ATIF JAVED: Hey, 3016.
Welcome to my final video project for the class.
My presentation will be on Modeling and Energy
Analysis of Liquid Crystals using Mathematica.
So liquid crystals pervade many areas
of technology in our world today,
including the screen you may be watching this on.
They're used in LCD displays, thermometers, surfactants,
polymers, detergents, and even Kevlar.
So the versatility of this phase of matter
is what definitely piqued my interest
and made me especially enthusiastic about this project
as an opportunity also to explore
this subject as a scientist would,
imposing my own questions and then
finding ways to answer them.
So let's move on to the subject matter.
Liquid crystals are the state of matter
which retains properties from the conventional crystalline
solid and the isotropic liquid state.
So here you see an example of the crystalline solid.
LC molecules are anisotropic, meaning
that they exhibit properties with different effects
when oriented in varied directions.
And that orientation implies a certain free energy.
And that allows for a structure with interesting properties,
such as let's say birefringence.
So here we see the entire disorder
of an isotropic liquid.
And as we move right, we're removing degrees of freedom
to eventually form nematic LCs that have orientational order,
and eventually smectic LCs, which
have both orientational and translational layering
order, which, by the way, is growing as a tool in research,
especially in nanotechnology and high-performance materials,
which I thought was pretty cool.
Everything ultimately references back to structure.
And here we can see the increasing degrees of freedom
and the characteristic decreasing order parameter
value as we transition to a liquid.
So the transitional degree of freedom
but restricted rotational freedom
is what will guide our assumptions
in developing a strong model.
All three subdivisions of the liquid crystal phase
can exhibit phase transitions, because of temperature
fluctuations.
So this will motivate my analysis
of the energy and entropy, which are directly
based on the order of the system,
and whose order is different based on the temperature.
So I will be using calamitic, or rod-shaped, liquid crystals,
and finish my analysis with a statistical modeling
of the energy and entropy over 100,000 trials,
and examine the graphical representation.
So since the molecules are calamitic,
I can think of them basically as straight rods.
Thus, the orientation is only based on the beginning
and the end of the rod.
So because we'll be looking at trends,
we can simplify the problem by first approaching it
in two dimensions.
And after a ton of trial and error,
I was able to put together a graphic of these rods
to shape the order of our liquid crystal system,
and demonstrate the type of phase that we're in.
And it's defined by the average theta
angle from the order director.
And this is the random generator for our LC
system, where we can change the theta value for our director.
And from this, we can work on extrapolating energies in them.
So here, we show three primary energy scenarios
for non-polar crystals.
And what can be considered their relative energy
is given by this equation at the top.
So by setting these conditions, when molecules are parallel
that gives a relative interaction energy
of negative 1; and when perpendicular,
a net energy of 0l and when angled
45 degrees to the director, yields an energy
of about negative 0.5.
So we can keep the negative sign as we
take the dot product of the two directions
to represent the fact that a lower free energy corresponds
to greater stability in either translation or orientation.
But we'll keep things positive for our graphics in the future.
So we can consider the molecules to be on average equidistant
from one another.
And so we can grid our liquid crystals in 2D space, like so.
Then we can visualize how the nearest neighbors, particularly
the ones in directly adjacent cells,
affect the energy of the system.
The energy effects of molecules farther
than these adjacent ones can be considered
negligible in comparison to these closer ones.
So we further our model by taking
the summation of energies in the necessary locations,
ultimately assigning a total energy to each crystal that
encompasses the energy of that crystal resulting
from orientation.
Or other energy contributions could exist, let's say,
such as magnetic.
But we'll focus on the energy of order and the results
from bonding and lowering of entropy.
Now subsequently, we can create a collection
of data angles produced from 100,000
trials of our random liquid crystal generator,
and obtain their energies from the relationships described
earlier.
So from this, we can develop a histogram,
as shown here in the top right-- in the top left.
And we clearly see it is relatively uniform Gaussian
distribution, as expected.
So moving forward, we want to look
to the third law of thermodynamics
to relate what we are seeing to entropy
using a fundamental law.
And so we'll graph our energy against entropy
using the natural level omega multiplied by the Boltzmann
constant.
Omega is the number of microstates, i.e.,
the number of configurations, basically,
that are still macroscopically a liquid crystal.
So by taking the log of our first graph,
we can obtain the second one plotting energy versus entropy.
And we see the curve flattens by taking
the natural log of the states.
And so after normalizing, we get the same graph,
except slightly wider and flatter.
And we can see this contrast in the lowest picture.
But now for the most interesting part of the problem,
we know from the Gibbs free energy equation
and the first law of thermodynamics
that for an isolated system such as this,
that dU equals TdS, where dU is the internal energy.
And so rewriting this, we see the relationship
between temperature and changing energy and entropy.
And since we have developed this information
from our previous graph, all we have to do
is rotate the axes of S against E,
making the slope of the curve equal to our temperature
for whatever generated liquid crystal.
Although I was unable to create the manipulate command that
would demonstrate the slope of dE versus dS
as the temperature was decreased, we would expect
a narrower distribution of beta angles, in other words
an increase of the order of the system
by being more aligned with the director,
to compress the histogram and result in a lower slope for T.
And this is pretty cool.
In my mind, it's a pretty cool thermodynamic proof,
that the narrower distribution of rods
equates to a lower temperature, and conversely,
a wider distribution of rods around energies
represents greater randomness, and thus higher temperature.
And so with confidence, we can say
that the order and temperature are mutually dependent.
And so I think the beauty of this problem
really stems from the simplicity derived
from taking appropriate steps and realizing where
the need for comparison and trend study
supersedes an unnecessarily complex setup.
I'm certain that that kind of technique
I'll be applying this in the future in DMSC courses.
And hopefully, I can continue that in my track,
in studying nanomaterials as well.
So to me, that's pretty awesome.
And my thanks go out to all of you, 3016 Fall 2012,
for making the class like quite an experience, granted
in retrospect.
But nonetheless, a really cool endeavor.
And my greatest appreciations go out to Ray, Esther,
and of course Professor Carter for their tireless efforts.
So thank you, guys.