 Gen. izgleda drugi in drugi tradisjačnj, da jaz na svetu traktivnja, v deltavo J-snega je, da je bi vsega, skupim, z numerator in denuminator. Mislim, že je log 4 divide by 3, in tjavej J-skega je 1 3. Zdaj se kaj je oplikujem terzojno tradisjačnja, vzhotnjating, videlj, da je zelo tudi cel, začo dobro je zelo temočno zelo nekaj, da je dobro zelo temočno zelo, zelo prez stat, je odelo cema. Kered prez stat, je odelo cema, zelo temočno zelo kod natrupe. In da je zelo cema... Zato, kako sem izgleda ferro in parastate, ferrostate je vse vse, vse parastate je dve energije za ferrostate delta minus 1.5 in za ferrostate zero. In kako sem izgleda dve energije za ferrostate, kako sem zasi dve energije za ferrostate delta minus 1.5 in je nekako, o čemu se vse završi na vse zatelje kako sem zasi završi na zvonku, za spenu 1 sistem. Vsmešaj sem se vse konšistenskega sezi, ki se priče več. Zato se najbolj na vsem zatišnjo vse, tenč o beta x, beta m, iko m, ti padlji temova, kaj se vzivo del desired. To je začela izhoditi različko sečjava na zetej vza. To je pravid, če je so kaj vzodil očetina vzela z 0vči. Ej začela vzela vzela vza na zetej. V medelji. Vzlušaj, da je tukaj kretične ljene, tu vse odporno zelo srečenje v m, in v q, občasno, počutno v m v q. Vzlušaj, da se kaj tukaj, kaj je tukaj srečenje, vzelo srečenje. Tukaj je tukaj kretične ljene. OK. Čekaj, ki se počutno... This are trivialities for those who know the problem. But I just, to be clear, I draw the curve. So the free energy will be flat here, and we'll develop a minimum around m equal to zero. Auto pushed for the so it's the same of cha, it's a sign change in diff. D to F DM to. K, the derivative of the free energy with respect to the lambda of order parameter. So if there is a sign change, then you locate the critical point and you get the line the critical line thatconnected disk airplane line which can be obtained analytically you just revolution you get so li for the coefficient and then you look where the coefficient is 0 and you get deline for the first order is more complicated because what happens is different so you know that at first order phase transitions a minimum first appear far from there is a symmetry OK, so m minus m, taj je m, je m axis, in je vse m minus m, in vzatvoj skupnjenj počak je extra minima, ki je metastable, OK. In se zelo v temprstvu, da pri temprstvu složite, in vsek pa se zelo vzajte vzaljaj minimo in zero in to je zelo početka na primer trončnji trončnji. In kako jaz všetne temprso, to je metastabil, minimo in zero in tudi zelo pričo. Vzelo početka na primer trončnji trončnji. Nisem tudi je to početka v numerikali. because there is a limiting point between these and these, and that's the point where the fourth order derivative of M is zero. And it's called the tri-critical point. So, the sixth order is positive and the fourth order is zero. So, for the tri-critical v britvenj počri so tem počer. Niske občutek v svoji hodnje energije in drugi hodnje energije v svoju broadly poče za njih odvijitov. Vsega nasya hodnje drža je pozitiva, ker je za njih nekaj energije odvijszene. Sledaj, občutek poče za svoje hodnje. Vsega ne primerja vznalo vznalo vznje vzneli svoj čas način, kaj je zelo soluzija na kronika ensample, ki je zelo čas na vaši vsega, in na vsega, ok. Zato sem postojnj, da da se zelo soluzija na naši kronika ensample. Zato vzela, da je to, da je vsega, prej energi for this model, and you see the second of the phase transition, as I've shown with the two minima, lateral minima in x, and the first star of the phase transition, this is the point of where the first star of the phase transition occurs, okay, these are the two cases, and in the middle you have a situation where the fourth order derivative is zero, okay, so let's go to the micro canonical, which is the where kersen 1 would say I should not go, but because it's not straight forward, it's not straight forward, but it's not too complicated in this case, because it's a mean field model, so let's draw a configuration, and the typical configuration will be a string of pluses, a string of minuses, and string of zeros, so this is a typical configuration, and since the interaction is mean field, if I swap the pluses, the minuses, and the zeros, I don't change the energy, so for a given energy, so with short range the problem to swap, but with long range you don't change the energy, okay, so the energy is fixed once you fix the number of pluses, the number of minuses, and the number of zeros, so how do you count in this case, you just swap all the spins, and this is n factorial, and then you get the same state, if you interchange the pluses, the minuses, and the zeros, so this is the volume of this model, okay, the Boltzmann factor, let's call omega, which Boltzmann call it w instead of omega, okay, and I don't know, I was trying to write w on a board yesterday, and came out omega, and that's because that's a Greek symbol on the, omega is a Greek symbol on the board of the computer, so you get omega if you press w in the board, check, so it's the same, omega and w, it's on the grave of Boltzmann in Vienna, and there is K log w without b in the constant, because we call it Boltzmann constant, but it has no b in the grave, okay, so this is the Boltzmann constant, so in what you do, you, okay, did you understand this counting, it's the same counting that you do for the Ising model, for the Ising model you don't have the zeros, okay, try to repeat it for the Ising model, for the Ising model you say, okay, the mean field Q revised model, I get a different state by swapping the spins, and then I get the same state if I swap the pluses and the minuses, so I have to divide by the, by the interchange of pluses and minuses, and I get the volume for the free, for the mean field Q revised model that you find in books of statistical mechanics, so what I'm doing, I'm just taking the same for, and, oh, the, disappeared disaster, the, I need the power, so maybe, since I have this limit, instead of waiting for my computer to restart, which will take time, I might give a solution to the exercise, one of the solutions to the exercise, to yesterday's exercise, okay, so do you agree? I will devote 20 minutes to the solution of, okay, so tomorrow we will, we will see the phase diagram of, so this obliges, ah, it's back, I don't know, it's back, but intermittently back, I don't know why, okay, but it's, I think it's a good point, because in this way you will have time to look into the solution in the micro canonical ensemble, and in the canonical ensemble you will begin to think of the, of the proposal that I gave you for the, for the, okay, let us solve the exercise, okay, so, so I will have to compute, I will do it for, with different variables in such a way that you will see the role of number of particles, because in the perfect gas it's easy, so, oh, this is something you find in, in the one, in the book of Wang, but it's, it doesn't do the, the legend, don't know why, it could have done, but, okay, so, okay, so this is the formula for, so, I will spend some time for, so, perfect gas, perfect gas, and this is the formula for the volume, the volume, so, I have the energy surface, okay, so, the Hamiltonian, the Hamiltonian is the one of the perfect gas, so, is the sum over pi square, pi is vector over 2m, okay, plus the potential, and the potential is infinite if you are outside of the gas, of the vol, volume is here, so, and the u, the potential here is infinite, and here is 0, okay, because if you are inside the volume, the, the potential is 0, if you are outside the volume is infinite, it is a perfect wall, okay, the gas will be obliged to stay inside, and then what you do, you estimate the states with energy, with energy h below e, okay, so, this is one type of entropy, you get similar, so, what I propose if you try also with the entropy defined on the, on the surface, okay, it will give you similar results, okay, so, then omega, the vol, which is the number of states, the number of states with h less than e, at fixed v and fixed n, has this formula, omega of e, v and n is k, I will tell you what is k, to 3n over 2, the correct gives factor, n factorial times gamma of 1 plus 3n over 2, if you want, I can derive this, but, okay, you find it in books, but if you have troubles, I, I derived it last night, before going to sleep, v to the n and e to the 3n over 2, this is the volume in, so, okay, the partition sum, so, so, this is the number of states with h less than e, kb is 1, kb is 1 and s of e, v and n is log of omega, so, Boltzmann was using log base 10 and omega was w, okay, don't worry, so, it's the same, okay, so, and then the partition sum, k, what is k? k is 2m pi over h square, h square is not the plan constant, omega is dimensional, because omega, omega formally is integral dp, dq1, okay, there are 3 components, okay, dqn, dp1, dpn, delta of, sorry, theta of e minus h, okay, this is omega, so, it's dimensional, you see, it's dimensional because it's pq, okay, and pq has h as dimension pq, okay, in order to make omega a dimensional and, so, it's inside the logarithm, so, it should be a dimensional, okay, should be dimensionless omega, this is h, and then in tjani, it is naturally the plan constant, that's why it's called h, and what is z? z of beta, v and n is k to the 3n over 2 over n factorial, v to n, this is much simpler to compute, even for the perfect gas, this is minus here, minus 3n over 2, okay, okay, this is the setting, so, now, okay, to solve z, you just, okay, so, let us do it now on the extensive variable, I did it, I don't know why, I used the other definition of the Laplace transform of the Legendre transform, so, log of omega minus beta e, okay, this is the Laplace transform of the entropy, sorry, the Legendre transform of the entropy, the soup over log omega minus beta e is the other version with extensive variables, don't worry, okay, so, now, we want to prove, we will see that the two are different, so, if I do this, so, to compute the soup, I have to derive with respect to e, so, I have 1 over omega, the derivative of omega with respect to e, which is better, okay, so, I have to solve this equation, but if I, omega is this, so, all that is a constant, okay, the only derivative is here with respect to e, and I have to do d omega over omega, so, the constant goes away, so, what I have, I have 3n over 2 times e to 3n over 2 minus 1, divided e to 3n over 2, which comes from the omega in the denominator, okay, so, I can eliminate these, okay, and what I get, I get the equipartition theorem, it's very interesting, I get the equipartition theorem because you see that you get beta equal 3n over 2, okay, so, the energy, so, if you write in the other way, so, the energy of the system will be the number of quadratic terms in the Hamiltonian, okay, the number of quadratic terms in the Hamiltonian is 3n, and so, you get the equipartition theorem, okay, and it's interesting if you replace this here, no, because you have to replace the minimum, okay, the soup into here, okay, you will kill the e, you see that you killed the e, and you remain only with constant 3n over 2, okay, and the other will be log of omega, now, if you write log z, and you compare with this, with the legend function, so, you have to write log z, log z, and you have to compare with this soup, okay, and you get a difference, where is it, yes, and the difference, you can check, let's give a name to this, let's call it, I don't know, psi, okay, psi of beta, psi of beta, so, so, I'm, I'm, plot, I'm writing now, psi of beta, or, yes, okay, psi of beta is a function of beta v and n, minus log of z, minus log of z, okay, and what I get is 3n over 2 log 3n over 2 minus, minus 3n over 2 minus log of gamma, this gamma, this is the gamma function, 1 plus 3n over 2, so, you see that they are not the same, so, the Legend transform, as I was telling you, there is no dependence here on beta, v, it's only dependence on n, okay, it's only dependence on n, but they are not the same, at finite n, so, the Legend transform of the, of the logarithm of omega of the perfect gas does not, is not equal to log of z, there are finite n corrections, but if you use a stealing formula for for the gamma function, gamma function is a factorial, so, you can use the stealing formula, you can check that in the large n, for n goes to infinity, this goes to zero, okay, that this difference goes to zero in the, in the, in the large n limit, no, because in the, there is this term here, which proportional to n, I, I, I, I, you mean, is, no, no, no, no, no, but these are, this is finite n, so, I have to divide by n then, so, all constants, so, this is the first order leading order in n, in the right, okay, the, then there is order constant, order constant go to zero if you divide by n, and then, but they are constants, you agree? Yeah, okay, okay, there is subleading term log n from stealing, and I didn't take into account, it was midnight, so, I took just the first, it was enough for me to see, to look at the leading term in n, and then I went to bed, okay, but you are right, yes, there is a subleading term in the stealing formula, and it will go to zero if you divide by n, so, you can repeat the, the calculation, you can repeat the calculation for the skin of the, of the sphere, which is different, there is different finite n corrections, the skin is, instead of taking the theta, you take the delta, okay, which is the derivative of the theta, so, you can simply take the theta as in the one book, you derive by e, by r, and you get the skin of the, of the hypersphere, no, of the hypersphere, and you will get different finite n corrections, I didn't check, because again it was midnight, but you can check, I think it will be different, but the leading order should be the same, so that they cancel if you use the stealing approximation, I would just like to comment, you will hear lectures by my good friend, Angelo Ulpiani, in the last week, so there have been a big, big debate in the literature, whether we should use the skin or the, or the volume, of course if the two goes to infinity, and I am not aware how the debate finished, but I know that the discussion was around the fact that if you take the theta, you will never get negative temperature in the infinite volume limit, because of course you are increasing, it's a never increasing function of e, while if you get the skin, of course you get the negative temperature, the discussion was are there or not negative temperature, and I think the conclusion is yes, there are negative temperature, but I don't want to enter in this debate, in the case of the perfect gas, there is no negative temperature, there are negative temperature for the Blume-Capelle model, you will see, okay, but there are negative temperature also for the Curivice model, because you have, at some point the entropy will go down, so the temperature will become negative. Question is if you can measure negative temperatures in real systems and so on, but I don't want to enter in that, so, but I propose you to do the calculation only also for the skin, which is easy to do. Okay, about the other exercise, three minutes, so what you have to do to argue for a non-existence of phase transition in one dimension for the Ising model, you have to take a typical configuration, which is pluses and minuses, separated by what are called defects, and you would try to, you have to try to estimate, so suppose you have dispersed defects in a configuration, okay, and you should try to argue what's the entropy and what's the energy of the defect, and then you will see that the energy is proportional to the number of defects, which if it is a small compared to the number of sides, it will not increase with the sides, while the entropy of such a configuration, you can see it also from the calculation, is proportional to n, so the energy will never be able to compete with entropy, and so the state will remain disordered in one dimension. Then I have another calculation and the exercise that I can propose to you, suppose I take the Dyson model, okay, suppose I take the Dyson model, and I take this configuration, this is an interesting calculation, I take, by the way, the proposal is in a paper by Taules, so I take a configuration in which the spins are flipped like this, okay, I flip some number of spin to the left, okay, and then I try to compute the change in energy. The change in energy, you will see that is able to compete with the entropy. That's a sign that the Dyson model is able to give a phase transition in one dimension, also in the region of weak long range forces, so in the region of sigma larger than zero, okay. Yes? Yes, this is the type of power, power argument that indeed fails in two dimensions because the change in energy is able to compete with entropy. The dynamical argument says that these defects, you can prove that these defects move like random walks, and there is no attraction among the defects. So if you wait long enough, the system will always create defects, and there is no way for the system to build a domain. By, the defects will arise because as temperature is finite, there will be flips, okay. Of course it is not true at zero temperature, zero temperature, in fact you can form a large domain and at zero temperature domain will be the full size of the system. It's sort of kinetic argument that these defects move like random walks and they will be free to arise at finite temperature and they will destroy the order in one dimension. So this is the dynamical. So this is a fire alarm, it's a test, but this is a nice point to stop and to run.