Okay, so we can start this morning session with Professor Los from Georgia Tech and his second lecture is entropy and the cac master equation. Thank you very much. Is the speakers are okay? Good.  So what was it? We start with an initial condition F0 in L1 of Sn minus 1 radius squared of n. Okay? So again, remind you this is the sphere in Rn. So it's an n minus 1 dimensional sphere and the radius is square root of n. You see, the way I talked about it last time, n is the energy of the system and we assume that the energy per particle is 1. So that's a reasonable number. Okay? Then what is the evolution of the master equation is the following equation, DTF equals n times qn minus the identity F and the initial condition F of V and 0 is F0 of V. Okay? And let me remind you V, I call V1 up to Vn. These are the velocities, huh? What is qn? qn was this operator here. By the way, can you read my handwriting? Okay, good. It's a sum of collision operators, if you like, over all pairs. So there are n choose two terms here, right, and I normalize it with n choose two. And the Rijs are given in the following way. You apply it to some function F. And what is it?  And these are just the post-collisional velocities. Let me remind you Vi star of theta, I mean, yeah, okay, let me write it down.  Okay? Now, rho of theta is a probability distribution. The best thing, it's not going to matter very much. Let's just assume that rho of theta is uniform, right? So that makes life easier. And also, it makes it easier for me to write things down. Okay. So, this is the model, right?  Remember, this depends on n. You could, right? You could let the n go to infinity. And when you assume that this is a chaotic sequence, that means that the F zero becomes roughly independent when n gets larger and larger, right? Then this kind of property of chaoticity is actually preserved under this time evolution, right? And we also know that the single particle marginal satisfies the nonlinear Boltzmann equation. So in other words, this ties this model, this Katz model, which is a model about collisions together with the Boltzmann equation, which is also a model about collisions. Katz model is in very high dimensions, the Boltzmann equation is in very low dimension, but nonlinear, okay? So that's the framework. Good. Now, one motivation of Katz to investigate this model was because you can study approach to equilibrium, okay? So let's talk about approach. So what is this? Now you notice this Qn, what it really is, it's an averaging operation, right? It averages things. So for example, when you stick an F to be the constant function, by the way, I should say the measure here, sigma n, is normalized. In other words, whenever I integrate over the sphere, I have to divide by the surface area of the sphere, right? That's my convention. And the advice is good because you see right away that the function one is a probability distribution on this space. You also notice when you stick in one in here and you average, you get one. So this operation preserves the function one, okay? In other words, when you stick it in here, you see when you stick in initial condition function one, this is zero, the function one is an equilibrium state. All right. So now you ask yourself, well, what are all the possible equilibrium states? And there's a little lemma.  I have to explain what this means. This equals the integral minus pi to pi d theta. Now here you put F of v1 vi star of theta j star of theta minus F of v1 up to vn squared, okay? So what is this? You notice by the way, I haven't told you yet, this distribution is of course a special property. First of all, it's constant, but it's also symmetric, right? Namely, rho of minus theta is equal to rho of theta. And what this means, it means that when you look at this Qn as an operator on L2 of this sphere of radius squared of n, again with the same measure, then this operator is self-adjoint, okay? And we call this microscopic reversibility, that's an important consideration if you want to use any kind of spectral estimates or anything of that sort. You see, as soon as you violate this condition, this thing is not self-adjoint, things get a little bit tricky and hairy, right? Okay, good. So now, what is this? This bracket here is just the inner product on my Hilbert space, right? Just the inner product on this L2 space. And what this is saying, so this is a simple computation, right? It's an extreme elementary computation. Namely, what you see here is that when you square this out, the F with itself, because this is a rotation, you can undo the rotation because the L2 space is invariant on the rotation. Then you get twice the integral of F squared minus the cross term, but the cross term, sorry, I forgot to sum on i and j, the cross term gives you just precisely the Qa, okay? So now what you see here, well, let me stick in the row theta for good measure, is this, if F is an equilibrium state. In other words, if this vanishes, then this side has to vanish too, which means that if this row of theta is halfway reasonable, I mean, what do I mean by halfway reasonable? For example, when you have your 0 and your theta, you would like to have a distribution like this, so you have lots of rotations. You see, if you put in a delta function at the origin, then of course you just have identically 0 and nothing happens. So now, make a little distribution. Now, what does it mean that this thing vanishes? This means that there's a ton of rotations in each plane for which this function is rotation invariant, which means, of course, that the function is constant, okay? That's a consequence, all right? So QnF equals F implies F equals 1, okay? All right. So now, what does this mean? Well, this means that there is approach to equilibrium. So maybe, let me leave this here, let me wipe this out. Maybe I'll do this over here, I wrote this here. So e to the n, Q minus i and converges to 1s. Well, how do we see that? You can go into the spectral representation of Qn. After all, we know it's a self-adjoint operator, right?  This was an argument which is sort of set in words by using spherical harmonics. Again, let me maybe repeat it. Since this operator is made up of rotations here, these rotations commute with the Laplacian on the sphere, the Laplace Beltrami. Therefore, the Q also commutes with the Laplace Beltrami. And therefore, you know that the subspaces consisting of spherical harmonics is an invariant subspace for Q, okay? And therefore, when you restrict your Q to each subspace, you have a finite dimensional problem. And therefore, you have this discrete spectrum, okay? That's easy. But you see, as I said last time, these subspaces, they can get large, right? Huge. So what do you know? Inside each subspace, the Laplacian is just the identity. The Q can be anything, okay? You don't know much. So now here, let me just look at the spectrum. So the first eigenvalue is a zero. We know zero is a simple eigenvalue. I just proved this, right? Because of the function one. And now let me order them in this way. And now what you can do is you can write down in L2 the following quantity minus the function one. L2, no, squared, right? So you evolve the function. You subtract the one. You ask yourself, does this go to zero? And the answer is yes. What you do is you write it out. Let me call the eigenfunctions uj. j equals zero to infinity. I can function. In fact, we know what these eigenfunctions are, right? They are spherical harmonics. But of course, I mean, it's not really saying much. Okay, so these are the eigenfunctions. And then you can expand this in terms of spectral representation. Now, you notice the eigenvalue that the eigenfunctions u zero is the constant one. This one doesn't show up because it's subtracted here. So therefore, you just have e to the minus lambda j t. j runs from one to infinity. And then you have, what is it, f in the product with uj. And now you notice that this is certainly lesser equals when you pull out the next one, the smallest one, which is not zero, right? And then what you left over is you just have to sum lift over. And that's of course nothing but the Fourier expansion of f minus one because I have a start at one. So that's what this is. And I forgot to do, put the two here because after all, I have the square here, right? So that's, so you see, nice. This goes to zero exponentially fast if you know what this lambda one is. The lambda one is positive. We know that, right? Okay. Good. So how big is it? So let me just check that it didn't cheat on anything so far. No. All right. So here comes the first conjecture of cuts. I think I can erase this lemma now. So cuts, so let me now call, okay, delta n, the spectral gap, one minus qn, okay? So in other words, this delta n was my lambda one, okay? But I call it delta n because I want to emphasize this thing can depend on n, okay? So let me write down the definition. Delta n is what? Well, it's the inner product of fn identity minus qn, f. You take the infimum over this quantity and you take the infimum over all normalized states in two and you want to make sure that f is not the function one. So you assume that this is perpendicular to the function one, okay? That's what it is. That's the gap. Yeah. Yeah, there's an n here. No, there's too many n's. No, thank you. And the norm is one, yes. Thank you. Okay? So that's the gap. And cuts, cuts as conjecture. That's, he did this in 1956. Delta n, okay. So let me draw a little picture here, right? So the story is this, here's zero. By the way, I forgot to say this operator q at one minus the identity is always greater than zero. It's a positive operator, right? That's easy to see. But this is always a positive now. Great or equal to zero, okay? It can never be negative, all right? So now you have here this delta n and so on and then you have other items, okay? And you see delta n is of course precisely the gap means what? There is no other eigenvalue in between. Okay? His conjecture was that this is greater or equal to some fixed number positive c independent of n. That's his conjecture. And what is interesting when you look into his paper, I mean he really wastes a lot of time and energy on proving this but he never succeeds. And then, so now you can of course say that the conjecture was proved in 1999. So you can actually have two takes on this. Either the conjecture is so difficult or the problem was so uninteresting, right? So it was proved in 1999 by Elise Jean Rez. Proved Jean Rez, okay? And the proof is essentially a version, uses a method which was invented by Hong Zoriao named the Martingale method, okay? That's, it's a complicated thing. Now when you think a little bit about it, in 2000 delta n equals one half n plus two over n minus one. So in other words, you can actually compute this explicitly. Now you see, just, I'm not going to prove it. It's actually not very hard but I will make some remarks. You see, it's fairly easy actually to figure out what the corresponding eigenfunction is. And the corresponding eigenfunction looks like this. It's a sum of vj fourth minus a constant which depends on n. And this constant is chosen in such a way that this function is orthogonal to the constant. It's an easy, I forgot the number. It doesn't really matter, okay? And this is a function which you can write down because you know more or less what this function is supposed to be on the level of the Boltzmann equation, right? So you write this down. And then you compute with this eigen, you compute the eigenvalue which, of course, it turns out it's an eigenfunction and so you compute the eigenvalue of this operator. You can do this explicitly. It's not a big deal. It's a little bit complicated, lots of indices. And what you get is this number. Of course, that doesn't tell you really whether this is the gap or not. And then you have to work, okay? And that's sort of the main part of the proof. It's usually quite easy to guess what these things are supposed to be. It's a little bit harder to make this, really show that this is a gap. In other words, there's nothing, no eigenvalue between zero and this number, right? But you see it's pretty convincing what turns out when n goes to infinity, this converges to a half. By the way, one half is precisely the eigenvalue which corresponds to the gap in the linearized Boltzmann equation, okay? The eigenfunction here when you take the first marginal corresponds precisely to the eigenfunction for the linearized Boltzmann equation, right? I mean you have this complete correspondence there, right? So I would say this is now in a very satisfactory way. I should also mention Maslin, David Maslin. He actually computed many, many more eigenvalues of this Katz operator using representation theory of permutations. Okay, so this is the story. Now, and by the way, I should also mention this of course is always assuming that rho is 1 over 2 pi. Because when rho is not 1 over 2 pi, you know, then things get much more complicated. All right. So now, I'm not going to prove this. And however, I would like to criticize this a little bit, the gap. You see, if you have a function in L1 on the sphere, let me just write L1, I mean otherwise, okay, make it clear. After all, it's a course. And let's assume that this function is chaotic. In other words, it's more or less a product of functions f of vj. It's approximate. And then you know, because it's a probability distribution that since the f is 1, and remember, the f should always be symmetric, I forgot to say. It's reasonable. Because this operator here respects symmetry and we do not distinguish the particles, right? Maybe I'll write this down here. F symmetric. Okay. So, now, you see, when you write this down, well, what you get? You get roughly the integral of f to the power m, right? And this, therefore, integral f should be 1. Now, let's look at the L2 norm. Integral f squared. Again, it's a little bit hand-waving, right? But I think it gets the idea across. Integral f squared, what is that? Well, it's more or less, morally, the integral of f squared to the power m.  Okay? So, in other words, what this is, it's a constant bigger than 1 to the power n. And imagine n, you have to think, is the number of the order 10 to the 26, right? Like Avogadro's number. So, these numbers are longers, right? So, what I'm saying is, when you look at the Katz conjecture, the L2 norm, the Hilbert space, L2, is not really a good way of looking at this problem. Why? Because the L2 norm initially is e c to the n c, a bigger number bigger than 1. And then you multiply by e to the minus delta n t. And this is roughly of the order 1 half. You have to wait astronomical times until this thing comes down to a reasonable quantity, right? Okay? This is not the way to look at this problem. Okay? So, therefore, we need a replacement. We need a replacement for this L2. So, therefore, let me get rid of the spectral gap. That's not the appropriate way. It's kind of embarrassing, right? You work hard to compute this gap because everybody tells you to do so and then finally you realize it's actually useless, right? Okay. But anyway, don't get discouraged by these things. At least you'll learn something, right? So, I did at least. So, let's see. What we have to look at is entropy. So, now, the relative entropy, I call it S of F. Really, I should write it in this way because it takes the entropy with respect to the equilibrium state, which is 1, is by definition the integral. Notice it's actually the negative entropy. It's F log F. I'm a little bit pedantic. I hope you don't mind, but I want to make sure that this is understood that this is a relative entropy as n minus 1 squared of n. Okay? That's what that is. Okay? Now, what's the advantage of a relative entropy or entropy over the gap? Because if you assume that your function F is roughly product function, okay? Then you see that here you get the integral of this product, but here this turns into a sum. So, you see right away that the sum is of the order of n, right? Or approximate product functions S of F out of the order of n times constant. So, in other words, it's extensive, right? So, the entropy per particle is actually over the 1, and now you can ask yourself, well, can you have approached equilibrium entropy? So, now, the point here, of course, is again, can we come up with a reasonable exponential decay in entropy with a rate which doesn't really depend very much on the number of particles? This is known as church and Janis conjecture. Let me make sure that I have not. Now, you have to tell me whether I screw up the church and Janis. Is this right? Okay. This was a conjecture which was originally formulated by Church and Janis about the Boltzmann equation. I mean, the full three-dimensional Boltzmann equation, but of course it carries over to this model, too. And the conjecture is the following S of F of t. So, what is F of t for me is just the evolution of F0 under this Katz evolution, right? I denote it by F of t. It's lesser equals e to the minus ct S of F0. You notice when F is equal to the constant function, of course, then you get 0, right? So, this is a reasonable conjecture. And c should be essentially independent. So, that's this conjecture, okay? So, now here we come exactly into the same situation as we learned from Laurent de Villette, right? We have to sort of discover what is called an entropy structure for our problem. We have an entropy structure. So, let's see to that. What do we need? Well, the first thing we need is that we have a unique equilibrium state, right? And we do. I just proved this before. We have this unique equilibrium state, okay? The next thing is that you need the dissipation of this entropy, right? So, in other words, what you compute is the derivative ds dt, which is d dt of this interval here. And now allow me that I forget about the one, right? So, and this depends on t. And when you do that, you get two terms, right? The one term here is n times q minus the identity f log f. That comes from the derivative here. And then you get the derivative from the other term.  And then, and then what you get is n times q minus the identity. Of course, the q always depends on n, right? And what you notice is I'm not so eloquent as Juan. How did you say this? Jack or something like this, right? Okay, good. So, you cancel this out. And you see now this term, of course, is zero, right? Because that's the mass preservation. You can also see differently, you can think of it as 1 n q minus the identity f. And since q is half a joint, you can push it over. And on the one function, you get zero, right? So this is just mass preservation. So this is equals to zero. And let me put the minus out. All right. So this, let me make it really clear. Okay. So, so this is what we call minus t of f to keep the notation of Laurent de Villette. Okay? So, and now what is the next question? The next question is you need a relation between the entropy production and the entropy. Okay? And let me write down a theorem. So this is the following definition. I call it gamma n. What do I do? I take the ratio of the entropy of this dissipation. I divide it by the entropy. And then I take the infimum over all f's, which, of course, are normalized, right? Oh, let me, let me say it out there. Put it out there. So d of f divided by s of f. And then you take the infimum over all f's whose integral is 1. That's called this gamma n. And this gamma n is actually what you call the entropy production, right? It's, it's how much entropy gets produced relative to the entropy you have. And this is a notion which was popularized by MacKean in 1965. And I should also mention Carlin and Carvalho, who were the one, I think, who made in the 90s, 1990s. They had a whole bunch of papers that made a huge amount of use of this entropy production idea. So, so what would you think? What one should have now is a lower bound and gamma n, which hopefully is independent of n? Okay, so let me start out with the first theorem, theorem, maybe I'll call it theorem one. And that's due to Cedric Villani, I think 2003. And he showed that gamma n is greater or equals 2 over n minus 1. And of course I always do this for rho equals 1 over 2 pi.  So problem is that this is actually more or less correct. Well, I wouldn't say sharp, but let me tell you what it is. Enough, when did you do it? I think this was, I think it was 2014 or something like, I forgot, I forgot, and it was my student, and I forgot, I mean, it's terrible. So he showed gamma, for any eta positive, there exists C eta so that gamma n less or equals C eta divided by n to the 1 plus C n independent of n. So in other words, I should say, what is it? No, minus. So in other words, what this, what this result says, Villani's estimate is essentially sharp. And I couldn't even imagine that even the constants are sharp. Okay, so nevertheless, I mean, you could say, well, this is a bad result or it's not useful. But what is, you see what it does, it points you to something. Namely, when you think about how Amit showed this, it's a complicated computation. It's very complicated. Because what you have to do is you have to find a reasonable trial state. And the way it does is you take 50% of the particles with very low energy and 50% of the particles have almost all the energy. And then you start letting them collide. And what these particles with high energy have to do is they have to hit these particles with low energy and start pumping them up so that they go towards an equilibrium, right? Okay? So now imagine when I talk about sort of the room in here, if you would take all these particles, all these gas particles and pack them in one corner and have a few of those left who bombard those particles with a lot of energy and hope that this really equilibrates within the reasonable amount of time. Time of order one, right? That's what you would like to have. I think this is very unlikely. That's my personal feeling. It will take a huge amount of time. And I think this points to the general problem about this area, namely, you see, these are phenomenological equations. You can write down lots of functions. But most of the functions which you write down don't make any physical sense. So what is it? Which functions do make physical sense? So, nevertheless, what I would like to do is I would like to give you a proof of Wilan's theorem because it involves some interesting technology and it raises a conjecture which actually Cedric made himself. Yeah. Just to understand, because you're mentioning this idea that half the particle with a lot of energy, but they're not localized. So this is homogenous in space, right? It's spatially uniform. Exactly. But still, you see what this means in this sphere, you have some very strange states which... Yeah, yeah, that's definitely because it has to be spatially homogenous with the particles somehow, half of them with a lot of energy. It's very, very unequally distributed on this sphere, right? And then you have to, what you simply showed, that you can actually push the gamma and very low, very small, okay? Makes sense. By the way, also let me mention something else. This is another interesting point which is related to the talk of Anton Arnold. You see, what does this telling you? It tells you that the entropy, when I draw a picture, so here's time, and here is the entropy S of t. What it says is that the entropy for very large N is very, very flat here, extremely flat. But you see, this doesn't really tell you that maybe after a certain amount of time say order one, that this thing suddenly can go like that. It doesn't tell you that. You would have to have other types of estimates, and it would love to see a function where you actually can see that the entropy decay is slow all the way to the end. I have never found an example. But the entropy production at the beginning, that you can make small. But it doesn't tell you what happens globally, right? So this is another problem which I have no clue how to deal with. Okay, good. So now let's start thinking about proving Villani's theorem. Make sure that I, yeah. So the first thing, let's do a little computation. And remember, I will always assume now that the rho of theta is equals uniform. It's 1 over 2 pi. Makes life just easier. Now let's take N equals 2. So what is the approach? I mean, I give you a proof which I worked out with Eric Carlin and Maria Cavallo, which is just based on induction. It's very elementary in a certain sense, and brings up some interesting mathematical issues afterwards. So let's start out with N equals 2. And then you know that df dt for two particles, what is it? It's twice qf minus the identity. But you see what is qf? f is a function which lives on the circle, right? Because it's two particles. And what does this q do? It just averages over the circle. Now when you average over the circle with a weight 1 over 2, with a weight 1, that's just the integral of f. But that's 1. So this is equals 2 minus twice f. And that's a simple differential equations which you can solve. And what is the solution? The solution is f of t is equals 1 minus e to the minus 2t plus e to the minus 2t0. It's completely elementary. I can just check. OK. Now you see, when you then look at the entropy at time t, you notice these two weights add up to 1. We know the entropy, the function. I don't have it here anymore. Let me write it down here again. The function f log f is a convex function. It looks like this. At 0, it's going to define 0 and here is 1. So that's x log. It's a convex function. So therefore, you can say, well, since this is a convex combination of the function 1 and f0, you can use Jensen's inequality and you see that this is 1 minus e to the minus 2t bounded below times s of 1 plus e to the minus 2t as f of 0. That's, of course, is equals to 0. So you learn that the entropy production, the dissipation, let me say this way. What am I doing? This is wrong. It's the other way. Inequality can only go one way. So this here is less or equals. This is 0. It's just a minus 2s f of 0 in the limit as t goes to 0, right? Because what you get on the right hand side when you subtract s of f0, you get e to the minus 2t minus 1. That gives you this kind of derivative here, right? Or put it differently, the gamma 2 is greater or equals. Well, what you have to do, you have to change the sign. So you get, and then you have to divide by s of f0. So you get 2t. That's completely elementary. All right. So now what you do, improving Wilhelm's theorem, how you pronounce Wilhelm? I have to Swiss German pronunciation. I hope you don't mind. So to prove this, you start doing an induction. Okay? Let me try to set this up. This one, I don't need it either. Yeah. So the first observation you have to make in this business, and that's also what we did with computing the gap, you have to use the symmetry. So when you look at qn, you can write this as 1 over n, the sum k equals 1 up to n, qn minus 1k. And I have to explain what this is. You see, what you do here, let me remind you, or have it here. This is the sum here, right? Now what is the qn minus 1k? I remove one particle, what I do, I remove the kth particle. I just remove it. Okay? So in other words, what you do now is, in the qnk, you have a sum i less than j between 1 and n minus 1, I mean, without this i k, the kth particle without k. And you have to divide by 1 over n minus 1 choose 2. Now you can easily check each term in here certainly shows up here, and to show up with the same weight. And you see right away that this is an average, and this is also an average, so this is a true equation, right? It's a little bit combinatorics. It's elementary. Okay. So now with that, you see, and this helps you to reduce the gamma n, that's the goal, to gamma n minus 1. But how do we do that? Let me write down df at the nth level. And that's n times the integral 1 minus qn. You see, I have to keep the n now, that's important, right? And that I can write as 1 over n minus 1 sum k equals 1 to n, the integral n minus 1, identity minus qn minus 1 of k, applied to f. By the way, so it's pretty clear that this is true. You see, this n minus 1, I stick in here, why? Because I want to have here the time evolution for one particle less. That's what this is. Therefore, I put an n minus 1 here, right? If you remember, nqn minus the identity, change the sign of course. So I put an n minus 1 here. And of course, I have to divide again. This n disappears against that, okay? Good. So now what you would like to do is you would like to use the induction assumptions about the entropy production for this operator. But you see there's a little bit of a problem. The particle vk is missing, okay? So now what you say, ah, no, that's very easy. What I do now is I split the integral into an integral d mu of vk times the integral of sn minus 2 squared of n minus vk squared d sigma n minus 2, n minus 1. And this is the uniform probability measure on that sphere, okay? This is of course simply what's left over in order to integrate things to get the total sphere. Namely, the sphere as n minus 1 over 8 squared of n, right? Okay? Good. And then you have what in here? n minus 1, ah, 1 minus qn minus 1k f log f, right? That's what you have. That's the way I think about it. And what's the problem? The problem is that this function f, when they restrict it to that sphere, is not a probability distribution. This integral is not 1, okay? You need that because otherwise the entropy estimates don't make any sense, right? So therefore what you do is you start normalizing this, okay? And how do I normalize it? So normalize as n minus 2 squared of n minus vk squared, okay? So how do I normalize it? So what you have to do is you simply integrate f over this sphere and divide, right? But there's a better way of thinking about it. You see, what is this integral over this sphere? You can think of it in the following way. I take the axis vk. Just, I think of it as the north pole axis, okay? And then I take my function and average it over all possible rotation which fixed that axis, okay? That gives you exactly the normalization, right? Because what it is is you integrate over all variables except the vk. You keep that fixed, okay? And this kind of, this kind of operation of averaging your function over all rotations which keep the vk axis fixed produces your projection. And I call this projection pk. So let's write this down here. Tkf equals, I'm not going to write down the formula, I could, but this is not, it's not the way to think about it. Equals the average f over all rotations of the sphere fix that keep the v axis, the vk axis, okay? So that's, that's, that's the normalization. That's the way I think about the normalization of f on this sphere. I'll put it differently. Gk equals f divided by pkf is normalized. And it's easy to see pk is a self-adjoint projection. Yeah, why is it a projection? Because what does this mean? The resulting function is a function which tends, depends only on the vk variable, right? And when you apply it once more nothing happens, right? Because it's already invariant on all rotations, okay? So this is, this is a good way of thinking about the normalization, okay? Good. So now let's go on and set up the induction. So let me just copy this down. I'm going to write, let me write down df again. n integral 1 minus, what did I say here? This one equals 1 up to n. Now I have here n minus 1, sorry, identity minus, okay? Next I can put here a gk times pkf, right? Because that's the definition of gk. Now you see, this is a function which depends only on vk what does this guy do? Well this guy only acts on functions which do not depend on vk, which are constant in vk. You see, this operator doesn't see that function. It's a constant. On this sphere, when I fix the vk squared it's a constant. So therefore I can take this function and push it over to here. Can you follow me? I think it's clear, right? So here's your g. Now here I don't have the g so I put it in and correct for it. So I divide by pkf but then I have to add it back on and what do I get? Let me write it out. It's kind of stupid but I do it anyway. k log pkf, right? Now look, this operator acts on this whole on this function here only. I have to really think of it this way. So what you're doing, you're taking this function and you integrate it against pk log pkf, right? That's the way I should think about it. Now look on the sphere, sn minus 2 squared of n minus vk squared. This operator here is self-adjoint. So therefore I can push it back onto. So let me write this in this way as if you like the inner product of d mu vk. Here I have pkf log pkf commutator n minus 1 1 minus q n minus 1 k gk. And this is the inner product on the sphere sn minus 2 square root of n minus vk squared, right? That's the way I can look at this. Now you see this is a function which on this sphere is just a constant. This function is just a constant. So therefore when you push this guy over, it's self-adjoint, it gets zero. So this whole term, this correction term is just not there, all right? So let me clean up the board. So this is therefore zero, zero. That's important, okay? So we get a fairly clean formula. Let me erase this now. We get a fairly clean formula, namely that this d of f can be written in this fashion. Now we can apply induction. So this we don't need anymore. So induction hypothesis, pf 1 over n minus 1 sum k equals 1 up to n. Integral pkf, this gadget is greater equals what? Well, I think of it again, sorry, as d mu of vk. I split the integral up. pkf, here I have the integral over sn minus 2 square root of n minus vk squared. And what do we have in here? By the induction hypothesis, this is greater equals gamma n minus 1 times the entropy of gk. That's the induction hypothesis. So this is gamma n minus 1 gk log. And now I can put things back again. What do I get? I get the gamma n minus 1, 1 over n minus 1 sum k equals 1 to n. And what is here? Well, here I have the integral again over my whole sphere. I put these things together. What is pkf times gk? That's my f. And now I have here a log gk, which are right as f divided by pk. That's what you do. It's fairly elementary so far. You see, if I would not have this guy here, this would be fantastical already, right? But I have this pkf down here. And that's sort of the killer in this business. What we get? We get that the dissipation is greater equals. Now let's see what we get.  That's your gamma n minus 1 n over n minus 1 s of f. That's clear, right? That's only the first term. But now you have this pkf, right? What are these? And it's a minus. You get a gamma n minus 1. You get a 1 over n minus 1. And then you have a sum k equals 1 up to n integral f log, okay? Now notice pkf is a function which depends only on vk. So now we integrate here over the whole sphere. That's the same as saying that I can integrate here over all rotations which keep the vk axis fixed. So I can actually for free stick in here. Okay, so let's stare at this. Maybe we should do this next time. What would be a reasonable estimate for that? A reasonable estimate, when you think of probability distributions, rn is the integral of f log f for the whole sum, okay? That would be nice. That would be absolutely wonderful. Because you see, if this were the case, if this here is less or equals to integral f log f, then of course you would be in fantastic shape because here you get an n minus 1 divided by that's a 1. That's gamma n minus 1. So you would get that gamma n is greater or equals gamma n minus 1.  It would be greater or equals 2. And you would be happy. It turns out if this integral is an integral over rn, it's true. I will prove this to you next time. This integral, however, is an integral over the sphere and then it's wrong. However, what is true is less than twice f log f. And that's the bad news. And that has to do with the sphere. That's a new kind of young's inequality on the sphere which I'm going to try to explain next time. And it's sharp. You cannot improve it, these two. And you see when you accept these two here, what you get, you get that df is greater or equals gamma n minus 1 times n minus 2 divided by n minus 1 times s of f. We'll put it differently. We learn that gamma n, my n is greater or equals 2, n minus 2 over n minus 1, gamma n minus 1. And you see, now you take this recursion with this initial condition, which is solved. It is extremely easy. You get precisely Villani's theorem. By the way, I should say Villani didn't have this. So in other words, this implies gamma n greater or equals 2 over n minus 1. I mean, Cedric didn't have this constant 2n minus 1. It just had a constant divided by n. But this is what you get. So let's talk about this. These are interesting new topics. Let's talk about this next time. It's also related to Braskam Liebbing equality, which I will try to explain, and it will be very helpful for afterwards when we go on. Okay? Thank you for attention, and I'm sorry. I was a little bit over time.


Welcome to our weekly roundup of blockchain news. I'm Ben. Throughout this video, we'll be telling you about everything that's been happening in a blockchain universe this week. Please, we ask questions about the videos, so don't be shy to let us know what you think in the comments below. The biggest story this week is, of course, the Segwit 2X. In the calm after the storm, Segwit 2X was found dead in the water. The controversial update to increase the size of Bitcoin's blocks that could have been led to Bitcoin forking itself has been put to rest by the creators themselves. Do you think that this was the right move? Alright, now let's talk about tomatoes. And in Italy, people have come under fire after an investigation revealed the exploitation of immigrant workers in the country's south, leading to debates over whether to implement blockchain technology. One side is arguing that blockchain technology will allow consumers to know the source and quality of their food products. And the opposition is arguing that blockchain technology can be limited in such a field, and sometimes you just don't want to know where it comes from. Okay, so our question of the week comes out of China. And it's this, does China's government have as tight a hold on the crypto market as people thought? The short answer is no, because people are people and people will always find a way to make money. The more spiritual Chinese investors seem to be following the ancient Confucian proverb, the man who moves a mountain begins by carrying the smallest stones away from China. It seems that, in spite of the government's hard stance against ICOs, Bitcoin, and the subsequent regulations imposed on the market, Chinese investors are finding ways to wriggle their way out of the government's back pocket. Investors are doing things as small as dropping the government-controlled Messenger app WeChat for a more secure, Russian-based telegram, as well as bigger changes, such as taking the crypto mining operations into less regulated areas. What is the future of Bitcoin in China? The United States Postal Service is one of the newest of what is sure to become one of the many companies and industries outside of the cryptoverse to embrace blockchain technology. Announcing a partnership with Bita, the consortium claimed that UPS will be implementing blockchain in systems used to track or monitor packages and facilitate payments between shipping parties. But this is only the beginning. The possibilities for blockchain are seemingly endless for UPS, and they could lead to improvements in efficiency and transparency in every aspect of the business, except the boxes themselves. Let's not make those transparent. It seems that UPS is one of the many companies stepping up to bring us greater transparency, even when it hurts. And now with a story out of Russia. Russia has announced plans to integrate blockchain technologies into its tourism sector within the next five to ten years. Russia, where its warm beaches and big smiles have attracted tourists for decades, want to use blockchain technology so that tourists will be able to connect straight to service providers without the need for third parties or intermediaries. They believe this will lead to more responsible practices by tourism agencies such as paddle boat rental services, beach umbrella vendors, and tiki torch manufacturers. Now we have just another reason to move to Cyprus. The SYCYSEC plans to integrate a DLT into its ELT after signing a partnership agreement with the BARC. What does this mean? The Cyprus Securities and Exchange Commission has announced that it will be integrating blockchain into its electronic payment system. It is their hope that this will lead to a more efficient, secure, and simplified financial industry. This is just another reason to go to Cyprus, especially if you have an ICO startup in mind. So should we be getting our financial advice from celebrities? Well, the SEC says that's your business. But they warn that celebrities like Paris Hilton and Jamie Foxx, who have been touting different ICOs, could be illegal if they're part of an undisclosed paid promotion. They stress that it's in your best interest to do your own research and make decisions based on hard facts over celebrity endorsements. And also, they'd like to remind folks that don't eat raw chicken, and if you plan to go out in a snowstorm, wear boots. Now we have Visa, who has decided to be the first major credit card to jump into the crypto pool. Starting with Singapore, card holders will now be able to use their Visa-backed cryptocurrency debit cards to make purchases at over 44 million locations worldwide. Parents of students at the Montessori schools in Flatiron and Soho recently asked the chairman, Marco Ciosa, if they could pay their children's tuition with Bitcoin, to which he replied, yes, and then very quickly added please. Though the school does not hold on to the bitcoins themselves, it has offered a more affordable, reliable, and efficient way for schools to process tuition payments, especially when you factor in the processing fees of most major credit cards. And out of the US, in a year rife with conflicts regarding transparency in US government, especially with the past election, blockchain technology might be on its way to save the day. The blockchain firm Consensus has partnered with the US Department of State to explore all the potential applications of blockchain technology. Do you think that this could be a new age of transparency in the US government? And really, do you want to know? And that's all the news for this week. Upcoming next week, we'll be bringing you daily news. Please don't forget to subscribe down below and leave us any comments with any suggestions or answers to any of the questions posed today. This being our first crack at a weekly news update, we want to know what kind of stories you want to hear about. Do you want to hear more about finance or do you want to hear more about the different advancements in blockchain and all different types of industries? Please let us know what was interesting to you and what you would like to see more of in the future. Thank you.


The recipient of the 2018 McLean Center Prize in clinical ethics is Dr. William Fagy, an epidemiologist and infectious disease physician. Dr. Fagy is best known for his work to eradicate smallpox in the 1960s and 1970s when he served as an epidemiologist for the CDC and later as the chief of the CDC's smallpox eradication program. Dr. Fagy developed and applied a new strategy for delivering smallpox vaccination, the strategy that would eventually rid the world of this dreaded disease. Because of limited time and supplies during a smallpox outbreak in rural Nigeria, Dr. Fagy and his colleagues created the surveillance containment strategy of vaccination, an alternative to the then prevailing strategy of mass vaccination. The containment strategy enabled health workers to vaccinate a much more limited population than did the mass immunization programs. The smallpox vaccine was given only to those who were in the radius of the outbreak or at its periphery. At the time, the population of eastern Nigeria was 12 million, but Dr. Fagy and his colleagues stopped an outbreak in the region by vaccinating only 750,000 people. It was a remarkable achievement. Before it was eradicated, it is estimated that 300 million people died of smallpox in the 20th century alone. Since 1980, there have been zero deaths from this terrible disease, and it has begun to fade into history. The eradication of smallpox from the entire world is surely one of the most stunning achievements in the history of modern medicine and public health. After the successful eradication of smallpox, Dr. Fagy went on to serve as the director of the CDC in Atlanta from 1977 to 1983. There he oversaw the response to major health problems in the U.S. and abroad, including the beginning of the HIV AIDS crisis. In 1984, Dr. Fagy became the head of the task force for child survival, formed by the Rockefeller Foundation, UNICEF, the World Health Organization and the World Bank. There he helped raise the global immunization levels from approximately 20 percent in 1984 to 80 percent by 1991. Today the task force reaches populations in 157 countries. Jim Grant, the head of UNICEF, called the vaccination program success the greatest peacetime achievement in history. From 1999 to 2011, Dr. Fagy worked as an advisor to the Bill and Melinda Gates Foundation, helping develop its childhood immunizations as part of its larger global health program. To learn more about the complex public health campaign to eradicate smallpox, I highly recommend Dr. Fagy's 2011 book, House on Fire, The Fight to Eradicate Smallpox. His most recent book published this year is Fears of the Rich, The Needs of the Poor, My Years at CDC. He was honored in 2001 with the Mary Laska Award for Public Service, in 2007 with the Jimmy and Roslyn Carter Award for Humanitarian Contributions to the Health of Humankind, and also received the Presidential Medal of Freedom from President Obama in 2012. Dr. Fagy's contributions to clinical medical ethics are extraordinary. He applied ethical standards to save lives, to eliminate health disparities, to give protective immunizations to many of the world's children, and to promote global health equity. As Dr. Fagy notes, the smallpox eradication program shows that humanity does not have to live in a world of plagues, conflict, and uncontrolled health risks. The coordinated action of a group of dedicated people can plan and bring about a better future. The fact of smallpox eradication remains a constant reminder that we should settle for nothing less. Today, Dr. Fagy will present his keynote lecture entitled Global Health, Ethical Challenges. Please join me in congratulating Dr. William Fagy on this extraordinary career and in welcome him and his family to the University of Chicago. Dr. Tolton. First, let me thank the Dean for that nice introduction, and second, let me apologize to the audience because it's all downhill from here on. What a golden standard program you have here. I was wondering yesterday and today, what would happen to American medicine if we had a McLean Center in every medical center, and what would happen if Mark was asked to train all of those directors of those programs? It's really very, very overwhelming. I'm impressed with so many of the things that I've heard over the last day and a half, stimulating things. Creativity, we know from the studies that creativity usually involves two different fields of knowledge coming together that were not put together before, and I kept thinking of that as I heard about surgery and prevention. Those two words are not heard very often, they were heard yesterday. I thought about that with clinical medicine and philosophy. I thought about that with every discipline plus ethics, and when I heard they talk on violence this morning, I realized how violence is one of the things that over the centuries has detracted from happiness. Violence combined with depression, addiction, fatalism, poverty have been problems forever. So many good points, when Valerie talked about the informed consent yesterday and the fact that we have to think about how do we protect patients as well as physicians. In the area of immunization we spent so much time on informed consent, but it eventually led to a program with a surge charge on vaccines put into a program run by the government so that people who have adverse reactions from immunizations now have recourse to be able to get compensation. When I heard the malaria program I was reminded of the fact that malaria is the first disease that we've ever had a national surveillance program for in this country, 1950 and 1951. The second one happened to be in 1955 because of the cutter incident with polio vaccine. The third in 1957 with influenza and now we have dozens and dozens of national surveillance programs. Yesterday when I heard the question of why do we underfund WHO and then expect them to do more, I thought back to when WHO was formed about 70 years ago and the role of the United States and other countries in making that organization such a difficult organization to run. We insisted on strong regional offices because we were trying to protect the Pan American Health Organization. We made them so strong that regional directors could always undermine the director in Geneva and that continues to happen so it makes it very difficult to have a global program. We insisted on having a board of governors, all the ministers of health in the world, 195 ministers of health on the board, many of them in place for only two or three years at a time. They're not really invested in WHO but they love going to Geneva every May for the World Health Assembly. And then third every year we asked them to reduce their budget. There were some years ago when the U.S. was not providing its fair share, its dues and I should tell you that we save more money each year in this country because of smallpox eradication than our dues to WHO and still we were not paying our dues. I wrote an editorial for the American Public Health Association and I quoted Dolly Parton. She once said you'd be surprised how much it costs to look this cheap. In 2014 Ebola in West Africa showed us how much it costs to look that cheap. Ethics is not abstract. It should drive us to solutions to benefit our ultimate bosses. Who are our ultimate bosses? One who will be born in the future because they have given their proxy to us and they're depending on us to make the right decisions. Will Durant said immortality in his view was the absorption of one's soul in deathless acts. You're all involved in deathless acts and I thank you for that.


Letton Tea and Letton Soup present Inner Sanctum Mysteries, starring Boris Karloff. The Inner Sanctum? This is your host. Welcome again through the squeaking door to another session of mystery, murder and madness. Oh, excuse me if I don't get up but I'm all worn out. Yes, I've had a hectic few days with an old friend who just blew into town. He's one of those earnest souls who insists on doing everything for himself. Consultations with the monument makers, the grave diggers, fittings of the coffin maker. Yes, quite a busy body. But then we only die once, you know. Ah, these friends of yours, they're such unhappy people. They never seem to enjoy life. Never seem interested in any of the quiet, peaceful, good things of life. For instance, what's the use of telling one of your spooky characters about Lipton Tea? They wouldn't like it. But other people enjoy that brisk Lipton flavor. They settle back in an easy chair and say to themselves, hmm, Lipton certainly has a rich, hearty flavor. Never the least bit wishy-washy. No sirree. But would a ghost appreciate Lipton's? Indeed he would not.  Or else it wouldn't be the world's largest selling brand of tea. Mary, you've been very hard on my friends. Very. And they won't like it. But then most live folks don't enjoy being scared to death. And that's just what's going to happen to you tonight. Our story is called The Corridor of Doom. It's an original radio play written by Robert Newman. And our star tonight is a man who gives even me the shakes. The famous star of stage, screen and radio, Boris Karloff.  Not the fact that it's inevitable that it must come to all of us someday, but rather how it will come. Do you think of it as a sleep and a waking, of a sudden transition from one state of being to another, or to a state of non-being? John Clay was one of those people who never thought about it at all, until he found himself walking down that dim and endless passage which, but suppose we let Boris Karloff in the role of John Clay tell you about it himself.  When I woke up, I had no idea of where I was or how I'd gotten there.  There was a dull pain in my abdomen. Touching it tentatively, I felt a bandage. So that was it. An operation, but for what? And where was I? At that moment, the door opened. And she came in the good afternoon. Or is it evening? Whichever you prefer. It doesn't matter. My name's Clay, John Clay. Yes. And yours? You can call me Nadar. Exactly. Where am I? In what hospital? It has no name. What?  I have to speak to Dr. Rogers. If you'll get him for me, please. There is no Dr. Rogers. At least not here. Then who operated on me? And for what? Listen, I'm not a well man. I have a very bad heart. Will you get someone who can talk to me? If you wish. I'll call Dr. Stone. Chill crept through my bones. It wasn't cold. It was fear. Unreasoning and abysmal fear. The door opened again. And there stood a heavy set man. His hair flaked with gray. And with him, my son-in-law, Alex Bartlett. Alex, I can't tell you how glad I am to see you. Hello, Father. Why are you standing out there? Why don't you come in? I shouldn't advise it, Mr. Clay. And why not? And why? Was it you who operated on me? Yes. I'm Dr. Stone. Why wasn't Dr. Rogers called in? He's taken care of me for years. There wasn't time. It happened during the night. Acute appendicitis. And even as it was. Even as it was? What? And why are you dressed that way, Alex? All in black. Well, it's customary. After all, you are my father-in-law. Of course I am, but... Now, look, Alex. You've got to stop being so mysterious. You know about my heart, what any sudden shock will do. I don't think you need to worry about it any more, Mr. Clay. And as far as the mystery is concerned, this initial period of adjustment is always a little difficult. Difficult? Do you realize what it's like lying here, helpless, completely isolated, as if I were all alone in the world? Or isn't there someone I can talk to? Some of the other patients? Not just yet. When the time comes, you'll meet them. But look, Doctor, I can't stand much more of this. I can't. If I don't find someone who really cares about me, who treat me like a normal human being... My dog. How about my dog? What do you think, Doctor? Yes, that's possible. We'll see what we can do, Mr. Clay. Come along, Bartlett. Yes, Father?  I don't know. I'll try, but it's difficult. Very difficult. Then don't go, Alec. Don't leave me here all alone. Come back. Come back! I waited and watched. Watched and waited. Then the door opened and there was the Doctor again.  My dog! You brought my dog. All right, Martin. Give it to him. Yes, sir. Come on, carry.   Wake up. Why, what's the matter? Carry! He's not asleep. He's dead. You wanted him, Mr. Clay, but... but why didn't you tell me? When did he die? How old was he? Eleven and a half. Maybe twelve. Pretty old for a dog. That's probably why he could come. What do you mean? What are you trying to do to me, don't you realize?  Easy, easy. I won't take it easy. I won't stay here another minute. I'm leaving right now. Oh, well, we'll see about that. You're getting yourself all upset for no reason, Mr. Clay, making it very difficult for the rest of us. Martin. You'd better let me have some of that bottle there. About 10 cc's. The red medicine? Yes. I don't want any medicine. I won't take it. Now, please, Mr. Clay. I won't, I tell you. No, I don't want the... Oh, it's awful. Salty. It tastes like... Yes, but I think you'll find that it will make things much easier for you. Very much easier. Yeah, you're doping me out. That's what you're doing. Putting me to sleep. You... You think that when I wake up, I'll forget about everything. Yes, Mr. Clay. You'll forget about everything. Everything. Somewhere deep down under the earth. It was a passageway, stone flagged and with stone walls, and I was walking slowly down it in my bare feet. I could feel the chill of the cold stones through the thick layer of dust. The passageway stretched ahead of me endlessly, and suddenly, I noticed that there were doors set into the walls on either side, closed doors, and on each door there was a name, Abel, Abercrombie, Abington. Where was I? What was this place? What was behind those awful ominously closed doors? Something seemed to be drawing me on, on down the terrible passageway, Addison, Agar, I could feel the cold creeping up my legs higher and higher, my heart pounding faster and faster, and suddenly I knew knew where I was and where I was going. Knew what was waiting for me there ahead of me down the passage. No, it's hurting all my will. I turn, cry to go back, with a worry in my ears. I was falling through the darkness. Falling, falling. In my eyes, I was in that cold, white room again, clutching the blankets with trembling hands. How do you feel now, Mr. Clay? You cried out, sir, as if... A dream, the most awful, horrible nightmare I ever had. A dream? The doctor will be very interested. Would you care to tell me all about it? Oh, I don't even want to think about it. It was about your former life? Former life? I shouldn't have said that I meant. Here, where are you going? You can't go, you can't, Martin, help me. Oh, don't you realize that if I do stay here, if I dream that dream again, listen, I was in a passageway, an endless, eternal passageway like a corridor of doom. It stretched on and on to infinity with doors, closed doors on either side, and on each one of the doors in alphabetical order, there was a name. A name of all those who had died since the beginning of time. I was walking down that corridor on my bare feet and... Why? Why are you looking at me that way? You mustn't talk about that. You mustn't, do you hear? But you ask. You didn't dream that dream, you couldn't have. And you've got to get it out of your mind. We will help you. We'll give you a massage. That should make you relax. The alcohol, Martin, right over there. A massage, you think that'll help? A stone. Try something else. I see. Now what? What are you staring at? Your feet. And the soles. Dust. Gray dust. Dust, like the dust in the passage. The corridor of doom. And that means... It wasn't a dream. It means... I was really there. Dirty feet on those nice, clean sheets. No wonder our friend the nurse seems so upset. Or was that the reason? Maybe she was just disappointed that he still hadn't told her about his operation. That always has them in stitches. Great, big stitches. Like the ones they take in a shroud. Mr. Host, I'm afraid I just can't believe this story. I can't believe that it really happened. You said so, Mary. You think Mr. Clay got that gray dust in his feet because he has feet of clay? There you go again. Always looking on the discouraging side of things. I really do believe I'd rather talk to cheerful folk like those nice young men and women who sang that new Lipton T song when I was at the studio. Sing a song of Lipton's. Yes, that's the way people ought to feel about tea.  Yes, brisk means that Lipton's is never wishy-washy. No, no, no, sirree, as they say in the song. So when you've had a busy afternoon or maybe when friends drop in for a little talk, pour yourself a cup of cheer with brisk flavored Lipton T. It's got a special flavor that always tastes like home. And it always tastes like more, too. Well, now I think it's about time to take another little walk. Yes, down the corridor of doom. Will our star Boris Karlov?  That's the nice part of a trip like that. You don't have to worry about coming back. I lay there staring down at my feet. No, it had not been a dream. But there on my feet was the thick, heavy dust from the corridor of doom I had been there, walking down its awful silence, not in my mind, but in the flesh. My heart clenched like a dicey fist that I threw the blankets aside, started to get up. Mr. Clay, what are you doing? Where are you going? Can I go?  You can't leave. Oh, let me go. I can click help me. Please, Mr. Clay. Oh, I haven't sleep. Let me go. Don't you realize what this means? If it wasn't a dream, and if I stay here, go down to that horrible place again. We've got to make it quiet down. Some more of that medicine, Martin. Another 10 C C. Right. Oh, no, no more of that. It's here, Mr. Clay. You must take it. You must. It will make you sleep. Sleep so soundly that you won't be able to go down there again. No. All right. Give it to me. You stay here, Martin. I'll go get Dr. Stone and tell him. Better, Mr. Clay. I don't know. Dark red taste. It's like, yes, I know. And he makes me sleep. My eyes get heavy. Have you been here for a long time, Martin? No, not long. What is it like outside of this room? It's strange. Like no place else. And the other patients. What are they like? They're strange, too. Listen, Martin, I'm a rich man. You're the only friend I've got here. You've got to help me. Whether you're rich or poor doesn't matter. As for helping you, that's what I'm here for. You've got to stay here. Watch me. Every minute. My heart. I don't think it'll stand like that. My first sensation was one of cold, numbing cold, creeping up my limbs. I reached for the blankets, couldn't find them. And I opened my eyes, and I was there again. Back there in that awful endless passage, that corridor of doom. The same stone wall, stone floor, covered with a thick layer of dust. The same doors with a name on each one on both sides of it. But now. Now I was up to the bees. That one there, Baba, next to it, Babbit, and then Backer. I tried to cry out, but I couldn't utter a sound. I tried to stop, check myself. My muscles didn't respond. Slowly, heavily, I continued walking on down that endless passage. Past Badger, Baffin, Bagley, past the bees and towards the seas. Towards a door with my name on it. My heart, pulse, pounding with terror. My breath, rust in my throat, convulsing me, I clutched at the walls, forced myself completely around. Then, as if I were fighting against a roaring gale, I drove myself back. One step I took, two, three, and I stumbled. I was falling again, falling through darkness, complete, absolute unending. Before I opened my eyes, I knew what I was. Back in my room, the sheets crumpled in my sweating hand. I lay there for a moment. I knew that this was my last chance. I couldn't decide knew what I was. Back in my room, the sheets crumpled in my sweating hand. Slipped out of bed, dipped over the door of the room, opened it a crack and peered out to hospital corridor. And sitting at a desk half way down at the nurse. Could I slip past her? Then, on a table next to the door, I saw the telephone. A telephone! Someone would save me, take me out of this place. Beating it up quietly, I dialed my daughter's number. Bartlett's wife. Oh, thank you. Hello? Jane, it's your father. Listen, you've got to help me. You've got to come and get me. I'm at the hospital. Alex knows where... Hello, is anyone there? Yes, can't you hear me? Didn't you hear what I said? It's your father. Jane, Jane! He heard her, but she couldn't hear me. Something wrong with the phone, her phone. I've got to get hold of somebody, somebody, but who? Dr. Rogers? Oh, I might be out. Come in while I'm phoning. Oh, I know, of course. Hello, police? This is John Clay of Riverside Road. I'm at the hospital. I don't know where, but... Can't you hear me, officer? For heaven's sake, listen. It's a matter of life and death. John Clay at a hospital. My son-in-law, Alec Bartlett, can tell you where. No! Officer, officer, listen, don't hang up. Don't! Officer, officer, hello! Anything the matter, Mr. Clay? Uh, just tone. Your telephone is... There's something wrong with it. No, Mr. Clay, there's nothing wrong. Not with the telephone. But I tried to make two calls. Two different numbers, and... I know. And you should have known. Nurse, all of them. You should have known what. Why couldn't they hear me when I could hear them? Why? Yes, Dr. Stone. We'll put Mr. Clay back into bed. I'm awfully sorry, sir. I only went out for a minute. Come on, Mr. Clay. No, no, leave me alone, please. Mr. Clay, please. You know that, don't you? Yes, I know. Doctor, I won't have to go back down there again, will I? Down to the collider? That's not up to me. All right, nurse. I think we're ready for another dose. The final one. Yes, doctor. No, Dr. Stone. No, not that red medicine. Not again. I'm sorry, but you've had a lot of time. All the time we can give you. No, I won't take it. You know what it means, doctor. I'll go back down there again to the corridor. It'll be to the letter C. To the place where my name is. If you won't take it by himself, perhaps you'd better help her, Martin. Yes, sir. No, no, no, I won't, Dr. Clay. Again, I knew what I was before I opened my eyes. I could feel the dust under my bare feet and through the dust the biting chill of the coals stones. I was there, back in the corridor, walking down its silent length past the blank closed doors. But the names of the doors now they were all seized. Cabot, cat and cahoon. On I walked, the beating of my heart, the pounding of my pulse loud in my ears, on down the passage. No longer even trying to fight against what I knew was in it. On past Cameron, Chelsea, Chisick and then, suddenly, terribly one door. A door with my name on it, gaping, waiting for me. I tried to stop to turn, but my legs carried me forward up as but two doors away. I could see into it now, see that it contained nothing, absolutely nothing, not even a coffin, just stone walls and a flat stone stab. I was turning, turning to step over the threshold. I made a last massive effort, but she won't have it to stop me. All right, Martin, pick them up. Yes, sir. Is it all over? Hello, Bartlett. In at the death, eh? I don't see if there's any pulse, of course, but I should think it is all over. It is stone, but not the way you think. Well, Clay, he's not dead. No thanks to you. All right, get them up, both of you. Here, Mr. Clay, let me help you. All right, Martin, I'll be fine from now on. But how? Don't look so surprised, Alec. Mr. Martin is a detective. I hired him some time ago. You see, I had a feeling that something was wrong when that railing broke accidentally and I took that bad fall, so I decided I should investigate. You can't prove it. You can't prove anything. The first results of Martin's investigation showed me what bad financial shape you were in, and it was then that I realized that you had actually been trying to murder me to get hold of my money, so I faked that story of having a very bad heart. You mean that... I thought it would give you the idea of trying something more subtle and less dangerous, and it certainly did. But you still can't prove anything, not a thing. No, don't you worry about that. Come back here, Bartlett. You'll never have a chance to prove it. Come back, Bartlett. You shot him, killed him. Well, I couldn't have. I fired up in the air to get him to stop. Come on. I don't understand it. I've got a mark on him. But he is dead. He was the one who had the bad heart. That's what gave me the idea of pretending. Good heavens, look. At what? This hallway was supposed to be the corridor of doom. When I reached the door with my name on it, I was supposed to die. Look. Look at the name on that door there. The one right next to him. Bartlett. His name. So what? Nothing, Martin. Nothing at all. Now, where do you think old Dr. Stern got the idea for that little alphabetical graveyard? That's right, for me. Huh? You don't believe me? Then come on home with me tonight, and I'll show you the one in the cellar of my house. What's more, I'll show you a door and a neat stone slab with your name on it. Nonsense, Mr. Host. Mr. Clay just explained that the whole thing was a hoax. And I'm not going to sit here and hear you say otherwise. Well, then don't sit. Stand up and we'll take a walk, Mary Bennett. Yes, back to your name. Back to the bees. Baker Bartlett Bennett. You don't scare me. Yeah? Well, how would you like it if we went to the elves and found a door marked Lipton? Oh, well, that's fine. Inside, we'd find a wonderful, friendly beverage, Lipton's, the tea with the brisk flavor, the tea that's welcome at all hours of the day. Yes, the largest selling brand of tea in the whole world, Lipton tea. And now a word of advice. If you should wake up tonight with a sudden chill, find yourself walking barefoot down a dusty stone corridor with doors on both sides of it. Don't get excited. Just insist on a doom with a view. By the way, this one's in a sanctum mystery novel is The Whistling Legs by Roman McDougal. Yes, and next week's inner sanctum story directed by Hyman Brown and brought to you by Lipton Tea and Lipton Soup. Next week's story is about women.  They'll love you for it. And who do you think is going to be their boyfriend? Hmm? That's right. Boris Karlov. Boris Karlov will be with us again next week because who else could love such women? Oh, if you're in a tender mood, tune in next Tuesday. Until then, good night. Pleasant dream? Looks, it's wonderful how quick and easy cooking can be these days. I guess some of you remember when it used to take half a day to make a pot of chicken noodle soup. But now we have Lipton's noodle soup mix. You might say Lipton's takes no time at all to prepare, and yet it has a fresh cooked chickeny taste, a real old-fashioned homemade flavor. Yes, and it's brimful of tender golden-aid noodles. Lipton's is grand for quick meals, and it's also a perfect beginning for the most elaborate dinner. So don't forget to serve Lipton's noodle soup. And don't forget to tune in next Tuesday night for another inner sanctum mystery. This is CBS, the Columbia Broadcasting System.


Hey, welcome to Drinks Made Easy. Today I'm gonna show you how to make the Blue Hawaiian, a Pina Colada derivative that is often confused for the Blue Hawaii, a daiquiri derivative. Now, Harry Yee, the originator of the Blue Hawaii, tried to jokingly address the confusion by saying that the only Blue Hawaiian that he was aware of was one that had drowned. Now on that note, let's make some drinks. To make your Blue Hawaiian, the first thing you're gonna do is you're gonna take one ounce of white rum and add it to a blender with about two cups of ice in it, add another ounce of blue curacao, one ounce of coconut cream or cocoa lopez, which is a brand name for coconut cream, and two ounces of pineapple. Cap it off and blend. Pour into your hurricane glass, a cherry, a straw, and an umbrella. And there you have the Blue Hawaiian. Hey, thanks so much for watching. And if you haven't already subscribed, go ahead and do so and hit the bell icon to be notified anytime we make a new video. And always remember to drink responsibly because everybody likes to make friends, but no one likes to make them in a prison shower.


Third category of pranayams are called tranquilizing pranayams. Tranquilizing practices of pranayams are designed to relax the body increasing while simultaneously increasing the pranic capacity and conscious awareness. These pranayams stimulate the parasympathetic nervous systems and draws the awareness within. Some bring more, some bring greater psychic sensitivity while others cool down the system. Tranquilizing techniques are usually practiced after Nadi Shodhan which balances the sympathetic and parasympathetic nervous system. Tranquilizing practices are also done through the both nostrils together and in some cases through the mouth as well. We are going to look at these practices in a short while. These practices should be avoided by the persons who are excessively introverted, over sensed to or psychically unbalanced as they may exerbate these. So, shithili, shithkari and these kind of pranayams should be avoided by excessively introverted or over sensed to or psychically unbalanced situations. Shithili and shithkari pranayams are these two examples of the tranquilizing pranayams. These two practices were designed to reduce the body heat and very effective during hot weather. The root word shith, not shith, it is not an English shith, it is shith that means cold while shithal means calm, passionless and unemotional. So, these practices cool both body and mind actually that is true with all the pranayam practices. Shithili and shithkari are cooling practices because the inhalation takes place through mouth rather than nose. When the breath is drawn in through the mouth, evaporation of the moisture on the tongue and the inner surface of the mouth cool the air. This cooled air then cools the blood vessels throughout the lungs which gradually eliminates excess blood heat. The technique and effects of shithili and shithkari are the same with only difference in the inhalation and we are going to look at it. First let us look at the technique of shithkari pranayam. Sit in comfortable meditative posture, close eyes, relax the body, hold teeth tightly together, separate lips, exposing teeth and then the tongue may be kept flat or folded against the soft palate in the khachiri mudra. We have not discussed khachiri mudra but that is a very powerful mudra and you can learn that online Jaggi Vasudevji or Swami Ramadevji. So, with the flat or folded palate, inhale slowly and deeply through the teeth. At the end of the inhalation, close the mouth, exhale slowly through the nose in the controlled manner. This is one round, we can have 11 rounds of practice and that can be increased to the 21 rounds. Shithili pranayam is slightly different where we need not to close our mouth by keeping the teeth lines touching each other. So, we can keep the mouth open, put the tongue outside of the mouth, it can be flat or it can be folded that is the variation in the shithili pranayam from the shithkari. Another pranayam which is very common and suddenly immensely benefited and suddenly immensely beneficial is a brahmari pranayam. Brahmar is the name for the bee. This practice is called because we imitate the same deep and low-pitched humming sound of the black bee. It is used in nathayog to awaken awareness of the inner psychic sound. Technique is that sit in comfortable meditation asanas with the hands resting on the knees in the gyan or chenamudra. Close the eyes, relax whole body, jaw should be relaxed with the lips gently closed and teeth slightly separated. This allows the sound vibration to be heard and felt more distinctly. Raise the arms sideways and bend the elbows bringing the hand to the ears. That is the first step towards the main important steps of the pranayam. Then use the index finger to close the flaps of the ears. Bring the awareness to the center of the head where agnya chakra is located and keep body absolutely still. Inhale through the nose while exhaling slowly in controlled manner produce a deep steady humming sound like that of the black bee. Be aware of the continuous humming sound within the head. The humming should be smooth even for the duration of the exhalation. The sound should be soft and mellow making the front of the skull reverberate. At the end of exhalation hands may remain in the unpriced position or may can be returned to the knees and then raised again for the next round. This is one round and 11 round practice is the with which we can start and it can be increased to 21 rounds. So these are some of the pranayams we discussed in this session. These are most commonly practiced pranayams. But there are multiple types of pranayams. There are pranayams which are practiced in conjunction with specific asanas, specific mudras. So there are there is a whole range of pranayams starting from very simple like conscious breathing to very complicated where you have to remember multiple steps and there is a particular protocol as it is taught in the Shakti Chalankriya by the Inner Engineering Foundation. So there can be multiple ways and different practices depending on our emotional, physiological and spiritual needs we can pick up appropriate practices. Our gurus, the mentors and the teachers of yoga must be respected and we must take up any practice under the guidance and particularly the practice of pranayam must be taken under the guidance of a trained yoga teacher or yoga master. And we can also gradually enhance the complexity of the practice. We need to be also conscious what is my core need? Is my need mostly a physiological, emotional, relational well-being or my needs are beyond these and also at the level of spiritual awakening and spiritual evolution. So accordingly we need to pick up the practice. This is the sense in this session we just try to give the overview of the pranayam and how this can be excellent intervention for the positive psychological outcome. So the final thoughts of pranayams can be summarized in the words of Stephen Parker. We must remember that focusing awareness on the breath activates all the integrative functions of middle, prefrontal cortex and that is the basic process in all kinds of pranayams. So that is a great benefit. Second aspect is that pranayam allows individuals to be more objective in the more objective space. Because awareness on the breath prepares our mind to remain more objective stage and that awareness brings the emotional awareness and we can control our emotions better. Yoga asserts that relaxing through the pause in the breath helps the mind to maintain its attention and concentration on a single object. Breath awareness and especially focused attention on exhalation phase of the breath activates parasympathetic relaxation response. The resultant release of oxytocin into the blood stream that elevates the mood problems and counters the stress as well as it promotes human bonding and healing. Many of you might know that oxytocin is also called cuddling hormone. It promotes human bonding and result into healing all of which contributes to our emotional health. With our anxiety in check our me, you or we maps as identified by the impersonal neurobiology are likely to become much more accurate. Our distinction based on ego can come down and our connection based on our shared experience of life and emotions can increase with the practice of pranayama.