Retrospect
● Last lecture (Underlying principles behind life)
● We have learned the concept of thermal motion and others in lecture 2.
● We'll try to understand: How does the thermal motion control the sub-cellular world? Lecture 3 Molecular dance: Brown motion & Random walks
Zhanchun Tu ( 涂展春 )
Department of Physics, BNU
Email: [email protected] Homepage: www.tuzc.org Main contents
● Probability
● Understanding the laws of ideal gas
● Brown motion & Random walks
● Friction and diffusion
● Configuration of polymers
● Biological applications of diffusion §3.1 Probability Discrete distribution
experiment
total times 1 2 3 4 5 6 12 0.250 0.167 0.000 0.083 0.250 0.250 120 0.183 0.158 0.225 0.175 0.125 0.133 1200 0.163 0.169 0.172 0.146 0.189 0.161 12000 0.170 0.161 0.168 0.165 0.169 0.169 120000 0.166 0.167 0.168 0.166 0.166 0.167
dice ( 骰子 ) ∞ 1/6 1/6 1/6 1/6 1/6 1/6
Suppose some variable X can take only discrete values x , x , ..., x . 1 2 M Suppose we have measured x on N unrelated occasions, finding X takes x on N occasions, takes x on N occasions, and so on. 1 1 2 2 =lim / ≤ ≤ Definition: P xi N i N 0 P x 1 N ∞ i
M ∑ = / = / = Normalized condition: P xi N 1 N 2 ... N M N N N 1 i=1 Continuous distribution
Suppose variable X can take on any value in a continuous interval a ≤ x ≤ b
frequence dx= lim ∣b−a∣/ M M ∞ d N x
a x dx b
Definition of distribution function: d N x p xdx=lim [ ] N ∞ N
Normalized condition: ∫b = a p x dx 1 ● Probability of finding X in interval (x ,x ) ⊂ 1 2 a ,b
x2 x 2 x P =∫ p xdx 2≤ x1 P 1 x1 x1 ● Uniform distribution in interval (a,b) p(x)
1 ∈ px={ − , x a ,b b a 0, x∉a ,b a b x ● Gaussian distribution (bell curve, normal distribution) p(x) − − 2 / 2 = x x0 2 p x A e σ 1 σ <σ <σ 1 2 3
∞ σ ∫ = ⇒ = 1 2 −∞ p x dx 1 A σ 2 2 3 x x 0 Mean and variance ● Mean (expectation value) If f is a function of x ∑ ∑ 〈 〉= xi P xi , discrete f x P x , discrete x { i 〈 〉={ i i i ∫ f x p x dx , continuous ∫ f x p xdx , continuous
● Variance ( 方差 ) variancex=〈x−〈 x〉2 〉=?〈x2 〉−〈 x〉2
Standard deviation (root-mean-square deviation or RMS deviation) s=variance x
Problem: If p(x) is a gaussian distribution, please prove that s=σ. Addition and multiplication rules ● Addition rule
For a discrete distribution, the probability that the next measured value of X is either x or x (i ≠ j) equals i j P(x ) + P(x ). i j
For a continuous distribution, the probability that the next measured value of x is either in (a,b) or (c,d) equals ∫b ∫d a p x dx c p x dx provided the two intervals don't overlap. ● Multiplication rule
Suppose we measure two independent quantities X and Y. The probability that X takes value x and Y takes i value y , simultaneously, is j P (x ,y )=P (x )P (y ). joint i j X i Y j
Note: For two connected events (for example, the chance of rain and wind) we don’t get such a simple relation. ● More complicated example
Wind gives random shifts to the X component of your arrows' arrival locations, and independent random shifts to the Y component. Shooting arrows = 1 − x2 / 2 2 = 1 − y2/ 2 2 Assume: p X x e pY y e into a distant target 22 2 2 Y Problem: Find the probability, p(r)dr, that an arrow lands a distance between r and r + dr from the center. = × (x+dx,y+dy) Multiplication rule p x , y dxdy p x dx p y dy XY X Y (x,y) ⇒ = 1 −x 2 y 2/2 2= 1 −r2/ 2 2 r+dr pXY x , y e e 2 2 2 2 r X =∑ p(r) Addition rule p r dr pXY x , y dxdy 1 −r 2/ 2 2 = e ×2 r dr 2 2
r − 2 / 2 ⇒ pr= e r 2 2 r §3.2 The laws of ideal gas Thermal equilibrium state
● A state of a system of a huge number of particles ● In this state – all past history is forgotten – all macroscopic quantities (such as P, V, T, ...) cease to change with time – probability distributions of microscopic physical quantities (such as r, v, ...) don’t change with time Temperature reflects the average kinetic energy of thermal motion ● Law of ideal gas P=nk T Note: here P is the pressure! B Problem: please check the dimension! ● The origin of P
= / (a round trip) t 2L vx = / = 2/ f 1 2 mv x t mvx L
= / 2= 2/ 3 p1 f 1 L mv x L
=〈∑ 〉=∑ 〈 〉= / 3 〈 2〉= 〈 2〉 P p1 p1 N L m vx nm vx ● Interpretation of temperature
= P nk T 1 2 1 1 2 1 2 B }⇒ m〈v 〉= k T = m〈v 〉= m 〈v 〉 = 〈 2〉 2 x 2 B 2 y 2 z P nm v x
The average energy per DOF is k T/2. B
1 3 v2=v2v2 v2⇒ m〈v2〉= k T x y z 2 2 B
Problem: prove that 〈v2 〉≃500 m/s for air moleculesat room temperature Maxwell-Boltzmann distribution ● Distribution of molecular velocities
dN v ,v ,v x y z ≡ pv ,v ,v dv dv dv N x y z x y z
Distribution function Maxwell distribution
2 −mv /2k T 2 2 2 2 = B v =v v v p v x ,v y ,vz Ae x y z
Problem: find A? / ∞ ∞ ∞ 3 2 ∫ ∫ ∫ = ⇒ = m −∞ dvx −∞ dv y −∞ dvz p vx ,v y , vz 1 A 2 k B T The proportionate number of molecules with velocities between u and u+du
=∣ ∣= 2 2 2 u v vx v y vz
dN u ≡ pudu=∑ p v ,v , v dv dv dv N x y z x y z 3/ 2 −mu2 / 2k T = m B × 2 e 4 u du 2 k BT p(u)
1 / 2 3/ 2 − 2 / (T/m) = 2 m 2 mu 2k BT 1 p u u e k T (T/m) B 2
Question: what will happen when T→0? (T/m) Problem: please prove 3 − / u ∝ E K k B T p EK E K e (T/m) <(T/m) <(T/m) 1 2 3 Stern distribution [Zeits. f. Physik 2 (1920) 49]
Qestion: What's the number of molecules that can escape from a slit in a given time t with velocities between u and u+du? 0
× dN escape = A udt dN u total Volume
1/2 3 / 2 − 2/ dN u = 2 m 2 mu 2 k B T u e du N k B T
−mu2 / 2k T = 3 B C is a constant dN escape Cu e du
Related problem: the escaping of RNA from the nuclear pores. Experiment [Miller & Kusch, Phys. Rev. 99(1955)1314]
L = / = / /= / u L T 0 L 2 L 2 = −/ d T 0 2
Angle corresponds to slots ∝ Problem: please prove du u ?
− 2 / = 3 mu 2k B T dN escape Cu e du
− 2 / = 4 mu 2k B T dN detect u C 1 u e
= 4 −2u 2 dN detect u C 2u e = / Maxwell distribution is correct! u u m 4k B T ● Boltzmann distribution −E state/ k T pstate=e B
What's the state and the E(state) for N particles? { }={ } r ; v x1 , y1 , z1 ,... , xN , yN , zN ;v1x ,v1y ,v1z ,... ,v Nx ,v Ny ,v Nz ={ } r1 ,... ,r N ; v1 ,..., vN N { }= ∑ 1 2 E r ;v U r1 ,... ,r N mi vi i =1 2
Potential energy Kinetic energy Configuration space ( 构形空间 ) ≡{ } configuration r1 ,..., r N Chemical reaction
● Reaction coordinate & activation barrier Reaction coordinate ( 反应坐标 ): the resultant fast reaction path from reactant to resultant in the configuration space
Potential reactant energy barrier Configuration space
ΔE a
reactant ΔE : activation energy a resultant Reaction 活化能 coordinate Potential curve along the reaction coordinate ● Reaction rate in simple reaction The reaction coordinate can be [Science 319(2008)183] simply taken as
Assume the collision induces the chemical reaction, only the DOF in the direction of collision plays role in.
1 2 1 1 E = m v2 m v2 K 2 1 1 2 2 2 ≥ Intuitively, the molecules of reactant with E K E a can escape from the barrier. Thus the reaction rate dN E ≥ E r ∝ K a N ≥ Problem: please prove dN E K E a − E /k T =e a B N v 2 2 2= dN(E ≥ΔE )= number of molecules outside the ellipse m1 v1 m2 v2 2 Ea K a
− 2 2/ ∝ m1 v1 m2 v 2 2k BT v p v1, v2 dv1 dv2 e dv1 dv2 1
− 2 2/ ∝ m V 1 V 2 2 k B T p V 1, V 2 dV 1 dV 2 e dV 1 dV 2
V V =V 2V 2 2 1 2 −mV 2/ 2k T pV dV ∝V e B dV
V ∞ 1 ≥ ∫ dN E E 2 E / m p V dV − / K a = a = Ea k B T ∞ e 2 2= N m V 1 V 2 2 Ea ∫ p V dV 0
Hard Question: What is − / ∝ Ea k B T missed in our discussion? r e Arrhenius rate law ● How to increase the reaction rate?
− / = E a k BT Arrhenius rate law r r0 e r r
r r 0 0
T ΔE fixed ΔE fixed T a a
Two ways to increase the reaction rate: (1) increase temperature (heat); (2) decrease the activation barrier (enzyme) Relaxation to equilibrium
● Our experience: when a system is left to itself under constant external conditions for a long time, it usually arrives at a thermal equilibrium state. This process is called relaxation to equilibrium. ● Example 1: gas expansion p(x) Disorder increases!
Initial state 1/L 1
1/L Final state 2 Thermal bath (constant T) x L L 1 2 ● Example 2: high energy molecules injection
N particles is at an equilibrium state in an isolated box. A fast molecule is suddenly injected in the box.
Gas is not so idea, molecules have a size! Problem: Prove that the fast one transfers a lot of its E to the slow one when they collide elastically. K Now we have two medium-fast molecules; each is closer to the average than it was to begin with. the p(u) equilibrium distribution will be gradually approached with the more collisions.
The directed motion (mechanical energy) of the original molecule has degraded to the average random motion (thermal energy) of all molecules.
Friction: E -->E u K T §3.3 Brown motion & Random walks Brief history of Brownian motion
● Brown (1828) – Pollen grains (1μm) suspended in water do a peculiar dance – The motion of the pollen never stopped – Totally lifeless particles do exactly the same thing
● Einstein (1905) – Quantitative explanation from thermal motion of water molecules Random walks
● Drunkard's walk ( 醉汉走路 )
Are there any rules on the position R of the drunkard?
Note: R is a vector ● 1D random walk
L L L L L L L L L L L L L L L x 0 L---Length of each step x =0---start point 0 x ---position after the n-th step n k L---displacement of the n-th step with P(k =1)=P(k =-1)=1/2 n n n = xn xn−1 k n L
Problem: prove that
2= 2= 2 2 2 xn xn−1 k n L xn−1 2Lk n xn−1 k n L
2=± 2= k n 1 1 1/2 〈 〉= − − = k n xn−1 1 xn−1 P 1 1 xn−1 P 1 0
〈 2 〉=〈 2 2 2〉=〈 2 〉 2 xn xn−1 2 Lkn xn−1 kn L xn−1 L
⇒ 〈 2 〉= 2 ⇔ 〈 2 〉= xN NL x N 2 Dt
D=L2/2t diffusion constant ● 2D random walk y = = rn xn , yn xn−1 , yn−1 k xn ,k yn L L P k xn 2= 2 2 L k 1 0 -1 L yn rn xn yn L x 1 0 1/4 0 L Problem: prove that L 0 1/4 0 1/4 L L L L L L -1 0 1/4 0 〈 2 〉= r N 4 Dt
● 3D random walk = = rn xn , yn , zn xn−1 , yn−1 , zn−1 k xn ,k yn ,k zn L
P±1,0,0=P0,±1,0=P 0,0,±1=1/6; 0 for others.
Problem: prove that 〈 2 〉= r N 6 Dt ● Diffusion law 〈[r t]2〉=2 mDt
Spacial dimension
D=L2/2t
Length of each step Time of each step
Diffusion constant Microscopic parameters in brown motion Macroscopic measurable quantity §3.4 Friction & diffusion Einstein relation ● A ball in viscous fluid (macroscopic analysis) = f f f friction v friction ext = − / = − / a f ext f friction m f ext v m v /⇒ ⇒ ⇒ / i :v f ext a 0 v v f ext /⇒ ⇒ ⇒ / t ii : v f ext a 0 v v f ext
≡ / ⇔ = vdrift f ext f ext f friction The process to reach the drift velocity in micrometer scale / = = − / = − / ⇒ dv dt a f ext v m vdrift v m
= [ − ] − t /m v t vdrift v 0 vdrift e The larger ζ/m, v reaches more quickly to v drift
Stokes formula: =6R η--viscosity − / t c η ≈10-3kg m-1s-1 R ≈ 1μm e water pollen 0.35 0.3 − 0.25 ≡ /≃ 7 0.2 m 10 s 0.15 c 0.1
0.05 t/τ 2 3 4 5 c v→v very quickly such that we can omit the process! drift ● A ball in viscous fluid (microscopic analysis) Assumption: (1) The collisions occur exactly once per Δt f ext (2) In between kicks, the ball is free of random influences, so a=f /m ext (3) v is the starting value just after a kick 0 (4) Each collision obliterates ( 抹去 ) all memory of the previous step − ⇒ = / 2= / / 2 1 3 x v0x t 1 2 a t v0x t 1 2 f ext m t ⇒ 〈 〉= 4 v0x 0 〈 〉 x f ext v ≡ = = / drift t 2m t ● Einstein relation D=L2 /2t 2 = L D m =2m/t t In our discussion, we have confined L and ∆t to be constant. In fact, they are also stochastic quantities. Strict derivation gives
2 L D=m〈 〉=m〈v2〉 t x = 1 1 m〈v2〉= k T D k B T 2 x 2 B
Hypothesis Testable prediction Heat is disordered molecular motion ! Einstein relation More rigorous treatment ● Mean field thought
The collisions of water molecules are simplified as a white noise x 〈 t〉=0, 〈 tt '〉=g t−t '
/ =− (Langevin equation) Newton's law==> m dvx dt v x t
Solution: −t / 1 t −t−s / v =v e c ∫ e c sds , ≡m/ x 0 m 0 c
t −t / 1 t −t −s/ xt=∫ v dt=v [1−e c ] ∫ [1−e c] s ds 0 x 0 c 0 − / t − − / 2 2 v − / t − − / 2= 2 2t c 1 [∫ t s c ] 0 t c∫ t s c v x v0 e e s ds e e s ds m2 0 m 0
Assume that v is independent of Γ(s): 〈v s〉=〈v 〉〈 s〉=0 0 0 0
2 − / t − − / 〈 2〉=〈 2 〉 2t c 1 〈[∫ t s c ] 〉 vx v0 e e s ds m2 0
− / g 2 g 2 t = 〈v 〉− e c 2 m 0 2m
The observed time scale ≫ t c
〈v2〉=g/ 2m x / = 1 1 g 2 k B T m〈v2 〉= k T 2 x 2 B − / t − − / 2 2 2 t 2 1 t s 2 x t=v [1−e c] {∫ [1−e c ] sds} 0 c 2 0 v − / t − − / 0 c [ − t c ]∫ [ − t s c ] 2 1 e 0 1 e s ds
− / t − − / 2 2 2 2 t 2 1 t s 〈 x t〉=〈v 〉 [1−e c ] 〈{∫ [1−e c] sds} 〉 0 c 2 0
2 g 2 −t / 2 g −t / =〈v 〉− [1−e c] [t 1−e c] 0 2 m c 2 c ≫ The observed time scale t c 〈 x2t〉= g /2t ≡2Dt ⇒ ⇒ = / 2 D g 2 = / = D k B T g 2 k B T Diffusion equation ● What's meaning of the average?
〈 2 〉= e.g. r t 4 Dt (2-dimensional random walk)
r (t) y 1 y y r (t) 2 x ...... x x
r (t) M M 〈r 2t〉=lim ∑ r2t = ∞ i 4Dt M i=1 Equivalent to release simultaneously M particles at origin and then let them do independently random walk (diffuse to other places). ● Fick's law Assumption-- ● Large number particles ● Uniform in y, z direction ● Nonuniform in x direction ● Jump distance L per time Δt ● Same jump probability to left and right
N(x,t): the number of particles in the box between x-L/2 and x+L/2 at time t. (1/2)N(x,t) particles will pass through x-L/2 to the left in Δt.
Simultaneously, (1/2)N(x-L,t) particles pass through x-L/2 to the right.
1 Density of number Flux [N x−L ,t −N x ,t] 2 ∂c j= =−D N x ,t × ∂ cx ,t= t A x A×L D=L2 /2t Question: why is there a flux since Reason for net flow: there are each particle has the same jump more particles in one slot than in probability to left and right? the neighboring one.
● Continuity equation
L L N x ,t=[ jx− − j x ] A t 2 2
∂ j =− L×A×t ⇒ ∂ x ∂ ∂ c =− j ∂t ∂ x ● Diffusion equation and its solution ∂c j=−D ∂ ∂ ∂2 x c = c ∂ ∂ ∂ D (Diffusion equation) c =− j t ∂ x2 ∂t ∂ x Pulse solution ∂ ∂2 c = c D (Diffusion equation) 2 ∂ ∂ 2 1 −x /4 Dt t x cx ,t= e 4 Dt cx ,0=x (initial condition) c(x,t)
0.5 t=0.25/D 0.4 Problem: Does the solution satisfy the 0.3 diffusion equation and initial condition? 0.2 t=1/D 0.1 t=10/D x -6 -4 -2 2 4 6 ● Fluctuation Molecular thermal motion (random) ==> Diffusion equation
Question: Why is diffusion equation deterministic?
Answer: Average on ensemble (collection of large number repeated systems). Available to describe a system of large number particles.
Finite particle system: its behavior deviated from the prediction of diffusion equation. This deviation is called statistical fluctuation.
Few particle system: Statistical fluctuations will be significant, and the system's evolution really will appear random, not deterministic.
Nanoworld of single molecules: need to take fluctuations seriously §3.5 Configurations of polymers Simple polymer
● Example
PE: polyethylene ( 聚乙烯 )
Long-chain molecule
Bond length
Bond angle
Thermal motion does NOT excite the DOFs of bond length and angle! Rotational DOF
l -120° 0° 120° 0 C l , θ fixed! C C 0 C gauche trans( 反式 ) gauche( 旁式 ) ● Flexibility ( 柔性 ) Locally static flexibility ( 局部静态柔性 ) P gauche − /k T =e B ≃1 for k T P trans B Gauche/trans conformations will be found in similar frequency in the local part of a polymer. Thus the local part of a polymer appears as a random coil.
P gauche − /k T =e B ≃0 for ≫k T P trans B Gauche conformations is seldom found in local part of a polymer. Only trans conformations. Locally like a rigid rod. Persistence length ( 驻留长度 ) and Globally static flexibility
/ k T = B Question: what's physical meaning of l ? l p l0 e p − / k B T Probability of gauche conformation e between near neighbor bonds.
How many bonds will occur 1 gauche conformation?
/ k T 1 = B = / 1 gauche conformation can − /k T e l p l0 e B occur in persistence length
(1) if total length < l , the polymer seems a rigid rod. p
(2) if total length >> l , many gauche conformations p occur in the polymer. The whole chain is random coil. Locally rigid while Globally flexible. Persistence time ( 驻留时间 ) & Dynamic flexibility ( 动态柔性 )
/ = E k B T p 0 e
τ ~10 ps 0
Transition time from trans to gauche
gauche trans gauche
(1) t <τ , polymer is frozen in one configuration. Dynamically rigid. observe p
(2) t >>τ , polymer transits in different configurations. observe p Dynamically flexible.
Now, we only consider the case length >> l and t >>τ p observe p ● End-to-end distance ( 首末端距 )
R : End-to-end vector z N R N length >> l p Polymer transits between a y t >>τ large number of configurations observe p x Thus, R is a stochastic variable! N
〈 〉= 〈 2 〉= = R N ?; R N ?; p R N ? Typical models for polymers
● Freely jointed chain r N L: segment length
r : vector for the i-th segment, random R i N L distribution vector with constant length = ⋯ r R r r r r 2 N 1 2 N O 1
Problem: prove that 〈 〉= 〈 2 〉= 2 R N 0 R N NL Discuss: equivalent to random walk Question: valid condition for polymer? (L>l ) p Problem: find the distribution function of R N ∣ ∣− We have known that = r n L ⋯ =∏ p rn ; p r1, ,r N p r n 4 L2 =∑ R N rn
=∫ 3 ⋯∫ 3 −∑ ⋯ p R N d r1 d r N RN r n p r1, ,r N (a classic math formula)
⋅ − 1 3 3 3 i k R ∑ r = ∫ d k∫ d r ⋯∫ d r e N n pr ⋯,r 23 1 N 1, N N − ⋅ ⋅ 3 i k rn sin kL 1 i k R sin kL ∫ = ⇒ = ∫ 3 N d r n e p rn p RN d k e kL 23 kL sin x sin x = = 1 1, x 0 For N>>1, only kL≈0 is important! x x 0.8 2 2 ∣sin x∣ sin kL≈ − k L ≈ −k 2 L2/ 6 0.6 1, x 0 1 e x kL 6 0.4 2 / sin x x 4 3 2 = − − 2 / 2 0.2 1 O x 3 3R 2 NL x 6 p R = e N x N 2 2 4 6 8 10 2 NL -0.2 ● Freely rotating chain
rn E==0⇒ freely rotating rn− 1 ∣ ∣= rn l0 bondlength
= ∣ Problem: prove that p r p r r − p r − n n n 1 n 1 ∫ 3 ∣ = d r n p rn r n−1 1 −∣ ∣− ∣ = n rn l0 p rn rn −1 〈 〉=〈 〉 2 − 2 l 0 sin rn rn 1 cos = ⋯ R N r1 r2 r N
Problem: prove that − N 〈 〉= 1 cos 〈 〉 R r =0 for random r n 1−cos 1 1 Problem: Prove that 〈 ⋅ 〉=〈 ⋅ 〉 = 2 n−m r n rmn rn−1 rm n cos l0 cos
Problem: Prove that for θ≠0,
− N N n 1 N − n 〈 2 〉=∑ 〈 2〉 ∑ ∑ 〈 ⋅ 〉= 2 2 ∑ cos cos R N rn 2 r n r m Nl0 2l0 − n=1 n=1 m=1 n=1 1 cos 1cos 〈 R2 〉=Nl2 when N ≫1 N 0 1−cos 〈 2 〉= 2 2 Question: what will happen if θ=0? R N N l0 Discuss: relationship between this model and random walk
(1) θ=π/2, random walk without memory (2) 0<θ<π/2, random walk with memory of last step's direction Problem: find the distribution function of R N The exact result has not yet obtained for finite N! N>>1, divide: N=N +ν+N p R =∫ d 3 R p R p R − R 1 2 N N 1 1 N 1 2 N N 1
⋅ =∫ 3 i k RN R G N ,k d RN e p RN N2 N = − G N 1 , k G N N 1 ,k
RN = 1 G N 1 N 2 ,k G N 1 ,k G N 2 , k = − f kN N1 G N ,k e ≫ ⇒ = 2 4 N 1 requires f k 0 f k ak O k ν −i k⋅R = 1 ∫ 3 N 0 p R d k e G N ,k N 23 −i k⋅R − 2 ≃ ≫ ν bonds is a point in = 1 ∫ 3 N ak N N 1 N 2 d ke e macroscopic scale. 23
ν large enough, such that 3 − 2 / RN 4 aN 〈 ⋅ 〉= 2 = / r r l cos 0 1 4 aN e N 1 N 1 0 Problem: find the positive number a
3 −R2 / 4aN = / N ⇒ 〈 2 〉=∫ R 2 = p R N 1 4 aN e R N d N RN p R N 6aN
1cos 2 〈 2 〉= 2 = l 0 1 cos R N Nl0 − a 1 cos 6 1−cos
/ −3 1−cosR2 − 3 2 N 3 1 cos 2 Nl 21cos pR =[ ] e 0 N 2 (Freely rotating chain, N>>1) 2 Nl0 1 cos
3/ 2 3 −3R2 /2 NL2 = N p R N e (Freely jointed chain, N>>1) 2 NL 2
Thus, the Gaussian form is universal for the long chain polymer! ● Ideal chain (Gaussian)
Definition: a chain of finite length whose distribution of end- to-end vector satisfies
3/ 2 3 −3R2 / 2 Nb2 = N p R N e (b: characteristic length) 2 Nb2
Equivalent model: (N+1) beads connected by N spring with 2 force constant 3kBT/b = R0 0 N ⋯ = 3k B T − 2 ∑ − U R1 , , R N 2 Rn Rn 1 2b n=1 Boltzmann distribution N 3 2 − ∑ − − − / 2 Rn Rn 1 ⋯ ∝ U k B T= 2 b n=1 p R1 , , R N e e Problem: find the distribution function of end-to-end vector
=∫ 3 ⋯∫ 3 ⋯ p R N d R1 d R N −1 p R1 , , RN Path-integral form
N N − 3 ∑ − 2 − / 2 ∫ / 2 Rn Rn−1 3 2 b dn d R dn 2 = n = 3 ⋯ 3 2b n 1 n=1 ∫ ∫ − =∫ [ ] C d R1 d RN 1 e D R n e
Correspondence principle
⇒ ∂ /∂ = / 2∇2 / 3 R2 p N 3 2b p 3 2 − N 3 2 ⇒ pR = e 2Nb = N 2 Initial condition p R0 N 2 Nb Problem: please derive p(R ) directly from N
N 3 2 − ∑ − − 2 Rn Rn 1 = ∫ 3 ⋯∫ 3 2b n=1 p R N C d R1 d RN −1 e
Hints: − 2 − Rn2 Rn − [ − 2 − 2 ] ∫ 3 Rn2 Rn1 Rn1 Rn ∝ 2 d Rn1 e e
N R − R 2 R2 −∑ − 2 − N 0 N Rn Rn−1 ∫ 3 ⋯∫ 3 n=1 ∝ N = N d R1 d R N −1e e e
2 − 3RN 2Nb2 = / 3 R2 p R N C e 3 2 − N 3 2 p R = e 2Nb N 2 ∫ 3 = 2 Nb d R N p RN 1 ● s d r s Worm-like chain ts= ds 2 s d ts' U = ∫ [ ] ds' z 2 0 ds' r(s)
Correspondence principle=> y 0 ∂ pts x = ∇2 pt s ∂ s t ≡ / ∣ ∣= ⇒ ∇2 k B T 2 t s 1 t is 2D Laplace operator
∞ l − =∑ ∑ l l 1 s [ 王竹溪，特殊函数论 ] p t s 2l 1 e Y lm , l=0 m=−l − ∣ − ∣ 〈 〉= 2 s2 s1 t s1 t s2 e − NL NL − 2 NL 〈 2〉=∫ ∫ 〈 〉= NL − 1 e R ds1 ds2 t s1 t s2 0 0 22 / − 2/ pR=3/2 NL3 2 e 3 R 2 NL ; N ≫1 [Daniels (1952)] Entropic elasticity ● Elasticity of crystals (change of internal energy)
f f
Diamond Undergo small deformations = − Fcryst U T S The crystal still have good order, the entropy is almost unchanged. ⇒ F =U Change of internal energy resists the deformation induced by external force! ● Elasticity of polymers (entropy)
f f 0 0
When a force is applied on the end of a polymer, both bond length and angle keep almost constant. Thus internal energy is unchanged.
The number of configurations (i.e. entropy) changes. ⇒ F =−T S
Change of entropy resists the deformation induced by external force!
Problem: The entropy of the polymer decreases when an extension is applied. Is the 2nd law violated? (discussion!) §3.6 Biological applications of diffusion Limit on bacterial metabolism ● Idealized model
O2 is dissolved in the water, with
Lake a concentration c0:
bacterium ∞= c c0 R Bacterium immediately gobbles 吞噬 up ( ) O2 at its surface: cR=0
Problem: Find the concentration profile c(r)
and the maximum consumed number of O2 per time ● Solution
I(r): the number of O2 per time passing through the fictitious ( 假想的 ) spherical shell
0 Assume: quasi-steady state, oxygen does 累积 r R not pile up ( ) anywhere: I r=I independentof r
= / 2 cr=c 1−R/r Flux of O2: j r I 4 r 0 =− / Fick's law: j r D dc dr
BCs: = & ∞= = c R 0 c c0 I 4 Dc0 R
maximum consumed number of O2 per time ● Limit on the size of the bacterium
The number of O2 per time that a living body needs ˙ ∝ 2/3∝ 2 Q M R Metabolic rate (ref. Lecture 2)
The maximum consumed number of O2 per time ˙ = Q I 4 Dc0 R I ˙ Q≤I
R≤R R max Rmax
There is an upper limit to the size of a bacterium! Permeability of membranes ● Ion channel for passive transport
Fick's law − dc c c j=−D =−D L 0 dx L ≡− D/ L c
P s L permeation constant ( 渗透系数 ) ● Idealized model: permeation of cell membrane
Assumptions: (1) c is almost constant because out the environment is infinite =− =− − (2) jout in P s c P s cin cout = / cin N in V ,cout cin / = / = / / d c dt dcin dt 1 V dN in dt / = =− dN in dt jout in A P S A c
/ =− / (relaxation of a concentration jump) d c dt P S A V c ⇒ = −t / ≡ / c c0 e V P s A (decay constant) Problem: analysis that V/P A has the dimension of time. s Membrane potentials ● Drift and diffusion: ionic solution under E-field =− / = =− / E V l f drift qE q V l + x + V + + + = /=− / vdrift f drift q V l =− Dq V vdrift = lk B T D k B T = =− Dq V jdrift c vdrift c lk B T ∂ ∂ =− c q V =− c j D[ ∂ c] jdiffusion D x lk T ∂ x B = (Nernst-Planck formula) j jdrift jdiffusion ● = = Equilibrium state j 0 at V V eq
∂ q V q V =− [ c eq ]= ⇒ d c =− eq j D ∂ c 0 c x lk B T d x lk B T
⇒ d ln c =−q V eq ⇒ ln c =−q V eq
d x lk B T l lk BT =−q V eq ln c (Nernst relation) k BT Membrane potentials maintains a concentration jump in equilibrium!
Discuss: Nernst relation and Boltzmann distribution. Electrical conductivity of solutions
∂c q V q V j=−D[ c] =− I ∂ x lk T jq D c ΔV B lk BT (Nernst-Planck formula) for constant c
+ + - j 2 Dq c A + - + - =− = + - I q jq qA V + - - k B T l 0 l 2 De2 c A = = (Ohm's law) I I e I −e V k B T l ∝c Electrical conductivity κ { ∝1/T §. Summary & further reading Summary ● Probability – Discrete and continuous distributions – Mean and variance – Addition and multiplication rules
● Understanding the laws of ideal gas – Relation between temperature and
Spacial dimension ● Friction and diffusion
– = Einstein relation D k B T ∂c – j=−D Fick's law ∂ x ∂ ∂2 – Diffusion equation c = c ∂ D 2 t ∂ x ● Configuration of polymers – Flexibility: Persistence length, Persistence time – End-to-end distance – Typical models ● Freely jointed chain, Freely rotating chain ● Ideal chain (Gaussian) ● Worm-like chain – Entropic elasticity
● Biological applications of diffusion – Bacterial metabolism – Permeability of membranes, membrane potentials – Electrical conductivity of solutions Further reading
● L. E. Reichl, A Modern Course in Statistical Physics ● P. G. de Gennes, Scaling concepts in polymer physics ● M. Doi & S. F. Edwards, The theory of polymer dynamics