text:
\section{Introduction}
\label{intro}
The calculation
of binding energies, masses of heavy quarkonia $q {\bar q}$ and
another physical parameter
using  renormalon method has attracted the attention recently. The calculations, based on perturbative expansions,
are primarily due to the knowledge of up to ${\rm N}^2{\rm LO}$
term ($\sim\!\alpha_s^3$) of the static quark-antiquark
potential $V(r)$ \cite{Peter:1996ig,Schroder:1998vy}
and partial knowledge of the ${\rm N}^3{\rm LO}$ term there,
and the ultrasoft gluon contributions to a corresponding
effective theory ${\rm N}^3{\rm LO}$ Hamiltonian
\cite{Kniehl:2002br,Brambilla:1999qa,Kniehl:1999ud};
and the knowledge of  the pole mass $m_q$ up to order
$\sim \alpha_s^3$ \cite{Gray:1990yh,Chetyrkin:1999qi}.
Another impetus in these calculations was given by
the observation of the fact that the contributions of the
leading infrared (IR) renormalon singularities (at $b=1/2$)
of the pole mass $m_q$ and of the static potential $V(r)$
cancel in the sum $2 m_q + V(r)$
\cite{Hoang:1998nz,Brambilla:1999xf,Beneke:1998rk}.
Consequently, this cancellation effect must be present also
in the total quarkonium mass $M = 2 m_q + E_{q \bar q}$
\cite{Pineda:2001zq,Lee:2003hh}, or more precisely,
in $M(s) = 2 m_q + E(s)$
where $E(s)$ is the hard$+$soft part of the binding energy,
i.e., the part which includes the contributions
of relative quark-antiquark momenta $|k^0|$, $|{\bf k}| \geq
m_q \alpha_s $, i.e., soft/potential scales (predominant)
and higher hard scales (smaller contributions).
In addition, the binding energy has contribution $E_{q \bar q}(us)$
from the ultrasoft momenta regime
$|k^0|$, $|{\bf k}| \sim m_q \alpha^2_s $. The ultrasoft
contribution is not related to the $b=1/2$ renormalon
singularity, since this singularity has to do with the
behavior of   theory in the region which includes the hard
($\sim m_q$) and soft/potential ($\sim m_q \alpha_s$) scales.

This contribution is based on ref. \cite{CCG:2004} and we present
  numerical calculation of
the binding ground energies $E_{b \bar b}$
(separately the $s$ and the $us$ parts)
and the mass $(2 m_b + E_{b \bar b})$ of the heavy $b {\bar b}$
system, by taking into account the leading IR
renormalon structure of $m_b$ and $E_{b \bar b}(s)$,
and combining some features as:
(a) the mass that we use in the perturbation expansions
is a renormalon-free mass
\cite{Beneke:1998rk,Pineda:2001zq,Bigi:1994em,Hoang:1998ng,Yakovlev:2000pv}
which we choose
to be the ${\overline {\rm MS}}$ mass
${\overline m}_q \equiv {\overline m}_q(\mu = {\overline m}_q)$;
(b) Borel integrations \cite{Lee:2003hh} are used to perform
resummations. Before resummations we perform
separation of the soft/potential ($s$) and ultrasoft ($us$)
part of the binding energies, and apply the renormalon-based
Borel resummation only to the $s$ part. The renormalization
scales used in the Borel resummations are $\mu_h \sim m_q$ (hard scale)
for $2m_{q}$, and $m_{q}\alpha_{s} \leq  {\mu}_{s} < m_{q}$ for $E_{q \bar q}(s)$.
The term corresponding to $E_{q \bar q}(us)$ is evaluated
at $\mu_{us} \sim m_q \alpha^{2}_{s}$ whenever perturbatively possible.
Further, the Borel resummations are performed
in three different ways: (a) using a slightly extended version
of the full bilocal expansion  Refs.~\cite{Lee:2002sn,Lee:2003hh};
(b) using a new ``$\sigma$-regularized'' full
bilocal expansion introduced in the ref. \cite{CCG:2004};
(c) using the form of the Borel transform where the leading IR
renormalon structure is a common factor of the transform
\cite{Caprini:1998wg,Cvetic:2001sn} (we call it $R$-method).
The Borel integrations for both $m_q$ and $E_{q \bar q}(s)$
are performed by the same prescription (generalized principal value PV
\cite{Khuri:bf,Caprini:1998wg,Cvetic:2001sn,Cvetic:2002qf})
so as to ensure the numerical cancellation of the
renormalon contributions in the sum $2 m_q + E_{q \bar q}(s)$.
Furthermore, we demonstrate numerically that in the latter
sum the residues at the renormalons are really
consistent with the renormalon cancellation
when a reasonable factorization scale parameter for the $s$-$us$
separation is used, while they become incosistent with
the aforementioned cancellation when no such separation is used.
The  numerical results allow us to extract the
mass ${\overline m}_b$ from the known $\Upsilon(1S)$
mass of the $b {\bar b}$ system.

\section{Pole mass}
\label{mass}

Here we   calculate   the pole mass $m_q$
in terms of the ${\overline {\rm MS}}$ renormalon-free mass
${\overline m}_q \equiv {\overline m}_q(\mu = {\overline m}_q)$
and of $\alpha_s(\mu,{\overline {\rm MS}})$, using
elements of the renormalization group approach of Ref.~\cite{Pineda:2001zq}
and the bilocal expansion method  Refs.~\cite{Lee:2002sn,Lee:2003hh}.
The ratio $ m_q/{\overline m}_q $ has, divergent, perturbation expansion
in ${\overline {\rm MS}}$ scheme which is at present known
to order $\sim\!\alpha_s^3$
(Ref.~\cite{Gray:1990yh} for $\sim\!\alpha_s^2$;
\cite{Chetyrkin:1999qi} for $\sim\!\alpha_s^3$. Let us define
\label{Sm}
\begin{eqnarray}
S \equiv \frac{m_q}{{\overline m}_q} - 1 & = & \frac{4}{3} a(\mu)
\left[ 1 + a(\mu) r_1(\mu) + a^2(\mu) r_2(\mu) + {\cal O}(a^3) \right]
\ ,
\label{Smexp}
\\
r_1(\mu) & = & \kappa_1 + \beta_0 L_m(\mu)
\ ,
\label{r1}
\\
r_2(\mu) & = & \kappa_2 + ( 2 \kappa_1 \beta_0 + \beta_1) L_m(\mu)
+ \beta_0^2 L_m^2(\mu) \ ,
\label{r2}
\\
(4/3) \kappa_1 & = & 6.248 \beta_0 - 3.739 \ ,
\label{k1}
\\
(4/3) \kappa_2 &= &   23.497 \beta_0^2 + 6.248 \beta_1
+ 1.019 \beta_0 - 29.94 \ ,
\label{k2}
\end{eqnarray}
where $L_m = \ln(\mu^2/{\overline m}_q^2)$, while
$\beta_0 = (11 - 2 n_f/3)/4$ and $\beta_1 = (102 - 38 n_f/3)/16$
are the renormalization scheme independent coefficients
with $n_f = n_{\ell}$ being the number of light active
flavors (quarks with masses lighter than $m_q$).
The natural renormalization scale here is $\mu = \mu_h \sim m_q$
(hard scale).

Let us consider $S$ as   a  function of the running coupling
$a(Q)=\alpha (q)/\pi$, and  the  perturbation
expansion for $S$ reads:
\begin{equation}
S(a(Q))=\sum_{n=0}^{\infty} s_{n}a^{n},
\end{equation}
which has to be summed. Then, one defines the Borel transform as
\begin{equation}
B(b,Q)=\sum_{n=0}^{\infty} s_{n}\frac{b^{n}}{n!}.
\end{equation}
This series has a finite radius of convergence in the $b$-plane,
and we introduce the Borel function $\widetilde{S}(a)$ (Borel Integral) corresponding to
$S(a)$ as
\begin{equation}
\widetilde{S}(a)=\int_{0}^{\infty} db B(b,Q) e^{-b/a}
\end{equation}
 and in our case $B_S(b)$ is known to order $\sim\!b^2$
\begin{equation}
B_{S}(b; \mu) = \frac{4}{3} \left[ 1 + \frac{r_1(\mu)}{1! \beta_0} b +
\frac{r_2}{2! \beta_0^2} b^2 + {\cal O}(b^3) \right] \ .
\label{BSm1}
\end{equation}
It has renormalon singularities at
$b = 1/2, 3/2, 2, \ldots, -1, -2, \ldots$
\cite{Bigi:1994em,Beneke:1994sw,Beneke:1999ui}.
The behavior of $B_S$ near the leading IR renormalon
singularity $b=1/2$ is determined by the resulting
renormalon ambiguity of $m_q$ which has to have the dimensions
of energy and should be renormalization scale and
scheme indepenent -- the only such QCD scale being
$const \times \Lambda_{\rm QCD}$
\cite{Beneke:1994rs}.
The Stevenson scale
$\widetilde \Lambda$ \cite{Stevenson:1981vj}  can be obtained in
terms of the strong coupling parameter
$a(\mu; c_2, c_3, \ldots) = \alpha_s(\mu; c_2, c_3, \ldots)/\pi$,
where $c_j = \beta_j/\beta_0$ ($j \geq 2$) are the
parameters characterizing the renormalization scheme,
by solving the renormalization group equation (RGE)
 \cite{Stevenson:1981vj}
\begin{eqnarray}
\frac{d a(\mu)}{d \ln \mu^2 }&=& - \beta_0 a^2(\mu)(1 + c_1 a(\mu)
+ c_2 a^2(\mu) + \cdots ) \quad \Rightarrow
\label{RGE}
\\
\ln \left( \frac{{\widetilde \Lambda}^2}{ \mu^2 } \right)
&=& \frac{1}{\beta_0} \int_0^{a(\mu)} dx \
\left[ \frac{1}{x^2(1 + c_1 x + c_2 x^2 + \cdots )}-
\frac{1}{x^2(1 + c_1 x)} \right]  \nonumber\\
& &  - \frac{1}{\beta_0 a(\mu)} +
\frac{c_1}{\beta_0} \ln \left( \frac{ 1 + c_1 a(\mu) }{c_1 a(\mu)}
\right) \ \Rightarrow
\label{solRGE}
\\
{\widetilde \Lambda} &=& \mu \
\exp \left( - \frac{1}{2 \beta_0 a(\mu)} \right)
\left( \frac{1 + c_1 a(\mu)}{ c_1 a(\mu)} \right)^{\nu} \nonumber\\
& &
\exp\left[ -  \frac{1}{2 \beta_0} \int_0^{a(\mu)} dx \
\frac{( c_2 + c_3 x + c_4 x^2 + \cdots )}{
( 1 + c_1 x ) ( 1 + c_1 x + c_2 x^2 + \cdots )} \right] \,
\label{tL1}
\end{eqnarray}
where $\nu = c_1/(2 \beta_0) = \beta_1/(2 \beta_0^2)$. Expansion of expression (\ref{tL1}) in powers of
$a(\mu)$ then gives
\begin{equation}
{\widetilde \Lambda} = \mu \exp \left( - \frac{1}{2 \beta_0 a(\mu)}
\right) a(\mu)^{- \nu} c_1^{- \nu} \left[
1 + \sum_{k=1}^{\infty} \ {\widetilde r}_k a^k(\mu) \right] \ ,
\label{tL2}
\end{equation}
where
\begin{eqnarray}
{\widetilde r}_1 &=& \frac{ ( c_1^2 - c_2) }{2 \beta_0} \ ,
\qquad
{\widetilde r}_2 = \frac{1}{8 \beta_0^2} \left[
( c_1^2 - c_2 )^2 - 2 \beta_0 ( c_1^3 - 2 c_1 c_2 + c_3 ) \right] \ ,
\nonumber \\
{\widetilde r}_3 & = & \frac{1}{48 \beta_0^3}  [
( c_1^2 - c_2 )^3 - 6 \beta_0 (c_1^2 - c_2 )(c_1^3 - 2 c_1 c_2 + c_3 )
\nonumber  \\
& & + 8 \beta_0^2 ( c_1^4 - 3 c_1^2 c_2 + c_2^2 + 2 c_1 c_3 - c_4 )] \ .
\end{eqnarray}
The singular part of the Borel transform $B_S(b)$ around $b=1/2$
must have the form
\label{BStc}
\begin{equation}
B_S(b; \mu) =  N_m \pi  \frac{\mu}{ {\overline m}_q }
 \frac{1}{ ( 1 - 2 b)^{1 + \nu} } \left[ 1 +
\sum_{k=1}^{\infty} {\widetilde c}_k ( 1 - 2 b)^k \right]
+ B_{S}^{\rm (an.)}(b; \mu) \ ,
\label{BSrenan}
\end{equation}
\begin{equation}
{\widetilde c}_1 = \frac{ {\widetilde r}_1 }{ (2 \beta_0 ) \nu } \ ,
\quad
{\widetilde c}_2 = \frac{ {\widetilde r}_2}{ (2 \beta_0 )^2 \nu (\nu -1) }
\ , \quad
{\widetilde c}_3 = \frac{ {\widetilde r}_3}
{ (2 \beta_0 )^3 \nu (\nu -1)(\nu - 2) } \ ,
\label{tc}
\end{equation}
and $B_{S}^{\rm (an.)}(b; \mu)$ is analytic on the disk $|b| < 1$.
The ${\overline {\rm MS}}$ coefficients for $ n_f\!=\!4$: $c_1= 1.5400$, $c_2=3.0476$ and $c_3=15.0660$ are
already known \cite{Tarasov:au,vanRitbergen:1997va},
but for $c_4$ we have only
estimates \cite{Ellis:1997sb, Elias:1998bi} obtained
by Pad\'e-related methods. We consider
 $c_4 =40 \pm 60$. Thus, ${\widetilde c}_j$ can be obtained: ${\tilde c_1} = -0.1054 $, ${\tilde c_2} = 0.2736
$ and ${\tilde c_3} = 0.01 \pm 0.17$.

The  bilocal method \cite{Lee:2002sn} consists of taking
in the expansion (\ref{BSrenan}) for the analytic part
$B_{S}^{\rm (an.)}$ a polynomial in powers of $b$, so that
the expansion of $B_S$ around $b\!=\!0$ agrees with
expansion (\ref{BSm1}). For that, the residue parameter
$N_m$ in Eq.~(\ref{BSrenan}) has to be determined.
Using  Refs.~\cite{Cvetic:2003wk}:
\begin{equation}
N_m = \frac{{\overline m}_q}{\mu} \frac{1}{\pi} R_S(b=1/2) \ ,
\label{Nmform}
\end{equation}
and according to (\ref{BSrenan})
\begin{equation}
R_S(b; \mu) \equiv  (1 - 2 b)^{1 + \nu} B_{S}(b; \mu) \ .
\label{RSm}
\end{equation}
Then, the $N_m$  is estimated (see Ref.~\cite{Cvetic:2003wk}),
using $R_S(b)$  TPS
and Pad\'e approximation $[1/1]$:
 \begin{equation}
N_m(n_f\!=\!4) = 0.555 \pm 0.020 \ ,
\label{Nmnf}
 \end{equation}
The bilocal expansion (\ref{BSrenan}) has then for the
analytic part the polynomial
\label{BSan}
\begin{eqnarray}
B_{S}^{\rm (an.)}(b; \mu) & = & h^{(m)}_0 + \frac{h^{(m)}_1}{1! \beta_0} b
+ \frac{h^{(m)}_2}{2! \beta_0^2} b^2 \ ,
\label{BSanexp}
\\
h^{(m)}_k & = & \frac{4}{3} r_k  - \pi N_m \frac{\mu}{{\overline m}_q}
(2 \beta_0)^k \sum_{n=0}^3 {\widetilde c}_n
\frac{ \Gamma ( \nu + k + 1 - n) }{ \Gamma(\nu + 1 - n) } \ ,
\label{hms}
\end{eqnarray}
where, by convention, $r_0 = {\widetilde c}_0 = 1$.
Then, the bilocal formula, is
\begin{equation}
B_S(b; \mu)^{\rm (biloc.)}   =   N_m \pi  \frac{\mu}{ {\overline m}_q }
 \frac{1}{ ( 1 - 2 b)^{1 + \nu} } \left[ 1 +
\sum_{k=1}^3{\widetilde c}_k ( 1 - 2 b)^k \right]
+ \sum_{k=0}^2 \ \frac{h_k^{(m)}}{k! \ \beta_0^k} \ b^k \ .
\label{BSbiloc}
\end{equation}
Applying the (generalized) principal value (PV)
prescription for the Borel integration
\begin{equation}
S(a) = \frac{1}{\beta}_0 {\rm Re}
\int_{\pm i \varepsilon}^{\infty \pm i \varepsilon} \ db
\ \exp \left( - \frac{b}{\beta_0 a(\mu)} \right) \ B_S(b; \mu) \ ,
\label{BSint}
\end{equation}
we obtain the pole mass $m_q$ in terms of the mass ${\overline m}_q$.
The numerical integration is performed, using the
Cauchy theorem ( Refs.~\cite{Cvetic:2001sn}).

In Figs.~\ref{mqmu.fig} (a) we present the resulting (PV) pole
masses of the $b$, as function of the renormalization
scale $\mu$. The spurious $\mu$-dependence is very weak.
In addition, results of another method (``R''-method )
are presented in Figs.~\ref{mqmu.fig} (a),  with the $\mu$-dependence
stronger in the low-$\mu$ region ($\mu/{\overline m}_q < 1$).
The R-method (Refs.~\cite{Caprini:1998wg,Cvetic:2001sn}) consists in
the Borel integration of the function (\ref{RSm})
\begin{equation}
S = \frac{1}{\beta_0} {\rm Re}
\int_{\pm i \varepsilon}^{\infty \pm i \varepsilon} \ db
\ \exp \left( - \frac{b}{\beta_0 a(\mu)} \right)
\ \frac{R_S(b; \mu)}{(1 - 2 b)^{1 + \nu}} \ ,
\label{BRSint}
\end{equation}
where for $R_S(b)$ the corresponding (NNLO) TPS is used.
When we take ${\overline m}_b = 4.23$ GeV  and we vary the values of the residue parameter $N_m$, the bilocal method gives,
at $\mu/{\overline  m}_q = 1$, variation
$\delta m_b = \mp 3$ MeV.
When the central values of $N_m$  are used, the
variation of the obtained values of $m_q$ with $\mu$,
when $\mu/{\overline m}_q$ grows from 1.0 to 1.5, is
about $5$ MeV  for $m_b$
(for $R$-method: $4$ MeV). When $c_4$ is varied, the variation is about $\mp\!2$ and $\mp\!1$ MeV
for $m_b$. The uncertainty in $\alpha_s$
can be taken as
$\alpha_s(m_{\tau}) = 0.3254 \pm 0.0125$ \cite{Cvetic:2001ws},
corresponding to $\alpha_s(M_Z) = 0.1192 \pm 0.0015$.
This uncertainty is by far the major source in the variation
of the pole masses: ($\delta m_b)_{\alpha_s} = ^{+135}_{-148}$ MeV
for bilocal method ($^{+137}_{-150}$ MeV for $R$-method).

The natural renormalization scale $\mu$ here is
a hard scale $\mu\!\sim\!{\overline m}_q$, and will be
denoted later in this work as $\mu_m$ in order to distinguish
if from the ``soft'' renormalization scale $\mu$ used in the
analogous renormalon-based resummations of the
(hard$+$)soft binding energy $E_{q \bar q}(s)$
(${\overline m}_{q} > \mu \geq {\overline m}_{q} \alpha_s$)
in Sec.~\ref{Eqq}. The fact that the two renormalization scales
are different does not affect the mechanism of the
($b\!=\!1/2$) renormalon cancellation in the bilocal calculations of
the meson mass $(2 m_q\!+\!E_{q \bar q}(s))$, because the
renormalon ambiguity in each of the two terms is
renormalization scale independent $\sim\!{\widetilde \Lambda}$,
as seen by Eqs.~(\ref{tL2})--(\ref{BStc}).
On the other hand, if $R$-type methods (\ref{BRSint})
[cf.~also Eq.~(\ref{Nmform})] are applied
for the resummations of $2 m_q$ and $E_{q \bar q}(s)$, the
renormalon ambiguities are renormalization scale independent
in the approximation of the one-loop RGE running, and the
renormalon cancellation is true at this one--loop level.

\section{Separation of the soft and ultrasoft contributions}
\label{sep}

The perturbation expansion of the (hard $+$ soft $+$ ultrasoft)
binding energy $E_{q \bar q}$ of the $q {\bar q}$ heavy quarkonium
vector ($S=1$) or scalar ($S=0$) ground state ($n=1, \ell=0$)
up to the ${\rm N}^3 {\rm LO}$ ${\cal O}(m_q \alpha_s^5)$
was given in \cite{Penin:2002zv}
The reference mass scale used was the pole mass $m_q$.
The ground state energy expansion has the form
\begin{eqnarray}
E_{q \bar q} & = & - \frac{4}{9} m_q \pi^2 a^2(\mu)
{\Big \{} 1 + a(\mu) \left[ k_{1,0} + k_{1,1} L_p(\mu) \right] \nonumber\\
&& +
a^2(\mu) \left[ k_{2,0} + k_{2,1} L_p(\mu) + k_{2,2} L_p^2(\mu) \right]
\nonumber\\
&& + a^3(\mu) \left[ k_{3,0} + k_{3,1} L_p(\mu) + k_{3,2} L_p^2(\mu)
+ k_{3,3} L_p^3(\mu) \right] + {\cal O}(a^4) {\Big \}} \,
\label{Eqqexp}
\end{eqnarray}
where
\begin{eqnarray}
L_p(\mu) &=& \ln \left( \frac{\mu}{\frac{4}{3} m_q \pi a(\mu) } \right) \ .
\label{Lp}
\end{eqnarray}
The   coefficients
$k_{i,j}$ of perturbation expansion (\ref{Eqqexp})   of the quarkonium ($n=1$; $\ell=0$; $S=1$ or $0$)
are given below. \cite{Titard:1993nn,Czarnecki:1997vz,Melnikov:1998ug,Penin:1998zh}, ~\cite{Penin:2002zv} --
For  $N_c=3$:
\label{kij}
\begin{eqnarray}
k_{1,1} & = & 4 \beta_0 \ ,
\qquad
k_{1,0} =  \left( \frac{97}{6} - \frac{11}{9} n_f \right) \ ,
\label{k1s}
\\
k_{2,2} & = & 12 \beta_0^2 \ ,
\qquad
k_{2,1} = \frac{927}{4} - \frac{193}{6} n_f + n_f^2  \ ,
\label{k2221}
\\
k_{2,0} & = & 361.342 - 40.9649 \ n_f + 1.16286 \ n_f^2  - 11.6973 \ S (S+1) \ ,
\label{k20}
\\
k_{3,3} & = & 32 \beta_0^3 \ ,
\qquad
k_{3,2} = \frac{4521}{2} - \frac{10955}{24} n_f + \frac{1027}{36} n_f^2 -
\frac{5}{9} n_f^3 \ ,
\label{k3332}
\\
k_{3,1} & = & 7242.3 - 1243.95 \ n_f + 69.1066 \ n_f^2  - 1.21714 \ n_f^3 \nonumber\\
&&
+ \frac{\pi^2}{2592} ( - 67584 + 4096 \ n_f ) \ S (S+1) \ ,
\label{k31}
\\
k_{3,0} & = & {\bigg [} \left( 7839.82 - 1223.68 \ n_f + 69.4508 \ n_f^2
- 1.21475 \ n_f^3 \right)
\nonumber \\
&& + (- 109.05 + 4.06858 \ n_f) \ S(S+1) \nonumber \\
&&
- \frac{\pi^2}{18}
\left( -1089 + 112 \ S(S+1) \right) \ln \left( a(\mu) \right)
+ 2 \frac{a_3}{4^3}
{\bigg ]} \ ,
\label{k30}
\end{eqnarray}
Here, $a_3$  have been estimated  in  Ref.~\cite{Cvetic:2003wk},
obtained from the condition of renormalon cancellation
in the sum $(2 m_q + V_{q \bar q}(r))$
\label{a3nf}
\begin{equation}
\frac{1}{4^3} \ a_{3}(n_f\!=\!4)   \approx
86. \pm 23. \ .
\end{equation}
The coefficients $k_{i,j}$ in the expansion (\ref{Eqqexp})
originate from quantum effects from various scale regimes
of the participating particles:
(a) the hard scales ($\sim\!m_q$);
(b) the soft and potential scales
where the three momenta are $|{\bf q}|\!\sim\!m_q \alpha_s$
($|q^0|\!\sim\!m_q \alpha_s$ in the soft
and $|q^0|\!\sim\!m_q \alpha_s^2$ in the potential regime);
(c) ultrasoft scales where
$|q^0|$ and $|{\bf q}|$ are both $\sim\!m_q \alpha_s^2$.
The coefficients are dominated by the soft scales;
the hard scales start contributing at the NNLO \cite{Kniehl:2002br}
and are numerically smaller. For this reason, we will usually refer to the combined soft and hard
regime contributions to the binding energy as simply
soft ($s$) contribution $E_{q\bar q}(s)$.
Strictly speaking, it is only the pure soft regime that
contributes to the $b=1/2$ renormalon.
However, for simplicity,
we will resum the hard$+$soft contributions $E_{q \bar q}(s)$
together. This will pose no problem,
since the hard regime, being clearly perturbative,
is not expected to deteriorate the convergence properties
of the series for $E_{q \bar q}(s)$. The natural
renormalization scale $\mu$ in the resummations of $E_{q \bar q}(s)$
is expected to be closer to the soft scale
($m_q \leq  \alpha_s  \mu < m_q$).

On the other hand, the ${\rm N}^3 {\rm LO}$ coefficient
$k_{3,0}$ obtains additional contributions from the
from the ultrasoft ($us$) regime.
The leading ultrasoft contribution comes from the exchange of
an ultrasoft gluon in the heavy quarkonium
\cite{Kniehl:1999ud,Brambilla:1999xf}. It consists of two parts:
\begin{enumerate}
\item
The retarded part, which cannot be interpreted in terms of
an instantaneous interaction
\begin{equation}
\frac{1}{\pi^3} k_{3,0}(us,{\rm ret.}) =
- \frac{2}{3 \pi} \left( \frac{4}{3} \right)^2 L_1^E \approx + 41.014 \ ,
\label{k30usr}
\end{equation}
where $L_1^E \approx - 81.538$ is the QCD Bethe logarithm
-  see Refs.~\cite{Kniehl:1999ud,Kniehl:2002br}.
\item
The non-retarded part can be calculated as expectation value
of the $us$ effective Hamiltonian ${\cal H}^{us}$ in the
Coulomb (i.e., leading order) ground state $|1 \rangle$,
where ${\cal H}^{us}$ (in momentum space) was derived in
Refs.~\cite{Kniehl:1999ud,Kniehl:2002br}. Direct calculation of the
expectation value, here in coordinate space, then gives:
\label{k30usnr}
\begin{eqnarray}
\frac{1}{\pi^3} k_{3,0}(us,{\rm nonret.})&=&
- \frac{9}{4 \pi^5}  \frac{1}{m_q a^5(\mu)}
\langle 1 | {\cal H}^{us} | 1 \rangle =
\frac{2}{ \pi^5 m_q a^4(\mu)}  [ \nonumber \\
& &
\frac{1}{2} \ln \frac{\mu_f^2}{(E_1^C)^2}
  + \frac{5}{6} - \ln 2  ]}
 \times {\Big \{
- \frac{27 \pi^3}{8} a^3(\mu) \langle 1 | \frac{1}{r} | 1 \rangle  \nonumber \\
& &
- 17 \pi^2 \frac{a^2(\mu)}{m_q} \langle 1 | \frac{1}{r^2} | 1 \rangle
 + \frac{4 \pi^2}{3} \frac{a(\mu)}{m_q^2}
\langle 1 | \delta({\bf r}) | 1 \rangle  \nonumber \\
& & +
3 \pi \frac{a(\mu)}{m_q^2} \langle 1 | \{ \Delta_{\bf r}, \frac{1}{r}
\} | 1 \rangle {\Big \}} \ ,
\label{k30usnr1}
\\
 & =&  - 14.196 \left[ \ln \left( \frac{ \mu_f}{m_q \alpha_s^2(\mu)}
\right) + 0.9511 \right] \ .
\label{k30usnr2}
\end{eqnarray}
Here, $E_1^C = -(4/9) m_q \alpha_s^2(\mu)$ is the Coulomb
energy of the state $|1 \rangle$, and $\mu_f$ is the factorization
energy between the soft ($\!\sim m_q \alpha_s$) and ultrasoft
($\!\sim m_q \alpha^2_s$) scale.
\end{enumerate}

The $s$--$us$ factorization scale $\mu_f$ can be
estimated as being roughly in the middle between
the $s$ and $us$ energies on the logarithmic scale
\cite{Cvetic:2003wk}
\begin{equation}
\mu_f \left[ \approx (E_{\rm S} E_{\rm US})^{1/2} \right] = \kappa \ m_q
\alpha_s(\mu_s)^{3/2} \ ,
\label{muf}
\end{equation}
where $\kappa \sim 1$ and $\mu_s \approx E_{\rm S}$ ($\!\mu$).
Therefore, the ultrasoft part of the ${\rm N}^3 {\rm LO}$
coefficient $k_{3,0}$ can be rewritten, by Eqs.~(\ref{k30usr}),
(\ref{k30usnr}) and (\ref{muf}), in terms of the
$s$--$us$ parameter $\kappa$ as
\begin{equation}
\frac{1}{\pi^3} k_{3,0}(us) = 27.512 + 7.098 \ln (\alpha_s(\mu_s))
- 14.196 \ln ( \kappa ) \ .
\label{k30us}
\end{equation}
The soft scale $\mu_s$ appearing here will be fixed
by the condition $\mu_s = (4/3) {\overline m}_q \alpha_s(\mu_s)$.

The formal perturbation expansions for the separate
soft and ultrasoft parts of the ground state binding energy (\ref{Eqqexp})
are then
\label{Eqqsus}
\begin{eqnarray}
E_{q \bar q}(s) & = & - \frac{4}{9} m_q \pi^2 a^2(\mu)
{\Big \{} 1 + \sum_{i=1}^2 a^i(\mu) \sum_{j=0}^i k_{i,j}
L_p(\mu)^j\nonumber \\
& &
+ a^3(\mu) \sum_{j=1}^3 k_{3,j} +
a^3(\mu)  \left[ k_{3,0}\!-\!k_{3,0}(us) \right] + {\cal O}(a^4)
{\Big \}} \ ,
\label{Eqqs}
\\
E_{q \bar q}(us) & = & - \frac{4}{9} m_q \pi^2 a^2(\mu)
{\big \{}  a^3(\mu) k_{3,0}(us)  + {\cal O}(a^4)
{\big \}} \ .
\label{Eqqus}
\end{eqnarray}
The energy $E_{q \bar q}(s)$ (\ref{Eqqs})
contains the leading IR renormalon effects,
and $E_{q \bar q}(us)$ (\ref{Eqqus}) does not.
In these expressions, the common factor is the soft scale
$\mu_p(\mu) = (4/3) m_q \alpha_s(\mu)$ which is also present
as the reference scale in the logarithms
$L_p(\mu) = \ln (\mu/\mu_p(\mu) )$ appearing with the coefficients
$k_{i,j}$ (when $j \geq 1$) in Eqs.~(\ref{Eqqexp}), (\ref{Lp}).

We will re-express $m_q$ everywhere in $E_{q \bar q}$ with the renormalon-free
mass ${\overline m}_q$, and will
consider the dimensionless soft-energy quantity
$E_{q \bar q}(s)/{\overline m}_q$.

Thus, we will divide the soft binding energy
with the quantity ${\overline \mu}({\widetilde \mu})
= (4/3) {\overline m}_q \alpha_s({\widetilde \mu})$, where
${\widetilde \mu}$ can be any soft scale.
We will fix this scale by the condition
${\widetilde \mu} = (4/3) {\overline m}_q  \alpha_s({\widetilde \mu})$
($\Rightarrow {\widetilde \mu}  = \mu_s$).
Further, in the logarithms $L_p(\mu)$ we
express the pole mass $m_q$ in terms of ${\overline m}_q$
and powers of $a(\mu)$ (cf.~Sec.~\ref{mass}),
and the powers of logarithms $\ln^k[a(\mu)]$ we
re-express in terms of $\ln^k [a(\widetilde \mu)]$.
This then results in the following soft binding energy
quantity $F(s)$ to be resummed
\begin{eqnarray}
F(s) & \equiv &  - \frac{9}{4 \pi^2}
\frac{E_{q \bar q}(s)}{ {\overline m}_q a(\widetilde \mu) }
= a(\mu) \left[ 1 + a(\mu) f_1 + a^2(\mu) f_2 + a^3(\mu) f_3
+ {\cal O}(a^4) \right] \ ,
\label{Fs}
\end{eqnarray}
where the coefficients $f_j$ depend on $\ln a(\widetilde \mu)$
and on three scales: the renormalization
scale $\mu$ ($\geq m_q \alpha_s$),
the (fixed) soft scale $\widetilde \mu$,
and ${\overline m}_q$. The coefficient $f_3$ depends,
in addition, on the parameters $\kappa$ (\ref{muf})-(\ref{k30us}),
$\mu_s$, and $a_3$ (\ref{a3nf}).
The coefficients $f_j$ are written explicitly in Appendix \ref{app:sbe}.
The $b=1/2$ renormalon in the quantity $F(s)$ is then of the type
of the renormalon of the pole mass $m_q$ discussed in
the previous Sec.~\ref{mass}.

However, if we divided in Eq.~(\ref{Fs}) by $m_q$ instead of
${\overline m_q}$ and at the same time used in the
resulting $f_j$-coefficients $\ln m_q$, the numerical
resummations of $F(s)$ by methods of Sec.~\ref{Eqq} would
give us values for $E_{q \bar q}(s)$ different usually by
not more than ${\cal O}(10^1 {\rm MeV})$ (we checked this
numerically). We will briefly refer to these approaches later in
this Section as ``pole mass'' approaches. A version of such
pole mass bilocal approach was applied in Ref.~\cite{Lee:2003hh}
for resummation of the unseparated $E_{q \bar q}(s\!+\!us)$.

The ultrasoft part (\ref{Eqqus}), on the other hand,
has no $b = 1/2$ renormalon. The mass scale used there
should also be renormalon free (${\overline m}_q$).
The renormalization scale $\mu$ there should be adjusted downward
to the typical $us$ scale of the associated process
$\mu \mapsto \mu_{us}$ ($\sim\! m_q \alpha_s^2$)
in order to come closer to a realistic estimate
\begin{eqnarray}
E_{q \bar q}(us) & \approx & - \frac{4}{9} {\overline m}_q \pi^2 k_{3,0}(us) a^5(\mu_{us}) \ .
\label{Eqqus2}
\end{eqnarray}

\section{
Mass ${\overline m}_b$ from the known mass of the vector $b {\bar b}$
[$\Upsilon (1S)$]}
\label{Eqq}

The soft binding energy
quantity to be resummed  is $F(s)$ of Eq.~(\ref{Fs}).
However, in the ${\rm N}^3{\rm LO}$ coefficient $f_3$ we have
dependence on   $a_3$, and on the $s$-$us$ factorization scale
parameter $\kappa\!\sim\!1$. Then, the value of  $\kappa$ (\ref{muf})
is obtained requiring that the residue parameter values
be reproduced from the Borel transform of the soft binding
energy quantity $F(s)$ of Eq.~(\ref{Fs}).

Similarly as in Eq.~(\ref{BSrenan}), we have
\begin{equation}
B_{F(s)}(b; \mu) = N_m \frac{9}{2 \pi}
\frac{\mu}{ {\overline m}_q a(\widetilde \mu) }
\frac{1}{ ( 1\!-\!2 b)^{1 + \nu} } \left[ 1 +
\sum_{k=1}^{\infty} {\widetilde c}_k ( 1\!-\!2 b)^k \right]
+ B_{F(s)}^{\rm (an.)}(b; \mu)\,
\label{BFrenan}
\end{equation}
where the factor in front of the singular part
was determined by the condition of renormalon cancellation
of the sum $2 m_q + E_{q \bar q}(s)$. We now define in analogy
with Eq.~(\ref{RSm})
\begin{equation}
R_{F(s)}(b; \mu; \mu_f) = (1 - 2 b)^{1 + \nu} B_{F(s)}(b;\mu; \mu_f) \ .
\label{RFs}
\end{equation}
Here we denoted,  explicitly the dependence
on the factorization scale $\mu_f$ and
\begin{equation}
N_m = \frac{2 \pi}{9} \frac{ {\overline m}_q a(\widetilde \mu) }{\mu}
R_{F(s)}(b;\mu;\mu_f) {\big |}_{b=1/2} \ .
\label{NmEqq}
\end{equation}
Theoretically, $R_{F(s)}(b)$ should be a function
with only a weak singularity (cut) at $b=1/2$, and the
nearest pole at $b=3/2$ ( \cite{Aglietti:1995tg}).
Resummations such as Pad\'e approximations (PA's)
are applied, then
  $R_{F(s)}[2/1](b)$ has physically acceptable
pole structure $|b_{\rm pole}| \geq 1$ for most of the
values of $\mu \geq m_q \alpha_s$ and $\kappa \sim 1$.
and the residue parameter $N_m$ is
reasonably stable under the variation of $\mu$.

In Fig.~\ref{Nmbb.fig}(a) we show the
dependence of $N_m$ on $\kappa$, at a typical
$\mu$ value $\mu\!=\!3$ GeV, for the $b \bar b$ system.
The  know value  of $N_m$ is
obtained by the $R_{F(s)}[2/1](b\!=\!1/2)$ expression at
$\kappa \approx 0.59$. In Fig.~\ref{Nmbb.fig}(b) we
present, for $\kappa = 0.59$, the dependence of
calculated $N_m$ on the renormalization scale $\mu$.
There, we include also the ($[2/1]$-resummed) curve for
the case when no separation of the $s$ and $us$ parts
of the energy is performed. In that case, the
obtained values of $N_m$ are unacceptable.
The other values of the input parameters are chosen
to have the $b \bar b$ ``central'' values:
$a_3/4^3 = 86$; ${\overline m}_b = 4.23$ GeV;
${\widetilde \mu} = 1.825$ GeV ($\approx \mu_s$)
and $\alpha_s(\widetilde \mu; n_f\!=\!4) = 0.3263 (\approx
\alpha_s(\mu_s; n_f\!=\!4) = 0.326$)
[from: $\alpha_s(m_{\tau}; n_f\!=\!3) = 0.3254$, i.e.,
$\alpha_s(M_Z) = 0.1192$ \cite{Cvetic:2001ws}.
For the RGE running, we   use four-loop ${\overline {\rm MS}}$
$\beta$-function (TPS).

Variation $N_m = 0.555 \pm 0.020$ [$b \bar b$]
implies $\kappa = 0.59 \pm 0.19$.
If, on the other hand, $a_3$ parameter is varied,
 then for $b \bar b$ $\kappa = 0.59 \mp 0.06$.
Thus, the value of $s$-$us$ factorization scale parameter $\kappa$
is influenced largely by the allowed values
of the renormalon residue parameter, and significantly less
by the allowed values $a_3$
of the ${\rm N}^3{\rm LO}$ coefficient of the static $q \bar q$
potential. Therefore, we will consider the variations
of $N_m$  and of $\kappa$ to be
related by a one-to-one relation, while the variations
of $a_3$  will be considered as independent.

In this way, we have the following value  for the $s$-$us$
factorization scale parameter:
\begin{equation}
N_m = 0.555 \pm 0.020 \ \Rightarrow \
\kappa   =   0.59 \pm 0.19
\quad (n_f = 4, \ S = 1) \
\end{equation}
and thus we obtain the ${\rm N}^3{\rm LO}$ TPS (\ref{Fs})
for the soft part of the ground binding energy.
Now with  the value of $\kappa$,
we can perform the resummation of the soft part of the
ground binding energy.
The full bilocal method \cite{Lee:2002sn,Lee:2003hh}
can be performed as in Sec.~\ref{mass},
Eqs.~(\ref{BSbiloc}) and (\ref{BSint}).
Therefore
\begin{equation}
B_{F(s)}^{(\rm biloc.)}(b;\mu)   =
N_m \frac{9}{2 \pi}
\frac{\mu}{ {\overline m}_q a(\widetilde \mu) }
\frac{1}{ ( 1\!-\!2 b)^{1 + \nu} } \sum_{k=0}^3 {\widetilde c}_k (1 - 2 b)^k
+ \sum_{k=0}^3 \ \frac{h_k}{k! \ \beta_0^k} \ b^k \ ,
\label{Fsbiloc}
\end{equation}
where the coefficients
$h_k$ in the expansion of the analytic part are     known
up to order $k=3$
\begin{equation}
h_k = f_k - N_m \ \frac{9}{2 \pi} \
\frac{\mu}{ {\overline m}_q a({\widetilde \mu}) } \
(2 \beta_0)^k \sum_{n=0}^3 \
{\widetilde c}_n
\frac{ \Gamma ( \nu + k + 1 - n) }{ \Gamma(\nu + 1 - n) }
\quad (k=0,1,2,3) \ .
\label{hss}
\end{equation}
  The result
  have   spurious $\mu$-dependence, and  for the of the Borel transformation
  turns out to have a problematic behavior:   the obtained pole of  $B_{F(s)}^{\rm (an.)}[2/1](b)$ is
  unacceptably small in size:
$|b_{\rm pole}| \leq 1/2$. Theoretically, $B_{F(s)}^{\rm (an.)}(b)$
should have the nearest pole at $b=3/2$ \cite{Aglietti:1995tg}.
The reason for this problem
appears to lie in the specific truncated form of the
singular part taken in the bilocal method.
While the latter part describes well the behavior of the
transform near $b=1/2$, it influences apparently strongly
the coefficients $h_k$ and thus the analytic part,
so that no reliable resummation of that part (apart from
TPS) can be done.   This problem can be alleviated by
introducing   a ``form'' factor
which suppresses that part away from $b \approx 1/2$,
but keeps it unchanged at $b \approx 1/2$. We choose
  a Gaussian type of function, and the ``$\sigma$-regularized''
bilocal expressions for the Borel transform is
\begin{eqnarray}
B_{F(s)}^{(\sigma)}(b;\mu) & = & N_{m}   \frac{9}{2 \pi} \frac{\mu}{ {\overline m}_{q} a( \widetilde \mu )} \
\frac{1}{ ( 1\!-\!2 b) ^{1 + \nu } }  [1 + {\widetilde c}_{1} (1\!-\!2b)  \nonumber  \\
&  & +   ( {\widetilde c}_{2} + \frac{1}{8 \sigma^{2}}   )
(1\!-\!2b)^{2}
+  ( {\widetilde c}_{3} + \frac{{\widetilde c}_{1}}{8 \sigma^{2}}  )(1\!-\!2b)^{3}   ]  \nonumber  \\
&  & \times \exp \left[ - \frac{1}{8 \sigma^{2}} (1\!-\!2 b)^2 \right]
+
\sum_{k=0}^{3} \ \frac{1}{k! \ \beta_{0}^{k}} \ h^{(\sigma)}_{k}  b^{k}  \ .
\label{Fssig}
\end{eqnarray}
The corrective terms $1/(8 \sigma^2)$ and
$ {\widetilde c}_1/(8 \sigma^2)$ in the
coefficients of Eq.~(\ref{Fssig}) appear to ensure the correct known behavior
of the renormalon part up to order $\sim\!(1 - 2b)^{-\nu + 2}$.
Numerical analysis indicate that $\sigma$ is between zero and one.
Namely, when $\sigma$ decreases from $\sigma\!=\!\infty$ to about
$\sigma \approx 0.3$-$0.4$, the
value of the pole of the $[2/1]$ Pad\'e-resummed analytic part
$B_{F(s)}^{({\rm an.} \sigma)}(b)$ of Eq.~(\ref{Fssig})
gradually turns acceptable ($|b_{\rm pole}| > 1$) and
rather stable when the renormalization
scale $\mu$ varies in the interval $[m_q \alpha_s, m_q]$
(except close to $\mu \approx m_q \alpha_s$). When the value of $\sigma$
falls below $0.3$, the analytic part starts showing
erratic behavior again and the Borel resummation
gives significantly differing results with the TPS-
and the Pad\'e-evaluated analytic parts.
On these grounds,
the obtained optimal $\sigma$ turn out to be
 \label{sig}
\begin{equation}
\sigma  =  0.36 \pm 0.03 \quad (n_f\!=\!4, \ S=1) \ .
\end{equation}

In Fig.~\ref{Eqqsig.fig}(a) we present the   Borel-resummed
soft part of ground state energy for the bottonium ($S=1$),
as a function of the $\sigma$ parameter of method (\ref{Fssig}).

Finally, the results for the soft binding energy $E_{b \bar b}(s)$
of the ground state of
bottonium using  the
$R$-method \cite{Caprini:1998wg,Cvetic:2001sn}, where
we resum the function $R_{F(s)}(b;\mu)$ (\ref{RFs})
and then employ the (PV) Borel resummation as written in
Eq.~(\ref{BRSint}) (with $R_{F(s)}$ instead of $R_S$ there)., as functions of the renormalization scale $\mu$,
are presented in Fig.~\ref{Ebbmu.fig}(a).
We observe from the Figure that the bilocal  ``$\sigma$-regularized'' method (\ref{Fssig}) ($\sigma=0.36$)
gives the TPS and PA results closer to each other.
The methods $\sigma$-TPS, $\sigma$-PA, and $R$-PA give
similar results in the entire presented $\mu$-interval.
$R$-TPS appears to fail at low $\mu$ ($\approx m_b \alpha_s
\approx 1$-$2$ GeV).
In Fig.~\ref{Ebbmu.fig}(b) we include, for comparison,
the simple TPS evaluation of $E_{b \bar b}(s)$, according to
formula [cf.~Eq.~(\ref{Fs})]
\begin{equation}
F(s)^{\rm (TPS)}  \equiv -\frac{9}{4 \pi}
\frac{1}{{\overline m}_b \alpha_s(\widetilde \mu)} E_{q \bar q}(s)
= a(\mu) \left[ 1 + a(\mu) f_1 + a^2(\mu) f_2 + a^3(\mu) f_3 \right]
\ ,
\label{FsTPS}
\end{equation}
where for ${\rm N}^2{\rm LO}$ TPS case we take $f_3=0$.
In Fig.~\ref{Ebbmu.fig}(b) the same input parameters are used
as in Fig.~\ref{Ebbmu.fig}(a).
We see that the perturbation series shows strongly divergent behavior
already at ${\rm N}^3{\rm LO}$.
In this Figure, we also included the ``perturbative''
ultrasoft part $E_{b \bar b}^{\rm (p)}(us;\mu)$ calculated
according to [see Eqs.~(\ref{k30us}) and (\ref{Eqqus})].
\begin{equation}
F^{\rm (p)}(us) \equiv -\frac{9}{4 \pi}
\frac{1}{{\overline m}_b \alpha_s(\widetilde \mu)}
E_{q \bar q}^{\rm (p)}(us;\mu) = k_{3,0} a^4(\mu) \ .
\label{Fusp}
\end{equation}

\subsection{Extraction of bottom mass}
\label{mb}

The estimate of the perturbative part
is given in Eq.~(\ref{Eqqus2}), where it was
essential to take for the renormalization scale
a $us$ scale $\mu \sim \mu_{us} \sim m_q \alpha_s^2$.

For the bottonium case, this scale is below $1$ GeV,
the energy at which we cannot determine perturbatively
$\alpha_s(\mu)$. This indicates that in the bottonium
the $us$ part of the binding energy has an appreciable
nonperturbative part. The lowest energy at which we can
still determine perturbatively $\alpha_s$ is
$\mu \approx 1.5$-$2.0$ GeV,
giving $\alpha_s(\mu) \approx 0.30 - 0.35$.
Although this is a soft scale for $b \bar b$, we will
use this also as an ultrasoft scale. Then by Eq.~(\ref{Eqqus2})
\begin{equation}
E_{b \bar b}(us)^{\rm (p)} \approx
- \frac{4}{9} {\overline m}_q \pi^2 k_{3,0}(us) a^5(\mu_{us})
\approx (- 150 \pm 100) \ {\rm MeV} \ .
\label{Ebbusp}
\end{equation}
The nonperturbative contribution coming from the
gluonic condensate is given by \cite{Voloshin:hc}
\begin{equation}
E_{b \bar b}(us)^{\rm (np)} \approx
{\overline m}_b
\pi^2 \frac{624}{425}
\left( \frac{4}{3} {\overline m}_b \alpha_s(\mu_{us}) \right)^{-4}
\langle a(\mu_{us}) G_{\mu \nu} G^{\mu \nu} \rangle
\approx  (50 \pm 35) \ {\rm MeV} \ ,
\label{Ebbusnp}
\end{equation}
where we used ${\overline m}_b = 4.2$ GeV, and the value
of the gluon condensate $\langle (\alpha_s/\pi) G^2 \rangle =
0.009 \pm 0.007 \ {\rm GeV}^4$  \cite{Ioffe:2002be}.
Eqs.~(\ref{Ebbusp}) and (\ref{Ebbusnp}) give
\begin{equation}
E_{b \bar b}(us)^{\rm (p+np)} \approx (-100 \pm 106) \ {\rm MeV} \ ,
\label{Ebbus}
\end{equation}
The finite charm mass contributions
has been calculated in Ref.~\cite{Brambilla:2001qk}
(Refs.~\cite{Gray:1990yh,Eiras:2000rh,Hoang:2000fm}).
The contribution to the mass
$M_{\Upsilon}(1S) = (2 m_b + E_{b \bar b})$ is
\begin{equation}
\delta M_{\Upsilon}(1S, m_c \not=0) \approx 25 \pm 10 \ {\rm MeV} \ .
\label{mcnot0}
\end{equation}
The estimates (\ref{Ebbus}), and (\ref{mcnot0})
then give a rough estimate of the $us$ and $m_c\!\not=\!0$ contributions
to the bottonium mass
$\delta M_{\Upsilon}(1S; us+m_c) \approx  (- 75 \pm 106)$ MeV.
The mass of the $\Upsilon(1S)$ vector bottonium ground
state is well measured $M_{\Upsilon}(1S) = 9460$ MeV
with virtually no uncertainty \cite{Hagiwara:fs}.
Therefore, the pure perturbative ``soft'' mass is
\begin{equation}
M_{\Upsilon}(1S; s) = 2 m_b + E_{b \bar b}(s) =  9535 \mp 106 \ {\rm MeV}
\ ,
\label{MUs}
\end{equation}
where the uncertainty  is
dominated by the uncertainty of the $us$ regime contribution.
Our numerical results for $E_{b \bar b}(s)$
and for $m_b$   allow us,
by varying the input value of ${\overline m}_b$, to adjust the
sum $2 m_b\!+\!E_{b \bar b}(s)$ to the value given in
Eq.~(\ref{MUs}). For the soft binding energy
we apply the ``$\sigma$-regularized bilocal methods
$\sigma$-TPS and $\sigma$-PA, and $R$-TPS and $R$-PA,
with the  input parameters:
$\alpha_s(M_Z) = 0.1192$; ${\widetilde \mu} = 1.825$ GeV
($\approx \mu_s$), thus $\alpha_s(\widetilde \mu, n_f\!=\!4) =
0.3263$ [$\alpha_s(\mu_s) = 0.326$]; $N_m = 0.555$; $\kappa = 0.59$;
$\sigma = 0.36$; $a_3/4^3 = 86$; $c_4({\overline {\rm MS}}) = 40$.
For $2 m_b$ we apply the bilocal-TPS and $R$-TPS
method, with renormalization scale $\mu_m/{\overline m}_b = 1$,
both methods giving us very similar results
[cf.~Fig.~\ref{mqmu.fig} (a)].
The bilocal-TPS method is applied for $2 m_b$ when $\sigma$-TPS and
$\sigma$-PA are applied for $E_{b \bar b}(s)$; the $R$-TPS
is applied for $2 m_b$ when $R$-TPS and $R$-PA are applied for
$E_{b \bar b}(s)$.
The extracted values
of ${\overline m}_b \equiv
{\overline m}_b(\mu\!=\!{\overline m}_b)$ are then
\begin{eqnarray}
\label{mbb}
{\overline m}_b & = & 4.225 \pm 0.054 \ {\rm GeV} \qquad
(\sigma\!-\!\textrm{TPS}) \ ,
\\
& = & 4.220 \pm 0.056  \ {\rm GeV} \qquad
(\sigma\!-\!\textrm{PA}) \ ,
\\
& = & 4.243 \pm 0.080  \ {\rm GeV} \qquad
(R\!-\!\textrm{TPS}) \ ,
\\
& = & 4.235 \pm 0.068  \ {\rm GeV} \qquad
(R\!-\!\textrm{PA}) \ .
\end{eqnarray}
Finally, the  average
of the central values  gives us
\begin{equation}
{\overline m}_b = 4.231 \pm 0.068 \ {\rm GeV}  \ .
\label{mbbav}
\end{equation}

\section{Comparisons and conclusions}
\label{summ}

Our result for the mass ${\overline m}_{b}=4.231 \pm 0.068 \ \rm GeV$,
is compared in Table \ref{table7}  with   recently values of ${\overline m}_b$.
The only input parameter common to all these
methods is $\alpha_s$.
Our central value was $\alpha_s(m_{\tau}) = 0.3254$
[$\Rightarrow \alpha_s(M_Z) = 0.1192$] since such \cite{Cvetic:2001ws},
or similar \cite{Geshkenbein:2001mn,Cvetic:2001sn},
values follow from the
(nonstrange) semihadronic $\tau$ decay data which are
very precise \cite{Barate:1998uf}. On the other hand, the
world average as of September 2002 is
$\alpha_s(M_Z) = 0.1183 \pm 0.0027$ \cite{Bethke:2002rv}.
Most of the authors during the last four years used
central value $\alpha_s(M_Z) \approx 0.118$.
Therefore, for comparisons, we convert our
results (\ref{mbb}) to this central value of $\alpha_s$ --
more specifically, from $\alpha_s(M_Z) = 0.1192 \pm 0.0015$
to $0.1180 \pm 0.0015$. This can be easily done by
inspecting in Table \ref{table2} the column
under $\alpha_s$,
an increase in the central values of ${\overline m}_b$ by $11$, $12$, $8$ and $10$ MeV,
respectively. This gives the average $10$ MeV
higher than in Eq.~(\ref{mbbav})
\begin{equation}
{\overline m}_b = 4.241 \pm 0.068 \ {\rm GeV} \quad
 \textrm{average \  when:} \  \alpha_s(M_Z) = 0.1180 \pm 0.0015   \ .
\label{mbbav2}
\end{equation}

There are two important numerical effects in our result.
The first is the separate evaluation of the ``perturbative''
ultrasoft energy part at the corresponding low renormalization energy
($ 1.5 - 2$ GeV), Eqs.~(\ref{Eqqus2}) and (\ref{Ebbusp}).
If we had not separated the (``perturbative'') ultrasoft from
the soft part of the binding energy,
the use of the common renormalization
energy scale $\mu$ ($\approx 3$ GeV) in the resummation then
would have given us the central value of $E_{b \bar b}(us)$
by about $+100$ MeV higher. Then the
extracted value of ${\overline m}_b$
would have gone down by about $46$ MeV, giving the
value ${\overline m}_b \approx 4.195 \pm 0.068$,
with the central value close to that of L03 in Table \ref{table7}.
On the other hand, that latter value is quite clearly
lower than the value PS02 in Table \ref{table7}, by about $150$ MeV,
principally because of the $b=1/2$ renormalon effect which
were taken into account here and in Ref.~\cite{Lee:2003hh}.
Thus, the renormalon effect brings down the extracted central
value of ${\overline m}_b$ by about 150 MeV, but the separate
evaluation  of the ultrasoft contribution brings
it up by about $50$ MeV.

\section*{Acknowledgments}
I would like  to thank G. Cvetic for helpful and discussions. This work was supported  by  Project USM-110321 of the UTFSM.

redpajama_set_name: RedPajamaArXiv


text:
Želatovice (Duits: Schelatowitz) is een Tsjechische gemeente in de regio Olomouc, en maakt deel uit van het district Přerov.

Želatovice telt 553 inwoners (2006).

Gemeente in Přerov

redpajama_set_name: RedPajamaWikipedia


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text:
The Wire, "React Quotes": You will believe a man can fly!
Spoilers for "The Wire" episode 5, "React Quotes," coming up just as soon as I order some pepper steak...
"He made another call?"
In a run that's never gotten the credit it deserves for being funny (or for any of the other things it does so brilliantly), the second meeting between our two fabulists -- and Jimmy's attempt to conceal a grin at the realization of just how full of it Scott really is -- is up there with the funniest "Wire" moments ever, whether it's Stringer chewing out Shamrock for taking notes on a criminal conspiracy, Omar's pajama-clad trip to the store, or Tommy and Norman sprinting to their car after Rawls turned the corner.
Nearly a year ago, undercover blogger and sometime "Wire" scribe David Mills reported from a "Wire" writers summit that this would be the funniest season ever: "We're talking the 'Dr. Strangelove' of police procedurals here." Mills wrote the script for "React Quotes," and the "Strangelove" comparison is dead-on. This is end of the world farce here, a portrayal of a system that's so ossified and self-destructive that you have to laugh at how bad things have gotten -- and how easy it is to game that system -- or else you'd want to blow your brains out. The only thing the season is lacking at this point is Landsman barking out, "You can't solve murders here -- this is the Homicide room!"
It was inevitable that McNulty and Templeton would cross paths, but the genius of how it happens is that (at first, at least) neither guy realizes how much he's helping the other one. Scott's laziness and fondness for making up the details -- reinforced when he transforms religious nut Nathan Levi Boston into dedicated family man Nathan Levi Boston -- makes him the perfect tool for Jimmy's attempt to turn the money tap back on, but it isn't until Scott expresses such surprise at the news of a second call that Jimmy realizes just how amazing his luck is.
I know there's been some debate in the On Demand thread about who knows what after that meeting at the Sun offices. While trying not to spoil things from future episodes, I will say that my reaction as I first watched that scene -- and as I've rewatched it several times since, often having to rewind because I'm laughing too hard -- is that Jimmy absolutely, positively figures out what's happening, while Scott has no clue. Just look at Jimmy's complete change in posture, the barely-constrained grin on his face, etc., while Scott just looks as befuddled as usual. I want to say more but this is one of those situations where your present knowledge of my future knowledge makes it hard for me to avoid spoilers, so I'll just shut up now.
Now, as Jimmy and Lester were discussing the details of their master plan -- to acquire intel on Marlo via their illegal wiretap and then credit the information to a phony CI -- I groaned (not for the show, but for what's become of these two cops), because what they're doing is just a more elaborate, deranged version of Herc and Carver's Fuzzy Dunlop scam from season two. If Herc and Carver (at the time) were symbols of all that's wrong with the Baltimore PD, what does it say about these two top investigators that they're trying something similar?
Or maybe the question should be this: What does it say about the Baltimore PD?
The scene where an indignant Lester argues with Daniels for more manpower -- "That's the cell number of the motherfucker who put 22 bodies on us!" -- and Daniels gets even more indignant in acknowledging how dry the well has become demonstrates exactly why Jimmy and Lester are stooping to this terrible, illegal level. (It's in some ways the equivalent of the "Strangelove" scene where Mandrake doesn't have enough change to call the President and asks Guano to shoot the lock off a Coke machine.) When Rawls and Burrell shut down the MCU, it was with the promise that they could get back up on Marlo if they acquired significant new information. I would say Marlo getting back up on cell phones qualifies, but of course that promise turns out to be as empty as every other one made in Baltimore. If you're Lester, seeing what's been done to this department and this city, and if you knew you could outsmart this mass-murdering SOB, wouldn't you perhaps go off the deep end and inadvertently ape those two knucklehead junior cops you once failed to school?
If you've read enough of these "Wire" reviews, you know how fond I am of pointing out the daisy chain of events that leads to momentous changes, both good and bad. (Think of all the things that had to happen for Randy to wind up back in the group home, or all the miscalculations Prop Joe had to make to wind up Marlo's latest victim.) In this round of "It's all connected," we have the bizarre story of Marlo's phone number. Vondas gives Marlo a phone because they need a means of communication. (I won't say a word about exactly what they're doing that so impresses Marlo and so confuses Lester; you'll find out eventually.) Marlo gives the number to Levy, to whom he was introduced by Joe (the final bit of wisdom Marlo absorbed from the man before killing him and replacing him in Vondas' organization). Herc, who works for Levy because of Marlo (and, no doubt, because of his role in the first Barksdale investigation), steals the number, both to get revenge on Marlo and to apologize to Carver for the Randy thing. And Carver, having finally grown into a good po-lice, takes it to exactly the right person: Lester. (Had he exercised such judgment with Randy instead of caving in to Herc's pleas for help with the camera, imagine how differently things would have gone.) I'm not saying how valuable this number will or won't be, but it's bizarre and yet totally "Wire"-esque that all these things had to happen for Lester to get ahold of it.
Getting back to what I said before about the danger of knowing too much about what's coming, I want to say as little as possible about Omar's superheroic leap to nowhere from his firefight with Chris, Snoop and Michael. All I'll give you is that when I found out what really happened there, I wasn't dissatisfied. As for the firefight itself, though its scale and choreography was far larger than life than the show usually gets, the show has definitely gone to this level before (think the "Let's bang out" gunfight at the Barksdale stash house in season three), always with Omar, the one larger-than-life character (other than maybe Brother Mouzone) on the show. There's a lot of colors in this show's palette; Omar just happens to be the most vivid shade of red, you know?
Beyond that, the point of the shootout -- of all the scenes with Omar, Donnie and Marlo's people in this episode -- was to show that, for the first time in this series, our man with the shotgun is up against a foe he may not be able to beat. For all that guys like the Barksdale crew and their hangers-on throw around the word "soldier," the cold hard fact is that most of the boys on those corners have no idea what to do with a gun, or with the kind of well-planned ambush that Omar specializes in. Chris, on the other hand, either has military training himself or was trained by someone who was ex-military, because he and his people are vastly better at marksmanship, firing patterns and tactics than your average bunch of slingers. Against Cheese or Wee-Bey or Stinkum, all Omar needed was a plan and the boldness to carry it out; against Chris and Snoop, he was lucky to make it to that window. (And, of course, Chris had a plan of his own, which we realized when Monk showed up at the rim shop wearing a bulletproof vest, which made the later surveillance scenes with Omar and Donnie extra-chilling. I just wanted to yell at the TV and tell him to peel on out of there already.)
While Omar has always seemed so indestructible because he exists outside of any institution, beholden to no one and nothing but his own code, Clay Davis invulnerability has come from his skill at mastering all the shortcuts within his institution. Yet by the end of the episode both men have been confronted with the limits of their power. Omar finally goes up against someone he can't intimidate or outshoot, and Clay finally seems caught in a political trap he can't talk his way out of. Both the major Clay scenes this episode were brilliant and hilarious. His argument with Neresse -- featuring the first "Sheeeeeeeitttttt!" of the season, and maybe the longest one ever (Mills is a fan of Clay's catchphrase) -- evoked the moment in season one when Brianna Barksdale convinced D'Angelo to fall on his sword for the sake of the family. Clay's whispered conversation with ex-mayor Royce (note the Afro-centric tie, as the man no longer has to worry about appeasing his handful of white constituents) at the courthouse rally offered up exactly what we imagine goes on away from the microphones at every one of Al Sharpton's photo ops, didn't it?
And with Omar in mortal danger, we have two other characters who may not be beholden to an institution, but who aren't exactly enjoying the benefits of that. Bubbs has finally kicked his addiction but is still empty and insistent on punishing himself for Sherrod and every other bad thing he did in his junkie days. Dukie is off the corner but at a loss for what to do with his life. And both get good-intentioned but not necessarily useful advice from street veterans Waylon and Cutty. (Cutty looking very different; I can't tell if Chad Coleman bulked up or if the shorter haircut just makes it look like he did.)
Dukie doesn't belong in The Game. However far gone Michael might be, even he knows that and wants to find another avenue for his friend. The problem is that Dukie's greatest strength is his brain, and by dropping out of school, it's going to be hard for him to find a realistic outlet for that. I know some people have suggested that Dukie and Michael are supposed to parallel the young Stringer and Avon, and while I can see Michael as Avon (they even share a boxing background, though of course Avon was raised in a family of kingpins), I can't (or maybe don't want to) see Dukie as growing up to be Stringer. Maybe Stringer was this sweet and innocent as a teenager, but Dukie doesn't seem to have the moral coldness that would allow him to be a Stringer (or a Joe, even though both share a fondness for fixing fans). Or maybe, like Cutty, all I've got for Dukie is hope and wishes, and it'll turn out that The Game is the only place for him. God, I hope not. I'm not expecting a Namond-style liberation from the streets, but I'd like to see the poor kid get some kind of upbeat but realistic ending.
Some other thoughts on "React Quotes":
Though the more obvious Dukie parallel in this episode is with Bubbs, you'll note that Omar would have killed Snoop had he not forgotten to jack a round into the chamber -- the same mistake Dukie makes during target practice with Michael.
Bad Dad award, not surprisingly, goes to Jimmy. As many terrible things he's done, both this season and in the past, I'm not sure I ever felt embarrassed for McNulty in the way I did when he showed up at Elena's house with no excuse for missing the play and absolutely no connection to his kids (who have come to expect their dad being pathetic and/or absent). Elena and Beadie are both pleading for Jimmy to clean his ass up, but I don't see how he pulls out of this spiral. He's enjoying himself way too much with this fake serial killer thing, which in turn fuels all the other self-destructive behavior.
Good Dad award, semi-surprisingly, goes to Chris, who has some Norman Rockwell-looking family stashed away in a nice part of town. On the one hand, Chris is an ice-blooded killer. On the other, outside of the deaths of Butchie and Bug's dead, he's always been shown to be a gentle killer, as these types go, no doubt a result of the abuse he suffered as a kid. So I can almost see him being a decent, if not always present, father. Still, bizarre to see such big smiles on his face and Marlo's. (I had forgotten Jamie Hector had teeth.)
Though Gus can smell the BS coming off of Scott, it's nice to see that he's not omniscient. He caught Fat Face Rick's name in the city council minutes in the premiere, but had no idea the importance of hearing Prop Joe's name as a murder victim on the police blotter.
One of the perils of doing a show about the media and current events that films months in advance is that sometimes you outdate yourself. The sports editors' discussion of baseball's steroid problem in the news meeting seems awfully quaint in the wake of the Mitchell Report and the Clemens press conference.
Fans of "The Corner" may have noticed that the nurse drawing Bubbs' blood for his HIV test was played by Fran Boyd. (If you didn't read the book, Khandi Alexander played her in the miniseries.)
Another great little touch: Michael and Cutty nod at each other, but Michael won't cross the entrance into the gym. He knows and respects what Cutty's about now, but he also knows that by joining Marlo's crew, he has no place there anymore. (Not that it's likely, but his presence there could jeopardize what Cutty's trying to do.)
Throughout this season, there have been these little throwaway mentions of violent images from popular culture: the "Boyz N the Hood" drive-by shootings, the media obsession with Natalee Holloway, and here Landsman referring to McNulty as "Clarice" (as in Starling, as in "Silence of the Lambs"). The media and the public lap this stuff up, these mentions seems to say, so why shouldn't McNulty be able to get away with his plan?
Lines of the week:
"How's it feel, Clay? Not much fun on the ass end, is it?" -Carcetti
"If Marlo Stanfield is using a cell phone, it is just a matter of time until we are up to our asses in pretrial motions mitigating a wiretap case. Joe gave him to us just in time." -Levy
"Where am I gonna find homeless people?" -Scott
"Not at home, I'd imagine." -Gus
"Whatever else I did to piss you off, remember that I also did this." -Herc
"What the fuck's wrong with this city?" -McNulty
"You make an appointment? Sorry. I'm booked up all afternoon." -homeless guy to Scott
"Do you believe Satan walks the earth in a fleshly form?" -Nathan Levi Boston (the real one)
"They look kinda hot with their clothes on." -Landsman, changing up his reading materials
After reading arguments for both sides and taking a realistic look at my schedule, I've decided to keep the posting arrangement as is: open thread for the On Demand episode on Monday morning, full episode review on Sunday night. I may not be able to resist coming in early with a review of the finale, but until then, status quo. So talk about this episode and the ones before it here; if I see any spoilers for the On Demand episode or down the road, they'll be deleted.
What did everybody else think?
Labels: The Wire, The Wire Season 5
It's worth pointing out that Herc & Carver got away with their Fuzzy Dunlop scam in season 2. It can be done.
10:08 PM, February 03, 2008
I may not be able to resist coming in early with a review of the finale, but until then, status quo.
Hate to break it to you, but I remember reading that HBO won't be making the finale available early On Demand. I guess they want the series finale to have a more traditional viewing experience.
which made the later surveillance scenes with Omar and Donnie extra-chilling. I just wanted to yell at the TV and tell him to peel on out of there already.
Same here. I kept expecting Chris & co. to ambush them in the car.
I read the scene at the paper the same way you did: Jimmy knows he's got an unwitting ally, and Scott knows nothing. If Scott had half a brain, he'd have caught on to Jimmy's scheme and made it work for him; luckily for Jimmy, Scott's a freakin' dufus. I hope Scott's unmasked as the fraud he is (conversely, I hope Jimmy and Lester get away with their effed-up scheme).
What the hell is going on with that phone? Computer line? Fax? Mutant code? Is it next week yet? ACK!
I thought the moment when Gus didn't recognize Prop Joe's name was a quiet tribute to Joe. His involvement in "the game" finally killed him, but Joe was tremendously successful and lasted for a long time in part because he avoided attracting too much attention. Keeping his career as low-key and "boring" as possible served Joe well--I haven't seen any other 60-year-old drug dealers on The Wire . . .
It seems obvious to me that Scott knows McNulty is lying. Scott says that the killer mentioned 12 victims, and then McNulty then says that the other call from the killer also mentioned 12. So, from Scott's point of view, he either got extremely lucky in guessing the 12 victims part, or McNulty just lied about the other call.
12:17 AM, February 04, 2008
Alan do you think you would be getting the final 3 episodes before they air?
I don't believe the issue is whether or not Scott has enough clues to realize that McNulty is lying (he certainly does), but I still think he doesn't get it. McNulty and Scott have completely different looks on their faces in the scene; McNulty is giddy as a schoolgirl at the fact that he doesn't have to fabricate another piece of the puzzle (someone else has done it for him), while Scott is just bewildered at what just happened.
Withnail said...
But what about in episode six when McNulty xxxxxxxxx with the xxxxxxx guy and then Marlos says xxxxxxxxxx to xxxxxxxx -
to avoid more posts like these, put up the On Demand Thread. It's been 1 hr and 16 minutes since it came on line.
Scott has no clue. But of course he thinks the police are telling him the truth. Now McNutty knows he can steer this wherever he wants.
It seems to me Mike did not come into cutty's Jim because while he respects him he is still wary of male adults
alex is right Joe prided himself on being low key. I think this is a final shout out to Joe. reminds me of the story i think told by Bunny about a dealer who was real low profile.
Upon reviewing the scene it looks like Omar hit the tree. Still loving the connections and zingers they lace throughout the show. An example - Levy being thankful that Joe steered Marlo his way.
Landsman - "they look good with the clothes on"
Chris Littmann said...
One of those great moments in the details: Lester placing the fake call to verify that it was Marlo's cell.
"You all still offer that carry out! I want a pepper steak."
Also, is there any truth to what was posted -- that the finale won't be shown on OnDemand first? Link?
Yeah, it's confirmed.
http://www.hollywoodreporter.com/hr/content_display/news/e3i35ab2fc2fb51e6e2eff4f5d6d94e9bb1
The Dukie storyline in this episode is one we've seen in basically every season (maybe not season 3) where someone who isn't made for the game is caught up in it and doesn't know how to get out or doesn't realize that they should, i.e., Wallace, Ziggy, Johnny, Namond, etc. But we also saw an echo of Randy's and Wallce's storyline where they turn to someone for help who doesn't know how to or can't help him. In Cutty, though, I saw a kind of deep failure, possibly a moral one, in his acceptance that he doesn't know how to get from Baltimore to the rest of the world. Unlike Bodie (why in the hell would anyone ever want to leave Baltimore), Dukie genuinely wants out and not just because he can't cut it. It seems that Dukie is the only character left other than Bubbles who actively empathizes with those around him. Shouldn't Cutty in acknowledging that he doesn't know how to help Dukie do something beyond telling him that. He seems to have left Dukie in a state of fatal aporia, instead of going to any of the people he knows who could help Dukie find out what he needs to know. Off the top of my head that would include the Deacon, Grace and Bunny. On Dukie's part, it would be nice to see him reach out to Prez, but for a 15 year old that seems to be asking too much. And Prez, for his part, drove by Dukie on the corner. As a viewer, this type of storyline is the one that the Wire has used to the greatest emotional effect on me. The lows of Wallace and Randy compared to the high of Namond, the latter always with a guilty twinge because of the feeling that he didn't deserve it as much as Randy or Dukie. That twinge, strikes me as precisely the point from the show, wherein as a viewer you become willing to write off one kid, who has done nothing to deserve his fate, for the sake of another who is similarly innocent but put upon in a different way. Like the city that we rail against for its compromises in the way it uses it's limited resources, instead of trying to find another way. The show of course annihilates those that try to change the system (Stringer and Bunny)and seems to reward those who game the system from within, a lesson McNutty and Lester seem to have taken to heart, a lesson aphorised by Marlo--you want it to be one way, but it's the other way.
In Cutty, though, I saw a kind of deep failure, possibly a moral one, in his acceptance that he doesn't know how to get from Baltimore to the rest of the world.
Exactly. Cutty's only alternative is fighting, and Dukie's not a fighter. He did point out that Dukie is smart, but with Cutty's limited experience/viewpoint, he doesn't know how Dukie can use his brain to get out. I kept hoping Dukie would mention his love of computers and that it would, in turn, spark an idea between the of them, but no. This is where Prez could do the most good, if only Dukie would let him, since Prez recognizes Dukie's intelligence and could steer him toward a path that would lead out of Baltimore and the life. Unfortunately, as we've seen, that kind of happy ending is rare :(
Great post by "Ben" by the way.
I'm having a hard time with Marlo, who is SO careful with all of his dealings to the point of compulsion, is OK with his lawyer hiring a cop who had been pursuing him previously and still has a reason to hold a very serious grudge (a situation which played out in the most recent
episode).
It seemed more convenient plotwise than it did logical to the characters we know so well now.
Siddartha
In terms of Marlo I agree with your basic premise that it's out of character for him. He's always been portrayed as the one who takes no chances, even moreso than Omar. Marlo has had people killed for less than Levy hiring Herc. As a viewer I tend to give the writers of the Wire the benefit of the doubt, so I'm looking for what's different about this situation. The main things that occur to me are: (1) Marlo is completely lost in the world away from the street (e.g., the bank) and is relying too much on what he learned from Prop Joe because in this arena he lacks the confidence to make decisions, after all if he fires Levy, how will he find another lawyer; and (2) Marlo misunderstands the nature of attorney-client privilege and wrongly believes it to be much more of a shield than it is (something I've seen as a lawyer).
I wonder what Chris meant at the start, when he said to Marlow, "you made your move, now I am going to make mine" . . . .. and if Prop Joe had a will or Slim knows where all the clean money is?
Anonymous 11:39 AM -
I could be wrong but I think that Chris' move was to consult with Levy on getting his and Snoop's gun charges dropped.
Thanks for the response, Ben. That assuages me a bit. I can't help but feel that if this season were longer, Herc wouldn't be in the office when Marlo comes in and then Levy sends Herc on an errand that clues Herc into the fact that Marlo is Levy's client and he has his cell number.
I also think that Levy purposely put Marlo's # in his rolodex with Herc watching knowing that Herc would eventually steal it.
^I agree, siddhartha. Marlo getting brought up on charges is good for Levy's business, as he more or less said to Herc after Marlo left.
Chris' move is setting up the sting on Omar.
With the caveat that I haven't watched subsequent episodes yet and don't wish to hear spoilers, I have to come down on the side that McNulty has no idea Templeton is lying.
Why would he assume that a reporter, in front of his all of his bosses, would fabricate possible evidence in a serial killer investigation? Crank calls to newspapers about high-profile cases are extremely common. I read McNulty's expression as simply one of "hey, here's a crank call I can turn to my advantage."
If McNulty knew Templeton was lying, then he must also have known he'd be exposing himself as a liar by corroborating the fake call. McNulty is certainly no stranger to risky behavior, but this seems a stretch to me.
His grin, to me, represents the (mistaken) thought "here's the perfect opportunity," rather than any great insight into Templeton's motives. And how like The Wire to put a grin on a character's face as he commits the error that could seal his fate...
"I thought the moment when Gus didn't recognize Prop Joe's name was a quiet tribute to Joe."
Totally agree with this assessment. All Joe wanted to do is buy for a dollar and sell for two and leave all the B.S. for street rep and corners for the young-ins.
My favorite thing about the McNulty/Templeton grand fabrication is that both men are helping the other in such a hge way, but based on what we know about them, they will never actually be in league together. Neither one will ever accuse the other of making things up, because it would reveal his own lie. Plus, neither would admit their own lie to the other in order to work together. It is a fantastic bit of plot in every way.
I agree with sr--I don't think Jimmy knows that Templeton is lying. Isn't it common for wackos to call newspapers claiming to be killers? At least, it seems to happen all the time in procedural movies and television.
I also suspect that if Templeton isn't sharp enough to have figured out that Jimmy is lying (and who wouldn't be? His including the reference to '12 bodies' gives him away) he will, soon. In fact, Templeton could end up being Jimmy's undoing. Talk about a scoop--veteran homicide detective fabricates serial killer and desecrates dead bodies? Grade-A Pulitzer bait.
^Except Templeton the Rat would have to cover his own lies and I don't think he's smart enough to pin it all on Jimmy. And if anyone else meets the real Nathan Levi Boston (esp. Gus), Scott's cooked for sure. Poor boy just ain't that bright.
BTW, does anyone else think Gus has some culpability in Scott's various frauds? Scott's obviously hungry for better assignments, and Gus just as obviously continues to thwart him. Granted, Scott's not a great writer, and Gus suspects him of hinky behavior, but instead of giving him assignments where Gus could excersize greater control over Scott's sourcing, etc., he gives him the react quotes which cause Scott to chaff.
Also, Scott seems to come in with these great scoops out of thin air--why isn't Gus more suspicious? Is he so dissatisfied with how things are being handled as the Sun is down-sized that he doesn't care anymore?
Dez,
What would be the alternative? Take a mediocre writer who has made no attempt to hide the fact that he views the paper as a stepping stone, and give him plum assignments and higher profile stories, while better, more experienced (or at least more enthusiastic like Alma) reporters take a back seat? Gus' job is not to coach the Mighty Ducks to Pee-Wee Hockey supremacy and more self0esteem. His job is to put out the best possible paper. A newsroom is a meritocracy, and Templeton is finding ways around that. Gus knows it but can't prove it as long as the higher-ups dig the crap he slings. The solution is not to give him a free pass to the Washington Post. Templeton's crimes are no one elses fault but his own. You can blame Gus about as much as you can blame Daniels for what McNulty is doing.
Gus has suspicions, but doesn't follow up on them. How easy would it have been for him to call Neresse himself when Scott came up with that quote about Daniels and attributed it to her? We know Gus is conscientious enough to call the paper in the middle of the night to make sure he got his numbers right in a story; why not take a few minutes to check with Neresse? At the very least, he can assign him to stories where he can cause a lot less damage than he is now.
Gus certainly could have confirmed with Nareese if he had wanted to, except Templeton said that had gotten the quote from her as an anonymous source and only gave Gus her name when he implied he wasn't comfortable with it.
As to his other stories, Gus tried to find info about "EJ" and said he had enough reservations to hold the story, but the higher-ups pushed it through. He said he thought the call from the killer was BS, but McNulty seemed to confirm it, and I don't know how he could go about confirming a quote from a homeless man...
Truth is, the only reasonable solution for what Gus should do with Templeton is fire him, but the higher-ups dig him, so Gus's hands are tied. Gus can't send him to Obituaries or some unimportant bureau for the same reason. The bosses like his stuff, so he stays.
And as far as the Baltimore Nibbler goes, Templeton wrote himself into the story. Gus has no choice on that now.
Basically, Gus is in a no-win situation with Scotty Boy. He can't fire him, can't bury him, won't promote him, and doesn't have the time or man-power to check up on everything he does. 15 years ago, during the paper's heyday, it would be different, but right now, Scott has found a loophole to exploit at the perfect time to exploit it. Gus is a victim of the same broken system that Scott is taking advantage of. He is an embodiment of "Doing more with less."
I haven't seen my two favorite lines from this episode mentioned anywhere, so here goes (yes, both Herc lines...what are the odds?!?):
"Don't ask, don't tell, it's like one of those guys sucking dick in the army".
"And Carv, when you put the bracelets on that bitch, remind him again of my fuckin' camera."
As for Clay Davis' extra long "sheeeeeet". He also had a long one ("sheeeet", that is) at the start of the fourth season when Sydnor served him with papers. I would love to know which one was longer....
That's it for this strangely profane post (apologies, Alan).
"In Cutty, though, I saw a kind of deep failure, possibly a moral one, in his acceptance that he doesn't know how to get from Baltimore to the rest of the world."
I disagree with this assessment. How many of us are willing to really look at our limitations and weaknesses and simply admit them as honestly as Cutty did. And he's done it twice. The first time to Avon who he served 14 years in prison for and who was in the midst of a street war and who was counting on him to step up and be a soldier. But like he said "it ain't in me no more". He couldn't kill anyone not even someone who put a gun in his face not because he didn't have heart but because he did.
And then the second time to a lost confused boy, 15 to 20 years his junior, he admitted that he's not smart enough or clever enough to help him. He hasn't seen or known enough of life of outside his own limited experience in it to know what to do to help Dukie. To a grown man in his late 30s that's got to be as sobering an admission as one could possibly make. But he not only did he have the wherewithal to make it but he told Dukie as plainspokenly and with as much warmth as he could I wish I could help you son but I just don't know how to. I think that speaks to how honest Cutty is with himself and to what kind of man he is than any failing he has.
And let's not forget the season isn't over and we haven't seen what Cutty will do. We've already seen with Justin and Michael that he doesn't give up on anyone easily so Grace, Prez, the Deacon, and Bunny none of whom we've seen yet may yet be paid a visit by Mr. Wise on Duquan's behalf.
Naresse is seriously gangster
Good to see Royce-- he and Davis both wearing their shit-eating grins.
Interesting to hear to legitimate people talk about "standing tall" just like the street folks.
Two moments in the show I love
Gus saying that a while ago he thought it was all bullshit.
Donnie briefly tuning to the radio show clay davis was on while sitting with omar.
Everyone talks about if we should give david simon and the writers the benefit of the doubt and follow it through to the end because the show was so great for four seasons. I think that makes no sense. The show still contains more truth than any other on television. It still provides inspiration and anger, and the occasional moment of beauty.
I just posted this @ Tim goodmans and thought I would also add it here for your thoughts - I hope this is not considered a spoiler cause I have no advanced info. no Ondemand (just directv here) but I think the answer to what the phone from Vondas does is in the opening credits. There is a samsung smart phone with a clock on it. I wonder If it syncs time with others or just sends a txt to others with the time?
How I fear that McNulty becomes the Homicide's Bayliss and that Omar become Oz's Siad (spelling)..but am I ever enjoying the ride! Kudos to HBO for NOT showing the Finale On Demand!!!
Okay, Andrew, you've convinced me :-)
It's probably too late for anyone to read this, but I need to get it off my chest. This season of The Wire isn't working for me. I still love it, but something is off. I think it's because season four ended with a cliffhanger that season five promised to pay off, but it's doing so with only one foot in reality. It's hard to take this seriously as a critique of the Baltimore Police Department when, one hopes, this isn't how people are working around the lack of funding. For whatever reason, Hamsterdam rang true for me in a way that this doesn't.
But none of that bothered me all that much until Omar jumped off the building. I don't want Omar to be a superhero. It cheapens everything that comes later. He should've died in that room and if the writers didn't want him to, they shouldn't have put him there. I want him to concoct a brilliant plan (in the five hours that are left) and bring his version of justice to Marlo, et al, but if we're throwing all of the "gritty realism" out the window, then it ends up being a different show - a much more enjoyable Godfather part Three. Great in its own right, but not a fitting end to the greatest show ever created.
So I'm going to try to enjoy what's left, and maybe I'll look back on this episode as the one I got too worked up over, and maybe they'll finish it spot on, but if they start throwing out reality, then it won't hurt as much when/if Omar, Dukie, Michael, Lester and all of the other great characters don't get their happy endings. And I'm okay with the unhappy endings, as long as a larger point is still being made.
I have to confess, I'm still not sold on this season of the Wire. Here is one example:
OK, Omar has always been bigger than life, but not in a completely unbelievable way. When he attacks, he knows the odds, and plays them to his advantage: either with cunning, intimidation or numbers (usually with a mix of all three). So how could he be so stupid as to perform essentially a frontal attack, with a handgun (as opposed to a more-efficient shotgun for close range), knowingly out manned, against an opponent who is ready for him, in a place of the opponent's choosing, slightly after the opponent "went to bed" ...oh, and he forgets to put one in the chamber before he subtlety kicks in the front door.
I served in Iraq and know a little bit about urban combat. In a situation where we had that much intelligence, we would come in with at least two squads (one through the front door and one through the back/balcony). It would be sightly before 4 in the morning. We would all be in full body armor with shotguns and automatic rifles. The power would be cut 1-2 minutes before entry and we would all have night vision. I'm not saying Omar has access to all the modern warfare equipment of the US Marines, but some cunning or precaution would have been preferred.
Is this an indication of a breakdown of Omar's previously-proved acumen, or of writer laziness? I would have been more impressed if Omar had attempted something smarter and still been foiled (perhaps by a "one-in-the-chamber lapse or something more elaborate).
Andrew - Spot on analysis of Templeton there.
johnever--
My point wasn't about Cutty being dishonest. I agree that he's honest and does understand his limitations. It's precisely because he understands those limitations that I have a problem with him simply saying he doesn't know the answer. He knows people who could help Dukie or send him on to someone else. Cutty has set himself up as a teacher and as an alternative to the game. Just like the Deacon told Colvin that he had to do more than provide a free zone, that he had to provide clean needles, etc., I think Cutty should extend himself a bit more. Perhaps I'm asking too much of him, and as you note the seasons not over and I don't know what will happen.
Following up on the allusions to movies in Alan's post, twice in ep 55 it was said to Dukie, "It's not like in the movies." The first was from Cutty in the boxing ring; the second, from Michael, was at the creek "firing range." Both times, each character was trying to explain to Dukie that just being able to effectively hit someone or carry/know-how-to-use a weapon wasn't enough--there were the consequences, the aftermath. The "street" would just keep coming at him, challenging him. No one-two punch followed by the "Rocky" theme, no sexy shoot-out settling matters of respect once and for all.
Does anyone know why Oscar calls Freamon "Socks?"
Is this an indication of a breakdown of Omar's previously-proved acumen, or of writer laziness?
I think it's two things: 1) Omar's judgment is clouded by his emotions for Butchie; 2) he doesn't fully know or understand Marlo's ruthlessness. Actually, three things because he doesn't have the same amount of "staff" he had in the past. He's outmanned, outgunned, and outmaneuvered. Almost like he's on "Survivor: Baltimore"!
Was anyone surprised that Lester called Marlo with his own cell phone? What if Marlo got suspicious and called back? If voicemail had picked up, it would say it was Lester Freamon (maybe even Detective Freamon)- at the very least, it wouldn't be the same phony accent that he put on.
lungfish, I assumed Freamon's phone blocks his number on outgoing calls. I have friends whose cell numbers show up as "Restricted" or "Unknown" when they call, so I don't think Marlo being able to call back would be a concern.
Re: marlo calling back. How many people call back a "wrong number" I sure as heck dont.
Mcnulty and Scott. I think mcnulty knows that dude is making it up not only because of the sly grin and change in posture but also because at the first meeting Scott asks for more details to spice up a mass murderer with a sexual connection. He sees that this is a writer with a penchant for making a story 'sexy'
On omar. If he honestly had been staking that place out for days, weeks even and we assume he has done this before based on his past why would he not believe he had Monk outgunned. Also, the assumption is that when he kicks inthe door he is coming to kill when the reality is probably that he is coming to get info on who killed butchie, THEN kill. No point in straight up killing monk if he doesnt get all info he needs. Of course emotion has clouded his judgement somewhat but I also dont think he has any idea who marlo is. He doesnt know he is up against the same dude he told at the card game "you must mistake me for a man who repeats himself"
To Anonymous
Omar knows exactly who Marlo is, he found that out last season when he was staking out the Co-Op and stole the shipment. Also, Omar knows who had Butchie killed because remember when they were torturing Butchie and after they killed him, they left the Big Dude alive to tell the story, "Chris telling the Big Dude to tell Omar everything that happened". The only info Omar needs is how to find Marlo which at this point his plan "was" to off as many of Marlo's people until he showed himself.
I can't simply go with the theory that Omar's emotions are clouded, because his emotions were clouded by Brandon's killing in the first season, and while his attack there wasn't entirely successful, it wasn't the unmitigated disaster of this one.
Also,Omar's been studying Marlo intently for two seasons. He knows exactly who and what Marlo is...I think...I need to review all those stake out sessions to understand if he understands Snoop and Chris.
I'm going to chalk this up to Simon and crew trying to bring Omar down a notch into the realm of realism while they are raising the rest of the characters (McNulty, etc.) into farse-like fictional archetypes.
The only character consistently mired in the real world of Baltimore seems to be Bubs...so maybe they are trying to distance themselves from all of the ups and downs of cops and crack-slingers, and get back to the message of "The Corner"; drugs are a disease that destroys the lives of everyone around them and all that is constant is the addict and the pain.
Maybe I need to change my mind about this season. Maybe Simon and co. are doing something even more grand than I ever imagined by building up characters over 4 seasons and spending the final season completely deconstructing (and destroying) them. If they can pull this together into something cohesive and meaningful, this might be a perfect ending to a near perfect show.
I thought Dukie's question to Cutty--"But how do I get from here to the real world?"--was one of the most touching lines of the series, and the question that every public official should be addressing when facing the problems the series depicts. As someone above noted, Dukie is merely the latest in a series of characters to ask that question in one way or another, or to want to pursue the goal of getting from the street to the "real world," but no one has ever presented them with the tools to do so. It's heartbreaking.
Nerese's speech to Clay was perfect and true--all I could think of was Marion Barry. Besides thinking of Brianna's speech to D'Angelo (I believe the phrasing--"Now you got to carry the weight"--was identical), of course. It also underscored just how complicit every damn person is in the system. Right down to that DJ who interviewed Clay.
On Jimmy and Scott--I think Scott known Jimmy is lying, but can't figure out why. Jummy suspects that Scott is lying, or too stupid to recognize a crank caller, but doesn't care why; he's just happy to have another piece he can use in building the Big Lie.
9:02 AM, September 08, 2009
Michael and Dukie seem more like D'Angelo and Wallace to me. Both are essentially decent guys, and the one who is deeper in the game is trying to keep the other one out of it.
Andy Hutchins said...
Nice musical touches: "Everybody Wants to Rule the World" playing in the bar as McNulty meets with Gutierrez and Templeton; "Just My Imagination" playing just after Templeton and McNulty/Freamon come up with their newest tall tales. What "My Girl Has Gone" means (it's played just after the McNulty-Templeton Bullshit-Off) is anyone's guess; same with "Gypsy Woman."
And when Beadie goes to Bunk, he says he won't snitch, and can't "speak out on that." Seemed very street to me.
Alex K said...
A nice shout-out to the book "Homicide": Jay telling Holley, who's on midnight shift, that Detective Worden will be coming in to relieve him.
I'm obviously way late to the Wire, but have thoroughly enjoyed Alan's work retroactively. Just watched this ep on HBOGO (who knew when this aired that that would be a thing?) and I've got to vehemently disagree, in the strongest possible terms, that this is playing as a Dr. Strangelove-ian farce. It's incredibly disappointing to see two characters veering so wildly put of character, Lester in particular. The show had already done a fantastic job of illustrating the dysfunction of the many institutions involved. Playing it like this, unfortunately, just feels like a betrayal of the first four seasons.
Brian Dunbar said...
So how could he be so stupid as to perform essentially a frontal attack, with a handgun
Several reasons
You had a lot of institutional knowledge informing your tactics. Practice at a MOUT facility, trainers, after-action reports, veterans from previous deployments fine-tuning tactics .. all devoted to informing a platoon sergeant of the best way to take a door.
Omar has himself. And what he's been doing has worked because, for the first time, his opposition was laying for him _and_ they had actual training.
Heck - I bet Chis has FM 90-10 on his bookshelf at home.
In Treatment week five thread
Strike Survival TV Club: Cupid, "Botched Makeover"
Lost: The time-traveler's girlfriend
Scrubs lives... on ABC?
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Sepinwall on TV: 'Unhitched,' 'Dirt' reviews
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American Idol: Men's semis, week two
Strike Survival TV Club: Cupid, "The Children's Hour"
Mad Men redux: Jews cruise
Okay, now this is getting out of hand
All TV: 'quarterlife' review
Oscar post-mortem
The Wire week 9 thread for the On Demand'ers
Breaking Bad episode 5 open thread
The Wire, "Clarifications": A kid's game
Academy Awards open thread
Sepinwall on TV: 'A Raisin in the Sun' review
Lost: Gone baby gone
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In Treatment, week four open thread
Mad Men redux: I don't want to be a Dick
Alan Sepinwall's week off
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In Treatment, week three: Paul and Gina
Double threats?
Lost: Sayid agent man
Strike Survival TV Club: Cupid, "Bachelorette Party"
Sepinwall on TV: 'Dexter' on CBS review
Survivor Micronesia: Love is all around
In Treatment, week three: Jake & Amy
8,000 degrees of John Munch?
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NBC firms up some plans
American Idol: Send in the pros from Dover
CBS announces which shows are coming back and when
In Treatment, week three: Alex
Terminator, "Queen's Gambit": All in the family
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Strike Survival TV Club: Cupid, "Grand Delusions"
Sepinwall on TV: 'Jericho' season two review
Early ABC pick-ups for next season, plus other str...
Mad Men redux: Mo' old money, mo' problems
Writers strike almost over?
Breaking Bad open thread
The Wire, "The Dickensian Aspect": Those left behind
Torchwood, "To The Last Man": No time for love
Friday Night Lights, "May the Best Man Win": Give ...
In Treatment, week two: Paul and Gina
Survivor Micronesia: No, they're not as stupid as ...
Strike Survival TV Club: Cupid, "A Great Personality"
Lost, "Confirmed Dead": Who you gonna call? Miles!
In Treatment, week two: Jake & Amy
All TV: 'Lipstick Jungle' review, 'Survivor' preview
In Treatment, week two: Sophie
House, "Don't Ever Change": Amber is slammin'
In Treatment, week two: Alex
What's in a nickname?
Strike Survival TV Club: Cupid, "Hung Jury"
In Treatment, week two: Laura
Mad Men redux: Slappy birthday
House, "Frozen" open thread
LET'S GO GIANTS!!!!!
The Wire, "React Quotes": You will believe a man c...
Sepinwall on TV: Big Blue fans give Aikman, Buck a...
Torchwood, "Sleeper": A farewell to arms
Friday Night Lights, "Leave No One Behind": KHAAAA...
In Treatment, week one: Paul & Gina
Matt Damon: Good sport, or greatest sport?
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