 The Langevin equation that we discussed is basically a dynamical equation, an equation of a Newtonian type to which a noise term is added. It is a simulation will give individual trajectories. Matter of practical interest is always a distribution in stochastic descriptions, what matters is to find the probability density or probability distribution that we can assign with respect to certain occupancy or certain being in certain position or velocity acquiring certain velocity. So, that job we did earlier in the random work context via developing a differential equation and we called it as a Fokker-Planck equation. When a similar exercise we do with the including now velocity fluctuations not necessarily a position fluctuation, we obtain a family or a new type of equation called the Klein-Kramer's equation. It is actually a Fokker-Planck equation in phase space as they say that is it is in the space of both velocity and position together. So, the position space FPE that we derived earlier, the position space FPE that is the Fokker-Planck equation for short. If you remember it is of the type dW by dt equal to some diffusion coefficient, let us say it is constant with respect to space minus some drift term arising out of some steady motion of the particle, steady velocity V let us say. One can even generalize it, this is in 1d for completeness we can always generalize it to d del square W minus divergence of V W say in 3D same. We did not solve any problem in 3D, but this is the way the diffusion equation is generalized. So, this we called as a Fokker-Planck description or in basically you can even call it as a convective diffusion equation if V is prescribed and given to you. So, what is this, what will we arrive at if we carry out the process of converting a dynamical equation with noise to a equation for probability density when velocity fluctuations are included. That exercise we do now because that completes the description of Langevin equation that we took up. For that we have to again take recourse to the Chapman-Kolmogorov description. So, to proceed we note that the average displacement in position due to velocity fluctuation delta x bar that will be always the instantaneous velocity into delta t. In a small time delta t this is the transition in transition in position in small time delta t, but there is a time in which basically the velocity also has undergone a small change and that change delta v bar we showed that in general it could be dependent on the friction of course and the external force term also f by m if it exists delta t and f is a function of space only not of velocity that is the main point. And most importantly the random force caused mean square change in velocity which is of the order of gamma delta t where gamma is the noise coefficient of the random fluctuations. So, we had derived several relationships one of them is being gamma equal to 2 beta k t by m in a thermal system. So, it is always related to the friction coefficient or rather beta there is a reciprocal relaxation time. We make use of this fundamental transition transition variables these are we assign basically that this is the mean transition value mean value of transition and this is the mean square value of transition and there exists a transition function characterized by these two parameters only and higher order transition coefficients will be of the order of delta t square and higher. This is true of course of Gaussian noise. So, basically the simplest assumption is a Gaussian noise Gaussian transition. So, with that we now end over to set up an equation for the probability of finding a particle at a position x having a velocity v at time t. In other words we are actually generalizing into hyperspace of in a one dimensional problem a one dimension of x and one dimension of v strictly speaking it will be a 6 dimensional problem there will be x y z and u v w there will be 3 components of velocity and you therefore, have to find what is the probability that a transition occurs from point x v to x plus delta x v plus delta v in this space and how the distribution of function evolves in this two dimensional space considering one dimension for space and one dimension for velocity. So, this is a kind of phase space that is why it is called phase space description. To do this we need to take recourse to Chapman Kolmogorov equation because we have seen that this equation gives a continuous variable formulation of the Markov processes Chapman Kolmogorov equation for continuous Markov processes we will call it as CKE whenever if we require further. Now, this equation we generally we can recall it that if in two dimensions for example, in the present context we can say that the probability of finding a particle at x and u velocity is say some v in a time small time change delta v that should be there will be transitions of both x and v now there are two variables. So, there will be a two integrals one over the elementary changes in x that take place and elementary changes in u. So, to come to x in the previous instant it must have been at a position w x minus delta x it must have had a velocity v minus delta v at time t the probability that you find a particle with this variables into transition probability in the remaining time delta t by a value delta x and this probability could depend on the previous coordinates. So, by an amount delta v in a time delta t we can formally write like this integrated over d delta x into d delta v. So, for all possible values of transitions, transition lengths and one defines therefore, a transition probability. Before we proceed further we now note one important thing about position transition. Now, position transition in a short time is primarily an advection process. In our Langevin model we do not have random assignment of transition to position we have random assignment of transition to velocity and position is basically d x by d t equal to u or in other words delta x that is any transition delta x should be v into delta t that instantaneous value of velocity coarse grained velocity not the impinging molecular velocity, but the coarse grained Brownian particle velocity into delta d that should be delta x. In other words the transition probability therefore, in the delta x context should be delta function. I must note that in this integral here I just missed p should also carry a delta x here by delta v delta x in a time delta t. So, it should be corrected at this. So, which we have done and integrated over all delta x and delta v. So, once we assume that the transition probability p at least has it will be p of x v delta v delta t, but it will be a delta function the direct delta function here as far as position transition is concerned that is the point. In velocity transition we will have to describe it, but position transition probabilities straight away settled by noting this point. So, that makes it much easier because this transition we already have worked out. Now we go to writing the Chapman Kolmogorov equation instead of x here we advance this distance a little because we want to avoid this x minus delta x and we note that delta x is of the order of v delta t. Hence we write that the probability that the particle is at a position x plus v delta t and has a velocity v in time t plus delta t. We apply C k equation to this it will now be a single integral because already the delta function we average out. So, that is the probability that the particle was at position x had a velocity v minus delta v only velocity transitions out we counted now at time t and the probability that at that position x and at that velocity v minus delta v it executed a transition by an amount delta v in time delta t integrated over all possible values of delta v. So, this is basically an integral now over all possible values of delta v. So, this way of writing Chapman Kolmogorov equation is the key conceptual advancement we make for arriving at a differential equation from Langevin dynamics. How do you proceed further from this? At this is the point where probably we can make some simplicity of I mean we can make some change for simplicity in writing like delta v and integration d delta v becomes a bit cumbersome to write. So, we just denote the quantity delta v by a new symbol psi we define for convenience it is no conceptual change we define for convenience of notation I would say psi equal to delta v. Then the C k equation can be written in the following form we can go back and see that here wherever delta v comes v will remain as v it is going to be psi. So, left hand side will remain the same that is W x plus v delta t v t t plus delta t because the right hand side is at an advanced next instant of time that should be equal to now the integral v over psi it will be W x v minus psi at time t and the transition probability at x v minus psi by an amount to psi in a time delta t d psi over all accessible values of the velocity. This will be the procedure or this will be the way to go forward we need to at this point in time we need to know the characteristics of the transition probability p. Broadly we say that it is a Gaussian and since it basically is a distribution over psi we say that it has characteristics such that the mean velocity or the mean velocity at any instant of time small time. So, delta the change of it is minus of beta v minus f by m delta t. So, this is one can p should be such that the mean arising out of it will be this. So, how does one get this basically implies integral p psi d psi should be equal to psi bar. So, that is the way it is defined. Similarly, psi square bar will be gamma delta t. So, this basically arises from the fact that psi square p of all this d psi this is psi square bar. So, these are the 2 important quantities that we require. Now, let us say we denote this by equation 1 now. So, the equation 1 LHS of equation 1 we can tailor expand both with the both in in both around x and t. So, there are 2 terms here left hand side there are 2 terms x plus v delta t which is a small term delta t is a small time and t plus delta t and we know how to do 2 variable expansions. So, that for example, can verify it it will take the form LHS equal to. So, we will write it like this LHS when we expand it will be dw by dt plus v dw by dx delta t that will be the expansion around this point this will be the 2 terms will be there first derivative will be dw by dt delta t let us redo it. So, the left hand side expansion will be first w x v then first with respect to t if we do it will be dw by dt delta t then if you do around x it will be v dw by dx delta t and plus higher order terms which right now a we do not intend to use it later. So, we stop it at this now the main thing is to how do we do the expansion of the right hand side. So, if you look at the right hand side within the integral of RHS if you look at RHS within its integral we have terms like w x v minus xi which is a small quantity xi is a small quantity delta v actually then there is a transition probability which also has v minus xi and xi in a small time delta t. So, we have to and of course, there is an integration being carried out over xi. So, this if we expand to the order xi and xi square this will give 3 terms so there will be 3 terms here also and we can then combine them. For example, we will say that if we expand the first term here w x v minus xi t it is going to be w x v t minus xi dw by dv plus xi square by 2 d2 w by dx dv square. So, that is one similarly if we write p if expand x v minus xi xi t this will be like same way p x v xi delta t. So, I will not write all these indices in the next part, but we know that it is going to be minus xi dp by dv plus again xi square by 2 d2 p by dv square. So, when we multiply these 2 and then combine it will lead to the following terms. Hence RHS within the integral will be will be equal to it will be w x v t w x v t into p x v xi minus of then their common term with the xi and that will be w dp by dv plus p dw by dv. You can see that these 2 terms will be added and then there will be a xi square by 2 term and there are 4, but 2 of them will add. So, you will have 3 terms that is p d2 dv square plus 2 dw by dv dp by dv plus w d2 p by dv square. So, we have term then of course, there will be higher order terms. So, now, when we integrate so, RHS as such is going to be the first term let us understand what I am saying. The first term is going to be an integral over w integral over p because I am integrating over xi and xi is in p and that is going to be 1 because it is a normalized thing. So, we are going to get just w. Second one if you carefully see it is going to be here second one is xi p. So, it is going to be xi bar, but d by dv of xi bar this one. Here if you come it is basically simply xi bar. So, this will give you xi bar and d by dp. Similarly, if you come here this will be xi square p integral which is xi square bar. Here it will be dw by dv and d by dv of xi square p bar. So, it is basically a derivative of xi square. Here similarly, this will be also second derivative of xi square. So, we write these terms carefully by noting these things then the RHS will become after the integration is carried out w x vt minus of systematically it will be d xi bar by dv plus xi bar dw by dv will be the first. Then the second derivative term will be half of d 2 w by dv square xi square bar plus twice d w by dv into d xi square bar by dv plus w d 2 xi square bar by dv square. It will have these many terms. Now, we note that again we come back to the fact that note xi bar was minus of beta v minus f by m delta t. So, whenever xi bar is a proportional to delta t, but xi bar derivative will be beta. So, considering these aspects we have therefore, very important result derivative of xi square bar by dv will be 0. So, this is because xi square bar is only a function of delta t, does not depend on velocity. So, we combine all this. So, we will then be left with the RHS will come to there will be a w there will be a delta t d by dv of beta v f by m of w then there will be delta t then gamma by 2 of d 2 w by dv square and we have already shown that our LHS was the first order derivative it had a w, it had a delta t and d w by dt and then we it had a delta t and v d w by dx higher order terms we are neglecting only up to the order delta t. That is the reason why we expand it here also up to xi square because we knew that higher orders are going to be in a Gaussian noise sense delta t square and higher. Now, when RHS is equal to LHS when we equate them it basically means these terms will cancel and also then the remaining delta t's also will cancel leading to the equation which can be written in the following form d w by dt plus v d w by dx plus external force divided by m d w by dv that will can be written as d by dv of gamma by 2 d w by dv plus beta v w. This is now the new differential equation this is called the Klein Kramer's equation or it is a Fokker Planck equation for the probability density or FPE for evolution of probability density in phase space. We have derived it basically by going through Klein Chapman Kolmogorov equation assume underlying assumption being the Markovian nature of the process. Thank you.