WEBVTT 00:00:00.070 --> 00:00:01.780 The following content is provided 00:00:01.780 --> 00:00:04.019 under a Creative Commons license. 00:00:04.019 --> 00:00:06.870 Your support will help MIT OpenCourseWare continue 00:00:06.870 --> 00:00:10.730 to offer high-quality educational resources for free. 00:00:10.730 --> 00:00:13.340 To make a donation or view additional materials 00:00:13.340 --> 00:00:17.217 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:17.217 --> 00:00:17.842 at ocw.mit.edu. 00:00:20.510 --> 00:00:21.320 PROFESSOR: OK. 00:00:21.320 --> 00:00:24.300 Let's start. 00:00:24.300 --> 00:00:26.460 So let's review what we've been trying 00:00:26.460 --> 00:00:29.980 to do in the past few lectures. 00:00:29.980 --> 00:00:33.710 It's to understand a typical situation involving something 00:00:33.710 --> 00:00:37.130 like a gas, which, let's say, initially 00:00:37.130 --> 00:00:39.770 is in one half of the container. 00:00:39.770 --> 00:00:45.020 And then expands until it occupies the entire container, 00:00:45.020 --> 00:00:49.070 going from one equilibrium state to another equilibrium state. 00:00:49.070 --> 00:00:52.570 Question is, how to describe this 00:00:52.570 --> 00:00:57.050 from the perspective of microscopic degrees of freedom, 00:00:57.050 --> 00:00:59.130 getting the idea that eventually you 00:00:59.130 --> 00:01:01.950 will come to some form of equilibrium. 00:01:01.950 --> 00:01:06.320 And actually, more seriously, quantifying the process 00:01:06.320 --> 00:01:11.040 and timescales by which the gas comes to equilibrium. 00:01:11.040 --> 00:01:15.560 So the first stage was to say that we will describe 00:01:15.560 --> 00:01:21.500 the process by looking at the density, which describes 00:01:21.500 --> 00:01:28.640 the number of representative examples of the system, whose 00:01:28.640 --> 00:01:31.540 collections of coordinates and momenta 00:01:31.540 --> 00:01:35.710 occupy a particular point in the 6N-dimensional phase space. 00:01:35.710 --> 00:01:42.210 So basically, this is a function that depends on 6N coordinates. 00:01:42.210 --> 00:01:45.300 And of course, it changes as a function of time. 00:01:45.300 --> 00:01:48.340 And we saw that, basically, if we 00:01:48.340 --> 00:01:53.980 were to look at how this set of points 00:01:53.980 --> 00:01:58.400 were streaming in phase space, and this 00:01:58.400 --> 00:02:01.100 can be mathematically represented 00:02:01.100 --> 00:02:05.350 by looking at the direct time dependence 00:02:05.350 --> 00:02:07.560 and the implicit time dependence of all 00:02:07.560 --> 00:02:12.010 of the coordinates captured through the Poisson bracket. 00:02:14.740 --> 00:02:17.950 And the answer was that it is a characteristic 00:02:17.950 --> 00:02:24.710 of the type of Hamiltonian evolution equations, 00:02:24.710 --> 00:02:27.590 that the flow is such that the phase space does not 00:02:27.590 --> 00:02:28.810 change volume. 00:02:28.810 --> 00:02:32.460 And hence, this quantity is 0. 00:02:32.460 --> 00:02:34.230 We said, OK, all of that is fine. 00:02:34.230 --> 00:02:39.040 But what does that tell me about how this gas is expanding? 00:02:39.040 --> 00:02:40.990 So for that, we said that we need 00:02:40.990 --> 00:02:45.590 to look at descriptions that are more appropriate to thinking 00:02:45.590 --> 00:02:50.310 about the volume and density of a gas expanding. 00:02:50.310 --> 00:02:52.770 And we really want to capture that 00:02:52.770 --> 00:02:55.200 through the one-particle density. 00:02:55.200 --> 00:02:59.380 But along the way, we introduced s particle densities, 00:02:59.380 --> 00:03:06.980 normalized up to something, to correspond to, essentially, 00:03:06.980 --> 00:03:09.810 integrating over all of the coordinates 00:03:09.810 --> 00:03:12.190 that we're not interested in. 00:03:12.190 --> 00:03:19.270 Say coordinates pertaining to numbers s plus 1 to N, 00:03:19.270 --> 00:03:21.850 corresponding 6N-dimensional phase 00:03:21.850 --> 00:03:26.470 space and the entire [INAUDIBLE] of this density or probability. 00:03:29.450 --> 00:03:35.460 Then the next stage was to be more specific about what 00:03:35.460 --> 00:03:39.230 governs the evolution of these gas particles, i.e., 00:03:39.230 --> 00:03:41.160 what is the Hamiltonian. 00:03:41.160 --> 00:03:46.590 And we said, let's focus on an N particle Hamiltonian that 00:03:46.590 --> 00:03:50.570 is composed, by necessity certainly, 00:03:50.570 --> 00:03:53.740 of one-body terms that is the kinetic energy 00:03:53.740 --> 00:03:58.320 and the potential energy, the latter describing the box, 00:03:58.320 --> 00:03:59.880 for example. 00:03:59.880 --> 00:04:02.580 And we added a two-particle interaction. 00:04:05.750 --> 00:04:09.780 qi minus qj, which, for the sake of simplicity, 00:04:09.780 --> 00:04:14.280 let's imagine is spherically symmetric-- 00:04:14.280 --> 00:04:18.790 only depends on the relative position of these particles. 00:04:18.790 --> 00:04:22.019 Then we say that if I look at the evolution 00:04:22.019 --> 00:04:25.490 of this s particle density, I start 00:04:25.490 --> 00:04:28.140 by writing an equation that kind of looks 00:04:28.140 --> 00:04:31.910 like the full Liouville equation. 00:04:31.910 --> 00:04:37.830 There is a Hamiltonian that describes this set of particles 00:04:37.830 --> 00:04:42.820 interacting among themselves, just replacing n by s. 00:04:42.820 --> 00:04:47.080 And so if these were the only particles in the universe, 00:04:47.080 --> 00:04:49.030 this is what I would get. 00:04:49.030 --> 00:04:52.900 Except that we know that there are other particles. 00:04:52.900 --> 00:04:56.940 And since we have two-body collisions possible, 00:04:56.940 --> 00:05:00.880 any one of the particles in the set that I'm describing here, 00:05:00.880 --> 00:05:03.790 composed of s particles, can potentially 00:05:03.790 --> 00:05:08.450 collide with a particle from this set 00:05:08.450 --> 00:05:10.420 that I don't want to include. 00:05:10.420 --> 00:05:13.900 And that will change the volume of phase space. 00:05:13.900 --> 00:05:17.290 This incompressibility condition was valid 00:05:17.290 --> 00:05:21.770 only if the entire thing of the Hamiltonian was considered. 00:05:21.770 --> 00:05:25.060 Here, I'm only looking at the partial subset. 00:05:25.060 --> 00:05:27.040 And so then what did I have here? 00:05:27.040 --> 00:05:31.870 I had the force exerted from particle s plus 1 00:05:31.870 --> 00:05:39.600 on particle n changing the momentum of the particle n. 00:05:39.600 --> 00:05:42.680 But the likelihood that this happens 00:05:42.680 --> 00:05:50.010 depends on finding the other particle also. 00:05:50.010 --> 00:05:56.910 So I had this dependence on the density that was s plus 1. 00:05:56.910 --> 00:06:00.490 So this was the BBGKY hierarchy. 00:06:00.490 --> 00:06:04.830 Then we said, let's take a look at the typical value 00:06:04.830 --> 00:06:08.050 of the terms that we have in this equation. 00:06:08.050 --> 00:06:15.640 And for s that is larger than 2, this Hamiltonian 00:06:15.640 --> 00:06:18.270 that pertains to s particles will 00:06:18.270 --> 00:06:21.300 have a collision term among the particles. 00:06:21.300 --> 00:06:25.130 And that collision term will give you some typical inverse 00:06:25.130 --> 00:06:30.080 timescale here, which is related to the collision time. 00:06:30.080 --> 00:06:34.320 Whereas on the right-hand side, for this collision 00:06:34.320 --> 00:06:39.430 to take place, you have to find another particle 00:06:39.430 --> 00:06:42.670 to interact with in the range of interactions 00:06:42.670 --> 00:06:44.700 that we indicated by d. 00:06:44.700 --> 00:06:48.450 And for density n, this was smaller 00:06:48.450 --> 00:06:53.850 by a factor of nd cubed for dilute gases. 00:06:53.850 --> 00:06:56.680 Therefore, the left-hand side is much larger 00:06:56.680 --> 00:06:58.290 than the right-hand side. 00:06:58.290 --> 00:07:01.250 And within particular approximation, 00:07:01.250 --> 00:07:04.500 we said that we are going to approximately set 00:07:04.500 --> 00:07:07.470 this right-hand side to 0. 00:07:07.470 --> 00:07:09.490 OK? 00:07:09.490 --> 00:07:12.730 Except that we couldn't do this for the equation 00:07:12.730 --> 00:07:20.990 that pertained to s equals to 1, because for s equals to 1, 00:07:20.990 --> 00:07:24.620 I did not have the analog of this term. 00:07:24.620 --> 00:07:30.940 If I were to write the precise form of this Poisson bracket, 00:07:30.940 --> 00:07:33.730 I had the partial derivative with respect 00:07:33.730 --> 00:07:41.110 to time, the momentum divided by mass, 00:07:41.110 --> 00:07:45.460 which is the velocity driving the change in coordinate. 00:07:45.460 --> 00:07:48.350 And the force that comes from the external potential, 00:07:48.350 --> 00:07:51.490 let's call it f1, we arrive in the change 00:07:51.490 --> 00:08:03.360 in momentum of the one-particle density, which is a function-- 00:08:03.360 --> 00:08:10.410 let's write it explicitly-- of p1, q1, and t. 00:08:10.410 --> 00:08:16.720 And on the other side, I had the possibility 00:08:16.720 --> 00:08:22.100 to have a term such as this that describes 00:08:22.100 --> 00:08:25.280 a collision with a second particle. 00:08:25.280 --> 00:08:27.840 I have to look at all possible second particles 00:08:27.840 --> 00:08:30.750 that I can collide with. 00:08:30.750 --> 00:08:36.182 They could come with all variety of momenta, 00:08:36.182 --> 00:08:38.679 which we will indicate p2. 00:08:38.679 --> 00:08:41.049 They can be all over the place. 00:08:41.049 --> 00:08:43.600 So I have integrations over space. 00:08:43.600 --> 00:08:46.700 And then I have a bunch of terms. 00:08:46.700 --> 00:08:53.440 So for this, we said, let's take a slightly better look 00:08:53.440 --> 00:08:59.400 at the kind of collisions that are implicitly described 00:08:59.400 --> 00:09:04.010 via the term that we would have here as f2. 00:09:04.010 --> 00:09:07.890 And since the right-hand side of f2 we set to 0, essentially, 00:09:07.890 --> 00:09:11.260 f2 really describes the Newtonian evolution 00:09:11.260 --> 00:09:14.580 of two particles coming together and having a collision. 00:09:17.530 --> 00:09:21.160 And this collision, I can-- really, 00:09:21.160 --> 00:09:23.630 the way that I have it over there 00:09:23.630 --> 00:09:27.510 is in what I would call a lab frame. 00:09:27.510 --> 00:09:30.010 And the perspective that we should have 00:09:30.010 --> 00:09:33.850 is that in this frame, I have a particle that 00:09:33.850 --> 00:09:35.800 is coming with momentum, let's say, 00:09:35.800 --> 00:09:41.920 p1, which is the one that I have specified 00:09:41.920 --> 00:09:44.240 on the left-hand side of the equation. 00:09:44.240 --> 00:09:46.240 That's the density that I'm following 00:09:46.240 --> 00:09:49.260 in the channel represented by p1. 00:09:49.260 --> 00:09:55.660 But then along comes a particle with momentum p2. 00:09:55.660 --> 00:09:59.340 And there is some place where they 00:09:59.340 --> 00:10:04.880 hit each other, after which this one goes off with p1 prime. 00:10:04.880 --> 00:10:07.970 Actually, let me call it p1 double prime. 00:10:07.970 --> 00:10:11.980 And this one goes off with p2 double prime. 00:10:14.810 --> 00:10:17.470 Now, we will find it useful to also look 00:10:17.470 --> 00:10:21.000 at the same thing in the center of mass frame. 00:10:30.480 --> 00:10:34.120 Now, in the center of mass frame-- oops-- 00:10:34.120 --> 00:10:44.050 my particle p1 is coming with a momentum that 00:10:44.050 --> 00:10:50.930 is shifted by the p of center of mass, 00:10:50.930 --> 00:10:53.360 which is really p1 plus p2 over 2. 00:10:53.360 --> 00:10:56.720 So this is p1 minus p2 over 2. 00:10:56.720 --> 00:11:01.180 And my particle number two comes with p2 minus center 00:11:01.180 --> 00:11:08.950 of mass, which is p2 minus p1 over 2. 00:11:08.950 --> 00:11:10.280 So they're minus each other. 00:11:10.280 --> 00:11:11.780 So in the center of mass frame, they 00:11:11.780 --> 00:11:16.270 are basically coming to each other along the same line. 00:11:16.270 --> 00:11:21.330 So we could define one of the axes of coordinates 00:11:21.330 --> 00:11:24.280 as the distance along which they are approaching. 00:11:24.280 --> 00:11:26.720 And let's say we put the center of mass 00:11:26.720 --> 00:11:30.690 at the origin, which means that I 00:11:30.690 --> 00:11:33.740 have two more directions to work with. 00:11:33.740 --> 00:11:39.660 So essentially, the coordinates of this second particle 00:11:39.660 --> 00:11:48.230 can be described through a vector b, 00:11:48.230 --> 00:11:52.380 which we call the impact parameter. 00:11:52.380 --> 00:11:55.290 And really, what happens to the collision 00:11:55.290 --> 00:12:01.460 is classically determined by how close they approach together 00:12:01.460 --> 00:12:03.580 to make that. 00:12:03.580 --> 00:12:06.650 To quantify that, we really need this impact vector. 00:12:06.650 --> 00:12:10.060 So b equals to 0, they are coming at each other head on. 00:12:10.060 --> 00:12:14.730 For a different b, they are coming at some angle. 00:12:14.730 --> 00:12:18.560 So really, I need to put here a d2b. 00:12:22.780 --> 00:12:27.690 Now, once I have specified the form of the interaction that 00:12:27.690 --> 00:12:32.730 is taking place when the two things come together, 00:12:32.730 --> 00:12:35.960 then the process is completely deterministic. 00:12:35.960 --> 00:12:38.670 Particle number one will come. 00:12:38.670 --> 00:12:43.960 And after the location at which you have the interaction, 00:12:43.960 --> 00:12:52.610 will go off with p1 double prime minus center of mass. 00:12:52.610 --> 00:12:56.210 And particle number two will come and get 00:12:56.210 --> 00:13:01.865 deflected to p2 prime minus-- or double prime minus p 00:13:01.865 --> 00:13:03.650 center of mass. 00:13:03.650 --> 00:13:04.150 OK? 00:13:07.800 --> 00:13:13.760 And again, it is important to note that this parameter 00:13:13.760 --> 00:13:19.560 A is irrelevant to the collision, in that once I have 00:13:19.560 --> 00:13:28.890 stated that before collision, I have a momentum, 00:13:28.890 --> 00:13:32.530 say, p1 and p2. 00:13:35.300 --> 00:13:40.490 And this impact parameter b is specified, 00:13:40.490 --> 00:13:45.850 then p1 double prime is known completely 00:13:45.850 --> 00:13:48.810 as a function of p1, p2, and b. 00:13:48.810 --> 00:13:52.600 And p2 double prime is known completely 00:13:52.600 --> 00:13:59.710 as a function of p1, p2, and b. 00:13:59.710 --> 00:14:05.030 And everything in that sense is well-defined. 00:14:05.030 --> 00:14:10.320 So really, this parameter A is not relevant. 00:14:10.320 --> 00:14:14.820 And it's really the B that is important. 00:14:14.820 --> 00:14:19.260 But ultimately, we also need to get these other terms, 00:14:19.260 --> 00:14:21.700 something that has the appropriate dimensions 00:14:21.700 --> 00:14:23.550 of inverse time. 00:14:23.550 --> 00:14:27.500 And if I have a density of particles coming 00:14:27.500 --> 00:14:32.520 hitting a target, how frequently they hit 00:14:32.520 --> 00:14:36.050 depends on the product of the density as well 00:14:36.050 --> 00:14:37.810 as the velocity. 00:14:37.810 --> 00:14:41.070 So really, I have to add here a term that 00:14:41.070 --> 00:14:44.100 is related to the velocity with which these things are 00:14:44.100 --> 00:14:45.630 approaching each other. 00:14:45.630 --> 00:14:51.030 And so I can write it either as v2 minus v1, or p2 minus 1 00:14:51.030 --> 00:14:51.690 over m. 00:14:54.690 --> 00:14:59.940 So that's the other thing that I have to look at. 00:14:59.940 --> 00:15:07.510 And then, essentially, what we saw 00:15:07.510 --> 00:15:11.390 last time is that the rest of the story-- if you 00:15:11.390 --> 00:15:13.920 sort of think of these as billiard balls, 00:15:13.920 --> 00:15:17.450 so that the collision is instantaneous and takes place 00:15:17.450 --> 00:15:20.690 at a well-specified point, instantaneously, 00:15:20.690 --> 00:15:23.910 at some location, you're going with some momentum, 00:15:23.910 --> 00:15:27.110 and then turn and go with some other momentum. 00:15:27.110 --> 00:15:31.430 So the channel that was carrying in momentum p1, which 00:15:31.430 --> 00:15:35.240 is the one that we are interested in, suddenly 00:15:35.240 --> 00:15:39.766 gets depleted by an amount that is related 00:15:39.766 --> 00:15:44.880 to the probability of having simultaneously particles 00:15:44.880 --> 00:15:49.580 of momenta p1 and p2. 00:15:53.580 --> 00:15:59.460 And the locations that I have to specify here-- well, 00:15:59.460 --> 00:16:03.360 again, I'm really following this channel. 00:16:03.360 --> 00:16:07.790 And in this channel, I have specified q1. 00:16:07.790 --> 00:16:12.110 And suddenly, I see that at the moment of collision, 00:16:12.110 --> 00:16:15.710 the momentum changes, if you are thinking about billiard balls. 00:16:15.710 --> 00:16:18.050 So really, it is at that location 00:16:18.050 --> 00:16:20.720 that I'm interested for particle one. 00:16:20.720 --> 00:16:24.050 For particle two, I'm interested at something 00:16:24.050 --> 00:16:27.580 that is, let's say, q1 plus, because it depends 00:16:27.580 --> 00:16:32.150 on the location of particle two shifted from particle one 00:16:32.150 --> 00:16:35.230 by an amount that is related by b and whatever 00:16:35.230 --> 00:16:38.190 this amount slightly of A is that is related 00:16:38.190 --> 00:16:41.531 to the size of your billiard ball. 00:16:41.531 --> 00:16:42.030 OK? 00:16:46.910 --> 00:16:52.550 Now, we said that there is a corresponding addition, which 00:16:52.550 --> 00:16:58.580 comes from the fact that it is possible suddenly to not 00:16:58.580 --> 00:17:04.140 have a particle moving along, say, this direction. 00:17:04.140 --> 00:17:07.240 And the collision of these two particles 00:17:07.240 --> 00:17:10.740 created something along that direction. 00:17:10.740 --> 00:17:13.119 So the same way that probabilities 00:17:13.119 --> 00:17:16.010 can be subtracted amongst some channels, 00:17:16.010 --> 00:17:18.569 they can be added to channels, because collision 00:17:18.569 --> 00:17:21.589 of two particles created something. 00:17:21.589 --> 00:17:28.900 And so I will indicate that by p1 prime, p2 prime, 00:17:28.900 --> 00:17:34.140 again, coordinates in t, something like this. 00:17:34.140 --> 00:17:38.950 So basically, I try to be a little bit more careful here, 00:17:38.950 --> 00:17:43.200 not calling these outgoing channels p prime, 00:17:43.200 --> 00:17:44.970 but p double prime. 00:17:44.970 --> 00:17:47.860 Because really, if I were to invert 00:17:47.860 --> 00:17:51.260 p1 double prime and p2 double prime, 00:17:51.260 --> 00:17:54.690 I will generate things that are going backward. 00:17:54.690 --> 00:17:58.320 So there's a couple of sign issues involved. 00:17:58.320 --> 00:18:02.220 But essentially, there is a similar function 00:18:02.220 --> 00:18:05.780 that relates p1 prime and p2 prime. 00:18:05.780 --> 00:18:07.900 It is, basically, you have to search out 00:18:07.900 --> 00:18:13.390 momenta whose outcome will create particles 00:18:13.390 --> 00:18:15.550 in the channels prescribed by p1 and p2. 00:18:19.290 --> 00:18:22.920 So as we said, up to here, we have 00:18:22.920 --> 00:18:26.240 done a variety of approximations. 00:18:26.240 --> 00:18:30.130 They're kind of physically reasonable. 00:18:30.130 --> 00:18:33.280 There is some difficulty here in specifying 00:18:33.280 --> 00:18:38.680 if I want to be very accurate where the locations of q1 00:18:38.680 --> 00:18:43.440 and q2 are within this interaction size 00:18:43.440 --> 00:18:45.760 d that I am looking at. 00:18:45.760 --> 00:18:48.870 And if I don't think about billiard balls, where 00:18:48.870 --> 00:18:51.810 the interactions are not quite instantaneous where 00:18:51.810 --> 00:18:56.470 you change momenta, but particles that are deformable, 00:18:56.470 --> 00:18:59.250 there is also certainly a certain amount of time 00:18:59.250 --> 00:19:01.710 where the potential is acting. 00:19:01.710 --> 00:19:05.530 And so exactly what time you are looking at here 00:19:05.530 --> 00:19:09.413 is also not appropriate-- is not completely, no. 00:19:12.830 --> 00:19:16.620 But if you sort of make all of those things precise, 00:19:16.620 --> 00:19:20.430 then my statement is that this equation still 00:19:20.430 --> 00:19:25.290 respects the reversibility of the equations of motions 00:19:25.290 --> 00:19:27.550 that we had initially. 00:19:27.550 --> 00:19:30.110 But at this stage, we say, well, what can 00:19:30.110 --> 00:19:31.950 I do with this equation? 00:19:31.950 --> 00:19:34.750 I really want to solve an equation that 00:19:34.750 --> 00:19:40.320 involves f of p1, q1, and t. 00:19:40.320 --> 00:19:42.310 So the best that I can do is, first 00:19:42.310 --> 00:19:46.300 of all, simplify all of the arguments here 00:19:46.300 --> 00:19:48.330 to be more or less at the same location. 00:19:48.330 --> 00:19:52.210 I don't want to think about one particle being shifted 00:19:52.210 --> 00:19:55.100 by an amount that is related by b plus something that 00:19:55.100 --> 00:20:00.390 is related to the interaction potential size, et cetera. 00:20:00.390 --> 00:20:06.020 Let's sort of change our focus, so that our picture size, 00:20:06.020 --> 00:20:07.790 essentially, is something that is 00:20:07.790 --> 00:20:12.050 of the order of the size of the interaction. 00:20:12.050 --> 00:20:14.910 And also, our resolution in time is 00:20:14.910 --> 00:20:19.630 shifted, so that we don't really ask about moments of time 00:20:19.630 --> 00:20:21.770 slightly before and after collision, 00:20:21.770 --> 00:20:26.480 where the particle, let's say, could be deformed or whatever. 00:20:26.480 --> 00:20:30.180 So having done that, we make these approximations 00:20:30.180 --> 00:20:33.780 that, essentially, all of these f2's, 00:20:33.780 --> 00:20:37.910 we are going to replace as a product of f1's 00:20:37.910 --> 00:20:41.540 evaluated all at the same location-- 00:20:41.540 --> 00:20:44.600 the location that is specified here, 00:20:44.600 --> 00:20:48.950 as well as the same time, the time that is specified there. 00:20:48.950 --> 00:21:15.490 So this becomes-- OK? 00:21:15.490 --> 00:21:20.120 So essentially, the result of the collisions 00:21:20.120 --> 00:21:27.790 is described on the left-hand side of our equation 00:21:27.790 --> 00:21:34.740 by a term that integrates over all momenta and overall impact 00:21:34.740 --> 00:21:39.840 parameters of a bracket, which is a product of two 00:21:39.840 --> 00:21:42.100 one-particle densities. 00:21:42.100 --> 00:21:46.080 And just to simplify notation, I will call this 00:21:46.080 --> 00:21:51.460 the collision term that involves two factors of f1. 00:21:51.460 --> 00:21:54.780 So that's just a shorthand for that. 00:21:54.780 --> 00:21:55.280 Yes. 00:21:55.280 --> 00:21:57.516 AUDIENCE: Is that process equivalent 00:21:57.516 --> 00:22:01.250 of a similar molecule [INAUDIBLE]? 00:22:01.250 --> 00:22:03.600 PROFESSOR: That, if you want to give it a name, yes. 00:22:03.600 --> 00:22:05.670 It's called the assumption of molecular chaos. 00:22:11.150 --> 00:22:16.480 And physically, it is the assumption 00:22:16.480 --> 00:22:20.200 that the probability to find the two particles 00:22:20.200 --> 00:22:23.500 is the product of one-particle probabilities. 00:22:23.500 --> 00:22:27.110 And again, we say that you have to worry 00:22:27.110 --> 00:22:30.460 about the accuracy of that if you 00:22:30.460 --> 00:22:34.530 want to focus on sizes that are smaller than the interaction 00:22:34.530 --> 00:22:36.010 parameter size. 00:22:36.010 --> 00:22:38.500 But we change that as part of our resolution. 00:22:38.500 --> 00:22:41.073 So we don't worry about that aspect. 00:22:41.073 --> 00:22:44.572 AUDIENCE: So molecular chaos doesn't 00:22:44.572 --> 00:22:49.009 include eliminating this space? 00:22:49.009 --> 00:22:50.488 It's a separate thing. 00:22:53.848 --> 00:23:00.090 PROFESSOR: I've only seen the word "molecular chaos" applied 00:23:00.090 --> 00:23:01.710 in this context. 00:23:01.710 --> 00:23:04.440 I think you are asking whether or not 00:23:04.440 --> 00:23:07.490 if q1 and q2 are different. 00:23:07.490 --> 00:23:11.060 I can replace that and call that molecular chaos. 00:23:11.060 --> 00:23:14.680 I would say that that's actually a correct statement. 00:23:14.680 --> 00:23:16.760 It's not an assumption. 00:23:16.760 --> 00:23:18.730 Basically, when the two particles 00:23:18.730 --> 00:23:22.640 are away from each other to a good approximation, 00:23:22.640 --> 00:23:25.910 they don't really know about-- well, actually, 00:23:25.910 --> 00:23:29.370 what I am saying is if I really want to sort of focus 00:23:29.370 --> 00:23:32.305 on time dependence, maybe that is also an assumption. 00:23:32.305 --> 00:23:33.730 Yes. 00:23:33.730 --> 00:23:38.290 So they're really, if you like, indeed, two parts. 00:23:38.290 --> 00:23:42.230 It is the replacing of f2 with the product of two 00:23:42.230 --> 00:23:48.010 f2's, and then evaluating them at the same point. 00:23:48.010 --> 00:23:48.510 Yes. 00:23:48.510 --> 00:23:50.690 AUDIENCE: But based on your first line-- I mean, 00:23:50.690 --> 00:23:55.140 you could break that assumption of different distribution 00:23:55.140 --> 00:23:58.254 functions have different events that occur when they get close, 00:23:58.254 --> 00:23:59.795 and the framework still works, right? 00:23:59.795 --> 00:24:03.090 It would just become-- when you got down to the bottom line, 00:24:03.090 --> 00:24:06.900 it would just become a much larger equation, right? 00:24:06.900 --> 00:24:08.150 PROFESSOR: I don't understand. 00:24:08.150 --> 00:24:10.316 AUDIENCE: Like if you assume different interactions. 00:24:12.810 --> 00:24:16.640 PROFESSOR: The interactions are here 00:24:16.640 --> 00:24:20.160 in how p1 prime and p2 prime depend on p1 and p2. 00:24:20.160 --> 00:24:21.146 AUDIENCE: Right. 00:24:21.146 --> 00:24:23.653 What I'm suggesting is if you change H 00:24:23.653 --> 00:24:26.650 by keeping the top [INAUDIBLE]. 00:24:26.650 --> 00:24:29.340 That's a pretty simple H. Would this framework 00:24:29.340 --> 00:24:32.799 work for a more complicated H. But then your bottom two lines 00:24:32.799 --> 00:24:33.840 would become [INAUDIBLE]. 00:24:33.840 --> 00:24:36.820 PROFESSOR: What more can I do here? 00:24:36.820 --> 00:24:38.225 How can I make this more general? 00:24:38.225 --> 00:24:39.600 Are you thinking about adding it? 00:24:39.600 --> 00:24:40.503 AUDIENCE: No, no, no. 00:24:40.503 --> 00:24:42.336 I'm not suggesting you make it more general. 00:24:42.336 --> 00:24:44.590 I'm suggesting you make it more specific to break 00:24:44.590 --> 00:24:49.130 that assumption of molecular chaos. 00:24:49.130 --> 00:24:50.150 PROFESSOR: OK. 00:24:50.150 --> 00:24:52.850 So how should I modify if I'm not saying-- 00:24:52.850 --> 00:24:55.174 AUDIENCE: I didn't have something in mind. 00:24:55.174 --> 00:24:57.058 I was just suggesting [INAUDIBLE]. 00:24:57.058 --> 00:24:59.540 PROFESSOR: Well, that's where I have problem, 00:24:59.540 --> 00:25:02.930 because it seems to me that this is very general. 00:25:02.930 --> 00:25:05.890 The only thing that I left out is the body 00:25:05.890 --> 00:25:10.400 interactions may be having to do with this being spherically 00:25:10.400 --> 00:25:11.450 symmetric. 00:25:11.450 --> 00:25:14.130 But apart from that, this is basically as general 00:25:14.130 --> 00:25:16.380 as you can get with two bodies. 00:25:16.380 --> 00:25:19.430 Now, the part that is pertaining to three bodies, 00:25:19.430 --> 00:25:24.000 again, you would basically-- not need to worry about as long 00:25:24.000 --> 00:25:26.970 as nd cubed is less than 1. 00:25:26.970 --> 00:25:32.515 So I don't quite know what you have in mind. 00:25:32.515 --> 00:25:34.393 AUDIENCE: I guess I was thinking about three- 00:25:34.393 --> 00:25:35.538 and four-body interactions. 00:25:35.538 --> 00:25:38.490 But if you're covered against that, then what I said, 00:25:38.490 --> 00:25:41.830 it's not valid or relevant. 00:25:41.830 --> 00:25:42.680 PROFESSOR: Right. 00:25:42.680 --> 00:25:45.970 So certainly, three-body and higher body interactions 00:25:45.970 --> 00:25:48.220 would be important if you're thinking 00:25:48.220 --> 00:25:52.790 about writing a description for a liquid, for example. 00:25:52.790 --> 00:25:57.440 And then the story becomes more complicated. 00:25:57.440 --> 00:26:00.550 You can't really make the truncations 00:26:00.550 --> 00:26:02.620 that would give you the Boltzmann equation. 00:26:02.620 --> 00:26:06.070 You need to go by some other set of approximations, 00:26:06.070 --> 00:26:08.160 such as the Vlasov equation, that you 00:26:08.160 --> 00:26:10.710 will see in the problem set. 00:26:10.710 --> 00:26:15.610 And the actual theory that one can write down for liquids 00:26:15.610 --> 00:26:17.510 is very complicated. 00:26:17.510 --> 00:26:20.150 So it's-- yeah. 00:26:20.150 --> 00:26:22.122 AUDIENCE: Can it get more complicated yet 00:26:22.122 --> 00:26:24.581 if, based on the same framework, isn't this similar 00:26:24.581 --> 00:26:26.975 to, like, the formulation of the neutron transport 00:26:26.975 --> 00:26:30.570 equation, where if you have, in addition to scattering, 00:26:30.570 --> 00:26:32.460 you have particles that don't attract, 00:26:32.460 --> 00:26:35.573 but you have fission events and you have inelastic center? 00:26:35.573 --> 00:26:36.156 PROFESSOR: OK. 00:26:36.156 --> 00:26:36.230 Yes. 00:26:36.230 --> 00:26:36.530 That's good. 00:26:36.530 --> 00:26:38.630 AUDIENCE: And then your H would change dramatically, right? 00:26:38.630 --> 00:26:39.440 PROFESSOR: Yes. 00:26:39.440 --> 00:26:39.726 That's right. 00:26:39.726 --> 00:26:39.870 Yes. 00:26:39.870 --> 00:26:41.860 AUDIENCE: But my point is that it's the same framework. 00:26:41.860 --> 00:26:42.420 PROFESSOR: Yes. 00:26:42.420 --> 00:26:42.919 Yes. 00:26:42.919 --> 00:26:44.320 So basically, you're right. 00:26:44.320 --> 00:26:47.650 One assumption that I have here is that the number of particles 00:26:47.650 --> 00:26:49.320 is conserved. 00:26:49.320 --> 00:26:53.010 We say that if the particles can break or diffuse, 00:26:53.010 --> 00:26:56.850 one has to generalize this approximation. 00:26:56.850 --> 00:27:00.330 And there would be an appropriate generalization 00:27:00.330 --> 00:27:03.470 of the Boltzmann equation to cover those terms. 00:27:03.470 --> 00:27:11.040 And I don't have a problem set related to that, 00:27:11.040 --> 00:27:14.900 but maybe you are suggesting that I should provide one. 00:27:14.900 --> 00:27:16.800 [LAUGHTER] 00:27:17.750 --> 00:27:18.400 OK. 00:27:18.400 --> 00:27:19.260 That's a good point. 00:27:21.690 --> 00:27:22.190 OK? 00:27:24.770 --> 00:27:25.270 All right. 00:27:25.270 --> 00:27:29.510 So we have made some assumptions here. 00:27:29.510 --> 00:27:34.090 And I guess my claim is that this equation-- well, 00:27:34.090 --> 00:27:35.670 what equation? 00:27:35.670 --> 00:27:38.880 So basically, what I'm going to do, 00:27:38.880 --> 00:27:41.630 I call the right-hand side of this equation 00:27:41.630 --> 00:27:46.090 after this assumption of molecular chaos, CFF. 00:27:46.090 --> 00:27:52.070 For simplicity, let me call this bunch of first derivatives 00:27:52.070 --> 00:27:56.380 that act on f1 L for Liouvillian. 00:27:56.380 --> 00:27:58.680 It's kind of like a Liouville operator. 00:27:58.680 --> 00:28:04.010 So I have an equation that is the bunch of first derivatives 00:28:04.010 --> 00:28:07.880 acting on f1, which is essentially 00:28:07.880 --> 00:28:12.700 a linear partial differential equation in six or seven 00:28:12.700 --> 00:28:15.770 dimensions, depending on how you count time-- is 00:28:15.770 --> 00:28:21.890 equal to a non-linear integral on the right-hand side. 00:28:21.890 --> 00:28:23.190 OK? 00:28:23.190 --> 00:28:26.370 So this is the entity that is the Boltzmann equation. 00:28:39.950 --> 00:28:48.710 Now, the statement is that this equation no longer 00:28:48.710 --> 00:28:51.220 has time reversal symmetry. 00:28:51.220 --> 00:28:56.420 And so if I were to solve this equation for the one function 00:28:56.420 --> 00:29:00.030 that I'm interested, after all, the f1, which really tells me 00:29:00.030 --> 00:29:03.740 how the particles stream from the left-hand side of the box 00:29:03.740 --> 00:29:06.480 to the right-hand side of the box, 00:29:06.480 --> 00:29:11.820 that equation is not reversible. 00:29:11.820 --> 00:29:14.480 That is, the solution to this equation 00:29:14.480 --> 00:29:16.840 will indeed take the density that 00:29:16.840 --> 00:29:18.940 was on the left-hand side of the box 00:29:18.940 --> 00:29:23.450 and uniformly distribute it on the two sides of the box. 00:29:23.450 --> 00:29:25.780 And that will stay forever. 00:29:25.780 --> 00:29:29.450 That's, of course, a consequence of the various approximations 00:29:29.450 --> 00:29:33.140 that we made here, removing the reversibility. 00:29:33.140 --> 00:29:34.810 OK? 00:29:34.810 --> 00:29:37.820 And the statement to show that mathematically 00:29:37.820 --> 00:29:41.280 is that if there is a function-- and now, for simplicity, 00:29:41.280 --> 00:29:45.750 since from now on, we don't care about f2, f3, et cetera, 00:29:45.750 --> 00:29:51.250 we are only caring about f1, I'm going to drop the index one 00:29:51.250 --> 00:29:55.980 and say that if there is a density f, which is really 00:29:55.980 --> 00:30:02.760 my f1, that satisfies the Boltzmann-- that is, 00:30:02.760 --> 00:30:10.610 I have a form such as this-- then there is a quantity H that 00:30:10.610 --> 00:30:15.620 only depends on time that, under the evolution that 00:30:15.620 --> 00:30:22.490 is implicit in this Boltzmann equation will always decrease, 00:30:22.490 --> 00:30:29.030 where H is an integral over the coordinates 00:30:29.030 --> 00:30:39.060 and momenta that are in this function f of p, q, and t. 00:30:39.060 --> 00:30:43.250 And all I need to do is to multiply by the log. 00:30:45.960 --> 00:30:50.220 And then all of these are vectors if I drop it 00:30:50.220 --> 00:30:52.480 for saving time. 00:30:52.480 --> 00:30:55.350 And we said that once we recognize 00:30:55.350 --> 00:30:57.660 that f, up to a normalization, is 00:30:57.660 --> 00:31:01.060 the same thing as the one-particle probability-- 00:31:01.060 --> 00:31:04.440 and having previously seen objects such as this 00:31:04.440 --> 00:31:06.990 in probability theory corresponding 00:31:06.990 --> 00:31:10.470 to entropy information, entropy of mixing, 00:31:10.470 --> 00:31:15.590 you will not be surprised at how this expression came about. 00:31:15.590 --> 00:31:19.370 And I had told you that in each one of the four sections, 00:31:19.370 --> 00:31:23.700 we will introduce some quantity that plays the role of entropy. 00:31:23.700 --> 00:31:26.680 So this is the one that is appropriate to our section 00:31:26.680 --> 00:31:28.880 on kinetic theory. 00:31:28.880 --> 00:31:30.949 So let's see how we go about-- 00:31:30.949 --> 00:31:31.740 AUDIENCE: Question. 00:31:31.740 --> 00:31:32.460 PROFESSOR: Yes? 00:31:32.460 --> 00:31:35.862 AUDIENCE: When you introduced the-- initialized the concept 00:31:35.862 --> 00:31:40.722 of entropy of information of the message that's 00:31:40.722 --> 00:31:42.666 p log p, [INAUDIBLE] entropy. 00:31:42.666 --> 00:31:44.620 And as far as I understand, this concept 00:31:44.620 --> 00:31:47.630 came up in, like, mid 20th century at least, right? 00:31:47.630 --> 00:31:50.450 PROFESSOR: I think slightly later than that. 00:31:50.450 --> 00:31:51.550 Of that order, yes. 00:31:51.550 --> 00:31:52.522 AUDIENCE: OK. 00:31:52.522 --> 00:31:54.813 And these are Boltzmann equations? 00:31:54.813 --> 00:31:55.438 PROFESSOR: Yes. 00:31:55.438 --> 00:31:58.578 AUDIENCE: And so this was, like, 100 years before, at least? 00:31:58.578 --> 00:32:00.911 PROFESSOR: Not 100 years, but maybe 60 years away, yeah. 00:32:00.911 --> 00:32:01.411 50, 60. 00:32:01.411 --> 00:32:04.131 AUDIENCE: OK. 00:32:04.131 --> 00:32:07.562 I'm just interested, what is the motivation for picking 00:32:07.562 --> 00:32:12.128 this form of functionality, that it's integral of f log f 00:32:12.128 --> 00:32:14.190 rather than integral of something else? 00:32:14.190 --> 00:32:14.950 PROFESSOR: OK. 00:32:14.950 --> 00:32:17.845 I mean, there is another place that you've seen it, 00:32:17.845 --> 00:32:22.210 that has nothing to do with information. 00:32:22.210 --> 00:32:24.460 And that's mixing entropy. 00:32:24.460 --> 00:32:27.550 If I have [? n ?] one particles of one type 00:32:27.550 --> 00:32:29.910 and two particles of another type, et cetera, 00:32:29.910 --> 00:32:33.130 and mixed them up, then the mixing entropy, 00:32:33.130 --> 00:32:36.780 you can relate to a form such as this. 00:32:36.780 --> 00:32:39.050 It is some ni log ni. 00:32:43.370 --> 00:32:43.870 OK? 00:32:47.826 --> 00:32:48.325 Fine. 00:32:52.370 --> 00:32:54.770 It's just for discrete variables, 00:32:54.770 --> 00:32:58.500 it makes much more sense than for the continuum. 00:32:58.500 --> 00:33:00.370 But OK. 00:33:00.370 --> 00:33:05.060 So let's see if this is indeed the case. 00:33:05.060 --> 00:33:10.550 So we say that dH by dt is the-- I 00:33:10.550 --> 00:33:15.330 have to take the time derivative inside. 00:33:15.330 --> 00:33:19.450 And as we discussed before, the full derivative 00:33:19.450 --> 00:33:23.210 becomes a partial derivative inside the integral. 00:33:23.210 --> 00:33:28.060 And I have either acting on f and acting 00:33:28.060 --> 00:33:30.720 on log f, which will give me 1 over f, 00:33:30.720 --> 00:33:33.110 canceling the f outside. 00:33:33.110 --> 00:33:38.660 And this term corresponded to the derivative 00:33:38.660 --> 00:33:41.760 of the normalization, which is fixed. 00:33:41.760 --> 00:33:49.215 And then we had to replace what we have for df by dt. 00:33:51.850 --> 00:34:00.466 And for df by dt, we either have the Poisson bracket 00:34:00.466 --> 00:34:06.790 of the one-particle Hamiltonian with f coming from the L 00:34:06.790 --> 00:34:12.000 part of the equation, once we get d by dt on one side. 00:34:12.000 --> 00:34:15.980 And then the collision part that depends on two f's. 00:34:15.980 --> 00:34:19.250 And we have to multiply this by f. 00:34:19.250 --> 00:34:24.920 And again, let's remind you that f really has argument-- oops, 00:34:24.920 --> 00:34:29.940 log f-- that has argument p1. 00:34:29.940 --> 00:34:31.500 I won't write q1 and t. 00:34:36.330 --> 00:34:40.580 And the next thing that we did at the end of last lecture 00:34:40.580 --> 00:34:46.889 is that typically, when you have integral of a Poisson bracket 00:34:46.889 --> 00:34:50.670 by doing some combination of integrations by part, 00:34:50.670 --> 00:34:56.360 you can even eventually show that this contribution is 0. 00:34:56.360 --> 00:35:00.510 And really, the important thing is the integration 00:35:00.510 --> 00:35:04.640 against the collision term of the Boltzmann equation. 00:35:04.640 --> 00:35:09.510 So writing that explicitly, what do we have? 00:35:09.510 --> 00:35:13.290 We have that it is integral d cubed. 00:35:13.290 --> 00:35:17.060 Let's say q1. 00:35:17.060 --> 00:35:20.290 Again, I don't really need to keep 00:35:20.290 --> 00:35:24.090 track of the index on this dummy variable. 00:35:24.090 --> 00:35:26.290 I have the integral over q1. 00:35:26.290 --> 00:35:28.860 I have the two integral over p1. 00:35:28.860 --> 00:35:32.530 Let's keep it so that it doesn't confuse anybody. 00:35:32.530 --> 00:35:36.040 Then I have the collision term. 00:35:36.040 --> 00:35:39.980 Now, the collision term is a bunch of integrals itself. 00:35:39.980 --> 00:35:47.530 It involves integrals over p2, integral over an impact vector 00:35:47.530 --> 00:35:50.070 b, the relative velocity. 00:35:53.940 --> 00:36:01.910 And then I had minus f of p1, f of p2 00:36:01.910 --> 00:36:05.990 for the loss in the channel because of the collisions, 00:36:05.990 --> 00:36:10.690 plus the addition to the channel from the inverse collisions. 00:36:13.890 --> 00:36:16.060 And this whole thing has to be multiplied 00:36:16.060 --> 00:36:22.041 by log f evaluated at p1. 00:36:22.041 --> 00:36:24.480 OK? 00:36:24.480 --> 00:36:28.280 So that's really the quantity that we are interested. 00:36:28.280 --> 00:36:31.200 And the quantity that we are interested, we notice, 00:36:31.200 --> 00:36:36.880 has some kind of a symmetry with respect to indices one and two. 00:36:36.880 --> 00:36:40.920 Something to do with prime and un-prime coordinates. 00:36:40.920 --> 00:36:44.980 But multiplied this bracket that has all of these symmetries, 00:36:44.980 --> 00:36:49.780 with this function that is only evaluated for one of the four 00:36:49.780 --> 00:36:54.460 coordinates for momenta that are appearing here. 00:36:54.460 --> 00:36:59.420 So our next set of tasks is to somehow make this as symmetric 00:36:59.420 --> 00:37:01.070 as possible. 00:37:01.070 --> 00:37:06.250 So the first thing that we do is we exchange indices one 00:37:06.250 --> 00:37:10.920 and two, because really, these are simply dummy integration 00:37:10.920 --> 00:37:12.370 variables. 00:37:12.370 --> 00:37:16.750 I can call the thing that I was calling p1 p2, and vice versa. 00:37:16.750 --> 00:37:21.380 And the integration then becomes the integral over q. 00:37:21.380 --> 00:37:26.290 Maybe that's why it's useful not to have, for the q, any index, 00:37:26.290 --> 00:37:32.110 because I don't care to change the index for that. 00:37:32.110 --> 00:37:34.500 d cubed p2, p1. 00:37:34.500 --> 00:37:38.780 Doesn't really matter in what order I write them. 00:37:38.780 --> 00:37:43.330 d2b is a relative separation. 00:37:43.330 --> 00:37:47.120 Again, it's an impact parameter. 00:37:47.120 --> 00:37:50.200 I can call this v1 minus v2. 00:37:50.200 --> 00:37:51.990 But since it's an absolute value, 00:37:51.990 --> 00:37:54.030 it doesn't matter really. 00:37:54.030 --> 00:37:56.950 And so the only thing that happens 00:37:56.950 --> 00:37:59.985 is that I will have-- essentially, the product is 00:37:59.985 --> 00:38:00.485 symmetric. 00:38:08.410 --> 00:38:13.970 The only thing that was separately f 00:38:13.970 --> 00:38:16.410 of p1 before now becomes f of p2. 00:38:18.980 --> 00:38:22.420 And then what I can do is I can say, OK, those are really 00:38:22.420 --> 00:38:27.040 two forms of the same integral, and I can average them, 00:38:27.040 --> 00:38:33.310 and write the average, which is 1/2 the integral d cubed q, d 00:38:33.310 --> 00:38:45.200 cubed p1, d cubed p2, d2 b, relative velocity, minus f 00:38:45.200 --> 00:38:53.700 of p1, f of p2, f of p1 prime, f of p2 prime. 00:38:53.700 --> 00:39:03.420 And then I have a log of f of p1 plus log of f of p2. 00:39:03.420 --> 00:39:07.845 And I have divided through 1/2. 00:39:07.845 --> 00:39:08.345 OK? 00:39:14.070 --> 00:39:22.200 So what I will do is to now write the next line. 00:39:22.200 --> 00:39:27.320 And then we'll spend some time discussing what happens. 00:39:27.320 --> 00:39:32.500 So artificially, what will happen looks like I essentially 00:39:32.500 --> 00:39:37.800 move prime and un-prime things the same way that I exchanged 00:39:37.800 --> 00:39:40.270 one and two indices before. 00:39:40.270 --> 00:39:45.010 I will exchange the superscripts that are prime or non-prime. 00:39:45.010 --> 00:39:51.268 And if I were to do that, I will get an integral d cubed q, 00:39:51.268 --> 00:39:59.980 d cubed, let's say, p1 prime, d cubed p2 prime, d2b, v2 minus 00:39:59.980 --> 00:40:09.740 v1 prime, minus f of p1-- [SIGHS] 00:40:12.710 --> 00:40:22.790 --prime, f or p2 prime, plus f of p1, f of p2. 00:40:22.790 --> 00:40:25.470 And then here, we would have log of f 00:40:25.470 --> 00:40:31.820 of p1 prime, log of f of p2 prime. 00:40:31.820 --> 00:40:34.180 OK? 00:40:34.180 --> 00:40:40.290 So this is a statement that just saying, 00:40:40.290 --> 00:40:43.550 I moved the primes and un-primes, 00:40:43.550 --> 00:40:47.080 it's very easy to come up with this answer. 00:40:47.080 --> 00:40:52.080 What you have to think very much about what that means. 00:40:52.080 --> 00:40:55.660 Essentially, what I had originally was 00:40:55.660 --> 00:41:02.070 integration invariables p1 and p2. 00:41:02.070 --> 00:41:07.090 And p1 prime and p2 prime were never integration variables. 00:41:07.090 --> 00:41:09.650 So there was not symmetry at the level 00:41:09.650 --> 00:41:13.070 that I have written between one and two, which 00:41:13.070 --> 00:41:15.790 are integration variables, and one prime 00:41:15.790 --> 00:41:19.630 and two prime, which are the quantities that, 00:41:19.630 --> 00:41:24.750 through the collision operator, are related to p1 and p2. 00:41:24.750 --> 00:41:28.920 So essentially, p1 prime and p2 prime 00:41:28.920 --> 00:41:33.330 are the kind of functions that we have written here. 00:41:33.330 --> 00:41:35.970 Here, I had p1 double prime, p2 double prime. 00:41:35.970 --> 00:41:40.300 But up to sort of inverting some signs, et cetera, 00:41:40.300 --> 00:41:45.290 this is the kind of functional relationship that you have 00:41:45.290 --> 00:41:50.380 between p primes and p's. 00:41:50.380 --> 00:41:51.810 OK? 00:41:51.810 --> 00:41:56.080 So now we have to worry about a number of things. 00:41:56.080 --> 00:41:58.180 The simplest one is actually what 00:41:58.180 --> 00:42:05.170 happened to this factor here-- the relative velocity. 00:42:05.170 --> 00:42:08.200 Now think back about collisions that you 00:42:08.200 --> 00:42:10.360 have in the center of mass. 00:42:10.360 --> 00:42:12.590 In the center of mass, two particles 00:42:12.590 --> 00:42:17.500 come with velocities that are opposite each other. 00:42:17.500 --> 00:42:21.320 They bang onto each other, go different direction. 00:42:21.320 --> 00:42:23.620 The magnitude of the velocities does not 00:42:23.620 --> 00:42:25.510 change an inelastic collision, which 00:42:25.510 --> 00:42:27.460 is what we are looking at. 00:42:27.460 --> 00:42:28.720 Right? 00:42:28.720 --> 00:42:34.800 So after the collision, what we have is that p2 prime minus 00:42:34.800 --> 00:42:38.380 p1 prime is the same thing as p1 and minus 00:42:38.380 --> 00:42:39.735 p2 for elastic conditions. 00:42:42.770 --> 00:42:45.640 So as far as this factor is concerned, 00:42:45.640 --> 00:42:55.180 I could very well replace that with v1 minus v2 00:42:55.180 --> 00:42:56.220 without the primes. 00:42:56.220 --> 00:42:57.835 It really doesn't make any difference. 00:43:01.210 --> 00:43:06.060 Another thing that you can convince yourself 00:43:06.060 --> 00:43:12.640 is that from the perspective of the picture that I drew, 00:43:12.640 --> 00:43:17.071 there was some kind of an impact vector B. 00:43:17.071 --> 00:43:19.060 You say, well, what happens if I look 00:43:19.060 --> 00:43:22.780 at from the perspective of the products of the interaction? 00:43:22.780 --> 00:43:24.600 So the products of the interaction 00:43:24.600 --> 00:43:27.270 are going in a different direction. 00:43:27.270 --> 00:43:30.670 I would have to redraw my coordinates that 00:43:30.670 --> 00:43:34.820 would correspond to A prime and B prime. 00:43:34.820 --> 00:43:38.380 But what you can convince yourself that essentially, 00:43:38.380 --> 00:43:43.200 the impact parameter, as far as these products is concerned, 00:43:43.200 --> 00:43:46.110 is just a rotated version of the original one 00:43:46.110 --> 00:43:49.070 without changing its magnitude. 00:43:49.070 --> 00:43:53.220 So you have also that magnitude of b prime 00:43:53.220 --> 00:43:55.950 is the same thing as magnitude of b. 00:43:55.950 --> 00:43:59.610 There's a rotation, but the impact parameter 00:43:59.610 --> 00:44:01.090 does not change. 00:44:01.090 --> 00:44:03.320 If you like, that is also related 00:44:03.320 --> 00:44:08.190 to conservation of energy, because if it wasn't so, 00:44:08.190 --> 00:44:11.670 the potential energy would be different 00:44:11.670 --> 00:44:13.920 right after the collisions. 00:44:13.920 --> 00:44:16.620 OK? 00:44:16.620 --> 00:44:21.215 So I really was kind of careless here. 00:44:21.215 --> 00:44:24.280 I should have written d2 b prime. 00:44:24.280 --> 00:44:26.310 But I could have written also d2b. 00:44:26.310 --> 00:44:27.660 It didn't make any difference. 00:44:27.660 --> 00:44:30.660 It's the same. 00:44:30.660 --> 00:44:38.860 And then, of course, these functions-- essentially, 00:44:38.860 --> 00:44:47.330 you can see that you say, OK, I start with, here, p1 and p2. 00:44:47.330 --> 00:44:50.920 And these were functions of p1 and p2. 00:44:50.920 --> 00:44:52.730 And here, I have reversed this. 00:44:52.730 --> 00:44:58.490 p1 prime and p2 prime are the integration variables. 00:44:58.490 --> 00:45:02.950 p1 and p2 are functions of p1 prime and p2 prime 00:45:02.950 --> 00:45:07.130 that, in principle, are obtained by inverting 00:45:07.130 --> 00:45:10.034 this set of equations. 00:45:10.034 --> 00:45:12.500 Right? 00:45:12.500 --> 00:45:15.300 But it really, again, doesn't matter, 00:45:15.300 --> 00:45:20.630 because once you have specified what incoming p1, p2, 00:45:20.630 --> 00:45:24.520 and b are, you know what the outcome of the collision is. 00:45:24.520 --> 00:45:26.670 So it's really the same function. 00:45:26.670 --> 00:45:28.750 Whether you are inverting it, you 00:45:28.750 --> 00:45:32.000 will, again, presume the different location 00:45:32.000 --> 00:45:35.080 of the arguments, but you will have the same function that 00:45:35.080 --> 00:45:38.500 would tell you what the inverse relationship is. 00:45:38.500 --> 00:45:41.120 So really, these functional forms 00:45:41.120 --> 00:45:45.120 are the same as the functional forms that we had before. 00:45:45.120 --> 00:45:48.990 And finally, I claim that in order 00:45:48.990 --> 00:45:52.920 to go through this change of variable, 00:45:52.920 --> 00:45:57.670 I have to also deal with the Jacobian of the transformation. 00:45:57.670 --> 00:46:02.790 But that we have the d3p1d3p2 is the same thing 00:46:02.790 --> 00:46:08.140 as d3p1 prime d3p2 prime. 00:46:08.140 --> 00:46:11.480 And that, in some sense, you can sort of also 00:46:11.480 --> 00:46:14.810 think about what we were doing before with the Liouville 00:46:14.810 --> 00:46:18.730 equation maintaining the volume of phase space. 00:46:18.730 --> 00:46:23.490 So I have some volume of phase space prior to the collision. 00:46:23.490 --> 00:46:27.730 After the collision, I would need to have the same volume. 00:46:27.730 --> 00:46:30.130 There is, of course, in the volumes 00:46:30.130 --> 00:46:33.890 that we are considering for the case of the Liouville operator, 00:46:33.890 --> 00:46:38.740 in addition, products of dq1, dq2. 00:46:38.740 --> 00:46:41.410 But you can see that all of those things 00:46:41.410 --> 00:46:44.780 are really put into the same volume when 00:46:44.780 --> 00:46:46.840 we are considering the collision. 00:46:46.840 --> 00:46:49.270 So really, the only thing that is left 00:46:49.270 --> 00:46:53.560 is that the Liouvillian preservation 00:46:53.560 --> 00:46:56.440 of the volume of phase space would 00:46:56.440 --> 00:47:00.160 say that the Jacobian here, you can also ignore. 00:47:00.160 --> 00:47:00.660 Yes. 00:47:00.660 --> 00:47:03.782 AUDIENCE: So are you assuming that this transformation 00:47:03.782 --> 00:47:08.246 [INAUDIBLE] is [? an article, ?] because it's [INAUDIBLE]? 00:47:08.246 --> 00:47:10.110 PROFESSOR: Yes. 00:47:10.110 --> 00:47:14.010 So everything that we do is within the framework 00:47:14.010 --> 00:47:18.112 of a Hamiltonian evolution equation. 00:47:18.112 --> 00:47:19.040 OK? 00:47:19.040 --> 00:47:22.850 So if you like, these kinds of things 00:47:22.850 --> 00:47:25.260 that I'm saying here sometimes are 00:47:25.260 --> 00:47:29.730 said to be consequence of microscopic reversibility, 00:47:29.730 --> 00:47:34.720 in that I'm sort of using, in this panel over here, 00:47:34.720 --> 00:47:38.300 things that are completely Newtonian and microscopic. 00:47:38.300 --> 00:47:45.186 The place that I use statistical argument was over here. 00:47:45.186 --> 00:47:46.540 OK? 00:47:46.540 --> 00:47:50.550 So essentially, having done that, 00:47:50.550 --> 00:47:55.200 you can basically also rename variables. 00:47:55.200 --> 00:47:59.000 Things that I was calling p1 prime I can call p1. 00:47:59.000 --> 00:48:05.860 It's another renaming of some variable. 00:48:08.570 --> 00:48:15.120 And again, it really doesn't matter which relative velocity 00:48:15.120 --> 00:48:15.790 I write down. 00:48:15.790 --> 00:48:17.630 They're all the same thing. 00:48:17.630 --> 00:48:19.380 So what will happen is that I will 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. 00:48:28.940 --> 00:48:37.670 And the only thing that happened is that the logs-- oops. 00:48:37.670 --> 00:48:39.010 I should have maintained. 00:48:43.735 --> 00:48:47.030 I don't know why-- where I made a mistake. 00:48:47.030 --> 00:48:52.080 But I should ultimately have come up with this answer. 00:48:55.490 --> 00:48:58.840 So what I wanted to do, and I really 00:48:58.840 --> 00:49:02.850 have to maybe look at it again, is 00:49:02.850 --> 00:49:07.920 do one more step of symmetrization. 00:49:07.920 --> 00:49:17.180 The answer will come down to an integral over p1 and p2, d2b, 00:49:17.180 --> 00:49:18.720 at relative velocity. 00:49:23.060 --> 00:49:31.410 We would have minus f of p1, f of p2, plus f of p1 prime, 00:49:31.410 --> 00:49:32.970 f of p2 prime. 00:49:32.970 --> 00:49:36.286 So this is our collision operator 00:49:36.286 --> 00:49:38.650 that we have discussed here-- the difference 00:49:38.650 --> 00:49:42.240 between what comes in and what comes out. 00:49:42.240 --> 00:49:49.660 And from the first part here, we should get log of f of p1, 00:49:49.660 --> 00:49:52.220 plus log of f of p2, which is the same thing 00:49:52.220 --> 00:49:53.475 as the log of the product. 00:49:56.070 --> 00:50:02.140 And if I had not made errors, the subtraction 00:50:02.140 --> 00:50:05.460 would have involved f of p1 prime, f of p2 prime. 00:50:09.890 --> 00:50:16.510 And the idea is that if you think about the function 00:50:16.510 --> 00:50:24.650 log of something as a function of that something, 00:50:24.650 --> 00:50:28.710 the log is a monotonic function. 00:50:28.710 --> 00:50:33.600 So that if you were to evaluate the log at two 00:50:33.600 --> 00:50:38.440 different points, the sign of the log 00:50:38.440 --> 00:50:42.070 would grow in the same way as the sign of the two 00:50:42.070 --> 00:50:49.190 points, which means that here, you have the difference of two 00:50:49.190 --> 00:50:50.470 logs. 00:50:50.470 --> 00:50:57.310 This is like log of s1 minus log of s2. 00:50:57.310 --> 00:51:02.820 And it is multiplied with minus s1 minus s2. 00:51:02.820 --> 00:51:05.860 So if s1 is greater than s2, log s1 00:51:05.860 --> 00:51:07.190 would be greater than log s2. 00:51:07.190 --> 00:51:12.040 If s1 is less than s2, log s1 would be less than log s2. 00:51:12.040 --> 00:51:15.050 In either case, what you would find 00:51:15.050 --> 00:51:18.830 is that the product of the two brackets is negative. 00:51:18.830 --> 00:51:20.570 And this whole thing has to be negative. 00:51:23.870 --> 00:51:24.370 OK. 00:51:24.370 --> 00:51:31.020 So the thing that confused me-- and maybe we can go back 00:51:31.020 --> 00:51:40.130 and look at is that probably, what I was doing here 00:51:40.130 --> 00:51:47.510 was a little bit too impatient, because when 00:51:47.510 --> 00:51:57.640 I changed variables, p1 prime f of p2. 00:52:00.310 --> 00:52:01.580 I removed the primes. 00:52:06.416 --> 00:52:06.915 Log. 00:52:09.726 --> 00:52:10.350 Reduced primes. 00:52:14.430 --> 00:52:15.785 Actually, I had it here. 00:52:15.785 --> 00:52:17.831 I don't know why I was worried. 00:52:17.831 --> 00:52:18.330 Yeah. 00:52:18.330 --> 00:52:22.570 So the signs are, I think, right, that when I add them, 00:52:22.570 --> 00:52:24.770 I would get the subtraction. 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. 00:52:32.640 --> 00:52:38.410 After I did this change, they appear with the positive sign. 00:52:38.410 --> 00:52:43.540 So there is the sign difference between the two of them. 00:52:43.540 --> 00:52:45.380 OK? 00:52:45.380 --> 00:52:49.230 So we have a function, therefore, 00:52:49.230 --> 00:52:52.330 that, just like entropy throughout the process, 00:52:52.330 --> 00:52:53.570 will increase. 00:52:53.570 --> 00:52:56.820 This is like a negative entropy. 00:52:56.820 --> 00:53:00.630 If I were to really solve this set of equations, 00:53:00.630 --> 00:53:03.305 starting with the f that describes things that 00:53:03.305 --> 00:53:06.200 are distributed in this box, and then 00:53:06.200 --> 00:53:09.600 allow it to expand into the other box, 00:53:09.600 --> 00:53:12.380 it will follow-- this solution will 00:53:12.380 --> 00:53:17.910 follow some particular trajectory that will ultimately 00:53:17.910 --> 00:53:21.050 no longer change as a function of time. 00:53:21.050 --> 00:53:22.840 It will not go back and forth. 00:53:22.840 --> 00:53:26.360 It will not have the full reversibility 00:53:26.360 --> 00:53:30.490 that these set of equations does have. 00:53:30.490 --> 00:53:32.780 Now, that's actually not a bad thing. 00:53:32.780 --> 00:53:37.690 It is true, indeed, that these equations are reversible. 00:53:37.690 --> 00:53:41.730 And there is a theorem that if you wait sufficiently long, 00:53:41.730 --> 00:53:44.960 this will go back into here. 00:53:44.960 --> 00:53:48.130 But the time it take for that to happen 00:53:48.130 --> 00:53:53.085 grows something of the order of the size of the system divided 00:53:53.085 --> 00:53:53.585 these. 00:53:56.100 --> 00:53:58.940 The time, up to various pre-factors, 00:53:58.940 --> 00:54:02.630 would grow exponentially with the number of particles 00:54:02.630 --> 00:54:04.490 that you have. 00:54:04.490 --> 00:54:07.390 And remember that you have of the order of 10 00:54:07.390 --> 00:54:10.050 to the 23 particles. 00:54:10.050 --> 00:54:13.290 So there is, indeed, mathematically rigorously, 00:54:13.290 --> 00:54:19.510 a recursion time if you really had this box sitting in vacuum 00:54:19.510 --> 00:54:23.420 and there was no influence on the walls of the container. 00:54:23.420 --> 00:54:27.760 After many, many, many, many ages of the universe, 00:54:27.760 --> 00:54:31.270 the gas would go back for an instant of time 00:54:31.270 --> 00:54:34.120 in the other box, and then would go over there. 00:54:34.120 --> 00:54:39.690 But that's something that really is of no practical relevance 00:54:39.690 --> 00:54:44.310 when we are thinking about the gas and how it expands. 00:54:44.310 --> 00:54:45.500 OK? 00:54:45.500 --> 00:54:49.210 But now let's see, with these approximations, 00:54:49.210 --> 00:54:51.140 can we do better. 00:54:51.140 --> 00:54:53.840 Can we figure out exactly how long does it 00:54:53.840 --> 00:54:58.180 take for the gas to go from one side to the other side. 00:54:58.180 --> 00:55:01.670 And what is the shape of the streamlines and everything else 00:55:01.670 --> 00:55:04.550 as this process is taking place? 00:55:04.550 --> 00:55:06.704 Yes. 00:55:06.704 --> 00:55:07.620 AUDIENCE: [INAUDIBLE]. 00:55:23.800 --> 00:55:25.580 PROFESSOR: Yes. 00:55:25.580 --> 00:55:26.080 Yes. 00:55:26.080 --> 00:55:30.380 So if you like, we want to think of this 00:55:30.380 --> 00:55:32.835 in terms of information, in terms of entropy. 00:55:32.835 --> 00:55:33.730 It doesn't matter. 00:55:33.730 --> 00:55:37.610 It is something that is changing as a function of time. 00:55:37.610 --> 00:55:40.620 And the timescale by which that happens 00:55:40.620 --> 00:55:45.200 has to do with the timescales that we set up and imprinted 00:55:45.200 --> 00:55:47.017 into this equation. 00:55:47.017 --> 00:55:48.440 OK? 00:55:48.440 --> 00:55:51.410 So the next step that I have to answer 00:55:51.410 --> 00:55:54.260 is, what is that time step? 00:55:54.260 --> 00:55:58.270 And in particular, if I look at this equation by itself, 00:55:58.270 --> 00:56:01.330 I would say that it has two timescales. 00:56:01.330 --> 00:56:04.310 On the left-hand side, you have this tau 00:56:04.310 --> 00:56:06.990 that has to do with the size of the box. 00:56:06.990 --> 00:56:09.350 We said it's very long. 00:56:09.350 --> 00:56:13.100 On the right-hand side, it has the time 00:56:13.100 --> 00:56:15.290 that I have to wait for a collision 00:56:15.290 --> 00:56:20.590 to occur, which is certainly much shorter than this time, 00:56:20.590 --> 00:56:23.170 for an ordinary gas. 00:56:23.170 --> 00:56:27.260 And so we expect, therefore, that the right-hand side 00:56:27.260 --> 00:56:31.800 of this equation would determine things much more rapidly 00:56:31.800 --> 00:56:35.040 compared to the left-hand side of the equation. 00:56:35.040 --> 00:56:39.700 And we want to now quantify that and make that 00:56:39.700 --> 00:56:42.610 into something that is relevant to things 00:56:42.610 --> 00:56:47.670 that we know and understand about a gas expand. 00:56:47.670 --> 00:56:49.480 OK? 00:56:49.480 --> 00:56:55.350 So let's sort of think about, well, what would happen? 00:56:55.350 --> 00:57:01.070 We said that this H, as a function of time, 00:57:01.070 --> 00:57:04.140 is going in one direction. 00:57:04.140 --> 00:57:08.060 Presumably, after some time-- we don't know how long here, 00:57:08.060 --> 00:57:10.440 and that's a good question-- eventually, 00:57:10.440 --> 00:57:12.500 hopefully, it will reach an equilibrium, 00:57:12.500 --> 00:57:15.320 like, the entropy will cease to change. 00:57:15.320 --> 00:57:19.175 And the question is, what is this equilibrium? 00:57:19.175 --> 00:57:21.800 So let's see if we can gain some information about equilibrium. 00:57:30.380 --> 00:57:34.210 So we said that dH by dt presumably 00:57:34.210 --> 00:57:38.330 has to be 0, because if it is not 0, 00:57:38.330 --> 00:57:40.650 then it can continue to decrease. 00:57:40.650 --> 00:57:44.920 You are not quite where you want to be. 00:57:44.920 --> 00:57:50.630 Now let's look at what is the condition for dH by dt to be 0. 00:57:50.630 --> 00:57:56.050 H is this integral that I have to perform over 00:57:56.050 --> 00:58:00.010 the entire phase space. 00:58:00.010 --> 00:58:04.050 Now, each element of that integral evaluated 00:58:04.050 --> 00:58:07.200 at some particular location in phase space 00:58:07.200 --> 00:58:09.090 is by itself positive. 00:58:09.090 --> 00:58:12.070 This argument that we had before is not 00:58:12.070 --> 00:58:13.800 about the entire integral. 00:58:13.800 --> 00:58:17.010 It's valid about every little piece of the integration 00:58:17.010 --> 00:58:18.340 that I am making. 00:58:18.340 --> 00:58:20.830 It has a particular sign. 00:58:20.830 --> 00:58:26.530 So if the entire integral has to be 0, every little piece of it 00:58:26.530 --> 00:58:29.770 has to be individually 0. 00:58:29.770 --> 00:58:32.310 So that means that I would require 00:58:32.310 --> 00:58:39.190 that log of f at any q evaluated, 00:58:39.190 --> 00:58:47.680 let's say, for p1 and p2 should be the same thing 00:58:47.680 --> 00:58:53.230 as log of f evaluated at p1 prime and p2. 00:58:58.490 --> 00:59:03.680 This would be true for all q. 00:59:03.680 --> 00:59:04.180 Right? 00:59:09.040 --> 00:59:11.630 You look at this and you say, well, what? 00:59:11.630 --> 00:59:14.560 This seems kind of-- yes? 00:59:14.560 --> 00:59:17.015 AUDIENCE: What about the other one in parentheses, 00:59:17.015 --> 00:59:19.500 beside the integral? 00:59:19.500 --> 00:59:21.350 PROFESSOR: It is the same thing. 00:59:21.350 --> 00:59:24.300 So this is s1 minus s2. 00:59:24.300 --> 00:59:26.700 And the other is log of s1 minus s2. 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. 00:59:34.447 --> 00:59:38.360 When s1 equals to s2, both parentheses are 0. 00:59:38.360 --> 00:59:41.000 Although, even if it wasn't, it would have been sufficient 00:59:41.000 --> 00:59:42.620 for one parentheses to be 0. 00:59:42.620 --> 00:59:48.210 But this is necessary, because the sign is determined by this. 00:59:48.210 --> 00:59:49.566 OK? 00:59:49.566 --> 00:59:51.450 Now, you look at this and you say, well, 00:59:51.450 --> 00:59:53.780 what the he-- what does this mean? 00:59:53.780 --> 00:59:59.220 How can I-- and the functions p1 prime and p2 prime I 00:59:59.220 --> 01:00:02.300 have to get by integrating Newton's equations. 01:00:02.300 --> 01:00:04.920 I have no idea how complicated they are. 01:00:04.920 --> 01:00:08.670 And you want me to solve this equation. 01:00:08.670 --> 01:00:10.410 Well, I tell you that the answer to this 01:00:10.410 --> 01:00:12.580 is actually very simple. 01:00:12.580 --> 01:00:17.080 So let me write the answer to it in stages. 01:00:17.080 --> 01:00:21.615 Log of f of p and q. 01:00:21.615 --> 01:00:24.390 So it's a function of p and q that I'm 01:00:24.390 --> 01:00:28.090 after that have to be evaluated at p1, p2, 01:00:28.090 --> 01:00:31.320 and then should be equal to p1 prime and p2 prime 01:00:31.320 --> 01:00:35.702 for any combination of p1, p2, et cetera, that I choose. 01:00:35.702 --> 01:00:41.980 Well, I say, OK, I will put any function of q that I like. 01:00:41.980 --> 01:00:46.810 And that will work, because it's independent of p. 01:00:46.810 --> 01:00:48.940 So I have minus alpha, minus alpha 01:00:48.940 --> 01:00:52.030 is the same as minus alpha, minus alpha. 01:00:52.030 --> 01:00:52.890 OK? 01:00:52.890 --> 01:00:57.800 Then I claim that, OK, I will put a minus gamma dot p. 01:00:57.800 --> 01:01:00.360 Gamma could be dependent on q. 01:01:00.360 --> 01:01:02.720 Dot p. 01:01:02.720 --> 01:01:07.590 And I claim that that would work, because that term would 01:01:07.590 --> 01:01:11.555 give me gamma dot p1 plus p2, gamma 01:01:11.555 --> 01:01:14.550 dot p1 prime plus p2 prime. 01:01:14.550 --> 01:01:16.150 What do I know for sure? 01:01:16.150 --> 01:01:18.240 Momentum is conserved. 01:01:18.240 --> 01:01:24.280 p1 plus p2 is the same thing as p1 prime plus p2 prime. 01:01:24.280 --> 01:01:26.760 It's just the conservation of momentum. 01:01:26.760 --> 01:01:30.670 I know that to be true for sure. 01:01:30.670 --> 01:01:34.790 And finally, kinetic energy is conserved. 01:01:34.790 --> 01:01:39.440 Once I'm away from the location where the interaction is taking 01:01:39.440 --> 01:01:43.580 place, I can choose any function of q 01:01:43.580 --> 01:01:46.970 multiplied with p squared over 2m. 01:01:46.970 --> 01:01:48.820 And then on the left-hand side, I 01:01:48.820 --> 01:01:51.910 would have the sum of the incoming kinetic energies. 01:01:51.910 --> 01:01:53.410 On the right-hand side, I would have 01:01:53.410 --> 01:01:56.740 the sum of outgoing kinetic energies. 01:01:56.740 --> 01:01:58.800 So what have I done? 01:01:58.800 --> 01:02:02.090 I have identified quantities that 01:02:02.090 --> 01:02:04.260 are conserved in the collision. 01:02:04.260 --> 01:02:06.960 So basically, I can keep going if I 01:02:06.960 --> 01:02:10.290 have additional such things. 01:02:10.290 --> 01:02:13.640 And so basically, the key to this 01:02:13.640 --> 01:02:18.020 is to identify collision-conserved quantities. 01:02:21.440 --> 01:02:26.030 And so momentum, energy, the first one corresponds 01:02:26.030 --> 01:02:27.970 to the number of particles. 01:02:27.970 --> 01:02:30.740 These are quantities that are conserved. 01:02:30.740 --> 01:02:33.610 And so you know that this form will work, 01:02:33.610 --> 01:02:38.450 which means that I have a candidate for equilibrium, 01:02:38.450 --> 01:02:43.580 which is that f of p and q should have a form that is 01:02:43.580 --> 01:02:49.540 exponential of some function of q that I don't-- at this stage, 01:02:49.540 --> 01:02:51.830 could be quite arbitrary. 01:02:51.830 --> 01:02:59.150 And some other function of q times p squared minus 2m. 01:02:59.150 --> 01:03:03.750 And let me absorb this other gamma term. 01:03:03.750 --> 01:03:06.690 I can certainly do so by putting something 01:03:06.690 --> 01:03:11.400 like pi of 1 squared over 2m. 01:03:15.900 --> 01:03:18.726 OK? 01:03:18.726 --> 01:03:23.030 And this certainly is a form that 01:03:23.030 --> 01:03:31.660 will set the right-hand side of the H theorem to 0. 01:03:31.660 --> 01:03:36.130 And furthermore, it sets the right-hand side 01:03:36.130 --> 01:03:39.220 of the Boltzmann equation to 0, because, again, 01:03:39.220 --> 01:03:40.840 for the Boltzmann equation, all I 01:03:40.840 --> 01:03:43.190 needed was that the product of two f's 01:03:43.190 --> 01:03:46.590 should be the same before and after the collision. 01:03:46.590 --> 01:03:49.030 And this is just the logs. 01:03:49.030 --> 01:03:53.300 If I were to exponentiate that, I have exactly what I need. 01:03:53.300 --> 01:03:57.690 So basically, this also sets the right-hand side 01:03:57.690 --> 01:04:00.000 of the Boltzmann equation to 0. 01:04:00.000 --> 01:04:02.650 And this is what I will call a local equilibrium. 01:04:08.420 --> 01:04:10.600 What do I mean by that? 01:04:10.600 --> 01:04:13.250 Essentially, we can see that right now, I 01:04:13.250 --> 01:04:16.510 have no knowledge about the relationship 01:04:16.510 --> 01:04:18.920 between different points in space, 01:04:18.920 --> 01:04:22.700 because these alpha, beta, and pi are completely arbitrary, as 01:04:22.700 --> 01:04:26.770 far as my argumentation so far is concerned. 01:04:26.770 --> 01:04:33.280 So locally, at each point in q, I can have a form such as this. 01:04:33.280 --> 01:04:36.480 And this form, you should remind you 01:04:36.480 --> 01:04:40.750 it's something like a Boltzmann weight of the kinetic energy, 01:04:40.750 --> 01:04:44.180 but moving in some particular direction. 01:04:44.180 --> 01:04:49.950 And essentially, what this captures 01:04:49.950 --> 01:04:54.220 is that through relaxing the right-hand side 01:04:54.220 --> 01:04:57.930 of the Boltzmann equation, we randomize 01:04:57.930 --> 01:05:03.170 the magnitude of the momenta, so that the magnitudes 01:05:03.170 --> 01:05:08.890 of the kinetic energy are kind of Boltzmann-distributed. 01:05:08.890 --> 01:05:11.910 This is what this term does. 01:05:11.910 --> 01:05:19.990 If I say that this term being much larger than this term 01:05:19.990 --> 01:05:21.440 is the more important one. 01:05:21.440 --> 01:05:24.020 If I need neglect this, then this is what happens. 01:05:24.020 --> 01:05:27.620 I will very rapidly reach a situation 01:05:27.620 --> 01:05:31.005 in which the momenta have been randomized 01:05:31.005 --> 01:05:32.410 through the collisions. 01:05:32.410 --> 01:05:35.040 Essentially, the collisions, they 01:05:35.040 --> 01:05:39.550 change the direction of the momenta. 01:05:39.550 --> 01:05:43.880 They preserve the average of the momenta. 01:05:43.880 --> 01:05:45.940 So there is some sense of the momentum 01:05:45.940 --> 01:05:49.660 of the incoming thing that is left over still. 01:05:49.660 --> 01:05:53.290 But there is some relaxation that took place. 01:05:53.290 --> 01:05:56.450 But clearly, this is not the end of the story, 01:05:56.450 --> 01:05:58.940 because this f that I have written here-- 01:05:58.940 --> 01:06:01.650 let's say local equilibrium. 01:06:01.650 --> 01:06:11.464 f local equilibrium does not satisfy the Boltzmann equation. 01:06:11.464 --> 01:06:13.080 Right? 01:06:13.080 --> 01:06:17.490 It set the right-hand side of the Boltzmann equation to 0. 01:06:17.490 --> 01:06:19.470 But it doesn't set the left-hand side. 01:06:19.470 --> 01:06:23.990 The left-hand side completely is unhappy with this. 01:06:23.990 --> 01:06:27.260 For the left-hand side to be 0, I 01:06:27.260 --> 01:06:32.640 would require the Poisson bracket of H1 and f to be 0. 01:06:32.640 --> 01:06:35.300 After all, the Liouvillian operator 01:06:35.300 --> 01:06:38.690 had the Poisson bracket of H1 and f. 01:06:38.690 --> 01:06:41.310 And we have seen functions that satisfy 01:06:41.310 --> 01:06:49.500 this have to be of the form f1 of H. And in our case, 01:06:49.500 --> 01:06:54.240 my f1 is simply the kinetic energy of one particle, 01:06:54.240 --> 01:06:56.385 plus the potential due to the box. 01:06:59.390 --> 01:07:01.560 OK? 01:07:01.560 --> 01:07:08.640 So you can see that the only way that I can make this marriage 01:07:08.640 --> 01:07:18.670 with this is to have a global equilibrium, where 01:07:18.670 --> 01:07:28.390 f for the global equilibrium is proportional, let's say, 01:07:28.390 --> 01:07:35.200 to exponential of minus beta p squared over 2m plus U of q. 01:07:39.592 --> 01:07:41.060 OK? 01:07:41.060 --> 01:07:47.430 So I identify a piece that looks like this p squared over 2m. 01:07:47.430 --> 01:07:51.740 I have to make pi to be 0 in order to achieve this. 01:07:51.740 --> 01:07:54.940 I have to make beta to be independent of q. 01:07:54.940 --> 01:07:58.120 I have to make alpha of f marriage with beta 01:07:58.120 --> 01:08:00.870 to give me the potential. 01:08:00.870 --> 01:08:02.440 OK? 01:08:02.440 --> 01:08:06.240 So essentially, what happens is this. 01:08:06.240 --> 01:08:11.400 I open this hole. 01:08:11.400 --> 01:08:15.120 The gas starts to stream out. 01:08:15.120 --> 01:08:18.899 And then the particles will collide with each other. 01:08:18.899 --> 01:08:21.910 The collision of particles over some timescale that 01:08:21.910 --> 01:08:24.470 is related to this collision time 01:08:24.470 --> 01:08:30.470 will randomize their momenta so that locally, suddenly, 01:08:30.470 --> 01:08:35.420 I have a solution such as this. 01:08:35.420 --> 01:08:39.859 But then the left-hand side of the Boltzmann equation 01:08:39.859 --> 01:08:41.440 takes over. 01:08:41.440 --> 01:08:44.220 And over a timescale that is much longer, 01:08:44.220 --> 01:08:47.740 it will then change the parameters of this. 01:08:47.740 --> 01:08:51.109 So basically, these parameters should really 01:08:51.109 --> 01:08:56.210 be thought of as functions of time, whose evolution changes. 01:08:56.210 --> 01:09:00.479 So initially, maybe locally over here, 01:09:00.479 --> 01:09:04.080 there is a stream velocity that goes towards the right. 01:09:04.080 --> 01:09:07.220 So I should have an average at that time that 01:09:07.220 --> 01:09:09.210 prefers to go to the right. 01:09:09.210 --> 01:09:11.049 I wait sufficiently long. 01:09:11.049 --> 01:09:13.090 Presumably, the whole thing comes to equilibrium. 01:09:13.090 --> 01:09:16.120 And that average goes to 0 [INAUDIBLE]. 01:09:16.120 --> 01:09:18.790 So presumably, the next thing that I 01:09:18.790 --> 01:09:22.020 need to do in order to answer the question, 01:09:22.020 --> 01:09:24.439 how does this thing come to equilibrium, 01:09:24.439 --> 01:09:29.510 is to find out how the left-hand side of the equation 01:09:29.510 --> 01:09:33.819 manipulates these parameters alpha, beta, and pi. 01:09:33.819 --> 01:09:36.910 But that's not quite consistent either. 01:09:36.910 --> 01:09:40.620 Because I first assume that the right-hand side is 0, 01:09:40.620 --> 01:09:43.569 and then said, oh, that's not a good solution. 01:09:43.569 --> 01:09:46.300 Let's go and see what the left-hand side does. 01:09:46.300 --> 01:09:49.120 But then if the left-hand side is changing things, 01:09:49.120 --> 01:09:53.279 I can't be over here where dH by dt is 0. 01:09:53.279 --> 01:09:56.240 So I have not done things consistently yet, 01:09:56.240 --> 01:09:59.480 although I have clearly captured the whole lot of what 01:09:59.480 --> 01:10:02.620 I want to describe eventually. 01:10:02.620 --> 01:10:03.600 OK? 01:10:03.600 --> 01:10:05.971 So let's see how we can do it systematically. 01:10:05.971 --> 01:10:06.762 AUDIENCE: Question. 01:10:06.762 --> 01:10:07.600 PROFESSOR: Yes. 01:10:07.600 --> 01:10:13.420 AUDIENCE: When you [? write ?] the condition for [INAUDIBLE], 01:10:13.420 --> 01:10:15.845 [? did you ?] write f1 of Hamiltonian. 01:10:15.845 --> 01:10:17.785 And ends up Hamiltonian [INAUDIBLE] 01:10:17.785 --> 01:10:19.105 includes interaction terms. 01:10:19.105 --> 01:10:20.980 PROFESSOR: It's the one-particle Hamiltonian. 01:10:20.980 --> 01:10:21.924 AUDIENCE: Oh. 01:10:21.924 --> 01:10:22.870 PROFESSOR: Right. 01:10:22.870 --> 01:10:27.840 So by the time I got to the equation for f1, I have H1. 01:10:27.840 --> 01:10:31.490 H1 does not have a second particle in it to collide with. 01:10:36.290 --> 01:10:39.105 OK? 01:10:39.105 --> 01:10:42.360 AUDIENCE: Then equilibrium system's distribution 01:10:42.360 --> 01:10:44.760 of particles in container does not 01:10:44.760 --> 01:10:50.418 depend on their interactions after the [INAUDIBLE]? 01:10:50.418 --> 01:10:53.681 PROFESSOR: In the story that we are following here, yes, it 01:10:53.681 --> 01:10:54.180 does not. 01:10:54.180 --> 01:10:57.810 So this is an approximation that actually 01:10:57.810 --> 01:11:00.680 goes with the approximations that we 01:11:00.680 --> 01:11:04.410 made in neglecting some of the higher order terms, with nd 01:11:04.410 --> 01:11:07.310 cubed being much less than 1, and not 01:11:07.310 --> 01:11:10.710 looking at things at resolution that is of the order of d. 01:11:10.710 --> 01:11:13.740 And we would imagine that it is only at the resolutions-- 01:11:13.740 --> 01:11:17.930 AUDIENCE: [INAUDIBLE] neglected [INAUDIBLE] atomized gas 01:11:17.930 --> 01:11:19.842 conjoined onto the molecules. 01:11:19.842 --> 01:11:20.798 Something like that. 01:11:20.798 --> 01:11:23.830 PROFESSOR: That would require a different description. 01:11:23.830 --> 01:11:24.658 AUDIENCE: Yeah. 01:11:24.658 --> 01:11:25.486 It would require [INAUDIBLE]. 01:11:25.486 --> 01:11:26.152 PROFESSOR: Yeah. 01:11:26.152 --> 01:11:27.080 So indeed. 01:11:27.080 --> 01:11:30.910 Eventually, in the description that we are going to get, 01:11:30.910 --> 01:11:33.400 we are looking at the distribution 01:11:33.400 --> 01:11:35.720 that would be appropriate to a dilute gas 01:11:35.720 --> 01:11:38.251 and which interactions are not seen. 01:11:41.620 --> 01:11:43.800 OK? 01:11:43.800 --> 01:11:44.300 All right. 01:11:44.300 --> 01:11:46.220 So let's see how we can do it better. 01:11:46.220 --> 01:11:49.170 So clearly, the key to the process 01:11:49.170 --> 01:11:56.620 is that collisions that take place frequently at timescale 01:11:56.620 --> 01:12:00.240 tau x randomize a lot of things. 01:12:00.240 --> 01:12:06.790 But they cannot randomize quantities that are conserved 01:12:06.790 --> 01:12:07.870 in collisions. 01:12:07.870 --> 01:12:11.057 So those are the quantities that we have to look at. 01:12:11.057 --> 01:12:13.140 And again, these are quantities that are conserved 01:12:13.140 --> 01:12:16.060 in collision, as opposed to quantities 01:12:16.060 --> 01:12:20.270 that are conserved according to H1. 01:12:20.270 --> 01:12:24.390 Like, momentum is conserved in the collision. 01:12:24.390 --> 01:12:26.720 Clearly, momentum was not conserved 01:12:26.720 --> 01:12:30.400 in H1 because of the size of the box 01:12:30.400 --> 01:12:34.400 and did not appear in the eventual solution. 01:12:34.400 --> 01:12:35.340 OK. 01:12:35.340 --> 01:12:36.797 Let's see what we get. 01:12:47.030 --> 01:12:51.100 So we are going to make a couple of definitions. 01:12:51.100 --> 01:12:53.495 First, we define a local density. 01:13:01.000 --> 01:13:04.560 I say that I take my solution to Boltzmann equation. 01:13:04.560 --> 01:13:06.170 And now I don't know what it is. 01:13:06.170 --> 01:13:09.070 But whatever it is, I take the solution 01:13:09.070 --> 01:13:12.520 to the Boltzmann equation as a function of p and q and t. 01:13:12.520 --> 01:13:16.120 But all I am interested is, at this particular location, 01:13:16.120 --> 01:13:19.400 have the particles arrived yet, what's the density. 01:13:19.400 --> 01:13:24.720 So to get that, I essentially integrate 01:13:24.720 --> 01:13:27.760 over the momenta, which is the quantity that I am not 01:13:27.760 --> 01:13:30.390 interested. 01:13:30.390 --> 01:13:30.890 OK? 01:13:30.890 --> 01:13:33.050 Obvious. 01:13:33.050 --> 01:13:37.160 I define-- maybe I need the average kinetic energy 01:13:37.160 --> 01:13:38.810 at that location. 01:13:38.810 --> 01:13:44.520 So maybe, in general, I need the average or some other quantity 01:13:44.520 --> 01:13:46.860 that I define at location q and t. 01:13:51.990 --> 01:14:03.830 What I do is I integrate at that position f of p, and q, and t, 01:14:03.830 --> 01:14:11.510 and this function O. Actually, I also define a normalization. 01:14:11.510 --> 01:14:17.420 I divide by n of q and t. 01:14:17.420 --> 01:14:17.920 OK? 01:14:22.820 --> 01:14:26.090 Now, I implicitly used something over there 01:14:26.090 --> 01:14:29.180 that, again, I define here, which 01:14:29.180 --> 01:14:43.620 is a collision-conserved quantity, 01:14:43.620 --> 01:14:54.320 is some function of momentum-- I could put q also-- 01:14:54.320 --> 01:15:03.020 such that when evaluated for p1 plus p2 for any pair of p1's 01:15:03.020 --> 01:15:08.630 and p2's that I take is the same thing as when evaluated 01:15:08.630 --> 01:15:13.120 for functions of p1 and p2 that would correspond 01:15:13.120 --> 01:15:18.990 to the outcomes of the collision from 01:15:18.990 --> 01:15:22.070 or giving rights to p1, p2. 01:15:22.070 --> 01:15:22.570 OK? 01:15:26.510 --> 01:15:28.420 And then there's a following theorem 01:15:28.420 --> 01:15:31.130 that we were going to use, which is 01:15:31.130 --> 01:15:40.800 that for any collision-conserved quantity, what I have is 01:15:40.800 --> 01:15:46.645 that I will define a quantity J, which is a function of q and t 01:15:46.645 --> 01:15:53.000 in principle, which is obtained by integrating 01:15:53.000 --> 01:15:57.830 over all momenta. 01:15:57.830 --> 01:16:02.915 Chi-- let's say, evaluated at p1; 01:16:02.915 --> 01:16:10.360 it could be evaluated at q-- times this collision operator-- 01:16:10.360 --> 01:16:14.775 C of f, f. 01:16:21.550 --> 01:16:25.680 And the answer is that for any collision-conserved quantity, 01:16:25.680 --> 01:16:31.175 if you were to evaluate this, the answer would be 0. 01:16:31.175 --> 01:16:33.460 OK? 01:16:33.460 --> 01:16:37.610 So let's write that down. 01:16:37.610 --> 01:16:44.500 So J of q and t is an integral over some momentum p1. 01:16:47.350 --> 01:16:50.396 Because of the collision integral, what do I have? 01:16:50.396 --> 01:16:53.450 I have the integration over momentum. 01:16:53.450 --> 01:16:56.430 I have the integration over b. 01:16:56.430 --> 01:17:01.480 I have the relative velocity v2 minus v1. 01:17:01.480 --> 01:17:06.420 And then I have the subtraction of interaction 01:17:06.420 --> 01:17:11.320 between p1 and p2-- addition of the interactions 01:17:11.320 --> 01:17:15.310 between p1 prime, p2 prime. 01:17:15.310 --> 01:17:19.380 And this collision integral has to be 01:17:19.380 --> 01:17:23.150 multiplied with chi evaluated at p1. 01:17:26.150 --> 01:17:28.390 OK? 01:17:28.390 --> 01:17:28.890 Fine. 01:17:32.200 --> 01:17:37.470 So then I will follow exactly the steps 01:17:37.470 --> 01:17:42.720 that we had in the proof of the H theorem. 01:17:42.720 --> 01:17:45.360 I have some function that only depends 01:17:45.360 --> 01:17:49.900 on one of four momenta that are appearing here. 01:17:49.900 --> 01:17:52.500 And I want to symmetrize it. 01:17:52.500 --> 01:17:58.590 So I change the index one to two. 01:17:58.590 --> 01:18:05.510 So this becomes 1/2 of chi of p1 plus chi of p2. 01:18:10.510 --> 01:18:14.670 I do this transition between primed and un-primed. 01:18:14.670 --> 01:18:18.940 And when I do that, I pick this additional factor 01:18:18.940 --> 01:18:26.808 of minus chi of p1 prime, chi of p2 prime. 01:18:26.808 --> 01:18:32.030 And this becomes division by 2 times 2. 01:18:36.690 --> 01:18:42.210 But then I stated that chi is a collision-conserved quantity, 01:18:42.210 --> 01:18:48.360 so that this term in the bracket is 0 for each point 01:18:48.360 --> 01:18:49.610 that I'm integrating. 01:18:49.610 --> 01:18:51.130 And hence, the answer is 0. 01:18:54.630 --> 01:18:55.630 Yes. 01:18:55.630 --> 01:18:59.130 AUDIENCE: In the [INAUDIBLE] theorem, 01:18:59.130 --> 01:19:05.252 in the second [INAUDIBLE], do you 01:19:05.252 --> 01:19:10.686 want to put minus sign of the--? 01:19:10.686 --> 01:19:13.410 PROFESSOR: There was a minus sign 01:19:13.410 --> 01:19:15.772 worry that I had somewhere. 01:19:15.772 --> 01:19:22.660 AUDIENCE: I think in front of 1/2, at the right term. 01:19:22.660 --> 01:19:27.040 PROFESSOR: In front of the 1/2, minus f of p1. 01:19:27.040 --> 01:19:28.580 You say I have to do this? 01:19:28.580 --> 01:19:29.568 AUDIENCE: No, no, no. 01:19:29.568 --> 01:19:31.544 The next term. 01:19:34.508 --> 01:19:35.496 PROFESSOR: Here? 01:19:35.496 --> 01:19:36.154 AUDIENCE: Yeah. 01:19:36.154 --> 01:19:41.918 I mean, if you put minus sign in front of the 1/2, 01:19:41.918 --> 01:19:43.894 then [INAUDIBLE]? 01:19:48.834 --> 01:19:49.610 PROFESSOR: OK. 01:19:49.610 --> 01:19:53.830 Let's put a question mark here, because I 01:19:53.830 --> 01:19:55.997 don't want to go over that. 01:19:55.997 --> 01:19:57.580 AUDIENCE: [INAUDIBLE] but you switched 01:19:57.580 --> 01:20:00.310 the order of the subtraction [? 1/2. ?] 01:20:00.310 --> 01:20:01.620 So it's minus [INAUDIBLE]. 01:20:01.620 --> 01:20:03.950 PROFESSOR: But I changed also the arguments here. 01:20:03.950 --> 01:20:05.202 AUDIENCE: Oh, OK. 01:20:05.202 --> 01:20:05.910 PROFESSOR: Right? 01:20:05.910 --> 01:20:07.545 So there is some issue. 01:20:07.545 --> 01:20:12.830 But this eventual thing is correct. 01:20:12.830 --> 01:20:16.271 So we will go back to that. 01:20:16.271 --> 01:20:16.770 OK. 01:20:19.300 --> 01:20:25.020 So what is the use of stating that 01:20:25.020 --> 01:20:29.040 for every conserved quantity, you have this? 01:20:29.040 --> 01:20:33.020 It will allow us to get equations 01:20:33.020 --> 01:20:36.820 that will determine how these parameters change 01:20:36.820 --> 01:20:39.190 as a function of time? 01:20:39.190 --> 01:20:39.690 OK. 01:20:39.690 --> 01:20:41.251 So let's see how that goes. 01:20:51.750 --> 01:20:57.530 If then I have that Lf is C, f, f, 01:20:57.530 --> 01:21:04.130 and I showed that the integral of chi against C, f, f is 0, 01:21:04.130 --> 01:21:07.690 then I also know that the integral 01:21:07.690 --> 01:21:23.971 of chi against this L operator on f is 0. 01:21:23.971 --> 01:21:24.470 OK? 01:21:24.470 --> 01:21:27.550 And let's remind you that the L operator is 01:21:27.550 --> 01:21:32.460 a bunch of first derivatives-- d by dt, d by dq, d by dp. 01:21:32.460 --> 01:21:36.790 And so I can use the following result. 01:21:36.790 --> 01:21:41.460 I can write the result as this derivative acting on both chi 01:21:41.460 --> 01:21:42.890 and f. 01:21:42.890 --> 01:21:46.755 And then I will get chi prime f plus f chi prime. 01:21:46.755 --> 01:21:54.220 Chi f prime I have, so I have to subtract fL acting on chi. 01:21:58.950 --> 01:22:00.850 OK? 01:22:00.850 --> 01:22:08.640 And proceeding a little bit. 01:22:08.640 --> 01:22:10.260 One more line maybe. 01:22:10.260 --> 01:22:12.390 d cubed p. 01:22:12.390 --> 01:22:19.100 I have dt plus-- well, actually, let 01:22:19.100 --> 01:22:21.843 me do the rest next time around.