Post on 14-Dec-2015
transcript
Microcanonical, canonical and
grand canonical pains with the Hagedorn spectrum
Luciano G. Moretto, L. Ferroni, J. B. Elliott, L. Phair
UCB and LBNL Berkeley
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( ) = dEρ E( )e−E T∫ =T0T
T0 −TeS0
€
ρ E( ) = eS = eS0 +
E
T0
€
S = S0 +ΔQ
T= S0 +
E
T0
T = Tc = 273Kor
0 ≤ T ≤ 273K ?
Summary
The true canonical temperature
The true microcanonical temperature (are they the same?)
The grand canonical mass spectrum
The gas of bags
Fluctuations and criticality. Is there a critical point?
Do bags have surface energy?
The (too) many ways of obtaining the Hagedorn spectrum( given the experimental evidence!!)
1. Bootstrap
1. Mit Bag
1. Regge Trajectories
1. Fractal shapes ( if no surface energy)
1. -----------
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
€
ρ m( ) = exp S( )∝ expm
TΗ
⎛
⎝ ⎜
⎞
⎠ ⎟
€
ρH m( )∝m0
m
⎛
⎝ ⎜
⎞
⎠ ⎟x
expm
TΗ
⎛
⎝ ⎜
⎞
⎠ ⎟
€
TH = B90
gπ 2
⎛
⎝ ⎜
⎞
⎠ ⎟
1
4
?
mm0
ln ρ
(m
)
Bag equilibrium with Fermionic obligatory constituents
• with
• For P0 = B the bag is stabilized by the quantum pressure and TH = 0.
€
aT 4 + P0 T = 0,n( ) + F T,n( ) = B
€
P0 = kn5 3
m
n
T
€
n0 =Bm
k
⎛
⎝ ⎜
⎞
⎠ ⎟
3
5
Fractal Shapes
Fisher Model
3/23/23/23/2 )11()(),( ATTCTACKATAC crsss eeeAgTAP −−+− ≅≅=
g(A)ln
g(A
,s)-C
ss/T
Ln
g(A
,s)
s
If Cs =0 S = KA
ρ(M) = exp KA
For 3d animals
A) 2exp(2.1198 A 0.120705)( 386751.-Ag =
Regge Trajectories
M2
I
M2 = kI
M2 = kn h
In how many ways can one write ∑<
=nn
nn1'
'
Euler’s Partitio Numerorum
€
P(n) ≈1
4n 3eπ 2n 3 ≈ ebM = P(M)
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( ) = dEρ E( )e−E T∫ =T0T
T0 −TeS0
€
ρ E( ) = eS = eS0 +
E
T0
€
S = S0 +ΔQ
T= S0 +
E
T0
T = Tc = 273Kor
0 ≤ T ≤ 273K ?
Thermal equilibriumSystem A System B
No thermostat: any temperature
One thermostat: one temperature
Two thermostats: no temperature
S
E
S
ES
E
S
ES
E
S
E
€
∂S1
∂E1
=∂S2
∂E2
∴ 1
T1
=1
T2
Equilibrium with Hagedorn bags:Example #1: the one dimensional harmonic oscillator
•BEWARE! A linear dependence of S on E spells danger to the unaware!•Beware of the canonical and grand canonical ensembles!•When in doubt, Go MICRO-canonical !
•For a one dimensional harmonic oscillator with energy ε in contact with a Hagedorn bag of energy E:
•The probability P(ε) is:
€
P ε( ) = ρ E −ε( ) = expE −ε
TΗ
⎛
⎝ ⎜
⎞
⎠ ⎟= exp
E
TΗ
⎛
⎝ ⎜
⎞
⎠ ⎟exp −
ε
TΗ
⎛
⎝ ⎜
⎞
⎠ ⎟
E ε
ln Pε
€
slope =1
TH•The most probable value of ε:
•
•For E : ε TH
€
ε =TH 1−E TH
exp E TH( ) −1
⎛
⎝ ⎜
⎞
⎠ ⎟
The resistible troubles with microcanonicity
Harmonic Oscillator ( 1 dim.) ctd
The microcanonical partition
)1()()( / −==Ω ∫ HTEH eTdPE εε
HTEH eTE /1
11ln−−
=∂Ω∂
)1( )/ HTEH eTT −−= : is this a Temperature ?
E
1.0
T/TH
Let us look at the spectra
ln P
(ε)
ε E1 E2 E3
Slope= 1/TH
ln P
(ε)
ε
Slope= 1/TH
Identical results for an ideal gas
Conclusion: Beware of microcanonical temperatures
)2
1()1(0
2/1
0
2/30 MM
MeK HT εεεε
−−=−
Microcanonical quirks for a gas of bags
Microcanonical Partition
==Ω−
επε
ε
d
dpm
m
pe
h
VE HT
M
22
4),(
2
3
0
m= M0 - ε
€
lnΩ(E,ε)
ε
⎛
⎝ ⎜
⎞
⎠ ⎟
ε
• 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 #2: an ideal vapor of N particles of mass m and energy ε
€
P E,ε( ) = ρH E −ε( )ρ iv ε( ) =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
+ lnV
N
mTΗ
2π
⎛
⎝ ⎜
⎞
⎠ ⎟
3
2 ⎡
⎣
⎢ ⎢
⎤
⎦
⎥ ⎥= 0⇒
N
V=
mTΗ
2π
⎛
⎝ ⎜
⎞
⎠ ⎟
3
2exp −
m
TΗ
⎛
⎝ ⎜
⎞
⎠ ⎟
ρ (E)
ideal vapor ρiv• 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. πs) 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.
ρH(E)
The story so far . . .
Stability of the Hagedorn bag against fragmentation
• If no translational or positional entropy, then the Hagedorn bag is indifferent to fragmentation.
ρH(m)ρH(mk)
ρH(m3)ρH(m2)
ρH(m4)
ρH(m5)
ρH(m5)
ρH(m1)
ρH(m6)
€
expm
TΗ
⎛
⎝ ⎜
⎞
⎠ ⎟= exp
mi
i=1
k
∑TΗ
⎛
⎝
⎜ ⎜ ⎜ ⎜
⎞
⎠
⎟ ⎟ ⎟ ⎟
indifferent
Resonance gas - A gas without pressure
No intrinsic energy and/or entropy penalty for aggregation
How many particles?
1 ≤ N ≤ Nmax
Ideal gas law:
€
p =N
VT
Resonance Gas Cont’d
€
A ⇔ nB
€
cBn
cA
=qB
n
qA
=qA
n−1
n3n / 2
€
qB =2πmAT
h2n
⎛
⎝ ⎜
⎞
⎠ ⎟3 / 2
=1
n3 / 2qA
€
limn→∞
cB = limn→∞
cA1/ nqA
n−1
n
n3 / 2= 0
€
%dissoc =1
n5 / 2q
n−1
ncA
1/ n
c0
%
diss.
n
n
HTCp 0
n -5/2
1.0
p almost constant !
n -3/2
• 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 #2: an ideal vapor of N particles of mass m and energy ε
€
P E,ε( ) = ρH E −ε( )ρ iv ε( ) =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
+ lnV
N
mTΗ
2π
⎛
⎝ ⎜
⎞
⎠ ⎟
3
2 ⎡
⎣
⎢ ⎢
⎤
⎦
⎥ ⎥= 0⇒
N
V=
mTΗ
2π
⎛
⎝ ⎜
⎞
⎠ ⎟
3
2exp −
m
TΗ
⎛
⎝ ⎜
⎞
⎠ ⎟
ρ (E)
ideal vapor ρiv• particle mass = m• volume = V• particle number = N• energy = ε
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
The EoS for the Gas of Hagedorns
For the mass spectrum
Is the most probable concentration of H with mass m
Energy conservation defines the maximal mass of Hagedorns mmax and fixes the number of Hagedorns Ntot and pressure P of the system
Maximal number of Hagedorns
The EoS for the Gas of HagedornsEnergy conservation defines the maximal mass of Hagedorns mmax
And gives EoS as P(E/V) : for M = 15.9 GeV to fit Ntot of all resonances with masses < 1.85 GeV
For plots E = 2 GeV
Is speed of sound square
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 as, we know the sign (+). The consequences of a surface term:
•
•
€
M = E ≅ H = aV A + aSA2 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
Is the surface energy temperature dependent?
•
•From the bag stability condition:
€
cs = cs0 1−
T
Tc
⎛
⎝ ⎜
⎞
⎠ ⎟
Tc
€
2c0V−2 3
•For V very small T Tc
€
σTH
4 = 3 B +2
3cs
0 1−T
Tc
⎛
⎝ ⎜
⎞
⎠ ⎟V
−2 3 ⎡
⎣ ⎢
⎤
⎦ ⎥
T€
3B
V
TT
c
Tindep
Tdep
The perfect “Granulator”
Tc
TH
Phase transitions from Hadronic to Partonic Worlds L. G. Moretto, K. A. Bugaev, J. B. Elliott and L. Phair
Lawrence Berkeley National LaboratoryNuclear Science Division
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 MeVρc ≈ 0.53 ρ0
pc ≈ 0.41 MeV/fm3
Phase transitions in the partonic world•Q. G. P. . . . Finite size effects?
Fractal Shapes
Fisher Model
3/23/23/23/2 )11()(),( ATTCTACKATAC crsss eeeAgTAP −−+− ≅≅=
g(A)ln
g(A
,s)-C
ss/T
Ln
g(A
,s)
s
If Cs =0 S = KA
ρ(M) = exp KA
For 3d animals
A) 2exp(2.1198 A 0.120705)( 386751.-Ag =
Conclusion1. Any system with a Hagedorn-like spectrum
is a perfect thermostat.
2. If such a system evaporated non-Hagedorn particles they constitute a saturated vapor in physical and chemical equilibrium.
3. A vapor of Hagedorn bags is indifferent to aggregation/fragmentation P=0.
4. Surface energy further destabilizes the vapor. Smaller bags are at higher temperature.
ConclusionsNuclear dropletsIsing lattices
• Surface is simplest correction for finite size effects (Rayleigh and Clapeyron)
• Complement accounts for finite size scaling of droplet
• For ground state droplets with A0<<Ld, finite size effects due to lattice size are minimal.
• Surface is simplest correction for finite size effects(Rayleigh and Clapeyron)
• Complement accounts for finite size scaling of droplet
• In Coulomb endowed systems, only by looking at transition state and removing Coulomb can one speak of traditional phase transitions
Bulk critical pointextracted whencomplement
takeninto account.
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
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