Discussion topic for week 3 : Entropic forces

• In the previous lectures we have stressed that random motion is

the dominant feature of cellular dynamics.

Why don't cells exploit the long-range Coulomb interactions to

accelerate enzyme reactions, binding of ligands to proteins, etc?

Thermodynamics in cells (Nelson, chap. 6)

See the Statistical Physics notes for detailed descriptions of Entropy,

Temperature and Free Energy.

Entropy is a measure of disorder in a closed system. It is defined as

where Ω is the number of available states in the phase space

Statistical postulate: An isolated system evolves to thermal equilibrium.

2nd law: S 0, equilibrium is attained when the entropy is at maximum.

Entropy of an ideal gas:

),,(ln NEkS

'ln)2(ln 2/32/3 CVENkmECVkS NN

Example:

Temperature is defined from entropy as

For the ideal gas:

11

dEdS

TordEdS

T

kTNE

ENk

CENkdEd

T 23

23

]")ln([1 2/3

02lnln)2ln(

)2(ln

ln

2/3

2/3

kNVVkS

CEVkS

CEVkS

NN

NNf

NNi

To create order, we have to do work on a system

volume change

work done

xFfT

xPAfT

xL

PVf

Tx

LNkT

fTV

xANkxf

T

VV

NE

ENkVEkS

P

kin

kinNNkin

11

111

23

lnln 2/3

xfWE

xfW

xAV

kin

x

To push the piston f > FP, then S > 0. At the equilibrium S = 0

Open systems:

Consider the same system but now

immersed in a heat bath which

keeps the temp. constant at T.

As the spring expands

Consider the total entropy of the system a+B

0

aaBaBatot

Batot

ESTESTSTSTST

SSS

LxNkS

EEEEEE

NkTE

a

aBasprkina

kin

0,0

23

a

Bremains const.

Helmholtz free energy of a system in contact with a thermal bath at T

System comes to an equilibrium when the total entropy is maximum

or when its free energy of is minimum.

How much work can we extract from such a system?

[This suggests that we can define an entropic force from free energy]

Maximum possible work:

dx

dF

x

Ff aaa

aaaaa TSEETSF )(

PALPVLNkTfF

xLNkTfLxTkNxfSTEF

a

aaa

0

(min)aa FFW

Assuming the initial length is Li, it will expand until

Internal kinetic energy remains the same

The free energy changes by

Introduce

Work done on the load is

i

f

i

f

iifffia

L

LNk

V

VNk

EVEVNkLLS

lnln

lnln),( 2/32/3

NkTEE fi 23

NkTxxfLLLfW fif )(

fNkTL f

i

faaa L

LNkTSTSTEF ln

xLLLLLx fifif 1)(

32)1ln(

32 xxxNkTxNkTFa

The difference between free energy and work done is wasted as heat!

In order to extract maximum possible work

Quasi-static processes provide the most efficient way for extracting work

If we bring this system into

contact with a heat bath at

T2 < T1 and recycle this process

we obtain a heat engine.

32

32 xxNkTWFa

fPAL

PVL

NkTLLx iF 1

T2

T1

Microscopic systems:

Consider a microscopic system in contact

with a heat bath which keeps the temp.

constant at T.

According to the statistical postulate

all the allowed states in the joint system

have the same probability P0

Probability of a particular state in “a”

kTEkESa

B

BatotBatotBBB

BBBBBBa

atotB eePEP

dE

dSEESEESES

PkESPEPEEP

)(0

000

)(

)()()(

/)(exp)(lnexp)()(

BaBatot EEEEE ,

Boltzmann distribution

Two-state system:

Consider a system, which is in thermal equilibrium with a reservoir and

has only two allowed states with energies E1 and E2

The probabilities of being in these states follow from the Boltzmann dist.

EaseeP

P

eee

eP

EEEeee

eP

eeCPP

CePCeP

kTEkTEE

kTEkTEkTE

kTE

kTEkTEkTE

kTE

kTEkTE

kTEkTE

0

1

1

,1

1

1

,

)(

1

2

2

121

121

21

12

21

2

21

1

21

21

Activation barrier:

In a more typical situation, there is also an energy barrier in the forward

direction (i.e. E2 E1), which is called the activation barrier, E‡

Rate constant: (probability of transition per unit time)

For N particles, the rate of flow is given by N.k

kTECek

The forward and backward rate constants and flux are given by

At equilibrium,

Approach to equilibrium

kTEEkTE

kTEkTEEkTE

eCNkNeCNkN

ekkCekCek

)(1122

)(

,

,

1111

1221

121

122

)()(

,

,

NkkNkNkNNkdt

dN

NNNNNN

NkNkdt

dNNkNk

dt

dN

tottot

tottot

kTEeNNkNkN 0

10212 (cf. Nernst pot.)

At equilibrium,

To solve the DE, introduce

Here is called the relaxation (or decay) time to equilibrium.

It critically depends on the activation barrier,

If E >> kT, k 0, and transition probability becomes

kk

NkN

dt

dN tot01

1 0

kTEkTE

ttkk

tot

eCekk

enentn

nkkNnkkNkdtdn

11

)0()0()(

)())((

)(

01

011 NNn

kTEe

tkt ekekP

)12(

Free energy in microscopic systems:

In a microscopic system, fluctuations in energy can be large, so we need

use the average energy. Given the probability of each state as Pi

Average energy

Entropy (from Shannon’s formula)

By analogy, we define the free energy of a microscopic system as

We need to show that this free energy is minimum at the equilibrium

and leads to the Boltzmann distribution

To find the minimum of Fa, set

i iia EPE

1,0

ii

i

a PP

Fto subject

i iia PPkS ln

aaa TSEF

where Z is the partition function

Substitute in Fa

Z

ePP

ePPkTEP

F

PPkTEPF

kTE

ii i

kTEiii

i

a

iii

iiia

i

i

1

01ln

ln

ZkT

ZkTEPkTEP

Z

ePkTEPF

iii

iii

kTE

ii

iiia

i

ln

ln

ln

i

kTEieZ

Complex two-state systems:

Because biological macromolecules are flexible, they do not have unique

structures. Rather in a given configuration, there are an ensemble of

states whose energies are very close.

Consider two such configurations I and II, with multiplicities N I and NII

If all the states in I had the same energy EI and in II, EII, then

More generally, they are different, so we need to use a Boltzmann dist.

Using

IIIkTE

I

IIkTE

I

kTEII

I

II EEEeN

N

eN

eN

P

PI

II

,

kTFIII

IeZZkTF ln

Z

Ze

ZP

Z

Ze

ZP II

N

Ni

kTEII

IN

i

kTEI

II

I

iI

i

11

1,

1

The relative probability becomes

Similarly, the transition rates between two complex states becomes

Note: when the volume changes in a reaction, the quantity to consider is

the Gibbs free energy

This is important for reactions involving the gas phase but can be

ignored in biological systems, where reactions occur in water

IIIkTF

kTF

kTF

I

II

I

II FFFee

e

Z

Z

P

PI

II

,

PVTSEPVFG

IIIkTF

III

III FFFek

k

,

Example: RNA folding as a two-state system (Bustamante et al, 2001)

Single molecule experiment using optical tweezers

Increasing the force on the bead triggers unfolding of RNA at ~14.5 pN

DNA handle

Reversible folding of RNA

Black line: stretching

Gray line: relaxing

z = b-a ~ 22 nm

When f = 14.1 pN, half of the RNA

in the sample is folded and half

unfolded

f.z = 14.1 x 22 = 310 pN.nm = 76 kT

F = 79 kT

Entropic Forces (Nelson, chap. 7)

Entropic forces are in action when a system in a heat bath does work

e.g., expansion of gas molecules in a box of volume V

VNkT

VF

P

constVNkTVCkTF

VCCV

epdpdrdrdCZ

ZkTF

N

NN

mkTNN

N

.ln'ln

')(

ln

2)(31

331

3 221

pp

dV

dFP

Adx

dF

A

f

dx

dFf

Ideal gas law applies equally to solutions

Osmotic pressure:

Sugar solution rises in the vessel until it reaches an equilibrium height zf

gzA

gAz

Amg

P ff

m

ckTV

NkTP

Amg

equil

Osmotic flow

Reverse osmosis

mg

van’t Hoff relation

Calculate the work done by osmosis when the piston moves from Li Lf

The sugar concentration changes as

For maximum efficiency, assume that we operate near equilibrium

This is the same expression we calculated for the free energy change

in an expanding volume of gas molecules.

Dilute solutes in water behave the same way as gas molecules.

FL

LNkTdx

x

NkTfdxW

x

NkTkTA

Ax

NckTAPAf

i

fL

L

L

L

f

i

f

i

ln

AxNxVNxc )()(

Osmotic pressure in cells:

Cells are full of solutes; proteins occupy roughly 30% of the cell volume.

Volume of a protein:

Protein concentration:

Note: c(mole/liter)=c(SI)/(103 NA), so c = 0.12 mM

From van’t Hoff relation

32438 104103

4m

pV

3221073.0 mccVp

Pa300101.4107 2122 ckTPeq (cf. 1 atm = 105 Pa)

RPdRRPPdV

RdRdAdW

RdRdAdAAAAdRRR

equilib 21

.24

8

8,,

Laplace’s formulafor surface tension

Thus 0.5 x 10-5 x 300 = 1.5 x 10-3 N/m = 1.5 pN/nm, rupture tension!

Osmotic pressure and depletion force

When small and large molecules coexist, large ones tend to aggregate.

The short range force driving this process is called the depletion force.

When d < 2R, depletion forces act to reduce the volume inaccessible

to smaller molecules.

RAckTFckTPVF

2

d < 2Rd > 2R

Electrostatic forces in solutions

Review of electrostatics:

Coulomb’s law:

Electric field:

Gauss’ law:

Differential form:

Central field:

Electric potential: & energy

rdUrd

qrdda

rd

qr

qq

VV

tot

VS

V

33

0

02

0

0

3

0

33

0

2112312

21

012

)()(21

,'')'(

41

)(

,0

1

'''

)'(4

1)(

41

rrrrr

r

EE

E

nE

rrrr

rrE

ErF

Dielectrics:

In a dielectric medium charges are screened: q q/ /

where is the dielectric constant of the medium.

Electric forces, fields and potentials are screened by the same amount.

Water has a large dipole moment and hence a large (= 80

Proteins and lipids have a much lower (≈ 2 - 5

Boundary between two media with different forms a polarizable interface

Eex

Induced surface charge density

nE ˆ221

210 ex i

n̂

Examples:

1. Charge near a membrane

Estimates of reaction force for an ion near a membrane (q’ ≈ q/84)

Useful quantities to remember

vacuum water

kTkTe

780560

,560)(14

1 2

0

= 80

= 2

q

d

d q’

RRR

RRR

qUd

q

qEFd

qE

qq

21

0

20

121

21

,2

'

4

1

,)2(

'

4

1

'

image charge

pNd

F

kTdd

U

R

R

)(

72

)(7.1

)(484560

22

2. Charge near a protein (spherical)

Note that the monopole term (l = 0) vanishes.

The dipole term gives

= 80

= 2

q

r

22

21

21

10 )1(

)(

4

1)(

l

lR r

a

ll

l

a

qr

No simple image charge solution

Series solution gives:

30

)(

7.1

24

1

21

4

10

arkTa

qU

raa

q

RR

R

a

q

3. Charge inside a channel (infinite cylinder)

Series solution gives (too complicated!)

The cylinder results are very different because, unlike the other cases,

the charge is inside a closed boundary (cf. induced charge is 40 times

larger).

= 80= 2

a

kTaa

qqU

aq

a

q

qqq

RR

indR

ind

)(

47

124

1

641

341

2

2

021

00

12

21

Poisson-Boltzmann (PB) equation:

Mobile ions in water respond to changes in the potential so as to

minimize the potential energy of the system.

Instantaneous potential is given by the Poisson eq’n.

At equilibrium, the mobile ions assume a Boltzmann distribution

Substituting this density in the Poisson eq’n gives the PB eq’n

Because of the exp dependence, the PB equation is very difficult to solve.

mobfix ,)(0 r

kTqmob ecq )(

0r

kTqfix ecq )(

00 )( rr

Gouy-Chapman solution of the PB eq’n for a charged surface:

1-1 electrolyte solution (e.g. NaCl)

surface charge density,

no fixed charges for x > 0

Boundary cond.:

Solution (see the Appendix)

where is the inverse Debye length,

0

)0()0(

dxd

E

kTeec

eeec

dx

d kTekTe

sinh2

0

0

0

02

2

x

xx

x

eekTe

ekTe

e

kTx

00

0

)4/tanh(1

)4/tanh(1ln

2)(

0

0002 ,2 kTce

Linearized PB equation:

Expand the Boltzmann factor in the ion density as

The first term vanishes from electroneutrality; substituting the second

term in the PB eq’n, we obtain the linearized PB eq’n

This is much easier to solve, e.g. for the previous problem we have

Boundary condition:

xexkT

ce

dx

d

02

0

02

2

2

)(2

kTqcqecq kTqmob )(10

)(0 rr

)()( 02

0 rr

kT

cqfix )( kTe

00

00

)0( dx

d

Charge density:

Debye length controls the thickness of the charge cloud near a charged

surface (at x = 1/, it drops to about 1/3 of its value at x = 0)

For a 1-1 salt solution at room temperature

Rule of thumb: for a 90 mM solution, 1/ = 10 Å

Although the approximate solution is obtained for

it remains valid up to

For nonlinear effects take over and the potential decays

faster than the exponential.

kTe

0

200 32//1 ceckT where c0 is in moles

mVekTkTe 25

kTe

xkTekTe

kTekTe

ekTceeeec

eeceec

02

0

00

2

,

Energy stored in a diffuse layer:

Potential energy due to an external field:

Energy per unit area for a charged surface

Estimate the energy for a unit charge per lipid head group

For a cell of radius 104 nm,

Ion clouds are permanent fixtures of charged surfaces.

220

2

2

41102

)(14 nm

kT

nmnm

eU

rdU 3

0

2

0

0

20

0

21

21

)()(

dxedxxxAU x

2nme

kT nm 2 998 104,10104 UA

Debye-Huckel solution of the linearized PB eq’n for a sphere:

Ion clouds modify the Coulomb interaction between macromolecules.

We can solve the PB equation for a sphere to describe this effect.

Represent the total charge q of the molecule with a point charge at r = 0

The coefficients c1 and c2 are determined from the boundary conditions

at r = a, namely, and E are continuous across the boundary

re

cr

arrdr

dr

arcrq

r

ar

out

in

)(

2

22

22

10

)(

,1

,4

1)(

a

q

Using the second boundary condition gives

Again we have an exponential “Debye” screening of the Coulomb potent.

To find the screening charge use

a

e

r

qr

a

qc

aac

a

q

dr

d

dr

d

ar

out

a

out

a

in

14

1)(

14

1

1

4

1

)(

0

02

2220

r

e

a

q ar )(22

02

02

14

)(2

2

1)(4)( arre

a

qrrrg

maximum at r = 1/

Total screening charge

Energy stored in the diffuse layer

For

This is too small, but charges on a protein are on the surface and q >> e

Coulomb interaction is mediated by the ion clouds around proteins.

qdrrea

qdrrr

a

ar

a

)(2

2

14)(

kTUaeq 02.0162

1.07,10/1,30,

20

2

)(22

2

0

22

)1(24

)1(44)()(

a

q

drea

qdrrrrU

a

ar

a

Why is water special?

Large dipole moment (p = 1.8 Debye) large dielectric const. ( = 80)

Tetrahedral structure of water

molecules in ice

Molecular structure of H2O

Covalent O-H bond: 1 Å

H-bond distance: 1.7 Å

O-O distance: 2.7 Å

q ~ 0.4 e

q

-2q

q

1 Å

104

1 Å

Energy scales involved in binding of molecules:

~ 100 kT single covalent bonds, e.g. C-C, C-N, C-O, O-H

~10 kT H-bonds, e.g. O---H, N---H (recall -helix and DNA)

~1 kT van der Waals attraction between neutral molecules

Note that the above values are for two molecules in vacuum.

In water, as with all other interactions, the H-bond energy is also reduced

to about 1-2 kT. Thus, H-bond energy is comparable to the average K.E.

of 3kT/2, and it can be relatively easily broken.

In ice, there are 4 H-bonds per water molecule.

In water at room temperature, the average number of H-bonds per water

molecule drops to ~3.5. That is, H-bond network is maintained to ~90% !

*** Dynamics of water involves making and breaking of H-bonds ***

Solvation of molecules in water:

Ions gain enough energy from solvation so that salt crystals dissolve in

water to separate ions. Born energy for solvation

Energy gain when a monovalent ion goes from vacuum into water:

This energy is larger than the binding energy of an ion in crystal form.

In water, ions have a tightly bound hydration shell around them.

Typically there are 6 water molecules in the first hydration shell.

Solventberg model: conductance of ions in water increases with temp.

2,24

1 2

0B

BB r

r

qU

for most ions

kTrq

UvacuumwaterB

B 1154

5601124

1 2

0

Solvation of nonpolar molecules and hydrophobic interactions:

When nonpolar molecules such as hydrocarbon chains are mixed in

water, they disrupt the H-bond network, costing free energy.

For small molecules, water molecules can form a ‘clathrate cage’ around

the intruder, which almost maintains the H-bond in bulk water.

But such an ordered structure costs entropy, i.e. free energy still goes up.

Size dependence of solubility

of small nonpol. molecules:

Butanol, C4H9OH (top)

Pentanol, C5H11OH

Hexanol, C6H13OH

Heptanol, C7H15OH (bottom)

Appendix: Gouy-Chapman solution of PB eq.

Solve the diff. eq.

Subject to the boundary condition:

First introduce the dimensionless potential,

and the inverse Debye length,

The diff. eq. becomes,

Multiply both sides by and integrate from ∞ to xdxd2

kTeec

dx

d

sinh

2

0

02

2

0

)0()0(

dxd

E

1cosh2cosh2 22

22

dxd

dxd

dxd

dxd

kTe /

kTce 0022

sinh22

2

dx

d

Take the square root of the diff. eq. using

Integrate [0, x]

Substitute

xx

u

u

ekT

e

kTe

e

xuu

xu

du

4tanh

4tanh

4tanh

4tanh

2tanh2tanhln

,sinh

00

0

0

2/,2/ dduu

2/sinh21cosh 2

x

dxd

dxd

0

2)2/sinh(

2/sinh2

0

(reject the + solution)

Using the identity

We can write

Boundary conditions: 0, x

x

x

x

x

vv

v

vv

vv

ekTe

ekTe

e

kTx

e

ee

z

zez

e

e

ee

eev

)4/tanh(1

)4/tanh(1ln

2)(

)4/tanh(1

)4/tanh(1

1

1

1

1tanh

0

0

0

02/

22

2

0

0

0

0

0

0

0

42sinh

2sinh

8)0(

eckT

e

kT

ekTc

dxd

Approximate solution for

Boundary condition:

Charge dist.:

x

kTekTe

kTekTe

ekTce

kTeeceeec

eeceec

02

00

00

2

sinh2

,

x

xx

x

x

xkTxe

ex

ekT

e

ekTe

ekTe

kT

xe

ekTe

ekTee

0

0

0

0

0

02)(

)(

21

41

41

2

)(1

)4tanh(1

)4tanh(1

00

00

)0( dx

d

kTe