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Transcript of Theoretical Biophysics
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Theoretical Biophysics
Stephan W. Grill 1,2 and Frank Julicher 2
transcribed by Falk Hoffmann and Fabian Rost
1Max-Planck Institute for Molecular Cell Biology and Genetics,
Pfotenhauerstr. 108, 01307 Dresden, Germany 2Max-Planck-Institute for the Physics of Complex Systems,
N othnitzerstrasse 38, 01187 Dresden, Germany
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Contents
A. Lecture: Theoretical Biophysics 4
I. Biopolymers 5
A. Freely jointed chain 5
1. General facts 5
2. Comparison with the diffusion of a Brown particle 6
3. Statistical mechanics 8
4. Freely jointed chain with external force 12
B. Bead-and-Spring Model - Gaussian chain 14
1. Small forces 14
2. Large forces 15
3. Effect of self-avoidance 17
C. Experiments 19
1. Magnetic Tweezers 19
2. Optical Tweezers 20
D. Semiexible Polymers 21
1. Stiff/Semiexible polymers 22
2. Effective stiffness 27
II. Biomembranes 29
A. Differential geometry (geometry of surfaces) 30
1. Curves in R2 30
2. Surfaces in R3 31
3. Curvature of a surface 32
4. Formalism of vectors and tensors 33
5. Tensor of curvature 34
6. Additional surface elements that are independent of parametrization: 35
B. Bending energy of a uid membrane 36
1. Expand bending energy 36
2. Theorem of Gauss-Bonnet 37
3. Biological implications 40
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C. Monge-representation 40
1. Tensor of curvature in Monge-representation 42
D. Persistence length of a membrane 44
1. Partition function 45
2. Continuum limit 47
3. Correlation functions 48
4. Periodic boundary conditions 50
5. Persistence length of a membrane 51
E. Stretching elasticity of a membrane 52
III. Dynamics of biological molecules 56
A. One-dimensional harmonic oscillator with uctuations 591. Linear response function 59
2. Correlation function 65
3. Fluctuation-dissipation-theorem 66
B. Stochastic motion in periodic potentials 66
1. Periodic boundary conditions 67
2. Kramers rules 68
3. Transition state picture 69
4. Detailed balance 70
5. Effective diffusion 70
6. Effective mobility 71
C. Experiments 72
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A. Lecture: Theoretical Biophysics
Cytoskeleton: actin, microtubules
Organells: nucleous, golgi, mitochondria, en-
dosoms, endoplasmatic reticulum(ER)
Molecular building blocks of the cell
small molecules
ions Lipids DNA/RNA Proteins
H2O cell membrane chromosomes enzymes
Ca++ Golgi, ER ribosomes actin, tubulin
ATP mRNA motor proteins
microRNAs ion channels
Statistical physics of biomolecules exible
semiexible
Statistical physics of membranes
Dynamics, non-equilibrium processes Langevin equation, Fokker-Planck equation
detailed balance, uctuation dissipation theorem active processes, dynamic uctuations
motor proteins
Provide the basic tools required to work in theoretical biophysics
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I. BIOPOLYMERS
A. Freely jointed chain
1. General facts
Biopolymers are long chain molecules. An easy model to describe them is
the freely jointed chain. In two dimensions it is similar to a bike chain.
N : angle to the horizontal
a: segment length
RN = R0 + a
N
i=1
cosi
sin i
R = RN R0 = aN
i=1
cosi
sin i
The probability for the conguration i has to be normalized
20 d1 . . . 20 dN P (1, . . . , N ) = 1No angle is preferred, so the probability is
P (1, . . . , N ) = 1(2)N
The rst expectation value for the end-to-end-distances:
< R > = 20 d1 . . . 20 dN RP = a 20 d1 . . . 20 dN N i=1 cosisin i 1(2)N = 0because
2
0d cos =
2
0d sin = 0
The next expectation value is
< ( R)2 > = a2 20 d1 . . . 20 dN N i,j =1 (cos i cos j + sin i sin j ) 1(2)N < ( R)2 > = Na2 20 d1 . . . 20 dN 1(2)N = Na2
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Diffusion equationP t
= D P
We can solve this equation with a Greens function. In two dimensions we get the following
solution:P (R, t ) =
14Dt
exp (R R0)2
4DtA repetition how to solve a gaussian integral:
dx e A2 x2 = 2A dx x2e A2 x2 = 2A 1A
The total probability must be normalized
dx dy P (R, t ) = 14Dt dx exp (x x0)24Dt dy exp (y y0)24Dt = 1With A = 12Dt we get
< (R R0)2 > = 1
4Dt
dx (x x0)2 exp x x04Dt
2A
dy (y y0)2 exp y y04Dt
2A
+ dx exp (x x0)24Dt dy exp (y y0)24Dt< (R R0)2 > =
2A
= 4 Dt (3)
Now we look to the distribution of the end-to-end-distance
R = RN R04Dt a
2N
P N ( R) = 1a 2N
e( R ) 2
a 2 N
2.2. Three dimensions:
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< ( R)2 > = Na2
P N ( R) = 3
2a 2N
32
e3( R ) 2
2a 2 N
P (R, t ) = 1
4Dt
32
exp (R
R0)2
4Dt
< R2 > = 6Dt (4)
6Dt = Na2
4Dt = 23
Na 2
Now we want to discuss the situation with an external force.
Therefore we need statistical mechanics.
3. Statistical mechanics
3.1. Basics
Microscopical conguration: n
Microcopic state energy: E n
The Micro state energy underlays a probability distribution.
Boltzmann distribution: P n = 1Z e E n
Partition function: Z = n e E n = 1kB T Free Energy: F =
kB T ln Z
Inner Energy: U = < E > = 1Z n E n e E n
F = U T S dF = dU T dS SdT
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The First Law of thermodynamics:
dU = dW + TdS
dQ
dF = dW SdT Entropy: dW = 0: S = F T Therefore:
S = F T
= kB ln Z kB T
Z T
n
E n e E n
Z
ln Z = 1Z
S = kB ln Z
F T + 1
TZ n
E n e E n
U T TS = F + U
This comes from information theory. The expectation value of ln P says something about
the entropy. The idea is:
S = kB n P n ln P nP n =
1Z
e E n
S = kBn
e E n
Z (E n ln Z )
S = kB ln Z + 1ZT
n
E n e E n = F T
(5)
3.2. Statistical mechanics with external force
Steric expansion in micro condition nLn Force-Ensemble
Partiton function:
Z =n
exp( (E n fL n ))
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Now we look to the effective stiffness. Lets assume a linear stiffness(spring):
F (T, < L > ) 12
K (< L > L0)2 (10)with L0 = < L > |f =0Lets calculate spring constant from the partition function!
F < L >
= f = K (< L > L0)
2F
< L > 2 = f
< L > = L0= K
f < L > = L 0
= < L >
f f =0
1
= 2H f 2 f =0
1
2H f 2 f =0
= 1K
(11)
We can use the thermodynamic potential H to calculate K.
H =
kB T ln Z (f )
H f
= 1Z n
Ln exp( (E n fL n ))
2H f 2
= Z 2 n
Ln exp( (E n fL n ))2
Z n
L2n exp( (E n fL n )) 2H f 2
= (< L 2 > < L > 2) f =0=
1K
< L 2 > < L > 2= < (L< L > )2 > = kB T 2H f 2
Temperature and stiffness set the scale of uctuation (Kubo-relation) For f = 0:
< L 2 > < L > 2= kB T
K (12)
The Equipartition theorem:
12
K (< L 2 > < L > 2) = 12
kB T (13)
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4. Freely jointed chain with external force
n = 1, . . . , N
E n = 0
L = ai
cos i
We write the partition function
Z = 20 d1 . . . 20 dN exp af i cos i= 20 d eaf cos N
I (A) = 2
0d eA cos
A = af
Z = I (af )N
H (T, f ) = kB T N ln I (af )dI dA
= I = 20 d cos eA cos d2I dA2 = I =
2
0 d cos2
eA cos
We look at f=0 A=0: Relaxed state!I (0) = 2
I (0) = 0
I (0) =
< L > = H f
= kB TNaI I
= 0(= L0)
2H f 2 f =0
= kB TN 2a2I I
= 12
Na 2
kB T
< R 2x > = < L2 > = kB T
2H f 2
= 12
Na 2
< R2 > = < R 2x > + < R2y > = N a
2
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So we have the following results for equipartition:
K = 2kB T
Na 2 (14)
< L 2 > = kB T
K =
1
2Na 2 (15)
That is the entropic elasticity.
4.1. Effective stiffness
Entropic feather:
K = 1 2 H f 2
= 2kB T
Na 2
F (T, < L > ) 12K < L >
2
F = U
U =0 T S
S = F T
= K T
< L > 2
dQ = T dS
S
kB
Na2 < L >
2
Rubber elasticity: Network of entropic feathers
T K contractionT K expansionthermical expansion coefficient is negative
f K (< L > L0) (16)
stretching leads to cooling!
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B. Bead-and-Spring Model - Gaussian chain
1. Small forces
contour length s = aN = anM
Rn = x
yN = nM
Entropic elasticity k = 2k
BT
na 2
Z = d2R1 . . . d2RM exp k2kB T i (R i+1 R i)2 + i (R i+1 R i)f Z = d2 R1 . . . d2 RM exp k2 i ( R i)2 + i R if Z = dx dy exp k2 (x2 + y2) + xf M
Again have to solve a gaussian integral:
dx exp A2 x2 + Bx = 2A eB 22AZ = 2k eM f 22k = 2kB T k e M 2kk B T f 2H =
M 2k
f 2 + H 0
1 2 H f 2
= K = kM
= 2kB T
Na 2 (17)
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1.1. Effective stiffness
Bead-and-Spring model is only correct for small forces.
L2 a2N 2
L f K = Na2f
2kB T aN af
2kB T 1a =
2kB T f
f 2kB T
a
Blob picture
L = M
2 = na 2
N = nM
n = 2
a2
L = N n
= Na2
=
fNa 2
2kB T =
f K
2. Large forces
f kB T
a (18)
af 1
Z = 20 d eaf cos N This integral can be explicitely solved with the Bessel function:
I nAi
= 12i n 20 d exp[(A cos() + in )]
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with n = 0 and A = af :
20 d eA cos = 2 I 0(A)Z = 2I
0
af
i
N
Sidenote about the Bessel function
The Bessel function looks like
I n (x) = 12i n 20 d exp[i(x cos() + n)]
For non-integer n it is combined by the following expressions:
J n (x) =x2
n
m =0
(
1)m
m!(m + n + 1)x2
2m
J n (x) =x2
n
m =0
(1)mm!(m n + 1)
x2
2m
I n (x) = c1J n (x) + c2J n (x)
with complex c1 and c2.
For integer n J n (x) and J n (x) are not independent. So we need the Neumann function
N n (x) = J n (x)cos n J n (x)sin nxand the Hankel functions
H (1)n (x) = J n (x) + iN n (x)
H (2)n (x) = J n (x) iN n (x)I n (x) = c1H (1)n (x) + c2H
(2)n (x)
with complex c1 anc c2.
But we want to look for an easier way to solve the integral by using a saddle-point
approximation.
With the gaussian integral:
e x2 = 16
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Saddle-Point-Approximation(large A, small ):
20 d eA cos d exp A 1 22
d exp A 1 2
2 eA
2A
Z 2af
N 2
eaNf
H kB T N af 12
ln af
2 = Naf +
12
kB T N ln af
2
< L > = H f Na
kB TN 2f
(19)
3. Effect of self-avoidance
For the gaussian chain: = 12 . That are normal uctuations in the absence of force.
Effective Free Energy F :
dF = d W S dT isothermal :dT =0= d W
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System can perform work until F at minimum (d F =0)
F = kB T ( R Na )2
Na 2F
R =
2kB T
Na2 ( R
Na )
2F R2
= K = 2kB T
Na 2
System will progress until F R = 0
R2 = N a2
Add an interaction potential V int ( R)
F ( R) = kB T ( R Na )2
Na 2 + V int ( R)
V int ( R) ddx (x)2We have introduced the density (x) = N ( R )d .
F ( R) = kB T ( R Na )2
Na 2 +
N 2
( R)d
We minimize the free energy:
F R
= 2kB T
Na 2 ( R Na ) d
N 2
( R)d+1 = 0
Scaling Arguments:
RN N
12 N 2
Rd+1 0 R N
12
N 3
Rd+1 0 Rd+2 N
12 Rd+1
0 for d small
N 3
0 for d large
0
d small:
Rd+2 N 3 R N
3d +2
Flory : F = 3d + 2
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d large:
Rd+2 N 12 Rd+1
R N 1
2
Gaussian chain : = 12
d F
1 1 1
2 343 35 0,588
4 1
212
5 3712
. . . . . . . . .
C. Experiments
Now we want to have a look of some experiments with our models Gaussian chain, Freely
jointed chain and Bead and spring.
Measure elasticity of DNA.
Magnetic Tweezers, Optical Tweezers(Lasertrap)
1. Magnetic Tweezers
Paramagnetic beads, several mTurn on force by turning on magnetic eld and measure the length the molecule of DNA
attains.
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Measured elastcity of single-stranded DNA:
Gaussian chain, s=100nt 30nm, a=1nt 0.3nm (N=sa =100)
k = 2kB T
Na 2 =
2kB T sa
8 pN nm30nm 0.3 nm 0.8
pNnm
Measured: 0.3 pNnma =
2kB T sk
8 pN nm2
30nm 0.3 pN 0.8 nm 2.5ntA hinge every 2-3 bases!
DNA is a semiexible polymer!
D. Semiexible Polymers
E [i] =N 1
i=1
k2
i+1 i2
L = aN
i=1
cos i
Z = 20 d1 . . . 20 dn exp N 1i=1 k2(i+1 i)2 af N
i=1
cos i
Continuum limit:
E [i] =N 1
i=1
a ka2
K (i+1 i)2a2
i (s); s = iaE [(s)] = Lmax0 ds K 2 2
= dds
K. . .Bending rigidity
s. . .Arc length
R i R(s)dRds
s = R; R2 = s2
(dRds
)2 = 1
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1. Stiff/Semiexible polymers
s. . .Arc length
c. . .curvature
R = t; t = cn; n = ct; t = cos
sin ; n = sin
cos ; c =
Bending Energy:
E = Lmax0 ds K 2 c2R(s) =
s
0ds t(s) =
s
0ds cos(s )
sin (s )
Z = D(s) exp Lmax0 ds K 2 2Z = D(s) exp Lmax0 ds K 2 2 + f cos
Correlation function:
D(s s ) = < t (s)t(s ) > = < cos ((s) (s )) >D(s s ) =
1Z D(s) cos((s) (s ))exp K 2kB T Lmax0 ds 2
We have to solve a gaussian path integral
Gaussian path integral:
I (B) = Ds(x) exp 12 dD x dD y s(x)K (x, y)s(y) + dD x B(x)s(x)Discreetion: x xi ; s i = s(xi); B i = a
D
B(xi); K kl = a2D
K (xk , x l)
I [B i] = i ds i exp 12 kl skK kl s l k Bksk
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K kl . . . symmetric matrix
eigen vectors u(i)k and eigen values (i)
l
K kl u(i)l =
(i)u(i)k
l
u(i)l u( j )l = ij
i
u(l)k u(i)l = kl
Variable transformation sk = l ukl s l; ukl = u(k)lUnitar matrix l U kl (U T )lm = S km ; det U kl = 1I [B i] =
i
ds i det sk
s l 1exp
1
2 k (k)s 2k +
k
skBk
Bk =l
ukl B l
I [B i] =k dsk exp 12 (k)s 2k Bksk
ds e 2 s2 + Bs = 2 eB 22I [B i] =
k 2
(k)e
B 2k2 ( k )
12
k
Bk1
(k)Bk =
12
lm
B lK 1lm Bm
I [B i] = (2)
N 2
det K ij exp12
ij
B iGij B j
Gij
= ( K 1)ij
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Continuum limit: Bi = aD B(xi); K ij = a2D K (xi , x j ); Gij = G(xi , x j ); l K kl Glm = km
dD x K (y, x)G(x, z ) = (D )(y z )aD
l
K (xk , x l)G(xl, xm ) = kl
aD
I [B] = N det K exp
12
P ij B i G ij B j
dD x dD y B(x)G(x, y)B(y)
I [B] = N det K exp
12 dD x dD yB (x)G(x, y)B(y)
dD x K (y, x)G(x, z ) = (D )(y z )
det K ij =k
(k) = expk
ln (k) = eTr(ln K ij )
Continuum limit: det K ij exp dk (k)I [B]I [0]
= exp12 dD x dD y B(x)G(x, y)B(y)
D(s1 s2) = < cos((s1) (s2)) > = < exp(i((s1) (s2))) >D(s1 s2) =
1Z D(s) exp 12 ds K kB T 2 cos((s1) (s2))
D(s1 s2) = 1Z D(s) exp Lmax0 ds K 2 2 + Lmax0 ds B (s)(s)
I [B] = D(s) exp Lmax0 ds K 2 2 Lmax0 ds B (s)(s)B(s) = i (s s1) i (s s2)
D(s1 s2) = I [B]I [0]
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I [B] = D(s) exp 12 ds ds (s)K (s, s )(s ) + ds B (s)(s)K (s, s ) =
K kB T
2s (s s )
ds K (s, s )G(s , s ) = (s s )I [B] = N det K exp 12 ds ds B(s)G(s, s )B(s )I [B]I [0]
= exp12 ds ds B(s)G(s, s )B(s )
G(s, s ) = G(s s ) = dq 2 G(q )exp( iq (s s ))K (s, s ) =
dq
2K (q ) exp(iq (s
s ))
G(q ) = ds G(s, 0)e iqs (s) =
12 dq eiqs
(q q ) = 12 ds exp(i(q q )s )
ds
dq 2
K (q )exp( iq (s s ))
K (s,s ) dq
2G(q )exp( iq (s s ))
G(s,s )=
dq
2 exp(iq (s s ))
12 dq dq (q q )K (q )G(q )exp( i(qs q s )) = dq 2 exp(iq (s s ))
K (q )G(q ) = 1
K (q ) = ds K (s, 0)e iqs = ds K kB T (s)e iqs = Kq 2kB T G(q ) =
kB T
Kq 2
G(s s ) = dq 2 kB T K exp(iq (s s ))q 2
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J = 12 ds ds B(s)G(s, s )B(s )
J = ( i)212 ds ds ( (s s1) (s s2))G(s, s )( (s s1) (s s2))
J =
1
2(G(s1, s1)
G(s1, s2)
G(s2, s1) + G(s2, s2))
J = 12
kB T K dq 2 2 exp(iq (s1 s2)) exp(iq (s1 s2))q 2
J = kB T K dq 2 1 cos
u
qsq 2J =
kB T K |s| du2 1 cos(u)u2
12J =
kB T 2K |s|
< t (s1)t(s2) > = I [B]I [0]
= e J = exp kB T 2K |s1 s2|
With persistence length = 2K kB T and K = 12 kB T ([K] = energy * length):
< t (s1)t(s2) > = e |s1 s2 |
Explicit values are:
DNA: 50nmActin: 16 mMicrotubuley: 2 mm
Now we calculate the end-to-end-distance: R = R(s) R(0) = s
0 ds t(s )
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< R2 > = s0 ds s0 ds < t (s )t(s ) >=
s
0ds
s
0ds exp |
s s |
= 2 s
0ds
s
0ds exp
s s
= 2 s0 ds exp s s s0= 2 s0 ds (1 e s )
< R2 > = 2|s|+ 2 2(e |s | 1) (20)
|s
|:
< R2 > 2|s| a2N N =
sa 2a = a
22 = a
|s| :< R2 > |s|2 + O
s3
< R2 >
2
|s
|+ 2 2 1
|s|
+ |s3|
33 + . . .
1
= 12 |s| 21|s|+ |s|2 13|s|3
+ . . . |s|2 O |s|3
2. Effective stiffness
F (T, < L > ) 12
K (L L0)2L0 = < L >
|f =0
K = 2H f 2 f =0
1
2H f 2
= (< L 2 > < L > 2)
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small forces, freely jointed boundaries:
1K
= 2
< R2 >
K = kB T
< R2
> =
kB T
2|s|+ 2 2(e |s |
1)(21)
compare with K f jc = 2kB T Na 2
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II. BIOMEMBRANES
Amphiphile moleculespolar head (hydrophilic)
chain (hydrophobic)
spontaneous formation of micelles beyond a critical concentration
spherical
cylindrical
Lipidspolar head
two chains
formation of micelles is sterically hindered
spontaneous formation of a bilayer instead membrane
Hydrophobic chains are in contact to water at
boundary: Energetic cost!
E = l
with the line tension and the length of
boundary lFormation of closed structures: Vesicles
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Holes close spontanously
E = 2r
Therefore: Lipid membranes are extremely stable!
Biological membranes
Lipid bilayer + proteins More than 200 types of lipidsBilayer is attached to the cytoskeletonAway from thermal equilibrium
Physical properties: morphology, mechanical properties, thermal uctuations
Fluid membranes
Relative position of lipids is not xed
Rapid Diffusion (D 1 10 m2
s )
Number of molecules is constant
surface is constant
Incompressible surface, no shear stress within bilayer ( water)Bending of the plane surface costs energy
Only important degree of freedom is the geometry (or shape) of surface Theory of uc-tuating geometries
Physics is invariant with respect to the choice of parametrisation of the surface
A. Differential geometry (geometry of surfaces)
1. Curves in R 2
t: tangent vector
n: normal vector
R(s); s: arc length
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Measure a distance:
( R)2 = ( R(s2) R(s1))2 dRds
s2
= t2( s)2
s is an arc length if t2 = 1
R2 = s2; t = R; t2 = R2
= 1
t = cnn = ct
c: curvature
t n = cn20 c; c = t n = R
2nd derivative
n
Circle:
x = R cos sRy = R sin sR
t = R = sin sR
cos sR; t2 = 1: parametrization is correct
t = R = 1R
c
cos sRsin sR
n
2. Surfaces in R 3
ui : parametrization (compare to arc length s)
Problem: There is no arc length parametrization for a surface!
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innitesimal displacement d ui
dR = Ru i
dui = eidui
dui as contravariant vectors, i covariant
tangent vector (not normalized)
ei = Ru i
= iR
distance on a surface
ds2 = (d R)2 = ei e j duidu jgij = ei e j
Metric tensor
We can now measure arc length using the metric tensor!
Local reference system ei = iR; e1 and e2; that is non-orthogonal
a = a ieinotation will become clear
later...Some things depend on parametrization chosen, some things do not.
Physics (e.g. Energy, curvature) cannot depend on the parametrization.
What are the parametrization-independent objects in this formalism?
3. Curvature of a surface
Tangent plane e1e2Normal vector n
How does the normal vector/tangent plane change with a displacement d ui?
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in = C ji e j (22)
ie j = D ij n + kij ek (23)lives in tangent plane
lives everywhere
Note C ji and D ij describe a curvature
C ki gkj = D ij related to the tensor of curvature
Christoffel-symbol (23): kij = ek ( ie j ) igki Divergence of a vectorRelationship between C ji and D ij
n ei = 0 i(n e j ) = ( in) e j + ( ie j ) n = 0(22) : ( in) e j = C ki ek e j = C ki gkj
(23) : ( ie j ) n = D ij n2 + kij ek n = D ij
C ki gkj = D ij
C ki is the tensor of curvature
D ij = ( ie j ) ne j = j R
D ij =
n
i j R (24)
curvature: 2 nd derivative!
How can we extract the curvature from C ji and D ij ?
Information is there: Geometry of the surface is locally characterized by C ji (and gij to
dene a metric).
4. Formalism of vectors and tensors
eis are not normalized and not orthogonal
Introduce a conjugate base e j
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ei e j = i jTwo coordinate systems: a = ai
contravariantei =
a i
covariant
ei
Coordinate transformation:
a i = a ei = a j e j ei = a j ji = aia i = a ei = a j e j ei = a j i j = a i
but also
a i = a ei = a j e j ei = a j g jiThe metric tensor transforms coordinates from contra- to covariant bases!
a i = a j g ji ; a j = aigij ; gij = ( gij ) 1
Scalar product:
a b = aib j ei e j = aib j gij = aib j ei e j = aib j ji = aibiParametrization-independent objects
Scalar: a b = aibiVector: a = aiei = a iei
Tensor: a = aij
(eie j ) = a
i j (eie
j
)Example: T a = T i j (eie j ) a = T i j (eie j akek) = T i j a j ei is a vector
5. Tensor of curvature
C ji = n gkj
symmetric
i k
symmetric
R
C ji is symmetric and can be diagonalized
2 Eigenvalues c1 and c2
2 orthogonal Eigenvectors
independent of parametrizationTwo scalars of curvature
(i) Mean curvature H = 12 T r(C ji ) =
12 (c1 + c2)
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(ii) Gaussian curvature K = det( C ji ) = c1c2
c1 and c2 the principal curvatures. The corresponding two axes of principal curvature are
orthogonal.
C ii = 2 H ; C ji C i j = (2 H )2
2K
c1 > 0
c2 > 0
H > 0
K > 0
c1 < 0
c2 < 0
H < 0
K > 0
c1 > 0
c2 < 0
K < 0
saddle point
6. Additional surface elements that are independent of parametrization:
Normal vector n
n = e e2| e1 e2 | ; n2 = 1
Surface element d A
ndA = ( 1Rdu1) ( 2Rdu2) = ( 1R 2R)du1du2 = ( e1 e2)du1du2
(ndA)2 = d A2 = ( e1
e2)2(du1du2)2 = ( g11g22
g12g21)
[(e1 e2 )2 = e21 e22 (e1 e2 )2 ](du1du2)2
(ndA)2 = d A2 = det gij (du1du2)2
g = det gij
dA = gdu1du2 independent of parametrization!A = gdu1du2
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B. Bending energy of a uid membrane
1. Expand bending energy
E = dA f (C ji (R))where C ji is the tensor of curvature and f (C
ji ) is a scalar, which is independent of the
parametrization. We now develop f to second order terms:
f a0 + a1C kk + a2(C kk )2 + a3C ki C ik + O (C ji )3
So we now have 4 parameters and 2 scalars of curvature H and K :
C kk = 2H
C ki C ik = (2 H )
2 2K With a0 = + 2 (c0)
2, a1 = c0, a2 = + 2 and a3 = 2 , f is typically written as:f = +
2
(2H c0)2 + K 2 Scalars of curvature:
H = 12 C kk mean curvature
K = det C ji gaussian curvature
4 Parameters:
bending rigidity
gaussian rigidity
surface tension
c0 spontaneous curvature
With that the bending energy looks like
E = A
constant+ dA 2(2H c0)2 + K discuss nextThe rst term A is constant, when A is constant. We are now going to discuss the last
term, which leads us to the Theorem of Gauss-Bonnet.
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2. Theorem of Gauss-Bonnet
A
dA K = 4 n (25)
The integrated surface must be closed. n = 1g, whereg is the topological genus (or the number of holes) of
the surface.
Gauss interpreted it in the following way: As above d A is:
dA n = ( 1R 2R)du1du2dA = g du1du2
g = det ( iR j R) = det gijNow lets look at the surface created by n(u1, u2). If the surface you integrate over is closed,
the surface created by all vectors n(u1, u2) is a sphere with radius 1. In anology to d A we
determine the surface element of this unit sphere d As2 :
dAs2 n s2 = ( 1n 2n)du1
du2
dAs2 = k du1du2k = det ( in j n)
One can see that
A
dAs2 n = 4n (26)
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Be aware that n is the normal vector and n is the one from equation (25). We now determine
k:
in j n = C ki ek C l j el = C ki gkl C l jk = det ( C ki gkl C l j ) = K 2 g
We used equations (22) and (23). So now:
4n = A
dAs2 n = A
k du1du2 = A
g du1du2 dAK =
A
dA K
A closed integral of gaussian curvature is topological invariant.
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Making equation (26) plausible
Lets look at a torus as an example. There exist two parallel normal vectors n and n for
each direction of n. So the surfaces created by n are two spheres with radius 1.
But it is not clear if d As2 is parallel
or antiparallel to n. We will check
this for the two normal vectors n and
n and the parametrization shown in
the gure on the left.
dAs2 = ( 1n 2n)du1du
2
nu 1
du1 = n(u1 + d u1, u2) n(u1, u2)
We have two different situations for n and n :
n and n s2 parallel n and ns2 antiparallel
So one sphere gives a surface area of 4 whereas the other gives 4:
dAs2 n = 4 4 = 0
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3. Biological implications
E = A
constant
+
A
dA2
(2H c0)2 +
A
dA K
constantIn a typical biological reaction, the number of holes in a given closed membrane does notchange and A dA K is constant. Exceptions are the budding or fusion of vesicles, wherethis term needs to be considered. Furthermore we assume A to be constant, so the rst termis constant, too.
E = A
dA2
(2H c0)2
For asymmetric bilayers c0 = 0:
solvent asymmetric composition asymmetric
In all other cases c0 = 0
E = A dA 2 (2H )2we conclude that the bending energy E only depends on a single parameter .
C. Monge-representation
Now lets choose a paramatrization!
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z = h(x, y); u1 = x; u2 = y; R =
x
y
h(x, y)
The base vectors therefore are:
ei = iR
ex =
1
0
xh
; ey =
0
1
yh
The normal vector is:
n = e1 e2|e1 e2| = 1 1 + (h)2
x h
yh1
Where:
h x h
yh; (h)
2 = ( x h)2 + ( yh)2
The metric tensor is:
gij = ei e jgij =
1 + ( x h)2 ( xh)( yh)
( xh)( yh) 1 + ( yh)2 = ij + ( ih)( j h)
Also
g = det gij = 1 + ( x h)2 1 + ( yh)2 ( x h)2( yh)2= 1 + ( x h)2 + ( yh)2 = 1 + (h)
2
The conjugate metric tensor gij can be written as follows:
gij = ( gij ) 1 = 1g
1 + ( yh)2 ( x h)( yh)( x h)( yh) 1 + ( x h)2
= ij ( ih)( j h)1 + (h)
2
Where we used:a b
b c
1
= 1
deta b
b c
c bb a
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We now calculate the conjugate base vectors:
e j = gij ei ex = g11ex + g
12ey = 1g
1 + ( yh)21
0
x h
1g
( xh)( yh)
0
1
yh
ex =
1g
1 + ( yh)2
( x h)( yh)
x h
ey = g21ex + g22ey = 1g
( xh)( yh)1 + ( x h)2
yh
Now we proove that ei e j = i jex ey =
1g
(( x h)( yh) + ( xh)( yh)) = 0ex ex =
1g
(1 + ( yh)2 + ( x h)2)
1+( h)2 = g= 1
1. Tensor of curvature in Monge-representation
With equation (24):
C ij = n i j R = 1
g x h yh
1
0
0
i j h
= i j h g
C ji = C ik gkj =
i kh g kj
( kh)( j h)g
= i j h g +
1g
32
k
( i kh)( kh)( j h)
Bending Energy
E = d2x 2(2H )2 2H = Tr(C ji)We now assume weak uctuations: ( h)
2 1.
g = 1 + O (h)2 2H = h 1 + O (h)2= 2x +
2y
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E = dx dy 2 ( h)2Partition function of a symmetric bilayer:
Z = DRVol( g) exp 2 dA (2H )2DR(u1, u2) Sum over all surfaces and parametrizations of each surfaceVol(g) Measure of all parametrizations of a surface conrmation, volume of
the group of reparametrizations
Z is invariant with respect to the
parametrization!
In Monge-representation it is:
Z = Dh exp 2 d2x ( h)2
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Sidenote about partial integration
b
a
u v dx = b
a
v u dx + [uv]ba
1st note:
dx 2
( 2x h)
u( 2x h)
v=
dx 2
( 3x h)
v( x h)
u+ ( xh)
u( 2x h)
v
=
dx 2
h( 4x h) + ( x h)( 2x h) h 3x h
0, if h( )=0=
dx
2h( 4x h)
2nd note:
d2y 1kB T ( y y + m) (x y) h(y) 4xP .I.= 1kB T ( x x + m)h(x) dx (x)f (x) = dx f (x) (x) = f (0)
D. Persistence length of a membrane
Compare to the persistence length of a polymer = 2K kB T calculated from the correlation
function < t (s1)t(s2) > .
To be able to calculate this quantity, we will place the membrane between two surfaces
and later move the surfaces apart from each other.
eliminates translation in h-
direction
Introduce a repulsive potential V(h):
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V tot = V (h + d) + V (d h)
12
mh 2 + const. m = 2
2
h 2V (h)
h= d
E = d2x
2( h)2 + m
2 h2
2xP .I.= d2x h 2 + m2 hThis is valid for periodic boundary conditions.
1. Partition function
Z (B) = Dh exp d2x 12h Kh + BhB(x) is an external force density. Periodic boundary conditions:
K = kB T
+ mkB T
With:
K (x, y) = kB T
y y + mkB T
(x y)One can write K as:
Kh = d2y K (x, y)h(y)Therefore:
Z (B) = Dh exp d2x d2y h(x)K (x, y)h(y) d2x B(x)h(x)45
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Reminder: Gaussian Path Integral (discrete case):
Z (B1, , Bn ) =n
i=1 dh i exp 12h iK ij h j B j h j=
2
N
det K ij exp 12BiGij B j
Gij = ( K 1)ij
Free Energy:
H (B1, , Bn ) = kB T ln Z =
kB T 2
ln (det K ij ) kB T
2 BiGij B j + const
Mean height (compare with Freely jointed chain: < L > = H F ):< h i > =
H B i
= kB T
2 Gij B j
= 0, if B = 0
Correlation function:
< h ih j > = 1kB T
2H B iB j
= Gij
Gij determines the correlations.H B i is a functional derivative in the continuum case.
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Functional Derivative Example: Minimize the bending energy of a surface
E [h(x)] = d2x 2 ( x h(x))2E h(x) + h(x)
E [h(x)] + E
E = d2x E h(x) h(x)Discrete:
E =i
E h i
h i
E [h(x) + h(x)] = d2x 2 [ x {h(x) + h(x)}]2= d
2x
2( x h(x))
2
E [h(x)]+ d
2x [ x h(x)] [ x h(x)] E E = d2x [ x h(x)] [ x h(x)]
2xP .I.= d2x ( x x h(x)) Eh ( x ) h(x)
E h(x)
= x x h(x)
For the surface with the minimal bending energy:
E h(x)
= 0
2. Continuum limit
< h (x) > = H B(x) B =0
= kB T 2 d2y G(x, y)B(y) B =0 = 0
< h (x)h(y) > = (2) H
B(x)B(y) B =0= G(x y)
Determine the propagator G = K 1
kB T
x x + mkB T
G(x y) = (x y)
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Fourier space:
G(x) = 1(2)2 d2q G(q )e iqx
(x) = 1
(2)2 d2q e iqx
kB T
q 4 + mkB T
G(q ) = 1
G(x) = kB T (2)2 d2q e iqxq 4 + m
If m = 0, G diverges for q 0. This is why we placed the membrane between two surfaces!
3. Correlation functions
< h (x)h(0) > = G(x) = kB T (2)2 d2q e iqxq 4 + m
< h(x) h(0)2 > = C (x) = < h (x)2 + h(0)2 > 2 < h (x)h(0) >
= < h (x)h(x) > + < h (0)h(0) > 2G(x)C (x) = 2 G(0) 2G(x)
What does C (x) look like?
So now lets calculate C (x). Therefore we introduce the lenth scale :
= m
14
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G(x) = kB T
1
(2)2 d2q e iqxq 4 + 4G(x) =
kB T
1(2)2
0
dq q 2
0
d e iqx cos
q 4 + 4
= kB T
1
2
0
dq qJ 0(qx)q 4 + 4
= kB T 2
2kei
x
J o(y) is a Bessel-function, kei(y) is the Thompson-function. There are approximations for
the Thompson-function:
kei (x) 4
+ 4
(a + bln x)x2 x 0kei(y)
1y
12
e y 2 sin
y 2 y
< h (x)2 > = kB T
8 2
C (x) = 2 G(0) 2G(x)= < h(x) h(0)
2 > = kB T 22 d2q 1e iqxq 4 + 4
C (x) = kB T
4a + bln
x
x2 x
C (x) = kB T
4 2 x
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C (x)L2
x
L
L kB T
12
L3
4. Periodic boundary conditions
Finite system size:
q min = 2L
q max = 2
a
a: microscopic length, thickness of the membrane, size of molecules. a4nm.
Because undulations on lengths smaller than the size of the microscopic constituent are
unphysical, we integrate from q min to q max .
G(x) = k
BT
(2)2 Q d2q
e iqx
q 4
< h 2 > = G(0) = kB T (2)2 qmaxqmin dq q 1q 4 = kB T 2 12q 2
2a
2L
= 1(2)3
kB T
(L2 a2)
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5. Persistence length of a membrane
< t (s1)t(s2) > = e |s 1 s 2 | p
p: units of length.
At which length do normal vectors de-correlate?
< (h)2 > 1 ?
Under what conditions is < (h)2 > 1, if we change the system size? This will give us
some kind of persistence length!
< h(x)h(y) > = xy < h (x)h(y) >
= 2x G(x y)=
kB T (2)2 Q d2q q 2e iq(x y)q 4
< h(x)2 > =
kB T (2)2 Q d2q 1q 2
= kB T 2 qmaxqmin dq 1q
= kB T 2
ln La
We need the mircroscopic cut-off! Gradient grows with larger system size L and will become
larger than one at some size. Persistence length is estimated as L = p, where < (h)2 > = 1.
p ae2
k B T
kB T 10 for typical lipids, a4nm p ae60 = 4 1017m astronomical number!
Persistence length of membranes is nite but very, very large: Membranes are at.
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But: Do not take thermodynamic limit (do not make system size innite as people
typically do in the statistical thermodynamic limit), because the membrane will fold!
E. Stretching elasticity of a membrane
: isotropic tension
dA = g dx dyg = 1 + (h)
2
A = d2x 1 + (h)2 A0 + 12 d2x (h)2A0 = L2: projected area.
Remember polymers:
Z (f ) = DR exp ds 2c2 + Lf H (f ) = kB T ln Z (f )
< L > = H f
Legendre transformation:
F (< L > ) = H + < L > f
f = F
< L >
= f
< L > =
2F < L > 2
= 2H f 2
1
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Do the same for membranes!
E = d2x 2( h)2 A0 A0 = A const 12 d2x (h)2
=
d2x
2( h)2 +
2(
h)2
A
Z ( ) = Dh exp d2x 2 ( h)2 + 2(h)2 + AH ( ) = kB T ln Z ( )
H
= kB T
Z Z
= A + < d2x 12(h)2 >
d2x
A 0< 12 ( h)
2 >
= < A 0 > average of the projected area
F (< A 0 > ) = H + Z < A 0 >F
< A 0 > =
Stretch molecules:
= < A 0 > Z
< A 0 > (units of energy / surface area)
= < A 0 > 2F
< A 0 > 2 =
< A 0 > 2 H 2
< A 0 > = A
1
2
A = < A 0 > 1 + 12
< (h)2 >
< A 0 > A 1 12
< (h)2 >
Total area A is larger then the projected are < A 0 > .
< (h)2 > =
kB T (2)2 d2q q 2q 4 + q 2 = kB T (2)2 qmaxqmin dq q 1q 2 +
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q min = 2L
q max = 2
a
< A 0 > = A 1 kB T 8
ln 4
2
a 2 +
4 2L 2 +
: Energy
area : Energy
L2
: Energy
area
< A 0 >Z =0
= A 1 kB T 4
ln La
For L 2 a 2 :
< A 0 > A 1 kB T 8
ln 4
2
a 2 +
4 2L 2 +
= A 1 + kB T 8
ln a2
4 2
Grows logarithmically with !
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A0 dlP V = A0
V 4R 2 R A 8R R
= RP
2
Measure A0 and and plot ln vs. < A 0 > . This should be a line with slope kB T 8 determined!
1 = 1
< A 0 >
< A 0 >
=0=
kB T
2 qmax
qmin
d2q 1
(2)2
q 4
(q 4
)2
= 323 2
kB T 1
L2 a2Stiffness at = 0: diverges when L = a, because there are no uctuations then.
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III. DYNAMICS OF BIOLOGICAL MOLECULES
Time-dependent stochastic processes.
Diffusion: Ficks second law:
t P = D P With:
=
2
R 2
P = P (R, t )
Solution for 2 dimensions:
P (R, t ) = 14Dt exp
(R
R
0)2
4Dt
Compare to gaussian polymer (freely jointed chain):
P (R, N ) = 1a 2N
exp (R R0)2
a2N
Random path of a diffusing particle:
a = v0 t; N = t t
4Dt a2
N
RN = R0 + aN
i=1
cosi
sini
R(t i + t) = R(t i) + v0cosi
sini
i t
i = random angle, i = random velocity.
< i > = v0 20 di2 cosisini = 0< i k > =
v2022 20 di 20 dk (cosi cos k + sin i sin k)
= v20 ik
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Continuum-limit: stochastic differential equation. t 0, D = const .dRdt
= (t) for Diffusion
What are the stochastic properties of (t)?
< (R R0)2 > = N a2 = t t
v20 t 2 = 4 Dt
D = 14
v20 t v20 =
4D t
a2 = v20 t2 = 4 D t
< (t i) (tk) > = 4 D ik t
Continuum-limit:
< (t) (t ) > = 4 D (t t )dRdt
= (t)
< (t) > = 0
< (t) (t ) > = 4D (t
t )
Vector components in any dimension:
< (t) (t ) > = 2 D (t t )Diffusion in a potential energy landscape:
W (R); =
: Mobility, : Random force
dRdt
= W R
+ < (t) (t ) > = 2D (t t )dRdt
= W R
Drift+
Diffusion< (t) (t ) > = 2
D2
(t t )
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Back to probability P (R, t ):
t P + R
J = 0
J =
D
P
Diffusion
RW P
DriftWith that you get the Smoluchowsky-Equation: t P = D P
Diffusion+
R
R
W P
DriftStationary State: t P = 0 R J = 0Steady state: J = 0
P (R) = N e
W ( R )D
Compare to the Boltzmann-Distribution:
P (R) = N e W (R ) =
1kB T
= D
This gives the Einstein-Relation :
D = kB T
Equation of Diffusion:
t P = D P
Stochastic differential equation:
t R = (t) < (t) > = 0
< (t) (t ) > = 2 D (t t )
Stochastic motion in a potential energy landscape, with random force (Langevin-Equation):
t R = W + (t) < (t) (t ) > = 2kB T
(t t )
Smoluchowsky-equation describes diffusion in a potential energy landscape:
t P = D P + (P W )
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Einstein-relation:
D = kB T
External force:
W (x) = V (x) f ext xHarmonic Oscillator with force:
W (x) = 12
kx2 f ext x
A. One-dimensional harmonic oscillator with uctuations
m 2t x + 1 t x + kx = + f ext < (t)(t ) > = 2kB T 1 (t t )
Correlation function
< x (t)x(t ) > = C (t t )
1. Linear response function
The response-function (t) is convoluted with the external force to give < x (t) > .
< x (t) > = t dt (t t )f ext (t ) + O(f 2ext ) t dt (t t )f ext (t )= dt (t t )f ext (t )
Where (t) = (t)(t).
Fourier-representation of (t) and f ext (t):
(t) = 12 d ()e it
f ext (t) = 12 d f ()e it
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With that < x (t) > is:
< x (t) > = dt d2 d2 ()e i(t t ) f ( )e i t=
d
2 d () f ( )e it
dt
2 e( )t
( )= d2 () f () =x()e it
< x (t) > = d x()e it x() = () f ()We look at the mean response < x (t) > , which should be independent of uctuations, = 0.
With that and the fourier-representation of < x (t) > the equation of motion for < x (t) >
looks like:
d2 m2 i 1 + k () f ()e it = d2 f ()e it () =
1
m2 i 1 + k () =
1m( 1)( 2)
Solve:2 +
im
km
= 0
Solution:
1/ 2 = i2m
12 4 km 1(m)2
With:
= 12 4 km 1(m)2
= 12m
One can write the solutions as:
1 = i2 = i
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(t) = d
2 ()e it
t < 0 : (t) = 0
t > 0 : (t) = (t) = d2 ()e itRemember that (t) = (t)( t).
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Note about Residuals:
dx f (x) = 2 i Res[f (x); a]a single pole:
Res[f (x); a] = limxa(x a)f (x)a multiple pole, order m:
Res[f (x); a] = 1
(m 1)! limxa
dm 1
dxm 1(x a)m f (x)
Only the rst term of a Laurent expansion survives, i.e. expand f (x) around x = a to order1x .
Explanation: Lets look at the pole 1zn :
dz 1zn
z = ei
dz = ieid
dz 1z n = 20 d ieie in = 20 d ie i(n 1)
This integral is always zero, except for n = 1:
if n = 1 : dz 1z n = i 12
0= 2 i
t > 0 : (t) = + i 1
m(1 2)e 1 t +
1m(2 1)
e 2 t
Note the sign change because of clockwise integration instead of counter clockwise (mathe-
matically positive).
(t) = i 1
2m e
1t
+ 12m e
i2
t
= i2m
e t i t e t+ i t
Note: ex+ iy = ex(cos y + i sin y)
(t) = i2m
e t [cos( t) + i sin( t) cos( t) i sin( t)]
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(t) = 1m
sin( t)e t for t > 0
(t) = 1m
sin( t)e t (t) t
(t) = (t) + i (t)
(t) real (t) imaginaryIn principal there are two ways to write (t):
1)
(t) = (t)antisymmetric, imaginary
(t) = (t)symmetric, real
2)
(t) = (t)symmetric, imaginary
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(t) = (t)antisymmetric, real
We now check if both ways are possible:
() = dt (t)eit ; (t) real() = dt (t)e it = () () complex
If (t) = (t) () = () : () realIf (t) = (t) () = () : () imaginary
We know: () = 1 m 2 i + k has real and imaginary part.
We still dont know which one is symmetric and which one is antisymmetric.
could it be that (t) is symmetric and (t) is antisymmetric?
Lets do the inverse transform:
() = ()
real: (t )= ( t )+ i ()
imaginary: (t)= (t )For (t) = (t) + i (t) to be real, (t) needs to be imaginary!
(t) symmetric, (t) antisymmetric.
Given a particular response function, causality
invokes a specic dissipation function ().
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2. Correlation function
f ext = 0
C ( ) = < x (t)x(t + ) >
C ( ) = d2 C ()e i C ( ) = C ( ) real
C () = C () = C () real and symmetric
More precisely:
C ( ) 12T
T
T dt x(t)x(t + )
= 12T T T dt d2 d2 x()x( )e it i (t+ )
= 12T T T dt d2 d2 x()x( )e it (+ ) i lim
T
12T T T dt e it (+ ) = 2 ( + )
C ( ) = d2 C ()e i With C () = < x()x() > . Now:
x() = ()( + f ext ) f ext = 0
x() = ()()
Therefore:
C () = < x()x() >= < ()() ()() > () = ()= < ()() > ()()= < ()() > | ()|2 < ()() > = 2
kB T
= 2kB T
1
|k m2 i |2
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We now calculate ():
() = 12i
[ () ()]=
1
2i
1
k m2 i
1
k m2 + i
=
1
|k m2 i |2This leads to the:
3. Fluctuation-dissipation-theorem
C () = 2kB T ()
Relationship between the complex part of the linear response function (the dissipation func-
tion) and the correlation function!
(t) = (t)kB T
dC (t)dt
B. Stochastic motion in periodic potentials
One-dimensional:
W (x + a) = W (x)
Smoluchowsky-Equation:
t P + x J = 0
J = 1
x P + ( x W f ext )P P denotes the probability density for the particle to be at position x at time t; P (x, t )
= (W f ext x)J 1 = x P P x
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Periodic system:
P (0) = P (a)
a
0dx P = 1
Nonequilibrium steady state (stationary state):
J = const. t P = 0
P (x) = N e ( x) J
e ( x) x0 dy e( y)
a0 dx P = 1 1 = N ao dx e ( x) J a0 dx e ( x) x0 dy e( y)N =
1 + J a0 dx e ( x) x0 dy e( y)
ao dx e ( x)
1. Periodic boundary conditions
P (0) = P (a)
Ne (0) = Ne ( a) J
e ( a) a0 dy e( y)
= ( a)
(0) =
f ext
a
N (e 1) = J
a
0dy e( y)
Solve for J:
(e 1) 1 + J
a0 dx e ( x) x0 dy e( y) = J a0 dy e( y) a0 dx e ( x)(1 e ) =
J a0 dx e( x) x0 dy e( y)e x0 dy e( y) + a0 dy e( y)
I (x)
I (x) = x+ a
ady e
x
0dy e +
a
0dy e =
x+ a
0dy e
x
0dy e =
x+ a
xdy e
J =
1 e
a0 dx e ( x) x+ ax dy e( y)v(x) = J P (x)
< v > = a J
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2. Kramers rules
< v > = v = akB T
a
0 dx e ( x)
x+ a
x dy e( y)
(1 e )
Denition:
r = kB T
a0 dx e ( x) x+ ax dy e( y)r + = r
r = re f ext ak B T
v = ar (1 ef ext ak B T )
v = a(r + r )
E (x) = W (x) f ext x; = E Saddle point approximation:
x+ ax dy e+ E (y) e+ E B dx e 2 |E B | (x xB )2 = 2 |W B
|
eE B
W B = E B . This integral is over one period and does not depend on x in this approximation!
a0 dx e E (x) e E A dx e 2 E A (x xA )2 = 2W A e E AW A = E A . Stationary rate:
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r 2
W A |W B |12
e E (f ext )
E = E B E A
r depends on second derivatives at minimum/maximum and on E (f ext )!
3. Transition state picture
Expand E (f ext ) to rst order in f ext :
E (f ext ) E 0 af ext + O(f 2
ext ) E 0 = W f ext =0
0 1
v0 = ar 0e W 0k B T
Where we dened r 0:
r0 = 2 (W A |W B |)
12
= 0 : v = v0 1 e f ext ak B T
= 12
: v = v0 ef ext a2k B T e
f ext a2k B T
= 1 : v = v0 ef ext ak B T 1
Motion in a continuous free energy landscape discrete hopping modelSystem spends most of the time at minima and moves to adjacent minima according to
Kramers Rules. This is not only true for periodic energy landscapes.
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4. Detailed balance
At equilibrium:
P N = N e W N k B T
or:
P ArAB = P B r BA
Therefore:r ABr BA
= eW A W B
k B T
Periodic potential:
v = a(r + r ) r+
r = e
f ext ak B T
5. Effective diffusion
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t P n = (r + + r )P n + r + P n 1 + r P n +1=
1a
(J n +1 J n ) xJ J n = a(r P n r + P n 1)
Detailed balance: J n = 0 at equilibrum!
J n = vP n + P n 1
2 DP n P n 1
a vP D x P
ar = v2
Da
ar + = v2
+ Da
v = a(r+ r )
D = 12
a2(r + r )
v a2
(W A |W B |)12 e W 0 (1 e f ext a )
D
a2
4 (W
A |W
B |)
12 e W 0 (1 + e f ext a )
6. Effective mobility
0 = vf ext f ext =0
D0 = Df ext =0
0 = a2
2kB T
(W A
|W B
|)
12 e W 0 =
D0
kB T Einstein Relation:
0kB T = D0
D0 Fluctuations Correlation function0 friction 1 Dissipation Response function
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C. Experiments
Single molecule, one protein at a time.
Example: RNA polymerase.
Copies DNA into messanger RNA. mRNA is then translated into protein by the ribosome.
Optical trap:
A bead in a laser trap can be described by an overdamped harmonic oscilator with uctua-
tions:
Overdamped: m = 0
1 t x + kx = f extNeed to determine the stiffness of the trap for calibrated measurements!
Measure at f ext = 0: thermal uctuations of bead.
Determine: C ( ) = < x (t)x(t + ) > and C () from experimental data.
Spectral Density C ()
C ()m =0
= 2kB T
1
k i 2
= 2kB T
1
k2 + 22D = kB T
= 2D
2k2 + 2
Lorentz-function:
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We know and D, so we can very accurately determine k by tting the data to this function!
Stiffness of DNA:
Semiexible polymer: K DN A = kB T 2 |s |+2 2 (e|s | 1) for small forces.
Motion of the polymerase along the DNA:Interaction potential is periodic with a period of 1 basepair!
Periodic Potential
Discrete hopping model with Kramers Rules
and external force!
THE END