1. 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

2. 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. + -

3. Type of Interaction Interaction Energy, w(r)
Summary
Type of Interaction Interaction Energy, w(r)
In vacuum: e=1
Charge-charge
Coulombic
Dipole-charge
Dipole-dipole
Keesom
Charge-nonpolar
Dipole-nonpolar
Debye
Nonpolar-nonpolar
Dispersive

4. 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.

5. L-J Potential for Ar (boiling point = 87 K)
London Constant calculated to be C = 4.5 x Jm6
Guessing B = Jm12
wmin  -5 x J
Compare to: (3/2)kTB= 2 x J
Actual s ~ 0.3 nm
(Guess for B is too large!)
(m)

6. Intermolecular Force for Ar (boiling point = 87 K)
F= dw/dr
(m)

7. Intermolecular Force for Ar (boiling point = 87 K)
F= dw/dr

8. 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.

9. Tip/Sample Interactions: Function of Distance
Physical contact between tip and surface

10. Tips for Scanning Probe Microscopy
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.
Radius of curvature ~ 10 nm
Ideally, one of the atoms at the tip is slightly above the others.
AFM tips from NT-MDT. See

11. Measuring Attractive Forces at the Nano-Scale
A = approach
B = “jump” to contact
C = contact
D = adhesion
E = pull-off
Tip deflection  Force C C A B E D
Vertical position

12. Creation of a New Surface Leads to an Adhesion Force
Surface area increases when tip is removed.
F  3pRG
G is the surface tension (energy) of the tip and the surface - assumed here to be equal.
Work of adhesion:

13. F The Capillary Force F  4pRG cosq
Pressure is required to bend a surface with a surface tension, G F
F  4pRG cosq
Max. Capillary Force:
With G = N/m for water and R = 10 nm, F is on the order of N!

14. 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.

15. Surface Force Apparatus
Mica has an atomistically smooth surface.
A piezoelectric moves the arm up by a known amount. Force on the mica is determined by measuring the distance between the mica.
Distance between mica sheets is measured with interferometry.

16. 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
w(r) = 4e[( )12 -( )6]
The molecular diameter in the gas state is s. Note that when r = s, then w = 0.
e is a bond energy (related to the London constant), such that w(r) = - e when r is at the equilibrium spacing of r = ro.

17. Lennard-Jones Potential for Molecular Pairs in a Crystal+
w(r) -e ro r s -

18. 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. [ ] -
We can solve this expression for r to find the equilibrium spacing, ro:
To find the minimum energy in the potential, we can evaluate it when r = ro: ) [ [(
- ( ] -

19. Variety of Atomic Spacings in Cubic Crystals
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.
8 nearest neighbours; 6 2nd nearest; 12 3rd nearest
12 nearest neighbours; 6 second nearest; 24 3rd nearest
6 nearest neighbours; 12 second nearest
Image from:

20. 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 Sr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 = and A6= There are different values for BCC, SC, etc.

21. Cohesive Energy of Atoms in a Molecular Crystal
w(r) = 4e[( )12 -( )6]
So, for a pair we write the interaction potential as:
For each atom in a molecular crystal, however, we write that the cohesive energy is:
W = 2e[A12( )12 -A6( )6]
From the first derivative, we can find the equilibrium spacing for an FCC crystal: ( )
We notice that the molecules are slightly closer together in a crystal compared to when they are in an isolated pair (ro=1.12 s).

22. Cohesive Energy of Atoms in a Molecular Crystal
We can evaluate W when r = ro to find for an FCC crystal: W -
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!

23. F Elastic Modulus of Molecular Crystals
We 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. F ro ao ao
The tensile stress st is defined as a force acting per unit area, so that:  -
The tensile strain et is given as the change in length as a result of the stress: -

24. Connection between the atomic and the macroscopicst et F A L Y
The Young’s modulus, Y, relates tensile stress and strain:
Y can thus be expressed in terms of atomic interactions: -
What is k?

25. Elastic Modulus of Molecular CrystalsW + -
-8.6e ro s r F + - ro
F = 0 when r = ro

26. [ ] [ ] ( ) ( ) Elastic Modulus of Molecular Crystals -
Force to separate atoms is the derivative of the potential: [ ] -
So, taking the derivative again: [ ] -
But we already know that: ( )
So we see that: ( )
We will therefore make a substitution for s when finding k.

27. [ ] [ ] Elastic Modulus of Molecular Crystals - -
To find k, we now need to evaluate dF/dr when r = ro.
Combining the constants to create new constants, C1 and C2, and setting r = ro, we can write: [ ] -
Finally, we find the Young’s modulus to be: -
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.

28. Bulk Modulus of Molecular Crystals
The definition of the bulk modulus, B, is: -
We recall the thermodynamic identity: dU = TdS - PdV
This identity tells us that: -
So B can be written as: -
If we neglect the kinetic energy in a crystal, then U  W(r).
After writing V in terms of s, and differentiating W, we obtain for an FCC molecular crystal:

29. Theory of Molecular Crystals Compares Well with Experiments
w(ro)

30. Response of Condensed Matter to Shear Stressy A
How does soft matter respond 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)

31. Elastic Response of Hookean Solids
The shear strain gs is given by the angle q (in units of radians). A F Dx y q A
The shear strain gs is linearly related to the shear stress by the shear modulus, G:
No time-dependence in the response to stress. Strain is instantaneous and constant over time.

32. Viscous Response of Newtonian Liquids
The top plane moves at a constant velocity, v, in response to a shear stress: v A y F Dx
There is a velocity gradient (v/y) normal to the area. The viscosity h relates the shear stress, ss, to the velocity gradient.
h 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:
Units of s-1
The viscosity can thus be seen to relate the shear stress to the shear rate:

33. Hookean Solids vs. 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”.

34. 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: t
Elastic response
Viscous response
(strain increases over time)
(strain is constant over time)
The shear strain, and hence the shear modulus, both change over time: gs(t) and

35. Response of Soft Matter to a Constant Shear Stress: Viscoelasticity
Slope: t
t is the “relaxation time” t
We see that 1/Go  (1/h)t
Hence, viscosity can be approximated as h  Got

36. Example of Viscoelasticity

37. Physical Meaning of the Relaxation Time
Constant strain applied
time s
Stress relaxes over time as molecules re-arrange
time
Stress relaxation:

38. Typical Relaxation Times
For solids, t is exceedingly large: t 1012 s
For simple liquids, t is very small: t  s
For soft matter,t takes intermediate values. For instance, for melted polymers, t 1 s.

39. Viscosity of Soft Matter Often Depends on the Shear Ratess
Newtonian:
(simple liquids like water) h ss h
or thickening:
Shear thinning h

40. An Example of Shear Thickening
Future lectures will explain how polymers and colloids respond to shear stress.

41. Problem Set 2 t = to exp(B/T-To),
1. 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 = Jm6 and B = 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 h 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, n.
Temp (C)
h (10-4 Pa s)
Temp (C)
h (10-4 Pa s)
3. In poly(styrene) the relaxation time for configurational rearrangements t follows a Vogel-Fulcher law given as
t = to exp(B/T-To),
where B = 710 C and To = 50 C. In an experiment with an effective timescale of texp = 1000 s, the glass transition temperature Tg of poly(styrene) is found to be C. If you carry out a second experiment with texp = 105 s, what value of Tg would be obtained?