 Okay, let's start. So in the last class, we discussed geodesic motion. In the last class, we discussed geodesic motion in an arbitrary space-time. And I'll just remind you of our main result. Once we define d by d tau, d by d tau is equal to dx mu by d tau times d mu. It's a definition of this covariant derivative with respect to tau along the path. Then our geodesic equation took the form d by d tau of dx mu by d tau is equal to zero. It's like Newton's law. In the absence of a force, acceleration is zero, except this acceleration is the covariant acceleration. Okay, we also saw that a particle here obeys the Hamilton Jacobi equation. The equation was g mu nu, d mu s d nu s is equal to m square. And we discussed how you can solve this equation. You remember the way of solving it is to find s is equal to s of x and alpha, x mu and alpha mu, where alpha mu are some integration constants, and then set del s by del alpha mu is equal to beta mu, where beta mu is some constant. This gives you as many equations as variables. So this was the way of solving for particle motion general relativity using the Hamilton Jacobi formulation. And then right at the end, in the last class, we also talked about the motion of massless waves. The motion of massless waves. Actually, our discussion was for massless waves, but we could equally have well-studied massive waves. We would come to exactly the same conclusion with the same Hamilton Jacobi equation. Okay, but in particular, we looked at a massless wave with something that obeyed the equation del squared phi is equal to zero. We said phi is equal to e to the power i psi. And then we found that in an appropriate limit, the limit in which the wavelength of a particle changed by a small fractional amount in one wavelength of propagation, we found that the psi obeyed the equation del psi, the whole thing squared is equal to zero. So very much like the Hamilton Jacobi equation, but with m squared set equals zero. And then we talked about how we would solve for wave packet motion in this problem. We would take, suppose we had a deep parameter set of solutions, psi of x mu alpha mu. What we would do is to take d alpha mu times some wave packet g alpha mu times e to the power i psi x mu alpha mu. And then we understood that the center of the wave packet, the center of the wave packet moved according to the equation del psi by del alpha mu is equal to some constant. Where this constant was the derivative of the phase of this g function with respect to alpha mu. Okay, so we got the same motion, same solution. We had to solve in both cases in wave motion and in Hamilton Jacobi motion. We had to solve the same equation. And in both cases we had to take that solution and do the same thing in order to deduce our particle moves. Okay, so this is a very clear demonstration. We try to do, we try to say some of this in our classical mechanics course. But we didn't have the time to say it as well as we have the time now. Okay, so this is a very clear demonstration of how wave motion reduces to particle like motion. Okay, and makes it clear how the, what the relationship between the Hamilton Jacobi formulation of classical mechanics and the Schrodinger formulation of quantum mechanics or wave motion is. Okay, it also makes it clear that a photon must satisfy the Geodc equation but in a massless way. Okay, and right at the end of the last class, well at some point in the last class we, we talked about the equation that the photon must satisfy. It was d by d, d by d let's say lambda of dx mu by d lambda is equal to zero. Where lambda is some appropriate parameterization of the path of the photon path and we're going to talk more about that now. Okay, so just to finish this discussion, what I want to do is to think or, is to have a way of thinking of, have a way of describing massless motion directly from the action for, for a particle moving so that we can think of massless motion as a limit of mass motion. Now in the action that we used that was not, that was not possible because the action was proportional to m. You said m to zero you just have no action. So we want to do something better and we're going to use a trick that by the way is, apart from the result, now we're pretty, you know, we've convinced ourselves of this result in one other way but apart from the result we're going to use this trick because it's a useful trick that comes up many times. Okay, so the action for massive motion was m square root of g, sorry, square root of g mu nu dx mu by d lambda dx nu by d, with a minus. Now I'm going to try to rewrite this action in a way that admits a reasonable m goes to zero limit. So let me try. What I'm going to do is the following. I'm going to both rewrite this action that, allowing for a reasonable m goes to zero limit also simplifies this action. And what I'm going to do is the following. Look, I'm going to say consider the following action. Consider the action g mu nu dx mu by d lambda dx nu by d lambda and then one by e of, you want to call this e, okay, e of lambda, which I don't know yet, times e of lambda d lambda. Then I say consider this action. Now what is this e of lambda? No, there's no root over anything. Okay, I may have to fix this a little bit but not with roots. That'll be part of the point that will produce the roots from dynamics. e of lambda is a new field I've added to the action. Okay, so this action at the moment has nothing to do with this action. It's just a new field that I've added to the action. Okay, now I'm going to ask, what are the equations of motion for this new, for this new action? So let me compute the equation of motion. Let's call this guy, this number that I don't yet know. Let's call it a. Okay, let me compute the equations of motion for this field e. Notice that the field e I've introduced without any derivatives. It's a Lagrange multiplier field. It's not really a dynamical field. Okay, the equations of motion with respect to e will give you algebraic equations. So what equations do we get? The equation we get is g mu nu dx mu by d lambda dx mu by d lambda 1 by e squared. Now this came with a minus sign but I'll take away the minus sign by putting the other term on the right hand side. It's equal to a. It's equal to a. This equation does is it gives us a solution for e. e squared is determined in terms of, is determined in terms of the, is determined in terms of the, of x as a function of lambda just by this or e is equal to square root of g mu nu dx mu by d lambda dx nu by d lambda. Now since this is just a Lagrange multiplier field. Okay, we can find the effective motion for the x field just by plugging in the solution for what e is. So let's do that. So what is the action for this, this whole thing become? I'll put also b. What is the action for this whole thing become when I just plug in the solution? Oh and I've forgotten one by a. Thank you. What is the action for this whole thing become when I just plug this in? Well I get s is equal to b times, now I get this square root business because square root cancels square root but I've got a square root a here. Okay, plus and this a and square root a cancel each other. I'll assume a is positive. So once again I get square root a times this square root business. Square root is just d mu nu. I mean this square root we have here. You are in this case because it's just a Lagrange multiplier. We can do it without that substitution if you wanted to. But you see what we're doing, suppose imagine for a moment that you were doing quantum mechanics. Okay, do the integral over e of lambda because there are no derivatives. The integral will just set e to be whatever according to the solution and you'll be left with a path integral over x of lambda with the substitution put in. And you can just substitute that because it's totally non-local. Yes, yes. And in quantum mechanics there could be a major factor that you have to worry about in doing that integral but classically you can forget. We can do this more, we can do it without it. It's just a bit easier. I'm saying suppose we were, I'm giving you a quick argument that it's correct using quantum language. You see, so suppose you were doing the path integral of quantum mechanics. You have to do the path integral over e and the path integral over x. Now because this has no derivatives, the path integral is over e is point by point. Over e is point by point. There are no derivatives in e. So path integral is an integral over each point in the path. You do that integral point by point. It's just doing an integral at each point. Okay, that gives you something. Now the thing that gives you the classical limit is just a substitution. All the rest is major factors that we ignore from quantum mechanics. And then you do the path. This is an unnecessarily complicated argument. We could say it in classical terms. But it's okay in this case. You can check it without doing this. In fact, we will find the equation of motion for x with this action and show you that it gives you the right equation of motion. I'll do that soon. It's fine. Okay, so now what we've got here is equal to b times 2b times square root a times this thing. Now I want to write this action in a way that it doesn't become singular and the limit m goes to 0. Okay, so I've got some freedom. I'll choose this freedom to, I don't know, let's say for instance that b equals half and a is equal to m squared, b equals minus half and a is equal to m squared. Okay, now this gives us the action m times square root with the minus sign. Okay, so what we've concluded is that s is equal to minus half into g mu nu dx mu by d lambda dx nu by d lambda plus m over 1 by e plus m squared e d lambda. Now this thing is the same dynamical system as the original action. Yes. No, yeah, it's not unique. You see what we're doing, trying to do is to produce another way of writing the same classical dynamical system. Any solution of this action will be a solution of this action. Okay, there could be many other ways of doing it. We're trying to find some useful way of writing the dynamics in a way that will allow us to take the m goes to 0 limit conveniently. What? How did someone write this action? Well, you know, the way you think of this actually is the following. Look, the original action was very carefully chosen to be reparameterization invariant under lambda goes to f of lambda. Now that requires the square root. Okay, the square root is unpleasant. We want to get rid of it. So we want to write something more natural looking like this, but this is not reparameterization invariant because there are two d by d lambdas here, but just one d lambdas. So what? Now, normally, in ordinary, if you were writing down the propagation of a scalar field not in one dimension along the world line, but in higher dimensions, you have d by d x mutes, two factors. And how do you make that reparameterization invariant? By putting a factor of inverse metric on your field theory. So what you do is to imagine that you've got an intrinsic metric on the world line of the particle. Okay? That metric is square root of e. Then the, sorry, that metric is e. Then the inverse of the metric is 1 by e. Okay? And then this is a cosmological constant term. This is a cosmological constant term in this thing. You see, so if we want this whole thing to be reparameterization invariant, this would be, if you assign appropriate reparameterization transformation properties to the z. So what I'm saying is that suppose under, no, I got this wrong actually. This, the metric, sorry, let me say this correctly. The metric is actually equal to, you know, what you would put here is g inverse times square root of g to make the whole thing reparameterization invariant. So g inverse times square root of g on the world line of the particle should be identified with 1 by e, which tells you that g is equal to e square, because then this would be e times 1 by e square, which is 1 by e. Okay? So the interpretation of this e here is that e square is the metric on the world line of the particle. Okay? And then this factor here is square root of g. So it's a cosmological constant kind of term on the world line of the particle. If these words are not making sense, it's okay. It's just to respond to Prathu's question. What I'm saying is that this thing is a reparameterization invariant action on the world line of the particle to write it in a way that does not involve the square root. You need to introduce a new field, which is the metric on the world line of the particle. How do you identify the metric on the world line? Okay. How do you identify this as the metric on the world line? You see, for this action to be reparameterization invariant, how must e transform? Okay? So the transformation of e is as follows. Let's look at this term. That's really simple. Okay? e of lambda, d of lambda must be must be invariant. Okay? But d lambda goes to d lambda by d lambda prime, d lambda prime. Okay? So it must be that e prime of lambda, e prime times d lambda prime is equal to e times d lambda. Substituting in, we find that e prime must be equal to d lambda by d lambda prime times e. Let's square this. e prime squared must be equal to d lambda by d lambda prime squared times e squared. Now this is exactly how a metric transforms in one, in one dimension. Okay? The same transformation property makes this guy invariant as well. Okay? The last five minutes have been, are not necessary. It's just to respond to Prathish's question about how, how, how this would be, how you would be inspired to think of it. Okay? In the end, we're going to do algebra, which doesn't care about how you, where you got it from. But just to give you a sense from where this came. Yeah? So this is now just a minimally coupled scalar equation on the world line of the particle with a particle metric. For those of you who are going to study string theory, this is the kind of thing we do all the time on the world sheet of the string, which is why. Okay? Okay. So now, so now just to, to come back, making these choices, we've got this, this action here, which gives us the same dynamical system as we had here. Okay? Now, as we just discussed, this action has a gauge invariance, a reparameterization invariance freedom. Okay? Now, one way of dealing with this action, or trying to understand the physics of this action is to set a gauge. Suppose I set the gauge, E is equal to 1. Use my reparameterization freedom to do this. This I can always do, but I have to be careful because after I've done this, I no longer have the equation of motion that I, that arose from varying E. Set whatever gauge I want. I can set E to whatever value I want. If I am careful to impose as an additional condition on dynamics, the equation that I would have got from varying E. But we know what the equation that you would have got from varying E is. It's this one. And now I've set E equals 1. Okay? Let me put the A here and write it as M squared. So the equation that I have to, so after, so it's equivalent to work with the dynamics of this action. Setting E equals 1. But then imposing the condition dX mu nu dX mu by d lambda, dx nu by d lambda is equal to, so what have we concluded? We've concluded that there is a simpler, that governs the motion of geodesic motion of both massive and massless particles. That simple action is simply s is equal to minus half. Now after I've set E equals 1, this is a constant term that I can just ignore. Right? So g mu nu dX mu by d lambda, dX nu by d lambda, d lambda. But you have to supplement this action, whatever you get from this action, with this condition. This plus the condition g mu nu dX mu by d lambda, dX nu by d lambda is equal to M squared. These two things, this, whatever, the variation, whatever you get from this action, coupled with this condition, is equivalent to the dynamics from this action. That's right. Because there's a re-parameter, I can choose whatever lambda I want. And by using the transformation law that we have written down, E prime is equal to d lambda by d lambda prime times E. I can always choose a lambda prime to set E prime equals 1. So I do that. Okay? Once I've done that, that I can always do, but I'm doing something else. I'm also trying to write down the action ignoring the fact that E was a dynamical field at all. Just a minute. That's only correct if we, in addition, impose the equation of motion for E. You see, suppose we were treating both as dynamical fields, X and E. I would get the equation of motion for X and the equation of motion for E. That's a full dynamical system. On that full dynamical system, I can set E to anything I want by using re-parameterization. So that's what I'm doing. I'm setting E to whatever I want. I get, I still have, I get what I get from the equations of motion from X, but I need to impose the equations of motion from E because I'm ignoring the fact that E is a dynamical field. Is this clear? So solving the full system in this particular choice of gauge is equivalent to just working with this action and additionally imposing this condition. Is this clear? Okay. So the second one already gives you the equation of motion. No, no. Second one just tells you that dx mu by d lambda, I mean, up to normalization is a four velocity. For instance, when the case m squared is zero, it'll tell you that a light ray is propagated in a null direction. So this is going to change into dx mu by d lambda? Yeah. What this is telling you for a massive particle is it's defining lambda. Because if lambda was tau, you would have this equation with one, with the right hand side being one. Sometimes this tells you that it's going to be the height of all of this. You know, what it's telling you is it's defining what your coordinate lambda is. You know, suppose you had a path, then if you chose proper time as your parameter along the path, then dx mu by d tau, dx mu by d tau, g mu nu would be equal to one. So what this is telling you is that lambda is proper time divided by m. The choice we made to fix our reparameterization invariance is to choose to parametrize the path, not by proper time along the path, but proper time divided by m. Rescaling of proper time. Is this clear? We have chosen some other value for it, other number. Yes, we would have got a further rescaled. Make basically no difference. But you see, the interesting thing of this way of working is now we can set m equals 0 in a completely smooth way. This now gives us the condition, you know, this now gives us the condition that dx mu by d lambda when m is equal to 0 is a null vector. And you just get the same equation of motion that you would have got otherwise, but from this action. Just to close the circle, let's derive the equation of motion that you get from this action. Just to make sure that we haven't done something wrong and make sure that we still get the geodesic equation that we got from the original square root action. It's really simple, let's do it. So now, when we differentiate with respect to this action, what do we get? We get terms of the variation from here. Variations from here. These are symmetric. So you get twice the variation from here and a variation from here. The variation from here comes with the opposite sign because of the integration by path. So let's do it quickly. So what do we get? We get so let's first do the variation from here. That's really easy. If we vary alpha, x alpha again. I'm looking at the thing multiplying delta x. The variation of the action. Okay. Now the variation from here is what? It's minus d by d lambda of g mu nu g alpha nu tx nu by d lambda and with a 2. So what do we get? We get delta x alpha into, now let's expand this out. Take the derivatives. So one term is just this delta alpha g mu nu. And then we get another term. Okay, let me write it down in steps. d x mu by d lambda. We get another term by differentiating this out. So we get minus 2 when d by d x theta let's say g alpha nu d x theta by d lambda d x nu by d lambda. And then we get the last term which is this guy. Just minus 2 g alpha nu d 2 by d lambda squared d x. Okay. Now you see both of these terms are something multiplying this quadratic factor in x. So I change dummy variables so as to make the quadratic factor always d x mu d x nu. Here we already have d x nu so that requires me to change theta to 0. Okay. Then so the equation of motion is that this whole thing is 0. So the equation of motion is that first let's identify this object. So this object here is del alpha g mu nu minus g alpha nu minus del nu g alpha mu. I can be thought of as this just by switching dummy variables mu and nu because this is multiplying something symmetric. So instead of writing it as twice this I can write it as this plus mu interchange with nu. Is this clear? So I can write this object here as this thing multiplying d x mu by d lambda d x nu by d lambda minus 2g alpha nu d 2x nu by d lambda squared. Okay. So now you recognize this. This is the combination that occurs in the Christopher symbols. Okay. So the equation of motion is and it occurs with the opposite sign the Christopher symbol has two pluses and a minus. This has two minuses and a plus but there's also a minus here. So we needed actually with a half but this has a two. We needed a g inverse but this has a g. Okay. So the full equation is simply d x mu by d lambda d mu of d x mu by d lambda is equal to 0. This is this. If you just write it out. Setting this to 0 is the same thing then. Setting this to 0. Now we got this equation with respect to lambda not with respect to tau as we had previously but the constraint equation tells us that lambda and tau are related up to an overall scale. Okay. So this equation is true any number time it is also true. So this is once again the GD C equation. Okay. It's very important we have this constraint equation because this equation is not true for arbitrary parameterizations of the curve. This equation is no longer reparameterization invariant under lambda. If you reparameterize lambda here you'll get some d lambda by d lambda prime kind of term here. You differentiate again you'll get a second derivative. It's a mess under reparameterization. So this equation is only true for a specific parameterization. Namely the parameterization in which this is true for in which this is true. For massive particles this is the same as parameterizing it by the proper time up to an overall scale. But for massless particles it's something new. The parameterization in which that we've hit upon for massless particles a parameterization which obeys this equation which obeys this equation and under which the equation of motion takes this compact form is called an affine parameterization. You see what the affine parameter is. You can take a massive particle obeys this GD C equation which tells you that it obeys this equation for dx mu by d lambda being a unit norm or not unit norm but m squared norm same as unit norm scale invariance. Unit norm time like vector. The GD C equation for null particles is the same equation where dx mu by d lambda is a null vector. Now you could approximately approach a null trajectory by taking a massive particle and make it move very fast. And you can ask how am I getting the null solution as a limit of working with the massive particle and boosting it to very high speeds. Now you see when you take a massive particle and boost it to very high speeds the proper time elapsed as a particle goes from here to there is almost zero because of Leven's contraction. So if you worked with that proper time coordinate it's going bad. It's not really parameterizing the path very well because it doesn't change by much even though you travel over very long distances. This way of thinking of the action makes it clear what to do. What you should do is work with a coordinate which is proper time divided by mass on lens scales of interest and then when you take m goes to zero that coordinate continues to be a good one. That coordinate is what's called an affine parameter along null gd6. We'll encounter affine parameterization in null gd6 much more as we continue our course especially in our study of black hole physics but okay. So this is the last thing I wanted to say about gd6 motion in general relativity. We spent perhaps more time on it than we should have but just to see it from many angles we'll have a couple of problem sets in this the problem set will have a couple of problems in this as well. Any questions or comments about gd6, Hamilton Jacobi wave motion in an arbitrary direction. Though we've not done any examples you now know everything you need to know to understand how waves move in the space or particles move in space or light rays move in the space. If you could g mu is equal to eta mu nu and lambda is equal to tau by m then we finally get the equation e square minus e square is equal to x. So is it as well as it's not that? Yes. You see that's absolutely right. This constraint is the mass shell condition. The mass shell condition that relates e squared and p squared to m squared. That's exactly right. It's also the constraint that came about that was the assertion of the Hamilton Jacobi equation which is also e squared minus p squared is equal to m squared. All these things are more or less the same set in different ways. Good point. Okay. Other questions or comments? Yes. We cannot say what? If I'm just going to 0, we don't know what g is. Yeah. So what are we doing is the following. We're looking at the action which is well defined at arbitrary m which is non-zero. Arbitrary non-zero m. I'm trying to find a way of taking the m goes to 0 limit of this action. But this e is not well defined when it comes to this. e is not well defined when... e is not well defined when m goes to 0. That's correct. That's correct. It's not well defined at m equals 0. But the limit is completely well defined. You see. What are we trying? We've got an action that doesn't make sense at m equals 0. What we're trying to do is to view it in a way so that the limit as m goes to 0 makes sense. Okay. So whatever we do will be singular at m equals 0 because the original action just was nonsensical at m equals 0. But what we're trying to do is to look at it in a way so that at arbitrarily small m it makes sense and then take the limit m goes to 0. So if you want, you can think of this as the correct action which gives rise to this guy when m is non-zero but also makes sense when m is zero. You see this is a way of formulating dynamics so that the m goes to 0 limit is censored. Right. That's the way to say it. We know it's correct by the way. You know you can ask, okay you've done some jugglery. How do you know it's the correct thing to do? The correct thing to do comes from this Hamilton Jacobi way of thinking because m equals 0 motion is really the motion of a massless wave. Okay. If you're looking at a photon, a photon is governed by Maxwell's equations. That is the absolutely correct thing to do to solve Maxwell's equations. And we've analyzed that problem. We've analyzed how to solve Maxwell's equations in a scalar form of Maxwell's equations in the Eichner approximation and we get the same answers. So this is the correct thing to do. That's the first principle is we're deriving it. This you can think of it just as a trick. You know a way of uniformly generating the equations of motion both for m equals 0 and m equals 0 together. Okay. Great. But Landau-Liffschitz calls this guy dx mu by d lambda. He calls it k mu. So when you see k mu in Landau-Liffschitz for null motion this is what he means. One last comment about affine parametrization. You see affine parametrization for a massive particle is completely defined once you set some normalization condition. That dx mu by d lambda is equal to m squared. The square of that is equal to m squared. But when you add m equals 0 if you've got one affine parameter you can generate another by multiplying by a constant. Okay. Because both this equation and this equation are scalar invariant. So unlike for massive particles where there's a completely natural normalization namely for instance proper time or square into m squared. For massive particles there's no natural normalization for this affine parameter. Affine parametrization of a null geodesic is defined only up to an overall rescaling of the parameter. For massive particles also there's no particularly good way of saying what the parameter is. But then if you can adopt a natural convention for instance if I adopt this convention then that fixes what lambda is. There's no longer scalar invariance. You can change the mass units and okay but suppose there was somebody else who decided to choose proper time as affine parametrization. It's true once you change units you can change anything. But let's say we've got fixed units you've got a fixed meter stick and then you can say proper time is my affine parametrization. That's a natural and unambiguous choice. There's no such unambiguous choice. No such natural unambiguous choice for massless particles. Okay great. Sorry we've probably gone through this in more detail than we more laboriously than we should have. But anyway let's move on. Any other questions or comments about geodesic wave motion in curve spaces? Why the word affine? That I don't know. There's probably some deep reason with a Greek and a Latin root and God knows what. We should put. That means the other word affine is used here? Affine is called something without a natural origin. Yeah. Okay I have no clue. Now before we proceed we're hoping in the next lecture to start writing down Einstein's equations. The equations that give you the dynamics of the metric field. But before we proceed to looking at the dynamics of gravity there are a few things we want to discuss here. The first thing that we started our lectures saying that this way of thinking has a chance of generating Newtonian gravity in an appropriate limit. Particle motion in an arbitrary metric like this has a chance of generating Newtonian gravity. It didn't have a chance of generating Maxwell's Coulomb law because that doesn't respect the equivalence principle. But it had a chance of generating Newtonian gravity because Newtonian gravity respects the equivalence principle. So now let's, now that we understand everything well, we can ask more precisely can this really generate Newton, you know does do these equations really reduce to Newton's laws in with some assumptions in any limit. Okay so now let's look at our action again. Our action is minus m square root of g mu nu dx mu by d lambda dx nu by d lambda d lambda. And since we want to try to make contact with Newtonian physics we'll do what we did for the Hamilton-Jacobie discussion. We'll choose lambda to be times. So that's equal to minus m square root of g mu nu dx mu by dt dx mu by dt times dt. This is our action. Now what was our idea? Our idea was that suppose we took g mu nu to be equal to eta mu nu plus h mu nu, where this guy is small, where h mu nu is somehow small. It's stuck of course to the non-relativistic limit. So that dx i by dt is also small, much much less than 1. And h mu nu is also much much less than 1. How these two are related we don't yet know. We'll soon find out. Now in this limit what does our action reduce to at leading order in small misses of both quantities together. Now many possibilities many things are possible but as we will see when we write down Einstein's equations all h's with all indices will be of roughly the same magnitude in a weak field limit. Which is the biggest term that involves an h in this action. You know terms without an h are bigger than terms with an h in general. What is the of all terms that involve an h in this action which is the biggest term. What? h squared but with what indices? So h could have h 0 0, h 0 i or h i j. Clearly time is special but not space. So we're not breaking rotations in anything. So which term will be most important? h 0 0. h 0 0, why? Because x dash 0 will be t. Exactly. Perfect. You see there are two sources of small misses. The fact that we're expanding g and the fact that we have these small velocities. If we take one smallness namely from h the leading contribution we want if such a term exists is not to have any other factor of the other smallness. And that we achieve by looking at the 0 0 component where both these terms are 1. Let me suppose that g 0 0 is equal to I think 1 plus I think I probably want to go 2 phi. I may change this normalization in a bit. So that h 0 is 2 phi. Okay. And ignore the contribution of all other h's to start with. Because those will be higher degrees of smallness. There will be h's multiplying velocities. Too small this is rather than 1. Okay. Now what does my action become? My action becomes minus m. Now so and then so g 0 i is equal to 0 g i j is equal to n times with a minus. Okay. So this is what I'm plugging in to have an approximate action. Okay. So then what do I get from here? From here I get 1 plus 2 phi minus d x i by d t the whole thing square. Where the square is now with just ordinary summation. So I've plugged these values of g into my action. That's what the action reduces to. Now this is small. That's our assumption. This is also small. So we'll expand to first order in both smallnesses. So what does our action become? Okay. This is equal to so I just expand the square root. Square root of 1 plus a is 1 plus half a square. Sorry 1 plus half a so that becomes s is equal to m by 2 d x i by d t let me put the m here. Square minus m phi. We'll put the by 2. What is this action? Somebody? The kinetic energy minus potential energy. So if phi was the Newtonian potential then this is exactly the non-relativistic action that Newton would have written down. This little exercise also allows us to judge what the relative smallnesses of this guy and velocities are in self consistent portion. In Newtonian motion what can you say about the relative magnitudes of the potential and the kinetic term? Same order. Because you can convert potential energy into kinetic energy. And you do in going around. So this phi here should be thought of as the same order as d x i by d t the whole thing squared. Okay. So now you see that we've done the right thing. If d x i by d t is thought to be of smallness order epsilon. So that d x i by d t the whole thing squared is of smallness order epsilon squared. Then phi is also order epsilon squared. Terms like h 0 i would have multiplied at least one velocity and therefore would have been of smallness order epsilon cube. So to leading non-trivial order and smallness we've done the correct thing by ignoring everything else. Though we've recovered Newton's laws. I mean we've recovered gravity in a Newtonian potential provided phi is the Newtonian potential. Okay. That we can't know by what we've said now. Because that requires the analog of Newton's equations. That requires the analog of the equation. The equations for dynamics for the gravitational field. We haven't yet written those down. But if we have some sensible equations such that those equations will reproduce the Newton's law for this potential in some appropriate situation then we will have recovered the Newtonian element. So this is really encouraging. This is really encouraging because it tells us that there's a chance the whole thing will work in much more detail than what we'd seen before. Okay. Great. Now there's a couple of other things that I want to immediately talk about before we move on to discussion of Einstein's actions. One thing we want to talk about is motion and what's called a static metric. So suppose we've got a metric in which, okay, let's just simplicity, g0i is equal to 0 and g0z all metric components are independent of time. The fact that g0i is equal to 0 is a statement about our coordinate system. You can always, you have coordinate invariance. You can always choose a choice of coordinates to make that true. But the fact that once you've done that all metric components are independent of time is not a statement about coordinates, it's a physical statement. As we will see as we go along, it's actually the geometric statement that there exists a time like killing vector in our geometry. We will understand this further as we go along. Just to say though that this is not a statement about coordinates, it's a statement about the kind of space time that we're studying. By time I mean x0. By time I mean x0. Suppose we're in a situation where all this is true. Okay. Now I want to tell you about an immediate, you know, that there is an immediate physical thing concerning time dilation in gravitational fields that I want to immediately discuss. Okay. Let's suppose that this is space. This is a cut of space, parametrized by the coordinate space. Okay. The key point is that nothing depends on what. None of the metric components depend on time. So it could be any cut, any constant time cut. Yes. In general, one thing, I mean, if Gij is not time dependent, one frame cannot be time dependent. Yes, it could. This condition is for all time. No, no. If the condition is that you see, if you can choose a coordinate system in which this is true, that's all required. The point is that not for every space time can you choose such a model. Okay. We will understand this. This is a geometric condition and we will understand it in more detail as we go along in our study. Okay. But at the moment we are, we are, you know, suppose this is true for whatever reason. This will be true for the metric of the sun. It will be true for the metric of the earth, some approximation. Sir. Okay. So it will be true of. Okay. This G00, the G, is a function of xi. It changes as we move around in space. It's not a function of time. Now, let's imagine that you are an observer. Okay. You start off at xi. There are two points, xi and yi. So let's suppose that you start off here. Okay. You walk slowly to the point yi. It takes a certain amount of time. You live at yi for a very long time. Let's say that you live at yi for proper time dy. Then you walk back to the point xi. Okay. And we're going to assume that the time that you live at yi is much larger than the time of transit. And xi is here. Yi is in the lecture theater, let's say. Okay. I take five minutes to go down by lift. I run down the stairs. I go to the lecture theater. Then I live there for 50 years. Okay. And then I come back. You guys on the other hand have decided to make A30 for your home. So you've just lived here. And then we meet 50 years later. Okay. The question I want to ask is, is one of us older than the other? Okay. So it's like the twin paradox kind of thing. Okay. You've got two twins. One of them journeys lives for a long time. Come back. Okay. And I want you to ignore the transit time as in small effect. I mean the actual answer will depend on the details of how you go and so on. But ignore that for now. Let's see. Business is over time and x is over time. Exactly. Exactly. I want some sort of space time diagram. This is x. This is t. One of our observers, both observers work together at the origin. One of them just lived all his life at the origin. The other one went here and we're ignoring this part of it. Okay. Lived a long time here and went here. This is much bigger than this. So we ignore those edge effects. Okay. Both, they come and meet. So both observers agree that the amount of time, the amount of coordinate time they've spent between parting and meeting is coordinate time t. Okay. Let's suppose this point was x. I'll give it a better name. Let's call this guy a and this point b. a i and b i. This is a special location. Such proper time as the observer has the observer at a i spent. Can somebody tell me? So let's say that this whole thing is coordinate time capital t. What is the proper time that is elapsed for? How much older is the observer at a i? What is his watch rate? ds squared is equal to g mu nu dx mu dx nu. Now how much? What we want to get the proper time is to integrate the square root of proper time is defined as integral ds along the path. So what should I put in for dx mu? What should I put in for dx nu? What should I put for g mu? For space I should put 0. For time I should put let's say for the whole path, so t. And then I have to take a square root. Okay. So I get the time g 00 at a times t. This is the proper time elapsed for the guy at a. What about the guy at b? There are segments here and here that depends on how he moves. But we have decided to ignore those, because that's not universal. What about ignoring those? What's the answer for him? Exactly. Square root g 00 at b times t. Now g 00 at b and g 00 at a are not the same. So one of the two is older than the other. The guy who spends most of his time at a place where g 00 is small is younger. Because less proper time will have elapsed for him. Okay. As we will see when we start studying Einstein's equations, g 00 becomes smaller and smaller the nearer you get to a massive object. In fact it goes to 0 at a special point in some massive objects called black holes called the event horizon. We will see this in detail. So the nearer you are to a massive object, the slower time goes. Okay. So I will have come back from the lecture theater younger than you guys. Not very, because the lecture theater is low. Not very much younger in this particular case. Okay. But you know, if you did this experiment near a very massive object like near a, it goes significantly near the event horizon of a black hole I could have come back much younger than you. But there is no limit to how much younger I could have come back. Okay. And this is a general relativistic version of the twin paradox of special relativity. You know, things moving goes, time runs slower when you move. Time also runs slower when you are near a massive object. By the way, we already know enough from what we've seen to see that g00 smaller near a massive object. Can somebody explain to me why in the Newtonian approximation we know this is true? Remember it was 1 plus 2 phi. Phi is, the most important thing is that it's what sign? Negative. And then as you go nearer and nearer, it becomes larger and larger. So as you go nearer and nearer to a massive object, g00 becomes smaller and smaller. Okay. So in the Newtonian approximation, the difference in ages would have been 2 times delta phi. Actually, no 2 because of the square root. Delta phi times t. In the Newtonian approximation, this would be the difference in ages between us. Okay. Excellent. There's been a problem set to compute in numbers what the difference would be between. What? G00 actually can be less than 0. It's just that no objects can sit there. This happens inside a black hole. We will understand this in great detail. One third of our course will be on black holes. Okay. Great. Okay, good. Now there's another experiment. So this is time dilation in gravitational fields. Okay. And it's a very simple phenomenon. It's a phenomenon that has been tested experimentally. It's actually tested in the shaft of the physics building at Harvard. And how was it tested? Well, you know, in order to do this testing, it's not like you're going to actually take someone and move them there, right? I mean, in order to do this testing, let's look at them more. Because we've got some point at this guy position A and another guy at position B. This guy sends pulses of light with a certain frequency. A certain proper frequency. So suppose what you're doing is generating light from a particular transition in the hydrogen atom. And you know exactly what the proper frequency of that is from physics. This guy here sits and compares the frequency of what he receives with what he adds by generating the same line from hydrogen atoms in his lab. Will he see that it's exactly the same frequency or not? Now in special relativity, both observers were at rest. The answer would be yes. Both observers would see exactly the same frequency. Because if one was motion, you'd see Doppler shift. One was motion, you'd see Doppler shift. But we're not in motion. We're at rest. In general relativity, what will we get? How do we answer this question? This question is very easy to answer without doing too much work by thinking as follows. What do we have to do to solve wave motion in a general relativistic background? We have to solve the equation del squared phi is equal to 0. Now, del squared phi is del squared including Christoffel symbols. Christoffel symbols come from the metric since no components of the metric depend on time. No components of the Christoffel symbols will depend on time. So this equation here has a symmetry. There's no time in the equation. So that's an easy way to solve this equation. The easy way to solve the equation is to say that there is a solution where the time dependence is e to the power i omega t. Remember t is the coordinate. Depending on details, the rest of the solution which is a function, this g which is a function of omega, might be some complicated function of the exercise. But I don't care about that. We isolated out the time dependence of the solution to the wave equation immediately. There's a solution where you get e to the power i omega t times some spatial dependence. That always solves the equation because the equation is independent of time. This guy here emits things at proper frequency, let's say omega p. Proper frequency. Now, what is frequency? Frequency is like 1 divided by a time interval. Omega p is the proper frequency then up to some 2 pi's which we don't care about. The time between 2 crests of the wave, the proper time between 2 crests of the wave is delta tau is equal to 1 over omega p. That's the proper time between 2 periods of the wave up to some 2 pi. Now, what is the coordinate time between 2 crests of the wave? Well, we've seen the formula delta tau is equal to square root g 0 at a times delta t delta let's call it. Therefore delta t is equal to delta t is equal to 1 by square root g 0 0 at a times omega p. So let me call this omega p concluded. We've concluded that omega. To get omega now that's the time between 2 troughs of the wave in the coordinate time. So we've concluded that omega is equal to square root g 0 0 of a times omega p. Since omega is a constant for wave motion the wave everywhere behaves like e to the power i omega t where t is coordinate time. The same calculation will go through for the point b and omega is the same omega. See that's the point of the statement. No this is omega. You see because it's in the coordinate time t that the wave equation doesn't depend on t. So this behaves like e to the power i omega t where t is coordinate time not proper time. No it's not that's the point because the wave equation is independent of time there is a solution here where omega is independent of time. Is this clear? Is this clear? So omega is also equal to square root of g 0 0 of b times omega p at b. We come to the conclusion that g 0 0 times omega squared is a constant omega p squared is a constant. For the same wave it's proportional to 1 by let's say squared g 0 0. What does this mean? Who has small g 0 0? It's nearer to a massive object. See as things go very fast. See as a large omega. Whereas somebody who has g 0 0 near 1 is far away from a massive object sees a correspondingly smaller omega. Omega increases as you fall down and decreases as you fall up. Roughly you can think of this as picking up energy as you fall down and losing energy as you go up. It's hard for the photon to go up. It doesn't express this hardness in slowing down because photons can't slow down. But it expresses this hardness in losing energy by becoming less by oscillating less rapidly. And the formula is precisely this one. Notice that omega is the same square root g factor as the time dilation which of course is intuitive. You can understand this also in the following way. Suppose you've got your guy here and he's knitting something every second. Time is going slower for this guy. So he sees these things coming not every second but let's say every half second. And therefore he sees a higher omega. It's completely intuitive. It gives you the analog of a Doppler shift kind of effect. The frequency of a photon, we're already familiar with the fact that in special relativity the frequency of an observed photon depends on the state of motion of the unit. This we're familiar. This all of you are familiar with I presume. I suppose you. Now we're seeing something else in gravity. The frequency of the observed photon depends also on the relative, approximately on the relative differences of the gravitational potential of the emitting and observed point. More precisely on the ratio of g 0 0 of the emitting and observed point. This is how this effect was actually detected in this experiment at Harvard. In the main physics building there's a big shaft. So they set up, I don't know whether it is up to down or down to up. They set up, let's say emitting up, measuring down and they measure the difference in frequency between locally generated lines in both places.