 Right, so now that we had this little excursion through linear response theory, let us try and make contact with the original model we started with. We started with the Langeva model and let us try to see if the results of the Langeva model match with those of linear response theory and if there is consistency between the two. And after we do that, we will then go a little deeper into the Langeva model itself or the generalizations of this model, look at a couple of physical situations and then take up the next topic. So the immediate program is to see how consistent the Langeva model is or its results are with those of linear response theory, which was the whole formalism that we developed. Now the Langeva model for a particle of Brownian particle in a fluid, we wrote down a stochastic differential equation, made some assumptions about the noise term in it and then derived quantities like the velocity, auto correlation, the equilibrium correlation functions. Linear response theory on the other hand was entirely in terms of correlation functions, equilibrium correlation functions. So the question is, is there a connection between these two? In the process, we should also take note of the fact that while we did the quantum case also simultaneously with the classical case in linear response theory, the Langeva model was completely classical. We never talked anything, said anything about operators or anything like that. If time permits later, we will look at a quantum Langeva equation but for the moment we are looking at it at a much more elementary level and let us try to make contact with the simplest Langeva model that we looked at. So let us call this, this has to do with the mobility of a Brownian particle inside a fluid. So recall that we looked at a single Cartesian component and quickly to recollect what we have done, we said a particle of mass m satisfies a stochastic differential equation which is V dot plus gamma V was equal to 1 over m eta of t. But this was a white noise, Gaussian white noise with 0 mean and a delta function correlation. So eta of t was equal to 0 and eta of t, eta of t prime was equal to some constant gamma delta of t minus t. Now we have been talking about what happens when you perturb a system. Now for the analogous equation would be to apply an external force to this, to this particle. So some external, arbitrary external force, so let us say 1 over m f external of t. So we have definitely taken the system out of equilibrium by applying this force here and we would like to know what the response of the system is in terms of what is the velocity, average velocity, averaged over some given initial condition. So we computed that, we found that V of t average was equal to V naught e to the minus gamma t. That was my initial condition plus 1 over m, a term involving the average of eta of t which was 0, so let us write that down explicitly, integral 0 to t dt prime, e to the minus gamma t minus t prime and then eta of t prime and the average of this quantity was 0. So once I took average it is this term went away but this term persists now. So you still have plus 1 over m, integral 0 to t dt prime, e to the minus gamma t minus t prime, f external or applied of t prime. There is nothing to average over it, it is a given force from outside and that is the result here, that is the explicit result for it. Now one of the questions we want to know is what happens if I apply a constant force over a lengthy period of time. Now normally we will switch on the perturbation at minus infinity, that was what we did in the formalism of response theory but now I want to simplify matters a little bit and say let us switch it on at t equal to 0 and let us do the simplest case which is a constant force at t equal to 0. So f external of t equal to f naught equal to a constant, what happens if I put a constant force on the system? Then this average v bar of t, v of t bar is v naught e to the minus gamma t plus f naught divided by m e to the minus gamma t, that is this portion and then inside the integral you have e to the gamma t minus 1 over gamma, if I do this trivial integral. This portion goes away so we can simplify it and write it in a little simpler form. So this is 1 minus e to the minus gamma t. Now we want the steady state response of the system. In other words, since we switched on the force at t equal to 0 we should let t go to infinity to get the steady state response, not the transient. This is the transient included here otherwise I should switch it on at minus infinity and see what happens at any finite time. It is easier to just let t go to infinity in this so what happens is that v of t divided by f naught per unit applied force, the average velocity or velocity response per unit applied force divided by this f naught, this quantity in the limit t tends to infinity is finite, is finite and it is equal to 1 over m gamma. Now you would expect that because physically what is it that you have? You have a fluid with some viscosity in it, some friction given by this gamma. You are going to apply constant force. If force keeps acting, constant force just like a particle falling under gravity, this particle you start with reaches terminal velocity and then it does not accelerate anymore because the drag force on it is exactly balanced by the external force. That is exactly what happens here. So that is why it tends to a constant because this is now going to tell you what is the terminal velocity per unit applied force, it is given by 1 over m gamma. There is a name for it, this quantity, definition of this mu mobility, the mobility which in this model happens to be equal to 1 over m gamma. In general if I have an arbitrary particle, an arbitrary model for it, we do not care what it is, I apply an external field then the average or the steady state average velocity per unit external force is called the mobility of this particle by definition. Now we have throughout a linear response theory applied forces which are time dependent, which are sinusoidal with some arbitrary frequency omega and so on. So this corresponds to no frequency at all, zero frequency because I put this f, when I put in frequencies then I have e to the minus i omega t etc. So one should really call it the static part of it or the omega equal to zero limit of it. So let me call this of zero in anticipation of the fact that this is happening at omega equal to zero because pretty soon now next I am going to ask what happens if the force were not constant in time but was sinusoidal with some fixed frequency omega. That would be the dynamic mobility as opposed to the static mobility or the frequency dependent mobility which by now you would have guessed is a kind of susceptibility, this is really what is happening. So what we are going to actually look at is the dynamic mobility but the static part is one over M gamma in this model, in this model. What we are going to do, linear response theory presumably will give us a general formula for this. So we have to go a little back and see how does this tie up with a general formula. Let us do the following. By the way, we can straight away make one connection here in this model and that is the following. That connection is, we know that the diffusion constant of this particle is K Boltzmann T over M gamma. So this is directly connected to this mobility by this relation and as you would have guessed by now, this relation between D and the static mobility has nothing to do with the Langevin model. In the Langevin model, it happens to be D happens to be this and mu happens to be one over M gamma but this relation between D and that should be more general, should really be more general and we will see why in a minute. Meanwhile, let us look at the dynamic mobility. What could be dynamic counterpart of this P? Well, we go right back and say now let the force on the particle f of t equal to f naught e to the minus i omega t for some given omega. So let f of t be equal to this. All I have to do is to put that into this equation here and do the same integral. It is again an exponential integral. So what happens now? v bar of t, this is again f naught over M integral again e to the gamma t and integral 0 to t dt prime e to the power because the force is e to the minus i omega t. I put that in here. This is f of t prime and I do this integral which is equal to v naught e to the minus gamma t plus f naught over M e to the power gamma t, sorry e to the minus gamma t here and outside and then inside is e to the i gamma minus i e to the power gamma minus i omega this is minus t minus 1 over gamma minus i omega and this simplifies. So let us put that in. This is minus e to the minus i e to the minus gamma t and this fellow becomes e to the power minus i omega t. Well, okay, one should really if you have a real force the response should be real. So one should say real part of this and I go through the whole real part but when we define the susceptibility we defined it by putting a sinusoidal external force f naught e to the minus i omega t and I said when you are going to compute physical quantities you are going to take the real part on both sides, etc. So we will do that after we do this integral. So now what happens? Now you can see that v of t over f naught limit as t tends to infinity. Of course the oscillatory part has no limit. This fellow will remain e to the minus i omega t because that is cosines and sines that do not have any definite limits they keep oscillating but this term will go to 0 and so will this and we do the same thing as before per unit amplitude, this quantity. This by definition is the frequency dependent mobility mu of omega by definition. And what is that equal to? f naught e to the minus i omega t, alright. So one should write this, let me write this in the following way, 10. So let us write this as t tends to infinity 1 over m gamma minus i omega times f naught e to the minus i omega t. So that makes it much clearer, right. So instead of just gamma you have gamma minus i omega here. Now recall what happened when we did linear response theory. We said that if you start with the Hamiltonian h naught and then you go to h naught minus a f of t and this f of t was f naught e to the minus i omega t, then the change in any other physical quantity, the perturbation delta b of t, this quantity was from when you switched on the force etcetera, etcetera minus infinity to t dt prime d, but how did we write this? Up to t dt prime f of t prime phi a b of t minus t prime. This is what we had and in the case when f of t was equal to f naught e to the minus i omega t, in that case for a single frequency we discovered that this quantity delta b of t was equal to chi of omega f naught e to the minus i omega t, chi a b. That is how we define this chi a b and then we went on to say this chi a b is integral from 0 one sided Fourier transform of a response function phi a b. But the first step was to say that if you apply a steady force at a given frequency, the steady response is also going to be at the same frequency, not a transient and it is attenuating the external force, the amplitude by this complex number chi. So if the force is the real part of this fellow, the response is the real part of this whole thing, the real part of a times b and a and b are complex numbers is not real a times real b. There are interference terms between the two. There will be a term which has a phase lag and so on, right, lag or lead depending on the kind of response. So that is what is happening here. What we have identified is exactly that. We have said that the velocity response, this should be v bar, v bar is this function of omega multiplied by the applied force. So it is clear that the dynamic mobility mu of omega is equal to 1 over m gamma minus i omega in this model, in this model and what kind of susceptibility is it? This is a generalized susceptibility but the question is what are the variables? Chi what should I write? Well the force applied is a mechanical force. So A, this operator A in the present case is x because if you take this as a potential and you differentiate with respect to x, the force is f of t minus the derivative with respect to x gives you f of t. So that is the reason for my putting a minus sign match with this case, right. So it is clear that in the present problem A is equal to x and what is B equal to? v because I am measuring the velocity response. So it is clear that what we have computed in this specific model is x v and that has turned out to be this quantity. At 0 frequency when you put omega equal to 0, this is the static mobility which is connected to the diffusion constant in this manner. We have to check that out also as a special case but we have all sorts of formulas in linear response theory for this guy. So the question is is that going to match? Is it going to give us exactly the same result but what was our general formula for the susceptibility and this is going to be the consistency check. If you did not know anything about the Langevin model but we did linear response theory then you are going to say chi x v of omega equal to integral from 0 to infinity d tau d tau that symbol e to the i omega tau phi x v of tau. That is what linear response theory defines in generalize susceptibility as we are doing classical physics. So what is phi x v of tau got to be? Remember that phi a b of tau was equal to beta times a dot of 0 b of tau in equilibrium. We derived that because we started with Poisson brackets and so on, played around with it, used the canonical ensemble and then we discovered in the classical case phi is just the correlation of this product but not a but a dot because there is a Poisson bracket with H naught which appeared here. So if you apply that blindly it is clear that this must be equal to beta times integral 0 to infinity d tau yes d tau times e to the i omega tau times a dot is x dot, x dot of 0 v of tau in equilibrium just applying that formula which is equal to 1 over k Boltzmann t times integral 0 to infinity d tau e to the i omega tau v of 0 v of tau in equilibrium but we want to compare with an ensemble obtained from the Langevin model. So we have to put in here whatever we got from the Langevin model from the Langevin equation without any external force but we already found an expression for this quantity. This thing here was equal to k Boltzmann t over m e to the minus gamma t for t positive gamma tau pardon me oh this is chi x of omega the k t cancels the 1 over m comes out and this is precisely equal to 1 over m with checks with checks with what we have here. So this is the magic of linear response theory this we got by solving the Langevin equation in the presence of the force directly solve for the average value blah blah blah etc and got this but this we got without any reference to any external force at all. We just said if you want this susceptibility which will tell you how the systems velocity will respond to some potential external force if you when you apply it it is given in terms of an equilibrium autocorrelation function in the absence of the external force but that the Langevin model gave an answer for and you plug it in and you get exactly the same result. So this checks the consistency of the Langevin model vis-a-vis linear response theory. Now one can go one step further we know that at 0 frequency this fellow was equal to that but remember independent of the Langevin model we computed explicitly what the mean square displacement was and we discovered that asymptotically for a Langevin particle all we did was to exactly compute the mean square displacement in this case and we discovered that we had to have d equal to integral 0 to infinity dt v of 0 v of t in equilibrium and that is just an integral of e to the minus gamma t nothing more than that. So this directly gave us k Boltzmann t over m gamma but if I multiply this by e to the i omega t then I get the mobility automatically apart from the beta factor. So this relation is a special case of the relation which defines the generalized mobility in terms of the velocity autocorrelation and what is that general relation? It is precisely this quantity that it is precisely the identification that chi x v is equal to mu of omega and in the general case this quantity here is precisely what we have computed namely this is equal to beta times an integral from 0 to infinity d tau e to the i omega tau v of 0 v of tau in equilibrium the general relation for the mobility independent of any Langevin model or anything like that. Because of the fact that the Langevin model itself was linear right? Yeah of course otherwise you would have had other terms in the exact calculation that you did for the model. Linear in what? The Langevin model does not have terms like v square. Ah okay it has got a good point this is a deep point. It is true that the Langevin model does not have any term it goes like v square but tell me where could it have had a v square term at which place could it have had a v square term at all? It is a good point where could the v square term have been and yes it is true that specific linearity of the Langevin equation is what is making it consistent in this case with linear response theory but where could it have been where could you have got a quadratic term or any other power? Exactly in the friction part you could have ended up in the friction part with a non-linear term in that and then it is so not obvious at all to do this. But we should still be able to say something about it and that we will when we look at it as a stochastic differential equation with a drift term specific drift term and relate it to a corresponding Fokker-Planck equation. So we can say more we will do that shortly. But yes at the moment yes there is another generalization of the Langevin equation which retains linearity and which should therefore be consistent with this guy completely consistent with this not as simple a model and that is the following. You see if you throw up and saying that this Brownian particle is much more massive than the fluid particles in the molecules themselves. If you make the mass of the Brownian particle smaller getting closer to the molecular region itself what would you expect? You would expect that the velocity would start remembering more and more what its past was. You would start building correlations in this velocity which are missing completely in the Langevin model. We have just said there is a gamma here and that is the time scale of this particle but now you are going to much smaller time scales so you would actually start remembering collisions and you would have to now include 2 particle, 3 particle correlations 2 time, 3 time correlations higher order correlations you would have to include in any case you would have to do that. One effective way of doing that is to say I will bury all that once again by saying that the particle has some memory the systematic part has some memory in it. So one way to generalize the Langevin equation would be to say v dot plus not gamma times v but let us say some integral over its entire history dt prime gamma of t minus t prime times v of t prime with some memory kernel here. Again dependent only on the lapse time this equal to 1 over m a dot t plus external force with n etc. That is one possibility. We have to examine whether it is consistent or not. It will turn out it is not strictly consistent with the white noise here. The moment you have this but you can still treat this as a model you can put an external force here and ask what is the mobility. You can ask what is the generalized susceptibility in this case. What would you expect will happen? Suppose you just took Fourier transforms and did this what would you expect would be the mobility in this case. You would have a gamma of omega you would certainly have a 1 over m times gamma of omega I will come to the notation in a minute minus i omega as before definitely. This is not gamma of t some kind of transform but because it is minus infinity to t the whole thing will be like a convolution. So what should this what will this be? A one sided Fourier transform precisely it will be a one sided Fourier. So let us call give it a name I do not want to call it a tilde because that looks like a Fourier transform. This follows anyway defined only for t greater than t prime it is getting cut off here. So let us call it gamma bar of omega where this is defined as integral 0 to infinity d t e to the i omega t gamma of t a one sided Fourier transform. So let me write this as exercise show that this thing here by the way is called the generalized it is called the generalized large y equation it is still linear but it is an integral differential equation and it will lead to dynamic mobility which looks like this but you know we have been saying this mobility is susceptibility and with our Fourier transform convention this object whatever be the mobility cannot have a singularity in the upper half plane in frequency we have to check that out. Now in this case when we found out that mu of omega so in the case when the ordinary Langevin equation this was 1 over m gamma minus i omega where is the singular as a function of omega omega is minus i gamma this is 1 over i so bring it to the right side so in other words in the omega plane you have a singularity here minus i m which is fine which is what you expect a causal susceptibility retarded susceptibility to be now you do not have to do the same thing here and show that this fellow the root of this the singularities of this denominator have to happen only at omega equal in the lower half plane it is not too difficult to show but you do not know the form of this function except assume some reasonable thing that it dies down sufficiently fast at infinity and also positivity you need positivity here so assume there is a monotonically decaying function of its argument positive non-negative and then it is sufficient to show that this singularity is actually in the lower half plane as it should be so at this level at least we are satisfied that there is consistency in this whole business now let us try a slightly harder problem rather than just the free particle look at the harmonically bound particle and see what happens again there should be complete consistency so we looked at a case of harmonically bound particles let us this was the oscillator in a fluid now we did this problem by not explicitly saying it is a Langevin equation we just wrote down the equation for the average well or average position this is a second order differential equation now so we wrote x double dot of t the average value plus plus gamma x dot of t this manner plus omega naught squared x of t I have already taken averages over the eta over the white noise so what is left is 1 over n f external of t and what did we do we found the green function for this problem and solve this inhomogeneous equation using that green function that was like a response function or that was like a susceptibility so if you recall we found that in this case chi of omega was minus 1 over m times omega squared plus i omega gamma minus omega naught squared and the poles were in the lower half plane this is an imaginary part which was essentially minus i gamma over 2 and then there was some part which depended on whether you had over damped or under damped oscillator or whatever what susceptibility is this by the way I should put subscripts here well we again apply a mechanical force that is how we got it which is x so a is x in this problem and b was also equal to x so we have essentially found chi xx of omega and we inverted this transform to get the response function right so if you recall we inverted this by saying the corresponding this fellow by the way was the Fourier transform of g it is the Fourier transform of the green function exactly and what was the green function itself g xx of omega tilde of t what is g xx of t was equal to a theta function of t times phi xx of t the retarded green function it is got a theta function in psi and the Fourier transform of that product is what gave us this guy so what we have to do is to take the inverse Fourier transform of this fellow and that gives us it should give us a theta function as well as this so this says theta of g this quantity here is equal to 1 over 2 pi an integral minus infinity to infinity d omega e to the minus i omega t g xx let us put that in chi xx of omega and you know how to do this integral because in the omega plane there are two singularities at these two points this is omega minus this is omega plus the contour of integration goes like this when t is positive you are compelled to close this in the negative lower half plane so that this contribution would go to 0 and you get a contribution when t is negative you are compelled to close it in the upper half plane and there is no singularities so the answer is 0 so it pulls out the theta function automatically for you so this thing here is equal to theta of t automatically and then there is 1 over m sitting there all the time and if you recall what we got we got the solution of the under damped harmonic oscillator in the presence of an external force right essentially so this involves e to the minus gamma t over 2 sin omega in the under damped case where omega under damped was equal to square root of omega not squared minus gamma squared over 4 so that is what we got so we can read off the phi xx of t now therefore in this problem phi xx of t is equal to e to the minus gamma t over 2 sin omega t over omega u we put a tau here if you like it makes you happy it does not matter looking at it for positive values of t anyway divided by m just a quick check we already saw that the time reversal property of this guy was dependent on the time reversal property of the 2 operators sitting here this is even or odd well we computed this only for t greater than 0 we should go back and compute for t less than 0 we get a mod here nothing is going to happen here at this point so it is actually an odd function in this case as it should be as it should be because the parity of this fellow depends on epsilon x epsilon x with a minus sign because it is a dot and a these do not change and this changes sign so it is got to be an odd function which it is so that part checks now what we need however for the mobility and the diffusion constant we need the xv because you are looking at the velocity response right so what is this going to be phi xv of t is there a quick way of finding out what the xv is you are going to put one more dot there right which means you are going to essentially differentiate once with respect to t so you are absolutely guaranteed that this thing is equal to the derivative of this so it is e to the minus gamma t over 2 over m omega u times first you differentiate this guy so it is omega u cos omega ut in this fashion so that is this portion that you differentiate this guy so it is minus gamma over 2 sin that is phi xv of t now what is the mobility what is the diffusion constant therefore what does linear response theory give you for this according to linear response theory this is got to be equal to beta times x dot of 0 v v of t by linear response theory right so this is got to be equal to 1 over k Boltzmann t times the velocity order correlation in equilibrium what is the diffusion constant so what is the diffusion constant this will imply that d equal to integral 0 to infinity dt v of 0 v of t in equilibrium which must be equal to therefore k Boltzmann t over m so let us get rid of this omega u let us divide here this goes this goes so that this is dimensionless kt over m so I have to do the integral 0 to infinity dt e to the minus gamma t over 2 times cos omega u t minus gamma over 2 omega u sin omega u t I have to do this integral so for harmonically bound particle this is the exact expression of the diffusion coefficient what does this integral give you so it is k Boltzmann t over m times the cosine just gives me gamma over 2 divided by this fellow here right so it is gamma over 2 over omega u square plus gamma square over 4 and the sin part minus there is another gamma over 2 and then an omega u divided by omega u times omega u square plus gamma square and this cancels against this which is identically 0 you should be surprised you are not surprised you should not be surprised so what is happening what is happening this particle is diffusing on the x axis and because there is a potential it is not translation invariant the point 0 is special it is in a potential half m omega square x square so it is really sitting inside this potential there is a spring which is bringing it back to the origin so the mean square displacement will not diverge it should not diverge remember that the diffusion constant is defined as x of t minus x of 0 whole square is supposed to go well to 2 d d so if you take the limit as t tends to infinity this guy this should be equal to d that is the definition of the diffusion constant in one dimension the means this place should go asymptotically like 2 d t or if you divide by 2 t that should give you the coefficient of diffusion that is turning out to be 0 so what do you conclude there is no long range diffusion in this problem the mean square displacement does not diverge goes to a finite constant and what finite constant can it go to so this is out the symptom that d equal to 0 this thing is a symptom of the fact that the mean square displacement does not go like t asymptotically it could go like a lower power of t but it is not doing that either it is actually going to a constant what would it go to what constant can let us take it in equilibrium so it does not matter what your starting point is what is this quantity for an oscillator yeah because we know this this is a quadratic term in the Hamiltonian the free Hamiltonian so we know that half m omega naught squared average must be equal to half k t per degree of freedom this is true so it implies that x squared average in equilibrium must go to k Boltzmann t over m omega naught square it does not diverge as a function of that is what the show showing up here it is explicitly showing up so we just sort of did this backwards right we computed the green function these solve the problem with an external force for the average position and then my little simple trick I differentiated it once to get the correlation between the velocity the velocity auto correlation function I did not directly compute it so this was useful trick notice that the differential equation that you have is for x and it is a second order differential equation so we did not really mess around with it we did not write if I had written an equation for v directly I could have done this but then the restoring force would have involved an integral the omega naught squared x would have involved an integral and I have had a different integral differential equation which is a bit of a mess so I circumvented that by actually computing the xx chi xx and then I argued or phi xx and I argued that phi xv is just one derivative I leave you to go back and find out what the corresponding chi xxv is so go back and find out in this case what the cross correlation is position velocity cross correlation so remember that you differentiate is multiplication by i omega and integration is division by i omega so in this problem as well there is complete consistency with now we should be able to get these results not each time appealing to the stochastic differential equation but by directly looking at what the probability distributions are like so we need a connection between the two and that is going to be the theme that we take up next I will not do that today but let me just mention the result here and then we will take it up in some detail next time onwards you see all the stochastic differential equations that we have been looking at all the stochastic equations have been in one variable for example there have been equations of the form some xi dot equal to some given function of xi could be linear could be constant we do not care so some function of xi it could even be non stationary although we have not looked at such cases this is the systematic part of the force that we are writing down here that we are writing here plus something which involves white noise 8 of t and the cases we looked at where 8 of t had a correlation which was gamma times delta of t minus t prime we can make it a little more general so let me write it in that language you could have a force a stochastic force which is of the form g of xi t some other function of the variable as well as time itself times let us call the zeta of t and this is a Gaussian white noise such that zeta of t equal to 0 and zeta of t zeta of t prime equal to delta function of t minus t prime I will subsume the constant in those in the cases where this is the constant in this function here so I have a slightly more general case than what we had so far general in several ways first of all this need not be a linear function or a constant in xi it could be time dependent itself this force what is multiplying at the amplitude of this white noise could itself be variable depend dependent on the driven variable and it could be a non-stationary so you have a very general situation here such a process is called by mathematicians diffusion processes when this is a Gaussian white noise these are called diffusion processes the usual kind of diffusion we are talking about of physical particle is a very very special case of it and relation I am going to talk about next would be to start to go from this thing here to an equation for the conditional density probability density of the variables I so there is can I write an equation for this quantity the conditional probability density function on the way we will make an assertion that this defines a special kind of Markov process which is then completely determined once you tell me this quantity here we write a differential equation for this called the Fokker-Planck equation and then we will study its properties okay so let me stop here