WEBVTT 00:00:00.000 --> 00:00:02.385 align:middle line:90% [SQUEAKING] 00:00:02.385 --> 00:00:04.293 align:middle line:90% [RUSTLING] 00:00:04.293 --> 00:00:07.155 align:middle line:90% [CLICKING] 00:00:07.155 --> 00:00:09.902 align:middle line:90% 00:00:09.902 --> 00:00:11.610 align:middle line:84% GIAN PAOLO BERETTA: So where do we stand? 00:00:11.610 --> 00:00:14.260 align:middle line:90% 00:00:14.260 --> 00:00:22.490 align:middle line:84% We've gone a long way, not yet done heat, 00:00:22.490 --> 00:00:24.890 align:middle line:90% but we've done temperature. 00:00:24.890 --> 00:00:30.380 align:middle line:84% And we introduced this diagram, energy versus entropy, 00:00:30.380 --> 00:00:35.780 align:middle line:84% that allows us to represent most of the results 00:00:35.780 --> 00:00:41.230 align:middle line:90% that we have discussed so far. 00:00:41.230 --> 00:00:49.420 align:middle line:84% So I want to do a little more with this diagram. 00:00:49.420 --> 00:00:53.590 align:middle line:84% Notice that it took us a while to get to temperature 00:00:53.590 --> 00:00:56.830 align:middle line:84% because in order to define temperature, 00:00:56.830 --> 00:01:01.210 align:middle line:84% we need the state principle and the fundamental relation 00:01:01.210 --> 00:01:03.080 align:middle line:84% for the stable equilibrium states. 00:01:03.080 --> 00:01:04.599 align:middle line:84% Because it's only for those states 00:01:04.599 --> 00:01:07.780 align:middle line:84% that we can define the slope of this curve, which 00:01:07.780 --> 00:01:10.210 align:middle line:90% is the temperature. 00:01:10.210 --> 00:01:14.300 align:middle line:84% In other words, states that are inside the region, 00:01:14.300 --> 00:01:20.560 align:middle line:84% these are non-stable, in general, non-equilibrium states. 00:01:20.560 --> 00:01:23.520 align:middle line:84% There is no temperature defined there. 00:01:23.520 --> 00:01:26.070 align:middle line:84% Although, we will see as we go on 00:01:26.070 --> 00:01:35.340 align:middle line:84% that we may be able to represent the system as a composite 00:01:35.340 --> 00:01:39.990 align:middle line:84% of several smaller systems that are close to equilibrium, so 00:01:39.990 --> 00:01:42.760 align:middle line:84% that somehow, we can assign temperatures. 00:01:42.760 --> 00:01:44.350 align:middle line:84% That's what we do in fluid mechanics. 00:01:44.350 --> 00:01:47.675 align:middle line:84% That's an interesting application of what we're doing. 00:01:47.675 --> 00:01:52.205 align:middle line:90% 00:01:52.205 --> 00:01:55.030 align:middle line:84% We have seen that with this diagram, 00:01:55.030 --> 00:01:58.450 align:middle line:84% if I have a generic state here and I 00:01:58.450 --> 00:02:01.540 align:middle line:84% want to represent what I can do-- the best 00:02:01.540 --> 00:02:04.990 align:middle line:84% I can do in a weight process, best 00:02:04.990 --> 00:02:10.240 align:middle line:84% in the sense of trying to extract as much energy as 00:02:10.240 --> 00:02:14.290 align:middle line:84% possible from our system in such a way 00:02:14.290 --> 00:02:20.440 align:middle line:84% that we make something useful, like lifting a weight, 00:02:20.440 --> 00:02:24.160 align:middle line:84% the maximum we can do in a weight process 00:02:24.160 --> 00:02:27.580 align:middle line:84% is the adiabatic availability of that state, which 00:02:27.580 --> 00:02:33.870 align:middle line:84% is represented here by the difference in energy 00:02:33.870 --> 00:02:36.400 align:middle line:90% on the vertical line. 00:02:36.400 --> 00:02:41.600 align:middle line:84% And in fact, the final state, when we do this maximum, 00:02:41.600 --> 00:02:43.660 align:middle line:84% we have done the process reversibly. 00:02:43.660 --> 00:02:45.610 align:middle line:90% Otherwise, you get less. 00:02:45.610 --> 00:02:49.430 align:middle line:84% And in a reversible weight process, 00:02:49.430 --> 00:02:50.870 align:middle line:90% the entropy remains constant. 00:02:50.870 --> 00:02:55.150 align:middle line:84% We've seen that as one of the theorems that we had cited. 00:02:55.150 --> 00:03:00.810 align:middle line:84% And so the process that extracts-- 00:03:00.810 --> 00:03:04.680 align:middle line:84% the weight process that extracts the adiabatic availability, 00:03:04.680 --> 00:03:10.410 align:middle line:84% moves vertically down until you reach the lowest energy 00:03:10.410 --> 00:03:11.890 align:middle line:90% state with that entropy. 00:03:11.890 --> 00:03:15.840 align:middle line:84% You cannot go below, so you cannot extract more energy 00:03:15.840 --> 00:03:18.890 align:middle line:84% because there are no states below that one. 00:03:18.890 --> 00:03:23.800 align:middle line:90% 00:03:23.800 --> 00:03:30.360 align:middle line:84% We've discussed the fact that the thermal reservoir 00:03:30.360 --> 00:03:31.870 align:middle line:90% has a constant temperature. 00:03:31.870 --> 00:03:34.670 align:middle line:90% 00:03:34.670 --> 00:03:40.740 align:middle line:84% And you actually had in the homework this discussion 00:03:40.740 --> 00:03:45.030 align:middle line:84% to try to reproduce it, the definition of temperature 00:03:45.030 --> 00:03:45.780 align:middle line:90% for a reservoir. 00:03:45.780 --> 00:03:48.450 align:middle line:84% So for the reservoir, the stable equilibrium state curve 00:03:48.450 --> 00:03:51.330 align:middle line:90% is just a straight line. 00:03:51.330 --> 00:03:54.690 align:middle line:84% The allowed states are on the left. 00:03:54.690 --> 00:03:58.050 align:middle line:84% Also, reservoirs can have non-equilibrium states. 00:03:58.050 --> 00:04:01.500 align:middle line:84% But most of the time, we are interested in 00:04:01.500 --> 00:04:06.140 align:middle line:90% their stable equilibrium ones. 00:04:06.140 --> 00:04:10.590 align:middle line:84% And then we also constructed the available energy. 00:04:10.590 --> 00:04:15.890 align:middle line:84% And I want to somehow do this again on the blackboard, 00:04:15.890 --> 00:04:23.990 align:middle line:84% just so that we can see together this construction. 00:04:23.990 --> 00:04:27.400 align:middle line:84% And I'm going to do it in a slightly different way, 00:04:27.400 --> 00:04:31.620 align:middle line:90% so that we can-- 00:04:31.620 --> 00:04:34.830 align:middle line:90% 00:04:34.830 --> 00:04:39.716 align:middle line:84% so, say that this is our system A, 00:04:39.716 --> 00:04:43.730 align:middle line:84% and this is its E versus S diagram. 00:04:43.730 --> 00:04:47.630 align:middle line:90% 00:04:47.630 --> 00:04:51.790 align:middle line:90% And this time, let's do-- 00:04:51.790 --> 00:04:55.870 align:middle line:84% you see in the viewgraph, state A1 00:04:55.870 --> 00:05:00.400 align:middle line:84% is to the left of the final mutual equilibrium state 00:05:00.400 --> 00:05:03.300 align:middle line:90% with reservoir R, which is AR. 00:05:03.300 --> 00:05:10.370 align:middle line:84% So I'm going to do it just for a change differently. 00:05:10.370 --> 00:05:16.210 align:middle line:84% So I assume now that the state is here, A1, 00:05:16.210 --> 00:05:21.770 align:middle line:84% and that the final state, when I will 00:05:21.770 --> 00:05:24.440 align:middle line:84% reach mutual equilibrium with the reservoir, 00:05:24.440 --> 00:05:25.360 align:middle line:90% it's going to be here. 00:05:25.360 --> 00:05:28.260 align:middle line:90% 00:05:28.260 --> 00:05:40.070 align:middle line:84% That means that the temperature of the reservoir is this one. 00:05:40.070 --> 00:05:42.420 align:middle line:84% Because when they are in mutual equilibrium, 00:05:42.420 --> 00:05:44.890 align:middle line:84% the system and the reservoir have the same temperature. 00:05:44.890 --> 00:05:49.580 align:middle line:84% We haven't proved that yet, but we will do it, hopefully, 00:05:49.580 --> 00:05:51.325 align:middle line:90% today just after this example. 00:05:51.325 --> 00:05:54.080 align:middle line:90% 00:05:54.080 --> 00:06:01.460 align:middle line:84% So let me draw below here, also, the E versus S 00:06:01.460 --> 00:06:02.970 align:middle line:90% diagram for the reservoir. 00:06:02.970 --> 00:06:07.730 align:middle line:90% 00:06:07.730 --> 00:06:11.720 align:middle line:84% So you see that in order to go from here-- from this state, 00:06:11.720 --> 00:06:20.150 align:middle line:84% to this state, the system must decrease its entropy. 00:06:20.150 --> 00:06:25.260 align:middle line:90% 00:06:25.260 --> 00:06:27.840 align:middle line:84% Definitely, it cannot do it in a weight process, 00:06:27.840 --> 00:06:31.410 align:middle line:84% because in a weight process, the most you can do is keep 00:06:31.410 --> 00:06:32.940 align:middle line:90% the entropy constant. 00:06:32.940 --> 00:06:35.980 align:middle line:84% But if you're not doing things reversibly, 00:06:35.980 --> 00:06:37.210 align:middle line:90% you generate entropy. 00:06:37.210 --> 00:06:40.350 align:middle line:84% So you're going to move to the right, not to the left. 00:06:40.350 --> 00:06:48.890 align:middle line:84% So in order to do that, this is the reason why I wanted to-- 00:06:48.890 --> 00:06:53.810 align:middle line:84% we can use the reservoir because we are considering-- 00:06:53.810 --> 00:06:58.530 align:middle line:84% we are starting the adiabatic availability. 00:06:58.530 --> 00:07:00.650 align:middle line:84% The available energy is the adiabatic availability 00:07:00.650 --> 00:07:04.050 align:middle line:84% of a composite system in which there is a reservoir. 00:07:04.050 --> 00:07:06.970 align:middle line:90% So my initial state is here. 00:07:06.970 --> 00:07:10.530 align:middle line:84% I have a reservoir, and what I want to do 00:07:10.530 --> 00:07:15.065 align:middle line:84% is lift the weight as much as possible. 00:07:15.065 --> 00:07:23.340 align:middle line:90% 00:07:23.340 --> 00:07:28.630 align:middle line:84% So the reservoir is available to do this job. 00:07:28.630 --> 00:07:35.870 align:middle line:84% We need to remove entropy from the system. 00:07:35.870 --> 00:07:43.370 align:middle line:84% So we can transfer that entropy from system A to the reservoir. 00:07:43.370 --> 00:07:52.340 align:middle line:90% 00:07:52.340 --> 00:08:09.080 align:middle line:84% So the system must decrease its entropy from S1 to SR. 00:08:09.080 --> 00:08:13.060 align:middle line:90% The reservoir can take it. 00:08:13.060 --> 00:08:18.350 align:middle line:84% OK, so the reservoir has a stable equilibrium state 00:08:18.350 --> 00:08:19.530 align:middle line:90% with this slope. 00:08:19.530 --> 00:08:21.130 align:middle line:90% So it's this. 00:08:21.130 --> 00:08:27.340 align:middle line:90% 00:08:27.340 --> 00:08:33.650 align:middle line:84% It must take entropy so its entropy will increase. 00:08:33.650 --> 00:08:37.490 align:middle line:84% So the initial state will be here, R1, 00:08:37.490 --> 00:08:42.120 align:middle line:84% and the final state will be here, R2. 00:08:42.120 --> 00:08:51.345 align:middle line:84% And the entropy of the reservoir will go from S1R to S1, to S2R. 00:08:51.345 --> 00:08:56.220 align:middle line:90% 00:08:56.220 --> 00:09:02.610 align:middle line:84% Now, so you would like to give just entropy to the reservoir. 00:09:02.610 --> 00:09:07.480 align:middle line:90% 00:09:07.480 --> 00:09:09.550 align:middle line:90% Because, look, so far-- 00:09:09.550 --> 00:09:12.610 align:middle line:84% suppose you consider here this is the initial energy 00:09:12.610 --> 00:09:21.560 align:middle line:84% of A, and this is the final energy of A. And we call omega 00:09:21.560 --> 00:09:27.050 align:middle line:84% 1R the energy that we transfer to the weight. 00:09:27.050 --> 00:09:43.470 align:middle line:84% So if we could do that process without the need 00:09:43.470 --> 00:09:47.175 align:middle line:84% of the reservoir, it would be a weight process for system A. 00:09:47.175 --> 00:09:50.445 align:middle line:84% And we would extract this much energy. 00:09:50.445 --> 00:09:53.100 align:middle line:90% 00:09:53.100 --> 00:09:56.790 align:middle line:84% But we cannot do it in a weight process, 00:09:56.790 --> 00:10:01.180 align:middle line:84% because we cannot subtract entropy. 00:10:01.180 --> 00:10:04.078 align:middle line:90% We cannot destroy entropy. 00:10:04.078 --> 00:10:05.870 align:middle line:84% So we need to transfer it to the reservoir. 00:10:05.870 --> 00:10:07.520 align:middle line:84% So we need to give it to the reservoir. 00:10:07.520 --> 00:10:13.170 align:middle line:84% And the reservoir, in order to accept entropy-- 00:10:13.170 --> 00:10:16.470 align:middle line:84% you see, we want to move this initial state of the reservoir 00:10:16.470 --> 00:10:17.080 align:middle line:90% to the right. 00:10:17.080 --> 00:10:18.300 align:middle line:84% But there are no states to the right, 00:10:18.300 --> 00:10:19.717 align:middle line:84% because that's the maximum entropy 00:10:19.717 --> 00:10:22.980 align:middle line:90% for that value for the energy. 00:10:22.980 --> 00:10:24.995 align:middle line:84% You need to give it also some energy. 00:10:24.995 --> 00:10:29.805 align:middle line:90% 00:10:29.805 --> 00:10:38.620 align:middle line:84% And that energy is the difference between E2 and E1 00:10:38.620 --> 00:10:41.078 align:middle line:90% for the reservoir. 00:10:41.078 --> 00:10:52.495 align:middle line:90% So E2, you must subtract that. 00:10:52.495 --> 00:10:58.180 align:middle line:90% 00:10:58.180 --> 00:11:02.230 align:middle line:84% And so you take this segment, and you put it up here, 00:11:02.230 --> 00:11:04.150 align:middle line:90% if you want. 00:11:04.150 --> 00:11:05.775 align:middle line:90% And it's this much. 00:11:05.775 --> 00:11:12.470 align:middle line:90% 00:11:12.470 --> 00:11:18.050 align:middle line:84% It's not precise in scale, but this much is that much. 00:11:18.050 --> 00:11:23.180 align:middle line:84% So you subtract from this energy difference, this energy 00:11:23.180 --> 00:11:26.630 align:middle line:84% difference, and what you are left with is, 00:11:26.630 --> 00:11:35.740 align:middle line:84% again, you see from this point to that point. 00:11:35.740 --> 00:11:38.710 align:middle line:90% And that's omega 1R. 00:11:38.710 --> 00:11:42.780 align:middle line:84% Notice that this point is equal to a geometrical construction. 00:11:42.780 --> 00:11:44.800 align:middle line:84% It doesn't correspond to a state, 00:11:44.800 --> 00:11:48.170 align:middle line:84% because it's below the stable equilibrium state curve. 00:11:48.170 --> 00:11:53.610 align:middle line:84% It's just the construction of bringing from-- 00:11:53.610 --> 00:11:56.890 align:middle line:84% translating this energy difference into there. 00:11:56.890 --> 00:11:59.250 align:middle line:84% So that I subtract it from the segment 00:11:59.250 --> 00:12:03.990 align:middle line:84% that represents the difference in energy for the rest. 00:12:03.990 --> 00:12:07.590 align:middle line:84% And then all I have to do is remember 00:12:07.590 --> 00:12:12.590 align:middle line:84% that the stable equilibrium state curve for a reservoir 00:12:12.590 --> 00:12:14.400 align:middle line:90% is a straight line. 00:12:14.400 --> 00:12:23.730 align:middle line:84% So I can rewrite this as TR times S2R minus S1R. 00:12:23.730 --> 00:12:26.960 align:middle line:90% 00:12:26.960 --> 00:12:31.130 align:middle line:84% And then I also remember that the amount of entropy 00:12:31.130 --> 00:12:36.050 align:middle line:84% that the reservoir must take is equal to the decrease in entropy 00:12:36.050 --> 00:12:38.280 align:middle line:90% that I need for my system. 00:12:38.280 --> 00:12:56.630 align:middle line:84% So S2 minus S1 is equal to S1A minus SRA. And so we 00:12:56.630 --> 00:13:02.900 align:middle line:84% finally get the expression for the available energy 00:13:02.900 --> 00:13:05.760 align:middle line:90% that we have-- 00:13:05.760 --> 00:13:09.176 align:middle line:84% we started with, we wanted to represent graphically. 00:13:09.176 --> 00:13:15.490 align:middle line:90% 00:13:15.490 --> 00:13:15.990 align:middle line:90% OK. 00:13:15.990 --> 00:13:23.070 align:middle line:90% 00:13:23.070 --> 00:13:31.560 align:middle line:84% So this brings us to a very important concept 00:13:31.560 --> 00:13:34.860 align:middle line:84% for applications, for applications in engineering, 00:13:34.860 --> 00:13:39.053 align:middle line:84% where you want to extract, convert energy from systems. 00:13:39.053 --> 00:13:43.950 align:middle line:90% 00:13:43.950 --> 00:13:48.780 align:middle line:84% And often, the systems you start with 00:13:48.780 --> 00:13:52.950 align:middle line:84% are in stable equilibrium states, maybe hot, but still 00:13:52.950 --> 00:13:56.010 align:middle line:90% stable equilibrium states. 00:13:56.010 --> 00:13:57.700 align:middle line:90% For example, when you-- 00:13:57.700 --> 00:14:03.610 align:middle line:84% in a steam power plant, you just heat up water. 00:14:03.610 --> 00:14:06.480 align:middle line:84% You make steam, and that steam is 00:14:06.480 --> 00:14:12.810 align:middle line:84% some sort of a high temperature stable equilibrium state. 00:14:12.810 --> 00:14:15.270 align:middle line:90% So here is the state. 00:14:15.270 --> 00:14:19.120 align:middle line:84% And you would like to extract as much energy as possible. 00:14:19.120 --> 00:14:20.560 align:middle line:90% So you would like to go down. 00:14:20.560 --> 00:14:26.490 align:middle line:84% But going down on the diagram, it would be an impossibility, 00:14:26.490 --> 00:14:29.350 align:middle line:84% I mean, a perpetual motion machine of the second kind. 00:14:29.350 --> 00:14:31.620 align:middle line:90% So it's impossible. 00:14:31.620 --> 00:14:33.960 align:middle line:84% In order to go down in energy, you 00:14:33.960 --> 00:14:39.040 align:middle line:84% need to subtract also some entropy. 00:14:39.040 --> 00:14:46.960 align:middle line:84% But then you need a system that takes up that entropy. 00:14:46.960 --> 00:14:50.650 align:middle line:84% And usually, that role is played by the environment, or maybe 00:14:50.650 --> 00:15:00.010 align:middle line:84% the nearby river, or lake, or the sea, or the air. 00:15:00.010 --> 00:15:07.990 align:middle line:84% That is the system in which you dump the entropy that you 00:15:07.990 --> 00:15:10.380 align:middle line:90% must extract from your system. 00:15:10.380 --> 00:15:13.070 align:middle line:90% 00:15:13.070 --> 00:15:18.350 align:middle line:84% So it's like you use the environment as an entropy sink. 00:15:18.350 --> 00:15:21.030 align:middle line:84% But the environment has its own needs. 00:15:21.030 --> 00:15:25.670 align:middle line:84% It does not take entropy for free. 00:15:25.670 --> 00:15:29.030 align:middle line:84% It takes entropy only if you give it, also, 00:15:29.030 --> 00:15:33.320 align:middle line:84% enough energy so that it can climb in the stable equilibrium 00:15:33.320 --> 00:15:33.850 align:middle line:90% state curve. 00:15:33.850 --> 00:15:38.800 align:middle line:90% 00:15:38.800 --> 00:15:41.830 align:middle line:84% So that energy that you are giving 00:15:41.830 --> 00:15:47.650 align:middle line:84% to the environment in a typical energy systems application 00:15:47.650 --> 00:15:53.030 align:middle line:90% is not a waste. 00:15:53.030 --> 00:15:55.140 align:middle line:84% Sometimes, you say, well, I'm wasting energy. 00:15:55.140 --> 00:15:58.310 align:middle line:90% No, no, it's not waste. 00:15:58.310 --> 00:16:01.770 align:middle line:84% You're using the environment for a very important function. 00:16:01.770 --> 00:16:03.900 align:middle line:90% You're using the energy there. 00:16:03.900 --> 00:16:07.010 align:middle line:84% That's not where the inefficiency of your power plant 00:16:07.010 --> 00:16:08.490 align:middle line:90% is. 00:16:08.490 --> 00:16:12.450 align:middle line:84% The inefficiencies come from the internal irreversibilities 00:16:12.450 --> 00:16:15.860 align:middle line:90% that are in your machinery. 00:16:15.860 --> 00:16:23.240 align:middle line:84% Instead, the role of the reservoir 00:16:23.240 --> 00:16:29.900 align:middle line:84% here is important because otherwise, you wouldn't 00:16:29.900 --> 00:16:33.470 align:middle line:90% be able to do anything useful. 00:16:33.470 --> 00:16:36.890 align:middle line:84% We'll come back to this when I do 00:16:36.890 --> 00:16:43.730 align:middle line:84% a quick review of heat engines and exergy stuff 00:16:43.730 --> 00:16:48.400 align:middle line:90% in a couple of lectures. 00:16:48.400 --> 00:16:57.130 align:middle line:84% But today, I want to go into some details on how 00:16:57.130 --> 00:17:04.200 align:middle line:84% we show that for systems to be in mutual equilibrium, 00:17:04.200 --> 00:17:05.750 align:middle line:84% you need equality of temperatures 00:17:05.750 --> 00:17:07.470 align:middle line:90% when they can exchange energy. 00:17:07.470 --> 00:17:11.400 align:middle line:84% You need equality of pressures when they can exchange volume. 00:17:11.400 --> 00:17:14.000 align:middle line:84% You need equality of chemical potentials 00:17:14.000 --> 00:17:17.240 align:middle line:84% of the various constituents for those constituents 00:17:17.240 --> 00:17:22.010 align:middle line:84% that can be transferred from one system to the other. 00:17:22.010 --> 00:17:29.750 align:middle line:84% And the way we prove this is by means of the maximum entropy 00:17:29.750 --> 00:17:30.500 align:middle line:90% principle. 00:17:30.500 --> 00:17:39.770 align:middle line:84% You remember that we proved that the stable equilibrium state is 00:17:39.770 --> 00:17:43.190 align:middle line:84% the maximum entropy one among those states that 00:17:43.190 --> 00:17:46.470 align:middle line:84% have the same energy and amounts and parameters. 00:17:46.470 --> 00:17:48.960 align:middle line:84% I'm not going to repeat amounts and parameters. 00:17:48.960 --> 00:17:52.670 align:middle line:84% Let's keep them fixed for the entire day. 00:17:52.670 --> 00:18:01.200 align:middle line:90% So how does the proof work? 00:18:01.200 --> 00:18:04.695 align:middle line:84% You say, all right, let's start from two systems, 00:18:04.695 --> 00:18:09.910 align:middle line:84% A and B, that are in mutual equilibrium. 00:18:09.910 --> 00:18:11.950 align:middle line:84% Therefore, the composite system C, 00:18:11.950 --> 00:18:15.660 align:middle line:84% this is the E versus S diagram of the composite system 00:18:15.660 --> 00:18:16.300 align:middle line:90% of the two. 00:18:16.300 --> 00:18:18.910 align:middle line:90% 00:18:18.910 --> 00:18:27.650 align:middle line:84% So the composite system is on the stable equilibrium state 00:18:27.650 --> 00:18:32.220 align:middle line:84% curve because by definition of mutual equilibrium, 00:18:32.220 --> 00:18:33.750 align:middle line:90% it's when the two systems-- 00:18:33.750 --> 00:18:36.406 align:middle line:84% the composite, is in stable equilibrium. 00:18:36.406 --> 00:18:39.120 align:middle line:90% 00:18:39.120 --> 00:18:43.990 align:middle line:84% So now what we do is, all right, let's construct another state. 00:18:43.990 --> 00:18:53.850 align:middle line:84% We'll call it C1, with the same energy, but a different state. 00:18:53.850 --> 00:18:56.500 align:middle line:84% Being a different state with the same energy, 00:18:56.500 --> 00:19:01.280 align:middle line:84% it necessarily has lesser entropy. 00:19:01.280 --> 00:19:06.050 align:middle line:84% We have many choices for this other state C1. 00:19:06.050 --> 00:19:09.950 align:middle line:84% So we are going to choose one that is easy for us 00:19:09.950 --> 00:19:10.615 align:middle line:90% in some sense. 00:19:10.615 --> 00:19:13.290 align:middle line:90% 00:19:13.290 --> 00:19:24.810 align:middle line:84% So for example, here is my system A. Here is my system B. 00:19:24.810 --> 00:19:26.790 align:middle line:84% In the mutual equilibrium state-- 00:19:26.790 --> 00:19:30.240 align:middle line:90% so now these are the partial-- 00:19:30.240 --> 00:19:36.810 align:middle line:84% the local energy-entropy diagrams. 00:19:36.810 --> 00:19:41.100 align:middle line:84% Sometimes I forget to write the ES because I give it 00:19:41.100 --> 00:19:45.020 align:middle line:90% for granted, but I shouldn't. 00:19:45.020 --> 00:19:50.510 align:middle line:84% So let's say that this is temperature T0 of system A, 00:19:50.510 --> 00:19:58.190 align:middle line:84% and this is temperature T0 of system B. And we want to prove, 00:19:58.190 --> 00:20:01.760 align:middle line:84% so I'll build it in already in this construction, 00:20:01.760 --> 00:20:04.800 align:middle line:84% that the two temperatures must be equal. 00:20:04.800 --> 00:20:13.910 align:middle line:84% So I start from for them being in a stable equilibrium state. 00:20:13.910 --> 00:20:15.940 align:middle line:84% And then I consider another state, 00:20:15.940 --> 00:20:23.200 align:middle line:84% which is this one, with the same energy for the composite, 00:20:23.200 --> 00:20:26.350 align:middle line:90% but different for the parts. 00:20:26.350 --> 00:20:28.930 align:middle line:90% So let's see. 00:20:28.930 --> 00:20:35.410 align:middle line:84% Let me arbitrarily choose a small amount of energy, dE, 00:20:35.410 --> 00:20:41.270 align:middle line:84% and therefore, assume that my system A-- 00:20:41.270 --> 00:20:44.380 align:middle line:84% I'll make it bigger so as to represent it-- 00:20:44.380 --> 00:20:52.890 align:middle line:84% has an energy higher by an amount dE. 00:20:52.890 --> 00:20:59.870 align:middle line:84% And B has an energy smaller by the same amount, 00:20:59.870 --> 00:21:01.580 align:middle line:84% so that since energy is additive, 00:21:01.580 --> 00:21:03.070 align:middle line:84% the total energy remains constant. 00:21:03.070 --> 00:21:04.810 align:middle line:84% And therefore, the composite system 00:21:04.810 --> 00:21:08.040 align:middle line:90% is still at the same energy. 00:21:08.040 --> 00:21:12.640 align:middle line:84% Plus, I mean, I have many choices on how to do this, 00:21:12.640 --> 00:21:16.860 align:middle line:84% but I'm going to choose the easy way, easy for our proof, 00:21:16.860 --> 00:21:22.860 align:middle line:84% so that both system A and system B change their energies, 00:21:22.860 --> 00:21:24.790 align:middle line:84% but remain in stable equilibrium states. 00:21:24.790 --> 00:21:28.900 align:middle line:90% 00:21:28.900 --> 00:21:36.270 align:middle line:84% So this is how I construct state A1 and B1 so as 00:21:36.270 --> 00:21:37.955 align:middle line:90% to make up the composite C1. 00:21:37.955 --> 00:21:43.510 align:middle line:90% 00:21:43.510 --> 00:21:49.400 align:middle line:84% Since I want to do, also, equality of pressure, 00:21:49.400 --> 00:21:51.790 align:middle line:90% I want to prove also that. 00:21:51.790 --> 00:21:57.840 align:middle line:84% Think also of the volume that changes. 00:21:57.840 --> 00:22:00.390 align:middle line:84% So this becomes a three-dimensional surface. 00:22:00.390 --> 00:22:01.640 align:middle line:90% I'm not going to represent it. 00:22:01.640 --> 00:22:03.980 align:middle line:90% Otherwise, it gets too messy. 00:22:03.980 --> 00:22:09.950 align:middle line:84% But imagine that in addition to changing the energy, 00:22:09.950 --> 00:22:13.620 align:middle line:84% I'm also going to change the volume. 00:22:13.620 --> 00:22:22.030 align:middle line:84% So the partition between system A and B is movable. 00:22:22.030 --> 00:22:26.530 align:middle line:84% And let's move the partition slightly to the right 00:22:26.530 --> 00:22:29.620 align:middle line:84% so that A gains a little volume dV, 00:22:29.620 --> 00:22:34.030 align:middle line:90% and B yields a little volume dV. 00:22:34.030 --> 00:22:34.810 align:middle line:90% OK. 00:22:34.810 --> 00:22:39.510 align:middle line:84% So now we constructed this state C1, 00:22:39.510 --> 00:22:41.880 align:middle line:84% that has the same energy and also, the same volume 00:22:41.880 --> 00:22:54.270 align:middle line:84% for system C. And what we want to make use of 00:22:54.270 --> 00:22:57.503 align:middle line:90% is the fact that-- 00:22:57.503 --> 00:22:58.920 align:middle line:84% the maximum entropy principle-- so 00:22:58.920 --> 00:23:05.850 align:middle line:84% that the entropy of C in state 0 is greater than that 00:23:05.850 --> 00:23:07.470 align:middle line:90% in, say, in C1. 00:23:07.470 --> 00:23:11.820 align:middle line:90% So that I can write it this way. 00:23:11.820 --> 00:23:16.050 align:middle line:84% S1C minus S10 must be less than 0. 00:23:16.050 --> 00:23:17.730 align:middle line:90% So we start. 00:23:17.730 --> 00:23:19.320 align:middle line:90% We want to compute this. 00:23:19.320 --> 00:23:26.780 align:middle line:84% And we end up imposing that it must be negative. 00:23:26.780 --> 00:23:33.290 align:middle line:84% Then the next steps are simply using the additivity of entropy. 00:23:33.290 --> 00:23:36.120 align:middle line:90% 00:23:36.120 --> 00:23:39.690 align:middle line:84% So in state initial and final state, the entropy of C 00:23:39.690 --> 00:23:46.830 align:middle line:84% is the sum of the entropies of A and B. We just rearrange, 00:23:46.830 --> 00:23:50.990 align:middle line:84% so that it's the additivity of the entropy differences. 00:23:50.990 --> 00:23:53.100 align:middle line:84% And then because of our construction, 00:23:53.100 --> 00:23:58.910 align:middle line:84% because we selected this particular state that still 00:23:58.910 --> 00:24:02.340 align:middle line:84% belongs to the stable equilibrium state curve, 00:24:02.340 --> 00:24:06.860 align:middle line:84% we can use a Taylor series expansion 00:24:06.860 --> 00:24:08.750 align:middle line:84% of the fundamental relation, which 00:24:08.750 --> 00:24:11.720 align:middle line:84% describes this curve, to represent 00:24:11.720 --> 00:24:14.510 align:middle line:84% how the entropy changes for the subsystems. 00:24:14.510 --> 00:24:18.560 align:middle line:90% 00:24:18.560 --> 00:24:24.350 align:middle line:90% You remember-- let's see-- 00:24:24.350 --> 00:24:32.790 align:middle line:84% yeah, we had S for stable equilibrium 00:24:32.790 --> 00:24:35.040 align:middle line:90% is a function like that. 00:24:35.040 --> 00:24:47.720 align:middle line:84% And if I want to compute the value of-- 00:24:47.720 --> 00:24:51.190 align:middle line:90% so this represents this curve. 00:24:51.190 --> 00:24:59.360 align:middle line:84% So if I want to compute the difference of, say, S2 minus S1, 00:24:59.360 --> 00:25:04.760 align:middle line:84% this being state 1 and this being state 2-- 00:25:04.760 --> 00:25:15.240 align:middle line:84% or actually, sorry, let's make it 0 and 1, so S1 minus S0. 00:25:15.240 --> 00:25:16.270 align:middle line:90% That's what we want. 00:25:16.270 --> 00:25:16.770 align:middle line:90% Yeah. 00:25:16.770 --> 00:25:21.450 align:middle line:90% 00:25:21.450 --> 00:25:25.020 align:middle line:84% I can do a Taylor series expansion 00:25:25.020 --> 00:25:28.200 align:middle line:84% around the initial state, the initial point 00:25:28.200 --> 00:25:29.820 align:middle line:90% of the function, so around E0. 00:25:29.820 --> 00:25:32.700 align:middle line:90% 00:25:32.700 --> 00:25:39.550 align:middle line:84% So let me do it only as if it were only a function of energy. 00:25:39.550 --> 00:25:43.200 align:middle line:84% So forget about these ones for the moment. 00:25:43.200 --> 00:26:01.700 align:middle line:84% So that would be S of E0 plus S prime of E0 times E1 minus E0 00:26:01.700 --> 00:26:09.650 align:middle line:84% plus S double prime divided by 2, or 2 factorial, whatever-- 00:26:09.650 --> 00:26:10.820 align:middle line:90% it's the same-- 00:26:10.820 --> 00:26:16.310 align:middle line:84% times E1 minus E0 squared plus dot, dot, dot. 00:26:16.310 --> 00:26:17.990 align:middle line:90% And the series continues. 00:26:17.990 --> 00:26:20.660 align:middle line:90% 00:26:20.660 --> 00:26:26.750 align:middle line:84% And of course, in the limit, as E1 minus E0 is small, 00:26:26.750 --> 00:26:29.570 align:middle line:84% the dominant terms are the first ones 00:26:29.570 --> 00:26:31.540 align:middle line:90% that you find in the expansion. 00:26:31.540 --> 00:26:33.880 align:middle line:84% So that's what we are doing here, 00:26:33.880 --> 00:26:38.770 align:middle line:84% with the notion that this S prime is what we 00:26:38.770 --> 00:26:40.430 align:middle line:90% call the inverse temperature. 00:26:40.430 --> 00:26:42.960 align:middle line:90% 00:26:42.960 --> 00:26:44.670 align:middle line:84% And when you do this, also, in terms 00:26:44.670 --> 00:26:48.150 align:middle line:84% of the second variable, which is the volume, 00:26:48.150 --> 00:26:52.940 align:middle line:84% that's what we call the pressure over temperature. 00:26:52.940 --> 00:27:01.080 align:middle line:84% So here is that expansion for system A. 00:27:01.080 --> 00:27:08.670 align:middle line:84% The second derivatives, let me write them as d2 00:27:08.670 --> 00:27:11.430 align:middle line:84% because it contains several terms. 00:27:11.430 --> 00:27:16.340 align:middle line:84% Since I'm doing that in terms of two variables, 00:27:16.340 --> 00:27:19.620 align:middle line:84% subsequent slide will have the details. 00:27:19.620 --> 00:27:23.270 align:middle line:84% And we'll do the same for system B. 00:27:23.270 --> 00:27:26.180 align:middle line:90% Then we use the fact-- 00:27:26.180 --> 00:27:28.500 align:middle line:90% well, then we combine terms. 00:27:28.500 --> 00:27:34.550 align:middle line:84% And here is what we get, 1 over T minus 1 over T for A 00:27:34.550 --> 00:27:40.058 align:middle line:84% and B times dE, P over T minus P over T, A and B, 00:27:40.058 --> 00:27:45.540 align:middle line:84% dV, plus the second differentials of S plus dot, 00:27:45.540 --> 00:27:46.040 align:middle line:90% dot, dots. 00:27:46.040 --> 00:27:51.150 align:middle line:90% 00:27:51.150 --> 00:27:52.720 align:middle line:90% So what can we infer? 00:27:52.720 --> 00:27:58.280 align:middle line:84% And this must be a negative, strictly negative. 00:27:58.280 --> 00:28:03.110 align:middle line:84% Of course, notice and remember, that when you do a series 00:28:03.110 --> 00:28:08.060 align:middle line:84% expansion, the terms here, like this one, 00:28:08.060 --> 00:28:11.220 align:middle line:84% are evaluated at the initial state. 00:28:11.220 --> 00:28:17.450 align:middle line:84% So they do not depend on how much you move around. 00:28:17.450 --> 00:28:23.200 align:middle line:84% So these terms in parentheses, and that term in parentheses, 00:28:23.200 --> 00:28:29.430 align:middle line:84% is fixed by the initial state, state 0. 00:28:29.430 --> 00:28:34.630 align:middle line:84% It's independent of how you move around to another state. 00:28:34.630 --> 00:28:36.817 align:middle line:84% So it's independent of dE and dV. 00:28:36.817 --> 00:28:40.610 align:middle line:90% 00:28:40.610 --> 00:28:45.160 align:middle line:84% On the other hand, you see here, I 00:28:45.160 --> 00:28:51.030 align:middle line:84% could have chosen another state of A with higher energy, 00:28:51.030 --> 00:28:52.740 align:middle line:90% but also, one with lower energy. 00:28:52.740 --> 00:28:56.750 align:middle line:84% So I could have chosen dE to be positive or negative, 00:28:56.750 --> 00:29:01.040 align:middle line:84% large or small, positive or negative, whatever, as long, 00:29:01.040 --> 00:29:06.450 align:middle line:84% of course, B gets the opposite, equal and opposite. 00:29:06.450 --> 00:29:12.760 align:middle line:84% So that means that dE and dV can be 00:29:12.760 --> 00:29:17.530 align:middle line:84% chosen arbitrarily, positive or negative. 00:29:17.530 --> 00:29:20.890 align:middle line:84% And if these terms in parentheses 00:29:20.890 --> 00:29:25.760 align:middle line:84% were different than 0, suppose they were positive, 00:29:25.760 --> 00:29:27.950 align:middle line:84% then I could choose a positive dE, 00:29:27.950 --> 00:29:32.880 align:middle line:84% and this would be a dominating term in the series, 00:29:32.880 --> 00:29:37.750 align:middle line:84% violating the fact that that sum must be negative. 00:29:37.750 --> 00:29:40.890 align:middle line:84% So the only possibility, due to the arbitrariness 00:29:40.890 --> 00:29:45.360 align:middle line:84% of the choice of dV and dE is that these terms in parentheses 00:29:45.360 --> 00:29:46.360 align:middle line:90% be equal to 0. 00:29:46.360 --> 00:29:48.920 align:middle line:90% 00:29:48.920 --> 00:29:53.750 align:middle line:84% And of course, then the burden to keep the inequality satisfied 00:29:53.750 --> 00:29:56.990 align:middle line:84% is going to be shifted to the second order terms 00:29:56.990 --> 00:30:02.320 align:middle line:84% in the expansion, which are going to be negative. 00:30:02.320 --> 00:30:05.570 align:middle line:84% So, essentially, what we proved here 00:30:05.570 --> 00:30:08.540 align:middle line:84% is that in order for that system here 00:30:08.540 --> 00:30:12.320 align:middle line:84% to be in a stable equilibrium state, 00:30:12.320 --> 00:30:16.710 align:middle line:84% the two temperatures must be equal 00:30:16.710 --> 00:30:18.680 align:middle line:84% and also, the two pressures must be equal. 00:30:18.680 --> 00:30:25.030 align:middle line:90% 00:30:25.030 --> 00:30:28.270 align:middle line:84% So above here is the same as before. 00:30:28.270 --> 00:30:32.890 align:middle line:84% I simply wrote more explicitly the expression 00:30:32.890 --> 00:30:38.580 align:middle line:84% for the second order differential, 00:30:38.580 --> 00:30:42.390 align:middle line:90% which must be non-positive. 00:30:42.390 --> 00:30:47.610 align:middle line:90% So it's either 0 or negative. 00:30:47.610 --> 00:30:50.700 align:middle line:84% If it is 0, it's the third order terms and fourth 00:30:50.700 --> 00:30:58.710 align:middle line:84% and go on, that must obey, must take up 00:30:58.710 --> 00:31:00.520 align:middle line:90% and satisfy this inequality. 00:31:00.520 --> 00:31:05.830 align:middle line:84% For example, see, if I have a reservoir, 00:31:05.830 --> 00:31:09.830 align:middle line:84% the second derivative is equal to 0 00:31:09.830 --> 00:31:11.390 align:middle line:90% of the fundamental relation. 00:31:11.390 --> 00:31:20.290 align:middle line:84% So for this reservoir, I need that somewhere along the line, 00:31:20.290 --> 00:31:24.550 align:middle line:84% there is a derivative which is negative. 00:31:24.550 --> 00:31:32.340 align:middle line:84% And since a straight line isn't like that, 00:31:32.340 --> 00:31:36.000 align:middle line:84% it means that this straight line in practice 00:31:36.000 --> 00:31:37.960 align:middle line:90% is really an approximation. 00:31:37.960 --> 00:31:43.230 align:middle line:84% So in other words, a real reservoir with a true straight 00:31:43.230 --> 00:31:47.250 align:middle line:84% line as a stable equilibrium straight curve cannot exist. 00:31:47.250 --> 00:31:48.690 align:middle line:90% It would violate this. 00:31:48.690 --> 00:31:54.720 align:middle line:84% So the reservoir is an approximation, extremely good 00:31:54.720 --> 00:31:59.640 align:middle line:84% approximation if you make it like water at the triple point, 00:31:59.640 --> 00:32:01.540 align:middle line:90% with sufficient amount of water. 00:32:01.540 --> 00:32:07.050 align:middle line:84% But if you go to small, very small amounts, 00:32:07.050 --> 00:32:12.940 align:middle line:84% then you may have difficulties constructing a real reservoir. 00:32:12.940 --> 00:32:19.324 align:middle line:90% 00:32:19.324 --> 00:32:22.720 align:middle line:84% So we proved equality of temperature and equality 00:32:22.720 --> 00:32:26.950 align:middle line:84% of pressure, and we proved, also, 00:32:26.950 --> 00:32:29.050 align:middle line:84% the sign of the second derivatives 00:32:29.050 --> 00:32:33.100 align:middle line:90% of the fundamental relation. 00:32:33.100 --> 00:32:38.320 align:middle line:84% The reason why I have drawn the E versus S diagram this way 00:32:38.320 --> 00:32:43.420 align:middle line:84% and not the other way, so with the concavity upward, 00:32:43.420 --> 00:32:46.092 align:middle line:90% is now proved here. 00:32:46.092 --> 00:32:48.130 align:middle line:84% The second derivative of S with respect to E 00:32:48.130 --> 00:32:54.600 align:middle line:90% is negative or 0, negative or 0. 00:32:54.600 --> 00:32:57.580 align:middle line:90% The same, you can prove. 00:32:57.580 --> 00:32:59.770 align:middle line:84% Since this form, differential form, 00:32:59.770 --> 00:33:06.010 align:middle line:84% must be negative for all possible values, 00:33:06.010 --> 00:33:12.250 align:middle line:84% you need that it's a quadratic form, 00:33:12.250 --> 00:33:14.830 align:middle line:84% and you need that the coefficient, the matrix 00:33:14.830 --> 00:33:17.170 align:middle line:90% of coefficients, be positive-- 00:33:17.170 --> 00:33:19.750 align:middle line:90% I mean negative definite-- 00:33:19.750 --> 00:33:23.150 align:middle line:84% or actually, non-positive definite, meaning that, 00:33:23.150 --> 00:33:26.240 align:middle line:84% for example, not only the terms on the diagonal, 00:33:26.240 --> 00:33:30.310 align:middle line:84% which are these two, but also, the determinant must be 00:33:30.310 --> 00:33:31.390 align:middle line:90% non-negative-- 00:33:31.390 --> 00:33:38.270 align:middle line:84% I mean non-positive, which gives us, also, this expression here. 00:33:38.270 --> 00:33:39.320 align:middle line:90% OK. 00:33:39.320 --> 00:33:47.450 align:middle line:84% Now, we can do, also, but more quickly the proof for when, 00:33:47.450 --> 00:33:56.150 align:middle line:84% instead of putting the movable piston-- 00:33:56.150 --> 00:34:00.110 align:middle line:90% I must change this drawing. 00:34:00.110 --> 00:34:08.030 align:middle line:84% So this should be replaced with a porous material, non-movable. 00:34:08.030 --> 00:34:12.380 align:middle line:84% So this is going to be a porous material that 00:34:12.380 --> 00:34:18.050 align:middle line:84% would allow the transfer of constituent i, 00:34:18.050 --> 00:34:22.409 align:middle line:84% so of molecules or atoms of type i, only one type. 00:34:22.409 --> 00:34:27.969 align:middle line:90% 00:34:27.969 --> 00:34:32.770 align:middle line:84% So they are in mutual equilibrium in state 0. 00:34:32.770 --> 00:34:38.110 align:middle line:84% And we construct another state with the same energy 00:34:38.110 --> 00:34:46.929 align:middle line:84% by shifting some energy from A to B and also, some amounts. 00:34:46.929 --> 00:34:52.500 align:middle line:84% So I take A with some more particles of that type 00:34:52.500 --> 00:34:55.170 align:middle line:90% and B with less particles. 00:34:55.170 --> 00:34:59.920 align:middle line:84% So it's like moving them from B to A. 00:34:59.920 --> 00:35:03.400 align:middle line:84% Well, exactly the same procedure, 00:35:03.400 --> 00:35:08.560 align:middle line:84% I'm going to exploit the fact that the entropy of S1 00:35:08.560 --> 00:35:11.770 align:middle line:90% must be less than that of S0. 00:35:11.770 --> 00:35:16.430 align:middle line:84% Only that when I do the series expansion for the energy, 00:35:16.430 --> 00:35:17.890 align:middle line:90% it's still the same. 00:35:17.890 --> 00:35:22.100 align:middle line:84% But now I'm not I'm not changing the volume. 00:35:22.100 --> 00:35:24.460 align:middle line:90% I'm changing the amount. 00:35:24.460 --> 00:35:29.350 align:middle line:84% So instead of P over T, the partial derivative 00:35:29.350 --> 00:35:32.920 align:middle line:84% with respect to amounts is minus the chemical potential 00:35:32.920 --> 00:35:40.480 align:middle line:90% over T. Combine in the same way. 00:35:40.480 --> 00:35:45.180 align:middle line:84% And this term by the arbitrariness 00:35:45.180 --> 00:35:52.680 align:middle line:84% of our choice of state C1, so the amounts dni, 00:35:52.680 --> 00:35:57.310 align:middle line:84% the difference are arbitrary, like the energy. 00:35:57.310 --> 00:35:59.370 align:middle line:84% And therefore, we find the same conclusion, 00:35:59.370 --> 00:36:01.460 align:middle line:84% that these coefficients must be 0. 00:36:01.460 --> 00:36:05.290 align:middle line:90% 00:36:05.290 --> 00:36:13.470 align:middle line:84% And of course, also, that second order differential 00:36:13.470 --> 00:36:15.900 align:middle line:90% must be non-positive. 00:36:15.900 --> 00:36:19.590 align:middle line:84% We prove, therefore, that we need equality of temperature 00:36:19.590 --> 00:36:22.680 align:middle line:84% and equality of the chemical potentials 00:36:22.680 --> 00:36:28.220 align:middle line:84% of that constituent in the two systems for mutual equilibrium. 00:36:28.220 --> 00:36:32.700 align:middle line:84% And we prove some other convexity 00:36:32.700 --> 00:36:34.666 align:middle line:90% or concavity relations. 00:36:34.666 --> 00:36:38.160 align:middle line:90% 00:36:38.160 --> 00:36:43.660 align:middle line:84% In fact, if you do it for several variables, 00:36:43.660 --> 00:36:51.910 align:middle line:84% you end up with the fact that all these second derivatives 00:36:51.910 --> 00:36:53.085 align:middle line:90% must be 0. 00:36:53.085 --> 00:36:55.770 align:middle line:90% 00:36:55.770 --> 00:36:59.760 align:middle line:84% But more than that, this matrix, which 00:36:59.760 --> 00:37:07.230 align:middle line:84% is the S matrix of partial, all the possible second order 00:37:07.230 --> 00:37:13.570 align:middle line:84% partial derivatives, must be negative definite or negative 00:37:13.570 --> 00:37:14.760 align:middle line:90% semi-definite. 00:37:14.760 --> 00:37:17.860 align:middle line:90% 00:37:17.860 --> 00:37:22.065 align:middle line:84% And therefore, you can apply all that you 00:37:22.065 --> 00:37:25.350 align:middle line:84% know about positive definite matrices, 00:37:25.350 --> 00:37:30.660 align:middle line:84% and so on, and obtain a number of inequalities. 00:37:30.660 --> 00:37:32.180 align:middle line:84% For example, from some of these-- 00:37:32.180 --> 00:37:34.710 align:middle line:90% 00:37:34.710 --> 00:37:37.830 align:middle line:84% one of these inequalities that is well-known 00:37:37.830 --> 00:37:41.640 align:middle line:84% is that the specific heat of a substance, 00:37:41.640 --> 00:37:43.860 align:middle line:84% which is related to the second derivative of S 00:37:43.860 --> 00:37:47.768 align:middle line:84% with respect to E, must be positive. 00:37:47.768 --> 00:37:51.360 align:middle line:90% 00:37:51.360 --> 00:37:56.240 align:middle line:84% But as you see here, that's only one result. 00:37:56.240 --> 00:37:58.665 align:middle line:90% You could obtain several. 00:37:58.665 --> 00:38:04.750 align:middle line:90% 00:38:04.750 --> 00:38:10.860 align:middle line:84% Now, by the same technique to prove the inequalities, 00:38:10.860 --> 00:38:15.460 align:middle line:84% we can also prove another interesting relation, 00:38:15.460 --> 00:38:19.500 align:middle line:84% which is known as the Young-Laplace equation 00:38:19.500 --> 00:38:21.540 align:middle line:90% for surface tension. 00:38:21.540 --> 00:38:27.610 align:middle line:84% So let me do this because it's also 00:38:27.610 --> 00:38:33.970 align:middle line:84% useful for those of you who are in fluid mechanics 00:38:33.970 --> 00:38:36.030 align:middle line:90% or mechanics of interfaces. 00:38:36.030 --> 00:38:49.050 align:middle line:84% For example, take the system which contains a drop-- 00:38:49.050 --> 00:38:51.660 align:middle line:90% so D stands for drops, say-- 00:38:51.660 --> 00:39:02.875 align:middle line:84% of one phase immersed or surrounded by another phase. 00:39:02.875 --> 00:39:05.630 align:middle line:90% 00:39:05.630 --> 00:39:07.830 align:middle line:84% And there are many examples here. 00:39:07.830 --> 00:39:09.360 align:middle line:90% I just put three of them. 00:39:09.360 --> 00:39:12.380 align:middle line:90% 00:39:12.380 --> 00:39:14.110 align:middle line:90% Let's start with the last one. 00:39:14.110 --> 00:39:18.550 align:middle line:84% It could be liquid and vapor, so a drop of liquid, 00:39:18.550 --> 00:39:21.450 align:middle line:84% and outside, you've got vapor, like water vapor 00:39:21.450 --> 00:39:23.470 align:middle line:90% or a vapor of any substance. 00:39:23.470 --> 00:39:27.045 align:middle line:90% 00:39:27.045 --> 00:39:35.600 align:middle line:84% Or you could have air on the outside, which is partially 00:39:35.600 --> 00:39:36.860 align:middle line:90% dissolved. 00:39:36.860 --> 00:39:41.120 align:middle line:84% Here is a drop of water in which there is some air dissolved. 00:39:41.120 --> 00:39:46.130 align:middle line:84% Or if it's not air, like CO2 or some other substance. 00:39:46.130 --> 00:39:48.890 align:middle line:84% Or you could have two liquids, like oil and water, 00:39:48.890 --> 00:39:50.840 align:middle line:90% that are not miscible. 00:39:50.840 --> 00:39:58.670 align:middle line:84% So in the bulk of each one, you don't find any of the other. 00:39:58.670 --> 00:40:03.710 align:middle line:84% But in the interfacial region, they 00:40:03.710 --> 00:40:09.656 align:middle line:84% must live together, these different molecules. 00:40:09.656 --> 00:40:12.380 align:middle line:90% 00:40:12.380 --> 00:40:16.040 align:middle line:84% So in this region, the properties 00:40:16.040 --> 00:40:19.400 align:middle line:84% of this layer, which is generally very 00:40:19.400 --> 00:40:27.530 align:middle line:84% thin, and therefore, these properties of the substance 00:40:27.530 --> 00:40:33.740 align:middle line:84% feel the different forces, intermolecular forces, 00:40:33.740 --> 00:40:35.850 align:middle line:90% than any of the two bulks. 00:40:35.850 --> 00:40:37.850 align:middle line:84% Because here, you have just forces 00:40:37.850 --> 00:40:41.700 align:middle line:90% between the same gray molecules. 00:40:41.700 --> 00:40:46.160 align:middle line:84% Here, you have just intermolecular forces 00:40:46.160 --> 00:40:48.210 align:middle line:90% of the dark molecules. 00:40:48.210 --> 00:40:52.350 align:middle line:84% And here, you have some and some. 00:40:52.350 --> 00:40:54.245 align:middle line:84% So they are really different systems. 00:40:54.245 --> 00:40:58.460 align:middle line:90% 00:40:58.460 --> 00:41:01.190 align:middle line:84% And the fact that the interface is-- 00:41:01.190 --> 00:41:04.400 align:middle line:84% or interfacial region is generally very, very 00:41:04.400 --> 00:41:07.590 align:middle line:90% thin, allows you to model it. 00:41:07.590 --> 00:41:11.800 align:middle line:84% You can model it, of course, in a continuous way. 00:41:11.800 --> 00:41:15.190 align:middle line:84% But here, we're going to do it like when you 00:41:15.190 --> 00:41:17.650 align:middle line:90% do a Young-Laplace equation. 00:41:17.650 --> 00:41:20.900 align:middle line:84% You model it in a much simpler way, 00:41:20.900 --> 00:41:32.440 align:middle line:84% as if it was just another very thin system, which can exchange 00:41:32.440 --> 00:41:37.260 align:middle line:90% molecules on both sides. 00:41:37.260 --> 00:41:42.140 align:middle line:84% But on the right, it can exchange only dark molecule, 00:41:42.140 --> 00:41:45.650 align:middle line:84% on the left, only gray molecules. 00:41:45.650 --> 00:41:50.360 align:middle line:84% Whereas, a situation in which you have coexistence of the two 00:41:50.360 --> 00:41:58.400 align:middle line:84% can only move along the surface in a 2D geometry. 00:41:58.400 --> 00:42:04.680 align:middle line:84% So it's only by moving along this circle here, 00:42:04.680 --> 00:42:09.380 align:middle line:84% if it is a spherical geometry, that you can 00:42:09.380 --> 00:42:14.106 align:middle line:90% find this coexisting situation. 00:42:14.106 --> 00:42:17.980 align:middle line:90% 00:42:17.980 --> 00:42:21.400 align:middle line:84% And it's the same kind of idea when 00:42:21.400 --> 00:42:26.950 align:middle line:84% you have phase separation, when the system in order 00:42:26.950 --> 00:42:38.710 align:middle line:84% to reach a stable equilibrium, doesn't find the maximum entropy 00:42:38.710 --> 00:42:42.710 align:middle line:84% state by remaining homogeneous, but it separates. 00:42:42.710 --> 00:42:47.860 align:middle line:84% So molecules like to bump to each other 00:42:47.860 --> 00:42:49.490 align:middle line:90% and become a liquid on one side. 00:42:49.490 --> 00:42:51.490 align:middle line:84% And on the other side, you've got a water vapor. 00:42:51.490 --> 00:42:54.600 align:middle line:90% 00:42:54.600 --> 00:42:58.290 align:middle line:84% The interface, interfacial region, 00:42:58.290 --> 00:43:03.990 align:middle line:84% is a very, very thin region in which the density changes 00:43:03.990 --> 00:43:05.650 align:middle line:90% drastically. 00:43:05.650 --> 00:43:09.030 align:middle line:84% Now, the factor of 1,000 is the density difference between 00:43:09.030 --> 00:43:13.860 align:middle line:84% the liquid and the vapor, like, for example, for water. 00:43:13.860 --> 00:43:20.150 align:middle line:84% So the density drops in a matter of few-- 00:43:20.150 --> 00:43:23.760 align:middle line:90% 00:43:23.760 --> 00:43:24.490 align:middle line:90% what is it? 00:43:24.490 --> 00:43:27.100 align:middle line:90% 00:43:27.100 --> 00:43:28.030 align:middle line:90% Nanometers. 00:43:28.030 --> 00:43:31.750 align:middle line:90% 00:43:31.750 --> 00:43:37.060 align:middle line:90% It drops by a factor of 1,000. 00:43:37.060 --> 00:43:49.180 align:middle line:84% So if we have a system like that, it's confined on a layer. 00:43:49.180 --> 00:43:53.440 align:middle line:84% So its extension is not measured in terms of thickness, 00:43:53.440 --> 00:43:57.820 align:middle line:84% but it's only in terms of the surface area 00:43:57.820 --> 00:44:02.070 align:middle line:90% where it can develop. 00:44:02.070 --> 00:44:04.350 align:middle line:84% So for that system, the parameter 00:44:04.350 --> 00:44:12.120 align:middle line:84% of the geometrical container, or let's say, 00:44:12.120 --> 00:44:18.000 align:middle line:84% mental container, where those molecules are confined, 00:44:18.000 --> 00:44:23.690 align:middle line:84% is not the volume, but it's the surface area. 00:44:23.690 --> 00:44:25.680 align:middle line:84% Or if you want, it is still the volume, 00:44:25.680 --> 00:44:28.100 align:middle line:84% but the volume can be split in the surface 00:44:28.100 --> 00:44:30.100 align:middle line:90% area and a very thin layer. 00:44:30.100 --> 00:44:34.200 align:middle line:90% 00:44:34.200 --> 00:44:39.350 align:middle line:84% So area of the surface becomes a parameter 00:44:39.350 --> 00:44:46.900 align:middle line:84% for that system, which plays a parallel role of that played 00:44:46.900 --> 00:44:51.740 align:middle line:90% by volume for bulk systems. 00:44:51.740 --> 00:44:58.610 align:middle line:84% And like we define the partial derivative of fundamental 00:44:58.610 --> 00:45:02.780 align:middle line:84% for the stable equilibrium state with respect to volume, 00:45:02.780 --> 00:45:10.640 align:middle line:84% we call it P over T, for this particular system, 00:45:10.640 --> 00:45:12.720 align:middle line:84% the derivative of S with respect to A, 00:45:12.720 --> 00:45:17.850 align:middle line:90% we call it minus sigma over T. 00:45:17.850 --> 00:45:22.110 align:middle line:84% And if we invert the function, we 00:45:22.110 --> 00:45:25.330 align:middle line:84% see that you can see, for the bulk system, 00:45:25.330 --> 00:45:27.000 align:middle line:84% the pressure is minus the derivative 00:45:27.000 --> 00:45:34.920 align:middle line:84% of the energy with respect to volume of the energy form 00:45:34.920 --> 00:45:37.020 align:middle line:90% of the fundamental relation. 00:45:37.020 --> 00:45:43.110 align:middle line:84% And for the interface system, it's sigma-- 00:45:43.110 --> 00:45:44.640 align:middle line:90% we call it surface tension-- 00:45:44.640 --> 00:45:46.980 align:middle line:84% is the partial of the energy with respect to A 00:45:46.980 --> 00:45:48.790 align:middle line:90% without the minus sign. 00:45:48.790 --> 00:45:51.440 align:middle line:84% That's why one is a pressure, the other is a tension. 00:45:51.440 --> 00:45:54.120 align:middle line:90% 00:45:54.120 --> 00:45:57.940 align:middle line:84% OK, so that is for the definition. 00:45:57.940 --> 00:46:08.150 align:middle line:84% So now we can play the same role, the same trick, 00:46:08.150 --> 00:46:10.940 align:middle line:84% of applying the maximum entropy principle, 00:46:10.940 --> 00:46:14.150 align:middle line:84% to see what happens at mutual equilibrium between those three 00:46:14.150 --> 00:46:22.790 align:middle line:84% phases, those three systems, B, D, and the interface C. 00:46:22.790 --> 00:46:27.270 align:middle line:84% And the trick is to construct another state. 00:46:27.270 --> 00:46:28.490 align:middle line:90% We call it F1. 00:46:28.490 --> 00:46:31.620 align:middle line:84% Because now I'm running out of letters. 00:46:31.620 --> 00:46:36.510 align:middle line:84% So F is the composite system of BCD. 00:46:36.510 --> 00:46:43.410 align:middle line:84% So this overall system is F. I'm not using letter A, because now. 00:46:43.410 --> 00:46:44.930 align:middle line:90% A is the area. 00:46:44.930 --> 00:46:49.250 align:middle line:90% So that's used up. 00:46:49.250 --> 00:46:53.850 align:middle line:84% So we need to construct another state. 00:46:53.850 --> 00:46:57.810 align:middle line:84% And the way we do it is by changing-- 00:46:57.810 --> 00:47:00.930 align:middle line:90% we do it in a particular way-- 00:47:00.930 --> 00:47:09.670 align:middle line:84% by changing the shape, perturbing 00:47:09.670 --> 00:47:12.955 align:middle line:90% the shape of the original drop. 00:47:12.955 --> 00:47:18.560 align:middle line:90% 00:47:18.560 --> 00:47:24.690 align:middle line:84% And then we proceed in the same way. 00:47:24.690 --> 00:47:30.340 align:middle line:84% We want to prove that this thing is negative-- 00:47:30.340 --> 00:47:32.860 align:middle line:90% 00:47:32.860 --> 00:47:38.110 align:middle line:84% yeah here is the minus, the negative sign. 00:47:38.110 --> 00:47:39.800 align:middle line:90% We apply additivity. 00:47:39.800 --> 00:47:40.995 align:middle line:90% We now have three systems. 00:47:40.995 --> 00:47:45.590 align:middle line:90% 00:47:45.590 --> 00:47:53.040 align:middle line:84% So the one, B and D, are like we've done before. 00:47:53.040 --> 00:47:57.840 align:middle line:84% But now we have, also, system C, for which the Taylor series 00:47:57.840 --> 00:48:00.960 align:middle line:84% expansion in the first terms in the expansion 00:48:00.960 --> 00:48:06.340 align:middle line:90% contain the surface tension. 00:48:06.340 --> 00:48:11.500 align:middle line:84% Of course, it's the value, like the pressure evaluated 00:48:11.500 --> 00:48:16.300 align:middle line:84% at the stable equilibrium state F0. 00:48:16.300 --> 00:48:20.595 align:middle line:84% Here the dots represent many things. 00:48:20.595 --> 00:48:21.970 align:middle line:84% I should have written also, but I 00:48:21.970 --> 00:48:25.690 align:middle line:84% didn't have space enough on the viewgraph, the derivative 00:48:25.690 --> 00:48:30.315 align:middle line:84% with respect to the amounts and the second-order terms. 00:48:30.315 --> 00:48:33.440 align:middle line:90% 00:48:33.440 --> 00:48:39.080 align:middle line:84% But in order to focus on the interesting part, new part, 00:48:39.080 --> 00:48:47.470 align:middle line:84% let me choose a particular situation in which-- 00:48:47.470 --> 00:48:49.410 align:middle line:90% this very particular value-- 00:48:49.410 --> 00:49:00.130 align:middle line:84% state F1, in which I have already 00:49:00.130 --> 00:49:05.930 align:middle line:84% satisfied the temperature equality condition. 00:49:05.930 --> 00:49:12.550 align:middle line:84% I already satisfied the chemical potential equality. 00:49:12.550 --> 00:49:20.750 align:middle line:84% And so all I am interested in is changing the volume. 00:49:20.750 --> 00:49:23.660 align:middle line:84% So that allows us to, essentially, 00:49:23.660 --> 00:49:27.990 align:middle line:84% concentrate on these three terms, this one, that one, 00:49:27.990 --> 00:49:29.050 align:middle line:90% and that one. 00:49:29.050 --> 00:49:32.390 align:middle line:90% 00:49:32.390 --> 00:49:34.230 align:middle line:90% And by getting rid of-- 00:49:34.230 --> 00:49:37.210 align:middle line:84% yeah, there shouldn't be 1 over T here that disappeared. 00:49:37.210 --> 00:49:46.183 align:middle line:84% I have to fix also that on the viewgraph 00:49:46.183 --> 00:49:49.180 align:middle line:84% because these temperatures are assumed to be equal. 00:49:49.180 --> 00:49:52.540 align:middle line:90% So I'm left with this equation. 00:49:52.540 --> 00:50:00.700 align:middle line:84% And this equation has to be negative, strictly negative. 00:50:00.700 --> 00:50:05.450 align:middle line:84% But as we will find out, is that, again, 00:50:05.450 --> 00:50:10.630 align:middle line:84% because of the arbitrariness of the choices 00:50:10.630 --> 00:50:19.010 align:middle line:84% of the change in the geometry, in the volume, 00:50:19.010 --> 00:50:22.640 align:middle line:84% it will turn out to have to be 0. 00:50:22.640 --> 00:50:25.130 align:middle line:84% And so the burden for the inequality 00:50:25.130 --> 00:50:31.225 align:middle line:84% will be carried by the extra additional terms. 00:50:31.225 --> 00:50:37.350 align:middle line:90% 00:50:37.350 --> 00:50:41.020 align:middle line:84% So yes, again, also here, I have to fix the viewgraph. 00:50:41.020 --> 00:50:43.805 align:middle line:84% So this is just in order to make room for something else. 00:50:43.805 --> 00:50:47.200 align:middle line:90% 00:50:47.200 --> 00:50:50.580 align:middle line:84% So now there is an easy way of proceeding here, 00:50:50.580 --> 00:50:52.630 align:middle line:90% is assume a geometry. 00:50:52.630 --> 00:50:57.710 align:middle line:84% If we assume a geometry, just to see what comes out-- 00:50:57.710 --> 00:51:05.690 align:middle line:84% so we assume, for example, spherical geometry of the drop, 00:51:05.690 --> 00:51:10.770 align:middle line:84% and we assume that we change the volume. 00:51:10.770 --> 00:51:16.430 align:middle line:84% So we consider a perturbed state with a volume slightly increased 00:51:16.430 --> 00:51:18.680 align:middle line:90% by an amount. 00:51:18.680 --> 00:51:21.830 align:middle line:84% If R was the radius of the drop, it 00:51:21.830 --> 00:51:26.448 align:middle line:84% becomes R plus dR or d epsilon, epsilon to stress 00:51:26.448 --> 00:51:27.490 align:middle line:90% the fact that it's small. 00:51:27.490 --> 00:51:30.290 align:middle line:90% 00:51:30.290 --> 00:51:31.850 align:middle line:90% OK. 00:51:31.850 --> 00:51:37.280 align:middle line:84% So the area of the interface of the surface of the sphere 00:51:37.280 --> 00:51:38.485 align:middle line:90% is 4 pi R squared. 00:51:38.485 --> 00:51:42.530 align:middle line:90% 00:51:42.530 --> 00:51:44.720 align:middle line:90% The volume that is-- 00:51:44.720 --> 00:51:49.360 align:middle line:90% 00:51:49.360 --> 00:51:52.570 align:middle line:84% if I increase this sphere from this solid line 00:51:52.570 --> 00:51:55.870 align:middle line:84% to the dashed line, the increase in volume 00:51:55.870 --> 00:52:04.100 align:middle line:84% is equal to the surface area times the shift in-- 00:52:04.100 --> 00:52:06.040 align:middle line:90% so times d epsilon. 00:52:06.040 --> 00:52:10.480 align:middle line:84% So that's dV, and that's what happens to what appears 00:52:10.480 --> 00:52:12.195 align:middle line:90% in this part of the equation. 00:52:12.195 --> 00:52:14.980 align:middle line:90% 00:52:14.980 --> 00:52:19.190 align:middle line:84% Also, since I am going from small to large, 00:52:19.190 --> 00:52:21.990 align:middle line:90% also the area changes. 00:52:21.990 --> 00:52:27.440 align:middle line:84% Now, the area changes because the radius has changed. 00:52:27.440 --> 00:52:35.738 align:middle line:84% So I go from 4 pi R squared to 4 pi R plus d epsilon squared. 00:52:35.738 --> 00:52:39.950 align:middle line:90% 00:52:39.950 --> 00:52:44.230 align:middle line:84% And if I forget about the d epsilon square 00:52:44.230 --> 00:52:46.270 align:middle line:84% because it's small with respect to d 00:52:46.270 --> 00:52:51.570 align:middle line:84% epsilon put into the others, what I'm left is-- 00:52:51.570 --> 00:52:54.090 align:middle line:84% so one of these cancels with that. 00:52:54.090 --> 00:52:57.005 align:middle line:90% I'm left with 8 pi R d epsilon. 00:52:57.005 --> 00:53:00.130 align:middle line:90% 00:53:00.130 --> 00:53:05.940 align:middle line:84% And this is the change in area that I put here. 00:53:05.940 --> 00:53:11.280 align:middle line:84% Still, the d epsilon is arbitrary. 00:53:11.280 --> 00:53:15.210 align:middle line:84% So that expression now becomes this expression. 00:53:15.210 --> 00:53:18.750 align:middle line:90% And it should be equal to 0-- 00:53:18.750 --> 00:53:22.280 align:middle line:84% I mean, negative, for arbitrary values of the epsilon. 00:53:22.280 --> 00:53:24.030 align:middle line:84% I could have chosen, also, d epsilon 00:53:24.030 --> 00:53:27.870 align:middle line:84% negative instead of positive, and still that 00:53:27.870 --> 00:53:29.250 align:middle line:90% should be satisfied. 00:53:29.250 --> 00:53:31.830 align:middle line:84% So essentially, it's just what multiplies 00:53:31.830 --> 00:53:35.510 align:middle line:90% d epsilon that has to be 0. 00:53:35.510 --> 00:53:41.450 align:middle line:84% So the 4 pi R cancels, and you get this expression here. 00:53:41.450 --> 00:53:47.130 align:middle line:84% There is a 2 that emerges from the 8 divided by 4. 00:53:47.130 --> 00:53:50.670 align:middle line:84% And so you find that the condition for mutual equilibrium 00:53:50.670 --> 00:53:53.640 align:middle line:84% between those three systems is that the pressure 00:53:53.640 --> 00:54:02.900 align:middle line:84% inside the drop must be greater than the pressure of the outside 00:54:02.900 --> 00:54:10.060 align:middle line:84% by the surface tension multiplied by 2 over R. 00:54:10.060 --> 00:54:17.156 align:middle line:84% Now, if you repeat that for a cylindrical shape, 00:54:17.156 --> 00:54:19.430 align:middle line:90% we just skip that. 00:54:19.430 --> 00:54:22.590 align:middle line:90% But here are the details. 00:54:22.590 --> 00:54:28.970 align:middle line:84% You get a similar expression, only with 1 over R. 00:54:28.970 --> 00:54:37.150 align:middle line:84% But a true story is that we don't know this shape. 00:54:37.150 --> 00:54:41.970 align:middle line:84% So we shouldn't, in principle, assume that shape, 00:54:41.970 --> 00:54:49.770 align:middle line:84% because in principle, who tells us 00:54:49.770 --> 00:54:54.240 align:middle line:84% that the shape should be spherical or geometry or 00:54:54.240 --> 00:54:55.830 align:middle line:90% cylindrical? 00:54:55.830 --> 00:55:02.650 align:middle line:84% That is also a result of this maximum entropy principle 00:55:02.650 --> 00:55:04.630 align:middle line:90% and the Young-Laplace. 00:55:04.630 --> 00:55:08.610 align:middle line:90% 00:55:08.610 --> 00:55:12.450 align:middle line:84% It's a bit complicated, so it took me days 00:55:12.450 --> 00:55:16.330 align:middle line:90% to construct these pictures. 00:55:16.330 --> 00:55:19.450 align:middle line:84% So I'm not going to spend too much time on them. 00:55:19.450 --> 00:55:22.980 align:middle line:84% But simply, I assume that you know 00:55:22.980 --> 00:55:26.100 align:middle line:84% that if I have a surface in space, 00:55:26.100 --> 00:55:31.030 align:middle line:84% I can define principal radii of curvature. 00:55:31.030 --> 00:55:38.110 align:middle line:84% And like in a sphere or in an ellipsoid, 00:55:38.110 --> 00:55:42.360 align:middle line:84% you could have the centers of curvature 00:55:42.360 --> 00:55:46.800 align:middle line:90% on the same side of the surface. 00:55:46.800 --> 00:55:50.170 align:middle line:84% But in a saddle point of your surface, 00:55:50.170 --> 00:55:53.040 align:middle line:84% you could have the centers of curvature 00:55:53.040 --> 00:55:55.140 align:middle line:90% on the two opposite sides. 00:55:55.140 --> 00:55:57.455 align:middle line:84% So for example, suppose this is the drop. 00:55:57.455 --> 00:56:00.520 align:middle line:90% 00:56:00.520 --> 00:56:03.700 align:middle line:84% In this situation, you have one center 00:56:03.700 --> 00:56:07.640 align:middle line:84% of curvature inside the drop and one outside. 00:56:07.640 --> 00:56:09.885 align:middle line:84% Whereas here, they are both inside. 00:56:09.885 --> 00:56:15.214 align:middle line:90% 00:56:15.214 --> 00:56:19.450 align:middle line:84% But, you see, this is a local picture. 00:56:19.450 --> 00:56:23.200 align:middle line:84% If you have an arbitrary shape, if you have a drop of arbitrary 00:56:23.200 --> 00:56:26.890 align:middle line:84% shape, like, for example, when you 00:56:26.890 --> 00:56:32.752 align:middle line:84% have a drop of water in gravity field, 00:56:32.752 --> 00:56:33.990 align:middle line:90% you know this is the shape. 00:56:33.990 --> 00:56:37.670 align:middle line:90% 00:56:37.670 --> 00:56:45.860 align:middle line:84% So the local curvature and the surface 00:56:45.860 --> 00:56:48.470 align:middle line:84% must be represented somehow by running 00:56:48.470 --> 00:56:53.280 align:middle line:84% over the entire surface with local variables and integrating. 00:56:53.280 --> 00:56:55.770 align:middle line:84% So in order to calculate the entire area, 00:56:55.770 --> 00:57:03.000 align:middle line:84% you have to do an integral of these local areas. 00:57:03.000 --> 00:57:06.900 align:middle line:84% So here is a lot of expressions that I hope 00:57:06.900 --> 00:57:09.270 align:middle line:90% are correct that allows you-- 00:57:09.270 --> 00:57:14.820 align:middle line:84% that describe these geometrical situations, 00:57:14.820 --> 00:57:20.430 align:middle line:84% one for the system when the two centers are on the two 00:57:20.430 --> 00:57:24.090 align:middle line:84% sides of the surface and one for this, the case in which they 00:57:24.090 --> 00:57:27.400 align:middle line:90% are both on the same side. 00:57:27.400 --> 00:57:32.510 align:middle line:84% The difference, you see, locally, 00:57:32.510 --> 00:57:35.330 align:middle line:84% the purpose of this formula here is 00:57:35.330 --> 00:57:43.780 align:middle line:84% to compute the ratio of the area change to the volume change. 00:57:43.780 --> 00:57:46.750 align:middle line:84% As you shift to C, remember that we were 00:57:46.750 --> 00:57:48.140 align:middle line:90% talking about this situation. 00:57:48.140 --> 00:57:53.590 align:middle line:84% We want to perturb the shape of the drop 00:57:53.590 --> 00:57:58.960 align:middle line:90% and go to another state. 00:57:58.960 --> 00:58:02.900 align:middle line:84% And we want to compute the volume and the area. 00:58:02.900 --> 00:58:06.220 align:middle line:84% So what is important is the ratio 00:58:06.220 --> 00:58:10.330 align:middle line:84% of the area change over the volume change. 00:58:10.330 --> 00:58:15.570 align:middle line:84% And that's why here, we compute it for locally. 00:58:15.570 --> 00:58:20.930 align:middle line:84% So as I perturb, move the surface, 00:58:20.930 --> 00:58:25.940 align:middle line:84% that entails a change in volume, delta V, and a change 00:58:25.940 --> 00:58:31.190 align:middle line:84% in surface area, delta A. And this ratio 00:58:31.190 --> 00:58:33.120 align:middle line:90% is related to the inverse. 00:58:33.120 --> 00:58:38.780 align:middle line:84% It's called the mean local curvature. 00:58:38.780 --> 00:58:40.800 align:middle line:84% It's the sum of two contributions, 00:58:40.800 --> 00:58:42.290 align:middle line:90% but it is an algebraic sum. 00:58:42.290 --> 00:58:44.540 align:middle line:90% Because one of the-- 00:58:44.540 --> 00:58:48.000 align:middle line:84% when the radius is on the other side, 00:58:48.000 --> 00:58:50.635 align:middle line:90% it has a negative in front. 00:58:50.635 --> 00:58:53.286 align:middle line:90% 00:58:53.286 --> 00:58:55.620 align:middle line:84% So if we go back to our picture here 00:58:55.620 --> 00:59:03.570 align:middle line:84% and try to compute these areas and these change in volume, 00:59:03.570 --> 00:59:07.980 align:middle line:84% we have to integrate somehow over the entire surface. 00:59:07.980 --> 00:59:11.980 align:middle line:84% That's why you need a double integral over some variables. 00:59:11.980 --> 00:59:20.100 align:middle line:84% So theta 1 and theta 2 maybe is a way to describe local angles. 00:59:20.100 --> 00:59:23.670 align:middle line:90% 00:59:23.670 --> 00:59:31.290 align:middle line:84% And then if you go on and do all the math, you calculate, 00:59:31.290 --> 00:59:35.490 align:middle line:84% again, the change in area, it's, again, R plus 00:59:35.490 --> 00:59:41.070 align:middle line:84% d epsilon and R. Now it's plus or minus d epsilon 00:59:41.070 --> 00:59:46.470 align:middle line:84% because when you move the surface on one side, 00:59:46.470 --> 00:59:54.980 align:middle line:84% in this case, one radius of curvature increases, 00:59:54.980 --> 01:00:01.430 align:middle line:84% and the other decreases by the amount d epsilon. 01:00:01.430 --> 01:00:11.700 align:middle line:84% And so in the end, you get a double integral over the surface 01:00:11.700 --> 01:00:13.410 align:middle line:90% coordinates. 01:00:13.410 --> 01:00:19.680 align:middle line:84% And the d epsilon, now you can choose arbitrarily 01:00:19.680 --> 01:00:24.540 align:middle line:84% the d epsilon at every location in the surface. 01:00:24.540 --> 01:00:25.970 align:middle line:90% And that's arbitrary. 01:00:25.970 --> 01:00:31.550 align:middle line:84% So that integral must be equal to 0 01:00:31.550 --> 01:00:34.370 align:middle line:84% for arbitrary choices of how you want 01:00:34.370 --> 01:00:35.725 align:middle line:90% to distribute the d epsilons. 01:00:35.725 --> 01:00:38.420 align:middle line:90% 01:00:38.420 --> 01:00:43.130 align:middle line:84% Which says that, therefore, it's the integrand that 01:00:43.130 --> 01:00:44.230 align:middle line:90% must be equal to 0. 01:00:44.230 --> 01:00:48.260 align:middle line:90% 01:00:48.260 --> 01:00:53.185 align:middle line:84% And so you find that you can rewrite it this way. 01:00:53.185 --> 01:00:58.100 align:middle line:90% 01:00:58.100 --> 01:01:01.040 align:middle line:84% And you find that therefore, the sum 01:01:01.040 --> 01:01:05.750 align:middle line:84% of the inverse of the radii of curvature, so 01:01:05.750 --> 01:01:09.740 align:middle line:84% the mean curvature, which, in principle, could 01:01:09.740 --> 01:01:15.050 align:middle line:84% be different locally from point to point, 01:01:15.050 --> 01:01:18.650 align:middle line:84% must be equal to this ratio of difference 01:01:18.650 --> 01:01:23.980 align:middle line:84% in pressure and the surface tension. 01:01:23.980 --> 01:01:27.840 align:middle line:84% Which is something that doesn't depend 01:01:27.840 --> 01:01:29.200 align:middle line:90% on location of the surface. 01:01:29.200 --> 01:01:33.560 align:middle line:90% 01:01:33.560 --> 01:01:41.930 align:middle line:84% So this result means that the average curvature 01:01:41.930 --> 01:01:45.800 align:middle line:84% must be uniform, must be equal everywhere on the surface 01:01:45.800 --> 01:01:54.580 align:middle line:84% when you are in mutual equilibrium, like in a sphere, 01:01:54.580 --> 01:01:57.010 align:middle line:90% like in a cylindrical situation. 01:01:57.010 --> 01:02:02.080 align:middle line:84% There are some other interesting surfaces-- 01:02:02.080 --> 01:02:08.320 align:middle line:84% if you look up in the Wikipedia, Surfaces 01:02:08.320 --> 01:02:13.420 align:middle line:84% of constant mean curvature, you find 01:02:13.420 --> 01:02:20.040 align:middle line:84% that somebody has found a very interesting shape where 01:02:20.040 --> 01:02:26.650 align:middle line:84% it's more complicated than a sphere, plus, of course-- 01:02:26.650 --> 01:02:29.490 align:middle line:84% so this is the main result, the uniformity 01:02:29.490 --> 01:02:31.860 align:middle line:90% of the mean curvature. 01:02:31.860 --> 01:02:36.420 align:middle line:84% And this is the standard Young-Laplace equation 01:02:36.420 --> 01:02:43.330 align:middle line:84% for a surface with the two different radii of curvature. 01:02:43.330 --> 01:02:52.890 align:middle line:90% 01:02:52.890 --> 01:02:59.640 align:middle line:84% So now we are ready for the next move. 01:02:59.640 --> 01:03:06.030 align:middle line:84% We want to prove the Clausius statement of the Second Law. 01:03:06.030 --> 01:03:08.700 align:middle line:84% We want to capitalize now on the fact 01:03:08.700 --> 01:03:20.005 align:middle line:84% that systems can exchange energy and entropy. 01:03:20.005 --> 01:03:26.290 align:middle line:90% 01:03:26.290 --> 01:03:29.030 align:middle line:84% And we would like to consider this particular situation, 01:03:29.030 --> 01:03:35.340 align:middle line:84% two systems A and B, initially in stable equilibrium states. 01:03:35.340 --> 01:03:39.030 align:middle line:90% 01:03:39.030 --> 01:03:41.160 align:middle line:84% For simplicity, for the moment, we just 01:03:41.160 --> 01:03:43.205 align:middle line:84% consider fixed volume, fixed amounts. 01:03:43.205 --> 01:03:48.100 align:middle line:90% 01:03:48.100 --> 01:03:52.485 align:middle line:84% And we want to transfer some energy from one to the other. 01:03:52.485 --> 01:03:55.480 align:middle line:90% 01:03:55.480 --> 01:03:58.770 align:middle line:84% The question is, Is it always possible, 01:03:58.770 --> 01:04:00.005 align:middle line:90% or when is it possible? 01:04:00.005 --> 01:04:05.190 align:middle line:90% 01:04:05.190 --> 01:04:08.060 align:middle line:84% Do I need to transfer, also, some entropy 01:04:08.060 --> 01:04:11.250 align:middle line:90% in order to make it possible? 01:04:11.250 --> 01:04:16.910 align:middle line:84% So that's the question answered by the Clausius-- 01:04:16.910 --> 01:04:24.810 align:middle line:84% the inequalities that we will prove right now. 01:04:24.810 --> 01:04:28.990 align:middle line:84% So, look, it's, again, the same story as we started with. 01:04:28.990 --> 01:04:33.750 align:middle line:84% If my system A starts in a stable equilibrium state here, 01:04:33.750 --> 01:04:38.910 align:middle line:84% and I want to extract the energy so that I give it to system B, 01:04:38.910 --> 01:04:40.950 align:middle line:90% I need to go down in energy. 01:04:40.950 --> 01:04:43.060 align:middle line:84% But in order to go down in energy, 01:04:43.060 --> 01:04:44.945 align:middle line:84% I also need to go down in entropy. 01:04:44.945 --> 01:04:50.070 align:middle line:90% 01:04:50.070 --> 01:04:56.920 align:middle line:84% So already from the picture, we see a geometrical proof. 01:04:56.920 --> 01:05:01.860 align:middle line:84% We will have viewgraphs that do that more in detail, 01:05:01.860 --> 01:05:03.730 align:middle line:90% but also more complicated. 01:05:03.730 --> 01:05:07.030 align:middle line:90% 01:05:07.030 --> 01:05:10.550 align:middle line:84% But first, let's do it with equations. 01:05:10.550 --> 01:05:12.410 align:middle line:90% So how do we model a system? 01:05:12.410 --> 01:05:14.740 align:middle line:84% We make energy balances and entropy balances. 01:05:14.740 --> 01:05:18.260 align:middle line:90% 01:05:18.260 --> 01:05:24.490 align:middle line:84% And let's consider a small change in-- 01:05:24.490 --> 01:05:28.460 align:middle line:84% a small amount of energy transferred. 01:05:28.460 --> 01:05:31.930 align:middle line:84% So if I make the energy balance for system 01:05:31.930 --> 01:05:34.150 align:middle line:90% A simply says that the rate-- 01:05:34.150 --> 01:05:36.430 align:middle line:84% I mean that the change in the energy of A 01:05:36.430 --> 01:05:43.750 align:middle line:84% is equal to minus the amount that I gave to B, that I 01:05:43.750 --> 01:05:47.680 align:middle line:84% extracted from A. And for system B, 01:05:47.680 --> 01:05:53.610 align:middle line:84% its energy change is equal to the amount that it received. 01:05:53.610 --> 01:05:54.800 align:middle line:90% Notice the arrows. 01:05:54.800 --> 01:05:57.740 align:middle line:90% 01:05:57.740 --> 01:06:04.360 align:middle line:84% The arrows tell you, when these symbols are positive, 01:06:04.360 --> 01:06:08.015 align:middle line:84% it means that the flow has been in the direction of the arrow. 01:06:08.015 --> 01:06:12.860 align:middle line:90% 01:06:12.860 --> 01:06:13.660 align:middle line:90% Entropy balance. 01:06:13.660 --> 01:06:17.020 align:middle line:90% 01:06:17.020 --> 01:06:26.110 align:middle line:84% The change in the entropy of system A as it goes to the left 01:06:26.110 --> 01:06:28.310 align:middle line:90% is equal to two contributions. 01:06:28.310 --> 01:06:34.530 align:middle line:84% One is the entropy that has been transferred out of system A, 01:06:34.530 --> 01:06:37.710 align:middle line:84% and the other, if any, is the entropy 01:06:37.710 --> 01:06:42.000 align:middle line:84% that has been produced by irreversibility within system A, 01:06:42.000 --> 01:06:45.720 align:middle line:90% if the process is irreversible. 01:06:45.720 --> 01:06:48.690 align:middle line:84% If you do it reversibly, this one isn't there. 01:06:48.690 --> 01:06:53.580 align:middle line:84% But if you do it irreversibly, this one, this term, 01:06:53.580 --> 01:06:56.400 align:middle line:90% must be greater than 0. 01:06:56.400 --> 01:06:58.990 align:middle line:84% For system B, the entropy balance 01:06:58.990 --> 01:07:03.420 align:middle line:84% says that the change in the entropy of B 01:07:03.420 --> 01:07:11.260 align:middle line:84% is equal to the entropy that it received from A plus, if any, 01:07:11.260 --> 01:07:18.650 align:middle line:84% the entropy that was generated within B by irreversibility. 01:07:18.650 --> 01:07:30.990 align:middle line:90% Now, again-- this state A1-- 01:07:30.990 --> 01:07:37.600 align:middle line:90% 01:07:37.600 --> 01:07:38.100 align:middle line:90% Yeah. 01:07:38.100 --> 01:07:41.980 align:middle line:90% 01:07:41.980 --> 01:07:48.850 align:middle line:84% And we extracted this much energy. 01:07:48.850 --> 01:08:01.990 align:middle line:84% I call it delta EA out of A and goes into B. 01:08:01.990 --> 01:08:09.560 align:middle line:84% I do not necessarily have to go to the stable equilibrium state. 01:08:09.560 --> 01:08:18.380 align:middle line:84% I could go to any of these states, in principle. 01:08:18.380 --> 01:08:21.770 align:middle line:84% Well, maybe I cannot go all that far. 01:08:21.770 --> 01:08:23.600 align:middle line:90% But that will depend. 01:08:23.600 --> 01:08:25.770 align:middle line:84% Because if I want to go this far, 01:08:25.770 --> 01:08:28.939 align:middle line:84% I need a system on the other side. 01:08:28.939 --> 01:08:32.029 align:middle line:84% Since I reduce the entropy, I need a system 01:08:32.029 --> 01:08:33.750 align:middle line:90% that picks up that entropy. 01:08:33.750 --> 01:08:38.240 align:middle line:84% So it has to have the ability to do that. 01:08:38.240 --> 01:08:44.630 align:middle line:84% So at a certain stage, we will see that we cannot go too far 01:08:44.630 --> 01:08:46.850 align:middle line:90% to the left. 01:08:46.850 --> 01:08:52.310 align:middle line:84% But in any case, these are the possible changes 01:08:52.310 --> 01:09:01.930 align:middle line:84% of state of system A. What I know 01:09:01.930 --> 01:09:07.330 align:middle line:84% is that the stable equilibrium state for the new value 01:09:07.330 --> 01:09:09.130 align:middle line:84% of the energy is the one that has 01:09:09.130 --> 01:09:13.810 align:middle line:84% the highest entropy, so again, maximum entropy principle. 01:09:13.810 --> 01:09:18.525 align:middle line:84% So the entropy S2, I'll define it. 01:09:18.525 --> 01:09:23.958 align:middle line:84% The entropy of state A2, suppose that's this one, 01:09:23.958 --> 01:09:28.529 align:middle line:84% is less than the maximum entropy corresponding 01:09:28.529 --> 01:09:30.180 align:middle line:84% to that value of the energy, which 01:09:30.180 --> 01:09:36.010 align:middle line:84% is the entropy of the stable equilibrium state corresponding 01:09:36.010 --> 01:09:42.149 align:middle line:90% to the new value of the energy. 01:09:42.149 --> 01:09:50.600 align:middle line:84% So here is the inequality, and here the fundamental relation, 01:09:50.600 --> 01:09:57.500 align:middle line:84% for which we can do, as usual, a Taylor series expansion 01:09:57.500 --> 01:10:01.720 align:middle line:90% around state 1. 01:10:01.720 --> 01:10:04.690 align:middle line:90% It will have a linear term. 01:10:04.690 --> 01:10:11.380 align:middle line:84% And let's keep, also, the second order term, plus dots, dots. 01:10:11.380 --> 01:10:20.960 align:middle line:90% 01:10:20.960 --> 01:10:23.580 align:middle line:90% So you see this expression. 01:10:23.580 --> 01:10:33.980 align:middle line:84% It's S2, S1, but S2 is S1 plus dS. 01:10:33.980 --> 01:10:38.150 align:middle line:84% So we can take the S1 on the left-hand side here. 01:10:38.150 --> 01:10:39.720 align:middle line:90% It becomes dS. 01:10:39.720 --> 01:10:42.330 align:middle line:90% And here is the rewriting. 01:10:42.330 --> 01:10:50.740 align:middle line:84% So dS is equal to dE over T plus the second derivative plus dots, 01:10:50.740 --> 01:10:52.530 align:middle line:90% dots, dots. 01:10:52.530 --> 01:10:55.870 align:middle line:84% What do we know about the second derivatives? 01:10:55.870 --> 01:11:01.286 align:middle line:84% Well, we just proved that they are negative. 01:11:01.286 --> 01:11:08.230 align:middle line:84% So if this is a negative number, and this is a squared number, 01:11:08.230 --> 01:11:09.750 align:middle line:90% this thing will subtract. 01:11:09.750 --> 01:11:12.280 align:middle line:90% 01:11:12.280 --> 01:11:20.490 align:middle line:84% Therefore, this will be less than just dE over T. 01:11:20.490 --> 01:11:23.370 align:middle line:90% There is an exception, though. 01:11:23.370 --> 01:11:27.480 align:middle line:84% If the system is a reservoir, that second derivative 01:11:27.480 --> 01:11:28.340 align:middle line:90% is equal to 0. 01:11:28.340 --> 01:11:33.540 align:middle line:90% 01:11:33.540 --> 01:11:39.180 align:middle line:84% So here, we would have an equal sign if A is a reservoir. 01:11:39.180 --> 01:11:48.860 align:middle line:84% And the same identical treatment goes for system B. 01:11:48.860 --> 01:11:51.653 align:middle line:84% So now we can put these all together. 01:11:51.653 --> 01:11:54.310 align:middle line:90% 01:11:54.310 --> 01:11:58.490 align:middle line:84% Just to shuffle things around, up to here, 01:11:58.490 --> 01:12:00.310 align:middle line:84% it's the same that we've seen so far. 01:12:00.310 --> 01:12:04.750 align:middle line:84% Now let me combine those equations. 01:12:04.750 --> 01:12:12.930 align:middle line:90% 01:12:12.930 --> 01:12:17.400 align:middle line:90% We have 1, 2, 3, 4, 5, 6. 01:12:17.400 --> 01:12:20.640 align:middle line:84% And we can eliminate four variables 01:12:20.640 --> 01:12:24.250 align:middle line:90% and remain two equations. 01:12:24.250 --> 01:12:28.520 align:middle line:90% 01:12:28.520 --> 01:12:30.300 align:middle line:90% I'll let you do the algebra. 01:12:30.300 --> 01:12:36.140 align:middle line:90% It's easy enough. 01:12:36.140 --> 01:12:39.350 align:middle line:84% And since I want to concentrate on the energy and entropy 01:12:39.350 --> 01:12:43.340 align:middle line:84% transfers, so the delta E and the delta S, 01:12:43.340 --> 01:12:45.370 align:middle line:84% these are the variables that I keep. 01:12:45.370 --> 01:12:48.370 align:middle line:84% I eliminate the other four, the changes. 01:12:48.370 --> 01:12:51.570 align:middle line:90% 01:12:51.570 --> 01:12:57.480 align:middle line:84% So this is for system A, from the balances from system A, 01:12:57.480 --> 01:13:02.100 align:middle line:90% and this is from system B. 01:13:02.100 --> 01:13:07.510 align:middle line:84% They both contain delta S, so I can combine 01:13:07.510 --> 01:13:12.670 align:middle line:84% these two in a single train of inequalities, 01:13:12.670 --> 01:13:20.370 align:middle line:84% where delta S, the entropy exchanged, is at the center. 01:13:20.370 --> 01:13:26.670 align:middle line:84% And it's less than if I add entropy 01:13:26.670 --> 01:13:28.420 align:middle line:90% produced by reversibility. 01:13:28.420 --> 01:13:32.820 align:middle line:84% And this must be less than delta, the energy over T1B. 01:13:32.820 --> 01:13:37.650 align:middle line:84% And on the other side, this is greater 01:13:37.650 --> 01:13:40.530 align:middle line:84% than if I subtract something positive, 01:13:40.530 --> 01:13:46.180 align:middle line:84% and it is, in turn, greater than the energy over T1A. 01:13:46.180 --> 01:13:49.270 align:middle line:90% 01:13:49.270 --> 01:13:58.540 align:middle line:84% So this is interesting by itself because it says-- 01:13:58.540 --> 01:14:01.470 align:middle line:84% all right, remember, we started from our two systems 01:14:01.470 --> 01:14:07.440 align:middle line:84% in two different stable equilibrium states 01:14:07.440 --> 01:14:09.150 align:middle line:90% with different temperatures. 01:14:09.150 --> 01:14:13.990 align:middle line:90% So T1A and T1B are different. 01:14:13.990 --> 01:14:19.230 align:middle line:90% 01:14:19.230 --> 01:14:24.320 align:middle line:84% And here, we find that if we forget about-- 01:14:24.320 --> 01:14:27.180 align:middle line:84% if you look only at the central and the two 01:14:27.180 --> 01:14:33.840 align:middle line:84% extremes of these inequalities, it says that definitely, 01:14:33.840 --> 01:14:36.815 align:middle line:90% if you want to have a finite-- 01:14:36.815 --> 01:14:40.700 align:middle line:84% I mean, not a finite, but a non-zero transfer of energy 01:14:40.700 --> 01:14:50.480 align:middle line:84% from A to B, you need an entropy change, 01:14:50.480 --> 01:14:55.620 align:middle line:84% an entropy transfer, with some value 01:14:55.620 --> 01:15:01.380 align:middle line:84% that stays within this interval of the energy transfer divided 01:15:01.380 --> 01:15:05.333 align:middle line:84% by the two temperatures of the two systems. 01:15:05.333 --> 01:15:15.650 align:middle line:90% 01:15:15.650 --> 01:15:17.350 align:middle line:90% So let me do it. 01:15:17.350 --> 01:15:21.710 align:middle line:90% 01:15:21.710 --> 01:15:28.810 align:middle line:90% So it says, What is it? 01:15:28.810 --> 01:15:32.970 align:middle line:84% I could write that inequality this way. 01:15:32.970 --> 01:15:39.930 align:middle line:90% 01:15:39.930 --> 01:15:43.710 align:middle line:84% Take things on the left-hand side. 01:15:43.710 --> 01:15:46.685 align:middle line:90% And am I right if I do this? 01:15:46.685 --> 01:15:50.940 align:middle line:90% 01:15:50.940 --> 01:15:51.960 align:middle line:90% Yeah. 01:15:51.960 --> 01:15:58.060 align:middle line:84% I'm taking this all the way to the other side. 01:15:58.060 --> 01:16:01.540 align:middle line:90% So what does that tell me? 01:16:01.540 --> 01:16:08.940 align:middle line:84% If I want delta EA positive, in other words, 01:16:08.940 --> 01:16:11.300 align:middle line:84% if I want to transfer energy from A 01:16:11.300 --> 01:16:17.540 align:middle line:84% to B, if this is positive, that has to be negative. 01:16:17.540 --> 01:16:22.100 align:middle line:84% If this has to be negative, it means that the second term must 01:16:22.100 --> 01:16:24.010 align:middle line:90% be bigger than the first one. 01:16:24.010 --> 01:16:26.990 align:middle line:90% 01:16:26.990 --> 01:16:31.640 align:middle line:84% So that means that the denominator must be smaller. 01:16:31.640 --> 01:16:39.990 align:middle line:84% So that implies T1A greater than T1B. 01:16:39.990 --> 01:16:41.480 align:middle line:90% Is that what we expected? 01:16:41.480 --> 01:16:43.960 align:middle line:90% Yeah. 01:16:43.960 --> 01:16:51.010 align:middle line:84% We can transfer energies only in the direction from the hot 01:16:51.010 --> 01:16:54.210 align:middle line:90% to the cold. 01:16:54.210 --> 01:16:57.180 align:middle line:84% That's Clausius's statement of the Second Law. 01:16:57.180 --> 01:17:00.120 align:middle line:84% Of course, if you want this negative, 01:17:00.120 --> 01:17:02.680 align:middle line:84% it can do it only in the opposite situation. 01:17:02.680 --> 01:17:14.500 align:middle line:90% 01:17:14.500 --> 01:17:20.932 align:middle line:84% So by looking at the two end parts, 01:17:20.932 --> 01:17:23.870 align:middle line:84% we proved the Clausius statement of the Second Law, 01:17:23.870 --> 01:17:29.560 align:middle line:84% that energy cannot flow spontaneously. 01:17:29.560 --> 01:17:32.620 align:middle line:84% And spontaneously is, well, while there 01:17:32.620 --> 01:17:35.785 align:middle line:90% are no other external effects. 01:17:35.785 --> 01:17:41.280 align:middle line:84% You're considering a system in which this composite A and B is 01:17:41.280 --> 01:17:43.790 align:middle line:90% isolated. 01:17:43.790 --> 01:17:50.260 align:middle line:84% So, it can only occur from hot to cold, 01:17:50.260 --> 01:17:52.280 align:middle line:84% from high temperature to low temperature. 01:17:52.280 --> 01:17:56.920 align:middle line:90% 01:17:56.920 --> 01:18:07.220 align:middle line:84% If the two systems initially are at different temperatures, 01:18:07.220 --> 01:18:11.800 align:middle line:84% It says that there must be also an entropy transfer 01:18:11.800 --> 01:18:15.070 align:middle line:90% between the two. 01:18:15.070 --> 01:18:20.860 align:middle line:84% And this entropy transfer must be contained between these two 01:18:20.860 --> 01:18:21.360 align:middle line:90% extremes. 01:18:21.360 --> 01:18:24.080 align:middle line:90% 01:18:24.080 --> 01:18:28.950 align:middle line:84% It doesn't have to have a specific value as long as it 01:18:28.950 --> 01:18:30.080 align:middle line:90% is within that interval. 01:18:30.080 --> 01:18:36.490 align:middle line:90% 01:18:36.490 --> 01:18:45.650 align:middle line:84% And clearly, if the temperatures are positive, 01:18:45.650 --> 01:18:51.230 align:middle line:84% as they are for most systems, that entropy transfer 01:18:51.230 --> 01:18:55.420 align:middle line:84% occurs in the same direction as the energy transfer. 01:18:55.420 --> 01:19:02.480 align:middle line:84% So the sign of delta S, I mean, if this is positive, 01:19:02.480 --> 01:19:03.740 align:middle line:90% temperature is positive. 01:19:03.740 --> 01:19:06.850 align:middle line:84% So this is a positive number, and delta S must be greater 01:19:06.850 --> 01:19:08.000 align:middle line:90% than a positive number. 01:19:08.000 --> 01:19:11.220 align:middle line:90% So it's positive, too. 01:19:11.220 --> 01:19:15.810 align:middle line:84% So I need, also, to transfer entropy, which is also 01:19:15.810 --> 01:19:17.360 align:middle line:90% the geometrical idea here. 01:19:17.360 --> 01:19:22.063 align:middle line:90% 01:19:22.063 --> 01:19:23.230 align:middle line:90% AUDIENCE: I have a question. 01:19:23.230 --> 01:19:24.220 align:middle line:90% PROFESSOR: Yeah? 01:19:24.220 --> 01:19:27.290 align:middle line:84% AUDIENCE: Under the inequalities, what do a 2A, 3A, 01:19:27.290 --> 01:19:30.310 align:middle line:90% 1A, 1B means at that? 01:19:30.310 --> 01:19:31.390 align:middle line:90% PROFESSOR: Yes. 01:19:31.390 --> 01:19:33.110 align:middle line:90% Yeah, let's discuss that. 01:19:33.110 --> 01:19:35.980 align:middle line:90% 01:19:35.980 --> 01:19:39.680 align:middle line:84% This is just a way to identify the inequalities. 01:19:39.680 --> 01:19:42.640 align:middle line:84% Because I'm interested, also, in, When 01:19:42.640 --> 01:19:47.870 align:middle line:84% are the strict equalities valid in these various situations? 01:19:47.870 --> 01:19:50.710 align:middle line:90% 01:19:50.710 --> 01:19:55.300 align:middle line:84% And so I kept trace of where they came from. 01:19:55.300 --> 01:19:58.900 align:middle line:90% For example, 1A comes from here. 01:19:58.900 --> 01:20:04.290 align:middle line:84% If system A undergoes a reversible process, 01:20:04.290 --> 01:20:10.114 align:middle line:84% so that there is no entropy produced in A, 01:20:10.114 --> 01:20:15.245 align:middle line:90% this becomes a strict equality. 01:20:15.245 --> 01:20:21.560 align:middle line:84% And so here's 1A, so this becomes a strict equality when 01:20:21.560 --> 01:20:27.840 align:middle line:84% this is 0; that also for system B. 01:20:27.840 --> 01:20:35.640 align:middle line:84% Then we have 3A-- and I'm sorry, 2A is here. 01:20:35.640 --> 01:20:39.570 align:middle line:84% When is the strict equality valid? 01:20:39.570 --> 01:20:45.990 align:middle line:84% OK, remember, there is where we did the Taylor series expansion. 01:20:45.990 --> 01:20:49.410 align:middle line:90% 01:20:49.410 --> 01:20:53.980 align:middle line:84% So if we remain on the curve for the stable equilibrium state, 01:20:53.980 --> 01:20:59.850 align:middle line:84% so if state 2 of system A is stable equilibrium, 01:20:59.850 --> 01:21:04.890 align:middle line:84% we have strict equality because we are in the maximum entropy 01:21:04.890 --> 01:21:06.420 align:middle line:90% state here. 01:21:06.420 --> 01:21:11.850 align:middle line:84% If you, instead, go somewhere inside, 01:21:11.850 --> 01:21:15.960 align:middle line:84% then you have an inequality, a strict inequality. 01:21:15.960 --> 01:21:17.580 align:middle line:84% So in other words, that inequality 01:21:17.580 --> 01:21:20.200 align:middle line:84% measures how distant you are from equilibrium. 01:21:20.200 --> 01:21:23.560 align:middle line:90% 01:21:23.560 --> 01:21:29.470 align:middle line:84% So that means that if system A ends up in a stable equilibrium 01:21:29.470 --> 01:21:36.340 align:middle line:84% state, you have the equality here in 2A, idem for 2B. 01:21:36.340 --> 01:21:39.526 align:middle line:90% 01:21:39.526 --> 01:21:49.650 align:middle line:84% 3A and 3B are, when do I have equality there? 01:21:49.650 --> 01:21:50.840 align:middle line:90% We said it. 01:21:50.840 --> 01:21:53.120 align:middle line:90% If we have a reservoir. 01:21:53.120 --> 01:21:57.470 align:middle line:84% If we have a reservoir, the second derivatives are 0. 01:21:57.470 --> 01:22:00.230 align:middle line:90% And therefore, this is equal. 01:22:00.230 --> 01:22:11.000 align:middle line:84% So in order for dSA to be strictly equal to dEA over T1, 01:22:11.000 --> 01:22:16.470 align:middle line:84% I need both these inequalities to be strict equality. 01:22:16.470 --> 01:22:19.860 align:middle line:84% So I need system A to end in a stable equilibrium state 01:22:19.860 --> 01:22:30.620 align:middle line:84% and system A to be a reservoir, idem for B. 01:22:30.620 --> 01:22:37.210 align:middle line:84% Now, it's clear, also, that since these two extreme values 01:22:37.210 --> 01:22:43.380 align:middle line:84% are different, if I started from different temperatures, 01:22:43.380 --> 01:22:52.180 align:middle line:84% it's impossible that all the strict equalities apply at once. 01:22:52.180 --> 01:22:56.190 align:middle line:84% Otherwise, I would have a situation in which one number is 01:22:56.190 --> 01:22:57.470 align:middle line:90% equal to a different number. 01:22:57.470 --> 01:23:03.040 align:middle line:90% 01:23:03.040 --> 01:23:09.040 align:middle line:84% But for example, suppose system A is a reservoir. 01:23:09.040 --> 01:23:13.190 align:middle line:84% And suppose it does end in a stable equilibrium state. 01:23:13.190 --> 01:23:15.740 align:middle line:84% And suppose that there is no entropy produced 01:23:15.740 --> 01:23:19.730 align:middle line:84% by irreversibility within A. That 01:23:19.730 --> 01:23:23.510 align:middle line:84% is enough for this strict equality to apply 01:23:23.510 --> 01:23:25.510 align:middle line:84% and that strict equality to apply. 01:23:25.510 --> 01:23:32.600 align:middle line:84% So that means that the amount of entropy that is extracted from A 01:23:32.600 --> 01:23:36.260 align:middle line:84% is equal to the amount of energy extracted divided 01:23:36.260 --> 01:23:43.140 align:middle line:90% by the temperature of A. 01:23:43.140 --> 01:23:50.040 align:middle line:84% And we are converging to the idea of a heat interaction. 01:23:50.040 --> 01:23:53.040 align:middle line:84% Because that will be a heat interaction. 01:23:53.040 --> 01:23:58.590 align:middle line:84% When the ratio of the energy to the entropy exchange 01:23:58.590 --> 01:24:04.110 align:middle line:84% is equal to the temperature of the system 01:24:04.110 --> 01:24:07.030 align:middle line:90% that gives the energy. 01:24:07.030 --> 01:24:12.340 align:middle line:90% 01:24:12.340 --> 01:24:19.000 align:middle line:84% So if A is a reservoir and ends up in an equilibrium state, 01:24:19.000 --> 01:24:24.600 align:middle line:84% then either this one or the other one 01:24:24.600 --> 01:24:27.190 align:middle line:90% must not be strictly equal. 01:24:27.190 --> 01:24:28.510 align:middle line:90% It cannot be strict equality. 01:24:28.510 --> 01:24:35.670 align:middle line:84% So if system B, if you insist that system B also 01:24:35.670 --> 01:24:38.440 align:middle line:84% ends in a stable equilibrium state, 01:24:38.440 --> 01:24:42.120 align:middle line:84% and maybe it's also a reservoir, so now we are having 01:24:42.120 --> 01:24:48.120 align:middle line:90% two reservoirs, it means-- 01:24:48.120 --> 01:24:49.830 align:middle line:90% so this one is equal. 01:24:49.830 --> 01:24:53.190 align:middle line:84% So it means that this one cannot be equal. 01:24:53.190 --> 01:24:55.980 align:middle line:84% And therefore, it means that you need 01:24:55.980 --> 01:25:02.760 align:middle line:84% to generate some entropy by irreversibility in system B, 01:25:02.760 --> 01:25:15.105 align:middle line:90% which we can picture this way. 01:25:15.105 --> 01:25:18.754 align:middle line:90% 01:25:18.754 --> 01:25:21.460 align:middle line:90% Oh, no, sorry, reservoirs. 01:25:21.460 --> 01:25:24.690 align:middle line:90% 01:25:24.690 --> 01:25:38.950 align:middle line:84% We'll do it here, system A, system B. So just say A and B. 01:25:38.950 --> 01:25:43.110 align:middle line:84% And you know what we mean, energy versus entropy. 01:25:43.110 --> 01:25:52.750 align:middle line:84% So if I extract energy from A, it 01:25:52.750 --> 01:26:01.110 align:middle line:84% ends up in a stable equilibrium state and no entropy production 01:26:01.110 --> 01:26:05.020 align:middle line:90% by irreversibility here. 01:26:05.020 --> 01:26:08.655 align:middle line:84% Now, system B must take that same energy. 01:26:08.655 --> 01:26:11.990 align:middle line:90% 01:26:11.990 --> 01:26:15.625 align:middle line:84% So somehow, I should take a segment of this match. 01:26:15.625 --> 01:26:19.360 align:middle line:90% 01:26:19.360 --> 01:26:26.540 align:middle line:84% It receives an entropy, which is equal to this. 01:26:26.540 --> 01:26:34.070 align:middle line:90% So it goes here. 01:26:34.070 --> 01:26:37.210 align:middle line:84% So the effect of the interaction with A 01:26:37.210 --> 01:26:44.170 align:middle line:84% is to transfer this much energy and this much entropy to B. B 01:26:44.170 --> 01:26:47.290 align:middle line:84% ends up in a non-equilibrium state. 01:26:47.290 --> 01:26:51.220 align:middle line:84% And if you wanted to go to an equilibrium state, 01:26:51.220 --> 01:26:55.412 align:middle line:84% it needs to generate this entropy by irreversibility. 01:26:55.412 --> 01:27:06.280 align:middle line:90% 01:27:06.280 --> 01:27:10.130 align:middle line:84% So this is a complicated picture. 01:27:10.130 --> 01:27:15.490 align:middle line:84% It's another couple of days of playing with PowerPoint. 01:27:15.490 --> 01:27:29.660 align:middle line:84% But it does provide a geometrical explanation 01:27:29.660 --> 01:27:35.880 align:middle line:84% of this train of inequalities, essentially 01:27:35.880 --> 01:27:44.420 align:middle line:84% showing that the entropy change, or actually the entropy 01:27:44.420 --> 01:27:49.680 align:middle line:84% transferred is between a minimum and a maximum value that 01:27:49.680 --> 01:27:55.260 align:middle line:84% is related to the slopes, the local slopes 01:27:55.260 --> 01:28:02.890 align:middle line:84% of the stable equilibrium states of the two systems A and B. 01:28:02.890 --> 01:28:06.810 align:middle line:84% And I think there is, also, an even more complicated one. 01:28:06.810 --> 01:28:12.030 align:middle line:90% 01:28:12.030 --> 01:28:21.760 align:middle line:90% You see, Clausius's statement. 01:28:21.760 --> 01:28:28.305 align:middle line:84% We proved it for infinitesimal amounts of energy transferred. 01:28:28.305 --> 01:28:31.190 align:middle line:90% 01:28:31.190 --> 01:28:34.130 align:middle line:84% But what about you want to extract 01:28:34.130 --> 01:28:35.965 align:middle line:90% a finite amount of energy? 01:28:35.965 --> 01:28:41.150 align:middle line:90% 01:28:41.150 --> 01:28:54.430 align:middle line:84% Well, finite amount of energy, again, we can reason this way. 01:28:54.430 --> 01:28:58.660 align:middle line:84% I'm starting from system A in state A1, 01:28:58.660 --> 01:29:02.055 align:middle line:84% and I want to extract this much energy. 01:29:02.055 --> 01:29:05.180 align:middle line:90% 01:29:05.180 --> 01:29:09.395 align:middle line:84% So as I said, it could go in any of these states. 01:29:09.395 --> 01:29:12.140 align:middle line:90% 01:29:12.140 --> 01:29:14.905 align:middle line:90% System B starts from here. 01:29:14.905 --> 01:29:18.990 align:middle line:90% 01:29:18.990 --> 01:29:20.510 align:middle line:90% It receives that energy. 01:29:20.510 --> 01:29:23.890 align:middle line:90% 01:29:23.890 --> 01:29:27.255 align:middle line:84% And in principle, it can go in any of these states. 01:29:27.255 --> 01:29:32.330 align:middle line:90% 01:29:32.330 --> 01:29:36.200 align:middle line:84% So that means that if you give to B this much energy, 01:29:36.200 --> 01:29:41.850 align:middle line:84% it is able to accommodate only this much entropy. 01:29:41.850 --> 01:29:44.510 align:middle line:84% It cannot go off to the right, because there are no states 01:29:44.510 --> 01:29:45.010 align:middle line:90% there. 01:29:45.010 --> 01:29:50.460 align:middle line:84% So this is the maximum amount of entropy that we can accommodate. 01:29:50.460 --> 01:29:56.550 align:middle line:84% And therefore, that's the maximum amount of entropy 01:29:56.550 --> 01:29:59.970 align:middle line:90% that A can decrease. 01:29:59.970 --> 01:30:08.490 align:middle line:84% So that means that somehow, A can only go down to here. 01:30:08.490 --> 01:30:13.300 align:middle line:84% And that is the meaning of these constructions. 01:30:13.300 --> 01:30:17.970 align:middle line:84% So there is a limited range of possible values. 01:30:17.970 --> 01:30:24.270 align:middle line:84% And here is the analytic expression 01:30:24.270 --> 01:30:26.620 align:middle line:90% for the Clausius inequality. 01:30:26.620 --> 01:30:31.450 align:middle line:84% Let's call it Clausius inequality for this. 01:30:31.450 --> 01:30:37.110 align:middle line:90% 01:30:37.110 --> 01:30:39.660 align:middle line:90% OK. 01:30:39.660 --> 01:30:40.700 align:middle line:90% We have five minutes. 01:30:40.700 --> 01:30:43.840 align:middle line:90% 01:30:43.840 --> 01:30:45.080 align:middle line:90% I think it's enough. 01:30:45.080 --> 01:30:46.470 align:middle line:90% Oops. 01:30:46.470 --> 01:30:48.820 align:middle line:90% No, next, yeah, sorry. 01:30:48.820 --> 01:30:53.300 align:middle line:84% I decided that I couldn't get this far. 01:30:53.300 --> 01:30:55.180 align:middle line:84% So I don't have viewgraphs anymore. 01:30:55.180 --> 01:30:58.730 align:middle line:84% So next time, we'll do work and heat. 01:30:58.730 --> 01:31:02.470 align:middle line:90% But let me anticipate the idea. 01:31:02.470 --> 01:31:09.980 align:middle line:84% First of all, we define work as an interaction in which 01:31:09.980 --> 01:31:12.355 align:middle line:90% there is no entropy transfer. 01:31:12.355 --> 01:31:15.700 align:middle line:90% 01:31:15.700 --> 01:31:21.640 align:middle line:84% See that we shifted the order of typical presentation, 01:31:21.640 --> 01:31:22.700 align:middle line:90% usually we define-- 01:31:22.700 --> 01:31:27.910 align:middle line:90% 01:31:27.910 --> 01:31:29.895 align:middle line:84% so here, we are using entropy to define work. 01:31:29.895 --> 01:31:34.650 align:middle line:90% 01:31:34.650 --> 01:31:41.140 align:middle line:84% And the other, anything that transfers entropy 01:31:41.140 --> 01:31:43.040 align:middle line:90% is a non-work interaction. 01:31:43.040 --> 01:31:45.570 align:middle line:90% 01:31:45.570 --> 01:31:52.120 align:middle line:84% And if there is an entropy transfer, 01:31:52.120 --> 01:31:55.800 align:middle line:84% it's not necessarily always heat. 01:31:55.800 --> 01:32:01.120 align:middle line:84% We're going to call heat a very special limiting situation. 01:32:01.120 --> 01:32:03.190 align:middle line:84% Essentially, that limiting situation 01:32:03.190 --> 01:32:07.990 align:middle line:84% that is captured by this train of inequalities 01:32:07.990 --> 01:32:16.190 align:middle line:84% here, when you squeeze the interval in which 01:32:16.190 --> 01:32:21.440 align:middle line:84% the entropy transfer values are confined, 01:32:21.440 --> 01:32:23.900 align:middle line:84% and you squeeze that interval, it 01:32:23.900 --> 01:32:27.848 align:middle line:84% becomes just a single possibility. 01:32:27.848 --> 01:32:29.390 align:middle line:84% Now, how do you squeeze that interval 01:32:29.390 --> 01:32:35.410 align:middle line:84% is by having the two systems begin at temperatures 01:32:35.410 --> 01:32:36.970 align:middle line:90% close and close. 01:32:36.970 --> 01:32:39.880 align:middle line:84% Or actually, in the limit in which the two systems have 01:32:39.880 --> 01:32:46.430 align:middle line:84% the same temperature, the entropy transfer 01:32:46.430 --> 01:32:48.840 align:middle line:90% has a precise value. 01:32:48.840 --> 01:32:50.730 align:middle line:84% It doesn't have any more a range, 01:32:50.730 --> 01:32:53.180 align:middle line:84% just a precise single value, which 01:32:53.180 --> 01:32:55.700 align:middle line:84% is equal to the energy transferred 01:32:55.700 --> 01:32:59.410 align:middle line:84% divided by the temperature of the two systems. 01:32:59.410 --> 01:33:01.220 align:middle line:90% We call it heat. 01:33:01.220 --> 01:33:05.000 align:middle line:84% And so energy, we will call it Q, 01:33:05.000 --> 01:33:09.500 align:middle line:84% so Q over T. So the entropy transfer 01:33:09.500 --> 01:33:13.460 align:middle line:84% will be the heat over the temperature, which 01:33:13.460 --> 01:33:20.810 align:middle line:84% I'm sure is the way you have seen probably 01:33:20.810 --> 01:33:23.500 align:middle line:84% the definition of entropy in previous courses. 01:33:23.500 --> 01:33:28.640 align:middle line:90% 01:33:28.640 --> 01:33:34.050 align:middle line:84% So, yeah, I think we can stop here for today. 01:33:34.050 --> 01:33:38.000 align:middle line:90%