Physics of other forces, other things in a curved background, in a curved background space. But we haven't yet understood what action sets the dynamics for the gravitational field itself, for the metric that sets itself. That's what we're going to discuss now. Okay. So, in order to do this, what we need is some expression built out of the metric field and its derivatives that is generally co-variant. So, we can add this expression to the action and generate the equations of motion for action. Okay. Now, how are we going to make an expression in terms of the metric field and its derivatives that is generally co-variant? Well, what we're going to do is to appeal to geometry to help us make such expressions. Okay. So, remember that we had this notion of parallel transport. We had this equation with delta A mu was equal to gamma mu alpha theta A theta dx alpha. Do you remember that a small change in one form when we parallely transported from x to x to x plus dx was given by this formula? You can ask the following question. Suppose I take this little one-form field and transport it in a little loop so that you come back to yourself. Does this process of parallel transport bring the vector back to itself or is that a change? If there's a change, what is the change? Okay. So, the answer is going to be in general one, bring it back to itself. There will be a change and we want to compute what it is. Okay. So, now let's suppose that the size of this loop that we're looking at is of order epsilon. We're looking at a loop whose length scale is of order epsilon. We're going to do our calculation to the first non-trivial order, the first place where we see an effect. And that first non-trivial order turns out to be at order epsilon squared. Okay. So, we're going to do this calculation of this chain, take this vector around a little loop, bring it back to itself. The loop size is order epsilon. At order epsilon squared, we'll see that we don't come back exactly to ourselves, but do something else and we're going to compute what that effect is. Now, how do we do this? So, we've got a differential equation. It tells us how A changes as we go around a little infinitesimal. We have to integrate this differential equation, come back to ourselves along the path that comes back to ourselves, and that gives us the answer. There's a formal solution to such differential equations in terms of path-ordered exponentials, if you want to do it in some exact way. But such complicated things are not for us at the moment. We can just work simply because we're working toward epsilon squared. So, you see, as we start, we start at a point and we've got a little loop. Now, A, let's say that at this first point, at the starting point, A mu is equal to A mu zero. Let's say we're starting at the point zero for convenience. Okay. Now, A changes as we move along this path. This integral here has an integral over the length of the path. That integral, this integrated dx will give you a factor of epsilon. So, if we are interested in changes in A at order epsilon squared, we only need to worry about, I mean, the change in A as we come around the path at order epsilon squared. On the right-hand side of this expression, we only need to correctly input the value of A corrected at order epsilon. Is this clear? Remember, is that we can't just do this integral treating A as a constant because A changes along the path by this expression. A is A zero here. There's something else here, something else here, something else there. So, in order to compute this change along this path, we're going to need what A is along the path. But the great way of working in this order-by-order way is that we only need to know what A is at order epsilon to get the change right at order epsilon squared. Is this clear? So, firstly, let's first compute A at order epsilon. Well, so if we use the same recursive logic to get A at order epsilon, we only need to input A at order zero into this expression. Okay? So, at order, so let's suppose that A mu is equal to A mu zero plus epsilon times A mu one plus epsilon squared times one. We want to know what A mu one is. Well, A mu one, A mu one obeys the equation, so A mu one is equal to gamma mu alpha theta at zero. So, if we take, you know, the Taylor expansion of gamma, the first term in the Taylor expansion is at order epsilon. Okay? So, to compute A mu one, that's an overkill because we're already doing an integral, which is another order epsilon. Okay? Gamma mu theta alpha at zero times A theta at zero dx alpha. Since these are both constants, the solution is obvious. The solution is A mu one is equal to gamma mu alpha theta of zero A theta zero times x alpha.  Okay? This is order zero. This is all, I mean, epsilon to the zero. This is order epsilon to the one. Everything higher will be epsilon to the two. Yeah, epsilon. Yeah, epsilon. Epsilon is a formal parameter. You can, in the end, set epsilon to one. It's just order counting parameter. Epsilon's there just for you to... In here. Epsilon's in your head. Okay? It's x or delta x that carries the power. Okay? In order to get this right to quadratic order, we also need gamma corrects to order epsilon. So, once again, we will get... That is simple. That we'll just take an experiment. Okay? So, gamma mu alpha theta is equal to gamma mu alpha theta This is an x. This is also an x. Okay? Alpha theta. It's called a del phi. X phi. Sorry. First the zero, please. Plus del phi gamma mu alpha theta. All derivatives evaluated at zero. Okay? I'm not going to indicate that explicitly. Just makes it to come to some. Times X phi. Plus higher order things at order epsilon squared. So, we don't need this. All we have to do is take this and plug it. So, let's do that. So, at order epsilon squared, we get that delta mu is equal to... Well, first we should ask what happens at order epsilon. Okay? So, at order epsilon, well, it's just a1 at zero minus a1 at zero. Which is zero.  Pass independent solution. So, you just get zero. What I'm saying is that the integral of dx over any closed curve is zero. Obviously. So, at order epsilon, there's no change. So, the first change is at order epsilon squared. Okay? So, what's the change at order epsilon squared? Well, we plug this in. So, we get one term is this term. So, del phi of gamma mu alpha theta, a theta x phi b x alpha. And the second term is in putting a1 here. Okay? So, plus gamma mu alpha theta. Now, then we want to put this with mu replaced by theta. So, gamma theta. We'll use something else for alpha. It's called a beta. And we'll use something for theta. Let's call it zeta. Now, we have to remember. There's a cheat sheet, mu went to theta, alpha went to beta, and theta went to zeta. This was an a theta. So, according to a cheat sheet, that becomes a zeta. And this was an x alpha. So, according to this, that was a beta. Okay? That became the x. And yes, dx alpha. While it's obviously true that the integral over a closed path of dx alpha is zero, as we got for the first order variation of a, is it true that the integral of x dx with lots of indices thrown in is always zero? No, it's not. Okay? This observation was made by many people, I'm sure, but, you know, it's essentially the observation of Stokes. This is, you know, Stokes theorem that tells you how you can take a vector field and what you get if you integrate it over a closed path. The answer is that it's the integral of the curl of the vector field over the area. We will say this in a slightly better way in just a moment. Okay? Now, at first order, we got zero because the vector field was a constant. So, it's curl with zero. It's the derivative of c. Okay? At second order, we don't get zero because the vector field is proportional to x. So, it's curl is not zero. The curl is constant, but it's not zero. So, all we have to do is to look at, to use Stokes theorem on this answer. Okay? So, in order to do that, let's rewrite this thing with indices. So, both of these have dx, alpha, dx with a dummy variable, alpha. But in this case, the dummy variable for x was beta. Well, in this case, it was phi. And here, we don't have a phi. So, we'll just replace beta by phi. We want to apply Stokes theorem. So, let me remind you what Stokes theorem says. Stokes theorem says that if you've got some integral like this, m alpha dx alpha over a closed loop, this is equal to the integral of del theta m alpha minus del alpha n theta dx theta dx alpha minus dx alpha dx theta divided by 2. What does this mean? You've got a little area element. You break it up into little boxes. The sides of the boxes have the two vectors dx1 and dx2. And you take this parallelepiped form from these two little vectors to make up this element. Okay? And sum over all the boxes. Compact that with this sum over all little boxes. Okay? This is, if you rewrite it in the way you're more familiar with, it's curl dot normal. Okay? Because the curl has an epsilon. Is this with an epsilon? And then taking that epsilon on this makes it the normal vector. Okay? But this is a better way to write Stokes theorem because it generalizes beyond three dimensions. It generalizes beyond. It's the natural way of writing the theorem. Okay? This is the wedge product notation for, this is the integral of a form for those of you who know what that means. We'll know what it means by the end of this course. Okay? Okay? Fine. So, now we're going to apply this, the byte 2 is because there are two terms here. So, when you dualize to the normal, you'll become twice normal and cancel the byte 2. Okay? Fine. So, now we're going to use this theorem here. We're going to use this theorem. We just have to identify what M alpha is. Well, so let's see. There are many indices around here. But most of them go along for the right. Like this mu index, as far as, as far as Stokes theorem is concerned, is going along for the right. See, what's important is that on the right-hand side here, we've got an alpha index contracted with the X alpha. Whatever that is, whatever else that is, there is on that alpha index, there will be some index, indices dressing these M. And those indices will just go along for the right. Is this clear? Please. It's just 1550, Stokes theorem. Yes. So, if we look for two dimensions, dx, d1, what is d1? Yeah. This is really dx theta 1 dx alpha 2 minus dx alpha 2 dx 1 1 2. We've got these two little vectors here. Okay? This is vector 1. This is vector 2. It's the area element along this little panel. Sorry, I shouldn't have. Okay. Excellent. That was a good question. Okay. So, this, by the way, this is what Landau-Liffchitz refers to as F12. It's the volume form on the hypercells. Okay. So, when you see F12 written in Landau-Liffchitz, this is what, or F alpha beta on our surface, this is what it means. Okay? Fine. Now, what we want to do is to compute this guy. So, what is M alpha for us? Just all the other indices go along for the right, but the alpha index was alpha there. Okay? We want to take the derivative with respect to theta. So, let me write this as phi so that the indices match. Yeah? The derivative of an expression linear and x just drops the x. Okay? So, this M alpha, del phi M alpha is simply this where we dropped the x. Okay? So, in our context, this del phi M alpha minus del alpha M phi is simply del phi of gamma mu alpha theta eta, but where I write eta is 0 theta. Okay? Plus gamma theta alpha gamma theta zeta phi. And then, according to Stokes' theorem, I'm supposed to subtract out this part. So, I'm supposed to subtract out phi going to alpha minus del alpha gamma mu phi theta minus gamma mu theta phi gamma theta alpha zeta zeta. Okay? So, the answer? Sir, please. Please. That M alpha dx alpha. Yes. We are matching with this, with this thing. Yes. Okay. So, then we are going to write the term for del phi M alpha minus del alpha M phi. Exactly. So, but here this phi thing is being contracted over there. It's contracted with an x, with x phi. So, now take derivative of this with respect to phi. Okay. Then, it just kills that. Okay. Fine. Fine. Now, this expression, this is our final answer. This integrated over the area. Okay? And this expression here, this expression here has a name. So, we will now get the conventions the same way. It's called the curvature, or let me say it more precisely. I'm going to give you a definition. So, let's define r iklm to be gamma i minus del gamma ikl plus gamma inl gamma nkm minus gamma inl gamma nkm. Okay? Then with this definition, if we've done everything right, it should be that delta a is ak. I'll write what lambda leaf should say, and then we'll check if we've got the same answer. Half r iklm flm by 2. Oh, and where's the a? Sorry, there's the ai. Okay. Let's check if that's working. So, firstly, this i index should be this i index here. So, that's this theta index here. Wait, something's... And then there's l here. Okay. Okay, let's check that this is right. Okay, let's do it properly. So, let's see. So, let me first write this guy in that form, and then we'll... Okay. So, what do we have? Our conclusion was... So, this is land-down issues. Okay? What did we conclude from here? We concluded that delta a... We were doing delta a mu, right? Yeah. So, we concluded that delta a mu was equal to... Okay. Now, gamma... Now, so what we should do is every time we have an a, we should contract with the same index. That was going to kill us, right? So, this... There was a theta here, and there was a zeta here. Zeta is not used here. So, let's make theta zetas. Okay? Zeta, zeta. Zeta, zeta. Okay. So, now we'll write zeta and a. So, now all the indices are good. So, let's write this as mu alpha plus... Write this. Minus del alpha gamma zeta mu phi plus gamma theta mu alpha gamma zeta theta phi minus gamma theta mu phi gamma zeta theta alpha e zeta. By 2 f, we had our phi alpha. That should be lm. So, I'm just going to replace in Landau-Liffchitz things. l and m by phi and alpha. Phi alpha, phi alpha... Sorry, sorry, sorry. lm. Phi... Now, which is an l and which is an i? Upper is i. Yes. Phi alpha, phi alpha. And then I should replace i and k, if I want to compare here, with zeta and mu. So, i and k is zeta and mu, zeta and mu. Zeta k is mu. i is zeta k is mu. i is zeta k is mu. i is zeta k is mu. Okay? And if we've done everything right, this should just match this now. Del phi mu alpha minus del alpha gamma zeta mu phi plus... The theta is something... is n. That's okay? Yeah, it's fine. Okay, we've written... So, we have the same expression as Landau-Liffchitz. Okay. So, what have we concluded? We've concluded that the change in this one form, as we go around this little loop, is given by this expression where this object here is what's called the curvature tense. Okay? So, one of the things we've learned is that parallel transport cannot be unambiguously used in general to define a vector field or a one-form field from a given vector at a point. You know, suppose I've got a given vector at one point or given one form at one point. Now, I try to say, let me define a field parallely transporting this vector to every other point in space. That process is ambiguous. It depends on what path you use for the parallel transport. In general, because different paths give you different answers as we've just seen. Okay? So, parallel transport less universal than you might have thought. It depends. It's path dependent. Now, the fact that there is curvature also tells you that covariant derivatives don't commute. Oh, something I should say, by the way, that this curvature here is obviously a tensor. Why is it a tensor? It's because this object in just the byproduct of two infinitesimals is obviously a tensor with upper indices. And curvature and delta ak is the difference between two vectors, two one-forms at the same point. So, that transforms like a one-form. Only way that can work is that this transforms like a tensor with all its indices, you know, with its indices indicated. Curvature is built out of gammas. Gammas contain derivatives of the metric field. So, we've succeeded in identifying a tensor that is built out of derivatives of the metric field. Okay? So, we've succeeded in our task. And we're going to use this R to build Einstein's action, to build the action for the gravitational field. But before we actually do that, there's some more gymnastics I'm going to do. Yes? So, is there a point considering the higher order terms in delta? Yes. Okay. The life can be iterated. If you want to get the whole answer, it can be written as a local expression. Meaning, the answer is complicated. But all the information is there in local curvatures. You see, it's like an integral. Okay? Suppose you want to know how much something changes from here to here. Okay? So, it changes over an infinitesimal little bit. Then you just add it up and you get the answer. It's the same thing here. What we've determined is how much something changes over a little area element. Now, you want the change in a big loop. You just add up the changes over the little area elements. The only complicated thing is that you're going to have to use the appropriate value of A. It doesn't matter what you're going to get. A loop can get close by itself. Yeah. It doesn't matter when you use the appropriate value of A. Okay. So, let's address this question in a little more detail. Suppose you actually wanted to compute. Suppose you actually wanted to compute what you would get over a big path. What you would actually do is take that original differential equation we wrote down and try to solve it. So, the original differential equation was delta A mu is equal to gamma alpha mu phi. So, one way of proceeding is to take this and just solve it along the path. Solve that differential equation along the path. Now, we could apply Stokes theorem. You see, this thing here also has the form integral of one form along a path. So, we could apply Stokes theorem to this whole thing here. Okay. And yeah. So, the complicated thing there is knowing what to put in the middle. Because you're going to have to find a field and A field. Gamma, of course, is defined everywhere. But A is not defined everywhere. You see, A was just defined at a point. So, what you're going to have to do is to find an A field. That has the correct values on the boundary. And the values on the boundary are determined by this differential equation. Once you do that, having done that, what you will get is the same expression locally done. But with the local value of this A. Do you understand what I mean? So, suppose I've got some A field, whatever it is. You find an A field that agrees with the values on the boundary. Agrees with the value that you would get by solving this differential equation on the boundary. Suppose you have any A field that does that. Yes, yes. The A field is totally unique. Sorry. No, no, no. The A, the boundary is unique. The filling up is up to you, I think. But then you're going to change the R. No, don't change R. R is a geometric quantity. What you will have to argue is that all these ambiguities drop out of the final answer. But just one way of doing it. Just one way in which we'd already tell you something important. So let me say it. One way of doing it is find an A field that fills up the surface, agrees with the right values of the boundary. And then with that A field, we just apply Stokes theorem. So we would get this with an integral as a finite expression, I think. I'm going to have to think about this. Let me think about this and give you a better answer. One thing that I do want to say, what I claim though is that if R is 0, then this will be 0 even for finite parts. This is obvious, right? Because for any little bit, it's 0 and so it adds up to 0. To get the actual answer will be complicated. But your question is willing to identify new geometric invariants. I don't think so. It'll just be in terms of R and maybe derivatives of R. R as a local field. OK, good question. I'll give you a better answer to that question. OK, excellent. Fine. So we've succeeded in our program of trying to, you know, the question. We'll talk about your question. D of something meaning what? Exterior derivative of something. No, it's not an exterior. R is not an exterior derivative. You know, exterior derivative applies only to forms, only to objects with completely anti-symmetric indices. OK. This is not something of. No, so it's not an exterior derivative. Yes, that is a Bianchi identity, which we will come to. Yes. Yes, yes. There is a sense in which you're right that it's an exterior derivative. We'll come to this in this class, hopefully. OK, excellent. Fine. So this has already given us a quantity that is built out of derivatives of R, out of derivatives of the metric that we can use to build Einstein's actions, and we will. But before we do that, there are a few more algebraic gymnastics that we want to do in order to identify more clearly what role this R plays. The next question I'm going to ask is, suppose we done this exercise with a vector field rather than a one-form field. OK. What would that answer be? OK. Now, if you remember when we started discussing parallel transporter vectors, we deduced how a vector would be parallel transporting in terms of the definition of how a one-form must be parallel transported, just from the observation that a scalar cannot receive no parallel transport. It doesn't change under parallel transport. OK. So we're going to use the same idea. Suppose you've got bi times ai, or b mu times a nu to stick w to. Under parallel transport, around a little loop, this is obviously 0. Scalar doesn't change under parallel transport. But 0 must be made up of two parts. 0 must be equal to delta b mu times a mu plus b mu times delta a mu. Delta a mu we already know. It's given by this expression. So let's plug that in. So we get that minus delta b mu a mu is equal to, OK, and now we've got this li business. So let's, this is delta a k. k, k, k, k, k, minus delta a k times a k is equal to what? Is equal to b and this now we'll call something else, some k prime. Delta a k prime.  This one was a k. Here we had an i. There's already a k here. What is this k? That is k prime. OK. So let's make this k.  Excellent. Degrees with lambda solutions. OK. So we've deduced not just how one form goes over a little loop but also vector changes. Now we can do this for arbitrary tensors. We can do it like we did in our discussion of parallel transport. We're using the fact that any tensor can be built as a sum of outer products of vectors and one forms. OK. The answer is obvious and is listed in lambda-olive sheets. I'm going to ask you to check that. It's just the sum of things. OK. Change of the first tens. Index by index. Just like it was for parallel transport. OK. Not even going to bother to write it down because it's a big expression. Is this OK? You can look it up in lambda-olive sheets. Excellent. Fine. So we understand how everything changes under loops of parallel transport. Fine. Now the next question I'm going to ask is, is there any other important significance of this curvature? And that is, you see, what is the point? The idea is that if you take a vector, you move it parallel here, here, here, here. You don't come back to yourself. But moving a vector like this has something to do with differentiating a vector coherently because that parallel transport there appears in the definition of the differential in this direction. Whereas moving a vector like this has something to do with differentiating it in this direction because differentiating it here involves moving it parallel in this way. Now this appears in the derivative that way. Whereas this will appear in the derivative the other way. Something like the inverse derivative. So this loop here has to do with what you will get by differentiating this way and differentiating this way minus the other way round. You see. This way and that way. OK. This loop can be thought of as this minus this. And this way is on the x-axis minus then y-axis. This way is y-axis then x-axis. OK. So just thinking about this in this geometrical way makes it seem like this curvature will have something to do with what you get from the commutator of covariant derivatives. I mean derivative derivative minus derivative derivative. Two different directions. We're going to now try to make that precise. OK. This I emphasize is important. Ordinary derivatives commute. Del x del y is same as del y del x. But covariant derivatives in general do not commute as we will see because the covariant derivative involves this gamma business. OK. And what are we going to get? So let's just work it out. I guess we're going to work it out really honestly. And then we'll use further calculations. We'll use tricks. So let's work with one form. Suppose I look at del alpha of a mu. What was that? This was equal to d mu of a alpha minus gamma mu theta zeta a z. The alpha of a mu minus alpha zeta mu. Now let's compute del beta of del alpha of a mu. What is this quantity? So this quantity by itself is an object with two indices down. So we have to use a formula like this, but with two such terms. Is this clear? So what are we going to get? So I'll write it down in two steps just to be... So those steps are totally clear. We get del beta of del alpha a mu minus... Now let's take the term where the a mu goes along for the right. So that's minus gamma beta alpha zeta del zeta a mu. And then there's the term where the alpha goes along for the right. So minus del beta of del alpha... That's for the gamma. Gamma beta zeta and this mu del alpha a z. This is what this quantity is. And now painfully we plug in what this is. So once we're going to do it in complete detail. From the next time we'll use some tricks. So let's plug it in. So that is equal to del beta of del alpha a mu minus gamma alpha mu zeta a zeta. Minus gamma zeta beta alpha into del zeta a mu minus zeta mu pi. Okay good. Minus gamma zeta beta mu into del alpha a zeta. So del alpha a zeta minus gamma zeta alpha a pi. Painful enough. That's what the expression is. Now what we're interested in is looking at this covariant derivative minus the opposite. So this expression that we've written down is this object. What we're interested in is this minus what we get from alpha and beta exchanged. So let's look at these terms. This term, the term with del beta del alpha is symmetric. So under interchange we cancel. This term here, anything that we get from this whole expression here is also symmetric under beta to alpha. Because gamma is symmetric with slow indices. So this object also cancels. And this term here is also symmetric under beta to alpha. Some dummy variable flip. So that also cancels. So the terms that we have to keep what remains. So the terms that we have to keep are this. Let me circle the terms we need to keep in orange. This one we need to keep. These we don't need to keep. This one we need to keep. Okay. And both of these nicely have the dummy index for a being zeta. So the terms that we're keeping are minus of del beta gamma zeta alpha mu gamma zeta beta mu. Wait, what happened to my... Sorry, I messed up on my... Sorry, I messed up on something. Just a minute, just a minute. What happened to the terms that... This is del zeta alpha. This is del zeta alpha? Zeta alpha. Is that right? I'm going to need to check that one minute. And I think I messed up on something. Let me just write what I had before. Del alpha zeta. Okay, just one minute. Just one minute, just one minute. So this del beta here... Whatever I've done. I've not kept the del beta acting on zeta. That's what I've not done. Okay. That's what I've not done. And this is coming with a minus sign. Sorry, people. Just one minute. So in this, of course, there's a term where the... Where... I'm sorry, I'm sorry, I'm sorry. I'm getting confused just one minute. Let me just step back. I messed up. So what did we do? Del alpha a mu is del alpha a mu minus this. This is correct. Now we took the covariant derivative of this object. This expression as far as I can see is okay. And then we plugged in what this was in each of these. Now it's certainly true that this cancels. There's no reason... That this is symmetric. No reason for this to be symmetric. This is certainly symmetric so that goes, right? The last term was not symmetric. Maybe the last term was not symmetric. Sorry. Let's look at that. So this was... Let's look at that. This term was a zeta. Yeah, I'm sorry. You're absolutely right. This term was not symmetric. You see, it was not symmetric. It was one of these gammas especially. Because one of them has an index that contracts with the... Okay. This term minus that term is going to give us the gamma, gamma minus gamma, gamma and curvature. But now this term and that term should have cancelled. The term that we had with derivative acting on just the e. What's funny me is that we're getting the same sign. So firstly, is it the right form? So we've got del beta a zeta. That is alpha del beta a. So the term coming from this... You're saying that this will make it symmetric. That's probably true. No, this will not make it symmetric. You have to write alpha. This is only del beta del alpha a. Then you have to subtract del alpha del beta a. I understand. So you're saying that the sum should vanish. It will vanish if this is symmetric. If the sum of this and this is symmetric in alpha beta. Let's check that. Thank you. Yeah. Sorry. So I've kept the right term, I think. But what I've not kept is this guy acting on this guy. And this guy acting on this guy. Let's check that. So that's... What I've not kept is del beta minus of del beta a zeta. Gamma zeta alpha mu. Del beta a zeta. Plus gamma zeta beta mu. Del alpha a zeta. Okay, it's perfect. As you said, it's symmetric. So under subtracting alpha goes to beta. That kills that. Okay? So when I circle this, I'll circle it in a more clever way. I'll circle that. So we'll have the derivative acting only on this. Okay? This part. Again, I uncircled this. But I circled that. Now we're in business. Thank you. Thank you. Okay. So now we're in business. And so what's our answer? We got... The term we wanted was... Okay, we might as well subtract alpha to beta immediately. So minus del alpha gamma zeta beta mu. Okay? And here we will get minus gamma beta... Minus is not there because overall outside. Minus beta mu zeta. Oh, minus into minus is plus. And I've taken a minus outside. So this is minus. Minus beta mu zeta gamma zeta alpha phi. Now we wanted the index which contracted A to be zeta. There's no phi here. So we'll make that phi appear. The whole thing will multiply. Okay? And then we'll take the other way around. Gamma alpha mu zeta gamma beta zeta phi. Okay. Now this is claimed to be... What we should have got is... Okay. So del beta del alpha of a mu minus del alpha del beta a mu should have been... beta z... My cheat sheet beta is equal to l. l is equal to beta. k is equal to alpha. mu is equal to i. Should have been equal to a m. r... Okay, we've got a phi. phi r phi... i is beta. k is alpha. And... Sorry, just one minute. l. That's l. l and... Sorry, sorry. i k l. i was mu. k and l are alpha and beta. Okay. So Landau-Liefschitz claims that this should have been our answer. If that's correct, it must be true. This is the answer if this expression here is this. That is... And with what sign? With what sign? The way I've written it, with a plus, it should have been... So if this is right, we should have that r phi mu alpha beta is equal to del alpha r phi beta mu minus del beta a pi plus gamma zeta beta alpha gamma pi alpha zeta minus zeta alpha gamma pi beta zeta. Okay. And now I'm going to check the definition and see if this works. The one with this index should have... Yeah, okay. I'm getting it. I'm getting both of them to the right side. Okay. This is... This is correct. Okay. So we found the right formula. We found the right formula. One of these indices should be mu. One of the... Oh yeah. There's something wrong with the indices here. How is that to go? This term came from here. First term beta beta. Zeta beta mu. Thank you. And this term came from here. This is it. I checked that... It's clearly the sum of these two terms and these two terms. I think I checked just now that I got the same signs as slant additions. Okay. So we've got... I'll urge you to check this carefully. Okay. You do it right in a second. Okay, excellent. Fine. So we've concluded that the difference in covariant derivatives, difference in the anti-symmetric part of two covariant derivatives, acting on an object that transforms, okay, is not zero, but is given by this formula. Okay. This is another way, of course, of seeing that this R is a tensor, because the covariant derivative was built to be a tensor. Difference between two covariant derivatives is also a tensor. A is a tensor. Therefore, this must be a tensor which transforms in the right way. Is this clear? Excellent. Fine. So now we understand many things about the curvature. Now, this curvature tensor, this curvature tensor is the basic geometric object that measures how much space-time deviates from flatness. Let's try to understand that statement. Firstly, suppose space-time was flat, then clearly curvature is zero, because it is possible in flat space to choose the usual coordinates, x, y, z, d. All gammas are zero in those coordinates. Therefore, derivatives of our gammas are zero. Therefore, curvature is zero in that coordinate system, but curvature is a tensor. So it's zero in one coordinate system and zero in every coordinate system. Okay. What about the converse? Suppose the curvature is zero everywhere, then is it possible to choose coordinates in that space-time, at least locally in the finite local patch. There could be identifications. There could be a circle. You could have things that are locally flat, but not globally identical to flat space, but ignoring such subtleties, in a little finite patch, is it possible to choose coordinates such that in that little local patch, the metric is just eta mu nu. Okay. Claim, yes. Let me give you the proof. It's a proof that can be given just in words without any equations, more or less. Okay. As we discussed, when curvature is non-zero, it is not possible to take a vector or a one-form at a point and define a vector field or a one-form field by parallel transport, because what you get going to another point depends on the path that you take. However, when curvature is zero, then all parallel transports through different paths gives you the same answer, but you don't go through some topological obstruction, just locally at least, some local patch. Therefore, the procedure of defining a vector or a one-form field by parallel transport away from a point is well-defined. Okay. So, now let me consider a manifold in which the curvature is zero. At a point, I go to coordinates such that the metric is eta mu nu at that point. So, I've got a special point here. I move to coordinates such that the metric is eta mu nu, the second one we do. Okay. And in this coordinate system at that point, I look at the one-form field. Let's say Mu is equal to, for instance, one, zero, zero. One in the time direction. Zero in the space directions. Let me start with this graph. I look at the one-form at that point. And then I define a one-form field by taking this one-form and parallely transporting it all over space. Okay. This gives me a mu and I'll call this T as a function of x, some one-form field. The T is to remind me that I started with time zero, zero, zero. In a similar manner, I define the one-form fields a mu of x1 as a function of x by parallely transporting zero, one, zero, zero. Okay. I define a mu of x2 as a function of x and a mu of x3 as a function of x. So, I've got these nice fields defined all over my new code. Now, I want to set up a coordinate system. Okay. At any point. Okay. So, I have my special point which is going to be my origin. Okay. And now, given any point, I associate to this point the coordinates T, x1, x2, x3 as follows. T is equal to integral a mu of T, dx mu from zero to that point. Similarly, x1 is equal to, is equal to integral a mu of x1 dx mu from zero to that point and so on. Now, some of you are asking but along what path? Answer doesn't matter because the curvature vanishes. Coordinate system. It's a coordinate system in which these one-form fields, a mu of T is just 1, 0, 0, 0 in this coordinate system. The expression for a mu of T is just 1, 0, 0, 0 everywhere. The reason for that is that 1, 0, 0, 0 dotted with dx mu has the property that that's dt. But our definition of T was that the change in T from one point to another is the integral of a mu of T dotted with dx mu. So, putting those two statements together, it follows that a mu T as a field in this coordinate system is the expression 1, 0, 0, 0 everywhere. Okay. Now, similarly, a mu of x1 is 0, 1, 0, 0 and so on. Now, remember that the a mu field was defined by parallel transport. By definition, a vector field defined by parallel transport satisfies the condition that the covariant derivative of this vector field is 0. You see, because the covariant derivative is take this guy, parallely transport it back. Suppose I want to define the covariant derivative of a field. What I have to do is to go to some other point, parallel transport this back. But, you know, we're just undoing the parallel transport that we did to get here. So, we just get 0 by taking this. So, this vector field, these one-form fields that I've defined by definition are covariantly constant. But in this coordinate system, the derivatives are also constant. Okay. So, that tells us that gamma, so let's work it out for this guy. Okay. So, saying is that of a mu T is equal to 0. We know this from construction. D mu of a mu of T is equal to D mu, now we work in the special coordinate system, is equal to D mu of a mu of T, but that's 0, plus gamma mu alpha theta a theta, a theta is just the time component. So, that's a T. Okay. Since we know that this is 0, that tells us that this is 0. But this argument was true for a mu of T, a mu of x1, a mu of x2, a mu of x3. That tells us that all gammas are 0. So, we've gone to a coordinate system in which all gammas are completely 0. Okay. And in this coordinate system, it's also true, I probably didn't even need to do this. It's probably even easier. So, it's probably obviously true that the metric is just eta mu nu. Yeah, that was probably the easiest thing to do. Sorry, Sagan? Gammas are 0, but if you remember, we got these gammas by... Okay, good question. It's a question of symmetry. Now, you see, we got these gammas by inverting this. Let's do the counting. How many independent gammas are there? Okay. The number of independent gammas are 10 into 4. 10 because symmetric down and 4 up. How many independent derivatives of the metric are there? Okay. 4 into 10 is the 10 components of the metric. And if you remember how we found the expression for gamma, we first had expressions for derivatives of metric written in terms of gamma, and then we inverted those equations. So, if gammas vanish, the derivatives of metric also vanish. Okay. Okay. Just the last step to complete the argument, I want to argue that... Just when there's one step left to complete this argument. And... Okay. There's something more direct to say, but anyway, now that we've shown that all gammas are 0, the metric is constant. That follows because... Okay. And if the metric is constant, we know what it is because it was eta mu nu at one point, and therefore it's eta mu nu everywhere else. There's probably a more direct way to argue that last statement. Anyway, we've shown that. Okay. So, if we have a coordinate system in which... I mean, a space time in which the curvature vanishes at least over an open neighborhood, then we can set up a coordinate system over that open neighborhood in which the metric is just eta mu nu. Okay. So, that tells us that curvature, this curvature tensor measures the obstruction for a space time to be flat, to setting up a coordinate system in space in which the metric is eta mu nu. If the curvature vanishes, the space is flat. If the space is flat, the curvature vanishes. So, now... Now... Okay. Okay. We've gone slower than I had hoped. Okay. Now, how do you think about... How useful do you find it for me to do algebra and class? If I... It's cheating. Okay, I wonder that. There's some temptation to say it can be shown that. We don't want that, right? Okay. Okay. No, no. We don't want that. Fine. So, let's move on. What I'm trying to head towards is building Einstein's action. But we need a few more definitions and a few more things about this curvature tensor. Well, the first property that we want to show about... So, there are two things we want to understand, two additional things we want to understand about this curvature tensor. The first is the one that Arashan talked about, namely, it's Bianchi identity. And the second, which we'll actually talk about first, it's something that's even more basic, even more elementary, that is the symmetry properties of the curvature tensor. Let's start talking about that immediately, though we're not going to get very far in this discussion. Let me tell you what the properties are, firstly. I'll tell you what. We've got five minutes or three minutes left. We're not going to finish the discussion of the symmetry of the curvature tensor in this much time. Well, instead I'm going to do the proof of the symmetry properties. What instead I'm going to do is just tell you the properties, which we'll prove next time. And then use that to do some counting, which we can do quite simply. So, there are three properties of the curvature tensor that are important and interesting. Then don't involve derivatives of the tensor, of this curvature tensor. So, we have this tensor R, A, B, C, D. Now, it's convenient in order to discuss... You know, it makes no sense to talk about symmetry properties between an upper and a lower index, because that's not coordinate invariant. Upper indices transform differently from lower indices. So, if you say there's an interchange symmetry, that can only be true in one special component system. But it makes sense to talk about symmetry properties of two lower indices. So, the symmetry of this object is maximally revealed when we lower this index as well. So, let's lower and look at the object R, A, B, C, D. You know what this is in terms of this, the metric look at it. Now, the thing we're going to do to start our class off next time is to show that this object, R, A, B, C, D, is anti-symmetric in interchange of C and D. Anti-symmetric in interchange of A and B, but symmetric in interchange of the pair A, B with C, D. So, let's write that down as an equation. So, our A, B, C, D is equal to minus R, B, A, C, D is equal to minus R, A, B, D, C is equal to R, C, D, A, B. And then, of course, by playing around, yeah, you can generate more identity. But the thing to remember is that it's anti-symmetric in its last two indices, anti-symmetric in its first two indices, anti-symmetric in the interchange of the pair, okay? And there's one more thing that we're going to prove. There's one more thing that we're going to prove in the next class. And that is that if we freeze one of these indices, let's say we freeze this index A, and then we cyclically permute the B, C, D index. R, A, B, B, C, D plus R, A, B, B, C plus R, A, C, D, D, B. This is equal to zero. And this is true for arbitrary indices, arbitrary choices of ABC. These three things we will prove in the next class using some algebra.  But using these properties, I'm going to ask the following question. I'm going to ask the question, how many independent components are there in the curvature tensor? So how many independent functions do you have in the curvature tensor? So firstly, if you ask the question, how many independent functions were there in the metric, the answer to that question was 10, right? 4 into 5 by 2. Okay? Now, let's try to get... Can somebody help me to do the counting? First, ignore this relation. If I ignore this relation, what would the answer be? Not quite. Unless you're in a different dimension. Tell... We can talk it aloud. Okay, so I'm going to break this question up. How many independent anti-symmetric pairs are there? No. How many independent components in an anti-symmetric 4x4 matrix? Six. Six, right? How many ways can you choose two pairs? 4 into 3 by 2. So there are six of these. How many here? So that was why he said 36. But the symmetry between interchange. He's not accounted for that. How do you account for a symmetric? 21 by 2. Well, it's good. What's the answer? 21. 21, fantastic. Six in a 7 by 2. The number of elements in a symmetric 6x6 matrix. Six in a 7 by 2. So that would be 21 elements. Okay? Now, what cut-down do we get from this relation? That's not so obvious, so let me help you work it out. The first thing is to understand when this relationship gives you something new beyond what you already knew. So this is true for all choices of indices. Let me see. Suppose I choose B, C and D all to be the same index. Then I'm claiming that there's no content in this thing. This is obvious. Why? Because these two are the same index, but it's anti-symmetric in this index. So it's zero. Great. Now, the next thing I'm going to do is to choose B, C and D to have two indices common. Let's say C and D are the same index. Let's see what I get. The first term is obviously zero because C and D are the same. Okay? What's the second term? Second term is R, A, B. Sorry, R, A, C, B, C. Third term is R, A, C, C, B, D, C. Anti-symmetric of this one, this makes this obvious. So this identity, when any two of the indices of B, C and D are the same, is not a new identity. Okay? In order to potentially get something we didn't already know, we haven't already accounted for an accounting, we have to choose B, C and D to be all distinct. Now, let's suppose we choose A equals one of these. Let's say B. Let's see what we get. So we get R, B, B, C, C, D, that's zero. Plus R, B, plus R, B, C. Everyone, what can you say about this pen? It's zero. Because this is equal to R, D, B, B, C is equal to minus R, B, D, B, C. So this is also automatic. We haven't learned anything new from this new identity that we didn't already know from the symmetry properties. If B, if this A index is one of these three indices. So the only time we learn something new is if all four indices are distinct. How many ways are there of choosing four distinct indices out of a set of four? One. So the number of independent components in the curvature is therefore 20, 20 minus one is 21. In the problem set, which will actually appear soon, we'll give you a problem in which we'll ask you to do this in D dimensions. I'll ask you to do the counting for how many independent components of the curvature tensor that are in D dimensions. Very good. Excellent. So the number of components in the curvature tensor is 20.  Oh, so there are two, three more definitions. We've now got this nice four index object, our ABCD. By the way, what order in derivatives? What is the highest derivative of G that appears in the curvature? Second order. Because gamma has a derivative and there are expressions which are derivative of gamma. So it's a two derivative term. That's ideal. In classical mechanics, we like actions with two derivative terms. There is lots of experience so far. It's a structure of phase space, two time derivatives, position momentum. It feeds nicely into quantum mechanics. So that's sort of ideal. So what we want to do is to take something built out of R and put it into the action. You might worry that having two derivative terms in the action is not what we want. That we want one derivative and one derivative squared. You see, the net total number of derivatives in R is never more than two. It's not like you have two derivatives of a metric multiplying one derivative of a metric, which would give you a three derivative to term the equation of motion. I mean, the equation of motion will obviously be a two derivative equation of motion, which is that if you build an action out of R, as we will see in more detail, which is what we want. But of course, R is not a scalar. So what we want to do is to build some scalar out of R and we can add it to the action. How do we do that? Well, clearly what we want to do is contract. Now, let's first see, can we get a two index object by contracting R? Can you think of a way of getting a two index object by contracting R? Are there many ways? What if I contract A and B? What do I get? Zero. Because this is an anti-symmetry. I contract C and D, I get zero. So what are my options? B and C and which one I choose doesn't matter because up to a minus sign, they'll all be the same. So I make another definition. This object, I define the Ritchie scale or whatever, Ritchie tensor. Rik is equal to glm, Rli mk. Let me make this A and B. R, A, B, G, B. Okay. You contract in Landau-Liefschitz conventions, you contract the first and the third index. And you think of this as a function of the second and the fourth indices. Okay. Now, what symmetry properties does this Ritchie tensor? Symmetric. Symmetric. Because if I make, if I do kA, I'll do, where's my k here? I'll do kA here. And then this is just dummy variable. Interchangeant. Okay. So this Ritchie tensor is the unique two index tensor you can get by contracting indices in the Riemann tensor. And there's a symmetric tensor in its indices. How many independent components in Ritchie? There are no further identities for Ritchie. How many independent components? Ten. Ten. Four into five by two. Excellent. Now, out of this two index objects,


What do you think it is for Herbert this upcoming season that we should be looking forward to? Yeah, it certainly won't be the same.  That kind of drives him to be the player that he is. I think you're seeing more of that from Justin this year, more input in the offense and just a better field. But he's still a young player in this league. He's going to keep getting better. I don't see him leveling off anytime soon. Youngest in the division. You've got to be so f***ing happy, Tom. I'm sorry, go ahead AJ. It will follow up on that. I was going to say, like, how nice is it to know that you have your franchise QB in place? You drafted, right? It really starts with a quarterback. Honestly, doesn't it? It feels like the Chargers have a lot of momentum and impopularity.


Hi everyone, Joe for Jaspy's CaseBreaks.com coming at you with 2014 Prism Football Salo Pack Break Number 23. Last filler we got to do here, nine of those hip parade football dual case break spots being given away at the end of the video. That'll be a different dice roll. The first dice roll will be for the break. Big thanks to this group for making it happen. I appreciate it. There are the teams. I appreciate the teams. Four and a four. Eight, the hard way for names and teams. Two, three, four, five, six, seven, and eight. Four and a four. Eight, the hard way for the teams. Two, three, four, five. All right, here is the first half of the list right there. No trade windows and the second half of the list right over there. All right, a little blast from the past here. 2014. Here's the final planner right here. 2014 is when Jaspy's first started getting group breaks going. So I've got a little nostalgia for some of these 2014. Probably Blake Bortles was a big chase. There's rookie Aaron Donald right here. There's Shaq Evans, JJ Watt. There's Jets Brett Farr. Looks like they're going to get Aaron Rodgers now too. That's exciting for Jets fans. Andy with the Rams gets the Aaron Donald. Now we've got Red, White, and Blues. Here's Andre Ellington, Heartline, and Julian Edelman. All right, nice little blast from the past. Now, let's gather everybody's names again, one through 32. And let's see who's going to win those nine Hit Parade dual case spots, top nine after 10. Six and a four, good luck. One, two, three, four, five, six, seven, eight. Nine and 10th and final time. So from 10 on down, no go unfortunately. Lee, close, but not quite. Appreciate you and everyone trying really, 10 through 32. Thank you. Top nine though, you're in, you're happy times. Cause you're in the Hit Parade break. Evan is happy. So is Edward, Kenny, Aaron, Chris, Alan, Chad, Kenny, and Eugene. Congrats, top nine. We'll see you in the next video for the break itself at jaspyscasebreaks.com.


breaking news, a source has told them Robbie Anderson has agreed to a deal with the Panthers, so the Jets do not re-sign Robbie Anderson. Yikes. Well, I remember a radio show very recently, that's a little bit of a shocker. I remember a radio show that I listened to every day from three to seven saying just yesterday they expected Robbie Anderson to be back. But with this total agreement. Two year, twenty million dollar deal including twelve million a year won with the Panthers according to Adam Sheff. Not crazy. That's not a huge deal. No. So that tells me when we talked to Riddick yesterday about that this is such a deep wide receiver draft that maybe the Jets feel like that that's the better way to go. But that tells me that they really weren't interested in bringing him back because that's not an outrageous amount of money to pay for him. No. No. For the amount of time he's been there. Well, is his former college coach. I'm sure he'll do well. Let's be honest, Robbie Anderson could have been really good. He never quite took the step into special. Can we just be honest? I mean, I would agree. His best year is twenty seventeen nine hundred forty one yards. Seven touchdowns, which is TD high. The last two years, seven seventy nine seven fifty two for yardage five and six touchdowns. He was if you had a great team, you know, he's your second wide receiver. He's never been an NFL prime time first wide receiver to defend our theory, or at least my theory that he was going to make it back to the Jets was that he was going to price himself out of the market, which if you look at the totally God, he probably did. It just that the Jets just that still was too much too rich for their blood. Well, they better get this kid some weapons. I mean, they're they're improving there. They better get him some weapons. I wonder if that says that Sam wasn't overly fond of him because don't you think if Sam was like, no, no, no, I love Robbie Anderson. He's a huge part of what we were going to do. Don't you think Sam could have fought for him and had an impact? I think I think it's all Gase. Yeah, I guess Gase isn't feeling it. I mean, he could have said his part, but if Gase didn't think he was the answer, let's go to all.


opening up the heart space and breathing into the heart as we sit in sacred circle and just feeling the essence of gratitude beginning to develop and grow starting in the root and the feet feeling that rootedness to earth and just expressing gratitude to the Mother Earth giving birth to us we're nurturing us for her patience and her compassion with us as humanity moves along in this learning process that we're in. Feeling this gratitude moving up the ankles the legs the knees the thighs rounding in the root chakra opening the roots and opening opening opening from this root area all the way down to the core of the earth the very crystal core of the earth very beautiful white fire opening and then allowing this white fire gratitude to move on up through the roots opening into the sacral blossoming open like a great ball of light in these two areas these two chakra areas centers merging as it moves up merging the roots and the sacral opening and just feeling the light feeling this incredible incredible column of light moving up light from within light from within each energy center opening and gratitude and empowerment moving up into the sacral moving up into the solar plexus and feeling the solar plexus blossoming open emerging with the chakra below this is gratitude light whose upwards into the area of the hearts and feeling the heart blossoming open and gratitude and light giving thanks for all that is giving thanks for our beloved divine mother and our divine father and opening now to the energy of light above welcoming in the energy of the divine father to our midst and sensing this great brilliant golden white lights opening up above the entire heavens opening up the great solar brilliance is brilliant to the sun becoming this portal for the divine essence of solar light welcoming that now coming down opening the crown chakra to receive this light very gently descending a great white gov atras got energies of both the dove and the albatross descending down into the crown it's brilliant brilliant white gold light feeling that crown chakra open and feeling also the the third eye chakra opening and feeling these two chakras just merging this light very slowly descends down filling the entire head and pineal area brilliant white light very gently gently open to descend this chakra begins to merge with the one below the stroke chakra opening up to receive this light this sense of joy beginning to form in the hearts as this beautiful light from above approaches the heart the throat chakra becomes completely open continues to peel open top to bottom very gently the high heart opening receive this light merging with the throat above feeling this beautiful brilliant pulse of light in the high heart in the primus chakra receiving with gratitude joy infinite peace this chakra fully opens and the chakra below the heart begins to open and receive this light from above these two energies merge in this beautiful dance feeling a pulse of deep intense joy recognition gratitude and love feel this unified love energy beginning to pulse in the heart just in rhythm with the heartbeat in rhythm with the beat the heart of Mother Earth in rhythm with the beat of the heart of the central Sun just take a few breaths feeling this pulse of the universe within the heart body in the heart bold and in the heart now becoming aware of how this heart is the central light the central son the light a spirit above duplicated in the heart or holographically projected in the heart it is one with the great source of light one in each of our hearts and in our collective heart opening the collective heart of this group and contributing to the opening of the collective heart of humanity and of the planet earth where they are inseparable radiance atomic state of being and feeling the power and the joy of this you may feel it duplicated in each of your energy centers all around the body every energy center every cell carrying a duplicate of this hologram this is the light of source the light that no darkness can exist within the light that when it shines brings all to perfect unity the light of remembrance of pure love and pure unity being throughout your being the aspect of your being every aspect of your physical body each cell radiating this light accepting and receiving this light we are worthy we claim our birthright as divine human beings upon a divine earth as divine and sovereign souls partaking in a divine and sovereign experience also collectively claiming our birthright of joy of true peace happiness infinite abundance beautiful connection vibrant health in all aspects of being claiming this is our truth here and now across all lifetimes across every level of existence we claim this now in one this and sovereignty let it be so let it be so let it be so for ourselves for our loved ones in connections for our communities for a beautiful mother earth for all of humanity all of life all of cosmic consciousness acknowledging that this is here free for all who are willing to partake of it knowing that this is the ultimate destiny of every soul and expressing great love and compassion for those who do not choose to partake at this time surrounding them with infinite love understanding and peace waiting for them whenever they are ready giving thanks to the directions to Father divine Father divine mother mother earth to our higher selves to our ancestors to all of our guiding spirits the spirits of the land the air waters and the fire and asking that all connections between us be those of love and light only we are sovereign beings in service to the light thank you thank you thank you so it is


Thank you for your loving kindness. Oh God. Man's will is not your will, Father God. Yes, Lord, we give you all the glory. We give you all the honor. Thank you, Father. Thank you, Lord. Thank you, Jesus. I would leave. I just want to thank you, people. For the spirit of love, oneness, resilience, commitment, support, and everything that you have shown to me, my husband, my children, the entire family, and indeed Zambia as a whole. I know that many people out there have been praying, and people have been praying and praying for years from the time we left office. And Father God is listening. Your prayers are not in vain. May we continue living united, loving one another so that peace may prevail forever and ever. May we not do this just for once, but may we live to support one another. Thank you to the PF, thank you to the members of our parliament who have come to rescue me from the gallows. I'm so grateful to all of you that have been there. You spent the whole day toiling. You tiled, you labored to have me released. It's not easy. This spirit should continue. The greatest reward will come from God for all that you are doing. Thank you so much. God bless you. What should I say that I wish I could repeat it went for a while, but you had a long day today. All of you are looking forward to some rest. Suffice to say may God bless each one of you, and may this not be the last. And for those of you who have doubts in the power of the Lord, you have sinned for yourself. And when people pray, God answers. God has got his own style of answering. Do not feel weakened because you think God has delayed. There is a purpose why God might not answer immediately. Sometimes he answers immediately. Like today he has answered immediately, but he has chosen to answer tomorrow. But God is the valet. It was on time. So mark my words. We should always look to God regardless of the situation in which we are. Everything has a purpose. And if the purpose is to enrich God's glory, so we should be grateful for it. The suffering we went through is purely to show the might of God. The loss of power by the PA, what I say today, was meant to show us how God works, and His working place would consider it. It is the video lives at home, community life, and indeed the whole nation. God bless you.


Meaning 14 and welcome all to this session of discourse. So, far we are discussing the show normal shop. And in all those discussions we have made this actually at locations. While the fluid flowing and across the shop there were these年的 In the fluid properties that we have seen. Now, there are situations where the shock is moving sometimes in a flow relative to a flow, flow may be rest at rest, flow also may be moving which are observed in certain practical cases like in air compressors which we discussed in turbo machines, sometimes in the exhaust and inlet manifold of internal combustion engines which may occur due to explosion, a shock wave propagates through a medium which may occur in case of fluid flowing in a pipeline, we just close or open the valve at the downstream, so a shock wave is generated which is propagated or moving to the in the fluid medium. In those cases it is easy to analyze the mathematical relations or to analyze the problem with respect to a coordinate system attached to the shock moving with uniform velocity, then the entire thing appears to be a stationary shock wave across which there is a change of fluid properties, but how these analysis are being made and results are being obtained, we have to see. Most important thing here in this context is to find out the velocity imposed by the motion of the shock wave to the fluid medium, so to understand this let us consider a case like this, let us consider a case like this rather I write here moving normal shock, moving normal shock, moving normal shock wave, shock wave. Let us consider a case like this where in a duct a shock wave, this is the shock, this is the shock wave, this is the shock wave, let us consider a situation like this that a shock wave is moving in a duct with a velocity of u s constant velocity of u s and the gas behind the shock that means downstream part of the shock it is at rest there is no velocity v is 0 and pressure is p 1 and temperature is t 1, whereas this shock has created a velocity in the fluid through which the shock has passed, let this velocity is create a velocity v, let us consider a velocity v, let us consider a velocity v which is being and it has created a pressure p 2 and temperature t 2, this is the velocity of flow v that means due to the shock passing over this fluid a velocity is imposed or induced by the shock. Now, this problem if you analyze in a way that this is a picture with respect to a coordinate frame attached to the wall that means a coordinate frame static frame of reference which is at rest. Now, if we attach the coordinate frame on the shock that means if we see this thing relative to the shock it is just like a stationary shock wave normal shock wave that means shock is at rest here that means then it will look like that the fluid velocity u s and a pressure p 1 and temperature t 1 approaches the shock and while it goes this side it is velocity is reduced u s minus v usually this u s is higher than v and therefore there is a deceleration in the same direction and this is u s minus v and the pressure is p 2 and t 2. So, therefore with respect to the coordinate axis attached to the shock wave this is the picture.  So, now if this is the thing now if we define the Mach number at the inlet that is upstream of the shock then what is the Mach number now here I can write the Mach number m 1 as u s by a 1 where a 1 is the acoustic speed sorry a 1 is root over gamma r t 1 well a 1 is equal to root over gamma r this is known as this m 1 is u s again I am writing a 1 is known as m s this is known as this is just a definition shock Mach number shock Mach number a shock Mach number m s. Now, what is m 2 now that means with respect to a stationary wave.  So, this v is the v generated due to the passing of the shock wave, but in a steady condition with reference to a coordinate frame attached to the shock wave that means if you make the shock wave stationary by imposing a velocity u s in the opposite direction to the entire system this is the situation. So, I am writing in that case the m 2 is the velocity that means the Mach number after the shock that is the stream velocity u s minus v. So, this become this one that means this is equal to m s into a 1 by a 2 minus this is v by a 2 minus v by a 2 minus a 2 is written as m 2 dash that means m 2 dash is nothing, but v by a 2 that means this v by a 2 is written as m 2 dash this is an important parameter. So, therefore, this relationship you have to first of all remember or you have to write that relations that means it happens that a fluid with a Mach number m 1 given by u s by a 1 and with a velocity u s and p 1 t 1 is passing through a stationary shock wave and its velocity is reduced that is Mach number rather is reduced to m 2 which is defined as u s minus v by a 2. Now, this quantities are with respect to this problem this is an unsteady problem where the shock is moving and these velocities are defined with respect to the coordinate frame static coordinate frame which is attached to the wall coordinate frame attached to the wall. Now, if we accept this now next part is that if we recall the analysis we if we now recall the shock relationships just I am writing this in earlier classes in few earlier classes we have discussed this thing that the suffix 2 is always the so now I am writing the shock relations normal shock relation normal shock just recapitulation normal shock relations recapitulation of normal shock relations this equal to this p 2 by p 1 that means the suffix 2 as you know is a usual symbol is after the shock and this is before the shock usually in shock more useful relationships are the ratios of pressure temperature density in terms of the inlet Mach number inlet Mach number is the input parameter. So, flow is usually specified by inlet Mach number so I am recollecting this formula this formulas that we already already deduce gamma plus 1 similarly rho 2 by rho 1 has an expression which is gamma plus 1 m 1 square divided by 2 plus gamma minus 1 m 1 square similarly we can write t 2 by t 1 which was all deduced earlier this is this 2 plus gamma minus 1 m 1 square if you remember this relationship 2 gamma m 1 square minus gamma minus 1 and divided by gamma plus 1 whole square m 1 square now this t 2 by t 1 can be written also equals to what is t 2 by t 1 a 2 by a 1 whole square why because the a 2 at the section 2 is root over gamma r t 2 and a 1 is root over gamma r t 1. So, ratio of temperature is the ratio of the acoustic speeds at that state corresponding states a 2 by a 1 square another very important relation is there that is the Mach number after the shock which is the subsonic Mach number gamma minus 1 m 1 square plus 2 2 gamma m 1 square minus gamma minus 1. So, at present I recall only this form this relationships again I tell what are those relationships this is the change in the fluid property in terms of the ratio pressure ratio the temperature t 2 by t 1 rho 2 by rho 1 the density and the Mach number after the shock in terms of the initial Mach number. Now, this relationship are exploited to find out a chart where against a Mach number 1 that is m 1 against m 1 against a Mach number m 1 we can found we can find all these properties. Another thing you have to remember with this or the recapitulation which has been already discussed earlier from the consideration of entropy change as a corollary of the second law we always tell that m 1 has to be greater than 1 for a shock to occur and m 2 will always be less than 1. That means a shock always decelerates the fluid approaching the shock from a supersonic one to a subsonic one after the shock.  So, this is our m m 2 now what you do if you write the same this replace this you get this.  So, according to the present nomenclature I write this one and t 2 by t 1 will be t 2 by t 1 will be again same that m 1 is replaced by m s m 1 is replaced by m s that means 2 plus gamma minus 1 m s square into 2 gamma m s square minus gamma minus 1. So, this is also equals to a 2 by a 1. So, this a 2 a 1 related to t 2 t 1 this t 2 t 1 p 2 p 1 these are scalar quantity they are not dependent on the coordinate transformations.   So, this is our m 2 at the present moment m s a 1 by a 2 minus m 2 dash now what happens if we take this equation and if we use the a 1 by a 2 then we get a very important relationship of m 2 dash let us do that.   So, you just see that this is the from your note that as you have taken I am writing then m 2 dash is equal to m s into a 1 by a 2 what is a 1 by a 2 is under root of that.  Now, the denominator will be under root gamma plus 1 whole square m s square that means this gamma plus 1 whole square that means gamma plus 1 into m s and denominator it will be m s very good denominator it is m s very good denominator it will be m s very correct so denominator it will be m s all right. So, m 2 dash is m s into this minus if you do that you will get gamma minus 1 m s square plus 2 to the power 0 plus 1 whole square 0.5 divided by 2 gamma m s square minus gamma minus 1 to the power 0.5 you can see it very correctly then this becomes equal to what then this becomes is equal to m s square gamma plus 1 so this will be the denominator 2 plus gamma minus 1 m s square to the power 0.5 let me write this 2 gamma m s square minus gamma minus 1 to the power 0.5 then minus so this 1 is 2 gamma that means this 1 2 plus gamma minus 1 m square 2 plus that means this will be simply 2 plus gamma minus 1 m s square that means the same thing 0.5 0.5 so if you now clear this thing then the denominator will be what denominator will be this big thing 2 plus gamma minus 1 m s square it is little tedious I know but you just follow that whether I am doing any mistake or not m square minus gamma minus 1 to the power 0.5 and now m s square gamma plus 1 minus 2 gamma minus 1 so gamma m square gamma m square is cancelled so this is m square and then gamma now if you write this gamma here I make a calculation gamma m s square plus m s square minus what is that 2 plus gamma minus minus gamma m square plus m square that means this cancels that means 2 into m square minus 1 this become 2 into m square minus 1 so this is 1 very important deductions that is here I can write m 2 dash I think you can see equals to which is very important m 2 dash why I am writing it is v by a 2 it is not m 2 m 2 is this 1 m s a 1 a 2 minus m 2 dash this is m 2 but I am finding out an expression of m 2 dash in terms of m s now you see what interesting results we have achieved now you see here what it is given that it is given not the downstream Mach number that means Mach number after the shock with respect to the shock the coordinate respect to the shock but it is given the ratio of the velocity induced because of the shock in actual case divided by a 2 m 2 dash in terms of the inlet or shock Mach number here m s you remember is u s by a 1 that is is equal to actual m 1 in this case now is a very interesting result it gives now what interesting result that if this if we increase the Mach number let us consider the inlet Mach number is very very high inlet Mach number is very very high that means the flow approaches with a very high supersonic velocity and as you know the strength of a shock as I told you earlier the is expressed by the pressure ratio that means the ratio of pressure after the shock to that before the shock and it is a direct function of Mach number I tell you is a direct function of inlet Mach number more is the Mach number more is this ratio that means the strength of the shock is usually expressed by this ratio which is a function of inlet Mach number or approaching Mach number more is the Mach number more is the strength of shock so therefore a Mach number approach a shock a fluid approaching with high Mach number to a shock wave is told as the shock is of very high strength so therefore when the shock Mach number is very high we usually call that this is a shock of high strength now whatever high strength shock comes there is a limiting value for m 2 dash physically now look it mathematically is there any limiting value when m s tends to 0 if I ask this question that when m s tends to sorry infinity what is the value of what is the limit rather m s tends to infinity what is the limit that limit I will now show you now m s tends to infinity m 2 dash can be written as so this type of thing as you know elementary mathematics at school level when m s tends to infinity let it expressed in terms of 1 by m s that means both numerator and denominator should be divided by m s square this is the standard technique see that whether the limit is there or not so m s square means 1 m s I can take here 1 m s I can take here and when it goes within the bracket again m square that means this will be 2 by m square that means this gives directly an indication that there is a finite limiting value the term is becoming free of m s m s by this m s we are dividing numerator and denominator by m s square this m s goes inside so therefore m s square here also 2 gamma minus gamma minus 1 so beautiful that means this give us an indication what we want that there is a finite limiting value so when m s tends to infinity it is 0 it is 0 and it is 0 so the limiting value there is a limiting value is root over 2 gamma into gamma minus 1 this is gamma minus 1 2 and 2 under root and this is 2 so therefore 1 2 under root will go up so therefore you see the limiting value that I was telling that limit m 2 dash or equal to m s square tends to infinity limit v by a 2 as m s that means the shock mach number tends to infinity is not infinity it is square root of 2 by gamma into gamma minus 1 this is a beautiful result and you see that for air for air if we take gamma 1.4 this m 2 dash is 1.8 that means you can say that in air whatever strength may be of the shock even a infinite strength shock very high strength shock can only induce a finite velocity which will be always less than this corresponding mach number 1.89 the actual velocity will depend upon the temperature we have to multiply this with a 2 which is root over gamma r t 2 that means velocity is usually specified in compressible flow by the mach number so you can tell for example in air whatever strong mach number may be at the upstream of a moving shock this can or whatever rather this way you tell whatever strong shock may be a moving shock whatever strong the shock may be whatever velocity the shock may be moving that the m s u s by a it cannot induce a flow whose mach number will be more than 1.89 in case of air so this is the limiting mach number in case of air now next is next we will rather now after this I think we will solve some problem so that your all these things will be made more clear by solving the problem before I go to a this reflection of the shock let us solve some problems. Now problem one example one a this is the first problem example say shock wave across which just you see this thing problem which the pressure ratio is 1.25 this is not a this is a moving shock a shock wave across which the pressure 1.25 is moving yes it is written is moving into steel air at a pressure of 200 kilo Pascal s and a temperature of 15 degree Celsius this is the steel air find the velocity pressure and temperature of air behind the shock wave that means it is simply the application of that how to do it now let us again solve the problem the shock is moving with a velocity is moving into steel air and we do not know its velocity let u s similarly that u s and this p 1 t 1 the same problem and then it induces velocity v and p and p 2 p 2 p 2 so with respect to the shock the problem is with respect to the shock the problem becomes steady with a velocity u s p 1 t 1 and it goes with the velocity u s minus v and p 2 t 2 what are given in the problem p 1 p 1 is not given rather p 2 by p 1 is 1.25 what are the given t 1 is given 15 degree Celsius p 1 is given 100 kilo Pascal what we have to find out velocity pressure air behind the shock that means we have to find out v we have to find out pressure p 2 and temperature t the first one is the school level thing p 2 because the pressure ratio is given 1.25 I know the pressure p 1 so p 2 will be 100 into 1.25 125 kilo Pascal straight way but we do not know t 2 how to calculate now I have told you earlier that in shock all those relationship we have derived that means for example if p 2 by p 1 is known I can find out m 1 because p 2 by p 1 is expressed in terms of m 1 again I tell you that just also at the beginning of this class I told you that all these relationship which was derived earlier for example this important relationship here one can find out algebraically p 2 by p 1 I know I can find out m 1. So if I know m 1 I can find out rho 2 by rho 1 t 2 by t 1 over this algebraic calculation sometimes are tedious. So therefore a table is drawn based on this calculation as usually drawn for different cases that table as I now tell you earlier also I told which is known as normal shock tables. So this is there just you see how to see a normal shock tables a normal now I not will do here I will rather do here a normal shock table we just how does it look like you can see in any book normal shock table just I show you probably I told you earlier also normal shock table looks like that in the left hand column these are the different column m 1 the corresponding nomenclature you know that is the mach number approaching the shock with respect to a stationary shock this is the after shock mach number then this is the ratio of pressure 2 is the section after the shock and this is all the ratios of the static pressure temperature the density and there is a ratio of stagnation pressure also because of the losses shock is an across the shock the process is irreversible. So therefore p o 2 by p o 1 now these there are number of mach numbers different values starts from 1.00 why because shock does not occur when the approach mach number is subsonic. So therefore it goes some for example 5.00 so corresponding number all these things are there. So then if I know the p 2 by p 1 now you have to see that what are the quantities given in this problem we know p 2 by p 1 we know p 2 by p 1 is 1.25. So if we know p 2 by p 1 I can find out everything m 2 t 2 by t 1 that means any one of these is known other things are known through the equations. So therefore it is not that always m 1 has to be known anything has to be known we can find out from the chart. So from this chart if you read with p 1 p 2 by p 1 1.25 you will get a mach number of m 1 which is which from normal shock table you will always get m 1 and m 2 m 1 means inlet mach number m 2 means outlet that means the downstream after the shock mach number 9 and t 2 by t 1 you get 1.0662. Now when you get t 2 by t 1 straight t 2 is t 1 is given t 1 is 15 that means 288 into 1.0662 so you get t 2 so t 2 is 307 in this case.  So therefore u s is the mach number I have got from the table into root over gamma that is 1.4 into r is 287 into the t 1 t 1 is 15 degree means 288.  Now you have to find out the velocity behind the shock that means this velocity it is very simple to do this what you have to do you have to write the expression for m 2 in your case in your case m 2 is what is m 2 u s minus v which you have to find out by a 2. So therefore you can write v is equal to a 2 a 2 u s minus m 2 a 2 u s is always I know that this is found 1.102 into root over 1.4 into 287 is the value of r into 288 minus m 2 which I have found out 0.9103 into again a 2 a 2 is root over 1.4 into sorry 287 into root over 1 into t 2 t 2 I have already calculated because I know t 2 by t 1 from the table. So t 2 is calculated 307. So most important parameter is this we want to know this this comes out to be 55.2 as far as this calculation I know the answer this is the answer. So this is a straight use of this formula which we derived the analysis which we have made for a moving shock. Now another interesting problem you see this is a relatively more interesting problem this problem what does it say what does it say a normal shock wave across which the pressure ratio is 1.17 moves down a duct into steel air at a pressure of this is almost the same and a temperature of 30 degrees rest part is little interesting find the temperature pressure and velocity of air behind the shock wave this part is routine. That means you can say that said this is the total repetition of the earlier problem with probably difference in the data. Now next one this shock wave passes over a small cylinder circular cylinder it is actually it will be here circular cylinder as shown in figure that I will show you the figure assume that the shock is unaffected by this small cylinder fine pressure acting at this stagnation point on the cylinder after the shock has passed over it. Now this is a problem here now this problem if you solve then how you what is the problem problem is this one that this is the shock shock at this is a small cylinder. So, shock has started from somewhere else in the upstream it has passed different sections at any instant as the problem tells when the shock has passed over the cylinder and shock remains unaffected what is the condition. That means in the entire upstream part of the shock through which the shock has already passed the fluid this has attained some induced velocity v now let us find this induced velocity v. So, therefore our case is similar to this then our case is similar to this that let this is the cylinder our case is similar to this. So, u s same thing u s minus v this is 1 and this is 2 and let this is p 2 t 2. So, p 1 t 1. So, I can find everything, but first of all I have to see with respect to this stationary shock what is the thing in pressure ratio same thing that means I am having pressure ratio that means I can immediately go to this shock table normal shock table and see against this pressure ratio what is my value m 1 m 2. So, this pressure against this pressure ratio the m 1 as found from the shock table is 1.07 m 2 as found from the shock is 0.936 and t 2 by t 1 as found is 1.046. So, therefore since t 2 by t 1 is known I can immediately calculate the temperature value temperature behind the shock that means this temperature t 2 I can find out t 1 is what t 1 is 30 degree that means t 2 is 1.046 into 273 plus 30 that means 303. So, I can find out this is this equals to 316.9. So, I know this and p 2 I can find out how p 2 is equal to this is the school level thing 1.17 it is already given 1.17. So, 1.17 into 105 kilo Pascal. So, I can find out and this becomes is equal to 122.9 kilo Pascal. Now, what I have to find out next assume that the shock is unaffected before that I have to find out the velocity of air behind the shock this is again the same thing that m 2 I know that means m 2 is 0.936 is equal to what u s minus v divided by a 2. Now, a 2 is what a 2 is equal to root over gamma r t 2 that means 1.4 gamma r 287 into t 2 because t 2 is 316.9.   So, that is very simple 3 16.9 you understand 3 16.9. So, therefore, I get the value of v that this v becomes equals to some 33 this becomes equal to v becomes equal to 39.38 meter per second I think it is absolutely all right. So, it is just the repetition of the earlier problem now the question is this that we have to find out the stagnation pressure now what happens if you see this picture now this cylinder actually here this is the model physically translated, but actual problem is this. This is the problem as we look with reference to a coordinate frame attached to the moving shock and flow becomes steady and this becomes stationary. So, actually this is the problem this v we are getting which is 39.8 that means now it is a problem that a cylinder circular cylinder is exposed here another thing is told stagnation point assuming that the shock is unaffected by this small cylinder find pressure acting on this stagnation point on this cylinder after the shock has passed over it and there is another thing which has to be told that consider this flow to be isentropic consider the flow the pressure acting at this stagnation point on this cylinder assuming that the flow is isentropic. Now, if we assume the flow to be isentropic then what happens there is an isentropic flow past a cylinder whose velocity is 39.38 meter per second. So, we can find out the stagnation pressure before that I have to tell you that if you recall the if you recall this isentropic flow relations isentropic table that isentropic flow relations are unnecessary without going to all algebraic calculations through the algebraic equations of isentropic flow relations I rather isentropic flow isentropic flow table that is normal shock table it is given in a tabular form similar to normal shock table that one column the extreme left column is m then these are the stagnation because here most important part of calculation is the ratio of stagnation temperature to static temperature stagnation to static all stagnation property respect of density to the corresponding static is the function of the mach number of flow. So, therefore, if we know the mach number of flow at that particular location we can find out all this thing similarly a 0 by a what is that a acoustic speed at the stagnation condition to the local acoustic speed a by a star probably you recall this things a star is the area where the sonic velocity is rich that is the minimum area known as throat area corresponding to m 1 and this a is the area local area at which we are concerned with and another parameter is theta I told probably earlier that does not come into picture here that will come when we will discuss the expansion wave this is the angle related to the expansion wave this is not coming at that there also the problems are treated as an isentropic flow that is why in the isentropic flow table itself this theta is given this is not required now. So, now if we know the mach number at a particular location in an isentropic flow this here mach number start 0.01 0.04 like that very small because all subsonic supersonic all levels are there. So, there will be somewhere 1 when the value of a by a star will be 1 this things probably I have discussed earlier. So, now if we know the local mach number I can find out the p 0 by p t 0 by t I can read that everything. So, here I have to find out the local mach number now how to find out because I know now the actual problem the cylinder is exposed to this uniform velocity 39.3. So, local mach number is m for example here is 39.38 by the local acoustic speed which is root over 1.4 gamma into r into this temperature that means this is 316 that means it is nothing, but a 2 it is nothing, but in our this problem this is a 2.  So, therefore, we take a value of m this m is equal to this is a v 2 and a pressure the mach number is 0.11 just I see the calculations. So, that you can check 0 point it has been calculated by me earlier. So, 0 point see that thing. So, therefore, in the table if you see this mach number you can find out the value of p 0 by p and that p 0 by p if you see from the table this also I tell you for this problem is this. So, therefore, you can very well find out p 0 is 1.0085 into the static pressure there which is p 2 122.9 because the condition prevailing is this p 2. So, this becomes is equal to 123.9. Let me see the value 123.9 as I have calculated already kilo Pascal. So, this is the value. So, it is clear that you can find out the value of the mach number and you can find out the value of the stagnation point pressure. So, this is one very interesting problem. Now, I will tell you something regarding the reflection of a shock wave regarding the reflection of a shock wave. What is reflection of a shock wave? So, what is reflection of a shock wave? Let me tell you what is reflection of a shock wave. This happens when the shock wave is moving. So, moving shock wave analysis we have done reflection of a shock wave. Now, reflection of a shock wave if we have to understand then just how does it happen? I tell you the physical picture now if there is a cylinder we have so far considered that it is moving with u s and it is creating a velocity v and a pressure p 2 t 2 and this side the pressure is p 1 and initial velocity v is 0 and temperature t 1 that was our thing that it is moving in this direction. So, with respect to these things are happening. Now, this is the actual problem physical problem as we are discussing. Now, what happens if this thing is closed at the end? That means there is a closed end or the pipe was open suddenly the valve is closed. That means the shock wave is moving and the pipe is closed at the end or is being closed by the closure of a valve at the end. So, what will happen? Physically you try to understand what will happen? This velocity will be induced so long the shock moves. Now, try to understand what happens in reality now when the shock comes here the velocity will be imposed v, but after that what happens this is the closed end. So, at this solid end the velocity will be 0 there has cannot be any velocity the velocity cannot penetrate into the solid like that. So, there will be no normal velocity. So, therefore, this velocity will be 0 and slowly in a compressible fluid the entire fluid will attain a velocity 0. For an ideally incompressible fluid this will be instantaneously 0, but for a compressible fluid the velocity will first fall here 0 and it will take some time which may be infinitely small depending upon the compressibility of the fluid that the entire velocity will again be 0. So, this thing is perceived in a way as if the shock wave after reaching there is being reflected back in a way that it creates a 0 velocity here is reflected back in a way that usually the velocity it created in this direction v again it created a velocity v in this direction in the opposite direction. So, that the final velocity is 0 that means the strength of the shock will be such for the reflected way. So, that it can create a 0 velocity while passing through it. So, that is the philosophy of the shock reflection of the shock. Now, let us do that reflection of the shock let us do that the reflection now if reflection of the shock is to be understood let us consider not the initial one the shock is being reflected now. The shock is being reflected now and we can just consider a case that when the shock is reflected and this is the reflection that reflected shock velocity and here let 1 and 2 if we see that 1 sorry this is why 1 and 2 I am telling that this can be expressed as we did earlier in this way that this is coming with us p 1 t 1. So, this 1 and this is 2 this is p 2 t 2 and this is going with us minus v 2. So, now what happened? So, this 2 will now prevail here that is here 2 that is p 2 t 2. So, now this is p 2 t 2 I am sorry now this will be sorry this when you will come here. So, entire thing was then p 2 t 2 that means 2 when it reaches when it reached here the entire fluid was at 2. So, when it is coming here this side is 2 that is p 2 t 2 I think you have understood and this side is 3 that is p 3 t 3 and velocity is 0. So, this is the reflected shock that means the reflected shock if we analyze you will analyze this way. Now with the help of the shock tables we can calculate things that means shock will again be reflected back here. So, this is a typical reflection and under certain circumstances shock will go back again. So, there will be a repeated movement of the shock. So, one such reflection of shock problem we will now solve that how a shock moving ultimately moving in this direction and a particular direction ultimately reaches the closed end and back when it reaches the closed end automatically the velocity becomes 0 this is the reality that means this is conceived as a movement of a reflected shock wave with velocity us are such that this velocity becomes 0 and the condition which is being created is p 3 t 3 again it is changed that means from p 1 t 1 this is p 2 t 2 this is with respect to the shock then the entire thing will become p 2 t 2 here also when it reaches here the entire thing becomes p 2 t 2 pressure p 2 temperature then again it will be changed to p 3 t 3 when the reflected shock will pass through that. So, this is the scenario physical scenario I think that this will be more easily understood if we solve a problem and this time we cannot I cannot continue because the time is up. So, in the next class I will solve a problem in relation to the reflection of a shock wave. Thank you.


news update. Welcome folks. We have the now industrials trading up 144 and Aztec is up 185, S&P is up 35. Gold contract down $3.90 trading at 18 up $3.90 traded at 1802 an ounce. We have silver down 5 cents, $24.13 an ounce late sweet crude flat $82.65 a barrel notes and bonds. The tenure note down 13 ticks trading 130, 24 the 30 year trade. We're also off 13 at 160 $18 and King dollar, King dollar down 453 ticks trade 93, 350. Now we have divergence in a must huge way here, folks. Okay we had the wide price spread on the dollar. It broke into the lower range, but yet gold didn't basically take the bait and go topside my take, especially the way that the dollar has gone down. The price has gone down. The price has gone down. The price has gone down. The price has gone down. The yen has also got stronger. So I expect what we're going to see coming into Friday. We're probably going to get some action here inside of the gold market. If we take a look at the NBX 100, it's going to get really intriguing watching this close. And the reason being is this. So the last two days it we sold down coming into the close and I expect we're going to have the same thing today. Now this is really $45.7 million the day before. $47 million. So it was $2 million less. Big deal. That's nothing. That's only 5%, right? Today, however, you're $25 million. So the real kicker is going to be, does it fail on price? Right now you're over the price. It hasn't failed on price yet. Bottom line, I expect we're going to see it because intraday what we have already had out here, the seller is out here. Whoever the seller is, bottom line, they're out here in spades, too, man. I mean, they came into this market and every time they're coming into this market, they're coming in the futures market and they're coming in with a big auto man because they're just smoking it. They smoked it. The first time they smoked it it was only a really, it was really small at 11.50 this morning, okay? That was their first footprint. Next footprint was much larger. The next footprint had 111,000 contracts in the NDX. When you see this folks, okay, bottom line most of these monsters don't stop. That's the real bottom line because what you've had is this.   They sell it down. They know right now there's no orders to buy above that market. When that happens, just a matter of time when it will work, when they sell it down. Stay right there folks. Come right back.