Lecture 6.

Preliminary and simple applications

of statistical mechanics

Zhanchun Tu (涂展春 )

Department of Physics, BNU

Email: tuzc@bnu.edu.cn

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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