>> Okay let's go ahead
and get started now.
So today we're going to talk
about how you can
determine the ways
in which molecules can vibrate
and so the first
thing we might start
out with is a little motivation.
So, why is it that we're
interested in the study
of molecular vibrations?
Does anybody want to
volunteer a motivation?
Yes?
>> Spectroscopy?
>> Spectroscopy.
Yes. So, as you've
learned probably mostly
in organic chemistry, molecule
vibrations often absorb
radiation in the infrared region
of electromagnetic spectrum
and if you shine infrared
light on a molecule,
you'll then excite transitions
between vibrational levels,
quantum mechanical levels, and
this gives you an opportunity
to find out basically what
are the energies involved
and it turns out that
chemical bonds, you know,
have very distinct signatures
in the infrared spectrum.
So you can tell a carbon-carbon
single bond by where it appears
in the infrared spectrum or
carbon-carbon double bond
or different types of
carbon-oxygen bonds,
et cetera, et cetera, et cetera.
So, if you can measure
the vibrational spectrum
of the molecule and assign
it meaning label the peaks
according to what kind of
bonds give rise to those peaks
and also bending modes, those
various kinds of motions
that show up in the infrared
spectrum then this is one
of the many tools that
you have at your disposal
to determine the
structure of the molecule.
In addition, the frequencies
at which the vibrations show
up in the infrared
spectrum tell you something
about the strengths
of the bonds.
So, stronger bonds have higher
characteristic frequencies
and weaker bonds have lower
characteristic frequencies.
So vibrational spectroscopy is
also a way to gain deep insight
into chemical bonding
in general.
So it's both an analytical tool
and then also a very nice tool
for getting fundamental
information on chemical bonding.
Now if you want to use a
vibrational spectrum in order
to do some kind of structure
determination, then you need
to know what are the frequencies
associated with different types
of chemical bonds and there's
various ways to do that.
One is to just put in molecules
of known structure and then sort
of heuristically figure
out where they are.
Another is that you may
use a theoretical model
to help you understand
the spectrum
and so that's what we're
going to talk about today
and the theoretical model
that we will discuss and work
out in some detail
it'll take at least all
of this class is what's known
as normal mode analysis.
So in order to illustrate what
normal model analysis gives
for you I figured up a little
Spartan calculation here
where I sketched in
C02, carbon dioxide,
and I minimized the energy
with respect to the positions
of the nuclei and
then I asked Spartan
to do a normal mode analysis
for me or vibrational analysis
and what I get out is a list
of 4 different vibrations
with their corresponding
frequencies.
It's not really a frequency
it's called the wave number,
inverse centimeters, which is
a common spectroscopic unit
for vibrational spectra.
So, what I want to
start out with today is
to show you what these
things correspond to
and then we'll talk about
how you actually get this
information and it's going
to be a long exercise mostly
for your entertainment,
but it'll be comprehensive
in the sense that you'll
be using a lot of the tools
that we have learned this
quarter so it'll be kind
of a review where we use a lot
of the tricks that we've learned
over the quarter to solve a
long and complicated problem.
I won't ask you to solve such
problems but I hope that you'll
at least having seen this
once have an appreciation
for where these modes come from
that you find in your textbook
and that you can calculate
yourself using a program
like Spartan.
Okay so if I go over here
to this little table here,
which reports the
frequencies of the vibrations,
the first thing I notice
is that there are 4,
there are 4 vibrations
here and it looks like 2
of them have essentially
the same frequency.
If I click on the checkbox here,
I see that the one corresponding
to about 526 wave numbers
is actually a bending motion
of the CO2 molecule.
Okay. It turns out
that the bending motion
in CO2 is degenerate because
I can bend in the plane
of the screen or I can bend
out of the plane of the screen.
Those have essentially the same
frequencies and you can see
when I click on the other
one that one was in and out
of the plane of the screen.
So there's two bending motions
and then there's this
higher frequency motion.
Let's see what that looks like.
That's what we call
the symmetric stretch
where you can see the
oxygen atoms are moving away
from the carbon away
and toward but they're
in phase both of those.
Then the highest frequency
motion is what's called the
asymmetric or anti-symmetric
stretch where you see
as 1 oxygen moves out
the other one moves in.
So these are what we
call the normal modes
of vibration for CO2.
Now first a little
comment on the number.
So if I have N atoms in a
molecule and each atom can move
in X, Y and Z directions,
so 3 directions,
what that means is the
total number of degrees
of freedom are ways of moving
the atoms around relative
to one another and with respect
to our reference frame would
be 3 times the number of atoms.
Now, in vibrational
analysis, we're not interested
in overall translations and
rotations of the molecule.
So actually the number of
vibrations is less than 3N
by a number that
depends on whether
or not the molecule is linear.
In the case of linear
molecule like CO2,
there's 3 times the number of
atoms minus 5 unique vibrations.
So, for CO2 we have 3 atoms and
3 times 3 is 9 minus 5 is 4.
So we expect to have 4 normal
modes and, in fact, we do.
In the case of non-linear
molecules,
the number is 3N minus 6.
So a molecule like
water has 3 atoms,
3 times 3 is 9, 9 minus 6 is 3.
So water has a bend like CO2,
it's got a symmetric stretch
where the hydrogens
move in and out together
and it's got an anti-symmetric
stretch.
Those are the characteristic
modes
and you've probably seen these
in your books when you talked
about molecular vibrations.
So today what we're going
to do is we're going to see
where that comes from.
I don't know if you ever
wondered where those things come
from but today we'll
see where they come
from using a harmonic model
meaning we're going to assume
that the vibrations and the
molecule can be described
as simple springs.
Now in order to formulate the
problem I have to give you a lot
of background information
all right so I'm going
to spend a little bit of time
over at the boards sketching
for you what we need
to do if we want
to determine these [inaudible].
They look really simple but
as you'll see where they come
from is not necessarily
so simple.
All right so what we're going
to do is we're actually going
to do the case of
CO2 but we're going
to simplify it a little bit
because we want to keep it
as simple as possible so
we can see how it works
without getting carried away.
So what I'm going to do is
I'm going to suppose that all
of the atoms in CO2 are on
the X axis and I'm going
to have 1 oxygen here, a
carbon here and an oxygen here.
So this is 0C0 and I'm going to
label this atom A, this one B
and this one C. Now, one way
that I can describe the
vibration so the changes
of the bond length say
so maybe I didn't say it.
We're going to do
this 1 dimensional CO2
where the atoms are only allowed
to move along the X axis.
So we're not going to be
able to describe the bends,
but we will be able
to get the stretches.
We're going to start out
by describing the stretches
in terms of the lengths
of the bonds.
So we'll call this RAB
and this will be RBC.
Okay. Now initially we're
going to assume that the energy
that describes how, the
energy as a function
of bond length we're
going to describe it
by what's called
the Morse potential.
We've seen that earlier
in the quarter.
So we're going to have a
Morse potential for this bond
and a Morse potential
for this bond.
So, what that means then
is we're going to have
for the first bond potential
energy vAB, which is going
to be a function of this RAB
and that looks like a number D,
which is the dissociation
energy times E
to the minus 2 A times RAB
minus B. Now this B here is the
preferred value, the equilibrium
bond length if you like.
Then we have minus 2 E to
the minus A, RAB minus B.
As we saw earlier in the
quarter what this looks
like something like this.
So it's a typical
bond stretching energy
and this here is the
dissociation energy the position
of the minimum is B MMA
determines the curvature
of this thing.
So we're going to
have a potential
like this describing the
deformation of this bond
and then we're going
to have another one
that describes this bond, 2
more, and then we're going
to say that the total
potential energy which is going
to be a function of both of the
bond lengths is equal to vAB
as a function of RAB plus
VBC as a function of RBC.
So that's going to describe
the total potential energy
associated with stretching
or compressing those bonds.
All right so what I want to do
now is we're going to go ahead
and type these potentials in
and then we'll have a look
at the total potential energy
and then what we're going
to do is we're going to
generate harmonic approximation
so you've done this in
one of your homeworks
for 1 Morse potential.
We'll do it for the
2-dimensional
or the 3-dimensional
potential energy.
All right.
Then we're going to change
the coordinate system
but let's get started by having
a look at the potential energy
and its harmonic approximation.
Okay. So we're going
to get rid of Spartan
and bring up Mathematica.
[ Pause ]
Okay. All right.
So first is vAB so I'm going to
call that little v capital AB
and it's going to be a function
of little i capital AB
underscore colon equals.
Then that's going to be D times
parentheses EXP bracket minus 2
times A times RAB minus B. Okay
and so then we have minus
2 times EXP minus A times
parentheses R AB minus
B and then parentheses.
Okay. Now we can copy
this in and put it back
and then just change
all the ABs to BCs.
There's 1, there's another,
another and another.
All right.
Now I'm going to define
the total potential energy
and I'll just call that
v bracket and it's going
to be a function of RAB and
underscore and RBC underscore,
underscore, colon
equals so that'll be vAB
of R AB plus vBC of RBC.
Okay. Now let's do
a couple of things.
So the first thing is let's
just plot one of these guys
so let's do plot of
the first one vAB
so just remind ourselves what
the Morse potential looks like.
So this will be vAB of RAB and
we use replacement rules to put
in some values so for CO2 D here
is going to be 7.65 and this is
in units of electron
volts not curly.
A is going to be 2.5 and B CO
bond length is 1.162 angstroms
and now we'll plot
that for RAB going
between 0.5 angstrom
and 3 angstrom.
All right so let's have a look
and we don't get anything.
All right.
[ Inaudible response] Oh,
I haven't entered that?
Okay, there we go.
I thought I had them all
in 1 cell but I guess not.
There you go.
Thank you very much.
Okay, so as advertised,
you know,
this is the Morse potential
you've seen it before
and so this would be
the energy associated
with deforming the AB bond away
from its equilibrium bond length
where the energy is minimum.
So you see as you compress
it the energy goes up
and as you stretch it, the
energy goes up and then
at some point the
bond would dissociate.
Okay, now let's have a look
at the 3-dimensional potential
which is a functionale of
both of the bond coordinates
so this will be the total
potential energy of vibration
for this 1-dimensional
CO2 molecule.
So now what we'll do is we'll
say plot 3D V total of RAB
and RBC and we can mouse
in the same parameters
and put them here.
Okay. We'll go ahead and plot
this for RAB goes from 0.5
to 2 angstrom and
the same for RBC.
Okay. I'm also going to put
in a plot range for the energy
so it'll be arrow
minus 16 to 10.
All right.
Let's have a look.
Whoops.
[ Inaudible response ]
I don't have?
Oh, arrow here, yeah.
Thank you.
Okay. So let's have a look.
So here now is a plot of the
potential energy as a function
of 2 of these coordinates.
So one of the bond
lengths is along this axis
and the other one
is along this axis.
There's a minimum where
both RAB and RBC are
at their equilibrium values
and then the energy increases
as you move away from
the equilibrium value
and it got cut off here
because I set the plot range
to go only up to 10.
Okay. All right.
So this now would be the
total potential energy
if we described both bonds
with the Morse potential.
Now, to do a normal
mode analysis,
the first approximation
you bring
in when you do normal mode
analysis is that you assume
that near the minimum, the
potential energy minimum,
that the system behaves
as a harmonic system.
So now what we're going
to do is we're going
to use the series command
to generate an approximation
to this 3-dimensional
Morse potential it's going
to be a tailor expansion around
the minimum up to second order.
You did that already in your
homework for 1-dimensional Morse
and now we're going to do it
for the 2-dimensional Morse
and that will be
a starting point
for doing the normal
load approximation,
normal load analysis.
Okay, so to do that I'm going to
go ahead and since we're going
to plot it in just a
minute I'll go ahead
and start a new line here.
I'm going to call this
approximate one v tailor.
So it's going to be a
tailor series expansion.
That's going to be normal
of series of VRAB RBC
and we have 2 variables
that we're expanding around.
So we'll say that RAB is being
expanded around the minimum, B,
equilibrium bond length,
and we're keeping 2 terms
harmonic approximation
and then the same thing for RBC.
So RBCB 2.
All right.
So let's go ahead
and enter that.
We need 1 more bracket here.
Okay. So notice now
we have this equation.
So there's this constant term
that involves the
dissociation entry
and then notice there's 2 terms
that are quadratic
in the bond lengths.
So that's the form you get
when you make the
harmonic approximation.
Okay so let's go ahead now and
plot that approximation along
with the actual function so
we can see what we're doing
when we make the
harmonic approximation.
So to do that I'm just going
to put a curly bracket
here and another one here.
Wait, let's not put
that one there.
Let's put I here.
So now we'll add to
the list here v tailor.
All right let's see
if that works.
Okay. So it looks kind of
funky, but it's good enough
to make the point
that I wanted to make.
So, just to remind you
here's our original Morse
and you can sort of see
it from the side it looks
like the 1-dimensional
one from this angle
and then this bowl here that's
basically a parabola that's been
spun around an axis drawn
through the minimum this is
the harmonic approximation.
So whenever you talk about
the harmonic approximation
to potential energy what you
will generally observe is
that close to the
minimum the harmonic,
very close to the minimum,
the harmonic approximation
is a very good representation
of the actual potential and
the assumption that we make
when we invoke the
harmonic approximation is,
and it's a reasonable
one in this case,
is that under normal
circumstances the amount
of deviation of the
bond lengths away
from the minimum energy
configuration is pretty small.
So, we really are only hanging
out down near the bottom.
Now why do we make the
harmonic approximation?
Well, the reason we do that is
because we can't
solve the problem
of determining the
vibrations of Morse potentials
without really using some
sophisticated analysis,
but if we can represent
the Morse potential
by this harmonic approximation,
the problem still will look
pretty hard but it's much, much,
much easier and then we
get out the normal modes.
So now I'm going to take
a little pause and go back
to the board because it also
turns out to be the case
that these variables that we're
using here the relative, well,
the separations of the atoms,
this is not the most
convenient set of coordinates
for doing our normal
mode analysis.
So now I'm going to come up with
a version of this formula here
for the potential energy
that's going to be a function
of the Cartesian coordinates of
the atoms, the X coordinates.
So we're going to move
from 2 variables describing the
separations between the atoms
to 3 variables defining the
positions of each of the atoms.
So I have to show you now
how we're going to do that.
[ Pause ]
Okay so first thing I'll
do is I'll just write
down what we just did.
So what we just did was
generated our harmonic
approximation to
the Morse in terms
of the distances
between the atoms.
So essentially what we said
was we have now an approximate
potential that looks like this.
[ Writing on board ]
Okay. Now, next thing is I want
to notice that this variable,
RAB, for a 1-dimensional CO2
this is just the difference
between the X coordinate of
atom B and the X coordinate
of atom A. Now I'm going to
define some displacement away
from equilibrium position
in Cartesian coordinates.
We have a similar one for RBC.
So I'm going to say Delta X sub
A is equal to the actual value
of X sub A minus its
preferred equilibrium position.
So we can call that X A 0.
I can similarly define Delta
X sub B. Now what I want
to do is rewrite this guy
essentially but bringing
in the Xs and getting
rid of the Rs.
So now the next thing
I'm going to notice is
that B equilibrium bond
length is the difference
between the preferred
positions of atoms B and A.
So that's XB0 minus XA0.
Each of these 2 guys has
their equilibrium position.
The difference between those
equilibrium positions is the
bond length.
All right.
So now I can write XA is
equal to Delta XA plus XA0
so that's just rearranging this.
XB is equal to Delta
XB plus XB0.
All right.
So now if I put these guys
into here I see that RAB,
the thing I want to eliminate,
can be written in terms
of the Deltas as follows.
That'll be Delta XB minus
Delta XA plus XA0, sorry, XB0,
minus XA0, but this is just B.
So now this thing RAB minus B
is just Delta XB minus Delta XA.
Okay.
[ Writing on board ]
And I can do exactly the same
thing for B and C and find
that RBC minus B is
Delta XC minus Delta XB.
All right.
So what that means now is that
I can rewrite this potential
but now as a function of these
displacements of the atoms
from the equilibrium
positions as follows.
B, and now there's 3 of them,
Delta XA, Delta XB and Delta XC.
That's going to be equal
to minus 2 times D plus A
squared D times Delta XB minus
Delta XA squared and then plus
A squared D times Delta XC minus
Delta XB squared.
Okay so that was a little bit
of effort to go from 2 variables
to a more convenient
set as we'll see 3.
Now, to make this look like
the problem that you solve
in physics class when you
talk about harmonic motion,
you presumably did that, what
I'm going to do is I'm going
to say that this is equal to V0
so some energy offset plus
1/2 times a force constant, K,
so that's the K that
appears in Hooke's Law,
times Delta XB minus
Delta XA squared,
and then plus a half
times the same K,
Delta XC minus Delta
XB quantity squared
and then just comparing
these 2 I see that that means
that V0 is equal to minus 2D
and K is equal to 2A squared D
in terms of our Morse
parameters.
Okay, now the next thing
we want to do is we want
to actually come up with
essentially Newton's equations
of motion for these 3
dynamical variables.
So we're just going to do a
classical mechanics treatment
and find out what the
equations of motion are
for these 3 variables
in this potential.
All right.
And the easiest way
to do that is a way
that you probably didn't
learn in your physics class
and I'm just going
to briefly summarize
for you what we're going to do
and if you ever have
the pleasure
of taking an advanced
mechanic's class,
then you'll have the
pleasure of learning what I'm
about to tell you
but if it looks
like something you don't want
to be bothered with,
you can ignore it.
All right so now we
have potential energy
and once we have potential
energy and we know how to write
down kinetic energy, then
there's a nice fancy way
that we can come up with
the equations of motion.
All right.
This fancy way is
what's referred
to as the Lagrangian Formalism
of Classical Mechanics.
Has anybody heard of that?
Probably not.
Have you asked?
Okay. Well, then I'll remind you
or you can help me if you want.
Okay, so according to the
Lagrangian Formulation
of Classical Mechanics
it's a sophisticated
and more extended version of
Newton and what you get out,
what we're going to get out is
essentially Newton's equations
for these 3 variables, but one
of the nice things
is Mathematica knows
about the Lagrangian,
essentially the Lagrangian
Formalism and so we'll be able
to crank out the equations
of motion using Mathematica.
Okay, but the basic idea here
and this is the part you
can ignore if you don't care
about it, is there's a
function that we can write
down called the Lagrangian
and it's a function of the set
of all the positions,
which I'm just going
to use X. This is general,
this is way more general
than the problem
we're doing here.
So this is just a
set of positions
and then it's also a function
of the time derivatives
of the positions, which are
related to the velocities
and just to, this dot is
just a nice shorthand way
of saying DXI IDT.
So it's a function of positions
and velocities essentially.
The function is defined
as the difference
between the kinetic
energy, which is a function
of the velocities and
the potential energy,
which we're going to assume
here is just a function
of the positions.
So this is a function defined
as kinetic energy
minus potential energy.
Here I would range from 1 to the
number of objects in our system.
So for us it's going to be 3.
Okay. Now, skipping lots
and lots and lots and lots
of details the beauty
of the Lagrangian Formulism
is once you write this guy
down there's an equation that
tells you what the equations
of motion are and you get
those equations of motion
by plugging your Lagrangian
into the so-called
Euler-Lagrange equations.
What those look like in
this notation is derivative
with respect to T of the
derivative of the Lagrangian
with respect to the positions,
with respect to velocity, sorry,
and then subtract off from that
the derivative of the Lagrangian
with respect to the positions
and set that equal to 0.
Now, looks kind of scary,
but actually it's not.
It's a very nice way of
deriving the equations of motion
and like I say if you take a
more advanced mechanics class
that's something
you would learn.
So the idea here is
then what we want to do
for our system here we have
the potential energy sitting
right here.
The kinetic energy is just going
to be one-half M times the
time derivative of the Deltas
and we know how to
write that down.
So we're going to
construct a Lagrangian
for our 3-atom vibration system
problem and then we're going
to use a function of
Mathematica which will calculate
or determine the
Euler-Lagrange equations for us.
Then we're going to rearrange
those equations of motion
and we're going to see
that we can write them
down as a matrix equation
that is an eigenvalue problem.
Then we're going to see
that the eigenvalues
of that eigenvalue
problem are related
to the vibrational frequencies
and the eigenvectors will tell
us how the atoms actually move.
In other words, they're going
to give us the normal modes
like what we just animated
for the CO2 molecule.
All right so I hope I haven't
lost you yet, but okay.
So, what I'm going to do right
now is we'll go ahead and get
to this point here and then I'll
need to come back to the board
and give a little more
information before we turn it
into an eigenvalue problem.
Okay. Any questions?
Does that look familiar?
Okay, good.
All right.
So, the first thing we're going
to do is type in the version
of this guy here that
was derived over there
in Cartesian coordinates.
So I'm going to call
that V harm cart.
So meaning the harmonic
potential energy
and Cartesian coordinates.
Then I'm going to say that's
equal to V0 plus K divided by 2,
force constant divided by 2,
times and then I'm not
going to put the Deltas in.
I'm just going to call Delta X,
X. So the first term is X
capital B of T minus X capital A
of T and then the
whole thing squared.
Then we'll have the second term
K over 2 times and now we put
in X capital C of
T minus X capital B
of T. Not a curly, squared.
All right so that's
our potential energy.
Now the kinetic energy
I'll call kinetic E equals
so this is just the sum of
1/2 MV squared so for A,
B and C. So we're going to
have 1/2 times and I'm going
to call the mass of
atom A little m sub a
and then I'm going to multiply
that by XA prime of T squared
because according
to our formulism
over there we write the velocity
as the time derivative
of the position.
Then we want to just put in
analogous terms for B and C. So,
plus, 1/2 times MB times
parentheses XB prime
of T squared.
Then finally 1/2 times MC
times XC prime T squared.
So now we have our kinetic
energy and our potential energy
and now I can form
the Lagrangian.
The Lagrangian equals
kinetic, E, minus V harm cart.
Okay. So let's go
ahead and enter that.
Now, as I told you, the
Mathematica knows how
to generate Euler-Lagrange
equations
if we give it a Lagrangian, but
the command that you need to do
that is in a package
that's not normally loaded.
So we're going to
load the package
that contains the
appropriate command
and it's called variational
methods.
So we say less than,
less than, variational
and then capital M methods.
Whoops, have to spell it right.
Then left single quote
and we can enter.
All right.
The command that
we're going to use
that generates the
Lagrange equations is called
Euler equations.
So Euler equations, whoops,
have to spell it right,
and then we put in the
Lagrangian and then we put
in the variables that
we want the equations
for so that's going to be X,
we need a curly here, XA of T,
XB of T, XC of T and then
we put in the time variable
which is going to be T and
then a bracket and then I know
from experience the
equations will look better
if I put a post fix simplify.
Another thing I'm going to do
is I'm going to stash these
in a variable, which
I'm going to call EOM
for equations of motion.
All right.
So let's go ahead
and let that rip.
So notice what we
get; three equations.
This one has only XA and notice
it's got acceleration and, no,
actually sorry it's got
the B in it also, right,
because A and B are bonded.
In any case, these are
equations of motion,
but I want to rearrange
them a little bit
to make them look more
amenable to writing as a matrix.
So, I'm going to put
these in the form XA
of T equals something,
something;
XB of T equals something,
something;
and XC of T equals
something, something.
The way we can do
that as follows.
I can say solve EOM and so
if I want XA that's going
to be bracket, bracket
1, bracket, bracket;
the first element in that list.
Solve it for XA double
prime of T. All right.
And if we do that, we
get something that looks
like acceleration of XA, the
second derivative with respect
to time, is equal to force
on XA divided by mass
of A. Then we can do the
same thing for B and C.
So we can just mouse this in
and put in 2 and solve for B
and then we can do
the same thing for C.
[ Pause ]
All right.
Now, why the heck did I do that?
Well, the reason I did
that I'm going to go
over to the board and show you.
All right.
Okay, just to make it clear does
everybody understand what we
have here?
These are basically
Newton's equations now
for the displacement of
the X coordinate of A,
B and C. That's what
we've generated there.
Euler equations took
our Lagrangian
and did all this for us.
All right.
Now, let's have a look
at what we've got here.
[ Pause ]
So if I write those
3 guys as a vector.
[ Pause ]
Let's have a look
at the first one.
So the first one minus
K over MA times XA
of T. Then we have plus K
over MA times XB of T and then
to sort of hint of where we're
going here we have plus 0 times
XC of T. And then for B, so
these are actually Deltas,
actually all of these are
Deltas in our original notation.
So the next one is K over MB
times Delta XA of T minus 2K
over MB times Delta XB
of T and then plus K
over MB times Delta
XC of T. Then
for XC double prime we have
0 times Delta XA of T plus K
over MC times Delta XB
of T and then minus K
over MC times Delta XC T. Now,
those of you who have worked
with matrices maybe
can recognize
that we can write this
array of equations here
as a matrix times a vector, a
3X3 matrix times a 3X1 matrix.
So, if I put here Delta XA
of T and then Delta XB of T
and Delta XC of T then I can get
these 3 equations by multiplying
by a matrix that has
the following terms.
They're just the
coefficients of those
in each of these equations.
So I have minus K over MA plus
K over MA, 0, and then I have K
over MB minus 2K over
MB and plus K over MB.
Then I have 0 K over MC
and then minus K over MC.
You see that?
So this is a concise way of
writing our equations of motion.
Now, it's not quite an
eigenvalue problem yet
and to turn it into an
eigenvalue problem now we invoke
one of the key assumptions
of the
so called normal mode analysis.
To kind of explain
that assumption,
I just want to remind
you for the case
of a single oscillating
coordinate X,
the harmonic isolater what
we normally do is we write it
as follows in physics.
We have our equation of
motion, which is MX double prime
of T is equal to minus KX of
T. That's just Hooke's law
for the force and this is
mass times acceleration.
I guess I could put
Deltas in here
to make it look more
like what we have.
And we can write this, this is
what's often been when you talk
about the harmonic oscillator
is Delta X of T is equal
to minus omega squared X of T
where omega now is the
characteristic frequency
of the oscillator and it's
equal to we can call it omega 0.
It's equal to the square root
of the force constant
divided by the mass.
We're interested in frequencies
when we do normal
mode analysis, right?
We want to know for
a given configuration
of a molecule that's
at an energy minimum
what are the frequencies
and modes of vibration.
So in order to do that, what
we're going to do is we're going
to suppose that for
each vibrational mode,
which will correspond to, you
know, particular set of dynamics
for the 3 atoms, there's going
to be all of the atoms are going
to be oscillating at
the same frequency
and so what we're
going to say now is
that we can write this each of
these equations in this form,
which means we're
going to say it's equal
to minus omega squared times
XA of T Delta, Delta XB of T
and Delta XC of T.
So for one mode
of the vibration we would
have 1 frequency omega
and what's hopefully apparent
here is I've just written this
down as an eigenvalue problem.
Matrix times vector
equals number times vector.
As we did before, we can
take the, so there will be 3
such equations with
3 frequencies
and we can collect them all
into 1 equation which I'm going
to write as the following.
Matrix, that's this
matrix here, M. I'm going
to collect the 3 modes
of vibration in vectors
in an eigenvector matrix
call that X. Then we know
that will be X times lambda
where now this is a diagonal
matrix whose elements are minus
omega squared.
So we've transformed this
problem of determining the modes
and frequencies of
the vibrations
in this 1-dimensional triatomic
CO2 molecule into the problem
of finding the eigenvalues
and eigenvectors of
this matrix here.
That we know how to do; we
just have to type this in.
Now, what are we
going to get out?
We're going to get out
the eigenvectors are going
to give us amplitudes
of oscillatory motion
and the eigenvalues will be
related to the frequencies.
So let's go ahead and get those
and then the last thing we're
going to do is I'm going
to show you how to
animate the eigenvectors.
So we'll recover
the usual symmetric
and anti-symmetric stretch,
and then we'll check
to see how good the frequencies
are for a particular set
of parameters, A, B and D.
We'll actually see by comparing
to experimental infrared
spectrum of CO2
that this actually
works very well.
Okay, so that's what we're
going to do and that's the end
of the mathematics part.
All right.
So let's go ahead and
type in our matrix.
I'll call it matrix equals
it's going to be 3X3.
This first element
minus K over MA
and then we have
K over MA and 0.
Second row K over
MB minus 2 times K
over MB and then K over MB.
Then we have for the third row
0 K over MC and minus K over MC.
Now let's have a look at that in
matrix form to see if it looks
like what's on the board.
Does that look right?
It looks good.
Okay. Now, let's
plug in some numbers.
So for the masses I'm going
to use atomic mass units.
So MA and oxygen is equal to 16.
B is a carbon so that's 12.
Then C is an oxygen
so that's 16.
Then I'm going to put
in A equals 2.5 is our
curvature parameter; B,
the equilibrium bond length
is 1.162 angstrom units;
D the dissociation
energy is 7.65
and this is in electron volts.
We need to remind ourselves
what is K. We determined earlier
that K is equal to 2 times A
squared times D. All right.
Now we're ready to get the
frequencies and eigenvectors.
So the frequencies equal the
square root of minus eigenvalues
of the matrix because
the eigenvalues are minus
omega squared.
If we want omega, we take square
root of minus eigenvalues.
Okay. Now, I know from having
done this example before
that we will benefit by putting
in the chop here to get rid
of the tiny values that
don't mean anything.
Chop here and here.
Okay, oh, I want to
take off the semicolon.
Okay, now, we get 3 numbers.
So there's 3 modes of motion.
One has a frequency of 4.68,
another has 2.44
and the other has 0.
We'll talk about
the 0 in a minute.
These are in funny units because
energy is in electron volts,
distance in angstrom units and
we're going to convert those
to wave numbers so we can
compare experimental data.
That's probably going to wait
until Tuesday, but in any case,
don't be worried about the
numbers because they're
in fun units but we do
see one higher frequency,
one lower frequency and one 0.
Now let's get the amplitudes of
the motion, the eigenvectors,
and we'll call that modes
equals and we're going
to chop the eigenvectors,
of matrix.
All right.
And if you enter that,
you get a list of 3 lists.
This is the eigenvector
telling us how the atoms move
for the vibration that
has this frequency.
This is the eigenvector
that tells us how the atoms
move at this frequency.
This is the eigenvector
for the 0 frequency mode.
So we can already see what
these vibrations are just
by looking at these numbers.
This one tells us
that for this mode
that has 0 frequency
all the atoms are moving
in the same direction.
They have exactly
the same amplitude.
So that's just the
relation of the molecule
as a whole; it's
not a vibration.
You see that?
So that's not interesting so
we don't care about that one.
Look at this one.
This one says that the
carbon atom doesn't move
and the 2 oxygens are moving in
opposite directions with respect
to the carbon so that's
our symmetric stretch.
Then this one here we have
carbons moving in one direction
and this oxygen is
moving to the left
and the other oxygen
is moving to the left;
that's our anti-symmetric
stretch.
All right?
So we can already see that
without doing any animation,
but let's go ahead and do
just for fun an animation
of the 2 modes and
just bear with me.
I'm going to type in the
long commands and then
if you want you can study
what they're actually doing.
Okay, now to actually
animate the modes we have
to do a couple of things.
What we're basically going
to do is we're going to step
over cycles, we're going to
step over a couple of cycles
of oscillation and
we're going to use
as the oscillatory component.
These are just amplitudes.
This is not the oscillatory
motion;
it's just the coefficient
that goes in front
of this oscillatory motion.
We're going to use a sign motion
for the oscillatory motion
to animate the modes and I'm
going to use a little time step,
which I'm going to call
DT and I'll set it to 0.1
and then I'm going to use
this variable here I mode
to indicate the number of
the mode that I'm animating
so this first one will be the
first mode and then I'm going
to say W is going to be a
frequency so that's going
to be frequencies of I mode
so I'm just going to pick
out from the list the frequency
corresponding to I mode.
I'm going to use a
variable called this mode
to equal the eigenvector.
So modes of I mode and then I'm
going to use the variable dim
to tell me how many elements
there are in the eigenvector
so that's length of this mode.
Then this number here
is just a scaling factor
to make my picture look nice
and I'm going to call it 0.25.
Now I'm going to
generate a bunch of images
that show the atoms moving
around and the atoms are going
to be represented
as black disks.
So I'm going to say
points equals graphics
so it'll allow me to drive
disks, and then I'm going
to specify them in a table.
So I have disk of I
plus scale times time
so here's our oscillatory
motion, W times T,
and then times the amplitude
which is this mode bracket,
bracket I and then
put in a 0, curly,
and then I think that's
the size of the disk.
All right.
Then I'm generating the table
for I goes from 1 to dim,
which is 3, the length
of our eigenvector.
So it's going to give me a bunch
of pictures where there's going
to be 3 disks drawn,
but they're going to be
in different positions depending
on where we are along
the oscillation
and I'm missing something here.
[ Pause ]
Curly bracket here.
Okay. All right now I'm
going to say plot range arrow
and these are some values that
I determine by trial and error
to make a nice looking plot.
Minus dim over 4 comma
dim plus 0.75 curly comma
and then minus dim over 4 to dim
over 4 curly square
bracket semicolon.
You're going to be underwhelmed
by how beautiful this,
how not so beautiful this
picture is going to be
but it's fun to be
able to do it.
Next thing we'll do is we'll
say graphs equals table points
and then we're going to have
time variable, T, going from 0
to 2 times pi divided by
W. Two times pi divided
by W is the period
of the oscillation
that has frequency W. We're
stepping through an interval DT.
Okay. So we can put
a semicolon there.
We need instead of a
comma here there we go.
Okay, then finally list
animate, graphs, 8,
animation running arrow false.
Okay. Let's see if it works.
All right.
We've got something.
Notice we have a little box.
It's got our C02 molecule in it
and now what we're going to do
by pressing play is
we're actually going
to see what's the
mode of vibration
of the first normal
mode that we calculated.
Sure enough as we surmised
from the eigenvector,
that's the anti-symmetric strip.
Pretty cool, huh?
You did it all by
yourself and now you know
where that comes from.
Okay, let's go ahead and
do the second one now
that we've done all
of this hard work.
We just mouse and need to
change a couple of things
and what we need to change is
only I mode because we set it
up to be nice and general.
We enter and get the same
picture except now we see
that the second mode is
the symmetric stretch.
Pretty cool, huh?
Any questions?
Okay so let's just
reiterate what we did today.
So first of all I just want
to tell you that the objective
of this whole exercise, well,
there are a few objectives.
So one was to see how we
could put together a lot
of the various things we've
learned through the quarter
to work on a pretty complicated
problem and we needed one
or two little extra things
like the Euler equations.
So that's one objective.
The other objective is that I
wanted to allow you to explore
where those normal modes come
from that you've probably seen
on at least a few
occasions in your textbooks.
Then also to have you appreciate
a little more what's involved
if you actually want to use
theoretical calculations
of vibrational modes to assign
actual vibrational spectra.
The normal mode analysis or
normal mode method is a root
to being able to do that.
Now, to convert the units
is actually a little bit
non-trivial so we're
going to do that next time
because what I want to do,
what I want to show you is
that these numbers we
calculated, let me go back
to the big screen here, so we
have these frequencies, right?
I told you they're in some funny
units because we used a mixture
of units in our calculations.
So what we're going to do next
time is we're actually going
to convert the first two
numbers so this is the frequency
of the anti-symmetric stretch
and the symmetric stretch,
we're going to convert
those to wave numbers
and then we'll go look
up experimental data
for the anti-symmetric stretch
and symmetric stretch
in CO2 and compare.
What we'll see is actually that
the agreement is quite good
and what that tells us is that
the Morse model that we started
with and the harmonic
approximation
that we made along the way
were actually pretty reasonably
accurate for being able
to identify these
normal modes in CO2.
All right so that's something
to look forward to on Monday
and I guess that will
conclude our class today.
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