 OK. OK, now it's better. OK, so sorry for those of you who, I think, in the master in Paris have done this exercise already. But OK, if you're bored, you can ask a lot of questions. No, no, no, no, shall we give it to them? Oh, I took one because they're numbered. No, maybe I took it. Here you are. Is this number four? Yeah. OK, OK. So our only handles up to now are the frequencies of the oscillators, which I am free to choose. There will be many. So this is something that I cannot choose because we want a bath, which is, by definition, something that it's infinite, potentially. And so we have the frequencies and the strength of the interactions. I am free to choose them. And the result does and does not depend, you will see. But this is, for the moment, the freedom I have. Then I will take another freedom that you will see. OK? So now it's very easy what we have to do. We have to solve the bath so that we can get rid of it. So let's get rid of the bath. So OK, the equations for the system are Newton's equations. I can even skip. Sorry, sorry, don't copy anything. This time I'm, oh, yeah, yeah. Let's skip one step for brevity. So acceleration equals force. And then we have the equations for the bath. OK, so the trick is the following. This I chose one degree of freedom because, as I told you, conceptually it's the same to do one or more. So my system, what I call my system, could have five degrees of freedom, whatever. But it's not a bath, it's a system. And you, I cannot solve this equation for sure because the potential could be anything. So this is not something I can solve analytically. The trick is that the equation for the bath, you see it's linear, except that it has this thing which plays the role of a field. So you can think of this as a harmonic oscillator. They're all decoupled. And this is a term that depends on time. It comes from the other equation, but forget it for a second. And this plays a role of a field that is entering this equation. So the equation for each oscillator of the bath is simply the equation of a forced oscillator. And this we know how to solve. And the solution is this thing I'm going to write, which is really ugly, but it's an oscillator that has a past in which it has been forced. This is solvable with the initial condition of which, more in a second, the moment at time zero. And this term that comes from precisely, this is just the generic solution which you can look in Wikipedia, of a harmonic oscillator that is being forced. And this is the forcing term. Of course, it is forced by my system. And OK. And now the strategy is very simple. We have this guy now. So now we have solved for the bath. We can put it in here. The equation we didn't know how to solve. And we will end up with a gigantic monstrous equation for q, which could have been. And this is going to be our effective equation, having gotten rid of the bath. OK. So if now I get this thing and I plug it in here, we are done. And the equation we get is the following. Oh, I'm sorry. I should have put a mass here. Forgot it. Sorry. Initial equation. Then there is a term, which I have to tell you what it is. And then there is another term, which I also have to tell you what it is. This is a memory of what's going on. And we have another term. The other definition that we're missing. OK, so now copy this and then we will stop and think for a second. And what I did is completely elementary. You can follow those steps. I'm going to give you a reference. Sorry, what's wrong? Is it q dot inside the integral there, or q? Because upstairs is? It's q dot. There has been an integration by parts. It's just for convenience. We could have integrated the other way around, and this would be. But it's nicer to write it this way. So yes, this is a good question. Here there would be q, because it is q somewhere. But because it is in an integral, we integrate ones by part. Because we prefer the derivative here for reasons that will be clear in a second. No. No, this is the one that belongs to the other sum. So I'm going to make it a bit shorter. So what is qA of 0? And the parenthesis there, maybe it's just q of 0. Thank you. Mahesh, you have to keep up to his standards when he goes away now. If there are doubts, please ask me. We cannot let Matteo do all the asking. He's already graduated. OK, so now let us think five minutes. This is, I think, a lovely exercise. So let's go back in time. You remember Brownian motion? So Mr. Brown discovered that particles of pollen sort of jitter in the water. And then at the beginning, they thought that perhaps they were alive. Then they decided that no. And then there was this jittering motion. And then Einstein understood what was going on. I mean, the essence of the problem. And it's the following. You look at a little piece of pollen. And you expect it to have an energy that is compatible, a kinetic energy, that is compatible with the temperature at which the water is. Because at his time, atoms were not so consolidated, as you would think. Even at the beginning of the 20th century, the idea of atoms was still. But thermodynamics, they did know. And they expected this to be in equilibrium with a bath. So have the energy compatible because it's been sitting there for hours. OK, so this is doing a jittery motion. And of course, Einstein already, he believed in atoms and importance of atoms. And so he understood immediately, I don't know if he was the first, that the jittery motion comes from, you are being bombarded. You pollen are being bombarded all the time by molecules of the water, which are kicking you in directions. Because the molecules have a finite size, they are not infinitely small, you are pushed by a discrete number of quantized objects that have a given mass. So even though on average, you are being kicked in all directions, because the molecules are finite, notice that this is already telling you something about the size of the molecules. Because they are finite, you are being pushed by relatively big guys that are pushing you. And this is why it's visible. And this already tells you this parenthetic remark that the size of the pollen is not so enormously larger than the size of a molecule. Yes, it is 10 to the something. But it is not 10 to the 24, the size of a single molecule. And then you have two things. But because you are being kicked in this way, why don't you simply gain energy? No, because if I kick something all the time, it should heat up, get more energy on average. And the reason why, and this is what Einstein understood, this is the heart of the matter, is that the water does not only give you the kicks, but it also gives you some friction, which stops you. And now the point that he understood, and this is crucial, is that you see that friction and noise, noise is this kicking, have to be in an appropriate balance so that you have the right energy corresponding to the temperature you have. So miracle, miracle, he discovered, or he realized, that friction and noise are not independent. They are tied to you by the necessity of being in thermal equilibrium. This is an extremely deep insight. I mean, it's one of the big things he did. In our modest little measure, we are discovering a bit the same thing. This, which I will call noise for the moment, it is not random. But it is an additive term that is shaking you. And, well, it is a function, of course, of the initial condition of the oscillators, which I will have to this one. On the other hand, this term, you have to interpret it as a term of friction. What is friction? Friction is something that is proportional to the velocity. When you want to move fast, 0. So this is the initial condition of the, again, sorry. And also the other one? Yes, this is very strange. The nodes have a mistake. It's right, you're right, sorry. This is the initial conditions of that. You're still losing towards Matteo, sorry. It's amazing the printed nodes were wrong. OK, yes, of course. Thank you. So why do I call this friction? Friction is, in general, something that depends on velocity. It doesn't want you to move. So if I'm standing still, I don't feel friction. I mean, I don't feel it as a force. And then when I try to move, the friction stops me. So this is a term with friction. But it's a bizarre friction because it's a friction that has memory of the past. So it's a friction with memory. It's a friction that doesn't depend only on the velocity you have now, but it depends on the velocity you had in the past. This is generic about friction. Now we will say a lot about what this function gamma looks like, so don't worry. But OK, so this is noise, and this is memory. Sorry, this is a friction term. So we have the two elements here. One is like, in this case, it's not molecules. It's harmonic oscillators that are bothering you. And the friction is stopping you. And this balance between bothering you and stopping you has to exist and give you something to do with, we will see, the temperature of the bath because this is what we want. But we are not there yet. So now comes the up to now everything is still absolutely Hamilton-Newton's equation. If you give me the initial condition of the oscillator and the initial condition of my particle in the potential, I can follow the thing, no problem. And it's completely deterministic. And it's only that I partially solved it, but that's only the only thing I did. So the memory that you speak of, in the example of Brownian motion, that would be of very short times, case of order of 10 to the minus 13 or minus 14. Absolutely. The memory term is the memory that, of course, if a molecule is stopping me now, so I am touching a molecule, and then comes another molecule which touches me in another direction, there is some interval between them because they occupy space. This is how he estimated the time to replace one molecule by the other. In the case of a liquid, it takes you some time. And this is a characteristic time of your bath. We will have here a complete freedom to choose the time scales you will see. So the principle I am telling you of balance between noise and friction is called the fluctuation dissipation theorem. And we will say a lot about this because it's very important. So now we have to say what the bath is like. Remember that we already chose arbitrarily this and this. And we will play with the choices. This was up to now our choice. But I didn't tell you how hot the bath was. I have to tell you what is the temperature of the bath. So I have to give you the energy of the bath, for example. Yes, OK. So now we make a crucial step is that we are going to assume, and Matteo yesterday wasn't happy with it, but even so we will go on, that we are going to choose that the pi a's and the x a's are in equilibrium, microcanonical equilibrium at temperature T. What Matteo didn't like is that ideally I should do microcanonical because I am now including the canonical information which I ask you to believe. But OK, it's really only simplifies life a bit. So I want to ask them to be some energy compatible with equilibrium at temperature T. I'm telling you how hot my oscillators are. This is very important. But then I'm going to do a bit more. And here comes the enormous step, which is very easy, but it is very big also, which is to say, yes, I'm going to ask this, but I'm going to ask them to be taken at random. So they will be taken at random with probability e to the minus beta h of bath divided by the normalization. But randomly, the energy. So I will throw a problem with a lot of these variables now are random. Random and with the typical energy in equilibrium, it's very easy because it's a harmonic oscillator. So it's just the equipartition of a harmonic oscillator. So what is the energy of a harmonic oscillator at temperature T? A half kT for the potential energy, a half kT for the kinetic and so on. So we are going to replace the initial values, but we are going to consider them a random variable. So this is the handle number 2, 3 we have of the problem. Handle number 1 was what the oscillator's frequencies are. This will survive to the end of our problem. Handle number 2 is the interactions between the oscillators of my system, handle number 2. And now the temperature of the initial conditions of the bath. The bath is so big that it doesn't care very much about you. So these we are going to choose so that on average, they have the right energy. And what is the consequence of this? That we are choosing a temperature. This is where temperature enters in our problem, in saying what is the distribution of the oscillators at the beginning of time. There is another point that is very important that now we have at this moment introduced randomness in the initial condition of the oscillators. The future life of their oscillators has entered in the initial condition and the initial condition is random. So what was a deterministic problem now is becoming a stochastic problem. And this has dramatic consequences because it seems like an innocent step and you will see that it takes one second to do it. But it has completely dramatic consequences because, for example, remember I told you that the volume of a drop of ink, let's say, initial conditions in phase space, preserves its volume in Hamilton's equation. Now the equations we will obtain for the system in contact with this bath do not do this and tend to equilibrium and everything is nice and easy. So we are going through all this detour and at the end of the day, at the end of today, you will see that we will end up in a stochastic equation which, in spite of appearances, is infinitely easier than doing Hamilton. And there is an entire branch of physics that is enormous that is dedicated to this and then they call it stochastic thermodynamics. You will see that although mathematically you have to deal with a random noise, in practice, all issues concerning ergodicity become trivial. Do you have something? Yeah, so no, actually, from what I see, so the dynamics is still deterministic here, in the sense that, at this stage, at least, you are just saying the initial conditions are random. But after that, it's like you are saying that I have my system and butter in a bigger butter up to time t. No, no, no, no, no. And then after that I take out and there is isolated and they go on deterministically. OK, yes. But not the system and bath. I have only the bath. I prepare my bath with a good temperature. God knows how, but I couldn't do it micro-canonically. We will be the same. But the calculation here will be a bit nastier. And then it's deterministic. Only that, as you will see, we will study trajectories on average with respect to the initial conditions. And that's what stochasticity enters. OK? It's an ensemble of possible things. Yes, yes, yes. But this is where irreversibility comes in. OK, so now, we have to compute this. And with this, we really and finally get rid of the bath. Get rid of the oscillators. This is a function. It's OK. I don't have much to say. Now let me say something about the noise. You see that it has signs and cosines. So on average, you believe me, it's zero. So it's something that is zero. And now I'm telling you that it's random via these numbers. OK? So now, what is it that I have to ask? Well, I have to ask for the correlation. So I have to compute, you know, when a variable. So now, because this is going to be a sum of many things, what happens with a sum of many things, it's going to be a Gaussian, but then correlate it in time. So now, we have to take this expression, multiply two of these, yes, of t and of t prime. And then remember that the average here is an average over initial condition. This is where randomness came in. OK? So I think I'm tempted to leave this as an exercise for you. You take this thing at t. You take this thing at t prime. And then the only thing you have to remember is that every time that pA squared appears, you are considering that this is the only random thing, pA of 0, sorry. And every time xA of 0 appears, this is the average over initial conditions. And when pA times of 0 times xA of 0 appears, this is 0. This is a harmonic oscillator. It's this way. And in a harmonic oscillator in equilibrium, this is a half kT. So you make the product in the product terms that have pA squared or have xA squared have to be replaced by the temperature, taking into account the temperature with the Boltzmann constant, of course. But as I told you, it's 1. If there is a term like this, pA, pB, or pA, sorry, xA, xB, et cetera, all these are 0 on average because the oscillators are completely independent, these are averages, remember, over initial conditions, and so on and so forth. Is it clear? I just don't want to fill the blackboard, but I just have to write this at T, write at T prime, and then collect all the terms that are pairs and replace the expectation values by the initial thing. And the consequence of this is that what I get, miracle, miracle, and it's not a very hard thing to do. This is not factorial, it's just an admiration sign. And of course, the average of eta is 0 at any time. And now we have finally gotten rid completely of the bath, and now we are left with, OK, so we manage. Yes, please. So the temperature is fixed, so the condition is isothermal, so can we make it to other variables like changing T to some pressure or volume? Yes, yes, yes, yes, yes. Yes, so, yes, for example, so what he is asking is can I have a bath that instead of keeping temperature, keeps pressure, OK? So one way to do it and in simulations, you take your system and then you can put a piston that is in contact with the harmonic oscillator and is pushing your system. So it fluctuates, so it's not the energy that fluctuates, it is the depth of the piston, and it has a, you can do it in the computer. In fact, it could be a nice exercise for one of the groups we will make. And this keeps, you can do it at constant energy, so it's conserved, but it's interacting with the piston, and you might make it in such a way that both pressure and temperature are maintained. You can do it with any variable. In practice, one way to do it, which is super nice, is imagine you have a gas of spheres. Let's say hard spheres, no, boing, boing, boing, boing. And you want to fix the pressure. So what you can do is you can make the radius of the sphere a dynamic variable, which is a bit like using a piston because when they inflate, it's as if they were more dense. If you do that and you make a kind of spring that doesn't want the radius to change, you can fix the pressure in your Monte Carlo procedure. So yes, you can do whatever. You can fix magnetization with a little bit of imagination. You can do whatever. Thank you. The idea of being always the same. OK, so here appears the thing that is so important. Remember what the thing I said about my pollen particle and the realization that the intensity of the kicks and the intensity of the friction have to be matched. And the matching, the dictionary matching, is precisely the temperature. So if you are at a big temperature, your noise is very strong because your molecules where you're leaving are kicking you like mad because they have high velocities, even if your friction is relatively smaller. So when you, as a pollen particle, are swimming in this water, you receive a lot of energy from the bath and relatively are taken away by friction and more modest quantity, and you are hotter yourself, which is good because this guarantees that you have the right energy in equilibrium with the bath. So fluctuation dissipation, which is the fluctuation dissipation, this one is called of the first kind. When we do the one of the second kind, you will see that it's of the second kind. Flactuation dissipation is precisely the statement that the properties of the noise, the variance, and the properties of the friction, the memory kernel of the friction, are exactly equal except for a temperature. They are not similar. They are exactly equal. You underlined the same function. Properties of the noise, gamma. What did I underline? Here. Yes. Yes. So OK. So now, forget the oscillators forever, almost forever. But if for some reason, the gamma and the gamma are not the same function, you could imagine that I write this equation and I choose a random Gaussian noise with a gamma and a different gamma here. I can do it mathematically. Then it does not correspond to an equilibrium bath. And you are in trouble. You're in trouble. What does it mean? No, you have a dynamics that is legitimate. But don't expect the system to reach any micro canonical or canonical or anything. You don't know what it will reach. So if the gamma here and the gamma here do not match. They are not proportional. Then you have a system that is inherently an unequilibrium system. Inherently means that if I started, even an equilibrium system, when I started, I put water here in this room, which is at, I don't know, 30 degrees. What was this room? 25 degrees. You put it a glass of water, which is at 5 degrees. It will take some time to equilibrate, after which it will equilibrate. A system that is in contact with a decent bath with fluctuation dissipation, meaning that the gamma here and the gamma here match, will do that. It will take some time, but it will equilibrate. It will become an equilibrium system. And from there onwards, I can apply canonical measure for everything. If instead this gamma and that gamma do not match, they will reach a stationary measure because you have the mechanism of pumping heat in and pumping heat out by noise and friction. But the measure it will reach is not the canonical measure because the bath in itself was not an equilibrium bath. So what measure does it reach? You don't know. You have to solve the entire system. There is no quick rule to discover that. Once the bath is a bad bath, then there is no general rule for anything. You have to solve the dynamics and see what you get. You cannot get rid of time, for example. So strictly speaking, essentially, so this noise here remains a deterministic function, the eta. If somebody gives you the initial positions of yours. The argument is that essentially you compute the average of eta, eta, eta of the d prime. And then you can say, OK, so this given the initial condition of Gaussian, this is a linear combination. So the noise will be Gaussian throughout. Exactly. And then you say, I can substitute this eta with, say, white noise with the same correlation. We will come to the white noise in a second. OK. So this is as far as we get. I'm so sorry. You should also have that this gamma is finite, right? So this imposes when, I mean, so you should have some limit when the number of oscillators goes to infinity to have this type of molecular chaos type of. Sure, your noise is a sum of oscillations. If you want the noise to look like a noise, you need many oscillators. We have many oscillators. Now, what happens if you're so cruel? OK, now we're coming to this now. All right, so now that we are here, OK, we're here. And OK, so now let me cancel this and say, OK, there's only one function gamma. And now you have to choose it. Yes? When the body is not good, when you say that you cannot assume the ensemble of the system anymore, it means that you cannot put the canonical or micro canonical here. And how will you introduce? At the beginning, yes. How will you introduce the randomness? You could, for example, because you decided that, for example, the initial condition is non-equilibrium one. For example, the initial condition, you can have half of the bath at temperature T1 and half of the bath at temperature T2. And with different omegas, we will play with this. So the high frequencies of this thing are hotter than the low frequencies. We will do this. And then your system doesn't know what temperature because there is no single temperature. We will do this, because as you will see when you do models of active matter, which is very important subject now, precisely it's as if you have a bath that is not at a temperature, but it's giving you energy, pumping in and then pumping out, but not with this precise, delicate equilibrium, which is called detail balance, by the way. Or while depending on the context, we will come back to it. But fluctuation, dissipation, theorem. And you will see that it bleeds you to detail balance. So you will see how this comes about, yeah? I know we cannot know that as a priori. We have to just try what works. There is no once your bath is not a decent equilibrium bath, you're on your own. You have to solve the dynamics. And I don't know of any problem that is not purely harmonic. Because when it's purely harmonic, then you can do everything. But a problem that is nonlinear and at the same time has a bath that is not a good bath, let's say. There is no like a Boltzmann weight, a bit modified, that helps your life. There's nothing like that. There is no generic rule to calculate what the measure will be. You will get a measure that will replace Boltzmann measure, but there is no nothing. You have to solve your entire problem. When it happens that you can, it's because it can be mapped. It's a hidden equilibrium one. Because you can remap it into equilibrium. But if not, there is nothing. So for example, when you have active matter and you ask what is the probability distribution of velocities, these are motorized particles. You know that, have some little motor that moves them and they lose energy when they crash. So you ask, well, what is the distribution of velocity? Is it the Maxwell distribution? You don't know. Simply you don't know. The only thing you have to do is do the simulation, take the histogram and record it. There is nothing that can be done. OK. Yes? What's the distribution function you are taking here to calculate this averages? Sorry? What's the probability distribution? The probability here is the one that comes from having, so the averages here are the averages over the initial velocities. No, I'm asking what's the kind of distribution you are taking, Gaussian or probability? Oh, the distribution of a task? Yeah. What is it? Yeah. So it's a distribution when you have many, many oscillators. It becomes a Gaussian with correlated in time in this way. With this one, it's a Gaussian, so with that, you completely have specified it. OK. So now we can choose whatever we want from gamma. You see playing around with your oscillators and their amplitudes. Remember that these were the quantities I now forgot. Now all this in blue, what I told you was one of my handles. And handle number one, handle number two, and handle number three was the temperature. OK. So these I can choose as I want. And so these cosines I can choose as I want. But then you see that this is the Fourier transform, more or less, of this. So I can choose the Fourier transform the way I like. So basically, basically, I can choose this and I have a lot of freedom to choose it. But then this gamma, I don't have any freedom anymore if I want my system to correspond to an equilibrium path. But whatever the gamma I choose, with very few conditions of positivity, because you see this is a correlation. So the Fourier transform has to be positive definite. OK. But this thing, I am very free. So because I am very free, let me erase it and tell you what this means. So OK, this one is obvious, so I erase it. So this is what is called a generalized Langevin equation. I think that Langevin himself wrote this one. Usually this is called generalized, because now you're going to see the one that is not generalized. So just to be clear, this is gamma. And this memory kernel affects your past and multiplies your velocity in the immediate past. And you are free to choose, provided, of course, the fluctuation dissipation relation holds. And so now what people do is take the easiest of the easiest thing, which is the following. It's if we take for gamma a very peaked function, a delta. So this would be a delta, delta of t minus t prime, to be proper. And here it's a half of the delta, because you see I'm integrating up to t, so it's only the half of the delta that I'm choosing. So in this case, you end up with this lovely equation. Why lovely? Because now the memory kernel doesn't go to the past and still respects this important thing. So now this equation in that case becomes what is more often called the Langevin equation. Q dot, gamma q dot, because this is a delta, it evaluates, or half a delta in reality, it evaluates q dot just in the time in the immediate past, so proportional to some gamma. Eventually, I didn't use it before, but we could ask, we could add a force that doesn't derive from a potential if you want. The eta i's now are independent, and I am now putting it with many variables so that it's nicer. So when gamma is very peaked in the recent past, you get this equation where the noise is now white. It's only correlated at very short times. And delta ij means that different noises corresponding to different variables are completely independent. So every variable of the q's, remember that here I have only one q, here I made many, but the proof is the same. Every degree of freedom q has a private bath of its own, which is completely independent, has a different independent set of oscillators that talk to him. So this is a bit rare in the literature. You find it, but you find it a bit less. I'm sorry, there's something I don't like here. My god. There has to be a gamma here. If not, fluctuation dissipation is not satisfied. Yes, there has to be a gamma here. Because that relation between the intensity of the noise and the friction has to be respected. Because if not, it's a bad bath. So you see that, again, this t here is this t here. So now you recognize the friction of a true friction. And the gamma that appears here has to appear here, because if the friction is stronger, and I want to keep the temperature fixed, I need to make the noise stronger. So if you're, yeah. Ah, OK. You want something whose Fourier, so you have to, I erased it, but if you go to the expression I did for gamma, you see that it's a sum of cosines with coefficients. You want that to be, to give you a delta. So take Fourier transform right and left. The Fourier transform of the delta is a constant. And then the Fourier transform of the right side is the sum of cosines that you had, because it's a Fourier transform of cosines. The sum of spikes at the values, so you see my point of the cosines. And then check that your density of oscillators gives you this constant. I'm not telling you it's flat, because there is a 1 over omega squared that you have to take into account and so on. But yes, yes. There is a condition. And then there is another condition that is very famous and gives you a noise that is not white, which is called the Debye condition, which corresponds to, you are in a solid, and the oscillators are in fact the sound modes within this sample. And these are distributed in a way that is a Debye way, and this is the one you use for solid state. So there are several famous, the omic, that one is the one that is called omic, low bomb, and so on. So there are some famous things. For the white one, you can. In order to make this generalization to many particles, does we have to assume that none of them has interaction with each other? They have interaction with each other, but the baths are independent. The oscillators of each other, Q1 talks to some oscillators, Q2 talks to other oscillators. But Q1 and Q2 can talk with each other, yes. If not, this wouldn't be very useful. And if not, they wouldn't be in equilibrium. Well, in the end, they wouldn't, yeah. OK, so this is the second, as I said, most well-known Langevin equation. So gamma, if you are in air, gamma is very small. Air at 20 degrees, gamma is small, so you have little friction, but you also have little noise. If you are in water, the gamma is going to be much larger. You have much more friction, but the noise you will get will also be much larger. And if you are in honey at 20 degrees, this is going to be very big. And this is going to be the same thing, very big. And the proportionality is always t, so there are three things that are at the same temperature. So the viscosity of your system where you're moving is decided by this gamma, at least in comparison with the inertia you have. This is the inertia of your problem, the mass. So the idea is that if gamma is large and M is small, maybe you can forget the inertia. Think of, for example, a Brownian particle in water. The inertia of a particle of pollen is nothing, or the inertia of a bacterium in water is nothing. So we have the temptation of throwing this away sometimes. Now, this brings a lot of headaches, but the Langevin equation most of you will find in life is without this term. So this is general. This is medium general with inertia. And the one you find most often is this one without this term. And when you don't have this term, then you can absorb gamma into time, absorb it here in the derivative, and absorb it here in the delta. And you get a Langevin equation that is only like this. Eventually, there could be some forcing that, of course, will drive the system out of equilibrium. And this equation is called Langevin, or sometimes Moluchov. I think the correct name should be Smoluchovsky. So regarding this neglecting the mass, I mean, if I have a forcing term, there can be trouble if I neglect the mass or? Well, if your forces sometimes are very impulsive and you want to see how inertia reacts, yes, you have tried. So here we've got three levels. And so, sorry, and this has to be supplemented with gamma now went away, the same with a two, but without the gamma. So gamma has been eliminated by absorbing it in time. And there is a temperature after so much saying that it was important. OK. So, and what we will do is we will talk about an ensemble of trajectories. OK, finished copying. Does that mean in overdamp Langevin equation there is no memory? There is no memory in the sense that this one is completely Markovian, yes. Markovian means that it depends on the immediate past, this one. This one is a bit Markovian, but it has an acceleration term, which a bit remembers if you want the past because you cannot do immediate changes. And this one is the generic one where you have all the memory. Thank you. So we've gone a long way. And now we have a situation where we have this equation of motion, which would be, sorry, not this one, this equation of motion, which would be Newton or Hamilton. Then we have a random noise. And you have a memory that is deterministic, but it's very much related to the noise by way of its variance. Here we lost this one, so it's not Newton anymore. So we only have the motion due to the potential. Eventually a force that doesn't derive from a potential, which, of course, will make the system an unequilibrium system. I put it here also, and then the noise. And why is the noise random? Well, if you go back and back and back and back, you will see that it's random because the initial conditions were random. And those are of the bath. Those are the ones that decide the randomness. And when we will study the dynamics of this and of this, we will find that we are not going to do a trajectory, or sometimes we will do a trajectory. But a trajectory is for given realization of noise. But many times, we will be interested in the average over all the possible noises that are chosen with the appropriate rule. They are Gaussian noises. And we do the average over trajectory. Even in your computer, when you want to simulate a problem, you run it once with random numbers that are the ietas that are chosen intelligently. And then you make a dynamics, and you record the values your particle took at given times. And then you run it again with another seed for the random numbers. And then again, and then again, and then again. And you make a histogram. So at each time, the location of a particle is a distributed quantity. You consider the ensemble of trajectories given the ensemble of noises. And all this comes back to, if you want to go backwards, in the original problem, to the ensemble of initial conditions of the bath, which is where the randomness was born. OK? Good. I think we are almost done for today. The only thing that I want to say now is that what we will do next time is that if I have an equation like this one, or like, oops, not this one, this one, I have the possibility, for example, of studying a trajectory. This is q with a given eta. Or I can, with another eta, have another trajectory, which will be slightly different. OK? And then if I consider all the possible eta's, I will have a probability distribution for q. It could be a vector. OK? This is one description. One description is to solve the problem for a given realization of eta, even in your computer. And another possibility is to study at each time what the probability cloud of this distribution, how it is evolving. OK? So if I start here in this is a free motion, so let's say this is 0, this is 0, and I only have eta, one particle will do like this, another one will do like this, and so on and so forth. And I will have, if I consider my envelope, all these. And at a longer time, of course, I will have this. And at a shorter time, I will have this. So I can do two things. I can study one trajectory, and there's a lot of developments that you do for a single trajectory. And you can instead study how this potato of probability is expanding. So for studying this thing, you have this. But you want to ask yourself, can I have an equation for p of q directly without doing individual trajectory? And the answer is yes. And this is the Smoluchowski equation. Fokker Planck, Smoluchowski, sorry. Maybe Smoluchowski doesn't apply here. Let's leave it for Langevin. But we will find that there is an equation for p of q and t. There is an equation that tells me, given that my initial things evolve like this one or this one, there are two different equations. How do these probabilities in space evolve? And this is going to be what we're going to do tomorrow. And with this, we complete, if you want, the passage from a world where everything was deterministic to a world where we have introduced randomness. So all this subject of adding randomness is very old. But in the last 25 years, there has been an enormous revival of activity here under the name. It is thermodynamics because you have a big bath at the end of the story. So even if you have one degree of freedom, remember there is a bath involved. And there have been incredibly, for something that is so elementary and so old, a lot of developments in the last 25 years on this enormously many. We will try to see a few of these depending time permitting. But there's a lot, a lot of things one can say. I think that I will stop here. So can we derive entropy of the system by calculating this PQ, this probability of Q? Yes. Yes. Yes. We will do the exercise. So we will do P of Qt log of P of Qt, which is the sensible thing to do. And we will minus. And then this quantity, which we will call entropy by analogy with the usual entropy, this is a kind of dynamical entropy. We will see that this quantity minus the energy to get a free energy is always going down with time. Going up, sorry, the entropy is going up with time. So is this branch interesting because we are getting thermodynamics from single particle dynamics? Yes. But it's fake single particle because there is a hidden bath and there is a hidden average you took. So there is a coarse-graining involved. And I told you that always when you have second principle, there is a coarse-graining. There is a coarse-graining involved in the fact that you're not considering one trajectory, but many, meaning that you are not looking at the details of the thermal bath. You're taking an average. So yes, you recover thermodynamics. Thanks to that. Thank you. One thing that I will insist is that unless I am very much mistaken in the books of thermodynamics, sometimes it is sort of implied that the Boltzmann equation in the old books somehow proves irreversibility. And I remember the books with which I studied and Huang. And the answer is no. I mean, it doesn't. What makes irreversibility for you in a Hamiltonian system is coarse-graining. It so happens that Boltzmann's equation has coarse-graining implicit. This procedure also has coarse-graining where in the fact that I don't give you the details of the bath, I give you only an average. So it is coarse-graining that gives you irreversibility, not a miracle of Boltzmann equation. That's historically people to be faithful to Boltzmann. They over, I think, represent the value of that equation. Maybe they were over-correcting for all the attacks he received. Yes, and that poor man ended his days a few, an hour walk from here. OK. Sorry, other questions? OK, so let's thank. And so I think now there is a coffee break. And Matteo has just asked me to tell you to not take the other coffee break from, I mean, because I understood. I mean, do not act like grasshoppers, you know. Thank you. See you in half an hour. We will meet in the info lab.