 So, today what we will do is to look at this in a more general context and write on multidimensional Fokker-Planck equations and apply it to the case of particles moving in phase space itself. So to recall to you what the equations were, we had an equation that said x dot minus v was equal to 0 that just says the velocity is the derivative of the position and then v dot plus gamma v this is the friction part the systematic part of the random force on it. This was equal to square root of gamma over m eta of t where this was Gaussian white noise. In other words it is a delta correlated Gaussian process stationary Markov process which is a delta correlated with unit strength. So, eta of t equal to 0 and eta of t, eta of t prime equal to delta function of t minus t prime. Now the question is what is the corresponding Fokker-Planck equation satisfied by the probability conditional probability density in both x and v jointly. So, we are asking for the equation satisfied by p of x v t given that it started from x naught and v naught at time 0. It is important to remember that earlier we had equations only in v and of course the equation satisfied by v was the Fokker-Planck by v was the Langevin equation. We got a Fokker-Planck equation for it and the solution turned out to be the Onstein-Ulenbeck distribution and we discovered also that the velocity process was exponentially correlated and that it was a stationary random process. So, that much we have already seen. Now the question is what does this look like? What equation is satisfied by this quantity here corresponding to the set of equations here. Now it turns out that if you have a general process, a multidimensional process of this kind, if you put them together in some vector form and you call that a vector xi say xi dot plus if you got a linear drift or xi, this is some constant matrix here acting on this column vector here. If this is equal to on the right hand side you have the usual noise of some kind. So, we have gamma over m and then it has a vector of 8 of t and this stands this quantity xi stands for x and e and this 8 of t stands for 0 8 of t. There is no noise here in this equation but there is a noise driving noise on this side due to the collisions with the other particles in the fluid. If you have a Langevin equation a stochastic differential equation of this kind corresponding to it this quantity xi its conditional probability density satisfies a certain Fokker-Planck equation and that Fokker-Planck equation is given by the following it is given by delta over delta t and let me call it rho. Let me call this rho so that you know that it is a phase space density this is equal to on the right hand side you have r i j delta over delta xi i xi j rho that is the drift term the first term plus d i j d to rho over delta xi i delta xi j where this matrix d i j is in this case a 2 by 2 matrix which is 0 0 0 and then whatever was the matrix which came out from one half the square of this guy here but remember that consistency required that gamma was equal to 2 m gamma k Boltzmann t that was the fluctuation dissipation theorem this is what kept the velocity in equilibrium in the Maxwellian was retained in time then one half the square of this guy the 2 goes away there is an m on top comes and that m cancels out here with this m squared and give you a factor m in the denominator so you are going to get gamma k Boltzmann t over m this is what this guy is and that is the Fokker-Planck equation okay all we have to do and there is a summation over repeated indices i and j run over 1 and 2 okay now in this problem what is r well from these two equations it follows that r in our problem here is equal to 0 minus 1 0 and gamma so r 1 2 is minus 1 r 2 1 is 0 there is no x here and then r 2 2 is gamma and that is it so that is the Fokker-Planck equation all we have to do is to write it out explicitly and you immediately see that delta rho this thing here is equal to in this particular problem r 1 2 is a minus sign so this is minus delta over delta i is equal to 1 so there is an delta over delta psi i r 1 2 is what we need so this is equal to 1 that gives me a minus delta over delta x and this is 2 so that is a v times rho out here and that is it there is one more term which is r 2 2 which is delta over delta v gamma v rho that is the 2 2 part plus the only term that survives in this term here is d 2 2 that is the only nonzero element here and then it becomes delta v 2 so plus gamma k Boltzmann p over m d 2 rho over delta v 2 now of course v is independent of x because it is a dynamical variable which is independent of x so this is equal to minus v delta over delta x rho plus gamma times delta over delta v v rho plus gamma k Boltzmann t over m this is the Fokker-Planck equation satisfied by rho it is a fairly complicated partial differential equation this portion was already there in the Fokker-Planck equation satisfied by the conditional density of the velocity itself so was this but there is an extra term here out here and this is like a convective derivative because it is v delta over delta it is like v dot del three dimensions it is a one dimensional analog of that so if you bring it to the left hand side you have d by dt d over dt of rho the full derivative equal to whatever is on the right hand side with this linear drift and this extra diffusive term here now the question is what is the solution well depends on the initial condition and we are talking about this conditional density so at t equal to 0 this satisfies rho of x v 0 at x naught v naught is delta of x minus x naught delta of v minus v naught so with that initial condition if you solve that partial differential equation with the boundary condition that these things vanish as x and v tend to infinity you get Gaussian's as a solution finally we are not going to do that here but you end up with some complicated Gaussian solution a bivariate Gaussian both in x and v so it will have terms like e to the minus x squared minus e to the minus v squared and e to the minus plus x v there be cross terms the interesting question to ask is what happens to this as t tends to infinity in this problem t tending to infinity means t much much greater than the velocity correlation time gamma inverse so it turns out that rho of x v t given x naught v naught tends as t becomes much much bigger than the correlation time it tends to the Maxwellian in v and the Maxwellian in v is of course m over 2 pi k Boltzmann t to the power half e to the minus v squared over 2 k Boltzmann t so the velocity thermalizes and then you have the positional part and that of course is 1 over 4 pi d t to the power half e to the minus x minus x naught whole squared over 4 d t you recognize this to be the limit of the Antstein-Ohlenberg distribution as t tends to infinity that is just the equilibrium Maxwellian distribution and that is the solution to the ordinary diffusion equation the question is what is this capital D and we saw by comparison of the Langevin equation with the diffusion equation in the long time limit we also saw that the other the other consistency condition was d is k t over m gamma so is there an equilibrium distribution no because as t tends to infinity this vanishes out here so it says that there is no stationary distribution for ordinary diffusion at all everything goes to 0 the whole Gaussian goes down flattens out to 0 the velocity however thermalizes the velocity is a stationary random process it thermalizes one could ask alright this is the phase space density what happens if I integrate over one of the variables I should get the distribution in the other variable right so if you read this if you took the exact solution if you take integral over dx rho of x v t x naught v naught over the full space what should this give you what would you expect it to give it should give you well it should be still a function of time because all you have done is to take the joint distribution in x and v and you were integrated over the position so it should give you the conditional density in the velocity alone and what is that distribution it is not the Maxwellian that is the limit distribution the on-strain ohlinbeck distribution so it will end up with exactly the on-strain ohlinbeck distribution so this thing here will become m over 2 pi k Boltzmann t 1 e to the minus 2 gamma t that is the variance to the power half exponential of minus v minus the mean value which is e to the minus gamma t remember this is conditional on this initial condition always so it is this squared divide m times that over 2 k Boltzmann t 1 minus e to the 2 gamma t this is the on-strain ohlinbeck distribution so indeed it will turn out after you integrate over the position you end up with this okay what would you expect if you integrate over the velocity instead what do you think you will get if you did minus infinity to infinity dv rho of x v t x naught what should you get what would you expect would it be this would it be this what does your intuition tell you would it be that distribution I got rid of the velocity altogether so do you think it would be that distribution that is only valid in the diffusion limit that is only valid for t much much greater than gamma inverse we are not saying anything like that here we just integrating over v so what would you expect it is the positional distribution no doubt about it it is the positional conditional distribution density in the position it will be a Gaussian because you have got a bivariate Gaussian you are integrating over one variable you are going to get a Gaussian in the other variable no doubt but much more complicated than that simple diffusion equation solution it is some Gaussian you need to know the exact solution of this equation in order to be able to do that I am not going to write it down here I will give it to you in writing elsewhere it is a complicated formula but that distribution will tend as gamma t tends to infinity will tend to that diffusion equation solution okay and interestingly enough there is no simple formula here whatever distribution you get here that does not obey any simple master equation okay so although the phase space density itself obeys this master equation the Fokker-Planck equation and the velocity process itself obeys a Fokker-Planck equation the position does not obey such an equation simple equation at all except in the diffusion limit and you go to this long distance and in fact several ways of seeing this that the position is not stationary random process we will talk about the Wiener process very shortly and then you will see how complicated the position actually can become okay and what Brownian motion means and so on and so forth okay so much for this that this distribution has an asymptotic limit which does this and when you integrate it over the position you end up with the Onstein-Ohlenbeck distribution now we can generalize this a little bit go a little further we can ask what happens if I had an external force on the problem what would then be the effect of an external force one way to approach this is to ask is to look at a slightly different model instead of a free particle let us look at the problem of an oscillator a simple harmonic oscillator now what we are talking about is a particle on the x axis here is x equal to 0 and this is a simple harmonic oscillator in the absence of any external force but we now imagine it is in a fluid with viscosity and this particle is being hit randomly by the other particles in the fluid okay and we want a Langevan type model for its motion here so there is a white noise exactly as in this case but there is also a systematic part of the force which corresponds to putting in the harmonically bound particle namely this thing is in a potential now this potential is going to do something interesting the actual harmonic oscillator potential is half m omega naught squared x squared as you know now the first effect of this potential is that the translation symmetry on this x axis is lost because this point becomes a special point is a potential here the other thing is your it is as if this particle is connected by a spring harmonic spring to the origin and therefore there is a strong restoring force if the displacement is very far away from the origin so you might expect okay there is going to be two effects here one of them is the random force due to the other particles which is tending to diffuse move this particle away and make its mean square displacement diverge as a function of time the other effect is this restoring force which is always acting the question is what is going to happen to this particle will it diffuse or will it not will the variance of the displacement diverge as t tends to infinity if it does so linearly you know that it is diffusive if it goes to finite constant then you know it is not no long range diffusion at all so the question is what is going to happen so that is the physical question we can answer it in the following way I can write this down by adding to this the restoring force and that of course is minus m omega squared x and I have divided out by m so there is also a term v dot plus gamma v plus omega naught squared x let me call it omega naught the unperturbed frequency of this oscillator so this is the case of a harmonically bound and that is the equation everything else is unchanged exactly as it is now of course the velocity once again thermalizes and there is an equilibrium velocity distribution which is again the Maxwellian distribution before but there is also this potential energy term coming from the potential in which the particle is okay so now what happens exactly as before I can write down the same problem I have rho of x v t and we could start it from zero for example just to make sure that it is a simplest condition but whatever it is given some x naught given some v naught at zero the question is what does this particle do what is the density satisfy again the same Fokker Planck equation as before same as this exactly but whereas in the previous case this fellow was equal to zero r was equal to zero minus one zero gamma what is it now remember there is an extra term here omega naught squared and this is appearing with a coefficient x in the equation for v so it means there is a term here which is omega naught squared that is the only difference between the two everything else goes through exactly as before so there is going to be here another term which is equal to plus there is a term which is equal to delta over delta and what would this be it is in the v equation okay so r 2 1 is what contributes so this is delta over delta v omega naught squared x rho the term of that kind because it is going to contribute to a delta over delta v because it is r 2 and this is a 1 so there is an x inside omega naught squared x that is the only change and we have an exact answer for the Fokker Planck equation satisfied by a harmonically bound particle in phase space and of course omega naught squared is constant so let us write that out so let us write this as equal to minus v delta rho over delta x plus omega naught squared times x times delta over delta v delta rho over delta because this time x is independent of v so that comes out just as out here v was independent of x so it came out and that is the Fokker Planck equation okay now the solution to this is very different from the solution in the absence of omega naught altogether what do you think will happen physically what do you think will happen think back to equilibrium statistical mechanics if you have a free particle free gas the position is irrelevant in the position is equally spaced distributed entirely in the container then it is only the velocity that is relevant okay so gas and thermal equilibrium classical gas you only talk about velocity distribution at a given temperature the position is taken to be uniform throughout but now there is a harmonically bound particle so it is clear that the energy of this particle there is a contribution which is half m omega squared x squared to the potential energy of this particle and we know that the relative probability of any value of the energies e to the minus epsilon over kt right so you would expect to get in equilibrium the distribution would be biased towards x equal to 0 obviously most probable value of x would be 0 in this case and what is the actual distribution it is the Boltzmann distribution right so what is the actual equilibrium distribution row equilibrium of x and v or the stationary distribution from equilibrium statistical mechanics what would this be it is e to the minus the energy over kt and that is it normalized appropriately normalized so this guy would obviously be e to the minus mv squared over 2k Boltzmann t that is the kinetic energy part minus m omega naught squared x squared over 2k Boltzmann t that is the potential energy part right times normalization factors e to the minus ax squared where a is a positive constant if you integrate you end up with square root of pi over a so it is square root of a over pi is a normalization right this thing here equal to square root of m over 2 pi k Boltzmann t square root of m omega naught squared over 2k Boltzmann t this is the normalized distribution such that if you integrate over x and b you are going to get unity so the distribution factors into something in x Gaussian in x and a Gaussian in v velocity you expect this on physical grounds this is going to happen and that is it so this exact solution will reflect that will reflect precisely this now tell me does the variance of the mean displacement diverges a function of time do you think or no it cannot it cannot because the limit value limiting distribution in position is a Gaussian and Gaussian has finite variance completely so there is no long range diffusion the behavior is not diffusive standing to a Gaussian it is going to do exactly what the velocity does except for change of constant here and so on notice dimensionally everything is okay because the Maxwellian distribution in velocity must have physical dimensions 1 over the velocity so that p of v dv is equal to 1 when you integrate right so there is an m here and there is an m k t which is m l square t to the minus 2 and the m cancels and you got a l square t to the minus 2 square root which is l t inverse so this whole thing the velocity is in fact 1 over velocity 1 over velocity in physical dimensions and similarly this is going to be 1 over length because the t to the minus 2 cancels here okay so this is perfectly correct dimensionally and you end up with distributions which are stationary distributions in this case so the mean square displacement will not diverge what will it be actually the mean square displacement what do you think this will be what is what do you think v square will be in equilibrium what should this be k t k t over m yeah because half m v square will have half k t so this is k t this is k b p over m x squared in equilibrium what will this be no long range diffusion and that is reflected by the solution to this guy again this is a bivariate Gaussian this fellow here is a bivariate Gaussian which will now tend in as t tends to infinity it will tend to this equilibrium distribution here but there is an interesting wrinkle here what do you think is the relaxation time for the position and the velocity you can tell what this is by looking at these equations here and in a minute you will see what the difference is in a minute you see what happens is that the velocity thermalizes over a time scale gamma inverse that is the correlation time of the velocity but the position does not thermalize with that well for instance if it is an under damped oscillator it is clear that it is going to reach equilibrium values in an oscillatory fashion always if it is over damped it is going to go monotonically and so on right so the omega not and gamma together will provide two time scales in this problem and what is the criterion for over damping of an oscillator given given this current a natural frequency omega not and a damping constant gamma both of which have time scale dimensions of frequency so you know that in this general expression you end up with gamma over 2 whether that is bigger than omega not or smaller than omega not etc this is over damped and of course less than omega not is under damped we do not care what it is this equation is valid in general whatever the value of omega not and gamma b this equation is true in general and there is a complicated solution in bivariate Gaussian solution but the relaxation times of the two are different from each other we will see in a minute show you what the relaxation effective relaxation time is for the case of the position because you can also ask in this problem in which I am telling you I am asserting even without writing the general Gaussian solution down that x does not diffuse the behavior of the there is no diffusion long range diffusion here and that there is a stationary distribution both in x and v this case both x and v are stationary Gaussian processes processes in this case so here if you integrate over the velocity you will end up with the distribution for the position whose solution is some kind of Gaussian and describes a stationary random process it is the distribution conditional density is a stationary that of a stationary Markov process similarly for the velocity as well okay. Now we could ask can we quickly see what the position itself does how do how do we get at it assuming that the velocity has thermalized what would one do in this case well this is what one would write down at start with the second equation and write it in this form x double dot plus gamma x dot plus omega not squared x equal to square root of gamma over m 8 of t okay and then I look at the over damped oscillator case in other words very high friction so this is called the high friction limit there is a systematic way of doing this high friction I have divided by m but imagine putting the m here and saying this term dominates very high friction gamma is very large then the inertia term can be neglected okay the effect of the mass is supposed to be like the friction is supposed to be very high and then m can be neglect this original term and then it becomes a Langevin equation just like the velocity Langevin equation for a free particle right except a slight difference here so now you got an equation which says x dot plus omega not squared over gamma x let us put it on the right hand side so x dot equal to minus this guy plus square root of gamma over m gamma 8 of t that is the Langevin equation for the position in this limit okay so let us simplify it a little bit and what does this give you this is equal to minus omega not squared over gamma x plus remember this was 2m gamma whatever it was right and there is an m gamma out here so this is square root of 2k Boltzmann t over m gamma 8 of t in this limit in this high friction limit so that is a Langevin equation compare this with the Langevin equation for the velocity for a free particle what did we get we found that v dot compare with v dot equal to minus gamma v plus square root of gamma over m that was equal to square root of 2 gamma k Boltzmann t over m compare with that so apart from this thing gamma here being replaced by omega not squared over gamma and then this constant changing to this other constant it was exactly the same in structure right so what is the Fokker Planck equation for this p here so what is this going to be p of x t assuming that you start from some origin at p equal to 0 x not we do not care satisfies delta p over delta t equal to and what is the first term on the right hand side this is a simple Langevin equation for which you can write the Fokker Planck equation immediately right so this is equal to omega not squared over gamma delta over delta x because everything is in x x p that is the drift part plus one half the square of this guy whatever this was plus k Boltzmann t over m gamma d 2 p over d x that is the Fokker Planck equation for p of x comma t right do you recognize k t over m gamma what is that that is the diffusion constant that is the diffusion constant so what we have got here is what happens in the high friction limit to harmonically bound particle as it diffuses so there is a correction if you like due to the potential to the diffusion term this is d capital D without this omega not it would be delta p over delta t is d times d 2 p over d x so that is the plane diffusion equation and you have long range diffusion but now you have got this extra term here as soon as you have that what does that imply it says something very interesting it says immediately that a stationary distribution would exist p of x as t tends to infinity and it should satisfy an equation in which this term is 0 delta p over delta t is 0 right it must satisfy stationary distribution p stationary of x must satisfy d over d x so let us write it down of k Boltzmann t over m gamma d p stationary over d x plus omega not squared over gamma p stationary equal to 0 what I have done is to say this term is a function of x alone now so this goes away and I have just put that in and this is it or there is an x d over d x x times p stationary which implies of course that this comes out and then if I take this down there it says m omega not squared over k Boltzmann t therefore this is a constant independent of x okay that is the current and if you say the current is 0 at infinity p stationary vanishes and so on and derivatives vanish then the constant is 0 everywhere so we can get rid of this in infinite medium and is equal to 0 implies p stationary equal to e to the power minus m omega not squared x squared over 2 k Boltzmann t all I have to do is integrate x and that is x squared over 2 with a minus sign and then a normalization constant which is m omega not squared over 2 k Boltzmann t square root 2 pi and that is the solution which is exactly what you get from equilibrium statistical mechanics. So you have got the Gaussian stationary Gaussian there is no long range diffusion etc but the general equation is this this equation is the equation satisfied by a diffusing particle in the presence of an external force a potential in this case a harmonic potential in this case it is called a Smolukowski equation this this equation here is an example that term is generally applied to the diffusion equation in the presence of an arbitrary external force we are going to apply to other situations as well for instance if you put a magnetic field the particle is charged and you put a magnetic field it moves in 3 dimensions and the question is what kind of probability density to the position and velocity and the phase space density have that is a question we are going to answer but this is an the simplest example of a Smolukowski equation okay what can generalize this a little bit and say it does not have to be a harmonic potential it could be any function any force you like we need that because we are soon going to do the case of a magnetic field so what do these equations look like so particle let us write in a potential let us say v of x here let us write the phase space equation down and then see what happens so again let us do it in one dimension first and then we will generalize this to higher dimensions so you have x dot minus v equal to 0 but now let us put the v on the right hand side just dot equal to v and m v dot equal to minus m gamma v that is the systematic part and then there is a term which is minus v prime of x that is the force on the particle plus whatever is the random force plus so let us divide through by this guy as always 1 over m plus square root of gamma over m and now this v prime of x may be nonlinear in x that makes the equation extremely hard to solve because you no longer have a linear drift now we got a really hard problem on our hands so the question is what does this solution look like what does this thing look like in general what is the phase space density Fokker Planck equation look like okay so for this we need a slight generalization of the linear drift case where the drift matrix was given by R now it is some complicated nonlinear function in general so let us say as usual that we have an equation of the form psi dot equal to some vector valued function F of psi I have in mind putting this psi as the x's and v's depending on how many dimensions there are all the dynamical variables plus a multiplicative noise which is let us say g of psi on this a dot and let us be completely general we do not know what the dimensionality of the psi is if it is a particle moving in one dimension then there is an x and a v if it is moving in three dimensions then there are three x's and three v's and so on and so forth and we have so far been looking at the case where the noise is just in that single component of the velocity but there could be different noises on different components they could be completely uncorrelated noises for instance if the particle moves in three dimensions you resolve it into three Cartesian coordinates there is no reason to expect that all the forces in all the directions are identical they are completely uncorrelated in fact what we will do and end up doing is to say that this force eta i of t eta j of t prime expectation where i and j are Cartesian components would be uncorrelated with each other if i is not equal to j so I am going to say this is equal to delta i j delta of t minus t prime for instance we are going to do things like that so we need to keep that in mind so let us be completely general and say that this is an n dimensional object so I write it as a column vector in other words n by 1 object column vector of this kind so is this and this noise could be in some of the components it may not have any noise at all in some of the others for example there is no noise here no explicit noise okay so let us say this fellow here is nu by 1 so it is nu dimensional white noise there are nu of these guys eta 1 eta 2 up to eta sub nu then this g in general this would be an n times nu matrix so that when it acts on the nu times 1 it is going to give you an n by 1 okay that is the general situation then corresponding to this equation this implies by this correspondence we have between Markov process the Langevin type equation for a Markov process and the corresponding Fokker-Planck equation for a for a diffusion process and the line Fokker-Planck equation we have delta rho over delta t equal to in this case delta over delta psi i psi i times rho okay and summed over the components i sorry f i times rho plus delta 2 over delta psi i delta psi j d i j which could in general be a function of this psi because this guy is times rho where d i j equal to one half as you would expect g g transpose i is a transpose that is the generalization okay remember that g is an n by nu matrix so g transpose is a nu by n so this whole thing turns out to have the right dimensions and that is the general Fokker-Planck equation and what we need to do is to apply to this case out here the reason I need this generalization is because this is not linear in the oscillator case this was linear and I just identified that our matrix in the matter was over but now this is not linear so the solutions in general would not be Gaussian's or anything like that fairly complicated things at all and at the moment we do not know there is a stationary distribution or not at all so what does this give us it says in this case delta rho over delta t equal to and first we got to do this term here so exactly as before you are going to have F in this problem so let us write down the F as I in this problem is just X and V F in this problem is equal to there is a V and there is a minus V prime of X over M minus V d ij mercifully is easy enough it is exactly as before because again this is just constant out here this is 0 out here what is new in this problem n is 2 little n is 2 because it is a 2 dimensional object what is little new just 1 just 1 so it is a very trivial problem in this case so d ij once again this matrix in this problem is again 0 0 0 gamma k volts may be half g g transpose etc so we can write down what this equation is in one dimension delta rho over delta t equal to minus V delta over delta X rho that is one portion that comes from here because there is a minus sorry minus I took F on the right hand side so there is a minus sign okay and then plus gamma delta over delta V V rho that term will always remain and then there is a term which is delta over delta V so this is equal to plus V prime of X over M delta over delta V rho that is this term plus the usual plus gamma k volts many over M d to rho over delta this is the one dimensional phase space density conditional density for particles moving on the X axis alone one just one component and it is got this extra term here due to the external force okay or the potential applied potential everything else is familiar as it stands this equation is called the Kramer's equation we will write down generalizations of this to 3 dimensional problems etc there it is fairly very straightforward to write it down in 3 dimensions in fact I urge you to do this as an exercise do this in general when you write this as a vector and this equation too as a vector now this gamma arises from a fluid and it is related to the viscosity of the fluid which will take to be isotropic so it is the same in all the directions does not matter for all the Cartesian components and this of course will be replaced by the gradient minus the gradient of the potential and then for the noise take this to be eta sub whatever I corresponding to V I and eta I is delta correlated such that different Cartesian components are not correlated to each other okay and write the general Kramer's equation down in 3 dimensions for the phase space density as a function of capital vector R vector V and T so that is a fairly straightforward generalization what would happen if you put this particle in a magnetic field what kind of force does it see that is an interesting case I mean we are going to do this problem but just to anticipate what is going to happen what do you think will happen so I put a magnetic field in some direction like this and you have this particle moving in 3 dimensions inside a container fluid and it is diffusing and there is also this magnetic field what do you think will happen well the particle will tend to do cyclotron motion around the direction of this field right on the other hand it is also diffusing what is the force applied by this magnetic field on the particle it is the Lorentz force right so there is a Q times V cross B so there is going to be an extra term on the right hand side which is Q over M V cross B that force is not derivable from a potential from a scalar potential which is a function of position it is a velocity dependent force but the problem is still very tractable and solvable why is that what is it about this force that makes the problem solvable completely it is linear it is linear in V so as soon as we have a linear problem this drift matrix becomes R which is a constant matrix and then I have a problem of exponentiating this constant matrix the green functions can be found etc etc may be complicated but in principle but in practice it is doable completely so the fact that it is linear is very very crucial so this problem can be solved it is a velocity dependent force of course it is going to look like this term it is going to have a V but the difference of course is that V 1 dot will involve V 2 and V 3 whereas this term is just V 1 in EB 1 dot etc so it will mix it up and secondly this gamma is represents the effect of dissipation as you know if you put a particle in a magnetic field a free particle its energy does not change kinetic energy does not change so there is no loss of energy at all so that portion of the drift term will be reversible in time but this portion will not be this is what leads to things at tending to an equilibrium etc now tell me suppose I have a gas of charge particles and there is overall neutrality maintained by some background and then these particles are diffusing at some finite temperature clearly if I put a magnetic field that field does no work on these particles whatsoever so will it change the Maxwellian distribution at all if it is in equilibrium it should not change this Maxwellian distribution the temperature will remain exactly the same nothing is going to happen but will the diffusion constant get affected what will happen what do you think is going to happen well we know that otherwise if it is one dimensional motion we know that this d was equal to k Boltzmann t over m gamma that by the way remains true no matter how many dimensions the particle moves in because each Cartesian component the variance goes like 2 dt and if you have 3 components in the r squared goes like 6 dt that is all that happens so the d is exactly the same k t over m gamma but now I put a magnetic field what do you think is going to happen this isotropy is broken there is a specific direction single doubt by the magnetic field will the diffusion along this direction be affected at all it would not be affected at all because there is no force in this direction at all on the other hand in the perpendicular plane there are 2 other directions in this plane the diffusion is inhibited because when it tries to diffuse there is a cyclotron motion kicking in trying to make it curve its path back again so we are going to discover that this diffusion tensor d i j is not going to be a constant times a unit matrix it is going to be such that the x and y components are not as large as the z z component so d 3 3 if you put the magnetic field in the 3 direction is going to be bigger and in fact we expected to be just k t over m gamma but in the other 2 directions the diffusion constant is going to be inhibited we will see explicitly how that comes about but that is just a physical argument we will write it down explicitly and see what happens here. So this equation here this basic equation the Kramers equation is a starting point for discovering whether the system has a stationary distribution or not and so on what we are going to do now next is to first take care of the problem of the magnetic field and after that I am going to go back and say let us look at the diffusion process itself the simple diffusion equation itself a little more carefully and ask what sort of process is the process x where the density of x the probability density of x obeys delta p over delta t equal to d d 2 p over dx 2 the original diffusion equation in one dimension and the corresponding Langevin equation which in this case was just x dot equal to square root of 2 d times 8 over t in the diffusion limit this means that you have gone to time scales much bigger than gamma inverse and then the particles is essentially as if the velocity is uncorrelated it is delta correlated it is a noise in this case in this approximation you have x dot is just white noise on the other side and corresponding density obeys the ordinary diffusion equation this is called Brownian motion mathematical Brownian motion and this process x this x process is called a Wiener process and it is the integral of white noise because formally what is happening this will imply that formally x of t minus x of 0 equal to integral 0 square root of 2 d times integral 0 to t dt prime 8 of t prime so what we have is a process that is the integral of white noise and it is much less singular than the white noise itself which has this delta function kind of correlation and we are going to ask what is this x of t it is called a Wiener process and we are going to study it and its sample parts in some detail so that will be the next program because this thing is what acts as a paradigm the very model for random process as random as you can get in some sense and it has a lot of interesting properties once we do that I will come back and make a connection between this and the Onstein-Ulenbeck process which as I told you is a unique continuous stationary Gaussian Markov process it is a unique process and we will see that there is a theorem which will tell you that in some sense if you studied this process the Wiener process you studied all Gauss-Markov processes we will see how this is done very mapping okay so that will be the next article thank you.