 And then I will start the statistical field theory for charge systems. Okay, so, yes? Yes, okay, yeah, we can do it. So maybe I just recap, I just do once again the cylindrical case. I'll spend few minutes on the spherical case and then it's a big split because then we start the statistical field theory which is complete. And then we can review all the things which are not, is it okay? Okay, so I just try to make a little bit more compact what we discussed last time about cylinder, charge cylinder, which charge density is sigma, negatively charged in the presence of counter ions, so in presence of just plus ions. So the Poisson Boltzmann equation is just d2 phi by dr square plus r d by d, plus 1 over r d by dr phi equals 2C0 over epsilon E sinh beta E phi. Okay, so C0 is the particle concentration of the counter ion. With the boundary, so the boundary condition is just that d phi dr which is minus the electric field is equal to minus sigma over epsilon where sigma is the surface charge density. So since you have this symmetry and this is the r axis if you want, the radial coordinate, so it is convenient to introduce linear charge, so sigma is the surface charge density of the cylinder. So it's convenient to define tau equals 2 pi A times sigma, which is the linear charge density. So tau is the charge per unit length, per one unit length of the cylinder. Right, this is L equals 1. Tau is the charge, the total charge of one length, and in fact it's convenient to introduce a dimensionless linear charge density which is tau, so tau times LB over E. Right, tau is a charge divided by a length, so the natural length is the beer room length, LB, and E is the charge, and let's, I assume that all the counter ions are monovalent so they have a charge plus E. So this quantity is the dimensionless linear charge density, and if you remember there was a critical point when tau was E over LB, so which corresponds to tau bar equals 1, so this is what we will see now. Okay, so, sorry, it's not a cinch, I don't know why I write a cinch, it's a, it's a, cinch is when it's a salt, but this is not a salt, it's E to the minus beta E5, it's counter ions only, okay. Cinch is when you have both plus and minus charges. Okay, so the solution of this equation, and yes, so the solution of this equation is essentially, sorry, yes, and there is no, okay, so I'm sorry because I took my notes for the salt case, so this is the correct equation. Okay, so the solution now for this is, depends on the value of tau bar, so, and this is, so the derivation of this solution, as you saw, you get it by writing a first integral, you have to make change of variable using, using a variable U equals log R over A, et cetera, but the final result is that the solution is 2 tau bar log R over A if tau bar is smaller or equal to 1, and the solution is 2 times log R over A plus log 1 plus tau bar minus 1 log R over A if tau bar is larger or equal to 1. You see that for tau bar equals 1, you get, of course, the same, the same solution, but this is for tau, so this is for weak charge on the cylinder, for weak linear charge, and this is for strong linear charge, so now, so the boundary condition as we, as we saw, since you have a finite amount of charges, one of the boundary conditions is that phi has to go to infinity at infinity when R goes to infinity, phi goes to plus infinity, and therefore we choose a boundary condition, the electrostatic potential you can choose as you want, the origin of, so we choose phi of A equals 0, and if we choose phi of A equals 0, it means that the concentration of counter ions at A is equal to C0. Okay, so this is the solution, and now the question is what is the charge, what are the ions which are bound, what is everything? So in order to get that, what you do is you do the change of variable, you have a first integral, and then you have in your, you have, when you do the integral, you have an integral which has a denominator with a square root of something plus a constant, and this constant is positive or negative depending whether tau bar is smaller or larger than one, and this determines, this gives a totally different behavior of the solution depending on the sign of C, right? This is what we saw last time, it's a bit, it's a bit complicated calculation, but I can, so it comes, all boils down to an integral which is, it's an integral which is dv divided by 2k square e to the v plus c, and then depending whether c is positive or negative, you will have the two different behaviors of this, yes? This is not the concentration, this is, okay, so this is where I come to now. Okay, so now the concentration c of r of counter ions, so when you have this, if you remember, in fact the c of r is c0 e to the minus beta e phi, and in fact, if you remember, this is equal to e to the minus beta e phi divided by some factor z, right? Because the integral of c of r should be one, or should be n, the total number of particles if you want, total number of counter ions, right? If we integrate so with, right? This is c of r, this is in Poisson Boltzmann, this is really the approximation. So now what I will show you is that when we are in this regime, the partition function is divergent, and therefore c of r is zero, which means that the concentration of counter ions, when the volume goes to infinity, the concentration goes to zero, which means that the counter ions are just filling up the whole space, whereas when you are in the limit of strong charge on the cylinder, the integral is finite, and therefore it means that there is a finite concentration at the wall. Okay, so the point is that this partition function is just if, so because there is translational invariance in the z direction, so I can put a factor, so if I look at one, one unit length in the z direction, this is just two pi integral d r, r e to the minus beta e phi of r from a to plus infinity, right? This is for one unit length of the cylinder in space. So now in this case, z will be equal to two pi integral from a to plus infinity d r, r, and then e to the minus beta e times two tau bar log r over a. Okay, I just replay, so is it clear that this is the normalization of this? The concentration of counter ions in the system is the Boltzmann weight. This is the Boltzmann weight of the counter ions. The counter ions are just charges which are floating around in a potential phi. So it's e to the minus beta e phi, that's the Boltzmann weight, and then you have to normalize it so that the total number is equal to n. So you divide by this so that if you integrate over r, this is one and this is n, no? What's, yes? Yes, because I said you take it over one unit length because you have this. So if you want, if you call l, call l the total length of the system, you have this factor l. But it's, the l is irrelevant because it's an infinite cylinder. So n itself, the n which is here will be also proportional to l. So I take l equals one, it doesn't matter, right? Okay, put l and l is equal to one, the unit length of the cylinder. Yes, you had another? It was the same as that question. Okay, yes, okay, so it's just, it's clear? Okay, so this, now this is just, proportional to r to the power one minus two beta e tau bar. Okay, so you know that now this integral at infinity, so which you write as one over r to the two beta e tau bar minus one. So this integral at infinity will, so at a equals zero, no problem, everything converges. But when r goes to infinity, this integral will converge. If and only if two beta e tau bar minus one is larger or equal to one. Sorry, without, no, for one, it's the logarithmic, the divergent. Sorry, for one, if it's equal, converges, yes, you're right, sorry. Yes, for equal one, it's divergent, it's logarithmically divergent, yes. Absolutely, okay, so the solution will be, so there is a critical tau bar which we saw which is beta e tau bar equals one. And, okay, so there is a mix up, yeah, so this, just a second, I see that, okay, there is a mix up in the units as usual, and the over e. So there, okay, so I'm sorry I made a mistake in the pre-factors, but the result should be tau star bar equals one and there is some factors that I, yes, yes, actually, oh, yes, yes, yes. Oh, yes, there is no beta e, absolutely, because it's beta e phi, thank you. That's the, so it's one minus two tau bar, okay. This is much better, yes, yes. Okay, so it converges if two tau bar minus one larger than one, and therefore there is a critical tau bar equals one. And since here we are in a situation where tau bar is smaller or equal to one, so converges only if tau bar is larger than one. But here we are in the situation where tau bar is smaller than one, and therefore this integral goes to plus infinity. This is what I wanted to do, okay? Because we are, this is the solution with tau bar smaller than one. So if z goes to infinity, it means that the c of r z is infinite. So c of r in this case is equal to zero, which means, doesn't mean that there is suddenly no ions in the solution. But the ions, if you have a big box like this, where here you have n over v ions, where n is fixed, because n is the total number of counter ions, it balances exactly the charge on the cylinder. So when v goes to infinity, this goes to zero. If there is no, everywhere, because there is no ions which are bound on the surface of the ion. So they are just floating around. It's just a gas of particle which are unbound to the cylinder. Now in the case here, is it clear? In the case here, the integral that we have to deal with is integral from a to plus infinity. So I forget all the factors, but the equivalent of the partition function integral is just the integral of e to the minus beta e phi, so it's drr times e to the minus 2 log r over a, doesn't matter, minus 2 log 1 plus tau bar minus 1 log r over a, which is just integral from a to infinity dr of r divided by 1 over r. This is r over r square. So it's dr1 over r and 1 over 1 plus tau bar minus 1 log r over a to the square. I use all the time that e to the log x equals x. So this is just, and now I don't make the mistake of putting back beta e phi. So it's e to the minus beta e phi. There's e to the minus 2 log. It's r over a if you want, but the a is irrelevant. It's just a constant. Minus 2 log 1 plus. So it's dr over r times this. There is a square here. So if you put u equals log r over a, the integral here, so dr over r is just du. So it's just an integral for r equal a. It's 0 to plus infinity du. This is du. And then it's 1 over 1 plus tau bar minus 1 times u square. And therefore, when u goes to infinity at large distances, this is finite. It converges because it goes like 1 over u square. You can do the integral, but the important thing is that it's smaller than plus infinity. So z is finite. And if z is finite, then the concentration here is finite, which means that C0 is finite. And C0 is just the concentration of mobile iron at the surface of the cylinder. Is it clear? OK, so from this, you can calculate a certain number of quantities, which is what I showed you last time. So which means that in this case, C0 is equal to 0. And here, C0 not equal to 0. OK, so the interesting thing is that the number of bound ions, so which is the total number of bound ions, is really the integral in all space of C of r. So in fact, the system, you can view it as a two fluid. You have a fraction of the counter ions which are bound to the cylinder. So the fraction below this, the fraction is 0. There is no, and then when you increase the charge of the cylinder, suddenly the counter ions start condensing on the cylinder. This is the Manning condensation. And they condense more and more as you increase the charge on the cylinder. So actually, it's increasing the charge on the cylinder or decreasing the temperature because the quantity, it's tau bar equals 1, and tau bar equals 1. It's E over LB. I mean, the critical charge is tau bar equals 1, which means tau in charge of the linear charge density, tau star equals E over LB. And E over LB is just 4 pi epsilon over E beta over beta. LB is beta E squared over 4 pi. Yes. So the charge density is 4 pi epsilon over E times T, temperature. So this is how it goes. That's the critical point. OK, so the fraction of bound ions is the fraction tau bar minus 1 over tau bar times tau over E. And OK, so if you look now the, so the interesting thing is if you plot the density or the concentration of bound ions as a function of the surface charge as a linear charge density, so it is 0. And then it goes like this, right? This is the, this is as a function of tau bar. And this is the fraction of bound ions. So it is 0 up to tau bar equals 1. And then it grows linearly according to this. Yes? How do you calculate this? This is just the result of the integral of C of r. The number of bound ions is just, the total number of bound ions is just per unit length. It's always per unit length. It's integral from A to plus infinity dr of r 2 pi. So this is the volume element, C of r. So as we saw, if you are below the threshold, the number of bound ions is 0. And if you are above the threshold, this integral is not 0. And you can calculate it exactly because essentially it's an integral something like that. And you can just do the calculation itself. And this is what you get. And the interesting thing now, which is the main result, as I said last time, is that so you have your cylinder which is charged like this with minus charges. And it will attract and bind a certain number of ions, of counter ions around it. But then there is a certain number of counter ions which are unbound. So there is only a finite fraction of all the counter ions which are bound. And if you look at all the apparent charge of the cylinder, which is the total charge of the cylinder minus the charge of the counter ions which are bound, so that you put your cylinder in the fluid with the counter ions, a certain number of counter ions. When you are above the critical point, a certain number of counter ions will condense. And the interesting result, a quite amazing result, is that the number of the ions will condense, the counter ions, in such a way that the total apparent surface charge of the cylinder remains equal to one, to the critical value. So if you write here, and of course the apparent charge or residual charge of the cylinder, so the residual charge of the cylinder is the total charge of the cylinder minus nb times b, if you want. The total charge of the cylinder is just related to the original linear charge density. And the number of bound is given by this. So if you do the difference of the two, the plot will look something like this, so you have the function of tau. So this is one. And the thing, so once you reach the critical point, the residual charge of the cylinder, which is total charge minus the charge of the counter ion cloud, remains constant equal to one, for tau bar smaller than one. Because the quantity that I write here, it's OK. So the charge, it's because nb is 0. OK. I write it as a function. I will write it. It's absolutely trivial. This is just, on this axis, it's just tau bar minus tau bar plus. So tau bar is the reduced linear charge density of the system. And tau bar plus is the charge of the bound ions. OK. So here, so let me plot it. So what I plot, this second plot is really tau bar minus tau bar plus as a function of tau bar. So tau bar is the linear charge density rescaled. And tau bar plus is the linear density, if you want, of the bound charges. So when you are below 1, tau bar plus is 0. There is no bound charges. So it's just tau bar as a function of tau bar. So it's just a linear straight line, right? And then it turns out that when you hit tau bar equals 1, then this starts to increase and increases in such a way that at this point it remains constant. Whatever you charge, so you charge your cylinder, you're just getting more and more counter and condensing. It's like when you have a liquid gas transition or so, you know, you have this plateau in the PV plane. You have something like this, right? So in this region, you have condensation and you are transforming the vapor into liquid. And the more you decrease the volume, the more you create liquid. So it's exactly the same kind of phenomenon. As you increase the charge on the surface or as you decrease the temperature, there is more and more ions which are condensing on the system. OK, so this phenomenon of ion condensation doesn't survive when you have salt. This is what I was trying to show you in the case of Debye Huckel. So in the case when you have salt, so now if you have your cylinder which is charged and you have salt plus minus, then it turns out that you have a screening length, lambda d, Debye length. And this Debye length is finite. It goes like 1 over the square root of the salt concentration, if you remember. And therefore, the interaction, the electrostatic potential, is never, the potential goes as we saw as e to the minus if you want r over lambda d divided by square root of lambda, the square root of r. So the potential decays exponentially fast. So you cannot, certainly the electrostatic potential doesn't propagate beyond lambda d, essentially. There is a scale lambda d beyond which the potential is not strong enough. And therefore, you cannot have this phenomenon of condensation because the phenomenon of condensation is really related to the fact that the potential is long range and is infinite range because it's logarithmic. In both cases, the potential goes logarithmically. And as a result, you don't have a condensation which is clear cut like this. But what you have, if you look at the concentration as a function of r, so both positive and negative ions, there is a kind of bulk. So if you're in a big system, there is a bulk concentration, let's say, which I call C0. And the counter ions will have a small overshoot like this. And the co-ions will go like this. So there is a small part which is attracted to the cylinder. And one can calculate within the bifurcated theory what is the typical size of the adsorbed layer and things like that. But it's not at all like here. So this is so far for the cylindrical case. I will spend a few minutes about the spherical case, the case of colloids. And then I take questions about it. And then that will conclude completely this Poisson Boltzmann by Huckel kind of theme. So spherical case, which is the case of colloids. So colloids are usually spheres with a fairly large radius, a. And they carry a charge q, which can be big because the object is quite big. So there is a fairly big charge. And this, of course, you can have many colloidal particles. And they are sitting in a sea of salt. So the big difference, the first big difference between these spherical particles and the cylindrical or planar case is that these objects are finite. The two others, the plane or the cylinder, are infinite objects. They have an infinite size. So they carry an infinite charge altogether. Whereas here, everything is finite. OK. So if we look at the charge, the interaction, let's say, of the salt particle with the colloid. So the interaction is phi. The potential is q over 4 pi epsilon r. And therefore, beta e phi, if you want, is q. So let me call it q. The total charge is q e. Let me call it q e rather than q, just for q e. So it's q e squared over 4 pi epsilon r. So it's q times lb over r, where q is the total number of charges on the, OK. And it can be plus or minus. So the partition function for one of these ions in the system is just integral d3r of e to the plus or minus q lb over r. So I write plus or minus because it depends if it's the partition function for plus or minus ion. And therefore, you see that the concentrations, so this function for any sign, when r goes to infinity, this goes to 1 because 1 over r. So this is infinite. Therefore, when you have a colloidal particle, you have a sphere. No particle can be bound to the sphere. So all the ions are floating around, but they are not bound. However, it turns out that if the charge q is large, so if q is much larger than 1, what happens is that there is a small layer. So if the q is negatively charged, if the colloidal particle is negatively charged, what you have is that you have a small layer of size d where you have counter ions, the ions opposite to this, to the colloidal particle, which are mostly localized, with very few because the charge is big. All the negative ions are expelled. And this means that if you look far away, there is a renormalization of the charge. And as a result, the Debye-Huckel, or the large distance electrostatic potential, which usually would look like phi of r created by q in an environment like that, should be q times e to the minus kappa d r divided by r. And this is what we saw that if you are in a system in a salt, the electrostatic potential created by a charge is this. So because of this effect that the charge q is large because it's a colloidal particle, so it's a particle. Colloidal particle is typically a fraction of micrometers. So it's of the order of 10 to the minus. It's of the order from 10 nanometers to 1 micrometer. Of course, you don't see it with the eye unless you have very, very good eyes. And it carries a charge which is quite big because compared to atomic dimensions, which are angstroms, it's quite big. So because of this large charge, in fact, this expression is modified. And phi of r takes the form when r goes to infinity for large r. So it's a q effective divided by 4 pi epsilon. And people like to put, so because the radius is, they put e to the minus k d r minus a over r, and they put an additional factor 1 plus k d a. And so q e, so this is, if you want the, it's the, this is just the by Huckel potential countered from r equals a. So anyway, this is just conventional, this form. And this defines q e. And q e is called the renormalized charge of the colloid. And it is the charge that you have to use when you start looking at phase diagrams of mixtures of colloids or things like that. You cannot use the real bear charge of the colloid. It doesn't work experimentally when you make experiments. If you try to fit with this q, it doesn't work. So the q has a cloud of counter ions surrounding it, although you're in salt. The cloud is important because the charge of the colloid is large. And therefore, it has an effective charge. And this effective charge is called the renormalized charge. So this is all I want to tell you about. Yes? Yes. Oh, yes, okay. Yes, yes. So the notation is stupid. I'm sorry. Call it q effective. Sorry. Or, yes, I'm very sorry. It's the effective. And, yeah, actually, it's even worse than that. It's, because as I defined it, it's q, so it should be qe here because q is the number of terms. And then it's qe times e. Sorry. So this is just the Debye-Huckel. So this is what you would expect if the theory, if a linear theory was okay, everything was okay. But it turns out that the linearized theory is not okay. If you look, if you do even simulations, numerical simulations, or if you do experiments, you see that asymptotically, for large distances, the potential created by a colloid is not given in this form. It has the right exponential decay, et cetera. It's just the pre-factor which is here, which is not correct. So the pre-factor is redefined in such a way. I mean, this is just, you see that the functional form e to the minus k dr over r is unchanged. So there is just this, which is modified, and this is just a kind of renormalization of the charge in the sense that your sphere is surrounded by opposite ions, which screen part of the charge. And at large distance, you don't see the bare charge of the thing, but you see the renormalized charge. Yes? Why? In this case, in here, because, okay, so I'm interested in the behavior, so this z plus or minus, so because of spherical symmetry, it's just 4 pi integral from a to plus infinity dr r squared, that's the d3r, e to the plus or minus q lb over r. So what I'm looking at is the behavior when r goes to infinity because all the rest is finite. So when r goes to infinity, this goes to 1. So it's the integral dr r squared. That's why it diverges like the volume if you want. Yes? Sorry? Yes? Oh, the size of the colloid, yes. But no, but this doesn't matter because of Gauss theorem. You know that if you have any object here, if you look far away because of the Gauss theorem, the divergence of e is equal to the total charge inside, so it's the total charge. The size of this charge doesn't matter, so you don't care about the size of it. I mean, if you start compressing and things like that, then, of course, you will need to have the apparent size, but if you just look what is the potential of interaction in a diluting of solution, then the size of the colloid, the apparent size of the colloid is not going to play any role. It's only if you start compressing that when the clouds will start to overlap. OK. So if you had some questions, maybe we can... Yes? Yes? So I assume that I have a system which contains, let's do it with one species, OK? So I have n ions with charge q, let's say, in a certain volume. So the probability, the Boltzmann weight, is the probability to find the particle at... So the p of r, probability to find one of these particles at point r, is e to the minus beta q phi of r divided by integral d3r e to the minus beta q phi of r. That's the definition, that's not the definition, so this is the probability. How do you go from probability to concentration? Yes. So this probability, it's... If you take a unit volume, you take the unit volume, it's the... Exactly. So then what I say is that c of r, yes, is just n e to the minus beta q phi of r divided by integral d3r e to the minus beta q phi of r. Up to now I guess that the integral of c of r in the whole volume should give us n. Exactly. No, yes, in the whole volume is n, exactly. No, I say that. I say that, and this is true if phi of r goes to zero at infinity. Or if your system is... If you have your system, let's say in a kind of membrane, so you have your fluid all over the place, and let's say the system is localized, then it's just that this integral, essentially if phi of r goes to zero, so of course this is true if you have finite objects and things like that, then this is just the volume. Right, there is correction from the finite regions where phi is nonzero, but it's... Yes, yes, and then you forget all this. We will see, actually, so this is really a mean field approximation. When we do field theory we will see that it's quite different, that what comes into the game is not, so it's not the concentration. So c, the concentration is n divided by integral d3r e to the minus beta q phi of r. But we'll see that in fact the natural thing is to work in grand canonical ensemble and everything is in terms of fugacity rather than... Exactly, yes. Any other questions? Yeah, go ahead. You wrote the entropy as a function of the concentrations. Yes, I didn't get the argument. I think I saw the same result for a simple system but I couldn't find it. Okay, so let me do it again. Yes. So actually this is a very important thing so we better... So the entropy in this kind of system, let's do it for the ideal gas. So the ideal gas, the partition function, if you have n particles it's one over... So there is something which I always, or not always but often forget, which is one over factorial n and this one over factorial n is because you have n identical particles and you should count each configuration in the configuration counting only once. Right, if you exchange particles it's the same question. So integral d3r... Oh, sorry, product over i equals one to n because you have n particles, d3ri and then e to the minus beta e, which is a function of all the ri's but if you have an ideal gas this is equal to zero. So the partition function is particularly simple to write because this is zero. If the volume of the system is v this is just... Each integral will give you a factor of v. So it's one over factorial n, v to the n. That's the partition function of an ideal. So this, you can write it... So sterling formula. So everybody knows sterling formula, which is the asymptotic expansion for factorial n. It's just n to the n e to the minus n and then... I mean, it's not equal. And then there is a square root of 2 pi n. There are corrections here, which I don't care. So you write it like this. So if I... So v, zn is just one over n to the n, v to the n, e to the plus n, which I write as e to the minus. So this term will give me n log n minus n log v minus one, minus n if you want. So this is just e to the minus n log n over v. So n over v is just a concentration. I write it as c and plus n. So this is when you have globally something and so. So now assume that your system... So now what you want to do is some kind of density functional theory. You want to coarse-grain the system and define locally a concentration. So assume that the concentration is not constant, but it's slowly varying in space. So it's not c, but it's a certain c of r. And the variations of c of r are shorted as... are slow enough that you can... OK. So from this, you can make a local theory. Actually, this is important because it goes well beyond the Coulombic system. It's true for... Here it's true for any fluid, any gas of particle. It's even true for polymers and things like that. And this, of course... OK. So zn, by the way, zn is e to the minus beta f. And f is u minus ts. So beta f is u. And u is zero in this case. Of course, there is no... Because the energy is zero. So the free energy is only related to the entropy. So beta f, it's minus k minus s over k, where k is the Boltzmann factor. So you get from this, this is just e to the plus s over k. So I will write that s over k equals minus n log c plus n. OK. So now I go to a local theory. So the local theory, instead of having a uniform c, I will have a c of r. And this concentration, c of r, will be such that the integral d3r of c of r will be equal to n, because that's the total number of particles of my system. But now I have a problem because initially this c, yes, OK. The problem I had already before. The problem is that c has a dimension. It has a dimension of a volume. So you don't like to write things like that in terms of dimensional quantities. So you want to write... So just a second. Before going to that, I'll write that s over k is equal to minus integral d3r c of r log c of r plus integral dr c of r. OK. So this is when I do things locally. This is exactly what I would write. And this is essentially what I wrote, except that now you need a reference state because when we looked at this, the entropy is always defined with respect to a state. So what I want to write is not the absolute entropy of the system, but it's the excess entropy with respect to a reference state. So the reference state will be a state with a constant concentration. And therefore I can subtract anything that I want, which is a constant. And so I will write that s of k is equal to minus integral d3r c of r log c of r divided by c0 plus integral d3r c of r minus c0. So when I do that, what I have done, what I have done is I have subtracted a term which is minus integral d3r c of r log c0 and a term integral d3r c0. So this term, it's a constant because it's just n log c0, but n is fixed so there is nothing. So I can add or subtract any constant. It becomes a reference. And this term is just c0v. It's also a constant. So when I insert this reference state, I have just shifted this total entropy by a constant so that if c of r is a constant equal to c0, the system has zero entropy. So that's the zero of entropy. And then if my c of r starts making variations or wiggling with respect to the c0, then I will have a delta s if you want with respect to this reference state. So this is what you use usually in the soft condensed matter as a... This is an... Yes? This one? This one? Sorry? Which one? Yes? Yes? Yes. Yes. But... Which is here? I think I understand because if we just substitute, we should have the integral of c r times the logarithm of... Yes. No, no. What I'm saying, it's not exact. It's just from this equation I try to make up a local entropy density which has the right properties, which in particular if c of r is constant, if c of r is equal to c, we'll give you back this. So this is the most natural way to do it. If you have... It means that if you're... If you have a big system, right, you have your big system and you look locally in a small volume here with the usual, you know, which is where the dimension d is much larger than microscopic dimension but much smaller than the total size of the system, within this small object, it's exactly like if you are somehow at equilibrium and if there is no interaction, et cetera, you can use this kind of formula within this and that's how you... Okay? I mean, you break it into small pieces if you want. You call this i, so it's sum over i of c i v, et cetera. So this is fairly straightforward and then you have this normalization by using a reference point where the entropy is zero. And so this is the fluctuation. It's a delta S if you want with respect to the completely uniform situation. Okay? Any other question? Yes? Because if it's an ideal case, the interaction energy between particles is zero and the internal energy is just the average of the, right? In the statistical physics, the internal energy is just the average of e over all configurations. So in an ideal gas, of course, in an ideal gas the free energy is purely entropic. It's the same with polymers. It's the same in many systems in soft matter. So this is why you have very often this balance between entropy and internal energy, which makes the system interesting. Yes? So the kinetic energy is just a... Okay. So the kinetic energy, of course... Okay. So if you want to go to... In detail, we can... Okay. If you want to really include the free energy, the kinetic energy, so the best way to do it is to go to the grand canonical ensemble. So the, okay. So the real partition function, as it should be written, is Zn equals one over factorial of n. Integral product over i equals one to n. D3 Ri, D3 Pi over 2 pi h bar. So is it to the 3 half or... I never know if it's 3 half or... Or 3, no? Sorry? 3. 3, yeah. Okay. And then the energy, the configuration of energy is 0, but then you have e to the minus beta over 2 Pi square over n. So actually, this can be an exercise for you. So the integral over Pi can be done. And this is just one over factorial n. So the integral over Ri is unchanged. It's still v to the n. And the integral over Pi will give you 2 Pi m over beta to the 3 half. And then for Pi square h Pi square to the 3n over 2. So it's m over 2 Pi beta h bar square to the 3n over 2 times v to the n over factorial n. Now, if you go to the grand canonical ensemble, to the grand canonical ensemble, so the partition function, if I define Z lambda equals e to the beta mu. So mu is the so-called chemical potential of the particles or whatever. So the partition function, the grand partition function Z of lambda should use a different notation, but it's by definition sum n equals 0 to infinity of lambda n by factorial n Zn. So let me call this as lambda 0 to the n, vn over factorial n with lambda 0 equals m over 2 Pi beta h bar square to the 3 half. Right? If I write it like this, then so I define lambda as e to the beta m times lambda 0, where lambda 0 is this. So lambda 0 is really the contribution from the kinetic energy of the particle, of the free of the ideal gas. So mu is the chemical potential. So this is my partition function. And then if I use Zn, so it's sum n equals 0 to infinity lambda n over factorial n. And this we saw was just v to the n. OK, with my notation, there is no factorial n, right? By definition, the factorial n is included here. So this is just e to the lambda v. And the grand potential, so Z of lambda as we saw, is e to the minus beta the grand potential. So the grand potential is minus lambda kT times v. Now, what is lambda? So lambda is such that the average number of particles in the system is lambda d by d lambda log Z of lambda. Yes? Sorry? Yes, OK, yes, absolutely. OK, so you're absolutely right. So let me write lambda e to the n beta mu lambda 0 to n Zn. OK? This is the chemical potential. This is lambda 0. So the definition of mu, you see that if I take the derivative of log Z of mu, let's say, either lambda or mu doesn't matter, by d mu, right? It's beta, so it's n. The average n is one over beta d by d mu of n. Do you agree? And this, yes? No, I look at this. This is the chemical potential. This is the lambda 0 which comes. So all this is lambda to the n divided by factorial n. Look, OK, what I want to say is that the chemical potential or the lambda, if you want, so if I look at it here, you see that we have also that expectation value of n is just lambda d by d lambda log Z. And this, so usually in a system what you have is the n, so usually it's divided by the volume if you want. But what is fixed is the average number of particles. So what you do is you solve this for lambda as a function of n, of average n. This is the n average. So this defines your lambda. So whether the lambda has this form or any other form, it's total value, so you cannot dissociate mu and lambda 0 because it all enters in the same together as lambda. And it's the product of the two which is determined by this equation. You see what I mean? So why did I say that? It's because in fact these chemical potentials and the effect of the kinetic energy in the partition function is very irrelevant. You don't care about it because once, because the fugacity is modified by this kinetic term, but eventually you write it like this. And the equation which determines this fugacity, this renormalized fugacity is this one. And whether it is separated as this times this or any other way doesn't matter. The only thing which you can determine physically is the product of the two. So just to say that you don't care about the kinetic energy. Yes. When you take, if you are interested in the, yes, in Boltzmann weights they disappear, if you are interested in distribution as a function of r, but if you are interested in distribution of r and velocities then you need both factors. But right, if you are interested in the probability to find a particle at a specific point and don't care about the velocity then the factors disappear and they are decoupled and they just don't get it. But it goes even beyond, even because you could think that the kinetic energy will play a role in the, I don't know, in the pressure or things like that, but it doesn't play any role. It's already contained in the configurational entropy of the system. Okay? Any other questions or, sorry? Sorry? Yes, so. If you have no energy, no. Even if you have energy because they are just decoupled. It doesn't, it never plays a role in the, at equilibrium, at equilibrium that's for sure. It never plays a role because it's just factorized. And the entropy really comes from the configurational entropy. It doesn't come from the velocities. It's funny because when you do kinetics of gas, if you take an ideal gas then you see that the pressure comes from the shock of the particles on the wall or on the piston. And so you would have the idea that it comes from a transfer of the kinetic energy of the particles of the gas to the wall. But in fact when you do the, at equilibrium, thermodynamic equilibrium, the entropy is purely configurational. That's a fact of life. Look, given that it's, yeah, any other question? OK. If sometime you have other questions, since it's maybe a bit dense, you can ask me and we can take sometimes, sometimes to review some of the questions. So now I will switch completely and show you a different type of description of these systems in terms of statistical field theory for Coulomb system. How many of you have studied field theory, quantum field theory or things like that? Yes? One, two. Most have never, most of you have never seen quantum field theory or statistic, you know? Don't be shy. It's not shameful. It's not shameful. OK. OK. So essentially this is what we will do during all this week, since I don't know if it's true, but it seems that many of you have never seen field theory. So if you have any questions or if I go too fast, you stop me and I'll try to explain. So the idea is that, OK, if you consider a set of charges qi at points ri, the partition function will be something like integral product over i d3 ri e to the minus beta over 2 sum over i not equal to j of qi 1 over 4 pi epsilon qi qj divided by ri minus rj. And if you have, in addition, you can have some fixed, so you have the charges floating around, you can have some fixed charges rho f of r being the charge density of these particles. So the interaction energy of the charges with the fixed charges will be minus beta sum over i integral d3 r of 1 over 4 pi epsilon qi rho f of r ri minus r. And you can have also, you can maybe take into account the interaction of the fixed charges with themselves, the electrostatic, total electrostatic interaction. So it will be minus beta integral dr dr prime 1 over 4 pi epsilon rho f of r rho f of r prime divided by r minus r prime. And with a factor two, with a factor two because it's the self-interaction of the fixed charges in the system. So that's the partition function. This is the total, and by the way, this quantity is a pure constant. It comes out of the configuration in integral because it doesn't depend on ri. It's fixed charges and these fixed charges have an interaction energy which doesn't depend on the position of the particle. So I will use this notation that vc of r minus r prime, let's say, is 1 over 4 pi epsilon 1 over r minus r prime, which is the Coulomb interaction between two charges one. And an important result of that is that Laplacian, and this is the Gauss equation. OK, so this is the partition function of the system in terms of particles, in terms of coordinates of particles. Now, we know that there is an electric field which is a field which exists in all space. There is also a charge density rho of r, which is, let's say, sum over i. So it's rho f of r plus sum over i of qi delta of r minus ri. So instead of having, of formulating the system in terms of integral over particle position, it would be nice to write the system, the partition function of the system, as an integral over all possible configurations of the electrostatic field or of the electrostatic density of particles. Right, and that's the idea of field theory that instead each configuration of particles here creates a certain electrostatic potential and a certain charge density. So the idea is that instead of integrating the exponential of the energy over all particle configuration, it would be to write the partition function as an integral over all electrostatic potential configurations or all charge density configurations. So there are essentially two ways to do it. I will show you one, and we will get to an expression of the partition function as a field theory over charge density or electrostatic potential or both. OK, so the first thing is to recognize if you look at the electrostatic energy, which is here. You can see easily that in terms of this charge density rho of r, it is just one half integral d r d r prime, rho of r vc of r minus r prime, rho of r prime. How do you see that? Well, it's very simple. You just replace. So this is my definition of rho of r. So rho of r is the charge density. So it is the charge density due to fixed charges plus the charge density due to particles. And you just plot it in here. You have the kernel interaction. So for instance, if there was, you see, you just replace. OK, so indeed, you see that if you replace, so the first term, for instance, this times this is just this term. Now, this times this plus this times this, they just did exactly the same value. And you reconstruct this term. And the last term, which you have here, so tells you that here you replace r by ri. Here you replace r prime by rj. And you reconstruct qi, qj. In fact, there is a small subtlety, which is related to this chemical potential thing, to this fugacity. When you write it like this, so this expression is really the energy up to one thing, which is that in the real free energy, in the real energy the summation runs on i not equal to j. But when you write it like this, you see that in this summation, it's really the last term that you get is one half of sum over i and j of qi, qj, vc of ri minus rj. And there is no restriction i not equal to j. So if we want really to have the correct free energy, we should subtract the term corresponding to i equal j, so minus one half sum over i qi square vc of zero. And vc of zero, of course, is divergent, but forget it for a moment. It's not a problem. It's just a constant which will be multiplying everything. So in principle, the correct expression should be that the total energy is this, minus one half sum over i qi square vc of zero. So to make things a little bit simpler, I will assume that there is only one species of ion. So one, and we will generalize very easily afterwards. So in that case, the energy is just one half integral dr rho of r v of r minus r prime, rho of r prime minus n over 2, n species of ion of charge q, q square vc of zero. So the partition function can be written as z equals. So all this is equal to this. So it's integral. So this is zn, product over i equals one to n. I have n particles, d3 r i of e to the minus beta over 2 integral d3 r, d3 r prime, rho of r. And then in front of here, I will write e to the beta over 2 q square vc of zero to the power n. So where rho of r is defined in terms of the r i as sum over i as q sum over i delta of r minus r i. That's the definition of n plus rho f. So how do I go? Now what I want to do is instead of having an integral over r, I want to have an integral over rho over the charge density configurations, fluctuations of rho. So what I have to do is to do a change of variable. So what I will do is I will enforce this as a constraint at each point in space. So zn is e to the beta over 2 q square vc of zero to the n. So this factor, I will not write it later. So it's integral product over i equals 1 to n d3 r i. And then I will write product over r at each point r of space, d rho r. So at each point of space, I introduce a variable delta of rho r minus this. So minus q sum over i delta of r minus r i minus rho f of r. And then e to the minus beta over 2 integral d r, d r prime, rho of r vc of r minus r prime, rho of r prime. Here I have done absolutely nothing. I just introduced one because integral d rho r of this is one. So at each point of space, I introduce a new variable, rho r, which I will later write as a rho of r. But just to, I put it as an index now. But in fact, it's like this. Everybody follows. OK, so then I will use the Fourier representation of the delta distribution. So I will use that for each point delta of rho r minus q sum over i delta of r minus r i minus rho f of r. So I will use the Fourier representation. So the Fourier of the delta function is, for instance, delta of x equals integral dk over 2 pi of e to the i kx. Right? Yes? Yes? If you have a configuration of r i, the configuration of points, then this defines the variable rho r. Yes? Yes, so yes, of course. You just integrate over rho r, it gives one. So this is obvious and I could put, so I will use it in a slightly different form. You see that it's integral dk over 2 pi e to the i beta kx if I put beta here. OK, I can put any factor in front. It's not going to make any difference. So then I introduce, so k is called the conjugate variable of x. So I introduce a variable phi r to define this Fourier transform. So it's a product over r, integral d phi r over 2 pi, so with a factor beta. As you will see, all these pre-factors are totally irrelevant because anyway they diverge. And then I have e to the i beta phi r rho r minus i beta phi r q sum over i delta of r minus r i minus rho f, plus rho f of r. OK, so I just use the Fourier representation of the delta function and this is a product over r, so this is true at all points in space. Now, as I told you, you can put, in fact, I put a factor of beta here, but you can put any factor you want in front of the delta function. So if I assume that I have, that I discretized my space, the total volume of the system, to define the various points r at which you put things, and if I call A the lattice spacing, instead of writing it like this, I could have written it as product over r beta A cubed, integral d phi r over 2 pi e to the i beta A cubed phi r rho r minus i. Right, I just, I mean the new variable instead of being phi r is beta A cubed. I mean it's just a simple rescaling due to the, you can do this rescaling, of course, because all these integrals which define the Fourier transform of, the Fourier transform representation of runs from minus infinity to plus infinity. So any rescaling of K is not making any difference. Is it clear up to here? No? Okay. Sorry? Here, yes. Yes, this is one. The integral, integral dx delta of x minus x0 is equal to one. So when I add this term, I don't change anything. The integral of a rho r is just equal to one. So if I do the integral, I still have a rho of r, remember it's given by this. So it's not going to make any difference. It's a way to force a change of variable in the integral. No, there is no Jacobian because I express, I have explicitly rho r as a function of the other variables. It's not a function of rho r. It's really the definition of rho r. It's really like this. What you're thinking if I wrote dx delta of some function f of x, then I would have a Jacobian. But here I don't have a Jacobian because I use it really in this form, right? The new variable is rho r defined as a function of the other ones. So the integral d rho r is one. No question. Is it okay? Clear? Okay, so now this is a product over r. The partition functions z. I want to, okay, no, before writing it like this, I want to concentrate on this term still. So this term, there is a certain constant, which is these products of terms like this. Now I take, I have factors of a cube and sum over r, right? Because it's a product at each point. So you have a cube sum over r of phi r rho r, and here a sum over r of phi r, et cetera. So when you have a cube sum over r, this is really identical to integral d3r in the limit when a goes to zero. So I have discretized the system to be able to make this change of variable at each point. Now I take the limit, the continuous limit, a going to zero. And this term, which is there, is just equal. And you write it like this. It's d phi of r. This means so there is some constant in front of it, but it's really all, so you can view this, when you do this integral, each configuration is a certain configuration with a given value of the field in each space, in each point of space. So this integral is an integral over all configurations of the field, of the electrostatic field. Is it clear? Right? If you take at each point a certain, this integral is an integral. It's a sum over all configurations of the field where the field phi r takes any value at each point of space. So then this expression, which is there, can be written as e to the i beta. So since it's a product over r here, it means that everywhere here is a sum over r. So it's e to the i beta integral d3r phi of r minus i beta integral, sorry, i phi r rho of r. So the rho r becomes a function in all space when you're in continue. So it's i integral phi of r rho of r minus i beta integral d3r phi of r times qi sum over i delta of r minus r i plus rho f of r. This is just the delta function, which is here. When you go to the continuous limit, because it's a product over r here, you can make a sum in your exponent. So you have i a cubed sum over r of phi r rho r, which gives you the first integral. And then here you have minus i beta a cubed sum over r of this, and this gives you this. Yes? Yes? It's product of integrals, but there is nothing which couples the integral. So it's like if you have product over r integral d phi r. Let me take just the first term, e to the i beta a phi r rho r. This is just, if I make the product of all these integrals, the integral product over r d phi r, and then e to the i beta a sum over r phi r rho r. OK? So this is what I call the curly d phi. This is an integral. For each configuration you have a field which has different values. You have to integrate over all possible configurations of the field phi of r in a 3D space. And then this sum times a, it's a cubed. This sum becomes the integral in the limit a going to zero becomes the integral d 3 r of phi of r rho of r. So this is a very, very standard technique to go from a particle description in statistical physics to a field description to a field theory in statistical physics. This really, it's very general. You can do it with any theory, any system. OK. And this can be further simplified as, so this d phi of r is equal by definition. There is some constant factors like beta a cubed 2 pi at each point of space. So it's a product on each point of space of this factor, beta a cubed over 2 pi. But this is just a constant which in general is going to be an infinite constant. But as you know, you don't really care because the partition function is just a normalization. It's always divided. So this multiplicative constant that you have in the partition function just cancel out when you calculate Boltzmann weights. So this is not a problem. So let me say there is a normalization, infinite normalization constant. And then product at each point of space of d phi r, the value of the field which varies from minus infinity to plus infinity at each point of space. Is it clear? So we are almost done. The problem is that I will need, OK. So I continue here. So there is this factor. I will come back to it later. So I have, so this term is really what I wrote here. And this product over r d rho r is also in the continuous, if I take a continuous limit. So I'll write something to tell me if you agree. So there is this factor which I don't write. This is d rho of r d phi of r e to the i integral d3 r phi of r times. So I will group this term and this term. And then I have this term and this term. So I can do the r integral. So there is still this integral product over i equals 1, 2, n d3 r i. And this term, if I integrate it, is just e to the minus i beta q sum over i phi of r i, right? So this is e i beta q and the integral d3 r. It gives me just the value of phi at point r i. And then I have this term, which is e to the minus beta over 2 integral d r, d r prime of rho of r vc of r minus r prime rho of r. So we are almost done. You all seem, is it complicated? It's clear. OK. Very good. So this is just going from here to there. So this integral now, there is this specific integral, which is completely decoupled on all r i. So this is exactly the same. Where should I write it? It's this integral. It's just like integral d3 r e to the minus i beta q phi of r to the power n, right? Because it's n times the same integral because it's a product. So I get the final result, which I will write here, which is that the partition function. So this is zn with n particles. So zn is an integral d rho r d phi r. So an integral over all field configurations, all charge density configurations, all electrostatic potential configurations. This is not the electrostatic potential, as we will see, but it's related to it. So I write it like e to the minus beta over 2 integral rho of r vc of r minus r prime rho of r prime, plus I forgot, I'm sorry, I forgot the beta here. I forgot the beta. This is n visible from here. Ah, OK, OK, OK. I'll try to make it visible. I'm a magician. OK, this one. This one was the, yes. So z equals integral d rho d phi e to the minus beta over 2 integral d 3 r plus I beta times integral d 3 r e to the minus I beta q phi of r to the power n. Is it readable? You can move over. OK, and now last thing, one more step. So this is the canonical partition function. But what we will use always in the following is not the canonical partition function, but the grand canonical partition function because it has a much better form. And the grand canonical partition function is expressed as z of lambda where lambda is the fugacity sum from n equals 0 to infinity of lambda to the n over factorial n zn. And so when you have this expression, so if you take this expression, you multiply by lambda to the n divided by factorial n. You see that this just amounts to exponentiate this term. And therefore the partition function z of lambda, and I will stop here, is just an integral d rho d phi e to the minus beta over 2 integral rho of r vc r minus r prime rho of r prime plus i beta integral phi of r rho of r minus rho f of r. And plus lambda integral d3r e to the minus i beta q phi of r. It comes from the fact that sum from n equals 0 to infinity of lambda n over factorial n times this integral e to the minus i beta q phi of r is equal to e to the lambda integral. And this is, OK, so I will start again from there next time. But this is one of the first basic forms for the partition function of a Coulomb gas expressed as a field theory. Any question about, is it difficult? There are many steps, but they are very straightforward. There is no subtle, OK, so next time, which is tomorrow, we continue on this and show how this can be simplified and the form it takes for various systems, et cetera.