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1. ATOMIC PHYSICS HOMEWORK HELP
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2. Problem
This week’s lectures contained an introduction into continuum mechanical concepts. In addition to more traditional engineering
applications, continuum theory is also extremely valuable in analyzing the mechanics of small-scale materials such as individual molecules,
thin films. It is also an essential part of fracture many fracture theories.
As pointed out in the lecture notes, the significance of elasticity problems goes far beyond simply studying reversible deformation. For
example, a beam bending problem similar as the one studied in the lecture can be used to carry out coupling from atomistic to mesoscopic
scales within a hierarchical multi-scale modeling framework.
This first assignment is focused around basic continuum mechanics and introductory material on molecular dynamics simulation.
PARTA
In the first and second lecture, we have discussed a beam bending problem with the following geometry:
Length
rectangular
cross-section
A=b x h
q =gA
In class, we have obtained the section force and moment distribution (Fz and My), as well as the distribution of curvatures. The purpose of this
first problem set is to determine the distribution of displacements along the x-direction using the differential beam equilibrium equations (see
Section 2.5.5 in the lecture notes).
1) Write out the equilibrium equations for this case.
2) Write out all boundary conditions (displacements, moments, forces, rotations etc.).
3) Solve the differential equilibrium equations so that you obtain the displacement distribution uz and the rotation distribution y, all as a
function of x.
Introduction To Atomistic Modeling Techniques
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3. Compare the results for Fz and My with those obtained earlier.
PART B – Basic molecular dynamics
1. Many materials failure processes occur at extremely short time scales. Molecular modeling can provide important information about
how a crystal undergoes deformation, including all atomic details and their temporal resolution, often beyond the capabilities of
current experimental techniques.
Is Monte Carlo (MC) or Molecular Dynamics (MD) advantageous for such instability problems? Describe why; give keywords
only.
2. Write the differential equation you solve in molecular dynamics and explain how it relates to Newton’s laws explained in the first
lecture.
PART C – Interatomic potentials
1. A popular potential to describe the interaction between atoms is the Morse function (named after physicist Philip M. Morse), a pair
potential that describes the energy stored in the bond between pairs of atoms (here i and j ) as:
ij ij m
2
(r ) D
1 exp Br r
.
(1)
i
rij
j
The Morse potential has three parameters, rm , D and B .
What are the units of these three potential parameters (e.g. length, energy, …)?
Write the total energy of a system of N particles, assuming only pair wise interactions between atoms, without any cutoff
radius, as summations over particles and energy expressions, in terms of (rij ) .
2. By taking the first derivative of (rij ) (denoted by '(rij ) ) with respect to rij (proportional to the force between particles i and
j ) and setting it to zero, calculate the equilibrium position between pairs of atoms, denoted by r0 .
Discuss all possible solutions that yield '(rij ) 0, and which specific terms are
required to be zero for each case. Express the corresponding atomic separation as a function of the potential parameters, for
each solution.
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4. Calculate the limiting value of (rij ) for rij and for rij 0.
From these results, calculate the energy stored in each bond, as a function of potential parameters.
Discuss the dependence of these properties on the parameter B . What influence does the parameter B have?
Hint: Consider that the second derivative of (rij ) corresponds to the force constant k ''(rij r0 ) ; without considering the actual
derivatives you can see
that the second partial is proportional to B with some exponent. This force constant approximates the potential behavior in the vicinity of
the equilibrium position r0 , i.e. if the bonds are soft or stiff, within a harmonic approximation
ij 0
2
(r ) ~ k(r r ) .
3. Using the above results, sketch the potential shape, drawing (rij ) as a function of the distance between two particles rij , indicating the
value of specific potential parameters rm and D in the plot.
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5. PARTA
A1)
Load only in z direction, therefore only consider the equation for My. With
and
we arrive at the equilibrium equation:
A2) uz (x 0) 0
M x M z 0
M y (x L) 0
y (x 0) 0
y ( x 0) 0 Qz (
x L) 0 uz ( x
0) 0
A3) Solution after integration:
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Solutions
6. Results for Qz and My match those obtained in class.
PART B
B1) MD is better suited since it can describe nonequilbrium processes; and provides full information about the dynamics. MC is suited for
equilibrium processes.
d 2
r
rj j
B2) m j
U(r ) , for j 1..N ,
dt2
which is a system of coupled 2nd
order nonlinear differential equations that can be solved by discretizing the equations in time.
PART C – Interatomic potentials
C1) Units: [rm ] = Angstrom, [ B ] = 1/Angstrom, and [ D ] = Energy.
These units can easily be derived based on the following arguments: The potential itself expresses the energy as a function of distance;
the argument of the exponential needs to be unit less.
The total energy of the system is given by the sum over the energetic contributions over all pairs of atoms in the system.
This can be written using two summations, with a prefactor of 1/2 to account for the fact that bonds are double counted (e.g. one
accounts for bond 2-3 and the energy of this bond is counted again when the bond 3-2 is considered):
N N
1
U ij
| (r )
total 2
i1 j1 i j
This expression sums the energy contributions of the pair-wise interactions of all atoms, without considering the interaction of atoms
with itself (i j ).
C2) The force between pairs of atoms goes to zero either at the equilibrium distance between pairs of atoms - given by rij r0 rm(
exp(B(rij rm )) 1), or for rij
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7. ( exp(B(rij rm )) 0 ). If no potential cutoff is introduced, two atoms will always assume the equilibrium position r0 . Note that r0
only depends on rm , and on no other potential parameters.
ij ij ij ij m
2
For r , (r ) D , and for r 0, (r ) D(1 exp(Br )) (repulsion).
For rij r0 , (rij ) 0 . Thus, the potential energy stored in each bond is given by D
(also denoted as “Morse potential [energy]”).
The potential parameter B changes the shape of the potential (wider or more narrow) around the equilibrium separation.
Not required: The parameter B controls the width of the potential and is equal to the square root of half the force constant k
''(rij r0 ) divided by the Morse potential,
B k
2D
.
C3) Sketch of the Morse potential, indication the parameters rm and D :
D
Ener
gy
rm
Interatomic separation (r)
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