3sms potential energy in condensed matter and the response to mechanical stress 29 january, 2007...

3SMS

Potential Energy in Condensed Matter and the Response to

Mechanical Stress

29 January, 2007

Lecture 3

See Jones’ Soft Condensed Matter, Chapt. 2; Ashcroft and Mermin, Ch. 20

Last Lecture:

• Discussed polar molecules and dipole moments (Debye units) and described charge-dipole and dipole-dipole interactions.

• Discussed polarisability of molecules (electronic and orientational) and described charge-nonpolar and dispersive (London) interactions.

• Summarised ways to measure polarisability.• Related the interaction energy to cohesive energy and

boiling temperatures.

+ +- +

+

-

-

SummaryType of Interaction Interaction Energy, w(r)

Charge-charge rQQ

o421 Coulombic

Nonpolar-nonpolar 62

2

443

r

hrw

o

o

)(_=)(

Dispersive

Charge-nonpolar 42

2

42 rQ

o )(_

Dipole-charge24 r

Qu

ocos_

42

22

46 kTruQ

o )(_

Dipole-dipole

62

22

21

43 kTruu

o )(_

Keesom

321

22

21

4 rfuu

o ),,(_

Dipole-nonpolar

62

2

4 ru

o )(_

Debye

62

22

4231

ru

o )()cos+(_

In vacuum: =1

Lennard-Jones Potential

• To describe the total interaction energy (and hence the force) between two molecules at a distance r, a pair potential is used.

• The pair potential for isolated molecules that are affected only by van der Waals’ interactions can be described by a Lennard-Jones potential: w(r) = +B/r12 - C/r6

• The -ve r -6 term is the attractive v.d.W. contribution• The +ve r -12 term describes the hard-core repulsion

stemming from the Pauli-exclusion. 12 is a mathematically-convenient exponent with no physical significance!

• The two terms are additive.

L-J Potential for Ar (boiling point = 87 K)

London Constant calculated to be C = 4.5 x 10-78 Jm6

Guessing B = 10-134 Jm12

wmin -5 x 10-22 J

Compare to: (3/2)kTB= 2 x 10-21 J

Actual ~ 0.3 nm

(Guess for B is too large!)

(m)

Intermolecular Force for Ar (boiling point = 87 K)

F= dw/dr

(m)

Intermolecular Force for Ar (boiling point = 87 K)

F= dw/dr

(m)

(m)

Weak Nano-scale Forces Can be Measured with an Atomic Force Microscope

The AFM probe is exceedingly sharp so that only a few atoms are at its tip!

Sensitive to forces on the order of nano-Newtons.

Tip/Sample Interactions: Function of Distance

Physical contact between tip and surface

AFM tips from NT-MDT. See www.ntmdt.ru

Tips for Scanning Probe Microscopy

Radius of curvature ~ 10 nm

Ideally, one of the atoms at the tip is slightly above the others.

The tip is on a cantilever, which typically has a spring constant on the order of k = 10 N/m.

Modelled as a simple spring:

F = kz

where z is the deflection in the vertical direction.

F

Measuring Attractive Forces at the Nano-Scale

A = approach

B = “jump” to contact

C = contact

D = adhesion

E = pull-off

Tip deflection Force

Vertical position

AB

C

D

EC

R

Creation of a New Surface Leads to an Adhesion Force

F

F 3R is the surface tension (energy) of the tip and the surface - assumed here to be equal.

Work of adhesion:

Surface area increases when tip is removed.

∫= dAW

121≈]

11[=

P

RdPAF

2≈=1

Pressure is required to bend a surface with a surface tension,

F

Max. Capillary Force: F 4R cos

With = 0.072 N/m for water and R = 10 nm, F is on the order of 10-8 -10-9 N!

The Capillary Force

Imaging with the AFM Tip

The AFM tip is held at a constant distance from the surface - or a constant force is applied - as it scanned back and forth.

www.fisica.unam.mx/liquids/tutorials/surface.html

Distance between mica sheets is measured with

interferometry.

A piezoelectric moves the arm up by a known amount. Force on the mica is determined by measuring the distance between the mica.

Surface Force Apparatus

Mica has an atomistically

smooth surface.

L-J Potential in Molecular Crystals

Noble gases, such as Ar and Xe, form crystalline solids (called molecular crystals) that are held together solely by the dispersive energy.

In molecular crystals, the pair potential for neighbouring atoms (or molecules) is written as

The molecular diameter in the gas state is . Note that when r = , then w = 0.

is a bond energy (related to the London constant), such that w(r) = - when r is at the equilibrium spacing of r = ro.

w(r) = 4[( )12 -( )6]r

r

Lennard-Jones Potential for Molecular Pairs in a Crystal

rw(r)

+

-

-

ro

L-J Potential in Molecular Crystals

The minimum of the potential is found from the first derivative of the potential. Also corresponds to the point where F = 0.

7

6

13

12 61240

rrF

drdw

+=== [ ]-

We can solve this expression for r to find the equilibrium spacing, ro:

1212 61

.==or

)

21

41

422

426

61

12

61

61

===)(w [[( ) - ( ] ]- -

To find the minimum energy in the potential, we can evaluate it when r = ro:

Variety of Atomic Spacings in Cubic Crystals

Image from: http://www.uccs.edu/~tchriste/courses/PHYS549/549lectures/figures/cubes.gif

8 nearest neighbours; 6

2nd nearest; 12 3rd nearest

12 nearest neighbours; 6 second nearest; 24

3rd nearest

6 nearest neighbours; 12 second nearest

The particular crystal structure (FCC, BCC, etc.) defines the distances between nearest neighbours, 2nd neighbours, 3rd neighbours, etc., and it defines the number of neighbours at each distance.

Potential Energy of an Atom in a Molecular Crystal

• For each atom/molecule in a molecular crystal, we need to sum up the interaction energies between all pairs (assuming additivity of the potential energies).

• The total cohesive energy per atom is W = 1/2 r

w(r) since each

atom in a pair “owns” only 1/2 of the interaction energy.

• As shown already, the particular crystal structure (FCC, BCC, etc.) defines the numbers and distances of neighbours, 2nd neighbours, 3rd neighbours, etc., and it defines the number of neighbours at each distance.

• This geometric information that is determined by the crystal structure can be described by constants, known as the lattice sums: A12 and A6 (where the 12 and 6 represent the two terms of the L-J potential.)

• For FCC crystals, A12 = 12.13 and A6= 14.45. There are different values for BCC, SC, etc.

Cohesive Energy of Atoms in a Molecular Crystal

w(r) = 4[( )12 -( )6]r

r

So, for a pair we write the interaction potential as:

We notice that the molecules are slightly closer together in a crystal compared to when they are in an isolated pair (ro=1.12 ).

From the first derivative, we can find the equilibrium spacing for an FCC crystal:

0912 6

1

6

12 .==AA

ro ( )

For each atom in a molecular crystal, however, we write that the cohesive energy is:

W = 2[A12( )12 -A6( )6]r

r

Cohesive Energy of Atoms in a Molecular Crystal

We can evaluate W when r = ro to find for an FCC crystal:

682 12

26 .==

AA- -

This expression represents the energy holding an atom/molecule within the molecular crystal. Its value is only 8.6 times the interaction energy for an isolated pair.

This result demonstrates that the dispersive energy is operative over fairly short distances, so that most of the interaction energy is contributed by the nearest neighbours. In an FCC crystal, each atom has 12 nearest neighbours!

W

Elastic Modulus of Molecular CrystalsWe can model the intermolecular force using a spring with a spring constant, k. The force, F, to separate two atoms in the crystal is: F = k(r - ro). At equilibrium, r = ro.

The tensile stress t is defined as a force acting per unit area, so that:

2AF

o

ot r

rrk )(=

-

Fro

The tensile strain t is given as the change in length as a result of the stress:

o

o

ot r

rrrr

==

-

oo

FA

L

The Young’s modulus, Y, relates tensile stress and strain:

ttY

=

Connection between the atomic and the macroscopic

Y

t

t

Y can thus be expressed in terms of atomic interactions:

oo

o

o

o

tt

rk

rrr

rrrk

Y =)(

)(

==2

-

-What is k?

rW

+

-

-8.6ro

F = 0 when r = ro

orrdr

dFk ==

Elastic Modulus of Molecular Crystals

rF

+

-

ro

drdW

F =

Elastic Modulus of Molecular Crystals

Force to separate atoms is the derivative of the potential:

7

66

13

1212 612

2r

A

r

AF

drdW

+== [ ]-

So, taking the derivative again:

8

66

14

1212 671213(

2r

A

r

AdrdF

)()

= [ ]-

But we already know that: 61

6

122AA

ro =( )

So we see that: 61

12

6

2AA

ro= ( )We will therefore make a substitution for when finding k.

Elastic Modulus of Molecular Crystals

To find k, we now need to evaluate dF/dr when r = ro.

8

66

14

1212 671213(

2r

A

r

AdrdF

)()

= [ ]-

Combining the constants to create new constants, C1 and C2, and setting r = ro, we can write:

)(== 221

8

62

14

121 22

oo

o

o

o

r

CC

r

rC

r

rCk [ ]- -

Finally, we find the Young’s modulus to be:

3212

oo r

CCrk

Y)(

== -

As ro3 can be considered an atomic volume, we see that the

modulus can be considered an energy density, directly related to the pair interaction energy.

Bulk Modulus of Molecular Crystals

We recall the thermodynamic identity: dU = TdS - PdV

The definition of the bulk modulus, B, is: TVP

VB )(=

-

This identity tells us that: SVU

P )(=

-

If we neglect the kinetic energy in a crystal, then U W(r).

So B can be written as:TSTS V

UV

VU

VVB ,)(+=])([= 2

2

- -

After writing V in terms of , and differentiating W, we obtain for an FCC molecular crystal:

325

12

6123

754

=)(= /

AA

AB

Theory of Molecular Crystals Compares Well with Experiments

w(ro)

Response of Condensed Matter to Shear Stress

When exposed to a shear stress, the response of condensed matter can fall between two extremes: Hookean (solid-like) or Newtonian (liquid-like)

How does soft matter respond to shear stress?

A

A

y

F

AF

s =

Elastic Response of Hookean Solids

No time-dependence in the response to stress. Strain is instantaneous and constant over time.

The shear strain s is given by the angle (in units of radians).

The shear strain s is linearly related to the shear stress by the shear modulus, G:

Gs

s

=

A

A

y

FAF

s =

yx

s

~=

x

Viscous Response of Newtonian Liquids

AF

s =

A

A

y

Fx

tx

v

=

There is a velocity gradient (v/y) normal to the area. The viscosity relates the shear stress, s, to the velocity gradient.

ytx

yv

s

==

The viscosity can thus be seen to relate the shear stress to the shear rate:

====tty

xyt

xs

The top plane moves at a constant velocity, v, in response to a shear stress:

v

has S.I. units of Pa s.

The shear strain increases by a constant amount over a time interval, allowing us to define a strain rate:

t

= Units of s-1

Hookean Solids vs. Newtonian Liquids

Hookean Solids:

G=Newtonian Liquids:

=Many substances, i.e. “structured liquids”, display both type of behaviour, depending on the time scale.

Examples include colloidal dispersions and melted polymers.

This type of response is called “viscoelastic”.

Response of Soft Matter to a Constant Shear Stress: Viscoelasticity

When a constant stress is applied, the molecules initially bear the stress. Over time, they can re-arrange and flow to relieve the stress:

The shear strain, and hence the shear modulus, both change over time: s(t) and

)(=)(

ttG

s

s

t

s

s ttG

)(=

)(1

Elastic response

Viscous response

(strain is constant over time)

(strain increases over time)

Response of Soft Matter to a Constant Shear Stress: Viscoelasticity

t

s

ttG

)(=

)(1

oG1

Slope:

1

==)(s

s

s

s

dtd

We see that 1/Go (1/)

is the “relaxation time”

Hence, viscosity can be approximated as Go

Example of Viscoelasticity

Physical Meaning of the Relaxation Time

time

Constant strain applied

Stress relaxes over time as molecules re-arrange

time

teGt =)(Stress relaxation:

Typical Relaxation Times

For solids, is exceedingly large: 1012 s

For simple liquids, is very small: 10 -12 s

For soft matter, takes intermediate values. For instance, for melted polymers, 1 s.

Viscosity of Soft Matter Often Depends on the Shear Rate

Newtonian:

(simple liquids like water)

s

Shear thinning or thickening:

s

s s

s s

An Example of Shear Thickening

Future lectures will explain how polymers and colloids respond to shear stress.

Problem Set 21. Calculate the energy required to separate two atoms from their equilibrium spacing ro to a very large distance apart. Then calculate the maximum force required to separate the atoms. The pair potential is given as w(r) = - A/r6 + B/r12, where A = 10-77 Jm6 and B = 10-134 Jm12. Finally, provide a rough estimate of the modulus of a solid composed of these atoms.

2. The latent heat of vaporisation of water is given as 40.7 kJ mole -1. The temperature dependence of the

viscosity of water is given in the table below. (i) Does the viscosity follow the expectations of an Arrhenius relationship with a reasonable activation energy?(ii) The shear modulus G of ice at 0 C is 2.5 x 109 Pa. Assume that this modulus is comparable to the instantaneous shear modulus of water Go and estimate the characteristic frequency of vibration for water, .

Temp (C) 0 10 20 30 40 50(10-4 Pa s) 17.93 13.07 10.02 7.98 6.53 5.47

Temp (C) 60 70 80 90 100(10-4 Pa s) 4.67 4.04 3.54 3.15 2.82

3. In poly(styrene) the relaxation time for configurational rearrangements follows a Vogel-Fulcher law given as

= o exp(B/T-To),

where B = 710 C and To = 50 C. In an experiment with an effective timescale of exp = 1000 s, the glass transition temperature Tg of poly(styrene) is found to be 101.4 C. If you carry out a second experiment with exp = 105 s, what value of Tg would be obtained?