 theorem, let me make some remarks about the fluctuation theorem and about looks like a lot. Okay, so one thing that is nice is that we saw that delta S total of gamma is ln of P, well, let's call it probe to keep the notation. Probability of gamma over probability of gamma reversed. Now I can define a new probability which is probability reversed of gamma is equal to probability of gamma reversed, okay. If I use this new function, okay, so if I use this new function, then I can write that thing as probe of gamma over probe reversed of gamma. There are different probabilities in a sense that they do they give different ways to the trajectories. Now if I want to write the average delta S total, that will be a sum over all gamma probability of gamma ln of the probability of gamma over probability of gamma reversed, right. And if you know Kubak library divergences, that's exactly the Kubak library divergences between probe and probe reversed, okay, which is like a very explicit relationship between the production and you know the fact that the production is related to the difference between four trajectories and reverse trajectories, okay. And the Kubak library divergences is something that shows up a lot in the information theorem. It's not really a distance because it's not symmetric. So you know, if I change, if I change these two things here of the position of these two things, I get a different thing. So it's not symmetric, but it's always larger than zero, right. It's a kind of very nice formula, okay. So you know, if I write the average introduction, I can write the average production this way, which can be very useful in different circumstances. And because you know there are lots of properties that this Kubak library divergences must fulfill. So if you write it like this, you can use properties that are well known in information theory, okay. So that's a very nice formula. When people talk about the relationship between entropy and irreversibility or you know, the fact that the fourth trajectory must be somehow different from the reverse trajectory, more likely than the reverse trajectory. I mean, that's really the formula that expresses this idea quite explicitly, okay. The other thing is, you know, we did, we did this fluctuation theorem here, right, to the power of minus delta s total equal to one. That's the one we did. But there is another one, which is I can think about the probability that delta s total is equal to x, right. That would be the sum overall trajectories, the probability of the trajectory plus I must sum overall trajectory such that delta s total of gamma is equal to x, right. So if I sum x is a number here, okay, so let's say x equal to 10. And I wonder what's the probability that the interaction is equal to 10. I have to sum the weight of all these trajectories such that the delta s total of that trajectory is equal to 10, okay. Okay. Now, an important property is that if delta s total of a trajectory gamma is equal to x, then the delta s total of the reverse trajectory is going to be equal to minus x, right. From the definition of delta s total, I remember delta s total was equal to ln of, sorry, delta s total was equal to gamma was equal to ln of probability of gamma over probability of gamma reversed, right. And so, you know, if I do the task of gamma reversed, and of course, each trajectory has only one reverse, right. It is a one to one nappy. So if I do the delta s of gamma reversed, I'll get ln of probability of gamma reversed divided by ln of probability of gamma of gamma, which is the reverse of gamma reversed, right. So, yeah, so that's the properties I would be interested in. Now, okay, so if I do these two probabilities, probability of delta s total equals to x, probability of delta s total equals to minus x. So I'm doing the ratio of these two things. Then I'm going to get the sum of gamma, overall gamma, probability of gamma of delta s total, s total equals to x divided by probability of sum over gamma with delta s equals to x probability. I could see gamma reversed here. That works also. Okay. So, you know, I can change this thing here by probability of gamma reversed, e to the power of x, right. Because e to the power of delta s total is going to be probability of gamma divided by probability of gamma reversed, okay. So if I look at these two fractions, each term in the sum that is this is multiplies into the power of x, so the power of x can go out of the sum, and then that's going to be equal to the power of x, okay. So this is another version of the fluctuation theorem. I mean, we could say it's more general than the other one. So this is called the integral fluctuation theorem. This is sometimes called detailed fluctuation theorem, but a little bit tricky calling detailed fluctuation theorem depends on the situation. But, you know, they are two different versions. So, you know, there's just this quality, the very well-known jazzynski quality is of this form here, and what's called the crux relation is of this form here, okay. But they are more or less equivalent, okay. It's not, the both of them are true. One could say this one here is even more a little bit more general, but, you know, again, it's just a different version of the fluctuation theorem, okay. Another remark I should make is that, I mean, we could also consider a case where the transition rates like W, I, J depend on time, okay. And, you know, if we're talking about the crux relation or the jazzynski relation, they are such that the transition rates depend on time. What does it mean for the transition rates to depend on time? It means that some thermodynamic parameter is time dependent. Like, for example, the temperature, the energy, so that is like a, that's what we call a protocol, okay. When you hear the word protocol, it simply means that there is some thermodynamic parameter that's a function of time. It could be the energy, could be the, could be, could be the temperature, some chemical potential, some affinity, whatever, okay. And this, you know, mathematically, it simply means that the transition rates would depend on time. Now, if you have a protocol that's time dependent, okay, you always have to consider the probability of the forward trajectory. And then you have to consider the probability of the reverse trajectory, but this probability of the reverse trajectory must be, you must have a star on it, because you also have to reverse the protocol, okay. So, I mean, let me give you a physical example, okay. Let's say you have a particle in a harmonic trap, okay. And you can do a Langevin equation for that. You can also do a master equation, which will be discrete, okay. There's a single particle, a single particle in a harmonic trap, okay. And let's say, let's call x0 as the position of the, the minimum of this harmonic trap, okay. But let's say I make x0 depend on time, okay. So, let's say I have x0, I'm using x for a lot of stuff, but maybe I don't call this x0. Let's call this y0, okay. But y is, okay, this x here has nothing to do with the first state in a trajectory, okay. That's just the position of the minimum of the harmonic trap when I have a colloidal part, okay. The black thing is a colloidal particle, and the other thing is a harmonic trap, okay. Let's say, I don't know, I don't experiment, but I increase the position of the harmonic trap linearly with time, okay. I do something like this, okay. I started some position, let's say, I don't know, it starts at 10, whatever the unit is, let's say it's micrometers or whatever, and I go until 100, okay. I don't know, the time is initially zero, and the final time is a final time tf, okay. That's a protocol, right. That's a way of changing things. And you know, if I was to transition rates in a discrete setting, or if I was to do a continuum thing, I would have transition rates that change with time, okay. So what does it mean to do the forward protocol? The forward protocol means I start at 10 and I increase linearly up to 100. The reverse protocol would be I start at 100, and I decrease linearly to 10, okay. So you know, if I was to do the fluctuation theorem for a situation like this, I would have to consider four trajectories with the four protocol going from 10 to 100. And then I would have to consider the reverse trajectory not with the forward protocol, but with the reverse protocol, okay. I would have to start at 100 and go to 10, okay. So that's kind of an important difference. If you have some protocol or some time dependent transition rates, not only have to reverse the trajectory, but you also should consider the reverse of the protocol, okay. And if you do that, you are going to get the fluctuation theorem, okay. That's kind of a technical note. Okay. The other thing is, you know, I discretize the time, and here I would like to give a little bit of sophistication. So, you know, why this great time. So if you remember, I defined the matrix M, you know, and the diagonal elements of this matrix look like 1 minus delta R i. And the off diagonal elements of this matrix look like delta W ij. That would be Mji, okay. All right. And that would be Mji, okay. Now, if delta is very small, the diagonal elements are close to 1, and the off diagonal elements are close to 0, okay. What does it mean a trajectory? You know, trajectory means that if I have a trajectory that starts at x0 goes to x1, and then at some point it's going to be at some point that's quite xm. And then keep going until the final point of trajectory. Now, if I do a trajectory, then, you know, the most likely scenario, let's say I start at state i, okay. Let's say this initial state x0 is just, I call it i. If I'm at state i, then the most likely scenario that after a jump, I will remain in state i because, you know, the probability of not changing my state is much higher than the probability of changing state. So what I could ask is, let's say xm here is defined as the first, the point where the state changes for the first time in this case, right. That's the first change of the state. And let's say I want to calculate the probability, what's the probability that it takes m jumps, m transitions for the state to change, right. Well, that's pretty easy to calculate. That's going to be one minus delta ri to the power of m, right. That's the probability that I do not change my state for m jumps, right. And I have to normalize this probability in order to normalize it. I have to, if I sum over all m, I must guess one. So to normalize it, I just multiply it by delta ri, okay. So that's going to be, so if now, if I did that, if I sum p m, p m from zero to infinity, I got one, okay. Okay, so that's the probability that that's kind of my waiting time probability is how long, how long it's going to take me for me to, I mean, it could take one jump, it could take me 10 transitions, it depends on the particular model, but, you know, that's kind of a probability of, that's kind of a waiting time probability, right. So how many jumps do I need? How many transitions do I need to jump in? What I'm going to do is just take the continue limit now. So let's say I have time t, that's the time it takes for me to, that's the waiting time, okay. How long does it take me to leave the state where I was? And so this time is just going to be m delta, right. Because again, delta is the size of the thing. So I can write p m as one minus, then the m, I write it as t over delta. And then this one, I write as t. And then I divide by t over delta, right. Okay, so that's it. And then I have a delta r i outside, okay. Okay, so, you know, if I take the limit, limit of delta going to zero, this thing here is simply going to be each of the power of minus r i t, okay. That's the definition of an exponential. And if I, you know, if I transform from the variable m, discrete m to continuous. So instead of p, let's use some other letter. Let's just call the waiting time distribution q. And let's not use a function. Let's just use a subscript because m is discrete, okay. So I'm calling this q because I don't want to use p anymore. Okay, so I mean, if I do that, then I will recover the exponential. And now if I go from, from q m, discrete q m to continuous q of t, then, then I must divide by delta. And so my q of t from that formula that is simply going to be r i, if the power of minus r i t, okay. And, you know, if I do a continuous time trajectory, as I told you, I'm going to have this waiting time, okay. So then the continuous time trajectory, I have a waiting time. So I stay in a trajectory for a certain amount of time. And then I do a transition. So the reason to use discrete time is that, you know, the trajectory is just a product of this m transpose x n. Let me write this here. So for the discrete time, I wrote the trajectory as m transpose x n, x n plus one. If I want to write the weight of a trajectory in continuous time, I have to keep this waiting times, okay. So it will be a product of the transition rate, W ij. So W from x to the next x. But I will also have this waiting times trajectory, which makes the trajectory more complicated. It makes the number of possible trajectories equal to infinite. So, you know, it's something that is mathematically a little bit harder to deal with, okay. Though it's completely unnecessary, and I don't know why people still do it in stochastic thermodynamics, but because if something is true for, for discrete time, then it must be true for continuous time, because continuous time is a particular case of discrete time, okay. While the other way around is not true. If you prove something for continuous time, it's not necessarily true for discrete time, okay. Or for all discrete time cases. Discrete time is a little bit more general than continuous time. Okay. So that's sort of an explanation to why I have used discrete time and not continuous time. It's just that I wanted to avoid using waiting times distributions when I write the probability of a trajectory, which kind of makes things a little bit harder. Okay. So with this, we finish the fluctuation theorem. And I want to start the thermodynamic uncertainty relation, okay. So, you know, the way I advertised things was that there are like these two main results in stochastic thermodynamics. One was the fluctuation theorem that was in the mid 90s. This one is way more recent. It's 2015. And this was, this one was done by myself and Ludwos Eifert. And I mean, what they have in Cuomo again is that they both, they are both about the fact that fluctuations cannot be anything, right. For the fluctuation theorem, what we saw is that fluctuation cannot be anything. Either you say the probability distribution of entropy has the symmetry. It can't be anything, but it is, it must fulfill the symmetry. Or, you know, you can, you can do the integral fluctuation theorem, one of the two. And the thermodynamic uncertainty relation is another constraint of possible fluctuations in thermodynamic quantities. Okay. So in order to do the fluctuation theorem, I'm going to consider a specific example. And the specific example is the same one from before is the case of a, that is the enzyme. Okay. That will be substrate us. And then there will be product P. Okay. And again, concentrations of S and P are fixed. And you know, we are in a steady state, meaning that the rate at which we burn substrate S is fixed. Okay. I call it stationary. I think I don't remember anymore, but I think I called it stationary, but stationary or steady, it doesn't really matter. Stay stationary state. Okay. That's the same situation as before. Now let's think about the random variable X, which is the number. And again, I'm using X a lot, but this X is not trajectory anymore, trajectory anymore, but it is the number of consumed S by E. Okay. So, you know, that's how if I could measure an experiment, I just, you know, whatever E goes to the suit, maybe I see some sort of blinking in my experiment. And then I can measure how many, how many ATP has my enzyme burn it after 10 minutes. Okay. And then I can measure it's a number, it might be 100, it might be two, it might be minus five, right? It could be negative because sometimes it's the chemical reaction in the reverse direction. And so, you know, this X is going to be a random variable, right? Okay. And, you know, the average of X divided by T is going to be the current that we saw in the previous part of the lecture, right? I'm going to simplify the version of the model that you had before. It's going to be even simpler than the true state model. But, you know, that's what we saw. And we also saw that, you know, in equilibrium, the average X is going to be zero, right? In equilibrium, I have no current. So, you know, I do not, on average, I do not consume substrate sometimes I consume product, but you know, I can go either left to right or right to left, right? There is no preferred direction to chemical reaction. So, you know, on average, if I'm in equilibrium, my average X is going to be zero. Now, I would like to think about epsilon square X square minus X square, which gives like the precision of the variable X, right? And, you know, what I can see is that in equilibrium, epsilon goes to infinity. So, I have no precision in equilibrium in a sense, okay? I just, you know, since X is average zero, I could say that I have no precision in a sense. Now, if I want to have a finite epsilon, so let's say I want a precision of one percent, that would mean I want epsilon to be equal to ten to the power of minus two, okay? The question is, what is the cost of precision in X, okay? You could say X is the output of the chemical reaction, right? X is the product that I produce or the substrate that I consume, whatever, okay? But what's the cost of precision, right? I know that I must be out of equilibrium in order to have some precision, but the question is, is there a minimal amount of energy I must pay? So, let's say I want a precision of one percent, how much do I have to pay for that, okay? That's the kind of, and I mean cost here simply means the amount of energy that I'm burning, right? So, if I burn 180p, I would burn like 20 kBT, right? In physiological conditions. So, how much substrate do I have to consume in order to have a certain precision? That's the question, am I? So, how much energy do I have to burn to have a certain precision? And, you know, there is a minimal cost to that, that's the term, and that's for the term that I'm considering, tell us what is the cost of this thing? Okay, so now I'm going to I'm going to simplify the model for that. I'm just going to say that I have like a bias random walk, okay? And if the bias random walk goes from x into x plus one, that will be the same as doing this chemical reaction here, e plus s goes into s, goes into e p, goes into e plus p. And if the bias random walk goes from x into x minus one, I'm going to say that e plus p went into e p. So, it's an even more coarse-grained version of the model that we had before. I don't want to describe all three states. I just want to say that if I do one chemical reaction, it's a jump forward in a bias random walk. And if I do the other chemical reaction, it's a dump backward in the other direction. Okay, so that's what I mean by a jump forward and that's what I mean by a jump backward. Now, this will happen at a rate k plus, okay? And this one here will happen if I rate k minus, okay? So, basically, from the generalized state balance we have that k plus over k minus is going to have to be each of the power of beta delta mu, okay? That was generalized state balance condition. Before I had three rates because I was doing rates for each one of these transitions, but now I'm just saying that, you know, I have a bias random walk. If I jump from x to x plus one, that's the same as doing the chemical reaction in the forward direction. And if I jump from x to x minus one, that's the same as jumping, as going the backward direction of the chemical reaction. And I just use k plus, k minus as effective transition rates, which is called that. And, you know, from what we had before, this initial rates must fulfill this relation here, okay? Okay, so let's see how the master equation will look like to look like p of x t dt is equal to k plus p of x minus one t plus k minus p of x plus one t minus k plus plus k minus p of x t, right? That's the master equation for a bias random walk. And then I can jump into a state x from x minus one if I rate k plus, I can jump into x from x plus one if I rate k minus, and then I can jump out of x either k plus or k minus, k plus would be, I jump to x plus one and k minus would be that I jump to x minus one, okay? That's my master equation. And now I want to solve this equation. Basically, what I want to do is calculate the x square minus x average square over x average square, okay? That's what I want to calculate. That's the first thing I want to calculate, and then I want to calculate the cost, okay? So, first time I'm going to calculate that. Then I'm going to calculate the cost. What I assure is that there will be a trade-off between, not bad, one could say a trade-off maybe, there will be some relationship between both of them, and this relationship between both of them is going to give, is going to answer me this question here, okay? The question I want to answer is this one, what's the cost of precision? And that's the question that the thermodynamic and central relation answers, okay? Okay, so let's calculate that. Okay, so that's not a very hard calculation. The way you solve an equation like this is a linear equation, okay? You do a Laplace transform that will make your life much easier. And so I will define the p tilde z t as the sum from x. x goes from minus infinity to plus infinity, okay? e to the power of z x p of x t, okay? Now I'm going to multiply both sides by e to the power of z x. When I say both sides, I mean of this equation here, okay? I'm going to multiply both sides by e to the power of z x, and I'm going to sum over all x in both sides of the equation, okay? So, I get d p tilde z d t is equal to the sum in x e to the power of z x p. So, for this one, I have p x minus 1, so k plus p x minus 1 t plus sum in x e to the power of z x k minus p of x minus 1 t. Then I have minus k plus plus k minus, and in this case, I just get p tilde of z and t, right? Now, this sum here can be written as e to the power of z, okay? Let me do this one explicitly so that you get the idea, sum over x e to the power of z x plus 1 x minus 1 t, sorry. It's minus 1 not plus 1, okay? Which of course, you know, sum over all x and sum over all x minus 1 is the same thing, so that thing here is just going to be p tilde of z and t, okay? Okay, so basically this term here is going to be e to the power of z p tilde of z and t, and this term here is going to give me a k plus e to the power of minus z, not k plus k minus, sorry, the power of minus z p tilde of z and t, okay? So, you know, if I want to rewrite that equation there, I've got d p tilde z t dt is equal to e to the power of z k plus plus e to the power of minus z k minus minus k plus plus k minus, and everything multiplies p tilde t. Now, it's very easy to solve the equation, p tilde of z and t is just going to be exponential of e to the power of z k plus plus e to the power of minus z k minus minus k plus plus k minus, and then I multiply by t, okay? That's the solution of the equation. Now, I assume that initially, so, you know, p of x zero is equal to delta x zero. Let's just say that initially I started zero, okay? But it's just the initial condition, and that will tell me that p tilde of z and zero is just equal to one, okay? And that you can show by just using this relationship here, right? So, you know, if at t equals to zero, p of x zero is just delta of x and zero, p tilde of z and t at t equals to zero is going to be equal to one, okay? Okay, so that's consistent with this initial condition is consistent with what I got, because, you know, I can determine this function up to an initial condition, but if I impose this simple initial condition of starting at x equals to zero, meaning initially I have neither burn it nor I have done nothing initially, okay? And so that's the solution of the equation. I mean, that's the Laplace transform, but the Laplace transform will have everything. Now, what we want to do is to calculate the average x and the average x square minus x square, okay? Okay, for that I will define a lambda equals to ln of p tilde of z and t, okay? And that's just going to be whatever is inside the exponential there, so that's just k plus e to the power of z plus k minus e to the power of minus z, then minus k plus, minus k minus t, okay? That's my lambda of z and t, which, you know, is called the scalar cumulon generating function. Typically, I would have to divide by t, but anyway. Okay, so that's my lambda of z and t. Now, let's, so lambda, another useful formula here will be that lambda of z and t is the ln of sum, sum in x, p, x, t e to the power of z, x, yeah, that's it. So if I do d lambda d z, I get sum in x, x, p, x, t e to the power of z, x over sum in x, p, x, t e to the power of z, x, okay? And, you know, the lambda d z at z equals to zero is simply going to be the average x, right? I hope that's clear to everybody. I mean, the whole idea is that I had to solve the question for p of x in order to do that, I do a Laplace transform. And now what I really want to calculate are the moments of distribution average x and average x square. In order to do that, I just take derivatives at z equals to zero of what's called the generating function or more generally the skillet cumulon generating function, okay? That's the name of z. That's the idea for generating function probability. It's just it's simply easier to calculate than the full probability itself. Though, you know, they contain the same information. They are just expressed in a different way. Okay, so if I do d two lambda d z two, I don't want to do this explicitly, but I can do that that's going to be x square minus x average square, okay? At z equals to zero, okay? Again, it's a fairly easy calculation to do with everything I have done already. So, I mean, it's pretty easy to see from this expression that that's what's going to happen. Okay, so, you know, I have lambda. And, you know, if I take a derivative with respect to lambda, it's with respect to z, sorry, it's very simple. These two things here, this is simply going to disappear after I take a derivative. And if the derivative is odd, then I'll get something like k plus minus k minus because of this sign here. And if the derivative is even, I'll get k plus plus k minus, okay? So, basically, this is going to be equals to k plus plus k minus multiplied by t. Remember that I have a t here, okay? So, you know, whatever I do, I'll get a t. And if I do that for x, x is going to be k plus minus k minus multiplied by t also, okay? So, again, all I just solved the master equation for a bias random walk by doing a Laplace transform, which something you should be able to find in several different books, okay? All right. So, that's the solution of the equation. So, that's the solution of the problem. So, we calculated epsilon square. Epsilon square is equal to k plus plus, well, let's write it explicitly. So, it's x average square minus over. So, that's going to be k plus plus k minus multiplied by t divided by k plus minus k minus square. And then I have one t and t square below. So, I have one over t. One over t. Now, what can I see in this expression? So, equilibrium would mean k plus equals to k minus. That would mean the delta mu is zero. Remember that k plus over k minus must be e to the power of beta delta mu, right? So, in equilibrium, this is going to infinity if k plus is equal to k minus my epsilon diverges. That's something you already know. And it's consistent with this result we got. And the other important thing is the t here, okay? So, the longer I run my chemical reaction, the more precise I get, okay? So, you know, naively you might say, okay, so you ask it, you know, what's the cost of precision, but you know, I can get any precision I want because I can just run the chemical reaction forever. And the more I run, the smaller the epsilon I'm going to get. Okay, so if I want to get epsilon equals to one percent, I just run the chemical reaction for a long time. And when the time is big enough, I just stop. That is true, but the longer you run, so the longer t, the higher the cost. And if I should write like this, so longer t implies higher cost, okay? So, you know, if I'm burning ATP, constantly burning ATP when I'm running the chemical reaction, if I keep running for longer, the cost is also longer. So, I do not want to know only about the precision, but I also worried about the cost. And the question I'm asking is if I want to get a certain precision, what's the cost? And running the chemical reaction for a longer time is not really a very good answer because I'm going to increase my cost. Now, let's think about the cost. The cost is going to be very simple to calculate, right? The cost is simply going to be the average x, what's the average x? That's the average number of ATP that the chemical reaction has burned. And I just have to multiply this by the delta mu, okay? So, every time there is an x, it means that I took a mu s and I transformed it into a mu p, okay? So, the energy that I use for every single substrate I take is just delta mu, right? Equals to mu s minus mu p. So, if I want to calculate the average cost of the chemical reaction, that's just the average x multiplied by delta mu. It's a pretty simple problem, okay? Which is, you know, again, that's the average rate of anti-production multiplied by t, okay? So, my cost is simply going to be equals to k plus minus k minus delta mu t. Now, what you see is what I was discussing, while the uncertainty gets better the longer I run, the cost gets worse the longer I run, okay? So, while the uncertainty goes with 1 over t and that's very general, okay? That would be true for any thermodynamic flux in any circumstance. The cost grows with t, which else is going to be true for pretty much any sort of steady state you can imagine, okay? So, running the thing for a longer time does not really sort this problem. And the interesting thing happens when I multiply both things. So, let's multiply both things. Let's do cost multiplied by epsilon square. What do I get when I multiply these two things? Well, so the cost is going to be k plus minus k minus, that is a delta mu. The cost has a t and then the uncertainty was k plus plus k minus over k plus minus k minus square and I have a 1 over t. Now, the t will disappear when I look at this product, okay? There is no time anymore. This square cancels out with this and so that's going to be equal to delta mu k plus plus k minus over k plus minus k minus. That's the product, cost multiplied by epsilon square. Now, I can use the fact that k plus over k minus must be equal to e to the power of beta delta mu. So, this cost is going to be delta mu e to the power of beta delta mu plus 1 over e to the power of beta delta mu minus 1. Now, this is actually an increasing function of delta mu. So, if you plot this function here, it's going to be an increasing function of delta mu. I mean, it's easy to see, right? For largely left delta mu, it just grows near you with delta mu, okay? But it is an increasing function of delta mu. So, the minimum is going to be reached when delta mu goes to zero, okay? For delta mu going to zero, but not strictly zero. Limit of delta mu going to zero. Strictly zero, I get no uncertainty, but in the limit of delta mu going to zero is when I'm going to reach the minimum of this function. And if I do this calculation for delta mu going to zero, this expression here becomes delta mu multiplied by 2 over beta delta mu, right? Again, all I did here is a clear expansion, okay? So, if I write exponential as 1 plus beta delta mu for both exponentials, I can do that, right? I'm going to get delta mu multiplied by 2, then the next order term beta delta mu is not going to matter. In the term up, in the term down, you know, the one cancels out, okay? So, this cancels out, that is actually 2 k B T, right? So, basically, what I'm saying is that the cost multiplied by epsilon square must be larger than the limit of delta mu going to zero of that expression, because again, this is an increasing function of delta mu, which is 2 k B T. Okay? And this relation here is what's called the thermodynamic uncertainty relation. It tells me, what does it tell me? Let's say I want a precision of 1 percent. So, let's say epsilon, I want epsilon equals to 10 to the power of minus 2. It means that the minimal cost of 1 percent is 20,000 k B T, okay? There is no way you can have a chemical reaction or whatever you can have. I mean, in general, x can be seen as a thermodynamic flux, okay? There is no way you can have a thermodynamic flux with better than 1 percent precision if you're not willing to pay 20,000 k B T, which is, you know, a good amount of energy. 1 percent is not that precise if you think about things that happen in biology, for example. Okay? So, that's a thermodynamic uncertainty relation. That's what the product tells you. The product says, if you want to get an uncertainty epsilon, you must dissipate at least 2 divided by epsilon square k B T. Anyway, the reason you should look at the product, I mean, the motivation that we had was simply that the T would disappear, okay? So, the T disappears, then you should get something, and it turns out that this something will give you a limit. Of course, this is true for everything, so, you know, I did an example here for this enzyme, okay? I did the calculation explicitly and showed that the inequality is true, but it's inequality is true for everything. By everything, I mean, you take a stochastic process, okay? You pick a current, so, you know, x would be some current. So, again, that's the thermodynamic uncertainty relation. Again, proving it in general is a little bit sophisticated. It's not something with these lectures, okay? There are different ways of proving it in general. One of them is by using what's called level 2.5 large deviations. The other one, you have to use some inequalities from information theory. And as far as I know, there are these two methods pretty much that you can use to prove this relation. Again, the proof is a little bit involved, a little bit more involved in the proof from the fluctuation theorem. But at least for this example, it's simple to demonstrate. And again, in general, okay, in general, let's say I have a general Markov process in stochastic thermodynamics, then I can define some current x. It could be any current you want, okay? It can be a single current, and typically there'll be many different currents in your problem, in your model, so, you know, let's say I look at a single x, and then I can define the diffusion coefficient associated with that current as x square minus x. And typically people divide this by 2, okay? That's kind of a definition of a diffusion coefficient. And then I define the current as the average x part t. And you know, the cost is going to be my entire production sigma multiplied by t, okay? And now the uncertainty relation, I mean, that I derive there will simply be expressed as 2b divided by j square multiplied by sigma must be larger than 2, okay? That's pretty much the same thing that I did before. That's a general way of writing. And again, my sigma here is going to be that formula that we had before, right? Sum of your i, pi, wij, ln of wij over wji, okay? That's just the introduction. Again, that's a very non-trivial relation because j, again, sigma might involve the sum of many different currents, and j can be any single current you want in your model, okay? Again, for this simple example, it's very simple. It's just one current, so it's much simpler. But you know, if you had something more complicated, then these things are way less trivial. And again, the proof is a little bit involved, okay? I mean, when we conjectured it, it was in 2015, we did not really have a proof. It was a conjecture, I think one year after it was the first proof modulus. Okay, so again, that's the thermodynamics relation. It tells you about what's the minimal cost of precision. Again, it's not like the quantum mechanics uncertainty relation. You know, it's not about the precision of two different things. It's just about precision, the relationship between precision and thermodynamic cost, okay? Okay, so even that we now know what's the thermodynamic incident relation, let's talk about some applications of this relation. So thermodynamic uncertainty relation, okay? All right, so let's say we have a molecular motor, okay? So the first application would be for a molecular motor. And let's say I have an experiment with a molecular motor, and I can measure x is the position of the motor, okay? Now the motor is like a machine that, you know, I guess Edgar told you about that. But the motor is a machine that burns chemical work, burns ATP, use chemical work to do some mechanical work, maybe to push some colloidal particle. But, you know, if we were to write like the second law for that load, there's the entire production for this thing would look like I have some chemical work minus the force multiplied by the current, where this current here is just the position of the motor divided by the time, right? It's just the velocity of the motor, okay? So j here would be just be the velocity of the motor, okay? Now again, I can measure x, I can measure the position of the motor, but I cannot measure the amount of ATP the motor is burning, okay? I don't really know what's the chemical reaction scheme for the burning of the ATP, which is, you know, can be the case in many different experiments, okay? So different motors have different chemical reaction schemes, these pathways can be a little bit more complicated or a little bit simpler. But I mean, I'm not really able to monitor how much ATP the motor is burning, but I can monitor the position of the motor, that's not easy to monitor in the experiment, okay? Okay, here I'm assuming that kb equals to t equals to 1, okay? So there is no temperature that I'm going to write here, it's just for simplicity, okay? So I can only measure x and I would like to know about the efficiency of the motor, which is just, you know, this thing here would be the mechanical work, right? That's quite w mechanical. So, you know, it's just the velocity of the motor, not the mechanical, this is the mechanical power, okay? And this is the power, it's work per unit of time, and that's mechanical work per unit of time, okay? So they are both power. And, you know, if I wanted to calculate that, that would be like my f, f being the external force, okay? So again, in this notation, f is the external force and j is the velocity of the motor, okay? So that would be just fj divided by the chemical work, right? Again, that's not something I can calculate because I cannot calculate the chemical work, okay? I don't know what's the ATP. But I would like to infer the efficiency, at least a bound on the efficiency, by simply measuring x. And how can I do that? Well, I can do that by using the thermodynamical symmetry relation, okay? So let's say I can measure x, okay? Well, I can measure j equals to x over t. And I can also measure the diffusion coefficient of the motor, where it's going to be x square minus x over u square over 2t, okay? And again, the 2 here is just some conventional people typically define diffusion coefficient dividing by 2, okay? Okay, so I can measure these two things. And what we know from the thermodynamical symmetry relation is that 2d sigma over j square must be larger than 2, okay? Okay, now all I have to do is, so d and j, so if I can measure sigma, I cannot measure, but I can find a bound on sigma. If I can find a bound on sigma, that's the same as finding a bound on efficiency. So what you do here is you use these two equations, w chemical minus fj. Again, f is the force, okay? f is like the force that I have to do to drag a colloidal part or something like that. And j is the velocity of the motor, okay? And I use the equation eta equals to fj over dub cam. So my j can be written as, my sigma can be written as fj one over eta mean the efficiency minus one, right? That's my sigma. Okay, now I'm going to throw this equation here. The both just can console, okay? So I get d. Again, this stuff I can measure is just a diffusion coefficient of the position of the motor. And I get fj over j square, then one over eta minus one is larger or equal than one, okay? So d over j f is larger or equal, then eta over one minus eta, right? So one minus eta, well, okay. So if you keep doing that, I don't want to do all steps. That's a bit pedantic. So you are going to find that eta is less or equal than one or fd over j plus fd. Again, j is just the average velocity of the motor and d is just a diffusion coefficient of the motor. So I mean, even if you don't measure anything about the ATP, you know nothing about the ATP, you can still infer the efficiency of the motor, at least a bound on the efficiency of the motor by measuring the position of the motor or the average position of the motor and the fluctuations of the motor, okay? So the fluctuations can tell you something about the efficiency, okay? Which is not really trivial. So, you know, the thermodynamic interpolation can be think as the minimal cost of precision, but can also be seen as an inference too, okay? You can always infer anthroproduction, and in this case, efficiency by simply measuring the position of the motor and fluctuations of this position of the motor, okay? So that would be one application of the thermodynamic interpolation. All right, I will think about the second one, a similar one, but here it's not really about measuring field. So let's now think about steady state heat engine. Again, so the first application I talked about was simply, I can infer the efficiency of the motor by simply look at the position. I have to know nothing about how much, about the chemical work, about how exactly the motor is burning ATP or about how much ATP it's burning. Without any knowledge about that, I'm able to bound the efficiency of the motor, okay? That's the first thing. I just have to measure the position of the motor. And again, there are very old experiments by very old, I mean, mid 90s, where people were, that's when single model experiments portal has started, right? And there are experiments from the 90s where people do measure this J and this D, but you know, at least at that time, they didn't know anything about the efficiency of the motor. Okay, so let's think now about the steady state heat engine. Now what's a heat engine? So the engine will take heat from a hot reservoir and the heat will do some work and release, I don't know if release is a good word, but let's leave it's release heat into a cold reservoir. That's a heat engine and imagine a steady state heat engine, okay? It's operating at steady state. So, you know, probably I should have done a picture. That's a better idea. So I have a hot reservoir. I have my system, which is my engine. So hot reservoir, temperature hot, that is heat going to the system, then some of the heat go to the cold reservoir, okay? And this could be a work reservoir. I don't know if I should go, I can quite work as a wife, you want, and then, you know, they do deliver some work, okay? Okay, so for this problem, if you write the second law, the enterprise sigma is going to be the rate of heat dissipated in the hot reservoir plus the heat of heat dissipating the cold reservoir, and that must be multiplied by a bit. Now QH here is dissipated, Q is always dissipated heat, okay? So heat minus QH will be heat taken from hot reservoir, okay? So the idea here is that minus QH is positive, QC is positive, and let's call this W as work per unit of time. So all these things are going to be positive, okay? So that's the second law, so that's much bigger than or equal to zero. And my first law is that the work or the power, because here is all per unit of time, so it's enterprise production, the power is going to be minus QH minus QC, okay? Remember that minus QH is heat taken from the hot reservoir, and minus QC would be heat taken from the cold reservoir, which is negative, so QC is positive, QH is negative, okay? So this is the second law, and this is the first law, okay? Now I'm not using the strength relation yet, but what is known here, and maybe I can derive that, let's see, yeah, I have time for that. So, it's difficult, yes? Maybe you can do a written break for us, if somebody has a question. Okay. Can you hear me? Yeah. Like in the previous exercise, but we bound the efficiency. Yeah. How good is that bound? Like if you see the real efficiency of a motor? It depends. It can be okay, it can be quite bad, it can be even larger than one, so of course, when it's larger than one, then it doesn't tell you anything. So I remember Patrick did that, Patrick was a student with Voodoo, now he's a postdoc at the MPI in Dresden, so it's Patrick Petsonka, the full name. So I remember he did that for some experimental data, and some points he would get like 0.5, and some points he would get something larger than one. I mean, typically, if you were to bound the sigma, the bound would be quite bad, because you are far from equilibrium, and typically when you're very far from equilibrium, the bound is quite far, but for the efficiency when you do all these ratios, I mean, you know, it's, I don't really know how good it is also because in these experiments, they don't know what's the efficiency of the motor, because they don't know how much heat you burn. But, you know, if you know nothing about the efficiency of the motor, just the fact that you know, for example, that the bound is, you know, the efficiency must be below 0.5 is already something in a sense, but in general, there are better methods probably to infer the efficiency of the motor, that would be a more direct one. But it can be, it can be okay, it can be bad, it depends. Okay, thank you. Are there ways to improve the bound, including the difference in the chemical potential? Is there a general way of doing this? Yes, yes, we have, so, you know, I talked about the uncertainty relation here. We do have a version of the uncertainty relation where we find the inequality that takes into account the affinity, the thermodynamic force. So, that is a, that is a bound that we have found. I mean, we found this already, if you read the original paper of the uncertainty relation, the bound is there a little bit, then we wrote papers about that. We wrote papers about how to infer things considering the thermodynamic force, which is the delta mu in this case, into account. But there is a bound inequality that does take delta mu into account, and that would be more, it would give you a definitely a much tighter bound. I don't know if you can do a bound on efficiency, that would be a little bit more complicated, but there are bounds that do take the delta mu into account, and they are better. And I mean, we did that as a conjecture pretty much when we wrote the first paper. And then after, I don't know, it took us maybe one or two years after you have a proof of this kind of bound that does take the delta mu into account. Yes, in some time, there is some question in the chat, I think. You want that I read for you? Sure, okay, I can't read the chat right now. Okay, you can count or you can? I cannot. Okay, I will read for you. Okay, the first question is nothing. The second question, the limit k plus go to k minus, and take go to infinity, is it commuted to limit? Is what? Okay, they commute. I mean, k plus going to k minus, it depends on what you're calculating, I guess. I mean, in this case, I don't really have to care too much. I mean, for the bias random walk, really, even for finite t, it works. I don't have to take the go to infinity. So for this calculation I did, I don't really take the limit to go to infinity. But if you were to do a calculation, first you would take the go to infinity, then you would think about k plus the limit of what's called linear response theory, okay, the limit where the delta mu is small, okay? That's the limit that it would do after t go to infinity. Whether both limits commute, I'm not sure, it depends on the situation. But I don't think it's really relevant for this thing. Typically, you just take the go to infinity. After that, you deal with linear response theory, making k plus close to k minus. Okay, so I ask this for people which are online, as Su Jung, which asks this question. You can interrupt Andre when you want, because no, we don't see your question. After maybe it was one hour ago. So it's better for the next time that you interrupt the speaker by telling, I have a question. This is good. We are here for a question. So do you listen to me, Su Jung? Yes, yes, I will do that. Thank you. Okay, perfect, perfect, okay. Alexander, no question. Okay, so Andre, you have again 20 minutes. Nobody have other question? Okay, so Andre, you have again 20 minutes today. I will try to close the chat. Okay, so let's go back to the case of the engine. Okay, so the engine is taking heat from our hot reservoir, delivered to the closed reservoir, and doing some work. Now, you know, if I use these two equations here, I mean, I'm going to do something that you probably did in your thermal physics course. So, you know, let's make qh disappear. So qh, no, let's make qc disappear, not qh. qc is equal to minus qh minus w, right? So I get sigma is equal to beta hqh plus beta c minus qh minus w. And that must be larger than zero, right? Of course, we remember the efficiency of the heat engine is going to be w over minus qh, okay? That's just the efficiency of the engine. It's the work divide. And it is not work, okay, w here is power, not work. It's work period of time because sigma is interproduction, it's rate of interproduction. So this is all heat period of time and work period of time, okay? So it's power, okay? So w is my power. I have my efficiency. I have my sigma. So sigma is going to be beta c minus beta h minus qh. Again, remember that the minus qh is the positive one minus beta c. W is the extracted work or extracted power, sorry. That's larger than zero, okay? So that pretty much means that my efficiency, you know, if I divide everything by beta c, I will get that the efficiency eta equals to w over minus qh. That's just a reminder, okay? It's going to be smaller than the Carnot efficiency, okay? Which is one minus, you cannot do it with betas or with t. So let's do it with betas. It's beta h over beta c, okay? That is the c over th, if you like, okay? And that's the c here means Carnot, okay? Carnot, that's the Carnot efficiency. And you know, what is known here is that the efficiency is smaller than the Carnot efficiency. That's very well known. That's just the second law. But now I would like to use the certainty relation to find something better than this, okay? I would like a relation that involves w power, that involves the official coefficient of the power, and that involves the efficiency, okay? Why, okay? The efficiency is about the Carnot efficiency. The result is known for hundreds of years. This result does not really tell us much about power, okay? What happens at finite time, okay? Now I can do that by looking at sigma. So sigma can be written as beta c minus beta h. Then I have this minus qh minus beta cw, right? So that's what I wrote for sigma. Now this qh here can be written as w divided by the efficiency eta, right? Remember that the efficiency was defined as w over minus qh, okay? Okay, that's the first relation I need, okay? I'm writing sigma in terms of efficiency and w. The other relation I would need was the uncertainty relation, which would be 2dw sigma over w square, must be larger than 2. And you know, I will do kb equals to 1 and tc equals to 1 also for the reminder of this relation. Okay, so if you do that, then you manipulate this equation. So from these two equations, if we just throw this sigma here, you are going to find this relation here, w over dw divided by eta, Carnot efficiency minus efficiency to the power of minus 1 must be smaller than equal to 1, okay? That's a very nice relation, okay? So the first relation we had, again, that's just the second law. The second law does not tell us anything about power, okay? Now, if you remember thermodynamics, you must do quasi-static things, okay? If the system is quasi-static, it means that the time it takes to complete a cycle is infinite, okay? If this time is infinite, it means my power is zero, okay? So for an ideal engine that does reach the Carnot efficiency, my power has to be zero. That's incomplete agreement with this equation here. So if I have eta c going to eta, going to eta c, this number here becomes infinity, but I must be smaller than 1. For that to be the case, my w has to be zero, okay? So if I want to approach the Carnot efficiency, I need a power that goes to zero, okay? But that's the question is very nice because if I want to get a certain efficiency, there is a trade-off between power and efficiency, okay? That's a very explicit trade-off between power and efficiency. And I mean, there was nothing like that, I guess, before this relation was out. And again, it's a direct application of the thermodynamics of this relation. Now, another interesting point about this relation is that there is another way of making the Carnot efficiency or an efficiency approach to the Carnot efficiency. And this way is to make the w very small, no, very large. So if I have a very large w, then I can make my efficiency approach to the Carnot efficiency. So if I have a finite power, but my power has huge fluctuations, then my efficiency can come very close to the Carnot efficiency, okay? That's what it would be. And there are examples in the literature of people finding something approaching the Carnot efficiency, but at the cost of having a very large, not the cost, but you just need really large fluctuations. So that's what might happen, close to a critical point where you can have large fluctuations. And yeah, so that's another application. Again, the idea here is not really that I want to infer the efficiency, that's not the idea. I just want to trade off between efficiency, power, and fluctuations of the power of the work, okay? And that's the relation, yeah. Okay, could you repeat what was dw? Sorry, I don't know. dw is just a diffusion coefficient of the work. So if x, if my w is x over t, x being the work, okay, so x would be whatever is the work in this engine you are dealing with, then the d is just going to be x square dw. I didn't call it d now. I should call it d as before, but yeah. It's just the same as before. Whatever my x is, x would depend on what the model is, but x is just the variable that's counting the work, okay? x would be the total work that I do, okay? If I divide it by time, I get my power, and the dw would just be the diffusion coefficient. So if you think about the molecular model, my x is not really the position of the model, but would be the position multiplied by the force f. That's what would give me my work, okay? Something like that. So again, that goes a little bit beyond the second law, which just tells you that the efficiency, the maximum efficiency of the carbon efficiency here, it tells you something a little bit stronger that there is a trade-off between power and efficiency. And you know, you can get a certain power with a certain efficiency, but this trade-off also involves fluctuations of the work. Okay, so I guess I finished the lecture here. If there are more questions, I'm happy to answer them. Is there a question? I have a question. You say that you can approach the carbon limit by having an HD, no? Yeah. Okay, so you can approach carbon efficiency like having extreme fluctuations in the model. Something like that, yeah. I mean, there is a paper about that, but it's not really a model. It's like a nice model. And what they do is that if you make if you're like close to criticality, where fluctuations go like a power law, they show that you can approach carbon efficiency. But I should say that there is no limit to how close you can come to the carbon efficiency. Okay, I mean, strictly speaking, efficiency equal to the carbon efficiency, typically it's only this quasi-static limit, but coming close to the carbon efficiency is always possible. Okay, so if you have an efficiency that is extremely close to the carbon efficiency by 99% close, that can always happen. Okay, and that happens a lot in many different models. So there is no... I mean, the second law simply tells you that the efficiency is smaller than the carbon efficiency. Okay, but it doesn't tell you how close you can come. Okay, it simply tells that there is a strict case where I really become equals on its quasi-static limit. But this is a question a little bit more explicit telling that typically if you are close to the carbon efficiency, you're probably either getting a low power or a large fluctuations, one of the two. Thank you. All right. Maybe this question is a bit stupid, but what came to mind is there a way to connect this with the fluctuation dissipation theorem? This what? The thermo-dynamic incident relation, or the thermo-dynamic incident relation, you mean? The standard fluctuation dissipation theorem connecting it to kind of transport efficient fluctuations in the system. I mean, so the fluctuation dissipation theorem is a relation between fluctuations and the response function, right? I mean, the fluctuation dissipation theorem is about a response function. So I mean, there is a connection with the fluctuation theorem. Okay, you can derive the fluctuation dissipation theorem or a version of the fluctuation dissipation theorem from the fluctuation theorem, that's possible. Okay, now with the incident relation, I mean, if you do, that is not really, I mean, first the philosophy is very different because again, in the fluctuation dissipation theorem, I'm comparing fluctuations with a response function. In the thermo-dynamic incident relation, I'm comparing fluctuations with cost, okay? There is no response function in the fluctuation, in the thermo-dynamic incident relation. There is no response function there. So it's sort of physically different. Now, if you do a linear response, if you do the thermo-dynamic incident relation in linear response theory, okay, then you kind of will find that that bound B equal to two is the same as the fluctuation dissipation relation, but that's more like a coincidence. It only happens if you are doing linear response theory and if you are doing like a unicycle, for example, okay? But that's really, I would say it's more a coincidence because those other coefficients must be equal to the diffusion coefficient. But again, in general, they are different relations. I mean that the idea of the thermo-dynamic isolation has nothing to do with, there is no response function there, okay? But the term that the fluctuation dissipation theorem is connected to the fluctuation theorem, though. That one, you can derive the FTD from the FTD, that's possible, in principle. Hi, my name is Elena and I was wondering, you said that you could increase the diffusion constant and then you would have fluctuations at transition point. So in general, will all kind of processes become reversible at transition? Because usually with the Kano cycles, you have reversible... I mean, yeah, so what I said is a little bit more complicated, but in the Kano cycle, it's not... This would not be a reversible cycle. This would be a cycle that operate, I mean this kind of model they did was a model that operates at finite time. There is interproduction, that the production is not zero, so it's not reversible in this sense. It's not zero interproduction. It's not quasi-static like the typical Kano cycle, okay? And they have... And they use the fact that they operate close to criticality in a sense to kind of try to achieve an efficiency that's very close to the Kano efficiency. But I mean, it's a bit... I wasn't actually be careful to compare this relation, it's not so simple. But yeah, in the Kano cycle, this kind of paper I'm talking about, it was a paper that people did, I should probably give reference in the last lecture tomorrow, so I'll try to give some references. But there it's not like the Kano cycle, okay? It's not quasi-static, which is a very important point. And it's not reversible in the sense of what I'm saying. The reversible simply means interproduction is zero. It's not zero interproduction, it is non-zero. Okay, so it moves towards the Kano efficiency, but it doesn't resemble the Kano cycle. It moves to the Kano efficiency and different from the Kano cycle, it's not finite. It's not infinite time, okay? It's not quasi-static, it's finite time. And it's not, it's not reversible also. Right. But again, this is not... I mean, you can always come very close to the Kano efficiency, okay? That's not... I mean, this was known even before. It's not... I mean, you shouldn't be so surprised if you see, I don't know, 99% of the Kano efficiency. What is probably true is that being equal to the Kano efficiency you know, equality should probably be only reached at the... I mean, there might be a case where these fluctuations really diverge and you take a thermodynamic limit and you might reach an equality. That could be the case. But equality typically is reached in this Kano cycle, this quasi-static reversible limit. Thank you. I have a question. Get closer because... Sorry, can you hear me? Yes. Okay. Do you reach the... Like, do you saturate the inequality at the equilibrium, no? The thermodynamic concentrated relation? Not at equilibrium and in linearity. At equilibrium, it's strictly speaking that the epsilon is infinite and the sigma is zero, so... But close... In the linear response regime, close to equilibrium, I would say. Not at equilibrium, I would say. At equilibrium, things are not really defined, I would say. Okay, but it means also that every time that you are like close to saturate the inequality, you are so close to equilibrium also. Typically, when I saturate the inequality... I mean, if I'm close to equilibrium, I might not saturate the inequality. It's possible to be very far from the inequality, even if you are close to equilibrium. But if you are close to the inequality, you must be close to equilibrium. Okay. In diffusion process, you can saturate also the equilibrium. Right? Andrei is talking about Markov-Tiamov process. So you can have a diffusion. In diffusion, you can... You can saturate in diffusion. But that's because diffusion is kind of a linear response thing. But yeah, sure. I mean, you can kind of... When you do this diffusion, limit your diffusion process, it's like you are in a linear response regime, model-less. But yeah, that's what Edgar says. But diffusion is kind of a linear response thing, but okay. Yeah. Okay, thank you. There's a question online. Do you see? No, please read it for me. Okay, I'll tell you. When the waiting time distribution is power law, would we model the system using Markovian process? It's the best. I mean, yes and no. So if... I mean, nothing is really a power law because there is always a cut-off. Okay, so if it's a finite system, your power law must have a cut-off at some point. So if there is a cut-off, then you can. If there is no cut-off, but in reality, there must be some cut-off. At some point, your power law must stop, okay? So if there is no cut-off, probably not. If there is a cut-off, then yes. That would be my question. My answer, sorry. But for example, power laws that show up in like biophysics, they will have some cut-off, okay? They come... I mean, you know, this Markovian or Markovian thing might be a problem if you're doing like quantum stuff. So if you do a quantum open system, then things will be truly no Markovian. There is no way out of it. But if you think about things in biophysics, biomolector stuff that, you know, there is no quantum effects there, at least at some level of description, the Markov description must be a good one. At some point, it must be. I don't know any examples where, you know, it's never going to be. But for quantum open systems, where there is like a quantum open no Markovian process, then the no Markovian shadow character is impossible to sort it out. It is really no Markovian, and then it's kind of a problem. But yeah. It is a remark. Nine, but other questions? Well, if there are no more questions, we thank Andrei again. And we continue this afternoon with... Next lecture on the rest of it. Right. It's tomorrow. And the afternoon, there is a lecture of one. Exactly. This afternoon is one. With... Oh, the afternoon for what? Exercise. Okay. The important part is... 215. Okay, so we... Now we go to... We close the system here. We go all to the rest of the lunch. All right. Now we go by...