 So, we had got to the stage last time where I had mentioned that if you took a more realistic equation of motion for the Brownian particle namely the equation of motion m v dot equal to minus m gamma v plus 8 of t where this was noise then we might now get a more physical result for the mean and the mean square which is as follows. You solve this equation this implies of course that v of t is v not e to the minus gamma t plus 1 over m an integral from 0 to t dt prime 8 of t prime or let us write it as t 1. So, that is the formal solution to this first order differential equation with the initial condition that v of 0 is v not some prescribed v not and then we took the average. So v of t average was v not e to the minus gamma t which was very good because it says that if you wait for long enough after the initial instant then the mean value does indeed the conditional mean does indeed vanish as the equilibrium mean already we know vanishes okay. The more important test was what happens to the mean square so we computed v square of t and that was v not squared e to the minus 2 gamma t and let us just compute the average value v square plus the cross term vanishes because the average of eta is 0 plus 1 over m square an integral from 0 to t dt 1 0 to t dt 2 and then the average of eta of t 1 eta of t 2. Now this quantity we assume was delta correlated because the time scale on which its correlation decays is much much smaller than the time scale on which this object is moving. By the way there is a dissipation time scale already introduced into the problem by introducing the constant gamma, gamma inverse is the physical dimensions of time. So it is roughly representing already we can see from the structure of this equation that it is representing some kind of dissipative time scale the time scale on which the motion of this particle relaxes but we are going to make that concrete. So if I put in the fact that eta of t 1 eta of t 2 average the same as the equilibrium average of this quantity and this is equal to some gamma delta of t 1 minus t 2 if I put that in here and do this integration then you immediately see that v square of t is equal to v0 square e to the minus 2 gamma t plus 1 over gamma over m square and then I do the delta function integration here and I just set t 1 equal to t 2 except that this is the wrong solution this is e to the minus gamma t minus t 1 you have the integrating factor here because of this term in the differential equation here. So that sits in here times e to the minus gamma this multiplied by e to the minus gamma t minus t 1 minus gamma t minus t 2 that is sitting in here. So if I keep track of that this is gamma over m square and integral from 0 to t dt dt 1 e to the minus 2 gamma t minus t 1 because the delta function fires and puts t 1 t 2 equal to t 1 and this factor gives you a factor 2 here yeah yeah I am going to come back to it. So I am saying that the moment I put in a friction like this a term like this here then we see this this is dv over dt so this immediately means that this gamma inverse has the physical dimensions of time and this is a damped oscillator in the oscillator case if you did this for x except here it is for v here you are already familiar from the equation of motion for a damped simple harmonic oscillator that a first order term will represent some kind of dissipation the coefficient here will represent some relaxation time we are going to make that concrete here okay. So this is what the equation the solution is out here and all we have to do is to do this integral so this is v naught squared e to the minus 2 gamma t plus gamma over m square and there is a 2 m square gamma because this factor comes down and then you have e to the minus 2 gamma t times e to the 2 gamma t minus 1 and if I put this take this factor in then it is 1 minus e to the minus 2 gamma t or if I combine the t dependences here this is equal to v naught squared minus gamma over 2 m square gamma little gamma e to the minus 2 gamma t plus a constant so that is the value of the conditional mean mean square of the velocity of this particle which starts out with the deterministic velocity v naught at t equal to 0 and this is what happens to the average velocity as t for any arbitrary t greater than 0 okay. Now two things can be done one of them is to say this particle this Brownian particle which started off with some given fixed initial condition which I imposed on it so what I did was to pick out that particle or that set of particles which had initial velocity v 0 and I am following the path of one of these particles the subsequent history and I average over all such particles and this is the answer here. Now one possibility is to say if t becomes very large compared to 0 much much larger than 0 t tends to infinity then this term drops out and this goes to a constant and the expectation would be it should go to that constant which is what the equilibrium value of the mean square velocity is. So if you expect if v square of t tends as t tends to infinity to v square in equilibrium the whole thing is in equilibrium in any case here then if you let t go to infinity it is clear that this portion goes away and you get this answer here right. So this must be equal this would imply then gamma over 2 m square gamma must be equal to the equilibrium value of this mean square value thing in one dimension which is k t over m from the Maxwell distribution of velocities. So I would expect this must be equal to k Boltzmann t over m which would imply that capital gamma is 2 m gamma k Boltzmann t. So it gives you a relation between capital gamma which is the strength of this white noise if you like the delta correlated noise because we said this is some gamma times delta of t 1 minus t 2 on the one hand. So this eta is driving fluctuations into the velocity of this Brownian particle and this gamma measures the strength of these fluctuations in some sense okay. That gets related to this constant gamma which represents the systematic part of the random force on this particle here this part is also random because v is a random variable and that systematic part has a constant m gamma sitting here and the two are related to each other okay and you can see for instance that this v square of t is relaxing with this time constant here. So it is related to gamma the gamma inverse is related to the relaxation time of this average to whatever value it attains in equilibrium. So there is a relation between the fluctuations on the one hand and the relaxation on the other hand or the dissipation and this is called a fluctuation dissipation relationship. It is the simplest of the fluctuation dissipation relationships we will come across several more. It is sometimes called a fluctuation dissipation theorem but we will see in what sense this yes. Sir when you ensemble average on the left hand side the v square of t is tending towards the full average the right hand side you are assuming kind of like all these t. I am assuming that the system remains in equilibrium yes definitely I am assuming that the system remains in equilibrium and there is a unique equilibrium state which is what I started with and I have not done anything to the object and system. I have not perturbed it in any way it is just a beaker containing some colloidal particles beaker of water and there it is a whole thing is in thermal equilibrium. It is just that I started by looking at a sub ensemble of particles with some given initial velocity v naught and I am asking what is the history of these guys as time goes on and since the system remains in equilibrium throughout this conditional average must tend to the equilibrium average. But if you do not like that argument this is an even better argument which is let us average this fellow over v naught. So let us find so that is the alternative or let us take v square of t and find this quantity which by definition is equal to the integral over rho equilibrium of v naught v square of t conditional average d v naught from minus infinity to infinity by definition now I go from the conditional average to the full average. So I complete the procedure by integrating over all possible initial conditions v naught remember this is a function of v naught here. So all I have to do is to take this fellow and use the Maxwellian distribution of velocities for v naught and average and do the average here. So this will then give you automatically there is a v naught sitting here and that gets average and the answer is k t over m. So this is equal to k Boltzmann t over m minus gamma over 2 m square gamma e to the minus 2 gamma t plus gamma over 2 m square gamma. But this cannot depend on time because the system is in equilibrium you have done a full average it has to be the Maxwellian distribution once again. So you are compelled to say that this must be equal to k t over m there is no choice there is only one way in which that can happen in which this time dependence vanishes and this becomes k t over m and that is if this condition is satisfied. So you have no choice consistency demands the only consistent solution is that the capital gamma the constant capital gamma is related to the little gamma by this relationship there is no alternative okay that the noise is thermal yes absolutely. We are looking at a problem in which the noise is entirely thermal fluctuations yeah no other external source experimental error none of those things we are just looking at a system in which the temperature the fact that the system is at finite temperature is what is driving fluctuations into the system absolutely no quantum fluctuations for example here at all we will we will at some stage if time permits talk about the quantum Langevin equation and see what happens there but this is at this very elementary level of just thermal fluctuations mind you we have not really addressed anything which is truly non equilibrium because we are still talking about a system in thermal equilibrium but we are asking some time dependent questions of this system here I am saying one time simple time independent if you make a measurement at some instant of yes of your time particle yes how that is going to how it is going to reach the equilibrium phase you get hold of one particle it is moving with velocity v naught that of course is drawn from the Maxwellian distribution some number v naught but now I follow that particle and ask what happens to it this velocity of this particle as time goes on well it merges into the distribution and the way it does it is by this relaxes in this fashion okay. So now this is completely consistent what we found is that this this will automatically imply that v of t full average is equal to 0 because I average over the Maxwellian in v naught by symmetry this vanishes this is an odd function and the mean square comes out to be the right answer here provided this consistency condition is met and now what we are saying is I take one particular particle any particle they are all equivalent to each other I take one of these particles and I discover that its velocity initially is some number v naught I make this measurement and then I ask what happens to it as I follow it and so on well it gets hit by the other particles etc etc and if I take an ensemble average over all these objects and wait long enough you are going to get the equilibrium values for the average so this is all I am saying or better still I do a second average over the initial conditions and I get the same equilibrium that is the meaning of an equilibrium average in any case right. So this is now as you can see you can appreciate now the fact that this is a consistency condition otherwise you run into inconsistencies earlier we ran into a very bad inconsistency which said that the mean square value of the velocity is spontaneously increased linearly with time we fix that problem by saying no no there must be some kind of damping term here systematic portion which on physical argument to say there must be a term of this kind but the moment you put in a term of this kind that problem that this runs away as a function of time does not happen anymore it tends to a constant but then what constant can it tend to consistency demands that it tend to the equilibrium constant value right. So I like to say this is a very important relationship it is a consistency condition and it is called the fluctuation dissipation we will come across many of these such relationships yeah. The spectrum the spectrum of the noise as it acts on this part it is not some absolute yeah you could have in principle done that but that is a good nice question why should it is why should the strength of this noise depend on on the particular particle if I change that particle I get another value of the strength of this noise good question but it is not the strength of noise on what strength of what noise the strength of the noise on that particle right. So it is not surprising that it has some dependence on that particle I change the mass of the particle I get a different strength of course because this is a strength of the effective noise due to all the other molecules as it acts on that large particle. So there must be some dependence on that particle and that is this case but we will see some similar magic is going to happen very soon when we talk about the diffusion coefficient you go back to the original equation right and divide everything by n yeah and then you define it at tilde which is no yeah in principle you could do that yeah in that case but it is it is making the question it is essentially the same thing it is finally the same you the physical answer is simply that this is the force as it acts on that particular particle right okay but having done this now let us go on and ask what happens to the velocity correlation time that is crucial. So let us see at the level of the mean square we are all happy but let us see what happens if I did this I want to now compute I would like to now compute what is the average value of v of t v of t prime where t prime is not the same as t in general and let us find the average value of this quantity here eventually I want to take the full average but let us find the average value of the conditional average first this is v0 squared e to the minus gamma t plus t prime that is the first term the cross term vanishes as before but then plus gamma over m squared integral 0 to t d t 1 0 to t d t 2 e to the minus e to the minus gamma times t minus t 1 minus gamma times t minus t 2 t t prime minus t 2 yes because it is a t prime here and then a delta function of t 1 minus t 2 but this time I cannot close my eyes and write t 2 equal to t 1 inside the integration because the limits of integration over these two this is t prime the limits of integration are different for the two integrals right. So if I draw a little picture here is t 1 and here is t 2 I am integrating t 1 up to the point t but t 2 only up to the point t prime in the case when t is greater than t prime if t is less than t prime the rectangle is height is greater than the width. So without loss of generality let us take t prime to be less than t the other cases trivial by interchanging t and t prime in the answer and what is the delta function constraint it says t 1 is equal to t 2 which is the 45 degree line. So as you integrate for a fixed value of t 1 as you integrate t 2 the delta function fires as long as t is as long as t 1 is restricted to the range 0 to t prime the moment it goes outside there is nothing to integrate the delta function does not fire because t prime t 2 is cut off at t prime okay. So this means that you can replace t 2 by t 1 but the t 1 integration is now restricted to the range 0 to t prime the lesser of the 2 right. So this is immediately equal to v naught squared e to the minus gamma t plus t prime plus gamma over m squared integral 0 to t prime smaller of the 2 and then this goes through. So this is e to the minus gamma t plus t let us take that out of integral e to the minus t plus t prime and integral 0 to t prime d t 1 e to the power gamma t 1 gamma t 1 2 gamma t 1 I do the integration once again 2 m squared gamma comes down and then it is e to the 2 gamma t prime minus 1 that is the integration right. So as you can see the first term is going to be e to the minus gamma t and then e to the plus and then the plus t prime because there is a plus twice here and the second term is going to be e to the minus just this fellow here same thing. So therefore v of t v of t prime average is equal to v naught squared minus take this guy here by the way okay let us write it in 2 steps gamma over 2 m squared gamma e to the minus gamma t plus t prime that takes care of the second term and the first term is plus gamma over 2 m squared gamma e to the minus gamma t minus t that is this portion but we already know that capital gamma over 2 m squared gamma must be equal to kt over m that was our consistency condition. So we have to put that in and if you do that this immediately gives you v naught squared minus k Boltzmann t over m e to the minus gamma t plus t prime plus k Boltzmann t over m e to the minus gamma t minus. Now what do you expect once again if I argue that I let t become infinite t prime become infinite such that the difference t minus t prime is finite any finite value this term goes away and that survives. On the other hand if I do an average over the initial value of the velocity average over v naught squared then the average of v naught squared is in fact kt over m and this cancels and there is nothing to average there because v naught this term is independent of v naught here. So whichever way you look at it it gives you an answer for the correlation time. So now we immediately follow v of t v of t prime is equal to this therefore is equal to k Boltzmann t over m e to the minus gamma and we did this for t greater than t prime had we done this for t less than t prime t and t t and t prime just have been interchanged. So this is simply replaced by the modulus which is a very important result. So now we understand finally what I meant by the relaxation time it says the velocity correlation time is gamma inverse because it relaxes to 0 this thing becomes 0 if t minus t prime tends to infinity it gets decorrelated starting from if t equal to t prime you set this equal to 0 the e to the 0 is 1 it is kt over m which is what you expect. So this quantity here as a function of t minus t prime has this following behavior it starts at this kt over m which is the mean square value and it decays exponentially to 0 with the characteristic time scale gamma inverse. Of course since we have a modulus here we can also ask what happens for t minus t prime less than 0 and it just has this it is symmetric function but that is a very important result it is saying something very crucial about this velocity process which we have not talked about yet about its distribution we will discover later that this velocity process under the assumption that this 8 of t is a Gaussian white noise namely a noise which is got equal power in all frequencies delta correlated Gaussian Markov process stationary etc then under those conditions this equation actually implies that the driven process v the velocity process remains Gaussian in nature it is Markov it continues to be Markov it is stationary but it is not delta correlated it is exponentially correlated. So the driving noise is delta correlated it is no correlation at unequal times it is just correlation function vanishes but the driven process is much slower than that in the sense that it retains some memory of where it was by this formula here which tells you exactly how it decorrelates as time increases okay that is a crucial lesson we will see that this v is called is satisfies its probability distribution satisfies that of an Onstein-Ohlenbeck process but it is the very very paradigm of a Gaussian stationary Markov process okay but I have not said any of those things yet I am just using those words I have not established any of those things yet but we will do so okay. In fact the only Gaussian stationary Markov process that is exponentially correlated is called the Onstein-Ohlenbeck process there is no other such process at all so it is sort of fundamental process random process it is kind continuous process yeah I should also add that it is the v is a continuous process because this is continuous noise and continuous not a jump process. So the only continuous stationary Gaussian Markov process which is exponentially correlated is the Onstein-Ohlenbeck process okay. So we have some powerful results here let us put in some numbers and see whether this whole thing makes any sense or not okay. Now I said that these particles are much bigger than the molecules themselves okay. So let us assume they are spherical particles the simplest assumption they are like pollen grains or rather the original experiments were done on Brownian motion by Robert Brown around 1827 or something like that and he actually used crushed pollen grains. So if you take pollen grains and you crush it fine to a fine powder then you get micron sized particles of some kind and then they float around in water they do a jiggly dance called Brownian motion and they are described by this model very adequately. Let us see if that makes any sense what the time scale this gamma is going to turn out to be like well if you put this in water and let us assume that the radius of a particle is of the order of a micron so 10 to the minus 6 meters and let us assume that its mass is the same density as water so we can compute what its mass is water is a density of 10 to the 3 kilograms per meter cube I find this standard international units rather inconvenient to use for such small objects but anyway so the mass of this particle is of the order of 10 to the 3 kilograms per meter cube so that is the density multiplied by the volume which is a cube so times 10 to the minus 18 is of the order of 10 to the minus 12 kilograms. Similarly it is even smaller 10 to the minus 15 which is okay now we need to find out what this gamma is right since these are particles which are suspended in water and they are floating around we said the effect of gravity is essentially nullified by the viscosity they are at terminal velocity in that sense so we have Stokes formula for the terminal velocity of a spherical particle dropping under gravity and whose gravity the force of gravity is exactly balanced by the viscosity viscous drag so what is that formula yeah so it says M gamma is equal to 6 pi a times eta by the way this is viscosity it is not that eta okay sorry for this notation it is not this noise here it is the viscosity of water the dynamic viscosity not the kinematic viscosity which is this divided by the density okay. So that tells you that gamma equal to 6 pi a eta over M which is equal to which is of the order of 6 pi is of the order of 10 and then a was 10 to the minus 6 divided by 10 to the minus 15 right and then eta what is the viscosity of water like Newton's law of viscosity what one normally uses and viscosity is measured in the CGS unit is but this standard international unit is Pascal second Newton second per meter squared okay and for water at room temperature it is easy to remember at 20 degree Celsius or something it is of the scale gamma is of the order of 10 to the minus 6 or 10 to the minus 7 seconds microsecond so you see we started out with something which had motion on the order of picoseconds the molecules themselves the atoms and all of a sudden out has come this time scale which is much larger and it is of the order of microseconds. So that is the time scale on which this Brownian motion particles velocity memory the velocity will be lost many orders of magnitude greater than the molecular motion time scales okay. So that is one of the reasons why I said this theory does not apply to individual molecules themselves several reasons why that it fails at that level starting with correlated motion but this is now in the right ballpark at least in the right ballpark okay. So what next well we could ask we did this in one dimension what happens in three dimensions that is something we can straight away take care of so let us do that so in 3D what we need to do is to write v dot of t equal to minus gamma v of t plus 1 over m 8 of t vector of t. Now all we need to ask is what are the statistical properties of this vector noise here well we resolve it into Cartesian components in the fluid is a homogeneous fluid isotropic fluid in all directions so it is very reasonable that each Cartesian component satisfies exactly the same equation with a different noise but essentially the same statistical properties right. So now what you do is to again assume that eta i of t average equal to 0 i equal to 1 2 or 3 for the Cartesian components and eta i of t eta j of t prime the correlation in these two fellows equal to a gamma times a delta of t minus t prime but the different Cartesian components are completely uncorrelated with each other so all you need to do is to put the delta i j the chronicle delta and you go through exactly the same thing as before you will get precisely the same results nothing will change what will happen to the velocity correlation time well now I have to write v i v j and there is going to be delta i j sitting out there. So this will imply v i of t v j of t prime equal to delta i j k Boltzmann t over m e to the minus gamma that is all that is going to happen if the different Cartesian components were correlated with each other for some reason then this would no longer be true at all they are completely independent so mind you we are talking about Cartesian components okay if I were talking about more complicated objects like what is happening what is the kind of stochastic equation satisfied by the magnitude of the velocity of the speed that is a more difficult problem we will say a little bit of it when we about it when we come to Brownian motion but this is what this thing does so or if you like write it in simpler language v of t dot product v j of v of t prime equal to k Boltzmann t over m e to the minus gamma times what we have to take the trace of this tensor so it is and what is this equal to equal to zero unequal components are uncorrelated right can you think of a situation where the different velocity components get correlated I mean that has to show up here itself something that says the acceleration in the direction one is related to the velocity in direction two if I put a magnetic field that is going to happen there is going to be immediately the Lorentz force on it and this part of it will depend now on the other components here and now you can see what will happen is that there is going to be some correlation between the different components of the velocity and you have to solve this problem all over again so however certain things will not change certain things won't change at all because the magnetic field does not change the energy of a particle just causes the part to curve around it does not do any work on this particle at all it does not change the energy so the interesting thing that will happen is that while diffusion in the direction perpendicular to the magnetic field will be inhibited simply because system starts moving in circles before it is disturbed by thermal noise certain other things won't change like the velocity correlation this this relation here and so on we will see what happens to it okay all right now since I talked about displacement let us compute that too we need to do that as well again let us for simplicity go back to a one-dimensional case well I can always compute the displacement by simply using the fact that x of t minus x of 0 is by definition integral 0 to t dt1 v of t that is it so the statistical properties of this displacement from whatever initial this value positive the position had are given by the statistical properties of this integral where this is a random variable here and we know everything about this at least we know reasonable amount about it such as its equilibrium correlation function so let us try to compute what happens to this displacement as a function of t so let me use another symbol let us call x of t equal to little x of t minus x not wherever it was this is the definition the displacement okay x is the position and that is the displacement right and then let us compute x squared of t and take its average so this is equal to integral 0 to t dt1 integral 0 to t dt2 times the average of v of t1 v of t2 well I am going to go straight for the full average I could do the conditional average first write down the formulas and then do this but we are interested in this thing here full average we need to compute this quantity some simplification can be done directly immediately because this quantity we already know its value we already know that this is equal to k Boltzmann t over m e to the minus gamma modulus t1 minus t2 we already know that so all we need to do is to put this in and do this integration and the integration is over a symmetric region 0 to t it is a square so if you go back here here is t t1 s t2 and integration is over this over this square so we are integrating over this full square but this is a symmetric function under the interchange of t1 and t2 which means if I reflect about the 45 degree line then the integrants value does not change in other words here is the 45 degree line the value of the integrant here is the same as the value at this point the value at this point is the same as the value reflected at that point and so on now in this triangle t1 is bigger than t2 and in this triangle t2 is bigger than t1 it does not matter which triangle I choose so this thing here is equal to twice the integral from 0 to t d t1 from 0 to t1 so I integrate here I am going to choose this integral I start with the value of t between t1 between 0 and t I charge this value but t2 is integrated only up to that point and I am going to scan this in this fashion so it is 0 to t1 I stop at t1 for the integration over t2 dt2 k Boltzmann t over m e to the minus gamma t1 minus t2 I can get rid of the modulus sign because t1 is bigger than t2 this integration what is this equal to this is 2 k Boltzmann t over m integral 0 to t dt1 e to the minus gamma t1 times e to the gamma t2 from 0 to t1 right that is equal to e to the gamma t1 minus 1 over gamma and integrating that that is a trivial integral so this is equal to so it says that the displacement x square of t is equal to 2 k t over m gamma times this integral is the integral of 1 which is just t itself t and then the integral of e to the minus gamma t1 which is equal to plus so it is equal to minus 1 over minus gamma e to the minus gamma t minus 1 which is equal to 2 k Boltzmann t over m gamma square and then inside you have gamma t minus 1 plus e to the minus gamma t3 minus is minus 1 yeah so that is the answer I have put the taken the gamma out so that this thing becomes dimensionless everything inside this dimensionless that is the mean square displacement and you can see this got the physical dimension which are right this is got to be the square of a length and indeed that is energy divided by mass that is t to the minus 2 so it is just square of a length so that is the exact expression now notice that as t becomes very small now we can ask what do I mean by small time and large time there is a time scale in the problem which is gamma inverse so when I say small time I mean t much much less than gamma inverse and large time is t much greater than gamma inverse or gamma t turning to infinity now what happens is that there is some slope and this term dominates at very long times and at very short times if I expand this the 1 cancels the gamma t cancels it starts with gamma squared t squared in fact what does it do this goes as gamma t tends to 0 this fellow tends to there is a gamma squared t squared so that cancels to the two cancels so k Boltzmann t over m times t squared but remember that this fellow here is v squared of b squared equilibrium times v squared so it is as if this Brownian particle is undergoing ballistic motion but with a velocity which is the root mean square velocity but more significant here is what happens at very long times so at very long times gamma t much much greater than unity this goes to two k Boltzmann t over m gamma times t it grows linearly with time so if I plot it this fellow is going to do something like this and the asymptotic slope is twice this constant and that is called the diffusion constant because as you know in diffusive processes the mean square displacement goes linearly with time twice dt and that coefficient d in each dimension is the diffusion coefficient in three dimensions r squared will go like six times this coefficient so we have actually ended up with a formula for the diffusion coefficient and if you put m gamma equal to 6 pi a eta then you have a relation which says this diffusion coefficient is k Boltzmann t over m gamma equal to k Boltzmann t over 6 pi a eta this is the viscosity so independent of the Brownian particle there is nothing in the last formula which has any reference to the Brownian particle at all you are finding the diffusion coefficient of this particle this was Einstein's great achievement this formula is due to Einstein and of course it uses the Stokes relation so it is called the Stokes Einstein formula and Einstein brilliantly derived this by very different argument all together this part of his thesis in the famous papers in 1905 but coincidentally a couple of months earlier this physicist called Sutherland from Australia also derived the same thing published it so I should really call it Stokes Sutherland Einstein relation but it is a very famous one and it helped Einstein compute now the size of a molecule the size of a Brownian particle from here in fact what I did not mention was that essentially it tells you what is the value of Avogadro's number which I will mention how that comes about little later but this was a fundamental relationship which was found by different argument but we see how this images at this stage I will stop here today and next time we will take this up a little deeper we will go a little further into whether this model can be generalized where you can look at more complicated systems and how they relax and what is the relation between the relaxation and the response of the function so let me stop here today.