 Hello? Okay. So what Federico Iceta-Senghi will tell you about is the statistical physics of these systems both equilibrium and non-equilibrium. And so what I wanted to do with you today and what we agreed with him is that we will review some of the basic stuff in particular statistical mechanics of the easy model actually even simpler than mean field easy model. Okay. So I imagine most of you have seen this who have not seen the mean field easy model. No one. Okay. At least four people have never seen the mean field easy model others have seen it but maybe this will help us to be on the same page on this for this course. So what is the easy model? Well the easy model is a model for a ferromagnet. Okay. So we think of a sample of a ferromagnet and we think it being composed of atoms. Okay. And each atom is modeled as a with a magnetic moment that can be either up or down. So these are extremely simplified picture. So for each atom so there are for each atom we have a spin variable sigma i which can be either plus one or minus one. Okay. And this i goes from one to n it labels the atoms the different atoms the n atoms usually it's over there as proportional to the volume. Okay. And so these are the variables the dynamical variables in our system. Okay. And in order to discuss the physics of the system we need to specify what is the what is the energy or the Hamiltonian corresponding to any configuration C by C I will mean a configuration sigma one sigma n of spins. Okay. So C will label the configurations. How many different configuration do I have? Two to the end. Two to the end. Okay. Very good. And so usually this configuration this energy is something that is written in the following form. There is an interaction term which involves nearest neighbors and then there is an external field. There is an external field which acts on each spin. Okay. This sum here means that this is a sum that runs on all the pairs i and j which are nearest neighbors in physical space. Okay. So this happens to be a very difficult model to study the system. And it's mostly because of the underlying structure of the interactions. So the fact that these interactions are over a three dimensional lattice in this case. And this makes calculation very hard. Okay. So the idea is to study instead of this system to study the mean field that is in model which assigns to this same configuration an energy which is just this one. Okay. So the external field is the same. But now this sum runs on all the pairs of spins. Okay. It runs on all i and on all j which are larger than i. Okay. And you may notice also that this j has been substituted with j divided by n. And this is because we want the energy to be extensive. Okay. So if you have here, so a sum over i of whatever is expected to be something which is of order n. It's proportional to n. Okay. Because you are summing over n terms. Okay. A sum over i and j will be proportional to n squared. Okay. So if you want this term to be of order n, you should divide by n. Okay. So this is the idea. If you don't do that, then you will be in trouble when you take the limit n going to infinity. And we will see this. Okay. So essentially this is the problem that we will study. We will study the property of a system, the statistical physics of a system that has this Hamiltonian. Yes. Sorry? Sorry. You mean this should be n squared. So if this is n squared, then this term will be of order 1. And this will be of order n, which is a problem. So this will outweigh this term. So in general, if you can consider even a more, say, a complicated model, probably you will see this, a p-spin model. Okay. But the energy of a configuration c is something like minus j divided by n to the p minus 1 sum over i1 less than i2 less than ip of sigma i1 sigma i2 sigma ip minus h sum over i sigma i. Okay. So probably you will also see models of this type in the course of Federico-Ritchett thing. And you see, in order to make these two terms to be of the same order in the thermodynamic limit, you need to have n to the p minus 1 here. Okay. Other questions? Very good. So what do we do with this energy? Well, first of all, let's see what is the ground state? So what is the configuration of minimal energy that minimizes this energy? So the configuration that minimizes the energy is clearly the configuration where all the spins are parallel. So this is equal to 1 for all the pairs. And all the spins are parallel to the sign of H. Okay. So the configuration of the ground state is 1 where, say, sigma i are only equal to, say, plus 1, sorry, the sign of H. Okay. But this I mean that if H is positive, this is plus 1. If H is negative, this is minus 1. Okay. So this is if H is different from 0. And what happens if H is equal to 0, then I have two ground states. Okay. For H equal to 0, there are both, say, C0 plus and minus are both ground states. So there are two ground states. Okay. And so the opposite limit, as you will see, is the limit where, say, the energy is maximal or if you want of infinite temperature. And that is the situation where, essentially, you don't care at all about the energy. You just pick a configuration at random. Okay. So let's say the situation at the temperature equal to infinity is a configuration where you think you consider a probability distribution which is uniform over, which assigns the same probability to any configuration. Okay. Then given this probability distribution, the average energy, well, the average energy is 0. Can you tell me a simple argument why it should be 0? A simple argument, a very simple argument, why the energy should be 0? Why should it be positive? Sort of. Exactly. So if you think actually the argument is a little bit, but yes, roughly because, essentially, for any up configuration you have an equivalent down configuration. And so, essentially, since, you see, the sum of i less than j of sigma i sigma j, these can be written as one half the sum of i and j, which go from 1 to n of sigma i sigma j. Right? So if I take the sum of i and j that go from 1 to n, both go from 1 to n, I will have twice, I will have, in this sum there is the sum of i less than j over i less than j, but there is also the sum over j less than i. And on top of that, there is also a term which is with i equal to j. Okay? So if I take this sum, take out all the terms where i are equal to j, then this is twice the sum over i less than j. Are you okay with this? Okay? Now, this guy, since i is equal to j, this is equal to 1, okay? So this thing is j is, and this thing, this guy here is noting that, but the sum over i from 1 to n of sigma i squared. Okay? Okay? So the sum over i and j, I can take sigma, I can take the sum over i sigma i times the sum over j sigma j, and this is just the sum over i sigma i squared. Okay? Very good. So you see now the interesting fact of this Hamiltonian, that indeed it can be written as something like, so if I call, yes, minus h sum over i minus plus j over 2, okay? So it can be written just as a function of the magnetization. The magnetization is the sum over i of all the spins, okay? And now, if all configurations have equal probability of having spins up and down, okay? So we expect that the magnetization will be equal to 0. Okay? Yes, please? This one? Yes. Yes, here, well, here essentially I'm writing, so I could write this like this, okay? Okay? So the diagonal term in this sum is a sigma, sum over i sigma i squared, and sigma i squared is equal to 1. Okay? Ah, yes, so no, this, okay, so you're saying this guy can be either positive or negative. And whereas this one, apart from this term here, can only be positive. And yes, no, that tells you, this one, yes, so the, so this fact, this, what this tells you is that this term here is plus n over 2 will be positive, is positive, okay? Typically it is positive, okay, unless this thing is equal to 0, okay, which is the magnetization is equal to 0, okay? And yes, are you convinced? No? Not really? Eh? Yes, so, yes, so this is a sort of sigma squared essentially, so it's, it's, it's is a squared object, so it is more, more positive than negative, okay? Okay, so, so because of this property, so you expect that in this, in this state, the, the expected value of the magnetization of this thing will be 0. And so the expected value of the, of the energy will essentially be typically 0, okay? Because these and these are 0. Of course I mean there are these, there are corrections to this, so this will be 0 only when n goes to infinity, okay? So, so you see this is the main reason why one studies this problem because you have two different situations. So one situation where if you take h equal to 0, you have a two-gram state and one where if you take, if you, if you say, take the opposite situation where you don't care about the energy or say you take infinity temperature, you get one single ground state, okay? And so the issue is how you go from this situation to this situation, so the, which is what we are interested in. And there are two different ways of discussing this system. One is if you think this physical system as an isolated system, if it is an isolated system that then it does not exchange heat with the environment and then you will study this with the micro canonical. Otherwise, you think at least as a, say, as a system which is in contact with a larger system or with the universe and is exchanging energy with this, so it is, say, a system in equilibrium with the environment and then you are going to use the canonical ensemble. So do you remember what, I mean, everybody knows what is a canonical ensemble, what is a micro canonical ensemble, okay? So let's look at the micro canonical ensemble, what is the important quantity, thermodynamic quantity, what is the state function, entropy, okay? So what is the entropy? Let's say we have to study the entropy as a function of energy, right? Because energy is fixed, okay? Energy is constant, it's a constant of the motion, okay? So it's fixed. So now if you look at this equation here, so what you see is that, as we said, this energy is a function of magnetization, okay? It's a function of m squared minus h times m, where m is sum over i from 1 to n of sigma i, okay? There is also the j over 2, but this factor j over 2 is going to be, first is a constant and so the energy, you don't care about constants in the energy. And second is also sub-extensive, so it's not extensive. So I'm going to neglect this in the rest, okay? Okay, so what we ask is what is, so what is the entropy? So what is the entropy of configurations which have a certain value of m, okay? So this is my function em, okay? So what I have to do, the entropy is to compute the number, the log of the number of configurations that have this particular energy here, okay? And this is simple because essentially, so this is the log, the number of configurations that have this particular energy is, so all configurations of this type will have n plus m over 2 sigma i, which are plus 1, n minus m over 2 sigma i that are minus 1, doesn't matter how, I mean any configuration where I choose n plus m over 2 spins to be plus 1 and the rest is minus 1 will have this energy, okay? So the number of configuration with this energy is, yes, it's n, choose n plus m over 2, okay? Very good. So and so now if we said, if we say that m is, we introduce lower case m as the average magnetization or also the magnetization per spin. And if we use a Stirling formula, which says that n factorial is essentially e to the n times, e to the minus n times n to the n times a constant which also, okay, which we neglect. So if we take this approximation, then this becomes minus n times 1 minus m over 2, let me log 1 minus m over 2, 1 plus m over 2, okay? So this is the micro canonical entropy and I will write this as n times this small, let me use small s, okay, as a function of, but this is a function of m in this case, okay? Now if you, so the energy for this configuration is written here, so the energy density is written here and it is equal to minus j over 2 m squared and minus h times m, okay? So this is the energy density. So in the micro canonical description, what you have is that as a function of the entropy, you have the entropy, okay? And so you see you can plot this curve parametrically. So when m is equal to 0, the entropy is 1 half log, 1 half minus 1 half log, 1 half, so it is equal to log 2 and when m is equal to 0, this is 0, okay? So you start from this point and then if h is equal to 0, then you have just one curve, okay? One curve that goes when m equals, goes to plus or minus 1, you go to minus j divided by 2 and essentially you will follow something like a curve like this, okay? This is h equals 0. If h is not equal to 0, then you will end up in two points. One is minus j over 2 minus h. If you follow a branch where the magnetization has the same sign of h, if you follow a branch where the magnetization has opposite sign to h, you will end up in minus j over 2 plus h, okay? And essentially if you plot this function, you will get something like this that again goes through this point and then does like this. So I have a red chalk so I can do some artwork, okay? So this is h different from 0, okay? Now if you have a certain energy, you will see that given the energy if h is different from 0, you will have this particular entropy and that's it, okay? And so if you have a finite energy, you start having two branches, okay? Which means that there is a state with higher entropy which is the stable state and it is a state of a lower entropy which is a metastable state, okay? It is, so they have, so just the number of configurations which are here is much higher than the number of configurations which are here, okay? And so in this sense, we say this is the stable state because if we pick a state at random, then it will most likely with probability 1 be here and goes to infinity, okay? Still there is this metastable state. The other interesting thing about this, this system in this particular energy is that if you look at the configurations which have this entropy and this entropy, these correspond to two different values of the magnetization which means that they correspond to configurations which are very far apart. So in order to go from one of these configurations to the other, you need to change at least a fraction of the spins, okay? So this means that essentially if you have your, we will discuss the dynamics, but if you have your picture of the state of configurations, then this situation here corresponds to a situation where you have many, many configurations somewhere and few configurations somewhere else. And these two set of configurations are disconnected, which means that if you have a dynamics that starts in one of these states and that moves from one state to states which are close by, which are very similar, then you will never jump from one state, set of states to the other which means that you have ergodicity breaking, okay? And of course, say, I think a lot of the work, a lot of the course of Federico usage, it will be about ergodicity breaking or studying systems where at a given energy the configuration space is divided into many, many minima or many, many different connected set and study the consequences of this, okay? Okay, so now if you go to the canonical ensemble instead, so of course, this picture here that I just erased, we will discuss this later, so no, let's discuss this later. Okay, so in the canonical ensemble instead, what we say is that our system is not, you know, the system is not, you know, the system is isolated, but it is in contact with the external environment, okay? And since it is in contact with the external environment, it exchanges heat and so the energy is not fixed, but essentially at the equilibrium, the average energy will be fixed, okay? The expected value of the energy will be fixed, okay? So, and this means that essentially what matters is, so the scripture of this system will be one where every configuration C has a, okay? So, what happens is that the probability which is given by the Boltzmann distribution, okay? So where does this Boltzmann distribution comes from? Yes, okay, so this comes from a calculation where essentially you say, well, the energy, this function here is all that matters to these spins, everything else is not important, so essentially what I will do is I will look at maximum energy or maximum entropy configuration which has a given value of the expected energy, okay? And then the solution of this problem will give you this probability distribution, okay? The other equivalent way of seeing this is to say, okay, let's say a priori all configurations are equally probable, but among this, so I have a distribution over the space of configuration by want to find out that distribution that also satisfy the fact that the average energy must be fixed and must be, and so this makes, this is a large deviation problem and if you solve this large deviation problem, you get this, okay? Very good. Now, so the problem here is to compute this partition function, in order to compute the partition function, you have to sum of all the configurations, it's just a normalization constant, okay? But what we have seen is that in the end that this function only depends on the magnetization, the energy, only depends on the magnetization. So what I can do is to say, let me sum over all configurations such that the magnetization of this configuration is equal to m, and let me sum over all values of m from minus n to n, right? In steps of 2, minus m, minus m, okay, so it's clear that the values, the possible values of the magnetization is minus m, minus m plus 2, minus n plus 4, et cetera, et cetera, okay? It's in step of 2, okay? And then here we have e to the minus, this energy of m divided by t, okay? And now, again, what is this object here? This is just the entropy, e to the entropy, okay? So it's just the sum m from minus n to n, n, choose n plus m over 2 e to the minus e over m, e over m divided by t, okay? Very good. And now what you can do is to, is to essentially go from the capital M to the lower case m, so and write this as, and you, you can turn the, the sum over m into an integral from minus 1 to 1 in lower case m, okay? And then you here you will have e to the minus n times a function of m divided by t, and this function of m is, is essentially what is called free energy, and it's energy minus t times entropy, okay? So in the end, so this f of m is what you read from here, it's minus j over 2 m squared minus h times m, and then there is minus t times the entropy that I wrote before. Okay, you remember 1 minus 1, 1 minus m divided by 2 log 1 minus m divided by 2 minus 1 plus m over 2, log 1 plus m over 2. Okay, so and, so, and then how, when you have integrals of this type, we then go into infinity, you do this by, by, you do it by? Third point, okay? So this is say, you write it like this to say that we only care about the exponential terms f of m star divided by t, where m star is such that the derivative of f with respect to m at m star is equal to 0. Okay? So let's take a break of five minutes, then we'll continue from here, okay? No, because you go from 1, from m equal to minus m to minus m plus 2 plus 4 plus 8, etc. Okay? It is, if n is very large, the difference between, yeah, taking essentially this term and taking the integral from minus m to n is essentially the same, okay? And now if you make a further change of variable where m is equal to big n times small m, then you will get this, okay? So, I mean, the difference between, so if you have a small m, you have a function, then, you know, the sum will be just something like this, the area under this curve and the integral will be just the integral, okay? So the difference, if n is very, very large, then the difference between the sum and the integral will be relatively very small. Okay, so let's go back to this object here, and now what we want to do is to study the behavior of this system, and essentially if you take this derivative and set it to 0, what you will find out is that your m star satisfies this equation, and then once you have solved this equation, which is an implicit equation for m star, then you plug it back here and you have your partition function, and given your partition function, you have the probability distribution over your configurations. So let's ask, what is the probability that I will find a particular value, let's say, what is the probability that we write this explicitly? What is the probability that my magnetization will be equal to m star? It will be equal to some value of m, okay? So in order to compute this probability, what you have to do is to sum these, to sum over all configurations such that the magnetization of this configuration is equal to n times small n, okay? And then you will have the probability of these configurations, okay? Now we have seen that, so this z is e to the minus n times f of m star divided by t, so this is e to the n divided by t of f of m star, this is 1 over z, and then what I have is minus n divided by t times the energy, okay, which is minus j over 2 m squared minus h times n, okay? And again, this is the probability that sum will give me e to the n times the entropy, okay? So in the end, when you do this calculation, the leading term will be e to the minus n over t times the free energy at m minus the free energy at m star, okay? So this is the probability with which you will see a particular value of the magnetization. Okay, now what is the shape of this? So you see, if this difference is finite, then this probability is exponentially small in n, okay? So this probability will be non-zero only when f of m is close to f of m star, or when m is close to m star, okay? So what you can do is you expand here f of m around f of m star, and what you get is that this is equal to the first derivative at m star times m minus m star plus one-half minus m star, the second derivative of f at m star times m minus m star squared plus higher-order terms, okay? Now this by definition of m star, this is equal to zero, okay? By definition of m star, this is going to be equal to zero, so that what you have here is e to the minus n over t times one-half, say, second derivative of f m star m minus m star squared plus higher order terms, okay? So now what you can do for exercise is to compute this second derivative or the derivative and to check that it is negative, okay? So this second order derivative is related to a susceptibility, but you can check that this is negative, okay? So what you get by this result is that the distribution, so the magnetization is essentially, okay, this magnetization one over n is essentially a Gaussian variable with mean m star and variance given by this object here, okay? And variance given by t divided by the absolute value of d squared f d m squared at m star, okay? So what it means is that the fluctuations, so the fluctuations of this object will be about one over square root n, okay? So they will be very small. So what this means is that when n goes to infinity, the probability to find a magnetization which is different from m star will be essentially zero, okay? Okay. So maybe we should do this too now. Okay, the other important point which one has to make in this respect. So essentially if we think about the, if we think about the limit, thermodynamic limit when n goes to infinity, essentially the magnetization per spin does not fluctuate. It's just a constant, okay? Which is equal to m star is given by the solution of the equation. Okay, so now, so then what we expect is that this m star should be equal to the average magnetization, right? Now, take the case where h is equal to zero. When h is equal to zero, essentially the probability distribution is symmetric for positive and negative magnetization, okay? So that we expect that this must be equal to zero, okay? Indeed, if you, if you put h equal to zero, okay? So h goes to zero here, m star equal to zero is a solution, okay? However, what happens is that, so I actually say also, so if you take the limit as h goes to zero of the average magnetization, so by this I mean the expected value over this distribution, okay? And then I take the limit, so if I take this limit as h goes to zero of the average magnetization, this is going to be equal to zero for a finite system. And this is just because in a finite system, I mean this is, this is just the sum of all configurations of one over z e to the minus energy of configuration divided by t times one over n sum over i sigma i, okay? So this is what it is, okay? And since this part is symmetric for sigma going to minus sigma and this part is anti-symmetric, this sum is equal to zero. Everybody is okay with this? You okay? Okay. Then, okay, so for any finite system, this is going to be equal to zero, then essentially also if you take the limit as h goes to infinity, then this is going to be equal to zero, okay? Now, however, what happens is that if you take the limit in the opposite order, things might be different, okay? So if you first take the limit as n goes to infinity of the average magnetization and then take the limit as h goes to zero, then this may be different from zero, okay? So let's see why, okay? Okay, so when you take the limit n going to infinity, we have seen that the magnetization just becomes a constant, which is m star, okay? So we have just to study m star as a function of h, okay? And m star is the solution of that equation there. Let me do it graphically, okay? So on the left hand side, you just have m star. I'm just plotting as a function of m star the left hand side and then the right hand side. So let's take a small positive h positive, okay? Let's say that h goes to zero from the positive side, okay? So if h is positive, then this function is something like this. Going to from, sorry, it should go across zero at some point here, okay? So this, when, so the hyperbolic tangent will be equal to zero when m star is equal to minus h divided by j, right? Okay? So if h is positive, this function here is displaced, okay? So this is the solution of that equation, okay? Now let's look at what happens when h goes to zero, okay? So when h goes to zero, essentially this curve moves to the right, okay? So it will move like this and like this, et cetera, et cetera. When h is equal to zero, it will just pass through zero, okay? So you see that, so this is the corresponding solution will converge to zero, okay? Okay, so this is true if the slope of this function here at, when it crosses the origin is less than one, okay? But if we have another situation, let me redraw the thing. If we have another situation where the curve is like this, so then the solution will be up here. But if you move the curve to the right, you see that the point will change very little. And at the end, when it reaches zero, okay? h reaches zero, then you will have an m star which is different from zero, okay? So you see the solution of this equation, when h goes to zero from a positive or I could have done from the negative side, actually if I had done from the negative side, you would have guessed that the intersection would have been on this other side, okay? So what you get in this case is that the limit when h goes to zero plus is different from zero. And this happens only when the derivative of this function in a positive or negative side, okay? So h equal to zero and m star equal to zero is larger than one. So essentially it happens only when j divided by t is larger than one. Or when t is less than a critical temperature which is equal to j, okay? So in other words, if you take, so what you find out is that if it's a function of, say, what you can do is the following. So as a function of h, you can plot m star, okay? So when m is positive, you will find, so this is plus one, this is minus one, okay? When h is positive, you will find a curve that goes to zero plus sorry, as a function of temperature, okay? So when h is positive, you will find a curve that goes from essentially one to essentially zero when the temperature goes to infinity or from minus one to zero, okay? But if you take the limit when h goes to zero, if you decrease h, then this branch will not converge to zero. So this will converge to zero above a certain tc, but the limit of this curve will not be zero, it will be something like this, okay? And this is essentially, this is what comes out of the function of h, okay? So solving that equation, okay? So taking the limit in this order, okay? Now what is the right, yes, please? Sorry? Why? Because you see, so after I have taken the limit h goes to zero, essentially this equation here is completely symmetric between m and minus m. So the only solution of this equation, say, so the, okay, so when you are here, you will find that what happens to that equation is that essentially you have three solutions, one, two, and three. Okay, sorry for my drawing. So you have three solutions, one, two, and three, okay? So these are different from zero. So this is, let's call it plus h, okay? So this is, let's call it m minus m and zero, okay? So because the probability of this is equal to the probability of this in equilibrium, when h is equal to zero, then if you take the average minus m plus m, then you get zero, okay? So this is why you get this result. For any finite m, n, you get that this limit will give you zero because the probability of the positive magnetization is equal to the probability of the negative magnetization, okay? Now if you take, however, if n is very, very large, then n, if your h is not zero but is positive, one of these two states, say if h is positive, the state with m positive will be exponentially more probable than the other one, okay? So for any positive h, for any positive h, this limit will give zero weight to this solution and will give weight equal to one to this, okay? So the probability is different, okay? So it's a different calculation, okay? And actually so physically what happens is it's very interesting because essentially so it's like you have, this is what you see also physically, so you have a ferromagnetic system, you polarize it into one direction, okay? So this is spontaneous magnetization, okay? So it's a positive finite system. In the end, I mean it's a finite sample, okay? So you are not in this situation here, you are in this situation here, okay? So what happens is that, well, so the, your finite sample, if you wait long enough and you take an average long enough, then the magnetization should be zero. So what will happen is that at some point it will spontaneously re-magnetize in the opposite direction. So probably say maybe the time you have to wait for this to happen may be longer than the time of the universe, but it will happen, okay? So, and actually we will see this when we will discuss, we will discuss the dynamics that essentially much of the phenomenology that you are going to see in this course has to do with the fact that system may get trapped into metastable states, into states which are, may not be equilibrium states for a very, very long time, for a time which is practically infinite, okay? So the infinities model is a, say, simplest example where you have this phenomenon of spontaneous symmetry breaking. And okay, there are other quite interesting things I want to recall about this solution and you will receive lecture notes on this part, but one thing which is interesting to notice is that essentially because this is a mean field model, what you have is that, so you can compute these object which is the correlation between two spins. And so since the, so in the limit when n goes to infinity, the probability distribution only depends on the average magnetization, then effectively the correlation between these two, between two spins is very weak, is about 1 over n because it goes through how much sigma j changes the average magnetization, which determines the probability of sigma j. So, and the change when you change just one spin is 1 over n, okay? So essentially what you will find is that the limit as n goes to infinity of this correlation is equal to essentially the m star squared, which is essentially equal to the average of sigma i times the average of sigma j. So spins essentially in this mean field easy model behave as independent, okay? So this is a particular feature of mean field models and which is not true for final dimensional systems where instead correlation may decay with distance. The other interesting thing is that you can characterize this point here, which is called critical point, which is the critical point. And there is a lot of interesting features that happen at the critical point. And these are related to the singularity that emerges at this point. In particular, you can ask yourself what is the magnetization as a function of t when you are close to this point here? And so this magnetization is, and how do you find this? Well, essentially you have to take this equation here, okay? So you take this equation here and you consider that if you are close to this point, then m star is small, it's very small. So you can expand this, you can expand the hyperbolic tangent in powers, okay? So the first term will be j m divided by t, then you will have a cube term, then you will have a fifth order term. So you do this, please do this exercise. And what you will check is that essentially if you are at t larger than t c, t is equal to 0. If you are at t less than t c, then this is equal to square root of 3 t c minus t divided by t. t is less than t c, okay? So which means that the magnetization as a square root singularity goes to 0 as a square root is not, you see this function here is not differentiable at t c, okay? So the derivative goes to infinity, which means that there is a singularity. And also what I told you is that the fluctuations of the hyperbolic magnetization are Gaussian, which means they are of order 1 over square root n. But if you are precisely at this point, this is no longer true. You can repeat the calculation that we have done and check that when you are precisely at t c, in the expansion, the exponential probabilities, you compute this probability that we computed that 1 over n. So what you will see is that this is no longer, sorry, the value we computed this as e to the minus n over t f of m minus f of m star. So when you are at t c, the first order term here is not the quadratic term, but it is a quadratic term. So this will be proportional to e to the minus n over times a constant times m to the fourth, okay? This tells you that the typical fluctuation, the average magnetization of order n to the one fourth, okay? And n to the minus one fourth, okay? So the fluctuation in the total magnetization will be of order n to the three fourth. You have much more fluctuations at the critical point than away from the critical point, okay? So these anomalous fluctuations are essentially at the heart of this non-trivial behavior of systems that we have here. At the critical point, okay? So I think it's way beyond the time to stop. So I hope it was not too simple for all of you. Any questions?