
There we have the exact solution for the background fields.
If we put back this
solution in the sigma model action to evaluate $\gamma$, we recover the
value $\gamma=0$ \cite{cons},
only in the limiting case $t\rightarrow -\infty$ which is
same as $a\rightarrow a_{0}$.
In other words, our exact solution is not simultaneous solution
of background as well as string equations of motions for time not
exactly $-\infty$. Moreover, we cannot dispense with
 the position coordinates in the expression for pressure and
energy density. To get a simple expression for $\gamma$
as a function of time
we assume, following Gasperini etal \cite{NET},
that the source term represents an ensemble of classical strings.
Initially all the strings are within the horizon, hence pressure and $\gamma$
are zero. The two types of initial
distribution of the lengths
of the strings, we consider, will give different expressions
for $\gamma$. Here, first we review the basic elements of the
model, then derive the expression for $\gamma$ from the model to be used
in the iteration procedure, to obtain the
background fields. Then we utilize these solutions to evaluate the
corresponding holographic ratios and subsequently compare and
contrast the features
of the two types of distributions.
\par
Now let us briefly recapitulate
some of the essential features of
the model. For further details we refer the
interested reader to the ref.\cite{NET}.
 We consider an ensemble of large number, say N,
of classical strings. Moreover, let the length of
i-th string at an instant of time $t$ be $L^{i}(t)$,
and number density of strings of length $L$ at that instant be $n(L, t)$.
Then $N=\int_{l_{s}}^{\infty}n(L, t)dL$.
 Here, the differential equations satisfied by $L^{i}(t)$ and $n(L, t)$ are,
\begin{eqnarray}
\dot{L}^{i}(t)=H(t)L^{i}(t)\theta(L^{i}(t)-D(t))\nonumber\\
a\partial_a{n}+\partial_{L}\left[n(L,a)L\theta(L-D)\right]=0
\end{eqnarray}
where, $D=H^{-1}$ is the Hubble length and $H$ is the Hubble parameter.
The energy density of stable and unstable strings (i.e. length
less or greater than the horizon) are given by
\begin{eqnarray}
E_{s}=\int d^{3}x \sqrt{g}\rho_{s}
 =\frac{1}{\pi \alpha^{\prime}}\int Ln(L,a)\theta(D-L)dL\\
E_{u}=\int d^{3}x\sqrt{g}\rho_{u}
=\frac{1}{\pi \alpha^{\prime}}\int Ln(L,a)\theta(L-D)dL
\end{eqnarray}
Note that in the far past when all the strings lie within the
horizon, $\gamma=0$ and $E_{u}=0$. On the otherhand, $\gamma=-\frac{1}{3}$
corresponds to the situation $E_{s}=0$.
So, $\gamma$ could be approximated by
\begin{equation}
\gamma(t)=-\frac{1}{3}\frac{E_{u}}{E_{u}+E_{s}}
\end{equation}
From the definition(3.3) it follows that
$\tilde{E}_{u}= \pi \alpha^{\prime}E_{u}$ satisfies a
differential equation
\begin{equation}
\partial_{D}{\tilde{E}_{u}}-\frac{1}{D}\frac{\partial{\ln{a}}}{\partial{\ln{D}}}
\tilde{E}_{u}=
-D n_{-\infty}(D)
\end{equation}
where, $n_{-\infty}(D)=n(D,t=-\infty)$ expresses the
number density of strings of size of the horizon at time $t$ according
to the distribution of number density at the far past. Note that this
is the differential equation for evaluating $\tilde{E}_{u}$ at
any time and
we will be using it again and again. On the contrary,
once we assume the
form of the distribution of number density at the remote past,
$E_{s}$ can be
directly evaluated from the definition(3.2). So
let us discuss the two types of distributions,
\begin{eqnarray}
n_{-\infty}(L)&=&\Lambda^{2}L^{-3},\\
n_{-\infty}(L)&=&\frac{N'}{L_{0}}exp(-\frac{L}{L_0})
\end{eqnarray}
where, $\Lambda$ and $N'$ are related to the total number of strings.
The mean length of a string in the two cases are respectively $2l_{s}$
and $l_{s}+L_{0}$.

Now for the first type of number distribution
$\gamma$ goes to $-\frac{1}{3}$ as $t_{c}$ is of the order
$l_{s}$ whereas for the
second type $t_{c}$ can vary.
Let us see how this happens a la Gasperini etal \cite{NET}. First we note
that when $\gamma$ becomes $-\frac{1}{3}$ or, almost all strings
cross the horizon, $a(-t)\sim (-t)^{\alpha}$\cite{DPISC},
$\alpha$ being some
negative fraction.
Then we find
\begin{eqnarray}
\tilde{E}_{u}&=&\Lambda^{2}\frac{H}{(1+\alpha)},\nonumber\\
\tilde{E}_{s}&=&\Lambda^{2}[(l_s)^{-1}-H],\nonumber\\
\gamma&=&-\frac{1}{3}\frac{1}{\frac{1+\alpha}{Hl_{s}}-\alpha}
\end{eqnarray}
So, $\gamma$ tends to $-\frac{1}{3}$ only when $D\sim l_{s}$.
\newline
Let us now consider the exponential distribution.
The differential equation for $\tilde{E}_{u}$ near
$\gamma=-\frac{1}{3}$ yields,
\begin{equation}
\tilde{E}_{u}=N'L_0
(\frac{D}{L_0})^{\alpha}\Gamma\left[2-\alpha,\frac{D}{L_0}\right]
\end{equation}
whereas,
\begin{equation}
\tilde{E}_{s}=N'L_{0}[(1+\frac{l_s}{L_0})exp(-\frac{l_s}{L_0})-
(1+\frac{D}{L_0})exp(-\frac{D}{L_{0}})]
\end{equation}
and as a result,
\begin{equation}
\gamma=-\frac{1}{3}
\frac{(\frac{D}{L_0})^{\alpha}\Gamma\left[2-\alpha,\frac{D}{L_0}\right]}
{[(1+\frac{l_s}{L_0})exp(-\frac{l_s}{L_0})-
(1+\frac{D}{L_0})exp(-\frac{D}{L_{0}})]
+(\frac{D}{L_0})^{\alpha}\Gamma\left[2-\alpha,\frac{D}{L_0}\right]}
\end{equation}
where, $\Gamma\left[2-\alpha,\frac{D}{L_0}\right]$ is incomplete
gamma function.
 So if $L_0$ is such that it is greater than $l_s$ as well as
$t_{cl}$ (i.e. $L_{0}>l_{s}$ and $L_{0}> t_{cl}$)
then $\gamma$ becomes $-\frac{1}{3}$ at the time $t_{cl}$
 for a particular value of $L_0$, say $\frac{2f_{m}}{\sqrt{3}}t_{cl}$.
On the contrary, if $L_{0} \sim l_{s}$ then
$\gamma$ becomes $-\frac{1}{3}$ when $t_{c}\sim l_{s}$.
We note here that for
the exponential distribution of mean length $l_{s}$, $\gamma$ falls off
faster than that of power law.\\
Again as we are considering the situation almost in the far past
i.e. near $t\rightarrow-\infty$ where the horizon
is very large, almost all strings of any size are within the horizon.
It should not matter much whether we take strings of mean length
$l_{s}$ or, larger and we take power law distribution of number of
strings or, exponential distribution. We will see how far this is true
in the following. Before we move on to estimate the holographic
ratio, let us describe
the way we calculate the
background fields, the prescriptions for the
iterations and how we put the holographic ratio
in proper form to compute it at each stage of iteration.
\par
In order to obtain the background field configurations order by order
through iteration, it is
suitable to use a varaible
$Y=\ln{\frac{a}{a_{0}}}$, instead of $x$, along the line of
ref.\cite{NET}.
 In terms
of the variable $Y$,
eqn.s (2.13)-(2.15) are rewritten as
\begin{eqnarray}
2 l_{s}^{2}\bar{\varrho}(Y)=(\frac{\Gamma}{l})^{2} e^{\phi_{0}}
[(w^{\prime})^{2}-\frac{3}{4}w^{2}]\\
\bar{\phi}(Y)=\phi_{0}+2\ln{w(Y)}\\
w^{\prime\prime}+\frac{\Gamma^{\prime}}{\Gamma}w^{\prime}-\frac{3}{4}w=0
\end{eqnarray}
where, $w(Y)$ is an auxiliary function introduced through the
relation
\begin{equation}
x+x_{0}=-2\Gamma\frac{w^{\prime}}{w}
\end{equation}
Our main aim will be to solve the $w$ equation using the expression
for $\Gamma$.
Note that
putting $\gamma=0$ or, $\Gamma=\Gamma^{0}$ in the differential
equation for $w$ we get
\begin{equation}
w=\frac{4l}{T}e^{-\phi_{0}}sinh{\frac{\sqrt{3}}{2}Y}
\end{equation}
We use the expression(3.16) for $w$ to
find the first order corrected $\gamma$. Moreover,
for nonconstant $\gamma$, $\Gamma$ is given by,
\begin{equation}
\Gamma=\Gamma^{0}+\int_{0}^{Y}\gamma(x)\frac{dx}{dY}dY.
\end{equation}

Let us now explain the procedure for iteration as we carry it out
for power law distribution.
In that case, putting the zeroth order expression for $w$ we get $\gamma$
which involves
the parameter $\epsilon$.
Once we put the first order corrected $\gamma$
in the differential equation we obtain
$w$ corrected to order $\epsilon$.
This procedure is repeated to get higher order corrections in $w$.
For exponential distribution procedure is similar. Due to the
presence of $Exp(-\frac{1}{f Y^{2}})$, there is no analog of the parameter
$\epsilon$;
though at the first sight
it might appear that $\frac{1}{f}$ or, $f$ can be used as expansion
parameter depending on whether $f$
is greater or less than one. There, improvement on the
zeroth order result through iteration is relevent
in the powers of $Y$ only.
Whatever be the distribution,
at each stage we
utilize the $w$ to evalute the holographic ratio. To achieve that we
write the holographic ratio in a proper form. Moreover,
let us keep in mind that
we are considering the Hubble horizon throughout
( We mention in passing that in the remote past the event horizon goes
as square root of the Hubble horizon and for small time both converge).
Now in the string frame, $Planck length$, $l_p(t)$, is time-dependent
and is given by\cite{dine}
\begin{equation}
l_p(t)=\frac{l_s}{\sqrt{\frac{V_6}{l_{s}^{6}}}}exp(\phi/2)=l_s exp(\phi/2)
\end{equation}
We have taken
$\frac{V_6}{l_{s}^{6}}$ to be one in
eqn(1.1).
The ratio of entropy contained within the Hubble Horizon to
the horizon area is given by
\begin{eqnarray}
l_p^{2}(t)\frac{S}{A}&=&l_s^{2}
exp(\bar{\phi})\sqrt{g}\frac{S}{A}\nonumber\\
&=&l_s^{2}
exp(\bar{\phi})
\frac{\bar{\varrho}(1+\gamma)}{T^{\prime}}
\frac{V_{H}}{A_{H}}\nonumber\\
&=&\l_s^{2}
exp(\bar{\phi})
\frac{\bar{\varrho}(1+\gamma)}{T^{\prime}}\frac{D}{3}\nonumber\\
&=&l_s^{2}
exp(\phi_0)
\frac{\bar{\varrho}(1+\gamma)}{T^{\prime}}
\frac{w(Y)^{2}}{3\frac{dY}{dt}}\nonumber
\end{eqnarray}
where, $T^{\prime}$ is defined to be the temperature.
It follows from the covariant conservation of $T_{\mu\nu}$
in the string frame
that the time development of string sources is adiabatic.
So $\gamma$ changes with time
keeping entropy per comoving volume constant.
Consequently,
\begin{eqnarray}
l_p^{2}(t)\frac{S}{A}&=&l_s^{2}
exp(\phi_0)
\frac{\bar{\varrho}_{ind}f(Y)}{T^{'}_{ind}f(Y)}
\frac{w^2}{3\frac{dY}{dt}}\nonumber\\
&=&l_s^{2}
exp(\phi_0)
\frac{exp(\phi_{0})}{(2l_{s}^{2})(4l^2 \beta)}
\frac{w^2}{3\frac{dY}{dt}}\nonumber\\
&=&
\frac{1}{\sqrt{3}\beta T}
\frac{w_{Y}^{2}}{3(\frac{dY}{dt})_{Y}}\nonumber\\
&=&\frac{1}{\sqrt{3}\beta T}
\frac{w_{Y}^{2}}{3(\frac{dY}{dt})_{Y}}
\end{eqnarray}
where,$\bar{\varrho}_{ind}$, $T^{'}_{ind}$ are factors independent of
$Y$ in $\bar{\varrho}$, $T^{'}$ respectively. On the otherhand,
we write $Y$-dependent factors of $w$ and $\frac{dY}{dt}$
as $w_{Y}$
and $(\frac{dY}{dt})_{Y}$.
$\beta$ is the temperature at $Y=0$, or equivalently at $t=-\infty$.
\par
Let us estimate
the holographic ratio in the zeroth order. This comes out as
\begin{equation}
l_p^{2}(t)\frac{S}{A}=
\frac{1}{\sqrt{3}\beta T}
\end{equation}
Now $\beta$ and $T$ are constants. Hence the holographic ratio also, in
the zeroth order, is constant, for all time.
So entropy per unit Planck area of the horizon
is also constant. It is given by the above
expression. This result is valid when the horizon is finite. Now
the same solution describes the Universe in
the early phase i.e. when $t \rightarrow -\infty$
or, $a \rightarrow a_{0}$; the horizon tends to infinite then. Hence we
infer, as a limiting procedure,
 that the entropy per unit Planck area of the
horizon is given by the expression(3.20) when
the Universe is
flat. Since the area is tending to infinity,
entropy within the Hubble horizon
also must be very large. As the Universe is cold at the
begining, it may appear contradictory.
But the energy density per unit comoving voulme is
also constant. Hence entropy per comoving volume is constant. As the
number of comoving volumes within the horizon
tends to infinity, entropy
within the Hubble horizon also tends to infinity.

This ratio has good physical implication. If this zeroth order ratio is one
(as is taken in time $-T$ in the paper\cite{venezia}), or,
at least bounded from above, then $\beta T = finite\quad number$.
Again in the PBB cosmology (i) $T$ which is the
 duration of dilaton driven phase
 is very large,
(ii) $\beta$, the temperature at the begining of the Universe, is very low.
Hence
product, $\beta T$ is also a finite number. So two
independent considerations lead to the same conclusion.
Recently Veneziano
has shown$\cite{venezia}$
that the ratio assumed of the order one,
explains the entropy budget of the Universe from the PBB cosmology
quite accurately upto some
numerical factors.
Therefore it is quite reasonable to assume
that the ratio is of the order
one. In other words,
the Universe in the flat beginning seems to show holography as in
AdS spaces \cite{leny}.
\section{Holographic ratio near the remote past}
After discussing the features associated with the zeroth
order holographic ratio
 being bounded let us see how the ratio changes
with time as the Universe evolves from the remote past. The evolution
will depend on the number distribution of strings at the remote past.
As a result the holographic ratio will also be getting modified
differentially.
Now the
holographic ratio
$l_{p}^{2}\frac{S}{A}$ is roughly
$e^{\phi} \frac{D}{\sqrt{g}}$ . As the
Universe evolves the first factor increases whereas the
second factor decreases. Hence the holographic ratio will increase or
decrease depending on whether the relative increament of the first
factor is more or less compared to relative decreament of the second
factor. We will see in our study below that in the case of power law
distribution the scale factor dominates whereas in the case of exponential
distribution it is the coupling constant which initially
dominates, though very weakly, for a
short interval of time before being overtaken by the scale factor.
\subsection{Power Law Distribution}
Let us now first consider the power law distribution,
$n_{-\infty}(L)=\Lambda^{2}L^{-3}$\cite{NET}. Then equation(3.5) becomes
\begin{equation}
\partial_{Y}{\tilde{E}_{u}}-\tilde{E}_{u}=\Lambda^{2}\partial_{Y}{D^{-1}}
\end{equation}
and on integration we get,
\begin{equation}
\frac{\tilde{E}_{u}}{\Lambda^{2}}=H+e^{Y}\int_{0}^{Y}He^{-Y}
\end{equation}
On the otherhand, if we directly evaluate eqn.(3.2) we arrive at,
\begin{equation}
\frac{\tilde{E}_{s}}{\Lambda^{2}}=\frac{1}{l_{s}^{2}}-H
\end{equation}
Hence, in this distribution, the expression
for $\gamma$ takes the following form,
\begin{equation}
\gamma=-\frac{1}{3}\frac{H+e^{Y}\int_{0}^{Y}He^{-Y}}
        {\frac{1}{l_{s}}+e^{Y}\int_{0}^{Y}He^{-Y}}
\end{equation}
where,
$H$ is the Hubble parameter of the Universe given by
\begin{equation}
H = \frac{dY}{dt}
\end{equation}

Now from eqn.s(3.12)-(3.15) we get
\begin{eqnarray}
\frac{dY}{dt}=\frac{1}{\sqrt{3}T}(cosh{\sqrt{3}Y}-1)\\
\frac{dx}{dY}=\frac{3}{2}\Gamma^{0}(cosech{\frac{\sqrt{3}}{2}Y})^{2}
\end{eqnarray}
where we have used the zeroth order expression for $w$,
\begin{eqnarray}
w=\frac{4l}{T}e^{-\phi_{0}}sinh{\frac{\sqrt{3}}{2}Y}\nonumber
\end{eqnarray}
and then substituting eqn.(4.6) in eqn.(4.4), we get $\gamma$ to
order $\epsilon$ as,
\begin{equation}
\gamma=-\frac{\epsilon}{2}[cosh{\sqrt{3}}Y-e^{Y}
+\frac{1}{\sqrt{3}}sinh{\sqrt{3}Y}]+o(\epsilon)^{2}
\end{equation}
Here, integration of the R.H.S. of the expression (3.17) for $\Gamma$ is not
possible if we keep terms upto all orders in $Y$. So to start with,
in the first step of iteration, we keep terms upto $Y^{2}$, for the
sake of simplicity and we get,
\begin{equation}
\Gamma=\Gamma^{0} [1-\frac{\epsilon}{2}Y+o(Y)^{3}o(\epsilon)^{2}]
\end{equation}
where, $\epsilon=\frac{l_{s}}{\sqrt{3}T}$.
Now, in the remote past $Y$ is small.
So whether $T$ is large or small, correction
term in the above expression is small compared to one.
Zeroth order solution acts as good perturbative vaccuum near
the remote past. Moreover, if $T$ is large, zeroth order solution acts as
good perturbative vaccuum for all $Y$\footnote{By all $Y$ we mean that
for all $Y$ for which power series converges}.

To carry on iteration, let us assume the $\Gamma$ obtained in eqn.(4.9)
to hold good for all orders in $Y$.
We put this $\Gamma$ in the background equation to get
$w$ with $o(\epsilon)$ correction,
\begin{equation}
w= \frac{4l}{T}e^{-\phi_{0}}
(1+\frac{\epsilon}{2}Y)sinh{\frac{\sqrt{3}}{2}Y}
\end{equation}
Consequently,
\begin{eqnarray}
\frac{dY}{dt}&=&\frac{1}{\sqrt{3}T}(cosh{\sqrt{3}}Y-1)\\
\frac{dx}{dY}&=&\frac{3}{2}\Gamma^{0}(cosech{\frac{\sqrt{3}}{2}Y})^{2}
                     [1-\epsilon Y+\frac{\epsilon}{\sqrt{3}}sinh{\sqrt{3}Y}]
\end{eqnarray}
Then, the model which takes care of string equations of
motion, gives
\begin{equation}
\gamma=-\frac{\epsilon}{2}\frac{cosh{\sqrt{3}Y}-e^{Y}
+\frac{1}{\sqrt{3}}sinh{\sqrt{3}Y}}
{1+\frac{\epsilon}{2}[\sqrt{3}sinh{\sqrt{3}Y}+cosh{\sqrt{3}Y}+2-3e^{Y}]}
\end{equation}
Let us determine $w$ to order $\epsilon^{2}$.
Note that when we keep terms upto $Y^{4}$
the expression for $\gamma$ does not come with $\epsilon^{2}$.
Therefore, retaining terms upto $Y^5$
we find,
\begin{equation}
\Gamma=\Gamma^{0}[1-\epsilon (Y+\frac{Y^2}{6} +\frac{Y^3}{36}
-\frac{Y^4}{240}-\frac{7 Y^5}{3600})+\frac{\epsilon^{2}}{40}Y^{5}
+o(Y)^{6}o(\epsilon)^{3}]
\end{equation}
Let us use the expression for $\Gamma$ in the background
equations of motion (3.12)-(3.14) to find the
$w$ and the corresponding fields upto $\epsilon^{2}$.
Assuming that this
 $\Gamma$ holds for all order in $Y$ we get the
expression for $w$ as in below:
\begin{eqnarray}
w&=& \frac{4l}{T}e^{-\phi_{0}}\nonumber\\
&&[[1+\frac{\epsilon}{2}Y]sinh{\frac{\sqrt{3}}{2}Y}
+\epsilon[\frac{\sqrt{3}}{2}cosh{\frac{\sqrt{3}}{2}Y}
(-\frac{Y}{10}-\frac{2Y^2}{135}+\frac{Y^3}{180}+
\frac{7Y^4}{2160})\nonumber\\
&&+sinh{\frac{\sqrt{3}}{2}Y}(\frac{1}{10}+\frac{2}{135}Y+
\frac{3}{40}Y^2+\frac{1}{135}Y^3-\frac{1}{480}Y^4-
\frac{7}{7200}Y^5)]\nonumber\\
&&+\epsilon^{2}[-\frac{\sqrt{3}}{2}cosh{\frac{\sqrt{3}}{2}Y}
(\frac{119303}{437400}Y
+\frac{1}{324}Y^2+\frac{52883}{874800}Y^3-
\frac{403}{9720}Y^4\nonumber\\
&&-\frac{6997}{5832000}Y^5-
\frac{953}{3888000}Y^6 -\frac{677}{27216000}Y^7+
\frac{7}{1296000}Y^8\nonumber\\
&&+\frac{49}{34992000}Y^9 )+
 sinh{\frac{\sqrt{3}}{2}Y} (\frac{119303}{437400}
+ \frac{1}{324}Y + \frac{269423}{583200}Y^2\nonumber\\
&&+\frac{829}{19440}Y^3+\frac{68243}{2332800}Y^4
-\frac{16301}{1296000}Y^5-
\frac{14111}{23328}Y^{6}-\frac{11}{108000}Y^{7}\nonumber\\
&&-\frac{31}{3456000}Y^{8}+
\frac{7}{3456000}Y^{9}+
\frac{49}{103680000}Y^{10} )]
\end{eqnarray}
We can use this result to get $\gamma$ and $\Gamma$ for
third order or, upto $\epsilon^3$ and higher order in $Y$.
In other words, this procedure, in principle, could
be used to determine
higher order terms in $\epsilon$.
 However, we compute terms upto
$\epsilon^{2}$.
Now,
we can use the
first and second order expressions for $w$
to estimate the holographic ratio.
\par
Let us evaluate the holographic ratio
for the first order i.e. including the $\epsilon$ correction
\begin{equation}
l_p^{2}(t)\frac{S}{A}=
\frac{1}{\sqrt{3}\beta T}(1+\epsilon Y)
\end{equation}
The holographic ratio in the second
order(i.e. when we include terms upto $\epsilon^{2}$) is
\begin{eqnarray}
l_p^{2}(t)\frac{S}{A}&=&
\frac{1}{\sqrt{3}\beta T}\nonumber\\
&&[1-\epsilon(Y+\frac{Y^2}{6}+\frac{Y^3}{36}
-\frac{Y^4}{240}-
\frac{7Y^5}{3600}\nonumber\\
&&-\frac{-720-240Y-60Y^2 +12Y^3
+7Y^4}{360\sqrt{3}}sinh{\sqrt{3}Y})\nonumber\\
&&-\epsilon^{2}[-\frac{139}{405}-\frac{521Y}{2430}
-\frac{5233Y^2}{4860}
-\frac{16297Y^3}{48600}\nonumber\\
&&-\frac{5429Y^4}{72900}
-\frac{1501Y^5}{64800}+\frac{1877Y^6}{388800}\nonumber\\
&&+\frac{Y^7}{1296}+\frac{23 Y^8}{388800}-\frac{7Y^9}{432000}
-\frac{49Y^{10}}{12960000}\nonumber\\
&&-\frac{139(-720-240Y-60Y^2 +12Y^3 +7Y^4)
cosh{\sqrt{3}Y}}{145800}\nonumber\\
&&-(1601280+1000320 Y+358080Y^2 + 9312Y^3\nonumber\\
&&-41344Y^4 - 12888Y^5 - 1500Y^6 + 504Y^7 + 147Y^8)
\frac{cosh{2\sqrt{3}Y}}{4665600}\nonumber\\
&&+(-2Y-Y^2 -\frac{Y^3}{3}-\frac{55Y^4}{216}+
\frac{73Y^5}{1800}+
\frac{133Y^6}{14400}\nonumber\\
&&+\frac{47Y^7}{43200}-\frac{7Y^8}{32000}-
\frac{49Y^9}{864000})
\frac{sinh{\sqrt{3}Y}}{\sqrt{3}}\nonumber\\
&&+(\frac{1}{10}+\frac{17Y}{270}+
\frac{Y^2}{648}
-\frac{287Y^3}{16200}\nonumber\\
&&-\frac{31Y^4}{4320}-\frac{53Y^5}{48600}+
\frac{49Y^6}{129600}
+\frac{49Y^7}{388800})\frac{sinh{2\sqrt{3}Y}}{\sqrt{3}}]]
\end{eqnarray}
We observe
that coefficient of $\epsilon$ in the above equation(4.17)
is of opposite sign compared to that of first order eqn.(4.16).
This sign difference arises due to the following reason.
In obtaining the expression(4.9) for $\Gamma$ we have kept terms,
upto $Y^2$. But
while calculating $w$, eqn.(4.10), we have kept terms upto
all orders in $Y$.
Then we have used this expression (4.10)
for $w$ in the holographic formula.
It gives rise to the
apparent discrepency.
If we had strictly kept terms upto $Y^2$, coefficient of
$\epsilon$ would have been absent in the first order
holographic ratio.
Again in the second order
formula(4.17) there is no term in
the coefficient of $\epsilon$ upto $Y^{2}$.
Hence the second order result
is consistent with the first order one.
Moreover, if we had kept terms in the expression for $w$ upto
$Y^5$, strictly, the
holographic ratio with $\epsilon^{2}$ correction
would have contained terms upto $Y^3$.
Now in the expression for the holographic ratio(4.17),
coefficient of $\epsilon$ decreases and that of $\epsilon^{2}$
increases with $Y$, if terms upto $Y^3$
are considered. This feature will be manifest from the plots below.
Again as a next step, if we go on to consider
the third order in $\epsilon$,
 to start with we have to keep terms higher order than
$Y^{5}$ in the expression for $\Gamma$. And this will not
change the coefficients of terms upto $Y^{5}$ in the
expression for $w$ and the coefficients of terms upto $Y^{3}$
in the expression for holographic ratio.
Hence, we conclude, that as the Universe
evolves form $t=-\infty$, the holographic ratio decreases,
at least upto the time terms from $Y^{6}$ onwards
are negligible.\\
The plots of the coefficients
of $\epsilon$ and $\epsilon^{2}$ in the holographic ratio,
for small values of $Y$, are shown in Fig. 2 and 3 respectively.
\begin{figure}[h]
\centerline{\leavevmode\epsfysize=5truecm \epsfbox{epsilon1.eps}}
\vspace{0.2in}
\caption{ The plot of $Y-dependent$ coefficient
of $\epsilon$ in $r=l_{P}^{2}\frac{S}{A}$
against $Y$.}
\label{a}
\end{figure}
\begin{figure}[h]
\centerline{\leavevmode\epsfysize=5truecm \epsfbox{epsilon2.eps}}
\vspace{0.2in}
\caption{ The plot of $Y-dependent$ coefficient of $\epsilon^{2}$ in
$r=l_{P}^{2}\frac{S}{A}$ against $Y$.}
\label{a}
\end{figure}
We observe from the plots that the coefficient of $\epsilon$ decreases
whereas the coefficient of $\epsilon^{2}$ increases. Note that in both
the cases the coefficients are indeed small.
As a result the
net holographic ratio decreases as stated in the previous paragraph.

\subsection{Exponential distribution}
After discussing all about the power law distribution, let us go over
to the second case, namely exponential one.
The differential equation satisfied by $\tilde{E}_{u}$ for the exponential
distribution takes the form,
\begin{equation}
\partial_{Y}{\tilde{E}_{u}}-\tilde{E}_{u}=
-N'\partial_{Y}{D} \frac{D}{L_{0}}exp(-\frac{D}{L_0})
\end{equation}
which on integration gives,
\begin{equation}
\tilde{E}_{u}=N'L_{0}[\frac{D}{L_{0}}exp(-\frac{D}{L_0})-exp(Y)\int_{0}^{Y}
\partial_{Y}{(exp(-Y)\frac{D}{L_{0}})}exp(-\frac{D}{L_0})dY]
\end{equation}
Since, $H=D^{-1}=\frac{dY}{dt}=\frac{\sqrt{3}}{2T}Y^2$ upto order of $Y^2$
we get,
\begin{equation}
\gamma=-\frac{1}{3}\frac{1}{fY^{2}}exp(-\frac{1}{fY^{2}})
\end{equation}
Use of the eqn.(3.17) leads to
\begin{equation}
\Gamma=\Gamma^{0}
[1-\frac{1}{3}(\frac{exp(-\frac{1}{fY^{2}})}{Y}-
\frac{\sqrt{\pi f}}{2}(Erf(\frac{1}{\sqrt{f}Y})-1))]
\end{equation}
where, $Erf(\frac{1}{\sqrt{f}Y})$ is the error function\cite{grad}.
Now to iterate we assume that the expression(4.21) for
 $\Gamma$ to be correct to all order
in $Y$. We use this $\Gamma$ in the differential equation(3.14) and
obtain $w$. Here, we keep terms upto $Y^{5}$ in the expression for $w$
and get
\begin{equation}
w=\frac{4l}{T}exp(-\phi_0)[sinh(\frac{\sqrt{3}}{2}Y)+c(Y)]
\end{equation}
where,
$c(Y)$ is given by
\begin{eqnarray}
c(Y)&=&\frac{1}{3}sinh(\frac{\sqrt{3}}{2}Y)\left[(\frac{1}{Y}+\frac{3Y}{8f})
exp(-\frac{1}{fY^{2}})
-\sqrt{f\pi}(Erf(\frac{1}{\sqrt{f}Y})-1)
(1+\frac{3}{4f}-\frac{3}{8f^2})\right]\nonumber\\
&&-\frac{1}{3}cosh(\frac{\sqrt{3}}{2}Y)
\left[\frac{\sqrt{3}}{2}
(1+\frac{3}{40}\frac{Y^2}{f})exp(-\frac{1}{fY^{2}})
-\frac{\sqrt{3}}{4f}
(1-\frac{3}{40f})ExpIntegralEi(-\frac{1}{fY^{2}})\right]\nonumber\\
&&
\end{eqnarray}
where, $ExpIntegralEi(-\frac{1}{fY^{2}})$ is the exponential integral
function \cite{grad}.

Then the holographic ratio becomes,
\begin{equation}
l_p^2(t)\frac{S}{A}=
\frac{1}{\sqrt{3}\beta T}[1+COR(Y)]
\end{equation}
with $COR(Y)$ is given by,
\begin{eqnarray}
COR(Y)&=&exp(-\frac{1}{fY^{2}})[\frac{8}{3\sqrt{3} f^2 Y^{7}}+
\frac{2}{\sqrt{3}f^2 Y^{5}}-\frac{4}{\sqrt{3}fY^{5}}-\nonumber\\
&&\frac{4}{3\sqrt{3}Y^{3}}+\frac{1}{2\sqrt{3} f^2 Y^{3}}+
\frac{1}{3fY^{3}}-\frac{5}{\sqrt{3}fY^{3}}+\frac{1}{2Y}-\nonumber\\
&&\frac{1}{2\sqrt{3}Y}+\frac{1}{80 f^2 Y}-\frac{3\sqrt{3}}{80 f^2 Y}+
\frac{1}{6fY}-\frac{\sqrt{3}}{2fY}+\frac{Y}{40f}\nonumber\\
&&-\frac{3\sqrt{3}Y}{40f}+
\frac{1}{2}coth(\frac{\sqrt{3}}{2}Y)(1-\frac{\sqrt{3}}{4}+
\frac{3Y^2}{40f}-\frac{3\sqrt{3}Y^2}{160f})\nonumber\\
&&-cosh(\frac{3\sqrt{3}}{2}Y)cosech(\frac{\sqrt{3}}{2}Y)
(\frac{1}{8\sqrt{3}}+\frac{\sqrt{3}Y^2}{320f})]\nonumber\\
&&-exp(-\frac{1}{f Y^{2}})[-\frac{1}{4\sqrt{3}}
+\frac{3}{40f}+\frac{1}{8\sqrt{3}f}-\frac{4}{3 f^2 Y^{6}}\nonumber\\
&&-\frac{1}{20 f^3 Y^{4}}-
\frac{2}{3 f^2 Y^{4}}
+\frac{2}{fY^{4}}+\frac{2}{3\sqrt{3}fY^{4}}+
\frac{1}{3Y^2}+\frac{1}{3\sqrt{3}Y^2}\nonumber\\
&&-\frac{3}{40 f^2 Y^2}
+\frac{5}{6fY^2}+\frac{1}{2\sqrt{3}fY^2}
-\frac{\sqrt{3}Y^2}{160f}]sinh(\sqrt{3}Y)\nonumber\\
&&-cosh(\sqrt{3}Y)[exp(-\frac{1}{fY^{2}})(\frac{8}{3\sqrt{3} f^2 Y^7}
+\frac{2}{\sqrt{3} f^2 Y^5}-\frac{4}{\sqrt{3}f Y^5}\nonumber\\
&&-\frac{4}{3\sqrt{3} Y^3}+\frac{1}{2\sqrt{3} f^2 Y^3}
-\frac{1}{3fY^3}-\frac{\sqrt{3}}{fY^3}+\frac{1}{6Y}\nonumber\\
&&-\frac{1}{2\sqrt{3}Y}-\frac{1}{80 f^2 Y}
-\frac{\sqrt{3}}{80 f^2 Y}-\frac{1}{6fY}-\frac{1}{2\sqrt{3}fY}\nonumber\\
&&-\frac{Y}{40f}-\frac{\sqrt{3}Y}{40f})
+\frac{\sqrt{f\pi}}{12}(1-\sqrt{3})(1-Erf(\frac{1}{\sqrt{f}Y}))]\nonumber\\
&&+(\frac{(3-\sqrt{3})(3-40f)}
{480f^2}coth(\frac{\sqrt{3}}{2}Y)
ExpIntegralEi(-\frac{1}{f Y^{2}}))
\end{eqnarray}
To see how the holographic ratio
 changes with $Y$, we plot the
correction to ratio
against $Y$ for f equal to 10 in figure 4.:
\begin{figure}[h]
\centerline{\leavevmode\epsfysize=5truecm \epsfbox{f1.eps}}
\vspace{0.2in}
\caption{ The plot of correction to holographic
ratio $l_{P}^{2}\frac{S}{A}$ denoted
as $r$, against $Y$ for exponential distribution with $L_{0}=fT$ and $f=10$.}
\label{a}
\end{figure}
We note that the holographic ratio, in this case,
deviates slightly upward from the zeroth order result.
Let us check that this deviation is not spurious. For this purpose
we plot the correction to holographic
ratio for $w$ upto $Y^{2}$ order also.
We get an identical looking curve.
We recall that we have kept terms upto $Y^{2}$
in the expression for $H$. This means
that the region of $Y$ taken in the plot is the region where higher order
terms are negligible compared to $Y^{2}$. Hence this little upward deviation
from the zeroth order result will remain if we
go to higher order in $Y$.
Actually this little upward deviation is generic of exponential
distribution. Again, to come to this
conclusion, we plot the ratio for $f=1$. This is shown in the figure 5:
\begin{figure}[h]
\centerline{\leavevmode\epsfysize=5truecm \epsfbox{f2.eps}}
\vspace{0.2in}
\caption{ The plot of correction to holographic
ratio $l_{P}^{2}\frac{S}{A}$ denoted
as $r$ against $Y$ for exponential distribution with $L_{0}=fT$ anf $f=1$.}
\label{a}
\end{figure}
We see that the domain of the
initial plataue or, rectangular shaped region
 in the above figure, $FIG. 5$, shifts with respect to the previous
plot, $FIG. 4$, to slightly larger value
of $Y$. Hence we infer that if we go on decreasing $f$ to $\epsilon$
this nature of the curve will remain, only the
domain of the curve will shift to larger value of $Y$.
This plataue occurs due to first order correction.
So the initial rise will remain if we include
the higher order corrections also.
Note the magnitude of smallness in the initial rise in the holographic
ratio. We mention in passing that for $\gamma=-\frac{1}{3}$ solution to be
valid, $f=\frac{L_{0}}{T}=\frac{t_{cl}}{T}$ has to be much greater
than $\frac{l_{s}}{T}$ but less than one, as $t_{cl}<T$.

Uptill now we were concentrating on how the holographic
ratio evolves over small time interval near the remote
past. In this regime, only few strings cross the horizon.
Perturbative techniques are useful also. But we don't
have an easy way to study the holographic behaviour when,
arbitrary number of strings say, half
of the total number of strings are more than the size of the horizon. Obvious
possibility is to go to the other extreme
i.e. when all the strings became nondynamical.
This is what we are going to do in the next section.
\section{Holographic ratio for $\gamma=-\frac{1}{3}$}
The holographic ratio is given by (for
$\gamma=-\frac{1}{3}$ or, $t\sim -t_{cl}$)
\begin{equation}
l_p^{2}(t)\frac{S}{A}=
\frac{1}{3\beta_{t_{cl}}t_{0}}
(\frac{-t}{t_{0}})^{-\frac{1}{2}}
\end{equation}
where, $\beta_{t_{cl}}$ is the temperature at the time
$t=-t_{0}$.
So the holographic ratio at the time $t=-t_{cl}$ is given by
$\frac{1}{3\beta_{t_{cl}} (-t_{cl})^{\frac{1}{2}}(t_{0})^{\frac{1}{2}}}$.
Hence the ratio of the holographic ratios, denoted by $R_{H}$, at the time
$-t_{cl}$ and $-\infty$ is
\begin{equation}
\frac{R_{H}(-t_{cl})}{R_{H}(-\infty)}=
\frac{3\sqrt{3}}{4}\frac{T}{(-t_{cl})^{\frac{1}{2}}}
(\frac{e^{\phi_{0}}}{3l})^{\frac{1}{2}}
\end{equation}
where we have used the constancy of comoving volume entropy.
Now assuming that the holographic ratio is one at the remote past we
find that the ratio at the time $-t_{cl}$ is also one provided the
parameters of the theory $\phi_{0}$ and $l$ are
constrained by the relation
$(\frac{e^{\phi_{0}}}{3l})^{-\frac{1}{2}}=
\frac{3\sqrt{3}}{4}\frac{T}{(-t_{cl})^{\frac{1}{2}}}$.
Note that $T$ is larger than $|t_{cl}|$ (see the Fig.1).

\centerline{\bf{Comparison with the Einstein frame ratio}}

Let us study another feature of the holographic ratio. This is whether
the holographic ratio is the
same in both the string and Einstein
frames. In the PBB cosmology, qualitative features of the Universe
like blue shift, shrinking of the horizon
etc. \cite{MPL}
are same in both the frames. So the question whether similarly
 the holographic
ratio is equal in the two frames motivates us to study the
corresponding ratio in the Einstein frame.
The metrics in the Einstein frame and the string frame
are related by the following conformal
transformation
\begin{equation}
8\pi g^{s}_{\mu\nu}=e^{\phi-\Phi_{0}}g^{E}_{\mu\nu}
\end{equation}
$\Phi_{0}$ is the present day value of the dilaton and we set
$16 \pi \l_{s}^{2}e^{\Phi_{0}}=l_p^{2}$.
Then if we remain in the synchronous gauge i.e. if we set
\begin{equation}
ds_{E}^{2}=dt_{E}^{2}-a_{E}^{2}(dx^{i})^{2}
\end{equation}
we get
\begin{eqnarray}
dt_{E}&=&\sqrt{8\pi}e^{-\frac{1}{2}(\phi-\Phi_{0})}dt\\
a_{E}(t_{E})&=&\sqrt{8\pi}e^{-\frac{1}{2}(\phi-\Phi_{0})}a(t)\\
\varrho^{s}_{E}(t_{E})&=&\frac{e^{2(\phi-\Phi_{0})}}{(8\pi)^{2}}
\varrho^{s}(t)
\end{eqnarray}
as,
$(T_{\mu}^{\mu})_{E}=\frac{\sqrt{g_{s}}}{\sqrt{g_{E}}}(T_{\mu}^{\mu})$
\cite{MPL}.
In the Einstein frame the equations of motion satisfied by the fields
are
\begin{eqnarray}
R_{\mu}^{\nu} -\frac{1}{2}R g_{\mu}^{\nu}&=&\frac{l_{p}^{2}}{2}T_{\mu}^{\nu}\cr
\nabla_{\mu}\nabla^{\mu}\phi&=&\frac{l_{p}^{2}}{2}(T^{S})_{\mu}^{\mu}
\end{eqnarray}
where, in the above,
$T_{\mu}^{\nu}=(T^{D})_{\mu}^{\nu}+(T^{S})_{\mu}^{\nu}$
and $(T^{D})_{\mu}^{\nu}$ is the stress-energy tensor corresponding
to the dilaton.

These equations of motions
have been derived from an action $S^{E}$ which in turn
has been obtained from the effective
action in the string frame by conformal transformation and is given as
\begin{equation}
\hbar^{-1}S^{E}=-\frac{1}{l_{p}^{2}}\int{ d^{4}x \sqrt{-g_{E}}
       (R-\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi)}\\
      +S_{\sigma}
\end{equation}
where,
\begin{equation}
S_{\sigma}=\frac{1}{4\pi\alpha^{'}}
           \int{d^{2}\sigma \partial_{\alpha}X^{\mu}
           \partial^{\alpha}X^{\nu}g_{\mu\nu}}e^{\phi}.
\end{equation}
After this short preliminary about the conversion of string frame
to Einstein frame in $(1+3)$-dimension, let us
proceed to compute the ratio in the Einstein frame corresponding to
$\gamma=-\frac{1}{3}$ in the string frame.
We note that in the Einstein frame\cite{DPISC},
\begin{eqnarray}
a_{E}(t_{E})&=&\frac{\sqrt{8\pi}}
{e^{\frac{\phi_{0}^{'}}{2}}}
              (\frac{-t_{E}}{t_{0E}})^{\frac{2}{5}}\\
\phi(t_{E})&=&\phi_{0n}-\frac{6}{5}\ln(\frac{-t_{E}}{t_{0E}})\\
\varrho^{s}(t_{E})&=&\frac{6}{25 L_{p}^{2}}\frac{1}{(-t_{E})^{2}}
\end{eqnarray}
where, in the above, $\varrho^{s}(t_{E})$ is the energy density of the
string source in the Einstein frame and $L_{p}$ is the present
day value of the Planck length. Again in the Einstein frame,
the dilaton also contributes to the energy density and the pressure.
 Consequently, the net
eneregy density and the pressure in the Einstein frame are
\begin{equation}
\varrho=\frac{24}{25 L_{p}^{2}}\frac{1}{(-t_{E})^{2}},\quad
p=\frac{16}{25 L_{p}^{2}}\frac{1}{(-t_{E})^{2}}
\end{equation}
The effective equation of state is
\begin{equation}
p=\frac{2}{3}\varrho
\end{equation}
The holographic ratio turns out to be
\begin{eqnarray}
L_{p}^{2}\frac{S}{A}&=&L_{p}^{2}\frac{S^c}{\sqrt{g}}\frac{D}{3}\nonumber\\
&=&L_{p}^{2}\frac{1}{\sqrt{g}}\frac{\bar{\varrho}(1+\gamma)}{T^{'}}
\frac{1}{3|\frac{\dot{a}}{a}|}\nonumber\\
&=&\frac{4}{3 \beta_{t_{cl}}^{E}}
\frac{(-t_{E})^{-\frac{1}{5}}}{(t_{0E})^{\frac{4}{5}}}
\end{eqnarray}
where, $\beta_{t_{cl}}^{E}$ is the temperature at the time $t_{0E}$.
Now, the string frame time and the Einstein frame time are related by
\begin{eqnarray}
t_{0}&=&\frac{5}{2\sqrt{8\pi}}e^{\frac{\phi_{0}^{'}}{2}}t_{0E}\nonumber\\
-t&=&\frac{5}{2\sqrt{8\pi}}e^{\frac{\phi_{0}^{'}}{2}}(t_{0E})^{\frac{3}{5}}
       (-t_{E})^{\frac{2}{5}}
\end{eqnarray}
we get,
\begin{equation}
\frac{(-t_{E})^{-\frac{1}{5}}}
{(t_{0E})^{\frac{4}{5}}}=\frac{5}
{2\sqrt{8\pi}}e^{\frac{\phi_{0}^{'}}{2}}
\frac{1}{t_{0}}(\frac{-t_{}}{t_{0}})^{-\frac{1}{2}}
\end{equation}
As a result the ratio, in the two frames eqn.(5.1) and eqn.(5.16)
 have exactly the same time dependences.
Again the entropy per comoving volume is the same in both the
string and the Einstein frames. Hence
\begin{equation}
\beta_{t_{cl}}^{E}=\frac{10}
{(8\pi)^{\frac{1}{2}}}e^{\frac{\phi_{0}^{'}}{2}}
\beta_{t_{cl}}
\end{equation}
where, $\phi_{0}^{'}=\phi_{0n}-\Phi_{0}$ and $\Phi_{0}$ is the present
day value of the dilaton.
Therefore the holographic ratios in both the frames are exactly the same.
\par
Uptill now we have been in $(1+3)$-dimension, came across nice features
like constancy of holographic ratio in the remote past,
identical ratio in the string and Einstein frames
in the recent past,i.e. $|t|=t_{cl}=t_{c}$. Let us now move
on to diverse dimensions
looking for the similar interesting features.
\section{Holographic ratio in diverse dimension}
Here we write down the string effective action in $(1+d)$-dimension and
the equations of motions derived from it.
The effective action in $(1+d)$-dimension is
\begin{equation}
\hbar^{-1}S^{s}=-\frac{1}{2 l_{s}^{d-1}}
      \int{d^{d+1}x \sqrt{-g} e^{-\phi}(R+
      \partial_{\mu}\phi \partial^{\mu}\phi)}
      +S_{\sigma}
\end{equation}
The resulting equations of motions \cite{PBBISC} are
\begin{eqnarray}
\dot{\bar{\phi}}^{2}-2\ddot{\bar{\phi}}+ mH^{2}+ nF^{2}&=&0\nonumber\\
\dot{\bar{\phi}}^{2}-mH^{2}-nF^{2}&=&2l_{s}^{d-1}\bar{\varrho}
e^{\bar{\phi}}\nonumber\\
2(\dot{H}-H\dot{\bar{\phi}})&=&2
\l_{s}^{d-1}\bar{p}e^{\bar{\phi}}\nonumber\\
2(\dot{F}-F\dot{\bar{\phi}})&=&2
\l_{s}^{d-1}\bar{q}e^{\bar{\phi}}
\end{eqnarray}
First let us keep in mind that we are writing $d=m+n$,
where, (i) $m$ is the
number of expanding spatial dimensions and (ii) $n$ is the number of
contracting spatial dimensions in the string frame.
Scale factor of a contracting dimension will be
denoted as $a_{con}$ and it will be just reciprocal of
$a_{ex}$. This is the only posssibility compatible with non-zero
energy density \cite{PBBISC}.
Here, now onwards we will be
confining ourselves to consider mainly time dependences
of quantities.
Moreover we will look into the following aspects,
(i) whether the holographic ratio is
constant or, not in different dimensions in the remote past,
(ii) whether the holographic ratios in the remote past are the
same in both the string and the
Einstein frames in $(1+3)$ as well as in the
diverse dimensions and
(iii) how the time dependences of the holographic ratios in the recent past,
in the string frame, go together with the corresponding ones in the
Einstein frames in various dimensions.
\subsection{Remote Past}
Let us now start with the case when we have nine isotropically
 expanding dimensions.
Then we have\cite{NET}
after solving the above set(6.2) of equations
as in the $(1+3)$-dimensional case with $p=q=0$,
\begin{eqnarray}
a&=&a_{0}(1-\frac{2T}{t})^{\frac{1}{3}}\nonumber\\
e^{\bar{\phi}}&=&\frac{16 l^{2} e^{-\phi_{0}}}{|t(t-2T)|}\nonumber\\
\bar{\varrho}&=&\frac{e^{\phi_{0}}}{8 l^{2}l_{s}^{8}}
\end{eqnarray}
And we know \cite{dten} in the ten dimension, $Planck length$, is
given by,
\begin{equation}
(l_{p}^{10})^{2}= g_{s}^{\frac{1}{2}}l_{s}^{2}=e^{\frac{\phi}{4}}l_{s}^{2}
\end{equation}
So, the holographic ratio is given by,
\begin{eqnarray}
(l_{p}^{10})^{8}\frac{S}{A}&=&l_{s}^{8}\frac{e^{\phi}}{\sqrt{g}}
\frac{\bar{\varrho}}{\beta_{9e}}\frac{V_{H}}{A_{H}}\nonumber\\
&=&\frac{3}{2k}\frac{1}{\beta_{9e}T}
\end{eqnarray}
where, $k$ is a numerical factor coming through
 $\frac{V_{H}}{A_{H}}=\frac{1}{k H}$
and $\beta_{9e}$ is the temperature in the remote past.
\par
Next let us consider the case when three dimensions are expanding
and six other dimensions are contracting. Then, observing that
equations of motions in terms of SFD
invariant variables remain same as in the previous case, we get
\begin{eqnarray}
a_{ex}&=&a_{0ex}|1-\frac{2T}{t}|^{\frac{1}{3}}\nonumber\\
a_{con}&=&a_{ex}^{-1}\nonumber\\
e^{\bar{\phi}}&=&\frac{16 l^{2} e^{-\phi_{0}}}{|t(t-2T)|}\nonumber\\
\bar{\varrho}&=&\frac{e^{\phi_{0}}}{8 l^{2}l_{s}^{8}}
\end{eqnarray}
And
$(l_{p}^{10})^{2}$ is written as in the nine expanding case. So, the
holographic ratio
\begin{equation}
(l_{p}^{10})^{8}\frac{S}{A}=l_{s}^{8}\frac{e^{\phi}}{\sqrt{g}}
\frac{\bar{\varrho}}{\beta_{3e}}\frac{V_{H}}{A_{H}}
\end{equation}
Again, the ratio of the Hubble volume to Hubble area is given by
\begin{eqnarray}
\frac{V_{H}}{A_{H}}
&=&\frac{\sqrt{\pi}}{36(\frac{\sqrt{\pi}}{2}|H| + \frac{16}{15}|F|)}\nonumber\\
&=&\frac{\sqrt{\pi}}{24}\frac{1}{\frac{\sqrt{\pi}}{2} + \frac{16}{15}}
\frac{t}{T}(t-2T)
\end{eqnarray}
where, $|H|$ and $|F|$ are the Hubble parameters of the expanding and
the contracting dimensions.
Consequently,
\begin{equation}
(l_{p}^{10})^{8}\frac{S}{A}=\frac{\sqrt{\pi}}{12}
\frac{1}{\frac{\sqrt{\pi}}{2} + \frac{16}{15}}\frac{1}{\beta_{3e}T}
\end{equation}
Therefore,
we see, in both the cases as $t$ tends to $-\infty$, the holographic ratio
 becomes constant as in the $(1+3)$-dimension.
Let us examine whether this happens in any arbitrary dimensions.

We note that for $(1+d)$-dimensional world,
\begin{eqnarray}
a_{ex}&=&a_{0ex}(1-\frac{2T}{t})^{\frac{1}{\sqrt{d}}}\nonumber\\
e^{\bar{\phi}}&=&\frac{16 l^{2} e^{-\phi_{0}}}{|t(t-2T)|}\nonumber\\
\bar{\varrho}&=&\frac{e^{\phi_{0}}}{8 l^{2}l_{s}^{d-1}}
\end{eqnarray}
And as far as time dependence is concerned, in the large $|t|$ limit
(i.e. $|t|>>T$) we have
\begin{eqnarray}
l_{p}^{d-1}(t) \frac{S}{A}&\sim&
\frac{V_{H}}{A_{H}}\frac{e^{\phi}}{\sqrt{g}}\nonumber\\
&\sim&\frac{e^{\phi}}{\sqrt{g}}\frac{1}{|H|}\nonumber\\
&\sim& |t|^{0}
\end{eqnarray}

Again, in the $(1+d)$-dimension, the Einstein frame scale factor and time
are related to the string frame ones\cite{MPL} as
\begin{eqnarray}
a_{E}(t_{E})&=&e^{-\frac{\phi}{d-1}} a(t)\nonumber\\
|t_{E}|&=& |t|^{\frac{d+1}{d-1}}
\end{eqnarray}
Thus the scale factor in the Einstein frame is, in the large
$|t|$ limit, in terms of the string frame time is expressed as
\begin{equation}
a_{E}(t_{E})= |t|^{\frac{2}{d-1}}
\end{equation}
Note that in this limit the Universe is isotropic even if
in the string frame it is mixed isotropic. In
$(1+3)$-dimension, in particular, the equation of state is
\begin{equation}
p=\frac{1}{3}\varrho
\end{equation}
 Then we arrive at the following holographic ratio in the Einstein
frame, in the $(1+d)$-dimension
\begin{equation}
L_{p}^{d-1}\frac{S}{A} \sim \frac{1}{\sqrt{g_{E}}}
|\frac{a}{\frac{da}{dt_{E}}}| \sim |t_{E}|^{\frac{2-d}{2}}
          \sim |t|^{-1}
\end{equation}
where, $|t_{E}|$ corresponds to the time in the Einstein frame.
So the time dependences of the holographic ratios
in the two frames in this large $|t|$ limit do not match
in any dimension. But we have seen that in the recent past the
ratio is same in both the frames in the four dimension. Therefore, let us
check, in the recent past, whether
in arbitrary dimensions time dependences of the
holographic ratio match in both the frame.
\subsection{Recent Past}
Now in the $(1+d)$-dimensional spacetime near the time $-t_{cl}$,
$\gamma=-\frac{1}{d}$
and the relevent field configurations, namely
 the scale factor of the expanding
dimension and the SFD invariant dilaton\cite{MPL} are
\begin{eqnarray}
a({t})&=&|t|^{-\frac{2}{1+d}} \nonumber\\
\bar{\phi}&=&dln|a(t)|
\end{eqnarray}
As a result the holographic ratio in the string frame
becomes
\begin{equation}
l_{p}^{d-1}\frac{S}{A} \sim |t|^{\frac{1-d}{1+d}}
\end{equation}
\newline
But now the Einstein frame time $t_{E}$ is related to the string frame
time by
\begin{equation}
|t_{E}|\sim |t|^{1+\frac{4m}{d^2 -1}}
\end{equation}
and the scale factors in the Einstein frame,
corresponding to the expanding and contracting
dimensions in the string frame, are
\begin{eqnarray}
a_{Em}(t_{E})\sim |t_{E}|^{2\frac{m+1-n}{d^2 +4m -1}}
\sim |t|^{-2\frac{n-m-1}{d^2 -1}}  \nonumber\\
a_{En}(t_{E})\sim |t_{E}|^{2\frac{2m-1+d}{d^2 +4m -1}}
\sim |t|^{2\frac{d-1+2m}{d^2 -1}}
\end{eqnarray}
Consequently, the corresponding ratio in the Einstein frame in terms
of the string time is given by
\begin{equation}
L_{p}^{d-1}\frac{S}{A} \sim
|t|^{\frac{1-d}{1+d}}
\end{equation}
Hence in the recent past the holographic
ratios in the two frames have the same time dependences in all dimensions.
We note here that time
dependences of holographic ratio as in eqn.(6) was arrived
at\cite{bak}
from different consideration.

\section{Discussion}
In this paper, we
have set out to compute the holographic ratio. First, we
obtained the holographic ratio in the zeroth order and
discussed the implication of the ratio to be bounded. Then
for the power law distribution,
we obtained the
correction to the ratio
in the first and second order in $\epsilon$ following an iteration
procedure.
We found that the correction depletes the ratio, albeit by small amount.
For the exponential distribution, the
correction increases the value of the ratio by very small
amount ($10^{-308}$),
towards the beginning of evolution, over a small range of time.
This is due to the domination of coupling constant over scale factor
during that period.
Again, we computed the ratio in the other extreme, when all the strings
crossed the horizon, assuming the mean length of the classical strings
is much larger than $l_{s}$. We then showed that the holographic ratio,
has the same time dependences in both the string and Einstein frames
in this limit, and taking care of all the factors in one particular
case, namely
in the four dimension we found that the ratio is identical.
This gives another
evidence that qualitatively
the two frames are equivalent in the PBB cosmology and Hubble horizon
is a good choice in both the frames.
On the other hand, in the remote past the ratio
in the string frame is constant
in all dimensions.
However,
in the Einstein frame, it goes as inverse of the string time.
The ratio is different in the two frames.
We have checked that solutions obtained by conformal trasformation
do satisfy the equations of motions in the Einstein frame in this
limit when coupling constant goes to zero.
We may mention that if one follows the proposal of ref\cite{bak},
then also one gets the corresponding holographic ratio which is
inversely proportional to the string time.
Thus we find that the holographic ratio in the remote past is not on
similar footing as other physical quantities.
We have studied the holographic behaviour in the presence of
string sources almost in two extreme cases in $(1+3)$-dimensions.
It is possible to generalise the exercise of getting
 the correction by iterative procedure to
$(1+d)$-dimension.
Again, we applied
pertubative method to go away from the extremality ($\gamma=0$).
It will be
very interesting to study the holographic ratio explicitly in
the intermediate region.
It is also desirable to see how the zeroth order holographic ratio
gets modified in presence of higher dimensional branes.
Moreover it is very much of relevence to
ask about the mechanism of holography in this case.

\centerline{Acknowledgement}

We have benifitted from discussions with
Professors R. Brustein, M. Gasperini
and especially with Professor A. Sen.
We are grateful to Professor G. Veneziano
for many valuable discussions
at various stages of the work
and this investigation is outcome of his suggestions
to explore consequences of holography
in presence of the string sources.
We are thankful to D. Goswami for drawing the first figure on behalf of us.

redpajama_set_name: RedPajamaArXiv


Surf's up dude! Get your boards waxed and ready in this surf shack by the beach and later your little surfer can go catch some gnarly waves. If the sun becomes too hot, they can get some shade inside and maybe earn a seashell or two renting out some boards. When the burnt sun starts to recede behind the salty blue ocean, let your beach bum climb the ladder to the bed above so that they can dream of catching the perfect wave while they look at the clear starry skies above them.


text:
Home · Oscars
Quvenzhane Wallis and Denzel Washington Earn Oscar Nominations
Wallis is nominated for Best Actress for her role in Beasts of the Southern Wild.
By Derrick Bryson Taylor	· Updated October 28, 2020
Actress Quvenzhane Wallis and renown actor Denzel Washington have much to celebrate this morning! Both stars were nominated for Academy Awards.
At 9 years old, Wallis is officially the youngest actress in Oscar history to be recognized. She's nominated for Best Actress for her role in Beasts of the Southern Wild—which also earned a Best Picture nomination. She will go up against Jessica Chastain who was nominated for her role in Zero Dark Thirty, Jennifer Lawrence for SIlver Linings Playbook, Emmanuelle Riva for Amour, and Naomi Watts for The Impossible.
Washington knows a thing or two about taking home a little gold statue. This marks his fourth Best Actor nomination. He drew critical acclaim last year when he played drug addicted and sex-crazed pilot William 'Whip' Whitaker in Flight. The competition is stiff —Bradley Cooper is nominated for his work in Silver Linings Playbook, Daniel Day-Lewis is nominated for his role in Lincoln, Hugh Jackman is nominated for Les Misérable and Joaquin Phoenix for The Master.
Quentin Tarantino's controversial film Django Unchained earned a Best Picture nomination and will go up against Amour, Argo, Les Misérable, Life of Pi, Lincoln, Silver Linings Playbook, Zero Dark Thirty and the aforementioned Beast of the Southern Wild.
The awards show will take place live in Los Angeles on February 24.
Who are your rooting for?
TOPICS: Denzel Washington

redpajama_set_name: RedPajamaCommonCrawl


Andrija Maurović is considered the father of Croatian and Yugoslav comics. Born in Muo near Kotor, Austria-Hungary (now Montenegro), he spent his childhood in Cracow, Poland and Dubrovnik. He broke off his studies at the Academy of Arts in Zagreb during the first year to become an independent illustrator for books, weekly and daily newspapers. For several years, he made caricatures for the satirical magazine Koprive, and he also produced artwork for Jutarnji list, Novosti, Zenski Svijet and Kulisa. He also made illustrations based on literary works and often worked in cooperation with prominent writers like Franjo Fuis.
Maurovic created his first comic in May 1935, 'Vjerenica Maca' ('Bride of the Sword'), which was published in Novesti, a Zagreb newspaper. That same year Maurovic co-launched Oko, the first Yugoslav comics magazine. Until 1968, he created nearly 150 stories for the magazine, including 'Stari Macak', 'Podzemma Carica' and 'Ljubavnica's Marsa'.
Inspired by big literary works, Maurovic adapted several works by Jack London, Max Brand , Tolstoy, Henryk Sienkiewicz and H.G. Wells to comics. Among his own creations were 'Dan', 'Old Tom-cat' and 'Radoslav', all of which are now a part of the Sudac Collection. He was specialized in historical comics, for which he didn't use any documentation, just his large imagination.


text:
Film & TV »
What The Flux?
Author Topic: What The Flux? (Read 3593 times)
Batman's Superior Cousin
What's sup MAN?
Re: What The Flux?
Short Answer: No
Long Answer: Well… Episode Three is atrocious, so I would say no. I'll get back to you once I see the fourth episode.
I can't help but feel that Godpleton's avatar/icon gets more appropriate everyday... - TordelBack
Texts from Last Night
Leigh S
I think the common factor here is Chibnall. Imagine being asked to perform the scene where your door i smashed down by an axe wielding figure and your reaction is "whats it with you lot tonight?"
Or your miniaturised house is handed to you and you have to say "How am I meant to fit in that?"
The greatest actors would (and do!) struggle with his insipid, unnatural dialogue and paper thin plotting
Quote from: pauljholden on 17 November, 2021, 02:18:17 PM
Is it just me or is John Bishop appalling - not his character but his acting. It's sort of flat and utterly unconvincing, maybe he's capable of a good performance but they haven't got it from him yet, but just ugh...
pauljholden
2000AD Creator
PJ Holden
I dunno, everyone else has to work their way through awful dialogue, it strikes me those lines were written with the hope a Bishop would make them funny based on his normal dry delivery, but Bishop's delivery is pretty flat.
La Placa Rifa
That's a fair point, though I do think a big factor is how hard it is to embue human rections to such unnatural dialogue.
As for this weeks, the Maxine Alderton factor failed to materialise... that said, has there been a decent Weeping Angels story outside Blink?
AS originally conceived they just dont work in anything but very very narrow scenarios, so various bells and whistles keep getting added until they are just a playground monster that has random powers to keep the plot moving - place such a monster in a Chris "anything goes if it keeps the plt moving" Chibnall story and we get this mess...
I did laugh when 13 wandered off early on to "investigate" a signal with her sonic - Jay Exci's video 100% accuracy rate maintained
Beast Code WWW
I didn't even realise Dr Who was back on until I accidentally stumbled across it tonight.
I'd stopped watching around the end of Jodi W's first year. Tonight's episode was much better than anything I'd seen since Capaldi's last season. Apparently I've missed three episodes of this story, but it was still easy to just pick up most of what was happening as it went along (it helped that I already knew who the angels were). I thought the angels were scary enough, and that the characters were facing enough threat to keep me interested. The two time zones colliding and the little girl talking to her future self were brilliant, and the edge of the village just turning into empty space looked spectacular.
I've no doubt that the criticism of earlier episodes in this thread is justified, but this episode was good fun and Im looking forward to next week's.
Proudhuff
Lorne; it's a square thing
Well that was an improvement, they should have started with this one.
DDT did a job on me
I strongly suspect that what we have here is a clever time-twisty Maxine Alderton story about angels in a village, with proper human characters and dialogue, that Chibnall has hammered into his Division plot. The former sections in the village were great, the rest ... not so much.
IndigoPrime
« Reply #37 on: 23 November, 2021, 11:46:02 AM »
I thought it was an improvement. I cared a bit about the characters. It looked very good. The acting wasn't bad. The script was coherent, despite, as Dan said, the Division thing being smashed into this with all the subtlety of a sledgehammer.
Mrs IP summed it up though: "I'm happy to give shows a chance, but I shouldn't have to wait for episode four of Doctor Who, a show I've been watching for years, for it to get good."
I'm just glad it was better than the finale of The Morning Show (which is the worst piece of television I've watched in a long, long time).
Hawkmumbler
'Uncontrollable urge to make a Scouse joke'
von Boom
As it has been said this was an improvement from last week, however, last week lowered the bar to such a degree I'm waiting to see if they can manage two episodes in a row before making any judgements on this one.
Colin YNWA
testing testing...
I was half watching the third episode as the family watched it and from that perspective it worked quite well. It was a decent character piece - even with the poweeerrr offff luuuuvvvv stuff. Reading the comments here I think I'll not bother to watch it any closer and skip ahead to the 4th!
Expecting some kind of clever twist to magic us out of the fan fiction was perhaps a bit much - it's not as if Chibnall has ever shown he hs any finesse as a writer.
But that was a painful 50 minutes or so - even if it had been "better", not reliant on a plot skeleton of a 100% coincidences, who would this appeal to? Casual viewers surely just want "adventures in Time and Space". I;d guess most fans want the Doctor's mystery preserved, but eeven if they all were gagging for the Doctor to meet her mom, what % of the audience is that?
How the Beeb let that snotty nosed knob goblin from "Open Air" take control of the show...thats some Long game. somewhere, I see Michael Grade pulling his rubber mask off to reveal the Master - Zoom shot! SCREEEEEE_ WUDDLE_U_DUM_DE WUDDLE_DUM...
abelardsnazz
Forget it Jake, it's Chinatown.
Really disappointed with this week's episode, I've only enjoyed 2 and 4 so far. Hoping that the even number success continues with the finale. Bah.
For a full review of this weeks episode. Let's go to the Simpsons:
Lisa: Perhaps there is no moral to this story.
Homer: Exactly! It's just a bunch of stuff that happened.
« Reply #44 on: 03 December, 2021, 11:32:37 AM »
I finally watched episode 5 last night and thought it was bollocks. Now, that could be entirely my own fault for missing the first three episodes, but on the other hand I could still follow and enjoy episode 4. So I suspect that the real problem is what Dan said:
Quote from: Dandontdare on 22 November, 2021, 09:57:40 PM


Home » The sea as a good cause » Projects and grants philanthropy » Support the WoRMS Editors!
WoRMS, the World Register of Marine Species, is the list of all life in the ocean. Currently, more than 200 editors across the world, each with their expertise in a certain taxonomic group, are working on this Register voluntarily. Each year up to 2,000 newly described species are added to the Register.
WoRMS, the World Register of Marine Species aims to capture the names of all species described since the pioneering work of Carl Linnaeus and is continuously updated and improved by a group of taxonomic experts ('the editors').
Completing and correcting WoRMS requires an enormous continuing effort and is entirely dependent on the expertise and time of the editors. On top of that, it is a race against time as species are at risk of disappearing due to changing maritime conditions (such as warming, pollution and acidification), before they are discovered.
Our support: coordination and dissemination of funds to the WoRMS editors.
Your gift is used to support the work of the WoRMS editors: including filling gaps in coverage, participation in international workshops and meetings, expanding the content and enhancing the quality of taxonomic databases, engaging personnel for verification of taxonomic information, and purchase of scientific literature.


text:
Garden Centipede: Plant Damage and Control
Garden centipedes, also called garden symphylans, are not insects, but members of the class Symphyla. Species of this class are common soil arthropods worldwide. Symphyla are small, whitish "centipede-like" creatures ranging from less than 1/8 inch up to about 5/8 inch (or 1/3 inch for garden symphylans). Other names for this pest are garden symphylan and greenhouse centipede. Despite its name, garden centipede is not a true centipede at all, but only resembles that group, because centipese feed on other insects, but garden centipedes are pests that eat plant roots. One way to tell them apart is by the number of legs they have: symphylans have 12 pairs, while centipedes have at least 15. Garden centipedes live about four years.
Garden centipedes are quite small rarely exceeding 1/3 of an inch (8.5 mm). You need a magnifying glass to distinguish them from other soil inhabiting creatures with which they may be confused. These include springtails, young millipedes, and centipedes. Symphylans are very active and fragile. Though their color may vary, depending on what they have eaten, garden symphylans are generally whiter and smaller than true centipedes.
The female lays about a dozen white eggs in the soil and remains with them until young emerge. The first instar nymphs emerge from the eggs with 6 pairs of legs and 6 antennal segments. The second instar looks more similar to the adult. Each of the six subsequent molts results in the addition of a pair of legs.4
By Soniamartinez (Own work) [GFDL (http://www.gnu.org/copyleft/fdl.html) or CC BY-SA 4.0-3.0-2.5-2.0-1.0 (http://creativecommons.org/licenses/by-sa/4.0-3.0-2.5-2.0-1.0)], via Wikimedia Commons
Garden centipedes can potentially damage a wide range of vegetables, small fruits, and flowers. Most serious damage occurs to belowground parts of plants. The damage caused to roots may initiate root rot inflicted by bacteria and fungi. Some damage signs are specific to the crop. The leaves of tomato, for example, take on a bluish tinge and the plant becomes very stunted. Lettuce plants do not develop a heart. Garden centipedes are also decomposers that feed on decaying matter. High rates of raw to partially decomposed organic matter additions are the main stimulant to these pests.1,2,4
To determine if the damage is caused by garden centipedes, lift poorly growing plants and quickly submerge them in a bucket of water. Centipedes will come out of the soil and float up to the surface. Ten or more centipedes around the roots of a single plant are enough to signal concern.
Control can be extremely difficult due to symphylans' vertical movement in the soil, the comlexity of sampling, and the lack of simple, effective control methods. 5 Tilling the soil is still one of the most effective method of control. By physically crushing garden symphylans, the populations can be reduced substantially, although it may also harm key garden symphylan predators such as true centipedes and predaceous mites. Pesticides may have the effect of both killing garden symphylans and repelling them from the surface soil.3 Crop rotation may help reduce the populations of symphylans. Populations decrease significantly in potato crops. They are also lower after a spring oat winter cover crop. Manure applications tend to increase the populations.5
Garden insects of North America: the ultimate guide to backyard bugs By Whitney Cranshaw
The gardener's guide to common-sense pest control By William Olkowski, Sheila Daar, Helga Olkowski
Symphylans: Soil Pest Management Options
Garden Symphylans in High Tunnel Production (HortReport, July 2002)
Brown Stink Bug
Buffalo Treehopper
California Red Scale
Garden Centipede
Garden Webworm
Pests of Perennials
Vegetable Weevil
Tarnished Plant Bug


The Recycle for Merseyside website uses Google Analytics, a web analytics service provided by Google, Inc. Google Analytics sets cookies in order to evaluate your use of the site and compile activity reports to enable us to improve the site and gather visitor statistics.
Google stores the information collected by its cookies on servers in the United States. Google may also transfer this information to third parties where required to do so by law, or where such third parties process the information on Google's behalf. Google will not associate your device with any other data held by Google. By using the Recycle for Merseyside site you consent to the processing of data about you by Google in the manner and for the purposes set out above.
The Recycle for Merseyside website is constructed to be as easy and enjoyable to use as possible. Cookies are used to identify your visits to the site and to enable the efficient display of content in a useful and attractive way.
The website stores information about your visits privately and securely on our server and this information is inaccessible to anyone other than MRWA and its appointed web design and hosting company.
Please be aware that restricting cookies may impact on the functionality of the Recycle for Merseyside website.


text:
Excel 2019/365: The VLOOKUP Function
· Learn how to use the VLOOKUP function to look up data in a table.
The VLOOKUP (vertical lookup; Excel also has an HLookup—horizontal lookup) function searches for a value in the left-most column of a table (called the table_array in Excel). When it finds the value, it returns the value in the same row using the column_index_number to determine which column to select the value from.
Syntax for the VLOOKUP function is:
VLOOKUP(lookup_value, table_array, column_index_number, range_lookup)
VLookup arguments
· Lookup_value is the value to search for in the first column of the table_array.
· Table_array is a range (rectangle) of data whose first column is usually sorted in ascending order (there is an exception; see below).
· Column_index_number is the column number in table_array from which the matching value must be returned. The first column is 1.
· Range_lookup determines if you are looking for an exact match for the lookup_value. Enter FALSE to find an exact match. Enter TRUE to find an approximate match, which means that if an exact match if not found, then the VLOOKUP function will look for the largest value that is less than the lookup_value.
Possible errors:
· If column_index_number is less than 1, the VLookup function will return the error #VALUE!.
· If column_index_number is greater than the number of columns in table_array, the VLookup function will return the error #REF!.
· If you enter FALSE for the range_lookup parameter and no exact match is found, then the VLookup function will return #N/A.
Use the samples workbook for some examples.
Assume the following table is in A1:B12.
This table can be used to look up letter grades if we know a percentage score.
If the score is 95%, we would look up the letter grade using the following formula:
=VLOOKUP(95%,$A$2:$B$12,2)
· The first argument is the value we want to look up: 95%. Note that this would most likely be a cell reference rather than a constant as in this example. See the green arrow below.
· The second argument is the table. Note:
· The headings are not included in the reference.
· An absolute cell reference is used to refer to the table. This is only necessary if the formula is going to be copied (as they usually are – if you're not going to copy the formula, there's probably no need to go to the trouble of setting up the lookup table).
· The third argument is the column that is to be used to retrieve the value ("A"). See the blue arrow below.
Again, assume we are using the same lookup table:
In this example, the lookup_value is not an exact match for any of the value in the Score column.
The arguments are similar to those in example 1, except that the 86% does not match a value in the Score column. In this case, Excel searches through the Score column until it finds the first value that is larger than 86% (this would be the 87% in row 8), then it backs up one row (to the 85% in row 7). Then it retrieves the value from the second column of the table (the Grade column): B-
If the numbers in the first column of your lookup table are not sorted, you can still do a lookup, but the value you are looking for must have an exact match in the table. It is also not necessary for the data in the first column of the lookup table to be numbers; it could also be text. Consider the following table:
If the data in the first column of your lookup table isn't sorted, you must include the fourth argument to the VLOOKUP function, and its value must be FALSE, which means that it will look for an exact match. For example, to look up Ben's phone number (the name "Ben" is in E2), use the following formula:
=VLOOKUP(E2,$A$2:$B$12,2,FALSE)
Be careful with text. Spaces on the end of a name can mess everything up!


text:
Putin has congratulated the actress Inna Churikova on anniversary
The President of Russia Vladimir Putin has sent the congratulatory telegram to the honored actress of the USSR Inna Churikova on the occasion of her anniversary. The text of the telegram on Friday is published on the official site of the Kremlin. In the congratulation the head of state has written: "You - the bright, original actress who has decorated with the talent a set of remarkable movies and theatrical performances - are sincerely loved by millions of people in our country". Also Putin has noted that admirers of talent of Inna Churikova appreciate her feeling of internal advantage and goodwill and also with great respect treat works of the actress. Besides, the president has wished to the heroine of the anniversary of inspiration, wellbeing and health. Inna Churikova was born in the republic Bashkiria in the city of Belebey on October 5, 1943. After the termination of Shchepkinsky school in 1965 she has started a theatrical career in the Moscow Theatre for Young Audience, now Churikova plays in Moscow Lenkom Theatre. She has begun to act in movies still being a student in 1961 and since then has acted more than in forty pictures. For the creative career Inna Churikova has been conferred a set of awards and awards.
of fapnews Anatoly Vasilyev Today, 10:47 0
URL: http://wellnews.us/articles/putin_has_congratulated_the_actress_inna_churikova


Julius, Frederick Gilder. b. 28 Feb 1811, 4th s of Dr George Charles Julius of Richmond, Surrey. Day Boy, Jan 1823 - Sep 1826. M.R.C.S. and L.S.A. 1833; M.D. (Lambeth) 1851; F.R.C.S. 1853: practised at Richmond. m. (1) 1839 (sic) Ellen Hannah o.d. of William Smith of Nottingham; (2) 1871 Sarah Hannah, widow of Hon. Judge Charles James Hargrave, Judge of Landed Estates Courts, Ireland. d. Richmond 4 Jan 1886.
Frederick went to Charterhouse School with his brothers. He took his Medical Degree in Edinburgh. Dr Frederick's career was equally as distinguished as his father's before him and he too was under the patronage of the Royal Family. He was a Medical Officer of the Royal Hospital, the opening of which was chiefly due to him. Originally known as the Richmond Infirmary , it enjoyed the patronage of Queen Victoria and by her command  was hence-forth known as the "Royal Hospital". After his retirement Dr Frederick usually spent the winter months travelling in the Mediterranean region, visiting on different occasions the Azores, Spain, Italy, Corsica, Algiers, Egypt, Palestine and Constantinople. Finally he 'fixed his abode upon Richmond Green', where he died from inflammation of the lungs, in his 75th year, after a very few days illness. He was buried in the Richmond Cemetery, thus ending a family association with that village which had lasted three quarters of a century.
Julius Jottings, January 1901. No 4.
He practiced for a few years at Bognor, where his eldest sister, Emily, kept house for him. He then fell to bad health, and had to travel abroad. On his return he went into partnership with his father Dr George C Julius snr. (note he was apprenticed as an Apothecary in 1827) He lived most of his life in the Old Palace, Richmond, but retired from practice when about 60 years and travelled extensively.
He was a fine looking man, about 5' 11 ", with fair complexion, blue eyes and a long white beard. He was very witty, clever, and amusing; very sarcastic and alarming to those who did not know him very well; but very kind and gentle in sickness.
Two of his daughters, Anne Ellen and Katherine Pauline married doctors. Dr. William Bateman and Dr. Edward Fenn respectively. These two doctors carried on their father-in- laws practice in Richmond until 1912, when the practice, which had been in the hands of the JULIUS FAMILY since it was first established by Dr. George Charles Julius in 1814, 98 years before, passed into other hands.
Richmond Census 1841: Frederick Julius, 30yrs, Surgeon. Ellen Julius, 27 yrs.
Frederick and his father appear to have had a close association with Arthur George Onslow the 3rd Lord Onslow (1777 - 1870). Onslow had no surviving heir, it is clear Frederick's youngest son was named after Onslow, who was his godfather.
It is unclear what the note below refers to.
This note relates to a French snuff box left to Frederick presumably at Onslow's death.
"I leave the snuffbox to my good and excellent friend Mr Frederick Julius; in token of my high regard for him, and under the full appreciation of those truly Christian qualities which so are not ostentatiously adorned his private character"
An apprentice to George Charles Julius of Richmond Surrey Apothecary for 5 years. Indenture dated 6th Feb 1827.
June 13 1833 Mr. Frederick Gilder Julius of full age candidate for certificate to practice apothecary.
2 courses on Materia Medica and Botany ?
2 courses on Principles and Practice of Medicine ?
Examined by Mr Bacot and approved.
M.D. University of Edinburgh 1825.
Member Royal College Surgeons England 1826.
M.D. by Doctorate granted by Archbishop of Canterbury 1851.
Registered 1 Jan. 1859. Julius Frederick Gilder Richmond Surrey surgeon.
Member 1833, Fellow 1853, Royal College Surgeons England.
Registered 17 June 1859 Julius George Charles Horwood Bideford Devon.
NOTES ON FREDERICK'S EDUCATION AND WORK.
NOTABLE BRITISH TRIALS - Dr Thomas Smethurst 1859 - L.A.Parry. Penguin Books Famous Trials, edited by James H. Hodge - Dr Thomas Smethurst 1859 by L.A. Parry.
Miss Bankes was until her marriage [to Dr Thomas Smethurst] in reasonably good health, though subject to bilious attacks. Soon after, she was taken ill and for the first few days her husband attended her. As she got no better it was decided to call in another doctor, and on the advice of the landlady, Dr Julius a well known I Richmond practitioner, was sent for. He first attended on 3rd April. When he was told by Smethurst that his wife was suffering from violent diarrhoea and vomiting -------- As the drugs tried led to no improvement, Dr Julius began to suspect that some irritant poison was being given to her. He asked his partner, Dr Bird, to see Miss Banks------- lie came to the same conclusion as his partner, but only after Dr Julius had mentioned his own fear to him.
2nd Trial 15 August 1859.
Dr Julius gave evidence about Miss Bankes's illness and her treatment.
He was cross - examined by Mr Sergeant Parry, for the defence.
Mr Sergeant Parry: Are you a Doctor of Medicine?
Mr. Parry: Is yours a London Degree?
Mr Parry: What Degree is it?
Dr Julius: It is the Archbishop of Canterbury's degree.
Mr Parry: What ! Can he make a doctor of medicine.
The Lord Chief Baron: Yes; and he can also make a Master of Arts.
Mr Parry:Did you take your degree as a matter of course?
Mr Julius:Oh, dear, no ! It is a very uncommon thing. I had to get a certificate from 2 members of the College of Physicians, Stating that they had known me for a length of time, and that I was a proper person to have a degree?
Mr Parry: And having that certificate you got your degree?
Mr Julius: Yes; but it only entitles me to call myself "doctor"
The Lord Chief Baron: But you are a member of the College of Surgeons, and a member of the Society of Apothecaries?
Note by the Editor, L.A. Parry. This power to grant medical degrees was a remnant of Papal authority, which was reserved to the Archbishop of Canterbury by a statute passed in 1533 in the reign of Henry VIII. The Medical Act 1857 abolished any qualifications to practice with these degrees unless they were granted prior to the passing of the Act.
Evidence given by Master Julius.
I am the son of Dr Julius. On 30 April 2 sealed bottles were given to me by Dr Bird and I took them to Mr Buzzard's in Great Marlborough. . . . .
It was noted that Dr Julius had bought the case to the attention of the magistrates, after a statement had been made to the effect that Drs Julius and Bird grounded their conviction of the guilt of the person on the evidence of Dr Taylor, who found arsenic in certain evacuations of Miss Bankes.
Dr Smethurst was found guilty of murdering Miss Bankes, whom he had married bigamously. He was however given a Royal Pardon as impure reagents had been used in the tests for arsenic. He was then charged with bigamy and found guilty and was sentenced to one year's hard labour.
THE TIMES Wednesday - Sept. 7 1859. Pg 7 col. f.
Dr Julius and Mr Bird waited on the Home Secretary with the view of undoing their evidence against Dr Smethurst. They represented to the right hon. gentleman that they chiefly grounded their conviction of the guilt of the prisoner on the evidence of Dr Taylor relative to the finding arsenic in certain evacuations of Miss Banks. ; whereas it is now ascertained that the arsenic so discovered came from the impure reagents which were employed as tests of the presence of poison in the lady's body. We further understand that the two medical gentlemen in question are most anxious to relieve themselves, by this course, from the moral responsibility they had incurred by the criminating nature of the evidence they gave on Dr Smethurst's trail.
In reply to the above, I beg to state that I was requested to attend at the office of Messrs Symes, Teesdale & Co. the solicitors for the prosecution, to answer certain questions, I believe suggested by the Home Secretary, and my answers were corroborative of the evidence I gave on trial ; and I can truly say that neither Mr Bird nor myself feel any desire to be removed from a responsibility which has unavoidably been incurred by the discharge of a most painful duty to society and to ourselves as conscientious men. And I would also add that our opinions as to the cause of Miss Banks death have not been in the slightest degree shaken".
Old Palace Richmond Sept. 6.
Some 2 years after his release Dr Smethurst bought an action to prove the will of the woman whom he had been convicted of poisoning and actually won it.
With sincere regret, we record the death of Dr Frederick Gilder Julius, which took place on Monday last, at his residence on Richmond Green, after a very few days illness, from inflammations of the lung. The deceased gentleman, like his eldest brother and father before him, practised for many years in Richmond, with which town the name of Julius has now been closely connected for nearly three quarters of a century. Dr F.G. Julius was born at Bristol in February, 1811, immediately after the arrival of his parents from India. His father shortly afterwards joined Sir David Dundas, at that time Sergeant-Surgeon to George III., in practice at Richmond, taking up in his residence in the Old Palace in 1814, where he and his son Dr Frederick Julius after him continued to reside until the latter retired from practice in the summer of 1871. After his retirement, Dr F.G. Julius usually spent the winter months travelling in the neighbourhood of the Mediterranean, visiting upon different occasions the Azores, Spain, Italy, Corsica, Algiers, Egypt, Palestine, and Constantinople. Finally he fixed his residence upon Richmond Green again, where the first 60 years of his life had been passed, and where his career has now ended in the 75th year of his age.
The funeral is to take place at the Richmond Cemetery today, at 12 o'clock.
Julius, Frederick Gilder (1811-1886) M.R.C.S., April 22, 1833; F.R.C.S., June 2, 1853; M.D. Lambeth, 1851; L.S.A., 1833. Born in Bristol; his father soon afterwards removed to Richmond and joined Sir David Dundas in practice, Sir David being then Sergeant Surgeon to George III. Here he was afterwards joined by his two sons, George C. Julius and Frederick Gilder Julius. The letter received his professional training at St George's Hospital, and his Lambeth degree was granted by the Archbishop of Canterbury. "my degree is not the London degree" he said, "it is the Archbishop of Canterbury's; he has the power of granting it; I did not take it out as a matter of form, it is a very uncommon thing. In order to obtain it, I had not merely to pay the fees, I had to go to to Fellows of the Royal College of Physicians, who stated that they had known me a certain length of time, and then I was a person upon whom a doctor's degree might be conferred; upon that you are enabled to call yourself a doctor, that is all; that was my object."
Julius was for many years ago a leading practitioner in Richmond, Surrey, and in 1859 was the principal medical witness in the trial known as "The Queen against Thomas Smethurst", which began on August 15 of that year and lasted several days. Dr Smethurst was accused of the murder by poison of a Miss Isabella Banks, whom he had bigamously married - he having as wife an older lady, alive at the time. Miss Banks, when suffering from severe illness, involving constant diarrhoea and vomiting, had made a will in Smethurst's favour on May 1. Julius who was called in and was in constant attendance upon her suspected that some drug of an irritant nature was being administered to her in addition to those prescribed by him. She died on May 4. His suspicions having been aroused, Julius had applied to a magistrate sometime before her death, but Smethurst was not arrested till the death occurred. Many important medical witnesses were called both for the prosecution and for the defence, and eventually Smethurst was found guilty of murder, but was reprieved owing to insufficiency of evidence, and in the end, after litigation, succeeded to Miss Banks fortune. Julius, however, had acted with entire propriety in the whole matter, as had his partner, Samuel Dougan Bird MRCS. He retired in 1871 and travelled much in countries bordering on the Mediterranean. His death occurred at Richmond on January 4, 1886.
Ref: Plarr's Lives of the Fellows July Pg 640.
Julius Frederick Gilder Esq. M.D. 8 Feb 1886.
The Will with codicil of Frederick Gilder Julius M.D. late of 4 Portland Ter., parish of St Mary, Richmond who died 4 Jan 1886 at 4 Portland Terrace was proved 8 Feb 1886 by Arthur Onslow Julius of 15 Finsbury Circus the son, and William Adolphus Frederick Bateman of Rothsay House, Parish of St Mary Surgeon Exors at L6023 15s 2d Resworn Aug 1886 L7126 15s 5d.
Awdrey Julius's notes, and notes from the family Bible of George Charles Julius put Frederick born at Bristol. Baptism date is doubtful.
Mar Qtr 1839 Nottingham shown for marriage is unchecked.
The above Bible is a 1723 version of the King James, printed by John Baskett.
655. Frederick G Julius Arm.
Arms: As above, but the stars Gules, and on an escutcheon, Or, a chevron cotised between three demi-griffins Sable [Smith].
Crest and Motto of Julius.
The following concerns a pivotal case at law in which Frederick was somewhat of a pawn.
THE TIMES - Friday March 5 1880. pg. 4 col. a.
Details of an appeal to The House of Lords against a decision of the Court of Appeal on a complaint against the Bishop of Oxford concerning the practice of the Rev Thomas Thelluson Carter Rector of the Parish of Clewer.
"I now reside and have for many years have resided in the Parish of Clewer Berks and am a parishioner of the said parish of St Andrew."
The Oxford Movement was the cause of concern to some of the Parishioners of Clewer as it had influenced the Rev Thomas Carter and his form of worship. The Movement began in Oxford in 1833, and had reawakened spiritual and devotional life of the Church of England. However some people feared it might lead the Established Church towards Roman Catholicism.
Carter had on accepting the living of Clewer in 1844, began introducing "order and decency" into the worship of the run down parish.
After two petitions by some Parishioners backed by the Church Association were rejected by Carter's bishop they found a new complainant in Frederick Julius. Whilst legally a parishioner, Frederick rarely resided within the parish and did not attend the church; indeed the affidavit dated December 28th 1878, was sworn at Her Majesty's Consular Court in Cairo. The Bishop was minded to veto the application as before. But now the Church Association endeavoured to force the Bishop's hand by challenging the right of veto.
Mixing water with the wine at the offertory.
Standing by the Altar with his back turned to the congregation.
Bowing during the Consecration and elevating the bread and wine.
The outcome of the Bishop's appeal was a reversal of the Queen's Bench decision, but this was not the end. Frederick took the case to the House of Lords. Once more attention was focused not on the original charges but on the bishop's right of veto. The Lords upheld the Court of Appeal, Frederick had finally lost his case and the bishop's veto was never questioned again.
The case of Julius v the Bishop of Oxford made legal history in establishing the principle that if a party has a discretion in a matter, i.e. they may act in a certain way, it does not necessarily mean they must do so. The Clewer case is cited when this point of law is at issue and has been used as recently as July 1986.
Carter's daughter Jane, records in her biography of her father how Carter and Frederick met for the first time at a house on Clewer Hill. Frederick was charmed by him and expressed regret at having commenced litigation. The two men shook hands and parted as friends. In 1887 Mary Isabel Julius  Fredericks neice was admitted as a Clewer Sister where she was known as Sr Mary Isabel. She was professed in 1889 and in 1890 went to India where she served until 1894. She died on March 3rd 1898.
Given by Valerie Bonham on 4th May 2001at St Andrew's Church, Clewer Village.
England: Canterbury - Index to the Act Books of the Archbishops of Canterbury, 1663-1859 (A-K) Lambeth. County: General Country: England Julius, Frederick Gilder (of Richmond, Surrey, M.R.C.S.), 1851 : Degree Doctor of Physic; 17, 73.
1. Moore's Royal Prophetic Almanack: The Book of Fate, 1840, Thought to Have Belonged To Dr Frederick G Julius.
3. Lambeth Degree Dr of Physic, 24 Jun 1851. Transcription of Grant of Doctor of Physic.
I Frederick Gilder Julius the person to be admitted to the Degree of Doctor of Physic by the Most Reverend Father in God John Bird by Divine Providence Lord Archbishop of Canterbury Primate of all England and Metropolitan do sincerely promise and swear that I will be faithful and bear true allegiance to her Majesty Queen Victoria - So Help Me God.
J.H.T. Manners Sutton - Registrar.
Registered on the twenty-sixth day of June in the year of our Lord One Thousand Eight Hundred and Fifty One by the Clerk of her Majesty for the Faculties in Chancery.
George Humphreys and Josiah Humphreys.


text:
PDL Schedule Released; FC Miami City Set To Kickoff In May
By Omar Moubayed on Saturday, February 3, 2018
It seems like there's a lot of dust in the air surrounding the beautiful game here locally. While every other league's schedule is still "under construction", the PDL released their schedule recently. "PDL?", you ask. Yes FC Miami City, the only club in the PDL from the magic city, will return to Tropical Park for the 2018 PDL season.
FC Miami City will look to pick up right where they left off in 2017. Finishing second in the Southeast Division, City amassed 33 points (10-3-1) and an outstanding plus-29 goal differential, falling just short of division champions SIMA Aguilas by a point. Qualifying for the playoffs after missing out in 2016, City defeated the OKC Energy 1-0 before losing in penalties to Mississippi Brilla F.C. after 120 minutes of scoreless play. However, City did earn a spot in the U.S. Open Cup joining fellow magic city clubs Red Force, Miami United, and FC Kendall.
Miami City will begin their 2018 campaign with a four game home-stand on Saturday May 19th against the Lakeland Tropics. The PDL schedule is a challenge as teams play every few days. City will have a back-to-back on May 23rd and 24th against defending division champions SIMA, followed by IMG Academy. The new look North County United (formerly South Florida Surf) will end the home-stand on Saturday June 2nd.
All home games are scheduled to kick off at 6PM. For more information about F.C. Miami City, visit their website by clicking here. You'll find coverage of City throughout the season here, at your home for all things soccer in Miami-Dade County.
2018 PDL FC Miami City FCMC PDL
FC Miami City USL League Two
FC Miami City Beat Weston; Move To Second In Table
Early First Half Equalizer Gives FCMC A 1-1 Draw
FC Miami City Overhaul Roster & Staff Seeking A Return To Normal
Pingback: Another Dade Derby as The Miami FC 2 and FC Miami City to face off in U.S. Open Cup - Magic City Soccer


I have printed the prototype because i was to late to buy the pnp from kickstarter .
But when i look at screen shots of the official pnp. The food number of the cards seems so different ?
Is the official more balanced out ? or does it not really matter ?
The original PnP prototype is a great game, but the balance of the final cards set is better. The benefits of the final version are incremental, not transformational, so I wound't worry if I were you.
The final card set has Defensive Herding and Intelligence instead of Unstable DNA, Camouflage, and Good Eyesight.
The final set has 7 copies of 16 different trait cards and 17 copies of the Carnivore card (129 total cards). The original PnP prototype has 6 copies of 17 different trait cards and 15 copies of the Carnivore card (117 total cards). The extra cards allow for a better 6 player game.
The food distribution of the final set has a slightly greater range than what you'll find in the original PnP version. This increases the tactical options of the food play, leads to more surprises, and increases game play variety. It creates a game environment that is a little less stable and a little more dynamic.
But if you want the final art PnP files for $5, let me know and I'll work something out for you. Cheers!
"The moment you doubt whether you can fly, you cease for ever to be able to do it."
Would you be willing to share the latest food distribution, so we can just edit the PnP numbers? I guess it's also important to know which numbers go onto which cards.
Really man? Come on you know that's not fair to the backers or to North Star Games.
Dom just wrote that he would, if you'll get him the $5 that the other backers paid for that info.
I would't mind releasing the food distribution numbers for free ($5 doesn't affect our financials in any meaningful way), but I don't want to be unfair to our backers. So if you want the updated PnP files, let me know by geekmail and we'll sell it to you for $5. Sorry for the bad news.
On the other hand, the free PnP files is a great game without any changes. So if you're tight on money, just play it as is. I doubt you'll be disappointing.
I would't mind releasing the food distribution numbers for free ($5 doesn't affect our financials in any meaningful way), but I don't want to be unfair to our backers.
I think I'll take the opportunity to practice some patience, and I'll try and wait for the official release. Thanks though!
Edit: to be clear, I'm sorry if I offended anyone. I did not follow the kickstarter and am not aware of any promises made to the backers. I just fiddled around with the PnP files I found here on the geek.
Yes, the original PnP, but aren't those the one without the newly revised cards (as in content, not just updated art?)? Or am I confused?
I thought the new cards are only found in the paid PnP. I was inquiring about possible final cards but with original artwork.
No, we have not paid for a graphic artist to make the file you are asking for. You can either pay $5 for the final PnP file, or use the one on the BGG Evolution page.