 So today let us look at some of the consequences of the Boltzmann equation vis-a-vis relaxation. One of the things that primary points to be emphasized with regard to the Boltzmann equation is that it gives you an evolution equation for the phase space density in mu space for systems which are not necessarily in equilibrium, in general out of equilibrium and then the question arises as to how the system could relax to equilibrium under suitable conditions. For instance if we switch off the external force how it would eventually go to the Maxwellian distribution of velocities and the distribution in space would be uniform if you did not have any external potential acting on the system. So the question is how does this relaxation proceed and this equation should presumably give you an exact answer. So let us look at such phenomena, let us call it relaxation to equilibrium from the Boltzmann equation but of course the full Boltzmann equation is too difficult to solve. Even if you linearize it it is still too difficult to solve. You need a systematic approximation procedure for in powers of some correction to the distribution from the equilibrium distribution and that is a long-winded story. So the question is can we make a quick crude approximation and get some 0th order idea of how relaxation occurs in various situations. So this is what we are going to do today, talk about different situations where you have relaxation to equilibrium in a very simple, particularly simple mechanism. The first of these before I do that let us write the Boltzmann equation down just for you to refresh your memory. So it is Delta F over Delta T and that is in general a function of R, V and T plus V dot Del with respect to R of F plus if there is an external force it is F over M dot gradient with respect to V of F and that is equal to the collision integral Delta F over Delta T collision. So all the physics is contained here. This is basically kinematics but this is where all the physics is. Now of course we know that it is not horribly complicated but even after you make a linear approximation it is still as I said very messy so let us make the simplest possible approximation. We have seen already that if you switch off the external force there is no force at all then this thing here tends to an equilibrium distribution which is F, do not remember the notation I called it F naught but let me call it F equilibrium, F equilibrium of V was equal to well recall on first we should write our normalization. The normalization of this F was integral d 3 R integral d 3 V F of V T equal to N the total number of particles in the system. If you integrate it over V alone so if you integrate d 3 V F of R how should I write this R V T V alone then you get a function of R and T which is equal to well it is the number but it changes with the number density but it changes as a function of position and time so this is by definition equal to N of R and T right. If you integrate it over R alone d 3 V F of R sorry whatever I done over R V T how did I write the normalization down did I mess up the normalization so this is okay as it stands but how did I get the number density oh yes so we wrote integral d 3 V F equilibrium of V was equal to N over V equal to N that is the normalization right thank you. So let me define this non-uniform density by just integrating over the velocity alone this distribution function and in equilibrium there is no T dependence there is no R dependence if I integrate over d 3 V I should get N over V okay so that is those are the normalizations. Now what we would like to say is that remember remember that F I need a little more notation F equilibrium of V was equal to N this thing here in space for the spatial part and then the velocity part was the Maxwellian distribution so the normalized distribution was m over 2 pi k Boltzmann T to the 3 halves e to the minus m v square over 2 k Boltzmann T so that is the equilibrium distribution this is a constant now this function is going to keep appearing all the time the Maxwellian distribution in the velocity alone okay so let us give it a name let us call this equal to identically equal to N times let us call this W of V to show that this is the Maxwellian distribution because in equilibrium everything is in the absence of an external force the distribution in space is uniform so we have uniform density number density N to unit volume and then multiplied by a velocity distribution which is the Maxwellian distribution and this is normalized to unity so we know that integral d 3 V W of V equal to 1 that is the reason for this factor 2 pi k T whatever it is right now what the approximation does the simplest approximation to the right hand side is to say that collisions cause if you have a small departure from equilibrium collisions cause you to go towards equilibrium they help you to equilibrium so if your distribution F of R V, T is a little bit away from F equilibrium then if you make the single relaxation time approximation the so-called single relaxation time approximation so let us write that down single then if you start with the uniform density the system is completely in equilibrium and you ask if I have this perturb it a little bit away from equilibrium the velocities are not thermalized they are little bit away from equilibrium how would these velocities equilibrate yes 0 force case I will do the system with the force present but 0 force case so that is the reason whenever this happens whenever this is a function of R you know that there is probably a force present but even in the absence of a force instantaneously there could be density fluctuations and there are so in general I would call this N of R and T R, T but when you have reached equilibrium in space and time and in velocity and position then the whole thing is just a constant here okay now how does the velocity relax that is the first question you want to ask so let us call 1 within the single relaxation time approximation relaxation of the velocity that is given by the following now there is no space dependence because it is uniform throughout the velocity is supposed to relax to equilibrium now we already know there is a model for this this was a Langevin model when I wrote down the Langevin equation and then I argued that if there is white noise which causes collisions and so on or the effect of collisions is mimicked by white noise Gaussian white noise etc then you are a Fokker Planck equation and the solution to that was the Onstein-Ohlenbeck distribution and then that constant gamma the friction constant determined how things relax to equilibrium and the velocity correlation time was an exponential single decaying exponential with the relaxation time gamma inverse that was that model okay but now we do not have that model at all there is no stochastic force or anything like that this is a completely self consistent system and we are not saying that the mass of this particle is different from the masses of the rest of it they are not saying that at all no such assumption but we made a huge approximation to this in this single relaxation time approximation so now this F is going to satisfy Delta F of V t over Delta t no external force the distribution is uniform in space so this term is missing and you only have this and this is equal to the collision integral and what the single relaxation time approximation says is that this is equal to there is a relaxation time brought in a parameter tau just one of them so it is minus 1 over tau and inside is just F of V t minus F equilibrium of V t so that is the approximation it replace by saying the deviation at any instant of time of the velocity distribution from the Maxwellian distribution is negligible is very very small and then what you have is exactly like in radioactive decay very similar to that this difference divided by tau and it is got the right dimensions it is the dimensionality of Delta F over Delta t except this F is discretized and said that is the difference between the distribution instantaneous distribution and the equilibrium one divided by the time scale tau so this is an extremely simple model what does it predict now we can solve this is a trivial equation to solve the first order equation so it is trivial to solve and of course what one should do is to write set when we can either write down the integrating factor or else set F of V t equal to e to the minus t over tau obviously want to relax with that characteristic time scale tau multiplied by something else so let me call this psi of V then the equation on the left hand side is the first term is minus 1 over tau times this whole business which will cancel against that on the right hand side and it says e to the minus t over tau Delta psi over Delta t equal to the first term cancels so it is equal to 1 over tau F equilibrium of moving this to the right hand side but you have to tell me what the initial condition is you have to tell me what is the initial velocity distribution so let us put that in so initial condition F of V t equal to some initial distribution some prescribed initial distribution so that immediately implies psi implies psi of 0 is F initial because this term becomes 1 at t equal to 0 so what is the solution to this thing it says psi of V t minus F initial of V is solving this differential equation you have to integrate this from 0 to t put t prime and integrate 0 to t so that integral becomes the t tau cancels as you can see so it becomes e to the t over tau minus 1 times F equilibrium of V which implies that F of V t equal to I move this to the right hand side and multiply this by e to the minus t over tau so it gives me e to the minus t over tau F initial of t plus I multiply this by e to the minus t over tau so this is 1 minus e to the minus t over tau F equilibrium that is the solution which is exactly what you would expect at t equal to 0 this thing is equal to the initial value because that vanishes and as t tends to infinity this term goes away that term goes away and you are hitting B F equilibrium exactly so this will work as long as this is very close to equilibrium so that is the whole point that you have replaced this entire collision integral by saying that the system is very close to max value in distribution pardon me whatever you prescribed there is no reason why that should be max value for consistency of the approximation this too should be fairly close to the max value in distribution of course no that is not the point there is a characteristic time scale tau we do not know anything about it we have simply said this entire collision integral has been replaced by this discretized version we have not had do not have any handle here about what this tau is at all what we need to do is to find some way of finding out how good this approximation is is to find some way of measuring this tau if it exists now in practice that is not going to happen but you can see what is what we can do what we should let me indicate how one should go about it in the general problem in the general case we are looking at a function of V this quantity is a function of V and T now we know that the equilibrium distribution is a max value is this one of here okay now any arbitrary function of V where we run each component of V runs minus V infinity to infinity can be expanded as Gaussian Gaussian and V times Hermite polynomials so in principle you can say this quantity here is a superposition of all the Hermite polynomials times the max value distribution with coefficients which are possibly time dependent which are certainly time dependent and that is how the equilibrium will happen now this is the crudest approximation to that it says that in some sense it just says the polynomial talking about is 1 okay you have taken the constant there is no Hermite polynomials anywhere there is no function of V anywhere you just have this function of V and that is it this is already the max value okay so there is a systematic way of justifying I mean finding out what this approximation means but in physical terms it says that if you replace this entire collision integral by saying that we do not care what is happening in the inner mechanism there is some effective time term over which the system equilibrates then this is what happens as far as the algebra is concerned but now how far this is good how good a good an approximation this is can we get an handle on tau can we measure it and so on are not answered questions within this framework as yet we are going to find out we are going to find out of this tau appears anywhere else and then discover how ask how what tau could possibly be okay certainly yeah it will not be a single molecular collision time because there are many many time scales involved here so this is some effective time scale on which the velocity is thermalized we already know from the Lanjama model that the time scale on which the velocity thermalizes in that model seems to have something to do with the viscosity of the medium but that was dependent on the assumption that the Lanjama equation was valid which is itself true only if the mass of that particle is much much higher that is not obvious here it is not obvious here so it is clear there is a spectrum of relaxation times and we have taken the leading contribution is it really what has happened now the question is can this also could this result have been produced in any other argument any other way is there a simple model which will produce exactly the same result some stochastic model which will produce the same result it is clear this is not the on-steen Ohlenberg distribution at all so it is not certainly the Lanjama model at all it is very different from that the Lanjama model has this on-steen Ohlenberg distribution which is again a Gaussian where the system tends to the Maxwellian distribution exponentially with this time scale gamma inverse but it is very different looking from this here it is just either they this is and incidentally this if I start with the particle like we did in the Lanjama model with a given velocity v0 this of course would be delta of v minus v0 this would be replaced by delta function but I am allowing a little more general solution by saying that it could be a distribution in itself not all fixed at one velocity is there some way of producing this thing here by a simple model assumption on the random on the nature of the random velocity we do this at all not the Lanjama model here now the answer is yes but it must be a much simpler model than the Lanjama equation because that involves the Fokker-Planck equation which we solved this thing here looks like it is a much more trivial thing to solve right there is an equation for the distribution function which is first order in time and involves the same thing out here so it just looks like a Markov process we have just assumed it to be a Markov process but it looks like it is a discrete Markov process well it is continuous because v is continuous but it looks like v is changing through a jump process of some kind so let us see if that is validated or not suppose v underwent suppose v is given by a Markov process suppose so what I am trying to do is to argue that there is an effective model where v is just taken to a Markov process by which you will reproduce this exact result so that is another way of understanding what is the meaning of this single relaxation time approximation suppose v is a Markov collision will take v to some other value each component of v varies continuously minus infinity to infinity the moment you say it is a Markov process then its probability density delta I need to use another symbol for it since I am reserving F for the symbol we use in the Boltzmann equation the distribution in new space let us just call it p so delta p of v and t divided by delta t must be equal to on the right hand side we need to write a rate equation because it is a Markov process so in general you can write this since it is a continuous process d3 v prime on this side times a gain term and a loss term that is what you do for any Markov process for which you can define a transition probability per unit time right so times the probability that you reached the velocity v prime at time t multiplied by the transition rate per unit time that you hit v given v prime starting from v prime so this guy here is the v prime to v that is the gain term and the loss term in the rate equation is minus w you jump out that is the Markovian master equation now the question is can I produce this solution by writing a model for w that is the question being asked detail balance must obtain in equilibrium so what sort of function should I put in here such that in equilibrium this will become p equilibrium that becomes p equilibrium what sort of function should I put in here such that detail balance will obtain well let us look at the physical process you have an initial velocity v in this transition rate you have a velocity v and it is getting knocked out into any other velocity v prime now what is the effect of these collisions one possible approximation is to say the collisions are extremely weak so whatever you started with that changes very slightly in a given collision in the limit of no collisions it will not change at all but in the limit of weak collisions it will change very small by a very small amount delta v to start with v a collision will connect you only to those velocities which are within a range delta v of the initial that is the weak collision approximation it turns out that approximation leads to the Fokker-Planck equation assuming that there is no memory in these collisions and it is a Markov process I am not going to do that but there is a way of systematically deriving from this master equation the Fokker-Planck equation by making the so-called weak collision approximation there is another possibility and that is a very strong collision approximation it says look each collision is such that it thermalizes it at once in other words it says that this quantity lambda v prime the transition rate from any velocity v prime to any other velocity v does not depend on the initial value at all the collisions are so strong that in a given collision it is immediately thermalized that is one possibility that the transition rate is independent of the initial state it depends only on the final state so suppose that were true suppose but equal to we need this is a transition rate so you need a constant of dimensions time inverse let us call it lambda times some function let me I should not use the function psi I should go to use some other function what is a good what is a good psi of v independent of v prime suppose so the transition rate every time there is a collision and the average rate of collisions is lambda the velocity changes whatever be the initial velocity it goes to a final velocity depends only on the final velocity suppose this is true and now you impose detail balance in equilibrium then what will it imply this will imply detail balance implies then with this assumption if you let t tend to infinity this term is 0 the time derivative it goes to equilibrium this becomes p equilibrium of v prime and this becomes lambda times psi of only the final state of v p equilibrium of v prime must be equal to lambda times psi of v prime p equilibrium of v and there is only one solution to that which is that this is p equilibrium of v itself so it is obviously the strong collision approximation so in one shot yeah so what it is saying is the following you have an equilibrium distribution in every collision the velocity is reset from whatever its value to some value drawn from the equilibrium distribution with those probabilities so if your distribution is a Gaussian and you are at this point you have this initial velocity after collision the velocity new velocity is reset to be one of these velocities with this distribution with this probability it is chosen okay so that is the implication of this statement here random process in which a Markovian process continuous process in which the transition rate is independent of the initial state and is drawn from the final state alone and is a function of the final state value alone and moreover there is detail balance this is called a Kubo Anderson process so what we have shown so far is that if you assume a Markov process for this guy and the Kubo Anderson process with detail balance then this equation simplifies enormously and look at what happens to it exactly apart from that factor n there is a factor n for our phase space distribution this is only in the velocity so I have distinguished it by writing p equilibrium separately so look at what happens to this equation you have d3 v prime this is a function is i of v p equilibrium of v and then you have d3 v prime p of v prime t which is equal to 1 because it is got to be normalized so this immediately says delta p over delta t equal to lambda times this fellow here is p equilibrium of v in this side minus lambda times and now this is an integral d3 v prime of p equilibrium of v prime which is 1 in this side times p of v and t so it is clear that with the identification lambda equal to 1 over tau you get the single relaxation time approximation it is exactly the same so the solution is exactly the same so one way of interpreting the single relaxation time approximation in the weak in the Boltzmann equation with a linearized Boltzmann equation is to say it is equivalent to saying that the velocity v is just a Markov process a Kubo Anderson process nothing more than that this gives exactly the same solution well you can ask well can I tweak this a little bit can I try to improve this model by saying look each collision need not take you to the velocity drawn from the equilibrium distribution so there could be one limit in which you have a very weak effect of collisions this is a strong collision limit in which it immediately thermalizes in some sense so is there anything in between I leave it as an exercise to you to show the following rather simple exercise that if you say independent of this if you say that this W is not this but it is equal to lambda times let us say some gamma times I should gamma is a bad number alpha times well alpha times a function of Xi this is p equilibrium so let me write it like that plus 1 minus alpha times at delta of v minus v prime you could do that this interpolates when alpha is 1 you have the Kubo Anderson process when alpha is 0 this goes away and you have no collision at all it does not do anything this will not equilibrate at all it will remain where it is but this here with alpha equal to 1 equilibrates with the time constant lambda inverse so now figure out what happens you can easily satisfy yourself that this again satisfies the detail balance condition so it is a kind of interpolation model between the strong collision limit and the 0 no collision there alpha is any number between 0 and 1 it does not equal if that depends on lambda this depends on lambda right so it does not equilibrate in one collision it simply says that the transition rate is drawn from the equilibrium distribution independent of where you started yes you can but this is an arbitrary parameter this is an arbitrary yes there are models for collision broadening in gases and so on where you use a strong collision limit seems to be the correct limit work with yes there are physical systems which display this behavior so the point I am trying to make is if you did not have this you have the strong collision limit it is a rather trivial solution it is equivalent to the single relaxation time approximation in the Boltzmann equation in the linearized Boltzmann equation on the other hand if you did not have this but you had this alone there is no collision there is no physics it is no equilibration at all but if you tweak this and made this a weak collision limit then you get the Fokker-Plank equation okay so it is possible to derive the Fokker-Plank equation from the Boltzmann equation by making a single relaxation time approximation and a weak collision approximation okay I am not going to do that that is a little bit of machinery but I am not going to do that I just wanted to point this out okay but now you could choose this alpha to be between 0 and 1 what do you think will happen this model also satisfies the detail balance condition for any alpha between 0 and 1 so what do you think will happen to the solutions again it should be solvable completely because there is a delta function here all it does is to rescale time it does not it just buys you some time it does not do anything the second term does not do very much yeah not entirely but this one here so it just changes lambda alpha and then because it is lambda alpha inverse as you can see if alpha was 0 that means collisions do not change the velocity at all so there be no equilibration which is equivalent to saying the relaxation time must become infinite so it is not surprising that it appears here in this place so the reason purpose of introducing this was to show you that relaxation phenomena can be modeled from the in a systematic way from the Boltzmann equation now let us look at the other thing we did with the Langevin model which absolutely absolutely yeah it is not the collision time very emphatically no not the time between collisions no in fact we will see what this tau is in the single collision approximation this was just a model the lambda was completely arbitrary here but let us go back to the single collision thing and look at the other phenomenon which was we have a gas in which you have a non-uniform initial distribution the velocity has thermalized and now the system diffuses that was our famous diffusion regime where the mean square displacement went like linearly with time and the velocity correlation time had long died down and you are looking at longer time scales let us see how that comes out in this language in the Boltzmann equation so this is the smoothing out of a non-uniform initial distribution in other words diffusion self diffusion this time because the particles of the medium are themselves no external force as before and the Boltzmann equation is delta F over delta T plus V dot gradient with respect to R of F this term is very much present because this is a function of R V and T that is equal to the single relaxation time approximation and what would you write it as this time it is 1 over tau F of R V and T minus what would you write it this time as there is an R distribution there is definitely an R distribution right we are trying to okay but the velocity has thermalized so you write it as N of R, T times W of V and recall that this fellow here is integral V 3 V F of R V by definition non-uniform distribution the velocity has thermalized it is the maximum we want to know how this guy relaxes how do you solve an equation like this well it is got both space and time derivatives so clearly you are going to do the plus transform with respect to time derivative time and Fourier transform with respect to space so you are going to define an F tilde of K V and S to be integral 0 to infinity DT 0 to the minus ST integral D 3 R 0 to the minus I K dot R F of I do not want to put 2 tilde I mean it stands for okay this stands for Fourier transform with respect to R Laplace transform with respect to time of this we plug that in here then let us write the equation down directly oh by the way I need to tell you what is the initial value of this fellow what is the initial distribution so I have to tell you what the initial distribution is so let us put N F of 0 is equal to N initial of R whatever be that initial non-uniform distribution that you plugged in times W of V of course so what is the equation you get this let us call this F tilde when I do the Fourier Laplace transform so F tilde S times F tilde of K V S minus this guy here that is the formula for the Laplace transform of the time derivative but I have got to do a Fourier transform with respect to space let us call that N tilde of initial of K W of V just sits there as a spectator this plus V dot gradient what is this guy going to do gradient with respect to R pulls on minus I K so we got minus I K dot V times F tilde S that is the transform of this fellow here this is equal to on the right hand side equal to 1 over tau minus 1 over tau F tilde of K V and S it should be plus so I expand this in terms of I expand this fellow yeah I expand this fellow and then it is F tilde I leave it you to put in all the 2 pi factors and stuff like that okay yeah yeah when I go to the Fourier transform in every one of these terms there is a 1 over 2 pi in the inverse transform that cancels out but this is a plus because what I am doing is to write this fellow in terms of as 1 over 2 pi D 3 K this guy it was a plus I K dot R so I K dot is equal to this this term plus 1 over tau the Fourier transform of this plus 1 over tau N tilde of K and S W now this looks like a hopeless task because we do not know what this is and we do not know what this is what should I do this is presumably given to me this is known function here I integrate both sides with respect to V what happens then so what happens to integral D 3 notice that integral D 3 V F of R V T integrated over V is N of R if I take Laplace transform this becomes N of R S if I do Fourier transform it becomes N tilde of K, S so I integrate both sides with respect to V in which case you are going to get this thing here and this is going to be integrated over with respect to V on this side but you got to be a little careful in doing this this N initial will move to the right hand side and there is this guy so we should not yet integrate over V you need to pull various things and then do the integral over V let us write all the F guys together so there is S that is this term plus I K dot V plus I K dot V so this term is gone this term is gone plus 1 over tau plus 1 over tau times F tilde of K V and S is equal to so this term is gone one I have written single written one extra term I think I have an extra term somewhere this is equal to let us see where this takes us N initial tilde of K times W of V that is certainly there plus last term plus 1 over tau where is the problem no no what did I do how did was this quite right how did this come about this is equal to this is F of R V it is F tilde of K V and 0 is it not this term is F tilde of K V and 0 and which we wrote as the initial distribution in R times the equilibrium distribution in V and then the Fourier transform with respect to that which was this guy so this is okay this is okay plus N tilde of yeah so that part is alright 1 over tau N tilde yes K and S W of V so this is gone that is the equation right now let us move this to the right hand side so this is divided by S plus I K dot V plus 1 over tau S plus I K dot V plus 1 over tau I am going to leave the rest of the completion of this algebra to you so I move this there and I integrate over V then if I do this on both sides D 3 V of this whole thing of the entire thing both the left and right hand side then what is this quantity equal to I have integrated over this so this is N tilde of K comma S so I get an equation which says N tilde of K S equal to the integral of W of V alone is 1 but you have got this in the denominator so you cannot do anything with it it is equal to N initial tilde of K integral D 3 V W of V over S plus I K dot V plus 1 over tau this fellow sitting there that is some number it depends on S and it depends on K it is some function of S and K plus the same number once again because there is W of V over this guy 1 over tau N tilde of K comma S times the same integral D 3 V W of V blah blah blah call that integral something or the other and bring it to the left hand side and you have an equation now for N tilde of K comma S you invert with this there are plus transform you get N tilde of K comma T you invert the Fourier transform you get N of R comma in principle now doing it analytically is formidable but you have a closed equation for it completely and what we will do next time is to take this and see how we are going to get the diffusion coefficient in the so called hydrodynamic limit first of all you have a relaxation time tau here we have seen that the velocity is relaxing with the relaxation time tau we know the diffusion regime is when you are at time scales much bigger than the relaxation time right so in terms of the Laplace variable we need not even invert the Laplace variable the diffusion regime would mean S times tau this guy here would be very very small this is small compared to 1 over tau that corresponds to long times small s is large t okay so this guy here should be equal much much less than 1 and K should be very very small also because you are looking at hydrodynamic modes you are looking at long wavelength fluctuations not on very short length scales long time scales long length scales should give us then the diffusion coefficient okay we will see how that emerges from here systematically we get actually more information from this equation but the basic trick is the following the basic trick is you write the single relaxation approximation by saying the collision integral is some minus 1 over tau times the difference between the distribution function and the asymptotic form or whatever and then you solve that equation in a self consistent way in this case the trick was to integrate over v and then get a self consistent equation for n tilde of K comma s which is what we are trying to find n of r comma t and what is done is to find this Laplace Fourier transforms first into this okay so we will complete this we will do this meanwhile try this out and then we will see how a systematic approximation procedure will give us the diffusion coefficient okay then there also remains the case of what happens if you apply a uniform but time dependent force on the system how will it take it out of equilibrium so you start with an initial condition that spatially uniform and the velocity is thermalized and then say I am going to switch on a force which does not depend on the position in the simplest instance but on time how is the system going to go out of equilibrium again in the single relaxation time approximation so we will look at that as well and those are things which these are the things which will help you to find things like the viscosity the diffusion coefficient the thermal conductivity and so on as I said before okay so let me stop here.