 from the Fokker-Planck equation that is the Claim Cramer's equation, we can determine the evolution of the probabilities of Brownian particles under any situation including external forces in time and in space as well as in velocity. For that sometimes it is quite useful to carry out some non-dimensional non-dimensionalization of the space, time and velocity variables from within the existing parameters. As you can see we derived this following equation in the previous lecture d W by dt plus V the velocity d W by dx plus the external force d W by dV equal to gamma by 2 d 2 W by dV square plus beta d by dV of V W. We in fact can combine these two in by taking the d by dV out, but here we write explicitly. So, specifically let us say fx equal to 0 no external force it is a free particle it is not necessary to do that, but this is just as in illustration. Then the equation takes the form d W by dt plus V d W by dx equal to gamma by 2 d 2 W by dV square plus beta d by dV of V W. This is a second order equation both in velocity and a first order in space somewhat interesting because if you go back to the Fokker-Planck equation that was second order in space and that second order term accounted for diffusion. Here the second order is in velocity and the therefore, it is basically referring to diffusion in velocity space that is what Brownian motion is that diffusion in velocity space is going to introduce randomness in excess space via a first order coupling. So, mathematically it is somewhat not so obvious how one gets a Fokker-Planck equation for example, from this and it is a bit of an advanced kind of an exercise to perform that, but right now we work with how we can make this equation look simpler. So, for that we note that when we have put force equal to 0 we have basically 2 parameters gamma and beta and they are not all unrelated and we have this thing to note that beta is basically kt by f as we said beta is sorry beta is f by m and gamma equal to 2 beta kt by m where m is the mass of the particle kt is the temperature f is the friction coefficient etcetera. Beta has a dimension of reciprocal time and hence 1 by beta provides a natural time scale for the system. So, we define natural time length and velocity scales that is the way we always non-dimensionalize. First we should look for what are the natural scales. One thing is obvious that the tau relaxation or the characteristic time let us say Tc, c always represents the characteristic scale that can be 1 by beta. The for the velocity v it is a characteristic velocity is the thermal velocity it is always root kt by m that provides the natural velocity. Now the once we have a characteristic time and a characteristic velocity we can generate a characteristic length. Hence a characteristic length lc that is going to be it is a product of velocity into time vc Tc meter per second and per second. So, it is automatically generated. So, with that we can scale these are characteristic scales. With that we can generate for example, the term gamma can now be expressed because it is 2 beta kt by m which is going to be vc square by Tc and now we define dimensionless variables. Normally we write it as a x star right. So, you can say for example, velocity let us start with the velocity first. Velocity dimensionless velocity is true velocity divided by the characteristic velocity. But somehow to avoid tediousness in writing we do not use a star, but let us say I decide we use u when we say u which means it is dimensionless velocity v. Similarly now we define a dimensionless length you can call it x star which is x divided by characteristic length which we already defined. But again for ease of writing I call it as xi. Similarly T star the characteristic time the scaled time dimensionless time is a true time divided by the characteristic time scale. Again for ease of writing I call it as tau. So, these are dimensionless variables of the problem. If we substitute these things, if we substitute these characteristic times eliminate beta v x t gamma by 2 exactly in terms of all these variables all the hanging the parameters cancel out. The entire equation is an exercise that you should do to get the feeling the entire equation takes a form free of any parameter it is becomes an easy to remember in the dimensionless u the probability density expressed in terms of these quantities instead of x we said it is going to be xi which is the dimensionless length scale that becomes d 2 w by d u square plus d w d by d w of u w. It is a very elegant looking one dimensional equation for free particle you know this cannot be done very generally or universally when there is a force field. If you know the force field it is dependence on x again of course, the same scaling you can introduce and you can write down, but it may also have its own length scales. So, there will be new length scales will develop. So, that is a matter of detail which we do not take up now. So, this is non-dimensionalized KKE non-dimensional claim climbers equation specifically for f x equal to 0 it has no hanging parameters now everything is in non-dimensional form. For a free particle we should specify the domains or xi domain could be the real physical space domain the velocity domain of course, it will be again minus infinity to infinity and of course, the time is always positive. So, these are also an additional thing with that we can note. Being a second order in space as being a second order in velocity we one may wonder that we may need boundary conditions two boundary conditions on velocity, but by virtue of the fact that velocity the domain of velocity includes infinity and generally in most problems velocity cannot be controlled from outside its domain cannot be regulated and then the velocity at large values cannot occur very frequently and they should decay down hence W will be assumed to tend to 0 when velocity tends to either plus infinity or minus infinity. So, a natural this equation therefore, has a natural boundary condition in the velocity space all that one might have to provide are the initial conditions or the boundary conditions in the position and time space. Let us develop some simple solutions most of the solutions we have attained in some way we have obtained in our exercises earlier, but this is basically reassertion of the same via the Klein Kramer's equation. So, let us consider case one solutions some solutions to Klein Kramer's equation. Let us consider case one equilibrium solution. So, when we say a system is in equilibrium there should be no flux of any type and there should be no change in time and hence equilibrium by definition means d by dt or for that matter d by d tau in our new variable this should be 0. Similarly, there should be no space change which basically implies that d by d psi should be 0. So, when we do that our equation becomes very simple it becomes d by d u of d w by d u plus u w equal to 0. So, first integral can be easily executed. So, that leads us d w by d u plus u w equal to some constant let us say K. Now what is the how do you determine this constant? Here we invoke the fact that I just mentioned that the domain of velocity includes minus infinity and plus infinity and since this is supposed to be constant across all velocities the if you determine the value at one velocity that should be sufficient and we know that when velocity is infinity when u tends to plus minus infinity both w and d w by d u should tend to 0 and in fact, w should tend to 0 much faster than any power when that is required because we need to have moments of w that is why. So, in other words even u w will go to 0. Once we invoke this condition it implies hence K has to be 0. So, this brings us to the next level of simplicity one has this equation d w by d y d u u w equal to 0. The first order equation I simply write the solution w equal to w u now we can call it as w infinity maybe subscript w infinity u some constant e to the power minus u square by 2 the non-dimensional way and let us recall u was a non-dimensional expression for the true velocity in terms of the characteristic velocity and characteristic velocity was root KT by m where vc was root KT by m. Hence w infinity in terms of the true velocity will be somewhere constant a e to the power minus m v square by 2 KBT. It is always put KB for the Boltzmann constant because not to confuse with many case we use. So, this basically means now a will be determined a determined by normalization of the probability by integral w infinity d v should be 1 which of course will give you an expression in terms of KT and m. So, with that we can so that leads us to w infinity v after you do that basically will be root of square root of m by 2 pi KBT e to the power minus m v square by 2 where T is the temperature. So, this equation there is nothing new in this this only guarantees that the equation we derived is consistent with the physics that we started because this is the well known Maxwell Boltzmann distribution and also is consistent with the principle of equipartition of energy etcetera. It will reproduce all the results that we had used from basic equilibrium principles in obtaining the second moments etcetera of the Langevin dynamics. Let us however go a little further let us take to a now a little more complicated problem that is where now particle starts with an initial velocity. It is not an equilibrium solution it is a non equilibrium situation with an initial velocity v naught or in the u space u naught, but it is homogeneous in space. You can imagine a situation that particles are distributed homogeneous in space, but all of them have a particular velocity that is we continue to hold that d by dx is 0, but w therefore it is a function of v at a time 0 is delta let us say we are working with the u variable. So, we can straight away write it for u variable as that u minus u naught particles is projected at some velocity u naught, but homogeneous in space. So, the equation now is a same equation, but it does have time dependence d w by d tau has the right hand side d by d u of u w with the initial condition that w u at tau equal to 0 is delta u minus u naught. This equation looks similar to the Fokker-Planck equation in position space or what we call as diffusion equation, but it is not. So, mainly because of this term this adds a certain level of complexity and it is definitely of theoretical interest to develop skills to solve this kind of linear equations. It has been already done, but just to illustrate how such things are done, we quickly go through the method of Fourier transforms which is very useful. So, as usual apply Fourier transform simply call it as f t. So, by definition we know that there is a conjugate variable for velocity. For velocity we do the Fourier transforming time remains the same. So, velocity goes over and by definition it is integrated over u and conjugate to u is going to be k and my distribution function becomes transformed to 1 with a carrot now. So, with this definition the thing that we are familiar is that the Fourier transform of the second order term is well known. We have seen that Fourier transform of the second derivative it is going from u to k that is what we are going to do. This Fourier transform is simply minus k square Fourier transform of the density itself which is a function of k and tau. This is known. So, the question or some exercise that one has to do is what is the Fourier transform of the second term that is d by d u of u w. So, this term requires some exercise for that we note that this is basically minus infinity to infinity by definition i k u then d by d u of u w integrated over all u. Do an integrating factor method integrate this and also integrate the second term and differentiate this term then apply the limits for the first part and those limits will disappear because limits will lead to the function to go to 0 and then one is left for left with only the second term which is minus of when you differentiate you are going to get i k here and then the integral will be e to the power i k u and this integral is going to be u w d u. So, this part is quite fine, but how do I do this because it is not just w it is u w for that. So, to do that we note that if we differentiate the definition since w carrot is given as integral e to the power i k u w d u if I differentiate with respect to k I am allowed to do because it is a function of k then I will have by d k for example then u will come here and i will come out. So, it is going to be e to the power i k u u w d u which is i which is a Fourier transform of i into f t of which is what you want to this is f t of u w. So, f t of u w is the minus i d w by d k and when we put all of them together in the Fokker-Planck equation we get a nice looking first order equation. So, let me let us complete this exercise hence we have f t of hence f t of d by d u of u w is going to be minus i k and this will be 1 by i. So, i square is minus and then there will be a minus eventually be minus k d w by d k. So, when we combine all that we obtain in the transformed space a first order equation that is the beauty of Fourier transforms it is going to be minus k square w cap minus k d w by d k or I can take this to the LHS right d w by d tau plus k d w by d k equal to minus k square w. So, it is the partial differential equation now it has become first order partial differential equation from the second order. So, there is a advantage this first order equations are generally solved by the method of characteristics and well to do that we need some initial condition and in view of the fact that we have this initial condition we can apply it over to the transformed variable which is nothing but integral e to the power i k u delta u minus u naught d u which is e to the power i k u naught. So, we have this system now which is completely determined to which we have to find a solution for that we use method of characteristics or a new variable we will find out or transformations in many ways the first order. The simplest way is to go to a new set of transformations say new variables in from the old variables k and tau where the old variables we go to new variables eta and tau prime. So, if you choose this eta is this and of course, the tau prime let us say is the same as tau, but just because it is a new coordinate system we put a prime to it with this these are the new variables. Then if you do a transformation for example, then use things like this we need d by d tau. So, d by d tau is going to be d by d tau prime d tau prime by d tau plus in terms of the new variables it is d by d eta d eta by d tau etcetera. Similarly, d by d k is going to be d by d tau prime d tau prime by d k partial plus d by d eta d eta by d k. And one thing we notice this is 0 and this is 1 and this of course, can be easily expressed by differentiating eta with respect to tau which is going to be k e to the power minus tau itself and minus sign will come. So, it will be minus eta this will become minus eta. So, likewise we can we need d eta by d k and d eta by d k is simply e to the power minus tau prime this will be e to the power minus tau or tau prime. So, when we put all these transformations the equation actually cancels many terms cancels one gets a very simple equation d w by d tau prime. Now in terms of this variable simply minus of eta square e to the power 2 tau prime w cap in terms of eta tau variables. So, we can integrate this noting that this is a function of eta and tau and it is a partial derivative. So, when we integrate this we will get w eta tau prime is going to be some unknown function it can be a function of eta can easily verify and it will be minus eta square by 2 e to the power 2 tau prime or go back to the original variables go back to k tau and then this attains the form w carat k tau will be the unknown function c will be a function of k e to the power minus tau e to the power minus k square by 2. Here c is unknown function. Now the unknown function we determined by the initial condition since we know that since w cap k 0 is e to the power i k u naught by putting tau equal to 0 we simply obtain c k therefore, is going to be e to the power i k u naught plus k square by 2 or c k e to the power minus tau which is what we want is going to be e to the power i u naught k e to the power minus tau plus k square by 2 e to the power minus 2 tau with all that. So, finally, our solution therefore, is going to be k tau equal to e to the power i u naught k e to the power minus tau minus minus k square by 2 into 1 minus e to the power minus 2 tau. Now formally the solution is obtained in the k tau space we have to do inversion from k to real velocity u. So, since it is a Gaussian integral inversion we know and we leave it to that. So, when we do that final inversion we get the probability density distribution w u tau equal to 1 by square root of 2 pi into 1 minus e to the power minus 2 tau e to the power minus of u minus u naught e to the power minus tau whole square divided by twice of 1 minus e to the power minus 2 tau. So, this is a Gaussian, but with the shifted mean and if you carefully see the mean is actually the one we had already obtained u naught e to the power minus tau and this is the variance which we had seen as 1 minus e to the power minus 2 tau by actually by actually working with the Langevin equation. But this gives you the confirms that the equation that we developed is consistent with the Gaussian approximation that we carried out and if one has to plot initially we had a delta function like this particularly that had some velocity at tau equal to 0 some velocity u naught. But before it attains equilibrium it becomes quite broad and the peak shifts with the time at any time to tau this is to any time tau peak shifts to a value e naught e to the power minus tau and the standard deviation here which is becomes square root of 1 minus e to the power minus 2 tau. So, this is the essence of the free particle problem solution to the free particle problem from the claim Cramer's equation. One can actually increase the level of complexity this has been done long ago and it is very well known and the classic book of S. Chandrasekhar reviews of modern physics in 1943 gives variety of solutions. Solutions for a projectile for example, where we do not assume that it is distributed uniformly in space particle is localized in velocity as well as in position solution is of course, apparently possible same double Fourier transform one has to use both for velocity as well as for position we get a very complicated multi Gaussian bivariate Gaussian distribution you will get and that has been done. One can further work with greater complexities of particles having some force field specially harmonic forces. So, its Brownian motion of bound particles all that can be done via the claim Cramer's equation. The claim Cramer's equations great advantages it enables us to understand how particles surmount barriers, why chemical reactions take place and many many other aspects. All this is relatively more advanced subject and we relegate all this to future. For now more or less we have examined ourselves starting from random walk model to Fokker Planck equation model to Cramer's model the kind of stochastic phenomena that a real physical systems encounter and how they can be described. Thank you.