 Right, so let us get started just to recap what where we ended last time we ended with the statement that if you looked at the velocity of a particle satisfying the Langevin equation the stochastic differential equation which recall was V dot is minus gamma V plus square root of gamma over m eta of t and this was a Gaussian white noise a Gaussian white noise which was delta correlated and had mean 0 then we discovered this was entirely equivalent to and this was a statement made it was entirely equivalent to the Fokker-Planck equation for the conditional probability density of the velocity which was delta over delta t P of V t given some initial condition V naught this was equal to gamma times delta over delta V V times P plus gamma k t over m d 2 P over d V 2 this is the Fokker-Planck equation and this was the Langevin equation this was a stochastic differential equation of the first order with a white noise term here and it said that the corresponding probability density for some specified initial condition satisfied this Fokker-Planck equation second order partial differential equation. I have put in the fact that the consistency requires that the system if it is in thermal equilibrium consistency requires that this quantity be related to little gamma according to gamma is 2 m gamma k bolts we will put that consistency condition in all the time because it implies that the system is in thermal equilibrium. So in a sense what this probability density describes is the approach of some arbitrary initial condition to the equilibrium distribution the Maxwellian distribution. So it is really telling you something about the manner in which equilibrium is approached and how does it equilibrate? It equilibrates due to collisions due to thermal collisions right alright. So given these two they form a pair out here one can also ask what is the correlation of the velocity itself like what is the solution to this Fokker-Planck equation like and so on and I mentioned that the solution is itself a Gaussian once again just as the noise is. So this was a driving force and it is Gaussian and so is the output variable V it is a Gaussian and the solution to the Fokker-Planck equation solution to the Fokker-Planck equation was the Onstein-Ohlenbeck process which read P of V, T V0 equal to there is a normalization factor and then it was a Gaussian which is the exponential of minus V minus V0 e to the minus gamma T whole square that is the mean V0 e to the minus gamma T the conditional mean divided by 2 K Boltzmann T minus M times that into 1 minus e to the minus 2 gamma T and then there is a normalization factor which was M over 2 pi K Boltzmann T 1 minus e to the minus 2 gamma T this to the power of half times that exponential it is a Gaussian in which the mean slowly drips from V0 to 0 and the variance slowly broadens till it hits the value at equilibrium the thermal equilibrium value of KT over N. So we know everything about the velocity from this now the question is what is the correlation time of the velocity and this requires you can do this in many ways we could start by the Langevin equation here and have an issue of compute the correlation time my correlation function autocorrelation by asking what is V of T 1 V of T 2 average over all realizations of this noise and I left that as an exercise to you we found the mean square and I said now find what the correlation looks like and if you have done that then you know that the expectation value of V of T 1 V of T 2 equal to K Boltzmann T over M e to the power minus gamma mod T 1 minus T 2. So it is exponentially correlated now there is a little theorem which says and this theorem is due to do it is called doob's theorem and it says the following in effect it says that the only continuous this is a continuous process this V is a continuous random process the only continuous stationary Gaussian process is the onstein-ohlenbeck process it is a Markov process and it is a the onstein-ohlenbeck process in particular. So note note the conditions the only the only stationary Gaussian exponentially correlated process I forgot to say that earlier so all the conditions are important you want stationarity this process is certainly stationary V is a stationary random process as is it the noise here it is Gaussian we see that the distribution function is a Gaussian I hear the density function it is exponentially correlated out here and the only such process is Markov and more over on top of it the onstein-ohlenbeck process okay all the conditions are needed if you drop for instance Gaussian and said what about a stationary Markov process which is exponentially correlated the only continuous it should be continuous because otherwise we have directly before us we have already seen an example of a process which is stationary which is Markov which is exponentially correlated and that is the dichotomous Markov process the dichotomous Markov process had all these conditions but it was a discontinuous process it jumped from one value to another and certainly not Gaussian or anything like that you could ask are there continuous stationary non Gaussian processes which are exponentially correlated and which are Markov also yes there is a whole family of such processes the many many such processes but they can be classified into 5 families in all for instance just as a matter of infumer curiosity let me mention that these are examples of processes which are continuous stationary Markov and exponentially correlated but which are different from the onstein-ohlenbeck process because they are not Gaussian okay and such processes are those for which this quantity p of xi t xi not this asymptotically as t tends to infinity tends to the stationary probability density p of xi where this p of xi satisfies satisfies the equation dp over dx equal to say some a of x over b of x p of x and this follows at best linear of the form a 1 x p of x let us just use the symbol x for this whole thing and this guy is quadratic so consider those processes for which the stationary distribution satisfies this differential equation these are called Fisher processes where a is at best a linear function of x and b is at best a quadratic function and of course these are constants whose values can be adjusted so for instance you could have a situation where this these to go away and this is a constant here and so on you can easily see the conditions under which this p of x is going to become a Gaussian you want to basically have an x there and nothing more and then it becomes a Gaussian but if you have any of these possibilities you have other other kinds of processes random processes which are continuous stationary exponentially correlated and Markov but not Gaussian and doves theorem does not apply in those cases so there are such processes and they are useful in certain context but we are focusing here on this theorem here there are other processes which are discontinuous like the daikon Thomas Markov process there is another process called the Kubo Anderson process and yet another called the kangaroo process I will talk about these little later when we come to some applications and they too will be exponentially correlated in certain special cases but you will see they are not Gaussian and they are Markov but they are not Gaussian so this is a very useful theorem this guy here the velocity is exponentially correlated in the large my equation okay now where does that get us what we need to see is what happens to the other piece of information that we know and we found that out by taking the continuum limit of a random walk itself so let us look at the position variable of this diffusing particle and we know that the position variable in some limit satisfies its probability density satisfies the diffusion equation so we have a statement which says delta P I will use the same symbol P it should be confusing for the probability density of the position of X, T given some starting point say X0 over delta T equal to D times D to P this is the ordinary diffusion equation and we know this X of T is also a Markov process continuous this should imply a stochastic differential equation of what kind this implies immediately by the rule that we had earlier this should be the corresponding quote unquote Langevin equation in this case should be X dot equal to there is no drift term this term is missing and you just have a diffusion term and by this correspondence I talked about its square root of 2 D times eta of T but this again is Gaussian white noise which is unit delta correlated how does this link up with that after all X is a position variable corresponding to the integral of the velocity so the question is how is this this connect up with this result here in what limit is this is this result going to go over into whatever this says mind you this says that the velocity this guy here X dot is a velocity is delta correlated on the other hand we know from this Langevin equation that the velocity is exponentially correlated right under what conditions is that exponent going to become a delta function if gamma tends to infinity if gamma tends to infinity or if you like if gamma T is much much bigger than 1 right so we keep that in the back of our minds that on time scales on which T is much much bigger than gamma inverse it is as if the velocity is delta correlated after all what is the graph of this function look like if I plot this correlation function as a function of T and say C of T is K B T over M e to the minus gamma mod T this guy looks like this it is an e to the minus gamma mod T looks like that this is symmetric function and you can see that when gamma becomes very large this function is going to be 0 everywhere unless T is 0 so it is going to become like a delta function spike so that gives us a hint that says this must be the limit of either gamma going to infinity or more physically the situation where T is much much bigger than gamma inverse therefore gamma inverse can be neglected essentially set to equal to 0 so let us see how that comes about where this comes about for that let us go back and compute what is the exact mean square displacement because what is the thing that this implies this diffusion equation immediately implies that the variance that the expectation value of X of T minus X of 0 whole squared this quantity this guy equal to 2 D T and that is trivial to do because we know the solution of this the free solution of this it is the Gaussian e to the minus X minus X naught squared over 4 D T you put that in integrate over X squared and you get this result right this is diffusive behavior it says the mean square displacement goes like linearly in the time that is pure diffusive behavior right now let us see if that is exact or it is an approximation from what we did you talked about so far it is clearly an approximation so the question is what is the quote unquote exact result in this case and that is easily found because all we have to do is to go back and say X of T minus X of 0 squared this guy equal to X of T minus X of 0 is the integral of the velocity 0 to T D T 1 V of T 1 and I want to square it and then take the average so it is clear this equal to 0 to T D T 1 0 to T D T 2 average value of V of T 1 V of T 2 full average over the initial condition as well because I put these bars here and we need to compute this number this quantity but now we have the explicit expression for this correlation function and this is by definition true this thing is equal to that by definition by the very definition of velocity this is true so this is equal to K B T over M integral 0 to T D T 1 integral D T 2 e to the minus gamma mod T 1 minus T 2 now what is the region of integration in the variables T 1 and T 2 the region of integration is 0 to T in each case 0 to T in each of these cases right this is a symmetric function of T 1 minus T 2 so if I reverse the sign of T 1 minus T 2 the function does not change and you are going to integrate over this unit square over the square from 0 to T in this range the value at any point is equal to the value of the function at this point it is completely symmetric so the integral can be written as equal to twice 0 to T D T 1 0 to T 1 D T 2 so I integrate in T 2 only up to T 1 which means in this function I integrate like this up to that point and that is the way I scan it I fix a T 1 between 0 and T I integrate in T 2 up to T 1 and I go on to the next value of T 1 and integrate and it gives me this now since T 2 T 1 is always bigger than T 2 I can get rid of this and write it as T 1 minus T 2 and I put the factor 2 outside which is equal to twice k Boltzmann T over M integral 0 to T D T 1 e to the minus gamma T 1 and then I integrate e to the gamma T 2 from 0 to T 1 so that is trivial this equal to e to the gamma T 1 minus 1 over gamma but this I can write as 1 minus e to the minus gamma T 1 so which is equal to twice k Boltzmann T over M gamma and then I have to do this integral which of course is equal to T the first term is T and then this minus cancels against that so it is a 1 over gamma e to the minus gamma T minus 1 that is the integral right so let us pull out this gamma squared here and write this as gamma T minus 1 plus e to the minus gamma T that is the exact expression for the mean square displacement for a particle obeying the Langevin whose velocity obeys the Langevin equation what does it do for very small gamma for small T this is really not a very good model for T much much less than gamma inverse but it is a good model for T much of the order of and bigger than gamma inverse certainly it is a good model what does it do just for fun what does it do for T much much less than gamma inverse so gamma T tending to 0 say what is the leading term well I have to expand this the 1 cancels the gamma T cancels so you left with gamma squared T squared over 2 factorial right so this becomes x of t minus x of 0 whole squared is equal to this whole thing and this tends as gamma T tends to 0 the leading behavior is k b t over m the 2 cancels with the 2 factorial the gamma squared cancels and then T square which is of course equal to v squared in thermal equilibrium T squared because the mean square display velocity is in a fluid in thermal equilibrium at temperature T this is equal to precisely k t over m because half m v squared must be half k t by the equipartition theorem so this is certainly true from the Gaussian so v squared average is found by averaging over the Maxwellian distribution and it gives you this in other words the root mean square displacement is the root mean square velocity multiplied by the time it is almost ballistic but with this effective velocity which is the root mean square velocity as you would physically expect but what is interesting here is a long time limit so what happens to this as gamma T tends to infinity what happens in that limit well this term is certainly negligible that is negligible this guy is becoming extremely large so you end up with one of the gammas cancels so it is twice k b t over m gamma times T but that is precisely the result we got from the diffusion equation which said that the mean square displacement goes like twice the diffusion constant times T you get exactly the same thing and now as a bonus you get a formula for the diffusion constant earlier the diffusion constant and the diffusion equation was in input it was just a parameter you put in but now you related it to a microscopic parameter so this implies this thing implies D is k Boltzmann T over m gamma that is a fundamental result this is the result that was used by Einstein to determine Avogadro's number in this early work on Brownian motion because what he did was to argue that if you looked at these particles spherical particles of radius a or something like that then m gamma times v is a retarding force that must be by Stokes law equal to 6 pi a eta v right so that comparison immediately tells you that D is k Boltzmann T over 6 pi a eta of course this quantity here is Avogadro's the gas constant divided by Avogadro's number so it gives you a formula for Avogadro's number if you know all the other parameters in the system does this mean by the way that the diffusion constant in a fluid is directly proportional to the temperature you would expect things would diffuse faster at higher temperatures right would it go linearly what would it do the viscosity is very strongly dependent on the temperature by the RNA's formula so it increases much more rapidly than that so there is a huge dependence due to this is in fact irrelevant practically compared to this activation form that is the RNA's form that is sitting here in eta okay but anyway it gives us a formula for the diffusion constant in fact you can go further we use the fact that in the Langevin model the velocity was exponentially correlated we do not have to do that we do not really have to do that what we need is the following we need to recognize that this guy is equal to integral 0 to T dT1 integral 0 to T dT2 times the velocity correlation function right so let me call that C of modulus T1 minus T2 and this fellow here stands for V of T1 V of T all we need is a statement that this V is a stationary process so that that C of T1 minus T2 the velocity correlation is a function of the modulus the time difference and the modulus of the difference okay so all we need is that piece of information and then we are done actually in here because now you will immediately get a formula for the diffusion constant in general even outside the Langevin model because as soon as you have this this function here put here which is a symmetric function you can write this as twice the usual things so write this as twice 0 to T dT1 integral 0 to T1 dT2 C of T1 minus T2. Now let us put T let T1 minus T2 equal to T prime C I want to change from T2 to T prime okay so this becomes equal to twice integral 0 to T dT1 and then what do I get when T2 equal to T1 the limit is 0 and when T2 equal to 0 the limit is T1 but there is a minus sign between dT2 and dT prime so it again gives you 0 to T1 dT prime C of T prime again gives you exactly the same thing okay but what does this look like now the region of integration here is T1 and here is T prime and it says integrate up to this point so it says this is T is 0 is also T and we are integrating over this triangle for each T1 you are integrating up to T1 in T prime instead of doing that way we could as well flip the integration and integrate in this direction in this fashion okay so what does that become this is equal to twice integral again T prime must from run from 0 to T that is clear from this figure 0 to T dT prime and since this guy does not depend on T1 I can bring it out of the integral so C of T prime and then an integral over dT1 and what should that run over T prime ran from 0 to T1 T1 was always bigger than T prime right so if you are going to scan it this way then it is clearly running from T prime up to T so this integral is from T prime up to T one way to remember it is that T1 is bigger than T prime therefore T prime is smaller than T okay so instead of going raster going vertically you go horizontally but that is a trivial integral to do so it says this quantity is twice integral 0 to T dT prime T minus T prime C of T prime that is exact we have not made any assumption except that the velocity is a stationary random process and that the velocity autocorrelation function is a symmetric function of its argument in even function of its argument right that actually follows from the time reversal property of the velocity I have not proved that here explicitly but we have seen that in our specific example that this is true so this is it this is general thing now we ask what does this guy do as T becomes very large there is no guarantee that C of T prime is such that this integral converges when T becomes infinite no guarantee as of now but if it converges then what does it mean this means that as T tends to infinity the large T limit this factor in general is going to die down as the upper limit increases it is a correlation it is going to die down as the argument increases so in this factor if T becomes very large this is negligible compared to this so you can pull it out of the integral but you might argue oh that may not be correct because what about the contribution from those regions of integration where T prime is comparable to T when you cannot throw out T prime relative to T but those are large values of T prime and for those this takes care of the convergence goes down you can say this much more rigorously using this dominated convergence theorem but the fact is that as long as this integral over C of T prime is finite this is guaranteed to be twice T integral zero to infinity DT prime C of T prime provided provided this converges if this integral of the velocity correlation function over zero to infinity does not converge that is a symptom that this behavior is not going to be proportional to T asymptotically but if it converges there is no doubt that it is equal to this guy here but we know that asymptotically the diffusion constant is defined as the limit of the mean square displacement divided by 2 T as T tends to infinity that is the definition of the diffusion constant so what does it tell us finally it says in general I might as well write it as V of 0 V of T drop the primes write it in this fashion so I shift the origin to zero this is a stationary call autocorrelation function and the integral of it from zero to infinity gives me the diffusion constant this is called a Kubo green formula it is an example of what is called a Kubo green formula where various response functions susceptibilities and so on and so forth are given as autocorrelation integrals over autocorrelations in equilibrium it is part of a more general definition where you subject this particle to a sinusoidal force and ask what is its steady state response going to be depends on the frequency dependent mobility and that will have a Fourier Laplace transform here e to the i omega t and at omega equal to zero the static susceptibility is called the real related to the diffusion constant okay well that takes us into non-equilibrium statistical mechanics I do not want to get into that here but just to tell you that the specific case of the Langevin model where this guy was kt over m e to the minus gamma t is going to immediately give you d is kt over m gamma but this is the more general formula okay okay so now we understand where this diffusion thing came from it came in the limit when gamma became very very large or in the regime in which gamma t was much much bigger than unity okay then the diffusion equation is a good approximation to the position of this variable the next question that arises is can I not look at the position and velocity together in phase space and try to write down a formula or function expression for an equation for the distribution not of x and v independently but for the conditional distribution in x and v together the joint distribution yes indeed we can but we need to write down now the corresponding Langevin equation for this and what does that look like well let us write it down and see what happens so you have x dot and that is of course equal to v by definition and v dot equal to minus gamma v in our model plus square root of gamma over m eta of t that was the Langevin equation of this guy here so let's regard these two together as a vector Langevin equation I should worry about dimensions but let's assume these are all in dimensionless various units and let's introduce a xi vector which is equal to x v this fashion then we have a vector Langevin equation for a two-component vector which looks like this it says xi dot equal to well let's bring this v to this side so plus some matrix times xi because these two guys are linear in the coordinate and position in fact x doesn't appear here at all so let's say there's some matrix R which takes care of the dissipation this guy here equal to square root of gamma over m a vector valued eta of t and this fellow stands for 0 the usual eta of t remember that capital gamma is little to m gamma k t we're going to put that in always as a consistency condition what is this vector with this matrix R yeah it's going to be 0 minus 1 0 gamma to take care of these two guys bring them to this side and then gamma becomes positive this becomes negative acting on x comma v is going to give you precisely these equations now just as we solve the ordinary Langevin equation by using this integrating factor e to the minus gamma t I need to now use the integrating factor e to the minus R t and so 2 by 2 matrix so I need to exponentiate this matrix to find that green function right but that's not very difficult to do because you notice that R squared this implies that R squared if I multiply twice together is going to give me a minus gamma on that side and a gamma squared in the denominator so this equal to gamma times R so once you have that the exponential is trivial because then R cubed is gamma R squared which is gamma squared R and so on so R to the power n is just gamma to the n minus 1 times R and therefore you can compute the exponential so in principle you can write the solution down and take averages and so on as we did earlier so in principle you can find all the correlations you can find v of 0 v of t x of 0 v of t we can do all these things but you must remember that x is not a stationary random variable it's the integral of the velocity in the velocity is a stationary random variable when you integrate this stationarity property is lost but it doesn't matter you can specify initial conditions x 0 v 0 etc and compute this whole thing but you would now ask just as I have a Langevin equation here with white noise is this not equal to a Fokker-Planck equation equivalent to some Fokker-Planck equation a matrix Fokker-Planck equation this time will involve some matrices etc but for the joint distribution of x and v and indeed that is so this implies the following equation so this guy implies the following implies the following Fokker-Planck equation for the density of x and v together so let's not use the symbol p for it let's use the symbol rho of x v t given x 0 and v 0 at t equal to 0 so let's specify that at the initial instant of time you had some initial position some initial velocity and this is the one particle density phase space density rho or probability conditional probability density in this case okay it implies the following equation delta rho over delta t equal to okay and then because this thing is a drift term is linear in this case it's R ij delta over delta xi i xi j times rho it's got to be scalar so it's just contracted plus d ij d2 rho over delta xi i where I have to explain what these matrices are this guy here this d ij is just 0 0 0 gamma k t over m because the noise is only in the second equation in principle we could have had a noise here two separately but then x dot is v by definition in this example there's no noise there and the noise is in the velocity because it's the force that's random so it appears in the equation for the acceleration okay and the summation over repeated indices is implied xi 1 by definition equal to x xi 2 by definition is v so it implies this equation again by this equivalence between the stochastic differential and Langevin equation and the Fokker-Planck equation for a two-dimensional diffusion process in this case and what does that look like what does this equation look like now if you simplify it this says delta rho over delta t equal to R ij now R 11 is 0 and R 12 is got a minus 1 out here so it's R 12 delta over delta x and then you have v times rho and what's R 12 minus minus delta over delta x v rho is there any other term R 22 is not 0 right so plus gamma times delta over delta v v rho because xi 2 was equal to v and i and j are both equal to 2 and then the only term that survives in this term thing here is plus gamma k Boltzmann t over m d2 rho over delta in this case all the other terms are 0 this is the equation satisfied by the phase space conditional density in position and velocity together but what does this tell us this says delta rho over delta t plus I bring it to this side and since v and x are independent variables in phase space this says v delta over delta x rho equal to gamma the usual thing that appeared on the right-hand side what does this remind you of it's a v dot del so the left-hand side is like a convective derivative this term is like a convective derivative you have partial delta over delta t plus v dot del that's the total derivative so it says just a very physical thing it tells you as soon as you have a flow which goes like this this will automatically emerge it will emerge automatically that there will be this term delta over delta t plus v dot del will appear automatically and then that acts on row to give the total time derivative and that will have the dynamics on the right-hand side okay so this is the Fokker Planck equation in the case when you have a phase space when you when you look at the Langevin equation as an equation in phase space in both x and b the question is what sort of solutions does this have again the initial condition is obvious in this case our initial condition is rho of x v 0 x naught v naught 0 is equal to delta of v minus v naught delta of x minus x naught and you can do this in several ways you get a bivariate Gaussian in x and v so it will have terms like e to the minus something or the other times v squared and then there's an x v term and then the minus x squared okay you can get some Gaussian of this form which can be written down we can actually write the solution down I am not going to do that here but what's of interest to us is what does this solution become what does the solution become if you reduce this to one of the two variables if you reduce to the v variable for which we already have a relation so I want p of v t v naught this should be found by actually integrating over all x the other guys so I integrate d x minus infinity to infinity rho of x v t whatever and it turns out this gives you the on strain ohlin by distribution as it should which is exactly what you expect what would you have what would happen if I integrated over v instead and get an expression for x right I get an expression for x but it will not be the solution to the diffusion equation it won't be that Gaussian because that's only true when gamma t is much much bigger than 1 so it's some complicated expression which will in fact involve v naught as well not just x naught but v naught as well showing that we x is not a stationary random variable that's not a much interest we can compute it it's not a much interest but we can ask what does this guy do when t becomes very large what would you expect it to do so I take this solution take this guy here with this initial condition and ask what is rho of x v t x naught v naught 0 as t tends to infinity or gamma t is much much bigger than 1 what would it tend to what would you expect now I'd expect the velocity to have thermalized completely and then what would be the distribution of the velocity the Maxwellian I'd expect a Maxwellian in velocity it would multiply a distribution in x and what would that is distribution be the solution of the diffusion equation because the limit is the diffusion limit right so this also is true that this becomes in this limit to Maxwellian in v times times e to the minus x minus x naught squared over 4 d t over square root so let's write the Maxwellian as well it would be equal to m over 2 pi k Boltzmann t to the power half e to the minus m b squared over so the memory of v naught is gone is lost because this tends to the equilibrium distribution this guy there's no equilibrium distribution in position we saw in the diffusion equation everything goes to 0 at all points in an infinite medium again exactly what you would expect on physical grounds so we won't go further into this but we could ask for one more generalization which is physically very relevant so out I've assumed there's no external force on this particle I've said there's an internal random force and there's a friction for consistency what if I impose an external force on it from some potential b of x then this equation here is going to get modified because I'm going to have minus out here the potential v prime of x over m that term is going to be there minus gamma v plus root gamma over m so I've got an extra term here and that's a complicated function of x depending on the potential when it's linear function of x then I argued that the whole thing is a linear drift and I could extend this formalism but when this is a nonlinear function of x then I'm in trouble so we look at what happens next time tomorrow we look at what happens in the general case here we'll first look at a linear problem see what happens very drastic change is going to happen depending on what kind of force this is for instance what is physically happening is that without this force this particle is diffusing clearly the variance of the position is becoming unbounded with time linearly now the question is suppose we put this in a potential such that it doesn't let it go too far cost too much energy I expect the diffusion may be curtailed it may not be able to fluctuate that far in other words asymptotically it's possible the variance of the displacement is going to be bounded it's not going to diverge and we'll see explicitly how this happens so we'll do that tomorrow