 So, let us continue with our study of the Fokker Planck equation and for the next lecture or two we are going to talk about the physical model of particle in a fluid undergoing random kicks due to the molecular collisions and at the same time possibly be subject to some external force but let us look at the case first where you just have a free diffusion of a particle in a fluid. I would not specify at the moment how big this particle has to be. We will come a little later and distinguish between different timescales and we will see under what conditions this whole thing applies but right now we take a very naive approach and say alright suppose I have a small particle inside a fluid and this is undergoing collisions random collisions due to the agitation thermal agitation of the molecules of the fluid what kind of equation of motion can we write for this system here. Recall that we made a correspondence we said we stated that there was a correspondence between certain kinds of stochastic differential equations describing diffusion processes and the corresponding Fokker Planck equation. So, what I am about to do now is a physical example of that in a very simple case in which the drift term will be a linear linear in the variable. So let us look at this little bit of detail again to refresh your memory I said that if you had a stochastic equation in some random variable X I use the symbol X but now for this application I am going to use X for the Cartesian coordinate of the particle. So, let us be neutral and call it Xi or something like that. So, if you have a stochastic equation of the form Xi dot equal to some f of Xi plus g of Xi times a white noise and this was white noise Gaussian white noise in fact satisfying eta equal to 0 and the autocorrelation T eta of T prime equal to delta of T minus T prime under those conditions this was equivalent to a Fokker Planck equation for the conditional density of this Xi with some initial condition of the form delta over delta T P of Xi T for a given Xi not say this is equal to minus delta over delta Xi f of Xi times P plus one half delta 2 over delta Xi 2 g square of Xi times P this was the Fokker Planck equation whereas that was the stochastic differential equation for this random variables I now let us look at some the simplest example we even wrote this down I said if the position of a particle undergoing diffusion on a line satisfies X dot equal to square root of 2 D times eta of T this immediately implied that P of X, T with some initial point X not satisfies the diffusion equation delta P over delta T equal to D D 2 P over DX 2 that was the correspondence we had made but now let us be a little more detailed and ask look whatever force the particle is subject to is going to cause an acceleration. So let us say that the equation of motion that you should write down for this particle moving in one dimension or one Cartesian coordinate of it let us call the velocity V and we should really say well MB dot that is the acceleration should be equal to whatever force it is subject to and in the simplest instance you would say this force is a completely random force it is due to all these molecular collisions I do not know anything about it. So in the simplest instance you would say this is equal to eta of T itself in this fashion but of course dimensional reasons and as well as the fact that this eta of T need not have unit delta function strength but some arbitrary number. So let us call it equal to square root of some gamma times eta of T where this is some constant which we may or may not be able to determine in a self consistent way to start with okay. Now what does this imply this immediately implies a Fokker-Planck equation for the quantity of interest namely the velocity distribution function density function but instead of writing this down let us take this stochastic equation seriously and ask whether it makes any physical sense or not before we do this. Now this will of course mean V dot is this and you can formally solve this equation this is a stochastic differential equation but it is very very simple and we can formally in principle we can solve it. So this will of course immediately imply that V of T equal to some initial condition whatever it be some initial value V naught plus square root of gamma over M integral from 0 to T dt prime eta of T prime. Now remember our physical context we have a fluid we imagine a fluid in which we are looking at one Cartesian component of the velocity of some tagged particle and this fluid is taken to be a thermal equilibrium at some temperature T okay. Then it says the velocity is equal to this the instantaneous velocity but what we are interested in is averages always so what is the average value of V of T now when I say averages I got to be a little careful average over what ensemble we have already specified an initial condition. So it is an average over all those particles whose initial value of the velocity is given to be some number V0 right it is not an average over all possible initial velocities as well that will come a little later. So we have 2 classes of averages one is over a sub collection of particles whose initial velocity is V naught and then we say look let us average over V naught as well over some initial distribution or since I said already that the fluid is in thermal equilibrium at temperature T over say the Maxwellian distribution at temperature T that is what I should do really. So to distinguish between these 2 things these 2 kinds of averages let me put an overhead bar to denote averages over a given initial condition V naught right and then a subsequent average over these V naughts will give me a final average for which I will use angular brackets. So this will immediately imply that V of T bar is equal to well average of V naught but V naught is a deterministic given number of course V naught plus the average of this integral but the integral is essentially a summation over different values of 8 of T prime. So since 2 different sums commute in either order this overhead bar is the same as putting it inside the integral right. So this is becomes plus square root of gamma over M integral 0 to T T T prime 8 of T prime an average over all realizations of this 8 of T prime. Now the physical assumption is that you are looking at one particle that is with a small number of degrees of freedom inside a huge collection of particles at in thermal equilibrium and the heat bath which is providing the fluctuations on the of the velocity of this particle that is not going to be affected by what this particle does. So whether I fix the initial velocity of the particle or not is not going to affect the average value of the random force at all. So as far as eta is concerned whether I put a bar or take a full average it does not matter at all we have already assumed that it is got 0 average the Gaussian white noise. So this quantity is actually 0 this integral and this becomes equal to V0 since the average value of this random force is 0 in any case. So so far so good it just says the average value remains this which is physically expected you are saying that the force can be as much acts as much to the right as to the left on the average the velocity component is 0 the average is 0 it remains whatever it was initially. But what does this tell you what is V squared of T if you compute the square of the velocity then I have to find the square of this guy and find the average. So it is evident that there is a term which is V0 squared and then there is twice V0 times this and then I want to take an average. So let us put that average in there there is nothing to average here and then I have V0 times eta of T prime averaged but V0 comes out of the averaging and it is just the average of eta of T prime again. So the cross terms average again goes to 0 but then I have a term which is plus gamma over m squared when I square this term here and then I have to take 0 to T dt let us call it dt 1 so that I do not mess around with primes and then again 0 to T dt 2 eta of T 1 eta of T 2 with an average out there but that quantity is not the product of averages because it has a random variable with the delta correlation this guy. So I have to put that delta correlation there and then it says that V squared of T average is V0 squared plus gamma over m squared an integral over this with delta function did I put the yes a delta of T 1 minus T 2 in here okay and that contributes as long as T 1 equal to T 2 and that always contributes for all values of T 1 between 0 and T that is easy to see because if you draw a little picture here is T 1 here is T 2 and in each case you are going to integrate from 0 to T in this fashion and the delta function constraint tells you T 1 equal to T 2 out there. So it says if you do the T 2 integration first which is the way I have written it here then it is clear that no matter what T 1 I have between 0 and T there is a value of T 2 as you scan this at which the delta function fires. So I can therefore remove the T 2 integral and replace wherever T 2 appears I replace it with T 1 okay and that gives you 0 to T d T 1 and this integral is gone the T 2 integration is gone which therefore gives V naught squared plus gamma T over n squared. So it gives us this rather unphysical result which says that the square of the average value of the square of the velocity increases without bound as T increases okay if you identify the temperature with half m B squared average this means that if you leave this particle untouched you keep a beaker of fluid then the average kinetic energy of any particle in there increases without bound okay the effective temperature increases without bound. So it is completely unphysical complete so this cannot be right this model cannot be right this equation cannot be right because there is nothing else that has gone can go wrong here okay you might say oh this perhaps this is unphysical perhaps this is not correct I should not use a delta correlation I should use an exponentially decaying correlation with some finite correlation time but even if you did that you would still get an unphysical answer and I leave you to check this out even if I took this quantity to be some e to the minus T over tau for some very small value of tau and computed what this number is explicitly you would still get an unpleasant answer here it would still be unphysical and this is not right. So the only thing that can possibly be incorrect would be the initial model itself this model cannot be right okay and what have I left out of it I have left out the fact that there is a systematic component in the random force this force it has completely random uncorrelated and so on that is fine but however if this tag particle starts moving in one direction at a velocity higher than the average velocity it gets hit back by the sister friction in the problem by the viscosity the very same molecules that cause fluctuations in its velocity will also damp out these fluctuations by having more collisions from the front than from the back if you are moving in this direction okay. So this means that I have to modify this model and this is not correct let us see what the correct model is the correct model would be mb dot equal to there is an eta of t so there is a square root of gamma times eta of t that part certainly exists that was the original model but there is a portion which says there is a viscous damping and for small velocities Newton's law of viscosity tells you the viscous drag is proportional to the force but with an opposite sign right. So this is equal to minus m gamma v plus this but gamma is a quantity which has dimensions of 1 over time so that it matches this on this side it is a friction coefficient I took out the m explicitly because it is easier that way. So this is my model this is by the way the simplest example of what is called a Langevin equation but this is again Gaussian white noise but there is a systematic component to the random force a model again it is a model and we have to see whether it makes any physical sense or not. Now look at what is going to happen we repeat exactly what we did before and I divide through by m then average v of t equal to the solution now by the way we can write down the solution first let us do that so the solution is v of t equal to v not but that is multiplied by e to the minus gamma t because there is this term here plus square root of gamma over m integral from 0 to t dt prime e to the minus gamma t minus t prime a eta of t prime because this is of the form dy over dx plus p of x times y equal to q of x and you have the standard formula for solving the first order differential equation of that kind and inhomogeneous equation and this gives you the explicit solution that is already telling us we are on the right track because it immediately tells us that v of t average is v not e to the minus gamma t because whether this factors present or not when you take the average of eta it is again 0 so you have this feature here already getting us on the right track because it says that if t becomes very large compared to gamma inverse this says the average velocity goes to 0 which is exactly what you would expect there is no reason why the memory of the initial condition should remain forever when you have viscosity in the system it is going to be damped out so this is a good feature that it exponentially vanishes as t increases and what does the square do it does exactly what happened earlier except for that extra factor so this is now v not squared e to the minus twice gamma t because that factor remains plus gamma over m squared once again integral 0 to t dt 1 0 to t dt 2 e to the minus gamma t minus t 1 e to the minus gamma t minus t 2 and then an eta of t 1 eta of t 2 whose correlation average value is delta of t 1 minus t 2 exactly the same way as before we can now do the t 2 integration and replace t 2 with t 1 everywhere so this integral is easy to do it is v not squared e to the minus 2 gamma t plus gamma over m squared integral 0 to t dt 1 and then there was already an e to the minus 2 gamma t sitting here and then you had e to the gamma t 1 e to the gamma t 2 but now t 2 is equal to t 1 right so e to the 2 gamma t 1 and that is the integral e to the 2 gamma t comes out of the integration and then you have to do this fellow here so this is v not squared e to the minus 2 gamma t plus gamma over 2 m squared gamma because when I do this integration that factor comes out downstairs and then e to the 2 gamma t 1 from 0 to t here the first term will cancel out and give you a 1 and the second one gives you e to the minus 2 gamma because this integration is e to the 2 gamma t 1 minus gamma t minus 1 so that is the result out here and this has no exponential blow up this linear blow up this does not increase unboundedly as t goes along because this exponential factors cancel and go to 0 and it looks like it is going to some constant which is what you should expect because if you are in thermal equilibrium it should remain fixed in thermal equilibrium right you could also rewrite this as equal to gamma over 2 m squared gamma plus v not squared minus gamma over 2 m squared gamma e to the minus 2 gamma this is for a given v 0 for some over the sub ensemble of particles with a given v 0 this is what you get and now what happens to it as t tends to infinity well gamma t tends to infinity this becomes independent of the initial condition it is forgotten the initial condition and it tends to some fixed limit okay. So remember that this average is being taken over a conditional density the condition being that the initial condition is v 0 so it is being taken really that average is being taken over this p of v t given a v 0 this thing is given t we have not yet found this we have not yet even written down the Fokker-Planck equation for this process but already it is telling us that the mean square value has this structure better have this structure from the Langevin equation itself okay and now if you insist that in equilibrium this whole thing is independent of time as t tends to infinity now if you insist that this should be true at all instance of time because this fellow here is in thermal equilibrium another way of saying it is let us compute what v squared average is over a full average so I compute v squared of t and what can that be how do I compute it from v squared bar of t I should now average over all v 0 right over what ensemble should I average this it is in thermal equilibrium so over the Maxwell distribution right so I need to compute I need to compute v squared of t equal to integral v squared of t bar over p of v 0 where this thing here is the equilibrium or stationary distribution and that is of course m over 2 pi k Boltzmann t to the power of half e to the minus m v square over 2 k Boltzmann t in v 0 so I have to do the average over this distribution right and then I get v squared average here let us do that and look at what happens this is equal to well this fellow is a number there is nothing to average and p of v 0 is a normalized distribution so you are back with this this is got this as it is so this is equal to gamma over 2 m squared gamma plus this quantity is averaged over you need to average over this definitely but what is the value of v 0 squared average over the Maxwell distribution but the average kinetic energy is half k t it is only one degree of freedom that is just the Gaussian when you once you put that in and do a Gaussian integral you discover that the half half m v 0 squared average is equal to half k t right so v 0 squared average is k t over m so this immediately tells us this is k Boltzmann t over m minus gamma over 2 m squared gamma e to the minus 2 but this cannot depend on time the system is in thermal equilibrium right it cannot depend on time and the only way that can happen is if this is equal to that so if this is equal to that if gamma is such that this quantity is equal to that the time dependence goes away because the system is in equilibrium at no time does half m v squared average change it remains k t over m because the system is in thermal equilibrium right and that is completely consistent with the fact that if this goes away you get exactly k t over m here once again you need that because if this constant had been different from that constant you are in trouble the fact that you insist that this should be equal to that automatically gives the right value here also so it says that consistency requires we must have gamma over 2 m squared gamma equal to k Boltzmann t over m or capital gamma equal to 2 m little gamma k Boltzmann t this is required by consistency now what does it mean physically well if you go back back to the Langevin equation that equation was v dot equal to gamma v with a minus sign equal to minus gamma v plus square root of gamma over m eta of t now this measured the viscous damping in the medium the viscosity in the medium which damped out fluctuations this measured the strength of these fluctuations how far does it get pushed out in some sense how strongly does it get kicked and now we are saying that the two are not independent parameters the larger the viscosity the larger the fluctuations here or conversely the larger the fluctuations the larger the viscosity must be to damp out those fluctuations and maintain thermal equilibrium there is therefore a connection between the source of fluctuations and the dissipation in the system okay and this is the first example of it the simplest example of it it is called the fluctuation dissipation relation or theorem if you like some case will come across more further examples of this but this is the simplest of the lot you are already familiar with this in the context of thermal noise in a resistor electrical resistor you know if you have an electrical resistor due to the Brownian motion of the electrons in it voltages are set up at the two ends and there is a current which fluctuates and flows in this resistor and this current you can ask this fluctuating current what is the power spectrum of this fluctuating current how much is the power carried by it in some given frequency window it will talk about power spectrum of noise a little later but we know that there is a relationship which connects the dissipation in the resistor measured by the resistance to this quant the temperature on the right hand side so we know that the power spectrum of the fluctuations in the response of the system is related to the resistance multiplied by the temperature this is called the Nyquist relation this is the Nyquist relation in this context it is exactly the Nyquist relation for thermal noise or Johnson noise whatever okay but physically what it means is the same the same fluctuations that give rise to while oscillations or which give rise to randomness in the velocity are the ones also responsible for the dissipation in the system and there is a consistency condition between the two you cannot have one unboundedly growing without independent of the other in this context it is required for thermal equilibrium this is required to maintain thermal equilibrium so we will put that in henceforth and now notice that once you put it in this thing here becomes equal to KBT over M independent independent of T so it is already starting to tell us that perhaps this V of T is really going to be a stationary Markov process we started with that assumption we already put that in I have not explicitly shown it here but we already when we wrote this Langevin equation to cut a long story short once you have a Langevin equation of this kind then what it means is if this is a Gaussian white noise in other words it is a stationary Gaussian Delta correlated Markov process then you are guaranteed that this V the output or driven variable is going to be a stationary Gaussian Markov process but not Delta correlated it will have a finite correlation time what do you think will be the correlation time of this velocity or what time scale is this thing losing its or what time scale is this average value going to 0 it is going e to the minus gamma T so there is only one such time scale in the problem which is little gamma inverse so indeed it will turn out little gamma inverse is the velocity correlation time with that information there which you have got there we can actually write down the solution completely to the Fokker Planck equation fully but we will not quite do that as yet we will just see what the Fokker Planck equation is before we do this but we are going to use this connection in school now let us write that Fokker Planck equation down immediately using this correspondence between the Langevin equation and the Fokker Planck equation we write it down then look at what its solution is but mean the meantime will compute the velocity correlation auto correlation function so now that we know what this little gamma is this is equal to minus gamma V plus this fellow is 2 m little gamma k T with a square root over m so let us take that over n that is the Langevin equation where we put in this consistency condition and what does that imply at once it at once implies that delta over delta T P of V, T V0 must satisfy the drift term is this but remember by looking at our general rule here it is minus so the minus cancels gamma is a constant delta over delta P V times P the same P plus one half the square of this guy in the half this kills that so you have gamma k Boltzmann T over m D2 P over delta V2 this was the original Fokker Planck equation the first one written down we use that term in general for the second order master equation with up to second derivative but this was the original one with a linear drift term now of course you can take this equation and ask what is its solution we need the initial condition and that is obvious here the initial condition is P of V T V0 equal to delta of V minus this is the initial condition of course what is the stationary distribution is there a stationary distribution in this problem unlike the diffusion equation where everything went to 0 the question is is there a stationary distribution that would be found by putting this equal to 0 and then asking what happens to this stationary distribution what should we expect as a stationary distribution the Maxwellian I should expect the Maxwellian once again right because I should expect that limit T tends to infinity P of V T V0 should be equal to P equilibrium of V0 V the memory of the initial condition should be raised and you should have the equilibrium distribution again if this process is a stationary random process. So let us see if that happens well the stationary distribution in is now does not have any T dependence so I write ordinary derivatives with respect to V and it must be this quantity must be equal to 0 of course gamma is a constant so let us get rid of that and then it says D over D V of gamma gamma goes away k Boltzmann T over M DP equilibrium over DV plus V times P equilibrium equal to 0 that is the equation right if it exists and it should be normalized to unity and so on. So what does it say it says this quantity in the in the bracket should be a constant independent of V so erase this and write this I want this P equilibrium to be a normalizable distribution so P equilibrium must vanish as mod V tends to infinity that is a necessary condition otherwise it is not normalizable you are going to integrate minus infinity to infinity so the function at better vanish at the end point sufficiently rapidly. I want all moments of this also to be finite I want the mean square for example to be finite I want it to be equal to k T over M right so you want this also to go to 0 you want this quantity to go to 0 at infinity faster than any power of V because you want all these moments to be finite therefore its derivative also will go to 0 faster than any power of V the value of the constant therefore is 0 because it is independent of V and when V is plus tens towards infinity the value is 0 therefore it is the value everywhere is that is that clear so this constant is 0 but there is another way to write this down to look at it you know if you write this Fokker-Planck equation down you can also write it like a continuity equation you can write delta P over delta T plus delta J over delta V equal to 0 this is del dot J in one dimension with V being the independent variable and what is J it is just this fellow but without the equilibrium without the equilibrium time dependent with the time dependent density that is the current this J is the probability current you do not want this current to be finite at infinity you do not want any flux at infinity of probability so this guy must be 0 at infinity but in the case of the stationary distribution its 0 everywhere for all values of V because it is got to be a constant pardon me in that case scenario yeah at infinity the flux yes is the boundary condition at infinity will be such that this quantity delta KT over M delta P over delta V plus V times P where this is time dependent the conditional density this will tend to 0 as V tends to infinity mod V tends to infinity every problem it will become 0 not necessarily we are going to do finite problems where this may not be 0 okay for instance if I had not a velocity but it is a diffusing particle say and on one side I have a barrier where if it hits that barrier it bounces back and forth on the other side I have a barrier where it absorbs like a sponge for instance then this boundary condition on the right hand side it is not true it is an absorbing barrier it is only saying the flux is 0 so it is equivalent to saying that there is a reflecting boundary condition but here it is at infinity so this is a natural boundary condition that this whole thing vanishes at infinity otherwise you do not even have a normalizable density okay so I used a physical argument to say that I want this equilibrium distribution to have finite moments in particular I want to finite variance so that I can relate it to the average kinetic energy and so on okay so this thing immediately tells me if I solve this equation it is an ordinary first order differential equation of course it immediately says m v over k Boltzmann t times this is 0 and that of course automatically implies that p equilibrium apart from a normalization constant is e to the minus m v squared over 2 k Boltzmann t all I have to do is to integrate this move it to the right hand side separate variables and that is it okay and that is your Maxwellian distribution back again so we now know that this equation this Fokker-Planck equation is consistent with the equilibrium distribution the Maxwellian distribution can we write a solution of this equation down a time dependent solution which satisfies this initial condition what key input would you need for that we should be able to solve this equation that is one thing but it is a hard equation to solve at least it is not a trivial equation to solve and you have to integrate this second order partial differential equation it is first order in time second order in space and so on and so forth we could do the following we could ask what is the mean square value mean value etc for a given initial condition which we derive from the Langevin equation but the question is can we do it directly from the Fokker-Planck equation yes we can if we indeed can suppose I multiply both sides by v and integrate then this is the rate of change of the mean value of v and that satisfies an ordinary differential equation because if I integrate I multiply by v all I have to do is to integrate by parts to bring these derivatives out of this p and write it as some averages so you can get an equation for v average bar v squared average bar by multiplying the v squared and ask can those be and do they form a close set of equations or not in this case they do and you can solve them so you would have the variance and the mean the conditional variance and the mean and then what else would be needed what assumption would you need to say that that is sufficient what kind of process would it continuous process are you familiar with in which a knowledge of the variance in the mean is sufficient to write a Gaussian yeah if this were Gaussian that would be it all higher cumulants are gone and then we could write the solution down explicitly yeah. Only there is a correspondence between the Langevin equation and the Fokker-Planck equation yeah and we already use the fact that v not is maxillian in the Langevin equation that is right so is it just the reflection of that yes absolutely so this should be consistent otherwise I would be very very surprised but we have not yet said that this is the solution of this we have not said is a Gaussian that is not so obvious as yet if it were then this would be immediately true I could write the solution down explicitly all I need to know is what is the variance to and what does the mean do and I can write this down immediately provided it were a Gaussian because I know how to write a Gaussian down once you will give me the mean and the variance in fact let us go back there and ask what is the variance of this guy what what does the variance look like so we need to compute we go right back at this stage and say we have an expression for v squared we got an expression for v average let us see what the conditional variance looks like so that is a good exercise to do we have v of t bar is v not e to the minus gamma t and we have v squared of t average equal to v not squared e to the minus 2 gamma t plus gamma over 2 m squared gamma 1 minus e to the minus 2 that was what v squared is so the variance v of t so let us say for given v not we should always remember that this is with a conditional ensemble then we given v not this is equal to this fellow minus this squared so this cancels out and you end up with gamma over 2 m squared gamma 1 minus e to the minus 2 notice that v not gets rid of is gone there is no dependence on v not at all whatever be the initial v not some given v not the variance of the velocity does not reflect it at all it is gone and now if you put in the fluctuation dissipation relation which just says the system remains in equilibrium then this fellow here is k t k Boltzmann t over m 1 minus e to the minus 2 gamma t if I now assert without proof at the moment that the solution to this with this initial condition is a Gaussian if the solution is a Gaussian is a Gaussian then we can write it down p of v t v not must be equal to e to the power minus v minus v of t bar square over twice the variance whatever it is divided by times a normalization factor times this guy you know explicitly write it down well let us put those factors in and see what it is so it looks big but it is actually very straight forward p of v t v not equal to now the variance is given to you it is this guy here so it is 1 over root 2 pi sigma squared and sigma squared is here so it is equal to m over 2 pi k Boltzmann t 1 minus e to the minus 2 gamma t this guy here and the whole thing is to the power a half 1 over square root of 2 pi sigma squared and then e to the power exponent minus x minus v minus v not e to the minus gamma t that is the mean squared divided by 2 sigma squared and that is minus m and then there is a 2 sigma squared is 2 k t 1 minus e to the minus 2 gamma t looks complicated but it is actually very very simple in structure okay again you have to check that as t tends to infinity it goes to the Maxwellian the right Maxwellian and indeed it does because as t tends to infinity this goes away the exponent that goes away this goes away and you have m v squared over 2 k t which is precisely the max value as it should be what happens as t goes to 0 it becomes singular it become you have to be very careful taking limits it becomes singular because this becomes this t goes to 0 this factor goes away you have e to the minus v minus v not whole squared and then this fellow here vanishes so intuitively it is clear that the only contribution will come from v equal to v not so that the numerator also vanishes and it should in fact go to the delta function so it starts as a singular not square integrable or anything it is a singular delta function distribution and then smooth smooth becomes smooth becomes a Gaussian so in pictures what it does is the following if I plot v here then initially it is a spike at v equal to v not a delta function spike and asymptotically it is a Gaussian the equilibrium distribution in velocity and as time increases the mean value which is also the peak in this case shifts gradually to the left like v not e to the minus gamma t so after a little bit of time it looks like this and then this peak shifts it comes down so that the area under the curve remains one always and it finally settles at the Maxwellian distribution which is just what you would expect this thing would do and we need to prove of course that the solution is a Gaussian that takes a little bit of doing but it is not very hard in this context we can actually compute the other cumulants and discover that they are all 0 and then it is exactly this there are other ways of solving this and we will not do that right now we will come back and if time permits we will talk about other ways of solving this equation okay because I want to also introduce another stochastic equation for the position of a brown harmonically bound particle which would look exactly like this and we know all about the harmonic oscillators so we can use that knowledge to solve this equation but you can see physically what is remaining in this context is to ask what is the velocity correlation itself to the correlation of the velocity do and then there are several questions that arise which you will all answer successfully namely what about the position of the particle we made this model we solve the equation of motion we found v of t we found its average distribution and so on what about the position and then a much deeper question should we not really look at this particle in phase space namely both x and v together and then should we not write down a stochastic differential equation for this quantity and then find this distribution or conditional density in phase space for both position and velocity together that is really what we should do because then we could put external forces on the particle and write the correct Langevin equation down and solve it in phase space because dynamics happens in phase space so we will do that we will write that that will mean a multi-dimensional two-dimensional Fokker-Planck equation but we can do that without much difficulty and you will see how the physics goes in just one remark here and that is to find the following quantity we found already v squared of t let us do v of t v of t prime and find the average here where t and t prime are both positive numbers but different numbers I leave this as an exercise to you because you would have exactly the same thing as before again each of these is v naught e to the minus gamma t etc. So there is going to be a v naught squared e to the minus 2 minus gamma t plus t prime those would be the first terms then there be one term where you have a v naught term multiplying an eta and a eta of t 2 say and then a v naught multiplying a eta of t 1 those averages go away and then you are left with plus gamma over m squared integral 0 to t dt 1 0 t prime dt 2 this fashion e to the minus gamma t minus t 1 minus gamma t prime minus t 2 this case times eta of t 1 eta of t 2 average that is a delta function. So you have a delta function now you have got to be careful okay so this much is straight forward but now in removing this delta function do the integral you have to be a little careful. So let us for example see in pictures what happens here is t 1 here is t 2 this fellow is integrated up to t the other guy is integrated up to t prime let us suppose t prime is smaller than t we also have to look at the case where it is larger but this whole thing is completely symmetrical in t 1 and t 2 so we could actually interchange after we find the result. So let us suppose this is t prime and what is the constraint on the integration the delta function and where does that fire on a line which is at 45 degrees so it clearly fires on this line this is the line t 1 equal to t 2 this is the case t greater than t prime otherwise the rectangle is upwards. Now what does that tell you it says that you are going to fix a t 1 and scan t 2 so you fix a t 1 and you scan in t 2 in this fashion and of course you hit this delta function you fix the next t 1 and scan you hit the delta function and you can do this till a t 1 hits t prime and after that you get 0 the answer so the integration is get it is cut off at this point here so this thing reduces to integral 0 to t prime d t 1 e to the minus gamma t minus t 1 minus gamma t prime minus t 1 once again you can set t 1 equal to t 2 inside the integrand by using the delta function constraint but the t 1 integration is constrained to stop at t prime okay and then you have to do this integral etc etc if t were greater than t prime do the t 2 integration later do the t 1 first not surprisingly what answer would you expect finally of this what kind of function of t and t prime would you expect I would expect it to be symmetric under t goes t and t prime getting exchange with each other right but we also know that if it is a stationary process I would expect this to be a function of the time difference and I want it to be symmetric so what would you expect you expect more t minus t prime I would expect the answer to be more t minus t prime so show that this fellow reduces to e to the minus gamma times more t minus t prime if it suffices to do it for t greater than t prime and then use this symmetry okay and you will see therefore that the velocity is exponentially correlated okay and there is a very powerful theorem which says there is only one process which is Gaussian which is continuous process which is a one dimensional process which is Gaussian stationary and Markov and exponentially correlated and that is this process there is nothing else and various other processes can be reduced to it by changes of variables reparameterization and so on so that is the reason why this is worth studying in such great detail in addition to this particular example here okay. So we will take it up from this point and I will point out how we can extend this to phase space and see what exactly where the diffusion approximation comes and so on we have got to understand the role of this gamma a little harder a little better okay so we take that up on Monday.