1 00:00:00,060 --> 00:00:01,780 The following content is provided 2 00:00:01,780 --> 00:00:04,019 under a Creative Commons license. 3 00:00:04,019 --> 00:00:06,870 Your support will help MIT OpenCourseWare continue 4 00:00:06,870 --> 00:00:10,730 to offer high quality educational resources for free. 5 00:00:10,730 --> 00:00:13,330 To make a donation or view additional materials 6 00:00:13,330 --> 00:00:17,217 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:17,217 --> 00:00:17,842 at ocw.mit.edu. 8 00:00:20,345 --> 00:00:22,570 PROFESSOR: So the question is, how do we 9 00:00:22,570 --> 00:00:26,800 describe the various motions that are taking place? 10 00:00:26,800 --> 00:00:28,780 And in principle, the Boltzmann equation 11 00:00:28,780 --> 00:00:30,460 that we have been developing should 12 00:00:30,460 --> 00:00:33,690 be able to tell us something about all of that. 13 00:00:33,690 --> 00:00:37,750 Because we put essentially all of the phenomena 14 00:00:37,750 --> 00:00:41,940 that we think are relevant to a dilute gas, such as what's 15 00:00:41,940 --> 00:00:46,180 going on in this room, into that equation, 16 00:00:46,180 --> 00:00:49,700 and reduced the equation to something that 17 00:00:49,700 --> 00:00:53,660 had a simple form where on the left hand side 18 00:00:53,660 --> 00:00:59,430 there was a set of derivatives acting on the one particle, 19 00:00:59,430 --> 00:01:04,160 probability as a function of position, and momentum, 20 00:01:04,160 --> 00:01:06,440 let's say, for the gas in this room. 21 00:01:06,440 --> 00:01:11,970 And on the right hand side, we had the collision operator. 22 00:01:11,970 --> 00:01:17,180 And again, to get the notation straight, 23 00:01:17,180 --> 00:01:22,050 this set of operations on the left hand side 24 00:01:22,050 --> 00:01:26,775 includes a time derivative, a thing 25 00:01:26,775 --> 00:01:32,900 that involves the velocity causing variations 26 00:01:32,900 --> 00:01:39,420 in the coordinate, and any external forces causing 27 00:01:39,420 --> 00:01:40,585 variations in momentum. 28 00:01:43,870 --> 00:01:51,910 I write that because I will, in order to save some space, 29 00:01:51,910 --> 00:01:56,960 use the following notation in the remainder of this lecture. 30 00:01:56,960 --> 00:02:00,490 I will use del sub t to indicate partial derivatives 31 00:02:00,490 --> 00:02:02,170 with respect to time. 32 00:02:02,170 --> 00:02:07,950 This term I will indicate as p alpha over m d alpha. 33 00:02:07,950 --> 00:02:11,560 So d sub alpha stands for derivative with respect 34 00:02:11,560 --> 00:02:15,450 to, say, x, y, and z-coordinates of position. 35 00:02:15,450 --> 00:02:20,290 And this summation over the repeated index is implied here. 36 00:02:20,290 --> 00:02:25,230 And similarly here, we would write F alpha d by dp alpha. 37 00:02:25,230 --> 00:02:27,630 We won't simplify that. 38 00:02:27,630 --> 00:02:33,100 So basically, L is also this with the summation implicit. 39 00:02:36,030 --> 00:02:38,090 Now, the other entity that we have, 40 00:02:38,090 --> 00:02:40,810 which is the right hand side of the equation, 41 00:02:40,810 --> 00:02:43,530 had something to do with collisions. 42 00:02:43,530 --> 00:02:54,240 And the collision operator Cff was an integral 43 00:02:54,240 --> 00:03:00,240 that involved bringing in a second coordinate, 44 00:03:00,240 --> 00:03:04,830 or second particle, with momentum p2, 45 00:03:04,830 --> 00:03:12,010 would come from a location such that the impact parameter would 46 00:03:12,010 --> 00:03:15,340 be indicated by B. We would need to know 47 00:03:15,340 --> 00:03:17,115 the flux of incoming particles. 48 00:03:17,115 --> 00:03:20,070 So we had the relative velocities. 49 00:03:20,070 --> 00:03:23,940 And then there was a term that was essentially throwing you 50 00:03:23,940 --> 00:03:27,180 off the channel that you were looking at. 51 00:03:27,180 --> 00:03:32,230 And let's say we indicate the variable here rather than say p 52 00:03:32,230 --> 00:03:34,180 by p1. 53 00:03:34,180 --> 00:03:38,780 Then I would have here F evaluated at p1, 54 00:03:38,780 --> 00:03:44,960 F evaluated at p2, p2 being this momentum that I'm integrating. 55 00:03:44,960 --> 00:03:49,770 And then there was the addition from channels 56 00:03:49,770 --> 00:03:53,210 that would bring in probability. 57 00:03:53,210 --> 00:03:55,950 So p1 prime and p2 prime, their collision 58 00:03:55,950 --> 00:03:59,240 would create particles in the channel p1, p2. 59 00:03:59,240 --> 00:04:01,450 And there was some complicated relation 60 00:04:01,450 --> 00:04:03,690 between these functions and these functions 61 00:04:03,690 --> 00:04:06,580 for which you have to solve Newton's equation. 62 00:04:06,580 --> 00:04:09,260 But fortunately, for our purposes, 63 00:04:09,260 --> 00:04:11,980 we don't really need all of that. 64 00:04:11,980 --> 00:04:16,598 Again, all of this is evaluated at the same location as the one 65 00:04:16,598 --> 00:04:17,556 that we have over here. 66 00:04:21,020 --> 00:04:24,370 Now let's do this. 67 00:04:24,370 --> 00:04:33,276 Let's take this, which is a function of some particular q 68 00:04:33,276 --> 00:04:37,020 and some particular p1, which is the things that we 69 00:04:37,020 --> 00:04:40,160 have specified over there. 70 00:04:40,160 --> 00:04:46,230 Let's multiply it with some function that depends on p1, q, 71 00:04:46,230 --> 00:04:52,090 and potentially t, and integrate it over P1. 72 00:04:57,220 --> 00:05:01,870 Once I have done that, then the only thing that I have left 73 00:05:01,870 --> 00:05:05,630 depends on q and t, because everything else 74 00:05:05,630 --> 00:05:07,630 I integrated over. 75 00:05:07,630 --> 00:05:12,550 But in principle, I made a different integration. 76 00:05:12,550 --> 00:05:14,430 I didn't have the integration over q. 77 00:05:14,430 --> 00:05:19,170 So eventually, this thing will become a function of q. 78 00:05:19,170 --> 00:05:21,610 OK, so let's do the same thing on the right hand 79 00:05:21,610 --> 00:05:23,240 side of this equation. 80 00:05:23,240 --> 00:05:29,370 So what I did was I added the integration over p1. 81 00:05:29,370 --> 00:05:32,740 And I multiplied by some function. 82 00:05:32,740 --> 00:05:36,770 And I'll remember that it does depend on q. 83 00:05:36,770 --> 00:05:39,790 But since I haven't written the q argument, 84 00:05:39,790 --> 00:05:40,890 I won't write it here. 85 00:05:40,890 --> 00:05:44,220 So it's chi of p1. 86 00:05:44,220 --> 00:05:47,640 So I want to-- this quantity j that I 87 00:05:47,640 --> 00:05:51,580 wrote is equal to this integral on the right. 88 00:05:54,120 --> 00:05:57,140 Now, we encountered almost the same integral 89 00:05:57,140 --> 00:06:00,960 when we were looking for the proof of the H-theorem 90 00:06:00,960 --> 00:06:03,200 where the analog of this chi of p1 91 00:06:03,200 --> 00:06:07,470 was in fact log of f evaluated at p1. 92 00:06:07,470 --> 00:06:11,200 And we did some manipulations that we can do over here. 93 00:06:11,200 --> 00:06:14,210 First of all, here, we have dummy integration variables 94 00:06:14,210 --> 00:06:15,790 p1 and p2. 95 00:06:15,790 --> 00:06:22,668 We can just change their name and then essentially average 96 00:06:22,668 --> 00:06:23,876 over those two possibilities. 97 00:06:27,340 --> 00:06:30,510 And the other thing that we did was this equation 98 00:06:30,510 --> 00:06:34,000 has the set of things that come into the collision, 99 00:06:34,000 --> 00:06:36,760 and set the things that, in some sense, 100 00:06:36,760 --> 00:06:39,860 got out of the collision, or basically things that, 101 00:06:39,860 --> 00:06:45,210 as a result of these collisions, create these two. 102 00:06:45,210 --> 00:06:49,400 So you have this symmetry between the initiators 103 00:06:49,400 --> 00:06:51,400 and products of the collision. 104 00:06:51,400 --> 00:06:53,440 Because essentially the same function 105 00:06:53,440 --> 00:06:57,100 describes going one way and inverting 106 00:06:57,100 --> 00:06:59,800 things and going backwards. 107 00:06:59,800 --> 00:07:01,840 And we said that in principle, I could 108 00:07:01,840 --> 00:07:04,890 change variables of integration. 109 00:07:04,890 --> 00:07:08,590 And the effect of doing that is kind of moving the prime 110 00:07:08,590 --> 00:07:12,160 coordinates to the things that don't have primes. 111 00:07:12,160 --> 00:07:16,200 I don't know how last time I made the mistake of the sign. 112 00:07:16,200 --> 00:07:19,470 But it's clear that if I just put the primes from here 113 00:07:19,470 --> 00:07:22,960 to here, there will be a minus sign. 114 00:07:22,960 --> 00:07:26,300 So the result of doing that symmetrization 115 00:07:26,300 --> 00:07:32,190 should be a minus chi of p1 prime minus chi of p2 prime. 116 00:07:32,190 --> 00:07:35,935 And again, to do the averaging, I have to put something else. 117 00:07:41,170 --> 00:07:45,460 Now, this statement is quite generally true. 118 00:07:45,460 --> 00:07:49,310 Whatever chi I choose, I will have this value of j 119 00:07:49,310 --> 00:07:52,930 as a result of that integration. 120 00:07:52,930 --> 00:07:56,660 But now we are going to look at something specific. 121 00:07:56,660 --> 00:07:59,490 Let's assume that we have a quantity that 122 00:07:59,490 --> 00:08:02,340 is conserved in collision. 123 00:08:02,340 --> 00:08:06,315 This will be 0 for collision conserved quantity. 124 00:08:15,920 --> 00:08:19,620 Like let's say if my chi that I had chosen here 125 00:08:19,620 --> 00:08:23,030 was some component of momentum px, 126 00:08:23,030 --> 00:08:26,760 then whatever some of the incoming momenta 127 00:08:26,760 --> 00:08:29,060 will be the sum of the outgoing momenta. 128 00:08:29,060 --> 00:08:32,450 So essentially anything that I can 129 00:08:32,450 --> 00:08:36,210 think of that is conserved in the collision, this function 130 00:08:36,210 --> 00:08:39,450 that relates p primes to p1 and p2 131 00:08:39,450 --> 00:08:43,712 has the right property to ensure that this whole thing will 132 00:08:43,712 --> 00:08:45,280 be 0. 133 00:08:45,280 --> 00:08:48,350 And that's actually really the ultimate reason. 134 00:08:48,350 --> 00:08:50,560 I don't really need to know about 135 00:08:50,560 --> 00:08:53,420 all of these cross sections and all of the collision 136 00:08:53,420 --> 00:08:54,500 properties, et cetera. 137 00:08:54,500 --> 00:08:57,220 Because my focus will be on things 138 00:08:57,220 --> 00:08:59,350 that are conserved in collisions. 139 00:08:59,350 --> 00:09:03,530 Because those are the variables that are very slowly relaxing, 140 00:09:03,530 --> 00:09:06,040 and the things that I'm interested in. 141 00:09:06,040 --> 00:09:08,930 So what you have is that for these collision conserved 142 00:09:08,930 --> 00:09:15,000 quantities, which is the things that I'm interested in, 143 00:09:15,000 --> 00:09:16,357 this equation is 0. 144 00:09:20,480 --> 00:09:24,660 Now, if f satisfies that equation, 145 00:09:24,660 --> 00:09:31,702 I can certainly substitute over here for Cff Lf. 146 00:09:36,340 --> 00:09:40,930 So if f satisfies that equation, and I 147 00:09:40,930 --> 00:09:43,710 pick a collision conserved quantity, 148 00:09:43,710 --> 00:09:47,160 the integral over p1 of that function of the collision 149 00:09:47,160 --> 00:09:50,700 conserved quantity times the bunch 150 00:09:50,700 --> 00:09:53,690 of first derivatives acting on f has to be 0. 151 00:09:56,890 --> 00:09:59,140 So this I can write in the following way-- 152 00:09:59,140 --> 00:10:03,330 0 is the integral over p1. 153 00:10:03,330 --> 00:10:05,220 Actually, I have only one momentum. 154 00:10:05,220 --> 00:10:08,080 So let's just ignore it from henceforth. 155 00:10:08,080 --> 00:10:10,800 The other momentum I just introduced in order 156 00:10:10,800 --> 00:10:13,830 to be able to show that when integrated against 157 00:10:13,830 --> 00:10:17,180 the collision operator, it will give me 0. 158 00:10:17,180 --> 00:10:23,720 I have the chi, and then this bunch of derivatives dt plus p 159 00:10:23,720 --> 00:10:30,280 alpha over m p alpha plus f alpha d by dp alpha 160 00:10:30,280 --> 00:10:33,346 acting on f. 161 00:10:33,346 --> 00:10:34,995 And that has to be 0. 162 00:10:34,995 --> 00:10:37,370 STUDENT: So alpha stands for x, y, z? 163 00:10:37,370 --> 00:10:41,360 PROFESSOR: Yes, alpha stands for the three components 164 00:10:41,360 --> 00:10:45,120 x, y, and z throughout this lecture. 165 00:10:45,120 --> 00:10:49,008 And summation over a repeated index is assumed. 166 00:10:52,920 --> 00:10:56,250 All right, so now what I want to do 167 00:10:56,250 --> 00:11:00,970 is to move this chi so that the derivatives act 168 00:11:00,970 --> 00:11:02,826 on both of them. 169 00:11:02,826 --> 00:11:06,460 So I'll simply write the integral of dqp-- 170 00:11:06,460 --> 00:11:09,040 if you like, this bunch of derivatives 171 00:11:09,040 --> 00:11:15,160 that we call L acting on the combination chi f. 172 00:11:15,160 --> 00:11:19,440 But a derivative of f chi gives me f prime chi. 173 00:11:19,440 --> 00:11:23,870 But also it keeps me chi prime f that I don't have here. 174 00:11:23,870 --> 00:11:25,550 So I have to subtract that. 175 00:11:33,480 --> 00:11:36,170 And why did I do that? 176 00:11:36,170 --> 00:11:39,950 Because now I end up with integrals 177 00:11:39,950 --> 00:11:43,845 that involve integrating over f against something. 178 00:11:47,050 --> 00:11:50,520 So let's think about these typical integrals. 179 00:11:50,520 --> 00:11:58,100 If I take the integral over momentum of f of p and q 180 00:11:58,100 --> 00:12:03,320 and t-- remember, f was the one particle density. 181 00:12:03,320 --> 00:12:06,640 So I'm integrating, let's say, at a particular position 182 00:12:06,640 --> 00:12:10,510 in space over all momentum. 183 00:12:10,510 --> 00:12:12,870 So it says, I don't care what momentum. 184 00:12:12,870 --> 00:12:15,930 I just want to know whether there's a particle here. 185 00:12:15,930 --> 00:12:17,350 So what is that quantity? 186 00:12:17,350 --> 00:12:20,520 That quantity is simply the density at that location. 187 00:12:26,364 --> 00:12:28,990 Now suppose I were to integrate this 188 00:12:28,990 --> 00:12:31,830 against some other function, which 189 00:12:31,830 --> 00:12:34,685 could depend on p, q, and t, for example? 190 00:12:38,030 --> 00:12:43,440 I use that to define an average. 191 00:12:43,440 --> 00:12:48,480 So this is going to be defined to be the average of O 192 00:12:48,480 --> 00:12:50,870 at that location q and t. 193 00:12:50,870 --> 00:12:54,700 So for example, rather than just calculating whether or not 194 00:12:54,700 --> 00:12:56,620 there are particles here, I could 195 00:12:56,620 --> 00:12:59,500 be asking, what is the average kinetic energy 196 00:12:59,500 --> 00:13:01,330 of the particles that are here? 197 00:13:01,330 --> 00:13:04,760 Then I would integrate this against p squared over 2m. 198 00:13:04,760 --> 00:13:09,750 And this average would give me the local expectation value 199 00:13:09,750 --> 00:13:13,020 of p squared over m, just a normalization n 200 00:13:13,020 --> 00:13:15,650 so that it's appropriately normalized. 201 00:13:19,150 --> 00:13:22,960 So with this definition, I can write the various terms 202 00:13:22,960 --> 00:13:24,120 that I have over here. 203 00:13:24,120 --> 00:13:27,540 So let me write it a little bit more explicitly. 204 00:13:27,540 --> 00:13:28,636 What do we have? 205 00:13:28,636 --> 00:13:32,780 We have integral d cubed p. 206 00:13:32,780 --> 00:13:44,780 We have this bunch of derivatives acting on chi f 207 00:13:44,780 --> 00:13:54,840 minus f times this bunch of derivatives acting on chi. 208 00:13:58,610 --> 00:14:03,020 So let's now look at things term by term. 209 00:14:03,020 --> 00:14:05,220 The first term is a time derivative. 210 00:14:05,220 --> 00:14:09,470 The time derivative I can take outside the integral. 211 00:14:09,470 --> 00:14:13,090 Once I take the time derivative outside the integral, 212 00:14:13,090 --> 00:14:14,580 what is left? 213 00:14:14,580 --> 00:14:19,060 What is left is the integral of chi f, 214 00:14:19,060 --> 00:14:21,230 exactly what I have here. 215 00:14:21,230 --> 00:14:23,590 O has been replaced by chi. 216 00:14:23,590 --> 00:14:26,780 So what I have is the time derivative 217 00:14:26,780 --> 00:14:33,190 of n expectation value of chi using this definition. 218 00:14:33,190 --> 00:14:35,620 Let's look at the next term. 219 00:14:35,620 --> 00:14:38,180 The next term, these derivatives are all over position. 220 00:14:38,180 --> 00:14:40,350 The integration is over momentum. 221 00:14:40,350 --> 00:14:42,460 I can take it outside. 222 00:14:42,460 --> 00:14:45,390 So I can write it as d alpha. 223 00:14:45,390 --> 00:14:50,150 And then I have a quantity that I'm integrating against f. 224 00:14:50,150 --> 00:14:55,220 So I will get n times this local average of that quantity. 225 00:14:55,220 --> 00:14:56,520 What's that quantity? 226 00:14:56,520 --> 00:15:00,170 It's p alpha over m times chi. 227 00:15:03,850 --> 00:15:06,060 What's the third term? 228 00:15:06,060 --> 00:15:09,860 The third term is actually an integral over momentum. 229 00:15:09,860 --> 00:15:12,770 But I'm integrating over momentum. 230 00:15:12,770 --> 00:15:16,430 So again, you can sort of remove things to boundaries 231 00:15:16,430 --> 00:15:18,920 and convince yourself that that integral will not 232 00:15:18,920 --> 00:15:20,090 give you a contribution. 233 00:15:22,910 --> 00:15:26,910 The next bunch of terms are simply directly this-- f 234 00:15:26,910 --> 00:15:29,170 integrated against something. 235 00:15:29,170 --> 00:15:31,700 So they're going to give me minus n 236 00:15:31,700 --> 00:15:36,570 times the various averages involved-- d t chi minus n, 237 00:15:36,570 --> 00:15:41,250 the average of p alpha over m, the alpha chi. 238 00:15:41,250 --> 00:15:47,340 And then f alpha I can actually take outside, minus n f alpha, 239 00:15:47,340 --> 00:15:51,330 the average of d chi by d p alpha. 240 00:15:51,330 --> 00:15:54,670 And what we've established is that that whole thing 241 00:15:54,670 --> 00:15:58,939 is 0 for quantities that are conserved under collisions. 242 00:16:05,510 --> 00:16:09,100 So why did I do all of that? 243 00:16:09,100 --> 00:16:13,220 It's because solving the Boltzmann equation 244 00:16:13,220 --> 00:16:17,560 in six dimensional phase space with all of its integrations 245 00:16:17,560 --> 00:16:21,490 and derivatives is very complicated. 246 00:16:21,490 --> 00:16:25,460 But really, the things that are slowly relaxing 247 00:16:25,460 --> 00:16:27,610 are quantities that are conserved 248 00:16:27,610 --> 00:16:32,480 collisions, such as densities, average momentum, et cetera. 249 00:16:32,480 --> 00:16:36,880 And so I can focus on variations of these 250 00:16:36,880 --> 00:16:40,670 through this kind of equation. 251 00:16:40,670 --> 00:16:44,160 Essentially, what that will allow me to do 252 00:16:44,160 --> 00:16:51,750 is to construct what are known as hydrodynamic equations, 253 00:16:51,750 --> 00:16:59,390 which describe the time evolution of slow variables 254 00:16:59,390 --> 00:17:01,710 of your system, the variables that 255 00:17:01,710 --> 00:17:05,690 are kind of relevant to making thermodynamic observations, 256 00:17:05,690 --> 00:17:07,400 as opposed to variables that you would 257 00:17:07,400 --> 00:17:08,910 be interested in if you're thinking 258 00:17:08,910 --> 00:17:09,924 about atomic collisions. 259 00:17:12,780 --> 00:17:16,849 So what I need to do is to go into that equation 260 00:17:16,849 --> 00:17:21,569 and pick out my conserved quantities. 261 00:17:21,569 --> 00:17:31,220 So what are the conserved quantities, 262 00:17:31,220 --> 00:17:36,480 and how can I describe them by some chi? 263 00:17:36,480 --> 00:17:40,950 Well, we already saw this when we were earlier on trying 264 00:17:40,950 --> 00:17:45,220 to find some kind of a solution to the H by dt equals 0. 265 00:17:45,220 --> 00:17:47,615 We said that log f has to be the sum 266 00:17:47,615 --> 00:17:49,780 of collision conserved quantities. 267 00:17:49,780 --> 00:17:53,190 And we identified three types of quantities. 268 00:17:53,190 --> 00:17:56,830 One of them was related to the number conservation. 269 00:17:56,830 --> 00:17:59,610 And essentially, what you have is that 1 plus 1 270 00:17:59,610 --> 00:18:01,060 equals to 1 plus 1. 271 00:18:01,060 --> 00:18:02,750 So it's obvious. 272 00:18:02,750 --> 00:18:04,345 The other is momentum. 273 00:18:07,520 --> 00:18:12,670 And there are three components of this-- px, py, pz. 274 00:18:12,670 --> 00:18:15,892 And the third one is the kinetic energy, 275 00:18:15,892 --> 00:18:18,320 which is conserved in collisions. 276 00:18:18,320 --> 00:18:21,480 In a potential, clearly the kinetic energy of a particle 277 00:18:21,480 --> 00:18:24,140 changes as a function of position. 278 00:18:24,140 --> 00:18:28,040 But within the short distances of the collisions 279 00:18:28,040 --> 00:18:30,330 that we are interested in, the kinetic energy 280 00:18:30,330 --> 00:18:31,330 is a conserved quantity. 281 00:18:35,390 --> 00:18:41,360 So my task is to insert these values of chi 282 00:18:41,360 --> 00:18:46,840 into that equation and see what information they tell me 283 00:18:46,840 --> 00:18:50,690 about the time evolution of the corresponding conserved 284 00:18:50,690 --> 00:18:52,460 quantities. 285 00:18:52,460 --> 00:18:56,820 So let's do this one by one. 286 00:18:56,820 --> 00:19:01,030 Let's start with chi equals to 1. 287 00:19:01,030 --> 00:19:03,690 If I put chi equals to 1, all of these terms 288 00:19:03,690 --> 00:19:07,991 that involve derivatives clearly needed to vanish. 289 00:19:07,991 --> 00:19:13,700 And here, I would get the time derivative of the density. 290 00:19:13,700 --> 00:19:21,330 And from here, I would get the alpha n expectation value of p 291 00:19:21,330 --> 00:19:23,300 alpha over m. 292 00:19:23,300 --> 00:19:24,560 We'll give that a name. 293 00:19:24,560 --> 00:19:27,310 We'll call that u alpha. 294 00:19:27,310 --> 00:19:33,170 So I have introduced u alpha to be the expectation 295 00:19:33,170 --> 00:19:36,640 value of p alpha over m. 296 00:19:36,640 --> 00:19:40,940 And it can in principle depend on which location in space 297 00:19:40,940 --> 00:19:42,090 you are looking at. 298 00:19:42,090 --> 00:19:44,030 Somebody opens the door, there will 299 00:19:44,030 --> 00:19:45,445 be a current that is established. 300 00:19:45,445 --> 00:19:48,110 And so there will be a local velocity 301 00:19:48,110 --> 00:19:50,520 of the air that would be different from other places 302 00:19:50,520 --> 00:19:52,260 in the room. 303 00:19:52,260 --> 00:19:53,920 And that's all we have. 304 00:19:53,920 --> 00:19:56,710 And this is equal to 0. 305 00:19:56,710 --> 00:19:58,640 And this is of course the equation 306 00:19:58,640 --> 00:20:00,860 of the continuity of the number of particles. 307 00:20:03,800 --> 00:20:07,680 You don't create or destroy particles. 308 00:20:07,680 --> 00:20:10,340 And so this density has to satisfy 309 00:20:10,340 --> 00:20:13,800 this nice, simple equation. 310 00:20:13,800 --> 00:20:17,780 We will sometimes rewrite this slightly in the following way. 311 00:20:17,780 --> 00:20:20,060 This is the derivative of two objects. 312 00:20:20,060 --> 00:20:25,720 I can expand that and write it as dt of n plus, let's say, 313 00:20:25,720 --> 00:20:29,405 u alpha d alpha of n. 314 00:20:29,405 --> 00:20:31,120 And then I would have a term that 315 00:20:31,120 --> 00:20:34,140 is n d alpha u alpha, which I will 316 00:20:34,140 --> 00:20:35,936 take to the other side of the equation. 317 00:20:43,460 --> 00:20:46,940 Why have I done that? 318 00:20:46,940 --> 00:20:52,760 Because if I think of n as a function of position and time-- 319 00:20:52,760 --> 00:20:58,300 and as usual, we did before define a derivative 320 00:20:58,300 --> 00:21:01,670 that moves along this streamline-- 321 00:21:01,670 --> 00:21:05,560 you will have both the implicit time derivative and the time 322 00:21:05,560 --> 00:21:08,820 derivative because the stream changes position 323 00:21:08,820 --> 00:21:12,950 by an amount that is related to velocity. 324 00:21:12,950 --> 00:21:17,550 Now, for the Liouville equation, we have something like this, 325 00:21:17,550 --> 00:21:20,040 except that the Liouville equation, the right hand side 326 00:21:20,040 --> 00:21:20,590 was 0. 327 00:21:20,590 --> 00:21:23,990 Because the flows of the Liouville equation, 328 00:21:23,990 --> 00:21:27,620 the Hamiltonian flows, were divergenceless. 329 00:21:27,620 --> 00:21:29,990 But in general, for a compressable system, 330 00:21:29,990 --> 00:21:34,230 such as the gas in this room, the compressibility 331 00:21:34,230 --> 00:21:37,431 is indicated to a nonzero divergence of u. 332 00:21:37,431 --> 00:21:39,764 And there's a corresponding term on the right hand side. 333 00:21:44,322 --> 00:21:48,280 So that's the first thing we can do. 334 00:21:48,280 --> 00:21:52,370 What's the second thing we can do? 335 00:21:52,370 --> 00:21:59,840 I can pick p-- let's say p beta, what I wrote over there. 336 00:21:59,840 --> 00:22:02,510 But I can actually scale it. 337 00:22:02,510 --> 00:22:05,200 If p is a conserved quantity, p/m 338 00:22:05,200 --> 00:22:06,575 is also a conserved quantity. 339 00:22:09,450 --> 00:22:12,010 Actually, as far as this chi is concerned, 340 00:22:12,010 --> 00:22:15,700 I can add anything that depends on q and not p. 341 00:22:15,700 --> 00:22:19,135 So I can subtract the average value of this quantity. 342 00:22:24,950 --> 00:22:27,340 And this is conserved during collisions. 343 00:22:27,340 --> 00:22:31,060 Because this part is the same thing as something 344 00:22:31,060 --> 00:22:32,690 that is related to density. 345 00:22:32,690 --> 00:22:35,490 It's like 1 plus 1 equals to 1 plus 1. 346 00:22:35,490 --> 00:22:38,900 And p over beta is conserved also. 347 00:22:38,900 --> 00:22:42,150 So we'll call this quantity that we 348 00:22:42,150 --> 00:22:47,220 will use for our candidate chi as c beta. 349 00:22:47,220 --> 00:22:54,020 So essentially, it's the additional fluctuating speed, 350 00:22:54,020 --> 00:22:56,790 velocity that the particles have, 351 00:22:56,790 --> 00:23:02,060 on top of the average velocity that is due to the flow. 352 00:23:02,060 --> 00:23:04,790 And the reason maybe it's useful to do 353 00:23:04,790 --> 00:23:08,670 this is because clearly the average of c is 0. 354 00:23:08,670 --> 00:23:12,040 Because the average of p beta over m is u beta. 355 00:23:12,040 --> 00:23:14,990 And if I do that, then clearly at least the first thing 356 00:23:14,990 --> 00:23:17,070 in the equation I don't have to worry about. 357 00:23:17,070 --> 00:23:19,510 I have removed one term in the equation. 358 00:23:22,810 --> 00:23:26,870 So let's put c beta for chi over here. 359 00:23:26,870 --> 00:23:28,580 We said that the first term is 0. 360 00:23:28,580 --> 00:23:31,170 So we go and start with the second term. 361 00:23:31,170 --> 00:23:31,812 What do I have? 362 00:23:31,812 --> 00:23:37,110 I have d alpha expectation value of c beta. 363 00:23:37,110 --> 00:23:39,970 And then I have p alpha over m. 364 00:23:39,970 --> 00:23:44,710 Well, p alpha over m is going to be u alpha plus c alpha. 365 00:23:44,710 --> 00:23:49,165 So let's write it in this fashion-- u alpha plus c 366 00:23:49,165 --> 00:23:52,980 alpha for p alpha over m. 367 00:23:52,980 --> 00:23:54,920 And that's the average I have to take. 368 00:23:59,980 --> 00:24:03,450 Now let's look at all of these terms that involve derivatives. 369 00:24:06,170 --> 00:24:09,730 Well, if I want to take a time derivative of this quantity, 370 00:24:09,730 --> 00:24:13,020 now that I have introduced this average, 371 00:24:13,020 --> 00:24:16,120 there is a time derivative here. 372 00:24:16,120 --> 00:24:20,130 So the average of the time derivative of chi 373 00:24:20,130 --> 00:24:23,600 will give me the time derivative from u beta. 374 00:24:23,600 --> 00:24:26,160 And actually the minus sign will cancel. 375 00:24:26,160 --> 00:24:31,210 And so I will have plus n, the expectation value of d-- well, 376 00:24:31,210 --> 00:24:37,190 there's no expectation value of something like this. 377 00:24:37,190 --> 00:24:39,690 When I integrate over p, there's no p dependence. 378 00:24:39,690 --> 00:24:41,230 So it's just itself. 379 00:24:41,230 --> 00:24:45,085 So it is n dt of u beta. 380 00:24:49,320 --> 00:24:51,560 OK, what do we have for the next term? 381 00:24:51,560 --> 00:24:53,290 Let's write it explicitly. 382 00:24:53,290 --> 00:24:56,690 I have n. 383 00:24:56,690 --> 00:25:03,020 p alpha over m I'm writing as u alpha plus c alpha. 384 00:25:03,020 --> 00:25:08,530 And then I have the position derivative of c beta. 385 00:25:08,530 --> 00:25:10,880 And that goes over here. 386 00:25:10,880 --> 00:25:16,560 So I will get d alpha of u beta with a minus sign. 387 00:25:16,560 --> 00:25:17,830 So this becomes a plus. 388 00:25:23,260 --> 00:25:29,880 The last term is minus n f alpha. 389 00:25:29,880 --> 00:25:33,210 And I have to take a derivative of this object with respect 390 00:25:33,210 --> 00:25:35,090 to p alpha. 391 00:25:35,090 --> 00:25:37,310 Well, I have a p here. 392 00:25:37,310 --> 00:25:40,140 The derivative of p beta with respect to p alpha 393 00:25:40,140 --> 00:25:42,690 gives me delta alpha beta. 394 00:25:42,690 --> 00:25:47,810 So this is going to give me delta alpha beta over m. 395 00:25:47,810 --> 00:25:49,700 And the whole thing is 0. 396 00:25:54,600 --> 00:25:59,710 So let's rearrange the terms over here. 397 00:25:59,710 --> 00:26:04,130 The only thing that I have in the denominator is a 1/m. 398 00:26:04,130 --> 00:26:08,080 So let me multiply the whole equation by m 399 00:26:08,080 --> 00:26:10,920 and see what happens. 400 00:26:10,920 --> 00:26:12,930 This term let's deal with last. 401 00:26:12,930 --> 00:26:20,685 This term, the first term, becomes nmdt of u beta. 402 00:26:23,310 --> 00:26:29,250 The change in velocity kind of looks like an acceleration. 403 00:26:29,250 --> 00:26:31,540 But you have to be careful. 404 00:26:31,540 --> 00:26:35,490 Because you can only talk about acceleration 405 00:26:35,490 --> 00:26:37,360 acting for a particle. 406 00:26:37,360 --> 00:26:40,990 And the particle is moving with the stream. 407 00:26:40,990 --> 00:26:45,020 OK, and this term will give me the appropriate derivative 408 00:26:45,020 --> 00:26:48,820 to make it a stream velocity. 409 00:26:48,820 --> 00:26:52,930 Now, when I look at this, the average of c that appears here 410 00:26:52,930 --> 00:26:56,030 will be 0. 411 00:26:56,030 --> 00:26:58,110 So the term that I have over there 412 00:26:58,110 --> 00:27:00,070 is u alpha d alpha u beta. 413 00:27:00,070 --> 00:27:01,850 There's no average involved. 414 00:27:01,850 --> 00:27:04,231 It will give me n. 415 00:27:04,231 --> 00:27:11,960 m is common, so I will get u alpha d alpha u beta, 416 00:27:11,960 --> 00:27:12,730 which is nice. 417 00:27:12,730 --> 00:27:16,770 Because then I can certainly regard this 418 00:27:16,770 --> 00:27:19,720 as one of these stream derivatives. 419 00:27:25,100 --> 00:27:28,230 So these two terms, the stream derivative 420 00:27:28,230 --> 00:27:34,460 of velocity with time, times mass, mass times 421 00:27:34,460 --> 00:27:37,720 the density to make it mass per unit volume, 422 00:27:37,720 --> 00:27:39,200 looks like an acceleration. 423 00:27:39,200 --> 00:27:42,800 So it's like mass times acceleration. 424 00:27:42,800 --> 00:27:45,515 Newton's law, it should be equal to the force. 425 00:27:45,515 --> 00:27:47,400 And what do we have if we take this 426 00:27:47,400 --> 00:27:48,940 to the other side of the equation? 427 00:27:48,940 --> 00:27:52,040 We have f beta. 428 00:27:52,040 --> 00:27:58,620 OK, good, so we have reproduced Newton's equation. 429 00:27:58,620 --> 00:28:03,090 In this context, if we're moving along with the stream, 430 00:28:03,090 --> 00:28:07,450 mass times the acceleration of the group of particles moving 431 00:28:07,450 --> 00:28:09,540 with the stream is the force that 432 00:28:09,540 --> 00:28:13,777 is felt from the external potential. 433 00:28:13,777 --> 00:28:15,110 But there's one other term here. 434 00:28:17,830 --> 00:28:20,190 In this term, the term that is uc 435 00:28:20,190 --> 00:28:24,780 will average to 0, because the average of u is 0. 436 00:28:24,780 --> 00:28:28,620 So what I will have is minus-- and I 437 00:28:28,620 --> 00:28:31,650 forgot to write down an n somewhere here. 438 00:28:31,650 --> 00:28:32,658 There will be an n. 439 00:28:36,860 --> 00:28:40,800 Because all averages had this additional n 440 00:28:40,800 --> 00:28:43,550 that I forgot to put. 441 00:28:43,550 --> 00:28:46,370 I will take it to the right hand side of the equation. 442 00:28:46,370 --> 00:28:52,620 And it becomes d by dq alpha of n. 443 00:28:52,620 --> 00:28:56,610 I multiply the entire equation by m. 444 00:28:56,610 --> 00:28:59,787 And then I have the average of c alpha c beta. 445 00:29:06,700 --> 00:29:09,160 So what happened here? 446 00:29:09,160 --> 00:29:14,080 Isn't force just the mass times acceleration? 447 00:29:14,080 --> 00:29:18,700 Well, as long as you include all forces involved. 448 00:29:18,700 --> 00:29:27,240 So if you imagine that this room is totally stationary air, 449 00:29:27,240 --> 00:29:33,240 and I heat one corner here, then the particles 450 00:29:33,240 --> 00:29:36,310 here will start to move more rapidly. 451 00:29:36,310 --> 00:29:38,120 There will be more pressure here. 452 00:29:38,120 --> 00:29:40,680 Because pressure is proportional to temperature, if you like. 453 00:29:40,680 --> 00:29:43,110 There will be more pressure, less pressure here. 454 00:29:43,110 --> 00:29:45,370 The difference in pressure will drive the flow. 455 00:29:45,370 --> 00:29:47,920 There will be an additional force. 456 00:29:47,920 --> 00:29:49,610 And that's what it says. 457 00:29:49,610 --> 00:29:54,800 If there's variation in these speeds of the particles, 458 00:29:54,800 --> 00:29:58,830 the change in pressure will give you a force. 459 00:29:58,830 --> 00:30:04,990 And so this thing, p alpha beta, is called the pressure tensor. 460 00:30:12,580 --> 00:30:15,140 Yes. 461 00:30:15,140 --> 00:30:19,000 STUDENT: Shouldn't f beta be multiplied by n, 462 00:30:19,000 --> 00:30:22,260 or is there an n on the other side of that? 463 00:30:22,260 --> 00:30:24,670 PROFESSOR: There is an n here that I forgot, yes. 464 00:30:27,980 --> 00:30:31,295 So the n was in the first equation, somehow got lost. 465 00:30:34,160 --> 00:30:37,160 STUDENT: So the pressure is coming 466 00:30:37,160 --> 00:30:39,160 from the local fluctuation? 467 00:30:39,160 --> 00:30:40,302 PROFESSOR: Yes. 468 00:30:40,302 --> 00:30:42,310 And if you think about it, the temperature 469 00:30:42,310 --> 00:30:44,372 is also the local fluctuation. 470 00:30:44,372 --> 00:30:48,090 So it has something to do with temperature differences. 471 00:30:48,090 --> 00:30:50,120 Pressure is related to temperature. 472 00:30:50,120 --> 00:30:52,950 So all the things are connected. 473 00:30:52,950 --> 00:30:56,670 And in about two minutes, I'll actually evaluate that for you, 474 00:30:56,670 --> 00:30:59,120 and you'll see how. 475 00:30:59,120 --> 00:30:59,870 Yes. 476 00:30:59,870 --> 00:31:02,684 STUDENT: Is the pressure tensor distinct from the stress 477 00:31:02,684 --> 00:31:05,030 tensor? 478 00:31:05,030 --> 00:31:07,800 PROFESSOR: It's the stress tensor 479 00:31:07,800 --> 00:31:10,880 that you would have for a fluid. 480 00:31:10,880 --> 00:31:13,920 For something more complicated, like an elastic material, 481 00:31:13,920 --> 00:31:18,061 it would be much more complicated-- not much more 482 00:31:18,061 --> 00:31:19,477 complicated, but more complicated. 483 00:31:22,890 --> 00:31:25,020 Essentially, there's always some kind 484 00:31:25,020 --> 00:31:27,450 of a force per unit volume depending 485 00:31:27,450 --> 00:31:30,020 on what kind of medium you have. 486 00:31:30,020 --> 00:31:31,611 And for the gas, this is what it is. 487 00:31:31,611 --> 00:31:32,110 Yes. 488 00:31:32,110 --> 00:31:34,054 STUDENT: So a basic question. 489 00:31:34,054 --> 00:31:39,886 When we say u alpha is averaged, averaged over what? 490 00:31:39,886 --> 00:31:41,360 Is it by the area? 491 00:31:41,360 --> 00:31:43,740 PROFESSOR: OK, this is the definition. 492 00:31:43,740 --> 00:31:47,940 So whenever I use this notation with these angles, 493 00:31:47,940 --> 00:31:50,920 it means that I integrated over p. 494 00:31:50,920 --> 00:31:52,400 Why do I do that? 495 00:31:52,400 --> 00:31:57,020 Because of this asymmetry between momenta and collision 496 00:31:57,020 --> 00:32:01,400 and coordinates that is inherent to the Boltzmann equation. 497 00:32:01,400 --> 00:32:03,880 When we wrote down the Liouville equation, 498 00:32:03,880 --> 00:32:06,870 p and q were completely equivalent. 499 00:32:06,870 --> 00:32:09,390 But by the time we made our approximations 500 00:32:09,390 --> 00:32:11,970 and we talked about collisions, et cetera, 501 00:32:11,970 --> 00:32:15,730 we saw that momenta quickly relax. 502 00:32:15,730 --> 00:32:18,725 And so we can look at the particular position 503 00:32:18,725 --> 00:32:22,320 and integrate over momenta and define averages 504 00:32:22,320 --> 00:32:25,440 in the sense of when you think about what's happening 505 00:32:25,440 --> 00:32:29,240 in this room, you think about the wind velocity 506 00:32:29,240 --> 00:32:33,530 here over there, but not fluctuations in the momentum 507 00:32:33,530 --> 00:32:38,140 so much, OK? 508 00:32:41,500 --> 00:32:46,530 All right, so this clearly is kind 509 00:32:46,530 --> 00:32:52,430 of a Navier-Stokes like equation, if you like, 510 00:32:52,430 --> 00:32:55,795 for this gas. 511 00:32:58,970 --> 00:33:02,930 That tells you how the velocity of this fluid changes. 512 00:33:05,690 --> 00:33:13,600 And finally, we would need to construct an equation that 513 00:33:13,600 --> 00:33:16,520 is relevant to the kinetic energy, which 514 00:33:16,520 --> 00:33:20,910 is something like p squared over 2m. 515 00:33:20,910 --> 00:33:25,960 And we can follow what we did over here 516 00:33:25,960 --> 00:33:28,590 and subtract the average. 517 00:33:34,320 --> 00:33:36,780 And so essentially, this is kinetic energy 518 00:33:36,780 --> 00:33:43,210 on top of the kinetic energy of the entire stream. 519 00:33:43,210 --> 00:33:45,460 This is clearly the same thing as mc 520 00:33:45,460 --> 00:33:49,400 squared over 2, c being the quantity that we 521 00:33:49,400 --> 00:33:51,720 defined before. 522 00:33:51,720 --> 00:34:03,640 And the average of mc squared over 2 523 00:34:03,640 --> 00:34:06,030 I will indicate by epsilon. 524 00:34:06,030 --> 00:34:07,130 It's the heat content. 525 00:34:11,350 --> 00:34:13,280 Or actually, let's say energy density. 526 00:34:13,280 --> 00:34:14,425 It's probably better. 527 00:34:24,350 --> 00:34:31,210 So now I have to put mc squared over 2 in this equation for chi 528 00:34:31,210 --> 00:34:34,719 and do various manipulations along the lines of things 529 00:34:34,719 --> 00:34:36,520 that I did before. 530 00:34:36,520 --> 00:34:39,330 I will just write down the final answer. 531 00:34:39,330 --> 00:34:48,679 So the final answer will be that dt of epsilon. 532 00:34:48,679 --> 00:34:50,489 We've defined dt. 533 00:34:50,489 --> 00:34:52,179 I move with the streamline. 534 00:34:52,179 --> 00:34:56,380 So I have dt plus u alpha d alpha 535 00:34:56,380 --> 00:35:02,650 acting on this density, which is a function of position 536 00:35:02,650 --> 00:35:04,790 and time. 537 00:35:04,790 --> 00:35:10,280 And the right hand side of this will have two terms. 538 00:35:10,280 --> 00:35:16,910 One term is essentially how this pressure kind 539 00:35:16,910 --> 00:35:25,080 of moves against the velocity, or the velocity and pressure 540 00:35:25,080 --> 00:35:27,940 are kind of hitting against each other. 541 00:35:27,940 --> 00:35:31,980 So it's kind of like if you were to rub two things-- 542 00:35:31,980 --> 00:35:33,550 "rub" was the word I was looking. 543 00:35:33,550 --> 00:35:35,810 If you were to rub two things against each other, 544 00:35:35,810 --> 00:35:37,930 there's heat that is generated. 545 00:35:37,930 --> 00:35:40,710 And so that's the term that we are looking at. 546 00:35:40,710 --> 00:35:43,290 So what is this u alpha beta? 547 00:35:43,290 --> 00:35:47,180 u alpha beta-- it's just because p alpha beta 548 00:35:47,180 --> 00:35:48,730 is a symmetric object. 549 00:35:48,730 --> 00:35:53,900 It doesn't make any difference if you exchange alpha and beta. 550 00:35:53,900 --> 00:35:56,770 You symmetrize the derivative of the velocity. 551 00:36:03,360 --> 00:36:06,020 And sometimes it's called the rate of strain. 552 00:36:13,480 --> 00:36:18,950 And there's another term, which is 553 00:36:18,950 --> 00:36:25,420 minus 1/n d alpha of h alpha. 554 00:36:25,420 --> 00:36:30,880 And for that, I need to define yet another quantity, this h 555 00:36:30,880 --> 00:36:44,240 alpha, which is nm over 2 the average of 3 c's, c squared 556 00:36:44,240 --> 00:36:46,160 and then c alpha. 557 00:36:46,160 --> 00:36:48,336 And this is called the heat transport. 558 00:37:00,890 --> 00:37:05,200 So for a simpler fluid where these 559 00:37:05,200 --> 00:37:08,490 are the only conserved quantities that I have, 560 00:37:08,490 --> 00:37:12,540 in order to figure out how the fluid evolves over time, 561 00:37:12,540 --> 00:37:15,470 I have one equation that tells me 562 00:37:15,470 --> 00:37:17,490 about how the density changes. 563 00:37:17,490 --> 00:37:21,440 And it's related to the continuity of matter. 564 00:37:21,440 --> 00:37:30,540 I have one equation that tells me how the velocity changes. 565 00:37:30,540 --> 00:37:33,430 And it's kind of an appropriately generalized 566 00:37:33,430 --> 00:37:37,560 version of Newton's law in which mass times acceleration 567 00:37:37,560 --> 00:37:40,820 is equated with appropriate forces. 568 00:37:40,820 --> 00:37:44,420 And mostly we are interested in the forces that are internally 569 00:37:44,420 --> 00:37:48,630 generated, because of the variations in pressure. 570 00:37:48,630 --> 00:37:54,360 And finally, there is variations in pressure 571 00:37:54,360 --> 00:37:57,170 related to variations in temperature. 572 00:37:57,170 --> 00:38:01,640 And they're governed by another equation that tells us 573 00:38:01,640 --> 00:38:07,470 how the local energy density, local content of energy, 574 00:38:07,470 --> 00:38:09,016 changes as a function of time. 575 00:38:12,590 --> 00:38:18,450 So rather than solving the Boltzmann equation, 576 00:38:18,450 --> 00:38:21,660 I say, OK, all I need to do is to solve 577 00:38:21,660 --> 00:38:23,455 these hydrodynamic equations. 578 00:38:27,100 --> 00:38:29,383 Question? 579 00:38:29,383 --> 00:38:32,811 STUDENT: Last time for the Boltzmann equation, 580 00:38:32,811 --> 00:38:33,311 [INAUDIBLE]. 581 00:38:53,000 --> 00:38:58,500 PROFESSOR: What it says is that conservation laws are much more 582 00:38:58,500 --> 00:38:59,720 general. 583 00:38:59,720 --> 00:39:02,260 So this equation you could have written for a liquid, 584 00:39:02,260 --> 00:39:05,020 you could have written for anything. 585 00:39:05,020 --> 00:39:07,290 This equation kind of looks like you 586 00:39:07,290 --> 00:39:09,870 would have been able to write it for everything. 587 00:39:09,870 --> 00:39:12,370 And it is true, except that you wouldn't 588 00:39:12,370 --> 00:39:16,450 know what the pressure is. 589 00:39:16,450 --> 00:39:19,460 This equation you would have written on the basis of energy 590 00:39:19,460 --> 00:39:22,760 conservation, except that you wouldn't 591 00:39:22,760 --> 00:39:27,230 know what the heat transport vector is. 592 00:39:27,230 --> 00:39:31,370 So what we gained through this Boltzmann prescription, 593 00:39:31,370 --> 00:39:33,600 on top of what you may just guess 594 00:39:33,600 --> 00:39:36,060 on the basis of conservation laws, 595 00:39:36,060 --> 00:39:40,700 are expressions for quantities that you would need in order 596 00:39:40,700 --> 00:39:44,380 to sort these equations, because of the internal pressures that 597 00:39:44,380 --> 00:39:48,558 are generated because of the way that the heat is flowing. 598 00:39:48,558 --> 00:39:56,180 STUDENT: And this quantity is correct in the limit of-- 599 00:39:56,180 --> 00:39:58,550 PROFESSOR: In the limit, yes. 600 00:39:58,550 --> 00:40:04,756 But that also really is the Achilles' heel 601 00:40:04,756 --> 00:40:09,070 of the presentation I have given to you right now. 602 00:40:09,070 --> 00:40:13,680 Because in order to solve these equations, 603 00:40:13,680 --> 00:40:16,290 I should be able to put an expression here 604 00:40:16,290 --> 00:40:21,670 for the pressure, and an expression here for the h. 605 00:40:21,670 --> 00:40:24,690 But what is my prescription for getting 606 00:40:24,690 --> 00:40:28,150 the expression for pressure and h? 607 00:40:28,150 --> 00:40:31,780 I have to do an average that involves the f. 608 00:40:31,780 --> 00:40:34,440 And I don't have the f. 609 00:40:34,440 --> 00:40:40,470 So have I gained anything, all right? 610 00:40:40,470 --> 00:40:45,040 So these equations are general. 611 00:40:45,040 --> 00:40:49,860 We have to figure out what to do for the p and h in order 612 00:40:49,860 --> 00:40:52,380 to be able to solve it. 613 00:40:52,380 --> 00:40:52,898 Yes. 614 00:40:52,898 --> 00:40:54,397 STUDENT: In the last equation, isn't 615 00:40:54,397 --> 00:40:56,966 that n epsilon instead of epsilon? 616 00:41:00,170 --> 00:41:05,885 PROFESSOR: mc squared over 2-- I guess 617 00:41:05,885 --> 00:41:09,240 if I put here mc squared over 2, probably it is nf. 618 00:41:15,386 --> 00:41:17,719 STUDENT: I think maybe the last equation it's n epsilon, 619 00:41:17,719 --> 00:41:21,298 the equation there. 620 00:41:21,298 --> 00:41:22,589 PROFESSOR: This equation is OK. 621 00:41:27,970 --> 00:41:33,240 OK, so what you're saying-- that if I directly put chi here 622 00:41:33,240 --> 00:41:36,970 to be this quantity, what I would need on the left hand 623 00:41:36,970 --> 00:41:39,430 side of the equations would involve 624 00:41:39,430 --> 00:41:41,700 derivatives of n epsilon. 625 00:41:41,700 --> 00:41:46,450 Now, those derivatives I can expand, write them, let's say, 626 00:41:46,450 --> 00:41:52,000 dt of n epsilon is epsilon dt of n plus ndt of epsilon. 627 00:41:52,000 --> 00:41:54,850 And then you can use these equations 628 00:41:54,850 --> 00:41:57,350 to reduce some of that. 629 00:41:57,350 --> 00:42:00,195 And the reason that I didn't go through the steps 630 00:42:00,195 --> 00:42:02,660 that I would go from here to here 631 00:42:02,660 --> 00:42:04,277 is because it would have involved 632 00:42:04,277 --> 00:42:05,610 a number of those cancellations. 633 00:42:05,610 --> 00:42:09,398 And it would have taken me an additional 10, 15 minutes. 634 00:42:16,230 --> 00:42:19,320 All right, so conceptually, that's 635 00:42:19,320 --> 00:42:20,960 the more important thing. 636 00:42:20,960 --> 00:42:25,110 We have to find some way of doing these things. 637 00:42:25,110 --> 00:42:31,130 Now, when I wrote this equation, we 638 00:42:31,130 --> 00:42:33,980 said that there is some kind of a separation of time 639 00:42:33,980 --> 00:42:37,470 scales involved in that the left hand 640 00:42:37,470 --> 00:42:40,780 side of the equation, the characteristic times 641 00:42:40,780 --> 00:42:43,550 are order of the time it takes for a particle 642 00:42:43,550 --> 00:42:46,720 to, say, go over the sides of the box, 643 00:42:46,720 --> 00:42:50,970 whereas the collision times, 1 over tau x, 644 00:42:50,970 --> 00:42:55,500 are such that the right hand side is 645 00:42:55,500 --> 00:42:57,710 much larger than the left hand side. 646 00:43:00,830 --> 00:43:08,570 So as a 0-th order approximation, 647 00:43:08,570 --> 00:43:12,650 what I will assume is that the left hand side 648 00:43:12,650 --> 00:43:18,380 is so insignificant that I will set it to 0. 649 00:43:18,380 --> 00:43:21,680 And then my approximation for the collision 650 00:43:21,680 --> 00:43:25,710 is the thing that essentially sets this bracket to 0. 651 00:43:25,710 --> 00:43:30,680 This is the local equilibrium that we wrote down before. 652 00:43:30,680 --> 00:43:36,980 So that means that I'm assuming a 0-th order approximation 653 00:43:36,980 --> 00:43:41,030 to the solution of the Boltzmann equation. 654 00:43:41,030 --> 00:43:43,410 And very shortly, we will improve up that. 655 00:43:43,410 --> 00:43:47,100 But let's see what this 0-th order approximation gives us, 656 00:43:47,100 --> 00:43:49,610 which is-- we saw what it is. 657 00:43:49,610 --> 00:43:54,470 It was essentially something like a Gaussian in momentum. 658 00:43:54,470 --> 00:43:59,350 But the coefficient out front of it was kind of arbitrary. 659 00:43:59,350 --> 00:44:03,560 And now that I have defined the integral over momentum 660 00:44:03,560 --> 00:44:07,360 to be density, I will multiply a normalized Gaussian 661 00:44:07,360 --> 00:44:09,970 by the density locally. 662 00:44:09,970 --> 00:44:13,060 And I will have an exponential. 663 00:44:13,060 --> 00:44:18,480 And average of p I will shift by an amount 664 00:44:18,480 --> 00:44:21,870 that depends on position. 665 00:44:21,870 --> 00:44:27,220 And I divide by some parameter we had called before beta. 666 00:44:27,220 --> 00:44:32,030 But that beta I can rewrite in this fashion. 667 00:44:32,030 --> 00:44:35,840 So I have just rewritten the beta 668 00:44:35,840 --> 00:44:38,350 that we had before that was a function of q and t 669 00:44:38,350 --> 00:44:40,500 as 1 over kBT. 670 00:44:40,500 --> 00:44:43,180 And this has to be properly normalized. 671 00:44:43,180 --> 00:44:48,020 So I will have 2 pi mkBT, which is a function of position 672 00:44:48,020 --> 00:44:48,520 to the 3/2. 673 00:44:53,040 --> 00:44:56,850 And you can check that the form that I have written here 674 00:44:56,850 --> 00:45:00,410 respects the definitions that I gave, 675 00:45:00,410 --> 00:45:02,750 namely that if I were to integrate it 676 00:45:02,750 --> 00:45:05,885 over momentum, since the momentum part is 677 00:45:05,885 --> 00:45:09,810 a normalized Gaussian, I will just get the density. 678 00:45:09,810 --> 00:45:14,210 If I were to calculate the average of p/m, 679 00:45:14,210 --> 00:45:16,810 I have shifted the Gaussian appropriately so 680 00:45:16,810 --> 00:45:22,860 that the average of p/m is the quantity that I'm calling u. 681 00:45:22,860 --> 00:45:25,390 The other one-- let's check. 682 00:45:25,390 --> 00:45:28,720 Essentially what is happening here, 683 00:45:28,720 --> 00:45:33,570 this quantity is the same thing as mc squared over 2kT 684 00:45:33,570 --> 00:45:37,790 if I use the definition of c that I have over there. 685 00:45:37,790 --> 00:45:40,140 So it's a Gaussian weight. 686 00:45:40,140 --> 00:45:41,740 And from the Gaussian weight, you 687 00:45:41,740 --> 00:45:43,930 can immediately see that the average of c 688 00:45:43,930 --> 00:45:48,720 alpha c beta, it's in fact diagonal. 689 00:45:48,720 --> 00:45:50,390 It's cx squared, cy squared. 690 00:45:50,390 --> 00:45:52,950 So the answer is going to be delta alpha beta. 691 00:45:55,630 --> 00:45:58,120 And for each particular component, 692 00:45:58,120 --> 00:45:59,871 I will get kT over m. 693 00:46:05,770 --> 00:46:09,740 So this quantity that I was calling epsilon, 694 00:46:09,740 --> 00:46:13,790 which was the average of mc squared over 2, 695 00:46:13,790 --> 00:46:17,570 is essentially multiplying this by m/2 696 00:46:17,570 --> 00:46:20,630 and summing over delta alpha alpha, 697 00:46:20,630 --> 00:46:22,490 which gives me a factor of 3. 698 00:46:22,490 --> 00:46:26,340 So this is going to give me 3/2 kT. 699 00:46:26,340 --> 00:46:31,840 So really, my energy density is none other 700 00:46:31,840 --> 00:46:34,610 than the local 3/2 kT. 701 00:46:34,610 --> 00:46:36,916 Yes? 702 00:46:36,916 --> 00:46:40,518 STUDENT: So you've just defined, what is the temperature. 703 00:46:40,518 --> 00:46:44,107 So over all previous derivations, 704 00:46:44,107 --> 00:46:47,586 we didn't really use the classical temperature. 705 00:46:47,586 --> 00:46:51,065 And now you define it as sort of average kinetic energy. 706 00:46:51,065 --> 00:46:55,450 PROFESSOR: Yeah, I have introduced a quantity T here, 707 00:46:55,450 --> 00:46:57,710 which will indeed eventually be the temperature 708 00:46:57,710 --> 00:46:58,610 for the whole thing. 709 00:46:58,610 --> 00:47:00,200 But right now, it is something that 710 00:47:00,200 --> 00:47:03,190 is varying locally from position to position. 711 00:47:03,190 --> 00:47:07,310 But you can see that the typical kinetic energy at each location 712 00:47:07,310 --> 00:47:09,646 is of the order of kT at that location. 713 00:47:13,460 --> 00:47:18,140 And the pressure tensor p alpha beta, 714 00:47:18,140 --> 00:47:22,440 which is nm expectation value of c alpha c beta, 715 00:47:22,440 --> 00:47:30,740 simply becomes kT over m-- sorry, nKT delta alpha beta. 716 00:47:30,740 --> 00:47:32,450 So now we can sort of start. 717 00:47:32,450 --> 00:47:34,980 Now probably it's a better time to think 718 00:47:34,980 --> 00:47:37,080 about this as temperature. 719 00:47:37,080 --> 00:47:41,000 Because we know about the ideal gas type of behavior 720 00:47:41,000 --> 00:47:44,570 where the pressure of the ideal gas is simply density times kT. 721 00:47:47,480 --> 00:47:51,300 So the diagonal elements of this pressure tensor 722 00:47:51,300 --> 00:47:54,440 are the things that we usually think about 723 00:47:54,440 --> 00:47:57,080 as being the pressure of a gas, now 724 00:47:57,080 --> 00:47:59,600 at the appropriate temperature and density, 725 00:47:59,600 --> 00:48:02,910 and that there are no off diagonal components here. 726 00:48:05,700 --> 00:48:08,710 I said I also need to evaluate the h alpha. 727 00:48:11,240 --> 00:48:14,980 h alpha involves three factors of c. 728 00:48:19,300 --> 00:48:24,090 And the way that we have written down is Gaussian. 729 00:48:24,090 --> 00:48:26,090 So it's symmetric. 730 00:48:26,090 --> 00:48:29,220 So all odd powers are going to be 0. 731 00:48:29,220 --> 00:48:31,950 There is no heat transport vector here. 732 00:48:36,270 --> 00:48:41,964 So within this 0-th order, what do we have? 733 00:48:41,964 --> 00:48:49,150 We have that the total density variation, which 734 00:48:49,150 --> 00:48:58,040 is dt plus u alpha d alpha acting on density, 735 00:48:58,040 --> 00:49:01,640 is minus nd alpha u alpha. 736 00:49:01,640 --> 00:49:06,110 That does not involve any of these factors that I need. 737 00:49:06,110 --> 00:49:08,200 This equation-- let's see. 738 00:49:08,200 --> 00:49:10,490 Let's divide by mn. 739 00:49:10,490 --> 00:49:13,225 So we have Dt of u beta. 740 00:49:16,480 --> 00:49:21,810 And let's again look at what's happening inside the room. 741 00:49:21,810 --> 00:49:25,140 Forget about boundary conditions at the side of the box. 742 00:49:25,140 --> 00:49:26,940 So I'm going to write this essentially 743 00:49:26,940 --> 00:49:29,610 for the case that is inside the box. 744 00:49:29,610 --> 00:49:32,040 I can forget about the external force. 745 00:49:32,040 --> 00:49:35,220 And all I'm interested in is the internal forces 746 00:49:35,220 --> 00:49:37,870 that are generated through pressure. 747 00:49:37,870 --> 00:49:44,570 So this is dt plus u alpha d alpha of u beta. 748 00:49:44,570 --> 00:49:47,210 I said let's forget the external force. 749 00:49:47,210 --> 00:49:48,580 So what do we have? 750 00:49:48,580 --> 00:49:52,780 We have the contribution that comes from pressure. 751 00:49:52,780 --> 00:49:57,110 So we have minus the alpha. 752 00:49:57,110 --> 00:49:59,480 I divided through by nm. 753 00:49:59,480 --> 00:50:04,630 So let me write it correctly as 1 over nm. 754 00:50:04,630 --> 00:50:06,520 I have the alpha. 755 00:50:06,520 --> 00:50:14,950 My pressure tensor is nkT delta alpha beta. 756 00:50:14,950 --> 00:50:18,780 Delta alpha beta and this d alpha, I can get rid of that 757 00:50:18,780 --> 00:50:20,971 and write it simply as d beta. 758 00:50:25,210 --> 00:50:31,070 So that's the equation that governs the variations 759 00:50:31,070 --> 00:50:34,540 in the local stream velocity that you 760 00:50:34,540 --> 00:50:38,290 have in the gas in response to the changes in temperature 761 00:50:38,290 --> 00:50:42,200 and density that you have in the gas. 762 00:50:42,200 --> 00:50:48,080 And finally, the equation for the energy density, I 763 00:50:48,080 --> 00:50:53,490 have dt plus u alpha d alpha. 764 00:50:53,490 --> 00:50:59,040 My energy density is simply related to this quantity T. 765 00:50:59,040 --> 00:51:04,610 So I can write it as variations of this temperature 766 00:51:04,610 --> 00:51:06,780 in position. 767 00:51:06,780 --> 00:51:10,040 And what do I have on the right hand side? 768 00:51:10,040 --> 00:51:13,770 I certainly don't have the heat transport vectors. 769 00:51:13,770 --> 00:51:19,740 So all I have to do is to take this diagonal p alpha beta 770 00:51:19,740 --> 00:51:25,150 and contract it with this strain tensor u alpha beta. 771 00:51:25,150 --> 00:51:28,280 So the only term that I'm going to get after contracting 772 00:51:28,280 --> 00:51:31,620 delta alpha beta is going to be d alpha u alpha. 773 00:51:31,620 --> 00:51:35,350 So let's make sure that we get the factors right. 774 00:51:35,350 --> 00:51:42,440 So I have minus p alpha beta is nkT d alpha u alpha. 775 00:51:51,740 --> 00:51:56,910 So now we have a closed set of equations. 776 00:51:56,910 --> 00:52:05,050 They tell me how the density, temperature, and velocity 777 00:52:05,050 --> 00:52:08,620 vary from one location to another location in the gas. 778 00:52:08,620 --> 00:52:10,070 They're completely closed. 779 00:52:10,070 --> 00:52:13,980 That's the only set of things that come together. 780 00:52:13,980 --> 00:52:17,620 So I should be able to now figure out, 781 00:52:17,620 --> 00:52:20,570 if I make a disturbance in the gas in this room 782 00:52:20,570 --> 00:52:25,630 by walking around, by talking, by striking a match, 783 00:52:25,630 --> 00:52:29,630 how does that eventually, as a function of time, 784 00:52:29,630 --> 00:52:32,710 relax to something that is uniform? 785 00:52:32,710 --> 00:52:37,870 Because our expectation is that these equations ultimately 786 00:52:37,870 --> 00:52:39,610 will reach equilibrium. 787 00:52:39,610 --> 00:52:43,450 That's essentially the most important thing 788 00:52:43,450 --> 00:52:45,560 that we deduce from the Boltzmann equation, 789 00:52:45,560 --> 00:52:47,950 that it was allowing things to reach equilibrium. 790 00:52:51,220 --> 00:52:52,145 Yes. 791 00:52:52,145 --> 00:52:54,522 STUDENT: For the second equation, that's alpha? 792 00:52:54,522 --> 00:52:56,105 The right side of the second equation? 793 00:53:00,520 --> 00:53:02,270 PROFESSOR: The alpha index is summed over. 794 00:53:02,270 --> 00:53:04,174 STUDENT: The right side. 795 00:53:04,174 --> 00:53:09,410 Is it the derivative of alpha or beta? 796 00:53:09,410 --> 00:53:10,124 Yeah, that one. 797 00:53:10,124 --> 00:53:11,040 PROFESSOR: It is beta. 798 00:53:11,040 --> 00:53:14,410 Because, you see, the only index that I have left is beta. 799 00:53:14,410 --> 00:53:17,620 So if it's an index by itself, it better be beta. 800 00:53:17,620 --> 00:53:21,990 How did this index d alpha become d beta? 801 00:53:21,990 --> 00:53:24,870 Because the alpha beta was delta alpha beta. 802 00:53:24,870 --> 00:53:28,020 STUDENT: Also, is it alpha or is it beta? 803 00:53:28,020 --> 00:53:30,920 PROFESSOR: When I sum over alpha of d alpha delta alpha beta, 804 00:53:30,920 --> 00:53:31,780 I get d beta. 805 00:53:35,628 --> 00:53:36,590 Yes. 806 00:53:36,590 --> 00:53:39,910 STUDENT: Can I ask again, how did you come up with the f0? 807 00:53:39,910 --> 00:53:42,070 Why do you say that option? 808 00:53:42,070 --> 00:53:45,110 PROFESSOR: OK, so this goes back to what 809 00:53:45,110 --> 00:53:47,090 we did last time around. 810 00:53:47,090 --> 00:53:49,830 Because we saw that when we were writing 811 00:53:49,830 --> 00:53:53,390 the equation for the hydt, we came up 812 00:53:53,390 --> 00:53:56,300 with a factor of what that was-- this multiplying 813 00:53:56,300 --> 00:53:58,950 the difference of the logs. 814 00:53:58,950 --> 00:54:01,800 And we said that what I can do in order 815 00:54:01,800 --> 00:54:04,930 to make sure that this equation is 816 00:54:04,930 --> 00:54:12,050 0 is to say that log is additive in conserved quantities, 817 00:54:12,050 --> 00:54:15,880 so log additive in conserved quantities. 818 00:54:15,880 --> 00:54:18,630 I then exponentiate it. 819 00:54:18,630 --> 00:54:21,700 So this is log of a number. 820 00:54:21,700 --> 00:54:23,840 And these are all things that are, 821 00:54:23,840 --> 00:54:26,500 when I take the log, proportional to p squared 822 00:54:26,500 --> 00:54:29,175 and p, which are the conserved quantities. 823 00:54:33,950 --> 00:54:36,930 So I know that this form sets the right hand 824 00:54:36,930 --> 00:54:39,290 side of the Boltzmann equation to 0. 825 00:54:39,290 --> 00:54:41,890 And that's the largest part of the Boltzmann equation. 826 00:54:51,510 --> 00:54:56,200 Now what happens is that within this equation, 827 00:54:56,200 --> 00:54:58,860 some quantities do not relax to equilibrium. 828 00:55:01,960 --> 00:55:04,750 Some-- let's call them variations. 829 00:55:04,750 --> 00:55:08,704 Sometimes I will use the word "modes"-- 830 00:55:08,704 --> 00:55:13,650 do not relax to equilibrium. 831 00:55:18,040 --> 00:55:23,240 And let's start with the following. 832 00:55:26,710 --> 00:55:33,740 When you have a sheer velocity-- what do I mean by that? 833 00:55:33,740 --> 00:55:39,500 So let's imagine that you have a wall that 834 00:55:39,500 --> 00:55:42,510 extends in the x direction. 835 00:55:42,510 --> 00:55:50,750 And along the y direction, you encounter a velocity field. 836 00:55:50,750 --> 00:55:55,590 The velocity field is always pointing along this direction. 837 00:55:55,590 --> 00:55:58,020 So it only has the x component. 838 00:55:58,020 --> 00:56:00,820 There's no y component or z component. 839 00:56:00,820 --> 00:56:02,980 But this x component maybe varies 840 00:56:02,980 --> 00:56:05,490 as a function of position. 841 00:56:05,490 --> 00:56:08,990 So my ux is a function of y. 842 00:56:08,990 --> 00:56:12,200 This corresponds to some kind of a sheer. 843 00:56:16,200 --> 00:56:21,010 Now, if I do that, then you can see 844 00:56:21,010 --> 00:56:26,770 that the only derivatives that would be nonzero 845 00:56:26,770 --> 00:56:29,710 are derivatives that are along the y direction. 846 00:56:33,530 --> 00:56:38,740 But this derivative along the y direction 847 00:56:38,740 --> 00:56:44,410 in all of these equations has to be contracted typically 848 00:56:44,410 --> 00:56:46,990 with something else. 849 00:56:46,990 --> 00:56:50,070 It has to be contacted with u. 850 00:56:50,070 --> 00:56:55,100 But the u's have no component along the y direction. 851 00:56:55,100 --> 00:56:59,030 So essentially, all my u's would be of this form. 852 00:56:59,030 --> 00:57:05,680 Basically, there will be something like uy. 853 00:57:05,680 --> 00:57:09,450 Something like this would have to be 0. 854 00:57:09,450 --> 00:57:12,340 You can see that if I start with an initial condition 855 00:57:12,340 --> 00:57:18,640 such as that, then the equations are that dt of n-- 856 00:57:18,640 --> 00:57:23,800 this term I have to forget-- is 0. 857 00:57:23,800 --> 00:57:26,920 Because for this, I need a divergence. 858 00:57:26,920 --> 00:57:31,450 And this flow has no divergence. 859 00:57:31,450 --> 00:57:38,260 And similarly over here what I see as dt of the temperature 860 00:57:38,260 --> 00:57:40,030 is 0. 861 00:57:40,030 --> 00:57:42,560 Temperature doesn't change. 862 00:57:42,560 --> 00:57:50,180 And if I assume that I am under circumstances 863 00:57:50,180 --> 00:57:57,290 in which the pressure is uniform, 864 00:57:57,290 --> 00:58:01,070 there's also nothing that I would get from here. 865 00:58:01,070 --> 00:58:06,790 So essentially, this flow will exist forever. 866 00:58:06,790 --> 00:58:09,060 Yes. 867 00:58:09,060 --> 00:58:12,840 STUDENT: Why does your u alpha d alpha n term go away? 868 00:58:12,840 --> 00:58:14,122 Wouldn't you get a uxdxn? 869 00:58:16,905 --> 00:58:19,620 PROFESSOR: OK, let's see, you want a uxdxn. 870 00:58:22,720 --> 00:58:24,980 What I said is that all variations 871 00:58:24,980 --> 00:58:27,764 are along the y direction. 872 00:58:27,764 --> 00:58:29,680 STUDENT: Oh, so this is not just for velocity, 873 00:58:29,680 --> 00:58:30,590 but for everything. 874 00:58:30,590 --> 00:58:32,750 PROFESSOR: Yes, so I make an assumption 875 00:58:32,750 --> 00:58:34,460 about some particular form. 876 00:58:34,460 --> 00:58:37,240 So this is the reasoning. 877 00:58:37,240 --> 00:58:42,970 If these equations bring everything to equilibrium, 878 00:58:42,970 --> 00:58:46,110 I should be able to pick any initial condition 879 00:58:46,110 --> 00:58:49,140 and ask, how long does it take to come to equilibrium? 880 00:58:49,140 --> 00:58:52,510 I pick this specific type of equation 881 00:58:52,510 --> 00:58:56,290 in which the only variations for all quantities 882 00:58:56,290 --> 00:58:57,780 are along the y direction. 883 00:58:57,780 --> 00:58:59,280 It's a non-equilibrium state. 884 00:58:59,280 --> 00:59:01,160 It's not a uniform state. 885 00:59:01,160 --> 00:59:03,340 Does it come relax to equilibrium? 886 00:59:03,340 --> 00:59:06,272 And the answer is no, it doesn't. 887 00:59:11,152 --> 00:59:14,568 STUDENT: What other properties, other than velocity, 888 00:59:14,568 --> 00:59:16,832 is given [INAUDIBLE]? 889 00:59:16,832 --> 00:59:18,290 PROFESSOR: Density and temperature. 890 00:59:18,290 --> 00:59:20,950 So these equations describe the variations 891 00:59:20,950 --> 00:59:23,750 of velocity, density, and temperature. 892 00:59:23,750 --> 00:59:28,350 And the statement is, if the system is to reach equilibrium, 893 00:59:28,350 --> 00:59:32,170 I should be able to start with any initial configuration 894 00:59:32,170 --> 00:59:34,750 of these three quantities that I want. 895 00:59:34,750 --> 00:59:39,710 And I see that after a while, it reaches a uniform state. 896 00:59:39,710 --> 00:59:40,540 Yes. 897 00:59:40,540 --> 00:59:43,420 STUDENT: But if your initial conditions aren't exactly that, 898 00:59:43,420 --> 00:59:44,860 but you add a slight fluctuation, 899 00:59:44,860 --> 00:59:51,280 it is likely to grow, and it will eventually relax. 900 00:59:51,280 --> 00:59:53,950 PROFESSOR: It turns out the answer is no. 901 00:59:53,950 --> 00:59:58,850 So I'm sort of approaching this problem from this more 902 00:59:58,850 --> 01:00:02,600 kind of hand-waving perspective. 903 01:00:02,600 --> 01:00:04,610 More correctly, what you can do is 904 01:00:04,610 --> 01:00:08,740 you can start with some initial condition that, let's say, 905 01:00:08,740 --> 01:00:13,160 is in equilibrium, and then do a perturbation, 906 01:00:13,160 --> 01:00:17,650 and ask whether the perturbation will eventually relax to 0 907 01:00:17,650 --> 01:00:19,960 or not. 908 01:00:19,960 --> 01:00:23,340 And let's in fact do that for another quantity, 909 01:00:23,340 --> 01:00:24,640 which is the sound mode. 910 01:00:34,130 --> 01:00:37,350 So let's imagine that we start with a totally 911 01:00:37,350 --> 01:00:40,070 nice, uniform state. 912 01:00:40,070 --> 01:00:43,740 There is zero velocity initially. 913 01:00:43,740 --> 01:00:45,130 The density is uniform. 914 01:00:45,130 --> 01:00:47,490 The temperature is uniform. 915 01:00:47,490 --> 01:00:50,740 And then what I do is I will start here. 916 01:00:50,740 --> 01:00:58,650 And I will start talking, creating a variation that 917 01:00:58,650 --> 01:01:02,430 propagates in this x direction. 918 01:01:02,430 --> 01:01:10,210 So I generated a stream that is moving along the x direction. 919 01:01:10,210 --> 01:01:13,810 And presumably, as I move along the x direction, 920 01:01:13,810 --> 01:01:18,290 there is a velocity that changes with position and temperature. 921 01:01:22,240 --> 01:01:26,250 Now initially, I had the density. 922 01:01:26,250 --> 01:01:27,390 I said that was uniform. 923 01:01:30,280 --> 01:01:35,780 Once I make this sound, as I move along the x direction, 924 01:01:35,780 --> 01:01:39,320 and the air is flowing back and forth, what happens 925 01:01:39,320 --> 01:01:47,640 is that the density will vary from the uniform state. 926 01:01:47,640 --> 01:01:50,020 And the deviations from the uniform 927 01:01:50,020 --> 01:01:55,260 state I will indicate by mu. 928 01:01:55,260 --> 01:02:01,410 Similarly, the third quantity, let's assume, 929 01:02:01,410 --> 01:02:05,870 will have a form such as this. 930 01:02:05,870 --> 01:02:09,770 And currently, I have written the most general form 931 01:02:09,770 --> 01:02:14,464 of variations that I can have along the x direction. 932 01:02:14,464 --> 01:02:16,130 You could do it in different directions. 933 01:02:16,130 --> 01:02:18,890 But let's say for simplicity, we stick with this. 934 01:02:18,890 --> 01:02:22,830 I haven't told you what mu theta and u are. 935 01:02:22,830 --> 01:02:28,110 So I have to see what they are consistent with the equations 936 01:02:28,110 --> 01:02:30,990 that we have up there. 937 01:02:30,990 --> 01:02:35,900 One thing that I will assume is that these things 938 01:02:35,900 --> 01:02:41,440 are small perturbations around the uniform state. 939 01:02:41,440 --> 01:02:46,350 And uniform-- sorry, small perturbations typically 940 01:02:46,350 --> 01:02:49,570 means that what I intend to do is 941 01:02:49,570 --> 01:02:52,473 to do a linearized approximation. 942 01:02:57,020 --> 01:03:01,200 So basically, what I will do is I will essentially 943 01:03:01,200 --> 01:03:04,040 look at the linear version of these equations. 944 01:03:04,040 --> 01:03:08,320 And again, maybe I didn't emphasize it before. 945 01:03:08,320 --> 01:03:10,630 Clearly these are nonlinear equations. 946 01:03:10,630 --> 01:03:13,010 Because let's say you have u grad u. 947 01:03:13,010 --> 01:03:15,100 It's the same nonlinearity that you have, 948 01:03:15,100 --> 01:03:17,280 let's say, in Navier-Stokes equation. 949 01:03:17,280 --> 01:03:20,300 Because you're transporting something and moving along 950 01:03:20,300 --> 01:03:22,120 with the flow. 951 01:03:22,120 --> 01:03:25,300 But when you do the linearization, then 952 01:03:25,300 --> 01:03:30,050 these operators that involve dt plus something like u-- 953 01:03:30,050 --> 01:03:33,120 I guess in this case, the only direction that is varying 954 01:03:33,120 --> 01:03:39,970 is x-- something like this of whatever quantity that I have, 955 01:03:39,970 --> 01:03:43,330 I can drop this nonlinear term. 956 01:03:43,330 --> 01:03:44,440 Why? 957 01:03:44,440 --> 01:03:48,940 Because u is a perturbation around a uniform state. 958 01:03:48,940 --> 01:03:53,340 And gradients will pick up some perturbations 959 01:03:53,340 --> 01:03:55,000 around the uniform state. 960 01:03:55,000 --> 01:03:57,430 So essentially the linearization amounts 961 01:03:57,430 --> 01:04:01,690 to dropping these nonlinear components 962 01:04:01,690 --> 01:04:05,240 and some other things that I will linearizer also. 963 01:04:05,240 --> 01:04:08,490 Because all of these functions here, 964 01:04:08,490 --> 01:04:13,690 the derivatives act on product of n temperature over here. 965 01:04:13,690 --> 01:04:18,960 These are all nonlinear operations. 966 01:04:18,960 --> 01:04:21,920 So let's linearize what we have. 967 01:04:21,920 --> 01:04:28,860 We have that Dt of the density-- I guess when I take the time 968 01:04:28,860 --> 01:04:34,010 derivative, I get n bar the time derivative of the quantity 969 01:04:34,010 --> 01:04:35,980 that I'm calling mu. 970 01:04:35,980 --> 01:04:36,770 And that's it. 971 01:04:36,770 --> 01:04:39,820 I don't need to worry about the convective part, 972 01:04:39,820 --> 01:04:41,570 the u dot grad part. 973 01:04:41,570 --> 01:04:43,180 That's second order. 974 01:04:43,180 --> 01:04:45,901 On the right hand side, what do I have? 975 01:04:45,901 --> 01:04:48,956 I have ndu. 976 01:04:48,956 --> 01:04:52,780 Well, divergence of u is already the first variation. 977 01:04:52,780 --> 01:04:56,400 So for n, I will take its 0-th order term. 978 01:04:56,400 --> 01:04:59,180 So I have minus n bar dxux. 979 01:05:04,020 --> 01:05:09,970 The equation for ux, really the only component that I have, 980 01:05:09,970 --> 01:05:13,090 is dt of ux. 981 01:05:13,090 --> 01:05:18,210 Actually, let's write down the equation for temperature. 982 01:05:18,210 --> 01:05:19,610 Let's look at this equation. 983 01:05:19,610 --> 01:05:28,750 So I have that dt acting on 3/2 kB times 984 01:05:28,750 --> 01:05:31,670 T. I will pick up T bar. 985 01:05:31,670 --> 01:05:35,880 And then I would have dt of theta. 986 01:05:35,880 --> 01:05:38,340 What do I have on the right hand side? 987 01:05:38,340 --> 01:05:39,750 I have a derivative here. 988 01:05:39,750 --> 01:05:43,200 So everything else here I will evaluate at the 0-th order 989 01:05:43,200 --> 01:05:47,505 term, so n bar k T bar dxux. 990 01:05:53,670 --> 01:05:58,900 So I can see that I can certainly divide through n bar 991 01:05:58,900 --> 01:06:00,110 here. 992 01:06:00,110 --> 01:06:06,350 And one of my equations becomes dt of mu is minus dxux. 993 01:06:10,520 --> 01:06:18,750 But from here, I see that dxux is also 994 01:06:18,750 --> 01:06:28,780 related once I divide by kT to 3/2 dt theta. 995 01:06:28,780 --> 01:06:30,080 And I know this to be true. 996 01:06:30,080 --> 01:06:34,720 And I seem to have an additional factor of 1 over n bar here. 997 01:06:34,720 --> 01:06:42,310 And so I made the mistake at some point, 998 01:06:42,310 --> 01:06:46,295 probably when I wrote this equation. 999 01:06:49,010 --> 01:06:51,880 STUDENT: It's the third equation. 1000 01:06:51,880 --> 01:06:55,500 PROFESSOR: Yeah, so this should not be here. 1001 01:06:58,030 --> 01:07:02,000 And that should not be here means 1002 01:07:02,000 --> 01:07:07,380 that I probably made a mistake here. 1003 01:07:07,380 --> 01:07:10,870 So this should be a 1/n, sorry. 1004 01:07:10,870 --> 01:07:15,595 There was indeed a 1/n / here. 1005 01:07:15,595 --> 01:07:19,209 And there is no factor here. 1006 01:07:23,700 --> 01:07:32,720 So we have a relationship between the time derivatives 1007 01:07:32,720 --> 01:07:39,650 of these variations in density and dx of ux. 1008 01:07:39,650 --> 01:07:44,200 Fine, what does the equation for u tell us? 1009 01:07:44,200 --> 01:07:54,880 It tells us that dt of ux is minus 1 over m n bar. 1010 01:07:54,880 --> 01:07:59,180 Because of the derivative, I can set everything 1011 01:07:59,180 --> 01:08:00,960 at the variation. 1012 01:08:00,960 --> 01:08:02,950 And what do I have here? 1013 01:08:02,950 --> 01:08:13,360 I have d by dx of n bar kB T bar. 1014 01:08:13,360 --> 01:08:17,930 And if I look at the variations, I have 1 plus mu plus theta. 1015 01:08:17,930 --> 01:08:20,649 The higher order terms I will forget. 1016 01:08:20,649 --> 01:08:21,453 Yes. 1017 01:08:21,453 --> 01:08:25,074 STUDENT: Shouldn't that be plus 3/2 dt theta? 1018 01:08:25,074 --> 01:08:27,224 PROFESSOR: It should be plus, yes. 1019 01:08:27,224 --> 01:08:28,847 There is a minus sign here. 1020 01:08:28,847 --> 01:08:30,146 And that makes it plus. 1021 01:08:36,979 --> 01:08:41,770 So the n bar we can take outside. 1022 01:08:41,770 --> 01:08:53,040 This becomes minus kT over m at space variations 1023 01:08:53,040 --> 01:08:54,176 of mu plus theta. 1024 01:08:59,569 --> 01:09:04,729 Now, what we do is that what I have here 1025 01:09:04,729 --> 01:09:09,310 is information about the time derivatives of mu and theta. 1026 01:09:09,310 --> 01:09:12,170 And here I have space derivatives. 1027 01:09:12,170 --> 01:09:13,800 So what do I do? 1028 01:09:13,800 --> 01:09:19,590 I basically apply an additionally dt here, 1029 01:09:19,590 --> 01:09:21,950 which we'll apply here. 1030 01:09:21,950 --> 01:09:23,850 And then we can apply it also here. 1031 01:09:30,960 --> 01:09:39,044 And then we know how dt of mu and dt of theta 1032 01:09:39,044 --> 01:09:42,380 are related to dx of ux. 1033 01:09:42,380 --> 01:09:45,120 The minus signs disappear. 1034 01:09:45,120 --> 01:09:51,144 I have kB T bar divided by m. 1035 01:09:51,144 --> 01:09:55,540 I have dt of mu is dxux. 1036 01:09:55,540 --> 01:09:58,130 dt of theta is 2/3 dxux. 1037 01:09:58,130 --> 01:10:04,540 So I will get 1 plus 2/3 dx squared of ux. 1038 01:10:10,330 --> 01:10:15,470 So the second derivative of ux in time 1039 01:10:15,470 --> 01:10:22,900 is proportional to the second derivative of ux in space. 1040 01:10:22,900 --> 01:10:25,890 So that's the standard wave equation. 1041 01:10:25,890 --> 01:10:29,090 And the velocity that we have calculated 1042 01:10:29,090 --> 01:10:36,048 for these sound waves is 5/3 kB the average T over m. 1043 01:10:41,050 --> 01:10:43,750 So that part is good. 1044 01:10:43,750 --> 01:10:49,930 These equations tell me that if I create these disturbances, 1045 01:10:49,930 --> 01:10:52,510 there are sound waves, and we know there are sound waves. 1046 01:10:52,510 --> 01:10:57,190 And sound waves will propagate with some velocity that 1047 01:10:57,190 --> 01:11:02,120 is related up to some factors to the average velocity of the gas 1048 01:11:02,120 --> 01:11:04,780 particles. 1049 01:11:04,780 --> 01:11:08,570 But what is not good is that, according 1050 01:11:08,570 --> 01:11:12,480 to this equation, if my waves, let's say, 1051 01:11:12,480 --> 01:11:15,710 bounce off perfectly from the walls, 1052 01:11:15,710 --> 01:11:18,950 they will last in this room forever. 1053 01:11:18,950 --> 01:11:22,480 So you should still be hearing what I was saying last week 1054 01:11:22,480 --> 01:11:24,320 and the week before. 1055 01:11:24,320 --> 01:11:27,250 And clearly what we are missing is 1056 01:11:27,250 --> 01:11:31,310 the damping that is required. 1057 01:11:31,310 --> 01:11:35,340 So the statement is that all of these equations are fine. 1058 01:11:35,340 --> 01:11:38,670 They capture a lot of the physics. 1059 01:11:38,670 --> 01:11:41,190 But there is something important that 1060 01:11:41,190 --> 01:11:45,510 is left out in that there are some modes-- 1061 01:11:45,510 --> 01:11:49,680 and I describe two of them here-- that basically last 1062 01:11:49,680 --> 01:11:52,590 forever, and don't come to equilibrium. 1063 01:11:52,590 --> 01:11:56,960 But we said that the Boltzmann equation should eventually 1064 01:11:56,960 --> 01:12:00,220 bring things to equilibrium. 1065 01:12:00,220 --> 01:12:03,490 So where did we go wrong? 1066 01:12:03,490 --> 01:12:05,760 Well, we didn't solve the Boltzmann equation. 1067 01:12:05,760 --> 01:12:09,690 We solved an approximation to the Boltzmann equation. 1068 01:12:09,690 --> 01:12:11,230 So let's try to do better. 1069 01:12:16,079 --> 01:12:18,953 STUDENT: I'm sorry, but for the last equation, 1070 01:12:18,953 --> 01:12:25,660 you took another derivative with respect to t. 1071 01:12:25,660 --> 01:12:30,180 PROFESSOR: Yes, I took a derivative with respect to t. 1072 01:12:30,180 --> 01:12:32,790 And it noted that the derivative with respect 1073 01:12:32,790 --> 01:12:36,310 to t of these quantities mu is related 1074 01:12:36,310 --> 01:12:39,741 to derivative with respect to x or u. 1075 01:12:43,590 --> 01:12:45,890 And there was one other derivative with respect 1076 01:12:45,890 --> 01:12:48,942 to x already, making it two derivatives. 1077 01:12:55,690 --> 01:13:01,865 So this is the kind of situation that we are facing. 1078 01:13:07,590 --> 01:13:08,780 Yes. 1079 01:13:08,780 --> 01:13:16,750 STUDENT: Is the 5/3 k in any way related to the heat capacity 1080 01:13:16,750 --> 01:13:17,960 ratio of [INAUDIBLE] gas? 1081 01:13:17,960 --> 01:13:19,964 PROFESSOR: Yes, that's right, yes. 1082 01:13:19,964 --> 01:13:23,050 So there are lots of these things that 1083 01:13:23,050 --> 01:13:24,780 are implicit in these questions. 1084 01:13:24,780 --> 01:13:32,360 And actually, that 3/2 is the same thing as this 3/2. 1085 01:13:32,360 --> 01:13:34,740 So you can trace a lot of these things 1086 01:13:34,740 --> 01:13:37,900 to the Gaussian distribution. 1087 01:13:37,900 --> 01:13:41,030 And they appear in cp versus cv and other things. 1088 01:13:41,030 --> 01:13:41,920 Yes. 1089 01:13:41,920 --> 01:13:43,880 STUDENT: Just clarifying something-- 1090 01:13:43,880 --> 01:13:49,130 this v is different from the mu in the top right? 1091 01:13:49,130 --> 01:13:52,040 PROFESSOR: Yes, this is v, and that's mu. 1092 01:13:52,040 --> 01:13:54,450 This v is the velocity of the sound. 1093 01:13:54,450 --> 01:13:58,760 So I defined this combination, the coefficient relating 1094 01:13:58,760 --> 01:14:01,720 the second derivatives in time and space 1095 01:14:01,720 --> 01:14:03,010 as the sound velocity. 1096 01:14:03,010 --> 01:14:17,020 So let's maybe even-- we can call it vs. All right? 1097 01:14:17,020 --> 01:14:19,000 STUDENT: And how did you know that that 1098 01:14:19,000 --> 01:14:25,450 is the [INAUDIBLE] oscillation, the solution that you got? 1099 01:14:25,450 --> 01:14:31,950 PROFESSOR: Because I know that the solution to dx 1100 01:14:31,950 --> 01:14:38,360 squared anything is v squared-- sorry, 1101 01:14:38,360 --> 01:14:44,390 v squared dx squared anything is dt squared anything, is 1102 01:14:44,390 --> 01:14:47,872 phi is some function of x minus vt. 1103 01:14:47,872 --> 01:14:51,620 That is a pulse that moves with uniform velocity 1104 01:14:51,620 --> 01:14:53,237 is a solution to this equation. 1105 01:15:05,180 --> 01:15:06,320 So we want to do better. 1106 01:15:09,350 --> 01:15:14,980 And better becomes so-called first order solution. 1107 01:15:14,980 --> 01:15:18,340 Now, the kind of equation that we 1108 01:15:18,340 --> 01:15:22,680 are trying to solve at the top is something 1109 01:15:22,680 --> 01:15:24,950 that its algebraic analog would be 1110 01:15:24,950 --> 01:15:28,830 something like this-- 2 times x. 1111 01:15:28,830 --> 01:15:31,830 It's a linear on the left hand side, 1112 01:15:31,830 --> 01:15:34,000 is quadratic on the right hand side. 1113 01:15:34,000 --> 01:15:37,080 Let's write it in this form-- except 1114 01:15:37,080 --> 01:15:39,970 that the typical magnitude of one side 1115 01:15:39,970 --> 01:15:42,240 is much larger than the other side, 1116 01:15:42,240 --> 01:15:45,890 so let's say something like this. 1117 01:15:45,890 --> 01:15:48,720 So if I wanted to solve this equation, 1118 01:15:48,720 --> 01:15:52,860 I would say that unless x is very close to 2, 1119 01:15:52,860 --> 01:15:56,280 this 10 to the 6 will blow things up. 1120 01:15:56,280 --> 01:15:59,832 So my x0 is 2. 1121 01:15:59,832 --> 01:16:01,040 And that's what we have done. 1122 01:16:01,040 --> 01:16:02,748 We've solved, essentially, the right hand 1123 01:16:02,748 --> 01:16:05,130 side of the equation. 1124 01:16:05,130 --> 01:16:09,310 But I can get a better solution by taking 2 and saying 1125 01:16:09,310 --> 01:16:13,260 there's a small variation to that that I want to calculate. 1126 01:16:13,260 --> 01:16:16,140 And I substitute that into the original equation. 1127 01:16:16,140 --> 01:16:22,250 On the left hand side, I will get 2 times 2, 1 plus epsilon. 1128 01:16:22,250 --> 01:16:25,600 On the right hand side, I will get 10 to the sixth. 1129 01:16:25,600 --> 01:16:28,660 And then essentially, I subtract 2 1130 01:16:28,660 --> 01:16:31,640 plus 2 epsilon squared from 5. 1131 01:16:31,640 --> 01:16:32,460 What do I get? 1132 01:16:32,460 --> 01:16:39,620 I will get 4 epsilon plus 4 epsilon squared. 1133 01:16:43,040 --> 01:16:45,040 Then I say that epsilon is small. 1134 01:16:45,040 --> 01:16:48,870 So essentially, I linearize the right hand side. 1135 01:16:48,870 --> 01:16:50,970 I forget about that. 1136 01:16:50,970 --> 01:16:52,980 I say that I keep the epsilon here, 1137 01:16:52,980 --> 01:16:55,550 because it's multiplying 10 to the 6. 1138 01:16:55,550 --> 01:16:59,130 But the epsilon on the other side is multiplying nothing. 1139 01:16:59,130 --> 01:17:00,850 So I forget that. 1140 01:17:00,850 --> 01:17:06,560 So then I will have my epsilon to be roughly, I don't know, 1141 01:17:06,560 --> 01:17:12,430 2 times 2 divided by 4 times 10 to the 6. 1142 01:17:12,430 --> 01:17:18,647 So I have gotten the correction to my first 0-th order solution 1143 01:17:18,647 --> 01:17:19,480 to this first order. 1144 01:17:22,140 --> 01:17:24,360 Now we will do exactly the same thing, 1145 01:17:24,360 --> 01:17:29,640 not for our algebraic equation, but for our Boltzmann equation. 1146 01:17:29,640 --> 01:17:37,700 So for the Boltzmann equation, which was Lf is C of ff, 1147 01:17:37,700 --> 01:17:41,620 we said that the right hand side is larger by a factor of 1 1148 01:17:41,620 --> 01:17:45,100 over tau x compared to the left hand side. 1149 01:17:45,100 --> 01:17:49,350 And so what we did was we found a solution f0 1150 01:17:49,350 --> 01:17:51,590 that when we put in the collision integral, 1151 01:17:51,590 --> 01:17:52,610 the answer was 0. 1152 01:17:55,180 --> 01:17:59,830 Now I want to have a better solution that I will call f1. 1153 01:17:59,830 --> 01:18:01,650 Just like I did over there, I will 1154 01:18:01,650 --> 01:18:05,250 assume that f0 is added to a small function 1155 01:18:05,250 --> 01:18:06,290 that I will call g. 1156 01:18:08,920 --> 01:18:14,230 And then I substitute this equation, this thing, 1157 01:18:14,230 --> 01:18:16,910 to the equation. 1158 01:18:16,910 --> 01:18:23,010 So when I substitute, I will get L acting on f0 1 plus g. 1159 01:18:28,720 --> 01:18:31,750 Now, what did I do over here? 1160 01:18:31,750 --> 01:18:36,390 On the left hand side, I ignored the first order term. 1161 01:18:36,390 --> 01:18:39,020 Because I expect the first order term to be already small. 1162 01:18:39,020 --> 01:18:42,016 And the left hand side is already small. 1163 01:18:42,016 --> 01:18:45,290 So I will ignore this. 1164 01:18:45,290 --> 01:18:51,030 And on the right hand side, I have to linearize. 1165 01:18:51,030 --> 01:18:59,430 So I have to put f0 1 plus g, f0 1 plus g. 1166 01:18:59,430 --> 01:19:03,460 Essentially what I have to do is to go to the collisions 1167 01:19:03,460 --> 01:19:09,590 that I have over here and write for this f0 1 plus g. 1168 01:19:09,590 --> 01:19:11,742 There are four of such things. 1169 01:19:11,742 --> 01:19:14,610 Now, the 0-th order term already cancels. 1170 01:19:14,610 --> 01:19:20,500 Because f0 f0 was f0 f0 by the way that I constructed things. 1171 01:19:20,500 --> 01:19:22,510 And then I can pull out one factor 1172 01:19:22,510 --> 01:19:25,000 of f0 out of the integration. 1173 01:19:25,000 --> 01:19:27,470 So when I linearize this, what I will get 1174 01:19:27,470 --> 01:19:31,310 is something like f0 that goes on the outside. 1175 01:19:31,310 --> 01:19:37,950 I have the integral d2p2 d2b v2 minus v1. 1176 01:19:37,950 --> 01:19:43,790 And then I have something like g of p1 plus g of p2 minus 1177 01:19:43,790 --> 01:19:47,490 g of p1 prime minus g of p2 prime. 1178 01:19:53,120 --> 01:19:56,610 So basically, what we have done is 1179 01:19:56,610 --> 01:20:01,090 we have defined a linearized version of the collision 1180 01:20:01,090 --> 01:20:04,970 operator that is now linear in this variable g. 1181 01:20:11,390 --> 01:20:15,820 Now in general, this is also still, 1182 01:20:15,820 --> 01:20:19,740 although a linear operator, much simpler 1183 01:20:19,740 --> 01:20:22,400 than the previous quadratic operator-- 1184 01:20:22,400 --> 01:20:25,500 still has a lot of junk in it. 1185 01:20:25,500 --> 01:20:30,275 So we are going to simply state that I 1186 01:20:30,275 --> 01:20:34,290 will use a form of this linearized approximation that 1187 01:20:34,290 --> 01:20:38,390 is simply g over tau, get rid of all of the integration. 1188 01:20:38,390 --> 01:20:41,730 And this is called the single collision time approximation. 1189 01:20:55,910 --> 01:20:58,560 So having done that, what I have is 1190 01:20:58,560 --> 01:21:05,590 that the L acting on f0 on the left hand side 1191 01:21:05,590 --> 01:21:14,110 is minus f0 g over tau x on the right hand side. 1192 01:21:14,110 --> 01:21:17,170 And so I can immediately solve for g. 1193 01:21:17,170 --> 01:21:21,540 Because L is a first order derivative operator. 1194 01:21:21,540 --> 01:21:28,320 My g is minus tau x, the Liouville operator 1195 01:21:28,320 --> 01:21:39,230 acting on f0-- sorry, log of f0. 1196 01:21:39,230 --> 01:21:41,540 I divide it through by f0. 1197 01:21:41,540 --> 01:21:45,770 So derivative of f0 divided by f0 1198 01:21:45,770 --> 01:21:48,690 is the derivative acting on log of f0. 1199 01:21:48,690 --> 01:21:52,410 So all I need to do is to essentially do 1200 01:21:52,410 --> 01:21:56,390 the operations involved in taking these derivatives. 1201 01:21:59,300 --> 01:22:01,420 Let's say we forget about the force. 1202 01:22:01,420 --> 01:22:04,280 Because we are looking in the middle of the box, 1203 01:22:04,280 --> 01:22:08,230 acting on log of what I had written before. 1204 01:22:08,230 --> 01:22:17,260 So what I have is log of n minus p minus mu squared over 2mkT. 1205 01:22:17,260 --> 01:22:21,960 Remember, ukTn are all functions of position. 1206 01:22:21,960 --> 01:22:24,230 So there will be derivatives involved here. 1207 01:22:30,750 --> 01:22:37,150 And I will just write down what the answer is. 1208 01:22:37,150 --> 01:22:41,760 So the answer becomes minus tau x. 1209 01:22:41,760 --> 01:22:45,730 You would have, once you do all of these derivatives 1210 01:22:45,730 --> 01:22:48,070 and take advantage of the equations 1211 01:22:48,070 --> 01:22:50,630 that you have written before-- so there's 1212 01:22:50,630 --> 01:22:53,150 some lines of algebra involved. 1213 01:22:53,150 --> 01:23:07,540 The final answer is going to be nm c alpha c beta minus-- 1214 01:23:07,540 --> 01:23:10,990 I should really look at this. 1215 01:23:19,930 --> 01:23:32,580 m over kT c alpha c beta minus delta alpha beta over 3 1216 01:23:32,580 --> 01:23:40,800 c squared u alpha beta, and then mc 1217 01:23:40,800 --> 01:23:56,560 squared over 2kT minus 5/2 c alpha over T d alpha of T. Yes. 1218 01:23:56,560 --> 01:23:58,730 STUDENT: Sorry, what's the thing next to the c 1219 01:23:58,730 --> 01:24:01,290 squared, something alpha beta? 1220 01:24:01,290 --> 01:24:03,188 PROFESSOR: Delta alpha beta, sorry. 1221 01:24:07,700 --> 01:24:11,790 So there is a well-defined procedure-- 1222 01:24:11,790 --> 01:24:16,140 it's kind of algebraically involved-- by which 1223 01:24:16,140 --> 01:24:18,460 more or less in the same fashion that you 1224 01:24:18,460 --> 01:24:22,480 can improve on the algebraic solution, 1225 01:24:22,480 --> 01:24:25,000 get a better solution than the one 1226 01:24:25,000 --> 01:24:30,380 that we had that now knows something 1227 01:24:30,380 --> 01:24:33,420 about these relaxations. 1228 01:24:33,420 --> 01:24:36,890 See, the 0-th order solution that we had 1229 01:24:36,890 --> 01:24:38,560 knew nothing about tau x. 1230 01:24:38,560 --> 01:24:42,270 We just set tau x to be very small, 1231 01:24:42,270 --> 01:24:44,510 and set the right hand side to 0. 1232 01:24:44,510 --> 01:24:46,740 And then nothing relaxed. 1233 01:24:46,740 --> 01:24:51,950 Now we have a better solution that involves explicitly tau x. 1234 01:24:51,950 --> 01:24:54,490 And if we start with that, we'll find 1235 01:24:54,490 --> 01:24:58,750 that we can get relaxation of all of these modes 1236 01:24:58,750 --> 01:25:05,510 once we calculate p alpha beta and h alpha with this better 1237 01:25:05,510 --> 01:25:06,640 solution. 1238 01:25:06,640 --> 01:25:09,330 We can immediately, for example, see 1239 01:25:09,330 --> 01:25:13,020 that this new solution will have terms that are odd. 1240 01:25:13,020 --> 01:25:15,500 There is c cubed term here. 1241 01:25:15,500 --> 01:25:18,540 So when you're evaluating this average, 1242 01:25:18,540 --> 01:25:20,640 you will no longer get 0. 1243 01:25:20,640 --> 01:25:24,520 Heat has a chance to flow with this improved equation. 1244 01:25:24,520 --> 01:25:27,530 And again, whereas before our pressure 1245 01:25:27,530 --> 01:25:30,140 was diagonal because of these terms, 1246 01:25:30,140 --> 01:25:32,680 we will have off diagonal terms that 1247 01:25:32,680 --> 01:25:34,950 will allow us to relax the shear modes. 1248 01:25:34,950 --> 01:25:37,471 And we'll do that next.