 So, the V coordinate is the V minus r coordinate. So, with the center blackboard R by V T will be greater than 0. So, if we have a V amount 0 it is greater than 0. So, you can have all the way that. And the V coordinate lines are possibly greater than 6 lines are constant or this is the line constant. Now, please. What do you say? This is going to be r by V T like how V r by V T along the unicycle like no. So, again max V V by V T is like the velocity that we will be talking about. And sorry V r by V T is usually the velocity that we will be talking about for a part. You are talking about whether something exceeds the speed of light or not? Yeah. I mean V r by V T. You know the question of whether it exceeds the speed of light or not is just a question of whether it goes and stays at its future light. Everything always stays inside its future light. Is different relations between different you know coordinate. But the light code is the geometric concept and everything always stays with it. But the light code can turn like for example, there is a diagram in the non-stick. That's right. The light code can turn in those coordinates. In this picture the light code is always true. Right because that's how light this is an in going light this is not going. The light code is the thing that is that is enclosed between in going and the outside. Now this is because this is a nice well drawn diagram in real coordinates. Now if you draw on some other coordinates the light codes can get it. If we are drawing R P coordinate. You want to draw the light codes in R P coordinates? So that we have enough of you. I will have to think of that. So let's see. So how do the light codes work? A R P coordinates. So this is the question of what do the R P coordinates look like in this system. So see basically this is the point. Inside the black hole R becomes a light light variable. And T is the space magnet. Going to smaller R is going to future time. So of course the light code coordinates we can just solve by setting ds by equal to 0. So ds squared is equal to 1 minus 2 R by m by R T squared minus d R squared over. So this is the light code. Outside. So first let's look at the motion of the light trace. So how do the light trace move? They move by d R by d T. As you put a plus minus 1 minus 2 by m by R. So this was squared. This was squared. So we get the plus minus. Let's... These numbers are negative inside the... Inside the... Inside the... Inside the calculus. Okay. So now the future directly light ray. Always go... So suppose you want to draw the two light codes that are pointing towards the future. In the article suppose I draw R and T here. The future directly light ray is always towards smaller R. Okay. And then it's done. So it's always once when... This is R is equal to m. It's always pointing like this. Okay. And how much T changes is given by this one. So one of them is smaller than one. Whereas outside here, the future direct light code is always towards larger T. Okay. So the light code took me... So here the light code looks like this. Here you can put someone like this suppose. I suppose at this point it's sort of fancy. And here it didn't start. So the numerical value then... If I do the... It would be more than... Yes. The numerical value has no meaning. We only have that... Yes. You have to... You have to... If the question is... Spirit of light is a question of... Is the ray in space light? Not the light. The numerical value changes when it changes. It's got to do. Okay. This is all the orientation. Okay. Okay. That's all. Okay. So this is our last lecture. And there's a lot of things I would like to... If you guys are gay, we could have these extra classes after December 15th on A, the theory of inflation, and B, some more details about that. Okay. I'm related to the cross. But just in this lecture... I have a book called... In this lecture there are many things about light code physics that we should talk about and that I will sort of touch on. I work on all details of everything just so that I can tell you about many things. Okay. The first thing about light codes that I wanted to tell you about was this notion of the eventualism. Okay. So in this solution that we've talked about, in this particular solution that we've talked about, we see that there is this surface and we are equals to m beyond which things look different. I mean, they don't look globally different. They look globally different. Okay. And now you can ask, how do you invariantly characterize this difference? I mean, what is the clear statement? And the difference is this, you see, there are some regions of space time from which a light ray can reach that's its coordinates. But there are other regions of space time from which it can. Okay. So if some regions of space time are causally connected to space time, do you have an observer here? You can receive signals. Or let's say you have an observer here. Some large boxes down. You can receive signals from the sky if it's here, but not from there. Okay. So now when there are some regions of space time that have a property and other regions that don't, there's a boundary between these two. Regions. And the boundary between the regions of points that can communicate with the scribe class. And those that cannot. As a name, it's called the event release. So in this diagram, the vaporizer is the surface. Now this, of course, is a very simple solution. Okay. Sorry. In this diagram, if you really write scribe class in this one, which scribe class is it? Okay. So if you put scribe class prime, then the event release for that would be this. But let's say we choose the vaporized device in this scribe class. That would give this interest. Okay. Because anything to the left of it cannot communicate. Okay. Now, of course, having these two scribe class and so it's an artifact of dealing with this internal domain. As we discussed. Okay. In the tenduous diagram of lack of form, due to colants, which looks like this. There's only one. Okay. There's this confusion of that. There's something. Okay. So this already brings you to a point that the notion of the event release is a little, then this particular static launch. In particular, suppose we have the situation that we talked about at the end of last class, because let's say there's a shell of light that is imploded, or a shell of dust that implodes. Okay. To start off with, space time could be very, you know, near flat everywhere. But this implosion can cause, can cause the creation of black. As you see, it's a tremendous thing. So in this full space time, including the full dynamical space time, with all the dynamical details of the formation of the black hole. Okay. There's still this notion, there's still a separation, the big one, of space time that can communicate with survivors and those that cannot. And there's a boundary between these two points. There's a boundary between the imaginables. So the event horizon is well defined with a complicated dynamical situation that involves black formation or involves scattering of black. Okay. Now, there are many things about the event horizon that are very interesting. The first thing about it is, that is what it's called now the event horizon. So the event horizon is a manifold, it's a sub-manifold of big black space time in four dimensional space time, it's a three dimensional manifold. And, any co-dimension once of a sub-manifold can be, you know, it's conveniently described by some equation of the four measures equal to zero where it's a sub-manifold. Okay. Now, even a manifold like this, there's a notion of the normal to the manifold. So what is the normal to such a manifold? The normal to a manifold is a one-fold. It was composed in such del mu of s. Take the function s, take del mu of s, that gives you one-fold because there's lower matrices. So the normal to a manifold is characterised by one-fold. You can also characterise by vector, by basing, because naturally it's characterised by one-fold. Now, what is the defining property of this one-fold? The characteristic property is that, look, suppose s is equal to zero on this manifold. Then you start from a point on the manifold. You take del mu of s and multiply it by dx mu. So then dx mu stays on the manifold. That takes you to another point on the manifold where s is equal to zero everywhere. So this is equal to zero. Okay? So the normal one-fold is called the normal, because it is orthogonal to all vectors that lie in the tangent space of this manifold, of the sub-manifold. In Euclidean space, the sub-manifolds are very easy to understand. You can cut a normal vector, and now you're enticed to then multiply it. Now, even in the tangent space, something else counterintuitive happens. If counterintuitive, if you're thinking of Euclidean space. And the counterintuitive thing that happens is this one. You see, since we've got a manifold with, let's try to characterise these manifolds. It's clear, since there are three kinds of one-folds, three kinds of vectors, qualitatively three kinds of vectors, you know, one-fold we have a space like, like, like or not. Okay? So it's clear that locally there are three distinct kinds of manifolds. Suppose you've got a space like, suppose you're looking at a manifold with normal and specific. You can always boost the good coordinates in the local little batch so that your manifold is x i equal to 0 in a local, uh, non-existent space. Okay? So the normal is given by tangent. It's given by in the coordinates t, x 1, x 2, and x 3 as the components 1, 0, 0, 0. So one-fold will be the components 1, 0, 0. Okay? So that manifold is like space locally in some way. So to keep it out of the space, it's such that a manifold with a space like locally always looks like that. All right. On the other hand, a manifold could be time like. In some appropriate coordinate system, the coordinates on the manifold could be t, x 1, x 2. The normal could be x 3. So that's a time like that. A manifold where the normal thread is space like, the manifold includes two directions of space somewhere like that. Okay? The most interesting kind of sub-manifold in the red since space, a sub-manifold where the normal is null. Okay? So, uh, just to get some intuitions with this, by appropriate coordinate changes locally, you can always take any null vector and put it in the form 1, 0. 1, 1, 0, 0. Let's see what vectors lie what local, uh, what local tangent vectors lie with this method. So that's everything that's rounded to this. So, of course, 0, 0, 1, 0 and 0, 0, 0, 1 lie with the manifold. But in addition, one more vector lies with the condition of normality. It's one condition of four-dimensional vector space. It gives you three-dimensional vector space. Can you tell me what the third vector is? No. 1, 1. 1, 1, 0, 0. Because this vector is null. So it's in particular orthogonal to itself. Null manifold. The null manifold is spanned by two space lines and one null direction. And the null direction is the normal direction itself. So the null manifold, the normal vector lies in the manifold. So it's normal in the manifold, but also it's an tangent to it. It's really counter-intuitive. I mean, even if you go normal in this direction, it stays on the manifold. In this direction, it stays on the manifold. Okay. Yes, so this is a tangent vector. It's both normal as well as tangent. Sir, is there only one such manifold? One such manifold. Locally, of course. No, but there are many nulls. No, I was saying flat-space. In flat-space, right? Let's see. Basically, any s such that it's gradient is null. Defines an untenable. Okay. So what does that mean? What we want is m or s and m or s to be 0. Now, let's look at it as an example of such a manifold. Yeah. So for instance, suppose we take a manifold that is where s is equal to s of x minus d. This will do it. But if we take s as equal to s of r minus d, that will also do it. Of course, x minus d, y minus d are related to each other. The r minus d is not related to each other. Okay. Now, at least if you're not interested, it's not one of your possible singularities. Two-dimensional polarity. All I have to do is I have to say it. Excellent. Now, why do I tell you all of this? I told you all of this because we just introduced the event horizon. Okay? Now, the event horizon is such a manifold and we don't understand what time it is. Okay? And of course, you know, you know, there's a particularly interesting surface and a particularly interesting kind of manifold. It's going to be a lot of intervals. So what I want to prove to you is that the event horizon of any black hole is obviously not. We'll show this by contradiction. Suppose we were going to append to a black hole. Yes, that's right. Okay. It's almost obvious. Let me prove it. Suppose at some point, the manifold was a space like. I mean, at a space like norm. Okay? So then, some local coordinate system, we could think of this manifold as including time and two special directions. These two special directions are found in this coordinate system. This time is the one direction. One normal special direction and the two special directions are found. Okay? This is supposed to be out. This is supposed to be. This is claimed to be the surface of things that separate causal relations. So inside can never be described as but outside can. That's the claim. This is obviously the question. Why? Because suppose the claim is that this point is positively disconnected from the scribe plus, but this point is positively connected to the scribe plus. Well, this point is positively connected to this point. Let's say I've got a long line trade-off. A lot sometimes. These two points are positively connected. So, a signal can go from here to here and then to scribe plus. If a signal can go from here to scribe plus. If a signal can go from here to scribe plus. It can also go from here to scribe plus. So we've reached the conclusion. It's impossible that this surface that divides points that can and cannot communicate with scribe plus has a space like has a space like? No. Similarly, I want to now show that it's impossible that the surface that divides points that can and cannot communicate with scribe plus has a time like? Because suppose it was true. Suppose we have some situation here. So suppose we have, you know, something meteorized here on the surface. And I go to coordinates where I make this look flat. Which I understand. We make up a normal vector for time management. All the vectors are going to this space. This point is supposed to be able to communicate with scribe plus. Though these points can locally we can draw such a signal. So any curve that's obeys causality. All the curves that are obeying causality from this point eventually end up somewhere in this region. Any curve within the local little local light corner of this point eventually ends up in this region. So if these points cannot communicate with scribe plus causality. Because, you know, as an eventualizer this globally is a very complicated surface. But locally it's very simple. If you buy this, causality is even better. So any event that emerges from here and goes in a time like a light light project eventually ends up here. Why is it true that none of these points can communicate with scribe plus? Try to look at this point therefore. And therefore it's a contradiction. It's impossible that this is outside the event horizon. Why is this? Is this here? So neither of these possibilities both these possibilities raise a contradiction. So what possibility do we have? Namely that the event horizon is null. So it must be that the event horizon is null. Let's just say that if the event horizon is null. And neither of the two equations that we looked at occur. Okay? Of course the event horizon is null so that this 45 degree angle is null. The first contradiction was that suppose this was inside and this was outside. Then we could draw some causally connected line from here to the other side and get out. So if this 45 degree curve so the first half goes away. Similarly, the second argument was that every curve emerging from here ends up inside here. But if it's 45 degrees it's not true. The objections that we raised don't apply if the manifold is null. It's the only possibility left. So the event horizon isn't null. Remember what that means? What that means is that it's normal is null. It's easy to convince yourself that in any non-manifold there is a unique null. Suppose in this local coordinate system that we talked about before there's a unique combination of these three that are null, namely this one. Anything else would be specific. So for any non-manifold there are no time like that. And a unique null. What does this mean more physically? More physically it means that if you shine at a real light and you want to propagate along with another map this ray has only one way. That is along the normal which also happens to be attached. It's a digital manifold but anyway let me just assert it and maybe it's intuitive to you. D, suppose you take you do the following. You take the normal vector and you use that as a tangent vector to make the following assertion. And we make the following assertion which I've not really tried to prove in the moment. We are just because we're in the last lecture I'm not going to say if we lose out these sessions in December we go through all this. So suppose you take the Kevin. On a null manifold there is an any null manifold that any null manifold can be foliate by okay so suppose we're in a three-dimensional of a gamma-threading any null manifold can be foliated by two parameters that was null gld6 such that the tangents to this gld6 at every point go itself up to a normalization to the normalization. So how do we add there are all lengths 0 There are all lengths 0 So how do we actually go from one point to another? You can go from one point to another universe of lengths. Length is a formula just we choose to call them Let's say that there is a two-parameter set of null gld6 which completely well I don't know the right word it's completely that the manifold is the union of a two-parameter set of null gld6 then we draw that where there is such that tangential of the gld6 at every point up to normalization coincides with the normalization Both these universes are the same No, no, wait Let's not bring this up What I mean is that there is a two-parameter set of gld6 and looking at null manifolds that are co-dimension one and three-plus one dimensions So the null manifold itself is two plus zero two plus zero okay so there what I'm claiming is this that you can take another manifold okay you can think of it as the as so this null manifold is a three-dimension out of any ray you cover one dimension okay a parameter set of such a ray which completely constitute the null manifold the null manifold passes through one of these one of these gld6 one of these guys now these two-parameter sets are age gld6 null gld6 okay and in particular if that is true it's obvious but anyway I'm saying it's obvious because there is a unique null vector at any point that tangentials coincide with all the left and right actually I don't need to say that it's obvious because the tangent vector to null gld6 is null since these gld6 lie on the manifold at that point it's a null vector on the manifold and therefore it must be proportional to the null is the claim here now this formation is unique is unique you see because it's a gld6 it's obvious take any point solve the gld6 equation through that you know light ray starts along the manifold solve the gld6 equation no no it's a unique null vector that points over the null vector this is not true for normal null variables what? this is not true they cannot be non-unique for you sure sure sure we give a physical interpretation something like what we say for null variables okay so look on the null manifold it's a unique null vector at any given point but the tangent to the gld6 can only be that point and the gld6 equation for a null the null gld6 equation is completely characterized by extension two parameters because this runs through one dimensional line now we start with another point forget about the gld6 but two plus one is three so to make up the whole null manifold so gld6 so all the points one of my null gld6 will be described by the same set set of parameters set of parameters that describe the gld6 of course I was going to be telling you about you can try to parameterize the manifold by affine parameter long gld6 and with gld6 so if you want to parameterize that manifold just check this is the one single parameter we are telling you that it was m squared or something that it was zero yes it's not zero yes we have two directions you see suppose it was not zero suppose we had a manifold in time meaning by which I mean a space right now time is part of the okay the span of this manifold by gld6 what you can do is imagine that you have a dust particle for this manifold you choose an initial surface and you keep the arbitrary velocities of the instrument many different ways in which you can do that all of which will give you gld6 that make up the manifold okay you have to lie firstly it's not even clear that you see there are many things about it firstly in arbitrary sub-manifold there won't even exist gld6 that will stay on the manifold in arbitrary manifold gld6 will leave the manifold there are various things about it so it's not even clear that it's possible to orient it okay when it's possible it's not clear that it's unique do that the number of parameters yes this is a very general thing that applies to two dimensions okay it's a 10 dimensional space you would have a 9 dimensional eventurized okay but you have a 8 parameter set of countries in the context of the eventurized what are gld6 these gld6 are the rays of light that are trying to make it out to scratch the but never quite making it you see outside the eventurized and go to scratch this inside the eventurized it's nothing that goes to scratch this all the light rays fall into the city light but on the eventurized there are these light rays that are at the edge they don't quite make it to scratch this neither do they fall into the city light they just stay suspended in particular in our particular these light rays are staying here so these light rays so these are the light rays they are vanquishing you try to shine your torch out to it because as far out as it can which constitutes neither going out nor away and these are the gld6 that make up your this making the eventurized is the gld6 that we want not parallel in the sense that they can expand area crystal but yes it's a good point that if your space time is I am going to describe that okay is this clear any questions about so if a dust particle or something penetrates the event horizon and that light which is suspended at event horizon gets scattered by it it can leave it cannot but there is another another direction suppose you have a day of life it's going global and then you somehow scatter it it can go inwards but it cannot go out you see all scattering happens within your light and all the eventurized scattered here are not out because another thing prevents you from it very hot excellent now I should say that these light rays are no fictional and nobody is claiming that okay what if some matter falls into an r equal to m increases whether the manner falls and r equals to m increases then that tells you that in such a space time r equals to m is not quite there it's a very good question so the question was without ascending into can you take this as the last problem here your problem says consider ds squared equal to minus d squared into 1 minus m to v of by r plus 2 dv this is the very diametric that describes back hole formation due to things falling there is a kind of thing that we have in mind because m is a function of v okay because m is a function of v it's changing now if m was a constant the event horizon manifold the event horizon manifold would just be would just be precisely m is equal to 2m is equal to r is equal to 2 okay the exercise is very but slowly so that m dot is smaller compared to whatever should be smaller bigger m squared let's think more carefully m is an L so I want to dimension this condition m dot maybe it's just 1 but if you need to assume that small compared to something else m dot is very small okay compute the event horizon of this okay given this and given also that m dot as v goes to infinity and it tends to this faster than anything that you let's say that the technique is exponential at large times it approaches m dot very fast given these two things compute the event horizons in a power series expansion this is an answer question okay all you need to do in order to do this is to remember that the event horizon is a null answer but just in order to answer this question try to write down the manifold manifold is described by some s of r r is equal to z because there is no angle to the problem it's going to be null and it has to at large late times tend to r equals to m these two things completely define the manifold it's not a difficult exercise it's an interesting exercise so what you will find is that at time v the leading order in your derivative expansion at any given time v r will be equal to m of v but there will be connections to this suppressed by derivatives find the first number should we take as a definition the non-standard space to remember that so null manifold that separates curves and communicates with scratch class from those that they don't remember a space line scratch class is well defined some points can communicate with scratch class some points can it's a purely grammatical thing can you draw a causal curve that goes from this point to scratch class if you can it's insane it's not hard we have non-manifold for a lot I mean that problem is when you lack of some for one direction at some point how would you know what happens to every s so the investorized manifold becomes single it's always true what happens this curve is that the event horizon topology changes in one case we have this condition there will be a zero in every case there will be two disconnections and when you bring them up it will certainly become one look at this for instance suppose you look at the following equation suppose you look at the equation x square minus y square is equal to alpha let's plot these curves when alpha is positive okay so y is equal to zero x this is alpha positive let's plot the curve when alpha is negative symmetric like this what happens when alpha is zero it's x plus y and x minus y it becomes like this and alpha positive alpha negative we've got smooth curves it's a manifold it's not a manifold so this is what's going to happen okay you have two event horizons there's some point of touch in some singular way you know on a large amount there's no right way to do that yes at that point at the time that it touches it's not a manifold it's a manifold by definition every locally patches isomorphic to a manifold at that point it's not so you can use these manifold notions just before and just after this is a singular singular thing to the event horizon no claim to singularity in physics just this concept auxiliary concept that you define it values the singularity okay I want to tell you more about more about more about these okay and okay just so that we finish what we're talking about now I'm going to prove things very lightly beautiful discussion in terms of chapter six of these so for those of you that I strongly recommend this class is going to be chapter six the thing I'm going to try to I'm going to tell you about is one of the one of the classic and most important terms of general relativity namely Hawking's area at least there okay so to do this we're going to start the collection of geodesics that foliate a manifold it's called a concoct it's just a name so collection that foliates a manifold use the word concoct so what we have to do is that any null manifold admits a null concoct null geodesic the congruence of current geodesics okay now what we're going to try to do is to understand how these how these congruences evolve in time evolve in more precisely affine parameter a lot of these things but I'll never try to tell you precisely okay so now let's look at any given value and any given value has its tangent vector I'll call it B okay this tangent vector in some local coordinates we'll look at 1, 1, 0 so what we're going to do is suppose we had a congruence of time like curves in particular curve there's a three-dimensional there's a there's a distinguished three-dimensional subspace spatial subspace there's a spatial subspace that's to the tangent that we want however suppose you've got a null congruence okay or just a null geodesic there's a null geodesic and I say let me look at the orthogonal subspace okay the space of vectors that is orthogonal to this this geodesic at any point this of course as we've seen includes includes the geodesic itself again a three-dimensional space but includes the geodesic itself because the geodesic is null now what I want to do is somehow define a spatial direction that is orthogonal to this subspace the kind of thing that I want to do is to focus on these two directions I want to define I want to focus on the x-y directions for what reasons it's okay so okay yeah so what I want to do is to focus on the x-y directions of this okay so in order to do that how could I do that but one way of doing this is the problem look the vector of one this one is zero this is a vector number called t because another vector is going to be called n okay in the square I have this spatial vector one minus one now let's look at the algebraic relations okay pi and t firstly of course t squared is 0 t is okay I'm going to put t dot n okay t dot n is right so t dot n is 2 so let me put a one so t dot n is 0 so what I'm going to do is the following along my now I've got this null junction suppose I've got a vector a such that t dot n is equal to one at some point and then you take the vector and parallel transported along along the gdc parallel transport doesn't change scales we know that the tangent vector along the gdc changes by parallel transport if n also changes by parallel transport in particular t squared n squared and t dot n all don't change okay so given one such vector which I've chosen if there's any immediately on the gdc and then define like a field such vectors or one parallel set of such vectors okay along the gdc I defined it by parallel transport so my definition of ways in particular basically the condition that that t dot v of n be the same because the n view field is defined by a parallel transport that is what changes the parallel transport okay now given this p and this n okay for any gdc I define the projector or subject to t and n okay we do it but here in this case what would we do we would want e w view okay we would want minus e w view in this case negative okay and then I would want to get rid of the one zero plus so I want to do something like plus t new n new plus t new in and now we check what this works okay so this thing is minus one one point one this guy let's look at zero zero zero zero gives one no yeah this thing with a minus is minus one one and this thing zero zero with t new and new would give one one one both ways would give one one the one by square root two would give half yeah so this works okay well with the space guy that way it works so this would just become zero zero something you can check so this operator gives you a projector onto a two dimensional space like plane orthogonal to both the tangent and it's null counterpart okay it's projecting onto a two dimensional spatial another part is not any right yeah let's let's think about let's let's work around here you have to assume it doesn't assume it doesn't matter just one so of course what you could do is choose one and any normal that would be other and then by normalizing this appropriately you could get the other product right so this is not unique it's not unique and it's not going to matter what we choose what is important is that once you make one choice you then make the same choice all through and that's this question that is once you make one choice of n mu the n mu's along the geodesic are then defined by parallelism okay so that's what I'm saying so we got these you take some slice let's say some space like slice and you have to cut it on the space like slice you get to make a choice of this n one for each time it cuts you get to choose this but after that so there's ambiguity but after that how it evolves along the geodesic it's unproduced okay that's all we will need to do and then there's something okay yeah now I'm going to tell you about various things that you can very simply prove maybe the first of these are true for you and then a very small algebraic exercise is going to do that are very simple to prove they're all very important, very detailed in terms of geodes I'll prove one again for you so one thing is at the end of these algebraic exercises what is it? okay so let me start so the first thing that you can prove is suppose p times beta is usually at any one point so it's losing a little vector in the tangent space in the one point around any geodesic this vector lies in your orthogonal direction so this projector is your orthogonal projector okay so suppose p times eta is equal to eta that is you've got a vector that is in your two dimensional set special sets whatever you're finding think I'm going to try to prove with the following that if p times eta is equal to eta that p at one point lies with inverse then you have to follow the equation d eta mu where d means derivative of the tangent d mu is equal to p mu lambda p lambda rho p and where p is is equal to at times eta theta the index of the equation is mu is equal to no lambda rho rho nu I'm going to try to prove it for you in the rest of the lecture I will prove it right out of point let's prove this one it's the most important just to show you how easy it is okay but what is this equation saying this equation is saying that evolution of eta along the geodesic it's given by some linear transformation on eta okay so that if you try to you try to parallel to transport this this vector eta it's it rotates in some way that depends on eta but the important point of course this model that I talked about anyway the important point is that you got these projectors so the change in eta is always in the projected actions so what I mean is that suppose you take some eta that lies in this two-dimensional subspace orthogonal to both these two-dimensional okay then you take this eta and transport it by parallel transport along the geodesic it remains orthogonal to this null subspace more or less obvious but it remains orthogonal and remains orthogonal by this rule then you get a particular rule for how this thing moves along okay let me try to show you this this process this way let's start by t dot d of eta this is very important this is equal to t dot d of p mu mu eta mu t dot d on n and p dot d of t so p mu nu so what is eta what is p mu nu? p mu nu is eta mu nu okay plus n mu t nu plus d of t nu but t dot t of both m and t t dot d of t is equal to d of t is a geodesic question and t dot d of n is equal to 0 this is not a final equation of this feeling so as far as this operation t dot d is concerned both t mu and n mu amongst the vectors so so so we can you know add this p mu nu basically for free because you can just pull the vectors outside okay let's write this out this is t dot d of eta mu nu eta mu plus n mu t nu acting on eta mu give 0 of this you just take this on this side and that's 0 this is just because p eta is equal to eta is equal to p eta this is a triviality sorry sorry but now let's see so the next step is we will write this as this is equal to now I can bring the p mu nu out because t dot del commutes to p mu so I can write this as p mu nu times t dot del on eta mu okay we can do this okay and now we have this definition of v when we had v nu rho is equal to what was it del del rho so now you write that using that sorry we write v alpha v is equal to v alpha one more thing that we need here that is that is that the other thing that I didn't tell you about is the following look see suppose I have my suppose I have my components okay suppose I have my components here and I choose coordinates for my null manifold to be a coordinate which is which coordinates and a coordinate which is wound up which tells you which goes around this affect how we develop the genesis okay now my these null coordinates is n mu okay can be thought of as pointing from one one genesis to the next one you see these n mu yeah so that my null manifold okay remember that my null manifold was basically spanned by this at any given point there was this null direction and then there was space like direction okay so if I wanted to point from one guy to the next guy okay this is some linear combination of a space like vector and this null vector okay and this condition that I am choosing let me say suppose I have some arbitrary y's okay if I draw little vectors from one genesis to the next these little vectors are not unique if I could draw this guy or this guy okay but the non-uniqueness is by is by shifts in t shifts in t remember t has minus one of this a vector basically suppose I suppose I take a little vector pointing from one genesis to the next a little vector pointing from one genesis to the next okay and I choose this vector to belong to this orthogonal subspace the orthogonal subspace that we were talking about that picks up a unique direction okay because we shifted by anything proportional to t that would spoil the orthogonal so it would not have to be orthogonal to and it would continue to be orthogonal to t but not to n so it would not have to be orthogonal to this whole, it would belong to this whole protected subspace okay good so you can look at little vectors you can look at these n vectors as being a choice of vectors that go from from one genesis to the next okay and use this to define use these vectors to define the zeros of lambda okay you know you can you can find parameters along each of these genesis now these applied parameters can be shifted by anything independently genesis equation okay now what I tell you is to take these little orthogonal vectors that are normal to n okay and define that if lambda is equal to zero here lambda will be zero along along the line everywhere okay so that gives you like that gives you a surface of constant lambda this defines your surface of constant lambda now with this definition this is the surface of constant lambda clearly these vectors these n vectors are the vectors del by del yf at constant lambda or the proportional to that because you're going from one genesis to the other but at constant lambda okay so these n vectors can be thought of as proportional to del by del yf and these these these stuff we thought of as proportional to del by del yf okay but now del by del yf at constant lambda commutes with del by del del by del lambda at constant yf but del by del yf sorry lambda at constant yf is the tangent vector so these vectors here can be chosen to commute with the tangent vector and that is actually what we need for this so let me you see suppose yeah so suppose but the eta vectors I mean del by del yf okay I mean del by del yf then we have that these vectors commute and therefore g mu g mu on okay let me sorry I should have okay let me show you this okay suppose I'm sorry about this so t is this vector that we've talked about before and del eta is the vector del by del yf at constant yf and these vectors commute d commutes with both eta okay so that tells you that d mu of actual displacement on this sorry d nu d nu of eta mu alpha minus eta mu alpha del mu of d mu is equal to 0 by definition of the vector's commute okay now because this is true it's also true that del nu of eta mu alpha minus eta mu nu alpha del nu t mu is equal to 0 and the reason that this goes to this is that the two terms cancel because of the energy you see the way they differ is by a term that is d nu gamma mu nu theta eta alpha and this one is eta mu alpha gamma mu nu theta d nu theta but these are the same okay the point is that this the only nonzero index is here okay so these two are now the same but this quantity is what we call d okay and this quantity is what we had d dot d times eta is what we had here okay so you can replace this by an eta dot b and that gives us this okay sorry about that sorry about that so let me let me let me let me let me summarize again okay so the eta's for which we are doing this are the del by del y of this okay because these eta's commute with rectity this goes over to to this and we've got this equation and once we've got this equation we can rewrite this as p b p eta because p has it okay so we prove this equation that we want sorry about that what is important here is the fact that t n get and eta commute t n get eta commute I'm trying to find a surface I'm trying to find how these normal vectors evolve as we move and the normal vectors define these normal vectors define the sort of coordinates of that surface I'm trying to see how the surface that's evolving how it shares compresses and moves okay now we've just got like a few minutes left let me let me very quickly so from this equation okay we've got this important quantity b and b here so b is an effectively 2 cross 2 matrix because it's orthogonal to the other directions it's a matrix it's really a 4 cross 4 matrix but it's projected on both sides so it's effectively like a 2 cross 2 matrix okay and there's some standard stuff you can do you can define theta is equal to this b hat mu so b hat mu v v v okay so theta is equal to b hat mu mu sigma hat mu nu is equal to b hat mu mu select rise minus the trace point and omega hat mu nu is equal to b hat mu so this is the answer so in terms of this of the of these components b hat can be written as is equal to half theta mu nu plus sigma hat mu nu plus omega hat so what I've done is I've taken a probability 2 cross 2 matrix and write it as symmetric traces plus anti-symmetric plus trace matrix okay so it's just that you can do my matrix okay now one thing you can prove we'll find this now to the null GeoD6 that go around the event matrix okay we haven't yet here used that this congruence of null GeoD6 is the congruence that is normal to a null matrix that's just any bunch of matrix okay it's normal to that this congruence is the congruence of the normals of another manifold allows you to show and I will try to do it now that omega hat now you define the following thing in this little congruence of null GeoD6 you define the following area okay you define a is equal to epsilon mu nu epsilon with the square root okay so epsilon mu mu rho sigma t mu eta mu that's n n mu and then eta 1 rho eta 2 what we're doing we're taking two little two little vectors in this normal space it doesn't matter which ones so two little vectors in this normal space these vectors span an area just a two-dimensional area okay and just the usual standard definition of this two-dimensional area is written as follows basically you know projecting on the useless one zero and one one and one minus one direction and computing just the area of the spatial two-dimensional space so you so suppose you've got this congruence of GeoD6 and between any so we've got this two parameters out of GeoD6 we've got another GeoD6 here and another GeoD6 here and any given time given outside parameter you've got these GeoD6 have a set of two-dimensional associated with them in the sense of this form okay now another thing you can quite easily show is that dA by dT is equal to T why is the why is it you know areas always how do you define the area of a triangle in this sense yeah it's a one eight to minus eight together so whenever you try to measure a volume it's measured by it's actually by a form the duality you know you dualise it to a scalar that's this epsilon and then measure that scale you also associate okay so something else you can prove I'm sorry I'm not going to prove all these things but it's very nicely done in this section okay is that dA by dT is equal to T so what's the picture with you suppose you've got an event horizon surface and you cut the event horizon by some spatial two-dimensional spatial map you can associate an area with this an actual horizon a few bits of area between each of the two between neighboring geodesics now as these geodesics move around these area elements change okay what's happening is it's like a fluid as the geodesics move around you can think of the geodesics following the fluid particles okay so the fluid particles may move away from each other or move towards each other is this the same thing the fluid particles can also shear that is you know move like this and do funny things without changing the area that shear is associated with that shear is this thing and the fluid particles can also have some vorticity okay and vorticity is this over here for the case of non-manifold there are only vortices here so it's vorticity free flow and vorticity free particles are moving so as to generate essentially vorticity free flow with just shear and compression okay great so all this is very nice but the most important thing comes is what the equations that are of the salt that we describe that again describe very nicely in these lecture notes right you're doing form the boundary equation you found an equation for the derivative as you move along as you move along okay if I'm the equation the derivative equation then d d theta by d lambda is minus theta squared by 2 minus sigma mu mu sigma mu plus omega mu mu omega mu mu minus or mu mu d mu t and basically all you you start by d theta in order to derive this you start by d theta by d lambda d dot l is just a definition of d mu mu sigma theta is the trace of speed and then you will use various manipulations you do various manipulations all the time using the using the gudc equation and the fact that the ends are transported there with a few random manipulations you can start with this equation and reach this beautiful you have to note at the bottom this thing here is negative definite this thing here is negative definite you might wonder is it really negative definite because both its squares and relativity can be either positive or negative depending on whether this makes like a time limit but theta of course is the second one and sigma is the tensor completely in the space-life direction so this kind of is negative definite this guy here is negative definite this guy here is positive but its just zero for the case of interest that is not there and then you also have this this term the thing that is multiplying r mu mu in this term is is an algebraic let me put that into vectors along these genesis so now lets use Einstein equations Einstein equations relate r mu mu to T mu mu and to T times G mu mu but G mu mu the G mu mu part vanishes because the T's are not T G G is there so you can basically take this and replace this by T mu mu with the right hand