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

- free (noninteracting and deconfined) gas of massless quarks and gluons
- 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)
- 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.
- \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- Note +B in E, since it costs extra energy to be in this phase.
- Note -B in P, since the "bubble" collapses if no quarks to provide balancing pressure.
- E = -P for vacuum part follows from
*Lorentz invariance*of the vacuum:- energy-momentum tensor T^{\mu\nu} = C g^{\mu\nu} from Lorentz invariance (nothing else to build a tensor from!)
- But T^{00} = E and T^{11} = p, so E = C g^{00} = C and p = C g^{11} = - C, so P=-E

- P = 1/3 E for massless particle part, from
*scale invariance*- scale invariance implies trace of energy-momentum tensor for this part is zero: g_{\mu\nu} T^{\mu\nu} = 0.
- So T^{00} - T^{11} - T^{22} - T^{33} = 0, or E - 3P = 0 (rest frame, so 11, 22, 33 elements the same)

- free gass of
*massless*pions - \mu=0 Energy density: E_II = 3 (Pi^2/30) T^4

\mu=0 Pressure: P_II = 3 (Pi^2/90) T^4- Note that E = 3P again, as expected for massless particles

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:

- equal temperatures T_I = T_II
- equal chemical potentials (baryon \mu's here) \mu_I = \mu_II
- 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:

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

- Setting P_II(T,\mu=0) = P_I(T,\mu=0) gave

37 (Pi^2/90) T^4 - B = 3 (Pi^2/90) T^4- So T_c is 144 MeV with B^(1/4) = 206 MeV
- Only changes to T_c = 140 MeV if pion part neglected!

- Plugging back in to the energy density:

E_I = 37 (Pi^2/30) T^4 + B = 920 MeV/fm^3

E_II = 3 (Pi^2/30) T^4 = 55 MeV/fm^3 (tiny!) - What does this tell us?
- first-order phase transition because
*latent heat* - cf., water boiling --> heat until 100 degrees, then sit there putting more energy in, in a mixed phase of liquid and vapor, until all vapor, and then the temperature increases again
- a signal that the
*entropy*is very different between the phases.

- first-order phase transition because
- Now do T=0, \mu_B <> 0 [g_q = 3 colors x 2 spins x 2 flavors = 12]
- E_I = (g_q/8 Pi^2) (\mu_B/3)^4 + B with \mu_q = \mu_B/3

P_I = (g_q/24 Pi^2) (\mu_B/3)^4

E_II = P_II = 0 (no contribution from pions at T=0) - P_I = P_II --> \mu_B = 1.3 GeV
- Plugging in to find the energy densities:

E_I = 940 MeV/fm^3 and E_II=0. - Again, a sign of latent heat. Almost the same number as above because E = 4B to good approximation in the QGP phase. Most of the energy density is nonperturbative vacuum energy.

- E_I = (g_q/8 Pi^2) (\mu_B/3)^4 + B with \mu_q = \mu_B/3

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

- Energy: E = E(S,V,N)
- S = entropy, V = volume, N = number of particles (could be baryon number or strangeness as well)
- 1st law: dE = TdS - PdV + \mu dN
- T=(dE/dS)_V,N

P=-(dE/dV)_S,N

\mu=(dE/dN)_S,V - The last expression defines the chemical potential

- 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.
- Helmholtz free energy: F = F(T,V,N) = E - TS
- Gibbs free energy: G = G(T,P,N) = E - TS + PV
- So we can also calculate the chemical potential from

\mu = (dE/dN)_T,V or \mu = (dG/dN)_T,P - Thermodynamic potential: \Omega = \Omega(T,V,\mu) = F - \mu N
= E - TS -\mu N
- d\Omega = -SdT - PdV - Nd\mu
- S=-(d\Omega/dT)_V,\mu

P=-(d\Omega/dV)_T,\mu

N=-(d\Omega/d\mu)_T,V

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.

- Consider a
*scale change*close to one: \lambda = 1 + \epsilon - 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.
- So we have:

\lambda E = E(\lambda S,\lambda V,\lambda N) - 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 ) - 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

- In terms of energy densities,

e = T\sigma - P + \mu\rho - At \mu=0, we have

e = T\sigma - P - In phase equilibrium, this implies

e_I - e_II = T(\sigma_I - \sigma_II) = latent heat/volume - Solving for the entropy, \sigma = (P+e)/T, which implies that P=-e for the vacuum (as before)
- Plugging in numbers for Phase I and Phase II, we find \sigma_I >> \sigma_II. So there is a much higher entropy density in the QGP phase.
- \sigma is proportional to T^3, and so is n_pi, so in the hadron phase the entropy/pion = 3.6 independent of temperature and pressure!
- At T=0, equal pressures and chemical potentials imply that the baryon densities in the two phases are very different.

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:

- Z = Tr[e^(-\beta(H-\mu N))], where H is the hamiltonian and N the relevant number operator (like baryon number).
- \Omega(T,V,\mu) = -(1/\beta) ln Z
- Z can be calculated for QCD on a lattice! (next module)

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.

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

- Recall that the chemical potential \mu can be thought of as a
*Lagrange multiplier*--- a way to include a constraint in the problem. - Example: given f(x,y,z) to minimize subject to the constraint
g(x,y,z)=0.
- Construct H(x,y,z) = f - \lambda g
- Set dH/dx = dH/dy = dH/dz = 0
**and**dH/d\lambda = 0 - If two constraints, then two chemical potentials:

g(x,y,z)=0 and h(x,y,z)=0 ==> H = f - \lambda g - \mu h - Find point on plane 2x-3y+5z=10 nearest origin.

f = x^2+y^2+z^2 and g = 2x-3y+5z-19

Solving simultaneous equations from partial derivative equations gives answer: \lambda=1, x=1, y=-3/2, z=5/2.

- \mu serves to enforce the constraint that a conserved charge is, in
fact, conserved (in canonical ensemble). We have one for each conserved
charge, such as baryon number and strangeness
- If we chose to, we could include one for electric charge as well.

- P. Braun-Munzinger and J. Stachel, nucl-th/9606017 and references therein
- Assume that in the hadronic fluid every local rest frame is a grand canonical ensemble of free (noninteracting) fermions and bosons in thermal and chemical equilibrium at freeze-out temperature T.
- The general expression for the particle number density \rho_i of the
i'th hadron is:

\rho_i = g_i \int (d^3p/(2 Pi)^3) 1/( e^(E_i-\mu_B B_i - \mu_S S_i)/T +/- 1)

\equiv g_i \int (d^3p/(2 Pi)^3) n_i(p)- g_i => spin-isospin density
- E_i => relativistic energy: sqrt{p^2 + M_i^2)
- B_i => baryon number
- S_i => strangeness
- \mu_B => baryon chemical potential (same for all! only B_i changes)
- \mu_S => strangness chemical potential (same for all!)
- + for fermions; - for bosons

- Include all baryons up to 2 GeV and all mesons up to 1.5 GeV
- More massive hadrons have little impact for T < 180 MeV

- Include corrections for the finite size of the system and an excluded volume correction for baryons (see paper)
- Sample hadrons:
- pion: g_\pi=3, E_\pi=p_\pi (if massless), B_\pi=0, S_\pi=0
- proton: g=4, E=\sqrt{p^2+M_p^2}, B=1, S=0
- anti-proton: g=4, E=\sqrt{p^2+M_p^2}, B=-1, S=0
- Lambda: g=2, E=\sqrt{p^2+M_\Lamba^2}, B=1, S=-1

- Total energy density, pressure, and entropy density found from
summing contributions from all hadrons. Contributions from the i'th
hadron are:
- energy density e_i = g_i \int (d^3p/(2 Pi)^3) E_i n_i(p)
- pressure P_i = g_i \int (d^3p/(2 Pi)^3) 1/3(p^2/E_i) n_i(p)
- entropy density from thermodynamic relation between e_i and P_i

- Pressure can also be found from

P_i = -\Omega_i/V = -/+ g_i/\beta g_i \int (d^3p/(2 Pi)^3) ln[1 +/- e^{-\beta(E_i-\mu_B B_i - \mu_S S_i) } ]

with a partial integration (try it!)

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.

- Figure 1 shows the P vs. T plot at fixed \mu_B for each phase, used to find the temperature of phase coexistence
- Figure 3 shows the resulting phase boundary graphs with T versus chemical potential and also baryon density. Note that the baryon densities for the two phases are not equal!
- The other Figure 3 shows that the latent heat is about 1 GeV/fm^3, similar to the numbers we found above.
- Next time we'll go over the other figures.

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Copyright © 1997,1998 Richard Furnstahl and James Steele.