\documentstyle[preprint,epsf,aps]{revtex}
\begin{document}
\preprint{EHU-FT/9803}
\draft
\title{The speed of cool soft pions}
\author{J. M. Mart\'\i nez Resco \thanks{\tt wtbmarej@lg.ehu.es}, \\ 
        M. A. Valle Basagoiti \thanks{\tt wtpvabam@lg.ehu.es}}

\address{Departamento de F\'\i sica Te\'orica, \\ 
 Universidad del Pa\'\i s Vasco, Apartado 644, E-48080 Bilbao, Spain}
 \date{\today}
\maketitle
\date{today}
\begin{abstract}
The speed of cool pions in the chiral limit is analytically computed at 
low temperature within the imaginary time formalism to two loop order. 
This evaluation shows a logarithmic dependence in the temperature where 
the scale within the logarithm is very large compared to the pion decay 
constant.
\end{abstract}   


\pacs{11.10.Wx, 12.39.Fe, 14.40.Aq} 


%\newpage

The dynamics of pions in a thermal bath has been studied intensively in the 
last few years \cite{GyL,go,schenk,kap,Pi,toublan} 
because its absorptive and 
dispersive properties can determine the termalization processes ocurring in
the aftermath of high-energy nuclear collisions. 

In the limit of exact chiral symmetry, pions are true Goldstone bosons and as
such, they travel at speed of light in the vacuum. At non zero temperature, 
however, relativistic invariance is lost, the thermal bath providing a 
priviliged rest frame and pions travel at less than the speed of light. 
Accordingly, the pion dispersion relation as a function of momentum must be 
modified. In chiral perturbation theory to leading order in low temperature, 
this modification first shows up at two loops, $\sim T^4/F_{\pi}^4$, but in a 
linear $\sigma$ model the effect already appears at one loop order, 
$\sim T^4/F_{\pi}^2 \, m_{\sigma}^2$ \cite{Pi}.

Only very recently \cite{toublan}, the speed of thermal pions has been 
explicitly computed. This has been obtained from the 
axial current two-point correlator in the real time formalism within 
chiral perturbation theory. 
There, some integrals had to be numerically computed due to its complexity. 
The final result shows a logarithmic dependence in the temperature,
\begin{equation}
v^{2}=1-\frac{T^{4}}{27\,F_{\pi}^{4}}\log\left(\frac{\Lambda}{T}\right) \,,    
\end{equation}
where the scale obtained is $\Lambda \approx 1.8$ GeV.

In this letter, we follow  a rather different approach to 
the evaluation of the speed
of the pion. We will work in the imaginary time formalism and use the background
field method to compute the required piece of the effective action. As we shall 
see below, only diagrams having external zero momentum and energy are required. 
Thus, we take advantage of the methods developed by Arnold an Zhai \cite{az} to 
analytically compute  integrals associated with higher-loop vacuum bubbles 
of the partition function in gauge theories. We find a result which is in 
agreement with the one previously given by Toublan \cite{toublan} except for the
scale within the logarithm. Here we find $\Lambda \approx 4.3$ GeV. 
In spite of this difference, the behavior of the pion speed is not greatly 
modified in the range of temperature where this calculation can be trusted, 
$T < F_{\pi}$.

In the limit of exact chiral symmetry $SU(2)_{L} \times SU(2)_{R}$ which is 
spontaneously broken to $SU(2)_{V}$,  the pion interaction at 
small energies is known from the effective chiral lagrangian  
with an increasing number of derivatives corresponding to an expansion 
in powers of momentum \cite{dsm},   
\begin{eqnarray}
{\cal L}_{eff}&=&\frac{F_{\pi}^{2}}{4} 
  {\mbox{Tr}}[\partial_{\mu}U \partial^{\mu}U^{\dagger}] \nonumber \\ 
   &&+L_{1}({\mbox{Tr}}[\partial_{\mu}U \partial^{\mu}U^{\dagger}])^{2}+
  L_{2} {\mbox{Tr}}[\partial_{\mu}U \partial_{\nu}U^{\dagger}]{\mbox{Tr}}
   [\partial^{\mu}U \partial^{\nu}U^{\dagger}]  +
    {\cal L}_{6}+\ldots 
\end{eqnarray}
Fortunately, the explicit form of ${\cal L}_{6}$, giving the tree level 
counterterms to be used in a two loop computation, is not required because 
its contribution at tree level is independent of the temperature. 
A very efficient technique which organizes the graphs to be computed is provided
by background field perturbation theory. Here, the $U$ field is parametrized by 
$U=\xi(x) h(x) \xi(x)$ where $\xi(x) = \exp (i \pi_{cl}/2 F_{\pi})$ is a 
classical backgroung field and $h(x) = \exp({i\tilde{\pi}}/F_{\pi})$ is a
fluctuating quantum field. When the lagrangian is written  in terms of 
$\xi$ and $h$ it becomes ${\cal L}({\tilde{\pi}}, A, V)$. This lagrangian 
has an exact local chiral $SU(2)$ invariance implemented by the connection 
$V_{\mu}= (\xi^{\dagger}\partial_{\mu}\xi + \xi\partial_{\mu}\xi^{\dagger})/2$.
Such a field is not propagating and the remaining 
fields ${\tilde{\pi}}, 
A_{\mu}= (\xi^{\dagger}\partial_{\mu}\xi - \xi\partial_{\mu}\xi^{\dagger})/2$, 
transform covariantly under $SU(2)$ transformations. 
Thus, the covariant derivative
on quantum fluctuations is 
\begin{equation}
 D_{\mu}{\tilde{\pi}} = \partial_{\mu}{\tilde{\pi}} + [V_{\mu},{\tilde{\pi}}]\,.  
\end{equation}

It is our aim here to compute the effective action $\Gamma[\pi_{cl}]$ in order
to obtain the pion inverse propagator from two functional derivations.  
This can be computed from the following equality \cite{abb} 
\begin{equation}
\Gamma[\pi_{cl}]= \left.\tilde{\Gamma}[\tilde{\pi}_{cl}; A[\pi_{cl}], 
V[\pi_{cl}] ]\right|_{\tilde{\pi}_{cl}=0} \,,
\end{equation}
where $\tilde{\Gamma}[\tilde{\pi}_{cl}; A[\pi_{cl}], V[\pi_{cl}] ]$ is calculated restricting ourselves to diagrams with no external 
$\tilde{\pi}_{cl}$ lines. These diagrams are the one-particle-irreducible 
subset with only internal $\tilde{\pi}$ lines and external 
$V_{\mu}, A_{\mu}$ lines. Since the required piece of the effective action 
is quadratic in $\pi_{cl}$, and the expansion of background fieds is 
\begin{equation}
A_{\mu}=\frac{i}{2 F_{\pi}}\partial_{\mu}\pi_{cl} + {\cal O}(\pi_{cl}^3)\,, 
\;\;\;\;\;
V_{\mu}=\frac{1}{8 F_{\pi}^{2}}[\pi_{cl},\partial_{\mu}\pi_{cl}] + 
  {\cal O}(\pi_{cl}^4) \,, 
\end{equation}
we only need to consider the first few terms of the expansion
\begin{eqnarray}
%\Gamma[\pi_{cl}]
\tilde{\Gamma}[0; A[\pi_{cl}],  V[\pi_{cl}]]
&=&
\int d^{4}x V_{\mu}^{a}(x)\langle V_{\mu}^{a}(x)\rangle _{1PI}\nonumber
\\&&+
\frac{1}{2}\int\!\!
\int d^{4}x\,d^{4}y  \ A_{\mu}^{a}(x)
\langle A_{\mu}^{a}(x) A_{\nu}^{b}(y)\rangle_{1PI}A_{\nu}^{b}(y) + \ldots
\end{eqnarray}   
 
Expanding $h$ up to order $1/F_{\pi}^{4}$ we get for 
${\cal L}_{2}$
\begin{eqnarray}
{\cal L}_{2}(\tilde{\pi}; A, V)&=& 
 -F_{\pi}^{2}{\mbox{Tr}}[A_{\mu}A^{\mu}]+\frac{1}{4}{\mbox{Tr}}[D_{\mu}\tilde{\pi}D^{\mu}\tilde{\pi}]-
\frac{1}{4}{\mbox{Tr}}[[A_{\mu},\tilde{\pi}]^{2}] \nonumber \\
&&-i F_{\pi}\, {\mbox{Tr}}[\partial_{\mu}\tilde{\pi} A_{\mu}]
  +i F_{\pi}{\mbox{Tr}}[\tilde{\pi}[V,A]] +
  \frac{2 i}{3 F_{\pi}}{\mbox{Tr}}[(\vec{\tilde{\pi}}\vec{\tilde{\pi}}
  \partial_{\mu}\tilde{\pi}-
   \vec{\tilde{\pi}}\partial_{\mu}\vec{\tilde{\pi}}\tilde{\pi}) A_{\mu}] \nonumber \\
&&-\frac{1}{6 F_{\pi}^{2}}{\mbox{Tr}}[(\vec{\tilde{\pi}}\vec{\tilde{\pi}})
    (V-A)\tilde{\pi}(V+A)\tilde{\pi}] + 
    \frac{1}{6 F_{\pi}^{2}}{\mbox{Tr}}[(\vec{\tilde{\pi}}
    \vec{\tilde{\pi}})^{2}(V-A)(V+A)] \nonumber \\
&& -\frac{5i}{12 F_{\pi}}{\mbox{Tr}}\left[
(\vec{\tilde{\pi}}\vec{\tilde{\pi}})[V,A]\pi \right] +
\frac{1}{6 F_{\pi}^{2}}{\mbox{Tr}}\left[ 
(\vec{\tilde{\pi}}\vec{\tilde{\pi}})[\pi,\partial_{\mu}\pi]V^{\mu}\right] \,.
\end{eqnarray}
For ${\cal L}_{4}$ we write only the terms that will contribute 
at tree and one loop order,
\begin{eqnarray}
{\cal L}_{4}
 &=&
L_{1}\left[-\frac{8}{ F_{\pi}^{2}}{\mbox{Tr}}[A_{\mu}A^{\mu}]{\mbox{Tr}}[\partial_{\mu}\tilde{\pi}\partial^{\mu}
\tilde{\pi}]-\frac{16}{F_{\pi}^{2}}{\mbox{Tr}}[A_{\mu}\partial^{\mu}\tilde{\pi}]
{\mbox{Tr}}[A_{\nu}\partial^{\nu}\tilde{\pi}]\right]\nonumber \\
&&+
L_{2} \left[ -\frac{8}{F_{\pi}^{2}}{\mbox{Tr}}[A_{\mu}A_{\nu}]{\mbox{Tr}}[\partial^{\mu}\tilde{\pi}\partial^{\nu}
\tilde{\pi}]\right. \nonumber \\   
&&- \left. \frac{4}{F_{\pi}^{2}}{\mbox{Tr}}[A_{\mu}\partial_{\nu}\tilde{\pi}+A_{\nu}\partial_{\mu}
\tilde{\pi}]{\mbox{Tr}}[A^{\mu}\partial^{\nu}\tilde{\pi}+
A^{\nu}\partial^{\mu}\tilde{\pi}] \  \right]
 +\ldots
\end{eqnarray}
The diagrams we have to compute from \ ${\cal L}_{2}$ \ are in Fig.1. 
The first gives a tree level contribution 
$\langle A_{\mu}^{a}A_{\nu}^{b}\rangle
=-4 F_{\pi}^{2}\delta^{ab}\delta_{\mu\nu}$. 
The fourth, fifth and the sixth diagrams  vanish 
trivially due to the form of the vertices involved and we 
have $\langle V_{\mu}^{a}\rangle_{1PI}=0$ up to two loops. 
The second, third and the seventh diagrams are necessarily proportional to 
$\delta^{\mu\nu}$. In consequence they give a contribution proportional 
to $P^{\mu}P^{\nu}\delta_{\mu\nu}$ which vanishes on shell. 
Then, only the last diagram at zero momentum remains to be computed.  
With the short-hand notation 
\begin{equation}
{\mbox{Tr}}_{K} \longrightarrow \mu^{2\epsilon}\!{\displaystyle\sum_{k_{0}=2\pi nT}}
\int \frac{d^{3-2\epsilon}k}{(2\pi)^{3-2\epsilon}}\,, 
\end{equation}
it reads
\begin{eqnarray}
\label{ab} 
P^{\mu} &\langle& A_{\mu}^{a}A_{\nu}^{b}\rangle P^{\nu}=\nonumber \\
& & -\frac{1}{3!} \, \frac{64}{ F_{\pi}^{2}}\delta^{ab}\, 
{\mbox{Tr}}_{Q}{\mbox{Tr}}_{K}\,[(\, (PK)^{2}+(PQ)^{2}+PK\,PQ )\,
\Delta(Q)\Delta(K)\Delta(K+Q) ]\,,
\end{eqnarray}
where we have approximated 
the euclidean thermal propagator 
$\Delta(K+Q-P)\simeq\Delta(K+Q)$ 
because the  required quadratic dependence in $P$ (see Eq.~(\ref{re}) below) 
is already present in  
the derivatives of 
$A[\pi_{cl}]$. Hence, it suffices to evaluate the sunset graph 
at zero external momentum.  Some details for this calculation are given in 
the Appendix. The end result for the sunset contribution takes the form at 
$P^2=0$,  
\begin{eqnarray}
\label{sunset}
\frac{1}{4 F_{\pi}^{2}} P^{\mu} &\langle& A_{\mu}^{a} A_{\nu}^{b}\rangle P^{\mu} 
=  \nonumber \\
& & \frac{1}{F_{\pi}^{4}}\delta^{ab} p^{2} 
\left( -\frac{T^{4}}{54}\, \frac{1}{\epsilon} + 
{\frac{\large{\gamma}_{E}\,{T^4}}{81}}-\frac{2\,T^{4}}{27}\log(\frac{\mu}{T}) 
 -\frac{46\,{T^4}}{1215}+\frac{T^{4}}{81}\log(16\pi) \right. \nonumber \\
& & + \left. \frac{80\,T^4}{27}\,\zeta'(-3) + 
 \frac{8\,{T^4}}{9}\,\zeta'(-1) + \frac{20\,{T^4}}{9\,\pi^{4}}\,\zeta'(4) +
{\cal O}(\epsilon) \right) \,,
\end{eqnarray}
where $p$ denotes the three momentum.  
The only contribution from ${\cal L}_{4}$ is (see Fig.2),
\begin{eqnarray}
& &\frac{1}{4 F_{\pi}^{2}} 
P^{\mu} \langle A_{\mu}^{a}A_{\nu}^{b}\rangle P^{\nu}=
-\frac{1}{4 F_{\pi}^{2}}\,
 \frac{1}{2!}\frac{256\,\delta^{ab}}{F_{\pi}^{2} }(L_{1}+2 L_{2}) \ 
{ \mbox Tr }_{Q}(PQ)^{2}\Delta(Q) \nonumber \\ 
& & = -\frac{64 \pi^{2}}{45}\ \frac{p^{2}\, T^{4}}{F_{\pi}^{4}}(L_{1}+2 L_{2})
\left[ 1+ \epsilon\,\left( 240\,\zeta'(-3)-\gamma_{E}+
\log(\frac{\mu^{2}}{4\pi T^{2}})
+\frac{13}{6}\right) + {\cal O}(\epsilon^2)  \right] \,.
\end{eqnarray}

The renormalization at finite temperature is the same as for $T=0$. 
This means that the  use of the renormalization prescription \cite{dsm}
\begin{equation}
L_{i}=L_{i}^{r}(\mu) - 
\,\frac{\gamma_{i}}{32 \pi^{2}}\left(\frac{1}{\epsilon}+\log(4 \pi)+
1-\Huge{\gamma}_{E}\right)\,,
\end{equation}
with $\gamma_{1}=1/12$ and $\gamma_{2}=1/6$ \ for $SU(2)$ must cancel the 
pole terms.  
Indeed, putting together the
contributions coming from \ ${\cal L}_{2}$ and ${\cal L}_{4}$ the poles cancel 
as they should.  

Now we can easily obtain the inverse propagator
\begin{equation}
\Delta_{ab}^{-1}(x,y)= \left.
\frac{\delta^{2}\Gamma[\pi_{cl}]}{\delta\pi_{cl}^{a}(x)\delta
\pi_{cl}^{b}(y)} \right|_{\pi_{cl} = 0}  
 =-\frac{1}{4 F_{\pi}^{2}}\partial_{x}^{\mu}\partial_{y}^{\nu}
\langle A_{\mu}^{a}(x)A_{\nu}^{b}(y)\rangle_{1PI} \,. 
\end{equation}
In momentum space $\Delta_{ab}^{-1}(P)=
\delta^{ab} \left( \Delta^{-1}_{0}(P)+\Sigma(P) \right)$. Thus,  
we can read the dispersion relation from the retarded proper 
self-energy $\Sigma^{\rm{ret}}(\omega,p)$ which can be 
obtained from the euclidean 
counterpart by analytic continuation as  $\Sigma^{\rm{ret}}(\omega,p) = 
\Sigma( p_{0}= -i \omega + 0^{+}, p)$. If the retarded inverse propagator 
has a zero located at $\omega= \omega(p) - i \gamma(p)$ with 
$\gamma(p) > 0$ we get
\begin{eqnarray}
\label{re}
v^{2}&=&1 + \frac{ {\rm{Re}}\Sigma^{\rm{ret}}(\omega = p)}{p^{2}}\,, \\ 
\gamma(p) &=& -\frac{ {\rm{Im}}\Sigma^{\rm{ret}}(\omega = p)}{2 p}\,. 
\end{eqnarray}
Then the speed of the pions in the chiral limit is   
\begin{eqnarray}
\label{speed}
v^{2}&=& 1-\frac{T^{4}}{F_{\pi}^{4}} 
\left(\frac{1}{27}\log\left(\frac{\mu}{T}\right)+
\frac{64 \pi^{2}}{45}(L_{1}^{r}(\mu)+
 2 L_{2}^{r}(\mu)) \right. \nonumber \\
&& - \left. \frac{101}{4860}+\frac{2 \Huge{\gamma}_{E}}{81}-
\frac{1}{81} \log(16 \pi) 
- \frac{8}{9}\,\zeta'(-1)-\frac{200}{27}\zeta'(-3)-
\frac{20}{9 \pi^{4}}\zeta'(4) \right) \,+{\cal O}(p) \,. 
\end{eqnarray}

This is the main result of this letter. The scale $\mu$ enters the calculation 
in such a way that, had we used a different scale $\mu^\prime$ but kept the speed 
invariant, we would have had
\begin{equation}
L_{i}^{r}(\mu^\prime) = L_{i}^{r}(\mu) -
\frac{\gamma_{i}}{16 \pi^{2}}\log \left(\frac{\mu^\prime}{\mu}\right)\,,
\end{equation}
which is the correct result for the running chiral coefficients. 
At the scale of the mass of the pion, $\mu = m_{\pi}$, 
we take \cite{wanders,toublan}
$32\pi^{2}(L_{1}^{r}+2 L_{2}^{r})\simeq 1.66$ to yield  
\begin{equation}
v^{2}=1-\frac{T^{4}}{27 F_{\pi}^{4}}\log\left(\frac{\Lambda}{T}\right)\,,
\end{equation}
with $\Lambda \simeq 4300$ MeV. Notice that we have obtained an analytic
expression 
\begin{equation}
- \frac{101}{4860}+\frac{2 \Huge{\gamma}_{E}}{81}-
\frac{1}{81} \log(16 \pi) 
- \frac{8}{9}\,\zeta'(-1)-\frac{200}{27}\zeta'(-3)-
\frac{20}{9 \pi^{4}}\zeta'(4) \simeq -0.0538 \,, 
\end{equation}
for the constant in Eq.~(\ref{speed}). 
This constant was computed by numerical integration in \cite{toublan}, where
the value $-0.0202$ is given. This discrepancy explains the difference in 
the scale $\Lambda$ from
$1.8$ GeV to $4.3$ GeV. However, this difference between logarithmic 
scales has a moderate impact 
on the quantitative behavior of the pion speed in the range of temperature 
where this calculation can be trusted. Thus, the change produced in $v^2$ is 
less than $6\%$ up to $T \sim 100$ MeV.      
   
The pion damping rate can also be computed easily within the 
imaginary time formalism at leading order. The result is 
$\gamma(p) = p^2 T^3 \log(T/p)/(18 \pi F_{\pi}^4) + {\cal O}(p^3)$ 
in agreement with the
one previously reported in \cite{go,smilga}. 

How do extend these results to the case of non-zero quark masses? In 
principle, this question could be answered by considering the 
new graphs coming from the breaking part of the effective Lagrangian 
${\cal L}_{breaking}= 
 F_{\pi}^2/4\,{\mbox{Tr}}[\chi U^\dagger + \chi^\dagger U] 
 +\ldots$, where $\xi = 2 B m$, with $m$ the quark mass matrix.
However, the use of background field perturbation theory 
which in the chiral limit greatly reduces the number 
of graphs to be computed, does not seem to  
simplify the computation now, since the Lagrangian 
${\cal L}(\tilde{\pi},\xi, \chi)$ obtained 
after the splitting of $U$ has lost its 
(local) invariance under chiral transformations. 
Moreover, the integrals involved 
are complicated by the pion mass, making their 
analytical computation more difficult. 
We hope to address this question in the future.             



\subsection*{Acknowledgements}

Discussions with M. A. Go\~ni and J. L. Ma\~nes are gratefully acknowledged. 
J.M.M. is supported by a UPV/EHU grant. 
This work has been also partially supported 
by CICYT under project No. AEN96-1668 and the University of the Basque Country 
grant UPV-EHU 063.310-EB225/95.  

\newpage

\appendix

\section {}
\label { pi appendix}

To do our calculations we follow Arnold and Zhai\cite{az} which allows 
us to evaluate the sum-integrals analytically.
We can rewritte Eq.~(\ref{ab}) (without the factors in front of it) as :
\begin{equation}
 \frac{1}{4}\ {\mbox{Tr}}_{Q}\,\Delta(Q) [P^{\mu}P^{\nu}\Pi^{\mu \nu}(Q) +
 3 (PQ)^{2}\Pi(Q) + 2 P^{2}\,{\mbox{Tr}}_{K} \Delta(K) ] \,,
\end{equation}
where
\begin{equation}
\Pi^{\mu \nu}(Q)={\mbox{Tr}}_{Q}[ (2K+Q)^{\mu}(2K+Q)^{\nu} 
 \Delta(K)\Delta(K+Q)]
 -2\,\delta^{\mu \nu}\,{\mbox{Tr}}_{K}\Delta(K) \, ,
\end{equation}
and
\begin{equation}
\Pi(Q)={\mbox{Tr}}_{K}\,\Delta(K)\Delta(K+Q)\, .
\end{equation}
We start with the term proportional to 
$\Pi(Q)$. As the result is independent of the direction of \,$\vec{p}$ \, 
we average over its angles in $d-1$ dimensions, 
\begin{equation}
\label{angles}
{\mbox{Tr}}_{Q}\,(p_{0}q_{0}\, +\vec{p}\vec{q}\,)^{2}\Delta(Q) \Pi(Q)=
\left(p_{0}^{2}-\frac{p^{2}}{d-1}\right)
{\mbox{Tr}}_{Q}\,q_{0}^{2}\,\Pi(Q)\,\Delta(Q) 
+\frac{p^{2}}{(d-1)}{\mbox{Tr}}_{Q}\,\Pi(Q)\,.
\end{equation}
Following \cite{az} we split $\Pi(Q)$ into two terms
\begin{equation}
\Pi(Q)=\Pi^{(0)}(Q)+\Pi^{(T)}(Q)\,,
\end{equation}
where $\Pi^{(0)}(Q)$ is the zero-temperature contribution and  
$\Pi^{(T)}(Q) = {\tilde{\Pi}}^{(T)}(Q) + 
 \Pi^{UV}(Q) $ is the temperature dependent piece. Although $\Pi^{(T)}(Q)$ 
is UV finite, its large-$Q$ asymptotic behaviour 
$\Pi^{UV}(Q)\sim 1/Q^2$,  
gives rise to 
a UV divergence in 
${\mbox{Tr}}_{Q} \,q_{0}^{2}\,\Pi^{(T)}(Q)\,\Delta(Q)$. Thus, it is useful 
to evaluate separately the three contributions,         
\begin{eqnarray}
{\mbox{Tr}}_{Q}\,q_{0}^{2}\Delta(Q)\,\tilde{\Pi}^{(T)}(Q)&=&
\frac{T^{4}}{1080}(7+3 \gamma_{E}+
1440\,\zeta'(-3)+180\,\zeta'(-1)) + {\cal O}(\epsilon)\,, \\ 
{\mbox{Tr}}_{Q}q_{0}^{2}\Delta(Q)\,\tilde{\Pi}^{UV}(Q) &=&  
-\frac{19 T^{4}}{2160}-\frac{T^{4}}{720}\frac{1}{\epsilon}-
\frac{T^{4} \Huge{\gamma}_{E}}{360}-
\frac{T^{4}}{360}\log(\frac{\mu^{2}}{2\,T^{2}})+
\frac{T^{4}}{4\,\pi^{4}}\zeta'(4) + {\cal O}(\epsilon)\,,  \\
{\mbox{Tr}}_{Q}\,q_{0}^{2}\Delta(Q)\Pi^{(0)}(Q)&=&
-\frac{T^{4}}{480} \frac{1}{\epsilon}-
\frac{T^{4}}{80}+\frac{T^{4}\,\gamma_{E}}{240}+
\frac{T^{4}}{240}\log(4\,\pi) \nonumber \\
& & -T^{4}\,\zeta'(-3)-\frac{T^{4}}{120}\,\log(\frac{\mu}{T})+
{\cal O}(\epsilon).
\end{eqnarray}
The remaining integral in Eq.~(\ref{angles}) gives
\begin{equation}
 {\mbox{Tr}}_{Q}\,\Pi(Q)=\left({\mbox{Tr}}_{K}\frac{1}{K^{2}}\right)^{2}=
\frac{T^{4}}{144}\ + {\cal O}(\epsilon) \,.
\end{equation}

Now we have to evaluate 
${\mbox{Tr}}_{Q}\Delta(Q)\,P^{\mu}\,P^{\nu}\,\Pi^{\mu\,\nu}(Q)$. 
We can use the orthogonality property  
$Q^{\mu} \Pi^{\mu \nu}(Q)=0$ to transform it into (after the average over 
the angles of $\vec{p})$
\begin{equation}
{\mbox{Tr}}_{Q}P^{\mu}P^{\nu}\Pi^{\mu\nu}(Q)\Delta(Q) = 
 -\frac{5}{432}p^{2}T^{4}+\left(p_{0}^{2}-
\frac{p^{2}}{(d-1)}\right){\mbox{Tr}}_{Q}\Pi^{00}\Delta(Q)\,, 
\end{equation}
where has been used ${\mbox{Tr}}_{Q} \Delta(Q) = T^2/12$. 
Following the same steps as for the case of $\Pi(Q)$ we get:
\begin{eqnarray}
{\mbox{Tr}}_{Q}\tilde{\Pi}^{00}_{T}\,\Delta(Q)&=&
-\frac{T^{4}}{1080}(1-21\Huge{\gamma}_{E}-2880\zeta'(-3)
-540\zeta'(-1))+ {\cal O}(\epsilon) \,, \\
{\mbox{Tr}}_{Q}\Pi^{00}_{UV}\,\Delta(Q)&=&
-\frac{7 T^4}{720}\frac{1}{ \epsilon } -
\frac{7 \Huge{\gamma}_{E} T^{4} }{360} + 
\frac{17 T^{4} }{2160}+\frac{7 T^4}{4\,\pi^{4} }\zeta'(4)-
\frac{7 T^4}{360}\log(\frac{\mu^{2} }{2T^{2} } )+ {\cal O}(\epsilon) \,, \\
{\mbox{Tr}}_{Q}\Pi^{00}_{0}\,\Delta(Q)&=&
-\frac{T^{4} }{1440}\frac{1}{ \epsilon } -
\frac{T^{4} }{3} \zeta'(-3) + \frac{\Huge{\gamma}_{E} T^{4} }{720}-
\frac{11 T^{4} }{2160}-
\frac{T^{4} }{720}\log(\frac{\mu^{2} }{4 \pi T^{2} })+{\cal O}(\epsilon) \,. 
\end{eqnarray}
Adding up all contributions we finally obtain the result given in 
Eq.~(\ref{sunset}).

The contribution from the counterterms is proportional to  
${\mbox{Tr}}_{Q}(PQ)^{2}\Delta(Q)$. This is required up to order 
$\epsilon$ because of the pole in $(L_{1}+2 L_{2})$, 
\begin{equation}
{\mbox{Tr}}_{Q}(PQ)^{2}\Delta(Q)=\left(p_{0}^{2}-
\frac{p^{2}}{(d-1)}\right)\,{\mbox{Tr}}_{Q}
\left(\frac{q_{0}^{2}}{Q^{2}}\right)\ +
\frac{p^{2}}{(d-1)}\ {\mbox{Tr}}_{Q}\,1 \,,
\end{equation}
where the last term is zero in dimensional regularization. 
The other term gives
\begin{equation}
{\mbox{Tr}}_{Q}(PQ)^{2}\Delta(Q)=\frac{4\pi^{2} p^{2}}{90}
\left(1+\epsilon\,(240 \zeta'(-3)-
 \gamma_{E}+\log(\frac{\mu^{2}}{4\pi T^{2}}) +
\frac{13}{6}\right) \,.
\end{equation}


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\end{references}

\begin{figure}
\centering
\leavevmode
\epsfbox{fig1.eps}
%\caption{Primera figura}
\label{fg:fig1}
\end{figure}

\begin{figure}
\centering
\leavevmode
\epsfbox{fig2.eps}
%\caption{Segunda figura}
\label{fg:fig2}
\end{figure}


\end{document}


