 So, let us take a look at another aspect of the Langevin model. We have seen that there is a connection between the general diffusion type of stochastic equation for a diffusion process and the Fokker-Planck equation corresponding Fokker-Planck equation. Before I go on, let us take care of one of the questions that was raised. The question is as follows. In a multi-dimensional process, if you have an equation of the form x dot equal to some vector valued function of all the x's plus a multiplicative noise, so this was g of x times as zeta of t where this is set of noises, independent noises and the dimensionality of this was nu. So remember that this was nu times 1 column vector whereas this was an n times nu matrix, rectangular matrix and the idea was that if x is this quantity is an n dimensional noise represented by an n dimensional column vector then the question being asked is these components are independent of each other, they are different noise components, 1 to nu but because of the presence of this, does it not mix up the various noise components? Yes, in general that is the most general possibility because even in the most trivial of instances when nu was 1 and n was equal to 2, remember that you had an equation of the form x dot equal to v and v dot equal to minus gamma v, the drift term on this side plus a noise which was square root of gamma over m times the white noise, Gaussian white noise. Now the question is is it not true that this noise affects the x itself? Sure it does. So there is already a coupling, it is a couple set of equations in any case. So what this is saying is that the most general possibility is that the strength of each of these noises is dependent on the current values of all the random variables which are anyway coupled to each other dynamically. So there is nothing inconsistent about it, it is perfectly reasonable. What is true is that these noise components are specified independently. So the statistics of this noise is taken to be quite independent of what is happening to the output variables, the driven variable x. You have to prescribe for me all the statistical properties of this multi-dimensional noise before I can tell you what the dynamics of the driven variables is. That is certainly true. And the whole assumption in this kind of modelling is that the driven variables do not affect that noise. So there is no feedback on to the noise from the driven variables. We take this in the simplest lines of our cases. We took the noise to come from the heat bath and I said that the effect of the motion of a single Brownian particle on the heat path is negligible. It does not throw it out of equilibrium or anything like that. So it is in that philosophy. That is the very general philosophy that you prescribe for me an external noise of some kind, statistics of this noise is prescribed independent of what is happening and then the statistics of the driven variable x is dependent on solving this set of equations, coupled equations. So that is the philosophy behind this stochastic differential equation approach with the driving noise, okay. The different components of this are uncorrelated with each other. In the simplest case, we take them to be Cartesian components for instance. They are uncorrelated with each other. But there is no reason why the coupling should not be such that this instantaneous strength of this noise as it acts on any one of the variables depends on all the other variables. And that is where this g comes in here. So is this clear? Okay. Now let us go back and let me point out what is wrong with the Langevin model. We will see where some serious problems crop. And we will do this in the simplest context which is that of a tag particle, a Brownian particle moving in a fluid. And again the essential physics is already contained in the one-dimensional case. So let us go back to that. Let me call this inconsistencies of the Langevin model. So first I will point out the inconsistency and later on we will fix it on physical grounds. We will so to speak derive or deduce an improved Langevin model which will take care of this immediate problem of what main problems flaws are in this model. So we call once again that for the velocity process, velocity of the Brownian particle, we had an equation of the form v dot is equal to minus gamma v plus square root of gamma over m zeta of t where this was Gaussian white noise, zero mean and a delta correlation. And this was taken to be Gaussian, stationary, Markov and delta correlated. And what was the output process? It was the Anstein-Rohlenbeck process. It was also Markov. It was also Gaussian. It was also stationary but it did not have a delta correlation of some kind. But we certainly proved that it was stationary. In particular we proved this relation. We proved the fact that v of t not, v of t not plus t in equilibrium, this quantity, we calculated what it was from this using the consistency condition that asymptotically this correlation will factor the product of averages and that the mean square value will remain in thermal equilibrium at the Maxwellian value. So this was found to be independent of t not and it was k t over m e to the minus gamma modulus t. It is a stationary correlation. Now if you go back to the general linear response theory that whose formalism we worked out, this is actually a response function. And in the language of the response function, what we have is phi AB in this problem, A was equal to x and B was equal to v because we perturbed the system by applying a mechanical force minus x times this f of x, f of t and then we measured the velocity, the average velocity. So the observed variable is the velocity and the variable which is cos coupling to the force is x, the dynamical variable. And this came about as a response function. So in a sense this quantity is in fact phi x v. Recall that in classical apart from some factor of k t or something like that. Recall that in classical physics, in the classical case we showed that this response function is beta times the equilibrium average value of A dot at 0 with B at time t. And A dot at 0 is v of t not. So this is what we derived here. And this is independent of t not by stationarity. But look at its consequence. If you took that seriously, if you took this thing seriously, it immediately follows that if I differentiate this function, this quantity with respect to t not, the answer should be 0 because it is independent of t not, right. But what happens if you differentiate it? You end up with v dot of t not, v of t not plus t average plus v of t not, v dot of t not plus t equilibrium should be equal to 0 because there is no t not dependence on the right hand side by stationarity. Now set t equal to 0 from above or t equal to going to 0, approaching 0 from above, right. Then it immediately says that at any instant of time v of t not, v dot of t not equilibrium must be 0, must vanish by stationarity. So what we have to do is to let t tends to 0, let t tend to 0 and immediately you get this. And they are classical variables, so it does not matter in which order they appear. So what it says is that the velocity at any instant of time is not correlated with the acceleration at the same instant of time. That is consistent, certainly with our assumption that the force drives the acceleration and not the velocity. So it is sort of reasonable that the velocity is not correlated with the acceleration at the same instant of time, okay. And in any case it is demanded by stationarity. But if I do that here, if I differentiate with respect to t and then set t equal to 0, it is like differentiating this guy here. So let us take t to be positive. If t is positive this goes away, I differentiate both sides with respect to t and then set t equal to 0. And that is going to give you v of t not, v dot of t not in equilibrium equal to minus gamma k Boltzmann t over m which is not equal to 0. All I do is to differentiate this with respect to t and put t equal to 0 from above. And you get a non-zero answer. So it says the velocity at any instant of time t not in equilibrium is correlated to the acceleration by this number here. The equal time correlation is not 0 identically but some finite number and yet if we stationary which we showed explicitly, this must vanish by stationarity. So there is obviously an inconsistency somewhere in this model, okay. So one or the other must be wrong. Either this or that must be wrong, right. Now the stationarity, this thing by the way, we derived this by making an assumption when we computed this correlation function. We assume that the velocity is uncorrelated with the force at later instance of time including that particular instant of time itself. So we made that assumption and that now leads to a result which is inconsistent with that assumption. So there is something wrong somewhere. Stationarity on the other hand we proved under very general considerations for from linear response theory we showed the stationarity of this correlation of the response function and yet this is happening. Let us see if there are any other problems. So you agree that there is a serious problem here. The equal time correlation between variable and its derivative should be 0 if the variable is a stationary random variable but according to this model it is not, alright. Any other problem? Well let us again look at this response function. It is phi x v so let me drop the subscripts for a moment and argue in the following way. I would like to find out what is the mean square velocity is k T over m in thermal equilibrium. What is the mean square acceleration? That should be some finite number. It is a physical quantity. It should be a finite number. So if I compute v dot on both sides I should get a finite answer. Now let us call this correlation phi of t minus t prime equal to v of t prime v of t in equilibrium. That was our call our response function which is e to the minus gamma mod t minus t prime in this model, right. Now the Fourier transform of this is a spectral function and the moments of the spectral function can now be computed because we have the spectral function explicitly. So what is that equal to? What is the spectral function equal to? Phi tilde of omega equal to integral 0 to infinity d t t to the i omega t times the phi of t. That is the definition of the spectral function, right. In this model, in the Langevin model we know what it is. We know explicitly what this thing is. So this is equal to, well let us compute, let us compute the power spectral density of this process, the v process which means we take its autocorrelation and we find its Fourier transform, okay. So what we need is the Fourier transform which is minus infinity to infinity. This is the power spectral density and what is this equal to? Well, we need a slightly different symbol for this whole thing. I call this the spectral function. It was, how did we define the spectral function? It was the Fourier transform of the response function, right. So it was this. I am just trying to get my factors right. This is in fact equal to phi tilde of omega. The susceptibility was the one sided Fourier transform. The spectral function is just this guy here. Now in engineering, the power spectral density is defined in various ways, apart from some factor of 2 or something like that, sometimes 2 pi, sometimes 2 times the Fourier transform of the autocorrelation function. So modulo of that, let us just define it to be the Fourier transform itself and let us compute what this number is in the case of the velocity process, yeah. So this is equal to k Boltzmann t over m integral minus infinity to infinity d t e to the i omega t e to the minus gamma modulus t because recall that for t negative, this is e to the minus modulus t. It is a symmetric function of t, e to the minus gamma modulus t and why should it be symmetric? Why should this quantity be symmetric? Because under time reversal, v goes to minus v. Both these fellows change signs, so this fellow does not change sign. Or yet another way of looking at it is, it is the time parity of a dot times the time parity of b and a dot is x dot, which is also a velocity. So it is completely consistent. But this is a trivial integral to do. This is twice the integral from 0 to infinity because it is a symmetric function and the sign part of it vanishes. So you have got 0 to infinity with a 2 factor k Boltzmann t over m, 0 to infinity e to the minus gamma t cos omega t, which is of course gamma over gamma square plus omega square. It is a Lorentzian, okay. The Fourier transform of the e to the minus x is a Lorentzian in k. So that is all that we are saying here. But this has a physical meaning, this quantity here because you can differentiate this on this side. Each time you differentiate with respect to t, you pull out an i omega from this, okay. So let us do that. Let us write this. So you have v of t prime, v of t, this is the correlation function. This is equal to 1 over 2 pi integral minus infinity to infinity d omega e to the minus i omega t minus t prime phi tilde of omega by definition, that is the inverse Fourier transform, right. And we know what phi tilde of omega is. So now what I do is v dot of t prime, v of t in equilibrium, again in equilibrium is with the i omega pull down and if I differentiate with respect to t as well, it is with the minus i omega pull down, right. So immediately you get this, 1 over 2 pi integral minus infinity to infinity d omega e to the minus i omega t minus t prime omega square times phi tilde of omega. The second moment apart from this time dependent factor. Now I set t prime equal to t. So that tells me v dot of t square in equilibrium. It is stationary, it is got to therefore be independent of time. The time dependence disappears from this thing and this is equal to 1 over 2 pi integral minus infinity to infinity d omega omega square phi tilde of but that is equal to this fellow here. So if I put it in, it is omega square over omega square minus square apart from constants. So it is equal to 1 over 2 pi, 2 k Boltzmann t over m times this fellow. But this is a finite physical quantity, it says mean square value of the acceleration but that is infinite. That has turned out to be infinite because this is not dying down fast enough at infinity. It goes like 1 over omega square and therefore goes to a constant as omega tends to infinity on either side. So this is tends to infinity. It is clearly wrong, which is clearly wrong. So again we are faced with a serious problem and of course once the second moment goes to infinity, the higher moments will all go to infinity, omega 4, omega 6, etc. will all blow up. So the velocity process therefore will not have, its derivatives will not have finite mean square values, which is clearly unphysical. Now what is the origin of this disease? Where could we have gone wrong in this business? It is after all a very very simple model. So where could this have been wrong? There was one assumption made in deriving this model and the assumption was that the model is valid as long as you are looking at time scales, which is not at the level of the molecular time scale, the collision time between molecules but may be 5 orders of magnitude higher. So time scales of the order of gamma inverse and greater. If it becomes very much greater than gamma inverse it goes to the diffusion regime but certainly it is not valid on time scales going towards 0 on the other side. So right there we made an assumption which is not valid and now notice that all these diseases arose when you took the time difference to go to 0 and in this case when you took the frequency to go to infinity, so infinite frequency is like 0 time, very small time in the Fourier language, high frequency is like small time and vice versa. So that is where the problem was, how do we fix it? In the context of this model itself how would we fix it? We would still like to have retarded response, we would still like to have causal response and we would like to stay within the regime of linear response. So clearly the one place where we made an assumption was that the acceleration of this particle we wrote mv dot is equal to the force on the particle, so there could be possibly some external force on the particle which could be time dependent, we do not care and then there was a random force on the particle but this random force, so let us write it as f random of t and we said this random force has two components to it, two parts to it, one part was a systematic part and the other part was the truly random part coming from molecular collisions. So this thing we took to be square root of gamma times zeta of t with this to be white noise and this part we took to be minus gamma m gamma v of t itself. Now this friction arose saying that the friction is proportional to the the frictional force is proportional to the instantaneous velocity assumes that this frictional force acts instantaneously, the reaction of the medium acts instantaneously on the particle that is not true, that is not tenable not mathematically instantaneously because the medium itself has sometimes scale in it, the molecules they cannot act instantaneously on this. So this model does not go all the way to t equal to z to instantaneous response, okay. So this is the point that had to be fixed, this model is untenable at very very short time intervals, how will we fix it? Staying within causal, retarded and linear response, how can we fix it? Well we will we will do the following, we will say look it must depend on the velocity but not on the instantaneous velocity but on the previous, the history of the velocity earlier velocity, right. So the way to fix that model is to replace this by minus m gamma and integral from wherever, from the infinite past, okay, minus infinity up to t because we certainly would not want this part to go beyond t, we are looking at the acceleration at time t that is not going to depend on what happened to the velocity at later times, right, times an integral dt prime and it must be a linear functional so v of t prime must be there, it must be causal, it must be retarded that means it must only depend on the time elapsed since t prime happened, so this must be some function gamma of t minus t prime, so this thing here is a function, it is a kind of memory kernel, oh there is no gamma here, so I make this gamma a function of time, defined only for positive values of its argument, non-negative values of its argument and what other property would you like of this gamma? Well, we would certainly expect the following, we would like gamma, let us call it gamma of tau to be neutral, gamma of tau memory kernel, we just want to write some general properties of it, we do not want to make specific modeling of gamma but some general properties, gamma of tau, we should take it to be identically 0 for tau less than 0, does not make sense to go, that fix is fixed by causality in any case but by the fact that I cut this off here but you could formally say this gamma is defined only for positive values of its argument, okay. Then I expect the effect to keep going down as the elapsed time increases, so I would certainly expect that gamma of tau decreasing function, so as tau increases I want this effect to go down, okay. Anything else you want of gamma? Well, we would like it to go to 0 at infinity sufficiently fast, if it does not wear in trouble then we have to fix it, we will do that, okay. And right now I am not saying how fast it should go to 0, as tau tends to infinity but what else do you require? There is one more very important criterion we need, one more condition, it represents frictional force, right, so you want its sign to be positive, you do not want it to be negative, right. So non-negative, non-negative decreasing function, let us put that in and see what happens, so my Langema equation now reads m v dot of t plus m, I put this m here purely for dimensional reasons so that this gamma whatever it is still has the dimensionality of 1 over time, so minus infinity to infinity d t prime, oh sorry, no, no, no, there is an extra time factor, we will fix that in a minute. I want to put this m here so that this m cancels out on this side completely, this we will see what dimensionality this gamma has but from this equation itself t minus t prime, v of t prime equal to any force you exert on the system, some applied force, deterministic force plus the random part which is the eta of the force we denoted by eta of t earlier, so let us leave that as it is but important not necessarily white noise, in fact I am going to show it cannot be white noise, it is noise, okay, it is random but we cannot assume, we cannot, we have to prove that it is consistent to assume that it is delta correlated, we will have to see, we do not know at the moment. So pardon me, oh I am sorry because I said already 0 so it cuts off but okay 0 to t, by the way you can recover the original version by saying if gamma of tau equal to gamma delta of tau where this is a constant implies implies the Langevin equation, Langevin model. So the physical dimensionality of this kernel is gamma divided by time, there is an extra 1 over t so it is 1 over t squared is the dimensionality because there is an extra t here and when you hit this you get a 1 over time and then it has the same dimensions as this, okay. So this eta of t at the moment is just noise but not the systematic part of it, the noise from the medium, okay and what are the assumptions we have permitted to make of this eta of t, well exactly the same physical assumptions namely that on the average it is 0, with or without the external force because again the assumption is you are in a heat bath and the fact that you applied an external force does not change anything, does not change thermal equilibrium, remains as it is and it certainly does not get affected by what is happening here. So we think, we do not know for sure as yet but this is a harmless assumption here. So we have no clue at the moment what sort of noise is consistent, the whole idea is to make this model consistent with causality because it looked like we are violating causality and to make it consistent with stationarity, it looked like that too was getting violated, okay. So this is this thing here is called the generalized Langevin equation and we need to fix this model in such a way that it will give us the results that we know are already true namely this velocity should finally the probability density should turn to the Maxwellian, should remain in equilibrium, that it should have a stationary autocorrelation function and that the mobility of the system is governed by the velocity correlation, that was a very general theorem that we deduced and that cannot be violated. So that result has to be checked to do this. So the whole idea will be to fix this model in such a way, precise way that is the only way we can fix it such that it will be consistent with stationarity, causality and lead to the correct Kubo formulas for the mobility and so on. In the process we will discover the fluctuation dissipation theorems. Now remember always keep this at the back of your mind, when this eta of t was Gaussian white noise with a strength square root of gamma that capital gamma and this little gamma here were related to each other by this relation. So we had this consistency condition gamma was equal to 2 m gamma k Boltzmann t. We call this the fluctuation dissipation theorem because this measured the strength of the fluctuations and that measured the dissipation in this system. We need to find out what is the new fluctuation dissipation theorem. So that will that also tell us something about eta? Yes, yes. So that is going to tell us something about the correlation of eta here because ultimately if this fact that the system is in equilibrium is happening because thermal fluctuations throw you out of equilibrium, the dissipation brings you back. The noise itself has to play a role to bring you back to equilibrium. So there has got to be a consistency condition between these two. Precisely what it is we do not know at the moment. But what we do know but what I can assert and we will have to prove this is the following that it will be inconsistent to assume this to be white noise. It will show that it cannot be white noise. It has got to have a finite correlation time. Not surprisingly that correlation time will be related to this memory function. They have to be related to each other. So there will be a relation between them. That will be the replacement of this theorem. That will change this. So let us see if this is going to work. By the way, this kind of business is used in understanding the movement of particles in liquids but it is a much more intricate formalism than this simple stochastic equation formalism. But this is a standard model. It is next step to improving the Langevin model before we go to the full glory of something like the BBJQI hierarchy or something like that. But this thing here will already tell you a great deal of, give you a great deal of information, okay. So now let us do the usual game. We are going to find average quantities. So let us find the mobility for instance. And we do this by the standard trick of saying well the average value is 0 and what we need for the mobility is for unit applied force, what is the average velocity. And so the average velocity per unit force, applied force is called the mobility of the system. And we also define the dynamic mobility mu of omega by saying suppose I apply a sinusoidal force with frequency omega, what is the response look like, average look like. So a short way of doing that, I am not going to start from the beginning like we did for the Langevin equation, is to simply say alright, let us take this thing, apply a portfolio transform on both sides and see what happens. So we are going to define V of T equal to as usual, our convention was 1 over 2 pi minus infinity to infinity d omega e to the minus i omega t V tilde of omega and the same for eta, the same for f, etc, etc. So any function of T is written in Fourier language in this fashion here. And then let us see what happens to this equation, what does it give? And we will also take averages so that the average value of this goes to 0 anyway and where does that get us? Well if I substitute that, let us write down the equation for the average value of V of T directly or rather the average value of the Fourier transform of V of T directly here. So I apply this, I put this in on both sides and see what happens. So the first term gives me m times minus i omega because there is a dot, right? And I am going to equate Fourier coefficients on both sides. So minus i omega, the average value of V of omega, this term here plus something or the other plus and let us do this properly, m divided by 2, but I am going to read out Fourier coefficients, okay. Let us do this slowly, pardon me, it is a convolution here, I am not okay since it is forced on me, this is a convolution, it stops at T and we can write down what this is going to be. It is going to be the one sided Fourier transform of this V of this gamma because the integral gets cut off and then you equate Fourier coefficients. So I hope you see that directly and the 2 pi is cancelled out everywhere here, here, here, etc. And you end up with this V tilde of omega times minus i omega plus I need a symbol for this quantity integral 0 to infinity d tau e to the i omega tau times gamma of tau. That is going to be the integral over dT prime. When I change variables from T minus T prime to tau, this integral will run from 0 up to infinity, it will get inverted and the minus sign will go away, right? I replace T prime by tau which is T minus T prime. Then this becomes gamma of tau and the e to the i omega minus i omega T which came in in the Fourier transform is going to give you e to the minus i omega T times e to the i omega tau. So let us give this a name, it is not gamma tilde because this is a one sided Fourier transform. It is if you define it to be... But I do not, I want to keep the notation straight so that I do not want to be confused that sometimes I define it that way and sometimes I do not. We have said gamma of tau is defined for positive values of its non-negative values of its argument. So only this thing, this definition, let us call it gamma bar of omega. For the response function I use chi, I just said that is the susceptibility because that is the standard symbol. For this, as far as I know there is no standard symbol, let us just call it gamma bar. I have used star for complex conjugation, not a bar. So it is gamma bar of omega minus i omega in this fashion here is equal to f tilde external of omega, m times. There is nothing to average in the deterministic applied force. So that is just f tilde. And the eta tilde goes away because the average is 0. So if I divide this by v tilde of omega, so therefore, pardon me, yeah, I have done that. I have done that, sorry. Thank you. Yes, of course, of course. It is the average. It is the average, why do not I write equilibrium outside this average? Why, why have not I put a subscript equilibrium? I am not in equilibrium. I have applied an external force. So this is the perturbed system. And that is why the system, the average velocity is not 0 because it is moving. In the absence of this external force, the average velocity is 0. It does not go anywhere. But I will still retain this notation. I will still retain the notation, this bracket here because I am doing a full average. There is an average over the initial condition v naught and then there is an average over all possible v naught. So I do not care. I am not even going to specify what that ensemble is. This is and the average force is 0 and that immediately gives me this. But we know that this fellow divided by this f tilde external of omega, we know this is by definition the dynamic mobility which in this model is 1 over m gamma bar of omega minus i omega. This is the dynamic mobility. It is a susceptibility. And what kind of generalized susceptibility is? Is it? It is chi x v of omega. Because I am applying a force which is purely mechanical, minus x times f external is the change in the Hamiltonian of the system. And therefore you get this. But there is no reference to Hamiltonians or anything like that here. Once I put in friction, it is not a Hamiltonian system anyway. But I do not care. The model still works. This still works. So in the sense of generalized susceptibilities, this is what the mobility is. But in simple and physical terms, it is the average velocity per unit applied force. And when you apply a force which is sinusoidal, you do this division for each frequency component and you call it the dynamic susceptibility or mobility here. We know there is a general property of this guy which says that the singularities in the omega plane cannot be in the upper half plane, not with this Fourier transform convention. So I am going to leave it as an exercise to you to show that this, the singularities of this function, by the way, this is not a constant. If it were a constant, the matter is trivial because then you know that the pole is in minus i gamma. But now you have to show that the singularities of this are in the lower half plane or better still, there is no singularity in the upper half plane. So any root of this omega equal to i times gamma bar omega must be in the lower half plane. That follows from this representation. So that is the susceptibility, the general in this form, in this model. We still have to see what it does in terms of the velocity autocorrelation function. We still have to compute that explicitly. And we have to show that this quantity is, in fact, the velocity autocorrelation function apart from some KT factor and then e to the i omega t integrated over t. We still have to show this. So show that u of omega has no singularities in the upper half plane. And of course you know that in the case when this fellow is constant, when the memory kernel is just gamma times delta of tau, this thing becomes gamma, comes out and then this is the gamma minus i omega. It was already the dynamics susceptibility in the large of my equation. We had already found that. So it goes to that correctly. But now we have to know, we have to see what is going to happen when you put in this memory kernel here. So now we set this aside and let us go back and compute the velocity autocorrelation in this equation in the absence of the external force because the whole point of linear response theory is that the response in the presence of the perturbation to first order in the perturbation is given by some physical quantity which is a response function dependent entirely on equilibrium fluctuations or correlations in equilibrium with no reference to the external force at all. So that is the crux of linear response theory and we now need to show that. So now in the absence of this quantity of this external force, this is the quantity that we have. So let us put this as 1 over m and remove this m. I want to find the autocorrelation function. So instead of trying to solve this integral differential equation, which is by the way a mess because one way to do this is to do what we did, namely do Fourier transforms and compute formally a solution. That is not going to help us find autocorrelation functions very easily. So we will cut a long story short and anticipate the fact that V is a stationary process and directly calculate the autocorrelation function in equilibrium. In other words, in the absence of the external force and we will check if this assumption of stationarity was right or not post-facto. So let us write this equation down with an arbitrary time. One point I wanted to make was yes, as an equation, as an equation, mathematical equation, it is an integral differential equation and as you know when you have an integral differential equation of this kind, in principle you can convert it to a differential equation but it will be an infinite order differential equation in general. So as it stands, there is no guarantee that, there is absolutely no guarantee that you have a Markov process as the output because for a Markov process I would expect a Fokker-Planck equation which has only the second derivative in the velocity etc. But this looks like there is infinite memory in the problem. So yes, it is indeed true that this does not lead to a Markov process. This output process V is not Markov even if this were Markovian which we do not know right now. So it is not, it goes beyond the usual Markov process. The next question, if it is not a Markov process, will its conditional density satisfy some kind of Fokker-Planck equation? In general, you do not expect it at all but in this case it turns out that it does. It satisfies a Fokker-Planck like equation very similar to it okay. And what is needed for that? I am just trying to motivate my sort of, heuristically what the answers would be, eta should do something, yeah. Now if you are familiar with Markov processes, you know that when you have a driving force of this kind in general, if this were a white noise then you have that Langevin type equation and sure enough you have a Fokker-Planck equation. That is a rigorous mathematical equivalence. Otherwise, the partial derivative order on the right hand side in the variable goes to infinite order. It is called the Kramers-Moyal expansion in general. It gets cut down to second order very crudely and roughly when higher order cumulants vanish. Now what sort of random variable has all cumulants beyond the second equal to 0? Gaussian random variable. So if this is a Gaussian noise, then it turns out that this output process in spite of this infinite memory here still satisfies a Fokker-Planck like equation, okay. It is not Markov but it still satisfies such an equation, a great simplification. We are not going to do that. We want the more physical aspect of this namely we want to fix the problem at stationarity, causality and to find the mobility. We would like to find that here. So let us do that. So let us write this equation at T naught plus. But T naught is an arbitrary number and T is any positive of this plus integral from minus infinity to T naught plus T dt prime gamma t naught plus. T naught is an arbitrary point and I am writing this equation at T naught plus T where T is any positive number. And now what would be the way to find the correlation function in equilibrium? It would be to multiply this by T naught on this side and then do an integral from 0 to infinity after multiplying by e to the i omega t. That will give me what the left hand side of the cobalt formula has, right? So I do that. I am going to do this here. But in the normal derivation for the ordinary Langevin equation, we could have done that and we did do that. We found the correlation function directly by multiplying by V of T naught, taking averages and arguing that V of T naught is uncorrelated to eta of T naught plus T, put that side equal to 0 on the average by saying that the force at a later time cannot govern the velocity at the present time. And then we ran into this problem with stationarity Gaussian, with causality and so on. We certainly want this property. We certainly want V of T naught, V dot of T naught in equilibrium to be equal to 0 because we would like this thing to be stationary, certainly. It turns out that the only way to do this with this equation here with this model is to argue that V of T naught, the correlation between V of T naught and the random force is not V of T naught with eta of T naught plus T correlation equal to 0 but rather a portion of this memory should be taken out on the right hand side and called that is the effective force on the particle. Now the question is what portion? What portion? That is left arbitrary at the moment but we know the result that we are aiming for and we want consistency here. So the argument is the following. Look at what happens if I break this integral up V dot of T naught plus T plus integral T naught to T naught plus T, dt prime gamma of T naught plus T minus T prime V of T prime is equal to 1 over m eta of T naught plus T minus this fellow here from minus infinity to T T naught dt prime V of T naught plus minus T prime sorry gamma of T naught plus T minus T prime. There are many ways of motivating this but the simplest is going to be the following. I am going to multiply both sides of this equation with V of T naught e to the i omega t and do a different on both sides of the equation after I average. So if I average this guy here that is going to be the average that exists in the cobalt formula for the dynamic mobility. If this is stationary if I put T equal to 0 this part of it here should become independent right? It will be equal time and should vanish this fellow should vanish. When will that happen? If I put T equal to 0 this integral vanishes and I am going to get this correlation equal to 0 provided the V of T naught is uncorrelated with this guy. Not with eta of T naught plus T but this portion of the velocity history subtracted out and that is the only way you can split this to achieve this point. We will see as we go along that how it becomes consistent right? And then we will justify it. So let us call this something else. Let us call this noise some h of it is a function of T naught plus T but it is also a function of T naught because there is a T naught sitting here and there is a T naught plus T sitting here. So it is dependent on both the starting point T naught and the current time T naught plus T and it is not stationary. This fellow is not stationary by any means. It is a function of T naught plus T and T naught separately so therefore it is a function of T and T naught separately. Now we will assume and we have to see if this is consistent or not that V of T naught is not correlated with h of. This is for any positive people okay. So that will be the next step. I have to stop here today since we have run out of time but we will take it from this point next night. We would like this to be true. We would like this to be true so that we do not have the problem which we had in the Langevin model where on the one hand stationary said it should be true but the exact computation of the velocity correlation said it is not true. We definitely like it to be 0 and that is achieved by saying look let this thing trivially vanishes if T is equal to 0. So I split that portion and the rest of it but it has got a physical meaning. There is on the time axis there is some arbitrary instant T naught and this is T naught plus T and we are writing the velocity or acceleration at this instant of time by saying that this is affected by a friction which operated all the way from the infinite past up to here but that is broken up into two pieces part of the velocity history which operated till here and the rest of it. And this portion of it I am subtracting from the noise at this instant of time and arguing that the effective force, random force with which the velocity here is uncorrelated is in fact this portion, this subtracted portion. That is the only way in which I can ensure that if T goes to 0 this correlation here is going to vanish identically so it preserves stationarity but now that is fine as a kind of fix but we need to see if it gives you the correct Kubo formula and it gives you the correct fluctuation dissipation theorem. By the way there are two fluctuation dissipation theorems here. One of them is the statement that when you apply an external force the response to first order depends on equilibrium fluctuations. So it is a kind of Kubo formula for the susceptibility. The formula for the susceptibility itself in terms of the response function is called the second fluctuation dissipation theorem because historically the fluctuation dissipation theorem was written in the context of the Langevin model where you have a specific stochastic model Langevin equation or model and then there was a connection between the dissipation constant and the strength of this random noise which you put in by hand and that was called the first fluctuation dissipation theorem. So the second one is in fact the consequence of linear response theory directly the first one is specific to a stochastic model but the second one must be valid as well clearly otherwise linear response theory does not make sense okay. So since our Langevin model is a linear model we expect that the linear response theory formulas will also be true it is linear causal retarded and therefore it satisfies the conditions required for the linear response theory formulas. So I hope this is clear that there are two of these theorems and we will write them both down explicitly.