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PROFESSOR: So the
question is, how do we

00:00:22.570 --> 00:00:26.800
describe the various motions
that are taking place?

00:00:26.800 --> 00:00:28.780
And in principle, the
Boltzmann equation

00:00:28.780 --> 00:00:30.460
that we have been
developing should

00:00:30.460 --> 00:00:33.690
be able to tell us
something about all of that.

00:00:33.690 --> 00:00:37.750
Because we put essentially
all of the phenomena

00:00:37.750 --> 00:00:41.940
that we think are relevant to
a dilute gas, such as what's

00:00:41.940 --> 00:00:46.180
going on in this room,
into that equation,

00:00:46.180 --> 00:00:49.700
and reduced the equation
to something that

00:00:49.700 --> 00:00:53.660
had a simple form where
on the left hand side

00:00:53.660 --> 00:00:59.430
there was a set of derivatives
acting on the one particle,

00:00:59.430 --> 00:01:04.160
probability as a function
of position, and momentum,

00:01:04.160 --> 00:01:06.440
let's say, for the
gas in this room.

00:01:06.440 --> 00:01:11.970
And on the right hand side,
we had the collision operator.

00:01:11.970 --> 00:01:17.180
And again, to get the
notation straight,

00:01:17.180 --> 00:01:22.050
this set of operations
on the left hand side

00:01:22.050 --> 00:01:26.775
includes a time
derivative, a thing

00:01:26.775 --> 00:01:32.900
that involves the velocity
causing variations

00:01:32.900 --> 00:01:39.420
in the coordinate, and any
external forces causing

00:01:39.420 --> 00:01:40.585
variations in momentum.

00:01:43.870 --> 00:01:51.910
I write that because I will,
in order to save some space,

00:01:51.910 --> 00:01:56.960
use the following notation in
the remainder of this lecture.

00:01:56.960 --> 00:02:00.490
I will use del sub t to
indicate partial derivatives

00:02:00.490 --> 00:02:02.170
with respect to time.

00:02:02.170 --> 00:02:07.950
This term I will indicate
as p alpha over m d alpha.

00:02:07.950 --> 00:02:11.560
So d sub alpha stands for
derivative with respect

00:02:11.560 --> 00:02:15.450
to, say, x, y, and
z-coordinates of position.

00:02:15.450 --> 00:02:20.290
And this summation over the
repeated index is implied here.

00:02:20.290 --> 00:02:25.230
And similarly here, we would
write F alpha d by dp alpha.

00:02:25.230 --> 00:02:27.630
We won't simplify that.

00:02:27.630 --> 00:02:33.100
So basically, L is also this
with the summation implicit.

00:02:36.030 --> 00:02:38.090
Now, the other
entity that we have,

00:02:38.090 --> 00:02:40.810
which is the right hand
side of the equation,

00:02:40.810 --> 00:02:43.530
had something to
do with collisions.

00:02:43.530 --> 00:02:54.240
And the collision operator
Cff was an integral

00:02:54.240 --> 00:03:00.240
that involved bringing
in a second coordinate,

00:03:00.240 --> 00:03:04.830
or second particle,
with momentum p2,

00:03:04.830 --> 00:03:12.010
would come from a location such
that the impact parameter would

00:03:12.010 --> 00:03:15.340
be indicated by B.
We would need to know

00:03:15.340 --> 00:03:17.115
the flux of incoming particles.

00:03:17.115 --> 00:03:20.070
So we had the
relative velocities.

00:03:20.070 --> 00:03:23.940
And then there was a term that
was essentially throwing you

00:03:23.940 --> 00:03:27.180
off the channel that
you were looking at.

00:03:27.180 --> 00:03:32.230
And let's say we indicate the
variable here rather than say p

00:03:32.230 --> 00:03:34.180
by p1.

00:03:34.180 --> 00:03:38.780
Then I would have here
F evaluated at p1,

00:03:38.780 --> 00:03:44.960
F evaluated at p2, p2 being this
momentum that I'm integrating.

00:03:44.960 --> 00:03:49.770
And then there was the
addition from channels

00:03:49.770 --> 00:03:53.210
that would bring in probability.

00:03:53.210 --> 00:03:55.950
So p1 prime and p2
prime, their collision

00:03:55.950 --> 00:03:59.240
would create particles
in the channel p1, p2.

00:03:59.240 --> 00:04:01.450
And there was some
complicated relation

00:04:01.450 --> 00:04:03.690
between these functions
and these functions

00:04:03.690 --> 00:04:06.580
for which you have to
solve Newton's equation.

00:04:06.580 --> 00:04:09.260
But fortunately,
for our purposes,

00:04:09.260 --> 00:04:11.980
we don't really
need all of that.

00:04:11.980 --> 00:04:16.598
Again, all of this is evaluated
at the same location as the one

00:04:16.598 --> 00:04:17.556
that we have over here.

00:04:21.020 --> 00:04:24.370
Now let's do this.

00:04:24.370 --> 00:04:33.276
Let's take this, which is a
function of some particular q

00:04:33.276 --> 00:04:37.020
and some particular p1,
which is the things that we

00:04:37.020 --> 00:04:40.160
have specified over there.

00:04:40.160 --> 00:04:46.230
Let's multiply it with some
function that depends on p1, q,

00:04:46.230 --> 00:04:52.090
and potentially t, and
integrate it over P1.

00:04:57.220 --> 00:05:01.870
Once I have done that, then
the only thing that I have left

00:05:01.870 --> 00:05:05.630
depends on q and t,
because everything else

00:05:05.630 --> 00:05:07.630
I integrated over.

00:05:07.630 --> 00:05:12.550
But in principle, I made
a different integration.

00:05:12.550 --> 00:05:14.430
I didn't have the
integration over q.

00:05:14.430 --> 00:05:19.170
So eventually, this thing
will become a function of q.

00:05:19.170 --> 00:05:21.610
OK, so let's do the same
thing on the right hand

00:05:21.610 --> 00:05:23.240
side of this equation.

00:05:23.240 --> 00:05:29.370
So what I did was I added
the integration over p1.

00:05:29.370 --> 00:05:32.740
And I multiplied
by some function.

00:05:32.740 --> 00:05:36.770
And I'll remember that
it does depend on q.

00:05:36.770 --> 00:05:39.790
But since I haven't
written the q argument,

00:05:39.790 --> 00:05:40.890
I won't write it here.

00:05:40.890 --> 00:05:44.220
So it's chi of p1.

00:05:44.220 --> 00:05:47.640
So I want to-- this
quantity j that I

00:05:47.640 --> 00:05:51.580
wrote is equal to this
integral on the right.

00:05:54.120 --> 00:05:57.140
Now, we encountered
almost the same integral

00:05:57.140 --> 00:06:00.960
when we were looking for
the proof of the H-theorem

00:06:00.960 --> 00:06:03.200
where the analog
of this chi of p1

00:06:03.200 --> 00:06:07.470
was in fact log of
f evaluated at p1.

00:06:07.470 --> 00:06:11.200
And we did some manipulations
that we can do over here.

00:06:11.200 --> 00:06:14.210
First of all, here, we have
dummy integration variables

00:06:14.210 --> 00:06:15.790
p1 and p2.

00:06:15.790 --> 00:06:22.668
We can just change their name
and then essentially average

00:06:22.668 --> 00:06:23.876
over those two possibilities.

00:06:27.340 --> 00:06:30.510
And the other thing that
we did was this equation

00:06:30.510 --> 00:06:34.000
has the set of things that
come into the collision,

00:06:34.000 --> 00:06:36.760
and set the things
that, in some sense,

00:06:36.760 --> 00:06:39.860
got out of the collision,
or basically things that,

00:06:39.860 --> 00:06:45.210
as a result of these
collisions, create these two.

00:06:45.210 --> 00:06:49.400
So you have this symmetry
between the initiators

00:06:49.400 --> 00:06:51.400
and products of the collision.

00:06:51.400 --> 00:06:53.440
Because essentially
the same function

00:06:53.440 --> 00:06:57.100
describes going one
way and inverting

00:06:57.100 --> 00:06:59.800
things and going backwards.

00:06:59.800 --> 00:07:01.840
And we said that in
principle, I could

00:07:01.840 --> 00:07:04.890
change variables of integration.

00:07:04.890 --> 00:07:08.590
And the effect of doing that
is kind of moving the prime

00:07:08.590 --> 00:07:12.160
coordinates to the things
that don't have primes.

00:07:12.160 --> 00:07:16.200
I don't know how last time I
made the mistake of the sign.

00:07:16.200 --> 00:07:19.470
But it's clear that if I
just put the primes from here

00:07:19.470 --> 00:07:22.960
to here, there will
be a minus sign.

00:07:22.960 --> 00:07:26.300
So the result of doing
that symmetrization

00:07:26.300 --> 00:07:32.190
should be a minus chi of p1
prime minus chi of p2 prime.

00:07:32.190 --> 00:07:35.935
And again, to do the averaging,
I have to put something else.

00:07:41.170 --> 00:07:45.460
Now, this statement is
quite generally true.

00:07:45.460 --> 00:07:49.310
Whatever chi I choose, I
will have this value of j

00:07:49.310 --> 00:07:52.930
as a result of that integration.

00:07:52.930 --> 00:07:56.660
But now we are going to
look at something specific.

00:07:56.660 --> 00:07:59.490
Let's assume that we
have a quantity that

00:07:59.490 --> 00:08:02.340
is conserved in collision.

00:08:02.340 --> 00:08:06.315
This will be 0 for collision
conserved quantity.

00:08:15.920 --> 00:08:19.620
Like let's say if my chi
that I had chosen here

00:08:19.620 --> 00:08:23.030
was some component
of momentum px,

00:08:23.030 --> 00:08:26.760
then whatever some of
the incoming momenta

00:08:26.760 --> 00:08:29.060
will be the sum of
the outgoing momenta.

00:08:29.060 --> 00:08:32.450
So essentially
anything that I can

00:08:32.450 --> 00:08:36.210
think of that is conserved in
the collision, this function

00:08:36.210 --> 00:08:39.450
that relates p
primes to p1 and p2

00:08:39.450 --> 00:08:43.712
has the right property to ensure
that this whole thing will

00:08:43.712 --> 00:08:45.280
be 0.

00:08:45.280 --> 00:08:48.350
And that's actually really
the ultimate reason.

00:08:48.350 --> 00:08:50.560
I don't really
need to know about

00:08:50.560 --> 00:08:53.420
all of these cross sections
and all of the collision

00:08:53.420 --> 00:08:54.500
properties, et cetera.

00:08:54.500 --> 00:08:57.220
Because my focus
will be on things

00:08:57.220 --> 00:08:59.350
that are conserved
in collisions.

00:08:59.350 --> 00:09:03.530
Because those are the variables
that are very slowly relaxing,

00:09:03.530 --> 00:09:06.040
and the things that
I'm interested in.

00:09:06.040 --> 00:09:08.930
So what you have is that for
these collision conserved

00:09:08.930 --> 00:09:15.000
quantities, which is the
things that I'm interested in,

00:09:15.000 --> 00:09:16.357
this equation is 0.

00:09:20.480 --> 00:09:24.660
Now, if f satisfies
that equation,

00:09:24.660 --> 00:09:31.702
I can certainly substitute
over here for Cff Lf.

00:09:36.340 --> 00:09:40.930
So if f satisfies
that equation, and I

00:09:40.930 --> 00:09:43.710
pick a collision
conserved quantity,

00:09:43.710 --> 00:09:47.160
the integral over p1 of that
function of the collision

00:09:47.160 --> 00:09:50.700
conserved quantity
times the bunch

00:09:50.700 --> 00:09:53.690
of first derivatives
acting on f has to be 0.

00:09:56.890 --> 00:09:59.140
So this I can write
in the following way--

00:09:59.140 --> 00:10:03.330
0 is the integral over p1.

00:10:03.330 --> 00:10:05.220
Actually, I have
only one momentum.

00:10:05.220 --> 00:10:08.080
So let's just ignore
it from henceforth.

00:10:08.080 --> 00:10:10.800
The other momentum I
just introduced in order

00:10:10.800 --> 00:10:13.830
to be able to show that
when integrated against

00:10:13.830 --> 00:10:17.180
the collision operator,
it will give me 0.

00:10:17.180 --> 00:10:23.720
I have the chi, and then this
bunch of derivatives dt plus p

00:10:23.720 --> 00:10:30.280
alpha over m p alpha plus
f alpha d by dp alpha

00:10:30.280 --> 00:10:33.346
acting on f.

00:10:33.346 --> 00:10:34.995
And that has to be 0.

00:10:34.995 --> 00:10:37.370
STUDENT: So alpha
stands for x, y, z?

00:10:37.370 --> 00:10:41.360
PROFESSOR: Yes, alpha stands
for the three components

00:10:41.360 --> 00:10:45.120
x, y, and z throughout
this lecture.

00:10:45.120 --> 00:10:49.008
And summation over a
repeated index is assumed.

00:10:52.920 --> 00:10:56.250
All right, so now
what I want to do

00:10:56.250 --> 00:11:00.970
is to move this chi so
that the derivatives act

00:11:00.970 --> 00:11:02.826
on both of them.

00:11:02.826 --> 00:11:06.460
So I'll simply write
the integral of dqp--

00:11:06.460 --> 00:11:09.040
if you like, this
bunch of derivatives

00:11:09.040 --> 00:11:15.160
that we call L acting on
the combination chi f.

00:11:15.160 --> 00:11:19.440
But a derivative of f
chi gives me f prime chi.

00:11:19.440 --> 00:11:23.870
But also it keeps me chi prime
f that I don't have here.

00:11:23.870 --> 00:11:25.550
So I have to subtract that.

00:11:33.480 --> 00:11:36.170
And why did I do that?

00:11:36.170 --> 00:11:39.950
Because now I end
up with integrals

00:11:39.950 --> 00:11:43.845
that involve integrating
over f against something.

00:11:47.050 --> 00:11:50.520
So let's think about
these typical integrals.

00:11:50.520 --> 00:11:58.100
If I take the integral over
momentum of f of p and q

00:11:58.100 --> 00:12:03.320
and t-- remember, f was
the one particle density.

00:12:03.320 --> 00:12:06.640
So I'm integrating, let's
say, at a particular position

00:12:06.640 --> 00:12:10.510
in space over all momentum.

00:12:10.510 --> 00:12:12.870
So it says, I don't
care what momentum.

00:12:12.870 --> 00:12:15.930
I just want to know whether
there's a particle here.

00:12:15.930 --> 00:12:17.350
So what is that quantity?

00:12:17.350 --> 00:12:20.520
That quantity is simply the
density at that location.

00:12:26.364 --> 00:12:28.990
Now suppose I were
to integrate this

00:12:28.990 --> 00:12:31.830
against some other
function, which

00:12:31.830 --> 00:12:34.685
could depend on p, q,
and t, for example?

00:12:38.030 --> 00:12:43.440
I use that to define an average.

00:12:43.440 --> 00:12:48.480
So this is going to be
defined to be the average of O

00:12:48.480 --> 00:12:50.870
at that location q and t.

00:12:50.870 --> 00:12:54.700
So for example, rather than
just calculating whether or not

00:12:54.700 --> 00:12:56.620
there are particles
here, I could

00:12:56.620 --> 00:12:59.500
be asking, what is the
average kinetic energy

00:12:59.500 --> 00:13:01.330
of the particles that are here?

00:13:01.330 --> 00:13:04.760
Then I would integrate this
against p squared over 2m.

00:13:04.760 --> 00:13:09.750
And this average would give
me the local expectation value

00:13:09.750 --> 00:13:13.020
of p squared over m,
just a normalization n

00:13:13.020 --> 00:13:15.650
so that it's
appropriately normalized.

00:13:19.150 --> 00:13:22.960
So with this definition, I
can write the various terms

00:13:22.960 --> 00:13:24.120
that I have over here.

00:13:24.120 --> 00:13:27.540
So let me write it a
little bit more explicitly.

00:13:27.540 --> 00:13:28.636
What do we have?

00:13:28.636 --> 00:13:32.780
We have integral d cubed p.

00:13:32.780 --> 00:13:44.780
We have this bunch of
derivatives acting on chi f

00:13:44.780 --> 00:13:54.840
minus f times this bunch of
derivatives acting on chi.

00:13:58.610 --> 00:14:03.020
So let's now look at
things term by term.

00:14:03.020 --> 00:14:05.220
The first term is
a time derivative.

00:14:05.220 --> 00:14:09.470
The time derivative I can
take outside the integral.

00:14:09.470 --> 00:14:13.090
Once I take the time derivative
outside the integral,

00:14:13.090 --> 00:14:14.580
what is left?

00:14:14.580 --> 00:14:19.060
What is left is the
integral of chi f,

00:14:19.060 --> 00:14:21.230
exactly what I have here.

00:14:21.230 --> 00:14:23.590
O has been replaced by chi.

00:14:23.590 --> 00:14:26.780
So what I have is
the time derivative

00:14:26.780 --> 00:14:33.190
of n expectation value of
chi using this definition.

00:14:33.190 --> 00:14:35.620
Let's look at the next term.

00:14:35.620 --> 00:14:38.180
The next term, these derivatives
are all over position.

00:14:38.180 --> 00:14:40.350
The integration
is over momentum.

00:14:40.350 --> 00:14:42.460
I can take it outside.

00:14:42.460 --> 00:14:45.390
So I can write it as d alpha.

00:14:45.390 --> 00:14:50.150
And then I have a quantity
that I'm integrating against f.

00:14:50.150 --> 00:14:55.220
So I will get n times this
local average of that quantity.

00:14:55.220 --> 00:14:56.520
What's that quantity?

00:14:56.520 --> 00:15:00.170
It's p alpha over m times chi.

00:15:03.850 --> 00:15:06.060
What's the third term?

00:15:06.060 --> 00:15:09.860
The third term is actually
an integral over momentum.

00:15:09.860 --> 00:15:12.770
But I'm integrating
over momentum.

00:15:12.770 --> 00:15:16.430
So again, you can sort of
remove things to boundaries

00:15:16.430 --> 00:15:18.920
and convince yourself that
that integral will not

00:15:18.920 --> 00:15:20.090
give you a contribution.

00:15:22.910 --> 00:15:26.910
The next bunch of terms are
simply directly this-- f

00:15:26.910 --> 00:15:29.170
integrated against something.

00:15:29.170 --> 00:15:31.700
So they're going
to give me minus n

00:15:31.700 --> 00:15:36.570
times the various averages
involved-- d t chi minus n,

00:15:36.570 --> 00:15:41.250
the average of p alpha
over m, the alpha chi.

00:15:41.250 --> 00:15:47.340
And then f alpha I can actually
take outside, minus n f alpha,

00:15:47.340 --> 00:15:51.330
the average of d
chi by d p alpha.

00:15:51.330 --> 00:15:54.670
And what we've established
is that that whole thing

00:15:54.670 --> 00:15:58.939
is 0 for quantities that are
conserved under collisions.

00:16:05.510 --> 00:16:09.100
So why did I do all of that?

00:16:09.100 --> 00:16:13.220
It's because solving
the Boltzmann equation

00:16:13.220 --> 00:16:17.560
in six dimensional phase space
with all of its integrations

00:16:17.560 --> 00:16:21.490
and derivatives is
very complicated.

00:16:21.490 --> 00:16:25.460
But really, the things
that are slowly relaxing

00:16:25.460 --> 00:16:27.610
are quantities
that are conserved

00:16:27.610 --> 00:16:32.480
collisions, such as densities,
average momentum, et cetera.

00:16:32.480 --> 00:16:36.880
And so I can focus on
variations of these

00:16:36.880 --> 00:16:40.670
through this kind of equation.

00:16:40.670 --> 00:16:44.160
Essentially, what that
will allow me to do

00:16:44.160 --> 00:16:51.750
is to construct what are known
as hydrodynamic equations,

00:16:51.750 --> 00:16:59.390
which describe the time
evolution of slow variables

00:16:59.390 --> 00:17:01.710
of your system,
the variables that

00:17:01.710 --> 00:17:05.690
are kind of relevant to making
thermodynamic observations,

00:17:05.690 --> 00:17:07.400
as opposed to variables
that you would

00:17:07.400 --> 00:17:08.910
be interested in
if you're thinking

00:17:08.910 --> 00:17:09.924
about atomic collisions.

00:17:12.780 --> 00:17:16.849
So what I need to do is
to go into that equation

00:17:16.849 --> 00:17:21.569
and pick out my
conserved quantities.

00:17:21.569 --> 00:17:31.220
So what are the
conserved quantities,

00:17:31.220 --> 00:17:36.480
and how can I describe
them by some chi?

00:17:36.480 --> 00:17:40.950
Well, we already saw this
when we were earlier on trying

00:17:40.950 --> 00:17:45.220
to find some kind of a solution
to the H by dt equals 0.

00:17:45.220 --> 00:17:47.615
We said that log f
has to be the sum

00:17:47.615 --> 00:17:49.780
of collision
conserved quantities.

00:17:49.780 --> 00:17:53.190
And we identified three
types of quantities.

00:17:53.190 --> 00:17:56.830
One of them was related to
the number conservation.

00:17:56.830 --> 00:17:59.610
And essentially, what
you have is that 1 plus 1

00:17:59.610 --> 00:18:01.060
equals to 1 plus 1.

00:18:01.060 --> 00:18:02.750
So it's obvious.

00:18:02.750 --> 00:18:04.345
The other is momentum.

00:18:07.520 --> 00:18:12.670
And there are three components
of this-- px, py, pz.

00:18:12.670 --> 00:18:15.892
And the third one is
the kinetic energy,

00:18:15.892 --> 00:18:18.320
which is conserved
in collisions.

00:18:18.320 --> 00:18:21.480
In a potential, clearly the
kinetic energy of a particle

00:18:21.480 --> 00:18:24.140
changes as a
function of position.

00:18:24.140 --> 00:18:28.040
But within the short
distances of the collisions

00:18:28.040 --> 00:18:30.330
that we are interested
in, the kinetic energy

00:18:30.330 --> 00:18:31.330
is a conserved quantity.

00:18:35.390 --> 00:18:41.360
So my task is to insert
these values of chi

00:18:41.360 --> 00:18:46.840
into that equation and see
what information they tell me

00:18:46.840 --> 00:18:50.690
about the time evolution of
the corresponding conserved

00:18:50.690 --> 00:18:52.460
quantities.

00:18:52.460 --> 00:18:56.820
So let's do this one by one.

00:18:56.820 --> 00:19:01.030
Let's start with
chi equals to 1.

00:19:01.030 --> 00:19:03.690
If I put chi equals to
1, all of these terms

00:19:03.690 --> 00:19:07.991
that involve derivatives
clearly needed to vanish.

00:19:07.991 --> 00:19:13.700
And here, I would get the time
derivative of the density.

00:19:13.700 --> 00:19:21.330
And from here, I would get the
alpha n expectation value of p

00:19:21.330 --> 00:19:23.300
alpha over m.

00:19:23.300 --> 00:19:24.560
We'll give that a name.

00:19:24.560 --> 00:19:27.310
We'll call that u alpha.

00:19:27.310 --> 00:19:33.170
So I have introduced u
alpha to be the expectation

00:19:33.170 --> 00:19:36.640
value of p alpha over m.

00:19:36.640 --> 00:19:40.940
And it can in principle depend
on which location in space

00:19:40.940 --> 00:19:42.090
you are looking at.

00:19:42.090 --> 00:19:44.030
Somebody opens the
door, there will

00:19:44.030 --> 00:19:45.445
be a current that
is established.

00:19:45.445 --> 00:19:48.110
And so there will
be a local velocity

00:19:48.110 --> 00:19:50.520
of the air that would be
different from other places

00:19:50.520 --> 00:19:52.260
in the room.

00:19:52.260 --> 00:19:53.920
And that's all we have.

00:19:53.920 --> 00:19:56.710
And this is equal to 0.

00:19:56.710 --> 00:19:58.640
And this is of
course the equation

00:19:58.640 --> 00:20:00.860
of the continuity of
the number of particles.

00:20:03.800 --> 00:20:07.680
You don't create or
destroy particles.

00:20:07.680 --> 00:20:10.340
And so this density
has to satisfy

00:20:10.340 --> 00:20:13.800
this nice, simple equation.

00:20:13.800 --> 00:20:17.780
We will sometimes rewrite this
slightly in the following way.

00:20:17.780 --> 00:20:20.060
This is the derivative
of two objects.

00:20:20.060 --> 00:20:25.720
I can expand that and write
it as dt of n plus, let's say,

00:20:25.720 --> 00:20:29.405
u alpha d alpha of n.

00:20:29.405 --> 00:20:31.120
And then I would
have a term that

00:20:31.120 --> 00:20:34.140
is n d alpha u
alpha, which I will

00:20:34.140 --> 00:20:35.936
take to the other
side of the equation.

00:20:43.460 --> 00:20:46.940
Why have I done that?

00:20:46.940 --> 00:20:52.760
Because if I think of n as a
function of position and time--

00:20:52.760 --> 00:20:58.300
and as usual, we did
before define a derivative

00:20:58.300 --> 00:21:01.670
that moves along
this streamline--

00:21:01.670 --> 00:21:05.560
you will have both the implicit
time derivative and the time

00:21:05.560 --> 00:21:08.820
derivative because the
stream changes position

00:21:08.820 --> 00:21:12.950
by an amount that is
related to velocity.

00:21:12.950 --> 00:21:17.550
Now, for the Liouville equation,
we have something like this,

00:21:17.550 --> 00:21:20.040
except that the Liouville
equation, the right hand side

00:21:20.040 --> 00:21:20.590
was 0.

00:21:20.590 --> 00:21:23.990
Because the flows of
the Liouville equation,

00:21:23.990 --> 00:21:27.620
the Hamiltonian flows,
were divergenceless.

00:21:27.620 --> 00:21:29.990
But in general, for a
compressable system,

00:21:29.990 --> 00:21:34.230
such as the gas in this
room, the compressibility

00:21:34.230 --> 00:21:37.431
is indicated to a
nonzero divergence of u.

00:21:37.431 --> 00:21:39.764
And there's a corresponding
term on the right hand side.

00:21:44.322 --> 00:21:48.280
So that's the first
thing we can do.

00:21:48.280 --> 00:21:52.370
What's the second
thing we can do?

00:21:52.370 --> 00:21:59.840
I can pick p-- let's say p
beta, what I wrote over there.

00:21:59.840 --> 00:22:02.510
But I can actually scale it.

00:22:02.510 --> 00:22:05.200
If p is a conserved
quantity, p/m

00:22:05.200 --> 00:22:06.575
is also a conserved quantity.

00:22:09.450 --> 00:22:12.010
Actually, as far as
this chi is concerned,

00:22:12.010 --> 00:22:15.700
I can add anything that
depends on q and not p.

00:22:15.700 --> 00:22:19.135
So I can subtract the average
value of this quantity.

00:22:24.950 --> 00:22:27.340
And this is conserved
during collisions.

00:22:27.340 --> 00:22:31.060
Because this part is the
same thing as something

00:22:31.060 --> 00:22:32.690
that is related to density.

00:22:32.690 --> 00:22:35.490
It's like 1 plus 1
equals to 1 plus 1.

00:22:35.490 --> 00:22:38.900
And p over beta
is conserved also.

00:22:38.900 --> 00:22:42.150
So we'll call this
quantity that we

00:22:42.150 --> 00:22:47.220
will use for our
candidate chi as c beta.

00:22:47.220 --> 00:22:54.020
So essentially, it's the
additional fluctuating speed,

00:22:54.020 --> 00:22:56.790
velocity that the
particles have,

00:22:56.790 --> 00:23:02.060
on top of the average velocity
that is due to the flow.

00:23:02.060 --> 00:23:04.790
And the reason maybe
it's useful to do

00:23:04.790 --> 00:23:08.670
this is because clearly
the average of c is 0.

00:23:08.670 --> 00:23:12.040
Because the average of
p beta over m is u beta.

00:23:12.040 --> 00:23:14.990
And if I do that, then clearly
at least the first thing

00:23:14.990 --> 00:23:17.070
in the equation I don't
have to worry about.

00:23:17.070 --> 00:23:19.510
I have removed one
term in the equation.

00:23:22.810 --> 00:23:26.870
So let's put c beta
for chi over here.

00:23:26.870 --> 00:23:28.580
We said that the
first term is 0.

00:23:28.580 --> 00:23:31.170
So we go and start
with the second term.

00:23:31.170 --> 00:23:31.812
What do I have?

00:23:31.812 --> 00:23:37.110
I have d alpha expectation
value of c beta.

00:23:37.110 --> 00:23:39.970
And then I have p alpha over m.

00:23:39.970 --> 00:23:44.710
Well, p alpha over m is going
to be u alpha plus c alpha.

00:23:44.710 --> 00:23:49.165
So let's write it in this
fashion-- u alpha plus c

00:23:49.165 --> 00:23:52.980
alpha for p alpha over m.

00:23:52.980 --> 00:23:54.920
And that's the average
I have to take.

00:23:59.980 --> 00:24:03.450
Now let's look at all of these
terms that involve derivatives.

00:24:06.170 --> 00:24:09.730
Well, if I want to take a time
derivative of this quantity,

00:24:09.730 --> 00:24:13.020
now that I have
introduced this average,

00:24:13.020 --> 00:24:16.120
there is a time derivative here.

00:24:16.120 --> 00:24:20.130
So the average of the
time derivative of chi

00:24:20.130 --> 00:24:23.600
will give me the time
derivative from u beta.

00:24:23.600 --> 00:24:26.160
And actually the minus
sign will cancel.

00:24:26.160 --> 00:24:31.210
And so I will have plus n, the
expectation value of d-- well,

00:24:31.210 --> 00:24:37.190
there's no expectation value
of something like this.

00:24:37.190 --> 00:24:39.690
When I integrate over p,
there's no p dependence.

00:24:39.690 --> 00:24:41.230
So it's just itself.

00:24:41.230 --> 00:24:45.085
So it is n dt of u beta.

00:24:49.320 --> 00:24:51.560
OK, what do we have
for the next term?

00:24:51.560 --> 00:24:53.290
Let's write it explicitly.

00:24:53.290 --> 00:24:56.690
I have n.

00:24:56.690 --> 00:25:03.020
p alpha over m I'm writing
as u alpha plus c alpha.

00:25:03.020 --> 00:25:08.530
And then I have the position
derivative of c beta.

00:25:08.530 --> 00:25:10.880
And that goes over here.

00:25:10.880 --> 00:25:16.560
So I will get d alpha of
u beta with a minus sign.

00:25:16.560 --> 00:25:17.830
So this becomes a plus.

00:25:23.260 --> 00:25:29.880
The last term is
minus n f alpha.

00:25:29.880 --> 00:25:33.210
And I have to take a derivative
of this object with respect

00:25:33.210 --> 00:25:35.090
to p alpha.

00:25:35.090 --> 00:25:37.310
Well, I have a p here.

00:25:37.310 --> 00:25:40.140
The derivative of p beta
with respect to p alpha

00:25:40.140 --> 00:25:42.690
gives me delta alpha beta.

00:25:42.690 --> 00:25:47.810
So this is going to give
me delta alpha beta over m.

00:25:47.810 --> 00:25:49.700
And the whole thing is 0.

00:25:54.600 --> 00:25:59.710
So let's rearrange
the terms over here.

00:25:59.710 --> 00:26:04.130
The only thing that I have
in the denominator is a 1/m.

00:26:04.130 --> 00:26:08.080
So let me multiply the
whole equation by m

00:26:08.080 --> 00:26:10.920
and see what happens.

00:26:10.920 --> 00:26:12.930
This term let's deal with last.

00:26:12.930 --> 00:26:20.685
This term, the first term,
becomes nmdt of u beta.

00:26:23.310 --> 00:26:29.250
The change in velocity kind
of looks like an acceleration.

00:26:29.250 --> 00:26:31.540
But you have to be careful.

00:26:31.540 --> 00:26:35.490
Because you can only
talk about acceleration

00:26:35.490 --> 00:26:37.360
acting for a particle.

00:26:37.360 --> 00:26:40.990
And the particle is
moving with the stream.

00:26:40.990 --> 00:26:45.020
OK, and this term will give
me the appropriate derivative

00:26:45.020 --> 00:26:48.820
to make it a stream velocity.

00:26:48.820 --> 00:26:52.930
Now, when I look at this, the
average of c that appears here

00:26:52.930 --> 00:26:56.030
will be 0.

00:26:56.030 --> 00:26:58.110
So the term that
I have over there

00:26:58.110 --> 00:27:00.070
is u alpha d alpha u beta.

00:27:00.070 --> 00:27:01.850
There's no average involved.

00:27:01.850 --> 00:27:04.231
It will give me n.

00:27:04.231 --> 00:27:11.960
m is common, so I will get
u alpha d alpha u beta,

00:27:11.960 --> 00:27:12.730
which is nice.

00:27:12.730 --> 00:27:16.770
Because then I can
certainly regard this

00:27:16.770 --> 00:27:19.720
as one of these
stream derivatives.

00:27:25.100 --> 00:27:28.230
So these two terms,
the stream derivative

00:27:28.230 --> 00:27:34.460
of velocity with time,
times mass, mass times

00:27:34.460 --> 00:27:37.720
the density to make it
mass per unit volume,

00:27:37.720 --> 00:27:39.200
looks like an acceleration.

00:27:39.200 --> 00:27:42.800
So it's like mass
times acceleration.

00:27:42.800 --> 00:27:45.515
Newton's law, it should
be equal to the force.

00:27:45.515 --> 00:27:47.400
And what do we have
if we take this

00:27:47.400 --> 00:27:48.940
to the other side
of the equation?

00:27:48.940 --> 00:27:52.040
We have f beta.

00:27:52.040 --> 00:27:58.620
OK, good, so we have
reproduced Newton's equation.

00:27:58.620 --> 00:28:03.090
In this context, if we're
moving along with the stream,

00:28:03.090 --> 00:28:07.450
mass times the acceleration of
the group of particles moving

00:28:07.450 --> 00:28:09.540
with the stream
is the force that

00:28:09.540 --> 00:28:13.777
is felt from the
external potential.

00:28:13.777 --> 00:28:15.110
But there's one other term here.

00:28:17.830 --> 00:28:20.190
In this term, the
term that is uc

00:28:20.190 --> 00:28:24.780
will average to 0, because
the average of u is 0.

00:28:24.780 --> 00:28:28.620
So what I will have
is minus-- and I

00:28:28.620 --> 00:28:31.650
forgot to write down
an n somewhere here.

00:28:31.650 --> 00:28:32.658
There will be an n.

00:28:36.860 --> 00:28:40.800
Because all averages
had this additional n

00:28:40.800 --> 00:28:43.550
that I forgot to put.

00:28:43.550 --> 00:28:46.370
I will take it to the right
hand side of the equation.

00:28:46.370 --> 00:28:52.620
And it becomes d
by dq alpha of n.

00:28:52.620 --> 00:28:56.610
I multiply the
entire equation by m.

00:28:56.610 --> 00:28:59.787
And then I have the
average of c alpha c beta.

00:29:06.700 --> 00:29:09.160
So what happened here?

00:29:09.160 --> 00:29:14.080
Isn't force just the
mass times acceleration?

00:29:14.080 --> 00:29:18.700
Well, as long as you
include all forces involved.

00:29:18.700 --> 00:29:27.240
So if you imagine that this
room is totally stationary air,

00:29:27.240 --> 00:29:33.240
and I heat one corner
here, then the particles

00:29:33.240 --> 00:29:36.310
here will start to
move more rapidly.

00:29:36.310 --> 00:29:38.120
There will be more
pressure here.

00:29:38.120 --> 00:29:40.680
Because pressure is proportional
to temperature, if you like.

00:29:40.680 --> 00:29:43.110
There will be more pressure,
less pressure here.

00:29:43.110 --> 00:29:45.370
The difference in pressure
will drive the flow.

00:29:45.370 --> 00:29:47.920
There will be an
additional force.

00:29:47.920 --> 00:29:49.610
And that's what it says.

00:29:49.610 --> 00:29:54.800
If there's variation in these
speeds of the particles,

00:29:54.800 --> 00:29:58.830
the change in pressure
will give you a force.

00:29:58.830 --> 00:30:04.990
And so this thing, p alpha beta,
is called the pressure tensor.

00:30:12.580 --> 00:30:15.140
Yes.

00:30:15.140 --> 00:30:19.000
STUDENT: Shouldn't f
beta be multiplied by n,

00:30:19.000 --> 00:30:22.260
or is there an n on
the other side of that?

00:30:22.260 --> 00:30:24.670
PROFESSOR: There is an n
here that I forgot, yes.

00:30:27.980 --> 00:30:31.295
So the n was in the first
equation, somehow got lost.

00:30:34.160 --> 00:30:37.160
STUDENT: So the
pressure is coming

00:30:37.160 --> 00:30:39.160
from the local fluctuation?

00:30:39.160 --> 00:30:40.302
PROFESSOR: Yes.

00:30:40.302 --> 00:30:42.310
And if you think about
it, the temperature

00:30:42.310 --> 00:30:44.372
is also the local fluctuation.

00:30:44.372 --> 00:30:48.090
So it has something to do
with temperature differences.

00:30:48.090 --> 00:30:50.120
Pressure is related
to temperature.

00:30:50.120 --> 00:30:52.950
So all the things are connected.

00:30:52.950 --> 00:30:56.670
And in about two minutes, I'll
actually evaluate that for you,

00:30:56.670 --> 00:30:59.120
and you'll see how.

00:30:59.120 --> 00:30:59.870
Yes.

00:30:59.870 --> 00:31:02.684
STUDENT: Is the pressure
tensor distinct from the stress

00:31:02.684 --> 00:31:05.030
tensor?

00:31:05.030 --> 00:31:07.800
PROFESSOR: It's
the stress tensor

00:31:07.800 --> 00:31:10.880
that you would have for a fluid.

00:31:10.880 --> 00:31:13.920
For something more complicated,
like an elastic material,

00:31:13.920 --> 00:31:18.061
it would be much more
complicated-- not much more

00:31:18.061 --> 00:31:19.477
complicated, but
more complicated.

00:31:22.890 --> 00:31:25.020
Essentially, there's
always some kind

00:31:25.020 --> 00:31:27.450
of a force per unit
volume depending

00:31:27.450 --> 00:31:30.020
on what kind of medium you have.

00:31:30.020 --> 00:31:31.611
And for the gas,
this is what it is.

00:31:31.611 --> 00:31:32.110
Yes.

00:31:32.110 --> 00:31:34.054
STUDENT: So a basic question.

00:31:34.054 --> 00:31:39.886
When we say u alpha is
averaged, averaged over what?

00:31:39.886 --> 00:31:41.360
Is it by the area?

00:31:41.360 --> 00:31:43.740
PROFESSOR: OK, this
is the definition.

00:31:43.740 --> 00:31:47.940
So whenever I use this
notation with these angles,

00:31:47.940 --> 00:31:50.920
it means that I
integrated over p.

00:31:50.920 --> 00:31:52.400
Why do I do that?

00:31:52.400 --> 00:31:57.020
Because of this asymmetry
between momenta and collision

00:31:57.020 --> 00:32:01.400
and coordinates that is inherent
to the Boltzmann equation.

00:32:01.400 --> 00:32:03.880
When we wrote down the
Liouville equation,

00:32:03.880 --> 00:32:06.870
p and q were
completely equivalent.

00:32:06.870 --> 00:32:09.390
But by the time we
made our approximations

00:32:09.390 --> 00:32:11.970
and we talked about
collisions, et cetera,

00:32:11.970 --> 00:32:15.730
we saw that momenta
quickly relax.

00:32:15.730 --> 00:32:18.725
And so we can look at
the particular position

00:32:18.725 --> 00:32:22.320
and integrate over momenta
and define averages

00:32:22.320 --> 00:32:25.440
in the sense of when you
think about what's happening

00:32:25.440 --> 00:32:29.240
in this room, you think
about the wind velocity

00:32:29.240 --> 00:32:33.530
here over there, but not
fluctuations in the momentum

00:32:33.530 --> 00:32:38.140
so much, OK?

00:32:41.500 --> 00:32:46.530
All right, so this
clearly is kind

00:32:46.530 --> 00:32:52.430
of a Navier-Stokes like
equation, if you like,

00:32:52.430 --> 00:32:55.795
for this gas.

00:32:58.970 --> 00:33:02.930
That tells you how the
velocity of this fluid changes.

00:33:05.690 --> 00:33:13.600
And finally, we would need
to construct an equation that

00:33:13.600 --> 00:33:16.520
is relevant to the
kinetic energy, which

00:33:16.520 --> 00:33:20.910
is something like
p squared over 2m.

00:33:20.910 --> 00:33:25.960
And we can follow
what we did over here

00:33:25.960 --> 00:33:28.590
and subtract the average.

00:33:34.320 --> 00:33:36.780
And so essentially,
this is kinetic energy

00:33:36.780 --> 00:33:43.210
on top of the kinetic
energy of the entire stream.

00:33:43.210 --> 00:33:45.460
This is clearly the
same thing as mc

00:33:45.460 --> 00:33:49.400
squared over 2, c being
the quantity that we

00:33:49.400 --> 00:33:51.720
defined before.

00:33:51.720 --> 00:34:03.640
And the average of
mc squared over 2

00:34:03.640 --> 00:34:06.030
I will indicate by epsilon.

00:34:06.030 --> 00:34:07.130
It's the heat content.

00:34:11.350 --> 00:34:13.280
Or actually, let's
say energy density.

00:34:13.280 --> 00:34:14.425
It's probably better.

00:34:24.350 --> 00:34:31.210
So now I have to put mc squared
over 2 in this equation for chi

00:34:31.210 --> 00:34:34.719
and do various manipulations
along the lines of things

00:34:34.719 --> 00:34:36.520
that I did before.

00:34:36.520 --> 00:34:39.330
I will just write
down the final answer.

00:34:39.330 --> 00:34:48.679
So the final answer will
be that dt of epsilon.

00:34:48.679 --> 00:34:50.489
We've defined dt.

00:34:50.489 --> 00:34:52.179
I move with the streamline.

00:34:52.179 --> 00:34:56.380
So I have dt plus
u alpha d alpha

00:34:56.380 --> 00:35:02.650
acting on this density, which
is a function of position

00:35:02.650 --> 00:35:04.790
and time.

00:35:04.790 --> 00:35:10.280
And the right hand side of
this will have two terms.

00:35:10.280 --> 00:35:16.910
One term is essentially
how this pressure kind

00:35:16.910 --> 00:35:25.080
of moves against the velocity,
or the velocity and pressure

00:35:25.080 --> 00:35:27.940
are kind of hitting
against each other.

00:35:27.940 --> 00:35:31.980
So it's kind of like if you
were to rub two things--

00:35:31.980 --> 00:35:33.550
"rub" was the word
I was looking.

00:35:33.550 --> 00:35:35.810
If you were to rub two
things against each other,

00:35:35.810 --> 00:35:37.930
there's heat that is generated.

00:35:37.930 --> 00:35:40.710
And so that's the term
that we are looking at.

00:35:40.710 --> 00:35:43.290
So what is this u alpha beta?

00:35:43.290 --> 00:35:47.180
u alpha beta-- it's just
because p alpha beta

00:35:47.180 --> 00:35:48.730
is a symmetric object.

00:35:48.730 --> 00:35:53.900
It doesn't make any difference
if you exchange alpha and beta.

00:35:53.900 --> 00:35:56.770
You symmetrize the
derivative of the velocity.

00:36:03.360 --> 00:36:06.020
And sometimes it's called
the rate of strain.

00:36:13.480 --> 00:36:18.950
And there's another
term, which is

00:36:18.950 --> 00:36:25.420
minus 1/n d alpha of h alpha.

00:36:25.420 --> 00:36:30.880
And for that, I need to define
yet another quantity, this h

00:36:30.880 --> 00:36:44.240
alpha, which is nm over 2 the
average of 3 c's, c squared

00:36:44.240 --> 00:36:46.160
and then c alpha.

00:36:46.160 --> 00:36:48.336
And this is called
the heat transport.

00:37:00.890 --> 00:37:05.200
So for a simpler
fluid where these

00:37:05.200 --> 00:37:08.490
are the only conserved
quantities that I have,

00:37:08.490 --> 00:37:12.540
in order to figure out how
the fluid evolves over time,

00:37:12.540 --> 00:37:15.470
I have one equation
that tells me

00:37:15.470 --> 00:37:17.490
about how the density changes.

00:37:17.490 --> 00:37:21.440
And it's related to the
continuity of matter.

00:37:21.440 --> 00:37:30.540
I have one equation that tells
me how the velocity changes.

00:37:30.540 --> 00:37:33.430
And it's kind of an
appropriately generalized

00:37:33.430 --> 00:37:37.560
version of Newton's law in
which mass times acceleration

00:37:37.560 --> 00:37:40.820
is equated with
appropriate forces.

00:37:40.820 --> 00:37:44.420
And mostly we are interested in
the forces that are internally

00:37:44.420 --> 00:37:48.630
generated, because of the
variations in pressure.

00:37:48.630 --> 00:37:54.360
And finally, there is
variations in pressure

00:37:54.360 --> 00:37:57.170
related to variations
in temperature.

00:37:57.170 --> 00:38:01.640
And they're governed by
another equation that tells us

00:38:01.640 --> 00:38:07.470
how the local energy density,
local content of energy,

00:38:07.470 --> 00:38:09.016
changes as a function of time.

00:38:12.590 --> 00:38:18.450
So rather than solving
the Boltzmann equation,

00:38:18.450 --> 00:38:21.660
I say, OK, all I need
to do is to solve

00:38:21.660 --> 00:38:23.455
these hydrodynamic equations.

00:38:27.100 --> 00:38:29.383
Question?

00:38:29.383 --> 00:38:32.811
STUDENT: Last time for
the Boltzmann equation,

00:38:32.811 --> 00:38:33.311
[INAUDIBLE].

00:38:53.000 --> 00:38:58.500
PROFESSOR: What it says is that
conservation laws are much more

00:38:58.500 --> 00:38:59.720
general.

00:38:59.720 --> 00:39:02.260
So this equation you could
have written for a liquid,

00:39:02.260 --> 00:39:05.020
you could have
written for anything.

00:39:05.020 --> 00:39:07.290
This equation kind
of looks like you

00:39:07.290 --> 00:39:09.870
would have been able to
write it for everything.

00:39:09.870 --> 00:39:12.370
And it is true, except
that you wouldn't

00:39:12.370 --> 00:39:16.450
know what the pressure is.

00:39:16.450 --> 00:39:19.460
This equation you would have
written on the basis of energy

00:39:19.460 --> 00:39:22.760
conservation, except
that you wouldn't

00:39:22.760 --> 00:39:27.230
know what the heat
transport vector is.

00:39:27.230 --> 00:39:31.370
So what we gained through
this Boltzmann prescription,

00:39:31.370 --> 00:39:33.600
on top of what
you may just guess

00:39:33.600 --> 00:39:36.060
on the basis of
conservation laws,

00:39:36.060 --> 00:39:40.700
are expressions for quantities
that you would need in order

00:39:40.700 --> 00:39:44.380
to sort these equations, because
of the internal pressures that

00:39:44.380 --> 00:39:48.558
are generated because of the
way that the heat is flowing.

00:39:48.558 --> 00:39:56.180
STUDENT: And this quantity
is correct in the limit of--

00:39:56.180 --> 00:39:58.550
PROFESSOR: In the limit, yes.

00:39:58.550 --> 00:40:04.756
But that also really
is the Achilles' heel

00:40:04.756 --> 00:40:09.070
of the presentation I have
given to you right now.

00:40:09.070 --> 00:40:13.680
Because in order to
solve these equations,

00:40:13.680 --> 00:40:16.290
I should be able to
put an expression here

00:40:16.290 --> 00:40:21.670
for the pressure, and an
expression here for the h.

00:40:21.670 --> 00:40:24.690
But what is my
prescription for getting

00:40:24.690 --> 00:40:28.150
the expression for
pressure and h?

00:40:28.150 --> 00:40:31.780
I have to do an average
that involves the f.

00:40:31.780 --> 00:40:34.440
And I don't have the f.

00:40:34.440 --> 00:40:40.470
So have I gained
anything, all right?

00:40:40.470 --> 00:40:45.040
So these equations are general.

00:40:45.040 --> 00:40:49.860
We have to figure out what to
do for the p and h in order

00:40:49.860 --> 00:40:52.380
to be able to solve it.

00:40:52.380 --> 00:40:52.898
Yes.

00:40:52.898 --> 00:40:54.397
STUDENT: In the
last equation, isn't

00:40:54.397 --> 00:40:56.966
that n epsilon
instead of epsilon?

00:41:00.170 --> 00:41:05.885
PROFESSOR: mc squared
over 2-- I guess

00:41:05.885 --> 00:41:09.240
if I put here mc squared
over 2, probably it is nf.

00:41:15.386 --> 00:41:17.719
STUDENT: I think maybe the
last equation it's n epsilon,

00:41:17.719 --> 00:41:21.298
the equation there.

00:41:21.298 --> 00:41:22.589
PROFESSOR: This equation is OK.

00:41:27.970 --> 00:41:33.240
OK, so what you're saying--
that if I directly put chi here

00:41:33.240 --> 00:41:36.970
to be this quantity, what I
would need on the left hand

00:41:36.970 --> 00:41:39.430
side of the equations
would involve

00:41:39.430 --> 00:41:41.700
derivatives of n epsilon.

00:41:41.700 --> 00:41:46.450
Now, those derivatives I can
expand, write them, let's say,

00:41:46.450 --> 00:41:52.000
dt of n epsilon is epsilon
dt of n plus ndt of epsilon.

00:41:52.000 --> 00:41:54.850
And then you can
use these equations

00:41:54.850 --> 00:41:57.350
to reduce some of that.

00:41:57.350 --> 00:42:00.195
And the reason that I
didn't go through the steps

00:42:00.195 --> 00:42:02.660
that I would go
from here to here

00:42:02.660 --> 00:42:04.277
is because it
would have involved

00:42:04.277 --> 00:42:05.610
a number of those cancellations.

00:42:05.610 --> 00:42:09.398
And it would have taken me
an additional 10, 15 minutes.

00:42:16.230 --> 00:42:19.320
All right, so
conceptually, that's

00:42:19.320 --> 00:42:20.960
the more important thing.

00:42:20.960 --> 00:42:25.110
We have to find some way
of doing these things.

00:42:25.110 --> 00:42:31.130
Now, when I wrote
this equation, we

00:42:31.130 --> 00:42:33.980
said that there is some
kind of a separation of time

00:42:33.980 --> 00:42:37.470
scales involved in
that the left hand

00:42:37.470 --> 00:42:40.780
side of the equation,
the characteristic times

00:42:40.780 --> 00:42:43.550
are order of the time
it takes for a particle

00:42:43.550 --> 00:42:46.720
to, say, go over the
sides of the box,

00:42:46.720 --> 00:42:50.970
whereas the collision
times, 1 over tau x,

00:42:50.970 --> 00:42:55.500
are such that the
right hand side is

00:42:55.500 --> 00:42:57.710
much larger than
the left hand side.

00:43:00.830 --> 00:43:08.570
So as a 0-th order
approximation,

00:43:08.570 --> 00:43:12.650
what I will assume is
that the left hand side

00:43:12.650 --> 00:43:18.380
is so insignificant
that I will set it to 0.

00:43:18.380 --> 00:43:21.680
And then my approximation
for the collision

00:43:21.680 --> 00:43:25.710
is the thing that essentially
sets this bracket to 0.

00:43:25.710 --> 00:43:30.680
This is the local equilibrium
that we wrote down before.

00:43:30.680 --> 00:43:36.980
So that means that I'm assuming
a 0-th order approximation

00:43:36.980 --> 00:43:41.030
to the solution of the
Boltzmann equation.

00:43:41.030 --> 00:43:43.410
And very shortly, we
will improve up that.

00:43:43.410 --> 00:43:47.100
But let's see what this 0-th
order approximation gives us,

00:43:47.100 --> 00:43:49.610
which is-- we saw what it is.

00:43:49.610 --> 00:43:54.470
It was essentially something
like a Gaussian in momentum.

00:43:54.470 --> 00:43:59.350
But the coefficient out front
of it was kind of arbitrary.

00:43:59.350 --> 00:44:03.560
And now that I have defined
the integral over momentum

00:44:03.560 --> 00:44:07.360
to be density, I will
multiply a normalized Gaussian

00:44:07.360 --> 00:44:09.970
by the density locally.

00:44:09.970 --> 00:44:13.060
And I will have an exponential.

00:44:13.060 --> 00:44:18.480
And average of p I
will shift by an amount

00:44:18.480 --> 00:44:21.870
that depends on position.

00:44:21.870 --> 00:44:27.220
And I divide by some parameter
we had called before beta.

00:44:27.220 --> 00:44:32.030
But that beta I can
rewrite in this fashion.

00:44:32.030 --> 00:44:35.840
So I have just
rewritten the beta

00:44:35.840 --> 00:44:38.350
that we had before that
was a function of q and t

00:44:38.350 --> 00:44:40.500
as 1 over kBT.

00:44:40.500 --> 00:44:43.180
And this has to be
properly normalized.

00:44:43.180 --> 00:44:48.020
So I will have 2 pi mkBT,
which is a function of position

00:44:48.020 --> 00:44:48.520
to the 3/2.

00:44:53.040 --> 00:44:56.850
And you can check that the
form that I have written here

00:44:56.850 --> 00:45:00.410
respects the
definitions that I gave,

00:45:00.410 --> 00:45:02.750
namely that if I
were to integrate it

00:45:02.750 --> 00:45:05.885
over momentum, since
the momentum part is

00:45:05.885 --> 00:45:09.810
a normalized Gaussian, I
will just get the density.

00:45:09.810 --> 00:45:14.210
If I were to calculate
the average of p/m,

00:45:14.210 --> 00:45:16.810
I have shifted the
Gaussian appropriately so

00:45:16.810 --> 00:45:22.860
that the average of p/m is the
quantity that I'm calling u.

00:45:22.860 --> 00:45:25.390
The other one-- let's check.

00:45:25.390 --> 00:45:28.720
Essentially what
is happening here,

00:45:28.720 --> 00:45:33.570
this quantity is the same
thing as mc squared over 2kT

00:45:33.570 --> 00:45:37.790
if I use the definition of
c that I have over there.

00:45:37.790 --> 00:45:40.140
So it's a Gaussian weight.

00:45:40.140 --> 00:45:41.740
And from the
Gaussian weight, you

00:45:41.740 --> 00:45:43.930
can immediately see
that the average of c

00:45:43.930 --> 00:45:48.720
alpha c beta, it's
in fact diagonal.

00:45:48.720 --> 00:45:50.390
It's cx squared, cy squared.

00:45:50.390 --> 00:45:52.950
So the answer is going
to be delta alpha beta.

00:45:55.630 --> 00:45:58.120
And for each
particular component,

00:45:58.120 --> 00:45:59.871
I will get kT over m.

00:46:05.770 --> 00:46:09.740
So this quantity that
I was calling epsilon,

00:46:09.740 --> 00:46:13.790
which was the average
of mc squared over 2,

00:46:13.790 --> 00:46:17.570
is essentially
multiplying this by m/2

00:46:17.570 --> 00:46:20.630
and summing over
delta alpha alpha,

00:46:20.630 --> 00:46:22.490
which gives me a factor of 3.

00:46:22.490 --> 00:46:26.340
So this is going
to give me 3/2 kT.

00:46:26.340 --> 00:46:31.840
So really, my energy
density is none other

00:46:31.840 --> 00:46:34.610
than the local 3/2 kT.

00:46:34.610 --> 00:46:36.916
Yes?

00:46:36.916 --> 00:46:40.518
STUDENT: So you've just defined,
what is the temperature.

00:46:40.518 --> 00:46:44.107
So over all previous
derivations,

00:46:44.107 --> 00:46:47.586
we didn't really use the
classical temperature.

00:46:47.586 --> 00:46:51.065
And now you define it as sort
of average kinetic energy.

00:46:51.065 --> 00:46:55.450
PROFESSOR: Yeah, I have
introduced a quantity T here,

00:46:55.450 --> 00:46:57.710
which will indeed eventually
be the temperature

00:46:57.710 --> 00:46:58.610
for the whole thing.

00:46:58.610 --> 00:47:00.200
But right now, it
is something that

00:47:00.200 --> 00:47:03.190
is varying locally from
position to position.

00:47:03.190 --> 00:47:07.310
But you can see that the typical
kinetic energy at each location

00:47:07.310 --> 00:47:09.646
is of the order of
kT at that location.

00:47:13.460 --> 00:47:18.140
And the pressure
tensor p alpha beta,

00:47:18.140 --> 00:47:22.440
which is nm expectation
value of c alpha c beta,

00:47:22.440 --> 00:47:30.740
simply becomes kT over m--
sorry, nKT delta alpha beta.

00:47:30.740 --> 00:47:32.450
So now we can sort of start.

00:47:32.450 --> 00:47:34.980
Now probably it's a
better time to think

00:47:34.980 --> 00:47:37.080
about this as temperature.

00:47:37.080 --> 00:47:41.000
Because we know about the
ideal gas type of behavior

00:47:41.000 --> 00:47:44.570
where the pressure of the ideal
gas is simply density times kT.

00:47:47.480 --> 00:47:51.300
So the diagonal elements
of this pressure tensor

00:47:51.300 --> 00:47:54.440
are the things that
we usually think about

00:47:54.440 --> 00:47:57.080
as being the pressure
of a gas, now

00:47:57.080 --> 00:47:59.600
at the appropriate
temperature and density,

00:47:59.600 --> 00:48:02.910
and that there are no off
diagonal components here.

00:48:05.700 --> 00:48:08.710
I said I also need to
evaluate the h alpha.

00:48:11.240 --> 00:48:14.980
h alpha involves
three factors of c.

00:48:19.300 --> 00:48:24.090
And the way that we have
written down is Gaussian.

00:48:24.090 --> 00:48:26.090
So it's symmetric.

00:48:26.090 --> 00:48:29.220
So all odd powers
are going to be 0.

00:48:29.220 --> 00:48:31.950
There is no heat
transport vector here.

00:48:36.270 --> 00:48:41.964
So within this 0-th
order, what do we have?

00:48:41.964 --> 00:48:49.150
We have that the total
density variation, which

00:48:49.150 --> 00:48:58.040
is dt plus u alpha d
alpha acting on density,

00:48:58.040 --> 00:49:01.640
is minus nd alpha u alpha.

00:49:01.640 --> 00:49:06.110
That does not involve any of
these factors that I need.

00:49:06.110 --> 00:49:08.200
This equation-- let's see.

00:49:08.200 --> 00:49:10.490
Let's divide by mn.

00:49:10.490 --> 00:49:13.225
So we have Dt of u beta.

00:49:16.480 --> 00:49:21.810
And let's again look at what's
happening inside the room.

00:49:21.810 --> 00:49:25.140
Forget about boundary conditions
at the side of the box.

00:49:25.140 --> 00:49:26.940
So I'm going to write
this essentially

00:49:26.940 --> 00:49:29.610
for the case that
is inside the box.

00:49:29.610 --> 00:49:32.040
I can forget about
the external force.

00:49:32.040 --> 00:49:35.220
And all I'm interested
in is the internal forces

00:49:35.220 --> 00:49:37.870
that are generated
through pressure.

00:49:37.870 --> 00:49:44.570
So this is dt plus u
alpha d alpha of u beta.

00:49:44.570 --> 00:49:47.210
I said let's forget
the external force.

00:49:47.210 --> 00:49:48.580
So what do we have?

00:49:48.580 --> 00:49:52.780
We have the contribution
that comes from pressure.

00:49:52.780 --> 00:49:57.110
So we have minus the alpha.

00:49:57.110 --> 00:49:59.480
I divided through by nm.

00:49:59.480 --> 00:50:04.630
So let me write it
correctly as 1 over nm.

00:50:04.630 --> 00:50:06.520
I have the alpha.

00:50:06.520 --> 00:50:14.950
My pressure tensor is
nkT delta alpha beta.

00:50:14.950 --> 00:50:18.780
Delta alpha beta and this d
alpha, I can get rid of that

00:50:18.780 --> 00:50:20.971
and write it simply as d beta.

00:50:25.210 --> 00:50:31.070
So that's the equation
that governs the variations

00:50:31.070 --> 00:50:34.540
in the local stream
velocity that you

00:50:34.540 --> 00:50:38.290
have in the gas in response
to the changes in temperature

00:50:38.290 --> 00:50:42.200
and density that
you have in the gas.

00:50:42.200 --> 00:50:48.080
And finally, the equation
for the energy density, I

00:50:48.080 --> 00:50:53.490
have dt plus u alpha d alpha.

00:50:53.490 --> 00:50:59.040
My energy density is simply
related to this quantity T.

00:50:59.040 --> 00:51:04.610
So I can write it as
variations of this temperature

00:51:04.610 --> 00:51:06.780
in position.

00:51:06.780 --> 00:51:10.040
And what do I have on
the right hand side?

00:51:10.040 --> 00:51:13.770
I certainly don't have the
heat transport vectors.

00:51:13.770 --> 00:51:19.740
So all I have to do is to take
this diagonal p alpha beta

00:51:19.740 --> 00:51:25.150
and contract it with this
strain tensor u alpha beta.

00:51:25.150 --> 00:51:28.280
So the only term that I'm
going to get after contracting

00:51:28.280 --> 00:51:31.620
delta alpha beta is going
to be d alpha u alpha.

00:51:31.620 --> 00:51:35.350
So let's make sure that
we get the factors right.

00:51:35.350 --> 00:51:42.440
So I have minus p alpha
beta is nkT d alpha u alpha.

00:51:51.740 --> 00:51:56.910
So now we have a closed
set of equations.

00:51:56.910 --> 00:52:05.050
They tell me how the density,
temperature, and velocity

00:52:05.050 --> 00:52:08.620
vary from one location to
another location in the gas.

00:52:08.620 --> 00:52:10.070
They're completely closed.

00:52:10.070 --> 00:52:13.980
That's the only set of
things that come together.

00:52:13.980 --> 00:52:17.620
So I should be able
to now figure out,

00:52:17.620 --> 00:52:20.570
if I make a disturbance
in the gas in this room

00:52:20.570 --> 00:52:25.630
by walking around, by
talking, by striking a match,

00:52:25.630 --> 00:52:29.630
how does that eventually,
as a function of time,

00:52:29.630 --> 00:52:32.710
relax to something
that is uniform?

00:52:32.710 --> 00:52:37.870
Because our expectation is
that these equations ultimately

00:52:37.870 --> 00:52:39.610
will reach equilibrium.

00:52:39.610 --> 00:52:43.450
That's essentially the
most important thing

00:52:43.450 --> 00:52:45.560
that we deduce from
the Boltzmann equation,

00:52:45.560 --> 00:52:47.950
that it was allowing things
to reach equilibrium.

00:52:51.220 --> 00:52:52.145
Yes.

00:52:52.145 --> 00:52:54.522
STUDENT: For the second
equation, that's alpha?

00:52:54.522 --> 00:52:56.105
The right side of
the second equation?

00:53:00.520 --> 00:53:02.270
PROFESSOR: The alpha
index is summed over.

00:53:02.270 --> 00:53:04.174
STUDENT: The right side.

00:53:04.174 --> 00:53:09.410
Is it the derivative
of alpha or beta?

00:53:09.410 --> 00:53:10.124
Yeah, that one.

00:53:10.124 --> 00:53:11.040
PROFESSOR: It is beta.

00:53:11.040 --> 00:53:14.410
Because, you see, the only
index that I have left is beta.

00:53:14.410 --> 00:53:17.620
So if it's an index by
itself, it better be beta.

00:53:17.620 --> 00:53:21.990
How did this index d
alpha become d beta?

00:53:21.990 --> 00:53:24.870
Because the alpha beta
was delta alpha beta.

00:53:24.870 --> 00:53:28.020
STUDENT: Also, is it
alpha or is it beta?

00:53:28.020 --> 00:53:30.920
PROFESSOR: When I sum over alpha
of d alpha delta alpha beta,

00:53:30.920 --> 00:53:31.780
I get d beta.

00:53:35.628 --> 00:53:36.590
Yes.

00:53:36.590 --> 00:53:39.910
STUDENT: Can I ask again, how
did you come up with the f0?

00:53:39.910 --> 00:53:42.070
Why do you say that option?

00:53:42.070 --> 00:53:45.110
PROFESSOR: OK, so
this goes back to what

00:53:45.110 --> 00:53:47.090
we did last time around.

00:53:47.090 --> 00:53:49.830
Because we saw that
when we were writing

00:53:49.830 --> 00:53:53.390
the equation for
the hydt, we came up

00:53:53.390 --> 00:53:56.300
with a factor of what that
was-- this multiplying

00:53:56.300 --> 00:53:58.950
the difference of the logs.

00:53:58.950 --> 00:54:01.800
And we said that what
I can do in order

00:54:01.800 --> 00:54:04.930
to make sure that
this equation is

00:54:04.930 --> 00:54:12.050
0 is to say that log is additive
in conserved quantities,

00:54:12.050 --> 00:54:15.880
so log additive in
conserved quantities.

00:54:15.880 --> 00:54:18.630
I then exponentiate it.

00:54:18.630 --> 00:54:21.700
So this is log of a number.

00:54:21.700 --> 00:54:23.840
And these are all
things that are,

00:54:23.840 --> 00:54:26.500
when I take the log,
proportional to p squared

00:54:26.500 --> 00:54:29.175
and p, which are the
conserved quantities.

00:54:33.950 --> 00:54:36.930
So I know that this
form sets the right hand

00:54:36.930 --> 00:54:39.290
side of the Boltzmann
equation to 0.

00:54:39.290 --> 00:54:41.890
And that's the largest part
of the Boltzmann equation.

00:54:51.510 --> 00:54:56.200
Now what happens is that
within this equation,

00:54:56.200 --> 00:54:58.860
some quantities do not
relax to equilibrium.

00:55:01.960 --> 00:55:04.750
Some-- let's call
them variations.

00:55:04.750 --> 00:55:08.704
Sometimes I will use
the word "modes"--

00:55:08.704 --> 00:55:13.650
do not relax to equilibrium.

00:55:18.040 --> 00:55:23.240
And let's start
with the following.

00:55:26.710 --> 00:55:33.740
When you have a sheer velocity--
what do I mean by that?

00:55:33.740 --> 00:55:39.500
So let's imagine that
you have a wall that

00:55:39.500 --> 00:55:42.510
extends in the x direction.

00:55:42.510 --> 00:55:50.750
And along the y direction, you
encounter a velocity field.

00:55:50.750 --> 00:55:55.590
The velocity field is always
pointing along this direction.

00:55:55.590 --> 00:55:58.020
So it only has the x component.

00:55:58.020 --> 00:56:00.820
There's no y component
or z component.

00:56:00.820 --> 00:56:02.980
But this x component
maybe varies

00:56:02.980 --> 00:56:05.490
as a function of position.

00:56:05.490 --> 00:56:08.990
So my ux is a function of y.

00:56:08.990 --> 00:56:12.200
This corresponds to
some kind of a sheer.

00:56:16.200 --> 00:56:21.010
Now, if I do that,
then you can see

00:56:21.010 --> 00:56:26.770
that the only derivatives
that would be nonzero

00:56:26.770 --> 00:56:29.710
are derivatives that are
along the y direction.

00:56:33.530 --> 00:56:38.740
But this derivative
along the y direction

00:56:38.740 --> 00:56:44.410
in all of these equations has
to be contracted typically

00:56:44.410 --> 00:56:46.990
with something else.

00:56:46.990 --> 00:56:50.070
It has to be contacted with u.

00:56:50.070 --> 00:56:55.100
But the u's have no component
along the y direction.

00:56:55.100 --> 00:56:59.030
So essentially, all my
u's would be of this form.

00:56:59.030 --> 00:57:05.680
Basically, there will
be something like uy.

00:57:05.680 --> 00:57:09.450
Something like this
would have to be 0.

00:57:09.450 --> 00:57:12.340
You can see that if I start
with an initial condition

00:57:12.340 --> 00:57:18.640
such as that, then the
equations are that dt of n--

00:57:18.640 --> 00:57:23.800
this term I have
to forget-- is 0.

00:57:23.800 --> 00:57:26.920
Because for this, I
need a divergence.

00:57:26.920 --> 00:57:31.450
And this flow has no divergence.

00:57:31.450 --> 00:57:38.260
And similarly over here what
I see as dt of the temperature

00:57:38.260 --> 00:57:40.030
is 0.

00:57:40.030 --> 00:57:42.560
Temperature doesn't change.

00:57:42.560 --> 00:57:50.180
And if I assume that I
am under circumstances

00:57:50.180 --> 00:57:57.290
in which the
pressure is uniform,

00:57:57.290 --> 00:58:01.070
there's also nothing that
I would get from here.

00:58:01.070 --> 00:58:06.790
So essentially, this
flow will exist forever.

00:58:06.790 --> 00:58:09.060
Yes.

00:58:09.060 --> 00:58:12.840
STUDENT: Why does your u
alpha d alpha n term go away?

00:58:12.840 --> 00:58:14.122
Wouldn't you get a uxdxn?

00:58:16.905 --> 00:58:19.620
PROFESSOR: OK, let's
see, you want a uxdxn.

00:58:22.720 --> 00:58:24.980
What I said is
that all variations

00:58:24.980 --> 00:58:27.764
are along the y direction.

00:58:27.764 --> 00:58:29.680
STUDENT: Oh, so this is
not just for velocity,

00:58:29.680 --> 00:58:30.590
but for everything.

00:58:30.590 --> 00:58:32.750
PROFESSOR: Yes, so
I make an assumption

00:58:32.750 --> 00:58:34.460
about some particular form.

00:58:34.460 --> 00:58:37.240
So this is the reasoning.

00:58:37.240 --> 00:58:42.970
If these equations bring
everything to equilibrium,

00:58:42.970 --> 00:58:46.110
I should be able to pick
any initial condition

00:58:46.110 --> 00:58:49.140
and ask, how long does it
take to come to equilibrium?

00:58:49.140 --> 00:58:52.510
I pick this specific
type of equation

00:58:52.510 --> 00:58:56.290
in which the only variations
for all quantities

00:58:56.290 --> 00:58:57.780
are along the y direction.

00:58:57.780 --> 00:58:59.280
It's a non-equilibrium state.

00:58:59.280 --> 00:59:01.160
It's not a uniform state.

00:59:01.160 --> 00:59:03.340
Does it come relax
to equilibrium?

00:59:03.340 --> 00:59:06.272
And the answer is
no, it doesn't.

00:59:11.152 --> 00:59:14.568
STUDENT: What other properties,
other than velocity,

00:59:14.568 --> 00:59:16.832
is given [INAUDIBLE]?

00:59:16.832 --> 00:59:18.290
PROFESSOR: Density
and temperature.

00:59:18.290 --> 00:59:20.950
So these equations
describe the variations

00:59:20.950 --> 00:59:23.750
of velocity, density,
and temperature.

00:59:23.750 --> 00:59:28.350
And the statement is, if the
system is to reach equilibrium,

00:59:28.350 --> 00:59:32.170
I should be able to start
with any initial configuration

00:59:32.170 --> 00:59:34.750
of these three
quantities that I want.

00:59:34.750 --> 00:59:39.710
And I see that after a while,
it reaches a uniform state.

00:59:39.710 --> 00:59:40.540
Yes.

00:59:40.540 --> 00:59:43.420
STUDENT: But if your initial
conditions aren't exactly that,

00:59:43.420 --> 00:59:44.860
but you add a
slight fluctuation,

00:59:44.860 --> 00:59:51.280
it is likely to grow, and
it will eventually relax.

00:59:51.280 --> 00:59:53.950
PROFESSOR: It turns
out the answer is no.

00:59:53.950 --> 00:59:58.850
So I'm sort of approaching
this problem from this more

00:59:58.850 --> 01:00:02.600
kind of hand-waving perspective.

01:00:02.600 --> 01:00:04.610
More correctly,
what you can do is

01:00:04.610 --> 01:00:08.740
you can start with some initial
condition that, let's say,

01:00:08.740 --> 01:00:13.160
is in equilibrium, and
then do a perturbation,

01:00:13.160 --> 01:00:17.650
and ask whether the perturbation
will eventually relax to 0

01:00:17.650 --> 01:00:19.960
or not.

01:00:19.960 --> 01:00:23.340
And let's in fact do that
for another quantity,

01:00:23.340 --> 01:00:24.640
which is the sound mode.

01:00:34.130 --> 01:00:37.350
So let's imagine that
we start with a totally

01:00:37.350 --> 01:00:40.070
nice, uniform state.

01:00:40.070 --> 01:00:43.740
There is zero
velocity initially.

01:00:43.740 --> 01:00:45.130
The density is uniform.

01:00:45.130 --> 01:00:47.490
The temperature is uniform.

01:00:47.490 --> 01:00:50.740
And then what I do
is I will start here.

01:00:50.740 --> 01:00:58.650
And I will start talking,
creating a variation that

01:00:58.650 --> 01:01:02.430
propagates in this x direction.

01:01:02.430 --> 01:01:10.210
So I generated a stream that is
moving along the x direction.

01:01:10.210 --> 01:01:13.810
And presumably, as I move
along the x direction,

01:01:13.810 --> 01:01:18.290
there is a velocity that changes
with position and temperature.

01:01:22.240 --> 01:01:26.250
Now initially, I
had the density.

01:01:26.250 --> 01:01:27.390
I said that was uniform.

01:01:30.280 --> 01:01:35.780
Once I make this sound, as I
move along the x direction,

01:01:35.780 --> 01:01:39.320
and the air is flowing back
and forth, what happens

01:01:39.320 --> 01:01:47.640
is that the density will
vary from the uniform state.

01:01:47.640 --> 01:01:50.020
And the deviations
from the uniform

01:01:50.020 --> 01:01:55.260
state I will indicate by mu.

01:01:55.260 --> 01:02:01.410
Similarly, the third
quantity, let's assume,

01:02:01.410 --> 01:02:05.870
will have a form such as this.

01:02:05.870 --> 01:02:09.770
And currently, I have
written the most general form

01:02:09.770 --> 01:02:14.464
of variations that I can
have along the x direction.

01:02:14.464 --> 01:02:16.130
You could do it in
different directions.

01:02:16.130 --> 01:02:18.890
But let's say for simplicity,
we stick with this.

01:02:18.890 --> 01:02:22.830
I haven't told you what
mu theta and u are.

01:02:22.830 --> 01:02:28.110
So I have to see what they are
consistent with the equations

01:02:28.110 --> 01:02:30.990
that we have up there.

01:02:30.990 --> 01:02:35.900
One thing that I will
assume is that these things

01:02:35.900 --> 01:02:41.440
are small perturbations
around the uniform state.

01:02:41.440 --> 01:02:46.350
And uniform-- sorry, small
perturbations typically

01:02:46.350 --> 01:02:49.570
means that what
I intend to do is

01:02:49.570 --> 01:02:52.473
to do a linearized
approximation.

01:02:57.020 --> 01:03:01.200
So basically, what I will
do is I will essentially

01:03:01.200 --> 01:03:04.040
look at the linear version
of these equations.

01:03:04.040 --> 01:03:08.320
And again, maybe I didn't
emphasize it before.

01:03:08.320 --> 01:03:10.630
Clearly these are
nonlinear equations.

01:03:10.630 --> 01:03:13.010
Because let's say
you have u grad u.

01:03:13.010 --> 01:03:15.100
It's the same nonlinearity
that you have,

01:03:15.100 --> 01:03:17.280
let's say, in
Navier-Stokes equation.

01:03:17.280 --> 01:03:20.300
Because you're transporting
something and moving along

01:03:20.300 --> 01:03:22.120
with the flow.

01:03:22.120 --> 01:03:25.300
But when you do the
linearization, then

01:03:25.300 --> 01:03:30.050
these operators that involve
dt plus something like u--

01:03:30.050 --> 01:03:33.120
I guess in this case, the
only direction that is varying

01:03:33.120 --> 01:03:39.970
is x-- something like this of
whatever quantity that I have,

01:03:39.970 --> 01:03:43.330
I can drop this nonlinear term.

01:03:43.330 --> 01:03:44.440
Why?

01:03:44.440 --> 01:03:48.940
Because u is a perturbation
around a uniform state.

01:03:48.940 --> 01:03:53.340
And gradients will pick
up some perturbations

01:03:53.340 --> 01:03:55.000
around the uniform state.

01:03:55.000 --> 01:03:57.430
So essentially the
linearization amounts

01:03:57.430 --> 01:04:01.690
to dropping these
nonlinear components

01:04:01.690 --> 01:04:05.240
and some other things that
I will linearizer also.

01:04:05.240 --> 01:04:08.490
Because all of these
functions here,

01:04:08.490 --> 01:04:13.690
the derivatives act on product
of n temperature over here.

01:04:13.690 --> 01:04:18.960
These are all
nonlinear operations.

01:04:18.960 --> 01:04:21.920
So let's linearize what we have.

01:04:21.920 --> 01:04:28.860
We have that Dt of the density--
I guess when I take the time

01:04:28.860 --> 01:04:34.010
derivative, I get n bar the
time derivative of the quantity

01:04:34.010 --> 01:04:35.980
that I'm calling mu.

01:04:35.980 --> 01:04:36.770
And that's it.

01:04:36.770 --> 01:04:39.820
I don't need to worry
about the convective part,

01:04:39.820 --> 01:04:41.570
the u dot grad part.

01:04:41.570 --> 01:04:43.180
That's second order.

01:04:43.180 --> 01:04:45.901
On the right hand
side, what do I have?

01:04:45.901 --> 01:04:48.956
I have ndu.

01:04:48.956 --> 01:04:52.780
Well, divergence of u is
already the first variation.

01:04:52.780 --> 01:04:56.400
So for n, I will take
its 0-th order term.

01:04:56.400 --> 01:04:59.180
So I have minus n bar dxux.

01:05:04.020 --> 01:05:09.970
The equation for ux, really
the only component that I have,

01:05:09.970 --> 01:05:13.090
is dt of ux.

01:05:13.090 --> 01:05:18.210
Actually, let's write down
the equation for temperature.

01:05:18.210 --> 01:05:19.610
Let's look at this equation.

01:05:19.610 --> 01:05:28.750
So I have that dt
acting on 3/2 kB times

01:05:28.750 --> 01:05:31.670
T. I will pick up T bar.

01:05:31.670 --> 01:05:35.880
And then I would
have dt of theta.

01:05:35.880 --> 01:05:38.340
What do I have on
the right hand side?

01:05:38.340 --> 01:05:39.750
I have a derivative here.

01:05:39.750 --> 01:05:43.200
So everything else here I will
evaluate at the 0-th order

01:05:43.200 --> 01:05:47.505
term, so n bar k T bar dxux.

01:05:53.670 --> 01:05:58.900
So I can see that I can
certainly divide through n bar

01:05:58.900 --> 01:06:00.110
here.

01:06:00.110 --> 01:06:06.350
And one of my equations
becomes dt of mu is minus dxux.

01:06:10.520 --> 01:06:18.750
But from here, I see
that dxux is also

01:06:18.750 --> 01:06:28.780
related once I divide
by kT to 3/2 dt theta.

01:06:28.780 --> 01:06:30.080
And I know this to be true.

01:06:30.080 --> 01:06:34.720
And I seem to have an additional
factor of 1 over n bar here.

01:06:34.720 --> 01:06:42.310
And so I made the
mistake at some point,

01:06:42.310 --> 01:06:46.295
probably when I
wrote this equation.

01:06:49.010 --> 01:06:51.880
STUDENT: It's the
third equation.

01:06:51.880 --> 01:06:55.500
PROFESSOR: Yeah, so
this should not be here.

01:06:58.030 --> 01:07:02.000
And that should
not be here means

01:07:02.000 --> 01:07:07.380
that I probably
made a mistake here.

01:07:07.380 --> 01:07:10.870
So this should be a 1/n, sorry.

01:07:10.870 --> 01:07:15.595
There was indeed a 1/n / here.

01:07:15.595 --> 01:07:19.209
And there is no factor here.

01:07:23.700 --> 01:07:32.720
So we have a relationship
between the time derivatives

01:07:32.720 --> 01:07:39.650
of these variations in
density and dx of ux.

01:07:39.650 --> 01:07:44.200
Fine, what does the
equation for u tell us?

01:07:44.200 --> 01:07:54.880
It tells us that dt of ux
is minus 1 over m n bar.

01:07:54.880 --> 01:07:59.180
Because of the derivative,
I can set everything

01:07:59.180 --> 01:08:00.960
at the variation.

01:08:00.960 --> 01:08:02.950
And what do I have here?

01:08:02.950 --> 01:08:13.360
I have d by dx of
n bar kB T bar.

01:08:13.360 --> 01:08:17.930
And if I look at the variations,
I have 1 plus mu plus theta.

01:08:17.930 --> 01:08:20.649
The higher order
terms I will forget.

01:08:20.649 --> 01:08:21.453
Yes.

01:08:21.453 --> 01:08:25.074
STUDENT: Shouldn't that
be plus 3/2 dt theta?

01:08:25.074 --> 01:08:27.224
PROFESSOR: It
should be plus, yes.

01:08:27.224 --> 01:08:28.847
There is a minus sign here.

01:08:28.847 --> 01:08:30.146
And that makes it plus.

01:08:36.979 --> 01:08:41.770
So the n bar we
can take outside.

01:08:41.770 --> 01:08:53.040
This becomes minus kT
over m at space variations

01:08:53.040 --> 01:08:54.176
of mu plus theta.

01:08:59.569 --> 01:09:04.729
Now, what we do is
that what I have here

01:09:04.729 --> 01:09:09.310
is information about the time
derivatives of mu and theta.

01:09:09.310 --> 01:09:12.170
And here I have
space derivatives.

01:09:12.170 --> 01:09:13.800
So what do I do?

01:09:13.800 --> 01:09:19.590
I basically apply an
additionally dt here,

01:09:19.590 --> 01:09:21.950
which we'll apply here.

01:09:21.950 --> 01:09:23.850
And then we can
apply it also here.

01:09:30.960 --> 01:09:39.044
And then we know how dt
of mu and dt of theta

01:09:39.044 --> 01:09:42.380
are related to dx of ux.

01:09:42.380 --> 01:09:45.120
The minus signs disappear.

01:09:45.120 --> 01:09:51.144
I have kB T bar divided by m.

01:09:51.144 --> 01:09:55.540
I have dt of mu is dxux.

01:09:55.540 --> 01:09:58.130
dt of theta is 2/3 dxux.

01:09:58.130 --> 01:10:04.540
So I will get 1 plus
2/3 dx squared of ux.

01:10:10.330 --> 01:10:15.470
So the second
derivative of ux in time

01:10:15.470 --> 01:10:22.900
is proportional to the second
derivative of ux in space.

01:10:22.900 --> 01:10:25.890
So that's the standard
wave equation.

01:10:25.890 --> 01:10:29.090
And the velocity that
we have calculated

01:10:29.090 --> 01:10:36.048
for these sound waves is
5/3 kB the average T over m.

01:10:41.050 --> 01:10:43.750
So that part is good.

01:10:43.750 --> 01:10:49.930
These equations tell me that
if I create these disturbances,

01:10:49.930 --> 01:10:52.510
there are sound waves, and we
know there are sound waves.

01:10:52.510 --> 01:10:57.190
And sound waves will propagate
with some velocity that

01:10:57.190 --> 01:11:02.120
is related up to some factors to
the average velocity of the gas

01:11:02.120 --> 01:11:04.780
particles.

01:11:04.780 --> 01:11:08.570
But what is not good
is that, according

01:11:08.570 --> 01:11:12.480
to this equation, if
my waves, let's say,

01:11:12.480 --> 01:11:15.710
bounce off perfectly
from the walls,

01:11:15.710 --> 01:11:18.950
they will last in
this room forever.

01:11:18.950 --> 01:11:22.480
So you should still be hearing
what I was saying last week

01:11:22.480 --> 01:11:24.320
and the week before.

01:11:24.320 --> 01:11:27.250
And clearly what
we are missing is

01:11:27.250 --> 01:11:31.310
the damping that is required.

01:11:31.310 --> 01:11:35.340
So the statement is that all
of these equations are fine.

01:11:35.340 --> 01:11:38.670
They capture a lot
of the physics.

01:11:38.670 --> 01:11:41.190
But there is something
important that

01:11:41.190 --> 01:11:45.510
is left out in that
there are some modes--

01:11:45.510 --> 01:11:49.680
and I describe two of them
here-- that basically last

01:11:49.680 --> 01:11:52.590
forever, and don't
come to equilibrium.

01:11:52.590 --> 01:11:56.960
But we said that the Boltzmann
equation should eventually

01:11:56.960 --> 01:12:00.220
bring things to equilibrium.

01:12:00.220 --> 01:12:03.490
So where did we go wrong?

01:12:03.490 --> 01:12:05.760
Well, we didn't solve
the Boltzmann equation.

01:12:05.760 --> 01:12:09.690
We solved an approximation
to the Boltzmann equation.

01:12:09.690 --> 01:12:11.230
So let's try to do better.

01:12:16.079 --> 01:12:18.953
STUDENT: I'm sorry, but
for the last equation,

01:12:18.953 --> 01:12:25.660
you took another derivative
with respect to t.

01:12:25.660 --> 01:12:30.180
PROFESSOR: Yes, I took a
derivative with respect to t.

01:12:30.180 --> 01:12:32.790
And it noted that the
derivative with respect

01:12:32.790 --> 01:12:36.310
to t of these
quantities mu is related

01:12:36.310 --> 01:12:39.741
to derivative with
respect to x or u.

01:12:43.590 --> 01:12:45.890
And there was one other
derivative with respect

01:12:45.890 --> 01:12:48.942
to x already, making
it two derivatives.

01:12:55.690 --> 01:13:01.865
So this is the kind of
situation that we are facing.

01:13:07.590 --> 01:13:08.780
Yes.

01:13:08.780 --> 01:13:16.750
STUDENT: Is the 5/3 k in any
way related to the heat capacity

01:13:16.750 --> 01:13:17.960
ratio of [INAUDIBLE] gas?

01:13:17.960 --> 01:13:19.964
PROFESSOR: Yes,
that's right, yes.

01:13:19.964 --> 01:13:23.050
So there are lots
of these things that

01:13:23.050 --> 01:13:24.780
are implicit in these questions.

01:13:24.780 --> 01:13:32.360
And actually, that 3/2 is
the same thing as this 3/2.

01:13:32.360 --> 01:13:34.740
So you can trace a
lot of these things

01:13:34.740 --> 01:13:37.900
to the Gaussian distribution.

01:13:37.900 --> 01:13:41.030
And they appear in cp
versus cv and other things.

01:13:41.030 --> 01:13:41.920
Yes.

01:13:41.920 --> 01:13:43.880
STUDENT: Just
clarifying something--

01:13:43.880 --> 01:13:49.130
this v is different from
the mu in the top right?

01:13:49.130 --> 01:13:52.040
PROFESSOR: Yes, this
is v, and that's mu.

01:13:52.040 --> 01:13:54.450
This v is the
velocity of the sound.

01:13:54.450 --> 01:13:58.760
So I defined this combination,
the coefficient relating

01:13:58.760 --> 01:14:01.720
the second derivatives
in time and space

01:14:01.720 --> 01:14:03.010
as the sound velocity.

01:14:03.010 --> 01:14:17.020
So let's maybe even-- we
can call it vs. All right?

01:14:17.020 --> 01:14:19.000
STUDENT: And how did
you know that that

01:14:19.000 --> 01:14:25.450
is the [INAUDIBLE] oscillation,
the solution that you got?

01:14:25.450 --> 01:14:31.950
PROFESSOR: Because I know
that the solution to dx

01:14:31.950 --> 01:14:38.360
squared anything is
v squared-- sorry,

01:14:38.360 --> 01:14:44.390
v squared dx squared anything
is dt squared anything, is

01:14:44.390 --> 01:14:47.872
phi is some function
of x minus vt.

01:14:47.872 --> 01:14:51.620
That is a pulse that moves
with uniform velocity

01:14:51.620 --> 01:14:53.237
is a solution to this equation.

01:15:05.180 --> 01:15:06.320
So we want to do better.

01:15:09.350 --> 01:15:14.980
And better becomes so-called
first order solution.

01:15:14.980 --> 01:15:18.340
Now, the kind of
equation that we

01:15:18.340 --> 01:15:22.680
are trying to solve at
the top is something

01:15:22.680 --> 01:15:24.950
that its algebraic
analog would be

01:15:24.950 --> 01:15:28.830
something like this-- 2 times x.

01:15:28.830 --> 01:15:31.830
It's a linear on
the left hand side,

01:15:31.830 --> 01:15:34.000
is quadratic on the
right hand side.

01:15:34.000 --> 01:15:37.080
Let's write it in
this form-- except

01:15:37.080 --> 01:15:39.970
that the typical
magnitude of one side

01:15:39.970 --> 01:15:42.240
is much larger than
the other side,

01:15:42.240 --> 01:15:45.890
so let's say
something like this.

01:15:45.890 --> 01:15:48.720
So if I wanted to
solve this equation,

01:15:48.720 --> 01:15:52.860
I would say that unless
x is very close to 2,

01:15:52.860 --> 01:15:56.280
this 10 to the 6
will blow things up.

01:15:56.280 --> 01:15:59.832
So my x0 is 2.

01:15:59.832 --> 01:16:01.040
And that's what we have done.

01:16:01.040 --> 01:16:02.748
We've solved, essentially,
the right hand

01:16:02.748 --> 01:16:05.130
side of the equation.

01:16:05.130 --> 01:16:09.310
But I can get a better
solution by taking 2 and saying

01:16:09.310 --> 01:16:13.260
there's a small variation to
that that I want to calculate.

01:16:13.260 --> 01:16:16.140
And I substitute that into
the original equation.

01:16:16.140 --> 01:16:22.250
On the left hand side, I will
get 2 times 2, 1 plus epsilon.

01:16:22.250 --> 01:16:25.600
On the right hand side, I
will get 10 to the sixth.

01:16:25.600 --> 01:16:28.660
And then essentially,
I subtract 2

01:16:28.660 --> 01:16:31.640
plus 2 epsilon squared from 5.

01:16:31.640 --> 01:16:32.460
What do I get?

01:16:32.460 --> 01:16:39.620
I will get 4 epsilon
plus 4 epsilon squared.

01:16:43.040 --> 01:16:45.040
Then I say that
epsilon is small.

01:16:45.040 --> 01:16:48.870
So essentially, I linearize
the right hand side.

01:16:48.870 --> 01:16:50.970
I forget about that.

01:16:50.970 --> 01:16:52.980
I say that I keep
the epsilon here,

01:16:52.980 --> 01:16:55.550
because it's
multiplying 10 to the 6.

01:16:55.550 --> 01:16:59.130
But the epsilon on the other
side is multiplying nothing.

01:16:59.130 --> 01:17:00.850
So I forget that.

01:17:00.850 --> 01:17:06.560
So then I will have my epsilon
to be roughly, I don't know,

01:17:06.560 --> 01:17:12.430
2 times 2 divided by
4 times 10 to the 6.

01:17:12.430 --> 01:17:18.647
So I have gotten the correction
to my first 0-th order solution

01:17:18.647 --> 01:17:19.480
to this first order.

01:17:22.140 --> 01:17:24.360
Now we will do exactly
the same thing,

01:17:24.360 --> 01:17:29.640
not for our algebraic equation,
but for our Boltzmann equation.

01:17:29.640 --> 01:17:37.700
So for the Boltzmann equation,
which was Lf is C of ff,

01:17:37.700 --> 01:17:41.620
we said that the right hand
side is larger by a factor of 1

01:17:41.620 --> 01:17:45.100
over tau x compared
to the left hand side.

01:17:45.100 --> 01:17:49.350
And so what we did was
we found a solution f0

01:17:49.350 --> 01:17:51.590
that when we put in
the collision integral,

01:17:51.590 --> 01:17:52.610
the answer was 0.

01:17:55.180 --> 01:17:59.830
Now I want to have a better
solution that I will call f1.

01:17:59.830 --> 01:18:01.650
Just like I did
over there, I will

01:18:01.650 --> 01:18:05.250
assume that f0 is added
to a small function

01:18:05.250 --> 01:18:06.290
that I will call g.

01:18:08.920 --> 01:18:14.230
And then I substitute
this equation, this thing,

01:18:14.230 --> 01:18:16.910
to the equation.

01:18:16.910 --> 01:18:23.010
So when I substitute, I will
get L acting on f0 1 plus g.

01:18:28.720 --> 01:18:31.750
Now, what did I do over here?

01:18:31.750 --> 01:18:36.390
On the left hand side, I
ignored the first order term.

01:18:36.390 --> 01:18:39.020
Because I expect the first
order term to be already small.

01:18:39.020 --> 01:18:42.016
And the left hand
side is already small.

01:18:42.016 --> 01:18:45.290
So I will ignore this.

01:18:45.290 --> 01:18:51.030
And on the right hand
side, I have to linearize.

01:18:51.030 --> 01:18:59.430
So I have to put f0 1
plus g, f0 1 plus g.

01:18:59.430 --> 01:19:03.460
Essentially what I have to
do is to go to the collisions

01:19:03.460 --> 01:19:09.590
that I have over here and
write for this f0 1 plus g.

01:19:09.590 --> 01:19:11.742
There are four of such things.

01:19:11.742 --> 01:19:14.610
Now, the 0-th order
term already cancels.

01:19:14.610 --> 01:19:20.500
Because f0 f0 was f0 f0 by the
way that I constructed things.

01:19:20.500 --> 01:19:22.510
And then I can
pull out one factor

01:19:22.510 --> 01:19:25.000
of f0 out of the integration.

01:19:25.000 --> 01:19:27.470
So when I linearize
this, what I will get

01:19:27.470 --> 01:19:31.310
is something like f0
that goes on the outside.

01:19:31.310 --> 01:19:37.950
I have the integral
d2p2 d2b v2 minus v1.

01:19:37.950 --> 01:19:43.790
And then I have something like
g of p1 plus g of p2 minus

01:19:43.790 --> 01:19:47.490
g of p1 prime minus
g of p2 prime.

01:19:53.120 --> 01:19:56.610
So basically, what
we have done is

01:19:56.610 --> 01:20:01.090
we have defined a linearized
version of the collision

01:20:01.090 --> 01:20:04.970
operator that is now
linear in this variable g.

01:20:11.390 --> 01:20:15.820
Now in general,
this is also still,

01:20:15.820 --> 01:20:19.740
although a linear
operator, much simpler

01:20:19.740 --> 01:20:22.400
than the previous
quadratic operator--

01:20:22.400 --> 01:20:25.500
still has a lot of junk in it.

01:20:25.500 --> 01:20:30.275
So we are going to
simply state that I

01:20:30.275 --> 01:20:34.290
will use a form of this
linearized approximation that

01:20:34.290 --> 01:20:38.390
is simply g over tau, get rid
of all of the integration.

01:20:38.390 --> 01:20:41.730
And this is called the single
collision time approximation.

01:20:55.910 --> 01:20:58.560
So having done
that, what I have is

01:20:58.560 --> 01:21:05.590
that the L acting on f0
on the left hand side

01:21:05.590 --> 01:21:14.110
is minus f0 g over tau x
on the right hand side.

01:21:14.110 --> 01:21:17.170
And so I can
immediately solve for g.

01:21:17.170 --> 01:21:21.540
Because L is a first
order derivative operator.

01:21:21.540 --> 01:21:28.320
My g is minus tau x,
the Liouville operator

01:21:28.320 --> 01:21:39.230
acting on f0-- sorry, log of f0.

01:21:39.230 --> 01:21:41.540
I divide it through by f0.

01:21:41.540 --> 01:21:45.770
So derivative of
f0 divided by f0

01:21:45.770 --> 01:21:48.690
is the derivative
acting on log of f0.

01:21:48.690 --> 01:21:52.410
So all I need to do
is to essentially do

01:21:52.410 --> 01:21:56.390
the operations involved in
taking these derivatives.

01:21:59.300 --> 01:22:01.420
Let's say we forget
about the force.

01:22:01.420 --> 01:22:04.280
Because we are looking
in the middle of the box,

01:22:04.280 --> 01:22:08.230
acting on log of what
I had written before.

01:22:08.230 --> 01:22:17.260
So what I have is log of n minus
p minus mu squared over 2mkT.

01:22:17.260 --> 01:22:21.960
Remember, ukTn are all
functions of position.

01:22:21.960 --> 01:22:24.230
So there will be
derivatives involved here.

01:22:30.750 --> 01:22:37.150
And I will just write
down what the answer is.

01:22:37.150 --> 01:22:41.760
So the answer
becomes minus tau x.

01:22:41.760 --> 01:22:45.730
You would have, once you
do all of these derivatives

01:22:45.730 --> 01:22:48.070
and take advantage
of the equations

01:22:48.070 --> 01:22:50.630
that you have written
before-- so there's

01:22:50.630 --> 01:22:53.150
some lines of algebra involved.

01:22:53.150 --> 01:23:07.540
The final answer is going to
be nm c alpha c beta minus--

01:23:07.540 --> 01:23:10.990
I should really look at this.

01:23:19.930 --> 01:23:32.580
m over kT c alpha c beta
minus delta alpha beta over 3

01:23:32.580 --> 01:23:40.800
c squared u alpha
beta, and then mc

01:23:40.800 --> 01:23:56.560
squared over 2kT minus 5/2 c
alpha over T d alpha of T. Yes.

01:23:56.560 --> 01:23:58.730
STUDENT: Sorry, what's
the thing next to the c

01:23:58.730 --> 01:24:01.290
squared, something alpha beta?

01:24:01.290 --> 01:24:03.188
PROFESSOR: Delta
alpha beta, sorry.

01:24:07.700 --> 01:24:11.790
So there is a
well-defined procedure--

01:24:11.790 --> 01:24:16.140
it's kind of algebraically
involved-- by which

01:24:16.140 --> 01:24:18.460
more or less in the
same fashion that you

01:24:18.460 --> 01:24:22.480
can improve on the
algebraic solution,

01:24:22.480 --> 01:24:25.000
get a better
solution than the one

01:24:25.000 --> 01:24:30.380
that we had that
now knows something

01:24:30.380 --> 01:24:33.420
about these relaxations.

01:24:33.420 --> 01:24:36.890
See, the 0-th order
solution that we had

01:24:36.890 --> 01:24:38.560
knew nothing about tau x.

01:24:38.560 --> 01:24:42.270
We just set tau x
to be very small,

01:24:42.270 --> 01:24:44.510
and set the right
hand side to 0.

01:24:44.510 --> 01:24:46.740
And then nothing relaxed.

01:24:46.740 --> 01:24:51.950
Now we have a better solution
that involves explicitly tau x.

01:24:51.950 --> 01:24:54.490
And if we start with
that, we'll find

01:24:54.490 --> 01:24:58.750
that we can get relaxation
of all of these modes

01:24:58.750 --> 01:25:05.510
once we calculate p alpha beta
and h alpha with this better

01:25:05.510 --> 01:25:06.640
solution.

01:25:06.640 --> 01:25:09.330
We can immediately,
for example, see

01:25:09.330 --> 01:25:13.020
that this new solution will
have terms that are odd.

01:25:13.020 --> 01:25:15.500
There is c cubed term here.

01:25:15.500 --> 01:25:18.540
So when you're
evaluating this average,

01:25:18.540 --> 01:25:20.640
you will no longer get 0.

01:25:20.640 --> 01:25:24.520
Heat has a chance to flow
with this improved equation.

01:25:24.520 --> 01:25:27.530
And again, whereas
before our pressure

01:25:27.530 --> 01:25:30.140
was diagonal because
of these terms,

01:25:30.140 --> 01:25:32.680
we will have off
diagonal terms that

01:25:32.680 --> 01:25:34.950
will allow us to
relax the shear modes.

01:25:34.950 --> 01:25:37.471
And we'll do that next.