Luciano G. Moretto LBNL-Berkeley CA Fights for hadronic – partonic phase transitions.
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Transcript of Luciano G. Moretto LBNL-Berkeley CA Fights for hadronic – partonic phase transitions.
![Page 1: Luciano G. Moretto LBNL-Berkeley CA Fights for hadronic – partonic phase transitions.](https://reader035.fdocuments.net/reader035/viewer/2022070400/56649e755503460f94b762ff/html5/thumbnails/1.jpg)
Luciano G. MorettoLBNL-Berkeley CA
Fights for hadronic – partonic phase transitions
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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?
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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?
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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”?
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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?
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Negative heat capacities in infinite mixed phase
b)
T
P
T
p
(q/ T) < 0
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Negative heat capacities in a single phase
T
P
Negative heat capacity here!!!!
a)
€
CX = CP −Vm
∂p
∂T X
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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
⎛
⎝ ⎜
⎞
⎠ ⎟
⎛
⎝ ⎜
⎞
⎠ ⎟
€
→
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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 ≈
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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.
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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
=
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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!
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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)?
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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 −
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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)
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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
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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
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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
)
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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
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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 ?
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• 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
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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 . . .
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Now to the gas of bags …
(Gas of resonances? )
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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
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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
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T < TH T = TH T = TH
Non saturated gas of π etc.
Gas of bags +
saturated gas of π etc.
One big bag
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T
€
εT 4
TH
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Bags have no surface energy :
What about criticality?
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€
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.
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Lattice Animals
///
/// ///
/// ///
///
///
///
How many animals of size A ?
Fisher guesses
To my knowledge nobody knows exactly why .
€
lnP(A) = −τ ln A + KA2
3
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A) 2exp(2.1198 A 0.120705)( 386751.-Ag =
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……. 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
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For 3d animals
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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
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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
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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
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Conclusions ctd..
3) The lack of surface energy entropically drives bag to fractal shape
4) Addition of surface energy makes drops non isothermal.