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PROFESSOR: So we've
been wondering

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00:00:23,090 --> 00:00:28,180
how a gas, such as
the one in this room,

10
00:00:28,180 --> 00:00:32,220
with particles following
Newtonian equations of motion,

11
00:00:32,220 --> 00:00:34,710
comes to equilibrium.

12
00:00:34,710 --> 00:00:39,270
We decided to explain that
by relying on the Boltzmann

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00:00:39,270 --> 00:00:41,120
equation.

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00:00:41,120 --> 00:00:46,190
Essentially, we said, let's look
at the density, or probability,

15
00:00:46,190 --> 00:00:50,180
that we will find particles
with momentum p at location q

16
00:00:50,180 --> 00:00:56,190
at time t, and we found that
within some approximations,

17
00:00:56,190 --> 00:01:00,540
we could represent
the evolution of this

18
00:01:00,540 --> 00:01:04,379
through a linear operator
on the left-hand side

19
00:01:04,379 --> 00:01:07,340
equal to some
collision second order

20
00:01:07,340 --> 00:01:11,090
operator on the right-hand side.

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00:01:11,090 --> 00:01:16,255
The linear operator is
a bunch of derivatives.

22
00:01:16,255 --> 00:01:22,190
There is the time derivative,
then the coordinate moves

23
00:01:22,190 --> 00:01:26,580
according to the velocity,
which is P over m.

24
00:01:26,580 --> 00:01:31,540
And summation over the index
alpha running from 1 to 3,

25
00:01:31,540 --> 00:01:34,600
or x, y, z, is assumed.

26
00:01:34,600 --> 00:01:38,940
And at this stage,
there is symmetry

27
00:01:38,940 --> 00:01:40,800
within the
representation in terms

28
00:01:40,800 --> 00:01:42,880
of coordinates and momenta.

29
00:01:42,880 --> 00:01:48,420
There is also a force that
causes changes of momentum.

30
00:01:48,420 --> 00:01:50,860
Actually, for most
of today, we are

31
00:01:50,860 --> 00:01:53,110
going to be interested
in something

32
00:01:53,110 --> 00:01:56,910
like the gas in this room
far away from the walls,

33
00:01:56,910 --> 00:02:00,290
where essentially there
is no external potential.

34
00:02:00,290 --> 00:02:04,870
And for the interest of thinking
about modes such as sound,

35
00:02:04,870 --> 00:02:09,729
et cetera, we can remove that.

36
00:02:09,729 --> 00:02:17,940
The second order operator-- the
collision operator-- however,

37
00:02:17,940 --> 00:02:22,350
explicitly because we were
thinking about collisions

38
00:02:22,350 --> 00:02:24,390
among particles that
are taking place

39
00:02:24,390 --> 00:02:27,810
when they are in close
contact, breaks the symmetry

40
00:02:27,810 --> 00:02:31,960
between coordinate and momenta
that were present here.

41
00:02:31,960 --> 00:02:35,320
So this collision
operator was an integral

42
00:02:35,320 --> 00:02:39,640
over the momentum of
another particle that

43
00:02:39,640 --> 00:02:42,590
would come at some
impact parameter

44
00:02:42,590 --> 00:02:46,050
with relative
velocity, and then you

45
00:02:46,050 --> 00:02:53,680
had a subtraction due to
collisions, and then addition

46
00:02:53,680 --> 00:02:55,940
due to reverse collisions.

47
00:03:03,770 --> 00:03:07,500
Now, what we said is
because of the way

48
00:03:07,500 --> 00:03:12,100
that this symmetry is
broken, we can distinguish

49
00:03:12,100 --> 00:03:17,550
between averages in
position and in momentum.

50
00:03:17,550 --> 00:03:21,530
For example, typically we
are interested in variations

51
00:03:21,530 --> 00:03:24,050
of various quantities in space.

52
00:03:24,050 --> 00:03:27,860
And so what we can do is
we can define the density,

53
00:03:27,860 --> 00:03:30,280
let's say at some
particular point,

54
00:03:30,280 --> 00:03:34,330
by integrating over momentum.

55
00:03:37,550 --> 00:03:40,400
So again, I said I don't
really need too much

56
00:03:40,400 --> 00:03:43,630
about the dependence
on momentum,

57
00:03:43,630 --> 00:03:45,770
but I really am
interested in how

58
00:03:45,770 --> 00:03:48,940
things vary from
position to position.

59
00:03:48,940 --> 00:03:53,700
Once we have defined density, we
could define various averages,

60
00:03:53,700 --> 00:03:57,880
where we would multiply this
integral by some function of P

61
00:03:57,880 --> 00:04:03,630
and q, and the average was
defined in this fashion.

62
00:04:03,630 --> 00:04:07,900
What we found was
that really what

63
00:04:07,900 --> 00:04:13,510
happens through this collision
operator-- that typically

64
00:04:13,510 --> 00:04:17,420
is much more important, as
far as the inverse time scales

65
00:04:17,420 --> 00:04:20,010
are concerned, than the
operators on the left,

66
00:04:20,010 --> 00:04:23,170
it has a much bigger
magnitude-- is

67
00:04:23,170 --> 00:04:31,890
that momenta are very rapidly
exchanged and randomized.

68
00:04:31,890 --> 00:04:36,560
And the things that are
randomized most slowly

69
00:04:36,560 --> 00:04:38,853
are quantities that are
conserved in collision.

70
00:04:45,000 --> 00:04:49,870
And so we focused on quantities
that were collision conserved,

71
00:04:49,870 --> 00:04:53,540
and we found that for each
one of these quantities,

72
00:04:53,540 --> 00:04:59,000
we could write down some kind
of a hydrodynamic equation.

73
00:04:59,000 --> 00:05:03,770
In particular, if we looked
at number conservation--

74
00:05:03,770 --> 00:05:05,750
two particles come
in, two particles

75
00:05:05,750 --> 00:05:09,480
go out-- we found that the
equation that described that

76
00:05:09,480 --> 00:05:18,230
was that the time derivative of
the density plus moving along

77
00:05:18,230 --> 00:05:24,480
the streamline had
a special form.

78
00:05:24,480 --> 00:05:32,250
And here, I defined the quantity
u alpha, which is simply

79
00:05:32,250 --> 00:05:38,077
the average of P alpha
over m, defined in the way

80
00:05:38,077 --> 00:05:41,660
that averages defined over here.

81
00:05:41,660 --> 00:05:45,520
And that this is the
number of particles,

82
00:05:45,520 --> 00:05:48,820
how it changes if I move
along the streamline

83
00:05:48,820 --> 00:05:52,300
And variations of this,
if you are incompressible,

84
00:05:52,300 --> 00:05:58,140
comes from the divergence
of your flow velocity.

85
00:05:58,140 --> 00:06:03,340
And actually, this operator we
call the total, or streamline,

86
00:06:03,340 --> 00:06:03,840
derivative.

87
00:06:08,040 --> 00:06:09,780
Now, the next thing
that we said is

88
00:06:09,780 --> 00:06:13,930
that OK, number is conserved,
but momentum is also conserved.

89
00:06:13,930 --> 00:06:19,460
So we can see what equation
I get if I look at momentum.

90
00:06:19,460 --> 00:06:22,180
But in fact, we put
the quantity that

91
00:06:22,180 --> 00:06:27,770
was momentum divided by mass
to make it like a velocity,

92
00:06:27,770 --> 00:06:31,190
and how much you
deviate from the average

93
00:06:31,190 --> 00:06:33,630
that we just calculated.

94
00:06:33,630 --> 00:06:38,040
And this quantity
we also call c.

95
00:06:38,040 --> 00:06:42,020
And when we looked at the
equation that corresponded

96
00:06:42,020 --> 00:06:45,290
to the conservation of this
quantity in collisions,

97
00:06:45,290 --> 00:06:51,060
we found that we had
something like mass times

98
00:06:51,060 --> 00:06:53,310
acceleration along
the streamline.

99
00:06:53,310 --> 00:06:56,160
So basically this
operator multiplied

100
00:06:56,160 --> 00:07:01,770
by mass acting on this
was external force.

101
00:07:01,770 --> 00:07:04,500
Well, currently we've
said this external force

102
00:07:04,500 --> 00:07:07,290
to be zero-- so when
we are inside the box.

103
00:07:07,290 --> 00:07:10,140
Here there was an
additional force

104
00:07:10,140 --> 00:07:16,940
that came from variations
of pressure in the gas.

105
00:07:16,940 --> 00:07:22,410
And so we had a term here that
was 1 over n d alpha of P alpha

106
00:07:22,410 --> 00:07:29,330
beta, and we needed to define
a pressure tensor P alpha

107
00:07:29,330 --> 00:07:36,997
beta, which was nm
expectation line of c alpha c.

108
00:07:41,400 --> 00:07:45,020
Finally, in collisions
there is another quantity

109
00:07:45,020 --> 00:07:48,720
that is conserved, which
is the kinetic energy.

110
00:07:48,720 --> 00:07:53,470
So here the third quantity
could be the kinetic energy--

111
00:07:53,470 --> 00:07:56,970
or actually, we chose
the combination mc

112
00:07:56,970 --> 00:08:02,210
squared over 2, which is the
additional kinetic energy

113
00:08:02,210 --> 00:08:03,260
on top of this.

114
00:08:03,260 --> 00:08:07,000
And it's very easy to check that
if kinetic energy is conserved,

115
00:08:07,000 --> 00:08:09,630
this quantity is also conserved.

116
00:08:09,630 --> 00:08:13,045
We call the average of
this quantity in the way

117
00:08:13,045 --> 00:08:18,850
that we have defined
above as epsilon.

118
00:08:18,850 --> 00:08:21,630
And then the
hydrodynamic equation

119
00:08:21,630 --> 00:08:26,536
is that as you move
along the streams--

120
00:08:26,536 --> 00:08:29,800
so you have this derivative
acting on this quantity

121
00:08:29,800 --> 00:08:35,049
epsilon-- what we are going to
get is something like minus 1

122
00:08:35,049 --> 00:08:41,059
over n d alpha of a
new vector, h alpha.

123
00:08:41,059 --> 00:08:50,430
h alpha, basically, tells me how
this quantity is transported.

124
00:08:50,430 --> 00:08:55,900
So all I need to do is to
have something like mc squared

125
00:08:55,900 --> 00:09:01,690
over 2 transported
along direction alpha.

126
00:09:01,690 --> 00:09:04,330
And then there was
another term, which

127
00:09:04,330 --> 00:09:09,040
was P alpha beta-- the
P alpha beta that we

128
00:09:09,040 --> 00:09:13,060
defined above--
times u alpha beta.

129
00:09:13,060 --> 00:09:20,100
U alpha beta was simply the
derivative of this quantity,

130
00:09:20,100 --> 00:09:20,600
symmetrized.

131
00:09:28,300 --> 00:09:32,050
So the statement
is that something

132
00:09:32,050 --> 00:09:38,270
like a gas-- or any
other fluid, in fact-- we

133
00:09:38,270 --> 00:09:42,020
can describe through
these quantities that

134
00:09:42,020 --> 00:09:45,265
are varying from one
location to another location.

135
00:09:45,265 --> 00:09:47,650
There is something
like a density,

136
00:09:47,650 --> 00:09:50,030
how dense it is
at this location.

137
00:09:50,030 --> 00:09:53,160
How fast particles are
streaming from one location

138
00:09:53,160 --> 00:09:54,700
to another location.

139
00:09:54,700 --> 00:09:57,830
And the energy content
that is ultimately related

140
00:09:57,830 --> 00:09:59,400
to something like
the temperature,

141
00:09:59,400 --> 00:10:00,681
how hot it is locally.

142
00:10:03,330 --> 00:10:05,770
So you have these equations.

143
00:10:05,770 --> 00:10:11,340
Solving these equations,
presumably, is equivalent,

144
00:10:11,340 --> 00:10:13,990
in some sense, to solving
the Boltzmann equation.

145
00:10:13,990 --> 00:10:16,690
The Boltzmann equation, we
know, ultimately reaches

146
00:10:16,690 --> 00:10:19,870
an equilibrium, so we
should be able to figure out

147
00:10:19,870 --> 00:10:22,700
how the system, such as
the gas in this room,

148
00:10:22,700 --> 00:10:26,260
is if disturbed, comes to
equilibrium, if we follow

149
00:10:26,260 --> 00:10:30,930
the density, velocity,
and temperature.

150
00:10:30,930 --> 00:10:33,940
Now, the problem with the
equations as I have written

151
00:10:33,940 --> 00:10:37,240
is that they are not closed
in terms of these three

152
00:10:37,240 --> 00:10:43,080
quantities, because I need
to evaluate the pressure,

153
00:10:43,080 --> 00:10:48,540
I need to evaluate the
heat transfer vector.

154
00:10:48,540 --> 00:10:53,040
And to calculate
these quantities,

155
00:10:53,040 --> 00:10:56,420
I need to be able to
evaluate these averages.

156
00:10:56,420 --> 00:11:01,920
In order to evaluate these
averages, I need to know f.

157
00:11:01,920 --> 00:11:05,380
So how did we proceed?

158
00:11:05,380 --> 00:11:10,690
We said well, let's try to find
approximate solutions for f.

159
00:11:10,690 --> 00:11:13,240
So the next task
is to find maybe

160
00:11:13,240 --> 00:11:19,740
some f, which is a function
of p, q, and t, and we

161
00:11:19,740 --> 00:11:23,650
can substitute over there.

162
00:11:23,650 --> 00:11:26,340
Now, the first
thing that we said

163
00:11:26,340 --> 00:11:33,300
was, OK, maybe what I can do
is I can look at the equation

164
00:11:33,300 --> 00:11:35,050
itself.

165
00:11:35,050 --> 00:11:37,740
Notice that this
part of the equation

166
00:11:37,740 --> 00:11:41,780
is order of 1 over the
time it takes for particles

167
00:11:41,780 --> 00:11:44,430
to move in the gas and
find another particle

168
00:11:44,430 --> 00:11:48,430
to collide with, whereas the
left-hand side is presumably

169
00:11:48,430 --> 00:11:52,510
something that is
related to how far I

170
00:11:52,510 --> 00:11:57,640
go before I see some variation
due to the external box.

171
00:11:57,640 --> 00:12:03,970
And we are really thinking about
cases where the gas particles

172
00:12:03,970 --> 00:12:06,520
are not that
dilute, in the sense

173
00:12:06,520 --> 00:12:09,380
that along the way to go
from one side of the room

174
00:12:09,380 --> 00:12:11,920
to another side of the
room, you encounter

175
00:12:11,920 --> 00:12:13,930
many, many collisions.

176
00:12:13,930 --> 00:12:18,290
So the term on the right-hand
side is much larger.

177
00:12:18,290 --> 00:12:20,720
If that is the case,
we said that maybe it's

178
00:12:20,720 --> 00:12:25,030
justifiable to just solve this
equation on the right-hand side

179
00:12:25,030 --> 00:12:26,970
as a zeroth order.

180
00:12:26,970 --> 00:12:32,180
And to solve that, I really
have to set, for example,

181
00:12:32,180 --> 00:12:37,490
the right-hand side that
I'm integrating here, to 0.

182
00:12:37,490 --> 00:12:39,400
And I know how to do that.

183
00:12:39,400 --> 00:12:46,070
If log f involves collision
conserved quantities,

184
00:12:46,070 --> 00:12:50,310
then ff before is
the same as ff after.

185
00:12:50,310 --> 00:12:53,200
And the solution
that I get by doing

186
00:12:53,200 --> 00:13:00,350
that has the form of exponential
involving conserved quantities,

187
00:13:00,350 --> 00:13:02,480
which are the quantities
that I have indicated

188
00:13:02,480 --> 00:13:05,350
over here-- let's
say, such as c.

189
00:13:05,350 --> 00:13:07,590
And so log of that
would be something

190
00:13:07,590 --> 00:13:10,440
that involves--
I can write as mc

191
00:13:10,440 --> 00:13:14,210
squared over 2 with
some coefficient

192
00:13:14,210 --> 00:13:18,534
that, in principle, varies from
location to another location.

193
00:13:21,200 --> 00:13:25,980
I want to integrate this and
come up with the density,

194
00:13:25,980 --> 00:13:30,460
so I put the density out here,
and I normalize the Gaussian.

195
00:13:35,890 --> 00:13:39,870
And so this is a
reasonable solution.

196
00:13:39,870 --> 00:13:44,400
Indeed, this is the
zeroth order solution

197
00:13:44,400 --> 00:13:45,940
for f-- I'll call that f0.

198
00:13:49,530 --> 00:13:52,810
So once you have the
zeroth order solution,

199
00:13:52,810 --> 00:13:55,850
from that you can calculate
these two quantities.

200
00:13:55,850 --> 00:14:01,220
For example, because the zeroth
order solution is even in c,

201
00:14:01,220 --> 00:14:04,620
the heat vector will
be 0, because it

202
00:14:04,620 --> 00:14:07,540
is symmetric in the
different components.

203
00:14:07,540 --> 00:14:12,710
The pressure tensor will be
proportional to delta alpha

204
00:14:12,710 --> 00:14:15,040
beta.

205
00:14:15,040 --> 00:14:16,480
OK?

206
00:14:16,480 --> 00:14:18,170
We started with that.

207
00:14:18,170 --> 00:14:21,740
Put those over
here, and we found

208
00:14:21,740 --> 00:14:24,640
that we could get some
results that were interesting.

209
00:14:24,640 --> 00:14:30,320
For example, we could see that
the gas can have sound modes.

210
00:14:30,320 --> 00:14:32,570
We could calculate
the speed of sound.

211
00:14:32,570 --> 00:14:35,030
But these sound modes
were not damped.

212
00:14:35,030 --> 00:14:38,130
And there were other modes,
such as sheer modes, that

213
00:14:38,130 --> 00:14:41,760
existed forever,
confounding our expectation

214
00:14:41,760 --> 00:14:44,630
that these equations
should eventually

215
00:14:44,630 --> 00:14:47,400
come to an equilibrium.

216
00:14:47,400 --> 00:14:50,810
So we said, OK, this
was a good attempt.

217
00:14:50,810 --> 00:14:54,660
But what was not good enough to
give us complete equilibrium,

218
00:14:54,660 --> 00:14:57,820
so let's try to find
a better solution.

219
00:14:57,820 --> 00:15:00,960
So how did we find
the better solution?

220
00:15:00,960 --> 00:15:04,320
We said that the
better solution-- let's

221
00:15:04,320 --> 00:15:08,330
assume that the
solution is like this,

222
00:15:08,330 --> 00:15:14,770
but is slightly changed
by a correction.

223
00:15:14,770 --> 00:15:17,570
And the correction comes
because of the effect

224
00:15:17,570 --> 00:15:21,690
of the left-hand side,
which we had ignored so far.

225
00:15:21,690 --> 00:15:24,620
And since the left-hand
side is smaller

226
00:15:24,620 --> 00:15:27,970
than the right-hand side by
a factor involving tau x,

227
00:15:27,970 --> 00:15:32,490
presumably the correction
will involve this tau x.

228
00:15:32,490 --> 00:15:32,990
OK.

229
00:15:32,990 --> 00:15:34,690
Good?

230
00:15:34,690 --> 00:15:37,960
We said that in
order to see that

231
00:15:37,960 --> 00:15:40,990
as a correction,
what I need to do

232
00:15:40,990 --> 00:15:44,760
is to essentially
linearize this expression.

233
00:15:44,760 --> 00:15:52,976
So what we did was we replaced
this f's with f0's 1 plus g,

234
00:15:52,976 --> 00:15:56,850
1 plus g, and so forth.

235
00:15:56,850 --> 00:16:00,920
The zeroth term, by
construction, is 0.

236
00:16:00,920 --> 00:16:04,680
And so if we ignore terms
that are order of g squared,

237
00:16:04,680 --> 00:16:08,553
we get something
that is linear in g.

238
00:16:08,553 --> 00:16:09,940
OK?

239
00:16:09,940 --> 00:16:13,190
Now, it's still an
integral operator,

240
00:16:13,190 --> 00:16:16,920
but we said let's
approximate that integration

241
00:16:16,920 --> 00:16:31,710
and let's do a linearized one
collision time approximation.

242
00:16:31,710 --> 00:16:35,050
What that approximation
amounted to was

243
00:16:35,050 --> 00:16:41,010
that, independent of what this
deviation is, if we are relaxed

244
00:16:41,010 --> 00:16:44,510
to the zeroth order
solution over time scale

245
00:16:44,510 --> 00:16:48,280
that is the same, and that
times scale we'll call tau x.

246
00:16:48,280 --> 00:16:54,150
So essentially we wrote this
as minus f0, essentially

247
00:16:54,150 --> 00:16:59,830
g, which is the difference
between f and f0,

248
00:16:59,830 --> 00:17:02,670
divided by tau x.

249
00:17:05,876 --> 00:17:07,140
f0, this was g.

250
00:17:12,829 --> 00:17:13,450
No.

251
00:17:13,450 --> 00:17:15,109
I guess we don't have this.

252
00:17:15,109 --> 00:17:18,810
We wrote it in this fashion.

253
00:17:18,810 --> 00:17:21,300
It's just writing the
way that I wrote before.

254
00:17:21,300 --> 00:17:26,329
We wrote this as g
over tau x, where

255
00:17:26,329 --> 00:17:30,590
g is the correction that
I have to write here.

256
00:17:30,590 --> 00:17:34,120
And you can see that the
correction is obtained

257
00:17:34,120 --> 00:17:41,200
by multiplying minus tau x with
l acting on f0 divided by f0,

258
00:17:41,200 --> 00:17:45,210
which is l acting on log of f0.

259
00:17:45,210 --> 00:17:50,600
So I would have
here 1 minus tau x--

260
00:17:50,600 --> 00:17:53,420
let's make this curly
bracket-- and then

261
00:17:53,420 --> 00:18:03,740
in the bracket over here, I
have to put the action of l

262
00:18:03,740 --> 00:18:04,770
on the log of f0.

263
00:18:09,020 --> 00:18:12,200
So I have to-- this is my f0.

264
00:18:12,200 --> 00:18:13,320
I take its log.

265
00:18:13,320 --> 00:18:15,600
So the log will have
minus mc squared

266
00:18:15,600 --> 00:18:19,900
over 2 kt and the log
of this combination,

267
00:18:19,900 --> 00:18:26,370
and I do d by dt plus P alpha
over m acting on this log,

268
00:18:26,370 --> 00:18:29,630
and then a bunch of
algebra will leave you

269
00:18:29,630 --> 00:18:31,450
to the following answer.

270
00:18:31,450 --> 00:18:35,900
Not surprisingly, you're
going to get factors of mkT.

271
00:18:35,900 --> 00:18:39,280
So you will get m over kT.

272
00:18:39,280 --> 00:18:44,380
You get c alpha c beta
minus delta alpha beta

273
00:18:44,380 --> 00:18:50,750
c squared over 3 acting on
this rate of strength tensor

274
00:18:50,750 --> 00:18:53,160
that we have defined over here.

275
00:18:53,160 --> 00:18:54,760
And then there are
derivatives that

276
00:18:54,760 --> 00:18:58,170
will act on temperature--
because temperature is allowed

277
00:18:58,170 --> 00:19:01,155
to vary from position to
position-- so there will be

278
00:19:01,155 --> 00:19:06,990
a term that will involve the
derivative of temperature.

279
00:19:06,990 --> 00:19:12,740
In fact, it will come in the
form over T c alpha multiplying

280
00:19:12,740 --> 00:19:21,530
another combination which is
mc squared over 2kT minus 5/2.

281
00:19:21,530 --> 00:19:25,625
Let me see if I got all of
the factors of 1/2 correct.

282
00:19:30,430 --> 00:19:30,930
Yeah.

283
00:19:35,061 --> 00:19:35,560
OK.

284
00:19:35,560 --> 00:19:37,280
So this is the first order term.

285
00:19:37,280 --> 00:19:40,660
Presumably, there will be
high order corrections,

286
00:19:40,660 --> 00:19:48,650
but this is the improved
solution to the Boltzmann

287
00:19:48,650 --> 00:19:53,450
equation beyond the zeroth
order approximation.

288
00:19:53,450 --> 00:19:58,800
It's a solution that involves
both sides of the equation now.

289
00:19:58,800 --> 00:20:01,380
We relied heavily on
the right-hand side

290
00:20:01,380 --> 00:20:05,920
to calculate f0, and we
used the left-hand side--

291
00:20:05,920 --> 00:20:09,530
through this log l
acting on log of f0--

292
00:20:09,530 --> 00:20:13,300
to get the correction
that is order of tau x

293
00:20:13,300 --> 00:20:14,730
that is coming to this equation.

294
00:20:17,490 --> 00:20:23,310
So now, with this
improved solution,

295
00:20:23,310 --> 00:20:27,390
we can go back and re-check
some of the conclusions

296
00:20:27,390 --> 00:20:29,830
that we had before.

297
00:20:29,830 --> 00:20:35,100
So let's, for example, start
by calculating this pressure

298
00:20:35,100 --> 00:20:39,550
tensor P alpha beta, and see
how it was made different.

299
00:20:39,550 --> 00:20:46,050
So what I need to do is
to calculate this average.

300
00:20:48,620 --> 00:20:51,720
How do I calculate that average?

301
00:20:51,720 --> 00:20:56,130
I essentially multiply this f,
as I have indicated over there,

302
00:20:56,130 --> 00:21:01,700
by c alpha c beta, and then
I integrate over all momenta.

303
00:21:01,700 --> 00:21:04,110
Essentially, I have
to do integrations

304
00:21:04,110 --> 00:21:07,700
with this Gaussian weight.

305
00:21:07,700 --> 00:21:10,690
So when I do the
average of c alpha

306
00:21:10,690 --> 00:21:15,230
c beta with this Gaussian
weight, not surprisingly,

307
00:21:15,230 --> 00:21:18,640
I will get the delta
alpha beta, and I

308
00:21:18,640 --> 00:21:21,220
will get from here kT over m.

309
00:21:21,220 --> 00:21:26,340
Multiplying by nm
will give me mkT.

310
00:21:26,340 --> 00:21:33,520
So this is the diagonal form,
where the diagonal elements

311
00:21:33,520 --> 00:21:34,995
are our familiar pressures.

312
00:21:37,890 --> 00:21:39,510
So that was the
zeroth order term.

313
00:21:39,510 --> 00:21:42,320
Essentially, that's
the 1 over here,

314
00:21:42,320 --> 00:21:47,340
multiplying c alpha c
beta before I integrate.

315
00:21:47,340 --> 00:21:51,130
So that's the 1 in this bracket.

316
00:21:51,130 --> 00:21:54,340
But there will be order
of tau x corrections,

317
00:21:54,340 --> 00:21:58,970
because I will have to
multiply two more factors of c

318
00:21:58,970 --> 00:22:00,770
with the c's that
I have over here.

319
00:22:03,650 --> 00:22:06,200
Now, none of these
terms are important,

320
00:22:06,200 --> 00:22:09,710
because these are,
again, odd terms.

321
00:22:09,710 --> 00:22:15,270
So when I multiply two
c's with three or one c

322
00:22:15,270 --> 00:22:17,030
the average will be zero.

323
00:22:17,030 --> 00:22:21,651
So all of the averages are
really coming from these terms.

324
00:22:21,651 --> 00:22:26,170
Now, these terms involve
four factors of c's.

325
00:22:26,170 --> 00:22:26,670
Right?

326
00:22:26,670 --> 00:22:29,240
There's the two c's
that I put out here--

327
00:22:29,240 --> 00:22:31,560
and actually, I
really have to think

328
00:22:31,560 --> 00:22:37,086
of these as different indices,
let's say nu nu, nu nu.

329
00:22:37,086 --> 00:22:42,410
Summation convention
again assumed.

330
00:22:42,410 --> 00:22:47,380
And then when I multiply by
c alpha c beta, I will have,

331
00:22:47,380 --> 00:22:49,930
essentially, four indices
to play with-- c alpha

332
00:22:49,930 --> 00:22:53,070
c beta, c nu c nu.

333
00:22:53,070 --> 00:22:57,970
But it's all done with the
Gaussian weight out here.

334
00:22:57,970 --> 00:23:01,520
And we showed and
discussed how there

335
00:23:01,520 --> 00:23:05,550
was this nice fixed theorem
that enabled you to rapidly

336
00:23:05,550 --> 00:23:11,340
calculate these Gaussian
weights with four factors of c,

337
00:23:11,340 --> 00:23:14,340
or more factors of c--
doesn't really matter.

338
00:23:14,340 --> 00:23:18,130
And so in principle you
know how to do that,

339
00:23:18,130 --> 00:23:22,250
and I'll skip the corresponding
algebra and write the answer.

340
00:23:22,250 --> 00:23:25,860
It is proportional
to minus tau x.

341
00:23:25,860 --> 00:23:29,960
And what it gives you, once
you do these calculations,

342
00:23:29,960 --> 00:23:37,730
is a factor of u alpha beta
minus delta alpha beta over 3

343
00:23:37,730 --> 00:23:40,040
nu gamma gamma.

344
00:23:40,040 --> 00:23:44,270
And again, let me make sure
I did not miss out anything.

345
00:23:47,100 --> 00:23:49,110
And apparently I missed
out the factor of 2.

346
00:24:02,280 --> 00:24:05,370
So really, the only
thing that's happened,

347
00:24:05,370 --> 00:24:11,060
once we went and included this
correction, we added this term.

348
00:24:11,060 --> 00:24:13,540
But the nice thing
about this term

349
00:24:13,540 --> 00:24:17,500
is that potentially, it
has off-diagonal terms.

350
00:24:17,500 --> 00:24:21,280
This matrix initially
was completely diagonal.

351
00:24:21,280 --> 00:24:23,760
The corrections that
we have calculated

352
00:24:23,760 --> 00:24:27,150
are potentially
off-diagonal coming

353
00:24:27,150 --> 00:24:33,600
from this term that corrected
the original Gaussian weight.

354
00:24:33,600 --> 00:24:35,610
So where is that useful?

355
00:24:35,610 --> 00:24:38,490
Well, one of the problems
that we discussed

356
00:24:38,490 --> 00:24:43,210
was that I can imagine a
configuration of velocities,

357
00:24:43,210 --> 00:24:46,600
let's say close to a
wall-- but it does not

358
00:24:46,600 --> 00:24:48,980
have to be a wall,
but something where

359
00:24:48,980 --> 00:24:53,730
I have a profile
of velocities which

360
00:24:53,730 --> 00:25:00,060
exist with components only
along the x direction,

361
00:25:00,060 --> 00:25:03,860
but vary along the y direction.

362
00:25:03,860 --> 00:25:06,780
So this ux is
different from this ux,

363
00:25:06,780 --> 00:25:08,780
because they correspond
to different y's.

364
00:25:08,780 --> 00:25:12,680
So essentially, this is a
pattern of sheering a gas,

365
00:25:12,680 --> 00:25:14,910
if you like.

366
00:25:14,910 --> 00:25:19,660
And the question is,
well, hopefully, this

367
00:25:19,660 --> 00:25:23,850
will come to relax and
eventually give us, let's say,

368
00:25:23,850 --> 00:25:28,160
uniform velocity, or zero
velocity, even better--

369
00:25:28,160 --> 00:25:32,630
if there's a wall, and
the wall velocity is zero.

370
00:25:32,630 --> 00:25:34,570
So how does that happen?

371
00:25:34,570 --> 00:25:38,485
Well, the equation
that I have to satisfy

372
00:25:38,485 --> 00:25:42,290
is this one that involves u.

373
00:25:42,290 --> 00:25:45,380
So I have m.

374
00:25:45,380 --> 00:25:46,705
Well, what do I have to have?

375
00:25:46,705 --> 00:25:55,140
I have du by dt plus
something that acts on ux.

376
00:25:58,690 --> 00:26:02,710
ux only has variations
along the y direction.

377
00:26:02,710 --> 00:26:06,210
So this term, if
it was there, had

378
00:26:06,210 --> 00:26:12,370
to involve a derivative along
the y direction multiplying uy,

379
00:26:12,370 --> 00:26:14,295
but it's not present,
because there's no uy.

380
00:26:16,970 --> 00:26:21,270
On the right-hand side of the
equation, I have to put minus 1

381
00:26:21,270 --> 00:26:24,040
over n.

382
00:26:24,040 --> 00:26:27,560
Again, the only variations
that I'm allowed to have

383
00:26:27,560 --> 00:26:29,110
are along the y direction.

384
00:26:29,110 --> 00:26:32,300
So I have, when I do the
summation over alpha,

385
00:26:32,300 --> 00:26:35,920
the only alpha that
contributes is y.

386
00:26:35,920 --> 00:26:39,910
And so I need the
pressure y, but I'm

387
00:26:39,910 --> 00:26:43,010
looking for velocities
along the x direction,

388
00:26:43,010 --> 00:26:47,042
so the other index better be x.

389
00:26:47,042 --> 00:26:49,020
OK?

390
00:26:49,020 --> 00:26:52,490
So the time course of
the velocity profile

391
00:26:52,490 --> 00:26:57,290
that I set up over
here is determined

392
00:26:57,290 --> 00:27:00,960
by the y derivative of the
yx component of the pressures

393
00:27:00,960 --> 00:27:02,960
tensor.

394
00:27:02,960 --> 00:27:05,400
Now, previously our
problem was that we stopped

395
00:27:05,400 --> 00:27:08,270
at the zeroth order,
and at the zeroth order,

396
00:27:08,270 --> 00:27:10,250
the pressure tensor
was diagonal.

397
00:27:10,250 --> 00:27:12,970
It didn't have a yx component.

398
00:27:12,970 --> 00:27:17,300
So this profile
would stay forever.

399
00:27:17,300 --> 00:27:22,880
But now we do have a
yx component, and so

400
00:27:22,880 --> 00:27:24,250
what do I get?

401
00:27:24,250 --> 00:27:30,690
I will get here
minus 1 over n dy dy.

402
00:27:30,690 --> 00:27:35,130
The yx component will
come from minus 2 tau

403
00:27:35,130 --> 00:27:41,900
x multiplying nkT and
then multiplying-- well,

404
00:27:41,900 --> 00:27:44,320
this term, again, is diagonal.

405
00:27:44,320 --> 00:27:46,210
I can forget about that term.

406
00:27:46,210 --> 00:27:50,360
So it comes from the uxy.

407
00:27:50,360 --> 00:27:51,815
What is uxy?

408
00:27:51,815 --> 00:27:57,480
It is 1/2 of the x derivative
of uy that doesn't exist,

409
00:27:57,480 --> 00:27:59,968
and the y derivative
of ux that does exist.

410
00:28:03,952 --> 00:28:05,950
OK?

411
00:28:05,950 --> 00:28:10,420
So what we have,
once we divide by m,

412
00:28:10,420 --> 00:28:14,580
is that the time
derivative of ux

413
00:28:14,580 --> 00:28:19,880
is given by a bunch
of coefficients.

414
00:28:19,880 --> 00:28:23,720
The n I can cancel
if it does not vary.

415
00:28:23,720 --> 00:28:29,370
What I have is the 2's cancel.

416
00:28:29,370 --> 00:28:31,420
The n's cancel.

417
00:28:31,420 --> 00:28:39,140
I will have tau x kT over m.

418
00:28:41,890 --> 00:28:44,260
tau x kT over m, that's fine.

419
00:28:44,260 --> 00:28:49,790
And then the second derivative
along the y direction of ux.

420
00:28:57,940 --> 00:29:02,170
So suddenly, I have
a different equation.

421
00:29:02,170 --> 00:29:06,870
Rather than having the time
derivative does not change,

422
00:29:06,870 --> 00:29:09,650
I find that the time
derivative of ux

423
00:29:09,650 --> 00:29:11,310
is proportional to Laplacian.

424
00:29:11,310 --> 00:29:13,430
It's a diffusion equation.

425
00:29:13,430 --> 00:29:15,970
And we know the solution
to the diffusion equation,

426
00:29:15,970 --> 00:29:20,110
how it looks qualitative if I
have a profile such as this.

427
00:29:20,110 --> 00:29:23,340
Because of the
fusion, eventually it

428
00:29:23,340 --> 00:29:27,900
will become more and
more uniform in time.

429
00:29:27,900 --> 00:29:30,410
And the characteristic
time over which

430
00:29:30,410 --> 00:29:34,570
it does so, if I assume that,
let's say, in the y direction,

431
00:29:34,570 --> 00:29:39,080
I have a pattern that has some
characteristic size lambda,

432
00:29:39,080 --> 00:29:44,950
then the characteristic
relaxation time for diffusion

433
00:29:44,950 --> 00:29:48,160
will be proportional
to lambda squared.

434
00:29:48,160 --> 00:29:50,460
There's a proportionality here.

435
00:29:50,460 --> 00:29:53,560
The constant of
proportionality is simply

436
00:29:53,560 --> 00:29:55,350
this diffusion coefficient.

437
00:29:55,350 --> 00:29:57,920
So it is inversely.

438
00:29:57,920 --> 00:30:02,731
So it is m kT tau x.

439
00:30:07,920 --> 00:30:10,840
Actually, we want to
think about it further.

440
00:30:10,840 --> 00:30:14,790
kT over m is roughly the square
of the terminal velocities

441
00:30:14,790 --> 00:30:16,610
of the particles.

442
00:30:16,610 --> 00:30:20,890
So lambda squared
divided by v squared

443
00:30:20,890 --> 00:30:24,490
is roughly the time that
you would have ballistically

444
00:30:24,490 --> 00:30:28,760
traveled over this line
scale of the variation.

445
00:30:28,760 --> 00:30:31,930
The square of that
time has to be provided

446
00:30:31,930 --> 00:30:34,740
by the characteristic
collision time,

447
00:30:34,740 --> 00:30:37,060
and that tells
you the time scale

448
00:30:37,060 --> 00:30:42,090
over which this kind
of relaxation occurs.

449
00:30:42,090 --> 00:30:43,071
Yes?

450
00:30:43,071 --> 00:30:46,999
AUDIENCE: So if x
depends linearly, why?

451
00:30:46,999 --> 00:30:49,340
Would still get 0
on the-- why can't--

452
00:30:51,960 --> 00:30:52,780
PROFESSOR: OK.

453
00:30:52,780 --> 00:30:57,000
So what you are setting
up is a variation

454
00:30:57,000 --> 00:31:00,450
where there is some kind
of a sheer velocity that

455
00:31:00,450 --> 00:31:01,930
exists forever.

456
00:31:01,930 --> 00:31:05,730
So indeed, this
kind of pattern will

457
00:31:05,730 --> 00:31:08,530
persist, unless you say
that there's actually

458
00:31:08,530 --> 00:31:10,740
another wall at the other end.

459
00:31:10,740 --> 00:31:14,500
So then you will have
some kind of a pattern

460
00:31:14,500 --> 00:31:15,660
that you would relax.

461
00:31:15,660 --> 00:31:20,140
So unless you're willing to send
this all the way to infinity,

462
00:31:20,140 --> 00:31:21,900
is will eventually relax.

463
00:31:26,365 --> 00:31:26,864
Yes?

464
00:31:26,864 --> 00:31:29,648
AUDIENCE: Can you just
say again the last term

465
00:31:29,648 --> 00:31:32,281
on the second
line, the 1/2 term.

466
00:31:32,281 --> 00:31:33,322
Where did that come from?

467
00:31:33,322 --> 00:31:35,140
Can you just explain that?

468
00:31:35,140 --> 00:31:36,576
PROFESSOR: This term?

469
00:31:36,576 --> 00:31:37,201
AUDIENCE: Yeah.

470
00:31:37,201 --> 00:31:38,420
The last half of it.

471
00:31:38,420 --> 00:31:39,711
PROFESSOR: The last half of it.

472
00:31:39,711 --> 00:31:41,210
So what did we have here?

473
00:31:41,210 --> 00:31:52,690
So what I have is
that over here,

474
00:31:52,690 --> 00:31:56,210
in calculating
the pressure, when

475
00:31:56,210 --> 00:31:59,760
I'm looking at the xy
component, I better

476
00:31:59,760 --> 00:32:03,530
find some element here
that is off diagonal.

477
00:32:03,530 --> 00:32:05,620
What's the off
diagonal element here?

478
00:32:05,620 --> 00:32:10,220
It is u xy What is u xy?

479
00:32:10,220 --> 00:32:14,160
u xy is this, calculated for.

480
00:32:14,160 --> 00:32:19,700
So it is 1/2 of dx uy,
which is what I don't have,

481
00:32:19,700 --> 00:32:22,150
because my uy is 0.

482
00:32:22,150 --> 00:32:24,090
And the other half or
it, of symmetrization,

483
00:32:24,090 --> 00:32:26,595
is 1/2 the value of x.

484
00:32:26,595 --> 00:32:28,050
Yes?

485
00:32:28,050 --> 00:32:31,990
AUDIENCE: [INAUDIBLE]
derivative [INAUDIBLE] for ux.

486
00:32:31,990 --> 00:32:36,240
You're starting out
the second term uy uy.

487
00:32:36,240 --> 00:32:41,360
Also, shouldn't there
be a term ux dx?

488
00:32:41,360 --> 00:32:42,240
PROFESSOR: Yes.

489
00:32:42,240 --> 00:32:42,740
Yeah.

490
00:32:42,740 --> 00:32:43,960
So this is a-- yes.

491
00:32:43,960 --> 00:32:46,810
There is a term that
is ux dx, but there

492
00:32:46,810 --> 00:32:49,760
is no variation along
the x direction.

493
00:32:49,760 --> 00:32:52,440
I said that I set up
a configuration where

494
00:32:52,440 --> 00:32:56,900
the only non-zero derivatives
are along the y direction.

495
00:32:56,900 --> 00:32:57,900
But in general, yes.

496
00:32:57,900 --> 00:32:58,483
You are right.

497
00:32:58,483 --> 00:33:00,195
This is like a divergence.

498
00:33:00,195 --> 00:33:03,460
It has three terms, but
the way that I set it up,

499
00:33:03,460 --> 00:33:06,140
only one term is non-zero.

500
00:33:06,140 --> 00:33:10,120
AUDIENCE: Also, if you have
[INAUDIBLE] one layer moving

501
00:33:10,120 --> 00:33:13,280
fast at some point, the other
layer moving slower than that.

502
00:33:13,280 --> 00:33:15,921
And you potentially can
create some kind of curl?

503
00:33:15,921 --> 00:33:19,470
But as far as I understand, this
would be an even higher order

504
00:33:19,470 --> 00:33:19,970
effect?

505
00:33:19,970 --> 00:33:22,369
Like turbulence.

506
00:33:22,369 --> 00:33:23,660
PROFESSOR: You said two things.

507
00:33:23,660 --> 00:33:27,560
One of them was you
started saying viscosity.

508
00:33:27,560 --> 00:33:30,240
And indeed, what
we've calculated here,

509
00:33:30,240 --> 00:33:33,150
this thing is the
coefficient of viscosity.

510
00:33:33,150 --> 00:33:36,450
So this is really the
viscosity of the material.

511
00:33:36,450 --> 00:33:41,980
So actually, once I
include this term,

512
00:33:41,980 --> 00:33:44,960
I have the full the
Navier-Stokes equations

513
00:33:44,960 --> 00:33:46,640
with viscosity.

514
00:33:46,640 --> 00:33:49,890
So all of the
vortices, et cetera,

515
00:33:49,890 --> 00:33:53,000
should also be present
and discussed, however

516
00:33:53,000 --> 00:33:55,880
way you want to do it with
Navier-Stokes equation

517
00:33:55,880 --> 00:33:57,850
into this equations.

518
00:33:57,850 --> 00:34:02,890
AUDIENCE: So components
of speed which

519
00:34:02,890 --> 00:34:06,730
are not just x components,
but other components which

520
00:34:06,730 --> 00:34:09,757
are initially zero, will
change because of this, right?

521
00:34:09,757 --> 00:34:10,965
PROFESSOR: Yes, that's right.

522
00:34:10,965 --> 00:34:11,464
That's right.

523
00:34:11,464 --> 00:34:12,462
AUDIENCE: You just
haven't introduced them

524
00:34:12,462 --> 00:34:13,460
in the equation?

525
00:34:13,460 --> 00:34:14,085
PROFESSOR: Yes.

526
00:34:14,085 --> 00:34:17,449
So this is I'm looking at the
initial profile in principle.

527
00:34:17,449 --> 00:34:19,300
I think here I have
set up something

528
00:34:19,300 --> 00:34:23,020
that because of symmetry will
always maintain this equation.

529
00:34:23,020 --> 00:34:24,850
But if you put a
little bit bump,

530
00:34:24,850 --> 00:34:26,489
if you make some
perturbation, you

531
00:34:26,489 --> 00:34:28,966
will certainly generate
other components of velocity.

532
00:34:34,433 --> 00:34:35,929
OK?

533
00:34:35,929 --> 00:34:40,730
And so this resolved one of the
modes that was not relaxing.

534
00:34:40,730 --> 00:34:45,710
There was another mode that we
were looking at where I set up

535
00:34:45,710 --> 00:34:50,350
a situation where temperature
and density were changing

536
00:34:50,350 --> 00:34:52,980
across the system,
but their products--

537
00:34:52,980 --> 00:34:55,500
that is, the
pressure-- was uniform.

538
00:34:55,500 --> 00:34:58,780
So you had a system
that was always there

539
00:34:58,780 --> 00:35:01,970
in the zeroth order, despite
having different temperatures

540
00:35:01,970 --> 00:35:03,886
at two different points.

541
00:35:03,886 --> 00:35:08,000
Well, that was partly
because the heat transport

542
00:35:08,000 --> 00:35:09,912
vector was zero.

543
00:35:09,912 --> 00:35:17,290
Now, if I want to calculate
this with this more complicated

544
00:35:17,290 --> 00:35:21,090
equation, well, what I
need-- the problem before

545
00:35:21,090 --> 00:35:24,110
with the zeroth order was
that I had three factors of c,

546
00:35:24,110 --> 00:35:25,870
and that was odd.

547
00:35:25,870 --> 00:35:30,240
But now, I have a term in the
equation this is also odd.

548
00:35:30,240 --> 00:35:34,700
So from here, I will get
a term, and I will in fact

549
00:35:34,700 --> 00:35:39,790
find eventually that
the flow of heat

550
00:35:39,790 --> 00:35:48,020
is proportional to
gradient of temperature.

551
00:35:48,020 --> 00:35:51,630
And one can compute
this coefficient, again,

552
00:35:51,630 --> 00:35:54,940
in terms of mass,
density, et cetera,

553
00:35:54,940 --> 00:35:57,820
just like we calculated
this coefficient over here.

554
00:35:57,820 --> 00:36:00,460
This will give you relaxation.

555
00:36:00,460 --> 00:36:04,310
We can also look at the
sound modes including this,

556
00:36:04,310 --> 00:36:07,130
and you find that the
wave equation that I

557
00:36:07,130 --> 00:36:09,920
had before for the
sound modes will also

558
00:36:09,920 --> 00:36:12,770
get a second derivative
term, and that

559
00:36:12,770 --> 00:36:15,130
will lead to damping
of the modes of sound.

560
00:36:15,130 --> 00:36:18,650
So everything at
this level, now we

561
00:36:18,650 --> 00:36:23,180
have some way of seeing
how the gas will eventually

562
00:36:23,180 --> 00:36:24,820
come to equilibrium.

563
00:36:24,820 --> 00:36:29,460
And given some knowledge of
rough parameters of the gas,

564
00:36:29,460 --> 00:36:31,360
like-- and most
importantly-- what's

565
00:36:31,360 --> 00:36:33,570
the time between
collisions, we can

566
00:36:33,570 --> 00:36:38,510
compute the typical relaxation
time, and the relaxation manner

567
00:36:38,510 --> 00:36:39,250
of the gas.

568
00:36:53,230 --> 00:36:56,820
So we are now going to
change directions, and forget

569
00:36:56,820 --> 00:36:59,360
about time dependents.

570
00:36:59,360 --> 00:37:02,330
So if you have questions
about this section,

571
00:37:02,330 --> 00:37:04,573
now may be a good time.

572
00:37:04,573 --> 00:37:05,072
Yes?

573
00:37:05,072 --> 00:37:07,087
AUDIENCE: [INAUDIBLE]
what is lambda?

574
00:37:07,087 --> 00:37:07,670
PROFESSOR: OK.

575
00:37:07,670 --> 00:37:11,420
So I assume that
what I have to do

576
00:37:11,420 --> 00:37:15,940
is to solve the equation
for some initial condition.

577
00:37:15,940 --> 00:37:18,860
Let's imagine that that
initial condition, let's say,

578
00:37:18,860 --> 00:37:22,380
is a periodic pattern of
some wavelength lambda.

579
00:37:22,380 --> 00:37:24,460
Or it could be any
other shape that

580
00:37:24,460 --> 00:37:26,880
has some characteristic
dimension.

581
00:37:26,880 --> 00:37:30,750
The important thing is
that the diffusion constant

582
00:37:30,750 --> 00:37:35,580
has units of length
squared over time.

583
00:37:35,580 --> 00:37:39,440
So eventually, you'll
find that all times are

584
00:37:39,440 --> 00:37:41,740
proportional to
some lengths where

585
00:37:41,740 --> 00:37:44,590
times the inverse of
the diffusion constant.

586
00:37:44,590 --> 00:37:47,380
And so you have to look
at your initial system

587
00:37:47,380 --> 00:37:51,830
that you want to relax, identify
the longest length scale that

588
00:37:51,830 --> 00:37:54,590
is involved, and then
your relaxation time

589
00:37:54,590 --> 00:37:56,473
would be roughly of that order.

590
00:38:05,611 --> 00:38:06,110
OK.

591
00:38:06,110 --> 00:38:11,340
So now we get to
the fourth section

592
00:38:11,340 --> 00:38:14,620
of our course,
that eventually has

593
00:38:14,620 --> 00:38:16,280
to do with
statistical mechanics.

594
00:38:22,170 --> 00:38:26,080
And at the very,
very first lecture,

595
00:38:26,080 --> 00:38:32,150
I wrote the definition for
you that I will write again,

596
00:38:32,150 --> 00:38:42,620
that statistical mechanics
is a probabilistic approach

597
00:38:42,620 --> 00:38:58,270
to equilibrium-- also
microscopic-- properties

598
00:38:58,270 --> 00:39:05,750
of large numbers of
degrees of freedom.

599
00:39:16,690 --> 00:39:21,650
So what we did to
thermodynamics was

600
00:39:21,650 --> 00:39:28,390
to identify what equilibrium
microscopic properties are.

601
00:39:28,390 --> 00:39:33,520
They are things such as
identifying the energy, volume,

602
00:39:33,520 --> 00:39:36,480
number of particles of the gas.

603
00:39:36,480 --> 00:39:40,240
There could be other things,
such as temperature, pressure,

604
00:39:40,240 --> 00:39:45,460
number, or other
collections of variables

605
00:39:45,460 --> 00:39:48,830
that are independently
sufficient to describe

606
00:39:48,830 --> 00:39:50,570
a macrostate.

607
00:39:50,570 --> 00:39:58,480
And what we're going to do is
to indicate that macroscopic set

608
00:39:58,480 --> 00:40:02,160
of parameters that
thermodynamically characterize

609
00:40:02,160 --> 00:40:07,030
the equilibrium
system by big M. OK?

610
00:40:10,000 --> 00:40:14,150
Clearly, for a
probabilistic approach,

611
00:40:14,150 --> 00:40:18,000
you want to know something
about the probability.

612
00:40:18,000 --> 00:40:26,700
And we saw that probabilities
involving large numbers--

613
00:40:26,700 --> 00:40:29,930
and actually various things
involving large numbers--

614
00:40:29,930 --> 00:40:34,095
had some simplified character
that we are going to exploit.

615
00:40:37,200 --> 00:40:39,850
And lastly, in the
last section, we

616
00:40:39,850 --> 00:40:45,040
have been thinking about
microscopic description.

617
00:40:45,040 --> 00:40:47,620
So these large number
of degrees of freedom

618
00:40:47,620 --> 00:40:50,560
we said identify
some point, let's

619
00:40:50,560 --> 00:40:55,540
say, for particles in a six-n
dimensional phase space.

620
00:40:55,540 --> 00:41:02,450
So this would for gas particles
be this collection p and q.

621
00:41:02,450 --> 00:41:07,260
And that these
quantities we know

622
00:41:07,260 --> 00:41:10,860
are in fact subject
to dynamics that

623
00:41:10,860 --> 00:41:14,810
is governed by some Hamiltonian.

624
00:41:14,810 --> 00:41:19,810
And we saw that
if we sort of look

625
00:41:19,810 --> 00:41:23,900
at the entirety of the
probability in this six

626
00:41:23,900 --> 00:41:26,640
n dimensional phase
space, that it

627
00:41:26,640 --> 00:41:29,590
is subject to
Liouville's question that

628
00:41:29,590 --> 00:41:35,953
said that dp by dt is a
Poisson bracket of H and p.

629
00:41:41,850 --> 00:41:47,180
And if the only thing that
we take from equilibrium

630
00:41:47,180 --> 00:41:52,220
is that things should not
change as a function of time,

631
00:41:52,220 --> 00:41:55,280
then requiring this
probability in phase space

632
00:41:55,280 --> 00:41:58,910
to be independent
of time would then

633
00:41:58,910 --> 00:42:05,240
require us to have a p which
is a function of H, which

634
00:42:05,240 --> 00:42:09,046
is defined in p and
q and potentially

635
00:42:09,046 --> 00:42:10,170
other conserved quantities.

636
00:42:15,850 --> 00:42:21,270
So what we are going to do
in statistical mechanics

637
00:42:21,270 --> 00:42:26,830
is to forget about how things
eventually reach equilibrium.

638
00:42:26,830 --> 00:42:30,230
We spent a lot of time
and energy thinking

639
00:42:30,230 --> 00:42:33,060
about how a gas
reaches equilibrium.

640
00:42:33,060 --> 00:42:36,320
Having established
what it requires

641
00:42:36,320 --> 00:42:40,690
to devise that solution
in one particular case,

642
00:42:40,690 --> 00:42:44,450
and one of few cases where
you can actually get far,

643
00:42:44,450 --> 00:42:45,640
you're going to ignore that.

644
00:42:45,640 --> 00:42:49,250
We say that somehow my system
reached this state that

645
00:42:49,250 --> 00:42:52,270
is time independent
and is equilibrium,

646
00:42:52,270 --> 00:42:55,390
and therefore the
probability should somehow

647
00:42:55,390 --> 00:42:57,280
have this character.

648
00:42:57,280 --> 00:43:01,460
And it depends,
however, on what choice

649
00:43:01,460 --> 00:43:04,380
I make for the macrostate.

650
00:43:04,380 --> 00:43:09,160
So what I need here-- this was
a probability of a microstate.

651
00:43:09,160 --> 00:43:12,820
So what I want to do
is to have a statement

652
00:43:12,820 --> 00:43:15,740
about the probability
of a microstate given

653
00:43:15,740 --> 00:43:18,820
some specification
of the macrostate

654
00:43:18,820 --> 00:43:21,570
that I'm interested in.

655
00:43:21,570 --> 00:43:22,690
So that's the task.

656
00:43:26,840 --> 00:43:33,750
And you're going to do that
first in the ensemble--

657
00:43:33,750 --> 00:43:37,062
and I'll tell you again
what ensemble means shortly.

658
00:43:37,062 --> 00:43:38,270
That's called microcanonical.

659
00:43:48,110 --> 00:43:54,830
And if you recall, when we
were constructing our approach

660
00:43:54,830 --> 00:43:59,580
to thermodynamics, one of the
first things that we did was we

661
00:43:59,580 --> 00:44:02,580
said, there's a whole bunch
of things that we don't know,

662
00:44:02,580 --> 00:44:07,440
so let's imagine that our
box is as simple as possible.

663
00:44:07,440 --> 00:44:10,170
So the system that
we are looking

664
00:44:10,170 --> 00:44:14,070
is completely isolated from
the rest of the universe.

665
00:44:14,070 --> 00:44:14,910
It's a box.

666
00:44:14,910 --> 00:44:17,010
There's lots of
things in the box.

667
00:44:17,010 --> 00:44:20,940
But the box has no contact
with the rest of the universe.

668
00:44:20,940 --> 00:44:23,630
So that was the case
where essentially there

669
00:44:23,630 --> 00:44:28,080
was no heat, no work that
was going into the system

670
00:44:28,080 --> 00:44:30,810
and was being exchanged,
and so clearly

671
00:44:30,810 --> 00:44:32,670
this system has a
constant energy.

672
00:44:35,600 --> 00:44:43,520
And so you can certainly
prescribe a particular energy

673
00:44:43,520 --> 00:44:47,110
content to whatever
is in the box.

674
00:44:47,110 --> 00:44:53,450
And that's the chief identity
of the microcanonical ensemble.

675
00:44:53,450 --> 00:44:56,650
It's basically a
collection of boxes

676
00:44:56,650 --> 00:44:59,290
that represent the
same equilibrium.

677
00:44:59,290 --> 00:45:04,420
So if there is
essentially a gas,

678
00:45:04,420 --> 00:45:06,110
the volume is
fixed, so that there

679
00:45:06,110 --> 00:45:08,940
would be no work
that will be done.

680
00:45:08,940 --> 00:45:10,630
The number of
particles is fixed,

681
00:45:10,630 --> 00:45:13,550
so that there is no chemical
work that is being done.

682
00:45:13,550 --> 00:45:18,140
So essentially, in general,
all of the quantities

683
00:45:18,140 --> 00:45:22,070
that we identified
with displacements

684
00:45:22,070 --> 00:45:28,100
are held fixed, as well as
the energy in this ensemble.

685
00:45:28,100 --> 00:45:31,540
So the microcanonical
ensemble would

686
00:45:31,540 --> 00:45:38,310
be essentially E, x, and N
would be the quantities that

687
00:45:38,310 --> 00:45:40,660
parametrize the
equilibrium state.

688
00:45:40,660 --> 00:45:42,700
Of course, there's
a whole collection

689
00:45:42,700 --> 00:45:45,210
of different microstates
that would correspond

690
00:45:45,210 --> 00:45:49,310
to the same macrostate here.

691
00:45:49,310 --> 00:45:59,590
So presumably, this bunch of
particles that are inside this

692
00:45:59,590 --> 00:46:04,560
explore a huge
multidimensional microstate.

693
00:46:04,560 --> 00:46:09,320
And what I want to do is to
assign a probability that,

694
00:46:09,320 --> 00:46:13,540
given that I have
fixed E, x, and n,

695
00:46:13,540 --> 00:46:17,210
that a particular
microstate occurs.

696
00:46:17,210 --> 00:46:18,150
OK?

697
00:46:18,150 --> 00:46:20,210
How do I do that?

698
00:46:20,210 --> 00:46:27,940
Well, I say that OK, if the
energy of that microstate

699
00:46:27,940 --> 00:46:33,350
that I can calculate is not
equal to the energy that I know

700
00:46:33,350 --> 00:46:36,320
I have in the box,
then that's not

701
00:46:36,320 --> 00:46:38,890
one of the microstates that
should be allowed in the box.

702
00:46:41,810 --> 00:46:47,030
But presumably there's a
whole bunch of micro states

703
00:46:47,030 --> 00:46:50,530
whose energy is
compatible with the energy

704
00:46:50,530 --> 00:46:52,980
that I put in the box.

705
00:46:52,980 --> 00:46:57,715
And then I say, OK, if there is
no other conserved quantity--

706
00:46:57,715 --> 00:47:02,850
and let's assume that there
isn't-- I have no way a priori

707
00:47:02,850 --> 00:47:05,200
of distinguishing between them.

708
00:47:05,200 --> 00:47:08,000
So they are just like
the faces of the dice,

709
00:47:08,000 --> 00:47:10,770
and I say they're
all equally likely.

710
00:47:10,770 --> 00:47:14,220
For the dice, I
would give 1/6 here.

711
00:47:14,220 --> 00:47:16,470
There's presumably some
kind of a normalization that

712
00:47:16,470 --> 00:47:23,420
depends on E, x, and n,
that I have to put in.

713
00:47:23,420 --> 00:47:26,420
And again, note
that this is kind

714
00:47:26,420 --> 00:47:31,050
of like a delta
function that I have.

715
00:47:31,050 --> 00:47:33,870
It's like a delta of H
minus E, and it's therefore

716
00:47:33,870 --> 00:47:37,310
consistent with this
Liouville equation.

717
00:47:37,310 --> 00:47:41,410
It's one of these
functions that is related

718
00:47:41,410 --> 00:47:45,250
through H to the probability
on the microstate.

719
00:47:48,560 --> 00:47:51,550
Now, this is an assumption.

720
00:47:51,550 --> 00:47:56,840
It is a way of
assigning probabilities.

721
00:47:56,840 --> 00:48:06,200
It's called assumption of
equal a priori probabilities.

722
00:48:06,200 --> 00:48:09,140
It's like the subjective
assignment of probabilities

723
00:48:09,140 --> 00:48:11,420
and like the faces
of the dice, it's

724
00:48:11,420 --> 00:48:13,750
essentially the
best that you can

725
00:48:13,750 --> 00:48:18,140
do without any
other information.

726
00:48:18,140 --> 00:48:24,710
Now, the statement is that once
I have made this assumption,

727
00:48:24,710 --> 00:48:28,090
I can derive the three
laws of thermodynamics.

728
00:48:28,090 --> 00:48:28,590
No, sorry.

729
00:48:28,590 --> 00:48:33,220
I can derive two of the
three laws of thermodynamics.

730
00:48:33,220 --> 00:48:36,970
Actually, three of the four
laws of thermodynamics, since we

731
00:48:36,970 --> 00:48:37,990
had the zeroth law.

732
00:48:37,990 --> 00:48:40,550
So let's proceed.

733
00:48:40,550 --> 00:48:41,530
OK?

734
00:48:41,530 --> 00:48:46,585
So we want to have a
proof of thermodynamics.

735
00:49:04,650 --> 00:49:07,350
So the zeroth law
had something to do

736
00:49:07,350 --> 00:49:10,610
with putting two
systems in contact,

737
00:49:10,610 --> 00:49:13,560
and when they were
in equilibrium,

738
00:49:13,560 --> 00:49:16,520
there was some empirical
temperature from one

739
00:49:16,520 --> 00:49:20,270
that was the same as what
you had for the other one.

740
00:49:20,270 --> 00:49:25,350
So basically, let's
pick our two systems

741
00:49:25,350 --> 00:49:32,375
and put a wall between them that
allows the exchange of energy.

742
00:49:35,500 --> 00:49:40,140
And so this is my
system one, and they

743
00:49:40,140 --> 00:49:43,950
have an spontaneous energy bond.

744
00:49:43,950 --> 00:49:47,160
This is the part that is
two, it has energy E2.

745
00:49:49,830 --> 00:50:01,360
So I start with an initial state
where when I look at E2 and E2,

746
00:50:01,360 --> 00:50:07,350
I have some initial value of
E1, 0, let's say, and E2, 0.

747
00:50:10,170 --> 00:50:15,660
So my initial state
is here, in these two.

748
00:50:15,660 --> 00:50:20,760
Now, as I proceed in time,
because the two systems can

749
00:50:20,760 --> 00:50:25,320
exchange energy, E1
and E2 can change.

750
00:50:25,320 --> 00:50:31,080
But certainly, what I have is
that E1 plus E2 is something E

751
00:50:31,080 --> 00:50:36,590
total that is E1,0 plus E2,0.

752
00:50:36,590 --> 00:50:41,700
Which means that I'm
always exploring the line

753
00:50:41,700 --> 00:50:45,380
that corresponds to E1 plus
E2 is a constant, which

754
00:50:45,380 --> 00:50:47,961
it runs by 45-degree
along this space.

755
00:50:53,260 --> 00:50:59,750
So once I remove the
constraint that E1 is fixed

756
00:50:59,750 --> 00:51:02,050
and E2 is fixed,
they can exchange.

757
00:51:02,050 --> 00:51:05,830
They explore a whole
bunch of other states

758
00:51:05,830 --> 00:51:08,030
that is available to them.

759
00:51:08,030 --> 00:51:12,480
And I would probably
say that the probability

760
00:51:12,480 --> 00:51:18,610
of the microstates is the same
up to some normalization that

761
00:51:18,610 --> 00:51:28,360
comes from E1.

762
00:51:28,360 --> 00:51:32,640
So the normalization-- not
the entirety 1 plus 2--

763
00:51:32,640 --> 00:51:37,210
is a microcanonical ensemble
but with energy E total.

764
00:51:37,210 --> 00:51:41,320
So there is a
corresponding omega

765
00:51:41,320 --> 00:51:44,740
that is associated with
the combined system.

766
00:51:44,740 --> 00:51:48,170
And to obtain that,
all I need to do

767
00:51:48,170 --> 00:51:52,540
is to sum or integrate
over the energy, let's

768
00:51:52,540 --> 00:51:55,220
say, that I have
in the first one.

769
00:51:55,220 --> 00:51:58,230
The number of states
that I would have,

770
00:51:58,230 --> 00:52:00,020
or the volume of
phase space that I

771
00:52:00,020 --> 00:52:06,980
would have if I
was at E1, and then

772
00:52:06,980 --> 00:52:12,850
simultaneously multiplying by
how many states the second part

773
00:52:12,850 --> 00:52:15,390
can have at the energy
that corresponds

774
00:52:15,390 --> 00:52:17,492
to E total minus E1.

775
00:52:22,220 --> 00:52:27,770
So what I need to do is
to essentially multiply

776
00:52:27,770 --> 00:52:31,560
the number of states that I
would encounter going along

777
00:52:31,560 --> 00:52:34,790
this axis, and the
number of states

778
00:52:34,790 --> 00:52:37,690
that I would multiply
going along the other axis.

779
00:52:37,690 --> 00:52:40,980
So let's try to sort of
indicate those things

780
00:52:40,980 --> 00:52:45,250
with some kind of
a color density.

781
00:52:45,250 --> 00:52:50,100
So let's say that the
density is kind of low here,

782
00:52:50,100 --> 00:52:54,500
it comes kind of high here,
and then goes low here.

783
00:52:54,500 --> 00:52:59,360
If I move along this axis,
let's say that along this axis,

784
00:52:59,360 --> 00:53:04,740
I maybe become high here,
and stay high, and then

785
00:53:04,740 --> 00:53:06,070
become low later.

786
00:53:06,070 --> 00:53:07,610
Some kind of thing.

787
00:53:07,610 --> 00:53:11,370
So all I need to do is to
multiply these two colors

788
00:53:11,370 --> 00:53:13,640
and generate the
color along this--

789
00:53:13,640 --> 00:53:17,980
there is this direction, which
I will plot going, hopefully,

790
00:53:17,980 --> 00:53:19,190
coming out of the board.

791
00:53:19,190 --> 00:53:21,630
And maybe that product
looks something like this.

792
00:53:25,280 --> 00:53:33,013
Stating that somewhere there
is a most probable state,

793
00:53:33,013 --> 00:53:40,250
and I can indicate the most
probable state by E1 star

794
00:53:40,250 --> 00:53:41,955
and E2 star, let's say.

795
00:53:46,940 --> 00:53:49,170
So you may say, OK,
it actually could

796
00:53:49,170 --> 00:53:52,010
be something, who says it
should have one maximum.

797
00:53:52,010 --> 00:53:54,520
It could have multiple
maxima, or things like that.

798
00:53:54,520 --> 00:53:56,510
You could certainly allow
all kinds of things.

799
00:53:59,080 --> 00:54:03,440
Now, my claim is that
when I go and rely

800
00:54:03,440 --> 00:54:10,770
on this particular
limit, I can state

801
00:54:10,770 --> 00:54:15,710
that if I explore
all of these states,

802
00:54:15,710 --> 00:54:20,300
I will find my system in the
vicinity of these energies

803
00:54:20,300 --> 00:54:23,070
with probability that in
the n goes to infinity,

804
00:54:23,070 --> 00:54:26,590
limit becomes 1.

805
00:54:26,590 --> 00:54:30,230
And that kind of
relies on the fact

806
00:54:30,230 --> 00:54:33,340
that each one of
these quantities

807
00:54:33,340 --> 00:54:37,480
is really an exponentially
large quantity.

808
00:54:37,480 --> 00:54:41,620
Before I do that, I
forgot to do something

809
00:54:41,620 --> 00:54:44,250
that I wanted to do over here.

810
00:54:44,250 --> 00:54:47,130
I have promised that as
we go through the course,

811
00:54:47,130 --> 00:54:50,870
at each stage we will
define for each section

812
00:54:50,870 --> 00:54:53,620
its own definition of entropy.

813
00:54:53,620 --> 00:54:57,050
Well, almost, but
not quite that.

814
00:54:57,050 --> 00:55:00,700
Once I have a probability,
I had told you

815
00:55:00,700 --> 00:55:05,080
how to define an entropy
associated with a probability.

816
00:55:05,080 --> 00:55:09,950
So here, I can say that
when I have a probability p,

817
00:55:09,950 --> 00:55:15,920
I can identify the average of
log p, the factor of minus p

818
00:55:15,920 --> 00:55:21,870
log p, to be the entropy
of that probability.

819
00:55:21,870 --> 00:55:27,180
Kind of linked to what we
had for mixing entropy,

820
00:55:27,180 --> 00:55:31,260
except that in
thermodynamics, entropy

821
00:55:31,260 --> 00:55:33,690
had some particular units.

822
00:55:33,690 --> 00:55:37,260
It was related to
heat or temperature.

823
00:55:37,260 --> 00:55:41,210
So we multiplied
by a quantity kb

824
00:55:41,210 --> 00:55:44,440
that has the right
units of energy

825
00:55:44,440 --> 00:55:47,360
divided by degrees Kelvin.

826
00:55:47,360 --> 00:55:51,340
Now, for the choice of the
probability that we have,

827
00:55:51,340 --> 00:55:55,510
it is kind of like a
step function in energy.

828
00:55:55,510 --> 00:56:01,260
The probability is
either 0 or 1 over omega.

829
00:56:01,260 --> 00:56:04,870
So when you're at 0, this
p log p will give you a 0.

830
00:56:04,870 --> 00:56:08,730
When you're over here, p log
p will give you log of omega.

831
00:56:08,730 --> 00:56:14,630
So this is going to give
you kb log of omega.

832
00:56:14,630 --> 00:56:20,980
So we can identify in the
macrocanonical ensemble,

833
00:56:20,980 --> 00:56:25,610
once we've stated
what E, x, and n are,

834
00:56:25,610 --> 00:56:27,930
what the analog
of the six that we

835
00:56:27,930 --> 00:56:32,430
have for the throwing
of the dices, what's

836
00:56:32,430 --> 00:56:37,340
the number of microstates that
are compatible with the energy.

837
00:56:37,340 --> 00:56:40,580
We will have to do a little bit
of massaging that to understand

838
00:56:40,580 --> 00:56:42,880
what that means in
the continuum limit.

839
00:56:42,880 --> 00:56:44,200
We'll fix that.

840
00:56:44,200 --> 00:56:47,380
But once we know that
number, essentially the log

841
00:56:47,380 --> 00:56:50,110
of that number up
to a factor would

842
00:56:50,110 --> 00:56:53,670
give something that would be
the entropy of that probability

843
00:56:53,670 --> 00:56:56,520
that eventually we're
going to identify

844
00:56:56,520 --> 00:57:00,130
with the thermodynamic
entropy that

845
00:57:00,130 --> 00:57:02,470
corresponds to this system.

846
00:57:02,470 --> 00:57:04,620
But having done
that definition, I

847
00:57:04,620 --> 00:57:12,860
can rewrite this quantity
as E1 e to the 1 over kb S1

848
00:57:12,860 --> 00:57:17,520
plus 1 over kb S2.

849
00:57:17,520 --> 00:57:20,280
This one is evaluated at E1.

850
00:57:20,280 --> 00:57:26,029
This one is evaluated at E2,
which is E total minus E1.

851
00:57:33,020 --> 00:57:37,450
Now, the statement that we are
going to gradually build upon

852
00:57:37,450 --> 00:57:43,600
is that these omegas are
these types of quantities

853
00:57:43,600 --> 00:57:47,650
that I mentioned that
depend exponentially

854
00:57:47,650 --> 00:57:49,900
on the number of particles.

855
00:57:49,900 --> 00:57:53,620
So let's say, if you just think
about volume, one particle

856
00:57:53,620 --> 00:57:55,520
then, can be anywhere in this.

857
00:57:55,520 --> 00:57:58,550
So that's a factor of v.
Two particles, v squared.

858
00:57:58,550 --> 00:58:00,050
Three particles, v cubed.

859
00:58:00,050 --> 00:58:02,640
n particles, v to the n.

860
00:58:02,640 --> 00:58:05,640
So these omegas
have buried in them

861
00:58:05,640 --> 00:58:08,130
an exponential dependence on n.

862
00:58:08,130 --> 00:58:12,150
When you take the log, these are
quantities that are extensing.

863
00:58:12,150 --> 00:58:14,500
They're proportionate
to this number

864
00:58:14,500 --> 00:58:17,450
n that becomes very large.

865
00:58:17,450 --> 00:58:20,380
So this is one of
those examples where

866
00:58:20,380 --> 00:58:23,760
to calculate this
omega in total,

867
00:58:23,760 --> 00:58:29,080
I have to evaluate an integral
where the quantities are

868
00:58:29,080 --> 00:58:31,590
exponentially large.

869
00:58:31,590 --> 00:58:34,440
And we saw that
when that happens,

870
00:58:34,440 --> 00:58:38,220
I could replace this
integral essentially,

871
00:58:38,220 --> 00:58:39,690
with its largest value.

872
00:58:39,690 --> 00:58:50,000
So I would have 1 over kb S1
of E1 star plus S2 of E2 star.

873
00:58:57,140 --> 00:59:03,540
Now, how do I identify where
E1 star and E2 star are?

874
00:59:03,540 --> 00:59:09,220
Well, given that I scale along
this axis or along that axis,

875
00:59:09,220 --> 00:59:12,400
essentially I want
to find locations

876
00:59:12,400 --> 00:59:15,540
where the exponent
is the largest.

877
00:59:15,540 --> 00:59:17,520
So how do I find
those locations?

878
00:59:17,520 --> 00:59:20,490
I essentially set
the derivative to 0.

879
00:59:20,490 --> 00:59:23,840
So if I take the derivative
of this with respect to E1,

880
00:59:23,840 --> 00:59:24,470
what do I get?

881
00:59:24,470 --> 00:59:28,722
I get dS1 by dE1.

882
00:59:28,722 --> 00:59:32,790
And from here, I
get dS2 with respect

883
00:59:32,790 --> 00:59:37,670
to an argument that is
in fact has a minus E1,

884
00:59:37,670 --> 00:59:43,500
so I would have minus dS2 with
respect to its own energy.

885
00:59:43,500 --> 00:59:45,600
With respect to its
own energy argument,

886
00:59:45,600 --> 00:59:48,070
but evaluated with
an energy argument

887
00:59:48,070 --> 00:59:51,030
that goes opposite way with E1.

888
00:59:51,030 --> 00:59:54,410
And all of these are
calculated at conditions

889
00:59:54,410 --> 00:59:56,980
where the corresponding
x's and n's are fixed.

890
01:00:02,570 --> 01:00:05,360
And this has to
be 0, which means

891
01:00:05,360 --> 01:00:10,420
that at the maxima, essentially,
I would have this condition.

892
01:00:13,850 --> 01:00:21,020
So again, because of the
exponential character,

893
01:00:21,020 --> 01:00:25,100
there could be multiple
of these maxima.

894
01:00:25,100 --> 01:00:28,940
But if one of them
is slightly larger

895
01:00:28,940 --> 01:00:33,050
than the other in
absolute terms,

896
01:00:33,050 --> 01:00:35,800
in terms of the
intensive quantities,

897
01:00:35,800 --> 01:00:38,850
once I multiply by
these n's, it will

898
01:00:38,850 --> 01:00:40,905
be exponentially
larger than the others.

899
01:00:40,905 --> 01:00:44,090
We sort of discussed that when
we were doing the saddle point

900
01:00:44,090 --> 01:00:48,690
approximation, how essentially
the best maximum is

901
01:00:48,690 --> 01:00:51,610
exponentially larger
than all the others.

902
01:00:51,610 --> 01:00:55,400
And so in that sense, it is
exponentially much more likely

903
01:00:55,400 --> 01:00:58,320
that you would be here
as opposed to here

904
01:00:58,320 --> 01:01:00,500
and anywhere else.

905
01:01:00,500 --> 01:01:04,970
So the statement, again, is that
just a matter of probabilities.

906
01:01:04,970 --> 01:01:08,530
I don't say what's the dynamics
by which the energies can

907
01:01:08,530 --> 01:01:11,230
explore these two axes.

908
01:01:11,230 --> 01:01:13,990
But I imagine like
shuffling cards,

909
01:01:13,990 --> 01:01:19,350
the red and the black cards
have been mixed sufficiently,

910
01:01:19,350 --> 01:01:21,180
and then you ask
a question about

911
01:01:21,180 --> 01:01:23,200
the typical configuration.

912
01:01:23,200 --> 01:01:25,160
And in typical
configurations, you

913
01:01:25,160 --> 01:01:27,960
don't expect a run of ten
black cards, or whatever,

914
01:01:27,960 --> 01:01:31,370
because they're
exponentially unlikely.

915
01:01:31,370 --> 01:01:34,000
So this is the same
statement, that once you

916
01:01:34,000 --> 01:01:38,150
allow this system to equilibrate
its energy, after a while

917
01:01:38,150 --> 01:01:40,990
you look at it, and
with probability 1

918
01:01:40,990 --> 01:01:44,960
you will find it
at a location where

919
01:01:44,960 --> 01:01:47,980
these derivatives are the same.

920
01:01:47,980 --> 01:01:49,800
Now, each one of
these derivatives

921
01:01:49,800 --> 01:01:54,340
is completely something that
pertains to its own system,

922
01:01:54,340 --> 01:01:55,930
so one of them could be a gas.

923
01:01:55,930 --> 01:01:57,220
The other could be a spring.

924
01:01:57,220 --> 01:01:59,170
It could be anything.

925
01:01:59,170 --> 01:02:02,260
And these derivatives could
be very different quantities.

926
01:02:02,260 --> 01:02:04,520
But this equality would hold.

927
01:02:04,520 --> 01:02:08,600
And that is what we
have for the zeroth law.

928
01:02:08,600 --> 01:02:11,710
That when systems come
into equilibrium, there

929
01:02:11,710 --> 01:02:14,990
is a function of
parameters of one

930
01:02:14,990 --> 01:02:18,300
that has to be the same as
the function of the parameters

931
01:02:18,300 --> 01:02:20,850
of the other, which we call
the empirical temperature.

932
01:02:27,420 --> 01:02:35,070
So in principal, we could define
any function of temperature,

933
01:02:35,070 --> 01:02:39,090
and in practice, for
consistently, that function

934
01:02:39,090 --> 01:02:40,415
is one over the temperature.

935
01:02:49,680 --> 01:02:55,670
So the zeroth law
established that exists

936
01:02:55,670 --> 01:02:58,490
this empirical function.

937
01:02:58,490 --> 01:03:01,340
This choice of 1
over T is so that we

938
01:03:01,340 --> 01:03:04,000
are aligned with
everything else that we

939
01:03:04,000 --> 01:03:06,310
have done so far, as
we will see shortly.

940
01:03:09,020 --> 01:03:10,390
OK?

941
01:03:10,390 --> 01:03:14,670
The next thing is
the first law, which

942
01:03:14,670 --> 01:03:18,690
had something to
do with the change

943
01:03:18,690 --> 01:03:20,280
in the energy of the system.

944
01:03:20,280 --> 01:03:22,460
When we go from one
state to another state

945
01:03:22,460 --> 01:03:28,240
had to be made up by a
combination of heat and work.

946
01:03:28,240 --> 01:03:31,900
So for this, let's
expand our system

947
01:03:31,900 --> 01:03:35,060
to allow some kind of work.

948
01:03:35,060 --> 01:03:39,280
So let's imagine that
I have something--

949
01:03:39,280 --> 01:03:45,740
let's say if it is a gas-- and
the quantity that would change

950
01:03:45,740 --> 01:03:49,020
if there's work done on
the gas is the volume.

951
01:03:49,020 --> 01:03:53,340
So let's imagine that there
is a piston that can slide.

952
01:03:53,340 --> 01:03:57,510
And this piston exerts
for the gas a pressure,

953
01:03:57,510 --> 01:04:00,540
or in general, whatever
the conjugate variable

954
01:04:00,540 --> 01:04:04,610
is to the displacement that's
we now allow to change.

955
01:04:04,610 --> 01:04:09,270
So there's potentially a
change in the displacement.

956
01:04:09,270 --> 01:04:12,730
So what we are doing
is there's work

957
01:04:12,730 --> 01:04:18,050
that is done that
corresponds to j delta x.

958
01:04:18,050 --> 01:04:27,440
So what happens if this
j delta x amount of work

959
01:04:27,440 --> 01:04:29,610
is done on the system?

960
01:04:29,610 --> 01:04:33,480
Then the system goes from
one configuration that

961
01:04:33,480 --> 01:04:36,270
is characterized by x to
another configuration that

962
01:04:36,270 --> 01:04:39,920
is characterized
by x plus delta x.

963
01:04:39,920 --> 01:04:48,040
And let's see what the change
in entropy is when that happens.

964
01:04:48,040 --> 01:04:51,090
Change in entropy, or the
log of the number of states

965
01:04:51,090 --> 01:04:52,780
that we have defined.

966
01:04:52,780 --> 01:05:05,860
And so this is essentially the
change starting from E and x

967
01:05:05,860 --> 01:05:14,430
to going to the case where
x changed by an amount dx.

968
01:05:14,430 --> 01:05:17,730
But through this
process I did work,

969
01:05:17,730 --> 01:05:22,500
and so the amount of energy that
is inside the system increases

970
01:05:22,500 --> 01:05:23,980
by an amount that is j delta x.

971
01:05:28,250 --> 01:05:33,610
So if all of these
quantities are infinitesimal,

972
01:05:33,610 --> 01:05:38,540
I have two arguments of S now
have changed infinitesimally,

973
01:05:38,540 --> 01:05:42,230
and I can make the corresponding
expansion in derivatives.

974
01:05:42,230 --> 01:05:48,190
There is a dS with respect
to dE at constant x.

975
01:05:48,190 --> 01:05:55,170
The corresponding change
in E is j delta x.

976
01:05:55,170 --> 01:06:02,660
But then there's also a
dS by dx at constant E

977
01:06:02,660 --> 01:06:05,370
with the same delta
x, so I factored out

978
01:06:05,370 --> 01:06:07,214
the delta x between
the two of them.

979
01:06:11,070 --> 01:06:15,840
Now, dS by dE at
constant x is related

980
01:06:15,840 --> 01:06:18,540
to the empirical
temperature, which we now

981
01:06:18,540 --> 01:06:24,980
set to be 1 over
T. Now, the claim

982
01:06:24,980 --> 01:06:30,110
is that if you have a situation
such as this, where you have

983
01:06:30,110 --> 01:06:32,170
a state that is sitting
in equilibrium--

984
01:06:32,170 --> 01:06:37,980
let's say a gas with a
piston-- then by definition

985
01:06:37,980 --> 01:06:41,900
of equilibrium, the system
does not spontaneously

986
01:06:41,900 --> 01:06:43,360
change its volume.

987
01:06:46,730 --> 01:06:49,850
But it will change its volume
because the number of states

988
01:06:49,850 --> 01:06:56,120
that is availability to it
increases, or S increases.

989
01:06:56,120 --> 01:06:58,920
So in order to
make sure that you

990
01:06:58,920 --> 01:07:02,210
don't go to a state that is
more probable, because there are

991
01:07:02,210 --> 01:07:04,940
more possibilities, there
are more microstates,

992
01:07:04,940 --> 01:07:09,160
I better make sure that
this first derivative is 0.

993
01:07:09,160 --> 01:07:12,620
Otherwise, depending on
whether this is plus or minus,

994
01:07:12,620 --> 01:07:14,590
I could make a
corresponding change

995
01:07:14,590 --> 01:07:18,380
in delta x that would
increase delta S.

996
01:07:18,380 --> 01:07:21,250
So what happens
here if I require

997
01:07:21,250 --> 01:07:25,460
this to be 0 is that
I can identify now

998
01:07:25,460 --> 01:07:32,690
that the derivative of this
quantity dS by dx at constant E

999
01:07:32,690 --> 01:07:43,170
has to be minus j
over T. Currently,

1000
01:07:43,170 --> 01:07:46,500
I have only these two
variables, and quite

1001
01:07:46,500 --> 01:07:54,040
generically, I can say that dS
is dS by dE at constant x dE

1002
01:07:54,040 --> 01:07:59,410
plus dS by dx at constant E dx.

1003
01:07:59,410 --> 01:08:02,550
And now I have identified
these two derivatives.

1004
01:08:02,550 --> 01:08:10,420
dS by dE is 1 over T, so I have
dE over T. dS by dx is minus

1005
01:08:10,420 --> 01:08:16,960
j over T, so I have
minus j dx over T.

1006
01:08:16,960 --> 01:08:28,295
And I can rearrange this, and I
see that dE is T dS plus j dx.

1007
01:08:32,390 --> 01:08:37,210
Now, the j dx I
recognize as before, it's

1008
01:08:37,210 --> 01:08:40,040
the mechanical work.

1009
01:08:40,040 --> 01:08:42,720
So I have identified
that generically,

1010
01:08:42,720 --> 01:08:47,340
when you make a transformation,
in addition to mechanical work,

1011
01:08:47,340 --> 01:08:52,040
there's a component that
changes the energy that

1012
01:08:52,040 --> 01:08:55,370
is the one that we can
identify with the heat.

1013
01:08:55,370 --> 01:08:56,229
Yes?

1014
01:08:56,229 --> 01:09:00,540
AUDIENCE: Could you explain
why you set that S to 0?

1015
01:09:00,540 --> 01:09:03,710
PROFESSOR: Why did
I set delta S to 0?

1016
01:09:03,710 --> 01:09:09,979
So I have a box, and this
box has a certain volume,

1017
01:09:09,979 --> 01:09:12,700
and there's a bunch
of particles here,

1018
01:09:12,700 --> 01:09:15,420
but the piston that
is holding this

1019
01:09:15,420 --> 01:09:19,270
is allowed to slide
to go up and down.

1020
01:09:19,270 --> 01:09:20,770
OK?

1021
01:09:20,770 --> 01:09:26,620
Now, I can ask what happens if
the volume goes up and down.

1022
01:09:26,620 --> 01:09:32,370
If the volume goes up and down,
how many states are available?

1023
01:09:32,370 --> 01:09:35,189
So basically, the
statement has been

1024
01:09:35,189 --> 01:09:38,729
that for each
configuration, there

1025
01:09:38,729 --> 01:09:42,460
is a number of states
that is available.

1026
01:09:42,460 --> 01:09:45,649
And if I were to change
that configuration,

1027
01:09:45,649 --> 01:09:47,630
the number of
states will change.

1028
01:09:47,630 --> 01:09:50,979
And hence the log of it, which
is the entropy, will change.

1029
01:09:50,979 --> 01:09:53,290
So how much does it change?

1030
01:09:53,290 --> 01:09:55,300
Well, let's see what
arguments changed.

1031
01:09:55,300 --> 01:09:58,530
So certainly, the
volume changed.

1032
01:09:58,530 --> 01:10:01,970
So x went to x plus dx.

1033
01:10:01,970 --> 01:10:05,390
And because I did
some amount of work,

1034
01:10:05,390 --> 01:10:11,100
pdv, the amount of energy that
was in the box also changed.

1035
01:10:11,100 --> 01:10:11,910
OK?

1036
01:10:11,910 --> 01:10:15,040
So after this transformation
with this thing going

1037
01:10:15,040 --> 01:10:19,600
up or down, what is
the new logarithm

1038
01:10:19,600 --> 01:10:21,140
of the number of states?

1039
01:10:21,140 --> 01:10:22,730
How much has it changed?

1040
01:10:22,730 --> 01:10:26,230
And how much of the
change is given by this?

1041
01:10:26,230 --> 01:10:28,570
Now, suppose that
I make this change,

1042
01:10:28,570 --> 01:10:31,700
and I find that suddenly I
have a thousand more states

1043
01:10:31,700 --> 01:10:32,850
available to me.

1044
01:10:32,850 --> 01:10:34,910
It's a thousand
times more likely

1045
01:10:34,910 --> 01:10:37,540
that I will accept this change.

1046
01:10:37,540 --> 01:10:43,440
So in order for me to be
penalized-- or actually,

1047
01:10:43,440 --> 01:10:46,150
not penalized, or
not gain-- because I

1048
01:10:46,150 --> 01:10:49,430
make this transformation,
this change better be 0.

1049
01:10:51,970 --> 01:10:55,840
If it is not 0, if
this quantity is,

1050
01:10:55,840 --> 01:10:58,380
let's say, positive,
then I will choose

1051
01:10:58,380 --> 01:11:03,510
a delta v that is positive, and
delta S will become positive.

1052
01:11:03,510 --> 01:11:06,135
If this quantity
happens to be negative,

1053
01:11:06,135 --> 01:11:08,710
then I will choose a
delta v that is negative,

1054
01:11:08,710 --> 01:11:11,880
and then delta S will
be positive again.

1055
01:11:11,880 --> 01:11:15,340
So the statement that
this thing originally

1056
01:11:15,340 --> 01:11:19,460
was sitting there by
itself in equilibrium

1057
01:11:19,460 --> 01:11:23,500
and did not spontaneously
go up or down

1058
01:11:23,500 --> 01:11:26,416
is this statement
this derivative 0.

1059
01:11:33,178 --> 01:11:37,540
AUDIENCE: That's only 0 when
the system is in equilibrium?

1060
01:11:37,540 --> 01:11:38,290
PROFESSOR: Yes.

1061
01:11:38,290 --> 01:11:39,260
Yes.

1062
01:11:39,260 --> 01:11:46,000
So indeed, this is
then a relationship

1063
01:11:46,000 --> 01:11:49,460
that involves parameters of
the system in equilibrium.

1064
01:11:49,460 --> 01:11:49,960
Yeah?

1065
01:11:49,960 --> 01:11:52,600
So there is
thermodynamically, we

1066
01:11:52,600 --> 01:11:58,810
said that once I specify
what my energy and volume

1067
01:11:58,810 --> 01:12:01,140
and number of particles
are in equilibrium,

1068
01:12:01,140 --> 01:12:03,610
there is a particular
S. And what

1069
01:12:03,610 --> 01:12:06,780
I want to know is if
I go from one state

1070
01:12:06,780 --> 01:12:11,550
to another state in equilibrium,
what is the change in dS?

1071
01:12:11,550 --> 01:12:13,880
That's what this statement is.

1072
01:12:13,880 --> 01:12:17,250
Then I can rearrange
it and see that when

1073
01:12:17,250 --> 01:12:21,190
I go from one equilibrium state
to another equilibrium state,

1074
01:12:21,190 --> 01:12:23,560
I have to change
internal energy, which

1075
01:12:23,560 --> 01:12:28,992
I can do either by doing
work or by doing heat.

1076
01:12:33,948 --> 01:12:34,448
OK?

1077
01:12:54,290 --> 01:12:56,110
Now, the second law
is actually obvious.

1078
01:13:00,620 --> 01:13:05,050
I have stated that I start
from this configuration

1079
01:13:05,050 --> 01:13:09,810
and go to that configuration
simply because of probability.

1080
01:13:09,810 --> 01:13:15,290
The probabilities are inverse
of this omega, if you like.

1081
01:13:15,290 --> 01:13:20,780
So I start with some location--
like the pack of cards,

1082
01:13:20,780 --> 01:13:22,655
all of the black
on one side, all

1083
01:13:22,655 --> 01:13:27,920
of the red on the other
side-- and I do some dynamics,

1084
01:13:27,920 --> 01:13:29,800
and I will end up
with some other state.

1085
01:13:29,800 --> 01:13:31,730
I will, because why?

1086
01:13:31,730 --> 01:13:36,990
Because that state has a much
more volume of possibilities.

1087
01:13:36,990 --> 01:13:38,880
There is a single
state that we identify

1088
01:13:38,880 --> 01:13:41,260
as all-black and
all-red, and there's

1089
01:13:41,260 --> 01:13:45,130
a myriad of states that we say
they're all randomly mixed,

1090
01:13:45,130 --> 01:13:47,120
and so we're much
more likely to find

1091
01:13:47,120 --> 01:13:50,780
from that subset of states.

1092
01:13:50,780 --> 01:13:57,520
So I have certainly stated
that S1 evaluated at E1 star,

1093
01:13:57,520 --> 01:14:02,480
plus S2 evaluated at
E2 star, essentially

1094
01:14:02,480 --> 01:14:11,140
the peak of that object, is much
larger than S1 plus E1 plus 0

1095
01:14:11,140 --> 01:14:12,616
plus S2 [INAUDIBLE].

1096
01:14:22,970 --> 01:14:28,270
And more specifically,
if we sort of follow

1097
01:14:28,270 --> 01:14:34,790
the course of the system
as it starts from here,

1098
01:14:34,790 --> 01:14:39,120
you will find that it will
basically go in the direction

1099
01:14:39,120 --> 01:14:46,490
always such that the derivatives
that are related to temperature

1100
01:14:46,490 --> 01:14:51,960
are such that the energy
will flow from the hot air

1101
01:14:51,960 --> 01:14:55,460
to the colder body,
consistent with what

1102
01:14:55,460 --> 01:14:56,710
we expect from thermodynamics.

1103
01:15:01,140 --> 01:15:04,920
The one law of thermodynamic
that I cannot prove from what I

1104
01:15:04,920 --> 01:15:07,565
have given you so
far is the third law.

1105
01:15:07,565 --> 01:15:10,280
There is no reason
why the entropy should

1106
01:15:10,280 --> 01:15:13,450
go to 0 as you go
to 0 temperature

1107
01:15:13,450 --> 01:15:14,519
within this perspective.

1108
01:15:21,510 --> 01:15:28,310
So it's good to look at our
canonical example, which

1109
01:15:28,310 --> 01:15:29,560
is the ideal gas.

1110
01:15:34,390 --> 01:15:37,365
Again, for the
ideal gas, if I am

1111
01:15:37,365 --> 01:15:39,740
in a microcanonical
ensemble, it means

1112
01:15:39,740 --> 01:15:42,880
that I have told you what
the energy of the box is,

1113
01:15:42,880 --> 01:15:45,250
what the volume of the box
is, and how many particles

1114
01:15:45,250 --> 01:15:48,190
are in it.

1115
01:15:48,190 --> 01:15:52,100
So clearly the microstate
that corresponds to that

1116
01:15:52,100 --> 01:15:55,720
is the collection of
the 6N coordinates

1117
01:15:55,720 --> 01:16:00,540
and momenta of the
particles in the box.

1118
01:16:00,540 --> 01:16:05,130
And the energy of
the system is made up

1119
01:16:05,130 --> 01:16:10,150
by the sum of the energies
of all of the particles,

1120
01:16:10,150 --> 01:16:17,700
and basically, ideal gas means
that the energy that we write

1121
01:16:17,700 --> 01:16:20,485
is the sum of the
contributions that you

1122
01:16:20,485 --> 01:16:23,540
would have from
individual particles.

1123
01:16:23,540 --> 01:16:26,960
So individual particles
have a kinetic energy

1124
01:16:26,960 --> 01:16:30,810
and some potential
energy, and since we

1125
01:16:30,810 --> 01:16:33,900
are stating that we
have a box of volume v,

1126
01:16:33,900 --> 01:16:39,450
this u essentially represents
this box of volume.

1127
01:16:39,450 --> 01:16:43,450
It's a potential that is 0
inside this volume and infinity

1128
01:16:43,450 --> 01:16:44,321
outside of it.

1129
01:16:47,550 --> 01:16:49,230
Fine.

1130
01:16:49,230 --> 01:16:53,990
So you ask what's
the probability

1131
01:16:53,990 --> 01:16:56,220
of some particular
microstate, given

1132
01:16:56,220 --> 01:16:59,790
that I specified E, v, and n.

1133
01:16:59,790 --> 01:17:02,290
I would say, well, OK.

1134
01:17:02,290 --> 01:17:09,820
This is, by the construction
that we had, 0 or 1

1135
01:17:09,820 --> 01:17:16,910
over some omega, depending on
whether the energy is right.

1136
01:17:16,910 --> 01:17:18,910
And since for the
box, the energy

1137
01:17:18,910 --> 01:17:21,250
is made up of the
kinetic energy,

1138
01:17:21,250 --> 01:17:32,910
if sum over i p i squared
over 2m is not equal to E,

1139
01:17:32,910 --> 01:17:37,755
or particle q is
outside the box.

1140
01:17:42,710 --> 01:17:51,350
And it is the same value if
sum over i p i squared over 2m

1141
01:17:51,350 --> 01:17:56,490
is equal to v and q i's
are all inside the box.

1142
01:18:07,921 --> 01:18:14,230
So a note about normalization.

1143
01:18:14,230 --> 01:18:17,690
I kind of skipped on
this over the last one.

1144
01:18:17,690 --> 01:18:23,400
p is a probability in this
6N dimensional phase space.

1145
01:18:23,400 --> 01:18:32,820
So the normalization is that the
integral over all of the p i's

1146
01:18:32,820 --> 01:18:42,470
and qi's should be equal to 1.

1147
01:18:42,470 --> 01:18:46,640
And so this omega,
more precisely,

1148
01:18:46,640 --> 01:18:49,920
when we are not talking
about discrete numbers,

1149
01:18:49,920 --> 01:18:53,530
is the quantity
that I have to put

1150
01:18:53,530 --> 01:18:58,890
so that when I integrate this
over phase space, I will get 1.

1151
01:18:58,890 --> 01:19:03,940
So basically, if I write this
in the form that I have written,

1152
01:19:03,940 --> 01:19:11,830
you can see that omega is
obtained by integrating over

1153
01:19:11,830 --> 01:19:18,850
p i q i that correspond
to this accessible states.

1154
01:19:25,290 --> 01:19:27,670
It's kind of like
a delta function.

1155
01:19:27,670 --> 01:19:32,240
Basically, there's huge
portions of phase space

1156
01:19:32,240 --> 01:19:38,060
that are 0 probability,
and then there's

1157
01:19:38,060 --> 01:19:43,990
a surface that has
the probability that

1158
01:19:43,990 --> 01:19:48,610
is 1 over omega and it is
the area of that surface

1159
01:19:48,610 --> 01:19:52,000
that I have to ensure
is appearing here,

1160
01:19:52,000 --> 01:19:54,730
so that when I integrate
over that area of one

1161
01:19:54,730 --> 01:19:59,070
over that area, I will get one.

1162
01:19:59,070 --> 01:20:03,230
Now, the part that corresponds
to the q coordinates

1163
01:20:03,230 --> 01:20:06,190
is actually very
simple, because the q

1164
01:20:06,190 --> 01:20:09,340
coordinates have to
be inside the box.

1165
01:20:09,340 --> 01:20:12,820
Each one of them has a volume
v, so the coordinate part

1166
01:20:12,820 --> 01:20:16,041
of the integral
gives me v to the m.

1167
01:20:16,041 --> 01:20:16,540
OK?

1168
01:20:19,330 --> 01:20:24,170
Now, the momentum
part, the momenta

1169
01:20:24,170 --> 01:20:26,480
are constrained by
something like this.

1170
01:20:30,640 --> 01:20:38,040
I can rewrite that as sum over
i p i squared equals to 2mE.

1171
01:20:41,500 --> 01:20:47,290
And if I regard this 2mE as
something like r squared,

1172
01:20:47,290 --> 01:20:49,680
like a radius
squared, you can see

1173
01:20:49,680 --> 01:20:52,310
that this is like
the equation that you

1174
01:20:52,310 --> 01:20:54,620
would have for
the generalization

1175
01:20:54,620 --> 01:20:57,600
of the sphere in 3N dimensions.

1176
01:20:57,600 --> 01:21:00,090
Because if I had
just x squared, x

1177
01:21:00,090 --> 01:21:03,030
squared plus y squared
is r squared is a circle,

1178
01:21:03,030 --> 01:21:05,140
x squared plus y
squared plus z squared

1179
01:21:05,140 --> 01:21:07,700
is r squared is
a sphere, so this

1180
01:21:07,700 --> 01:21:15,410
is the hypersphere
in 3N dimensions.

1181
01:21:18,160 --> 01:21:22,740
So when I do the
integrations over q,

1182
01:21:22,740 --> 01:21:32,880
I have to integrate
over the surface of a 3N

1183
01:21:32,880 --> 01:21:45,270
dimensional hypersphere of
radius square root of 2m.

1184
01:21:47,850 --> 01:21:49,744
OK?

1185
01:21:49,744 --> 01:21:54,140
And there is a simple
formula for that.

1186
01:21:54,140 --> 01:21:59,800
This surface area,
in general, is

1187
01:21:59,800 --> 01:22:03,460
going to be
proportional to r raised

1188
01:22:03,460 --> 01:22:07,880
to the power of
dimension minus 1.

1189
01:22:07,880 --> 01:22:12,210
And there's the generalization
of 2pi r over 4pi r squared

1190
01:22:12,210 --> 01:22:14,940
that you would have in
two or three dimensions,

1191
01:22:14,940 --> 01:22:17,900
which we will discuss next time.

1192
01:22:17,900 --> 01:22:22,170
It is 2 to the
power of 3n over 2

1193
01:22:22,170 --> 01:22:26,980
divided by 3n over 2 factorial.

1194
01:22:26,980 --> 01:22:29,950
So we have the formula
for this omega.

1195
01:22:29,950 --> 01:22:34,150
We will re-derive it and
discuss it next time.