General Relativity | Lecture 3
Susskind, Leonard
(October 8, 2012) Leonard Susskind continues his discussion of Riemannian geometry and uses it as a foundation for general relativity. This series is the fourth installment of a six-quarter series that explore the foundations of modern physics. In this quarter, Susskind focuses on Einstein's General Theory of Relativity.
 Stanford University. General relativity has a reputation for being very difficult. I think the reason is that it's very difficult. It's kind of calculationally intensive. Indices and symbols. There are ways people have invented to express things in sort of more condensed, very, very neat notations. But just learning them in itself is a task that would take some time. Things like theobines and forms, spinners and twisters, and all sorts of different kinds of mathematical objects. You could call many of them notation, notational if you like. And they do simplify the equations. So I sometimes feel as if in presenting these things the way I do, that it's sort of like Maxwell who wrote down every single equation of his, how many Maxwell equations are there? I think eight. Eight, eight all together. We don't usually write all components of the equations. We put them together into vector notation and so forth. And if we're smart, we can even avoid the indices by inventing symbols like curl and so forth. And the same thing can be done to some extent for general relativity. But as I said, the computational methods and techniques are harder. And I'm not an expert in them. I'm by no means an expert in them. I'm not an expert in general relativity. I know how to use it. But quite honestly, when I go to do something using general relativity, I go back to the indices. I can follow when somebody else uses the fancy methods. But I can't do them myself. It's not just that I can't do them myself. It's not natural to me. I wind up going back to the same thing that we're doing here. Indices, derivatives, the whole works. And my feeling is it's the right way to see it for the first time. But it is a nuisance. It's sort of indices all over the place, notations all over the place, too much of it. Can't help it. So some aspect of it, you have to get over. You have to get over that hump of tensor notation, the indices and so forth. And then it goes easier. Then it goes more smoothly until you come to do a calculation. And then right back to the mess again. Now, if you're really interested in doing computations in general relativity, there are packages. There are, you know, you just put in the metric as a function of position, the metric tensor. And out it will spit various tensors and Riemann tensors, this kind of tensor, that kind of tensor, Einstein tensors. And then you can say, without even looking at the result, you can say, okay, please, Mr. Computer, set the Riemann tensors, set the Einstein tensor equal to the energy momentum tensor and tell me what comes out. So, yeah, computers can do a lot better. But as I said, general relativity is hard. And part of the hardness is just the complexity of the notations that are necessary. Incidentally, the subject that we're studying up till now is Riemannian geometry. Riemannian geometry, the geometry of curved spaces, the geometry of smooth, curved metric spaces, they're called. And I suppose we have Riemann to blame for the complexity of the notation. Actually, he simplified it and Einstein simplified it further. But nevertheless, it is notationally and computationally intensive. The basic concepts are really not so hard, but we have to go through the mess before we get to the concepts. Or before we can express those concepts. If there are no questions, yeah. I know very well, yeah. Well, yeah, there are several. The most active areas of research, the ones which are really very, very strongly active among theoretical physicists, is the quantization of general relativity, the quantization of black hole physics. That's probably the most active area. There are astrophysical areas where, again, black holes are very important. Astrophysical areas are entirely classical general relativity. There's numerical relativity. And usually for the purposes of solving astrophysical problems, numerically putting in various initial situations, a pair of black holes orbiting around each other, and simply running the computer to find out what happens by solving Einstein's equations. And it's a very subtle business, incidentally. It is not that you just put in the equations and grind. It's much more subtle than that. And numerical relativity is also a very hard subject. The purpose, for example, being to find out how much gravitational radiation and other kinds of things is entirely classical. Gravitational radiation is classical radiation when two black holes fall into each other, bang. A lot of gravitational radiation is produced, gravitational waves. Why are we interested in them? Well, among other reasons, because we have a chance of detecting them, anything you can detect is a window into phenomena that you've done a new phenomena. So gravitational wave detectors may be able to detect gravitational waves from the collision of two black holes. It is estimated that there are a few collisions every year of black holes that can be seen by gravitational radiation. So there's a lot of different areas. It's an extremely active area. The equations of general relativity, just the equations of general relativity, and the equations of black holes have been found to be useful, extremely useful, and condensed matter physics and fluid dynamics for having nothing to do with black holes and gravity. Same equations. And so there's a heavy amount of activity in the areas of applying the equations of general relativity to other areas of physics. It's probably, I would say, probably the most active. It went over the last, over my career in physics from being a real backwater in physics. Nobody studied general relativity very much in the 1960s when I was a student, except for the people who were really especially interested and were focused on questions of general relativity. The bulk of theoretical physicists were not interested. They just started to become interesting in astrophysics, and eventually it took over questions of fundamental theoretical physics. Why? Because of a clash of gravity in general relativity. Anyway, for whatever reason, it is probably now more important than the day-to-day workings of theoretical physicists than almost any other subject, that and quantum field theory, and the connection between them. Okay, let's, today will be the last day in which we'll be studying Riemannian geometry as such without really discussing gravity. Next time we will really get into gravity. What does all of this have to do with gravity? I told you a little bit in the first lecture that the problem of finding out if there's a real gravitational field as opposed to just some funny coordinates in which you're expressing space and time, which leads to curved trajectories, but not really curved trajectories. The problem of finding out there really is a gravitational field, is a problem which is mathematically, I would say very similar, but I think I would go further than that. I would say is identical to the problem of finding out if a certain metric geometry, geometry characterized by a metric tensor, is flat or not. A geometry which is flat, let's say a two-dimensional geometry to begin with, that means a surface of some sort. A flat geometry is one for which Euclid's postulates are correct. There are parallel lines, all the stuff you learned in Euclidean geometry is called flat. And it's characterized in terms of a metric tensor by the statement not that the metric has a very special simple form. That's not the right statement, the special form being the chronic or delta. That's not the right statement that the metric has that form. The right statement is that coordinates can be found such that with those coordinates, the metric has a simple flat form. But more generally, just given an arbitrary metric tensor which varies from place to place, assume that it's smooth, assume that it has all good differential properties and so forth, the problem of determining if that space is flat or not is a somewhat difficult problem. What you have to do is look at, here's the bad way to do the problem. The bad way to do the problem is to search through all possible coordinate systems, transform the metric from the given form, however it's given, the way that it's presented to you, transform the metric to the new coordinate system that you have constructed, calculate the new metric and ask whether it looks like a flat metric. Now, of course, that will take an infinite amount of time, you'll never get there, so you need a better technique. And the better technique is to search for diagnostic quantities, quantities which are built out of the metric, and its derivatives and so forth, things that you can calculate which, if they are zero everywhere throughout the space, that says the space is flat, if they're nonzero any place, it says that the space is curved there. The diagnostic quantity that does that is called the curvature tensor. Today's goal is to get to the curvature tensor of a Riemannian geometry. And it's a complicated goal, I warn you, you've got to pay attention or else you'll fall asleep. This is a very superific subject. I'll try from time to time to tell a joke or whatever to keep you awake. But on the other hand, if you can follow it, you've learned a lot and it is very interesting. I think you'll find it very interesting. I'll try not to make it too boring. Okay. I told my friend Barack Obama, never apologize, and here I am apologizing already for being boring. Okay. What do we start with? We start with a space, and a space means, first of all, a number of dimensions. The number of dimensions in Riemannian geometry can be anything. In principle, you can even have a zero dimensional space, but that's pretty damn boring. It's just a point. A one dimensional space is either an infinite line or a closed curve. If it's a closed curve, what is it characterized by? It's characterized by only one thing, the total length of the curve. Every curve is equivalent to every other curve of the same length. In other words, just think about it for a moment. Take a piece of rope that closes on itself, it has a certain length, and wiggle it in any kinds of ways, or curve it in any kind of ways. It can always be mapped or put on top of another piece of rope of exactly the same length in a precise way without stretching it. One dimensional spaces are all equivalent to each other, not quite. They're characterized only by their length. They're either infinite, or they have a length L, and every one of them have given length L is identical to every other one in its intrinsic properties. A bug walking along a little piece of rope cannot distinguish one geometry from the other geometry. All he can do is count the number of steps that it takes him to go around the rope. He leaves a marker someplace, goes around the rope until he comes back to the marker, counts the number of steps, and that's the only thing about that space. Two dimensional spaces begin to become more complicated. There are flat ones. There are curved ones. Flat ones are plain. Curved ones could be a sphere. It could be just a space with lumps, the surface of the earth, including the mountains and the valleys and so forth. It can even have weird topology, the surface of a donut, a torus. Or you can poke another hole in the torus and make a torus with two holes and so forth. So by the time you get to two dimensional spaces, they're pretty complicated. Nevertheless, they are characterized by the metric tensor. And the question of whether they're flat is a computation, as a first question, find something which distinguishes whether they're flat. Find a quantity g mu nu, g mu nu g m n, the metric tensor. If it is not equal to zero, that does not mean the space is not flat. It just means you're maybe in the wrong coordinates. So you can't diagnose the space by asking whether g m n is zero or not. You want to find a quantity. Let us call it r. At the moment, I'm not going to tell you how many indices it has. Some quantity. But as best as should be a tensor, because if it's a tensor and it's nonzero, it's nonzero in every frame of reference. We like to find some quantity called r. r stands for Riemann incidentally. Riemann was the first person who asked this question and answered it. We want to find a thing called Riemann such that if Riemann is equal to zero, that implies that the space is flat and that there exist coordinates such that, sorry, did I say g m n not equal to zero? I mean not equal to delta m n. g m n is an interesting space. It's not very big. Yeah, g m n equal delta m n is what I meant. And that's what we're going to search for. The curvature tensor. It is a tensor. Let me just remind you about a few properties of the metric tensor and what you do with it. First of all, g m n is by definition the thing that's a function of position in general. It was not a function of position incidentally. If it was a constant matrix, a constant set of coefficients, then it's very easy to find coordinates in which g m n is delta m n. You simply have to straighten out the angles. So if g m n is not dependent on position, if it's just a set of coefficients, a set of numerical coefficients, then the space is definitely flat. So g m n of x is the metric tensor and what do you do with it? You take a little differential distance, d x m, multiply it by d x n. Here's your little differential displacement. It's characterized by a set of differential coordinate shifts, d x m, d x n. You multiply them together and that gives you the square of the length between the tail and the arrow and the head of the vector. That's equal to d s squared, s standing for spatial distance, I suppose. All right, now, the metric tensor is a tensor. I think we proved that last time and the way you see that is just by saying d s squared, that's an invariant quantity. All coordinate, let's call them coordinate systems, all coordinate systems should agree about the length of a given little vector and if we take that same d s squared and write it in y coordinates instead of x coordinates, the length has to be the same and that's sufficient to prove that g m n really is a tensor and that it transforms like a tensor. Okay, next property of g m n is it has an inverse. g m n is a matrix. It's a matrix and it has an inverse. The inverse satisfies the following equation. g, it's written again as a matrix. When I say it has an inverse, I mean it has an inverse as a matrix. g m n is a matrix with entries and it has an inverse. g m n, which is written with upper indices, they are in fact, contravariant indices and the properties of g m n with upper indices are such that if you multiply it by g, let's call it n r and sum over n, but I don't have to sum over n because Einstein tells me that an upper and a lower index like this are automatically to be summed over. This is the sum. This actually corresponds to taking the matrix g m n and multiplying it by the matrix g n r and the resulting matrix is the unit matrix. Why? Because I told you that g m n is the inverse matrix to this one here and the product then is the identity matrix and the identity matrix is the Kronecker delta and Kronecker delta in general relativity or in Riemannian geometry is always written as a thing with one upper index and one lower index. At least if it's being treated as a tensor, the Kronecker delta, this is a thing to prove, prove that if you take the Kronecker delta and you transform it pretending it's a tensor, you just get back the Kronecker delta again. That's the thing that you can prove. So these are tensor equations. This is the meaning of saying the g m n is the inverse matrix of g n r. And once you have that, oh, there's another thing that is important. In Riemannian geometry, as opposed to Einsteinian geometry, in Riemannian geometry all lengths are positive. All lengths are positive, squares of all lengths. More important, squares of all lengths are positive and all lengths are also positive. But the square of all lengths are positive and that says something about the properties of the metric tensor. What it says is that there are no directions along which if you evaluate this you get anything which is not positive. That's equivalent to a statement about the eigenvalues of g m n. It says all of the eigenvalues of it are positive. But for our purposes now, all lengths are positive. It's the same thing. And it does say something about the metric tensor. It says it's positive definite. That's the same thing as saying all of its eigenvalues are positive. That will change when we go to the theory of real gravity. That will go when we go to Einsteinian geometry. All right, the metric allows you to do something. It allows you to raise and lower indices. If you have a contravariant vector, or better yet, you have a vector which has contravariant components v m, contravariant means upstairs. And I put it into correspondence with another vector with covariant indices. And it's really the same vector. We talked about this a little last time. I told you the difference between contravariant and covariant indices. And the way you construct the covariant components, components of the words I'm looking for, the covariant components of a vector giving its contravariant components is just to multiply it by the metric tensor g m r. And, whoops, let's call it g m n, and sum over m. This is now g sub n, or v sub n. Likewise, this is something that you need to prove. Given a v m with upper indices, this, if you like, can be thought of as the definition of the lower of the covariant component of a vector. How do you go back? Well, you have to solve this equation, but it's very easy to solve, and I'll tell you what the solution is if you know the inverse matrix with upper indices. Namely, v m is equal to g m n v n. So if you know the upper indices, if you know the covariant, the contravariant components of a vector, this is the way you build its covariant components, and if you know the covariant components, this is the way you build the contravariant components. There's a little bit of a proof here that you have to do. So, if you have a question, you can ask me. You can see what we have to do. You'll have to check that these two are compatible with each other, but they aren't. Okay. To what extent can you force the metric to look like a flat metric? To what extent? Assume the metric is not necessarily a flat metric. I will mean not only is the space flat, but that the coordinates are such that g m n is equal to delta m n. If that's the case, if that's the case, this space is definitely flat. But to what extent, given any space, can you find coordinates such that g m n might be, for example, the karmic of delta over some limited region, or approximately over some limited region equal to delta m n? All right, so here's a theorem. There are coordinates at every point. These coordinates are coordinates which are good in the immediate vicinity of a point. They're not global coordinates. They don't cover the whole space. They're just good within the immediate vicinity of a point, which means they're good up to some order in little displacements away from this point, namely second order. All right. Coordinates called Gaussian normal coordinates. Here's what you do. Just to first think of it intuitively, we have some curved surface. We go to some specific point, and we draw coordinate axes. We draw coordinate axes, and we keep those coordinate axes as straight as we possibly can. Later we'll learn that the term for straight as possibly can curve is a geodesic, but you make a straight curve as you can. You have a little bug. He moves on this surface, and he moves following his nose going as straight ahead as he possibly can, and that defines one coordinate axis, then he comes back to a point and goes off in right angles, does some surveying to figure out where right angles to his little line are, and goes off in another direction as straight as he can, and then builds coordinates based on those... The theorem says that first of all, at every point, at every point you can choose coordinates, as I said, Gaussian normal coordinates, such that, first of all, at that point, gmn is equal to delta mn at x equals x0. This point is x0. You can do that in more than one way. If you find coordinates in which it's true, you can rotate the coordinates. You can rotate the coordinates, and if you rotate them, that just corresponds to rotating the coordinates in different axes and so forth. So there's more than one way you can do it. Next, you can choose your coordinates. You can straighten out these lines to a degree where the derivative of the metric with respect to any direction... Well, let's put an xr here. Any direction, the derivative of any component of the metric with respect to any direction in space can be said equal to zero. That you do by choosing these curves to be as straight as possible. I'll tell you how you prove it in a minute. It's actually very easy to prove. It's just a counting problem of counting how many variables there are and how many equations there are, but I'll show you in a minute. Okay, finally... Yeah? Say it again? No, only at x equals zero. At... Sorry, not at x equals zero. Could be x equals zero. At x equals x zero. If you try to extend that same coordinates away from that point, no matter how you try to extend them, then in general, unless the space is flat, if you try to extend the coordinates away from that point, you'll find out that the derivatives, the higher derivatives of gmn are not equal to zero, and in general, for the general generic situation, the second derivatives, let's say with respect to some XR and some other XS, those in general cannot all be chosen equal to zero, except, of course, if the space is flat. Okay? So, at a point... At a point, there is no content, really, in saying that the metric can be chosen to be flat-like, let's call this flat-like, even up to first derivatives that can always be done. It's in these second derivatives that somehow the fact that the space is either flat or not flat starts to show up. All right, first of all, how do you prove this? Well, this is actually not hard. Let's take our point of interest to be the origin. Let's just take it to be the origin. Let's set it equal to zero. Now, supposing we have some general metric and some coordinates Y in which the metric has some form, not of this form here. Let's look for some XS, which we're going to build, which are going to be functions of the Y. Let's choose it in the following way. At the place where X is equal to zero, in other words, at our origin over here, let's also assume that Y is equal to zero. So, the two sets of coordinates have the same origin. That means that X will start out... We're just equal to Y. XM equal to YM. And then, in the next order, we'll have some set of coefficients. I'm simply expanding X in powers of Y. The next order will be something quadratic in Y. So, let's put something quadratic in Y here. We'll call that CM. And let's put here N and R and put XNXR. That's the most general quadratic thing that I can write. Oops. Sorry, Ys. YN, YR. And then, things may get more complicated later. Okay. How many such... Let's pick a number of dimensions of space. Let's pick a number of dimensions of space. Let's pick four dimensions of space. How many such coefficients are there? Well, how many combinations do we have in Y in YR? Are there? The answer is ten. Try counting them. There's Y1 squared, Y2 squared, Y3 squared, Y4 squared, Y1, Y2, Y1, Y3, Y1, Y4, Y2, Y3, Y2, Y4, and Y3, Y4. Did I miss any? I think I have every possible thing here, which is quadratic in Y. And there's one, two, three, four, five, six, seven, eight, ten, ten of them. So there are ten possibilities here. How many components of the metric are there? There are ten YN, YRs. So that means N and R give you ten possibilities. But then for each N and R, you have an M. That means there are 40 independent coefficients here. 40 independent coefficients. The most general X that you can build to quadratic order has 40 independent coefficients here. Now, how many independent components of G are there? The answer is ten. So how many equations does this correspond to? Ten components. And setting all of their derivatives equal to zero means that this is 40 equations. 40 equations, 40 unknowns. And that allows you to be sure that you can solve these equations to quadratic order. Solving them to quadratic order is saying the same thing that you can solve these equations, but you'll fail at the next order. You'll fail at the next order, and here's the next order. So it is always possible to choose a metric that has the property that to quadratic order, in other words, the first derivative of the metric is equal to zero and the metric is equal to a flat metric at a point. Excuse me, I'm making something crucial here. It's very beginning. We have that equation with X and normal left and Y is on the right. Does that represent an attested coordinate transformation? Yeah, that's a coordinate transformation. Sure. Yeah, that's a coordinate transformation. Yeah, this is a coordinate transformation which we're going to try to choose these C's in such a way as to make the derivatives of the metric zero. You can always do it. So every metric in that sense is flat over a sufficiently small distance. Is that the same as it's locally flat? That's saying it's locally flat. That's the statement that it's local. That's what it being locally flat means, yeah. Intuitively, it just says that any smooth surface can be approximated over a small distance by a plane that's tangent to it. Right, that's what that says. Mm-hmm. That can be solved. That can be put, Gmn is invertible. Yeah. You can't always solve for the equation necessarily. Say it again? You can't always solve for the equation before the unknowns, right? Yeah, I think in this case you can. I think in this case you can. Yeah. Yeah. Yeah, in this case you can. There's, try it out. Do it with two by two matrices. Do it in two dimensional case. And there's nothing special about four dimensions. You'll find you can do it. All right, I don't want to try to do it on the blackboard. It's, all right, in any case it's a theorem. It's a theorem that Gaussian normal coordinates exist at every point, at every point, and they have this property. All right, now the reason we introduce that is because we want to give some meaning to derivatives of vectors. We've talked about some, of tensors in general, derivatives of tensors. We've talked about various procedures, various operations we can do on tensors to make new tensors. We can add them, we can subtract them, we can multiply them. We can do all kinds of things. Can we differentiate them? In particular, if we have a tensor which is a function of position, that means all of its components are functions of position, can we construct another tensor which is in some appropriate sense the derivative with respect to position of the tensor that we have? So let me show you why this is a little bit tricky. You could just say, well, okay, take the components of the tensor, let's say covariant components, and differentiate them, and that gives you a new set of components, which are the derivatives of the components. That gives you the derivative of the vector. It's true you can do that, but what you get will not be a tensor. Okay? And let me show you why. Yeah. The equation you have is C. There are no first order terms. This is first order. You mean zero-th order? Yeah. Yeah, that's right. It's not that there are no zero-th order terms, it's just I said let's choose the origins of the two coordinate systems to be at the same point. So that says wherever y is equal to zero, x is equal to zero. It was arbitrary, but it doesn't cost anything. Okay. All right, let's look at a vector field, a vector field, a vector at every position, and let's start with coordinates which either the space is flat, in which case I'll just use flat ordinary Cartesian coordinates, or the space is curved, in which case these coordinates here represent Gaussian normal coordinates, as straight as possible, so to speak. And now let's take another coordinate system and superimpose it on top of that. These are not Gaussian normal coordinates. They may be the coordinates that we started with. I'll represent those in green. And they're curved relative to these coordinates. For example, the coordinate lines might look like that, one of the coordinate lines, the other one might look like this, or whatever. All right, now, before we ask how we differentiate a vector field, vector field means a vector in every position here, let's ask what it means, what it means for the vector field to be constant in space. The problem here is because the space is curved, it becomes hard to compare the vector at one point and another point. The coordinates, if they can't be chosen to be everywhere flat, then what exactly do you mean by saying a vector over here is equal to a vector over here? It doesn't really mean anything, because to compare a vector over here with a vector over here, unless you have nice flat coordinates, there's no unique way to do that. So let's look at it for a moment. If the space is flat, really flat, then I think I know what it means for a vector over here to be the same or to be equal to a vector over here. It means it's equal, it points in the same direction and has the same length. On the other hand, does it say that I'm thinking about the green coordinates now? Let's think about the green coordinates. In the black coordinates, the components of this black vector and this black vector are clearly the same in the nice Cartesian coordinates. But what about the coordinates of this vector here and this vector over here along the green axes? Let's take a special case. Let's take this vector to be vertically up so that we know that it only has a Y component in the flat coordinates. Let's assume that this vector over here is equal to this vector it points in the same direction in flat space. Are its coordinates along the green axes, are they the same as the coordinates along the black axes? No. This coordinate along the green axis here is different than the coordinate along the black axis here. And in fact, this vector would have both X and Y components along the green axes, whatever the green axes happen to be, has both kinds of components. So it's clear that even though the vectors are the same, the components are not the same. So you cannot judge when you have curvy linear coordinates and in general, you will have curvy linear coordinates. You can't judge by whether the components, either covariant or contravariant, you can't judge whether the vector is constant in space. Another way to say the same thing is that in one coordinate system, the derivatives, let's say with respect to some coordinate XR of the mth component of a vector, might be zero in one coordinate system and not zero in another coordinate system. It may even be the case that all of the derivatives of Vm are zero in one coordinate system and not zero in the other coordinate system because the coordinates shift, because the coordinate direction, not because the vector is changing, but because the coordinates are changing. So this statement that the derivatives of a vector are equal to zero is not an equation which is likely to be the same in every reference frame. Therefore, it cannot be a tensor equation. It's not a tensor equation. And the derivative of a vector, the components of a vector with respect to coordinates, are not themselves components of a tensor. They're not components of a tensor because they can be zero all of them in one frame of reference and not in another frame of reference. If they were components of a tensor, we would think of this as a rank 2 tensor, an m index and an r index. We might say, look, could this be a tensor Tmr? Well, if it's a tensor Tmr, then if it's zero in one frame or in one coordinate system, it's zero in every coordinate system. But that's simply not true of the derivatives of components of vectors, not because the vector may change from point to point, but because the coordinates, the representation of the coordinates change. So we need a better definition, we need a better definition of the derivative of a vector than just differentiating its components. We need something which, if it's zero in one frame, is zero in every frame. Okay, so here's what we'll do. Here's how we'll define the derivative of a vector. First of all, to define the derivative at a point, you only need to look at nearby points, right? You only need to look at nearby points. So the first thing you'll do is to construct Gaussian normal coordinates near that point. Gaussian normal coordinates, they're as straight as possible. Then, let's put this vector a little bit closer just to indicate that it's all like that. Then, pretending that the Gaussian normal coordinates are really nice flat coordinates, we shift this vector so that its origin is the same as the starting vector over here. In other words, shift it, keeping its components fixed and constant in Gaussian normal coordinates. So in the Gaussian normal coordinates, I keep the components of this vector fixed, and I compare it with this vector over here. Just shift it down to here, and the difference between these two is the thing that I will use to define the derivative. The derivative will be the difference between these two divided by the little separation along this axis here. In other words, to differentiate a vector, differentiate it in Gaussian normal coordinates. They are the ones which are closest of all to Cartesian coordinates. The derivatives in the Gaussian normal coordinates define the derivative. Now, when you differentiate a vector in general coordinates, you'll get two terms. The two terms come, one term comes because the vector may be changing, and the other term comes because the coordinates are changing. Coordinates are rotating out from under you, even if the vector doesn't change, and so forth. If you work out this prescription, take the derivative, just the ordinary derivative in Gaussian normal coordinates, and then re-express it in arbitrary coordinates. Here's what you will find. You will find that the derivative, which will now be called capital D along the R axis, of the components of a vector Vm in arbitrary coordinates are not just the derivative with respect to R of Vm, but there's another term, and the other term has to do with the way the coordinates themselves are changing from point to point. What it looks like is something that's proportional to V. If you double the size of V, you'll double this term here, obviously. So it's proportional to V. It will have a V in it. It does not have a derivative because this term doesn't come from the fact that the vector is changing. It's coming from the fact that the coordinates are changing. Let's put some indices and try to see what kind of thing it could be. It might have the T component here. On the left-hand side, we have indices R and M. That means we have to have some kind of object over here that has index R and M, just to match the indices on the left-hand side. On the other hand, it's going to be proportional to the components themselves of V, so it's going to have some VT here. This equation doesn't quite make sense. There are two directions here. There are three here. You've got to soak them up. I have to put another index into this object here. In fact, the standard convention is to put a minus sign there. This, in fact, is what you will get if you take a vector differentiated in Gaussian normal coordinates and then transform it to other coordinates. In any other coordinate system, what you will get is the derivative minus a thing, what's this called a thing for the moment, times components of V themselves. This is summed over T, and the index structure on the left-hand side is just R and M. That holds for an arbitrary coordinate system, right? That holds now for an arbitrary coordinate system, but I haven't told you what gammas are, what these gamma objects are. Question. What does that formula become if it's a Gaussian normal? No, in Gaussian normal coordinates, it's just this. Okay, so the gammas come in because of this? Yeah, the gammas, yeah. Right, we're going to find that the gammas are proportional to derivatives of the metric. They have to do with the metric changing from place to place. They're going to be proportional to the derivatives of the metric in Gaussian normal coordinates to derivatives of the metric of zero. Okay, that's the trick. Use Gaussian normal coordinates to differentiate because to first order the derivatives of the metric of zero. Once you've differentiated it, you can then transform the object as if it were a tensor to any frame of reference, and the form will always be like this. It will always be like this. It will contain indeed a derivative, but it will also contain another term which will not be differentiated, and it has to do with the, as I said, it has to do with the fact that not the vector is changing from point to point, but the fact that the coordinates are changing from point to point. These objects have various names. This is called a connection. It connects neighboring points and tells you how to differentiate something from one point to another. It tells you basically how to compare vectors at different points. It tells you how to compare vectors even though the coordinate system may be changing. This is called a covariant derivative, not because its index is downstairs. It's called a covariant derivative for, just because it's called a covariant derivative. It's called a covariant derivative, and it is a tensor. It's a tensor in that if it is zero in any frame, it's zero in every frame. Is that a generalization of the convective derivative? You know, I think it is. I never thought about it that way, but I think it is. What do you mean? Whether it's connected with a convective derivative, no doubt it is. No doubt it is. I would have to sit down and think about it, but there certainly are covariant derivatives in fluid flow. Right, okay, so, yeah, this is called a covariant derivative, and these symbols here go under two names usually. They're either called the connection coefficients or the Christoffel symbols. Christoffel. I, Christ. They've also been known on occasion as Christoffel symbols, and they are, they're not that complicated, but they're complicated enough. They're unlikeable enough. Okay, now, one fact about them that follows, I'm not going to prove every single thing I say, because they're just too many little pieces, but they're easy to check. It follows from the definition, namely the definition is you differentiate in Gaussian normal coordinates, and then you treat the object as a tensor and transform it as a tensor with two indices. It follows as a theorem, it's an easy theorem, but it's a little too intricate to do on the blackboard, that this Christoffel symbol has a symmetry. The symmetry of the Christoffel symbol is the gamma t, r, m, is equal to gamma t, m, r. It doesn't matter which order you put m and r, they're the same. That follows, there are generalized Riemannian geometries. They're called geometries with torsion. The symmetry is not true, but those geometries are not widely in use in ordinary gravitational theory. Ordinary gravitational theory is Riemannian geometry, it follows from the principles that I laid out, and gamma r, m is equal to gamma m, r. You know, So it's just the gamma that the Christoffel symbol, not gamma. The gamma, the gamma, the gamma. You know, doing the derivative in these Gaussian normal coordinates, which are almost flat, or as flat as can be, and then treating it as an object on its own right, it is very, very similar to what we do in gravitational theory when we evaluate something in a freely falling frame. And then, for example, in a freely falling frame, we calculated how light moved across an elevator. And then we transformed it to the frame of reference in which the elevator was accelerating. Okay. That's closely related to the operations we're doing here. We calculate something because we know how to do it in coordinates which are as flat as possible. That would be a freely falling frame in general relativity. And then we transform it to any coordinates we like, accelerated coordinates, or anything we like. And we translate the statement from one coordinate system to another. Here, the notion of variation of a vector from point to point is done first in Gaussian normal coordinates, and then it's transformed. And this is the form that you'll get for the corresponding tensor. We're not going very fast tonight, so I want to move along. All right. Now, for higher derivative tensors, we will need this tonight. Sorry, for higher rank tensors, supposing we have a tensor with more than one index, m, let's say, n, and we want to differentiate it, co-variantly differentiate it, again, in this case, along the s-axis. I don't know why I chose this instead of r. Here, I used r, but I want to differentiate it along the s-axis. Does everybody know what I mean d sub r here? That means derivative with respect to xr. Derivative with respect to xr. d sub r. That's a Gaussian normal. No, no, no, that's in any coordinate system. d sub r will always mean d by dxr. No, but I mean, in this equation, this is a Gaussian. Which question? The one here, you mean error routing. If you want to even talk about the rest of the equation. This one here? Yeah. No, that's in any coordinate system. In Gaussian normal coordinates, this term isn't there. In Gaussian normal coordinates, you just differentiate. Gaussian normal coordinates, everything is easy and ordinary. The derivative of a vector is just the derivative. In arbitrary coordinates, there's another term. You can work that out by transforming, just by doing transformations. Yeah? Good thing, when you say that in Gaussian normal coordinates, the polar derivative is just the derivative. That's only at the point. Yes, yes, yes, yes. Only at that point. Right. So, the question, are the Christoffel's coordinates, are those, is that a 2, 1 tensor or whatever? No, it is not a tensor. Not a tensor. This is intended to be a tensor. We're going to try to choose gamma so that this is a tensor. This we know is not a tensor, and therefore this cannot be a tensor. Right, no. Gammas are not tensors. In fact, they're built up out of derivatives of the metric. In a coordinate system in which the derivatives of the metric are zero, the Christoffel symbols are zero, but a tensor, if it's zero and one coordinate system, is zero in every coordinate system. So, they can't be tensors. And they're not. Okay. Let's just, before we go and figure out what the gammas are, we're going to work them out. They're not that hard to work out. There's a trick. Let's write this down. It's the analog of this, except for every index in the tensor, there is a term like this. There is a term for every index, and I'll show you how it works. There is a Christoffel term for every index. So, for example, we start out by hiding in. Let's hide in and pretend that this was just a vector. Then we would write that this is the derivative of t with respect to xs, that's the first term here, and then we would put a term gamma s m. Let's see what I used here. s m t t m. But now I expose the n here, and I put it in on both sides. Now I'm not finished. I have to do exactly the same thing, covering up m. Cover up m. I won't try to do it, but I have another term here where m is passive, n is active, and so we get gamma s s n t t m t. Look at the form of it for a minute. Oh, minus signs. Minus signs are wrong. Same as here. Minus signs as here. Notice if I covered up the second index, t m n. If I covered up every place the second index, it would look exactly like the covariant derivative of a vector. If I covered up the first index, it would look exactly like the covariant derivative of a vector. Say it again. You're right. This should be m t. Thank you. Yeah. Paralleling this over here. All right. Notice in one of the terms, it occurs in the first place, and in the second term it occurs in the second place. So that's the form that the covariant derivative of a tensor would take. The rule is the same. You go back to Gaussian normal coordinates, and you just differentiate the tensor, just derivative, none of this junk. And then you transform it to arbitrary coordinates, assuming that the result here is a tensor, and this is what you get. So it allows you to differentiate any tensor. At the moment, we're just dealing with tensors with covariant indices. We'll come to tensors with contravariant indices at some point. All right. In particular, let's apply this to the metric tensor itself. But there's something special about the metric tensor. In Gaussian normal coordinates, its derivatives are all zero. Yeah. I'm sure you must have said this one. It seems like a really fundamental thing. What's the covariant derivative for? It's for differentiating things. It's for comparing things at different points. What are we going to use it for? Field equations. Field equations. Field equations are going to be differential equations, which represent how fields change from one place to another. But we want them to be the same equations in every reference frame. We don't want to write down equations which are special to some peculiar reference frame. We want to write down equations which are general, so that if they're true in one frame, they'll be true in all frames. That means they have to be tensor equations, and that means we have to know how to differentiate tensors to get other tensors. So normally you have a derivative if you multiply it by corresponding to the decimal, you get the change. This is the partial derivative. This is the partial derivative with respect to the coordinate xs. I'm trying to throw me for the covariant derivative, but is there some differential you would multiply times the covariant? Yes, yes, absolutely, absolutely. Then what is that? dx. If you wanted to know how a thing changes from one point to a neighboring point, you would just multiply this by dxs. But is that what the partial derivative is? No, no, no, no, no. I say it again. The point is that the components of the vector may change even if the vector doesn't. Or that, right, we have to keep that in mind. So this is, right, we'll see some examples about how it works. This is representing effectively a distortion of the, an out of flat distortion of the coordinates. Yeah, that's right. It's an out of flat distortion of the coordinates. It's there even in flat space if you choose funny coordinates. That's an important point. It's there even in flat space if you choose funny coordinates. In fact, if you choose any coordinates in which the derivatives of the GMNs are not zero in which the coordinates vary from point to point in funny ways, this will be there. This is not really a feature necessarily of curved spaces. It's a feature of curved coordinates. Now, okay. So because the metric tensor is constant, or at least has zero derivative in Gaussian normal coordinates, and the definition of covariant derivative is to differentiate first and Gaussian normal coordinates and then take what you get and treat it as a tensor, it must mean that the covariant derivative of the metric tensor is zero. That's what's going to allow us to compute the Christoffel symbols. All right, so let's write down that equation. Here is the form for the different derivative, the covariant derivative of any tensor with two lower indices. So let's apply it to the metric tensor itself. dS of GMN is equal, first term, dGMN by dxS minus gamma TSMGTN minus gamma TSNGMT. All I've done is rewritten this equation, and now I'm going to add that the covariant derivative of the metric tensor is equal to zero. Why? Because the ordinary derivative is zero in Gaussian coordinates, in Gaussian normal coordinates. All right, now, let's write the same equation except permuting the indices, interchanging S and M and so forth. This is the equation for differentiating with respect to S of GMN by inspection, why am I doing this? I'm doing this because I want to get as much information out here about things like this as I can, and then try to see if there's any way that I can manipulate the equations to get to isolate and figure out what gamma is. So it's, you know, people who are experienced at this sort of thing look at it and say, I know what to do. If you don't know what to do, I'll tell you what to do. You do it once, and then when your friend comes and asks you, how do you find the Christoffel symbols, you'll say, I know what to do. I'm experienced. Okay, so you do the same thing except now I'm going to interchange S and M. Just permute S and M, and that means we're going to write down dM of GSN. How do I do it? I just do it by interchanging or replacing S by M and M by S, and that gives me derivative of GSN with respect to XM minus, now interchanging S and M here doesn't change anything because S and M, yeah, from the symmetry here, so this is the same thing, gamma SMT, GT, let's see. Yeah, we're just interchanging S and M, GTN, and then over here, so we're interchanging S and M, so this is gamma MNT GST, and of course for exactly the same reason, this is also equal to zero. And finally, there's one last possibility. Here, S is the odd man out, M and N go together. Here, M is the odd man out, S and N go together. And the last one is dN of GSM, and let's see, I think I'll just copy it down. It equals d by dxN of GSM minus gamma SMT, GTN, GTM minus gamma MNT GST, and that's equal to zero. Now you stare at this for a few minutes. Now how can I either air these, subtract these, or do something clever that will isolate only one of these terms with the gammas so that I don't have a horrible bunch of equations but basically only one equation? So I'll tell you what to do. Call this equation three, call this equation two, and call this equation one, and add equation three to two, and subtract one. And we'll see what happens. We'll see what happens. First of all, of course, we will get these derivatives. We'll get this derivative plus this derivative minus this derivative. So let's put them down on one side of the equation. I'm going to transpose all of this junk to the other side. OK, so on the left side of the equation we will get dN, dN GSM plus dN GSM, no, dN MG, the GSN M, that's this one, and the last one is minus GMN, sorry, derivative GMN. That's this plus this minus this, or this plus this minus this. Now, how about on the other side? Let's see, there may be some things which cancel. Let's hope this thing which cancel. This comes with a minus sign, this comes with a plus sign, this comes with a plus sign. So this and this are the same, but they cancel because of the opposite sign. No, listen, this are gone. They are gone, what else is gone? GTSN, ah, here it is, this one and this one come with opposite signs. So they are gone. We're left with two terms, but they're the same. So we're in luck. We can write that this is equal to twice, we need over to the right hand side, twice gamma MN T GST. Well, we still have, we're still not there, we would like to get gamma MN by itself. Trying to find out what the gamma, what the Christoffel symbol is all about, that's our goal, figure out what they are in terms of derivatives of the metric. We're almost there, first of all we can see that if all the derivatives of the metric are zero, the Christoffel symbol must be zero. But how are we going to get rid of this GST? And the answer is GST has an inverse. Basically, you can just take this over to the other side, but replace it by the inverse matrix. So first of all, let's put a one half. And now, put a bracket around this and take this GST over to the other side, except in its inverse form. And this is the correct formula. This is the formula for the Christoffel symbol. It looks pretty simple. We'll look at it for a moment. The M and N are symmetric. Look at it, M, N, if you want to change these two, and here it's symmetric again, has one negative term. Oops, yeah, one negative term, two positive terms, not very complicated. The problem is there's a boatload of them. When you think about four dimensional space and you let S and T and, sorry, and let all of these coefficients range over from one to four, there's just a lot of Christoffel symbols. And it's that which makes doing calculations in general relativity a very tedious business. Intrinsically, there's nothing hard about it. But in fact, doing a calculation in some sort of general relativity context usually fills page after page after page of nothing more complicated than just computing these derivatives and putting them together. Yeah. I'm sorry, I just want to say it again. G is constant, has zero derivatives in Gaussian normal coordinates. At that point. Oh, yeah, at that point. All of this is just at that one point. Yeah, but once it's true at one point, it's true at every point. To derive it at any point, you go to Gaussian normal coordinates at that point. So you say, I want a formula appropriate to a particular point. I go to that point, use Gaussian normal coordinates, and find out that this is what gamma is. But once you know it at any point, it's true at every point. So this is the connection between gamma and the derivatives of G at every point. The thing is that this is not a tensor. It's not a tensor, and so it can be zero in one frame of reference and not zero in others. For example, in Gaussian normal coordinates at the point, at that point, at that point, this is equal to zero, but in some other coordinate system it's not. Even more, take ordinary flat space. In flat space, you can find coordinates in which the G's are constant everywhere, in flat space. But you can also find other coordinates in which they're not constant. Polar coordinates, the G's are not constant. In Cartesian coordinates, they are constant. The Christoffel symbols can be nonzero even if the space is flat. In fact, they generally are. They are in polar coordinates. Yeah? Just to clarify, you can represent the metric that G has a 2 by 2 matrix in this... Not a 2 by 2, an n by n, or 4 by 4, or whatever it is. Whatever it is, but here you're on the... Well, whatever it is, it's 4 by 4, n by n. What is the Christoffel symbol, so it will also come right in? No, it's n by n by n. It's got three indices. n times n, now it's n by n by n, but there's some symmetry. So there's 40 components to it. It has 40 components. Everything always has 40 components. M and n, if you fix t, fix t, then gamma mn has 10 components. Same 10 components for a symmetric tensor. And then you let t run over all four components, and that gives you 40. So there are 40 independent Christoffel symbols. Excuse me. Yeah. I just want to make sure I understand what you said. I agree that you proved that at that one particular point, and it's true in all points, but for each point that you do it, you have to use a different G. Do you use the G that's appropriate for the Gaussian normal coordinates? No, no, no, no, no, no, no. You have some coordinates which are general coordinates all over the place. And I want to know how to differentiate at that point. So at that point, I go to Gaussian normal coordinates. I do my differentiating, and then I relate it back to the coordinates that are general coordinates which cover the whole space. At each point, at each point, this is what you do with this formula. No. You take your curved coordinates, whatever they are, and then you say, I want to know how to compare vectors at neighboring points. So I go to Gaussian normal coordinates at that point, and it tells me how to compare them at that point. Here's what it tells me to do. And those G's are the G's associated with Gaussian coordinates at that point? No. They're the G's that are associated with any coordinates. With Gaussian normal coordinates, these are zero. They're all zero at that point. The derivatives are all zero at that point. So no, these are the coordinates that you began with. You transformed the Gaussian coordinates, derived all of this, but then you finished with it. You finished with the Gaussian normal coordinates. Now you're finished with the Gaussian normal coordinates, and you go on. This was just a trick to, the Gaussian normal coordinates were just a trick to eliminate locally at each point the variation of the metric pencil with position. Right, good. Is this equation, the equation that tells you that a space is flat? It tells you what? That your space is flat? No. No, no, no, no, no. This does not tell you that the space is flat. This is just, which just tells you what gamma is. If gamma is, I assure you, if gamma is zero everywhere, it not only means the space is flat, but it also means you've chosen Cartesian coordinates. Didn't we get here by setting the derivative, the covariant derivative of the metric tensor equal to zero? And so that by itself is not enough to tell you the flight length. No, the covariant derivative of the metric tensor is always zero. That's not the same as saying the metric is constant over space, the metric tensor. The statement that the metric tensor itself is constant would be that the ordinary derivative is zero. Right, but the covariant derivative of the metric is always zero. Why? Because in Gaussian coordinates it's zero. And if a tensor is zero in one coordinate frame, it's zero in every coordinate frame. And one more thing, that covariant derivatives are tensors. So this is conceptually tricky. But at the end of the day, the rule is calculate your Christoffel symbols and replace ordinary derivative in many contexts, replace ordinary derivatives with covariant derivatives. You can write your equations in Gaussian normal coordinates, then they just involve ordinary derivatives. But if you want the same equations in general coordinates, then you replace the ordinary derivatives by covariant derivatives. That's the rule. Think about it. Look, this requires sitting down and thinking about it and understanding what we're doing. We have not got the curvature yet, so let's try to get to curvature. We have the definition of... Now, let's talk about what curvature is. It's easiest to start with two-dimensional curvature. How do we probe for curvature? Well, let's draw... Let's begin by drawing a space which is curved. This sphere is obviously curved, but I don't want to deal with a sphere. I want to deal with a cone, but not a cone, but a cone with a round point at the top. The top of a mountain, but with the sides being nice and flat. So that if you're away from the top of the volcano up here, I don't know, it's not a volcano, it's just a rounded cone, on the sides here it looks cone-like. At the top it's smoothed off, it's been shaved off and made round. The first statement is such a cone is flat, everywhere is except up near the top. To see that, just take the same space down here, but continue it so that it really does form a genuine cone. Okay, so now we have a cone. It only differs from the space that we're talking about up in this vicinity up here. Now, take the cone and slice it. Slice it down here and open it up. This is a cone, slice it down here and open it up. It's clearly flat. It's clearly a flat space, but it's a kind of question. Okay, open it up and what does it look like? It looks like, depending on the angle of the cone up here, it looks like a flat space, but when you cut it and open it up, there was a piece cut out of it. Missing, a missing piece. I don't have a scissor, I should have brought a scissor. I didn't bring a scissor. Take a piece of paper, cut it, cut out a wedge of a given angle, and then glue it together and you make a cone. The cone has a tip up here. The tip is clearly not flat, but the rest of the cone is flat. You cannot flatten this like a coffee filter. You can't flatten it down on the table without tearing it, but if you cut it anywhere, if you cut it and open it up, you can lay it out and make it flat. Scissor? Hmm? Scissor? Scissor. Oh, a scissor. A scissor, oh. Okay, but now I have to cut my notes. If you have a piece of paper, somebody have a piece of paper, you can do it. Alright, I've cut out a certain angle, and now when I sew them together or glue them together, in this case it made a kind of flat cone. I could wear it as a hat. Okay? If I cut out a bigger angle, I'll make a more acute angle at the top of the cone. For example, I can cut out a large angle. I cut out a larger angle, and now when I sew it together, it makes a pointy cone, except I can't really do it because, yeah, I think you get the idea though, huh? Yeah, it makes a pointier cone. So the bigger the angle that you think you, the bigger the angle that you cut out, the bigger the deficit angle, it's called the deficit angle or the conical deficit, the bigger the conical deficit, the pointy of the cone. If the conical deficit is a very small acute angle, it means the cone is very flat on top. Okay? All right, now let's think about, let's think about a vector, a vector field on the space here which is constant. Let's take a vector field which is constant. What does it mean to be constant? It means that it's pointing in the same direction everywhere, and for simplicity, let's take it to be pointing in this direction. Something funny over here, over here it's along this line, over here it's practically perpendicular to this line, but remember when the cut is identified like this, the line is the same line. Let's do it. Let's take a vector pointing over here, here, here, here, and here, and now when we sew it together, we discover that there's a funny jump in its direction. It's pointing along the cut here, and it's pointing in some funny direction here. Even though when you open it up, you see they're all in the same, did you see that or was I hiding it there? As you go around, starting here, pointing in the same direction until you get around to the other side of the cut, in which case you find that it's not in the same direction. As the arrow was over here. Nevertheless, when I open it up, they're all parallel. So the tip of the cone is such that if you carry a vector around, when you get around to the other side, it has been rotated. It has been rotated due to the conical deficit at the top. Exactly the same will be true. In this geometry over here, if I take a vector along a given direction and I carry it around the mountain in such a way that it's always pointing in the same direction if I opened it up, by the time I get back to the other side, it'll be pointing in some other direction. That is because the top of the cone is curved. That's the effect of curvature. The effect of curvature, a region of curvature has the property that if you take a vector pointing in the same direction as much as you can do, keep it pointing in the same direction and take it around the region of curvature, the vector will rotate. The vector will rotate when it goes around any region of curvature, even though you took pains to make sure that at every point you were moving it parallel to itself. In other words, you took pains to make sure that the covariant derivative of that vector was zero, but nevertheless when you came around to the other side, it had shifted. There's another way to say this, which is actually more useful. Here's the other way to say it. It's equivalent. It's equivalent. If you have a curved space in some curvature here, in some curvature here, and you take a vector field and differentiate it along one axis, we're going to move it and differentiate it, differentiate it along one axis, and then differentiate it along a second axis. We take the vector, displace it a little bit, and then displace it along some second axis. We will... the vector will change. And how will the vector change? It will change typically by differentiating the vector along these two axes in sequence. You first differentiate the vector along one axis, then differentiate it along the second axis, and there's a small change in the vector due to these two derivatives. So if I go along here, in other words, if I take the difference of the vector over here and over here, it consists of two changes, a change to go from here to here and a change to go from here to here. And that change, the total change in that vector is proportional to a second derivative. It would be true in any coordinates. It would be true in any coordinates that if you carry the vector from here to here and then from here to here and then compare it with the vector at this point over here, what you would be calculating is the second partial derivative of the vector, let's say derivative of Vn, call this the Xs direction and the Xr direction. Then this would be, let's see, which derivative would it be? It would be derivative with respect, it would be covariant derivative with respect to r of the covariant derivative with respect to s of the vector, and let's say the mth component of the vector. This operation says you move the vector a little bit along the s direction, and this one then says you take the difference, the little derivative, and move it along the r direction, and the net result is the change in the vector going from here to here. Done covariantly, which means that you could have done the same thing in Gaussian normal coordinates with just ordinary derivatives. In Gaussian normal coordinates, you would just be doing this derivative here, or two ordinary derivatives. You could have also gone in the other direction. You could, and that is to say, you could have gone first in the r direction and then in the s direction and calculated the way the vector changes from here to here. That would be the opposite order of differentiation. Ordinarily, and in flat space in general, these two are equal to each other. This is just the fact that derivatives, it doesn't matter which order you take them, but it's not true in curved space. In curved space, this net, the difference between these two operations can be thought of as taking the vector around the closed loop. The difference between these two operations, taking the vector from here to here and then from here to here, minus the opposite direction, is essentially taking the vector around the little closed loop. Taking it around the little closed loop, we discovered that when there's curvature at the tip of the cone here, you don't get back to what you started with. In curved space, covariant derivatives are not interchangeable. In flat space, they are interchangeable. That's the way we test for whether a space is flat or not. We test whether differentiating tensors, in particular vectors, in opposite order, give zero. If the answer is yes, everywhere is in the space. For any vector, then the space is flat. If we discover that there are places in the space where the order of differentiation gives different answers, then we know the space has some kind of defect in it, or it has curvature. It has curvature analogous to the top of the mountain. So all we have to do, and it's a little bit complicated, but it's not that complicated, is compute the second derivative of a vector, the second covariant derivative of a vector, in opposite order and compare them. I'll tell you what to do. You're going to do it. I'm not going to do it on the blackboard. It's too late. It's easy. It's really quite easy. It's mechanical. It's a mechanical operation. I'll tell you what the steps are, and I'll tell you the answer. And it's entirely mechanical. It's just pure plug-in. It's too late to carry it out. What we're interested in is taking a vector, the Vn, the components, covariant components of the vector, computing the covariant derivatives in the r direction, and then differentiating it in the s direction. Later on, we will interchange s and r and subtract. But let's just do this. This is equal to ds. This object over here is a tensor, trn. It's a tensor, and we have the rule for calculating the derivative of a tensor. We have a contribution from each index, and I'm going to write it down. I'm going to write down the answer. I'll write down the, let's see. ds of ds of dRVn minus gamma Rnt Vt. I'm just stuck in here the covariant derivative of Vn. So that's what's in this box here, just this thing over here. Now we have to differentiate it again using this formula. We have a two-indexed object, and let's see if I can, let's see. I'm going to leave it to you to calculate. Let's see. All right, I'll write down one further equation. It's ds of dRVn minus gamma Pn. Oh, brother. No, it's getting too late. I don't want to try to write it down. I'm going to leave it for you to do. Just pure plug-in. Pure plug-in. You plug-in for Tmn, this object over here. Put it right into there, or calculate the second derivative, and then compute the following. ds dR of Vn minus dR ds Vn. That corresponds to the difference between the two ways that you could get from here to here by going through this direction or that direction. I'll tell you what you get. You get something called R. It's a tensor with three covariant indices, one contravariant index, and it multiplies Vt, where this is pretty easy, except that R is a little bit complicated, and I'll tell you what R is. So, ds dR sRnT equals dR gamma snT minus ds gamma rnT gamma snP gamma prT minus, one more term, gamma rnP gamma pst. Two terms involving derivatives of the Christoffel symbols and two terms involving products of Christoffel symbols. There's a sum over p here and a sum over p here. The indices, take a look at it, the indices match. It's a complicated expression, and it's even more complicated when you remember what the gammas are. The gammas themselves are complicated derivatives of g's and the curvature tensor. This is the curvature tensor, the Riemann curvature tensor. This Riemann curvature tensor is itself another derivative and also products of Christoffel symbols. Let's see what it has in it. When you differentiate, the Christoffel symbol involves a derivative of g. Differentiating again gives you second derivatives of g. Remember, the second derivatives of g are the things that you cannot generally set equal to zero. The first derivatives of g, you can find frame of reference where they're equal to zero. The second derivatives of g, you can't. So by the time you finish calculating this thing here, second derivatives of g have come into it, and it is testing out and feeling out the geometry a little more thoroughly than the first derivatives would feel it out. It's feeling out the second derivatives, feeling out the coordinate systems to quadratic order. So it has second derivatives of Christoffel symbols, and it has squares or quadratic things involving first derivatives. It's a complicated thing, and if I were to actually write it in terms of the Christoffel symbols, or I were to try to calculate it for a given metric, it can fill up pages, and it will fill up pages. But conceptually, what it's doing is just calculating the difference in a vector if you transport it around here, keeping it parallel to itself until it comes around all the ways around. The little change in the vector in going in a loop. This is the curvature tensor. The curvature tensor, as I said, it's complicated to calculate, but you can calculate it. You can put it into a computer, and you can ask the computer, is it zero? In fact, if the computer, it's good if you have a computer that can do algebra, and if you have the metric in some algebraic form, or you have the metric in some mathematical form, you can do these operations and then test out whether this curvature tensor is zero everywhere. If the curvature tensor is zero everywhere, all components of it, the space is flat. Now, we'll study the curvature tensor a little more, but as I said, it's a complicated thing, but the main thing about it is that it tells you whether the space is flat, and if not, how unflatted it is. It is closely related to a quantity in gravitational physics. Can you guess what it is? It's a local quantity which tells you that the space is not flat. So it must be something telling you whether there's really a gravitational field present or not. Tidal forces? Tidal forces. Tidal forces. It's exactly related to tidal forces. Those things which in a gravitational field squeeze one way, stretch the other way, that's what these things are, and in fact, they really are the tidal forces. Tidal forces are represented by the curvature tensor. Now, you can sort of see something like that if you were to imagine, yeah, imagine you have, again, this kind of conical geometry. It doesn't have to be conical geometry, but something which is flat far away. Yeah, let's say it's flat far away. Nice and flat far away. And somewhere in the center it has a bulge so that it's curved. Now you take some object out here. It's a thing living in flat land. It's two-dimensional. You take some object that might be composed of a bunch of links, there might be a bunch of triangles connected together, links, and then you start moving it. As you move it in the flat region, nothing happens to it. It's perfectly happy. It doesn't get stretched. It doesn't get distorted. It doesn't get deformed. It doesn't get squeezed one way and stretched the other way. But now what happens when you try to move it into the curved region? When you try to move it into the curved region, it simply can't follow the curvature without having to stretch its links. It has to follow the metric properties of the curved space. And in particular, if you go around a little closed loop in this space of links here, what you're doing is sampling this double covariant derivative here. And what you're going to find out is that when you go around a little closed loop, some angle shifts from what it would have been in the flat space. The angle shift, the length shifts when you try to push this up into here. It gets stressed. It gets deformed. It gets stressed. It gets stretched and so forth. And the measure of how much it gets stressed locally is exactly this curvature tensor. This curvature tensor is telling you how the system gets deformed as it moves up into the region of curvature. That's why it's connected with tidal forces. So it really is an important property. If you're in a region where this curvature, you feel it. You feel it as a distortion. If you're in a region of a gravitational field, let's say a uniform gravitational field, uniform gravitational fields don't have curvature. And that's why in free fall, in a perfectly uniform gravitational field, you simply feel nothing. OK, so the next time we will enter into gravity land, we will talk about what you have to do to change from Riemannian geometry to Einstein geometry. And then we will study a particular example. The particular example is the Schwarzschild geometry. The Schwarzschild geometry is the geometry of a black hole or a star or any gravitating mass. And see what we learn. We've left out one subject that was in my notes for tonight. And it has to do with geodesics. But I guess we'll have to do that. But we'll do that after we start thinking about gravity, real gravity a little bit. OK, I wish I had a way to make this fewer indices, but I don't. Yeah. And the first question says that the differences between an accelerating object and an opposite gravitational field is not always the same. No. So the difference between looking at an object in freefall from an accelerated coordinate system or accelerated frame of reference, and a non-accelerated frame of reference, or looking at a system which has a genuine gravitational field. In other words, you can simulate what looks like a gravitational field for many purposes by just an accelerated frame of reference. But it won't cause tidal forces. If you have a real gravitational field made by real matter out there, then it produces tidal forces. If there are tidal forces, it means that there's something real going on. It's not just a coordinate transformation. So you can tell the difference between the accelerating reference frame and the gravitational field. Yeah. Except for the uniform field. Well, there are no uniform gravitational fields. The real gravitational field of the Earth looks like this. It's not uniform. It produces tidal forces. It varies from place to place. There's no such thing as a real uniform gravitational field. It's an approximation. An approximation near the surface of a very big planet over a relatively small solid angle, the gravitational field looks approximately uniform. Now, there's another thing to say about it. Tidal forces have much less effect on small objects than on large objects. You can see that also as true of curvature. If I have a very, very small object here, and I move it around, if it's much, much smaller than the scale on which things vary, then it won't experience any great deformation. If you have a large object and you move it around, it will experience deformation because different parts of it get stressed differently. It's the same thing. Tidal forces have very small effects on very small objects. A bacteria falling into the surface of the Earth would experience, essentially no, as long as it was free-forming, would experience no tidal force, what it's speaking about. Whereas the 2000-mile man would be crushed by the tidal force as he falls toward the Earth. So when you speak about the strength of tidal forces, you have to discuss how big an object you're talking about. The same thing is true here. When you speak about the effect of curvature on an object moving around in this space, again, it depends on the size of the object that you're talking about. And this correspondence is not accidental. It's literally the same thing. Do you have to calculate the curvature of the temperature at every point? Is it technically still an infinite problem? Yeah, it is. But very often you'll be presented with the metric tensor in some analytic form. There'll be even a formula for it. And with a formula, you can differentiate the formula and everything becomes some analytic function that you can calculate when you look at it. If it's zero, it's zero. You're right. If you were just given numerically the metric tensor, you were just given a bunch of numbers at every point, you know, a huge, big spreadsheet filled with loads and loads of numbers, you would have to check the curvature tensor at every point, and that is, you know, as complicated as anything else. So where this is useful, the situations where you know the metric in some analytic sense and which you can calculate the derivatives, subtract them, do all the things you have to do, and then look at the analytic answer and see if the analytic answer is zero. Yeah. Right. So in that sense, it's the problem is it's not as bad as having to look at every possible metric and see whether you can make it. You have to look at every possible metric at every possible point. That's this is this is this is better. Not as hard as every possible metric. Every possible coordinate system is what I meant. Every possible coordinate system at every possible point until you find one which expresses the metric as being flat. This is easier. You calculate this one thing and you compare it from point to point. Okay. For more, please visit us at stanford.edu.
 Stanford University.  General relativity has a reputation for being very difficult.  I think the reason is that it's very difficult.  It's kind of calculationally intensive.  Indices and symbols.  There are ways people have invented to express things in sort of more condensed, very, very neat notations.  But just learning them in itself is a task that would take some time.  Things like theobines and forms, spinners and twisters,  and all sorts of different kinds of mathematical objects.  You could call many of them notation, notational if you like.  And they do simplify the equations.  So I sometimes feel as if in presenting these things the way I do,  that it's sort of like Maxwell who wrote down every single equation of his,  how many Maxwell equations are there?  I think eight.  Eight, eight all together.  We don't usually write all components of the equations.  We put them together into vector notation and so forth.  And if we're smart, we can even avoid the indices by inventing symbols like curl and so forth.  And the same thing can be done to some extent for general relativity.  But as I said, the computational methods and techniques are harder.  And I'm not an expert in them.  I'm by no means an expert in them.  I'm not an expert in general relativity.  I know how to use it.  But quite honestly, when I go to do something using general relativity,  I go back to the indices.  I can follow when somebody else uses the fancy methods.  But I can't do them myself.  It's not just that I can't do them myself.  It's not natural to me.  I wind up going back to the same thing that we're doing here.  Indices, derivatives, the whole works.  And my feeling is it's the right way to see it for the first time.  But it is a nuisance.  It's sort of indices all over the place, notations all over the place, too much of it.  Can't help it.  So some aspect of it, you have to get over.  You have to get over that hump of tensor notation, the indices and so forth.  And then it goes easier.  Then it goes more smoothly until you come to do a calculation.  And then right back to the mess again.  Now, if you're really interested in doing computations in general relativity,  there are packages.  There are, you know, you just put in the metric as a function of position,  the metric tensor.  And out it will spit various tensors and Riemann tensors, this kind of tensor,  that kind of tensor, Einstein tensors.  And then you can say, without even looking at the result,  you can say, okay, please, Mr. Computer, set the Riemann tensors,  set the Einstein tensor equal to the energy momentum tensor and tell me what comes out.  So, yeah, computers can do a lot better.  But as I said, general relativity is hard.  And part of the hardness is just the complexity of the notations that are necessary.  Incidentally, the subject that we're studying up till now is Riemannian geometry.  Riemannian geometry, the geometry of curved spaces, the geometry of smooth,  curved metric spaces, they're called.  And I suppose we have Riemann to blame for the complexity of the notation.  Actually, he simplified it and Einstein simplified it further.  But nevertheless, it is notationally and computationally intensive.  The basic concepts are really not so hard, but we have to go through the mess before we get to the concepts.  Or before we can express those concepts.  If there are no questions, yeah.  I know very well, yeah.  Well, yeah, there are several.  The most active areas of research, the ones which are really very, very strongly active among theoretical physicists,  is the quantization of general relativity, the quantization of black hole physics.  That's probably the most active area.  There are astrophysical areas where, again, black holes are very important.  Astrophysical areas are entirely classical general relativity.  There's numerical relativity.  And usually for the purposes of solving astrophysical problems,  numerically putting in various initial situations, a pair of black holes orbiting around each other,  and simply running the computer to find out what happens by solving Einstein's equations.  And it's a very subtle business, incidentally.  It is not that you just put in the equations and grind.  It's much more subtle than that.  And numerical relativity is also a very hard subject.  The purpose, for example, being to find out how much gravitational radiation and other kinds of things is entirely classical.  Gravitational radiation is classical radiation when two black holes fall into each other, bang.  A lot of gravitational radiation is produced, gravitational waves.  Why are we interested in them?  Well, among other reasons, because we have a chance of detecting them,  anything you can detect is a window into phenomena that you've done a new phenomena.  So gravitational wave detectors may be able to detect gravitational waves from the collision of two black holes.  It is estimated that there are a few collisions every year of black holes that can be seen by gravitational radiation.  So there's a lot of different areas.  It's an extremely active area.  The equations of general relativity, just the equations of general relativity,  and the equations of black holes have been found to be useful, extremely useful,  and condensed matter physics and fluid dynamics for having nothing to do with black holes and gravity.  Same equations.  And so there's a heavy amount of activity in the areas of applying the equations of general relativity to other areas of physics.  It's probably, I would say, probably the most active.  It went over the last, over my career in physics from being a real backwater in physics.  Nobody studied general relativity very much in the 1960s when I was a student,  except for the people who were really especially interested and were focused on questions of general relativity.  The bulk of theoretical physicists were not interested.  They just started to become interesting in astrophysics,  and eventually it took over questions of fundamental theoretical physics.  Why? Because of a clash of gravity in general relativity.  Anyway, for whatever reason, it is probably now more important than the day-to-day workings of theoretical physicists  than almost any other subject, that and quantum field theory, and the connection between them.  Okay, let's, today will be the last day in which we'll be studying Riemannian geometry as such without really discussing gravity.  Next time we will really get into gravity.  What does all of this have to do with gravity?  I told you a little bit in the first lecture that the problem of finding out if there's a real gravitational field  as opposed to just some funny coordinates in which you're expressing space and time,  which leads to curved trajectories, but not really curved trajectories.  The problem of finding out there really is a gravitational field, is a problem which is mathematically,  I would say very similar, but I think I would go further than that.  I would say is identical to the problem of finding out if a certain metric geometry,  geometry characterized by a metric tensor, is flat or not.  A geometry which is flat, let's say a two-dimensional geometry to begin with,  that means a surface of some sort.  A flat geometry is one for which Euclid's postulates are correct.  There are parallel lines, all the stuff you learned in Euclidean geometry is called flat.  And it's characterized in terms of a metric tensor by the statement not that the metric has a very special simple form.  That's not the right statement, the special form being the chronic or delta.  That's not the right statement that the metric has that form.  The right statement is that coordinates can be found such that with those coordinates,  the metric has a simple flat form.  But more generally, just given an arbitrary metric tensor which varies from place to place,  assume that it's smooth, assume that it has all good differential properties and so forth,  the problem of determining if that space is flat or not is a somewhat difficult problem.  What you have to do is look at, here's the bad way to do the problem.  The bad way to do the problem is to search through all possible coordinate systems,  transform the metric from the given form, however it's given, the way that it's presented to you,  transform the metric to the new coordinate system that you have constructed,  calculate the new metric and ask whether it looks like a flat metric.  Now, of course, that will take an infinite amount of time, you'll never get there,  so you need a better technique.  And the better technique is to search for diagnostic quantities, quantities which are built out of the metric,  and its derivatives and so forth, things that you can calculate which,  if they are zero everywhere throughout the space, that says the space is flat,  if they're nonzero any place, it says that the space is curved there.  The diagnostic quantity that does that is called the curvature tensor.  Today's goal is to get to the curvature tensor of a Riemannian geometry.  And it's a complicated goal, I warn you, you've got to pay attention or else you'll fall asleep.  This is a very superific subject.  I'll try from time to time to tell a joke or whatever to keep you awake.  But on the other hand, if you can follow it, you've learned a lot and it is very interesting.  I think you'll find it very interesting.  I'll try not to make it too boring.  Okay.  I told my friend Barack Obama, never apologize, and here I am apologizing already for being boring.  Okay.  What do we start with?  We start with a space, and a space means, first of all, a number of dimensions.  The number of dimensions in Riemannian geometry can be anything.  In principle, you can even have a zero dimensional space, but that's pretty damn boring.  It's just a point.  A one dimensional space is either an infinite line or a closed curve.  If it's a closed curve, what is it characterized by?  It's characterized by only one thing, the total length of the curve.  Every curve is equivalent to every other curve of the same length.  In other words, just think about it for a moment.  Take a piece of rope that closes on itself, it has a certain length,  and wiggle it in any kinds of ways, or curve it in any kind of ways.  It can always be mapped or put on top of another piece of rope of exactly the same length  in a precise way without stretching it.  One dimensional spaces are all equivalent to each other, not quite.  They're characterized only by their length.  They're either infinite, or they have a length L,  and every one of them have given length L is identical to every other one in its intrinsic properties.  A bug walking along a little piece of rope cannot distinguish one geometry from the other geometry.  All he can do is count the number of steps that it takes him to go around the rope.  He leaves a marker someplace, goes around the rope until he comes back to the marker,  counts the number of steps, and that's the only thing about that space.  Two dimensional spaces begin to become more complicated.  There are flat ones.  There are curved ones.  Flat ones are plain.  Curved ones could be a sphere.  It could be just a space with lumps, the surface of the earth, including the mountains and the valleys and so forth.  It can even have weird topology, the surface of a donut, a torus.  Or you can poke another hole in the torus and make a torus with two holes and so forth.  So by the time you get to two dimensional spaces, they're pretty complicated.  Nevertheless, they are characterized by the metric tensor.  And the question of whether they're flat is a computation,  as a first question, find something which distinguishes whether they're flat.  Find a quantity g mu nu, g mu nu g m n, the metric tensor.  If it is not equal to zero, that does not mean the space is not flat.  It just means you're maybe in the wrong coordinates.  So you can't diagnose the space by asking whether g m n is zero or not.  You want to find a quantity.  Let us call it r.  At the moment, I'm not going to tell you how many indices it has.  Some quantity.  But as best as should be a tensor, because if it's a tensor and it's nonzero,  it's nonzero in every frame of reference.  We like to find some quantity called r.  r stands for Riemann incidentally.  Riemann was the first person who asked this question and answered it.  We want to find a thing called Riemann such that if Riemann is equal to zero,  that implies that the space is flat and that there exist coordinates such that,  sorry, did I say g m n not equal to zero?  I mean not equal to delta m n.  g m n is an interesting space.  It's not very big.  Yeah, g m n equal delta m n is what I meant.  And that's what we're going to search for.  The curvature tensor.  It is a tensor.  Let me just remind you about a few properties of the metric tensor and what you do with it.  First of all, g m n is by definition the thing that's a function of position in general.  It was not a function of position incidentally.  If it was a constant matrix, a constant set of coefficients,  then it's very easy to find coordinates in which g m n is delta m n.  You simply have to straighten out the angles.  So if g m n is not dependent on position,  if it's just a set of coefficients, a set of numerical coefficients,  then the space is definitely flat.  So g m n of x is the metric tensor and what do you do with it?  You take a little differential distance, d x m,  multiply it by d x n.  Here's your little differential displacement.  It's characterized by a set of differential coordinate shifts, d x m, d x n.  You multiply them together and that gives you the square of the length  between the tail and the arrow and the head of the vector.  That's equal to d s squared, s standing for spatial distance, I suppose.  All right, now, the metric tensor is a tensor.  I think we proved that last time and the way you see that is just by saying d s squared,  that's an invariant quantity.  All coordinate, let's call them coordinate systems,  all coordinate systems should agree about the length of a given little vector  and if we take that same d s squared and write it in y coordinates instead of x coordinates,  the length has to be the same and that's sufficient to prove that g m n really is a tensor  and that it transforms like a tensor.  Okay, next property of g m n is it has an inverse.  g m n is a matrix.  It's a matrix and it has an inverse.  The inverse satisfies the following equation.  g, it's written again as a matrix.  When I say it has an inverse, I mean it has an inverse as a matrix.  g m n is a matrix with entries  and it has an inverse.  g m n, which is written with upper indices, they are in fact,  contravariant indices and the properties of g m n with upper indices  are such that if you multiply it by g, let's call it n r  and sum over n, but I don't have to sum over n because Einstein tells me  that an upper and a lower index like this are automatically to be summed over.  This is the sum.  This actually corresponds to taking the matrix g m n  and multiplying it by the matrix g n r  and the resulting matrix is the unit matrix.  Why? Because I told you that g m n is the inverse matrix to this one here  and the product then is the identity matrix  and the identity matrix is the Kronecker delta  and Kronecker delta in general relativity or in Riemannian geometry  is always written as a thing with one upper index and one lower index.  At least if it's being treated as a tensor, the Kronecker delta,  this is a thing to prove, prove that if you take the Kronecker delta  and you transform it pretending it's a tensor,  you just get back the Kronecker delta again.  That's the thing that you can prove.  So these are tensor equations.  This is the meaning of saying the g m n is the inverse matrix of g n r.  And once you have that, oh, there's another thing that is important.  In Riemannian geometry, as opposed to Einsteinian geometry,  in Riemannian geometry all lengths are positive.  All lengths are positive, squares of all lengths.  More important, squares of all lengths are positive  and all lengths are also positive.  But the square of all lengths are positive  and that says something about the properties of the metric tensor.  What it says is that there are no directions  along which if you evaluate this you get anything which is not positive.  That's equivalent to a statement about the eigenvalues of g m n.  It says all of the eigenvalues of it are positive.  But for our purposes now, all lengths are positive.  It's the same thing.  And it does say something about the metric tensor.  It says it's positive definite.  That's the same thing as saying all of its eigenvalues are positive.  That will change when we go to the theory of real gravity.  That will go when we go to Einsteinian geometry.  All right, the metric allows you to do something.  It allows you to raise and lower indices.  If you have a contravariant vector, or better yet,  you have a vector which has contravariant components v m,  contravariant means upstairs.  And I put it into correspondence with another vector with covariant indices.  And it's really the same vector.  We talked about this a little last time.  I told you the difference between contravariant and covariant indices.  And the way you construct the covariant components,  components of the words I'm looking for,  the covariant components of a vector giving its contravariant components  is just to multiply it by the metric tensor g m r.  And, whoops, let's call it g m n, and sum over m.  This is now g sub n, or v sub n.  Likewise, this is something that you need to prove.  Given a v m with upper indices,  this, if you like, can be thought of as the definition  of the lower of the covariant component of a vector.  How do you go back?  Well, you have to solve this equation,  but it's very easy to solve,  and I'll tell you what the solution is  if you know the inverse matrix with upper indices.  Namely, v m is equal to g m n v n.  So if you know the upper indices,  if you know the covariant, the contravariant components of a vector,  this is the way you build its covariant components,  and if you know the covariant components,  this is the way you build the contravariant components.  There's a little bit of a proof here that you have to do.  So, if you have a question, you can ask me.  You can see what we have to do.  You'll have to check that these two are compatible with each other,  but they aren't.  Okay.  To what extent can you force the metric to look like a flat metric?  To what extent?  Assume the metric is not necessarily a flat metric.  I will mean not only is the space flat,  but that the coordinates are such that g m n is equal to delta m n.  If that's the case, if that's the case,  this space is definitely flat.  But to what extent, given any space,  can you find coordinates such that g m n might be, for example,  the karmic of delta over some limited region,  or approximately over some limited region equal to delta m n?  All right, so here's a theorem.  There are coordinates at every point.  These coordinates are coordinates which are good  in the immediate vicinity of a point.  They're not global coordinates.  They don't cover the whole space.  They're just good within the immediate vicinity of a point,  which means they're good up to some order in little displacements  away from this point, namely second order.  All right.  Coordinates called Gaussian normal coordinates.  Here's what you do.  Just to first think of it intuitively,  we have some curved surface.  We go to some specific point, and we draw coordinate axes.  We draw coordinate axes, and we keep those coordinate axes  as straight as we possibly can.  Later we'll learn that the term for straight as possibly can curve  is a geodesic, but you make a straight curve as you can.  You have a little bug.  He moves on this surface, and he moves following his nose  going as straight ahead as he possibly can,  and that defines one coordinate axis,  then he comes back to a point and goes off in right angles,  does some surveying to figure out where right angles to his little line are,  and goes off in another direction as straight as he can,  and then builds coordinates based on those...  The theorem says that first of all, at every point,  at every point you can choose coordinates,  as I said, Gaussian normal coordinates, such that,  first of all, at that point,  gmn is equal to delta mn at x equals x0.  This point is x0.  You can do that in more than one way.  If you find coordinates in which it's true,  you can rotate the coordinates.  You can rotate the coordinates,  and if you rotate them, that just corresponds to rotating the coordinates  in different axes and so forth.  So there's more than one way you can do it.  Next, you can choose your coordinates.  You can straighten out these lines to a degree  where the derivative of the metric  with respect to any direction...  Well, let's put an xr here.  Any direction, the derivative of any component of the metric  with respect to any direction in space  can be said equal to zero.  That you do by choosing these curves to be as straight as possible.  I'll tell you how you prove it in a minute.  It's actually very easy to prove.  It's just a counting problem  of counting how many variables there are  and how many equations there are,  but I'll show you in a minute.  Okay, finally...  Yeah?  Say it again?  No, only at x equals zero.  At... Sorry, not at x equals zero.  Could be x equals zero.  At x equals x zero.  If you try to extend that same coordinates away from that point,  no matter how you try to extend them,  then in general, unless the space is flat,  if you try to extend the coordinates away from that point,  you'll find out that the derivatives,  the higher derivatives of gmn are not equal to zero,  and in general, for the general generic situation,  the second derivatives, let's say with respect to some XR  and some other XS,  those in general cannot all be chosen equal to zero,  except, of course, if the space is flat.  Okay?  So, at a point...  At a point, there is no content, really,  in saying that the metric can be chosen to be flat-like,  let's call this flat-like,  even up to first derivatives that can always be done.  It's in these second derivatives  that somehow the fact that the space is either flat or not flat  starts to show up.  All right, first of all, how do you prove this?  Well, this is actually not hard.  Let's take our point of interest to be the origin.  Let's just take it to be the origin.  Let's set it equal to zero.  Now, supposing we have some general metric  and some coordinates Y  in which the metric has some form,  not of this form here.  Let's look for some XS,  which we're going to build,  which are going to be functions of the Y.  Let's choose it in the following way.  At the place where X is equal to zero,  in other words, at our origin over here,  let's also assume that Y is equal to zero.  So, the two sets of coordinates have the same origin.  That means that X will start out...  We're just equal to Y.  XM equal to YM.  And then, in the next order,  we'll have some set of coefficients.  I'm simply expanding X in powers of Y.  The next order will be something quadratic in Y.  So, let's put something quadratic in Y here.  We'll call that CM.  And let's put here N and R and put XNXR.  That's the most general quadratic thing that I can write.  Oops. Sorry, Ys.  YN, YR.  And then, things may get more complicated later.  Okay.  How many such...  Let's pick a number of dimensions of space.  Let's pick a number of dimensions of space.  Let's pick four dimensions of space.  How many such coefficients are there?  Well, how many combinations  do we have in Y in YR?  Are there?  The answer is ten.  Try counting them.  There's Y1 squared, Y2 squared,  Y3 squared, Y4 squared,  Y1, Y2, Y1, Y3,  Y1, Y4,  Y2, Y3,  Y2, Y4, and Y3, Y4.  Did I miss any?  I think I have every possible thing here,  which is quadratic in Y.  And there's one, two, three, four, five, six, seven, eight,  ten, ten of them.  So there are ten possibilities here.  How many components of the metric are there?  There are ten YN, YRs.  So that means N and R give you ten possibilities.  But then for each N and R, you have an M.  That means there are 40 independent coefficients here.  40 independent coefficients.  The most general X that you can build to quadratic order  has 40 independent coefficients here.  Now, how many independent components of G are there?  The answer is ten.  So how many equations does this correspond to?  Ten components.  And setting all of their derivatives equal to zero  means that this is 40 equations.  40 equations, 40 unknowns.  And that allows you to be sure that you can solve these  equations to quadratic order.  Solving them to quadratic order is saying the same thing  that you can solve these equations, but you'll fail at the  next order.  You'll fail at the next order, and here's the next order.  So it is always possible to choose a metric that has the  property that to quadratic order, in other words,  the first derivative of the metric is equal to zero  and the metric is equal to a flat metric at a point.  Excuse me, I'm making something crucial here.  It's very beginning.  We have that equation with X and normal left and Y is on the  right.  Does that represent an attested coordinate transformation?  Yeah, that's a coordinate transformation.  Sure.  Yeah, that's a coordinate transformation.  Yeah, this is a coordinate transformation which we're going  to try to choose these C's in such a way as to make the  derivatives of the metric zero.  You can always do it.  So every metric in that sense is flat over a sufficiently  small distance.  Is that the same as it's locally flat?  That's saying it's locally flat.  That's the statement that it's local.  That's what it being locally flat means, yeah.  Intuitively, it just says that any smooth surface can be  approximated over a small distance by a plane that's  tangent to it.  Right, that's what that says.  Mm-hmm.  That can be solved.  That can be put, Gmn is invertible.  Yeah.  You can't always solve for the equation necessarily.  Say it again?  You can't always solve for the equation before the unknowns,  right?  Yeah, I think in this case you can.  I think in this case you can.  Yeah.  Yeah.  Yeah, in this case you can.  There's, try it out.  Do it with two by two matrices.  Do it in two dimensional case.  And there's nothing special about four dimensions.  You'll find you can do it.  All right, I don't want to try to do it on the blackboard.  It's, all right, in any case it's a theorem.  It's a theorem that Gaussian normal coordinates exist  at every point, at every point, and they have this property.  All right, now the reason we introduce that is because we want  to give some meaning to derivatives of vectors.  We've talked about some, of tensors in general,  derivatives of tensors.  We've talked about various procedures, various operations  we can do on tensors to make new tensors.  We can add them, we can subtract them, we can multiply them.  We can do all kinds of things.  Can we differentiate them?  In particular, if we have a tensor which is a function of position,  that means all of its components are functions of position,  can we construct another tensor which is in some appropriate sense  the derivative with respect to position of the tensor that we have?  So let me show you why this is a little bit tricky.  You could just say, well, okay, take the components of the tensor,  let's say covariant components, and differentiate them,  and that gives you a new set of components,  which are the derivatives of the components.  That gives you the derivative of the vector.  It's true you can do that, but what you get will not be a tensor.  Okay?  And let me show you why.  Yeah.  The equation you have is C.  There are no first order terms.  This is first order.  You mean zero-th order?  Yeah.  Yeah, that's right.  It's not that there are no zero-th order terms,  it's just I said let's choose the origins of the two coordinate systems  to be at the same point.  So that says wherever y is equal to zero, x is equal to zero.  It was arbitrary, but it doesn't cost anything.  Okay.  All right, let's look at a vector field,  a vector field, a vector at every position,  and let's start with coordinates which either the space is flat,  in which case I'll just use flat ordinary Cartesian coordinates,  or the space is curved, in which case these coordinates here  represent Gaussian normal coordinates,  as straight as possible, so to speak.  And now let's take another coordinate system  and superimpose it on top of that.  These are not Gaussian normal coordinates.  They may be the coordinates that we started with.  I'll represent those in green.  And they're curved relative to these coordinates.  For example, the coordinate lines might look like that,  one of the coordinate lines,  the other one might look like this, or whatever.  All right, now,  before we ask how we differentiate a vector field,  vector field means a vector in every position here,  let's ask what it means,  what it means for the vector field to be constant in space.  The problem here is because the space is curved,  it becomes hard to compare the vector at one point and another point.  The coordinates, if they can't be chosen to be everywhere flat,  then what exactly do you mean by saying a vector over here  is equal to a vector over here?  It doesn't really mean anything,  because to compare a vector over here with a vector over here,  unless you have nice flat coordinates,  there's no unique way to do that.  So let's look at it for a moment.  If the space is flat, really flat,  then I think I know what it means for a vector over here  to be the same or to be equal to a vector over here.  It means it's equal, it points in the same direction  and has the same length.  On the other hand, does it say that I'm thinking about the green coordinates now?  Let's think about the green coordinates.  In the black coordinates, the components of this black vector  and this black vector are clearly the same in the nice Cartesian coordinates.  But what about the coordinates of this vector here  and this vector over here along the green axes?  Let's take a special case.  Let's take this vector to be vertically up  so that we know that it only has a Y component in the flat coordinates.  Let's assume that this vector over here is equal to this vector  it points in the same direction in flat space.  Are its coordinates along the green axes,  are they the same as the coordinates along the black axes? No.  This coordinate along the green axis here  is different than the coordinate along the black axis here.  And in fact, this vector would have both X and Y components  along the green axes,  whatever the green axes happen to be,  has both kinds of components.  So it's clear that even though the vectors are the same,  the components are not the same.  So you cannot judge when you have curvy linear coordinates  and in general, you will have curvy linear coordinates.  You can't judge by whether the components,  either covariant or contravariant,  you can't judge whether the vector is constant in space.  Another way to say the same thing  is that in one coordinate system,  the derivatives, let's say with respect to some coordinate XR  of the mth component of a vector,  might be zero in one coordinate system  and not zero in another coordinate system.  It may even be the case that all of the derivatives of Vm  are zero in one coordinate system  and not zero in the other coordinate system  because the coordinates shift,  because the coordinate direction,  not because the vector is changing,  but because the coordinates are changing.  So this statement that the derivatives of a vector are equal to zero  is not an equation which is likely to be the same  in every reference frame.  Therefore, it cannot be a tensor equation.  It's not a tensor equation.  And the derivative of a vector,  the components of a vector with respect to coordinates,  are not themselves components of a tensor.  They're not components of a tensor  because they can be zero all of them in one frame of reference  and not in another frame of reference.  If they were components of a tensor,  we would think of this as a rank 2 tensor,  an m index and an r index.  We might say, look, could this be a tensor Tmr?  Well, if it's a tensor Tmr,  then if it's zero in one frame or in one coordinate system,  it's zero in every coordinate system.  But that's simply not true of the derivatives  of components of vectors,  not because the vector may change from point to point,  but because the coordinates,  the representation of the coordinates change.  So we need a better definition,  we need a better definition of the derivative of a vector  than just differentiating its components.  We need something which, if it's zero in one frame,  is zero in every frame.  Okay, so here's what we'll do.  Here's how we'll define the derivative of a vector.  First of all, to define the derivative at a point,  you only need to look at nearby points, right?  You only need to look at nearby points.  So the first thing you'll do is to construct  Gaussian normal coordinates near that point.  Gaussian normal coordinates, they're as straight as possible.  Then, let's put this vector a little bit closer  just to indicate that it's all like that.  Then, pretending that the Gaussian normal coordinates  are really nice flat coordinates,  we shift this vector so that its origin  is the same as the starting vector over here.  In other words, shift it,  keeping its components fixed and constant  in Gaussian normal coordinates.  So in the Gaussian normal coordinates,  I keep the components of this vector fixed,  and I compare it with this vector over here.  Just shift it down to here,  and the difference between these two  is the thing that I will use to define the derivative.  The derivative will be the difference between these two  divided by the little separation along this axis here.  In other words, to differentiate a vector,  differentiate it in Gaussian normal coordinates.  They are the ones which are closest of all to Cartesian coordinates.  The derivatives in the Gaussian normal coordinates  define the derivative.  Now, when you differentiate a vector in general coordinates,  you'll get two terms.  The two terms come, one term comes  because the vector may be changing,  and the other term comes because the coordinates are changing.  Coordinates are rotating out from under you,  even if the vector doesn't change, and so forth.  If you work out this prescription,  take the derivative,  just the ordinary derivative in Gaussian normal coordinates,  and then re-express it in arbitrary coordinates.  Here's what you will find.  You will find that the derivative,  which will now be called capital D along the R axis,  of the components of a vector Vm in arbitrary coordinates  are not just the derivative with respect to R of Vm,  but there's another term,  and the other term has to do with the way the coordinates themselves  are changing from point to point.  What it looks like is something that's proportional to V.  If you double the size of V,  you'll double this term here, obviously.  So it's proportional to V.  It will have a V in it.  It does not have a derivative  because this term doesn't come from the fact  that the vector is changing.  It's coming from the fact that the coordinates are changing.  Let's put some indices  and try to see what kind of thing it could be.  It might have the T component here.  On the left-hand side, we have indices R and M.  That means we have to have some kind of object over here  that has index R and M,  just to match the indices on the left-hand side.  On the other hand, it's going to be proportional  to the components themselves of V,  so it's going to have some VT here.  This equation doesn't quite make sense.  There are two directions here.  There are three here.  You've got to soak them up.  I have to put another index into this object here.  In fact, the standard convention is to put a minus sign there.  This, in fact, is what you will get  if you take a vector differentiated in Gaussian normal coordinates  and then transform it to other coordinates.  In any other coordinate system,  what you will get is the derivative minus a thing,  what's this called a thing for the moment,  times components of V themselves.  This is summed over T,  and the index structure on the left-hand side is just R and M.  That holds for an arbitrary coordinate system, right?  That holds now for an arbitrary coordinate system,  but I haven't told you what gammas are,  what these gamma objects are.  Question.  What does that formula become if it's a Gaussian normal?  No, in Gaussian normal coordinates, it's just this.  Okay, so the gammas come in because of this?  Yeah, the gammas, yeah.  Right, we're going to find that the gammas are proportional to derivatives of the metric.  They have to do with the metric changing from place to place.  They're going to be proportional to the derivatives of the metric  in Gaussian normal coordinates to derivatives of the metric of zero.  Okay, that's the trick.  Use Gaussian normal coordinates to differentiate  because to first order the derivatives of the metric of zero.  Once you've differentiated it, you can then transform the object  as if it were a tensor to any frame of reference,  and the form will always be like this.  It will always be like this.  It will contain indeed a derivative,  but it will also contain another term which will not be differentiated,  and it has to do with the, as I said, it has to do with the fact  that not the vector is changing from point to point,  but the fact that the coordinates are changing from point to point.  These objects have various names.  This is called a connection.  It connects neighboring points and tells you how to differentiate something  from one point to another.  It tells you basically how to compare vectors at different points.  It tells you how to compare vectors even though the coordinate system may be changing.  This is called a covariant derivative,  not because its index is downstairs.  It's called a covariant derivative for,  just because it's called a covariant derivative.  It's called a covariant derivative, and it is a tensor.  It's a tensor in that if it is zero in any frame,  it's zero in every frame.  Is that a generalization of the convective derivative?  You know, I think it is.  I never thought about it that way, but I think it is.  What do you mean?  Whether it's connected with a convective derivative,  no doubt it is.  No doubt it is.  I would have to sit down and think about it,  but there certainly are covariant derivatives in fluid flow.  Right, okay, so, yeah, this is called a covariant derivative,  and these symbols here go under two names usually.  They're either called the connection coefficients or the Christoffel symbols.  Christoffel.  I, Christ.  They've also been known on occasion as Christoffel symbols,  and they are, they're not that complicated,  but they're complicated enough.  They're unlikeable enough.  Okay, now, one fact about them that follows,  I'm not going to prove every single thing I say,  because they're just too many little pieces,  but they're easy to check.  It follows from the definition,  namely the definition is you differentiate in Gaussian normal coordinates,  and then you treat the object as a tensor and transform it as a tensor with two indices.  It follows as a theorem, it's an easy theorem,  but it's a little too intricate to do on the blackboard,  that this Christoffel symbol has a symmetry.  The symmetry of the Christoffel symbol is the gamma t, r, m,  is equal to gamma t, m, r.  It doesn't matter which order you put m and r, they're the same.  That follows, there are generalized Riemannian geometries.  They're called geometries with torsion.  The symmetry is not true,  but those geometries are not widely in use in ordinary gravitational theory.  Ordinary gravitational theory is Riemannian geometry,  it follows from the principles that I laid out,  and gamma r, m is equal to gamma m, r.  You know,  So it's just the gamma that the Christoffel symbol, not gamma.  The gamma, the gamma, the gamma.  You know, doing the derivative in these Gaussian normal coordinates,  which are almost flat, or as flat as can be,  and then treating it as an object on its own right,  it is very, very similar to what we do in gravitational theory  when we evaluate something in a freely falling frame.  And then, for example,  in a freely falling frame,  we calculated how light moved across an elevator.  And then we transformed it to the frame of reference  in which the elevator was accelerating.  Okay.  That's closely related to the operations we're doing here.  We calculate something because we know how to do it  in coordinates which are as flat as possible.  That would be a freely falling frame in general relativity.  And then we transform it to any coordinates we like,  accelerated coordinates, or anything we like.  And we translate the statement from one coordinate system to another.  Here, the notion of variation of a vector from point to point  is done first in Gaussian normal coordinates,  and then it's transformed.  And this is the form that you'll get for the corresponding tensor.  We're not going very fast tonight, so I want to move along.  All right.  Now, for higher derivative tensors, we will need this tonight.  Sorry, for higher rank tensors,  supposing we have a tensor with more than one index,  m, let's say, n,  and we want to differentiate it,  co-variantly differentiate it, again,  in this case, along the s-axis.  I don't know why I chose this instead of r.  Here, I used r, but I want to differentiate it along the s-axis.  Does everybody know what I mean d sub r here?  That means derivative with respect to xr.  Derivative with respect to xr.  d sub r.  That's a Gaussian normal.  No, no, no, that's in any coordinate system.  d sub r will always mean d by dxr.  No, but I mean, in this equation, this is a Gaussian.  Which question?  The one here, you mean error routing.  If you want to even talk about the rest of the equation.  This one here?  Yeah.  No, that's in any coordinate system.  In Gaussian normal coordinates, this term isn't there.  In Gaussian normal coordinates, you just differentiate.  Gaussian normal coordinates, everything is easy and ordinary.  The derivative of a vector is just the derivative.  In arbitrary coordinates, there's another term.  You can work that out by transforming, just by doing transformations.  Yeah?  Good thing, when you say that in Gaussian normal coordinates,  the polar derivative is just the derivative.  That's only at the point.  Yes, yes, yes, yes.  Only at that point.  Right.  So, the question, are the Christoffel's coordinates, are those,  is that a 2, 1 tensor or whatever?  No, it is not a tensor.  Not a tensor.  This is intended to be a tensor.  We're going to try to choose gamma so that this is a tensor.  This we know is not a tensor, and therefore this cannot be a tensor.  Right, no.  Gammas are not tensors.  In fact, they're built up out of derivatives of the metric.  In a coordinate system in which the derivatives of the metric are zero,  the Christoffel symbols are zero,  but a tensor, if it's zero and one coordinate system,  is zero in every coordinate system.  So, they can't be tensors.  And they're not.  Okay.  Let's just, before we go and figure out what the gammas are,  we're going to work them out.  They're not that hard to work out.  There's a trick.  Let's write this down.  It's the analog of this,  except for every index in the tensor,  there is a term like this.  There is a term for every index,  and I'll show you how it works.  There is a Christoffel term for every index.  So, for example, we start out by hiding in.  Let's hide in and pretend that this was just a vector.  Then we would write that this is the derivative of t  with respect to xs,  that's the first term here,  and then we would put a term gamma s m.  Let's see what I used here.  s m t t m.  But now I expose the n here,  and I put it in on both sides.  Now I'm not finished.  I have to do exactly the same thing, covering up m.  Cover up m.  I won't try to do it,  but I have another term here where m is passive,  n is active,  and so we get gamma s s n t t m t.  Look at the form of it for a minute.  Oh, minus signs.  Minus signs are wrong.  Same as here.  Minus signs as here.  Notice if I covered up the second index, t m n.  If I covered up every place the second index,  it would look exactly like the covariant derivative of a vector.  If I covered up the first index,  it would look exactly like the covariant derivative of a vector.  Say it again.  You're right.  This should be m t.  Thank you.  Yeah.  Paralleling this over here.  All right.  Notice in one of the terms,  it occurs in the first place,  and in the second term it occurs in the second place.  So that's the form that the covariant derivative of a tensor would take.  The rule is the same.  You go back to Gaussian normal coordinates,  and you just differentiate the tensor,  just derivative, none of this junk.  And then you transform it to arbitrary coordinates,  assuming that the result here is a tensor,  and this is what you get.  So it allows you to differentiate any tensor.  At the moment, we're just dealing with tensors with covariant indices.  We'll come to tensors with contravariant indices at some point.  All right.  In particular, let's apply this to the metric tensor itself.  But there's something special about the metric tensor.  In Gaussian normal coordinates,  its derivatives are all zero.  Yeah.  I'm sure you must have said this one. It seems like a really fundamental thing.  What's the covariant derivative for?  It's for differentiating things.  It's for comparing things at different points.  What are we going to use it for?  Field equations.  Field equations.  Field equations are going to be differential equations,  which represent how fields change from one place to another.  But we want them to be the same equations in every reference frame.  We don't want to write down equations which are special  to some peculiar reference frame.  We want to write down equations which are general,  so that if they're true in one frame, they'll be true in all frames.  That means they have to be tensor equations,  and that means we have to know how to differentiate tensors  to get other tensors.  So normally you have a derivative if you multiply it by corresponding  to the decimal, you get the change.  This is the partial derivative.  This is the partial derivative with respect to the coordinate xs.  I'm trying to throw me for the covariant derivative,  but is there some differential you would multiply times the covariant?  Yes, yes, absolutely, absolutely.  Then what is that?  dx.  If you wanted to know how a thing changes from one point to a neighboring point,  you would just multiply this by dxs.  But is that what the partial derivative is?  No, no, no, no, no.  I say it again.  The point is that the components of the vector may change  even if the vector doesn't.  Or that, right, we have to keep that in mind.  So this is, right, we'll see some examples about how it works.  This is representing effectively a distortion of the,  an out of flat distortion of the coordinates.  Yeah, that's right.  It's an out of flat distortion of the coordinates.  It's there even in flat space if you choose funny coordinates.  That's an important point.  It's there even in flat space if you choose funny coordinates.  In fact, if you choose any coordinates in which the derivatives of the GMNs  are not zero in which the coordinates vary from point to point in funny ways,  this will be there.  This is not really a feature necessarily of curved spaces.  It's a feature of curved coordinates.  Now, okay.  So because the metric tensor is constant,  or at least has zero derivative in Gaussian normal coordinates,  and the definition of covariant derivative is to differentiate first  and Gaussian normal coordinates and then take what you get and treat it as a tensor,  it must mean that the covariant derivative of the metric tensor is zero.  That's what's going to allow us to compute the Christoffel symbols.  All right, so let's write down that equation.  Here is the form for the different derivative,  the covariant derivative of any tensor with two lower indices.  So let's apply it to the metric tensor itself.  dS of GMN is equal, first term,  dGMN by dxS minus gamma TSMGTN minus gamma TSNGMT.  All I've done is rewritten this equation,  and now I'm going to add that the covariant derivative of the metric tensor is equal to zero.  Why? Because the ordinary derivative is zero in Gaussian coordinates,  in Gaussian normal coordinates.  All right, now, let's write the same equation  except permuting the indices, interchanging S and M and so forth.  This is the equation for differentiating with respect to S of GMN  by inspection, why am I doing this?  I'm doing this because I want to get as much information out here  about things like this as I can,  and then try to see if there's any way that I can manipulate the equations  to get to isolate and figure out what gamma is.  So it's, you know, people who are experienced at this sort of thing  look at it and say, I know what to do.  If you don't know what to do, I'll tell you what to do.  You do it once, and then when your friend comes and asks you,  how do you find the Christoffel symbols, you'll say, I know what to do.  I'm experienced.  Okay, so you do the same thing except now I'm going to interchange S and M.  Just permute S and M, and that means we're going to write down dM of GSN.  How do I do it? I just do it by interchanging or replacing S by M and M by S,  and that gives me derivative of GSN with respect to XM minus,  now interchanging S and M here doesn't change anything  because S and M, yeah, from the symmetry here,  so this is the same thing, gamma SMT, GT, let's see.  Yeah, we're just interchanging S and M, GTN,  and then over here, so we're interchanging S and M,  so this is gamma MNT GST, and of course for exactly the same reason,  this is also equal to zero.  And finally, there's one last possibility.  Here, S is the odd man out, M and N go together.  Here, M is the odd man out, S and N go together.  And the last one is dN of GSM, and let's see,  I think I'll just copy it down.  It equals d by dxN of GSM minus gamma SMT, GTN,  GTM minus gamma MNT GST, and that's equal to zero.  Now you stare at this for a few minutes.  Now how can I either air these, subtract these,  or do something clever that will isolate only one of these terms  with the gammas so that I don't have a horrible bunch of equations  but basically only one equation?  So I'll tell you what to do.  Call this equation three, call this equation two,  and call this equation one, and add equation three to two,  and subtract one.  And we'll see what happens.  We'll see what happens.  First of all, of course, we will get these derivatives.  We'll get this derivative plus this derivative minus this derivative.  So let's put them down on one side of the equation.  I'm going to transpose all of this junk to the other side.  OK, so on the left side of the equation we will get dN,  dN GSM plus dN GSM, no, dN MG,  the GSN M, that's this one,  and the last one is minus GMN, sorry, derivative GMN.  That's this plus this minus this, or this plus this minus this.  Now, how about on the other side?  Let's see, there may be some things which cancel.  Let's hope this thing which cancel.  This comes with a minus sign, this comes with a plus sign,  this comes with a plus sign.  So this and this are the same, but they cancel because of the opposite sign.  No, listen, this are gone.  They are gone, what else is gone?  GTSN, ah, here it is, this one and this one come with opposite signs.  So they are gone.  We're left with two terms, but they're the same.  So we're in luck.  We can write that this is equal to twice,  we need over to the right hand side, twice gamma MN T GST.  Well, we still have, we're still not there,  we would like to get gamma MN by itself.  Trying to find out what the gamma, what the Christoffel symbol is all about,  that's our goal, figure out what they are in terms of derivatives of the metric.  We're almost there, first of all we can see that if all the derivatives of the metric  are zero, the Christoffel symbol must be zero.  But how are we going to get rid of this GST?  And the answer is GST has an inverse.  Basically, you can just take this over to the other side,  but replace it by the inverse matrix.  So first of all, let's put a one half.  And now, put a bracket around this and take this GST over to the other side,  except in its inverse form.  And this is the correct formula.  This is the formula for the Christoffel symbol.  It looks pretty simple.  We'll look at it for a moment.  The M and N are symmetric.  Look at it, M, N, if you want to change these two,  and here it's symmetric again, has one negative term.  Oops, yeah, one negative term, two positive terms, not very complicated.  The problem is there's a boatload of them.  When you think about four dimensional space and you let S and T and,  sorry, and let all of these coefficients range over from one to four,  there's just a lot of Christoffel symbols.  And it's that which makes doing calculations in general relativity  a very tedious business.  Intrinsically, there's nothing hard about it.  But in fact, doing a calculation in some sort of general relativity context  usually fills page after page after page of nothing more complicated  than just computing these derivatives and putting them together.  Yeah.  I'm sorry, I just want to say it again.  G is constant, has zero derivatives in Gaussian normal coordinates.  At that point.  Oh, yeah, at that point.  All of this is just at that one point.  Yeah, but once it's true at one point, it's true at every point.  To derive it at any point, you go to Gaussian normal coordinates at that point.  So you say, I want a formula appropriate to a particular point.  I go to that point, use Gaussian normal coordinates,  and find out that this is what gamma is.  But once you know it at any point, it's true at every point.  So this is the connection between gamma and the derivatives of G at every point.  The thing is that this is not a tensor.  It's not a tensor, and so it can be zero in one frame of reference  and not zero in others.  For example, in Gaussian normal coordinates at the point, at that point,  at that point, this is equal to zero,  but in some other coordinate system it's not.  Even more, take ordinary flat space.  In flat space, you can find coordinates in which the G's are constant everywhere,  in flat space.  But you can also find other coordinates in which they're not constant.  Polar coordinates, the G's are not constant.  In Cartesian coordinates, they are constant.  The Christoffel symbols can be nonzero even if the space is flat.  In fact, they generally are. They are in polar coordinates.  Yeah?  Just to clarify, you can represent the metric that G has a 2 by 2 matrix in this...  Not a 2 by 2, an n by n, or 4 by 4, or whatever it is.  Whatever it is, but here you're on the...  Well, whatever it is, it's 4 by 4, n by n.  What is the Christoffel symbol, so it will also come right in?  No, it's n by n by n.  It's got three indices.  n times n, now it's n by n by n, but there's some symmetry.  So there's 40 components to it.  It has 40 components.  Everything always has 40 components.  M and n, if you fix t, fix t,  then gamma mn has 10 components.  Same 10 components for a symmetric tensor.  And then you let t run over all four components, and that gives you 40.  So there are 40 independent Christoffel symbols.  Excuse me.  Yeah.  I just want to make sure I understand what you said.  I agree that you proved that at that one particular point, and it's true in all points,  but for each point that you do it, you have to use a different G.  Do you use the G that's appropriate for the Gaussian normal coordinates?  No, no, no, no, no, no, no.  You have some coordinates which are general coordinates all over the place.  And I want to know how to differentiate at that point.  So at that point, I go to Gaussian normal coordinates.  I do my differentiating, and then I relate it back to the coordinates that are general  coordinates which cover the whole space.  At each point, at each point, this is what you do with this formula.  No.  You take your curved coordinates, whatever they are, and then you say,  I want to know how to compare vectors at neighboring points.  So I go to Gaussian normal coordinates at that point,  and it tells me how to compare them at that point.  Here's what it tells me to do.  And those G's are the G's associated with Gaussian coordinates at that point?  No.  They're the G's that are associated with any coordinates.  With Gaussian normal coordinates, these are zero.  They're all zero at that point.  The derivatives are all zero at that point.  So no, these are the coordinates that you began with.  You transformed the Gaussian coordinates, derived all of this,  but then you finished with it.  You finished with the Gaussian normal coordinates.  Now you're finished with the Gaussian normal coordinates, and you go on.  This was just a trick to, the Gaussian normal coordinates were just a trick  to eliminate locally at each point the variation of the metric pencil with position.  Right, good.  Is this equation, the equation that tells you that a space is flat?  It tells you what?  That your space is flat?  No.  No, no, no, no, no.  This does not tell you that the space is flat.  This is just, which just tells you what gamma is.  If gamma is, I assure you, if gamma is zero everywhere,  it not only means the space is flat, but it also means you've chosen Cartesian coordinates.  Didn't we get here by setting the derivative, the covariant derivative of the metric tensor equal to zero?  And so that by itself is not enough to tell you the flight length.  No, the covariant derivative of the metric tensor is always zero.  That's not the same as saying the metric is constant over space, the metric tensor.  The statement that the metric tensor itself is constant would be that the ordinary derivative is zero.  Right, but the covariant derivative of the metric is always zero.  Why?  Because in Gaussian coordinates it's zero.  And if a tensor is zero in one coordinate frame, it's zero in every coordinate frame.  And one more thing, that covariant derivatives are tensors.  So this is conceptually tricky.  But at the end of the day, the rule is calculate your Christoffel symbols  and replace ordinary derivative in many contexts,  replace ordinary derivatives with covariant derivatives.  You can write your equations in Gaussian normal coordinates,  then they just involve ordinary derivatives.  But if you want the same equations in general coordinates,  then you replace the ordinary derivatives by covariant derivatives.  That's the rule.  Think about it.  Look, this requires sitting down and thinking about it and understanding what we're doing.  We have not got the curvature yet, so let's try to get to curvature.  We have the definition of...  Now, let's talk about what curvature is.  It's easiest to start with two-dimensional curvature.  How do we probe for curvature?  Well, let's draw...  Let's begin by drawing a space which is curved.  This sphere is obviously curved, but I don't want to deal with a sphere.  I want to deal with a cone, but not a cone,  but a cone with a round point at the top.  The top of a mountain, but with the sides being nice and flat.  So that if you're away from the top of the volcano up here,  I don't know, it's not a volcano, it's just a rounded cone,  on the sides here it looks cone-like.  At the top it's smoothed off, it's been shaved off and made round.  The first statement is such a cone is flat,  everywhere is except up near the top.  To see that, just take the same space down here,  but continue it so that it really does form a genuine cone.  Okay, so now we have a cone.  It only differs from the space that we're talking about  up in this vicinity up here.  Now, take the cone and slice it.  Slice it down here and open it up.  This is a cone, slice it down here and open it up.  It's clearly flat.  It's clearly a flat space, but it's a kind of question.  Okay, open it up and what does it look like?  It looks like, depending on the angle of the cone up here,  it looks like a flat space, but when you cut it and open it up,  there was a piece cut out of it.  Missing, a missing piece.  I don't have a scissor, I should have brought a scissor.  I didn't bring a scissor.  Take a piece of paper, cut it, cut out a wedge of a given angle,  and then glue it together and you make a cone.  The cone has a tip up here.  The tip is clearly not flat, but the rest of the cone is flat.  You cannot flatten this like a coffee filter.  You can't flatten it down on the table without tearing it,  but if you cut it anywhere, if you cut it and open it up,  you can lay it out and make it flat.  Scissor?  Hmm?  Scissor? Scissor.  Oh, a scissor. A scissor, oh.  Okay, but now I have to cut my notes.  If you have a piece of paper, somebody have a piece of paper,  you can do it.  Alright, I've cut out a certain angle,  and now when I sew them together or glue them together,  in this case it made a kind of flat cone.  I could wear it as a hat.  Okay?  If I cut out a bigger angle,  I'll make a more acute angle at the top of the cone.  For example, I can cut out a large angle.  I cut out a larger angle,  and now when I sew it together,  it makes a pointy cone,  except I can't really do it because,  yeah, I think you get the idea though, huh?  Yeah, it makes a pointier cone.  So the bigger the angle that you think you,  the bigger the angle that you cut out,  the bigger the deficit angle,  it's called the deficit angle or the conical deficit,  the bigger the conical deficit,  the pointy of the cone.  If the conical deficit is a very small acute angle,  it means the cone is very flat on top.  Okay?  All right, now let's think about,  let's think about a vector,  a vector field on the space here which is constant.  Let's take a vector field which is constant.  What does it mean to be constant?  It means that it's pointing in the same direction everywhere,  and for simplicity,  let's take it to be pointing in this direction.  Something funny over here,  over here it's along this line,  over here it's practically perpendicular to this line,  but remember when the cut is identified like this,  the line is the same line.  Let's do it.  Let's take a vector pointing over here,  here, here, here, and here,  and now when we sew it together,  we discover that there's a funny jump in its direction.  It's pointing along the cut here,  and it's pointing in some funny direction here.  Even though when you open it up,  you see they're all in the same,  did you see that or was I hiding it there?  As you go around, starting here,  pointing in the same direction until you get around  to the other side of the cut,  in which case you find that it's not in the same direction.  As the arrow was over here.  Nevertheless, when I open it up,  they're all parallel.  So the tip of the cone is such  that if you carry a vector around,  when you get around to the other side,  it has been rotated.  It has been rotated due to the conical deficit at the top.  Exactly the same will be true.  In this geometry over here,  if I take a vector along a given direction  and I carry it around the mountain in such a way  that it's always pointing in the same direction  if I opened it up,  by the time I get back to the other side,  it'll be pointing in some other direction.  That is because the top of the cone is curved.  That's the effect of curvature.  The effect of curvature,  a region of curvature has the property  that if you take a vector pointing in the same direction  as much as you can do,  keep it pointing in the same direction  and take it around the region of curvature,  the vector will rotate.  The vector will rotate when it goes around  any region of curvature,  even though you took pains to make sure  that at every point you were moving it parallel to itself.  In other words,  you took pains to make sure that the covariant derivative  of that vector was zero,  but nevertheless when you came around to the other side,  it had shifted.  There's another way to say this,  which is actually more useful.  Here's the other way to say it.  It's equivalent.  It's equivalent.  If you have a curved space in some curvature here,  in some curvature here,  and you take a vector field  and differentiate it along one axis,  we're going to move it and differentiate it,  differentiate it along one axis,  and then differentiate it along a second axis.  We take the vector,  displace it a little bit, and then displace it along some second axis.  We will...  the vector will change.  And how will the vector change?  It will change typically by differentiating the vector  along these two axes in sequence.  You first differentiate the vector along one axis,  then differentiate it along the second axis,  and there's a small change in the vector  due to these two derivatives.  So if I go along here,  in other words, if I take the difference of the vector  over here and over here,  it consists of two changes,  a change to go from here to here  and a change to go from here to here.  And that change, the total change in that vector  is proportional to a second derivative.  It would be true in any coordinates.  It would be true in any coordinates  that if you carry the vector from here to here  and then from here to here  and then compare it with the vector at this point over here,  what you would be calculating  is the second partial derivative of the vector,  let's say derivative of Vn,  call this the Xs direction and the Xr direction.  Then this would be, let's see,  which derivative would it be?  It would be derivative with respect,  it would be covariant derivative with respect to r  of the covariant derivative with respect to s  of the vector, and let's say the mth component of the vector.  This operation says you move the vector a little bit  along the s direction,  and this one then says you take the difference,  the little derivative, and move it along the r direction,  and the net result is the change in the vector  going from here to here.  Done covariantly, which means that you could have done  the same thing in Gaussian normal coordinates  with just ordinary derivatives.  In Gaussian normal coordinates,  you would just be doing this derivative here,  or two ordinary derivatives.  You could have also gone in the other direction.  You could, and that is to say,  you could have gone first in the r direction  and then in the s direction  and calculated the way the vector changes from here to here.  That would be the opposite order of differentiation.  Ordinarily, and in flat space in general,  these two are equal to each other.  This is just the fact that derivatives,  it doesn't matter which order you take them,  but it's not true in curved space.  In curved space, this net,  the difference between these two operations  can be thought of as taking the vector around the closed loop.  The difference between these two operations,  taking the vector from here to here  and then from here to here,  minus the opposite direction,  is essentially taking the vector around the little closed loop.  Taking it around the little closed loop,  we discovered that when there's curvature  at the tip of the cone here,  you don't get back to what you started with.  In curved space, covariant derivatives  are not interchangeable.  In flat space, they are interchangeable.  That's the way we test for whether a space is flat or not.  We test whether differentiating tensors,  in particular vectors, in opposite order, give zero.  If the answer is yes, everywhere is in the space.  For any vector, then the space is flat.  If we discover that there are places in the space  where the order of differentiation gives different answers,  then we know the space has some kind of defect in it,  or it has curvature.  It has curvature analogous to the top of the mountain.  So all we have to do,  and it's a little bit complicated, but it's not that complicated,  is compute the second derivative of a vector,  the second covariant derivative of a vector,  in opposite order and compare them.  I'll tell you what to do. You're going to do it.  I'm not going to do it on the blackboard. It's too late.  It's easy. It's really quite easy. It's mechanical.  It's a mechanical operation.  I'll tell you what the steps are, and I'll tell you the answer.  And it's entirely mechanical. It's just pure plug-in.  It's too late to carry it out.  What we're interested in is taking a vector,  the Vn, the components, covariant components of the vector,  computing the covariant derivatives in the r direction,  and then differentiating it in the s direction.  Later on, we will interchange s and r and subtract.  But let's just do this.  This is equal to ds.  This object over here is a tensor, trn.  It's a tensor, and we have the rule for calculating  the derivative of a tensor.  We have a contribution from each index,  and I'm going to write it down.  I'm going to write down the answer.  I'll write down the, let's see.  ds of ds of dRVn minus gamma Rnt Vt.  I'm just stuck in here the covariant derivative of Vn.  So that's what's in this box here, just this thing over here.  Now we have to differentiate it again using this formula.  We have a two-indexed object, and let's see if I can, let's see.  I'm going to leave it to you to calculate.  Let's see.  All right, I'll write down one further equation.  It's ds of dRVn minus gamma Pn.  Oh, brother.  No, it's getting too late.  I don't want to try to write it down.  I'm going to leave it for you to do.  Just pure plug-in.  Pure plug-in.  You plug-in for Tmn, this object over here.  Put it right into there, or calculate the second derivative,  and then compute the following.  ds dR of Vn minus dR ds Vn.  That corresponds to the difference between the two ways  that you could get from here to here  by going through this direction or that direction.  I'll tell you what you get.  You get something called R.  It's a tensor with three covariant indices,  one contravariant index, and it multiplies Vt,  where this is pretty easy, except that R is a little bit  complicated, and I'll tell you what R is.  So, ds dR sRnT equals dR gamma snT minus ds gamma rnT  gamma snP gamma prT minus, one more term,  gamma rnP gamma pst.  Two terms involving derivatives of the Christoffel symbols  and two terms involving products of Christoffel symbols.  There's a sum over p here and a sum over p here.  The indices, take a look at it, the indices match.  It's a complicated expression, and it's even more complicated  when you remember what the gammas are.  The gammas themselves are complicated derivatives of g's  and the curvature tensor.  This is the curvature tensor, the Riemann curvature tensor.  This Riemann curvature tensor is itself another derivative  and also products of Christoffel symbols.  Let's see what it has in it.  When you differentiate, the Christoffel symbol  involves a derivative of g.  Differentiating again gives you second derivatives of g.  Remember, the second derivatives of g are the things  that you cannot generally set equal to zero.  The first derivatives of g, you can find frame of reference  where they're equal to zero.  The second derivatives of g, you can't.  So by the time you finish calculating this thing here,  second derivatives of g have come into it,  and it is testing out and feeling out the geometry  a little more thoroughly than the first derivatives would feel it out.  It's feeling out the second derivatives,  feeling out the coordinate systems to quadratic order.  So it has second derivatives of Christoffel symbols,  and it has squares or quadratic things involving first derivatives.  It's a complicated thing, and if I were to actually write it  in terms of the Christoffel symbols,  or I were to try to calculate it for a given metric,  it can fill up pages, and it will fill up pages.  But conceptually, what it's doing is just calculating  the difference in a vector if you transport it around here,  keeping it parallel to itself until it comes around all the ways around.  The little change in the vector in going in a loop.  This is the curvature tensor.  The curvature tensor, as I said, it's complicated to calculate,  but you can calculate it.  You can put it into a computer, and you can ask the computer,  is it zero?  In fact, if the computer, it's good if you have a computer  that can do algebra, and if you have the metric in some algebraic form,  or you have the metric in some mathematical form,  you can do these operations and then test out whether this curvature tensor  is zero everywhere.  If the curvature tensor is zero everywhere, all components of it,  the space is flat.  Now, we'll study the curvature tensor a little more,  but as I said, it's a complicated thing,  but the main thing about it is that it tells you whether the space is flat,  and if not, how unflatted it is.  It is closely related to a quantity in gravitational physics.  Can you guess what it is?  It's a local quantity which tells you that the space is not flat.  So it must be something telling you whether there's really a gravitational field present or not.  Tidal forces?  Tidal forces.  Tidal forces.  It's exactly related to tidal forces.  Those things which in a gravitational field squeeze one way,  stretch the other way, that's what these things are,  and in fact, they really are the tidal forces.  Tidal forces are represented by the curvature tensor.  Now, you can sort of see something like that if you were to imagine,  yeah, imagine you have, again, this kind of conical geometry.  It doesn't have to be conical geometry, but something which is flat far away.  Yeah, let's say it's flat far away.  Nice and flat far away.  And somewhere in the center it has a bulge so that it's curved.  Now you take some object out here.  It's a thing living in flat land.  It's two-dimensional.  You take some object that might be composed of a bunch of links,  there might be a bunch of triangles connected together, links,  and then you start moving it.  As you move it in the flat region, nothing happens to it.  It's perfectly happy.  It doesn't get stretched.  It doesn't get distorted.  It doesn't get deformed.  It doesn't get squeezed one way and stretched the other way.  But now what happens when you try to move it into the curved region?  When you try to move it into the curved region,  it simply can't follow the curvature without having to stretch its links.  It has to follow the metric properties of the curved space.  And in particular, if you go around a little closed loop in this space of links here,  what you're doing is sampling this double covariant derivative here.  And what you're going to find out is that when you go around a little closed loop,  some angle shifts from what it would have been in the flat space.  The angle shift, the length shifts when you try to push this up into here.  It gets stressed.  It gets deformed.  It gets stressed.  It gets stretched and so forth.  And the measure of how much it gets stressed locally is exactly this curvature tensor.  This curvature tensor is telling you how the system gets deformed as it moves up  into the region of curvature.  That's why it's connected with tidal forces.  So it really is an important property.  If you're in a region where this curvature, you feel it.  You feel it as a distortion.  If you're in a region of a gravitational field, let's say a uniform gravitational field,  uniform gravitational fields don't have curvature.  And that's why in free fall, in a perfectly uniform gravitational field,  you simply feel nothing.  OK, so the next time we will enter into gravity land,  we will talk about what you have to do to change from Riemannian geometry to Einstein geometry.  And then we will study a particular example.  The particular example is the Schwarzschild geometry.  The Schwarzschild geometry is the geometry of a black hole or a star or any gravitating mass.  And see what we learn.  We've left out one subject that was in my notes for tonight.  And it has to do with geodesics.  But I guess we'll have to do that.  But we'll do that after we start thinking about gravity, real gravity a little bit.  OK, I wish I had a way to make this fewer indices, but I don't.  Yeah.  And the first question says that the differences between an accelerating object and an opposite gravitational field is not always the same.  No.  So the difference between looking at an object in freefall from an accelerated coordinate system or accelerated frame of reference,  and a non-accelerated frame of reference, or looking at a system which has a genuine gravitational field.  In other words, you can simulate what looks like a gravitational field for many purposes by just an accelerated frame of reference.  But it won't cause tidal forces.  If you have a real gravitational field made by real matter out there, then it produces tidal forces.  If there are tidal forces, it means that there's something real going on.  It's not just a coordinate transformation.  So you can tell the difference between the accelerating reference frame and the gravitational field.  Yeah.  Except for the uniform field.  Well, there are no uniform gravitational fields.  The real gravitational field of the Earth looks like this.  It's not uniform.  It produces tidal forces.  It varies from place to place.  There's no such thing as a real uniform gravitational field.  It's an approximation.  An approximation near the surface of a very big planet over a relatively small solid angle, the gravitational field looks approximately uniform.  Now, there's another thing to say about it.  Tidal forces have much less effect on small objects than on large objects.  You can see that also as true of curvature.  If I have a very, very small object here, and I move it around, if it's much, much smaller than the scale on which things vary, then it won't experience any great deformation.  If you have a large object and you move it around, it will experience deformation because different parts of it get stressed differently.  It's the same thing.  Tidal forces have very small effects on very small objects.  A bacteria falling into the surface of the Earth would experience, essentially no, as long as it was free-forming, would experience no tidal force, what it's speaking about.  Whereas the 2000-mile man would be crushed by the tidal force as he falls toward the Earth.  So when you speak about the strength of tidal forces, you have to discuss how big an object you're talking about.  The same thing is true here.  When you speak about the effect of curvature on an object moving around in this space, again, it depends on the size of the object that you're talking about.  And this correspondence is not accidental. It's literally the same thing.  Do you have to calculate the curvature of the temperature at every point? Is it technically still an infinite problem?  Yeah, it is.  But very often you'll be presented with the metric tensor in some analytic form.  There'll be even a formula for it.  And with a formula, you can differentiate the formula and everything becomes some analytic function that you can calculate when you look at it.  If it's zero, it's zero.  You're right.  If you were just given numerically the metric tensor, you were just given a bunch of numbers at every point, you know, a huge, big spreadsheet filled with loads and loads of numbers,  you would have to check the curvature tensor at every point, and that is, you know, as complicated as anything else.  So where this is useful, the situations where you know the metric in some analytic sense and which you can calculate the derivatives,  subtract them, do all the things you have to do, and then look at the analytic answer and see if the analytic answer is zero.  Yeah.  Right.  So in that sense, it's the problem is it's not as bad as having to look at every possible metric and see whether you can make it.  You have to look at every possible metric at every possible point.  That's this is this is this is better.  Not as hard as every possible metric.  Every possible coordinate system is what I meant.  Every possible coordinate system at every possible point until you find one which expresses the metric as being flat.  This is easier.  You calculate this one thing and you compare it from point to point.  Okay.  For more, please visit us at stanford.edu.
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Stanford University. General relativity has a reputation for being very difficult. I think the reason is that it's very difficult. It's kind of calculationally intensive. Indices and symbols. There are ways people have invented to express things in sort of more condensed, very, very neat notations. But just learning them in itself is a task that would take some time. Things like theobines and forms, spinners and twisters, and all sorts of different kinds of mathematical objects. You could call many of them notation, notational if you like. And they do simplify the equations. So I sometimes feel as if in presenting these things the way I do, that it's sort of like Maxwell who wrote down every single equation of his, how many Maxwell equations are there? I think eight. Eight, eight all together. We don't usually write all components of the equations. We put them together into vector notation and so forth. And if we're smart, we can even avoid the indices by inventing symbols like curl and so forth. And the same thing can be done to some extent for general relativity. But as I said, the computational methods and techniques are harder. And I'm not an expert in them. I'm by no means an expert in them. I'm not an expert in general relativity. I know how to use it. But quite honestly, when I go to do something using general relativity, I go back to the indices. I can follow when somebody else uses the fancy methods. But I can't do them myself. It's not just that I can't do them myself. It's not natural to me. I wind up going back to the same thing that we're doing here. Indices, derivatives, the whole works. And my feeling is it's the right way to see it for the first time. But it is a nuisance. It's sort of indices all over the place, notations all over the place, too much of it. Can't help it. So some aspect of it, you have to get over. You have to get over that hump of tensor notation, the indices and so forth. And then it goes easier. Then it goes more smoothly until you come to do a calculation. And then right back to the mess again. Now, if you're really interested in doing computations in general relativity, there are packages. There are, you know, you just put in the metric as a function of position, the metric tensor. And out it will spit various tensors and Riemann tensors, this kind of tensor, that kind of tensor, Einstein tensors. And then you can say, without even looking at the result, you can say, okay, please, Mr. Computer, set the Riemann tensors, set the Einstein tensor equal to the energy momentum tensor and tell me what comes out. So, yeah, computers can do a lot better. But as I said, general relativity is hard. And part of the hardness is just the complexity of the notations that are necessary. Incidentally, the subject that we're studying up till now is Riemannian geometry. Riemannian geometry, the geometry of curved spaces, the geometry of smooth, curved metric spaces, they're called. And I suppose we have Riemann to blame for the complexity of the notation. Actually, he simplified it and Einstein simplified it further. But nevertheless, it is notationally and computationally intensive. The basic concepts are really not so hard, but we have to go through the mess before we get to the concepts. Or before we can express those concepts. If there are no questions, yeah. I know very well, yeah. Well, yeah, there are several. The most active areas of research, the ones which are really very, very strongly active among theoretical physicists, is the quantization of general relativity, the quantization of black hole physics. That's probably the most active area. There are astrophysical areas where, again, black holes are very important. Astrophysical areas are entirely classical general relativity. There's numerical relativity. And usually for the purposes of solving astrophysical problems, numerically putting in various initial situations, a pair of black holes orbiting around each other, and simply running the computer to find out what happens by solving Einstein's equations. And it's a very subtle business, incidentally. It is not that you just put in the equations and grind. It's much more subtle than that. And numerical relativity is also a very hard subject. The purpose, for example, being to find out how much gravitational radiation and other kinds of things is entirely classical. Gravitational radiation is classical radiation when two black holes fall into each other, bang. A lot of gravitational radiation is produced, gravitational waves. Why are we interested in them? Well, among other reasons, because we have a chance of detecting them, anything you can detect is a window into phenomena that you've done a new phenomena. So gravitational wave detectors may be able to detect gravitational waves from the collision of two black holes. It is estimated that there are a few collisions every year of black holes that can be seen by gravitational radiation. So there's a lot of different areas. It's an extremely active area. The equations of general relativity, just the equations of general relativity, and the equations of black holes have been found to be useful, extremely useful, and condensed matter physics and fluid dynamics for having nothing to do with black holes and gravity. Same equations. And so there's a heavy amount of activity in the areas of applying the equations of general relativity to other areas of physics. It's probably, I would say, probably the most active. It went over the last, over my career in physics from being a real backwater in physics. Nobody studied general relativity very much in the 1960s when I was a student, except for the people who were really especially interested and were focused on questions of general relativity. The bulk of theoretical physicists were not interested. They just started to become interesting in astrophysics, and eventually it took over questions of fundamental theoretical physics. Why? Because of a clash of gravity in general relativity. Anyway, for whatever reason, it is probably now more important than the day-to-day workings of theoretical physicists than almost any other subject, that and quantum field theory, and the connection between them. Okay, let's, today will be the last day in which we'll be studying Riemannian geometry as such without really discussing gravity. Next time we will really get into gravity. What does all of this have to do with gravity? I told you a little bit in the first lecture that the problem of finding out if there's a real gravitational field as opposed to just some funny coordinates in which you're expressing space and time, which leads to curved trajectories, but not really curved trajectories. The problem of finding out there really is a gravitational field, is a problem which is mathematically, I would say very similar, but I think I would go further than that. I would say is identical to the problem of finding out if a certain metric geometry, geometry characterized by a metric tensor, is flat or not. A geometry which is flat, let's say a two-dimensional geometry to begin with, that means a surface of some sort. A flat geometry is one for which Euclid's postulates are correct. There are parallel lines, all the stuff you learned in Euclidean geometry is called flat. And it's characterized in terms of a metric tensor by the statement not that the metric has a very special simple form. That's not the right statement, the special form being the chronic or delta. That's not the right statement that the metric has that form. The right statement is that coordinates can be found such that with those coordinates, the metric has a simple flat form. But more generally, just given an arbitrary metric tensor which varies from place to place, assume that it's smooth, assume that it has all good differential properties and so forth, the problem of determining if that space is flat or not is a somewhat difficult problem. What you have to do is look at, here's the bad way to do the problem. The bad way to do the problem is to search through all possible coordinate systems, transform the metric from the given form, however it's given, the way that it's presented to you, transform the metric to the new coordinate system that you have constructed, calculate the new metric and ask whether it looks like a flat metric. Now, of course, that will take an infinite amount of time, you'll never get there, so you need a better technique. And the better technique is to search for diagnostic quantities, quantities which are built out of the metric, and its derivatives and so forth, things that you can calculate which, if they are zero everywhere throughout the space, that says the space is flat, if they're nonzero any place, it says that the space is curved there. The diagnostic quantity that does that is called the curvature tensor. Today's goal is to get to the curvature tensor of a Riemannian geometry. And it's a complicated goal, I warn you, you've got to pay attention or else you'll fall asleep. This is a very superific subject. I'll try from time to time to tell a joke or whatever to keep you awake. But on the other hand, if you can follow it, you've learned a lot and it is very interesting. I think you'll find it very interesting. I'll try not to make it too boring. Okay. I told my friend Barack Obama, never apologize, and here I am apologizing already for being boring. Okay. What do we start with? We start with a space, and a space means, first of all, a number of dimensions. The number of dimensions in Riemannian geometry can be anything. In principle, you can even have a zero dimensional space, but that's pretty damn boring. It's just a point. A one dimensional space is either an infinite line or a closed curve. If it's a closed curve, what is it characterized by? It's characterized by only one thing, the total length of the curve. Every curve is equivalent to every other curve of the same length. In other words, just think about it for a moment. Take a piece of rope that closes on itself, it has a certain length, and wiggle it in any kinds of ways, or curve it in any kind of ways. It can always be mapped or put on top of another piece of rope of exactly the same length in a precise way without stretching it. One dimensional spaces are all equivalent to each other, not quite. They're characterized only by their length. They're either infinite, or they have a length L, and every one of them have given length L is identical to every other one in its intrinsic properties. A bug walking along a little piece of rope cannot distinguish one geometry from the other geometry. All he can do is count the number of steps that it takes him to go around the rope. He leaves a marker someplace, goes around the rope until he comes back to the marker, counts the number of steps, and that's the only thing about that space. Two dimensional spaces begin to become more complicated. There are flat ones. There are curved ones. Flat ones are plain. Curved ones could be a sphere. It could be just a space with lumps, the surface of the earth, including the mountains and the valleys and so forth. It can even have weird topology, the surface of a donut, a torus. Or you can poke another hole in the torus and make a torus with two holes and so forth. So by the time you get to two dimensional spaces, they're pretty complicated. Nevertheless, they are characterized by the metric tensor. And the question of whether they're flat is a computation, as a first question, find something which distinguishes whether they're flat. Find a quantity g mu nu, g mu nu g m n, the metric tensor. If it is not equal to zero, that does not mean the space is not flat. It just means you're maybe in the wrong coordinates. So you can't diagnose the space by asking whether g m n is zero or not. You want to find a quantity. Let us call it r. At the moment, I'm not going to tell you how many indices it has. Some quantity. But as best as should be a tensor, because if it's a tensor and it's nonzero, it's nonzero in every frame of reference. We like to find some quantity called r. r stands for Riemann incidentally. Riemann was the first person who asked this question and answered it. We want to find a thing called Riemann such that if Riemann is equal to zero, that implies that the space is flat and that there exist coordinates such that, sorry, did I say g m n not equal to zero? I mean not equal to delta m n. g m n is an interesting space. It's not very big. Yeah, g m n equal delta m n is what I meant. And that's what we're going to search for. The curvature tensor. It is a tensor. Let me just remind you about a few properties of the metric tensor and what you do with it. First of all, g m n is by definition the thing that's a function of position in general. It was not a function of position incidentally. If it was a constant matrix, a constant set of coefficients, then it's very easy to find coordinates in which g m n is delta m n. You simply have to straighten out the angles. So if g m n is not dependent on position, if it's just a set of coefficients, a set of numerical coefficients, then the space is definitely flat. So g m n of x is the metric tensor and what do you do with it? You take a little differential distance, d x m, multiply it by d x n. Here's your little differential displacement. It's characterized by a set of differential coordinate shifts, d x m, d x n. You multiply them together and that gives you the square of the length between the tail and the arrow and the head of the vector. That's equal to d s squared, s standing for spatial distance, I suppose. All right, now, the metric tensor is a tensor. I think we proved that last time and the way you see that is just by saying d s squared, that's an invariant quantity. All coordinate, let's call them coordinate systems, all coordinate systems should agree about the length of a given little vector and if we take that same d s squared and write it in y coordinates instead of x coordinates, the length has to be the same and that's sufficient to prove that g m n really is a tensor and that it transforms like a tensor. Okay, next property of g m n is it has an inverse. g m n is a matrix. It's a matrix and it has an inverse. The inverse satisfies the following equation. g, it's written again as a matrix. When I say it has an inverse, I mean it has an inverse as a matrix. g m n is a matrix with entries and it has an inverse. g m n, which is written with upper indices, they are in fact, contravariant indices and the properties of g m n with upper indices are such that if you multiply it by g, let's call it n r and sum over n, but I don't have to sum over n because Einstein tells me that an upper and a lower index like this are automatically to be summed over. This is the sum. This actually corresponds to taking the matrix g m n and multiplying it by the matrix g n r and the resulting matrix is the unit matrix. Why? Because I told you that g m n is the inverse matrix to this one here and the product then is the identity matrix and the identity matrix is the Kronecker delta and Kronecker delta in general relativity or in Riemannian geometry is always written as a thing with one upper index and one lower index. At least if it's being treated as a tensor, the Kronecker delta, this is a thing to prove, prove that if you take the Kronecker delta and you transform it pretending it's a tensor, you just get back the Kronecker delta again. That's the thing that you can prove. So these are tensor equations. This is the meaning of saying the g m n is the inverse matrix of g n r. And once you have that, oh, there's another thing that is important. In Riemannian geometry, as opposed to Einsteinian geometry, in Riemannian geometry all lengths are positive. All lengths are positive, squares of all lengths. More important, squares of all lengths are positive and all lengths are also positive. But the square of all lengths are positive and that says something about the properties of the metric tensor. What it says is that there are no directions along which if you evaluate this you get anything which is not positive. That's equivalent to a statement about the eigenvalues of g m n. It says all of the eigenvalues of it are positive. But for our purposes now, all lengths are positive. It's the same thing. And it does say something about the metric tensor. It says it's positive definite. That's the same thing as saying all of its eigenvalues are positive. That will change when we go to the theory of real gravity. That will go when we go to Einsteinian geometry. All right, the metric allows you to do something. It allows you to raise and lower indices. If you have a contravariant vector, or better yet, you have a vector which has contravariant components v m, contravariant means upstairs. And I put it into correspondence with another vector with covariant indices. And it's really the same vector. We talked about this a little last time. I told you the difference between contravariant and covariant indices. And the way you construct the covariant components, components of the words I'm looking for, the covariant components of a vector giving its contravariant components is just to multiply it by the metric tensor g m r. And, whoops, let's call it g m n, and sum over m. This is now g sub n, or v sub n. Likewise, this is something that you need to prove. Given a v m with upper indices, this, if you like, can be thought of as the definition of the lower of the covariant component of a vector. How do you go back? Well, you have to solve this equation, but it's very easy to solve, and I'll tell you what the solution is if you know the inverse matrix with upper indices. Namely, v m is equal to g m n v n. So if you know the upper indices, if you know the covariant, the contravariant components of a vector, this is the way you build its covariant components, and if you know the covariant components, this is the way you build the contravariant components. There's a little bit of a proof here that you have to do. So, if you have a question, you can ask me. You can see what we have to do. You'll have to check that these two are compatible with each other, but they aren't. Okay. To what extent can you force the metric to look like a flat metric? To what extent? Assume the metric is not necessarily a flat metric. I will mean not only is the space flat, but that the coordinates are such that g m n is equal to delta m n. If that's the case, if that's the case, this space is definitely flat. But to what extent, given any space, can you find coordinates such that g m n might be, for example, the karmic of delta over some limited region, or approximately over some limited region equal to delta m n? All right, so here's a theorem. There are coordinates at every point. These coordinates are coordinates which are good in the immediate vicinity of a point. They're not global coordinates. They don't cover the whole space. They're just good within the immediate vicinity of a point, which means they're good up to some order in little displacements away from this point, namely second order. All right. Coordinates called Gaussian normal coordinates. Here's what you do. Just to first think of it intuitively, we have some curved surface. We go to some specific point, and we draw coordinate axes. We draw coordinate axes, and we keep those coordinate axes as straight as we possibly can. Later we'll learn that the term for straight as possibly can curve is a geodesic, but you make a straight curve as you can. You have a little bug. He moves on this surface, and he moves following his nose going as straight ahead as he possibly can, and that defines one coordinate axis, then he comes back to a point and goes off in right angles, does some surveying to figure out where right angles to his little line are, and goes off in another direction as straight as he can, and then builds coordinates based on those... The theorem says that first of all, at every point, at every point you can choose coordinates, as I said, Gaussian normal coordinates, such that, first of all, at that point, gmn is equal to delta mn at x equals x0. This point is x0. You can do that in more than one way. If you find coordinates in which it's true, you can rotate the coordinates. You can rotate the coordinates, and if you rotate them, that just corresponds to rotating the coordinates in different axes and so forth. So there's more than one way you can do it. Next, you can choose your coordinates. You can straighten out these lines to a degree where the derivative of the metric with respect to any direction... Well, let's put an xr here. Any direction, the derivative of any component of the metric with respect to any direction in space can be said equal to zero. That you do by choosing these curves to be as straight as possible. I'll tell you how you prove it in a minute. It's actually very easy to prove. It's just a counting problem of counting how many variables there are and how many equations there are, but I'll show you in a minute. Okay, finally... Yeah? Say it again? No, only at x equals zero. At... Sorry, not at x equals zero. Could be x equals zero. At x equals x zero. If you try to extend that same coordinates away from that point, no matter how you try to extend them, then in general, unless the space is flat, if you try to extend the coordinates away from that point, you'll find out that the derivatives, the higher derivatives of gmn are not equal to zero, and in general, for the general generic situation, the second derivatives, let's say with respect to some XR and some other XS, those in general cannot all be chosen equal to zero, except, of course, if the space is flat. Okay? So, at a point... At a point, there is no content, really, in saying that the metric can be chosen to be flat-like, let's call this flat-like, even up to first derivatives that can always be done. It's in these second derivatives that somehow the fact that the space is either flat or not flat starts to show up. All right, first of all, how do you prove this? Well, this is actually not hard. Let's take our point of interest to be the origin. Let's just take it to be the origin. Let's set it equal to zero. Now, supposing we have some general metric and some coordinates Y in which the metric has some form, not of this form here. Let's look for some XS, which we're going to build, which are going to be functions of the Y. Let's choose it in the following way. At the place where X is equal to zero, in other words, at our origin over here, let's also assume that Y is equal to zero. So, the two sets of coordinates have the same origin. That means that X will start out... We're just equal to Y. XM equal to YM. And then, in the next order, we'll have some set of coefficients. I'm simply expanding X in powers of Y. The next order will be something quadratic in Y. So, let's put something quadratic in Y here. We'll call that CM. And let's put here N and R and put XNXR. That's the most general quadratic thing that I can write. Oops. Sorry, Ys. YN, YR. And then, things may get more complicated later. Okay. How many such... Let's pick a number of dimensions of space. Let's pick a number of dimensions of space. Let's pick four dimensions of space. How many such coefficients are there? Well, how many combinations do we have in Y in YR? Are there? The answer is ten. Try counting them. There's Y1 squared, Y2 squared, Y3 squared, Y4 squared, Y1, Y2, Y1, Y3, Y1, Y4, Y2, Y3, Y2, Y4, and Y3, Y4. Did I miss any? I think I have every possible thing here, which is quadratic in Y. And there's one, two, three, four, five, six, seven, eight, ten, ten of them. So there are ten possibilities here. How many components of the metric are there? There are ten YN, YRs. So that means N and R give you ten possibilities. But then for each N and R, you have an M. That means there are 40 independent coefficients here. 40 independent coefficients. The most general X that you can build to quadratic order has 40 independent coefficients here. Now, how many independent components of G are there? The answer is ten. So how many equations does this correspond to? Ten components. And setting all of their derivatives equal to zero means that this is 40 equations. 40 equations, 40 unknowns. And that allows you to be sure that you can solve these equations to quadratic order. Solving them to quadratic order is saying the same thing that you can solve these equations, but you'll fail at the next order. You'll fail at the next order, and here's the next order. So it is always possible to choose a metric that has the property that to quadratic order, in other words, the first derivative of the metric is equal to zero and the metric is equal to a flat metric at a point. Excuse me, I'm making something crucial here. It's very beginning. We have that equation with X and normal left and Y is on the right. Does that represent an attested coordinate transformation? Yeah, that's a coordinate transformation. Sure. Yeah, that's a coordinate transformation. Yeah, this is a coordinate transformation which we're going to try to choose these C's in such a way as to make the derivatives of the metric zero. You can always do it. So every metric in that sense is flat over a sufficiently small distance. Is that the same as it's locally flat? That's saying it's locally flat. That's the statement that it's local. That's what it being locally flat means, yeah. Intuitively, it just says that any smooth surface can be approximated over a small distance by a plane that's tangent to it. Right, that's what that says. Mm-hmm. That can be solved. That can be put, Gmn is invertible. Yeah. You can't always solve for the equation necessarily. Say it again? You can't always solve for the equation before the unknowns, right? Yeah, I think in this case you can. I think in this case you can. Yeah. Yeah. Yeah, in this case you can. There's, try it out. Do it with two by two matrices. Do it in two dimensional case. And there's nothing special about four dimensions. You'll find you can do it. All right, I don't want to try to do it on the blackboard. It's, all right, in any case it's a theorem. It's a theorem that Gaussian normal coordinates exist at every point, at every point, and they have this property. All right, now the reason we introduce that is because we want to give some meaning to derivatives of vectors. We've talked about some, of tensors in general, derivatives of tensors. We've talked about various procedures, various operations we can do on tensors to make new tensors. We can add them, we can subtract them, we can multiply them. We can do all kinds of things. Can we differentiate them? In particular, if we have a tensor which is a function of position, that means all of its components are functions of position, can we construct another tensor which is in some appropriate sense the derivative with respect to position of the tensor that we have? So let me show you why this is a little bit tricky. You could just say, well, okay, take the components of the tensor, let's say covariant components, and differentiate them, and that gives you a new set of components, which are the derivatives of the components. That gives you the derivative of the vector. It's true you can do that, but what you get will not be a tensor. Okay? And let me show you why. Yeah. The equation you have is C. There are no first order terms. This is first order. You mean zero-th order? Yeah. Yeah, that's right. It's not that there are no zero-th order terms, it's just I said let's choose the origins of the two coordinate systems to be at the same point. So that says wherever y is equal to zero, x is equal to zero. It was arbitrary, but it doesn't cost anything. Okay. All right, let's look at a vector field, a vector field, a vector at every position, and let's start with coordinates which either the space is flat, in which case I'll just use flat ordinary Cartesian coordinates, or the space is curved, in which case these coordinates here represent Gaussian normal coordinates, as straight as possible, so to speak. And now let's take another coordinate system and superimpose it on top of that. These are not Gaussian normal coordinates. They may be the coordinates that we started with. I'll represent those in green. And they're curved relative to these coordinates. For example, the coordinate lines might look like that, one of the coordinate lines, the other one might look like this, or whatever. All right, now, before we ask how we differentiate a vector field, vector field means a vector in every position here, let's ask what it means, what it means for the vector field to be constant in space. The problem here is because the space is curved, it becomes hard to compare the vector at one point and another point. The coordinates, if they can't be chosen to be everywhere flat, then what exactly do you mean by saying a vector over here is equal to a vector over here? It doesn't really mean anything, because to compare a vector over here with a vector over here, unless you have nice flat coordinates, there's no unique way to do that. So let's look at it for a moment. If the space is flat, really flat, then I think I know what it means for a vector over here to be the same or to be equal to a vector over here. It means it's equal, it points in the same direction and has the same length. On the other hand, does it say that I'm thinking about the green coordinates now? Let's think about the green coordinates. In the black coordinates, the components of this black vector and this black vector are clearly the same in the nice Cartesian coordinates. But what about the coordinates of this vector here and this vector over here along the green axes? Let's take a special case. Let's take this vector to be vertically up so that we know that it only has a Y component in the flat coordinates. Let's assume that this vector over here is equal to this vector it points in the same direction in flat space. Are its coordinates along the green axes, are they the same as the coordinates along the black axes? No. This coordinate along the green axis here is different than the coordinate along the black axis here. And in fact, this vector would have both X and Y components along the green axes, whatever the green axes happen to be, has both kinds of components. So it's clear that even though the vectors are the same, the components are not the same. So you cannot judge when you have curvy linear coordinates and in general, you will have curvy linear coordinates. You can't judge by whether the components, either covariant or contravariant, you can't judge whether the vector is constant in space. Another way to say the same thing is that in one coordinate system, the derivatives, let's say with respect to some coordinate XR of the mth component of a vector, might be zero in one coordinate system and not zero in another coordinate system. It may even be the case that all of the derivatives of Vm are zero in one coordinate system and not zero in the other coordinate system because the coordinates shift, because the coordinate direction, not because the vector is changing, but because the coordinates are changing. So this statement that the derivatives of a vector are equal to zero is not an equation which is likely to be the same in every reference frame. Therefore, it cannot be a tensor equation. It's not a tensor equation. And the derivative of a vector, the components of a vector with respect to coordinates, are not themselves components of a tensor. They're not components of a tensor because they can be zero all of them in one frame of reference and not in another frame of reference. If they were components of a tensor, we would think of this as a rank 2 tensor, an m index and an r index. We might say, look, could this be a tensor Tmr? Well, if it's a tensor Tmr, then if it's zero in one frame or in one coordinate system, it's zero in every coordinate system. But that's simply not true of the derivatives of components of vectors, not because the vector may change from point to point, but because the coordinates, the representation of the coordinates change. So we need a better definition, we need a better definition of the derivative of a vector than just differentiating its components. We need something which, if it's zero in one frame, is zero in every frame. Okay, so here's what we'll do. Here's how we'll define the derivative of a vector. First of all, to define the derivative at a point, you only need to look at nearby points, right? You only need to look at nearby points. So the first thing you'll do is to construct Gaussian normal coordinates near that point. Gaussian normal coordinates, they're as straight as possible. Then, let's put this vector a little bit closer just to indicate that it's all like that. Then, pretending that the Gaussian normal coordinates are really nice flat coordinates, we shift this vector so that its origin is the same as the starting vector over here. In other words, shift it, keeping its components fixed and constant in Gaussian normal coordinates. So in the Gaussian normal coordinates, I keep the components of this vector fixed, and I compare it with this vector over here. Just shift it down to here, and the difference between these two is the thing that I will use to define the derivative. The derivative will be the difference between these two divided by the little separation along this axis here. In other words, to differentiate a vector, differentiate it in Gaussian normal coordinates. They are the ones which are closest of all to Cartesian coordinates. The derivatives in the Gaussian normal coordinates define the derivative. Now, when you differentiate a vector in general coordinates, you'll get two terms. The two terms come, one term comes because the vector may be changing, and the other term comes because the coordinates are changing. Coordinates are rotating out from under you, even if the vector doesn't change, and so forth. If you work out this prescription, take the derivative, just the ordinary derivative in Gaussian normal coordinates, and then re-express it in arbitrary coordinates. Here's what you will find. You will find that the derivative, which will now be called capital D along the R axis, of the components of a vector Vm in arbitrary coordinates are not just the derivative with respect to R of Vm, but there's another term, and the other term has to do with the way the coordinates themselves are changing from point to point. What it looks like is something that's proportional to V. If you double the size of V, you'll double this term here, obviously. So it's proportional to V. It will have a V in it. It does not have a derivative because this term doesn't come from the fact that the vector is changing. It's coming from the fact that the coordinates are changing. Let's put some indices and try to see what kind of thing it could be. It might have the T component here. On the left-hand side, we have indices R and M. That means we have to have some kind of object over here that has index R and M, just to match the indices on the left-hand side. On the other hand, it's going to be proportional to the components themselves of V, so it's going to have some VT here. This equation doesn't quite make sense. There are two directions here. There are three here. You've got to soak them up. I have to put another index into this object here. In fact, the standard convention is to put a minus sign there. This, in fact, is what you will get if you take a vector differentiated in Gaussian normal coordinates and then transform it to other coordinates. In any other coordinate system, what you will get is the derivative minus a thing, what's this called a thing for the moment, times components of V themselves. This is summed over T, and the index structure on the left-hand side is just R and M. That holds for an arbitrary coordinate system, right? That holds now for an arbitrary coordinate system, but I haven't told you what gammas are, what these gamma objects are. Question. What does that formula become if it's a Gaussian normal? No, in Gaussian normal coordinates, it's just this. Okay, so the gammas come in because of this? Yeah, the gammas, yeah. Right, we're going to find that the gammas are proportional to derivatives of the metric. They have to do with the metric changing from place to place. They're going to be proportional to the derivatives of the metric in Gaussian normal coordinates to derivatives of the metric of zero. Okay, that's the trick. Use Gaussian normal coordinates to differentiate because to first order the derivatives of the metric of zero. Once you've differentiated it, you can then transform the object as if it were a tensor to any frame of reference, and the form will always be like this. It will always be like this. It will contain indeed a derivative, but it will also contain another term which will not be differentiated, and it has to do with the, as I said, it has to do with the fact that not the vector is changing from point to point, but the fact that the coordinates are changing from point to point. These objects have various names. This is called a connection. It connects neighboring points and tells you how to differentiate something from one point to another. It tells you basically how to compare vectors at different points. It tells you how to compare vectors even though the coordinate system may be changing. This is called a covariant derivative, not because its index is downstairs. It's called a covariant derivative for, just because it's called a covariant derivative. It's called a covariant derivative, and it is a tensor. It's a tensor in that if it is zero in any frame, it's zero in every frame. Is that a generalization of the convective derivative? You know, I think it is. I never thought about it that way, but I think it is. What do you mean? Whether it's connected with a convective derivative, no doubt it is. No doubt it is. I would have to sit down and think about it, but there certainly are covariant derivatives in fluid flow. Right, okay, so, yeah, this is called a covariant derivative, and these symbols here go under two names usually. They're either called the connection coefficients or the Christoffel symbols. Christoffel. I, Christ. They've also been known on occasion as Christoffel symbols, and they are, they're not that complicated, but they're complicated enough. They're unlikeable enough. Okay, now, one fact about them that follows, I'm not going to prove every single thing I say, because they're just too many little pieces, but they're easy to check. It follows from the definition, namely the definition is you differentiate in Gaussian normal coordinates, and then you treat the object as a tensor and transform it as a tensor with two indices. It follows as a theorem, it's an easy theorem, but it's a little too intricate to do on the blackboard, that this Christoffel symbol has a symmetry. The symmetry of the Christoffel symbol is the gamma t, r, m, is equal to gamma t, m, r. It doesn't matter which order you put m and r, they're the same. That follows, there are generalized Riemannian geometries. They're called geometries with torsion. The symmetry is not true, but those geometries are not widely in use in ordinary gravitational theory. Ordinary gravitational theory is Riemannian geometry, it follows from the principles that I laid out, and gamma r, m is equal to gamma m, r. You know, So it's just the gamma that the Christoffel symbol, not gamma. The gamma, the gamma, the gamma. You know, doing the derivative in these Gaussian normal coordinates, which are almost flat, or as flat as can be, and then treating it as an object on its own right, it is very, very similar to what we do in gravitational theory when we evaluate something in a freely falling frame. And then, for example, in a freely falling frame, we calculated how light moved across an elevator. And then we transformed it to the frame of reference in which the elevator was accelerating. Okay. That's closely related to the operations we're doing here. We calculate something because we know how to do it in coordinates which are as flat as possible. That would be a freely falling frame in general relativity. And then we transform it to any coordinates we like, accelerated coordinates, or anything we like. And we translate the statement from one coordinate system to another. Here, the notion of variation of a vector from point to point is done first in Gaussian normal coordinates, and then it's transformed. And this is the form that you'll get for the corresponding tensor. We're not going very fast tonight, so I want to move along. All right. Now, for higher derivative tensors, we will need this tonight. Sorry, for higher rank tensors, supposing we have a tensor with more than one index, m, let's say, n, and we want to differentiate it, co-variantly differentiate it, again, in this case, along the s-axis. I don't know why I chose this instead of r. Here, I used r, but I want to differentiate it along the s-axis. Does everybody know what I mean d sub r here? That means derivative with respect to xr. Derivative with respect to xr. d sub r. That's a Gaussian normal. No, no, no, that's in any coordinate system. d sub r will always mean d by dxr. No, but I mean, in this equation, this is a Gaussian. Which question? The one here, you mean error routing. If you want to even talk about the rest of the equation. This one here? Yeah. No, that's in any coordinate system. In Gaussian normal coordinates, this term isn't there. In Gaussian normal coordinates, you just differentiate. Gaussian normal coordinates, everything is easy and ordinary. The derivative of a vector is just the derivative. In arbitrary coordinates, there's another term. You can work that out by transforming, just by doing transformations. Yeah? Good thing, when you say that in Gaussian normal coordinates, the polar derivative is just the derivative. That's only at the point. Yes, yes, yes, yes. Only at that point. Right. So, the question, are the Christoffel's coordinates, are those, is that a 2, 1 tensor or whatever? No, it is not a tensor. Not a tensor. This is intended to be a tensor. We're going to try to choose gamma so that this is a tensor. This we know is not a tensor, and therefore this cannot be a tensor. Right, no. Gammas are not tensors. In fact, they're built up out of derivatives of the metric. In a coordinate system in which the derivatives of the metric are zero, the Christoffel symbols are zero, but a tensor, if it's zero and one coordinate system, is zero in every coordinate system. So, they can't be tensors. And they're not. Okay. Let's just, before we go and figure out what the gammas are, we're going to work them out. They're not that hard to work out. There's a trick. Let's write this down. It's the analog of this, except for every index in the tensor, there is a term like this. There is a term for every index, and I'll show you how it works. There is a Christoffel term for every index. So, for example, we start out by hiding in. Let's hide in and pretend that this was just a vector. Then we would write that this is the derivative of t with respect to xs, that's the first term here, and then we would put a term gamma s m. Let's see what I used here. s m t t m. But now I expose the n here, and I put it in on both sides. Now I'm not finished. I have to do exactly the same thing, covering up m. Cover up m. I won't try to do it, but I have another term here where m is passive, n is active, and so we get gamma s s n t t m t. Look at the form of it for a minute. Oh, minus signs. Minus signs are wrong. Same as here. Minus signs as here. Notice if I covered up the second index, t m n. If I covered up every place the second index, it would look exactly like the covariant derivative of a vector. If I covered up the first index, it would look exactly like the covariant derivative of a vector. Say it again. You're right. This should be m t. Thank you. Yeah. Paralleling this over here. All right. Notice in one of the terms, it occurs in the first place, and in the second term it occurs in the second place. So that's the form that the covariant derivative of a tensor would take. The rule is the same. You go back to Gaussian normal coordinates, and you just differentiate the tensor, just derivative, none of this junk. And then you transform it to arbitrary coordinates, assuming that the result here is a tensor, and this is what you get. So it allows you to differentiate any tensor. At the moment, we're just dealing with tensors with covariant indices. We'll come to tensors with contravariant indices at some point. All right. In particular, let's apply this to the metric tensor itself. But there's something special about the metric tensor. In Gaussian normal coordinates, its derivatives are all zero. Yeah. I'm sure you must have said this one. It seems like a really fundamental thing. What's the covariant derivative for? It's for differentiating things. It's for comparing things at different points. What are we going to use it for? Field equations. Field equations. Field equations are going to be differential equations, which represent how fields change from one place to another. But we want them to be the same equations in every reference frame. We don't want to write down equations which are special to some peculiar reference frame. We want to write down equations which are general, so that if they're true in one frame, they'll be true in all frames. That means they have to be tensor equations, and that means we have to know how to differentiate tensors to get other tensors. So normally you have a derivative if you multiply it by corresponding to the decimal, you get the change. This is the partial derivative. This is the partial derivative with respect to the coordinate xs. I'm trying to throw me for the covariant derivative, but is there some differential you would multiply times the covariant? Yes, yes, absolutely, absolutely. Then what is that? dx. If you wanted to know how a thing changes from one point to a neighboring point, you would just multiply this by dxs. But is that what the partial derivative is? No, no, no, no, no. I say it again. The point is that the components of the vector may change even if the vector doesn't. Or that, right, we have to keep that in mind. So this is, right, we'll see some examples about how it works. This is representing effectively a distortion of the, an out of flat distortion of the coordinates. Yeah, that's right. It's an out of flat distortion of the coordinates. It's there even in flat space if you choose funny coordinates. That's an important point. It's there even in flat space if you choose funny coordinates. In fact, if you choose any coordinates in which the derivatives of the GMNs are not zero in which the coordinates vary from point to point in funny ways, this will be there. This is not really a feature necessarily of curved spaces. It's a feature of curved coordinates. Now, okay. So because the metric tensor is constant, or at least has zero derivative in Gaussian normal coordinates, and the definition of covariant derivative is to differentiate first and Gaussian normal coordinates and then take what you get and treat it as a tensor, it must mean that the covariant derivative of the metric tensor is zero. That's what's going to allow us to compute the Christoffel symbols. All right, so let's write down that equation. Here is the form for the different derivative, the covariant derivative of any tensor with two lower indices. So let's apply it to the metric tensor itself. dS of GMN is equal, first term, dGMN by dxS minus gamma TSMGTN minus gamma TSNGMT. All I've done is rewritten this equation, and now I'm going to add that the covariant derivative of the metric tensor is equal to zero. Why? Because the ordinary derivative is zero in Gaussian coordinates, in Gaussian normal coordinates. All right, now, let's write the same equation except permuting the indices, interchanging S and M and so forth. This is the equation for differentiating with respect to S of GMN by inspection, why am I doing this? I'm doing this because I want to get as much information out here about things like this as I can, and then try to see if there's any way that I can manipulate the equations to get to isolate and figure out what gamma is. So it's, you know, people who are experienced at this sort of thing look at it and say, I know what to do. If you don't know what to do, I'll tell you what to do. You do it once, and then when your friend comes and asks you, how do you find the Christoffel symbols, you'll say, I know what to do. I'm experienced. Okay, so you do the same thing except now I'm going to interchange S and M. Just permute S and M, and that means we're going to write down dM of GSN. How do I do it? I just do it by interchanging or replacing S by M and M by S, and that gives me derivative of GSN with respect to XM minus, now interchanging S and M here doesn't change anything because S and M, yeah, from the symmetry here, so this is the same thing, gamma SMT, GT, let's see. Yeah, we're just interchanging S and M, GTN, and then over here, so we're interchanging S and M, so this is gamma MNT GST, and of course for exactly the same reason, this is also equal to zero. And finally, there's one last possibility. Here, S is the odd man out, M and N go together. Here, M is the odd man out, S and N go together. And the last one is dN of GSM, and let's see, I think I'll just copy it down. It equals d by dxN of GSM minus gamma SMT, GTN, GTM minus gamma MNT GST, and that's equal to zero. Now you stare at this for a few minutes. Now how can I either air these, subtract these, or do something clever that will isolate only one of these terms with the gammas so that I don't have a horrible bunch of equations but basically only one equation? So I'll tell you what to do. Call this equation three, call this equation two, and call this equation one, and add equation three to two, and subtract one. And we'll see what happens. We'll see what happens. First of all, of course, we will get these derivatives. We'll get this derivative plus this derivative minus this derivative. So let's put them down on one side of the equation. I'm going to transpose all of this junk to the other side. OK, so on the left side of the equation we will get dN, dN GSM plus dN GSM, no, dN MG, the GSN M, that's this one, and the last one is minus GMN, sorry, derivative GMN. That's this plus this minus this, or this plus this minus this. Now, how about on the other side? Let's see, there may be some things which cancel. Let's hope this thing which cancel. This comes with a minus sign, this comes with a plus sign, this comes with a plus sign. So this and this are the same, but they cancel because of the opposite sign. No, listen, this are gone. They are gone, what else is gone? GTSN, ah, here it is, this one and this one come with opposite signs. So they are gone. We're left with two terms, but they're the same. So we're in luck. We can write that this is equal to twice, we need over to the right hand side, twice gamma MN T GST. Well, we still have, we're still not there, we would like to get gamma MN by itself. Trying to find out what the gamma, what the Christoffel symbol is all about, that's our goal, figure out what they are in terms of derivatives of the metric. We're almost there, first of all we can see that if all the derivatives of the metric are zero, the Christoffel symbol must be zero. But how are we going to get rid of this GST? And the answer is GST has an inverse. Basically, you can just take this over to the other side, but replace it by the inverse matrix. So first of all, let's put a one half. And now, put a bracket around this and take this GST over to the other side, except in its inverse form. And this is the correct formula. This is the formula for the Christoffel symbol. It looks pretty simple. We'll look at it for a moment. The M and N are symmetric. Look at it, M, N, if you want to change these two, and here it's symmetric again, has one negative term. Oops, yeah, one negative term, two positive terms, not very complicated. The problem is there's a boatload of them. When you think about four dimensional space and you let S and T and, sorry, and let all of these coefficients range over from one to four, there's just a lot of Christoffel symbols. And it's that which makes doing calculations in general relativity a very tedious business. Intrinsically, there's nothing hard about it. But in fact, doing a calculation in some sort of general relativity context usually fills page after page after page of nothing more complicated than just computing these derivatives and putting them together. Yeah. I'm sorry, I just want to say it again. G is constant, has zero derivatives in Gaussian normal coordinates. At that point. Oh, yeah, at that point. All of this is just at that one point. Yeah, but once it's true at one point, it's true at every point. To derive it at any point, you go to Gaussian normal coordinates at that point. So you say, I want a formula appropriate to a particular point. I go to that point, use Gaussian normal coordinates, and find out that this is what gamma is. But once you know it at any point, it's true at every point. So this is the connection between gamma and the derivatives of G at every point. The thing is that this is not a tensor. It's not a tensor, and so it can be zero in one frame of reference and not zero in others. For example, in Gaussian normal coordinates at the point, at that point, at that point, this is equal to zero, but in some other coordinate system it's not. Even more, take ordinary flat space. In flat space, you can find coordinates in which the G's are constant everywhere, in flat space. But you can also find other coordinates in which they're not constant. Polar coordinates, the G's are not constant. In Cartesian coordinates, they are constant. The Christoffel symbols can be nonzero even if the space is flat. In fact, they generally are. They are in polar coordinates. Yeah? Just to clarify, you can represent the metric that G has a 2 by 2 matrix in this... Not a 2 by 2, an n by n, or 4 by 4, or whatever it is. Whatever it is, but here you're on the... Well, whatever it is, it's 4 by 4, n by n. What is the Christoffel symbol, so it will also come right in? No, it's n by n by n. It's got three indices. n times n, now it's n by n by n, but there's some symmetry. So there's 40 components to it. It has 40 components. Everything always has 40 components. M and n, if you fix t, fix t, then gamma mn has 10 components. Same 10 components for a symmetric tensor. And then you let t run over all four components, and that gives you 40. So there are 40 independent Christoffel symbols. Excuse me. Yeah. I just want to make sure I understand what you said. I agree that you proved that at that one particular point, and it's true in all points, but for each point that you do it, you have to use a different G. Do you use the G that's appropriate for the Gaussian normal coordinates? No, no, no, no, no, no, no. You have some coordinates which are general coordinates all over the place. And I want to know how to differentiate at that point. So at that point, I go to Gaussian normal coordinates. I do my differentiating, and then I relate it back to the coordinates that are general coordinates which cover the whole space. At each point, at each point, this is what you do with this formula. No. You take your curved coordinates, whatever they are, and then you say, I want to know how to compare vectors at neighboring points. So I go to Gaussian normal coordinates at that point, and it tells me how to compare them at that point. Here's what it tells me to do. And those G's are the G's associated with Gaussian coordinates at that point? No. They're the G's that are associated with any coordinates. With Gaussian normal coordinates, these are zero. They're all zero at that point. The derivatives are all zero at that point. So no, these are the coordinates that you began with. You transformed the Gaussian coordinates, derived all of this, but then you finished with it. You finished with the Gaussian normal coordinates. Now you're finished with the Gaussian normal coordinates, and you go on. This was just a trick to, the Gaussian normal coordinates were just a trick to eliminate locally at each point the variation of the metric pencil with position. Right, good. Is this equation, the equation that tells you that a space is flat? It tells you what? That your space is flat? No. No, no, no, no, no. This does not tell you that the space is flat. This is just, which just tells you what gamma is. If gamma is, I assure you, if gamma is zero everywhere, it not only means the space is flat, but it also means you've chosen Cartesian coordinates. Didn't we get here by setting the derivative, the covariant derivative of the metric tensor equal to zero? And so that by itself is not enough to tell you the flight length. No, the covariant derivative of the metric tensor is always zero. That's not the same as saying the metric is constant over space, the metric tensor. The statement that the metric tensor itself is constant would be that the ordinary derivative is zero. Right, but the covariant derivative of the metric is always zero. Why? Because in Gaussian coordinates it's zero. And if a tensor is zero in one coordinate frame, it's zero in every coordinate frame. And one more thing, that covariant derivatives are tensors. So this is conceptually tricky. But at the end of the day, the rule is calculate your Christoffel symbols and replace ordinary derivative in many contexts, replace ordinary derivatives with covariant derivatives. You can write your equations in Gaussian normal coordinates, then they just involve ordinary derivatives. But if you want the same equations in general coordinates, then you replace the ordinary derivatives by covariant derivatives. That's the rule. Think about it. Look, this requires sitting down and thinking about it and understanding what we're doing. We have not got the curvature yet, so let's try to get to curvature. We have the definition of... Now, let's talk about what curvature is. It's easiest to start with two-dimensional curvature. How do we probe for curvature? Well, let's draw... Let's begin by drawing a space which is curved. This sphere is obviously curved, but I don't want to deal with a sphere. I want to deal with a cone, but not a cone, but a cone with a round point at the top. The top of a mountain, but with the sides being nice and flat. So that if you're away from the top of the volcano up here, I don't know, it's not a volcano, it's just a rounded cone, on the sides here it looks cone-like. At the top it's smoothed off, it's been shaved off and made round. The first statement is such a cone is flat, everywhere is except up near the top. To see that, just take the same space down here, but continue it so that it really does form a genuine cone. Okay, so now we have a cone. It only differs from the space that we're talking about up in this vicinity up here. Now, take the cone and slice it. Slice it down here and open it up. This is a cone, slice it down here and open it up. It's clearly flat. It's clearly a flat space, but it's a kind of question. Okay, open it up and what does it look like? It looks like, depending on the angle of the cone up here, it looks like a flat space, but when you cut it and open it up, there was a piece cut out of it. Missing, a missing piece. I don't have a scissor, I should have brought a scissor. I didn't bring a scissor. Take a piece of paper, cut it, cut out a wedge of a given angle, and then glue it together and you make a cone. The cone has a tip up here. The tip is clearly not flat, but the rest of the cone is flat. You cannot flatten this like a coffee filter. You can't flatten it down on the table without tearing it, but if you cut it anywhere, if you cut it and open it up, you can lay it out and make it flat. Scissor? Hmm? Scissor? Scissor. Oh, a scissor. A scissor, oh. Okay, but now I have to cut my notes. If you have a piece of paper, somebody have a piece of paper, you can do it. Alright, I've cut out a certain angle, and now when I sew them together or glue them together, in this case it made a kind of flat cone. I could wear it as a hat. Okay? If I cut out a bigger angle, I'll make a more acute angle at the top of the cone. For example, I can cut out a large angle. I cut out a larger angle, and now when I sew it together, it makes a pointy cone, except I can't really do it because, yeah, I think you get the idea though, huh? Yeah, it makes a pointier cone. So the bigger the angle that you think you, the bigger the angle that you cut out, the bigger the deficit angle, it's called the deficit angle or the conical deficit, the bigger the conical deficit, the pointy of the cone. If the conical deficit is a very small acute angle, it means the cone is very flat on top. Okay? All right, now let's think about, let's think about a vector, a vector field on the space here which is constant. Let's take a vector field which is constant. What does it mean to be constant? It means that it's pointing in the same direction everywhere, and for simplicity, let's take it to be pointing in this direction. Something funny over here, over here it's along this line, over here it's practically perpendicular to this line, but remember when the cut is identified like this, the line is the same line. Let's do it. Let's take a vector pointing over here, here, here, here, and here, and now when we sew it together, we discover that there's a funny jump in its direction. It's pointing along the cut here, and it's pointing in some funny direction here. Even though when you open it up, you see they're all in the same, did you see that or was I hiding it there? As you go around, starting here, pointing in the same direction until you get around to the other side of the cut, in which case you find that it's not in the same direction. As the arrow was over here. Nevertheless, when I open it up, they're all parallel. So the tip of the cone is such that if you carry a vector around, when you get around to the other side, it has been rotated. It has been rotated due to the conical deficit at the top. Exactly the same will be true. In this geometry over here, if I take a vector along a given direction and I carry it around the mountain in such a way that it's always pointing in the same direction if I opened it up, by the time I get back to the other side, it'll be pointing in some other direction. That is because the top of the cone is curved. That's the effect of curvature. The effect of curvature, a region of curvature has the property that if you take a vector pointing in the same direction as much as you can do, keep it pointing in the same direction and take it around the region of curvature, the vector will rotate. The vector will rotate when it goes around any region of curvature, even though you took pains to make sure that at every point you were moving it parallel to itself. In other words, you took pains to make sure that the covariant derivative of that vector was zero, but nevertheless when you came around to the other side, it had shifted. There's another way to say this, which is actually more useful. Here's the other way to say it. It's equivalent. It's equivalent. If you have a curved space in some curvature here, in some curvature here, and you take a vector field and differentiate it along one axis, we're going to move it and differentiate it, differentiate it along one axis, and then differentiate it along a second axis. We take the vector, displace it a little bit, and then displace it along some second axis. We will... the vector will change. And how will the vector change? It will change typically by differentiating the vector along these two axes in sequence. You first differentiate the vector along one axis, then differentiate it along the second axis, and there's a small change in the vector due to these two derivatives. So if I go along here, in other words, if I take the difference of the vector over here and over here, it consists of two changes, a change to go from here to here and a change to go from here to here. And that change, the total change in that vector is proportional to a second derivative. It would be true in any coordinates. It would be true in any coordinates that if you carry the vector from here to here and then from here to here and then compare it with the vector at this point over here, what you would be calculating is the second partial derivative of the vector, let's say derivative of Vn, call this the Xs direction and the Xr direction. Then this would be, let's see, which derivative would it be? It would be derivative with respect, it would be covariant derivative with respect to r of the covariant derivative with respect to s of the vector, and let's say the mth component of the vector. This operation says you move the vector a little bit along the s direction, and this one then says you take the difference, the little derivative, and move it along the r direction, and the net result is the change in the vector going from here to here. Done covariantly, which means that you could have done the same thing in Gaussian normal coordinates with just ordinary derivatives. In Gaussian normal coordinates, you would just be doing this derivative here, or two ordinary derivatives. You could have also gone in the other direction. You could, and that is to say, you could have gone first in the r direction and then in the s direction and calculated the way the vector changes from here to here. That would be the opposite order of differentiation. Ordinarily, and in flat space in general, these two are equal to each other. This is just the fact that derivatives, it doesn't matter which order you take them, but it's not true in curved space. In curved space, this net, the difference between these two operations can be thought of as taking the vector around the closed loop. The difference between these two operations, taking the vector from here to here and then from here to here, minus the opposite direction, is essentially taking the vector around the little closed loop. Taking it around the little closed loop, we discovered that when there's curvature at the tip of the cone here, you don't get back to what you started with. In curved space, covariant derivatives are not interchangeable. In flat space, they are interchangeable. That's the way we test for whether a space is flat or not. We test whether differentiating tensors, in particular vectors, in opposite order, give zero. If the answer is yes, everywhere is in the space. For any vector, then the space is flat. If we discover that there are places in the space where the order of differentiation gives different answers, then we know the space has some kind of defect in it, or it has curvature. It has curvature analogous to the top of the mountain. So all we have to do, and it's a little bit complicated, but it's not that complicated, is compute the second derivative of a vector, the second covariant derivative of a vector, in opposite order and compare them. I'll tell you what to do. You're going to do it. I'm not going to do it on the blackboard. It's too late. It's easy. It's really quite easy. It's mechanical. It's a mechanical operation. I'll tell you what the steps are, and I'll tell you the answer. And it's entirely mechanical. It's just pure plug-in. It's too late to carry it out. What we're interested in is taking a vector, the Vn, the components, covariant components of the vector, computing the covariant derivatives in the r direction, and then differentiating it in the s direction. Later on, we will interchange s and r and subtract. But let's just do this. This is equal to ds. This object over here is a tensor, trn. It's a tensor, and we have the rule for calculating the derivative of a tensor. We have a contribution from each index, and I'm going to write it down. I'm going to write down the answer. I'll write down the, let's see. ds of ds of dRVn minus gamma Rnt Vt. I'm just stuck in here the covariant derivative of Vn. So that's what's in this box here, just this thing over here. Now we have to differentiate it again using this formula. We have a two-indexed object, and let's see if I can, let's see. I'm going to leave it to you to calculate. Let's see. All right, I'll write down one further equation. It's ds of dRVn minus gamma Pn. Oh, brother. No, it's getting too late. I don't want to try to write it down. I'm going to leave it for you to do. Just pure plug-in. Pure plug-in. You plug-in for Tmn, this object over here. Put it right into there, or calculate the second derivative, and then compute the following. ds dR of Vn minus dR ds Vn. That corresponds to the difference between the two ways that you could get from here to here by going through this direction or that direction. I'll tell you what you get. You get something called R. It's a tensor with three covariant indices, one contravariant index, and it multiplies Vt, where this is pretty easy, except that R is a little bit complicated, and I'll tell you what R is. So, ds dR sRnT equals dR gamma snT minus ds gamma rnT gamma snP gamma prT minus, one more term, gamma rnP gamma pst. Two terms involving derivatives of the Christoffel symbols and two terms involving products of Christoffel symbols. There's a sum over p here and a sum over p here. The indices, take a look at it, the indices match. It's a complicated expression, and it's even more complicated when you remember what the gammas are. The gammas themselves are complicated derivatives of g's and the curvature tensor. This is the curvature tensor, the Riemann curvature tensor. This Riemann curvature tensor is itself another derivative and also products of Christoffel symbols. Let's see what it has in it. When you differentiate, the Christoffel symbol involves a derivative of g. Differentiating again gives you second derivatives of g. Remember, the second derivatives of g are the things that you cannot generally set equal to zero. The first derivatives of g, you can find frame of reference where they're equal to zero. The second derivatives of g, you can't. So by the time you finish calculating this thing here, second derivatives of g have come into it, and it is testing out and feeling out the geometry a little more thoroughly than the first derivatives would feel it out. It's feeling out the second derivatives, feeling out the coordinate systems to quadratic order. So it has second derivatives of Christoffel symbols, and it has squares or quadratic things involving first derivatives. It's a complicated thing, and if I were to actually write it in terms of the Christoffel symbols, or I were to try to calculate it for a given metric, it can fill up pages, and it will fill up pages. But conceptually, what it's doing is just calculating the difference in a vector if you transport it around here, keeping it parallel to itself until it comes around all the ways around. The little change in the vector in going in a loop. This is the curvature tensor. The curvature tensor, as I said, it's complicated to calculate, but you can calculate it. You can put it into a computer, and you can ask the computer, is it zero? In fact, if the computer, it's good if you have a computer that can do algebra, and if you have the metric in some algebraic form, or you have the metric in some mathematical form, you can do these operations and then test out whether this curvature tensor is zero everywhere. If the curvature tensor is zero everywhere, all components of it, the space is flat. Now, we'll study the curvature tensor a little more, but as I said, it's a complicated thing, but the main thing about it is that it tells you whether the space is flat, and if not, how unflatted it is. It is closely related to a quantity in gravitational physics. Can you guess what it is? It's a local quantity which tells you that the space is not flat. So it must be something telling you whether there's really a gravitational field present or not. Tidal forces? Tidal forces. Tidal forces. It's exactly related to tidal forces. Those things which in a gravitational field squeeze one way, stretch the other way, that's what these things are, and in fact, they really are the tidal forces. Tidal forces are represented by the curvature tensor. Now, you can sort of see something like that if you were to imagine, yeah, imagine you have, again, this kind of conical geometry. It doesn't have to be conical geometry, but something which is flat far away. Yeah, let's say it's flat far away. Nice and flat far away. And somewhere in the center it has a bulge so that it's curved. Now you take some object out here. It's a thing living in flat land. It's two-dimensional. You take some object that might be composed of a bunch of links, there might be a bunch of triangles connected together, links, and then you start moving it. As you move it in the flat region, nothing happens to it. It's perfectly happy. It doesn't get stretched. It doesn't get distorted. It doesn't get deformed. It doesn't get squeezed one way and stretched the other way. But now what happens when you try to move it into the curved region? When you try to move it into the curved region, it simply can't follow the curvature without having to stretch its links. It has to follow the metric properties of the curved space. And in particular, if you go around a little closed loop in this space of links here, what you're doing is sampling this double covariant derivative here. And what you're going to find out is that when you go around a little closed loop, some angle shifts from what it would have been in the flat space. The angle shift, the length shifts when you try to push this up into here. It gets stressed. It gets deformed. It gets stressed. It gets stretched and so forth. And the measure of how much it gets stressed locally is exactly this curvature tensor. This curvature tensor is telling you how the system gets deformed as it moves up into the region of curvature. That's why it's connected with tidal forces. So it really is an important property. If you're in a region where this curvature, you feel it. You feel it as a distortion. If you're in a region of a gravitational field, let's say a uniform gravitational field, uniform gravitational fields don't have curvature. And that's why in free fall, in a perfectly uniform gravitational field, you simply feel nothing. OK, so the next time we will enter into gravity land, we will talk about what you have to do to change from Riemannian geometry to Einstein geometry. And then we will study a particular example. The particular example is the Schwarzschild geometry. The Schwarzschild geometry is the geometry of a black hole or a star or any gravitating mass. And see what we learn. We've left out one subject that was in my notes for tonight. And it has to do with geodesics. But I guess we'll have to do that. But we'll do that after we start thinking about gravity, real gravity a little bit. OK, I wish I had a way to make this fewer indices, but I don't. Yeah. And the first question says that the differences between an accelerating object and an opposite gravitational field is not always the same. No. So the difference between looking at an object in freefall from an accelerated coordinate system or accelerated frame of reference, and a non-accelerated frame of reference, or looking at a system which has a genuine gravitational field. In other words, you can simulate what looks like a gravitational field for many purposes by just an accelerated frame of reference. But it won't cause tidal forces. If you have a real gravitational field made by real matter out there, then it produces tidal forces. If there are tidal forces, it means that there's something real going on. It's not just a coordinate transformation. So you can tell the difference between the accelerating reference frame and the gravitational field. Yeah. Except for the uniform field. Well, there are no uniform gravitational fields. The real gravitational field of the Earth looks like this. It's not uniform. It produces tidal forces. It varies from place to place. There's no such thing as a real uniform gravitational field. It's an approximation. An approximation near the surface of a very big planet over a relatively small solid angle, the gravitational field looks approximately uniform. Now, there's another thing to say about it. Tidal forces have much less effect on small objects than on large objects. You can see that also as true of curvature. If I have a very, very small object here, and I move it around, if it's much, much smaller than the scale on which things vary, then it won't experience any great deformation. If you have a large object and you move it around, it will experience deformation because different parts of it get stressed differently. It's the same thing. Tidal forces have very small effects on very small objects. A bacteria falling into the surface of the Earth would experience, essentially no, as long as it was free-forming, would experience no tidal force, what it's speaking about. Whereas the 2000-mile man would be crushed by the tidal force as he falls toward the Earth. So when you speak about the strength of tidal forces, you have to discuss how big an object you're talking about. The same thing is true here. When you speak about the effect of curvature on an object moving around in this space, again, it depends on the size of the object that you're talking about. And this correspondence is not accidental. It's literally the same thing. Do you have to calculate the curvature of the temperature at every point? Is it technically still an infinite problem? Yeah, it is. But very often you'll be presented with the metric tensor in some analytic form. There'll be even a formula for it. And with a formula, you can differentiate the formula and everything becomes some analytic function that you can calculate when you look at it. If it's zero, it's zero. You're right. If you were just given numerically the metric tensor, you were just given a bunch of numbers at every point, you know, a huge, big spreadsheet filled with loads and loads of numbers, you would have to check the curvature tensor at every point, and that is, you know, as complicated as anything else. So where this is useful, the situations where you know the metric in some analytic sense and which you can calculate the derivatives, subtract them, do all the things you have to do, and then look at the analytic answer and see if the analytic answer is zero. Yeah. Right. So in that sense, it's the problem is it's not as bad as having to look at every possible metric and see whether you can make it. You have to look at every possible metric at every possible point. That's this is this is this is better. Not as hard as every possible metric. Every possible coordinate system is what I meant. Every possible coordinate system at every possible point until you find one which expresses the metric as being flat. This is easier. You calculate this one thing and you compare it from point to point. Okay. For more, please visit us at stanford.edu.
this kind of tensor, that kind of tensor, physics. Why? Because of a clash of gravity in general relativity or in Riemannian geometry is as a tensor and transform it as a tensor with two formula. This is the formula for the Christoffel