text:
\section{Introduction}

Since the precise measurement of the Gottfried sum rule
\cite{gottfried67} by the New Muon Collaboration (NMC) \cite{nmc91} in
1991, the possibility that $\overline u (x) \neq \overline d (x)$ in the
proton has been extensively studied. The reason for this attention is that, in
the quark-parton model, the Gottfried sum rule is expressed as:

\begin{equation}
S_G = \int_0^1 [F_2^{p}(x) - F_2^{n}(x)]\frac{dx}{x} = \frac{1}{3}
+ \frac{2}{3}\int_0^1 [\overline{u} (x) - \overline d (x)],
\label{0}
\end{equation}
where charge symmetry between the proton and the neutron was used.
If SU(2) flavor symmetry, $\overline u = \overline d$, is also assumed,
$S_G$ is reduced to 1/3.

In order to test the Gottfried sum rule experimentally, the NMC \cite{nmc91} performed
deep inelastic muon scattering on hydrogen and deuterium and measured
the deuteron/proton cross-section ratio. To extract the proton
and neutron structure functions, they used the simple relation
$F_2^p (x) - F_2^n (x) = 2F_2^d (x) (1 - F_2^n (x)/ F_2^p (x))/
(1 + F_2^n (x) / F_2^p (x))$, with $ F_2^n (x) / F_2^p (x)= 2F_2^d (x)
 / F_2^p (x) -1$. A parametrization \cite{nmc91,nmc92} for $F_2^p (x)$
based on NMC, SLAC and BCDMS data was used.
The NMC reported the following value for the Gottfried sum
rule:

\begin{eqnarray}
S_G(0.004\leq x \leq 0.8) &=& 0.221\pm 0.008(stat)\pm 0.019(syst),
\nonumber \\*
S_G &=& 0.235\pm 0.026 ,
\label{00}
\end{eqnarray}
where the contribution for $x>0.8$ was estimated using a smooth
extrapolation of $F_2^n / F_2^p$ and for the region $x<0.004$
a Regge-like behavior ($ax^b$ with $a=0.2\pm 0.03$ and
$b=0.59\pm 0.06$) was assumed. It is worth noticing that in the extraction of
$F_2^n / F_2^p$, target mass corrections and nuclear effects were
not included, nor were higher twist corrections included
in the sum rule as a whole. However, a detailed examination of binding,
shadowing and meson exchange (anti-shadowing)
corrections \cite{wally93} does not greatly change
their conclusions.

The experimental result from the NMC yields:

\begin{equation}
\int_0^1 dx [\overline u(x) - \overline d (x) ]= -0.147 \pm 0.039.
\label{000}
\end{equation}

Theoretically, there has been some
indication for $\overline u \neq \overline d$
since the work of Feynman and Field \cite{feynman77}, where the Pauli principle
was invoked to justify that $u\overline u$ pair creation is suppressed relative
to $d\overline d$. Later, Ito et al. \cite{ito} measured continuum dimuon production and
determined the sea quark distribution
in the context of the Drell-Yan description
of dilepton production. From an analysis of the
logarithmic derivative of the measured
cross section, they inferred that the Drell-Yan model, assuming a symmetric
sea, underestimated the measured slope. This fact was interpreted
as an indication of broken symmetry in the sea,
with $\overline u < \overline d$.
Somewhat later, Thomas \cite{tony83}, following Sullivan's suggestion
\cite{sullivan} that the pion cloud can contribute to the nucleon
structure function, realized that pion dressing of the nucleon
naturally predicts an excess of $\overline d$ over $\overline u$. This is
because, for instance, the proton has a greater probability to
emit a $\pi^+$ than a $\pi^0$ or a $\pi^-$. In fact, this observation was
used \cite{tony84} to interpret the Ito et al. results on dilepton production.
Since then, the idea of pions producing an
asymmetric sea has been widely explored
\cite{find} as a possible interpretation of the NMC results.
Further theoretical evidence for asymmetry in the sea was found in
the calculation of quark distributions of Signal and Thomas \cite{signal89}.
The matrix elements of the quark distribution involve intermediate
states where the photon scatters on a valence quark and also intermediate
states where the photon oscillates into a $q\overline q$ pair and a quark or
antiquark is inserted in the nucleon. Again, following the idea of Field and
Feynman, it is easier to insert a $d$ than a $u$ quark into the proton.

Even with all this evidence, until the NMC experiment, the nucleon
continued to be seen as having a symmetric sea. There is a simple explanation
for this. The whole set of data available could be described by
parametrizations of parton distributions where it was assumed that
$\overline u(x) = \overline d(x)$, with the price of a slightly odd
behaviour of the valence quark distribution when $x\rightarrow 0$ \cite{mrs93a}.
In this line of thought, it was also proposed that the NMC result
could still be reproduced with flavor and charge symmetry but
with a modified behaviour for $F_2 (x)$ at small $x$, in such way that
the integral $(F_2^p (x) - F_2^n (x))/x$ saturates the naive
expectation 1/3 \cite{mrs90}. However, the NA51 collaboration \cite{na51}
recently carried out an experiment proposed by Ellis and Stirling \cite{ellis91}
to distinguish between a symmetry breaking effect
and an odd small $x$ behaviour of the parton distributions.
The NA51 results indicate
a strong flavor symmetry breaking at $x=0.18$, thus reinforcing the
idea that the sea is indeed flavor asymmetric. In a recent reanalysis
\cite{steffens96} of
the NA51 result, this conclusion was reinforced, although the
exact size of the flavor asymmetry is dependent on the extent of
the charge symmetry breaking between the proton and the neutron. Finally,
calculations based on meson cloud models \cite{harald} are also able to produce
an asymmetry compatible with what is measured by the NA51 collaboration.

We then see that the Pauli principle and pion dressing of the
nucleon are two powerful tools to understand flavor symmetry breaking.
If a bare proton is seen as having three valence quarks,
two up and one down, then
all the antiquarks have their origin in gluonic and/or
pionic effects. For instance,
pair creation through gluon emission of one of the valence quarks will
produce a intermediate state of four quarks and one antiquark. The flavor
of the antiquark is certainly restricted by the valence structure of the proton
via the Pauli principle. However, one should also account for
the antisymmetrization between the various quarks. In this case, the valence
structure will also interfere in the number of antiquarks created.
In order to understand the interplay of these effects, we study in some
detail the relation between the
Pauli principle and quark antisymmetrization in pair creation inside
a nucleon.

In the pionic case the same reasoning applies. Of course, the pion is
complicated by the fact that it is a pseudo-Goldstone boson. Thus its
internal structure will be quite complicated and the probability to find
just a $q-\bar{q}$ pair with the quark
in a 1-s state, would be expected to low. As
anti-symmetrization with the valence quarks in the nucleon will only
effect quarks in the 1-s state (of the nucleon bag) we expect that
anti-symmetry should have a rather small effect on pion loops.
Nevertheless, it is important to quantify this prejudice and that is the
main purpose of the present calculation.

\section{The Pauli Principle in the Proton Sea}

The most natural idea to account for the observed discrepancy between
theory and
experiment is to invoke the Pauli principle. Field and Feynman
\cite{feynman77} were the first ones to use this idea: if the
proton is composed of two valence $u$ quarks and one valence
$d$ quark then the creation of a quark - antiquark pair
through gluon emission will tend to give more $d\overline{d}$ pairs
than $u\overline{u}$ pairs. This is easy to understand
because there are five empty states
for the $d$ quark and four for the $u$ quark. Although this
idea is attractive and essentially correct, there is one other effect
to consider. We will see that we also have to
consider graphs containing interference between the sea quarks
generated in the gluon emission and the remaining quarks in the nucleon.
This effect will, in fact, hide the excess of $\overline{d}$
over $\overline{u}$ due to the Pauli principle.

To illustrate our discussion, we begin by reviewing the pair creation through
gluon exchange, following the calculations of Donoghue and
Golowich \cite{donoghue}. The assumption made is that the bare proton is
composed of two valence $u$ quarks, one valence
$d$ quark and its color, flavor and spin wave function is given by:

\begin{equation}
|p\rangle_0 = \frac{1}{\sqrt{18}}\epsilon^{\alpha \beta \gamma}[b^{\dagger} (u,\uparrow,\alpha)
b^{\dagger} (d,\downarrow,\beta) - b^{\dagger} (u,\downarrow,\alpha)
b^{\dagger} (d,\uparrow,\beta)] b^{\dagger} (u, \uparrow, \gamma) |0\rangle .
\label{1}
\end{equation}
The sea quarks are generated through a quark gluon interaction
given by:

\begin{equation}
H_I (x)=g\overline{\psi}(x)\gamma^{\mu}\frac{\lambda^a}{2}\psi (x){\it A}^{a}_{\mu} (x),
\label{2}
\end{equation}
where $g$ is the coupling constant, $A^a$ are the gluon fields and $\psi$ the
quark field. We are not going to worry about the form of the spatial part
of these operators but will concentrate only on
the color, spin and flavor part of the proton dressed with a quark - antiquark
pair:

\begin{equation}
|p\rangle = |p\rangle_0 +  \frac{1}{E_0 - H_0}H_I
\frac{1}{E_0 - H_0}H_I |p\rangle_0 + ...,
\label{3}
\end{equation}
with $H_0$ the free Hamiltonian. (Note that we have omitted the term corresponding to
a single gluon with the three valence quarks.)

The vector coupling between quarks and gluons allows for vertices
where the quark that emits the gluon either changes its spin or not.
Another feature of the pair creation process is that, since a particle
has opposite intrinsic parity to an antiparticle, at least one quark (or
the antiquark) or the two quarks and the antiquark have to be in a state of odd parity
in order to conserve the parity of the proton.
The proton wave function is then written as:

\begin{equation}
|p\rangle = |p\rangle_0 + C_s |\psi_s\rangle + C_v |\psi_v\rangle ,
\label{4}
\end{equation}
with $|\psi_v\rangle$ the wave function for the case where the
quark that emits the gluon can change its spin and $|\psi_s\rangle$
the wave function for
the case that the quark emitting the gluon does not change its
spin. The factors $C_s$ and $C_v$ depend on the particular form
for the spatial wave functions, according to Eq. (\ref{3}).
We then have:
\begin{eqnarray}
\langle p|p\rangle &=& 1 + |C_s|^2 \langle \psi_s|\psi_s\rangle
+ |C_v|^2 \langle \psi_v|\psi_v\rangle  \nonumber \\*
     &+& C_s^* C_v \langle \psi_s | \psi_v \rangle + C_v^* C_s \langle \psi_v | \psi_s \rangle .
\label{5}
\end{eqnarray}
The wave function involved in the vector coupling is calculated
from Eqs. (\ref{2}) and (\ref{3}):

\begin{eqnarray}
|\psi_v\rangle &=& (-1)^{\tilde{n} -1/2}b^{\dagger}(s,n,\rho)
\sigma^{l}_{n\tilde{n}}\lambda^a_{\rho\tilde{\rho}}
d^{\dagger}(\overline s , -\tilde{n},\tilde{\rho}) \nonumber \\*
             && b^{\dagger}(v,m,\delta)\sigma^{l}_{m \tilde{m}}
\lambda^a_{\delta \tilde{\delta}} b(\tilde{v},\tilde{m},\tilde{\delta})|p\rangle_0
\nonumber \\*
            &=&\frac{(-1)^{\tilde{n} -1/2}}{\sqrt{18}}\sigma^{l}_{n\tilde{n}}
\sigma^{l}_{m \tilde{m}}\lambda^a_{\rho\tilde{\rho}}\lambda^a_{\delta \tilde{\delta}}
\nonumber \\*
  &&\{\epsilon^{\alpha\beta\gamma}(
\delta_{\tilde{v}u}\delta_{\tilde{\delta}\alpha}A^{\dagger}
- \delta_{\tilde{v}d}\delta_{\tilde{\delta}\beta} B^{\dagger}
+ \delta_{\tilde{v}u}\delta_{\tilde{\delta}\gamma} C^{\dagger})|0\rangle \},
\label{6}
\end{eqnarray}
where $\sigma^l$ are the Pauli spin matrices, $\lambda^a$ are the Gell-Mann
color matrices and the spin and flavor indices are summed. We also
have:

\begin{eqnarray}
A^{\dagger} &=&b^{\dagger}(s,n,\rho)d^{\dagger}(\overline s , -\tilde{n},\tilde{\rho})
[b^{\dagger}(v,m,\delta)b^{\dagger}(d,\downarrow,\beta)\delta_{\tilde{m}\uparrow}
-b^{\dagger}(v,m,\delta)b^{\dagger}(d,\uparrow,\beta)\delta_{\tilde{m}\downarrow}]
b^{\dagger}(u,\uparrow,\gamma) \nonumber \\*
B^{\dagger} &=&b^{\dagger}(s,n,\rho)d^{\dagger}(\overline s , -\tilde{n},\tilde{\rho})
[b^{\dagger}(v,m,\delta)b^{\dagger}(u,\uparrow,\alpha)\delta_{\tilde{m}\downarrow}
-b^{\dagger}(v,m,\delta)b^{\dagger}(u,\downarrow,\alpha)\delta_{\tilde{m}\uparrow}]
b^{\dagger}(u,\uparrow,\gamma) \nonumber \\*
C^{\dagger} &=&b^{\dagger}(s,n,\rho)d^{\dagger}(\overline s , -\tilde{n},\tilde{\rho})
b^{\dagger}(v,m,\delta)\delta_{\tilde{m}\uparrow}[b^{\dagger}(u,\uparrow,\alpha)
b^{\dagger}(d,\downarrow,\beta)
-b^{\dagger}(u,\downarrow,\alpha)b^{\dagger}(d,\uparrow,\beta)].
\label{7}
\end{eqnarray}
For the scalar coupling  the wave function is the same except that
the Pauli spin matrices are omitted and $n=\tilde n$, $m=\tilde m$.

The calculation of the wave function overlap is very long and tedious.
The results are displayed in Table \ref{tb21}, where $v$ refers to the
state of the valence quark after the gluon emission, $s$ corresponds to the state
where the quark in the sea is created and $g$ refers to the ground state.
Notice that there is no interference between the second and the third lines of
the table because of angular momentum conservation.
These calculations agree with the results of Donoghue and Golowich \cite{donoghue}.

\begin{table}
\begin{center}
\begin{tabular}{|c|c|c|c|}
State & $\langle \psi_s|\psi_s\rangle$ & $\langle \psi_v|\psi_v\rangle$ &
$\langle \psi_s|\psi_v\rangle$ \\ \hline
v=s=g & 0 & 4608 + 1792$\delta_{su}$ - 1024$\delta_{sd}$ & 0 \\
v=g,s$\neq$g & 0 & 4608 & 0 \\
v$\neq$g, s=g & 1152 + 640$\delta_{su}$ + 320$\delta_{sd}$ &
3456 + 1280$\delta_{su}$ - 320$\delta_{sd}$ & -320$\delta_{su}$ - 640$\delta_{sd}$ \\
v=s$\neq$g & 1152 + 128$\delta_{su}$ + 64$\delta_{sd}$ &
3456 - 384$\delta_{su}$ - 192$\delta_{sd}$ & 384$\delta_{su}$ + 192$\delta_{sd}$ \\
v$\neq$s, v$\neq$g, s$\neq$g & 1152 & 3456 & 0 \\
\hline
\end{tabular}
\end{center}
\caption{Amplitudes of probabilities to find a quark in the sea for the various
possible states of the sea quark and the valence quark that
emits the gluon. These probabilities are based only on the
spin-flavour-colour wave function.}
\label{tb21}
\end{table}

The most striking result to be read from Table \ref{tb21} is that
the probability to find the sea with a $u\overline u$ pair
is bigger than the sea composed of a $d\overline d$ pair.
This conclusion is the opposite of the experimental result collected
by the $NMC$ \cite{nmc91} where the measured
Gottfried sum rule is smaller than 1/3 - a result that implies
$\overline d > \overline u$.
How can we then understand that the intrinsic sea in the
proton generated by gluons favours $u\overline u$ pairs over
$d\overline d$ pairs? In principle, this result also contradicts the
intuitive picture introduced by Field and Feynman \cite{feynman77},
based on the Pauli principle. For each flavor there are
6 empty states (2 from spin and 3 from color). As the bare proton has
2 $u$ quarks and one $d$ quark, there are four available states for
insertion of $u$ quarks and 5 available states for insertion of $d$ quarks.
Although this reasoning is correct, one also cannot forget that
antisymmetrization between quarks is an additional complication.
The same excess of $u$ valence quarks that prevents $u\overline u$ pair
creation in comparison with $d\overline d$ pair creation, also
produces extra contributions because of antisymmetrization between the
$u$ quarks that does not exist for the $d$ quarks.

To understand this effect in more detail, we consider the case
where the quark that emits the gluon does not change its spin
but goes to an excited state and the sea is
created in the ground state or in a P state, according to
parity conservation. This case was chosen because it is
the simplest, as can be seen from table \ref{tb21}.
In Fig. \ref{figb2} we show the graphs containing $u\overline u$ pairs
and in Fig. \ref{figb3} the graphs containing $d\overline d$ pairs.

\begin{figure}[htb]
\vspace{8cm}
\centering{\
    \epsfig{angle=270,figure=uubar.eps,height=7.5cm}  }
\caption{Graphs containing $u\overline u$ pairs - in the case where the valence
quark emitting the virtual gluon goes to an excited state (i.e. $s=g, v\neq g$).}
\label{figb2}
\end{figure}

\begin{figure}[htb]
\vspace{8cm}
\centering{\
    \epsfig{angle=270,figure=ddbar.eps,height=7.5cm}  }
\caption{Graphs containing $d\overline d$ pairs - in the case where the valence
quark emitting the virtual gluon goes to an excited state (i.e. $s=g, v\neq g$).}
\label{figb3}
\end{figure}

The diagrams (a), (c) and (d) for $u\overline u$ are the analogues
of the diagrams (a), (c) and (d) for $\overline d$. Their values are
respectively the same because the created sea does not mix with the
valence quarks and, for this reason, it is not possible to distinguish
the sea flavor. Thus, in the case where the sea contracts with itself,
the values of these graphs are the same for any flavor. Of course, the
heavier the quark the smaller its contribution through these diagrams,
but this is an effect related solely to mass and not to spin statistics.
The computed value is
$\langle \psi_s|\psi_s\rangle=384$ for (a), $\langle \psi_s|\psi_s\rangle=
128$ for (c) and $\langle \psi_s|\psi_s\rangle=64$ for (d).
The diagram (b) for $\overline u$ is the analogue of diagram
(b) for $\overline d$ and they only exist when $u\overline u$ and
$d\overline d$ pairs are created. Their values are $\langle \psi_s|\psi_s\rangle=
4\times 160/3$ for (b) of Fig. \ref{figb2} and $\langle \psi_s|\psi_s\rangle
=6\times 160/3$ for (b) of Fig. \ref{figb3}. Generally speaking,  the set of
diagrams just mentioned are the only ones which are comparable. From
them, one easily sees the Pauli principle in action and
from it an excess of $\overline d$ over $\overline u$ as expected.
However, we also have to include graphs
(e) and (f) of Fig. \ref{figb2} for $\overline u$. They appear because
there is an excess of $u$ valence
over $d$ valence quarks.
If the graphs (e) and (f) are included, as they should
be, then the results change because these graphs give positive
contributions: $\langle \psi_s|\psi_s\rangle=320$ for (e)
and $\langle \psi_s|\psi_s\rangle=2\times 160/3$ for (f). With these
graphs included the probability to find a $\overline u$ antiquark in the proton
is bigger than the probability to find a $\overline d$ in the same proton.

One could doubt this interpretation of the relation between the
Pauli principle and antisymmetrization effects if one were to count only those
graphs where a $d$ quark emits a gluon. In this case, where a
$d$ quark emits a gluon,
we would have 6 empty states for a $d$ quark and only 4
empty states for the $u$ quark (if the sea is created in the
ground state). On the other hand, from Fig. \ref{figb2}
we see that graphs (b), (c) and (d) (also graph (f) if we
include the one coming exclusively from the excess of $u$ valence
quarks) give a larger contribution than the corresponding graphs
(c) and (d) of Fig. \ref{figb3}. As the graphs (b), (c) and (d), in principle, are not
related to the excess of $u$ valence quarks, there is an apparent
contradiction with the simple counting of states and the whole argument
of the role of the interference graphs presented by us before.
In reality, this contradiction is only apparent.

To better understand what is happening, consider a proton made of only
one $u$ and one $d$ quark. We consider again the case where the quark
that emits the gluon goes to an excited state.
In this case, if a $d$ quark emits a gluon, we would
have in principle 6 empty states for the insertion of a $d$ quark and $5$
for the insertion of a
u quark. The possible graphs would be the analogue of (b) and (c) from
Fig. \ref{figb2} for a $u\overline u$ in the sea and the analogue
of graph (c) from Fig. \ref{figb3} for a $d\overline d$ in the sea.
Again, we would have more $u\overline u$ pairs than $d\overline d$ pairs
and now it is clear why that happens: this is because there is one free
valence $u$ quark that can be exchanged with the sea and there is
no such free valence $d$ quark to be exchanged (in the case of a $d\overline d$
sea). The opposite situation happens when the $u$ quark emits the
gluon such that the sum of all diagrams, gluon emission from
$u$ and $d$ valence quarks, renders an equal probability for a
$u\overline u$ and $d\overline d$ pair creation, as expected in
a proton containing only one quark of each flavor. The lesson is that
we cannot treat the gluon emission in the proton from different
flavors separately, and expect the Pauli principle to work free
of any other effects.

For the combined result, the probability to find a $\overline u$
is bigger than the probability to find a  $\overline d$.
One then can say that interference terms overcome the
naive expectation of the Pauli principle. This result is extremely important
because it also says that if we have to antisymmetrize the sea
quarks with valence quarks in the case where the sea quarks are
generated through pions, then the whole set of conclusions about
the importance of the pions to the Gottfried sum rule might need to
be revised. In section III we investigate whether
this is in fact the case.

\subsection{Discussion of the role of anti-symmetry}

Before discussing pions it is of interest to relate the findings which
we have just reported to a similar issue in lattice QCD.
Figures 2(a), 2(c) and 2(d) are a very simple example of the
class of diagrams known in lattice QCD as disconnected insertions (DI)
\cite{leinweber96}.
While the DI's in lattice calculations include an infinite number of
gluon exchanges, the simple example which we have just presented
illustrates an important principal. The intermediate 4q states in
Figs. 2(a), 2(c) and 2(d) are unphysical in the sense that their wave function
is not totally anti-symmetric. Only when the exchange process, shown
in Fig. 2(b), is included do we have a physically meaningful result.
Within the lattice terminology the latter are connected insertions (CI).
Thus we find that neither the CI, nor the DI are physically meaningful
alone. This must be taken into account when one is trying to interpret
the physical mechanisms behind lattice results for flavour breaking
in the nucleon sea, the $\sigma$ commutator and so on.

\section{Antisymmetrization of the Quarks of the Pion}

We now wish to investigate the effect of anti-symmetrization on the
internal structure of the pion. As already noted in the introduction,
the pion (as a pseudo-Goldstone boson) is expected to have a complicated
internal structure. The effect of anti-symmetrization on this internal
structure can only effect components of the wavefunction where a quark
is in the 1-s state of the nucleon bag. We consider the case where the
pion couples to just one $q-\bar{q}$ pair with the extra quark in the
1-s state. In order to estimate the amplitude for this five-quark
configuration we use the cloudy bag model in which the pion-quark
coupling is dictated by chiral symmetry.
Motivated by the findings in the gluon case, we wish to check whether the
$\overline u$ antiquark is favoured over the $\overline d$
antiquark. If this turns out to be true, the asymmetry
in favour of $\overline d$ (from the process $u\rightarrow d\pi^+$)
over $\overline u$ (from  $d\rightarrow u\pi^-$) may be at risk.

The relevant diagram to study antisymmetrization
in relation to the pion cloud is the one shown in
Fig. 3b, involving two loops. However, we will be mainly interested
in the relative importance of the two loop process in comparison with
that involving only
one loop. To this end we also need to compute
diagram (a) of Fig. \ref{figb4}, which we do as a warm-up exercise.
The calculations are going to be done at the quark level
($q\rightarrow \pi q$), which means
that we need an interaction Hamiltonian between quarks and pions.
To fix the calculation in a specific scheme, we shall use the
interaction between quarks and pions as given by the
Cloudy Bag Model \cite{theberge80,thomas84}, where the interaction is
totally specified by chiral symmetry:

\begin{figure}[htb]
\vspace{5cm}
\centering{\
    \epsfig{angle=270,figure=pion1.eps,height=5cm}  }
\caption{The one and two loop graphs involved in pion emission.}
\label{figb4}
\end{figure}

\begin{equation}
H_I = \frac{i}{2f}\int d^3 x \overline\psi (x) \gamma_5
\tau^i \psi (x) \phi^i \delta(r - R),
\label{8}
\end{equation}
with $R$ the bag radius and

\begin{eqnarray}
\vec\phi (\vec x) &=& \frac{1}{(2\pi)^{2/3}}\int \frac{d\vec k}
{(2\omega_k)^{1/2}}(\vec a_{\vec k} e^{i\vec k \cdot \vec x} +
\vec a_{\vec k}^{\dagger} e^{-i\vec k \cdot \vec x}), \nonumber \\*
\psi^s (\vec x)&=&\frac{N}{(4\pi)^{1/2}}\sum\left\{
\left(\matrix{ij_0 (\omega r/R)\cr
              -j_1 (\omega r/R)\vec\sigma\cdot\hat r \cr}\right)\chi_m b(f,m,\alpha) +
\left(\matrix{-ij_1 (\omega r/R)\vec\sigma\cdot\hat r \cr
              j_0 (\omega r/R)\cr}\right)\chi_m d^{\dagger}(f,m,\alpha)\right\} \nonumber \\*
\psi^p (\vec x)&=&\frac{N}{(4\pi)^{1/2}}\sum\left\{
\left(\matrix{ij_1 (\omega r/R)\vec\sigma\cdot\hat r \cr
              j_0 (\omega r/R)\cr}\right)\chi_m b(f,m,\alpha) +
\left(\matrix{ij_0 (\omega r/R)\cr
              j_1 (\omega r/R)\vec\sigma\cdot\hat r \cr}\right)\chi_m d^{\dagger}(f,m,\alpha)\right\}.
\label{9}
\end{eqnarray}
Here $N^2=\omega^3/(2R^3 (\omega -1)sin^2 \omega)$, $\omega$ is the frequency
associated with a given principal quantum number and orbital angular momentum
and $j_0$ and $j_1$ are Bessel functions subject to the condition
$j_0 = j_1$ at the bag surface, $r=R$. We wrote the
explicit forms for the $s$ and $p$ waves for a quark inside a cavity because
they are going to be used later on.

Three vertices are relevant to our diagrams: $q\rightarrow \pi q$,
$\pi\rightarrow q\overline q$ and $0\rightarrow \pi q\overline q$.
As in the gluon case, there are some restrictions
due to the conservation of parity, that can be helpful. As is well known,
for fermions a particle has
opposite intrinsic parity to the antiparticle. By convention, a particle has parity
$+ 1$ and an antiparticle parity $- 1$. Also, the parity of one particle
relative to a set of others particles is given by $(-1)^l$, with $l$ the
orbital angular momentum
of the particle in question. If $\overline q$ has parity -1 then one
of the quarks (or the same antiquark) has to be in a state $l=1$, or a p-wave,
so that the proton parity is conserved. In general, the system must always be
in a state of even parity.

We now write down the explicit form for the interaction Hamiltonian for
both vertices. It happens that their form, besides the
creation operator for a quark or for a antiquark, is the same for both
processes $q\rightarrow \pi q$ and $0\rightarrow \pi q\overline q$:

\begin{eqnarray}
H_I^{q^{\tilde f} \rightarrow \pi q^f}&=& H_I^{0\rightarrow \pi q^f
\overline{q}^{\tilde f}} =
\frac{i}{2f(2\pi)^{3/2}}\left(\frac{\omega^a \omega^b}{(\omega^a -1)
(\omega^b - 1)}\right) \nonumber \\*
&&\int \frac{d^3 k}{(2\omega_k)^{1/2}}\chi^{\dagger}_m \vec\sigma\cdot\vec k \chi_{\tilde m}
\frac{j_1 (kR)}{kR} b^{\dagger}(f,m,\alpha)\tau^i_{f\tilde f}
    \matrix{b(\tilde f, \tilde m, \tilde\alpha)\cr
           d^{\dagger}(\tilde f, \tilde m, \tilde\alpha )\cr} a^{\dagger i}_{\vec k}.
\label{10}
\end{eqnarray}
For the case $\pi\rightarrow q\overline q$ the interaction is just
slightly different:

\begin{eqnarray}
H_I^{\pi\rightarrow q^f \overline{q}^{\tilde f}} &=&
\frac{-i}{2f(2\pi)^{3/2}}\left(\frac{\omega^a \omega^b}{(\omega^a -1)
(\omega^b - 1)}\right) \nonumber \\*
&&\int \frac{d^3 k}{(2\omega_k)^{1/2}}\chi^{\dagger}_m \vec\sigma\cdot\vec k \chi_{\tilde m}
\frac{j_1 (kR)}{kR} b^{\dagger}(f,m,\alpha)\tau^i_{f\tilde f}
 d^{\dagger}(\tilde f, \tilde m, \tilde\alpha ) a^{i}_{\vec k}.
\label{101}
\end{eqnarray}
In expressions (\ref{10}) and (\ref{101}) the indices $a$ and $b$
refer to the spatial wavefunctions of the quarks, or/and antiquarks,
involved in a specific reaction.

Once we have the interaction Hamiltonian, it is easy to
calculate the quantities in which we are interested. As announced, we
first calculate the probability to find a pion in the nucleon.
To be specific, we set $\omega =\omega^a = \omega^b$ in Eq. (\ref{10}),
which means that the quark remains in the ground state after it
emits the pion. In this case we get:

\begin{eqnarray}
\frac{1}{E_0 - H_0}H_I |p\rangle_0 &=& \frac{-i}{2f(2\pi)^{3/2}}\frac{\omega}{(\omega -1)}
\int \frac{d^3 k}{(2\omega_k)^{1/2}}\frac{1}{- \omega_k}
\chi^{\dagger}_m \vec\sigma\cdot\vec k \chi_{\tilde m}\frac{j_1 (kR)}{kR} \nonumber \\*
&&b^{\dagger}(v,m,\alpha)\tau^i_{v\tilde v}
      b(\tilde v, \tilde m, \tilde\alpha) a^{\dagger i}_{\vec k} |p\rangle_0 ,
\label{11}
\end{eqnarray}
where we used $(E_0 - H_0)(H_I |p\rangle_0) = (E_0 - (E_0 + \omega_k))
(H_I |p\rangle_0)$.
The probability to find a pion is then given by:

\begin{eqnarray}
P_{\pi} &=& \left(_0\langle p|H_I\frac{1}{E_0 - H_0}\right)\left(\frac{1}
{E_0 - H_0}H_I |p\rangle_0\right) \nonumber \\*
&=&\frac{\pi}{6f^2(2\pi)^3}\frac{\omega^2}{(\omega - 1)^2}\int_0^{\infty}
\frac{k^4 dk}{\omega_k^3}\frac{j_1^2 (kR)}{(kR)^2} \nonumber \\*
&& \times _0\langle p|b^{\dagger}(\tilde{v}^{\prime},\tilde{m}^{\prime}
\tilde{\delta}^{\prime})
\tau^i_{\tilde{v}^{\prime}v^{\prime}}\sigma^j_{\tilde{m}^{\prime}m^{\prime}}
b(v^{\prime}, m^{\prime}, \delta^{\prime}) b^{\dagger}(v,m,\delta)\tau^i_{v\tilde v}
\sigma^j_{m\tilde m}b(\tilde v, \tilde m, \tilde\delta)|p\rangle_0 .
\label{12}
\end{eqnarray}
After some calculation, the expectation value between bare proton states is found
to be 57. In terms of the pion components, this result reads 57 =$22\delta_{\pi^+}
+ 19 \delta_{\pi^0} + 16\delta_{\pi^-}$. This means that, as expected, the
probability to find a $\overline d$ in the nucleon is bigger than to
find a $\overline u$.
Using the identity $\omega/(f(\omega - 1)) = \sqrt{4\pi}18 f_{\pi NN}/
5m_{\pi}$, where $m_{\pi}$ is the physical pion mass and
$f_{\pi NN}$ the pion nucleon nucleon coupling constant, expression (\ref{12})
is rewritten as:

\begin{equation}
P_{\pi} = \frac{57}{25}\left(\frac{f_{\pi NN}}{m_{\pi}}\right)^2\frac{3}{\pi}
\int_0^{\infty} \frac{k^4 dk}{\omega_k^3}\left(\frac{3j_1 (kR)}{(kR)}\right)^2.
\label{13}
\end{equation}

\begin{figure}[htb]
\vspace{5cm}
\centering{\
    \epsfig{angle=270,figure=pion2.eps,height=5cm}  }
\caption{The two loop graph in time order.}
\label{figb5}
\end{figure}

The next step is the evaluation of the two-loop graph. Its calculation
is a straightforward application of Eqs. (\ref{3}), (\ref{10}) and (\ref{101}).
We start with graph (a) of fig. \ref{figb5}. The process is pion creation
with subsequent decay into a $q\overline q$ pair:

\begin{eqnarray}
\frac{1}{E_0 - H_0}H_I^{\pi\rightarrow q\overline q}\frac{1}{E_0 - H_0}H_I^{q\rightarrow\pi q}
|p\rangle_0 &=& \frac{1}{(2f)^2(2\pi)^3}\sqrt{\frac{\omega_f \omega_o \omega^s \omega^p}
{(\omega_f - 1)(\omega_o - 1)(\omega^s - 1)(\omega^p - 1)}} \nonumber \\*
   && \int d^3 k^{\prime} \int d^3 k \frac{1}{(2\omega_{k^{\prime}}
2\omega_k)^{1/2}}
\frac{R}{\omega_o - \omega_f - R\omega_k}\frac{R}{\omega_o - \omega_f -
\omega^s - \omega^p} \times \nonumber \\*
   &&\chi^{\dagger}_n \vec\sigma\cdot\vec{k}^{\prime} \chi_{\tilde n}
\frac{j_1 (k^{\prime} R)}
{k^{\prime}R}b^{\dagger}(s,n,\rho)\tau^j_{s\tilde s}
d^{\dagger}(\tilde s, \tilde n, \rho) a^j_{\vec{k}^{\prime}}\times \nonumber \\*
   &&\chi^{\dagger}_m \vec\sigma\cdot\vec k \chi_{\tilde m}\frac{j_1 (kR)}{kR}
b^{\dagger}(v,m,\delta)\tau^i_{v\tilde v}
b(\tilde v, \tilde m, \delta) a^{\dagger i}_{\vec k} |p\rangle_0 .
\label{14}
\end{eqnarray}
The energies of the intermediate states are clearly indicated on the
figure. The notation is the following: $\omega_0$ and $\omega_f$ are,
respectively, the quark frequencies before and after the pion emission
or absorption, $\omega^s$ and $\omega^p$ are, respectively, the frequencies of the s wave
quark and of the $p$ wave antiquark.

Similarly to case (a), we calculate the contribution from graph (b), where
a pion and a $q\overline q$ pair is created and the pion is
subsequently absorbed by a quark:

\begin{eqnarray}
\frac{1}{E_0 - H_0}H_I^{q\pi\rightarrow q}\frac{1}{E_0 - H_0}H_I^{0\rightarrow\pi q\overline q}
|p\rangle_0 &=& \frac{1}{(2f)^2(2\pi)^3}\sqrt{
\frac{\omega_f \omega_o \omega^s \omega^p}
{(\omega_f - 1)(\omega_o - 1)(\omega^s - 1)(\omega^p - 1)}} \nonumber \\*
   &&\int d^3 k^{\prime} \int d^3 k \frac{1}{(2\omega_{k^{\prime}} 2\omega_k)^{1/2}}
\frac{R}{\omega_o - \omega_f -\omega^s - \omega^p}\frac{R}{-\omega^s - \omega^p -
R\omega_k}\times \nonumber \\*
   &&\chi^{\dagger}_n \vec\sigma\cdot\vec{k}^{\prime} \chi_{\tilde n}
\frac{j_1 (k^{\prime} R)}
{k^{\prime}R}b^{\dagger}(v,n,\delta)\tau^i_{v\tilde v}
b^(\tilde v, \tilde n, \delta) a^i_{\vec{k}^{\prime}}\times \nonumber \\*
   &&\chi^{\dagger}_m \vec\sigma\cdot\vec k \chi_{\tilde m}\frac{j_1 (kR)}{kR}
b^{\dagger}(s,m,\rho)\tau^j_{s\tilde s}
d^{\dagger}(\tilde s, \tilde m, \rho) a^{\dagger j}_{\vec k} |p\rangle_0 .
\label{15}
\end{eqnarray}
The total amplitude for the two loop process is then given by the sum of graphs (a)
and (b). To better deal with this sum, we rewrite the product of operators in
Eq. (\ref{15}) as:

\begin{eqnarray}
&&b^{\dagger}(v,n,\delta)\tau^i_{v\tilde v}
b^(\tilde v, \tilde n, \delta) a^i_{\vec{k}^{\prime}}
b^{\dagger}(s,m,\rho)\tau^j_{s\tilde s}
d^{\dagger}(\tilde s, \tilde m, \rho) a^{\dagger j}_{\vec k} |p\rangle_0 \nonumber \\*
 &&= b^{\dagger}(s,n,\rho)\tau^j_{s\tilde s}d^{\dagger}(\tilde s, \tilde n, \rho)
b^{\dagger}(v,m,\delta)\tau^i_{v\tilde v}b(\tilde v, \tilde m, \delta)
a^i_{\vec{k}^{\prime}}a^{\dagger j}_{\vec k}|p\rangle_0 \nonumber \\*
 && + \delta_{\tilde v s}\delta_{\tilde n m}\delta_{\delta\rho}b^{\dagger}(v,n,\delta)
\tau^i_{v\tilde v}\tau^j_{s\tilde s}d^{\dagger}(\tilde s, \tilde m, \rho)
a^i_{\vec{k}^{\prime}}a^{\dagger j}_{\vec k})|p\rangle_0.
\label{16}
\end{eqnarray}
We then use $a^i_{\vec{k}^{\prime}}a^{\dagger j}_{\vec k}|0\rangle = \delta^{ij}
\delta(\vec{k}^{\prime} - \vec{k})|0\rangle$ to write the sum of the graphs (a) and (b)
as:

\begin{eqnarray}
  &&\frac{1}{E_0 - H_0}H_I^{\pi\rightarrow q\overline q}\frac{1}{E_0 - H_0}H_I^{q\rightarrow\pi q}
|p\rangle_0 + \frac{1}{E_0 - H_0}H_I^{q\pi\rightarrow q}\frac{1}{E_0 - H_0}H_I^{0\rightarrow\pi q\overline q}
|p\rangle_0 \nonumber \\*
  &&=\frac{1}{(2f)^2(2\pi)^3}\sqrt{\frac{\omega_f \omega_o \omega^s \omega^p}
{(\omega_f - 1)(\omega_o - 1)(\omega^s - 1)(\omega^p - 1)}} \times \nonumber \\*
  &&\int d^3 k \frac{(\vec\sigma\cdot\vec k)_{n\tilde n}(\vec\sigma\cdot\vec k)_{m\tilde m}}
{2\omega_k}\frac{j_1^2 (kR)}{(kR)^2} \frac{R^2}{\omega_o - \omega_f - \omega^s - \omega^p}
\times \nonumber \\*
  &&\left\{
\left(\frac{1}{\omega_o - \omega_f - R\omega_k} + \frac{1}{-\omega^s - \omega^p - R\omega_k}\right)
b^{\dagger}(s,n,\rho)\tau^i_{s\tilde s}d^{\dagger}(\tilde s, \tilde n, \rho)
b^{\dagger}(v,m,\delta)\tau^i_{v\tilde v}b(\tilde v, \tilde m, \delta) \right. \nonumber \\*
  &&\left. + \frac{1}
{-\omega^s - \omega^p - R\omega_k}\delta_{\tilde v s}\delta_{\tilde n m}\delta_{\delta\rho}
b^{\dagger}(v,n,\delta)\tau^i_{v\tilde v}\tau^i_{s\tilde s}d^{\dagger}(\tilde s, \tilde m, \rho)
\right\}|p\rangle_0 .
\label{17}
\end{eqnarray}

We are now ready to compute the probability, $P_{\pi q\overline q}$, to find a
$q \overline q$ pair in the nucleon. It is given by the square of the
amplitude (\ref{17}):

\begin{eqnarray}
P_{\pi q\overline q} &=& \left(\frac{1}{(2f)^2(2\pi)^3}\right)^2 \frac{\omega_f \omega_o \omega^s \omega^p}
{(\omega_f - 1)(\omega_o - 1)(\omega^s - 1)(\omega^p - 1)}\left(\frac{4\pi}{3}\right)^2 \nonumber \\*
  &&\int_o^{\infty} \frac{k^4 dk}{2\omega_k}\frac{j_1^2 (kR)}{(kR)^2}
\int_o^{\infty} \frac{k^{\prime 4} dk^{\prime}}{2\omega_{k^{\prime}}}
\frac{j_1^2 (k^{\prime}R)}{(k^{\prime}R)^2}\left(\frac{R^2}
{\omega_o - \omega_f - \omega^s - \omega^p}\right)^2 \times \nonumber \\*
  && \left\{\left(\frac{1}{\omega_o - \omega_f - R\omega_k} +
\frac{1}{-\omega^s - \omega^p - R\omega_k}\right) \left(\frac{1}{\omega_o - \omega_f -
R\omega_{k^{\prime}}} + \frac{1}{-\omega^s - \omega^p - R\omega_{k^{\prime}}}\right) \times
\right. \nonumber \\*
  && \left. \hspace{1em} _0\langle p| b^{\dagger}(\tilde{v}^{\prime},\tilde{m}^{\prime},\delta^{\prime})
\tau^i_{\tilde{v}^{\prime} v^{\prime}} \sigma^j_{\tilde{m}^{\prime} m^{\prime}}
b(v^{\prime}, m^{\prime}, \delta^{\prime})d(\tilde{s}^{\prime},\tilde{n}^{\prime},\rho^{\prime})
\tau^i_{\tilde{s}^{\prime} s^{\prime}} \sigma^j_{\tilde{n}^{\prime} n^{\prime}}
b(v^{\prime}, m^{\prime}, \delta^{\prime}) \right. \nonumber \\*
  && \left. \hspace{7em}
b^{\dagger}(s,n,\rho)\tau^k_{s\tilde s}\sigma^l_{n\tilde n}d^{\dagger}(\tilde s, \tilde n, \rho)
b^{\dagger}(v,m,\delta)\tau^k_{v\tilde v}\sigma^l_{m\tilde m} b(\tilde v, \tilde m, \delta)|p\rangle_0
\right. \nonumber \\*
  && \left. \hspace{1em} + \frac{1}{-\omega^s - \omega^p - R\omega_k}\frac{1}{-\omega^s - \omega^p -
R\omega_{k^{\prime}}}\times \right. \nonumber \\*
  &&\left. \hspace{1em} _0\langle p|d(\tilde{s}^{\prime}, \tilde{m}^{\prime}, \rho^{\prime})
\tau^i_{\tilde{s}^{\prime} s^{\prime}}\tau^i_{s^{\prime} v^{\prime}}
\sigma^j_{\tilde{m}^{\prime} m^{\prime}}\sigma^j_{m^{\prime} n^{\prime}}
b(v^{\prime}, n^{\prime}, \rho^{\prime})
b^{\dagger}(v,n,\rho)\tau^k_{vs}\tau^k_{s\tilde s}\sigma^l_{n m}
\sigma^l_{m \tilde m} d^{\dagger}(\tilde s, \tilde m, \rho) |p\rangle_0 \right\} .
\label{18}
\end{eqnarray}
The last quantities to be calculated are the expectation values between the bare
proton states.
It is a very long but straightforward calculation. We calculated only the case
where the valence quark that emits or absorbs the pion remains in the same orbital state.
The results are:

\begin{eqnarray}
  && _0\langle p| b^{\dagger}(\tilde{v}^{\prime},\tilde{m}^{\prime},\delta^{\prime})
\tau^i_{\tilde{v}^{\prime} v^{\prime}} \sigma^j_{\tilde{m}^{\prime} m^{\prime}}
b(v^{\prime}, m^{\prime}, \delta^{\prime})d(\tilde{s}^{\prime},\tilde{n}^{\prime},\rho^{\prime})
\tau^i_{\tilde{s}^{\prime} s^{\prime}} \sigma^j_{\tilde{n}^{\prime} n^{\prime}}
b(v^{\prime}, m^{\prime}, \rho^{\prime}) \nonumber \\*
  && \hspace{6em}
b^{\dagger}(s,n,\rho)\tau^k_{s\tilde s}\sigma^l_{n\tilde n}d^{\dagger}(\tilde s, \tilde n, \rho)
b^{\dagger}(v,m,\delta)\tau^k_{v\tilde v}\sigma^l_{m\tilde m} b(\tilde v, \tilde m, \delta)|p\rangle_0
= \nonumber \\*
  && = 684 - 110\delta_{\tilde s \overline u} - 181 \delta_{\tilde s \overline d} \nonumber \\*
  && _0\langle p|d(\tilde{s}^{\prime}, \tilde{m}^{\prime}, \rho^{\prime})
\tau^i_{\tilde{s}^{\prime} s^{\prime}}\tau^i_{s^{\prime} v^{\prime}}
\sigma^j_{\tilde{m}^{\prime} m^{\prime}}\sigma^j_{m^{\prime} n^{\prime}}
b(v^{\prime}, n^{\prime}, \rho^{\prime})
b^{\dagger}(v,n,\rho)\tau^k_{vs}\tau^k_{s\tilde s}\sigma^l_{n m}
\sigma^l_{m \tilde m} d^{\dagger}(\tilde s, \tilde m, \rho) |p\rangle_0 \nonumber \\*
  && = 972 - 54\delta_{\tilde s \overline u} - 81 \delta_{\tilde s \overline d} .
\label{19}
\end{eqnarray}

The results contained in expressions (\ref{19}) are quite surprising.
They say that, if the quark structure of the pion is important and if the
quarks from the pion are allowed to antisymmetrize with the
quarks from the parent proton, then the probability
for the antiquark in the pion to be a $\overline u$ is bigger than $\overline d$.
The above result is independent of particularities of the given model, in the
sense that expressions (\ref{19}) are a direct consequence of the bare proton
wave function, Eq. (\ref{1}). They are also a consequence
of assuming a pion quark interaction.

To better determine how important these second order effects could be, we will calculate
the ratio of probabilities $P_{\pi q\overline q}/P_{\pi}$. To this end we need to
perform the integrals over $k$ and $k^{\prime}$ in Eqs. (\ref{13}) and (\ref{18}).
These integrals are dependent on the particular value of the bag radius
and here we will display the results for $R=0.6,\; 0.8$ and $1\; fm$. We set
$\omega_0 = \omega_f$ and define the following integrals:

\begin{eqnarray}
I_1&=& \frac{1}{2}\int_0^{\infty}\frac{dx}{m_{\pi}^2 R^2 + x^2}
\frac{\omega^s + \omega^p + 2(m_{\pi}^2 R^2 + x^2)^{1/2}}
{\omega^s + \omega^p + (m_{\pi}^2 R^2 + x^2)^{1/2}}
\left(\frac{sin^2 x}{x^2} + cos^2 x - \frac{sinx cosx}{x}\right),
\nonumber \\*
I_2&=& \frac{1}{2}\int_0^{\infty}\frac{dx}{(m_{\pi}^2 R^2 + x^2)^{1/2}}
\frac{1}{\omega^s + \omega^p + (m_{\pi}^2 R^2 + x^2)^{1/2}}
\left(\frac{sin^2 x}{x^2} + cos^2 x - \frac{sinx cosx}{x}\right),
\nonumber \\*
I_3&=& \int_0^{\infty}\frac{dx}{(m_{\pi}^2 R^2 + x^2)^{3/2}}
\left(\frac{sin^2 x}{x^2} + cos^2 x - \frac{sinx cosx}{x}\right),
\label{20}
\end{eqnarray}
where we use $x=kR$. The numerical values of these integrals are
displayed in Table \ref{tb22}.
\newpage

\begin{table}
\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
Integral & $R=0.6\; fm$ & $R=0.8\; fm$ & $R=1\; fm$  \\ \hline
$I_1$ & 1.6848 & 1.24065 & 0.980306 \\
$I_2$ & 0.1946 & 0.1734430 & 0.157762 \\
$I_3$ & 4.92548 & 2.65299 & 1.63279 \\
$P_{\pi q\overline q}/P_{\pi}$ & 0.0217 & 0.0124 & 0.0081 \\
\hline
\end{tabular}
\end{center}
\caption{The two to one loop ratio for various bag radii.}
\label{tb22}
\end{table}
With these definitions, we rewrite expression
(\ref{18}) as:

\begin{eqnarray}
P_{\pi q\overline q} &=& \left(\frac{1}{(2Rf)^2(2\pi)^3}\right)^2 \frac{\omega_o^2}
{(\omega_o - 1)^2}\frac{\omega^s \omega^p}
{(\omega^s - 1)(\omega^p - 1)}\left(\frac{4\pi}{3}\right)^2
\left(\frac{1}{\omega^s + \omega^p}\right)^2\nonumber \\*
  &&\left\{I_1^2 \; (684 - 110\delta_{\tilde s \overline u} -
181 \delta_{\tilde s \overline d}) + I_2^2 \; (972 - 54\delta_{\tilde\delta \overline u} -
81 \delta_{\tilde\delta \overline d}) \right\}.
\label{21}
\end{eqnarray}
The ratio $P_{\pi q\overline q}/P_{\pi}$ is then easily expressed:

\begin{eqnarray}
\frac{P_{\pi q\overline q}}{P_{\pi}}&=&
\frac{1}{(2Rf)^2(2\pi)^3} \frac{\omega^s \omega^p}
{(\omega^s - 1)(\omega^p - 1)}\frac{4\pi}{3}
\left(\frac{1}{\omega^s + \omega^p}\right)^2 \nonumber \\*
  &&\left\{\frac{I_1^2}{I_3}\frac{(684 - 110\delta_{\tilde s \overline u} -
181 \delta_{\tilde s \overline d})}{57} + \frac{I_2^2}{I_3}
\frac{(972 - 54\delta_{\tilde\delta \overline u} -
81 \delta_{\tilde\delta \overline d})}{57} \right\} .
\label{22}
\end{eqnarray}
The value of this ratio for different sizes of the bag is also displayed
in Table \ref{tb22}, where we used $m_{\pi} = 140\; MeV$ for the pion mass and
$f=93\; MeV$ for the pion decay constant. There is a strong dependence of
the calculated ratio on the bag size but, even for the worst scenario
(the case of a small bag), the size of the two-loop contribution is just
$2\%$ of the one loop term.

\section{Final Remarks}

Our primary interest has been to investigate the effect of anti-symmetry
between the valence quarks in the nucleon and the internal structure of
the pion. Within the cloudy bag model, in which the coupling of the pion
to quarks is dictated by chiral symmetry, we found that it
appears to be safe to neglect possible antisymmetrization effects
between pion quarks and the nucleon valence quarks. Of course, when writing
the nucleon wave function one would have to add the contributions from all
possible states:
\begin{equation}
|N> = Z^{1/2} [|N>_0 + |N\pi>_0 + \sum_{q\; states}|N\pi q\overline q>_0 + ...],
\label{23}
\end{equation}
where the sum of all quark states (1s, 2s, etc...) was particularly
emphasized. If the quark that emits the pion remains in the ground state
the only contribution from antisymmetry is the one in Table \ref{tb22}. As
we have to sum over all other possible states for this quark, it turns
out that the antisymmetry effects are further diluted. However, we
will also have some other contributions even in this case, because if the
quark in the pion is in the ground state it can antisymmetrize with the
two spectator valence quarks. Also, if the quark in the pion is in an excited
state, it can combine with a particular excited state of the valence
quark that emitted the pion in the first place, as in the gluon example.
We did not make these calculations because of their level of complexity.
Our goal
was to examine the behaviour of the dominant contribution, displayed
in Table \ref{tb22}, and our results indicate that this number is
itself already very small.

Of course, as the sum over quark states in Eq. (\ref{23}) is infinite,
we can not rule out the possibility that eventually the antisymmetry
graphs could be important. However, we believe that this is
improbable, as the contributions from graphs which would be unaffected by
antisymmetry would grow even faster.
In summary, the major purpose of the present calculation was
to check whether the antisymmetrization effects in pion emission can be safely
disregarded - they can.
\newline
\newline
\newline
We would like to thank to S. Gardner, D. Leinweber, W. Melnitchouk, K-F.
Liu, A. I. Signal and A. G. Williams for helpful discussions.
This work was supported
by the Australian Research Council and by CAPES (Brazil).

\addcontentsline{toc}{chapter}{\protect\numberline{}{References}}

redpajama_set_name: RedPajamaArXiv


text:
The Wellmen
Guardians of Civilization
Author: Faelinn Shadowmoon
Guardians of Civilization - the Benevolent Order of Wellmen
By Faelinn Shadowmoon
The Wellmen are a mysterious but very old organization. As with all secret societies, not much is known about them. This is by design as the Order sees it as their duty to use secrecy to better aid them in their task: ensuring the survival of civilization.
We must stand together to fight the Infinite Dark to stand any chance of survival.
The time for secrecy has come to a close.
Early history in many cultures is not written down and what tales remain from that time are handed down through oral traditions. The earliest written mention of the Benevolent Order of the Wellmen is the document entitled "On the Sealing of Bedlam" speculated to be written in what is believed to be the year 56 of the common Realms calendar. This document chronicles the "Great Work" done by the Wellmen which was the sealing of the plane of Bedlam.
It is speculated that the Wellmen have existed in one form or another since the dawn of mortal civilization, particularly of that of the humans. Civilization, knowledge, and society are things generally attributed only to higher sentient creatures. In a discussion with a member of the Order, he claimed that civilization was the only thing separating men from beasts.
The Wellmen are organized by family, tracing their lineage through biological relatives. Cousins, children and even those adopted into the Order through a patron become one of the line. Should they choose to exercise their right by blood or by patron, any of the line can become a member of the Order. There were twelve original families of Wellmen, though some were thought to be lost of all time. Even if one family member exists, the line can be rekindled and reconnected to the Order of Wellmen.
The Wellmen are governed by the council known as the Conclave of the Mind. This decision-making body answers to the nominal leader of the Wellmen known as the Soul. It is believed that the position of the Soul is inherited through a specific bloodline, documented by hints in Wellmen documents about Bedlam. A position known as the Mouth speaks for the Conclave and disseminates orders to the Wellmen.
The line of Banecroft has the strongest claim to the Soul for they are scions of Rubis Banecroft, heir to the Soul of his generation who was destroyed by an Eater of Memories. The line continued through siblings, cousins and marriages for generations. The most well know of this line are Baron Sir Diamond Banecroft, Magus Savyne Banecroft and their son.
Who are the Wellmen?
Who is a member of this secret society? That is hard to say. Members have been know to keep their involvement with the order a secret even from their husbands and wives. Wellmen have been known to carry their secrets to the grave rather than divulge information about the order and who else may be a part of it.
As stated before, there are families of Wellmen each tracing their lineage to before the first fight against Bedlam. Some families have been well documented in papers pertaining to Bedlam such as the Banecroft and Hamofel lines. Others remain obscured and well-hidden unless you know what to look for, but they have also been documented. Samuel Cowperthwaite has become the "face" for the Wellmen in dealing with the adventurers of the Realms. He was first met in Chimeron in 1008 posing as a merchant with a fabulous treasure dungeon. Since then he has been the known contact person in dealing with most things Wellmen related. A friendly face for the Wellmen to engage with the adventurers of the Realms.
The other well known Wellman is a woman known as Maudlin. She was first encountered in 1005 on the expedition to open Bedlam. Many of the initial documents on Bedlam and information about the Wellmen were found from her as she had "borrowed" documents from the Order. Prior to the expedition, Maudlin leaked a serious of private missives from the Wellmen message courier system. This line of communication became known as the "Maudlin line" and was monitored after the opening of Bedlam. Though Maudlin hasn't been seen since 1005, there is at least one agent of the Wellmen who is looking for those who have used it in the past. Anyone who gained access to those private missives and those who ventured with Maudlin from Perdian to open Bedlam should beware of a man named Dinas Kingsman. No further warning was given about who this man is or what his intentions are.
Presently known Wellmen operatives:
Samuel Cowperwaite
Cover: merchant
Base of Operations: Chimeron
Gene Fitzgibben
Cover: author
Base of Operations: unknown
Iapeth
Base of Operations: City of Ivory
Cover: unknown
Status: unknown
Suspected Wellmen and Operatives:
Dinas Kingsman
Erastus
Deceased Wellmen (First War with Bedlam)
Gerald Harden
Base of Operations: unknwon
Status: deceased
Rodic Hamofel
Phiranos Hamofel
Ruis Banecroft
Bases of Operation
n the fall of 1010, the adventurers were allowed into a complex known as Wellmen Central. The means of getting there were obscured as were the means for leaving the complex afterwards. While this complex might have been a central location, there remain satellite bases in operation in both Chimeron and the City of Ivory.
The Wellmen have been allowed to operate unmolested in the City of Ivory and the lands of Ivory by securing their missions through legitimate merchants and guilds. When their presence was made known the the City Council, it was decided so long as they worked for the betterment of the City and did not violate any laws they were allowed to continue operations.
Places of learning and knowledge such as churches, libraries and universities or schools can be homes to Wellmen and their agents and have been confirmed as such. The Library of Ivory as well as the Library in Chimeron have been havens for Wellmen agents in their goal to preserve knowledge. Merchant guild, caravans, and even adventuring groups have been used to conceal and transport Wellmen operatives as they make their way across the Realms.
Known Operations of the Wellmen:
Project Rosebud
Purpose: unknown - Gerald Harden was assigned to it
Project Gatekeeper
Purpose: unknown - last status known 1005/pre-reopening Bedlam
Project Bookwalker
Purpose: Utilize the Book of the Moon artifact to combat Bedlam abroad.
Project Nightshade
Purpose: Bypass possible Skinwalker interference for full disclosure.
While the Wellmen cannot be everywhere at once, they keep track of information and monitor situations through agents. These agents can be petitioners to the Order, family members, or adventures either with or with out the knowledge of who they are truly working for. Wellmen have been known to monitor their operatives through magical scrying as well as warding meeting between agents..
During the first battle with the Infinite Dark, the Wellmen created weapons to combat the growing darkness. Weapons such as the Light Cannon were constructed to better combat bois. Caches of weapons were stored in secret locations, such as the tomb of Tristan the LightKeeper, were scattered all over the Realms. Some were lost to Bedlam or the memories of the Order. Others are suspected to remain in controll by the Wellmen and have yet to be fully utilized.
As well as physical weapons, the Order of the Wellmen have another powerful tool at their disposal: faith. The worship of Aurora, Goddess of Light, is held in high regard among many of the Order. Samuel Cowperwaite attended a meeting of the Church of Aurora in December 1010 to keep his fellow Wellmen of faith appraised of the situation. Samuel admitted that there are some who have harnessed more arcane and dangerous practices of Auroran faith. Those Aurorans seek to utilize whatever tools they have at their disposal, including personal sacrifices and blood magic, and may have fallen to the despair of the Path of Shadow. There is an old Auroran saying "The Light is always brightest just before it goes out". It is feared that some have given the ultimate sacrifice to fight Bedlam even if just to gain a small temporary advantage.
The Heralds of Destiny
Not much is known about the Heralds of Destiny. They were first mentioned in a document recovered from the Black and White Masquerade in the fall of 1009, but first "came into play" so to speak at the annual Feast of Chimeron in 1009.
The Heralds of Destiny are dangerous creatures, heralding a time when the world becomes unstable and the future is no longer clearly divined. They are great boons in the fight against the Infinite Dark, but as more come into play the harder and more impossible the task to defeat the Dark becomes. The Heralds of Destiny are the safeguards that civilization has put into place in order to preserve itself against the flowing tides of Bedlam. The Wellmen have hinted that the appearance of the Heralds is a signal that the civilization and society is in danger.
The document recovered in the vault at the Black and White in 1009 was translated as follows.
"After a visit from my young friend. I am sorry that Emerauld had to leave so soon. Her young company is a welcome joy to these tired bones. She is still but a girl, though, and she would not be delayed. I have to wonder, is Emerauld truly a daughter of Banecroft? A descendant of Wayland? Is she a lost connection to two of the first? Regardless, her fate is clearly entwined with our own. I wish my divinations proved more clear.
Gem stems from warrior
Twice must it fall
Risen the third, the sunset is marked
Creation does battle
One falls or all
The shadow selves must know to follow
The Emerald Path is only for them.
Further thoughts: could Emerauld be a Herald of Destiny?
Clan Defender – she is no warrior.
Empowering Soul – possible but she is still young.
Giver of Life – she has greed, so this cannot be her.
Kindred Spirit – not until she knows herself.
Light Keeper – she does not seek to protect knowledge, but seek it – there are better options.
Lucid Dreamer – the most likely for she seeks information which was lost.
Mercy Bringer – her heart will harden beyond this, or all. That much I can see.
Spear of the Just – can she kill so willingly?
Taker of Death – what man can take this role?
Voice of Reason – not so early in life.
Weaver of Names – she shows no control over creation.
Emerauld's story has only begun. Would that it began sooner, so I could see it through to the end. If only I could meet the followers of The Emerald Path in-person, I have so many questions for them. On my honor, a Wellman, I will set the remainder of my spells to be of aid once I may help no longer."
The author of this work is unknown, but was a friend to the young woman Emeraude Banecroft, the cousin of Rubis Banecroft. At the time of writing this (June 1011), two of the Heralds of Destiny have been activated.
In the creation of the ZulaBois, Zula Darkwillow became tainted with a darkness that threatened to consume him utterly and create a living gate of darkness. After brainstorming and failed ideas, it was decided to utilize an artifact known as a Tear of Aurora to cleanse the darkness within Zula. This came with the side effect that Zula would forever be changed and become of the Light. Zula was ritually infused with the essence of the Tear and, unknowingly, the Herald of Destiny known as the Light Keeper was activated. This took place at the Feast of Chimeron in 1009 and the first of the Heralds was put into play out of order bypassing the "unlocking" mechanism. Zula's duty as Light Keeper is to preserve knowledge so that it may be kept from the Infinite Dark and not be consumed.
The second Herald of Destiny to be activated is Clan Defender. This"unlock" condition was the only one known for the Heralds. This was unwittingly met, for none of the Heralds can be "forced" into play. While searching for a Lost Rune in Paradise, Laurante D'Rhian of Folkestone gave shelter to a small girl by using a shield. The consequences of the "unlock" being met on that snowy day weren't realized until weeks later. At the assault on Wayland fortress and the final battle against the smith, the Church of Vandor opted to stay behind to fight off the sweeping tides of Bedlam so that others could escape. In making that choice, the Church of Vandor was activated as the Clan Defender. The Church stood their ground and prevented Bedlam from claiming the live of those fleeing.
The Emerald Path
It is unknown what exactly the Emerald Path is exactly. It is thought to be named after Emeraude Banecroft, cousin of Rubis Banecroft, who walked a path of discovery long ago.
As the heroes from another realm embark on the journey of discovery along the Emerald Path, the information gained transfers to a copy of the book located at the Broken Harp Tavern in Perriden.
Not much is known about the heroes from another world, but they can sometimes "see" though our eyes and "know" what we know. A trip to Perriden might reveal more information to anyone curious. There in the Broken Harp Tavern, Emeraude's journal is being revealed.
Author's final notes
I sometimes pretend or imply that I know more than I actually do. This time, I am sad to admit I don't know everything we need to win this war. I know enough. I know that the knowledge cannot be allowed to be take from those who wish to fight the Infinite Dark and that it cannot be allowed to die with one person. In my three documents On the Infinite Dark, The Many Woven Branches, and Guardians of Civilization I have tried to find how Bedlam affects others, who might be our enemies and allies, and how we could combat this threat..
How do you combat the threat of a Skin Walker taking all the information and keeping it from those who would fight? You put the information for all to see. Information can be a powerful tool or a deadly weapon. The time for secrecy has passed. If you have read my documents, you know what I do. I suspect there will be some who will not be pleased at the revelation of secrets and my life might be forfeit in their eyes. I would give those of the family business the honor of introducing themselves first but time is of the essence.
You are part of my extended family, but times are changing and cooperation is needed. I have done your work hand in hand with my own for the preservation of knowledge and life. I've taken the advice and doing the family's work again, but in my own way.
If we are to survive, we must do so in the light.
I pray that those shrouded in mystery will come forward to all and share what they know.
I pray that it will be enough.
In closing I leave you with the words of Gerald Harden, a former Wellmen on his deathbed, to think about.
"I am a Wellman to the end, and I did what I did to save humanity. I can think of no nobler goal. We were right to tell them to fear us, Stewart. We are the enemy - more so than any king, or demon, or monster. We have lost our light and been consumed by the shadows we sought to protect others against. This was the only solution."
~Faelinn Shadowmoon
High Priestess of Aurora
Lady of the City of Ivory
Knight of the Crown
First Lady of Faerie
Lady of Darkvale
Magus of the Realms
Kathrani-To
Ichigiritorrennin-Ma
Last Modified by Faelinn Shadowmoon (Leanne Micciche) at 06-11-11 08:31 PM

redpajama_set_name: RedPajamaCommonCrawl


text:
Universities: the neoliberal agenda
Bob Brecher
Centre for Applied Philosophy, Politics and Ethics
Research output: Chapter in Book/Conference proceeding with ISSN or ISBN › Chapter
The neoliberal attack on universities needs to be understood for what it is: an explicit and deliberate attack on the very idea of knowledge and of human agency in its creation. To understand why in turn requires an understanding of what neoliberalism is, and in particular of how it differs from traditional forms of liberalism. I therefore offer an abbreviated account of neoliberalism, based on Mirowski's analysis, that situates it in its political context. I emphasize Hayek's antipathy (however contradictorily expressed) to knowledge, as opposed to information, and thus to the traditional liberal notion of the individual as epistemic agent, before going on to elaborate how and why universities thus cannot but be anathema to the neoliberal project of "engineering the human soul".
Interrogating the Neoliberal Lifecycle
B. Clack, M. Paule
Accepted/In press - 30 Jul 2018
Mirowski
Brecher, B. (Accepted/In press). Universities: the neoliberal agenda. In B. Clack, & M. Paule (Eds.), Interrogating the Neoliberal Lifecycle London: Palgrave Macmillan.
Brecher, Bob. / Universities : the neoliberal agenda. Interrogating the Neoliberal Lifecycle. editor / B. Clack ; M. Paule. London : Palgrave Macmillan, 2018.
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AB - The neoliberal attack on universities needs to be understood for what it is: an explicit and deliberate attack on the very idea of knowledge and of human agency in its creation. To understand why in turn requires an understanding of what neoliberalism is, and in particular of how it differs from traditional forms of liberalism. I therefore offer an abbreviated account of neoliberalism, based on Mirowski's analysis, that situates it in its political context. I emphasize Hayek's antipathy (however contradictorily expressed) to knowledge, as opposed to information, and thus to the traditional liberal notion of the individual as epistemic agent, before going on to elaborate how and why universities thus cannot but be anathema to the neoliberal project of "engineering the human soul".
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Palma Beach Resort is a no-frills offering just off the main thoroughfare and a stone's throw from Nipah Beach.
The intimate grounds and stucco white buildings are impressive-looking but they border on spooky at night as the walkways are poorly lit. On a good day a friendly, smiling front man will present himself; otherwise the non-smiling 2IC will be there to extend the warmth of an ice cube. Nonetheless, location and reasonable rates make Palma a decent pick for budget travellers. With 27 clean and basic rooms available, there is a choice of upstairs or downstairs and although furnishings have seen better days, air-con, hot-water shower and private veranda makes up for it. Their large open-air reception area offers a hot and cold-water dispenser, a few tables and chairs and free WiFi. Discounted rates may be negotiable for longer stays.


Roughly 350 students at South Lakes High School took part in a walkout at noon today to remember the 17 people killed last week at Marjory Stoneman Douglas High School in Parkland, Florida.
Students left the school and stood outside for 17 minutes before returning for class or lunch, school officials said.
A similar walkout happened at Langston Hughes Middle School today. Around 250 students participated in that event.
"I am proud of how the students conducted themselves, including a moment of silence that was very moving," said SLHS Principal Kim Retzer in a statement.
Our teachers, administrators and staff continue to reinforce a sense of positive school community and we ask for your partnership in working with your child to discuss meaningful actions that they can take to engage in their community.
Another walkout is planned for National School Walkout Day on March 14. Schools across the country are expected to walk out to demand action on gun violence and school shootings.


CB William Gay was benched last week in favor of CB Greg Toler. This is a situation to monitor closely as it was just two seasons ago when Toler put up 82 solos. Could end up being a very sneaky pickup.
Eagles waived LB Brian Rolle. Look for Jamar Chaney and/or Akeem Jordan to assume the WLB duties. Not much fantasy value here at all.
Troy Polamalu looks like he will play this week. Solid DB2.
Redskins SS Brandon Meriweather will be out "at least a week" with a left MCL/PCL injury. Reed Doughty started in Meriweather's place following his pre-game collision Sunday.
Giants SS Kenny Phillips has been diagnosed with a sprained right MCL, and is week-to-week.
Will Witherspoon rotated heavily with Zac Diles at linebacker last week. Colin McCarthy looks like he will be back this week, but Witherspoon should be dumped either way.
Studs such as Jared Allen, Cameron Wake, Charles Johnson, and George Wilson all had very nice weeks and are back on the path to IDP stardom.
Bobby Wagner had a nice game, but he still isn't playing in subpackages. Until he does, I would avoid him as anything other than a LB4+.
With the Jets losing Santonio Holmes for the year with a foot fracture, they could very well end up being the worst offense in the league. This could mean a big spike in tackle opportunity for LB David Harris. He's been outstanding the last couple of weeks and has a top tier tackle matchup this week against Houston. He has LB1 potential ROS. I would also upgrade LaRon Landry and Yeremiah Bell as well.
RDE Cameron Wake (MIA) - I've been preaching patience with Wake the last couple of weeks. Leading up to last week, he led the league in QB hurries/hits, however had no sacks to show for it. That changed last week when he destroyed the Arizona Cardinals with 4.5 sacks. He's a DL1 moving forward and should be owned in all formats. If someone was dumb/impatient enough to drop him, scoop him up now!
RDE John Abraham (ATL) - Abraham leads all current NFL players in sacks. However, Father Time is starting to catch up to the 13 year veteran and he doesn't receive nearly the amount of snaps he use to in his younger days. His limited snap count is keeping him more fresh however and he has 3 sacks so far this season. He has a nice matchup against the Redskins this week and is a top 10 DL play in balanced and sack-heavy scoring formats.
DT Henry Melton (CHI) - Melton has consistently been beating interior linemen off the ball and last week against Dallas was no exception. He's had a sack in 3 of the first 4 weeks and I expect that trend to continue this week against the Jacksonville Jaguars. He's a solid DL3 option.
LDE Everson Griffen (MIN) - Griffin doesn't receive enough playing time to be counted on as anything more than a long shot DL3, deep league DL4 type player at this point. But he has had 3 sacks so far this season and with Jared Allen gaining more effectiveness, Griffin should see more one-on-one matchup opportunities in the interior. Not a superb play, but you could do worse this week.
WLB Vontaze Burfict (CIN) - Ever since Thomas Howard went on IR after Week 1, I've been keeping a close eye on the Cincinnati WLB position. I think Burfict is the real deal and has a chance to be the best fantasy linebacker on the team come year's end. He's an instinctive player who has a knack for being around the ball. The coaches like him and most importantly, he has little competition for tackles. He's worth a LB4 roster spot in 10 and 12 team leagues.
WILB Pat Angerer (IND) - Angerer looks like he will make his season debut this week against the Green Bay Packers. There's no reason why he shouldn't be on your roster if he is still sitting on the waiver wire (unless you are already stacked at LB). He was a stud last year and will be a stud this year as long as he's healthy.
MLB Colin McCarthy (TEN) - This is an obvious one. Coaches said McCarthy looked great in practice and he is looking like a good bet to start against the Vikings this week. If by some chance, he's still on your waiver wire, pick him up. LB1 with top 5 potential.
LOLB James Harrison (PIT) - Similar to Angerer, Harrison will likely see his first game action of the season this week. He goes up against Michael Vick and the Eagles. Although he's suffered some significant injuries over the last couple of seasons, Harrison is as tough as they come and has shown that he can play quite effectively even when not at 100%. He's a guy that I always like to take a chance on if I need big production out of my LB3/4 spot and am willing to get a bit risky.
SILB Demorrio Williams (SD) - This is going to be a one week waiver wire pickup suggestion. Williams is not a starer. However, he plays in all of San Diego's nickel packages and with the Saints on the schedule this week, he will figure to see 80-85% of the snaps. Historically, he's produced when he's played. If you are looking for a LB4 in super deep leagues to start this week, Williams is a nice option.
SS Jordan Babineaux (TEN) - Babineaux has been on this list before, so I won't expel too much about him. In short, he's a mediocre player in a great situation. The Titans are receiving a ton of tackle opportunity and will likely continue to receive a ton as the season wears on. He's a solid DB2+, especially in tackle-heavy leagues.
FS Matt Giordano (OAK) - Due to the injuries at the CB position, the Raiders have temporarily moved Michael Huff from free safety to cornerback. Giordano has taken his place at safety and his put up good numbers during that time. We've seen similar production from Giordano in short starter stints last season. He should be counted on for 5-7 solos per game.
FS Rahim Moore (DEN) - Moore, a 2nd round pick out of UCLA last season, has started to emerge as one of the better young players on the Broncos roster. He's still 22 years old and still suffers from consistency problems, but his upside is high enough in certain matchups to use him as a bye week/injury DB3 filler. He has the Chargers and Patriots in his next two matchups and has a good chance of exceeding expectations in those weeks and perhaps beyond.
FS Malcolm Jenkins (NO) - For most of his career, Jenkins has failed to make a big splash in the tackle column on an extended basis. However, the Saints' defense has been on the field for larger spans of time this season which has led to an increased amount of tackle opportunity. Add in a weaker front seven and you have a recipe for a significant increase in tackle production. So far through four games, Jenkins has registered 33 total tackles. He has a very good matchup this week against the Chargers and should be started with confidence.


text:
> Fed's Kocherlakota: Fear of Another Crisis Slows Recovery
Fed's Kocherlakota: Fear of Another Crisis Slows Recovery
ForexLive
Tuesday, 10/04/2012 | 18:20 GMT-0
–Expects More Accommodation, Making MonPol Inappropriately Easy
By Denny Gulino
WASHINGTON (MNI) – Fear of another crisis like 2008 is a headwind
that is adding years to recovery but monetary policy will still have to
adjust to what improvement occurs, even to the extent of raising rates
late this year, Minneapolis Federal Reserve Bank President Naryana
Kocherlakota said Tuesday.
In remarks prepared for delivery in a dairy in Nicollet, Minnesota,
Kocherlakota underlined his own outlook, that the current level of
accommodation will become increasingly inappropriate soon.
He also repeated that should more accommodation become necessary,
he would support buying more Treasuries and Fannie Mae and Freddie Mac
MBS.
"I expect that the FOMC's policy will be even more accommodative
than I would see as appropriate," he said. "I expect that the core and
headline personal consumption expenditure inflation rates will be around
2 percent his year and rise to 2.3% in 2013."
For now, "Low household net worth and wealth will continue to
represent significant headwinds for demand," and so "I would view a
highly accommodative monetary policy as being appropriate."
Kocherlakota will be on the FOMC in 2014. His appearance in
Nicollet, a community of about 1,000, is part of a "two-way
communication between the residents of Main Street and the Federal
Reserve," he said.
Although the economy has been improving at a moderate pace, he
said, the crisis cost it trillions of dollars in wealth, years worth of
growth and a productive capacity much depressed from where it would be
otherwise. "Real output has recovered to 2007 levels, but remains well
below what one would expect it to be in light of historical growth
patterns."
Employment is still "well below 2007 levels" and restraining the
kind of progress that would help the economy catch up faster is that,
"Households and firms now feel that they must stay prepared for the kind
of financial market shock they experienced in 2008."
"Existing firms' fear of a 2008 financial market shock keeps them
from hiring workers whom they might have to fire if 2008 recurred," he
The fear makes it "harder for destroyed jobs to be replaced by
created jobs," he said.
The fact that inflation is "remarkably close" to the Fed's target
and not much less — or even lapsed into the kind of deflation that
occurred in the Great Depression — suggests demand for goods is not
"significantly below the productive capacity of the United States."
It has been the Fed's highly accommodative policy that has
prevented deflation even considering the "fall in demand and damage to
the productive capacity of the country," he said. At around 2.5% to 3%
growth in each of the next two years, "output will remain well below
what we might have expected it to be back in 2007."
The rate of employment growth will be similarly slowed, "to be only
slightly higher than the rate of population growth, he said. "I expect
that the unemployment rate will continue to rall slowly, to around 7.7
percent at the end of this year and to around 7 percent by the end of
Answering his own question whether the FOMC should reduce
accommodation, Kocherlakota said, "Yes. From the point of view of the
dual mandate, the outlook is better than a year ago — and so we should
have less accommodation in place."
While that doesn't mean "we should be raising rates anytime soon,"
he went on to repeat that, "My own belief is that we will need to
initiate our somewhat lengthy exit strategy sometime in the next six to
nine months or so, and that conditions will warrant raising rates
sometime in 2013 or, possibly, late 2012."
"I see no need for additional accommodation at this time, and I
believe that conditions will warrant raising rates well before the end
of 2014," he reiterated.
** MNI Washington Bureau **
[TOPICS: M$U$$$,MMUFE$,MGU$$$,MFU$$$]


I read this article and found it very interesting, thought it might be something for you. The article is called 'Municipal burglary' challenges local democracy in UK and is located at https://www.bangthetable.com/news/municipal-burglary-challenges-local-democracy/.
Peter Latham calls for the remaking of local government – and local democracy – as he confronts the UK Conservative Government's devolution agenda in his book Who stole the town hall?: The end of local government as we know it. Drawing on examples from around the UK, Latham argues that the agenda hides a vision with major implications for British democracy: the privatisation of public services and local government.
Latham traces the role of the central government and private sector in diminishing local government, and projects the consequences of its reorganisation. He proposes the mending and remodelling of the tax system with reforms for taxes on land, wealth, and income to sustain critical public services and feed the revival of local and national democracy.
In his foreword to Who stole the town hall, Rodney Bickerstaffe, former general secretary of Unison, warns of the decline of local government and the rise in civic disengagement and distrust. Bickerstaffe points to the pressing need to improve public services and save local government as a vital democratic institution. He identifies 'municipal burglary' and corporate profiteering through government as a major challenge for democracy, and presents Latham's research and analysis as a diagnosis of the current situation.
The book builds from Latham's earlier work The state and local government: Towards a new basis for 'local democracy' and the defeat of big business control (2011), in which he calls for the revival of grassroots political networks and structures to address privatisation and reorient government. .
Sociologist Peter Latham, former researcher at the London School of Economics, is an official in the University and College Union, member of the Labour Land Campaign, the Economic Committee of the Communist Party of Britain, and a delegate to Croydon Trades Union Council. He was Treasurer and then Secretary in the Labour Campaign for Open Local Government (1999-2006).


text:
Sell Your Cards Now
22 Yogi Berra Baseball Cards You Need To Own
Items of Yankee legends are always popular among sports collectors and Yogi Berra baseball cards are no exception.
One of the most well-known names in baseball history, he also became a cultural icon for his funny sayings like "the future ain't what it used to be" and "I never said most of the things I said".
Born Lawrence Berra, he earned the nickname "Yogi" from a childhood friend who thought he looked similar to an Indian yogi he saw during a movie.
Yogi went on to be one of the best (if not the best) catchers and most feared hitters of all-time. A 3-time MVP, 15-time All-Star and member of 10 World Series championship teams, Berra's place in baseball history is forever cemented.
Here is a list of the 22 best Yogi Berra cards that any serious collector should pursue.
Ross Uitts - Owner
Are you selling sports cards that were produced from 1868 - 1975?
If so, then please fill out the form below and I'll be in touch right away. Or, feel free to call/text me at 305-684-6680!
1948 Bowman #6 Yogi Berra Rookie Card
The 1948 Bowman #6 is the only official Yogi Berra rookie card in the hobby. The small 2 1/16" by 2 1/2" black and white card shows a very young Berra looking up after a swing. Mint condition cards are almost impossible to come by, and will easily run for thousands of dollars. By 1948, Berra had been playing with the Yankees for two seasons. Over the next 14 years, a season would not pass that Yogi was not called up to play for at least 100 games. Along with the Warren Spahn and Stan Musial rookie cards, this one is the key to the set.
1949 Bowman #60
1949 would be the first of five times in a row that the Yankees would win the World Series with Yogi on the field. His ability to hit low pitches into home runs made him a huge asset to the team's success. He was once again called up to the All-Star team for the second year in his young career. The bright red background, while the rest of the colors remain muted, give the picture a surreal look. His name does not appear anywhere on the front of the card as he was part of the first half of the 1949 Bowman print run which didn't include player nameplates.
Before Topps arrived on the scene and took over the world of baseball cards, Bowman was the go to for kids collecting cards. The 1950 set was especially amazing because for the first time ever it featured full color, hand painted renditions of real photos of the players. Any collector would want to have a piece of what many have called one of the most important sets in baseball card history. Yogi is shown squatting and arm cocked, ready to gun out a would-be base runner.
1951 Bowman #2
As the Yankees went on to win the World Series for a third consecutive time, Yogi would be awarded his first title of American League Most Valuable Player. The card commemorating this great year shows a close up of Berra smiling in his pinstripe jersey. Also of note is the use of his real name "Larry" along the bottom of the card. It's a beautiful card but centering and other condition challenges are still an issue.
Yogi followed up his amazing '51 season with 98 RBI in and 30 home runs, which would be his all-time career high. He was once again an All-Star and World Series champ. This card often suffers from toning problems, staining and centering issues but it's still a very attractive issue. The artist's rendition features a focused Berra, ready at the plate, along with his autograph across the image. It would be one of two mainstream cards on which Berra appeared that year.
1952 Topps #191
This card is the most valuable of Yogi's career even though it was issued several years after his rookie season--something you don't see too often in the hobby. The iconic '52 Topps set features full color photos of candid shots of each player. Yogi's name and autograph are surrounded by stars with a Yankees logo nearby. The set is one of the most difficult to collect as there are varying numbers depending on if you're trying to collect the basic set or the true master set. Pair that with the outrageously high prices of many of the cards, and you've got yourself one of the hardest sets to own. The Mantle and Mays cards easily outshine every other player in the set but Berra's is still key. Tilting and centering are typical challenges with this card.
1953 Bowman Color #121
Bowman stepped up their game in 1953 by offering full-color, large-sized cards featuring the 160 best major league players. Nothing else appears on the front of these cards, with all focus on the picture of the player. Yogi's card in this set is one of the entire collection's most important cards. It is incredibly difficult to find in good condition, as many specimens feature print lines and poor centering. Relaxing in the dugout, it's easy to envision Yogi laying out one of his famous sayings in this picture.
It would be the last of the five year streak of World Series wins for the Yankees. Yogi would be called up to the All-Star game once again for his stellar performance during the season. The card features a portrait of Berra with his cap on and a determined look on his face. His name, position and team are found in a red box at the bottom left corner. A beautiful card, no doubt, it's still one of the few of his mainstream issues where he's not smiling. Chipping along the red bottom border is a common concern for collectors.
1954 Bowman #161
In 1954, Yogi received his second of three American League Most Valuable Player titles. The card has an excellent full color photo of Yogi smiling on it. Underneath in a salmon colored box, Yogi's signature appears. For such a great looking card, the price tag isn't as steep as many Berra cards. The backs of the cards in this set are graced with trivia questions, humorous anecdotes and player records.
1954 Topps #50
Topps embraced a whole new design for their '54 set. Yogi's card has a bold green background with a black and white cut-out action shot of him swinging the bat. A larger, cut-out color photo of Yogi about to make a catch steals the focus of the card. His signature trails across the bottom of the card while his name, team and position are recorded at the top. These cards are prone to rounded corners and centering issues, which devalues them drastically.
Bowman clearly put some creative thought into their 1955 set. Photos of players are in borders that are meant to look like color TVs. Berra has his bat over his shoulder, ready to hit whatever comes to him. The only problem with the awesome TV borders is that the dark brown color chips so easily. Only his last name "Berra" accompanies him on this card.
The Yankees faced the Brooklyn Dodgers in the 1955 World Series, a series in which one of the most iconic moments in the game's history took place. Down 4-6 in the top of the 8th of Game 1, Jackie Robinson famously stole home plate--something Berra remained upset about for years thereafter.
The match-up would end up being a 7 game battle, one which the Yankees would inevitably lose. In game 7 of the championship, Yogi hit a fly ball in the sixth inning that had the potential to change the game, but Dodgers' left fielder Sandy Amorós caught it, which sealed the Yankees' fate. The horizontal design, a first for Topps, and the bright yellow borders make this card one of the more attractive Yogi Berra baseball cards. Centering and print blotches in the yellow background are common problems.
The Yankees would have their chance at redemption in 1956, as once again they would end up facing the Dodgers in the World Series. This time, the Yankees brought the heat. Even though Berra helped catch in Don Larsen's perfect, no hit game, the teams still pushed the battle to a game 7. Much to the credit of Yogi, who hit a homerun and batted 4 runners in, the Yankees smashed the Dodgers in a 9-0 domination. The whole season was a great one for Berra, giving him many of his personal bests which still hold to this day. Another great horizontal issue, Berra is is depicted in a close up head shot along the left and in a small action shot on the right.
1957 Topps #2
The 1957 Topps set was fairly large in size, at 407 cards. While Yogi's teammate, Mickey Mantle, is the real star of the set, Berra's card still sells close to $3,000 in top condition. The focus in '57 was elsewhere in New York, as it would be the last season for the Brooklyn Dodgers and the New York Giants before the teams relocated to the west coast. His '57 Topps issue is a sharp-looking card with a full on headshot of the Yankee legend on center stage. Centering is probably the number one condition issue that affects this card.
Topps decided to punch up the color factor in the 1958 set. Each card has one of nine different brightly colored backgrounds. The backs of these cards featured small cartoons about the players. Yogi's card is electric orange with a cutout photo of him smiling over top. A blue banner at the bottom reads his position and team name. It was an excellent season for Yogi. He played 88 perfect games landing him one of several 1.000 fielding averages in his career. Centering and print blotches are typically the biggest challenges for this card.
Berra was the first catcher to keep one finger out of the glove while catching. Many other catchers followed suit in the years to come. Yogi would once again make it to the All-Star team in 1959. The Topps set from that year was bold and totally new in design. A circular photo of Berra holding the bat with his autograph across it lays in front of an orange background. His name trails diagonally upwards at the the top of the card.
The Yankees would find themselves once again in a World Series, this time facing the Pittsburgh Pirates. In what would be a devastating Game 7 loss, the Pirates' Bill Mazeroski would hit a homerun that would win the game 10-9. The Yankees both outhit and outscored the Pirates, but one bad pitch would doom the game. It's the last time Berra was featured on a horizontally-designed mainstream card. The iconic 1960 Topps card design displays a small black and white action shot of Berra on the left and a larger head shot on the right. Print dots can affect these cards but they're still great looking.
The 1961 Topps set is considered a classic despite its somewhat plain appearance. Print bubbles can appear in the colored blocks at the bottom but for the most part this is one of the easier sets in the '60s to find in high grade. Yogi's card features a posed photo of him smiling, bat in hand. The backs of the card had no humorous anecdotes, just the plain old stats about each player. The Yankees would win yet another World Series in '61, this time against the Cincinnati Reds. This was their 19th championship win out of 39 seasons. They won the match-up in just five games, completely crushing the Reds.
Once again the Yankees found themselves in a World Series Championship going head to head with Willie Mays and the San Francisco Giants. The match went right to game 7, with the Yankees stealing the victory by just one run. Yogi was 37 years old by this point, but he still played with stamina, even catching the entirety of a 22 inning game against the Detroit Tigers, which lasted seven whole hours. The famous wood-grain borders of the 1962 Topps set present chipping and wear challenges. Centering is also a common problem.
This card features two photos of Berra, one small black and white shot of him holding the bat in a bright yellow circle, and a much larger full color photo of him smiling. The larger photo really shows his age. This would be the year that Yogi would retire as a player for The Yankees, and take his turn as a manager and coach for the team. Centering is a severe challenge and the red bottom borders also easily show wear and chipping.
Yogi was able to lead the Yankees to the American League pennant against the St Louis Cardinals as a first year coach. The Yankees lost the matchup over seven games. Berra was having difficulty convincing the team that he was ready to manage, and he eventually was fired from the Yankees, leading to over a decade of bad blood between Yogi and the team. One of the easier sets to find in terms of print quality, 1964 Topps Berra cards are much easier to find in top condition. It's also one of two Yogi Berra baseball cards to show him as a manager.
In 1965, Berra officially ended his Yankees career by signing with the New York Mets as a coach. He would appear occasionally as a catcher throughout that season, mostly just for show. He would continue to be a coach in the years that followed, eventually leading the team to a World Series title. Tilt and centering are usually the more common issues with this card.
Yogi Berra Baseball Cards Wrap-up
Berra caught for the Yankees in each of his 19 years professional playing career and left a great series of baseball cards to collect.
Many have dubbed Yogi as the greatest catcher of all time. It's not hard to understand why, as he caught 173 shutouts during his baseball career, an all-time high in the MLB.
Inducted into the Baseball Hall of Fame in 1972, the nation's love for Yogi is obvious, as he has both a museum and a university stadium named in his honor. Yogi's baseball career is the stuff legends are made of. Two plaques exist to commemorate Berra, one at the Baseball Hall of Fame and one in Monument Park. Berra would not enter The Yankees stadium for years after this honor, until the man responsible for firing him made a formal apology.
On September 22, 2015, Berra passed in his sleep at the age of 90. All of New York mourned the loss, with homage to Berra being paid as the Empire State Building was lit in Yankees pinstripes the following day.
After his death in 2015, President Obama awarded Yogi the Presidential Medal of Freedom for contributing to America's strength as a nation. Yogi was married to his wife Carmen for over 65 years, until she passed away just one year before him. Yogi left behind a story that is worth being told, having had numerous films, plays and documentaries being made about him. As you collect the cards honoring this major icon in American history, be sure to tell yourself, "it ain't over till it's over.
Ross Uitts
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Ross is the founder of Old Sports Cards and has been collecting sports cards for over 30 years. He also loves to write about the hobby and has written for Beckett, Topps, SABR and of course, this website. Need help buying or selling cards or have a general question about the hobby? Contact him at [email protected]
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Copyright © 2020 by Old Sports Cards, LLC


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