 We will try to motivate a very important equation in kinetic theory namely the Boltzmann equation which is along lines somewhat different to the ones that we have been following so far. We looked at the Langevin model, the generalized Langevin model, a lot of linear response theory etc. But then the question arises as to whether there is a more basic way of deriving densities in phase space, in particular the one particle density say, the simplest of these such densities. Well this was a problem of kinetic theory and a great deal of work was done in the 19th century as well as the 20th century on this. A lot of problems still remain to be solved in this area and I will try to give you a flavour of how even in the simplest of instances the problem reduces to something which is fairly complicated and is fairly non-linear as you will see. I am not going to give a rigorous derivation of the Boltzmann equation but I will point out what the physical assumptions are, very reasonable physical assumptions, justifying which is going to be much more complicated but even without that just the mechanism involved is something which could be explained in reasonably simple terms. So let us look at that, let us look at what the problem is and how it is tackled. This problem arose initially in the context of extremely dilute gases when people were trying to derive from first principles things like the Maxwell distribution of velocities etc. At the very foundations of statistical physics Boltzmann worked out what is now known as the Boltzmann equation in one form or the other and he motivated much of what happened later on and this continues to be of great importance in many many applications including plasma physics, fluid dynamics etc etc. Now what is the problem that we are going to look at? We have in mind a gas, a very dilute gas of particles which of course interact by collisions among the particles. To understand the collisions themselves one would probably have to look at the quantum nature of these particles which is outside the purview of what we are going to do now but given the fact that particles scatter off each other, there is some collision cross section which you can compute in principle, the question is what does it say about single particle densities? So the rest of the argument has to do with just physical arguments on how phase space densities evolve, okay. Suppose you have n particles, a very dilute gas of n particles, we are not going to get into things like quantum statistics or anything like that with these particles, a very classical kind of gas, very dilute gas and this gas is in thermal equilibrium. The question then arises if you start with some initial conditions which of course means that you are not talking, you are talking about special initial conditions, the question is how does the gas equilibrate or if you start by disturbing the gas a little bit by applying a field or something like that, how does it go back after you switch it off, how does it go back to the equilibrium distribution? So this is the question asked. Now of course to actually deduce what is happening to all the particles, every one of the particles in this gas is a formidable task. We need to know conditional probabilities or densities of given a particle here, what is the probability of a particle there and the probability of yet another particle there and so on, that is formidable but we could say alright, all these particles are in equal footing and let us say, let us look at the phase space of any one of these particles, a given particle say. Now this phase space is described by 3 spatial coordinates and of course 3 momenta or 3 velocities. Since we are going to talk in terms of collisions, let us look at velocities instead, that is the conventional way it is done. So we will say that a particle is described by specifying its position and velocity at any given time t. Now this space of all possible r's and v's and r is inside the volume of the container if you like, that space, the phase space of a single particle is known as mu space. So the space occupied by these quantities is called mu space and the idea is that this mu space is the same for all the particles, any one of the particles will have the same set of possible values of r, possible values of t. So the single particle phase space is what I call mu space. Then the particles of course are of a discrete nature but we would like to have some kind of continuum description of these particles. So let us look at some numbers. At normal temperature and pressure in a dilute gas, you have about 10 to the 25 molecules per cubic meter. This space if you know coarse-grain it, if you try to make it into small cells, if you break up this space into cells, so let us suppose schematically you have the velocity in that direction and the coordinate in this direction schematically, pardon me, the coarse-grain in mu space, the space, the phase space of a single particle, one particle phase space. If I coarse-grain it in some fashion, then well since in physical volume, I have already said that each cubic meter has 10 to the 25 particles, if you look at the physical volume of say 10 to the minus 15 meter cubed, that is 10 to the minus 5 meters on each side a small cube, 10th of a millimeter on each side, 100th of a millimeter on each side, you still have about 10 to the 10 particles in it. So in that sense, these particles form practically a continuum even though you are looking at a point, a point as small as a volume of 10 to the minus 15 meter cube which is pretty much like a point but even then you have a very large number of particles in it. So let us suppose that you coarse-grain it in this fashion and this thing here is what I call mu sub i, that is the ith cell in mu space, okay and I label it by the points of its center. Then I can ask at any given instant of time t, what is the number of particles in this cell here? Let us call that number of particles f of r, v and t where r and v are the coordinates here in this point of the center of that cell. So this is the number of particles in this volume at time t when multiplied by d3r, d3v. So let me call this, give it a shorthand, let us call this delta mu i, that is the measure, the phase space measure of the ith cell, okay. So this is like a density, number density if you like and it is a local quantity, okay. And the idea is that you do not have very sharp variations of this f as you move from one cell to the neighboring cell, etc. So you can treat this as a continuous variable here. And now the question is what is the total number of particles in the gas? The total number, remember that each particle, each of the n particles is represented by some somewhere in this space, in this mu space because each of them has an r and a v and therefore is some kind of point in this space. And when you add up all these numbers, you are going to get the total number, right. So it is clear that if I integrate d3r, d3v, f of r, well if I take this and sum it, I sum over i, f of r, v, t, delta mu i, summed over i, all the cells i, this must be equal to n and in the limit in which these volumes go to 0, in which these, this measure goes to 0, this goes to an integral f of r, v, t, d3r, d3 velocity and let me call this d, the short end for that I call d, right. And this must be equal to n. If I integrate this quantity over r alone, right, what happens then? I get a volume, if I integrate over r, I get a volume and that will give me n over v, okay. But whatever, this is the constraint, this is the normalization of this single particle density and the whole point of kinetic theory is to find out what this f is. So that is the task to calculate this f of r, v, t because from that it is like the density for phase space density for a single particle, from that I should be able to find all kinds of physical properties, transport properties and so on. I am not given anything at the moment except this condition and I have tried to find this Abbein issue, okay. It will sort of satisfy some kind of differential equation and I have to tell you what it is and of course the standard problem would be given the value of f of r v, t at t equal to 0, initial condition, find out what it does at later times. That would be the typical problem. So the whole target is to find an equation for this. Just as when we did density matrices for instance in classical or quantum mechanics, we found there was an equation for the density operator, the Louisville equation. It says delta rho over delta t was equal to the Poisson bracket or commutator of rho with h or h with rho. We want a comparable equation for this f here. Now we are going to assume that the gas is dilute, okay. What that means is that it is so dilute, it is spread apart that the De Broglie wavelength of each of these particles is much smaller than the thermal wavelength. That of course as you know is a standard criterion for non-degenerate in the quantum mechanical sense for non-degeneracy in the quantum sense. Namely that the thermal wavelength, so schematically what it means is you have little fuzzy objects which are the individual particles and these follows should be at a distance considerably greater than the fuzziness in the position of each of these particles, right. So you want to say that lambda thermal which is equal to Planck's constant divided by the average momentum but in a free gas the kinetic energy is proportional to k t, p square over m is proportional to k t, so p is proportional to square root of m k t roughly, right. So you want this square root of m k Boltzmann t, this fellow here must be much much smaller than the inter-particle separation. Now the volume per particle is V over n and that is like a 3 dimensional, it is the cube of the inter-particle separation, so you want this to be much much less than V over n to the power one-third. That is the condition under which classical statistical mechanics will apply. Shouldn't I have a, yeah because this is saying that once you prescribe for me how the particles are distributed in this phase space and that is the end of it, right. Yes, if you like, yeah I already said that R and V are the coordinates of this point here, the middle point, yeah, yeah sure, okay. So if I bring it to this side it says the number density if I define n equal to n divided by V, number of particles per unit volume, it says n h cube over m k Boltzmann t to the 3 halves much much greater than 1. That is the condition for classical statistics as you know. Low densities, high temperatures and then this becomes much much less than 1. For the gas in this room for instance, nitrogen gas at normal temperature and pressure, this quantity will be of the order of 10 to the minus 6 or minus 7 or something like that and then you are in very good shape, okay. On the other hand if you go to extremely low temperatures and higher densities then you are in a very degenerate situation, right. If you go to something like a Newton star, this quantity becomes enormous. Then at even relatively high temperatures you would still have, this would get reversed in the other direction, okay. And then you have to do full quantum statistics, okay. So we are going to work in this regime, in this classical, completely classical regime, okay. Now the matter is very simple if you did not have collisions at all. First of all throughout we are going to neglect to start with collisions with the walls, okay because now we have to then deal with the atomic structure of the walls and ask what the interaction is. You are going to ignore that and look at what happens in the bulk first and then this is what happens to this F. If you consider what happens to F of R V T, D Mu, okay, D 3 R, D 3 V, etc. and ask what happens to these as T goes to T plus D T, okay. Then the particle in this cell at this position would maybe get out of this cell, it would acquire a new position and a new velocity depending on what force there is on this particle. Let us suppose you impose an external force on the particle, not collisions, an external force on this particle, some F, okay capital F and that force could vary with time but very slowly compared to the time scale of molecular motion. It could change with position, we do not care whatever. Then this quantity, all the particles here would move into new positions, coordinates would be equal to F of R plus V Delta T and V becomes V plus F over M Delta T and the time is of course T plus Delta T, that is the density multiplied by D Mu prime where Mu prime refers to the measure of the cell corresponding to these coordinates. So this is clearly conservation of number, nothing has happened, these people, these molecules have moved into some other cell in this phase space, okay. We will assume Delta T to be quite small and the cells to be adequately small, the limit, we are going to take limit in which all these things become point objects but that is what this conservation law is to start with in the absence of collisions, no collisions. But now, yeah, that is why it is changing with time, sure. No, no, no, it is a different cell, all I am saying is this set of particles moves out, right. This is the way it evolves in time. In a sufficiently small interval Delta T, the velocity changes by V plus F over M Delta T, this is the acceleration, right. This is the change of momentum divided by mass, that is your change of velocity, okay. So that is the whole point, may be a complicated equation of motion but we are looking at a very infinitesimal interval of time. So that is all it is in the absence of collisions. But we also know that if this happened in phase space, imagine for a moment we are working in phase space in Hamiltonian dynamics in R and P, then we know the motion is such that phase space volume elements are preserved under time, it is like an incompressible fluid, okay. But we are working with velocities here but this is still true, this is still true and if you look at Huang's book for instance in statistical mechanics where a nice derivation is given of this Boltzmann equation, he points out that this is still true even in terms of the velocities. Essentially in this problem it is kind of trivial that it is true because what happens is if in the volume element you have this, initial volume element in R and P, you have this, then since R changes by R plus V dt and V changes by a similar kind of thing, this goes over into something that looks like the origin is shifted, it looks like this and you can show this area is equal to this area, my simple geometry which I am not going to do now, okay. I have stopped all collisions, I said we do not have collisions, the problem will arise when we are going to include collisions which you have to but I am trying to see what happens purely geometrically without collisions, if you ignore collisions. So let us put it another way, this is in the absence of collisions which is equivalent to saying this is in the case in which the particles can go right through each other without affecting each other, okay. Imagine they float through each other, they do not know about each other as present, so the scattering cross section is 0, okay, then this would be true but we have to include the scattering cross section which we will do in a moment, okay. So since d mu equal to d mu prime, it implies that this is equal to that explicitly but what does that mean? We can now go to the limit in which delta t goes to 0 implies this f is equal to this f in the limit in which delta t goes to 0, okay. And let us expand, Taylor expand this, the first term cancels and retain to first order in the infinitesimals, right. That immediately tells you delta f over delta t if there is explicit t dependence and then when you differentiate with respect to this, you get the convective derivative part gradient r with respect to rf and when I differentiate this, I get plus f over m dot gradient with respect to v f equal to 0 in the absence of collisions but now there are collisions, there certainly are collisions, so whole physics is because there are collisions, equilibration occurs because there are collisions, so this is wrong and what is left out is precisely due to collisions, this differs from this precisely by the extra term due to collisions, so this is, so that is our first equation and everything is sitting here, everything is in this term here, pardon me, pardon me, normal derivative with respect to what? This is that the, why? This is the way in which the f changes in a time, the change in f in a time delta t due to collisions is delta f over delta t times delta t, I have just cancelled out the delta t on both sides, it is the contribution to this f due to collisions, we do not know what is happening to this f due to collisions, so now we are going to ask what is going to happen right, so you had a certain number of particles in this d mu in here, here is a molecule, in the time between t and t plus delta t, a molecule can come and hit this and throw this out of this cell, so it will deplete it, on the other hand there may be a collision here between 2 molecules and in the time between t and t plus delta t, this fellow may get in here, that will add to this f and you therefore have to take into account both these fellows out here, things which get in will increase this f and things which get out will decrease this f right, so this is equal to r bar minus r, this is the standard notation and I will explain what this r and r bar are exactly, 2 or n body we do not know, we do not know as yet okay, we do not know what these collisions are like, we are now going to start systematically making approximations okay, so r delta t equal to number of particles that oh yes, we now make an assumption, we now make a crucial assumption that the cell is so small that any collision takes it out of the cell, we are going to count the number of collisions, so we need to know what these collisions do, you may have a collision which still leaves it in the cell, a very mild collision but we are going to say the cell size is so small in the limit of 0 volume that any collision takes you out of the cell because finally what is happening, we are trying to find a local differential equation, completely local equation at each point, so some kind of continuum approximation is being made and the idea is that any collision kicks you out of the cell, so number of particles that leave d mu of our cell at the point r at the volume, at the velocity v, so that is r delta t and similarly r bar number that come in due to collisions in this time interval, that is a very standard kind of assumption that you say that these collisions are such that any collision gets you out of it but there could also be some collisions that bring you into it, this is like these assumptions which in the case of Markov processes there is a gain and there is a loss term in any rate equation, it is pretty much in the same spirit here, we have still not said anything about dilute, we have to explain where this dilute business comes in, we have only assumed that any collision, the cells are sufficiently small that collisions knock you out of it, right. Now the question is what sort of collisions are we talking about? In the simplest instance let us assume these are elastic collisions, so the energy is conserved in these collisions, there is no absorption, when two molecules collide we come out with the total kinetic energy is the same before and after the collision, okay. And we have an elaborate theory of scattering in quantum mechanics to describe scattering processes but the simplest instance we will assume that the collisions are binary collisions, then we only two at a time, the idea is that the probability of three particles scattering of each other at the same instant of time is 0, effectively 0 which is a very good approximation, yeah. There will be fluctuations but the assumption is that look at the cell size that we are talking about, it is sufficiently large that you are in the continuum approximation as far as the number of particles is concerned, so it is not as if you are looking at a cell size in which all of a sudden 10 to the 15 particles will make become 10 particles for instance, that is not going to happen ever, okay. The probability of such a large number fluctuation is negligibly small, okay, it is always true, I mean even in the gas in this room for instance it is a binomial distribution in thermal equilibrium, so the probability of something which is very far away from the mean is going to be extremely small. It should move out of the cell, yeah. So that would imply that the mean path length of molecule should be larger than the cell size itself. Yeah. So that after the collision it does not actually move out of the cell. Yes. So then you will need an intersection of the fact that the cell size needs to be large enough so that. Yes. So is that? Yeah, that is built in, yeah, yeah, absolutely, we will see where this gets us, yes. So the question is, is this going to be exact when the answer is no, it is not an exact equation. But we know the precise conditions under which the approximation is valid and for dilute gases it will turn out to be an extremely good one. Now the crucial assumption is this business of binary collisions, that the collision events are all two particle collisions, okay. Now what happens in a two particle elastic collision, very simple, if you have a particle with velocity v1 colliding with a particle with velocity v2 then these particles go to new particle, the same particles go out with velocities v1 prime and v2 prime for instance. So some collision event happening in which you have v1 coming in, v2 coming in there and then this goes off v1 prime and v2 prime in this fashion. Then the conservation of momentum in this case, they all assume to be equal mass particles is of course that v1 prime plus v2 prime, that is the first equation and the second one says the energy is conserved, right? So it says v1 squared plus v2 squared equal to v1 prime squared. Now in this situation it is obviously sensible to go to a frame of reference in which you are in the centre of mass, frame of reference so that it looks like two fellows coming in and two fellows going out in this fashion. So in the centre of mass frame of reference, the collision would not look like this at all in the CM frame, it looks like two particles are coming in in this fashion with equal and opposite velocities. So this comes in say with some u over 2 and this is minus u over 2 and then after collision they go off in this fashion. So this is u prime over 2 and this fellow is minus u prime. And how is that achieved? Well, you go to a frame of reference in which the capital V, the centre of mass has a velocity capital V which is half v1 plus v2 to start with. So let us convert to v equal to half v1 plus v2 and u is v2 minus v1. That is the relative velocity. Similarly for the prime coordinates, so v prime equal to half v1 prime plus v2 prime and u prime equal to v2 prime. Then what is the content of these two equations? Well, the first one of course is that capital V equal to v prime, that is trivial. Total initial velocity here is equal to the same thing out there and this thing here if you substitute for little v1, little v2 etc in terms of capital V and u, it becomes the cross term v dot whatever cancels out and you are left with u equal to u prime. The magnitude of the relative velocity is the same, is unchanged in an elastic collision. So for all practical purposes, you can view this as scattering of particles of velocity u in a fictitious potential, whatever that potential be. And we know how to do scattering theory in that case, right? So you have a center of attraction, center of force for example, some potential and then you have an incident beam of particles coming in with a velocity u. If you treat this quantum mechanically, then particles scatter off in all possible directions and one of the fundamental problems of, the fundamental problem of quantum scattering theory is to find out the flux of particles in any solid angle that depends on the nature of this potential. It depends on the, whether there is spherical symmetry or not, it etc, depends on the potential. But it is a one-body problem, if I do it. Classically too, we can solve the problem. You give me the potential and I tell you what the scattering in different directions is. The other force scattering for example from a Coulomb potential will tell you exactly what the differential cross section is in various directions. So that is a problem in scattering theory and it essentially says that at the end of the day, if you ask what is, given an incident flux in some direction, what is the number of particles, what is the flux of particles that is coming out in a solid angle d omega at an azimuthal angle phi and polar angle theta with respect to this incident direction. So this problem is a problem in scattering theory. And what is, how is that solved? That is solved by saying this number of particles, the flux, let us say, number of particles scattered into the cone of solid angle d omega per unit time is equal to what? Well, it will depend on the incident flux. The number of particles coming in per unit area, per unit time to start with. So let me call that incident flux equal to incident flux multiplied by the scattering cross section, the differential scattering cross section, right? So this is equal to d sigma which depends on omega over d omega times d omega because I am asking for the number in d omega and this is the differential cross section. In this two-particle scattering geometry, a lot of interesting things happen. First of all, it is easy to show that the measure does not change, that it is, this is simple exercise. I want you to show this from these algebraic equations. You can show that d3 v1, d3 v2 equal to d3 v, d3 u, if I change variables, that is equal to d3 v prime, d3 u prime which in turn is equal to d3 v1 prime, d3 v2 prime. So this volume measurement in phase, in velocity space does not change under the scattering, okay. The rest is just, it is just simple algebra, it simply says counting, just counting because we need to find out the following. Yes, let us look at the term r first, we have r bar minus r delta t, let us look at what gets out of this phase space cell, okay, in a time delta t, okay. What is that going to depend on? It is going to depend on the incident flux. So let us pretend that we, we know this quantity by solving the scattering problem and let us pretend that v1, we fix v1 at some value and ask what are all those v2's which will cause this scattering out into some cone or the other, okay. What are all those v2's? So for a fixed v1, we want all the v2's which will hit against this v1 in that phase space cell in time t to t plus delta t and kick it out, okay. So it is equal to the incident flux multiplied by this scattering cross section. But the incident flux is provided by what? It is provided by f of r v2 because that is what is hitting against the v1 at time t multiplying, multiplying the differential cross section and any other factor, well you have to integrate over all these fellows here, d3 v2 and integral. We do not care what angle it gets scattered into. Before we get to this, we have not finished with the incident flux yet. We need to have all those particles which are within the mod u of it because per unit time. So mod v2 minus v1 and then d sigma. So this is going to be the r for a fixed v1. Yes, no the condition on v2 comes from, this is going to depend, this is a function of u. It will be a function of the energy and of course the angles because what is happening is that if you have a scattering center here in this geometry, in this geometry, I ask how many particles are going to go here? That is going to depend on what the initial velocity is or initial energy is because this elastic scattering, u prime is equal to u. So it will definitely depend on mod u and then relative to this direction of u, it will depend on what this scattering angle is and it will also depend on what the phi is in this direction. I have included both of them in writing omega. If it is a spherically symmetric potential, then it will not depend on the phi but it will still depend on theta and of course on the incident energy which is measured by mod u, right. So that is there, that is sitting here. This is providing the factor in the flux, the number per unit time depends on how many of them can reach this point, the particle v1 to be scattered out, okay. So this is going to be the r term, okay but that is for one particle but all those which are sitting in this phase space within the volume element d3 r, r1 are going to get scattered out, right. So you must multiply this once again by f of r v1, that is the number of scattering centers. So r is going to be proportional to this, oh we do not care in what direction this particle gets scattered out. So there is also an integral over this. So I should bring that fellow down here and then write this. Now let us put this in but there is an integral over d3 v2. All the molecular details of the actual scattering process, actual collision etc is sitting here. So the computation of this is a separate story altogether but in a very rough sense because of this binary approximation you have 2 of these fellows sitting here, okay. And mod v1 minus v2, absolutely. That is what couples that too. So this is what r is proportional to, it is essentially the r. But what is r bar which is those which are getting scattered outside to get into this phase space, okay. How are we going to find that? Well we argue in the following way. If you have a scattering process, it is equal to, it is actually equal to. If you have a scattering process in which v1 plus v2 goes to v1 prime plus v2 prime then assuming that the dynamics is time reversal invariant, the scattering process in which v1 prime plus v2 prime combined to give you v1 plus v2 as the final products is also possible and has the same probability. So by symmetry, time reversal symmetry it turns out the scattering cross section for that process is the same as the scattering cross section for the original, for this process, okay. It is exactly the same. So there are some intermediate steps in between where you have to show this is really true and so on and so forth. But we will take this, let me just make a statement that owing to time reversal symmetry plus the overall rotation symmetry of this and inversion, reflection symmetry of this scattering process, the other scattering cross section for v1 prime v2 prime scattering into v1 and v2 is exactly the same, right. So therefore we will introduce some notation. So it is conventional to call f1, f1 of r, f of r v1 t is conventional to call it f1 v1 prime t equal to f1 prime and similarly f of v2 prime equal to f2 prime. Then r bar is exactly the same thing except that you have the primes here, sitting here. It turns out that this is exactly the same factor with suitable changes. Remember that mod v1 prime minus v2 prime was the same as mod v1 minus v2, okay. So that is also put in, is used and you end up with an equation which essentially says that delta f over delta t of r v1 t, that is our fixed velocity to start with plus v1 dot del r, let us put the f outside, f, okay, t plus the external force f over m dot gradient v1 f of v1 on the right hand side is equal to dv2 d3 v2 integral d sigma differential cross section and then inside f1 prime f2 prime minus f1 f2 mod v1. And notice what these quantities are, f1 is this quantity, f2 stands for this quantity here and that is integrated over, the v2 is integrated over. So all those v2s such that the relative velocity is v1 minus v2, it is integrated with this weight factor and this quantity depends on that modulus, on this modulus here, it is a function of that as well as the angles involved. Similarly for the prime quantities where these primes stand for v1 prime v2 prime such that v1 prime minus v2 prime modulus is equal to v1 minus v2 modulus. So those quantities v1 prime v2 prime are determined in terms of v1 and v2. But ultimately these are the fellow scattering into the volume, these are the fellow scattering out of the volume, that is the measure of these quantities here and the binary business comes here. But now let us stand back and look at this. The moment you see an equation like this, first of all it is an integral differential equation to start with. This is the famous Boltzmann equation in this language but what is the most striking thing about it? All these quantities are very similar to the kind of things we had in wrote down various Fokker-Planck equations and so on. But now this is nothing to do with that, there is no randomness or anything like that. We are just saying we are going to look at all deterministic processes such that the phase space density changes due to collisions. But the most striking thing of all is that this is non-linear in this F. So it is a non-linear integral differential equation, extremely formidable to solve mathematically or even to show the existence of solutions under suitable conditions. So already even with the simplest of assumptions with binary collisions and so on and so forth, even if you knew the dynamics, even if you knew this scattering cross-section here, you still have the problem that you got to solve this integral differential equation which is highly non-linear. So that is the root of the source of the problems, the source of the difficulty. Now where did this question of dilute come in? Well, we assume that these two things are independent of each other. So we assume that two particles which collide have no correlation at all, no memory, no correlation at all. Those who collide once are gone forever, after that they do not see each other again. And moreover, that it is uniform across R, independent of R, the same geometry applies. So in that sense, we are really using this so-called molecular chaos assumption of Boltzmann, that there is no correlation of this kind and there is independent of position space, the velocity distributions act in this multiplicative manner. Now of course if you have correlations between particles, this is not true anymore. So if you go to dense gases, this is not true. Even with binary collisions, this kind of assumption is not true anymore, okay. And there are many cases where it is spectacularly not true at all. For instance, if you imagine gas and elastic gas of equal mass particles on a line, one dimension, then you know that in one dimension when you have equal mass particles, when two particles collide they just exchange velocities, right. So if you have one particle A in between two particles B and C, then all the collisions of A are always with B or with C. So the number of recollisions, all collisions are recollisions in some sense. At best there is one more particle with which it can collide in between two collisions with one neighbour on one side. So it is the opposite extreme, it is highly correlated, it is extremely correlated, controlled entirely by recollisions but the Boltzmann equation assumes there are no recollisions at all. It neglected it, right. And in three dimensions that pretty much it is clear that it is pretty much valid, this assumption is valid. Two fellows collide with each other, they are gone after that. So this is the most elementary way of sort of motivating the derivation of the Boltzmann equation. I have not gone through all the steps but very roughly this is what it entails in this case. The rest has to do with approximations. How to linearise this, what are the assumptions one makes to linearise it etc. For instance in plasma physics there is a very famous approximation called the Blasov equation which follows from the Boltzmann equation etc. But that gets us into kinetic theory and that is not my purpose really. One problem that one should tackle with this is all the transport coefficients. So there is an elaborate theory which will tell you how to compute the viscosity of this gas dilute gas using starting with the Boltzmann equation in this fashion making suitable approximations. But I thought I should mention this thing here because earlier assumptions were all stochastic in nature but now we are dealing with a situation where we do not have a tagged particle that is much more massive than the rest of the particles. All of them are in equal footing completely, right. So that is one of the reasons why the equation is not expected to be as simple as in the case where you have a very heavy particle and distinct from all the rest from the heat path, okay. So maybe I will stop here today and any questions and comments we will take it next time.