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\begin{document}

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\begin{flushright} 
DAMTP-HEP-98-7 \\ 
{\tt hep-ph/9801197} \\
%\\ 
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\vskip 2.5cm


\begin{center}{\Large\bf Hydrostatic  Pressure of the $O(N)$ 
$\phi^{4}$ Theory in the Large $N $ Limit} \vskip 1.2cm {\large 
P.Jizba\footnote{E-mail:
{\tt P.Jizba$@$damtp.cam.ac.uk}}}\\ {\em DAMTP, University of Cambridge,
Silver Street, Cambridge, CB3 9EW, UK} \end{center}

\addtocounter{footnote}{2} 

\vskip 2.0cm
\begin{center} {\large\bf Abstract}
\end{center}


\begin{minipage}{13.8cm} {\small We present a self-contained calculation
of the hydrostatic pressure in $\lambda \phi^{4}$ theory with $O(N)$
internal symmetry. In the large $N$ limit the pressure can be expressed in
a resummed form. Our calculations are based on composite-operator
renormalization. As a byproduct, the renormalized (or improved)
energy-momentum tensor is obtained. The Dyson-Schwinger equations are used
to discuss both the mass and coupling constant renormalization. We also
calculate the high-temperature expansion for the pressure in $D=4$. The
latter is done by means of the Mellin transform technique. The pressure
computed is in complete agreement with the recent calculations of the
thermodynamic pressure \tsecite{ID1}.} \end{minipage}


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\section{Introduction}

A significant quantity of physical interest that one may want to calculate 
in field theory at finite temperature, either at equilibrium or out of 
equilibrium, is a pressure. In thermal quantum field theory (both in the 
real and imaginary time formalism) where one usually deals with systems 
in thermal equilibrium there is an easy prescription for a pressure 
calculation. The latter is based on the observation that for thermally 
equilibrated systems the grand canonical partition function $Z$ is given as

\begin{equation}
Z = e^{-\beta \Omega} = Tr(e^{-\beta(H-\mu_{i}N_{i})}),
\tselea{1}
\end{equation}

\vspace{2mm}
\noindent where $\Omega$ is the grand canonical potential, $H$ is the
Hamiltonian, $N_{i}$ are conserved charges, $\mu_{i}$ are corresponding
chemical potentials, and $\beta$ is the inverse temperature: $\beta = 1/T$
($k_{B}=1$). Using identity $\beta \frac{\partial}{\partial \beta}= -T
\frac{\partial}{\partial T}$ together with (\tseref{1}) one gets

\begin{equation}
 T \left( \frac{\partial \Omega}{\partial T} \right)_{\mu_{i},V} =  
\Omega -E + \mu_{i} N_{i},
\tselea{2}
\end{equation} 

\vspace{2mm}
\noindent with $E$ and $V$ being the averaged energy and volume of the
system respectively. A comparison of (\tseref{2}) with a corresponding
thermodynamic expression for the grand canonical potential \tsecite{LW,GM,
Cub} requires that entropy $S=- \left(\frac{\partial \Omega}{\partial T}
\right)_{\mu_{i}, V}$, so that

\vspace{5mm}

\begin{equation}
d\Omega = -SdT - pdV -N_{i}d\mu_{i}\; \Rightarrow \; p= -\left( 
\frac{\partial \Omega}{\partial V} \; \right)_{\mu_{i}, T}.
\tselea{3}
\end{equation}
 
\vspace{2mm} 

\noindent For large systems one can usually neglect surface effects so $E$
and $N_{i}$ become extensive quantities. Eq.(\tseref{1}) then immediately
implies that $\Omega$ is extensive quantity as well, so (\tseref{3})
simplifies to

\begin{equation}
p = -\frac{\Omega}{V} = \frac{\mbox{ln}Z}{\beta V}.  
\tselea{4}
\end{equation}

\vspace{3mm}
\noindent The pressure defined by Eq.(\tseref{4}) is so called thermodynamic 
pressure. 


\vspace{3mm}

\noindent As $\mbox{ln}Z$ can be systematically calculated summing up all
the loop diagrams \tsecite{LW, DJ}, the pressure calculated via (\tseref{4})
enjoys a considerable popularity \tsecite{ID1, LW, LB1, ID}. 
Unfortunately, the latter procedure can not be extended to out of
equilibrium as there is, in general, no definition of the partition
function $Z$ nor grand canonical potential $\Omega$ away from an
equilibrium. 

\vspace{3mm}

\noindent Yet another, alternative definition of a pressure not hinging on
thermodynamics can be provided; namely the hydrostatic pressure which is
formulated through the energy-momentum tensor $\Theta^{\mu \nu}$. The
formal argument leading to the hydrostatic pressure in $D$ space-time
dimensions is based on the observation that $\langle \Theta^{0 j}(x)
\rangle$ is the mean (or macroscopic) density of momenta ${\vect{p}}^{j}$
in the point $x^{\mu}$. Let ${\vect{P}}$ be the mean total
$(D-1)$-momentum of an infinitesimal volume $V^{(D-1)}$ centred at
${\vect{x}}$, then the rate of change of $j$-component of ${\vect{P}}$
reads


\begin{equation}
\frac{d{\vect{P}}^{j}(x)}{dt} = \int_{V^{(D-1)}}d^{D-1}{\vect{x}}' \; 
\frac{\partial}{\partial
x^{0}} \langle \Theta^{0 j}(x^{0},{\vect{x}}' ) \rangle = \sum_{i=1}^{D-1}\int_{\partial 
V^{(D-1)}} d{\vect{s}}^{i} \; \langle \Theta^{i j}
\rangle.
\tselea{5}
\end{equation}

\vspace{2mm}

\noindent In the second equality we have exploited the continuity equation
for $\langle \Theta^{\mu j} \rangle$ and successively we have used Gauss's
theorem \footnote{The macroscopic conservation law for $\langle
\Theta^{\mu \nu} \rangle$ (i.e. the continuity equation) has to be
postulated. For some systems, however, the later can be directly derived
from the corresponding microscopic conservation law \tsecite{DG}.}. The
$\partial V^{(D-1)}$ corresponds to the surface of $V^{(D-1)}$.

\vspace{3mm}

\noindent Anticipating  a system out  of equilibrium, we must assume a
non-trivial  distribution   of   the    mean  particle   four-velocity
$U^{\mu}(x)$ (hydrodynamic velocity). Now, a pressure is by definition
a scalar quantity.  This particularly means  that it should not depend
on the hydrodynamic velocity. We must thus go to  the local rest frame
and evaluate pressure there. However, in  the local rest frame, unlike
the  equilibrium, the  notion of  a  pressure  acting equally in   all
directions is   lost.   In order   to retain  the scalar  character of
pressure,  one customarily defines  the {\em pressure  at a point} (in
the following denoted as  $p(x)$) \tsecite{Bach}, which is  simply the
`averaged pressure' {\footnote {To  be  precise, we should  talk about
averaging the normal  components of stress  \tsecite{Bach}.}} over all
directions at  a given point.  In the local rest frame Eq.(\tseref{5})
describes $j$-component  of the  force  exerted by  the medium  on the
infinitesimal volume $V^{(D-1)}$  as, by  definition, there cannot  be
any contribution   to $\frac{d{\vect{P}}^{j}(x)}{dt}$  caused  by  the
particle   convection through  $V^{(D-1)}/;$.   Averaging the  LHS  of
(\tseref{5})        over    all   directions    of     the      normal
${\vect{n}}({\vect{x}})$, we get 
\footnote{The angular average  is standardly defined for scalars (say,
$A$) as; $\int A \;d\Omega({\vect{n}})/ \int d\Omega({\vect{n}})$, and
for  vectors  (say, ${\vect{A}}^{i}$) as; $\sum_{j}\int {\vect{A}}^{j}
\;       {\vect{n}}^{j}\;      d\Omega         ({\vect{n}})/      \int
d\Omega({\vect{n}})$. Similarly   we might   write  down  the  angular
averages for tensors of a higher rank.} 


\begin{eqnarray}
\mbox{$\frac{1}{\left(S^{D-2}_{1}\right)}$}\sum_{j=1}^{D-1}\int \; 
\frac{d{\vect{P}}^{j}(x)}{dt}\; {\vect{n}}^{j}
\; d
\Omega({\vect{n}}) &=& 
\mbox{$\frac{1}{\left(S^{D-2}_{1}\right)}$}\sum_{j,i=1}^{D-1}\int_{\partial 
V^{(D-1)}} 
ds\; \langle \Theta^{ij}(x') \rangle \; \int d\Omega({\vect{n}})\;  
{\vect{n}}^{i}{\vect{n}}^{j}\nonumber\\
&=& - \frac{1}{(D-1)}\sum_{i=1}^{D-1} \int_{\partial V^{(D-1)}}ds \; 
\langle \Theta^{i}_{\; i}(x')
\rangle,
\tselea{EMT22}
\end{eqnarray}

\vspace{3mm}

\noindent where $d\Omega({\vect{n}})$ is an element of solid angle about
${\vect{n}}$ and $S^{D-2}_{1}$ is the surface of $(D-2)$-sphere with unit
radius ($\int d\Omega({\vect{n}}) = S^{D-2}_{1} = 2
\pi^{\frac{D-1}{2}}/\Gamma (\mbox{$\frac{D-1}{2}$})$) . On the other hand,
from the definition of the pressure at a point $x^{\mu}$ we might write

\vspace{2mm}

\begin{equation}
\left(S^{D-2}_{1}\right)^{-1}\sum_{j=1}^{D-1}\int \; 
\frac{d{\vect{P}}^{j}(x)}{dt}\; {\vect{n}}^{j}
\; d
\Omega({\vect{n}}) = - p(x) \; \int_{\partial V^{(D-1)}}ds,
\tseleq{EMT23}
\end{equation}

\vspace{3mm}

\noindent here the minus sign reflects that the force responsible for a
compression (conventionally assigned as a positive pressure) has reversed
orientation than the surface normals ${\vect{n}}$ (pointing outward). In
order to keep track with the standard text-book definition of a sign of
a pressure \tsecite{Cub, Bach} we have used in (\tseref{EMT23}) the 
normal ${\vect{n}}$ in a contravariant notation (note, ${\vect{n}}^{i} = -
{\vect{n}}_{i}$). Comparing (\tseref{EMT22}) with (\tseref{EMT23}) we can
write for a sufficiently small volume $V^{(D-1)}$


\begin{equation}
p(x)= - \frac{1}{(D-1)} \sum_{i=1}^{D-1}\langle \Theta^{i}_{\;i}(x) 
\rangle. \tseleq{EMT24}
\end{equation}


\vspace{3mm}

\noindent We should point out that in equilibrium the thermodynamic
pressure is usually identified with the hydrostatic one via the virial
theorem \tsecite{LW, Zub}. In the remainder of this note we shall deal
with the hydrostatic pressure at equilibrium. We shall denote the
foregoing as ${\cali{P}}(T)$, where $T$ stands for temperature. We consider
the non-equilibrium case in a future paper. 

\vspace{4mm}

\noindent The plan of this paper is as follows. In Section 2 we review
the necessary mathematical framework needed for the renormalization of
the energy-momentum tensor.(For an extensive review see also
ref.\tsecite{LW, Collins, Brown}.)  The latter is discussed on the
$O(N)\; \phi^{4}$ theory. As a byproduct we renormalize
$\phi_{a}^{2}$, $\phi_{a}\phi_{b}$ and $\Theta^{\mu \nu}$
operators. The corresponding QFT extension of (\tseref{EMT24}) is
obtained. 

\vspace{3mm}

\noindent Resummed form for the pressure in the large $N$ limit,
together with the discussion of both coupling constant and mass
renormalization is worked out in Section 3. The discussion is
substantially simplified by means of the Dyson-Schwinger equations. 


\vspace{3mm} 

\noindent In Section 4 we end up with the high-temperature expansion of
the pressure. Calculations are performed for $D=4$ (both for massive and
massless fields) and the result is expressed in terms of renormalized
masses $m_{r}(0)$ and $m_{r}(T)$. The former is done by means of the
Mellin transform technique. 

\vspace{3mm}

\noindent The paper is furnished with two appendices. For the
completeness' sake we compute in Appendix A1 a renormalized coupling
constant at finite temperature. Finally in Appendix A2 we clarify some
mathematical manipulations needed in Section 3. 


\vspace{5mm}
\section{Renormalization}


If we proceed with (\tseref{EMT24}) to QFT this leads to the notorious
difficulties connected with the fact that $\Theta^{\mu \nu}$ is a
(local) composite operator.  If only a free theory would be in
question then the normal ordering prescription would be sufficient to
render $\langle \Theta^{\mu \nu} \rangle$ finite. In the general case,
when the interacting theory is of interest, one must work with the
Zimmerman `normal' ordering prescription instead. Let us demonstrate
the latter on the $O(N)\; \phi^{4}$ theory. (In this Section we 
keep $N$ arbitrary.) Such a theory is defined by the bare Lagrange
function

\begin{equation}
{\cali{L}}= \frac{1}{2}\sum_{a=1}^{N}\left( (\partial 
\phi_{a})^{2}-m_{0}^{2}\phi_{a}^{2} \right) - 
\frac{\lambda_{0}}{4!N}\left( \sum_{a=1}^{N} (\phi_{a})^{2} \right)^{2}.
\tseleq{6}
\end{equation}

\vspace{2mm}

\noindent The corresponding canonical energy-momentum tensor is 
given by


\begin{equation}
\Theta^{\mu \nu}_{C} = 
\sum_{a}\partial^{\mu}\phi_{a}\partial^{\nu}\phi_{a} - g^{\mu \nu} 
{\cali{L}}.
\tseleq{7}
\end{equation}

\vspace{2mm}


\noindent The Feynman rules for Green's functions with the energy-momentum
insertion can be easily explained in momentum space. In the reasonings to
follow we shall need the (thermal) composite Green's function


\begin{equation}
D^{\mu \nu}(x^{n}|y) = \langle T^{*}\left\{ \phi_{r}(x_{1}) \ldots 
\phi_{r}(x_{n}) \Theta^{\mu \nu}_{C}(y) \right\} \rangle.
\tseleq{8}
\end{equation}


\vspace{2mm}

\noindent Here the subscript $r$ denotes the renormalized fields in the
Heisenberg picture (the internal indices are suppressed) and $T^{*}$ is so
called $T^{*}$ product (or covariant $T$ product) \tsecite{N, RJ, CCR,
IZ}.  The $T^{*}$ product is defined in such a way that for fields in the
interaction picture it is simply the $T$ product with all differential
operators ${\cali{D}}_{\mu_{i}}$ pulled out of the $T$-ordering symbol,
i.e. 


\begin{equation}
T^{*}\{ {\cali{D}}_{\mu_{1}}^{x_{1}}\phi_{I}(x_{1})\ldots
{\cali{D}}_{\mu_{n}}^{x_{n}}\phi_{I}(x_{n})
\} = {\cali{D}}(i\partial_{\{\mu\}}) T\{\phi_{I}(x_{1})\ldots
\phi_{I}(x_{n})\},
\tseleq{TP1}
\end{equation}
\vspace{1mm}

 
\noindent where ${\cali{D}}(i\partial_{\{\mu\}})$ is just a useful
 short-hand notation for
${\cali{D}}_{\mu_{1}}^{x_{1}}{\cali{D}}_{\mu_{2}}^{x_{2}}\ldots
{\cali{D}}_{\mu_{n}}^{x_{n}}$. In the case of thermal Green's functions,
the $T$ might be as well a contour ordering symbol. It is the mean value
of the $T^{*}$ ordered fields rather than the $T$ ones, which corresponds
at $T=0$ and at equilibrium to the Feynman path integral representation of
Green's functions \tsecite{CCR, JS}. 


\vspace{3mm}

\noindent A typical contribution to $\Theta^{\mu \nu}_{C}(y)$ can be 
written as

\begin{equation}
{\cali{D}}_{\mu_{1}}\phi(y)
\; {\cali{D}}_{\mu_{2}}\phi(y) \ldots {\cali{D}}_{\mu_{n}}\phi(y),
\tseleq{CO1}
\end{equation}

\vspace{2mm}

\noindent so the typical term in (\tseref{8}) is

\begin{displaymath}
{\cali{D}}(i\partial_{\{\mu\}})\; \langle T^{*}\left\{ \phi_{r}(x_{1})\ldots 
\phi_{r}(x_{n})\phi(y_{1})\ldots \phi(y_{k}) \right\}\rangle \left. 
\right|_{y_{i}=y}= D_{\{\mu \}}(x^{n}|y^{k})\left. 
\right|_{y_{i}=y}.
\end{displaymath}

\vspace{3mm}

\noindent Performing the Fourier transform in (\tseref{8}) we get 


\begin{equation}
D^{\mu \nu}(p^{n}|p)= \sum_{k=\{2,4\}}\; \int  \left( \prod_{i=1}^{k} 
\frac{d^{D}q_{i}}{(2\pi)^{D}}\right) (2\pi)^{D}\; 
\delta^{D}(p-\sum_{j=1}^{k}q_{j})\; {\cali{D}}_{(k)}^{\mu \nu}(q_{\{ \mu 
\}}) \; D(p^{n}|q^{k}),
\tseleq{11}
\end{equation}

\vspace{3mm}

\noindent where ${\cali{D}}_{(k)}^{\mu \nu}(\ldots)$ is a Fourier
transformed differential operator corresponding to the quadratic and
quartic terms in $\Theta^{\mu \nu}_{C}$. Denoting the new vertex
corresponding to ${\cali{D}}_{(k)}^{\mu \nu}(\ldots)$ as $\otimes$, we can
graphically represent (\tseref{8}) through (\tseref{11}) as



\begin{figure}[h]
\begin{center}
\leavevmode
\hbox{%
\epsfxsize=7.5cm
\epsffile{N1.eps}}
\caption{\em The graphical representation of $D^{\mu \nu}(p^{n}|p)$.}
\label{fig1}
\end{center}
\begin{picture}(10,10)
\put(50,100){$D^{\mu \nu}(p^{n}|p)=$}
\end{picture}
\end{figure}
\vspace{3mm}
 

\noindent For the case at hand one can easily read off from (\tseref{7}) 
an explicit form of the bare composite vertices, the foregoing are


\begin{figure}[h]
\begin{flushleft}
\leavevmode
\hbox{%
\epsfxsize=2cm
\epsffile{N2.eps}}
\end{flushleft}
\begin{picture}(20,5)
\put(65,105){$\sim \;{\cali{D}}_{(2)}^{\mu \nu}(q_{\{ \mu \}}) = 
\frac{1}{2}\;\delta_{ab}\;\{ 2(q_{1}-p)^{\mu}q_{1}^{\nu}-g^{\mu 
\nu}((q_{1}-p)_{\lambda}q_{1}^{\lambda}-m_{0}^{2})\}$}
\put(65,50){$\sim \;{\cali{D}}_{(4)}^{\mu \nu}(q_{\{ \mu \}}) = \frac{g^{\mu 
\nu} \lambda_{0}}{4!N}\{ 
2(\delta_{ab}\delta_{cd}+\delta_{ac}\delta_{bd}+\delta_{ad}\delta_{bc}) 
- 5 \delta_{ab}\delta_{cd}\delta_{ac}\}$}      
\end{picture} 
\end{figure}

\vspace{2mm}

\noindent (For the internal indices we do not adopt Einstein's summation 
convention.) We have tacitly assumed in Fig.\ref{fig1} that the vacuum 
bubble diagrams present in the shaded blobs are divided out. We have also
implicitly assumed that summation over internal indices is understood.
Note that in the case of thermal composite Green's function, the new
vertices are clearly of type-1\footnote{For a brief introduction to the
real-time formalism in thermal QFT see for example \tsecite{PVL2}.} as the
fields from which they are deduced have all a real-time argument (type-1
fields). 

\vspace{7mm}
\noindent {\it Renormalization of $\phi_{a}(x)\phi_{b}(x)$}
\vspace{2mm}

\noindent Now, if there would be no $\Theta^{\mu \nu}_{C}$ insertion in
(\tseref{8}), the latter would be finite, and  so it is natural to define the
renormalized energy-momentum tensor $[\Theta^{\mu \nu}_{C}]$ (or Zimmermann 
normal ordering) in such a way that

\begin{displaymath}
D_{r}^{\mu \nu}(x^{n}|y) = \langle T^{*}\left\{ \phi_{r}(x_{1}) \ldots 
\phi_{r}(x_{n})\; [\Theta^{\mu \nu}_{C}] \right\} \rangle,
\end{displaymath}


\noindent is finite for any $n > 0$. To see what is involved, we
illustrate the mechanism of the composite operator renormalization on
$\phi_{a}(x)\phi_{b}(x)$. We shall use the mass-independent
renormalization (or minimal subtraction scheme - (MS)) which is
particularly suitable for this purpose.  In MS we can expand the bare
parameters into the Laurent series which has a simple form
\tsecite{Brown, IZ,
JS}, namely

\begin{equation}
\lambda_{0} = \mu^{4-D}\; \lambda_{r} \left( 1 + \sum_{k=1}^{\infty}
\frac{a_{k}(\lambda_{r}; D)}{(D-4)^{k}}\right)
\tseleq{CO2}
\end{equation}
\begin{equation}
m_{0}^{2} = m_{r}^{2} \left( 1 + \sum_{k=1}^{\infty}\frac{b_{k}(\lambda_{r}; 
D)}{(D-4)^{k}} \right).
\tseleq{CO3}
\end{equation}
    
\vspace{2mm}

\noindent Here $a_{0}$ and $b_{0}$ are analytic in $D=4$. The parameter
$\mu$ is the scale introduced by the renormalization in order to keep
$\lambda_{r}$ dimensionless. An important point is that both $a_{k}$'s and
$b_{k}$'s are mass, temperature and momentum independent. 

\vspace{3mm}

\noindent It was Zimmermann who first realized that the forest formula
known from the ordinary Green's function renormalization \tsecite{IZ,
Collins} can be also utilized for the composite Green's functions rendering
them finite \tsecite{Collins, Zimm}. That is, we start with Feynman
diagrams expressed in terms of physical (i.e. finite) coupling constants
and masses (this holds for the composite vertex as well). As we calculate
diagrams to a given order, we meet UV divergences which might be cancelled
by adding counterterm diagrams. The forest formula then prescribes how to
systematically cancel all the UV loop divergences by counterterms to all
orders. However, in contrast to the coupling constant renormalization, the
composite vertex need not to be renormalized multiplicatively. We shall
illustrate this fact in the sequel. Let us also observe that in the lowest
order (no loop) the renormalized composite vertex equals to the bare one,
and so to that order $A=[A]$, for any composite operator $A$. 


\vspace{3mm}

\noindent Now, from (\tseref{CO2}) and (\tseref{CO3}) follows that for any 
function $F= F(m_{r}, \lambda_{r})$ we have
\vspace{2mm}

\begin{displaymath}
\frac{\partial F}{\partial m^{2}_{r}} = \frac{\partial 
m_{0}^{2}}{\partial m_{r}^{2}}\; \frac{\partial F}{\partial m_{0}^{2}} = 
\frac{
m_{0}^{2}}{ m_{r}^{2}}\; \frac{\partial F}{\partial m_{0}^{2}} . 
\end{displaymath}

\vspace{2mm}

\noindent So particularly for

\begin{displaymath}
F= D(x_{1}, \ldots , x_{n}) = \langle T^{*} \{ \phi_{r}(x_{1}) \ldots 
\phi_{r}(x_{n}) \} \rangle,
\end{displaymath}

\noindent one reads

\begin{eqnarray}
&&m^{2}_{r}\; \frac{\partial}{\partial m^{2}_{r}} D(x_{1}, \ldots , x_{n}) 
= m^{2}_{0}\; \frac{\partial}{\partial m^{2}_{0}} D(x_{1}, \ldots , 
x_{n})\nonumber \\
&&\mbox{\hspace{1.5cm}}= \left( - \frac{i}{2} \right) {\cali{N}}\; \int 
d^{D}x\;\sum_{a=1}^{N}\; 
\int {\cali{D}}\phi\; \phi_{r}(x_{1}) \ldots \phi_{r}(x_{n})\; 
m_{0}^{2}\phi_{a}^{2}(x)\; \mbox{exp}(iS[\phi, T])\nonumber \\
&&\mbox{\hspace{1.5cm}}= \left( - \frac{i}{2} \right) \; \int 
d^{D}x\;\sum_{a=1}^{N}\;D_{a}(x_{1}, \ldots, x_{n}|x; m_{0}^{2}).
\tselea{CO5}
\end{eqnarray}

\vspace{2mm}

\noindent Here ${\cali{N}}^{-1}$ is the standard denominator of the path
integral representation of Green's function. We should apply the
derivative also on ${\cali{N}}$ but this would produce disconnected graphs
with bubble diagrams. The former precisely cancel the very same
disconnected graphs in the first term, so we are finally left with no
bubble diagrams in (\tseref{CO5}). In the Fourier space (\tseref{CO5})
reads


\begin{equation}
m^{2}_{r}\; \frac{\partial}{\partial m^{2}_{r}} D(p_{1}, \ldots ,p_{n}) = 
\left( - \frac{i}{2}  \right) \sum_{a=1}^{N}\; D_{a}(p_{1}, \ldots , 
p_{n}|0;m_{0}^{2}).
\tseleq{CO6}
\end{equation}

\vspace{2mm}

\noindent As the LHS is finite there cannot be any pole terms on the RHS
either, and so $\sum_{a} m^{2}_{0}\phi_{a}^{2}$ is by itself a renormalized
composite operator. We see that $m_{0}^{2}$ precisely compensates the
singularity of $\sum_{a=1}^{N} \phi_{a}^{2}$. 

\vspace{3mm}

\noindent Now, it is well known that any second-rank tensor (say $M_{a
b}$) can be generally decomposed into three irreducible tensors; an
antisymmetric tensor, a symmetric traceless tensor and an invariant
tensor. Let us set $M_{ab}=\phi_{a}\phi_{b}$, so the symmetric traceless
tensor $K_{a b}$ reads

\vspace{-2mm}
\begin{equation}
K_{ab}(x) = \phi_{a}(x)\phi_{b}(x)-\delta_{a
b}/N\;\sum_{c=1}^{n}\phi^{2}_{c}(x),
\tseleq{CO64}
\end{equation}

\noindent whilst the invariant tensor $I_{a b}$ is 

\begin{displaymath} 
I_{a b}(x) = \delta_{a b}/N \sum_{c=1}^{N}\phi_{c}^{2}(x). 
\end{displaymath}

\noindent Because the renormalized composite operators have to preserve a
tensorial structure of the bare ones, we immediately have that

\begin{equation}
K_{a b} = A_{1}[K_{a b}]\;\;\; \mbox{and}\;\;\; I_{a b} = A_{2}[I_{a b}],
\tseleq{CO66}
\end{equation}

\vspace{2mm}

\noindent where both $A_{1}$ and $A_{2}$ must have structure $(1 + \sum
(\mbox{poles}))$. The foregoing guarantees that to the lowest order $K_{a
b} = [K_{a b}]$ and $I_{a b}= [I_{a b}]$. As we saw in (\tseref{CO6}),
$m_{0}^{2}I_{a b}$ is renormalized, and so from (\tseref{CO66}) follows
that $m_{0}^{2}I_{a b} = C\; [I_{a b}]$. Here $C$ has dimension $[m^{2}]$
and is analytic in $D=4$.  We can uniquely set $C=m^{2}_{r}$ because only 
this choice fulfils the lowest order condition $I_{ab}=[I_{ab}]$ (c.f.
Eq.(\tseref{CO3})). Collecting our results together we might write

\begin{equation}
\sum_{c} \phi_{c}^{2} = Z_{\Sigma \phi^{2}}\; \left[ \sum_{c} 
\phi_{c}^{2} \right] = Z_{\Sigma \phi^{2}} \sum_{c} [\phi_{c}^{2}],
\tseleq{CO67}
\end{equation}

\noindent with $Z_{\Sigma\phi^{2}} = A_{2}= \frac{m^{2}{r}}{m^{2}_{0}}$. 
In the second equality we have used an obvious linearity \tsecite{Collins}
of $[\ldots]$. From (\tseref{CO64}) and (\tseref{CO67}) follows that


\begin{equation}
\phi_{a}(x)\phi_{b}(x) = A_{1}[\phi_{a}(x)\phi_{b}(x)] - 
\frac{\delta_{ab}}{N}(A_{1} - Z_{\Sigma \phi^{2}})\; 
\sum_{c=1}^{N}[\phi_{c}^{2}(x)]. \tseleq{CO57}
\end{equation}


\noindent So particularly for $\phi^{2}_{a}$ one reads

\begin{equation}
\phi^{2}_{a} = \frac{1}{N}\left( (N-1)A_{1} + Z_{\Sigma \phi^{2}} \right) 
\; [\phi^{2}_{a}] - \frac{1}{N} \left( A_{1} - Z_{\Sigma \phi^{2}} 
\right) \sum_{c \not = a } [\phi_{c}^{2}].
\tseleq{CO58}
\end{equation}


\vspace{3mm}


\noindent From the discussion above it does not seem to be possible to
obtain more information about $A_{1}$ without doing an explicit
perturbative calculations, however it is easy to demonstrate that $A_{1}
\not= Z_{\Sigma \phi^{2}}$. To show this, let us consider the simplest
non-trivial case; i.e. N=2, and calculate $A_{1}$ to order $\lambda_{r}$. 
For that we need to discuss the renormalization of the $n$-point composite
Green's function with, say, $\phi^{2}_{1}$ insertion. To do that, it
suffices to discuss the renormalization of the corresponding 1PI $n$-point
Green's function. The perturbative expansion for the composite vertex to
order $\lambda_{r}$ can be easily generated via the Dyson-Schwinger
equation \tsecite{PC} and it reads


\begin{figure}[h]
\begin{center}
\leavevmode
\hbox{%
\epsfxsize=12.3cm
\epsffile{N558.eps}}
%xdvi large\label{fig101}
\end{center}
\setlength{\unitlength}{1mm}
\begin{picture}(20,7)
\put(0,40){where}
\put(139,71){(2.16)}
\end{picture}
\end{figure}

\addtocounter{equation}{1}



\noindent Here cross-hatched blobs refer to (renormalized) 1PI
$(n+2)$-point Green's function, circled indices mark a type of the field
propagated on the indicated line, and uncircled numbers refer to thermal
indices (we explicitly indicate only relevant thermal indices). The
counterterms, symbolized by a heavy dot, are extracted from the boxed
diagrams (elementary Zimmermann forests).  In MS scheme one gets the
following results: 

\begin{figure}[h]
\begin{flushleft}
\leavevmode
\hbox{%
\epsfxsize=2cm
\epsffile{N559.eps}}
\end{flushleft}
\setlength{\unitlength}{1mm}
\begin{picture}(15,10)
\put(25,43){$= \frac{i\;\lambda_{r}\mu^{4-D}}{4}\int 
\frac{d^{D}q}{(2pi)^{D}} \left\{ \D_{11}(q)\D_{11}(-q) - 
\D_{12}(q)\D_{12}(-q) \right\}|_{\mbox{{\scriptsize MS pole term}}}$}
\put(25,32){$= - \frac{1}{4} \partial_{m_{r}^{2}}\left( 
\frac{\Gamma\left(1-\frac{D}{2}\right)}{(4 \pi)^{\frac{D}{2}}} \; 
\lambda_{r}\; \mu^{4-D}\; m_{r}^{D-2}\right)|_{\mbox{\scriptsize MS}} = 
- \lambda_{r} \mu^{4-D} /2\; (D-4)\; (4 \pi)^{2}$}
\put(25,17){$ = - \lambda_{r} \mu^{4-D} / 6\; (D-4)\; (6 \pi)^{2}$.}
\end{picture}
\end{figure}


\noindent Here $\D_{11}$ and $\D_{12}$ are the usual thermal propagators
in the real-time formalism \tsecite{LW, LB1, PVL2} (see also Section 3). 
From (2.16) we can directly read off that

\begin{displaymath}
[\phi^{2}_{1}] = \left(1 - \frac{\lambda_{r} \mu^{4-D}}{2\;(D-4)\; (4 
\pi)^{2}} + 
{\cali{O}}(\lambda_{r}^{2})\right)\;\phi_{1}^{2} + \left(- 
\frac{\lambda_{r} \mu^{4-D}}{6\; (D-4)\; (4 \pi)^{2}} +
{\cali{O}}(\lambda_{r}^{2})\right)\;\phi_{2}^{2}.
\end{displaymath}


\vspace{3mm}

\noindent As the coefficient before $\phi_{2}^{2}$ is not zero, we
conclude that $A_{1} \not= Z_{\Sigma \phi^{2}}$. It is not a great
challenge to repeat the previous calculations for the $\phi_{1}\phi_{2}$
insertion. The latter gives


\begin{displaymath}
 A_{1} = 1 - \frac{\lambda_{r} \mu^{4-D}}{3\; (D-4)\;
(4 \pi)^{2}} +
{\cali{O}}(\lambda_{r}^{2}).
\end{displaymath} 




\vspace{3mm}

\noindent Eq.(\tseref{CO58}) exhibits the so called operator mixing
\tsecite{IZ}; the renormalization of $\phi_{a}^{2}$ cannot be considered
independently of the renormalization of $\phi_{c}^{2}$ ($c \not= a$). The
latter is a general feature of composite operator renormalization. 
Note, however, that $\phi_{a}\phi_{b}$ ($a \not=b$) do not mix by
renormalization, i.e. they renormalize multiplicatively. It can be shown
that composite operators mix under renormalization only with those composite
operators which have dimension less or equal \tsecite{IZ, Collins, Zimm}. 

\vspace{3mm}

\noindent Unfortunately, if we apply the previous arguments to $n=0$, the
result is not finite; another additional renormalization must be performed.
The fact that the expectation values of $[\ldots]$ are generally UV
divergent, in spite of being finite for the composite Green's functions
\footnote{Also called the matrix elements of $[\ldots]$.}, can be 
nicely illustrated with the composite operator $[\phi^{2}]$ in the $N=1$
theory. Taking the diagrams for $D(0|0)$ and applying successively the
(unrenormalized) Dyson-Schwinger equation \tsecite{PC} we get

\begin{figure}[h]
\begin{center}  
\leavevmode
\hbox{% 
\epsfxsize=12cm
\epsffile{N556.eps}}
%xdvi large\label{fig101}
\end{center}
\setlength{\unitlength}{1mm}
\begin{picture}(20,7)
%\put(65,15){$ = m_{0}^{2}\;\delta_{ba} \delta_{ca}$}
\put(140,20){(2.17)}
\end{picture}
\end{figure}

\addtocounter{equation}{1} 

\noindent Eq.(2.17) might be rewritten as 

\begin{eqnarray}
D(0|0) &=& D(0|0)|_{\lambda_{r}^{0}}\nonumber \\
&& + \frac{1}{2} \; \int 
\frac{d^{D}q_{1}}{(2\pi)^{D}}\frac{d^{D}q_{2}}{(2\pi)^{D}}\; 
\delta^{D}(q_{1}+ q_{2})\; 
D^{amp}(q^{2}|0)|_{\lambda_{r}}\; D(q^{2})\nonumber \\
&& + \frac{1}{36} \; \int \prod_{i=1}^{6} 
\frac{d^{D}q_{i}}{(2\pi)^{D}}\; \delta^{D}(\sum_{j=1}^{6}q_{j})\; 
D^{amp}(q^{6}|0)|_{\lambda_{r}^{2}}\; D(q^{6}),
\tselea{CO8}
\end{eqnarray}

\vspace{3mm}

\noindent where $D^{amp}(q^{m}|0)|_{\lambda_{r}^{k}}$ is the
$m$-point amputated composite Green's function to order $\lambda_{r}^{k}$,
and $D(q^{m})$ is the full $m$-point Green's function.  The crucial point is
that we can write $D(0|0)$ as a sum of terms, which, apart
from the first (free field) diagram, are factorized to the product of the
composite Green's function with $n > 0$ and the full Green's function. 
(The factorization is represented in (2.17) by the dashed lines. ) 

\vspace{3mm}

\noindent Now, utilizing the counterterm renormalization to the last two
diagrams in (2.17) we get situation depicted in Fig.\ref{fig11}. Terms
inside of the parentheses are finite, this is because both the composite
Green's functions ($n \ge 2$ !) and the full Green's functions are finite
after renormalization. The counterterm diagrams, which appear on the RHS
of the parentheses, precisely cancel the UV divergences coming from the
loop integrations over momenta $q_{1}\ldots q_{i}$ which must be finally
performed.  The heavy dots schematically indicates the corresponding
counterterms. In the spirit of the counterterm renormalization we should
finally subtract the counterterm associated with the overall
superficial divergence \footnote{A simple power counting in the $\phi^{4}$
theory reveals \tsecite{IZ} that for a composite operator $A$ with
dimension $\omega_{A}$ the superficial degree of divergence $\omega$
corresponding to an $n$-point diagram is $\omega = \omega_{A}-n$.} related
to the diagrams in question. But as we saw this is not necessary; 
individual counterterm diagrams (Zimmermann forests) mutually cancel their
divergences leaving behind a finite result. 


\begin{figure}[h]
\begin{center}
\leavevmode
\hbox{%
\epsfxsize=14cm
\epsffile{N609.eps}}
%\epsffile{N600.eps}}
\caption{\em Counterterm renormalization of the last two diagrams in 
Eq.(2.17). (Cut legs indicate amputations.)} \label{fig11}
\end{center}
\setlength{\unitlength}{1mm}
\begin{picture}(10,7)
%\put(20,33){ $\sum_{i=2}^{\infty}$ }
\end{picture}
\end{figure}
\vspace{3mm}



\vspace{3mm}  

\noindent So the only UV divergence in Fig.(2.17) which cannot be cured by
existing counterterms is that coming from the first (i.e.  free field or
ring) diagram. The foregoing divergence is evidently temperature
independent (to see that, simply use an explicite form of the free thermal
propagator $\D_{11}$). Hence, if we define

\begin{equation}
\langle \phi^{2} \rangle_{\mbox{\footnotesize{renorm}}} = \langle [\phi^{2}] 
\rangle - \langle 0| [\phi^{2}] |0 \rangle,
\tseleq{Norm}
\end{equation}
\noindent or, alternatively
\begin{equation}
\langle \phi^{2} \rangle_{\mbox{\footnotesize{renorm}}} = \langle [\phi^{2}] 
\rangle - \langle [\phi^{2}] \rangle|_{\mbox{\footnotesize{free fields}}}, 
\tseleq{Norm2}
\end{equation}

\vspace{2mm}

\noindent we get  finite quantities, as desired. On the other hand, we 
should emphasize that

\begin{equation}
\langle \phi^{2} \rangle - \langle 0| \phi^{2} |0 \rangle = 
Z_{\phi^{2}}\left\{ \langle [\phi^{2}] \rangle - \langle
0| [\phi^{2}] |0 \rangle \right\} \not= \mbox{finite in $D$=4}. 
\end{equation}

\vspace{2mm}

\noindent An extension of the previous reasonings to any $N>1$ is 
straightforward, only difference is that we must deal with operator 
mixing which makes (\tseref{Norm}) and (\tseref{Norm2}) less trivial. 

\vspace{3mm}

\noindent The important lesson which we have learnt here is that the
naive ``double dotted'' normal product (i.e. subtraction of the vacuum
expectation value from a given operator) does not generally give
a finite result. The former is perfectly suited for the free theory 
($Z_{\Sigma \phi^{2}} = 1$) but in the
interacting case we must resort to the prescription (\tseref{Norm}) or 
(\tseref{Norm2}) instead.


\vspace{6mm}
\noindent{\it Renormalization of the energy-momentum tensor}
\vspace{3mm}

\noindent In order to calculate the thermodynamic pressure, we need to
find such $\langle \Theta^{\mu \nu}_{C}
\rangle|_{\mbox{\footnotesize{renorm}}}$ which apart from being finite is
also consistent with our derivation of the pressure ${\cali{P}}$ performed in
the introductory Section. In view of the previous treatment, we however
cannot expect that $\Theta^{\mu \nu}_{C}$ will be renormalized
multiplicatively. Instead new terms with a different structure than
$\Theta^{\mu \nu}_{C}$ itself will be generated during renormalization.
The latter must add up to $\Theta^{\mu \nu}_{C}$ in order to render
$D^{\mu \nu}(x^{n}|y)$ finite \footnote{In fact it can be shown
\tsecite{LW, Collins} that the Noether currents corresponding to a given
internal symmetry are renormalized, i.e $J^{a}= [J^{a}]$, however, this is
not the case for the Noether currents corresponding to external symmetries
(like $\Theta^{\mu \nu}_{C}$ is).}. 

\vspace{3mm}

\noindent Now, the key ingredient exploited in Eq.(\tseref{5}) is the
conservation law (continuity equation). It is well known that one can
`modify' $\Theta^{\mu \nu}_{C}$ in such a way that the new tensor
$\Theta^{\mu \nu}$ preserves the convergence properties of $\Theta^{\mu
\nu}_{C}$. Such a modification (the Pauli transformation) reads


\begin{eqnarray}
&&\Theta^{\mu \nu} = \Theta^{\mu \nu}_{C} +
\partial_{\lambda}X^{\lambda \mu \nu}\nonumber \\
&& X^{\lambda \mu \nu} = -X^{\mu \lambda \nu}.
\tselea{pp1}
\end{eqnarray}

\vspace{2mm}

\noindent For scalar fields (\tseref{pp1}) is the only transformation
which neither changes the divergence properties of $\Theta^{\mu \nu}_{C}$
nor the generators of the Poincare group constructed out of $\Theta^{\mu
\nu}_{C}$ \tsecite{LW, DG, RJ, IZ}. Because the renormalized (or improved)
energy momentum tensor must be conserved (otherwise theory would be
anomalous), it has to mix with $\Theta^{\mu \nu}_{C}$ under
renormalization only via the Pauli transformation, i.e. 


\begin{equation}
[\Theta^{\mu \nu}_{C}] = \Theta^{\mu \nu}_{C} + 
\partial_{\lambda}X^{\lambda \mu \nu}\nonumber.
\tseleq{ppp1}
\end{equation}

\vspace{2mm}


\vspace{3mm}

\noindent In order to determine $X^{\lambda \mu \nu}$, we should
realize that its role is to cancel divergences present in $\Theta^{\mu
\nu}_{C}$.  Such a cancellation can be, however, performed only by
means of composite operators which are even in the number of fields
(note that $\Theta^{\mu \nu}$ is even in fields and Green's functions
with the odd number of fields vanish).  Recalling the condition that
renormalization can mix only operators with dimension less or equal,
we see that the dimension of $X^{\lambda \mu \nu}$ must be $D-1$, and
that $X^{\lambda \mu \nu}$ must be quadratic in fields. The only
possible form which is compatible with tensorial  structure
(\tseref{pp1}) is

\begin{equation}
X^{\lambda \mu \nu} = \sum_{a, b = 1}^{N} c_{a b}(\lambda_{r};D)\; \left( 
\partial^{\mu} g^{\lambda \nu} - \partial^{\lambda} g^{\mu \nu} 
\right) \; \phi_{a}\phi_{b}.
\tseleq{ppp5}
\end{equation}

\vspace{2mm}

\noindent From the fact that $\Theta^{\mu \nu}_{C}$ is $O(N)$ invariant
(see Eq.(\tseref{7})), $\partial _{\lambda} X^{\lambda \mu \nu}$ must be
also $O(N)$ invariant in order to cancel divergences appearing in
$\Theta^{\mu \nu}_{C}$, so $c_{a b} = \delta_{ab} c$. Thus, finally we can
write

\begin{equation}
[\Theta^{\mu\nu}_{C}] = \Theta^{\mu \nu}_{C} + c(\lambda_{r};D)\; \sum_{a 
=1}^{N} \left( \partial^{\mu}\partial^{\nu} - g^{\mu 
\nu} \partial^{2}\right) \; \phi_{a}^{2},
\tseleq{ppp6}
\end{equation}

\vspace{3mm}

\noindent with $c = c_{0} + \sum (\mbox{poles})$, here $c_{0}$ is analytic
in $D$. Structure of $c(\lambda_{r};D)$ could be further determined,
similarly as in the $N=1$ theory, employing a renormalization group
equation \tsecite{Brown}. We do not intend to do that as the detailed
structure of $c$ will show totally irrelevant for the following
discussion, however, it turns out to be important in non-equilibrium case. 

\vspace{3mm}

\noindent Now, similarly as before, $[\Theta^{\mu \nu}_{C}]$ gives the
finite composite Green's functions if $n >0$ but the expectation value
$\langle [\Theta^{\mu \nu}_{C}] \rangle$ is divergent (discussion for the
$N=1$ theory can be found in Brown \tsecite{Brown}). The 
unrenormalized Dyson-Schwinger equation  for $D^{\mu \nu}(0|0)$ 
reads


\begin{figure}[h]
\begin{center}
\leavevmode  
\hbox{%
\epsfxsize=14cm
\epsffile{N601.eps}}
%xdvi large\label{fig101}
\end{center}
\setlength{\unitlength}{1mm}
\begin{picture}(20,7)
%\put(65,15){$ = m_{0}^{2}\;\delta_{ba} \delta_{ca}$}
\put(139,20){(2.26)}
\end{picture}
\end{figure} 

\addtocounter{equation}{1}

\noindent The structure of the composite vertices in (2.26) is that
described at the beginning of this Section. Note that the amputated
composite Green's functions in individual parentheses are of the same
order in $\lambda_{r}$. Performing the counterterm renormalization as in
the case of $\langle [\phi^{2}] \rangle$, we factorize the graphs inside
of parentheses into the product of the renormalized 2- (and 6-) point
composite Green's function and the renormalized full 2- (and 6-) point
Green's function. The latter are finite. The UV divergences arisen during
the integrations over momenta connecting both composite and full Green's
functions are precisely cancelled by the remaining counterterm diagrams.
Only divergence comes from the free-field contribution, more precisely
from the $T=0$ ring diagram. Defining

\begin{equation} \langle \Theta^{\mu \nu}_{C} \rangle
|_{\mbox{\footnotesize{renorm}}} = \langle [\Theta^{\mu \nu}_{C}] \rangle
- \langle 0| [\Theta^{\mu \nu}_{C}]|0 \rangle, \tseleq{ppp2}
\end{equation} \noindent or \begin{equation} \langle \Theta^{\mu \nu}_{C}
\rangle |_{\mbox{\footnotesize{renorm}}} = \langle [\Theta^{\mu \nu}_{C}]
\rangle - \langle [\Theta^{\mu \nu}_{C}] \rangle
|_{\mbox{\footnotesize{free field}}}, \tseleq{ppp3} \end{equation}

\vspace{2mm}

\noindent we get the finite expressions. Note that the conservation law is
manifest in both cases. In equilibrium (and in $T=0$) we can, due to
space-time translational invariance of $\langle \ldots \rangle$, write

\begin{equation}
\langle [\Theta^{\mu \nu}_{C}] \rangle = \langle \Theta^{\mu 
\nu}_{C}\rangle + 
\partial_{\lambda} \langle X^{\lambda \mu \nu } \rangle = \langle 
\Theta^{\mu \nu}_{C}\rangle.
\tseleq{ppp8}
\end{equation}

\vspace{2mm}

\noindent Using (\tseref{ppp2}) or (\tseref{ppp3}) we get either  the  
thermal interaction pressure or the interaction pressure, 
respectively. This can be explicitly written as

\begin{equation}
{\cali{P}}_{\mbox{\footnotesize{th.int.}}}(T) = {\cali{P}}(T) - 
{\cali{P}}(0) = 
-\frac{1}{(D-1)} \sum_{i=1}^{D-1}\left\{ \langle \Theta_{C\; i}^{i} 
\rangle - \langle 0| \Theta_{C\;i}^{i} | 0 \rangle \right\},
\tseleq{ppp9}
\end{equation}
\noindent or
\begin{equation}
{\cali{P}}_{\mbox{\footnotesize{int.}}}(T) = {\cali{P}}(T) - 
{\cali{P}}_{\mbox{\footnotesize{free 
field}}}(T) = -\frac{1}{(D-1)} \sum_{i=1}^{D-1} \left\{ \langle \Theta_{C\; 
i}^{i} \rangle -
\langle \Theta_{C\; i}^{i} \rangle|_{\mbox{\footnotesize{free field}}}
 \right\}. 
\tseleq{ppp10}
\end{equation}

\vspace{2mm}

\noindent In order to keep connection with calculations done by Drummond
{\em et al.} in \tsecite{ID1} we shall in the sequel deal with the thermal
interaction pressure only. If instead of an equilibrium, a non-equilibrium
medium would be in question, translational invariance of $\langle \ldots
\rangle$ might be lost, in that case either prescription (\tseref{ppp2})
or (\tseref{ppp3}) is obligatory, and consequently $c(\lambda_{r};D)$ in
(\tseref{ppp6}) must be precisely specified. 

   
\section{Hydrostatic pressure}

\renewcommand{\thefootnote}{\fnsymbol{footnote}}
\setcounter{footnote}{0}


In the previous section we have prepared ground for a hydrostatic
pressure calculations in the $O(N)\; \phi^{4}$ theory. In this section
we aim to apply the previous results to the massive $O(N)\;  \phi^{4}$
theory in the large $N$ limit. Anticipating an out of equilibrium
application, we shall use the real-time formalism even if the
imaginary-time one is more natural in the equilibrium context. As we
aim to evaluate the hydrostatic pressure in 4 dimensions, we use here,
similarly as in the previous Section, the usual dimensional
regularization to regulate the theory (i.e. here and throughout we
keep $D$ slightly away from the physical value $D=4$).

\vspace{3mm}

\noindent Let us start first with some essentials of our model system 
at finite temperature. 


\subsection{Mass renormalization}

In the Dyson multiplicative renormalization the fact that the complete 
propagator has a pole at the physical mass leads to the usual 
mass renormalization prescription \tsecite {IZ}:


\begin{equation}
m_{r}^{2}= m_{0}^{2} + \Sigma(m^{2}_{r}),
\tseleq{c1}
\end{equation}

\vspace{2mm}

\noindent where $m_{r}$ is renormalized mass and $\Sigma(m_{r}^{2})$ is
the proper self-energy evaluated at the mass shell; $p^{2}= m_{r}^{2}$. In
fact, Eq.(\tseref{c1}) is nothing but the statement that 2-point vertex
function $\Gamma^{(2)}_{r}$ evaluated at the mass-shell must vanish. The
Dyson-Schwinger equation corresponding to the proper self-energy reads
\tsecite{PC, EM, PJ}: 


\begin{figure}[h]
\begin{center}
\leavevmode
\hbox{%
\epsfxsize=8.5cm
\epsffile{N5.eps}}
\label{fig5}
\end{center}
\setlength{\unitlength}{1mm}
\begin{picture}(10,7)
\put(17,35.5){$\Sigma\; =$}
\put(141,35.5){(3.2)}
\end{picture}
\end{figure}    

\addtocounter{equation}{1}

\noindent where hatched blobs represent 2-point connected Green function
whilst cross-hatched blob represents proper vertex $\Gamma^{(4)}_{r}$ (i.e. 
1PI 4-point Green's function). A summation over internal indices is 
understood. In the sequel the following convention is accepted: 



\begin{figure}[h]
\begin{center}
\leavevmode
\hbox{%
\epsfxsize=2cm
\epsffile{N55.eps}}
%xdvi large0\label{fig5}
\end{center}
\setlength{\unitlength}{1mm}
\begin{picture}(10,7)
\put(87,20.5){$ = \Gamma^{(n)}_{r}$}
%\put(139,24.5){(3.2)}
\end{picture}
\end{figure}



\noindent The second term in (3.2) actually does not contribute in the
large $N$ limit. This is because each hatched blob behaves at most as
$N^{0}$ whilst $\Gamma^{(4)}$ goes maximally as $N^{-1}$ \footnote{ In the
$\phi^{4}$ theory there is a simple relation between the number of loops
($L$), vertices ($V$) and external lines ($E$); $4V = 2I +E$. Together
with the Euler relation for connected graphs; $L= I -V +1$ (here $I$ is
the number of internal lines), we have $L-V = \frac{2-E}{2}$. As each loop
carries maximally a factor of $N$ (this is saturated only for `tadpole'
loops) and each vertex carries a factor of $N^{-1}$, the overall blob
contribution behaves at most as $N^{L-V} = N^{\frac{2-E}{2}}.$}.
Consequently, various contributions from the first graph in (3.2)
contribute at most $N^{0}$, whereas in the second graph the contributions
contribute up to order $N^{-1}$. So the first diagram dominates, provided
we retain only such 2-point connected Green's functions which are
proportional to $N^{0}$ (as mentioned in the footnote, these are
comprised only of `tadpole' loops.). After neglecting the `setting sun'
graph, Eg.(3.2) generates upon iterating the so called superdaisy diagrams
\tsecite{ID1, EM, CJT}. 

\vspace{3mm}

\noindent Let us now define $\Sigma(m^{2}_{r}) = \lambda_{0}\; 
{\cali{M}}(m_{r}^{2})$. Because the `tadpole' diagram in (3.2) can be 
easily resummed we observe that 

\begin{equation}
{\cali{M}}(m_{r}^{2}) = \frac{1}{2}\int \frac{d^{D}q}{(2 \pi)^{D}} \; 
\frac{i}{q^{2}-m_{0}^{2}-\Sigma(m^{2}_{r})+i\epsilon} = \frac{1}{2}\int 
\frac{d^{D}q}{(2 \pi)^{D}} \;
\frac{i}{q^{2}-m^{2}_{r}+i\epsilon},
\tseleq{b45}
\end{equation}   

\vspace{4mm}

\noindent hence we see that $\Sigma$ is external-momentum independent. If
we had started with the renormalization prescription: 
$i\Gamma^{(2)}_{r}(p^{2}=0)= - m^{2}_{r}$, we would arrived at 
(\tseref{c1}) as well (this is not the case for $N=1$!). 

\vspace{5mm}


\noindent At finite temperature the strategy is analogous. Due to a
doubling of degrees of freedom, the full propagator is a $2\times 2$
matrix. The latter satisfies, similarly as at $T=0$, Dyson's equation

\begin{equation}
\D = \D_{F} + \D_{F} \left(-i{\vect{\Sigma}} 
\right) \D.
\tseleq{b5}
\end{equation}

\vspace{2mm}

\noindent An important point is that there exists a real, non-singular 
matrix $\M $ (Bogoliubov matrix) \tsecite{LW, LB1, PVL2} having a property 
that

\begin{equation}
\D_{F} = \M \left( \begin{array}{cc}
                                   i\Delta_{F} & 0 \\
                                        0      & -i\Delta^{*}_{F}
                                    \end{array} \right) \M
\;\; \; \; \mbox{and} \; \; \; \;
{\vect{\Sigma}} = \M^{-1} \left( \begin{array}{cc}
                                   \Sigma_{T} & 0 \\
                                        0      & -\Sigma^{*}_{T}
                                    \end{array} \right) \M^{-1}. 
\tseleq{m8} 
\end{equation} 
 
\vspace{2mm}

\noindent Here $\Delta_{F}$ is the standard Feynman propagator and `*'
denotes the complex conjugation. Consequently, the full matrix propagator
may be written as

\begin{equation}
\D = \M \left( \begin{array}{cc}
              \frac{i}{p^{2}-m_{0}^{2}-\Sigma_{T}+i\epsilon} & 0 \\
                             0 & 
\frac{-i}{p^{2}-m_{0}^{2}-\Sigma^{*}_{T}-i\epsilon}
                  \end{array} \right) \M.
\tseleq{m9}
\end{equation}

\vspace{2mm}

\noindent Similarly as in many body systems, the position of the
(real) pole of $\D$ in $p^{2}$ fixes the temperature-dependent effective
mass $m_{r}(T)$ \tsecite{LB1,FW}. The latter is determined by the equation

\begin{equation}
m_{r}^{2}(T) = m_{0}^{2} + \mbox{Re}\left(\Sigma_{T}(m^{2}_{r}(T))\right).
\tseleq{m10}
\end{equation}

\vspace{2mm}

\noindent From the explicit form of $\M$ it is possible to show that 
$\mbox{Re}{\vect{\Sigma}}_{11}=\mbox{Re}\Sigma_{T}$ \tsecite{LW, LB1}. As 
before, the structure of the proper self-energy can be deduced from the  
corresponding Dyson-Schwinger equation. Following the usual real-time 
formalism convention (type-1 vertex $\sim -i\lambda_{0}$, type-2 vertex 
$\sim i\lambda_{0}$ ), the former reads:

\begin{figure}[h] 
\begin{center}
\leavevmode   
\hbox{%
\epsfxsize=12.4cm
\epsffile{N6.eps}} 
\label{fig6} 
\end{center}
\setlength{\unitlength}{1mm}
\begin{picture}(10,7)
\put(20,67.5){${-i\vect{\Sigma}}_{11}=$}
\put(76,67.5){${-i\vect{\Sigma}}_{22}=$}
\put(0,47.5){where}
\put(141.5,67.5){(3.8)}
\end{picture}
\end{figure}

\addtocounter{equation}{1} 


%\begin{figure}[h]
%\begin{center}
%\leavevmode
%\hbox{%
%\epsfxsize=13.7cm   
%\epsffile{N65.eps}}
%\label{fig65}
%\end{center}
%%\setlength{\unitlength}{1mm}
%%\begin{picture}(10,7)
%%\put(20,25.5){${\vect{\Sigma}}_{11}=$}
%%\end{picture}
%\end{figure} 


\noindent In (3.8) we have omitted the `setting sun' diagrams because they
do not contribute in the large $N$ limit. Note that the latter fact
implies that the off-diagonal elements of ${\vect{\Sigma}}$ are zero.
Inspection of Eq.(3.8) reveals that

\begin{equation}
{\vect{\Sigma}}_{11}= \frac{\lambda_{0}}{2}\; \int 
\frac{d^{D}q}{(2\pi)^{D}} \; {\D}_{11}(q;T) \; \; \; \mbox{and} \; \; \; 
{\vect{\Sigma}}_{22}= -\frac{\lambda_{0}}{2}\; \int
\frac{d^{D}q}{(2\pi)^{D}} \; {\D}_{22}(q;T).
\tseleq{m11}
\end{equation}

\vspace{3mm}

\noindent It directly follows from Eq.(\tseref{m11}) that both
${\vect{\Sigma}}_{11}$ and ${\vect{\Sigma}}_{22}$ are external-momentum
independent and real \footnote{Reality of ${\vect{\Sigma}}_{11}$ can be
most easily seen from the largest-time equation \tsecite{PJ}. The LTE
states that ${\vect{\Sigma}}_{11} + {\vect{\Sigma}}_{22} +
{\vect{\Sigma}}_{12} + {\vect{\Sigma}}_{21} = 0$.  Because no `setting
sun' graphs are present, ${\vect{\Sigma}}_{12} + {\vect{\Sigma}}_{21} =
0$, on the other hand ${\vect{\Sigma}}_{11} + {\vect{\Sigma}}_{22} =
2i\mbox{Im}{\vect{\Sigma}}_{11}$ (see (\tseref{m8})).}. If we define
$\Sigma_{T}(m_{r}^{2}(T)) = \lambda_{0}\;  {\cali{M}}_{T}(m_{r}^{2}(T))$,
then Eq.(\tseref{m10}) through Eq.(\tseref{m11}) implies that

\begin{equation}
m_{r}^{2}(T) = m^{2}_{0} +  \lambda_{0}\;
{\cali{M}}_{T}(m_{r}^{2}(T)).
\tseleq{m12}
\end{equation}

\vspace{2mm}


\noindent A resummed version of $\D_{11}$ is easily obtainable from 
(\tseref{m9}) \tsecite{LW, LB1} and consequently  
(\tseref{m11}) yields 

\begin{eqnarray}
{\cali{M}}_{T}(m_{r}^{2}(T)) &=& \frac{1}{2}\; \int 
\frac{d^{D}q}{(2\pi)^{D}}\; \left\{ \frac{i}{q^{2} - m^{2}_{r}(T) + 
i\epsilon} \; \; +\; (4\pi)\; \delta^{+}(q^{2}-m^{2}_{r}(T))\; 
\frac{1}{e^{q_{0}\beta}-1} \right\}\nonumber \\
&=&- \int \frac{d^{D}q}{(2\pi)^{D}}\;  
\frac{\varepsilon 
(q_{0})}{e^{q_{0}\beta}-1}\; \mbox{Im}\frac{1}{q^{2}-m_{r}^{2}(T) 
+i\epsilon}. \tselea{m14}
\end{eqnarray}

\vspace{3mm}

\noindent Let us remark that (\tseref{m14}) is manifestly independent of
any particular real-time formalism version. This is because the various
real-time formalisms differ only in the off-diagonal elements of $\D$
\tsecite{LW, LB1}. 


\vspace{3mm}

\noindent In passing it may be mentioned that because
${\vect{\Sigma}}_{11}(m^{2}_{r})$ is momentum independent, the wave
function renormalization $Z_{\phi} = 1$. (The K{\"a}llen-Lehmann
representation requires the renormalized propagator to have a pole of
residue $i$ at $p^{2}=m^{2}_{r}$. The former in turn implies that
$Z_{\phi}=(1-{\vect{\Sigma}}_{11}'(p^{2})|_{p^{2}=m^{2}_{r}})^{-1} =1$.)
Trivial consequence of the foregoing fact is that
$\Gamma^{(2)}_{r}=\Gamma^{(2)}$ and $\Gamma^{(4)}_{r}=\Gamma^{(4)}$. 



\subsection{Coupling constant renormalization}

Let us start with $T=0$ first. By assumption the fields $\phi_{a}$ have
non-vanishing masses, so we can safely choose the renormalization
prescription for $\lambda_{r}$ at $s = 0$ ($s$ is the standard Mandelstam
variable). Thus we require that

\begin{equation}
\Gamma^{(4)}(s=0)= -\lambda_{r}/N.
\tseleq{m13}
\end{equation}

\vspace{2mm}

\noindent The formula (\tseref{m13}) clearly agrees with the tree level
value $\Gamma_{tree}^{(4)}(s=0)= -\lambda_{0}/N$. The structure of
$\Gamma^{(4)}$ is encoded in the Dyson-Schwinger equation which reads
\tsecite{IZ, PC}: 




\begin{figure}[h]
\begin{center}
\leavevmode  
\hbox{%
\epsfxsize=13.5cm
\epsffile{N7.eps}}
%\label{fig7}
\end{center}
\setlength{\unitlength}{1mm}
\begin{picture}(10,7)
%\put(2,49.5){$\Gamma^{(4)}\;=$}
\put(42.5,26){$\sum_{i=1}^{3}$}
\put(52,56){$\sum_{i=1}^{3}$}
\put(139.5,26){(3.13)}
\end{picture}
\end{figure}

\addtocounter{equation}{1}


\noindent The sum $\sum_{i=1}^{3}$ schematically represents a summation over
$s,t$ and $u$ scattering channels (as usual, summation over internal
indices on the RHS is understood). Similarly as before, we can argue that
both the third and fourth graphs contribute at most $N^{-2}$, whilst the
second (`fish') graphs may contribute up to order $N^{-1}$. So in the
large $N$ limit the last two diagrams may be neglected, provided we keep
in the $4$-point vertex function only graphs proportional to $N^{-1}$.
However, the former can be only fulfilled if we retain such a `fish' graph
where summation over internal index on the loop is allowed. Remaining two
graphs in the sum $\sum_{i=1}^{3}$ (i.e. $t$ and $u$-channel interactions)
are suppressed by the factor $N^{-1}$ as the internal index on the loop is
fixed. In this way we are left with the relation


\begin{eqnarray} 
\Gamma^{(4)}(s=0) &=& -\frac{\lambda_{0}}{N} -
\left. \frac{i\lambda_{0}}{2} \int \frac{d^{D}q}{(2\pi)^{D}}\; 
\Gamma^{(4)}(s) \; 
\frac{i}{\left( q^{2} - m_{r}^{2} + i\epsilon \right)}\; \frac{i}{\left(
(q -Q)^{2} - m_{r}^{2} + i\epsilon \right)}\right|_{s=0}  
\nonumber \\
 &=& - \frac{\lambda_{0}}{N} - \frac{\lambda_{0} \lambda_{r}}{2N} 
\int^{1}_{0}dx \int
\frac{d^{D}q}{(2\pi)^{D}}\;\left. \frac{i}{\left( q^{2} -
m_{r}^{2} + x(1-x)s + i\epsilon \right)^{2}} \right|_{s=0},\nonumber\\
\tseleq{m145}
\end{eqnarray}

\noindent (with $Q=p_{1}+p_{2}$ and $s=Q^{2}$, $p_{1}, p_{2}$ are 
the external momenta) or equivalently


\begin{equation}
\lambda_{r} = \lambda_{0} + \lambda_{0}\lambda_{r}\; {\cali{M}}'(m^{2}_{r}),
\tseleq{m146}
\end{equation} 

\vspace{2mm}

\noindent the prime means differentiation with respect to $m_{r}^{2}$; 
${\cali{M}}(m^{2}_{r})$ is defined by (\tseref{b45}).

\vspace{4mm}

\noindent As the renormalization of the coupling constant at finite 
temperature does not enter the following reasonings we shall postpone its 
discussion to Appendix A1.


\subsection{The pressure}

The partition function $Z$ has a well known path-integral 
representation at finite temperature, namely


\begin{eqnarray}
&&Z[T] = \mbox{exp}(\Phi[T]) = \int{\cali{D}}\phi\; 
\mbox{exp}(iS[\phi;T])\nonumber \\ &&S[\phi;T] = \int_{C}d^{D}x\; 
{\cali{L}}(x). \tselea{b1}
\end{eqnarray}

\vspace{2mm}

\noindent Here $\Phi= -\beta \Omega$ is so called Massieu function
\tsecite{LW} and $\int_{C}d^{D}x =
\int_{C}dx_{0}\int_{V}d^{D-1}{\vect{x}}$ with the subscript $C$ suggesting
that the time runs along some contour in the complex plane. In the
real-time formalism, which we adopt throughout, the most natural version
is the so called Keldysh-Schwinger one \tsecite{LW, LB1}, which is
represented by contour in Fig.\ref{fig18}

\begin{figure}[h]
\vspace{4mm}
\epsfxsize=11cm
\centerline{\epsffile{fig18.eps}}
\caption{\em The Keldysh-Schwinger time path.}
\label{fig18}
\vspace{4mm}
\end{figure}


\noindent Let us mention that the fields within the path-integral
(\tseref{b1}) are further restricted by the periodic boundary condition (KMS 
condition) \tsecite{LW,LB1, PVL2} which in our case reads:

\begin{displaymath}
\phi_{a}(t_{i}-i\beta, {\vect{x}})= \phi_{a}(t_{i}, {\vect{x}}).
\end{displaymath}
 

\vspace{3mm}


\noindent As explained in Section 2, we can use for a pressure
calculation the canonical energy-momentum tensor $\Theta^{\mu \nu}_{C}$. 
Employing for $\Theta^{\mu \nu}_{C}(x)$ its
explicit form (\tseref{7}) together with (\tseref{11}), one may write

\begin{equation}
\langle \Theta^{\mu \nu}_{C} \rangle = \frac{N}{2}\; \int 
\frac{d^{D}q}{(2\pi)^{D}}(2q^{\mu}q^{\nu}-g^{\mu 
\nu}(q^{2}-m^{2}_{0}))\; \D_{11}(q;T) 
\; \; + \frac{\lambda_{0}}{4!N} g^{\mu \nu} \left\langle \left( 
\sum_{a=1}^{N}\phi_{a}^{2}(0)\right)^{2} \right\rangle,
\tseleq{b2}
\end{equation}


\vspace{2mm}

\noindent where $\D_{11}$ is the Dyson-resummed thermal propagator 
\tsecite{LW, LB1}, i.e. 

\begin{equation}
\D_{11}(q;T) = \frac{i}{q^{2}-m^{2}_{r}(T)+i\epsilon} \; \; + (2\pi) \; 
\delta (q^{2}-m^{2}_{r}(T))\; \frac{1}{e^{|q_{0}|\beta}-1}.
\tseleq{b3}
\end{equation}

\vspace{2mm}

\noindent Note that we have exploited in (\tseref{b2}) the fact that
the expectation value of $\Theta^{\mu \nu}_{C}(x)$ is $x$
independent.On the other hand in (\tseref{b3}) we have used the fact
that $m^{2}_{r}$ is $q$ independent. In order to calculate the
expectation value of the quartic term in Eq.(\tseref{b2}), let us
observe (c.f. (\tseref{b1})) that the derivative of $\Phi$ with
respect to the bare coupling $\lambda_{0}$ (taken at fixed $m_{0}$)
gives


\begin{equation}
\frac{\partial\Phi[T]}{\partial \lambda_{0}}= - \frac{i}{4!\; N}\; \int_{C} 
d^{D} x \left\langle \left(\sum_{a=1}^{N}\phi_{a}^{2}(0) \right)^{2} 
\right\rangle \nonumber \nonumber,
\end{equation} 

\noindent which implies that 

\begin{equation}
\left\langle \left(\sum_{a=1}^{N}\phi_{a}^{2}(0) 
\right)^{2} \right\rangle 
= - \frac{N4!}{\beta V}\;\frac{\partial\Phi[T]}{\partial \lambda_{0}}.
\tselea{b4}
\end{equation}

\vspace{2mm}

\noindent The key point now is that we can calculate $\Phi[T]$ in a
non-perturbative form. (The latter is based on the fact that we know the
Dyson-resummed propagator $\D_{11}(q;T)$ (see (\tseref{b3}).) Indeed,
taking derivative of $\Phi$ with respect to $m_{0}^{2}$
(keeping $\lambda_{0}$ fixed) we obtain


\begin{eqnarray}
\frac{\partial \Phi[T]}{\partial m_{0}^{2}}&=& -\frac{iN}{2}\int_{C}d^{D}x 
\left\langle \phi^{2}(0) \right\rangle = -\frac{\beta V N}{2} \int 
\frac{d^{D}q}{(2\pi)^{D}}\; \D_{11}(q;T)\nonumber \\
&=& - \beta V N\; {\cali{M}}_{T}(m_{r}^{2}(T)),
\tselea{b5}
\end{eqnarray}
\noindent thus

\begin{equation}
\Phi[T;\lambda_{0}; m_{0}^{2}] = \beta V N \; \int_{m_{0}^{2}}^{\infty} 
d{\hat{m}}_{0}^{2}\;{\cali{M}}_{T}({\hat{m}}_{r}^{2}(T))\; \; + 
\Phi[T;\lambda_{0}; \infty].
\tseleq{b6}
\end{equation}

\vspace{2mm}

\noindent Let us note that $\Phi[T;\lambda_{0}; \infty]$ is actually zero
\footnote{To be precise, we should also include in Fig.\ref{fig3} an
(infinite) circle diagram corresponding to the free pressure \tsecite{LW,
EM}. However the later is $\lambda_{0}$ independent (although $m_{0}$
dependent) and so it is irrelevant for the successive discussion (c.f.
Eq.(\tseref{b4})).} because $\Phi[T;\lambda_{0};m^{2}_{0}]$ has the
standard loop expansion \tsecite{LW, PC}: 

\begin{figure}[h] 
\begin{center}
\leavevmode   
\hbox{%
\epsfxsize=11cm
\epsffile{N3.eps}}
\caption{\em First few bubble diagrams in the $\Phi$ expansion.}
\label{fig3}
\end{center}
\begin{picture}(10,10)
%\put(50,100){$\Phi[T;\lambda_{0};m_{0}]\; \sim$}
\end{picture}
\end{figure}


\noindent It is worth mentioning that in the previous expansion one must
always have at least one type-1 vertex \tsecite{LW}. The RHS of
Fig.\ref{fig3} clearly tends to zero for $m_{0} \rightarrow \infty$ as all
the (free) thermal propagators from which the individual diagrams are
constructed tend to zero in this limit. The former result can be also
deduced from the CJT effective action formalism \tsecite{CJT} or from a
heuristic argumentation based on a thermodynamic pressure \tsecite{ID1}.
Note that in the large $N$ limit the fourth and fifth diagrams in
Fig.\ref{fig3} must be omitted. 


\vspace{3mm}


\noindent The expectation value (\tseref{b4}) can be now explicitly 
written as 


\begin{equation}
\left\langle \left( \sum_{a=1}^{N}\phi_{a}^{2}(0) \right)^{2} 
\right\rangle = 
N^{2}4!\; \int_{m_{0}^{2}}^{\infty} d{\hat{m}}_{0}^{2}\; \int 
\frac{d^{D}q}{(2\pi)^{D}}\; \frac{\varepsilon(q_{0})}{e^{q_{0}\beta}-1}\; 
\mbox{Im} \left(\frac{\frac{\partial 
\Sigma_{T}({\hat{m}}_{r}^{2}(T))}{\partial 
\lambda_{0}}}{(q^{2} - {\hat{m}}^{2}_{r} +i\epsilon)^{2}}\right).
\tseleq{b65}
\end{equation}

\vspace{2mm}

\noindent In fact, the differentiation of the proper self-energy in 
(\tseref{b65}) can be carried out easily. Using (\tseref{m12}), we get

\begin{displaymath}
\frac{\partial \Sigma_{T}}{\partial \lambda_{0}} = 
\frac{\Sigma_{T}}{\lambda_{0}} 
+ \lambda_{0}{\cali{M}}_{T}' \; \frac{\partial \Sigma_{T}}{\partial 
\lambda_{0}} \;\; \Rightarrow \;\; \frac{\partial \Sigma_{T}}{\partial 
\lambda_{0}} = \frac{\Sigma_{T}}{\lambda_{0}(1-\lambda_{0}{\cali{M}}'_{T})}.
%\tseleq{b7}
\end{displaymath}

\vspace{2mm}

\noindent From Eq.(\tseref{m12}) it directly follows that

\begin{displaymath}
\frac{dm_{r}^{2}(T)}{dm_{0}^{2}} = \frac{1}{(1- 
\lambda_{0}\;{\cali{M}}'_{T})},
\end{displaymath}

\noindent which, together with the definition of ${\cali{M}}_{T}$, gives


\begin{eqnarray}
\left\langle \left( \sum_{a=1}^{N} \phi_{a}^{2}(0) \right)^{2} 
\right\rangle &=& N^{2}4!\; \int_{m_{r}^{2}(T)}^{\infty} 
d{\hat{m}}_{r}^{2}\; \int \frac{d^{D}q}{(2 \pi)^{D}} \; 
\frac{\varepsilon(q_{0})}{e^{q_{0}\beta} -1} \;
\mbox{Im}\frac{{\cali{M}}_{T}({\hat{m}}_{r}^{2})}{(q^{2}-
{\hat{m}}_{r}^{2} +i\epsilon)^{2}}\nonumber \\
&=& -\; N^{2}4! \; \int_{m_{r}^{2}(T)}^{\infty} d{\hat{m}}_{r}^{2}\; 
{\cali{M}}_{T}({\hat{m}}_{r}^{2})
\; \frac{\partial {\cali{M}}_{T}({\hat{m}}_{r}^{2})}{\partial 
{\hat{m}}_{r}^{2}}\nonumber \\
&=& \frac{N^{2}4!}{2}\; {\cali{M}}_{T}^{2}(m_{r}^{2}(T)).
\tselea{b8}
\end{eqnarray}

\vspace{2mm}

\noindent where we have exploited in the last line the fact that
${\cali{M}}_{T}^{2}(m_{r}^{2} \rightarrow \infty)=0$. Let us mention that
the crucial point in the previous manipulations was that $m_{r}$ is both
real and momentum independent. Collecting our results together, we can
write for the hydrostatic pressure per particle (cf. Eq(\tseref{ppp9}))


\begin{eqnarray}
{\cali{P}}(T)-{\cali{P}}(0) &=& -\; \frac{1}{(D-1)N}\; \left( \langle 
\Theta^{i}_{C \;i} \rangle - \langle 0| \Theta^{i}_{C\; i} | 0 \rangle 
\right)\nonumber \\
&=& +\; \frac{1}{2}\; \int \frac{d^{D}q}{(2 \pi)^{D-1}}\; 
\left( \frac{2 {\vect{q}}^{2}}{(D-1)}  \right)\;  
\frac{\varepsilon(q_{0})}{e^{q_{0}\beta}-1}\; 
\delta(q^{2}-m_{r}^{2}(T))\nonumber \\
&& -\; \frac{1}{2}\; \int \frac{d^{D}q}{(2 
\pi)^{D-1}}\;
\left( \frac{2 {\vect{q}}^{2}}{(D-1)} \right)\; 
\delta^{+}(q^{2}-m_{r}^{2}(0))\nonumber \\ 
&&+ \frac{1}{2\lambda_{0}}\left( \Sigma^{2}_{T}(m_{r}^{2}(T)) - 
\Sigma^{2}(m_{r}^{2}(0)) \right).
\tselea{b9}
\end{eqnarray}

\vspace{2mm}


\noindent Applying the Green theorem to the last two integrals and 
eliminating the surface terms (for details see Appendix A2) we find

\begin{eqnarray} {\cali{P}}(T) - {\cali{P}}(0) &=& \frac{1}{2}\; \int
\frac{d^{D}q}{(2 \pi)^{D-1}}\; \frac{\varepsilon(q_{0})}{e^{q_{0}\beta}-1}\; 
\theta(q^{2}-m^{2}_{r}(T))\nonumber \\ &&- \frac{1}{2} \; \int
\frac{d^{D}q}{(2 \pi)^{D-1}}\; \theta(q_{0}) \; 
\theta(q^{2}-m^{2}_{r}(0))\nonumber \\
&&+\frac{1}{2\lambda_{0}}\left( \Sigma^{2}_{T}(m_{r}^{2}(T))-
\Sigma^{2}(m_{r}^{2}(0))\right) \nonumber \\ &&\nonumber \\ &=&
{\cali{N}}_{T}(m_{r}^{2}(T)) - {\cali{N}}(m_{r}^{2}(0)) +
\frac{1}{2\lambda_{0}} \left( \Sigma^{2}_{T}(m_{r}^{2}(T))-
\Sigma^{2}(m_{r}^{2}(0))\right),\nonumber \\ && \tseleq{b10}
\end{eqnarray}

\vspace{2mm}



\noindent where we have introduced new functions
${\cali{N}}_{T}(m^{2}_{r}(T)$ and ${\cali{N}}(m^{2}_{r})$;


\begin{eqnarray}
{\cali{N}}_{T}(m^{2}_{r}(T)) &=& \frac{1}{2}\; \int
\frac{d^{D}q}{(2 \pi)^{D-1}}\; \frac{\varepsilon(q_{0})}{e^{q_{0}\beta}-1}\;
\theta(q^{2}-m^{2}_{r}(T)) \nonumber \\
{\cali{N}}(m^{2}_{r}) &=& \lim_{T 
\rightarrow
\; 0}{\cali{N}}_{T}(m^{2}_{r}(T)). \tselea{b101}
\end{eqnarray}


\noindent Eq.(\tseref{b10}) can be rephrased into a form which exhibits an
explicit independence of bar quantities. Using the trivial identity: 


\begin{eqnarray}
&&\frac{1}{2\lambda_{0}}\; \left( \Sigma^{2}_{T}(m_{r}^{2}(T))- 
\Sigma^{2}(m_{r}(0))\right)\nonumber \\
&& \mbox{\hspace{1.5cm}}= \frac{1}{2\lambda_{0}}\; \left( 
\Sigma_{T}(m_{r}^{2}(T))-\Sigma(m^{2}_{r}(0))\right)\left( 
\Sigma_{T}(m_{r}^{2}(T))+\Sigma(m^{2}_{r}(0))\right) \nonumber \\
&& \mbox{\hspace{1.5cm}}= \frac{\delta m^{2}(T)}{2}\;\left(
{\cali{M}}_{T}(m_{r}^{2}(T))+{\cali{M}}(m^{2}_{r}(0))\right).
\tselea{b111}
\end{eqnarray}

\noindent we get

\begin{equation} {\cali{P}}(T) - {\cali{P}}(0) = 
{\cali{N}}_{T}(m_{r}^{2}(T)) -
{\cali{N}}(m_{r}^{2}(0))+ \frac{\delta m^{2}(T)}{2}\;\left(
{\cali{M}}_{T}(m_{r}^{2}(T))+{\cali{M}}(m^{2}_{r}(0))\right),
\tseleq{b11}
\end{equation}

\vspace{2mm}

\noindent where $\delta m^{2}(T)= m_{r}^{2}(T)-m_{r}^{2}(0)$. The result
(\tseref{b11}) has been previously obtained by authors \tsecite{ID1} in the
purely thermodynamic pressure framework. 



\section{Hydrostatic pressure in $D=4$ (high-temperature expansion)}


In order to obtain the high-temperature expansion of the pressure in
$D=4$, it is presumably the easiest to go back to equation (\tseref{b9})
and employ identity (\tseref{b111}). Let us split this task into two
parts.  We firstly evaluate the integrals with potentially UV divergent
parts using the dimensional regularization. The remaining integrals, with
the Bose-Einstein distribution insertion, are safe of UV singularities and
can be computed by means of the Mellin transform technique. 

\vspace{3mm}

\noindent Inspecting (\tseref{b9}) and (\tseref{b111}), we observe that 
the only UV divergent contributions come from the integrals:

\begin{eqnarray}
&+& \frac{1}{(D-1)} \int \frac{d^{D}q}{(2\pi)^{D-1}} \; {\vect{q}}^{2}\; 
\left( \delta^{+}(q^{2}-m^{2}_{r}(T)) - \delta^{+}(q^{2}-m^{2}_{r}(0))
\right)\nonumber \\
&+& \frac{\delta m^{2}(T)}{4} \int \frac{d^{D}q}{(2\pi)^{D}} \; \left( 
\frac{i}{q^{2} - m^{2}_{r}(T) + i\epsilon } + 
\frac{i}{q^{2}-m^{2}_{r}(0)+i\epsilon}\right), 
\tselea{4.1} 
\end{eqnarray}


\vspace{2mm} 

\noindent which, if integrated over, give

\begin{eqnarray}
\mbox{(\tseref{4.1})} = &+& 
\frac{\Gamma(\frac{-D}{2})\Gamma(\frac{D}{2}+\frac{1}{2})}{(D-1) 
\Gamma(\frac{D-1}{2}) (4 \pi)^{\frac{D}{2}}} \left( 
(m^{2}_{r}(T))^{\frac{D}{2}} - 
(m^{2}_{r}(0))^{\frac{D}{2}}\right)\nonumber \\   
&+&\frac{\delta m^{2}(T)\; \Gamma(1-\frac{D}{2})}{4 
(4\pi)^{\frac{D}{2}}} \left( (m^{2}_{r}(T))^{\frac{D}{2}-1} + 
(m^{2}_{r}(0))^{\frac{D}{2}-1}\right).
\tselea{4.2}
\end{eqnarray}

\vspace{3mm}

\noindent Taking the limit $D=4-2\varepsilon \rightarrow 4$ and using 
expansions

\begin{eqnarray*}
\Gamma(-n + \varepsilon) &=& \frac{(-1)^{n}}{n!} \left(\frac{1}{ 
\varepsilon} + \sum_{k=1}^{n} \frac{1}{k} - \gamma +  
{\cali{O}}(\varepsilon) \right)\nonumber \\
a^{x+\varepsilon} &=&  a^{x} \left( 1 + \varepsilon \;\mbox{ln}a + 
{\cali{O}}(\varepsilon^{2}) \right),
\end{eqnarray*}

\noindent ($\gamma$ is the Euler-Mascheroni constant) we are finally left
with

\begin{equation}
\left.\mbox{(\tseref{4.1})}\right|_{D \rightarrow 4} =
- \frac{m^{2}_{r}(0) m^{2}_{r}(T)}{4 \;(4\pi)^{2}} \; \mbox{ln}\left( 
\frac{m^{2}_{r}(T)}{m^{2}_{r}(0)}\right)+ \delta m^{2}(T)\;(m_{r}^{2}(T) 
+ m^{2}_{r}(0))\; \frac{1}{128\; \pi^{2}}\;. \tseleq{4.3}
\end{equation}


\vspace{3mm}

   
\noindent The fact that we get finite result should not be 
surprising as entire analysis of Section 2 was made to show that 
${\cali{P}}(T)-{\cali{P}}(0)$ defined via $\Theta^{\mu \nu}_{C}$ is 
finite in $D=4$. 


\vspace{3mm}


\noindent We may now concentrate on the remaining terms in (\tseref{b9}), 
the latter read (we might, and we shall, from now on work in $D=4$)


\begin{equation}
\frac{1}{3} \int \frac{d^{4}q}{(2 \pi)^{3}} \; {\vect{q}}^{2}\; 
\frac{1}{e^{|q_{0}|\beta}-1}\; \delta(q^{2}-m_{r}^{2}(T))
+ \frac{\delta m^{2}(T)}{4} \int \frac{d^{4}q}{(2\pi)^{3}} \; 
\frac{1}{e^{|q_{0}|\beta}-1}  \delta(q^{2}-m_{r}^{2}(T)).
\tseleq{4.4}
\end{equation}

\vspace{2mm}

\noindent Our following strategy is based on the observation that the 
previous integrals have generic form:



\begin{eqnarray}
I_{2\nu}(m_{r}) &=& \int \frac{d^{4}q}{(2\pi)^{3}}\; {\vect{q}}^{2\nu}\; 
\frac{1}{e^{|q_{0}|\beta}-1}\; 
\delta(q^{2}-m^{2}_{r}),\nonumber\\
&=& \frac{m_{r}^{2+2\nu}}{2\; \pi^{2}}\; \int_{1}^{\infty} dx 
\;(x^{2}-1)^{\frac{1+2\nu}{2}}\; \frac{1}{e^{xy}-1},
\tseleq{c14}
\end{eqnarray}
 
\vspace{2mm}


\noindent with $\nu =0,1$ and $y = m_{r}\beta$.  Unfortunately, the
integral (\tseref{c14}) can not be evaluated exactly, however, its small
$y$ (i.e. high-temperature) behaviour can be successfully analyzed by means of
the Mellin transform technique \tsecite{LW, EM}. Before going further, let us
briefly outline the basic steps needed for such a small $y$ expansion. 

\vspace{3mm}

\noindent The Mellin transform $\hat{f}(s)$ is done by the prescription
\tsecite{LW, EM, B}:

\begin{equation}
\hat{f}(s)= \int_{0}^{\infty} dx\; x^{s-1}\; f(x),
\tseleq{c12}
\end{equation}

\vspace{2mm}

\noindent with $s$ being a complex number. One can easily check that the 
inverse Mellin transform reads

\begin{equation}
f(x)= \frac{1}{i(2\pi)} \int_{-i\infty + a}^{i\infty +a} ds\; x^{-s} \; 
\hat{f}(s), \tseleq{c13}
\end{equation} 

\vspace{2mm}

\noindent where the real constant `$a$' is chosen in such a way that 
$\hat{f}(s)$ is convergent in the neighbourhood of a straight line 
($-i\infty +a,\; i\infty +a$). So particularly if $f(x) = 
\frac{1}{e^{xy}-1}$ one can find (\tsecite{B}; formula I.3.19) that


\begin{equation}
\hat{f}(s) = \Gamma(s)\zeta(s) y^{-s} \mbox{\hspace{1cm}} (\mbox{Re}s > 1),
\tseleq{4.6}
\end{equation}    

\vspace{2mm}

\noindent where $\zeta$ is the Riemann zeta function ($\zeta(s)=
\sum_{n=1}^{\infty}n^{-s}$). Now we insert the Mellin transform of $f(x) =
\frac{1}{e^{xy}-1}$ to (\tseref{c14}) and interchange integrals (this is
legitimate only if the integrals are convergent before the
interchange). As a result we have

\vspace{2mm}

\begin{equation}
\int_{0}^{\infty} dx \; g(x)\ \frac{1}{e^{xy}-1} = \int_{-i\infty 
+a}^{i\infty + a} \frac{ds}{i(2\pi)} \; \Gamma(s)\zeta(s) y^{-s} 
\hat{g}(1-s), \tseleq{4.7}
\end{equation}

\vspace{4mm}

\noindent with $g(x) = \theta(x-1)\;(x^{2}-1)^{\frac{1+2\nu}{2}}$. Using 
the tabulated result (\tsecite{BB}; formula 6.2.32) we find

\begin{equation}
\hat{g}(1-s) = \frac{1}{2} B(-\nu -1 + 
\mbox{$\frac{1}{2}s$}; 
\mbox{$\frac{3}{2}$} + \nu) \mbox{\hspace{1.5cm}} (\mbox{Re}s >2+2\nu),
\tseleq{4.8}
\end{equation}

\vspace{2mm}

\noindent with $B(\;;\;)$ being the beta function. Because the integrand
on the RHS of (\tseref{4.7}) is analytic for $\mbox{Re}s > 2 + 2\nu$ and
the LHS is finite, we must choose such $a$ that the integration is
defined. The foregoing is achieved choosing $\mbox{a} > 2+2\nu$. Another
useful expressions for $\hat{g}(1-s)$ are (\tsecite{BB}; formula I.2.34 or
I.2.37)

\begin{eqnarray*}
\hat{g}(1-s) &=& B(\mbox{$\frac{3}{2}$}+\nu; -2 - 2\nu +s) \; 
{~}_{2}F_{1}[-\mbox{$\frac{1}{2}$} -\nu ; -2 - 2\nu +s; 
-\mbox{$\frac{1}{2}$} -\nu +s; -1] \nonumber \\
&=& 2^{\frac{1}{2} +\nu} \; B(\mbox{$\frac{3}{2}$}+\nu; -2 - 2\nu +s)\; 
{~}_{2}F_{1}[-\mbox{$\frac{1}{2}$} -\nu ; \mbox{$\frac{3}{2}$} +\nu; 
-\mbox{$\frac{1}{2}$} -\nu +s; \mbox{$\frac{1}{2}$}], 
\end{eqnarray*}
   
\vspace{3mm}

\noindent where ${~}_{2}F_{1}$ is the (Gauss) hypergeometric
function \tsecite{BB}.  Using identity 

\vspace{1mm}
\begin{displaymath}
\Gamma(2x) = \frac{2^{2x-1}}{\sqrt{\pi}}\; 
\Gamma(x)\Gamma(x+\mbox{$\frac{1}{2}$}),
\end{displaymath}
\vspace{-3mm}
\noindent we can write

\begin{equation}
\mbox{(\tseref{4.7})} = \frac{\Gamma(\mbox{$\frac{3}{2}$} +\nu)}{4 
\sqrt{\pi}}\; \int_{-i\infty + a}^{i\infty +a} \frac{ds}{i(2\pi)} 
\Gamma(\mbox{$\frac{1}{2}$}s) \zeta(s) \left(\mbox{$\frac{1}{2}$}y 
\right)^{-s}\Gamma(-\nu - 1 + \mbox{$\frac{1}{2}$}s).
\tselea{4.9}
\end{equation}

\vspace{3mm}

\noindent The integrand of (\tseref{4.9}) has simple poles in $s=-2n \;
(n=1,2,\ldots)$, $s=1$, $s=-2n +2\nu +2 \;(n= 0,1, \ldots, \nu)$ 
 and double pole in $s=0$. An important point in the former pole analysis
was the fact that $\zeta(s)$ has simple zeros in $-2m$ ($m>0$) and only
one simple pole in $s=1$. The former together with identity


\begin{displaymath}
\Gamma\left( \frac{x}{2} \right)\; \pi^{- \frac{x}{2}}\; \zeta(x) = 
\Gamma \left( \frac{1-x}{2} \right) \; \pi^{ \frac{x-1}{2}} \; \zeta(1-x),
\end{displaymath}

\noindent shows that no double pole except for $s=0$ is present in
(\tseref{4.9}). Now, we can close the contour to the left as the value of
the contour integral around the large arc is zero in the limit of infinite
radius (c.f. \tsecite{B} and \tsecite{GR};  formula 8.328.1). Using
successively the Cauchy theorem we obtain

\begin{eqnarray}
\lefteqn{\frac{4\sqrt{\pi}\;\mbox{(\tseref{4.7})}}{\Gamma(\mbox{$\frac{3}{2}$}+\nu)}}\nonumber \\ 
&=& \sum_{n=0}^{\nu} y^{2n - 2\nu -2} \; 
\frac{\pi^{-2n +2\nu +2}(-n+\nu)!\; (-1)^{n} |B_{-2n + 2\nu +2}|}{n! \; 
(-2n +2\nu +2)!\; 2^{4n -4\nu -4}}\nonumber \\
&+& \sum_{n=1}^{\infty} y^{2n} \; \frac{\pi^{-2n} \;(2n)! 
\;\zeta(1+2n)\; (-1)^{n + \nu +1}}{n!\; (n + 1 + \nu)! 
\;2^{4n-1}}\nonumber \\
&+& y^{-1} \; \frac{\pi\; (-1)^{\nu + 1} \; (\nu + 1)! \;2^{2\nu +3}}{(2\nu 
+ 2)!} + \frac{2\; 
(-1)^{\nu + 1}}{(\nu + 1)!}\left\{ \mbox{ln} \left( \frac{y}{4\pi}\right) + 
\gamma - \mbox{$\frac{1}{2}$} \sum_{k=1}^{\nu + 1}\frac{1}{k} 
\right\},\nonumber \\
\tselea{4.10}
\end{eqnarray}

\vspace{3mm}

\noindent where $B_{\alpha}$'s are the Bernoulli numbers. Let us mention
that for $\zeta(2n + 1)$ only numerical values are available. 

\vspace{3mm}

\noindent Inserting (\tseref{4.10}) back to (\tseref{4.4}), we get 
for ${\cali{P}}(T) - {\cali{P}}(0)$

\begin{eqnarray}
&&{\cali{P}}(T) - {\cali{P}}(0) = (\mbox{\tseref{4.3}}) + 
\frac{1}{3}\; I_{2}(m_{r}(T)) + \frac{\delta 
m^{2}(T)}{4} \;I_{0}(m_{r}(T))\nonumber \\
&&\mbox{}\nonumber\\
&& \mbox{\hspace{5mm}}=\frac{T^{4}\; \pi^{2}}{90} - 
\frac{T^{2}}{24} \left( m^{2}_{r}(T) - \frac{\delta m^{2}(T)}{2} 
\right) + \frac{T\; m_{r}(T)}{4\; \pi} \left( 
\frac{m^{2}_{r}(T)}{3} - \frac{\delta m^{2}(T)}{4} \right)\nonumber \\ 
&&\mbox{}\nonumber\\
&& \mbox{\hspace{5mm}} -\; \frac{m_{r}^{2}(T)\;m^{2}_{r}(0)}{32\; \pi^{2}}\; 
\mbox{ln}\left( \frac{m_{r}(0)\; 4\pi}{T}  \right) - 
\frac{m^{2}_{r}(0)}{128 \; \pi^{2}} \left( 3 m^{2}_{r}(T) - 
\delta m^{2}(T)\right) \nonumber \\
&&\mbox{}\nonumber\\
&& \mbox{\hspace{5mm}} - \;  
\sum_{n=1}^{\infty} \left( m^{2}_{r}(T) - \mbox{$\frac{(n+2)}{2}$}\; \delta 
m^{2}(T)\right) \; \frac{m_{r}^{2n+2}(T) \; \pi^{-2n-2}\; 
(2n)!\; \zeta (1+2n) \; (-1)^{1+n}}{T^{2n}\;n!\;(n+2)!\; 
2^{4n+4}}.\nonumber\\
\tselea{4.11}
\end{eqnarray}
 
\vspace{3mm}

\noindent Note that (\tseref{4.3}) cancelled against the same term in
$\frac{1}{3}\; I_{2}(m_{r}(T)) + \frac{\delta m^{2}(T)}{4}
\;I_{0}(m_{r}(T))$. One can see that (\tseref{4.11}) rapidly converges 
for large $T$, so that only first four terms dominate. The 
latter come from the poles nearby the straight line $(-i\infty + a, \; 
i\infty +a)$ (the more dominant contribution the closer pole). It is a 
typical feature of the Mellin transform technique that integrals of type 

\begin{displaymath}
\int_{0}^{\infty}dx \; g(x) \; \frac{1}{e^{xy}-1},
\end{displaymath}

\vspace{2mm}

\noindent can be expressed as an expansion which rapidly converges for 
small $y$ (high-temperature expansion) or large $y$ (low-temperature 
expansion)\footnote{By the same token we get the low-temperature 
expansion if the integral contour must be closed to the right.}.

\vspace{3mm}

\noindent Note that our discussion of the mass renormalization in
Section 3.1 can be directly extended to the case when $m_{r}(0) =
0$. This shows  that the original massless scalar particles acquire
the thermal mass $m_{r}^{2}(T)= \delta m^{2}(T) $ . From
(\tseref{4.11}) one then may immediately deduce the pressure for massless
fields $\phi_{a}$  in terms of $\delta m(T)$. The latter reads

\begin{eqnarray}
{\cali{P}}(T) - {\cali{P}}(0) &=& \frac{T^{4} \; \pi^{2}}{90}  - 
\frac{T^{2}\; (\delta m(T))^{2}}{48} + \frac{T\; (\delta m(T))^{3}}{48 
\; \pi} \nonumber \\
&& \mbox{} \nonumber \\
&+& \; \sum_{n=1}^{\infty} \frac{(\delta m(T))^{2n+4} 
\;\pi^{-2n-2}\; (2n)!\; \zeta(1+2n)\; (-1)^{n+1}}{T^{2n} \;(n-1)!\; 
(n+2)!\; 2^{4n+5}}. \tselea{4.12}
\end{eqnarray}

\vspace{2mm}

\noindent This result is identical to that found by Drummond {\em et al.} in 
\tsecite{ID1}. 

\vspace{3mm}

\noindent Noteworthy observation is that when the energy of a thermal
motion is much higher then the mass of particles in the rest, then the
massive theory approaches the massless one. This is justified in the
first (high-temperature dominant) term of (\tseref{4.11}) and
(\tseref{4.12}).  This term is nothing but a half of the black body
radiation pressure for photons \tsecite{GM, Cub} (photons have two
degrees of freedom connected with two transverse polarizations). One
could also obtain the temperature dominant contributions directly from
the Stefan-Boltzmann law \tsecite{LW, GM, Cub} for the density energy
(i.e. $\langle \Theta^{00} \rangle$). The formal argument leading to
this statement is based on the noticing that at high energy
(temperature) the scalar field theory is approximately conformally
invariant, which in turn implies that the energy-momentum tensor is
traceless \tsecite{CCR}.  Taking into account the definition of the
hydrostatic pressure (\tseref{EMT24}), we can with a little effort
recover the leading high-temperature contributions for the massive
case.


\section{Conclusions}

In the present article we have clarified the status of hydrostatic
pressure in (equilibrium) thermal QFT. The former is explained in terms of
the thermal expectation value of the `weighted' space-like trace of the
energy-momentum tensor $\Theta^{\mu \nu}$. In classical field theory there
is a clear microscopic picture of the hydrostatic pressure which is
further enhanced by a mathematical connection (through the virial theorem)
with the thermodynamic pressure. In addition, it is the hydrostatic
pressure which can be naturally extended to a non-equilibrium medium.
Quantum theoretic treatment of the hydrostatic pressure is however pretty
delicate. In order to get a sensible, finite answer we must give up the
idea of total hydrostatic pressure. Instead, thermal interaction pressure
or/and interaction pressure must be used (see (\tseref{ppp9}) and
(\tseref{ppp10})). We have established this result for a special case when
the theory in question is the scalar $\phi^{4}$ theory with $O(N)$
internal symmetry; but it can be easily extended to more complex
situations. Moreover, due to a lucky interplay between the conservation of
$\Theta^{\mu \nu}_{C}$ and the space-time translational invariance of an
equilibrium (and $T=0$) expectation value we can use the simple canonical
(i.e. unrenormalized) energy-momentum tensor. In the course of our
treatment in Section 2 we heavily relied on the counterterm
renormalization, which seems to be the most natural when one discusses
renormalization of composite Green's functions. To be specific, we have
resorted to the minimal subtraction scheme which has proved useful in
several technical points. 

\vspace{3mm}

\noindent We have applied the prescriptions obtained for the QFT
hydrostatic pressure to $\phi^{4}$ theory in the large $N$ limit. The
former has the undeniable advantage of being exactly soluble. This is
because of the fact that the large-$N$ limit eliminates `nasty' classes of
diagrams in the Dyson-Schwinger expansion. The surviving class of diagrams
(superdaisy diagrams) can be exactly resummed, because the (thermal)
proper self-energy ${\vect{\Sigma}}$, as well as the renormalized coupling
constant $\lambda_{r}$ are momentum independent. For the model at hand the
resummed form of the pressure was firstly derived (in the purely
thermodynamic pressure context) by Drummond {\em et al.} in \tsecite{ID1}.
We have checked, using the prescription (\tseref{ppp9}) for the thermal
interaction pressure, that their results are in agreement with ours. The
former is a nice vindication of the validity of the virial theorem even in
the QFT context. 

\vspace{3mm}

\noindent The expression for the pressure obtained was in a suitable form
which allowed us to take advantage of the Mellin transform technique. We
were then able to write down the high-temperature expansion for the
pressure in $D=4$ (both for massive and massless fields) in terms of
renormalized masses $m_{r}(T)$ and $m_{r}(0)$. (The corresponding mass
shift equation (gap equation) was already numerically solved in
\tsecite{ID1} and so we did not tackle this problem.) We have
explicitly checked that all UV divergences present in the individual thermal
diagrams `miraculously' cancel in accordance with our analysis of the
composite operators in Section 2. 





\section*{Acknowledgements}

I am indebted to P.V.Landshoff for reading the manuscript and for
invaluable discussion. I am also grateful to N.P.Landsman and H.Osborn for
helpful conversations. The work is supported in part by Fitzwilliam
College. 


\vspace{2cm}

\appendix 
\section{Appendix} 

\subsection{}

In this Appendix we shall derive resummed form for the renormalized
coupling constant at finite temperature. Extension to the finite
temperature however needs more care.  We can not simply require
$\Gamma^{(4)}_{1111}(s=0)= -\lambda_{r}(T)/N$.  The crux is that the heat
bath breaks the Lorentz invariance and hence $\Gamma^{(4)}$ is not
function of $s$ any more. Instead, $\Gamma^{(4)}$ is, on account of the
remaining rotational invariance, function of ${\vect{Q}}^{2}$ and $Q_{0}$.
In what follows it will prove useful to require the following
(non-physical) renormalization condition: 

\begin{equation}
\Gamma^{(4)}_{1111}(Q_{0}=0,{\vect{Q}}^{2}=0;T) = -\lambda_{r}(T)/N,
\tseleq{m15}
\end{equation}

\vspace{2mm}

\noindent (in the sequel we shall omit the argument of
$\Gamma^{(4)}$) the corresponding Dyson-Schwinger equation for
$\Gamma^{(4)}_{1111}$ reads: 



\begin{figure}[h]
\begin{center}
\leavevmode
\hbox{%
\epsfxsize=12.5cm
\epsffile{N8.eps}}
%\label{fig7}
\end{center}
\setlength{\unitlength}{1mm}
\begin{picture}(10,7)
%\put(2,49.5){$\Gamma^{(4)}\;=$}
%\put(42.5,26){$\sum_{i=1}^{3}$}
%\put(52,56){$\sum_{i=1}^{3}$}
\put(137.5,24){(A.2a)}
\end{picture}
\end{figure}
 

\noindent The `setting sun' diagrams are suppressed as before. In order 
to resolve (A.2a) we must furnish it with two further Dyson-Schwinger 
equations:


\begin{figure}[h]
\begin{center}  
\leavevmode
\hbox{%
\epsfxsize=11.5cm
\epsffile{N88.eps}}
%\label{fig7}
\end{center}  
\setlength{\unitlength}{1mm}
\begin{picture}(10,7)
%\put(2,49.5){$\Gamma^{(4)}\;=$}
%\put(42.5,26){$\sum_{i=1}^{3}$}
%\put(52,56){$\sum_{i=1}^{3}$}
\put(137.5,40){(A.2b)}
\put(137.5,14){(A.2c)}
\end{picture}
\end{figure}

\addtocounter{equation}{1}

\noindent From (A.2a) and (A.2c) we can directly read that


\begin{eqnarray}
\frac{\lambda_{r}(T)}{N} &=& \frac{\lambda_{0}}{N} -  
\frac{i\lambda_{0} \lambda_{r}(T)}{2N}\; \int 
\frac{d^{D}q}{(2 \pi)^{D}}\D_{11}(q)\D_{11}(-q)\nonumber \\
&+&\frac{i 
\lambda_{0}}{2}\; 
\Gamma^{(4)}_{2211}\int \frac{d^{D}q}{(2\pi)^{D}} \D_{12}(q)\D_{12}(-q), 
\tseleq{m165}
\end{eqnarray}

\vspace{2mm}

\noindent similarly (3.17b) and (3.17c) lead to 


\begin{equation}
\Gamma^{(4)}_{2211} = \frac{i\lambda_{0}}{2}\Gamma^{(4)}_{2211}\; \int 
\frac{d^{D}q}{(2\pi)^{D}} \D_{22}(q)\D_{22}(-q) - 
\frac{i\lambda_{0}\lambda_{r}(T)}{2N} \; \int \frac{d^{D}q}{(2 
\pi)^{D}}\D_{21}(q)\D_{21}(-q).
\tseleq{m166}    
\end{equation}


%\noindent In the foregoing two equations we have used the identity

%\begin{displaymath}
%\int \frac{d^{D}q}{(2\pi)^{D}}\; A(q)A(-q+Q)|_{Q_{0}=0; |{\vect{Q}}|=0} = 
%\int \frac{d^{D}q}{(2\pi)^{D}}\; A(q)A(-q),
%\end{displaymath}

\noindent 
%which is true for any function $A(q_{0}, {\vect{q}}) = f(q_{0},
%{\vect{q}}^{2})$.
Using the mass-derivative formulas \tsecite{LW, LB1}

\begin{eqnarray}
i\partial_{m^{2}_{r}}\D_{11}(q) &=& \D_{11}(q)\D_{11}(-q) - 
\D_{21}(q)\D_{21}(-q)\nonumber \\
-i\partial_{m^{2}_{r}}\D_{22}(q) &=& \D_{22}(q)\D_{22}(-q) -
\D_{12}(q)\D_{12}(-q),
\tseleq{m17}
\end{eqnarray}

\vspace{2mm}

\noindent we may write

\begin{equation}
\lambda_{r}(T) = \lambda_{0} + \lambda_{0} \lambda_{r}(T)\; 
{\cali{M}}'_{T} + N\; 
\Gamma^{(4)}_{2211}(1-\lambda_{0}{\cali{M}}'_{T}), \tseleq{m18}
\end{equation}

\vspace{2mm}

\noindent where differentiation is meant with respect to $m^{2}_{r}(T)$; 
${\cali{M}}_{T}$ is done by (\tseref{m14}). Using the explicit form of 
$\Gamma^{(4)}_{2211}$ from (\tseref{m166}), we can easily evaluate the 
imaginary part of $\lambda_{r}$. The former reads

\begin{equation}
\mbox{Im}(\lambda_{r}(T)) = \frac{-\lambda_{0}\; \int \frac{d^{D}q}{(2 
\pi)^{D}}\D_{12}(q)\D_{12}(-q)}{2(1+\lambda^{2}_{0}({\cali{M}}'_{T})^{2})},
\end{equation}

\noindent whilst the real part of $\lambda_{r}(T)$ is 

\begin{equation}
\mbox{Re}(\lambda_{r}(T)) = \lambda_{0} + 
\lambda_{0}\mbox{Re}(\lambda_{r}(T))\; {\cali{M}}'_{T}(m^{2}_{r}(T)).
\tseleq{m124}
\end{equation} 

\vspace{3mm}

\noindent Note that $\mbox{Im}(\lambda_{r})$ vanishes in the small $T$ limit 
because $\lim_{T \rightarrow 0}\D_{12}(q)\D_{12}(-q) = 0 $. 
Correspondingly, $\mbox{Re}(\lambda_{r})$ approaches $\lambda_{r}$ from
(\tseref{m146}). The very same results can be obtained using the
largest-time equation \tsecite{PJ}. 


\subsection{}

In this Appendix we give some details of the derivation of
Eq.(\tseref{b10}). We particularly show that the surface integrals arisen
during the transition from (\tseref{b9}) to (\tseref{b10}) mutually cancel
among themselves. As usual, the integrals will be evaluated for integer 
values of $D$ and corresponding results then analytically continued to a 
desired (generally complex) $D$.

\vspace{3mm}

\noindent The key quantity in question is

\vspace{2mm}

\begin{eqnarray}
&+& \frac{1}{2}\; \int \frac{d^{D}q}{(2\pi)^{D-1}} \left( 
\frac{2{\vect{q}}^{2}}{(D-1)} 
\right) \; \frac{\varepsilon(q_{0})}{e^{\beta 
q_{0}}-1}\;\delta(q^{2}-m^{2}_{r}(T))\nonumber \\
&-& \frac{1}{2}\; \int \frac{d^{D}q}{(2\pi)^{D-1}} \left( 
\frac{2{\vect{q}}^{2}}{(D-1)} \right)\; \delta^{+}(q^{2}-m^{2}_{r}(0)).
\tselea{a1}
\end{eqnarray} 

\vspace{2mm} 

\noindent Applying Green's theorem (i.e. integrating by parts with 
respect to ${\vect{q}}$) on (\tseref{a1}) one finds

\vspace{2mm}

\begin{eqnarray}
\mbox{(\tseref{a1})}&=& {\cali{N}}_{T}(m^{2}_{r}(T)) - 
{\cali{N}}(m^{2}_{r}(0))\nonumber\\
 &+& \lim_{R \rightarrow \infty} \; 
\frac{1}{2(D-1)} \int \frac{dq_{0}}{(2\pi)^{D-1}}\; \int_{\partial 
S^{D-2}_{R}} 
d{\vect{s}}\;{\vect{q}}\; \theta(q^{2}-m^{2}_{r}(T))\; \theta(q_{0})\; 
\left( \frac{2}{e^{\beta q_{0}}-1}+1 \right)\nonumber \\
&-&\lim_{R \rightarrow \infty} \; \frac{1}{2(D-1)} \int    
\frac{dq_{0}}{(2\pi)^{D-1}}\; \int_{\partial S^{D-2}_{R}}
d{\vect{s}}\;{\vect{q}}\; \theta(q^{2}-m^{2}_{r}(0))\; \theta(q_{0}).
\tseleq{a2}
\end{eqnarray}

\vspace{4mm}

\noindent As usual, ${\vect{a}}{\vect{b}} =
\sum_{i=1}^{D-1}{\vect{a}}_{i}{\vect{b}}_{i}$ and $S^{D-2}_{R}$ is a
$(D-2)$-sphere with the radius $R$. The expressions for ${\cali{N}}_{T}$ and
${\cali{N}}$ are done by (\tseref{b101}). 

\vspace{3mm}

\noindent With the relation (\tseref{a2}) we can show that the surface 
terms cancel in the large $R$ limit. Let us first observe that

\begin{eqnarray}
&&\lim_{R \rightarrow \infty} \; \int
\frac{dq_{0}}{(2\pi)^{D-1}}\; \int_{\partial S^{D-2}_{R}}
d{\vect{s}}\;{\vect{q}}\; \theta(q^{2}-m^{2}_{r}(T))\; 
\frac{2 \theta(q_{0})}{e^{\beta q_{0}}-1}\nonumber \\
&=& \lim_{R \rightarrow \infty}\; 
\frac{2\pi^{\frac{D-1}{2}}R^{D-1}}{\Gamma \left(\frac{D-1}{2}\right)} \; 
\int \frac{dq_{0}}{(2\pi)^{D-1}} \; \theta(q_{0}^{2}-R^{2}-m^{2}_{r}(T))\;
\frac{2 \theta(q_{0})}{e^{\beta q_{0}}-1}\nonumber \\
&=& \lim_{R \rightarrow \infty} 
\frac{\pi^{\frac{1-D}{2}}R^{D-1}}{2^{D-2}\Gamma 
\left( \frac{D-1}{2} \right)}\int_{\sqrt{R^{2}+m_{r}^{2}(T)}}^{\infty} 
dq_{0}\; \frac{2}{e^{\beta q_{0}}-1} \; = \;0.
\tseleq{a3}
\end{eqnarray}

\vspace{3mm}


  
\noindent In 2-nd line we have exploited Gauss's theorem and in the last line
we have used L'H{\^{o}}pital's rule as the expression is in the
indeterminate form $0/0$. The remaining surface terms in (\tseref{a2})
read

\begin{eqnarray}
&& \lim_{R \rightarrow \infty} \int
\frac{dq_{0}}{(2\pi)^{D-1}} \; \int_{\partial S^{D-2}_{R}}
d{\vect{s}}\;{\vect{q}}\; \left\{ \theta(q^{2}-m^{2}_{r}(T)) - 
\theta(q^{2}-m^{2}_{r}(0)) \right\}\; \theta(q_{0}) \nonumber\\
&=& \lim_{R \rightarrow \infty}\; 
\frac{\pi^{\frac{1-D}{2}}R^{D-1}}{2^{D-2}\Gamma \left( \frac{D-1}{2} 
\right)}\; \left\{ \int_{\sqrt{R^{2}+m_{r}^{2}(T)}}^{\infty} - 
\int_{\sqrt{R^{2}+m_{r}^{2}(0)}}^{\infty} \right\}dq_{0}\; = \; 0.
\tseleq{a4}
\end{eqnarray}

\vspace{2mm}

\noindent The last identity follows either by applying L'H{\^{o}}spital's
rule or by a simple transformation of variables which renders both
integrals inside of $\{ \ldots \}$ equal. Expressions on the last lines in
(\tseref{a3}) and (\tseref{a4}) can be clearly (single-valuedly) continued
to the region $\mbox{Re}D > 1$ as they are analytic there. We 
thus end up with the statement that


\vspace{2mm}

\begin{displaymath}
\mbox{(\tseref{a1})}= {\cali{N}}_{T}(m^{2}_{r}(T))- {\cali{N}}(m^{2}_{r}(0)).
\end{displaymath}


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\end{document}



