Lecture 6. Preliminary and simple applications of statistical mechanics

Transcription

1 Lecture 6. Preliminary and simple applications of statistical mechanics Zhanchun Tu ( 涂展春 ) Department of Physics, BNU tuzc@bnu.edu.cn Homepage:

2 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

3 6.1 Fundamental concepts & equations

4 Conventional thermodynamics Object Systems: large number of particles (~103) Short-range interaction between particles (ideal gas, vdw gas, plasma, polymer; gravity system, + or - charged system) Interaction potential r 3 for r An isolated system can reach thermal equilibrium through long enough but finitetime relaxation

5 Four thermodynamic laws 0th: it is possible to build a thermometer 1st: energy is conserved nd: not all heat can be converted into work du =d Q d W k TdS dq 3rd: 0K cannot be reached via a finite reversible steps Note in thermodynamics: (1) temperature, similar to mass, is an operational definition ( 操作性定义 ) () entropy can be defined as ds =d Q/T via reversible process

6 Free energy (Helmholtz) F =U TS 1st+nd+const. T: df =du TdS =d W k d Q TdS d W k d W k d F Free energy change = maximum work done by the system on the outer environment in isothermal process

7 Statistical mechanics Function N-particle system Statistical mechanics Newtonian mechanics Thermodynamics (Ensemble average) A bridge from microscopic motions of a large number of molecules to macroscopic behavior of the system consisting of these molecules.

8 Macrostate: thermodynamic EQ state e.g. PVT, HMT etc. Microstate: phase point (q,p) p q A macrostate 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!

9 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?

10 Measure disorder Event Probability P1 P PM x1 x... X Possible values xm A good quantity (I) to measure disorder satisfies Continuity: continuous respect to P1,P,...,PM 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

11 Shannon's information entropy (1948) M I = K P j ln P j ; K 0 j=1 Problem: prove that I max=k ln M if and only if P j =1/ M. (1) constraint: M P j =1 j=1 () Lagrange problem: 1 / K { } I / P j =0 P j =e M extremum I I j=1 (3) Max. or Min.?: P j 1 I P j Pk M I / =0 P j =1 P j =1 / M (const.) P j =1/ M j=1 = KM jk maximum!

12 Boltzmann Entropy Equilibrium of isolated system: const. E, N, V W(E,N,V): the number of allowed microstates W Measure of disorder: I = K P j ln P j j=1 Constraint: W P j =1 1 P j= W j=1 Each microstate is equally probable! kb S I max =k B ln W K I =I max (Boltzmann entropy) Boltzmann Entropy = constant X maximal value of disorder

13 Canonical system: A system in equilibrium state at constant N, V & <E> Allowed microsates: j { r 1, p1 ; r, p ; ; r N, p N } j with energy E j ; j=1,, I = K j P j ln P j Measure of disorder: Constraints: P j P E j Probability: j P j =1 j P j E j=const. Problem: prove that max{i/k} with the above constraints gives E j P j=e where β is a constant and /Z Z = j e E j ---Boltzmann distribution ---Partition function

14 T and F in Stat. Mech. Problem: if we interpret <E>=U, what's the physical meaning of β? E j P j =e / Z U = E = j E j e E j /Z ( ) β E β E kb e e S I max = k B j P j ln P j= k B j ln K Z Z j = k B j e β E j Z ( β E j ln Z )=k B β U +k B ln Z For constant N, V U =U E (Homework: proof equations) S =k B [U E ] S Thermodynamic relation in du =T ds isochoric ( 等容 ) reversal process U =1 /k B T S U N,V =k B 1 = T N,V j

15 Temperature in Stat. Mech. S and U for constant V and N can be calculated in Stat. Mech. We can define 1 S = T U N,V Then Stat. Mech. & thermodynamics are consistent.

16 Free energy in Stat. Mech. Z = j e P j =e E / Z E j j U = E = j E j e E j =1/k B T /Z kb S I max =k B U k B ln Z K Problem: prove that F = k B T ln Z Problem: prove that U = ln Z

17 Macrostate & microstate are relative One person s macrostate is another s microstate n1 p X 1 = m=1 1 pm = Z n1 e m m=1 1 G X e Z Z1 G ( X 1 )= k B T ln Z 1 1

18 6. Non-interaction ideal models

19 Ideal gas Law of ideal gas P=nkBT Note: here P is the pressure! Problem: please check the dimension! The origin of P t = L / v x (a round trip) x f 1= mv x / t=mv / L p 1= f 1 / L =mv x / L 3 P= p1 = p1 = N /L m v x =n m v x 3

20 Interpretation of temperature } P=n k B T m v x = k B T = m v y = m v z P=n m v x The average energy per DOF is kbt/. 1 3 v =v v v m v = k B T x y z <EK> of each molecule (3/)kBT! Problem: prove that v 500 m /s for air moleculesat room temperature

21 Maxwell distribution of molecular velocities dn v x, v y, v z p v x, v y, v z dv x dv y dv z N Distribution function Maxwell distribution m v / k BT p v x, v y, v z = Ae Problem: find A? dv x dv y dv z p v x, v y, v z =1 A= v =v x v y v z m kbt 3/ Problem: please confirm that <EK> of each molecule (3/)kBT

22 The proportional number of molecules with velocities between u and u+du u= v = vx v y v z dn u p u du= p v x, v y, v z dv x dv y dv z N 3/ m m u / k T = e 4 u du k BT B p(u) 1/ 3/ p u = m kbt m u / k BT u e (T/m)1 (T/m) Question: what will happen when T 0? (T/m)3 Problem: please prove u E K / k B T p E K E K e (T/m)1<(T/m)<(T/m)3

23 Chemical reaction - Reaction coordinate & activation barrier Reaction coordinate ( 反应坐标 ): the fast reaction path from reactant to resultant in the configuration space Potential energy resultant reactant Configuration space barrier ΔEa reactant resultant ΔEa: activation energy Reaction coordinate Potential curve along the reaction coordinate 活化能

24 - Reaction rate in simple reaction [Science 319(008)183] The reaction coordinate can be simply taken as Assume the collision induces the chemical reaction, only the DOF in the direction of collision plays role in E K = m1 v 1 m v Intuitively, the molecules of reactant with E K E a can escape from the barrier. Thus the reaction rate dn E K E a r N

25 Problem: please prove m1 v1 m v = E a v dn E K E a E =e N a /k B T dn(ek ΔEa)= number of molecules outside the ellipse v1 m1 v1 m v / k BT p v 1, v dv 1 dv e dv1 dv m V 1 V / k B T p V 1, V dv 1 dv e V V = V 1 V mv / k B T p V dv V e V1 Hard Question: What is missed in our discussion? dv dn E K E a E / m p V dv E / k = =e N p V dv a m V 1 V = E a dv 1 dv a B T 0 Ea / k B T r e Arrhenius rate law

26 - How to increase the reaction rate? E a /k BT r=r 0 e r r r0 r0 fixed ΔEa T Arrhenius rate law fixed T Two ways to increase the reaction rate: (1) increase temperature (heat) () decrease the activation barrier (enzyme) ΔEa

27 S of Ideal gas at constant E, N, V N E= i=1 N m m m vi = v i u i=1 u = E / m u is a 3N-dimensional vector u is located in (3N-1)-dimensional sphere with radius (E/m)1/ For N molecules at fixed positions, each vector u can be reached W E, N,V surface of thesphere E / m Each particle can stay any place in position space W E, N,V =C V N E 3 N / S =k B ln W E, N,V =k B 3 N 1 W E, N,V V (for large N) 3N ln C N ln V ln E Question: please find the dimension of constant C. N

28 U, S, F of a box of ideal gas at constant N, V N E = i=1 pi m Z= 1 e N! i pi / m 3 i 3 d r i d pi h 3 Why is there N!? N Δpx [ ] V m Z= N! h Δx N ln Z =ln 3 N / V m N! h 3 N/ 3N ln ln Z 3 N 3 N k B T U = = = U 3 k BT = N

29 ln N! N ln N N [ ] [ ] N F = k B T ln Z = k B T ln V m N! h 3 N / 3 / V mk B T = Nk B T 1 ln N h S= [ = Nk B T 1 ln U F =Nk B T [ ] V m N h 3/ 5 V mk B T ln N h ] 3/ Problem: what's ΔS if V V? S =Nk B ln Problem: find the entropy density for (dilute) ideal gas c = mk B T h 3/ 3 10 m 3 N S c For c= c, = k B c ln V V c

30 Dilute Solutions Free energy of solution & solute chemical potential Total lattice: N s N H O Total states: W= 3D lattice model S mix =k B ln Stirling approx. D lattice model

31 dilute solution==> N s / N H O 1 NH O Ns with c= V, c 0= V tot tot 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 ci 0 i 0= is k B T ln c0

32 Osmotic pressure Permeability of the membrane to water LH O= RH O,T L =T R=T equilibrium state Chemical potential of water H O T, p = Gtot NH O 0 = H O T, p T,p Ns k T NH O B Van't Hoff formula v: volume per water molecule LH O= 0H O T, p L R H O 0 HO 0 = H O Ns Ns 0 T, p R k T H O T, p L p R p L k T NH O B p NH O B Ns p= p R p L = k T V B

33 Estimate: Osmotic pressure in an E. coli cell Characteristic concentration of inorganic ions in cells is 100 mm N s = =6 10 m 3 3 V 10 m ions/e. coli Ns 5 1 Δ pion =( p i n p out )= k B T =(6 10 ) (4 10 ) Pa V = Pa=.4 atm Contributions of other molecules are much smaller than ions Ns Example: proteins = 106 ions/e. coli V p protein =0.08 atm Remark: the crowding in cell might change the above results

34 Entropic force L System a: gas+piston fext a What's meaning of fa? fa fext Effective force on the piston by the collisions of gas molecules S a =Nk B ln L const. 3 U a = f ext L Nk B T 3 F a= f ext L T S a Nk B T ds a min F a f ext =T f a dl 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

35 If fa>fext, gas pushes the piston to move and do work against the load Question: Where does the work come from? T is fixed ==> internal energy of the gas molecules is unchanged Δ U 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 Key point: The cost of upgrading energy from thermal to mechanical form is that we must give up order

36 6.3 Simple models of ligandreceptor binding & gene expression

37 Ligand-receptor binding Lattice model Number of ligand Number of lattice For L! L L! Binding probability = +

38 Binding curve c L/V tot c 0 /V tot Hill function with Hill coefficient n=1 Assume lattice constant ~1nm c0= M? 3 (1 nm) Question: which effect (energy or entropy) dominates at small c? Answer: entropy

39 Gene expression Genetic control in the central dogma simplify Very complicated controls, simplified as a problem as RNA polymerase binds at the promoter of the related gene

40 Lattice model Experiment reveals most cellular RNA polymerase molecules are bound to DNA Assumption: (1) all RNA polymerase molecules are bound to DNA () different binding energies for specific site (promoter) and non-specific sites (3) P RNA polymerases and NNS non-specific sites

41 Binding probability of RNA polymerase to its promoter P 1 N NS P 1 /k T e e P 1! = P P 1 N NS / k T N NS P 1 e e P! P 1! S pd with B NS pd B S pd NS pd /k B T / k BT S pd / k B T e Determine the shapes of bind curve

42 6.4 Law of mass action

43 Chemical reaction 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 Chemical potential dn B= B dn ref A= G NA T,p dn AB = AB dn ref B= G NB T,p A A B B AB AB =0 AB= G N AB T,p

44 General reaction i X i 0 equilibrium constant N N N i 0i c = c e i=1 i i i=1 i i0 i=1 K eq 1 Kd Law of mass action dissociation constant

45 Ligand Receptor Binding (again) Simple binding Law of mass action==> Probability of receptor occupied by ligand Compare lattice model v: volume of single lattice

46 Cooperative binding Hill function Assume: single occupied states are rare Hill function with n=

47 6.5 Discussions on irreversible process

48 Entropy is not conserved Clausius & Kelvin (nd 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

49 Free expansion of ideal gas S =Nk B ln Entropy increased spontaneously when we suddenly released a constraint. (Isolated system) The system forfeits ( 丧失 ) some order in the free expansion. 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)

50 Origin of irreversibility Increase of entropy However, Irreversibility in macroscopic process Molecular collisions happens in reverse! Microscopic equations of motion is reversible! Puzzle: Where does the irreversibility come from? Origin of irreversibility: a highly specialized initial state 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.

51 Why can all molecules not go back to the left side spontaneously? The chance that all molecules in the left side: N N/4 P= 1/ for N 10 3 Time scale that we can observe this happens: = s univ y 10 s τ0 can be taken as the characteristic time (ps) of molecular collisions

52 Efficiency of free energy transduction Initial pressure: p i= w1 w / A Initial height of piston: Ideal gas Li= Nk B T / Api Let w slide away from the piston, and relax gas Final pressure: p f =w1 / A Final height of piston: L f =Nk B T / 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 Wk =? F

53 Wk x = F ln 1 x w 0 for w =0 x= = w1 w 1 for w = { W k / F x W k / F 0 when w ; W k / F 1 when w 0 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!

54 Heat engine Ideal heat engine---carnot engine E cycle =0 W net =Q 1 Q Isothermal expansion (absorb) Adiabatic comp. S cycle =0 Adiabatic expansion (release) Isothermal compression Q1 Q =0 T1 T W net Q 1 Q T C = = =1 Q1 Q1 T1

55 Non-ideal Carnot-like heat engines E cycle =Q 1 Q W net =0 W net =Q1 Q Adiabatic Q=0 Adiabatic Q=0 Non-ideal => irreversible => spontaneous entropy generation ir S 0 Q1 Q ir S cycle = S =0 T1 T Q1 Q T1 T T Q T 1 Q1 W net Q1 Q Q T = = =1 1 C Q1 Q1 Q1 T1

56 Efficiency of an engine at maximum power (1975) T T1 CA =1 =1 1 C, with C =1 T1 T1

57 During the finite time, the working substance undergoes: (1): isothermal expansion, temperature T1w<T1 (bath) Absorb Heat : Q 1 = T 1 T 1w t 1 (4) (1) (3) () T1 (): adiabatic expansion, no heat exchange T1w (3): isothermal compression, temperature Tw>T (bath) Release Heat : Q = T w T t Tw T Total time assumption: (4): adiabatic compression, no heat exchange t total = t 1 t =>Power output: P= Q1 Q / t 1 t Endoreversible assumption: in the above cycle Q1 Q S =0 T = T 1w w max{p}: with x=t 1 T 1w, y=t w T Efficiency at maximum power: T w T P P = =0 = T 1w T w T 1w T1 Q 1 Q T w T CA= =1 =1 Q1 T 1w T1

58 Summary & further reading

59 Summary Classic thermodynamics: 0th, 1st, nd, 3rd laws Statistical mechanics for closed system Entropy and Internal energy S= k B j P j ln P j ; U = j P j E j Temperature 1 S = T U Free energy F =U TS = k B T ln Z Boltzmann distribution P j =e E / Z V j Z = j e E j 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.

60 Law of mass action N N N c = c i=1 i i i=1 i i0 i 0i e i=1 1 K eq Kd Applications Ideal gas U 3 k BT = N ci i= i 0 k B T ln ci 0 Ns p= p R p L = kbt V Dilute solution Ligand-receptor simple binding Gene express (simple model) Ligand-receptor cooperative binding

61 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 arrow, Physcis Today 46, 3