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