 So, from velocity now we come to the elephant in the room we analyze the behavior of the displacement of the particle because of the random forces. We know fundamentally that if x is the displacement it is very elementary knowledge that the dx by dt is related to the velocity v t. It is a very starting point in mechanics. So, which means just as formal integration would say x t equal to x 0 0 to t v t prime dt prime. If v t is imposed and known and is integrable then we can always express it as an integral over v. To simplify the matter let us say that x naught equal to say let us say that x naught equal to 0. The particle starts from the origin we have fixed our coordinate with respect to the original position of the particle and it becomes simple x t is just an integral over the velocity of the particle at all times up to since we have derived an expression for velocity we can calculate very easily the x bar at any time t which will be 0 to t v bar t prime dt prime and v bar dt prime is simply v naught v naught e to the power minus beta t prime dt prime. Well v bar is v naught e to the power minus beta t which leads to a simple result equal to v naught divided by beta into 1 minus e to the power minus beta t. So, all that it says is starting from 0 as a t tends to infinity the particle eventually gets a displaced from its original position through a distance of v naught by beta. So, x bar infinity is v naught by beta in some specific literature particle mechanics literature this is called as the stopping distance. This is the distance at which the particle stops a projectile stops within of course all those assumptions we made it should be a creeping flow low Reynolds number creeping flow it is very important concept in the study of for example aerosol systems. So, x bar is clear it is a simple exponential rising exponential behavior. So, we now go to the next important metric that is x square bar as a function of time. So, from the definition of x which was an integral over velocity x square bar is going to take the form it is an integral over t of the velocities v t 1 v t 2 over bar d t 1 d t 2 this is the axiomatic from the very definition, but we have derived the autocorrelation function v t 1 in the previous lecture that we wrote in a form v naught square minus gamma by 2 beta into e to the power minus beta t 1 plus t 2 plus a modulus term 2 beta e to the power minus beta into mod of t 2 minus t 1 regardless of whichever is whichever is larger. So, if we substitute this and integrate looks quite formidable and formidable, but actually a nice pattern emerges. First we see that when we substitute in this integral when we substitute this formula in this integral the first part is going to be an integral over e to the power minus beta t 1 from 0 to t and e to the power minus beta 2 from 0 to t and they are uncoupled integrals. So, they just are the same it is just a square of a single integral term. So, which can easily be identified. So, we again call this. So, contribution first contribution we call it as a contribution equal to v naught square minus gamma by 2 beta into the double integral that is 0 to t 0 to t e to the power minus beta t 1 d t 1 and e to the power minus beta t 2 d t 2 is simply going to be right here v naught square minus gamma by 2 beta into 1 minus e to the power minus beta t whole square divided by cos beta square. We just call a definitionally just define it as let us say a. We call it later. Now, so the first term we have done now comes the second term. So, in the second term we have once again we have to take care of the fact that we have a modulus function. So, that b integral let us define the b integral let here also this is let b be defined as gamma by 2 beta integral 0 to t 0 to t e to the power minus beta mod t 2 minus t 1 d t 1 d t 2. So, here since it is mod we have to remove this mod for integration. So, mod basically means if t 2 is greater than t 1 then this exponent is simply t 2 minus t 1. If t 2 is a smaller than t 1 then it will be t 1 minus t 2. So, we have to keep that thing in mind. So, with that let us write the beta term as as gamma by 2 beta then integration from 0 to t d t 1 let us keep out and that t 2 integral I will first evaluate up to t 1 then e to the power minus it will be t 1 minus t 2 because my t 2 is now running all less than t 1. This integral is all t 2 is less than t 1 then t 2 is less than t 1 it is always beta t 1 minus t 2 and the second part of the integral. So, here I will put another bracket. So, the second part of the integral will be t 1 to t because both are running up to t here it will be e to the power minus beta of t 2 minus t 1 d t 2 equal to bracket and the square bracket closed. So, the first integral this will be 0 to t e to the power minus beta t 1 in it will be actually e to the power beta t 1 minus 1 by beta it will be d t 1 now one integral we easily evaluate and the next one here 0 to t d t 1 this is also an exponential integral which can be easily evaluated and we note that e to the power beta t 1 we keep out only t 2 if you integrate it will have the form e to the power minus beta t 1 minus e to the power minus beta t divided by beta. In fact, both now can be one more integration can be easily carried out it is an exponential term this is a kind of a homework you can actually do and both the integrals become same leading to a formula for b which will be all this will come to twice gamma by 2 beta square into integral 0 to t 1 minus e to the power minus beta t 1 d t 1 it all comes to this. So, now when we compare which of course, is easily integrable. So, it comes to gamma by beta square this is a t minus 1 minus e to the power minus beta t by beta one should be careful to put the brackets at the right places. So, now if you combine b and a we had our a which was a square term here complete square and we have a b now. So, our result is a basically a and b together. So, finally, the x square t and some leverage is going to be v naught square minus gamma by 2 beta into 1 minus e to the power minus beta t by beta whole square one term and the second term is gamma by beta square into t minus 1 minus e to the power minus beta t by beta. Now since v naught square e 1 minus e to the power minus beta t is actually v bar. So, it becomes x bar square as we calculated if we go back to our x bar we see that x bar already involved v naught by beta into 1 minus e to the power minus beta t. So, it is a square is going to be that square term. So, we can write in a neater form sigma square x which is x square bar minus x bar square and that will have the form I will write with gamma by beta cube as out. So, it will be beta t minus 1 minus e to the power minus beta t minus half of 1 minus e to the power minus beta t whole. So, because of the gamma by 2 beta this half term comes. Now let us it is a kind of a neatly written form with the powers of 1 minus e to the power minus beta t. Now let us first explore what happens to sigma square x as t tends to infinity. So, then we see that it will be gamma by beta cube here there will be beta t and here there will be 1 and here there will be half and then all other terms will decay to 0 since e to the power minus beta t tends to 0 as t tends to infinity. Basically we can say that sigma square x decays as gamma by beta square into t minus some constant 3 by 2 into gamma by beta cube. This is just a small constant compared to the first term which is directly proportional to t. So, as we move to larger and larger times the second constant is just a displacement it is a small shift value we can neglect that and sigma square x can be written in transparent form as going as gamma by beta square t as t tends to infinity which basically now helps us to find an expression for the diffusion coefficient. Conventionally the mean square displacement we know that sigma square of x is written as twice d into t where d is diffusion coefficient. We did not have any theory for it we merely postulated a diffusion coefficient. In fact, in our random work model it was square of the jump length divided by the time and neither of them had any physical basis they were just some numbers we took or some concepts we introduced, but here now we compare and we see that upon comparing sigma square x upon comparing we see that gamma by beta square equal to twice d or since we already have since we have established from fluctuation dissipation theorem 1 that gamma by 2 beta equal to v square average which was kt by m from equi partition this implies we know combined with the previous one that d will be kt by m beta. But if you look at definition beta equal to f by m again. So, this leads to kt by m. So, we arrive at or we simply say get the Stokes Einstein relation d equal to kt by f. So, what this relation says is that the tendency or diffusion coefficient is reciprocally related to the friction coefficient. So, to put it in words diffusion coefficient is a reciprocally related to friction coefficient conceptually what it means is that the tendency to diffuse the tendency to disperse. So, that is what diffusion coefficient is about because of the random motion there is a tendency to disperse. There is a friction coefficient which tends in fact, if it tends to align the velocity into an average velocity a particle falling under gravity for example, because of the friction coefficient it attains a terminal velocity. So, there is a reciprocal connection between the tendency to align and move in a certain direction called the drift and the tendency to disperse. Although they look slightly opposing both of them come from the same underlying mechanics that is the mechanics of collisions. Because of this important conceptual connection between dispersity and friction which can be put the other way between the friction and the dispersity it has been given a very important terminology called the fluctuation dissipation theorem. Maybe we can call it 2, but this is the original fluctuation dissipation theorem introduced by Einstein without bringing in the concept of gamma. It can be proven in many ways and this is one way we arrived at it that is a relationship between the friction and the tendency to diffuse. This theorem of course, has been generalized and it is it forms the cornerstone of non equilibrium statistical mechanics today, but the Brownian example is still the classic example which illustrates this very meaningful relationship. As I mentioned there are other important ways of proving this of Stokes Einstein relation. We will revert to it some later time, but let us before that let us examine the small time behaviour of the sigma square x as a t goes to 0. So, we can also write if we use the notation let us say small time limit of sigma square x. So, define let us say dimensionless time tau equal to beta t yesterday definition. Then we have our exact expression sigma square x can be written as gamma by beta cube into tau minus 1 minus e to the power minus tau minus half of 1 minus e to the power minus tau square looks much neater now to confirm that this is indeed what we obtained we just refer to this expression. So, we have only put beta t equal to tau. Now, if we carry out the Taylor expansion of this which I am not going to do, but you can do it yourself e to the power minus tau is 1 minus tau etcetera etcetera. You can write this as actually this turns out to be gamma t cube into 1 third minus tau divided by 4 then there are higher order terms. So, essentially it means the short time limit it goes as gamma t cube it is a very interesting behaviour. So, while for large times the variance in the position goes as proportional to time at very early times the variance that is the difference between the actual displacement square minus the average displacement that will go as t cube it is not just t square it goes as t cube. So, this gives us an idea that we can derive the important relationship between the diffusion coefficient and the friction coefficient. So, often used in all engineering problems whether there it is in the colloids or in aerosols or many other thermal systems all that can be derived from Langevin equation under the assumption of a delta correlated acceleration. In order to develop this thought process into solvable situations one must pass from a deterministic type equation that is the Newton's equation to a stochastic or a differential equation. To carry out such a passage it is very important first to develop transition probabilities in both the velocity as well as the physical domains. In the next lecture we make an attempt to develop these transition probabilities and write the kind of differential equation that one arrives at based on them. Thank you.