 So, we arrived at a very important result in the last lecture called the Stokes Einstein relation which was derived by considering the autocorrelation of the velocity fluctuations. The Stokes Einstein relation is D, the diffusion coefficient equal to the thermal energy k T divided by the friction coefficient and within the Stokes regime friction coefficient is basically 6 pi eta a. In view of its importance it is a useful to see whether we can derive it in alternate ways. In fact, the original derivation from Einstein does not consider Langevin equation. You can derive it by thermodynamic equilibrium arguments using Fokker Planck classical Fokker Planck equation in position space itself. However, having started with the Langevin equation dynamics we can derive this relationship without using the autocorrelation functions and that is what we do now. For that let us start with the Langevin equation which says if the particle has an instantaneous velocity v and then when I say dot it means derivative with respect to time. So, v dot plus beta v T equal to a T is the Langevin equation which is where a T is random acceleration with the properties that a bar ensemble average is 0 and the autocorrelation function T minus T prime at two different times the autocorrelation functions at two different times a T and a T prime expectation equal to gamma delta T minus T prime is delta correlated where gamma is basically in intensity of fluctuation parameter which in turn we could relate it to the temperature the friction coefficient etcetera all that we did in the last lectures. Now, let us start with this equation 1 itself and see whether we can arrive at the diffusion law. For that it is a very important educator because from a velocity equation we will have to examine in the presence of random force we will have to examine how we get the position equation. So, for that of course, first thing we know that velocity itself is the time derivative of position. So, we can call it as x dot T. So, we will specify whenever we say x dot or dot represents derivative with respect to time. So, with this the Langevin equation 1 takes the form x double dot it is a function of time, but we will just for brevity we will omit it and it will be beta x dot velocity is x dot and the right hand side is going to be just a T. Now, let us multiply this equation let us say we call it as equation 2. Multiply equation dot by x T just there are a couple of steps multiply equation 2 by x T itself. So, it simply looks like x x double dot plus beta x x dot will be x T at all all referring to a single time now because x is a function of T x dot is a function of T and x double dot also will be a function of T double dot meaning second derivative d 2 x by dx dt square. Now, we note the following note this term x x dot can be written as half of d by dt of x square because if you differentiate x square you will get 2 x 2 cancels x into dx by dt which is x x dot. Similarly, you can say and this is a bit long winded, but you can do it x x double dot can be written as half d 2 by dt square of x square minus x dot square. You can verify it by expanding first you differentiate you will get the first derivative term then again you differentiate you will get x double dot only in fact, x dot square will cancel from both sides. So, this way of writing the advantage is now you go back and call this is equation 3 and this we call as let us say equation 4 a and equation 4 b then we can upon substituting these equations equation 4 a b in equation 3 then we will obtain the following equation because x x double dot gets replaced with this. So, you will have half d 2 by dt square of x square then x dot is v. So, we can just write it as minus v square then of course, beta x x dot becomes beta by 2 into d by dt of x square and the right hand side now will be x t a t let us call it as equation 5. It may be useful to all the time right instead of writing x square to just define a tentatively a quantity y just equal to x square just for reference define then it becomes looks like differential equation formulation we are quite familiar d 2 y by dt square and of course, I multiply by 2 throughout. So, this becomes a 2 v square v is the particle velocity at time t plus beta dy by dt will be twice x t and a t to see to confirm that it is true we just take a step back and look we have put x square equal to y here that is all we have done x square as y. So, you will have d 2 y by dt square minus v square beta minus 2 v square because 2 v multiplied. So, we will y and x t a t now we make very important assumption coming from physics right away. Supposing I take an average now ensemble average equation call it as 6 the previous one was 5. So, this will be 6 by ensemble average we again just to remind ourselves we mean it is an average over the replicas of a path of a given Brownian particle, but paths will not be the same of course, each time the randomness will make a different, but under identical host fluid conditions whose host fluid conditions is stationary it does not change with the time. So, every time I do an experiment the underlying fluctuation character remains the same, but particle trajectories will be different. So, you average over them when you do that this equation becomes d 2 say y bar by d 2 square minus twice it will be v square bar plus beta d y bar by dt that will be twice of x t a t. Now, we assume that we are basically working with the macroscopic timescale t. All the microscopic variations let us say have died down. So, as a result the system is almost near thermal equilibrium it is undergoing spontaneous fluctuations of course, and hence there is always an x associated with it, but the underlying fluctuations are coming from the fluid and the fluid is stationary. And as we have seen in our previous class asymptotically at a timescales much larger than 1 by beta or the tau relaxation time v square average reaches the equipartition equivalent of mean square velocity. So, assuming that the particle is in equilibrium particle gas equilibrium we can assume therefore, that v square average has reached the equilibrium value of k t by m that is one important assumption. So, this is assumption one another important assumption we make is that the ensemble average of the quantity x t a t will be 0. This is a little subtle and we can invoke a causal argument for it. After all a t is an acceleration coming from various impulses from the fluid x t is the displacement of the particle. Impulses first they affect the velocity and then it takes some time for it to affect the position of the particle. And hence the impulse that has been imparted at time t cannot be related to the position that the particle has at that time because the position at that time cannot be influenced by a t. It will always some delay effect is there. And that qualitative argument we can use here to justify x t a t equal to 0. You can also see it because we have already developed an expression for x t detailed expression. And then if you multiply by a t and take ensemble average it will also turn out to be 0. But so, we will not but then we do not want to depend on that proof. Hence independently we argue by virtue of a kind of causal relationship that the force that a particle receives now cannot influence or cannot be dependent or the position now cannot be dependent on the force that is it receives at that point in time. There is always a certain delay. Hence the ensemble average of position and imposed acceleration should be 0. So, with this the equation therefore takes the form you can write it in the way d by dt if you can take it out of dy bar by dt plus beta y bar equal to 2 kt by m. You can see that if v square average is kt by m then that we can take to the right hand side and this expectation is 0. And hence you will have these two terms on the left hand side and d by dt is common. So, I take it out that is what we have done. We can actually integrate this now step by step the first integral for example, can be obtained we can I will write down the solution for you the first integral is for dy bar by dt. So, dy bar by dt for example, from this we will have the solution 2 kt by m beta plus some constant of integration e to the power minus beta t. This you could do by taking dy bar by dt as z or something and then this will be dz by dt plus beta z equal to this it is a first order equation and then you can very easily integrate by integrating factor use the integrating factor then you will get back to this result. So, which is now further easy to integrate once again you can do the integration. So, you will have y bar equal to 2 kt by m beta t plus some c 1 by beta into 1 minus e to the power minus beta t where we know by definition that y bar is nothing, but x square bar because y was x square in the limit t tends to large exponential functions decay much faster of course, in fact, t much greater than when we need not really be infinity. We can even say that that is t much greater than 1 by beta that is sufficient. Then we will get x square bar equal to 2 kt by m beta into time plus some constant number c 1. C 1 is actually maybe the value that the system attains a certain displacement on an average, but when t is very very large this will tend compared to any initial value this will tend to t kt by m beta into t only as t tends to infinity because the constant number can be neglected with term which involves an ever increasing time. So, we now compare this expression compare with this definitional expression compare with the definition x square bar is twice dt. Then we get d equal to kt by m beta and note that beta was by definition f by m and hence m beta is f. So, we get kt by f. So, this is the desired result we have proved what we set out to do. So, if you work back in this we straight away rather than solve for v and then analyze the behavior for x y auto correlation functions by using two assumptions one that the Brownian particle eventually has its velocity fluctuation governed by thermal energy and the auto correlation function between the position instantaneous position and the instantaneous force it receives from the host gas is uncorrelated. Its instantaneous position is uncorrelated with the acceleration it receives we obtain this derivation. The idea of proving this is that there is a great underlying physics or underlying phenomena that relates the two that is the existence of thermodynamic equilibrium for the displaced or Brownian particle eventually that is underlying cause of this relationship between two otherwise very different quantities as which we discussed yesterday. So, the learning point that we should note in this why this relationship is further important comes from the following consideration. Supposing we have to discuss suppose we have to consider a Brownian motion of a particle in the presence of an external field supposing consider Brownian motion in the presence of an external field f x some external force f x we call it as f e x e for external. Then the Newton's law if we write that is a Langevin equation of course, we have d v by dt plus beta v equal to the systematic force this external force we are talking about is a very systematic force which is a function of position let us say plus of course, the random acceleration. So, since we are writing for v this will be f by m we have divided by mass throughout plus random acceleration at time t. Supposing we take an expectation over the various ensembles of the differential equation itself then the equation reduces to the form d v bar by dt plus beta v bar and since that expectation force does not depend on this expectation we can keep this term as such. But however, this will be 0 since a t bar is 0 when we do ensemble average. Now this equation if we solve for most forces if we consider the behavior of the average velocity after some time that is for times much greater than 1 by beta. Like for example, if 1 by beta is on my in micro seconds then we are interested in let us say milliseconds and more then the entire transient effects can be forgotten. So, then we can set d v bar by dt equal to 0 some steady state will be achieved and the velocity now will be virtually governed by the external force alone and that results in a formula. So, v bar therefore, in such a situation will be f external x divided by m beta. So, when you have an external field a steady state velocity that is system or Brownian particle here Brownian particle attains a steady velocity. Usually such steady velocities are expressed in non-equilibrium of physics as v bar equal to some mobility b into external force. This is we can call it as mobility coefficient. Sometimes the notation mu is also used for ions and for molecules mu is used. We are talking of particles Brownian particle and n's we use b. Now, if you compare this expression with what we obtained from the Brownian motion theory we can we will get b equal to 1 by m beta and we know beta equal to f by m. So, m beta is f so, 1 by f. So, mobility coefficient is inversely proportional to the friction coefficient. So, one can put it this way that if friction coefficient represents the concept of how difficult it is to move the mobility coefficient reciprocally answers same question of how easy it is to attain a velocity. Higher the b more will be the velocity attained whereas, friction coefficient would say that the higher the friction coefficient less will be the velocity attained. So, if they are reciprocal concepts. These two statements are of course, within classical fluid mechanics, but because of the Stokes Einstein relation now we have shown that f equal to is related to D. So, since D is kT by f we have the relationship using Stokes Einstein relation for f we obtain the relation B equal to since we can come back here since the Stokes Einstein relation tells us f equal to kT by D. So, we can replace f with the kT by D. So, 1 by f will be D by kT. So, B is D by kT. This is a very equally deep relationship which says mobility proportional to diffusion coefficient. To put it verbally the tendency for particles to scatter is what is represented by the diffusion coefficient. The tendency for particles to move with the smooth directed uniform velocity is what characterizes the mobility coefficient and there is a connection between these two apparently opposing concepts. And the underlying depth being underlying quality or a feature of the host gas being the random fluctuations the existence of random fluctuations both these originate from the existence of random fluctuation and thermodynamic of course, equilibrium of the system. From this point the question that is of certain interest is twofold. One we obtained the many several characteristics of the velocity and its variance of a Brownian particle. How does one go over to making predictions on the eventual evolution of the probability distributions? One we had the Fokker-Planck description for pure random walk. We moved over to a differential equation formalism, but there was no velocity involved. Now that we have brought in a concept of velocity what kind of equations are we going to get? So, that is one question. Second of course, how can we make use of all that we learnt for actually simulating? I want to do a numerical simulation of velocity fluctuations. How do I do that? So, these two aspects we will address in the coming lecture. Thank you.