 And if you recall the problems with the original Langevin equation were that it was inconsistent with stationarity of the output process and moreover the white noise assumption probably led to this inconsistency. We have to check this out and we also found that the power spectrum did not go to 0 fast enough and you ended up with the fact that physical quantities like mean square accelerations were infinite. So we also had some issues with causality which was rather subtle. We made this assumption that the velocity is not correlated to the noise at later times but what happens at equal times? We are not rather vague about this. So we need to fix all these problems and for that we imposed, we introduced this generalized Langevin equation and what I am going to show today is that it solves these problems. It is completely consistent. It is a consistent way of describing the velocity of a Brownian particle in terms of some memory of frequency dependent friction and it also provides correct consistent answers to this problem at stationarity with causality and so on and so forth. Let us see now systematically how this goes. So generalized Langevin equation and we looked at it in the case of a single velocity component of a Brownian particle. This equation said m v dot of t plus m times an integral from minus infinity to t dt prime some memory kernel t minus t prime v of t prime was equal to the noise on the right hand side whatever that was and right now I am just going to say it is a noise and that is it. We are not going to specify anything about it except to say it is a stationary noise. So some eta of t stationary noise with the average value equal to 0. That is it. That is the only assumption. We have to see in a consistent way what kind of statistics are tenable for this noise. Then if we took averages we computed the mobility of the particle. So we said that the average value of the velocity per unit applied force when I apply an external force this thing becomes plus f external of t. So this is the generalized Langevin equation and we said that the average velocity to first order in the external force in this case it is a linear equation so it is actually an exact relation. This quantity divided by f external for constant external force the limit in which t tends to infinity this quantity tends to what was called the mobility of the particle. The static mobility of the particle at 0 frequency. The dynamic mobility is the steady state response that you get when the force is sinusoidal with some fixed frequency some given frequency omega. So in that case we discovered that v tilde of omega average namely the Fourier component of the velocity at that frequency at which you are pushing the system divided by per unit force so this was f tilde of omega. That is the component of the external force corresponding to frequency omega. This quantity here was equal to the dynamic mobility and we had in our model we computed this quantity simply by taking Fourier transforms on both sides. There are some niceties about taking Fourier transforms of a random function I slurred over those niceties but one can make this rigorous the result is perfectly correct and this turns out to be 1 over m times if you recall gamma bar of omega minus i omega but this is the one sided Fourier transform of the memory kernel which is defined for non negative values of its argument. So that is the place to which we had got. Gamma bar of omega is equal to an integral from 0 to infinity dt into the i omega t gamma of t. The assumption is the memory kernel is such that dies down as t tends to infinity in such a way that this integral converges. If it does not then you have to take Laplace transforms and then analytically continue to minus i omega or something like that in the Laplace transform variable. So this is the place we had up to which we had come. Now the question is in linear response theory we know we have a formula for the mobility. We are applying a perturbation which essentially couples via the variable X and we are measuring the average velocity V. So in the formalism of linear response theory this quantity A is X and the quantity B is V, the velocity. So we should have compatibility with the formula which we get from linear response theory and the question is is that consistent with this model or not. So we need to make sure that the model correctly reproduces that expression, the general formula. Then we can assert that these two are consistent with each other. So I want to emphasize again that the general formula we have for the mobility from linear response theory is that mu of omega, so from linear response theory, mu of omega is just the same as chi of X V of omega and we are looking at a classical particle. So this quantity is equal to the Fourier transform, this is equal to integral 0 to infinity dt e to the i omega t phi X V of t. So it is a generalized susceptibility and the question is what is phi X V of t equal to and this is classical. If you recall it is equal to beta times the expectation value, the equilibrium expectation value of A dot with B, so this is X dot of 0 with V of t in equilibrium but of course this is V of 0. So linear response theory says on general grounds that this fellow had better be equal to 1 over k Boltzmann t integral from 0 to infinity dt e to the i omega t V of 0 V of t but V is supposed to be a stationary random variable in thermal equilibrium. So this would be some V of t0 V of t0 plus t, t0 is arbitrary completely because if V is a stationary random variable then this correlation function is a function only of the time difference of the two arguments and it is just a function of t. So this much the general linear response theory says independent of any Langevin model, okay, that is what we found in general in the classical case. Now the question is is the generalized Langevin equation compatible with this or not, can we derive this or not? So what we have to show is the expression you get for the mobility here in the generalized Langevin equation. So this is, this is equal to 1 over M gamma bar of omega minus i omega and the question is, is this equal to this quantity here in the same model? If it is then I assert that these two are compatible with each other, okay. So you see the logic in the generalized Langevin equation model we explicitly found the mobility which is computing the response straight away and we discovered by taking Fourier transforms or whatever, we discovered that it is equal to this. On the other hand general linear response theory says it should be equal to that. So we need to go back to the Langevin equation, compute this quantity and see if it is integral from 0 to infinity multiplied by this gives you this and if it does we are done. So let us see if this is true. Now the way to do this is as follows. First of all the B must turn out to be a stationary process and we must be able to show that quite rigorously and we will see that it consistently turns out to be 1, okay. So let us start since I am anticipating myself, since it will turn out that this V is a stationary process namely this correlator is only a function of t, let us start, let us take a shortcut, let us start with the Langevin equation at time t0 plus t, whatever be t0, okay. So I have m V dot of t0 plus t plus m integral from minus infinity to t0 plus t and let us break that up into 2 pieces for a reason which will become clear. Let us break it up into t0 t0 plus t dt prime gamma of t0 plus t minus t prime V of t prime equal to on the right hand side the noise at time t0 plus t because that is the instant at which I am writing my equation down minus the piece that I took away from here. So it is minus m integral minus infinity up to t0 dt prime gamma of t0 plus t minus t prime V of t prime, okay. So I have split the force in the the frictional force into 2 pieces, one depends on the velocity history after my initial time t0 whatever it is and the other is past history from way back when whenever and let us give this a name. So let us call this thing by definition some H, it is a kind of noise because this is a random variable, this is a random variable which is imposed from outside. So let us call it some H and what is it a function of? It has got to be a function of t0 as well as t separately. So since t0 plus t appears here and t0 appears here separately, let us write it out in that form, let me call it t0 plus t t0, it is a function of both and I define this H in this fashion here. Now there is a reason why I did this because you see I want the velocity auto correlation, I want the velocity to be a stationary process, right. So in particular I want the correlator V of t0, V of t0 plus t equilibrium average to be independent of t0. So in particular if I take dot, if I take derivatives on both sides V of t0, V dot of t0 must be 0 at the same instant. So if I left multiply by V of t0 here, I get m V of t0, V dot of t0 plus t and let us left multiply on both sides and take full averages. There is no external force, so everything is in equilibrium and let us multiply this by e to the power i omega t and in this way integrate from 0 to infinity with respect to t. I have to do the same thing everywhere but the reason for my splitting it up till t0 is that if I set t equal to 0, this integral vanishes out here and then I am guaranteed that V of t0, V dot of t0 is actually 0 provided V of t0 is not correlated with this provided. So let us impose that, impose causality by the condition of t0 h of t0 plus t0 equilibrium equal to 0 for all t greater than 0. Impose it, I am going to impose it from outside. Nively you would have said V of t0 with eta of t0 plus t it should be 0, that is causality says the force at a later time cannot affect the velocity at an earlier time but that is not consistent, that is the important thing. It turns out it is not consistent and you will see why. On the other hand we do not know this condition is going to work but it has one merit. If I do that then straight away if I multiply by V of t0 here and set t equal to 0 then this integral vanishes, this becomes V of t0 with V dot of t0 and that is equal to 0, it constructs automatically. So stationarity is imposed, it is automatically satisfied provided I can get away with this condition. At the moment it is an artifact, we have to see what its consequences are. Is this clear? The reason why I am doing this? You could have split it anywhere, you could have from its minus infinity up to this point, you could have split the integral anywhere but I can only split it at one point in order to maintain the fact that I want this, I require by stationarity, it is required and this condition, this imposition is going to achieve that once I break the integral of t0, oh by the way I should say right from the beginning in this equation itself, t is positive, t is positive. So when I say t0 plus t I mean an instant later than t0 always, t greater than equal to 0, I can take the limit t going to 0 but from above always, yes, yes, yes, yes, we will see, we will see. So it looks like we, the noise and the velocity are mixed up in some complicated way, we will see, we will see what happens, we will see, it should not contain any memory, okay, we will see what happens as we go along but right now the motivation as to why I am doing this is very simple, I want that stationarity and that is achieved automatically if this integral runs from t0 to t0 plus t because if I set t equal to 0 this integral vanishes, okay. But it looks like I am paying a stiff price for it because I am going to impose this which as he rightly points out means that if I multiply, if I take this term, multiply by v of t0 and multiply by v of t0 here and take averages, these two are related to each other, we will see. So meanwhile let us see what happens here. If I take this average here, by construction this condition has been imposed, so left multiplying by v of t0 taking averages and integrating respect to t with this weight factor says this term plus m times integral from 0 to infinity dt integral t0 to t0 plus t dt prime gamma of t0 plus t minus t prime, this is a lot of algebra but it is worth looking at it carefully to see what happens. So you have v of t0, v of t prime, average equilibrium multiplied by e to the i omega t and that must be equal to 0 because by construction v of t0 multiplying by h of t0 plus t average is 0 by assumption. So this term plus that term must be equal to 0, okay. Well let us do this integral. By the way this equal to 0 means m is removed, it is gone and this plus that is 0. Now what does this give us? Well this is d over dt here because this dot I can take to act on t, it is derivative with respect to t and so I do integration by parts, right. The first term gives me v of t0, v of t0 plus t average equilibrium times e to the i omega t at t equal to 0 and t equal to infinity. That is the first term and then minus since it is integration by parts, the derivative with respect to t of this guy which is minus i omega times that, so minus i omega integral 0 to infinity dt e to the i omega t times v of t0, v of t0 plus t equilibrium. So that is this portion gone plus this double integral and I now need to simplify this double integral so let me erase the board here, let me do it up here. I will just retain this term, this integral, let me see what it does, okay. First step of course the obvious thing to do is to change variables from t prime to t prime minus t0, clearly that is a sensible thing to do because this thing here will then become 0, okay. So let us put t1 equal to t0, t1 is t prime minus t0, okay. There are several ways of doing this, t prime is, alright, let us see where it gets us. So this integral becomes integral 0 to infinity dt integral from where to where 0 and then t prime is t0 plus t, so it is t, 0 to t and then is the dt prime is dt1, more plus signs, gamma of what, t, t minus t1 that is this portion and then v of 0, v of t prime, t prime is t0 plus t, t prime is t1 plus t1, sorry, t0 plus t1 equilibrium, e to the i omega t that still remains as it is, that is the integral. I need to somehow get this integral e to the i omega t to act here somehow, right. So the thing to do is to interchange the orders of integration and what does this become if I interchange the order of integration, 0 to infinity t and for each value of t, t1 goes up to t, so if I interchange right now t1 is less than t, so t is greater than t1, so this will become integral 0 to infinity dt1 integral t1 to infinity dt, gamma of t minus t1, v of t0, v of t0 plus t1 equilibrium e to the i omega t. Now let us put t minus t1 equal to tau because that is the obvious thing to do, okay, right. So can we get rid of this? The first two terms gave us this, so we must keep this, let us keep the last part of it, just this integral is being simplified and let us put t minus equal to tau, so dt equal to dt for a fixed t1, so this is equal to integral 0 to infinity dt1 integral 0 to infinity again d tau, gamma of tau, so you see, finally you have ended up with the memory kernel and integral and then this remains as it is, v of t0, v of t0 plus t1 equilibrium and then e to the i omega, t is t1 plus tau. So there is an e to the i omega t1, e to the i omega tau, so let us bring the e to the i omega tau here and this integral was just e to the i omega t1, let me bring that there and then this completely factors, integral 0 to infinity d tau, e to the i omega tau, gamma of tau, totally factors out and what did we call this? We call this gamma bar of omega, the one sided Fourier transform of this, the weighted with e to the i omega tau, this integral was gamma bar of omega, so function of omega alone, right. So this becomes gamma bar of omega and this already a minus i omega times the same integral, dt1 e to the i omega t1, v of t0, v of t0 plus t1, this v of t0, v of t0 plus t e to the i omega t, instead of t1 the integration variable is t, but it is the same integral. So it says this, so it says this times gamma bar of omega minus i omega of this integral, times this integral with a plus sign is equal to 0, but what is this integral? This term here, if you put in t equal to infinity, this becomes v of t0, v of infinity, that of course decorrelates your equilibrium, this t going to infinity, this becomes a product of averages and the average velocity is 0. So the upper limit is 0, this therefore goes away, minus whatever happens at the lower limit, that is equal to 0, right. At the lower limit, e to the i omega t is 1, as t is 0, this becomes v of t0 because t is 0, therefore you get a square, v squared of t0 in equilibrium and there was a minus sign. So I move it to the other side and this is equal to that. So this fellow divided by gamma bar of omega minus i omega is equal to this integral, but v squared in equilibrium is k t over m, that is the Maxwellian distribution. So this is k Boltzmann t over m, now let us retain the m here, I put 1 here and remove the k t to this side, 1 over k Boltzmann, but this is what we call mu of omega. So we have actually established directly from the Langevin equation that the mobility on the one hand is given by this, on the other hand it is the same equation says, it is also equal to this integral weighted, this correlation function weighted with e to the i omega t integrated from 0 to infinity, which is with the 1 over k t, which is exactly the linear response theory formula, okay. So the model is consistent with linear response theory, this is sometimes called the first fluctuation dissipation theory. It is different from the equation, capital gamma is 2 little gamma m k t that we got, that is the second fluctuation dissipation theorem, we will come back to that, we will come back to that. So this is just a, in that sense the formula for the generalized susceptibility in terms of the correlation function from which comes inside the response function is in fact the first fluctuation dissipation theorem. So this is one way of saying it is to say that chi A B of omega equal to integral 0 to infinity d t e to the i omega t phi A B but remember that phi A B in general was equal to beta times A dot of 0 semi colon B of tau of t in equilibrium. This is sometimes called the first, in a sense it is just a definition but it is more than that. It says the actual response function is given by this equilibrium expectation value and that is the consequence of all the dynamics that we went through both classically and quantum mechanically. Recall what this fellow was, it was just a product of these 2 operators, these 2 quantities in the classical case, in the quantum case it stood for integral 1 over beta d lambda etc etc. That is the exact formula but whatever it is it is some correlation function and this is completely consistent with that. So the assumption we made that V of t naught is correlated, uncorrelated with H of t naught plus t comma t naught that has led to stationarity being recovered, being maintained and the fact that the first fluctuation dissipation theorem which comes out of linear response theory is satisfied. What we need to do now is to go back and say alright we made this assumption about V and H, some correlator was equal to 0, what happens to that? What does that lead to? So let us see where it gets us and it will lead to a little bit of a surprise, not too surprising. It will turn out that it is no longer consistent to make A tau of t white noise, it will not be delta correlated noise at all but it has some finite correlation time and we will see what happens. So let us go back and examine the consequence of saying that V of t naught H of t naught plus t t naught equilibrium is equal to 0 implies that V of t naught A tau of t naught plus t equilibrium, this is the first term in this H and then there was a minus something so we are saying this equal to 0 so that is equal to m times an integral from minus infinity up to time t naught dt gamma of t naught plus t minus t prime V of t prime V of t naught equilibrium. So making this assumption is equivalent to saying this. Now what is the first thing we can do here? It is clear that you can immediately change variables here so that I get rid of this t naught here. So let us do that. Let us put t 1 equal to t naught minus t prime. So this implies that this quantity here is equal to m times integral so t prime equal to t naught minus t naught. So if t prime is t naught, t 1 is 0 and if t prime is minus infinity, t 1 is infinity and then there is a minus sign in the Jacobian. So this is 0 to infinity dt 1 gamma of t naught t minus t prime minus t naught what do we get? T naught plus t minus t prime so it is minus t naught plus t 1. So it is t plus t 1 gamma of t plus t 1. Correlation V of t prime but t prime is t naught minus t 1 V of t naught. We can simplify this a little bit because we know this is stationary. This process is stationary. We have explicitly checked it out now. So I can shift time arguments in this. What should I do? Add t 1 to both sides right. So this is equal to therefore V of t naught for whatever t naught you like, 8 of t naught plus t for positive values of t or non-negative we can take the limit as t goes to 0 from above equilibrium must be equal to m times an integral from 0 to infinity dt 1 gamma of t 1 plus t times correlation V of t naught V of t naught. We could choose t naught to be 0 it does not matter just as this is independent of t naught 2 stationary. So you can put t naught equal to 0 and then you get V of 0 V of t 1 that is the correlator gamma t 1 plus t integrated over t 1 must be equal to this. In particular in particular we can also it says for consistency it says for consistency you have no choice but to say that this V is indeed correlated with the noise via this formula. So this is true for all t greater than 0 and the coupling is happening because of the memory kernel here. So if you let t go to 0 from above so let t 0 from above then it says that V of t naught at any time 8 of t naught namely the velocity at any time the output process at any time multiplied by the input noise at the same time and you take the average value this quantity is not 0. But there is a correlation between the 2 which is precisely given by the integral of the memory kernel. So this is M times an integral from 0 to infinity d t 1 gamma of t 1 and then velocity here is V of t naught t naught plus t 1 so you could as well write it as V of 0 V of t 1 in equilibrium and you can in fact remove this integration variable call it t. It has to be so. So it says this random force is not all that random. You specify whatever random force you like stationary, stationary random force whatever you like then the only consistent way to describe the motion of this particle is to say that there is this correlation otherwise you violate the stationarity principle. So this at instance equal instance of time there is a correlation. So this maintains causality that is the whole point. This is the way causality is imposed in this model for consistency. Yes, exactly. So it is intricately linked. What is it you are doing? It is not any random force. It is the random force on this particle. So there is a characteristic of the particle that has come in and it is not surprising. It is not surprising at all for the following reason. Go back to the original Langevin equation. I have not yet come to the first second fluctuation dissipation theorem which I will in a second but it is not surprising because after all when you say m b dot so this was the original Langevin equation. So m b dot plus m gamma v equal to eta of t which was white noise. We wrote this as square root of gamma times eta of t but this was unit delta, unit correlation delta function as the correlation. Then one would think look this is completely arbitrarily specified. This is a property of the medium but the two are related. You know that is not independent of each other. You know gamma must be equal to 2 m gamma k Boltzmann t and we also discovered that even though you started by saying when we took averages that v was uncorrelated with eta at the same instant of time, after you computed things you discovered that was not really true. That v of t naught, eta of t naught was not 0 identically. But even this is the coupling between gamma and t naught. Exactly. So there is a coupling back here. So this is the consistency condition. There is a reaction on the medium. Yes, yes but that, that is because this one is also sitting here all the way up. So this is the only way in which you can have this consistency. It is a backward correlation because this t is increasing out here. No, no it looks counter intuitive but it is not. It is completely consistent. So this separation we had in our minds that this random force is, I said it very glibly that this random force is due to molecules that won't get affected by the motion of this particle and so on but it is affected. It is the random force on this particle and it has got to be self-consistently determined. You will see this more dramatically in a second. So there is a coupling. There is indeed a coupling but we are trying to make this coupling consistent with causality, stationarity and so on. So this is, this is a relation which says that the velocity and the force and so-called random force at the same instant of time are not uncorrelated with each other. The equal time correlation is some integral. This is a number by the way. It is a pure number because it has got no dependence on anything. It is a pure number and that is the value of this guy. In a sense it is measuring the strength of this but now let us see it in a more dramatic way. You could do the following now. You could say alright, let us start with, let us turn this around and start with, so let us keep that aside. So we note that there is a correlation between the V and A to keep that in mind and now let us turn this around and say what was the definition of this H of, H of t0 plus t, t0 was equal to A to of t0 plus t minus M and integral from minus infinity up to time t0 dt prime gamma of t0 plus t0 plus t minus t prime V of t prime, okay. That was my definition here. Now let us multiply this by H of t0 with t0. Just as I pre-multiplied by V of t0, I multiply with H of t0. When I found the velocity autocorrelation, I took V of t0 plus t or V dot from that matter from the Langevin equation pre-multiplied by V of t0 and took averages. Now I want to find the correlation of the noise with itself. So I have H of t0 plus t and H of t0 multiplied on this side. That is equal to this term here but out here for this term, this fellow here, I can substitute from the Langevin equation. I have done so for this quantity so it must be multiplied by H of t0 t0 on this side. But what is this fellow, this guy here? I go back to the Langevin equation and it is equal to M times V of t0 V dot plus M times an integral from t0 to what? t0. That portion went away. So it is M times V0 dot M times V of t0. I play the same trick as before. I take this, so I take averages like that over this whole thing, substituting this for that in here, multiply by e to the i omega t and integrate over t from 0 to infinity. But all the quantities on the right hand side, I do the same manipulations as before, integrate by parts, use this relation between V and Eta because that is going to be important, right? And simplify. This is a slightly messier calculation on the previous one. There are going to be terms proportional to M squared, etc. I have simplified the whole thing. Use the fact that we already know what mu of omega is. We know the integral over the velocity correlation is mu of omega, right? And that is 1 over gamma bar of omega minus i omega with an M. So use that and finally after simplification, after which I am going to leave to you, you get the following result. You get M times gamma bar of omega equal to 1 over k Boltzmann t, integral 0 to infinity dt h of t0 t0 e to the i omega t. That looks like a fluctuation. It looks like one of these theorems. It says the one-sided Fourier transform of some correlator is some, this time not the mobility but the memory kernel itself on this side. But you can say now look that is not quite, it is not simple enough because this still involves the velocity. It does not involve the noise Eta alone. This h is not a stationary noise. That is why you need both arguments here. But if you use the properties of this correlator which you derived from this equation, then one can show that this quantity is exactly equal to Eta of t0, Eta of t0 plus t. No, it is equal to it. It is equal, period. This portion is exactly equal to this. So you end up with this relation which says gamma bar of omega equal to 1 over mk Boltzmann t, integral 0 to infinity dt e to the i omega t, Eta of t0 Eta of t0 plus t. This is the second fluctuation theorem. Just for comparison, let us write the first one down. The first one was mu of omega equal to 1 over m, sorry, 1 over k t, integral 0 to infinity dt e to the i omega t, the output process, v of t0. So the dynamic mobility which measures the response of the system to an external force in linear response theory is this one sided Fourier transform of this correlator, the output process. On the other hand, this is specific to the Langevin model. This was general in linear response theory because in the Langevin model, we explicitly find a model for the velocity and solve the equation of motion, random equation. It says that the correlation of the force, stationary force is not arbitrary but it must be related to the friction kernel, the memory kernel in this fashion. This is the equivalent of the fluctuation dissipation theorem, the capital gamma, etc. Because how do you recover that? By saying that this memory kernel is just a delta function, then this gamma bar, this gamma of t minus t prime is just delta of t minus t prime times a constant gamma. This fellow would become gamma out here. This would be a delta function but half a delta function because you are integrating 0 to infinity. So you would get half here and capital gamma on top and you would get capital gamma is 2m gamma little k k Boltzmann t, which would give you the original theorem back again. But this is the general version of it. And what is the big lesson it is telling us? It is saying that nice in this generalized model, this eta here is stationary fine but it cannot be delta correlated because if it is delta correlated, there can be no omega dependence. And then this side is omega dependent. So it is not consistent. So this is immediately telling you that the moment you introduce a memory kernel, the noise cannot consistently be white noise. It has got to be colored noise. There is a correlation time. And what is that correlation time? You define the correlation time by putting omega equal to 0 here, right? And then dividing by the mean square value or whatever it is. So it gives you a quantity of dimensions time. And that in this case is just the memory kernel. So little gamma of t, little gamma of t dt from 0 to infinity is going to specify for your correlation time. Yes, yes, exactly. Yeah, it says in this generalized model you cannot independently, just as in the original model you cannot independently specify little gamma, the friction constant and the strength of the white noise. You cannot independently do it. In exactly the same way the more general statement is you cannot specify the memory kernel and the random forces correlation independently. They are related in this fashion. It is a consistency check on the model. And when that is satisfied, so you now in model building you do what you like. You tell me a random force which is stationary. I then say that it is consistent with the Langema equation with a specific memory kernel, not any old memory kernel. On the other hand in modeling empirically if you discover that this friction is modeled by a memory kernel which is maybe an exponential or some function of time decreasing function, that fixes for you the correlation of the noise. The manner in which this correlation behaves, it fixes for you through these relationships here. And that is what is more generally done. It depends on the model building that you have in mind. So you could say alright this thing here is exponentially correlated with some correlation time tau, single exponentially correlated. That is the first thing you do. It is a Markov process. We will assume it is Gaussian. We will assume it is stationary. And now you say it is not delta correlated but exponentially correlated, maybe an on-steen or unbacked process, maybe. Then it has a correlation which is exponential in time that fixes for you the memory kernel. It fixes gamma bar of omega and from which you can find what sort of gamma of t you should have had in order to have this. So that is an interesting exercise, simple exercise. Take this to be an exponentially correlated, e to the minus t over tau. Then figure out what is this guy going to be. And therefore what is gamma of t going to be? That would be the simplest model in this instance. So this kind of brings us to an end to this part of the program. We will, there are some loose ends to be tied up. I will mention them to the extent that I can remember. We will talk about them tomorrow and then go on from there.