text:
\section{Introduction}
In the recent years, reaction-diffusion problems have stimulated a large
body of work in many different directions~\cite{RITTEN,LEE}.
The simplest example is provided by the annihilation process $A+A
\rightarrow 0$, in which the $A$ particles diffuse  and react by pairs
on contact.

Usually such kinetic processes are described in terms of macroscopic
rate equations giving the time evolution for the local averaged
concentrations.  One assumes that the reaction
is completely described in terms of the local average densities, and that the
reaction introduces no correlations between the reacting species.  This
is reminiscent of a mean-field like approximation in statistical
physics. However, an important aspect of the problem is neglected,
namely the microscopic fluctuations and it is well known  that these
fluctuations play an important role in low dimensional
systems~\cite{WIL,LEBO,CCD}.

Indeed, for the annihilation process $A+A \rightarrow 0$, the rate
equation predicts that, in the long time regime, the concentration of
$A$ will decrease in time as $a \simeq {\cal A} t^{-1}$ while a
calculation taking into account the local microscopic fluctuations in the
particle density  gives~\cite{KANG} $a \simeq {\cal A} t^{\alpha}$,
with $\alpha={\rm Min}(1,\frac{d}{2})$, where $d$ is the dimensionality
of the system. Moreover, for a Poissonian initial state, the amplitude
${\cal A}$ is a universal quantity, which is in particular
independent on the initial density of the particles.  This nonclassical
power law behavior is called anomalous kinetics.

Thus to describe correctly these systems, it is crucial to work within
a formalism which is able to  keep track of the fluctuations. This is
a difficult task and this is the reason why exact analytical results are
scarce, but for one-dimensional systems. Complex reaction-diffusion systems
are present in nature~\cite{BORN}. However, the study of simple models,
like the diffusion-coagulation process (DC) $A+A \rightarrow A$, or the
diffusion-annihilation one (DA) $A+A \rightarrow 0$ provide a very useful
testing ground for new theoretical approaches.

Different methods have been explicitly developed for the one-dimensional
systems (see for example~\cite{RITTEN,RACZ,LUSH,BALD,SPOU,PRIV,GRYN}).
In most of them, one maps the reaction diffusion problem onto a different
model (quantum chain~\cite{RITTEN}, kinetic Ising model~\cite{RACZ},
invasion process~\cite{BALD}, etc), which is exactly solvable. A different
approach consists in finding the "good observables" for which an exact
equation of motion can be derived and solved. For the DC model, Doering
{\it et al}.~\cite{LIN} have shown that the system was best analyzed in
terms of the time dependent probability ${\cal E}(x,y,t)$ that the interval
between $y$ and $x$ is empty at time $t$. The particle density $a(x,t)$
is simply given by $a(x,t)=-\frac{\partial}{\partial x}{\cal E}(x,y,t)
\vert_{x=y}$. In the continuous limit, ${\cal E}(x,y,t)$ obeys to a simple
diffusion  equation which can be solved analytically. However, Doering
{\it et al.}\/\ were not able to find a similar quantity for the DA model.

In arbitrary dimension, it was soon recognized, based on dynamical
renormalization group arguments~\cite{PELITI_UNO,SAS}, that the DA and
DC models belong to the same universality class. The power law exponents
describing the decay of the particle densities are the same.
However, for the amplitude $\cal A$, the situation turns out to be more
subtle. Whereas for the DC model this amplitude seems to be strictly
independent of the initial state, this is no longer true for the
one-dimensional DA model. Indeed, as noticed by Family and Amar~\cite{FAMILY},
short-ranged  initial correlations can change the asymptotic amplitude in a
continuous way. In their analysis, the DA model is mapped onto a kinetic Ising
model at zero temperature~\cite{RACZ}. Explicit asymptotic results for the
average domain size, average magnetization squared and pair-correlation
function are derived for arbitrary initial conditions. For the case of an
initial magnetization $m_0=0$, the results for the DA model with Poissonian
initial conditions are reproduced. However, for $m_0 \not = 0$, the
particle density has a nontrivial dependence on the initial magnetization.
A nonzero value of $m_0$ means for the DA model that there are some
correlations among the particles in the initial state. What is the mechanism
responsible for this dependence and could a similar situation occur also in
the DC process for which there is no mapping onto a kinetic Ising model?

To answer this question we shall, in this paper, revisit the problem of
universality in the DC and DA model. The strategy is to study the
field theory which can be associated to a reaction-diffusion process,
following the method introduced by  Doi~\cite{DOI} and revisited by
Grassberger {\it et al}.~\cite{GRASS} and Peliti~\cite{PELITI_DUE}.
We shall consider a class of diffusion-coagulation-annihilation models
(called $\alpha$-models), allowing us to go continuously from the DC
process to the DA one.

The paper is organized as follows. In Section 2, the way to associate a field
theory to a reaction-diffusion process is briefly reviewed and the
$\alpha$-models are introduced. Within a functional integral formalism,
different $d$-dimensional $\alpha$-models are related one to the others.
In Section 3, the one-dimensional DC model is revisited. It is shown how the
equation for the time dependent probability ${\cal E}(x,y,t)$ that an
interval between $y$ and $x$ is empty at time $t$ can be obtained within the
field theoretical framework. In Section 4, we introduce the quantity
${\cal T}^{(\alpha)}(x,y,t)$ which is the natural generalization for the
$\alpha$-model of ${\cal E}(x,y,t)$. Its equation of motion can be derived
and solved exactly, leading to an exact expression for the asymptotic behavior
of the particle density. It is shown explicitly how and when the amplitude
universality can be violated. In particular, it is proved that for an initial
state in which particles are pairwise correlated, the violation of amplitude
universality is only possible for the DA model. Finally, some extensions of
this work are discussed in the conclusion.

\section{Field theoretical approach}

For the sake of completeness and to set up the notations,
we briefly sketch the main steps necessary to derive the field theory
associated to a d-dimensional  reaction-diffusion process. We follow the
method introduced by Doi~\cite{DOI} and revisited by Grassberger
{\it et al}.~\cite{GRASS} and Peliti~\cite{PELITI_DUE}.

We start from a description of the system in terms of the probabilities
$P_n(x_1,\dots,x_n;t)$ that a configuration with $n$ particles at points
$x_1,\dots, x_n$ is realized at time $t$. As we are not interested in
specifying which particle is where, we choose $P_n(x_1,\dots,x_n;t)$ to be
a symmetric function in the variables $x_1,\dots, x_n$. Note that, as the
particle number is not conserved, a state of the system  must be specified
by the entire set $\Phi(t)=\{P_n (t)\}_{n=0,\dots,\infty}$ with the
normalization:
$$\sum_n {1\over n!}\int dx_1\dots dx_n \ P_n(x_1,\dots,x_n;t)=1\quad .$$
The dynamics is defined by the master equation:
  \begin{equation}
\partial_t P_n(x_1,\dots,x_n;t) = [{\cal H} P]_n(x_1,\dots,x_n;t)\quad ,
  \end{equation}
where ${\cal H}$ is an operator acting on the set of the
probabilities~\cite{DOI,GRASS}.

Let us consider as an example the following DA process. The $A$ particles
diffuse
and annihilate each other with a reaction rate $V(\vert x-y\vert)$ depending
on the distance. The corresponding master equation reads:
  \begin{equation}
\partial_t P_n(x_1,\dots,x_n;t)=
D\sum_{i=1}^{n}{\partial^2 \over \partial x_i^2}P_n(x_1,\dots,x_n;t)
  \label{equ:master}\end{equation}
$$-{1\over 2}\sum_{i\ne j}V(\vert x_i-x_j\vert)P_n(x_1,\dots,x_n;t)
+{1\over 2}\int dydz\ V(\vert y-z\vert)P_{n+2}(y,z,x_1,\dots,x_n;t)\quad ,$$
where $D$ is the diffusion constant.

We introduce a Fock space representation for the states of the system.
Space-valued annihilation $\psi(x)$ and creation $\pi(x)=\psi^{\dag}(x)$
operators are introduced. The vacuum state $\vert0\rangle$ is defined by
$$\psi(x)\vert0\rangle =0$$
and the annihilation and creation operators obey bosonic commutation
relations:
$$[\psi(x),\pi(y)]=\delta(x-y)\quad .$$
The state specified by the set $\Phi=\{P_n\}_{n=0,\dots,\infty}$ reads:
  \begin{equation}
\vert \Phi\rangle =\sum_{n=0}^{\infty}{1\over n!}\int dx_1\dots dx_n\
P_n(x_1,\dots,x_n)\pi(x_1)\dots\pi(x_n)\vert 0\rangle
  \label{equ:vettore}\end{equation}
and the statistical average $
\langle A \rangle$ over the state $\Phi$ of an observable
$\{A_n(x_1,\dots,x_n)\}_{n=0,\dots,\infty}$ defined by
$$\langle A\rangle=\sum_n {1\over n!}\int dx_1\dots dx_n\
A_n(x_1,\dots,x_n)P_n(x_1,\dots,x_n)$$
takes the form of a scalar product:
$$\langle A\rangle=\langle\vert A\vert \Phi\rangle\quad ,$$
where $A$ is the corresponding operator of the Fock space, such that
$\langle 0\vert \psi(x_1)\dots\psi(x_n)A\vert \Phi \rangle=A_n(x_1,\dots,x_n)$
and $\langle\vert$ is a projection state given by
$$\langle \vert=\langle 0\vert e^{\int dz\ \psi(z)}\quad ,$$
with the property $\langle\vert\pi(x)=\langle\vert$. In particular,
the particle density is given by  $c(x)=\langle \psi(x)\rangle$ and
the n-point correlation functions by $\rho(x_1,\dots,x_n)=\langle
\psi(x_1)\dots\psi(x_n)\rangle$. The normalization of the state is expressed
by the condition $\langle\vert \Phi \rangle=1$. A correlated state,
characterized by the $p$-points cumulants:
$$\langle \psi(x_1)\dots\psi(x_p)\rangle_{cum}=g_{(p)}(x_1,\dots,x_p)$$
has the form:
  \begin{equation}
\vert \Phi \rangle= \exp \left[
\sum_{p=1}^{\infty}{1 \over p!}\int dx_1 \dots dx_p\
g_{(p)}(x_1 ,\dots,x_p)(\pi(x_1)-1)\dots(\pi(x_p)-1)
\right]\vert 0\rangle\quad .
  \label{equ:state}\end{equation}

It can be shown that the master equation can be written in
the Schr\"odinger-like form:
  \begin{equation}
{\partial \over \partial t}\vert \Phi(t)\rangle =H\vert \Phi(t)\rangle\quad .
  \label{equ:schroedinger}\end{equation}
For our annihilation model,  the non-Hermitian evolution operator $H$
is given by
$$H=H_0+H_a\quad ,$$
with
  \begin{equation}
H_0=D\int dz\ \pi(z)\nabla^2\psi(z)
  \label{equ:H_0}\end{equation}
and
  \begin{equation}
H_a=-\int dzdz^{\prime}\ V_a(\vert z-z^{\prime}\vert)[\pi(z)\pi(z^{\prime})-1]
\psi(z)\psi(z^{\prime})\quad .
  \label{equ:H_a}\end{equation}
A second example is provided by the diffusion-coagulation   process.
The interaction part of the evolution operator becomes:
  \begin{equation}
H_c =-\int dzdz^{\prime}\ V_c(\vert z-z^{\prime}\vert)[\pi(z)\pi(z^{\prime})-
\pi(z)]\psi(z)\psi(z^{\prime})
  \label{equ:H_c}\end{equation}
If now both coagulation and annihilation  reactions are simultaneously allowed,
one has:
 \begin{equation}
H_{a+c}=H_a +H_c =
-\int dzdz^{\prime}\ [(V_a +V_c)\pi(z)\pi(z^{\prime})-
V_c\pi(z)-V_a]\psi(z)\psi(z^{\prime})\quad .
  \end{equation}
Although not Hermitian, these evolution operators have the property to
preserve the state normalization: $\langle\vert H=0$. The evolution equation
of an observable $A$ is:
  \begin{equation}
{\partial\over\partial t}\langle A\rangle=\langle AH\rangle=\langle
[A,H]\rangle
  \quad .\label{equ:evoluzioneA}\end{equation}

It can be shown~\cite{PELITI_DUE}, that the correlation functions may
be expressed by functional integrals as:
  \begin{equation}
\langle \psi(x_1,t_1)\dots \psi(x_n,t_n)\rangle _{{\cal S},F_0} =
\int {\cal D}\psi {\cal D}{\bar \psi}\ \psi(x_1,t_1)\dots \psi(x_n,t_n)
e^{-{\cal S}} F_0 [{\bar \psi} (x,0)]\quad .
  \label{equ:integralefunzionale}\end{equation}
The action ${\cal S}$ appearing in the measure $e^{-\cal S}$ is related to
$H$ as follows:
$${\cal S}[{\bar\psi},\psi]
=\int dtdx\ {\bar\psi}\dot{\psi}-\int dt\ H[\pi={\bar\psi}
+1,\psi]\quad .$$
$F_0$ is a functional of the auxiliary field ${\bar \psi}$ at t=0,
and is related to the cumulants of the initial state by
  \begin{equation}
F_0[{\bar \psi}]=
{\rm exp}\left[\sum_{p=1}^{\infty}{1 \over p!}\int dx_1 \dots dx_p\
g_{(p)}(x_1 ,\dots,x_p;0){\bar \psi}(x_1)\dots
{\bar \psi}(x_p)\right]\quad .
  \label{equ:state2}\end{equation}

Let us now assume that $V_c$ is proportional to $V_a$. This leads us to
introduce
a class of model depending on a continuous real parameter $\alpha$ (called
the $\alpha$-models) and defined by the following action:
  \begin{equation}
{\cal S}_{\alpha}=
\int dtdz\ {\bar\psi}\left({\dot\psi}-D\nabla^2\psi\right)+
\int dtdzdz^{\prime}\ V(\vert z-z^{\prime}\vert)
[{\bar\psi}(z){\bar\psi}(z^{\prime})+\alpha{\bar\psi}(z)]
\psi (z)\psi(z^{\prime})\quad .
  \label{equ:Sac}\end{equation}
The simple coagulation and annihilation reactions correspond respectively to
$\alpha=1$ and $\alpha=2$ and will be labeled by the indices $c$ and $a$ in
the following. Thus the $\alpha$-models ($\alpha\in[1,2]$) interpolate
between the coagulation process and the annihilation one.

We are now in the position to relate dynamics with different values of $\alpha$
in a very simple way. Taking the coagulation case as reference,
eq.~(\ref{equ:Sac}) gives:
$${\cal S}_{\alpha} [{\bar \psi}, \psi]
={\cal S}_c [{{\bar \psi} \over \alpha},\alpha \psi]$$
and, by simply rescaling the functions in the functional integral, we get
the following relations between correlation functions:
  \begin{equation}
\langle \psi(x_1,t_1)\dots \psi(x_n,t_n)\rangle _{\alpha,F_0}
=\alpha^{-n} \langle \psi(x_1,t_1)\dots \psi(x_n,t_n)\rangle
_{c,F_0 ^{(1/\alpha)}}\quad ,
  \label{equ:equiv}\end{equation}
where:
$$F_0 ^{(1/\alpha)}[{\bar \psi}]
=F_0 [\alpha{\bar \psi}]\quad .$$
In particular this means that, if at $t=0$ we consider two states
$\vert \Phi_{\alpha}(0)\rangle$ and $\vert \Phi_c(0)\rangle$ such that:
$$\langle\vert \psi(x_1)\dots \psi(x_n)\vert \Phi_{\alpha}(0)\rangle
=\alpha^{-n} \langle\vert \psi(x_1)\dots \psi(x_n)\vert \Phi_c(0)\rangle
\quad ,$$
for all $n$, then this relation in conserved for all times, if
$\vert \Phi_{\alpha} \rangle$
evolves according to the dynamic of the $\alpha$-model and $ \vert \Phi_c
\rangle$ according
to the dynamic of the coagulation one. Moreover, if a set of equations
among correlation functions, or more generally among operators, is satisfied
for the coagulation case, then a similar set of relations exists for the
$\alpha$-model, providing that one rescales the fields according to
$\psi\rightarrow \alpha\psi$. Note that similar relations have been recently
derived in a different context by Henkel {\it et al}.~\cite{MALTE}.

For the particular case $\alpha=2$ and for homogeneous Poissonian initial
conditions for both $\vert \Phi_c \rangle$ and $\vert \Phi_a \rangle$, we have
the following relation
between the concentrations:
  \begin{equation}
c_c(t)=2c_a(t)\qquad\forall t>0\quad ,
  \end{equation}
if initially $c_c(0)=2c_a(0)$. This result has been already derived by
several authors (see for example~\cite{SPOU,PRIV,MALTE}). Moreover, the
relation among the actions ${\cal S}_c$ and ${\cal S}_a$ has been
already used in previous works~\cite{PELITI_UNO,SAS} for Poissonian initial
conditions. However, our results are valid for all the $\alpha$-models and
take care explicitly of arbitrary initial condition. In addition, from
eq.~(\ref{equ:equiv}), we clearly see that quantities involving correlation
functions of the same order are simply related. But more complicated
quantities as for example the interparticle distribution function (which
involves many correlation functions of different order), are no longer
simply related. This explain why the asymptotic interparticle distribution
function between DC and DA model are qualitatively different\cite{BEN}.

\section{The one-dimensional coagulation model revisited}

Let us define ${\cal E}(x,y,t)$, the time dependent probability that the
interval $(y,x)$ is empty at time $t$. Our goal is to rederive in the
framework of the field theory the equation of motion for ${\cal E}(x,y,t)$
obtained by Doering {\it et al}.~\cite{LIN}. This equation reads:
  \begin{equation}
{\partial {\cal E}(x,y,t) \over \partial t} =
D\left( {\partial ^2 \over \partial x^2}+{\partial ^2 \over \partial y^2}
\right) {\cal E}(x,y,t)\quad .
  \label{equ:charlie}\end{equation}
First, we have to identify the Fock space operator corresponding to the
probability ${\cal E}(x,y,t)$. This means that we are looking for a
time-independent operator $E(x,y)$ such that, given a state $\vert \Phi(t)
\rangle$, the probability that the interval between $y$ and $x$
is empty at time $t$ is given by
  \begin{equation}
{\cal E}(x,y,t)=\langle \vert E(x,y)\vert \Phi(t)\rangle \quad .
  \label{equ:defEdue}\end{equation}
This operator is:
  \begin{equation}
E(x,y)=e^{-\int_{y}^{x} dz\psi(z)}\quad .
  \label{equ:defE}\end{equation}
The easiest way to show this consists in using the explicit form of the
projection state
$\langle \vert=\langle 0\vert e^{\int_{-\infty}^{+\infty} dz\psi(z)}$ in
eq. (\ref{equ:defEdue}). Indeed, for a general state $\vert\Phi\rangle$,
we find:
$${\cal E}(x,y)=\langle 0\vert e^{(\int_{x}^{\infty} +\int_{-\infty}^{y})
dz\ \psi(z)}\vert \Phi\rangle\quad .$$
If we consider the state
$\vert \Phi^{\prime}\rangle=\pi(z_1)\dots \pi(z_n)\vert \Phi\rangle$
obtained by adding
particles in  $z_1 ,z_2 ,\dots,z_n$ to the state $\vert\Phi\rangle$
we obtain:
  \begin{equation}
\langle E(x,y)\rangle _{\Phi^{\prime}}=\cases{\langle E(x,y)\rangle_\Phi
&if\ $z_n \not\in (y,x)\ \forall n,$ \cr 0 &otherwise.}
  \label{equ:a}\end{equation}
This defines the probability ${\cal E}(x,y)$ introduced above. Moreover,
we have:
\begin{equation}
\psi(x)=-\left[{\partial \over \partial x}E(x,y)\right]_{x=y}\quad ,
  \label{equ:psi-E}\end{equation}
in agreement with Doering {\it et al}.~\cite{LIN}.

The basic commutation relation is:
  \begin{equation}
[E(x,y),\pi(z)]=-f(z,x,y)E(x,y)\quad ,
  \label{equ:commE}\end{equation}
where the function $f$ is given by:
  \begin{equation}
f(z,x,y)=\cases {1 &if\ $y<z<x,$ \cr 0 &otherwise.}
  \label{equ:f}\end{equation}
This commutation relation can be used to prove eq.~(\ref{equ:a}) without
any reference to the explicit form of the projection state.

We can now derive the equation of motion for ${\cal E}(x,y,t)$.
{}From eqs.~(\ref{equ:H_c},\ref{equ:evoluzioneA}) and the property
(\ref{equ:a}), we have:
  \begin{eqnarray}
{\partial \over \partial t}\langle E(x,y)\rangle&=&
\langle E(x,y)(H_0+H_c)\rangle\label{equ:evol}\\
&=&-D\langle E(x,y)[\psi^{\prime}(x)-\psi^{\prime}(y)]\rangle
+\int_{\Sigma} dzdz^{\prime}V(\vert z-z^{\prime}\vert)
\langle E(x,y)\psi(z)\psi(z^{\prime})\rangle\quad ,
  \nonumber\end{eqnarray}
where the domain of integration is
$$\Sigma=\{ (z,z^{\prime})\ \vert\ z\not\in (y,x),\ z^{\prime}\in (y,x)\}$$
To proceed, we have to specify  the interaction $V(\xi)$.
We choose the following form:
  \begin{equation}
V(\xi)=\cases {v &if $0<\xi<\sigma,$ \cr 0 &if $\xi>\sigma$}
  \label{equ:V}\end{equation}
and  we shall eventually take the $\sigma \rightarrow 0$ limit.
For this purpose let us define:
$$v\sigma =\lambda \quad\hbox{and}\quad v\sigma ^2 =\chi\quad .$$
Taking into account the form of the domain of integration
$\Sigma$, the expansion in $\sigma$ of eq. (\ref{equ:evol}) gives:
  \begin{eqnarray}
{\partial \over \partial t}\langle E(x,y)\rangle&=&
\langle E(x,y)(H_0+H_c)\rangle\label{equ:expansion}\\
&=&-D\langle E(x,y)[\psi^{\prime}(x)-\psi^{\prime}(y)]\rangle
+{1\over 2}\chi \langle E(x)[\psi^2 (x)+\psi^2 (0)]\rangle +O(\sigma ^3 )
  \nonumber\end{eqnarray}
{}From the definition of $E(x,y)$ (eq.~(\ref{equ:defE})), we obtain:
  \begin{equation}
({\partial^2 \over \partial x^2}+{\partial^2 \over \partial y^2})
\langle E(x,y)\rangle
=-\langle E(x,y)[\psi^{\prime}(x)-\psi^{\prime}(y)]\rangle
+\langle E(x,y)[\psi^2 (x)+\psi^2 (y)]\rangle\quad .
  \label{equ:derE}\end{equation}
Thus in the limit
  \begin{equation}
\sigma \rightarrow 0, \qquad v \rightarrow \infty \qquad
\hbox{such that} \qquad v\sigma^2 =\chi \rightarrow 2D\quad ,
  \label{equ:limite}\end{equation}
we recover the desired equation:
\begin{equation}
{\partial \over \partial t}\langle E(x,y)\rangle =
D\left( {\partial ^2 \over \partial x^2}+{\partial ^2 \over \partial y^2}
\right) \langle E(x,y)\rangle +O(\sigma^3)\quad .
\label{equ:charlie2}\end{equation}
The particular choice $\chi\rightarrow 2D$ has its physical motivation in the
relation between a model on a discrete lattice and its continuous
version~\cite{NEXT}. Moreover, our closed equation is valid only in the
limit~(\ref{equ:limite}), which corresponds to an infinite pointlike coupling:
  \begin{equation}
V(\vert z-z^\prime \vert)\rightarrow 2\lambda \delta(z-z^\prime ),
\quad\hbox{with}\quad\lambda=v\sigma \rightarrow \infty\quad .
  \label{equ:lambdainfinito}\end{equation}
This fact is not surprising, considering that such a coupling
corresponds to an instantaneous coagulation on contact: this is exactly the
condition which was used in all the exact
solutions~\cite{RACZ,LUSH,BALD,SPOU,PRIV,GRYN}.

This aspect can also be nicely understood in the renormalization group
framework (see for example~\cite{LEE}). The limit $\lambda\rightarrow\infty$
corresponds to approaching the non-trivial fixed point of the theory and thus
to be in
the asymptotic long time regime. In consequence, we can interpret the
result~(\ref{equ:charlie2}), valid in the limits~(\ref{equ:limite})
and~(\ref{equ:lambdainfinito}), as an equation for the asymptotic long time
regime for a diffusion-coagulation problem, with arbitrary reaction rate.

\section{Generalization  to $\alpha$-reactions}

Following the results obtained in section 2, on the correspondence between
the coagulation model and an $\alpha$-model, we can now find the quantity
corresponding to $E$ for the $\alpha$-model and its equation of motion.

Let us define the operator:
  \begin{equation}
T^{(\alpha)}(x,y)=e^{-\alpha\int_{y}^{x} dz\psi(z)}\quad .
  \label{equ:defR}\end{equation}
Thus, ${\cal T}^{(\alpha)}(x,y)=\langle T^{(\alpha)}(x,y)\rangle$
satisfies the closed equation:
  \begin{equation}
{\partial \over \partial t}{\cal T}^{(\alpha)}(x,y,t)=
D\left( {\partial ^2 \over \partial x^2}+{\partial ^2 \over \partial y^2}
\right) {\cal T}^{(\alpha)}(x,y,t) +O(\sigma^3)\quad .
  \label{equ:charlie3}\end{equation}
To see the physical meaning of this operator, we consider a state
$\vert \Phi\rangle=\pi(z_1)\dots \pi(z_n)\vert 0\rangle$. It is then easy
to show that:
  \begin{equation}
\langle T^{(\alpha)}(x,y)\rangle_{\Phi} =\prod_{i=1}^n a_i\quad\hbox{with}\quad
a_i=\cases {1 &if $z_i\not\in (y,x)$, \cr 1-\alpha &if $z_i\in (y,x)$. \cr}
  \label{equ:interpretazioneT}
\end{equation}
${\cal T}^{(2)}(x,y)$, which is relevant for the annihilation case,
is related to the probability of having an even number of particles
in the interval between $y$ and $x$. For a generic vector of the form
$\vert \Phi\rangle=F[\pi]\vert 0\rangle$, we also have:
  \begin{equation}
{\cal T}^{(\alpha)}(x,y)=F[1-f^{(\alpha)}(z)]\quad ,
  \label{equ:TsuF}\end{equation}
where the function $f^{(\alpha)}$ is the generalization of~(\ref{equ:f}),
{\it i.e.}\/,
$$f^{(\alpha)}(z)=\cases{\alpha &if $y<z<x$, \cr 0& {\rm otherwise.}\cr}$$

Eq.~(\ref{equ:charlie3}) has been solved by Doering {\it et al.}~\cite{BUR},
with appropriate boundary conditions at $x=y$ and $x=\infty$ and the initial
condition ${\cal T}^{(\alpha)}(x,y,0)={\cal T}_0^{(\alpha)}(x,y)$. The
solution of this equation is particularly simple if the system is homogeneous.
{}From translational invariance ${\cal T}^{(\alpha)}(x,y,t)=
{\cal T}^{(\alpha)}(x-y,t)$ and eq.~(\ref{equ:charlie3}) becomes:
  \begin{equation}
{\partial \over \partial t}{\cal T}^{(\alpha)}(x,t) =
2D{\partial ^2 \over \partial x^2}{\cal T}^{(\alpha)}(x,t)\quad ,
  \label{equ:charlieomogenea}\end{equation}
with the conditions:
$${\cal T}^{(\alpha)}(0,t)=1\ , \quad
\vert {\cal T}^{(\alpha)}(\infty,t)\vert \leq 1\quad\hbox{and}\quad
{\cal T}^{(\alpha)}(x,0)={\cal T}^{(\alpha)}_0 (x)\quad .$$
The second condition follows from the fact that $T^{(\alpha)}(x,y)$
is a stochastic variable which takes its value in the
interval $[-1,+1]$.
 To take care of the  first two boundary conditions, it is judicious
to define:
  \begin{equation}
t^{(\alpha)}(x,t) \equiv
{\partial ^2 \over \partial x^2}{\cal T}^{(\alpha)}(x,t)\quad ,
  \label{equ:deft}\end{equation}
whose equation of motion is:
  \begin{equation}
\left\{ \begin{array}{ll}
{\displaystyle \frac{\partial}{\partial t}t^{(\alpha)}(x,t)=
2D{\partial ^2 \over \partial x^2} t^{(\alpha)}(x,t)}\quad ,\\ [.4cm]
t^{(\alpha)}(0,t)=t^{(\alpha)}(\infty,t)=0\quad .
  \end{array}\right.
  \end{equation}
The solution reads:
  \begin{equation}
t^{(\alpha)}(x,t)=\int_0 ^{\infty}dz\ G(x,z,t)t^{(\alpha)}_0 (z)\quad ,
  \label{equ:soluzionet}\end{equation}
where $G(x,z,t)$ is the Green function:
  \begin{equation}
G(x,z,t)=(8\pi Dt)^{-1/2} \left[{\rm exp}\left(-{(x-z)^2 \over 8Dt}\right)
-{\rm exp}\left(-{(x+z)^2 \over 8Dt}\right)\right]\quad .
  \end{equation}

We can now study the behavior of the particle density. From
eq.~(\ref{equ:defR}), it follows that the (homogeneous) density is:
  \begin{equation}
c_{\alpha}(t)=\langle\psi(0)\rangle=-\alpha^{-1}\left[{\partial \over \partial
x}
{\cal T}^{(\alpha)}(x,t) \right]_{x=0}
  \label{equ:densita}\end{equation}
and then from~(\ref{equ:deft}):
  \begin{equation}
c_{\alpha}(t)=\alpha^{-1}\int_0 ^{\infty}dz\ t^{(\alpha)}(z,t)\quad .
  \end{equation}
Using eq.~(\ref{equ:soluzionet})
we obtain:
  \begin{eqnarray}
c_{\alpha}(t)&=&\alpha^{-1}\int_0 ^{\infty}dz\ {\rm erf}
\left({z\over (8Dt)^{1/2}}\right)t_0^{(\alpha)}(z)\\
&=&\alpha^{-1}(2\pi Dt)^{-1/2}\int_0 ^{\infty}d\xi\ e^{-\xi^2}\xi
\left[1-{\cal T}^{(\alpha)}_0(2\xi t^{1/2})\right]\quad ,
  \nonumber\end{eqnarray}
whose asymptotic behavior for large $t$ is:
  \begin{equation}
c_{\alpha}(t)\sim
(2\alpha)^{-1}(2\pi Dt)^{-1/2}\left[1-{\cal T}^{(\alpha)}_0(\infty)\right]\quad
{}.
  \label{equ:densitaasintotica}\end{equation}
This result shows that, besides the trivial dependence on the factor
$(2\alpha)^{-1}$ which follows from the rescaling $\psi\rightarrow\alpha\psi$,
the asymptotic density amplitude depends also on the initial value
${\cal T}^{(\alpha)}_0(\infty)$. The study of this quantity for a general
initial state is a very difficult task. From~(\ref{equ:TsuF})
and~(\ref{equ:state}) we have:
  \begin{equation}
{\cal T}_0^{(\alpha)}(x)={\rm exp}\left[
\sum_{p=1}^{\infty}{(-\alpha)^p \over p!}{\bar g}_{(p)}(x)\right]\quad ,
  \end{equation}
where
$${\bar g}_{(p)}(x)=
\int_{0}^{x}dx_1 \dots \int_{0}^{x}dx_p\
g_{(p)}(x_1 ,\dots,x_p)$$
and it is difficult to predict the value ${\cal T}_0^{(\alpha)}(\infty)$,
without expliciting the $g_{(p)}$'s.

Let us then consider a translational invariant state having the following
properties:
  \begin{equation}
g_{(1)}(x)=c\ ,\quad g_{(2)}(x_1,x_2)=g(\vert x_1-x_2\vert)\quad\hbox{and}
\quad g_{(p)}=0\ ,\ \hbox{for}\ p\geq 3\quad .
  \label{equ:statoparticolare}\end{equation}
Such a state can be easily built, as we shall see later. Thus, we have:
  \begin{equation}
{\cal T}_0^{(\alpha)}(x)={\rm exp}\left\{ -\alpha cx +{\alpha^2 \over 2}
{\bar g}_{(2)}(x)\right\}\quad ,
  \label{equ:tinit}\end{equation}
with
$${\bar g}_{(2)}(x)=2x\int_{0}^{x}dz\ g(z)-2\int_{0}^{x}dz\ z\,g(z)\quad .$$
If we define
$$A=2\int_{0}^{\infty}dz\ g(z)\quad\hbox{and}\quad
B=\int_{0}^{\infty}dz\ z\,g(z)\quad ,$$
which are finite for short range correlations, one sees that
in the limit $x\rightarrow\infty$ three cases have to be distinguished:
  \begin{eqnarray}
1)\ \ \alpha c-{\alpha^2 \over 2}A>0 & : & {\cal T}_0^{(\alpha)}(\infty)=0
    \nonumber \\
2)\ \ \alpha c-{\alpha^2 \over 2}A<0 & : & {\cal T}_0^{(\alpha)}(\infty)=\infty
\label{equ:condit}\\
3)\ \ \alpha c-{\alpha^2 \over 2}A=0 & : & {\cal T}_0^{(\alpha)}(\infty)
    =e^{-\alpha^2B} \nonumber
  \end{eqnarray}
The case $2)$ is unphysical, because the particle density must be positive.
Thus the value of $A$ is such that $\alpha A\leq 2c$ for all $\alpha\in[1,2]$,
which implies $A\leq c$. This constrain is a consequence of the assumption
$g_{(p)}=0$ for $p\geq 3$ and $B<\infty$. This also means that the case $3)$
is only realizable for $\alpha=2$, $A=c$. From~(\ref{equ:densitaasintotica})
it follows that for an initial state of the type~(\ref{equ:statoparticolare}),
we have the same asymptotic density amplitude for all the $\alpha$-models
with $1\leq \alpha <2$ and for the pure-annihilation case with $A<c$. But in
the pure-annihilation case with $A=c$, the amplitude depends on the value of
$B$.

We now show how to built  a state of the form
(\ref{equ:statoparticolare}) and understand its physical content.
Let us take a Poissonian state with density ${\hat c}$:
$$\vert P\rangle =e^{{\hat c}\int dz\ (\pi(z)-1)}\vert 0\rangle$$
and the state $\vert \Phi_{\beta}\rangle$ obtained by the replacement:
  \begin{equation}
\pi(z) \rightarrow \pi(z)\left
[\beta+{1 \over 2{\hat c}}\int dy\ \pi(z+y)g(y)\right]\ \ \ \ \ \
0\leq \beta \leq 1
  \label{equ:b}\end{equation}
The normalization $\langle \vert \Phi_\beta\rangle=1$ implies that:
$$A=2\int_{0}^{\infty}dy\ g(y)=2(1-\beta){\hat c}\quad .$$
The modification (\ref{equ:b}) means that we go from the state
$\vert P\rangle$, where particles
are randomly distributed, to the state $\vert \Phi_{\beta}\rangle$,
where both single particles and pairs are
randomly distributed.
The probability density that a pair have an extension
$\sigma$ is ${1 \over 2{\hat c}}g(\sigma)$.
If $\beta=0$ there are only pairs in the state $\vert \Phi_0\rangle$ .
A straightforward calculation gives:
$$\vert \Phi_{\beta}\rangle={\rm exp}\left\{ (2-\beta){\hat c}\int dz\
(\pi(z)-1)+{1 \over 2}\int dzdy\ g(z-y)(\pi(z)-1)(\pi(y)-1)\right\}\quad .$$
Thus we have a state of the type~(\ref{equ:statoparticolare}) with density
$c=(2-\beta){\hat c}$. For $\beta=0$, we have $A=c$. The mean extension of
a pair is:
$${\bar \sigma}={1\over 2{\hat c}}\int_0^{\infty}d\sigma\ \sigma g(\sigma)=
{(2-\beta)B\over 2c}\quad .$$
Our $\alpha$-models, with initial correlations can now be parametrized in
terms of $(\alpha,\beta,{\bar\sigma})\in [1,2]\times [0,1]\times [0,\infty)$.
The asymptotic density amplitudes are universal for all the values of the
parameters, except on the line $(\alpha=2,\beta=0,{\bar\sigma})$ where the
amplitude is proportional to $(1-e^{-4c{\bar\sigma}})$. In this case, the
asymptotic density amplitude tends to zero if ${\bar \sigma}\rightarrow 0$.
The physical reason is that when ${\bar \sigma}=0$, the two particles of a pair
are at the same place and immediately annihilate. This situation cannot occur
if
coagulation is present, because single isolated particles will remain.

It would be interesting to see how a more general initial state would affect
the amplitudes. Note finally, that the relations~(\ref{equ:condit}),
characterizing the initial state, are preserved by the dynamical evolution.
Indeed, every evolved state can be considered as a new initial state evolving
towards the same asymptotic solution.

\vskip 1truecm
\section{Conclusions}

We have studied a class of $d$-dimensional diffusion-reaction models
interpolating continuously between the diffusion-coagulation and the
diffusion-annihilation models. The field theoretical approach used leads
to exact relations between the observables of these different models. In one
dimension, it was shown how correlations in the initial state can lead to a
violation of the universality of the amplitude for the DC models.

Several extension of the present work concerning one-dimensional models with
reversible diffusion-reaction systems, or the presence of fronts in
inhomogeneous systems are under investigation. Moreover, one may expect
to construct ``good observables'' in $d$-dimensional systems for which a
closed equation of motion can be derived.

\section*{Acknowledgements}

We thanks Z. R\'acz for many useful discussions.

redpajama_set_name: RedPajamaArXiv


Sylvaner grape – a white wine variety grown in the region of Alsace.
Sylvaner grape is a white wine variety, grown in and around the regions of Alsace of Germany, where its official name is also called Gruner Sylvaner.
The grape is an old variety, with its DNA fingerprinting dating it back to as old as the middle ages, around the 1600s when its first ever written records have been found and uncovered.
Being an old grape, and going through a lot of historical events of all these years, Sylvaner suffered several setbacks as well as economic booms and golden ages throughout its existence.
It is now one of them most well known and popularly grown white grape varieties in the world and some of the biggest exporters include Australia, France, Germany and the US, including of course other notable regions like Slovakia, Romania and Croatia among others that produce Sylvaner to an appreciable number.
When fully ripened, the grape produces rather fine wines that can reach to almost blandness when not controlled as far as its yields are concerned.
However, during good yields, the grape gives off some good quality grapes that have excellent musts. Sylvaner, because of this reason, and its high acidity, is preferred and used to a considerable extent as a blending grape.
Sylvaner is an old variety that has been in cultivation in Central Europe at least since the ancient times, however, its first ever records go back to the middle ages.
Certain DNA analysis has given proofs that the grape was a cross of Traminer and Osterreich-Wiess, the 'Austrian white' variety.
For how long it has existed since then, we cannot be specific about it, however, since 'hunnic' varieties, which in this case is the Osterreich-Weiss, are ancient varieties, first being introduced by the Huns when they started settling down after pushing back Eastern Roman Empire's territories around Romania and Hungary, we can thus be quite certain that the grape is of ancient lineage.
Sylvaner's real history however, starts from the middle ages, around 1600s, during the existence of thirty years war that included a lot of Germanic states as its participating combatants.
During this time, since Austria was one of the main participants in the conflict, the grape seemed to have been transferred to such territories where it could be grown in comparatively safer environments.
It is thus thought, without any credible evidence really, that the grape came to Germany after the war ended, since there are records that state that the Country of Castell in Franconia, dated 5 April, 1659, cultivated the Sylvaner grape.
In the coming decades, the Sylvaner grape witnessed ups and downs through its German winemaking history. Being cultivated in a colder country than its southern European counterparts, the Sylvaner could never really be affected by the phylloxera epidemic that wrecked havoc with European vineyards in the 1880's.
However, the grape underwent some major economic declines in the 1970s, when, after being Germany's most planted wine variety; it got replaced by such varieties like the Muller-Thurgau.
The grape suffered a drastic decrease in its popularity during this time as other events started piling up; the grape lost its preference due to certain yields being over productive, resulting in the grape losing its characteristics and ultimately, changing tastes led to several of its wines being stopped arriving at a full stop.
Only some regions, however, like Franconia, still produce Sylvaner, its blending into Liebfraumilch being appreciated even today.
Ever since the disastrous 1970s, Sylvaner has, however, kept up its popularity index, simply because of the high quality characteristics it produces when cultivated in a proper manner and when its yields are set under control.
In recent decades, however, the Sylvaner has seen some resurgence in the Alsace region, especially with its varietal purpose being produced more rather than its blending one. In 2006, the grape was officially recognised as the first to be designated an Alsace Grand Cru.
Sylvaner wines can have certain differing traits and characteristics according to the region and soil the grape has been cultivated in.
Italian and Swiss Sylvaners typically have much lighter and crisper flavours than their Alsatian and Franken counterparts, and these can be paired up well with a lot of items from Germany and Italy.
An Alsatian onion tart or even simple desserts are things that should definitely be consumed well with a Sylvaner wine.


text:
Hanna Kim Trio & Ben Matthews presented by PJS
This event has passed!
PJS Presents a double bill performance from pianists Ben Matthews and Hanna Kim. Both pianists will perform a diverse set of original music. Hanna Kim Trio Hanna Kim was born in South Korea and started studying piano at the age of six.
She graduated with a Bachelor of Music degree in Jazz Piano Performance from Seoul Art College and performed regularly in Korea before moving to Perth in 2014. Recently, she formed Hanna Kim Trio and has begun composing music towards her own album. Hanna Kim Trio is comprised of Francisco Javier Munoz Melo on drums and Jonathan Chen on bass guitar. Hanna Kim Trio is a melting pot of different musical cultures; Korean composer and pianist, Chilean drummer and Malaysian bassist.
Experience the varied and beautiful music that the trio plays by blending their own musical backgrounds through jazz, influenced by artists such as Roy Hargrove, Yellow Jackets and Fourplay. Ben Matthews Ben Matthews was drawn to the piano at a young age, following a passion for artistic expression through improvisation and composition. He performs in a unique style that blends genres drawing influence from jazz legends like Herbie Hancock and Oscar Peterson, and from soundtrack titans like John Williams and Joe Hisaishi (Studio Ghibli) to other pianistic composers like Ludovico Einaudi and Ryuichi Sakamoto.
After a decade performing as both a sideman and bandleader with some of Australia's top-tier musicians, WAAPA graduate jazz pianist and contemporary keyboardist Ben Matthews turned his focus to composition in late 2015. In 2016, he wrote and recorded the following projects: an original jazz soundtrack to THQ Nordic's game 'This Is The Police'; an original jazz/instrumental neo-soul double-CD release, 'LiveWyre'; a collaborative E.P. of contemporary ballads with vocalist Edwina Masson entitled 'An Open Letter of Imperfections'; a seven track, entirely improvised solo piano album, 'My Story'; music for the short film 'Lightbulb'; and two improvised solo piano singles, 'Indulge in a Daydream' and 'Uncertain Times'. In 2017 a second solo piano album 'Memoir' was improvised.
In early 2018, Ben was flown to Norway, to compose and record for game title "This Is The Police 2", creating a rich and diverse soundtrack featuring a blend of nordic folk instruments, world music, 60's jazz music and hard fusion-rock. This project included collaborations with composer Kevin Penkin (Made in Abyss, Florence the game), producer Jia Lih, vocalist Holli Scott. Over the years, five of Ben's compositions have been nominated as finalists for Western Australian Music (WAM) Song of the Year awards (2011, 2015, 2016, 2018).
His latest release, 'This Too Shall Pass' (2018) is a third improvised solo piano album, exploring concepts of darkness, balance, gesture and storytelling through music. Ben's current aspiration is to create honest and emotional music that balances creativity and accessibility.
7.00 - 10.00pm Wednesday, 17 July
The Ellington Jazz Club
191 Beaufort St

redpajama_set_name: RedPajamaCommonCrawl


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Experience the life of Louisiana's most ancient residents- ALLIGATORS! See them in crystal clear water in the climate controlled environment of a working alligator ranch, home to more than 2,000 gators. Your guided tour gives you a first-hand account of the Louisiana Alligator Industry, from hatchling to handbag.


text:
FilmFracture
Movie News, Movie Reviews, and Features With Your Time in Mind
News and Curiosities
Cinema Fearité
What To Expect When You're Expecting
May 12, 2012 by Kathryn Schroeder
Synopsis: Inspired by the perennial New York Times bestseller of the same name, WHAT TO EXPECT WHEN YOU'RE EXPECTING is a hilarious and heartfelt big-screen comedy about five couples whose intertwined lives are turned upside down by the challenges of impending parenthood.
Over the moon about starting a family, TV fitness guru Jules and dance show star Evan find that their high-octane celebrity lives don't stand a chance against the surprise demands of pregnancy. Baby-crazy author and advocate Wendy gets a taste of her own militant mommy advice when pregnancy hormones ravage her body; while Wendy's husband, Gary, struggles not to be outdone by his competitive alpha-Dad, who's expecting twins with his much younger trophy wife, Skyler. Photographer Holly is prepared to travel the globe to adopt a child, but her husband Alex isn't so sure, and tries to quiet his panic by attending a "dudes" support group, where new fathers get to tell it like it really is. And rival food truck chefs Rosie and Marco's surprise hook-up results in an unexpected quandary: what to do when your first child comes before your first date?
Watch clips and more from the film here.
Release Date: May 18, 2012 MPAA Rating: PG-13
Genre(s): Comedy, Drama
Every film deserves a shot at appealing to a wide audience that includes the unintended viewer; What To Expect When You're Expecting never stood a chance. Following five women during their pregnancies already screams "chick flick," the inclusion of a dude's group helps to bring in the male demographic only slightly. What To Expect When You're Expecting is a movie about women, for women, and surrounded by women. What it is not is a fleshed out narrative for any one story in particular; each individual scenario is consistently cut short, forgotten, and then returned to in fits of time lapses to reveal again a brief snippet of said woman's experience with pregnancy, only to be looked over again and again.
The five women taking on pregnancy in What To Expect When You're Expecting all have different experiences, and the film does do a great job in outlining that no one trip towards motherhood is the same. The opening of the film introduces the viewer to one of the pregnant women in question, Jules (Cameron Diaz). It also sets the tone for the film and introduces the one connection all of the women will have throughout–reality television programming. The tone falls right in line with reality television, an inside look into impending motherhood without editing out the less than desirable side or the tragedy that can befall a pregnancy. What To Expect When You're Expecting takes the honesty route, from the women and the men's perspectives; those without children beware, the film may make you re-think parenthood. Jules' opening scene takes place during the final dance-off for the show "Celebrity Dance Factor," where her and her partner/father-to-be Evan (Matthew Morrison from "Glee") await the final decision on who will be crowned the winner. Jules is feeling nauseous, but the star of her own losing weight reality television show does not quit, and she performs her dance number regardless. They say one of the first signs of pregnancy is vomiting and Jules discovers this the hard way, and most embarrassing, on live television. The remaining four women will not suffer live-television humiliation but they will all struggle in their own way–besides one.
Holly (Jennifer Lopez) desperately wants to start a family with her husband Alex (Rodrigo Santoro) but her ovaries are not cooperative; adoption is the route being taken by Holly and Alex. While Holly may appear to be the character focused upon in this relationship it is actually Alex that has the stronger storyline. He is unsure about fatherhood, unprepared and afraid. The solution is to send him to "Dude's Group." Dude's Group is a group of father's, led by Chris Rock's Vic, who meet for a walk in the park each week to share their dad experiences. The second rule of Dude's Group is that you do not share with anyone what is said during Dude's Group–obviously they do not walk around praising their lives as fathers or husbands making for some key comedic moments. Shifting the greater focus of motherhood onto fatherhood in this particular relationship works well for the story overall. Holly is far too eager and excited to be a mother, and there are no bumps along the way in their adoption process to warrant any sort of story arc for the couple.
Brooklyn Decker's twenty-something Skyler wasn't trying for a baby with MUCH older husband Ramsey (Dennis Quaid). If there is one woman to dislike in What To Expect When You're Expecting it is Skyler, due to the fact that she is having the perfect pregnancy, sans any of the common ailments, including weight gain. At seven months Skyler is frolicking around in a teeny tiny bikini at her pool party baby shower. Her daughter-in-law, who is older than Skylar, naturally, is having the complete opposite pregnancy experience. Wendy (Elizabeth Banks) has been trying to conceive with her husband Gary (Ben Falcone) for the past two years; a frisky night full of wine and a movie in the park finally does the trick. But pregnancy is not the dream experience Wendy has been selling at her baby-themed store all these years, or in her children's book. It is a nightmare, and Banks' performance is what gives What To Expect When You're Expecting the majority of its comedic moments. There is humor in misery, and pregnancy has all sorts of avenues to address of the comedic sort.
Rounding out the five women is Rosie (Anna Kendrick), the poster child for why safe-sex is important. A one-night stand with Marco (Chace Crawford), whom she has known since high school, results in a pregnancy neither one of the food-truck runners were expecting. Rosie and Marco are the tragic story in What To Expect When You're Expecting, finding out that deciding to have a baby is just as difficult as losing one you never exactly wanted. Kendrick and Crawford do their absolute best in displaying the torn feelings over their situation given the soapy dialogue and awkward meetings between the two characters. Their story is far more engrossing even with its limited screen time than Diaz's Jules and Morrison's Evan fighting over whether to circumcise their son. The melodrama wins out over the bitchy type-A head-butting.
What To Expect When You're Expecting is full of memorable moments, showcasing the ups and downs and myriad of emotions for both men and women when expecting a child. It is the execution of the film that causes it to suffer; never clearly defining one story from the next and needlessly trying to intersect each one in unnecessary ways. The women are engaging, the men more than likely forgettable, but there is heart in the picture. The modern landscape of our times presents varying ways in which people have children, and very different approaches to parenthood for each individual. What To Expect When You're Expecting does its best to highlight them all but simply falls into the trappings of serial programming and soap-opera style execution, never letting the characters stories move beyond carefully selected glimpses of a journey.
Comedy Factor
Pregnancy is a beautiful thing, when you're not pregnant–this one statement could describe Elizabeth Banks' Wendy's experience with pregnancy. Thankfully for the viewer Wendy is having a terrible time being pregnant because her character provides nearly all of the memorable comedic moments in What To Expect When You're Expecting. Wendy is very honest about what is happening to her body, and a tad to much information is given repeatedly about her ordeal that gives way to laugh-out-loud scenes scattered throughout the film. Discussing peeing yourself a little, uncontrollable gas, backne, hemorrhoids, and all sorts of other inappropriate topics in conversation is what Wendy does, and Banks nails each and every line and even more so with her facial expressions showcasing her absolute misery and uncomfortableness. What To Expect When You're Expecting deals with all the types of pregnancies you can have, but it is the most miserable of them all that gets the most laughs, all thanks to Elizabeth Banks.
The other slightly humorous part of the film, besides the occasional with the women other than Wendy, comes in the form of the "Dude's Group." Led by Chris Rock's Vic the Dudes are not a group of men to be envied; they stroll through the park with strollers, kids strapped to their chests, bottles at the ready, and look ridiculous. They are a group of dad's who envy the single guy they run into all the time, Joe Manganiello's Davis, and worship at his feet. The stories the dad's tell about their lives with children are oftentimes funny, and make the Dude's Group scenes more than bearable. You would think Chris Rock would have more great lines than he does, or that Thomas Lennon's Craig would not be subjected to repeat the same joke over and over again about his son's name or the need to get a high-five from Davis. But the Dude's do have their comedic moments, more subtle than brass, but overall mildly entertaining.
Director(s): Kirk Jones
Producer(s): Mike MedavoyArnold MesserDavid Thwaites
Screenwriter(s): Shauna CrossHeather Hach
Cast: Cameron Diaz (Jules)Jennifer Lopez (Holly)Elizabeth Banks (Wendy) Chace Crawford (Marco)Brooklyn Decker (Skyler)Ben Falcone (Gary)Anna Kendrick (Rosie)Matthew Morrison (Evan)Dennis Quaid (Ramsey)Chris Rock (Vic)Rodrigo Santoro (Alex)Joe Manganiello (Davis)
Cinematographer: Xavier Perez Grobet
Production Designer(s):
Casting Director(s):
Music Score: Mark Mothersbaugh
Music Performed By:
Filed Under: Movie Review
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/// Kathryn Schroeder time to read: 6 min
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