HE3 Soft Matter Potential Energy in Condensed Matter and the Response to Mechanical Stress 25 February, 2010 Lecture 3 See Jones’ Soft Condensed Matter, Chapt. 2; Ashcroft and Mermin, Ch. 20

Interaction Potentials: w = -Cr -n If n 3, molecules interact only with the nearer neighbours. Gravity: negligible at the molecular level. W(r) = -Cr -1 Coulombic: relevant for salts, ionic liquids and charged molecules. W(r) = -Cr -1 van der Waals’ Interaction: three types; usually quite weak; causes attraction between ANY two molecules. W(r) = -Cr -6 Covalent bonds: usually the strongest type of bond; directional forces - not described by a simple potential. Hydrogen bonding: stronger than van der Waals bonds; charge attracting resulting from unshielded proton in H. In the previous lecture:

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, polar- nonpolar, and dispersive (London) interactions. Summarised ways to measure polarisability  u u u  u  

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

Cohesive Energy Def’n.: Energy required to separate all molecules in the condensed phase or energy holding molecules in the condensed phase. We found that for a single molecule, and with n>3: with  = number of molecules per unit volume   -3, where  is the molecular diameter. So, with n = 6: For one mole, E substance = (1/2)N A E E substance = sum of heats of melting + vaporisation. Predictions agree well with experiment! 1/2 to avoid double counting!

Boiling Point At the boiling point, T B, for a liquid, the thermal energy of a monoatomic molecule, 3/2 kT B, will exactly equal the energy of attraction between molecules. Of course, the strongest attraction will be between the “nearest neighbours”, rather than pairs of molecules that are farther away. The interaction energy for van der Waals’ interactions is of the form, w(r) = -Cr -6. If molecules have a diameter of , then the shortest centre-to-centre distance will likewise be . Thus the boiling point is approximately:

Comparison of Theory and Experiment Evaluated at close contact where r = . Note that  o and C increase with . Non-polar

Additivity of Interactions Molecule Mol. Wt. u (D) T B (°C) Ethane: CH 3 CH Formaldehyde: HCHO Methanol: CH 3 OH C=O H H C-O-H H H H C-C H H H H H H Dispersive only Keesom + dispersive H-bonding + Keesom + dispersive

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/r 12 - C/r 6 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 London Constant for Ar is calculated to be C = 4.5 x Jm 6 We can guess that B = Jm 12 w min  -5 x J Compare to: (3/2)kT B = 2 x J Actual  ~ 0.3 nm (Our guess for B is too large!) (m) Ar boiling point = 87 K

Intermolecular Force for Ar Pair F = dw / dr (m) Very short-range force!

Comparison of Force and Potential Energy for Ar Pair F= dw / dr (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.

AFM tips from NT-MDT. See 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

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

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 A B C DE C 0

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.

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 and knowing the arm’s stiffness. 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 = r o.

Lennard-Jones Potential for Molecular Pairs in a Crystal r w(r) + - -- roro 

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

Variety of Atomic Spacings in Cubic Crystals Image from: 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 tot = 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 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: A 12 and A 6 (where the 12 and 6 represent the two terms of the L-J potential.) For FCC crystals, A 12 = and A 6 = There are different values for BCC, SC, etc.

Cohesive Energy of Atoms in a Molecular Crystal 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 (r o =1.12  ). From the first derivative, we can find the equilibrium spacing for an FCC crystal: For each atom in a molecular crystal, however, we write that the cohesive energy is:

Cohesive Energy of Atoms in a Molecular Crystal We can evaluate W when r = r o to find for an FCC crystal: - - 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, but the equivalent of 8.6 of them contribute to the energy!) W TOT FCC

F A L The Young’s modulus, Y, relates tensile stress and strain: Connection between the macroscopic and the atomic Y tt tt How does the interatomic force, F, relate to the macroscopic Y?

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 - r o ). At equilibrium, r = r o and F = 0. The tensile stress  t is defined as a force acting per unit area, so that : F roro The tensile strain  t is given as the change in length as a result of the stress: oo oo roro What is k? Y can thus be expressed in terms of atomic interactions:

r W tot  roro  F = 0 when r = r o Elastic Modulus of Molecular Crystals r F + - roro The Young’s modulus is sometimes known simply as the elastic modulus.

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: Re-arranging, we see that: 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 = r o. Combining the constants to create new constants, C 1 and C 2, and setting r = r o, we can write: Finally, we find the Young’s modulus to be: As r o 3 can be considered an atomic volume, we see that the modulus can be considered an energy density, directly related to the pair interaction energy.

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

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: A A y F  xx

Viscous Response of Newtonian Liquids A A y F xx There is a velocity gradient (v/y) normal to the area. The viscosity  relates the shear stress,  s, to the velocity gradient. The viscosity can thus be seen to relate the shear stress to the shear rate: 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: Units of s -1

Hookean Solids vs. Newtonian Liquids Hookean Solids: Newtonian Liquids: Many substances, i.e. “structured liquids”, display both types 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 themselves and flow to relieve the stress: The shear strain, and hence the shear modulus, both change over time:  s (t) and t Elastic response Viscous response (strain is roughly constant over time) (strain increases over time)

Response of Soft Matter to a Constant Shear Stress: Viscoelasticity t Slope: We see that 1/G o  ( 1 /  )    is the “relaxation time” Hence, viscosity can be approximated as   G o 

Example of Viscoelasticity Video: Viscoelastic gel

Relaxation Time in a Creep Experiment In a “creep” experiment, a constant stress is applied for a fixed period of time and then released. The strain is recorded while the stress is applied and after it is released. The instantaneous (elastic) response is modeled as a spring giving an elastic modulus, Y. The time-dependent (viscous flow) response is modeled as a “dashpot” giving a viscosity, . Stress,  Time Strain,  Time  Y 0 0

Modelling of Creep and Relaxation Maxwell model: Spring and dashpot in series Instantaneous elastic deformation is followed by flow. Elastic deformation is recovered but not the flow.  Voigt-Kelvin model: Spring and dashpot in parallel  Anelastic deformation is followed by recovery with an exponential dependence on  =  / Y, which is called the relaxation time. Deformation :

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

Total Creep and Recovery Time Displacement Strain is described by the sum of the Maxwell and Voigt-Kelvin models. Elastic Anelastic Flow Constant stress applied Spring and dashpot acting both in series and in parallel.

Stress Relaxation under Shear time  Constant shear strain applied  Stress relaxes over time as molecules re-arrange time Stress relaxation :  is the relaxation time

Viscosity of Soft Matter Often Depends on the Shear Rate Newtonian: (simple liquids like water) ss   Shear thinning or thickening :  ss 

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

Problem Set 2 1. Calculate the energy required to separate two atoms from their equilibrium spacing r o 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/r 6 + B/r 12, where A = Jm 6 and B = Jm 12. 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 10 9 Pa. Assume that this modulus is comparable to the instantaneous shear modulus of water G o and estimate the characteristic frequency of vibration for water,. Temp (  C)  (10 -4 Pa s) Temp (  C)  (10 -4 Pa s) In poly(styrene) the relaxation time for configurational rearrangements  follows a Vogel-Fulcher law given as  =  o exp(B/T-T o ), where B = 710  C and T o = 50  C. In an experiment with an effective timescale of  exp = 1000 s, the glass transition temperature T g of poly(styrene) is found to be  C. If you carry out a second experiment with  exp = 10 5 s, what value of T g would be obtained?

R Creation of a New Surface Leads to a “Thermodynamic” Adhesion Force F  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.

S L  LS A L S  SV A  LV A When liquid (L) and solid (S) are separated, two new interfaces with the vapour (V) are created. W =  LV A +  SV A -  LS A Work per unit area, W: W  LV +  SV -  SL  SV   LV  SL Work per unit area, W: Young-Dupré Equation

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

Force for Polymer Deformation