1 00:00:00,070 --> 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 --> 00:00:13,340 To make a donation or view additional materials 6 00:00:13,340 --> 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 00:00:20,510 --> 00:00:21,320 PROFESSOR: OK. 9 00:00:21,320 --> 00:00:24,300 Let's start. 10 00:00:24,300 --> 00:00:26,460 So let's review what we've been trying 11 00:00:26,460 --> 00:00:29,980 to do in the past few lectures. 12 00:00:29,980 --> 00:00:33,710 It's to understand a typical situation involving something 13 00:00:33,710 --> 00:00:37,130 like a gas, which, let's say, initially 14 00:00:37,130 --> 00:00:39,770 is in one half of the container. 15 00:00:39,770 --> 00:00:45,020 And then expands until it occupies the entire container, 16 00:00:45,020 --> 00:00:49,070 going from one equilibrium state to another equilibrium state. 17 00:00:49,070 --> 00:00:52,570 Question is, how to describe this 18 00:00:52,570 --> 00:00:57,050 from the perspective of microscopic degrees of freedom, 19 00:00:57,050 --> 00:00:59,130 getting the idea that eventually you 20 00:00:59,130 --> 00:01:01,950 will come to some form of equilibrium. 21 00:01:01,950 --> 00:01:06,320 And actually, more seriously, quantifying the process 22 00:01:06,320 --> 00:01:11,040 and timescales by which the gas comes to equilibrium. 23 00:01:11,040 --> 00:01:15,560 So the first stage was to say that we will describe 24 00:01:15,560 --> 00:01:21,500 the process by looking at the density, which describes 25 00:01:21,500 --> 00:01:28,640 the number of representative examples of the system, whose 26 00:01:28,640 --> 00:01:31,540 collections of coordinates and momenta 27 00:01:31,540 --> 00:01:35,710 occupy a particular point in the 6N-dimensional phase space. 28 00:01:35,710 --> 00:01:42,210 So basically, this is a function that depends on 6N coordinates. 29 00:01:42,210 --> 00:01:45,300 And of course, it changes as a function of time. 30 00:01:45,300 --> 00:01:48,340 And we saw that, basically, if we 31 00:01:48,340 --> 00:01:53,980 were to look at how this set of points 32 00:01:53,980 --> 00:01:58,400 were streaming in phase space, and this 33 00:01:58,400 --> 00:02:01,100 can be mathematically represented 34 00:02:01,100 --> 00:02:05,350 by looking at the direct time dependence 35 00:02:05,350 --> 00:02:07,560 and the implicit time dependence of all 36 00:02:07,560 --> 00:02:12,010 of the coordinates captured through the Poisson bracket. 37 00:02:14,740 --> 00:02:17,950 And the answer was that it is a characteristic 38 00:02:17,950 --> 00:02:24,710 of the type of Hamiltonian evolution equations, 39 00:02:24,710 --> 00:02:27,590 that the flow is such that the phase space does not 40 00:02:27,590 --> 00:02:28,810 change volume. 41 00:02:28,810 --> 00:02:32,460 And hence, this quantity is 0. 42 00:02:32,460 --> 00:02:34,230 We said, OK, all of that is fine. 43 00:02:34,230 --> 00:02:39,040 But what does that tell me about how this gas is expanding? 44 00:02:39,040 --> 00:02:40,990 So for that, we said that we need 45 00:02:40,990 --> 00:02:45,590 to look at descriptions that are more appropriate to thinking 46 00:02:45,590 --> 00:02:50,310 about the volume and density of a gas expanding. 47 00:02:50,310 --> 00:02:52,770 And we really want to capture that 48 00:02:52,770 --> 00:02:55,200 through the one-particle density. 49 00:02:55,200 --> 00:02:59,380 But along the way, we introduced s particle densities, 50 00:02:59,380 --> 00:03:06,980 normalized up to something, to correspond to, essentially, 51 00:03:06,980 --> 00:03:09,810 integrating over all of the coordinates 52 00:03:09,810 --> 00:03:12,190 that we're not interested in. 53 00:03:12,190 --> 00:03:19,270 Say coordinates pertaining to numbers s plus 1 to N, 54 00:03:19,270 --> 00:03:21,850 corresponding 6N-dimensional phase 55 00:03:21,850 --> 00:03:26,470 space and the entire [INAUDIBLE] of this density or probability. 56 00:03:29,450 --> 00:03:35,460 Then the next stage was to be more specific about what 57 00:03:35,460 --> 00:03:39,230 governs the evolution of these gas particles, i.e., 58 00:03:39,230 --> 00:03:41,160 what is the Hamiltonian. 59 00:03:41,160 --> 00:03:46,590 And we said, let's focus on an N particle Hamiltonian that 60 00:03:46,590 --> 00:03:50,570 is composed, by necessity certainly, 61 00:03:50,570 --> 00:03:53,740 of one-body terms that is the kinetic energy 62 00:03:53,740 --> 00:03:58,320 and the potential energy, the latter describing the box, 63 00:03:58,320 --> 00:03:59,880 for example. 64 00:03:59,880 --> 00:04:02,580 And we added a two-particle interaction. 65 00:04:05,750 --> 00:04:09,780 qi minus qj, which, for the sake of simplicity, 66 00:04:09,780 --> 00:04:14,280 let's imagine is spherically symmetric-- 67 00:04:14,280 --> 00:04:18,790 only depends on the relative position of these particles. 68 00:04:18,790 --> 00:04:22,019 Then we say that if I look at the evolution 69 00:04:22,019 --> 00:04:25,490 of this s particle density, I start 70 00:04:25,490 --> 00:04:28,140 by writing an equation that kind of looks 71 00:04:28,140 --> 00:04:31,910 like the full Liouville equation. 72 00:04:31,910 --> 00:04:37,830 There is a Hamiltonian that describes this set of particles 73 00:04:37,830 --> 00:04:42,820 interacting among themselves, just replacing n by s. 74 00:04:42,820 --> 00:04:47,080 And so if these were the only particles in the universe, 75 00:04:47,080 --> 00:04:49,030 this is what I would get. 76 00:04:49,030 --> 00:04:52,900 Except that we know that there are other particles. 77 00:04:52,900 --> 00:04:56,940 And since we have two-body collisions possible, 78 00:04:56,940 --> 00:05:00,880 any one of the particles in the set that I'm describing here, 79 00:05:00,880 --> 00:05:03,790 composed of s particles, can potentially 80 00:05:03,790 --> 00:05:08,450 collide with a particle from this set 81 00:05:08,450 --> 00:05:10,420 that I don't want to include. 82 00:05:10,420 --> 00:05:13,900 And that will change the volume of phase space. 83 00:05:13,900 --> 00:05:17,290 This incompressibility condition was valid 84 00:05:17,290 --> 00:05:21,770 only if the entire thing of the Hamiltonian was considered. 85 00:05:21,770 --> 00:05:25,060 Here, I'm only looking at the partial subset. 86 00:05:25,060 --> 00:05:27,040 And so then what did I have here? 87 00:05:27,040 --> 00:05:31,870 I had the force exerted from particle s plus 1 88 00:05:31,870 --> 00:05:39,600 on particle n changing the momentum of the particle n. 89 00:05:39,600 --> 00:05:42,680 But the likelihood that this happens 90 00:05:42,680 --> 00:05:50,010 depends on finding the other particle also. 91 00:05:50,010 --> 00:05:56,910 So I had this dependence on the density that was s plus 1. 92 00:05:56,910 --> 00:06:00,490 So this was the BBGKY hierarchy. 93 00:06:00,490 --> 00:06:04,830 Then we said, let's take a look at the typical value 94 00:06:04,830 --> 00:06:08,050 of the terms that we have in this equation. 95 00:06:08,050 --> 00:06:15,640 And for s that is larger than 2, this Hamiltonian 96 00:06:15,640 --> 00:06:18,270 that pertains to s particles will 97 00:06:18,270 --> 00:06:21,300 have a collision term among the particles. 98 00:06:21,300 --> 00:06:25,130 And that collision term will give you some typical inverse 99 00:06:25,130 --> 00:06:30,080 timescale here, which is related to the collision time. 100 00:06:30,080 --> 00:06:34,320 Whereas on the right-hand side, for this collision 101 00:06:34,320 --> 00:06:39,430 to take place, you have to find another particle 102 00:06:39,430 --> 00:06:42,670 to interact with in the range of interactions 103 00:06:42,670 --> 00:06:44,700 that we indicated by d. 104 00:06:44,700 --> 00:06:48,450 And for density n, this was smaller 105 00:06:48,450 --> 00:06:53,850 by a factor of nd cubed for dilute gases. 106 00:06:53,850 --> 00:06:56,680 Therefore, the left-hand side is much larger 107 00:06:56,680 --> 00:06:58,290 than the right-hand side. 108 00:06:58,290 --> 00:07:01,250 And within particular approximation, 109 00:07:01,250 --> 00:07:04,500 we said that we are going to approximately set 110 00:07:04,500 --> 00:07:07,470 this right-hand side to 0. 111 00:07:07,470 --> 00:07:09,490 OK? 112 00:07:09,490 --> 00:07:12,730 Except that we couldn't do this for the equation 113 00:07:12,730 --> 00:07:20,990 that pertained to s equals to 1, because for s equals to 1, 114 00:07:20,990 --> 00:07:24,620 I did not have the analog of this term. 115 00:07:24,620 --> 00:07:30,940 If I were to write the precise form of this Poisson bracket, 116 00:07:30,940 --> 00:07:33,730 I had the partial derivative with respect 117 00:07:33,730 --> 00:07:41,110 to time, the momentum divided by mass, 118 00:07:41,110 --> 00:07:45,460 which is the velocity driving the change in coordinate. 119 00:07:45,460 --> 00:07:48,350 And the force that comes from the external potential, 120 00:07:48,350 --> 00:07:51,490 let's call it f1, we arrive in the change 121 00:07:51,490 --> 00:08:03,360 in momentum of the one-particle density, which is a function-- 122 00:08:03,360 --> 00:08:10,410 let's write it explicitly-- of p1, q1, and t. 123 00:08:10,410 --> 00:08:16,720 And on the other side, I had the possibility 124 00:08:16,720 --> 00:08:22,100 to have a term such as this that describes 125 00:08:22,100 --> 00:08:25,280 a collision with a second particle. 126 00:08:25,280 --> 00:08:27,840 I have to look at all possible second particles 127 00:08:27,840 --> 00:08:30,750 that I can collide with. 128 00:08:30,750 --> 00:08:36,182 They could come with all variety of momenta, 129 00:08:36,182 --> 00:08:38,679 which we will indicate p2. 130 00:08:38,679 --> 00:08:41,049 They can be all over the place. 131 00:08:41,049 --> 00:08:43,600 So I have integrations over space. 132 00:08:43,600 --> 00:08:46,700 And then I have a bunch of terms. 133 00:08:46,700 --> 00:08:53,440 So for this, we said, let's take a slightly better look 134 00:08:53,440 --> 00:08:59,400 at the kind of collisions that are implicitly described 135 00:08:59,400 --> 00:09:04,010 via the term that we would have here as f2. 136 00:09:04,010 --> 00:09:07,890 And since the right-hand side of f2 we set to 0, essentially, 137 00:09:07,890 --> 00:09:11,260 f2 really describes the Newtonian evolution 138 00:09:11,260 --> 00:09:14,580 of two particles coming together and having a collision. 139 00:09:17,530 --> 00:09:21,160 And this collision, I can-- really, 140 00:09:21,160 --> 00:09:23,630 the way that I have it over there 141 00:09:23,630 --> 00:09:27,510 is in what I would call a lab frame. 142 00:09:27,510 --> 00:09:30,010 And the perspective that we should have 143 00:09:30,010 --> 00:09:33,850 is that in this frame, I have a particle that 144 00:09:33,850 --> 00:09:35,800 is coming with momentum, let's say, 145 00:09:35,800 --> 00:09:41,920 p1, which is the one that I have specified 146 00:09:41,920 --> 00:09:44,240 on the left-hand side of the equation. 147 00:09:44,240 --> 00:09:46,240 That's the density that I'm following 148 00:09:46,240 --> 00:09:49,260 in the channel represented by p1. 149 00:09:49,260 --> 00:09:55,660 But then along comes a particle with momentum p2. 150 00:09:55,660 --> 00:09:59,340 And there is some place where they 151 00:09:59,340 --> 00:10:04,880 hit each other, after which this one goes off with p1 prime. 152 00:10:04,880 --> 00:10:07,970 Actually, let me call it p1 double prime. 153 00:10:07,970 --> 00:10:11,980 And this one goes off with p2 double prime. 154 00:10:14,810 --> 00:10:17,470 Now, we will find it useful to also look 155 00:10:17,470 --> 00:10:21,000 at the same thing in the center of mass frame. 156 00:10:30,480 --> 00:10:34,120 Now, in the center of mass frame-- oops-- 157 00:10:34,120 --> 00:10:44,050 my particle p1 is coming with a momentum that 158 00:10:44,050 --> 00:10:50,930 is shifted by the p of center of mass, 159 00:10:50,930 --> 00:10:53,360 which is really p1 plus p2 over 2. 160 00:10:53,360 --> 00:10:56,720 So this is p1 minus p2 over 2. 161 00:10:56,720 --> 00:11:01,180 And my particle number two comes with p2 minus center 162 00:11:01,180 --> 00:11:08,950 of mass, which is p2 minus p1 over 2. 163 00:11:08,950 --> 00:11:10,280 So they're minus each other. 164 00:11:10,280 --> 00:11:11,780 So in the center of mass frame, they 165 00:11:11,780 --> 00:11:16,270 are basically coming to each other along the same line. 166 00:11:16,270 --> 00:11:21,330 So we could define one of the axes of coordinates 167 00:11:21,330 --> 00:11:24,280 as the distance along which they are approaching. 168 00:11:24,280 --> 00:11:26,720 And let's say we put the center of mass 169 00:11:26,720 --> 00:11:30,690 at the origin, which means that I 170 00:11:30,690 --> 00:11:33,740 have two more directions to work with. 171 00:11:33,740 --> 00:11:39,660 So essentially, the coordinates of this second particle 172 00:11:39,660 --> 00:11:48,230 can be described through a vector b, 173 00:11:48,230 --> 00:11:52,380 which we call the impact parameter. 174 00:11:52,380 --> 00:11:55,290 And really, what happens to the collision 175 00:11:55,290 --> 00:12:01,460 is classically determined by how close they approach together 176 00:12:01,460 --> 00:12:03,580 to make that. 177 00:12:03,580 --> 00:12:06,650 To quantify that, we really need this impact vector. 178 00:12:06,650 --> 00:12:10,060 So b equals to 0, they are coming at each other head on. 179 00:12:10,060 --> 00:12:14,730 For a different b, they are coming at some angle. 180 00:12:14,730 --> 00:12:18,560 So really, I need to put here a d2b. 181 00:12:22,780 --> 00:12:27,690 Now, once I have specified the form of the interaction that 182 00:12:27,690 --> 00:12:32,730 is taking place when the two things come together, 183 00:12:32,730 --> 00:12:35,960 then the process is completely deterministic. 184 00:12:35,960 --> 00:12:38,670 Particle number one will come. 185 00:12:38,670 --> 00:12:43,960 And after the location at which you have the interaction, 186 00:12:43,960 --> 00:12:52,610 will go off with p1 double prime minus center of mass. 187 00:12:52,610 --> 00:12:56,210 And particle number two will come and get 188 00:12:56,210 --> 00:13:01,865 deflected to p2 prime minus-- or double prime minus p 189 00:13:01,865 --> 00:13:03,650 center of mass. 190 00:13:03,650 --> 00:13:04,150 OK? 191 00:13:07,800 --> 00:13:13,760 And again, it is important to note that this parameter 192 00:13:13,760 --> 00:13:19,560 A is irrelevant to the collision, in that once I have 193 00:13:19,560 --> 00:13:28,890 stated that before collision, I have a momentum, 194 00:13:28,890 --> 00:13:32,530 say, p1 and p2. 195 00:13:35,300 --> 00:13:40,490 And this impact parameter b is specified, 196 00:13:40,490 --> 00:13:45,850 then p1 double prime is known completely 197 00:13:45,850 --> 00:13:48,810 as a function of p1, p2, and b. 198 00:13:48,810 --> 00:13:52,600 And p2 double prime is known completely 199 00:13:52,600 --> 00:13:59,710 as a function of p1, p2, and b. 200 00:13:59,710 --> 00:14:05,030 And everything in that sense is well-defined. 201 00:14:05,030 --> 00:14:10,320 So really, this parameter A is not relevant. 202 00:14:10,320 --> 00:14:14,820 And it's really the B that is important. 203 00:14:14,820 --> 00:14:19,260 But ultimately, we also need to get these other terms, 204 00:14:19,260 --> 00:14:21,700 something that has the appropriate dimensions 205 00:14:21,700 --> 00:14:23,550 of inverse time. 206 00:14:23,550 --> 00:14:27,500 And if I have a density of particles coming 207 00:14:27,500 --> 00:14:32,520 hitting a target, how frequently they hit 208 00:14:32,520 --> 00:14:36,050 depends on the product of the density as well 209 00:14:36,050 --> 00:14:37,810 as the velocity. 210 00:14:37,810 --> 00:14:41,070 So really, I have to add here a term that 211 00:14:41,070 --> 00:14:44,100 is related to the velocity with which these things are 212 00:14:44,100 --> 00:14:45,630 approaching each other. 213 00:14:45,630 --> 00:14:51,030 And so I can write it either as v2 minus v1, or p2 minus 1 214 00:14:51,030 --> 00:14:51,690 over m. 215 00:14:54,690 --> 00:14:59,940 So that's the other thing that I have to look at. 216 00:14:59,940 --> 00:15:07,510 And then, essentially, what we saw 217 00:15:07,510 --> 00:15:11,390 last time is that the rest of the story-- if you 218 00:15:11,390 --> 00:15:13,920 sort of think of these as billiard balls, 219 00:15:13,920 --> 00:15:17,450 so that the collision is instantaneous and takes place 220 00:15:17,450 --> 00:15:20,690 at a well-specified point, instantaneously, 221 00:15:20,690 --> 00:15:23,910 at some location, you're going with some momentum, 222 00:15:23,910 --> 00:15:27,110 and then turn and go with some other momentum. 223 00:15:27,110 --> 00:15:31,430 So the channel that was carrying in momentum p1, which 224 00:15:31,430 --> 00:15:35,240 is the one that we are interested in, suddenly 225 00:15:35,240 --> 00:15:39,766 gets depleted by an amount that is related 226 00:15:39,766 --> 00:15:44,880 to the probability of having simultaneously particles 227 00:15:44,880 --> 00:15:49,580 of momenta p1 and p2. 228 00:15:53,580 --> 00:15:59,460 And the locations that I have to specify here-- well, 229 00:15:59,460 --> 00:16:03,360 again, I'm really following this channel. 230 00:16:03,360 --> 00:16:07,790 And in this channel, I have specified q1. 231 00:16:07,790 --> 00:16:12,110 And suddenly, I see that at the moment of collision, 232 00:16:12,110 --> 00:16:15,710 the momentum changes, if you are thinking about billiard balls. 233 00:16:15,710 --> 00:16:18,050 So really, it is at that location 234 00:16:18,050 --> 00:16:20,720 that I'm interested for particle one. 235 00:16:20,720 --> 00:16:24,050 For particle two, I'm interested at something 236 00:16:24,050 --> 00:16:27,580 that is, let's say, q1 plus, because it depends 237 00:16:27,580 --> 00:16:32,150 on the location of particle two shifted from particle one 238 00:16:32,150 --> 00:16:35,230 by an amount that is related by b and whatever 239 00:16:35,230 --> 00:16:38,190 this amount slightly of A is that is related 240 00:16:38,190 --> 00:16:41,531 to the size of your billiard ball. 241 00:16:41,531 --> 00:16:42,030 OK? 242 00:16:46,910 --> 00:16:52,550 Now, we said that there is a corresponding addition, which 243 00:16:52,550 --> 00:16:58,580 comes from the fact that it is possible suddenly to not 244 00:16:58,580 --> 00:17:04,140 have a particle moving along, say, this direction. 245 00:17:04,140 --> 00:17:07,240 And the collision of these two particles 246 00:17:07,240 --> 00:17:10,740 created something along that direction. 247 00:17:10,740 --> 00:17:13,119 So the same way that probabilities 248 00:17:13,119 --> 00:17:16,010 can be subtracted amongst some channels, 249 00:17:16,010 --> 00:17:18,569 they can be added to channels, because collision 250 00:17:18,569 --> 00:17:21,589 of two particles created something. 251 00:17:21,589 --> 00:17:28,900 And so I will indicate that by p1 prime, p2 prime, 252 00:17:28,900 --> 00:17:34,140 again, coordinates in t, something like this. 253 00:17:34,140 --> 00:17:38,950 So basically, I try to be a little bit more careful here, 254 00:17:38,950 --> 00:17:43,200 not calling these outgoing channels p prime, 255 00:17:43,200 --> 00:17:44,970 but p double prime. 256 00:17:44,970 --> 00:17:47,860 Because really, if I were to invert 257 00:17:47,860 --> 00:17:51,260 p1 double prime and p2 double prime, 258 00:17:51,260 --> 00:17:54,690 I will generate things that are going backward. 259 00:17:54,690 --> 00:17:58,320 So there's a couple of sign issues involved. 260 00:17:58,320 --> 00:18:02,220 But essentially, there is a similar function 261 00:18:02,220 --> 00:18:05,780 that relates p1 prime and p2 prime. 262 00:18:05,780 --> 00:18:07,900 It is, basically, you have to search out 263 00:18:07,900 --> 00:18:13,390 momenta whose outcome will create particles 264 00:18:13,390 --> 00:18:15,550 in the channels prescribed by p1 and p2. 265 00:18:19,290 --> 00:18:22,920 So as we said, up to here, we have 266 00:18:22,920 --> 00:18:26,240 done a variety of approximations. 267 00:18:26,240 --> 00:18:30,130 They're kind of physically reasonable. 268 00:18:30,130 --> 00:18:33,280 There is some difficulty here in specifying 269 00:18:33,280 --> 00:18:38,680 if I want to be very accurate where the locations of q1 270 00:18:38,680 --> 00:18:43,440 and q2 are within this interaction size 271 00:18:43,440 --> 00:18:45,760 d that I am looking at. 272 00:18:45,760 --> 00:18:48,870 And if I don't think about billiard balls, where 273 00:18:48,870 --> 00:18:51,810 the interactions are not quite instantaneous where 274 00:18:51,810 --> 00:18:56,470 you change momenta, but particles that are deformable, 275 00:18:56,470 --> 00:18:59,250 there is also certainly a certain amount of time 276 00:18:59,250 --> 00:19:01,710 where the potential is acting. 277 00:19:01,710 --> 00:19:05,530 And so exactly what time you are looking at here 278 00:19:05,530 --> 00:19:09,413 is also not appropriate-- is not completely, no. 279 00:19:12,830 --> 00:19:16,620 But if you sort of make all of those things precise, 280 00:19:16,620 --> 00:19:20,430 then my statement is that this equation still 281 00:19:20,430 --> 00:19:25,290 respects the reversibility of the equations of motions 282 00:19:25,290 --> 00:19:27,550 that we had initially. 283 00:19:27,550 --> 00:19:30,110 But at this stage, we say, well, what can 284 00:19:30,110 --> 00:19:31,950 I do with this equation? 285 00:19:31,950 --> 00:19:34,750 I really want to solve an equation that 286 00:19:34,750 --> 00:19:40,320 involves f of p1, q1, and t. 287 00:19:40,320 --> 00:19:42,310 So the best that I can do is, first 288 00:19:42,310 --> 00:19:46,300 of all, simplify all of the arguments here 289 00:19:46,300 --> 00:19:48,330 to be more or less at the same location. 290 00:19:48,330 --> 00:19:52,210 I don't want to think about one particle being shifted 291 00:19:52,210 --> 00:19:55,100 by an amount that is related by b plus something that 292 00:19:55,100 --> 00:20:00,390 is related to the interaction potential size, et cetera. 293 00:20:00,390 --> 00:20:06,020 Let's sort of change our focus, so that our picture size, 294 00:20:06,020 --> 00:20:07,790 essentially, is something that is 295 00:20:07,790 --> 00:20:12,050 of the order of the size of the interaction. 296 00:20:12,050 --> 00:20:14,910 And also, our resolution in time is 297 00:20:14,910 --> 00:20:19,630 shifted, so that we don't really ask about moments of time 298 00:20:19,630 --> 00:20:21,770 slightly before and after collision, 299 00:20:21,770 --> 00:20:26,480 where the particle, let's say, could be deformed or whatever. 300 00:20:26,480 --> 00:20:30,180 So having done that, we make these approximations 301 00:20:30,180 --> 00:20:33,780 that, essentially, all of these f2's, 302 00:20:33,780 --> 00:20:37,910 we are going to replace as a product of f1's 303 00:20:37,910 --> 00:20:41,540 evaluated all at the same location-- 304 00:20:41,540 --> 00:20:44,600 the location that is specified here, 305 00:20:44,600 --> 00:20:48,950 as well as the same time, the time that is specified there. 306 00:20:48,950 --> 00:21:15,490 So this becomes-- OK? 307 00:21:15,490 --> 00:21:20,120 So essentially, the result of the collisions 308 00:21:20,120 --> 00:21:27,790 is described on the left-hand side of our equation 309 00:21:27,790 --> 00:21:34,740 by a term that integrates over all momenta and overall impact 310 00:21:34,740 --> 00:21:39,840 parameters of a bracket, which is a product of two 311 00:21:39,840 --> 00:21:42,100 one-particle densities. 312 00:21:42,100 --> 00:21:46,080 And just to simplify notation, I will call this 313 00:21:46,080 --> 00:21:51,460 the collision term that involves two factors of f1. 314 00:21:51,460 --> 00:21:54,780 So that's just a shorthand for that. 315 00:21:54,780 --> 00:21:55,280 Yes. 316 00:21:55,280 --> 00:21:57,516 AUDIENCE: Is that process equivalent 317 00:21:57,516 --> 00:22:01,250 of a similar molecule [INAUDIBLE]? 318 00:22:01,250 --> 00:22:03,600 PROFESSOR: That, if you want to give it a name, yes. 319 00:22:03,600 --> 00:22:05,670 It's called the assumption of molecular chaos. 320 00:22:11,150 --> 00:22:16,480 And physically, it is the assumption 321 00:22:16,480 --> 00:22:20,200 that the probability to find the two particles 322 00:22:20,200 --> 00:22:23,500 is the product of one-particle probabilities. 323 00:22:23,500 --> 00:22:27,110 And again, we say that you have to worry 324 00:22:27,110 --> 00:22:30,460 about the accuracy of that if you 325 00:22:30,460 --> 00:22:34,530 want to focus on sizes that are smaller than the interaction 326 00:22:34,530 --> 00:22:36,010 parameter size. 327 00:22:36,010 --> 00:22:38,500 But we change that as part of our resolution. 328 00:22:38,500 --> 00:22:41,073 So we don't worry about that aspect. 329 00:22:41,073 --> 00:22:44,572 AUDIENCE: So molecular chaos doesn't 330 00:22:44,572 --> 00:22:49,009 include eliminating this space? 331 00:22:49,009 --> 00:22:50,488 It's a separate thing. 332 00:22:53,848 --> 00:23:00,090 PROFESSOR: I've only seen the word "molecular chaos" applied 333 00:23:00,090 --> 00:23:01,710 in this context. 334 00:23:01,710 --> 00:23:04,440 I think you are asking whether or not 335 00:23:04,440 --> 00:23:07,490 if q1 and q2 are different. 336 00:23:07,490 --> 00:23:11,060 I can replace that and call that molecular chaos. 337 00:23:11,060 --> 00:23:14,680 I would say that that's actually a correct statement. 338 00:23:14,680 --> 00:23:16,760 It's not an assumption. 339 00:23:16,760 --> 00:23:18,730 Basically, when the two particles 340 00:23:18,730 --> 00:23:22,640 are away from each other to a good approximation, 341 00:23:22,640 --> 00:23:25,910 they don't really know about-- well, actually, 342 00:23:25,910 --> 00:23:29,370 what I am saying is if I really want to sort of focus 343 00:23:29,370 --> 00:23:32,305 on time dependence, maybe that is also an assumption. 344 00:23:32,305 --> 00:23:33,730 Yes. 345 00:23:33,730 --> 00:23:38,290 So they're really, if you like, indeed, two parts. 346 00:23:38,290 --> 00:23:42,230 It is the replacing of f2 with the product of two 347 00:23:42,230 --> 00:23:48,010 f2's, and then evaluating them at the same point. 348 00:23:48,010 --> 00:23:48,510 Yes. 349 00:23:48,510 --> 00:23:50,690 AUDIENCE: But based on your first line-- I mean, 350 00:23:50,690 --> 00:23:55,140 you could break that assumption of different distribution 351 00:23:55,140 --> 00:23:58,254 functions have different events that occur when they get close, 352 00:23:58,254 --> 00:23:59,795 and the framework still works, right? 353 00:23:59,795 --> 00:24:03,090 It would just become-- when you got down to the bottom line, 354 00:24:03,090 --> 00:24:06,900 it would just become a much larger equation, right? 355 00:24:06,900 --> 00:24:08,150 PROFESSOR: I don't understand. 356 00:24:08,150 --> 00:24:10,316 AUDIENCE: Like if you assume different interactions. 357 00:24:12,810 --> 00:24:16,640 PROFESSOR: The interactions are here 358 00:24:16,640 --> 00:24:20,160 in how p1 prime and p2 prime depend on p1 and p2. 359 00:24:20,160 --> 00:24:21,146 AUDIENCE: Right. 360 00:24:21,146 --> 00:24:23,653 What I'm suggesting is if you change H 361 00:24:23,653 --> 00:24:26,650 by keeping the top [INAUDIBLE]. 362 00:24:26,650 --> 00:24:29,340 That's a pretty simple H. Would this framework 363 00:24:29,340 --> 00:24:32,799 work for a more complicated H. But then your bottom two lines 364 00:24:32,799 --> 00:24:33,840 would become [INAUDIBLE]. 365 00:24:33,840 --> 00:24:36,820 PROFESSOR: What more can I do here? 366 00:24:36,820 --> 00:24:38,225 How can I make this more general? 367 00:24:38,225 --> 00:24:39,600 Are you thinking about adding it? 368 00:24:39,600 --> 00:24:40,503 AUDIENCE: No, no, no. 369 00:24:40,503 --> 00:24:42,336 I'm not suggesting you make it more general. 370 00:24:42,336 --> 00:24:44,590 I'm suggesting you make it more specific to break 371 00:24:44,590 --> 00:24:49,130 that assumption of molecular chaos. 372 00:24:49,130 --> 00:24:50,150 PROFESSOR: OK. 373 00:24:50,150 --> 00:24:52,850 So how should I modify if I'm not saying-- 374 00:24:52,850 --> 00:24:55,174 AUDIENCE: I didn't have something in mind. 375 00:24:55,174 --> 00:24:57,058 I was just suggesting [INAUDIBLE]. 376 00:24:57,058 --> 00:24:59,540 PROFESSOR: Well, that's where I have problem, 377 00:24:59,540 --> 00:25:02,930 because it seems to me that this is very general. 378 00:25:02,930 --> 00:25:05,890 The only thing that I left out is the body 379 00:25:05,890 --> 00:25:10,400 interactions may be having to do with this being spherically 380 00:25:10,400 --> 00:25:11,450 symmetric. 381 00:25:11,450 --> 00:25:14,130 But apart from that, this is basically as general 382 00:25:14,130 --> 00:25:16,380 as you can get with two bodies. 383 00:25:16,380 --> 00:25:19,430 Now, the part that is pertaining to three bodies, 384 00:25:19,430 --> 00:25:24,000 again, you would basically-- not need to worry about as long 385 00:25:24,000 --> 00:25:26,970 as nd cubed is less than 1. 386 00:25:26,970 --> 00:25:32,515 So I don't quite know what you have in mind. 387 00:25:32,515 --> 00:25:34,393 AUDIENCE: I guess I was thinking about three- 388 00:25:34,393 --> 00:25:35,538 and four-body interactions. 389 00:25:35,538 --> 00:25:38,490 But if you're covered against that, then what I said, 390 00:25:38,490 --> 00:25:41,830 it's not valid or relevant. 391 00:25:41,830 --> 00:25:42,680 PROFESSOR: Right. 392 00:25:42,680 --> 00:25:45,970 So certainly, three-body and higher body interactions 393 00:25:45,970 --> 00:25:48,220 would be important if you're thinking 394 00:25:48,220 --> 00:25:52,790 about writing a description for a liquid, for example. 395 00:25:52,790 --> 00:25:57,440 And then the story becomes more complicated. 396 00:25:57,440 --> 00:26:00,550 You can't really make the truncations 397 00:26:00,550 --> 00:26:02,620 that would give you the Boltzmann equation. 398 00:26:02,620 --> 00:26:06,070 You need to go by some other set of approximations, 399 00:26:06,070 --> 00:26:08,160 such as the Vlasov equation, that you 400 00:26:08,160 --> 00:26:10,710 will see in the problem set. 401 00:26:10,710 --> 00:26:15,610 And the actual theory that one can write down for liquids 402 00:26:15,610 --> 00:26:17,510 is very complicated. 403 00:26:17,510 --> 00:26:20,150 So it's-- yeah. 404 00:26:20,150 --> 00:26:22,122 AUDIENCE: Can it get more complicated yet 405 00:26:22,122 --> 00:26:24,581 if, based on the same framework, isn't this similar 406 00:26:24,581 --> 00:26:26,975 to, like, the formulation of the neutron transport 407 00:26:26,975 --> 00:26:30,570 equation, where if you have, in addition to scattering, 408 00:26:30,570 --> 00:26:32,460 you have particles that don't attract, 409 00:26:32,460 --> 00:26:35,573 but you have fission events and you have inelastic center? 410 00:26:35,573 --> 00:26:36,156 PROFESSOR: OK. 411 00:26:36,156 --> 00:26:36,230 Yes. 412 00:26:36,230 --> 00:26:36,530 That's good. 413 00:26:36,530 --> 00:26:38,630 AUDIENCE: And then your H would change dramatically, right? 414 00:26:38,630 --> 00:26:39,440 PROFESSOR: Yes. 415 00:26:39,440 --> 00:26:39,726 That's right. 416 00:26:39,726 --> 00:26:39,870 Yes. 417 00:26:39,870 --> 00:26:41,860 AUDIENCE: But my point is that it's the same framework. 418 00:26:41,860 --> 00:26:42,420 PROFESSOR: Yes. 419 00:26:42,420 --> 00:26:42,919 Yes. 420 00:26:42,919 --> 00:26:44,320 So basically, you're right. 421 00:26:44,320 --> 00:26:47,650 One assumption that I have here is that the number of particles 422 00:26:47,650 --> 00:26:49,320 is conserved. 423 00:26:49,320 --> 00:26:53,010 We say that if the particles can break or diffuse, 424 00:26:53,010 --> 00:26:56,850 one has to generalize this approximation. 425 00:26:56,850 --> 00:27:00,330 And there would be an appropriate generalization 426 00:27:00,330 --> 00:27:03,470 of the Boltzmann equation to cover those terms. 427 00:27:03,470 --> 00:27:11,040 And I don't have a problem set related to that, 428 00:27:11,040 --> 00:27:14,900 but maybe you are suggesting that I should provide one. 429 00:27:14,900 --> 00:27:16,800 [LAUGHTER] 430 00:27:17,750 --> 00:27:18,400 OK. 431 00:27:18,400 --> 00:27:19,260 That's a good point. 432 00:27:21,690 --> 00:27:22,190 OK? 433 00:27:24,770 --> 00:27:25,270 All right. 434 00:27:25,270 --> 00:27:29,510 So we have made some assumptions here. 435 00:27:29,510 --> 00:27:34,090 And I guess my claim is that this equation-- well, 436 00:27:34,090 --> 00:27:35,670 what equation? 437 00:27:35,670 --> 00:27:38,880 So basically, what I'm going to do, 438 00:27:38,880 --> 00:27:41,630 I call the right-hand side of this equation 439 00:27:41,630 --> 00:27:46,090 after this assumption of molecular chaos, CFF. 440 00:27:46,090 --> 00:27:52,070 For simplicity, let me call this bunch of first derivatives 441 00:27:52,070 --> 00:27:56,380 that act on f1 L for Liouvillian. 442 00:27:56,380 --> 00:27:58,680 It's kind of like a Liouville operator. 443 00:27:58,680 --> 00:28:04,010 So I have an equation that is the bunch of first derivatives 444 00:28:04,010 --> 00:28:07,880 acting on f1, which is essentially 445 00:28:07,880 --> 00:28:12,700 a linear partial differential equation in six or seven 446 00:28:12,700 --> 00:28:15,770 dimensions, depending on how you count time-- is 447 00:28:15,770 --> 00:28:21,890 equal to a non-linear integral on the right-hand side. 448 00:28:21,890 --> 00:28:23,190 OK? 449 00:28:23,190 --> 00:28:26,370 So this is the entity that is the Boltzmann equation. 450 00:28:39,950 --> 00:28:48,710 Now, the statement is that this equation no longer 451 00:28:48,710 --> 00:28:51,220 has time reversal symmetry. 452 00:28:51,220 --> 00:28:56,420 And so if I were to solve this equation for the one function 453 00:28:56,420 --> 00:29:00,030 that I'm interested, after all, the f1, which really tells me 454 00:29:00,030 --> 00:29:03,740 how the particles stream from the left-hand side of the box 455 00:29:03,740 --> 00:29:06,480 to the right-hand side of the box, 456 00:29:06,480 --> 00:29:11,820 that equation is not reversible. 457 00:29:11,820 --> 00:29:14,480 That is, the solution to this equation 458 00:29:14,480 --> 00:29:16,840 will indeed take the density that 459 00:29:16,840 --> 00:29:18,940 was on the left-hand side of the box 460 00:29:18,940 --> 00:29:23,450 and uniformly distribute it on the two sides of the box. 461 00:29:23,450 --> 00:29:25,780 And that will stay forever. 462 00:29:25,780 --> 00:29:29,450 That's, of course, a consequence of the various approximations 463 00:29:29,450 --> 00:29:33,140 that we made here, removing the reversibility. 464 00:29:33,140 --> 00:29:34,810 OK? 465 00:29:34,810 --> 00:29:37,820 And the statement to show that mathematically 466 00:29:37,820 --> 00:29:41,280 is that if there is a function-- and now, for simplicity, 467 00:29:41,280 --> 00:29:45,750 since from now on, we don't care about f2, f3, et cetera, 468 00:29:45,750 --> 00:29:51,250 we are only caring about f1, I'm going to drop the index one 469 00:29:51,250 --> 00:29:55,980 and say that if there is a density f, which is really 470 00:29:55,980 --> 00:30:02,760 my f1, that satisfies the Boltzmann-- that is, 471 00:30:02,760 --> 00:30:10,610 I have a form such as this-- then there is a quantity H that 472 00:30:10,610 --> 00:30:15,620 only depends on time that, under the evolution that 473 00:30:15,620 --> 00:30:22,490 is implicit in this Boltzmann equation will always decrease, 474 00:30:22,490 --> 00:30:29,030 where H is an integral over the coordinates 475 00:30:29,030 --> 00:30:39,060 and momenta that are in this function f of p, q, and t. 476 00:30:39,060 --> 00:30:43,250 And all I need to do is to multiply by the log. 477 00:30:45,960 --> 00:30:50,220 And then all of these are vectors if I drop it 478 00:30:50,220 --> 00:30:52,480 for saving time. 479 00:30:52,480 --> 00:30:55,350 And we said that once we recognize 480 00:30:55,350 --> 00:30:57,660 that f, up to a normalization, is 481 00:30:57,660 --> 00:31:01,060 the same thing as the one-particle probability-- 482 00:31:01,060 --> 00:31:04,440 and having previously seen objects such as this 483 00:31:04,440 --> 00:31:06,990 in probability theory corresponding 484 00:31:06,990 --> 00:31:10,470 to entropy information, entropy of mixing, 485 00:31:10,470 --> 00:31:15,590 you will not be surprised at how this expression came about. 486 00:31:15,590 --> 00:31:19,370 And I had told you that in each one of the four sections, 487 00:31:19,370 --> 00:31:23,700 we will introduce some quantity that plays the role of entropy. 488 00:31:23,700 --> 00:31:26,680 So this is the one that is appropriate to our section 489 00:31:26,680 --> 00:31:28,880 on kinetic theory. 490 00:31:28,880 --> 00:31:30,949 So let's see how we go about-- 491 00:31:30,949 --> 00:31:31,740 AUDIENCE: Question. 492 00:31:31,740 --> 00:31:32,460 PROFESSOR: Yes? 493 00:31:32,460 --> 00:31:35,862 AUDIENCE: When you introduced the-- initialized the concept 494 00:31:35,862 --> 00:31:40,722 of entropy of information of the message that's 495 00:31:40,722 --> 00:31:42,666 p log p, [INAUDIBLE] entropy. 496 00:31:42,666 --> 00:31:44,620 And as far as I understand, this concept 497 00:31:44,620 --> 00:31:47,630 came up in, like, mid 20th century at least, right? 498 00:31:47,630 --> 00:31:50,450 PROFESSOR: I think slightly later than that. 499 00:31:50,450 --> 00:31:51,550 Of that order, yes. 500 00:31:51,550 --> 00:31:52,522 AUDIENCE: OK. 501 00:31:52,522 --> 00:31:54,813 And these are Boltzmann equations? 502 00:31:54,813 --> 00:31:55,438 PROFESSOR: Yes. 503 00:31:55,438 --> 00:31:58,578 AUDIENCE: And so this was, like, 100 years before, at least? 504 00:31:58,578 --> 00:32:00,911 PROFESSOR: Not 100 years, but maybe 60 years away, yeah. 505 00:32:00,911 --> 00:32:01,411 50, 60. 506 00:32:01,411 --> 00:32:04,131 AUDIENCE: OK. 507 00:32:04,131 --> 00:32:07,562 I'm just interested, what is the motivation for picking 508 00:32:07,562 --> 00:32:12,128 this form of functionality, that it's integral of f log f 509 00:32:12,128 --> 00:32:14,190 rather than integral of something else? 510 00:32:14,190 --> 00:32:14,950 PROFESSOR: OK. 511 00:32:14,950 --> 00:32:17,845 I mean, there is another place that you've seen it, 512 00:32:17,845 --> 00:32:22,210 that has nothing to do with information. 513 00:32:22,210 --> 00:32:24,460 And that's mixing entropy. 514 00:32:24,460 --> 00:32:27,550 If I have [? n ?] one particles of one type 515 00:32:27,550 --> 00:32:29,910 and two particles of another type, et cetera, 516 00:32:29,910 --> 00:32:33,130 and mixed them up, then the mixing entropy, 517 00:32:33,130 --> 00:32:36,780 you can relate to a form such as this. 518 00:32:36,780 --> 00:32:39,050 It is some ni log ni. 519 00:32:43,370 --> 00:32:43,870 OK? 520 00:32:47,826 --> 00:32:48,325 Fine. 521 00:32:52,370 --> 00:32:54,770 It's just for discrete variables, 522 00:32:54,770 --> 00:32:58,500 it makes much more sense than for the continuum. 523 00:32:58,500 --> 00:33:00,370 But OK. 524 00:33:00,370 --> 00:33:05,060 So let's see if this is indeed the case. 525 00:33:05,060 --> 00:33:10,550 So we say that dH by dt is the-- I 526 00:33:10,550 --> 00:33:15,330 have to take the time derivative inside. 527 00:33:15,330 --> 00:33:19,450 And as we discussed before, the full derivative 528 00:33:19,450 --> 00:33:23,210 becomes a partial derivative inside the integral. 529 00:33:23,210 --> 00:33:28,060 And I have either acting on f and acting 530 00:33:28,060 --> 00:33:30,720 on log f, which will give me 1 over f, 531 00:33:30,720 --> 00:33:33,110 canceling the f outside. 532 00:33:33,110 --> 00:33:38,660 And this term corresponded to the derivative 533 00:33:38,660 --> 00:33:41,760 of the normalization, which is fixed. 534 00:33:41,760 --> 00:33:49,215 And then we had to replace what we have for df by dt. 535 00:33:51,850 --> 00:34:00,466 And for df by dt, we either have the Poisson bracket 536 00:34:00,466 --> 00:34:06,790 of the one-particle Hamiltonian with f coming from the L 537 00:34:06,790 --> 00:34:12,000 part of the equation, once we get d by dt on one side. 538 00:34:12,000 --> 00:34:15,980 And then the collision part that depends on two f's. 539 00:34:15,980 --> 00:34:19,250 And we have to multiply this by f. 540 00:34:19,250 --> 00:34:24,920 And again, let's remind you that f really has argument-- oops, 541 00:34:24,920 --> 00:34:29,940 log f-- that has argument p1. 542 00:34:29,940 --> 00:34:31,500 I won't write q1 and t. 543 00:34:36,330 --> 00:34:40,580 And the next thing that we did at the end of last lecture 544 00:34:40,580 --> 00:34:46,889 is that typically, when you have integral of a Poisson bracket 545 00:34:46,889 --> 00:34:50,670 by doing some combination of integrations by part, 546 00:34:50,670 --> 00:34:56,360 you can even eventually show that this contribution is 0. 547 00:34:56,360 --> 00:35:00,510 And really, the important thing is the integration 548 00:35:00,510 --> 00:35:04,640 against the collision term of the Boltzmann equation. 549 00:35:04,640 --> 00:35:09,510 So writing that explicitly, what do we have? 550 00:35:09,510 --> 00:35:13,290 We have that it is integral d cubed. 551 00:35:13,290 --> 00:35:17,060 Let's say q1. 552 00:35:17,060 --> 00:35:20,290 Again, I don't really need to keep 553 00:35:20,290 --> 00:35:24,090 track of the index on this dummy variable. 554 00:35:24,090 --> 00:35:26,290 I have the integral over q1. 555 00:35:26,290 --> 00:35:28,860 I have the two integral over p1. 556 00:35:28,860 --> 00:35:32,530 Let's keep it so that it doesn't confuse anybody. 557 00:35:32,530 --> 00:35:36,040 Then I have the collision term. 558 00:35:36,040 --> 00:35:39,980 Now, the collision term is a bunch of integrals itself. 559 00:35:39,980 --> 00:35:47,530 It involves integrals over p2, integral over an impact vector 560 00:35:47,530 --> 00:35:50,070 b, the relative velocity. 561 00:35:53,940 --> 00:36:01,910 And then I had minus f of p1, f of p2 562 00:36:01,910 --> 00:36:05,990 for the loss in the channel because of the collisions, 563 00:36:05,990 --> 00:36:10,690 plus the addition to the channel from the inverse collisions. 564 00:36:13,890 --> 00:36:16,060 And this whole thing has to be multiplied 565 00:36:16,060 --> 00:36:22,041 by log f evaluated at p1. 566 00:36:22,041 --> 00:36:24,480 OK? 567 00:36:24,480 --> 00:36:28,280 So that's really the quantity that we are interested. 568 00:36:28,280 --> 00:36:31,200 And the quantity that we are interested, we notice, 569 00:36:31,200 --> 00:36:36,880 has some kind of a symmetry with respect to indices one and two. 570 00:36:36,880 --> 00:36:40,920 Something to do with prime and un-prime coordinates. 571 00:36:40,920 --> 00:36:44,980 But multiplied this bracket that has all of these symmetries, 572 00:36:44,980 --> 00:36:49,780 with this function that is only evaluated for one of the four 573 00:36:49,780 --> 00:36:54,460 coordinates for momenta that are appearing here. 574 00:36:54,460 --> 00:36:59,420 So our next set of tasks is to somehow make this as symmetric 575 00:36:59,420 --> 00:37:01,070 as possible. 576 00:37:01,070 --> 00:37:06,250 So the first thing that we do is we exchange indices one 577 00:37:06,250 --> 00:37:10,920 and two, because really, these are simply dummy integration 578 00:37:10,920 --> 00:37:12,370 variables. 579 00:37:12,370 --> 00:37:16,750 I can call the thing that I was calling p1 p2, and vice versa. 580 00:37:16,750 --> 00:37:21,380 And the integration then becomes the integral over q. 581 00:37:21,380 --> 00:37:26,290 Maybe that's why it's useful not to have, for the q, any index, 582 00:37:26,290 --> 00:37:32,110 because I don't care to change the index for that. 583 00:37:32,110 --> 00:37:34,500 d cubed p2, p1. 584 00:37:34,500 --> 00:37:38,780 Doesn't really matter in what order I write them. 585 00:37:38,780 --> 00:37:43,330 d2b is a relative separation. 586 00:37:43,330 --> 00:37:47,120 Again, it's an impact parameter. 587 00:37:47,120 --> 00:37:50,200 I can call this v1 minus v2. 588 00:37:50,200 --> 00:37:51,990 But since it's an absolute value, 589 00:37:51,990 --> 00:37:54,030 it doesn't matter really. 590 00:37:54,030 --> 00:37:56,950 And so the only thing that happens 591 00:37:56,950 --> 00:37:59,985 is that I will have-- essentially, the product is 592 00:37:59,985 --> 00:38:00,485 symmetric. 593 00:38:08,410 --> 00:38:13,970 The only thing that was separately f 594 00:38:13,970 --> 00:38:16,410 of p1 before now becomes f of p2. 595 00:38:18,980 --> 00:38:22,420 And then what I can do is I can say, OK, those are really 596 00:38:22,420 --> 00:38:27,040 two forms of the same integral, and I can average them, 597 00:38:27,040 --> 00:38:33,310 and write the average, which is 1/2 the integral d cubed q, d 598 00:38:33,310 --> 00:38:45,200 cubed p1, d cubed p2, d2 b, relative velocity, minus f 599 00:38:45,200 --> 00:38:53,700 of p1, f of p2, f of p1 prime, f of p2 prime. 600 00:38:53,700 --> 00:39:03,420 And then I have a log of f of p1 plus log of f of p2. 601 00:39:03,420 --> 00:39:07,845 And I have divided through 1/2. 602 00:39:07,845 --> 00:39:08,345 OK? 603 00:39:14,070 --> 00:39:22,200 So what I will do is to now write the next line. 604 00:39:22,200 --> 00:39:27,320 And then we'll spend some time discussing what happens. 605 00:39:27,320 --> 00:39:32,500 So artificially, what will happen looks like I essentially 606 00:39:32,500 --> 00:39:37,800 move prime and un-prime things the same way that I exchanged 607 00:39:37,800 --> 00:39:40,270 one and two indices before. 608 00:39:40,270 --> 00:39:45,010 I will exchange the superscripts that are prime or non-prime. 609 00:39:45,010 --> 00:39:51,268 And if I were to do that, I will get an integral d cubed q, 610 00:39:51,268 --> 00:39:59,980 d cubed, let's say, p1 prime, d cubed p2 prime, d2b, v2 minus 611 00:39:59,980 --> 00:40:09,740 v1 prime, minus f of p1-- [SIGHS] 612 00:40:12,710 --> 00:40:22,790 --prime, f or p2 prime, plus f of p1, f of p2. 613 00:40:22,790 --> 00:40:25,470 And then here, we would have log of f 614 00:40:25,470 --> 00:40:31,820 of p1 prime, log of f of p2 prime. 615 00:40:31,820 --> 00:40:34,180 OK? 616 00:40:34,180 --> 00:40:40,290 So this is a statement that just saying, 617 00:40:40,290 --> 00:40:43,550 I moved the primes and un-primes, 618 00:40:43,550 --> 00:40:47,080 it's very easy to come up with this answer. 619 00:40:47,080 --> 00:40:52,080 What you have to think very much about what that means. 620 00:40:52,080 --> 00:40:55,660 Essentially, what I had originally was 621 00:40:55,660 --> 00:41:02,070 integration invariables p1 and p2. 622 00:41:02,070 --> 00:41:07,090 And p1 prime and p2 prime were never integration variables. 623 00:41:07,090 --> 00:41:09,650 So there was not symmetry at the level 624 00:41:09,650 --> 00:41:13,070 that I have written between one and two, which 625 00:41:13,070 --> 00:41:15,790 are integration variables, and one prime 626 00:41:15,790 --> 00:41:19,630 and two prime, which are the quantities that, 627 00:41:19,630 --> 00:41:24,750 through the collision operator, are related to p1 and p2. 628 00:41:24,750 --> 00:41:28,920 So essentially, p1 prime and p2 prime 629 00:41:28,920 --> 00:41:33,330 are the kind of functions that we have written here. 630 00:41:33,330 --> 00:41:35,970 Here, I had p1 double prime, p2 double prime. 631 00:41:35,970 --> 00:41:40,300 But up to sort of inverting some signs, et cetera, 632 00:41:40,300 --> 00:41:45,290 this is the kind of functional relationship that you have 633 00:41:45,290 --> 00:41:50,380 between p primes and p's. 634 00:41:50,380 --> 00:41:51,810 OK? 635 00:41:51,810 --> 00:41:56,080 So now we have to worry about a number of things. 636 00:41:56,080 --> 00:41:58,180 The simplest one is actually what 637 00:41:58,180 --> 00:42:05,170 happened to this factor here-- the relative velocity. 638 00:42:05,170 --> 00:42:08,200 Now think back about collisions that you 639 00:42:08,200 --> 00:42:10,360 have in the center of mass. 640 00:42:10,360 --> 00:42:12,590 In the center of mass, two particles 641 00:42:12,590 --> 00:42:17,500 come with velocities that are opposite each other. 642 00:42:17,500 --> 00:42:21,320 They bang onto each other, go different direction. 643 00:42:21,320 --> 00:42:23,620 The magnitude of the velocities does not 644 00:42:23,620 --> 00:42:25,510 change an inelastic collision, which 645 00:42:25,510 --> 00:42:27,460 is what we are looking at. 646 00:42:27,460 --> 00:42:28,720 Right? 647 00:42:28,720 --> 00:42:34,800 So after the collision, what we have is that p2 prime minus 648 00:42:34,800 --> 00:42:38,380 p1 prime is the same thing as p1 and minus 649 00:42:38,380 --> 00:42:39,735 p2 for elastic conditions. 650 00:42:42,770 --> 00:42:45,640 So as far as this factor is concerned, 651 00:42:45,640 --> 00:42:55,180 I could very well replace that with v1 minus v2 652 00:42:55,180 --> 00:42:56,220 without the primes. 653 00:42:56,220 --> 00:42:57,835 It really doesn't make any difference. 654 00:43:01,210 --> 00:43:06,060 Another thing that you can convince yourself 655 00:43:06,060 --> 00:43:12,640 is that from the perspective of the picture that I drew, 656 00:43:12,640 --> 00:43:17,071 there was some kind of an impact vector B. 657 00:43:17,071 --> 00:43:19,060 You say, well, what happens if I look 658 00:43:19,060 --> 00:43:22,780 at from the perspective of the products of the interaction? 659 00:43:22,780 --> 00:43:24,600 So the products of the interaction 660 00:43:24,600 --> 00:43:27,270 are going in a different direction. 661 00:43:27,270 --> 00:43:30,670 I would have to redraw my coordinates that 662 00:43:30,670 --> 00:43:34,820 would correspond to A prime and B prime. 663 00:43:34,820 --> 00:43:38,380 But what you can convince yourself that essentially, 664 00:43:38,380 --> 00:43:43,200 the impact parameter, as far as these products is concerned, 665 00:43:43,200 --> 00:43:46,110 is just a rotated version of the original one 666 00:43:46,110 --> 00:43:49,070 without changing its magnitude. 667 00:43:49,070 --> 00:43:53,220 So you have also that magnitude of b prime 668 00:43:53,220 --> 00:43:55,950 is the same thing as magnitude of b. 669 00:43:55,950 --> 00:43:59,610 There's a rotation, but the impact parameter 670 00:43:59,610 --> 00:44:01,090 does not change. 671 00:44:01,090 --> 00:44:03,320 If you like, that is also related 672 00:44:03,320 --> 00:44:08,190 to conservation of energy, because if it wasn't so, 673 00:44:08,190 --> 00:44:11,670 the potential energy would be different 674 00:44:11,670 --> 00:44:13,920 right after the collisions. 675 00:44:13,920 --> 00:44:16,620 OK? 676 00:44:16,620 --> 00:44:21,215 So I really was kind of careless here. 677 00:44:21,215 --> 00:44:24,280 I should have written d2 b prime. 678 00:44:24,280 --> 00:44:26,310 But I could have written also d2b. 679 00:44:26,310 --> 00:44:27,660 It didn't make any difference. 680 00:44:27,660 --> 00:44:30,660 It's the same. 681 00:44:30,660 --> 00:44:38,860 And then, of course, these functions-- essentially, 682 00:44:38,860 --> 00:44:47,330 you can see that you say, OK, I start with, here, p1 and p2. 683 00:44:47,330 --> 00:44:50,920 And these were functions of p1 and p2. 684 00:44:50,920 --> 00:44:52,730 And here, I have reversed this. 685 00:44:52,730 --> 00:44:58,490 p1 prime and p2 prime are the integration variables. 686 00:44:58,490 --> 00:45:02,950 p1 and p2 are functions of p1 prime and p2 prime 687 00:45:02,950 --> 00:45:07,130 that, in principle, are obtained by inverting 688 00:45:07,130 --> 00:45:10,034 this set of equations. 689 00:45:10,034 --> 00:45:12,500 Right? 690 00:45:12,500 --> 00:45:15,300 But it really, again, doesn't matter, 691 00:45:15,300 --> 00:45:20,630 because once you have specified what incoming p1, p2, 692 00:45:20,630 --> 00:45:24,520 and b are, you know what the outcome of the collision is. 693 00:45:24,520 --> 00:45:26,670 So it's really the same function. 694 00:45:26,670 --> 00:45:28,750 Whether you are inverting it, you 695 00:45:28,750 --> 00:45:32,000 will, again, presume the different location 696 00:45:32,000 --> 00:45:35,080 of the arguments, but you will have the same function that 697 00:45:35,080 --> 00:45:38,500 would tell you what the inverse relationship is. 698 00:45:38,500 --> 00:45:41,120 So really, these functional forms 699 00:45:41,120 --> 00:45:45,120 are the same as the functional forms that we had before. 700 00:45:45,120 --> 00:45:48,990 And finally, I claim that in order 701 00:45:48,990 --> 00:45:52,920 to go through this change of variable, 702 00:45:52,920 --> 00:45:57,670 I have to also deal with the Jacobian of the transformation. 703 00:45:57,670 --> 00:46:02,790 But that we have the d3p1d3p2 is the same thing 704 00:46:02,790 --> 00:46:08,140 as d3p1 prime d3p2 prime. 705 00:46:08,140 --> 00:46:11,480 And that, in some sense, you can sort of also 706 00:46:11,480 --> 00:46:14,810 think about what we were doing before with the Liouville 707 00:46:14,810 --> 00:46:18,730 equation maintaining the volume of phase space. 708 00:46:18,730 --> 00:46:23,490 So I have some volume of phase space prior to the collision. 709 00:46:23,490 --> 00:46:27,730 After the collision, I would need to have the same volume. 710 00:46:27,730 --> 00:46:30,130 There is, of course, in the volumes 711 00:46:30,130 --> 00:46:33,890 that we are considering for the case of the Liouville operator, 712 00:46:33,890 --> 00:46:38,740 in addition, products of dq1, dq2. 713 00:46:38,740 --> 00:46:41,410 But you can see that all of those things 714 00:46:41,410 --> 00:46:44,780 are really put into the same volume when 715 00:46:44,780 --> 00:46:46,840 we are considering the collision. 716 00:46:46,840 --> 00:46:49,270 So really, the only thing that is left 717 00:46:49,270 --> 00:46:53,560 is that the Liouvillian preservation 718 00:46:53,560 --> 00:46:56,440 of the volume of phase space would 719 00:46:56,440 --> 00:47:00,160 say that the Jacobian here, you can also ignore. 720 00:47:00,160 --> 00:47:00,660 Yes. 721 00:47:00,660 --> 00:47:03,782 AUDIENCE: So are you assuming that this transformation 722 00:47:03,782 --> 00:47:08,246 [INAUDIBLE] is [? an article, ?] because it's [INAUDIBLE]? 723 00:47:08,246 --> 00:47:10,110 PROFESSOR: Yes. 724 00:47:10,110 --> 00:47:14,010 So everything that we do is within the framework 725 00:47:14,010 --> 00:47:18,112 of a Hamiltonian evolution equation. 726 00:47:18,112 --> 00:47:19,040 OK? 727 00:47:19,040 --> 00:47:22,850 So if you like, these kinds of things 728 00:47:22,850 --> 00:47:25,260 that I'm saying here sometimes are 729 00:47:25,260 --> 00:47:29,730 said to be consequence of microscopic reversibility, 730 00:47:29,730 --> 00:47:34,720 in that I'm sort of using, in this panel over here, 731 00:47:34,720 --> 00:47:38,300 things that are completely Newtonian and microscopic. 732 00:47:38,300 --> 00:47:45,186 The place that I use statistical argument was over here. 733 00:47:45,186 --> 00:47:46,540 OK? 734 00:47:46,540 --> 00:47:50,550 So essentially, having done that, 735 00:47:50,550 --> 00:47:55,200 you can basically also rename variables. 736 00:47:55,200 --> 00:47:59,000 Things that I was calling p1 prime I can call p1. 737 00:47:59,000 --> 00:48:05,860 It's another renaming of some variable. 738 00:48:08,570 --> 00:48:15,120 And again, it really doesn't matter which relative velocity 739 00:48:15,120 --> 00:48:15,790 I write down. 740 00:48:15,790 --> 00:48:17,630 They're all the same thing. 741 00:48:17,630 --> 00:48:19,380 So what will happen is that I will 742 00:48:19,380 --> 00:48:28,940 get minus f or p1, f of p2, plus f of p1 prime, f of p2 prime. 743 00:48:28,940 --> 00:48:37,670 And the only thing that happened is that the logs-- oops. 744 00:48:37,670 --> 00:48:39,010 I should have maintained. 745 00:48:43,735 --> 00:48:47,030 I don't know why-- where I made a mistake. 746 00:48:47,030 --> 00:48:52,080 But I should ultimately have come up with this answer. 747 00:48:55,490 --> 00:48:58,840 So what I wanted to do, and I really 748 00:48:58,840 --> 00:49:02,850 have to maybe look at it again, is 749 00:49:02,850 --> 00:49:07,920 do one more step of symmetrization. 750 00:49:07,920 --> 00:49:17,180 The answer will come down to an integral over p1 and p2, d2b, 751 00:49:17,180 --> 00:49:18,720 at relative velocity. 752 00:49:23,060 --> 00:49:31,410 We would have minus f of p1, f of p2, plus f of p1 prime, 753 00:49:31,410 --> 00:49:32,970 f of p2 prime. 754 00:49:32,970 --> 00:49:36,286 So this is our collision operator 755 00:49:36,286 --> 00:49:38,650 that we have discussed here-- the difference 756 00:49:38,650 --> 00:49:42,240 between what comes in and what comes out. 757 00:49:42,240 --> 00:49:49,660 And from the first part here, we should get log of f of p1, 758 00:49:49,660 --> 00:49:52,220 plus log of f of p2, which is the same thing 759 00:49:52,220 --> 00:49:53,475 as the log of the product. 760 00:49:56,070 --> 00:50:02,140 And if I had not made errors, the subtraction 761 00:50:02,140 --> 00:50:05,460 would have involved f of p1 prime, f of p2 prime. 762 00:50:09,890 --> 00:50:16,510 And the idea is that if you think about the function 763 00:50:16,510 --> 00:50:24,650 log of something as a function of that something, 764 00:50:24,650 --> 00:50:28,710 the log is a monotonic function. 765 00:50:28,710 --> 00:50:33,600 So that if you were to evaluate the log at two 766 00:50:33,600 --> 00:50:38,440 different points, the sign of the log 767 00:50:38,440 --> 00:50:42,070 would grow in the same way as the sign of the two 768 00:50:42,070 --> 00:50:49,190 points, which means that here, you have the difference of two 769 00:50:49,190 --> 00:50:50,470 logs. 770 00:50:50,470 --> 00:50:57,310 This is like log of s1 minus log of s2. 771 00:50:57,310 --> 00:51:02,820 And it is multiplied with minus s1 minus s2. 772 00:51:02,820 --> 00:51:05,860 So if s1 is greater than s2, log s1 773 00:51:05,860 --> 00:51:07,190 would be greater than log s2. 774 00:51:07,190 --> 00:51:12,040 If s1 is less than s2, log s1 would be less than log s2. 775 00:51:12,040 --> 00:51:15,050 In either case, what you would find 776 00:51:15,050 --> 00:51:18,830 is that the product of the two brackets is negative. 777 00:51:18,830 --> 00:51:20,570 And this whole thing has to be negative. 778 00:51:23,870 --> 00:51:24,370 OK. 779 00:51:24,370 --> 00:51:31,020 So the thing that confused me-- and maybe we can go back 780 00:51:31,020 --> 00:51:40,130 and look at is that probably, what I was doing here 781 00:51:40,130 --> 00:51:47,510 was a little bit too impatient, because when 782 00:51:47,510 --> 00:51:57,640 I changed variables, p1 prime f of p2. 783 00:52:00,310 --> 00:52:01,580 I removed the primes. 784 00:52:06,416 --> 00:52:06,915 Log. 785 00:52:09,726 --> 00:52:10,350 Reduced primes. 786 00:52:14,430 --> 00:52:15,785 Actually, I had it here. 787 00:52:15,785 --> 00:52:17,831 I don't know why I was worried. 788 00:52:17,831 --> 00:52:18,330 Yeah. 789 00:52:18,330 --> 00:52:22,570 So the signs are, I think, right, that when I add them, 790 00:52:22,570 --> 00:52:24,770 I would get the subtraction. 791 00:52:24,770 --> 00:52:32,640 So in one case, I had the f's of the p's with the negative sign. 792 00:52:32,640 --> 00:52:38,410 After I did this change, they appear with the positive sign. 793 00:52:38,410 --> 00:52:43,540 So there is the sign difference between the two of them. 794 00:52:43,540 --> 00:52:45,380 OK? 795 00:52:45,380 --> 00:52:49,230 So we have a function, therefore, 796 00:52:49,230 --> 00:52:52,330 that, just like entropy throughout the process, 797 00:52:52,330 --> 00:52:53,570 will increase. 798 00:52:53,570 --> 00:52:56,820 This is like a negative entropy. 799 00:52:56,820 --> 00:53:00,630 If I were to really solve this set of equations, 800 00:53:00,630 --> 00:53:03,305 starting with the f that describes things that 801 00:53:03,305 --> 00:53:06,200 are distributed in this box, and then 802 00:53:06,200 --> 00:53:09,600 allow it to expand into the other box, 803 00:53:09,600 --> 00:53:12,380 it will follow-- this solution will 804 00:53:12,380 --> 00:53:17,910 follow some particular trajectory that will ultimately 805 00:53:17,910 --> 00:53:21,050 no longer change as a function of time. 806 00:53:21,050 --> 00:53:22,840 It will not go back and forth. 807 00:53:22,840 --> 00:53:26,360 It will not have the full reversibility 808 00:53:26,360 --> 00:53:30,490 that these set of equations does have. 809 00:53:30,490 --> 00:53:32,780 Now, that's actually not a bad thing. 810 00:53:32,780 --> 00:53:37,690 It is true, indeed, that these equations are reversible. 811 00:53:37,690 --> 00:53:41,730 And there is a theorem that if you wait sufficiently long, 812 00:53:41,730 --> 00:53:44,960 this will go back into here. 813 00:53:44,960 --> 00:53:48,130 But the time it take for that to happen 814 00:53:48,130 --> 00:53:53,085 grows something of the order of the size of the system divided 815 00:53:53,085 --> 00:53:53,585 these. 816 00:53:56,100 --> 00:53:58,940 The time, up to various pre-factors, 817 00:53:58,940 --> 00:54:02,630 would grow exponentially with the number of particles 818 00:54:02,630 --> 00:54:04,490 that you have. 819 00:54:04,490 --> 00:54:07,390 And remember that you have of the order of 10 820 00:54:07,390 --> 00:54:10,050 to the 23 particles. 821 00:54:10,050 --> 00:54:13,290 So there is, indeed, mathematically rigorously, 822 00:54:13,290 --> 00:54:19,510 a recursion time if you really had this box sitting in vacuum 823 00:54:19,510 --> 00:54:23,420 and there was no influence on the walls of the container. 824 00:54:23,420 --> 00:54:27,760 After many, many, many, many ages of the universe, 825 00:54:27,760 --> 00:54:31,270 the gas would go back for an instant of time 826 00:54:31,270 --> 00:54:34,120 in the other box, and then would go over there. 827 00:54:34,120 --> 00:54:39,690 But that's something that really is of no practical relevance 828 00:54:39,690 --> 00:54:44,310 when we are thinking about the gas and how it expands. 829 00:54:44,310 --> 00:54:45,500 OK? 830 00:54:45,500 --> 00:54:49,210 But now let's see, with these approximations, 831 00:54:49,210 --> 00:54:51,140 can we do better. 832 00:54:51,140 --> 00:54:53,840 Can we figure out exactly how long does it 833 00:54:53,840 --> 00:54:58,180 take for the gas to go from one side to the other side. 834 00:54:58,180 --> 00:55:01,670 And what is the shape of the streamlines and everything else 835 00:55:01,670 --> 00:55:04,550 as this process is taking place? 836 00:55:04,550 --> 00:55:06,704 Yes. 837 00:55:06,704 --> 00:55:07,620 AUDIENCE: [INAUDIBLE]. 838 00:55:23,800 --> 00:55:25,580 PROFESSOR: Yes. 839 00:55:25,580 --> 00:55:26,080 Yes. 840 00:55:26,080 --> 00:55:30,380 So if you like, we want to think of this 841 00:55:30,380 --> 00:55:32,835 in terms of information, in terms of entropy. 842 00:55:32,835 --> 00:55:33,730 It doesn't matter. 843 00:55:33,730 --> 00:55:37,610 It is something that is changing as a function of time. 844 00:55:37,610 --> 00:55:40,620 And the timescale by which that happens 845 00:55:40,620 --> 00:55:45,200 has to do with the timescales that we set up and imprinted 846 00:55:45,200 --> 00:55:47,017 into this equation. 847 00:55:47,017 --> 00:55:48,440 OK? 848 00:55:48,440 --> 00:55:51,410 So the next step that I have to answer 849 00:55:51,410 --> 00:55:54,260 is, what is that time step? 850 00:55:54,260 --> 00:55:58,270 And in particular, if I look at this equation by itself, 851 00:55:58,270 --> 00:56:01,330 I would say that it has two timescales. 852 00:56:01,330 --> 00:56:04,310 On the left-hand side, you have this tau 853 00:56:04,310 --> 00:56:06,990 that has to do with the size of the box. 854 00:56:06,990 --> 00:56:09,350 We said it's very long. 855 00:56:09,350 --> 00:56:13,100 On the right-hand side, it has the time 856 00:56:13,100 --> 00:56:15,290 that I have to wait for a collision 857 00:56:15,290 --> 00:56:20,590 to occur, which is certainly much shorter than this time, 858 00:56:20,590 --> 00:56:23,170 for an ordinary gas. 859 00:56:23,170 --> 00:56:27,260 And so we expect, therefore, that the right-hand side 860 00:56:27,260 --> 00:56:31,800 of this equation would determine things much more rapidly 861 00:56:31,800 --> 00:56:35,040 compared to the left-hand side of the equation. 862 00:56:35,040 --> 00:56:39,700 And we want to now quantify that and make that 863 00:56:39,700 --> 00:56:42,610 into something that is relevant to things 864 00:56:42,610 --> 00:56:47,670 that we know and understand about a gas expand. 865 00:56:47,670 --> 00:56:49,480 OK? 866 00:56:49,480 --> 00:56:55,350 So let's sort of think about, well, what would happen? 867 00:56:55,350 --> 00:57:01,070 We said that this H, as a function of time, 868 00:57:01,070 --> 00:57:04,140 is going in one direction. 869 00:57:04,140 --> 00:57:08,060 Presumably, after some time-- we don't know how long here, 870 00:57:08,060 --> 00:57:10,440 and that's a good question-- eventually, 871 00:57:10,440 --> 00:57:12,500 hopefully, it will reach an equilibrium, 872 00:57:12,500 --> 00:57:15,320 like, the entropy will cease to change. 873 00:57:15,320 --> 00:57:19,175 And the question is, what is this equilibrium? 874 00:57:19,175 --> 00:57:21,800 So let's see if we can gain some information about equilibrium. 875 00:57:30,380 --> 00:57:34,210 So we said that dH by dt presumably 876 00:57:34,210 --> 00:57:38,330 has to be 0, because if it is not 0, 877 00:57:38,330 --> 00:57:40,650 then it can continue to decrease. 878 00:57:40,650 --> 00:57:44,920 You are not quite where you want to be. 879 00:57:44,920 --> 00:57:50,630 Now let's look at what is the condition for dH by dt to be 0. 880 00:57:50,630 --> 00:57:56,050 H is this integral that I have to perform over 881 00:57:56,050 --> 00:58:00,010 the entire phase space. 882 00:58:00,010 --> 00:58:04,050 Now, each element of that integral evaluated 883 00:58:04,050 --> 00:58:07,200 at some particular location in phase space 884 00:58:07,200 --> 00:58:09,090 is by itself positive. 885 00:58:09,090 --> 00:58:12,070 This argument that we had before is not 886 00:58:12,070 --> 00:58:13,800 about the entire integral. 887 00:58:13,800 --> 00:58:17,010 It's valid about every little piece of the integration 888 00:58:17,010 --> 00:58:18,340 that I am making. 889 00:58:18,340 --> 00:58:20,830 It has a particular sign. 890 00:58:20,830 --> 00:58:26,530 So if the entire integral has to be 0, every little piece of it 891 00:58:26,530 --> 00:58:29,770 has to be individually 0. 892 00:58:29,770 --> 00:58:32,310 So that means that I would require 893 00:58:32,310 --> 00:58:39,190 that log of f at any q evaluated, 894 00:58:39,190 --> 00:58:47,680 let's say, for p1 and p2 should be the same thing 895 00:58:47,680 --> 00:58:53,230 as log of f evaluated at p1 prime and p2. 896 00:58:58,490 --> 00:59:03,680 This would be true for all q. 897 00:59:03,680 --> 00:59:04,180 Right? 898 00:59:09,040 --> 00:59:11,630 You look at this and you say, well, what? 899 00:59:11,630 --> 00:59:14,560 This seems kind of-- yes? 900 00:59:14,560 --> 00:59:17,015 AUDIENCE: What about the other one in parentheses, 901 00:59:17,015 --> 00:59:19,500 beside the integral? 902 00:59:19,500 --> 00:59:21,350 PROFESSOR: It is the same thing. 903 00:59:21,350 --> 00:59:24,300 So this is s1 minus s2. 904 00:59:24,300 --> 00:59:26,700 And the other is log of s1 minus s2. 905 00:59:26,700 --> 00:59:30,823 So I'm saying that the only time it is 0 is s1 is equal to s2. 906 00:59:34,447 --> 00:59:38,360 When s1 equals to s2, both parentheses are 0. 907 00:59:38,360 --> 00:59:41,000 Although, even if it wasn't, it would have been sufficient 908 00:59:41,000 --> 00:59:42,620 for one parentheses to be 0. 909 00:59:42,620 --> 00:59:48,210 But this is necessary, because the sign is determined by this. 910 00:59:48,210 --> 00:59:49,566 OK? 911 00:59:49,566 --> 00:59:51,450 Now, you look at this and you say, well, 912 00:59:51,450 --> 00:59:53,780 what the he-- what does this mean? 913 00:59:53,780 --> 00:59:59,220 How can I-- and the functions p1 prime and p2 prime I 914 00:59:59,220 --> 01:00:02,300 have to get by integrating Newton's equations. 915 01:00:02,300 --> 01:00:04,920 I have no idea how complicated they are. 916 01:00:04,920 --> 01:00:08,670 And you want me to solve this equation. 917 01:00:08,670 --> 01:00:10,410 Well, I tell you that the answer to this 918 01:00:10,410 --> 01:00:12,580 is actually very simple. 919 01:00:12,580 --> 01:00:17,080 So let me write the answer to it in stages. 920 01:00:17,080 --> 01:00:21,615 Log of f of p and q. 921 01:00:21,615 --> 01:00:24,390 So it's a function of p and q that I'm 922 01:00:24,390 --> 01:00:28,090 after that have to be evaluated at p1, p2, 923 01:00:28,090 --> 01:00:31,320 and then should be equal to p1 prime and p2 prime 924 01:00:31,320 --> 01:00:35,702 for any combination of p1, p2, et cetera, that I choose. 925 01:00:35,702 --> 01:00:41,980 Well, I say, OK, I will put any function of q that I like. 926 01:00:41,980 --> 01:00:46,810 And that will work, because it's independent of p. 927 01:00:46,810 --> 01:00:48,940 So I have minus alpha, minus alpha 928 01:00:48,940 --> 01:00:52,030 is the same as minus alpha, minus alpha. 929 01:00:52,030 --> 01:00:52,890 OK? 930 01:00:52,890 --> 01:00:57,800 Then I claim that, OK, I will put a minus gamma dot p. 931 01:00:57,800 --> 01:01:00,360 Gamma could be dependent on q. 932 01:01:00,360 --> 01:01:02,720 Dot p. 933 01:01:02,720 --> 01:01:07,590 And I claim that that would work, because that term would 934 01:01:07,590 --> 01:01:11,555 give me gamma dot p1 plus p2, gamma 935 01:01:11,555 --> 01:01:14,550 dot p1 prime plus p2 prime. 936 01:01:14,550 --> 01:01:16,150 What do I know for sure? 937 01:01:16,150 --> 01:01:18,240 Momentum is conserved. 938 01:01:18,240 --> 01:01:24,280 p1 plus p2 is the same thing as p1 prime plus p2 prime. 939 01:01:24,280 --> 01:01:26,760 It's just the conservation of momentum. 940 01:01:26,760 --> 01:01:30,670 I know that to be true for sure. 941 01:01:30,670 --> 01:01:34,790 And finally, kinetic energy is conserved. 942 01:01:34,790 --> 01:01:39,440 Once I'm away from the location where the interaction is taking 943 01:01:39,440 --> 01:01:43,580 place, I can choose any function of q 944 01:01:43,580 --> 01:01:46,970 multiplied with p squared over 2m. 945 01:01:46,970 --> 01:01:48,820 And then on the left-hand side, I 946 01:01:48,820 --> 01:01:51,910 would have the sum of the incoming kinetic energies. 947 01:01:51,910 --> 01:01:53,410 On the right-hand side, I would have 948 01:01:53,410 --> 01:01:56,740 the sum of outgoing kinetic energies. 949 01:01:56,740 --> 01:01:58,800 So what have I done? 950 01:01:58,800 --> 01:02:02,090 I have identified quantities that 951 01:02:02,090 --> 01:02:04,260 are conserved in the collision. 952 01:02:04,260 --> 01:02:06,960 So basically, I can keep going if I 953 01:02:06,960 --> 01:02:10,290 have additional such things. 954 01:02:10,290 --> 01:02:13,640 And so basically, the key to this 955 01:02:13,640 --> 01:02:18,020 is to identify collision-conserved quantities. 956 01:02:21,440 --> 01:02:26,030 And so momentum, energy, the first one corresponds 957 01:02:26,030 --> 01:02:27,970 to the number of particles. 958 01:02:27,970 --> 01:02:30,740 These are quantities that are conserved. 959 01:02:30,740 --> 01:02:33,610 And so you know that this form will work, 960 01:02:33,610 --> 01:02:38,450 which means that I have a candidate for equilibrium, 961 01:02:38,450 --> 01:02:43,580 which is that f of p and q should have a form that is 962 01:02:43,580 --> 01:02:49,540 exponential of some function of q that I don't-- at this stage, 963 01:02:49,540 --> 01:02:51,830 could be quite arbitrary. 964 01:02:51,830 --> 01:02:59,150 And some other function of q times p squared minus 2m. 965 01:02:59,150 --> 01:03:03,750 And let me absorb this other gamma term. 966 01:03:03,750 --> 01:03:06,690 I can certainly do so by putting something 967 01:03:06,690 --> 01:03:11,400 like pi of 1 squared over 2m. 968 01:03:15,900 --> 01:03:18,726 OK? 969 01:03:18,726 --> 01:03:23,030 And this certainly is a form that 970 01:03:23,030 --> 01:03:31,660 will set the right-hand side of the H theorem to 0. 971 01:03:31,660 --> 01:03:36,130 And furthermore, it sets the right-hand side 972 01:03:36,130 --> 01:03:39,220 of the Boltzmann equation to 0, because, again, 973 01:03:39,220 --> 01:03:40,840 for the Boltzmann equation, all I 974 01:03:40,840 --> 01:03:43,190 needed was that the product of two f's 975 01:03:43,190 --> 01:03:46,590 should be the same before and after the collision. 976 01:03:46,590 --> 01:03:49,030 And this is just the logs. 977 01:03:49,030 --> 01:03:53,300 If I were to exponentiate that, I have exactly what I need. 978 01:03:53,300 --> 01:03:57,690 So basically, this also sets the right-hand side 979 01:03:57,690 --> 01:04:00,000 of the Boltzmann equation to 0. 980 01:04:00,000 --> 01:04:02,650 And this is what I will call a local equilibrium. 981 01:04:08,420 --> 01:04:10,600 What do I mean by that? 982 01:04:10,600 --> 01:04:13,250 Essentially, we can see that right now, I 983 01:04:13,250 --> 01:04:16,510 have no knowledge about the relationship 984 01:04:16,510 --> 01:04:18,920 between different points in space, 985 01:04:18,920 --> 01:04:22,700 because these alpha, beta, and pi are completely arbitrary, as 986 01:04:22,700 --> 01:04:26,770 far as my argumentation so far is concerned. 987 01:04:26,770 --> 01:04:33,280 So locally, at each point in q, I can have a form such as this. 988 01:04:33,280 --> 01:04:36,480 And this form, you should remind you 989 01:04:36,480 --> 01:04:40,750 it's something like a Boltzmann weight of the kinetic energy, 990 01:04:40,750 --> 01:04:44,180 but moving in some particular direction. 991 01:04:44,180 --> 01:04:49,950 And essentially, what this captures 992 01:04:49,950 --> 01:04:54,220 is that through relaxing the right-hand side 993 01:04:54,220 --> 01:04:57,930 of the Boltzmann equation, we randomize 994 01:04:57,930 --> 01:05:03,170 the magnitude of the momenta, so that the magnitudes 995 01:05:03,170 --> 01:05:08,890 of the kinetic energy are kind of Boltzmann-distributed. 996 01:05:08,890 --> 01:05:11,910 This is what this term does. 997 01:05:11,910 --> 01:05:19,990 If I say that this term being much larger than this term 998 01:05:19,990 --> 01:05:21,440 is the more important one. 999 01:05:21,440 --> 01:05:24,020 If I need neglect this, then this is what happens. 1000 01:05:24,020 --> 01:05:27,620 I will very rapidly reach a situation 1001 01:05:27,620 --> 01:05:31,005 in which the momenta have been randomized 1002 01:05:31,005 --> 01:05:32,410 through the collisions. 1003 01:05:32,410 --> 01:05:35,040 Essentially, the collisions, they 1004 01:05:35,040 --> 01:05:39,550 change the direction of the momenta. 1005 01:05:39,550 --> 01:05:43,880 They preserve the average of the momenta. 1006 01:05:43,880 --> 01:05:45,940 So there is some sense of the momentum 1007 01:05:45,940 --> 01:05:49,660 of the incoming thing that is left over still. 1008 01:05:49,660 --> 01:05:53,290 But there is some relaxation that took place. 1009 01:05:53,290 --> 01:05:56,450 But clearly, this is not the end of the story, 1010 01:05:56,450 --> 01:05:58,940 because this f that I have written here-- 1011 01:05:58,940 --> 01:06:01,650 let's say local equilibrium. 1012 01:06:01,650 --> 01:06:11,464 f local equilibrium does not satisfy the Boltzmann equation. 1013 01:06:11,464 --> 01:06:13,080 Right? 1014 01:06:13,080 --> 01:06:17,490 It set the right-hand side of the Boltzmann equation to 0. 1015 01:06:17,490 --> 01:06:19,470 But it doesn't set the left-hand side. 1016 01:06:19,470 --> 01:06:23,990 The left-hand side completely is unhappy with this. 1017 01:06:23,990 --> 01:06:27,260 For the left-hand side to be 0, I 1018 01:06:27,260 --> 01:06:32,640 would require the Poisson bracket of H1 and f to be 0. 1019 01:06:32,640 --> 01:06:35,300 After all, the Liouvillian operator 1020 01:06:35,300 --> 01:06:38,690 had the Poisson bracket of H1 and f. 1021 01:06:38,690 --> 01:06:41,310 And we have seen functions that satisfy 1022 01:06:41,310 --> 01:06:49,500 this have to be of the form f1 of H. And in our case, 1023 01:06:49,500 --> 01:06:54,240 my f1 is simply the kinetic energy of one particle, 1024 01:06:54,240 --> 01:06:56,385 plus the potential due to the box. 1025 01:06:59,390 --> 01:07:01,560 OK? 1026 01:07:01,560 --> 01:07:08,640 So you can see that the only way that I can make this marriage 1027 01:07:08,640 --> 01:07:18,670 with this is to have a global equilibrium, where 1028 01:07:18,670 --> 01:07:28,390 f for the global equilibrium is proportional, let's say, 1029 01:07:28,390 --> 01:07:35,200 to exponential of minus beta p squared over 2m plus U of q. 1030 01:07:39,592 --> 01:07:41,060 OK? 1031 01:07:41,060 --> 01:07:47,430 So I identify a piece that looks like this p squared over 2m. 1032 01:07:47,430 --> 01:07:51,740 I have to make pi to be 0 in order to achieve this. 1033 01:07:51,740 --> 01:07:54,940 I have to make beta to be independent of q. 1034 01:07:54,940 --> 01:07:58,120 I have to make alpha of f marriage with beta 1035 01:07:58,120 --> 01:08:00,870 to give me the potential. 1036 01:08:00,870 --> 01:08:02,440 OK? 1037 01:08:02,440 --> 01:08:06,240 So essentially, what happens is this. 1038 01:08:06,240 --> 01:08:11,400 I open this hole. 1039 01:08:11,400 --> 01:08:15,120 The gas starts to stream out. 1040 01:08:15,120 --> 01:08:18,899 And then the particles will collide with each other. 1041 01:08:18,899 --> 01:08:21,910 The collision of particles over some timescale that 1042 01:08:21,910 --> 01:08:24,470 is related to this collision time 1043 01:08:24,470 --> 01:08:30,470 will randomize their momenta so that locally, suddenly, 1044 01:08:30,470 --> 01:08:35,420 I have a solution such as this. 1045 01:08:35,420 --> 01:08:39,859 But then the left-hand side of the Boltzmann equation 1046 01:08:39,859 --> 01:08:41,440 takes over. 1047 01:08:41,440 --> 01:08:44,220 And over a timescale that is much longer, 1048 01:08:44,220 --> 01:08:47,740 it will then change the parameters of this. 1049 01:08:47,740 --> 01:08:51,109 So basically, these parameters should really 1050 01:08:51,109 --> 01:08:56,210 be thought of as functions of time, whose evolution changes. 1051 01:08:56,210 --> 01:09:00,479 So initially, maybe locally over here, 1052 01:09:00,479 --> 01:09:04,080 there is a stream velocity that goes towards the right. 1053 01:09:04,080 --> 01:09:07,220 So I should have an average at that time that 1054 01:09:07,220 --> 01:09:09,210 prefers to go to the right. 1055 01:09:09,210 --> 01:09:11,049 I wait sufficiently long. 1056 01:09:11,049 --> 01:09:13,090 Presumably, the whole thing comes to equilibrium. 1057 01:09:13,090 --> 01:09:16,120 And that average goes to 0 [INAUDIBLE]. 1058 01:09:16,120 --> 01:09:18,790 So presumably, the next thing that I 1059 01:09:18,790 --> 01:09:22,020 need to do in order to answer the question, 1060 01:09:22,020 --> 01:09:24,439 how does this thing come to equilibrium, 1061 01:09:24,439 --> 01:09:29,510 is to find out how the left-hand side of the equation 1062 01:09:29,510 --> 01:09:33,819 manipulates these parameters alpha, beta, and pi. 1063 01:09:33,819 --> 01:09:36,910 But that's not quite consistent either. 1064 01:09:36,910 --> 01:09:40,620 Because I first assume that the right-hand side is 0, 1065 01:09:40,620 --> 01:09:43,569 and then said, oh, that's not a good solution. 1066 01:09:43,569 --> 01:09:46,300 Let's go and see what the left-hand side does. 1067 01:09:46,300 --> 01:09:49,120 But then if the left-hand side is changing things, 1068 01:09:49,120 --> 01:09:53,279 I can't be over here where dH by dt is 0. 1069 01:09:53,279 --> 01:09:56,240 So I have not done things consistently yet, 1070 01:09:56,240 --> 01:09:59,480 although I have clearly captured the whole lot of what 1071 01:09:59,480 --> 01:10:02,620 I want to describe eventually. 1072 01:10:02,620 --> 01:10:03,600 OK? 1073 01:10:03,600 --> 01:10:05,971 So let's see how we can do it systematically. 1074 01:10:05,971 --> 01:10:06,762 AUDIENCE: Question. 1075 01:10:06,762 --> 01:10:07,600 PROFESSOR: Yes. 1076 01:10:07,600 --> 01:10:13,420 AUDIENCE: When you [? write ?] the condition for [INAUDIBLE], 1077 01:10:13,420 --> 01:10:15,845 [? did you ?] write f1 of Hamiltonian. 1078 01:10:15,845 --> 01:10:17,785 And ends up Hamiltonian [INAUDIBLE] 1079 01:10:17,785 --> 01:10:19,105 includes interaction terms. 1080 01:10:19,105 --> 01:10:20,980 PROFESSOR: It's the one-particle Hamiltonian. 1081 01:10:20,980 --> 01:10:21,924 AUDIENCE: Oh. 1082 01:10:21,924 --> 01:10:22,870 PROFESSOR: Right. 1083 01:10:22,870 --> 01:10:27,840 So by the time I got to the equation for f1, I have H1. 1084 01:10:27,840 --> 01:10:31,490 H1 does not have a second particle in it to collide with. 1085 01:10:36,290 --> 01:10:39,105 OK? 1086 01:10:39,105 --> 01:10:42,360 AUDIENCE: Then equilibrium system's distribution 1087 01:10:42,360 --> 01:10:44,760 of particles in container does not 1088 01:10:44,760 --> 01:10:50,418 depend on their interactions after the [INAUDIBLE]? 1089 01:10:50,418 --> 01:10:53,681 PROFESSOR: In the story that we are following here, yes, it 1090 01:10:53,681 --> 01:10:54,180 does not. 1091 01:10:54,180 --> 01:10:57,810 So this is an approximation that actually 1092 01:10:57,810 --> 01:11:00,680 goes with the approximations that we 1093 01:11:00,680 --> 01:11:04,410 made in neglecting some of the higher order terms, with nd 1094 01:11:04,410 --> 01:11:07,310 cubed being much less than 1, and not 1095 01:11:07,310 --> 01:11:10,710 looking at things at resolution that is of the order of d. 1096 01:11:10,710 --> 01:11:13,740 And we would imagine that it is only at the resolutions-- 1097 01:11:13,740 --> 01:11:17,930 AUDIENCE: [INAUDIBLE] neglected [INAUDIBLE] atomized gas 1098 01:11:17,930 --> 01:11:19,842 conjoined onto the molecules. 1099 01:11:19,842 --> 01:11:20,798 Something like that. 1100 01:11:20,798 --> 01:11:23,830 PROFESSOR: That would require a different description. 1101 01:11:23,830 --> 01:11:24,658 AUDIENCE: Yeah. 1102 01:11:24,658 --> 01:11:25,486 It would require [INAUDIBLE]. 1103 01:11:25,486 --> 01:11:26,152 PROFESSOR: Yeah. 1104 01:11:26,152 --> 01:11:27,080 So indeed. 1105 01:11:27,080 --> 01:11:30,910 Eventually, in the description that we are going to get, 1106 01:11:30,910 --> 01:11:33,400 we are looking at the distribution 1107 01:11:33,400 --> 01:11:35,720 that would be appropriate to a dilute gas 1108 01:11:35,720 --> 01:11:38,251 and which interactions are not seen. 1109 01:11:41,620 --> 01:11:43,800 OK? 1110 01:11:43,800 --> 01:11:44,300 All right. 1111 01:11:44,300 --> 01:11:46,220 So let's see how we can do it better. 1112 01:11:46,220 --> 01:11:49,170 So clearly, the key to the process 1113 01:11:49,170 --> 01:11:56,620 is that collisions that take place frequently at timescale 1114 01:11:56,620 --> 01:12:00,240 tau x randomize a lot of things. 1115 01:12:00,240 --> 01:12:06,790 But they cannot randomize quantities that are conserved 1116 01:12:06,790 --> 01:12:07,870 in collisions. 1117 01:12:07,870 --> 01:12:11,057 So those are the quantities that we have to look at. 1118 01:12:11,057 --> 01:12:13,140 And again, these are quantities that are conserved 1119 01:12:13,140 --> 01:12:16,060 in collision, as opposed to quantities 1120 01:12:16,060 --> 01:12:20,270 that are conserved according to H1. 1121 01:12:20,270 --> 01:12:24,390 Like, momentum is conserved in the collision. 1122 01:12:24,390 --> 01:12:26,720 Clearly, momentum was not conserved 1123 01:12:26,720 --> 01:12:30,400 in H1 because of the size of the box 1124 01:12:30,400 --> 01:12:34,400 and did not appear in the eventual solution. 1125 01:12:34,400 --> 01:12:35,340 OK. 1126 01:12:35,340 --> 01:12:36,797 Let's see what we get. 1127 01:12:47,030 --> 01:12:51,100 So we are going to make a couple of definitions. 1128 01:12:51,100 --> 01:12:53,495 First, we define a local density. 1129 01:13:01,000 --> 01:13:04,560 I say that I take my solution to Boltzmann equation. 1130 01:13:04,560 --> 01:13:06,170 And now I don't know what it is. 1131 01:13:06,170 --> 01:13:09,070 But whatever it is, I take the solution 1132 01:13:09,070 --> 01:13:12,520 to the Boltzmann equation as a function of p and q and t. 1133 01:13:12,520 --> 01:13:16,120 But all I am interested is, at this particular location, 1134 01:13:16,120 --> 01:13:19,400 have the particles arrived yet, what's the density. 1135 01:13:19,400 --> 01:13:24,720 So to get that, I essentially integrate 1136 01:13:24,720 --> 01:13:27,760 over the momenta, which is the quantity that I am not 1137 01:13:27,760 --> 01:13:30,390 interested. 1138 01:13:30,390 --> 01:13:30,890 OK? 1139 01:13:30,890 --> 01:13:33,050 Obvious. 1140 01:13:33,050 --> 01:13:37,160 I define-- maybe I need the average kinetic energy 1141 01:13:37,160 --> 01:13:38,810 at that location. 1142 01:13:38,810 --> 01:13:44,520 So maybe, in general, I need the average or some other quantity 1143 01:13:44,520 --> 01:13:46,860 that I define at location q and t. 1144 01:13:51,990 --> 01:14:03,830 What I do is I integrate at that position f of p, and q, and t, 1145 01:14:03,830 --> 01:14:11,510 and this function O. Actually, I also define a normalization. 1146 01:14:11,510 --> 01:14:17,420 I divide by n of q and t. 1147 01:14:17,420 --> 01:14:17,920 OK? 1148 01:14:22,820 --> 01:14:26,090 Now, I implicitly used something over there 1149 01:14:26,090 --> 01:14:29,180 that, again, I define here, which 1150 01:14:29,180 --> 01:14:43,620 is a collision-conserved quantity, 1151 01:14:43,620 --> 01:14:54,320 is some function of momentum-- I could put q also-- 1152 01:14:54,320 --> 01:15:03,020 such that when evaluated for p1 plus p2 for any pair of p1's 1153 01:15:03,020 --> 01:15:08,630 and p2's that I take is the same thing as when evaluated 1154 01:15:08,630 --> 01:15:13,120 for functions of p1 and p2 that would correspond 1155 01:15:13,120 --> 01:15:18,990 to the outcomes of the collision from 1156 01:15:18,990 --> 01:15:22,070 or giving rights to p1, p2. 1157 01:15:22,070 --> 01:15:22,570 OK? 1158 01:15:26,510 --> 01:15:28,420 And then there's a following theorem 1159 01:15:28,420 --> 01:15:31,130 that we were going to use, which is 1160 01:15:31,130 --> 01:15:40,800 that for any collision-conserved quantity, what I have is 1161 01:15:40,800 --> 01:15:46,645 that I will define a quantity J, which is a function of q and t 1162 01:15:46,645 --> 01:15:53,000 in principle, which is obtained by integrating 1163 01:15:53,000 --> 01:15:57,830 over all momenta. 1164 01:15:57,830 --> 01:16:02,915 Chi-- let's say, evaluated at p1; 1165 01:16:02,915 --> 01:16:10,360 it could be evaluated at q-- times this collision operator-- 1166 01:16:10,360 --> 01:16:14,775 C of f, f. 1167 01:16:21,550 --> 01:16:25,680 And the answer is that for any collision-conserved quantity, 1168 01:16:25,680 --> 01:16:31,175 if you were to evaluate this, the answer would be 0. 1169 01:16:31,175 --> 01:16:33,460 OK? 1170 01:16:33,460 --> 01:16:37,610 So let's write that down. 1171 01:16:37,610 --> 01:16:44,500 So J of q and t is an integral over some momentum p1. 1172 01:16:47,350 --> 01:16:50,396 Because of the collision integral, what do I have? 1173 01:16:50,396 --> 01:16:53,450 I have the integration over momentum. 1174 01:16:53,450 --> 01:16:56,430 I have the integration over b. 1175 01:16:56,430 --> 01:17:01,480 I have the relative velocity v2 minus v1. 1176 01:17:01,480 --> 01:17:06,420 And then I have the subtraction of interaction 1177 01:17:06,420 --> 01:17:11,320 between p1 and p2-- addition of the interactions 1178 01:17:11,320 --> 01:17:15,310 between p1 prime, p2 prime. 1179 01:17:15,310 --> 01:17:19,380 And this collision integral has to be 1180 01:17:19,380 --> 01:17:23,150 multiplied with chi evaluated at p1. 1181 01:17:26,150 --> 01:17:28,390 OK? 1182 01:17:28,390 --> 01:17:28,890 Fine. 1183 01:17:32,200 --> 01:17:37,470 So then I will follow exactly the steps 1184 01:17:37,470 --> 01:17:42,720 that we had in the proof of the H theorem. 1185 01:17:42,720 --> 01:17:45,360 I have some function that only depends 1186 01:17:45,360 --> 01:17:49,900 on one of four momenta that are appearing here. 1187 01:17:49,900 --> 01:17:52,500 And I want to symmetrize it. 1188 01:17:52,500 --> 01:17:58,590 So I change the index one to two. 1189 01:17:58,590 --> 01:18:05,510 So this becomes 1/2 of chi of p1 plus chi of p2. 1190 01:18:10,510 --> 01:18:14,670 I do this transition between primed and un-primed. 1191 01:18:14,670 --> 01:18:18,940 And when I do that, I pick this additional factor 1192 01:18:18,940 --> 01:18:26,808 of minus chi of p1 prime, chi of p2 prime. 1193 01:18:26,808 --> 01:18:32,030 And this becomes division by 2 times 2. 1194 01:18:36,690 --> 01:18:42,210 But then I stated that chi is a collision-conserved quantity, 1195 01:18:42,210 --> 01:18:48,360 so that this term in the bracket is 0 for each point 1196 01:18:48,360 --> 01:18:49,610 that I'm integrating. 1197 01:18:49,610 --> 01:18:51,130 And hence, the answer is 0. 1198 01:18:54,630 --> 01:18:55,630 Yes. 1199 01:18:55,630 --> 01:18:59,130 AUDIENCE: In the [INAUDIBLE] theorem, 1200 01:18:59,130 --> 01:19:05,252 in the second [INAUDIBLE], do you 1201 01:19:05,252 --> 01:19:10,686 want to put minus sign of the--? 1202 01:19:10,686 --> 01:19:13,410 PROFESSOR: There was a minus sign 1203 01:19:13,410 --> 01:19:15,772 worry that I had somewhere. 1204 01:19:15,772 --> 01:19:22,660 AUDIENCE: I think in front of 1/2, at the right term. 1205 01:19:22,660 --> 01:19:27,040 PROFESSOR: In front of the 1/2, minus f of p1. 1206 01:19:27,040 --> 01:19:28,580 You say I have to do this? 1207 01:19:28,580 --> 01:19:29,568 AUDIENCE: No, no, no. 1208 01:19:29,568 --> 01:19:31,544 The next term. 1209 01:19:34,508 --> 01:19:35,496 PROFESSOR: Here? 1210 01:19:35,496 --> 01:19:36,154 AUDIENCE: Yeah. 1211 01:19:36,154 --> 01:19:41,918 I mean, if you put minus sign in front of the 1/2, 1212 01:19:41,918 --> 01:19:43,894 then [INAUDIBLE]? 1213 01:19:48,834 --> 01:19:49,610 PROFESSOR: OK. 1214 01:19:49,610 --> 01:19:53,830 Let's put a question mark here, because I 1215 01:19:53,830 --> 01:19:55,997 don't want to go over that. 1216 01:19:55,997 --> 01:19:57,580 AUDIENCE: [INAUDIBLE] but you switched 1217 01:19:57,580 --> 01:20:00,310 the order of the subtraction [? 1/2. ?] 1218 01:20:00,310 --> 01:20:01,620 So it's minus [INAUDIBLE]. 1219 01:20:01,620 --> 01:20:03,950 PROFESSOR: But I changed also the arguments here. 1220 01:20:03,950 --> 01:20:05,202 AUDIENCE: Oh, OK. 1221 01:20:05,202 --> 01:20:05,910 PROFESSOR: Right? 1222 01:20:05,910 --> 01:20:07,545 So there is some issue. 1223 01:20:07,545 --> 01:20:12,830 But this eventual thing is correct. 1224 01:20:12,830 --> 01:20:16,271 So we will go back to that. 1225 01:20:16,271 --> 01:20:16,770 OK. 1226 01:20:19,300 --> 01:20:25,020 So what is the use of stating that 1227 01:20:25,020 --> 01:20:29,040 for every conserved quantity, you have this? 1228 01:20:29,040 --> 01:20:33,020 It will allow us to get equations 1229 01:20:33,020 --> 01:20:36,820 that will determine how these parameters change 1230 01:20:36,820 --> 01:20:39,190 as a function of time? 1231 01:20:39,190 --> 01:20:39,690 OK. 1232 01:20:39,690 --> 01:20:41,251 So let's see how that goes. 1233 01:20:51,750 --> 01:20:57,530 If then I have that Lf is C, f, f, 1234 01:20:57,530 --> 01:21:04,130 and I showed that the integral of chi against C, f, f is 0, 1235 01:21:04,130 --> 01:21:07,690 then I also know that the integral 1236 01:21:07,690 --> 01:21:23,971 of chi against this L operator on f is 0. 1237 01:21:23,971 --> 01:21:24,470 OK? 1238 01:21:24,470 --> 01:21:27,550 And let's remind you that the L operator is 1239 01:21:27,550 --> 01:21:32,460 a bunch of first derivatives-- d by dt, d by dq, d by dp. 1240 01:21:32,460 --> 01:21:36,790 And so I can use the following result. 1241 01:21:36,790 --> 01:21:41,460 I can write the result as this derivative acting on both chi 1242 01:21:41,460 --> 01:21:42,890 and f. 1243 01:21:42,890 --> 01:21:46,755 And then I will get chi prime f plus f chi prime. 1244 01:21:46,755 --> 01:21:54,220 Chi f prime I have, so I have to subtract fL acting on chi. 1245 01:21:58,950 --> 01:22:00,850 OK? 1246 01:22:00,850 --> 01:22:08,640 And proceeding a little bit. 1247 01:22:08,640 --> 01:22:10,260 One more line maybe. 1248 01:22:10,260 --> 01:22:12,390 d cubed p. 1249 01:22:12,390 --> 01:22:19,100 I have dt plus-- well, actually, let 1250 01:22:19,100 --> 01:22:21,843 me do the rest next time around.