Luciano G. MorettoLBNL-Berkeley CA

Fights for hadronic – partonic phase transitions

Phase transitions from Hadronic to Partonic Worlds

Phase transitions in the Hadronic world•Pairing (superconductive) Transition

finite size effects: correlation length•Shape transition

all finite size effects, shell effects

•Liquid-vapor (with reservations) van der Waals-like finite size effects due to surface

Tc ≈ 18.1 MeVc ≈ 0.53 0

pc ≈ 0.41 MeV/fm3

Phase transitions in the partonic world•Q. G. P. . . . Finite size effects?

Phase diagram via the liquid:Caloric curves

J. Pochodzalla et al., Phys. Rev. Lett. 75, 1040 (1995).

Excited nuclei treated as a heated liquid: Measure energy E and

temperature T E vs. T plotted as a

caloric curve: This curve is suggestive,

but some questions must be answered: For instance: what is the

volume and pressure of the system?

Caloric Curves and Heat Capacities …ctdCaloric Curves and Heat Capacities …ctd

A resistible temptation:

T

H

This is at constant pressure

T

∆E

This is at constant volume

But….Nuclei decay in vacuum …

so Heat capacities at constant “what”?

But….Nuclei decay in vacuum …

so Heat capacities at constant “what”?

Phase diagram via the liquid:Heat capacity

M. D'Agostino et al,Phys. Lett. B473 (2000) 219.

Excited nuclei treated as a heated liquid: Some measure of E is

partitioned Fluctuations in E compared to

nominal fluctuations: The results is interesting, but

some questions arise: How well was E

measured? How well were partitions of

E constructed? Fundamental questions of

the thermodynamics of small systems?

Negative heat capacities in infinite mixed phase

b)

T

P

T

p

(q/ T) < 0

Negative heat capacities in a single phase

T

P

Negative heat capacity here!!!!

a)

€

CX = CP −Vm

∂p

∂T X

Thermodynamic aside #1• Clausius-Clapeyron Equation:

– valid when:

vapor pressure ~ ideal gas Hevaporation independent of T

• Neither true as T Tc:– The two deviations compensate:

•

• Observed empirically for several fluids: “Thermodynamics” E. A. Guggenheim.

€

p = p0 exp −ΔH /T( )

€

p

pc

= expΔH

Tc

1−Tc

T

⎛

⎝ ⎜

⎞

⎠ ⎟

⎛

⎝ ⎜

⎞

⎠ ⎟

€

→

Thermodynamic aside #2• Principle of corresponding states:

Cubic coexistence curve. Empirically given by:

for liquid for vapor.

• Observed empirically in many fluids: E. A. Guggenheim, J. Chem. Phys. 13, 253 (1945).J. Verschaffelt, Comm. Leiden 28, (1896).J. Verschaffelt, Proc. Kon. Akad. Sci. Amsterdam 2, 588 (1900).D. A. Goldhammer, Z.f. Physike. Chemie 71, 577 (1910).

• 1/3 is critical exponent ≈

Or, why there are so few nuclear phase diagrams...The liquid vapor phase diagram – 3 problems:

Finite size: How to scale to the infinite system?

Coulomb: Long range force

No vapor in equilibrium with a liquid drop. Emission into the vacuum.

Saturated VaporSaturated Vapor ( V.d.W forces) and the phase diagram

Infinite system :the Clapeyron equation or Thermodynamic frugality

Hm≈ aV+p Vm≈ aV+T

Vm≈ Vm ≈ T/p

vap

Now integrate the Clapeyron equation to obtain the phase diagram p= p(T)

m

m

VT

H

dT

dP

=

Finiteness Effects : LiquidFiniteness Effects : LiquidShort Range Forces ( V.d.W.)

Finiteness can be handled to a good approximation by the liquid drop expansion ( A-

1/3)

EB= aVA + aSA2/3 +aC A1/3 ……. = A(aV +aSA-1/3+ aCA-2/3…..)

Liquid Drop Model in nuclei: • stops to 1st order in A-1/3 • good to 1% ( ≈ 10 MeV)• good down to very small A (A ≈ 20)

Extra bonus:• aV≈ -aS in all V.d.W systems

The binding energy/nucleon aV is essentially sufficient to do the job!

The saturated vapor is a non ideal gas. We describe it in terms of a Physical Cluster Model.

Physical Cluster Model: Vapor is an ideal gas of clusters in equilibrium

If we have n(A,T), we have the phase diagram:

P=T n(A,T)

= An(A,T)So

What is n(A,T)?

If we have n(A,T), we have the phase diagram:

P=T n(A,T)

= An(A,T)So

What is n(A,T)?

Fisher ModelFisher Model

n(A,T)= q0A- exp- T

Ac σε with c

c

T

TT −=ε

Where does this come from?

Example: Two dimensional Ising Model

n(A,T)=g(A)exp-T

Ac σ

Asymptotic expression for g(A)

g(A)~A- exp kAσ

Fisher writesFisher writes

n(A,T)= q0A- expT

Ac

T

Ac

c

σσ −

Finite size effects: Complement

• Infinite liquid • Finite drop

nA (T) g(A)exp ES (A)

T

nA (A0,T) g(A)g(A0 A)

g(A0)exp

ES (A0,A)

T

• Generalization: instead of ES(A0, A) use ELD(A0, A) which includes Coulomb, symmetry, etc.• Specifically, for the Fisher expression:

nA (T) q0

A A0 A

A0 exp

c0ε Aσ (A0 A)σ A0σ

T

Fit the yields and infer Tc (NOTE: this is the finite size correction)

Test Complement with Ising model

• 2d lattice, L=40, =0.05, ground state drop A=80

• Regular Fisher, Tc=2.07

• Tc = 2.32±0.02

to be compared with the theoretical value of 2.27...

• Can we declare victory?

A=1

A=10

Complement for excited nuclei

• Fisher scaling collapses data onto coexistence line

• Gives bulk

Tc=18.6±0.7 MeV

• pc ≈ 0.36 MeV/fm3

• Clausius-Clapyron fit: E ≈ 15.2 MeV

• Fisher + ideal gas:

p

pc

T nA T

A

T nA Tc

A

• Fisher + ideal gas:

v

c

nA T A

A

nA Tc A

A

• c ≈ 0.45 0

• Full curve via Guggenheim

Fit parameters:L(E*), Tc, q0, Dsecondary

Fixed parameters:, σ, liquid-drop coefficients

The partonic world (Q.G.P.)(a world without surface?)

• The M.I.T. bag model says the pressure of a Q.G.P. bag is constant:

• ; g: # degrees of freedom, constant p = B, constant .

• The enthalpy density is then

•

• which leads to an entropy of

•

• and a bag mass/energy spectrum (level density) of

• .

• This is a Hagedorn spectrum:

•

Partonic vacuum

Hadronic vacuum

€

p =gπ 2

90TΗ

4 = B

€

ε =H

V=

E

V+ p =

gπ 2

30TΗ

4 + B

€

S =δQ

T∫ =

dH

T0

H

∫ =H

TΗ

≡m

TΗ

€

m( ) = exp S( )∝ expm

TΗ

⎛

⎝ ⎜

⎞

⎠ ⎟

€

H m( )∝m0

m

⎛

⎝ ⎜

⎞

⎠ ⎟x

expm

TΗ

⎛

⎝ ⎜

⎞

⎠ ⎟

€

TH = B90

gπ 2

⎛

⎝ ⎜

⎞

⎠ ⎟

1

4

?

mm0

ln

(m

)

Origin of the bag pressure

•To make room for a bubble of volume V an energy E = BV is necessary.•To stabilize the bubble, the internal vapor pressure p(T) must be equal to the external pressure B.

•Notice that the surface energy coefficient in this example is not obviously related to the volume energy coefficient.

10 m

B = 1 atm

Can a “thermostat” have a temperature other than its own?

•

•

• Is T0 just a “parameter”?

•

• According to this, a thermostat, can

have any temperature lower than its

own!

Z T dE E e E T T0T

T0 TeS0

E eS eS0

E

T0

S S0 Q

TS0

E

T0

T = Tc = 273Kor

0 ≤ T ≤ 273K ?

• The total level density:

• Most probable energy partition:

• TH is the sole temperature characterizing the system:

• A Hagedorn-like system is a perfect thermostat.

• If particles are generated by the Hagedorn bag, their concentration is:

• Volume independent! Saturation! Just as for ordinary water, but with

only one possible temperature, TH!

Equilibrium with Hagedorn bags:Example : an ideal vapor of N particles of mass m and energy ε

€

P E,ε( ) = ρ H E −ε( )ρ iv ε( ) = g(m)V N

N!3

2N

⎛

⎝ ⎜

⎞

⎠ ⎟!

mε

2π

⎛

⎝ ⎜

⎞

⎠ ⎟

3

2N

expE − mN −ε

TΗ

⎛

⎝ ⎜

⎞

⎠ ⎟

€

∂ lnP

∂ε=

3N

2ε−

1

TΗ

= 0⇒ε

N=

3

2TΗ

€

∂ lnP

∂N V

= −m

TΗ

+ ln g(m)V

N

mTΗ

2π

⎛

⎝ ⎜

⎞

⎠ ⎟

3

2 ⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥= 0⇒

N

V= g(m)

mTΗ

2π

⎛

⎝ ⎜

⎞

⎠ ⎟

3

2exp −

m

TΗ

⎛

⎝ ⎜

⎞

⎠ ⎟

(E)

ideal vapor iv• particle mass = m• volume = V• particle number = N• energy = ε

1. Anything in contact with a Hagedorn bag acquires the temperature

TH of the Hagedorn bag.

2. If particles (e.g. s) can be created from a Hagedorn bag, they will

form a saturated vapor at fixed temperature TH.

3. If different particles (i.e. particles of different mass m) are created

they will be in chemical equilibrium.

H(E)

The story so far . . .

Now to the gas of bags …

(Gas of resonances? )

Stability of the Hagedorn bag against fragmentation

• If no translational or positional entropy, then the Hagedorn bag is indifferent to fragmentation.

H(m)H(mk)

H(m3)H(m2)

H(m4)

H(m5)

H(m5)

H(m1)

H(m6)

€

expm

TΗ

⎛

⎝ ⎜

⎞

⎠ ⎟= exp

mi

i=1

k

∑TΗ

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

indifferent

Equilibrium with Hagedorn bags:

€

N

V= g(m)

mTΗ

2π

⎛

⎝ ⎜

⎞

⎠ ⎟

3

2exp −

m

TΗ

⎛

⎝ ⎜

⎞

⎠ ⎟

(E)

ideal vapor iv• particle mass = m• volume = V• particle number = N• energy = ε

€

g(m) = em

TH

€

N

V=

mTΗ

2π

⎛

⎝ ⎜

⎞

⎠ ⎟

3

2

T < TH T = TH T = TH

Non saturated gas of π etc.

Gas of bags +

saturated gas of π etc.

One big bag

T

€

εT 4

TH

Bags have no surface energy :

What about criticality?

€

P(A) = Kg(A)e−

C s A2 3

T = KA−τ eC s A

23

Tcr e−

C s A2

3

T

€

g(A) ≅ A−τ eKA2

3

? ?

€

lng(A) = SurfaceEntropy = KA2

3 =Cs

Tcr

A2

3

This is predicated upon a nearly spherical cluster.

Lattice Animals

///

/// ///

/// ///

///

///

///

How many animals of size A ?

Fisher guesses

To my knowledge nobody knows exactly why .

€

lnP(A) = −τ ln A + KA2

3

A) 2exp(2.1198 A 0.120705)( 386751.-Ag =

……. Instead

€

lnP(A) = αA2

3 + βA

How to resolve this conundrum?….

ln P

(A,S

)

S

Sphere-likefractal

€

P(A,S)e−

S

T

€

P(A,T) = P(A,S)e−

S

T ds∫With increasing temperature ..

T

Fractal dimension goes from surface-like to volume-like

For 3d animals

A bag with a surface?•Remember the leptodermous expansion:

•

•Notice that in most liquids aS ≈ -aV

•However, in the MIT bag there is only a volume term

•

• Should we introduce a surface term? Although we may not know the magnitude of it, we know the sign (+). The consequences of a surface term:

•

•

€

M = E ≅ H = aV A + aS A2 3 + aC A1 3

€

εV = H = f T( ) + B[ ]V + aSV2 3 ?( )

V

TT

c

€

p =1

3f T( ) − B +

2

3aSV

−1 3 ⎛

⎝ ⎜

⎞

⎠ ⎟= 0 at equilibrium

€

T = f −1 3 B +2

3aSV

−1 3 ⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥

V

0

Cp=

0

V

εV

Stability of a gas of bags

Bags of different size are of different temperature. If the bags can fuse or fission, the lowest temperature solution at constant energy is a single bag. The isothermal solution of many equal bags is clearly unstable.

A gas of bags is always thermodynamically unstable.

A bag decays in vacuum by radiating (e.g. pions). As the bag gets smaller, it becomes HOTTER! Like a mini-black hole.

The decay of a bag with surface

Conclusions

1) The bag supports a 1st order phase transition

2) A gas of bags is entropically unstable towards coalescence

Bag Non Hagedorn particles ( pions?)

at a single TH

Hagedorn dropsBag

Conclusions ctd..

3) The lack of surface energy entropically drives bag to fractal shape

4) Addition of surface energy makes drops non isothermal.