 So in this lecture, what we are going to try to do is to understand black holes based on better than we understand them. So last time we did some intuitive calculations. I wanted to make these calculations more precise to get a more precise and better understanding of black holes based on. So the very important thing about a black hole is its causal structure. Okay? And so this reason why we are going to try to do is to understand how to characterize causal structures based on this. So let's first start with a very simple example. Let's start with flat space. The square is equal to d r squared plus d r squared plus d r squared plus r squared d omega 2 squared. So what we are trying to do is to define u is equal to or d minus r v is equal to d plus r. These are the flat space outgoing and outgoing coordinates. In the sense that u is constant along an outgoing g v square and v is constant along an outgoing g v square. Now let's rewrite the metric in terms of u and v. So you see that, so you see that d t squared minus d r squared is simply d u d v. So d s squared is equal to d u d v minus r squared d omega 2 squared. So let's pause for a moment to note that u and v are, it's manifest from this from the u and v unknown coordinates. Why is that? That's because if you keep v and the square points fixed, then the change of the interval is 0 because there's no d u squared. d v is 0. Okay? Similarly with you. So this manifest as we already knew. So staying on a line of constant u is another thing. Staying on a line of, the diagnostic for that is that there's no d v squared. Staying like on a line of constant v is another thing. The diagnostic for that is that there's no d u squared. Excellent. Now what is the range of variation of u and v? Okay? Now r is greater than 0. r is u plus v. Sorry, v minus u. So v minus u is greater than 0. Other than that, v and u are completely unconstrained. Okay? Now, so I'm going to draw that. Okay? So we put in these conditions in a moment. But imagine that it wasn't there. Then u and v would go in both range from minus infinity to infinity. And we have the space time in which u and v go from minus infinity to infinity. Now I want a visual representation of the space time. Pembro's inventive. A visual representation of the space time. Okay? Which faithfully captures all the causal relationships of the original space time. Okay? That's what I'm going to describe. Okay. So the way you do this is the what? See, v, if you want to draw a diagram of the space time, because v and u go from minus infinity to infinity, it's a diagram that you fill out of your board. And beyond. However, if I make a variable change, suppose I say v is equal to tan, v2 is equal to tan u. The fact that v is equal to tan is not important. It's just some one-to-one function that ranges over a compact value when u and v go from minus infinity. So if I ignore this, what would the range of u and v mean? We would have a u belongs to minus 5 by 2, 5 by 2. v belongs to minus 5 by 2, 5 by 2. This is very normal. Taking this as an account, since v and u are monotonic functions of same monotonic functions of u and v, the condition is simply that we got this range plus that v is greater than u. I'm going to draw a diagram of the space in which the u-axis and the v-axis will be depicted as tilted at 45 degrees to reflect the fact that they land. So the diagram, first I'm going to draw this diagonal. What are lines of constant u? Lines of constant u are outgoing values. These are lines of constant u. Is that the definition? These will be lines of constant u. Now in this diagram, time you should imagine is going upwards. Each of u and v increases time increases as you see. And space roughly speaking goes like this. So as you go up on this diagram, u increases, so this line here is u equals 5 by 2. This line here is u is equal to minus 5 by 2. This line here is v is equal to 5 by 2. This line here is v is equal to minus 5 by 2. Now space, I'm not yet implemented the condition that v is greater than u. v is greater than u. This condition, however, just restricts us to the right-hand side because this line is the line of constant u. It's the line r equals 0. To the right of this piece. So the diagram, the space that we're interested in in this v u plane goes like this. This is u equals minus 5 by 2. This is v is equal to 5 by 2. This is the line v is u. These are lines of constant u. These are lines of constant v. And my whole space time is different to u. Now let me write out the matrix of my space time. The matrix of my space time is what? ds squared is equal to sec squared u sec squared v into minus r squared cos squared u cos squared v. And r itself is a function of u. In order to work out what function, though we won't really need it. In order to work out what function, we know that v minus u, that r is more v minus u by 2. That's all. Can't be minus, can't be 0. Exactly. So v minus u is equal to 5 by 2 is equal to r. And therefore r is equal to 10 v minus 10 u by 2. So this matrix with r substituted with this formula is the original matrix of space time. This looks like an enormously complicated matrix. But notice that if we just dropped this factor, this overall factor, then as far as following radial light rays is concerned, this matrix, the u dv is the same as the original factor. So up to an overall factor, it's true that this thing looks pretty complicated. But if we're not interested in that, we are just looking at the radial motion. Up to an overall factor, this matrix is very simple. The matrix is just up to an overall factor. So in particular, light rays continue to move at 45 degrees in the capital U capital v plane, just like they live in the small use model. So lines of constant capital U and lines of constant capital V continue to be null geodesics. The range of capital U and capital V is compact. So this little diagram here gives you the complete depiction of your space time. That is, free sufficient to determine causal questions. Suppose you want to know, you've got some point in space. And you want to know, can it affect, if you set the center, can it reach here? Just a question of whether it's inside or outside this 45 degree line. Don't look down. You see that at any given point in space, for instance, you can receive signals from this point of view. What is this line? This line is just r equals z. It's the origin of your space time. What is this point? This is the infinite time. This point was minus infinity. And this point, if you think about it, was infinite space, almost infinity. Even txt. However, in this diagram, you're seeing that the boundary of space time has a lot more to it than you might have thought. Again, as far as causal relations are concerned, the points t goes to infinity given r fixed, the infinite future. Or r goes to infinity given t fixed. I just smoke up particular examples of ways of going to infinity. The other ways of going to infinity are going to infinity and t going to infinity. In the same way, keeping r minus t fixed. The way t infinity reaches upon the spender's diagram, which is called scribes. Scribers describes the future of all outgoing light rays. Any outgoing light ray in the future will eventually get scribes. Just because all outgoing null gd6 are like this. This bar is called scribes. And it represents the path of any outgoing null gd6. All outgoing null gd6 are come from this. I don't even know what this guy is called. I0 plus I0 points. These points here represent the path of all time-like observers. Because every time, no time-like observer. You go to the path, you can't go at 45 degrees. So every time-like observer has come from here. This is the future of all time-like observers. And this is just infinite distance at fixed time. This is a depiction of flat space. This is a form of flat space. It's not terribly frigid. But this whole way of thinking comes into its own when you start looking at black holes. So what we're going to try to do is draw an equilibrium diagram from black holes. Then you'll see what's interesting. But even here, suppose you're here and a bomb explodes. Letting out light. What parts of spacetime does it get an effect? Where does light go? It was destroyed plus here. This thing was a pendulum diagram of flat space. And I didn't have too many surprises. But anyway, what we're going to try to do is to understand this black hole in the pendulum by drawing a similar diagram. We're going to do this in six. So what we're going to do is to remember that we had these coordinates for the black hole. That we used, which was T minus integral dr by 1 minus 2 by r. T equals T minus integral dr or 1 plus 1 minus 2 by r. This integral, by the way, is often called r star. It often tells that people see this as our star. Whatever that is. The result, thank you. This wasn't easy integral. So let's compute r star. r star is equal to integral dr by 1 minus 2m by r. This is equal to r dr by r minus 2m. Exactly. Exactly. Yeah, exactly. So this is equal to r minus 2m plus 2m over r minus 2m over r minus 2m. So this is equal to r, as you said, plus 2m log of r. We see many useful things in this expression. But it's the large r. The leading behavior of this is just r. So the large r r star is a very different one. However, when r goes to 2m, which is a finite value, r star goes to minus 8 log of c. So r star varies from minus infinity to 8. As r varies from 2m to 8. Remember. We describe these null coordinates. We describe the black hole in either null, in going or null out going. Black hole is another set of coordinates. In in going coordinates, we used v and r. In in going, we used v and r as the coordinates we described. Outgoing coordinates, we used u and r. You might ask, I wonder, why we're using u and v? This would be the analog of what we did to draw the Vendor's diagram. Because these are the null coordinates. Let's draw it. Let's see what we get when we do du, du. So now, by definition, d of r star is 1 over 1 minus 2m. So this is equal to dv minus dr by h. So with the black hole space, then we start the graph. Apart from an overall factor, this is equal. So the formula is that 1 minus 2m by r into du d. For the rest of this discussion, we know the sphere part. The ds department. Please don't hold it in discussion. Is equal to the original black hole metric. d squared, you know, 1 minus 2m by r minus 3r squared over 1. You're right. This sort of similar to the analysis in flat space. It's sort of similar to the analysis in flat space. Except for one difference. There's an r here, which of course should be regarded as a function of u and v. Somehow, in these coordinates, the black hole metric is not manifestly non-sigma. It wasn't the ur coordinate. So it wasn't the vr coordinate. But the reason in these coordinates, because something funny happens in the metric of these coordinates. Namely, the full metric bandages. The determinant of the metric bandages of r goes to a singular. In these coordinates, we're just going to see the most metric in the whole system. So these aren't yet satisfactory coordinates. They're not yet satisfactory. So we can easily make them satisfactory. The idea is that if we're going to do a coordinate exchange here, then we're yet to do this here. Let us remember what r was as a function of u and v. So it's exactly like what we had before. r star is equal to v minus u by 2. Because v is equal to t plus r star. u is equal to t minus r star. So v minus u by 2 is r star. And r star, it was a function of r. If you want this, you can take this relationship, invert it to find some using this r star. Using this, you invert it to find some relationship that will give you r is equal to r of v minus u by 2. It's a complicated inversion. But in principle, you could invert this equation and find that expression. So here we'll be writing this r. We should think of it as a function of u, v and u, obtained by plugging v minus u by 2 to this and solving for r as function. The first important point is that r star is not a function of v and u independently. There is a function of v minus u. Now, what I'm going to do now is to try to get rid of this singularity. Look, it's in general a complicated function of r. But let me take this relationship and specialize it to the neighborhood of r equals 2n. Where this guy goes. Then what I get is 2n into log of r minus 2n, v minus u by 2. So that's 4n. The relationship occurs very easily. But because we get r minus 2n is equal to e to the power of v minus u by 4. There was a plus r plus 1. Is that right again? Yes. Perhaps suppose we go to the, you come to this. Suppose we go to the neighborhood of r near to it. This term dominates this. So the neighborhood of the evangelization, where things are going back. And we can forget about this. So everything interesting is happening from here. So the neighborhood of that evangelization, we get this equation, right? This suggests the following. This suggests the following change of events. Suppose we have v is equal to e to the power, exactly. Right. So you see, this is the kind of factor that we have out here. So this is the kind of factor that we have out here. So this is the kind of thing that we want to obtain from the change of events. So suppose I do 4n. Suppose I do v is equal to e to the power of v by 4n. And u is equal to e to the power minus u by 4n. So suppose I make this change of events. The metric, which was, right now we've got du, du, du. And du, du is equal to e to the power v minus u by 4n, du, du. So that du, du is this object. And there's the same kind of singularity as the physical metric. Like I say, since this physical metric has a 1 by r here, du, du by 2n of the event horizon becomes whatever this is, 2m. So it's approximately r minus 2m by 2n. Just approximately r minus 2m by r. 1 minus 2m by r. Which is the true metric, do you see? So do you see that the metric, the black hole metric in the day of event horizon is desingularized by moving to these capital V capital U coordinates. Because in this metric, expanding to reading order, it's 1 by r by 2n. Right. But the way I expand it is r equals 2m. So the only thing, as far as that's concerned, you see, what I'm saying is that these two differ by, this is 1 over r by 2n. And this is 1 over 2m by 2n. Yeah, r equals 2n. So this is between r and 2n by 2n. In every other place, we say r equals 2n. So now, what I'm saying is that this funny-looking metric in the little, little requirements becomes just a very simple metric in the capital V capital U. Now let's be a little more precise. We'll just work with these capital U and capital U coordinates and write down the forward black hole metric without making approximations. What we guarantee is that this metric will not have any similarities at the event horizon, because we've just got the expansion down the same, the event horizon. And we've seen that in the neighborhood of the event horizon, this metric is just, there's a motivation for making this change in there, doing the expansion near the event horizon and then seeing that this is the change of variables we need to desingularize it. Is this completely clear? Once we know what change of variables we want to make, of course, it's just algebra. This was a metric. You want to realize it in terms of these U and V coordinates, capital U and capital V coordinates. So let's do that. So what is a metric? It is 1 minus 2m by R into, now, dU is equal to d capital U into the power plus U by 4m. Plumbing area? We're in a factor of 4m. We're in a factor of 4m. Thank you. Let's get the factor of 4m. Thank you. Thank you. Thank you. So we want, this thing would have 1 over 4m. So we're going to have to process this to make it to the limit. Firstly, of course, everything here should be written in terms of capital U by capital V itself. So this is minus 4 in the working square into 1 minus 2m by R, dU vp. Next question. What is R? In terms of A as a function of capital V. Okay. So we need to write down the final answer. What is it supposed to get? Oh, definitely. So the usual definition is U is equal to minus 6. Let me just check that it is U and V at the same time. Yeah, U is T minus r star. V is T equal to plus r star. And here we are expecting U with a minus in the bar. So then I suppose the only difference here is that this becomes plus. But actually, it will become minus again because of U. I know. Minus 1 is minus 1. Then the minus is minus 2. Well, 2 minus is cancel. This minus was therefore the differentiation. Yeah. Okay. So here there was a minus. Okay. Another minus. One minus. One minus probably. Exactly. But this minus is minus 2. Right. So actually there was no minus. And then there you go. Okay. Now writing down what they claim that the answer is, let's write it down a little bit. Okay. So they say that we get, that you can write V as square as equals to, I will say it is 2 m 2 prime 4 e to the power minus r by m. V U V V. I mean the square part. Where U V is equal to minus r by m prime 2 m. Let's check that U V as we have defined the business property. That's obvious actually. You see U V was minus of e to the power V minus U by 4 m. Okay. Now V minus U was 2 r star. It was equal to minus e to the power r star by 2 m. But r star was r plus that law. Okay. It was equal to minus e to the power r by 2 m times e to the power and there was a log of r minus 2 m. So we get, and that's 2 m log of r minus 2 m. Now r star at 2 m times log of r minus 2 m. So 2 m is cancelled. So we get. 2 m is 2 m. Yeah. 2 m. Where do they have this extra 2 m? Where does that? Where do they have this integration? Ah, that's the, sorry. Sorry, sorry. In r star. I'm going to modify it. Okay. My r star was equal to r plus log of r minus 2 m plus constant. Okay. And I'm going to choose the constant to be log minus log of 2 m. So this becomes the dimensions. Yeah, that's how we're going to do it. Okay. Then you get the 2 m. And we're happy, right? So we've confirmed that u v is equal to this. Okay. So that's the relationship between r and u times v. Okay. And now we want to confirm that in these new coordinates, the metric takes that form. Okay. So let's see. So what was the metric? The metric was minus 4 m, the whole thing squared. 1 minus 2 m by r. Into this. And then we substitute, we substitute v minus u is 2 r star. So this is minus, this is e to the power minus 2 r star by, sorry, minus r star by 2 m. Do you agree? Then we make the same substitution. Okay. So let's make that substitution. So we get, this is just exactly the inverse of this. So we get minus 4 m whole thing squared. 1 minus 2 m by r. Divided by e to the power r by 2 m. Into r minus 2 m by 2 m. Okay. So that becomes what? Yeah. 1 minus 2 m. Yeah. But that's what remains is 2 m divided by r. So we get this 2 m by r. 1 m in the packet. Now what is 2 r? Then 1 minus 2 m by r. That's it. Right. So you get 2 m divided by r times 4 m squared. Times e to the power minus r by 2 m. Times 2 minus. So that's 4 into 4 into 2. That's 32 m into n squared. That's m cubed divided by r e to the power. So I'll write 2 as just because you did. Continents in which the black hole is manifest Q vector. At the regularization. These are the UV coordinates. Why is the black hole manifest Q vector at the regularization? Because now, you see, there's nothing funny happening at the regularization as I was going to do. Of course, the coordinate change is similar as I was going to do. There's nothing funny happening in the coordinate change. That's not right. One of these will go through it. Actually, either u goes to 0. Nothing. That's not true. It's just here. It's just 0. So we've got a manifestly nice and smooth coordinate system. Okay? Not a manifestly nice and smooth coordinate system for the backup. Now what I want to do is before proceeding, I want to understand what were the allowed ranges for the coordinates u and v. Firstly, what were the allowed ranges for the coordinates u and v? First, stick outside the event horizon. If I stick outside the event horizon, r is always greater than 12. What do u and v rate? It's like the flat space k as we consider except r star doesn't stop at 0. r star is not a positive guy. It goes through r and all that. So little u and v both belong to minus infinity to infinity outside the event horizon. Take an example of something. You see the little u, little v coordinates. As little u and v rate overall, they're possibly cool. It covers only part of the space time. The part of the space time that's outside the event horizon. We have already seen that there is a part of the space time that is inside the event horizon. We're going to these ingoing coordinates, coordinates in which we have little even hour or less than an hour. So we know that there is more of the space time to cover. But the little u and v coordinates are only covering part of it. How are the big u and big v coordinates different? So let's recall the definition of big u and v. Big u is equal to minus u to the power minus u by 4n. And big v is equal to u to the power v by 4n. So when little v went to minus infinity, big v went to change. v belongs to zero to infinity. What about big u? Where are you going to ask me? Think ranges from minus infinity to zero. u belongs to minus infinity to zero. Actually the reason I put this minus is so that capital u increases when little u increases. The space time in the little u and all of the packs that little u and little v saw is zero to infinity with v and minus infinity to zero in u. Now zero to infinity and minus infinity to zero in u. But we have seen that nothing funny happens to the metric in the u v coordinates as u and v go to zero. We can just take this metric and regard it as a true metric of a space time. Or regard it as a perfectly fine metric for arbitrary values of capital u and v. A capital u and capital v. This is a metric that manifestly agrees with outside the event horizon of the black hole in the region in these two regions. But it makes sense everywhere because nothing funny happens when capital u and capital v go to zero. All that happens is that r goes to v. We will tentatively say u belongs to minus infinity to infinity and v belongs to minus infinity. And what will happen? Is that really the full space time? You see while r equals to m is not a real singularity in the Schwarzschild space time. There is a genuine singularity in the Schwarzschild space time. That's at r equals zero. So if you go to r equals zero then it's easy to check. I haven't checked for you but it's easy to check. Curvature is actually not a real singularity in the Schwarzschild space time. They are not going to give you a space time past that curvature. It's not a true metaphor. So where does r equals zero sit on the space time? In order to find that, that's a beautiful formula. r equals zero is the equation u v is equal to minus one. u v is equal to r equals one. So r is equal to one. Let's do the other thing. So now all of space time. This way of seeing when we are using the relation where we are already on the limits of u v. Then we are using the relation like how we define the capital U of capital V. And that from there we are already on the limits of u v. We know that we are. Now this space time is supposed to be this metric with this relation. It's supposed to solve for r as a function of u v from the other side. But for this relation we already have a limit on what's the value of capital V. No, no, no. This is not the full space time. You see, we got this space time by looking at what it was in the u v patch. But now our idea is to take this space time and trust it everywhere we can. It's like an analytic continuation of a function. You know the function in a disk. And then you ask, can I use the same formulas to define the whole complex limit? It's like that. Because there's nothing stopping me. I'll do it. Fine. So we're going to be stopped when u v is equal to plus 4. Okay. I'm now going to try to draw a picture of this space time. Allow me to learn some of the general diagrams. All right. So let's do that. I'll keep this on the board. Okay. So. Yes. Sir, if it's small u and small v, how can we do it with the only word from ISE? Yeah. Actually, more technically what happens is that this space time is geodesically incomplete, which means, as we've seen, as we saw last class, that if you have an observer, you can reach the boundary of that space time and find that property. So that space time does not answer the question, what happens to the observer at 5 o'clock in the evening? And if what he reaches the boundary of the space time, then what? It makes no sense. I was able to write the metric in terms of small u and small v. What? Small u and small v is greater than v. And there's a relation between u, small u, small v and v. Yes. That relationship works outside the elections. So how is this different? What is this difference between small u, small v and capital u? In these two relations, I can say that I'm not in the whole space time. In these two, I can say I'm not in the whole space time. You see, what we're going to find is that this metric that we get by extending it to the maximum form in capital U and capital N is not geodesically. That is, for every observer, for infinite proper title along his world line, you lie within this full metric. You see, the more you take the short sheet metric as a metric that is defined only outside the entrance, that makes no sense. Because, as we just discussed, an observer drifting along in his rocket ship will reach the edge of the space time and find that proper time. Coverture is a fine and smooth thing. Nothing catastrophic is happening here. Now, we have to answer the question of what happens to the coordinates. So it makes no sense to do that. So what are the options? The options are the, but it causes a solution to Einstein's equations. If there's a solution in this coordinate system, it will be singularity. What are the options? The options are look. We found a solution to Einstein's equations, but somehow we can't use it in stupid coordinates. We found it in only one coordinate patch. Now, what we're trying to do is to find the solution. Who's one coordinate patch is the schochi's. Now, because we're looking at solutions of Einstein's gravity without any sources, the solution is going to be completely analytic. It's not going to be like there's a non-analytistic situation somewhere. The only way that could happen is if there's a non-analytic source, solutions without sources. So what we're looking for is a nice smooth space time that, an analytic function of coordinates, that in some region reduces to what we already have, 5 coordinates. So what I'm now going to solve, what I'm going to do for you is solve that in person. You take this space time. Imagine God is giving it to you. You know, somebody is giving it to you. You've got this space time. Now, in this space time, I'll find a patch which, under an appropriate coordinate change, becomes exactly the schochi's. But this space time has no pathologies. None of the pathologies are associated with the schochi's. So it's the true solution. We're going to use these coordinates that we found this in. You can later then check this in these coordinates. That's the sort of sense equation that we have. It's not a branch we can see where there is no singularity. There's no singularity except the part of the V. R is the zero singularity. That's what this should be all about. Yeah, and we're going to see this time. The diagram I draw will make it with V. Okay. Now, look, because there's U and V business going on, we've got the V. So now we're going to play the same game that we played in space time to draw a better space time. This is a perfectly smooth space time. The only thing that's not so nice when we draw, depicting it in our minds and our diagram, is that even we go to a very infinite space. So we've played the same thing that we did before. Maybe we let U as it would have done. I think I'm going to relate to this. I would have liked to use new U, but that's confusing. Think of some letter that reminds you of U. U brain. Is that good enough? Okay. U. I'll tell you that. Time-proud thing that reminds me of Dan. And V is equal to Dan V. Plugging this then gives you the metric in terms of this Dan, V till that, and the V till that, V till that. What else? The figure of the metric and all that, I'm very interested. The most important thing is what is the range of the book. Firstly, U is capital U and capital V range from minus infinity to infinity, just like in flat space. V till that, V till that will range from minus 5 to 5. However, this space time is going to be a couple of an awful line. Capital U, capital V is equal to 1. Okay. So first let's draw this. So first we have this diamond that we drew also from that space. Okay. This is U is equal to minus 5 by 2. This line is U is equal. No. U till that. U till that is equal to 5 by 2. This line is V till that is equal to 5 by 2. V till that is equal to minus 5 by 2. This diagram is a square. This is the 90 degrees. Okay. Now, the next question is what are the lines UB is equal to? Capital U, capital V is equal to 1. Which is the same as the question, what are the lines? Tan U till that, tan V till that is equal to 1. But now think about it. Suppose V till that was 5 by 2 minus U till, then tan V till that would be got U till that and this relation would be satisfied. So one way this can be satisfied is that V till that was U till is equal to 5 by 2. Is there any other way this can be satisfied? Well, what about V till that? A V till that is minus 5 by 2 plus U till that. This is one of the questions. 3 by 2 equals? 3 by 2 equals? 4 by 4. 3 by 2 which is the same as minus 5 by 2. Let me check this. I want V till that, tan of U till that. So one way is having sin and cos being interchanged. The other way is having sin and cos being interchanged with the minus 5. So I think that does that. So it's pi minus pi minus. Let me depict this first line here. We'll get the second one by symmetry. And then we'll check it. Let me find this first line. This first line here is V till that plus U till that is equal to pi by 2. So if we were to draw this square like this, this would have been a 45 degree line like this. And it would have been a 45 degree line that pass through 5 by 2. It's from the middle of this to the middle of this. And I'm claiming this also. Which means I suppose that I just want to replace pi by 2 by minus 5 by 2. Let me check that. So suppose V till that is equal to minus 5 by 2 plus U till that is minus cos. Because if tan V till that is equal to 1 upon tan U 1 till that is equal to 1 upon U till that. Perfect. That agrees with this. So it must work out now. Exactly see. And accept it. And he's right. It's clear that it works. Pi worked. Minus 5. Thank you, thank you. It worked minus before and minus now now. Thank you, thank you. And that's exactly what we wanted. Same slope, but different intercept. Here it's pi by 2, here it's minus pi by 2. So it's right in the middle of this equation. Draw this little better so that it looks like this. So this is what it was. And then we take this part and cut it off. This part is not there. This part is also not there. See, we brought this line with a movie. Because we see lines as opposed to represent singularities. This diagram is not geely simply complete because let's draw this. Now I'm going to give you a case. This thing here is called a singularity. How do we have scribes? Just like for flatness. It's where if you understand. This is right. This is another square plus and another square minus. These points are the spatial infinity. These points are the... Now let's go back to this diagram and look at it. In this diagram, where was the original patch that was outside our back hole? It was this. This was the patch that represented the little u and little v coins. Little u and little v going from minus infinity to infinity. This is outside the event horizon. The singularity point is also at zero point. It has one point. This is a bit misleading because you multiply by infinite factors here. This point is the whole space. It's a bit misleading to think of that. In this compactification. Now, the first thing we did in the past lecture is going to go to these in-going editor-in-game standards. In-going editor-in-game standard coordinates were constant along lines of v. And we found that the event horizon was nice and smooth in these coordinates. Now what are these in-going lines? These are these lines. First of all, let's have the coordinates in an out-going editor-in-game standard. Those were lines of constant u. As R went. What were those lines? Those were these lines. That the event horizon was smooth in these coordinates. Now look at this diagram and see that we've actually concluded the smoothness of different things. This was the event horizon that we saw in the in-going editor-in-game standard. This was the event horizon that we saw in the out-going editor-in-game standard. The point where these two diagrams meet on the Federal's diagram is the names of the bisonation. It's an important point when studying the properties of interval black holes would be better. There are several things I want to say. Look at this diagram. In this diagram, look at this upper right. Suppose you're here. Your future light goes here. So any signal you send from here will go ahead and hit the singular. Anything, no matter what you do, you will also go ahead and hit it. Suppose you're here. A signal you send is two options. One is to go ahead and hit the singular. The other is to go out. And the scribe-plus has a black hole that wasn't there. Just like in flat-nose. You yourself have two options. You can either go and fall into the black hole or out. It's up to you. What about if you're here? If you're here, every fast license, everything that has happened has in this space-time, in this space-time, we talk about this relation to actual black holes. This was called a neutral black hole. It's a mathematical idealization of a black hole. It's four-wheel drive, as we will discuss. But in this maximally analytical continued space-time, in this space-time, we have a region here with two distinct singularities. One of which is the past of everything, except light-rays that are emerging. And actually, there's also this. These points are also past. And oh-oh. Oh my, there's this dog. It's here. Okay, yeah, we're going to have to hit it. Right? So this region is called a white hole. Because everything in its past was a singularity. Whereas this region is called a black hole. Because everything in its future is this. Another really strange thing about this diagram is that there are in this diagram two asymptotic scripluses. And two asymptotic scribinuses. So this diagram depicts two universes. Two universes in which people can grow up, be happy, and die. They're both communicating with each other. Except through going into the same diagram. They've gone into the same diagram and they've talked for a while. Not too long. It's very strange. The space time is, when you look at it from here, it looks like a small deviation of that space, but a global structure. It's very different. Connected, described. Two mothers are connected by what's called an Einstein-Rosenbridge world. Okay. Now this is the structure of the space time. One last thing I want to say. Last thing I want to say about this was that in this diagram, what do lines of constant t, the original Schwarzschild time difference, in this region? Okay. Lines of constant time look the following. Okay. They look like this. This event horizon line, this line that separates between inside the blackboard from the outside of the blackboard, is a line of t equals infinity or t equals minus. If you look at it and make it supplementary. Okay. If you look at it, it's connected to the fact that this is described by the v-core of it. Because in the v-core that v is equal to what's it? Minus rc plus rc. Okay. Now, if v is held fixed and r star goes to minus infinity, then t must be equal to plus infinity. But r star doesn't go to minus infinity. Okay. So, this region here, the region explored by the in-go-ary that we could stand in coordinates are the line t equals infinity. There's all the line here that t equals infinity. The region explored by out-go-ary that we could stand in coordinates t is minus infinity. Actually, we're going to have to stop it in two minutes because this is the stop. But actually, there are many interesting things but, you know, I already imagine the step is diagonal. It's two minutes more. Step is diagonal to think about. Very interesting. Okay. The last two minutes, I have to make the boundary before the physical boundary. This sounds totally weird, you know. This sounds totally weird because blackouts in our universe can be formed by collapse. That's the one we talked about in our last lecture of this course. Monday would be the last lecture of this course. Because I have to go over it next week. It's basically time now. Okay. Two minutes to go. Okay. So, what we described in the last lecture of this course is more about blackout. This is going to be a formation of blackouts in other properties. But blackouts in our universe can be formed by collapse. You don't expect a little localized collapse process to give birth to a whole new universe. You know, black spaces don't have these regions. Because black holes we rarely have in the future. Now, the important thing is that it's space time for this is what is called an internal blackout. It's better for a black hole than it will always be there in the future and it has always been there in the past. The more you look at the metric for a black hole that is formed from collapse. That's very simple next week that, you know, it gives Indian physicists to vary the outlook. First of all, to describe black hole formed from the collapse of a short burst of null weight. It's like a shell of light. That's very simple. That implodes inwards, starts spherical and implodes inwards. The shell of light contains energy and once it becomes small enough can create a black hole. It's interesting. But essentially, the point is this. A shell of light carries some energy. Suppose this light has some thickness to it. Where the thickness of the light is much smaller than the Schwarzschild radius associated with that energy. The light just says into the Schwarzschild radius without interacting with itself. But now the metric outside the light becomes Schwarzschild. And once it's inside, it's a renderise. No matter what happens here, the black hole is formed. This is a very beautiful metric that we wrote down in a particular limit for several processes. And that is df squared is equal to 1 minus 2m of v by r. dv squared minus 2 dv v r plus the initial r squared dv. Where m of v is a function that interpolates from 0 at v equals minus infinity to constant. At some time, actually this metric is only accurate for this process. This metric is sourced by another, what they call null dust, lightless. It's only accurate when m of v varies rather quickly. The thickness of the pulse is not very much. Now for such an endpoint, you see if m was constant, it would be black hole in entry point. But if m is not constant, it starts from 0 and then goes to 0. Now you can ask, why does the Vendor's diagram have such a black hole? Not even to describe it for you in detail, but I'll just tell you the answer. The answer is this. So it has this part of the black hole. That's right. But not the white hole part and not the other one. Here's inside the black hole. Here's outside the black hole. So black holes that are formed by collapse, for instance, of the shell of that. I'll add one problem to your problem, just one more. To describe this process. These things have a crucial Vendor's diagram that looks like this. It's part of this diagram. Removing this part. It's part of this diagram. It has some of the elements of this diagram. But not the weird elements. So this diagram here describes a mathematically convenient space time. Imagine how interesting calculations that captures the physics of realistic situations where black holes are formed by collapse. That's all I want to say for today. Let's continue on. Once again, this lecture.