 Okay, so, next one. So, maybe two, three minutes of cosmology, to come to start my book. What's the... I'm sure it was between the two of you in the lecture. The pre-alien mandi was clear. But I thought that any... other than any questions you'd like to ask about that, we'll come back to that. In the seminar, for example, there's a part of the material that's moving the most in... in the city. Do you think that the future needs to put the role in the presentation? You deal with the preparation. So, say, suppose you've got to get part of the... the first question you ask is, suppose you've got to make a... you've got the universe that has some inflation, that Johnson's defined. Okay? Now, we know that Johnson... cost money. And you can ask for whatever reason. And that reason is a lecture between one of the fluctuations. So, for whatever reason, suppose there's a small fluctuation in the metric of everything. And you'd be like, how would it cost? Now, the problem to do fluctuation in the metric is really involved with the problem of this system. But there's a similar problem, which is just about the scale of it. Suppose you've got the scale of in addition to the metric. And the scale of it in the background value scale. Now, you're doing small fluctuations at the scale of it. First, you've unionized all the others at once. So, in some equations of motion, you take some small fluctuation changes in that. Now, on top of that, we mean nonlinear variations. The point is that, look, suppose you've got some sort of gravitational potential. So, in this case, it's like this. And that's a matter. Some of the people want more matter than I want in low-potential ranges. Okay? And I mean, suppose there's more matter in here, the potential goes into your mind. In fact, you need more matter. This is a very genetic factor of matter. But because gravity is an attraction to the universe, they attract it. Matting distributions, at least from large enough states, are basic units. So, if you've got small fluctuation out of this thing, you can grow it. But this is more of a fluctuation in terms of the small wavelengths of gravity first. There's a variation in the large wavelengths of gravity. And largely, there's a difference between the smaller scales and the larger scales of gravity. And that's what the structure of gravity is. Now, let's see if we can start. The C and B that should run under the structural load, gives you a picture of what the universe looked like very early. A few hundred thousand years ago. Because the universe became neutral after that, after a minute or so, gravity changed. In fact, it doesn't move much. Okay? And let's just restrain this first. So, a fluctuation, a fluctuation has to go very much. So, in that, about the universe, the universe of all of us, so it is an attraction, but at one point, and the deviations from this isotropy, you see these deviations in this puzzle. This was the deviations of the gravity. The gravity of gravity. It's not an effectant. It's a little better, but it's a photon. It's an austenite. Yes? Now, let's see if we can bring that down. Yes. How do we bring that down? If an age of the universe has an age in the framework of the universe. Yeah? So, suppose you take a part of that, a part of the universe. Forget that it ever stopped. Continue on the equals. In that, a part of the universe will make an influence. Actually, this universe might make an influence. Okay? But this is the age, the 1146 structure of the universe. Because before inflation, or during inflation, according to the theory of the universe, it was just homogeneism as well. So, if there was a puzzle like that, there's a star that is older than the universe. This is a puzzle. If the star was older than the FRW, it doesn't help those inflation. But it's a type of one. Because that was just homogeneism. Okay? Yeah. So, you know, in age, when you talk to the age of the universe, we have this diagram. Here's, this is the diagram. Here's the transition between FRW and inflation. And it's the amount of time from the age of the unknown to the age of the FRW. The size of the universe that you have to be able to calculate. So that's the thing that you can see in the point of view. Exactly. Exactly. Because how do you define that sum is a bit up to you. But whatever you do, what number you put to the size of the FRW, whatever you mean by that, is this. Now, whatever you want to measure to the age of, or yesterday's age of, you know, that's all. So, plus scale of measure. We try to fit it in with a plus scale equation. So, we try to fit it in with a plus scale equation. Actually, in behind the south, with plus scale. Yeah. Okay. Excellent. Any more questions? Thinking to cosmology. But the study of black hole physics is here. You know, study cosmology is perhaps the most single important piece of this course. Because this is the single most important use of chemical energy in real physics. The study of black holes, maybe the single most potentially important. Potentially important, okay. So, cosmology is very important. There's a beautiful theory that has been built there. And isn't there something by the three of them that we were able to understand only the sections of the theory of cosmology. Cosmologists, and I should say this I can't quite the way they think of it, you know, they find lots of distances is given by the effort of you, has three values, the effort, and then a more value is an indication, that if you respond to it, the effort of you, has an extra value of 4, plus an extra value of 1. Given an extra value of 1, extra value of 1, extra value of 1, extra value of 1, if you respond to it, you know basically everything that is given by you. You know in today, you know basically everything that is given by you, and for instance, what is given by you, you know everything that is given by you, everything you want to know, it is so easily done. You see the small, not the proper sense, it is a beautiful theory, but an inflation theory becomes even more beautiful. It continues to be quite simple. It is a pure theory in the beginning, whether it is a question or whether it is a question. Oh wait, it is the end. Okay, so now that is the one. Let us start with the topic of today's lecture. Next we need to make the backwards. So the topic is that just to mention that we had a refund from the first course, that it was here, it was minus 25 yard. D squared plus d r squared over minus 25 yard, it was r squared d. You remember become this is the unique, static solution of uncertainty made with normal stress sets. Okay? And in the next three lectures, we are going to take these and that. We are going to take the solution and similar solutions and analyze that. So if the state is given the face of the solution, it is something funny like this, an argument. Because at this point of r less than 2 r, this kind becomes negative so that overall becomes positive, but this kind becomes positive. The metric blows up. So if can this metric be of physical use of the metric? Okay? If r equals 2 r, the singularity of the solution. And if so, it is in the right side of the physical answer. So what is this question? There is no small singularity in the solution. But we are going to try to understand that. And have a really good basis. But first we are going to do this understanding. The first thing we are going to do is technically understand. You see, this metric looks singular because of 1 minus 2 m by r in the other. And you see this is a coordinate out of r. Rather than being a factor. In other words, we are going to try to do that. And we are going to try to redefine coordinates. In other words, we are going to redefine a new coordinate out of r. G is equal to v plus g f of r. That is the coordinate rate definition. Of this sort, so as to set the R-spectrum, this form, the new R-spectrum is the only R-spectrum plus, plus what? New R-spectrum equal to 1 over 1 minus 2 m by r minus g prime of r into 1 minus 2 m by r. This is here. You plug this in here. We get a dt, dt is dt plus g prime of r here. So the R-square has this thing as g prime of r-square along with this. So g prime of r is equal to plus minus 1 by 1 minus 2 m by r. What makes this look? The d R-square along with the matrices. What remains? Of course, this has to be established anyway because we are going to push the sickness of the matrix into some other coordinate. So what about the dv-square of the coordinate? So now, that's why we are doing this. Well, first of all, that's exactly what was dt-square. Because dt is dv minus or plus g prime of r. So the new matrix, the matrix is formed in the new coordinate. 1 minus 2 m by r dv-square by the zero. Then it's my construction. What about a dv here? dv by r w. That's how you do that. The good thing about dv by r w is dv by r. There are two pieces because then it's like a derivative. Actually, one times 2 m by r, it comes from here. In a dv plus g prime of r d r. So it takes square. You just need to take a dv after this. So what do we get? This gives us 2 dv by d r g prime of r into 1 minus 2 by square and 2 by square. Okay? So this is equal to 2 dv by d r d r plus r square dv by m by r w. This will work. I think something strange still happened at r w by 2 m. But it never was your square. Now there isn't also a dv square. Okay? You might think maybe this means that it's your matrix. And therefore the matrix is doing this. That's what you could equally represent. Let's look at the matrix in matrix form. Let's look at the matrix form. Let's look at the matrix in r v, r v. And of course this is the s to the power minus 2. Okay? This part is very simple. It's r square times square d r square by 5. This is 2 plus 2 times 2 plus 2. Now the r square times d prime. Now what about the dv d r prime? This is plus minus 1 minus 1. What about this term? This was 1 minus 2 by r w. And the term for this matrix. The term for this matrix is 1 into the term for this square root of g. The term for this matrix is equal to r over 4 sine square d r sigma. This is 2 everywhere. And it has no problem except at all. The matrix is perfectly smooth. Okay? This is an analytic measure. Works for the matrix of 5 hours. In the inverse of 5, this is no problem in the matrix. Now it's 2 everywhere. This already tells us the formula. This tells us that the value singularity of the matrix. And now it goes to n. Because in fact a coordinate element. At numbers that we were working in coordinates, we somehow went back. By an appropriate change of coordinates, we've gone to making this lecture. You can manifestly smoothen it. But look at this matrix. This is a characteristic interpretation. The first characteristic interpretation of this coordinate, v that we have. That's even solutions. We have g prime r is equal to 1 over 1 minus 2 m by the power. And if you want, you can take this matrix and integrate it. Without much trouble. But the only important thing that I want to point out here, is that g of r has a logarithmic singularity. Okay? Nya r equals 2 m. So it's like, 2 m, Nya r equals 2 m. Yes? Yes. It was because this takes place to 1. Because this minuses. Goes, blows up, Now it goes to n. Now you may be, well, well, well, well. Isn't that something you think? We've kind of thought that transformation that goes up, as r goes to n. 2 m goes to n. And the question is, what was the coordinate? We've got a singular metric. And then desingularize it by coordinate change. The only way to do that is to make a singular coordinate. The point is that the right formula for the metric, we should think of the all-important system as taking a perfectly fine metric in singular coordinates. It's the other one that we're talking about. And the reason that we've got something singular only is that we took this nice new metric and we did this strange thing, because of this coordinate, the coordinate change, to make it look singular. Why? It's true that it was a coordinate of some, but there are many things that we might want to say. One of the things we might want to say is that d by d t was a thing, which is true. d by d t is also a thing. The other thing we might want to say is that things that deal with the coordinate system of things that are also static on the spectrum of that group, whatever that means. And that, in some sense, is true. And what we're going to see is that the reason that that coordinate system is singular is that it's impossible for anything to be static and our list of perfectly-assigned coordinate systems reflects in some sense the principle of that, that it's impossible to do in an appropriate sense the static coordinate system for our list. Right? In the evidence, it's zero to zero. This would also give you flat space. It would just give you flat space time. But let's say the evidence is zero. Okay. So, let me take the n goes to zero. Okay? So, exercise. Now, if this is zero, we're going to be glad. In the n goes to zero to the limit, g prime of r is equal to 1. We integrate because there's a new integration concept. You choose to say zero. So, the new coordinate, t goes to equal to v plus minus r. The new matrix in the n goes to zero to the limit is equal to dv squared plus minus 2 dv dr plus r squared dv dr. Okay? So, this flat space, what is this v? Now, this v now, v is equal to t minus plus so, let's say this is the coordinate. The radial light source would be, I think, going up. Let's look at that radial light source that are out. Of the other hand, even more black. For instance, we have flat space. For instance, that remain constant along either that space in the n goes to zero to the limit. But it's slightly unusual coordinates. One of the coordinates is just usually a coordinate. Well, the other coordinate, instead of being time, is a coordinate that's chosen to remain constant. And along either input or input, output. Second is actually true even on the coordinate. Even minus in integral dr, what minus input by r, nudge v is dr minus 2 by r, is an input. Meaning, like when I say v is equal to this, I mean, whole v constant. So, now, input nudge v is 6. Sorry, output nudge v is 6. The light source would be in this case. The light source, input nudge v is 6. Now, just as we've distinguished that, people often do the following. They are hoping to name u to the light that's actually and v to the light. Yeah, the coordinate, v is constant along something to do with it. It's not a time, sir. Like, it's no time. Say what we said. We said that when the state goes towards just an accident, what I can do with the light source, the access becomes a lot more personal. But you know, in the new, if you think just a boost, I think you're going to get a boost. If you did just a boost in the new constant, you'll get the same magic in the new constant. Exactly. Okay, if you, at any given velocity, and hence in the limit, you'll get the same energy. But now, why do you do that? Because you see what we're doing. Like, a boost is like x goes to something here, a combination of x and v. The output goes to the new constant. What we're doing here is to take t to be something in the combination of r and v. But r is the data. As you see, in terms of why you love, we can write it naturally. Now, I'm going to argue that life is a constant between me and now. Because, say, I mean, when I say constant v, of course, I'm only thinking the constant. So constant angle and constant v is obvious, because v is that constant, and the angle is that angle. But there's no reaction like that, no? Uh, uh, it's clear that if you start with a particular angle, you know what things are angle, so this is now effectively a two-dimensional. And a two-dimensional space, every now and then, is a geolocation. If it's all geolocation, space is just a null angle. And that is your one-dimensional solution. Actually, this is a more general fact, isn't it? I'm saying now. Yes, what does that mean? Why? It would better happen, because you see, there is starting from every point, starting from every point, they should be a unique, outgoing null genes. And unique white human. Into dimensions, you should be able to put your thoughts out. Into dimensions. Into dimensions, right? So there's a left-overs and a right-overs focus on the right-overs. So there should be a unique, right-overs null genes. Because if you flash it out at that point, you'll go somewhere. But that's a unique null curve. We're going to the right side. A null gene, you see, is in particular a null. But the null curve into dimensions is unique, and going to the right. What? Oh, yeah. What? Null gene is 6. Then, non-null gene is 6. Because null gene, you see, it's always open speed of light. Into dimensions, it's always speed of light. It's always the same. I don't defend that thing like that. Then, the better we look at genes, the better we look at the equation. That's true. That's the point. It won't work for us. We'll stick it along in point and out point genes. Okay. Let's try to let's try to understand in the... Yes. Yes. What happened to this coordinate? It also... Now, the geometry of the matrix was not changing. It was always one negative. Which one was positive and which one was negative was changing. In this coordinate system, moving along R at constant speed is always a null thing. Changes from being finite outside. What does that mean? It means that if you were to try to hold something as fixed R outside the vacuum, that's possible. Because every physical object moves along a timeline or null in the extreme-phase null variation. Let's look at the area. Let's look at the object. Let's look at how it was constant and detailed. Just to be a safety. Okay. It's a timeline variation outside of that. It's a null variation of the event horizon. R equals to R. And the space stack inside of that, which means nothing can move can stay a fixed value. We've already just said one of the most important properties of this of this black hole space stack. That is, objects can be held in static outside the black hole, but cannot be held in static inside the black hole. If you wanted to try to hold something static R equals to R just outside the black hole, you could do that only by bringing extremely high force. It's actually the surface of the black hole that forces it infinitely. And inside the black hole, no force can be out of it. It's good. Okay, let's try to understand a little more what happens in this black hole space stack. So now to do this, I'm going to look at the equations for massive geodesies. We do this actually in both coordinates. Just to see how it works. So let's first, in the old coordinates, square is equal to 1 minus 2 in p squared minus 2. Now, in geodesies, we can always get constant angular position. We don't need to write this. But computing on one arm, there's actually an easier way to do it. And that is as follows. We all know that the motion of the particle and the equation of the motion of a particle are all being thrown to follow the action. They are being thrown from the action in integral square root of ds. In integral ds is equal to minus m in integral square root of g mu here ds mu here. And if you look at the equation of the motion from this form, in my day, it's actually that would be some better computing response. The action is equal to minus m into the original working system. So we have square root of 1 minus 2 and pi r into d squared minus d r squared into 1 minus 2. It's a very interesting motion to find out how this function is. You can write this as I said in integral square root of 1 minus 2 and pi r minus d r by dt of d squared into 1 minus 2. Now, how do we integrate the equations from such a from such a gradual you could of course try to find the equations in motion but of course this is simply the energies. So the simplest thing to do is to compute the Hamilton of the system and then equate it to the constant. The energies in itself, because there's fine numbers. So what's the Hamilton? We actually don't know the answer. Let's do it in this language. So we have a minus m then what we're supposed to do is to do p q dot minus n dot. So p is there granted by the energies by their velocity. The fact is two answers that we get are not by 1 minus 2 m by r into the square root of this system. That's p q dot r dot squared. So then that becomes simply minus m into 1 dot squared by 1 minus 2 m squared and then we open up the square root by 1 minus 2 m by r as well. So the square root was this. So it gives plus 1 minus 2 m by r double squared minus r dot squared into 1 minus 2 m by r squared cancels one factor of this by r dot squared minus 2 to the 0 h to the 1 this is 1 minus r dot square root of r dot squared it's a little bit of an extra by this side. There is a limit to that plus square root of m minus r by r. Now that's going to matter. There is a limit to n by d r by r. So it's coming up to minus r by r. Ah, we're saying this is going to be minus r. Okay. Wait, I thought this side and then guess it. Okay. So this side of the is going to be conserved. It is not squared. This is equal to 1 minus 2 m by r squared over 1 minus 2 m by r minus r dot squared minus 2 m by r is to solve r dot and equal to m. All this because that might be what's happening in the second row. Okay. All this because we put that in the process and it's a simple application of the formula but it will be a point. And it will make the difference. So r dot squared is not squared in 1 minus 2 m by r. Okay. It's not squared in 1 minus 2 m by r minus m squared in 1 minus 2 m by r squared. We have e naught squared r dot squared r dot is equal to let's try that again. Then we take the equation d r by square root of 1 minus 2 m by r into e naught squared in 1 minus 2 m by r minus m squared in 1 minus 2 m by r dot squared. What is this equation? How does this integral become 5 in the neighborhood of r equals 2 m? Okay. So this describes a particle following in this gravitational potential. Okay. And I'm interested in what this equation becomes like. Okay. In the neighborhood of r equals 2 m. So in the neighborhood of r equals 2 m by r, let's see. In that neighborhood every this is going to be 0. This is going to be 0 even fast. So you just ignore this. Then you got 1 minus 2 m by r and you got 1 minus 2 m by r. So this what does this integral have to do? Does it have to be 0? Exactly. That's what that said. Yeah, exactly. So this becomes d r divided by 1 minus 2 m by r in that axis. This is complete. This becomes the equation in the circle of 1 minus 2 m by r by r. But let me know what you're interested in. Okay. So this so the equation becomes either u equals constant or v is equal. Is that particles falling into the black hole in which they internalize them. Become lines of v equals constant. These are v equal constant. When it falls and goes it speeds up. As far as the observer is going and the event horizon moves to the speed of light. So it's trajectory comes in this equation from the last quote. But we know the light photons are working at a light of v equals constant. So it's near the horizon massive particles also move along the lines of v equals constant. The next thing we want to check is how long does it take for an observer because of course you're on a rocket ship and you're falling into a star and you're falling into a black hole. How long does it take for you to reach it? How does it take for you to reach it and then you fall and to reach anywhere but you reach the event horizon at a high light time or infinity power of the future as measured by a rocket ship. What? As measured by the data form. The data changes. And then you take an internal body and that's the body form and that's the plastic form. Again, you start with a finite distance but if you take how much I'm going to take there will be potential for an algorithm that says in the motion near the in the motion, near the event horizon. So we're going to decalculate this in v. What? In the decalculate in v. In the v, that's right. You will reach you will do it at t equals infinity but the question about So to take this trajectory to compute the proper distance is to compute the proper distance in this last bit of motion. Proper distance in the last bit of motion. Zero. Because it's very like an algorithm. You reach the you reach the event horizon at a finite proper time and in fact that last bit just whizzes by they almost don't to go through the end. Okay, so the next thing that you see is this that if in finite proper time once again you don't study the trajectory of light rays in this geometry that what we start with the trajectory of light rays in this geometry inside or outside okay, so inside the event horizon I want to look at what the what the outgoing genesis so suppose an inside event horizon and an arc less than 2 m outgoing non-genesis what for the outgoing non-genesis for non-genesis for arc greater than 2 m these multiple lines are usually u is equal to t is plus or minus minus in technical terms 1 minus 2 m let me be like this trajectory in terms of quantity easiest way togetherness, where is the easiest way to go back to the to the basic method either let's say in these quarters we have inside the event horizon our d r squared becomes a light light voltage that's all of this non-equation this non-equation is d r by t is equal to d r by t in the outgoing genesis we have smaller smaller arc in let me make the rest I'm sorry let me use let me use the following let me use d r squared to be squared 1 minus 2 m by r and then when I was using v it was plus or minus so the input line was minus right the guy was going to be the symmetric let me yes in just one this is the method now let's look at the 2 g v 6 the 2 g v 6 and then the 2 now see to get the 2 and 1 now what we have to do is to set v s squared the solution of course is just d v is equal to 0 light is constant v r now you know it what are the automations but the outgoing g v 6 are the ones that solve d v in the 1 minus 2 m by r is equal to where? no I want my definition was supposed to be v is constant I am going g v 6 u is constant and I am not sure if I have the right sign here d v in the 1 minus 2 m by r is equal to 2 m yeah I have the right sign 1 minus 2 m by r over 2 so this is the second g v 6 the second mother g v 6 the one that should be out for you now v is you know like that v increases that ok so as v increases r increases the nano g v 6 go out when we are out of the naturalization but even the second nano g v 6 goes inwards inside the naturalization both nano g v 6 go towards smaller arm the first one was the 2 and nano g v 6 zoom to smaller arm and also v the second one towards towards smaller arm and I should have given you the following picture of this picture here that is the solution which gives you the form but out from which nano g v because when other g v is transmitted then no g v is exactly which nano g v is and then this this this fact in your picture ok and the picture is in terms of what is called there is a drawing of the space standard in which the drawing does not fit fully represent the metric but fitfully represents the metric multiplying and overriding such that light rail and causality is because 45 degree lines between the path of light rail in this diagram as well as in the space standard in both the picture there is the predictor of the standard ok so what we are going to say is to this batch of the space standard we are going to say that the space standard this is the electrolytes the electrolytes itself are now at an extensive it is the line of all of this outside of all of this it is the region inside of all of this on this line it is the path of the finite on this diagram so if you are outside the electrolytes you can go in or you can go out inside the electrolytes you can try to go in in which case you go in it or out in it you can try to go out in which case it is still in that it is this region space in which space can form ok we understand this diagram ok thanks ok