Interactions in Condensed Matter: Lennard-Jones Potential & Atomistic Forces

1. HE3 SM Potential Energy in Condensed Matter and the Response to Mechanical Stress 5 February, 2009 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. Summary Charge-charge Coulombic Dipole-charge Dipole-dipole Keesom Charge-nonpolar Dipole-nonpolar Debye Nonpolar-nonpolar Dispersive Type of InteractionInteraction Energy, w(r) In vacuum:e=1

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. Actual s ~ 0.3 nm (Guess for B is too large!) wmin -5 x 10-22 J Compare to: (3/2)kTB= 2 x 10-21 J (m) L-J Potential for Ar (boiling point = 87 K) London Constant calculated to be C = 4.5 x 10-78 Jm6 Guessing that B = 10-134 Jm12

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

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

8. The AFM probe is exceedingly sharp so that only a few atoms are at its tip! Weak Nano-scale Forces Can be Measured with an Atomic Force Microscope Sensitive to forces on the order of nano-Newtons.

9. F 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 www.ntmdt.ru

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

11. C A B C E D Measuring Attractive Forces at the Nano-Scale A = approach B = “jump” to contact C = contact D = adhesion E = pull-off Tip deflection  Force 0 Vertical position

12. R Creation of a New Surface Leads to a “Thermodynamic” Adhesion Force F Surface area increases when tip is removed. Gis the surface tension (energy) of the tip and the surface - assumed here to be equal. Work of adhesion:

13. W = GLVA + GSVA - GLSA Work per unit area, W: W = GLV+ GSV - GSL GLV q GSV GSL Work per unit area, W: Young-Dupré Equation L L GLVA GLSA S GSVA S When liquid (L) and solid (S) are separated, two new interfaces with the vapour (V) are created.

14. The Capillary Force Pressure is required to bend a surface with a surface tension,G F F  4pRG cosq Max. Capillary Force: With G = 0.072 N/m for water and R = 10 nm, F is on the order of 10-8 -10-9 N!

15. Force for Polymer Deformation

16. 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.

17. 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 and knowing the arm’s stiffness. Distance between mica sheets is measured with interferometry. www.fisica.unam.mx/liquids/tutorials/surface.html

18. 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 s. Note that when r = s, then w = 0. eis a bond energy (related to the London constant), such that w(r) = - ewhen r is at the equilibrium spacing of r = ro.

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

20. 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 pair’s equilibrium spacing, ro: To find the minimum energy in the potential, we can evaluate it when r = ro:

21. 6 nearest neighbours; 12 second nearest 8 nearest neighbours; 6 2nd nearest; 12 3rd nearest 12 nearest neighbours; 6 second nearest; 24 3rd nearest 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. Image from: http://www.uccs.edu/~tchriste/courses/PHYS549/549lectures/figures/cubes.gif

22. 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 = 1/2Sr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 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.

23. Cohesive Energy of Atoms in a Molecular Crystal 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: 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=1.12s).

24. 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!

25. F ro The tensile stress st is defined as a force acting per unit area, so that:  The tensile strainetis given as the change in length as a result of the stress: - 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. ao ao ro

26. + W ro s r -8.6e - + F ro r - Elastic Modulus of Molecular Crystals F = 0 when r = ro

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

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

29. 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: - Elastic Modulus of Molecular Crystals [ ] - To find k, we now need to evaluate dF/dr when r = ro. 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.

30. F Response of Condensed Matter to Shear Stress A y 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. q The shear straingsis linearly related to the shear stress by the shear modulus, G: Elastic Response of Hookean Solids The shear strain gs is given by the angle q (in units of radians). A F Dx y A 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: A F Dx v y A There is a velocity gradient (v/y) normal to the area. The viscosityh relates the shear stress, ss, to the velocity gradient. The shear strain increases by a constant amount over a time interval, allowing us to define a strain rate: The viscosity can thus be seen to relate the shear stress to the shear rate: hhas S.I. units of Pa s. Units of s-1

33. Hookean Solids: Newtonian Liquids: 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 Viscous response Elastic response t The shear strain, and hence the shear modulus, both change over time:gs(t) and When a constant stress is applied, the molecules initially bear the stress. Over time, they can re-arrange and flow to relieve the stress: (strain increases over time) (strain is constant over time)

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

36. Example of Viscoelasticity

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

39. Typical Relaxation Times For solids,tis exceedingly large: t1012 s For simple liquids,tis very small: t10 -12 s For soft matter,ttakes intermediate values. For instance, for melted polymers,t1 s.

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

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

42. 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 = 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 waterhis 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)0 10 20 30 40 50 h(10-4 Pa s) 17.93 13.07 10.02 7.98 6.53 5.47 Temp (C)60 70 80 90 100 h(10-4 Pa s) 4.67 4.04 3.54 3.15 2.82 3. In poly(styrene) the relaxation time for configurational rearrangementstfollows a Vogel-Fulcher law given as t = toexp(B/T-To), where B = 710 C and To = 50 C. In an experiment with an effective timescale oftexp= 1000 s, the glass transition temperature Tg of poly(styrene) is found to be 101.4 C. If you carry out a second experiment withtexp = 105 s, what value of Tg would be obtained? Problem Set 2