 So, we had this detour of deriving a formal expression for the random force and its two time correlation function. So, originally the fundamental contribution of Langevin to stochastic phenomena is to introduce this equation where he combined Newton's law with the random acceleration term and as we saw in our earlier lectures, he formulated an equation of this type. Here we have divided by mass beta is the friction coefficient by mass of the particle. So, we no longer subscript it now because now the gas is a just a background the fluid is just a background we do not need its internal properties. We merely need to know that it has a certain temperature t fluid temperature t and it is in equilibrium where if my particle has been disturbed I we have imparted it a certain velocity some external means you know like a particle thrown into a liquid with certain velocity and we want to know what happens to this velocity because now molecular molecules are going to strike and a random force is going to be introduced and this equation describes the behavior. The unique property of these random forces is that its average is 0 and its two time average or two time correlation involving averaging over all random components is some constant gamma delta t minus t prime. When Langevin originally postulated a and its properties that is this delta correlation property gamma had was a parameter it was not known on what it would depend on, but he merely introduced as a coefficient to be determined later. We have a model via gas-coilitic theory, but that is only a specific model. When we wrote the Langevin equation we have not considered a specific model of the fluid we merely postulated the existence of random component of acceleration impinging on the particle. So, with this input let us see how to solve this equation. So, we start with an initial condition it is a first order equation with the time dependent force. So, the initial condition we impose is that v 0 equal to v naught particle is just thrown with some velocity. Now, what happens to it in the presence of this random force which has this delta correlation. So, for that we integrate this equation it is very easy it is a first order equation. I can easily say if you have an equation of the type say it is like if you have an equation of dy by dx plus some px into y equal to qx all of us study that y e to the power integral px prime dx prime is qx double prime in e to the power integral px double prime dx prime ok is x prime dx prime plus an constant c. So, this is called an integrating factor. So, we use this property from that we can easily write down that velocity v t formally will be v naught e to the power minus beta t plus e to the power minus beta t and we do not know the property of a t explicitly here this will be a t prime dt prime. So, it is the source term. So, we integrate with the integrating factor is e to the power beta t prime and we integrate this source term. This is a formal simple solution. Let me again emphasize that in the gas kinetic model we explicitly assumed a as a series of delta functions here no such assumption is made all that we know are the two point correlations about it. So, with this v t is of course, now a random velocity because a is random v values will change depending on the particular realization of a. So, that is not the matter of interest to us although we would need it at a later stage, but right now let us focus only on the average behavior. So, what is the average value? We call it as v bar. So, sometimes over barring is the same as ensemble averaging. So, for convenience of writing I sometimes over bar. So, we can note that the first term will not have any fluctuations it is a deterministic term we have given a very specific velocity. So, it will not depend on our averaging process. The second term the averaging process is basically an integration process or a summing process. So, I can take it inside because none of the quantities which depend on beta and t prime they are not statistical quantities the only statistical quantity is a t prime. So, the average can be taken inside the integral. It is an important step it comes from independence of the quantity on which we are averaging, but we know that this average is 0 by the property 1 of the random acceleration. On an average it does not have any contribution which means v bar t is going to be v naught e to the power minus beta t. What it means is the particle starting some velocity v naught if this is the velocity axis and the time axis necessarily its velocity will come to 0. This is a kind of deterministic decay of velocity. This is a deterministic result from Stokes law itself because it is basically an exponential function. The time scale that is the characteristic time what we call as a tau, tau relaxation we can simply call it as tau R for the particle that will be 1 by beta. So, if you throw a particle into a fluid with some velocity v naught it necessarily its average velocity will come to 0 and the time required for it to come or time required in the sense of an exponential decay 70 percent or of or so to lose will be 1 by beta or tau R. Just to give you a physical feeling for these numbers the expression for tau of a beta is a 6 pi a eta by m of particle or tau is the reciprocal of that tau equal to beta to the power minus 1. So, if we estimate for for example, for if you take air where eta is of the order of 1 1.8 into 10 to the power minus 5 Pascal second and particle of of the order of 50 nanometers and then corresponding mass and let us say density of particles you can take just for references 1000 kg per meter cube rho particle 1000 kg per meter cube. We will get tau the relaxation time of the order of the in air for particle well it should be for particle. So, let us not confuse that. So, we will call it as a tau relax equal to about 30 nanoseconds. So, that is even the our macroscopic times itself is of the order of few nanoseconds when we are talking of particles of the order of 50 nanometers or so. If we have larger particles it could be much larger. So, it goes as it will go as actually square of the diameter or radius of the particle, but we should still note that 30 nanoseconds is a much longer time as compared to the molecular collision times. You can actually collision times of the order of 1 by nu c tau collision this is of the order of 1 by nu c and you can estimate from gas kinetic theory this is of the order of pi a square n g into say g square average of c whatever mean thermal speed and it will come to about 10 to the power minus 14 seconds 1 4. So, you can imagine that the each of those kicks has occurred in extremely short time intervals in even in a gas and it would be far higher in a liquid. So, that it will be even shorter. So, in other words we have highly differentiated timescales in this process acting and we are justifiable in our assuming that in tau relaxation is in fact, a macroscopic timescale although it is in nanoseconds as compared to the molecular collision timescales which are close to much less than even picoseconds. So, that is we have tau collision is far smaller than tau relaxation. Hence, the Brownian motion is a visualization of microscopic phenomena in macroscopic timescales that is what actually the whole sum and substance of this exercise is. Now, we go to calculating a higher order average v bar did not actually depend on the random acceleration at all, but quantity of interest could be v square average v square t average it can be a bar sometimes or the angular brackets. Let us see how to estimate this quantity from the expression we have for v. So, this is average of or mean square average mean mean square velocity of the particle to obtain this we again note that the system is in thermal equilibrium the gas the host gas or the host fluid in thermal equilibrium with itself it has temperature t. So, all these molecules are colliding subject to the constraint that they have to maintain an average temperature t. So, Brownian particle is immersed in that into this system. So, since we have the expression for v t which we have seen. So, v square t can be written from that expression as just v square t before averaging we go back. So, we have this expression for v t here v naught e to the power minus beta t plus a term that involves random acceleration. So, we use that to construct v square t which will be v naught square we have to square that term first. So, there will be strictly yes a plus b whole square type. So, this will be minus 2 beta t the second will be 2 v naught e to the power minus beta t this will involve the a term beta t prime a t prime d t prime. And there will be a next term which will come as the square of the integral itself and that can be written as v e to the power minus 2 beta t. So, that square can be written as double integrals each of them running from 0 to t, but one of them the having the random not random the arbitrary index as a t prime the integration variable as t prime and for the other it can be as well t double prime. So, we maintain this distinction. So, that I can write the two accelerations within the integral as d t prime d t double prime. So, this is a small thing that we should always note that when we square an integral it can be written as double integrals, but the internal variables the dummy variables should be then differentiated cannot use the same index one will end up with mistakes. So, they are kept like this. So, when we now take an expectation when we do this v square d it will be null the expectation over the first term which of course has no random component in it. So, it will remain as such will be e naught square minus 2 beta t very interestingly the second term this term which involves a t here when we do the average that will go to 0 because expectation of a is 0. So, the second term will necessarily go to 0 and the third term will now appear as 0 to t 0 to t e to the power beta t prime plus t double prime ensemble average of the accelerations d t prime d t double prime. So, the last term is the involves the two time correlation functions naturally coming because of the double integral and we as we did earlier we took the ensemble average inside the integral because none of the parameters other than a and a prime other than a t prime and a t double prime depends on the averages over which the ensemble average is taken the parameters are independent basically. So, we now make use of the fact that a t prime a t double prime is gamma into delta t minus t prime. So, we replace this ensemble average. So, we get v square t average will be v naught square e to the power minus 2 beta t plus gamma I can take out e to the power minus 2 beta t was already there. So, this double integral 0 to t 0 to t over t prime and t double prime, but the in t grants add this function t prime and t double prime delta t prime minus t double prime as a two point correlation integrated over both the variables. Now, we see that both t prime and t if you see they vary from 0 to t, t prime or t double prime both t prime and t double prime vary between 0 to t. That means, anyone say if I first do the integration over t double prime it necessarily will cross t prime somewhere because t prime is also going to run from 0 to t. Hence the delta function necessarily will select the value corresponding to its 0 which will always be met because of the nature of the in double integral. Hence the selection will mean that the first integration over t double prime will contribute at t double prime equal to t prime and we can then substitute the value. So, that will lead to the expression v naught square e to the power minus 2 beta t plus gamma e to the power minus 2 beta t. So, when we do integration over t double prime first that will cancel, but t prime integration will remain. So, it will be 2 beta t prime dt prime delta has now selected at t double prime equal to t prime which is very easy to integrate. So, we get the answer v naught square e to the power minus 2 beta t plus gamma by 2 beta 1 minus e to the power minus 2 beta t. So, in writing this equation I have just skipped one step here first you do this integration it will have e to the power 2 beta t minus 1 by 2 beta, but then you multiply by e to the power minus 2 beta t the first term becomes 1 and the second term becomes 1 minus 2 beta t and it will have this term gamma by 2 beta. So, let us try to understand it at t equal to 0 at t equal to 0 the second term becomes 0. So, necessarily v square average had a value v naught square because v naught was a defined velocity it is perfect because when t equal to 0 this term is 1 and this is 0. However, when t tends to infinity and when I say t tends to infinity it is not some very long time it is just large as compared to the relaxation time 1 by beta. That means, within a few microseconds for example, beta t is going to become much more than 1 hence this term will decay. When we are talking of v square correlation the impact of the original velocity will be very short lift necessarily it will undergo a damping and as t tends to infinity e to the power minus 2 beta t tends to 0 and hence we will get or we obtain v square average t tending to infinity will be gamma by 2 beta only. We have arrived at a very interesting result that no matter what the initial velocity was the particle eventually will relax and attain a certain stationary value of its mean square velocity dictated by the ratio of the strength of the random accelerations and the relaxation rate. Although the individual velocity goes to 0 we showed that the previous derivation showed that the mean velocity went to 0 exponentially decayed here via beta minus t the mean square velocity tends to a finite value hence the particle has v square t or v square average now time has been taken to infinity asymptotically tends to the expression gamma by 2 beta the mean square velocity does not decay to 0 eventually it approaches a constant that is constant depending on the strength of the random force it is acceleration but we will just qualitatively call it force and there the viscous or relaxation rate of the particle it is a very important noteworthy finding which has to be kept in mind that the very presence of the random acceleration led this would not have been so if we did not introduce the random acceleration term. So, if you look back we introduce random acceleration with a property of two point correlation and that if you look back at the steps that we use that is the key result by which we obtain this. Not only that if we now take recourse to thermodynamics which states that given sufficient time any subsystem should also attain equilibrium at the same temperature as the host medium. So, a subsystem the Brownian particle we introduced it should attain equilibrium with the host fluid. So, now, we invoke thermodynamics the Brownian particle should attain equilibrium with the host gas with host fluid temperature t if it attains equilibrium it must have the mean energy associated with that per degree of freedom should be equal to half kT. Then equipartition law becomes valid for the particle law applies for particle also because it has become a macroscopic molecule in the host fluid that by implying that v square average in fact equipartition law says that half mv square average should be equal to half kT per degree of freedom. In this case we are now discussing x component. So, it will be half into half kT if you are talking of the vector v it will be 3 by 2 kT. So, likewise because there are 3 degrees of freedom if we include rotation then again even those degrees of freedom have to be considered. But in one dimensional problem this is always true and this leads to the result that v square average therefore, is kT by m one more result. Substituting back in the expression here that leads us to a relationship kT by m equal to gamma by 2 beta or we can estimate that unknown pre factor for the strength of random fluctuations as that is gamma that is gamma equal to 2 beta kT by m. Without one could do it without any model for random occurrences or random events taking place on to the particle. This approach of the Langevin by invoking fluid equilibrium was in fact, very informative because at the end it connects us back to the random coefficient of the random force and allows us to evaluate it. For us the interesting thing is if you go back to the previous lecture we had shown how the random collisions have led us to a formula for the pre factor of acceleration coefficients. So, from our gas kinetic model we obtained we called it as gamma one because it was the model dependent whereas, the gamma here is not model dependent. So, that gamma one we showed that was a 2 kT beta one we called it divided by mp. So, these two expressions are completely consistent. Numerically the beta one could be different from beta depending on whether one is talking of it coming from Navier stokes or it is coming from gas kinetic models. But formally the strength of the random fluctuation its dependence on the friction coefficient or the relaxation and particle masses and a temperature perfectly identical. The take away from this is particle necessarily reaches equilibrium and the strength of fluctuations that it experiences is dependent on its relaxation parameter. From this now we move over to actual diffusion of this Brownian particle in the physical space that is from the velocity space to physical space and that gives us another profound relation that is the called the Stokes Einstein relation. So, we will show how that is derived from within the Langevin equation formalism. Thank you.