 Right, so today what we will do is to look at one of the most important consequences of the equation for the density, probability, density function for diffusion in a potential. It is called the Smolukowski equation as you know and we will derive from it a very important consequence. That is something which has a wide range of applications. So to recall to you what the Smolukowski equation was, we are looking at a one dimensional variable X like the position of a diffusing particle on the X axis and then the Smolukowski equation was the equation for the probability density function P of X, t which is derived from the general Langevin equation in the high friction limit. So if you recall the equation in the high friction limit in the actual Langevin equation was Mx double dot plus M gamma x dot was equal to minus the force on the particle which is V prime of X and then there was a noise term which was essentially square root of 2 M gamma k Boltzmann t times the white noise, the delta correlated white noise with unit strength here and then this went over in the high friction limit, so gamma large, it went over to this equation X dot is minus 1 over M gamma V prime of X plus the square root of this divided by M gamma so it is twice k Boltzmann t over M gamma times Z of t. So that is the Langevin equation in one variable with this drift term and that diffusion term. So it immediately follows that this quantity P of X, t this conditional density some initial condition satisfies the corresponding diffusion equation in the presence of a potential which we call the Smolukowski equation. So this I think implied that delta P over delta t was equal to minus the drift term the derivative of it, so it is 1 over M gamma delta over delta X V prime of X times P that is the first term plus the diffusion term which is k Boltzmann t over M gamma. It is a square of this divided by 2, the G squared over 2 part times delta 2 P over delta X 2. This is the Smolukowski equation, this equation here is a Smolukowski equation. Okay, one of the questions, important questions is is there a steady state or equilibrium distribution at all? So let us start at that point. If at all there is such a distribution, so we are saying that P of X, t if tends as t tends to infinity tends to P equilibrium of X then this quantity P equilibrium of X must satisfy an ordinary differential equation with this set equal to 0 and that equation of course is 1 over is k Boltzmann t it is the following. Before that, before that before I write this notice that this equation because this is a derivative can be pulled out of on the right hand side could be written as equal to minus delta over delta X J of X, t some current. So this equation has the form of a continuity equation delta rho over delta t plus divergence of J equal to 0. So it has got exactly that form but this J of X, t equal to minus whatever is inside the d over dx here. So it is minus V prime of X over M gamma P inside there plus k Boltzmann t over M gamma delta P over delta X all I have done is to write out delta over delta X outside and then times this bracket here. So this if you like is the current, the probability current and now if if P of X, t tends to P equilibrium of X as t tends to infinity then it means delta P over delta t is 0 and therefore this quantity here d over dx of this right must be equal to 0. So this will imply that d over dx of this fellow is 0 so it says d over dx of minus V prime of X over M gamma minus d over dx of V prime of X plus k Boltzmann t. I am going to retain these factors for a minute as you will see why d P equilibrium over dx times P equilibrium. So let me write these factors out carefully P equilibrium of X plus k Boltzmann t over M gamma d P by dx must be equal to 0 which means that this quantity is constant and if it is a normalizable distribution then you would say the constant is 0, it is 0 at infinity so it is 0 everywhere and then you would get a simple solution for P equilibrium of X. On the other hand it is conceivable that we could look at this as a problem in which you have a steady flux of many particles coming in, standard flux and then there is some complicated potential, these particles are riding through under thermal agitation as well and then there is a flux outwards. So there is a stationary current so it is possible there is a stationary current in the problem which we do not know as yet. So the general solution is not that this quantity equal to 0 but that this quantity is equal to constant right. So this means that the general solution is that V prime of X over M gamma P equilibrium of X, let us write the derivative term first so k Boltzmann t over M gamma d P equilibrium over dx plus V prime of X P equilibrium divided by M gamma is equal to minus j stationary. So this is some stationary current because it is stationary it is a function of X alone just as these fellows are all functions of X alone that is the general solution to this equation right. So what does that imply? It says d P equilibrium over dx plus V prime of X over k Boltzmann t P equilibrium is equal to minus M gamma over k Boltzmann t j stationary and if it is a flux which is moving smoothly constant uniform flux that is a constant, the constant is some number there right. So you can pull that out and you have to solve this equation with this constant as some constant right and what is the general solution to this equation? Well it depends on the initial condition but suppose I have an arbitrary initial condition I will integrate this equation without specifying an initial condition with an integration constant put in there. So the general solution to this equation pretending for a minute that we know this j j stationary so pretending that we know this constant we will determine it in a self consistent way is the following. So it says P equilibrium of X e to the power you have to integrate e to the integral P dx this is function P which is V of X over k t so this says P equilibrium e to the V of X over k Boltzmann t is equal to some constant integration constant minus M gamma over k Boltzmann t j stationary whatever that number is times an integral dx e to the integral P dx so e to the power V of X by k Boltzmann t that is the solution and if I move this factor to the right hand side then it is this e to the minus V of X over k Boltzmann t. So notice we still have to do this integral and we do not know what this V of X is in general but formally we can write the solution down if you tell me and you measure a steady flux at some point in the x axis and then you say that that number is j stationary pardon me no I am saying that it is a stationary current it is a constant d over dx of something is equal to 0 which implies this something is equal to a constant and j stationary has to be a constant because it would have to satisfy the continuity equation right and the P corresponding P is 0 and derivative is 0 so j stationary has to be a constant so sorry this is wrong I should not have put the x there it is a constant so that is the formal solution and to write it down explicitly you have to give me some boundary conditions you have to tell me at x equal to whatever point some x naught or something you have to tell me what this is and then in terms of that I determine t put the same right okay. Now let us apply this solution to a very crucial problem and this problem is called Kramers escape rate and we are going to derive this formula. Now the formula itself and this are just the following suppose you have a particle moving in a potential and the potential has ups and downs and so on and at some point it encounters a minimum of the potential you would expect that the concentration of particles is very high the minimum very low in the maximum of the potentials and somewhere in between between extreme right. Now you could ask because there is thermal agitation particles have been kicked all the time is there a possibility that the particle might jump over the barrier and go to the other side of the of a maximum okay and if so at what rate is it doing so the general scenario and besides it is the classical thermally assisted process completely classical so it is the analog of quantum mechanical tunneling tunneling appears happens because of quantum physics there is no classical tunneling at all but the following can happen let us suppose that the v of x looks like this there is a minimum somewhere and then there is a maximum and then it goes off in this fashion so if I schematically say this is v of x versus x on this side you could ask if I start with a particle here at what rate will it have flow outwards across the maximum to the other side at what rate will it escape the barrier okay. Now you could ask why should it escape the barrier well if you put a particle here with this zero kinetic energy it is just going to remain here but remember it is being buffeted by thermal agitation so all the while there is kicks which are moving the particle and now the question is what is the escape rate going to be it is a hard problem if you have quantum mechanical tunneling and if your k t is so high this is an energy scale in the vertical side is so high that the barrier is not even visible to it but what happens if k t is much smaller than the height of the barrier so that is the question we are going to look at let us give some names to these points so let us say this point is x naught this point is x 1 some point here is x 2 and I want to know what is the rate of escape of particles if there is a steady flux coming in from one side from the region around x naught to a point like x 2 so this is the question we would like to answer yeah I am so I am not worried about what happens here I would like to know what is what is the rate at which particles escape the minimum can go across the maximum and get to the other side of it precisely where on the other side is an irrelevant detail I can compute it for any point the assumption is that v of x 1 minus v of x naught which I will call delta v this is the definition of the barrier height the maximum minus the minimum that is the height of the barrier the assumption is that is much bigger than the energy imparted by single thermal agitation by k t much greater than k t so delta v much much greater than k Boltzmann t that is the regime we are interested in now you can see why this is going to be a tricky problem because you have a particle here imagine it is called zero kinetic energy it gets kicked to the right by thermal agitation k t typically k t energy it goes up a little bit but then it gets kicked back and so on it will oscillate about this point move randomly about this point to get it across this barrier in some fashion you have to coherently kick it with a lot of k t is all in one direction so it is very improbable it is very improbable but it is a non-zero probability and the question is to calculate what it is yeah so we are going to make that we are going to say there is a constant flux of particles coming in so for some given constant flux what is it in fact we are going to do something better we are going to argue and we know this that the equation the diffusion equation obeyed by the concentration of particles is the same as the diffusion equation obeyed by the probability density for a single particle if they are not all interacting with each other it is the same diffusion equation so in a sense either I talk about the concentration of particles here or I talk about the probability mass of the particles here so I am going to ask a general question I take particles in some region A to B and I ask how many particles are there it is the same as the integral of the probability density equilibrium density from A to B and I ask what is the flux at a point like X 2 and the answer is that the flux at X 2 on the right hand side is equal to the probability mass around X naught multiplied by rate of escape of each particle so I find the constant rate of escape of each particle multiply that by the probability mass and that is the flux at I am interested in computing this rate of escape now you can see that this is a paradigm for a large number of processes any thermally assisted process which involves an activation barrier is precisely this problem for instance if you have molecules here and you have reaction chemical reaction and there is an energy barrier against this reaction so the molecule here has to hit the molecule there in order for the reaction to happen but there is an energy barrier in between then the rate of this reaction is going to depend on the rate at which this escapes the barrier so in that sense this is a very general formula now the question is how is it going to depend on this height and how is it going to depend on the temperature given the fact that this height is much much greater than the temperature so that is the question we are going to answer and that is going to give us a final answer for this rate of escape which is called the Kramer's escape rate formula so it surely has to do with the shape of the potential here the shape of the potential there and so on they are going to play a role the height of this potential barrier is also going to play a role it is biased in the sense that the moment you have a potential there is a bias at every point and that is the reason why in the Smolkovsky equation you got a drift term so the drift term is acting like an X dependent bias if you like there are times when it will be favourable to one direction favourable to the other depends on V of X V prime of X we will look at the problem of a constant bias like gravity we will do that at the end but here we got a much more general question and right now we are looking at the problem of an escape over a barrier so definitely the potential is assumed to have a minimum and a maximum and we would like to compute these quantities here so the rate of escape is what we want to compute but first we need to compute the probability mass so let us remember this formula the rate of escape lambda escape rate is equal to flux at X 2 that is the stationary current so this is J stationary divided by probability mass around the point X 0 so divided by an integral from A to B DX P equilibrium of X so this is what we want to compute and we are going to make a number of approximations in doing this because the formula is an approximate one and you will see how good the approximation is as we go along now first step we already found a formula for P equilibrium and you have to tell me what it is it was if I recall right it was C minus M gamma over K Boltzmann T no times and J stationary integral DX E to the power V of X over K Boltzmann T this is an indefinite integral so it is a function of X multiplying E to the minus V of X over K Boltzmann T I can erase this now but you see we need to do this to do to find this we need to know what is J stationary so this fellow is sitting right here in this ring here but now I am going to argue that this J stationary is small is negligible because it is a small flux you what do you how many particles you expect to come out here it is a highly improbable thing so to leading approximation except you cannot kill it here then of course there is nothing right but anything which comes from here goes up goes to higher order in J stationary so to leading order in the denominator we are going to kill this here so this is approximately equal to C E to the minus V of X over K Boltzmann T so I want you to understand clearly that this is self consistent this whole thing is an approximate formula defining the leading contribution under these conditions so this is going to give you higher order terms including it but in the leading order this is what it is but there is an unknown constant sitting here how are we going to determine that well if this is true then it is immediately clear that P equilibrium of X naught whatever that be whatever that be at the minimum of the potential is equal to the same constant C E to the minus V of X naught over K Boltzmann T I do not care I do not care you specify what A should be what B should be because you will see that this answer is going to become independent of A and B due to the nature of the integrant as you will see in a minute in just a minute so P equilibrium is this quantity and the ratio of the two C cancels out so you could also write this as equal to P equilibrium of X naught which is itself unknown times E to the power V of X naught minus V of X over K Boltzmann T so I got rid of this but in favour of this constant now I need to integrate here at any point so what is happening is the what you are envisaging is way out here you are going to insert particles at the stationary rate and they are going to come out of that stationary rate but the question is how is this barrier affecting that J stationary? Yeah it will depend on the it will depend on this flux it is a constant it is independent of X because that is how we defined J stationary it was the integration constant on the right hand side right we had D over DX this current state this DP whatever it is equal to 0 and therefore that whatever in the bracket is equal to constant and that has got the significance of a current some stationary current right. So the question is how is this barrier affecting it so essentially what we are asking is finally the final question is given that the part A particle starts here what is the probability that it is going to be thermally assisted to get out of it even if you wait for a long long long time what is going to be the probability all right. So we have a formula for P equilibrium but we need to integrate it from A to B let us do that therefore integral DX P equilibrium of X this is a constant so it comes out P equilibrium of X not E to the power V of X not comes out and then you have to integrate integral from A to B DX E to the minus V of X over K Boltzmann you have to do this integral. Now look at the nature of this integration you got E to the power function in a neighbourhood of the minimum at this point right so whenever V of X not V of X is large this integral dies down because E to the minus and the largest contribution will come from the minimum of the potential because that is where V of X out there as this integral is going to make sense it is going to be nearest E to the 0 which is 1 right all right. So this means you can write V of X equal to V of X not plus X minus X not V prime of X not which is 0 so this is now straight forward Gaussian integration that is 0 plus X minus X not whole square over 2 factorial V double prime of X not plus higher order terms and they are all sitting in the exponent so this means you can take this out write as P equilibrium of X not E to the minus the leading term is V of X not which cancels this goes away and then you have an integral A to B DX E to the minus X minus X not whole squared over 2 or there is a K Boltzmann T also so K Boltzmann T here and then V double prime at X not into 1 plus higher terms because I can write the next term the cube term by taking it out down here by expanding the exponent but the leading contribution comes from here that is like a Gaussian integral and in a Gaussian integral the bulk of the integration comes from the maximum of the potential in E to the minus A X squared the bulk of the contribution comes from the point near the origin right and the rest of it is exponentially down so this is a standard approximation it is called goes by many names it is the starting point of something called the method of steepest descent the method of the phase method and so on saddle point method it has got many many names we are looking at the simplest version of it okay and then in the same spirit you can actually extend this integral from minus infinity to infinity because again the contribution comes from just the center everything else is exponentially down so the dependence on A and B is gone out here and we know the formula for this E to the minus A X squared is squared of pi over A from minus infinity to infinity provided A is positive that it is KT is positive and this is at a minimum of the potential so the second derivative is positive be sure about that so this quantity here is greater than 0 so this is P equilibrium of X naught and then this exponential factor is gone square root of 2 pi K Boltzmann T divided by V double prime of X naught that is the denominator so this is equal to J stationery divided by P equilibrium of X naught square root of 2 pi K Boltzmann T over V double prime we set that aside it is going to come because we still have to find J stationery right we still have to find J stationery so how are we going to do that but we have to go right back now and put things in so how are you going to find J stationery we have a formula for we need a formula for this guy we have this fellow sitting here what are we going to do next how are you going to find J you have to remind me of this equation J stationery pardon me yeah 1 over M gamma this is a minus sign so this P equilibrium over DX so let us pull out yeah plus P equilibrium times there is another term V prime exactly so that was J stationery right which is equal to minus K B T over M gamma I will write this as a total derivative so I can write this as D over DX times D over DX of P equilibrium times P to the power V of X by K Boltzmann T because it looks like a total derivative if I do this I get D P over DX times exponential plus P equilibrium times V prime divided by K B T but I should now compensate for that E to the V power minus V of X so I rewrite it in this form therefore J stationery multiplied by E power V of X over K B T is therefore equal to minus K B T over M gamma D over DX P equilibrium E to the V of X over K if I get this right with all the factors and so on will be miracle because I want to be careful about the factors and not make a mistake somewhere because then it means the formula I know the final end product formula that is very easy to remember so I just want to make sure I get all the factors right okay. Now let us integrate this from remember the geometry like this and then went off this fellow was X naught this was the peak X 1 at this point and then you looked at some point X 2 here so let us integrate this from X naught to X 1 X 2 let us integrate DX from X naught to X 2 so this is integrated X naught to X 2 DX so remember this is some horrible function we do not know what it is but I am going to accept it has a shape like this and I am going to integrate it from X naught to X 2 what is appearing on this side is E to the V X with a plus sign so again this is going to contribute the maximum at the maximum of the potential else where it is going to drop down very very rapidly right. So now let us write this therefore as integral from X 1 to X naught to X 2 J stationary okay and then DX E to the V of X let us do an integration saddle point or Gaussian integration around the maximum of the potential X 1 so I will write this as E to the power V of X 1 and again the first derivative is 0 so the next term is X minus X 1 whole square over 2 factorial V double prime at X 1 plus higher order terms the whole thing divided by K T that is the left hand side which is approximately equal to J stationary E to the V of X 1 over K Boltzmann T times a Gaussian integral but it is got a plus out here however you are at a maximum and therefore the second derivative is guaranteed to be negative therefore I can write this as modulus and there is a minus sign here and that integral is equal to square root of pi over A in a formula square root of 2 pi K Boltzmann T divided by modulus of V double prime that must be equal to on the right hand side this integral out here from X naught to X 2 of a total derivative right so this is equal to minus K Boltzmann T over M gamma times all we have to do is to write now P equilibrium at X 2 E to the power V of X 2 over K Boltzmann T minus P equilibrium of X naught E to the power V of X naught over K Boltzmann is that a plus sign or a minus sign plus sign it is a plus sign as it stands not too thrilled by that yeah that is okay that is okay we have to go where it takes us right now the probability itself at any point is vanishing out here like X 2 most of the probability mass is sitting here out here therefore compared to this term this term is negligible potential is concerned they do not make a difference they are of same weight so these factors are of the same order but this thing here is much smaller than that is very improbable that you find something at X 2 so you neglect this and you get approximately this thing here is approximately K Boltzmann T over M gamma times P of equilibrium of X naught E to the power V of X naught over K Boltzmann T and we are nearly home so this is J stationary it took to take everything to the right hand side and this therefore is equal to K Boltzmann T over M gamma P equilibrium of X naught is still sitting there I move this factor to the right hand side it comes with a minus sign and this is a plus sign here so E to the power minus delta V at last over K Boltzmann T that is sitting there and then take this to the right hand side so square root of modulus V double prime of X 1 divided by K Boltzmann T divided by this guy the whole thing divided by P equilibrium of X naught and then this factor so I can again write this as a 2 pi right and then this goes up in the numerator once again so this is V double prime of X naught divided by 2 pi T Boltzmann T and notice P equilibrium of X naught cancels out we do not know this number we are not able to compute that number without actually solving the full equation and finding out what the steady state solution is but it mercifully cancels out the T cancels out as well so the K Boltzmann T cancels this and this so we finally get this lambda escape equal to square root of V double prime of X naught modulus V double prime of X 1 divided by 2 pi M gamma E to the minus delta V over K Boltzmann T this is the Kramers escape rate formula so it tells you that the rate at which this escape happens depends exponentially on the barrier height with a 1 over K T that is precisely of the Arrhenius form and as you know this is an extremely sensitive form of the temperature a small change in the temperature changes things enormously the rates of chemical reactions which are thermally assisted changes enormously with the temperature for a given barrier height similarly once you increase the barrier height this exponentially becomes more difficult it justifies post facto or assumption that the probability is actually pretty small the escape rate is pretty small but this is indeed non-perdibutive in the sense that it is not power series and delta V over K T or anything like that it is actually given exponential form in 1 over T that is why chemical reactions require very high temperatures in general unless of course you have enzymes otherwise just think about it every time you take in food your system actually breaks down these complex molecules and makes other complex molecules but when you cook food you do so at very very high temperatures that is how much energy it takes to break those bonds down right on the other hand you ingest this food and your stomach does it actually effortlessly apparently at 37 Celsius so it would have come down enormously the rate of reaction but it is working because there are enzymes but this temperature dependence of the Arrhenius formula this thermally activated processes is very typical multiplied by these factors here now geometrically of course these are the curvatures of the point of the potential at that point so you can write this in a very compact way that is the way it is normally written you could say well if I have potential like that around this point this curvature can be subsumed in the frequency of harmonic oscillations about this minima so let us call that frequency omega naught around the point X 1 X naught and similarly this is an inverted parabola so let us call the frequency of this inverted parabola the harmonic approximation let us call that frequency omega 1 then of course this curvature here V double prime this is equal to this thing here is M omega naught squared and this is M omega 1 squared and the M cancels that way so this gives you a nice little formula is omega naught omega 1 over 2 pi gamma e to the minus delta V over K volts this is the Kramers formula that is a simple derivation of this formula now we can put in all the various complications and so on for instance you could ask what happens if this is potential barrier height is brought down KT is made larger what happens if you have more singular potential and so on what happens if you have quantum tunneling assisting this thermal process these are complicated questions but they do have physical applications physical importance but in the simplest instance this is the way the Kramers formula is derived the ingredients in it are the curvature the minimum the curvature at the maximum of the potential at the height of the barrier at the bottom of the barrier and the energy difference potential difference with the temperature appearing here this is a very mild dependence on temperature powers and so on but the pre factor as it turns out in the leading approximation is not even temperature dependent it is gone yeah now we have assumed that things are curvature exist it is finite etc etc so you will have to do those case by case or if it is but this is the generic case you could ask what happens if V double prime is also 0 and it is a very flat 4th order potential and so on I leave you to play with those things and find out what happens but the trick essential trick is the Gaussian trick okay the next problem we are going to look at is what happens in a constant force feet I said we would do this and then go back to the small Kofsky equation that is a linear problem it is relatively simple again I call attention to the fact that what we have done is a self consistent calculation we assume that there is a stationary current which is nonzero stationary flux and then we kind of found it by a self consistent way by a sequence of approximations knowing beforehand that the problem equilibrium stationary equilibrium density at beyond the barrier is actually going to be quite small because of this barrier how small exponentially small by the height of this barrier and one should have expected this because the only energy scale in the problem is this and of course the height of the barrier so this is got to be some function of the ratio let us look at the problem of sedimentation again let us convert it to a one dimensional problem and now this time let us do it under gravity just so that we have our physical picture this is x equal to 0 and the x axis moves upwards and the potential is mg x so v of x in this problem with reference to the floor level and we have a particle diffusing in a column in one dimension up there in a big fluid column infinite say semi infinite and then we ask what will be the equilibrium distribution of the probability density or the density itself like a column of gas assuming the whole thing is at constant temperature that is not true of the real atmosphere but assuming that you are going to get an exponential distribution something with the probability of finding it at a height x is going to decay exponentially with the height that is called the barometric distribution and we can actually write it down it is very clear that you will have the equilibrium of x and x by the way 0 less than equal to x less than infinity so it is going in the vertical direction this is proportional to e to the power minus mg x over k Boltzmann t that is the Boltzmann factor the Hamiltonian is going to have p squared over 2 m plus mg x and e to the minus beta Hamiltonian the x part of it is precisely this one should expect that so let us write the exact equation down for the actual probability density using the Smolukowski equation so delta p over delta t x t is equal to 1 over m gamma delta over delta x v prime of x but v prime of x is just mg with a minus sign right minus v prime of x so minus right the force is in the downward direction times p p the full p there is no x because it got differentiated out so it is actually an easier problem than the Fokker Planck equation was for the velocity process or the OU process plus the usual thing k t over m gamma d 2 p I have already had a minus v prime of x right this will become of course let us see the dimensions are okay this is g over gamma and it should have dimensions of of a velocity l t inverse because then l here and t here which it does this is l t to the minus 2 and this is l t t to the minus 1 so it is l t inverse which is a speed so let us give it a name let g over gamma be equal to c it is got dimensions of velocity of speed let us call it c and this is our old friend d of course we can immediately write down what the equilibrium distribution is it is precisely this because I pull out this is equal to 0 I pull out a d over dx and whatever is in the bracket is the current and that current must be 0 at infinity it is a constant in any case and therefore you end up with this distribution you have to normalize it from 0 to infinity you integrate and the answer is some k t over m g whatever it is times the exponential so that is the barometric distribution but our interest here is in some some things like well I would like to get the full time dependence if possible but first let us see what this stationary thing actually is like notice that if I call so this is equal to c delta p over delta x plus d times d to p over dx where the c is given by that and d is k t over gamma as usual now you can form if you had for instance a finite column then there is one more microscopic length scale in the problem which is the length of this fluid column then you can form a dimensionless number so you can form l c over d if this column extended to height l from 0 to l for instance then this quantity has got dimensions of this is l and that is l t inverse so this l square t inverse this is l square t inverse so that is a dimensionless number and it will govern the motion fluid dynamics under a constant field of force it has got a name does anyone familiar with this name here all these dimensionless numbers in fluid dynamics have specific names like the Reynolds number which you are familiar with this is called the Peclet number and it plays a role in fluid dynamics and whatever so we have settled the equilibrium distribution it is this notice gamma cancels out in it as it should there is no role for it in that there is dimensionally it cannot exist in p equilibrium this in fact cancels out as you put it out of the bracket and set the rest equal to 0 then you just get k t over mg which is what happens here now what about the exact solution what would it look like I am not going to solve it here now but what would this actually look like it depends on the initial condition so we have to specify the initial condition and the boundary conditions so the initial condition p of x 0 equal to some delta of x minus x 0 same so if I plot this p of x p as a function of x initially it is a delta function at x 0 remember you cannot go to the left of x equal to 0 nor can p be negative as t increases and becomes infinite this thing goes to the asymptotic exponential distribution p of x 0 and this distribution as p of x infinity which is p equilibrium it is an exponential thing so the question is how does this delta function spike at t equal to 0 how does it spread out and become this exponential with a p cut 0 no matter what x 0 is so clearly sort of physically you can see that initially immediately it will start diffusing so it will do something like this and then the peak meanwhile start shifting to the left so it will start doing this but then there is a bounce back anything that hits this cannot go there so it will probability mass will bounce back and gradually as the shifts left it becomes more and more pronounced and this gets flatter and flatter and eventually it goes to this so clearly there is reflection at x equal to 0 which is obvious because under this diffusion process particles are going to hit the floor and they cannot go through the floor there is no leakage through the floor so what is the boundary condition for this so the boundary condition the p of x t tends as x tends to plus infinity this tends to 0 it is a normalizable distribution so the probability density must go to 0 as x goes to infinity what is the boundary condition at the origin at x equal to 0 so p of 0 t what is the condition on this it is not 0 it is not 0 at any arbitrary time this probability density is not 0 in fact at t going to infinity all the probability mass much of it is concentrated here but the current is 0 and what is the boundary condition and everything lies in the boundary condition if you have understood that then job is done so remember that the current here you pull out a d over dt and the gamma so you can write this as equal to this whole side as delta over delta x g over gamma p plus k Boltzmann t over m gamma dp delta p over delta x and minus this quantity is the current and you want the current at the floor to be 0 so it is a the boundary condition at bc at 0 is delta p over delta x the gamma goes away plus m over k Boltzmann mg over k Boltzmann tp equal to 0 at x equal to 0 for all t so sometimes that is a reflecting boundary condition if you discretize this and did it as a random walk problem it says the moment you come to 0 the particle cannot go to the left so anything which tries to go to the left has to go to the right so the probability of a jump to the right is plus 1 you have to impose that as a boundary condition at 0 so it is the current that goes to 0 not the probability density itself that does not do so so given that you have to solve this equation I am going to leave this as an exercise mention what the solution is it is messy to write down how would you do this well you can do the you solve the full partial differential equation we know the initial condition we know t runs from 0 to infinity we know x runs from 0 to infinity so take Laplace transforms with respect to time and then you get an ordinary second order differential equation this is a constant coefficient and so is this so you can find a solution for the Laplace transform you have to translate you know the initial condition out here we know the Laplace transform of the time derivative it is s times the transform of this minus the initial value it becomes a green function equation for whatever is the Laplace transform and then you have to invert that so there would be various error functions etc etc etc and then after that you have to invert the Laplace transform but the problem is doable it is exactly solved but we have extracted all the physical information from it now the rest of the day all right so let me stop here today