 Right, so let us now today look at the general correspondence between a stochastic differential equation for a diffusion process and the corresponding Fokker-Planck equation. So let us call this the general stochastic differential equation and the Fokker-Planck equation correspondence. Just to refresh your memories, let us go back and look at what we had come out with in the case of the ordinary Langevin equation. There we had V dot, a velocity, one component was minus gamma V plus a noise. This noise after I divided by m, etc., etc. This turned out to be square root of 2 gamma k T over m and then there was a zeta of T here. I have used the fluctuation dissipation theorem divided by m and so on. And then you have, you had a 2 m gamma k T divided by m square, I put it inside the square root, so this becomes an m. And this implied, if you recall, the original Fokker-Planck equation for the Ornstein-Ohlenbeck distribution. So this implied immediately that the conditional density of V satisfied this equation, P over V T divided by T is equal to gamma times delta over delta V, V times P plus gamma k T over m, this was k Boltzmann T over m, d 2 P over d v 2. And we have to solve this with some specific initial condition, V naught for instance. So the initial condition was P of V comma 0 equal to a delta function at V naught and this led to P of V T equal to the Ornstein-Ohlenbeck distribution. That was the first Fokker-Planck equation we came out with. We also saw what happened if you assume the velocity to be a delta correlated process. Then we were in the diffusion regime. So we saw that for T much, much greater than gamma inverse, which we could implement by actually going back to this equation and then throwing out the inertia term and retaining just this term here, in which case we had a term, high friction limit or high friction limit, we had x dot, that is V is equal to the same thing here, but square root of 2 k Boltzmann T over m gamma z dot, because this term was negligible and I took this on this side and brought the gamma down and you ended up with that thing which then implied here just as this automatically implied it. So similarly back here you have delta P of x T over delta T equal to k Boltzmann T over m gamma d 2 P over d x 2. This was the ordinary diffusion equation because this quantity was the diffusion coefficient. So these are special cases of a more general result which I wrote down which was that if you have an equation which says x dot plus some matrix x, this is a higher dimensional process and this is some, this quantity is an n by 1 column vector and the drift was linear in it. This here on the right hand side was equal to some coefficients, constant coefficients not a function or anything like that times a noise of some kind. So there was some coefficients, constant coefficients and we will make this precise in a minute and then there was a vector valued noise zeta of T. So these are all n by 1 quantities and this is an n by n drift matrix. So if you had a situation like this which is a special case which is a generalization of this fellow here, right. So the reason we want that is because we want to be able to write down Fokker-Plank equations for the phase space density, joint density in x as well as v together. So if you had a situation like this then this immediately implied that delta over delta t and whenever there is a phase space density namely a density in more than one variable I will use a symbol rho so that it is distinct from this little p. So this quantity delta over delta, this quantity here was equal to k ij delta over delta xi xj rho plus some matrix, constant matrix formed from these coefficients here and the natural thing to call it as a diffusion matrix, dij d2 rho over dx i dxj. This diffusion matrix, there is a definite prescription for finding it from these coefficients. We are going to write it in the more general case so I do not bother about writing this down in each special case or anything like that. Suffice it to say that just like in these cases this was a constant, this was a constant, etc. Essentially it is a square of this constant divided by 2. But I want to make it a little more systematic. So you end up with an equation like this. This is the Fokker-Plank equation. Now we can apply this to the situation which we did earlier namely the Langevin particle in one dimension and then we did this quickly for three dimensions. So in one dimension this was of the form x dot was equal to v so x dot minus v equal to 0 and v dot plus gamma v this thing here essentially we are writing this equation again is equal to whatever is there, square root of 2 gamma k t k Boltzmann t over m in this fashion zeta of t. So the noise, the vector noise is 0 in the first row and it is equal to this fellow in the second row. So the coefficient, the matrix of the column vector of coefficients is just this and this. But you actually need a matrix here. So in this particular problem let us write what these things were, k was the matrix 0 minus 1, let us stand, let us put this in here. By the way, the identification is x 1 equal to x, x 2 equal to, that is the identification which we are making to write this down. Then it is 0 minus 1 and then 0 here and a gamma here, this gamma. So it is a linear drift matrix. In fact it is more than that, it is even simpler than that, it is constants everywhere here and it is zeros along this, it helps. And d ij in this case, let me call that matrix d whose elements are d ij. I do not want to write d alone here without subscripts because I have used that for this symbol. I have used that symbol for this quantity and we will stick to that. This thing here is equal to 0, 0, 0, gamma k Boltzmann t over n. So that is the equation obeyed by the phase space density and we have done this already. So let us write that out and then we are on to new material. So that immediately implies that the phase space density satisfies delta rho over delta t equal to, now we need to put this in k ij so the first term is with the minus sign as you can see and then you have a k 1, 2 that is the only element which is present here which is minus 1 and therefore it is delta over delta x of v rho with the minus sign. But v is independent of x, x 2 is independent of x 1 so this is equal to minus v delta rho over delta x is the first term and then k 2 1 is anyway 0, k 2 2 delta over delta v times whatever is sitting here, whatever is sitting inside here which is gamma, okay. So the next term is plus gamma delta over delta v times v rho delta over delta x 2 rho and x 2 is v and then the only term present from the diffusion matrix is the 2, 2 term that is the only term that is non-zero out here so it gives us exactly what we had in the original Fokker-Planck equation, gamma k Boltzmann t over m v 2 rho delta v 2. This is the equation satisfied by the phase space density rho of v x and you are supposed to solve this equation with some given initial conditions so you are supposed to solve it with the initial conditions rho of x v 0 equal to delta of x minus x naught delta of v minus v naught. It is a little harder to solve than the Fokker-Planck equation for the velocity process alone which did not have this term and then we got the Anstein-Ulenberg distribution. Now you have to use this term as well and solve it, it is not such a trivial solution but the solution is a general Gaussian in both x and v, a joint Gaussian and x and v and that Gaussian has the property that as t tends to infinity this Gaussian goes to the p equilibrium in v, the Maxwellian distribution multiplied by the solution of the diffusion equation which of course vanishes as t tends to infinity strictly everywhere but we want a leading behaviour the asymptotic regime. This is p of x comma t and this satisfies delta p over delta t equal to d d 2 p over d x that is the regime in which the velocity has lost its memory gamma t is much much greater than gamma inverse. Gamma inverse is called the Smolukovsky time, the correlation time of the velocity this is gamma inverse is called so it is a time scale in the problem. There is also a length scale in this problem. There is no potential, there is no general external potential because if the particle were moving in a potential like a harmonic oscillator potential or something like that you could get a length scale in the problem but here there is no length scale at all from the potential no potential but there is still a length scale and what would that be? There is a characteristic time scale in the problem which is gamma inverse well there is a characteristic velocity in the problem and therefore there is a or a speed and therefore there is a length scale. What is the characteristic speed in this problem? There is one quantity of dimension speed the mean square in equilibrium from this right so this v square equilibrium is k t k Boltzmann t over m square root of this quantity gives you a speed characteristic speed therefore there is also a length scale what could you do? You divide by T inverse right so this fellow here we put this in is a characteristic length scale in this problem. This is L T inverse this is T inverse so it is an L. You can compute what it is you can compute what it is for the Brownian particles we are talking about we know what the masses are between 10 to the minus 12 and 10 to the minus 15 kilos. We know what the temperature is 300 Kelvin so we know k t. We know what gamma is because gamma is related to the viscosity of a fluid take water at room temperature for instance then gammas of the order of 10 to the minus 6 minus 7 seconds therefore we know what this quantities can compute what this length scale is it is indeed very tiny it is very very tiny alright but we keep at the back of our minds that there is such a length scale even in the absence of an external potential. So having got this far we would like to know what happens next what happens if I put the system under an external force under an applied potential like the harmonic bound for a particle suppose there is a potential present X dependent potential what would happen to these equations. Well the first thing that would happen is that your Langevin equation would become different. Oh incidentally we also saw what the 3 dimensional generalization of this was we wrote the 3 dimensional generalization this becomes V dot gradient with respect to R rho this becomes gradient with respect to V dot the divergence of V times rho with respect to V here and then this is the del square with respect to V so that part is over and in general the solution is some kind of Gaussian which would asymptotically become something like this writing it down in 3 dimensions is very messy but in one dimension anyone can write the explicit solution down okay. By the way one important point in the 3 dimensional case whatever we wrote here in terms of the drift matrix K and so on is still true but in that case what would happen is in 3D for example this matrix K would become 0 minus the identity matrix 0 gamma times the identity matrix where I is a 3 by 3 identity matrix so this is fairly straight forward this is 6 by 6 object and so the generalization is kind of trivial. The reason I am emphasizing this is because we would like to what did we do in the original Langevin case we actually solve the equation of motion Langevin equation and then started taking velocity averages and so on. Now you have a matrix equation to solve which will involve the exponentiation of this matrix you have to find e to the power K just as we found e to the gamma t you have e to the k t that would be your integrating factor if you like but e to the k t this matrix has interesting properties therefore you can find immediately the Eigen values are 0 and gamma so you can actually find the exponential quite explicitly and write the rest of the solution alright so now let us turn to what happens when you have a potential it is a Brownian particle in a potential and again let us do the one dimensional case first so you have as before first let us do the phase space distribution and then we will come back and do the positional distribution technically a little easier because you can see what are the approximations involved because we need to now know what is the diffusion regime we need to know that first so let us look at the phase space problem x dot equal to v as before and v dot is minus gamma v same model as before but now there is an external potential some f of x and I divided by m so it is minus 1 over m v prime of x this is some potential in a potential v of x capital v of x is that term present and then the last term is exactly as it was before plus root 2 gamma k Boltzmann t over m z dot this consistency condition to keep the system in equilibrium is going to remain in any case no matter what v of x does it would still remain because what happens is that the velocity will still thermalize and that is the external potential acting on the particle we do not care about it but the velocity will thermalize and of course the distributions themselves will become very different there is no reason why now if I solve if I write down the phase space density why the solution should be generalized Gaussian no reason at all the introduction of this v prime of x makes the equation of motion non-linear earlier in all the cases we looked at the equation of motion was linear in the dynamical variables there was an external noise but now that is gone unless v of x is either a constant or or is it either a linear function or a quadratic function if it is a linear function this becomes a constant like gravity for instance if this thing is a quadratic function it is like the oscillator problem but in all other cases the equation of motions themselves become non-linear in the dynamical variables and that will go to make it complicated. Now what is the way in which we write down the Fokker-Planck equation we cannot do what we did earlier because we have this fellow which is a non-linear term right so the general correspondence goes like this we do not care how many what the dimensionality of our phase space variables is could be n dimensional in general. So let us write the general Langevin equation general diffusion process has a stochastic equation which is x dot equal to on the right hand side some possibly non-linear function of the variables and this is also vector value. So this guy here x is an n by 1 column matrix and this is an n by 1 vector force there is a force for each of these components and that is in general a function of all the x's all the coordinates. We have still not taken the most general case where this could be explicitly time dependent I said we are not going to look at those problems at the moment plus a g of x this is multiplicative noise in general but you look at times the noise but look at what happened so this is some zeta of t in general this is some matrix acting on this column vector but look at what happened in the earlier case the equation x dot equal to v did not have any noise on the right hand side but v dot equal to something or the other had a noise because we wrote an equation for random force on the particle. So the dimensionality of this noise may be lower than the dimensionality of your phase space variables so we have to allow for that. So suppose this fellow is a nu by 1 column matrix so we have n equations for the components of this but only nu of them have noise on the right hand side and what is the smallest value that nu can have? 1 if it has 0 then everything is deterministic we do not even have a stochastic equation 1 less than equal to nu less than equal to n because all n of them might have noise we do not care. Then what sort of matrix has this got to be? This got to be an n by nu matrix got to be an n by nu matrix this on the other hand is an n by 1 this is n by 1 and therefore n by nu and then a nu by 1 gives you an n by 1 that is the most general diffusion process that we can write down. There is an n dimensional dynamic a set of dynamical variables and then there is a nu dimensional noise on the other hand and the noise is multiplicative in general and then value of the noise for each coefficient could depend on all the dynamical variables you can see this is very general. The question is what is the Fokker-Planck equation corresponding to this and I am just going to write it down so this implies and is implied by Fokker-Planck equation which is delta rho over delta t equal to minus delta over delta x i F i times rho it is like a generalized del dot whatever so it is a divergence term and then the next term is plus d 2 over d x i d x j d i j times rho there is a diffusion matrix but it is not a set of constant coefficients because you have this sitting here so the question is what is this d i j equal to. So it is clear that we have to be a little careful it is not just g squared as it would have been in the 1 dimensional case this matrix d is one half g matrix g transpose because this is n by nu and that follows nu by n so the product is n by n d is n by n diffusion matrix because the indices i and j run from 1 to n and there is a summation over repeated indices so in explicit form this implies that d i j equal to 1 half a summation from alpha equal to 1 to nu g i alpha g j alpha that is what is meant by transpose here clear. So this is the general Langevin equation general stochastic differential equation and this is the general Fokker plant notice that this set of coefficients is not necessarily constants because it depends on this function g inside here and it gets differentiated it is inside so this is fairly complicated here likewise no matter how non-linear the forces this drift term is it is inside here so given that let us try and ask what does our equation look like here for this situation in the 1 dimensional case. So we want an equation for delta rho over delta t where this rho is rho of x v and t so this is equal to well x 1 is x and x 2 is v so we want minus delta over delta x 1 and then f so let us first write this f matrix in this case f equal to this fellow is a column vector and the first one is just v and second portion is minus gamma v minus 1 over m v prime of x that is this column and what is d or the matrix d well it is the same as before because you can see that this thing here is acting in this problem what is n equal to 2 of course and what is nu equal to 1 that is really true only one of these equations as noise the other one does not so this is a very trivial case but the reason I wrote this down is because I am now going to require you next to do this for the 3 dimensional case when you have an external field potential as well as a magnetic field which will make a velocity dependent force so that is a little intricate but it is a 6 dimensional phase space but we can write the equations down given the general expressions. Now what is this matrix equal to it is obviously equal to 0 0 0 and half the square of this which is gamma k Boltzmann t and now we are all set all we got to do is to copy this and write the equation down for rho so this is therefore rho of x v t satisfies the Fokker Planck equation delta rho over delta t equal to minus there is a minus here and then the v comes out because the first is delta over delta x times v rho but v is independent of x as before so this term is still present minus v delta rho over delta x this is the convective derivative v dot del with respect to the coordinate and then there is another term here and what is that that is equal to plus gamma delta over delta v v rho this portion takes care of this and then there is a plus 1 over m delta over delta v of v prime of x times rho but v is independent of x so v prime of x comes out that term sits there and then finally there is only one diffusion term here which is plus gamma k Boltzmann t over m v to rho over dv2 so that is the equation for the phase space density the one particle phase space density in the presence of an external potential v of x for a Brownian particle that is the exact equation. There is no guarantee that the solution to this is a Gaussian a joint Gaussian and x and v because this term is a mess no guarantee at all in general it is some complicated nonlinear function of x and there is no guarantee of anything here this equation is called the Kramer's equation generalized to other potentials in 3 dimensions may be even velocity dependent forces so some generalization of this is a Kramer's equation this is the simplest form of the Kramer's equation now of course you can recognize the oscillator case very trivially because oscillator case goes back to the old problem so if oscillator case potential implies that this term is this is m omega not squared x so this term becomes omega not squared delta rho over delta sorry the not squared x delta rho this particular term it is a linear term so it could in fact be combined with all these fellows it reduces to the old case which is already been solved but this is a little harder to solve in this case because while the diffusion matrix remains the same this will imply that this matrix K that we had written down which was 0 minus 1 0 gamma in the free particle case what would it be in this case remember there is an extra force here this is minus omega not squared x if you bring it to the left it is omega not squared x x not v right so the drift would have one more term this is still 0 x dot is minus v but there is a term here it is omega not squared in this problem and the D would look exactly the same as before this is not so simple to exponentiate although it is any 2 by 2 matrix can be exponentiate it why do I say that because with the 0 here the eigenvalues of 0 and gamma and exponentiating this matrix would have been trivial because K squared would have been proportional to gamma K or something like that but now that is not going to happen there are 2 eigenvalues here which depend on both gamma and omega not squared not surprisingly they would be precisely the eigenvalues of the zeros of the susceptibility the poles of the susceptibility so they depend whether they are under damped over damped critically damped definitely have that so the solution will be in terms of trigonometric functions and damped exponential in the under damped case of hyperbolic functions in the over damped case so it is considerably more messy although you can exponentiate this so not hard but this is where it would change but we can write the formal equation now what do you expect the solution of such as an equation to become for long times I would still like to look at the diffusion regime and see what happens in the free particle case when this was not there recall that this row asymptotically went to a product of the equilibrium Maxwellian distribution in v multiplied by the solution of the free diffusion equation in x right now what do you think will happen will that happen in this problem as well well in general you have to look at what happens when t is much much bigger than gamma inverse the diffusion regime or gamma is very large the high friction limit so you have to ask what happens to the solution in that high friction limit but there is also a length scale in the problem so now the diffusion regime is defined as follows diffusion not only should you have t much much greater than gamma inverse but also the force v prime of x must not vary too rapidly within the characteristic length scale so it is as if the force did not exist at all in that length scale right or a constant or whatever so within v now v squared equilibrium to the half divided by gamma so that will give you a criterion you can compute this number and then ask check whether v prime or the force is rapidly varying spatially or not if it is a gentle potential of some kind it would not very significant then how would you go to this regime what would you do you do exactly what you did for the original free particle case namely you would say I am going to throw away the inertia term look at the high friction limit and ask what happens there so in that high friction limit let us write it down below this so what happens is that this row tends in the diffusion regime to p equilibrium of v multiplied by an equation for p of x, t where p of x, t satisfies the Fokker Planck equation corresponding to setting the friction term the setting the inertia term to 0 so essentially what you are doing it is to take m x double dot plus m gamma x dot equal to minus v prime of x plus a noise and the noise was precisely the noise that we had in the original so 2 m gamma k Boltzmann t times zeta of t this term and you are dropping this term the high friction limit so this term dominates so that gives us an equation which says x dot equal to minus 1 over m gamma v prime of x that is the drift term plus square root of I have to divide by m square gamma squared so it is twice k Boltzmann t over m gamma that is our old friend square root of 2 d appearing again times zeta of t corresponding to this stochastic equation and what is this stochastic equation corresponding to x dot equal to minus 1 over m gamma v prime of x plus square root of 2 k Boltzmann t over m gamma zeta of t that is a one dimensional equation with a non-linear drift term here but we know how to write the Fokker Planck equation down for it so what would that be in other words delta over delta t of p of x, t should be equal to with a minus sign so that goes away and then you have equal to 1 over m gamma sitting here d over dx now of v prime of x times p because that is inside you cannot do anything about it plus the second derivative half of the square of this which is k t over m gamma but that is just our old friend d plus d times this is as before k Boltzmann t that is the Fokker Planck equation this equation is called the Smolakowski equation so it gives you the diffusion equation in the presence of a potential but you see where it comes from it really comes from the Fokker Planck equation in phase space it gets reduced to this in the diffusion regime where you have to have an extra criterion now about the variation of v prime of x we have not taken we have not done the problem in the presence of a velocity dependent force I assume that the external force was a potential which depends only on the position so no velocity dependent forces no magnetic field etc has been included here as yet now the question is does this have does this have any equilibrium solution or not as t tends to infinity we know that if it is not this is not there it does not it is just the Gaussian which tends to 0 flattens out but now this will depend on what this v prime of x is we kind of expect that if you had a confining potential that prevents you from having long range diffusion then the mean square displacement won't diverge with t and therefore it might be in equilibrium case for instance the oscillator now let us look at the harmonic oscillator in the over damped highly over damped high friction limit case so highly over damped what is the Smolakowski equation for the positional position probability density of the highly over damped harmonic oscillator well this fellow is trivial to write down it is m omega not squared x and the m cancels right so you end up with delta p over delta t equal to omega not squared over gamma times delta over delta x of x p plus d times this is k t over m gamma that is the Fokker-Planck equation does it remind you of any other Fokker-Planck equation you have seen yeah the velocity process for a free particle what did that equation look like that looked like delta p over delta t equal to gamma delta over delta v times v times p plus gamma k Boltzmann t over m d to p over d x d v to where this p was p of v comma whereas this p is p of x comma it is not the same function and these identical apart from reinterpretation of constants the one over gamma the gamma here is replaced by omega not squared by gamma and same physical dimensions for both and the k t over m gamma is replaced by gamma k t over m this is the diffusion constant in velocity space for a free particle or for a particle in the presence of a potential it does not matter that is the diffusion constant position space what was the solution to this equation the on-synol and by distribution with the mean which went like e to the v naught e to the minus gamma t and a variance which started with the delta function with zero and then went to the variance of the equilibrium max value so there was an equilibrium distribution which was precisely the max value in distribution this is exactly the same mathematically exactly the same problem so that is why you would very often see the on-synol and by process described in some books as the oscillator process what they mean is that the conditional density of the simple harmonic oscillator a harmonically bound particle the Fokker-Planck equation satisfying that gives you the on-synol and by distribution just as for the free particle the velocity has this on-synol and by distribution in the over damped limit in the Smolukowski approximation this thing becomes precisely the on-synol and by distribution now how do you find the equilibrium distribution in this case you would set this to equal to 0 then you would find the current here would vanish at infinity you would get the Gaussian e to the minus m v naught squared over 2 k t what will you get here what would be that distribution so there is an equilibrium distribution in this case we have to look at the general case we will subsequently of when you have an equilibrium distribution but in this case there is no doubt at all that p of x t tends as t tends to infinity there is genuinely an equilibrium distribution here because that would correspond to setting this equal to 0 then this is d over dx comes out and becomes whatever is inside is a constant which can be set equal to 0 for normalizable distribution and then you get d p over dx plus omega squared over gamma times m gamma over k t times p equal to 0 so what would you get what is and there is an x so what kind of solution do you get it is a Gaussian and what would that Gaussian be constant times e to the power minus pardon me s m omega naught squared x square over right you expect that of course because that is the Boltzmann factor for the potential energy in equilibrium you expected to go to the Gibbs distribution the Maxwell Boltzmann distribution so it is e to the minus the Hamiltonian and the Hamiltonian as half m u squared plus half m omega naught squared x squared divided by k t is what is coming out so you want that as a consistency check otherwise it is really off right so you do expect that this will go to the Maxwell Gaussian distribution in the general case no guarantee because you would have to set this equal to 0 and then ask whether this ordinary differential equation has a solution or not and yes suppose it has a solution what would that look like so this becomes a dx d over dx p equilibrium this becomes d to p over dx to equilibrium what would this solution look like we want this to be equal to 0 now of course that simply means d over dx of this whole business the current could be constant it could be going to some stationary constant current then without knowledge of that stationary current you cannot solve the problem but let us assume that at t could plus minus infinity this thing vanishes then what you have is an ordinary differential equation it says d d p equilibrium over dx plus 1 over m gamma times v prime of x p equilibrium equal to 0 but d is k t over m gamma they cannot be any reference to the friction constant in the equilibrium positional distribution right so this goes away and this was a k t so this becomes k Boltzmann t and what would that tell you this tells you implies p equilibrium of x is proportional to e to the power minus v of x over k Boltzmann the integral of v prime x is v of x but that is just the Boltzmann factor which is what you expect in equilibrium that will be e to the minus the Hamiltonian over k t and the Hamiltonian has a kinetic part which you took all separately and then a potential which is precisely this so checks this thing checks there are the possibilities we have assumed the stationary current is not there but otherwise so this will happen whenever this v of x is bounded I mean it binds the point sort of stops long range diffusion then the particle is bound in some sense and it cannot diffuse out to infinity and you end up with an equilibrium stationary distribution no guarantee that it always exists but in these cases normal cases it would exist okay so the magnetic field case is a little more intricate than this and we will talk about this later on and then see how to generalize this stop