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visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Let's start. Are there any questions? We would like to have a
perspective for this really common observation that if you
have a gas that is initially in one half of a box, and
the other half is empty, and some kind of a partition
is removed so that the gas can expand, and it can
flow, and eventually we will reach another
equilibrium state where the gas occupies more chambers. How do we describe
this observation? We can certainly characterize
it thermodynamically from the perspectives
of atoms and molecules. We said that if I
want to describe the configuration of the gas
before it starts, and also throughout the expansion,
I would basically have to look at all sets
of coordinates and momenta that make up this particle. There would be some
point in this [? six ?], and I mention our
phase space, that would correspond to where
this particle was originally. We can certainly follow
the dynamics of this point, but is that useful? Normally, I could
start with billions of different types of
boxes, or the same box in a different
instance of time, and I would have totally different
initial conditions. The initial
conditions presumably can be characterized to a
density in this phase space. You can look at some volume
and see how it changes, and how many points
you have there, and define this
phase space density row of all of the
Q's and P's, and it works as a function of time. One way of looking at how it
works as a function of time is to look at this box
and where this box will be in some other
instance of time. Essentially then, we are
following a kind of evolution that goes along this streamline. Basically, the derivative
that we are going to look at involves changes both
explicitly in the time variable, and also increasingly
to the changes of all of the
coordinates and momenta, according to the Hamiltonian
that governs the system. I have to do, essentially,
a sum over all coordinates. I would have the
change in coordinate i, Qi dot, dot, d row by dQi. Then I would have Pi, dot-- I
guess these are all vectors-- d row by dPi. There are six end coordinates
that implicitly depend on time. In principle, if I am
following along the streamline, I have to look at
all of these things. The characteristic of evolution,
according to some Hamiltonian, was that this volume of
phase space does not change. Secondly, we could
characterize, once we wrote Qi dot, as dH by dP,
and the i dot as the H by dQ. This combination of derivatives
essentially could be captured, and be written as 0 by
dt is the Poisson bracket of H and [? P. ?] One of the things,
however, that we emphasize is that as far as evolution
according to a Hamiltonian and this set of
dynamics is concerned, the situation is completely
reversible in time so that some
intermediate process, if I were to reverse all of
the momenta, then the gas would basically come back
to the initial position. That's true. There is nothing to do about it. That kind of seems to
go against the intuition that we have from
thermodynamics. We said, well, in
practical situations, I really don't care about
all the six end pieces of information that
are embedded currently in this full phase
space density. If I'm really trying
to physically describe this gas expanding,
typically the things that I'm interested in are
that at some intermediate time, whether the particles
have reached this point or that
point, and what is this streamline velocity
that I'm seeing before the thing relaxes, presumably,
eventually into zero velocity? There's a lot of
things that I would need to characterize
this relaxation process, but that is still
much, much, much less than all of the information
that is currently encoded in all of these six end
coordinates and momenta. We said that for things that
I'm really interested in, what I could, for
example, look at, is a density that involves
only one particle. What I can do is to
then integrate over all of the positions and
coordinates of particles that I'm not interested in. I'm sort of repeating this to
introduce some notation so as to not to repeat all of
these integration variables, so I will call dVi the phase
place contribution of particle i. What I may be interested in
is that this is something that, if I integrate
over P1 and Q1, it is clearly
normalized to unity because my row, by definition,
was normalized to unity. Typically we may be
interested in something else that I call F1, P1
Q1 P, which is simply n times this-- n times
the integral product out i2 to n, dVi, the full row. Why we do that is
because typically you are interested or used
to calculating things in [? terms ?] of
a number density, like how many particles are
within some small volume here, defining the density so
that when I integrate over the entire volume
of f1, I would get the total number of
particles, for example. That's the kind of normalization
that people have used for f. More generally,
we also introduced fs, which depended on
coordinates representing s sets of points, or s
particles, if you like, that was normalized to be-- We said, OK, what I'm really
interested in, in order to calculate the properties of
the gases it expands in terms of things that I'm
able to measure, is f1. Let's write down the
time evolution of f1. Actually, we said,
let's write down the time evolution
of fs, along with it. So there's the time
evolution of fs. If I were to go
along this stream, it would be the fs
by dt, and then I would have contributions
that would correspond to a the changes in
coordinates of these particles. In order to progress
along this direction, we said, let's define
the total Hamiltonian. We will have a simple form,
and certainly for the gas, it would be a good
representation. I have the kinetic energies
of all of the particles. I have the box that confines
the particles, or some other one particle potential, if you like,
but I will write in this much. Then you have the interactions
between all pairs of particles. Let's write it as sum over i,
less than j, V of Qi minus Qj. This depends on n set of
particles, coordinates, and momenta. Then we said that for purposes
of manipulations that you have to deal with, since
there are s coordinates that are appearing here
whose time derivatives I have to look at, I'm going
to simply rewrite this as the
contribution that comes from those s particles,
the contribution that comes from the remaining
n minus s particles, and some kind of [? term ?]
that covers the two sets of particles. This, actually, I didn't quite
need here until the next stage because what I write
here could, presumably, be sufficiently general,
like we have here some n running from 1 to s. Let me be consistent
with my S's. Then I have Qn, dot, dFs by
dQn, plus Pn, dot, dFs by dPn. If I just look at the
coordinates that appear here, and say, following this
as they move in time, there is the explicit
time dependence on all of the implicit
time dependence, this would be the
total derivative moving along the streamline. Qn dot I know is
simply the momentum. It is the H by dPn. The H by dPn I have from
this formula over here. It is simply Pn divided by m. It's the velocity--
momentum divided by mass. This is the velocity
of the particle. Pn dot, the rate of
change of momentum is the force that is
acting on the particle. What I need to do is to take
the derivatives of various terms here. So I have minus dU by dPn. What is this? This is essentially the
force that the particle feels from the
external potential. If you are in the
box in this room, It is zero until you
hit the edge of the box. I will call this Fn to represent
external potential that is acting on the system. What else is there? I have the force that will
come from the interaction with all other [? guys. ?] I
will write here a sum over m, dV of Qm minus Qn,
by dQn-- dU by dQm. I'm sorry. What is this? This Is basically the
sum of the forces that is exerted by the n
particle on the m particle. Define it in this fashion. If this was the entire
story, what I would have had here is a group of
s particles that are dominated by
their own dynamics. If there is no other
particle involved, they basically have to
satisfy the Liouville equation that I have written, now
appropriate to s particles. Of course, we know that
that's not the entire story because there are
all these other terms involving the interactions
with particles that I have not included. That's the whole
essence of the story. Let's say I want to think
about one or two particles. There is the interaction
between the two particles, and they would be evolving
according to some trajectories. But there are all of
these other particles in the gas in this room
that will collide with them. So those conditions
are not something that we had in the
Liouville equation, with everything considered. Here, I have to
include the effect of all of those other particles. We saw that the
way that it appears is that I have to imagine that
there's another particle whose coordinates and
momenta are captured through some volume for
the s plus 1 particle. This s plus 1
particle can interact with any of the particles
that are in the set that I have on the other side. There is an index
that runs from 1 to s. What I would have here is the
force that will come from this s plus 1 particle, acting on
particle n the same way that this force was deriving
the change of the momentum, this force will derive the
change of the momentum of-- I guess I put an m here-- The thing that I
have to put here is now a density that also
keeps track of the probability to find the s plus 1 particle
in the location in phase place that I need to
integrate with both. I have to integrate
over all positions. One particle is moving along
a straight line by itself, let's say. Then there are all of the
other particles in the system. I have to ask, what is
the possibility that there is a second particle with
some particular momentum and coordinate that I
will be interacting with. This is the general set up
of these D-B-G-K-Y hierarchy of equations. At this stage, we
really have just rewritten what we had for
the Liouville equation. We said, I'm really, really
interested only one particle [? thing, ?] row one and F1. Let's focus on that. Let's write those
equations in more detail In the first equation, I
have that the explicit time dependence, plus the time
dependence of the position coordinate, plus the time
dependence of the momentum coordinate, which is driven
by the external force, acting on this one
particle density, which is dependent on
p1, q1 at time t. On the right hand
side of the equation. I need to worry about a second
particle with momenta P2 at position Q2 that
will, therefore, be able to exert a force. Once I know the position,
I can calculate the force that particle exerts. What was my notation? The order was 2 and
1, dotted by d by dP1. I need now f2, p1, q2 at time t. We say, well, this
is unfortunate. I have to worry about
dependence on F2, but maybe I can get
away with things by estimating
order of magnitudes of the various terms. What is the left hand
side set of operations? The left hand side
set of operations describes essentially one
particle moving by itself. If that particle has to cross
a distance of this order of L, and I tell you that the typical
velocity of these particles is off the order of V, then
that time scale is going to be of the order of L
over V. The operations here will give me
a V over L, which is what we call the
inverse of Tau u. This is a reasonably
long macroscopic time. OK, that's fine. How big is the right hand side? We said that the
right hand side has something to do with collisions. I have a particle in my system. Let's say that particle has
some characteristic dimension that we call d. This particle is moving with
velocity V. Alternatively, you can think of this
particle as being stationary, and all the other
particles are coming at it with some velocity V. If I say that the density
of these particles is n, then the typical time for which,
as I shoot these particles, they will hit this
target is related to V squared and V, the
volume of particles. Over time t, I have to
consider this times V tau x. V tau xn V squared should
be of the order of one. This gave us a
formula for tau x. The inverse of tau
x that controls what's happening on this
side is n V squared V. Is the term on the right
hand side more important, or the term on the
left hand side? The term on the
right hand side has to do with the two body term. There's a particle
that is moving, and then there's
another particle with a slightly different
velocity that it is behind it. In the absence of
collisions, these particles would just go along
a straight line. They would bounce off the
walls, but the magnitude of their energy,
and hence, velocity, would not change from
these elastic collisions. But if the particles
can catch up and interact, which is governed
by V2, V on the other side, then what happens is that the
particles, when they interact, would collide and
go different ways. Quickly, their velocities,
and momenta, and everything would get mixed up. How rapidly that happens depends
on this collision distance, which is much less than
the size of the system, and, therefore, the term that
you have on the right hand side in magnitude is
much larger than what is happening on
the left hand side. There is no way in
order to describe the relaxation of the gas
that I can neglect collisions between gas particles. If I neglect collisions
between gas particles, there is no reason why
the kinetic energies of individual particles
should change. They would stay
the same forever. I have to keep this. Let's go and look at the second
equation in the hierarchy. What do you have? You have d by dT, P1 over m d
by d Q1, P2 over m, P d by d Q2. Then we have F1 d
by d Q1, plus F2, d by d Q2 coming from
the external potential. Then we have the force
that the involves the collision between
particles one and two. When I write down the
Hamiltonian for two particles, there is going to be
already for two particles and interactions between them. That's where the
F1 2 comes from. F1 2 changes d by the
momentum of particle one. I should write, it's
2 1 that changes momentum of particle two. But as 2 1 is simply minus F1
2, I can put the two of them together in this fashion. This acting on F2 is
then equal to something like integral over V3, F3 1,
d by dP1, plus F3 2, d by dP2. [INAUDIBLE] on F3 P1
and Q3 [INAUDIBLE]. Are we going to do this forever? Well, we said, let's
take another look at the magnitude of
the various terms. This term on the
right hand side still involves a collision that
involves a third particle. I have to find that
third particle, so I need to have,
essentially, a third particle within some
characteristic volume, so I have something
that is of that order. Whereas on the
left hand side now, I have a term that
from all perspectives, looks like the kinds of terms
that I had before except that it involves the collision
between two particles. What it describes is the
duration that collision. We said this is of the
order of 1 over tau c, which replaces
the n over there with some characteristic
dimension. Suddenly, this term is very big. We should be able to use that. There was a question. AUDIENCE: On the left hand
side of both of your equations, for F1 and F2, shouldn't
all the derivatives that are multiplied by your
forces be derivatives of the effects of momentum?
[INAUDIBLE] the coordinates? [INAUDIBLE] reasons? PROFESSOR: Let's go back here. I have a function that
depends on P, Q, and t. Then there's the explicit
time derivative, d by dt. Then there is the Q dot here,
which will go by d by dQ. Then there's the P dot term
that will go by d by dP. All of things have to be there. I should have derivatives
in respect to momenta, and derivatives with
respect to coordinate. Dimensions are, of
course, important. Somewhat, what I write
for this and for this should make up for that. As I have written it now,
it's obvious, of course. This has dimensions of Q
over T. The Q's cancel. I would have one over
T. D over Dps cancel. I have 1 over P. Here,
dimensionality is correct. I have to just make sure
I haven't made a mistake. Q dot is a velocity. Velocity is momentum
divided by mass. So that should
dimensionally work out. P dot is a force. Everything here is force. In a reasonable coordinate-- AUDIENCE: [INAUDIBLE] PROFESSOR: What did I do here? I made mistakes? AUDIENCE: [INAUDIBLE] PROFESSOR: Why didn't
you say that in a way that-- If I don't
understand the question, please correct me before I
spend another five minutes. Hopefully, this is now
free of these deficiencies. This there is very big. Now, compared to the
right hand side in fact, we said that the
right hand side is smaller by a factor that
measures how many particles are within an
interaction volume. And for a typical
gas, this would be a number that's of the
order of 10 to the minus 4. Using 10 to the minus
4 being this small, we are going to set the
right hand side to zero. Now, I don't have to
write the equation for F2. I'll answer a question here that
may arise, which is ultimately, we will do sufficient
manipulations so that we end up with a
particular equation, known as the Boltzmann
Equation, that we will show does not obey
the time reversibility that we wrote over here. Clearly, that is built in to the
various approximations I make. The first question
is, the approximation that I've made here,
did I destroy this time reversibility? The answer is no. You can look at this
set of equations, and do the manipulations
necessary to see what happens if P goes to minus
P. You will find that you will be able to reverse your
trajectory without any problem. Yes? AUDIENCE: Given that it
is only an interaction from our left side
that's very big, that's the reason why we can
ignore the stuff on the right. Why is it that we
are then keeping all of the other terms that
were even smaller before? PROFESSOR: I will ignore them. Sure. AUDIENCE: [LAUGHTER] PROFESSOR: There was the
question of time reversibility. This term here has to do
with three particles coming together, and how
that would modify what we have for just
two-body collisions. In principle, there
is some probability to have three particles
coming together and some combined interactions. You can imagine some
fictitious model, which in addition to these
two-body interactions, you cook up some body
interaction so that it precisely cancels what
would have happened when three particles
come together. We can write a computer
program in which we have two body conditions. But if three bodies come
close enough to each other, they essentially become ghosts
and pass through each other. That computer program
would be fully reversible. That's why sort of
dropping this there is not causing any
problems at this point. What is it that you
have included so far? What we have is a situation
where the change in F1 is governed by a
process in which I have a particle that I
describe on the left hand side with momentum one, and
it collides with some particle that I'm integrating over, but
in some particular instance of integration, has momentum P2. Presumably they come
close enough to each other so that afterwards, the
momenta have changed over so that I have some P1 prime,
and I have some P2 prime. We want to make sure that we
characterize these correctly. There was a question about
while this term is big, these kinds of terms are small. Why should I basically
bother to keep them? It is reasonable. What we are following here are
particles in my picture that were ejected by the
first box, and they collide into each
other, or they were colliding in the first box. As long as you are away from
the [? vols ?] of the container, you really don't care
about these terms. They don't really
moved very rapidly. This is the process of
collision of two particles, and it's also the same process
that is described over here. Somehow, I should be able
to simplify the collision process that is going on
here with the knowledge that the evolution
of two particles is now completely deterministic. This equation by itself
says, take two particles as if they are the only
thing in the universe, and they would follow some
completely deterministic trajectory, that if you
put lots of them together, is captured through
this density. Let's see whether we can
massage this equation to look like this equation. Well, the force term, we
have, except that here we have dP by P1 here. We have d by dP
1 minus d by dP2. So let's do this. Minus d by dP2, acting on F2. Did I do something wrong? The answer is no, because I
added the complete derivative over something that
I'm integrating over. This is perfectly
legitimate mathematics. This part now looks like this. I have to find what is the
most important term that matches this. Again, let's think
about this procedure. What I have to make
sure of is what is the extent of the
collision, and how important is the collision? If I have one
particle moving here, and another particle off there,
they will pass each other. Nothing interesting
could happen. The important thing is how
close they come together. It Is kind of
important that I keep track of the relative
coordinate, Q, which is Q2 minus Q1,
as opposed to the center of mass coordinate, which
is just Q1 plus Q2 over 2. That kind of also
indicates maybe it's a good thing for me to
look at this entire process in the center of mass frame. So this is the lab frame. If I were to look at this
same picture in the center of mass frame,
what would I have? In the center of mass frame, I
would have the initial particle coming with P1 prime, P1
minus P center of mass. The other particle that
you are interacting with comes with P2 minus
P center of mass. I actually drew these
vectors that are hopefully equal and opposite,
because you know that in the center of
mass, one of them, in fact, would be P1 minus P2 over 2. The other would be
P2 minus P1 over 2. They would, indeed,
in the center of mass be equal and opposite momenta. Along the direction
of these objects, I can look at how close
they come together. I can look at some
coordinate that I will call A, which measures
the separation between them at some instant of time. Then there's another
pair of coordinates that I could put into a vector
that tells me how head to head they are. If I think about they're
being on the center of mass, two things that are
approaching each other, they can either
approach head on-- that would correspond
to be equal to 0-- or they could be slightly
off a head-on collision. There is a so-called
impact parameter B, which is a measure
of this addition fact. Why is that going to
be relevant to us? Again, we said that there
are parts of this expression that all of the
order of this term, they're kind of
not that important. If I think about the collision,
and what the collision does, I will have forces that
are significant when I am within this
range of interactions, D. I really have to look at
what happens when the two things come close to each other. It Is only when this relative
parameter A has approached D that these particles
will start to deviate from their straight
line trajectory, and presumably go, to say in
this case, P2 prime minus P center of mass. This one occurs
[? and ?] will go, and eventually P1 prime
minus P center of mass. These deviations will
occur over a distance that is of the order of
this collision and D. The important changes that
occur in various densities, in various
potentials, et cetera, are all taking place when this
relative coordinate is small. Things become big when the
relative coordinate is small. They are big as a function
of the relative coordinate. In order to get big
things, what I need to do is to replace these
d by dQ's with the corresponding
derivatives with respect to the center of mass. One of them would come
be the minus sign. The other would come
be the plus sign. It doesn't matter
which is which. It depends on the
definition, whether I make Q2 minus Q1,
or Q1 minus Q2. We see that the big
terms are the force that changes the momenta
and the variations that you have over these
relative coordinates. What I can do now
is to replace this by equating the two big
terms that I have over here. The two big terms
are P2 minus P1 over m, dotted by d by dQ of F2. There is some other
approximation that I did. As was told to me before,
this is the biggest term, and there is the
part of this that is big and compensates for that. But there are all these
other bunches of terms. There's also this d by dt. What I have done
over here is to look at this slightly coarser
perspective on time. Increasing all the equations
that I have over there tells me everything
about particles approaching each
other and going away. I can follow through
the mechanics precisely everything
that is happening, even in the vicinity
of this collision. If I have two squishy balls,
and I run my hand through them properly, I can
see how the things get squished then released. There's a lot of
information, but again, a lot of information
that I don't really care to know as far as
the properties of this gas expansion process is concerned. What you have done is to forget
about the detailed variations in time and space that
are taking place here. We're going to shortly make
that even more explicit by noting the following. This integration over
here is an integration over phase space of
the second particle. I had written before d
cubed, P2, d cubed, Q2, but I can change
coordinates and look at the relative
coordinate, Q, over here. What I'm asking is,
I have one particle moving through the gas. What is the chance that
the second particle comes with momentum P2, and the
appropriate relative distance Q, and I integrate over both the
P and the relative distance Q? This is the quantity
that I have to integrate. Let's do one more
calculation, and then we will try to give a
physical perspective. In this picture of the center
of mass, what did I do? I do replaced the coordinate, Q,
with a part that was the impact parameter, which
had two components, and a part that was
the relative distance. What was this relative distance? The relative
distance was measured along this line that was
giving me the closest approach. What is the direction
of this line? The direction of this
line is P1 minus P2. This is P1 minus P2 over 2. It doesn't matter. The direction is P1 minus P2. What I'm doing here is I am
taking the derivative precisely along this line of
constant approach. I'm taking a derivative, and
I'm integrating along that. If I were to rewrite the
whole thing, what do I have? I have d by dt, plus P1 over m,
d by dQ1, plus F1, d by dP1-- don't make a mistake--
acting on F1, P1, Q1, t. What do I have to write
on the right hand side? I have an integral
over the momentum of this particle with which
I'm going to make a collision. I have an integral over
the impact parameter that tells me the distance
of closest approach. I have to do the
magnitude of P2 minus P1 over n, which is
really the magnitude of the relative velocity
of the two particles. I can write it as P2
minus P1, or P1 minus P2. These are, of course, vectors. and I look at the modulus. I have the integral
of the derivative. Very simply, I will
write the answer as F2 that is evaluated
at some large distance, plus infinity minus F2
evaluated at minus infinity. I have infinity. In principle, I have
to integrate over F2 from minus infinity
to plus infinity. But once I am beyond
the range of where the interaction changes,
then the two particles just move away forever. They will never see each other. Really, what I should write here
is F2 of-- after the collision, I have P1 prime, P2
prime, at some Q plus, minus F2, P1, P2, at
some position minus. What I need to do is to
do the integration when I'm far away from the
collision, or wait until I am far
after the collision. Really, I have to just
integrate slightly below, after, and before the collision occurs. In principle, if I
just go a few d's in one direction or
the other direction, this should be enough. Let's see physically
what this describes. There is a connection
between this and this thing that I
had over here, in fact. This equation on the left
hand side, if it was zero, it would describe
one particle that is just moving by itself until
it hits the wall, at which point it basically
reverses its trajectory, and otherwise goes forward. But what you have on
the right hand side says that suddenly
there could be another particle with
which I interact. Then I change my direction. I need to know the
probability, given that I'm moving
with velocity P1, that there is a second
particle with P2 that comes close enough. There is this additional factor. From what does this
additional factor come? It's the same factor
that we have over here. It is, if you have a
target of size d squared, and we have a set of
bullets with a density of n, the number of collisions that
I get depends both on density and how fast these things go. The time between collisions, if
you like, is proportional to n, and it is also related to
V. That's what this is. I need some kind of a time
between the collisions that I make. I have already
specified that I'm only interested in the
set of particles that have momentum P2 for this
particular [? point in ?] integration, and that they
have this kind of area or cross section. So I replace this
V squared and V with the relative coordinates. This is the corresponding
thing to V squared, and this is really a
two particle density. This is a subtraction. The addition is
because it is true that I'm going with velocity P1,
and practically, any collisions that are significant
will move me off kilter. So there has to be a
subtraction for the channel that was described by P1
because of this collision. This then, is the
addition, because it says that it could be that
there is no particle going in the horizontal direction. I was actually coming along
the vertical direction. Because of the
collision, I suddenly was shifted to move
along this direction. The addition comes from having
particles that would correspond to momenta that somehow,
if I were in some sense to reverse this, and
then put a minus sign, a reverse collision
would create something that was along the
direction of P1. Here I also made
several approximations. I said, what is chief among
them is that basically I ignored the details of the
process that is taking place at scale the order of
d, so I have thrown away some amount of detail
and information. It is, again, legitimate
to say, is this the stage at which you
made an approximation so that the time
reversibility was lost? The answer is still no. If you are careful enough with
making precise definitions of what these Q's are before
and after the collision, and follow what happens if you
were to reverse everything, you'll find that the
equations is fully reversible. Even at this stage, I have
not made any transition. I have made approximations,
but I haven't made something to be time irreversible. That comes at the
next stage where we make the so-called
assumption of molecular chaos. The assumption is
that what's the chance that I have a particle
here and a particle there? You would say, it's
a chance that I have one here and one there. You say that if two of
any P1, P2, Q1, Q2, t is the same thing as the product
of F1, P1, Q1, t, F1, P2, Q2, t. Of course, this assumption
is generally varied. If I were to look
at the probability that I have two particles
as a function of, let's say, the relative
separation, I certainly expect that if
they are far away, the density should be the
product of the one particle densities. But you would say that if the
two particles come to distances that are closer than
their separation d, then the probability and
the range of interaction d-- and let's say the
interaction is highly repulsive like hardcore-- then
the probability should go to 0. Clearly, you can
make this assumption, but up to some degree. Part of the reason we
went through this process was to indeed make sure that
we are integrating things at the locations where
the particles are far away from each other. I said that the range of
that integration over A would be someplace
where they are far apart after the
collision, and far apart before the collision. You have an
assumption like that, which is, in
principle, something that I can insert into that. Having to make a distinction
between the arguments that are appearing in this equation
is kind of not so pleasant. What you are going to do is
to make another assumption. Make sure that everything is
evaluated at the same point. What we will eventually now have
is the equation that d by dt, plus P1 over n, d by dQ1,
plus F1, dot, d by dP1, acting on F1, on
the left hand side, is, on the right hand side,
equal to all collisions in the particle of
momentum P2, approaching at all possible cross
sections, calculating the flux of the
incoming particle that corresponds to
that channel, which is proportional to V2 minus V1. Then here, we subtract the
collision of the two particles. We write that as F1 of
P1 at this location, Q1, t, F1 of t2 at the
same location Q1, t. Then add F1 prime, P1 prime, Q1
t, F1 prime, P2 prime, Q2, t. In order to make the
equation eventually manageable, what you
did is to evaluate all off the coordinates that
we have on the right hand side at the same location, which
is the same Q1 that you specify on the left hand side. That immediately means
that what you have done is you have changed the
resolution with which you are looking at space. You have kind of washed
out the difference between here and here. Your resolution has to
put this whole area that is of the order of d squared
or d cubed in three dimensions into one pixel. You have changed the
resolution that you have. You are not looking at things
at this [? fine ?] [? state. ?] You are losing additional
information here through this change of
the resolution in space. You have also lost
some information in making the assumption that
the two [? point ?] densities are completely within always
as the product one particle densities. Both of those things
correspond to taking something that is very precise
and deterministic, and making it kind of vague
and a little undefined. It's not surprising then,
that if you have in some sense changed the precision of
your computer-- let's say, that is running the particles
forward-- at some point, you've changed the resolution. Then you can't
really run backward. In fact, to sort of precisely
be able to run the equations forward and backward,
you would need to keep resolution
at all levels. Here, we have sort of removed
some amount of resolution. We have a very good guess
that the equation that you have over here no longer
respects time reversal inversions that you
had originally posed. Our next task is to prove
that you need this equation. It goes in one particular
direction in time, and cannot be drawn
backward, as opposed to all of the predecessors that I
had written up to this point. Are there any questions? AUDIENCE: [INAUDIBLE] PROFESSOR: Yes, Q prime
and Q1, not Q1 prime. There is no dash. AUDIENCE: Oh, I see. It is Q1. PROFESSOR: Yes, it is. Look at this equation. On the left hand side,
what are the arguments? The arguments are P1 and Q1. What is it that I have
on the other side? I still have P1 and Q1. I have introduced
P1 and b, which is simply an impact parameter. What I will do is
I will evaluate all of these things, always
at the same location, Q1. Then I have P1 and P2. That's part of my story of
the change in resolution. When I write here Q1,
and you say Q1 prime, but what is Q1 prime? Is it Q1 plus b? Is it Q1 minus b? Something like this
I'm going to ignore. It's also legitimate,
and you should ask, what is P1
prime and Q2 prime? What are they? What I have to do, is I
have to run on the computer or otherwise, the
equations for what happens if I have P1
and P2 come together at an impact parameter
that is set by me. I then integrate
the equations, and I find that deterministically,
that collision will lead to some P1
prime and P2 prime. P1 prime and P2 prime are
some complicated functions of P1, P2, and b. Given that you know two
particles are approaching each other at distance d with
momenta P1 P2, in principle, you can integrate
Newton's equations, and figure out with what
momenta they end up. This equation, in fact, hides a
very, very complicated function here, which describes
P1 prime and P2 prime as a function of P1 and P2. If you really needed all of
the details of that function, you would surely be in trouble. Fortunately, we don't. As we shall see shortly, you
can kind of get a lot of mileage without knowing that. Yes, what is your question? AUDIENCE: There
was an assumption that all the interactions
between different molecules are central potentials
[INAUDIBLE]. Does the force of the
direction between two particles lie along the [INAUDIBLE]? PROFESSOR: For the things that
I have written, yes it does. I should have been more precise. I should have put
absolute value here. AUDIENCE: You have
particles moving along one line
towards each other, and b is some arbitrary vector. You have two directions,
so you define a plane. Opposite direction particles
stay at the same plane. Have you reduced-- PROFESSOR: Particles
stay in the same plane? AUDIENCE: If the two particles
were moving towards each other, and also you have
in the integral your input parameter,
which one is [INAUDIBLE]. There's two directions. All particles align,
and all b's align. They form a plane. [? Opposite ?] direction
particles [? stand ?] in the-- PROFESSOR: Yes, they
stand in the same plane. AUDIENCE: My
question is, what is [INAUDIBLE] use the
integral on the right from a two-dimensional
integral [? in v ?] into employing central symmetry? PROFESSOR: Yes, you could. You could, in principle, write
this as b db, if you like, if that's what you want. AUDIENCE: [INAUDIBLE] PROFESSOR: Yes, you
could do that if you have simple enough potential. Let's show that this equation
leads to irreversibility. That you are going to do here. This, by the way, is called
the Boltzmann equation. There's an associated
Boltzmann H-Theorem, which restates the following--
If F of P1, Q1, and t satisfies the above
Boltzmann equation, then there is a quantity H that
always decreases in time, where H is the integral over P
and Q of F1, log of F1. The composition of
irreversibility, as we saw in thermal
dynamics, was that there was a
quantity entropy that was always increasing. If you have calculated
for this system, entropy before for the half
box, and entropy afterwards for the space both boxes
occupy, the second one would certainly be larger. This H is a quantity like
that, except that when it is defined this
way, it always decreases as a function of time. But it certainly is very
much related to entropy. You may have asked,
why did Boltzmann come across such
a function, which is F log F, except that
actually right now, you should know
why you write this. When we were dealing
with probabilities, we introduced the entropy of the
probability distribution, which was related to something
like sum over iPi, log of Pi, with a minus sign. Up to this factor
of normalization N, this F1 really is a
one-particle probability. After this normalization
N, you have a one-particle probability,
the probability that you have occupation
of one-particle free space. This occupation of
one-particle phase space is changing as a
function of time. What this statement says is
that if the one-particle density evolves in time according
to this equation, the corresponding
minus entropy decreases as a function of time. Let's see if that's the case. To prove that, let's do this. We have the formula for H, so
let's calculate the H by dt. I have an integral
over the phase space of particle one, the particle
that I just called one. I could have
labeled it anything. After integration, H is
only a function of time. I have to take the
time derivative. The time derivative
can act on F1. Then I will get the F1
by dt, times log F1. Or I will have F1 times
the derivative of log F1. The derivative of log F1
would be dF1 by dt, and then 1 over F1. Then I multiply by F1. This term is simply 1. AUDIENCE: Don't you want to
write the full derivative, F1 with respect [INAUDIBLE]? PROFESSOR: I thought we
did that with this before. If you have something that
I am summing over lots of [? points, ?] and these
[? points ?] can be positioned, then I have S at location
one, S at location two, S at location three,
discretized versions of x. If I take the time
derivative, I take the time derivative
of this, plus this, plus this, which are
partial derivatives. If I actually take the
time derivative here, I get the integral d cubed P1,
d cubed Q1, the time derivative. This would be that
partial dF1 by dt is the time derivative
of n, which is 0. The number of particles
does not change. Indeed, I realize that 1
integrated against dF1 by dt is the same thing that's here. This term gives you 0. All I need to worry
about is integrating log F against the Fydt. I have an integral over P1 and
Q1 of log F against the Fydt. We have said that F1 satisfies
the Boltzmann equation. So the F1 by dt, if I
were to rearrange it, I have the F1 by dt. I take this part to the
other side of the equation. This part is also
the Poisson bracket of a one-particle H with F1. If I take it to
the other side, it will be the Poisson
bracket of H with F1. Then there is this
whole thing that involves the collision
of two particles. So I define whatever is
on the right hand side to be some collision
operator that acts on two [? powers ?] of F1. This is plus a collision
operator, F1, F1. What I do is I
replace this dF1 by dt with the Poisson bracket of H,
or H1, if you like, with F1. The collision operator I will
shortly write explicitly. But for the time being, let
me just write it as C of F1. There is a first
term in this sum-- let's call it number one--
which I claim to be 0. Typically, when you
get these integrations with Poisson brackets,
you would get 0. Let's explicitly show that. I have an integral over
P1 and Q1 of log of F1, and this Poisson
bracket of H1 and F1, which is essentially
these terms. Alternatively, I could write
it as dH1 by dQ1, dF1 by dt1, minus the H1 by
dt1, dF1, by dQ1. I've explicitly written this
form for the one-particle in terms of the Hamiltonian. The advantage of
that is that now I can start doing
integrations by parts. I'm taking derivatives
with respect to P, but I have integrations
with respect to P here. I could take the F1 out. I will have a minus. I have an integral, P1, Q1. I took F1 out. Then this d by dP1 acts on
everything that came before it. It can act on the H1. I would get d2 H1 with
respect to dP1, dQ1. Or it could act on the
log of F1, in which case I will get set dH1 by dQ1. Then I would have d
by dP acting on log of F, which would
give me dF1 by dP1, then the derivative of the
log, which is 1 over F1. This is only the first term. I also have this term, with
which I will do the same thing. AUDIENCE: [INAUDIBLE] The
second derivative [INAUDIBLE] should be multiplied
by log of F. PROFESSOR: Yes, it should be. It is Log F1. Thank you. For the next term, I have F1. I have d2 H1, and the other
order of derivatives, dQ1, dP1. Now I'll make sure I
write down the log of F1. Then I have dH1
with respect to dQ1. Then I have a dot product with
the derivative of log F, which is the derivative of F1 with
respect to Q1 and 1 over F1. Here are the terms
that are proportional to the second derivative. The order of the
derivatives does not matter. One often is positive. One often is negative,
so they cancel out. Then I have these
additional terms. For the additional
terms, you'll note that the F1 and the
1 over F1 cancels. These are just a product
of two first derivatives. I will apply the five
parts process one more time to get rid of the derivative
that is acting on F1. The answer becomes plus
d cubed P1, d cubed Q1. Then I have F1, d2 H1, dP1,
dQ1, minus d2 H1, dQ1, dP1. These two cancel each other
out, and the answer is 0. So that first term vanishes. Now for the second term,
number two, what I have is the first term vanished. So I have the H by dt. It is the integral
over P1 and Q1. I have log of F1. F1 is a function of
P1, and Q1, and t. I will focus, and make sure I
write the argument of momentum, for reasons that will
become shortly apparent. I have to multiply with
the collision term. The collision term
involves integrations over a second particle,
over an impact parameter, a relative velocity, once I
have defined what P2 and P1 are. I have a subtraction
of F evaluated at P1, F evaluated at
P2, plus addition, F evaluated at P1 prime,
F evaluated at P2 prime. Eventually, this whole thing
is only a function of time. There are a whole bunch of
arguments appearing here, but all of those arguments
are being integrated over. In particular, I have arguments
that are indexed by P1 and P2. These are dummy
variables of integration. If I have a function
of x and y that I'm integrating over x and
y, I can call x "z." I can call y "t." I would integrate
over z and t, and I would have the same answer. I would have exactly
the same answer if I were to call all of the
dummy integration variable that is indexed 1, "2." Any dummy variable
that is indexed 2, if I rename it and call it 1,
the integral would not change. If I do that, what do I have? I have integral
over Q-- actually, let's get of the
integration number on Q. It really doesn't matter. I have the integrals
over P1 and P1. I have to integrate over
both sets of momenta. I have to integrate over
the cross section, which is relative between 1 and 2. I have V2 minus V1, rather
than V1 minus V2, rather than V2 minus V1. The absolute value
doesn't matter. If I were to replace these
indices with an absolute value, [? or do a ?] V2 minus V1
goes to minus V1 minus V2. The absolute value
does not change. Here, what do I have? I have minus F of P1. It becomes F of P2, F of
P1, plus F of P2 prime, f of P1 prime. They are a product. It doesn't really matter in
which order I write them. The only thing
that really matters is that the argument was
previously called F1 of P1 for the log, and now it
will be called F1 of P2. Just its name changed. If I take this, and the
first way of writing things, which are really two ways of
writing the same integral, and just average them, I will
get 1/2 an integral d cubed Q, d cubed P1, d cubed P2,
d2 b, and V2 minus V1. I will have F1 of P1, F1 of
P2, plus F1 of P1 prime, F1 of P2 prime. Then in one term, I
had log of F1 of P1, and I averaged it
with the other way of writing things, which was
log of F-- let's put the two logs together, multiplied by F1. So the sum of the
two logs I wrote, that's a log of the product. I just rewrote that equation. If you like, I symmetrized It
with respect to index 1 and 2. So the log of 1,
that previously had one argument through
this symmetrization, became one half
of the sum of it. The next thing one has to
think about, what I want to do, is to replace primed and
unprimed coordinates. What I would
eventually write down is d cubed P1 prime, d cubed P2
prime, d2 b, V2 prime minus V1 prime, minus F1 of P1 prime,
F1 of P2 prime, plus F1 of P1, F1 of P2. Then log of F1 of P1
prime, F1 of P2 prime. I've symmetrized originally
the indices 1 and 2 that were not quite
symmetric, and I end up with an expression that has
variables P1, P2, and functions P1 prime and P2 prime, which
are not quite symmetric again, because I have F's evaluated
for P's, but not for P primes. What does this mean? This mathematical expression
that I have written down here actually is not correct,
because what this amounts to, is to change variables
of integration. In the expression
that I have up here, P1 and P2 are variables
of integration. P1 prime and P2 prime are
some complicated functions of P1 and P2. P1 prime is some complicated
function that I don't know. P1, P2, and V, for which I
need to solve in principle, is Newton's equation. This is similarly for P2 prime. What I have done
is I have changed from my original variables
to these functions. When I write things over here,
now P1 prime and P2 prime are the integration variables. P1 and P2 are supposed
to be regarded as functions of P1
prime and P2 prime. You say, well, what
does that mean? You can't simply
take an integral dx, let's say F of some function of
x, and replace this function. You can't call it
a new variable, and do integral dx prime. You have to multiply with the
Jacobian of the transformation that takes you from the P
variables to the new variables. My claim is that this
Jacobian of the integration is, in fact, the unit. The reason is as follows. These equations that have
to be integrated to give me the correlation are
time reversible. If I give you two momenta, and
I know what the outcomes are, I can write the
equations backward, and I will have the
opposite momenta go back to minus the original momenta. Up to a factor of minus, you
can see that this equation has this character, that P1, P2
go to P1 prime, P2 prime, then minus P1 prime, minus P2
prime, go to P1, and P2. If you sort of
follow that, and say that you do the
transformation twice, you have to get back up
to where a sign actually disappears to where you want. You have to multiply
by two Jacobians, and you get the same unit. You can convince yourself that
this Jacobian has to be unit. Next time, I guess we'll
take it from there. I will explain this
stuff a little bit more, and show that this implies what
we had said about the Boltzmann equation.