 Thank you for watching who's ever watching Yeah, so at the end of the lecture this morning I I presented Some topics in irreversible aggregation. I started with constant kernel aggregation And derived the generating function solution then I turned to deriving a scaling solution Which in some sense is the most general way to deal with problems Which are sufficiently complicated that you don't know even how to start and what I want to do today is show you the generating function solution for the product kernel aggregation, which is one of the Classic exactly soluble models the solution is very elegant and very pretty and it's a little bit advanced So I hope that I'm not going to blow anyone's mind here, but it's just it's it's just so beautiful It's it's worth trying to show So we're talking about Product kernel aggregation and that is where the reaction rate between a cluster of mass i and mass j is equal to i times j and one way one can motivate this Reaction rate is you say that we're dealing with say molecules that are multifunctional units And so when and there's like K, you know for a cluster of size i there's I functional units and when an I and a j come together, you know because there's like I grabby hands from one guy and j grabby hands from another guy the rate Which they'll actually react is proportional to ij and So we want to try and understand the dynamics of this process So let me write down the master the rate equation for the concentration of clusters of mass k. So it's equal to one half summation i plus j is equal to k I see I j c j minus K c k and the other term here I See I summed over all I by definition. That's equal to one because just that's just the total mass in the system Which we can set equal to one So I want to solve this set of equations and so the natural thing is to try the generating function now here there's a few almost Preliminaries that one should do to try and simplify the algebra so First thing is that you see that there's an I and there's a J sitting out in front here There's a case sitting out in front here, which normally would appear by differentiating the generating function with respect to its argument so You know if you didn't know any better and you just tried to do the original generating function You would see after you went through some of the steps They said oh geez it would have been a lot simpler if instead of doing the generating function per se I introduced a slightly different generating function. So let me call G of Yt is equal to summation overall k K times Ck times Zk So I I it just turns out to be convenient to put a K out in front But then well, there's a Z. There's a why what did I do here? And so another point is that because I'm doing differentiation if I were differentiating perspective Z You know there would be like one less power of k of Z here And it's a little bit clumsy to deal with it So it turns out that it would be better to instead not use Z But use e to the Y so I'm going to put it here to the Yk So that's the generating function. I'm going to use so it you know Conceptually it's no different than before it's just that now I'm saying that Z is e to the Y And I'm using Y as my basic variable and I put a K out in front for convenience and you'll see let's let's work it out And you'll see that it just makes a much simpler equation Okay, so I'm going to take my original equation I'm going to multiply by K e to the Yk and I'm going to sum Overall K and let's see what comes out. So on the left hand side I just have the time derivative of the generating functions. I have g t So again, I'm going to use subscripts to denote partial differentiation with respect to t Here Let me put a little worksheet over here. So I have i j and then I have i plus j C i C j and So Let's see what I have here, and then I have times e to the Y I e to the Y e to the Y J because I have e to the Yk but k is equal to i plus j Okay, so let's see what we can do with this so There's there's two terms and so there's a term which is like I squared, you know, I I squared j C i C j e to the Y i e to the Y j And so the J C j e to the Y j is a generating function The term I squared C i e to the Y i is just a generating function differentiated with respect to Y one time and that's why I use e to the Yk because it involves derivatives and I don't have to worry about losing a power So the very first term and then there's a one half here That's going to cancel the fact that there are two identical such terms here with i and j interchange so I have G times d g d y times two and the half and so I'm going to get here G d g d y so that's the first term In the second term so I have k squared C k e to the Y k So that's a derivative of the generating function with respect to y just brings down one more factor of k So this is just minus g y so I get So I'm going to get g y G G minus one and So if you stare at this for a second if there wasn't this it's just a good old linear wave equation now it's a linear wave equation, but with a Velocity that depends on the amplitude of the wave So it's a nonlinear wave equation and in fact, it's the burgers disquiet burgers equation in disguise So the fact that it's a nonlinear wave equation means that typically you have wave breaking phenomena And it turns out that the gelation transition corresponds to this wave actually breaking so that's one point second point is now we want to solve this equation and You know again, it's like one of these things that It's a first-order partial differential equation. It's soluble by the method of characteristics And if you haven't seen method of characteristics, it looks mysterious But it's like something that any sort of math major will take and it's very simple in principle So let me rewrite this equation as gt. Actually, let me be really pedantic. I'll write it as g d d d d d g d t plus one minus g d g d y Is equal to zero But let me also write this as dg the total derivative gt and And so if I do it this way then I can also write this as let me go back up over here So the second right is dg Because this is g is a function of y and t so I can write this as dg by dt plus dg by dy dy by dt Equal zero so the point here is that g is a fun is a two variable function of y and t It satisfies this partial differential equation I can think of it as a total time derivative if I understand, you know If I when I compute the total time derivative, I just get the partial time derivative plus the part that involves the derivative respect to y So the point is that on what's called a characteristic curve so on the curve where So so on the curve dy by dt Is equal to one minus g So if dy dt is one minus g then this equation is the same as what we have before and The curve is g is constant so on this curve Keep looking for my brush On this curve G equals constant So, you know This is very similar to what we saw in the case of the linear wave equation that you had that The function you had to solve for was some very simple function that you would match to initial condition Here we have more or less the same kind of situation occurring So What we have is that on this curve G is constant so we can integrate this equation and get y of t is equal to 1 minus g t Plus some constant and let me call that and so in particular at t equals zero y at t equals zero Which is equal to f and now we match that to the initial condition and once we do that then we can work this out and Solve for the generating function. So initially our generating function. So at t equals zero G of y and t equals zero and again for the monomer only initial condition So if there's only monomers then this sum at t equals zero is just one term one c1 e to the y Is there something wrong with that? Because you see that over here If if this is true D y dt is 1 minus g. So then the that's the equation D g dt plus 1 minus g. Yeah, and by definition that was equal to Zero so it's constant. So anyways, so for the monomer only initial condition We have you know c1 is equal to 1 so we have one c1. That's one e to the y so this is e to the y So at t equals zero y is equal to log g So finally y of t At any time is equal to 1 minus g t plus log g so You know looks like I haven't you know looks like I'm just going around in a circle But this is actually the solution to the problem. So let's exponentiate this. We'll get e to the y Is equal to g times e to the 1 minus g t or G e to the minus g is equal to e to the y minus t e g t here, so This is an implicit Solution for the generating function. So the thing is that you know We have this complicated function of g is like a is like some function of time And so in some sense we have time as a function of the generating function But what we want is the generating function is a function of time So we want to do something with this and invert this series or invert this function to figure out g is a function of time And this is where the black magic comes in Anyways, let me just do one more step here, which is let's call this g t is e times e to the minus g t So it's all involving g t is equal to t e to the y minus t And let me call this thing x and so then we have the equation that x e to the minus x and let me call all this stuff here equal to y equal y So we have y as a function of x But we want is x is a function of y if I could invert this turn this Function inside out then I'd have the solution to the problem But in fact, we don't need to turn the entire function inside out We just because g is is you know what we really want to the term the power series representation of g So we only want not the full function x but you just want its power series representation And so how do we do something with that and I keep losing my rags? So let's do something with that And so there's something and so the black magic here something called the Lagrange inversion formula and If you haven't seen it, it's so beautiful You know, it's just every time I see it like I'm always in awe of just like the wonders of complex analysis So what we want? The power series and but I just want to check that my notation agrees with what I have here So I won't confuse myself So I managed to do something exactly backwards Compared to my notes, I'm going to call this thing y I Want and I'm going to call this thing x sorry about this If I do it the other way around then I'll get halfway through it. I'll get myself confused Okay, so I have x as a function of y what I want is a y as a function of x. We want the power series for Y of x and so this is going to be some power series a n x to the n summation n equals, you know one to infinity So how do we formally compute the power series? Well, we'd say well if I want the nth term here what I can do is Convert this from a power series to a Laurent series by dividing through by x to the n plus one and doing a contra integral so a n It's equal to 1 over 2 pi i a contra integral of y divided by x to the n plus 1 dx But y is this Thing that we don't know But now here comes the black magic and it's incredibly simple, but very powerful which let's do the following Let's convert this integral in the complex x plane to an integral in the complex y plane by just doing this dx dy dy and So let's see what we get. So this is 1 over 2 pi i the contra integral Okay So X we know what it is. So we have here y divided by x to the n plus 1 so that's y to the n plus 1 I e to the minus n plus 1 y and dx dy well, I don't know what that is, but I know what because I have Or yeah, I do know so let's compute now dx by dy so that's going to be So that derivative is e to the minus y out in front and when I differentiate with respect to x There's one term which is just one and another term. I'm sorry. There's a term with y and a term with a minus Try again dx by dy I can differentiate the y and get a 1 or a differentiate up here They get a minus y so there's a 1 minus y So I just convert the integral in x to an integral in y Um, so let's see what else I can do with this So Almost done So what we're going to get then is that a n is equal to 1 over 2 pi i integral so I have So this cancels one of these guys So I'm going to get 1 minus y and then this cancels one of these guys and then I have divided by y to the n And then I have e to the n y dy and once again what I want is the The You know this is now the nth term in the power series representation of the function By the way, one other point I should mention is that notice that for y and x both very small I can forget about this so for small y or small x y and x are the same thing and because I'm doing the integral around a Small circle about the origin because x and y are linear functions of each other I don't do any violence to this integral by changing variables from x to y because for x and y going to zero They're both the same variable. So I didn't do any violence by this by this change of variables But the thing is that this function is now sufficiently simple that to evaluate its residue I don't need to do any integration I can just forget about it about doing the corner integral and I just evaluate the residue directly from here So what I would say then is a n is equal to the residue of the following thing so I have One over y to the n minus one over y to the n minus one and Then here that this has a simple power series representation This is a summation from k equals zero to infinity of n y to the power K Over k factorial So to find the residue I just have to find the coefficients of like one over you know y to the n plus one So this is equal to and so what I'm going to get here is that for this guy. I want The term n minus one will combine with this to give me a term which is of order Give me a term of order, you know the residue so there's going to be one term which is like When k is equal to n minus one so it's n to the n to the n minus one over n minus one factorial and then this guy is going to give me n to the n minus two over n minus two factorial so that's And then this and now allow me because this is the kind of thing I suck at let me just Tell you that this has a very simple Thing here. This is just equal to And to the n minus one over n factorial trust me that that's correct so finally Why? Which is equal to? gt so again, we were stuck here we had Gt was you know like here, but now we have solved for y as a function of x or at least found its power series representation This is the coefficient in the power series representation. So this is equal to summation a n x to the power n so this is equal to So let me write it Actually, let me write it as summation a k x to the power k k equals one to infinity and so what I get is so there's this k to the k minus one over k factorial and Then I have x Which is this guy here? So it's going to be t to the power k minus one e to the minus oh, sorry There's also there's I have to multiply by t so there's a t to the k minus one But then there's t to the k and then I have the exponent e to the k y e to the plus Sorry e to the e to the k y e to the minus k t So that is my generating function and This part and One power of k is I erased it crap in the definition of the generating function Remember that it was like c k c k e to the y k. So I take the cove if I take Get rid of the e to the y k divide by one power of k then I have the concentration So the final concentration is c k is equal to k to the power k minus 2 over k factorial t to the k e to the minus k t So that's the final answer Above so the n to the n minus 1 divided by n minus 1 factorial minus n to the n minus 2 So apparently that is n to the n minus 2 divided by n factorial Where sort of going from here to here. Yeah So No, the downs down here. Yeah, that that one. Yes, okay So because if you take the second term and multiply by n minus 1 factorial Then I don't know. Okay. Okay. I mean it's again. It's like it looks confusing and every time It takes me three times to get it right, but it's okay. Yeah, okay So anyways, that's the cluster size distribution. So again There's nothing here. That's like technically very difficult, but it's just that there's some Concepts here, which are probably unfamiliar to the average Student, but it's just that some of these methods are so beautiful I just wanted to like expose them Okay, so now that we have the Solution we can ask well, let's do something with it. Let's extract some physics out of all of this and So let's look at this thing. So we have ck is equal to K to the k minus 2 over K factorial T to the power k e to the minus Kt And so typically what we're interested in is in the limit of large k We like to look at the asymptotic form of the cluster size distribution. So asymptotically I'd say well let's use Stirling's approximation and Get rid of k factorial and for analytic functions. So k factorial is nothing more than k over e to the power k times square root of 2 pi k So we have k to the k minus 2 K over e to the power k times square root of 2 pi k that's Stirling's approximation T to the k e to the minus kt By the way, I'm using the last piece of Japanese chalk that was on the chalk drop, but it's like disappearing so Something is going to happen to my lecturing style very quickly. So anyways, I have here a square root of 1 over 2 pi There's So k to the k that cancels this and then there's k to the minus 2 with a root k So there's a 1 over k to the power of 5 halves and then I have t to the power k e to the minus So I've k t and then there's one more e to the k here. So it's e to the minus k t minus 1 And now the other thing is now what we need to do is so how does this function? This is the large k form of the distribution. How does this vary as a function of time? So we want to look what's happening as a function of time. So let me do one more step here Whatever the square root of 2 pi 1 over k to the 5 halves and then I have e to the minus So what have I got here? So I've got minus k t plus k minus k log t and so What's interesting is that if you stare at this for a second first of all notice that when t is equal to 1? When t is equal to 1 this is 0. This is 0. So e to the 1 so at t equals 1 This is a power law decay for all k Whereas for small times, it's There's an exponential here and so there's an exponential and to figure out the asymptotics of this exponential Let's suppose. Let's look at assume or let's try T is equal to 1 minus epsilon because we know that when t is equal to 1 the exponential goes away and so pure power loss So let's look close to this point and then What I have here is minus k t plus k Minus k log t is equal to minus k. So t is 1 minus epsilon plus k Minus k and then log t. So that's log of 1 minus epsilon. So it's minus epsilon plus epsilon squared Sorry t Let's get this straight. So here I'm going to write No, I guess that's fine Everything I was everything is fine. So log t. So Logarithm of 1 minus epsilon. So it's minus epsilon minus epsilon squared over 2 dot dot dot dot Yep, ask you have so plus k and then minus k log t Plus k Mine plus k and then minus k log. Thank you. Thank you try. Let me try this again Plus k and minus k log t. So this is this will make look a little bit better. So it's minus epsilon minus epsilon squared over 2 Okay, so This k cancels that k Plus k epsilon cancels this It seemed to have a sign mistake somewhere. No, I said so this is this is fine because minus minus it makes a plus Which is plus k? I should put plus k t here Okay, all right and Minus k. Yeah, I mean, this is the kind of thing. I really am terrible at so anyways trust me that everything cancels out except for the term minus K epsilon squared over 2. So the point is that this distribution Asymptotically looks like k to the minus 5 halves for t equals 1 and for t Less than but approaching one. This is k to the minus 5 halves e to the minus k Over 2 an epsilon is 1 minus t squared so this is the exponential decay and we can rewrite this as You know, I got to start with a bigger piece of chalk. This is not working anymore. So it's k To the minus 5 over 2 e to the minus k over k star Where this characteristic size k star? It's equal to 2 over 1 minus t quantity squared So this is the main result of this product kernel aggregation process that indeed the cluster size distribution is a power law At t equals 1 and for t less than 1 it's a power law with an exponential cutoff And if you remember from earlier the previous lecture, I showed I argued that for Type 1 kernels, which is where large large interactions are dominant Which is what happens in product kernel aggregation that the cluster mass distribution my plot it ck as a function of k So let's plot log of this thing. So it's a straight line and here There's a cutoff an exponential cutoff at a k star Which is partial to 1 minus 1 over t Square log of it. Okay Yeah, sorry, it's log ck versus log k. Okay For those of you who really remembered the the previous lecture you might ask me the following question We saw last time that if I have a homogeneous kernel that the Typical mass when the homogeneity index was equal to 2 that the typical mass should Blow up as 1 over 1 minus t And now I have a different mass 1 over 1 minus t squared So how come I have different masses and part of the reason I have different masses is that I have a power law Distribution and so if I were to compute the average mass the total mass in the system that should be something like the integral of K ck integrated up to this upper limit and So what is this? So this is the upper limit k ck, which is k to the minus five halves DK So this integral is like k to the minus three halves integrated I get k to the minus one half I get the upper limit. So this is like k star to the power One half and so even though the upper limit where the cutoff occurs is blowing up like 1 over 1 minus t Quantity squared the total average mass contained in this distribution is of the order of 1 over 1 minus t So that is the gelation time So the thing is that you you know you take your jello you put it in your bowl You mix it up you put it in the fridge and as time goes on you're getting larger and larger clusters and at t equals 1 in all of these units magically an infinite cluster appears and You have gelation So what do you see that the solution for t larger than one is on physical? No, I mean it's it's not that it's on physical and you know This is very much like what happens in the classic percolation problem. So in classic percolation at PC there's an incipient infinite cluster below PC of only finite clusters at PC you have an incipient infinite cluster Above PC of one cluster that contains a finite fraction of the total mass in the system and all that's left is finite clusters So the cluster size distribution is still well defined for percolation both above PC and below PC It's just that above PC There's this one giant cluster that kind of has most of the mass and you just have the rest in the finite clusters And the same exact same thing is happening here. Well, I started at 2 at 230, right? Okay okay, so that is the end of the story of Aggregation and now I want to turn to the next section of this class which is irreversible absorption kinetics and The point here is that Once again, I'm going to try and argue to you that by taking a kinetic approach one can solve a complicated Configurational problem in an easy way by using, you know judiciously defined master equations So first of all, let me define what the problem is so imagine you have a sticky substrate and Molecules from above are impinging on the surface and whenever they land on the surface They stick there forever and then the question is as time goes on the surface is getting more and more covered And how covered is the surface? So we're talking about irreversible adsorption and let me Just talk about this in one dimension as I'm going to mention later on The one-dimensional case can be really solved completely high dimensions, there's only like asymptotic results and you know in general for a lot of these non-equilibrium processes Solving a one-dimensional rate equation like a birth death where there's a sort of a one-dimensional basic variable like the location of a particle the Number of particles the fraction of surface covered these can all be solved exactly But as soon as they add another spatial dimension basically very little is known But so the way I define the problem is have a one-dimensional substrate And so these little holes here meant to represent adsorption sites And so from above our molecules that you know they float around and they come in and They when they hit one of these things they absorb forever So in the case where we're talking about monomer adsorption The problem is really easy to solve and so let me just solve it because to give you a flavor of what the basic degrees of freedom are So let me call V the density vacancy density and so particles are being impinging at a fixed rate So then V dot So the density of vacancies they go away with a rate proportional to V itself So V is just equal to e to the minus t and The coverage so let's call row the density that's equal to 1 minus e to the minus t. So as t goes to infinity The coverage goes to one so the surface is full So there's nothing interesting to talk about in the case of monomer adsorption But the first interesting case is suppose now instead of absorbing monomers you absorb dimers So you can think of these as rigid dumbbells So I have something like this now and it It does something it does whatever it does But then it comes down and it occupies two nearest neighbor sites on the surface But now you could imagine a situation where this another dumbbell up here Then lands here and Now you've got this site here that's never has any chance of being filled and so the interesting question to ask here Is what is the coverage at infinite time? It's obviously less than one, but how much less than one and You can sort of get some idea of a bound on what it's going to be because the smallest the coverage can be is that if For every dumbbell there's a vacancy next to it then the minimum coverage is two-thirds And the maximum coverage is one, but what is the coverage and Just from the point of view of history it turns out that this problem is first solved by Polymer chemist named Paul Flory in 1939 and the solution is published in journal chemical physics and If you're interested it's worth reading it because it's only about a two-page long paper, but it is such subtle and beautiful combinatoric analysis and you know when I read it I think Yeah, yeah, yeah if I was twice as smart as I was as I am and I work twice as hard as I did Yeah, I think I could figure out how to do this and you know There's a very famous quote by Mark Cots a famous mathematician who died in the middle 80s Who described the difference between a genius and a magician? And so a genius is when somebody does a calculation they look at and say yeah, yeah, yeah If I was twice or three times as smart as it was and I worked three times as hard as I should I could do this But then a magician is so many look at what they did and you said I have no idea how they did it and I've been privileged in my life to meet a few magicians And when you meet a magician, you know it right away and I could reveal to you who I think my magicians are for me But anyways, let's go back to the problem of solving this irreversible absorption problem So first thing we need to do is we've got to like define our basic variables So at some intermediate stage of coverage, you'll be some sites that are occupied some sites that are empty and so There's one basic variable that you might think of defining And it's like what I would call the vacancy density So let me define it this way Vm is the probability of Having the following configuration an occupied site M empty sites in a row an occupied site And then I don't have to worry about anything outside of that That seems to be like a very natural way to describe the state of the surface So if we use that description Let's now try and compute the time evolution of the vacancy densities starting from an initially empty substrate So initially all the vacancy densities are one because the lattice has nobody in it So all these probabilities for any length interval be equal to one and so let me try and write down a rate equation for the time evolution of Vm so First point is that if I'm dropping down dimers So I have like a you know a parking spot of size M and we have a car of size 2 How many different ways can I drop a car? Somewhere inside of that parking space and if I do that then I kill this This this vacancy of size M and clearly if I have a car of size 2 I could use the first two spots the middle two spots the last two spots Basically, it's M minus one spots that I could put my car and so there's a minus M minus one Vm So that would be the loss term But then there's a gain term because I can have a much bigger vacancy and I could park somewhere in the middle of the vacancy I guess it never happens Italy to have lots of parking spots empty in a row But if you did do that then you would have plus a gain term and so that would happen if I had a Vacancy of size M plus 2 let's I see so let's do it like J plus 2 Summation M equals J to infinity. So if I have a Parking spot which has is this huge then you know like so if I have a big empty spot here I could park my car here Making something of size M or I could park my car here Making something of size M over here So there's two different ways that I could drop a car into a huge empty area to create a gap of size M And so there's going to be a two out in front So this is the math rate equation for the vacancy density and again It has this generic character of that there's a loss term and there's a gain term and in fact there's another way you can view this equation because What's happening is that you're taking a gap and you're cutting it into two pieces and So if you take a gap and you cut it into two You lose something but if you take a really large gap and cut into two such a one of the smaller gaps Is the size that you're looking for M then you gain And so in fact you could use the same formulation to describe the kinetics of a fragmentation process that you're just breaking material into smaller and smaller clusters and in fact You know I could give a chapter which I'm not going to do but I could give a chapter on fragmentation connects where this would be my starting point but in the case of car parking density This turns out to be not the right way to deal with the problem because you know I just arbitrarily introduced this quantity visa been the vacancy density, but it turns out there's another quantity which is simpler Whose equations of motion are simpler in character. And so let me look instead at what I call the empty interval density Excuse me before you proceed any further. Sure. I ask you like Clarification of the quantity VM. Yes, so it's a probability to have at least something like you draw drew like with m vacancies Right, what exactly m vacancies exact and but since you have like you imagine this long chain Right, and this is the probability to have one of these structures or no So the thing is that you know, I have my my large infinite dimensional lattice I just count the number of suppose it m is 12. So I just count how many How many parking spots of size 12 exist in my entire lattice divided by the size of the lattice that would be v12 Okay, yeah, okay So let me now look at a different variable which I'm going to call the empty Interval density and this is e sub m and this is going to be the probability of The following configuration. It's m Sites in a row that are empty and then I I put an x here for I don't care So I imagine that I'm looking at my lattice of blinders that are of size m And I say do I see m empty sites in a row and here? Yes So that adds one to em I move over by one Do I see m empty sites in a row and if the answer is yes, I add to e one But if there's like a now I There's a car or particle at this site at the edge then it's we don't contribute but the crucial point here is that I don't pay attention to what's happening outside of this interval and in so doing as you're going to see it actually makes For a much easier description because notice that this rate equation is non local because it involves m and all sizes bigger than m and so it's a little bit harder to solve than if I have a local equation So let's look at the rate equation for em dot So there's the same loss term as before m minus 1 em But now the only other way that em can evolve. I mean again notice that if I have blinders on There's no way I can create From an em you know a cluster an empty interval of size em if it's if I have an interval of size em The only thing you can happen to it it can disappear because I'm either parking in the interior or I'm parking at the edge There's no way I can create by looking with blinders of this size an empty interval of size em So how do I also lose I can I can imagine that I have? You know here's my I guess Let's make it the same number of sites So I have my m interval and if this happens to be empty Over here and my car parks here Then that loses the em em the empty interval of size em over here And I could park the car at the right edge or at the left edge So there's again two ways that I can park but it's just em plus one So it's much simpler because it's a local equation and in fact you'll see that it's it's easily soluble So is it a minus sign? Minus, yeah, exactly. It's a minus. There's no way of create. There's no creation process here So that's another feature that this is now a monotonic process. And so it's a little bit easier to deal with Okay, so how do we solve this? Well, you know, you see that it kind of looks like I don't know You know, it's a first order equation. I could do the following thing I could write em dot plus m minus 1 em is equal to minus 2 em plus 1 Then then you say well, let's write this as a total derivative. And so if I write this as em e to the m minus 1 t Dot so I multiply through by e to the m minus 1 t then this plus this is just a time derivative of this combination This is equal to minus 2 e to the m plus 1 e to the m minus 1 t So this combination of variables is going to be very handy throughout. So let's call this thing equal to phi and so this is a function of time, you know some function of time and in fact What this is doing in some sense is that it's separating out the time dependence from the m dependence So by its definition so phi is equal to e to the m e to the m minus 1 t or Another or if I just turn it around at em is equal to phi of t e to the minus m minus 1 t So you see that em which depends on time It's a two variable function and now we separate it into a function of time only and then a function which involves Size only and it turns out to be very convenient So anyways, I'm running out of space here. So let's go back over here. So we now have phi dot So listen, yeah in the second process So that involves the probability that the Extrasight is also empty right the probability what in the in the process in the second The two em plus one process right that involves also the probability that the extra site is also empty That's right. That's why it's em plus one So here I in order to park in the middle of the interval I just have to have em empty sites in a row to park at the edge of the interval I need to have em plus one sites empty, but I don't care about em plus two or em plus three or anything like that Okay, so we have phi dots is equal to minus two and now let's rewrite this in terms of phi So, you know, here's this involves em plus one, but em plus one Is equal to phi e to the minus m t just I just change m to m plus one here And so on the right hand side I have two Phi e to the minus m t I just have e to the minus t Yeah, yeah, ask away Yeah, exactly. You don't care exactly so Yeah, you know and once again, it's like you have to play with these variables for a while to like feel comfortable like what really is a difference between these two variables Yeah, but yeah, I'm just saying you know that what's inside of this time derivative. I'm just calling that equal to phi No, no, no, but phi depends on em obviously, but if I turn it around then em so there's a part here which doesn't involve em Okay, well, let's let's go on because you see let's go over here. Where's em? There's no em here Let's let's let me just let me do one or two more steps and hopefully you'll be convinced So anyways the point is that now we have to solve this equation. So this is log of phi Dot is equal to minus two e to the minus t. So phi of t Is equal to phi of zero. Well, actually let's do log of phi of t minus phi of zero. Let's just do it Slowly, so that's the the time integral of this. So it's to the integral from zero to t e to the minus t v t So that's equal to minus two So I integrate So I get this and then I'm going to change it around. So it's going to be one minus e to the minus t So finally phi of t Is equal to e to the minus two one minus e to the minus t So I'm almost done because we've now found a function phi and let's now compute The coverage that's the thing that we really want is what is the coverage? So first of all we have in fact more than just phi. We have em of t. So that's equal to e to the minus two one minus e to the minus t Times minus em minus one t So this is the time-dependent coverage of any empty interval of size m. So in particular e one Just it equal to e to the minus two one minus e to the minus t e one is a probability that there's an empty interval of size one One minus that quantity is a probability that there's a full interval of size one So the density the coverage row of t is nothing more than one minus e one of t This is equal to one minus e to the minus two one minus e to the minus t So final result row at t equals infinity Just plug t equals infinity and so this goes away, and we have one minus e to the minus two That is the cool result. That was first arrived by Florian 1939 and this number here Is like point eight six four dot dot dot something or other Isn't that beautiful? Yeah, I mean I just I lever every time I do this I just I'm always in awe just because it's so pretty So that is the final coverage and as you see it's somewhere between two-thirds and one and it's sort of like halfway in between almost And so there is the coverage for irreversible dimer absorption Going back to Angela's question. So in the last two equations there So you use So fee in principle should also depend on M, right? So in the last equation you should have a fee of M plus two I'm sorry. I'm confused here. So yeah, so so that is okay. It's a definition of phi, right? So I didn't do anything here yet. You didn't do anything But then when you write the same equation for M equal to M plus one, right? Then the fee if it depends on M should be a different function Yeah, but so there is an assumption that fee does not depend on M And I'm wondering where this come from. I guess what I would say is that if I you know if I don't say anything What fee is I don't I haven't made any assumption. I just said oh, I've just sort of regathered things And I look at the equation of motion for fee. There's no M in it. I mean there's no M So no because the initial condition I start with an empty lattice So the initial condition is every everybody is is one all empty interval sizes are one Okay Right, but the other thing to emphasize here is that in addition to getting flories result We have the concentration at any time And so if you now go back and look at flories original paper and struggle through it And it would take you to take the average person several days to struggle through all the details I argue that you know one nice thing about doing the non equilibrium formulation is that You know with Semi-elementary methods one can calculate the full time dependence and then the final coverage drops out as a corollary Whereas in flories calculation one has to work very hard. So it's it's kind of cool Okay, so there's sort of excuse me. There's two natural Extensions that we can think about here One is what happens if you have longer molecules and then the other one Which is maybe more relevant for Italy is suppose that instead of having like discreet size molecules You have like little fiat Cinco Cincos that are continuous trying to fit into a parking spot so let's look at these two extensions of this basic process and The thing which is kind of cool. Oh, I should mention one more thing about this Which is that now you can ask how quickly does the coverage approach is limiting value and it turns out that if you're dealing with discreet Molecules impinging on the surface the approach is exponential But as soon as you have continuous size Particles or continuous parking spaces then the approach is power law. So actually let me just do is one more corollary here Let's look at row of T. I Actually want to do it the other way around row of infinity minus row of T What is this? So this is 1 minus e to the minus 2 1 minus e to the minus t Minus 1 minus e to the minus 2 So the ones cancel and so I have e to the minus 2 Minus e to the minus 2 e to the 2 e to the minus t And for a long time I can expand this in a power series So this is going to be e to the minus 2 minus e to the minus 2 1 Okay, so when t is large, this is small quantity so we can expand so it's going to be 1 plus 2 e to the minus t and so The leading to e to the minus 2 cancels and we're going to get e to the minus 2 e to the minus 2 so that's some number times e to the minus t So the point is that the approach to the final density is exponential in time Okay So let's now extend this to absorption of camers Here it's like everything follows Exactly the same every every step is more or less exactly the same but it's just that it becomes like a little bit more tricky to try and keep track of the books and So in fact, I'll refer to my notes because I always managed to get this messed up But if we're doing camer absorption, there's two possibilities first of all you can have Your parking spot bigger than the car or the parking spot smaller than the car in the case of dimer absorption There's only one parking spot smaller in the car And that was a single a single empty site where nothing could happen So we didn't really have to worry about that, but here we actually do have to worry about The different sizes of parking spots so let me just write down the equation and then it's it's and You're going to see it's going to very much parallel what we have for dimer absorption. So em is equal to minus M minus k plus 1 em minus 2 summation em plus j j equals 1 to k minus 1 and this equation is true for m Larger than or equal to k So that means that the parking spot is big and let me write down the equation for for for a k less than m so it's small parking spots and in fact once we know The equation for the first one we can get the equation for the second one very simply because if we're trying to So actually let me explain this equation a bit more and then I'll write down the equation for the other guy So let's understand like what are all these different terms here? So once again, this is where we're going to park in the middle of our parking spot so for example if my parking spot is size Seven for example and my cars are size three So my car could be here It could be here It could be here here or here So the thing is that in principle if a car parked in the very first spot the tail was K minus M But one more spot because you know the last car the last the end of the car is over here So the end of the car could be here here here here here, which is just m minus k plus one So that describes that term on The other hand if I want to look at like sort of parking at the edge well You know if my car is of size three then the car could have parked here or could have parked here and If it parks here, it's already like an interior parking event. So that doesn't happen So if I have a size three if this was you know a car of size three There's two ways that I could park in an interval of which is big enough to accommodate You know that part of the car and so that is the equation for Gaps of size M when you're parking cars of discrete size K In the case where K is less than M then all you have to do in some sense is you're just Interchanging the role of gap and car. So when I was parking like a car in the middle of the gap Like this for example So in some sense I said well, I have to fit my car inside the gap If instead my gap is size three and my car is like a stretch limousine of size five for example Then I just have to count the different ways that this gap fits in Exactly inside the car and all that means is just interchange the role of K and M So I can get the equation just by interchanging K and M and so I'm going to have K minus M plus one E K Minus two integral J integral summation J equals one to M minus one E K plus J and that turns out to be the equation of motion for the other case and You know, I if you if you spend like another half hour like sort of puzzling out all the different combinations With a lot of work. You'll get to the same thing, but just by interchanging the role of car and gap It's it's much easier to get the same answer. So now I'm going to solve the set of equations by exactly the same Technique as I did for the case of dimer absorption. So once again, let's look at You know this set of equations and again, we have like a similar structure. So let's write this as E M E to the M Minus K plus one T So I bring this to the left-hand side of the equation And so if I multiply through by E to the M plus minus K plus one T The left-hand side is exactly this And on the right-hand side we have On the right-hand side then we have Minus twice summation And so it's going to look just similar to what we have here. So we're going to have E M plus J E to the M minus K Plus one T and once again, I'm going to call this thing Phi and so That says that E M Is equal to Phi E to the minus M minus K plus one T Okay, so Let's let's do something with this. So what we're going to have then is this equation So on the left-hand side We have Phi dot and on the right-hand side, we have is equal to minus two. We have a summation J equals one to K minus one and Then we have E to the M and So I'm going to write so E to the M plus J is just Phi and Then it's going to be E to the minus M plus J Minus K plus one T and then I have this Yeah, I have And I first I multiplied through by E to the M minus K plus one T so I have to move so E to the M minus K plus one. I'm sorry that my writing has gotten a bit slow small here Okay, so the M's cancel the K's cancel The ones cancel and so finally get a very simple equation Phi dot is equal to minus two summation J equals one to K minus one Phi E to the minus J T. I forgot a T here So you see that it looks not so diso Ran out of blackboard space. I had a very similar looking equation for the case of dimer adsorption There's only one term in that series and then you could integrate it So it's the same thing here. It's just they've got J terms in the series rather than one term in the series So we have log of Phi Dot is equal to minus two Phi E to the sorry E to the minus J T J equals one to K minus one. So I integrate that I'm gonna get Phi of T And again using the initial condition that my system starts out empty. So Phi of zero is equal to one So we're gonna have EXP of Minus two and then I'm gonna have summation J equals one to K minus one one minus E to the minus J T Over J So we're like sort of 90% done. We have our Phi function and now we have to compute the coverage. Oh, yes, I'm sorry Thank you. Thank you. Thank you By the way, I feel I'm going very fast because you know, there's nobody, you know There's very few people in the classroom to say wait a minute. What did you just do? So, you know Like in the peanut gallery, just please shut out a question Anyways, so now we have to compute the coverage and so that's equal to roll of T Which you know is equal to one minus E of one of T so We have to look then at E of one and its equation of motion. So Let's see. Where is E of one? So this is for big parkings You know I'm looking at this and I'm realizing dyslexia struck again This is you know, I wanted to have the two different limits. This is the same thing. So please excuse me This is for M less than K Sorry So anyways, so here's the equation of motion for small gaps and in particular for E one E one dot is Minus K E K So look at the equation for E one dot So it's equal to minus K E K Which makes perfect sense if I have a gap of size one and a car of size K The car can park in any one of K places and cover a gap of size one and there's no Possibility of edge parking. So this term doesn't play any role. So this is the equation for Gaps of size one and so, you know E one of T Is equal to minus the integral K E K of T DT from zero up to T and The density row is just equal to one minus E one And so this is equal to one minus K the integral from zero to T E K of T DT But now E K well, you so you know, it's now going to look a little bit funky because it's one minus K Integral from zero to T. So E K is What it's going to be Phi So E K is Phi e to the minus T so So it's um Phi so it's e to the minus T prime and then times Phi and Phi is times exp of Minus two summation J equals one to K minus one of One minus e to the J T prime Over J So that's the answer So that is the coverage in the case of kmer absorption So one thing about this type of problem is that you have double exponential So it always looks funky and every time I see it like I'm always scratching my head like what does this mean? Where's the double exponential? but you know, it's like a simple analytic function and What you can also show is that in the limits of infinite long time that the approach to the final coverage is Exponential time and then you can numerically evaluate This integral for infinite time and get what the coverage is for each value of K And so in the case of dimer absorption the final coverage was like you know point eight six six for trimmers It's a little bit smaller for formers. It's a little bit smaller still if you go to the limit where K goes to infinity Which means that you have the continuous limit then it can be solved again And it's called the reny constant and the asymptotic density is like some number like point seven four seven something like that So the last part of the story which I'm not sure I'm gonna finish is let's now look at the absorption of cars so this is like You know what I saw in Rome when I was there is that creative parking is everywhere and people are squeezing in cars and like every possible spot and the point is that When you do that then There's a possibility that there's a car if your cars of length one Somewhere in the system is a car of length one plus epsilon And so eventually that gets filled and then then there's a car spot of size one plus epsilon over two and eventually that'll get filled and because There's that possibility of like just filling in a spot Which is just infinitesimally longer than the car the approach to the final state instead of being? Exponential in time is the power law in time. So let's now talk. Oh, I guess I don't want to erase this Kamer Absorption so let's keep that here because I'll write parallel equations for car absorption car And let me put this in quotes because there are one dimensional cars to begin with and they can fit into any spot Which is infinitesimally larger than the car to infinity and so now the basic observable So now let's replace em which is a discreet variable with e x of t so this will be the probability that Interval length x is empty But I don't care what's happening just outside and again. That's just the simplest way to deal with the problem And so now we want to write down exactly parallel equations for Kamer absorption for car absorption and here because fortunately I didn't erase the equation So I can sort of like almost cheat by like looking at these equations to determine the equations of motion for car absorption So let's first of all look at D e of xt dt for the case where The parking spots are ample in size so x is bigger than One so I should say now when I'm talking about car absorption. I'm going to assume length one cars I Can choose the length to be anything by rescaling length So I'll choose my cars of length one and so my parking spots have to be at least a size one to accommodate a car So if the parking spot is one or bigger then I can accommodate a car in two different ways I can park inside the spot or I can park like at the edge of the spot if the spot is actually a little bit bigger than than one So there are two terms here. There's minus x minus one e x I Hope that term is now kind of obvious because If you have a car of length one and a spot of size x then you know The car can sort of fit the front end of the car could be right at the beginning or it could be a distance x minus one Away from the beginning and that means the end of the car would be at the end of the parking spot So there's x minus one places You know quote-unquote places to put the front end of the car Let me move this a little further to the right And then there is minus two and then I and now instead of a sum I have an integral so I've an integral from like x to x plus one E y t dy Again, this is parking in the interior of the interval This is parking at the edge of the interval either in the left side of the right side And now my parking space has to be at least between size x and x plus one in order that the car I Can you know fit for one thing and that that the that the end of the car actually impinges on the original parking spot of size x and then there's a similar equation for Small parking spots x less than one and here all I have to do is just again interchange the role of x and one So here we have minus one minus x e1 t minus two x to x plus one I'm sorry one to x plus one e y t dy and So again, we could do a careful enumeration of all the different configurations But once you're used to it all you just say is you interchange the role of car and empty interval and you get The equation in this case so it's like now to solve these equations you do exactly the same as before So now I'm going to define You know, you know, let's bring over this thing onto the left-hand side of the equation And so we're going to get e x t e to the x minus one t The time derivative of all this thing is equal to minus two x to x plus one e Y t dy times e to the x minus one t So I'm calling what's inside of here. This is my function phi And to turn it around then that says that e x t Is equal to phi e to the minus x minus one t So notice that for x equals one E one t is just phi just a useful limit. So now, you know, everything is kind of the same as before and I wish I had another Panel, so let's see if I can squeeze this in somehow So we have phi dot So that's let this is the left-hand side is equal to minus two And then I've integral from x to x plus one So e y so that's equal to phi e to the minus y minus one t times e to the x minus one t dy So this is equal to minus two integral from x to x plus one so the phi comes out and then let's see so I have plus t minus t so that goes away and then I have Put leave stuff inside the integral that should be inside so it's equal to minus two Phi e to the x t and then it integral from x to x plus one of E to the minus y t dy and so this is equal to minus two Phi e to the x t and so I'm going to integrate this thing I'll get minus but then I'm going to put the lower limit first So it'll be back to a plus sign and so it's going to be e to the minus x t Minus e to the minus x plus one t and when I integrate dy there's a factor one over t So let's put the one over t over here and Finally, I'm going to get minus two phi One minus e to the minus t over t So that is my phi dot and so now I want to compute e x Well, okay, so you know you can see now. I'm almost at the solution. So let's just let's just finish this I guess I have time to do this. I didn't think I was going to get this far So I'm a little worried that I've gone way too fast You can tell me afterwards if it's too fast for future lectures Okay, so Well, I'm trying to Slow you down with stupid questions, but I hope the students will also help me. Okay. Okay All right. Anyways, um Where am I here? Okay, so so now what I have here so I have phi dot Is equal to minus two phi One minus e to the minus t over t So I'm going to have log phi Dot is equal to minus two one minus e to the minus t over t Terrible penmanship. I should flunk rate to again T over t And so what we're gonna so if you solve for that we'll get phi of t is equal to phi of zero Which is equal to one and I'm going to be XP of Minus two the integral of one minus e to the minus t prime Over t prime dt prime from zero to t So that's the phi of t And so in the last thing now is I want to compute the coverage and So what is the coverage? So let's look at the equation for So what is the coverage? Yeah, so let's look at the equation for Let me just remind myself of this Yeah, so what is row of t the coverage? Well in order if I look at the coverage that means I've got to look at an infinitesimal region and say well Is there a car park there? So that's nothing more than one minus the probability that an infinitesimal region is empty. So this is e of zero t So we've got to compute e of zero t so that means back here We have to look at this guy for x equals to zero. So if x is equal to zero we're going to have So I Need another little piece of blackboard space here. So from the equations of motion. We're going to have d e zero t dt So that's equal to minus So e zero So this term doesn't contribute and all we have here is for zero. So it's minus e one of t minus e of one t Sorry, so can't you take e zero t from the last equation there? Which equation here? Yeah, no because notice that what I was solving was this equation. This is for x bigger than one I didn't I didn't talk equation. So I've got it's a it's a separate equation that I have to deal with So anyways, I guess I guess I'm going to have to raise something here Don't need this anymore. So that says it e zero t is equal to minus the integral from zero up to t of e one of t prime dt prime and Row, which is one minus e zero is equal to one minus e zero So I'm a little bit stuck here, but I didn't want to refer to my notes But I'm gonna I'm gonna cheat and refer to my notes here because I have lost my train of thought Yeah, and I think I did something terrible because I just erased what I wanted to point out and so remember Way back when that I found that Phi Where is Phi? Yeah, so you remember I Would like to read what's on the blackboard, but I erased it too. Well, sorry about this But I'm almost I'm sort of two minutes from the very end here. So we had Let me try and remember what I had because like my notes are so brief. I can't even read them Yeah, so I had e x t is equal to e to the minus x minus one t times Phi So in particular e one t This is equal to Phi just by definition So So this is equal to one minus No, but I'm missing something. I'm missing something really trivial here, and I'm just confusing myself Yeah, okay, I know what my problem is Let's go back. Let's go back one step here So if I integrate this equation D e by dt is equal to minus e1. This is going to give me e of Zero t minus e of zero and t equals zero So this is equal to minus that guy, but this thing is equal to one. So I have e Okay, so now I have the right answer e minus one is equal to minus the integral zero to t e one of t prime dt prime And so we have so e zero Is equal to one minus Integral from zero to t and e one is just nothing more than Phi of t prime dt prime Finally and One minus e naught is just the density. So we're going to have the density is Equal to plus integral from zero to t Phi of t prime dt prime Okay, so finally I have the answer So finally the coverage is is equal to the integral from zero to t dt prime of exp and So it's going to look really kind of funky. So it's to integral from zero to t prime because we want Phi at t prime One minus e to the minus t double prime over t double prime dt double prime dt prime so that's my answer This is the final coverage And in the limit where t goes to infinity so I'm I think I don't want to do this now because this is Straightforward, but tedious. I'll certainly do the wrong answer, but the limit as t goes to infinity This thing goes to a Asymptotic value known as the Renyi constant, which is like 0.747 Dot dot dot dot and it's called the Renyi constant Because this problem is first solved in 1950 by Albert Renyi the Renyi of the Erdos Renyi random graph same guy did this solution and What I'm going to show I mean, you know, it's like I'm sort of three minutes away from finishing this topic But I don't feel ready to do it right now But it's just that we can also compute the approach to the asymptotic limit and it turns out that it approaches the Asymptotic limit as 1 over t rather than exponential in time So for next time I will finish this discussion. I'll also talk briefly about absorption and higher dimensions for which basically Nothing is known exactly however One can get the asymptotic behavior and there's a very simple scaling arguments or I wouldn't even call it scaling right sort of dimension Not even dimensional like Hand-waving argument that gives you the asymptotic approach to the final coverage in higher dimensions And that's really the only known result there is so that'll be for next time Okay, thank you. Do we have a question?