Last time we introduced a model for the hadronic phase consisting of all noninteracting hadrons. The energy density and pressure are found from the expressions given last time, adding the contributions from each hadron included in the model. This simple construction is possible because we neglect interactions.

- Checkpoint: What is the baryon density at T=0, with a strangeness chemical potential of zero and a baryon chemical potential of 500 MeV?

We can use this model to test the possibility that thermal and chemical equilibrium are reached in a high-energy heavy-ion collision by measuring the relative abundances of different hadrons detected. If equilibrium was reached, these abundances should be consistent with a single temperature and constant baryon and strangeness chemical potentials at "freeze-out". Freeze-out is the point in the evolution of the plasma when hadrons have been formed and then have stopped interacting with each other. So energy and momentum are no longer exchanged (as when in thermal equilibrium) but are "frozen". Comparison of predicted (given a fit to T and the \mu's) and measured abundances are given in the other figures from the Braun-Munzinger and Stachel paper.

- Figure 4 shows the comparison of predicted and experimental relative abundances for an AGS collision.
- Figure 5 shows the same comparison for a higher energy SPS collision.
- What can you imply from the figures about the baryon chemical potentials?

- The quality of fit suggests that equilibrium is reached!
- Figure 6 shows the fitted temperatures and chemical potentials on the phase boundary plot (free quarks and gluons plus the bag pressure is used on the deconfinement side). Looks like phase boundary is reached or crossed during the evolution to freeze-out!
- Next: The evolution of the QGP!

Now we want to step back and fill in the gap between the estimate of Bjorken of the initial conditions (energy density and time) for the QGP and the hadronization of the plasma. We assume the QGP has formed and is in local equilibrium at proper time \tau_0, with initial e_0 following Bjorken.

- The QGP should evolve according to relativistic hydrodynamics.
- The question we ask is: How do e and the other thermodynamic variables evolve with the proper time?
- The most relevent quantities are the energy density e, the pressure
P, the temperature T, and the four-velocity u^\mu = dx^\mu/d\tau.
- e and P are given in the "co-moving" frame (the frame that is the local rest frame of the fluid)
- the equation of state will relate e, P, and T

- T^{\mu\nu} is the momentum in the \mu-direction per unit 3-surface area perpendicular to the \nu-direction. That is, the momentum flux.
- Consider F*, the local reference frame where a fluid element is at rest
- T^00
- p^0 is the energy
- 3 surface perpendicular to 0th direction (time) is ordinary 3-d spatial volume
- So T*^00 = \Delta E/ \Delta x^1 \Delta x^2 \Delta x^3 = e, the energy density

- T*^11
- momentum in 1-direction is \Delta p^1
- 3-surface element is \Delta t \Delta x^2 \Delta x^3
- So T*^11 = \Delta p^1/\Delta t \Delta x^2 \Delta x^3 = force/area = pressure P

- T*^12
- \Delta p^1/\Delta t \Delta x^1 \Delta x^3 = 0
- spatial surface area is parallel to momentum, so no flux through it.
- So T*^ij = P \delta^ij
- In more general frames, there will be fluid flow that makes flux through the surface, so T*^12 is not zero.

- T^00
- Now consider a general frame F, with fluid four-velocity u^\mu
- If the three-velocity is
**v**, then u^\mu = \gamma (1,**v**), where \gamma = 1/sqrt{1-**v**^2}. - As expected for a four-velocity, u^2 = u_\mu u^\mu = 1.
- In F, T^\mu\nu = A g^\mu\nu + B u^\mu u^\nu by Lorentz covariance
(we must build T from available tensors and these are the only
symmetric combinations, since T^\mu\nu = T^\nu\mu).
- Specialize to F*, where u^\mu = (1,
**0**):

T*^00 = e = A + B

T*^11 = P = -A

=> P = -A, B = e + P - So T^\mu\nu = (e+P)u^\mu u^\nu - P g^\mu\nu
- Note that T^\mu_\mu = e - 3P

- Specialize to F*, where u^\mu = (1,

- If the three-velocity is

- Consider the collision of two identical nuclei at \sqrt{s} >= 100 GeV/nucleon (should be in baryon-free regime).
- The rapidity distribution dN/dy (although we might actually measure
dN_ch/d\eta) has a plateau in y around y=0
- We saw this even with p-p collisions. Even more so for nucleus-nucleus collisions.
- => dN/dy|y_L < y < y_R = dN/dy|y=0
- That is, dN/dy is the same for a range of y around y=0.
- This is particularly relevant since our estimate of the initial energy density depended on dN/dy|y=0.

- Recall that y transforms under a longitudinal boost simply
by changing by a constant:

y --> y' = y - y_\beta , with y_\beta = 1/2 log(1+\beta/1-\beta) - Therefore, the rapidity distribution in frame F' is the same as in frame F if the transformation of the rapidity keeps it in the central region.
- Bjorken's estimate for e'_0(F') can then be related to e_0(F):

e'_0(F') = [m_T/\tau_0 A] dN/dy'|y'=0 = [m_T/\tau_0 A] dN/dy|y=y_\beta

= [m_T/\tau_0 A] dN/dy|y=0 if y_L < y_\beta < y_R

= e_0(F) !!!- So e_0 is Lorentz invariant for a large range of y_\beta
- => simplifies hydrodynamics!

- Since invariant along longitudinal direction, thermodynamic variables
are functions of Lorentz invariants only => proper time \tau
- e = e(\tau)
- P = P(\tau)
- T = T(\tau)

- Energy and momentum conservation imply

\partial T^\mu\nu/\partial x^\mu = 0 - Count equations:
- \nu = 0,1,2,3 => 4 equations
- e,P,T,u^0,u^1,u^2,u^3 => 6 quantities to determine.
- Relation 1: e = e(P,T) [equation of state]
- Relation 2: u_\mu u^\mu = 1
- ==> sufficient!

- simplify the problem by considering only z and t
- Solve \partial T^\mu\nu/\partial x^\mu = 0 subject to Lorentz-invariance condistions that e,P,T depend only on \tau
- Two fewer equations from T^\mu\nu and two fewer quantities (don't need u^1 or u^2)
- Initial conditions:

e_0 = (m_T/\tau_0 A) dN/dy|y=0

z/t = v_z = tanh y - We can project out the equation we want by contracting u_\mu with \partial T^\mu\nu/\partial x^\mu after substituting the expression for T^\mu\nu in terms of e, p, and u.
- Some simplifications:
- x_\mu u^\mu = x_\mu dx^\mu/d\tau = 1/2 d(x^2)/d\tau = 1/2 d(\tau^2)/d\tau = \tau
- \tau^2 = x_\mu x^\mu => 2\tau d\tau/dx^\mu = 2 x_\mu => d\tau/dx^\mu = x_\mu/\tau
- So u_\mu d\tau/dx^\mu = \tau/\tau = 1
- u_\nu du^\nu/dx^\mu = 1/2 d(u^2)/dx^\mu = 0 since u^2=1

- Applying these simplifications we are left with:

\partial e/\partial \tau + (e+P)\partial u^\mu/\partial dx^\mu = 0 - The final ingredient is that

\partial u^\mu/\partial dx^\mu = 1/\tau- This follows since the initial velocity field satisfies u^\mu = x^\mu/\tau
- Taking the derivative (in 2 space-time dimensions) and using the fact that the general function can only depend on \tau gives the result.

- So the final result is:

\partial e/\partial \tau + (e+P)/\tau = 0

- Now consider a QGP modeled as massless quarks and gluons => P = e/3 is the equation of state.
- So the equation from the last section reduces to: de/d\tau = -4/3 e/\tau
- This is easily solved to give:
e(\tau)/e_0 = (\tau_0/\tau)^(4/3)

P(\tau)/P(\tau_0) = (\tau_0/\tau)^(4/3) - Since e \propto T^4, T \propto e^1/4

=> T(\tau)/T(\tau_0) = (\tau_0/\tau)^(1/3) - Since the number density of quarks n_q \propto T^3,

=> n_q(\tau)/n_q(\tau_0) = \tau_0/\tau

and similarly with the antiquark and gluon number densities

So what have we got?

- At time \tau=\tau_0, we have local thermal equilibrium with initial energy density e_0 and initial temperature T(\tau_0) \propto e_0^(1/4)
- As time evolves, e and P drop like \tau^{-4/3} while T drops only as \tau^{-1/3}
- The entropy density is (see last time) is \sigma = (e+P)/T,
so

\sigma(\tau)/\sigma(\tau_0) = \tau_0/\tau

or \sigma(\tau)*\tau = constant!- Since the volume element is dx_T \tau dy, we can write \sigma * \tau = dS/dx_T dy is a constant, so there is constant entropy per unit of rapidity

- The temperature has dropped to the critical temperature T_c
at time

\tau_c = (T(\tau_0)/T_c)^3 \tau_0

where the transition to the hadron phase takes place.- In the process there is a mixed phase.
- The system sits at temperature T_c while the entropy density is converted to the hadronic value
- The time it takes is roughly the ratio of the entropy
to the 3/4 power:

\tau_hadronization = (g_qg/g_h)^{3/4} \approx 6 \tau_c

- After hadronization, the temperature in the hadron phase continues to decrease until freeze-out, which is where thermal contact (interactions between hadrons that exchange energy) stops.

The problem set project for PS #3 is to study the phase coexistence curve
with a model for the hadronic side including a free gas of pions and a model
for *interacting* nucleons. On the QGP side we use the same bag
model used before, although now you must solve it for general T and \mu_B,
not just the limiting cases.

- The model is the so-called \sigma-\omega model.
- Besides pions, the degrees of freedom are nucleons (protons and
neutrons), and Lorentz scalar and vector mesons.
- The mesons only contribute to the energy density and pressure and density of the nucleons.
- With the parameters given in the handout, at T=0 the system has a minimum in the energy/baryon that is consistent with ordinary nuclei.

- Expressions for the thermodynamic quantities are given in the
handout, but note that the solution is nontrivial. That is because it
must be
*self-consistent*. - The nucleon has an
*effective mass*that has to be found by solving an equation that depends on the temperature and chemical potential. - General procedure at finite T and \mu:
- Given T and \nu (not \mu!), solve for M^*. This must be done numerically!
- Then find \rho_B, which then tells you \mu = \nu + (gv^2/mv^2)\rho_B. Nontrivial! Also a numerical problem.
- Then evaluate e and P.
- Given T and \mu, solve the QGP side and compare pressures to find the coexistence curve.

So far we have calculated the pressure and energy density of the QGP phase in the limit of noninteracting (massless) quarks and gluons. Of course, there are interactions. If the effective coupling is relatively small, we might hope to use perturbation theory (Feynman diagrams) as in QED.

- To calculate finite temperature and chemical potential corrections, we use the same sort of diagrammatic expansion as at T=\mu=0, but the Feynman rules for the quark and gluon propagators now include thermal contributions.
- We won't go into detail here, but just indicate the diagrams for the leading corrections (in class) and give the result.
- The first corrections are two-loop diagrams, proportional to \alpha_s, the strong version of the fine structure constant.
- The energy density to this order is:

E = [1 - (15/4 Pi) \alpha_s] 8 Pi^2/15 T^4 + \sum_{flavors} { [1 - (50/21 Pi) \alpha_s] 7 Pi^2/20 T^4

+ [(1 - 2/Pi \alpha_s] 3/Pi^2 \mu_i^2 [Pi^2 T^2 + \mu_i^2/2]} + B - For \alpha_s = 0.5, which is an appropriate value for a phase transition around T_C = 150-200 MeV, the corrections reduce the energy density by about a factor of two!

One can work to higher order in perturbation theory, but there are complications from infrared divergences. The only reliable calculations when nonperturbative physics is still important are numerical simulations on a lattice. We consider these next.

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Copyright © 1997,1998 Richard Furnstahl and James Steele.