Lecture 6: Interactions between
macroscopic bodies II
What did we cover in the last
lecture?
1) Dispersion interactions between atoms/molecules and macroscopic
bodies
2) Dispersion interactions between two macroscopic bodies
In this lecture…
1) Interactions between macroscopic bodies
continued
2) The effects of geometry on dispersion
interactions
3) Determining the energy scale of dispersion
interactions (Hamaker constant)
4) Range of interactions between macroscopic
bodies
5) Magnitude of forces between macroscopic
bodies
Interactions between a sphere and a
solid
Problem 2: Derive an expression for the dispersion forces
which act between a sphere of radius R (for separations
D>>R) and a semi-infinite solid if the number density of
atoms in the slab and sphere are n1 and n2 respectively.
Summary of results
Dispersion Interaction
energy between an
atom/molecule and a semiinfinite solid
Dispersion Interaction
energy between a flat slab
of area, S, and a semiinfinite solid
Dispersion Interaction
energy between a sphere
of radius, R, and a semiinfinite solid
U ( D)  
nC
6D
3
n1n2CS
U ( D)  
2
12 D
n1n2 CR
U ( D)  
3D
All of these are longer range than dispersion
interaction between isolated molecules
2
The Hamaker Constant
The strength of the dispersion interaction between
materials is often quantified in terms of the Hamaker
constant, A
A12  n1n2 C
2
Measured in
Joules
n1 and n2 are the number densities of atoms/molecules in
the two interacting materials (m-3)
C is the ‘strength’ of the interaction between an
atom/molecule from one material and an atom/molecule
from the other (Jm6)
A12 is the Hamaker constant for materials 1 and 2
Dispersion interactions and
Hamaker constants
We can rewrite our expressions for the dispersion interaction
energy between two macroscopic bodies in terms of the
Hamaker constant
Dispersion Interaction
energy between a flat
slab of area, S, and a
semi-infinite solid
n1n2CS
A12 S
U ( D)  

2
2
12 D
12D
Dispersion Interaction
energy between a
sphere of radius, R,
and a semi-infinite
solid
n1n2 2CR
A12 R
U ( D)  

3D
3D
Dispersion Forces and Hamaker
constants
We can also write the forces in terms of the Hamaker
constant
Dispersion force
between a flat slab of
area, S, and a semiinfinite solid
Dispersion force between
a sphere of radius, R,
and a semi-infinite solid
dU
A12 S
F ( D)  

dD
6D 3
dU
A12 R
F ( D)  

dD
3D 2
So for a given geometry we can determine the
strength of the dispersion interaction A12between two
materials by measuring forces
Some other important geometries
The geometry of the macroscopic bodies is important in determining
the separation dependence of the dispersion interaction
U ( D)  
Parallel cylinders U ( D)  
A12 R1 R2
3D R1  R2
A12 L  R1R2 


3 
R  R2 
12 2 D 2  1
Sphere-sphere
1
2
A12
U ( D)  
R1 R2
6D
L
Perpendicular
cylinders
How big are Hamaker constant
values?
P190 Surface and Intermolecular forces by J. Israelachvili
Values of A typically lie in the range 10-21 to 10-19 Joules
Note: kTroom = 4.41x 10-21J
What is the range of dispersion
forces between macroscopic bodies?
We can calculate the range of the interaction by comparing
dispersion and thermal energies i.e. |U(Drange)| ~ kT
For a sphere of radius R and a solid surface we have
A12 R
U ( D) 
 kT
3Drange
Hence
A12 R
Drange 
3kT
For values of T=300K, R=10 nm and A12=10-19J
Drange ~ 10R = 100 nm
Note: Drange(atoms) ~ 0.3 nm
How big are the dispersion forces
between macroscopic bodies?
We can also calculate the magnitude of the forces
exerted on macroscopic bodies due dispersion forces
Sticking with the sphere and surface we have that
dU
A12 R
F ( D)  

2
dD
3D
If R=10 nm, A12=10-19J
When D=10nm,
When D=1nm,
F= 3 x 10-12 N ~ 3 pN
F= 3 x10 -10 N ~ 0.3 nN
But can we measure forces this small?
Force between two surfaces
In the last lecture we showed that the force between a
perfectly flat slab and a semi-infinite solid surface is
A12 S
F ( D)  
3
6D
If the area of the slab is S=10-4m2 and A12 = 10-19J. What
is the force between the slab and the surface at a
separation of 0.3 nm?
Ans: ~20,000 Newtons This is a very large force!
So why don’t all surfaces stick together irreversibly?
Summary of key concepts
1) Dispersion interactions between
macroscopic bodies occur over
significantly longer ranges than for
isolated atoms/molecules
2) We can derive expressions for the
dispersion interaction energy and forces
between macroscopic objects
3) The strength of the dispersion
interaction can be quantified in terms of
the Hamaker constant A
4) Dispersion forces on nanoscale
objects can be on the pN to nN scales