 Okay, so let's continue. So just to remind you, in the last lecture we discussed the notion of covariant differentiation. We developed nice formulas for the covariant derivative including a nice formula for this object the Christoffel symbol gamma alpha beta theta in terms of the metric and I'll remind you of that formula since it's important. This is g alpha phi of r by 2 then g alpha phi g alpha phi beta plus g beta phi alpha minus g alpha beta phi. So this is a good formula to remember. It raising an index. So the easier formula to remember is the one with all three indices lowered. So gamma has two indices that are symmetric and one special one. The special one is going to be raised but the guy with three indices lowered has the two indices special. You know you get g of one of the one of those and the special one comma the other one. Other one in the pair plus g of the other one in the pair times the special one comma the first one in the pair minus g of the pair comma the special one and then there's a half. Okay this formula you'll use I mean if you if you use general to be used many times. It's a good thing to remember. Okay this formula was obtained assuming compatibility of the metric with our notion of covariant differentiation and assuming that gamma was a symmetric object had no anti-symmetric piece which were minimalistic assumptions and which are assumptions that will you know this is the formula that will appear in equations of physics. Okay then at the end of the next class we started we started a discussion of of the motion of a particle through an arbitrary metric. We consider the Lagrangian s is equal to m by you know just m square root of g alpha beta x alpha dx alpha by d lambda dx beta by d lambda d lambda and bear a minus sign. And at the end of the last class we worked out the equation of motion dual to this that followed from the Lagrangian and that equation of motion took the following form. It took the form that is sometimes written as d by d tau of dx alpha by d tau is equal to zero. So covariant version of acceleration is equal to zero and what do we mean by this object what we meant by this object more precisely was dx alpha by d tau of d alpha okay acting on dx alpha by d tau is equal to zero and what was tau tau was equal to the proper time along the so d tau more precisely was along the well line exactly so d tau is equal to g alpha beta dx alpha minus minus minus minus minus minus metric I should remember is minus plus plus plus plus plus yeah so over the minus yes no no no we know we're using Landau's funny conventions right so yeah so our metric is plus minus minus minus and therefore there's no there's no there's no minus okay excellent fine now in the first I would like in today's class to wrap up a discussion to wrap up our basic discussion of the motion of physics in curved spaces you'll get an assignment sometime next week in which you'll do many exercises to get familiar with this stuff but the there are several things we need to discuss in class the first thing I want to discuss is the is the motion of a free particle in curved space in a little more detail I want to first tell you about the Hamiltonian formulation of of this problem which is extremely unnatural and very ugly because it because Hamiltonians make a special choice of choice of time and break the beautiful covariance of this problem but then I will go from there to the Hamilton Jacobi formulation of motion of a particle in general relativity which which regains the covariance that is very beautiful we'll talk about motion of light in geometric optics and general relativity and briefly discuss electromagnetism in curve space so let's see how much of that we can do today okay so the first thing I want to do is to work out the canonical formulation of this action okay now we wrote this action this nice covariant form and there's actually a way of working up the canonical formulation also this covariant form where you take lambda to be your time parameter that's a nice discussion but not needed for what we're going to do here and we're not going to do it what we're going to do is to break the covariance of of the theory by choosing time especially okay so we'll choose lambda this parameter lambda to be time okay so s is equal to minus m g alpha beta square root of dx alpha by dt dx beta by dt dt now this alpha beta index runs over time and space so dx alpha by dt when alpha is equal to zero is just one it's dt by dt but when alpha so the alpha indices is equal to zero stands for time and I which stands for space okay and this formula is accurate you have to remember that x zero is equal to t so dx zero by dt is equal to one okay this is of course this funny thing is the breaking of the covariance not very pleasant but bear with bear with me for five minutes it's just going to be intermediate breaking of covariance now let's work out the canonical momenta so so now we now that we fixed time what are the coordinates of our problem it's clear the coordinates are xi i is equal to 1 to 3 if you're living in a four dimensional space diamonds okay as we will be for a while okay okay so we've got three coordinates so we want to find the three momenta corresponding to these three coordinates so what are the three canonical momenta so the three canonical momenta well what's the rule we differentiate this with respect to xi dot what do we get we get minus m times square root of this business times the derivative of this object and we should put a 2 because there's a half in differentiating square root but that I times the derivative of this object with respect to xi dot so I take the derivative of this object with respect to xi dot I kill this guy for instance there's a factor of 2 because the two guys I could have killed that index becomes an i the other index runs over all variables so I'm just writing down the answer it's g i alpha dx alpha by dt okay so pi is equal to and I should I forgot the factor of 2 because the two choices here so that 2 cancels this 2 in the special case that the metric was suppose we look at the special case where g i 0 suppose g i 0 was equal to 0 so the metric had mixed space elements time was by itself suppose that was the case then this guy would have would you know would not be some linear combinations of the velocities in the case that g i 0 was not 0 it would be a linear combination of the velocities plus some function of x's that function of x's being g i 0 so these are the canonical momentum the next thing I'm going to do is to work out the Hamiltonian okay so yeah so the next thing I'm going to do is to work out the Hamiltonian of the property okay so the Hamiltonian is what it's equal to dx i by d tau I'm sorry by dt pi minus the Lagrangian so let's work that out okay let's work it out up here yeah so let's work out the Hamiltonian so the Hamiltonian is this object so let's work it out so we get dx i by dt and then substitute for what pi was so minus m into g i alpha dx alpha by dt times the square root minus the Lagrangian now the Lagrangian was minus square root so that becomes plus square root okay so we'll write this as and the Lagrangian also have the m so we'll take the m out common so we've got m divided by square root into g alpha beta dx alpha by dt dx beta by dt minus dx i by dt g i alpha dx alpha by dt now you might think that this was zero yes but it's not zero because the i index runs only over space so all that remains is the time part okay so we get the formula h is equal to m by square root times this index alpha we keep was common but the beta index being zero is all that counts beta index being zero gives dx zero by dt which is one times g alpha zero okay so we get g alpha zero times dx alpha by dt now let's let's say this in a slightly in a slightly a slightly nice way look at this object the object here p alpha let's define this object p alpha which is equal to dx x alpha by square root g alpha beta g theta phi dx theta dx phi okay oh if you want you could divide by dt dt here and divide by dt there this is an obviously covariant object because it's built it this object transforms like a like a vector and this object is an invariant okay so it's obviously a covariant vector i mean what is called a vector with one index up maybe it's called contra variant contra variant okay and now let's look at the formula you remember this one right this was I'll just write this down gi alpha dx alpha by dt by square root look at the formula for the spatial momentum and for the Hamiltonian the momentum conjugate and for the Hamiltonian okay you see that the spatial momentum is the ith component of the object built by multiplying minus m times this object lowered okay so now so what i'm going to do is of course i define p alpha just by lowering the index here with the metric as usual okay and so to be specific it's g alpha beta the x beta by square root that's p alpha and then if we look at this formula and this formula we identify the following we identify that p i is simply equal to p i by which i this sounds like a totality but what i mean is that the momentum conjugate to the ith coordinate oh yeah with the minus okay let me define these p's with an m but let's not um do we want to define them with a minus that is our next question now we're in this brilliant convention such that this has a minus in it right so yeah i think it's the natural definition with a minus so that if there's a positive velocity it's a positive it's a question of whether you want upper or low i don't know okay okay let's i'm sorry let's let's just stick let's just stick to this this this definition and um fine so then p i where this p is okay momentum conjugate to to x i is equal to okay so just a bit now i'm going to a chronologically change the notation to make things uh to make things clear i'm going to call the momentum conjugate pi i okay again these vectors are called p i's the p alphas so we can clear that pi i is equal to minus m times oh no we put the m in this here so minus p i and h is equal to p zero okay but we know we know of course sort of more or less by construction that pi alpha transformed like a covariant for vector pi alpha however is as we've seen h times minus p i so we conclude that the object the forward four column object that you built by whose first element is h and the remaining three elements are the momentum conjugate to the ith coordinates transform like um transform like uh uh four dimensional vector uh covariant vector you know one form pi i thank you thank you and p i transform like a forward vector but h and minus pi i transfer thank you okay excellent so you know that this is sort of nice which we take this completely non-covariant formulation and we start seeing some sort of hidden covariance uh the hidden covariance is that if you take the Hamiltonian which in the end is the generator for motions and time and add it to the momentum which in the end are the generators for motion and space you put them together you get a nice covariant for vector not too crazy but nice okay now the real reason i'm telling you all of this is um uh is the following you might remember from our discussion of classical mechanics last year that one way to view classical mechanics is through this the so-called Hamilton Jacobi formulation okay and since it's been uh over a semester since all of you mastered the Hamilton Jacobi formulation i'll give you a two minute review of that you remember the so what i what i'm leading to is to write down the Hamilton Jacobi equations for the motion of a particle in arbitrary curve space they're very beautiful equations and they're often the most useful way to solve practical problems in general relativity the more elementary textbooks um i don't emphasize this but you know you know real master like Landau you know gets to the heart of the matter okay so um yes okay if you try to build a phase space where there are four coordinates and four canonical momenta you're going to land up in a constrained formulation of phase space you have an energy and p square relation that's right so you can do this and it's the analog of what we do in string theory i mean it's if you want to manifestly covariant quantization of this particle that's what you would do okay uh this is very useful for formal purposes it's useful when you're trying to make a description of quantum mechanics based on this classical formulation but we're not going to get into that now because for actually solving problems classical motion uh it's not particularly useful okay it's a good question and you can't do it and it has many interesting applications but we're not going to get into that uh though i'd be very happy to discuss it with anyone who wants to probably out of class um okay so um let's remember how the Hamilton Jacobi formulation of classical mechanics worked um suppose you've got some initial conditions at some initial time that's a xi and ti and you have the some final conditions that's a xf x initial and x final and tf there's a very starting point of the discussion of classical mechanics that fixing these initial and final conditions you can find an action that goes between them and you know the action is extremized over paths blah blah blah uh that's what you do in the first lecture in classical mechanics but about the 20th lecture in classical mechanics you return to this thing and say let's not worry about the action as a formal object anymore but let's see what we can find up for about the value of the action as a function of initial initial position final position initial time and final time okay and something that we derived and i will not try to go through again unless you're very uncomfortable um and i might have to look up the text to get the minus signs correct but i'll do that i'll get signs and i'll check them um was that um the uh the derivative of of this action with respect to tf was the Hamiltonian whereas the derivative of the action with respect to xfi was equal to pi the uh with the minus one of them has a minus sign which one i'll check two two which is two okay uh yes i'm assuming two in london two is the final time so i've got it almost then uh this is something you can quite easily check okay that you think of the action now as a function of this initial time initial position initial time final position final time a very similar relations hold for differentiation with respect to initial position initial time but with a with an extra minus sign okay but we will not care about that we're going to think of it for fixed initial position and time as a function of final position in final time what you what you can do uh is to turn this equation here into a differential differential equation a partial differential equation for the action and the way we do it is as follows uh we have the del s by del t now all times and all positions are going to be final so i'll drop the f subscript del s by del t was equal to minus of h but h is a function of positions and and coordinates sorry positions and momentum this formula to substitute in what pi i was in terms of the action okay so we get an equation that says that del s by del t is equal to minus h of xi del s by del and then if you know the form of the Hamiltonian for your problem as we do in our problem okay if you know the form of the Hamiltonian in your problem then um all you have to do but then you just plug in and you get a partial differential equation for this function s now as we discussed in our classical mechanics class uh a year ago a semester ago um this sounds like antithesis of progress right you take a problem which involves solving ordinary differential equations and convert into a problem solving partial differential equations sounds like a weird thing to do however there are many formal and sometimes practical uses of this and how do you use this practically when you use this practically in the following way i'll remind you because we're going to be doing this repeatedly what you try to do is to find let's now specialize the case the three position coordinates what you are supposed to do is to find a three parameter solution to this set of partial differential equations you remember the general solution of partial differential equations depends on an infinite number of parameters initial values of functions but we aren't interested in the general solution we just want a three a three parameter set of solutions if you have more fine set all the other coefficients to zero choose three parameters okay and what we were supposed to do was to find some action that satisfied this equation um was a function of x size um and of d and of these three parameters which we call nullify okay let's label the something so we call this f something very clever we used this s the thing that we found as the generating function of a canonical transformation regarding x size as what they were the old positions and alphas as the new as the new momentum i'll remind you that canonical transformations are generated by generating function which are a function of one's choice of new things one choice of all things could have new positions old positions new positions all momentum new momentum all positions in any of the four combinations allowed we made this one choice um the next thing we did was to say well um after finding this we found the new Hamiltonian now this h prime was equal to old Hamiltonian and i hope i'm getting the signs right del f by del t f by del t was the old Hamilton with the minus sign so that was equal to zero so the point of doing this canonical transformation was we moved to a new set of variables in terms of which the Hamiltonian vanishes therefore all new variables positioned as well as momentum were constants and we actually found solutions to the equations of motion by doing the following maneuver we said well what are the new momentum according to the rules of canonical transformations the new momentum that generating are the derivative of the generating function sorry the new position the new positions what are the new positions they're the derivatives of the generating function with respect to the old the new momentum so these new positions wherever they are we call them beta i only thing about them that we care is that they're constants because the Hamiltonian express in terms of new variables was was zero that was equal to del f by del alpha i now gave you three equations which you can use to solve for x i of t in terms of six unknown parameters the six unknown parameters being the three alphas and the three betas and that gives you the most general solution to the equations of motion because the most general solution of the equations of motion is in terms of six unknown parameters was this was this lightning review clear any questions or comments about it if you've completely forgotten about this stuff you're going to have to go back to our good old textbook from last semester and read about it again okay this is assuming you more or less remember to jog your memory okay great so this is how the Hamiltonian Hamilton Jacobi formulation works now i'm going to do is to work out the Hamilton Jacobi equations in general relativity okay so what we want what what what what is the what is the what's the idea the idea is to find this equation right this equation that some equation the partial differential equation that relates del s by del t and and of course the way of constructively doing it is as i described but in our case there's a bit of a shortcut which shortcut means saves you two lines of algebra it's not much of a shortcut and that's this what we remember is that what we remember is that p alpha is equal to h and minus pi i alpha satisfies a very simple identity the identity is that g alpha beta p alpha p beta is equal to well what is it equal to well it's equal to g alpha beta and let's remember the definitions now d alpha x alpha means dx alpha lowered with the matrix dx beta now this is the same as okay let me write it once again square root of g alpha theta phi dx theta dx phi the whole thing squared because each of these terms had a one over square root then there was an m so this becomes an m square okay now a great thing about indices is that indices once contracted you can raise one lower the other and this cancels this okay so we get the equation that g alpha beta g alpha beta uh p alpha p beta is equal to m square let's find a formula for p alpha in terms of the action that's simple we know we know that p alpha is equal to h minus pi i and we know that h is equal to del s by del t with a minus pi i is equal to del s by del xi with a plus substitute these two into that that gives you that p alpha is equal to minus is equal to minus del s by del t del s by del xi gotten the bigger the overall minus because now this is an manifestly covariant expression okay this is equal to minus del s by del x alpha and then the fact that this is true gives us the Hamilton Jacobi equation g alpha beta del s by del x alpha del s by del x beta is equal to m square so this is the beautiful Hamilton Jacobi equation that governs the motion of of a particle in an arbitrary space how do you use it remember the idea is to somehow find a three-parameter set of solutions to these to these equations and then differentiate those solutions with respect to the three parameters set those derivatives to constant that gives you the solutions to the equations of motion okay great before we actually start using it actually there will be some time before we start using this because we have to actually find what matrix Fourier's realistic situations are we don't yet know the rules for that you know the we haven't yet discussed the equivalence of Newton's laws okay we have to do that but before we actually start using it for practical purposes there are a couple of formal things about this that I wanted to say the first formal thing is the following we have discussed so far we have discussed the motion of so far we have discussed the motion of a massive particle in in an arbitrary curved space we will also be interested in discussing the motion of massless particles in the motion of light through a curved charge but the action we've written down so far has not been good enough to deal with that because if you take the action we've written down you set m to zero you get zero so in order to understand how to describe the motion of light through an arbitrary curved space let's go back to first principle so what is light light in the end classically is a solution to a wave equation so the thing that we want to do is to understand how to describe solutions to wave equations massless wave equation how to describe solutions to wave equations in arbitrary curved space now actually light comes from solving the equations of Maxwell electrodynamics as you very well know and these equations are wave equations but irritating complications associated with polarization okay i'm going to illustrate everything important in the simpler case of looking at a minimally coupled scalar wave equation okay so consider so the light will obey a relativistic wave equation which is slightly complicated because of the indices in the Maxwell field but let's as a toy model of this problem let's study the wave equation that comes from the Lagrangian this is the people sometimes for the two years at one both of us half this is the Lagrangian of so-called minimally coupled scalar field does this make you uncomfortable by the way to see Lagrangians for does this does a formula like this make you uncomfortable does it make anyone uncomfortable is this familiar have you seen formulas like Lagrangians integrated over all the space before no have you seen a Lagrangian for Maxwell electrodynamics so very similar what's the Lagrangian for Maxwell electrodynamics that's well in curve space it's integral square of g f alpha beta alpha beta well by four will be the natural normalization but okay doesn't matter so it's some Lagrangian involving fields over all of space and you get the equations of motion for your theory by varying the action with respect to the to the fields in the Lagrangian are you sure this is familiar this this is familiar so then this is an easier version of that here are the fields were a mu of x here are the fields of phi of x a mu of x is more complicated than phi of x because there are four m u's to keep track of indices phi is a little simpler that's why i'm dealing with the phi is rather the image okay oh the Lagrangian has no integration sign let's call this the action the Lagrangian actually have an integration sign over space but not time so so it's in the action let's say all right the point of this course will be to get familiar with these kind of Lagrangian Lagrangian for gravity will be of this general structure okay so now let's take this Lagrangian and vary it with with with respect to our scalar field to find the equation of motion for the scalar field okay so what's the general principle the general principle is that delta s must vanish on shell del so this by two goes del mu of delta phi del mu of phi g mu nu okay now as always you should you're allowed to set the coefficient of delta phi to zero once you've taken all derivatives and put on the other fields okay so that implies that this is equal to integral of del mu of square root minus g g mu nu del nu of phi delta phi the only way this can vanish is if we have we have the equation of motion del mu of square root minus g g mu nu del nu of phi is equal to zero now this is true it's also true that one over square root of g minus g takes this is equal to zero because I can multiply zero by one over square root of minus g now can those of you who've been alert in the last class or who have notes of the last class tell me if you find this formula familiar go very in the editor of what this is the same as the formula for d mu d mu of phi you see you take a covariant derivative and act on a scalar it's just the ordinary derivative so this first d mu is just ordinary derivative raised but when you act with the second covariant derivative you're acting with a covariant derivative on a vector that's not the ordinary derivative but now we're not acting most general covariant derivative on vector we're acting with a covariant derivative contracted with a vector and you remember we had a simple formula for that that simple formula evaluated to this so there is an object here this differential operator is called the Laplacian the covariant Laplacian or I don't know box the the alembush in I don't know the many many such names for this this thing it's the natural covariant generalization of del square n flat space okay and see how beautifully it came out of varying the action this illustrates another general thing why did we get a covariant action no covariant equation of motion it's because we started with a covariant action in the action you might wonder where was covariant okay firstly just to make sure everyone's understanding everything why did I put the square root g in this action exactly in an earlier class we had shown that d4x is not generally coordinate invariant but square root g d4x is so this thing is in this expression is generally coordinate invariant because the d4x goes with square root g then I had derivatives here why didn't I put covariant derivatives there I don't I don't know because they're scalars I meant covariant derivatives but in the special case of scalars covariant derivatives the same as ordinary derivatives okay and why did I contract with the upper metric to make it covariant that's the only way to get a scalar coordinate invariant out of two lower index vectors so at the level of the action was really easy to write down a generally coordinate invariant action the lines of algebra gives you an equation that doesn't look generally coordinate invariant there are no gammas in this equation all these funny square root g's but it is it had to be as we checked effectively last day that is written in terms of covariant derivatives okay now there were a lot of questions in earlier classes about for instance this metric compatibility condition of the covariant derivative in the metric you know where does it come from you see once we actually start doing physics these questions evaporate okay they they evaporate because physics is determined by Lagrangian you start with a Lagrangian like this you get an equation of motion this equation of motion can be written in terms of the covariant derivative with the metric compatible connection so we did some abstract theory at the beginning that's some blah blah blah okay it's blah blah blah that allows you to give good words to the equations you get from physics but there's no ambiguity in physics you know it comes from a nice Lagrangian okay so we should be careful about being being too much like mathematicians you know okay this was a nice equation of motion now you might wonder what's this guy doing um he was talking about the motion of particles and suddenly he started talking to a minimally coupled scalar fields what's what's the point we can't get into the point okay i wanted to talk about the motion of massless particles the point was that massless particles really are described by a minimally coupled scalar equation or some version there of a massless scalar equation so what we want to do is to see how solutions to this equation propagate in a limit where we can think of these solutions as particle light this is like the classical limit of quantum mechanics where wave motion becomes particle like motion it's how geometry the emergence of geometric optics the motion of line linear motion of light from Maxwell electrodynamics is in everywhere analogous in sense in in fact is more or less the same thing as the classical limit of quantum mechanics okay so since you guys have taken several high-powered quantum mechanics courses you know all about this but i'll just remind you anyway okay um i'll just remind you how this works um so for notation sometimes i will use the symbol del mu instead of d mu for the covariant derivative some textbooks use this symbol as and it's good to get used to both okay so what del mu for d mu no i the thing i think about is del square i'm saying d mu is del mu and d mu d mu is sometimes called del square right i'll also sometimes okay good so let's use that notation just to get it familiar so the equation of motion is del square phi is equal to zero now i am going to solve this equation of motion in a special circumstance okay the circumstance is the icono under the iconol or in quantum mechanics it's called wkb approximation okay and the way you're going to solve it is the following you see this is the basic idea the basic idea is that you know some exact solutions to these equations let's say we were in flat space for a moment you know some exact solutions to these equations just plane waves plane waves at any moment exact solutions to some equations of motion you've got a family of exact solutions um like this you can try to find approximate solutions that piecewise approximate these exact solutions but where the parameters of this approximation vary slowly in space so the idea is to try to find solutions which locally look like e to the pi p dot x but where p itself is a function of x such that the variation of the function p is on lens scales much larger than the wavelength set by p so that over a region considering of thousand wavelengths it looks like you've got just a plane wave kind of solution but as you go slowly varying you see that p changes try to make that more precise what what what we're trying to say is we're looking for solutions which look like this so phi is equal to e to the power i times psi such that the derivatives of psi are not small because psi you should secretly think of as and so that del psi by del x mu at any value of x we will in fact call it p mu maybe it's better to call k mu of x wave number of x just as a definition but you see what i'm getting at right this k mu can be anything as long as it's near some plane wave not necessarily as long wavelength plane wave no one came used to vary fast and the derivatives of psi are small so we're in the regime uh the regime in which we're functioning is one in which now we can make this precise if you like um you see the length scale associated what we want basically is that a wavelength the change of a wavelength over the distance associated with the fractional change of a wavelength over the distance associated with the wavelength is small now let me make that condition precise just just for doing it once so this gives you k so the associated wavelength so lambda is of order one by k so one by del psi by del x and just doing order of magnitude here so I'm omitting all indices okay so what's the condition we want the condition we want is the fractional change in k over a wavelength is small so that lambda uh lambda times del by del x of log of k is much much less than one but lambda is one by del psi by del x and del by del x of log of k is del by del x of k divided by k so that's like del 2 psi by del x squared over del psi by del x the whole thing squared is much less okay I'm being rough here I'm omitting indices so second derivatives of psi divided by first derivative squared note this is dimensionless object is small this is the condition under which the w k b i can all approximation works okay under this condition what we're basically saying is forget about all second derivatives okay so let's do that let's look at um let's look at phi is equal to by psi and look at the second look at del square of this now what is del square of this del square and phi is del mu del mu of phi which is equal to del mu del times del mu i times del mu psi times e to the pi psi which is equal to minus of another what one of these is yes now there are two terms I differentiate a sofa only one good what remains so one term is minus del squared psi e to the pi psi and the remaining term is this is plus plus i and then minus del psi the whole thing squared e to the pi okay this is exact an exact manipulation okay but effectively what we're doing in the regime of interest to us is uh neglecting this term with respect to this term so the final equation becomes del psi the whole thing squared is equal to zero okay I've I've done one rough and dirty thing I've ignored variations of amplitude in this thing or equivalently I've treated psi as a complex number what I've done is okay in the end though you may have to I won't go into the justifications to it okay so so we get the equation of geometrical optics a geometrical wave motion that is in the iconol approximation in the wkp approximation this wave motion is described by the equation del psi the whole thing squared is equal to zero now I'm getting to the point of this thing let's look at what what equation what you see you might think that wave motions very different from classical mechanics that's not true classical mechanics can be formulated in terms of wave motion the Hamilton Jacobi equation okay and let's recall the earlier wave of the wave motion associated with the particle moving at mass n that we just derived in the earlier part of this class it was del s the whole thing squared is equal to m squared these equations are remarkably like each other under the substitution m squared equals zero equations being equal could be a coincidence whether the the important question is not whether the equations are equal it's whether the solution to the problem physical problem or interest is the same okay so you can try to convince you of is that massless motion is like a limit of passive motion just in a way that I will try to explain to you so let's keep this at the this this similarity at the back of our mind and let's proceed to study just this wave motion so we've got this phase of the iconol of the wave equation that is varying according to the equation del psi squared is zero now we want to know the for answer to the following question suppose I produce okay now just as in the study of classical mechanics I'm going to ask you to do the following I'm going to ask you to take this equation and solve it that is I mean find psi which is xi t and alpha for some three parameters alpha however you manage to do it okay you know for instance these three parameters could label the the three momenta in at some initial time this is a roughly plane wave at some moment okay it's like in quantum mechanics this would be like a base a basis set of of wave functions like the plane like the plane waves last place okay imagine it doesn't matter whether you can actually do the solving or not I'm going to ask you to imagine that you've managed to do the solving okay and then what we're going to do is to try now this solution that we've got represents as we've said something that's a plane wave that over a long only long distances varies from plane waves this describes the wave motion what we're interested in geometrical optics we're interested in the motion of little wave packets that move more or less like particles and we want to follow how these particles move how in quantum mechanics do you make particle like motion from wave motion wave packets excellent so you consider a basis of wave motion which is plane waves and then you superpose these to get wave packets so what you do is exactly what we're going to do here you take let's imagine that you've got some g of alpha which i'll break up into a modulus times e to the power i phi of alpha so the alpha labels which wave a wave packet is summing over many different waves with different amplitudes okay so that's what we're doing times e to the power i psi of xi t and alpha such an object now represents a potentially localized thing uh how do you estimate where the center of this wave packet is and how do you estimate the width of this wave packet somebody stationary phase approximation exactly using the stationary phase approximation you know the point is that the variations in in the psi are much faster than the variations due to due to changes of yeah due to changes of k okay so that justifies the stationary phase approximation much like it does in quantum mechanics in the wkb approximation and in the stationary phase approximation this wave packet is dominated okay this wave packet receives dominant contributions from um those xi and t such that when xi and t are tuned to those values the value of the derivative of this of the integral with respect to alpha vanishes the center of the wave packet is given by the equation del phi alpha of alpha by del of alpha i is equal to where the minus is equal to del psi by del alpha i of xi t and alpha now we are going to work with the wave packet that is localized in alpha space around some alpha equals alpha naught just like in quantum mechanics you look at a wave packet as localized both in momentum space and in position space okay that is localized in alpha space about um uh alphas about alpha naught with some thickness alpha not this thickness is supposed to be small so here we can roughly speaking substitute the value of this derivative at alpha equals alpha naught so let me define beta i is equal to minus del phi by del alpha i at alpha equals alpha naught as beta i then this equation becomes let's call it minus yeah this this equation becomes beta i is equal to del psi by del alpha i of xi t and alpha i and this is this is the equation that determines the motion of the center of the wave packet how do you determine the motion of the center of the wave packet alpha is some fixed number some here you put in alpha naught i it's localized around there beta is some fixed numbers you've got three fixed numbers alpha three fixed number beta i in terms of these three fixed numbers you solve for xi as a function of t that determines the motion of the wave packet does this sound familiar it's exactly what we do in the hamilton jacob equation you take this equation solve it remember what we do for the hamilton jacob equation you take the equation solve it in terms of these three alphas and to find motion what you do is to differentiate the action with respect to alphas you set that even to a constant that gives you the motion of the particle okay so you see here in its clearest form a little between wave motion and particle motion in the iconol approximation is particle motion the wave the iconol wave equation once you identify the iconol wave equation in this case del psi squared is equal to zero but the hamilton jacob equation of the particle we conclude it we've concluded that two things we've concluded two things the first one is that if we're interested in starting the motion of mass of particles we obtain the equation motion that is the geodesic equation that we described then in order to study the equation of massive particles we also looked at this problem in another way by studying the hamilton jacobi equation for massive particles then we studied the motion of massless particles we didn't have in any clean way the analog of a geodesic equation we went back to first principles asked what mass massless particles were and they came out of studying massless weights and we realized that in the appropriate approximation the w k b iconol approximation we got exactly the same mathematical problem as you got for studying mass of particles upon setting m equals zero so we couldn't we couldn't do that we couldn't do that directly in the Lagrangian for describing the massless particle we suddenly could do this in the in the hamilton jacobi description of massless particles so the hamilton jacobi description unifies massive and massless motion just said m squared equals zero in the hamilton jacobi description and you get the motion of massless particles now in the exercises we will try to take you through now you remember that the should be doing the i think we'll do in the exercise because we're somehow not going fast enough in this course right so yeah so we'll do it in the exercises we'll take you through you say you remember that just at the level of the equation of motion the geolastic equation for a massive particle was was the following it was dx alpha by d tau of d alpha dx beta by d tau what we said about the hamilton jacobi equation implies the following and i will try to get you to convince yourself of that in the exercises there's almost no convincing to do but okay you think about it suppose you take k alpha which is del psi by del x alpha this was the thing that was the momentum in the hamilton jacobi description and was like the wave number wave number where ten massless motion okay maybe i can just at the level of an analogy it's all what i'm saying it's almost obvious that we'll make you do some work with this suppose i i rewrote this equation for a massive particle in terms of its momentum in terms of even better in terms of the it's yeah okay let's let's do it this way that makes it really clear let's write this as d by d dx alpha when i say derivative with respect to x alpha with the lower what i mean is derivative with respect to x alpha with an upper lower by the metric i mean i'll raise by the metric okay was equal to dx alpha by d tau up to a factor of n is proportionate this equation told you that by dx alpha d alpha d s by d it doesn't matter upper lower is equal to zero since this must be derived from the hamilton jacobi equations since both are true equations both follow from classical mechanics we've not done them here but you do that in the exercise you'll check that these two you just go from the wave equation to this thing frankly i've never done this exercise myself but you guys will do it and show me how it's done okay uh or i'll work it out okay uh uh the same thing the same thing the same thing uh must be true you know because the mathematics is the same when you replace by psi so d psi by dx alpha d alpha d psi by d x beta zero if we define d psi by d x beta to be equal to k beta as we have before in studying eye canal motion this becomes the geodesic equation this becomes the geodesic equation for the motion of a massless particle k alpha the alpha of k beta okay and uh thinking of the motion of a massless particle remember k square is equal to zero from the wave equation massless particle thinking of a motion of a massless particle by being k beta is equal to dx beta by d sum the definition this gives you the trajectory of this massless particle okay so what i'm saying is that in studying the motion of massless particles most often the most convenient ways to use this hamilton jacobi equation directly sometimes you might find it convenient to use the geodesic equation uh also the fact of the hamilton jacobi equation for massless particles is equal to the equivalent of the geodesic equation means that the same is true for massless particles where the geodesic equation is deter is a geodesic equation for the wave number for this wave vector k alpha okay these last few remarks may have seen the bit obscure we'll flesh them out in your assignment so you note that out okay excellent now we're going to be shoot out of this room pretty soon very unfortunately so let's let's move as fast as we can um excellent questions or comments about this before we proceed okay so what we've done is to understand particle motion and massless particle motion in a nice way in general okay there are two things two other things that i want to discuss before we start discussing the formulation of einstein's field equations and these two other things are the following the first of these is to try to give you more intuition uh the first of these is to try to give you more intuition uh for being curved space okay uh i remember when i was a student roughly at your uh your stage of your your education um i remember being very uh fine in general relatively very hard to understand because you know all the familiar notions distance and time and all seem to have evaporated remember asking my professor look i can read you know in journals that the atromeda galaxy is so many light years away from us what what you know what precisely does that statement mean whatever i mean if you make a statement in a journal the last time a precise meaning what is the precise meaning uh in general relatively this kind of stuff really confused me and um i remember that section 84 and land out reading section you know i i was making the big mistake of reading this big fat book this telephone directory miss nathan and wheeler which i think is an awful book so that shouldn't be said in front of the camera then i i started reading land out leaf sheets and suddenly always clear you say four o'clock seven five minutes more okay um and uh do i somehow when preparing for this lecture when i read it again i couldn't see exactly why again i remember i found section 84 extremely illuminating so i can't really see i can't anymore see the point of it so much but somehow when i was learning the subject it was very important to me so i'm going to tell you about it okay um this is a section and what the distances and times mean in general so um so the question that landaw asks is okay we've got these abstract coordinate systems we're sort of physicist and we want to think about your things these abstract coordinate systems are irritating of course uh so let me imagine the following let me imagine that every fixed x is the trajectory of an observer that we've got a three parameter set of observers filling space time these observers sit at xi the three spatial xi is equals constant and propagate along in time so given a coordinate system i will associate to it a three a three parameter set of observers now the coordinate system i will associate to that another three parameters set of observers this kind of thing that by the way i understand it a lot while trying to understand special relativity i don't know if you've read about all these these clocks and meter sticks and all that he was quite obsessed by yeah um so you know if you come from a special relativity background so naturally want to try to think of things this okay so uh though i emphasize it's not the way any practitioner today actually thinks such but still maybe it's useful to get into the subject okay uh so i've got this this let's say i've got one three parameter set of observers now i want to know is my neighboring observer from me i'll ask later if i can ask this question over finite distances but at least infinitesimal how far is the observer my neighboring observer from now you know one of the lessons of general relativity as in all of 20th century physics is that if you ask a question you should be able to imagine an experiment that can answer it otherwise the question may not have an answer so when you ask how far is that guy from me it sounds like a good question but you have to give meaning to the question okay now over infinitesimal distances let's say i'm here at x equals zero and i've got another observer friend who's at delta x i'm going to give meaning to this question as follows these two observers are in different frames no no same frame in a given frame you want to break up you want to understand you know you've got one observer labeled by x another observer labeled by x plus delta x these x and delta x are just x i's okay spatial x you want to know how far the guy at x plus delta x is away from you the answer was somehow be contained in the metric but you want a spatial distance not not an interval there's an interval between two events that's clear but that's not the thing that you want to know you want to know this guy is standing he's moving but how far is he away from here it sounds like an intuitively reasonable question you have to make it precise in order to get an answer one way of making it precise is this you are a world observer doing this your friend is doing that ask your friend to hold up a mirror shine a ray of light it goes to your friend bounces back measure how long it takes to come back divide that time by two assign that distance to your friend because light travels at speed one through this course speed of light is one okay restored by dimensional else okay sounds like a reasonable definition of distance okay at least infinitesimally sounds like a reasonable definition of distance do you agree with us we're going to work out what this definition gives us it sounds like a reasonable definition of space what please obviously it seems to be flat no no no no because light travels this is an infinitesimal region yeah so infinitesimally space is locally flat all right yes yeah I suppose that was your question yes the answer is yes in this region you know curvature is like an infinitesimal square effect as we will see as as we go along okay so how do we make this question precise well it's very easy remember that two points connected by a ray of light light travels along null rays as we've seen okay so two points connected by a ray of light are separated by zero spacetime interval so we send out our light ray from x equals zero at equal zero we want to know when it will have reached our friend okay we want to know the answer to when will it have reached our friend what is the answer to that question well we get that answer by solving a quadratic equation say that it reaches our friend at time delta t the equation we have to solve is g zero zero delta t squared plus two delta t g zero i delta x i plus g i j delta x i delta x j is equal to zero solving this quadratic equation will tell us how long it took so let's solve so we get delta t is equal to minus b so minus two g zero i delta x i minus b squared so we keep a factor of two out g zero i delta x i the whole thing squared minus four ac the four is this two two cancels the denominator uh a here was g zero zero so uh minus g zero zero g i j delta x i delta x j oh we've gone out of the line but okay fine now you see there are two light rays here that starting from here okay would do the job one that came from the past and then one that went to the future it's a point of view we do this calculation for this guy please are the ones that we actually want do you see it's just the difference between these two roots do you see this time difference okay oh my we really have to stop soon is equal to two square root g zero i delta x i the whole thing squared minus g zero zero g i j delta x i delta x j divided by g zero zero now uh this is the coordinate time difference but our observers are sophisticated they know that they should be measuring proper time then they can divide by divide this by two and get the right distance but the proper time is related to the coordinate time by factor square root g zero zero clearly so for the square root so square root g zero zero this delta tau is equal to this is the answer for distance square which gives me a spatial metric distance square oh and then then i should divide by two so this goes over so the distance squared is given by g i j the effective spatial matrix gamma i j this is g i j um minus minus g zero i g zero j divided by g zero zero um uh squaring this okay we through okay uh i think we're gonna have to postpone uh things to the next class just this is the punch line this is the effective distance see assigned to neighboring observers by uh by the definition that we just prescribed we'll come back to that to this next class okay uh please let's go out and i had one question to ask the people about scheduling um