 But in my found, when bearing with respect to g alpha beta was del alpha x mu del beta x mu is equal to half of g alpha beta. And then del alpha x mu del alpha x. The raising is raising with its word sheet metric. That's what you're doing, my friend. This is the equation of motion with respect to g. OK, now let's take determinants of both sides. OK, if I find minus signs at the moment, that's always tricky. But if I find the signs, we get the determinant of this object of del alpha x mu and beta x mu. Now what? This, as far as the matrix structure is concerned, is a constant. It's 2 by 2 matrix. So the determinant is multiplied by constant square. So g is equal to half square, so that's one fault, times del alpha x mu del alpha x mu of x square times delta g. Now we want to put the minus signs. You see, we want to minus here. So we also want to minus the delta g, which is minus g. And then we get square root of both sides. So we get square root of this is equal to half. We take a positive square root. We take a positive square root, and with the sign you can mention this is this often without square is negative delta g. So we get minus, at the square root of g. The minus sign is because our sign can be mentioned is delta g. So the time derivative is negative of the space derivative. But the time derivative always dominates the space derivative. The time derivative is an energy. Space derivative is momentum. Energy is always last. So in our sign, we can mention this quantity as written is a negative. So we've got a positive number. We're taking a square root. We're using the convention of the square roots. Where it was to be possible to add plus something on minus. Using the convention that we get positive number. Positive quantity. OK, that's what we mean. Not in a sense. So that's where we put the minus sign. Is this here? I mean, it's just a matter of prevention. So therefore, this thing, just to say it again, this is going to have a kinetic term. And the kinetic term in this action will be positive with this structure. So you would do it in the expression of this term. Otherwise, do you think this is going to be negative? Oh, OK. You're going to say no. I agree with that. OK. Therefore, we find that this action is the same thing. This half is this half. Minus of the sign is as that. That's my conclusion. Whether we want it, oh, yeah, do the minuses. Sorry. OK, good question, do the minuses. We have to do the minuses, don't mention it. So let's see, let's see, let's see, let's see. This, OK, sorry. OK, let's see, let's see. Do you want something that is positive? Minus sign, what do you have there, I guess? Yes, in minus sign, it won't be there, I guess. Possibly, I mean. Because we took a square root of the thing, so there are two signs possible. So the one with positive will make it, maybe be actually positive. Yes, not so clear to be actually, because of our sign convention, we have negative type there. Minus sign is in. OK, yeah, of course, intake the square root, we have this plus, minus, so it'll be usually. Sorry, sorry. Was this the right sign to start with? And certainly, this is the right. And certainly, we want to get this. Otherwise, it won't make sense. These two signs have to be the same. But what's really puzzling is that if by the square root we mean the positive square root, this looks like a negative definite object. Whereas this looks like a... All right, now this statement in the square root we had, plus, minus, and negative, which was one of the signs that works, right? I can't think of, at the moment, the rationale for this particular sign. It actually doesn't matter, because all we really care about is the equations of motion of the same, unless you add these there anyway. If there's something sleep-disabled while you get this sign, let's not do it online here. Let me try to get back to this. Next slide. OK, fine. So, more you know this sign, we found that this action is the same as this action. Is this good? OK. Now, all this is very slick, but it's giving something quite important. And this is the thing that I want to end. OK. The question we're going to ask is, is this system the same classical system as this? By which I mean, is there a one-to-one map between solutions, the solutions of the two theories, such that, you know, anything you compute with this solution is the same. You compute with that. In order to answer that question, it's interesting to do the following thing. You see, the structure looks to start with the structure of these equations that look the same. It looks like we've introduced three new variables, the G alpha. But we have three new constraints from the equations of motion with respect to four G alpha beta, which allow us to determine what G alpha beta word is in terms of the X's. Plugging that solution back in, you get some equations of motion purely for X's. Therefore, it looks like these, which would be the same as the equations of motion that you derive from the X theory, because we've got the same action, on your side, but that doesn't matter for equations of motion. Therefore, it looks like we've got the same classical systems. However, this is wrong. Does anyone know why it's wrong? Can anyone see why it's wrong? It's not true that all three of the new additional degrees of freedom that we've got from our system are determined. It's not true that all three G alpha betas are determined by these equations of motion. How do we see that? Well, one way of seeing that is by taking the trace of these equations. Taking this equation and multiplying it by the other side. On this side, what do we get? We get del alpha by X mu del alpha by X mu. And on this side, what do we get? This becomes two betas. Trace of two graphs to matrix. And this is the same thing. Del alpha by X mu, del alpha by X mu. So what we've got here is an identity. We have three equations here. We represent matrix, we have three equations. We have three equations from the two diagonal components and the one off diagonal component. So that's just three. But actually we see we have left two equations. Because one of these equations is an identity aspect. It's not giving you information about G. It's just a trivial identity. Okay? So now, you might think, okay, what's going on? But we've only got two new equations. And we've only got two equations. And the thing is we check with the other two equations aren't you? That we've introduced is determined by this constraint equation that we found. But the third one is not. Given that, how did it happen that we managed to get the same actions for the two sides? Well, what happened to that third degree of freedom? Since we got the same action, we got an action plugging in with this equation of motion that didn't depend on this third degree of freedom. It just doesn't enter the action at all. Okay? Let's see how that works. Let's look at this action here. Consider two different metrics. G till and up, which is equal to G alpha beta times G alpha beta times e to the power 5. The arbitrary function of your whole G coordinate. Okay? Now, under this transformation, what happens to all the fields that exist in this action? The square root minus G alpha beta till and up alpha beta is equal to e to the power 2 5 by 2. The 2 5 is a terminus case like a square and the by 2 is because we take the square root of square root G alpha beta but also G inverse so G alpha beta till and up is equal to minus 5 G till and up G alpha beta. Action has both the square root and the terminus G alpha. There are no derivatives anywhere. Metric. So you plug this in and you find you get exactly the same action for G and G till and up. So, there is a huge symmetry in this action. Okay? The symmetry has a name. It's called wireless symmetry. So this action is symmetric on the wireless different metrics that are related to each other by rescaling with a function in which coordinates give you same functional form for the action. It's because the action at this wireless symmetry varying with respect to this direction in field space gives you no no equation of motion and varying with respect to this direction in field space you can easily convince yourself at the same time to trace these equations because you're varying with respect to the direction in field space in which the whole metric is changing in an overall way which you can convince yourself is basically the traceness of this equation. This equation was in a way that was proportional to the metric. It would be G alpha beta the variation is proportional to delta G alpha beta is proportional to G alpha beta and therefore that traceness. The reason for our equations of motion being too few to solve for this new degree of freedom is that the action we wrote down was new fantasy action we wrote down and a big symmetry. Now what is the consequence of this? The consequence of this is that as we have seen as we have seen if we had given a solution to the equations of motion that come from this action that solution determines two of the three components of the metric G alpha beta but the third component is completely our metric as far as this action is concerned so strictly speaking as dynamic these two actions are not the same for every solution to the equation of motion here there is an infinite number of solutions to the equation of motion here in the solution with any given metric times all possible Y transformations so while a naive manipulation seems to imply this action of this action of the same classical systems that is not quite true an infinite number of more solutions to the equation of motion of this now it's not dynamical but if you try to do a quantization of the system you say quantization is quantization of phase S and you would find a huge additional you know a much larger phase in the second system but if your point is completely right it is not dynamical it is it is completely correct the comment made was that even though we have lots of new solutions these new solutions are related to each other as an action so how do we make these two dynamical systems the same we make these two dynamical systems the same if we supplement this theory by a statement the statement that we are going to take and gauge it take the symmetry and treat the solutions that are related by Y transformations as the same solution once we do that we make the two dynamical systems the same you see we have now increased the degree of gauge in various dimensions in the original action itself we had a reparameterization in dependence and we agreed last class that we should be thinking of the system and being gauged by this reparameterization solutions that differ only by reparameterization the same solution in space time same physical motion in space now once we've gone as view form of the action this form of the action we have inherited a third gauge in there the solution should be regarding the same physical solution if they differ by either reparameterization in dependence or by wireless gaming it is with this understanding that this action is the same as this action as a classical dynamics so is this understanding that we should keep in our mind now in order to try to make a quantum theory starting with the same action which is much easier job than stuff is there any questions or comments before we just clear everything and write it as a more systematic way now our plan for this lecture in the next oh sorry comment about schedule I realized that there were conflicts on Tuesdays Wednesday is a Friday schedule Wednesday is a Friday as written in stone so what we are going to do in the next in this lecture in the next one and perhaps the next one is perform an honest quantization is perform a quantization of this action, this string action we are going to do this in two steps in this lecture I am going to try to get to the result as fast as I possibly can in short techniques we are trying to be as honest as we can in the next two lectures we will go in the next few lectures we will go back and revisit the question and discuss it very carefully using advanced and interesting techniques of conformal field theory but first before we get you know before we get formalism let's try to study the answer in a slightly simpler way alright so what's our goal? our goal is to take this action minus t now I am going to switch to the dimensions used in string theory t, what was t? the question behind the action was the tension of the string now in conventional it's conventional to write t as if from 1 over 2 pi let's look at dimensions tension is mass per unit length mass and energy are all the same dimensions because h cross is 1 and mass and inverse are all the same dimensions because h cross is 1 so we have got mass per unit length which is 1 over length squared therefore this quantity of u of dimensions is length squared the root of alpha prime is a length it's a famous length it's the length associated with the size of strings it's called the strings okay at the moment this is just a change in location but this square root of alpha prime will be the length scale that will appear all to our standard okay so t was this but now our action was minus t by 2 is minus 1 by 4 pi alpha right 2 to minus g then alpha is mu okay the way to understand how to deal with this action treating and while transformations as gates and entries so what we have to do is in order to contest this theory in an honest way the fastest way that I can is use the mishpash of the methods that we use class to class which is very well for the case of the particle the first the first thing I'm going to do is to use an analog of method number 4 the last method if you remember was to take an action that had a gates symmetry fix gauge but be careful about dealing with the constraints that accompany the gauges that's it so what kind of gauges we've got reparameterization in various how many functions of reparameterization in variance in two dimensions this is how many functions worth of invariance two functions worth of invariance that's two and we've got wider than that how many functions worth of invariance one function three functions worth of invariance functions of the worksheet worth of invariance it should be possible to use these three functions to set any three other dynamical functions to be something specific up to possibly discrete ambiguities which we deal with carefully in the next few years okay so what we're going to try to do is to set this g-alpha-beta okay g-alpha-beta to the form e to the power 5 eta-alpha-beta using reparameterization in variance and then to set the same thing to eta-alpha-beta using white we should see questions of can you do it can you always take any matrix and use reparameterization in variance to set it to this form these are interesting questions which depend on many details it depends on the space in which you're in and so on we will analyze all these questions in detail in a few languages but for now we're going to forget some subtleties they affect small things but not gross and pretend we can always do it okay pretend we can always do it so this is the gauge that we're going to set okay once we have set our gauge our action becomes enormous extremes our action becomes s-equal to minus 1 by 4 pi alpha prime square root of well there's no minus g now so it's just del-alpha-x mu del-alpha-x mu using the flat space matrix the system of 3 goes on the two dimensions which any one can see in-chicks engage in-chicks engage there are two things you have to worry about okay the first thing we have to worry about is the constraints that were associated with the gauge okay so that's why now the constraints associated with the gauge-fixing were the equations that you're going to write for us that are the equations of motion with respect to the objects that we fence which was the alpha-x mu the beta-x mu is equal to square root of g but that's a nascent square root of g is just 1 minus g is just 1 minus half del-alpha-x mu del-alpha-x mu del-alpha-x mu alpha-beta del-alpha-x mu we have to do our quantization consistent with these constraints these are constraints all these constraints are it's useful to use a nice set of coordinates of the world-sheet of the string okay so the coordinates that I'm going to introduce are the lifelike coordinates of the world-sheet so we've been using these coordinates sigma and tau in terms of which we've chosen the metric in terms of which we've chosen the metric we mined quite a bit okay but let's now move to the new set of coordinates sigma plus sigma and sigma minus which is tau minus and let's write everything of the world-sheet string everything in action in terms of coordinates every derivative in terms of derivatives for sigma plus and sigma minus so we've got to transform this action into that form that's very easy so let's remind ourselves how things work firstly about the action this thing was just was minus z tau squared of x plus z sigma squared of x but there was a minus sign so this is plus and minus let's rewrite this into let's see if we can find a simple expression in terms of plus and minus there in terms of x that gives us the same thing well so let's see so del tau so let's use the chain rule of possible translation so suppose you've got del a this is the same thing as del x a by del y b d by d x a we know what x so this tells us for instance from this condition that del plus is equal to del tau plus del sigma sorry let's use the chain rule let's use it quite d y b by d x a d by d b let's choose a y b sigma plus sigma minus and our x is to be the tau so tau and sigma we have del tau is equal to plus plus del minus and del sigma is equal to del plus minus del because del sigma by del tau is 1 del sigma plus del tau is 1 that's it so let's take this and plot this into here to see that this is the same thing as del del minus x and I think it's like 4 this is the same thing as del tau x0 squared minus minus x0 yeah the 4 is because you get a factor of 2 when you take this square that's why you didn't do it so let's try to get all of that actually written in terms of plus and minus as the following interesting form 1 by 5 on the front the minus we absorbed and the 4 I just cancelled and plus x mu del minus x mu needed let's see very possible let's see if we put it in what yeah and now is that consistent with that yeah it is good let's see what our constraint looks like in a moment okay and this is the part of the action so nicely because it helped us solve the equation more than very simply at this moment but okay the real reason for going to these variables is that this constraint becomes one so let's look at what the constraint looks like okay so this constraint here was of this form in any set of variables so let's use it in the variables where alpha, beta, range over plus and minus okay so now you can use that I need to know where g alpha, beta is in terms of plus and minus okay but minus beta squared plus g sigma squared is just identically the same as minus of d sigma plus d sigma minus a plus b, b minus b is a squared minus b squared okay which tells us that g plus minus is equal to minus half there's a matter of half between what we've got here and what we've got here because the metric is symmetric there are two times because g plus minus is same as g minus plus which tells us that g plus minus if we need that is equal to minus of 2 and we invert okay so now let's compute let's compute let's compute this constraint equation let's compute this constraint equation all three compute at the point well the first thing we see is that the plus plus components minus minus components the plus plus components and minus minus components are very simple because in a new in a new set of components g plus plus is 0 g minus minus is 0 so we get no contribution from this step is that right so when it's a g plus minus half when it's extra supplemented by g plus plus it's equal to 0 do you know the constraints of an example the del plus of x x mu del plus of x mu is equal to 0 and del minus of x mu del minus of x mu is equal to 0 what is the constraint equation that we get I think plus and minus can somebody do it without doing the algebra we've already set the answer it's 0 so as Logan and I have pointed out by taking the plus minus component in this score that's estimating the trace of it of these equations we're already saying that that should just not give us an equation that should be an identity now it's a matter of keeping your wits about you it's a matter of keeping your wits about let's let's try to see that it actually works it's a matter of twos and minus signs let's see if we can do that g alpha beta has minus half so that's let's look at the plus minus equation the plus minus equation is del plus x mu del minus of x mu on the left hand side then we have minus of one fourth minus of half minus of half that's g alpha beta and then this quantity now this quantity was what it was del plus of x mu del minus of x mu multiplied by g plus minus such terms del plus of x mu del plus of x mu and del minus of x mu del plus of x mu so two such terms and g plus minus is minus two so we get into minus four times del plus of x mu then minus x mu not that I think we got it so minus four zero there are two rounds of the three constraints that's the plus plus constraint the minus minus constraint the minus plus constraint is nothing okay so let's summarize what we got so far we understood that what we want to do is to quantize the action that I've written down here subject to actually go ahead and solve this problem there's one more irritating but important point that I've created in this video and that's the point we discussed at the beginning we discussed at the beginning that there may be subtleties about reaching the stage okay and we talk about the effect of the subtleties as we get to that point but there is a more important question that has important effect to quantization so we must talk about it now and that's the question of once we have reached this gauge have we completely exhausted our gauge freedom or not do we have additional gauge redundancy left once we have reached this gauge we were asking the question in general we've taken a metric and made it flat somehow or the other but we asked are there coordinate changes are there coordinate changes that commute with it being flat and that are the only very trivial questions Lorentz transformations such small numbers of things would have been there remember our gauge freedom helps not just coordinate variations with coordinate reactions plus wide transformations okay so if we perform a coordinate change that doesn't leave the metric in bed the effect on the metric is to rescale the metric by an overall function then that rescaling of the metric will be also a gauge symmetry any change of coordinates any refarameterization of the theory that is to perform an overall rescaling of the metric okay in this gauge is a symmetry that we have not yet fixed such coordinate changes such coordinate changes have a mathematical name they in infinitesimal form they induce by what are called conformal killing vectors so a mathematical addition would ask for a mathematical addition we started the problem of finding all the conformal killing vectors in our system though the mathematical additions which generally try to take post conditions at infinity for these vectors we for the loaded questions we are asking now can you think of any refarameterization in two dimensions that changes the metric of course changes the black net only by over at molecular the overall function now I know several of you know the answer but for somebody who doesn't know the answer okay because you've read it before can you think of any such function any such way molecular is often such such as an example but the many more in order to help you let me rewrite metric in our nice words d sigma plus d sigma plus this is our image now can you think of any changes of coordinates that only multiply this metric by an overall characteristic feature of the metric characteristic feature is that that zero is here and then something here because it's symmetric that something has to be same all you want is some change of coordinates that will leave the structure intact by which I think there is no d sigma plus d sigma plus term no d sigma minus d sigma minus term can you think of what the change of this metric that achieves that there you go sigma plus being redefined to any function of sigma plus sigma minus being redefined to any function of sigma minus preserves this so there is an important function worth of ambiguity function worth of not yet having fixed our gauge freedom completely in our problem there are sigma plus is equal to some function that's called f or sigma plus and sigma minus is equal to some other function that's called g of sigma minus functions worth of re-parametricity in variance that we have not yet fixed you might think well well what happened to our argument we used three functions to fix three variables maybe there is some discreet ambiguity some small ambiguity but how are we left with functions worth is the is the question clear? sounds unreasonable to be left with as much freedom as we thought we had fixed what's the answer to that? this is an independent question just you're fixing the edge the answer to that is to be not left with as much freedom as we had fixed because these are functions of one variable our general coordinate transformations have as re-parametricity in variance three functions of two variables what we are left with is two functions of one variable which are function x so measure zero is that function of two variables but yet it might be physically important is this clear? so before we just embark on a mathematical program of quantization we have to deal with the fact that we have not yet fixed all our gates so let's put that as point three need to fix unfixed need to fix unfixed gates and entry sigma plus is equal to the f of sigma plus sigma minus okay in order to emphasize to you I wanted to do this fixing and also in order to emphasize to you how important this fixing is before proceeding to the quantization I'm going to remind you of the classical solutions the classical solution this problem okay? so how do we compute the equations of motion? suppose we compute the equations of motion of x mu we vary the action and we find that plus and minus of x mu is the zero to make the linear equation of motion it's a traveling wave solution in one dimension and we know that all the solutions to this equation of motion are x mu is equal to that of x mu plus of sigma plus minus of sigma plus as you can easily prove this is the second solution to this equation of motion now are these functions x plus and x minus completely arbitrary but remember we were dealing with with a string such that the cos sigma coordinate was periodic between 0 and 2 pi so it's not completely arbitrary these functions have to be periodic in sigma plus and sigma minus but periodicity 0 to 2 pi which tells us that if we wanted to we could expand them in a Fourier series with congruent correcting TGF okay? we'll come to that we'll come to the details of the quantization of the solution in a moment but why I wanted to say this immediately was to emphasize how important the Romanian coordinate is the Romanian gauge redundancy you see, suppose I take suppose I take this this gauge transformation and I apply it on a solution of course I must get the solution, you know I get sigma plus replaced by f of sigma plus sigma minus replaced by f of g of sigma plus if we were interested in looking just at the space of solutions we could use this coordinate redefinition of the QT then to say any particular one of these x's to something specific you see any function of sigma plus is as good as the definition of sigma plus and sigma plus it's that provided a coordinate redefinition of the QT in our problem sigma plus is only defined up to functions of itself sigma minus is only defined up to functions of itself now I want to share the solutions to the equations of motion these things are divided up into functions of sigma plus and functions of sigma plus so one of these coordinates could be used as a definition of sigma plus okay? to say it again, even before we impose these constraints the space of any particular classical solutions is not suppose we were in d range d space dimensions it's not d functions of sigma plus and d functions of sigma minus it's d minus one functions of sigma plus and d minus one functions is this clear? the constraints on the other hand remember the constraints have this nine property that we have one constraint for pluses and one constraint for minuses then plus a function of sigma minus is zero so the first constraint constrains x pluses and the second constraint constrains x minuses so the constraints on the other hand of one more equation of non-sigma plus could be used as one more solution so the space of classical solutions of this problem the space of classical solutions of this problem is d minus two is d minus two functions of x plus and d minus two functions of x minus of x minus if we forgot about the original redundancy we would have got the answer d minus one and d minus one which would have been totally wrong so it's a very important thing as we will see as we will see with sigma now can somebody give me a physical interpretation of d minus two so let me ask you the following question suppose I have a string, a long string stretched in our physical three dimensional space extension and I hit it a little bit the oscillations of the string the oscillations of the string can be thought of as some number of scalar fields on the on the string in physical space the camera for you it's two the transversed oscillations can oscillate that way or can oscillate that way you might function in three the three directions one of the directions in which the string is itself can oscillate that way in a physical string you might need to oscillate that way you might need to pressure that's not the kind of string this is the only way there's no sense about which part of the string is right it's just a word line occupied by the string there's no sense in oscillating in the direction of the string that's meaningless so the number of oscillations in our problem should be d minus two that's what we're finding and it's very good because nothing else would have made sense so you're assuming that the number of problem of spatial equations is three ways but in the little toy question I gave you that was the answer but you can easily in your mind now heighten that up to d space that way and the answer for d minus two you can't oscillate in time you can't oscillate in the direction of the string for many directions you can oscillate it's d minus two is this clear? so that's all I wanted to say about rough analysis of classical motion you and all the intuition we gather to try to perform a quantization of the string but questions are common to the problem the question is the physical argument the physical argument was this see suppose you've got a string occupying a certain position and you have to ask what they there are no impedances there are no impedances there's not no necessary a quantization in various around the string if you ask where can a little bit of the string move it can move in that direction or it can move in that direction if it moves here that is changing the shape of the string and so it's not counted as a new shape you see because only if you're afraid the variables of string theory are what path in space span the string occupies there's no sense of what the path of the string isn't made up of atoms we can label the question of which atom is where the only thing we can say about a string is what space span go and it's occupying so if you try to move the string along its own shape that's not changing that space span go that's the physical argument we've seen we've seen it formally from here as we can see more clearly I think this function that we have here is the consequence of this like a symmetry exactly in this particular teacher you could choose a crazy gauge in which these symmetries are completely different consequences for that teacher yes but look at the voice our comment was completely right you see we argued that coordinate transformations whose only effect was to rescale the matrix our symmetry of our unfixed re-parameterization plus wild symmetry of our problem there's a second name for coordinate transformations whose only effect is to rescale the matrix and they're called conformal transformations so one something that we conclude in a general general is that the theory of the world sheet of the string must be conformal theory in the gauge symmetry deal with the unfixed gauge symmetries the unfixed gauge symmetries of our problem okay so before we actually start doing it I want to give you one one bit of one bit of orientation you see the orientation has to do with the nature of face space in any classical system face space is something that you are familiar with as a space parameterized by positions and momentar of the dynamical variables of the system but that's just one way of describing face space is not the best in it so why are we interested in positions and canonical momentar we're interested in positions and canonical momentar because we constitute the full set you know once you take down all constraints that happen to be in the problem they constitute the full set of initial conditions for classical mathematics okay as you know that's by a coordinate and a velocity velocity is essentially moment but initial conditions are in wonderful correspondence with the solutions of the theory for every initial condition there's a situation so the best way to think of face space is to think of it as the space of classical solutions okay that's not the face space of the system it's the space of classical solutions okay and when you quantize face space where quantization is a procedure of associating an operator algebra timber space algebra with face space the connection between the classical motion of the system and the quantum motion of the system the connection in one way or the other find the same classical solutions that's face space and then quantize the face space okay so that's the procedure we've already seen that the face space of the system is made up of 24 functions of x plus and 24 functions of x minus so that's the quantization of that problem okay so let's now try to move forward the formal procedure goes like this we see that we have oh sorry now D minus 2 D minus 2 yes now let's move forward let's move forward more forward we now get fixed as gauging there okay so let's fix it by a particular slice of face space okay so all our space of solutions we've got these equivalents that we talked about and we want to fix this up, this equivalence so on the space of solutions we've got the equivalence and I'm going to fix it by using something that you might think is a little weird but it was hit upon by trial and error and it's a good thing okay so let me find x 0 minus x plus x D as equal to x plus and x 0 minus x D minus 1 is equal to x minus now this whole thing gets confusing because there are two things going the space time the place where the strain is moving and that's the working strain x is remember our coordinates for the space time sigma is our coordinates for the working strain the two are different okay we use light cone variables sigma plus sigma minus in space in on the working now we're going to do something a little weird we're going to break explicit Lorentz invariance in space time what we're doing now is analog of the second way of quantizing that we had yesterday the way of taking this nice Lorentz invariant action and saying well let me use tau as time this was very clear then but it wasn't explicitly Lorentz okay so in order to get at in order to do the fastest route to quantization I'm going to do that I'm going to break explicit Lorentz event to the action by first choosing one special station to what? the D minus 1 to 1 combining with x 0 and x plus x plus combining this with the minus sign and x plus x plus so far nothing to relate to coordinates but where I do something is saying let me choose the gauge to on the space of solutions fix my reparametralization of beauty let me choose the gauge x minus is equal to alpha prime p minus then this is the final gate I'm going to fix but I would say in a way that makes it sound less let's say what is p minus at the moment some undetermined number I will convince you in a moment that p minus cannot be fixed we don't have freedom let's say p minus that's not the answer so we need an ambiguous number that parameterizes difference p minus will give you different solutions physically different solutions but let me write this better as x minus plus the right moving part can be clear x minus plus is equal to alpha prime p minus into sigma plus space and x minus minus alpha prime p minus into sigma minus remember every solution to the equation of motion was some function of sigma plus and some function of sigma minus this segment already tau function of sigma plus and sigma minus tau is equal to sigma plus plus sigma minus by 2 so real tau equal to sigma plus by sigma minus by 2 and then I use that to identify why I made that so this is the equivalent to this same not like making this gauge where I've done this to say well remember x minus of x plus of minus was in general arbitrary function of sigma minus and I then choose my reparameterization freedom that function of sigma minus is now sigma minus times across is this clear my name no and reparameterization of variance just changing sigma sigma plus to any function of sigma plus okay but now on this base of classical solutions I have D different functions of sigma plus x1 is a function of sigma plus x2 is a function of sigma plus or maybe xd minus 1 in studying classical solutions I could choose the following gauge you know there were these 20 different functions of sigma plus but I could use my sigma plus reparameterization of variance to choose any one of those functions to be my new sigma minus once I've done that that new function is one well I've done it with x minus actually it's x0 plus xd minus 1 that's what I have to do with it I could have done it with x0 essentially I did it and said whatever that is that's my new definition of sigma minus up with this concept and whatever x minus was x there's a minus and a plus and a minus and a minus x minus plus for sigma plus x minus minus for sigma minus up with the concept okay I'm using this reparameterization immediately to do this now what is this clear this is P minus and D minus this is just some number which would be fine now why this number why would I just say it would be sigma plus the point is that our reparameterization wasn't complete because there's one thing about the sigma coordinate that we know namely there are all functions of periodic with periodicity periodicity to pi and you shouldn't know a coordinate should change you do that should change that so if in some variable if in some variable your functions were periodic with periodicity to pi if you make a coordinate redefinition that respects that periodicity then you're safe but if you make a coordinate redefinition that breaks the periodicity then you're not because it takes periodicity to pi periodicity to pi times the risk rate okay but if I were to try to if I were to try to change this constant here what I would be doing is changing each of these functions by constant but the variable of sigma is just sigma plus minus sigma minus so that would also be changed by constant so I don't have enough to soak up the zero more part the zero more part of what x minus was is this clear? let me say this way let's remember we said that the most general solution the most general solution okay of any of these x's and therefore of x minus because of this form let's forget about the x sigma minus in what I've said look at just sigma plus okay this was periodic solution what we have is an x mu class redefinition in terms of of Fourier so we could write this as n not equal to zero e to the power i n sigma plus times the number let's say beta to the power n in this issue of course we don't need n not equal to zero to be constant that's right some of it all right but there's one more thing we can do that also solves the equations of motion the point that we can do is to add a constant times sigma sigma plus what? now this doesn't respect the symmetry because when you take sigma sigma plus 2 pi yeah we can add this plus a constant times sigma plus doesn't respect the symmetry but remember we also had to say x minus when I add this so if we chose a constant in x plus and the constant in x minus would be the same then the sigma part cancels between the two you see x is then just a function of tau that's respect the symmetry but it's re-direction of sigma plus and sigma minus that respect the periodicity of sigma which allows us to say all these terms to see so this constant whatever it is can't be changed this is a physical object and I have all the p minus why all the p minus will become clear very soon another moment is just some number parametrizing different equivalence in times isn't this clear? so I have parametrized I have now looked at the phase space of my problem then in the space of all the equivalent classical solutions solutions of 20 by 20 d minus 1 d minus 1 between the functions of sigma plus and sigma minus because I have set one of these functions to something specific it's not quite something specific there is a number which can be parametrized in equivalent solutions isn't this clear? okay great now the next question I want to ask is now that I fixed this gauge now that I fixed this gauge in my therapy can I just plug it into the action or do I have to worry about the equivalent of the constraints that come from fixing this gauge let's see the constraints from fixing a gauge are always the equations of motion that would have followed had you not fixed this gauge so we want to know what are the equations that come from varying x minus can somebody tell me what are the equations that come from varying x minus in space the minus you get an equation for x plus exactly so the equation of motion that comes from varying to the state x minus as I will give you the check is there a plus equation then minus or x plus to plug in our solution into the action that would be fine if we were already also to impose this condition that x plus that's right to the function of x sigma plus sigma minus now where does it work because we don't independently have to specify this constraint because it follows from the earlier constraints remember that our earlier constraint was there plus maybe that's the thing that we accurately said maybe that's the thing but maybe it's more accurate to say the following suppose we do impose this as a constraint we say that x x plus is the sum of sigma plus and sigma minus that function of sigma plus and sigma minus that we get for x plus is completely determined by the other constraints why is that what are the constraints of that the constraints of that del plus x plus del plus x minus plus sum over i del plus of x i del plus of x i where i runs over all the remaining d minus two coordinates now we chose a fixed gauge such that del plus of x minus is equal to alpha p minus by capital therefore that del plus of x plus d equal to 2 by alpha p minus into sigma of i del plus of x i del plus of x i and the solution if we just treat this as the solution of what x plus is of what x plus is you know given what the other i's are doing obviously you raise the fact that it's function only of sigma plus the minus part of it is the fact that it's function of sigma minus okay now what have we done that we know that x plus must be function of sigma plus and sigma minus put in that fact plus the earlier constraint to completely solve the plus in this gauge completely solve this little inaccurate we have not solved the zero mode of x plus you see we only have solved that part of x plus that is that is visible in the derivative of expansion but x plus we have constant peace there's some constant peace that we have not solved before this part is almost apart from that one constant gel the independent of sigma plus we have completely solved for x plus in terms of the motion minus two points in this gauge going to ask a question now we are going to ask a question all this blah blah blah can we now write down a simplified action okay for a dynamical system that has the same solutions you know that is dynamically equivalent to the system that we started namely an action that has all the same solutions an action that has all the same solutions to the equations of motion as the one we want because solutions to the equations of motion are parameterized just by x i's and by this one number of p minus also where 0 on part okay if we were interested in writing down an efficient action if we were interested in writing down an efficient action that action should just not have x plus and x minus in it okay maybe another another way to say it is take the action of maybe another way to say it is take the action of the theory that we had and just plug in the gauge plug in the gauge gauge check to see whether varying the equations of motion that follow from that action the equations of motion that follow from that action all of the equations give us the solutions that we want instead of this confusing thing that we said again well the original action that we had and we plugged in our gauge condition in alpha prime x minus is equal to constant times p minus times if the equations of motion is the equations of motion that follow for x plus however putting in those equations of motion same that x plus and x plus of sigma plus and sigma minus independently plus the constraints allow us simply to determine x plus this whole process does not back-react on x times okay our natural system is an arbitrary left-moving, right-moving functions of x i and all of that the system does is determine what x plus is doing in terms of way so basically the conclusion we've reached is that it would be legitimate in this situation just to take the action we have in our gauge okay once we've done this we can even forget we can even forget about our constraint equations because the constraint equations don't constrain the dynamic of the degrees of freedom that remain just relates other degrees of freedom to these guys why don't we care okay there is one degree of freedom that we would be wrong in doing this well what should we do okay so let's move so what we can do so we have one by one by okay let me go back to the slide so our action was equal to one by minus one by four by okay that's right one by four by alpha prime and we had x x mu dot x mu dot minus x prime this was an action into this action we put the gauge we put the gauge condition we put the gauge condition okay into the action we put the gauge condition then I'll come back to zero once in a moment but that essentially eliminates what does it do it makes we put in a condition such that x mu so there's part of this that's x plus x minus that is x minus okay let's find the x x plus x minus part you see all that survives here in the x plus x minus part are zero more contributions you see suppose you took x plus and Fourier expand in it in modes of the circle and x minus and Fourier expand in modes of the circle we set all the non-trivial non-zero Fourier modes of x minus to zero but in this integral the only way you can get something non-zero is that a Fourier mode with momentum k clicks with a Fourier mode with momentum minus the action that remains is an action only for the zero modes of x plus and x minus so there's no action remaining for the oscillation modes of x minus that's by handling the gauge difference is zero or all the oscillation modes for x plus they just don't enter the action okay x i's are just not affected by what's going on so the final conclusion that I wanted to show is the if is see if you take the action one over four pi alpha prime if you take the action one over four pi alpha prime x plus dot x minus dot times two pi integral d t plus d two sigma x i x i dot square minus x i prime square square prime okay then this is the full action for the dynamical degrees of freedom left okay this is the full remaining you know full action for the degrees of freedom of interest okay what was this x dot and x minus these were zero modes so we took x and expanded this as just a function of sigma I'm not sigma I'm down we took x plus and expanded it as x x plus sum over n x plus into pi n sigma we already argued that all the modes with non-zero n drop out this is the mode with n equals zero the two pi is now doing the example over sigma yeah in two minutes okay correct okay so sorry okay so the next conclusion then is that the action 1 over 2 alpha prime x plus dot x minus dot plus 1 over 4 pi plus plus 1 over 4 pi alpha prime x i dot square minus x i prime square okay okay this action we supplemented with one remaining constraint the one constraint that was not that was not completely redundant when we did this whole thing that one constraint is the zero mode part of this constraint why is that? why is that that this one constraint is left behind okay the one constraint is left behind basically because of this you see as we will see when we do the quantization of the system this one solution couples to one degree of freedom in x plus okay so that part of x plus is not redundant then when you look at when you look at the subjective form that the Poisson brackets between degrees of freedom in our system x plus alpha oscillators are Poisson brackets with the x plus oscillators so but throwing away x plus mostly x plus degrees of freedom we are not doing anything there however the zero mode of x minus couples with the zero mode of x plus so if we want a quantum system we want to build a quantum system out of a classical system that has solutions parameterized by a number in the zero mode sector of x minus we need to include that quantum system the zero mode of x plus and the remaining constraint the one remaining constraint is going to tell gives us the constraint that links the dynamics of zero mode of x plus with z okay so we basically run out of time and I know this has been confusing it's been confusing for two reasons it's been confusing for two reasons one of which is that we are trying to take shortcuts we are trying not to give the full detailed justification of the result now that is of course very unsatisfactory way to predict and it won't remain in the next few lectures I'll give you a much better derivation a much better derivation of this action nonetheless the way physics is actually done is it's only 20 years after the work is originally done that people come up with good derivations of the results so it would be it is it would be silly to not pay attention to slightly shady but intuitively correct ways of proceeding because that's how you always do physics so what I would like is all of you to think through everything we've talked about today on its own terms couple of questions next class about this in the first 10 minutes of the next class I'll review the logic of what you went through and then we will start by using this to derive the spectrum of the work actually the second reason I think it's been confusing is that the logic of this procedure actually relies on a way of thinking of the system of the way of thinking of the phase space of the system that's slightly non-standard please think about this as carefully as you can in the first 10 minutes of the next class we'll review the logic in detail and you will attack it and I'll try to defend it ok, ready what is this actual the measure of anything this action supplemented by zero mode of distribution