 So today I will continue with various geometries for electrostatic systems. So I remind you that yesterday we studied the case of one or two plates. And we saw that the concentration of ions goes like c of z. There is a certain factor divided by z plus lambda g square. And there was a pre-factor which is 1 over 2 pi Lb. And lambda g is the GUI Chapman. And the GUI Chapman length is given by e over 2 pi Lb sigma if sigma is the charge density of the wall. Somebody asked me what is the interpretation of this lambda g. So I was a bit not very clear. So lambda g is the distance at which the electrostatic energy of a particle is equal to KBT. So if I have a charge density sigma, and if I assume that it's just a simple charge density, then the electric field is sigma over 2 epsilon. It's sigma over epsilon if you can go only on one side, but if you can go on both sides, sigma over 2 epsilon. So the force if I put a charge e at a certain distance here, the force is f equals sigma e over 2 epsilon. And the potential energy of the charge, so it's the work that you take to go from here to here. So it's just sigma e over 2 epsilon times lambda. And let me call this lambda g. Now if I write that this energy is equal to KBT, you see that it gives lambda g equals 2 KBT epsilon over sigma e, which is 2 epsilon over beta sigma e. And okay, so sigma is always absolute value of sigma of course. And this is just, so if I express it in terms of LB, it's exactly e over 2 pi LB sigma. I remind you that LB is defined by 4 pi LB equals beta e square over epsilon. So the scale which enters in the decay of the ion, of the counter ion concentration, the scale which enters is the scale at which the electrostatic energy of a particle is equal to essentially 2 KBT. And beyond that in decay. And of course the ionic profile as a function of z goes like 1 over z square at large distance. Okay, so today I want to study something a bit. So we saw the case of one plate with salt can be solved exactly, but if you have two plates with salt or you cannot solve it exactly. So now I go to another important geometry, which is the cylindrical geometry. So you have a cylinder which is let's say negatively charged in either counter ion or salt. So this of course in real life it's rare that you have cylinders like this, but in biology for instance if you take DNA, DNA is very negatively charged. And by the way DNA are the largest charges existing on Earth because a molecule of DNA can be billions of bases long, each base carries one electron charge. So all together it's a huge global charge for microscopic object and so it's the most charged molecules that exist. So because of these charges, so the polymer of course is an object which is a completely random Brownian makes a random coil, but DNA because of the charges along this, the charges repel each other and it gives a certain rigidity to the chain. So the chain has a tendency to be straight, cylindrical, and there is a notion of persistence length which is the length over which you can consider that the polymer is a straight line or is not crumpled, not okay. So anyway the geometry that we will study is this one. So before that I want to, yes? Yes. Sorry? No, no, because there are counter ions. It's neutral. I mean if you are in a solution as I said, in real life everything is neutral. So there are other ions which are neutralizing the charge of the DNA, otherwise your hair would be and we couldn't approach you. Sorry? There is all kind of salts in your cell. That's the purpose of this course. In biology everything is neutral so you have very strongly charged objects like DNA, like membranes and things like that, and you have ions which are floating around, which are neutralizing the charge of all this and all together everything is neutral. As I said yesterday you cannot have a macroscopic charge anywhere because it's completely unstable. So if you have a tendency for a system to have a certain charge it will suck charges from everywhere to make the system neutral. Okay? Okay. So I want to discuss rapidly a qualitative discussion of ion condensation in various geometries. So the first geometry is the charged sphere. So assume that you have a charged sphere with a certain charge, let's say z minus or with a charge q minus. So this charge is fixed and then I put charge plus e floating around. So the electrostatic potential is phi of r equals 1 over 4 pi lb q minus divided by r. And the probability to find this charge e at point r is given by e to the minus, sorry, e to the minus beta e q minus over 4 pi epsilon r. And this, of course, is to be divided by the partition function. And the partition function is just the normalization. So the z is the integral d3r with a r, let's say larger than a. If a is the radius of the sphere of e to the minus beta q minus e over 4 pi epsilon r. So this is because of spherical symmetry. This is 4 pi integral from a to plus infinity dr of r. d3r of r square e to the minus some constant alpha over r. So you see that this object is divergent. This object is divergent because when r goes to infinity, this exponential goes to 1. And therefore, this quantity essentially diverges like the volume of the cell which goes to infinity. So if this diverges, it means that p of r essentially goes to 0, which means that the probability to find your iron at any point goes to 0, which means essentially that the iron is unbound. So your iron, which is here, if you have a charged sphere, the iron will go around, will float, but there is no, the concentration of iron, if you take the volume to infinity, will go to 0, and essentially you have an iron floating around and not bound to the sphere. The other example is a charged plane which we studied yesterday. So I want to show you what happened in the case of a charged plane. And the reason, the physical reason, of course, why the particle is not bound is that essentially the attraction of the particle goes like 1 over r. So when r goes to infinity, it goes to 0. Whereas the entropy of the particle, so the particle is in the whole volume, so the entropy goes essentially like the log of r cubed, like the log of the volume. So the entropy, the larger the volume is, the entropy diverges and the entropy will always be much larger than the electrostatic energy and therefore the particle is not localized and it just goes away. Now, in the case of a charged plane, if I take a charged plane with, so what are my notations? With the density sigma, so the electric field is e equals sigma over epsilon, phi of z is equal to minus sigma z over epsilon. And therefore, if I do the same kind of calculation, if I look what is the probability to find my particle at point z in the system, it's just e to the minus beta e phi. So it's e to the plus beta sigma e z over epsilon divided by z. So sigma is negative. So sigma is negative. Let me write it rather like minus like this. And the partition function z is just the integral 0z. So from 0 to plus infinity dz of e to the minus beta sigma e z over epsilon. So this, of course, is completely finite. It's just epsilon over beta sigma e. And therefore the probability to find the particle at any finite distance z from the plane, this probability is finite. It's non-zero. And therefore the particle will be localized. It cannot escape to infinity. It has a certain probability always to be at a certain distance z from the center. So this is a normalized. So here p r equals 0 particle delocalized, whereas here the charge is localized near the plane. And here it's a charge look. And here the reason, thank you, the reasoning can be made the same. Here you see that when z increases, the electrostatic energy increases linearly, whereas the entropy will scale still like log z times the volume. And so you see that now z is always much larger than log z. So the attraction of the particle to the charged plane will always be much larger than the entropy. And therefore the particle will be bound to the plane. Of course it's the entropy which has a tendency to delocalize the particle. It's a balance. The particles are attracted and they want to fly away because of the entropy of the free volume. And so whichever wins tells you whether the particle is localized or not. OK, so this is simple enough. And now we come to the intermediate case, which is the most interesting case, which is the cylindrical case. So the charged cylinder. So you take the axis z, let's say, and radially coordinate. This is r. So I take charge sigma. So it's very simple to calculate the electric field. To calculate the electric field at distance r, you use the Gauss theorem. So you do that the flux, if you have a cylinder, whatever, so you take a cylinder like this and you calculate, so you have the flux. So the vector e is perpendicular to the surface because it's a radial field. And the flux of e is equal to the total charge q inside divided by epsilon. So the flux of e is e times the surface. The surface, if this is r, is equal to 2 pi r. It's the external surface. So it's 2 pi r times, let's say, l. And this is equal to the charge inside. So the charge inside is equal to sigma, sigma being the charge density, times the surface. The surface is pi a squared times l, which means that the electric field at point r divided by epsilon. So it's equal to a squared over 2 epsilon r. No, it's pi a, pi a. Is it correct? No, it's 2 pi a, 2 pi a times l, right? 2 pi a is the, I mean, if you open this, you have a rectangle like this. So this is 2 pi a and this is l. So the total charge is 2 pi a times l times sigma, et cetera. And no 2. OK, anyway. So the point is that in that case, the electrostatic potential, p of r, can be written as e to the minus, let me write it here, p of r is equal to e to the minus 2 e lb. So if I call rho is the linear charge density, so it's 2 pi a times sigma. Sigma is the surface charge density. So if I multiply by 2 pi a, this is the linear charge density, right? So it's e to the minus 2 pi, 2 e lb, rho log r over a. So this is the electron divided by z. Right? You see that essentially what you see is that the electrostatic potential is obtained by integrating this with respect to r, right? This is minus d phi dr. So this gives you that phi is equal to minus a over epsilon log r. And you need a reference of potential. So for instance, if you take the potential on the cylinder is zero, so you put it like this. OK, and when you go to, so then you multiply by beta, et cetera, this is what you get. And divided by the partition function. So now the partition function. The partition function is z equals integral d to r of e to the minus 2 e lb rho log r over a. So times, because of transitional invariance, I don't put the z coordinate. So it's 2 pi integral from a to infinity dr of r times e to the minus 2 e lb rho log r over a. So it's essentially 2 pi integral from a to infinity dr of r to the power 1. And this is just 1 over r to the, so 1 minus 2 e lb rho. Yes? Yes, absolutely. Yes, q, but sigma is here. Yes? Oh, OK. So it's a sigma, exactly. Yes. And that's how you get the lb and all these things. OK. OK, so this is the quantity. And now you see that this integral will diverge or not diverge depending on this exponent which is here. And this exponent, so if, so which you can write as 1 over r to the 2 e lb rho minus 1. So if 2 e lb rho minus 1, if this exponent is smaller or equal to 1, which means if rho is smaller or equal to 1 over lb, then the integral diverges and the ions are delocalized, are not localized, right? Because the integral diverges and therefore the probability p is equal to 0. On the other hand, if this is larger than 1, which means if rho is larger than 1 over e lb, I think there is a mistake. It should be e over lb. So I must have its lb times rho. Yes, I'm sorry, but it's probably lb times rho over e. OK. Yeah, so you have to check, but it's certainly like this because of dimensions, right? Rho is a charge per unit length. So rho over e is 1 over a length, and that has to be dimensionless. So I'm sorry that e is like this. lb over e. So if rho is larger than e over lb, it converges and the ions are localized near the cylinder. So this phenomenon is a phase transition. We will see that this approach is extremely naive. It doesn't correspond really to the reality. I will show you in detail the solution of the real Poisson Boltzmann equation, but essentially it means that there is a critical rho star, which is essentially one unit charge per beer length, one charge per beer length. So if you have a linear charge on a cylinder which is equal to this rho star, if you are below the threshold, below this rho star, all the ions are delocalized. They float around. And if you are above this threshold, the ions will condense on the surface. It condenses in the sense that they are localized. They don't fly away to infinity. I mean, the fraction of ions which go away to infinity is zero. So we will see in more detail how it goes. And this phenomenon is called Manning condensation. Did I write it somewhere? No? So this is Manning or sometimes Manning-Ozawa condensation. So we will study this phenomenon in detail. It's a bit reminiscent, if you have studied statistical physics. It's a bit reminiscent to the Kosterlitz-Towles transition in Coulombic systems or in a XY model. I mean, in 2D there is a specific kind of transition, which is the XY Kosterlitz-Towles transition, which is a bit like this. Yes? Yes? We are always in 3D. Yes, exactly. To marginalize the probability density. Yes. So actually I didn't say it here, but you see here the potential. So the electrostatic potential is logarithmic. It goes like log r. And the entropy is also logarithmic. So they can balance. So depending on the coefficients in front of these two terms, either the energy will win, in which case the particles, the ions will be localized, or the entropy will win, in which case the particles will be delocalized. For example, the charge sphere. The what? The radial probability. Yes? This is finite, not zero. No, because you have to normalize. You see the point is that it's the normalization. The normalization is dominated by the infinite volume. And I mean it's this balance between the entropy and the, I mean the, yes? Sorry? I also have a question. Yes? Yes? Sure. It's a 2.50? Yes, of course. And in fact, you can see that if you, of course, what I will, this is a very naive presentation, because there is only one ion. But in reality, in the real system, you have many ions. And this is what we are going to see now. We have many ions which repel each other. So the picture is a bit different. And I will show you what happens. But what people believe is that you have a whole critical phase in the condensed case. You have a whole critical phase. Sorry? Yes, it looks a little bit like the KT transition. But you know Coulomb gas is equivalent to XY model. And so, OK. So what I want to do now is to show you the solution. So this is just a kind of hand-waving phenomenological kind of approach. Yes? It's a characteristic length at, when you have two charges interacting, it's the length at which the interaction energy between the two particles is equal to KT. So LB, it's called the B-room length. So if you have two charges, this is LB. So it's beta E squared over 4 pi epsilon LB equals 1. That's the definition which means LB. It's another way to write, so LB is beta E squared over 4 pi epsilon. So it's a way to write the electrostatic energy. In other words, beta E phi of r, if phi of r is 1 over 4 pi epsilon r beta E, so beta E phi of r is LB over r. It's a way to write the electrostatic potential as a ratio of two lengths. Yes? It's just the fact that you have algebraic interactions and that in the KT transition it's the same. You have a balance between the entropy of pairing and the attraction of the two pairs. So the attraction of vortices or if you think in terms of a Coulomb gas, the attraction between the opposite charges is logarithmic, the entropy is logarithmic, and the electrostatic stylist transition is just a balance between the two, like here, so that's in this sense. And OK, it's a continuous transition. 2D is always very special because the entropy is logarithmic and the interactions are logarithmic. So you have all kinds of weird transitions in 2D. OK, so now I go to the more serious stuff, which is the exact solution of the Poisson-Boltzmann equation for the case of a cylinder. So it's a little bit more complicated. So what I will do is I will consider one cylinder in infinite volume, but the exact solution has been devised in the 50, and in fact the exact solution was done for a charged cylinder like this in a big container, cylindrical cylinder of size r, so the small cylinder has radius a, you have infinite cylinder like this, and the ions are confined here. So the general case is more complicated. The solutions are more complex. Here I will look at just one cylinder. I will assume that we are in infinite space. So I call A the radius of the small cylinder, of the charged cylinder. Sigma is the charge density. And again I can define a linear charge density. So the linear charge density is the charge per unit length of the cylinder. So it's 2 pi A times sigma, right? It's the rho. If it's the linear charge density, it's the number of charges on unit L. So the number of charges on the cylinder for size L is 2 pi A times L. That's the surface of the cylinder times sigma divided by L. So it's really this, right? OK, so this is the linear charge density. So the Poisson Boltzmann equation, which is Laplacian phi equals minus C0E over epsilon sinh. Sorry, I take the, so I look at the case with only counter ions, only counter ions, minus. So in that case you have only one species of ion, which is E to the minus beta E phi. So this is the equation. And the boundary condition now is that the surface charge, so phi, so d phi dr at r equals A, is equal to minus sigma over epsilon, OK? And at infinity I assume that the charge density goes, I mean at most the total number of charges that I will have will neutralize the charges which are here. So I will assume that, so C of r, of course, from this, right, we have the C of r. The density of counter ions is C0E to minus beta E phi. So I want that at infinity C of r goes to zero when r goes to infinity. Or equivalently, phi of r goes to plus infinity when r goes to infinity. OK, so there is a change of variable to do. OK, first of all, because we are in cylindrical coordinates, we can write phi in cylindrical coordinates. So you have z, r, so it's r theta z. r is the radial distance. z is the coordinate like this. And theta is the angle, the polar angle in a reference plane, projected on a plane. Now, because of the symmetry, this is, of the cylindrical symmetry, this is just a function phi of r independent of theta. It doesn't depend on the angle. It doesn't depend because you have an infinite cylinder. It doesn't depend on z where you are sitting on z. So then the Laplacian in polar coordinates, if your function is independent of r, is independent of theta and z, it is just d2 by dr r square plus 1 over r d by dr. And so the Poisson-Boltzmann equation, which is here, takes the form d2 by dr square plus 1 over r d by dr phi of r equals minus c0e over epsilon e to the minus beta e phi. Question? No? OK. So we do a change of variable. The change of variable is u. You define a new variable, u equals log of r over a, or equivalently r equals ae to the u. And you define a new function v of u equals minus beta e phi of r plus 2u, or in other words, beta e phi equals 2u minus v of u. So instead of solving for phi of r, I will write this equation as an equation, as a differential equation for v of u, as a function of the variable u. Now, there is a question with the boundary conditions. So the boundary conditions are that, so d phi dr is equal to, let's say, d phi du. OK. I will define psi equals beta e phi for, right? So psi is beta e phi just to simplify the notation. So d psi dr, so which way do I write it? Yes. So d psi dr is equal to d psi du du dr. So d psi psi of r equals minus v of u plus 2u. So d psi du. And so d and du dr equals 1 over r from this. And 1 over r is e to the minus u over a. OK. So this is just e to the minus u over a times d psi du. And d psi du is 2 minus v prime of u. So I will need d psi dr. I will need the second derivative also. So d psi dr square will be the derivative of, so it will be 1 over a d by du of e to the minus u 2 minus v prime of u times du dr. And du dr, you read it off, times e to the minus u over a. So it's e to the minus u divided by a square times minus, so the first term is minus e to the minus u 2 minus v prime, minus e to the minus u v second. So it's minus e to the minus 2u over a square times 2 minus v prime plus v second. Now I need also, so 1 over r d by dr. So I can multiply both sides by beta e. OK. So I write that d 2 psi by dr square plus 1 over r d psi dr is equal to minus e to the minus 2u over a square. 2 minus v prime plus v second plus 1 over r. So 1 over r is e to the minus u over a. That's the 1 over r. Psi prime of r. So psi prime of r, I have another e to the minus u over a times 2 minus v prime. So you can see, you can group these two terms. You see that the 2 minus v prime disappear. And you get just the very nice result that this is just minus e to the minus 2u over a square v second. OK. So now I can write the Poisson-Boltzmann equation with these new variables. So the Poisson-Boltzmann equation becomes, so psi, I remind you psi is beta e phi. So if I want to write it for psi, if I want to write this equation for psi, I should multiply by beta e. Or I divide all these by beta e. So by beta e, right? To get the equation. So this is multiplied by 1 over beta e. So then I have the equation is minus e to the minus 2u over beta e a square v second. That's this term. OK. Equals this minus c0e over epsilon. And e to the minus beta e phi. So this is beta e phi. So e to the minus beta e phi is equal to e to the minus 2u, e to the v of u minus. So it's e to the minus beta. So e to the minus 2u times e to the v of u. So finally, you get the equation v second. And this is where you, so you have to do the algebra yourself. I mean, I show you. So you get c0 beta e square over epsilon a square e to the u, e to the v of u, which is, so beta e square over epsilon is 4 pi lb. So it's 4 pi lb c0a square e to the v of u. That's the simplified form of the Poisson Boltzmann. You see that now it's a standard equation. So when it has, let me check that. OK. So at this level what you do is you look for a first integral. So to get a first integral what you do is you multiply by v prime on both sides. So you have v prime v second equals 4 pi lb c0a square e to the v of u times v prime of u. So this is the derivative of, so you have one half v prime square prime equals 4 pi lb c0a square e to the v prime. And so if you integrate this equation you have one half v prime square equals 4 pi lb c0a square e to the v plus a certain constant which I call, how did I call it? OK, I call it c. Why not? c is not a good notation because it looks like a concentration. So let me call it d. OK. So I erased in the middle, I erased the boundary condition. So the boundary condition if you remember because you have the cylinder with the charge density sigma. So close to here the electric field is sigma over epsilon. So you have phi prime d phi dr at r equals a should be equal to minus sigma over epsilon. Is it complicated? I mean it's just a standard algebra. It's kind of thing you have to know how to do. So d phi by dr you can do it here. So d phi by dr is equal. So psi is phi multiplied by beta e. So it's 1 over beta e times e to the minus u over a to minus v prime of u. And now u taken at r equals a. If you take u for r equals a it is equal to zero. So taking the boundary condition d phi by dr at r equals a is equivalent to taking this at u equals zero. Right? Because for r equals a u is equal to zero. OK. So what you have is that the boundary condition becomes, so this is 1. So 1 over beta e a times 2 minus v prime of zero is equal to minus sigma over epsilon. Which is, since sigma is negative it's sigma over epsilon. OK. And so I keep this. OK. So you have 2 minus v prime of zero equals beta e a sigma over epsilon. And if you remember rho over 2 pi equals sigma a. So this is beta e rho over 2 pi epsilon. And this is just, if I use the theorem length, it's 2 times rho over lb. So if I write, so I have this equation. So I just show you rapidly. I don't want to bother you too much. But essentially from this equation we have this equation which relates v prime square of u to exponential v of u. And the boundary condition, which is that v prime of zero equals v prime of zero equals 2 times 1 minus rho over lb e. So if I solve this equation I get v prime equals square root of 8 pi lb c0. Sorry. Yes. OK. No. I don't want to write it like this. If I write it to zero I have 1 half v prime zero square equals, so for u equals zero v of zero. No. I don't know. Sorry. So what am I doing? v prime. Ah, yes. OK. So one thing that I forgot to say, of course, is that phi of a is the origin of the potential. So on the surface of the cylinder I choose phi of a to zero, which means that phi of a equals zero, which means that for u equals zero I have v of zero equals zero. Yes. That's what was missing. So then I can write that v prime square of zero is equal to 4 pi lb c0 a square plus d. And if I replace you get that d. So this is where I want to get that d is. So if I define a parameter d, so I get equals v prime square. Sorry. It's plus 2d. So 2d equals v prime square of zero minus 4 pi lb c0 a square. So I define kappa square. Kappa square equals, it's a bit messy, but I want to get to some result which will be a little bit more understandable. So kappa square is 4 pi lb c0 a square. So 2d equals v prime zero square, which is kappa square. 2d is kappa square plus, no. Sorry. So v prime zero square is kappa square plus 2d. And here I have v prime zero square equals 4 times 1 minus rho over lb e. OK, so this parameter is just the density. It's the, right? No, there is a sign. There is a problem with the dimensions. I'm sorry. Yes, I think I made a mistake somewhere. The lb should be in the numerator. But I cannot trace it. It's rho lb over e for dimensional reasons. E prime of zero. So, OK. So when you redo the calculation, please check. There was a mistake here. It's rho lb. OK, so the result now, I will write you the result and you will see what happens. So if I call tau is lb rho over e. So lb rho over e, there was a modulus of rho. So the final result, once you use the boundary condition in this, the result is that v prime square is equal to 2 kappa square, where kappa square is written, where did I write kappa square? So kappa square is 4 pi lb c0 a square. Yes, it's here. So it's 2 kappa square e to the v plus 4 times 1 minus tau square minus kappa square over 2. OK, so from there you can integrate. So when you have this, you write that v prime is the square root of this equals dv by du equals square root of this. And then you write dv. So you have du equals dv divided by square root of 2 k square e to the v plus 4, et cetera. And you can integrate the equation and you get everything in terms of one integral. So it's a bit heavy. It's a bit complicated. But so I give you the final result. And you see that as you will see, there is a transition which occurs for tau equals 1. So I write you as a function of tau. OK, as a function of tau, we have two regimes. For tau, larger or equal to 1. So tau, I remind you, is rho lb over e. So there is a special point which is tau star equals 1. And tau star equals 1 means rho star lb over e equals 1, which means rho star equals e over lb, which means when you have a charge density, a linear charge density on the cylinder, which is one electronic charge per Bjergm length, there is a phase transition. The phase transition is the following. So for tau, larger or equal to 1, beta e phi of r is equal to 2 times log r over a plus log of 1 plus tau minus 1 log r over a. And for tau, smaller or equal to 1, beta e phi of r is equal to 2 tau log r over a. So this is the final result. Depending on whether you are above or below the threshold of one, of one unit charge per Bjergm length, if you are below, you recover this is the standard Coulomb potential created by a charged cylinder. tau is just the standard 2 pi a sigma. So it's just the standard linear charge density times the log r over a. So this is the expected result of the electrostatic potential created by a cylinder. And when you are above the threshold, you have a correction. So you have no more tau here, tau comes in there. And you have an additional term, which is a log of a log, which occurs here. If you look at the concentration of the ions, I see that I forgot part of the ions. So if you look at the ionic concentration in that case, okay, so the C of r is, so in fact, in that case you can see that this corresponds to C0 equals 0, which means that there is no charge on the, you see this is exactly the free potential created by a cylinder. And that means that the charges, all the ions are unbound, they are not bound, they don't modify the potential created by the cylinder, which means that if you take a box bigger and bigger, all the ions will fly away like a normal gas. Okay, and what I want to say is you can calculate the fraction of bound ions. So if you calculate the fraction of ions which are bound, so what happens is that here the ions unbound and here part of the ions are bound. So how do we get the fraction of ions which are bound? It's by just calculating NB, number of bound ions, equals integral from A to infinity, d2r of N of r or C of r. And that's just 2 pi sum from A to infinity dr of r C of r. Okay, so you can calculate this in both cases. I let you do the calculation in the case where tau is smaller than 1, so in the unbound case you find that NB is equal to 0. And in the case where tau, so when you are above the threshold, you have a fraction of the ions which are bound, and this fraction is given by is equal to tau minus 1 over LB. This is the number of ions which are bound per unit length of the cylinder. So you see that at tau equals 1 it is 0 and it goes continuously up to whatever all the ions getting bound. Now there is an interesting thing which I want to show you which is quite important and that's the whole essence. So the picture when you are above the condensation you have your cylinder and you have a certain profile of ions. And so a certain fraction of the ions will be bound to the cylinder and because this NB is not the total charge, it's only a fraction of the ions which are bound. So all the other ions are free to go in free space. So what I will show you is that, so this is negatively charged. Here you have plus charges floating around. So what I will show you is that if you look at the, so if you look from far away you see the cylinder which is with some positive charges sticking on it when you are above tau equals 1. And what I will show you is that the total number of charges which are stuck on the cylinder or which are bound to the cylinder is such that when you look far away the apparent charge of the cylinder is exactly the critical charge. So if you put your cylinder, let's say you put your cylinder in a counter ion solution and you increase the charge of the cylinder. So when the charge of the cylinder is weak, everything is unbound. When you start heating the critical tau equals 1, so by increasing the charge, some ions will start to bind to the cylinder. They will bind in such a way that the apparent charge of the cylinder remains critical equal to 1 always. So you will add more and more charges to the cylinder and eventually the total charge will just be 1. So this is related to this quantity. So this you can get by calculating C of r, the concentration of ion, and integrating it like this. It's not a trivial calculation but it's doable. And you get this simple result where tau, I remind you, is the linear density of charge times Lb over E. Now the total charge on the cylinder, so the total charge on the cylinder is the essentially tau E over Lb. The total charge on the cylinder is essentially its rho. It's rho, let's say, times L, if I mean L. So per unit length, the total charge on the cylinder is just rho. Now the number of bound charges on the cylinder per unit length is tau minus 1 over Lb. And so it's rho Lb minus 1, rho Lb over E. Lb minus 1 over Lb. So it's rho over E minus Lb. And so if I look at the apparent N charge, N effective charge of the cylinder when you are above the transition, so there is a factor of, sorry, there is always factors of E which are, actually it's the number of bound charges. So if it's in terms of charges, it's like this. So it's N total. So the apparent charge is the total number of charges that you have on your cylinder minus the charges which are bound. It's the difference between what is stuck here. And if you look here, what you see is the difference between the total minus what is bound. That's the effective charge of your cylinder in the solution. And so it's N total minus N bound. And this is just, OK, so actually the number of total charge, let me write it rather like this. It's rho over E. Right? Why I'm always confused with this factor of E, it depends if we talk in number of charges or in charge itself, the ratio, the factor E comes from that, right? The total charge is rho is the charge density. So rho over E is the number density. That's why I have always this stupid mistake. So if I talk in terms of charge density, so the total number of charges is rho over E. That's the total charge per unit length on the cylinder. The total number of charges bound to the cylinder per unit length is tau minus 1 over LB. This is what comes out from this calculation, which I don't do because you see it's a bit complicated calculation. So this is just rho over E minus 1 over LB. So now if I want to see what is the effective charge of the cylinder once all the ions are bound on it, you see that it's this minus this. So it's just 1 over LB. And 1 over LB is just like you have one charge per LB, per theorem length. So the system, for any value of tau, for any density above any concentration, any charge density above the threshold, the apparent charge density of the cylinder remains constant equals to the critical density 1 over LB. And that's a very interesting result. It's a kind of charge regulation. You cannot have, if you charge and charge your polymer or your cylinder in an environment of counter ions, you cannot charge it. It will just suck the counter ions on the surface so as to keep always a maximum charge density equals to 1 over LB, equal to the threshold density. Yes? No. No. It's only true for counter ions. Yes. But OK, so counter ions, of course, it doesn't really exist situations. But if you're in very, very low salt concentration, so there is very few co-ions. But this is observed experimentally. I mean, in DNA, it's very, very frequently you can see this occurring, this manning condensation, and this kind of charge regulation occurring in DNA when you vary the charge by changing the pH or things like that. You can control the charge on the polymers by changing the pH of the solution or things like that. And then you can see this charging, this saturation of the charge and constants of the charge experimentally. OK, so this is for the counter ion case. So as you can see, the calculations are a bit laborious. And so now in the case of salt, I will just spend a few minutes on the case of salt. So the case of salt can be solved exactly, but it's extremely complicated. It's a panlevé kind of equations. And there has been a solution which was found a few years ago. It's very complicated, and it involves the theory of integrable systems, actually. So I will not discuss it. What I will do is I will discuss just to give you a flavor, the case of salt, but in the Debye-Huckel framework. So in principle, you should write the Poisson-Boltzmann equation for a charged cylinder in presence of salt this time. And in presence of salt, your Poisson-Boltzmann equation again is the same. It's going to be Laplacian phi equals 2C0E over epsilon sinh beta E phi, where C0 is the bulk concentration of the salt far away in the bulk of the system. With the boundary condition, so the boundary condition as you can see is d phi dr equals minus equal sigma over epsilon. OK, so as usual, we linearize this equation, and it becomes Laplacian minus Laplacian phi plus Kpd square phi equals sigma equals 0 with phi prime of 0 of A equals sigma over epsilon. This boundary condition is that r equals A. So this is the linearized version of this, and the Kpd square is 2C0 beta E square over epsilon, and beta E square over epsilon is 4 pi Lb, so it's 8 pi Lb C0. And that's the Debye constant to the square, or 1 over the Debye length to the square. So again, you go to cylindrical coordinates, so the equation becomes minus phi second plus phi prime over r plus Kpd square phi equals 0 with the boundary condition, and the solution is in terms of Bessel function. So the solution involves some special functions, which are Bessel functions, and the Bessel functions are defined. So the solution, I will write you the exact solution. The solution is psi of r equals beta E phi equals minus 2 Lb rho over KdA, so Kd Kpd is the Debye constant, times K1, this is a Bessel function, K1 of KpdA times K0 of R, Kpd, so Kpd is here, and it's 1 over lambda d square, the Debye length. So K0 and K1 are Bessel functions of 0 and first order, so Bessel functions, it's special functions. I don't know if you have ever heard about that. The functions which are defined by certain properties, initially they were defined by some partial differential equation. I can give you several definitions of these Bessel functions. But you see, it's funny to see that such a simple linear equation involves these complicated functions, so these functions, the k alpha, K alpha Bessel is defined by x square d2 by dx square k alpha plus xd by dx k alpha minus x square plus alpha square K alpha equals 0, with the boundary condition that k alpha of 0 equals plus infinity. And there are some integral representation, which I can give you, which is k alpha of x equals integral from 0 to infinity dt of e to the minus x cosh t times cosh alpha t. So K0, for instance, you put alpha equals 0, so it's just this exponential. K1, you have cosh t. And so you see that, for instance, K1 of x. So see, if you have cosh t here, it's just minus d by dx. So K1 of x is just minus d by dx of K0 of x. So you have this function, which is like this. What you can show is that at large, that's a property of this. So these Bessel functions are tabulated. You can find them if you use Mathematica. You can find them in all books. It's very well studied. So if you take this property, this here, this shows that when r goes to infinity, the solution phi of r behaves like e to the minus r over lambda d or kappa dr divided by square root of kappa dr. So it decays exponentially at large distance. And there is an additional factor, 1 over square root of r, instead of 1 over r that you usually have in the byhooker. But you have this standard exponential decay, which is characteristic of salt. When you have salt, you have screening, and you have exponential decay. Because I'm supposed to stop, maybe I will stop here. I will finish this in a few next time. And then I will start the field theory part next time. Any questions, yes? Sorry, notion of what? Yes, OK. First of all, all this is mean field. The theory is mean field, because Poisson-Boltzmann is mean field. So the scaling is a bit trivial and not very interesting. There is no exact, as far as I know...