 So, let us resume our discussion of the velocity process for Brownian particle in a fluid and just to recapitulate very quickly what we discovered we found that when the particle obeys the Langevin equation then its velocity process this random process driven by white noise Gaussian white noise is exponentially correlated. So, we discovered that this quantity V of t naught V of t naught plus t was a function only of the time difference t between these two arguments and this was essentially equal to k B t over m e to the minus gamma t this fashion. So, I might as well set t naught equal to 0 and write it as V of 0 V of t equal to k B t over m e to the minus gamma t. Now we also saw I am not sure if you prove this but we also saw using stationarity that this very trivially implies if I subtract minus t from each of these arguments this also implies that V of minus t V of 0 I said t to minus t here is going to give us a modulus there. So, in general it is clear that this quantity satisfies this expression. So, it is a symmetric function in t and it dies down exponentially on either side of the t axis okay in equilibrium we further saw as a consequence of this by the way this implies that there is a time scale in the problem gamma inverse which we could actually estimate by using the fact that we put this particle in a fluid with viscosity eta for instance and it has a radius r then we saw that m times gamma was equal to 6 pi r eta where this is the viscosity of the fluid and this is the mass of the particle and if you estimate this mass to be of the order of 10 to the minus 15 kilograms gamma turns out to be of the order of 10 to the minus 6 or 10 to the minus 7 second inverse. So gamma inverse turns out to be about 10 to the 6 or 10 to the 7 r micron sized particle 10 to the minus 6 eta is 10 to the minus 3 in SI units standard international units Newton second per meter squared or something like that then we discovered that there is a time scale in the problem which is of the order of microseconds or tens of microseconds and you must compare it to the other time scales in the problem. The other time scale we have is the actual interaction time between molecules and that is of the order of 10 to the minus 15 seconds or less and then there is a time scale between collisions of particles that is of the order of picoseconds or less and this time scale is another 6 orders of magnitude higher. Now of course you can consider times much much greater than gamma inverse and that is the diffusion regime in which the mean square displacement of the particle goes linearly in time with a diffusion constant this coefficient D given by KT over M gamma the diffusion coefficient of this massive particle in the fluid not of individual molecules. We also discovered as a consequence of this expression we could actually write down what the displacement did the mean square displacement did and we found that if you define an x of t to be equal to little x of t minus x of 0 and you computed the mean square value of this quantity x squared of t not the conditional mean but the actual mean of this quantity you computed the full equilibrium average value of this quantity then this turned out to be KBT over M gamma squared times gamma t minus 1 plus e to the minus gamma t that was the exact expression and all we needed was to use the fact that the position is the integral of the velocity and that is it so with that we immediately got an expression which went like this this of course goes for gamma t much much greater than 1 to twice DT where D is given by this expression in other words that is the diffusion regime but for all time it is exactly equal to this in this model as far as this model goes now you can ask further questions of this you could ask what is this quantity itself do for instance and what is its average value and so on you can I am going to leave this as an exercise just one step per step for instance if you compute what is x bar of t this is the conditional average the conditional average for a given V naught and a given x of 0 then this is easy to find notice that a simple integration immediately gives you x of t is equal to an integral from 0 to t DT 1 V of t 1 and we know what V of t 1 does the conditional average this quantity V of t 1 bar is equal to V naught e to the minus gamma t plus a portion that depends on the initial depends on the random force the eta and that average is to 0 so if I compute this integral here this thing becomes equal to all I need to do is to plug this in V naught over gamma 1 minus e to the minus gamma t all I have done is to substitute this expression here and compute it here and this is equal to x bar of t so the average value the conditional average of the displacement is this quantity here of course if I take a full average now over V naught it will vanish as it should the displacement should vanish having got this you can now ask what is the variance of this quantity not of this quantity itself but the deviation of this quantity from its mean so the natural definition of delta x of t capital X of t would of course be X of t minus the average value the conditional average of X of t that is obviously the natural definition of the deviation from the mean of the displacement not the position but the displacement itself and then one can ask what is delta X of t whole squared the full average you have to take this quantity square it in the usual way and take the full average if you use this fact what we need is information about this X bar of t obviously in this expression and we use that information in then it is not hard to show that this becomes equal to d over gamma times oh incidentally we could also have written this as d over gamma times gamma t minus 1 plus gamma e to the minus gamma t I substitute d equal to k t over m gamma. So if I do the same thing here for this variance then this is equal to this turns out to be equal to not sure if I remember this expression completely but let us see if I can mentally write this out 2 gamma t minus 3 plus 4 e to the minus gamma t I remember that part of it minus e to the minus 2 gamma t which comes from squaring this fellow. So you get some expression like that for the variance of this displacement itself of course as t tends to infinity this will go to 2 dt as it should. So if you are interested in the displacement rather than the velocity we have this expression here. Now what is interesting is that there is no simple equation as the width will be for the velocity process there will be some kind of equation we will talk about this equation which will give the distribution and probability of the conditional density of the conditional well of probability density of the velocity we are going to write that down shortly but there is no such equation for this. However the very fact that the velocity is just the position is the integral of the velocity helps you to find these things these quantities here you could in fact go on to find delta x of t 1 and delta x of t delta x of t prime it is a messy expression of some kind. So we can play with this find all these moments explicitly but let us come back here backtrack a little bit and say alright this is very nice can we say something about the probability density of the actual distribution of this velocity as a function of time the conditional distribution starting from the fact that it at t equal to 0 it is a delta function at some v not and then t tends to infinity it should go to the maxwellian distribution can we write this distribution down we are going to do that a little later when I show you that there is a correspondence between the Langevin equation for the variable itself and an equation called the Fokker-Planck equation for the conditional probability density there is a complete one to one correspondence in certain cases and we will exploit that but right now I want to write the answer down and introduce you to this distribution which you would have seen in other context perhaps but let me show you what this distribution is we can simply write it down in this particular case and it is as follows. So remember what we know about this velocity we know that it is a stationary process in equilibrium because it is a function of t alone in fact you can show that it is stationary in the strict sense in other words all its densities joint densities are independent of the origin of time you can shift the time at this level it is only the correlation the two point correlation that is been shown to be so but this is true for all its joint distributions. We know for instance that v of t average is v not e to the minus gamma t the conditional average for a given v not we know that v squared of t average we had an expression for this quantity we know that it is v not squared 1 minus e to the minus 2 gamma t if I take its side this average if I squared if I take the average with respect to v not there was an extra term here you have to remind me what this term is there was an extra term which also had this plus what was the actual expression we found out otherwise I have to now go back and start working out the algebra what was the actual expression as t tends to infinity so well perhaps this was correct average of v not squared is k b t over n so that is okay that is what it was explicitly okay right but we also put in I am a little unhappy about this what I would like to know what I would like to do is write down what is the variance of this quantity the conditional variance what was the actual expression for this quantity is this correct as it stands yes one involving gap yes it was v not square minus gamma upon 2 m square gamma times e power minus 2 gamma v not squared minus v not squared e to the minus 2 gamma t minus yeah plus plus capital gamma upon 2 m gamma square m square gamma into 1 minus 2 m square gamma no v not square no v not square gamma upon 2 m square gamma into times 1 minus e power minus 2 gamma good that is so this is equal to we wrote this as by the way this quantity we know there is a fluctuation dissipation relation so it is k t over m so it is k Boltzmann t over m plus v not squared minus k Boltzmann t over m e to the minus 2 gamma good so that was the expression so let us kill this and this was v squared of t exactly exactly now we argued that if you let t go to infinity for a fixed e not start with for a fixed origin on time then this term goes away and it reaches the equilibrium value on the other hand if you average this quantity with respect to the Maxwellian distribution in v not then this gives you a k t over m and that kills this and there is nothing to average here so it remains k t over m this was our consistency condition right so now given these two we can compute what is v squared of t minus v of v of t average squared this is equal to conditional variance of v and what is this equal to what we need to do is take this and subtract from this is square of this quantity and that kills this term here so this is equal to k Boltzmann t over m 1 minus e to the minus 2 gamma so we have the conditional mean and we got the conditional variance now if and this is a big if this process is a Gaussian and remains Gaussian at all times we know at t equal to infinity it is a Gaussian it is the Maxwellian distribution right if it remains a Gaussian at all times then we can actually write the distribution down and what distribution would that be it is the conditional probability density so p of v t given v not this is the conditional p d f of velocity it is conditioned upon this specific initial condition starts with the delta function at v equal to v not as t tends to infinity it goes to the Maxwell distribution in v and now the question is what is it actually equal to well if it is a Gaussian if it is a Gaussian then the mean and variance determine the distribution completely a Gaussian is determined by its mean and variance completely right so if that is so then apart from a normalization factor if is a Gaussian and we have to show this but we will do so later on then p of v t v not is actually equal to apart from a normalization factor e to the power minus v minus v not e to the minus gamma t the whole squared that is the Gaussian up there divided by twice the variance so it is minus m over twice k Boltzmann t 1 minus e to the minus 2 gamma t that is the exponential and all we have to do is to normalize this exponential which is m over 2 pi k Boltzmann t 1 minus e to the minus 2 gamma t to the power a half exponential of this whole thing in brackets so let me write it out neatly this is exponential of minus m v minus v not e to the minus gamma t whole squared over k Boltzmann t twice k Boltzmann t 1 minus e to the that tells you that this distribution in v p of v t v not starts with the delta function at v not at any intermediate time it looks like this a peak at this point v not e to the minus gamma t and a width that is growing all the time because as t increases this quantity decreases till at equal to infinity it reaches its value of 2 k t largest value of 2 k t twice the variance and then it becomes a symmetric Gaussian up here so it eventually ends up with this at t equal to infinity this is called the Onstein it is the Onstein-Ohlenbeck probability density function and the general statement which I made as a statement is that if you have a continuous stationary Gaussian Markov process then it is exponentially correlated and then it looks like this in general we wrote this for the velocity based on the Langevin equation and the assumption that it remains a Gaussian but this is in fact a Gauss-Markov process a prototypical Gauss-Markov process okay. So with that statement in place and I tell you it is Markov then all moments are known everything is known about it completely later we will little later we will derive the Fokker-Planck equation or at least justify the Fokker-Planck equation from which this of which this is going to be the solution appropriate solution okay for that given initial condition yes absolutely. So it is essentially telling you how does the probability density function itself relax to the equilibrium value so that is the crucial point about this okay. Having done that and talked a little bit about the displacement let us do the following the next question to ask natural question will be to say if we push this Langevin model little further then we also need to be able to say what is the joint distribution of the position and the velocity in phase space or the position and the momentum but I am not going to do that right now because it is a lot easier to do that in terms of the distribution function itself for which I would need something called the Fokker-Planck equation corresponding to that 2 dimensional process. So I have not justified that yet and we have not come to that yet so meanwhile before we do that we are going to do a lot of other things so let me keep that in abeyance for the moment because it will become easier to understand later on and go back now and look at a 3 dimensional case just to show you that the velocity correlation function can actually mix up different components of the velocity if for instance you have a magnetic field. So let us look at this very simple problem it is actually quite a simple problem of a Langevin particle in a magnetic field that problem too can be solved quite exactly a constant magnetic field and I want to look at the simplest case where this particle is placed in a constant magnetic field in this fluid and everything else remains exactly the same again I say this model which is a Langevin model for the particle I write its equation of motion but this time including the Lorentz force on the particle the V cross B force okay. So the masses M the particle mass equal to M charge equal to Q and the magnetic field applied is some B which is in some arbitrary direction defined by some unit vector N okay we could without loss of generality take this N to be along the Z direction but there is no reason to let us just look at it for arbitrary N a little more algebra but it is helpful to separate than the transverse and the longitudinal components easily okay. Now I am going to cut this story short and do it in the following way we will use physical arguments here to make certain assumptions which are actually justifiable completely rigorously first of all the fluid is in thermal equilibrium at temperature T okay now this particle all the other assumptions in the absence of the field continue to hold good so the distribution of velocity of this particle in equilibrium is not going to be different from the Maxwellian in thermal equilibrium nothing happens because the magnetic field does no work on this particle at all it does not change its energy and if it does not change its energy it remains in the canonical ensemble the equilibrium distribution is still going to be exactly the Maxwellian distribution on the other hand if I start with some given initial velocity V naught in some arbitrary direction then there is a question as to how it relaxes to this equilibrium distribution what how do the components different components relax we will continue to assume without with some intuition that the velocity remains a stationary process the different Cartesian components of the velocity may be correlated we do not know yet we are going to find this correlation so let us compute the correlation function directly and let us do it in the case of the magnetic field assuming that the system remains in thermal equilibrium and moreover that this velocity is a stationary process this is all we need then we could have computed in the absence of the field by a slight shortcut so let me do that with the presence of the field and you will see how this calculation goes so the Langevin equation that I write for it is M times let us look at some given Cartesian component J of T VJ dot of T J is one of the Cartesian components runs over 1 2 3 this is equal to minus M gamma VJ of T the same gamma I assume the fluid is isotropic gamma depends on the viscosity of the fluids completely isotropic exactly the same in all Cartesian components and then there is a portion there is a random force as usual but there is also a term which is the V cross B term so there is a Q times epsilon Jkl Vk Bl which is nl but let us put B outside I assume you are familiar with the index notation and with this epsilon symbol which is a short way of writing the cross product V cross B this quantity is equal to plus 1 if Jkl are in the order 1 2 3 or permutations cyclic permutations there are minus 1 if they are not and any 2 indices are equal it is equal to 0 right so the technical way of saying it is epsilon Jkl is plus 1 if Jkl is an even permutation of 1 2 3 the natural order minus 1 if it is an odd permutation and 0 in all other cases okay plus there is a force here now which is 8 a g it is a vector force so I write it is Jkl and let us as usual divide by m so this is 1 over m you have QB over m this fashion this goes off this goes off it is convenient to write this as minus gamma Vj of t minus this quantity QB over m has a physical interpretation it is a quantity of dimensions frequency or inverse time same as gamma that is dimensionless this is dimensionless that is called dimensions velocity now what is what do you call this quantity it is a cyclotron frequency so omega c equal to cyclotron frequency QB over m so let us write this as minus omega c m epsilon K Jl nl vk vk is a function of t plus 1 over m 8 and J and let me define a matrix let us define a tensor of rank 2 you can write as a matrix if you like let us define m define m K Jl to be equal to epsilon K Jl times nl l is contracted so sorry so this m K J so the K J element of this tensor of rank 2 or matrix 3 by 3 matrix is defined to be K Jl nl therefore I can remove this and write this as m J times vk now to find the correlation function we have assumed that the velocity we are going to assume the velocity is stationary process then a quick way of finding the correlation function is to write let us do let us pre multiply both sides by vi of 0 vj and take average equilibrium average we can go through the rigmarole of solving this equation taking conditional average and then taking full averages etc showing it stationary and so on let us cut all that short just multiply by vi of 0 t is greater than 0 here vj dot of t equal to whatever this is equal to minus gamma and then I go ahead and take averages vi of 0 vj of t minus omega c average value of vi of 0 vk of t times m Jk notice that m is a constant matrix all its elements are constants plus average value vi of 0 eta J of t 1 over n but this is 0 by causality because for all t greater than 0 this thing is not vanishes identically for all t greater than 0 by causality for all t greater than 0 this thing here vanishes because the effect cannot precede the cause what the random force done does at some time t greater than 0 cannot affect what the velocity was at t before 0 so that is it this equation now unlike the previous case where you had a correlation function now you got a correlation matrix because there symbols ij the indices ij etc so let us define c ij of t to be equal to vi of 0 vj of t. In fact let us do the following we know that asymptotically asymptotically if i is equal to j this quantity here is going to die down exponentially and at t equal to 0 it is going to start with kt over m the average value is going to start with the average value of v not squared which is kt over m so let us divide this by kt over m we will define a normalized correlation matrix by dividing by this asymptotically this initial value kt over m so c ij of 0 is equal to 1 by definition so let us keep that in mind if i is not equal to j then these two are uncorrelated at t equal to the same time but if i is equal to j then you just get v squared one component and that is equal to kt over m so if I write this c of 0 as a c of t as a matrix call it some matrix with components c ij then at t equal to 0 the matrix is the unit matrix that is the delta function is just this is these are just the elements of the unit matrix so that is a useful thing to know so now we can solve this equation we can solve this equation because it simply says d over dt c ij of t is equal to minus gamma c ij of t minus omega c m we should be careful with commutation m ik c ik of t mj is that correct m m sorry we had mjk what happened here did I define I flipped it to say change the sign and then I brought this you can change your definition of mk I should do that if you really do that properly jkl is okay nl just define that as mjk yeah there is no error so let us did I get this right this was correct this part is okay so this is m kj oh yeah oh yeah yeah I wrote this term sorry I wrote this term as minus omega c epsilon kjl vk and I define mkj as epsilon kjl nl so this fellow here was an mkj that is what it was and this is mkj you could do it by defining it anywhere you like but now it is consistent yeah this is therefore mkj thank you okay or in matrix form this is the same as saying d over dt c of t where this is a matrix c of t is a correlation matrix is equal to minus gamma times the unit matrix plus omega c times the matrix m on is that correct no no no c times the whole thing so c times yeah minus c times gamma times the unit matrix plus omega c times the m matrix yeah now we are in good shape you should either have it on the left or right that is all otherwise it does not matter at all but I should be careful where I put this c so where I write the solution so this is minus c of t times this fellow this is a matrix m is a matrix this is a unit matrix is a matrix equation but this is a constant matrix there is no time dependence here this fellow here so this immediately implies that the solution is c of t is equal to e to the power minus is equal to c of 0 times e to the power minus gamma times unit matrix plus omega c times m times t I have to be consistent this c is on the left of this thing here so in the solution to it is on the left in the present case it does not matter why is that c of 0 is the identity matrix it commutes with everything so we do not really care but it is just good discipline it may not always be the case so we should just be careful doing that so that is the solution for c and all you got to do is to read off this matrix but you have the problem of exponentiating this so by the way if since this is equal to the identity matrix this implies that the whole thing is equal to this implies c of t implies c of t equal to e to the minus gamma t times e to the minus omega c minus m omega c e so we have to find the exponential of m times is a constant that is all we have to do and then the problem is solved now how we go about doing that this is a 3 by 3 matrix if it were a 2 by 2 matrix we could write it in the Pauley basis in terms of Pauley matrices and we can read off what the exponential is but it is a 3 by 3 matrix what should we do then yeah there are several tricks to do this one of them could be the following we could we could take this matrix m it is a 3 by 3 matrix right now it is clear that the exponential of a 3 by 3 matrix cannot you have to be should be able to sum this thing reasonably if this has got some physical meaning and so on but writing down an explicit formula may not be all that easy you can write this exponential down because it is characteristic equation after all must be a third order polynomial in m which means that m cube can be written in terms of the identity matrix m and m squared therefore m 4 can be written in terms of these fellows and so on and so forth but that may pain in the neck to try to combine the coefficients and compute what it is not doable in general but this matrix is so simple that is possible to do it well another way to do this could be find its eigenvalues therefore write its characteristic equation down the secular equation and replace lambda by the matrix itself by the Cayley Hamilton theorem and then maybe one can find out what m cube this in an easy way but there is an easy even easier way this is a rotation matrix I hope you recognize this a rotation matrix so m squared k j equal to m k j equal to epsilon k j l n l this is a unit vectors a component so m squared k j is equal to epsilon it is equal to m k k r m r j where r is another index so this is equal to epsilon k r l n l epsilon r j s n s that is what it is right so this is equal to epsilon k r l epsilon j r s n l n s with a minus sign I flip this once here and then I use this identity well known identity when you take this epsilon symbol and you contract one of the symbols in the same position then it is just a product of delta function so this is equal to minus of delta k j delta node what did I do no no no this is equal to minus epsilon r k l epsilon r j s n l n s so this is equal to a delta function of these two fellows k j delta function of l s minus delta function of k with s delta function of l with j acting on n l n s which is equal to what the first term you have a delta k j delta l s n l n n s is n l n l which is equal to 1 it is a unit vector and the other term is plus wherever l appears replace with j wherever s appears replace with k so n j n k so it is just n j n k we have taken got the k j element so let us write it symmetrically n k n j minus delta k j that is the square of this matrix because it is how many people know that it is a mutated rotation matrix they are not going to believe you or me it is it is a rotation matrix we know it is I what do you think its eigenvalues are it is a rotation matrix in 3 dimensions and it is got it is a rotation about the direction l the index l right so it must it is clear that n l must be an eigenvector of this matrix with eigenvalue equal to 1 got to be so if it is a rotation matrix then there are 2 other eigenvalues can either of them be real or must they be imaginary well suppose it is real then this means if this eigenvalue is real then there is a direction in space which is also left invariant by this rotation but in a 3 dimensional rotation there is only one axis that can at best be left invariant right so the other 2 eigenvalues must be complex it is a rotation matrix so this matrix is unitary it is orthogonal the elements are real therefore it is all unitary matrix with real elements it is an orthogonal matrix because it is a unitary matrix all its eigenvalues must lie on the unit circle 1 is already an eigenvalue minus 1 is not an eigenvalue of this is immediate why is that why is that yeah if my the product of all the 3 eigenvalues must be equal to the determinant of this matrix which must be plus 1 it is a proper rotation right so if minus 1 appears as an eigenvalue it must appear second time 2 because 1 is already an eigenvalue right so it must be a repeated eigenvalue it also means that there is again a real eigenvector real eigenvalue will imply I leave you to figure out why minus 1 cannot appear as an eigenvalue the only other possibility is some e to the i theta appears and e to the minus i theta 2 minus yeah they have to give an argument as to why 2 minus 1 do not appear here for a real rotation matrix in this case in this case in three dimensional rotation so the only other possibility is a pair of complex conjugate roots which lie on the unit circle in this case there will be pi over 2 minus pi over 2 e to the i pi over 2 e to the minus pi over 2 look at it this way once you have this direction nl regard that as a z direction then rotation about it is a rotation in the xy plane which is given by matrix of the form cos theta sin theta minus minus sin theta cos theta and its eigenvalues are e to the plus minus i theta so that is all this case anyway we are going to discover the same thing in so this is m m m squared k j so implies m cubed k j therefore equal to m squared k r m r j and that is equal to what m squared k r is n k n r minus delta k r times epsilon r j l n l what does that work out to what is the first term n k n r n l epsilon r j l why is it 0 it is symmetric and r and l n r n l but there is an epsilon here which is anti symmetric in the two so the first term is 0 so the second term tells you that m cubed k j equal to the second term is a delta where this is replaced here so it is epsilon k j l n minus m k j so this implies that m cubed that is very simple by the way that also tells you the characteristic equation right away if I the Kayley Hamilton equation would be m times m squared plus i times m equal to 0 that is the Kayley Hamilton equation for this case because it is a characteristic equation where you replace the eigenvalue by the matrix once I write by the way this is the statement that m cubed is minus m once I write it like this it means that lambda squared plus 1 times lambda equal to 0 which means lambda is equal to you have your three eigenvalues in this problem okay alright once we yeah once we have this the exponential is very easy so now let us go back and do what is the exponential this case and now we can do this very fast so this says e to the power minus m omega c t equal to i minus m omega c t plus m squared over two factorial omega c t whole squared minus m cubed over three factorial omega c t cube plus m four over four factorial omega c t to the power four minus dot this is equal to i let us collect all these terms together minus m omega c t that is this portion and then an m cubed but m cubed is minus m so this becomes a plus out here so this gives you m times omega c t and this becomes a plus so this has to be a minus out there that term is going to keep going this fashion and then you have to deal with this fellow plus m squared over two factorial omega c t whole squared plus m four and what is m four it is equal to m times m cubed but m cubed is minus m so this is minus omega c t whole to the power four over four factorial plus dot dot dot this keeps going if I had a one here then this would be a cosine so let us add and subtract the one this came with a plus sign right so let us put a minus one here and put a plus one this fellow is minus cosine of omega c t and there was a one here that is the identity matrix in this one so we have done the job this here now is equal to i plus one minus cos omega c t m squared but before let us write the m term first i minus m sin omega c t plus m squared that is the final answer therefore the correlation matrix itself the normalized correlation matrix itself c of t equal to e to the minus gamma t and notice as t goes to zero this disappears that disappears you left with this times the unit matrix this is one so let me stop here since we run out of time but let us take it from this point on what now I request you to write down what c i j is it will of course start with e to the minus gamma t and then you have to write down whatever it is inside what do you get from here delta i j what is this epsilon i j l or i j k if you like n k sin omega c t plus m squared but remember we have a formula for the components m squared this fellow here was n i n j minus delta i j 1 minus cos this delta i j will cancel against that delta i j this n i n j will sit as it is here and there is a delta i j which also multiplies this so what is this equal to we can also write this as e to the minus gamma t n i n j that is the first term and then so there is plus correct me if I am wrong here delta i j well work this out explicitly I do not want to make a mistake in writing this expression but you notice that there is a portion which is odd in time and a portion which is even in time we will interpret this we will interpret what this whole thing is all this is for t greater than 0 so there is certainly a part which mixes up the components the velocity this thing scrambles it up here and we will interpret each of these stones so we start with this expression next time and see where this gets us and also try to see what happens as t becomes less than 0 we can write it down from here using physical arguments then the next step would be to use the Kubo formula to get the diffusion tensor so we do that and see what the transverse and longitudinal displacement diffusion coefficients are after which we will take up linear response theory and then come back to this once we have studied what the general formulas in this little bit so let me stop here now.