 Alright today we will try to work out the problem of the diffusion of a free particle in a fluid in the presence of an external magnetic field. So this is diffusion in a magnetic field. I will assume a field is constant in space and time and there is a magnetic field B and we have a particle of charge Q and mass M in moving in this field. As usual the particle is imagined to be inside a heat bath at some fixed temperature T and the question is what is the diffusion look like what is the diffusion coefficient look like in this case. It is a simple exercise because as we know the Lorentz force on the particle due to the magnetic field is linear in the velocity and as soon as we have a linear problem the Langevin equation can be solved explicitly. So let us write it out step by step we look at the velocity distribution as well. Now let us start by saying that the particle has a velocity V and V dot is equal to minus gamma times V that is the usual friction that we put in plus Q over M that is the charge over the mass times V cross B some constant magnetic field B is applied and then there is a usual noise term which as usual we will write as root gamma over M eta of T and this is Gaussian white noise. So its properties are any component eta j of T as 0 value average and eta j of T eta k of T prime is delta of T minus T prime. So it is the usual Gaussian white noise and this is the extra term out here okay. In component form we need that so let us write it out V j dot of T equal to minus gamma V j of T plus Q over M and we need the j component of this cross product which of course is epsilon jkl Vk bl plus square root of gamma over M eta j of T for each Cartesian component we have this and the components are mixed up with each other because of the magnetic field okay. Now as always we are going to say gamma is 2m gamma k Boltzmann T that is the fluctuation dissipation relation that does not change and the first thing we got to ask is what do we expect we expect that when you have this magnetic field on the Maxwellian distribution of velocities of the particle in thermal equilibrium is not disturbed because the magnetic field does not do any work on these particles it does not accelerate them so as to change the kinetic energy merely changes the direction of the velocity for every particle but it does not do anything else. So physically I expect that the equilibrium distribution will be Maxwellian will continue to remain Maxwellian however the Fokker Planck equation which is the conditional denser equation satisfied by the conditional probability density of the velocity will of course show you show a presence of this field it will certainly have an effect but asymptotically as T tends to infinity I expect it will tend again to the Maxwellian distribution okay. So we will look at the velocity Fokker Planck equation we could also look at the phase space Fokker Planck equation the full one in R and V because we know how to do that we know how to write down the Fokker Planck equation for the phase space density rho as a function of R, V and T in the presence of a linear force of this kind. So we will do that as well and then we try to compute what is the distribution etc. So corresponding to this we can write down the Fokker Planck equation immediately at once so if you write it in this language for example this will imply the Fokker Planck equation for P of V, T given some V naught so given some initial velocity V naught we can write down what the Fokker Planck equation for this is and that is going to read delta P over delta T equal to you have a minus sign there so it is gamma times delta over delta VJ VJ P that is this part is taken care of plus you can write it in a number of ways but here is the most convenient way of writing this plus Q over M epsilon Jkl Bl that is a constant so that comes out and then delta over delta VJ of VK P in that fashion plus half this guy so the square of it divided by 2 in the square this factor will go away and then this gets squared so it is gamma KT over M K Boltzmann P over M delta JK D2P over delta VJ delta VK this is essentially the del squared operator as you can see in velocity space so that is the Fokker Planck equation satisfied by the conditional density here for a given velocity so the initial condition on this is just a delta function 3 dimensional delta function at V equal to V naught okay. Now we can write this in vector form as well so let us do that let us put del V equal to gradient operator with respect to velocity components so just for notation let us call that del V otherwise I write del it could be with respect to position as well so let us just call it del sub V and then what does this say it says delta P over delta T equal to gamma times del V dot VP it is just the divergence here as you can see because it is the same J and then this guy here gives you minus oh sorry you got to be careful with the signs this was a plus here so this equation has a minus the drift comes with a minus F of X as if you remember so minus Q over M we will be could write this down directly as it is so it is del V dot V cross B times P so it is the divergence of that that is the drift part of it plus gamma K Boltzmann T over M del V squared on P so that is the equation satisfied by that is the Fokker-Planck equation okay the task is to solve this equation subject to this initial condition okay by the way a little bit of simplification it is convenient as you can see I have taken the magnetic field to be in some arbitrary direction it is convenient to do that rather than specialized to the z direction or anything like that but you can see the structure of this whole thing coming out more interestingly if you keep a general direction so let us say that the magnetic field B equal to B times some unit vector N in some arbitrary direction right then what you have here is Q B over M so let us simplify this equation a little bit in fact let us simplify this equation so for BL I am going to write B times N sub L and then I put a B here N sub L VK and let us define a matrix M so let us define a matrix M JK to be equal to epsilon JK L NL a matrix whose elements JK are given by epsilon JK L NL okay it is clear that this is an anti-symmetric matrix because the epsilon symbol is totally anti-symmetric in any two indices so that is one fact we need to know right away with M JK equal to minus M KJ it is anti-symmetric then if you look at this for instance or in that language it does not matter which way you look at it this is going to have two derivatives one of which is going to be delta VK over delta VJ that is delta JK times P but delta JK with epsilon JK contracted will give you 0 because this is symmetric and that is anti-symmetric and the other term is VK times delta P over delta VJ and that of course survives right so this means that you can simplify this a little bit and write this as del V on P dotted with V cross B because the del V dot V cross B is 0 identically right so divergence of a vector times a scalar is a scalar times the divergence of the vector plus the vector dotted with the gradient of the scalar and the first term vanishes in this case so we could also write this as V cross B dotted with del V P so a little simplification now what does this do this thing is what we called M JK so let us call this M JK and Q B over M is a direct physical significance it is the cyclotron frequency for a charge particle of charge Q and mass M in a field of magnitude B so let us call this omega cyclotron so out there we have omega C okay now we could either try to solve this equation directly or we write this Fokker-Planck equation down and try to solve this does not matter either way we would like to get at the velocity correlation function and after that we try to use the Kubo formula to find the diffusion constant we will also find the diffusion constant by another method going to the high friction limit and seeing whether we get the same answer or not this is what we want to check out first before anything else this thing here what does it do once you have once you have a magnetic field applied in some constant direction and you have some velocity to start with V naught and I apply this field what happens to this V naught is that it starts processing around the direction of N the unit vector N right so this is all that happens I mean if you look at without any noise without a damping without an external without anything except the external field just a free particle and it satisfies an equation of this kind V dot equal to omega C M this is a matrix times V that is what this equation is this is equal to that all I have written is V dot is Q over M V cross B in this language okay now what is the solution to this equation for a given initial condition so the initial condition is V naught so what happens is that if you have this as the unit vector N and you have a velocity V naught here you know that at time progresses this V naught processes in a cone around this direction of N this means that the component of V along the field does not change whereas the components the perpendicular components the transverse components they perform circular motion with a phase difference of pi over 2 right so V X and V Y exchange roles you move in a circle and you have this processional motion so what is the solution to this guy here the formal solution to this this thing will imply that V of T equal to E to the M omega C T acting on V naught that is the formal solution to this guy right and what we have M is a rotation matrix as we will see in a minute and we want to exponentiate it right and the angle changes at a constant rate omega C so what we need is a finite rotation formula in this case many ways of writing this down I am sure you are familiar with this from familiar what happens if you have a given vector R and then you have some direction unit direction N and you make a rotation about this direction rotate the coordinate system about this point through an angle psi you get a finite rotation formula yeah it is actually the generator this guy is like a rotation matrix through an angle omega C T as you can see right now of course you need to do this you need to compute this quantity but that is not very hard to do computing this quantity is not hard to do because let us do that as a side exercise so e to the M omega C T equal to I plus M omega C T plus M squared over 2 factorial omega C T whole squared dot dot dot etcetera right now what is it about M that strikes you immediately because M squared is going to be interesting M has elements JK equal to epsilon JK L NL where this is a component of a unit vector and what is M squared JK equal to well you got to put two of these epsilon's and then use the fact that this is a unit vector etcetera and simplify it using this formula for what happens when you take two epsilon's and contract one index you get a product of delta functions etcetera that chronicle deltas right. So in this case this turns out to be equal to N J N K minus delta I J and what does M cube become well you put another N here you put another M here and use this fact here and sorry not I J M cube becomes equal to minus M itself check this out so M 4 is minus M squared and so on. So therefore the exponential collapses completely and you can write down a formula for this guy so this thing here is equal to if I exponentiate it fully it is I plus M sin omega ct plus M squared 1 minus cosine omega ct because you see the M squared guy starts with omega ct squared over 2 factorial so it is you have got to put that 1 in separately and it is of course cos cosine as minus t squared this got a plus here so it is 1 minus cos and therefore you can write this down explicitly write down fully what this thing is so it says this guy by the way if I change this plus or minus all that happens is t goes to minus t so this becomes plus or minus well that is an interesting little exercise to play with this and it will reproduce for you the finite rotation formula so let us write that down see what happens the formula in fact is that if you if you have a unit vector n and I take a certain vector a and rotate it about this direction in rotate the coordinate system through an angle psi then a goes to a prime which is precisely e to the M psi acting on this original vector a so it is going to be equal to a cos psi plus 1 minus cos psi times n dot a unit vector in the direction n so if I rotate this there are only 3 possibilities one of them is it is along the original vector a the new vector has a component along a it has a component along n and a component perpendicular to these two guys right so this plus sin psi times a cross this is called the finite rotation formula can establish this for the coordinates and after that of course a vector is a quantity which transforms exactly like the coordinates and therefore any vector would transform in this fashion so in fact we can write down the answer now to what the solution to the Fokker Planck equation is explicitly because I have already said we have given physical argument to show that the energy does not change at all all that happens is that the velocity starts processing every velocity processes yeah pardon me ah good question so I am rotating the coordinate system about this point about the direction of the unit vector n so if the original coordinate system had an x axis like that it now has a thing like this and the amount of rotation in that plane is psi the amount the angle through which you have rotated to go to another point on the tip of this cone okay in our problem this rotation is happening all the time the precession so psi is replaced by omega c t so what it is saying is that if you just had a magnetic field and you looked at just that original equation so the equation is v dot equal to omega c m acting on v this will imply that we at any time t at any time would be equal to a to the m omega c t acting on v naught and that can be written down this is equal to cos omega c t times v naught plus 1 minus cos omega c t times v naught dot n times n plus sin omega c t times v naught cross n so that is the explicit solution in general to what happens to any initial velocity v naught at time t this is what it does okay let us give this a name let us call this u as a function of time and of course the initial velocity v naught so let us define u to be this okay the magnitude of u is the same as a magnitude of v naught nothing happens it just precesses now what do you think will be the solution to this guy we can do this painfully but we can write this down on physical grounds v t v naught what do you think this will be in the absence of the magnetic field this is the onstein-ohlenbeck distribution okay and the onstein-ohlenbeck distribution has the variance goes like 1 minus e to the minus 2 gamma t and the mean value decays to 0 like e to the minus gamma t so the original thing distribution was e to the minus v minus v naught e to the minus gamma t whole squared now instead of v naught you are going to have u of t comma v naught that is all that will happen so we can write this solution down it is going to be equal to m over 2 pi k Boltzmann t 1 minus e to the minus 2 gamma t the whole thing to the power 3 halves because you got 3 components and then this 3 dimensional Gaussian e to the exponential of minus v minus u of t comma v naught that is a vector e to the minus gamma t whole squared divided by of course this and minus m over 2 k Boltzmann t 1 minus e to the minus 2 gamma t so that is the solution to the Fokker-Planck equation all that happens in the magnetic field is that this v naught gets rotated keeps rotating and the damping is exactly as before that happens because the Langevin equation has a minus gamma v on the right hand side it damps it out and the portion v cross b this portion the linear drift but it is a reversible part this is the reversible drift this is irreversible due to dissipation and of course that also represents the effect of dissipation so you can see from this you can check backwards that it satisfies the Fokker-Planck equation but you can see from this there as t tends to infinity this goes to the max value this part goes away it is m over 2 pi k t to the 3 halves e to the minus m v squared over 2 k t so it certainly goes to the max value in distribution as t tends to infinity that is the equilibrium distribution the max value in 3 dimensional max value in distribution our task is slightly different we want to actually find out what happens to the diffusion how does it diffuse right for that we already have a formula we have a formula which says that the integral of the velocity correlation from 0 to infinity is in fact the diffusion coefficient we explicitly showed that by looking at the long time behavior of the mean square displacement now we have indices to worry about so we got to be a little more cautious so let us see we can do that and we do not have a diffusion coefficient we have a diffusion tensor now because the coefficient here when you write the diffusion equation down since we have 3 components we are going to have a diffusion tensor in this business so let us see what this is let us first find what the correlation velocity correlation is so we will go back and use the fact that we know the velocity is a stationary random process here so since we know that we can do this a little heuristically let us write v j dot of t equal to minus gamma v j of t plus and then there was an omega c m and if I recall right it was m j k v k of t and then there was a portion which was plus square root of gamma over m eta j of t that was a Langevin equation right we would like to find out what this quantity is we are seeking c i j of t define this to be equal to v i of 0 v j of t this is the quantity we want this angular average is the full average in equilibrium now what is this quantity equal to what c i j of 0 equal to what is this guy equal to so it is the correlation between the Cartesian component ith component at a given time 0 with the jth component at the same time now independent different Cartesian components are independent of each other this correlation is 0 in equilibrium but when the two are equal it is equal to k t over m right so we know that this guy equal to k b t over m delta i j for each component it is k t over m for each i i i equal to j it is equal to k t over m just the mean square velocity any given component so let us try to compute that what we need to do is to multiply this on the left by v i of 0 and then take averages times this whole thing and then take averages so the left hand side says d over d t of c i j of t minus gamma c i j of t plus omega c m j k c i k of t this is a constant matrix and then c i v i of 0 v k of t is c i k at time t plus the average value plus square root of gamma over m the average value of v i of 0 with eta j of t what is this equal to that is 0 because of causality it says the force and for any t greater than 0 cannot dictate what the velocity was is the velocity at an earlier time is not dependent on the force at a later time even at equal times this correlation is 0 it is the acceleration that is correlated to the force right so this vanishes we are now computing this for t greater than 0 because that is really what we need for the diffusion coefficient when you do it for t less than 0 we can compute it you got to be a little careful you got to be a little careful because remember that without this magnetic field the formula you got for the velocity correlation was e to the minus gamma modulus t and now there are going to be terms depending on the magnetic field so when you take t to minus t you have to be a little cautious because although this dissipation term will be e to the minus gamma modulus t the fact is that as t goes to minus t you should also be careful to reverse the sign of the magnetic field okay. So for time reversal operation you when you have an external magnetic field time reversal is taking t to minus t and the applied field B to minus B and the physical reason for that of course is that if this field is imagined to be from due to some electromagnets some electric charge is moving if you reverse time the current flows in the opposite direction the field changes sign okay. So the time reversal property of B is actually B goes to minus B unlike e okay so this term is 0 now this looks almost like a matrix multiplication except that you have a little problem there is a jk and an ik but of course we know that mjk is minus mkj so we can write this as equal to minus gamma cij of t plus omega c cik put a minus sign m let us write this properly m so if I call c the matrix I have d over dt c d over dt of the matrix c of t this is a matrix now equal to minus c of t times gamma i plus omega c m on the right hand side because it mkj and what is the solution to this guy this implies that c of t equal to e to the power minus sorry the other way c of 0 with e to the minus gamma i plus omega c m it is on acting on the right so I have better be careful keep it there of course this is also equal to e to the minus unit matrix commutes with everything so I can just bring it out so it is equal to e to the minus gamma t c of 0 e to the minus omega c m minus m omega c t and that is it but what c of 0 yeah so this will of course imply that c of 0 equal to k Boltzmann t over m times the identity matrix delta ij is just the component of the identity matrix so this becomes equal to k Boltzmann t over m e to the minus gamma t and all we got to do now is find e to the minus m omega c but we did that already we have already done it sitting here so we are in good shape we actually have an expression for the correlation so this says that c of t is equal to e to the minus gamma k t over m e to the minus gamma t times i plus m squared 1 minus cos omega c t minus because we got a minus MCD m sin omega c t so we have the matrix explicitly correlation matrix and what we need is the ijth component of it so it says c ij of t equal to k Boltzmann t over m e to the minus gamma t what is the ijth component of i the Kronecker delta of course Kronecker delta delta ij plus m squared remember was n i n j minus delta ij 1 minus cos omega c t this guy here minus and recall this fellow was epsilon we want the ijth component so it is ij k n k sin omega c t so we have an explicit formula for the velocity correlation remember that if you had a free particle in the absence of the magnetic field then the different Cartesian components did not get mixed up as a function of time and they were uncorrelated they started uncorrelated they remained uncorrelated now because of the magnetic field it is crambling things up so you got all kinds of mixtures out here what we need to compute is the integral of this quantity over t from 0 to infinity but there is a little change in the formula here got to be little cautious here the Kubo formula for the diffusion constant is now a formula for the diffusion tensor it turns out that d which was equal to integral 0 to infinity dt v of 0 v of t originally now the correct formula is di j the coefficient because it is a tensor now diffusion tensor this is equal to dt ci j of t plus c j i with a half it is the symmetric part of this coefficient of this tensor which turns out to be the diffusion coefficient we can prove this without much difficulty I have not done linear response theory explicitly we have not talked about the Kubo formula and how to derive it for the multi component case but in a couple of minutes we will try to corroborate that this is indeed to so I will write down the diffusion constant directly by looking at the high friction limit so what we need actually is not the integral of this but the integral of the symmetric part of it right so this says di j equal to the symmetric part this part is symmetric this part is symmetric this part is anti symmetric in i n j and so when I take the symmetric part it vanishes does not contribute and there is a factor half I can get rid of that half by adding this twice dividing by 2 so this immediately becomes k Boltzmann t over m by the way a little bit of simplification here itself let us write this whole thing as equal to k Boltzmann t over m e to the minus gamma t and then there is an n i n j let us write it like this this thing here is n i n j 1 minus cos omega c t plus delta i j cos omega c t because delta i j times 1 and this cancels out so we have a small simplification here in this fashion so let us write this c i j of t equal to k Boltzmann t over m e to the minus gamma t times we rearrange terms a little bit n i n j plus delta i j minus n i n j cos omega c t minus this other guy epsilon i j k n k sin let us check if that is correct that is correct that is right now you will recognize that this is the longitudinal part and that is the transverse part of this tensor and that is the anti symmetric part the last term okay so if I compute now the diffusion coefficient this guy here is equal to k Boltzmann t over m e to the minus sorry got to integrate over that times I have an n i n j and I integrate e to the minus gamma t that is just a 1 over gamma right so it is n i n j over plus and there is a delta i j minus n i n j times the integral of e to the minus gamma t cos omega c t and that is of course gamma divided by gamma squared plus omega c squared that is it and the other part last part vanishes because it is anti symmetric this is of course the longitudinal part and that is the transverse part and you can see the 2 are slightly different because let us pull out the k t over m gamma and then it is this plus that so you see at once that in the transverse direction this remember that in the original problem d was equal to k t over m gamma this followed by consistency in the diffusion limit we already saw this this is the Einstein Sutherland formula and that is sure there in the longitudinal direction but in the transverse direction it is modulated by this factor which is less than 1 so this shows exactly how the fact that you have the cyclotron precession is inhibiting the diffusion constant it is still diffusive even in the transverse directions but it is a smaller diffusion constant instead of being a constant it is dropping off as a function of omega c when omega c becomes very large it will of course go to 0 so you can easily see few if you took a special case if you took n to be equal to 0 0 1 c so you have a field in the z direction okay then it is immediately clear that d becomes diagonal and you have d 1 1 equal to d 2 2 so you put n 1 equal to n 2 equal to 0 and then only this term contributes so you have k b t m gamma times gamma squared over whereas d 3 3 then you get the 1 cancels out here and you get just this portion which is k b t so this is equal to d transverse equal to this is equal to the longitudinal so it shows this very simple problem exactly how the diffusion is inhibited it is still diffusive it is mean square displacement still diverges but the fact that this particles constantly pulled back into a circle in the transverse direction lowers your defective diffusion constant by this amount okay this is the simplest model now of course you have a lot of other complications in a real plasma you have many more complexities but this tells you already how this thing happens the simplest instance can we see this directly from the Langevin equation yes because what you need to do is this and I am going to leave details as an exercise what you need to do is the following let us go back to the original equation the Langevin equation itself and write it in the following way we have v dot equal to minus gamma v plus q over m v cross p plus root gamma m theta of t in this fashion right and we wrote this in a somewhat simpler form right we wrote this as equal to minus gamma times unit matrix plus plus the matrix omega c m acting on acting on v we wrote it in this form now in the high friction limit the inertia term is negligible so you can throw that out and then you have a thing which says gamma i minus omega c m acting on v which is r dot is equal to square root of gamma over m theta of t where this is Gaussian white noise so this is a high friction limit or diffusion limit if you like but you can rewrite this this implies that r dot equal to gamma i minus omega c m inverse acting on time square root of gamma over m acting on eta of t does this inverse exist got to be a little careful here to see if this thing exists what are the eigenvalues of m itself because we know what m is the m ij is epsilon ijk nk where nk is the component of a unit vector right so we can find the eigenvalues of this m very easily what do you think of the eigenvalues of m it is a way we had a property for m remember that remember remember that m cube plus m was equal to 0 right that is the characteristic polynomial of m so lambda cube plus lambda is 0 what are the eigenvalues of m 0 and then plus or minus i so 0 is certainly an eigenvalue of m and therefore the inverse of m does not exist we do not care we want the inverse of gamma i minus some number times m so you can stay away from the eigenvalues depending on what this gamma is in fact it is easy to see that this inverse exists we will find it explicitly okay so there is no problem that exists but we got an equation which looks like x dot equal to something or the other some constant whatever matrix is this acting on white noise so remember our original equation we got x dot equal to square root of 2d times eta and we at once said this immediately implies that the probability density of this x satisfies d d2p over dx2 that was my identification of the diffusion constant all we needed was white noise and the integral of white noise is diffusion of course here stationary white noise this guy stationary this is a constant independent of t so this whole noise is still a stationary noise it is still delta correlated it is still delta correlated because you can see what this noise is like so let us call h of t equal to let us define this to be equal to all this garbage square root of gamma over m gamma i plus minus omega cm inverse this whole thing acting on this eta of t now what is h i of t is equal to it is equal to gamma over m gamma i minus omega cm inverse i k eta k agree and what is h j of t prime this is equal to square root of gamma over m gamma i minus omega cm inverse jl acting on eta l of t prime so we can now find out what is h i of t h j of t prime and take the average that is the noise correlation right that is equal to gamma over m square gamma i minus omega cm inverse i k gamma i minus omega cm inverse jl the correlation of eta k with eta l eta k of t with eta l of t prime what is that equal to that is a chronic delta delta kl times the delta function delta of t minus t prime so this guy is going to have a chronic a delta kl delta of t minus t prime right and of course you can finish off this immediately because this jl kl you can get rid of this and make this a jk so it is still delta correlated this is still a noise and it is going to be square root of 2 di j times a delta function on this side when I do this for our dot I am going to get a square root of 2 i j whatever it is so this is what we need since we have proved that our dot equal to h of t component by component right this will tell you immediately that this noise is like square root of 2 di j for each component acting on the appropriate component of h on that side on the right hand side so this will immediately imply d 2 o d dp over dt equal to di j so you can identify what this di j is from this guy here so this is what I would like you to do this is 2 di j remember k is cancelled out so this whole thing so you can identify what all you have to do is to find this inverse and that inverse how are you going to do that how are you going to find these inverses well once again we know that if gamma is sufficiently large which is what you have to do it is a high friction limit then of course you can do a binomial expansion of this in powers of m and use the fact again that m cubed is a minus m so it will again terminate once again okay if you do not like that find the Laplace transform because if you find the Laplace transform of e to the power so e to the power m omega c t and you find it is Laplace transform it is e to the power minus s t times this guy find the Laplace transform of m omega c t 0 to infinity dt this is equal to si minus m omega c inverse formally this is true we already have a formula for this guy in terms of sines cosines and so on so put that in and do those simple integrals and you are going to get something which depends on m something which depends on m squared that is equal to this guy and now set s equal to gamma and you have this inverse so there are many tricks for finding out the inverse you do not have to work extra to do the inverse it is already there then put that here put that here whatever expressions you get and that will and identify that with 2 d i j read of d i j it corroborates the fact that in the diffusion limit you can directly from the fact that the position that the velocity is delta correlated you can actually find out with some coefficient the coefficient is essentially the diffusion coefficient we can compute what the d tensor is in this case we can do this in several ways we also have you can also compute now the phase space density completely so you can find out while we found out p of v, t for a given v0 you can also find out what row of r v and t given r0 and v0 is explicitly and that is not very hard to do either because once you have this expression for the exponential of this m matrix directly then everything follows automatically so in a sense this completes what happens in the simplest instance of a particle moving in a magnetic field does exactly what we expected on physical grounds so let me stop here today and we will come back to noise next time.