8.08 Statistical Physics (II) – Lecture note 1
Xiao-Gang Wen, MIT
2019 Spring
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Statistical Physics
• Partial information to partial result
- 1 mole = NA = 6.02214076 × 1023 (Avogadro number)
≈ numbers of 1g of protons or neutrons or H atoms
- To specify the micro state of 2g of H2 gas, we need to specify
3 × 6.022 × 1023 real numbers for the positions and
3 × 6.022 × 1023 real numbers for the momenta. Then we can
compute the properties of the gas at a later time.
- If we only have partial information of the gas (a few real numbers)
how to compute a few properties of the gas at a later time?
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Statistical ensemble
• All possible states that satisfy the partial information appear
with an equal probability.
- Statistical Physics deals with an ensemble of the systems.
• Time average and ensemble average
- For one system, we may obtain an ensemble of the systems is form
by the system at different times → an ensemble that satisfies the
property that all possible states appear with an equal probability.
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Microcanonical ensemble
• The partial information is 1 real number E ± 21 δE , the total energy
of the system → temperature and heat capacity of the systems.
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Microcanonical ensemble
• The partial information is 1 real number E ± 21 δE , the total energy
of the system → temperature and heat capacity of the systems.
• Number of states – an example of N spin-1/2’s
Consider N spins in magnetic field. The energy for an up-spin is
E↑ = 0 /2 and for a down-spin E↓ = −0 /2.
How many states are there with a total energy
E = M 20 + (N − M) −2 0 ? (M up-spin, N − M down-spin)
N!
The answer is Γ(E ) = M!(N−M)!
• The above does not make sense since E is 0 × integer. Include
the effect of δE :
N!
δE
Γ(E ) =
M!(N − M)! 0
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Microcanonical ensemble
• The partial information is 1 real number E ± 21 δE , the total energy
of the system → temperature and heat capacity of the systems.
• Number of states – an example of N spin-1/2’s
Consider N spins in magnetic field. The energy for an up-spin is
E↑ = 0 /2 and for a down-spin E↓ = −0 /2.
How many states are there with a total energy
E = M 20 + (N − M) −2 0 ? (M up-spin, N − M down-spin)
N!
The answer is Γ(E ) = M!(N−M)!
• The above does not make sense since E is 0 × integer. Include
the effect of δE :
N!
δE
Γ(E ) =
M!(N − M)! 0
• Entropy
S(E ) = kB ln Γ(E )
where kB = 1.3807 × 10−16 erg/K is the Boltzmann constant
– the unit of entropy
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Calculating entropy
Using the Stirling approximation
ln(N!) = N ln N − N +
1
ln(2πN) + O(1/N)
2
We find
kB−1 S(E ) = ln
δE
N!
+ ln
M!(N − M)!
0
s
δE
2πN
≈N ln N − M ln M − (N − M) ln(N − M) + ln
0 2πM 2π(N − M)
s
M
N
N −M
δE
= − M ln( ) − (N − M) ln(
) + ln
N
N
0 2πM(N − M)
δE
1
− ln 2πf↑ f↓
=N(−f↑ ln f↑ − f↓ ln f↓ ) + ln
0
2
|
{z
}
N 0 ln N terms
f↑ ≡
M
N , f↓
≡1−
M
N
the probabilities for a spin to be up and down.
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Entropy and information
• In the thermodynamic limit N → ∞, S(E ) has an extensive term
1
N(−f↑ ln f↑ − f↓ ln f↓ ) and a N 0 ln N term ln δE
0 − 2 ln 2πf↑ f↓ .
- In this class, we always deal with the thermodynamic limit N → ∞.
We will ignore the sub leading term:
S(E ) = kB N(−f↑ ln f↑ − f↓ ln f↓ )
- Entropy (the leading term) is additive S1+2 = S1 + S2 .
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Entropy and information
• In the thermodynamic limit N → ∞, S(E ) has an extensive term
1
N(−f↑ ln f↑ − f↓ ln f↓ ) and a N 0 ln N term ln δE
0 − 2 ln 2πf↑ f↓ .
- In this class, we always deal with the thermodynamic limit N → ∞.
We will ignore the sub leading term:
S(E ) = kB N(−f↑ ln f↑ − f↓ ln f↓ )
- Entropy (the leading term) is additive S1+2 = S1 + S2 .
• Entropy per spin measures the uncertainty (lack of information)
X
s = −kB f↑ ln f↑ − kB f↓ ln f↓ = kB
−Pi ln Pi
i
i labels different possible states. Pi is the probability for state-i.
- For a system
probability, Pi = Γ−1 ,
Pwith Γ states with
Pequal
Γ
→ S = kB i −Pi ln Pi = kB i=1 −Γ−1 ln Γ−1 = kB ln Γ.
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Entropy and information
• In the thermodynamic limit N → ∞, S(E ) has an extensive term
1
N(−f↑ ln f↑ − f↓ ln f↓ ) and a N 0 ln N term ln δE
0 − 2 ln 2πf↑ f↓ .
- In this class, we always deal with the thermodynamic limit N → ∞.
We will ignore the sub leading term:
S(E ) = kB N(−f↑ ln f↑ − f↓ ln f↓ )
- Entropy (the leading term) is additive S1+2 = S1 + S2 .
• Entropy per spin measures the uncertainty (lack of information)
X
s = −kB f↑ ln f↑ − kB f↓ ln f↓ = kB
−Pi ln Pi
i
i labels different possible states. Pi is the probability for state-i.
- For a system
probability, Pi = Γ−1 ,
Pwith Γ states with
Pequal
Γ
→ S = kB i −Pi ln Pi = kB i=1 −Γ−1 ln Γ−1 = kB ln Γ.
• Information carried by a message 0110011000101001011 · · ·
information per bit = −P0 ln P0 − P1 ln P1
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Entropy as a function of energy
From E = M 20 − (N − M) 20 , we have f↑ = 12 + EE0 and
f↓ = 21 − EE0 where E0 = N0 , we find
h
1
E
1
E
1
E
1
E i
S(E ) =kB N − ( + ) ln( + ) − ( − ) ln( − )
2 E0
2 E0
2 E0
2 E0
Using S(E ) we can compute all kinds of thermodynamic
properties of the system, such as temperature
• From the definition, the number of states with energy E ± 21 δE is
determined by the entropy
Γ(E ) = eS(E )/kB .
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Thermal equilibration and temperature
Consider two systems, one with energy E1 ± 12 δE and another with
energy E2 ± 21 δE . The numbers of the states for the two systems:
Γ1 (E1 ) = eS1 (E1 )/kB ,
Γ2 (E2 ) = eS2 (E2 )/kB .
sys.−1
system−2
E1
E2
• The combined system has a total energy E = E1 + E2 ± 12 δE
X
Γtot (E ) =
Γ1 (E1 )Γ2 (E2 )
E1
P(E1 ) ∝ Γ1 (E1 )Γ2 (E − E1 )
= eS1 (E1 )/kB eS2 (E −E1 )/kB
1
0.8
P(E 1)/P max
where we view energy as
“discrete” values with
discretization δE .
- The probability for sys.-1,sys.-2
to have energies (E1 , E − E1 )
0.6
10
0.4
100
0.2
1000
0
−0.5
−0.4
10000
−0.3
−0.2
−0.1
0
E1 / |E |
N1 = 21 N2 = 10, 100, · · ·, E1 = −N1 0
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Thermal equilibration and temperature
In thermodynamic limit, the system-1 has a “definite” energy E1
which maximizes P(E1 ). The equilibration energy satisfies
∂E1 P(E1 ) = 0 →
∂S1 (E1 )
∂S2 (E − E1 )
=−
∂E1
∂E1
• Let us introduce the temperature
of the system via S(E ):
1/T = kB β = ∂S(E1 )/∂E
kB conversion between T and E
- The equilibration condition
becomes T1 = T2
β
Τ
0
Τ
• For our spin system ( = E /N)
−0.5
ε /ε 0
1
kB 12 E0 − E kB 21 0 − kB f↓
= βkB =
ln 1
ln 1
ln
=
=
T
0
0
0 f↑
2 E0 + E
2 0 + - When T = ∞? When T < 0?
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
β
0.5
Boltzmann distribution
• We also see that
f↑
= e−0 /kB T = e−(E↑ −E↓ )/kB T ,
f↓
E↑ − E↓ = 0
which gives us
1
f↑ =
e− 2 0 /kB T
1
1
1
,
f↓ =
e+ 2 0 /kB T
1
1
e− 2 0 /kB T + e 2 0 /kB T ‘
e− 2 0 /kB T + e 2 0 /kB T ‘
- The above is the probability distribution of a single spin-1/2 at
temperature T .
But how can a single spin-1/2
to have a temperature T ?
• In general Pi ∝ e−Ei /kB T or
e−Ei /kB T
Pi = P −E /k T
j
B
j e
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
.
Curie’s law – high temperature magnetic susceptibility
• For a spin-1/2 system in magnetic field B, 0 = g µB B. The total
magnetic moment is M = N(f↑ − f↓ ) g µ2 B . We find
!
g µB
e −0 /kB T
1
M=N
−
2
e −0 /kB T + 1 e −0 /kB T + 1
=
we have
g 2 µ2B N
M=
B
4kB T
We find magnetic susceptibility
g 2 µ2B N
χ=
∝ 1/T .
4kB T
0.008
experiment
Curie law
(emu/mole Cu)
kB T
g µB ,
Ca 2Y2 Cu5 O10
χ
• For B 0
Ng µB
g µB B
Ng µB
tanh
=
tanh
2
2kB T
2
2kB T
0
0
100
200
300
T (K)
• The deviation at low temperatures is due to spin-spin interaction.
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Irreversible and reversible processes
Consider two identical systems with total energy E1 + E2 = 2Ē
Γ( E 1 ) Γ( E 2 )
E1
2E
E1
Γ( E ) Γ( E )
E2
(a)
E
(b)
E
E
(c)
Irreveisible
Reversible
4 states
4 states
7 states
4 states
• Irreversible process:
E
Number of many-body
energy levels in δE increases
by a factor e#N .
• Reversible process:
Number of many-body energy levels in δE does not change up to a
factor N. (Example: change the magnetic field of the spin system)
Irreversible process entropy increase.
Reversible process entropy no change.
Xiao-Gang Wen, MIT
gas
gas
8.08 Statistical Physics (II) – Lecture note 1
Canonical ensemble
A system has states labeled by n = 1, 2, 3, · · ·
.
What is the probability for
Sys
the system to be in the state n
with energy En .
Canonical
ensemble
Heat
bath
Temperature
T
Micro canonical ensemble
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Canonical ensemble
A system has states labeled by n = 1, 2, 3, · · ·
.
What is the probability for
Sys
the system to be in the state n
with energy En .
Heat
bath
Temperature
• The ideal heat bath:
Canonical
T
ensemble
T independent of energy
−1
From T = ∂S(E )/∂E
Micro canonical ensemble
→ Sbath (E ) = E /T + const.
→ number of states of the bath: Γbath (E ) = const. × eE /kB T
- The heat = ST .
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Canonical ensemble
A system has states labeled by n = 1, 2, 3, · · ·
.
What is the probability for
Sys
the system to be in the state n
with energy En .
Heat
bath
Temperature
• The ideal heat bath:
Canonical
T
ensemble
T independent of energy
−1
From T = ∂S(E )/∂E
Micro canonical ensemble
→ Sbath (E ) = E /T + const.
→ number of states of the bath: Γbath (E ) = const. × eE /kB T
- The heat = ST .
• Boltzmann distribution:
- sys. + bath = micro canonical ensemble with a total energy Etot .
- If the system is in state n, the heat bath has energy Etot − En
- The number of states of the whole system:
Γtot (En ) = const. × e(Etot −En )/kB T
- The probability distribution of sys. Pn ∝ # of states ∝ e−En /kB T
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Partition function and free energy
- Using Boltzmann distribution Pn ∝ e−En /kB T , we can compute
various thermodynamic properties of the system.
- Using the partition function Z (T ) or free energy, we can compute
various thermodynamic properties of the system.
• Partition function = the total “probability”
# of states per energy
# of states of sys.+bath
z
}|
{
X
−En /kB T
e
Z (T ) ≡
Z
=
n
# of states within δE
Z
=
dE
zX
}|
{
δ(E − En ) e−En /kB T
n
sharply peaked
z }| {
z
}|
{
Z
Γsys (E ) −E /kB T
dE − k 1T (E −S(E )T )
dE
e
=
e B
δE
δE
- The system has an energy Ē that maximizes the free energy
F̃ (E , T ) ≡ E − S(E )T
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Partition function and free energy
F (T ) = Ē − S(Ē )T
• The minimal value
.
is the real free energy which is a function of T ). Here Ē satisfies
dS(E )
1
. Equilibration T = Tbath = Tsys
T =
E =Ē
| dE {z
}
1/Tsys
- e−F̃ (E ,T )/kB T ∝ the probabiltiy of canonical ensemble.
System wants to minimize the free energy to reach equilibration.
When T = 0, the sys. minimizes the energy to reach equilibration
- The real free energy F (T ) is also given by
X
F (T ) = −kB T ln Z (T ) = −kB T ln
E =δE ×int.
e
− k 1T (E −S(E )T )
B
E
range −F (T )/kB T
The reasoning ln e−F (T )/kB T < ln Z (T ) < ln δE
e
F (T )
F (T )
− kB T < ln Z (T ) < − kB T + # ln N
P − 1 F̃ (E ,T )
− 1 F̃ (Ē ,T )
- A maximal term ∼ whole: ln E e kB T
∼ ln e kB T min
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Compare micro-canonical and canonical ensemble
• Probabilty for the system to have an energy E :
∝ Γ(E ) = eS(E )/kB
• Probabilty for the system to have an energy E :
−
∝ Γsys.+bath ∝ Γsys. (E )e−E /kB T ∝ e
Xiao-Gang Wen, MIT
(E −S(E )T )
kB T
−
∝e
F̃ (E ,T )
kB T
8.08 Statistical Physics (II) – Lecture note 1
Thermodynamic quantities and relations
• Thermodynamic quantities:
Extensive E , S, F , V
Intensive T
• Micro canonical ensemble: from E , V and S(E , V ) to obtain
other thermodynamic quantities.
1
- T = ∂S/∂E
|V .
- Adiabatic expandsion:
∆E = −force × ∆x = −area × p × ∆x = −∆V × p
∂S
∂S
∂S
∂S
0 = ∆S = ∂E
∆E + ∂V
∆V = − ∂E
p∆V + ∂V
∆V
→ pressure p =
∂S/∂V |E
∂S/∂E |V
∂S
= T ∂V
Xiao-Gang Wen, MIT
E
8.08 Statistical Physics (II) – Lecture note 1
Thermodynamic quantities and relations
• Canonical ensemble: from T , V and F (T , V ) to obtain other
.thermodynamic quantities.
- From F̃ (T , V , E ) = E − S(E , V )T , the free energy F (T , V ) is
obtained by minimizing F̃ respect to E :
F (T , V ) = F̃ (T , V , Ē (T , V ))
-
∂F
∂T
=
∂ F̃ (T , V , E )
∂E {z
|
E =Ē
∂F
∂V
=
∂ F̃ (T , V , E )
∂E {z
|
E =Ē
+
∂ F̃ (T ,V ,E )
∂T
= −S
∂ Ē (T ,V )
∂V
+
∂ F̃ (T ,V ,E )
∂V
∂S
= −T ∂V
= −p
}
=0
-
∂ Ē (T ,V )
∂T
}
=0
∂F
Summary: S = − ∂T
V
∂F
p = − ∂V
,
T
.
∂F
- The internal energy: Ē = F + S(Ē , V )T = F − T ∂T
V
- Heat capacity
∂ Ē
∂F
∂S
∂S
CV = ∂T
= ∂T
+ ∂ST
= −S + S + T ∂T
= T ∂T
∂T
V
V
V
Xiao-Gang Wen, MIT
V
V
8.08 Statistical Physics (II) – Lecture note 1
Energy cost of information
Consider a flexible chain
.
- Each link may point to right +1 or left −1.
- Total length of the chain NR − NL .
- No interaction energy.
- Total entropy:
Stot = kB ln 2N = NkB ln 2
- The entropy with length = 0:
N!
≈ kB (N ln N − 2 N2 ln N2 ) = NkB ln 2
S0 = kB ln (N/2)!(N/2)!
(The principle of one maximal term ∼ whole)
- The entropy with length = N:
SN = kB ln 1 = 0
• What is energy cost to stretch the chain from length = 0 to N?
No internal energy → no energy cost.
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Energy cost of information
In the stretch process,
the chain entropy decrease: ∆Schain = −kB N ln 2.
the heat bath entropy increase: ∆Sbath = kB N ln 2.
the heat bath energy increase: ∆Ebath = T ∆Sbath = kB TN ln 2.
Stretching the chain costs an energy NkB T ln 2.
• The change of free energy:
∆F = FN − F0 = (−TSN ) − (−TS0 ) = TkB ln 2
is equal to the change of total energy (sys. + bath)
The chain wants to shrink to lower free energy
At a finite T , a system wants to minimize its free energy F ,
just like at T = 0, a system wants to minimize its energy E .
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Energy cost of information
• Reducing a random link (50% left, 50% right) to 100% right,
descreases entropy kB ln 2 and cost energy kB T ln 2.
• Setting a random bit (50% 0, 50% 1) on a hard drive to a
particular value descreases entropy kB ln 2 and cost energy at least
kB T ln 2.
To write a 1T bytes hard drive at least cost energy
8 × 109 × kB T ln 2 ∼ 10−4 erg ∼ 10−11 J
An hard drive uses 1W = 1J/sec → Could fully rewrite itself 1011
times in one seccond.
( kB = 1.3807 × 10−16 erg/K ).
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Maxwell distribution of the velocity of the particles in a gas
Consider a single particle of mass m:
− k 1T
Prob. to find p in volum d3 p: P(p) ∝ Cp e
B
p2
2m
d3 p
mv 2
− 2k
BT
v 2 dv
Prob. to find |v | in range dv : P(v ) ∝ Cv e
r
Z ∞
π
− m v2
Cv e 2kB T v 2 dv = Cv
(kB T /m)3/2 = 1
2
0
2
p
− mv
P(v ) = (m/kB T )3/2 2/πv 2 e 2kB T Maxwell distribution
- Mono-atomic gas
at T = 300K ≈ 25C ◦
- Speed of sound in air
vsound = 343m/s
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Equipartition of energy
Kinetic energy of x-motion
R
R
px2 −βp 2 /2m
px2 −βpx2 /2m
e
e
dp 3 2m
dpx 2m
px2
h
i= R
= R
2 /2m
−βp
−βp
3
2m
dp e
dpx e x2 /2m
s
Z
√
∂
∂
2m
11
1
2
=−
ln dpx e−βpx /2m = −
ln π
=
= kB T
∂β
∂β
β
2β
2
which is independent of the mass m of the particle.
In a gas , each degree of freedoms, regardless the directions
and kinds of atoms, has a thermal energy 12 kB T
• Each atom in a gas carries a thermal energy 23 kB T .
Heat capacity of any gas cV = 32 kB per atom.
Total heat capacity of mixed gas CV = 23 kB (N1 + N2 + · · · )
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Molar heat capacity of gases
kB = 1.38 × 10−23 J/K . One mole = NA = 6.022 × 1023 .
Gas constant R = kB NA = 8.31J/K .
12.5/8.31 = 1.50
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Partition function Z (T , V ) of classical ideal gas
Gas: A systems of particles (atoms)
.Classical: Use Newton theory to describe the partciles.
Ideal: Assume there is no interactions between the particles.
Z Y
N
pi2
d3 pi d3 xi − k 1T PNi=1 2m
1 V N
1
B
e
=
Z (T , V ) =
N!
h3
N! λ3T
i=1
- Sum over states = sum over phase space (px , x), etc
- An area of ∆px ∆x = h correspond to a state.
- Identical particles: exchange two particles corresponds to the same
state.
p
R dp − 1 p2
√
1
kB T 2m
=
e
2mk
T
π
=
mkB T /2π~2 = 1/λT ,
B
h
p 2π~
where λT = 2π~2 /mkB T is the thermal wave length.
√
R +∞
2
Used −∞ dx e−x /A = Aπ
h
- Physical meaning: λT ∼ √
. Kinetic energy ∼ kB T .
m×(kinetic energy)
p
m × (kinetic energy) ∼ ∆p is the momentum uncertinty
→ λT size of the wave-packet.
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Free energy F (T , V ) and equation of state
h
i
F (T , V ) = −kB T ln Z (T , V ) = kB T N ln N − N + N ln(λ3T /V )
h
i
= NkB T ln(nλ3T ) − 1
- The above free energy is extensive. If we did have the N! term in
the partition function, the resulting free energy
F (T , V ) = NkB T ln(λ3T /V ) would not be extensive.
• Equation of state:
p=−
∂F
∂(−NkB T ln V + · · · )
kB NT
=−
=
∂V
∂V
V
or pV = NkB T .
∂F
- Calculate the pressure: − ∂V
= kB T ∂Z Z/∂V =
n (V )
− ∂E∂V
is the pressure from the nth state. →
Xiao-Gang Wen, MIT
− ∂En e−En /kB T
P ∂V−E /k T
n B
n e
∂F
− ∂V = hpi.
P
n
.
8.08 Statistical Physics (II) – Lecture note 1
Interacting classical gas
1
Z (T , V ) =
N!
Z Y
N
d3 pi d3 xi − k 1T
e B
h3
pi2
i=1 2m +U(x1 ,··· ,xN )
PN
i=1
- Replace V (x1 , · · · , xN ) by a constant – its average:
Z
1X
1
U(x1 , · · · , xN ) =
u(xi − xj ) ≈
d3 x d3 y n2 u(x − y )
2
2
i,j
Z
1
1
= Nn d3 xu(x) = Nnū
2
2
Z Y
N
pi2
1
d3 pi d3 xi − k 1T PNi=1 2m
+ 21 Nnū
Z (T , V ) =
e B
N!
h3
i=1
1 V − Nv0 N − k 1T 21 Nnū
=
e B
N!
λ3T
where v0 =
4π 3
3 r0
is the volume of one particle.
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
The free energy and equation of state for interacting gas
• Free energy:
h
F (T , V ) = NkB T ln
i 1 N2
Nλ3T
−1 +
ū
V − Nv0
2 V
• Equation of state:
∂F
NkB T
1 N2
mRT
a
p=−
=
+
ū, or p =
− 2
2
∂V
V − Nv0 2 V
V −b V
or (van Der Waals equation of state)
(p −
a
1 N2
ū)(V − Nv0 ) = NkB T , or (p + 2 )(V − b) = mRT ,
2
2V
V
where a = − 21 N 2 ū, b = Nv0 , and m = N/NA .
a > 0 → attraction, b > 0 → a hardcore.
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Plot the equation of state
- Green T > Tc
- Blue T = Tc
- Red T < Tc
p=
mRT
V −b
−
a
ab
2
• V 3 − (b + RT
p )V + p V − p = 0
can have three real solutions for V ,
such as F,H,J.
At T = Tc and p = pK ,
the three solutions becomes one:
(V − VK )3 = 0 →
a
2
c
3VK = b + RT
pK , 3VK = pK ,
VK3 = pabK →
The case for
VK = 3b = 3Nv0 ,
a=0
a
−ū
=
,
b
=0
pK = 27b
2
54v 2
Tc =
8b
a
27b 2
R
0
=
8a
27bR
=
−4ū
27kB v0
Isothermal curves
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
a
V2
Phase and phase transition
Pressure Pascal = 1N/m2 .
1 atmosphere ∼ 1kg-weight/cm2 = 101.325kPa
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Gibbs energy G (T , p)
How to obtain the real isothermal p-V curve for water?
.
• Free energ F (T , V ) is convinient for calculating thermodynamic
properties for a system with given T , V .
To calculate thermodynamic properties for a system with given
T , p, we like to introduce Gibbs energy:
Gibbs energy of the system
= Free energy of the sys. + spring
Free energy of spring =pV + const.
G̃ (T , p, V ) = F (T , V ) + pV
− freek energy
T
• Note that e
B
Heat
bath
Sys
Spring
Canonical
ensemble
Temperature
T
Micro canonical ensemble
is the total (unormalize) probability, thus the
−
G̃ (T ,p,V )
probabilty distribution of V is given by P(V ) ∝ e kB T .
The volume of the system is given by V = V̄ that mininizes G̃ .
• Real Gibbs energy: G (T , p) = G̃ [T , p, V̄ (T , p)],
,V )
where ∂F (T
+ p V =V̄ (T ,p) = 0.
∂V
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Minimizing G̃ (T , p, V ) ≡ G̃T ,p (V )
• Find V̄ :
(p +
∂ G̃ (T ,p,V )
∂V
=0→
∂F (T ,V )
∂V
+ p = 0 → Equation of state:
a
ab
mRT 2 a
)V + V −
=0
)(V − b) = mRT or V 3 − (b +
V2
p
p
p
V̄ , that minimizes G̃ , satisfies the equation of state.
• For T > Tc ,
G̃T ,p (V ) has one minimum.
• For T < Tc and p1 < p < p2 ,
G̃T ,p (V ) has two minima
and one maximum.
R
~
∂ G̃
• ∆G̃ = dV ∂V
G
R
∂F
= R dV (p + ∂V
)
~
= dV (p − psys )
G
~
G
V
T>T c
T<T c
T=T c
V
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Singularity in Gibbs energy G (T , p) and phase transition
G
T>T c
T<T c
~
G
V
~
G
~
G
~
G
V
V
V
~
G
~
G
V
~
G
V
p
• Fix T and plot G (T , p)
as a function of p
V
~
G
V
• A p-T phase diagram (for H2 O).
- Water and vapor belong to the same phase.
- A critial point appear at Tc = 647K ◦ = 374C ◦ and pc = 218atm
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Thermodynamic properties from Gibbs energy G (T , p)
From G = F + pV , we have
dG = dF + d(pV ) = −S dT − p dV + V dp + p dV
∂G
∂T
p
= −S,
∂G
∂p
T
= V;
∂F
∂T
V
∂F
∂V
= −S,
T
= −P.
• Heat capacity CV and Cp :
dE = dHeat − dW = dHeat − p dV
For fixed V : CV =
dHeat
dT
=
∂E
∂T V
∂S
= T ∂T
V
For combined sys. + spring Esys.+spring = E + PV :
dEsys.+spring = dHeat
For fixed p: Cp =
dHeat
dT
=
∂(E +pV )
∂T
p
Xiao-Gang Wen, MIT
=
∂(G +ST )
∂T
p
∂S
= T ∂T
p
8.08 Statistical Physics (II) – Lecture note 1
Clapeyron relation
• Along the liquid-gas transition line,
Gliq = Ggas , or dGliq = dGgas
∂G
∂G
• dGliq = ∂Tliq dT + ∂pliq dp
= −Sliq dT + Vliq dp
∂G
dGliq
∂G
• dGgas = ∂Tgas dT + ∂pgas dp
= −Sgas dT + Vgas dp
→
dp
dT
=
∆S
∆V
dG gas
(Clapeyron relation)
• Usually, from low temerature
phase to high temperature phase,
dp
both V and S increase, and dT
>0
• But for H2 O, from ice to water S increase and V decrease!
dp
→ dT
<0
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1
Equation of state near critical point and universal scaling
• Near the critial point
G̃ = − 12 C1 t(V − Vtr )2 + 14 C2 (V − Vtr )4 + · · ·
Minimizing G̃p
:
V − Vtr = ± C1 t/C2 .
→ ∆V ∼
t 1/2 .
1/2 is a scaling exponent β.
It is material independent,
ie universal.
~
G
V
t
~
G
∆V
• But in real systems, such a
universal scaling exponent
has a different value
β = 0.326419(3) in 3-dimensions.
The scaling exponent β = 1/8 for 2-dimensions and β = 1/2 for
4-dimensions and higher.
Xiao-Gang Wen, MIT
8.08 Statistical Physics (II) – Lecture note 1