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PROFESSOR: OK, let's start.
So last time, having dealt with the ideal gas
as much as possible, we started with interacting systems.
Let's say interacting gas.
And the first topic that we did was
what I called an approach towards interaction
through the Cumulant expansion.
The idea is that we certainly solved the problem where
we had particles in a box and it was just a trivial system.
Basically, the particles were independent of each other
in the canonical ensemble.
And things become interesting if you put interactions
among all of the particles.
Very soon, we will have a specific form
of this interaction.
But for the beginning, let's maintain
it as general as possible.
And the idea was that we are trying
to calculate the partition function for a system that
has a given temperature.
The box has some volume V and there are N particles in it.
And to do so, we have to integrate over
the entirety of the phase space of N particles.
So we have to integrate over all of the momenta and coordinates
of e to the minus beta h.
But we said that we are going to first of all,
note that for identical particles,
we can't tell apart the phase space
if you were to make these in [? factorial ?] permutations.
And we also made it dimensionless
by dividing by h to the 3n.
And then we have e to the minus beta
h, which has a part that is e to the minus sum over i pi
squared over 2m, that depends on the momenta.
And a part that depends on the interaction U.
The integrals over the momenta, we can easily perform.
They give us the result of Gaussians
that we can express as lambda to the power of 3n
where we introduced this lambda to be h over root 2 pi m k t.
And actually, sort of keeping in mind
the result that we had for the case of the ideal gas,
let's rewrite this as V over lambda cubed to the power of N.
So I essentially divided and multiplied by V to the N
so that this part becomes the partition
function than I would have had if U was absent.
And so then the remainder is the integral over all of the q's,
but I divided by V. So basically,
with each integration, I'm uniformly
sampling all of the points inside the box with probability
density 1/V e to the minus beta U.
And we said that this quantity we
can regard as sampling e to the minus beta U
through 0 to all other probability
distribution in which the particles ideal-like are
uniformly distributed.
And we call this e to the minus beta U.
And the index 0 again to indicate
that we are dealing with a uniform 0
to order distribution with respect
to which this average is taken.
And then, this quantity we can write as Z0.
And this exponential we can formally
express as a sum over all, let's say,
l running from 0 to infinity minus beta
to the l divided by l factorial.
Then, expanding this, a quantity that depends on the coordinates
is U. So I could have U raised to the l power, this kind
of average taken with respect to that.
And immediately, my writing something like this
suggests that what I'm after eventually
is some kind of a perturbation theory.
Because I know how to solve the problem where U is absent.
That's the ideal gas problem.
Maybe we can start to move away from the ideal gas
and calculate things perturbatively
in some quantity.
And exactly what that quantity is
will become apparent shortly.
At this time, it looks like it's an expansion in beta U
having to be a small quantity.
Now, of course, for calculating thermodynamic functions
and behaviors, we don't really rely usually on Z,
but log Z, which can be related to the free energy,
for example.
So taking the log of that expression,
I have a term that is log of Z0.
And a term that is the log of that sum.
But we recognize that that sum, in some sense,
is a generator of moments of this quantity U.
And we have experienced that when
we take the logarithm of a generator of moments, what
we will get is a generator of Cumulants,
now starting with l equals to 1 minus beta
to the power of l divided by l factorial U
to the l0 with a subscript c.
Of course, this subscript c sort of captures all of the various
subtractions that you have to make in order
to go from a moment to a Cumulant.
But that's quite general and something
that we know essentially is relating log of the expansion
to the expansion itself.
But that's well-known.
So presumably, I can perturbatively calculate these.
And I will have, therefore, progressively
better approximations to log Z in the presence
of this interaction.
Now at this point, we have to start
thinking about a particular expression for U.
So let's imagine that our U is a sum over pairwise interactions.
So if I think about the particles,
there are molecules that are in the gas in this room.
Basically, the most important thing is when the two of them
approach each other and I have to put a pairwise interaction
between them of this form.
In principle, I could add three-point and higher-order
interactions.
But for all intents and purposes,
this should be enough.
And then again, what I have here to evaluate,
the first term would be minus beta,
the average of the first power, and the next one
will be beta squared over 2.
For the first Cumulant, it is the same thing
as the first moment.
The second Cumulant is the thing that is the variance.
So this would have been U squared 0 minus average
of U squared, and then there will be higher orders.
So let's calculate these first two moments explicitly
for this potential.
So the first term U in this 0 to order average, what is it?
Well, that's my U. So I have to do a sum over i
and j of V of qi minus qj.
What does the averaging mean?
It means I have to integrate this over all qi divided
by V. Let's call this qk divided V and this is [INAUDIBLE].
OK, so let's say we look at the first term in this series which
involves V of q1 minus q2.
Let's say we pick that pair, q1 and q2.
And explicitly, what I have here is integrals over q1, q2,
qn, each one of them coming with a factor of V.
Now, of course, it doesn't matter which pair I pick.
Everything is symmetric.
The weight that I start with uniformly is symmetric.
So clearly, my choice of this pair
was arbitrary and the answer when
I do the sum is the answer for 1 times the number of pairs,
which is N N minus 1 over 2.
Now, over here I have integrals over q3 or qn,
et cetera, that don't appear in the function
that I'm integrating.
So all of those integrals will give me factors of 1.
And really, the only thing that I am going to be left with
is these two integrals over q1 and q2.
But even then, the function that I'm integrating
is only a function of the relative coordinate
that I can call q.
And then I can, if I want, integrate over q1 and q.
The integral over q1 would also give me
1 when I divide by the volume.
So the answer is going to be N N minus 1 over 2,
which is the number of pairs.
The integral over the relative coordinate times
the potential-- there is a factor of 1 over V
that I can put out here.
So you tell me what the first pair potential is
and I tell you that this is the first correction that you will
get once you multiply by minus beta
to log Z from which you can calculate the partition
functions.
Yes.
AUDIENCE: How did you turn the product of two differentials
d q1 times d q2 into just one differential?
PROFESSOR: OK, I have the integral dq q1 dq q2,
some function of q1 minus q2.
I change variables keeping q1 as one of my variables
and q [INAUDIBLE] replace q2 with q minus q2.
So then I have the integration over q1.
I have the integration over q2.
Not q2, but q f of q.
And then everything's fine.
All right, so that's the first term.
Let's calculate the next term.
Next term becomes somewhat interesting.
I have to calculate this quantity as a Cumulant.
It's the variance.
V itself is a sum over pairs.
So to get V squared to U squared,
I have to essentially square this expression.
So rather than having sum over one pair ij,
I have the sum over two pairs ij and kl.
And then what I need to calculate
is an average that if I were to square this,
I will get v of qi minus qj, v of qk minus ql 0.
This will be the average of U squared.
But when I calculate the variance,
I have to subtract from it the average of U squared.
So I have to subtract V of qi minus qj V of qk minus ql.
Fine.
So this is just rewriting of that expression
in terms of the sum.
Now, each one of these is N N minus 1 over two terms.
And so this, when I sum over all possibility,
is the square of N N minus 1 over two possible terms that
will occur here in the series that I've written.
Now, let's group those terms that can occur in this as
follows.
One class of terms will be when the pair
ij that I pick for the first sum is distinct from the pair kl.
So huge number of terms as I go through the pairs
will have this possibility.
OK, then what happens?
Then when I calculate the average of qi minus qj qk minus
ql, which is this part, what I need to do is to do all
of these integral of this form over all q's.
And then I have this V of qi minus qj V of qk minus ql.
Now, my statement is that I have, let's say here,
q1 and q2 and somewhere else q7 and q8.
Or, q7 and q974.
It doesn't matter.
The point is that this integral that
involves this pair of variables has
nothing to do with the integral that
involves the other pair of variables.
So this answer is the same thing as the average of qi
minus qj average of qk minus ql.
Essentially, in the 0 order probability that we are using,
the particles are completely independently
exploring this space and these averages will independently
rely on one pair and independently
rely on another pair.
And once this kind of factorization occurs,
it's clear that this subtraction that I
need to do for the variance will get rid of this term.
So these pairs do not contribute.
No contribution.
If you like, this is this thing.
All right, so let's do the next possible term, ij and ik,
with k not equal to this k.
So basically, I make the pairs have one point in common.
So previously, I had a pair here, a pair here.
Now I joined one of their points.
So what do I get here?
I will get for this average qi minus qj qi minus ql.
You can see that essentially, the three integrals that I need
to worry about are the integrals that
involve these three indices.
All the others will give me 0.
So this is going to be the integral qi qj ql divided by V.
I have V of qi minus qj V of qi minus ql.
AUDIENCE: You switched [INAUDIBLE].
PROFESSOR: OK.
I think that's fine.
All right.
Now, I can do the same thing that I did over here.
I have three variables.
I will pick qi to stay as one of my variables.
I replace qj with this difference, q j i.
I replace ql with its distance to i.
And again, I can independently integrate
over these three variables.
So this again, becomes the same thing
as V of qi minus qj average V of qi minus ql average.
And again, there is no contribution.
Now, the first class it was obvious because this pair
and this pair were completely distinct.
This class is a little bit more subtle
because I joined one of the points together,
and then I used this and this as well as this point
as independent variables.
And I saw that the sum breaks into pieces.
And you won't be surprised if no matter
how complicated I make various more interactions over here,
I can measure all of the coordinates with respect
to the single point and same thing would happen.
This class of diagrams are called 1-particle reducible.
And the first part had to do with distinct graphs.
You didn't have to worry about them.
But if I were to sort of convert this expression--
and we will do so shortly-- into graphs,
it corresponds to graphs where there
is a point from which everything else is hanging.
And measuring coordinates with respect
to that point will allow the average to break into pieces,
and then to be removed through this subtraction.
So this is simple example of something that is more general.
So at the level of this second-order theorem,
the only type of thing that will survive
is if the pairs are identical, ij and other term
is also the same pair ij.
What do I get then?
I get V of qi minus qj squared minus V
of qi minus qj 0 order squared.
So it's the variance of a single one of these bond
contributions.
If I write that in terms of integrals,
this becomes the integral d cubed q1,
d cubed q2-- well, qi qj.
I will have factors of V. Then, I
would have V squared qi minus qj for the first term
and then the square of something like this for the second term.
So putting everything together up to this order,
what do I get?
I will get log Z, which is log of Z0.
Let me remind you, log of Z0 was N log
of V over lambda cubed N. That's on form such as this.
And then I have these corrections.
Note that both of the terms that have survived
correspond to looking at one pair.
So the corrections will be of the order of N N minus 1
over 2.
Because in both cases, I really look at one pair.
The first contribution was minus beta integral d cubed q v of q.
There was a factor of 1 over V. Actually,
I can take-- well, let's put the factor of 1 over V
here for the time being.
The next term will be beta squared over 2
because I am looking at the second-order term.
And what do I have?
I have the difference of integral d cubed q
over V V squared.
And the square of the integral of d cubed q over V V.
And presumably, there will be higher-order terms
as we will discuss.
Now, I'm interested in the limit of thermodynamics
where N and V are large.
And in that limit, what do I get?
I will get log Z is log of Z0.
And then I can see that these terms I can write as beta.
N N minus 1 I will replace will N squared.
The factors-- here I have a 1 over V. Here,
I have 1 over V squared.
So this factor is smaller by an amount that
is order of V in the large V-limit.
I will bring the factor of V outside,
so I have something like this.
And then, what do I have here?
Actually, let's keep this in this form,
minus beta integral d cubed q V plus beta
squared over 2 integral d cubed q V squared [INAUDIBLE].
So the kinds of things that we are interested
and we can measure are the energy of the system,
but let's say we focus on pressure.
And if you look through various things that we derived before,
beta times the pressure you can get
from taking a derivative of log Z with respect to V.
We can express the log Z in terms of the free energy,
and then the derivative of free energy with respect to volume
will give you the pressure of the various factors of beta
[? and signs, ?] we will come up with this.
So from the first term, N long V, we get our ideal gas result.
Beta p is the same thing as density.
And we see that we get a correction from here when
I divide by-- when I take a derivative with respect
to V of 1 over V, I will get minus 1 over V squared.
So the next term would be the square of the density.
And then, there is a series that we
will encounter which depends on the interaction potential.
Now, it turns out that so far I have calculated
terms that relied on only two points, ij.
And hence, they become at the end
proportional to density squared.
If I go further in my expansion, I
will encounter things that I will need triangular points.
For example, i j k forming a triangle.
And then, that would give me something
that would be of the order of the density cubed.
So somehow, I can also see that ultimately this series
in perturbation theory, as we shall see,
also more precisely can be organized in powers of density.
Why is that important?
Because typically, as we discussed right
at the beginning, when you look at the pressure of a gas
when it is dilute and over V is very small,
it is always ideal gas-like.
As you make it more dense, you start
to get corrections that you can express in powers of density.
And the coefficients of that are called Virial coefficients.
So in some sense, we have already
started in calculating the second Virial coefficient,
and there are higher-order Virial coefficients,
which together give you the equation of states relating
pressure to density by a power series.
AUDIENCE: Question.
PROFESSOR: Yes.
AUDIENCE: The Virial coefficients,
if they have integral over the whole volume,
then they will be functions of volume, right?
PROFESSOR: Yes.
But imagine that I'm thinking about the gas in this room
and the interaction of two oxygen molecules.
So what I am saying is you pick an oxygen molecule.
There is interaction and I have to integrate over
where the other oxygen molecule is.
And by the time it is tiny, tiny bit
[? away, ?] the interaction is 0.
It doesn't really matter.
AUDIENCE: If the characteristic volume of interaction
is much smaller than volume--
PROFESSOR: Of the space.
AUDIENCE: [INAUDIBLE], then the integrals just converge
to [? constant. ?]
PROFESSOR: That's right.
So if you want to be even more precise,
you will get corrections when your particles are
close to the wall and uniformity, et cetera,
is violated.
So there will, actually, be corrections to, say, log
Z that are not only proportional ultimately to number volume,
but the area of the enclosure and other [? subleading ?]
factors.
But in the thermodynamic limit, we ignore all of that.
So a lot of that is resolved by this statement here.
Again, you won't be surprised that if I were to go ahead,
then there will be a term in this series that
would be beta cubed over 3 factorial integral over q
of vq, et cetera.
And that's actually very good because as I have written
for you currently, this expression is totally useless.
Why is it useless?
Because let's think back about these two oxygen
molecules in the gas in this room
and what the potential of interaction between them
would look like.
Presumably, it is a function of the relative separation.
We can call it r or q, it doesn't matter.
And if you bring them closer than the typical size
of these molecules, their potential will go to infinity.
So basically, they don't want to come close to each other.
If you go very far away, typically you
have the van der Waals attraction
between the particles.
So out here, it is attractive.
The potential is negative and falls off,
typically, as 1 over r to the 6 related
to the polarizabilities of the particles.
And these, when you come very short distances,
the electronic clouds will overlap
and the potential goes to infinity.
And if you want to sort of think about numbers, typical scales
that we have here-- let's say here,
are presumably of the order of angstroms.
And the typical depths of these potentials
in units that make sense to statistical physics
are of the order of 100 degrees Kelvin.
So it's why typical gases liquidy
at range of temperatures that is of the order of 100
degrees Kelvin.
Now, if I want to take this potential and calculate this,
I have trouble.
Because the integral over here will give me infinity.
So this expression is, in general, fine.
But whenever you have perturbation theory,
you start to evaluate things and you
have to see whether the correction indeed
is sufficiently small.
And it seems we can't make it sufficiently small.
Yes.
AUDIENCE: If you're modeling interatomic or intermolecular
potential as a Lennard-Jones potential,
isn't the [INAUDIBLE] term at the--
or near the origin, a phenomenological choice?
PROFESSOR: If you are using as a formula
for this the Lennard-Jones potential,
people phenomenologically write something
like 1 over r to the 12th power, et cetera.
But I don't have to assume that.
AUDIENCE: Right.
Because I've seen forms of the Lennard-Jones potential
which is exponentially decaying at the origin.
So it reaches a finite, albeit probably high value
at the origin.
Wouldn't that take care of the issue?
PROFESSOR: If you believe in that kind of potential, yes.
So what you are saying is that if I
make this come to a finite value,
I will be able to find the temperature that
is sufficiently high compared to this,
and then things would be fine.
If we had that, then we would have fusion right here going
on, right?
So clearly, there is some better truth
to a potential that is really very, very high compared
to temperatures that we are looking at.
So we don't have to worry about that.
Well, we have to worry about this issue.
OK?
All right.
So you say, well, how about the following-- let's say
that I am something like the gas in this room
and I am sufficiently dilute.
I'm going to forget about these terms.
But since this potential becomes essentially large,
I can never ignore a particular term in this series.
And I keep going adding more and more terms in the series.
If I sort of think back about the picture
that I was generating before diagrammatically-- which again,
we will clarify later on-- this term
corresponds to taking a pair of points and a V
that connects them.
And this term corresponds to two V's
going between the same pair of points.
And you say, well, there will be terms
that will involve three V's, four V's, and they
will contribute in the series in a manner that I can recognize.
So why don't I add all of those terms together?
And if you like, we can call the resulting object something
else.
What is that something else?
Well, what I have done is I have minus beta V.
Well, all of them-- so what is this?
It is the integral d cubed q.
I have minus beta V plus 1/2 beta V squared.
The next term is going to be minus 1
over 3 factorial beta V cubed because it will come
from the third-order term in the expansion.
And I can go on.
And you say, well, obviously this whole thing
came from the expansion of e to the minus beta V
except that I don't have the 0 order term, 1.
So this quantity that I will call f of q rather than V of q
is obtained as e to the minus beta V of q minus 1.
So if I were to, in fact, add up an infinite number
of terms in the series rather than having
to integrate V V squared V cubed, each one of them
is divergence.
I have to integrate this f of q.
So let's see what this f of q-- or correspondingly, f of r
looks like as a function of r.
So in the range that I have my hard core
and V is large and positive, this is going to give me 0.
So the function is minus 1 down here.
At exactly whatever point it is that the potential is 0,
then f is also 0.
So basically, I will come to the same 0.
When the potential is negative, this will be larger than 1.
My f will be positive.
So presumably, it will have some kind
of a peak around where this peak is.
And then at large distances, it goes back towards 0.
So this is going to go back towards 1 minus 1, which is 0.
So actually, this end of the potential
is the same thing as minus beta.
And whereas I couldn't integrate any one term in this series,
I can certainly integrate the sum over all positions
and it will give me a number, which
will depend on temperature, the properties
of the potential, et cetera.
And it is that number that will tell me
what the density squared term in the expression for the pressure
is and what the second Virial coefficient is.
Yes.
AUDIENCE: Is this basically an excluded volume of sorts?
PROFESSOR: Yeah.
So this part if I sent it to infinity,
corresponds to an excluded volume.
So if you want really excluded volume, this would be minus 1
up to some point.
AUDIENCE: [INAUDIBLE].
PROFESSOR: Yeah.
But I'm talking about very general,
so I would say that excluded volume can
be very easily captured within this [INAUDIBLE].
All right.
So we want to sort of follow-up from here.
But since you will have a problem set,
I will mention one other thing.
That this reorganization of the series that I made here
is appropriate to the limit of low densities, where
I would have a nice expansion in powers of density.
The problem that you will deal with
has to do with plasmas, where the interaction
range is very large.
And you already saw something along those lines
when we had the Vlasov equation as opposed
to Boltzmann equation.
There was a regime where you had to reorganize
the series different ways.
In that case, it was the BBGKY hierarchy
whether you were looking at the dense limit or a dilute limit.
So this is the analog of where the Boltzmann equation would
have been inappropriate.
The analog of the regime where you
are dense and something like the Vlasov equation
would be appropriate.
So there is some kind of interaction range,
n d cubed if it is much larger than 1.
Then it turns out that rather than looking at diagrams that
have the fewest number of points-- in this case, 2--
you have to look at diagrams that
have the largest number of points.
Because each additional point will give you
an additional factor of N over V.
You can see that these factors of N N minus 1 over 2
came from the number of points that I had selected.
So if I had to select three points,
I would have N cubed and the corresponding V. So the more
points that I have in my series, I
will have more powers of the density.
So in that case, it turns out that rather
than looking at two points and all lines between them,
you organize things in terms of what
are called ring diagrams, which are
things such as this, this, this.
So basically, for a given number of lines,
the most number of points is obtained by creating your ring.
And so one of the problems that you have
is to sort of sum these ring diagrams and see what happens.
But it seems like what I'm telling you here
is while we calculated order of density squared,
but maybe I want to calculate order of density cubed.
And it makes much more sense rather than when
faced with potentials of this form arranging
the series in powers of the potential to arrange it
in powers of this quantity e to the minus beta V minus 1.
So let's go through a route in which I directly
expand everything in terms of this quantity.
And that's the second type of expansion
that I will call the cluster expansion.
So once more, what we want to calculate
is the partition function.
It depends on temperature, volume, number
of particles obtained by integrating over
all degrees of freedom.
The integration over momenta we saw is very simple.
Ultimately, will give us this N factorial lambda to the 3N.
This time, I won't pull out the factor of V
to the N at this point.
And then I have the integration over all of these q's.
I did not divide by V because I did not
multiply by V to the N. OK.
And then I have e to the minus beta U,
but this is my U. U is the sum of various terms.
So e to the minus beta U will be a product over all,
let's say-- let's call this k pairs ij.
e to the minus beta V of qi minus qj.
So basically, the only thing that I did
was e to the minus beta-- this quantity
I wrote as a product of contributions
from the different pairs.
Now, this quantity I have it in the line above.
This is also a relative position.
Clearly, I can think of this as 1
plus the f that would correspond to the distance between qi
and qj.
And again, I can either write it as Vf of qi minus qj,
or simplify of i-- my notation-- and write it as f i j.
So f i j means-- so let me maybe write it here.
f i j I have defined to be e to the minus beta
V of qi minus qj minus 1.
So what do I expect?
I saw that the first correction to something
that was interesting to me had one power of f in it
that I had to ultimately integrate.
So maybe what I should do is I should start organizing things
in terms of how many f's I have.
So an expansion in powers of f.
So what's going to happen here?
I would have 1 over N factorial lambda to the 3N.
I have a product of all of the integrations.
And I have all of these factors of 1 plus f1 2 times 1.
So basically, this is maybe-- it's really a product of 1
plus f1 2, 1 plus f1 3, 1 plus all of one of these things.
So the thing that has least number of f's is
when I pick the 1 from all of these brackets
that are multiplying each other.
The next term is to pick 1f from one pair
and all the others would be 1.
The next term would be sum over ij kl f i j f k l,
and then there will be diagrams that would progressively
have more and more factors of f.
Now, what I will do is to represent the various terms
that I generate in this series diagrammatically.
So first of all, I have to integrate over N points.
So I put points 1 through N. So I have 1, 2, 3, 4.
It doesn't matter how I put them-- N.
And then, this 1 would correspond
to just this diagram.
The next thing is I put a line for f i j.
So let's say that I picked in this series
the term that was f 2 3.
I will represent that by a line that goes between 2 and 3.
Later on in the picture, maybe I will
pick a term that is f 2 3 f 4 5.
So some second-order term could be something like this.
Maybe later on in the series, I will
pick a term that is connecting f3 and f4.
So this would be a third-order term in the series.
Maybe later on the series, I have some other pair over here.
So any one of the huge number of terms
that I have generated here has one diagram
that is appearing over here.
Now, the next step is, of course I have to integrate over all
of the q's.
I mean, these diagrams, what they represent is essentially
some f of q1 minus q3, some f of whatever.
And I have to integrate over all of these q's.
So the contribution to the partition function
that I will get would be some value associated
with one of these graphs obtained by doing
the integrations over q.
Now, the thing that I want you to notice
is that contribution of one of these graphs or diagrams
is the product of contributions of its linked clusters.
So here I have one particular diagram
that is represented here.
Well, let's say 2, 3, 4, 5 are linked together
and separate from whatever this is.
Let's say this is 7, 9.
So when I do the integrations over q2, q3, q4, q5,
I don't really rely on any of the other things.
So my integration here, we break off
into the integration that involves
this, the integration that involves this, as well
as integrations over all of these points
that are not connected to anybody.
All of those will simply give me a factor of v.
So is everybody happy with this simple statement
that the contribution-- if I think of these linked clusters
as some collection of points that are linked together,
the value that I would get for this diagram
would be the contribution-- product of the contributions
that I would have for this.
So yes?
AUDIENCE: Can you explain [INAUDIBLE]?
PROFESSOR: OK.
So let's pick a particular thing.
Let's say we do something like this,
where this is number 7, 245, 6, 5, 4.
This is a particular thing.
So what do I have to do?
There's also a whole bunch of points-- 1, 2, 3.
What I am instructed to do is to integrate over
q1, q2, q3, et cetera.
Now, the integral over q1 is just the integral of q1.
So this integral by itself will give me
V. Any number of points that are by themselves
will also give me V. So the first term
that becomes nontrivial is when I
have to do the integration over q4, q5, q6 of f 4 5, f 5 6,
f 6 4.
I don't know what this is, but there
is something that will come from here.
Then, later on I have to integrate over 7
and something else.
So then I have the integral over q7, q-- let's call this 8.
f of 7 8.
So this is something else.
So the overall value of this term in the perturbation theory
is the product of contributions from a huge number of one
clusters.
Here, I have a two cluster.
Here, I have a three cluster.
Maybe I will have more things.
Also, notice that if I have more of these two clusters,
the result for them would be exactly the same.
So if I have lots of these pairs, the same way that I
had lots of single points-- and it became V
to the number of single points-- it
will become whatever that y is to the number of pair
points, x, whatever that is, to the number of triplets
that I have in triangles.
So this is how this object is built up.
Yes?
AUDIENCE: How are we deciding how many triplets
and how many pairs we actually have?
PROFESSOR: At this point, I haven't told you.
So there is a multiplicity that we have to calculate.
Yes.
So at this point, all I'm saying is given
this, the answer is the product of contribution.
There is a multiplicity factor-- you're right.
We have to calculate that.
OK?
Anything else?
All right.
So I mean, think of this as taking a number
and writing it as the product of its prime factors,
factorizing into primes.
So you sort of immediately know that somehow prime factors,
the prime numbers, are more important.
Because everything else you can write as prime numbers.
So clearly, what is also buried in heart
of calculating this partition function is these clusters.
So let me define the analog of prime numbers as follows.
I define bl to be the sum over contributions
of all linked clusters of l-points.
So let's go term by term.
b1 corresponds essentially to the one point
by itself in the diagram that I was drawing down here.
And corresponds to integrating over
the coordinate that goes all over the space.
And hence, the volume is the same thing as the volume V.
b2 is a cluster of two points.
So it is this.
And so it is the integral over q1 and q2,
which are two endpoints of this, of f of q1 minus q2.
And this we have seen already many times.
It's the same thing as volume, the integral over q f of q.
It is the thing that was related to our second Virial
coefficient.
Now, this is very important for b3.
I note that I underlined here "all."
And this is for later convenience.
It seems that I'm essentially pushing complexity
from one place to another.
So there are a number of diagrams
that have three things linked together.
This is one of them.
Let's say, think of 1, 2, 3 connected to each other.
But then, I have diagrams.
At this stage, I don't really care
that this is one particle irreducible.
Here, I have no constraints on one particle irreducibility.
And in fact, this comes in three varieties
because the bond that is missing can be one of three.
So once I pick the triplets of three points, sum
over all clusters that involve these three points is this,
in particular with this one, appearing three times.
Yes.
AUDIENCE: Do we still count all the diagrams
when particles are identical?
PROFESSOR: Yes.
Because here, at this stage, there
is an N factorial out here.
All I'm really doing is transcribing
a mathematical formula.
The mathematical formula says you
have to do a bunch of integrations.
And I am just following translating
that mathematical formula to diagrams.
At this stage, we forget.
This has nothing to do with this N factorial or identity.
It's just a transcription of the mathematics
So at this stage, don't try to correct my conceptual parts
but try to make sure that my mathematical steps are correct.
OK?
Yes.
AUDIENCE: I didn't understand what
that missing links represents.
PROFESSOR: OK.
So at the end of this story, what I want to write down
is that the value of some particular term in this series
is related to product of contributions.
I described to you how productive contributions
comes along.
And each contribution is a bunch of points
that are connected together.
So what I want to do, in order to make my algebra easier
later on, is to say that-- let's say
these three points are part of a cluster.
They are connected together somehow.
So they will be giving me some factor.
Now, they are connected, if they are connected like this,
like this, like this, like that.
So basically, any one of these is
a way of connecting these three points so that they
will make a contribution together.
Yes.
AUDIENCE: So the one with all three bonds
comes with a different [INAUDIBLE]
in the series than the ones with only two bonds, right?
PROFESSOR: Yes.
Now, there is reason, ultimately,
why I want to group all of them together.
Because there was a question before about
multiplicity factors.
I can write down a closed form, nice expression
for the multiplicity factor if I group them together.
Otherwise, I have to also worry that there are three of these
and there is one of them, et cetera.
There is an additional layer of multiplicity.
And I want to separate those two layers of multiplicity.
So here, the thing that I have is I have these endpoints.
I want to say that the eventual result for endpoints
comes from clusters that involves 1's, clusters that
involve pairs, and cluster that involve triplets.
Now, I realize suddenly that if I made the multiplicity count
here, then for this triplet, I could have put all of them
together like this or like that.
And the next order of business when
I am thinking about four objects,
I can connect them together into a cluster in multiple ways.
And there is a separate way of calculating
the relative multiplicity of what
comes outside and the multiplicity of this quartic
with respect to entirety.
So you have to think about this a little bit.
So I'll proceed with this.
So this mathematically would be the integral q1, q2, q3, f
1 2, f 2 3, f 3 1 plus, say, f 1 2, f 2 3, plus f 1 3 f 3 2
plus f 2 3, f 3 1, something like this.
Well, 1, 2.
2 is repeated.
3 is repeated.
I should repeat 1.
OK.
Of course, the last three are the same thing,
but this is the expression for this.
Now, before would be a huge complex of things.
It would have a diagram that is of this form.
It would have a number of diagrams that are of this form.
It would have diagrams that are of this form, diagrams
that are like this, and so forth.
Basically, even within the choice
of things that have four clusters,
there is a huge number when you go
to further and further [? down. ?]
OK, so maybe this statement will now clarify things.
So what I have is that the partition function
is 1 over N factorial lambda to the 3N.
And then it is all of the terms that
are obtained by summing this series.
And so I have to look at all possible ways
that I can break N-points into these clusters.
Suppose I created a situation where I have clusters of size l
and then I have nl of them.
So I saw that the contribution that I would get from the 1
clusters was essentially the product
of the number of 1 clusters.
It was V to the number of points that
are not connected together.
So I have to take b1, raise it to the power of N1,
which is the number of 1 cluster.
Then, I have to do the same thing for b2, b3, et cetera.
And then I have to make sure that I
have looked at all ways of partitioning
N the numbers into these clusters.
So I have a sum over nl l has to add up to n.
And I have to sum over all nl's that are
consistent with this constraint.
But then there is this issue of the multiplicity.
Given that I chose some articular set of these,
if I were to reorganize the numbers,
I would get the same contribution.
So let's say we pick this diagram that
has precisely a contribution that
is V, which is b1 to the number of these one points.
This to the number of b2 to the power
of the number of these, this, et cetera.
But then I can take these labels-- 1, 2, 3, 4, 5,
6-- that I assigned to these things and permute them.
If I permute them, I will get exactly the same contribution.
So there is a large number of diagrams
that have exactly this same contribution because
of this permutation.
There was a question back there.
AUDIENCE: You've already answered it.
PROFESSOR: All right.
Actually, this is an interesting thing--
I don't know how many of you recognize this.
So essentially, I have to take a huge number, N,
and break it into a number of 1 clusters, 2 clusters, et
cetera.
The number of ways of doing that is
called a partition of an integer.
And last century, the [INAUDIBLE]
calculated what that number is.
So there is a [INAUDIBLE] theorem associated with that.
And actually, later in the course
I will give you a problem to calculate
the asymptotic version of the [INAUDIBLE].
But this is different story.
So given that you had made one of these partitions
of this integer, what is this degeneracy factor?
So let me tell you what this degeneracy factor is.
So given this choice of nl's, you
say-- well, the first thing is what I told you.
For a particular graph, the value
is independent of how these numbers were assigned.
So I will permute those numbers in all possible ways
and I will get the same thing.
So that will give me N factorial.
It's the number of permutations.
But then, I have over-counted things because within,
let's say, a 2 cluster-- if I count 7, 8, 8,
7 they are the same thing.
And by multiplying by this N factorial,
I have over-counted that.
So I have to divide by 2 for every one of my 2 clusters.
So I have to divide by 2 to the power of N2.
For the 3 clusters, I have the permutation of everything
that is inside.
So it is 3 factor here.
So what I have here is I have to divide
by the over-counting, which is l factorial-- labels
within a cluster-- to the number of clusters
that I have that are subject to this.
There is another thing that this pair is 100, 101 and this pair
is 7, 8.
Exchanging that pair of numbers with that pair of numbers
is also part of the symmetries that I have now over-counted.
So that has to be taken into account
because the number of these 2 clusters here is 2.
Here is actually 3.
I have to divide by 3 factorial.
In general, I have to divide by nl factorial.
OK.
Again, this is one of those things
that the best advice that I can give you is to draw five
or six points so your n is small.
And draw some diagrams.
And convince yourself that what I told you
here rapidly is correct.
And it is correct.
Yes.
AUDIENCE: Can you just clarify which is in the [INAUDIBLE]?
PROFESSOR: OK.
So the N factorial is outside.
And then-- so maybe write it better.
For each l, I have to multiply nl factorial and l
factorial to the power of nl.
So this is my partition function in terms of these clusters.
Even with all of these definitions
that I have up there, maybe not so obvious an answer.
And one of the reason it is kind of an obscure answer
is because here I have to do a constrained sum.
That is, I have all of these variables.
Let's call them n1, n2, n3, et cetera.
Each one of them can go from 0, 1, 2, 3.
But they are all linked to each other
because their sum is kind of restricted by some total value
l, by this constraint.
And constrained sums are hard to do.
But in statistical physics, we know
how to gets rid of constrained sums.
The way that we do that is we essentially
allow this N to go all the way.
So if I say, let's make-- it's very hard for me
to do this n1 from 0 to something,
n2 from 0 to something with all of this constrained.
My life would be very easier if I could independently
have n1 go take any value, n2 take any value,
n3 take any value.
But if I do that essentially for each choice of n1, n2, n3,
I have shifted the value of big N.
But there is an ensemble that I know
which has possibly any value of N,
and that's the grand canonical.
So rather than looking at the partition function,
I say I will look at the grand partition function
Q that is obtained by summing over
all N. Can take any value from 0 to infinity.
I have e to the beta mu N times the partition function that
is for N particles.
So that's the definition of how you would go from the canonical
where you have fixed N to grand canonical
where you have fixed the chemical potential mu.
So let's apply this sum over there.
I have a sum over N, but I said that if I allowed
the nl's to vary independently, it
is equivalent to varying that N, recognizing
that this n is sum over l l nl.
That's the constraint.
So that's just the first term.
I have rewritten this sum that was constrained and this sum
over the total number as independent sums over the nl's.
Got rid of the constraint.
Now, I write W. Oh, OK.
The partition function.
Now, write the partition function.
I have 1 over N factorial.
I have lambda cubed raised to this power.
So actually, let me put this in this fashion.
So both e to the beta mu over lambda
cubed that is raised to that power.
OK, so that's that part.
The sum I have gotten rid of.
I notice that my W has an N factorial.
So this is the N factorial that came from here.
But then there is an N factorial that is up here from the W.
So the two of them will cancel each other, N factorial.
And then I have a product over all clusters.
Part of it is this bl to the power of nl,
and then I have the contribution that is nl factorial l
factorial to the power of nl.
So this part cancels.
For each l, I can now independently
sum over the values of nl can be anything, 0 to infinity.
And what do I have?
I have bl to the power of nl.
I have division by nl factorial.
I have l factorial to the power of nl.
And then I have e to the beta mu over lambda cubed raised
to the power of l nl.
I recognize that each one of these terms in the sum
is 1 over nl factorial something raised to the power of nl,
which is the definition of the exponential.
So my q is a product over all l's.
Once I exponentiate, I have e to the beta mu
over lambda cubed raised to the power of l.
And then I have bl divided by l factorial.
Or, log of Q, which is the quantity that I'm interested,
is obtained by summing over all clusters 1
to infinity e to the beta mu over lambda cubed raised
to the power of l bl divided by l factorial.
So what does this tell us, which is
kind of nice and fundamentally important?
You see, we started at the beginning over here
with a huge number of graphs.
These graphs could be organizing all kinds of clusters.
And they would give us either the partition function
or summing over N, the grand partition function.
But when we take the log, I get only single connect objects.
And this is something that you had already
seen as the connection that we have
between moments and Cumulants.
So the way that we got Cumulants was
to look at the expansion of the log.
So the function itself was the generator of all moments
and we took the log.
And graphically, we presented that
as getting the moments by putting together
all kinds of clusters of points that corresponded to Cumulants.
And this is again, a representation
of the same thing.
That is, in the log you have essentially
individual contributions.
Once you exponentiated, you get multiple contributions.
Now, the other thing is, that if I am thinking about the gas,
then log of Q is e to the minus beta g.
And g is E minus TS minus mu N. Bur for a gas that
has extensive properties, E is TS mu N minus PV.
So this is the same thing as beta PV.
So this quantity that we have calculated here
is, in fact, related to the pressure
directly through this formula.
And the important part or the important observation
is that it says that it should be proportional to volume.
You can see that each one of my b's
that I calculated here will have one free integral.
If you like, its center of mass.
You can go over the entire volume.
So all of my b's are indeed proportionality to volume.
And you can see what disaster it would
have been if there was a term here that was not just a linked
cluster but product of two linked clusters.
Then I would have something that would go like V squared.
It's not allowed.
So essentially, this linked cluster nature is also
related to extensivity in this sense.
So in some sense, what I have established here-- again,
related to that-- is that clearly all of my bl's
are proportional to volume.
And I can define something that I
will call bl bar, which is divide the thing by the volume.
And then what we have established
is that the pressure of this interacting gas as an expansion
beta p, which takes this nice, simple form, sum over l
e to the beta mu divided by lambda cubed bl
bar-- the intensive part of the contribution of these cluster--
divided by l factorial.
This, of course, runs from 1 to [INAUDIBLE].
Yes.
AUDIENCE: [INAUDIBLE] is raised to the l?
PROFESSOR: Who's asking the question?
AUDIENCE: The term in the parentheses is raised to the l?
PROFESSOR: Exactly.
Thank you very much.
Yes.
AUDIENCE: [INAUDIBLE].
But they don't, actually.
Could that be true, for example [INAUDIBLE]?
PROFESSOR: OK.
So the triangle, I have it here.
Integral d q1, d q2, d Q3, f 1 2, f 2 3, f 3.
Let's write it explicitly.
It is f of q1 minus q2, f of q2 minus q3, f of q3 minus q1.
I call this vector x.
I call that vector y.
This vector is x plus y.
Yes.
AUDIENCE: Does the contribution from [INAUDIBLE]?
PROFESSOR: OK.
So you say that I have here an expansion for the pressure.
I had given you previously an expansion for pressure
that I said is sensible, which is this Virial expansion, which
is powers of density.
This is not an expansion in powers of density.
So whether or not the terms in this expansion
becomes smaller or larger will depend
on density in some indirect way.
So my next task, which I will do next lecture,
is within this ensemble I have told you
what the chemical potential is.
Once I know what the chemical potential is,
I can calculate the number of particles
as d log Q by d beta mu.
And hence, I can calculate what the density is.
And so the answer for this will also be a series in powers of e
to the beta mu.
And what then I will do is I will combine these two
series to get an expansion for pressure in powers of density,
and then identify the convergence of this series
and all of that via that procedure.