 So, let us continue with our discussion of the probabilistic aspects of coarse grain dynamics in a dynamical system. Just to recall to you what we are looking at we have a certain phase space gamma region of phase space possibly into which the system is stuck and then we partitioned that into cells of some given resolution we would look at a typical cell C and the complement of it is C tilde the rest of it. So we mark this cell and then we are trying to ask for the statistics of recurrences to this cell for this purpose we started with the following quantity we said there exists an invariant measure for each cell so mu of Cj for each j there exists such an invariant measure the total invariant measure is equal to 1 this is the a priori probability that if I just put a pencil dot on any given on the phase space I would be in the cell Cj occasionally I will use the notation p of Cj for the same guy the measure is a little more general because as I said there could be individual points in this cell which have a finite probability and so on so we are allowing for that then the quantity that we want to compute is the probability that you are going to come back to this cell after a certain time n and I call that r sub n for recurrence to this cell and this was nothing but the joint but the conditional probability that you are going to be back at cell C in time n and you are in the complement in the previous instance C tilde 1 given that you started in this cell at time 0 so that is the recurrence time distribution as a function of n and running 1 2 3 etc etc ok so this is the quantity we want to compute but because it is a conditional probability we can also write this as p of Cn C tilde n minus 1 all the way up to C tilde 1 and C0 and divide this whole thing by p of C because a product of this times this conditional density is equal to this probability here ok so we could always write it in this form the question is to compute this what we did was to say let us first define the probability W n tilde to show that it belongs to the complement which is the probability that you are in the complement at time n minus 1 all the way up to the complement at time 0 so let us just define this joint probability that you started in the complement at 0 and you remain in this complement of C till time n minus 1 so that is the definition of W n and it is immediately clear that W1 tilde equal to p of C tilde itself it is the invariant measure of the complement here because the whole thing is stationary so the time index for a single time probability does not exist we also defined for convenience W not tilde is 1 we define it in that fashion simply so that the formulas look very uniform our target is this but we could start by asking what is the probability of a sojourn in the complement for a given amount of time so let us define a sojourn probability h sub n tilde to be a sojourn in this in this in this complement so it is the probability of C tilde at time n C tilde at n minus 1 dot C tilde at time 1 given that you were in the complement at time 0 so all the normalizable probabilities will be those which are conditioned on something so I say I started in the complement at t equal to 0 and what is the probability that I stay here till time n that is the sojourn probability h sub n okay h because halting I halted in this complement here okay but this is equal to the joint probability with the C tilde 0 without the bar but a semicolon divided by this invariant measure p of C tilde but this is obvious that from here this is equal to W n plus 1 tilde over the probability of C tilde which is W 1 and I emphasize that there is no question of normalizing this h sub n they are all probabilities there is no question of adding different ends because they are not mutually exclusive events what we do know is that this set of numbers is actually a decreasing sequence as n increases the probability that you are going to stay out of C is going to decrease because it is ergodic sooner or later the point will move into C so as n tends to infinity I expect W n tilde to actually go to 0 okay but we can show this quite rigorously without anything else because I know that W n tilde is a non increasing it is a decreasing sequence bounded from below by 0 because it is a probability so it cannot be negative and there is a theorem in analysis called the Bolzano-Weierstras theorem which says that if you have a limit if you have a sequence a decreasing sequence which is bounded from below then the limit exists okay I mean there is no reason why there should be a limit at all for W n it could oscillate or something like that but the limit exists so we are guaranteed limit n tends to infinity W n tilde exists and this is the so called Bolzano it is a decreasing sequence but the limit need not exist at all but now we are guaranteed by this theorem that if it is a decreasing sequence bounded from below then the limit exists and it is bounded from below by 0 because it is a probability ergodicity then implies and this is a crucial point is actually 0 this probability has to go to 0 so that the probability of hitting C becomes non-zero going into C becomes non-zero so this is all the input we need that this is a decreasing sequence and its limit point is 0 on this side okay so this gives us this so zone probability and the next step is to ask what is the escape probability so the probability escape probability out of C tilde so let us put again again again A n tilde this is the probability that you are out of C tilde at some time so C n given that you were inside at all earlier times including C tilde 1 given that you were in C tilde at times 0 so you start just for reference let us keep this here so this is C and the rest of it is C tilde so you start by saying I start out here but at time n I am out in I am inside here okay I have escaped from this compliment having stayed there till time n minus 1 at time n I get in here okay so the question is what is this probability equal to okay well again it is a joint probability divided by the invariant measure of this guy so we can certainly write it in that form but we can do it in yet another way it is related and that is to consider H n tilde what is this equal to that is written down here so just for reference let me write it down just below that P of C tilde n and the rest of the arguments are all exactly the same now what happens if you sum these two probabilities these two conditional probabilities well you started given that you are in C tilde at 0 and then you are asking for the probability that you are in C tilde it will tell n minus 1 and what can happen in time n either it can go out and be in C or it must can stay in C tilde so the sum of these two must necessarily be the probability that you have had a sojourn till time n minus 1 so it is immediately obvious that E n tilde plus H n tilde must be the probability P of C tilde n minus 1 C tilde 1 given that you are in C tilde because it is a sum over possible final states next and that is the probability of whatever has happened till then till that point but this quantity here is equal to H n minus 1 so that gives us our first result which says that this escape probability E n tilde is equal to H n minus 1 tilde minus H n tilde so it is the first difference of the state probability this is like minus d over dt it is real in the continuous case okay and it is physically very clear what is all you have done is to conserve probability but we have got a formula for this and this therefore is equal to W H n was W n plus 1 so H n minus 1 is W n tilde minus W n minus n plus 1 divided by W 1 tilde so that gives us our first result it says if you compute these numbers W n and we define those W n's here if you compute this number then you have actually computed the escape probability in this case and now you can check that this fellow is normalized because you can either escape at time 1 or at time 2 or at time 3 whatever time so let us see what happens if you sum this E n well this is W 1 minus W 2 plus W 2 minus W 3 all tilde etc divided by W 1 tilde is equal to 1 because all the terms cancel out except the first term okay so this is indeed a normalized probability because that is the only thing that can happen we have assumed ergodicity so this immediately says that sooner or later things will escape but you allow for escape at any instant of time 1 onwards and you have to get one as the answer so this is normalized as it stands you could ask for the mean escape time that is n times E sub n and you have to sum over n and normalize it so you can that is a trivial thing to find out here what this n times this thing is going to be okay it may or may not be finite so let us find out what is the mean escape time mean time it is like asking what is the mean first passage time we already had experience with this we found that even on a straight line with diffusion mean first passage time was infinite even though the probability of hitting that point in this case exiting C tilde is 1 you may still have an infinite mean well let us see what that implies mean escape time which is equal to we have a formula here so it is equal to 1 over W 1 tilde times 1 times for this properly even as W 1 minus W 2 so it is plus twice W 2 minus twice W 3 plus thrice W 3 minus W 4 right which is equal to W 1 tilde's everywhere W which is equal to summation n equal to 1 to infinity because you can see that each time it is going to be one shot of complete cancellation so it is W n tilde 1 over W 1 it is equal to this guy what can we say about this we know the W n's all decrease it is a decreasing sequence and we know the limit n tends to infinity W n is 0 that is a necessary condition for this series to converge the n term must go to 0 otherwise the series will never converge but it has to converge go fast enough for it to converge it has to go faster than 1 over n to converge but there is no such guarantee no such guarantee that this series converges all we know is that W n tilde goes to 0 as n tends to infinity so the sequence decreases but if it goes like for instance 1 over n then this is going to diverge logarithmically as you know the harmonic series is infinite okay so we have no guarantee at all and therefore the mean escape time might still be infinite so I want you to carefully note this because it is going to lead to a very interesting result if you start in this place by the way the C tilde could have been any one cell you get similar results always if you start here you are guaranteed to escape into this at some time but the mean time to do so may be infinite may be infinite but it may be finite we do not care but that does not tell us anything more unless you actually compute what W n is and tell me what the large n behavior is I cannot say right hand right away whether it converges or not may or may not be infinite now let us look at recurrence and see what happens so the lesson here is that the escape out of C tilde is the first difference of the so zone in C tilde by this simple formula now let us ask what about the recurrence itself and there I define what the recurrence was we defined it by saying the recurrence probability for the cell C r sub n is P of C n given C tilde n minus 1 C tilde 1 C 0 notice now it is C so the statement is I start in I start inside here at t equal to 0 then I jump out and come back and the question is what is the probability of this happening in n time steps so I start with that probability here conditioned on this but I can also write this as P of C n C tilde n minus 1 dot dot dot C tilde 1 semicolon C 0 divided by P of C the invariant measure of the cell itself as always I write this conditional probability as a joint probability divided by whatever is the probability of the conditioning but this thing here this here is equal to 1 minus P of C tilde because the total probability measure is 1 for the full phase space right but that is equal to W0 tilde minus W1 because W0 tilde was defined to be 1 and W1 tilde was P of C tilde right so in this case the denominator is not W1 tilde but it is this and this is a decreasing sequence so this number is smaller than that it is positive that is guaranteed now what can we say about this P on top we need to write it out and the idea is to write it out in terms of these W's but W is referred to C tilde so we have to be a little careful and let us see how we can write this out explicitly so we want this joint probability C n C tilde n minus 1 dot dot C tilde 1 and C 0 this is equal to an integral over the phase space over the invariant measure d mu x you like write this as rho of x dx and x is the full over all the variables in the phase space multidimensional integral formally times times that set of points where at time 0 you are inside the cell and at time 1 2 3 up to n minus 1 you are outside and at time n you are back in there so the way to do this is to introduce theta functions which when the argument is positive is 1 and when the argument is negative is 0 and in general when you have a multidimensional object like that it is called a characteristic function very bad name but that is the way it is and it goes as follows so we define the so called indicator function or characteristic function but indicator function is better it is just a big name called sky subscript C of x and it is very simple it is equal to 1 if x is an element of C 0 if x is an element of C the complement just a theta function but it is a very useful device because it helps us to do these integrals with constraints of this kind then what is this fellow equal to this here you must integrate over all those points with this invariant measure which satisfy that so you have a chi C of x that takes care of this portion because this integral will automatically be 0 of x is outside so the integral is now running over the full phase space but this function sees to it that x must lie inside here and then the remaining points but remember that x evolves so what is happening is that x k is given by a time k is given by some operator t some time development operator t acting on x k-1 this is a time evolution operator we do not care what it is in a map it could be some very complicated non-linear function this is what I called f of x last time but it is formally some time operator and it is completely general this whole thing is completely general so even in a continuous time dynamical system you can take time slices and what happens to the any phase space point at time k depends on where it was a time k-1 and there is a map that carries you from k-1 to k let us call that map t and this is iterated each time the rule is not changing it is an autonomous system so this is equal to t squared x k-2 t to the n x 0 the initial starting point so it is actually tracing the orbit if you have a point here at some time and the next instant it is there next instant it is here next instant it is here and that is given by this map here okay so this time evolution maps the phase space onto itself but it takes different points and maps them in different places okay and its iterates give you how the time evolution occurs so we have ensured by this that the starting point is inside C but in the first step it should be outside C therefore the next factor is 1-kai C of t on x this is the indicator function of the complement of C so if x is inside if tx is inside C this is going to give you 0 because it is 1-1 gives you 0 if tx is outside C in C tilde this will be 0 and you have got a 1 so you got the full measure okay and that should be true all the way up to n-1 so this whole thing becomes a product k equal to 1 to n-1 of 1-kai C of t kx so just a compact way of writing something very complicated if you started writing it out in any arbitrary phase space but this takes care of the fact that so this is taken care of this is taken care of right up to here is taken care of okay and then you got to be back inside here so the next one has to be kai C of tn and that is the formal expression for this joint probability there is no conditioning there is no conditioning on the initial state on initial point except to say that it is we want the probability that it is in 0 it is in C not that given that it is in C we want the probability so we are integrating over it but notice that by this device of putting in this time operator here I have got rid of the need to introduce x1 x2 x3 these things as variables of integration and putting in delta functions all the time I have buried all that by saying this map is iterated several times to give you this so if the first map x1 is some nonlinear function of x0 x2 is a nonlinear function of the nonlinear function of x0 so it can be extremely complicated but we do not care notation wise this takes care of it completely the next step is to say alright I want to relate this to the W n's so what we have to do is to replace this by 1-kai C of x and replace this by 1-kai C of t to the nx if you did that then you added some terms and you have to subtract them out one term you have already got but you have to subtract out the remaining terms that you put in I leave the details to you it is just a couple of lines of algebra what happens is simple if you took this here and this and this this product then it says time 0 you are in the compliment time 1 you are in the compliment time n-1 you are in the compliment time n also you are in the compliment here so it is not hard to see that this guy here becomes equal to W n plus 1 tilde because from 0 to n you are in the compliment and that probability joint probability we call W n tilde and then the term which involves just the kai and kai here on both these sides that is also sitting there and what would that mean this when you when you multiply the 1 with the 1 two of these guys are gone and you are left with 1 to n-1 but the integration variable is x on the other hand the argument is t times x so you are in a little bit of trouble till you realize that this is the invariant measure so it immediately says in this is very very crucial invariant measure implies remember what we said about the Frobenius-Perron equation we said the density rho of x the invariant density does not change under one iteration at all that is the whole reason why we called it the invariant density so then formal terms this implies that d mu x equal to d mu of t so I change variables from x to y where y is this nonlinear function of x and the measure does not change the probability measure does not change at all so in the term which is 1 we change variables to tx okay and then it will look exactly like another w and the w it will look like is 2 less than this 0 to n which had n plus 1 of these points but now you got n-1 of these points so there is going to be a plus w n-1 tilde and then in between you have 2 terms where you have this 1 times this with a minus sign and minus this times 1 again with a minus sign and I use the same trick of changing variables to tx in each case and they will have the original thing had with the ones you had n-1 time arguments with the full brackets you had n plus 1 and with one of them firing and the other not firing you have n of them and the 2 are equal to each other by this invariance you have to check this out but I am just motivating it it is very obvious here in this form so you have a minus twice w n tilde but that is the second difference of the w's okay so we are back here we now can write down what this famous recurrence probability is or n therefore is w n plus 1 tilde minus twice let us write it the other way because this is a decreasing sequence w n minus 1 tilde minus twice w n tilde plus n plus 1 tilde divided by now w naught tilde minus w 1 tilde that was mu of C the invariant measure of the cell so we got a really interesting formula which says that the recurrence time distribution to the cell C is the second difference of so Joan probabilities in the complement of C the first difference gave you the escape the second difference gives you the return but to C okay we can write this back in terms of I divide and multiply by mu of C so this is equal to mu of C tilde divided by mu of C this guy here is mu of C multiplied by w n minus 1 and then a w n plus 1 what was this fellow here in terms of so Joan probabilities w n plus 1 was h n so this fellow is h n minus 2 minus twice h n minus 1 tilde plus h n so it is the second difference of the so Joan probabilities and that is normalized we have to check normalization we still have to do this so that is a closed form expression this guy here and the only assumption we made is ergodicity we just said this sequence w n is a decreasing sequence that is it with a limit point 0 and that immediately gives us this result for the actual distribution itself you can it is obviously the first difference of escape that is also clear because you can also show that this is like apart from something else it is E n minus 1 minus E n times some constants because we already showed that E n is h n minus 1 minus h n that is it so it is just E n minus 1 minus E n okay so that is an interesting physical way of looking at it the escape but out of the complement into the cell C occurs at time n with probability E n tilde and E n minus 1 minus E n gives you the recurrence probability the second that is a non-trivial result it is not at all obvious to start with but all we did was to use the invariance of the measure the dimensionality did not matter the kind of dynamics did not matter how complicated it is did not matter we did not assume it was chaotic we did not assume anything just said it is ergodic in some region of phase space so even if you have a dissipative system in which a system falls into what is called a strange attractor this will apply on the strange attractor on that level so all that you need is that the system is in some region of phase space and it stays there forever does not we do not care what kind of motion it performs inside there but it should be ergodic given enough time it should visit the neighborhood of all points in this part of phase space then this follows at once we need still to check normalization we still need to check normalization so let us do that then we need to find the mean time so the normalization would say that R n tilde summation n equal to 1 to infinity this is equal to 1 over the measure of C that is W naught minus W 1 tilde times R 1 is W naught let me just leave out the tilde minus 2 W 1 plus W 2 plus W 1 minus 2 W 2 plus W 3 plus W 2 minus 2 W 3 plus W 4 dot so it is clear that the W 2 is cancel the W 3 is cancel and so on and you are left with W naught minus W 1 which is in the denominator also is equal to 1 so this series telescopes everything cancels out except just the numerator and denominator and it is normalized to 1 so this shows that recurrence is a certain event sure event again ergodicity if we all parts of phase space are guaranteed to be visited given enough time by a representative point then recurrence to any cell is a guaranteed event probability is 1 so now we can ask what is the mean time to come back in the mean time mean recurrence time for recurrence to C for return to any C this is equal to 1 over W naught minus W 1 tilde times 1 times this guy so it is W naught minus 2 W 1 plus W 2 plus 2 times this guy so it is 2 W 1 minus 4 W 2 plus 2 W 3 plus 3 W 2 minus 6 W 3 plus 3 W 4 and so on so this cancels out 3 plus 1 is 4 2 plus 4 is 6 because this is next term as 4 W 4 and so on W 3 and so on so everything cancels out except W naught on top so this is equal to W naught tilde W 1 not tilde minus W 1 tilde but W naught tilde is 1 and this fellow is the invariant measure of the cell C so we have a very elegant result this is Poincare ergodic theorem which says that the mean recurrence time to a cell is the reciprocal of the invariant measure of that cell and that is exactly what we have got here okay and that is the recurrence theorem the ergodic recurrence there is a corresponding theorem for Markov chains and so on well we have already come across a trivial instance of it you know when you take a coin toss that is not a dynamical system because all the tosses are independent of each other but it is a Markov process because it is a Bernoulli process and then if you have n coin tosses this fellow at a geometric distribution for the first time you got a head or a tail or something like that so if you say the tail and the head are 2 cells in phase space and I ask what is the mean time it takes to get the head to get into the head for the first time it goes like 1 over P the probability of a head okay that is but the measure of this that the probability measure of the success is P and 1 over P was the mean time so that is actually an illustration of this theorem it is this theorem at work in the case of a Markov process very very simple Markov process but here even if it is completely correlated we do not care what kind of dynamics it is you are still guaranteed to get this okay so it is a remarkable result Poincare's got a theorem now it is got consequences it is got consequences for this problem problem of irreversibility in microscopic physics you may say well suppose you take an ideal case where you have classical dynamics operating and you have a statistical mechanical system say the gas in this room it is got a huge phase space right of the order of Avogadro's number in just the degrees of freedom and now I ask will it ever it is a chaotic system will it ever come back to any given cell in the initial if you start with some initial conditions will it come back to it or not what is the recurrence time going to be like what do you think will happen this is the famous there is many many models of this and the idea is the famous model is the checkerboard model you have a checkerboard on which you start with some fleas and the fleas are jumping from one square to another at random so there is a discrete time say every time step each flea jumps around around somewhere here so the question is when will this system return to its initial configuration it is an ergodic system so it certainly will return but the question is what is the mean time it takes to do so it is a finite system completely for a phase space is finite and so on so the question is now when is it going to happen will this happen at all as you can see if it is one flea and just one flea then all you are asking is how long does it take for it to come back to its initial point on the average one representative point but now you are asking when is the system going to come back to its initial state that means every flea must come back to the initial point that it started from it is getting less and less probable as you can see it is not hard to see in this case that what will happen now is that the mean time will go like the exponential of the number of degrees of freedom enormous because any given initial state has a very small measure compared to the rest of the states because a huge number of states possible and therefore this fellow will typically be of the order of e to the minus n if you have two particles in each of them can be in two states then the total number of accessible states is 4 if you have n particle it is 2 to the n so it is exponential in this and the measure of any one of those states is 1 over 2 to the n so if this is 1 over 2 to the n there is 2 to the n goes on top and the mean time becomes exponentially large in the number of degrees of freedom okay. So even in an ideal world with no other things happening the mean time of recurrence will go like the exponential of the number of degrees of freedom now even for the gas in this room that will be of the order of e to the power n Avogadro that is e to the 10 to the 23 it is so large a number that I do not bother to measure I do not bother to specify the time units right whether I call it picoseconds femtoseconds or ages of the universe it still does not matter the age of the universe only 10 to the 17 seconds and a picosecond is 10 to the femtosecond is 10 to the minus 15 so this is a factor of 32 orders of magnitude but we are here speaking of 10 to the 10 to the 23 okay. So it is such a large number that the system appears irreversible okay because many other things happen in between so this explains why in large systems you do not in daily life you do not see reversibility it is highly improbable and even if it is sure event it will not come exactly back to the same point come arbitrarily close to the initial state but it will take ages of the universe or even far more than the ages of the universe to come back so this kind of argument actually explains the reason for apparent irreversibility in observations okay it is simply that when you have so many possibilities very specific thing that you want is not going to happen and irreversibility happens it can be said in many many ways for instance when you park a car it is much easier to get out than to get in as you know and you have all seen this so assuming that people park normally in any normal country with human beings there are exceptions you have a spot like this you have cars everywhere you come along and then what you do is to go forward and reverse into this in this fashion you pull out in the same way but it is much easier to pull out than to pull in much easier okay the reason is you may say why cannot I just reverse my motion whatever I did to get out I just reverse it in exactly the same way the fact is that when you are getting out when you get out and you are either like this or you are like this etc it does not matter there is enough phase space many accessible many possible microstates that you can get into but when you are in here there is only one microstate you have to be like this if you are in any other direction you are going to hit somebody okay so even though you have reversibility theoretically you can just reverse the motion exactly the aim has to be perfect while going in but it does not have to be perfect while going out so that is the reason for the ease in coming out but the difficulty in going in and that is an explanation of the fact that you have irreversibility in this problem even though the dynamics is completely reversible it is again has to do with how many microstates are accessible and that is one manifestation of it is this kind of here now you might have one objection to this thing which is to say look all this is nice in discrete time but when I go to continuous time I am going to be in trouble because if I put a time step tau then this is tau over mu C and now if I let the time step go to 0 this says the mean time of recurrence is 0 that is not true in continuous time dynamics I still have recurrences but this gives you a wrong answer because it says you cannot go to 0 sampling time you did that you are in trouble the reason this this flaw was already noticed by small kofsky way back in 1916 and he proposed a modification to punk array which goes like this the physical idea is the following we included in recurrence R1 okay this corresponded to P of C1 C0 given this is not a recurrence you just stayed on so there was no question of you going out and coming back you did not do that you just stayed on and it is a fake recurrence we counted this as a recurrence and we needed it because I showed you that summation Rn is 1 provided you sum from 1 to infinity if you sum from 0 or 2 upwards then you do not get 1 right so I will quickly put it that and summed it up this thing here but this is not a recurrence should not be included as a recurrence so the way to do this is to say alright I have a tau up there but in the denominator I must exclude this so I must call Rn equal to the probability that I start at C0 I get out at C tilde 1 and I stay out and get back at Cn plus 1 C tilde now this is a true recurrence because you told you start here you get out at 1 and then you ask I come back at time n plus 1 so the time difference is n again and that is R sub n okay if you do that then you got to divide by this probability to find the joint probability you go through similar manipulations but what is this fellow going to be this guy here says that you start inside the cell you are outside the cell at time 1 so you put the characteristic function etc and work it out and you discover this is W1 minus W2 and not W0 minus W1 in the denominator and now you will sum over 1 to infinity and everything is going to be fine in this case so this modification because what happens is you have 1 over 2 in this case in the limit in which tau goes to 0 this fellow will also vanish like tau and so will the numerator and there is a finite limit okay and therefore you can go to a continuous so this modification was proposed by Smolukovsky as I said way back in 1916 and it is been looked at after that in several cases along with some collaborators I have a paper on this which actually tells you how to handle this thing in the continuous time limit and to extend it to stochastic systems we have not talked about them and talked about it here at all and then in the case of chaotic systems which have got phenomena like intermittency and so on then very strange things start happening there could be regions of a space where you are stuck for a very long time and then you jump out etc etc but now I want to pose the following paradox to you we saw that it was possible that the mean escape time out of C tilde so here is C and I said the mean time of escape out of C tilde could be infinite but now I am saying the time mean time to start here and to come back here is finite that sounds weird because to start here and come back here I go out and then I escape from out back in here okay so how come the mean recurrence time is guaranteed to be finite for ergodic systems by the Ponca ergodic theorem but the mean escape time to go from here to there could be infinite depends on the convergence of this sequence W and what do you think is a resolution to this paradox how do you think that is happening you get the point right because it looks like recurrence involves an escape out of C into C tilde and a return escape out of C tilde into C and this mean time could be infinite so could that be infinite but we are saying this round trip is finite mean time is finite how is that happening in mathematics is quite clear so there is no mistake in that the resolution is that when you say average time you are averaging over certain realizations of events okay the recurrence time probability is normalized so you are guaranteed that things come back and therefore you are averaging over all those realizations of this random process for which the return is guaranteed on the other hand in the escape case you are sampling over a different set of realizations you are sampling over all realizations when I escape from here to there is guaranteed that is it nothing more this is the bigger set and therefore the average over it could well be infinite so it is a question of what you are averaging over once I normalize the probability in that fashion then I have said that I am averaging over all those realizations where the return is guaranteed and then you have a finite answer for the return distribution but this is over a different set of events just over the escape events and it is possible there are some for which the W n is so slow decreases so slowly that it does not work but another crude way of saying it here is another crude way of saying it if W n is proportional to 1 over n for instance then sigma over W n is going to be fine infinite right okay suppose it goes like 1 over n to n squared what happens to sigma W n that is finite right if it goes like this it is certainly no to diverge in this case but remember that when finding the recurrence time you are taking the second difference you are differentiating this guy twice more so it is going like 1 over n cubed and then things will be fine etc. So each time you take a difference you have a decreasing sequence the first difference will go to 0 faster than the sequence itself and the second difference will go to 0 even faster by a power of 1 each time so it is entirely likely that you may have a finite recurrence time but an infinite escape time this is entirely possible within this framework here okay. So there is a lot more that one can say here about this but I thought that give you a little flavor of what this looks like one could ask that takes us into a different subject itself one could ask can we illustrate this in general theorems using a simple model a map or something like that the answer is yes but it again involves partitioning phase space so here is a simple model which will do the trick one dimensional phase space so we have X n equal to some function of X n minus 1 some nonlinear function and it is a one dimensional map of the unit interval for example 0 to 1 0 to 1 and the map is simplicity itself but it is completely chaotic and the rule is the following start with any X naught between 0 and 1 and find X 1 equal to twice X naught modulo 1 so this fellow says just double the number and if it is less than 1 keep it if it is greater than 1 throw away the integer that is it that is the rule it is called the Bernoulli shift this map here and what does its graph look like what does the graph of this guy look like f of X equal to 2 X modulo 1 says 2 X would be a graph like this slope 2 when X is 1 it is equal to 2 but you want modulo 1 so you cut that piece and put it back here so it would look like this and this is at a half that is the map and this is the rule therefore if you start with X naught X 1 is 2 X naught mod 1 X 2 is 2 X 1 mod 1 which is equal to 4 times X naught mod 1 and what does the iterated map look like it is got slope 4 and mod 1 so it goes like this and now the slope is 4 etc no fixed points are stable in this case and all fixed points are unstable all periodic orbits are unstable etc and a typical point if you start with an X naught will as time goes along under this fill the entire unit interval densely and uniformly you can show that the invariant measure this constant in this case a typical point but if you start with a rational point it will be part of a periodic orbit any rational point will be part of for instance start with one-third next time it becomes two-thirds next time it becomes four-thirds which is the same as one-third it is back to one-third two-thirds so it is a period two you start with one-fifth goes to two-fifths then it goes to four-fifths and then it goes to eight-fifths which is the same as three-fifths and then it goes to six-fifths which is the same as one-fifth and you are back to a four period four cycle and so on so any rational point will be a period two cycle of some finite period and those points will repeat themselves etc but the rationals form a set of measures zero among the in the unit interval and all irrational points are guaranteed to be part of non-periodic orbits they are part of this chaotic orbit completely okay so you can find the invariant density you can now partition the cell the unit interval into say 10 parts 15 parts call each one of them a cell and keep track of only where the point is in which cell it is each time and you can play this game you can find W n in this problem you can actually compute what it is so I suggest you do the following is the simplest case just do two cells so C 1 equal to 0 to half C 2 equal to half one is the same as 0 by this model so call one of them C call one of them C tilde and find out what happens to W n etc you can do this directly without too much difficulty it is called the Bernoulli shift because you know what is happening is that if I start with an X naught and it is a number between 0 and 1 so I can write it as 0 point a naught a 1 a 2 a 3 etc you can write it irrational number will have an infinite decimal point expansion which is not recurring and I will do this in binary so if you do this in binary each ai is either 0 or 1 that is it and what does this a naught stand for this number actually stands for a naught divided by 2 to the 1 plus a 1 divided by 2 to the 2 plus dot dot dot that is what this stands for in binary so each ai equal to 0 or 1 if this is X naught X 1 doubles it and doubling it means multiplying by 2 so this becomes a naught divided by 2 to the 0 that means it comes out of the decimal point here so at the next step implies X 1 equal to a naught point a 1 a 2 a 3 dot dot dot dot but if a naught is 0 then the number is less than half to start with and when I doubled it it was less than 1 but if a naught is 1 it means it exceeded 1 but I got to throw away the 1 so this is again 0 that is what makes the map nonlinear otherwise they have been just a linear map but now I brought it cut it down it is nonlinear this means that if you give me an X 1 there were 2 X naughts from which you could have emerged this got this X 1 if you give me an X 2 there were 4 X naughts from which you could have got the initial point so in the past they could have been 2 to the n points which are feeding into 1 point or another way of saying it is in the forward direction a difference of epsilon will become twice epsilon next time 2 squared epsilon the next time etc till it becomes 2 to the n epsilon where n increases without bound and since your module of 1 it means that 2 points which are infinitesimally apart initially could become unit interval apart after given enough time that is what chaos is okay so this is called the Bernoulli shift because you lose information about this a naught the next time you lose information about a 1 etc and that is what this multiple valued mapping is saying telling you that you cannot retrace the back past uniquely okay and it is called the shift because the decimal point is just shifted each time one place to the right and whatever is on the left is thrown out okay the map is fully chaotic it displays its as random as a coin toss because it has a lot of fantastic properties here but you can actually tell what is going to happen here and now interestingly what will happen you will discover is that if you break it up into these 2 cells and keep track of only the cell it is in call it left and right for example less than half it is left right greater than half it is right so you have a string sequence L R L R L R etc just a sequence of letters here then it will turn out that the mapping to go from at any given time to go from one C 1 C 2 to the next C 1 C 2 becomes a Markov process the two state Markov process so you can actually use the entire power of Markov processes to now talk about recurrences here once it does not always happen this is called a Markov partition but it is not necessarily true that will always happen it will happen if you break it up into 2 to the n pieces in this particular map. So, let me stop here and the questions we will take them up.