 Right, so we will tie up one of the loose ends that was left in our discussion of the Langevin equation and Fokker-Planck equation and related matters. And this had to do with the possibility that the force on the Brownian particle may be velocity dependent and while the formalism is quite general, we will do this in the specific case of the magnetic field because that is a directly accessible, experimentally accessible situation and we have studied it in the past as well. So let us look at how to obtain the phase-space density for a particle moving in a magnetic field using our correspondence between the Langevin and Fokker-Planck equations. So let us say particle in a magnetic field, this is just the regular Langevin equation and I am going to apply uniform steady magnetic field B throughout and put this particle in a thermal heat bath at temperature T and ask how the Brownian motion of it is affected by the magnetic field. So as usual, all we have to do is to write down the Langevin equation, the pair of Langevin equations for the phase-space variables. So we could start with R dot equal to V as before but V dot is equal to the friction term or usual friction term. So this is minus gamma times V on this side plus this time there is a force due to the magnetic field which is a Lorentz force. So plus q over m B cross B, that is the systematic part of this force and then there is a usual noise term. Now this noise term, we have been writing it as a white noise and that is the same noise we are going to assume but now let us ask ourselves physically, you put a set of particles, Brownian particles in a magnetic field, the field does no work on these particles at all because the instantaneous velocity and the field are perpendicular to each other and therefore as you know the kinetic energy of the particle does not change, only the direction of its velocity changes and it goes in a cyclotron orbit. So we expect that whatever be the phase-space distribution, we expect that asymptotically as T tends to infinity, the conditional probability density of the velocity or the phase-space density as far as the velocity part is concerned is going to remain Maxwellian at the same temperature. So given that, we ask what is the condition necessary for the Maxwellian distribution at the end. We know that the average kinetic energy must be half, 3 half kT or whatever it is in this situation and so we put that in right from the beginning. And how was that implemented? It was implemented by saying that the strength of this noise is related to this dissipation coefficient by this relation which we have already written down many times, gamma is equal to 2m gamma k Boltzmann T. So we will put that in as we have been doing in all our examples in which case this becomes plus square root of, if I put that in, 2 gamma k Boltzmann T divided by m because it was mb dot equal to whatever it was on this side and this was square root of capital gamma. So 1n cancels and you end up with this times this noise zeta of T and this zeta of T is a vector value white noise with 0 means, so its properties are zeta i of T equal to 0 and of course zeta i of T, zeta i j of T prime is equal to delta ij delta of T minus T prime. So each Cartesian component of this zeta is a delta correlated stationary Gaussian Markov process, okay. So this is the set of Lajman equations here. Now that immediately tells us what the Fokker Planck equation should be. The fact that you have a velocity dependent force here means that you cannot write it like a Smolkovsky equation with the potential and derivative of the potential and so on. But of course we can still take this into account because I call this part of the drift. This whole thing is deterministic, the noise is here nowhere else. So I have a general formalism which tells me what I should do when I have some arbitrary non-linear drift and some arbitrary multiplicative noise. But this is a much simpler case, the noise is still additive and the drift is linear even though its velocity dependent, does not matter at all. So given that we can identify all the quantities needed in the Fokker Planck equation. Let us write it out for the velocity alone. So let us look at the velocity distribution alone. This is going to be, now this equation of motion, we can simplify it a little bit, write it in components and we have already done this earlier. So if I write any one component vj dot equal to minus gamma vj plus and let us put the field in some given direction. So b equal to b times unit vector in some arbitrary direction n. So this becomes qb over m which is omega cyclotron frequency that takes care of this factor here and then v cross b which is epsilon jkl vkl nl. I have just written out the components, the jth component here and the cross product in terms of the epsilon symbol plus this fellow here 2 gamma k Boltzmann T over m zeta j of t this fashion. So that is the Langevin equation. This is the f part of it and that is the g part of it. And so we can now write down what the actual solution is of this equation because it is a linear equation. We can solve it explicitly or we can write the Fokker-Plancky equation as the case may be. Let us see what this looks like if I write it out in terms of the matrix m which is a rotation matrix. So if you recall we introduced this symbol mjk equal to epsilon jkl is the definition nl. It is the matrix that takes care of rotations in 3 dimensions because it says that if you have this velocity vector v, v not in some direction and this is the unit vector n then as you switch on the magnetic field all that happens is the component of v not along n does not change and the perpendicular components process around the direction of the field with the cyclotron frequency. And that is given as stated by saying that v of t in this single deterministic problem v of t is a rotation matrix acting on this v not such that at time t you have gone through omega c t as the angle. So this is equal to e to the power m omega c t v not and of course we know what this thing is we can exponentiate this matrix m. We have done that in the past. We know that m cube terms of the properties of m are very straight forward. Once you define it like this, this implies that m squared jk is equal to n i nj, sorry nj nk minus delta jk. So m squared is straight forward enough and m cubed equal to minus m. All you have to do is to multiply once again by m and you discover that m cubed is minus m. So that is the property of this matrix m. What are the eigenvalues of this m by the way? Well what are the eigenvalues of e to the power rotation matrix? I mean if I define a rotation matrix about the angle n through the angle psi to be equal to this turns out by the way to be equal to m psi. We have gone through all the angular momentum generators etc put them all in and finally you end up with the rotation matrix written in this form. This is the simplest way of remembering the rotation matrix in 3 dimensions. What are the eigenvalues of this? Certainly one must be an eigenvalue because if you rotate about the direction n nothing is going to happen to any vector, right, which is along n. So this means that n itself is one of the eigenvectors and those vectors along n do not change at all. So one is an eigenvalue. Can you have another real eigenvalue? No, because if you did it means that you have this direction of axis of rotation say this and then you rotate nothing along this axis changes but all other points vectors must change. If you had another eigenvector which was value which was real then it is bad news. First of all no eigenvalue can have a modulus other than 1 because the rotation means no distances are changed between particles. So it must lie on the unit circle and you cannot have any second real eigenvector at all. And the eigenvalues are plus or minus e to the plus or minus i psi. The 3 eigenvalues of this are 1 e to the i psi e to the minus i psi, okay, yeah, alright. So this is going to tell you what the rotation does, just precession around this point and of course if I expand this out I am going to get a piece using m cubed is minus m. I am going to get a piece which does not change, a piece along b naught, a piece along n and a piece along b naught cross n, okay. So that is what this v of t is going to look like should be a vector here. Now if you put this in a fluid this Brownian particle and ask what happens due to the collisions this velocity, average velocity goes to 0. So all that happens is that the average v of t in a fluid becomes e to the minus gamma t times e to the m omega c t v naught. By this I mean write this as a 3 by 3 matrix, write this as a column vector and then operate on it, okay. So this is all that happens for a Brownian particle, okay. We have already seen this thing go on. So let us see instead of trying to solve this problem explicitly let us see if we can get a quick answer. So this is mjk vk, pardon me, we want to write the fpe for this definitely. So what would the Fokker-Planck equation look like? This would be delta over delta t p of v t v naught. Let us forget the v naught for a moment. It is understood. The initial condition on this is v naught equal to, so what should I write on this side? Minus, this is the drift term, this is the f here, right? So minus delta over delta vj of minus gamma vj plus omega c mjk vk, this fellow on p. So that takes care of the drift part and then plus what should I write here? Well clearly I should, pardon me, what should I write? Yeah, it is not del squared v, it is not del squared because this thing is a Cartesian component, right? And this is a scalar, this quantity here. So what should it be there? It is the diffusion matrix now obviously. You will not have a diffusion matrix. So this will be some d, let us use, let us use d, d ij d to p over delta v, let us call it jk, jk vj. And what is djk? This quantity here, what should it be? We need to compute this quantity. We need to compute this quantity. We have, we are not home yet. We need to compute this quantity. So what you have to do is to write this thing in terms of g is g, g transpose, work out what that is and then compute this, right? Yeah, so I am using bad notation here. Let me for the moment, since we are in velocity space, let us use this. Let us, let us use d, jk. And what is this fellow here? This is gamma k Boltzmann t over m delta ij because the different Cartesian components of this noise are appearing with the same strength. So it is a diagonal matrix more than that. It is actually unit matrix times this. So half the square of this guy is what this is. This is what the diffusion matrix is in velocity space, in velocity space. Now it is a simple matter to see if there is an equilibrium distribution and stationary distribution etc. It will turn out to be the Maxwellian distribution once again. We can solve this equation by the way. This is like, it is a linear quantity here. So apart from the small complication that the different components are mixed up, this is the Einstein Uhlenbeck distribution is the solution for this problem and it is an Einstein Uhlenbeck distribution with a mean which is given by this. So it is e to the power minus v minus this vector whole square divided by the variance 1 minus e to the minus 2 gamma t and so on. Asymptotically it will go to the Maxwellian distribution. So that part is quite straightforward. The next question is, what is the phase space density going to look like? Rho of r, v. Well now it is a little more intricate. You have a 6 dimensional phase space. So these objects, this drift matrix and the diffusion matrix will become 6 dimensional etc. But no sweat, it really does not matter because you will have this matrix, 6 dimensional fellow will be of the form, may be a 0 here, 0 here and then it is going to be just this. That portion is going to have this d matrix here. So it is not no sweat. I mean you can write down the big solution in this case. The full solution can be written down, okay. But now we have an interesting question. What about position space? What happens in position space? What is the diffusion of the particle look like? We already know that when you start with the Langevin equation in phase space, if you go to the diffusion regime which means t much much greater than gamma inverse or take the high friction limit in which you neglect the inertia term in favour of the dissipation term and then pretend this is the Langevin equation with a velocity that is essentially white noise, delta correlated then you can ask what does this Fokker-Planck equation look like? Look like and it gives you a diffusion equation with a diffusion tensor, okay. There was another way in which we did this. We solved this entire problem. I want to retain some of this stuff. I do not need this. We solved the same problem in another way. We said the displacement X, we called it capital X, X of t, X square of t average goes as t tends to infinity. We said that X square, we called it R square if I remember in three dimensions. This goes for t much much greater than gamma inverse. This goes like the diffusion constant. This fellow here will have some D ij, etc, etc. But we found that the diffusion constant was different in the X and Y direction, XY direction as opposed to the Z direction here. Now how do we get that? Well it obviously arises from the following. It arises from the fact that the positional probability density P of r, t if you like delta over delta t, this fellow here satisfies a diffusion equation which is not a homogenous, which is not an isotropic diffusion equation but must be of the form. Now let us use the ordinary D, D ij, D 2 P over D X i, D X j and then you are guaranteed that R square goes like the sum of X square goes like D 1 1 times 2 D 1 1 t plus 2 D 2 2 2 t and so on, okay. We need to find this. What did we do? We used the Kubo formula in this case and argued that this quantity D ij was equal to an integral from 0 to infinity D t V i of 0 V j of t equilibrium and that promptly gave us what this diffusion tensor was. We discovered that in the longitudinal direction it was not affected, it was K t over M gamma but in the transverse directions it was modulated by a factor which depended on relative strength of gamma and omega c, the cyclotron frequency. I want to derive that result by starting here and just using this going to your Smolakowski equation, okay. So let us see how to do that. In the high friction limit, exactly, exactly, yes, okay. In the high friction limit you are going to neglect V dot neglected in comparison with gamma V. So we go back to this equation and say let us neglect this term compared to that, okay and I bring this to the left hand side over here. So what does it say and I write it as a vector equation. So it says gamma times the identity matrix minus omega c times M that is this matrix acting on V is equal to this guy is equal to on the right hand side square root of 2 gamma K Boltzmann t over M zeta of t, pardon me. So that is the matrix equation. There is no V dot it is gone but there is this crazy matrix acting on V and this part of it is coming from the magnetic field. So formally this says R dot equal to square root of 2 gamma K Boltzmann t over M times gamma i minus omega c M this inverse of this matrix acting on zeta of t. This is also noise. This guy is also noise and this part does not involve the velocity or the dynamical variables at all. So is this additive noise or multiplicative noise? If this depended on the dynamical variables it is multiplicative noise otherwise it is just additive noise. So it is no drift, no drift and additive noise. The only t dependence is from this white noise and this follows delta correlated. So if I call this whole thing some noise some eta of t it is also delta correlated. It is guaranteed to be stationary because that part of it comes from here, okay. So it is stationary delta correlated. It is not isotropic I mean there is this fellow sitting here so you have to be careful about it but if you call this G the F term is 0 if you call this G then we know immediately what the Fokker-Planck equation in this case looks like it is exactly this and what is d ij? So this equation implies this with d ij equal to 1 half G times G transpose and the ijth element of it. So this is equal to 1 half and the half cancels against the 2 so it is gamma k Boltzmann t over m this matrix times its transpose and you want the ijth element of it or jkth element whatever ijth element of it. So what is that equal to? This is gamma i minus omega cm inverse gamma i no, no, no, why do you say that? The next one. Oh yeah, yeah, sure but I have written just G I have to write G transpose. So the ijth element is this times ik and then the kjth element but of the transpose so this is again equal to gamma i minus omega cm inverse transpose jk but this matrix m is an anti symmetric matrix epsilon ijk and k is anti symmetric in i and j and what is the transpose of this guy? It just says take the transpose of m because the unit matrix does not have any change and that makes it a plus. So you permit that I can write this as a plus and I have taken the transpose so now we can simplify things. The whole thing is essentially the inverse of gamma i plus omega cm gamma i minus omega cm inverse and then the ijth element of it. I interchange the orders when it does not matter in this case this guys commute with each other but I interchange the order because this was on the right in the inverse first and now it brought to the left but m and i commute with each other and therefore this is just gamma squared i minus omega c squared m squared so this is equal to gamma k Boltzmann t over m gamma squared i minus omega c squared m squared inverse. Let us pull out this gamma squared. Oh incidentally what is the guarantee that this fellow has an inverse to start with? What is the guarantee? m does not have an inverse what is the m itself does not have an inverse because 0 is an eigenvalue but what is the guarantee that this fellow has an inverse? What is the guarantee that this inverse exists? Yeah you see you can treat this like gamma is an eigenvalue gamma over omega c is an eigenvalue of this fellow and there are no real eigenvalues so as long as gamma and omega c are real this fellow can never this inverse will always exist okay. There is no vector such that m acting on it will give you a real number times m therefore this quantity has an inverse okay. Is that do you agree? Alright so let us pull out this gamma squared and I will write it out as m gamma and this goes away over gamma squared and then I want the ijth element of it inverse. Now the matter is very straight forward so I have d ij equal to k Boltzmann t over m gamma notice this original diffusion constant has emerged here but there is going to be dependence on the indices so inside you have i plus omega c squared over gamma squared m squared plus omega c squared 4 over gamma 4 m 4 but m 4 is m cubed times m and m cubed is minus m so this becomes a minus m squared plus omega c 6 over gamma 6 m 6 but that is m cubed whole squared and that is m squared ijth element which is equal to k Boltzmann t over m gamma times there is an i of course that sits here plus omega c squared over gamma squared m squared we take that out and then it is 1 minus omega c squared over gamma squared plus omega c 4 over gamma 4 minus dot dot dot and then ij this alternating thing says it is 1 over 1 plus this quantity here so this is equal to equal to k Boltzmann t over m gamma i plus omega c squared over gamma squared and then 1 over 1 plus this squared which is gamma squared over omega c squared plus gamma squared m squared ijth element home this cancels out and m squared we know we know what the elements of m squared are so d ij equal to k Boltzmann t over m gamma that sits inside and here the ijth element of the unit matrix of course delta ij plus omega c squared over gamma squared plus omega c squared times n i nj minus delta ij and that is the diffusion constant you can simplify this a little bit so this tells you the transverse part and the longitudinal part the coefficient of this portion here projects along the direction n and the other 2 ways the rest of this tensor will project in the other direction so that is a quick method we do not have to do the velocity correlation or anything although it is completely equivalent to this so it is a check of the fact that the Kubo formula is correct so you do get this diffusion coefficient directly all it requires is an inversion inversion of this matrix here this rotation matrix so once again in the presence of the magnetic field the problem is completely solvable explicitly it is diffusive in the normal sense in spite of this velocity dependent force and the diffusion constant is not isotropic diffusion constant right so that was one illustration the answers which we will take up here there is another aspect which I wanted to mention and that is the following we have seen that in all these cases we started with the Langevin equation and went to the Fokker-Planck equation occasionally you may want to do the opposite in order to identify the kind of stochastic differential equation that the random variable will obey when you have only a probabilistic distribution description of it in terms of the corresponding conditional density okay so here is a classic example let us look at the problem of diffusion in more than one dimension spatial dimension you have no field and so on so the diffusion equation in D dimensions would look like this let us do it in 3 dimensions to be specific so you have Delta P diffusion in 3 dimension so you have Delta P of RT over Delta T equal to D Del squared P of R I am interested in asking what is the equation obeyed by the distance from the origin okay I know what the equations what the equation the Langevin equation in this case is for each Cartesian component but I would like to know what the distance from the origin does square root of x squared plus y squared plus z squared that looks fairly intricate but let us see how we could answer that so I look at the diffusion equation and I am going to look at initial conditions which are completely spherically symmetric so we start at the origin and start with P of R0 equal to Delta 3 of R and P goes to 0 as R goes to infinity in all directions so this is the standard diffusion problem in which the solution is known to us it is a Gaussian right so we already know that P in the absence of boundaries P of R and T is 1 over 4 pi DT to the power 3 halves e to the minus R squared over 4 DT this is the fundamental Gaussian solution in 3 dimensions to the diffusion equation from here it follows that R squared average is 6 DT and so on okay but I want to know what little R does the distance from the origin so that my random variable is this it runs from 0 to infinity and I want to know what its distribution what its stochastic equation is okay this is a fairly messy thing to do because while the noise is not correlated for different Cartesian components the moment you go to spherical polar coordinates things become much more complicated well the first thing I do is to ask what is rho of R and T this is the sorry little R and T this is the probability density of the vector R it is another matter that in this case there is no theta or phi dependence because I started with natural boundary conditions which are spherically symmetric and initial conditions which are spherically symmetric so the solution is also spherically symmetric because the equation is spherically symmetric the differential operator del squared the initial condition the boundary conditions they all have spherical symmetry so the solution also has spherical symmetry. Now I want to know the distance here and that is of course the integral of this P probability density function over all angles and that just gives me a 4 pi factor and then there is an R squared because you want not bother about the direction but only the distance so there is a phase space factor R squared which is crucial right so this is equal to 4 pi R squared over 4 pi DT to the 3 halves P of R sorry e to the minus this is equal to 4 pi R squared P of R and T that is what I meant to say this is guaranteed to be normalized from 0 to infinity in R so integrate from 0 to infinity R time DR times this and you get 1 okay. Now what is the equation satisfied by this rho what is the Fokker Planck equation satisfied by this rho you got to go back here to this and substitute for it so we will let us do that we are interested in only the spherical symmetric part of it so therefore this del squared I retain only the R dependent part of it right in which case the equation is delta P over delta T this fellow here is equal to D times 1 over R squared delta over delta R R squared delta over delta R of P that is the equation we sought to get this Gaussian solution or whatever it is this is the radial part of the del squared operator and now let us put P equal to rho divided by 4 pi R squared into this equation so this will imply an equation for rho which is precisely the Fokker Planck equation for rho this is going to imply that delta rho over delta T in the equation for this guy here by the way if I put that in here and take partial derivatives notice that if I put P equal to rho over R squared out here it is a partial derivative with respect to T so the R squared comes out there is nothing to differentiate there and then it will cancel against things over here so this thing becomes relatively simple it becomes minus and in this case I know the answer delta over delta R 2 D over R times rho plus D times D to rho over delta R 2 check this out I am not 100 percent sure but I believe it is correct whatever it is this term is crucial okay so what will this imply backwards now implies a Langema equation for little R for little R which is square root of that crazy thing this is going to be R dot equal to this is F and G right and there is this term here so you want minus this fellow here so this is 2 D over R plus the usual whatever it is plus square root of 2 D times wide noise so there is a drift there is a drift here which is standing to increase R and there is no external force but yet you have a drift where is this coming from in the for the P there is no such thing there is no drift there is no applied force no potential nothing and therefore this was it had no stable equilibrium distribution it just went to 0 but here I have done nothing I have just changed variables and if this correspondence between the Fokker Planck and the Langema stochastic differential equation is to be believed you have a drift term where did this come from and how do we interpret this it is telling you that independent of the noise and of course it is not an independent drift term so notice that it is not like an external force or anything like that because D is sitting in there is very much there it is the same D that is sitting here so these are not unrelated to each other we are probably just change variables but what is the interpretation of this term notice that you start at the origin this term is very large close to the origin and standing to push it away R dot is positive so it is standing to push it away this means that if you do it in 2 dimensions you will see this immediately in the XY plane this is like 2 Brownian motions composed in the X and Y directions the forces are uncorrelated to each other and you have this motion particle doing Brownian motion but if you start here somewhere in this place the tendency is to get pushed out because this term is very large it is 1 over 1 over R and the reason of course is very clear it says that if you start in a sufficiently small neighborhood of this origin then due to any fluctuation R can only increase if R is sufficiently small in any fluctuation R can only increase strictly speaking at the origin any fluctuation will push it away from it right but that is actually true even if you are further away from the origin because suppose you are here at this point you can go in any direction with equal probability and the forces that are uncorrelated are in the X and Y directions so if you draw a little square like this this is a heuristic hand waving argument right now all points on the square are equally likely so to speak but there are more points outside than inside if you go inside it means R decreases if you go outside R increases so that is the drift that is appearing here okay this is the reason why it appears correlated in this case so you should not otherwise a priori you should not have such a drift but you do because of this simple geometrical fact so while the noise has this Cartesian symmetry the variable you are looking at is spherically symmetric right so you have a kind of drift effect here even without an external force or potential similarly you could write down the equations for if you had the single variable just X and it is undergoing Brownian motion you can ask what happens to X squared what does this look like or in higher dimensions what happens to R squared so that is an exercise interesting exercise find out what the stochastic equation satisfied by R squared is so consider R equal to let R equal to R squared find the Fokker Planck equation and Langevin equation for R without a square root in this case so it is all a question of changing variables and again you would use the same trick you would start with the master the solution of the diffusion equation the Gaussian solution is spherical symmetric Gaussian solution and rewrite that in terms of the density of probability density function of this R keeping track of all the Jacobian and so on and so forth and then having got that equation then we have first order term as well a drift term go back and write the Langevin equation down for it in this case okay by the way this thing here is easily generalized to D dimensions what would happen if you are in D spatial dimensions where little D greater than equal to 2 this becomes D minus 1 this becomes D minus 1 here this factor this factor 4 pi will change it is the surface of the unit sphere or whatever it is in D dimensions so this will be some dimension dependent factor but it is irrelevant for our purposes here this fellow however becomes D minus 1 okay times a constant so let us write it as proportional to P proportional to this guy and what happens to the Laplacian this is going to be Delta over Delta R and then this is exactly so this factor becomes D minus 1 this guy here becomes sorry D into D minus 1 and this remains R I do not remember if this is going to change on orbit you can figure that out this constant but this 1 over R drift is still there but this depends on this this guy here okay and of course it should be like this because when D equal to 1 limit you just have this there is no drift where should be similarly you could ask about what happens to various functions of Brownian motion various functionals of Brownian motion like the exponential and so on and so forth and they have strong strong applications there are many such applications but since we are not discussing stochastic processes per say I am not going to discuss those here okay the next thing we have to do again to fill up a gap is the following I mentioned that throughout the Langevin equation is an approximation we made some statements about when it is valid and so on and correspondingly the Fokker Planck model which is equivalent to the Langevin equation is an approximation in the same sense okay can we go back and ask what a particle in a fluid is actually going to do this is a very very hard problem it turns out even classically one could make the problem much simpler and ask what is the particle in a very dilute gas to what happens if it is removed knocked out of equilibrium and can we follow it find out what its time dependent phase space distribution is even the one particle stage okay it turns out that you can derive an equation called the Boltzmann equation to describe this and then from there you can find out what various transport properties are you can make a connection between the Boltzmann equation and the Langevin dynamics that we have talked about here under specific assumptions but in a sense the Boltzmann description is more fundamental it exposes more clearly precisely what are the statistical mechanical assumptions here so we will try to fill that gap I will try to give a very short derivation of the Boltzmann equation for a dilute gas that will be the next topic.