 Right, so we started looking at the way in which a non-uniform distribution equalizes becomes uniform as a function of time essentially the diffusion problem in the context of the Boltzmann equation. And we have a whole lot of symbols on the board but let me go over it again and then we will complete where we were getting at. So if you recall, we want to look at the distribution, we want to look at F of R V T on time scales where the velocity is essentially thermalized. That is where the diffusion regime is, okay, although we are not explicitly saying so. So we would like to ask what happens if the velocity distribution is Maxwellian essentially but the positional distribution is a function of time starts with some initial given prescribed distribution and then it obviously is a function of time due to diffusion, it equalizes. We are trying to understand that process and understand the time scale which appears in the problem when you try to find out how it goes to this equilibrium situation. So the equation we wrote for F was Delta over Delta T plus there is a V dot Del and that is definitely present, this term Del R is definitely present because we are trying to find out how this function of R changes, okay. So this times F plus an external force but that is 0, there is no external force. So F over M dot grad V, this term acting on F, this term is absent and this is equal to on the right hand side minus 1 over tau times F of R V and T minus whatever it relaxes to, whatever the system relaxes to. Remember that when we looked at the Ranjima problem in phase space, one particle phase space, I said that beyond the diffusion time the joint probability density in phase space factorizes into a portion which was essentially Maxwellian times a portion which satisfied the diffusion equation. I did not set it equal to constant, it satisfies the diffusion equation and we are trying to see if that emerges here, okay. So that is the point that you raised and that is what we are looking for. We are trying to find out what happens, just to recall to you, we had a rho there, we had a rho of R V and T and we said that when T was much much greater than gamma inverse in the Ranjima model in this case, this went to some probability distribution R T times the Maxwellian so in the velocity. That is the one that, that is what we discovered in the diffusion limit and then we found out what is the equation satisfied by this quantity. So it is in that spirit that this is being done. So what we need here is an N of R, T times W of V which is the Maxwellian distribution. So just to set your normalizations, remember that F of R V, T, if you integrate this over R and N, V, you end up with the number density finally. So this quantity here, F equilibrium, this goes to F equilibrium of V, that is the absolute uniform distribution in space. So this is equal to N, a constant, it is just the number density at constant times W of V which is the Maxwellian, the Gaussian distribution in V with the variance which is proportional to K T. So that is what is sitting here except that it is not the equilibrium distribution we are going to. We are trying to find out what is this distribution here. So our target is actually to find out what this quantity is. Well if T tends to infinity in the solution of the diffusion equation, everything becomes 0 in an infinite volume. Here everything will become a uniform distribution, 1 over the volume. That is not very interesting, right. We are trying to find out what is the mode by which, what is the timescale, that is really what we are trying to do. So we put in a timescale here and we are trying to see how this timescale is going to play a role in controlling the timescale on which the thing equalizes, okay. But of course we must also look at another fact which is that when you have a non-uniform distribution and you do a Fourier transform in space, you have all wave numbers. So you can resolve it into sinusoidal fluctuations on all length scales, on all wave numbers, right. Now the question is how does that play a role? So clearly the relaxation time is going to depend on the wave number itself. So they cannot be one relaxation time, right. There has got to be a dependence on the wavelength. So a disturbance which is very long wavelength disturbance is going to take a different time to relax than one which is a very sharp wiggle. That is what should come out, right. So although we put in one constant here, you are going to get a whole family of constants, a continuous family of constants. So it is going to depend on the wave number of the fluctuation. And the question is how, okay. So yeah, that is right, exactly, exactly. Yes, absolutely, yeah. We already saw what is the timescale on which in this one, in this single mode thing, we already saw that this tau is the velocity relaxation time. So now we are looking at timescales much bigger than that tau. But it will depend on what k is, what the wave number is. So this is what we are going to discover, right. Essentially you are going to find a whole family of relaxation times, one for each k. But how does it depend on that k is the question, okay. So this is the target and now we are in business. Let me go through this quickly since we did this already last time. I do a Fourier transform with respect to space and Laplace transform with respect to time. So essentially we are going to write f of r v t equal to well and 1 over 2 pi whole cubed integral 0 to infinity dt integral d3r e to the minus st, that is the Laplace transform and then the Fourier transform was e to the minus ik dot r f tilde of k v ns. Oh, what have I done? d3k and our Fourier transform convention was to put a plus sign there for this k. I should write the inverse transform, let me not do that. So let us write f tilde of k v ns equal to integral from 0 to infinity dt integral d3r e to the power minus e to the power minus st e to the power minus ik dot r f of r v nt, okay. And the inverse transform will have a 1 over 2 pi cube e to the plus ik dot r and this is going to become e to the plus st over the melenne contour, whatever. So what is going to happen here? d over dt of this guy but this fellow if I write it down is going to pull an ik dot r so let us without further ado write this down. So we have s times f tilde of k v s minus the initial value. Now we made the assumption that the initial distribution f of r v t, this quantity is equal to some initial distribution in r at 0 multiplied by w of v that was the assumption we had made, right. So minus if I take the transform n initial tilde of k w of v that is the value of this guy after a Fourier transform in space and then there is this term it is going to pull up v dot whatever del r acting on this fellow is going to put plus ik dot r k dot v on f tilde of k v and s and this is equal to minus 1 over tau f tilde of k v and s on this side plus 1 over tau n tilde of k and s times w of v, okay. We got all the factors right now. So I bring this, this and this to the left hand side and I have s plus ik dot v plus 1 over tau acting on f tilde of k v and s is equal to so this term has been taken into account so has this and so has this I move this to the right hand side this is equal to n initial tilde of k plus 1 over tau n tilde of k and s times w of v. So that takes care of this term and this term. Shouldn't there be a what? Well we are finding the Laplace transform of the derivative time derivative so it is the transform of this function minus the value of the function in time at t equal to 0 which is all I have written but then I did a Fourier transform in space and so let us move this fellow down here this thing divided by s plus ik dot v plus 1 over tau and let us multiply both sides by tau so that it is tau times this plus just that is equal to 1 plus s tau plus this guy, i tau k dot v and this goes to that side. Now of course we need to close this set of equation, this equation so what we do is to integrate over v and integrate over v and I get precisely n tilde of k and s once I finish integration over v because f of r v t if I integrate over v I get n of r t by definition so what we have is integral d 3 v remember that so I integrate and Fourier transform so I am going to get f tilde of k is n tilde of k comma s if I take Laplace transform etc so this tells you that n tilde of k and s is equal to tau n initial tilde of k whatever that be plus n tilde of k and s here in this bracket times an integral and that integral is integral d 3 v w of v divided by 1 plus s tau plus i tau k dot v I have to integrate this over v along that it comes out so let us call this integral something let us call this i some integral and it is a function of k and s the v is gone so I have 1 minus i k and s times n tilde of k s equal to tau n tilde initial of k let me write it properly is equal to tau n initial tilde of k times i of k and s and I divide by this factor 1 minus i that is the solution now everything is known we know this it is a Maxwellian in principle it is something in the denominator it is some kernel so once you do this integral you know this function explicitly it sits in both places and you know this from the initial distribution so therefore I know the Fourier Laplace transform of n of r comma t of this one and that is the one I am trying to find out of course it cannot be inverted analytically this is a terrible mess this itself is bad and then on top of it you have to put that here and then invert both the Fourier and the Laplace transform it is an impossible task okay but what are we trying to get we are trying to find out what happens at long times and what happens on long length scales the scales on which diffusion occurs the modes the longest wavelength the long and the shortest wavelength the largest sorry smallest k which means longest wavelength k is the wave number okay which is where the diffusive modes are so one way to do this is to expand this whole thing in powers of k this is 1 plus something and you say k is going to be near 0 so expand this in systematically in powers of k using the binomial theorem and do the first order second order etc. Now when you do that you can see immediately what is going to happen here so let us try to expand i of k and s equal to let us pull this out of the denominator and then do a binomial expansion in powers of k so that is independent of v so this is 1 over 1 plus s tau and then an integral d3 v W of v times 1 I am going to pull this out and take this out in the denominator here so it is 1 plus something inverse therefore it is 1 minus i k dot v i tau yeah yeah absolutely yeah yeah yeah precisely before what do you say is the first order term and k will vanish all odd things will vanish yeah I just want to show you explicitly how that happens you can see so this divided by 1 plus s tau we must keep the next term because that is the one that is not going to be 0 so it is 1 plus x inverse which is 1 minus x plus x square but then there is a minus sign here because of this guy so minus tau squared k dot v the whole squared divided by 1 plus s tau squared plus dot dot dot now integral d3 v W of v is 1 because it is a normalized max value already so the first term is 1 plus now this term here has a k dot v obviously you should use polar coordinates such that the polar axis is along v then this becomes k v cos theta and you will do an integral on cos theta from minus 1 to 1 d cos theta so that is an odd function and it vanishes it is obvious from isotropic here so the first order term is 0 then this one is the first nonzero term and that is easily evaluated so you have 1 minus 1 over 1 plus sorry so this is equal to 1 over 1 plus s tau minus tau squared k squared divided by 1 plus s tau the whole cubed is 1 factor here and 2 more there times and integral d3 v W of v and then v squared out here times cos squared theta where theta is a polar angle I have assumed that we are going to polar coordinates spherical polar coordinates in v space and chosen the polar angle polar direction to be along that of k because that is the vector that is sticking out right so you are going to get v squared here and then a cos squared plus higher orders yeah yeah I just want yeah I just want to do this in spherical polar coordinates because I do not want to get into Cartesian and count different kinds of integral there is only one integral to be done here right in fact I am not even going to do the integral I am going to leave it to you as homework problem it is a trivial problem but what I want to do is to extract the temperature dependence of this term that is crucial this temperature sitting there in W of v on the Maxwellian I want to know what it looks like that is that is the main point I want to get at right so we should keep this but you see this term here what is it going to be this first of all in the in the phi direction there is a 2 pi so first of all this fellow is m over 2 pi k Boltzmann t to the 3 halves that is the normalization of this W of v and then there is a v square d v and then there is a 2 pi from the 2 pi from the phi integration this is v square and cos square theta and cos square theta is integrated from minus 1 to 1 so it is twice integral 0 to 1 and then it is twice integral from 0 to 1 divided by 3 so it is 2 over 3 times 2 pi is 4 pi over 3 that is why I mean it is obvious you are going to get 4 pi over 3 4 pi over 3 and all that is finished and then you are left with an integral 0 to infinity d v there is already a v square and a v square so it is v 4 e to the minus m v square over 2 k Boltzmann t and there is a Gaussian integral with the v 4 here so it is a trivial thing to do and change variables to m v square over 2 k t then you are going to get if you call that u or something like that there is going to be a 1 over square root of u and then a u square so u to the 3 halves so there is going to be a gamma of 5 halves but you can scale this fellow out completely v square scales like k t so this will go like k t whole square and then there is a 1 over square root of k t here and there is this fellow sitting here so this sum and substance is there is going to be a k t finally whatever be that constant some k t over m or something like that okay I leave that to you as an exercise whatever it is finally let me write the final answer down there is a very well known answer let me write this down this fellow after you put that in here and here and then keep to order k square because that is the order to which you cut it off everywhere except this so the coefficient or whatever this is correcting this initial distribution has a term which is of the order 1 and then order k square and so on and so forth so if you do that this term here becomes equal to it looks like the following an initial tilde of k divided by s plus a function of s let me call it d of s times k square where I have used this d of s deliberately because it has dimensions of the diffusion coefficient this fellow has got dimensions of time inverse and this fellow is length to the minus 2 so this whole thing this therefore must have dimensions l squared over t which is the dimensions of the diffusion constant so I called it d of but it is a function of s where d of s equal to and now it is not surprising you are going to get the temperature dependence from here so it turns out to be k B t tau over m 1 over 1 plus s tau you could have sort of guessed some of these things 1 over 1 plus s tau because that is appearing here goes with the k square so that is the solution now you have to tell me what this is and then I have to invert it with this weight but now let us see we are trying to ask how does the initial distribution whatever it is how does it diffuse out it depends on the poles of this because when I do an inverse the plus transform 1 over s plus lambda as the inverse transform e to the minus lambda t right so it is going to depend on k square out here therefore the leading relaxation is going to be of the form for s tau much much less than 1 because we want long time or small s going to 0 you can neglect this term so this is going to look like n initial of k divided by s plus d of 0 k square and now you know exactly how it is going to relax because I can invert the Laplace transform therefore trivial and it says finally so it says n of r t and n tilde of k and t let me leave the k still is equal to n tilde initial of k e to the minus d k square t where d is k Boltzmann t tau over n so you see it is typical of what it is exactly what we would expect remember in the large one model we got k t over m gamma and the velocity correlation time was tau gamma inverse well here the velocity correlation time is tau in this model and m is the mass of a molecule so the approximation has been sort of gross in that sense that has become essentially the mass of a molecule but this tau is completely arbitrary completely arbitrary we just put that in my hand so that is the diffusive mode as you can see and you can see it is a Gaussian solution because what is the inverse Laplace inverse Fourier transform of just this guy if you make this if you give me a start with the delta function for instance then this fellow here is a constant if you give me a delta function in r this n tilde of k is a constant and then the inverse Laplace transform of k e to the minus k square Gaussian is of course e to the minus r square again Gaussian now this Gaussian has a variance 1 over d t and in r it will have a variance proportional to d t so you have exactly the diffusive behaviour with this diffusion constant d but it is giving you a lot more information this thing is telling you much more it is telling you how other modes in lax in principle so even though it is a single relaxation time approximation because of this complicated way in which k appears and s appears it is actually telling you much more it is going beyond this diffusion approximation so this is how a space dependent relaxation appears so first we said the velocity itself how does it relax given a uniform distribution we found that was rather trivial it was like a kubo Anderson process then we said what happens if you have a non-uniform spatial distribution well it is diffusive this is called diffusive mode if you like now let us ask the other question suppose the distribution in space is uniform and I start with the equilibrium distribution but now I switch on an external force then what is going to happen and let us in the simplest case let us switch on a force which is time dependent but uniform so that the uniform distribution in space does not change however you put on a force external force and these particles are going to get dragged so I expect the Maxwellian distribution will be disturbed and if the force finally goes to a constant force as a function of time then the Maxwellian certainly will be distorted sort of intuitively in the direction of that force right so that direction should be singled out and let us see how that happens okay and we play the same game and practice an even easier thing so let us do that so till now we were looking at problems where you are a little away from equilibrium you relax to equilibrium but now I start with equilibrium and I push it out of it and ask what is the new distribution to which it goes so constant uniform so I have f of r v 0 the initial thing is the equilibrium distribution is equal to n times w and it is a uniform force so uniform force is not going to destroy in initially uniform spatially uniform distribution is going to remain spatially uniform so I do not have to worry about the gradient term with respect to r acting on this thing at all then what happens to the equation that we have so some force f no r dependence here so we have delta f over delta t and this is a function of r v and t plus f of t over m dot the gradient with respect to the velocity that is important of f of r v t and this guy is equal to minus 1 over tau f of r v t minus f equilibrium of v so I want to know how the f is moved away from the equilibrium distribution because that is what I started with at t equal to 0 so this is the difference in the single relaxation time approximation so this is the equation I have to deal with now clearly it is sensible to call this something else equal to some g of r v and t in particular I want to compare the answer we are going to get here with whatever we know from linear response theory remember in linear response theory we found the average velocity divided by the applied force the constant applied force per unit applied force was the mobility and in the case in which the force was a constant time independent we got the static mobility so we want to see what happens now in this case but this is completely kinetic we do not need to carry the r dependence everywhere because there is no term with respect to this guy so we forget about it completely it is uniform everywhere so let us make this f of v comma t f of v comma t and that is it no r dependence the statement is you have a force which is uniform time dependent but uniform you start with an initial uniform distribution there is nothing which is going to make the distribution non-uniform in space okay so this is what we got to work out this fellow here and now let us put that in here so we get delta g over delta t plus f of t over m dotted with gradient with respect to v of f equilibrium that is equal to f equilibrium v plus g of v and t because f is f equilibrium plus g by definition this is equal to minus 1 by tau g of v we want to compare now that is an exact equation right however we want to compare with what happens when the external force is weak that is the whole idea that you want to compare with linear response theory now it is clear that this g is going to depend on the external force there is no force this g is 0 identically you remain in equilibrium so g is going to have a first order term in f second order third order etc etc but if you are going to work to linear response level then this term f times g is already of second order in g right this is the correction which is proportional to f and higher powers and there is already an f here so I am going to drop this in preference to this okay and then we get delta g to first order t the equation is delta over delta t g that is this term plus f of t over m dot the gradient with respect to v of f equilibrium of v equal to minus 1 over tau now of course one brings this to this side and that is the same as saying that g of v comma t equal to e to the minus t over tau times something else let us call it h of v comma t just removing this integrating factor so the first term is minus 1 over tau times g itself and that cancels and then you have e to the minus t over tau delta h over delta t equal to minus f of t over m dot gradient v f equilibrium so move e to the t over tau to the right hand side now what is the initial condition well the initial condition on this f was that the start with it was in the equilibrium distribution so g of v comma 0 is 0 therefore h of v comma 0 is 0 this is the initial condition and we have a formal solution and what is the formal solution now it says h of v t is equal to minus this is the initial condition this we got 1 over m integral from 0 to t dt prime f of t prime dot gradient v times e to the we should not forget that e to the t prime over tau times f of t prime dotted with gradient with respect to v f equilibrium of v that is h so it says g therefore is e to the minus t over time tau times that so this is e to the minus t over tau times and that is the formal solution for g f however is f equilibrium plus g so it says f equal to f equilibrium of v minus this guy and that is the solution we would like to see what happens at long times if I put a constant force so what I need to do is to switch on the force by whatever means I like and the typical way to do this just to get an idea of what the asymptotic behavior is is to do the following it is called adiabatic switching and you do this in quantum mechanics for example when you do perturbation theory you switch on the force slowly and not suddenly so that all the energy levels do not get jiggled up but the same spectrum remains but gets slowly perturbed okay in the same way here is a way to switch on the force so let us suppose that on the time axis you have here f of t magnitude or whatever you want to make it go to a constant value you want to make it go to a constant value this value is f and let us say you start switching it on at t equal to 0 because we took at t equal to 0 the system to be in equilibrium and typically what would happen is that you would switch it on it will do this you can choose any force history you like but I would like to see what happens at long times here so a simple model would be to say that f of t equal to it should start at 0 and should end up with a capital F which I will put at the end inside so 1 minus e to the minus t over some time scale whatever you call it is tau 1 this is typically tau 1 the time scale on which this fellow essentially reaches 1 over e its saturation value or something like that so that is one way to do this you put that in here then you integrate what is that get us so this fellow here becomes this term this term alone let us look at what happens to this so let us look at the whole thing so this implies that f there is no r f of v t equal to f equilibrium of v and then what minus e to the t e to the minus t over tau that was the relaxation time over m and then the time integral is essentially integral 0 to t dt prime e to the minus t prime over tau 1 minus e to the minus t prime over tau 1 and then there is outside there is an f dot v f dot grad v of f equilibrium of v this fellow is independent of time it just comes out as I said trivial integral to do this whole business first term is e to the whatever it is a plus I think this is a plus plus t plus out there plus t plus out there so this first term is going to give you e to the t over tau minus 1 and when you multiply by this it becomes 1 minus e to the minus t over tau the second term has got a smaller term which is a positive term so it is 1 over tau minus 1 over tau 1 if tau 1 is much bigger than tau for instance then this term is going to be positive 1 over tau minus 1 over tau 1 but smaller than 1 over tau therefore when you hit it with this this integral that part will vanish exponentially so the sum and substance is the only thing that remains for t much much greater than tau and tau 1 is f equilibrium of v minus what remains this is going to give you an e to the t over tau leading term and it is going to kill this so you are just going to get 1 over m f constant dot grad v of f equilibrium of v what is that going to be because remember that f equilibrium of v equal to n times this m over 2 pi k t stuff multiplied by e to the minus m v square over 2 k Boltzmann t now what is the gradient of e to the minus r square in physical space in 3 dimensional space just d over dr is going to appear and in the direction of r so the gradient of r square is essentially r vector twice whatever it is now when you differentiate this is going to give you a twice and that twice will cancel here and you get to m over k t in the denominator looking starting to look like the mobility as you can see and then there is going to be a piece which is essentially v so this whole thing is going to look like f dot v apart from some constants there is some constants I leave you to figure out and then f equilibrium so it is going to become proportional to this pardon me constant is m over k b t yeah so there is a 1 over k Boltzmann t and a plus and a plus sign because you are differentiating a minus here so and a plus so we can even write this down plus so constant is m over k b t so this m goes away but remember that when I integrate e to the t prime over tau I get a tau on top so yes so there is going to be a tau over k Boltzmann t let us write it properly so that is how it gets dragged in the direction of the force leading correction to 1st order in f we know the answer should be proportional to f it has got to be a scalar and the only other velocity you have is v itself so it has got to be proportional to f dot v times this this gives you the constant of proportionality okay exactly precisely f dot v is the rate at which power is supplied to the system is the rate at which energy is supplied to the system a work is done on the system by the force so that is a sort of very straightforward interpretation now what is done in transport theory is apart from the complications of solving the Boltzmann equation in various situations is you remember the collision invariance we talked about 1 and then the velocity itself which gives a momentum current and then the energy current itself so what one does is to take each of these currents and from that one can extract transport coefficients such as the shear modulus the thermal conductivity the electrical conductivity etc diffusion coefficient we have already extracted so you have to construct you can even construct the heat current the thermal conductivity because there is a mechanical portion which is the moment of all these things depend on taking this fellow here an integral of v times f of r v t times some something here some function of v as I put it these are the currents and if this fellow is 1 you got the momentum current if this was kinetic energy half m e squared you got the energy current and so on so you take half m e squared minus the average value 3 half k t or something like that and that gives you the heat current you can go one step further and say I will find out what thermal conductivity is what is the heat conduction equation you can derive Fourier's law for the heat conduction equation here within linear response theory by using this local equilibrium approximation you say that the temperature is different at different places but in a very mild way so you write t as a function of r itself then the derivatives are going to spatial derivatives will act on that t and in that manner you will end up with Fourier's law of heat conduction which essentially says delta capital t over little t partial little t is minus the thermal conductivity times del square of t you get that you get a actually we get a formula for the heat current which says it is proportional to the temperature gradient that is what will emerge automatically from this okay and so on. So I am going to call a halt to this portion of it which is a kinetic theory really and we go on now to the very current topic well it has been current for a while but it is still a very crucial one namely we will do dynamic critical phenomena next for that we will start with some notions of equilibrium critical phenomena second continuous phase transitions talk about mean field theory and then we will see how time dependence comes in so that is the next thing we take up.