 We can see your screen. Now you can hear me, right? Yeah. I just forgot to unmute. All right. Thanks to the organizers for organizing this interesting workshop. I also would like then to start thanking the student who did the work, Dan Mulder, and also Tom Aldrich for the collaboration. So the topic is about copying. And of course, copying is fundamental to life. And perhaps the most striking example of copying is the self-replication of a single-celled organism. But also the molecular scale copying is very important, DNA replication, gene expression. But also in cell signaling, copying is a key ingredient. And to illustrate that, I would like to briefly highlight a recent result from our group, right? And so recently we've been interested in cellular prediction, this question of how accurately can cells predict future changes in the environment. And so the general setup is that shown over here. So here you have an input signal that fluctuates in time. This signal is relayed via some signaling system not shown to the output signal shown in blue. And then the system then needs to predict from the current output now what the future input will be. Now, and while the system needs to predict the future input, it can, of course, only sense the current input and remember the past input. And this immediately implies that there is a fundamental limit to how much information the system can obtain about the future, the so-called predictive information, which is set by the information, the amount of information that is extracted from the past signal. Now, this information bound does not depend on the design of the system. It only depends on the statistics of the input signal. Yet how close the system can come to this information bound, that does depend on the design of the system. In particular, to reach the information bound, the cell should extract those characteristics of the past signal that are most informative about the future signal. Moreover, the amount of information that the system extracts from the past signal or can extract, that is determined by how many resources are devoted to maintaining and operating the signaling system. So receptor proteins and readout molecules are needed. They have to be synthesized on a timescale set by the growth rate. And also, these signaling systems are typically driven out of equilibrium requiring a chemical energy cost. And these resource calls then put a bound on the amount of information that the system can extract from the past signal, which in turn will put an upper bound on the amount of information they can possibly have about the future signal. Now, what I would like to briefly highlight now is that in order to understand this, we also have to understand the energetics of copying. And more concretely, so we have recently looked at a Markovian input signal. We've also looked at non-Markovian input signals. But for now, briefly, I want to focus on the Markovian input signal. And here in black, you then see this information bound, this predictive information as set by the past information. And as I already mentioned, this is really determined. This bound is really determined by the temporal statistics of this Markovian signal. But here, highlighted in green, this region shaded in green, that is the region that is accessible to this push-pull network under a resource cost C, which in this particular example is simply given by the number of receptors and the number of readout molecules times this growth rate. And there are three points worthy of note. The first is that as you relax the resource constrained, as more readout molecules and receptor molecules can extract information, that then more information, more past information and predictive information can be obtained. That's not so surprising. But what's perhaps more interesting is that this push-pull network, it can reach this information bound. But what's perhaps even more interesting is that this system can increase both its predictive power and its past information by moving away from this information bound. And so why is this now? And to elucidate this, we can look at the design of the system along this information frontier. And remember that along this frontier, the sum of the number of receptor and readout molecules, that is constant. But what you see is that as the system moves away from the bound, the ratio of the number of readout molecules over the receptor molecules increases as well as the integration time. So really what happens is that at this information bound where theta equals 0, there the integration time is 0. So that's the time to complete a cycle. The number of readout molecules equals the number of receptor molecules. So here at the information bound, the system is an instantaneous responder. It instantly copies the input signal into the output signal. But the system can do better by moving away from this bound by trading receptor molecules for readout molecules. And then what happens is illustrated here, it basically then takes more measurements per receptor. It stores receptor states in the past in the stable chemical modification states of these readout molecules. So it copies states of the receptor in the past. It copies these into these stable chemical modification states. So copying is really what makes it possible for the system to increase its predictive power. But what now about energy? So here in this example, I've ignored the operating cost. If I now include the operating cost, what you will see is that then the system moves away from this information bound. And that can be understood very intuitively because to reach this information bound, the integration time tower should be zero because only then can the system instantly copy the current input into the current output. But that's, of course, precisely the point where this operating cost diverges. So the operating cost diverges at the information bound. And in general, including it, moves then the system away from the information bound. So this example goes to illustrate that copying is fundamentally important for cell signaling. It's really what enables time integration. And copying requires not only molecules, proteins, but it also requires energy. So does copying or taking a measurement require energy and why? Of course, this is a very old question. So at least 100 years old, it goes back all the way to Maxwell's demon, this creature that tried to create an energy gradient without having to put in any work in a parent violation of the second law from the dynamics. And then Silar basically exercised this demon by realizing that you can only extract work if you do measurements, but measurements, taking a measurement, however, costs work. And then Bennett and Landauer, they carved this out more clearly. And at the end of the day, what happens is that taking a measurement or making a copy is all about correlating the state of the system of interest, let's say the data bed, with the state of the measurement device, the memory bid. Moreover, after the measurement has been taken, you typically want to separate your measurement device from the system that you just measured and creating correlations between two systems that are separated that do not directly interact that cost free energy as shown over here. And this extra cost is given by this mutual information. And so this means that work is needed to establish correlations. It also means that free energy is stored in these correlations, which means that work in principle could also be extracted. Now, where the work is put in to create these correlations, that is actually not of primary importance. Where you do the work is really an accounting exercise. So you could first reset or erase the memory bid. So here we have, let's say, the data bid that we want to copy inside a 0 or 1. We want to copy the state of the data bid into this memory bid, which initially is maybe 0 or 1. Then we can first reset the memory such that we start the copy process with the memory in a well-defined state. Then we need to put in work to reset the memory. The copy process as such can be then for free. That's one way of doing it. But you can also decide to just omit the reset step and directly copy. And then you need to put in work to carry out this copy operation. And yeah, so it's really work. At the end of the day, you need work because work input is required because you need to establish these correlations. Now, in principle, work can also be extracted. But that's typically not what you do in a typical copy process. And it's this failure to extract work that really underlines the thermodynamic irreversibility of a canonical measurement cycle. Now, these optimal protocols, they work really by tilting energy landscapes. While the cellular systems, they typically operate at a constant non-equilibrium driving force. So the cells do their best to maintain the concentrations of ATP, ADP, and PI at constant levels. So the driving force is typically constant. But then the copying proceeds via enzymes, catalysts that reduce barriers. And this has marked implications for the thermodynamic trade-offs between energy and precision. And this is discussed in this paper shown over here. So for now, for the remaining 10 minutes or 50 minutes or so, I will now focus on these more canonical protocols that work by changing energy landscapes. And I will ask the question, what is the trade-off between work and precision when this copy now needs to be made in a finite time? Of course, there has been a lot of work on bit reset in finite time. But as I'll help to show you, copying really brings new and interesting physics into the picture. So what are the questions then that we're going to address? So if you think about it, then you may expect that there's a trade-off between minimizing the work to copy the two respective states of the data bits separately, which then immediately raises this question, what is the optimal protocol that minimizes the average work to copy the data bits? How does this protocol depend on the bias of the data bits and the speed and precision of the copy operation? And how does this optimal copy protocol depend on the steps in this whole protocol that are in time constraint? And I will elucidate that in due time. So what is the protocol that we consider? So here we have the potential of a memory bit. We then bring it into contact with the data bits. If the data bit is 0, then the sequence of steps in the top row will be followed, else the steps in the bottom row. And a typical protocol consists of first establishing the correlation between the memory bits and the data bit. And this is done by changing these energy levels. Then we need to decouple the memory bit from the data bits. This we do by raising the barrier between the two states of the memory bit, and then we can decouple. And then at some point, there will also be a decorrelation step. And I will discuss that in much more detail later. So we will model the memory bit as a two state system. The energy level difference is given by this local detail balance rule. And here we assume that when we change these respective rates, there's some remains constants. But this is really not of much importance. Another procedure where you keep the forward rate constant also only changes the backward rates. That gives very similar results. So we start now by considering copying of one state of the data bit. And the work to copy one state of the data bit is then given by these equations over here. So if you integrate by parts, then you get this term and this term. So the first term here, that's the change in internal energy. The second term is the heat. And then in the spirit of stochastic thermodynamics, this heat is the entropy change in the environment, which is minus the entropy change of the system plus the irreversible entropy production. So from this, you then obtain that the work to copy a given state of the data bit is given by the free energy change of the memory bit plus this irreversible entropy production term. And then we can derive the optimal protocol using a Lagrangian approach. We've actually used an approach that's very similar to that described by Deweyse et al. in this paper. And this problem has similarities to bit reset, although we also find some differences. But I will not go into these details here. So what do we then find? So we'll consider copying the state of one data bit. And we will assume here for now that the initial distribution of the memory bit is 50-50. The two states of the memory bit are equally likely. And this then means that the cost of copying the two states of the data bit separately is the same. And so here you see then the energy change as a function of time for different total copy times. And this dashed line, that's the energy difference that yields the desired accuracy in the quasi-static limit where you have an infinite time to carry out the copy process. That corresponds to the equilibrium distribution consistent with the required precision. And what you then see is that when the copy time gets shorter, that you have to push the system more, you have to drive it more out of equilibrium. And that's, I think, pretty intuitive. What you can see here on the right is that there's a minimal copy time, just the time at which this work diverges to reach a desired accuracy. And the larger the required accuracy, the larger this minimal copy time should be. And so copying with a desired position requires a minimum copy time. Now, what is now the average work to copy a data bit? So the average work to copy data bit where p prime is the probability that the data bit is state one. That is then just given by this simple weighted average. And if you work it out, you'll find that it's given by these three terms. So the first term on the right-hand side, that's the mutual information between the data bit and the memory bit after the copy step. So this depends on the bias, the statistics of the data bit. It also depends on the precision of the copy process, because the bias and the data bit and the precision of the copy voter, that determines the distribution of the memory after the copy step. So that's the first term. That's the mutual information. The second term, that's the Kubeck-Leibler divergence between the memory distribution after the copy process versus the memory distribution before the copy process. So this is really the non-equilibrium free energy stored in the memory, in isolation. And then the last term is this entropy, irreversible entropy production. Another point worthy of note is that this first term, as I mentioned, that's just set by the statistics of the data bit and the required precision. There's nothing to optimize here. But the second term, that depends on this final distribution. That is set by the statistics of the data bit and the precision. This cannot be optimized. But this initial distribution, the initial distribution of the memory bit, that can be optimized. The irreversible entropy production, that also depends on the initial distribution of the memory bit. And as I will show you shortly on the next slide, this leads to an interesting trade-off between these two terms. Moreover, while these two terms, the mutual information and the Kubeck-Leibler divergence, they do not depend on time. The last term, of course, does depend on time. And so what we will see is that we will get a trade-off between these last two terms that actually depends on time. So that makes it interesting. So this is then the main results. So here you see the average work to copy the data bit as a function of the initial distribution of the memory bit as quantified by the probability that initially at t equals 0, it is in state 1. So we look here at the average work to copy a data bit that is biased towards state 1. So the probability that the data bit is in state 1 is 0.9, and the required accuracy is 95%. This dark red line, that's this mutual information, which is independent on the initial distribution of the memory bit. This red line, this bright red line, that's this non-equilibrium free energy stored in the memory bit after the copy process, given by the Kuhlberg-Leibler divergence between the final and initial memory distribution. And that has a minimum around here. So this is the final distribution. This point is the final distribution as set by the data bit and the position. And when the initial distribution matches that, then this Kuhlberg-Leibler divergence is indeed minimized, then it is 0. And this offset here between this orange line and this red line, that's this irreversible entropy production when the copy time is relatively large. While this difference between the yellow line and the red line, that's the irreversible entropy production when the copy time is much shorter. And what you can now see is that indeed, when the copy time becomes much shorter, this irreversible entropy production becomes much larger. Moreover, you see that the optimal distribution of the memory bit goes down. And that's also illustrated here. So when the copy time is very, very large, then the optimal initial distribution of the memory bit basically mirrors the bias in the data bit. But then at some point, it rapidly drops to a half, which means that for short copy times, the optimal distribution of the memory bit is 50-50. And it is even actually shown that the optimal distribution for short copy times is 50-50 irrespective of the bias in the data bit. And that is because if you would not go to the symmetric distribution, then the cost of copying the unlikely state of the data bit becomes prohibitively large. You need a certain precision, which means that you need to cover a certain distance similar to a Wasserstein distance in your information space. And when the copy time just becomes very short, then you really want to minimize this distance for both states. And that's why for short copy times, this initial distribution goes to 50-50 irrespective of the bias in the data bit. Now, in the last two minutes or so, what is now the role of this decorrelation step? So far, we've looked at the work to copy, which is basically the work to go from state 1 till state 7. And what's now interesting is that of course, if you want to use the data bit, you can always reverse the protocol. You can run it backwards in time and extract work from that. And in quasi-equilibrium, you would recover all of the work. But that's typically not what you want to do in a measurement. So for example, in the cellular system, after the state of the receptor was copied in the chemical modification state of the readout, then the receptor, if then the ligand molecule unbinds, then the memory, the readout molecule, continues to remember what that state of the receptor, the data bit, was in the past. That's typically what you have. So you don't want to use the data bit again. But if you know the statistics of the data bit, which is different from, of course, knowing the state itself, but if you know the statistics and if you know that the statistics is not symmetric, so if it's a bias to 1, then you can exploit that. And you can basically extract the free energy that is stored in this state of the memory bit. And you can basically recover this Kuhlbeck-Leibler divergence. Because remember, so when the copy time was short, what you see here, then this Kuhlbeck-Leibler term, then indeed becomes important. But this you can recover in this decorrelation step by knowing statistics. And the last point, and I cannot go into that in any further detail, but if the full cycle, the full copy cycle is under a time constraint, then you basically continuously overwrite, then there is no Kuhlbeck-Leibler term. But then the overall work is higher because this irreversible entropy production becomes larger. But that's maybe too much for now to go into further detail. So let me wrap up. So there's a minimum required time that depends on the position of the copy step, the optimal initial distribution of the memory bit depends on the statistics of the data bits, the precision, and the copy time. For large copy times, the optimal distribution of the memory bit mirrors the bias in the data bit. But for short copy time, this optimal distribution becomes 50-50 irrespective of the bias in the data bit, which I think is an interesting and non-frivial result. Lastly, if you think about this, as I mentioned, the optimal distribution goes back to 50-50 for short copy times. And that is really because the cost of copying this unlikely state of the data bit becomes very large. But now you may ask, OK, this is maybe. What happens now if I now allow for asymmetric copy accuracy? So I say I now copy at constant mutual information. And then, of course, the system can give up on copying the unlikely state of the data bit in favor of copying the more likely state of the data bit more accurately. And then you can actually show that then this strategy of mirroring the bias in the data bit, which is the dominant strategy for large copy times, that that becomes useful again. OK, with that, I stop and thank you for your attention. Thank you, Peter-Rain. So we've got a couple of minutes for any questions. Rashid? Yeah, so I was wondering, since you have that at very fast copying this 50-50 being the optimal, it suggests that maybe the amount of work that you need to put into fast copying doesn't depend on anything, the distributions of anything. So is that true? Is there a very simple expression for the amount of work at a really high rate of copying? Yeah, right. So really, so what happens for short copy times when the copy speed is very high is right. So in our setup, so we have that the copy accuracy, required accuracy for the two states of the data bit is the same. And then if your memory bit would be biased towards this state, but now you copy, let's say, state 0, you basically have to tilt this landscape. And this corresponds to a certain distance. And when the copy time is just very short, you're just not able to move over this probability sufficiently rapidly. And then the best you can do is to really have your memory like this, 50-50, because then the distances for copying the two respective states becomes identical. I think it's correct. I think for his question, which is basically, does the copy cost become something that is independent of the input distribution? Yes. I think that's true. That's true. That's also what I said. Yeah. So this 50-50 becomes the optimal input distribution in this limit irrespective of the bias of the data bit. Yeah. Very quickly, Ruxiang. How? Well, OK. Can you hear me? Can you hear me, Professor? Thanks for a very interesting talk. I'm just wondering why there is a irreversible input production term in the optimal work. It is a little bit counterintuitive, since when you, if you quasi-statically copying data bit, the entropy production term means that it seems that the optimal work will diverge since you need an infinite amount of time when you do it quasi-statically. How do you explain that? So if you do quasi-statically, then the data bit is 50-50. The required accuracy is 100%. You can do it at a finite cost of KVT log 2. And really what happens, the way to understand it intuitively, you change this energy level at some, and then you're more and more likely to end up in the right state. You have less and less probability here, which means that it becomes cheaper and cheaper to raise this energy further. And so at some point, you've transferred all your probability. You can basically then raise this energy level to infinity, essentially for free, because you've already transferred your probability. And this is the reason why, in this quasi-static limit, you can do it, you get 100% accuracy for only KVT log 2. But if you do it, but if you need this desired accuracy and you have, you can only do it in a finite time, then you have to really push it out of equilibrium. You have to push harder. And the harder you have to push, the more work you have to put in. OK, very quickly from Udo then, and we will move on. Yeah, Peter Ryan, hello. Very nice. Along the first question, I mean, you pointed out that if you do it in shorter time, it will need more work. But did you look at the scaling in time? Because for many of these processes, when you want to reach something move from A to B, we and others have found that it typically goes like 1 over T for short times. Yes, we also found that. We looked at the scaling, yes, yes. OK. Yeah, so that, yes. Is it 1 over T minus some critical minimal time? Because it diverges at finite time, right? Yeah, no, but I think the 1 over T, I think that's the long time. Is that the short time limit or the long time limit? Usually it's the short. I think it's the long time limit, right? And then it's time. It depends. For continuous systems, it's actually kind of scaling, Mary. And so it's 1 over T in both limits. I think there's a fundamental thing that comes in because there's a discrete process with a we limit the maximum transition rate. And that's what gives the divergence. OK, OK. Anyway, let's thank Peter Ryan again and move on to the next speaker. Apologies that I am not being very precise with my timekeeping. Biswa does, I believe, work fluctuations and introduction of a trap-browning particle in an active-like viso-elastic bath. OK. Yeah. Yeah, may I do it? Yes, you are. Oh, yeah. So, hi.