The following
content is provided
under a Creative
Commons license.
Your support will help MIT
OpenCourseWare continue
to offer high quality
educational resources for free.
To make a donation or
view additional materials
from hundreds of MIT courses,
visit MIT OpenCourseWare
at ocw.mit.edu.
PROFESSOR: So the
question is, how do we
describe the various motions
that are taking place?
And in principle, the
Boltzmann equation
that we have been
developing should
be able to tell us
something about all of that.
Because we put essentially
all of the phenomena
that we think are relevant to
a dilute gas, such as what's
going on in this room,
into that equation,
and reduced the equation
to something that
had a simple form where
on the left hand side
there was a set of derivatives
acting on the one particle,
probability as a function
of position, and momentum,
let's say, for the
gas in this room.
And on the right hand side,
we had the collision operator.
And again, to get the
notation straight,
this set of operations
on the left hand side
includes a time
derivative, a thing
that involves the velocity
causing variations
in the coordinate, and any
external forces causing
variations in momentum.
I write that because I will,
in order to save some space,
use the following notation in
the remainder of this lecture.
I will use del sub t to
indicate partial derivatives
with respect to time.
This term I will indicate
as p alpha over m d alpha.
So d sub alpha stands for
derivative with respect
to, say, x, y, and
z-coordinates of position.
And this summation over the
repeated index is implied here.
And similarly here, we would
write F alpha d by dp alpha.
We won't simplify that.
So basically, L is also this
with the summation implicit.
Now, the other
entity that we have,
which is the right hand
side of the equation,
had something to
do with collisions.
And the collision operator
Cff was an integral
that involved bringing
in a second coordinate,
or second particle,
with momentum p2,
would come from a location such
that the impact parameter would
be indicated by B.
We would need to know
the flux of incoming particles.
So we had the
relative velocities.
And then there was a term that
was essentially throwing you
off the channel that
you were looking at.
And let's say we indicate the
variable here rather than say p
by p1.
Then I would have here
F evaluated at p1,
F evaluated at p2, p2 being this
momentum that I'm integrating.
And then there was the
addition from channels
that would bring in probability.
So p1 prime and p2
prime, their collision
would create particles
in the channel p1, p2.
And there was some
complicated relation
between these functions
and these functions
for which you have to
solve Newton's equation.
But fortunately,
for our purposes,
we don't really
need all of that.
Again, all of this is evaluated
at the same location as the one
that we have over here.
Now let's do this.
Let's take this, which is a
function of some particular q
and some particular p1,
which is the things that we
have specified over there.
Let's multiply it with some
function that depends on p1, q,
and potentially t, and
integrate it over P1.
Once I have done that, then
the only thing that I have left
depends on q and t,
because everything else
I integrated over.
But in principle, I made
a different integration.
I didn't have the
integration over q.
So eventually, this thing
will become a function of q.
OK, so let's do the same
thing on the right hand
side of this equation.
So what I did was I added
the integration over p1.
And I multiplied
by some function.
And I'll remember that
it does depend on q.
But since I haven't
written the q argument,
I won't write it here.
So it's chi of p1.
So I want to-- this
quantity j that I
wrote is equal to this
integral on the right.
Now, we encountered
almost the same integral
when we were looking for
the proof of the H-theorem
where the analog
of this chi of p1
was in fact log of
f evaluated at p1.
And we did some manipulations
that we can do over here.
First of all, here, we have
dummy integration variables
p1 and p2.
We can just change their name
and then essentially average
over those two possibilities.
And the other thing that
we did was this equation
has the set of things that
come into the collision,
and set the things
that, in some sense,
got out of the collision,
or basically things that,
as a result of these
collisions, create these two.
So you have this symmetry
between the initiators
and products of the collision.
Because essentially
the same function
describes going one
way and inverting
things and going backwards.
And we said that in
principle, I could
change variables of integration.
And the effect of doing that
is kind of moving the prime
coordinates to the things
that don't have primes.
I don't know how last time I
made the mistake of the sign.
But it's clear that if I
just put the primes from here
to here, there will
be a minus sign.
So the result of doing
that symmetrization
should be a minus chi of p1
prime minus chi of p2 prime.
And again, to do the averaging,
I have to put something else.
Now, this statement is
quite generally true.
Whatever chi I choose, I
will have this value of j
as a result of that integration.
But now we are going to
look at something specific.
Let's assume that we
have a quantity that
is conserved in collision.
This will be 0 for collision
conserved quantity.
Like let's say if my chi
that I had chosen here
was some component
of momentum px,
then whatever some of
the incoming momenta
will be the sum of
the outgoing momenta.
So essentially
anything that I can
think of that is conserved in
the collision, this function
that relates p
primes to p1 and p2
has the right property to ensure
that this whole thing will
be 0.
And that's actually really
the ultimate reason.
I don't really
need to know about
all of these cross sections
and all of the collision
properties, et cetera.
Because my focus
will be on things
that are conserved
in collisions.
Because those are the variables
that are very slowly relaxing,
and the things that
I'm interested in.
So what you have is that for
these collision conserved
quantities, which is the
things that I'm interested in,
this equation is 0.
Now, if f satisfies
that equation,
I can certainly substitute
over here for Cff Lf.
So if f satisfies
that equation, and I
pick a collision
conserved quantity,
the integral over p1 of that
function of the collision
conserved quantity
times the bunch
of first derivatives
acting on f has to be 0.
So this I can write
in the following way--
0 is the integral over p1.
Actually, I have
only one momentum.
So let's just ignore
it from henceforth.
The other momentum I
just introduced in order
to be able to show that
when integrated against
the collision operator,
it will give me 0.
I have the chi, and then this
bunch of derivatives dt plus p
alpha over m p alpha plus
f alpha d by dp alpha
acting on f.
And that has to be 0.
STUDENT: So alpha
stands for x, y, z?
PROFESSOR: Yes, alpha stands
for the three components
x, y, and z throughout
this lecture.
And summation over a
repeated index is assumed.
All right, so now
what I want to do
is to move this chi so
that the derivatives act
on both of them.
So I'll simply write
the integral of dqp--
if you like, this
bunch of derivatives
that we call L acting on
the combination chi f.
But a derivative of f
chi gives me f prime chi.
But also it keeps me chi prime
f that I don't have here.
So I have to subtract that.
And why did I do that?
Because now I end
up with integrals
that involve integrating
over f against something.
So let's think about
these typical integrals.
If I take the integral over
momentum of f of p and q
and t-- remember, f was
the one particle density.
So I'm integrating, let's
say, at a particular position
in space over all momentum.
So it says, I don't
care what momentum.
I just want to know whether
there's a particle here.
So what is that quantity?
That quantity is simply the
density at that location.
Now suppose I were
to integrate this
against some other
function, which
could depend on p, q,
and t, for example?
I use that to define an average.
So this is going to be
defined to be the average of O
at that location q and t.
So for example, rather than
just calculating whether or not
there are particles
here, I could
be asking, what is the
average kinetic energy
of the particles that are here?
Then I would integrate this
against p squared over 2m.
And this average would give
me the local expectation value
of p squared over m,
just a normalization n
so that it's
appropriately normalized.
So with this definition, I
can write the various terms
that I have over here.
So let me write it a
little bit more explicitly.
What do we have?
We have integral d cubed p.
We have this bunch of
derivatives acting on chi f
minus f times this bunch of
derivatives acting on chi.
So let's now look at
things term by term.
The first term is
a time derivative.
The time derivative I can
take outside the integral.
Once I take the time derivative
outside the integral,
what is left?
What is left is the
integral of chi f,
exactly what I have here.
O has been replaced by chi.
So what I have is
the time derivative
of n expectation value of
chi using this definition.
Let's look at the next term.
The next term, these derivatives
are all over position.
The integration
is over momentum.
I can take it outside.
So I can write it as d alpha.
And then I have a quantity
that I'm integrating against f.
So I will get n times this
local average of that quantity.
What's that quantity?
It's p alpha over m times chi.
What's the third term?
The third term is actually
an integral over momentum.
But I'm integrating
over momentum.
So again, you can sort of
remove things to boundaries
and convince yourself that
that integral will not
give you a contribution.
The next bunch of terms are
simply directly this-- f
integrated against something.
So they're going
to give me minus n
times the various averages
involved-- d t chi minus n,
the average of p alpha
over m, the alpha chi.
And then f alpha I can actually
take outside, minus n f alpha,
the average of d
chi by d p alpha.
And what we've established
is that that whole thing
is 0 for quantities that are
conserved under collisions.
So why did I do all of that?
It's because solving
the Boltzmann equation
in six dimensional phase space
with all of its integrations
and derivatives is
very complicated.
But really, the things
that are slowly relaxing
are quantities
that are conserved
collisions, such as densities,
average momentum, et cetera.
And so I can focus on
variations of these
through this kind of equation.
Essentially, what that
will allow me to do
is to construct what are known
as hydrodynamic equations,
which describe the time
evolution of slow variables
of your system,
the variables that
are kind of relevant to making
thermodynamic observations,
as opposed to variables
that you would
be interested in
if you're thinking
about atomic collisions.
So what I need to do is
to go into that equation
and pick out my
conserved quantities.
So what are the
conserved quantities,
and how can I describe
them by some chi?
Well, we already saw this
when we were earlier on trying
to find some kind of a solution
to the H by dt equals 0.
We said that log f
has to be the sum
of collision
conserved quantities.
And we identified three
types of quantities.
One of them was related to
the number conservation.
And essentially, what
you have is that 1 plus 1
equals to 1 plus 1.
So it's obvious.
The other is momentum.
And there are three components
of this-- px, py, pz.
And the third one is
the kinetic energy,
which is conserved
in collisions.
In a potential, clearly the
kinetic energy of a particle
changes as a
function of position.
But within the short
distances of the collisions
that we are interested
in, the kinetic energy
is a conserved quantity.
So my task is to insert
these values of chi
into that equation and see
what information they tell me
about the time evolution of
the corresponding conserved
quantities.
So let's do this one by one.
Let's start with
chi equals to 1.
If I put chi equals to
1, all of these terms
that involve derivatives
clearly needed to vanish.
And here, I would get the time
derivative of the density.
And from here, I would get the
alpha n expectation value of p
alpha over m.
We'll give that a name.
We'll call that u alpha.
So I have introduced u
alpha to be the expectation
value of p alpha over m.
And it can in principle depend
on which location in space
you are looking at.
Somebody opens the
door, there will
be a current that
is established.
And so there will
be a local velocity
of the air that would be
different from other places
in the room.
And that's all we have.
And this is equal to 0.
And this is of
course the equation
of the continuity of
the number of particles.
You don't create or
destroy particles.
And so this density
has to satisfy
this nice, simple equation.
We will sometimes rewrite this
slightly in the following way.
This is the derivative
of two objects.
I can expand that and write
it as dt of n plus, let's say,
u alpha d alpha of n.
And then I would
have a term that
is n d alpha u
alpha, which I will
take to the other
side of the equation.
Why have I done that?
Because if I think of n as a
function of position and time--
and as usual, we did
before define a derivative
that moves along
this streamline--
you will have both the implicit
time derivative and the time
derivative because the
stream changes position
by an amount that is
related to velocity.
Now, for the Liouville equation,
we have something like this,
except that the Liouville
equation, the right hand side
was 0.
Because the flows of
the Liouville equation,
the Hamiltonian flows,
were divergenceless.
But in general, for a
compressable system,
such as the gas in this
room, the compressibility
is indicated to a
nonzero divergence of u.
And there's a corresponding
term on the right hand side.
So that's the first
thing we can do.
What's the second
thing we can do?
I can pick p-- let's say p
beta, what I wrote over there.
But I can actually scale it.
If p is a conserved
quantity, p/m
is also a conserved quantity.
Actually, as far as
this chi is concerned,
I can add anything that
depends on q and not p.
So I can subtract the average
value of this quantity.
And this is conserved
during collisions.
Because this part is the
same thing as something
that is related to density.
It's like 1 plus 1
equals to 1 plus 1.
And p over beta
is conserved also.
So we'll call this
quantity that we
will use for our
candidate chi as c beta.
So essentially, it's the
additional fluctuating speed,
velocity that the
particles have,
on top of the average velocity
that is due to the flow.
And the reason maybe
it's useful to do
this is because clearly
the average of c is 0.
Because the average of
p beta over m is u beta.
And if I do that, then clearly
at least the first thing
in the equation I don't
have to worry about.
I have removed one
term in the equation.
So let's put c beta
for chi over here.
We said that the
first term is 0.
So we go and start
with the second term.
What do I have?
I have d alpha expectation
value of c beta.
And then I have p alpha over m.
Well, p alpha over m is going
to be u alpha plus c alpha.
So let's write it in this
fashion-- u alpha plus c
alpha for p alpha over m.
And that's the average
I have to take.
Now let's look at all of these
terms that involve derivatives.
Well, if I want to take a time
derivative of this quantity,
now that I have
introduced this average,
there is a time derivative here.
So the average of the
time derivative of chi
will give me the time
derivative from u beta.
And actually the minus
sign will cancel.
And so I will have plus n, the
expectation value of d-- well,
there's no expectation value
of something like this.
When I integrate over p,
there's no p dependence.
So it's just itself.
So it is n dt of u beta.
OK, what do we have
for the next term?
Let's write it explicitly.
I have n.
p alpha over m I'm writing
as u alpha plus c alpha.
And then I have the position
derivative of c beta.
And that goes over here.
So I will get d alpha of
u beta with a minus sign.
So this becomes a plus.
The last term is
minus n f alpha.
And I have to take a derivative
of this object with respect
to p alpha.
Well, I have a p here.
The derivative of p beta
with respect to p alpha
gives me delta alpha beta.
So this is going to give
me delta alpha beta over m.
And the whole thing is 0.
So let's rearrange
the terms over here.
The only thing that I have
in the denominator is a 1/m.
So let me multiply the
whole equation by m
and see what happens.
This term let's deal with last.
This term, the first term,
becomes nmdt of u beta.
The change in velocity kind
of looks like an acceleration.
But you have to be careful.
Because you can only
talk about acceleration
acting for a particle.
And the particle is
moving with the stream.
OK, and this term will give
me the appropriate derivative
to make it a stream velocity.
Now, when I look at this, the
average of c that appears here
will be 0.
So the term that
I have over there
is u alpha d alpha u beta.
There's no average involved.
It will give me n.
m is common, so I will get
u alpha d alpha u beta,
which is nice.
Because then I can
certainly regard this
as one of these
stream derivatives.
So these two terms,
the stream derivative
of velocity with time,
times mass, mass times
the density to make it
mass per unit volume,
looks like an acceleration.
So it's like mass
times acceleration.
Newton's law, it should
be equal to the force.
And what do we have
if we take this
to the other side
of the equation?
We have f beta.
OK, good, so we have
reproduced Newton's equation.
In this context, if we're
moving along with the stream,
mass times the acceleration of
the group of particles moving
with the stream
is the force that
is felt from the
external potential.
But there's one other term here.
In this term, the
term that is uc
will average to 0, because
the average of u is 0.
So what I will have
is minus-- and I
forgot to write down
an n somewhere here.
There will be an n.
Because all averages
had this additional n
that I forgot to put.
I will take it to the right
hand side of the equation.
And it becomes d
by dq alpha of n.
I multiply the
entire equation by m.
And then I have the
average of c alpha c beta.
So what happened here?
Isn't force just the
mass times acceleration?
Well, as long as you
include all forces involved.
So if you imagine that this
room is totally stationary air,
and I heat one corner
here, then the particles
here will start to
move more rapidly.
There will be more
pressure here.
Because pressure is proportional
to temperature, if you like.
There will be more pressure,
less pressure here.
The difference in pressure
will drive the flow.
There will be an
additional force.
And that's what it says.
If there's variation in these
speeds of the particles,
the change in pressure
will give you a force.
And so this thing, p alpha beta,
is called the pressure tensor.
Yes.
STUDENT: Shouldn't f
beta be multiplied by n,
or is there an n on
the other side of that?
PROFESSOR: There is an n
here that I forgot, yes.
So the n was in the first
equation, somehow got lost.
STUDENT: So the
pressure is coming
from the local fluctuation?
PROFESSOR: Yes.
And if you think about
it, the temperature
is also the local fluctuation.
So it has something to do
with temperature differences.
Pressure is related
to temperature.
So all the things are connected.
And in about two minutes, I'll
actually evaluate that for you,
and you'll see how.
Yes.
STUDENT: Is the pressure
tensor distinct from the stress
tensor?
PROFESSOR: It's
the stress tensor
that you would have for a fluid.
For something more complicated,
like an elastic material,
it would be much more
complicated-- not much more
complicated, but
more complicated.
Essentially, there's
always some kind
of a force per unit
volume depending
on what kind of medium you have.
And for the gas,
this is what it is.
Yes.
STUDENT: So a basic question.
When we say u alpha is
averaged, averaged over what?
Is it by the area?
PROFESSOR: OK, this
is the definition.
So whenever I use this
notation with these angles,
it means that I
integrated over p.
Why do I do that?
Because of this asymmetry
between momenta and collision
and coordinates that is inherent
to the Boltzmann equation.
When we wrote down the
Liouville equation,
p and q were
completely equivalent.
But by the time we
made our approximations
and we talked about
collisions, et cetera,
we saw that momenta
quickly relax.
And so we can look at
the particular position
and integrate over momenta
and define averages
in the sense of when you
think about what's happening
in this room, you think
about the wind velocity
here over there, but not
fluctuations in the momentum
so much, OK?
All right, so this
clearly is kind
of a Navier-Stokes like
equation, if you like,
for this gas.
That tells you how the
velocity of this fluid changes.
And finally, we would need
to construct an equation that
is relevant to the
kinetic energy, which
is something like
p squared over 2m.
And we can follow
what we did over here
and subtract the average.
And so essentially,
this is kinetic energy
on top of the kinetic
energy of the entire stream.
This is clearly the
same thing as mc
squared over 2, c being
the quantity that we
defined before.
And the average of
mc squared over 2
I will indicate by epsilon.
It's the heat content.
Or actually, let's
say energy density.
It's probably better.
So now I have to put mc squared
over 2 in this equation for chi
and do various manipulations
along the lines of things
that I did before.
I will just write
down the final answer.
So the final answer will
be that dt of epsilon.
We've defined dt.
I move with the streamline.
So I have dt plus
u alpha d alpha
acting on this density, which
is a function of position
and time.
And the right hand side of
this will have two terms.
One term is essentially
how this pressure kind
of moves against the velocity,
or the velocity and pressure
are kind of hitting
against each other.
So it's kind of like if you
were to rub two things--
"rub" was the word
I was looking.
If you were to rub two
things against each other,
there's heat that is generated.
And so that's the term
that we are looking at.
So what is this u alpha beta?
u alpha beta-- it's just
because p alpha beta
is a symmetric object.
It doesn't make any difference
if you exchange alpha and beta.
You symmetrize the
derivative of the velocity.
And sometimes it's called
the rate of strain.
And there's another
term, which is
minus 1/n d alpha of h alpha.
And for that, I need to define
yet another quantity, this h
alpha, which is nm over 2 the
average of 3 c's, c squared
and then c alpha.
And this is called
the heat transport.
So for a simpler
fluid where these
are the only conserved
quantities that I have,
in order to figure out how
the fluid evolves over time,
I have one equation
that tells me
about how the density changes.
And it's related to the
continuity of matter.
I have one equation that tells
me how the velocity changes.
And it's kind of an
appropriately generalized
version of Newton's law in
which mass times acceleration
is equated with
appropriate forces.
And mostly we are interested in
the forces that are internally
generated, because of the
variations in pressure.
And finally, there is
variations in pressure
related to variations
in temperature.
And they're governed by
another equation that tells us
how the local energy density,
local content of energy,
changes as a function of time.
So rather than solving
the Boltzmann equation,
I say, OK, all I need
to do is to solve
these hydrodynamic equations.
Question?
STUDENT: Last time for
the Boltzmann equation,
[INAUDIBLE].
PROFESSOR: What it says is that
conservation laws are much more
general.
So this equation you could
have written for a liquid,
you could have
written for anything.
This equation kind
of looks like you
would have been able to
write it for everything.
And it is true, except
that you wouldn't
know what the pressure is.
This equation you would have
written on the basis of energy
conservation, except
that you wouldn't
know what the heat
transport vector is.
So what we gained through
this Boltzmann prescription,
on top of what
you may just guess
on the basis of
conservation laws,
are expressions for quantities
that you would need in order
to sort these equations, because
of the internal pressures that
are generated because of the
way that the heat is flowing.
STUDENT: And this quantity
is correct in the limit of--
PROFESSOR: In the limit, yes.
But that also really
is the Achilles' heel
of the presentation I have
given to you right now.
Because in order to
solve these equations,
I should be able to
put an expression here
for the pressure, and an
expression here for the h.
But what is my
prescription for getting
the expression for
pressure and h?
I have to do an average
that involves the f.
And I don't have the f.
So have I gained
anything, all right?
So these equations are general.
We have to figure out what to
do for the p and h in order
to be able to solve it.
Yes.
STUDENT: In the
last equation, isn't
that n epsilon
instead of epsilon?
PROFESSOR: mc squared
over 2-- I guess
if I put here mc squared
over 2, probably it is nf.
STUDENT: I think maybe the
last equation it's n epsilon,
the equation there.
PROFESSOR: This equation is OK.
OK, so what you're saying--
that if I directly put chi here
to be this quantity, what I
would need on the left hand
side of the equations
would involve
derivatives of n epsilon.
Now, those derivatives I can
expand, write them, let's say,
dt of n epsilon is epsilon
dt of n plus ndt of epsilon.
And then you can
use these equations
to reduce some of that.
And the reason that I
didn't go through the steps
that I would go
from here to here
is because it
would have involved
a number of those cancellations.
And it would have taken me
an additional 10, 15 minutes.
All right, so
conceptually, that's
the more important thing.
We have to find some way
of doing these things.
Now, when I wrote
this equation, we
said that there is some
kind of a separation of time
scales involved in
that the left hand
side of the equation,
the characteristic times
are order of the time
it takes for a particle
to, say, go over the
sides of the box,
whereas the collision
times, 1 over tau x,
are such that the
right hand side is
much larger than
the left hand side.
So as a 0-th order
approximation,
what I will assume is
that the left hand side
is so insignificant
that I will set it to 0.
And then my approximation
for the collision
is the thing that essentially
sets this bracket to 0.
This is the local equilibrium
that we wrote down before.
So that means that I'm assuming
a 0-th order approximation
to the solution of the
Boltzmann equation.
And very shortly, we
will improve up that.
But let's see what this 0-th
order approximation gives us,
which is-- we saw what it is.
It was essentially something
like a Gaussian in momentum.
But the coefficient out front
of it was kind of arbitrary.
And now that I have defined
the integral over momentum
to be density, I will
multiply a normalized Gaussian
by the density locally.
And I will have an exponential.
And average of p I
will shift by an amount
that depends on position.
And I divide by some parameter
we had called before beta.
But that beta I can
rewrite in this fashion.
So I have just
rewritten the beta
that we had before that
was a function of q and t
as 1 over kBT.
And this has to be
properly normalized.
So I will have 2 pi mkBT,
which is a function of position
to the 3/2.
And you can check that the
form that I have written here
respects the
definitions that I gave,
namely that if I
were to integrate it
over momentum, since
the momentum part is
a normalized Gaussian, I
will just get the density.
If I were to calculate
the average of p/m,
I have shifted the
Gaussian appropriately so
that the average of p/m is the
quantity that I'm calling u.
The other one-- let's check.
Essentially what
is happening here,
this quantity is the same
thing as mc squared over 2kT
if I use the definition of
c that I have over there.
So it's a Gaussian weight.
And from the
Gaussian weight, you
can immediately see
that the average of c
alpha c beta, it's
in fact diagonal.
It's cx squared, cy squared.
So the answer is going
to be delta alpha beta.
And for each
particular component,
I will get kT over m.
So this quantity that
I was calling epsilon,
which was the average
of mc squared over 2,
is essentially
multiplying this by m/2
and summing over
delta alpha alpha,
which gives me a factor of 3.
So this is going
to give me 3/2 kT.
So really, my energy
density is none other
than the local 3/2 kT.
Yes?
STUDENT: So you've just defined,
what is the temperature.
So over all previous
derivations,
we didn't really use the
classical temperature.
And now you define it as sort
of average kinetic energy.
PROFESSOR: Yeah, I have
introduced a quantity T here,
which will indeed eventually
be the temperature
for the whole thing.
But right now, it
is something that
is varying locally from
position to position.
But you can see that the typical
kinetic energy at each location
is of the order of
kT at that location.
And the pressure
tensor p alpha beta,
which is nm expectation
value of c alpha c beta,
simply becomes kT over m--
sorry, nKT delta alpha beta.
So now we can sort of start.
Now probably it's a
better time to think
about this as temperature.
Because we know about the
ideal gas type of behavior
where the pressure of the ideal
gas is simply density times kT.
So the diagonal elements
of this pressure tensor
are the things that
we usually think about
as being the pressure
of a gas, now
at the appropriate
temperature and density,
and that there are no off
diagonal components here.
I said I also need to
evaluate the h alpha.
h alpha involves
three factors of c.
And the way that we have
written down is Gaussian.
So it's symmetric.
So all odd powers
are going to be 0.
There is no heat
transport vector here.
So within this 0-th
order, what do we have?
We have that the total
density variation, which
is dt plus u alpha d
alpha acting on density,
is minus nd alpha u alpha.
That does not involve any of
these factors that I need.
This equation-- let's see.
Let's divide by mn.
So we have Dt of u beta.
And let's again look at what's
happening inside the room.
Forget about boundary conditions
at the side of the box.
So I'm going to write
this essentially
for the case that
is inside the box.
I can forget about
the external force.
And all I'm interested
in is the internal forces
that are generated
through pressure.
So this is dt plus u
alpha d alpha of u beta.
I said let's forget
the external force.
So what do we have?
We have the contribution
that comes from pressure.
So we have minus the alpha.
I divided through by nm.
So let me write it
correctly as 1 over nm.
I have the alpha.
My pressure tensor is
nkT delta alpha beta.
Delta alpha beta and this d
alpha, I can get rid of that
and write it simply as d beta.
So that's the equation
that governs the variations
in the local stream
velocity that you
have in the gas in response
to the changes in temperature
and density that
you have in the gas.
And finally, the equation
for the energy density, I
have dt plus u alpha d alpha.
My energy density is simply
related to this quantity T.
So I can write it as
variations of this temperature
in position.
And what do I have on
the right hand side?
I certainly don't have the
heat transport vectors.
So all I have to do is to take
this diagonal p alpha beta
and contract it with this
strain tensor u alpha beta.
So the only term that I'm
going to get after contracting
delta alpha beta is going
to be d alpha u alpha.
So let's make sure that
we get the factors right.
So I have minus p alpha
beta is nkT d alpha u alpha.
So now we have a closed
set of equations.
They tell me how the density,
temperature, and velocity
vary from one location to
another location in the gas.
They're completely closed.
That's the only set of
things that come together.
So I should be able
to now figure out,
if I make a disturbance
in the gas in this room
by walking around, by
talking, by striking a match,
how does that eventually,
as a function of time,
relax to something
that is uniform?
Because our expectation is
that these equations ultimately
will reach equilibrium.
That's essentially the
most important thing
that we deduce from
the Boltzmann equation,
that it was allowing things
to reach equilibrium.
Yes.
STUDENT: For the second
equation, that's alpha?
The right side of
the second equation?
PROFESSOR: The alpha
index is summed over.
STUDENT: The right side.
Is it the derivative
of alpha or beta?
Yeah, that one.
PROFESSOR: It is beta.
Because, you see, the only
index that I have left is beta.
So if it's an index by
itself, it better be beta.
How did this index d
alpha become d beta?
Because the alpha beta
was delta alpha beta.
STUDENT: Also, is it
alpha or is it beta?
PROFESSOR: When I sum over alpha
of d alpha delta alpha beta,
I get d beta.
Yes.
STUDENT: Can I ask again, how
did you come up with the f0?
Why do you say that option?
PROFESSOR: OK, so
this goes back to what
we did last time around.
Because we saw that
when we were writing
the equation for
the hydt, we came up
with a factor of what that
was-- this multiplying
the difference of the logs.
And we said that what
I can do in order
to make sure that
this equation is
0 is to say that log is additive
in conserved quantities,
so log additive in
conserved quantities.
I then exponentiate it.
So this is log of a number.
And these are all
things that are,
when I take the log,
proportional to p squared
and p, which are the
conserved quantities.
So I know that this
form sets the right hand
side of the Boltzmann
equation to 0.
And that's the largest part
of the Boltzmann equation.
Now what happens is that
within this equation,
some quantities do not
relax to equilibrium.
Some-- let's call
them variations.
Sometimes I will use
the word "modes"--
do not relax to equilibrium.
And let's start
with the following.
When you have a sheer velocity--
what do I mean by that?
So let's imagine that
you have a wall that
extends in the x direction.
And along the y direction, you
encounter a velocity field.
The velocity field is always
pointing along this direction.
So it only has the x component.
There's no y component
or z component.
But this x component
maybe varies
as a function of position.
So my ux is a function of y.
This corresponds to
some kind of a sheer.
Now, if I do that,
then you can see
that the only derivatives
that would be nonzero
are derivatives that are
along the y direction.
But this derivative
along the y direction
in all of these equations has
to be contracted typically
with something else.
It has to be contacted with u.
But the u's have no component
along the y direction.
So essentially, all my
u's would be of this form.
Basically, there will
be something like uy.
Something like this
would have to be 0.
You can see that if I start
with an initial condition
such as that, then the
equations are that dt of n--
this term I have
to forget-- is 0.
Because for this, I
need a divergence.
And this flow has no divergence.
And similarly over here what
I see as dt of the temperature
is 0.
Temperature doesn't change.
And if I assume that I
am under circumstances
in which the
pressure is uniform,
there's also nothing that
I would get from here.
So essentially, this
flow will exist forever.
Yes.
STUDENT: Why does your u
alpha d alpha n term go away?
Wouldn't you get a uxdxn?
PROFESSOR: OK, let's
see, you want a uxdxn.
What I said is
that all variations
are along the y direction.
STUDENT: Oh, so this is
not just for velocity,
but for everything.
PROFESSOR: Yes, so
I make an assumption
about some particular form.
So this is the reasoning.
If these equations bring
everything to equilibrium,
I should be able to pick
any initial condition
and ask, how long does it
take to come to equilibrium?
I pick this specific
type of equation
in which the only variations
for all quantities
are along the y direction.
It's a non-equilibrium state.
It's not a uniform state.
Does it come relax
to equilibrium?
And the answer is
no, it doesn't.
STUDENT: What other properties,
other than velocity,
is given [INAUDIBLE]?
PROFESSOR: Density
and temperature.
So these equations
describe the variations
of velocity, density,
and temperature.
And the statement is, if the
system is to reach equilibrium,
I should be able to start
with any initial configuration
of these three
quantities that I want.
And I see that after a while,
it reaches a uniform state.
Yes.
STUDENT: But if your initial
conditions aren't exactly that,
but you add a
slight fluctuation,
it is likely to grow, and
it will eventually relax.
PROFESSOR: It turns
out the answer is no.
So I'm sort of approaching
this problem from this more
kind of hand-waving perspective.
More correctly,
what you can do is
you can start with some initial
condition that, let's say,
is in equilibrium, and
then do a perturbation,
and ask whether the perturbation
will eventually relax to 0
or not.
And let's in fact do that
for another quantity,
which is the sound mode.
So let's imagine that
we start with a totally
nice, uniform state.
There is zero
velocity initially.
The density is uniform.
The temperature is uniform.
And then what I do
is I will start here.
And I will start talking,
creating a variation that
propagates in this x direction.
So I generated a stream that is
moving along the x direction.
And presumably, as I move
along the x direction,
there is a velocity that changes
with position and temperature.
Now initially, I
had the density.
I said that was uniform.
Once I make this sound, as I
move along the x direction,
and the air is flowing back
and forth, what happens
is that the density will
vary from the uniform state.
And the deviations
from the uniform
state I will indicate by mu.
Similarly, the third
quantity, let's assume,
will have a form such as this.
And currently, I have
written the most general form
of variations that I can
have along the x direction.
You could do it in
different directions.
But let's say for simplicity,
we stick with this.
I haven't told you what
mu theta and u are.
So I have to see what they are
consistent with the equations
that we have up there.
One thing that I will
assume is that these things
are small perturbations
around the uniform state.
And uniform-- sorry, small
perturbations typically
means that what
I intend to do is
to do a linearized
approximation.
So basically, what I will
do is I will essentially
look at the linear version
of these equations.
And again, maybe I didn't
emphasize it before.
Clearly these are
nonlinear equations.
Because let's say
you have u grad u.
It's the same nonlinearity
that you have,
let's say, in
Navier-Stokes equation.
Because you're transporting
something and moving along
with the flow.
But when you do the
linearization, then
these operators that involve
dt plus something like u--
I guess in this case, the
only direction that is varying
is x-- something like this of
whatever quantity that I have,
I can drop this nonlinear term.
Why?
Because u is a perturbation
around a uniform state.
And gradients will pick
up some perturbations
around the uniform state.
So essentially the
linearization amounts
to dropping these
nonlinear components
and some other things that
I will linearizer also.
Because all of these
functions here,
the derivatives act on product
of n temperature over here.
These are all
nonlinear operations.
So let's linearize what we have.
We have that Dt of the density--
I guess when I take the time
derivative, I get n bar the
time derivative of the quantity
that I'm calling mu.
And that's it.
I don't need to worry
about the convective part,
the u dot grad part.
That's second order.
On the right hand
side, what do I have?
I have ndu.
Well, divergence of u is
already the first variation.
So for n, I will take
its 0-th order term.
So I have minus n bar dxux.
The equation for ux, really
the only component that I have,
is dt of ux.
Actually, let's write down
the equation for temperature.
Let's look at this equation.
So I have that dt
acting on 3/2 kB times
T. I will pick up T bar.
And then I would
have dt of theta.
What do I have on
the right hand side?
I have a derivative here.
So everything else here I will
evaluate at the 0-th order
term, so n bar k T bar dxux.
So I can see that I can
certainly divide through n bar
here.
And one of my equations
becomes dt of mu is minus dxux.
But from here, I see
that dxux is also
related once I divide
by kT to 3/2 dt theta.
And I know this to be true.
And I seem to have an additional
factor of 1 over n bar here.
And so I made the
mistake at some point,
probably when I
wrote this equation.
STUDENT: It's the
third equation.
PROFESSOR: Yeah, so
this should not be here.
And that should
not be here means
that I probably
made a mistake here.
So this should be a 1/n, sorry.
There was indeed a 1/n / here.
And there is no factor here.
So we have a relationship
between the time derivatives
of these variations in
density and dx of ux.
Fine, what does the
equation for u tell us?
It tells us that dt of ux
is minus 1 over m n bar.
Because of the derivative,
I can set everything
at the variation.
And what do I have here?
I have d by dx of
n bar kB T bar.
And if I look at the variations,
I have 1 plus mu plus theta.
The higher order
terms I will forget.
Yes.
STUDENT: Shouldn't that
be plus 3/2 dt theta?
PROFESSOR: It
should be plus, yes.
There is a minus sign here.
And that makes it plus.
So the n bar we
can take outside.
This becomes minus kT
over m at space variations
of mu plus theta.
Now, what we do is
that what I have here
is information about the time
derivatives of mu and theta.
And here I have
space derivatives.
So what do I do?
I basically apply an
additionally dt here,
which we'll apply here.
And then we can
apply it also here.
And then we know how dt
of mu and dt of theta
are related to dx of ux.
The minus signs disappear.
I have kB T bar divided by m.
I have dt of mu is dxux.
dt of theta is 2/3 dxux.
So I will get 1 plus
2/3 dx squared of ux.
So the second
derivative of ux in time
is proportional to the second
derivative of ux in space.
So that's the standard
wave equation.
And the velocity that
we have calculated
for these sound waves is
5/3 kB the average T over m.
So that part is good.
These equations tell me that
if I create these disturbances,
there are sound waves, and we
know there are sound waves.
And sound waves will propagate
with some velocity that
is related up to some factors to
the average velocity of the gas
particles.
But what is not good
is that, according
to this equation, if
my waves, let's say,
bounce off perfectly
from the walls,
they will last in
this room forever.
So you should still be hearing
what I was saying last week
and the week before.
And clearly what
we are missing is
the damping that is required.
So the statement is that all
of these equations are fine.
They capture a lot
of the physics.
But there is something
important that
is left out in that
there are some modes--
and I describe two of them
here-- that basically last
forever, and don't
come to equilibrium.
But we said that the Boltzmann
equation should eventually
bring things to equilibrium.
So where did we go wrong?
Well, we didn't solve
the Boltzmann equation.
We solved an approximation
to the Boltzmann equation.
So let's try to do better.
STUDENT: I'm sorry, but
for the last equation,
you took another derivative
with respect to t.
PROFESSOR: Yes, I took a
derivative with respect to t.
And it noted that the
derivative with respect
to t of these
quantities mu is related
to derivative with
respect to x or u.
And there was one other
derivative with respect
to x already, making
it two derivatives.
So this is the kind of
situation that we are facing.
Yes.
STUDENT: Is the 5/3 k in any
way related to the heat capacity
ratio of [INAUDIBLE] gas?
PROFESSOR: Yes,
that's right, yes.
So there are lots
of these things that
are implicit in these questions.
And actually, that 3/2 is
the same thing as this 3/2.
So you can trace a
lot of these things
to the Gaussian distribution.
And they appear in cp
versus cv and other things.
Yes.
STUDENT: Just
clarifying something--
this v is different from
the mu in the top right?
PROFESSOR: Yes, this
is v, and that's mu.
This v is the
velocity of the sound.
So I defined this combination,
the coefficient relating
the second derivatives
in time and space
as the sound velocity.
So let's maybe even-- we
can call it vs. All right?
STUDENT: And how did
you know that that
is the [INAUDIBLE] oscillation,
the solution that you got?
PROFESSOR: Because I know
that the solution to dx
squared anything is
v squared-- sorry,
v squared dx squared anything
is dt squared anything, is
phi is some function
of x minus vt.
That is a pulse that moves
with uniform velocity
is a solution to this equation.
So we want to do better.
And better becomes so-called
first order solution.
Now, the kind of
equation that we
are trying to solve at
the top is something
that its algebraic
analog would be
something like this-- 2 times x.
It's a linear on
the left hand side,
is quadratic on the
right hand side.
Let's write it in
this form-- except
that the typical
magnitude of one side
is much larger than
the other side,
so let's say
something like this.
So if I wanted to
solve this equation,
I would say that unless
x is very close to 2,
this 10 to the 6
will blow things up.
So my x0 is 2.
And that's what we have done.
We've solved, essentially,
the right hand
side of the equation.
But I can get a better
solution by taking 2 and saying
there's a small variation to
that that I want to calculate.
And I substitute that into
the original equation.
On the left hand side, I will
get 2 times 2, 1 plus epsilon.
On the right hand side, I
will get 10 to the sixth.
And then essentially,
I subtract 2
plus 2 epsilon squared from 5.
What do I get?
I will get 4 epsilon
plus 4 epsilon squared.
Then I say that
epsilon is small.
So essentially, I linearize
the right hand side.
I forget about that.
I say that I keep
the epsilon here,
because it's
multiplying 10 to the 6.
But the epsilon on the other
side is multiplying nothing.
So I forget that.
So then I will have my epsilon
to be roughly, I don't know,
2 times 2 divided by
4 times 10 to the 6.
So I have gotten the correction
to my first 0-th order solution
to this first order.
Now we will do exactly
the same thing,
not for our algebraic equation,
but for our Boltzmann equation.
So for the Boltzmann equation,
which was Lf is C of ff,
we said that the right hand
side is larger by a factor of 1
over tau x compared
to the left hand side.
And so what we did was
we found a solution f0
that when we put in
the collision integral,
the answer was 0.
Now I want to have a better
solution that I will call f1.
Just like I did
over there, I will
assume that f0 is added
to a small function
that I will call g.
And then I substitute
this equation, this thing,
to the equation.
So when I substitute, I will
get L acting on f0 1 plus g.
Now, what did I do over here?
On the left hand side, I
ignored the first order term.
Because I expect the first
order term to be already small.
And the left hand
side is already small.
So I will ignore this.
And on the right hand
side, I have to linearize.
So I have to put f0 1
plus g, f0 1 plus g.
Essentially what I have to
do is to go to the collisions
that I have over here and
write for this f0 1 plus g.
There are four of such things.
Now, the 0-th order
term already cancels.
Because f0 f0 was f0 f0 by the
way that I constructed things.
And then I can
pull out one factor
of f0 out of the integration.
So when I linearize
this, what I will get
is something like f0
that goes on the outside.
I have the integral
d2p2 d2b v2 minus v1.
And then I have something like
g of p1 plus g of p2 minus
g of p1 prime minus
g of p2 prime.
So basically, what
we have done is
we have defined a linearized
version of the collision
operator that is now
linear in this variable g.
Now in general,
this is also still,
although a linear
operator, much simpler
than the previous
quadratic operator--
still has a lot of junk in it.
So we are going to
simply state that I
will use a form of this
linearized approximation that
is simply g over tau, get rid
of all of the integration.
And this is called the single
collision time approximation.
So having done
that, what I have is
that the L acting on f0
on the left hand side
is minus f0 g over tau x
on the right hand side.
And so I can
immediately solve for g.
Because L is a first
order derivative operator.
My g is minus tau x,
the Liouville operator
acting on f0-- sorry, log of f0.
I divide it through by f0.
So derivative of
f0 divided by f0
is the derivative
acting on log of f0.
So all I need to do
is to essentially do
the operations involved in
taking these derivatives.
Let's say we forget
about the force.
Because we are looking
in the middle of the box,
acting on log of what
I had written before.
So what I have is log of n minus
p minus mu squared over 2mkT.
Remember, ukTn are all
functions of position.
So there will be
derivatives involved here.
And I will just write
down what the answer is.
So the answer
becomes minus tau x.
You would have, once you
do all of these derivatives
and take advantage
of the equations
that you have written
before-- so there's
some lines of algebra involved.
The final answer is going to
be nm c alpha c beta minus--
I should really look at this.
m over kT c alpha c beta
minus delta alpha beta over 3
c squared u alpha
beta, and then mc
squared over 2kT minus 5/2 c
alpha over T d alpha of T. Yes.
STUDENT: Sorry, what's
the thing next to the c
squared, something alpha beta?
PROFESSOR: Delta
alpha beta, sorry.
So there is a
well-defined procedure--
it's kind of algebraically
involved-- by which
more or less in the
same fashion that you
can improve on the
algebraic solution,
get a better
solution than the one
that we had that
now knows something
about these relaxations.
See, the 0-th order
solution that we had
knew nothing about tau x.
We just set tau x
to be very small,
and set the right
hand side to 0.
And then nothing relaxed.
Now we have a better solution
that involves explicitly tau x.
And if we start with
that, we'll find
that we can get relaxation
of all of these modes
once we calculate p alpha beta
and h alpha with this better
solution.
We can immediately,
for example, see
that this new solution will
have terms that are odd.
There is c cubed term here.
So when you're
evaluating this average,
you will no longer get 0.
Heat has a chance to flow
with this improved equation.
And again, whereas
before our pressure
was diagonal because
of these terms,
we will have off
diagonal terms that
will allow us to
relax the shear modes.
And we'll do that next.
