 We drifted on the expression for it last time. It was d2z, you know, apart from all these factors, a Cs2 and g to the fourth, had the delta function for momentum conservation. That's all the way to there. It was d2z, 1z to the power, what was it? Alpha prime u by 2 minus 4. And 1 minus z to the power of alpha prime t by 2 minus 4. And, OK, so this is the expression that I've been interested in studying more in detail. OK, so we're going to study many aspects of this relation. But before we actually study in detail, I suggest to remind you again that this expression is not viewed as an analytic function of u and t. It's not, firstly, so to remind you, alpha prime u plus t plus s is equal to minus 16. It's a kind of natural relationship, we found last time. And, OK, so viewed as an analytic expression in u and t, it's not everywhere convergent, actually. It wants to condition convergence. If it is in convergence, it's firstly that alpha prime u by 2 minus 4 is greater than minus 2. And alpha prime t by 2 minus 4 is greater than minus 2. But also that alpha prime u by 2 minus 4 plus alpha prime t by 2 minus 4 is that the sum of these two is less than minus 2. So it dies off at infinity. Let's quickly take. So what was it? So k1 plus k2, the whole thing squared plus k1 plus k3, the whole thing squared plus k1 plus k4 plus k2 squared, which is 3k1 squared. And then from that mathematical observation, we've got another minus 2k. So that's k1 squared minus alpha prime. What's my minus? k1 squared minus alpha prime plus k1 squared plus k1 squared plus k1 squared plus all of these are possible. All of these are possible. Right, so then maybe I'm wrong. OK, good. But now let's remind us that it's going down. So I'm going to introduce you to number 16 now. So each of these is 4 by alpha prime and 4 by k. And then I'll follow. Right. OK, good. So we've got plus 16, maybe not plus 16. Exactly. That's probably the same. Let me just check this. It's just getting easier to see. He has minus 16, but then he put some minus alpha prime u. His u is probably minus of k1 plus k2 squared. So let's maybe try to be consistent with this. So we've got a minus here, minus here. We've changed our definition to s is equal to minus of k1 plus k2 to the power of k2 squared, which is a good thing, actually. s will be positive at high n, at least, this way. That's as t is equal to minus of k1 plus k2, because of k2 squared, u is equal to minus of k1 plus k2 squared. And then this thing is right to the minus. OK, thank you. In order for this to convert, the sum of these two from these two relations has been greater than minus 1. It ought to be less than minus 2. Well, that's fine. So there's a window between minus 4 and minus 2 in which all three relations in this aspect. So there's a place where it's converted to be evaluated there, and evaluate the entire equation. In order to do this, we have to understand how to do this in 10. So we can understand 10 in this input. So in the 10. What if we do it to minus 1? Yeah. Now with this, this should be minus n minus c. This whole coefficient here has to be greater than minus 2 in order to get convergence at 0. The other coefficient has to be greater than minus 2 in order to get convergence at 1. And the sum of these two coefficients has to be less than minus 2 in order to get convergence in unit. But it's important that these two conditions can sign with the instant efficiency. So the integral that we want to perform is an integral of the following form. It's an integral of the form mod z to the power 2 a minus z to the power 2 b minus 2 b to the power z. a is equal to minus of alpha prime u by 4 plus 1 is equal to minus of alpha prime t by 4 plus 1. a plus from a plus b. You see, if we take a plus b, we find minus 2. And then from here, we have alpha prime by 4 minus alpha prime u by 4 minus alpha prime t by 4. But remember that the sum of alpha prime by 4 with a minus sign is plus 4. So this thing can be written as a plus b is equal to minus 2 plus 4 plus alpha prime. But if we define c is equal to alpha, so that's equal to 2 plus alpha prime s by 4. So if I say it's equal to 1 plus alpha prime s by 4, the symmetric definition. Then we see that we have a plus b plus c. And a is associated with u by, let's see, sorry. So we have a is associated with u by this formula, b with t by this formula, and c is associated with s by this same formula. So as far as this integral is concerned, I'm going to just simply define a c such that a plus b plus c is equal to 1. And write the answer in some symmetrical way in terms of a, b, and c. And we know what c is in terms of x. We can already express anything in terms of 2 where everything we want, but it'll look more symmetrical a bit less than 3. So now let's go ahead and read it. So this is the integral that we want. So how are we going to do this integral? No, I want a plus b plus c equal to 1. Let's do this integral. So we're interested in this integral here. And in order to do it, we're going to use a trick. Well, we're going to use the fact that mod z squared to the power a minus 1 is the same thing as e to the power of minus a. e to the power minus mod z squared b divided by 1 over gamma of now what is it? If you have t to the power n plus 1, we have now more a. So minus a can be written as minus a minus 1 of minus a. Just make sure I've got it all right? So you know that t to the power minus n, and t to the power n is factorial, which is the same thing as the gamma of n plus 1. So I should write 1 and more than the coefficient that appears in the urn. So I should write this is minus correct. So let's see. Why is this true? It's true by just doing the change of variables. Suppose we find a new variable t to the power of just mod z squared times b. We'll absorb a way to change to get the mod z squared times a minus 1, as you can see by the power of t. And then what remains is the definition of gamma of minus a. Good. So similarly, we put in 1 minus mod z mod squared to the power of b minus 1 over gamma of 1 minus b. And then technically u, u to the power minus b into the power minus 1 minus 1 z mod squared times u. Let's insert these identities into the integral expression that we want to compute. So what do we get? We get g to z, then we get dt divided by gamma of 1 minus a, gamma of 1 minus b times exponential of. And now let's write z as equal to x plus iy. So we get minus t into x squared plus 1 squared. And minus u in a 1 minus x over x squared plus 1 squared. We're using Polychinsky's convention for all numbers. So for Polychinsky, d to the z is equal to dz times dz bar, which is twice of dx and y, because it's bounded between z and z bar, x and y, that's what we've determined still. So coincidentally, now we're going to add ui times x times x squared, simplify the exponential. So we simplify the exponential by completing square s. So it's minus t plus u into x into, OK, let's first write it out. x squared plus y squared. Then there is the linear term in x, which is plus 2 x u. And that's the constant piece, which is minus u. So that's also the d to the bar minus a and u to the bar minus a. So we get that. d to the bar minus a, d to the bar minus x. So the next step what we do is to complete the square exponent. d to the bar minus a, u to the bar minus b, d, d, d, x, d y, exponential of, OK, so now when we complete the square, we get something. So some shifted x, OK, so it's d plus u into x minus what's the shifted, u by d plus u plus y squared. And then we have to do the subtraction here, which gives us minus u and then plus of where or what. So that's 2 d to the bar minus a, u to the bar minus b, d, d, u, d, x, d, y, exponential of minus d plus u into this square business. And then let's simplify this. That is equal to minus of d, u divided by d. So now the next step, because we do the, now see the integral to x in the shifted x and y. So that gets you to the x and y integral. And we pick up a factor of square into pi, the whole thing squared. So that's 2 pi, I mean we have 2 here, so that's 2 pi. That's the integral of d to the bar minus a, u to the bar minus b, d, u, d, d, exponential of minus d, u, by d plus. Do you think this is a poorly detailed, by the way, because this is, you know, one and the same. You know, one's first formula is still here, 6, 8, 9, 6, 9, 6. So now the next step, and the next step what we do with, well, we've got some simplification to reason what to entangle to do after we do it. So the next step what we do is to say, well, let's switch variables to an overall scale of the u space and the ratio. So let t is equal to v alpha x, and u is equal to 1 minus x, and so on. So let's do, let's look at the terms that you go through for this change of variables. So dt by the x is alpha, let's say dt by the alpha is x, dt by the x is alpha, d, u by the alpha is 1 minus x, and d, u by the x minus alpha. So this is the determinant of change of variables, which is equal to, whose modulus is equal to 1, and the x alpha is x. OK, so there's no Jacobian for the change of variables. Fine. This expression here has 2 pi. Then this becomes, there's the entangle over alpha as true, right? So we want to say entangle over alpha. So we have 2 pi alpha to the power minus, Jacobian is equal to alpha, the alpha. So alpha to the power minus u, minus a, minus b, we have plus 1, OK? Times, then we have x to the power a, 1 minus x to the power b, and then we have exponential of minus alpha in the x and the y. It doesn't, Jacobians are, you know, what do you mean? What do you mean by sine? Yeah, I mean, p over n. Now we do the entangle over alpha. So the entangle over alpha gives us what? Firstly, we get gamma of a plus, of minus a plus, and somewhere along the way, we lost 1 over gamma. We lost 1 over gamma, so we could come back. It's just broad spaces a little bit. So we have 2 pi over, let's go to back it now. So gamma of minus a minus b over gamma of 1 minus a, and the gamma of 1 minus b. Well, times, we get this factor of x to the power minus x from power, OK? x to the power minus x from power, OK? So for each factor of alpha, we should get a 1 over x to the power minus x, OK? So we get integral dx x to the power minus b plus 2. 1 minus x to the power minus a plus. Each factor of alpha has a 1 over alpha. Sorry. So we should, when we have an alpha, we should take it to x to the power minus 1, 1 minus x to the power minus 1. So we have this factor, but we also have an alpha plus 2. So this should be replaced by x to the power minus 1. 1 minus x to the power minus 1 into minus, so we should make those pluses, so e plus b minus 2. It's 2 because there's a d outside of these d alpha plus, OK? And we have x to the power a, 1 minus x to the power minus b. Or I have to say, let's see. That's important. Thank you. We should have got a factor of 1 over e plus b. Just 1 over root e plus b squared, because this is x squared, I think. Just a factor of 1 over d plus b. Yeah, so that could be another factor of alpha. Or I hope we can trace this to the length here. I mean, I don't think we should say something like that. Is this fair? Thank you, thank you, Bhuvan. So we have 1 over d plus b. Thank you. I think the coefficient of x squared. No, it's not that difficult. x squared and x minus shift. Yes, but this is a very big question. Why? It doesn't matter. If you write it as an x minus u, it's not going to be like this. Yeah. OK, so let's 1 over d plus u here. Maybe at some point, we should go to the answer. Let's try. OK, let's try. So what do we get? We had d to the power minus a, u to the power minus b. d, u, d, d, d. So what the powers of alpha can now get by probability, right? So it's alpha to the power minus a minus b plus 1, which I'm trying to know. There's no plus 1 here, but there'll be even d alpha. This just cancels the alpha we get from the chip. Right? Same. OK. So this gives us a gamma of plus 1. T to the n is gamma of a plus 1. Plus 1. OK. Let's see. Let's take it here. OK. So what do we have here? So the remaining integral is dx into x to the... Did I get this part right? a plus b minus 1. That's 1. We have x to the power minus 1. 1 minus x to the power b plus a minus 1. Let's see. Now, the next step is to use the formula. The next step is to use the formula for the beta. You know, if you had a... That's not the equation. Let's just go through. This thing, you know, there's this beta function formula which tells you that this thing is equal to gamma of 2 a plus b in the gamma of 2 b plus a in the gamma of the sum of... OK. So the... The ring of integral integrations for T and u is 0 to infinity. Alpha and... x goes from 0 to 1. And alpha goes from 0 to 2. I have messed up some... You know, I messed up some signings up. I just killed it in one and I'm sorry. But there are previously different variables and they're like a few with... You give me one more minute. This integral is 0. This is the... Somehow, I mean I should have done like these extra powers with a negative sign. I don't think we'll miss the others. Ah, these were the negative sign, right? This had minus a and this had minus b. Because it was... t to the power minus a and t was x vector. OK. Sorry. So next modification. So this is minus and minus. This part is correct. But this is from... So what do we get here? Here we get b minus 1 and here we get a minus 1. Ah, this is a minus 1. OK. So now this thing. Gamma of... Gamma of... You remember this classic formula. You remember you have once learned this classic formula. What do you know, beta function formula? Yeah, this is a derivative value in terms of gamma functions. In terms of gamma of b. We're finished. We have to multiply this with this and that's our answer. OK. So what's the answer? It's 2 pi. Now, we use the factor b plus a minus b is c. So b plus b plus c is equal to 1. OK. So it's 2 pi into gamma of a gamma of b gamma of c divided by... And then we use the factor b plus a is 1 minus c. Gamma of 1 minus a you know gamma of 1 minus b is gamma of 1 minus c. You just remember what a b and c were. So we written that down on the border. What was it? a was equal to... And so did it for b and c. s is equal to... Well, so when we have all these delta functions in here, times c s 2 times g tacky on to the power 4 times 2 pi times gamma of an a b c in the denominator in the numerator. So minus 1 minus alpha prime u plus 2 divided by 4 and the next line into gamma of minus 1 minus alpha prime u by 4 gamma of minus 1 minus alpha prime s by 4 gamma of minus 1 minus alpha prime t by 4. The whole divided by b to 1 minus a a that is divided by gamma of 2 plus alpha prime u by 4 gamma of 2 plus alpha prime that's 2 pi to the power 26 and the delta function in the sum of all the things that we can see. What was the aspect of that? You see that this claim that we made in the last class that you know it was obviously symmetric under these digital x, d and u. The answer is manifest in this form. The format. Okay. Any questions can proceed to study this. No, you see at the moment it's a function of our arbitrary complex number. It's like there's no claim about positive. So let's let's let's start and immediately address the question that you have. Okay. So the question that you might have is where this formula has has poles. Okay. So what do the poles mean? So let's first understand what the first pole is. So let's suppose we're working with a variable s. So it has it's completely symmetric with three variables. So any statement about the poles in one variable will be affected. Okay. So let's say we're working with a variable s. Okay. So the s the way we've done it is is something positive and gets increasingly positive as we go to higher and higher energies. Okay. Therefore this gamma function here gets increasingly negative as we go to higher and higher energies. So we keep running into four or more poles. Because gamma is a negative integer. So that's the lowest energy poles. Okay. So the first time we get a pole is when this object goes to see it. So we have minus one minus alpha prime s by four is equal to the pole in the variable s. That's the smallest value of s for which we get a pole. And that adds an s is equal to an s is equal to minus four by alpha prime. Now what does it mean to have pole in s is equal to minus four by alpha prime? Exactly. Which you see what's happening is that we've already seen that our theory has a three cation vertex in fact in cation. If you have two cations propagating with this cation you do a third one. We can then produce to what happens. So this diagram that I draw all lines here that this contribution you see this answer is the fuller answer for the estimate. It's not you know some effective action from which you have to build trees. It's the fuller answer for the estimate. So this tree level exchange of all particles is the method that you can get by the way you do that beyond this and running in a channel will contribute and each of these contributions will give you a pole. It can make you know what's the mass spectrum of string theory. But mass spectrum of string theory with the the spectrum of closed strings is at m squared equals 4 by r top right into n minus 1 where n is all possible intensity number operator of the of the oscillators and every value of s that is that takes the form for any integer n. Now let's check when do we actually have poles? When we have actually a poles we have poles when alpha prime s by 4 minus minus 1 is equal to minus n which tells you that's when alpha prime s by 4 is equal to n minus so you put the 4 by so s is equal to 4 by alpha prime n minus 1 which is exactly the same places. Okay. So while scattering out with you has poles but it has poles exactly where we should expect them to be for physics books. Of course it's the same with the t-channel or the u-channel it's the same thing. Now let me understand this okay the next thing we should do is to try to understand this a little better. So okay if you're so smart you know not to say it has a pole but check that the rest of you is the right thing. What we have we have bad things this scattering amplitude remember was gt q times cs the other one was the same thing so the pole gt to the sixth times cs2 squared should be equal to the rest. gt to the fourth times cs2 times times all of it where we should plug alpha prime which is probably s is equal to minus of 4 by alpha prime to this okay this would be like s is equal to minus of 4 by alpha prime to this what do we get well let's see firstly from here we get gamma of 1 which is 1 just start floating around but I click I click with the other stuff cancels let's see how that works it's actually easiest to see you know we have the formula gamma a gamma b gamma c for gamma of 1 minus a gamma of 1 minus b gamma of 1 minus c and where we go our first pole is when let's say a is equal to c okay so when a is equal to 0 b is 1 minus c and c is 1 minus b so this this gamma of b cancels the gamma of 1 minus c and gamma of c cancels the gamma of 1 minus c so you see the residue to this pole the residue here so so so I'm claiming that the structure it's the same thing here just harder to see because there are more there's more kish yeah so we set s to the same we set s equal to 0 and we get well what do we get we get that in the neighborhood of that so we set s equal to this thing equal to 0 we get that in the neighborhood of that we get this to the pole that cs2 that's 2 pi times times 1 over this quantity so 1 over s minus alpha prime by 4 into s minus 4 s plus 4 written in this quantity is more needed value in the appropriate way versus s okay so in the neighborhood of this pole this is what that function looks like the residue of the pole when the residue of the pole in s minus 4 by alpha prime so that's 8 pi by alpha prime times cs2 times 2 pi which times gd to the 4 which would be the same thing as gd to the 6 times cs2 the point is the one thing we didn't know with what number we should put each vertex of that's why we named this gd which was some 3 3 now this relationship in physics that has to be still gd so we cancel this we make this square we cancel one factor of cs2 and we find that gt squared equal to 8 pi 5 alpha prime divided by cs2 we want to change all these things seriously from unit energy you know for the you know for the for the pre fact that we have to put beyond behind the vacuum okay so there's no check here but it's a way of determining what the right answer is for the pre fact now of course once we return if we calculate 5 and 6 point functions that act here then there will be checks and there's a lot of work out we're not going to try to do that but we determined what's it for all and now every time you have an act here on operator this is what you have to put in is this here now that we have what gt squared is in terms of cs2 what gt squared is in terms of cs2 let's go back and look at our let's go back and look at our three point function calculation involving one radicon and two tachyons and see what we can say about the normalization remember from last time we had our thought prime by like k23 mu k23 mu at the amplitude for one radicon a radicon in particle one the scatter to tachyons which are particles 2 and 3 will be proportional to this object we have our prime by 4 in the factor of 2 in our basic formula for lx and there was a second factor of 2 because we took k2 and made it we took k2 and made it k2 minus d3 by 2 this was dressed with one with a gt squared and then a gmask so it was dressed with a cs so gt squared and cs2 we know it's 8 by 5 gm times okay I'm not giving you that sign in the read back factor so our thought prime by 2 times 5 times k2 by 3 mu k2 by 3 mu times d mu mu times g that was gm by 5 it's used for this normalization not actually putschitzky but putschitzky that is plugged in this relationship so that's why I got one out of our prime and this is not just this is not 2 tachyons at the ground up but it's still not clear what it is because we still don't know what the what gm is but g for the master state for the ground up so how do we do that well I'm not going to actually go through the calculation because like it's not like I'm not going to go through the numbers right but let me just see how it works okay once again we want to take this 4-point function calculation and factorize it on the pole but this time it worked with the pole that had s is equal to 4 by out of prime it would be the pole that had s is equal to 0 once it's in the pole we get 2 factors of this vertex divided by the pole so the residue should be 2 factors of this vertex divided by the pole now there's not one master's part but the many master's part because we're supposed to sum over all master's part in the game so we get factors of this another factor with another in prime here factors of momentum and then there's supposed to sum over all polarizations that sum over polarizations can be done giving you a completeness thing okay and so when you do that and you plug it all in you get some powers of t right because you have 2 factors of momentum in each vertex so you get 4 factors of momentum in the end in the end there's something in there so it's some power of t that is quadratic some power of t or t some schematically it's scalar of form that you cannot get on your pole now you can go back and ask is that kind of residue that you actually get from the forward so let's study just without putting your APCR for a moment so suppose we study for APCR suppose we took it gamma A gamma B gamma C over gamma 1-A gamma 1-B gamma 1-C we saw that if we looked at the pole and A equals 0 the answer is independent of B and C the residue to that that goes in there so let me ask is this way to continue if we look at the residue of the pole and A equals minus 1 is the answer still independent of C and the answer is no, at least so if we look at the pole and A equals let's say minus N okay B will be equal so now let's say that we can substitute since A plus B is equal plus C is equal to 1 okay since A plus B plus C is equal to 1 let's say in substitute for C in terms of B so C will be equal to 0 1 plus B if A is equal to minus N we will accept so what we will get is something of the form of 1 by A times N factorial times gamma of B gamma of 1 plus N minus B over gamma of 1 minus B times gamma of 1 minus so 1 plus B minus now when N is equal to 0 the 1 cancels this cancels this this cancels this and we will be very happy but N is not equal to 0 we don't have perfect cancellation however we can use properties of the gamma functions the fact that gamma of N plus 1 is N times gamma of N it gets a simple formula let's do it for N equals 1 so for N equals 1 what we have is gamma of B gamma of B over gamma of 1 minus B into gamma of B minus 1 now this thing is equal to B minus 1 into gamma of B minus 1 and this thing is equal to 1 minus B into gamma of 1 minus B so we get B minus 1 into 1 minus B ok into into 1 using the formula gamma of N N plus 1 is N times gamma ok you can check more generally that if we went you can easily work out the formula that you get at high end it's like B minus 1 double B squared B minus 2 double B squared ok but let's remember in our context what this ok I am going to pull you closer with me much more you can work this out in terms of STMU compare convert that to momenta see that matches what you get from this thing like keep on this that's something quadratic in the T's ok which is something quadratic in the momenta as you as you get to get to get to take from this and then you can try to work it out in more detail work out the numbers I am not going to I am not going to try to keep on here like what I said that when we do this of course you get all the momenta part matching but that's a number that's left ok and then the GM will just be fixed by the bias matching and when you do that what you find GM is 2 divided by 2 divided by you get GM is equal to G tax non times you see you have two factors in GM left and you have two you see you have the three parts Catholic was GT was had a GT square times GM matching that's a GT to the 4 so the GT is cancelled and then you get GM is but there's a CS2 that's left which you can re-express in terms of GT so in the end when you know you have GM is equal to GT 2 divided by there's actually another way of working out the relative ratio between these two numbers and that is through the state operator you see the logic for these vertex operators was they were the operator that corresponded to a given state now one can ask with what normalization should we put the operator but if you think of it through the state operator map there's no guarantee we should put the state with normalization 1 and work out the operator that is dual to that state with normalization 1 now in relation to all of that you may want to put an overall factor because you're competing in S&P but the relative normalization between the two operators are two different things should be fixed by this so when the industry is in excellent work out the relative normalization from the state operator map of the vertex operators for the master states compared to the vertex operators for the datum being careful to ensure that the states to which your operators are dual are unit normalized you'll find the same some sort of check so once you get confident about this logic you just use the state operator map to find the coefficient that you put for behind the vertex operators and pass over a few checks that's the last thing I wanted to the last two things I wanted to tell you about before we quickly come to this we need to prepare for the next talk it is about the high energy behavior it's about the high energy behavior for example actually call that I mean quickly say one thing we have a lot of fancy analysis involving media involving the exact answer in order to determine the poles and the rest of the use the first pole is actually very easy to directly from the first integral representation and that's we have z to the power minus alpha prime by 4 u minus minus 4 times that's regular at z equals 0 and we want to do this integral we want to do the integral of d2z so what do we do we want the neighborhood of that thing about to become an origin so we take this and convert it into a radial and we get 4 pi from the angular integral because our measure d2z is twice the x u then this becomes r dr r to the power minus alpha prime u by 4 minus 4 minus 1 2 minus alpha prime u by 2 minus we know the answer integral r to the power n has derivative so if we integrate r to the power n we get 1 over n 5 so we get 1 over alpha prime u by 2 plus radius minus 3 and that is plus 1 so that's 8 pi over alpha prime u plus 8 pi by alpha prime into u plus alpha prime by 4 this is the same formula this is the simplest way to get it just to check that we have done anything crazy now exactly what we should be doing plus plus things so that's exactly what we want to do the last thing I want to tell you about is that higher than your behavior so I need to realize the scattering amplitude in terms of s t and u I am going to ask for help for some of these data loads so what was it? it was 2 pi times product of 3 damas divided by product of 3 damas can you help me build up to your arguments of these minus 1 minus alpha prime minus 1 minus alpha prime s by 4 similarly minus alpha prime I am sorry the energy limits that are of interest that's of interest is the limit in which we take s goes to infinity t goes to infinity u goes to infinity you know with the modulus of s goes to infinity modulus of t goes to infinity modulus because x plus t plus u c is equal to 1 it's kind of all over the polarity infinity is equal to 7 so before working on the answer there let me remind you what that means suppose you've got some particles suppose we work in the center of mass and we have two particles colliding at each other in the x direction so we have particles whose energies come this is k 1 and k 2 is at 0 it's the center of mass frame but the energy is not at 0 so we've got say k 3 is what k 1 becomes after its deletion so it becomes square root s by 2 times 1 cos theta sin theta 1 so it's theta 0 and k 4 was square root s 1 minus cos theta minus sin theta 0 0 after its capture so this is the momentum diagram something that comes in like this sometimes capital is a natural detail it's really k 2 simple matter with this momentum configuration we calculate s d nu s d nu for instance s is s we set it up that way you add the two of these factor 2 goes then you take the square root s but now we want to calculate t basically the point is that suppose you keep the angle fixed suppose you keep theta to be some fixed angle and take s to infinity because everything is multiplying some fixed function of theta with this factor so i angle high energy fixed angle if the energy is very large studying what's going on the limit s d nu ok so the question that we want to ask is what happens to our formula in this high energy fixed angle it's captured it's very large we can just use the crudest approximation that there is for a term large that is gamma of x is equal to x to the power x plus thanks to corrections so x is x to the power x so here what do we have we can figure out what the 1s and 2s are so what do we have in s up here it's minus alpha prime s by 4 to the power minus alpha prime s by 4 and in the denominator we have alpha prime s by 4 to the power alpha prime the modulus of this if we take modulus this goes like alpha prime s by 4 to the power minus alpha prime s by 2 times ok the key point this long s is a slowly growing function so roughly speaking but this is even further down but roughly speaking this amplitude is exponential in damage at large and the thing that sets the dimensions for this exponential dampening is the strength here physical interpretation this is a very important point the strength is actually becomes very small at very high energies it's very important because it's the opposite of the behavior of gravity on gravity on spectrum in the eye standard gravity on gravity becomes very large apparently at high energies the effect of gravitational coupling involves powers at the moment ok exponential damping of high energy amplitudes which is a general feature of string theory ok is at the heart of the finite in the substrate we have to be assured that you know that string loop amplitudes are finite they don't need to be cut off the physical reason for this is the exponential dampening things don't interact very much they do interpretation of exponential dampening have you ever studied physics before in computer and exponential behavior of a scattering amplitude you know in a very important extent something to change physics at the turn of the last century before this discovered the structure of the atom was by taking electrons and throwing it in the air and he done the calculation that showed him that you know they were they knew that they were positive and negative charges of the atom one model was this I don't know plump pudding model of the atom there's this large soup of virus the model in which the charge the negative charge of the atom maybe the positive I don't even know what the charge of the atom is uniformly spread over the range the scattering the scattering of the electrons of the atom would have been very very small it's just a general feature of Fourier transformation like if you scatter that a smaller wavelength over a sauce that is very large that clicks to your Fourier transform and it's like if you have a lens scale that is very much larger than the lens scale associated with the scattering wavelength you get exponential suppression Fourier transform very high gain while you get these zines corresponding to the fact that the sauce that was doing the scattering was very small this is localized nucleus this is a totally important experiment turn of the last century now what we are saying is that when we scatter energies high compared to the string scale when we make our moment work we get this exponential damping so it's what Ruffberg had originally explained it tells you that string states are big fuzzy objects on string size on string scale and the fourth factor which is fancy way of saying Fourier transform of the source from which scattering occurs causes exponential damping as strings scattering makes each other very high the key to the exponential damping is that strings are big fuzzy objects or strings do we have to stop by death? yes we have to stop one moment which is one moment so that's what happens at high energy fixed angles there is another thing that we have to do that was historically it was a physically important instrument and that is high energy scattering not at fixed angles but when we take s large but at fixed then you can ask what the hell is that and we can check if I go into the formulae but that corresponds to taking s to infinity while keeping s and theta square fixed so you are looking at low angle scattering and how long the angle goes to 0 as s goes to 0 ok I am going to leave it as an exercise which I am really urge you guys to do it's 5 minutes but please do it when you calculate the approximation of this stuff in the large s-fixity limit remember if you take large s and fixity you must be taking large u as well because s plus t plus u is equal to 1 ok so large s-fixity limit what you are finding is s to the power s to the power which is the regular formula is an example