 In the previous lecture, we discussed the the question of equilibrium in the random work models. Within the framework of the random work equations that we had set up, especially the nearest neighbor jump type of random work, we postulated that supposing there exists a steady state, then the state should be invariant with respect to number of steps and that would then lead to a condition of a constant flux and then a special case of that flux becoming 0 is called an equilibrium state. In that equilibrium state, the problem reduces to a simpler system of equations. One obtains for example, an equation of the type q say m w infinity m equal to P m minus 1 w infinity m minus 1. What it means is a special situation in which if you denote the states m and m minus 1 in this order, say this will be an upper state m plus 1 will exist here, m minus 2 will exist here, but if you look at the states m and m minus 1, while the random work equation connects the 3 of them together m plus 1 m and m minus 1, under the equilibrium situation only the 2 states get connected that is the meaning of this equation that we arrived. What it says is under equilibrium the downward transition from m to m minus 1 that is called as q's. If you remember our q's are backward transition or in this sequence it is a downward transition that rate should exactly balance the upward transition from m minus 2 m, this is P m minus 1. Such a balancing between just about 2 states whereas, the process of stochastic evolution depends on 3 states is called detailed balancing condition. It is a very important condition in equilibrium statistical physics. So, this detailed balancing enabled us to develop a quick solution in the form of a product sum we call it as a sum and a product together providing a stationary solution exists. We illustrated this by considering a case of a biased random walk constant site independent bias, but in the presence of a reflector at 0. So, only one sided random walk what is the distribution that we would expect in equilibrium situation and an equilibrium does exist for this if the bias is towards the reflector itself. So, you would see an exponentially distributed occupancy probabilities as once goes to the higher sites away from the reflector. So, this was a very specific result obtained as an illustration. One can actually use the transition probabilities tuned to reflect various types of attractive forces around points such as harmonic potentials or other bound bounding potentials and obtain a discrete state distribution. We then proceeded to say that such equilibrium can also be discussed in connection with some finite state problems as an equilibrium solution and that we called ah we we mentioned in that connection about Ahrenfest's flea model because originally a dog flea model flea model we will say Ahrenfest's flea model flea model. In the present lecture we will spend some time on understanding this particular model of equilibrium again because it has some connection with the non-equilibrium system we already discussed. Deceptively similar solutions come, but this is to be interpreted as an equilibrium solution obtained after infinite number of steps have been taken or rather process has reached that kind of a state. So, we continue to ah ah discuss the Ahrenfest model. So, this is about equilibrium continued. So, going back to the Ahrenfest's flea model it was originally proposed as an equilibrium as a tendency to attain equilibrium or tendency to move towards a centrality even without explicit causes it is actually an early version of the model for diffusion. So, we consider in this two animals I am just representing the animals by a circle say animal A and there is another animal B and let us say they are kept in an animal house in close quarters within a few feet away. Now, let us say that somehow there are total of L fleas total of L fleas which rest either on A or B they cannot rest anywhere else. So, this L number of fleas partition themselves either on A or on B and in a random manner it is not that they want to be on A or on want to be on B it just happens by purely a random ah process. We ask a question in equilibrium commonsense tells us that if there is no preference of the total let us say L fleas L by 2 should be on A and L by 2 should be on B this is a very deterministic mean, but in reality we know that fleas will be jumping from one surface to another and it will be deviated from the L by 2. So, the natural question for us is how much would be the extent of deviation in such a system. We can ask of course, a deeper question of the kinetics of the whole process supposing A is fresh just had a bath and B is full of the L number of fleas how much time it would take for A to reach that equilibrium if at all does he reach the equilibrium. The answer is in a system of this type it will reach equilibrium and there is a unique answer. However, how much time it would take and what functional form it would take is a bit complicated not easy to solve within analytical framework that we have set out to do. So, we relegate this question for future at the moment we merely ask that steady state has been attained and what is the distribution. So, to proceed further on this first we have to postulate transition probabilities. So, the question is the what is the partitioning what is the distribution of fleas between A and B that is the kind of a question we are asking. So, we have to postulate transition probabilities. In the random walk model we learnt basically to use two concepts in one dimensional random walk the so called forward or upward or backward or downward the transition probability is P and Q. Most of the discussions we had are either P equal to Q or a P and Q could be different from each other, but constant across the entire lattice system. Here accordingly we have to first define the lattice equivalent and what is the lattice equivalent that is A can have let us say 0 fleas 1 flea or 2 flea let us focus on A because the reciprocally the similar thing will be happening with B. So, focusing on A either it can be 0 fleas or 1 flea or all the L fleas on it. So, our state space here consists of space consists of L equal to 0, 1 etcetera up to L fleas. So, we therefore, have to find a distribution for occurrence of L fleas given that there are total of L maximum possible fleas. So, to find a distribution for occurrence of L fleas on A. Now, if A has L fleas let us say a particular realization then necessarily B must be having L minus L fleas. Now, we must ask the transition probability the so called upward transition probability. What is upward transition probability? L raising to L plus 1 is the upward transition probability. So, we define E L it is the upward transition probability that is L going to L plus 1. Correspondingly we can define Q L as the downward transition probability that is L going to L minus 1. Let us argue how L will become L plus 1 first. So, let us argue how to find an expression for P L which is the probability of picking up 1 flea. We here have already made an assumption that at the most I take given step or a given instant only one flea can gets transferred simultaneously more than one flea does not get transferred from B to A. So, this is therefore, an equivalent of a nearest neighbor like random walk. So, this is equivalent to a nearest neighbor transition probability of L becoming L plus 1 becoming L plus 1. Now, if you have the two animals A and B and if this is L and this is L minus L assume that any one of those fleas could have landed here it is just a random process. If there is an intrinsic probability a very small probability epsilon let us say for a one particular fleet to just change its mind and comments it on A that probability is uniform is the same for all the L minus L fleas which are there. Hence, the tendency for L to become L plus 1 that is a P L basically which means transition from L to L plus 1 this tendency should be proportional to the number of fleas present on B should be proportional to the number of fleas present on B that is we write it as proportional to L minus L or to convert it into equality we can write as some constant k into L minus L. So, the upward transition probability is proportional to the number of fleas which is present on the other animal. Now, let us discuss what is Q L that is the probability of transiting from L to L minus 1. The mechanism is the same randomly a flea just leaves one animal and moves over to the other on its own wheel free wheel. So, if that is the model and it has no preference between A and B then the probability that one flea will leave A is the same epsilon which I mentioned for that being on B. Hence, since there are L such fleas the probability that any one of them will leave is L times that epsilon. So, basically it should be proportional to L the number of fleas present on A will virtually determine the probability that one of them will leave. So, converting the proportionality to a equation you can write it again it as k into L. So, we have now two relationships P L equal to k into L minus L Q L equal to k into L and we use the constraint equation P L plus Q L use P L plus Q L equal to 1. Every step one flea has to go either from A to B or other way. So, which basically means that k into L minus L plus L should be 1 implies k equal to 1 by L. Hence, we can summarize P L equal to L minus L by L and Q L equal to L by L are the upward and downward transition probabilities. With this and all of us to the random walk problem we can set up the random walk equation for the occupancy probabilities. So, let W N L be the occupancy probability of L fleas on A at the nth step. The whole process is being executed in steps nothing happens in between steps. So, we allow the process to continue in terms of steps and that is the basic philosophy we have been following in all the discrete version of the stochastic modeling. And in fact, in the previous lecture I introduced that now it is time for us to move forward to more realistic situations of continuous time random walk models which are what natural systems are. So, towards that end we complete this problem and set up an equation for the occupancy probability W N plus 1 L will be P L minus 1 W N L minus 1 plus Q L plus 1 W N L plus 1 and that is all it is a nearest neighbor problem only one flea at a time. So, contribution to L can come either from L minus 1 with the propensity P L minus 1 or it can come from L plus 1 with the downward transition propensity of Q L plus 1. So, the problem is well defined we can start with any initial condition for completeness sake that is W 0 at the 0th step L let us say it can be delta L say 0. So, begin with there was no fleas on it this L of course, is 0 1 L. Now, because my P's and Q's are not constants they depend on the sites especially from the relationships that we derived here these relationships which show some linear dependencies. It is not so easy to get a close form solution to this even by generating function methods. So, in fact, in a finite space generating function is not very useful the site specific generating functions will have several problems when you try to execute it even the step some generating functions are also not quite they are also quite complicated. So, we at the moment desist from the asking a question on what is the transient solution. Instead we straight away first ask the question on what is the equilibrium solution. So, as earlier we postulate that such an equilibrium could exist sufficient large number of steps the occupancy probabilities will be invariant with respect to the number of steps and that would be W infinity and then this equation is just going to be W infinity L will be T L minus 1 W infinity L minus 1 plus Q L plus 1 W infinity L plus 1 for all L equal to L. Now, this equation is not very difficult to solve already we have established a method for equilibrium and with that method we showed that the final solution is actually a product of piece divided by a product of Q's. So, we showed in our previous lecture that in general the equilibrium solution W infinity let us say to any m th state will be P naught to P m minus 1 divided by Q 1 to Q m W infinity 0 this was proved previous lecture. So, we make use of this result although you one can do a exercise by actually arriving at independently for this problem also. So, accordingly we say that for the present problem for the flea problem we have W infinity L is going to be W say W naught. So, W infinity 0 the ground state the blank state as we say multiplied by the corresponding P values. So, it will be now what is P naught we go here P naught is L by L that is 1. So, we can say it is just L because the denominator L is common for B's and Q's you can note here P's and Q's occur equal times both in the numerator and denominator and hence any constant factor will disappear and that leads to a solution W infinity 0 into it will be L into L minus 1 etcetera L minus L minus 1 divided by 1 into 2 because the denominator is proportional to L only into L and this is simply L factorial divided by L minus L factorial into L factorial into W infinity 0. So, as a result we now have to find out. So, we need to find out W infinity 0 will just come from normalization is obtained from the normalization we get W infinity 0 will be actually 2 to the power L it will come to the binomial theorem because sum of sigma L C L is going to be just 2 to the power L by binomial theorem. So, the finally, we get the solution W L infinity equal to 1 by 2 to the power L L C equilibrium solution to the free problem for all L. As I mentioned it deceptively close to the random walk model where if we map now L is the number of steps and L is the distribution with respect to the sites lying within the number of steps there is a slight difference it is only L by 2, but it is nearly similar. In fact, if we transform this equation with respect to the medium value then it comes even closer to the random walk model, but we must remember that there it was the number of steps here whereas, L is the number of sites. So, it makes a conceptual difference and this is an equilibrium solution. Now, we can find the generating function for this G z which by definition of course, we can superscript it with the steady equilibrium state will be L equal to 0 to L of W L infinity W let us say L infinity z to the power L that is how we define and it simply will come to 1 by 2 to the power L sigma L C L z to the power L L equal to 0 to L and this is simply binomial expansion of 1 by z to the power L comes exactly the same. So, the answer is going to be 1 by 2 to the power L 1 plus z to the power L. So, generating function for equilibrium state is very easily developed by summing. So, from that we can obtain the properties for example. So, the characteristics main characteristics of the distribution such as the mean and the variance the mean L bar as we know that mean L bar is basically G prime z at z equal to 1 we can differentiate 1 plus z. So, simply it will come to L by 2 we can do a similar exercises for the variance in fact, L square bar for this problem L square bar is going to be G double prime 1 plus L bar if you differentiate it will come to this and if you when you carry out the whole thing it will be L square by 4 minus L by 4 plus L by 2. So, it is to be L square plus L by 4. So, similarly we can we get already L bar square we know that is L square by hence we get the variance sigma square L square bar minus L bar square then L square by 4 terms will cancel you will get L by 4 or the standard deviation sigma is L to the power half by 2. So, important characteristics we have obtained by this the distribution for example, just to see how it would look for small L it is a discrete distribution. So, it will not look very continuous it will have this kind of a look. So, if this is for L equal to 10 it will move further and further away and more smoother and smoother for say let us say this is up to 10 this is 20 this is for L equal to 20 the mean value will be always half. So, if it is 20 it should be at 10, but just we can we can let us say not to the scale, but just to demonstrate it or we can sort of correct the whole thing we can say that for 10 it will come let us say this way with the mean at 5 now it agrees its end point will be 10 whereas, for 20 it will have an end point here. The sum and substance of this is there is a fluctuation around the mean and that fluctuation tends to a Gaussian as L increases and that Gaussian variance is going to be just the variance that we derived that is L by 4. So, one can summarize it in the form of a Gaussian distribution which has the form that is W infinity will tend to be 1 by sigma root 2 pi as we see e to the power minus L square that is L by 2 minus L whole square divided by 2 sigma square where sigma is equal to L by root L by 2 tends to a Gaussian. So, we demonstrated the existence of an equilibrium in a finite system tendency towards Gaussian distribution as a stable distribution as an equilibrium step invariant distribution that is the interesting thing about it. And there are many problems now which can be addressed by the logic that we developed depending on the ratio p's and q's. However, as I mentioned in my previous talk also vast class of problems of complex nature can be solved in a better fashion when we move to continuous variable formulation and that is what we will do in our next lecture. Thank you.