[SOUND] [MUSIC] So, let me clarify my point. Well, first of all. And with the use of the metric tensor and its inverse tensor, I can also lower and higher the indices of any tensor. For example, if I have a tensor like this mu nu, say, alpha beta. Then I can higher the index of alpha like this, and define corresponding with the corresponding number of indices, like this. And here, I stress that the order of indices is important. That's the first thing. So now, I'm ready to explain why transfers are important. For example, if I have such a quantity, say like this, T mu of mu Gamma and I multiply it by another tensor, S, say carrying indices alpha, nu, gamma. This quantity should transform also as a tensor because it carries one index, tensor index. So it should transform as a vector. And it does transform as a vector under the conditions that these guys transform appropriately. Because transformation of every such index is compensated by a transformation of every such index and the same is true for this, so the only index which remains is this. As a result, this quantity is transformed as a vector. So manipulations with are self consistent and have obvious properties under the coordinate transformations. And that's what we are seeking for. So, and also, it is obvious that a scale or product of two vectors, scalar product of two vectors is invariant under the coordinate transformations like this. That is this quantity. And one can write it like this. Sorry, this is all the same. This is a different ways of writing of the same quantity. So, tanzers as a tool to use to write equations in special and general theory of relativity, so metric is, of course, is a tanza. Because as we discussed during the previous lecture for the metric, we have the following consideration. That integral in one coroney system looks like this. And if we transform it to the other coordinate systems, and then it does like this. G bar of beta of x bar d x bar alpha d x bar beta. Because of these and after coordinate interval doesn't change, we have the falling law of transformation for the metric that g mu nu affects, is equal to g bar alpha beta x bar d x bar alpha d x mu d x bar beta d x nu. And this transformation law we have been using during the previous lecture at the very beginning, so metric is just a concrete example of a tenant. So, but then, after all these definitions, we encounter the following problem. That differentials, differentials are not transforming appropriately, suppose. Suppose we consider a differential of a tensor. I mean, suppose we consider a difference of A mu at the point (x+dx) and subtract from it A mu (x). To linear order in dx, this is equal to d nu A mu of x d x nu. And now I'm going to show that despite the fact that this quantity carries two indices, carries two indices, so nu and mu, but it doesn't transform appropriately as. Now I'm going to show that. I'm going to show that this quantity, this quantity d nu a mu, dx nu which is just to linear order by definition of the difference of these two factors. It is not a tensor with two indices, this by the fact that it carries two linear. Well, by the way, I'm going to use several different notations for the ordinary differential. Namely dAmu/dxnu is the same as dnu A mu and the same as A mu comma nu. So, why this quantity doesn't transform appropriately. Consider this thing. Consider A bar alpha comma beta function of x bar. Well, this is d over d x bar beta d of a bar alpha, by definition, by definition, but this guy, let us express this guy through that unbarred quantity. Well, this is a function of x bar, through that unbarred quantity, so this becomes e over ex bar dX bar alpha over dX mu, A mu of X. So this, I just used the expression for the transformed form of this stanza. And now, I have to understand that, X. X is a function of x bar. A mu is a complicated function of x bar, it's a function of x which is and so on, write as a function of x bar. So, now I use Leibniz rule to differentiate this. First, I differentiate this quantity with x bar. But this is the same as differentiating complicated functions, meaning that I first take the derivative. Well, let me write it and then explain. So x. Of course, I write this DX Alpha, DX Mu. And then, I differentiate. So, what I do here, I explain in a second. A mu, comma. Nu x plus, before writing the remaining quantity let me just stress what I did here. So I applied this differential to this function, so I differentiated it, this guy, with respect to x nu. And here, and it's written and then differentiated x nu as a function of x bar with respect to x bar. So if we would have had only this term, this term. Then this quantity would transform as a tender. But the problem is that we also have to apply the differential, this differential to this guy. And that is a problem, in fact, because, what we get is a falling stage. So, we differentiate this guy with respect to twice x bar alpha d x mu d x nu. And then, so we assume that this is a function of x unbarred. And then we differentiate unbarred x nu with respect to the bar beta. So, I mean, we assume that this differential, this derivative here is a function of x, unbarred x. Times a mu (x) exactly because of this term, this quantity with that simple standard differentiation, doesn't transform as intended. Notice that if we would have had only linear transformations, linear coordinate transformations, this term would be absent and then this quantity would be a quantity. That's the situations we encounter in special relativity, in flat space-time. But in curved space-time, and for generic coordinate transformations, this guy doesn't vanish. As a result, this quantity doesn't transform as a tensor, that complicates things. My goal here is to define co-variant differential, which does transform as an appropriate tensor quantity. Let me define it. The problem with this saying, that we subtract a vector at the point. At this point, from the vector at this point. To make appropriate quantity, we have to parallel transport this quantity to the point x plus [INAUDIBLE]. How it is done, I'm going to explain right now. Now the co-variant derivative of a mu is the following thing. It's a difference between this quantity, And the following thing. [A mu(x) + delta A mu(x)]. And this guy is exactly result of the parallel transport of this quantity to the point (x+dx). Who is this guy, and what it means parallel transport? I will explain a bit later, but at this moment I am going to define this quantity, this quantity. So I'm going to explain this, the meaning of parallel transport on some simple example But at this moment, let me just specify. So I have a point x, I have a nearby point x plus dx, I have some vector field here, so I have, say, for vector a mu at this point, and I have a full vector a mu at this point. So this is A(x+dx), and this is A(x). So, this guy doesn't coincide with the part of the transport of this quantity. It's some value of this field at this point, but then I parallel transport. According to some rule how parallel transport is done, I paratransport this quantity here, and I obtain some different vector, which is A plus delta A, at this. And then I subtract from this is and that's how I get this quantity. So, what is this thing? What is this thing? It is as follows, it is as follows. First of all, I assume that this interval is very short, so the dx is very small despite the fact that I draw it big. So this delta a should be delta a should be a result of some transformation, so there is some matrix M, which does some transformation on our vector A at this point. So who is this matrix? So it's some rotation of this vector. The result of this result is one plus, so unit matrix plus this short rotation given by this M. So, who is this guy? For small dx, it should be proportional to dx. So I define it. This is just the definition with the minus sign with some quantity gamma mu, nu alpha of x, dx alpha. A mew of x. So, this quantity is referred to as a connection for the parallel transport. And I'm going to explain it geometric meaning in a second. [MUSIC]
