 We want this to be an alternative. No, of course, one way to do it is to choose this to be the same. Let me say that. Sorry. What we want is until the range transformations. Sorry. So we've got f of x, x mu. And let's look at the infinity of the range transformation. Okay? So let's, at times, mu, the change in x, we've not done the infinity of the range transformation. Okay? So how does this, this object change? Yeah, we've already put translation invariants by saying that this object is mu. Okay? So then this is del f by del x dot times del times mu. Okay? Because the dot, lambda has nothing to do with the instructions. The dot just goes. Okay? So in that situation, this is the change in the, in the action. Now, of course, one thing that's good for me to do is to make f, a lorenzimate, in which case this thing is zero. That's what we did last time. But suppose you want to allow non-zero values to this. Okay? What you demand is that this object here is a total element. Now, the important thing is that this object here is either constant or a function of x mu dot. I mean, x is a function of x mu dot. So del f by l del x mu dot is a function of x mu dot could have happened to be the constant function. Now, if it happens to be constant, then this thing is a total element. There is no other choice that makes it a total element. Because, you see, as you were saying, f of x times the x dot is a total element. That's a total element of the entire order of action. But there's no function of x dot. There's no absolute function of x dot. Obviously, x dot times the x dot. That's a total element. The x dot square is not a total element. The x dot cube is not a total element. And yes, with del x by l, this is no function of x. But initially, we said we are taking it to be translational. Translational. If it is not with some small translation, then it becomes very changed by some time variable. Yeah. We have to recall it. But what I was trying to know was the analog of what we did for delinium boosts. You remember that when we were trying to study an action that was invariant under delinium boosts, that a grunge in itself was translational invariant. But it wasn't invariant under boosts. It was invariant only under delinium. So what Shant's worry was that maybe there would be something analogous there. Let's first check that. So what do we conclude from? We conclude that this is a total derivative. Only a f dot is a constant. The del f by del x p dot is a constant. In which case, the thing that we're adding to the action is linear x p dot. In which case, the addition to the action is itself a total element. So it's a real addition. So basically, there are no possibilities like that for your nc by nc. Now we could recurse for the two implement translational invariants also. So let's continue. Let me review what we talked about. Oh, somebody used to put that one. So let me remind you what we discussed last class. In the last class, we started our discussion. So what is it? Return others, please. Recording. Yeah, I should say REC. Recording. So the last class we discussed, we discussed that we wanted to find, we had two parallel motivational streams in our discussion class. The first one was a slight discomfort with this engine between the two. So the question that was being asked was could you have a Lagrangian that is itself not Lorentz invariant but whose variation under Lorentz transformations is a total element. So we were looking for a Lagrangian which is a function of x mu dot. We're building in translation invariance in terms of Lagrangian. So we're looking at the Lagrangian, suppose the Lagrangian is some f of x mu dot, d dot. Now I want the variation of this Lagrangian under Lorentz transformation to be a total element. Let me consider infinity as the Lorentz transformation and let the variation of x mu under a Lorentz transformation, any particular Lorentz transformation be delta x mu. Again, this is something interesting but I don't care what it is. You've got nothing. Under a second Lorentz transformation the variation of x mu dot delta of x mu dot is simply delta x mu for all three dots. These Lorentz transformations have nothing to do with Sri Lanka happening. So now let's look at how this Lagrangian varies under a Lorentz transformation. The change in the Lagrangian is what? It's the change in f. It's delta f by delta x mu dot times the change in x mu which is delta x mu of the dot. Now the important point is delta x mu dot is linear in x mu dot since then. So the Lorentz transformation is this linear transformation. So this, with some matrix structure is some m mu mu x mu where this m is the matrix Lorentz. Now you see what we want is that this object here is a total derivative. Now one option is that it's zero. So that the change in f under a Lorentz transformation is zero. That's the option that we implemented in the Lorentz transformation. So we are studying options in which this is not the case but this whole thing doesn't vanish. Now we have two cases. One case is that this object here which is some function of x mu dot is a constant function. In this case, this is a constant matrix times the total derivative and therefore the total derivative. So the variation of this action under the Lorentz transformation is the total derivative. This is the kind of thing that we were interested in. However this unfortunately is trivial because in this case f was itself a linear function of x mu dot. So the change in the Lagrangian itself was a total derivative. So when you change a Lagrangian by a total derivative you are not changing the equations of motion by a total derivative. So this is a trivial option. Basically no other option gives you a total derivative. Okay? What? How is f a linear function of x mu dot? If its variation to the respect of x mu dot is a constant, then it's linear. What is the expression for that? What is the actual expression? The actual expression isn't the inner part of x mu dot of x mu dot. That expression gives zero to this thing. We are looking at things where the variation under Lorentz is not zero. So the answer is something in varying under Lorentz transformations plus something that has to be of this form. But it's a total derivative. So that's the rule. Anything here that is nonlinear in x mu dot is a nonlinear function of x mu dot. And that is never a total derivative for arbitrary. For instance, x dot squared is not a total derivative. x dot y dot is not a total derivative for arbitrary functions of x mu dot. Corresponding to the Lagrangian for a derivative of x mu dot is not a total derivative. So this is not a total derivative. What is not a total derivative? The fact that the Lagrangian action for plus n particles is invariant under... You see, so that's a good question. And the point there is that what was available is changing. You see, x0 dot was also taken to be 11. A transformation that in xi which was proportionate to x0 in this action is treated as linear in the field of action. In the other action would not be treated as linear in the field of action because x0 is time. That gives the non-ohmogenius part and that's the issue there. So that's how this comes through. The difference here is that Lorentz transformations are linear in the field whereas Galilean boosts are not linear in the field. Because the fields are x i's and other Galilean boosts x i goes to x i minus b t. But t is not a field. It's a family derivative. That was the reason. Okay, other questions about this? Okay. So just to remind you what we discussed in the last class. Yes, any Lagrangian that generates equations of motion that are Lorentz invariant should have the following property. Should have the property that the variation of the action under Lorentz transformation is 0. And therefore the variation of the Lagrangian under Lorentz transformation gives rise to Lorentz transformation. Now we were discussing that in the last class we talked about how if we looked at the Lagrangian for a big particle we could pin it down entirely with the demands of Lorentz invariant. Now there was a question saying what if the Lagrangian itself was not Lorentz invariant but only the action? And I gave the argument that all such additional possibilities are true in this particular case. Okay, excellent. So let's recall what we were talking about last class. Last class we approached the standard of general relativity from two different streams. The first one was a slight discomfort with the theory of special relativity. The theory of special relativity or framework of special relativity is this nice framework in which different ocean observers aren't governed. Namely, the laws of physics take the same functional form when expressed in terms of variables and inertia. However, this nice democracy between inertia and observance is broken when you move to non-inertia. So this is like ancient Greece perfect democracy between the citizens. Most of the people there were slaves. So that seems somehow understanding and it leads you to the kind of questions which last a lot in the early days of special relativity. Who decides which observer is inertia? Okay? So this led us to asking the following question. Can we try to write laws of physics in a manner that is completely democratic among observance? Or coordinate systems describing physics give you the same laws in the sense of the same functional form of the equations of motion? Now in order to try to implement this philosophy what we did was we wrote down the Lagrangian for a particle, a free particle moving that moving special relativity. We motivated the Lagrangian there were lots of questions about it but now we all believe that the Lagrangian of x mu dot, x mu dot you termed it the lambda was equal to the x mu by d lambda and lambda was another derivative this action is re-parameterization in that we change lambda to the function of lambda and we get the same action. Okay, this was correct in an inertia field and we said well let's try to implement a program for extension of democracy so we moved to non-inertia field reference. So we went like we moved to new quote that's y mu such that x mu where x mu was some arbitrary x mu of y mu and we found that the action took the form y mu dot, y mu dot g mu nu d lambda by g mu nu was equal to x mu nu d alpha beta eta alpha beta 20 this is correct in any coordinate system but we noted the following thing that this g mu nu which was obtained by this coordinate transformation of the metric eta was not the most general metric model right now by metric we, I mean symmetric tensor as a function of all variables we noted that was not the most general metric between the write down because the most general metric is 10 different functions of four variables and these kind of metrics are obtained from a particular form with four arbitrary functions so clearly it's not the most general metric okay, the grand general may look more demographic but it is a attraction because in the ocean observers that's the special and they are the observers in which to matrix it now I mean I've changed that, you physically how about this way of thinking suggests a common possibility suppose that in the real world this metric g mu nu was really allowed to be an arbitrary that was the case then to think that there would be no coordinate system in which the metric was incoming so there would be no special inertia in the surface all observers would then be on the same footing okay and we would have furthered this principle of democracy because it happened in the principle of relative top of this wild idea we did actually was NASA in Pakistan shame this wild idea which was then then this Lagrangian for an arbitrary g mu nu describes the correct equations of motion in some physical situations some g mu nu could be coordinate transform to eta nu that just gives us free motion in the right coordinates that's boring you know but suppose we looked into a g mu nu that were like eta mu nu plus small fluctuations that could not be transform to eta nu but were eta mu nu plus small and there should be small fluctuations about free motion in the red and red space now we know that small deviations for free motion of the red and red space we describe this by same of article this is subject to a force okay so at least for small fluctuations around the red and red space if this crazy idea that this could actually describe the laws of physics in some situations correct then it must be that the effect of turning on an arbitrary metric field mimics the action of some force then the next thing we asked was what force of nature could it be well on that example we immediately came to the following conclusion we came to the conclusion that since then we then we got to the algebraic fact which I didn't prove in the last class but we will discuss but it is possible even though it's not possible to set an arbitrary metric to eta mu nu much more over the space at any given point it's possible to set g to eta and g dot to zero okay so while having g non-zero and all of space leads to some effect of force this force has to have a very special kind of property that's not the property that this force wherever it is is of the form that allows you to move to a particular action frame in a very small infinitesimal neighborhood of this in which you see no forces because g is eta and g dot is zero and g dot is eta will be important you might think well whenever I've got a particle that's exhibiting a force I can just move along with that particle and say no force, what's the big deal but the point is that this should happen simultaneously for all forces for all particles we have 20 particles in the internet we should be able to move to a frame of reference to reach the force due to this force being then we'll be coming here to the metronome that is just for all of that now that's not true in the electromagnet because if we've got an electron and a positron we've got an electric field the electron fields force this way the positron fields force this way the frame of reference for the electron fields no force you'll see the positron actually so this idea could be described in an electric field so if it describes a force it must describe a force that can be gauged away by moving to the right frame of reference in an international labor for all particles is that such a force of nature though it be gauged there is a force of gravity it's a famous fact that the equations of motion of a particle subject to the gravitational force are independent of its mass and they're called the same for all particles okay so the gravitational force has this basic character that lends itself to a description by this maneuver no other force of nature that we are aware of has this feature so if this maneuver describes some force of nature it must be the gravitational force it's not again far from clear from these works that it does describe the gravitational force but that of course is the purpose of this force yes what's the difference between no force for all the different particles due to the gravitational force in a very small region you're going to inform a little force and you're going to move we're making this experiment at one over a little time over a little additional space we're looking at the system over a little time over a little additional space we look at that system in a particular coordinate system the coordinate system that says g to eta and g dot to z in that coordinate system we know that you see no force but you see no force for everything that's in your region not just for one particle but for all of them that is not possible in the region that is true of that what if you bring in another degree of freedom that's a positively charged neutral yes so this would be a thing positively charged at neutral for instance this would be as big a contradiction neutral guy accelerating with a positive charge just to clarify I'm not saying that this principle disallows the force of electromagnetic of course not I'm only saying that this manoeuvre that is coupling the particle into the metric that itself if rise to an effective force that effective force has this property if you change the Lagrangian in other ways for instance by the way you can add this Lagrangian into an electromagnetic field with potential in you and that cannot be gauged away by going to the right coordinate system I'm saying that this is the only possible force in special range I'm just saying that the force that you obtain by replacing g and eta with g that is an effective force when g is near eta that force has this property and there is a force in nature that has this property it's the force of gravity this property was remarked on by from the beginning of the modern study of physics by Galileo for instance it's a striking property of gravitation and once you have this idea it's natural that perhaps this is the reason behind this gravity property this striking property that all particles are accelerating in the same way gravity or where gravity force of gravity is gauged away it's called the principle of equivalence okay so such a maneuver implements the principle of equivalence perhaps it is the correct maneuver for the study of gravity so this was our motivation for trying to study such theory and I said that we will come back to the study of particle motion in the presence of general electric in general the most general coordinate transformation we consider are Lorentz transformations in general we have to be equal to all of them that's right where by the the size of an observer is the size of a coordinate that's right you know you would say and observe is that right oh way of measuring things yes right when the next thing we did was to say well if we are going to try to make a theory whose fundamental motivation is the idea that we should be able to describe things by same equations of motion in every coordinate system we better learn how to transform between coordinates okay and we did some mathematics which I'll briefly remind you of just continue take tracks on fire wings with the take tracks on fire wings you know we don't have any fermions so right yes so at the moment the metric is not even with dynamic so whether you write the metric as a function or whether you write it as dot product of dead tracks it makes no difference it's only a way in the metric becomes a dynamical field that we need to worry about whether the equations of the dead tracks we don't think the equations of motion is expected during this time so as long as you get some gmu new function it doesn't matter how it doesn't matter whether you get it by or some other way you put in that function you get and you've got a new function at the moment we're not allowed to ask this question it's only when we think we've got a dynamical difference alright yes what do you consider how do you choose coordinates how do you choose coordinates how do you perform a measurement in a particular coordinate how do you perform a measurement in a particular coordinate well take some convention for labeling points of space it doesn't matter what this convention is with this convention you will have equations of motion that will tell you how are you bringing the values to the other convention if we have some other convention there are formulas to transform between your results so if you want to ask questions about where something happens that is by itself not a meaningful question but the relationship between two events and if you think about it that's what physics is about does this particle hit this detector yes this detector somewhere this particle somewhere when they go inside there's an absolutely absolutely every other observer will agree hold back on your question until we start doing calculations you know then we start actually trying answers to any physical question and fall through carefully it doesn't matter about how you set up your coordinates is that am I addressing what you asked if you're asking what I think you're asking it's very good question it's a question everyone who first studies it is very uncomfortable about it till we start addressing specific questions then he will say other questions so we started doing some mathematics of general relativity of quantum transformations so we had a manifold somebody asked me to say what a manifold is manifold as far as we're concerned is any space any such that in order to label where you are you need to specify what numbers so an example of a two-dimensional manifold is this black where I'm in the black world let's say the x-quad is in the y-quad you may do some different grid but however you do it to specify where you are in the black world you have to specify two numbers in which you need to specify where you are in the space this is not sufficient for a mathematician he has said this would be his requirements but we won't get into that except if we need it this is sufficient for our intuitive progress okay so when I use a manifold I just mean switch time and in order to specify where you are you need to specify what numbers that's all you don't need okay so so we have a manifold here and we have some coordinates x and a new set of coordinates x was a function we had we had two friends who were trying to describe the same objects of the manifold so we start in the scalar function phi which was a function of x and the other guy just tried to describe that as a function of y and the transformation rule for this was that phi tilde of y was phi of x then we looked at the scalar okay we looked at del mu of x del mu of y your friend had this there to this scalar field and we found that d by d so let's do that because d by d x mu of y d by v y alpha of phi tilde was equal to del x theta by del y alpha del by del y theta sorry x theta theta of phi of x y and then we get this guy named we call it a mu so the transformation rule became a mu tilde of y was equal to del y del x alpha by del y mu into a alpha of x alpha we had similar rules for upper indices for vectors we talked about vectors being little displacements on our manifold and we described how they were transformed between the rule was that a mu tilde a mu tilde is equal to del y mu by del x alpha a alpha of x alpha in order to get the transformation rules for objects with indices down we needed only to know x as a function in order to get transformation rules for objects with indices up we need to know both x as a function of y and y as a function of x so this rule works even though the map is not inverted this rule requires the map to be improved so projection does here if you write a sub manifold y described coordinates on a sub manifold you can take objects with indices down and pull them down to the y coordinates and objects with indices up now let me talk about tensors with arbitrary index structure an arbitrary tensor had mu 1 mu n mu mu 1 and mu k and again the transformation rule in this tensor it was very simple the rule went index by index this index transformed according to this rule this index transformed the model to the same one so the one factor for every index up one factor for every index down and then the final answer was partial of x we wrote the last class right at the end of the last class we talked about invariant tensors so we talked about delta mu nu this object which is one of the indices 0 on the right we showed that it was an invariant tensor meaning that if you take delta and find delta twiddle according to this rule its value is also it okay there's one more invariant tensor that will be of a extensive use to us so let's talk about it it is square root g that's epsilon mu 1 to mu t okay let's say with four directions so mu 1 to mu 4 so now let me define it so firstly what is epsilon mu 1 to mu 4 epsilon mu 1 to mu 4 is the symbol okay so within the four dimensional space so firstly it's an object with four indices all of which are down okay now by definition we will say that epsilon 0 1 2 3 is equal to 1 and that epsilon is otherwise completely anti-symmetric you make transpositions of indices in order to get to the form 0 1 2 3 the number of transpositions you have to make tell you whether that epsilon symbol is plus 1 or minus 1 okay since it's familiar to you with down to 4 you will see so I'm going to now look at how this object here transforms under coordinate transformations okay so under coordinate transformations so suppose I've got this as my index this is my index in the x coordinate this is my tensor in the x coordinate and now I'm going to ask what is my tensor why do I ask that question well I just use my rules the rules are I take square root g which is the square root of the oh and what's square root g actually what is square root minus g actually the metric you remember we talked about was a matrix as a matrix you can take its determinant okay so hg through this course we will be referring to as g the determinant of the metric tensor will be referred to as g okay the determinant of the inverse metric is 1 because because determinants have the product the property that if determinant of a times b determinant of a times determinant of b but determinant of metric times metric inverse is identity therefore determinant of metric times determinant of metric inverse is 1 therefore determinant of metric inverse is 1 over the derivative of g okay and I look at determinant of square root minus g is as we saw last class, the metric has negative derivative. We don't want to deal with imaginary objects. So I've got this tensor in my x-coordinate frame. And I want to find what it becomes when I apply my rules of coordinate transformations. So what does it become? What's it? Let's just apply it. We get minus square root g, which is the metric in the x-coordinate times del x half power 1 by del, OK. So let's give this whole thing a name. Let's call it a mu1 form. Let me say it's better. So that a fila mu1 different form, what it transforms to in the different way, OK. Is equal to, well, is equal to what? Is equal to del x alpha 1 by del y mu1, del x alpha 2 by del y mu2, del x alpha 3 by del y mu3, del x alpha 4 by del y mu4, OK. Times whatever we originally had, OK. So times what we originally had, which was a, but that was just square root of minus g, times epsilon alpha 1 to alpha 4. Now this is the same metric evaluated at the point x over x. But this so far looks like a pretty unnatural thing. So look at this product. Can anyone recognize this product? It's the determinant of the check of mu. Or more precisely, this object is equal to, by the rules of determinants, so a fila mu1 to mu4 is equal to j determinant of j of x by y. I'll tell you what I mean by that in a moment. Square root minus g epsilon, where I use the rule epsilon alpha 1 alpha 4, m alpha 1 mu1, m alpha 2 mu2, m alpha 3 mu3, m alpha 4 mu4 is equal to epsilon mu1 to mu4 times determinant of the rules from the mathematics of determinants, which you are all familiar with already, refreshment. So where the matrix, the alpha beta elements of the matrix are del x alpha by del y beta. But that matrix, the matrix goes alpha beta elements, del x alpha by del y beta as a name. It's called the, I don't know if this may be the determinant, it's called the Jacobian. Anyway, we call it the Jacobian. Here we call it the Jacobian matrix. And we notice that matrix by j x by y. To say it's the Jacobian with x in the numerator and y in it. So now we've got this rule. This still looks a little weird because this j x by y and because square root minus g in the x square matrix. This product is nice. Why is this product nice? Oops, I should have gotten the, it's in there. Why is the product nice? Let's remember, let's remember what g tilde was in terms of g. G tilde of alpha beta was equal to del x mu1 by del x mu by del y alpha, del x mu by del y beta, g mu mu. And I'm not writing, I'm not immediately writing the arguments inside just because of formulas without being to write. Anyone's confused at any point. Go ahead, you're evaluating things awesome. Or it's the same, the same matrix, del x by del y. Okay, now let's write this in matrix language. It's similar to g as a matrix. Okay, then this becomes the, when g tilde is equal to j, g, j transpose, then determinant of g tilde, using this rule for determinants, products of determinants that we talked about, is equal to net j times net g times net j transpose. Does anyone know what net j transpose is? Net j transpose is net j transpose. So this is net j square. So square root of minus g tilde is equal to net j times square root of minus g. And I take the square root of the positive sign because I assume that my coordinate transformation is, we are assuming that coordinate transformation is one to one, right? You have to assume both. One to one guarantees that it's a definite sign, okay? Well, it doesn't guarantee that it's determinant is positive. Okay, yeah, so strictly we've got exactly, so this object is, well, it should be over here. Exactly, this object is invariant only under coordinate transformations that respect this object, otherwise it picks up a minus sign. This is the fact that it's not a parity. Okay, so let's restrict for two minutes to coordinate transformations for which net j is positive. Okay, there's a discreet of the two positive signs. If we assume that it's positive, then this times square root of minus g is square root of minus g tilde. And therefore, a tilde is equal to square root of minus g tilde times epsilon. And therefore, the formula for the, for a tilde is the same formula in terms of tilde of variables at the formula for a. This was our definition of invariant tensor. So, square root of g times epsilon is an invariant tensor. What is the point of having invariant tensor? The point of having invariant tensor is that it's very easy to make coordinate invariant objects using invariant tensor. See, if you take an arbitrary four tensor and multiply it by an arbitrary product of four vectors, contract indices, that object is not coordinate invariant. Because when you transform the four vectors, that induces a transformation on this tensor and this tensor in general rotates. But if the tensor has the property, that is, the rotated tensor is the same as the original tensor just by moving to the new coordinate system. That's all expression is invariant under Lorentz transportation. So, invariant tensor are important objects that help us to build Lorentz invariants or in knowledge, coordinate invariant, general coordinate invariant objects. And when we use an extensor, we do this course. So, these two invariant tensor that we encounter, they'll be delta v mu and epsilon v of 1, v of 2, v of 3, v of 4. We'll be very useful for us through this course. Any questions or comments, please. The last thing I wanted to say about this invariant stuff or the gradient slope, sorry, the last thing I wanted to say was about an integration question. We will often need to take something and integrate it over all the space there. So, let's put it in D4x. I want to see what this becomes in terms of D4y. Well, we all know it becomes x. So, this D4x is equal to D4y by the Jacobian of x, y, y. This is the same determinant. I'm not sure whether the Jacobian prefers to determine the dimensions. We will talk about these dimensions. Let's make up. So, this integration measure D4x is not coordinate invariant. This is obvious. D4x is a measure that encloses the volume with unit volume to unit length for a unit cube in x. And something that has a unit cube measured in x, clearly does not have unit size of measured in five dimensions. So, it's obvious that this is not an invariant integration measure. Now, since we don't need to integrate things over all the space time, there's an immediate question. Can we find any other coordinate invariant integration measure? And the answer is obvious. It's different from the same algorithm. The answer is that integral g D4x is a coordinate, but minus 1. Integral minus g D4x is a coordinate invariant. Why is that? Because we just see it around the coordinate transformations. This goes to minus g tilde times, well, OK. Let's look the other way around since that's what you see. Let's check what integral minus g D4y is. This guy we just seen goes to that j times minus square root g. So it goes to square root of minus g at that j of x by y. And this thing transforms to that j of y by x. But this is the determinant matrix. This is the determinant of x inverse. So the product is to determine this one. Let's get this one. Once again, there's some modules Once again, we're assuming that the size, here it won't matter. Because it depends on whether you take D4h to put a module. If you do that, if you say that, good, you're right, you're right. If you think of just an unsigned volume, such as D4y, then you put a module. So the first point was that this is this with this. And this formula really comes out with these modules. We choose a branch of the square root. So then this will just cancel. I agree. Here it won't matter. We have a particular rule for transformation of square root of the unit. You have to choose the sign by taking the square root. You choose the right sign so that it's always positive then it is better. Good, thank you. So this is an invariant integration measure. And it will be used to generate coordinate invariant integral. This is all I want to say about basic manipulations with coordinate changes. We're not going to talk about coordinate invariant differentiation. But any questions or comments? If I'm going too fast, you should then. We have the advantage of having a beautiful book on the formula. Of course, as usual, the books by Landau Lyftschitz are the best books on the subject. This might be 50 years old. It's just quite incredible. A beautiful book. Everything's crystal clear. And we'll go ahead and read it. Ask me to slow down. Ask me questions. OK, great. So now let's move on. So we've discussed some basic manipulations of coordinate transformations. You won't understand how to take derivatives in a coordinate invariant. So what is the issue? Suppose I have, for instance, this object of one form. And I'll take one index now. And I want to study its derivative. So I want to study del by del x alpha of u. The question I'm asking is, how does this object transform under coordinate transfer? Does it transform in some way? So what do we want to do? What we want to do is to compute del by del y alpha, let's say, of a tilde mu. Now that's related to del by del x alpha of u. But this is not something for which we need to independently specify a transformation. Since we already know how it is going to transform. So this is equal to del by del y alpha of dx of x star by dy mu of a of theta. This object, there are two terms. This derivative could act as this object. Or it could act as this. Let's first look at the term that you get by taking the del to the left. So that gives us del x theta by del y mu into del by del y alpha of a theta. But then I use the change. So del by del y alpha of x theta, del by del x theta of a. Now this term is the same object, this object in the old x coordinate system. These two terms are exactly what you would expect to transform the same object were it attached over two lower ends. So when you take the derivative and act it on a theta, you get exactly the transformation. That part of it gives you the transformation properties that you would expect were this object to be attached. So I take del by del alpha of a theta, del by del y alpha of a theta. But I use the change rule to convert the del by del y alpha into del by del y alpha of x beta and del by del x beta. What is it going to do? Coming, coming. So we get the second term, which is del to the x theta by del y alpha del y mu. Transformation rule tells you that this object is not a dense because it does not transform or convert. It's a rule of dense. If we leave it as a transformation rule, it's not homogenous. Derivatives of y do not depend on y derivative to the A mu, they are not in x coordinate system. They do not depend linearly on y derivative to the A mu in the y coordinate system. There's also a piece that depends just on A rather than derivatives. So this object wasn't a dense. This sounds like, this is sounding pretty bad. It's sounding pretty bad because, you know, all the ground regions in physics are full of derivatives. Maxwell's equations, any reasonable derivative chip will have derivatives. And if derivatives transforms so badly have a coordinate transformation, it will seem like a miracle if we can make coordinate very naturally. Yes? Can it be said of the same thing? Or two derivative of greater or less? Yes, it's the same. It's the same manipulation. Because we've seen how one derivative transforms. It's the same. I get it. Okay, good. So it looks like, you know, either a theory based on coordinate transforms or coordinate invariance involving derivatives is either going to be very messy. Or we need a better way to think of these things. Of course, there are better ways to think of these things. And that's what I'm going to explain. I'm going to first start by doing a bit of a hack along. I'm going to slowly understand what the elements are in this hack. What I'm going to say is that, look, I have, like, someone trying to invent, wait, first thing we ask, physically what's a hack? Why was it that we differentiated something that looked nice and got something that looked messy? And the point is that any tensor, like this ideal, has a transformation law that depends on del x by del y at the point at which you were interested in looking at the object. But derivative doesn't involve looking at something at one point. A derivative is measuring something at one point, subtracting from a thing, the same thing at another point. This first guy transforms, he transforms to the revolution of tensor at this point. This second guy transforms according to the rules of tensor at the second point. So they're different transforms nicely. It doesn't transform nicely under either rule. It's an unnatural thing from the point of view of tensor and algebra to subtract vectors at different points. But unfortunately, that's what derivative is. It involves subtraction of vectors of one point and do different points. That's the reason we're lining up at this test. Now, you know, kind of thing. We encounter just that study of coordinate systems in two dimensions. So first we've got two dimensions. We've got x and y. We've got a vector field in these two dimensions. A x as function of x and y. A y as function of x and y. Differentiating the components of A x and A y is straightforward. And this gives us a notion of the derivative of vector field in the x-y coordinate system. We get the x component. We differentiate the component by component multiplied by the unit vector field. Okay, however, you suppose you use a less well suited coordinate in a less good coordinate system. Suppose you use the model. So you wrote A was equal to r hat times A r plus theta hat times A theta. You try to differentiate this vector field and you just differentiate these components. A r and A theta. You get the wrong answer for differentiating the vector field. Because you also have to differentiate these unit vectors. We were lucky in the first case that we were using constitutive. That's basically another way of saying what the problem is. If we're working, we're just differentiating the component in a different way, differentiating the coefficient, but not the unit vectors. So let's think a little bit more, you know. Suppose we got everything very easy when we're in flat space. So we have some notion of what constant vectors are. Then we can always move to that coordinate system, do the differentiation, there's no confusion. But we're in general relatively number, there's no flat space. All coordinate systems are equal. Okay, now what do we do? So we need to have a following notion. First we've got some vector field here and some vector field here. What we really want to do, we have some a little vector here and a little vector here, parts of the vector here. What we want to do is to move this object to this point so that its transport is parallel. You want to move this vector parallel to this point. Then do the subtraction. That would be the correct answer for the vector field. There's no sort of parallelity, of course. In flat space it's very clear. Parallel means having the same x component, same y component. What we're going to try to do now is to try to achieve the same thing. That is, try to achieve the process of parallel transport of vectors along infinitesimal lines in an arbitrary method. First by introducing new objects and then by putting physical principles on these new objects to the term. So let's, you know, in this case, does anyone know what this parallel transport would be? Suppose I've got a vector which has some a, r and n theta, and I wanted to move it in the r direction. What would parallel transport amount do? Something in the r of theta direction? The removal of the r direction remains the same r, same theta direction. But if you move it in the theta direction, what will the muscle be? Subtracting an infinitesimal rotation. A rotation is a, is, is linear in the original vector. An infinitesimal rotation is linear in the original vector. So this process of parallel transport in simple examples, okay, is the following. If you want to parallel transport some vector, we'll name it. Okay, I'm going to do it first with one form. Okay, with one form we'll name you. From x to x plus delta x. Now let me do it right, from x to x plus delta x. Then what happens? Then I'm going to assume that a mu parallel transport is equal to a mu between infinitesimals. So really that was, okay. That's a small direction. But the small direction is itself linearly, just like it was in all the rotation. Okay, plus gamma mu alpha theta n, that's delta. This is the parallel transport. This is the result. I'm going to assume that my space time is embedded with some object. It's a three index object. I call it gamma. It's called a crystallism. Okay. So that, and this object tells us what's the correct notion of parallel transporters. How to move a vector from x to x plus delta x. And this gamma is a function of x. From some way, someone has given us a notion of what the correct rules for parallel transporting vectors are. Now, we won't rely on someone giving us a notion for very long. We'll determine this question. But that will take 20 minutes. So for now, just somehow there's some notion of parallel transporters. Okay. And the rule is that under parallel transport, a mu from x to x plus delta x, a mu goes to x. This is my rule. And sometimes, right? This has delta a mu is equal to, change, gamma mu alpha delta, delta abstract method, you know. This is possible to try to see what I can say about the things. The way to ask is, suppose somebody has given me this gamma three index symbol. How to do vectors that is objects with any index objects transform a vector. Transform. That's very easy to see. Consider mu can take, this thing is a scalar. How does it transform a parallel transport? It doesn't. It's meant. So the transformation of these objects on a parallel transport should be zero. But delta of this is equal to delta b mu a mu plus b mu delta a mu. So the rule for a infinitesimal. So it's equal to delta b mu which I don't know. That's a mu plus b mu delta a mu which I don't know. Gamma mu, gamma mu alpha theta a theta delta x mu. That's right delta x. This object has to vanish for any two vectors x and b. So x and b and b. Now, there are a lot of dummy indices here. Everyone is comfortable with this convention of repeated indices being summed. So there are lots of dummy indices here. This dummy index, let me call it, what I want to do is to change the dummy index on the a sum here to be the same one as here. So I'll change this mu to a fine. So I'll just write the answer. So zero is equal to delta b mu mu plus b. And let me call this phi, gamma, phi, alpha. And let me call this mu delta x alpha. I'll just change the few dummy indices. But b. So what this tells you is that zero is equal to delta b mu plus b phi, gamma, phi, alpha, mu, delta x alpha. Enjoying it? But then this object here is some vector adopted with an arbitrary a mu of zero, which must be in the distance. Which tells us that delta b is equal to minus of b phi, gamma, phi, alpha, mu. But in next transforms, by this rule, under parallel transport, then another index transforms by this rule. Yes, give it a rest. So what is the rule here in between the changes of this form? Well, firstly it's infinitesimal. So that's a mu plus something. It's got to be proportionate to delta x. The assumption, maybe, is that it's also proportional to 8. Why is that? That's because it's true for every... So you write down an arbitrary coordinate system in flat space and do parallel transport. We're always talking about some effective location. So that's what we're doing locally. What we've got is some coordinate system that is not flat. And we want to move the vectors parallel. So that amounts to moving the vector plus a rotation as a change of coordinate system. And a change of coordinate system transforms ASD. So that's the only assumption. Because if you're not working anything about what gamma is, there are no further effects. It should depend on three indices, but it has to. So linear, this linear, this has to be okay. So just writing down the most general thing. You see this. What? Gamma is a function of x. It's also convenient as a function of y. Determine what gamma is as we go along. At the moment we don't know what it is. We're leading up to the process of doing it. The next thing we're going to say, the next thing that we're interested in is to try to work out the transport of an arbitrary tensor field. Okay? So suppose we look at a tensor field that was being built to a disease gun. Now one example of that tensor field to a disease gun is a product of two one-forms with indices down. Moreover, any tensor field to a disease gun can be written as a sum over two vector, two one-forms with indices down. So if we determine how two one-forms with indices down transform, we determine how to index object. But that's what we do here. Because, so suppose we've got one mu and we do mu, mu. That's what it is that is equal to. Well, that's how we want mu to be two mu plus. We want mu that is down to two mu just because it's infinitesimal. We have like a formula, which tells us it's equal to gamma of mu alpha theta p1 theta plus gamma of mu plus p1 mu gamma of mu theta alpha p2 theta. That's the whole thing. So that tells you that for an arbitrary two index object delta of p mu mu is equal to operation on first index layer and second one unchanged because operation on second index layer is first one. So gamma mu alpha theta p theta mu plus gamma mu alpha theta index by index. Do it for the first index layer, the second one unchanged do it for the second index again for the first one. Similarly, when we've got a tensor with many lower indices, many upper indices. This gives us a rule for parallel transport about the train tensor. Now, some of you are thinking what is all this parallel transport nonsense? I mean, this parallel transport is quite abstract concept. I'm interested in formulas. This is an attitude I sympathize with. All this blah blah blah research in formulas. What is the formula? What interesting formula? You know, not for some abstract notion of parallel transport, but for something interesting. Does this lead to a formula? There's a class here for a class to be in constraint. Okay. A seventh. Okay, so that's going to be a good idea. Okay. So the thing that we're really interested in, of course, is not all this parallel transport. What we're really interested in is derivatives. How do we use derivatives to define how to use parallel transport to help us define good derivatives? There will be derivatives that will transform like this. So now let me try once again to do what we failed to do last time. I want to define the good version of d by dx alpha of a nu. Now what should I really be doing? This object here was equal to a nu of x plus, let me multiply this by delta. Okay. x plus delta x minus a nu of x divided by, okay, just a minute. So suppose I really was interested in d by dx. Okay. So suppose I was interested in this, then this x, you know, just moving just the delta alpha x only in the alpha direction and zero in the other directions and divided by delta x. I'm sorry for the comparison notation, but I hope it's okay. I'm just defining an element. Okay. That's what this object was and it didn't transform well. What is the object that should transform well? What should transform well is a and x plus dx parallel transformed to x. It's a fact. Then it's a fact. Okay. Now the parallel transform from x to x plus dx involved a plus sum. And therefore from x plus dx to x involved a minus sum. Okay. So of course this term, which was this derivative will be there. But in addition we will have minus sum. Okay. So this is a fact. But what we said is that this is a good object. We expect to be the following. Minus gamma zero alpha theta. This object now is the difference between a and x plus dx parallel transformed to x minus a of x divided by delta x. If we've got a good notion of parallel transform, now this should give us a good notion of an element. Because now this term I can do it. There's other signs based on points. Each of them transform to x. So there is which transform. Okay. So this notion of parallel transport is a fancy set of words. It actually tells you that what we're interested in is defining such a derivative. Notice the parallel transformation, a transport. We have a good notion of differentiation. And by good we mean this object must transform. Like a tensor. As long as that blah blah blah that works, let's start going on with the formula. So what we want is that this object transforms like a tensor. Let's process what that means. The minus sign is because even the derivative, okay, what we want is so there are two options you have. In defining this differentiation, what you must either do let's take one of them first. Let's take the vector field at x plus, plus dx. Transport it to x. And then subtract from that the vector field at x. Now the parallel transport from x plus dx to x has a minus sign. Because the parallel transport from x to x plus dx has a plus sign. Now there's a second way of saying it. Suppose you want to say, look, we have devolved to square two points. From one to the other. I don't like that. I'll transport from here, yes. Free to do it. Let's take that. Then what we want to do is to take the field at x, parallel transport to x plus dx. That parallel transport involves plus sign. But then the result that you get has to be subtracted. The vector field at x plus dx. So that will give you the minus sign. Either way it will give you the minus sign. It is to take this object and deduce from it a rule of coordinate transformation from the symbol. So we've got this parallel transport symbol. Its role in life is to create tensors from differentiating tensors. So this object must create a tensor from differentiating tensor. Let's see what can make that work. So what do we have to do? Let's compute this in the Dillback coordinate system. Let's compute it in the original coordinate system. Demand that the transformation rule is better in life. That will give us a transformation rule for gamma. So let's go. For that part of the calculation before we do it again. So we have first let's compute this in the Dillback coordinate system. So that's d by dy alpha a to the mu of minus let's call that gamma to the theta mu alpha a to the t. Now I apply the rule for transformational behavior. Let's d by d y alpha of del x phi by del y mu e phi to the right of both terms minus gamma to the theta mu alpha a okay that's right. This time that's right. So that's del x phi by del y theta we will simplify it a little more as we did a little time ago by looking at the two terms which is del 2x phi del y alpha del y mu and then there's a second term. This of course is going to be the problem. But let's look at the border terms first. That's plus plus del x phi by del y mu del x theta by del y alpha del y del x theta of a phi minus gamma theta which we still don't know of theta mu alpha del x phi del y theta. We want this whole thing to be this times this times the covariant derivative. The covariant derivative to the x. This whole business by the way this combined derivative is called covariant derivative. We want this that if the answer was this factor that this factor times covariant derivative to a phi we would declare it to be in the absence of this term that could work exactly if gamma transforms like a tensor. Suppose gamma till that transforms like a tensor with this index structure then we would get these two factors from these two terms and this factor would cancel this factor. Now in the absence of this term gamma would just transform as a tensor. So suppose now what we are going to do is to try to cancel this. Oh. So suppose we take the rule that gamma till that theta mu alpha is what it would have been where it would have been a tensor namely del y theta by del x theta prime del mu prime by del y mu del x alpha prime by del x alpha gamma theta prime mu prime alpha prime and then whatever is needed to cancel this which is plus x pi d2 have only net d2x theta by d y mu d y alpha times something that will cancel this to make it you know del this is zeta del y zeta del y theta theta by del x now this transformation rule we till this term make this transform like a tensor and give us a covariant derivative transform okay so we have denied that this fancy parallel transform object whatever it is as to a variable structure okay that's it people sorry we have to run outside the class