 So, we agreed that today will not be additional lectures, but it will be an exercise session. So, we start by asking you if you want some particular thing explained understood, you know you can ask a question how do you do this and then I may be able to answer I may not be able to answer, but we can try yes oh I thought I will be swamped with all kinds of questions can you explain the very good here is the following in the BTW model or more generally in these kinds of sand pile models you add grains and there are some avalanches and eventually the grains leave the system and you keep on adding and they keep on leaving. So, we ask for what is the probability of there are s toplings some event of size s and what I would like to argue is that the expectation value of number of toplings has to diverge as the size of the system goes to infinity if this is true then it immediately so if then you define this is probability of s for a given size then probability of s l usually is a function like this initially this is log probability divided by log s initially it kind of decays usually it is like some power, but there is a cut off ok, but if you let l go to infinity then the power log goes on for much longer. So, we define probability of s l limit as l goes to infinity equal to probability of s if this limit exists usually it does you know like even if you take the last size limit the probability that you will get no toppling when you add something is finite probability that you will get exactly one toppling is finite. So, this probe s exists even for even when l is very large and it is finite for each finite s then the question is is it a power log. So, the claim is that the first movement of this has to be infinite because what is the proof the proof is that suppose you take very large number suppose you take l is l is a finite number 100 by 100 or some such thing and I keep on adding grains and they keep on leaving ok. So, I add one trillion grains one after another and in each case I measure the amount of toppling done then I can label the grains as eventually each grain will be added at some time and it will leave at some time. So, I ask how many topplings this grain undergo the number of topplings one particular grain undergoes is going to increase with l as at least as l or typically as l squared ok. So, if the number of topplings incurred by each grain is going to diverge then the average number of topplings per grain will also be the same it will also diverge ok. The proof is precisely this number of topplings per grain average number is equal to number of topplings per actually there is a factor here by 4 because each toppling causes in an avalanche when I count one toppling 4 grains count it I got toppled I got toppled I got toppled like that ok. So, if I take the sample of all the trillion grains I added I will get so many topplings were done ok then I say per grain how many topplings were done each toppling in an avalanche will count as each of the grains will say I got toppled I got toppled on that one right. So, the number of topplings per avalanche per grain added average number of topplings found by one grain that found number of topplings under go by a marked grain ok. So, this number is clearly infinite and hence this number is infinite once this number is infinite then the probability of the this probability s the first moment must diverge probe s s summation over s is equal to infinite. So, it is a power level ok clearly I have not used any properties of the particular toppling rule of the sand pipe. So, it is a very robust argument the only point which we have used is that if you add a grain it keeps on moving, but it can only leave at the boundary it cannot disappear in the middle if the grain disappears in the middle then this argument does not work because argument said that it has to move order L in order to disappear right that argument will not work if there is dissipation in the bulk not only dissipation at the boundary ok. In one avalanche? No in the till it leaves from the time it is added till the time it leaves no it is not leaving in one avalanche. So, it is a flip of the you are counting number of topplings, but in two different ways one is per grain and one is per avalanche, but you should get the same answer sorry. Perfect size. Yeah that went as L squared or you know the bound that this number is greater than order L is a much more robust bound it does not even use the fact that the parts of grains are random walks ok. So, you just add it anywhere it has to take order L steps in order to leave ok ok. So, we say so I had this pile and I keep on adding grain grain number one grain number two grain number three all grain grains are marked and all of them leave eventually. So, I add how many grains how many topplings were encountered by grain one grain two grain three. Let us call them E 1 plus E 2 plus E 3 plus E trillion whatever E total number of grains added, but this will be compared with S 1 plus S 2 plus S 3 which is the number of topplings on adding the first grain number of topplings on adding the second grain and so on. But these two numbers have to be equal to a multiplicative constant sorry I beg your pardon I think there is a mistake here it is multiplied by 4 not divided by 4 even E 2 is the number of topplings encountered by grain one till it leaves E 2 is the number of topplings encountered by grain two till it leaves and so on. And this is number of topplings on adding the first grain number of topplings on adding the second grain like that. So, I am just counting all the number of topplings in two different ways I am rearranging the submission and so that is what it is ok it is the proof ok yes ok. So, so there is some I guess for me right now I will write the configuration will be given by the heights know. So, the heights will be 3 2 1 1 2 whatever some heights and the heights are written in the centers of squares ok those are my cells usually we will discuss this in the you know there is a square and the vertices are where the height should be. So, I should write 3 here and 2 here and 1 here and not in the centers. I think this issue is important, but we will stick to this one the I like this easier to draw know because I write 3 in the middle of the square ok. So, then what we have agreed which we will not show now is that there is a burning test which says that if you try to imagine that outside is all burned and then you burn any site whose height is equal to or bigger than the number of unburned neighbors and if everything burns then you are in the recurrent state ok. So, I start by this that I this stuff is all burnt outside or the sink site outside is considered as burnt this is the sink site. Now, what does what can be burnt in this one let us take 3 too many burns, but ok now I burn this this site can be burnt because it has only 2 unburned neighbors and its height is 3. So, I would be typically burning this first and what else can I burn right now nothing in this graph nothing else can be burnt, but now as soon as I come here then this site has height 2, but it has 2 unburned neighbors. So, in the next stage I can burn this once I have burned this then I can burn this and then I can burn this right no no sorry I beg your pardon not this one 2 ok. So, I am stuck. So, this was not a recurrent configuration what is left behind is a forbidden configuration this one because it has 3 unburned neighbors. So, right now I cannot burn it ah ah sorry I beg your pardon you are right of course. So, I should burn this one once I burn this one then this will also burn and now this one below and the one above and then this one as well and then this one as well and then this one as well and then it seems like it yeah this one yeah ok this one this one in the second row in the left one again this one also all right no no very good. So, there depending on the configuration they will burn, but you see there is a sequence in which they burn. So, now I will draw the path the fire took in order to get there yes 0 to 3 ok very good. So, there was there has been sort of flip flop in the notation and so one convention is that the heights go from 1 to 4 and the other convention is heights go from 0 to 3 in this case the toppling rule is that if height is greater than 4 then topple and here the rule is that if height is greater than equal to 4 then topple ok. So, we agreed that these two are actually equal there is no big deal you just have to redefine h prime equal to h minus 1 and then both the rules are actually the same ok, but the burning rule can also be phrased either in terms of h prime or in terms of h ok. And in terms of h prime there is one burning rule in terms of h there is a slightly different burning rule one of them says that burn only if everything if the height is bigger than the number of unburned neighbors the other says if it is bigger than or equal to unburned neighbors ok. So, that is comment it is not just remember that different when suppose somebody defines the model like this then he will use one version if he defines the model like this he will use the other version ok. So, now in our case we found that the fire went like this and then it went here and then it went here and then it went here and yeah then it went here, this fire went from here to here and from outside it went here and it went here and here and then it went here and it went here and so on ok. So, once site is burnt then it can burn and the fire can propagate from there. So, if everything burns then I just draw this path of fire burning and it will burn everything. Let us not write these heights and it will burn everything in some way. So, the burning path defines a spanning tree. Is this, yes and from already burned squares. Yes, but only from one of them. Yeah. Yeah. Yeah. So, the bottom one is a top one. Yeah. It is not possible. Very good. No, no, no. So, the basic philosophy is that the burning path corresponds to a spanning tree. This much is true. Now, is there a one to one correspondence between burning paths and spanning trees? That is the question you have to ask. Each burning path will be a spanning tree because I only burned once. Each site is burned only once. So, it is a tree. All sites are burned. So, it is a spanning tree. Now, we have to ask is there a one to one correspondence between these? That is a little bit trickier and so, what we have to do is to realize that if this site burns, then the height could be 2 or 3 and it will still burn. So, maybe there is no one to one correspondence between the burning paths and trees, but no there is actually because you can burn either from here or from here. So, any site which burns, I can say that the fire came from one of its already burned neighbors, which one? We will have to choose. I have some choice, but the choice is exactly equal to the height allowed at that site. So, I make up some rules saying if the height is maximum, then burn from this side. If the height is less than that, burn from this side and so on. Then there is a unique height association with the burning path. Once I have chosen, decided that if the height is 3, then I will burn from here. If the height is 2, I will burn from here. Now, when I give you a burning path, you will immediately know what is the height configuration because of my extra rules, I had it. So, in the burning test, we did not actually tell you have to burn from this side, burn from that side. We just said burn, but if you want to make a one-to-one correspondence, you add this extra bit and you say burn from this side, burn from this side and then there is a one-to-one correspondence between the burning path and the height configuration. Is that clear? Anything else? Yes? No, because this path, you know the fire comes like this, fire comes like this. At this point, one site gets burned, but I will decide whether to say that it burned from top or from bottom, but from not from both. So, then there is no possibility of a loop. Yes, sir? Because given a configuration which I know burns, I just draw the path of the fire. There is a construction. See the point is there are two sets. They are equal in number. You can make any one-to-one correspondence between them in principle. No, actually we first set up the correspondence, then count the number of spanning trees and then say that O, but this must be equal to the other number. But in the end, I am saying the correspondence between the spanning trees and the height configurations. Take some stupid example. This is a 2 by 2 graph. There is a spanning tree like this, there is a spanning tree like this, there is a spanning tree like this, there is a spanning tree like this. I cannot make any other trees on a square. Now, the height configurations are whatever they are, 2, 1, 2 and so on. Now, what we have to do? What corresponds to which one? You can make any correspondence. I can say this height configuration corresponds to this one, this height configuration corresponds to this one, this one, whatever. Given two equal sets, you can make any one-to-one correspondence. We are making the one which makes our life sort of easy and makes, you know, because there are big sizes. You make an algorithm to make an assignment. It is a free choice and we chose the burning test to use the assignment. Yes, please. For example, 7 minus 4 minus relations between operators. For example, we have a 1 to the power of 7. We wrote 8 to the power delta i equal to product over j, 8 to the power minus delta ij. So, j is not equal to i. What is the proof of this relation? This is the question, right? No, no, no. I refreshed your question, but you asked first the original question. Delta is some phonemetrics. Yeah, precisely. 7 equals 8 to the power of 4. I think you see in the columns and write the algorithm. No, I did not write any columns. I did not write any algorithm. So, let us go through this proof. It is important to be a sure of the argument. So, this relation is true for all configurations with positive heights. So, suppose there is some configuration or the other. I do not know what it is. So, this is called c and I would like to know what is a i to the power delta i acting on c. c is any configuration. The statement is that of course, you can add one and then topple and then add another and then topple and do all kinds of stuff. But the result will be the same as first I add all of them together and then topple. In that one, actually c can have some sides which are unstable also. It does not matter. Now, I will say that oh, but I am allowed to topple any order I like. So, first I will topple at i. So, the statement is only requires that at side i, now that I have added delta ii, it must topple. Once it topples, this is the same as acting minus delta ij particles at j acting on c, product over j. End of proof. No, no, no. It should be clear that the argument is only one line argument and it impose it uses the abelian property. It uses the property that once I add delta ii particles at the side, it must topple. That is all it uses nothing else. How do you what? So, forbidden configurations, but which? Yeah. Yeah. Ok, very good. So, the statement, no, no, it is very, it is all right. Suppose I am in the configuration, suppose my c belongs to R, suppose assume, suppose there is a stronger relation. The stronger relation is that product. So, let us write it like this 2, 1, 1, 1, 2, 1, 1, 1, 0 acting on c is equal to c. It says that take whatever configuration you have, that is the recurrent configuration add to it this stuff 2 here, 1 here, 1 here, and topple and see what you get. Then you must get back the recurrent configuration. If I do not get back the recurrent configuration, then what I do is that suppose I get this acting on c is equal to c prime which is not recurrent. Then in c prime they must be, it is then I look at c and I try to burn it and you will stop, the fire will not reach everywhere. Then whatever remains is a forbidden configuration. The forbidden configurations were sort of found first and the burning test was invented later based on generalizing the forbidden configuration, but now you can use the burning test to define whatever the fire cannot burn is a forbidden configuration. Yes, please. Is that clear? Yes, please. There were four spanning trees that is equal to the number of recurrent configurations. Ah, ok, very good. So, I think one should distinguish between these two things. There is a spanning tree like this and there is a sink site which is a, let us draw the sink site like one site, but this site is connected to this one with two bonds and connected to this one with two bonds and connected to this one with two bonds and connected to this one with two bonds. Now, I have to draw spanning trees on this bigger graph. That number is 192. You have to draw the edges in the suitable way and yeah, everything can be modified. What is the, there is some Latin word for this which I have forgotten, but anyway, ok. So, the proof to, so if there is a forbidden configuration, so you imagine that there is fire outside everywhere and you try to burn it and you would not burn it. You would not be able to reduce any further. So, then it is a forbidden configuration. Yes, please. Yes, yes, yes. You mean you add some grains. Yeah, yeah, yeah. Is there a configuration? Yeah, see eventually, yeah, yeah. It is a recurrent configuration. Recurrent means comes back again and again. Yeah, yes, yes. Ok. I have explained it three times now. Let me see. Can we discuss it later? I am saying the proof of the forbidden configurations was that if you don't start with two 0's like this, you will never get there. You cannot create this configuration by toppling process. That was the argument. Either it is there in the initial configuration or it won't be there because if you try to add something, you will never get 0 0, ok. So, once you are in the recurrent, in the initial configuration, you may be starting with everything 0. That is allowed. But as you keep on adding grains, the number of adjacent 0's keeps on decreasing and this number can never increase. And so eventually, they get lost and they are never found again. So, in the recurrent states, they don't occur. If you start from recurrent state, then this is never, it's not there and it cannot be generated, ok. Yes, please. So, we wrote E alpha is equal to product over j A j to the power B alpha j, right. This is what you want to understand, ok. So, firstly, this is an operator. This much is clear. Now, I want to show that E alpha raised to the power D alpha equal to what? Show is identity operator on the set of recurrent configurations, ok. So, what is E alpha to the power D alpha? E alpha is equal to product over j A j to the power D alpha E alpha j, right. This can be written as equal to product over j alpha j A j. D alpha is a big D. We wrote delta is equal to A D B. So, D was a diagonal matrix. So, D B is a diagonal matrix and D B is the same as D alpha B alpha beta, ok. So, now, but D B is equal to A inverse delta from the previous equation. So, this is equal to product over j A j A inverse delta alpha j, ok. But this can be written as product over j A j to the power delta alpha j A in A j to the power delta beta j A inverse alpha beta, ok. I wrote on the right, but it is the same because I have written indices. So, now, I do not have to worry about matrix, but this stuff is identity and so, this becomes the whole thing becomes identity, ok, end of proof. Yeah, precisely. See, if I have a group element, what power do I raise it to to get identity? If you raise it to enough power sometime or the other you will get identity, but these have been constructed. So, that this one raised to the power D alpha will give me identity, but I have different choices of D alpha and I get different generators and then this generator of the different group, ok. Yes, please. Yeah, of course, the set of recurrent configurations where the group works, if they you know we are determining the structure of the recurrent group. So, it is only in the recurrent group, no problem, no other questions, very good. So, I made up some questions because I thought maybe you will like to have a sample of the kinds of questions you might be asked or you can test your understanding on what you have already understood so far by seeing if you can answer these questions. So, for each question I will ask you, do you think you understand how to answer this? Means, maybe you cannot figure out the answer in two, three seconds, but you know how to go around it, given ten minutes you will be able to work it out or not. If you can answer it then you say yes, otherwise you say no, then I will sort of have a feeling as to how much of my effort has been successful in communicating some ideas, ok. So, here is a problem, the problem will be written down here, ok. So, it says define, some of the stuff will be written on said only in words. So, I define a sand pile with the following rules, on toppling Z j goes to Z j plus Z j 1 plus Z j 2 plus Z j 3 plus Z j 4 by 5. I did not write it very well, let me say it like this. There are these, there is a site if it says if Z is bigger than 3, then you just take the average of all these five numbers and give it to each one, each one gets the average of the five heights. If the height is to be then you just spread it out equally amongst all the neighbors, the heights are real numbers not integers. Is the definition of the model clear? Yeah, yeah, yeah. In the toppling process the value of the neighbors is also modified, ok. Sorry, it is precisely the same, all of them get the same value. You keep on adding until some site becomes bigger than 3 then you topple and when something goes to the boundary then things will leave, ok. Is the model clear? It is one of like one of those sand pile models. I did not say it, but yeah, you work with positive numbers. Sorry, average and? Yeah, yeah, yeah. So, I should say Z j, I go to this, all the five neighbors go to that value. Now, yes sir. No, they are not integers, they are real numbers I said, ok. So, it is clear, the definition of the model is clear. They are real numbers, they are real numbers. It is not, yes sir. The neighbors are zeros? Yeah, it does not matter. You just average and give everybody the average. Yeah, yeah, yeah. So, it says on the borders this guy will get some number and it will leave. Initially the height was zero then it gets something, but it is reset to zero. Heights outside are always reset to zero. The definition of the model is clear, ok. The question is only this. Is this model a billion or not? They are not getting one. They were, neighbors are not getting one, but they might get something, you know. It is, ok. No, no, no. So, let us not jump to conclusions. How many people think they answer, they know the answer. Just raise your hand. Yeah, the question is, is this model as defined a billion or not? Do you know the answer? I do not know whether it is yes or no, but do you, can you figure out the answer or you have no clue about what is the answer? That is the question. If you think you know the answer, just raise your hand. A little bit higher so that I can see. I see a substantial class of, substantial set of people have not raised their hand, ok. So, then there is a problem. See, I define the model clearly. Then I am asking, is this model a billion or not, ok. So, we have to check, you know, if the, if you add at i, add at j, this is the same result as adding at j and at, at i, you know, that was the test or that was the characteristic. So, can I just check it, you know, can I add this, start with some configuration, add at i, add at j, see what happens. No, so the answer is, they don't come here. No, no, so you are able to see or you are able to check that this model is not a billion, that is the answer. We keep on adding, no, you know, in that model, in the sand pile, there is no depth of house and grains. You keep on adding one after another and you reach the steady state, you keep on adding and so on and so forth. Yes, but the node is unstable if I add four grains. No, no, so in the beginning, there are some configurations where you add nothing happens. You add here, nothing happens, you add here, nothing happens, so of course it is the same. But the a billion property requires, therefore, all possible configurations, the same thing happens. If I have, for example, this, four, zero, what happens? So, this one will topple now, by, oh, so we said, yes, so we will imagine that there is a zero here and zero here and zero here and then divide this four equally and then erase all the boundary sites. So, I said that it's a real number, we said it tries, you are not paying attention. Okay, no, it's okay, the heights are real numbers. So, it's four by five, the height is, later it becomes four by five. This is four by five, this is also four. Yeah, I asked if it was a fraction one. Yeah, yeah. So, all I have to do is I have to find one configuration where you first add here and then add here is different than when you add here and then add there. That's all. Okay, so I will take a graph like this, put this, we said if z is bigger than three, so this is 2.5 and this is two. And let us say all the neighbors are one-one. Please, can you, please, let us take this configuration. I'm just made it up right now. Consider a big lattice and I construct this configuration where these heights are 2.5 and 2 and the neighbors are some 1.1, whatever you like. Now, first I add here an average over everything and then I add here an average over everything and then I do it in the reverse order. You will not get the same value. Okay, and so then the model is not a million, end of proof. So, we say, yes, we said 1.1, 1.2, 2.1, 1. So, we decide to add one right here. Yes. So, we go there and we take the sum over m divided by 5, each one of these distributions. We get this is stable. So, we want to prove that it's not a billion, we then add. Yeah, you added i and g. Yes. Construct a c, i, j such that a, i, a, j, c is not equal to a, j, a, i, c. Once you have constructed one such counter example, then the result is not a billion, end of proof. No, because we, in the proof of a billion, no, it is because you can start with this configuration and a, i, a, j, c is not equal to a, j, a, i, c. This we can construct. And then, but for a billion property, it has to be true for all configurations. Okay. So, it's not true. Okay. So, you can check. Again, you can construct this. Yeah. Of course, it is different. The point was only to see that given a model, you can check does it have a billion property or doesn't have a billion property, full stuff. It's more or less the definition of what is a billion property. But we are applying it to a context which was not recently discussed in class. This will work. Yeah, yeah. Sorry, can you repeat? Hmm. No, no, no. I thought it is true for all configurations. That's what we proved. Okay. Yes, sir. We can find the inverse. No, no, no, no. It's too far. Inverse, we don't, the a billion property does not necessarily require an inverse to exist. Because sometimes we work in this extended space of configurations where heights can be negative and all kinds of stuff. Let us not go into the group. Of course, if all operators are commuting, then they will form a commutative group. If they don't commute, maybe they form a non-commutative group. What do I know? Yeah. Yeah. Yeah. Yeah. Yeah. All I can do for sure, given a configuration C, which is by definition set of z i's. I know what is a i c is equal to z i prime. This I can construct. So the operators are well constructed. You can check if they are commuting or not. So it's not clear that there is such a relation. It's not clear that there is such a relation. But this happens because we have a real number of brains that we can have on each side, right? Because if I am in chairs, I can always write a relation. No. Because the a billion property may not still hold. It may happen that, see we said that a i to the power 4 is equal to a i 1, a i 2. That property also need not hold in general, okay? I think these are all special cases. Okay. Is this end of this discussion for this question? Yes, please. Yes. Necessary condition is just a i a j equal to a j a i. That is necessary and sufficient. No, no, no. So I understand a i a j equal to 0. Now you can deduce this from some other condition. Yeah. If that other condition uses the toppling rules in some way, then okay, I mean you have shown that that particular toppling rule is commutative. Do you want to prove that? Sorry, just let me answer this one. So let all possible, you know, you don't even tell me the toppling rule and you want to deduce this condition. With toppling rule. With toppling rule you can just check. Oh, I don't know. Yeah, no, I understand. No, no, no. These questions, they can be set in some more general setting and you can think about them. They are supposed to a check for comprehension and allow you to extrapolate further. We which we are not doing. Yes, please. There was another question here. Yes. Yes. Okay. No, they all become equal. That was the rule. All the heights become 11 by 5. So people didn't even understand. See, I asked three times, have you understood the definition of the model and people didn't understand. Okay. You serve. Is this argument clear to you so far? Okay. Okay. Yes, sir. Depends upon your neighbor. Yeah. No, it's not a proof. Depends on how it depends on the neighbor, no? Maybe, maybe not. I don't know. I'm saying that without giving a toppling rule you cannot give a proof that everything is a billion. There is a counter example. For some set of toppling rules, yes, they are a billion. Other set of toppling rules, they are not a billion. Yes, sir. Sorry? When a system is finite, and it seems to write the relation between operations like toppling in one side or in the other. Yeah. We can say that once I found the identity operation of the current state, then the system is a billion for sure, right? Because I am moving in a torus in one dimension or in another. Yeah. I don't know if there is some identity operation which is a non-trivial identity operation. You know, if you don't touch anything, then of course it remains unchanged. We are doing something, we say add all this stuff and then you relax again and you will still get back the same stuff. I don't know for an arbitrary toppling rule this result is true. Okay? Yeah. Yes, that is what I want to do, but I am sort of not being allowed. Yeah. Anything else? Okay. So, can we go on to the next question? Yes. This is not a proof. This is a reference to authority. No, no, no. I am asking for a proof, not saying in some textbook or the other, this is written like that. Can you give the proof yourself? Yeah. Yeah, yeah, yeah. The statement there is correct. Looking for a counter example, I agree. Ah, so, okay. Everybody satisfied and happy? We can go to the next question. Okay. Okay. So, this one is still referring to my first lectures. So, I want to have a profile H of X. You know, this was this profile, like this mountain profile. And I want to construct H of X, H of X plus delta, average value. Actually, I will write like this. H of X plus delta minus H of X squared average goes like delta to the power y for all y, for all delta. Okay. You cannot see. Okay. Construct a profile H of X such that H of X, H of X plus R, let's write minus squared goes like R to the power y. So, this was, you know, in the beginning, we said that there is this mountain profile. And if you calculate height difference at two sides, it goes like power y. And can you construct? Can you write a Monte Carlo program which will generate these height profiles? Given y. I give you y, you generate a program and give me how to give the output. It will give random height profile, but it will have this property that on the average H of X minus H of X plus R goes squared, goes like R to the power y. Average over, it's an ensemble average, but if you feature long configuration, you can average over X. Yes, please. It can be, hurry. Yeah, yeah, yeah. Yeah, yeah, yeah. But it should give you a random answer each time and the ensemble should have this property. Yes, that is the question. Has everybody understood the question? No, no, no. So, I am giving, put of the program is a random profile which has this ensemble property. Can you generate a random profile which has this, from this ensemble which has this average property? Yeah, H of, I only give you y, for all R it should work. Yeah, yeah, yeah. Trial and error. I think that will be very tiring. You write a computer program, let the computer do the trial and error. No, because I want 1 million outputs and you are going to do trial and error, you will keep me waiting for very long and that is, I cannot afford that much time. It's a question clear. Okay, okay, whatever, next. Anybody else has a better idea? Yes, ma'am. I would like to say the origin and then discretize the space for the smallest value of delta through the new height according to that distribution. So, taking the square root is distributed as H of X plus R is equal to H of X plus R for y equal to 0.7, can you work, will your program work for y equal to 0.7? If it only adds incrementally, then some kind of central limit theorem usually works and it doesn't generate long range correlations of the type I am looking for. Yes, there was some other question. Only one point. This is varied for all R. Okay, yes, please. Yes. Yeah, precisely. That is the answer. Okay, so the answer is, this is all questions depend on some pre-knowledge. So, I assume that you know this or you know this or you don't know this and I also know that all of you actually don't end up knowing that pre-requisite, but this is a test of that pre-requisite. So, it is well known that if you want to generate something like this, a random profile, the way to do it is like this. Suppose there is an H of X, then you can generate H of X, H of X plus R average. This is called C of R, defined. We are giving you a particular form of C of R, but you can choose other forms if you like. Okay, but then how do you generate something with C of R? So, it says that given C of R, define tilde of K is equal to C of R Fourier transform. C tilde of K is the power spectrum of the thing in the wave number K. So, you take the signal, break it into Fourier mode and the power in the Kth mode is called C tilde of K and if the power in the Kth mode. So, then if I want to generate H tilde of K, is a random variable with variance tilde of K. So, you pick H tilde of K as a random variable, but its square average is C tilde of K. Then you get all the set of H tilde of K, then take Fourier transform, you get H of X, because these numbers are random. You will get random answers each time, different answers. This F H now will be Gaussian random variables. Okay, so I didn't insist, but you know I just said make a construction which will give me this and this one works. Okay, so that is an example of such functions and by you know because this C tilde of K can be chosen to be whatever I like. I can in this case it is chosen to be 1 upon K to the power some number which depends on y and then when you Fourier transform this will give you some function and now you adjust the beta so that the variance comes out right. That step is not being done here, but the point about the question is only this that the answer is that you construct a function with a given Fourier components and then add up the random components and then you get the full spectrum and the power of the Kth mode is pre-specified. Yes of course, yes. So beta is also known. I didn't work out the relation between beta and gamma that is left as an exercise. Okay, yes please. This depends this is a translational invariant system and it has a stationary Gaussian process. It's working. Yes, yes. So there is a well known theorem which is called the Wiener-Kinchin theorem. We say if you have a stationary process then the Fourier transform the correlation function is the power spectrum. So if I give you the correlation function I am giving you the power spectrum. If I am giving you the power spectrum you make a signal with that power and then Fourier transform it back and you get the answer. No, this is the expectation is in the so this Wiener-Kinchin theorem is not known to everybody in the Wiener-Kinchin theorem. It says for a stationary process h of x Fourier transform of correlation function is equal to power spectrum. Okay. This was assumed to be known. Okay, so you know the power spectrum of a process then you construct a random signal with that power means quiet is fixed but the values need not be fixed then take Fourier transform and you get the random signal. Sorry, yes average. Okay. Can I go to the next question? Okay, so very good. So h x minus h of x plus r full squared average is equal to h squared average plus h of x plus r squared average minus h h but this is equal to this and it is independent of x because it is a translational invariant process and so it is just a number. So, h h correlation function is equal to constant minus h minus h squared average. Okay, so it is up to an additive constant these two are the same. Okay, next question I will erase this part it says show that directed asm expectation value of s I will you know is equal to constant time say it is not goes as it is equal to let us define the model. So, the model was defined like this there was some square lattice with periodic boundary conditions this length is l, this length is m and when some particle leaves from here it throws two particles down. Okay. And you add only at top and leave at in the particles leave at the bottom and then the question is that what can you show that the number of top links per particle added will be constant time set. So, if you remember the proof it said that on the average you know every particle will travel length l but for a directed pile every particle will actually travel exactly length l in order to leave. So, the argument works without any change is this clear yeah, yeah of course limited value of s by definition is in the steady state and there is a conservation of particles and the particles are added at top and they are removed at the bottom see nothing this proof does not require anything beyond what you have already learned but it might just require restructuring your thought a little bit to get the answer exactly right. Okay. Okay. So, this one find the steady state probability different configurations compartment plus engine train. So, to recall the train model was defined like this there is a train and there is some spring in between and there is a compartment one and there is a spring and there is a compartment two and this engine moves at rate one every time step but if this spring becomes of length more than two it becomes three then it resets to either length one or one two and it pulls everything behind so it just moves this compartment but then this spring will become longer and then maybe it will also reset and that compartment will jerk forward and so then it stops at some stage and then the engine moves again and so on and so forth and the compartment behind keep on moving in jerky motion. So, then there is a steady state now the question is can you if I take the steady state and I look just before the engine has moved further are the after the configuration has relaxed before the engine has moved one more step then what are the allowed configurations of the train there are different configurations and what are the probabilities of different configurations in the steady state how many of you can do this problem one half hand ok ok no no no so firstly once everything has relaxed what are the lengths of springs they can be one or two that is what we said nothing else so a configuration is specified by l 1 equal to 1 l 2 equal to 1 configuration is specified by l 1 l 2 which is the length of spring 1 length of spring 2 and the allowed configurations are 1 1 1 2 2 1 2 2 is this point clear now in the steady state these configurations will occur with and we want to know what are the probabilities of these different configurations ok so how do I do this well I have to find what what happens if the configuration now is 1 1 and the engine moves then what will happen to it 1 2 actually maybe I should write it like l 2 l 1 l 1 is to the right and so it will become 1 2 l 1 will increase and the other configuration will remain there right this is all that will happen but suppose the configuration is 1 2 then what happens so it will become 1 3 but 1 3 has to reset so this will reset to 1 or 2 but depending on that value this next value will also change so 1 3 once the engine moves further will become very good it goes to 2 1 or 1 2 with what probabilities we said with equal this was our rule with equal probability this probability 1 by 2 this is also probability 1 by 2 what happens if the initial configuration is 2 1 and the engine moves 2 2 what happens if the initial configuration is 2 2 and the engine moves so very good so the point is the engine will move then the first spring will relax and the second spring will also relax and then you have different possible outcomes so with some probability this will happen with some probability this will happen there are transition probabilities of different configurations so you can write in the end that p 1 1 p 1 2 vector at time t plus 1 is equal to w times p 1 1 p 1 2 p 2 at time t then I know I know the matrix w I just worked it out then I find the largest eigenvector of this w and that is the steady state yes please with equal probability so the relaxation occurs from the engine side the first spring just changes and it pulls the compartment behind once it pulls the compartment behind then the next chain will become long then it will pull the compartment behind and so on there is the dynamics as defined okay so the matrix w is a 4 by 4 matrix so it can be worked out without too much hassle that is how you determine the steady state of this system yes yes of course once you define the model the matrix w is well defined okay yes yes I guess so but so in the definition of the model we said that the first spring relaxes the next one relaxes yeah okay that you know you can ask what is the definition of the model which order do springs relax and so on and so the answer is that it may happen that you know this spring relaxes but then this one relaxes this relaxes again so we said no no no start from the right end and then only propagate towards the left yes please sorry yes I hope so no no no I am asking is this fact well known to everybody that the steady state will be the largest eigenvector of a Markov matrix yeah at least some people know it know it is clearly not known to everybody but I am asking that I we actually use this property sometime you know we said that there is a evolution operator w which we wrote down w is this matrix it is the probability evolution matrix yes the largest eigenvalue is one in this problem because the steady state the vector does not change all other vectors have eigenvalue less than one that is a general property of Markov processes yes of course this w is also w gives you one step evolution in time and then if you want 10 step evolution you write w to the power 10 value will be the raise to power 10 but one raise to the power 10 is still one all other numbers are smaller and they decrease so the proof is that if you write a Markov process in which there is the matrix w then it has one eigenvalue at least which is one and all other eigenvalues have to have modulus less than or equal to one they cannot be bigger than one because if it is bigger than one you raise it to power 15 probabilities this will become divergent which is not allowed so the Markov matrix cannot have eigenvalues with modulus bigger than one that depends on the details of the Markov matrix so then it is a degenerate Markov matrix in which nothing evolves it remains the step put then the steady state is not unique if the eigenvalue of the Markov matrix one is not unique if the eigenvector corresponding to eigenvalue one is not unique then the steady state is not unique okay yeah it is not okay can I ask a question about the previous exercise do we have to find C or do we have to prove it is you have to determine the eigenvector that is what it says determine the probabilities of different configurations exercise the previous seven do we have to find C in a question about yeah yeah yeah you have to find it it is an arm what is it I would say one I would say half actually we gave the same please remember we gave the argument each particle of course takes L steps but when I ask how many toplings occur in each topling two particles move so there is a half if you don't put down a half you will not get full credit for some such question clearly okay yes I would only look for if you have got the general idea I will not look for a very detailed and careful proof so long as I see that you have got the idea I will give you credit okay this is a low test low order test in the sense I am not really distinguishing between a person who gets the Nobel prize and the person who doesn't get the Nobel prize we are looking for people who pass the course and don't pass the course so the threshold for checking is also lower I will not ask very subtle questions I will not ask please excuse me when I talk I prefer that other people don't talk okay so I am saying that in the exam we are not really it is not a very high grade exam so we are only checking for basic understanding like if I ask you this problem if you somehow I am able to convince myself that you understand what is the W matrix and you have tried to write it you have tried to write the eigenvector you made some mistake in the algebra determining it you will get partial credit clearly you will not get full credit but a person who doesn't even recognize that this is a Markov process and there is a transition matrix then of course that doesn't get any credit whatsoever okay you are saying something namely you will not bad one I see okay I hope not my attempt has been to make clear okay okay so here is a suppose I write the same pile on a 4 by a lattice okay so the sides are labeled I guess this one is called well we will count x this way so this is called 1, 1 this is called 1, 2 this is called 1, 3 4 this one is called 2, 1, 2, 2 2, 3, 2, 4 is this point clear the coordinates are defined x this way and y this way this is consistent with the quantum mechanics is done and operators are written and so on and so forth or Urdu or Arabic is written right so we count from right to left and so there is an operator called Axy or there is a set of operators called Axy and then there is this operator algebra we already wrote down I do not know a to the power 4 equal to a with some indices below okay so the question is use the operator algebra express Axy in terms of a x equal to 1 y okay can one do this Axy in terms of put x equal to 1 and y can take value 1 to 4 so can you write a to 1 in terms of a 1 1, a 1 2, a 1 3, a 1 4 you think that is nice but can you do it no this is an interesting question that I cannot okay I think I am going for okay so here is my problem these operators can be expressed in terms of a 1 1, a 1 2, a 1 3, a 1 4 they are them what about this one a 2 1 well I have an equation which says a 1 1 to the power 4 is equal to a 2 1 a 1 2 right so why just invert it I write a 2 1 is equal to a 1 1 to the power 4 a 1 2 to the power minus 1 they are the addition operators no they are these matrices of size 4 to the power n by 4 to the power n actually working on the recurrent set now a operators are the operators which were defined in the first lecture and they are the addition operators they act on the space of configuration give you a new configuration but think of them as matrices and then this equation holds a 2 1 is equal to a 1 1 to the power 4 a 1 2 to the power minus 1 what is a 2 2 one of you is brave enough to give the final answer yes yeah you speak out then I will write a 2 2 equal to a a 2 2 to the power 4 no I want to write a 2 2 in terms of other things a 1 1 inverse that's it change the order but this commutes that sounds good to me is this clear so I can write also a 2 3 and a 2 4 there will also be some powers of a 1 1 a 1 2 a 1 3 a 1 4 once I have done this then I can also write a 3 1 1 a 3 1 y in terms of these powers ok you always get integer powers of the basic operators positive or negative I can do at least one more problem ok then yeah so we said express x y in terms of x equal to 1 y so it's clear they are just integer powers of it so all these operators they are looking like many but they are only expressed in terms of these 4 basic operators then you can ask further there are these 4 operators the generators of the full group or not so that question is a little bit you know difficult to answer not immediately clear not difficult to answer not immediately clear so that's not part of my question the question stops here if you recognize the basic operator satisfied by the algebra basic equation satisfied by the operators write the rearrange them in some way you get the result it's not a big deal the only thing which is required is a little bit of ability to think about the problem directly ok question number 10 is like this that I have a sand pile like this which is L by L but L is big then I add at some site at the boundary you cannot see anything particle so I will make the boundary like this this is my pile and this is empty so it is half space x bigger than equal to 0 ok now I add a particle at the origin here a is 0 ok and then you calculate how many toplings occur at distance r I think we showed earlier that g of r which is the number of toplings at r went like 1 upon r to the power d minus 2 ok this was shown and for site adding site at the boundary boundary g goes like 1 upon r to the power d minus 1 so I am just saying that I add a particle here how many toplings will occur how does that vary with r it varies with the different power then if this was in the bulk if it was in the middle not away from the boundary so this power is changed from d minus 2 to d minus 1 then can we show that it goes as r to the power d minus 1 ok now the proof will be suppose you have a point charge and I calculate the potential due to a point charge potential due to a point charge in 2 d goes like log r but suppose the potential is near a conducting wall I am sorry suppose the charge is near a conducting wall then the corresponding potential behaves differently what is it one word one word ok does everybody agree so I think the only thing to realize is that we were doing the electrostatic problem all alone nothing very profound the propagator g is just the electrostatic potential we did the register problem where we got some log r we solve the Laplace's equation to get the potential the Laplace's equation is the solution of an electrostatic problem this is the discrete analog of the electrostatic problem but there is there still dipoles and the potential at large r will go like 1 by r in 2 d 1 by r squared in 3 d multiplied by cos theta there is a angle dependence also to the potential right that is all there is sorry sorry please one at a time basic operators are the operators a then what of the group corresponding to the strip problem not for the universe ok yes yeah ok can you repeat kind of a general argument I didn't get it no that is why I am asking you to repeat ok the power r 1 over r to the d minus 2 but we are adding particles to the whole boundary and say like it's an easy plane right so we can say just we can do like integration on the boundary yeah I cannot do that integral in my head no no no you can do the integral but is it clear what is the answer of the integration that is proportional to r to the 1 over r to the d to d minus 2 times r but it is times 1 by r it's not times r it is times 1 by r the answer the first problem was log r the new answer is 1 by r if the potential dies faster not slower 1 r to the power d minus in 3 d the ordinary potential goes like 1 by r the potential near the conducting wall goes like 1 by r squared it goes faster goes down faster your integration would suggest the opposite ok no no but the argument is actually good what you have to say is that if I want to solve the problem with phi equal to 0 on the some line then there is an image method of images you take a positive and you take the image charge which is negative and add them up and that gives you the dipole potential and dipole potential then as a 1 by r extra factor added this argument is continuing to work for the discrete Laplace's problem can you give me the the argument is that both of them satisfy the Laplace's equation with the point source no that is the end of the argument it's a correct what an rigorous argument see if there are two different problems one is an electrostatics problem one is a sand pile problem what is the relation between these two problems the connection is that they satisfy similar equations so I already showed that the g function which I define for my sand pile problem satisfies the Laplace's equation Poisson equation and so there is problem is like this electrostatic problem of some sort suppose you are doing a real experiment with sand then what do you expect you know all the electrostatics you have read all of Jackson then what it depends on how much you are able to connect different problems ok so if I am an experimentalist is not an excuse if you should be an experimentalist with some imagination and some head then you are able to make the connection putting the machinery very good when I add a particle here things leave and then some toplings occur and things leave from here so fewer toplings occur because particles leave ok is that clear I think we should stop because I am getting tired ok