Course Outline for Physics 880.05

III. A. Thermodynamics and Phase Transitions (Chs. 9,13) Part 3

Two-Phase Model for Hadronic Matter

Last time we made a simple two-phase model of "hadronic matter". Let's recap the model and look at some numerical results.

Phase I: QGP Phase

  1. free (noninteracting and deconfined) gas of massless quarks and gluons
  2. plus bag pressure B, which accounts for the nonperturbative vacuum energy relative to the T=0 vacuum (which is defined to have zero energy here)
  3. the system is uniform with very large volume V, so we can neglect surface effects. In a more realistic treatment we'd take into account the finite size of the plasma.
  4. \mu=0 Energy density: E_I = 37 (Pi^2/30) T^4 + B
    \mu=0 Pressure: P_I = 37 (Pi^2/90) T^4 - B

Phase II: Hadronic Phase

  1. free gass of massless pions
  2. \mu=0 Energy density: E_II = 3 (Pi^2/30) T^4
    \mu=0 Pressure: P_II = 3 (Pi^2/90) T^4

Thermodynamic Equilibrium between Phases

In each of the phases we can calculate thermodynamic quantities and then decide which phase "wins". The conditions for equilibrium between Phase I and Phase II are:

  1. equal temperatures T_I = T_II
  2. equal chemical potentials (baryon \mu's here) \mu_I = \mu_II
  3. equal pressures P_I = P_II

If we have as our independent thermodynamic variables T and \mu, we can calculate P(T,\mu) for each phase and see where they are equal. A procedure to find the phase boundary:

  1. fix \mu_I = \mu_II = \mu_B
  2. calculate P_I(T,\mu_B) and P_II(T,\mu_B) as functions of T
  3. find the intersection of the curve, which determines T_c on the phase boundary for that \mu_B
  4. find the baryon densities \rho^I_B(T,\mu_B) and \rho^II_B(T,\mu_B)
  5. repeat with a new \mu_B

Numbers from Last Time

Thermodyanmics Review (one component system)

We're thinking of a system in thermodynamic equilibrium in a sufficiently large volume V.

Thermodynamic Energies

  1. Energy: E = E(S,V,N)
  2. Use Legendre transforms to change to other energies that depend on different indpendent variables. Use the one that is appropriate to the physical system and situation being considered.
  3. Helmholtz free energy: F = F(T,V,N) = E - TS
  4. Gibbs free energy: G = G(T,P,N) = E - TS + PV
  5. So we can also calculate the chemical potential from
    \mu = (dE/dN)_T,V or \mu = (dG/dN)_T,P
  6. Thermodynamic potential: \Omega = \Omega(T,V,\mu) = F - \mu N = E - TS -\mu N

Explicit Expressions

It's tempting to think we can just integrate the differentials to obtain, for example, \Omega = -ST - PV - N\mu. But this doesn't work! However, we can do exactly that for the energy for a special reason: the independent variables are all extensive (which means they are different for different subvolumes of the system). Let's see how that works.

  1. Consider a scale change close to one: \lambda = 1 + \epsilon
  2. The intensive variables T,\mu,P don't change, but S,V,N go to \lambda S,\lambda V,\lambda N and so does E.
  3. So we have:
    \lambda E = E(\lambda S,\lambda V,\lambda N)
  4. Now substitute \epsilon and expand the right side in a Taylor expansion with partial derivatives:
    (1+\epsilon)E = E + \epsilon( S(dE/dS)_V,N + V(dE/dV)_S,N + N(dE/dN)_S,V )
  5. Equating the terms proportional to \epsilon and identifying the partial derivatives from above gives our results:
    E = TS - PV + \mu N
    F = -PV + \mu N
    G = \mu N
    \Omega = -PV

Grand Canonical Ensemble

Usually we want T and \mu and V as independent variables, so \Omega is the relevant potential. We can find it from the partition function:

Thermal Model

What about the other hadrons? Shouldn't we count those as well? Ok, let's do it! We'll include the other hadrons as additional components of our noninteracting gas.

Chemical Potential as Lagrange Multiplier

Now we will have strange hadrons as well. How does that work?

Details of Thermal Model (Hadronic Side)

The results of using the thermal model for the hadronic phase and a bag model (similar to what we used) for the QGP side are shown in the figures.

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Copyright © 1997,1998 Richard Furnstahl and James Steele.