lecture 6. preliminary and simple applications of...
TRANSCRIPT
Lecture 6.
Preliminary and simple applications
of statistical mechanics
Zhanchun Tu (涂展春 )
Department of Physics, BNU
Email: [email protected]
Homepage: www.tuzc.org
Main contents
● Fundamental concepts and equations
● Non-interaction ideal models
● Simple models of ligand-receptor binding
and gene expression
● Law of mass action
● Discussions on irreversible process
§6.1 Fundamental concepts & equations
Conventional thermodynamics
● Object
– Systems: large number of particles (~1023)
– Short-range interaction between particles (ideal gas, vdW gas, plasma, polymer; gravity system, + or - charged system)
– An isolated system can reach thermal equilibrium through long enough but finite-time relaxation
Interaction potential r−3 for r∞
● Four thermodynamic laws– 0th: it is possible to build a thermometer
– 1st: energy is conserved
– 2nd: not all heat can be converted into work
– 3rd: 0K cannot be reached via a finite reversible steps
dU =d Qd W k
TdSd Q
(1) temperature, similar to mass, is an operational definition (操作性定义 )
Note in thermodynamics:
(2) entropy can be defined as via reversible process dS=d Q /T
1st+2nd+const. T:
dF=dU−TdS=d W kd Q−TdS d W k
● Free energy (Helmholtz)
F=U−TS
−d W k−d F
Free energy change = maximum work done by the system
on the outer environment in isothermal process
Statistical mechanics
● Function
N-particle system
Newtonian mechanics ThermodynamicsStatistical mechanics
(Ensemble average)
A bridge from microscopic motions of a large number of molecules to macroscopic behavior of the system consisting of these molecules.
● Macrostate: thermodynamic EQ state– e.g. PVT, HMT etc.
● Microstate: phase point (q,p)
A macrostate
q
p
A macrostate
q
p
Each macrostate corresponds to many microstates!Each microstate occurs at some probability.
Task of stat. mech. is to find this probability distribution and then explain further the macroscopic quantities!
● Statistical Postulate (统计假说 )
– When an isolated system is left alone long enough time, it evolves to thermal equilibrium.
– Equilibrium is not one microstate, but rather that probability distribution of microstates having the greatest possible disorder allowed by the physical constraints on the system.
Key question: How to measure disorder?
Measure disorder
● A good quantity (I) to measure disorder satisfies
– Continuity: continuous respect to P1,P
2,...,P
M
– Zero for pure state: I=0 for {P1=1 and others=0}
– Maximum for most mixed state: max I for {Pj=1/M}
– Additivity: I(X+Y)=I(X)+I(Y) for independent X and Y
x1
x2
xM
...
EventPossible values
X
Probability
P1
P2
PM
● Shannon's information entropy (1948)
I=−K∑j=1
M
P j ln P j ; K0
Problem: prove that I max=K ln M if and only if P j=1/M .
(1) constraint: ∑j=1
M
P j=1
(2) Lagrange problem:
extremum{I≡I−∑j=1
M
P j−1}∂ I /∂P j=0⇒ P j=e−1/K (const.)
∂ I /∂=0⇒∑j=1
M
P j=1 P j=1/M
(3) Max. or Min. ?: ∂2 I
∂ P j∂Pk∣P j=1 /M
=−KM jkmaximum!
Boltzmann Entropy
● Equilibrium of isolated system: const. E, N, V
W(E,N,V): the number of allowed microstates
I=−K∑j=1
W
P j ln P jMeasure of disorder:
Constraint: ∑j=1
W
P j=1P j=
1W
Each microstate is equally probable!
(Boltzmann entropy)S≡k B
KI max=k B lnW
Boltzmann Entropy = constant X maximal value of disorder
I=I max
● Canonical system: A system in equilibrium state at constant N, V & <E>
Allowed microsates: j≡{r1, p1 ; r2, p2 ;⋯;r N , pN } j
Measure of disorder:
Constraints:
I=−K∑ jP j ln P j
Probability: P j≡P E j
∑ jP j=1 ∑ j
P j E j=const.
with energy E j ; j=1,2,⋯
Problem: prove that max{I/K} with the above constraints gives
P j=e− E j /Z
where β is a constant and Z=∑ je− E j
---Boltzmann distribution
---Partition function
Problem: if we interpret <E>=U, what's the physical meaning of β?
P j=e− E j /Z ⇒U=⟨E ⟩=∑ jE j e
− E j /Z
Thermodynamic relation in isochoric (等容 ) reversal process
dU =T dS⇒∂ S∂U ∣
N , V
=1T
S≡k B
KI max=−k B∑ j
P j ln P j=−k B∑ j
e−β E j
Zln(e
−β E j
Z )=−k B∑ j
e−β Ej
Z(−βE j−ln Z )=k BβU+k B ln Z
=1 /k B T
T and F in Stat. Mech.
For constant N, V ∂U∂
=U 2−⟨E2
⟩
∂ S∂
=k B[U2−⟨E2
⟩ ]
∂ S∂U ∣
N ,V
=k B(Homework: proof 2 equations)
● Temperature in Stat. Mech.
S and U for constant V and N can be calculated in Stat. Mech.
We can define
1T=
∂ S∂U ∣
N ,V
Then Stat. Mech. & thermodynamics are consistent.
Problem: prove that F=−k B T ln Z
● Free energy in Stat. Mech.
Problem: prove that U=−∂ ln Z∂
P j=e− E j /Z
S≡k B
KI max=k BUk B ln Z
=1/k B T
U=⟨E ⟩=∑ jE j e
− E j/Z
Z=∑ je−E j
Macrostate & microstate are relative● One person’s macrostate is another’s microstate
p X 1=∑m=1
n1
pm=1Z∑m=1
n1
e− m≡
1Z
e− G X 1
Z1
G (X 1)=−k B T ln Z 1
§6.2 Non-interaction ideal models
Ideal gas● Law of ideal gas
P=nkBT
Problem: please check the dimension!
● The origin of P
f 1=2 mv x/ t=mvx2/L
t=2 L /vx(a round trip)
p1= f 1/L2=mv x
2/L3
P=⟨∑ p1⟩=∑ ⟨ p1⟩=N /L3m⟨vx
2⟩=nm⟨vx
2⟩
Note: here P is the pressure!
● Interpretation of temperature
P=n k B T
P=n m⟨v x2⟩}⇒
12
m⟨vx2⟩=
12
k BT=12
m⟨v y2⟩=
12
m ⟨vz2⟩
v2=vx
2vy
2vz
2⇒
12
m⟨v2⟩=
32
k B T
<EK> of each molecule (3/2)k
BT!
The average energy per DOF is kBT/2.
Problem: prove that
⟨v2⟩≃500 m /s for air moleculesat room temperature
● Maxwell distribution of molecular velocities
dN v x , v y , vz
N≡p vx , vy , vzdv x dv y dvz
Distribution function
p v x , v y , vz=Ae−mv2
/2 k BT
Maxwell distribution
Problem: find A?
∫−∞
∞
dvx∫−∞
∞
dv y∫−∞
∞
dvz pvx , v y , vz=1⇒ A= m2 k B T
3/2
v2=vx
2v y
2vz
2
Problem: please confirm that <EK> of each molecule (3/2)k
BT
The proportional number of molecules with velocities between u and u+du
u=∣v∣=vx2v y
2vz2
dN uN
≡ pu du=∑ pvx , v y , vzdvx dv y dv z
= m2 k BT
3/2
e−mu2 /2 kB T
×4u2 du
p u= 2
1 /2
mk BT
3/2
u2 e−mu2/2k BT
u
p(u)
(T/m)1
(T/m)2
(T/m)3
(T/m)1<(T/m)
2<(T/m)
3
Question: what will happen when T→0?
Problem: please prove
p EK ∝E K e−E K / kB T
Configuration space
reactant
resultantReaction coordinate (反应坐标 ): the fast reaction path from reactant to resultant in the configuration space
活化能Reactioncoordinate
reactant
resultant
Potentialenergy
Potential curve along the reaction coordinate
barrier
ΔEa
ΔEa: activation energy
● Chemical reaction
- Reaction coordinate & activation barrier
Intuitively, the molecules of reactant with
can escape from the barrier. Thus the reaction rate
EK≥ Ea
r∝dN EK≥ Ea
N
The reaction coordinate can be simply taken as[Science 319(2008)183]
Assume the collision induces the chemical reaction, only the DOF in the direction of collision plays role in.
EK=12
m1 v12
12
m2 v22
1 2
- Reaction rate in simple reaction
Problem: please prove dN EK≥ Ea
N=e− E a /kB T
v1
v2
m1 v12m2 v2
2=2 Ea
dN(EK≥ΔE
a)= number of molecules outside the ellipse
p v1, v2dv1 dv2∝e−m1 v1
2m2 v2
2 /2k BT
dv1 dv2
p V 1, V 2dV 1 dV 2∝e−m V 1
2V 2
2 /2 kB T
dV 1 dV 2
V1
V2
mV 12V 2
2=2 Ea
V =V 12V 2
2
p V dV ∝V e−mV 2
/2 kB TdV
dN EK≥ Ea
N=∫2 E a/m
∞
pV dV
∫0
∞
p V dV=e− Ea / kB T
r∝e− Ea / kB T Arrhenius rate lawHard Question: What is missed in our discussion?
r=r0 e− E a /k BT Arrhenius rate law
T
r
r0
fixed ΔEa
r
ΔEafixed T
r0
Two ways to increase the reaction rate:(1) increase temperature (heat)(2) decrease the activation barrier (enzyme)
- How to increase the reaction rate?
● S of Ideal gas at constant E, N, V
E=∑i=1
Nm2
vi2=
m2∑i=1
N
v i2≡
m2
u2 u is a 3N-dimensional vector
∣u∣=2 E /m u is located in (3N-1)-dimensional sphere with radius (2E/m)1/2
Each particle can stay any place in position space W E , N ,V ∝V N
For N molecules at fixed positions, each vector u can be reached
W E , N ,V ∝surfaceof thesphere ∝ 2 E /m 3 N−1
W E , N ,V =C×V N E 3 N /2 (for large N)
S=k B ln W E ,N ,V =k Bln CN ln V 3 N
2ln E
Question: please find the dimension of constant C.
E=∑i=1
N p i2
2 m
Δx
Δpx
Z=1
N !∫ e
−∑ipi
2 /2m∏i
d 3 ri d 3 pi
h3
Z=V N
N ! 2 m
h2 3 N /2
Why is there N ! ?
ln Z=ln[V N
N ! 2m
h2 3 N /2
]−3 N2
ln
● U, S, F of a box of ideal gas at constant N, V
U=−∂ ln Z∂
=3 N2
=3 N k BT
2⇒
UN=
3 k BT
2
F=−k B T ln Z=−k B T ln[V N
N ! 2 m
h2
3 N /2
]=−Nk BT [1lnVN 2m
h2
3/2
]
⇒ S=U−F
T=Nk B[ 5
2ln
VN 2 mk B T
h2 3/2
]
ln N !≈N ln N−N
=−Nk B T [1lnVN 2 mkB T
h2 3 /2
]Problem: what's ΔS if V→2V?
S=Nk B ln 2
Problem: find the entropy density for (dilute) ideal gas
c∗= 2mk B T
h2 3/2
≃1032 m−3 For c=NV≪c∗ ,
SV=−k B c ln
cc∗
Dilute Solutions● Free energy of solution & solute chemical potential
3D lattice model 2D lattice model
Smix=k B ln ≈
Stirling approx.
W=
Total lattice:
Total states:
N sN H 2 O
dilute solution==> N s /N H 2 O≪1
c=N s
V tot
, c0=N H 2 O
V totwith
Problem: please prove that the chemical potential of ideal gas has the similar form.
Homework: please prove that the chemical potential for non-interaction system of multi-components is
c i 0=1 M ---reference state
i 0= isk B T lnci 0
c0
● Osmotic pressurePermeability of the membrane to water
equilibrium state H 2 OL = H 2 O
R ,T L=T R=T
H 2 OL = H 2 O
0 T , pL
H 2 OR
= H 2 O0
T , pR−N s
N H 2 O
k BT≈ H 2 O0
T , pL∂H 2O0
∂ p pR−pL−N s
N H 2 O
k B T
Chemical potential of water
H 2 OT , p= ∂Gtot
∂ N H 2 O T , p
=H 2O0
T , p−N s
N H 2O
k BT
v: volume per water molecule
p= pR−pL=N s
Vk B T
Van't Hoff formula
Estimate: Osmotic pressure in an E. coli cell
Characteristic concentration of inorganic ions in cells is 100 mM
N s
V=
100×10−3×6×1023
10−3 m3=6×1025 m−3
Δ pion=( p i n−pout)=N s
Vk BT=(6×1025
)×(4×10−21) Pa
=2.4×105 Pa=2.4 atm
≡6 ×107 ions/E. coli
Contributions of other molecules are much smaller than ions
Example: proteinsN s
V= 2 ×106 ions/E. coli
⇒ pprotein=0.08atm
Remark: the crowding in cell might change the above results
Entropic force
a
fext
L System a: gas+piston
Sa=Nk B ln Lconst.
U a= f ext L32
Nk B T
F a= f ext L−T Sa32
Nk B T
min F a⇒ f ext=TdSa
dL≡ f a
What's meaning of fa?
fext
fa
Effective force on the piston by the collisions of gas molecules
Formally (形式上 ), fa is the force
resisting the change of entropy
Called entropic force
Remark: osmotic pressure can also be interpreted as entropic force
If fa>f
ext, gas pushes the piston to move and do work against the load
T is fixed ==> internal energy of the gas molecules is unchanged
Question: Where does the work come from?
ΔU gas=W k+Q=0⇒Q=−W k>0 because W k<0
The gas system absorbs thermal energy from the reservoir and converts it into mechanical work
Question: Doesn’t that violate the Second Law?
The gas system sacrifices (牺牲 ) some order in the expansion
The cost of upgrading energy from thermal to
mechanical form is that we must give up order
Key point:
§6.3 Simple models of ligand-receptor binding & gene expression
Ligand-receptor binding● Lattice model
!−L!
≈L
For L≪
=
+
Binding probability
Number of ligand
Number of lattice
● Binding curve
c≡L/V tot
c0≡/V tot
Hill function with Hill coefficient n=1
Assume lattice constant ~1nm
c0=1
(1 nm)3≈0.6 M ?
Question: which effect (energy or entropy) dominates at small c?
Answer: entropy
Gene expression● Genetic control in the central dogma
Very complicated controls, simplified as a problem as
RNA polymerase binds at the promoter of the related gene
simplify
● Lattice model Experiment reveals most cellular RNA polymerase molecules are bound to DNA
Assumption: (1) all RNA polymerase molecules are bound to DNA (2) different binding energies for specific site (promoter) and non-specific sites (3) P RNA polymerases and
NNS
non-specific sites
● Binding probability of RNA polymerase to its promoter
=
N NSP−1
P−1!e−P−1 pd
NS/k B T
e− pd
S/k B T
N NSP
P !e− pd
S/ k BT
N NS
P−1
P−1!e−P−1 pd
NS/ k BT e− pd
S/ k BT
with Determine the shapes of bind curve
§6.4 Law of mass action
● Simple reaction
Equilibrium at fixed T and p
When the numbers of A and B molecules decrease by 1 there is a corresponding increase by 1 in the number of AB molecules.
Introduce stoichiometric coefficients
dN A=A dN ref dN B=B dN ref dN AB=AB dN ref
Chemical potential A= ∂G∂ N A T , p
B= ∂G∂ N B T , p
AB= ∂G∂N AB T , p
A A B B AB AB=0
Chemical reaction
● General reaction
i X i 0
∏i=1
N
cii=∏
i=1
N
ci 0 i e
−∑i=1
N
i i0
≡K eq≡1
K d
equilibrium constant
dissociation constant
Law of mass action
Ligand–Receptor Binding (again)
● Simple binding
Law of mass action==>
Probability of receptor occupied by ligand
Compare lattice model
v: volume of single lattice
● Cooperative binding—Hill function
Hill function with n=2
Assume: single occupied states are rare
§6.5 Discussions on irreversible process
Entropy is not conserved● Clausius & Kelvin (2nd law)
– When we release an constraint on an isolated
macroscopic system in equilibrium, it eventually
reaches a new equilibrium whose entropy is at least
as great as before
● Boltzmann (Statistical meaning of entropy)
– Entropy reflects the disorder of a macroscopic
system in equilibrium, when we have limited and
coarse knowledge of microstates of the system
● Free expansion of ideal gas
S=Nk B ln 2Entropy increased spontaneously when we suddenly released a constraint.
The system forfeits (丧失 ) some order in the free expansion.(Isolated system)
Question: Would the above process spontaneously happen in reverse?
Answer: To get it back, we have to compress the gas with a piston which do mechanical work on the system, heating it up. To return its original state, we then have to cool it (remove some thermal energy).
The cost of recreating order is that we must degrade some organized energy into thermal form. (Lecture 4)
Origin of irreversibility
Increase of entropy Irreversibility in macroscopic process
Molecular collisions happens in reverse!
Microscopic equations of motion is reversible!
However,
Puzzle: Where does the irreversibility come from?
The instant after the switch is opened, suddenly a huge number of new allowed states appear, and the previously allowed states are suddenly a part of the total allowed states. There is no work-free way (不做功的方式 ) to suddenly forbid those new states to appear.
Origin of irreversibility: a highly specialized initial state
● Why can all molecules not go back to the left side spontaneously?
The chance that all molecules in the left side:
P=1/2 N≈10−N /4
≈10−1022
for N≃1023
Time scale that we can observe this happens:
=1
10−1022 0
τ0 can be taken as the characteristic time (ps) of molecular collisions
≈101022
s≫ univ≈200×108 y≈1018s
Efficiency of free energy transductionInitial pressure: p i=w1w2/A
Let w2 slide away from the piston, and relax gas
Final pressure: p f=w1/A
Initial height of piston:
Ideal gas
Li=Nk BT /Api
Final height of piston: L f =Nk BT /Ap f
Mechanical work to lift w1: ∣W k∣=w1 L f −Li
Free energy change of gas: ∣ F∣=Nk B T ln L f /Li
∣W k
F ∣=?
∣W k
F ∣=− xln 1−x
x=w2
w1w2
={0 for w 2=01 for w2=∞ x
∣W k / F∣
0 1
1
∣W k / F∣ 0 when w2∞ ; ∣W k / F∣1 when w20
Free energy transduction is least efficient when it
proceeds by the uncontrolled release of a big constraint.
Free energy transduction is most efficient when it proceeds by
the controlled release of many small constraints step by step.
“quasi-static process”!
● Ideal heat engine---Carnot engine
Isothermal expansion
Adiabatic expansion
(absorb)
(release)Isothermal compression
Adiabatic comp.
Ecycle=0
W net=Q1−Q2
Scycle=0
Q1
T 1
−Q 2
T 2
=0
⇒C=W net
Q1
=Q 1−Q 2
Q1
=1−T 2
T 1
Heat engine
● Non-ideal Carnot-like heat engines
Ecycle=Q 1−Q2−W net=0
W net=Q1−Q2
Scycle=Q1
T 1
−Q2
T 2
S ir=0
AdiabaticQ=0
AdiabaticQ=0
Non-ideal => irreversible =>spontaneous entropy generation
S ir0
Q1
T 1
Q 2
T 2
T 2
T 1
Q2
Q1
⇒=W net
Q1
=Q1−Q2
Q1
=1−Q 2
Q1
1−T 2
T 1
≡C
● Efficiency of an engine at maximum power
(1975) 22CA=1−T 2
T 1
=1−1−C , with C=1−T 1
T 1
During the finite time, the working substance undergoes:
T1
T2
T1w
T2w
(1)
(2)(3)
(4)
(2): adiabatic expansion, no heat exchange
Absorb Heat : Q1=T 1−T 1w t1
(1): isothermal expansion, temperature T1w
<T1 (bath)
(3): isothermal compression, temperature T2w
>T2 (bath)
Release Heat : Q2=T 2w−T 2 t 2
(4): adiabatic compression, no heat exchange
Endoreversible assumption: in the above cycle
Total time assumption: t total= t1t 2 =>Power output: P=Q1−Q2/t 1t 2
S=0 ⇒Q1
T 1w
=Q2
T 2w
∂P∂T 1w
=∂ P∂T 2w
=0⇒T 2w
T 1w
= T 2
T 1
max{P}:
Efficiency at maximum power: CA=Q1−Q2
Q1
=1−T 2w
T 1w
=1−T 2
T 1
with x=T 1−T 1w , y=T 2w−T 2
§ Summary & further reading
Summary● Classic thermodynamics: 0th, 1st, 2nd, 3rd laws
● Statistical mechanics for closed system
– Entropy and Internal energy
– Temperature
– Free energy
– Boltzmann distribution
– Principle of Min free energy
– Free energy transduction is most efficient when it proceeds by the controlled release of many small constraints step by step.
S=−k B∑ jP j ln P j ; U=∑ j
P j E j
1T=
∂ S∂U ∣
V
F=U−TS=−k BT ln Z
P j=e− E j /Z Z=∑ je−E j
● Applications
– Ideal gas
– Dilute solution
– Ligand-receptor simple binding
– Gene express (simple model)
– Ligand-receptor cooperative binding
UN=
3 k BT
2 i= i 0k B T ln
ci
ci 0
● Law of mass action
∏i=1
N
cii=∏
i=1
N
ci 0 i e
−∑i=1
N
i i0
≡K eq≡1
K d
p= pR−pL=N s
Vk B T
Further reading● Phillips et al., Physical Biology of the Cell, ch6● Nelson, Biological Physics, ch6● Jaynes (1989) Papers on Probability and
Statistics● Lebowitz (1993) Boltzmann's entropy and time's
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