 Okay, we would like to, I would like to make some more comments about the solution to the random walk problem, the linear random walk on a linear lattice with a bias. There are certain points about this solution which are general in nature and I would like to start with that and then go on to some more properties of general birth and death processes, okay, together with hopefully some physical examples of what is happening. Now if you recall our random walk problem, so this was random walk on a linear lattice, if you recall we had these sites on the lattice, this was site J and this was J plus 1, J minus 1 and so on, all the way only the side over this full set of integers and the rule of the game was that from every site J you jumped to the site J plus 1 with a mean rate which was lambda times alpha and you jumped from J to J minus 1 with a mean rate which was lambda times beta such that alpha plus beta is 1 and then we discovered that if you start at some arbitrary origin say 0, the probability P of J, T starting from the origin, this had an explicit solution and the generating function for this was z to the J summation J equal to minus infinity to infinity was a quantity we could evaluate explicitly and this was e to the power minus well e to the power lambda T times alpha z plus beta over z minus 1 and that was it. We interpreted this in two different ways, one of them was to write a birth and death equation for it and solve it using a generating function or to say that this process, this random process J is the difference of two Poisson processes with rates lambda alpha and lambda beta and that immediately gave us this as a generating function, okay. Now from this the mean values could be written down, the solution itself could be written down etc and just to recall to your mind what happened, this was e to the minus lambda T and then we could write the rest as e to the 2 lambda T over 2 square root of alpha over beta times z plus square root of beta over alpha 1 over z and that helped us immediately identify P of J, T as a certain Bessel function, right. So this immediately gave us the solution P of J, T starting from the origin to be equal to e to the minus lambda T and then the Bessel function ij of 2 lambda T times square root of alpha beta and then there was this factor alpha over beta square root to the power J or this to the power J over 2. So that was the explicit solution, okay, right to this probability for this probability. Now from this we can read off all the moments etc, etc and interpret them in some sense. For instance the mean value J of T, this was equal to if I call this thing the generating function f of z comma T equal to this, this was the generating function, the mean value here is just delta f over delta z at z equal to 1 and that follows directly from here you differentiate this you pull down a lambda T and then an alpha minus beta if you put z equal to 1 this goes away the exponent goes away and you have this is equal to lambda T times alpha minus beta. So you see what is happening is that this bias alpha minus beta this differential preference to the right or the left depending on alpha being bigger or smaller than beta acts like a velocity. So this is like the mean position of displacement after time t is proportional to the time and of course it must be multiplied by a velocity in this drift velocity is essentially lambda times alpha minus beta. So lambda times alpha minus beta is like a drift velocity. You can now compute the mean, the mean square and so on and we know that the variance actually proportional to lambda T. So one should do this a little more elegantly and so going on differentiating this function here let us just find the cumulant generating function that will solve the problem in one shot. Now the cumulant generating function if you recall is the log of the moment generating function and the moment generating function so k of u is the cumulant generating function equal to the log of the moment generating function m of u but that was the same as f of e to the power u and that is equal to all we got to do is to take logs out here. So this is equal to lambda T times alpha e to the u plus beta e to the minus u minus 1 that is it. Therefore it immediately follows that the rth cumulant k r is just d r k of u over d u to the power r evaluated at u equal to 0 and what does that give us all you got to do is to differentiate this fellow and each time you differentiate e to the u you get the same thing e to the u and you differentiate this you get a minus or plus sign depending on whether it is even or odd. So this becomes equal to lambda T times alpha plus minus 1 to the power r the first moment first cumulant is of course the mean value itself. So it is clear that this k 1 equal to j of t and that is of course alpha minus beta which you already derived right but what is interesting is that the variance is independent of this bias. So the drift is cancelled out by the fact that you subtract the square of the mean when you compute the variance from this mean square and therefore k 2 equal to the variance of j this is equal to lambda T this is a crucial result it is typical of diffusive behavior we know that the root mean square displacement must go like the square root of the time and that is exactly what is happening here when you compute the variance you subtract out the systematic drift and then you end up with precisely diffusive behavior. We can of course examine and ask whether p of j comma t itself has a limiting distribution or not remember we are in an infinite lattice now what would you physically expect I would expect if I start with probability 1 of being at the origin at t equal to 0 as time goes along you are spreading you have an infinite amount of space to spread out on both sides. So although the particle is somewhere on this line the probability of being at each particular point vanishing should be vanishingly small so you would expect it should go to 0 and there should be no stationary distribution in this problem. So in a birth and death process when you have an infinite lattice when this variable is not is unbounded from both sides then the probability stationary distribution will not exist unless you put in some external condition like you would put in a current of particles from one side and let them jump out from the other side or something like that but that is an external condition without that in general it will turn out that this probability should vanish as t tends to infinity but that is that happens here we can check that out because it turns out that ij of whatever is inside ij of let us call it whatever variable you want to call it xi whatever it is this tends as mod xi tends to infinity this fellow tends to e to the power xi over square root of 2 pi xi that is the leading behavior of the modified Bessel function and it is independent of j the order j provided the order itself does not become infinite. So if you apply that here it is immediately clear that P of j, t on 0 tends for long times which means lambda t much much greater than 1 tending to infinity in fact because the time scale in the problem is set by lambda inverse that is important remember this tends to e to the minus lambda t times 1 minus twice square root alpha beta and then alpha beta to the j over 2 divided by square root of whatever square root of 2 pi so that is 4 pi lambda t root of alpha and this quantity vanishes exponentially rapidly in t as long as this is bigger than that and remember that this quantity is 1 minus 2 root alpha beta equal to 1 minus 2 root alpha times 1 minus alpha where 0 is less than alpha is less than 1 and when does this quantity become large the largest at alpha equal to half because otherwise on both sides it is a parabola it comes down on both sides at alpha equal to half it is maximum and then of course the 1 cancels out. So it this quantity dies down exponentially as long as the coin is biased as long as the random walk is biased either way we do not care but this will exponentially vanish on the other hand when alpha so P of j t 0 tends in the case alpha equal to beta equal to half it goes like a power law so now this factor cancels out and it goes like 1 over square root of 4 pi at 2 2 pi lambda t goes like 1 over t to the half that is typical of diffusion in one dimension linear problem instantly now that we have written this down we can actually write down the solution to the random walk problem on an arbitrary dimensional lattice square lattice or a hyper cubic lattice or a hyper cubic lattice there is no problem at all in writing this down because one can do this by inspection because all the jumps are independent of each other and then you can see what is going to happen so let us generalize this right away when we come back to this so suppose I have a d dimensional cubic lattice or hyper cubic and let us suppose that the coordinates of any point are labeled by j 1 j 2 j d now let us call this the vector j so the components are all integers and they label the lattice point I start with some arbitrary origin it is an infinite dimensional lattice and then suppose I put biases on both sides which you can always do so to move to the right there is a probability to move to the left there is a different probability to move up or down or move in or out and so on right so you can associate probabilities alpha alpha r beta r where r runs from 1 to D and the sum of all these alphas and betas equal to 1 they are all positive numbers then we can write down what the solution is for this quantity p of j t given that you started at the origin what would this be what would this be because the total rate of jumps out of a site is lambda and that is now split earlier it was split between lambda alpha and lambda beta on this site now it split between lambda alpha 1 beta 1 alpha 2 beta 2 and so on the whole thing adds up to 1 so what do you think will happen what do you think will be the generalization of this now it is not the difference of two Poisson processes there are several processes going on at the same time right and they are all independent of each other so one can actually write this down by inspection what do you think it will be well yes you certainly would let us write it down here this j you certainly have this factor as before this comes remember it came from the generating function because there was a mind you subtracted in the master equation you subtracted minus p of j, t with alpha plus beta added to 1 so that produces this e to the minus lambda t and then the rest of it is sitting here so what you would have is simply the same thing as this except this would be alpha r over beta r j sub r over 2 and this would be j subscript r alpha r beta r here e to the minus lambda t a product from r equal to 1 to j to d and that is the solution it immediately follows that this is the solution in a d dimensional hyper cubic lattice so once again one can ask what happens if t tends to infinity lambda t tends to much much greater than one so this tends lambda t much much greater than one they call this the diffusion limit very long time limit what would this go to all you have to do is to use this fact up there and then whenever there is a bias there is going to be an exponentially damped factor what happens if there is no bias at all what would you expect if there is no bias unbiased random walk what would this correspond to for each alpha and beta on a hyper cubic lattice yeah so you have for example from this point you have up you can go down you can go this way you can go this way you can go into the board you can come out of the board and so on so there is six nearest neighbors in three dimensions in d dimensions there are 2d nearest neighbors right so for an unbiased random walk alpha r equal to beta r equal to 1 over 2d so when you sum over all these 2d numbers you get unity then what would this become these factors disappear all these factors disappear this factor in the exponential there cancels out exactly and you are left with the product of these fellows and alpha or beta r or 1 over 2d so there is a 1 over 2d multiplying this gives you some constant etc but the crucial point is unbiased the crucial thing is the t dependence and what does the t dependence do there is this factor for each of the Bessel functions right and there are d such Bessel functions so this immediately becomes proportional to 1 over t to the d over so it is a power law dk but it is proportional to 1 over t to the half d over 2 where d is the dimensionality of the space when we solve the diffusion equation in continuous time and space then again you will discover this is Gaussian solution has a dk which is exponential which in general is 1 over t to the for an unbiased diffusive process it is 1 over t to the dimension over 2 this fashion so for 1 dimension it is 1 over square root of t which is what we found this now okay so you can begin you can see that there is going to be close connection between this and the diffusion process and we will explore that when we go to the continuous space case but already from the random walk on a lattice one can deduce these facts now of course this depends you would say on a in higher dimensions it would depend on the lattice the nature of the lattice and what are the nearest neighbor jumps etc etc but actually this this feature is very general although this solution is very specific to that kind of lattice this is very very general and it actually follows from much more general considerations so we want to get into that right now except to say that I can go to a much more general kind of random walk I can say that I am not on a lattice I take steps in different directions of different lengths they could be distributed they could themselves be drawn from some distribution as long as the step length has a fine is a distributed in such a way that it has a finite variance finite first two moments this this kind of feature will always persist these these general features will persist so it is independent of dimensionality it is independent of the nature of the walk etc as long as the walk is Markovian this independent steps the steps are not correlated with each other if on the other hand you have a real material in which you have some correlation then this whole thing goes out of the window if for example you jump from left to right and then there is a predilection to jump back to the vacant site back again then of course you have correlation between different steps all these things have to be modified or if you have a situation in which you remember what happened earlier and you never is visit the sites you already visited once and those have a lot of memory and they go completely outside the purview of these Markovian random box okay so Markov assumption is very crucial here yeah now let us try to generalize this a little bit and put it in a slightly more general framework and that is the framework of what I called birth and death processes they have other names as well in the semiconductor industry business they are called generation recombination processes and so on and so forth there are different names in different areas will stick to this birth and death business now what was the fundamental assumption the assumption here was that on a linear lattice you jumped either one step to the right or one step to the left or in state space if J labels the states of the system their integer value and then J changes by plus 1 or minus 1 this is all if you generalize this a little bit then once again not with any reference to a lattice or anything like that but if I just write down the states of the system some random variable which takes integer values and you are at some point here let us call it N not let us call the let us call these sites by J once again so here is site J and if the process is such that you change this the transitions occur the W matrix I am going to write down will only connect neighboring states then I want to model for this quantity W K J which if you recall was W K J the transition rate to go from an initial state J to an initial to a final state K if this thing only connects neighboring states then there is got to be something which is proportional to a Kronecker delta such that K can only be J minus 1 or J plus 1 so there is a transition probability this way and one that way from J and all the other transition probabilities are 0 then what is the most general thing you can have you could have this to go such that the final state is out here with some rate and in the random walk problem we said that is a constant rate lambda times beta or the time apart from that lambda factor let me just call it beta but this could be dependent on where you are so as you move down the states the rate at which it makes transitions you still assume it goes to neighboring states on either side but that could depend the rate could depend on where you are in which case this would be some beta J times this plus the final state could be J plus 1 you could jump out there and in difference to the random walk problem that we used alpha and beta for the bias factors let us continue to call it alpha but it could depend on J now there are three possibilities for the state space either the state space which is takes on integer values runs from 0 to n or 1 to n some finite set or it is infinite on one side goes 0 1 2 3 all the way to infinity or it runs over the full set of integers these three are three different classes of problems altogether because when we write equations down with finite boundaries then I have to change the rate equation at the boundaries to make sure that the space is confined to that range just as when we did the Poisson process I said you have the probability rate at which the probability of 0 decays in a time t changes that decreases with time and then 1 2 3 etc etc they are all functions of time this is a Poisson distribution but the rate equation for DP 1 over dt was different from that for DP 0 over dt we will write the rate equation down once again right. So you have to be careful about the boundaries and then you can put all sorts of boundary conditions on it but for the moment let us take J to be a general number integer then what does the W matrix look like once you have this suppose J runs from 0 1 2 3 in that case you would have a finite matrix this kind if it is bounded from the other side to otherwise it is infinite that way and what does it typically look like let us look at W 1 2 for instance so J so in this case K is 1 and J is 2 so this is 1 sorry this is beta 2 and then when K is 1 it fires so you certainly have something here and then you have beta 2 in this case and this term contributes 0 okay all the others are 0 0 0 etc and what happens to 2 1 so if you look at W 2 1 this term does not contribute but this contributes K is 2 and if J is 1 it contributes a beta and what an alpha and what does it contribute it contributes alpha 1 and everything else is 0 down here and when it comes here you have a beta 3 contribution because that could have jumped to the left on this side and when you come here this could have an alpha 2 contribution on this side and what would the diagonal element be it would of course be we know this is equal to alpha 2 plus beta 2 this alone because I started with 1 2 3 4 labeling things with 1 2 3 4 there is no state 0 here in this problem this alone would be minus alpha 1 that would ensure that this is a stochastic matrix the sum of each row each column must be equal to 0 but the general structure of this matrix is a triangular matrix with something on the diagonal elements negative elements and then the super diagonal has all these beta factors and the sub diagonal has all the alpha factors so typically at some point in the middle on the jth column jth column everything is 0 till you hit a beta j and then a minus alpha j plus beta j and then you hit an alpha j and on the right hand side you have a beta j plus 1 this diagonal keeps going this diagonal keeps going and on the left hand side you have an alpha j minus 1 that is the general structure of this matrix okay and if there are end points to this matrix then you got to be careful at those ends etc and what would the rate equation correspond to in this case what is the rate thing now let us get over get rid of this j because that had this connotation of a lattice point but the examples we looked at earlier like the Poisson process etc we looked at a random variable n some number n so what would the rate equation be so let us write that because we are going to look at cases where n is running from 0 to capital N or 0 to infinity or minus infinity to infinity and what does that look like corresponding to this W the rate equation dp over dt dp n over dt as a function of t what is that going to be equal to well it is W times this p you write this p down in this fashion and what is the first term that is going to appear again you can see what is going to happen the growth of any particular knowledge call this n the growth of the probability here depends on fellows jumping from in from here and people jumping in from here right and whatever is jumping in from here has got to be beta but dependent on the site n plus 1 and what is jumping in from here is alpha dependent on the site n minus 1 right so it is clearly alpha n minus 1 p n minus 1 of t plus beta n plus 1 p n plus 1 of t minus alpha n plus beta n those would be these 2 processes which go out so this is alpha n beta n here this fellow was an alpha n minus 1 and this was a beta n plus 1 jumping in and these 2 other ones that jump out and that is multiplied by p n that is the general equation which you have for a one step birth and death process in which the elementary transition probabilities only connect you from the state n to the state n plus minus 1 and these are in general functions these rates are general functions apart from the fact that they are positive or non-negative you cannot say very much more so this is the general equation general master equation or rate equation and we looked at special cases of course we looked at special cases of it so random walk bias random walk alpha n equal to alpha beta n equal to beta alpha plus beta equal to 1 right what about the Poisson process what happened to the Poisson process and incidentally n was the element of the set of integers what about Poisson what did that correspond to we had a certain rate lambda but n now ran from 0 upwards and the Poisson process equation was for n equal to 0 1 2 etc to infinity and in this case alpha equal to beta equal to lambda but the process was dp n over dt equal to lambda times p n minus 1 so you had finished n minus 1 jumps in time t and then in the time that interval delta t after that you did one more with probability lambda delta t but you might have already reached n and you are now jumping to n plus 1 dk so therefore it is a lost right so this equation was this minus p n and that was true for n greater than equal to 1 and for n equal to 0 you had dp 0 over dt was minus lambda p 0 it is clear this is a special case of this very general thing with suitable choices of alphas and betas right this is a pure birth process because the number n kept increasing with time it did not ever decrease beyond a certain value so the left words probabilities were all 0 and you only had increase on one side incidentally this was a stationary Poisson process stationary in the sense that the statistical properties of the system do not change with time which means this rate lambda does not change with time what would have happened by the way we know the solution to that right we know that we know completely for the Poisson process Poisson process we know that p n of t is e to the minus lambda t lambda t to the n over n factorial n equal to 0 1 2 we know we can write down the generating function for this and compute what the coefficient of z to the n is and that is the end of it we did that explicitly by the way what would happen if this Poisson process was non-stationary suppose it turns out that for some reason lambda equal to lambda of t and you have a non-stationary but Poisson what would have happened to the solution this equation is still true because all you are saying all you are saying in these in this entire birth and death businesses you are saying that in a time interval delta t sufficiently small time interval delta t either there is a transition to the left or to the right or no transition at all it does not say that cannot be transitions to further points further states it just says those are of order delta t whole squared and so on and so forth whereas a transition to the nearest neighbors is proportional to delta t itself and then we got that differential equation so now the question is what happens if it is non-stationary you have a nice function lambda of t but it systematically changes with time in some deterministic fashion what would happen to the solution I would get exactly the same rate equations as before except this that this becomes lambda of t on this side and the initial condition is once again at t equal to 0 n is 0 okay what would have happened to that solution well the way we got that solution was we multiply this by z to the power n and summed over n to write the generating function down right and we got an equation with said essentially delta f of z, t over delta t was equal to lambda times z minus 1 f of z, t we got a z because there is an n minus 1 here so when you multiply by z to the n you got remove a z and write it as z to the n minus 1 and sum okay so that is how the z came out now what is going to happen is this once again the boundary condition is the initial condition is the same as before we know that only p 0 has a value at t equal to 0 that is 1 so f of z, 0 equal to 1 this is the initial condition what is the solution to this equation with that initial condition this is not a constant so how do you solve what is the solution yeah it is e to the power z minus 1 integral from 0 to t dt prime lambda that is it okay you have to integrate both sides its separation of variables immediately gives you this solution that is the general solution in the case when lambda is constant that exponent becomes lambda t as you can see okay and it satisfies the boundary conditions we got a unique solution so if you define this quantity you define this to be some lambda of t call it that when we are ready to write down what the probability density it is a probability itself is once again it is just the coefficient of z to the power n so you would get exactly the same solution except that you now have e to the minus lambda of t lambda of t to the power n over n so this particular making it non stationary is quite trivial it is quite a way with the solution immediately okay this occurs in the problem of what is called time dependent short noise okay but it is a trivial extension of the original Poisson process what is much harder to do is solving the general equation this equation here this is not at all a trivial matter even though it is a one step process in which you only go from n to n minus 1 or n plus 1 even then if these coefficients these coefficients are sufficiently independent in a complicated way then writing down an explicit solution is out of the question for the set of couple differential equations okay constant no problem that was a biostrandom or problem when alphas and betas are independent well the next thing you would guess after that which is solvable would be to say suppose these are dependent on n linearly that would be the next natural thing to do linearly for instance as a physical problem in which you have a quantum mechanical harmonic oscillator say levels are all quantized and let us suppose it is put bed in radiation of the right frequency so that the oscillator can absorb a photon go to one excite a state higher than it or it can emit a photon of the same frequency and go to a state lower right and then you ask what is the population probability of population of the state n of this oscillator that problem would be exactly this kind of rate equation in which the alphas and betas are linearly dependent on n. In fact that is how Einstein actually derived that is how Planck originally got the Planck law in that case what happened was for the radiation problem turns out that alpha n plus 1 equal to some constant times n and beta n equal to some of the constant b times n alpha n minus 1 was proportional to n and beta n was proportional to n once again with some constants a and b okay. I am going to leave it as an exercise for you to solve this set of differential equations given this not altogether trivial but at least find the stationary distribution in this problem n runs 0 1 2 3 up to infinity they label the levels of the harmonic oscillator right. So n is and find P n stationary I will give the answer to it is a geometric distribution which depends on the ratio a over b okay. So try to derive that what you have to do is to put this in and then equate this put P stationary here everywhere no T dependence and then equate it to 0 okay and then you try to find out what the stationary distribution is in this case but actually when you have a linear process you can go a little further and that is interesting because this rate equation can give you some physical information and here is how it happens let us suppose for ease of algebra let us suppose for a moment that n runs over all the integers so that I do not have to worry about these boundary points otherwise I have to be very careful about the boundary conditions and put in appropriate conditions each time I might for example if run n runs from 0 to infinity I will have to define alpha minus beta minus 1 as identically 0 put in a formal beta and define it to be 0 and so on but let us just say that n runs over all integers suppose n element of I look at this and I ask what is the average value of this random variable n at any time T. So what I need to do is to compute n of T this is equal to summation over n n times P n of T over all the integers since I am summing over all integers whether the summation index is n or n plus 1 or n minus 1 does not matter it is over the same set I am going to freely shift things okay then what does what happens to this what is D over DT of n of T equal to I multiply both sides by n and sum over n over all the integers then the first term gives me an n but I make it an n minus 1 and remember to add that one later on so it is going to be a summation n minus 1 alpha n minus 1 P n minus 1 plus summation now in this fellow here I multiply by n so I make it n plus 1 and subtract the n later on. So summation n plus 1 beta n plus 1 P n plus 1 minus summation n alpha n P n minus summation n beta n P n and then let us be careful I multiplied by n I wrote it as n minus 1 so I got to add so there is a plus I add 1 so this is summation alpha n minus 1 P n minus 1 and then I have to subtract this fellow here no no no I had an n I wrote it as n minus 1 plus 1 I had an n I wrote it as n plus 1 so I subtract 1 so minus summation beta n plus 1 P n plus 1 that is overall n but I can shift the summation index to n minus 1 and it is over the same range and this term cancels this and this term cancels this. So we have an interesting result which says I do not care what these coefficients are but if they are linear whatever they be I do not care there is an equation which says d over dt the average value of n of t is equal to this quantity and that quantity but what is this equal to again I translate the summation index to n but I am got the weighted average of alpha sub n whatever function of n that be so this is equal to alpha sub n so I have an equation actually pardon me a minus minus that equation could be non-linear in n in general it could be complicated etc so it is not trivial to solve it but you have an explicit equation for the average value in terms of possibly higher moments in general but you have an equation similarly if I multiply by n squared and compute what is n squared of t here again this thing is equal to n squared P n and the trick is exactly the same as before write it as n minus 1 squared and n plus 1 squared shift it and that will cancel against these fellows here but there was a 2n plus 1 which you added you must subtract those fellows out add or subtract those guys out but remember each time whenever these fellows n minus 1 you better match that here so in the linear terms you got to be careful once again subtract etc and you get a closed equation once again an equation in terms of average values so in this case you do not remember what it is but one can work this out on the right hand side you are going to get things like average value of n times alpha n plus beta n and then there is possibly an alpha n minus beta n etc plus something times there is maybe there is a 2 here because there was a 2 remember each time etc so these factors can be worked out but this is the sort of equation that you would get okay now tell me if the process is linear in n what is the special feature that emerges at once suppose the alphas and betas are at best linear in n each of them is some constant times n plus some other constant what would what happens then to these 2 equations sorry this is d over dt what happens then what is the special feature that emerges well this fellow is also linear in n so at best it is going to involve this guy so you get a closed equation for it right away which you can in principle solve okay and then what happens here again the worst you get is a quadratic in n the average value so this set of equations is closed completely and you can solve it even if you are not able to solve the full time dependent problem for p of n comma t you can still solve the equation explicitly for the average value as a function of t and the variance completely provided the process is linear provided the coefficients are linear so when I say the whole differential equation these are linear equations in the piece no doubt about that but when I say linear nonlinear etc I mean these coefficients if they are linear in n or nonlinear in n you have a big difference if they are linear then not only do you get a close set of equations for the a for the mean and the variance but you can also solve these equations explicitly but if it is nonlinear then it depends changes from problem to problem okay we will I will give some physical examples of this next time but you have already seen that the simplest of cases like the random work problem and everything that is mapped on to the random work problem these coefficients are constant and then when it is linear you have the radiation is an example of this chemical reactions would have similar kinds of rate equations etc a very interesting question is is the existence of a stationary distribution and the rule was and we saw already explicitly that in the random work problem there is no stationary distribution because j ran from minus infinity to infinity but when you have finite boundaries you physically expect there would be a stationary distribution unless these boundaries are absorbing boundaries because the system to disappear and we will give examples of that so let me stop here today and take it up tomorrow.