So, in the last part of this lecture today, we shall look at something a little different. We shall actually now go back again to looking at these linear transformations in terms of matrices. Earlier, we have seen that any vector space, any vector in a finite dimensional vector space can be represented as an n-touple through this assignment of coordinates. But now we are going to see that the same sort of an idea, the idea of an ordered basis precisely allows us to do something similar basically to these linear operators too. So, what is it that allows us to do this? And again, pictures here, these are not like the pictures of sets and all. But this picture here, if you remember this, it is not really about remembering also. If you just visualize this in your minds, often you can do get away just fine with this. So, what is the picture? The picture is you have this vector space V, of course, finite dimensional vector space. And you have this phi which acts on objects in the vector space V and maps them to objects in the vector space W. So, let us say the dimension of this is n as we always do and the dimension of this is m. And suppose you have an ordered basis for this given by V1, V2 till Vn and an ordered basis for this given by W1, W2 till Wm. So, this is ordered basis and this is also an ordered basis. The moment you have an ordered basis, what I can simply do is, through an isomorphism, we proved it is an isomorphism, right? When we go from V to F to the n, where F is of course, a field over which both V and W are defined. So, when we go from here to here, the mapping that we have is, remember the dictionary? Yeah, that is what we have. We take each of the vectors in this basis, represent them in terms of, yeah, right? And then we know that this is basically the coordinate represent. In fact, for this object, this V1 maps to what exactly? 1 0 0 0, V2 maps to 0 1 0 0 like so, right? Does everybody understand what is going on here? See, I am saying coordinate assignment for V1 would be this. A coordinate assignment for V2 would look something like this, because what is V1 after all? It is just 1 times V1 plus 0 times V2 plus 0 times V3 plus 0 times V4. So, the base, the coordinate representation of V1 is this. By the same token, the coordinate representation of V2 is like this and the coordinate representation of Vn is going to be like this. So, when I am writing this map here, that is exactly what I am saying. That is what I am talking about, yeah? Clear so far, right? So, I can always do this coordinate assignment here. Exactly in the same manner, I can also take this, objects in this vector space w through some coordinate assignment with respect to this basis, I can take them to m tuples. So, here is the important thing. The question we are asking is, this object phi which maps vectors in the abstract vector space V, this abstract vector space could be the vector space of functions, vector space of polynomials, vector space of what have you, not necessarily Euclidean spaces, right? So, this is an abstract vector space, this is another abstract vector space, this is a linear transformation between two abstract vector spaces. What I have done is, I have stripped them off their fancy names, got them down to their coordinates because of the fact that they are finite dimensional, I am able to do that and assign them certain n tuples and m tuples they are in. And now, what I am claiming is that there will be a matrix A of size f n cross m which will capture exactly the action of phi. What do you mean by which will capture exactly the action of phi? It means that if instead of looking at this phi in a standalone fashion, I take any vector whose effect I want to study under the action of phi, right? How phi affects that vector? What I can do instead of just passing it as an argument of phi is as follows, I can simply say take that vector, get it down to its representation in terms of this ordered basis, let a matrix act on that n tuple that will lead me to an m tuple and then take this inverse mapping. Remember this is an isomorphism, so the inverse will exist. So, when I am going against the direction of flow here, it is of course, the inverse mapping. So, by bypassing this direct evaluation through phi, I can simply reduce it down to just computations of matrices and n tuples and m tuples. Which one sorry? Ah sorry, yeah m cross n, thank you, right. So, what I am saying is that instead of looking at the direct action of phi on objects in v, I can simply go via this root and it is going to be exactly the same. But of course, this requires this needs to be established through this diagram it is pretty apparent. Each of these actions are legitimate, but who is to guarantee that I will be able to find such an a? You see, that is what we are now going to try and see. So, you have this basis here, we know that in order to find out what a linear transformation does to any object in a vector space, it suffices to just investigate what it does to objects in any basis, right. So, let us look at the action of phi on v i. The moment I have phi of v i, where does it reside? Remember, v i resides in v, but after I have allowed phi to act on this v i, it now resides inside w. So, if it resides inside w, it can jolly well be represented as a linear combination of objects in this set, because this is the basis for w, agreed. So, now, I can go ahead and write this as, so I will use summation alpha i w i, there it is. But look at this fellow v i, this is just some special v i, right. So, if I remove this i here, this is still legitimate, right. This is true of any v, because if I take once I let phi act on this v, it is already gone there to w, and then it can be represented in this manner, right. So, now, if I say that phi v is representation, so suppose I had gone via the direct root here. So, I have gone from v to phi v, and then I represent phi v in terms of v w, that would be equal to alpha 1, alpha 2 till alpha m, agreed, right. But now notice, what about this object v here? What can I write v as? v is residing inside the vector space v, and therefore, can be written as a linear combination of objects inside this set v v, which is a basis for v, right. So, this is equal to summation, let us give it some other name, beta i or beta j v j, of course, with j going from 1 through n. Now, if I say that phi v, that is what, what is this? Summation of phi v, that is what I can push this, pull this summation out, because of the linearity of the operation you see. So, I can just write this as beta j phi v j summation j going from 1 through. But now each and every object such as this can be written as what? This sort of a linear combination. So, I did not want to do with a double subscript here, but it seems I will eventually have to use it here, right, because I will be using a double summation here. So, now, what is that v j? This phi v j is again a linear combination of objects inside w. So, this is summation beta j, j going from 1 through n, and this is summation. So, I have to use a double subscript now. I have to use over i, a i j, I think w i, right, i going from 1 through m, is that right? Yeah, because the j is fixed here, I am summing over i. So, the moment I have fixed this and then you run this j through 1 through m, is this correct? So, I can probably erase this part now, right, we will revisit this figure. So, what is this? Okay, let me just write that term, so that it is easier for you to see. So, phi of v then turns out to be, let us just write the first term, shall we?  Again, you see something very familiar. So, you will begin to see the pattern after a while as you ask, right. So, now, I am going to collect together terms related to w 1, because what I am interested in is a coordinate representation of phi v, yeah. That is what I am interested in, right. So, that is what this is. I am going to try and evaluate these alphas and try and understand what they are going to constitute of. What is this? If I just collect together terms related to w 1, what is the coefficient of w 1? a 1 1 beta 1 plus a 1 2 beta 2 plus, dot, dot, dot till a 1 n beta n plus a 2 1 plus a 2 1 beta 1 plus a 2 2 beta 2 plus. So, these three dots, the mark of ellipses, then a 2 n beta n plus a m 1 beta 1 plus a m 2 beta 2 plus until a 2 n. So, a m n beta n times, oh, of course, we have w 2 w m, is all right. So, now, if you try and equate each of these alphas with terms like the one sitting here, what do we get? What is it that this reminds you of? Yeah, because it is a linear transformation, you see, at the end of the, the change of basis was also a special case of linear transformation. That is all there is. But now we are seeing that any linear transformation manifests itself through some matrix multiplication at the end of the day, because this means that I will just erase this punch, because I want you to be able to compare those alphas with what we have now gotten here. So, this means that through this route, we already know that this is what the representation would be. We want to find out what these are and now we have exactly found the answer to what these are. So, what is this? So, now, from this, all of this work that we have done here, in terms of b w, this turns out to be what? a 1 1 a 1 2 a 1 m, sorry, a m, no, is it? Where is it? Where is it? a 1 n, yeah? a 2 1 a 2 2 a 2 n a m 1 a m 2 a m n and what do we have here? beta 1 beta 2 till beta n. What is this exactly? What are these betas representing? If you look at how we started with the betas, what were these betas, right? So, this is not this a basis representation for the vector v in the vector space big v and what are these objects exactly? If you look at this, what is this? Yeah, so this is a representation of phi v 1, right? So, what is this? What is this? The action of phi on the first vector in the order basis for v, that is phi v 1 represented in terms of b w. This is phi v 2. So, this, let me just mark that. This column here is phi v 1 represented in terms of b w. This by the same token is phi v 2 represented in terms of b w and so on until this is nothing but phi v n represented in terms of b w, yeah? And this object is nothing but the representation of v in terms of objects in b v. There is nothing really to remember. It is very obvious. You see v is residing inside vector space v. So, the only way to represent it through a coordinate is through a ordered basis in v. So, that is what you have done. And if you want to study the action of this linear transformation phi on an object in the vector space v, in the abstract sense we know we need to study its effect on the fellows in the basis. So, in the coordinate sense, what we are doing is exactly that, because these numbers, these m tuples of numbers are exactly capturing the coordinate representation of the effect of phi on every member in the basis set of v. So, you are studying phi v 1, look at its basis representation in terms of w, b w. Phi v 2, look at its basis representation in terms of b w. Phi v n, look at its basis representation in terms of b w. That is what your equivalent A matrix is. So, that figure which we have raised now because of one of space, that A matrix which we had written probably in pink and we are asking the question as to what is the representation of the A matrix that captures the action of phi is nothing but this. So, in other words, in a very shorthand kind of a notation, we can just say in summary that this phi, of course, most of it is already written here. This phi v, if you want to represent in terms of coordinates, in terms of the basis in w because phi v is an object residing in w, then that is given by this A matrix acting on a basis, a coordinate representation of v. And this A matrix, let me erase this part now. And this A matrix is now defined in the following manner. It is nothing but phi v 1 represented in terms of b w, phi v 2 represented in terms of b w until phi v n represented in terms of b w, right? That is exactly the matrix.  Given a coordinate, there is a 1 to 1 mapping, right? So, given a coordinate representation in f m, you can always map it back to the fellow in w. And that is the same action you see. So, that u shaped path going through the downward trend and the forward path, they are constituting essentially the same action, right? So, now we will make a very important observation. Let us see how much time we have got, ok good. We have about 5 minutes. So, now I am going to draw that figure again and ask you to use your imagination a bit, not by much, but just a bit. And just look at this v and w.    So, the underlying linear transformation that you have in mind is phi. And then you have found some a matrix corresponding to your choices of basis in v as well as w, right? Your friend on the other hand decides that I am not going to choose the same set of basis. So, your friend chooses another basis, which is b v 2. So, of course, your friend decides that I have had enough. I am not going to choose the same as this other fellow. So, I am going to choose some other basis b w 2. So, of course, your friend is going to land up with a different a matrix, maybe an a tilde. But you see that at the end of the day, as I have said, it is the idea that counts. I mean the choice of basis is completely arbitrary at your will, right? You have chosen something, your friend has chosen something that should not really matter. So, now for the moment I just decide I am going to overlook, I am going to overlook this part and just look at it like a story of matrices, yeah? So, I know, suppose this is my choice. So, I know a, I know the basis I have chosen and I know the basis my friend has chosen. The question is, what is the corresponding matrix that my friend has obtained? So, try and understand the question. It is very important to understand the question. Once you understood the question, the answer is very obvious, it is staring us here. What I am saying is the following. This is something that I have done, the lower part of this. In going from V to W, I have chosen a basis in V, a basis in W and accordingly I have found this pink colored A, right? Suppose I know that my friend has chosen a different basis and I also know that basis in V he has chosen and the basis in W that he or she has chosen, right? The question is, how am I going to obtain this a tilde based on this knowledge? Suppose I do not even know what the V and W are. I mean, I can, I am, suppose I am told that you just forget about it. Then how do you go about this? I just know the matrices in question.  But I am not going to do, what I am just going to do? Forget about this blue part that exists here. So, now if you look at this picture here, it should be very obvious. What do I have to do? I have to find out this A, probably I will run out of colors before long. So, I have to find out this which is equivalent to this entire action. And now just by looking at this figure, that is why this figure is so powerful. What is it that I can say? You see, this A tilde that my friend has done is nothing but what has my friend first done? Born from here to here. But that is just the inverse of this action. So, this is, first action is the inversion of this coordinate allocation. That would take me to this. Then I would have B V1. Then I would require A, okay, maybe I should have had a little more space. So, let me write it elaborately here. So, this is B V2 inverse. Then I let it be acted on by B V1. Then I use the A that I have cooked up. And then what? Again an inversion of this B W1. B W1 inverse. And then finally, B W2. So, that is exactly how this A tilde matrix can be obtained. Yeah, if you try to derive this using all those enormous matrices and those large summations and everything, you will see that it is a pretty cumbersome thing. But once you understand what is going on and you follow the path here, right, and you see you land up with the requisite thing here. Just the final observation before we conclude today's lecture, we will revisit this in much detail, in some detail in the next lecture, not for much long. But if you now think of this V and W to be the same vector space. If you think of this V and W to be the same vector space and B V1 to be the same as B W1, B V2 to be the same as B W2, just observe what happens to this. Yeah, B V1 and B V2. Sorry, B V1 and B W1 are the same. B V2 and B W2 are the same, right. So, I am not saying that these two are the same. These two are still one and two are different choices. One is your choice, two is your friend's choice. So, these two do not cancel. But what happens to the sizes in the first place? The sizes are n cross n. And then what happens when you look at this? See, what am I saying? B V1 is equal to B W1. B V2 is equal to B W2. Don't you think these fellows are inverses of one another? These are all square matrices then. Remember, first observation is these are all square invertible matrices. Yeah. So, when you hit this with this and B W2 is B V2. B W1 is B V1, which means that I have chosen the same basis in the domain and the codomain. Your friend has also chosen the same basis in the domain and codomain. So, this is like T inverse A T. Similarity transformation is what we call them. And that is the reason why we studied briefly, very briefly when we are talking about fields, whether a matrix is diagonalizable through a similarity transformation. Because then we are saying that essentially there exists some non-singular transformation or just a change of basis will capture the same operation. And when you have a diagonal operator sitting here, it has very nice properties because it is in some sense like decoupled. We will revisit this point in much greater detail when we study eigenvalues and eigenvectors. But for now, at least please remember this and try to draw up an analogous figure when B V and W are the same vector space and B V1 and B W1 and B V2 and B W2 are also same. So, B V1 is equal to B W1, B V2 is B W2. That is all for today. Thank you.


Right, we will today discuss the answers to this quiz that we had last week and what I would like to do is to also make comments. Ideally I should have given you the answer sheet so you could check against it but I will read out the questions and then you will be able to recollect what the question was about and we will try to give justifications for the various choices. So the first question was in 2 of all was that as follows, consider a quantum mechanical system with Hamiltonian H0, its normalized density operator in the canonical ensemble is given by e to the minus beta H0 over the trace of the same thing. So you are given this row equilibrium is e to the minus beta H0 divided by the trace of and the question asked is what does this tend to as t tends to 0 from above of course as you go to absolute 0. The statement is that it tends to the unit operator. Is that true or false? Tends to the unit operator as t tends to 0. What happens to the state at t tends to 0? That means beta tends to infinity, this thing goes to infinity here, right? So you cannot just put beta going to infinity because then this goes to 0. But then this denominator has to be taken into account. So if you took this system to have a discrete set of energy levels e0, e1, e2 etc with corresponding Eigen functions, so if you set that H0 on phi n equal to en on phi n, if you did this then these states, normalized states form a complete set, an orthogonal set, an orthonormal set of Eigen vectors. N now labels all the possible states, stationary states of the system, all the state, all the Eigen states of the Hamiltonian. There could be degeneracy in which case n will be more than one label. It will have more than one quantum number and we are just schematically writing everything down as with an n here. So this trace is over all the states, diagonal elements over all the states clearly. Now in this representation of these phi n's, it is clear that this operator itself has a representation which is summation over n e to the minus beta en times the portion that projects onto the ground, onto the state n. So this is phi n, phi n divided by summation over n e to the minus beta en. This is over the states n. And out here when you take phi n, phi n, it is normalized to unity and you get this here. So the question is what happens to this? Now we can say what happens to this as beta tends to infinity. Clearly if all the en's are positive, then the numerator is also 0 and the denominator is also 0. But if e is 0 is finite, we do not care if it is positive or negative. We really do not care where you start measuring the spectrum from. Wherever is a ground state, it should not be minus infinity. Wherever you have this ground state, you can pull out e to the minus beta e 0 as a common factor and then whatever is inside all the other exponentials will be the difference of e 0 minus en's and they will all go to 0 as beta tends to infinity. So this thing will tend as beta tends to infinity to just the contribution from the lowest energy level, the ground state which is equal to of course e to the minus beta e 0, phi 0, phi 0 divided by e to the minus beta e 0. Because all the other terms will have decaying exponentials will all go to 0 and this cancels out. So it tends not to the unit operator but to the projector of the ground state as it should because the system goes to its ground state at absolute 0 of temperature, right. The probability of it being in any higher state is 0 essentially and it is with probability 1 in the ground state. Therefore the density operator is not any longer, there is no uncertainty, thermal uncertainty here, it is just the ground state operator, projector, that is it. Which is not the unit operator, which is not the unit operator because recall that it is this quantity, this is equal to the unit operator. At t equal to infinity, you can ask what happens if the temperature becomes infinite, then all these Boltzmann factors e to the minus beta en's essentially all of them go to 1, everything becomes equally probable and then the system will end up with this density operator. You see, if these, if these are the energy levels of the system, they do not have to be equally spaced and this is the ground state e 0, what statistical mechanics is telling you is that once you have a certain finite temperature then there is an average energy and there is a variance, a scatter about it, okay. On the other hand, if the temperature becomes infinite then k t which is the thermal noise, the quantum of thermal noise becomes much larger than the gap between energy levels, so these gaps are irrelevant completely and the system will just freely move around from state to state and therefore there is no bias at all in this case and then every state is equally probable essentially in the limit. But as the temperature is lowered, the bias comes more and more strongly towards the ground state till at t equal to 0, everything sits in the ground state. So that is the fundamental property of equilibrium, statistical mechanics and the answer therefore is false, it is just the projector of the ground state, okay. The next statement is given that the x component of the velocity of a particle of mass M in a fluid satisfies a generalized Langevin equation with a memory kernel which is a decreasing function of its argument and A t of t is a stationary noise, all you are given is that this is a stationary noise, you are not given anything more. So the statement is m b dot of t plus m integral minus infinity to t d t prime, some memory kernel gamma of t minus t prime, v of t prime is equal to on the right hand side, random force A t of t and this is stationary. So we are told that this quantity here has 0 mean at any time and A t of t t naught, A t of t naught plus t equal to function of t, the origin does not matter, that is all you are given and you are also told that this quantity gamma is a decreasing function, positive decreasing function of its argument for all positive values of its argument, right. So that is all you are told, nothing more. And then the statement is v of t is a Markov process and its density function, conditional density pdf satisfies a Fokker-Planck equation, these two are false. It is false because this incredible memory sitting here, this is an integral differential equation, it is not a Planck-Jama equation at all, the usual type of Planck-Jama equation. If you insist on writing this as a differential equation, you will have derivatives of all orders in the time variable. So there is a lot of memory in this business and it is not a Markov process. So there is no chance that the probability density function pdf satisfies the Fokker-Planck equation, none whatsoever, it is not a Markov process. But there is actually an interesting theorem, it says that if this is stationary and Gaussian, which I have not said in this problem, but if this is stationary as well as Gaussian, then it turns out that the density function pdf p of vt v0, this is the conditional density, turns out in that case when this is also a stationary Gaussian process to obey an equation which looks very much like the Fokker-Planck equation, but it is not a genuine Fokker-Planck equation. In particular, in particular you have to recognise that if it is not a Markov process, then this 2 point conditional density alone does not determine the rest of the process. You have got all the joint densities on top of it. So the fact that this satisfies a certain Fokker-Planck like equation might help in calculating this quantity and therefore the autocorrelation, but it does not help as far as understanding the nature of this stochastic process. That is much more intricate than this. Well then one of the questions says that this is inconsistent for eta to be delta correlated. And we can see right away we will see in a minute that it is not, this cannot be delta correlated. So the question is, suppose I tell you this is a stationary process, it is not delta correlated, it has got some finite correlation time. What is it you can say about the output process here? Well for that you have got to take recourse to these fluctuation dissipation here which we will talk about in some detail. So you can make some statements about the autocorrelation of V of t, compute it, but in general to find out what is this process, what is its probability density function exactly, that is not doable from this as it stands. You need to know all the properties of eta of t and case by case you have to analyse it which is not a trivial task. So the statement is, it is false, the statement is false. Next thing says, the assumption that the autocorrelation of eta of t is proportional to a delta function leads to an inconsistency. Is that true or false? That is true. It definitely does as follows so we can see what the problem is. You see all you have to do is to ask what is the second fluctuation dissipation theorem in this case. Recall that the original Panjama equation which had this m v dot plus m gamma v, just a local function with a constant gamma, the ordinary Panjama equation, this was set equal to square root of gamma times eta of t. This was a Gaussian white noise with 0 mean and a delta function autocorrelation. Then consistency required that 2 m gamma k Boltzmann t had to be equal to capital gamma. That was the second fluctuation dissipation theorem relating the dissipation in this equation with the strength of the fluctuations driven by the noise. The question is, what is that for this process? Not very hard to do. All you have to do is to find out what is the correlation here and how does it get related to this quantity here. The corresponding relation here turns out to be m m k Boltzmann t gamma bar of omega equal to an integral from 0 to infinity dt e to the i omega t eta of t0 eta of t0 plus t. That is the second fluctuation dissipation theorem in this problem. Now you can see how this is going to reduce to this, okay. Because in the case where this quantity is integral e to the i omega t gamma of t, one sided Fourier transform. To go to the ordinary Panjama equation, you have to assume this to be a delta correlated quantity. So with capital gamma times a delta function, if you put delta of t, then t is equal to 0. This goes away and you get capital gamma which is on this side. And on this side you are going to get m k t gamma bar. Now the way to go to a constant is to put this memory kernel gamma of t, little gamma of t to be equal to constant gamma times a delta function. So when gamma of t goes to gamma times delta of t, no memory, then you get the ordinary Panjama equation from the generalized Panjama equation. If you put that in, this term here becomes gamma but with the factor missing because this quantity here when I say delta, it got to run from 0 to minus infinity to infinity to pick up the full contribution of the delta function.  So that is how from the general second fluctuation dissipation theorem, you end up with the special case for the ordinary Panjama equation. Now the question asked is, it is inconsistent in the general Panjama equation, generalized equation to have a delta function as a correlation of the noise, to have this delta correlated? Yes indeed, because if you put in a delta function of t here times a constant, this integral becomes just that constant but on this side it has got a function of omega and there is an immediate inconsistency. You cannot have a function of omega equal to a constant because certainly we know that if this is non-trivial, this is some function of omega and that is gamma bar. So on this side you have got a function of omega and on this side you have got a constant which is inconsistent. So this is how we see that delta correlated noise is inconsistent with the generalized Panjama equation. This has got to have a finite correlation time. In fact that is the reason for introducing the generalized equation in the first place because we said look, in real life no noise is purely delta correlated. There is always some finite correlation time. So that is the reason why we have to introduce a memory kernel for consistency. So the statement as it stands is true. The next question says, the dynamic mobility of the particle vanishes for infinite frequency as omega tends to infinity. So this too is a very straightforward question because from this equation, we derive the fact, this equation implies that the dynamic mobility in this problem is 1 over M times gamma bar of omega minus R. Whatever be this, it is clear that this quantity here becomes infinite as omega tends to infinity and this just vanishes and physically that is expected because the dynamic mobility gives you the average velocity amplitude for a given sinusoidal amplitude of some sinusoidal frequency omega. As omega tends to infinity, this becomes so large that the system cannot respond. It does not, its inertia prevents it from responding. Therefore it should go to 0 which is what is happening. Without or without this term, it still goes to 0. So this statement is certainly true. The next one is a little tricky. It says in this model, the generalized Langevin equation model, the equal time correlation between the velocity and eta, the noise eta is not equal to 0 but the equal time correlation V of t with the acceleration V dot of t is equal to 0 and the question asked is whether this is true or false. So again within the context of this equation, the question asked is does this imply that V of t V dot of t in equilibrium equal to 0 but V of t with eta of t in equilibrium is not equal to 0 and you are asked is this true or not. It is true, it is true because the whole point about fixing the causality and stationarity conditions in this problem, well the first problem was in the ordinary Langevin equation, while we discovered that the correlation of the velocity is stationary, the velocity is a stationary random process that V of 0 V of t in equilibrium is e to the minus gamma modulus t, we ran into the difficulty that stationarity was apparently violated because if you took V square of t average, that is equal to k t over m, Maxwellian and you differentiate both sides. The right hand side has no dependence on little t, the left hand side you differentiate and then set t equal to 0 or something like that. Then you immediately discover that this must be equal to 0. For a stationary process, any stationary process xi of t, xi of t xi dot of t must be equal to 0 on the average, that is the meaning of stationarity but the Langevin equation violated it. The question is whether it is true in this equation and the answer is yes, it is true, this thing can be established rigorously, this is certainly true but is this true or not? Now in the ordinary Langevin equation we took this quantity, the force on the right hand side to be uncorrelated to the velocity at the same time arguing that the force determines the acceleration at any given time and the velocity comes a little later after integration, so this quantity is certainly 0 by causality but that led to a problem, that led to an inconsistency with stationarity. To fix that we took this problem and said look, stationarity has to be rigorously obtained here because equilibrium tells you that this velocity is stationary but causality has to be written more carefully, that the effective random force is not this but a portion of it which comes from the history here and the assumption of causality was that the velocity is uncorrelated with that effective force at a later instant of time. When you did that carefully then it turns out this quantity is not 0 at all but rather it has got a finite value which also we computed I think, so somewhere along the line we had this here, so it turns out that in this model if you compute this thing here, this turns out to be equal to m times an integral from 0 to infinity dt prime, gamma of t prime, v of 0, v of t prime. So at a given instant of time, this correlator is not 0 but rather depends on the memory kernel in this fashion. So it is an integral over the velocity correlation at all previous instance from 0 to infinity weighted with this memory function and it is not identically 0. So the statement as it is made is actually true but this is a little trickier to evaluate but it is straightforward to evaluate it after you make the proper causality assumption here. Then finally the last one on this set of questions is in the same generalized Langevin equation as tau tends to infinity suppose the memory kernel decays to 0 like some power of tau, inverse power of tau. So you are given, you are given that gamma of tau goes to 0 as tau tends to infinity like 1 over tau to the power alpha where 0 is less than alpha less than equal to 1, I believe it is less than equal to 1. The statement is does the static, the static mobility of the particle vanishes in this case. So the question asked is is mu at omega equal to 0, equal to 0 as a consequence of this? Yes or no? Yes, it is true. The reason is very straightforward. We know that mu at omega equal to 1 over m gamma bar of omega minus i omega, so mu of 0 is just 1 over m gamma bar of 0 and this integral is integral 0 to infinity d tau gamma of tau. If this dies down like 1 over tau or 1 over square root of tau or any power of tau less than 1 in the denominator then of course this integral blows up, it diverges. So this tends to infinity and therefore mu of 0 vanishes, this quantity therefore tends to 0 because this memory is so strong, this dissipation, this is like the effective drag is so strong that the system does not move. That is what is meant by very slowly decaying memory, okay. So much for the generalized Langevin equation. The next set of questions has to do with linear response theory. Given that A represents a physical observable of a quantum mechanical system and you have asked, the statement made is phi A A of tau equal to minus phi A A of minus where recalled that this quantity stands for the expectation value in quantum mechanics of A of 0, A of tau over i h cross inequality over the canonical ensemble here. So true or false? The statement is true and why do you say that? Because under time reversal you are asking what happens under time reversal and we know in general you have phi A B of tau in general is equal to epsilon A dot epsilon B phi A B of minus tau where epsilon A dot is a time parity plus or minus 1 but you know yeah there is a dot here which is equal to minus epsilon A epsilon B phi A B of minus tau because A dot and A have opposite time parities like position and velocity or velocity and acceleration.  So this response function is an odd function of tau and that helps us answer the next part of it. It says that the corresponding generalized susceptibility and spectral function are related according to real chi A A of omega is equal to half phi A A tilde of omega. So you are asked is real chi A A of omega equal to 1 half phi A A tilde of omega. Is this true or false? Well because of this this thing implies that it is phi A A tilde of omega is actually twice 2 I times the imaginary part of chi A A of omega. Had this been a symmetric function this would have been twice the real part but because it is anti-symmetric it is twice the imaginary part. But now we know that the imaginary part of the susceptibility is an odd function of omega while the real part is an even function of omega and in this case the spectral function is proportional to the imaginary part. So it is clearly an anti-symmetric function and therefore this statement that this is not true instead you have this statement here. So that is the correct statement okay. So the spectral function is therefore an odd function because it is proportional to the imaginary part of chi. So this of course immediately implies phi A A. So that takes care of this part and then there is set of questions on the Boltzmann equation. The first of which says let f of r, v, t denote the phase space density in the one particle phase space mu space of a dilute classical gas. We will assume that the molecules of the gas only undergo binary collisions with each other. Then the statement is at every instant of time the gas can only be in a state in which f satisfies the Boltzmann equation. Is this true or false? We are not saying the gas is in equilibrium. We are saying that it satisfies, we are not saying there is f of r, v, t and in the absence of an external force this thing goes in equilibrium to f equilibrium of v which is independent of time. And in an absence of an external force this quantity is equal to n times the Maxwellian w of v. We have seen that. Even if there is an external force present which is dependent only on the position of the particle, a conservative force then this quantity here goes to the Maxwell Boltzmann distribution. So this will go in equilibrium as t tends to infinity to e to the power minus phi of r over k Boltzmann t times w of v, right? Apart from some normalization factor such that the whole thing is normalized. The integral of this over d3r, d3v should be equal to the number density. So this constant will fix that normalization, okay. Now the question is at every instant of time is the Boltzmann equation satisfied or not? We are not saying that at every instant of time the phase space density is the equilibrium density. We are not saying that at all. The equilibrium density could happen in between. It could occur. It is one of the possibilities. The gas could be in a state where it is exactly where this density is exactly the equilibrium density. And as t tends to infinity we know that it attains that, okay. And now the question is at every instant of time is the Boltzmann equation itself valid or not? Well, no. In general no because you see the Boltzmann equation is valid if and only if the assumption of molecular chaos is valid. If and only if in mu space we have said that at some point are inside the cell, the centre of a cell in mu space we have assumed that the probability that you have a particle in this r with velocity v, v1 and t, probability density is this. And the probability density that you have 2 of these guys with velocities r1 and velocities v1 and v2 is the product of these densities. What does that mean? That means there is no correlation in velocity space at all, okay. This would certainly not be true if there are recollisions. Something hits against this particle, it goes and hits another particle and comes back and hits this. Then this kind of assumption is not valid anymore. You can easily see that it builds up a correlation between these 2 particles, okay. That is immediately clear. Take the simplest possible example, put everything on a line, let us say the particles are equal mass and put a wall at one end. Particle A is sitting here, particle B comes and hits it, bounces back, hits the wall, comes back and hits A. Surely the velocity of the projectile the second time is dependent on what happened the first time. So this is no longer true. It is only true at some instance of time. And at those instance the Boltzmann equation is valid, okay. So the answer is false. The Boltzmann equation holds good, only at those instance where the molecular chaos assumption is valid. In between there are all kinds of states possible, which do not satisfy the assumption. No, it says let f of r, v, t denote the phase space density in the one particle phase space or new space of a dilute classical gas. We assume that the molecules of the gas only undergo binary elastic collisions with each other. That is it. That means no energy is lost. And then the statement is that every instant of time the gas can only be in a state in which f of r, v, t satisfies the Boltzmann equation and that is not true. That is not true. The next statement says since the Boltzmann equation is the first order partial differential equation in t, delta f over delta t equal to something or the other, it follows that r and v comprise a 6 dimensional Markov process. Because as you know for a Markov process the master equation says that the density, whatever be the conditional density, satisfies a first order differential equation in time. What that equation is depends on the actual process, the transition rates and so on. But it satisfies a first order process in time. And not only that it satisfies the chain condition. So it is even if you like a non-linear integral equation just like the Boltzmann equation. But is the Boltzmann equation a Markovian master equation? What do you think?  Because this sort of term appears even when you write down the Langevin equation in phase space. When we wrote down the Kramer's equation for instance, we got terms like this. We got v streaming terms plus f over m dot gradient to respect to v of f. This is equal to on the right hand side delta f over delta t collision. And if you recall this was a complicated integral which involved quadratic squares, I mean products of 2f's at 2f's minus final minus initial and so on. We had a very complicated collision term here. So this is by no means a Markov process of any kind. It is a much, much more complicated thing than that. Finding what the correlation is between the autocorrelations and so on in this case is very tricky because you have to essentially solve this Boltzmann equation. And notice there is no explicit randomness which is brought in anywhere here. In principle you are supposed to know this on the right hand side and solve itself consistently. So it is certainly not a Markovian master equation by any means. However if you recall when we looked at the so-called single relaxation time approximation, in the case when you linearize the Boltzmann equation, now we did not clearly explain what are all the assumptions which you need to do the linearization. But if you did, it is a small departure from equilibrium and on top of it you make the single relaxation time approximation. Then under suitable cases the velocity relaxed like the for like a Kubo-Anderson process. It relaxed like a Markov process. So that is a special case of a special case. In that case you can mimic it but that is a crude approximation to the actual Boltzmann equation. This is not a Markovian master equation. And it does not say how strong the collisions are. It just says v1 and v2 go to v1 prime v2 prime. It does not say put any bounds on how big this quantity should be. It does not put any bounds on how big the relative velocity v1 minus v2 should be at all. Could be drastically different. Yes, yes. No, detailed balance only involve time reversal in variance for the dynamics, right? So in general what you do is for Markov processes if detailed balance obtains then you can write down the equilibrium distribution by using the detailed balance condition just as we could do so here too by inspection of that collision integral. But there is nothing which says detailed balance should be there because in the Markov process I can specify the transition rates, probabilities between different states of the system quite arbitrarily subject to some ordinary conditions. But the fact is they are completely arbitrary. There is no reason why detailed balance should apply at all for a general Markov process, okay. Whereas here the physics is saying that it does apply because ultimately the collisions, the dynamical process of the collisions is time reversal invariant. Whatever is happening there is assumed to be time reversal invariant, okay. So that is when some physics has been put in about the molecular level process of scattering and then you get this detailed balance coming out, popping out, okay. And finally consider the gas in the absence of an external force. The statement is for any given distribution f of v, t, there is no dependence on r. The Boltzmann H function is defined as f log f integrated over v. Statement, H of t is a monotonically decreasing function of t, true or false. So let us be careful. You define the Boltzmann function as integral d3 v f of v, t log f of v, t. And is it true that this quantity satisfies d H over d t? Alright, less than equal to. Is this true? Exactly. You see the definition, this quantity, you give me any distribution and I will define this quantity, the H function. What we have proved is that if this satisfies the assumption of molecular chaos and therefore satisfies the Boltzmann equation, then you can show this. So what it is saying is that if at instant of time t, if at this instant of time, some particular instant of time, say t0, the Boltzmann equation is satisfied, then the slope, local slope at that point is negative. So this H function will go like this. But it is not saying that at all these instance of time the Boltzmann equation is valid. The local instantaneous slope to one side is negative. And then you can also show by clever argument that if at this instant of time the Boltzmann equation is going to be satisfied, then at an immediately preceding instant the slope is positive. So this function actually fluctuates in this crazy fashion. It is not monotonic at all. What we do know and what we can establish is that at some of these peaks, at local peaks, at points where the Boltzmann equation is satisfied, the H function is at a local peak. The converse is not true. We are not saying that at all such peaks the Boltzmann equation is satisfied. And certainly it is not satisfied here or here or here. So it is only satisfied in a statistical sense. Then there are other questions about how frequent are these points, how dense are they and so on. Those are subtler questions. Okay, there is a lot of very intricate stuff involved there. So the theorem is very clear. And the statement as it stands is false. H of t is not monotonically decreasing function of t. It is a rapidly fluctuating function of t. And finally consider the gas in absence of an external force. Let F equilibrium be the equilibrium phase density and V1 plus V2 goes to V1 prime, V2 prime as usual. It is an elastic scattering process between two molecules. Then statement, F equilibrium is unique and satisfies the detail balance condition. Such that F of V1, F of V2 is equal to the product F of V1 prime, F of V2 prime. True or false? True. This satisfies by the system is in equilibrium. Okay. So at this stage detail balance is valid and it is a necessary and sufficient condition for equilibrium density. So that condition actually as we saw determines the equilibrium distribution completely because what you do is to take logs and then it says log F of V1 plus log F of V2 is log F of V1 prime plus V2 prime. So log F of V must be something which is a function of all the quantities that are conserved in the collision process such as a constant, the velocity itself for equivalence particles, the sum of initial velocities and the sum of the kinetic energies. So that is what necessary and sufficient condition. For proving the necessary part of it, we have to go through the H theorem. The sufficient was inspection. Just by looking at the fact that this condition satisfies the collision integral vanishes and in the absence of an external force, F is time independent explicitly. So that part was straight forward. So the statement is true. Okay. Then we had a few fill in the blanks. The first of which said that we have a system satisfying the large y equation and you are asked to find the velocity correlation time in terms of the memory current. So what is integral 0 to infinity dt V of t0, V of t0 plus t, equilibrium divided by V square equilibrium. This quantity has dimensions of time. This is by definition tau, the velocity correlation time. And the question is in the ordinary Langevin equation this was gamma inverse. The question is what is it in the generalized Langevin equation? Pardon me? Yeah, this quantity is equal to 1 over gamma bar of 0. And the way you establish that is to go back to the first fluctuation dissipation theorem which essentially says that if you multiply this by e to the i omega t, then you get this quantity is equal to V square equilibrium which is kt over m as usual divided by, divided by gamma bar of omega minus i omega. All you have to do is to set omega equal to 0 and divide by this quantity. So that immediately tells you this implies that tau velocity is 1 over gamma as indeed you can see that in the case of constant friction this will go to 1 over gamma. But again I repeat in the generalized Langevin equation this quantity is not an exponential decaying exponential something much more complicated. You can easily see what it will be. Instead of this e to the i omega t imagine this to be the Laplace transform. So certainly this is true e to the minus st of this is gamma bar of s where this is the Laplace transform plus s. Therefore this quantity is the inverse Laplace transform of this expression. Had this been a constant that is just e to the minus that constant times t but this is some complicated function of s we do not know what it is in general. So the correlation function can be very messy, very very messy indeed. You have to invert this Laplace transform that is not a rational function of s in general. Then the next one was to find the Fourier transforms of A of 0 B of t commutator equilibrium the Fourier transform of this fellow divided by the Fourier transform of the same thing but with the anticommutator. Now we know that this thing here was equal to the spectral function on the right hand side phi tilde of A B e to the minus i omega whatever it is. So the Fourier transform was essentially i h cross times the spectral function. The Fourier transform of this had the same thing except there was a cot hyperbolic become minus sign. So this ratio will become minus tan hyperbolic beta h cross that is all you have to do is to read off the Fourier transforms in each case and it is equal to this. Then the equilibrium mean square value of A squared is related to an integral of the spectral function according to according to what? Now you have a formula for A of t B of t prime or A of t prime B of t whatever it is expectation in terms of the spectral function a spectral representation. All you have to do in that is to put A equal to B and t equal to t prime right. So you end up with A squared equilibrium therefore cannot be independent on time. This is equal to i h cross I do not know there is a 2 pi there must be 2 pi somewhere over 2 pi integral minus infinity to infinity d omega phi tilde of A of omega 1 minus e to the beta h cross that is it yeah yeah you get a apparently different expression. So what happens if you use A of t so you have one expression for A of t prime B of t then you have another expression for B of t A of t prime then you have a third expression for the sum of the 2 which is really the full return. In every one of these except in the commutator where if you put A equal to B and t equal to t prime you get 0 equal to 0 except for that all of them will give you expressions there is only one expression right. So what is happening yeah they will all give you exactly the same if you simplify this. If you simplify this thing here this I pull out a half h cross omega here and then this becomes minus sign hyperbolic or whatever it is this side right. So they will all give you exactly the same expressions. One of them has an e to the beta h cross omega on top this one does not have it at all but you can convert one to the other but be little careful you also have that e to the i omega whatever it is right. So in doing this conversion you may have to change omega to minus omega so it is just a little piece of algebra but you should satisfy yourself that all of them will give exactly the same expression. I can write that in several ways then they came these questions on diffusion which I think could might have been a little tricky because we did not spend too much time on this when I discussed this topic. So the first question says to find for diffusion in 3 dimensions we have the Gaussian's fundamental solution of the diffusion equation and then you want to find the probability distribution of R squared which is also a random variable. Now that is straight forward because all you have to do is first you find the probability distribution of little r which is 4 pi r square times the fundamental distribution and then so let us suppose that you have p of r, t equal to 1 over 4 pi dt to the 3 halves e to the minus r square over 4 dt. From here you go to let me call it phi of r, t the distribution of r itself this is just equal to 4 pi r square it is this density you integrate over angles and you get the distribution pdf of the radial distance okay. Then you want to go to rho of r but we know that rho of r dr equal to phi rho of r and t if when little r is between r and r plus dr capital r is between r and capital dr capital r then this equation obtains in this case both are positive so we have no problem it is being mapped only once so if you plot it is a one to one map okay then therefore there is no complication about folding and things like that. So this immediately implies that this quantity so it says take this quantity write eliminate little r and write it in terms of capital r and put this Jacobian factor in which is equal to 1 over 2 root r should not forget that and then you get something like square root of capital r e to the minus capital r by 4 dt so that is the density that part is trivial. The next question is what is the Fokker Planck equation satisfied right and the last part of it says what is the Langevin equation satisfied by it. Now there you have to be careful to find the Fokker Planck equation is not very trivial what would you do? We will discuss the answer after I have gone through the answer books just to find out what people have done. What would you do for what? We know the Fokker Planck equation for this that is just the diffusion equation. What should you do to get the Fokker Planck equation for this? So this quantity is delta p over delta t equal to d del square p that is the Fokker Planck equation just the diffusion equation. What should I do to get the Fokker Planck equation for this? In this equation put p equal to 5 over 4 pi r square and then solve write it as an equation for 5. So that is the Fokker Planck equation for 5. What should you do for the Fokker Planck equation for this? Change variables once again do the same thing. Now use this use this relation here and then get you can do it in one step of course you can do both these in one shot right. Crucial point is each time you have to keep track of the Jacobians. This one is the phase phase factor and this one is the Jacobian factor you got to keep track of that. So you got the Fokker Planck equation for rho and then you got to use this correspondence backwards in order to get the Langevin equation for capital R. That will have a drift term and it will have multiplicative noise. So it is not the ordinary diffusion equation. Would I have just done this by starting with the Langevin equation for little r and then going to capital R? After all if I get some equation do not ask me how equal to something on the other. Can I simply say r equal to r square implies, can I put this in and get the Langevin equation for this? No. No. This is where the tricky part of white noise or the Wiener process comes in okay when you have multiplicative processes. But whereas going through this route of this correspondence between the Fokker Planck and the Langevin equation systematically is a shortcut to get the Langevin equation for capital R or little r for that matter which are non-trivial statements because this so-called white noise that we talked about is not a well-defined object mathematically. You have to use what is called a Wiener process and then you have to use Ito calculus which is somewhat different from normal calculus. So we will not do that but I will mention after I go through your answer books we will come back and I will mention but the right answer here is okay. So that this last portion would have been a little time consuming the rest I think was very straightforward etc okay. So we will start next time with critical phenomena.


Ready or not, good movies coming. Ooh, that's a hot mug, guys. Hey guys, this is my review for Ready or Not, a film that came out last year, which admittedly I didn't get to see, but I kinda really wish I did, because I heard a lot of good hype about this movie, but now having seen it, even with that pre-notion of how good it supposedly is, I dug it. And I agree with all of the positive criticism that this film has gotten. It is a fun movie, it's a smart movie, and it's a very well-paced film. The film follows this woman who's getting married into this family in a very kind of bizarre sort of ceremony. The whole thing feels really off. All of the family members are very weird, especially Hatchet and over on the side here. But then it comes to the game, which is a tradition in the family. Sometimes it's chess, sometimes it's checkers, but there is once in a while where a certain card is pulled, and that's when things kind of get a bit bloody. The game is hide and seek, and the family has to hunt down the bride, but she ain't no sissy, she ain't going down without a fight.  There's those moments in horror films where you'll be yelling at the screen, yelling at the characters, do this thing, don't do that, you dummy. She essentially anticipates it every time before it actually happens. Kind of like how Cloverfield 10 Lane would keep on subverting you, it would lead to the question and the answer that you were already saying earlier on, and it gives you the answer, and it gives you a completely different route to go. By the time you've figured out that, it will then deviate again because it's a thinking ahead of you. And this film does the same thing. It is thinking ahead of you. It is giving you a very smart, well-laid-out survival film. And it reminds me a lot of Your Next, which was an Adam Wingard film directed, I think in the early 2010s. Funny enough that people aren't talking about that movie because Ready or Not is essentially a better version of Your Next. Your Next is commendable. It's very well done. And actually, it follows a lot of the same ideologies as this film does. This film just has it going the whole time. There's never downtime. You're always in the character's corner and we're finding out more about the family, their history, and why they are the way they are. And you're also getting a lot of surprising laughs. There's a lot of very funny moments that happen throughout this film that kind of help with the gravitose of the situation. The thrilling aspect is great. There's some horror elements that are very well done. There's not a lot of jump scares. Thank God, in this kind of movie, it doesn't devalue itself for that. I actually enjoyed all of the characters in this film. Even the groom, who is the one who is bringing this bride, unfortunately, into this family game. And we see an arc with him, which admittedly doesn't end, and I think in the best of ways, it just kind of ends for the sake of the film ending. But otherwise, that's kind of my one detractor of it.  Either way, Ready or Not is a fun film. It subverts your expectations in a good way. It's smart enough to think ahead of you. Every time you think something should happen, it's gonna do it. This was a well-written script. It's a well-executed film. I'm very excited to see what these guys do next. I'm excited to see what this actress does next because she's essentially a different version of Margot Robbie. I even think she's Australian. But I really enjoyed this film. I had a fun time watching it. I was consistently engaged the whole time. It had me in its grasp and I was rooting for the character the whole time, which is definitely a sign of a good film. I also liked the visual aspect of it too. Everything from the set design to the costume design too. Apparently her dress is a combination of five different dresses. It wasn't just something they bought, they actually physically made it. I always appreciate you don't see that too much. In the end, I'm gonna give Ready or Not a six out of seven. I really like this movie. Anyways, guys, that's all for me. I hope you enjoyed this review. If you did, leave a like and if you're interested in more, subscribe. Otherwise, see you guys next time. Thanks for watching the video. My name is Nitz and you might remember me from the animated cult classic TV show, Undergrads. It's been a while, but I'm happy to say the click is finally getting back together in an all new movie thanks to a successful Kickstarter campaign. But we are still asking for your support. To see any and all updates about the upcoming Undergrads movie, be sure to check out and like the Bring Back Undergrads Facebook page. And with any luck, we'll see you guys soon.


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You know, Trump's comments in the Postal Service, I don't know if this is unprecedented, I'm not a historian, but I think it's the most open I've ever heard voter suppression in America discussed historically, except maybe the Reconstruction Period where white supremacists is the 1870s now, you know, maybe the 1870s. So this is really a throwback to the most open, you know, history of voter suppression, especially the black community, which has a long history. I mean, in the wake of the Civil War, the emergence of black voters was a pivot question for a lot of political issues, and it really was the key issue for moving the country in a more progressive direction after the Civil War, as opposed to towards the monopoly capitalism that it became.  So there's a long history of this. In recent years, it's become a much bigger issue. Republicans, you know, really since the Civil Rights Movement, but significantly for some time have, you know, really tried to use as a major part of their politics, the politics of racial grievance. The idea that for white working class and middle class people to try to say, well, all of the problems of the neoliberal era are really because there's a limited pie, and blacks have gotten more of the pie as opposed to they used to, and you're an immigrant now and you are failing because of that. And so in the past, really, I would say since about 1998, Republicans have developed an extremely sophisticated system of voter suppression that has been primarily based around purging people from the voting rolls, creating a range of different, you know, maneuvers that basically mean that millions and millions of people have been subject to having their names removed. A lot of times they don't know that this has really happened or don't fully understand it until they show up to vote. Then you have to fill out something called a provisional ballot. Provisional ballots often aren't even fully counted. So you know, there's a very large number of votes. Reg Pallas is a journalist who's done much work on this since the year 2000, you know, of potentially millions and millions of people who've been denied the right to vote. And so Trump's saying that he wants to prevent the postal service from getting additional funds that will allow it to deal with many things, including the onslaught of mail-in ballots because of COVID-19. Most elections will probably be very heavily via mail, but we're not set up for that in this country. So the postal service needs more infrastructure as do the states. And Trump openly saying he wants to do this defunding of the postal service to prevent mail to prevent people from voting. I mean, it really is that simple. He's claiming that the votes are off fraudulent, but obviously that's that's false. Larry Kudlow, who is the top economist of the White House, also said same day as Trump says, I'm going to stop the mail-in ballots. He says that voting rights is part of a liberal wish list of Democrats that Republicans have no interest in.  And obviously if those people, you know, will come out and vote against the feeling that there's some, you know, potential far right white supremacist institution, they have to be able to limit that in order to get what they want. So here we're seeing, you know, yet again the unfolding of this terrible situation. But I mean, it's really miraculous. I mean, people who don't follow the United States closely, I think for who, you know, see what's on Voice of America or CNN or whatever, it would be very surprised to know literally millions and millions of people have been disenfranchised. And many of the top architects of this are the biggest supporters of Trump in the country and many of them are deeply ensconced in his campaign already.


Hello Earth citizens. In this video, I will show you a 10-minute routine to do for a practice in brain education called bowing. Bowing is actually a very ancient practice that is done across many religions, many spiritual practices, as a form of releasing the ego, as a form of surrender, as a form of humility. For example, Buddhist monks practice bowing to overcome their ego, to become connected to their pure essence. So in brain education, we also practice bowing as a form of going inside, as a form of meditation, to connect with my soul, to connect with the essence of my life, beyond my thoughts, beyond my emotions, beyond all the surface things that may be clouding my connection to my deep true self. So I will share with you the technique of bowing. I hope that you can use bowing to meditate. A lot of the times people think meditation is just about sitting still in a dark lit room and chanting om, but that is not true at all. You can use your body to move and in your movement, you can practice meditation by connecting deeply inside. So let me show you the basics of bowing. I'll show you from a side view because it's a little bit easier to understand from the side angle rather than a front angle. So if you can, please have a mat. It's best to do this on a mat. If you don't have a mat, you can lay out a bath towel, something where you can create some sacred space for you to do this bowing practice. Now, if you don't have a yoga mat, if you don't have a bath towel, then at least do it on carpet. You don't want to do this on hard floor because it will hurt your knees. Okay?  Feel your breathing.  This energy pillar comes down your spine into your sixth chakra, fifth chakra in your throat, fourth chakra in your chest, third chakra in your solar plexus, second chakra in your lower abdomen beneath your navel and first chakra, your tailbone. Now this energy pillar comes down from your tailbone, to your legs, all the way down to the bottom of your feet. Imagine from the bottom of your feet, roots come down, making you ground it and plant it into the earth. Can you feel this energy line? Stay connected to this energy line and as you stay connected now, gently, just ever so gently so you don't break this focus, gently open your eyes slightly just to see my movement. Stay connected with this energy line in your body.  If you cannot, just go all the way down as much as you can and now lift your palms up, as if you are receiving Heaven's blessing. Now come back, pull your spine one vertebrae at a time. That is one bow.  So the cycle of bowing is a cycle of letting go, being reborn, letting go and being reborn and by repeating this process you are purifying your energy, making your body, mind and spirit light and fresh. So we will do ten bows together. So since we did one, let's do nine more. I will count. Please try to follow my count. If you cannot, it's okay. Just do it at your own pace and it doesn't have to be perfect. Just try to get the movement of coming down and coming back up while feeling your body, staying connected to this energy line. Okay, so we did one. Let's do nine more. Number two, palms together, come down, feel the energy line, bend forward, come to your knees, forehead to the floor, palms up, receive Heaven's blessing, come back up, straighten your legs, slowly roll your spine up, keep feeling the energy line at the center of your body. That's very important. Number three, palms together, bring them down, bend your upper body, down to your knees, forehead to the floor, palms up, receive Heaven's blessing, bend to your palms, come back up, slowly roll your spine up, one vertebrae at a time, feel the energy line. That's number three. This is number four. Stay connected to the energy line.  Number six, as you repeat this movement, try to feel the flow of the movements. Can you feel your mind becoming quieter? Number seven, you feel the artistry, the poeticness of the movements. This act of humility, this act of sincerity to purify my body, mind, and spirits. Number eight, number nine, last one, number ten. Great job. Slowly open your eyes, feel your body. What do you feel in your body now? Does your mind feel a little bit quieter than before? Does your mind feel a little bit calmer than before? When you get deeply into this bowing routine and feel the rhythm of the movement up and down, up and down, it's a very powerful way for you to let go of what's making you heavy and then purify your energy just like cleaning the inside of your body. It's like cleaning your energy. So please don't think that bowing is you're bowing to any kind of God or bowing to an idol or bowing to something else outside of yourself.  So in brain education, we recommend doing at least 21 bells, 21 bells. If you can, as a daily practice, we do 103 bells. After you do 103, you will have a huge difference in your body, mind, and spirit. So if you're up to the challenge, I recommend doing 21 first to start and then building up to 103. You did a great job. To experience a full one-hour brain education-based class, please visit a local body and brain yoga studio near you.


A panel of immigrants from all over the world came to speak at Central Lakes College about their experiences in America. Their presentation was a part of Central Lakes College's Cultural Thursdays. Each presenter spoke about how they came to America and how they hoped the presentation would change the way some people think about immigration. I hope that the message can get out into the positive message that we do have a lot to share. I think in our diversity among our populace is very similar to diversity in the environment and we can help take care of one another and be stronger through the multi-faceted approach rather than a single-minded approach. The next CLC Cultural Thursday is set to be held in February. If you've enjoyed this segment of Lakeland News, please consider making a tax-deductible contribution to Lakeland Public Television.


Good evening, friends of the creaking door. This is your host to welcome you into the inn of Sanctuary. Come in. Come on in. Well, let me tell you about the loot we picked up in our family. Mother got a new broom. Yeah, she needed it ever since she wrecked the old one in a forced landing on the roof last month. Ha, ha, ha, ha. Uncle William got a fur-lined stray jacket and dear darling little Aunt Matilda got a new hat pin. Yeah, the last one got stuck in the skull of her fourth husband. Ha, ha, ha, ha, ha, ha, ha. Turn your lights low and your radio up and remember that all the actors getting killed on this program are being well-paid for it. The winter wind was moaned around the Gable House like a wandering band of wailing women. As Caroline Giles puzzled and frightened by something she's seen in the gathering desk makes her way into the living room. This is young Mrs. Giles' first night at the family mansion built by her husband's grandfather. Only this morning the attractive bride arrived after a honeymoon trip in Europe. When she enters the huge, shallow-filled room she notices the servant girl at the window. She's a young girl, the only other person in the house and Caroline can see that she's scarcely able to control the terror she feels. Mary? Yes, ma'am. What's going on out there? Nothing. When will Mr. Giles be home? He should be here within an hour. Why are those cars stopping out there? Cars, Mrs. Giles? Really, Mary? You must see them as well as I do. Three of them stopped within a few minutes. Look, there's another one. Maybe that's Mr. Giles. No, the car isn't coming into the driveway. It's stopping outside. Who are these people, Mary? I don't know. Well, why should they stop outside? What did they want? I don't know. I don't know. Mary, you don't know. You don't know. There's no reason to come and step on it. Mrs. Giles, come away from the window. Why? You don't know them. Please come away from the window. Mary, why not? I warned you. The flooring walks through the window. Get down, Mrs. Giles. Get down on the floor. You'll be hurt. But, Mary, please. Mrs. Giles, just pray that they don't come in here. Just pray that they don't. Wait a minute, Mary. I hear sirens. It's the police. Oh, the police. Oh, there ain't any evidence. Don't make them go away. Mrs. Giles, I'm Richard Croyton, the mayor of this town. I'm sorry we had to come here under these circumstances. What do you do? This is John Dent, our local chief of police. Mrs. Giles, I am very sorry about what happened tonight. The crowd is gone now. I'll do my best to see that it never happens again. I've also come to offer what apologies I can. Must have been a frightful experience for you on your first day here. Our people may appear civilized, you know, but in many ways we're a very primitive community.  No, he didn't. But David told me about you, Mr. Croyton. He said you were the oldest friend he has. I believe you lived down the road? Yes, yes, that's right. How I happened to see those cars coming and I phoned Chief Dent. But why did those cars come, Mr. Croyton? Why? Well, yes. Why should those people who hardly know me do a thing like this? Well, didn't David tell you? Tell me what? Where is David now? He went to town. I expect him in a minute. Well, perhaps he'll tell you what you want to know when he returns. Why can't you tell me? I'm not sure that I care to. Mr. Dent, perhaps you'll be kind enough to explain this. Well, them people weren't after you, ma'am. They were after the man you married. Did? Good for? Dent, if Mr. Giles hasn't told, I don't think it's your place to do so. Well, the woman has her right to know.  Some call him even worse than that. What do you mean? Mrs. Giles, two years ago a series of killings began in this here county. Girls, young girls, four of them were found on lonely roads, stabbed to death. Homicidal maniac, that's what the state police officer said. And my husband? Well, your husband was arrested for them crimes. He was tried right here, but they let him go. We didn't have enough evidence. David, can't believe it. Why should they suspect him? He knew one of them girls, and he had been sent away, ma'am. Sent away? What are you talking about? Everyone knows there was something wrong with him. He had to go to an asylum for a while. That's quite enough, Dent. Mrs. Giles, I must tell you this. David was discharged from the rest home, completely cured. It was nothing more than a nervous breakdown. Dent, you have no right to frighten her like this. I'm not frightened. And I'm very grateful to Mr. Dent for telling me what he did. I just want to know one thing. My husband was acquitted. Why should they still believe he's guilty? Because since he left town six months ago, there ain't been none of these murders. Not a single one. Now that he's back, they'll start again. That's what you believe, Mr. Dent? Oh, no, ma'am, but that's what the folks are saying. Thank you. And now, if you don't mind, I'd like to be alone. Mrs. Giles, if there's anything I could know. No, no, thank you. There's nothing. Good night, Mr. Croydon. Good night. Come along, Dent. Good night, ma'am. Good night. Operator, will you give me the number of the railroad station, please? No. Never mind, operator. Don't bother. Are you going to leave, Karala? David, you heard. And I heard everything else, my very old friend Mr. Dent told you. I was behind the other door. Then why didn't you come in? Why ruin their fun? Besides, I was curious to see just what your reaction would be. David, why did they have to tell me? Why couldn't you? You think this was an easy thing to tell to a newly married bride? I would have found out sooner or later. I did plan to tell you, Carolyn, in due time. That's why I ordered Mary not to say anything. Darling, it isn't true. This is all an insane dream, a nightmare. No, Carolyn. It's not a dream, but in many ways it is a nightmare. Were you going to run away from me, Carolyn? Is that why you phoned the station? Yes. Carolyn, if you want to leave me, I can't blame you. I realize it may be asking a little too much of any woman to be my wife and to live here. Why did you come back here, David?  And I was hoping you'd help me. I thought maybe you loved me enough for that. Carolyn, I'm terribly sorry it must end this way. I'm not going to leave, David. Carolyn, I hope you understand what you're saying. You may be risking your life by remaining here. I love you, David. I'll risk it. I want you to be certain. Because in the end I may fail and be put to death. Yes, there are going to be other murders. Don't ask me how I know this, but I'm almost positive that sometime tonight a girl in or near this community will be killed, brutally stabbed to death. I'll have to go out tonight and see if I can prevent it. Do you still want to remain? Why are you telling me this? Because I want to know if you really love me or if you said what you did to placate a madman. To quiet a homicidal maniac who may still murder you. Are you going to stay, Carolyn? Yes, I'm going to stay, David. There was this child. Mary, now just can't calm down. I saw them. I saw them. Mary, try to put yourself together. Tell me what you saw. There was a man out there. A man moving around. Are you sure of it, Mary? I don't see anyone. Well, maybe you can't see him from here, but I saw him from the window of my room. All right, let's look out the window of your room. I don't want you to think I'm seeing things. I'm not the kind of girl who sees things. I didn't say you were, Mary, but I... Well, I don't notice anything unusual from your window either. You... you don't see anything? Just the lawn, the tree. Mary, I know we've all been under strain today. I guess we're a little jumpy, aren't we? I... I guess so. But I wish Mr. Giles were here. I... I don't like us two being alone. I know, Mary, but it can't be helped. Now come, dear, I think you'd better go back to bed. But I'm afraid to be alone in here. If I can get in through the window... Then we'll close the shutters. Help me, Mary. All right, then. There we are. The outside doors are all closed. Yes, I... I locked them myself, but still, this is a very old house. They can get in. Well, there's a lock on your door that works quite well. You can lock the door from the inside, and then you'll be as safe as you can possibly be. All right. I'll do that when you go. Very well, Mary. I'll go back to my room. I'm... I'm sure you'll sleep now. Good night. I... I hope so. Good night, Mrs. Giles. Did you lock the door, Mary? Yes, I... I just did. I'm turning out the light now. I'm sorry I troubled you, Mrs. Giles. Good night. Good night. Oh, she's right. I... God must be hearing things. Mrs. Giles! So, this is your host again. You ready to go on with our story of married blitz? I... I mean blitz. You know, if Caroline had only listened to this program, she'd have realized that it's always a mistake to marry a man who has a weakness for murdering the domestic help. It not only makes house cleaning difficult, but creates a terrible servant problem. But there is one compensation in marrying a homicidal maniac. When you'd call your mate crazy, you'd be so right. Well, let's get on with our story. As you remember, Caroline couldn't break into Mary's room, and now she rushes to the telephone and dials the operator. Please send someone right away. I think there's been a murder. Hang up that phone. What's the matter with you? What didn't she do as I told you? I don't want the police to know about this until we're ready. That... that room! You just came out of Mary's room. Yes, I saw her.  I'm all right now. How did you get into her room? I heard screams when I passed the car. The window to her room was open, and I climbed through. I saw her on the bed, and then I unlocked the door and came in here. You found the window open? But I locked it myself. I bolted the shutter. Well, it's open now. When did you bolt the window? A little while ago. The murderer must have been in the room hiding. But she locked herself in. I guess he escaped through the window. Yes. Perhaps that's what happened.  I explained that. Yes, I told you this. Carolyn, now listen to me. If I killed her, do you think I'd be talking to you now? Here, look. Son! Hmm. If I'd murdered Mary, the next step would be to kill you. No. And you see how easy it is for me to do it. Now, Carolyn, you believed in me before. I'm asking you to believe in me again. I don't want the police to know I left this house tonight. Tense like the rest of them. He hates me. He'd be glad to accuse me of this crime. Now, Carolyn, I want you to say I was home all night with you. And if I don't? I've got this gun, Carolyn. If you let me down, nothing much will matter then. That might be the police, Carolyn. I'm gonna open the door. Hey, Mr. Giles. Good evening, Dent. You got here quickly. Maybe that's because I was kind of expecting to come. Oh, have you been watching my house? I ain't saying. When I got that radio call, I didn't lose any time. Mrs. Giles, you phoned in something about a murder? Yes, sir. Our girl, Mary. She's a mayor. Mr. Giles, are you accusing me of this, too, Dent? It's mighty funny. Right here in your own home. This time, I can easily prove that I had nothing to do with it. My wife and I were together all night. Yes, sir. Is this right, Mrs. Giles? That's... that's right, Mr. Dent. Oh, David, Mrs. Come in. I understand the Dent questioned you again this morning. Yes, but he didn't get anywhere. I think even he's beginning to get the idea that I'm not a murderer. I see. Oh, Carolyn dear. Oh, hello there. David, why didn't you warn me your friend was coming tonight? Well, I phoned Richard only a few minutes ago. You see, Carolyn, I have to go out. Go out? Yeah. I didn't want to leave you alone after everything that happened. I... I... No, but... Well, couldn't we come with you, David? No. I'm afraid not. Why not? Well, it might be dangerous. David, where are you going? I'm sorry, dear, but I can't tell you. It won't be a long trip. It's only about 20 miles or so. That's all the information I can give you. Has... has there something to do with these murders? Perhaps. Now, David, I wouldn't advise you to take matters into your own hands. I'm afraid I have to, Richard. You see, the police haven't gotten very far, and my life is at stake. Your life? Isn't it obvious? The people are stirred up about these crimes. Who can tell when one of them may take it into his head to murder me? It's becoming a matter of the killer's life of my own. Furthermore, there's another reason why I must solve this. A reason I can't divulge right now. Why not? Darling, please don't ask me. I think that after tonight all this will be over. Bye, Richard. Goodbye, Carolyn. Richard? Yes? What did he mean by another reason? Well, Carolyn, if he's right about the murderer, then the next logical victim will be a girl, an attractive girl who's very close to him. So the people will suspect him even more. And the most likely person is yourself. Did he tell you how long he'd stay away, Richard? No, no, he didn't. Shall I deal another round? That must be David. All right, darling. Well, when will you be home? All right, I'll tell Richard. Goodbye. David's on his way home. He'll be here in an hour. Where did he call from? He called Waverly. He says he knows who the killer is. That's impossible. How do you know? Because if he knew, he'd never have made that call. I don't understand. It was a perfect signal to the murderer. Murderer? He told him that he was safe, that he had all the time he needs, a full hour. You... Please, I must warn you not to do anything foolish. It will only make matters worse for yourself. What do you have in your hand? A knife. You see, I simply press this button, and it snaps open. Try to control yourself, Karen. Don't try to fight the inevitable. Take my advice and submit to what will come. It will be less painful. You'll see how futile it is to struggle. I have the knife at your throat now. I have only to move my arm, and it's all over. Why don't you? Why are you waiting? There's no rush. For once, I have a little time. Time? Time seems to extend the experience. I must warn you not to scream. The moment you do it will all come to an end, suddenly, violently. Screams are dangerous. One can never tell who hears them. Take the knife away. No. Please. Fear seems to do something to your features, Karen. You look extraordinarily beautiful. And I must compliment you on your self-control. You still haven't even raised your voice. You can't get away with this, Richard. David will come home to find you. That's not quite how it will happen. David will come home and find you. I shall have the rare pleasure of seeing his reaction. And then it will be his turn.  Yes, it will be a new experience. I've never murdered a man before. If that possibly works, you'll be caught. I plan to be. But I shall plead self-defense. If the one will believe me because I'll tell them he murdered you. It's very difficult not to scream, isn't it, Carolyn? They won't believe you. They'll never believe you. Oh, yes, they will. My police commissioner hates David because he was freed after Dent arrested him. And his position also depends upon me. Oh, I admire you, Carolyn. None of the others held out this long. Then you must almost be insane with fear and terror, aren't you? And yet... I won't scream, Richard. I won't. In the end, you will. I won't. You must. You really are extraordinary, Carolyn. I won't. Almost all evening. I phoned from a neighbor's house to make him force his hand. I got back just in time. Still not? Yes. Richard. David, did you know? You couldn't have been that clever. I suspected you, Richard, but I wasn't certain till tonight. It was a trap? And Carolyn, the bait? All a trick? I'm afraid that's what it was, Richard. Why did you make them think it was me? Why? Because you were a giles. Oh, it's a little too clever. Never quite one of us. But you'll die for this, David. You'll be blamed for this and the other murders. I've left letters accusing you of the crimes. And Dent will believe them. Even over your wife's testimony. You'll die for this. Dude, kid, kid, you... David, is it true what he said? Pop the letters? Probably. Dent will never believe me. Good evening, Mr. Giles. Dent.  When I heard the shots, I came running in. So you killed him? Yes, but I had to. Dent, he was going to murder my wife. That's true. He's a homicidal maniac you've been looking for. You've got to believe that, Mr. Dent. Of course I believe it, ma'am. I've been suspecting him for a long time. Ever since he's spread them stories to folks about Mr. Giles. I heard what he said before he died. They've got nothing to fear now, Mrs. Giles. You and your husband can live here and we'll be proud to have you. And some folks will be mighty ashamed about what they've been saying and doing. And I reckon that better call in. Get them to take this...this man away.  As the sclera said to the walnut tree, you're still not as nutty as people. And that brings us to the moral of this story, which is taken from a retired witch who once said, Why slice your husband's head off? A little start to take the bed.  One has been brought to you through the facilities of the United States Armed Forces Radio Service, The Voice of Information and Education.