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\centerline{SWISS CHEESE INSTABILITY OF THE QUARK-GLUON PLASMA}
\vskip\baselineskip
\centerline{Benjamin Svetitsky}
\vskip\baselineskip  {\it
\centerline{School of Physics and Astronomy}
\centerline{Raymond and Beverly Sackler Faculty}
\centerline{of Exact Sciences}
\centerline{Tel Aviv University}
\centerline{69978 Tel Aviv, Israel}  }

\vskip \baselineskip

{\narrower
\noindent 
ABSTRACT.  I argue that the quark-gluon plasma is unstable against the
formation of bubbles of confining vacuum.

\vskip  \baselineskip}

This workshop has seen some beautiful talks about perturbation theory
in the quark-gluon plasma, about the chiral properties of the plasma,
and about studies of the phase transition.  
One common assumption of
all this work is that there indeed exists a high-temperature phase of
QCD which is homogeneous in equilibrium.  
I wish to present the
possibility that the opposite is true---that the homogeneous plasma
is unstable, and that the equilibrium state is full of ``holes,''
that is, bubbles of vacuum containing a gas of hadrons.
The main hard input to this argument is the anomalously small surface
tension at the quark--hadron interface which has emerged from bag
model$^1$ and lattice$^2$ calculations;
the intuitive picture is based on models where all the surface
physics comes from the fact that the fields in either phase interact
with the interface via given, fixed boundary conditions.
A test of this intuition in QCD should come to fruition shortly
in lattice gauge theory$^3$.
 
Consider the usual theory of bubble nucleation at a first-order phase
transition, which you may look up in Landau and Lifshitz$^4$.
For concreteness, let us discuss boiling water, near the transition
temperature $T_0=373$~K at atmospheric pressure.
A bubble of vapor in the water has a free energy which depends on its
radius:
$$
F(R)=\Delta P\cdot {4\over3}\pi R^3 + \sigma \cdot 4\pi R^2 + \cdots .
\eqno(1)
$$
The first term is the volume term, whose coefficient $\Delta P$ is
the difference in pressure between the two phases: $\Delta P >0$ for
$T<T_0$ where the liquid phase is stable, and $\Delta P <0$ for
$T>T_0$ where the vapor phase is stable.
The second term is the surface term, $\sigma>0$ being the (isothermal)
surface tension.
 
$F(R)$ is shown schematically in Figure 1.
For $T<T_0$, both the volume and surface terms act to suppress bubble
formation;
any bubble formed by a fluctuation collapses to $R=0$.
For $T>T_0$, however, the volume term encourages growth of bubbles
while the surface term creates a barrier.
The maximum of $F(R)$ occurs at some $R=R_{cr}$.
A bubble formed in the superheated liquid with $R<R_{cr}$ will shrink
away, but a bubble formed with $R>R_{cr}$ will grow until the entire
system is vapor.
The probability of getting a bubble with sufficiently large radius
is proportional to $\exp\left(-F(R_{cr})/T\right)$, and hence so is
the nucleation rate$^4$.
 
\eject\ 
\vskip 5.5truein
{\narrower \baselineskip12truept
\noindent \smallr
\textfont1=\smalli \scriptfont1=\smalleri
FIGURE 1.
Schematic representation of bubble free energy in boiling water, retaining 
only volume
and area terms, for $T<T_0$ (solid) and $T>T_0$ (dashed).
\smallskip}
In Equation 1 I have indicated that there might be more terms beyond
the area term, having in mind an expansion in powers of $1/R$.
Conventional treatments, of course, stop at the area term as above.
The next term, which we write $\alpha\cdot 8\pi R$, is clearly
negligible in condensed-matter contexts:
By dimensional analysis, $\alpha/\sigma \sim \lambda$, where
$\lambda$
is a microscopic length, and hence $(\alpha R)/(\sigma R^2) \sim
\lambda/R$,
an exceedingly small number since the typical nucleating bubble is
much larger than either the inter-molecular spacing or the
correlation length.
The picture in Figure 1, based on volume and area terms, is then the
whole story.
 
Note that our picture will be substantially the same for liquid
droplets in the vapor as for vapor bubbles in the liquid, since
$\sigma$ is the same in the two cases.
This is because the surface tension may be measured in planar
interfaces, and it doesn't matter which phase is on which side.
 
Let us turn to QCD, a theory with only one scale.
Masses, radii, the transition temperature, the latent heat, $\ldots$,
are all on the order of 200 MeV or 1 fm.
This 1 fm is then the only ``microscopic'' size, and 1 fm should also
be the typical size of a bubble.
Thus one may not ignore additional terms in (1).
Let us explore the possibility that $\sigma R^2 \ll \alpha R$;
for simplicity, let $\sigma=0$.
Then
$$
F(R)=\Delta P \cdot {4\over3}\pi R^3 + \alpha \cdot 8\pi R + \cdots\ .
\eqno(2)
$$
Further, let us neglect (for now) any terms beyond what is written in
(2).
 
If (2) represents the free energy of a droplet of plasma in the
hadron phase, with $R$ its radius, then the form of $F(R)$ is
qualitatively the same as that shown in Figure 1.
So, of course, is the picture of the nucleation process; the only
change is that $R_{cr}$ and $F(R_{cr})$ have to be calculated in
terms of $\Delta P$ and $\alpha$, instead of $\Delta P$ and $\sigma$.
 
The essential difference between (1) and (2) appears if we turn to
the inverted case, that of a vacuum bubble (presumably containing a
hadron gas) in the plasma.
To apply (1) or (2) to this case, we shall argue that it is correct
to take $R\to-R$.
To justify this, consider a free-field model, such as the bag model,
wherein the free energy of species $i$
is a sum over free-particle states weighted
by statistical factors:
$$
F_i(R,T)=\mp\gamma T \int_0^\infty dk\,\rho_i(k,R)
\log\left(1\pm e^{-\omega(k)/T}\right)\ .
\eqno(3)
$$
Here $\gamma$ is a degeneracy factor, the upper (lower) sign is for
fermions (bosons), and $\rho$ is the density of states in the
presence of the boundary.
(For details see Ref.~1.)
If we want $F(R,T)$ as a power series in $1/R$, it behooves us to
express $\rho(k,R)$ as such a series; this is precisely the result
of a calculation employing the Multiple Reflection Expansion$^5$:
$$
\rho={V\over2\pi^2}k^2 + c_1 k \int dS + c_2 \int dS\,
\left({1\over R_1} + {1\over R_2}\right) + \cdots\ .
\eqno(4)
$$
Equation (4) is valid for a smooth surface of any shape.
The first term is a universal volume term;
the integral in the second term is the surface area;
and the integrand in the third term is the extrinsic curvature, with
$R_1$ and $R_2$ the principal radii of curvature at each point of the
surface.
The coefficients $c_1$ and $c_2$ depend on the nature of the field
in question and on the boundary conditions chosen.
Specializing to a spherical interface, we have
$$
\rho(k,R)={2\over3\pi}R^3k^2 + c_1 4\pi R^2 k + c_2 8\pi R +
\cdots\ .
\eqno(5)
$$
The fact that the extrinsic curvature appearing in (4) is a signed
quantity implies that so is $R$ in the third term in (5).
In other words, if $R$ is chosen positive for a plasma droplet, it
must be negative for a vacuum bubble in the plasma.
 
Let us then write (2) with the change $R\to -R$, as appropriate for
a vacuum bubble:
$$
F(R)=(-\Delta P)\cdot {4\over3}\pi R^3 - \alpha\cdot8\pi R\ \ .
\eqno(6)
$$
The sign change of the first term is trivial, and is consistent with
our identification of the stable phase for temperatures above and
below $T_0$.
The area term, if it were present, would be unchanged, in line with
what we said above concerning invariance of the surface tension under
exchanging the two phases.
In the absence of the area term, however, we are left with the
``curvature term,'' which has flipped sign!
 
The form of $F(R)$, as shown in Figure 2, is qualitatively different
from that shown in Figure 1.
For $T<T_0$, where the hadron phase is stable, there is {\it no}
barrier against the growth of bubbles.
This means that it is impossible to supercool the plasma---bubbles
nucleate immediately upon crossing $T_0$, and grow without hindrance.

%\eject\ 
\vskip 5.3truein
{\narrower \baselineskip12truept
\noindent \smallr
\textfont1=\smalli \scriptfont1=\smalleri
FIGURE 2.
Free energy of a hadron gas bubble in the plasma for 
temperatures near $\smalli T_0=$\ 150~MeV.
Top to bottom:
$T=$\ 155~MeV, 150~MeV ($=T_0$), 147~MeV.
Results from bag model calculation, Ref.~1.
\smallskip}
 
Even stranger is the form of $F(R)$ for $T>T_0$:
It has a minimum at $R_{min}\not=0$.
This means that a bubble of radius $R_{min}$ has {\it lower} free
energy than the plasma with no bubble at all!
Here, where we had thought to have a stable plasma, we find that it
is energetically favorable to grow bubbles to a fixed radius.
Thus the equilibrium plasma contains a population of bubbles,
the density of which is presumably fixed by some bubble--bubble
interaction.
This is the ``Swiss-cheese instability.''
 
Naturally, the instability doesn't stop at spherical bubbles.
Once the surface tension is negligible and the curvature term
dominates the free energy, it should be energetically favorable to
roughen the surface of a bubble.
Perhaps dendritic growth occurs; perhaps the bubbles link up and
percolate.
Any change in shape which increases the local curvature will occur,
limited only by what little surface tension remains, and by other
terms in $F$.
 
Speaking of other terms in $F$, what, indeed, justifies neglecting
them in (2)?
The answer is, nothing.
Equation (2) represents an expansion about $R=\infty$, and as $R$ is
made smaller, more terms will indeed come into play.
In fact, many terms should appear together, if there is really only
one scale in the problem.
We can only point to the bag model calculation, in which the linear
curvature term is clearly responsible for an interval where $F(R)<0$
before other terms contribute at smaller radii.
 
Let us discuss the nature of the phase transition.
The high-temperature phase consists of a froth of bubbles, and
possibly stranger beasts, with plasma in between.
As we lower the temperature, the bubbles grow, and perhaps merge, but
always stay around some characteristic size.
At $T=T_0$ they reach some maximum size, and then when $T$ is lowered
further they grow immediately without bound, until the entire volume
is in the hadron phase.
The structure of the high-temperature phase should leave some
signature in the resulting hadron gas, be it in the early
Universe or in ultrarelativistic heavy-ion collisions.
(This structure may take the place of that expected to result from
nucleation if the phase transition were a conventional first-order one.$^{6,7}$)
 
Heating the hadron gas through the transition is also interesting to
think about, though it may be less accessible experimentally.
The hadron phase is well-behaved and smooth, stable against the
plasma droplets which arise through fluctuations near the phase transition.
At $T>T_0$, the critical radius $R_{cr}$ comes into play:
Droplets born with $R<R_{cr}$ disappear, while those born with
$R>R_{cr}$ grow.
The latter grow in the usual hydrodynamic way until they meet and
coalesce, eventually filling all space except for bubbles of vacuum
which are shrinking.
In an ordinary phase transition, these bubbles would vanish entirely;
here, they shrink down to $R_{min}$ and stay there,
the while assuming whatever shapes the curvature term forces upon
them.
 
Let us return to our starting point.
Is the surface tension indeed small?
First indications came from the bag model.$^{8,1}$
It turns out that gluons and massless quarks contribute nothing to
$\sigma$; the surface tension arises only from the massive strange
quarks in the plasma.
This accident does {\it not} apply to the curvature coefficient---it gets
contributions from all quarks and gluons, and is not small.
 
The surface tension in the pure glue system has been calculated
directly in lattice gauge theory.$^2$
Again, it turns out to be anomalously small.
What does ``small'' mean?
One would expect, on dimensional grounds, that the surface tension
should be on the order of $B^{3/4}$, where $B$ is the bag constant.
The lattice result is smaller by almost two orders of magnitude.
A calculation of the curvature coefficient $\alpha$, and indeed of
the full form of $F(R)$, is in progress. $^3$
 
This lattice calculation should close the obvious loopholes in the
argument.
(Alternatively, the argument might be hanged by these loopholes.)
The main source of discomfort is that the intuitive argument depends
on a picture wherein the interface is thin.
One expects, however, that the quark--hadron interface will have some
thickness, as well as dynamics of its own.
Will this invalidate the expansion in $1/R$, giving an $F(R)$ where
the sign of the curvature term either doesn't flip, or is irrelevant?
Stay tuned.
 
\vskip\baselineskip
\leftline{\bf References}
\frenchspacing
\vskip\baselineskip

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\bye
