text:
\section{Introduction}

The absence of a quantum theory of gravitation is probably one of the
most important open
problems in theoretical physics. In fact all of our knowledge about
gravitation refers only
to its classical aspects which are well described by General Relativity
and other
generalizations of this theory with the same low-energy limit. On
the other hand, the other
interactions (strong and electroweak) are well described, even at
the quantum level,
by the Standard Model, as it has been confirmed in detail in the
last years at the Large
Electron-Positron Collider (LEP) and many other experiments. Moreover
it seems
reasonable to think that there may exist some physical regime where
the gravitational
field could be treated classically whereas matter is quantized.
In fact this is the only regime that can be studied by now without
making extra hypothesis
beyond what has already been checked experimentally.
In this very conservative approach one considers, on the one hand, the
quantum gauge and matter
fields propagating on a curved space-time and, on the other hand,
the dynamics of the
classical degrees of freedom associated to the space-time. This is
the so-called
 semi-classical
approximation. It is clear that space-time curvature or torsion
(see \cite{Hehl,shap3} for a review) can
affect the quantum
fields in different ways. In addition, space-time is also affected by
the presence of the
quantum matter fields, as it happens, for instance, in
 the electromagnetic field case, where fermion loops modify the
electromagnetic dynamics by means of the
vacuum polarization and other
effects.

In order to study these new effects due to matter fields loops,
it is specially appropriate the use of the gravitational effective action
(EA). This
EA is obtained after integrating out the matter fields. In general
the EA
will be a nonlocal and non-analytical functional of the metric and
the connection.
The exact expression for the EA, obtained after integrating out the
matter
 fields $W[g_{\mu\nu}]$
 is not known for arbitrary space-time geometries. However there
are several techniques
 that have been proposed for its approximate calculation, namely:
perturbation theory, in which the metric tensor is divided in
a flat metric and a
small fluctuation.
The main difficulty of this method is that general covariance is explicitly lost
\cite{thve,capp,deser}. The Schwinger-DeWitt proper time representation \cite{Schw,DeWitt}
allows us to obtain a covariant asymptotic expansion for the EA, but as far as it is local,
it does not allow
to describe particle creation processes.
These difficulties are intended to be solved by means of the partial
resummation
of the Schwinger-DeWitt series \cite{Avramidi}.
There are also perturbative methods that respect
general covariance,
such as the
so called covariant perturbation theory \cite{Vilkovisky}.
Finally the local momentum representation \cite{BunchParker} is
based on the
 Riemann normal
coordinate expansion. This technique has the advantage of combining
the usual flat space-time
methods with formally covariant expressions. It has allowed to calculate
the
divergent local parts of the one loop effective action for scalar and
fermionic theories. The main aim of the present work is  to find a
representation of the {\it nonlocal finite} parts of the effective action in
this formalism.

Once we know the EA, we have all the information concerning the
semi-classical gravitational evolution. The corresponding equations of
motion will modify
the Einstein field equations taking into account quantum effects. Moreover,
the EA could have
a non-vanishing imaginary part, which can be interpreted as the particle
production probability  \cite{Schw,art2}.

In this work we show our results concerning the computation of the one-loop
EA after
integrating out scalar and fermionic fields by using the
local momentum representation. Our computation includes not only the
divergences but
also the nonlocal finite terms that can lead to instabilities of the
classical
solutions by particle emission. As we will show, the use of normal coordinates
allow us to obtain a useful representation of  the nonlocal form factors.
Such representation has been succesfully applied in a recent work to
obtain particle production in different cosmological contexts in a
remarkably easy way \cite{art2}. In addition, the results will
shed some light about the boundary conditions on the metric tensor which
permit a definition of the effective action.

The work is organized as follows. In Section 2 we do a brief review of
the Euler-Heisenberg
effective lagrangian for Quantum Electrodynamics (QED). This model will
be a guide for the
calculation of the corresponding
gravitational EA. In Section 3 we present the method to generate the
 derivative expansions
of the EA by means of Riemann normal coordinates. Applying such method to
 the scalar theory
in the presence of gravitation, we obtain the divergences as well as the
finite nonlocal
pieces of the EA up to quadratic terms in the curvature. In Section 4 we
do the same with fermionic  matter fields, obtaining
the corresponding gravitational EA up to quadratic terms in the curvature and
in this case also in the torsion. In particular we apply our results to the
 Standard Model particle
 content. Finally
 Section 5 contains
the main conclusions of this work an a brief discussion on their possible
applications to the
computation of particle production probabilities.
We have also included an Appendix
containing
 the
dimensional regularization formulae and some normal coordinates expansions
 used in the text.

\section{The Euler-Heisenberg lagrangian}
The historical origin of the semi-classical EA can be traced back to the
Euler-Heisenberg lagrangian for QED \cite{EulerHeisenberg}. When the
momentum $p$ of
photons  is
much smaller than the electron mass $M$, the one-loop effects, such as
vacuum
polarization, can be taken into account by adding local non-linear terms
to the classical
electromagnetic lagrangian. Consider the QED EA given by:
\begin{eqnarray}
e^{iW[A]}=\int [d\psi][d\overline\psi]e^{-\frac{i}{4}\int d^4x
F_{\mu\nu}F^{\mu\nu}}
\exp\left(i\int d^4x\overline \psi (i\not{\!{\!D}}-M+i\epsilon)\psi\right)=
e^{ -\frac{i}{4}\int d^4x F_{\mu\nu}F^{\mu\nu}} \det (i\not{\!{\!D}}-M+i\epsilon)
\end{eqnarray}
where as usual
 $\not{\!{\!D}}=\gamma^\mu(\partial_\mu-ieA_\mu)$. Using dimensional
regularization (see Appendix), it is possible to find the following
expression up to quadratic
terms in the photon field:
\begin{eqnarray}
W[A]&=&-\frac{1}{4}\int d^4x F_{\mu\nu}F^{\mu\nu}-i\mbox{Tr} \log((i\not{\!{\!D}}-M
+i\epsilon))
=-\frac{1}{4}\int d^4x F_{\mu\nu}F^{\mu\nu}
+i\sum_{k=1}^{\infty} \frac {(-e)^k}{k}\mbox{Tr}[(i{{\not{\!{\!\partial}}}} -M)^{-1} \not{\!{\!A}}]^k
\nonumber \\
&=&
\int d^4x \left[-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}-\frac{e^2\Delta}{3(4\pi)^2}
F^{\mu\nu}F_{\mu\nu}
+\frac{2e^2}{(4\pi)^2}F^{\mu\nu}\left(\frac{2}{3}\frac{M^2}{\Box}\right.
\right.
+\left.\left. \frac{1}{6}\left(1-2\frac{M^2}{\Box}\right)F(-\Box;M^2)\right)
F_{\mu\nu}\right]
+{\cal O}(A^4)
\label{ehea}
\end{eqnarray}
where $\Delta=N_\epsilon-\log (M^2/\mu^2)$, with
$N_\epsilon=2/\epsilon+\log 4\pi -\gamma$
and $\gamma\simeq0.577$ is the Euler constant. We
have performed the formal Taylor expansion of the logarithm and used the
expression:
\begin{eqnarray}
F(-\Box;M^2)F_{\mu\nu}(x)=\int d^4y\frac{d^4p}{(2\pi)^4}e^{ip(x-y)}F(p^2;M^2)
F_{\mu\nu}(y)
\end{eqnarray}
with:
\begin{eqnarray}
F(p^2;M^2)=2+\int_0^1 dt \log\left(1-\frac{p^2}{M^2}t(1-t)\right)
\label{mand1}
\end{eqnarray}
In a similar way, the inverse operator $1/\Box$ can be defined with the usual
boundary
conditions on the fields as:
\begin{eqnarray}
\frac{1}{-\Box}F_{\mu\nu}(x)=\int d^4y\frac{d^4p}{(2\pi)^4}e^{ip(x-y)}
\frac{1}{p^2+i\epsilon}F_{\mu\nu}(y)
\end{eqnarray}
The expression (\ref{ehea}) for the EA is nonlocal and has a regular massless limit. In fact,
for
small $p$ compared with $M$, the Mandelstam function $F(p^2;M^2)$ behaves as:
\begin{eqnarray}
F(p^2;M^2)= -\log \left(\frac{M^2}{-p^2-i\epsilon}\right)+{\cal O}(M^2)
\label{mandaprox}
\end{eqnarray}
From (\ref{ehea}) we can see that the only contributions in the massless limit
are
those coming, on one hand from the  $\Delta$ factor and, on the other hand,
from the
Mandelstam function. Both logarithmic contributions equal, up to  sign,
so that they cancel each other and we obtain:
\begin{eqnarray}
W[A]=\int d^4x \left(-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}
-\frac{e^2}{3(4\pi)^2} F^{\mu\nu}\Gamma(\Box)F_{\mu\nu}\right)+{\cal O}(A^4)
\label{eheam0}
\end{eqnarray}
where we have used the following notation:
\begin{eqnarray}
\Gamma(\Box)=N_\epsilon-\log\left(\frac{\Box}{\mu^2}\right)
\label{gama}
\end{eqnarray}
to be understood as in the previous cases through the corresponding Fourier
transform, with
the $i\epsilon$ factor as shown in (\ref{mandaprox}).

The EA (\ref{ehea}) allows us to derive in an exact fashion the photon
two-point
one loop Green functions. This, in turn,
allows us to obtain for example the vacuum polarization. The EA can be
expanded
as a power series in $p^2/M^2$, and also in $A$ to obtain the well-known
Euler-Heisenberg local
lagrangian
\cite{EulerHeisenberg}

\section{Integration of matter fields in a gravitational background}

Consider a scalar field in a curved space-time. The corresponding classical
action is given
by:

\begin{eqnarray}
S[\phi]=-\frac{1}{2}\int d^4x\sqrt{g} \phi\left( \Box+m^2+\xi R \right) \phi
\label{acsc}
\end{eqnarray}
where
$\Box \phi=g^{\mu\nu}\nabla_{\mu}\partial_{\nu}\phi
=g^{-1/2}\partial_{\mu}\left(g^{\mu\nu}\sqrt{g}\partial_{\nu}
\phi\right)$.
The non-minimal term  $\xi R$ is included so that for
$m=0$ and $\xi=1/6$, the classical lagrangian is invariant under
local conformal transformations.

The EA for the gravitational fields that arises after integrating out
real scalar matter fields is given by the following expression in Lorentzian
 signature:
\begin{eqnarray}
e^{iW[g_{\mu\nu}]}=\int [d\phi]e^{iS[g_{\mu\nu},\phi]}=
  \int [d\phi]e^{-\frac{i}{2}\int d^4x\sqrt{g}\phi(\Box+m^2+\xi R
-i\epsilon)\phi}=
(\det O)^{-1/2}
\label{eaes}
\end{eqnarray}
where
$O_{xy}(m^2)=(-\Box_y-m^2-\xi R(y) +i\epsilon)\delta^0(x,y)$ with
$\delta^0(x,y)$ being
the covariant delta $\delta^0(x,y)=g^{-1/2}(x)\delta(x,y)$. Therefore we
can write:
\begin{eqnarray}
W[g_{\mu\nu}]=\frac{i}{2}\log \; \det O(m^2)=\frac{i}{2}\mbox{Tr} \;\log O(m^2)
\label{esceff}
\end{eqnarray}
Since in this expression we have only integrated the scalars out, the
gravitational field
is treated classically. Accordingly, this EA is analogous to the classical
action but including the quantum effects due to the matter fields. In
addition, (\ref{eaes})
is the generating functional of the Green functions containing only scalar
loops and
gravitational external legs.

Let us now consider a fermion field propagating in a curved space-time with
torsion
\cite{Hehl}. The corresponding classical action will be given by:

\begin{eqnarray}
S=\int d^4x \sqrt{g}\overline\psi(i\not{\!{\!D}} -M)\psi= \int d^4x \sqrt{g}
\overline\psi\left[i\gamma^{\mu}\left(\partial_{\mu}-\frac{i}{2}
\Gamma^{a\;b}_{\mu}\Sigma_{ab}+\frac{i}{8}S_{\mu}\gamma_5\right)-M\right]\psi
\label{lm}
\end{eqnarray}
where $S_{\mu} =
\epsilon_{\rho\nu\lambda\mu}T^{\rho\nu\lambda}$
is the torsion pseudotrace, $\Gamma^{a\;b}_{\mu}$ are the Levi-Civita
spin-connection
components and $\Sigma_{ab}$ are the Lorentz group generators.
The fermionic EA is given by:
\begin{eqnarray}
e^{iW[e,\hat \Gamma,A]}=
  \int [d\psi d\overline\psi]\exp \left(i\int d^4x \sqrt{g} \;\overline
\psi (i{\not{\!{\!D}}}
- M+i\epsilon)\psi\right)=\det (i\not{\!{\!D}}-M+i\epsilon)
\label{eafe}
\end{eqnarray}
where $e$ denotes the vierbein, $\hat \Gamma$ the full connection
with torsion and $A$ collectively denotes the possible gauge fields.
Accordingly:
\begin{eqnarray}
W[e,\hat \Gamma,A]=-i\log \; \det (i\not{\!{\!D}}-M+i\epsilon)=-i\mbox{Tr}\log(i\not{\!{\!D}}
-M+i\epsilon)
\label{fermion}
\end{eqnarray}
In the massless limit, $M=0$, the classical fermionic lagrangian is also
conformally invariant.

Since the above model does not posses self-interactions, the one-loop
calculation is exact.
This does not mean that  it is possible
to explicitly calculate
the EA for an arbitrary space-time geometry. In those cases in which
there is
a high degree of symmetry, as in maximally symmetric manifolds, or in
the so called
conformally trivial situations, i.e, conformally flat manifolds and
conformally invariant
theories, it is possible to find the explicit form of the modified
Einstein equations
coming from the EA \cite{BiDa}.

In order to consider more general geometries, we will use an approximation
scheme similar
to the one used for the Euler-Heisenberg lagrangian. It consists in treating
the curvature
as a small perturbation. When we integrate massive fields out, this is
equivalent to consider
that the Compton wavelength corresponding to the massive particle is much
smaller than the
characteristic length scale of the gravitational field. In this framework,
the expression for
the EA will be an expansion in metric tensor derivatives over the
particle mass. Such expansion will be generated by the normal coordinates
expansions of
the  $O(m^2)$ and
$(i{\not{\!{\!D}}}-M)$ operators.
In the massless case, or if we are interested in the high-energy regime,
it is possible
to obtain an alternative expansion in powers of the curvatures (Riemann
and Ricci tensors
and scalar curvature), generically denoted ${\cal R}$.

\subsection{Riemann normal coordinates}

Let $x^\alpha_0$
be the coordinates of $P$ in a given coordinate system and consider the
set of
geodesics passing through $P$, that we will write as $x^\alpha(\tau)$.
We will choose the $\tau$ parameter in such a way that
$x^\alpha(0)=x_0^\alpha$. Each of these geodesics will be characterized
by the tangent vector
at $P$,
\begin{eqnarray}
\xi^\alpha=\left.\frac{dx^\alpha}{d\tau}\right\vert_{\tau=0}
\end{eqnarray}
and each point $A$ on each geodesic by certain value of the $\tau$ parameter.
The Riemann coordinates $y^\alpha$ of $A$ are defined as
$y^\alpha=\xi^\alpha \tau$ \cite{Petrov}. In a  neighborhood of $P$
where any other point $A$
can be joined to $P$ by a unique geodesic (normal neighborhood), the
correspondence between
the $x^\alpha$ and $y^\alpha$ coordinates is one to one.

When torsion is present and it is completely
antisymmetric,
the geodesic equation agrees with that obtained using
the Levi-Civita connection and therefore, some given Riemann coordinates
respect to the Levi-Civita
connection will also be Riemannian respect to the connection with torsion
\cite{Eisenhart,CogZer}.
 By means of a linear real homogeneous
coordinate transformation, it is possible to write the metric tensor at $y_0$
in the Minkowski form
$\eta_{\mu\nu}$.
The new coordinates are also Riemannian
and they are known as Riemann normal coordinates.

Let us consider the components of some tensor field at $y$, that we
will assume to be
analytic functions in a neighborhood of the normal coordinates origin $y_0$.
From their Taylor expansion around $y_0$ we can obtain
an expression
that is written as a series in curvatures and their covariant derivatives.
In paricular for the metric tensor componentes we get \cite{Petrov}:
\begin{eqnarray}
g_{\mu\nu}(y)&=&\eta_{\mu\nu}+\frac{1}{3}R_{\mu\alpha\nu\beta}(y_0)y^\alpha
y^\beta-
\frac{1}{6}R_{\mu\alpha\nu\beta ;\gamma}(y_0)y^\alpha y^\beta y^\gamma
+
\left[\frac{1}{20}R_{\mu\alpha\nu\beta ;\gamma\delta}(y_0)
+\frac{2}{45}R_{\alpha\mu\beta\lambda}(y_0)
R^{\lambda}_{\;\gamma\nu\delta}(y_0)\right]y^\alpha y^\beta y^\gamma y^\delta
\nonumber \\
&+&{\cal O} (\partial^5)
\label{normal}
\end{eqnarray}
Here, ${\cal O} (\partial^5)$ denotes terms with 5 or more derivatives and
the indices in those tensors evaluated in
$y_0$ are raised and lowered with the flat metric $\eta_{\mu\nu}$.
In the Appendix, we have written the expansions corresponding to the metric
 determinant and
other useful expressions.
It is important to note that  (\ref{normal})
is a superposition of two kind of expansions: on one hand, terms are organized
 by the
number of metric derivatives, but on the other hand, those terms with a certain
 number of
derivatives can be classified according to the number of curvature tensors they
 have.

It will also be useful to introduce the following relation:
$2\sigma(x,x')=y_\alpha y^\alpha$,
where the biscalar $\sigma(x,x')$ represents half of the geodesic distance
between the
$x$ and $x'$ points and
$y^\alpha$ denotes the normal coordinates of the $x$ point with origin at $x'$.
On the other hand,
$\partial_\alpha^x\sigma(x,x')$ is a tangent vector at $x$
to the geodesic joining $x$ and $x'$, whose length equals the geodesic distance
between these two points and it is oriented in the
 $x'\rightarrow x$ direction. In turn, ${\partial_\alpha^x}'\sigma(x,x')$
is tangent to the same geodesic at $x'$, with the same modulus and oriented
in the opposite
sense.

In normal coordinates with origin at $x'$, according to the previous
expressions,
we can write:
\begin{eqnarray}
\sigma_\alpha(x,x')=\frac{\partial}{\partial x^\alpha}\sigma(x,x')=y_\alpha
\end{eqnarray}
i.e, $y_{\alpha}$ are components of a vector tangent at  the origin.

The use of normal coordinates, apart from being basic to obtain the
derivative expansions
of the EA, allows us to work in momentum space in a similar way
to the flat space-time. Let us consider some scalar function $f(x,y_0)$
with normal coordinates
with origin at $y_0$. We can define its covariant Fourier transform
through \cite{BunchParker}:
\begin{eqnarray}
f(x,y_0)=\int \frac{d^4k}{(2\pi)^4} \hat f(k,y_0) e^{-ikx}
\end{eqnarray}
In a similar way we can introduce the covariant Dirac delta:
\begin{eqnarray}
\delta^0(x,y_0)=\int \frac{d^4k}{(2\pi)^4} e^{-ikx}
\end{eqnarray}
As far as one of the delta arguments is the origin of coordinates, we will
have
$\delta^0(x,y_0)=\delta(x,y_0)g^{-1/2}(y_0)=\delta(x,y_0)$. In the general
case, (arbitrary
arguments) there is also a covariant definition whose expression in arbitrary
coordinates is
given by \cite{Avramidi}:
\begin{eqnarray}
\delta^0(x,y)=g^{1/4}(x)g^{1/4}(y)\Delta^{1/2}(x,y_0)\Delta^{1/2}(y,y_0)
\int\frac{d^4k}{(2\pi)^4}g^{-1/2}(y_0)
\exp(ik_{\mu}(\sigma^{\mu}(y,y_0)-\sigma^{\mu}
(x,y_0)))
\end{eqnarray}
where ${k_\mu}$, $\sigma_{\mu}(y,y_0)$ and $\sigma_{\mu}(x,y_0)$
are tangent vectors at $y_0$ and $\Delta(x,x')$ is the Van Vleck-Morette
determinant, defined as:
 $\Delta(x,x')=g^{-1/2}(x)\det(-\nabla_{\mu '}\nabla_\nu\sigma(x,x'))
g^{-1/2}(x')$.
If we take $y_0$ as origin, we will have
 $\Delta(x,y_0)=g^{-1/2}(x)$, this expression reduces then to:
\begin{eqnarray}
\delta^0(x,y)=g^{-1/2}(x)
\int\frac{d^4k}{(2\pi)^4}\exp(-ik_{\mu}(x^{\mu}-y^{\mu}))
\label{delta}
\end{eqnarray}
All these definitions are valid only in normal neighborhoods of the
origin in
which geodesics do not intersect.

\subsection{Derivative expansions}
In normal coordinates there is a privileged point around which we
perform the expansions.
In addition, the different curvature tensors are defined on the
tangent plane corresponding
to that point.
This fact, together with the general coordinate invariance of the
EA will
allow us to obtain a covariant derivative expansion for the effective
lagrangian around
the origin. In the following we will discuss in detail the scalar case,
 although the
procedure is the same for fields with different spin.

Let us start with the scalar EA (\ref{esceff}).
Using the normal coordinates expansion for the metric tensor it is
easy to split the operator $O_{xy}(m^2)=(-\Box_y-m^2-\xi R(y) +i\epsilon)
\delta^0(x,y)$
 in a free part, that coincides with the flat space-time Klein-Gordon
operator
\begin{eqnarray}
A_{xy}(m^2)=(-\Box_0^y-m^2 +i\epsilon)\delta^0(x,y)
\end{eqnarray}
with $\Box_0^y=\eta^{\mu\nu}\partial_\mu^y \partial_\nu^y$,
and the interaction part $B_{xy}$ that includes all the curvature dependence:
\begin{eqnarray}
B_{xy}&=&\left[-\frac{2}{3}R^\lambda_{\;\;\rho}(y_0)y^\rho
\partial_\lambda^y+\frac{1}{3}R^{\mu\;\;\nu}
_{\;\;\epsilon\;\;\beta}(y_0)y^\epsilon y^\beta \partial_\mu^y\partial_\nu^y
-\xi R(y_0)\right.
-\left(\frac{1}{20}R_{\alpha\;\; ; \beta\gamma}^{\;\;\nu}(y_0)
+\frac{1}{20}R_{\alpha\;\; ; \gamma\beta}^{\;\;\nu}(y_0)
-\frac{1}{20}R^{\mu\;\;\nu}_{\;\;\alpha\;\;\beta;\mu\gamma}(y_0)\right.
\nonumber \\
&-&\frac{1}{20}R^{\mu\;\;\nu}_{\;\;\alpha\;\;\beta;\gamma\mu}(y_0)
 -\frac{8}{45}R_{\alpha\lambda}(y_0)R^{\lambda\;\;\nu}
_{\;\;\beta\;\;\gamma}(y_0)
+\frac{1}{15}R_{\alpha\;\;\beta\lambda}^{\;\;\mu}(y_0)
R^{\lambda\;\;\nu}_{\;\;\mu\;\;\gamma}(y_0)
+\frac{4}{45}R_{\alpha\;\;\beta\lambda}^{\;\;\mu}(y_0)
R^{\lambda\;\;\nu}_{\;\;\gamma\;\;\mu}(y_0)
+\frac{1}{40}R_{\alpha\beta;\gamma}^{\;\;\;\;\;\;\;\nu}(y_0)\nonumber \\
&+&\left.\frac{1}{40}R_{\alpha\beta;\;\;\gamma}^{\;\;\;\;\;\nu}(y_0)\right)
y^\alpha y^\beta y^\gamma \partial_\nu^y
-\left(-\frac{1}{20}R^{\mu\;\;\nu}_{\;\;\rho\;\;\epsilon;\delta\kappa}(y_0)
+
\frac{1}{15}R_{\rho\;\;\epsilon\lambda}^{\;\;\mu}(y_0)
R^{\lambda\;\;\nu}_{\;\;\delta\;\;\kappa}(y_0)
\right) y^\rho y^\epsilon y^\delta y^\kappa
\partial_\mu^y\partial_\nu^y\nonumber \\
&-&\left.\xi\frac{1}{2}
R_{;\alpha\beta}(y_0)y^\alpha y^\beta\right]\delta^0(x,y)
+...
\end{eqnarray}
We have only written the two and four derivatives contributions since
terms with an odd
number of metric derivatives are shown to be irrelevant for the final
result.
Therefore we have:
$O_{xy}(m^2)=A_{xy}(m^2)+B_{xy}$

We will also assume that space-time is asymptotically  flat and this
will allows us to
discard total derivatives of the curvatures in the EA. We have
included the
covariant Dirac delta $\delta^0(x,y)$ in the definition of the
free operator so that
we can use the covariant integration measure $d^4x g^{1/2}(x)$.
Taking all this into account we can write the EA as:
\begin{eqnarray}
W[g_{\mu\nu}]=\int d^4x {\cal L}_{eff}(x)=\frac{i}{2}\mbox{Tr}\log O(m^2)
=\frac{i}{2}\mbox{Tr} \log(A+B)
\end{eqnarray}
As far as the calculations will be done in normal coordinates with
respect to the $y_0$ point,
any term in the effective lagrangian is evaluated in that point. For
that reason the
integration involved in  the $\mbox{Tr}$ symbol
cannot be done immediately. This, in turn, keep us from expanding the
logarithm,
which only makes sense inside the trace.
A way to avoid the problem consists in formally differentiating the EA
with respect to  $m^2$ \cite{Zakharov}:
\begin{eqnarray}
\frac{d}{dm^2}W[g_{\mu\nu}]=-\frac{i}{2}\mbox{Tr} \frac{1}{O(m^2)}=
-\frac{i}{2}\mbox{Tr}((A+B)^{-1})
=-\frac{i}{2}\mbox{Tr}(A^{-1}-A^{-1}BA^{-1}+A^{-1}BA^{-1}BA^{-1}-...)
\label{invdes}
\end{eqnarray}
In the last step the inverse of $A+B$ has been expanded. This can be
done without taking the
functional trace.

The lowest order term in $B$ is linear in the Riemann tensor and therefore
it contains
two metric derivatives.  The term in (\ref{invdes}) with two $B$ factors
will be ${\cal O}({\cal R}^2)$ and will contain at least four metric
derivatives, etc. Since we are only interested in the result up to
${\cal O}({\cal R}^2)$, it
will be enough to consider the first three terms in (\ref{invdes}).

\subsection{Divergent parts}

In order to calculate the divergent parts of the EA,
we consider the previous
expansions. Since normal coordinates allows us to do momentum space
integrations,
we can  use the flat space-time regularization techniques such
as dimensional
regularization.
Our calculation method is based on the
local momentum representation proposed in \cite{BunchParker}, but
adapted to the EA techniques. The results
agree with the well-known expressions in \cite{BiDa}.

First we find the scalar propagator $A^{-1}_{xy}$, given by:
$A_{xy}A^{-1}_{yz}=\delta^0(x,z)$,
where we have used the De Witt generalized summation convention for repeated
indices.
Using (\ref{delta}), we can write this equation as:
\begin{eqnarray}
\int d^4y g^{1/2}(y)(-\Box_0^y-m^2+i\epsilon)\frac{\delta(x,y)}{g^{1/2}(x)}
 g^{-1/2}(y)\int \frac{d^4k}{(2\pi)^4} e^{-ik(y-z)}\tilde G(k)=
\frac{\delta(x,z)}{g^{1/2}(x)}
\end{eqnarray}
The $G(k)$ function is easily obtained and finally the propagator reads:
\begin{eqnarray}
A^{-1}_{yz}=\frac{1}{g^{1/2}(y)}\int \frac{d^4k}{(2\pi)^4}
e^{-ik(y-z)}\frac{1}{k^2-m^2+i\epsilon}
\label{propagador}
\end{eqnarray}
The first term in the EA expansion (\ref{invdes})
can be immediately evaluated and it reads:
\begin{eqnarray}
A^{-1}_{y_0 y_0}=\int d\tilde k \frac{1}{k^2-m^2}=-\frac{i}{(4\pi)^{D/2}}
\frac{\Gamma(1-D/2)\mu^\epsilon}{(m^2)^{1-D/2}}
\end{eqnarray}
with the notation $d\tilde k=d^Dk\mu^\epsilon/(2\pi)^D$.
When $y_0$ appears as a repeated subindex it must be understood that the
integration in $y_0$
has not been done. This final integration, that corresponds to the trace
in (\ref{invdes}),
will be performed below in an explicit way. In the last step we have used
the equation (\ref{rdint}) from the Appendix.
Performing the  $m^2$ integration we have the lowest order term in
the effective lagrangian:
\begin{eqnarray}
{\cal L}_{div}^{(0)}(y_0)=\frac{1}{64\pi^2}m^4\left(\Delta+\frac{3}{2}\right)
\label{lef0}
\end{eqnarray}
where we have defined $\Delta=N_\epsilon+\log(\mu^2/m^2)$.
As is well-known, this first term will give rise to the cosmological
constant renormalization. The integration constant can be set to zero without
 loosing
generality since it can be absorbed in the renormalization procedure as we
will show below.

Up to two metric derivatives, only the second term in (\ref{invdes})
contributes:
\begin{eqnarray}
(A^{-1}_{y_0 y}B_{yz}A^{-1}_{zy_0})^{(2)}&=&\int d^4y g^{1/2}(y)d^4z
g^{1/2}(z)\int d\tilde k
\frac{e^{iky}}{k^2-m^2}\left(-\frac{2}{3}R^\lambda_{\;\;\rho}(y_0)
z^\rho\partial_\lambda^z
\right.
\nonumber \\
&+& \left. \frac{1}{3}R^{\mu\;\;\nu}_{\;\;\epsilon\;\;\beta}(y_0)
z^\epsilon z^\beta \partial_\mu^z\partial_\nu^z
-\xi R(y_0)\right)\frac{\delta(y,z)}{g^{1/2}(y)}g^{-1/2}(z)
\int d\tilde q\frac{e^{-iqz}}{q^2-m^2}
\label{2senc}
\end{eqnarray}
Integrating by parts and removing the coordinates $z$
through $z\rightarrow i\partial_q$, we can rewrite this expression as:
\begin{eqnarray}
(A^{-1}_{y_0 y}B_{yz}A^{-1}_{zy_0})^{(2)}&=&\int d^4z d\tilde p d
\tilde q \frac{e^{-ipz}}{q^2-m^2}
\left(\frac{\frac{1}{3}R(y_0)-\xi R(y_0)}{(q-p)^2-m^2}
-\frac{2}{3}\frac{R^{\mu\nu}(y_0)q_\mu q_\nu}{((q-p)^2-m^2)^2}\right)
\nonumber \\
&=&
\frac{i}{(4\pi)^{D/2}}\frac{\Gamma(2-D/2)}{(m^2)^{2-D/2}}\left(\frac{1}{6}
-\xi \right) R(y_0)
\end{eqnarray}
In the last step we have done the $z$ integration first and then those
corresponding to the
momenta. Finally integrating in $m^2$
we find the effective lagrangian contribution to order ${\cal O}(\partial^2)$:
\begin{eqnarray}
{\cal L}_{div}^{(2)}(y_0)=-\frac{1}{32\pi^2}m^2\left(\Delta+1\right)
\left(\frac{1}{6}-\xi\right)R(y_0)
\label{lef2}
\end{eqnarray}

Now we derive the divergences with four metric derivatives. In this case,
 there are contributions
from $A^{-1}BA^{-1}$ and $A^{-1}BA^{-1}BA^{-1}$:
\begin{eqnarray}
(A^{-1}_{y_0 y}B_{yz}A^{-1}_{zy_0})^{(4)}&=&\int d^4y g^{1/2}(y)d^4z
g^{1/2}(z)\int d\tilde k
\frac{e^{iky}}{k^2-m^2}\left[-\left(\frac{1}{20}
R_{\alpha\;\; ; \beta\gamma}^{\;\;\nu}(y_0)
\right.\right.
+\frac{1}{20}R_{\alpha\;\; ; \gamma\beta}^{\;\;\nu}(y_0)
-\frac{1}{20}R^{\mu\;\;\nu}_{\;\;\alpha\;\;\beta;\mu\gamma}(y_0)\nonumber \\
&-&\frac{1}{20}R^{\mu\;\;\nu}_{\;\;\alpha\;\;\beta;\gamma\mu}(y_0)
-\frac{8}{45}R_{\alpha\lambda}(y_0)
R^{\lambda\;\;\nu}_{\;\;\beta\;\;\gamma}(y_0)
+\frac{1}{15}R_{\alpha\;\;\beta\lambda}^{\;\;\mu}(y_0)
R^{\lambda\;\;\nu}_{\;\;\mu\;\;\gamma}(y_0)
+\frac{4}{45}R_{\alpha\;\;\beta\lambda}^{\;\;\mu}(y_0)
R^{\lambda\;\;\nu}_{\;\;\gamma\;\;\mu}(y_0)\nonumber \\
&+&\left.\frac{1}{40}
R_{\alpha\beta;\gamma}^{\;\;\;\;\;\;\;\nu}(y_0)
+\frac{1}{40}R_{\alpha\beta;\;\;\gamma}^{\;\;\;\;\;\nu}(y_0)\right)
z^\alpha z^\beta z^\gamma \partial_\nu^z
+\left(\frac{1}{20}R^{\mu\;\;\nu}_{\;\;\rho\;\;\epsilon;\delta\kappa}
(y_0)-
\frac{1}{15}R_{\rho\;\;\epsilon\lambda}^{\;\;\mu}(y_0)R^{\lambda\;\;\nu}
_{\;\;\delta\;\;\kappa}(y_0)
\right)\nonumber \\
&\times& z^\rho z^\epsilon z^\delta z^\kappa \partial_\mu^z\partial_\nu^z
-\left.\xi\frac{1}{2}
R_{;\alpha\beta}(y_0)z^\alpha z^\beta\right]\frac{\delta(y,z)}{g^{1/2}(y)}
g^{-1/2}(z)
\int d\tilde q\frac{e^{-iqz}}{q^2-m^2}
\label{2prod}
\end{eqnarray}
Removing the coordinates as before and integrating in positions and
momenta we find:
\begin{eqnarray}
(A^{-1}_{y_0 y}B_{yz}A^{-1}_{zy_0})^{(4)}&=&-\frac{i}{(4\pi)^{D/2}}
\frac{\Gamma(3-D/2)}{(m^2)^{3-D/2}}\left(-\frac{1}{180}
R^{\mu\nu\lambda\rho}(y_0)
R_{\mu\nu\lambda\rho}(y_0)\right.
\nonumber \\
&-&\left.\frac{1}{270}R^{\mu\nu}(y_0)R_{\mu\nu}(y_0)
+\frac{1}{6}\left(\frac{1}{5}-\xi\right)\Box R(y_0)\right)
\label{42}
\end{eqnarray}
Notice that this result is finite. The divergences will appear when doing
the
 $m^2$ integration. We add to this term the lowest order contribution from
the next one,
namely:
\begin{eqnarray}
(A^{-1}_{y_0 y}B_{yz}A^{-1}_{zt}B_{tu}A^{-1}_{uy_0})^{(4)}&=&
\int d^4yg^{1/2}(y) d^4zg^{1/2}(z) d^4tg^{1/2}(t) d^4ug^{1/2}(u)
\int d\tilde k d\tilde q d\tilde p \frac{e^{iky}}{k^2-m^2}
\nonumber \\
&\times &
\left(-\frac{2}{3}R^\lambda_{\;\;\rho}(y_0)z^\rho\partial_\lambda^z \right.
+ \left.\frac{1}{3}R^{\mu\;\;\nu}
_{\;\;\epsilon\;\;\beta}(y_0)z^\epsilon z^\beta \partial_\mu^z\partial_\nu^z
-\xi R(y_0)\right)\frac{\delta(y,z)}{g^{1/2}(y)}g^{-1/2}(z)\frac{e^{-iq(z-t)}}
{q^2-m^2}
\nonumber \\
&\times &
\left(-\frac{2}{3}R^\lambda_{\;\;\rho}(y_0)u^\rho\partial_\lambda^u \right.
+\left.\frac{1}{3}R^{\mu\;\;\nu}
_{\;\;\epsilon\;\;\beta}(y_0)u^\epsilon u^\beta \partial_\mu^u\partial_\nu^u
-\xi R(y_0)\right)\frac{\delta(t,u)}{g^{1/2}(t)}
g^{-1/2}(u)\frac{e^{-ipu}}{p^2-m^2}
\label{4prod}
\end{eqnarray}
In a similar fashion to the previous case we find the results for the
regularized
integrals:
\begin{eqnarray}
(A^{-1}_{y_0 y}B_{yz}A^{-1}_{zt}B_{tu}A^{-1}_{uy_0})^{(4)}&=&
-\frac{i}{(4\pi)^{D/2}}
\frac{\Gamma(3-D/2)}{(m^2)^{3-D/2}}\left(\frac{1}{2}\left(\frac{1}{6}
-\xi\right)^2 R(y_0)^2
-\frac{1}{108}R^{\mu\nu}(y_0)R_{\mu\nu}(y_0)\right)
\label{43}
\end{eqnarray}
Substracting  (\ref{42}) from (\ref{43}) and integrating in $m^2$
we obtain the divergent lagrangian up to  ${\cal O}(\partial^4)$:
\begin{eqnarray}
{\cal L}_{div}^{(4)}(y_0)=\frac{\Delta}{32\pi^2}
\left(\frac{1}{180}R^{\mu\nu\lambda\rho}(y_0)
R_{\mu\nu\lambda\rho}(y_0)-\frac{1}{180}R^{\mu\nu}(y_0)R_{\mu\nu}(y_0)
\right.
- \left.\frac{1}{6}\left(\frac{1}{5}-\xi\right)\Box R(y_0)
+\frac{1}{2}\left(\frac{1}{6}-\xi\right)^2 R(y_0)^2\right)
\label{lef4}
\end{eqnarray}
Comparing with the well-known Schwinger-DeWitt expansion, we see that
 ${\cal L}_{div}^{(2)}(y_0)$ is proportional to $a_1(O,y_0)$ and
${\cal L}_{div}^{(4)}(y_0)$ to $a_2(O,y_0)$.
As far as the above expressions are scalars they will have the same
form in any coordinate
system, not neccessarily geodesic. We can then perform a coordinate change
 and integrate to obtain
the corresponding EA:
\begin{eqnarray}
W[g_{\mu\nu}]_{div}=\int d^4x {\cal L}_{div}(x)
\end{eqnarray}
We have included in ${\cal L}_{div}(x)$ the $g^{1/2}(x)$ factor coming
from the integration measure. It is possible, in principle, that when
doing the coordinate
change, new non-covariant terms appear provided they vanish only for
geodesic coordinates.
However, in absence of gravitational anomalies the EA is scalar and
accordingly
such terms are not permitted. On the other hand, terms with an odd
number of derivatives
yield terms with an odd number of momenta in the numerator
which vanish in dimensional regularization.
The above are the only possible divergences, higher derivative terms
give rise to
momentum integrals with more momenta in the denominator that turn out
 to be finite.

Apart from the divergences, which are purely local contributions, the EA
also contains finite nonlocal pieces that are responsible for the pair
creation
processes. We have seen that in the
QED case (\ref{ehea}), the massless limit is well-defined due to the
cancellation
of the mass logarithmic dependence associated to the divergences with
that coming from the nonlocal pieces. This fact allows us to extract
some information about
the nonlocal terms from the knowledge of the divergences. In next
section we will profit
this connection to find part of the nonlocal structure of the
gravitational EA
by means of a point-splitting procedure in the divergences.

\subsection{Nonlocal contributions}

In the previous section we have obtained local contributions up to
order ${\cal O}({\cal R}^2)$.
If we continue the calculation to higher orders, we would get a
power series
with terms of the following type:
\begin{eqnarray}
\frac{\nabla^n {\cal R}^p(y_0)}{m^{2p+n-4}}
\label{serie}
\end{eqnarray}
where $\nabla$ denotes the covariant derivative.
This is a typical derivative (or adiabatic) expansion which is only
valid at low
energies.

All the terms in (\ref{serie})
can be classified by the number of curvatures they contain. Those terms
with a fixed number $n$
of curvatures will give rise to a series with an increasing number of
covariant
derivatives. If we could add all these terms together, we would obtain
nonlocal contributions
that would provide the exact n-point Green functions (with curvatures
in the external legs)
\cite{Ball,Avramidi,Vilkovisky}.
These Green functions will be valid for any value of the mass $m$,
as in the QED case (\ref{ehea}).

In this section we propose a method that effectively carries out
that resummation
for the quadratic operators, i.e we are interested in those terms
of the form $\nabla^{n} {\cal R}^2 $.
The basic idea is to perform a point-splitting in the quadratic parts
in curvatures in  (\ref{2prod}) and (\ref{4prod}), at the end of the
section we will argue that this procedure gives rise to the correct
resummation up to ${\cal O}(m^2 {\cal R}^2)$ terms.

As we have just commented, all the terms in (\ref{serie})
are local and finite. The reason why we have not obtained nonlocal
terms as in
(\ref{ehea}) is that the normal coordinates expansions are performed
around a
single point $y_0$.  In (\ref{2prod}) and (\ref{4prod})
there are products of curvatures evaluated at the same point, i.e,
 ${\cal R}(y_0){\cal R}(y_0)$.
Using again the normal coordinates expansion
we can rewrite these products as:
\begin{eqnarray}
{\cal R}(y_0){\cal R}(y_0)&=&{\cal R}(y_0){\cal R}(y)+({\cal R}(y_0)
\nabla{\cal R}(y_0)y+
{\cal R}(y_0)\nabla^2{\cal R}(y_0)yy+...)\nonumber \\
&+&({\cal R}(y_0){\cal R}(y_0){\cal R}(y_0)yy+
{\cal R}(y_0)\nabla{\cal R}(y_0){\cal R}(y_0)yyy+...)+...
\label{pointsplit}
\end{eqnarray}
This expression allows us to split the points, the price to pay is the
modification of the coefficients of the infinite higher order terms.
We will not
modify the linear part
in ${\cal R}$ in (\ref{2prod}). Let us first obtain the nonlocal
result and then
we will argue that the new terms $\nabla^n {\cal R}^2$ generated in
(\ref{pointsplit}) will exactly cancel those coming from the expansion
in (\ref{invdes}).

By means of the above point-splitting, the ${\cal R}^2$ contributions in
(\ref{2prod}) and (\ref{4prod}) will remain as:
\begin{eqnarray}
(A^{-1}_{y_0 y}B_{yz}A^{-1}_{zy_0})^{({\cal R}^2)}&=&\int d^4y g^{1/2}(y)
d^4zg^{1/2}(z)\int d\tilde k
\frac{e^{iky}}{k^2-m^2}
\left[\left(
\frac{8}{45}R_{\alpha\lambda}(y_0)
R^{\lambda\;\;\nu}_{\;\;\beta\;\;\gamma}(y)
-\frac{1}{15}R_{\alpha\;\;\beta\lambda}^{\;\;\mu}(y_0)
R^{\lambda\;\;\nu}_{\;\;\mu\;\;\gamma}(y)\right.\right.\nonumber \\
&-&\left.\left.\frac{4}{45}R_{\alpha\;\;\beta\lambda}^{\;\;\mu}(y_0)
R^{\lambda\;\;\nu}_{\;\;\gamma\;\;\mu}(y)\right)
z^\alpha z^\beta z^\gamma \partial_\nu^z
-
\frac{1}{15}R_{\rho\;\;\epsilon\lambda}^{\;\;\mu}(y_0)
R^{\lambda\;\;\nu}_{\;\;\delta\;\;\kappa}(y)
 z^\rho z^\epsilon z^\delta z^\kappa \partial_\mu^z\partial_\nu^z
\right]\frac{\delta(y,z)}{g^{1/2}(y)}g^{-1/2}(z) \nonumber \\
&\times &
\int d\tilde q\frac{e^{-iqz}}{q^2-m^2}+{\cal O}(\nabla^2{\cal R}^2)
\end{eqnarray}
and:
\begin{eqnarray}
(A^{-1}_{y_0 y}B_{yz}A^{-1}_{zt}B_{tu}A^{-1}_{uy_0})^{({\cal R}^2)}&=&
\int d^4yg^{1/2}(y) d^4zg^{1/2}(z) d^4tg^{1/2}(t) d^4ug^{1/2}(u)
\int d\tilde k d\tilde q d\tilde p\frac{e^{iky}}{k^2-m^2}
\frac{e^{-ipu}}{p^2-m^2} \nonumber \\
&\times &
\left(-\frac{2}{3}R^\lambda_{\;\;\rho}(y_0)z^\rho\partial_\lambda^z
+\frac{1}{3}R^{\mu\;\;\nu}
_{\;\;\epsilon\;\;\beta}(y_0)z^\epsilon z^\beta \partial_\mu^z
\partial_\nu^z
-\xi R(y_0)\right)\frac{\delta(y,z)}{g^{1/2}(y)}g^{-1/2}(z)
\frac{e^{-iq(z-t)}}{q^2-m^2}\nonumber \\
&\times&
\left(-\frac{2}{3}R^\lambda_{\;\;\rho}(t)u^\rho\partial_\lambda^u
+\frac{1}{3}R^{\mu\;\;\nu}
_{\;\;\epsilon\;\;\beta}(t)u^\epsilon u^\beta \partial_\mu^u\partial_\nu^u
-\xi R(t)\right)\frac{\delta(t,u)}{g^{1/2}(t)}g^{-1/2}(u)
+{\cal O}(\nabla^2{\cal R}^2)\label{r21}
\end{eqnarray}
where, as mentioned before, ${\cal O}(\nabla^2{\cal R}^2)$ denotes
local finite terms with two curvatures and an arbitrary even number of
derivatives. There are in principle different ways of performing the
splitting, depending on
the choice of the pair of points, but all them are equivalent up to
higher order terms
as can be seen from (\ref{pointsplit}).

Removing the explicit coordinates occurrences, we obtain:
\begin{eqnarray}
(A^{-1}_{y_0 y}B_{yz}A^{-1}_{zt}B_{tu}A^{-1}_{uy_0})^{(R^2)}&=&
\int d^4t d\tilde q d\tilde p e^{ip t}\left(\frac{\left(\frac{1}{3}
-\xi\right)R(y_0)}
{(q^2-m^2)^2}-
\frac{2}{3}R^{\mu\nu}(y_0)\frac{q_\mu q_\nu}{(q^2-m^2)^3}\right)
\nonumber \\
&\times& \left(\frac{-\xi R(t)}{((p+q)^2-m^2)}
+\frac{2}{3}\frac{R^{\mu\nu}(t)(p+q)_\mu (p+q)_\nu}{((p+q)^2-m^2)^2}
 \right)
+{\cal O}(\nabla^2{\cal R}^2)
\end{eqnarray}
Using the equations (\ref{rd1}), (\ref{rd2}) and (\ref{rd4}) from the
 Appendix,
neglecting higher order terms and
integrating in $m^2$, we obtain the following contributions to
the effective lagrangian:
\begin{eqnarray}
{\cal L}_{1}^{({\cal R}^2)}(y_0)
=\frac{1}{32\pi^2}\int d^4t d\tilde p e^{ip t}
\left(\Delta-F(p^2;m^2)\right)
\left[\frac{1}{2}\left(\frac{1}{6}-\xi \right)^2R(y_0)R(t)
-\frac{1}{108}R_{\mu\nu}(y_0)R^{\mu\nu}(t)\right]+{\cal O}(\nabla^2{\cal R}^2)
\label{l1}
\end{eqnarray}
$F(p^2;m^2)$ being given in (\ref{mand1}).

Apart from the local divergent and nonlocal finite terms, in the
dimensional regularization
procedure, finite local terms do arise. However, their coefficients
will be absorbed in the
definition of the renormalized parameters
and they will not be explicitly considered.

In the same form as before we obtain from (\ref{r21}):
\begin{eqnarray}
{\cal L}_{2}^{({\cal R}^2)}(y_0)
=\frac{1}{32\pi^2}\int d^4t d\tilde p e^{ip t}
\left(\Delta-F(p^2;m^2)\right)
 \left[\frac{1}{180}R^{\mu\nu\lambda\rho}(y_0)
R_{\mu\nu\lambda\rho}(t)
+\frac{1}{270}R^{\mu\nu}(y_0)R_{\mu\nu}(t)\right]+{\cal O}(\nabla^2{\cal R}^2)
\label{l2}
\end{eqnarray}
Adding both contributions  in (\ref{l1}) and
(\ref{l2}) and including the $0$ and $2$ derivatives divergent contributions
given in
 (\ref{lef0}) and (\ref{lef2}), we can write the nonlocal EA in a slightly
different notation:
\begin{eqnarray}
W[g_{\mu\nu}]&=&\frac{1}{32\pi^2}\int d^4x \sqrt{g}\left[
\frac{m^4}{2}\left(\Delta+\frac{3}{2}\right)
-m^2\left(\Delta+1\right)
\left(\frac{1}{6}-\xi\right)R(x)
+\Delta
\left(\frac{1}{180}R^{\mu\nu\lambda\rho}(x)
R_{\mu\nu\lambda\rho}(x)
\right.\right.\nonumber \\
&-&\left.\frac{1}{180}R^{\mu\nu}(x)R_{\mu\nu}(x)
+ \frac{1}{2}\left(\frac{1}{6}-\xi\right)^2 R(x)^2\right)
-\left(\frac{1}{180}R^{\mu\nu\lambda\rho}(x)F(\Box;m^2)
R_{\mu\nu\lambda\rho}(x)\right.
\nonumber \\
&-&\left. \left. \frac{1}{180}R^{\mu\nu}(x)F(\Box;m^2)R_{\mu\nu}(x)
+\frac{1}{2}\left( \frac{1}{6}-\xi\right)^2 R(x)F(\Box;m^2)R(x)\right)
+ {\cal O}(m^2{\cal R}^2)\right] +{\cal O} ({\cal R}^3)
\label{ear2}
\end{eqnarray}
Here the $F(\Box;m^2)$ operator action should be understood through
the expressions
 (\ref{l1}) y (\ref{l2}) as we did in the QED case (\ref{ehea}),
in addition we have neglected local finite
terms and total derivatives.
On the other hand, we see that the quadratic
divergences agree with those obtained in the previous section.

Consider now the massless limit of the EA in (\ref{ear2}).
For small masses compared with $p$, the Mandelstam function behaves as
shown in
(\ref{mandaprox}).
As in the QED case, the term  proportional to
 $\log (m^2)$, associated to the divergences in (\ref{ear2}),
has the same coefficient but with opposite sign as that coming from
the Mandelstam function.
Therefore they exactly cancel. This allows us to obtain the regular
massless limit:
\begin{eqnarray}
W[g_{\mu\nu}]&=&\frac{1}{32\pi^2}\int d x \sqrt{g}
\left(\frac{1}{180}R^{\mu\nu\lambda\rho}(x)\Gamma(\Box)
R_{\mu\nu\lambda\rho}(x)-\frac{1}{180}R^{\mu\nu}(x)\Gamma(\Box)
R_{\mu\nu}(x)
\right. \nonumber \\
&+& \left.
\frac{1}{2}\left(\frac{1}{6}-\xi\right)^2 R(x)\Gamma(\Box)R(x)\right)
+{\cal O} ({\cal R}^3)
\label{nolocal}
\end{eqnarray}
where $\Gamma(\Box)$ is given in (\ref{gama}).
This result is what one had expected on dimensional grounds \cite{Duff},
however we have got in addition an explicit representation, completely
analogous to the flat space-time case, for
the form factors.
In principle, the result in (\ref{ear2}) cannot be considered,
strictly speaking,  as the
 ${\cal O}({\cal R}^2)$ contribution in the curvature expansion since in
the calculation
we have not considered those terms with two curvatures and an arbitrary
number of covariant
derivatives, denoted by
${\cal O}(\nabla^2{\cal R}^2)$. We have obtained a nonlocal expression,
but the infinite ${\cal O}(\nabla^2{\cal R}^2)$ terms could in principle modify it.
However, we have seen in the previous section that the divergences
have a logarithmic dependence in $m^2$ (whose calculation is unambiguous).
In addition, we showed in Section 2 in the QED case that the coefficients of
the $\log(m^2)$ in the divergent part have to be the same as those
of the $\log(p^2)$ in the nonlocal pieces, in order to have a regular
masless limit. In (\ref{nolocal}) we have seen that our result indeed
posseses a regular masless limit. Since the ${\cal O}(\nabla^2{\cal R}^2)$ terms
are finite, they could only modify the finite local and nonlocal part,
but not the
divergences. However such a modification in the nonlocal part could
spoil the regularity of the masless limit. Accordingly, we conclude that
the ${\cal O}(\nabla^2{\cal R}^2)$ terms can only modify the local finite
pieces and those
proportional to $m^2$, but not the nonlocal ones containing $\log (m^2)$.
For that reason, the result in (\ref{ear2}) includes all the quadratic
curvature contributions
except for those  containing an arbitrary mass power denoted
${\cal O}(m^2{\cal R}^2)$.

A compatible result has been obtained using partial resummation of
the Schwinger-DeWitt expansion
\cite{Avramidi} and also by means of the so called covariant
perturbation theory
\cite{Vilkovisky}, in this case it was possible to derive the
cubic terms in curvatures for asymptotically flat manifolds.
Notice however that the boundary conditions that normal coordinates
impose on
the metric tensor are slightly more general  than the asymptotically
Minkowskian condition. In fact, we have only required that the curvatures
and their covariant derivative vanish at infinity. This condition
includes, for instance, those Robertson-Walker manifolds in which
the expansion rate $\dot a/a$ asymptotically vanishes
in the future,  although the scale
factor itself $a(t)$ does not tend to a constant. This kind of manifolds
are known to accept the definition of the so-called adiabatic
vacua \cite{BiDa}. In \cite{art2} the possibility of defining
such vacua was extremely useful for the interpretation of the
effective action as vacuum persistence amplitude.

To summarize, the procedure we have just presented makes it possible
a partial resummation
of the higher order terms in the EA, within the mentioned
limits. Neglecting  ${\cal O}({\cal R}^3)$  terms implies that
 (\ref{ear2}) will be  a good approximation for
 $\nabla \nabla {\cal R}>>{\cal R}^2$. In a similar fashion, when we
neglect
${\cal O}(m^2{\cal R}^2)$ terms we are assuming that
$\nabla \nabla {\cal R}>>m^2 {\cal R}$. However, if we are only interested
in the two point
Green functions with external curvature legs, the massless limit in
(\ref{nolocal}) is exact. The renormalization of the EA
can be done following the standard procedure in the classical references
(\cite{BiDa} and \cite{shap3}).
Here we will only mention that from the viewpoint of the Appelquist and
Carrazone decoupling
theorem \cite{AppCarr}, the scalar field does not decouple from
gravity since
there are new terms in the EA (\ref{ear2}) which are not present in the
Einstein-Hilbert action
and they are not suppressed by powers of the particle
mass  $m$.

\section{Integration of the Standard Model matter fields: the EA for torsion}

Up to now we have only worked with scalar fields. However, in order to
include the effect of the
matter content present in the SM in the gravitational EA, we have to
deal with the integration
of fermionic fields.  In this case we can proceed in a similar way
using
 (\ref{fermion}) in order to obtain the EA for the gauge fields and
gravitation
by integrating out the SM matter fields. The main novelty is that, in
addition to the gravitational
field (the vierbein), fermions couple also to the pseudotrace of the
torsion as it
was discussed at the beginning of the previous section.
The SM matter lagrangian in a curved space-time with torsion can be
written as
\cite{anom}:
\begin{eqnarray}
{\cal L}_M=\sqrt{g} \left(\overline {\cal Q}(i\not{\!{\!D}}^Q-M^Q) {\cal Q}+
\overline {\cal L}(i\not{\!{\!D}}^L-M^L) {\cal L}\right)
\label{materia}
\end{eqnarray}
where:
\begin{eqnarray}
\not{\!{\!D}}^Q  = \gamma^\mu D^Q_\mu= \gamma^{\mu}(\partial_{\mu}
+\Omega^{Q}_{\mu}
+S^{Q}_{\mu}\gamma_5), \;\;
\not{\!{\!D}}^L =  \gamma^\mu D^L_\mu=\gamma^{\mu}(\partial_{\mu}
+\Omega^{L}_{\mu}+S^{L}_{\mu}\gamma_5)
\label{dos}
\end{eqnarray}
with $M^Q$ and $M^L$  the mass matrices of quarks and leptons.
We have followed the notation in \cite{anom}.
In the following we will concentrate only in the gravitational
couplings, so that we have
neglected the gauge fields contributions in the operators.

In order to obtain the torsion contribution to the EA up to ${\cal O}(S^2)$,
we first consider a flat space-time with torsion and afterwards we will
include the effect
of curvature. From the above lagrangian we see that torsion behaves as
the electromagnetic
field in QED, without considering the  $\gamma_5$ coupling, which does
not affect
the calculation of terms with an even number of fields, in the massless
limit. Using the result in
(\ref{ehea}) and changing
$e\not{\!{\!A}} \rightarrow -1/8 {\not{\!{\!S}}} \gamma_5$, we obtain in this limit:
\begin{eqnarray}
W[S]=-\left(\frac{N_\nu}{2}+N_{Df}+N_c N_q\right)
\frac{N_f}{192(4\pi)^2}\int d^4x
 S^{\mu\nu}\Gamma(\Box)
S_{\mu\nu}+{\cal O}(S^4)
\label{sea}
\end{eqnarray}
where $\Gamma(\Box)$ is defined in
(\ref{gama}), $N_f$ is the number of families, $N_\nu$ the number of
neutrinos,
$N_{Df}$ that of Dirac fermions and $N_q$
the number of quark flavors.
$S_{\mu\nu}=\partial_\mu S_\nu-\partial_\nu S_\mu$.
Due to the absence of right neutrinos, there could be, in principle,
parity violating
 terms. However it can be shown that those terms  are total derivatives.

In order to introduce the space-time curvature, we recall that the
SM matter sector is
locally conformally invariant (for massless fermions), this is also
the case of the
counterterms \cite{duff},
 (see \cite{shap2,shap3} where the renormalization procedure
has been studied for manifolds with torsion).
When including the curvature, apart from those in  (\ref{sea}),
there could be quadratic terms in torsion in the generic form
${\cal R}S^2$. However such terms are not conformally invariant.
Therefore the ones obtained
above are the only possible divergences. Next we will confirm these
results with an explicit
calculation of the divergences in  a space-time with curvature.

The SM Dirac operators in Euclidean space are not Hermitian because of
the electroweak gauge couplings and the absence of the right
neutrinos. However,  in Euclidean space the EA
divergences are real
\cite{Ball}, thus it is enough to calculate:
$2\mbox{Re}\; W[e,\hat \Gamma]=-\log \det ({\cal O})$,
where ${\cal O}=(\not{\!{\!D}} +M)^\dagger (\not{\!{\!D}} +M)$ and $\not{\!{\!D}}$
and M denotes the joint Euclidean Dirac operator and mass matrix
for quarks and leptons.
The heat-kernel expansion together with dimensional regularization
\cite{BiDa,JaOs}
allows us to obtain:
\begin{eqnarray}
-\frac{1}{2}\mbox{Tr} \log ({\cal O})=\frac{\mu^{\epsilon}}
{2(4\pi)^{D/2}}\mbox{tr}\sum_{n=0}^{\infty}M^{D-2n}
\Gamma\left(n-\frac{D}{2}\right)\int d^4x \sqrt{g}\;\,
a_n({\cal O},x)
\end{eqnarray}
The well-known HMDS coefficients  $a_n$ are given for the above operators
by:
\begin{eqnarray}
a_0({\cal O},x)&=&1, \;\;
a_1({\cal O},x)= \frac{1}{6}R-X \nonumber \\
a_2({\cal O},x)&=&\frac{1}{12}[D_{\mu},D_{\nu}][D^{\mu},D^{\nu}]
+\frac{1}{6}
[D_{\mu},[D^{\mu},X]]+ \frac{1}{2}X^2-
\frac{1}{6}RX \nonumber\\
&-&\frac{1}{30}R_{;\mu}^{\;\;\mu}+\frac{1}{72}R^2+\frac{1}{180}
(R_{\mu \nu \rho
\sigma}  R^{\mu \nu \rho \sigma} -R_{\mu \nu} R^{\mu \nu})
\label{swdwdes}
\end{eqnarray}
where, when we neglect the gauge fields contributions, the operator can
be written as:
\begin{eqnarray}
{\cal O}=D_{\mu}D^{\mu}+X+M^2,\;\;\;  \mbox{with}\;\;\;
X=\gamma_5 S^{\mu}_{\; ;\mu}+2S_{\mu}  S^{\mu}-
\frac{1}{4}[\gamma^{\mu},\gamma^{\nu}][d_{\mu},d_{\nu}]\nonumber
\end{eqnarray}

\noindent and \centerline{$D_{\mu}=\partial_\mu+\Omega_\mu
-\frac{1}{2}\gamma_5[\gamma_{\mu},
\gamma^{\nu}]S_{\nu}
=d_{\mu}
-\frac{1}{2}\gamma_5[\gamma_{\mu},\gamma^{\nu}]S_{\nu}$}

Writing the result in Lorentzian signature we have (see also the previous
works \cite{shap2,shap3}):
\begin{eqnarray}
W_{div}[e,\hat \Gamma]&=&-\sum_f \sum_i N_i\frac{1}{32\pi^2}\int
d^4x \sqrt{g}
\left[M_i^4\left(\frac{\Delta_i}{2}+\frac{3}{4}\right)-M_i^2\left(\Delta_i
+1\right)
\left(-\frac{1}{12}R
+\frac{1}{32}S^2\right)\right.\nonumber\\
&+&\left.\Delta_i\left(\frac{1}{384}S^{\mu\nu}S_{\mu\nu}
-\frac{7}{1440}R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}
-\frac{1}{180}R_{\mu\nu}R^{\mu\nu}
+\frac{1}{288}R^2\right)\right]
\label{sdiv}
\end{eqnarray}
$\sum_f$ denotes sum over the different families  and  $\sum_i=\sum_{lept.}
+N_c\sum_{quarks}$ over the different flavors in each family, including
quarks and leptons.
On the other hand,
$N_i$ is the number of spinor components: 2 for neutrinos and 4 for the
 rest of fermions.
$M_i$ are the different fermion masses, $\Delta_i=N_\epsilon
+\log (\mu^2/M_i^2)$ and
$\mu$ the renormalization scale.

This formula is compatible with the flat space-time result in
 (\ref{sea}). There is no
 ${\cal R}S^2$ term, as commented before, and we have discarded
total derivatives.
Following similar steps as for the scalar case, i.e. including
$\log (\Box/M^2)$
factors and taking the massless limit, we obtain
the finite nonlocal contributions depending on curvature and torsion
 from the divergences.
In the mentioned limit we have:
\begin{eqnarray}
W[e,\hat \Gamma]&=&-\frac{N_f(8N_c+6)}{32\pi^2}\int d^4x \sqrt{g}
\left(\frac{1}{384}S^{\mu\nu}\Gamma(\Box)S_{\mu\nu}
+\frac{1}{288}R\Gamma(\Box)R
\right.\nonumber\\
&-&\left.\frac{7}{1440}R_{\mu\nu\rho\sigma}\Gamma(\Box)R^{\mu\nu\rho\sigma}
-\frac{1}{180}R_{\mu\nu}\Gamma(\Box)R^{\mu\nu}
\right)
+{\cal O}({\cal R}^3)+{\cal O}(S^3)
\end{eqnarray}
where $\Gamma(\Box)$ is given in (\ref{gama}) and ${\cal O}(S^3)$ denotes terms
with 3 or more torsion fields. Finally, due to the presence of chiral
fermions, the EA could contain
an abnormal parity sector, responsible for the gauge and gravitational
anomalies. However,
as far as the previous result has been derived from the real part of the EA
(with normal parity), this sector is not taken into account.

The renormalization procedure in this case will require, not only the
introduction of
quadratic terms and a constant, but also a kinetic and mass terms for
 the torsion field.
Accordingly, the starting classical action should be:
\begin{eqnarray}
S_{G}=\int d x \sqrt{g}\left(\frac{R-2\Lambda}{16\pi G}+a_2
R^{\mu\nu\lambda\rho}
R_{\mu\nu\lambda\rho}+a_3R^{\mu\nu}R_{\mu\nu}
+ a_4R^2
+a_5 S_{\mu\nu}S^{\mu\nu}+a_6S^2\right)
\end{eqnarray}
The divergences can be absorbed by constants redefinition:
\begin{eqnarray}
a_j &\rightarrow & a'_j=a^r_j(\mu)
-\frac{1}{32\pi^2}C_j N_\epsilon+\mbox{finite constants},\; j=0..6
\end{eqnarray}
where $a_0=-2\Lambda/(16\pi G)$ and $a_1=1/(16\pi G)$.
The values of
$C_j$ are shown in Table 3.1.

The scale dependence of the renormalized constants is derived from the
renormalization
group equation \cite{shap3,shap2}, in an analogous way to the scalar case.
As a consequence, if at a given
scale $\mu$,
the constants have certain values $a_i^r(\mu)$,
their values at a different scale $\mu'$ will be given by:
\begin{eqnarray}
a_i^r(\mu')=a_i^r(\mu)+\frac{C_i}{32\pi^2}\log\left(\frac{\mu^2}{\mu'^2}
\right)
\label{escala}
\end{eqnarray}
where the  $C_i$ constants are those appearing in Table 3.1. In particular,
the renormalized
Newton constant $G^r(\mu)$ has a scale
dependence and therefore
its value should be specified for a given $\mu$.
Thus  $G$ will depend on the size of the system we are
considering and this
could have an enormous importance in cosmology (see \cite{goldman}).
\begin{center}
\begin{tabular}{|c|c|c|c|} \hline
$C_0$&$C_1$&$C_2$&$C_3$\\  \hline
$-\sum_f \sum_i N_i  \frac{M_i^4}{2}$&
$-\sum_f \sum_i N_i  \frac{M_i^2}{12}$&$\frac{7}{180}N_f N_c
+\frac{7}{240}N_f$&
$\frac{2}{45}N_f N_c+\frac{1}{30}N_f$\\
\hline \hline
$C_4$&$C_5$&$C_6$& \\ \hline
$-\frac{1}{36}N_f N_c-\frac{1}{48}N_f$&
$-\frac{1}{48}N_f N_c-\frac{1}{64}N_f$&
$\sum_f \sum_i N_i \frac{M_i^2}{32}$& \\ \hline
\end{tabular}
\end{center}
\centerline{\footnotesize {\bf Table 3.1:} Renormalization constants
for the SM case}

In the renormalized EA we can generate a torsion kinetic term by
means of a $S_\nu$ finite normalization as follows: $S_{\nu}^r=
Z_3^{-1/2}(\mu)S_{\nu}$,
where $a^r_5(\mu)=-\frac{1}{4}Z_3^{-1}(\mu)$. Thus in the massless
limit we have:
\begin{eqnarray}
W^r[S^r]&=&\int d^4x \sqrt{g} \left(-\frac{1}{4}S_{\mu\nu}^r S^{r\mu\nu}
+\mbox{nonlocal terms} +
{\cal O}(S^3)\right)
\end{eqnarray}
As a consequence the physical torsion field
$S_\mu^r$ will behave as an abelian gauge field.
Therefore we have generated a kinetic term for
torsion even starting from a theory without propagating torsion.

\section{Conclusions and discussion}

In this work we have dealt with the computation of the low-energy
Effective Action (EA)
for gravity obtained when matter fields, both scalar and fermionic,
are integrated out.
As a much simpler exercise, we have started with by reviewing the
 low-energy
EA for the electromagnetic field obtained when the electronic field is
integrated out. Special attention has been paid to the nonlocal terms
which are related
to the particle production probabilities.

In order to integrate the scalar fields to compute the EA for gravity,
 we have used the
normal coordinate expansion. This method has
been used
previously to compute the divergent terms of the EA. By comparison of
this kind of
expansion around different space-time points, we have been able to
extend the
previous work to obtain also the nonlocal finite terms of the EA up
to quadratic terms
in the curvature. In particular we have arrived to the
conclusion that
the scalar field does not decouple from the EA in the large mass limit
in the
 Appelquist-Carrazone sense. Finally we have also discussed  the
 meaning of
the expansion in the massless limit.

In order to be able to consider the matter field content present in
 the Standard Model
we have also studied the case of fermionic matter. The main novelty
in this case is that
this kind of fields can also couple to the torsion pseudotrace in
addition to the
gravitational field. Thus we have
obtained the low-energy EA for the gravitational field and the
torsion including the
divergent and the nonlocal finite terms up to quadratic terms
in the curvature and the
torsion. We have discussed the renormalization of
this EA and we have found
that a kinetic term is
generated for the torsion pseudotrace. This result is quite
interesting since in the
standard Einstein-Cartan action \cite{Hehl}, the torsion does not propagate.

Finally we would like to stress that the results obtained in this
work can be useful
for the study of some interesting physical effects. In particular,
the low-energy EA
can be applied for the study of the quantum stability of classical
solutions of the
Einstein equations of motion. More specifically, the nonlocal terms
of the EA
which have been computed here, are complex in general.  Classical
solutions that give
rise to an imaginary part when the EA is evaluated on them, are
unstable by particle
radiation even if they are stable at the classical level. Moreover,
 the imaginary part
of the EA evaluated on classical solutions, can be used to compute
particle production
rates and spectra thus  providing an alternative method to the more
 commonly used based
on the Bogolyubov transformations. The expression that we have obtained
for the nonlocal form factors is particularly appropriate for
the calculation of particle production probabilities in cosmological
space-times. The boundary conditions imposed by the normal coordinates
expansion, allow to apply this method not only to asymptotically
flat manifolds, but also to some other cosmological manifolds where
the expansion rate asymptotically vanishes.
Work has been done in this
 direction and it  will be
presented elsewhere \cite{art2}.

\vspace{0.5cm}

{\bf Acknowledgments:}
We thank I.L.Shapiro for his comments and suggestions in several
points of the paper.
This work has been partially supported by the Ministerio de Educaci\'on y
Ciencia (Spain)(CICYT AEN96-1634). A.L.M. acknowledges support from
SEUID-Royal Society.

\section{Appendix}
\subsection*{Dimensional regularization formulae}
In dimensional regularization \cite{regdim}, the space-time dimension is
taken to be
$D=4-\epsilon$ and the poles are parametrized through
$N_\epsilon$ that was defined in Section II. We will use the notation
$d\tilde q=\mu^\epsilon d^Dq/(2\pi)^D$ where $\mu$ is the renormalization
scale.
Some useful expressions are:
\begin{equation}
\int d\tilde{q}\frac{(q^2)^r}{[q^2-R^2]^m}=
i\frac{(-1)^{r-m}}{(4\pi)^{D/2}}\frac{\Gamma(r+D/2)
 \Gamma(m-r-D/2) \mu^{\epsilon}}{\Gamma(D/2)\Gamma(m)(R^2)^{m-r-D/2}}
\label{rdint}
\end{equation}
\begin{eqnarray}
I_1(p,m^2)\equiv \int d\tilde q\frac{1}{(q^2-m^2)^2((q+p)^2-m^2)}
&=&
i\frac{(-1)^{D/2}}{(4\pi)^{D/2}}\Gamma(3-D/2)\int_0^1
dt (1-t)(-m^2+p^2t(1-t))^{D/2-3}
\nonumber \\
\int^{m^2} I_1(p,m^2)dm^2 &=&\frac{i\mu^{\epsilon}}{2(4\pi)^2}
\left(N_\epsilon+2-\log \frac{m^2}{\mu^2}
-F(p^2;m^2)\right)
\label{rd1}
\end{eqnarray}

\begin{eqnarray}
I^{\mu\nu}_2(p,m^2)&\equiv&\int d\tilde q\frac{q^\mu q^\nu}{(q^2-m^2)^3((q+p)^2
-m^2)}\nonumber \\
&=&i\frac{(-1)^{D/2}}{2(4\pi)^{D/2}}\Gamma(4-D/2)\int_0^1
dt (1-t)^2(-m^2+p^2t(1-t))^{D/2-4}
 \left(p^\mu p^\nu+g^{\mu\nu}\frac{-m^2+p^2
 t(1-t)}{2(3-D/2)}\right)\nonumber \\
\int^{m^2}I_2^{\mu\nu}(p,m^2)dm^2 &=&\frac{i\mu^\epsilon}{12(4\pi)^2}
\left(N_\epsilon+\frac{13}{6}
-\log \frac{m^2}{\mu^2}-F(p^2;m^2)\right)g^{\mu\nu}
+{\cal O}(p^2)
\label{rd2}
\end{eqnarray}

\begin{eqnarray}
I^{\mu\nu\alpha\beta}_3(p,m^2)&\equiv&\int d\tilde q
\frac{q^\mu q^\nu q^\alpha q^\beta}
{(q^2-m^2)^4((q+p)^2-m^2)}\nonumber \\
&=&i\frac{(-1)^{D/2}}{6(4\pi)^{D/2}}\Gamma(5-D/2)\int_0^1
dt (1-t)^3(-m^2+p^2t(1-t))^{D/2-3}
\frac{g^{\alpha\nu}g^{\beta\mu}+g^{\alpha\mu}
g^{\beta\nu}+g^{\alpha\beta}g^{\mu\nu}}
{(6-D)(8-D)}\nonumber \\
&+&{\cal O}(p^2)\nonumber \\
\int^{m^2}I_3^{\mu\nu\alpha\beta}(p,m^2)dm^2 &=&
\frac{i\mu^\epsilon}{96(4\pi)^2}\left(N_\epsilon+\frac{7}{3}
-\log \frac{m^2}{\mu^2}-\left(1-\frac{2m^2}{p^2}\right)F(p^2;m^2)-
\frac{4m^2}{p^2}\right)\nonumber \\
&\times &(g^{\alpha\nu}g^{\beta\mu}+g^{\alpha\mu}g^{\beta\nu}
+g^{\alpha\beta}g^{\mu\nu})+{\cal O}(p^2)
\label{rd4}
\end{eqnarray}
where $F(p^2;M^2)$ is given in (\ref{mand1}).
These expressions have been obtained in the Minkowski space-time.
We have not explicitly
included the $+i\epsilon$ terms accompanying the momenta in order
to avoid the
confusion with that coming from dimensional regularization.

\subsection*{Normal coordinates expansions}

We will show the Riemann normal coordinates expansions for
different objects up to
order ${\cal O}(\partial^4)$. We will take the point $y_0$ as the
coordinates origin.
For the inverse metric tensor we have:
\begin{eqnarray}
g^{\mu\nu}(y)&=&\eta^{\mu\nu}-\frac{1}{3}
R^{\mu\;\;\;\nu}_{\;\;\alpha\;\;\beta}(y_0)y^\alpha y^\beta
+\frac{1}{6}R^{\mu\;\;\;\nu}_{\;\;\alpha\;\;\beta ;\gamma}(y_0)
y^\alpha y^\beta y^\gamma
+
\left[-\frac{1}{20}R^{\mu\;\;\;\nu}_{\;\;\alpha\;\;\beta ;\gamma\delta}
(y_0)
+\frac{1}{15}R_{\alpha\;\;\;\beta\lambda}^{\;\;\mu}(y_0)
R^{\lambda\;\;\nu}_{\;\gamma\;\;\delta}(y_0)\right]
y^\alpha y^\beta y^\gamma y^\delta \nonumber \\
&+&{\cal O} (\partial^5)
\end{eqnarray}

The metric determinant $g=\vert\det g_{\mu\nu}\vert$ has the
following expansion:
\begin{eqnarray}
g(y)=1+\frac{1}{3}R_{\alpha\beta}(y_0)y^{\alpha}y^{\beta}
-\frac{1}{6}R_{\alpha\beta;\gamma}
y^{\alpha}y^{\beta}y^{\gamma}+\left(\frac{1}{18}R_{\alpha\beta}
R_{\gamma\delta}-
\frac{1}{90}R_{\lambda\alpha\beta\kappa}
R^{\lambda\;\;\;\;\kappa}_{\;\;\gamma\delta}
+\frac{1}{20}R_{\alpha\beta;\gamma\delta}\right)
y^{\alpha}y^\beta y^\gamma y^\delta
+{\cal O}(\partial^5)
\label{detmet}
\end{eqnarray}
For the square root of the metric determinant and its
inverse we have:
\begin{eqnarray}
g^{1/2}(y)&=&1+\frac{1}{6}R_{\alpha\beta}(y_0)y^{\alpha}
y^{\beta}-\frac{1}{12}
R_{\alpha\beta;\gamma}
y^{\alpha}y^{\beta}y^{\gamma}
+\left(\frac{1}{72}R_{\alpha\beta}R_{\gamma\delta}-
\frac{1}{180}R_{\lambda\alpha\beta\kappa}
R^{\lambda\;\;\;\;\kappa}_{\;\;\gamma\delta}
+\frac{1}{40}R_{\alpha\beta;\gamma\delta}\right)
y^{\alpha}y^\beta y^\gamma y^\delta\nonumber \\
&+&{\cal O}(\partial^5),
\label{detmet12}\\
g^{-1/2}(y)&=&1-\frac{1}{6}R_{\alpha\beta}(y_0)y^{\alpha}
y^{\beta}+\frac{1}{12}
R_{\alpha\beta;\gamma}
y^{\alpha}y^{\beta}y^{\gamma}+\left(\frac{1}{72}R_{\alpha\beta}
R_{\gamma\delta}+
\frac{1}{180}R_{\lambda\alpha\beta\kappa}
R^{\lambda\;\;\;\;\kappa}_{\;\;\gamma\delta}
-\frac{1}{40}R_{\alpha\beta;\gamma\delta}\right)
y^{\alpha}y^\beta y^\gamma y^\delta\nonumber\\
&+&{\cal O}(\partial^5)
\label{detmet-12}
\end{eqnarray}
\vspace{-.5cm}
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\end{document}

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A Happy Healthy Mummy: Can you read a food label?
What information are brands happy to share?
In fact, consumer groups have often argued that standards of food labelling are varied and confusing. At the moment, manufacturers are only obliged to give a breakdown of the nutritional content of the food if the product also makes a nutrition claim such as 'low-fat', or a health claim such as 'calcium helps build strong bones', or if vitamins or minerals have been added to the product. You might have heard of the 'traffic light' system, although even this has been criticised in the past as it allowed retailers and manufacturers to present the nutritional information in a variety of ways. From June 2013 however, the Department of Health launched a new 'front of pack' format using the red, amber and green colour-coding, with nutritional information and percentage reference intakes (RIs, formerly known as Guideline Daily Amounts) to display the amount of energy, fat, saturates, sugars and salt in food and drink products. This is of course still a voluntary system, but does aim to provide a more consistent format for UK consumers, and many of the leading supermarkets and manufacturers are signed up to it - perhaps because shoppers are becoming more savvy - or perhaps because the rising cost of treating diseases that result from obesity and poor nutrition are major issues facing our society (and government) today. A cynic might even suggest that if they know if don't do it voluntarily, it's likely they could be forced into following a system they didn't have a stake in developing. It's really worth scanning labels as you shop - if only to see which brands are happy to provide you with information. I'm guessing those that don't have a reason not to!
Rather more encouragingly, manufacturers will face tougher laws after December 2016, when under EU rules, back of pack nutrition labelling will be compulsory for all pre-packed foods. After that it'll be up to the consumer to educate themselves more about nutrition - although it's still easy to feel confused when new research often contradicts what we may be have been led to believe was correct in the past!
How do you feel about reading labels on the foods you buy? Do you bother? Do you feel overwhelmed? Do you have a basic idea of what you should and shouldn't be looking for? Let me know in the comment box below.
I don't buy a huge amount of packaged and processed food but I do read the labels when I do ... and I am not sure I always totally understand what is written there. Better to cook from scratch therefore if you can.
The moral of the story seems to be that you have to take responsibility for what you eat - harsh but true!!! Ideally everyone should cook from scratch, but I'd still like the powers that be to be more forceful with manufacturers - people have a right to know if food is unhealthy. I suspect that in the future processed food will be more regulated - just as tobacco and cheap alcohol became increasingly regulated as society had to deal with the fallout from that. Thank you for commenting!


text:
Russia Seeks Arrest of Hermitage Capital Executive
By William Mauldin
Updated March 29, 2010 12:30 pm ET
MOSCOW—Lawyers for Hermitage Capital Management received a decree from Russian authorities in March seeking the arrest of one of its executives, about five months after one of the firm's attorneys, Sergei Magnitsky, died in a Moscow jail, the fund said Monday.
Hermitage's Ivan Cherkasov is wanted by the Russian interior ministry for underpayment of taxes, but Hermitage said those accusations contradict a finding by the Federal Tax Service that the taxes were in order, the fund said in a statement from its London office.

redpajama_set_name: RedPajamaCommonCrawl


text:
Gunnfríður Jónsdóttir (* 26. Dezember 1889 auf Island; † 15. Mai 1968 in Reykjavík) war eine isländische Bildhauerin.

Leben
Gunnfríður Jónsdóttir war eine Tochter von Halldóra Einarsdóttir Andréssonar und Jón Jónsson. Ihr Großvater war der Dichter Einar Andrésson – er stand im Verdacht, Hexerei zu betreiben.

Sie studierte an der Kungliga Konsthögskolan in Stockholm bei dem Bildhauer Carl Milles. Von 1924 bis 1940 war Gunnfríður Jónsdóttir mit dem Bildhauer Ásmundur Sveinsson (1893–1982) verheiratet, die Ehe blieb kinderlos.

Literatur

 Islensk höggmyndalist 1900–1950. Listasafn, Reykjavík 1990, S. 32–41 (Digitalisat, Isländisch; PDF; 4 MB).
 Steingrím Davíðsson (Hrsg.): Listaverk Gunnfríðar Jónsdóttur. Reykjavík 1966.

Siehe auch
 Isländische Malerei und Bildhauerei

Weblinks
 Gunnfríður Jónsdóttir
 Gunnfríður Jónsdóttir, arktiv.is (Datenbank isländischer bildender Künstler)

Bildhauer (Island)
Isländer
Geboren 1889
Gestorben 1968
Frau

redpajama_set_name: RedPajamaWikipedia


text:
Listen: Artist Talk with Esther Teichmann
Artist talk presented by Photoworks and University of Brighton Photography
Using still and moving image, collage and painting, Esther Teichmann creates alternate worlds that blur autobiography and fiction. Central to her work lies an exploration of the origins of fantasy and desire and how these are bound to experiences of loss and representation.
In this artist talk, Esther will explain her approach to making work, the themes explored within it, her use of still and moving image and the techniques she uses for installation.
Esther Teichmann (b Germany,1980) has lived in London since she was eighteen and earned her MA and doctorate at the Royal College of Art. She has exhibited widely and a monograph of visual works, Fulmine, and a book of essays, Falling – On Loss, Desire and the Photographic, will be published by Stanley/ Barker in 2017. In 2013 Esther was a guest professor at the California College of the Arts in San Francisco and she is currently a Senior Lecturer at the University of the Arts London and visiting lecturer at the Royal College of Art. Her solo exhibition Heavy the Sea shows until May 2017 at Transformer Station, Cleveland.
For more of Esther's work, click here.
For more from our Watch & Listen section, click here.
Listen: Brighton Photo Biennial Artist Robin Maddock
Listen to Robin Maddock discuss his series 'Nothing We Can't Fix By Running Away' shown as part of Brighton Photo Biennial 2018.
Watch: Brighton Photo Biennial Artist Tereza Červeňová
Watch Tereza Červeňová discuss her series 'June' shown as part of Brighton Photo Biennial 2018.


text:
Q: SQlite Data on selection not present showing though data is inserted public boolean checkWhetherDataPresentorNot(String question_id, String selected_option, String session_id) {
    Log.d(" CHECKDATA ", "Question Id" + question_id);
    Log.d("CHECKDATA ", "Selected_option " + selected_option);
    Log.d("CHECKDATA", "Session ID " + session_id);

    db = this.getReadableDatabase();
    @SuppressLint("Recycle") Cursor cursor = db.query(TABLE_TEST_START, new String [] {QUESTION_ID_TEST_START, SELECTED_OPTION_TEST_START,TEST_START_ID}, "question_id_test_start = ? AND selected_option_test_start = ? AND test_start_id = ?",
            new String[] { question_id, selected_option, session_id },
            null,
            null,
            null);

    if (cursor.moveToFirst()) {
        Log.d("CHECKDATA ", "DATA IS PRESENT");
        return true;
    }else{
        Log.d("CHECKDATA ","DATA NOT PRESENT");
        return false;
    }

}

redpajama_set_name: RedPajamaStackExchange


As Sony's Venom starring Tom Hardy slowly approaches theaters this coming October, fans are already creating artwork, and in this case, sculptures dedicated to the host/symbiote. In the footage above, we see YouTuber Steven Richter begin the process of creating the bust, first building Hardy's Eddie Brock before adding Venom's details on one side of his face.
Everything is time-lapsed, from getting its color complexion correct, molding Venom's obnoxiously large mouth and teeth, as well as even getting Brock's lower wardrobe and hair to look identical to his outfit in the movie.
In related news, Alec Baldwin has exited DC's upcoming Joker movie.
The works of seven artists made entirely by computers.
Throwback screenprints and customized sneakers.


The process by which of a trade is entered into the books and records of the counterparts to a transaction. The settlement of currency trades may or may not involve the actual physical exchange of one currency for another.
Order type whereby an open position is automatically liquidated at a specific price. Often used to minimize exposure to losses if the market moves against an investor's position. As an example, if an investor is long USD at 156.27, they might wish to put in a stop loss order for 155.49, which would limit losses should the dollar depreciate, possibly below 155.49.
A technique used in technical analysis that indicates a specific price ceiling and floor at which a given exchange rate will automatically correct itself.Opposite of resistance.
The scanner was faster and simpler, however normally not quite as sharp as the digicam and tent!
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aid physicians in distinct specialties of medicine.
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your skin), therefore aiding your skin to reduce naturally.
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body from free-radical destruction, accountable for several uncomfortable signals of aging.
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One of the most noticeable consequences as it pertains to aging is how the lines appear on your skin and after several years of creasing and transferring, this is an exceptionally organic method.


text:
package com.thoughtworks.go.config;

import com.thoughtworks.go.config.remote.PartialConfig;
import com.thoughtworks.go.config.remote.RepoConfigOrigin;
import com.thoughtworks.go.server.service.EntityHashes;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.stereotype.Component;

import java.util.*;
import java.util.stream.Collectors;

@Component
public class PartialConfigHelper {
    private final EntityHashes hashes;

    @Autowired
    public PartialConfigHelper(EntityHashes hashes) {
        this.hashes = hashes;
    }

    /**
     * Tests whether two {@link PartialConfig} instances have identical structure and origin
     *
     * @param left  a {@link PartialConfig}
     * @param right a {@link PartialConfig}
     * @return whether or not the {@link PartialConfig}s are equivalent
     */
    public boolean isEquivalent(PartialConfig left, PartialConfig right) {
        return hasSameOrigins(left, right) && isStructurallyEquivalent(left, right);
    }

    /**
     * Determines the equivalence of to collections of {@link PartialConfig}s
     *
     * @param left  a {@link Collection<PartialConfig>}
     * @param right a {@link Collection<PartialConfig>}
     * @return whether or not the {@link Collection<PartialConfig>}s are equivalent
     */
    public boolean isEquivalent(Collection<PartialConfig> left, Collection<PartialConfig> right) {
        return Objects.equals(identities(left), identities(right));
    }

    private Set<String> identities(Collection<PartialConfig> collection) {
        if (null == collection) return Collections.emptySet();
        return collection.stream().map(this::identity).collect(Collectors.toSet());
    }

    private String identity(PartialConfig partial) {
        final RepoConfigOrigin origin = (RepoConfigOrigin) partial.getOrigin();
        return hash(partial) + ":" + origin.getMaterial().getFingerprint() + ":" + origin.getRevision();
    }

    /**
     * Tests whether two {@link PartialConfig} instances define structurally identical configurations.
     *
     * @param previous a {@link PartialConfig}
     * @param incoming a {@link PartialConfig}
     * @return whether or not the structures are identical
     */
    private boolean isStructurallyEquivalent(PartialConfig previous, PartialConfig incoming) {
        return Objects.equals(hash(incoming), hash(previous));
    }

    /**
     * Tests whether two {@link PartialConfig} instances share an identical
     * {@link com.thoughtworks.go.config.remote.ConfigOrigin}.
     *

     * This is needed because we need to update the origins of the generated {@link PipelineConfig} instances to match
     * the revisions of their {@link com.thoughtworks.go.domain.materials.MaterialConfig}s. If they don't, the pipelines
     * will not be scheduled to build.
     *

     * See {@link com.thoughtworks.go.domain.buildcause.BuildCause#pipelineConfigAndMaterialRevisionMatch(PipelineConfig)}.
     *
     * @param previous a {@link PartialConfig}
     * @param incoming a {@link PartialConfig}
     * @return whether or not the origins are identical
     */
    private boolean hasSameOrigins(PartialConfig previous, PartialConfig incoming) {
        return Objects.equals(
                Optional.ofNullable(previous).map(PartialConfig::getOrigin).orElse(null),
                Optional.ofNullable(incoming).map(PartialConfig::getOrigin).orElse(null)
        );
    }

    public String hash(PartialConfig partial) {
        // in case of a deserialization bug, a plugin may return an invalid config which might fail during serializing into xml
        // hence, in case of a serialization error, return null has hash
        //
        // it is safe to return null on a structurally invalid config, as once a the config is fixed, a hash will be computed.
        try {
            return digestPartial(partial);
        } catch (Exception e) {
            return null;
        }
    }

    /**
     * Computes a digest of a {@link PartialConfig}.
     *

     * NOTE: Don't move this to EntityHashingService; it will create a cyclic dependency. The dependency graph leading
     * here is pretty deep. Your head may explode.
     *
     * @param partial a {@link PartialConfig} to hash
     * @return a cryptographic digest that can be used for comparison.
     */
    protected String digestPartial(PartialConfig partial) {
        if (null == partial) {
            return null;
        }

        return hashes.digest(
                hashes.digest(partial.getGroups()),
                hashes.digest(partial.getEnvironments()),
                hashes.digest(partial.getScms())
        );
    }

}

redpajama_set_name: RedPajamaGithub


text:
There are several rivers named Urubu River in Brazil:

 Urubu River (Amazonas)
 Urubu River (Rio de Janeiro)
 Urubu River (Roraima), a river of Roraima
 Urubu Grande River


text:
Home Music Music Lists
August 28, 2017 12:00PM ET
MTV VMAs 2017: The Best, Worst and Most WTF Moments
From Katy Perry's spotty debut as host to Kendrick Lamar's victory to a Chester Bennington tribute, here's the evening's highs and lows
Elias Leight
Al Shipley
At last year's MTV Video Music Awards, performer Britney Spears had the considerable misfortune of following Beyoncé's masterful Lemonade medley performance. This year's VMAs had the misfortune of simply lacking Beyoncé or a Beyoncé-caliber moment – or even a Spears moment, for that matter. Nevertheless, the raucous evening, held inside the Forum in Inglewood, California, produced plenty of fodder to kvell, kvetch and tweet about: from Kendrick Lamar's absolute domination to host Katy Perry's iffy jokes to a touching Chester Bennington tribute to the thunder-stealing return (sort of) of Taylor Swift. Here, Rolling Stone recaps the night's highs and lows.
Kevin Winter/Getty
Best: Kendrick Lamar's Opener Is Lit, Literally
Kendrick Lamar opened the VMAs with another emphatic performance, blazing through "DNA" and "Humble" with taut focus. Both of the Damn. tracks were produced with club-shaking force by Mike Will Made It, but at the VMAs, Lamar reworked them as muscular funk-rock, full of ringing guitars and clobbering drum fills. He opened the performance clad head-to-toe in red and rapped from within a shifting cage of lasers; later, the lasers were replaced by impressive pyrotechnics: a burning, multistory latticework and a dancer who twirled while enveloped in flames. Lamar broke up the comforting flow of his Number One hit "Humble" with fierce, stop-start passages, while the dancers appeared to climb the firewall behind him. It was an impressive display from the evening's leading light: Lamar was nominated for eight awards at the VMAs, more than any artist in any genre, and won six, including Video of the Year, for "Humble." EL
Worst: Newcomers Get Sidelined
This VMAs broadcast had a frustrating tendency to relegate vital new artists to the pre-show or cut their performances short before they made it past their first hook. This was especially egregious in the case of Cardi B: her hit "Bodak Yellow" rose to Number Three on the Hot 100 last week, but the ascendant rapper was still stuck in the pre-show portion of the event. She delivered one of the night's most exciting performances anyway, employing an arsenal of sneers, snarls and stiff-arms as she strutted around MTV's studio. Also stuck in the pre-show was Khalid, who hopped from one foot to the other as he delivered cheerful renditions of his hit "Location" and his latest single, "Young Dumb & Broke." (He'd later win the fan-voted Best New Artist during the show.) And during the main event, singer-songwriter Julia Michaels was just settling into her string-slathered, heart-on-sleeve ballad "Issues" when MTV cut to commercial. All three deserved better. EL
WTF: Katy Perry's Iffy Comedic Chops
Without the timing of a proper comedian – or, for that matter, Cardi B – Katy Perry relied on weirdness and inside jokes to take her through the show, which began with a painfully long "I've been on the moon for a year" astronaut bit. Some jokes landed – there was a nice nod to the conspiracy theory that Perry is in fact JonBenét Ramsey and a riff about a dirty drafted tweet to Ed Sheeran – but she too often fell short: normal rather than normie. Ellen DeGeneres was seen nearly wincing when Perry doubled down on a fake-baby gag by shooting fake breast milk under her dress. Ouch. NM
Best: Fifth Harmony Hits the Ejector Button
There was a lot to take in during Fifth Harmony's performance of "Angel" and "Down:" Tightly coordinated movements from the group, a slew of silver-suited dancers, Normani Kordei wowing the crowd by dropping into a split onstage, a brassy, EDM-inspired breakdown and a cool-as-ice Gucci Mane ambling onstage to deliver a laid-back verse. But all of this was overshadowed by Fifth Harmony's pointed opening: the group began "Angel" cloaked and hooded on top of tall glass boxes, with a faceless fifth singer in place of former member Camila Cabello. Suddenly the fifth was thrown backwards as if shot, and she tumbled out of sight. After that ejection, Kordei, Ally Brooke, Dinah Jane and Lauren Jauregui threw back their hoods and began to sing.
Perhaps this will stop interviewers like Charlamagne the God from continuing to ask questions about Cabello's exit. During the pre-show, he asked Fifth Harmony, "If y'all win the choreography award, do you give one to the member who's no longer in the group?" "If we win for best choreography, we'll actually give it to our choreographer," Kordei replied slyly. "He deserves it." EL
Worst: Lorde's Wordless Interpretive Dance
Performing a deep cut at an awards show, while all your peers are parading their biggest hits, is the kind of power move usually reserved for VMA superstars. Lorde isn't quite at that level, which made it all the more gutsy that she took the VMA stage with "Homemade Dynamite" instead of one of the Melodrama's two singles. More surprising was that she opted to simply dance, not even lip syncing as the album track rang out. To be fair to Lorde, that may not have been her original plan; she tweeted shortly before show time that "i have the flu so bad i needed an IV #showgoeson #cutelilproppedupcorpse." Nonetheless, it was not the triumphant moment for Lorde's quirky choreography that it could've been. AS
WTF: Ed Sheeran Makes Friends with Lil Uzi Vert
At this point, Ed Sheeran has rapped and collaborated with rappers more than, really, anybody but Ed Sheeran could possibly ask for. But it was still a little surprising when SoundCloud rap's favorite rock star Lil Uzi Vert crashed the stage to rap over the end of Sheeran's ubiquitous trop house chart-topper "The Shape of You." And when Sheeran began singing Uzi's "all my friends are dead" emo rap hit "XO Tour Llif3," Cardi B's raised eyebrow was the audience reaction shot that summed up how pretty much everyone at home felt. AS
Best: Jared Leto Honors Chester Bennington
Current-era award shows are filled with enough tributes to keep Dave Grohl busy year-round. Yet few felt as profound as the words Jared Leto spoke for Linkin Park's Chester Bennington, who took his own life at age 41 in July. "I think about his heart, which he wore on his sleeve," said Leto. "I think about his band who were really his brothers, and I remember his voice, at one ferocious and delicate, which will live forever." In a program filled with messages about suicide prevention, Leto's message was especially resonant: "The absolute biggest breakthroughs in life lie just beyond the darkest days." NM
Worst: The Bored-AF Audience
During the VMAs, many of honorees and presenters were cordoned off from the rabble in an awkward lounge area, and when the camera cut to capture a celebrity reaction to a joke, acceptance speech, or performance, theyfrequently appeared bored: The Bleachers' Jack Antonoff stared into the distance and munched on a banana, seemingly unaware of his surroundings, Lorde was surprisingly nonplussed during Pink's powerhouse medley, Ellen DeGeneres remained impassive as host Katy Perry whiffed on some of her one-liners, and DeGeneres' wife Portia de Rossi seemed similarly unmoved. There were a few notable exceptions: Bebe Rexha, who exploded with excitement when Fifth Harmony's Normani Kordei dropped into a split during "Down," and P!nk's daughter Willow, who flashed a toothy smile at her mother after she accepted the Michael Jackson Video Vanguard Award. EL
WTF: DNCE, Rod Stewart's "Sexy" Vegas Collab
Rod Stewart was making tasteless synth-rock before the tasteless synth-rockers of DNCE were even born. Their VMA collaboration, however, went back even further, resurrecting "Do Ya Think I'm Sexy?" for a performance that no one asked for but a few people, somewhere, probably appreciated. DNCE handled the first verse on their own, frontman Joe Jonas (with a modified porn-stache) crooning the vocals in a zebra print suit, unafraid of what stains the set's black lights might uncover. Stewart, even bolder, strutted off an elevator in skin-tight white pants, then sang the lyrics with the confidence of a man who believes that each year leaves him sexier than the one before. Naturally, the performance was filmed on a B-stage atop a tower in Vegas. NM
Best: A "Younger" Miley's Chill VMAs Return
Did Miley Cyrus – she's Dolly Parton's goddaughter, after all – just go full-on country? Sort of. With a Patsy Cline meets Tom Petty vibe, the recently reinvented Cyrus showcased her strongest vocals yet performing "Younger Now." She wore cat-eye glasses and a poodle skirt, and danced past a row of old folks in leather jackets, which made more sense during the song's Link Wray bridge. There were also boy scouts on tricycles, which made little sense at all. The whole thing was weird and fun and low-key, especially compared to the universe-splitting, twerking bears of her 2013 VMAs. But she was so much older then. NM
Worst: 30 Seconds to Mars' Goggle Gimmick
Even beyond Jared Leto's Hollywood fame, 30 Seconds to Mars have always had a flair for visual spectacle in their videos and concerts – which justified their selection as the VMAs' token rock performance this year. But the band's performance of their new single "Walk On Water" was perhaps too ambitious, filmed entirely through heat vision goggles that ultimately looked like a more boring version of the Beastie Boys' video for "So What'cha Want." Leto wore and removed a ski mask as a mild attempt at having fun with the heat sensors, and Travis Scott wandered into the frame to shoehorn a few bars of his hit "Butterfly Effect" into the song, but the performance just never had a pulse. AS
Best: Pink's Refreshingly Excellent Speech
Pink's performance before accepting the Video Vanguard Award was a solid testament to the strength of her discography, if not her fairly unremarkable videography, and as usual she elided the R&B-tinged Can't Take Me Home era from her career narrative. It was all worth it, however, for Pink's touching, funny acceptance speech, which began with an anecdote about her 6-year-old daughter, Willow (seen watching in the audience). "She said, 'I'm the ugliest girl I know.' And I said, 'huh?' And she was like 'Yeah, I look like a boy with long hair.'" She proceeded to wax eloquent about androgynous rock stars and "artists who live their truth" like Prince and Annie Lennox who made it possible for people like Pink and her daughter to feel comfortable in their own skin. AS
Best: Things Get Woke
In the all-time ranking of memorable VMA moments, a speech by a descendant of Confederate general Robert E. Lee (introducing Charlottesville victim Heather Heyer's mother, Susan Bro) may not be able to beat, say, the fist-fight between Kid Rock and Tommy Lee. Yet in the context of the 2017 ceremony, it was undoubtedly a highlight, its tone consistent with a show in which artists repeatedly tried to figure out how they – and their fans – can effect change. Kesha reminded viewers that they're not alone in the world. Cardi B promised to stand with Colin Kaepernick.
Logic's performance of "1-800-273-8255," with Khalid and Alessia Cara, was particularly intense. The song's title, the number of the National Suicide Prevention Hotline, illuminated the stage, which was also occupied by rows of suicide survivors. "I don't give a damn if you're black white or anything in between. We're all born equal, but we're treated unequal," the rapper said in an emotional speech after the last verse. NM
Worst: Katy Perry's Meh Finale
The rebuttal that wasn't. Closing out the show, Perry reprised the basketball theme of her recently released music video for "Swish Swish" when she brought the single to the VMAs, but the performance was flat and lacking sparks – especially when compared with her rival Taylor Swift's "Look What You Made Me Do" song and video (more on that in a second). Perry soared through the air over the stage twice with the aid of a harness, but she had already done that to start the show, so the trick began to wear thin, and when basketball-like balloons fell from the ceiling, the hoops-theme started to seem overdone. Nicki Minaj stood still as she rapped her "Swish Swish" verse and let the camera move around her, so the energy level of the performance barely shifted. Perry finished the song on a high note by triumphantly dunking a basketball. But moments later, she was left floating awkwardly in the air once more. EL
WTF: Taylor Swift, in Absentia, Hijacked Entire Night
MTV has always relished the collisions of big names and big egos that other award shows might prefer to keep polite and proper. And the moment in 2009 that Kanye West grabbed the mic and interrupted a young Taylor Swift instantly birthed the VMAs' most enduring soap opera. Both returned to the show in 2010 to debut songs seemingly inspired by the incident, Swift presented an award to West in 2015 (the same year she squashed another nascent beef with Nicki Minaj), and in 2016, West was nominated in several major categories for "Famous," the song that finally turned their odd shared history into a full-on feud. So it's no surprise that the 2017 VMAs offered a continuation of the saga, with Swift – present only in the video which premiered 20 minutes into the show – poking fun at her image in a playful, bombastic clip for her disdainful new ode to West (and others), "Look What You Made Me Do." And of course, Swift's second most famous frenemy, Perry, hosted the whole night, mostly avoiding the fracas besides a few highly interpretable, wordless winks and nudges. AS
In This Article: 30 Seconds to Mars,	Ed Sheeran,	Fifth Harmony,	Katy Perry,	Kendrick Lamar,	Lorde,	Miley Cyrus,	MTV Video Music Awards,	Pink,	Rod Stewart,	Taylor Swift


Home » Blogs » Real World Research » Audio Books vs. Book Books: Which Does the Brain Prefer?
I've recently started listening to audio books. The idea never appealed to me much because I've never liked being read to. Reading is a solitary experience for me and being read to always seemed a little icky, though I couldn't tell you why.
Certainly being read to has a venerable history. At one time, all writing was meant to be read aloud, since few people could read. And reading aloud was family entertainment in the pre-radio, pre-TV days. And, of course, reading to children is both cozy and the first step towards their literacy.
So it's not like listening to books is anything new. But downloadable audio books are increasingly popular (though the growing popularity of ebooks is the headline news in publishing.) Fans of audio books even have their own magazine.
The first audio book I listened to was Bossypants, which is read by Tina Fey herself. Now I'm listening to Never Let Me Go, by Kazua Ishiguro, which is beautifully read by Rosalyn Landor, who strikes a tone as wistful as the book and conveys changes of character with just the slightest change in her voice. Narration, I realize, is an art form unto itself.
But I'm still not sure how I feel about the audio book. It might be seducing me, but I worry about whether I'm having the experience of the book the author originally intended. Do we lose something of a novel when we don't see the words spelled out in front of us? Is the medium integral to the message?
I didn't think about this with Bossypants, a chatty autobiography read by the author. But with novels, I wonder if the experience of listening to a book is the equivalent of animating a painting. Cool, but different from what it was. Could we label one "better"?
A series of experiments measured participants' response times to true or false questions, some of which required imagery. For example, "A grapefruit is bigger than a cantaloupe" requires you to actually imagine those fruits in order to respond.
What this research revealed perhaps argues for the superiority of the audio book: Reading interferes with imagery.
When you think about it, this makes sense because (as the researchers explain) reading and imagining both require visual representation. It appears that when the visual bits of our brain are busy taking in the written word, there's less of them available for creating an image of the content.
And indeed, I find that to create an image while I read, I often must pause, look away from the page, and imagine. With an audio book, the images come practically unbidden. Listening doesn't get in the way of that.
The skill of the narrator undoubtedly affects the perception of the listener with audio books—but so does the skill of a reader who is reading a book. The better you read, the better your understanding of what you've read.
Listening also requires you to hear every single word, unlike reading, in which some of us are likely to take in whole sentences at a time, swallowing books in big gulps. That must also have an effect on our experience of a book, and for the better. I may be a more thorough listener than I am reader.
So far, everything I've read speaks to the superiority of listening to rather than reading a book. Funny though: I can't imagine curling up in a big armchair and spending an afternoon listening to a book. I'm certain I would become restless; I do most of my listening while driving or walking the dog or cleaning house. For some reason, reading seems more of an activity than listening, even though both require the same amount of cognitive engagement, and the physical engagement of reading—holding a book and turning pages—is nominal.
Do any fiction writers out there have feelings about people reading vs. listening to their stories?