 So, we saw yesterday that in the simple Langevin model for diffusion of particles in a fluid the equation is m v dot plus m gamma v is equal to a white noise 8 of t if this is a Gaussian white noise a random process with a delta correlation a Markov process which is stationary and has a delta correlation then the velocity v also turns out to be a Markov process and its conditional density turns out to be the Alnstein-Ulenbeck distribution. So, we saw that this implied and you can show that this implies and is implied by the fact that the conditional density the velocity is v at time t given that it was v 0 at some instant of time t equal to 0 this quantity this was given the normalized distribution was given by 2 pi k Boltzmann t 1 minus e to the minus 2 gamma t this is a time dependent normalization factor multiplied by a Gaussian function which is minus v minus the average value whole squared divided by the variance which is twice k Boltzmann t 1 minus e to the minus 2 gamma t this was the Alnstein-Ulenbeck distribution which starts as a delta function at v equal to v 0 at t equal to 0 and asymptotically approaches the Maxwellian distribution the usual Gaussian distribution corresponding to thermal equilibrium as t tends to infinity now of course once you are given this and the fact that v is a Markov process which we have not proved but I am asserting that it is so then all probability distributions connected with v are known n time distributions are completely known because everything could be written for a Markov process in terms of this conditional density alone but the interesting thing is this density also obeys a differential equation which looks like the diffusion equation but has another name extra term and we will discuss that a little bit today and that equation and this stochastic differential equation are equivalent to each other the equation obeyed by this p is delta p over delta t and it is called the master equation for the probability conditional density of a Markov process in this case this master equation is a second order partial differential equation and it happens to read equal to gamma delta over delta v v times p plus gamma k b t over m d 2 p so this equation together with the initial condition so p which is a function of v and t so p of v and t with the initial condition p of v 0 equal to delta of v minus v 0 and a certain set of boundary conditions because its second order in space in v we need boundary conditions and natural boundary conditions namely p of v t goes to 0 as v tends to plus or minus infinity specifies a unique solution and that is this distribution so the unique solution of this second order partial differential equation with this initial condition and this set of natural boundary conditions happens to be the answer in no one back distribution one can prove this rigorously you can solve this equation explicitly by a variety of methods we are not interested in that at the moment but this is where the answer in the Olin back distribution appears from this particular equation the master equation for the conditional density this equation is the first example of what is called a Fokker-Planck equation this is a class of equations for a specific kind of Markov process which I will mention in a minute and it is perhaps the first example or one of the first examples of a Fokker-Planck equation this equation for the velocity was actually written down by Lord Rayleigh and it is sometimes called the Rayleigh equation but it is common parlance it has become known as the Fokker-Planck equation because it is an example of a class of equations to which are labeled as the Fokker-Planck equations now we found also from this that the velocity was correlated exponentially so we found that the velocity v at time 0 v of t average v of t1 v of t2 was a function only of mod t1 – t2 I can shift the origin call 0 one of the arguments here in equilibrium this turned out to be kBT over m e to the – ? mod t – t prime this is the way you define for a Markov process a correlation time what you do is to take this process if you have a stationary Markov process and let me denote ? for general stationary Markov process you take this quantity ? of 0 ? of t take its average value that is the correlation function provided the mean is 0 at all times the mean is not 0 you should subtract the mean at every instant of time so let us even do that let us write the most general formula down the actual correlation function C of t we defined as ? of 0 – average ? which is time independent if it is a stationary process the average of that multiplied by this deviation from the average this quantity here would actually be the correlation function but you can easily see it is easy to check that this is the same as the average value of ? of 0 ? of t – mod average ? whole square this thing here is in fact ? of 0 ? of t – average ? whole square so it is a generalization of the variance of the process that is all that the correlation does let us for the time being for the case of v this was 0 so let us simply drop it and then you can define a correlation time in the following way so you take this quantity ? of 0 ? of t you normalize it by ? square that is the value of t equal to 0 this function is guaranteed to start at 1 at t equal to 0 and then in general it would drop off to 0 as t tends to infinity could drop off very slowly but here it drops off exponentially fast not always guaranteed but in this problem it drops off exponentially fast so if you took this quantity and you integrated over all t you get a quantity of dimensions time so 0 to infinity of this if this integral converges you get a certain characteristic time in the problem and this is what you would call the correlation time of the process so that is the way pardon me if this is a random process if this is a stationary random process then this is the way of acquiring of extracting a time scale in the problem the time scale on which the system loses memory some observable yes and that is a good question this question is if I look at this problem and I look at the velocity correlation time is that the same as the correlation time for say the position or anything else we will see it is an interesting question we will see what happens to the position in a minute for the velocity it is so this is certainly going to give you a correlation time pardon me this quantity this is I squared average so that this is normalized to unity otherwise if this quantity has a physical dimension of the square of Xi whatever it be I want to extract time from it so I remove this this quantity gets rid of the physical dimensions of Xi and it is something which would typically look like this this function here would start at 1 and die off and I am saying this quantity here this area under the curve is actually got the physical dimensions of a characteristic time scale that is not easy it is not hard to prove that if you took this in the present instance this would imply a tau correlation time equal to gamma inverse to really follows that you simply get 1 over gamma as a correlation time of the velocity now the next question is very nice we have something for the velocity what about the position what happens to the position something very complicated happens to the position the fact is it turns out that I was really look at the motion in phase space as we know in dynamics you must look at things in phase space so since I am looking at one component of the velocity the x component say I should really look at the pair x, v and then I should not even be talking about p of vt I should be talking about the joint density in phase space of x vt given say some initial position v0 at t equal to 0 so it is this quantity that I should look at and I should ask how does that evolve as a function of time it turns out you can write a Fokker-Planck equation for that quantity that would correspond to writing an equation for the probability density corresponding to the pair of variables here so I would write x dot equal to v and v dot equal to this look like a dynamical system except that it is got random components here and then I could write a probability density of that kind and then ask what is the equation obeyed by it it also satisfies a Fokker-Planck equation it is somewhat more complicated but we could actually write it down quite simply pardon me yeah the statement is dynamics happens in phase space as we know from our experience with dynamical systems so we are talking now about a random velocity of an individual particle the x component of the velocity say I should really talk about the position and the velocity together because that is what constitutes a point in phase space right so I should really write down a couple set of equations and here is the couple set of equations x dot is v and v dot satisfies this equation which is random this eta is random corresponding to this stochastic differential equation I have a probability density in phase space which is given by rho of x vt this is the probability density or when multiplied by dx dv this gives you the probability that if a particle starts at x0 with velocity v0 at time 0 it is found in the range v to v plus dv in velocity and x to x plus dx in position at time t this quantity satisfies the Fokker-Planck equation also which is more complicated than this equation in fact we could write this equation down let me do so that is the Fokker-Planck equation in phase space and it is actually not much more complicated than this yeah I should write it down really I should write down not this but I should write down r I should write down v and I should write down r0 v0 so let us do this in several steps I should really finally aim at writing rho of r vt given r0 pardon me no the motion occurs I look at particles in this room they are moving in three dimensional space but they move each of them is moving in a six dimensional phase space one particle phase space it is six dimensional it is not a Hamiltonian system at all this thing is not a Hamiltonian system is right away gone from that there is no way you can produce this friction in a Hamiltonian system I put this model in phenomenologically and I put in a random component here so we move to a stochastic dynamical system now which is in fact dissipative but no I am not saying anything like that I am saying that every particle has the particle motion occurs in phase space that is all I am saying I need to have equations of motion which I write down should be for all the dynamical variables and the position and the velocity other position in the momentum they determine the state of the particle at any time so it is not Hamiltonian does not have to be Hamiltonian at all I am not writing any Hamiltonian down at all but the idea of phase space is more general than that of Hamiltonians as you know if this particle had other internal degrees of freedom then the dimensionality of the corresponding phase space would change but it is a point particle its position and momentum you mean why do not I have what else could I have here what is the k it is got it is got to be a physical observable pertaining to the point particle what are the physical observables pertaining to a point particle at any instant of time that be its position and momentum that is about it it is true that if I have this particle in if it turns out that there are velocity dependent forces it turns out there are other degrees of freedom interacting with it it turns out for example that its basic equation of motion is six dimensional or seven dimensional I would have to include those extra terms but as long as I have forces conventional kinds of forces we know that the position and momentum are sufficient or in this case the position and velocity so this is the sort of quantity I would look at but let us first do this for one dimension and then I write it down immediately in orbit in three dimensions the equation turns out to be delta rho over delta t plus v delta rho over delta x equal to whatever we wrote on the right hand side gamma delta over delta v v times rho this time plus gamma kvt over m t2 rho over dv2 this is the Fokker-Planck equation in phase space nothing much has happened nothing much has happened you added this term here in fact you can intuitively guess where this term comes from what does this remind you of yes in fluid dynamics yeah v dot del so it reminds you exactly of that it is a convective derivative so the left hand side is really the total time derivative of rho in that sense so that is all that happens this is a very complicated solution if I write this down with initial conditions p of x v0 is delta of v minus v0 delta of x minus x0 and I use natural boundary condition saying that rho vanishes as x tends to plus minus infinity v tends to plus minus infinity then the solution turns out to be a fairly complicated distribution turns out to be a Gaussian jointly in x and v so it is a multivariate Gaussian distribution I am not going to write that down right now now how could you generalize this to rho of r, v, t that is also obvious almost obvious let us say so what would the general equation be in 6 dimensional phase space how would it change this remains the same and not surprisingly this becomes plus v dot the gradient of rho gradient with respect to r and on the right hand side out here you actually have gamma times the gradient with respect to v dotted with v rho the divergence of this this require this means the components of this are the partial derivatives with respect to the three velocity components in this term here becomes del with respect to v squared of rho that is the full Fokker-Planck equation for free particles in the absence of an external force in phase space and the solution is a generalized Gaussian in all these variables you can use a matrix method to solve this equation but it is a little intricate and we will not do that here so you can generalize these two actual motion in any number of dimensions but our purpose now is not to do that we go back we go back to this set of equations the Langevin equation and ask what can we say about the position of the particle we know all we need to know about the velocity because it starts conditional density is given by the answer in Olin by distribution what can we say about the position of the particle how far does it move in a given time so let us see what happens so we could start with the particle moving from the origin at t equal to 0 or at any point x0 and then let us find out what it does so I know that x of t minus x of 0 is equal to an integral from 0 to t dt1 v of t1 this is a definition of the displacement and let us call this delta x of t now I would like to find out what is the mean square displacement so all I have to do is to take the square of this and take the average in equilibrium so what is that give us that gives us 0 to t dt1 0 to t dt2 and then you have a v of t1 v of t2 in equilibrium no external force but we already know what this is this is the exponential that I wrote down therefore you can compute this number right you can actually calculate this number very straightforwardly because all you have to do is to put in that exponential there and compute it so let us do that a little trick involved here because remember that this quantity was k Boltzmann t over m e to the minus gamma times modulus t1 minus t2 is a modulus there in other words in the t1 t2 plane you have to integrate 0 to t for t2 and a t1 here this is the line t1 equal to t2 and you have to integrate over this whole square here but the function here is symmetric with respect to the function there because it is a function of the modulus the value at any point here is equal to the value reflected on this line so we could as well write this as equal to twice the value in one of the triangles so I could write this as twice the value from 0 to t dt1 and this goes from 0 to t1 dt2 k Boltzmann t over m e to the minus gamma since t1 is bigger than t2 I could just write this as the t1 minus t2 without the modulus and which triangle I am integrating over I am integrating over t1 from 0 to t and each time t2 is restricted it goes from 0 up to this quantity so it just goes up till there that is the integral I am doing that is a trivial integral to do very simple and the result is the following so you end up with delta x of t whole square is equal to turns out to k Boltzmann t over m gamma square because I am going to do two integrals here and the k Boltzmann t over m comes from here and then it is an integral which gives you gamma t minus 1 plus e to the minus gamma and that is the result what can you say about the displacement now right away from this result this quantity is the mean square displacement and it is a function of the time what sort of process is the displacement now if the mean square value depends on time what can you say about this process is it stationary or it is non stationary exactly so that is the first problem it is not even a stationary process unlike the velocity whose statistical properties were independent of time and therefore the mean the mean square and so on were all independent of time this process is not even a stationary random process so its correlation function is going to depend on both time arguments and is not a function of this of the difference of the two alone so if I compute delta x of t1 delta x of t2 it is a function of both t1 and t2 and how does this function look as a function of t if I plot delta x of t whole square exactly so it is clear that asymptotically this term and this term are going to dominate over this and this is a straight line which intersects at t equal to 1 over gamma so it is clear there is a straight line of this kind with some slope and this function look at what happens for t very small gamma t much less than 1 this goes like 1 minus gamma t which cancels both these things and then you got a gamma square t square so it starts off like a parabola and then it asymptotically hits this value here eventually so it is clear that this function looks like this and this line here its ordinate is 2d 2 2 whatever it is 2k Boltzmann t over m gamma t minus 1 over gamma it is a straight line and that is the asymptote so it behaves linearly for sufficiently long times but at short times it is quadratic and then in between it has both the exponential as well as a linear term and it is not a stationary process x of t is not a stationary turns out it is not even Markov but there is one case where things simplify where have we come across the statement that the mean square displacement of a diffusing particle increases linearly with the time this is a famous diffusion equation prediction so if you write the diffusion equation down for the positional probability density function then we can prove directly from that that the mean square displacement increases linearly with time but this is saying something else it is saying it is not linear it is only linear for sufficiently long times this thing here is gamma inverse and if t is much bigger than gamma inverse which is the velocity correlation time then the mean square displacement increases linearly with time so it is giving us a very important input it says the diffusion equation which I would normally nively write down from fixed laws of diffusion a valid only at sufficiently long times how long times much longer than the velocity is memory time correlation time so that is telling you something beyond the diffusion equation so now let us backtrack and write down the diffusion equation and see what that implies and what sort of stochastic equation that implies again so let us go back to our phenomenological the usual phenomenological description of diffusion is in terms of two conditional densities so let me call I need to use another symbol for it so let me use the symbol P script there for the position of a particle at time t this is the positional probability density and what you say about this it is like if I take a diffusing species and I write the concentration as a function of position and time I write the two laws of thick the fixed laws for diffusion the first one is just conservation of matter and the second law tells you how things move so the equations if you recall for the concentration of a species at time t the equations fix laws are delta C over delta t equal to on the right hand side this quantity is equal to it is equal to minus the divergence of a diffusion current that is the continuity equation that is fixed first law it is just the conservation of matter continuity equation but the second law says something dynamic it says under suitable conditions the current J of RT is proportional to the negative gradient of the concentration so the current moves from the region of higher concentration to the region of lower concentration so it is equal to minus the gradient of the concentration and the coefficient here is called the diffusion coefficient and when you put both these guys together of course you end up with the diffusion equation which says delta C over delta t equal to D del square that is the famous diffusion equation the positional probability density of a single particle in one dimension now moving in one dimension satisfies the same equation this is almost by definition and therefore it implies that we have the diffusion equation delta p over delta t equal to D d 2 p over dx 2 that is the diffusion equation so we set aside the Langevin model we are not going to think about it for the moment we come to this we look at this diffusion equation and ask can we say something about the motion of the particle from here well I need to specify initial conditions and boundary conditions before I am through so let us start with the simplest initial condition let us assume that on the x axis the initial concentration is at x equal to 0 delta function so let us say that p of x 0 is equal to delta of x so t equal to 0 it starts at x equal to 0 I could shift the origin as I please but let us choose the origin to be at x equal to be at the initial position of the particle what are the boundary conditions you would like to put well I certainly like to have this I certainly like to have the total concentration to be finite and the equivalent of that here is minus infinity to infinity dx p of x, t equal to a finite quantity and if it is a probability density function then I is equal to 1 for any t greater than equal to 0 I certainly like to normalize this probability so it should be some non-negative function which is whose total area under the curve is unity and which starts as a delta function at t equal to 0 what is a necessary condition for this integral to exist it is an integral which runs from minus infinity to infinity the entire x axis so when would that exist well this function must be integrable of course but it must vanish at the end points otherwise this integral does not exist it must vanish sufficiently rapidly at the end points but certainly it must vanish so the natural boundary conditions natural BC of x, t tends to 0 as mod x tends to infinity as x tends to plus or minus infinity on both ends I would like it to vanish given this initial condition and those boundary conditions this equation has a unique solution and we know the solution the famous diffusion equation solution is in fact p of x, t is equal to the exponential of minus x squared over the exponential of 4 d t and there is a normalization constant which is square root of this is the famous Gaussian solution it simply says that as time goes on this becomes a Gaussian then it widens out and eventually dissipates down completely the area under the curve remains one at all times but this curve broadens finally p vanishes at every point but the total area is one that is the fundamental Gaussian solution to the diffusion equation or the heat conduction equation these are all identical equations well for one thing we can find the value of the mean value of x or the mean square of x etc. you can in fact do that without solving this equation but once you have it you can write it down how do you find the mean value of x how do you find x of t average from this equation without solving the equation what would you do well I would like to compute x of t equal to an integral d x x times p of x, if this is normalized to unity then this is the definition of the mean value so let us do that let us multiply both sides of this equation by x and integrate over t over all x so what happens I multiply this by x on both sides and integrate from minus infinity to infinity I have integral minus infinity to infinity d x x times delta over delta t p of x t this quantity is equal to d times an integral minus infinity to infinity d x x times d 2 p over d x 2 but remember the t dependence comes entirely from here so I can actually pull this out once it is integrated over x this is just a function of t alone and I can pull that out and since it is a function of t alone I can simply write it as d over dt but what is this quantity equal to by definition it is the mean value so it says d over dt x of t is equal to this and the obvious thing to do is to integrate by parts so if I integrate this by parts this gives me d times x delta p over delta x at x equal to minus infinity plus infinity minus d times an integral minus infinity to infinity d x times the derivative of this function with respect to x. Which is one and the integral of this which is delta p over delta x so it just gives me delta p over delta x but this is 0 because we would like all moments of this quantity of x I like the mean value the mean square value and so on to be finite which means that p of x t must not only go to 0 as mod x goes to infinity but must go to 0 faster than any negative power of x otherwise that moment would diverge therefore since this goes to 0 faster than any negative power of x as x goes to plus minus infinity this quantity goes to 0. So this is gone and what does this do this gives you just the surface terms p at plus infinity and p at minus infinity which are 0 so that is gone and therefore you find this is 0 which implies that this quantity is a constant but if I start at the origin at t equal to 0 this is 0 at t equal to 0 and since it must remain constant it is 0 at all times so just telling you physically the fact that if you start with a symmetric initial distribution and let diffusion occur without bias the average would still the distribution would remain symmetric at all times the average would remain 0 so this is identically 0 what happens to the mean square displacement that is not 0 of course that is non-trivial so I would like to find x squared of t which is this and I start by finding x squared on the side I multiply by x squared so I have d over dt of x squared of t on the right hand side this is what we are going to get I am going to put an x squared here and integrate over x the first term the surface term will have an x squared dp over dx which goes to 0 and the next term will be a minus d times an integral minus infinity to infinity the derivative of x squared which is 2x so there is a twice and then there is a dx and then an x delta p over delta x the derivative of x squared the integral of dp over dx which is this and now I integrate by parts once again so that gives me equal to minus 2d x times p at minus and plus infinity that vanishes once again and then a plus sign 2d times the derivative of this which is unity and the integral of this which is p itself integrated but the integral of p is 1 because it is normalized so it is just this it finally leaves you with just 2d which implies at once that x squared of t equal to 2d t plus a constant but at t equal to 0 this is 0 and so the constant drops out and you end up with this famous result that the mean square displacement in diffusion is linearly proportional to time and the coefficient of proportionality is 2d in each dimension of motion in three dimensions if I computed r squared it would be average x squared plus average y squared plus average z squared which would give me 6d t now how does that tally with the result there this thing here is certainly not a linear function of t except in the limit gamma t much bigger than 1 and these two must match therefore in the diffusion regime we could call this the diffusion regime t much much bigger than gamma inverse in that regime x squared average of t equal to 2d t and we got something even more interesting if that is the case I must then equate this phenomenological coefficient d which is put in by hand with what we got from the microscopic model here and that implies that 2d t here must be 2d t there and that immediately tells you it is 2 sitting here in the gamma cancels and you end up with d equal to the diffusion is not arbitrary it is related to the dissipation in the system and it is related in this particular fashion this is another expression of the fluctuation dissipation relationship that I talked about the last time the diffusion actually tells you how things spread out how fluctuations act on the system and gamma tells you how the dissipation in the system acts and there is a connection between them this is also related to the viscosity of the fluid so now you could go back and ask what does this mean what sort of stochastic equation does x satisfy in order for this to happen and for this we need to know a little bit about the connection between stochastic differential equations and the corresponding master equations the Fokker-Planck equations I will write that relationship down without proof and it is as follows in the relationship only yes that is what happened I mean that particular microscopic model no not at all that is only a model as it stands so we have no guarantee of this at all finally you have to test against experiment whether the model is successful or not so the statement made is the following and now by hindsight we can mention what in this particular problem what the resolution of this whole business is the time between collisions the time of interaction between two molecules is of the order of 10 to the minus 15 seconds that is the sort of electromagnetic interaction time the time between collisions is of the order of a picosecond or less the next time scale is the velocity correlation time which under normal circumstances for the kind of fluids we are talking about would be of the order of microseconds so diffusion the diffusion equation would be a good approximation to the transport phenomenon on time scales much greater than microseconds so there are lots and lots of time scales in the problem they are well separated from each other which is why this trick works and of course in a dense fluid under different conditions and so on things can get much much more complicated the velocity is not given by the simple range of my equation the point I want to make here from the point of view of dynamical systems is that for a given stochastic differential equation you can write down a Fokker-Planck equation for the conditional density and it is as follows so if you have a random variable xi and it is given by a first order stochastic differential equation of the kind xi dot is some function of xi we do not care what some function which gives you the drift of this thing plus a white noise of this kind 8 of t let me call it zeta of t multiplied by perhaps a function of xi once again and this is the stochastic differential equation where zeta of t where f and g are given functions and zeta of t is a Gaussian delta correlated stationary Markov process with zero mean so we will make that assumption zeta of t is a Gaussian white noise with zero mean zero and zeta of t zeta of t prime equal to delta function t prime I have absorbed whatever constant sits here in this g if this is the stochastic differential equation obeyed by the random variables zeta xi then this implies xi of t is a Markov process whose conditional density satisfies a Fokker-Planck equation of the following kind p of xi satisfies delta p over delta t equal to minus delta over delta xi f of xi p of xi plus one half g squared of xi this is the Fokker-Planck set equation satisfied by p this is rigorously proved so the moment you have a random variable driven in this fashion by white noise Gaussian white noise stationary Gaussian delta correlated Markov process is zero mean then you are guaranteed that the output process p is a Markov process whose conditional density satisfies this equation this is generally given the name the Fokker-Planck equation and what we had is a special case in that case the equation we had for v if you recall was v dot equal to minus gamma v plus a white noise eta over 1 over m times this so g was essentially a constant so once we have this let us go back and ask what about the diffusion equation what kind of process was that well the diffusion equation said delta p of xt over delta t was equal to d d2p of xt over delta x2 so it corresponded to situation where f was zero this term did not appear at all and this term here was a constant which is given by g so it is quite clear that the stochastic differential equation x dot is given by a white noise on the right hand side and it is just square root of 2d times 8 so if you put g equal to square root of 2d then half g square is in fact d that is what we have f equal to 0 g was just a constant and this tally's with what we already know if this is a white noise its correlation time is 0 and we already saw that the displacement behaves its mean square displacement behaves linearly in the diffusion regime when t is much much bigger than gamma inverse in other words when the time scale is much larger than the correlation time of the velocity or another way of saying it is on time scales compared to which the velocity correlation time is 0 and acts like a white noise so really these two things are not contradictory at all but there is a consistency condition which connects one to the other in fact that can be generalized it does not have to be and this is in the final thing I want to mention it does not have to be the Lanjima model at all you could in fact start from very general considerations and ask what is the diffusion equation what does the diffusion constant do and that is not hard to see that is the final thing I am going to talk about so all we have to do is to start with this statement that the displacement in a time t is in fact the integral of the velocity over that time and then if I compute the square of this and average this gives me this quantity 0 to t t 2 and then I have V of t 1 V of t 2 average and if the velocity is stationary in equilibrium that is all I want in other words the energy does not change at all everything is in thermal equilibrium then independent of what the form of this quantity is I could in fact simplify this whole thing so the first step is to write this as this is a symmetric function so it is a function of mod t 1 – t 2 so the first step is to write it as twice the integral from 0 to this of this guy and write it as f of t 1 – t 2 now what have we done well here is t 1 here is t 2 and this is 0 to t 0 to t and I am integrating in this fashion but I could equally well integrate by interchanging orders of integration in this fashion so if you interchange orders of integration here this becomes 0 to t d t 2 on this side and this integral t 1 is always bigger than t 2 so it runs from t 2 up to t 1 up to t 2 sorry up to t that is right d t 1 and then this function of t 1 – t and now I change variables to t 1 – t 2 at t mine at in insight to t 1 – t 2 what happens to the lower limit of integration this becomes a 0 and then this integral is trivial to do one of the integrations is trivial to do and you end up with a result which is twice an integral 0 to t d t 2 if you like or let me call it d t prime t – t prime in this fashion and then inside is v of 0 v of t prime this quantity is expected to start at 1 and decrease to 0 and the integral runs up to t what is the asymptotic t going to infinity behavior of this well this term is going to dominate and this limit will go up to infinity the parts where t prime becomes comparable to t would be damped because if they go down here so it is clear that the t going to infinity limit of this the asymptotic behavior of that is in fact this quantity for sufficiently long times goes to twice t times an integral from 0 to infinity d t prime v of 0 v of t prime but if you insist that this should be equal to 2 d t by definition it gives you a basic formula for the diffusion coefficient so it implies that eventually the diffusion coefficient is intimately connected to the velocity correlation and independent of the model d is integral 0 to t d t prime 0 to infinity I might as well drop the prime here v of 0 in equilibrium so it says the transport is determined by equilibrium fluctuations and thermal equilibrium the fluctuations of the velocity this has a name it is an example of what is called a Kubo green formula in the Langevin model this was kt over m e to the minus gamma t and of course that give you d is kt over m gamma but that was specific to the model but this is much more general all you need is that in this integral converges actually does not even have to converge you can show that the d if the process is diffusive and not super diffusive or anything like that then this d is actually the analytic continuation of the Laplace transform of this to s equal to 0 so you could put in an e to the minus st then it is a Laplace transform of the velocity correlation and if you after doing the integral go to s equal to 0 that is the diffusion constant so this is in fact the beginning of the subject of non equilibrium statistical mechanics but our interest here was from the point of view of dynamical systems and what I wanted to show you was that even if you include noise then the formalism changes a little bit it is more convenient to talk about phase space densities or densities now like the invariant measures we talked about earlier to talk about time dependent probability density functions write equations for them and solve them and you get all the averages that you need and of course you could ask more complicated questions like what happens if you have a combination of deterministic and random components to the dynamics things get harder and harder but the formalism is well laid out so the model of the story is eventually you have to use statistical methods you have to write down equations for from either from or from other inputs either from the dynamical equations or otherwise you have to write down equations for the distribution functions the probability densities or the measures and then after that find statistical averages of the quantities that you want and then there may be internal consistency conditions like the one we found for D here which you have to impose in general they are not if the things of the random is the noise is totally different it is an external source they may not be such consistency conditions so very few general lessons but there are some pointers as to what one should do in such cases that maybe I should stop here thank you.