 Let me know, oh yes, you have to circulate this page. Any question about last time? Okay, everything was clear. So last time I started discussing the case of one charged plane with counter ions. Before that, I see that to be complete I will come back to one thing, which is so we store how to derive the Poisson Boltzmann equation. So the Poisson Boltzmann equation tells you about the distribution of ions in a system at thermodynamic equilibrium. But there is an interesting question about transport and about dynamical processes in a solution. So what happens, for instance, if you have a current, if you have a flow, if the ions are mobile and you are not at equilibrium? And I want to discuss this rapidly just to show you the equation and it's called the Poisson Nernst. So this is really for the dynamics and I will discuss it in the simplest case where the fluid, you have a steady fluid which is not mobile and you have ions in your fluid plus or minus charges which are moving in any direction. So this is very relevant to batteries, cell fuels, fuel cells and also to transport of charges through ion channels in biology and things like that. OK, so the starting point for this is the Langevin equation for the ions. So the ions I will assume as usual that there are Nk charges Qk in the system. And if I look at the equation of motion in a fluid of charge of type K, so there are, let's say, M species like this, so the equation is given by the Langevin equation. So the Langevin equation is essentially Newton equation. So Newton equation is just Ma acceleration equals the force. Now you know that when you are in a fluid at a finite temperature, thermal equilibrium, there is friction and there is a gaussian noise. Did you see, did you study Langevin equation? OK, so if you are in a fluid which has a certain viscosity, I will write the equation for the viscous and for the overdamped case. And the overdamped case is when the mass term is very small compared to the friction term. So I will write the equation in the following way. It's exactly equivalent to this. So I will write the overdamped Langevin equation in the following form. I'll write it as drk by dt equals drk by dt d beta frk plus the noise eta k of t. So d is the diffusion coefficient and it can be dependent on the particle species. It doesn't matter. So dk is called the diffusion coefficient. And it's related to the friction coefficient gamma k by the so-called Einstein relation, which is essentially that gamma k equals 1 over beta dk. OK, so that's the relation between the friction coefficient and the diffusion coefficient. Yes? Does this f function have to be the same for all k? No, it can be. But f, OK, so f is the force. It's the force acting on particle k. Now, we will study ions in a fluid. So essentially what we will take for f is the electrostatic force, the electric force. And so it is the same nature, but at different points, right? But it could be, of course, this equation is valid. Each particle has a different force. And the force, of course, can depend on the position of all other particles. Everything is coupled together. But we will see that in the Nernst-Planck equation, it's a kind of mean field approximation, a dynamical mean field approximation equivalent to Poisson Boltzmann in the other case. So for the moment, I will assume nothing, just this. OK. OK. So you know that when you have, and of course, the noise eta k of t is a noise. It's a white noise. So it's a random force. So a random force which has the following property that the average of the force is 0, right? And the variance of the force, or let's say, is equal to 2 delta kl dk delta of t minus t prime. So this is really the standard, standard Langevin equation. The noise is a white noise. So it's a Gaussian noise. It has a Gaussian distribution, probability distribution with zero average and correlation function, which tells you that the noise for particle k and particle L are decoupled because the noise comes, of course, the noise which is here essentially represents the collisions of the particle with the particles of the fluid with the particle of the solvent. And the noise is not correlated on time, so you have this delta function like this. So this is really classical, I guess. You probably saw this in your classes of statistical physics. So the important point is that this is a stochastic equation. And therefore, rk, because it's a stochastic equation because eta is a random noise. It's a stochastic variable. And therefore, the position of rk is not really deterministic. It depends on the history of the noise. And therefore, the position of the particle rk is determined by there is a probability distribution p, probability to find the particle at point rk at time t. And this probability distribution function satisfies a Fokker-Planck equation, which is d by dt of p rk t equals, or rather than writing it like this, I would rather write it like this, pk of rt is the probability to find particle of type k at point t. So dpk by dt equals d, so d index k, d by dr of dpk by dr minus beta Fk of rk times pk. Fk of r. So this is the Fokker-Planck equation. This is classic. Have you studied this kind of dynamics, the Langevin equation and the Fokker-Planck? Yes. OK, good. So this is a kind of diffusion equation. And Fk, so Fk of r is the force acting on particle of type k at point r, yes? Not because of any other ion. Sorry? So are we assuming that when you have an ion of type k, that the random forces are only due to fluid not because of any other ion? Yes. But that's the standard thing, that when you have many different kinds of particles, they are not at the same position. So you assume that there is no correlation in distance and it's an assumption that there is no correlation for the noise at different points in space and no correlation of the noise at different times. And therefore the what? Data k between different types of particles. So there is no correlation between particles, between plus ions and minus ions or so, for the noise. It's about the noise. It's not about the... Exactly. No, no. The force, the force, it's an electrostatic force so it includes all the interactions between all the particles. But the random noise comes from the fluid, from thermal collisions and therefore if you have a particle here and a particle here, they feel forces which are uncorrelated. OK? Sorry? Why? Yes, but they accept if they are very close by, if they are at, let's say the correlation, the noise, the distance, the correlation distance of the noise is very small. So if your ions are not too concentrated, the typical distance between the ions is much larger in general that the correlation of the noise, the correlation distance of the noise. But usually it's an approximation that is quite satisfied. The concentration of ions is very low and the correlation of the noise is very, very small. Very small compared to the typical distance between the ions. But this is not, it's not a strong assumption. OK. So this is now where the approximation come into the game. The principle fk of r is the sum of all Coulomb interaction between the particle k and all the other particles. And this you make a kind of mean field approximation by imagining that fk of r is qk the charge of the particle times the field ek at point r. And what you, or the not ek, e at point r, and what you write is therefore that this is minus qk gradient phi at point r. And therefore you have this equation, the Fokker-Planck equation, which becomes d by dt pk of r and t is dk d by dr. d pk by dr plus beta qk gradient phi of r pk of r. So this is for the probability distribution function. And this equation is linear. So if I multiply pk by nk, where nk is the number of particles of type k, then you have this relation that the concentration of type k at point r times t, that's just nk times ek of r and t. And therefore you have the equation, which is the Nernst-Planck equation, which is that d ck by dt equals dk. So usually people write it like this as gradient ck plus beta qk ck gradient phi. So this is the Nernst-Planck equation. It tells you how the concentration of iron varies in time, in presence of charges in the fluid, et cetera. And this equation, of course, has to be supplemented by the Poisson equation. And the Poisson equation is epsilon Laplacian phi with this phi because, of course, this equation relates ck to phi. You want another relation. To close this, you want an equation which will relate phi to ck. And the equation which relates phi to ck is just the Poisson equation, which tells you that minus epsilon Laplacian phi is the concentration of fixed charges plus sum over k, sum over all charges of qk ck of r. So this is the second equation. So this closes the set of equation between ck and phi, phi and ck. And this set of two equations is called the Poisson-Nernst-Planck equations. Is it clear? So starting from a certain configuration of charges, ck, so you can assume that you, I don't know, you have a certain electric field, you have a certain initial configuration of charges, and then you can just solve this equation numerically. There are essentially no analytic cases which are exactly soluble except maybe some trivial cases without, but people use extensively this equation numerically to calculate currents and distribution of currents and distribution of ions in ionic fluids which are circulating or ionic liquids. OK, yes? So pk of r is the probability to have a particle at point r. If you multiply by the total number of particles of this type in the system, that will tell you how many particles of type k, the density of particles of type k at point r, right? If you have the probability, you multiply by nk, that will give you the concentration per unit volume. It's normalized. You see, if this probability, if you integrate it, it's equal to one. If you integrate over r, it should be equal to one. So concentration is the total number of particles that you have in the system. So the probability, if you want, it's one over n the concentration. Yeah, it's always a density of probability. Yes, of course. When you're in continuous phase, when I say probability, it's a probability density, of course, so that the integral of pk of r, d3r, is one at each time. It's a density of probability. And if you multiply by nk, of course, it will be nk, and that's the total concentration of particles of type k at point r. It's the local concentration. Is it clear? Okay, so this was just, of course, I don't want to dig too much into this. It's just because it's an important equation if you happen to work in electrochemistry or transport in membranes and things like that. Just a last remark, which I will write here or here, which is if you're at equilibrium. So this equation, of course, takes the form of a continuity type of equation, which is dck by dt plus divergence of jk equals 0, where the current jk is minus dk times gradient ck plus beta qk ck gradient 5. Okay, so there are arrows everywhere. So this is the form of the transplank equation. It's really like a continuity equation, a conservation equation. It's the conservation of charge, really. And this equation, if you're at equilibrium, at thermodynamic equilibrium, then thermodynamic equilibrium, by definition, is that you have no current in the system. So it's jk equals 0. So jk equals 0 means dck by dt equals 0, but it's even more general. You could have a constant current, which is a stationary regime, would give you dck by dt equals 0, but thermodynamic equilibrium is really jk equals 0. And jk equals 0 is d gradient ck equals minus beta qk ck gradient 5, which you can write if you divide by ck. So you can integrate this equation trivially. This is a log of ck. This is minus beta qk gradient 5, minus beta qk 5. And since you need a constant of integration, instead of writing ck of r, I write ck over ck0, where ck0 is a constant. And therefore the solution, the equilibrium solution for this equation, you get back that ck of r is ck0, which is a constant, times e to the minus beta qk 5 of r, which is just the Boltzmann distribution for the particle in the field. So this Nernst-Planck equation is consistent with the Boltzmann distribution. And the Boltzmann distribution is the equilibrium solution of the Nernst-Planck equation. OK, no more questions. So this equation can be generalized to the case when there is a flow of the fluid. So in other words, if you have, let's say, if the solvent is flowing in the pipe and you have ions inside, so there is a, the Nernst-Planck equation takes a more complicated form, which I am not going to discuss, but there is all kind of generalizations of this Nernst-Planck equation. OK, so this closes, this was just to show you how to do mean field theory for ionic fluids in mean field theory, the equivalent of Poisson Boltzmann. Yes? No. Then, of course, first of all, there is a, so the equation is quite changed because there is a term with the velocity of the fluid which enters explicitly. And usually the simpler form is that you give the flow of the fluid. So the fluid is defined by a certain velocity at each point in space, and then you include it in the equation. But then the solution is not Boltzmann, of course. Boltzmann is really an equilibrium, thermal equilibrium. OK, so now I go back to what we started last time, which was the planar case. So can I erase? Yes? OK, so the planar Poisson Boltzmann equation, as I told you, before, so you have a plane here and you assume that it is charged. It is impenetrable, so the ions cannot go through it. And it has a charge, sigma, negative charge. It could be positive, but I take it negative. And then you can have plus ions, minus ions, which are floating around, so it can be either a salt. But the simplest case that we'll start, that we'll study in the beginning is the case of counter ions. So counter ions is the case where you have only positive ions. No, it's not a salt, it's just ions which are opposite to this. So by definition, counter ions are ions opposite to the object that you're studying. So for instance, if you study polyelectrolytes or polymers, if they are positive, or if you take DNA, DNA is negatively charged, it's a negatively charged polymer. And when it is surrounded by protons, by H plus, it's H plus are the counter ions. OK, so if we write the Poisson Boltzmann equation for counter ions only, the equation takes the form Laplacian phi, or minus Laplacian phi, equal rho f over epsilon plus, so I will call it C0E over epsilon e to the minus beta E phi. There is only one species of ion, and C0 is some constant. OK, so the first thing, of course, if we have only counter ions, if we have only counter ions, what we see is that here you have a certain amount of charges on the surface, so if you have charge neutrality, it means that here you will have the same amount of counter ions to neutralize so that the total charge of the system is zero. As a consequence, it turns out that at infinity, so C plus, if I write C plus the count of r is equal to zero when r goes to infinity. Because if it went to a finite value, then by integrating C of r over, if it went to a certain C0, then if you integrate integral C plus of r, d3r, this would be proportional to the volume, and therefore it would be much larger than what you have on the surface. So this is just, but we will see it anyway, just to show you that as a result, if the C goes to zero at infinity, it means that phi must go to plus infinity at infinity, which is unusual. As I said, most of the time in a system at infinity phi goes to zero, but this is when the solution is neutral, but here the solution is not neutral. You have only counter ions, so you have plus ions, and therefore it's a different system, and we will see what happens. Okay, so just to announce what happens. So as I said before, if you look at this equation, the fact that phi goes to infinity, okay, so you understand that the concentration of counter ion has to go to zero at infinity. It cannot go to a constant because if it went to a constant, when you integrate, it would essentially, if it went to a finite value, essentially at the end, this would behave like this, or something like that. And V is much larger, the volume, it's the area, so it's the area of this, if you want, times a certain distance. So this would be, and the distance, L goes to infinity, so this would be much larger than the sigma A, than the total charge that you have on the wall. You agree on that? So C at infinity, C must go to zero. But C, if you remember, C is directly proportional to e to the minus beta e phi. So if it goes to zero, then it means that phi must go to infinity, that's one. But we will see that phi indeed goes to infinity. Okay, so because of the symmetry of the problem, so I will call this z, of course, and x, y. So this is a plane, really, and x, y are the two directions in the plane if you want. So this can be x or y, sorry, x, y, z. And this is only a function of z. And rho f of x, y, z is just sigma delta of z. So the Poisson-Boltzmann equation becomes a normal ordinary differential equation, which is minus d to phi by d z squared equals sigma over epsilon delta of z, plus C zero e over epsilon e to the minus beta e phi of z. Okay, and the boundary condition here, you obtain it by integrating this equation between two points which are very close to the wall, one on the left, but on this side you have phi prime equals zero. The electric field on this side, this is a metal, for instance, this or whatever you want, it sits at the equilibrium. So the electric field inside is zero. And so you have, if you integrate here between, let's say, minus a and plus a, and you let a go to zero, so if you integrate, this will give you minus phi prime of a minus phi prime of minus a equals, so the integral of this is sigma over epsilon, and this is of order a. So when a goes to zero, it's very small. And we have the boundary condition, which is, so this is zero, so I write it here, so the boundary condition, so the equation, I can write the equation as minus a phi second, let's say. And I forget about this because this is just a boundary condition term, so it's minus phi second equals c zero e over epsilon e to the minus beta e phi of z, and the boundary condition is, so this term is zero, so it's minus phi prime of a, so phi prime of a, and a goes to zero, it's phi prime of zero on the right side equals minus sigma over epsilon. OK, so at this stage, you can do the integration very easily by, so you multiply both sides by phi prime, so phi prime phi second is the derivative of phi prime square over two, so you can write that d minus sigma square over two, OK, the derivative of phi prime square over two is phi prime phi second, and equals c zero e over epsilon, so you multiply by phi prime, so phi prime e to the minus beta e phi of z, this is minus one over beta e, d by dz of e to the minus beta e phi, so minus beta over e, so it's a c zero over beta epsilon, d by dz of e to the minus beta e phi of z, so this equation can be integrated trivially to give minus phi prime square over two equals c zero over beta epsilon e to the minus beta e phi of z, plus a constant. Now, the constant is quite simple because if you go to infinity, if the system is stable at infinity, so it means that the electric field at infinity should be zero. Now, we saw that at infinity you cannot have charges at infinity, so this is the density of charges, the concentration of charges at infinity, so it's also zero plus a, which means that a is equal to zero. There is something which is a bit weird because this is a bit negative. I forgot the minus sign. I forgot the minus sign here because that was a bit annoying. There is a minus sign here, and therefore we have this relation which becomes that phi prime equals square root of two c zero over beta epsilon e to the minus beta e over two phi. When you have this kind of equation, you know that you write d phi equals d phi by d z, and so you write d phi on one side. I give you the solution. You just write this as d z square root of two c zero over beta epsilon d z equals e to the beta e over two phi d phi. So, equals two over beta e d of e to the beta e phi over two. And, sorry, was okay. And so you can again integrate it. I will just, okay, so we have the boundary condition here. So, you integrate the equation, and the solution is just square root of two c zero over beta epsilon times z minus z zero, let's say, equals two over beta e exponential beta e phi over two. There is a, I just, a question? I just realized there is something which is a bit annoying in the notation that e is the electronic charge, but it's also the exponential. The e which is here is not the same as this one. Be careful. I'm sorry. Yes. So, when you print, when it's printed this one, this one is in Roman character whereas this one are Italic because it's mathematical character, but I cannot do that on the blackboard. Okay. And, okay, so this is the equation which gives me, so if I put everything back, so e to the beta e phi over two equals square root of beta e square c zero over two epsilon times z minus z zero. And, I remind you what we define as the B-room length is beta e square over four pi epsilon. So, beta e square over epsilon is four pi s, is four pi m e. So, this is just square root of four pi lb over two. So, it's two pi lb c zero z minus z zero, which gives you that beta e phi over two equals one half log two pi lb c zero plus log z minus z zero. Okay. So, now, if I want to apply the boundary condition, so, or I'll write it as beta e phi equals this plus twice. Now, the boundary condition, so, there are two things to determine. The first thing, so, I write that beta e phi prime is equal to two over z minus z zero. And, if I go at here, so, this will imply that beta e, if I go at phi at z equals zero, so I get beta e phi prime of zero, which is minus sigma over epsilon beta e equals minus two over z zero. So, that gives me that z zero is equal to epsilon to two, sorry, two epsilon over beta sigma e. And, therefore, it is negative because I remind you that sigma is negative, so it's minus two epsilon over beta sigma e. So, this quantity z zero, sorry, with a, okay, it's just a matching of boundary condition. So, I define a quantity which is called lambda g, which is the GUI Chapman length from the name of the two physicists who found the solution of the planar case. So, the GUI Chapman length is given, so this z zero is a length, of course, because it's z minus z zero, so it's homogeneous to a length. And, this length, lambda g is equal to two epsilon over beta sigma e. Yes? This one? Yes. No, I just say that when I have this, this is a differential. So, I can integrate, yes, between whatever you want and whatever you want. So, there is a constant, if you want. The integral of this is z is square root of this plus a constant. So, the constant I call, I write it in this form, that's all. Yes, because there is only one constant in the end. Okay. First of all, you will see it here already, because that's the solution, right? Whatever z zero is, you see that when z goes to infinity, the potential phi will go exponentially, will go logarithmically to infinity. Yes, phi goes to infinity when z goes to infinity. The reason, the physical reason, as I say, is that this is essentially proportional to the concentration of ion, of counter ions at point z. This concentration must go to zero when z goes to infinity, because you want, you see, you have minus charges, and you want to neutralize exactly this. So, you see, if you have a salt with plus or minus charges, then you can have a finite concentration of salt at infinity, because the salt is neutral. So, there is, you see, the charge, we will see that when you have a salt in the solution, you can have, you can attract the plus ions around here. You have the minus, and then at infinity, you have plus and minus, and you can have a finite concentration of plus and minus. So, you have a finite bulk concentration of salt when you go far away. But if it's pure counter ions as it is here, necessarily it has to go to zero. Yes, exactly. So, the potential at infinity, in the case of counter ions, is just because you, you attract to the wall only the necessary quantity to neutralize the surface charge, and then you cannot have steel ions. Right? Exactly. Yes, because I think, no, it's not infinity. It's not zero. It's, the concentration is zero. The concentration of counter ions is zero. No, because if it was zero at infinity, it's zero at infinity. No, because if it was zero at infinity, the concentration of ion would be finite. The concentration of counter ions would be finite, and that's not possible, because you would have too much. You cannot, you see? You cannot compensate. If you have a salt, it's different, because in a salt you have plus and minus in the solution. So, they compensate each other exactly. So, you have no excess charge. Right? At infinity, if you go locally at infinity, the concentration of charge is zero in a salt. But if you are in a, in a counter ion. Yes. And less particles. Yes. One can think that you have a sum of one over half to take as a salt. But they, the distances between particles increases so it becomes zero. Yes, but if you want to say that a particle tends to go to a larger potential, to a less, less, smaller potential, then you can say zero and minus one minus two. No, but the point is that you see what you say is, if you take the function one over x, for instance, the integral of one over, one over x goes to zero at infinity, but the integral diverges. And the integral diverges how far you go to zero. So, this is exactly the case where the potential doesn't decrease fast enough. So, okay. Thanks. And actually, so this is it. And so, the potential takes the form that beta e phi is equal to log, I write it as z. So, minus z zero is this lambda g, right, from this. So, z plus lambda g times the plus log 2 pi Lbc zero. Yes, yes, I will come to that very soon. Yes, so this length actually, the physical meaning of this length is, we will see. Just a second, I'm looking at one thing. Okay, so I'll come back to this length in a minute. So, the solution is this. And therefore, if we look at the concentration of ions, of counter ions as a function of z, so the concentration c of z is equal to c zero e to the minus beta e phi. And you see that is equal to c zero. And e to the minus beta, sorry, I made a mistake. There is a factor of two which I forgot, right? I forgot this factor of two. Beta e phi is log of this plus 2 log z minus z zero. So, it's c zero and e to the minus beta e phi is e to the minus log of 2 pi Lbc zero minus 2 log z minus z zero. And the exponential of a log is a function itself. So, exponential log like this, so it's c zero divided by 2 pi Lbc zero. So, it's 1 over 2 pi Lbc of z. And this is one. And z zero is, I write it as plus lambda g. So, 1 over z plus lambda g square. So, this is the ionic profile of the charges near the wall. Yes? Yes? From the equation from the potential. Yes, this one? I write that no matter how few positive charges you put, nevertheless you get a positive potential at plus infinity. Is this right? I mean, it's like. What you mean as? I mean, if you put a very low concentration of positive ions, you still get a potential that goes to plus infinity at plus infinity. Yes, so in fact, yes? No, it turns out that the system will always, the system will put as much counter ions as necessary to neutralize. You see that the concentration which is here doesn't depend on c zero. It's totally independent on c zero. And the reason is, in fact, if you take the Poisson-Boltzmann equation, initially it was phi second equal over epsilon, let's say, equals what was it? C zero e to the minus beta e phi, right? You see that I can, and with the boundary condition, which was phi prime of zero equals minus sigma over epsilon. So you see that I can include this c zero. If I redefine phi, OK, if I write beta e psi equals beta e phi minus log c zero. So it's just a shift in phi, right? I define this new psi. So you see that the equation for psi would be minus psi second over epsilon equals e to the minus beta e psi, right? I absorb the phi, the zero. And the equation here would be psi prime of zero equals minus sigma of epsilon over epsilon, which means that the quantity c zero is totally irrelevant in the problem. And the system, so it's not, you see c zero in the way we introduce it in the reason the way we introduce the c zero was I say that it's e to the minus beta e phi divided by integral, if you remember. And I say that if phi goes to zero at infinity, this quantity is the volume. And there was a factor of n. And that's how the concentration came, the c zero came into the game. But here, this integral, as we will see, this integral is finite. It's a number. It doesn't diverge like v. So which means that what happens is that the system adjusts the number of counter ends so that it is the only way that this whole thing has a meaning is when the system is completely neutral, in which case c zero is complete. It's just a constant. It's just a reference point for the potential. Yes. And we will check that it works. We'll check it immediately. And the c zero, you see that, for instance, c zero comes into the, so the potential, I write the potential, the potential is phi of z equal, so beta e phi, I mean beta e phi is the dimensionless potential, is equal to log 2 pi lb c zero plus 2 log z plus lambda g. OK? So you see that the c zero enters, doesn't enter at all in the concentration of ions. It enters here in the, it's just a reference point for the potential. You know that the potential, the value of the electrostatic potential depends on the constant because what's defined is the electric field. So c zero is just a constant which defines the origin of potential. Now, one thing I want to show you is what happens, what is the integral of, OK. So if I call sigma the charge here, if A is the area of the surface, then the total charge of my plate is sigma A. I mean it's negative but it's modulus of sigma A. So let me calculate some from zero to infinity. So the total number of counter ions is A times, that's the, because it's uniform in the transverse direction, sum from zero to infinity d z c of z. So it's A over 2 pi lb sum from zero to infinity d z 1 over z plus lambda g square. So this integral, the primitive is 1 over z, is minus 1 over z plus lambda g. So it's A the area times 1 over 2 pi lb lambda g. And so 1 over, so what is 2 pi lb lambda g is 2 pi lb and lambda g. So lb is beta e squared over 4 pi and lambda g is 2 epsilon, beta e squared over 4 pi epsilon, and times 2 epsilon over beta sigma e. So you do the calculation. So the 2 times 2 pi and the 4 pi go away. The epsilon goes away, the beta go away, and okay, the e's go away. I mean, okay, there is a convention of whether it's, the density is sigma or sigma e. So I, anyway, you can see that what you get is that essentially this is equal to, so it's the inverse, so it's A times sigma, A times sigma over e. Okay? And sigma over e is the number of, this quantity is the total number of charges on the surface. This is the total charge of the surface. So if you want to know how many charges you have on the surface, it's sigma A over e. So you see that the total number of counterions in the system, which are bound in the system, are exactly the number of charges on the surface. Is it clear? Right? The integral of the constant, so the integral of the concentration gives you the total number of counterions in the system. And the total number of counterions of the system is the surface area A times sigma over e. That's from the integral. And now if you have a surface with a charge, surface charge sigma, the total charge is sigma A. And if you want the total number of charges, it's sigma A over e. So it matches. So that's the first thing. And the second thing, which I want to show you, is which will help in the interpretation of the, of this glitch happening. Yes? Is the solution neutral? Yes. I mean, the total charge is zero. The total charge is zero, but it does it automatically. No, you don't, when it's counterions, you cannot control the concentration because you don't have a reservoir. When you have a salt, you can control the concentration because it's at, because in the bulk it is neutral. But counterions, you have no control. Counterions are released by the charged object. And the charged object will release the amount of counterions necessary to neutralize the solution. It's a big difference between counterions and salt. And I will see that the solution and the physics is complete. For instance, in the cylindrical case, it's completely different. And here also, actually. And the last thing I want to show you, so, is, if I, that, I want to show you is that the, the grid Chapman length. So this lambda g is the length over which half of the, half of the counterions are present. In other words, if this is your wall, if this is lambda g. So here, the total number of counterions is half of the total number of counterions is the solution. The way to see it is just that you integrate this concentration instead of integrating it from zero to infinity. If you integrate it to lambda g, c of z. So it's one over two pi lb, some from zero to lambda g, d z one over z plus lambda g square. And so it's one over two pi lb times one over, so the integral is just one over z plus lambda g. And so it's one over lambda g minus one over two lambda g. So it's one over two pi lb times one over two lambda g. So you see that if it was one over two pi lb lambda g, it was neutralizing the whole surface. So here it's one over two pi lb times one over two lambda g. So it's neutralizing half of the surface, which means that half of the counterions are contained in a layer of thickness lambda g. And that's the physical interpretation of this lambda g. Another interpretation of the lambda g is it's the distance at which, so lambda g, half of the counterions are absorbed. Half of the total number of counterions or half of the total charge is neutralized. It's also lambda g is such that the lb over lambda g. So lb over lambda g is essentially the Coulomb energy at distance lambda g from the wall. And this is equal to kt. So it's the distance from the wall at which the Coulomb energy is equal to kt. These are the two interpretations you can give of this lambda g. Yes? Sorry? Half of the counterions are absorbed. So half of them over a distance lambda g, half of the counterions of the total number of counterions are present and neutralized, half of the charge of the distance at which the Coulomb energy times beta, of course, is equal to kt lambda. Sorry, you don't write it like this. Yes? E squared. OK, so this is it for the case of counterions. So now I will discuss another important case, which is still the planar case, but now with salt. So if you have salt, so it means you have your negative thing, you have plus and minus charges. I will assume that it's a valence one salt. So the boundary condition will be the same. If sigma is the charge density on the wall, you have phi prime of zero equals minus sigma over epsilon. That's the same boundary condition as we had before because there is nothing changed. And the equation will be phi second of z equals 2 C0 over epsilon C0e over epsilon sinh beta e phi. OK, so now C0 is really the concentration of the, so the big difference with counterions is that here at infinity you can have salt because salt is neutral globally. So if you integrate the charge up to infinity, it's OK because if you go far from the wall, here you have still plus and minus ions, so the total charge locally is zero. So it doesn't matter. And C0 in that case is really the bulk concentration of the salt far away from the wall. OK, so you can work out the solution as in the previous case by looking for, you know, whenever you have in 1D, so it's still one-dimensional. Whenever you have something like that, you multiply by phi prime on both sides, and then you integrate. That's the standard method. So I'm not going to do it. I just show you the shape, the form of the solution. So the solution is a bit more complicated than in the previous case. In this case, it's minus 2 over beta e, log, 1 plus gamma e to the minus z over lambda e, divided by 1 minus gamma e to the minus z over lambda d. This is the solution where lambda d is the divide length. OK, and gamma is some constant, and I will show you. So as an exercise, I don't do it because it's quite lengthy to do it, but you can do it. I think it's this afternoon if you have time. Just plug it in here and check that it works. It's a few lines of calculation, but I am not very eager to do it now. So actually, you can write it as minus 2 over beta e, log of e to the z over lambda d plus gamma, divided by e to the z over lambda d minus gamma, and if you write it like this, you see that now I will try to satisfy the boundary condition. So phi prime of z is minus 2 over beta epsilon, and if I take the derivative of this with respect to z, so I write it as log of the numerator minus log of the denominator, so I get 1, so I get a factor of e to the lambda d times 1 over lambda d, and then 1 over e to the z over lambda d plus gamma minus 1 over e to the z over lambda d minus gamma. And if I write it at z equals 0, I have the boundary condition, which is that minus sigma over epsilon equals minus 2 over beta epsilon lambda d times 1 over 1 plus gamma minus 1 over 1 minus gamma. Am I just putting z equals 0? Yes? What is? P. P? Yes. Did I write a P? You mean d maybe? D? Gamma. Gamma. Gamma. Ah, it's gamma. It's a constant. It's a constant which is to be determined by the boundary condition. What I write here is that the general solution has this form with a certain gamma, and this gamma will be determined by this boundary condition, which is this relation which I write here. OK, so you write plus, plus. OK, so there is some, a little bit of algebra. OK, and this is minus 2 over beta epsilon lambda d times, I have a sign mistake somewhere. OK, so here it seems that I have minus 2 gamma over 1 minus gamma squared. OK, so, OK, so if I really find, so I remind you that lambda g was 2 epsilon over beta sigma e, then you can work this out, and you see that if you define lambda, capital lambda equals lambda g over lambda d, so the Gwichapan length divided by the Dubai length, so the Dubai length is essentially the screening length, and you can see that indeed it is a screening length because it is the range over which the potential varies. So essentially you can see that gamma satisfies a second order, a second degree equation, which is gamma squared plus 2 gamma lambda equals 1. And the solution is therefore, so you get gamma equals minus lambda plus square root of lambda squared plus 1. OK, so as a consequence, the concentration of the concentration of plus or minus ions at point like this is equal to c0 e to the minus plus beta e phi, and if I take the expression for phi, which is here, you get that it is c0 1 plus or minus e to the minus z over lambda d divided by 1 over minus plus e to the minus z over lambda d to the square. So if you look now, so this is what you get by just replacing phi with the value of gamma that you get from there in the problem. OK, so if I look at the profile as a function of z, of course you see that when z goes to infinity, which means when z is much larger than lambda d, the exponentials decay to zero, and then c plus or minus of z is just equal to c0. So the concentration of positive and negative ion both converge to the same value, to the same bulk value, which is c0, and this is just charge neutrality. This is why you can have the potential going. So you see that when z goes to infinity, the potential goes to zero this time, right? Because this term goes to zero, this term goes to zero, so you have log one. So at infinity you have zero potential, the potential goes to zero, and you go to a finite concentration, which is the salt concentration in the system. Now, if you look at the plus concentration, so there is a certain c0 like this. And, of course, so the wall is charged negatively, so you will see that this is c plus and c minus. So there is that traction of the salt of the counterions, opposite of the positive ions are attracted to the wall, and then at infinity they go to this bulk value c0, whereas the co-ions, which have the same sign, the negative ions, are repelled from the wall and go to the same c0 at infinity. The total charge of the salt, y, but one is plus, the other one is minus. Is that the bulk of concentration of plus charges? Yes. I guess that the total charge is the integral of one core minus the other. Yes, and the difference is exactly the charge of the surface. Because the total system has to be neutral. So since you have an excess of negative charge on the surface, you must have an excess of positive charge in the bulk. And it's over a finite distance only, but it was to compensate exactly what you have on the surface. So again, I guess I can mention here and put some kind of salt, just to have the same amount of positive or negative charge. Of the salt? I mean, I can define an experiment that was exactly the same question as before. I thought that you can buy a glass, a metal wall connected to a battery and put salt in the same amount of positive and negative charge. So this problem cannot be solved here. No, you cannot have a system with a macroscopic charge imbalance, with a macroscopic charge. This is not possible. It's not possible because of the nature of the Coulomb interaction. Because if you have, let's say, an excess of charge, which is Ne in a solution. If you have a solution which is not neutral with a charge Ne, then the total energy of this, the Coulombic energy of this, would go like N squared E squared. And this is huge, so the system cannot exist. Yes, for an infinite size, of course. For a small system, you can have charge imbalance. This occurs very often, but in a macroscopic system, it's impossible to have non-neutral systems. They would just blow up, and it's not possible. OK, so this is, yes? They go to a finite value, which is C0, E to the minus or plus beta E phi of zero. And phi of zero is just minus two beta E, so you just put Z equals zero. So it's log one of plus gamma divided by log one minus gamma. It's just a number. Sorry? So this is a mistake. Oh, this is, yes, OK. It's a mistake because I forgot the gamma. I forgot in, let me see. Yes, I think I forgot the value. Yes, I forgot the value of gamma. Yes, you see there is this gamma here. So when you take E to the minus beta E phi, it's, yes, it's one plus, sorry, there is gamma. Thanks. So it's one plus gamma over one minus gamma or one minus gamma over one plus gamma. Sorry. And gamma being the solution of this equation. OK, so this is planar solution with salt. And OK, so now to conclude, I will do, I will discuss one case before discussing the cylindrical case, which is very interesting, which we'll see next time. I will go back to a simpler case, which is the planar Poisson Boltzmann, but now with two plates. So imagine that you have two symmetric plates with salt in the middle. So this is the origin. Here is L over two. This is minus L over two. So in the case of pure counter ions, you can solve exactly the system in the case of salt when you have two symmetric plates. So symmetric means that you have the same charge density here on both sides. The non-symmetric case is much more complicated, and I will not discuss it. And the symmetric case, I will discuss it only in the Poisson Boltzmann approximation. So just one thing I want to say. With counter ions, you can solve exactly the problem. It's not very complicated. In the case of salt, you cannot have an explicit analytic solution. You can express it in terms of elliptic functions. It's a bit complicated, and I will not discuss it. So I will discuss what happens with salt in the Debye-Huckel approximation. Is there any question about the previous thing before I start this? I mean, for the other case, you have to work out the solution by yourself if you're interested and see how it works. It's quite easy, but the calculations are, I mean, you have to do it yourself. So the Debye-Huckel approximation for that. So actually, if you write the symmetric two plate case, the Debye-Huckel equation is just 5 seconds of Z minus kappa D. So in fact, it's minus 5 seconds squared plus kappa D by y of Z equals sigma over epsilon delta of Z minus L over 2. So the charge density is delta of Z minus L over 2 plus delta of Z plus L over 2. Yes? What do you mean exactly by non-symmetrical case? Non-symmetric case is when you would have sigma and sigma prime. You would have two different charge densities on the two plates. Here the symmetric case is you have exactly the same charge density on both sides. So the boundary conditions are very simple. Again, you have 5 prime of 0 equals minus sigma over epsilon. That's the electric field near the wall. And no, not on zero, let's say on L over 2. L over 2. And obviously in the center at prime zero by symmetry, 5 prime of zero is equal to zero. So we can study. So the system is completely symmetric with respect to zero. And we can study it between zero and plus infinity with this boundary condition. And the question is just minus phi second of Z. So phi second of Z equals kappa square phi of Z where kappa D square is the Dubai constant to the square. So the solution of this is, of course, a linear combination. So phi of Z is A cos kappa D Z plus B sinh kappa D Z. If I take a derivative, I get a sinh for this and a cosh for this. And if I write the boundary condition that the derivative should be zero at zero, it means that this term doesn't exist. So this is the solution, right? It's trivial. And then if I write, so phi prime is equal to kappa D A sinh kappa D Z. So the boundary condition is just that minus sigma over epsilon equals kappa D A sinh kappa D L over 2. So this gives me A, and this minus sigma is absolute value of sigma because sigma is negative. So I have phi of Z equals A. So it's sigma over epsilon kappa D sinh kappa D Z divided by sinh. No, cosh. It's A, so it's cosh. kappa D Z divided by sinh kappa D L over 2. So that's very simple. And you can calculate the profile of all species around there. OK, so the last thing I want to do is to calculate, so when you have these two plates like this, to calculate what is the osmotic pressure of the ions between the two plates. So what is the pressure that the ions are exerting on the two plates? So for that, if you remember, we saw that the free energy F in the Debye-Huckel case is just the integral between minus L over 2 plus L over 2. So there will be a factor of A, which is where A is the area of this because it's an integral D3R in the whole volume. So it's A the surface times the integral DL times DZ times 5 prime square plus kappa D square phi square times epsilon over 2. OK, so the A is just the surface. And you know that the pressure, the osmotic pressure is related to the free energy by the relation that the pressure, osmotic pressure, is given by minus DF by DV, like the pressure is given by minus DF DV. And in this case, it's just minus 1 over A DF by DL, where L is the distance between the two plates. OK, so we have F of Z here. So F phi prime is sigma over epsilon kD times kD sinh kappa DZ over sinh kappa DL over 2. So this is just, so F over A equals epsilon over 2. Since everything is symmetric, there is a factor of 2. So it's 2 times sum from 0 to L over 2 dz of, so 5 prime square, I will have this and this. So it will be, yes? Sorry, F is free energy. It's the free energy in the Debye-Huckel approximation, right? We saw that last time. It's, this is the, the first term is really the electrostatic energy because this is E square if you want. It's epsilon E square over 2. And the second term is the expansion of the entropy. That's the result of the expansion of the entropy. That's, that's the final result which you see. You can see easily it's given by this. So, so there are some prefactors which are sigma square over epsilon kappa D square. And then it's 5 prime square. So it gives a kappa D square plus kappa D square. So it's like this. So, 1 over sinh square kappa D L over 2 times cosh square kappa DZ plus sinh square kappa D. And so the free energy per unit area, F over A, is equal to epsilon. So the kappa D square will disappear. So it's epsilon over epsilon square. Sigma square over epsilon. 1 over sinh square kappa L over 2. Yes. And some from 0 to L over 2 DZ. And the cosh square plus sinh square is the cosh of twice of 2 kappa DZ. So it's sigma square over epsilon sinh square kappa L over 2 times. So the integral is 1 over 2 kappa D times sinh 2 kappa D L over 2. So it's kappa D times L. And therefore you have the final result. So if you remember that sinh kappa D L is twice sinh kappa D L over 2 times cosh kappa D L over 2, you get that F over A equals sigma square over epsilon kappa D times cosh kappa D L over 2 divided by sinh kappa D L over 2. It's a fairly simple expression for the free energy of a salt between two plates in the Debye-Huckel approximation. So this is the contingent. And if you take, so pi equals minus D by D L of F over A according to this. So pi is equal to sigma square over epsilon kappa D. And then the derivative of this with respect to L will give only of kappa D over 2 times 1 minus cosh kappa D L over 2 divided by sinh kappa D L over 2 to the square. And therefore pi is just equal with a minus sign, right, minus D by D L. And so the result is pi equals sigma square over 2 epsilon 1 over sinh square kappa D L over 2. And this is the final result. OK, so now if you, so the pressure is positive, of course, and we can study two regimes. So it goes like, so first of all, it goes like sigma square, so which means that the pressure doesn't depend on the sign of the charges on the two walls. The walls can be, if they are both positive or negatively charged, it's the same. And you see that if you are in the regime where kappa D L much smaller than 1, which means that L much smaller than lambda D, so if the distance between the two walls is much smaller than the bi length, you see that the pressure pi essentially is sigma square over epsilon lambda D square over L square. So the pressure shoots up like 1 over L square. So when you try to press together at distances shorter than the bi length, there is a very strong electrostatic pressure which prevents to push the plates together. On the other hand, if kappa D L is much larger than 1, then the sinh kappa D L over 2 is essentially 1 half of E to the kappa D L over 2. And therefore the pressure scales like sigma square over epsilon E to the minus kappa D L inch square. And there is a factor of, so the pressure goes to zero exponentially fast with the distance. So essentially when you are beyond the bi-hooker length, there is essentially no more pressure between the plates because the charges are completely neutralized and the plates essentially don't see each other. So here it goes to infinity and here it goes to zero. So it goes like this, something like this. OK. And I think I will stop here. Yes? It's the pressure which is due to the ions in the system. Like when you define the pressure of a gas in a container, it's the pressure due to the, I mean, the osmotic pressure is the pressure exerted by the particles on the wall. So whether it's a piston or, I mean, here you have the solvent and you have the particles which are floating around. It's like a gas of ions. So it's a gas of particles and in a container. So it's the osmotic pressure. It's the pressure due to the ions moving in the container. It's the same, I mean.