It was near Halloween when that day scared me to the core. I was surfing online to kill time, but had the unusual idea to visit the place where no innocent should explore. The dark web. I opened up to her and got to work. An hour looking through the dark web, all that investigating and nothing scary or creepy. I hadn't ordered anything online that day, but I opened the door wide, then they left in a rush. After that visit, it put me on edge even more, leaving me with the box. So I did some digging on a creepy wax sealed box and many things came up. Most websites said it contained a demonic being sealed with a crystal inside to contain the demon's power in the box. After searching online, I put the box in a closet and locked it with a key which I keep around my neck. Although I kept the box sealed in the closet, I still had my suspicions because picture frames fell down, lights kept turning on and off, and the creepiest of them all, the closet door shook violently. I tried to do more research, but it didn't say much. I had to think of a better way to get rid of the box. I finally thought of a better idea for the box. I buried the box tonight, near a cemetery. I knew it was a bad idea, but I lived in an urban area, so where else should I have hidden it? Although these paranormal events stopped, I still saw that man looking through my window, but maybe my mind was playing tricks on me. Still on guard, so let's just say unusual things are still happening.


Let's review today's topic of chryscare analysis. We discussed categorical variable versus continuous variables and contingency tables. Now, let me ask you a few questions to see how well you're understanding these concepts. First of all, if I have five groups, let's say of different breeds of dogs, is this a categorical variable or continuous variable? Who can tell me the answer to this question and explain why? Thank you. Yes, that is the correct answer. You can have dogs that are terriers or a German shepherd or a colleague. These are groups of types of dogs and it's a categorical variable. Here's my next question. If I am studying how tall my students are and I measure them in centimeters, is there a categorical variable or a continuous variable and why? Yes, that's right. How does a continuous variable, it can take on any value between its minimum value and its maximum value. A student can be 152 centimeter or 200 centimeter or anything in between. Each centimeter is part of a continuum and it's continuous variable. Here is a final question. Let's say I created a task comprising 20 multiple choice questions and I'll demonstrate that task to 40 students. Then I counted the number of correct answers out of the 20 questions, like 18, 17 or some student got 15 out of the 20. Can I use a contingency table for data analysis in this situation? Yes, no, and why? The answer is yes. If you are counting the number of the right answers, you can certainly use a contingency table. But if you are calculating the percentage of the right answers like 98%, we cannot use a contingency table because you are treating it as a continuous variable. I know it's confusing, but there are different kinds of percentage. We're going to talk about the two different kinds of percentage next time. But for now, what we need to remember is that if you can count the number of categories, we can use a contingency table.


I would like to thank the organizers for inviting me to give these lectures. I will be talking about black holes and quantum information. And my talk will be centered around the black hole information paradox. So I will review the paradox. And I will explain that it is a fundamental conflict between general relativity and quantum mechanics. And I will also emphasize why this paradox is closely related to a somewhat different question, which is what happens to an observer crossing the horizon of a black hole. So at first sight, this seems to be two different questions. But they're actually closely related. I hope this will become clear during my lectures. And a point I want to emphasize already now is that this is an IR paradox. So it is a paradox that we can phrase within effect field theory. And in particular, while we expect that our understanding of space time may be modified by quantum gravity at very small scales, this paradox suggests that there may be modifications of space time even at macroscopic scales at a quantum level. In particular, as we discuss, it may have implication about notions such as locality over macroscopic scales. And by thinking about this paradox, we may be led towards identifying the fundamental principles of quantum gravity. Now, apart from the general comments, there is also a concrete technical question that I will try to address in my lectures, which is related to these questions. And this is the question of how do we describe the interior of a black hole in the context of AD safety correspondence? So as you probably know, AD safety has been very successful in explaining certain quantum aspects of black holes, for example, black hole entropy. However, this question of what happens behind the horizon of a black hole remains mysterious even today. And in particular, even if we had the complete solution of the quantum field theory in AD safety, so for example, if somebody gave you the exact correlators of the N equals 4 or the strong coupling, at the moment, we don't even know how to get that information and extract from it the question of what is the space time behind the horizon. So there's an important conceptual problem here, which is open and it's very interesting.  So it is an important question. Now, in my lectures, I will start by a review of the information paradox. Then I will explain how this connected the smoothness of the horizon. In my third lecture, I will try to formulate these problems in AD safety. And finally, I will discuss some proposals for resolutions and some other recent developments. In particular, I will talk about something called the mirror operators, the ERIPR proposal, and some recent results related to reversible wormholes, which can be used to address some of these questions. So my first two lectures will be a little bit more pedagogical and slow. And my last two lectures will be more technical. And I have decided to use a combination of slides and blackboard, but you can give me feedback. If you prefer that I switch off the slides and adjust to blackboard, we can do that. Just let me know after the first lecture. And also, I would like to encourage you to ask questions during my lectures. All right, so let's start. So we start by reviewing very quickly some basic classical aspects of black holes. So we will be mostly thinking about the Schwarzschild black hole, so a slightly symmetric uncharged black hole. But most of the results that I will be talking about can also be translated to black holes which include the charge rotation, as long as we are away from extremality. So for extremal or supersymmetric black holes, most of the statements I'm going to make have to be modified. And as you know, this solution has two special features, the singularity at r equals zero on the horizon at r is equal to 2gm, where we know that the horizon is a coordinate singularity. So in particular, if you calculate the curvature of a spacetime as a function of the radius r, you find this result. And if you evaluate the curvature of the spacetime at the horizon by taking small r equal to 2gm, you find that the curvature of the horizon is inversely proportional to a part of the mass. So for a very large black hole, the curvature of the horizon is very small. And thus, we expect that an infalling observer would not experience anything dramatic when crossing the horizon of a big black hole. Now, to make this more precise, we can consider a change of coordinates. For example, we can introduce crucial coordinates which allow us to extend the spacetime beyond the original coordinate range that we started with. And we get this diagram where we have the black hole interior, represented by this part of the spacetime.  Now, when we first learn about this diagram in general relativity, usually we say that the left region is some sort of mathematical idealization and the physical meaning of this region is not very clear. But nowadays, there is a better understanding which I will try to address in my later lectures.  So this is a black hole whose exterior is described by the Schwarzley metric for all times. But if you have a black hole from by-graphician collapse, at early times, the metric is going to be different. And in particular, you have to take into account of the geometry in the interior of the star which is from the black hole. So instead of working with this diagram, if we have a black hole from by-collapse, the diagram is truncated at some point, and this gray region represents the interior of the star. So the metric inside the star is not the Schwarzley metric. It depends on the details of the star, but the exterior is Schwarzley. And in particular, if you want to ask questions about the late-time questions, for example, what happens to an observer falling into the black hole at late times, you can see that we can approximate those questions by doing a calculation in the eternal black hole rather than the one which is from by-collapse. So I will also be using Penrose diagrams quite a lot. So just to remind you very briefly, these diagrams are a way of depicting the spacetime of mapping the infinity of space into a finite region on the blackboard. So we consider a conformal compacted scaling of the metric. You take the metric and you multiply it by an overall scalar factor which depends on the coordinates. And in this way, you do not modify the causal relationship between events. So two events which are time-like in the original metric are going to be time-like in the new metric, the rescaled metric. Those which were null will remain null and so on and so forth. So the causal structure of spacetime can be read off from this diagram, but at the same time, we're able to map the infinite region of flat space, the asymptotic region of flat space into a finite region. So these diagrams are very useful for understanding the causal relationship between events, but they do not have the information of the physical distance between points. So in particular, physical distances can be misleading when you look at the Penrose diagram. For example, if you look at this little corner here, it seems to be a finite region on the Penrose diagram, but actually represents an infinite region of flat space. So in these diagrams, we have, this is a two-dimensional diagram for a four-dimensional metric. So every point on this diagram, you should think of a two-dimensional sphere living over every point and the size of this sphere will depend on where you are on the diagram. In particular, as you move towards the bifurcation point, the sphere gets smaller and reaches the minimum size, exactly the bifurcation point, where the radius becomes equal to two GM, and then it starts growing again as you move towards the left. And this sphere, also as you can see from the metric, shrinks to zero on the singularity where the space time, well, becomes singular. In the case of a collapsing black hole, the Penrose diagram has this form. So here we see the star which collapses and forms a black hole. And this vertical line represents the origin of spherical coordinates. So the sphere shrinks to zero size, but in a smooth way. So this is a smooth region of space time. And in particular, light rays which move at 45 degrees on this diagram when they hit their line r equals zero, they get reflected in a smooth way. All right, so this is the classical story and of course things become interesting when we consider quantum mechanics. So I will start by reviewing for you the calculation of Hawking, which will be the starting point for us. So what Hawking considered is the classical geometry of a collapsing star, which a star collapses into a black hole, and he considered the quantization of the scalar field on the background, on this classical background. And what he found was that even if you start with a state with no particles for this field in the far past, you get a thermal flux of particles in the far future. And in particular, he found that this flux, the radiation that you get depends only on the mass of the black hole and not on the details of the star which form the black hole. So where this calculation is done is by working in a particular approximation of quantum field theory in curved space time which means that we, of course, we cannot do a calculation in full quantum gravity. We don't know the full theory of quantum gravity, so we want to work in an approximation, and that approximation is the one where we take the classical metric to be, the background metric to be classical and given, and then we take a quantum field on top of that metric and we try to quantize it. Now, the reason that this approximation is valid is because the back reaction of this quantum field or the classical metric is suppressed by a factor of this form where mp is the Planck mass and m is the mass of the black hole. So the back reaction of the quantum field on the classical metric is suppressed by this factor if you plug in the numbers for an astrophysical size black hole. This is incredibly small of the order of 10 to the minus 40 or so. So all gravitational interactions between these quantum particles are suppressed and in particular, we can keep the metric classical and fixed and then quantize the field, the fields on top of that metric. So we will want to consider, for simplicity, massless Klein-Gordon field and we write down the equation for this field and we try to quantize it using this metric as a background. Now, let me remind you that if you look at this metric, if you are outside the star, the metric is short-shelled. In the interior of the star, the metric is different. So we have to take it into account when we try to solve this equation. Now, if you want to quantize a field, this is a free field, so an easy way to quantize it is to expand it to a set of modes and one choice is to select the modes with represent particles coming in from minus, from the infinite past. So this is a massless field, so it describes massless particles which are coming from minus null infinity, from past null infinity and you can take the field and expand it into a set of modes where you have some wave functions and operators which play the role of creation and inhalation operators and these operators represent particles coming in from the past. So these operators obey the usual algebra that they're labeled by a frequency and angular momentum and they obey the usual oscillator algebra. So in this way, you can build up a focus space of particles representing excitations coming in from the past. Yeah, it's convenient to work in a base of plane waves. You don't have to do that, but it's, yes. You can work with plane waves or we can try to build up waves packets out of these plane waves. Yeah, this expansion is relevant for this Cauchy slice that I have drawn in blue on this diagram. So this blue line represents a Cauchy slice and you can expand the field in modes which are defined on this slice. However, we can imagine taking this slice and continuously deforming it towards the future up to the point where it becomes this purple line, part of which is along the horizon and the other part is on future null infinity. This is another Cauchy slice. So when we do quantum field theory in curved space time, there is not a unique way to expand the field in modes as you know, there are more than one ways and in particular, we can consider an expansion of the same quantum fields in modes which are relevant for this Cauchy slice. When we do that, we get an expansion which looks slightly different. So now we have two different sets of modes, the modes G and H, which this mode G which multiply the creation and inhalation of B represent particles going out in future null infinity while these modes represent particles falling into the horizon. So we have two different set of modes which describe particles falling into the black hole or flying out to infinity. And correspondingly, we have creation and inhalation operators B and C. Yeah, these arrows just represent particles going through this horizon or through this, or through, you know, future null infinity. We have the quantum field phi and we have expanded it into different bases. But of course, there should be a relationship between them because after all, it's one quantum field. So in principle, we should be able to rewrite, to find the change of basis between the modes F and the modes G and H. So I remind you, these are solutions of the wave equation on this background and this is a complete base of solutions. These two guys together is another complete basis. So in principle, you should be able to transform from one to the other. And if you find the transformation between the wave functions, F, G and H, you can also find the transformation of the operators A, B and C. Yes, so B and B dagger obey a similar algebra. So it's similar to this one. C and C dagger the same and B and C commute. Because by definition, we have taken the modes B to have support on future null infinity and the C's to have support on the horizon. So B and C commute. So we want to find that the transformation between these two basis of solutions and this will allow us to express the operators B, C in terms of A and A dagger. Now, as we said before, the modes in the past null infinity form a complete basis. So in particular, if you take this operator B in the far future, it should be possible to express it as a linear combination of the A's and A daggers. Now, here we have some coefficients, multiplying the modes. This indices i and j, you can think of them as denoting the frequencies omega m or perhaps you can go from wave packets which are labeled by some quantum number i and j. So in general, what this equation is telling you is that these modes B in the far future will be linear combinations of A's and what is perhaps a little bit surprising is that not only A's but also A daggers. So as a matter of principle, we will want to write down the most general transformation between these modes and these modes which will mix positive and negative frequencies. So in general, we will have the transformation of this form. And what Hawking found was that in this particular geometry, it turns out that both of these coefficients are known zero which in particular means that what is an annihilation operator for a mode in the far future is a linear combination of creation and annihilation operators of modes in the far past. Yeah, yeah. Well, these are... No, I mean, no, no, this surface, these surfaces are in the future of this one, right? But this one and that one are sort of space-like separators, yeah. So what Hawking did is that he actually calculated or he found a way of estimating these coefficients and he found that both of them are known zero. The important thing for us is that beta is known zero which means that if you define the vacuum of the quantum field as the state which has no particles in the far past, is how we define the vacuum at early times, and if you try to calculate now the expectation value of the number operator for particles in the far future, so b dagger b, if you use this equation and the fact that beta is not zero, you find that you get something which is known zero. So you get particles in the far future. So this is the idea of the calculation. And if you really want to do it in detail, what you need to do in principle is take the wave equation of this classical geometry, find the solution of the wave equation and relate the solutions at late times to solutions at early times, right? Yes, I'm sorry, I can't hear you very well. This solution, the solution, the modes F are defined to have a simple form at minus infinity, right? At minus null infinity. While these are defined, G and A's are defined in future null infinity and on the horizon. So it is one geometry which is however time dependent, right? It's a collapsing black hole. So it's a time dependent geometry. So the expansion at early times may be different from the expansion at late times. You do not put any boundary condition on the horizon in this calculation, right? This small CNC dagger go through the horizon. So you do not constrain them, right? You do not impose any condition on these guys. For example, if you start with a wave pocket in the far past, you can evolve it forward using the Klein Gordon equation and the result of this calculation will tell you which part of the wave falls into the horizon and which part flies out to infinity. You don't want to impose anything by hand. Yeah, yeah, yeah, good. So the last question was, do we impose any boundary conditions to the horizon? Like, do we impose the condition that I don't know what you had in mind, but I do not impose any boundary condition on the horizon in this calculation. Yes. I'm sorry, I have a cold and I cannot hear very well. Could you speak a bit louder? Yeah, yeah, okay, so the question was, part of the space time differs from the Schwarzschild geometry. The interior of the star is not Schwarzschild, right? The metric inside the star is not the Schwarzschild metric. So when you try to solve this wave equation, if you want to do it properly, you have to take that into account. But as we will see in a little bit, that does not have an effect, an important effect for the behavior of this coefficients which are relevant for the late-time radiation. So the radiation at late times after the black hole settles down. So the reason will probably become clear in the next slide. Yes. Well, that's not very important because what I want to, so the question was, how exactly do I define this, how do I select these basis of functions? The answer to this question is that it doesn't really matter because for this calculation, all I need to know is the relationship between B and A. I don't want to make any statement right now about the state of the quantum field on the horizon. I just want to find the relationship between B and A. So for that calculation, the choice of this base, the details of this basis is not important. Okay, so this is the idea and in principle, we have to solve this complicated wave equation problem, but that's really difficult. So I want to now give you some intuitive understanding of how Hawking was able to estimate the result without solving the wave equation. And let me, let me introduce this coordinate R star, which is called the tortoise coordinate. This is a coordinate which allows us to write the short schematic in this form. So we introduce a new radial coordinate R star in such a way that the coefficients in front of the T squared and the new coordinate, the R star squared are the same because in this new coordinate system, we can immediately write down the equation obeyed by null geodesic, but radial null geodesics. So in this coordinate system, in falling light rays obey T plus R star is constant and outgoing light rays T minus R star constant. And by looking at this equation, this form of the metric and comparing it to the short solution that we started with, you can actually calculate what is R star in terms of R and you find that R star is equal to R plus 2gm log R minus 2gm over 2gm. So this new coordinate R star, if you go very far from the black hole, this is the dominant term. The logarithm is also important. So very far away from the black hole, R star is more or less the same thing as R. But when you go near the horizon, you notice that this logarithm blows up, it goes to minus infinity. So near the horizon, R star goes to minus infinity. Okay, so we introduce this coordinate and then because of this property that in falling light rays obey very simple equation, we introduce light-con coordinates, u and v. So this is v and this is u. So light rays which come from past null infinity can be labeled by the parameter v and light rays in the far future can be labeled by the parameter u. Now, as you can see from this diagram, there's a very special light ray in the far past which is at a particular value of v called v naught which has the property that as it goes in and when it gets reflected at time equals zero, it actually becomes the horizon. So this is a very special light ray. Now, any light ray which has a smaller value of v, so any light ray that was emitted earlier than v naught is going to reach R equals zero before the horizon is formed and it's going to fly out to infinity. While any light ray which comes in with a value of v larger than v naught is going to reach R equals zero after the horizon has formed and will follow the singularity. In this way, we can see that these particles that we get as hocking radiation at very late times if we're sitting there are these particles which came from this special point v naught, very near this point. In particular, remember that two slides ago, I told you that the distances in this penrose diagram are somewhat misleading. So this corner here represents an infinite region of large space even though it's mapped into this finite region and what we see from this diagram is that all of the light rays which will ever get out of the black hole even at very, very late times, they all emerge from this very small region near v equals to v naught. So in that sense, this very small region in the parameter v is magnified by an enormous factor by the black hole and is mapped into this infinite region in the u parameter. So the black hole acts as a microscope which magnifies the structure of the moles in the uv and maps them into moles of long wavelength in the far future. The reason that this happens can be understood by the fact that there's a very large gravitational redshift for these light rays which travel very close to horizon. So the idea is the following.  So there was no time dependence, just a star. When a light ray comes from infinity and falls towards the star, it gets blue shifted because the star is pulling the light ray so it goes down the gravitational potential and it gets blue shifted. It reaches r equals zero and gets reflected and as it goes out, it gets red shifted. Now if the star is static and time independent, these two effects are symmetric equivalent so there is no net blue shift or red shift as the light ray goes through the star. But here we have a time dependent geometry which means that in the beginning these light rays approach the star and they get blue shifted. But when they try to get out, the star has collapsed a little bit so the gravitational field is stronger now and these light rays try to escape but the red shift is more important than the blue shift that they underwent. So there's a net red shift for the light rays that go through this collapsing geometry. So if you start with a light ray of some particular frequency, what you will get out is a light ray with a lower frequency. And in fact, you can think of this special light ray V naught as a light ray which suffers an infinite amount of red shift and thus is unable to get out to infinity. So these light rays which are very close to V naught have the property that they undergo a very significant red shift which means that if you want to study modes which have frequency omega in the far future and if you trace these modes back to where they came from you will find that they came from modes with a much higher frequency. So we start with modes with very high frequency they propagate the star, they get red shifted and they merge on the other side as modes of low frequency. These are the Hawking particles. The Hawking particles have relatively low frequencies but if you trace them back to see where they came from they came from modes of the quantum field which had extremely high energy. Now, this is very useful for us because if you have the modes of the quantum field of very high frequency it means that they have very short wavelengths and then you want to study the propagation of modes with short wavelengths on a time dependent geometry. But as you can imagine if the wavelengths of the particles is very small we can use an approximation to study the propagation of those waves which is called geometric optics approximation. So that is what we do with light. If you want to study the propagation of light in a geometry whose typical size is much larger than the wavelength you can think of electromagnetic waves as light rays and they just move on null geodesics and then you can calculate how they propagate without having to solve the wave equation which is much harder. So because these modes start off as modes of very, very high frequency we can use a geometric optics approximation to study how they propagate to the star and then the only thing we need to know is the relationship of the equation for null geodesics following these trajectories which you can determine very easily and the important thing is that you get a relationship of this form. You get that at late times you can calculate the parameter u of a geodesic in terms of the parameter v for the geodesic when it started at early times and you find that it's given by this equation just by solving the geodesic equation in this diagram and the important thing is that we get this logarithm which blows up as v goes to v naught. So this equation makes sense only for the light rays which have v less than v naught because these are the light rays that are able to escape to infinity. These light rays will form to the singularity. So you get this geodesic equation and this allows you to estimate the square root of this coefficient alpha and beta without having to solve the wave equation. Well, the equation was, okay, in the beginning the wavelength is very short so we can use the geometric optics approximation but as I explained these waves undergo gravitational redshift so the wavelength increases and then there will be a point where you can no longer apply the geometric optics approximation and the question was how do we deal with that issue? The point is that it is necessary for us to apply the geometric optics approximation up to the point where this light ray is out of this collapsing star because the geometry of this part of the space time is a Schwarzschild static and it's relatively simple. So we just want to make sure that the geometric optics approximation is reliable up to the point where this particle can escape from the star. And this is true for if you want to study particles at late times. Yeah, this equation is true only for very large u or for V-veg. No, in this limit does not depend on details. It doesn't play an important role for estimating these coefficients. So the only thing we want to do out of this calculation is to estimate the coefficients alpha and beta which I remind you where the Bogolubov coefficients relating the most at late times to the most at early times. Okay, so we can then use the geometric optics approximation to calculate these coefficients and this is what Hawking did and as we will discuss later this predicts thermal radiation coming out of the black hole. Now, I want to also give you another intuitive picture of how this computation works. Here I have sort of plotted the cartoon of a small wave at a personal infinity and we think of this wave as propagating inwards and then we notice that when it gets reflected it sort of splits in two parts. The part that makes it out to infinity and the part that falls into the horizon, into the singularity. Now, this picture highlights the idea of pair production so it highlights the idea that Hawking radiation can be thought of as pair production and an important aspect of this calculation that will be important later is that because we start with this wave these points are close together and if you look at these two excitations they turn out to be in an entangled state. So these two parts, if you think of them as two separate particles that are in a highly entangled state, that's what follows in the calculation Hawking and the intuition is the one I have shown on this diagram. So these purple and blue waves are, if you think of them as quantum mechanical particles they're entangled and one of them falls into the singularity and the other one flies out to infinity.  So in this way you get particles out at infinity which seem to be entangled with a black hole which is sitting there. So this entanglement will play a very important role in formulating the paradox in the next slides. So please try to remember this point. Also I will make it more precise later but I will mention at this point that if you look at an infalling observer going through these particles the infalling observer does not detect these particles on self-propagating particles. So the infalling observer thinks that the quantum field is in the ground state locally and for that to happen, for that to be possible that the infalling observer does not detect any particles it is very important that these two particles are highly entangled. We will illustrate this in more detail in Minkowski space later but yeah, I just want to mention it now. Okay, so what Hawking found after doing this calculation is that this coefficient beta is non-zero and then this allows one to calculate the expectation value of the number operator of moles in the far future and what you find is a thermal distribution so this is thermal radiation multiplied by a coefficient which slightly modifies the spectrum from being thermal these are called the gray body factors and these coefficients are basically the probability for a particle to get absorbed by the black hole so it depends on the frequency and the angular momentum of the modes. This will not have an important significance for what I'm going to say so from now on we will ignore this factor and we'll just assume that the spectrum of Hawking radiation is thermal. The inverse temperature beta is related to the massive black hole by this equation and I want to emphasize that when we say that the spectrum is thermal we mean that the expectation value of the number operator is this one but also the higher moments so for example B dagger B is thermal and also if you start calculating the higher moments for example the expectation value of B dagger B to the power of K if you calculate this object it's precisely the same thing as what you would have calculated by considering a harmonic oscillator of frequency omega and a temperature beta so this is going to be trace of e to the minus beta omega B dagger B times B dagger B to the power K by Z. So what this means is that we do not get a definite number of particles that are given modes but we get a thermal distribution precisely the same as what the one you would have if you take a harmonic oscillator and you place it at temperature beta. So this is true for every single mode and moreover what you find in the calculation of Hawking is that higher point functions between this modes B are actually factorized the product of two point functions. So there are no genuine higher point functions in the calculation. They're not connected higher point functions, they connect the higher point function as zero. So this means that the radiation that we get in the calculation of Hawking is thermal so the particles seem to be, the modes seem to be thermally populated and what is very important, they seem to be uncorrelated. There are no correlations between the outgoing particles. This follows from the fact that these higher point functions factorize. Yes. Yes, of course this, the question was, is this, does this follow from the fact that phi did not interact with itself? That was a question. Yes, if the field phi is interacting then you will get some correlations or if you take into account the back reaction of the field on the metric, I told you before that we ignore that effect because of this condition. But so if there are self interactions between the field phi or if you take into account the interactions with the gravitational field, this will introduce some correlations. But for reasons I will explain later, it is believed that these correlations that we introduced in that way are not sufficient in order to solve the paradox that I'm going to talk about. You can compute it, but for this, like for the statement we want to make now it doesn't have any significance. We just want to know what happens at future and infinity now. Okay, so then Hawking predicts that the Hawking particles are in a thermal state. More precisely it means that if you calculate the density matrix of the Hawking radiation it is thermal and which means in particular it is diagonal in the occupation basis of the modes at future and infinity. Now if we now take this result that the black hole emits thermal radiation it means that it will lose energy and its mass will decrease its function of time. So you can try to estimate the rate at which the black hole loses energy and you find that the time derivative of the mass of black hole is given by this formula. Now to get this result, if you want to do it properly you need to consider the details of this calculation including this gray body factor and integrate over all modes to calculate the power generated by the black hole. But it turns out that you get a correct order of magnitude estimate by thinking of the horizon as a black body at temperature T. So then the Stefan Boltzmann law says that the power radiated by this black body so D by DT is going to be given by the area of this surface times some constant which sigma which will depend on the number of fields that you have in your theory and so on and so forth times T to the fourth. And if you replace the area by so this proportional to G squared M squared the area is proportional to G squared M squared and the temperature is one over GM. So we get one over T to the fourth M to the fourth which gives us a scaling that we have on the slide. So the black hole radiates as black body at temperature one over eight pi GM and it loses mass according to this formula which you can integrate and you can estimate the evaporation time and you find that it scales with the third part of the mass and as you probably know if you plug in some numbers for let's say a solar sized black hole this timescale turns out to be incredibly large like 10 to the 60 years or something like that which means that the evaporation of a large black hole is an extremely slow process. In particular this timescale is much, much, much longer than the typical timescale which plays a role in the calculation of Hawking and because of this hierarchy of scales between the evaporation time and the timescale relevant for calculating the Hawking effect, it is a good approximation to treat the black hole as quasi-static and do the Hawking calculation for any given mass and without worrying about the fact that the geometry will change the function of time. I mean the previous calculation that I highlighted was done in a geometry where there's no evaporation but now if we take into account the buck reaction of the evaporation of the geometry the black hole will eventually evaporate so this is the new Penrose diagram where the black hole evaporates and we're left with Hawking radiation flying out to infinity. Now in this new Penrose diagram we noticed that at some point the black hole disappears. Now this little corner is not, we cannot study it reliably because the black hole has a very small mass and the curvature at the horizon becomes very large so we don't really know what happens there but this will not have any important implication for what I'm going to talk about. Okay, so now we are ready to formulate the paradox at least in the most basic version is the following. We found that Hawking predicts that the black hole emits thermal radiation and that is inconsistent with unitary and quantum mechanics because well according to Hawking a pure state evolves into a mixed state, raw thermal, while in quantum mechanics a pure state will always remain a pure state because it evolves unitary with Hamiltonian. So under unitary evolution you can never have a pure state evolving into a mixed state. So this is a paradox, a conflict between the calculation of Hawking and what we expect in unitary quantum mechanics. Another way of saying that is that according to calculation of Hawking the radiation that you get in the far future only depends on the mass of the black hole. So obviously you can form a black hole of given mass in many different ways and according to the calculation of Hawking you always get the same final state which seems that there is some loss of reversibility in physics or differently there is some information loss by looking at the final radiation you cannot reconstruct what was the star that formed the black hole. Entering? Yes. Well the question was there are also some particles which are falling into the black hole, let's say these guys. So the question was since we have these particles is it correct to think of the outgoing photons as a closed system? That was the question, right? Well, you're right. If the black hole did not evaporate your objection will be perfectly fine and there will be no paradox because it will not have been a closed system. There will be something inside the black hole something outside maybe the information is stored inside the black hole. So that wouldn't be a paradox. The problem is that if we take the back reaction of this into account of this radiation the black hole evaporates and completely disappears at some point. So if you look at the updated Penrose diagram where we take evaporation into account we notice that at very late times let's say if you draw this slice there is no black hole anymore, right? You just have flat space, empty flat space and Hawker radiation flying out towards the future and infinity. So there's no black hole. So these modes that you were talking about have disappeared. So we cannot use them to say that the information is stored there. That's precisely the problem, the paradox. So right, so we have this paradox and now you could ask a question here is could this calculation that Hawker did is reliable only to the extent that as I already mentioned M-plank over M is much smaller than one but as the black hole evaporates it will get smaller and there will be a point where the size of the black hole will be plankian. And at that point the calculation Hawker is no longer reliable. So you could say could it be that during the final stages of the evaporation where the black hole is almost plankian size could it be that the radiation is dramatically modified at that point and all the information comes out at once during the endpoint of the evaporation? Now I will try to argue that that is not possible that it's not consistent and to do that let me introduce one more idea which is the following. We consider the Hawking particles as they come out of the black hole and we take the let's say the first n Hawking particles and we calculate the reduced density matrix rho n of those particles and the full Neumann entropy of that matrix of that density matrix. Now I already explained in the previous slide that the black, these Hawking particles turn out to be thermal and uncorrelated. So this means that if you look at only one Hawking particle it will have some entanglement entropy S1 which is minus trace rho one, the reduced density matrix of one Hawking particle which is thermal log rho one. So this quantity is non-zero. Now if we take two Hawking particles the reduced density matrix of two Hawking particles is going to be the product of rho one times rho one. Why? Because we argued that these particles are uncorrelated. According to the calculation of Hawking every Hawking particle is independent of the previous Hawking particles. So if you take two of those the density matrix will be the product. So if you calculate the entanglement entropy of this direct product, tensor product you find that the entanglement entropy is actually two times the entanglement entropy of one particle. And if you do it n times if you consider n Hawking particles you find that the density matrix is rho one to power n so the entanglement entropy of n Hawking particles is n times the entanglement entropy one. In other words according to the calculation of Hawking the entanglement entropy of the Hawking radiation keeps increasing linearly with a number of particles. Now in the beginning that's fine. It's like having a heat bath or some if you burn a piece of paper of some other object that find a temperature and you look at radiation coming from it then the photons are almost independent in the beginning and indeed the entropy of those particles keeps increasing. But so there's no problem in the beginning of this process but we can ask what happens at very late times and at very late times if we want the theory to be unitary it must be that after the blackboard has completely evaporated the entanglement entropy of the radiation has to go to zero. Why? Well if the process is unitary and if you start with a pure state then the Hawking radiation after the evaporation of the black hole has to be in a pure state and a pure state has zero entanglement entropy. So if we want to if we believe that the process is unitary this curve that Hawking calculated has to be corrected into a curve which dives down and raises zero when the black hole completely evaporates. So the question I was addressing before is could it be that this curve entanglement entropy of the Hawking radiation as a function of the number of particles increases linearly and then at some point very near the end point where the calculation of Hawking is no longer reliable it goes down to zero during the last burst of radiation coming from the black hole. So this would be the scenario where the black hole emits all of the information during the last stages where the calculation of Hawking is no longer reliable. Now I want to explain that this is not consistent. So are there any questions about the question I'm trying to address? So as the black hole evaporates and becomes smaller then the curvature near the horizon becomes very large, right? Exactly, so then can we say that when a particle crosses the horizon it will be like before that there's like little change or something. What? Means so earlier we could assume that as the particle crosses the horizon there's not much change because the curvature is... Yeah, yeah, so near the end point of the evaporation we cannot assume anything because the curvature is very high. The curvature could be of the order of the Planck scale so we can no longer control the geometry and in particular we cannot trust the Hawking calculation anymore, right? That's precisely the point. So the Hawking calculation would tell you that this curve will just continue to increase, right? Now one way of resolving the paradox would be to postulate that this last part of the calculation is unreliable because the black hole is tiny and the calculation has to be replaced by some other calculation of quantum gravity which will allow all the information to get out of one like in a very short amount of time. But this is a bit problematic for the following reason. This entanglement entropy, if you think about the physical meaning of this entanglement entropy, the Hawking radiation it is the following.  Now we believe that black holes are quantum mechanical systems. We have a Hilbert space with finite dimensionality and we believe that the size of that Hilbert space is determined by the area of the horizon. So in the beginning the black hole is very big. The Hilbert space of the black hole is very large so there is no problem. Everything is fine. It can be entangled with Hawking radiation but as the black hole evaporates the size of the Hilbert space of the black hole decreases because the area decreases. So if I plot on the same diagram the logarithm of the dimensionality of the Hilbert space of the black hole we start at a very large value where this is A over 4G and as the black hole evaporates the size of the Hilbert space of the black hole decreases. Now there will be a point where it will cross the other line beyond which it means that the size of the Hilbert space of the black hole is smaller than the entanglement entropy that you need in order to purify the Hawking radiation by the area of the black hole. Now I remind you that the entanglement entropy of a system is bounded from above by the logarithm of the dimensionality of that system. So when the black hole becomes smaller than what this curve predicts then it no longer makes sense to expect that the Hawking radiation is entangled with the black hole because the amount of entanglement that you need is actually larger than the size of the Hilbert space of the black hole. So this means that this scenario where the entanglement entropy keeps increasing and only goes down to zero near the end of the black hole is inconsistent way before the time where the black hole has Planckian size. You run into a contradiction way earlier where you can estimate the size of the black hole at that point and it turns out to be macroscopic. So it's not consistent to assume that the radiation is thermal all the way to the last stage of the evaporation where it suddenly becomes unitary and all the information comes out in a very short amount of time. That doesn't make sense. It is not consistent with the idea that a black hole of size A has a Hilbert space which is determined by this area. Yes. Good. So the question is what about corrections? Let me continue because I'm going to talk about corrections now. So now okay, we have formulated this paradox and what are the possibilities for resolving it? Well, there are two possibilities that have been discussed in literature. One is that the information is fundamentally lost that was the original proposal of Hawking but for reasons that I will explain a little bit, we no longer believe that this is the right answer. Another possibility is that of remnants which is the idea that the black hole, the evaporation stops at some point, let's say near that point where the size of the black hole is Planckian and you're left behind with an object which is called a remnant with a stable object which has very small mass so it has like Planck mass and it has a very large Hilbert space so the Hilbert space must be large enough to be able to accommodate all this entanglement with Hawking radiation and that would be a possibility but it leads to other problems and as I will explain, it is also ruled out by ADS-50. So both of these two possibilities we're not going to discuss them a lot. In my opinion, the strongest argument against these possibilities is that in the ADS-50 correspondence, we do not have information lost because the boundary theory is manifestly unitary and we don't have remnants because in some situations we can actually calculate the spectrum of the CFT and we can see that there are no particles with these peculiar properties. So we will consider the third possibility that somehow the information of the black hole is encoded in small corrections to the calculation Hawking.  If you burn a piece of paper, you get radiation which seems to be thermal. So you would, well, but we don't worry about information loss. Why is that? The answer is that we don't worry about information loss when we burn a piece of paper because the photons appear to be thermal but in reality, they're not exactly thermal. There are small correlations between those photons and those correlations are sufficient in order to encode the information of the piece of paper. So the question is, could something similar be happening in the case of the Hawking calculation? Could small corrections to the calculation of Hawking resolve the paradox? Now, before I go on, let me make two clarifications. The first is that in what I'm going to say we are not going to be able to actually calculate these corrections. So what we will be able to, this will be a very difficult problem which will be equivalent to being able to calculate the exact S matrix in quantum gravity. If you really want to be able to calculate the exact quantum state of the particles and the future null infinity, that will be as hard as being able to calculate scattering experiments including all full quantum gravity effects. So that's clearly out of reach today. So instead, what we are trying to do when we try to resolve this paradox is we try to estimate how big those corrections have to be, what's the minimal size of those corrections and then we want to see whether those corrections are expected based on general arguments and whether they are consistent with effective theory. So we want to estimate the size of these corrections and then based on the result that we'll get, we will decide whether this is a reasonable size, it's whether we expect corrections of a type and whether these corrections are consistent with effective theory. Yeah. So are you assuming that there are a large, there's a large number of Hawking quanta so that the overall entanglement entropy, the contribution to the entanglement entropy by these correlations, small correlations, the overall contribution is such that it cancels the thermal computation. So I would, I mean, of course we have a lot of Hawking particles, right? I mean, just to remind everyone, if a black hole evaporates, you can estimate the number of particles that you get and the number of particles of the order of the entropy of the black hole. So it's a huge number of particles and as you say, these small corrections, there are many possible pairs you can write down and you can introduce small corrections to the correlations between all of these pairs or you can take also groups of these particles and introduce correlations among them. So because of the size of this Hilbert space, this is a very large Hilbert space, it is possible to introduce small corrections which will turn a mixed state into a pure state. That will be the point. So, yes, I am running out of time, right? Do I have five minutes or? Okay, just five minutes. Okay. So, okay, so actually, okay, I will try to close now and this afternoon I will try to explain the following results in more detail. So the claim I want to present this afternoon is that the unitarity can be restored in the Hawking calculation just by introducing exponentially small corrections in the entropy, so e to the minus s, to simple observables in effect field theory. So what this means is that we will see that you can think of the Hawking calculation as being a reliable computation up to exponentially small corrections provided that you focus your attention on what we call simple observables in effect field theory. So I will make it a little bit more precise this afternoon, what do we mean by that? But roughly speaking, simple observables are low point functions of the quantum fields on the background of a black hole, while a correlation function with s black hole insertions would be classified as a complicated observable. So the statement would be that the Hawking calculation is reliable for the computation of low point functions up to exponentially small deviations, but it may be completely unreliable for the computation of high point functions of local operators when a black hole evaporates. So, and I will try to argue that this claims relies on a very basic property of quantum statistical mechanics, which is the property that if you have a large system, a large quantum system, which many degrees of freedom, most pure states look exactly the same, or almost exactly the same as mixed states. So this statement I just made, I will quantify it, I will make it precise this afternoon and I will explain what it implies about the information paradox. Thank you.


Hi everyone, Joe for jaspyscasebreaks.com. It's the end of Tuesday night and we did it. We knocked out the break credit promo We just had to fill this requisite number of breaks right here this amount of breaks right here gather all those names into one big randomizer list and Then randomize it roll the dice randomize your names in top five. We'll get five hundred dollars of break credit each To jaspyscasebreaks.com I I have to be honest with you. I'm normally a pretty positive person pretty optimistic about About about everything about people are at will ask me. Hey, do you think this brings gonna fill? I'll be like, yeah, of course. Yeah always a little nervous today first half was really slow But then second half the show really started heating up So an excellent rally to get all of these breaks done. So let's just quickly do a spot check here We have to do the immaculate random number block that we did Already only the number block counts right there Clearly nine which was done over the weekend. There's clearly nine Donner's optic 18 which we did today. There's optic 18 skybox metal universe to 12 and 13 so we did There's 13 right there and Then 12 I didn't do but there is 12 Then we had to do hit parade 10 a dual case break. There it is plus the filler. There it is the prestige pack We had to do Merlin 7 there's Merlin 7 since it's right there. There's Merlin 7 and we had to do elite 6 And there's the elite 6 list right there nice. So that's a total of 249 names And remember we keep track of all these on the break schedule Click the break schedule and then click the next tab over you can see 9 to 20 promo You can always keep track of it there make sure I don't forget your break All right, so again 249 we're gonna randomize it Five and a six almost full 12 five and a six eleven times one two three four five six seven eight nine ten And 11th and final time. Good luck everybody There's 11 times. There's 11 right there a lot of sad faces here But hopefully you did well in the breaks that you were in Hopefully we're sending you some hits so all is not lost But either way I appreciate you being part of it ladies and gentlemen. Thank you But there can be only five Now these promos are great though because really you're doing what you would normally do join breaks These really should be considered You know, I really wouldn't go out of my budget to try to chase break credit here But just do what you would normally do Maybe you end up in the top five and that would be pretty awesome the not awesome part is spot six Mark I am so I appreciate you getting in but That's the worst spot right there the that really close spot just on the outside of that break credit But that sucks, but I appreciate you giving it a shot But there can be only five sad times for a lot of people but happy times for the top five Sean Maddock is happy Nico is happy Michael Losey is happy Nico Martin again is happy and Arianna Daniel is happy So congrats the top five here Nico since you were lucky enough to win twice I'll send that I'll send one break credit code for a thousand But hey Arianna Nico Michael Nico again and Sean. Thank you very much $500 of break credit going your way. It never expires. You don't have to use it all at once So it'll save your balance You can only use it on jasky's case breaks calm Think that's it and it will arrive to your email as a gift card code check your spam folders if you don't see it There you go ladies and gentlemen. I'm Joe for jasky's case breaks calm. Thanks for this excellent rally Keep your eye out for more break credit promo like this on jasky's case breaks calm and I'm Joe And I'll give away some more money with you next time. Bye. Bye


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I'm Staff Sergeant Tiffany McCoy. I'm a KC-135 Crew Chief here at the 171st. Each aircraft has a team of refuel aircraft maintainers, also known as Crew Chiefs, assigned to it. McCoy is assigned to KC-135 tail number 591468 or, as her team refers to it as, 46th grade, that's what we call it. McCoy has recently been upgraded to Assistant Crew Chief. This comes with the luxury of having your name added onto the side of the jet. It's pretty cool actually to have your name on an aircraft. It's something you don't think about coming out of tech school because it's more of an active duty type thing. So I thought and then once you become a technician it's pretty cool. A typical day as a Crew Chief varies. You get your assignments in the morning and then by lunchtime you could be on something completely different. For instance, we'll take yesterday, we ended up towing two aircraft and then by lunchtime we were putting panels back on a totally different aircraft. So it just depends on what they need for the day. Definitely a total different ball game. When it's winter though you have more operations to get aircraft ready. Definitely have to watch out for the ice snow. You have to clean the aircraft off before you can get them ready to fly. So it's definitely different than just a normal day if it's sunny outside. The part about being Crew Chief would definitely be traveling. I have never left home before basic so getting to travel and see new places is definitely cool and also getting to meet new people is pretty awesome as well. Crew Chiefs are unique breed of airmen. They feel that the most rewarding piece of the job is. Getting safe airplanes off the ground for sure. Just producing good aircraft for our pilots and our booms. Reporting from Pittsburgh, I'm Erin First Class, Zoe Walkenfuss.


Hello everyone, this is Amit Dan of amitdan.com. Today I will have another episode of drink video and I'll compare three different types of coffees from three different big brand names, McDonald's, Starbucks, and T-Martins. If this is your first time, I'll be a visit. Thank you for your support. On this channel, I'm always trying new and different ideas. If you appreciate variety, please go ahead and come back for more. If you're only looking for drink videos, there'll be links in the description box below. So I have done a few coffee and tea videos and I always wanted to do a comparison video between three different coffees from three different big brand names. So I have Tim Horton's coffee right here and the McDonald's coffee right here and the next to it I have Starbucks, which I have it right here. What I noticed that in the past, whenever I went to Starbucks, it's like self-service. So you have to put your milk and sugar or cream and sugar yourself. They wouldn't put it at the store. But this time I noticed that the guy asked me for how much sugar and milk I want. And once I told him that he put it by himself and I asked him actually whether this is something new. And then he mentioned that due to COVID, they're actually putting cream and sugar or milk and sugar at all Starbucks, especially this is in Canada. I'm not sure about other Starbucks in the world, but in Canada, he said they're putting cream and sugar at the stores due to the pandemic. But definitely when I was there before the pandemic, you had to do on your own. They wouldn't put cream and sugar. But it's different for McDonald's and Tim Horton's. They would always put cream and sugar, whatever you want. You don't have to do anything. So it's not self-service. You get a whole coffee all like 100% done. Another thing I noticed that only Starbucks gave me the sleeve. This is like just to protect you from heat if it's too hot. But for McDonald's and Tim Horton's, you have to ask for it or you have to get it on your own. They wouldn't give you. But at Starbucks, it was just there or the guy put it. I didn't have to ask for anything. If you look at the lid or cover, it looks like McDonald's has the best cover or lid. So I'm just going to open them up and I'm just going to show you one by one because it's going to be easier. Okay. So what's happening here? This is a McDonald's lid. So if you take a look at it, so it looks like this is the most friendly user friendly and most expensive lid. You can tell by looking at it and also you have to just push it under down and then you can sip it. And Tim Horton's have the second best one and you have to just kind of like break it right here. And the Starbucks looks like they're still giving the cheaper version of the lid. They haven't improved their lid. This is what I saw like many years ago. So definitely McDonald's has the best lid among these three. And these coffee actually they're all small. I asked for the small size and looks like Starbucks is small. It's actually bigger than McDonald's and Tim Horton's looks like slightly smaller in size than McDonald's. But I have a feeling that probably they hold the same amount. But definitely Starbucks has a little bit more coffee than other. And all this coffee they're one cream, two sugar. So if you take a look at it, let me just put it a little bit closer. Actually I'm going to just empty out like all three of them just to show you a little bit better. Okay so I had to take out coffee from all of them. And looks like this actually this cup has a mixture of coffee of all three. This is like Tim Horton's Starbucks and McDonald's. This is how Starbucks coffee looks like and this is how McDonald's coffee looks like. This is how Tim Horton's coffee looks like. So in terms of how it looks like, Starbucks looks a little bit darker and looks like you know McDonald's looks the lightest. And Tim Horton's is kind of like you know slightly in between. And so let's just start now tasting them. I'll just remove this one. Before removing this I just tasted this coffee and it tastes kind of like you know weird because it's a combination of a Starbucks, a McDonald's and Tim Horton's. Okay so we have now three coffee and let's start with the Starbucks. So I'm going to just you know try one by one and see how it goes. So here we have the Starbucks so I'll just test this first. Okay so let me just get the McDonald's which I have it right here and I'll just taste this now and here we have Tim Horton's so I'll just try this now. Okay so all of them have two sugars but I'm not sure the Starbucks coffee doesn't like you know taste like it has sugar so it could be that you know it's a bit bitter and McDonald's coffee actually tastes the sweetest. Always two sugar and Tim Horton's coffee is like you know somehow in the middle so now let me just sip them again. So the Starbucks coffee is a bit bitter and stronger so definitely you know this is the nature of a Starbucks coffee that you know they give the strongest coffee and I always find it a bit bitter and if you're not used to it then you're gonna like the taste of a Starbucks for that reason actually I would prefer to have McDonald's and Tim Horton's so this is a very strong coffee and even after like you know two cream it tastes a little bit bitter to me so I'll just you know get McDonald's again and let me just sip again. So this is a very balanced coffee I would say a well-balanced coffee it's not you know bitter it's not lighter so it's just somewhere in the middle and what I noticed about McDonald's coffee is that regardless which location you buy from they're always very consistent and the taste wouldn't fluctuate a lot you are going to get like you know exact same thing or almost you know same coffee everywhere and that's what I'm getting right now because I remember this taste and this is what I had it you know other McDonald's location and I like it this is you know so whenever you're buying coffee from McDonald's you can know or it can be assured that you're going to get what you're used to it's not going to fluctuate a lot let me just try Tim Horton's again so the problem with Tim Horton's coffee is that actually it's not the problem the taste differs based on where you're buying from sometimes you know they miss they give you very bad coffee sometimes if they're you know they make perfect coffee then you're gonna get the best coffee so this happened to me a lot I went to let's say one Tim Horton's and I didn't like the coffee then another time a few days later I went down to Tim Horton then I like the coffee so it depends on who's making the coffee and I have a feeling that you know they have their beans are from different sources or I don't know how they brew their beans but they're not like you know the same at every location they fluctuate so whenever you're buying coffee from Tim Horton's you know you're not gonna get the same thing it all depends where and when you buy and also you know probably the person who's making it it always fluctuates but Tim Horton sorry some McDonald's is more consistent in terms of you know taste and flavor and everything so you're gonna get the same coffee or almost same coffee at every location and Starbucks is like you know you always gonna get the slightly bitter coffee stronger coffee than the rest so if you are like you know looking for a small stronger caffeine boost or more caffeine then definitely you know Starbucks is the way to go and also this is more expensive this is kind of like in a close to three dollars and these two are close to two dollars so you're like you know paying almost a dollar more for Starbucks coffee and you're gonna get a stronger a bitter coffee now I'm liking McDonald's the most before I thought you know I would prefer Tim Horton's better than McDonald's and I never had them like you know side-by-side at the same time but today I'm gonna have to say that you know I'm liking McDonald's coffee more than Tim Horton's coffee at this time but in general I prefer Tim Horton's coffee more than McDonald's coffee or this is my most favorite coffee but today I can feel the taste and everything that you know this tastes slightly better than Tim Horton's but they're very on par they're very close to each other but you know Tim Horton's is going to fluctuate whenever or based on where you buy but McDonald's is more consistent so according to what I had today I'm going to say here that you know I had the best coffee from McDonald's and this is based on my taste and needs and Tim Horton's is very close and the Starbucks is just you know a little bit too dark so I'm not going to you know usually go for Starbucks unless I need heavy caffeine boost or heavy caffeine so in general I'm gonna prefer either McDonald's or Tim Horton's but in today's like you know testing I have a feeling that I have the best coffee and this is from McDonald's so that's my conclusion on this channel I'm always trying new and different ideas if you like what you see please go ahead and subscribe and I have more coffee and tea videos which will be in the description box below so thank you for watching this video there will be more videos coming up so I'll see you shortly thank you


Hey everybody this is Brian welcome to the seventh Android tutorial. Last time we did the basic ListView and you can see in one of the graphics here you've got the that's very small but you've got a picture, let's see if I can blow that up a little bit, you got a picture with some text. We're gonna actually do that today because like I said ListViews can be very very complex. I'm hoping this will be a bit of a quick tutorial because I do have plans later but we're gonna see what happens here. So let's start a new Android project. Let's call this app 7 and we're just gonna leave the defaults on everything here. We're gonna make an empty activity and Android will calculate the sun moon and stars. Sometimes I honestly wonder what it's doing under the hood. I really should read up on that but anyway it's gonna do a quick gradle build. It's a lot bigger than it should be. We're just gonna move this around a little bit. The snap windows as they're called is kind of messing me up a little bit here. There we go. All right so we've got just our bare bones basic activity here. We're gonna get rid of this Hello World because well we're beyond Hello World at this point and we are going to go find ListView. There you are. I had to lean forward for that. I have this unnatural habit. I had Lasik a while ago, well over a year ago and I'm still used to wearing glasses. I'm leaning forward. So there's the code right here and you can just see ListView is gonna take up the whole thing. We're gonna give this an ID of LST main. Whoops. There we go. As soon as we give it our ID and it creates it, it puts in a what's called a visual template here. That's actually not really what the items are gonna look like. We're gonna get to that. So let's crack open our layout folder here and you can see there is our main activity that we were just looking at. We want to make essentially what's called a list item. We're gonna go new and we are going to have to hunt and pick through here. Whoops. Layout resources file. Yeah, we really didn't have to hunt and pick too far. All right so this may take me a minute here. Not used to doing it this way. Forgive me. Let's call this list item. We're gonna give it linear layout and we're gonna put it in the layout directory. Honestly, these available qualifiers, I haven't read up on these too much so I'm not really sure what those mean. I'll have to read up on those but we're just gonna hit okay for now. Really not used to doing it that way. Forgive me. And as you can see it makes another layout. Now what we're gonna do, let's flip back to the main activity here, is you see these items, items 1, item 2, item 3, and 4. We're gonna make our own template that each item is going to follow and that's what this list item is. So let's flip into this guy and I want to just kind of look at the code here, make sure we're not missing anything. Yeah, we want to set this to horizontal. There we go. The reason why we do that is if it was vertical they would stack down. We want them to go left to right. So what we want to do here is we want to get, if I can find it through here, I see I miss where Cute Widget has this little search thing where you could search for them. We want to get an image view in here. Did I actually do it or didn't do it? Didn't do it. We're just gonna type it then. Forget it. Actually you know what? I like drag and drop. We're just gonna do IC launcher. Why not? See there it goes right there. I didn't want to do a lot of typing on XML because it's quite frankly bores me to death and then we're gonna get a text view and we're just gonna pop this text view over here and you can see how it's kind of got this 50-50 thing. So we're just gonna grab this guy and just kind of scale him over a little bit and then we'll grab this and we need to give these some identifiers here. So we're gonna give this AMG item. I don't know where I come up with the names for these things. I gotta do better at that I guess. And TXT item, short for text item. We're gonna modify this a little bit here.  Let's go 18. Let's go 30. Maybe not 30. Maybe 24. Why not? Alright. And we are going to look at this here. Where's the padding on this bad boy? This part I probably will write because I can't find it through here. Although I'm probably looking right at it. We'll just flip in here. And we're gonna go padding top. We want to pad this a little bit. Now if you don't understand what 5dp means, we'll explain that in a future video. Just right now, just understand that this is not a pixel measurement. You can see how it drops it down a little bit. We're gonna actually do 10dp. You can see how it drops down even further. So that is the rough template that each one of our items is going to follow here. Now it's very important that when you do this, you have an identifier for the picture and the text box. Sometimes people forget that. I forget that all the time. We're going to kill that. Whoops, we're gonna go ahead and kill that. Now what we're going to do is we're going to say, Alright, let's see here in that type hash map. There we go. Now you should remember what a hash map is from your intro to Java classes. And this is one of those cases where we're gonna actually use it.  So you're going to have to have a key and a value pair here. So to do this, we're going to say items equals, whoops, it's hash map, not hash. And we're gonna say new hash map. And we want to give this an integer. And remember, these cannot take primitive types. So you can't do int. You have to do the actual integer. So that's gonna be our actual items here. Now we're going to list view and we'll call this LST main, because that's the name of our list view. And we're going to say LST main, list view, find view by ID, r.id. This is kind of the boring part of the code. LST main. So we're just, you know, getting our resource to the list view and actually casting it into an actual control or an actual view, if you will. Now we're going to populate our map. We're just going to very simply say four and i equal zero. i is less than five plus plus. And we're just going to say items.put. And here we're going to say i because that's our key. Remember, we have an integer and then we want to string for the value. So we want to say item number plus i. So really, that's all we're doing. Now comes the interesting bit. Let's save this. Let's actually give this a good build. Make sure we haven't screwed up on this so far. And it looks like it built just fine. I've got some icons somewhere. Here we go. And we're just going to take these, and we're going to copy these. Very important that you copy, not cut. And we're going to go to our drawable folder. If you don't have one, you can actually just right click on res. Go show in files. And that'll bring open your resources where they're physically located on disk. You can just add a drawable folder. But anyways, you're going to right click on drawable. And you're just going to go to paste. And it's going to say copy specified files. And it's going to give you the actual path. And you're just going to say, okay. And that copies the actual icons that we got in here. Put some into your Android project and registers them so that when, I shouldn't say registered, it just puts them in there so that when you build, the r variable actually has those. So now you can say r dot drawable. Whoops. Dot. And then there's your icons. That is how we're going to actually access the icons. Now before we do any of that, let's run this. And I'll show you that it does absolutely nothing. I just want to get the emulator up and booting. Make sure it actually runs. It was giving me fits last night when I was doing the pre version of this. Great. It went come on emulator. You can do it. I have this truck. It's not a truck. It's a Jeep. And I love this thing to death. And it's making some noise. I'm like really afraid to drive it. Who's texting me? I'm doing a video go away. Sorry, my daughter was texting me. Alright, so as you can see, it does absolutely nothing. So what we're going to do is we're going to populate this now to get those items in so it shows an image and a text. We have to use what is called a you guessed it anybody. You have to use what's called an array adapter. So let's kill that application. Let's close that. And let's go here to Java and we're going to go to new Java class. We're going to call this custom adapter. Now this is going to be a fair amount of typing. So the first thing we're going to do is we're going to extend the array adapter. And you notice how this is templated array adapter type. So what we need to do here, and this is going to look a little funky if you're not used to it, is we want to actually give this a type of hash map. And a lot of the tutorials out there you'll find won't do that they'll do like a string and then you'll hand it an array. And there's a reason for that we want to actually take this hash map that we just populated and give it to the array adapter and you'll see what I mean here in a second. So we're going to say hash map. And there's our key value pair. We're gonna say integer string. And you notice how it's already getting cranky with a snow default constructor. Okay, well, let's make a default constructor here. First thing we're going to do is we're going to say private hash map give it our key value pair here are types actually sorry, string. Yeah, I really love that cheap. I mean, it's anybody who's got a vehicle they've ever just really loved. I mean, you understand what I mean. It's like your baby almost. Alright, so we're gonna have our hash map and our context. We haven't really talked about context too much. context. It's like having a conversation with somebody in what context are you talking about? You know, I was walking around in my underwear. Well, in what context? We walking around in your underwear in your house or in the middle of a supermarket. That's a context. So in programming, you have to know the context of which is pretty much translates to what view are you talking about or what activity are you talking about? Because everything could have its own context, which I think internally really relies on scope. If you remember your your scope classes from Java. Alright, so we're going to do patrol. Oh, and we're not going to do that actually. Sorry about that. We're going to go hover over this. Try get that little. There it is that little icon to pop up. And we're going to create constructor. And we're just going to go with this first one. context and resource. And then we're going to add this in here, because I'm really sick of typing this out, we're just going to grab this. And we are going to say boom, we're gonna call that map. So there's our constructor. And now we are going to just say items equal to map. There's a whole lot of memory copying going on here. That's one thing I'm not really thrilled about with some of these languages. Now, some of them internally, like I think Java actually does internally, we'll actually use a pointer instead of copying. I may be wrong about that. But I think in the olden days it copied and now it actually uses pointers under the hood. So if you know anything about C plus plus, you'll know exactly what I'm talking about. Alright, so now we have our context, which we're going to need later on and we actually have a reference to the map reference or pointer that's up for debate. Anyways, sorry, I'm thinking I miss C plus plus really bad. Get count. So we're going to add get count. We need that because the array adapter will not show any items unless we actually return account. So if you're following along and you skip the step and none of your items show, this is why. Okay, so now we're going to return items dot size. That'll ensure that our array adapter actually does what it is supposed to do. All right, now. This is where the actual magic happens. We need to go control o a new keyboard. And we're going to say get view. Now notice how this says in position convert view, which is a view group, and it returns a view. So what are these things? Well, remember, a view is pretty much anything on the screen. So what we're really going to be returning is this list items dot XML, this guy, we're going to basically create an instance of this and shove it back, or I should say return it back. To do that, we have to do a little bit of magic under the hood here. Why is this thing suddenly missing return statement? We'll get to that in a minute. All right, so we have to use what's called a layout inflator, which is kind of a complex topic. But basically, a layout is an XML file, which we've seen an inflator is just a bit of code or class that actually takes that turns it into an object that we can work with. So we're going to say inflator equal layout C layout inflator. What? Why are you doing that?  Copy and paste. Forget it. There. Con dot get system service. And we're going to say context dot layout inflator service. Now if this looks like some weird voodoo black magic code, it's because it kind of is basically what we're saying here is get a layout inflator, which is a class that's going to turn this XML file into something we can work with. And then we're going to cast it here. And then we're saying con, which is the context that we're going to hand this custom adapter, which the context is actually going to be this guy, our main activity. Confusing. I know I'm very sorry. I didn't create this Google did they're much smarter than I am. Anyways, so then we're going to take that context of the main activity and we're going to get a system service called the layout inflator service. What does all that mean? Basically, it means that there is a bit of code running in the memory of the device that has a pointer in the context that is called the layout inflator service. So it's not us actually creating an instance constantly. It's us getting the devices instance of it and then using that instance. Not 100% accurate, but I think it makes the point. Remember, these things are designed to save memory to save power. So inflator dot, and then we're going to actually inflate. Now what are we inflating our dot? I have to look at my notes. Sorry, lay out the list items. This was actually a challenging tutorial for me to write. I did not like doing this one. So I'm trying to do good by you guys here. So we're just going to do a little bit of magic there. We're going to say we want to inflate that list item thing. So we're going to turn that to usable code. Alright, now we want to actually make a text view call this txt. And this is going to be you guessed it, text, text view. And we're going to say, whoops, we have to do the row dot fine view by ID r dot and I'll explain this here in a second. Just let me finish typing this out for I lose my place here in my notes. txt item. So what are we really doing here? We're taking the inflator. We're getting a we're getting a reference to our pointer to the layout flaters inflator service that's running on the device. We are then grabbing a view remember reviews pretty much anything shown on the screen. And we're calling that row and we're saying inflator inflate this XML file this guy. Once we've quote unquote inflated that or turn that into something we can work with this guy, we're then saying grab the text view out of that thing. So we're taking list view fine view by ID and we want to get the txt item, which is this right here. If that's not confusing, I don't know what is. Okay, so let's let's continue here. Sorry. I literally once I figured that out, I had to laugh and went wait, what that makes no sense. But then when you start thinking in what I call Google language, then it actually makes a lot of sense. Sometimes I wish Google would hire me just so I could get smarter. Bro. Alright, fine view by ID, we're gonna say our dot ID dot item main. So now we've got that image view on there also. So and if you had other controls in here or controls are other views in here, you'd want to grab those as well. So basically we got, you know, reference to that image in that text box, or sorry, text view. So used to other languages. We're going to close that close that because we don't need those anymore. Alright, we're almost there. We have a little bit of magic that we have to do here. Alright, so we're going to say, yes, that was actually me rubbing my hands together because I'm a little excited because I want to I want to get this done items dot get and we want to get the position. Sorry, my phone two string. So basically what we're doing is we're taking that map. And we're saying we want to set the text here. And just for the sake of argument, we're going to return the row. Whoopsie. And we're going to build run this just so we can see if this thing actually works. And there it goes. And alright, so it built but it's not displaying anything. Let's fix that super quick here. Let's flip back into our main. And we're going to say whoops, how many times can I say whoops in a video? Custom adapter, adapt, adapt her. A new custom adapter. This resource zero because we don't really care. And then we want to give it the items. So what we're really doing here is we're creating an instance of the custom adapter that we've been working on here. And we're giving it this as the context. So when we think of context, you have to think of, you know, what perspective are you talking from where we're talking from the main activity. So when we're in here using context and you'll see context a lot in Android tutorials. The context is our main activity, whereas the view context and get view is a different context. So we want to kind of flip those back and forth. And if that gets very confusing, there's like a whole three hour class on YouTube about it. And then LST main set adapter. And then adapter. So by doing this, what we're really doing here is we're saying all right, make a map or I'm sorry, hash map, fill it with values. We need some way of rendering those values, we're going to use a custom adapter. And then we're going to say the list view set the adapter, we're going to set the adapter here. So let's actually stop run this again. And when this thing pops up, we won't see any graphics. But yeah, we'll see the default that we put. Now what we need to do is actually find these icons here. See how we got icon 01234. We want to actually put those in there. So there'll be all different icons. This takes a little bit of magic here. The problem we're trying to solve is we're trying to take the r dot drawable. And then whatever the number is. You notice how that's code, that's actual, you know, objects, we want to be able to get that from a string. And that kind of just, I'm just going to level with it just completely pissed me off. It took me like, I want to say 20 minutes, actually a lot longer than that. I don't want to lie. It took me about half an afternoon to figure this out. Because most of the tutorials I found on the internet were just dead wrong, or they were ancient and they just didn't work. And that's been a big, big problem I've had with Android is a lot of the tutorials out there are just really old. So anyways, we're going to make a context and we're going to say the image, the image view, we're going to get the image views context. We're going to say int ID because we want the identifier because that's what everything in the resources is an identifier. And we'll say context dot get resources dot get identifier. And then we're going to say, ICO plus position, we want to get a drawable item. Make sure you spell that correctly. I did that wrong one time. And we want to get this in the context dot get package name, sorry. And then finally, we want to image dot set image resource ID. So what are we doing here? Essentially, we're saying get the context of the image view convert the actual string icon, whatever the position is, the position is getting from here. So whether it's item one, item two, item three, actually, it'll be zero base item zero item one and you get it. We want the drawable resource. And then we want the context to get the package name now the package is this entire application you can actually get resources in other packages if your security allows it there's all sorts of other hurdles you got to jump through but you can do it. And then finally, we're going to say image set image resource ID. This is the exact same thing as saying our dot drawable dot icon zero or icon one or whatever. We're basically just converting the string into the ID the Android understands. So we'll stop our application, make sure it stopped save our work. And let's run this bad boy. See what happens here. Boom, there you go. So now we have our individual icons with our text. And from that you can modify this template however you need it. You can put buttons you can put pretty much anything in there that you want. It's kind of crazy that what you can do with these that shows you basic item templating and how to handle with it. And there, you know, it acts just like a normal thing, you could actually put in code to handle the on click and all that stuff like we done in the previous tutorial. That was a mouthful. Well, for the source code for this and other programs, other tutorials, you go to my website void realms dot com under tutorials. And then I got to alphabetize that under Android, I'll pop it in right here. And if you're still watching, go to the contact button. There is the void realms Facebook group. There is I think today almost 1200 or just over 1200 programmers all walks of life all different languages. And we all just help each other out. It's getting to the time of the year where I need to renew my website subscription. So if this helped you, the site is 100% funded by your donations. Anything above and beyond just paying for the website I donate to charity. So that's it. Thank you for watching. I hope you found this educational entertaining.


Hey, it's Anfa and I just got home and I got mail. I Have a little suspicion What could it be? I'm not sure and I just want to show you Let's see. I Don't have any tools for opening the envelopes right here Yep, oh aha Oh dear. Well, let's take a look Can you see this? There is a a Card from Matt Tittle saying thanks for supporting helm There is something that there's a number. I don't know what it is. It doesn't it's not an IP number And maybe it's a phone number. It's signed By him with his own hands and I've got Actually two Nice lovely helm stickers. Wow. These are thick And is there anything else? There's some paper Yep, there's some paper to make it more sturdy. You wouldn't like these to get all you know crushed in male delivery Yeah, so I sent five dollars to Matt Tittle the other day after I recorded Anfa vlog 12 with helm because they thought hey, that's a cool instrument. Let's let's support this guy And this is funny because I actually sent five dollars just From the top of my head and then I read through his website and I saw If you send me five bucks, I'm gonna send you a helm sticker and actually he sent me two of them Thanks Matt I'm gonna proudly expose at least one of this known the sonoi convention in november See you there. Thanks for watching. Bye