watch casio model

the followingcontent is provided under a creativecommons license. your support will help mitopencourseware continue to offer high qualityeducational resources for free. to make a donation, orview additional materials from hundreds of mit courses,visit mit opencourseware at ocw.mit.edu. professor: so we are goingto switch directions. rather than thinkingabout binary variables,

these ising variables,that were discrete, think again about a lattice. but now at each site we put aspin that has unit magnitude, but m component. that is, si hascomponents 1, 2, to n. and the constraint that thissum over alpha si alpha squared is unit. so clearly, if i-- let'sput it here explicitly. alpha of 1 to n, so that'swhen i look at the case of n

equals to 1, essentiallyi have one component. 2 squared has to be 1, soit's either plus or minus. we recover the ising variable. for n equals to 2,it's essentially a unit vector who'sangle theta, for example, can change in three dimensionsif we were exploring the surface of the cube. and we alwaysassume that we have a weight that tends to makeour spins to be parallel.

so we use, essentially, thesame form as the ising model. we sum over near neighbors. and the interaction,rather than sigma i sigma j, we put it as thissi dot sj, where these are the dotproducts of two vectors. let's call the dimensionlessinteraction in front k0. so when we want to calculatethe partition function, we need to integrateover all configurations of these spins of this weight.

now for each case, wehave to do n components. but there is a constraint,which is this one. now i'm going to be focusedon the ground state. so when t equals to 0, weexpect that spontaneously the particular configurationwill be chosen. everybody will be alignedto that configuration. without loss of generality,let's choose aligned state to point along thelast component. that is, all of thesi at t equal to 0

will be of the form 0,0, exceptthat the last component is pointing along someparticular direction. so if it was two components, they component would always be 1. it would be alignedalong the y direction. yes, question? audience: what dimensionalityis the lattice? professor: it can be anything. so basically we have twoparameters, as usual. n is the dimensionalityof spin, and d

would be the dimensionalityof our lattice. in practice for whatthe calculations that we are goingto be doing, we will be focusing ind that is close to 2. now if the odd fluctuationsat finite t, what happens is that the state of thevector is going to change. so this si at finitetemperature would no longer be pointing alongthe last component. it will start tohave fluctuations.

those fluctuations will changethe 0 from the ground state to some value i'll callpi 1, the next one pi 2, all of the big pi n minus 1. and since the whole entirething is a unit vector, the last component has toshrink to adjust for that. so we would indicate thelast component by sigma. so essentially this subspace offluctuations around the ground state is capturedthrough this vector pi that is n minus1 dimensional.

and this correspondsto the transverse modes that we're lookingat when we were doing the expansion of thelandau-ginzburg model around its symmetrybroken state. in this case, thelongitudinal mode, essentially, isinfinitely stiff. you don't have theability to stretch along the longitudinal modebecause of the constraint that we have put over here.

so if you think back,we had this wine bottle, or mexican hat potential. and the goldstonemodes corresponded to going along thebottom, and how easy it was to climbthis mexican hat was determined by thelongitudinal mode. in this case, the mexican hathas become very, very stiff to climb on the sides. so you don't have thelongitudinal mode.

you just have thesegoldstone modes. the cost to payfor that is that i have to be very careful incalculating the partition function. if i'm integratingover the n components of some particular spin,i have to make sure that i remember that this sumof all of these components is 1. so i have to integratesubject to that constraint. and the way that i havebroken things down now,

i'm integrating over the n minus1 component of this vector pi. and this additionaldirection, d sigma, but i can't do bothof them independently. because there's adelta function that enforces that sigma squaredplus pi squared equals to 1. the pi squared correspondsto the magnitude of this n minus 1component vector. and essentially, i can solvefor this delta function, and really replacethis sigma over here

with square root of1 minus pi squared. but i have to be a little bitcareful in my integrations. because this deltafunction i can write as a deltafunction of sigma plus or minus square rootof 1 minus pi squared. and there is a rule that ifi use this delta function to set sigma to be equal tosquare root of 1 minus pi squared, like ihave done over here, i have to be careful that thedelta function of a times x

is actually a delta functionof x divided by modulus of a. so essentially, i have tosubstitute something here. so this is, in fact, equalto the integration in the pi directions because of theuse of this delta function to set the valueof sigma, i have to divide by the squareroot of 1 minus pi squared. which actuallyshortly we will write this in the following way. i guess there's anoverall factor of 1/2

but it doesn't really matter. so yes? audience: so what doyou do with the fact that there are two places wherethe delta function is done? professor: i'mcontinuously connecting to the solution thatstarts at 0 temperature with a particular state. so i have removed thatambiguity by the starting points of my cube.

but if i was integratingover all possibilities, then i should reallyadd that on too. and really just makethe partition function with the sum of twoequivalent terms-- one around this ground state,one around another state. audience: the product issupposed to be for a lattice site-- for the integrationvariable, not for the-- professor: right. so i did something bad here.

so here i should have written--so this is an n component integration that i haveto do on each side. now let's pick one of the sides. so let's say wepick the side pi. for that, i have a smalln integration to do. what does it say? it basically saysthat if, for example, i am looking at thecase of n equals to 2, then i have startedwith a state that

points along this direction. but now i'm allowingfluctuations pi in this direction. and i can't simply say that theamount of these fluctuations, pi, is going let's say fromminus infinity to infinity. because how muchpi i have changes actually whether it issmall or whether it is large when i'm down here. and so there's constraintslet's say on how big pi can be.

pi cannot be larger than 1. and essentially, aparticular magnitude of pi-- how much weight doesit has, it is captured by this. so that's one thingto remember when we are dealing withintegration over unit spins, and we want to lookat the fluctuations. the other choice of notationthat i would like to do is the following. i said that mystarting [inaudible]

is a 0 sum over all nearestneighbors si dot sj. now in the statewhere all of the spins are pointing in one direction,this factor is unity. so the 0 temperature state getsa factor of 1 here on each one. let's say we are in ahyper cubic lattice. there are d bonds per site. so at 0 temperature, iwould have ndk0, basically-- the value of this ground state. and then if i havefluctuations from that state,

i can capture that asfollows, as minus k0 over 2. it's a reduction in this energy,as sum over ij si minus sj squared. and you can check that. if i square theseterms, i'm going to get 1, 1, 1/2, whichbasically reproduces this, which actuallygoes over her. and the dot product, minus 2 sidot sj, cancels this minus 1/2, basically givesyou this exactly.

so the reason i writeit in this fashion is because very shortly, i wantto switch from going and doing things on a lattice togoing to a continuum. and you can see thatthis form, summing over the difference betweennear neighbors, i very nicely can goto a gradient squared. so essentially that'swhat i want to do. whenever i have asum over a site, i want to replace it withan integral over a space.

and i guess to keepthings dimensionless, i have to divide by a to get d. so i can call that the densitythat i have to include, which is also the same thingas the number of lattice points in the body of the box. so my minus betah in the continuum goes over to whateverthe contribution of the completelyaligned state is. and then whatever thedifference of the spins

is, because of thesmall fluctuations, i will capture through anintegration of gradient of s and i call the originalcoupling that we're basically the strength of theinteraction divide by kdk0. clearly in orderto get the coupling that i have in thecontinuum, i have to have this fact of a to the d. but then in the gradient, ialso have to divide by distance. so there's something here, afactor of a to the 2 minus d

that relates these two factors. it should be theother way around. it doesn't matter. audience: question. professor: yes. audience: this step isonly valid for a cubicle-- so if it was somethinglike a triangular lattice, or something, there would besome miracle factors here. audience: but imean like writing

the difference ofthe spin squared as the gradient squared. like if it were atriangular lattice? professor: yeah,so the statement is that whateverlattice you have, what am i doing at thelevel of the lattice, i'm trying to keepthings that are close to each other aligned. so when i go to the continuum,how is this captured,

it's a term like agradient squared. now on the hyper cubiclattices, the relationship between what youput on the bonds of the hyper cubic lattice andwhat we've got in the continuum is immediately apparent. if you try to do it onthe triangular lattice, you still can. and you'll find thatat the end of the day, you will get the factorof square root of 3,

or something like that. so there's some miraclefactor that comes into play. and then at the endof the day, i also want to replace thesegradient of a squared naturally in terms ofessentially s has n components. n minus one of them are pi,and one of them is sigma. so this would be minusk/2 integral d dx. i have gradient ofthe pi component, and a gradient of thesigma component squared.

so after integrating sigmausing the delta functions, i'm going to thecontinuum limit. the partition functionthat we have to evaluate up to various non-singularfactors, such as this constant over here, is obtainedby integrating over all configurationsof our pi field, now regarded as acontinuously varying object under the dimensional lattice. and the weight,which is as follows,

there is a gradient ofpi squared, essentially this term over here. there is the gradientthe other term. the other term, however,if i use the delta function in square root of 1minus pi squared squared. and then there's thisfactor from the integration that i have to be carefulof, which i can also take to the exponent, and writeas, again, this density times log of 1 minus pi squared.

there's, i think, afactor of [inaudible]. so the weight that ihad started with, with s dot s was kind ofvery simple-looking. but because of theconstraints, was hiding a number of conditions. and if we explicitlylook at those conditions and ask what is theweight of the fluctuations that i have to putaround the ground state, thesegoldstone modes, that

is captured withthis hamiltonian. part of it is thisold contribution from goldstonemodes, the transfer modes that we had seen. but now being more careful, wesee that these goldstone modes, i have to be careful aboutintegrating over them because of the additional termsthat capture, essentially, the original full symmetry,full rotational symmetry, that was present inintegration over s. yes?

audience: the integration--the functional integration, pi should be linkedup to a sphere of 1. professor: this will keep. so i put thatconstraint over here. and it's not just that itis limited to something, but for a particularvalue of pi, it gets this additional weight. so if you like, once itry to take my integrals outside that region, thatfactor says the weight as usual.

so this entity is calledthe non-linear sigma model. and i never understood why theydon't call it a non-linear pi model. because we integrateimmediately with sigma. that's how it is. so what you're going to doif we had, essentially, stuff that the first term not includedany of the other things, we will have had theanalysis of goldstone modes that we had done previously.

the effect of thesethings, you can see if i start makingexpansion in powers of pi, is to generateinteractions that will be non-linear termsamong the parts. so these goldstone modesthat we were previously dealing with as independentmodes of the system, are actuallynon-linearly coupled. and we want to knowwhat the effect of that is on the behaviorof the entire system.

so whenever we're facedwith a non-linear theory, we have to do some kind ofa preservative analysis. and the first thing thatyou may be tempted to do is to expand thepowers of pi, and then look at the gaussian part,and then the higher order parts, etc. that's a way of doing it. but there's actually anotherway that is more consistent, which is to organizethe terms in this weight

according to powersof temperature. because after all, i startedwith a zero temperature configuration. and i'm hopingthat i'm expanding for small fluctuation. so my idea is to-- iknow the ground state. i want to see what happensif i go slightly beyond that. and the reason forfluctuations is temperature, so organize terms in thiseffective hamiltonian

for the pis in powersof temperature. and temperature, by this i meanthe inverse of this coupling constant, k,because even, again, if i go throughmy old derivation, you can see thati go minus beta h, so k0 should be inverselyproportional to temperature. k is proportion to k0. it should be inverselyproportional to temperature. so to some overallcoefficient, let's just

define temperature [inaudible]. now we see that at thelevel that we were looking at things before,from this term it's kind of like agaussian form, where i have something like k, whichis the inverse temperature pi squared. so just on dimensional grounds,up to functional forms, etc. we expect pi squared to beproportional to temperature at the 0 order, if you like.

because, again, iftemperature goes to 0, there's not goingto no fluctuations. as i go away from 0 temperature,the average fluctuations will be 0. average squared will beproportional to temperature. it all makes sense. so then if i lookat this term, i see that dimensionally,it is inverse temperature pi squared is of theorder of temperature.

so this is dimensionallyt to the 0. whereas if i startto expand this, this log i can start toexpand as minus pi squared plus pi to the 4th over 2 pi tothe 6th over 3, and so forth. you can see that subsequentterms in this series are higher and higherorder in this temperature. this will be the order oftemperature-- temperature squared, temperature cubed. and already we can seethat this term is small

compared to this term. so although thisis a gaussian term, and i would'vemaybe been tempted to put it in the 0order hamiltonian, if i'm organizingthings according to orders of temperature, my0-th order will remain this. this will be thecontribution to first order, 2nd order, 3rd order. and similarly, i canstart expanding this.

square root is 1 minuspi squared over 2. so then i take the gradientof minus pi squared over 2. i will get pi gradient of pi. you can see that the lowestorder term in this expansion will be pi gradient ofpi squared, and then higher order terms. and this is something thatis order of pi to the 4th, so it gives the order oftemperature squared multiplied by inverse temperatures.

so this is a term that iscontributing to order of t to 0, t to the 1. so basically, at order of tto 0, i have as my beta h0 just the integral d dx k/2gradient of pi squared. while at order oft the 1st power, i will have a correctionwhich has two types of terms. one term is this k/2 integrald dx pi gradient of pi squared, coming from what was thegradient of sigma squared. and then from here, iwill get a minus rho

over 2 integral d dx pi squared. and then there will be otherterms like order of t squared, u2, and so forth. so i just re-organized termsin this interacting hamiltonian in what i expected to bepowers of this temperature. now here we-- one of thefirst things that we will do is to look at this andrealize that we can decompose into modes by goingto previous space, i do a fourier transform.

this thing becomes k/2 integraldd q divided by 2 pi to the d q squared pi theta of q squared. so let's write it aspi theta [inaudible]. and again, as usual,we will end up needing to calculate averageswith this gaussian rate. and what we have here is thatpi alpha of q1 pi beta of q2, we get this 0-th ordered rate. the components haveto be the same. the sum of the twomomenta has to be 0.

and if so, i justget k q squared. now i can similarlyfourier transform the terms that i have over here. so the interactionsto one-- first one becomes rather complicated. we saw that when wehave something that's is four powers of a field. and when we go tofourier space, rather than having oneintegral over x, we

ended up withmultiple integrals. so i will have, essentially,fourier transform of four factors of pi. for each one of them iwill have an integration. so i will have ddq1, dd q2, dd q3. and the reason idon't have the 4th one is because of the integrationover x, forcing the four q's to be added up to 0. so i will have pi alphaof q1, pi alpha of q2,

now note that thishigh gradient of pi came from a gradientof pi squared, which means that the twopis that go with this, carry the same index. whereas for the nextfactor, pi gradient of pi, they came from different ones. so i have pi q3 i betaminus q1 minus q2 minus q3. now if i just writtenthis, this would've been the fourier transform ofmy usual 4th order interaction.

but that's not whati have because i have two additional gradients. and so for two ofthese factors actually i had to take thegradient first. and every time i take agradient in fourier space, i will bring a factor of i q. so i will have here i q1dotted with iq let's say 3. so the fourier transform ofthe leading quartic interaction that i have, is actually theform that i have over here.

there is a trivialterm that comes from fourier transforming. it's pi squared becausethen i fourier transform that, i get simply pialpha of q squared. yes? audience: does itmatter which q's you're pulling outas the gradient? professor: you can see thatthese four pis over here in fourier space appearcompletely interchangeably.

so it really doesn't matter, no. because by permutation andre-ordering these integration, you can move itinto something else. no, there is-- i shouldn't--i'll draw a diagram that corresponds to that that willmake one constraint apparent. so when i was drawinginteraction terms for m to the 4th tier forlandau-ginzburg, and i have somethingthat has 4 interactions, i would draw somethingthat has 2 lines.

but the 2 lines had 2 branches. and the branching was supposedto indicate that 2 of them were carrier 1index, and 2 of them were carrying the same index. now i have to makesure that i indicate that the branchesof these things additionally havethese gradients for the iq'sassociated with them. and i make aconvention the branch,

or the q, that has the gradienton it, i will put a line. now you can seethat if i go back and look at the origin ofthis, that one of the gradients acts on one pair ofpis, and the other acts of the other pairs of pis. so the other dashed line icannot put on the same branch, but i have to put over here. so the one constraint thati have to be careful of is that these iq' shouldpick one from alpha and one

from beta. this is the diagrammaticpresentation. so what i can do is to nowstart doing perturbation in these interaction. you want to do thelowest order to see what the first correctionbecause of fluctuations and interaction ofthese goldstone modes. but rather than dothings in two steps, first doing perturbation,encountering difficulty,

and then converting thingsto a normalization group, which we've already seen thathappen, that story, in dealing with the landau-ginzburgmodel, let's immediately do the perturbativerenormalization group of this model. so what i'm supposedto do things is to note that allof these theories came from someunderlying lattice model. i was carefully drawingfor you the first lattice

model originally. which means that thereis some cut off here, some lattice cut off. which means that wheni go to fourier space, there is always some kind of arange of wave numbers or wave vectors that i haveto integrate with. so essentially, mypi's are limited after i do a littlebit of averaging, if you like that there issome shortest wavelength,

and the correspondinglargest wave number, lambda, in [inaudible]. and the procedure forrg, the first one, was to think aboutall of these pi modes, and brake them into two pieces. one's that we're respondingto the short wavelength fluctuations that wewant to get rid of, and the ones that correspondto long wavelength fluctuations that we would like to keep.

so my task is as follows, thati have to really calculate the partition function overhere, which in it's fourier representation indicatesaveraging over all modes that's are in this orange. but those modes i'mgoing to represent as d pi lesser, aswell as d pi greater. each one of thesepi's is, of course, an n minus on component vector. and i have a ratethat i obtained

by substituting pi lesser andpi greater in the expressions that i have up there. and we can see already thatthe 0-th order terms, as usual, nicely separates outinto a contribution that we have for pi lesser,a contribution that we have for pi greater, and that theinteraction terms will then involve both of these modes. and in principle, icould proceed and include higher and higher orders.

now i want to get rid of allof the modes that are here. so that i have an effectivetheory governing the modes that are the longer wavelengths,once i have gotten rid of the short wavelengthfluctuations. so formally, once i haveintegrated over pi greater in this doubleintegral, i will be left with the integrationover the pi lesser field. and the exponential getsmodified as follows. first of all, if i wereto ignore the interactions

at the lowest order,the effect of doing the integration of the gaussianmodes that are out here, will, as usual,be a contribution to the free energy of thesystem coming from the modes that i integrated out. and clearly it alsodepends, i forgot to say, that the range of integrationis now between lambda over b lambda, where b is myrenormalization factor. audience: because you'recoming from a lattice, does

the particular shape ofthe brillouin zone matter more now, or still not really? professor: it is in noway different from what we were doing before inthe landau-ginzburg model. in the landau-ginzburg model,i could have also started by putting spins,or whatever degrees of freedom on a lattice. and let's say if i wasin hyper cubic lattice, i would've had brillouinzones, such as this.

and the first thingthat we always said was that integratingall of these things gives you an additionaltotally harmless component to the energy that hasno similar part in it. so we're always searchingfor the singularities that arise at the coreof this integration. whatever you dowith the boundaries, no matter how complicated shapesthey have, they don't matter. so going back to here.

if we had ignoredthe interactions, integrating overpi greater would've giving me this contributionto the free energy. and, of course, beta h0 ofpi lesser would've remained. but now the effect of havingthe interactions, as usual, it is like integratinginto the minus u with the rate over here. so i would have anaverage such as this. and we do the cumulantexpansion, as usual.

and the first term iwould get is the average of this quantity withrespect to the gaussian rate, integrating out the highcomponent modes, high frequency modes, and highorder corrections. audience: so right here you'redoing two expansions kind of simultaneously. one is you have non-linearmodel that you're expanding differentpowers and temperature. and then youfurther on expand it

to cumulants to be ableto account for that. professor: no, because ican organize this expansion in cumulants inpowers of temperature. so this u has an expansionthat is u1, u2, etc. organized in powersof temperature. audience: ok. professor: and then wheni take the first cumulant, you can see that the average,the lowest order term, will be--

audience: the first cumulantis linear in temperature, and that's what you want? so i'm being consistentalso with the perturbation that i had originally stated. actually, since i drew adiagram for the first term, i should state that this term,since we are now also thinking of it as a correctionin u1, i have to regard it as 2 factors of pi. so i could potentially representit by a diagram such as this.

so diagrammatically, my u1that i have to take the average is composed ofthese two entities. so what i need to do is to takethe average of that expression. so i can either dothat average over here. take the averageof this expression, or do it diagrammatically. let us go by thediagrammatic route. so essentially,what i'm doing is that every line that i see overthere that corresponds to pi,

i am really decomposinginto two parts. one of them i will draw as astraight line that corresponds to the pi lesserthat i am keeping. or i replace itwith a wavy line, which is the pi greater thati would be averaging over. so the first diagram ihad essentially something like this-- actually,the second diagram. the one that comesfrom rho pi squared. it's actually trivial, so let'sgo through the possibilities.

i can either have both ofthese to be pi lessers-- sorry, pi greaters. so this is pigreater, pi greater. and when i haveto do an average, then i can usethe formula that i have in red about theaverage of 2 pi greaters. and that wouldessentially amount to closing this thing down. and numerically, it would givesme a factor of minus rho over 2

integral d dk over 2 pi to thed in the interval between lambda over b, lambda . and i have theaverage of pi alpha pi alpha using a factorof delta alpha alpha. summing over alpha will giveme a factor of n minus 1. and the average would besomething like k k squared. so i would have to evaluatesomething like this. but at the end of the day,i don't care about it. why don't i care about it?

because clearly the result ofdoing this is another constant. it doesn't depend on pi lesser. so this is an additionto the free energy once i integrate modes betweenlambda over b to lambda, there is a contributionto the free energy that comes from this term. it doesn't changethe rate that i have to assign to configurationsof the pi lesser field. that's another possibility.

another possibility is ihave one of them being a pi greater, one of thembeing a pi lesser. clearly, when i try to getan average of this form, i have an averageof one factor of pi with a gaussianfield that is even. so this is 0. we don't have to worry about it. and finally, i will get aterm, which is like this. which doesn't involveany integrations,

and really amounts totaking that term that i have over there, and justmaking both of those pi to be pi lessers. so it's essentially the sameform that will reappear, now the integration beingfrom 0 to lambda over 2. so we know exactly what happenswith the term on the right. nothing useful, or importantinformation emerges from it. if i go and look atthis one however, depending on wherei choose to put

the solid linesor the wavy lines, i will have a numberof possibilities. one thing that isclearly going to be there is essentially i put pi lesserfor each one of the branches. essentially, when i writeit here like this one, it is reproducingthe integration that i have over there,except that, again, it only goes between 0 and lambda. and now i can startadding wavy lines.

any diagram that has one wavy,and i can put the wavy line either on that typeof branch, or i can put it on thistype of branch. it has only one factor of pi. by symmetry, it willgo to 0, like this. there will be thingsthat will have three factors of pi lesser. and all of these--again because i'm dealing with an oddnumber of factors of pi

greater that i'maveraging will give me 0. there's one other thingthat is kind of interesting. i can have all fourof these lines wavy. and if i calculatethat average, there's a number of ways ofcontracting these four pi's that will giveme nontrivial factors. but these are also contributionsto the free energy. they don't depend on thepi's that i'm leaving out. so they don't have to worryabout any of these diagrams

so far. now i dealt with the 0,1, 3, and 4 wavy lines. so i'm left with 2 wavylines and 2 straight lines. so let's go through those. i could have onebranch be wavy lines and one branch bestraight lines. and then i take theaverage of this object. i have a pi greater--a pi greater here, and therefore i cando an average of two

of those pi greaters. that average will give me afactor of 1 over k k squared. i have to integrate over that. but one of these branches hadthis additional dash thing that corresponds tohaving a factor of k. so the integralthat i have to do involves something like this. and then i integrate over theentirety of the k integration. this is an odd power, and sothat will give me a 0 also.

so this is also 0. and there's anotherone that's is like this where i go like this. and although i do thesame thing now with two different branches, the kintegration is the same. and that vanishes too. so you say, is thereanything that is left? the answer is yes. so the things that areleft are the following.

i can do something like this. or i can do something like this. so these are the two things thatsurvive and will be nontrivial. you can see thatthis one will be proportional topi lesser squared, while this one is going tobe proportional to gradient of pi lesser squared. so this one will renormalize,if you like, this coefficient. whereas this one, we've modifiedand renormalize our coupling

straight. so it turns outthat that is really the more important point. but let's calculatethe other one too. audience: why do we connect theones down here with the loops, but left all the endsfree in the ones. was that just a matter of thecase of how to write diagram, or does that signify something? professor: couldyou repeat that?

i'm not sure i understand. audience: so when we had twowavy line, both coming out one of the diagram, thisline, they just stop. we connected themtogether when we were writing the oneson the bottom line. professor: so basically, istart with an entity that has two solid linesand two wavy lines. and what i'm supposed to dois to do an integration-- an average of thisover these pi greaters.

now the process ofaveraging essentially joins the two branches. if i had the momentum here,q1, and a momentum here, q2, if i had anindex here, alpha, and an index here, beta,that process of averaging is equivalent to saying thesame momentum has to go through, the same indexhas to go through. there is no averaging that isbeing done on the solid lines, so there is-- meaninglessto do anything.

so this entitymeans the following. i have k/ 2 let's callit legs 1, 2, 3, and 4. integral q1 and q2, but q1and q2 you can see explicitly are solid. so these are the integrationfrom 0 to lambda over b. i have an integration overq3, which is over a wavy line. so it's between lambdaover b and lambda. if i call this branch alphaand this branch beta, from here i have actually pi lesser alphaof q1 pi lesser beta of q2.

i should have put themoutside the integration, but it doesn't matter. and then here i had pialpha of q3, pi beta of q4. but these also had theselines associated with them. so i have here actuallyan i q3, an i q4. again, q4 has to stand for minusq1 minus q2 minus q3 from this. and then i had the pipi here, which give me, because of the averaging,delta alpha beta. and then i willhave an integration

that forces q2 plus q4 to be 0. and then i have k q3 squared. now q3 plus q4 is the same thingas minus q1 minus q3, if you like, because ofthat constraint. so i can take thatoutside the integration. there's no problem. i have one integration left,which is 1 over k q3 squared, but these two thenbecome the same. these pi's i will take outside.

i note that because of thisconstraint, q1 and q2 being the same, these two reallybecome one integration that goes between 0and lambda over b. and these indices havebeen made to be the same. so i have pi alpha ofq, this q, squared. then i have the integrationfrom lambda over b to lambda. d d q3 2 pi to the d. here i have i q3 i q4. but q4 was said to be minus q3.

so the two i's and theminus cancel each other. and i will get afactor of q3 squared. and then here i have afactor of k q3 squared. so the overall thingis just that we can see that the k's cancel. i have one factor integraldd q 2 pi to the d. i have pi alpha of q squared. and these are q lessers. i integrated out this quantity.

the q3's vanish, so i reallyhave the integral of q3 over 2 pi to the d. now if i had done theintegral of q3 2 pi to the d, all the way from 0 tolambda, what would i have done? if i multiply thisvolume here, that would be the number of modes. so this is, in fact, n/v,which is the quantity that i have called the density. but what i'm doingis, in fact, doing

just a fraction of this integralfrom 0 to lambda over b. so if i do the fractionfrom 0 to lambda over b, then i will get 1minus b to the minus d. sorry, from lambdaover b to lambda. then if i had done all ofthe way from 0 to lambda, i will have had one,but i'm subtracting this fraction of it. so the answer is rho 1minus b to the minus d. the overall thing here getsmultiplied by rho 1 minus

b to the minus d. it just would correct thatfactor of density that we have. we'll see shortly it's notsomething to worry about. the next one is really themore interesting thing. so here we have this diagram,which is k/2 integral from 0 to lambda over b. essentially, i willget the same structure. this time let me writethe pi alpha lesser of q1 pi beta lesser of q2outside the last configuration.

i have the integral from lambdaover b to lambda dd of q3 2 pi to the 3d. and i will have the samedelta function structure, except that now these factorsof i q become i q1 times i q2. so i can put themoutside already. and then i have herethe delta function. so the only difference isthat previously the q squared was inside the integration. now the q squared isoutside the integration.

so the final answer willbe k/2 integral 0 to lambda over d d d q 2 pi to the d. i have a q squared. pi of q lesser squared. and the coefficientof that would look like what i had before,except without this factor. so it's the integralfrom lambda over b to lambda dd of q3 divided by 2pi to the d 1 over kq3 squared. so once i do explicitlythis calculation,

the answer is going to bea weight that only depends on this pi lesserthat i'm keeping and i would indicate thatby beta h tilde, as usual. that depends of this pi lesser. and we now have allof the terms that contribute to this beta h tilde. so let's write them down. there's a number ofterms that correspond to changes in the free energy.

so we have a v delta f p atthe 0-th order contribution of delta f p at the firstorder that are essentially a bunch of diagrams, both fromhere as well as from here. but we don't reallycare about them. then we have types ofterms that look like this. i can write them alreadyafter fourier transformation in real space. so let me do that. i have integral ddx in real space,

realizing that my cutoff has been changed to ba-- things that areproportional to gradient now gradient ofpi lesser squared is the things i infourier space becomes q squared pi lesser squared. and what is the coefficient? i had k/2, whichcomes from here. if i don't do anything, ijust have the 0-th order modes acting on these.

but i just calculateda correction to that that issomething like this. so in addition towhat i had before, i have this correction,k/2 times this integral. and i'm going to call theresult of this integral to be i sub d-- seethat this integral is proportional to 1/k. and when the integration--let's call the result of the integralid of b because it

depends on bothdimension of integration, as well as the factorv through here. so i have 1/k id of b. so that's one type ofthat i have generated. i started from this0-th order form. and i saw that once i make theexpansion of this to the lowest order, i will get acorrection to that. and actually, just tosort of think about it in terms of formulae,you see what

happened was that the firstterm that i had over here was pi gradient ofpi repeated twice. and what i did wasi essentially did an average of these two pi's andgot the correction to gradient of pi squared, which is what wascomputed here and [inaudible]. the next term that i haveis this term by itself, not calculated with the pilessers only-- so this object. so i will write it as k/2i gradient of pi squared. there's no correction.

the final term thati have is this term. so i have minus rho over2 pi lesser squared. and then the correction that ihave was exactly the same form. it was 1 minus 1 minusb to the minus b. the rho over 2 is going tobe hung onto both of them. you can see that there'sa rho and there's a 2. and so basically i havetwo add this to that. and to first order, this isthe entire thing of the thing that i will get.

so this however isjust a course gradient. it's the first step of the rg. and it has to be followed bythe next two steps of the rg. where here, youlook at you field and you can see that thefield is much coarser because the short distancecut off rather than being a, has been switched to ba. so you define your x primeto x/b, so that you shrink. you get the same course pixelsize that you had before.

and you also have todo a change in pi. so you replace pi lesserwith some factor zeta pi prime, so that thecontrast will look fine. so once i do that, youcan see that this effect of this transformationis that this coupling k will change to k prime. because of the changeof x with b x prime, from the integration,i will get b to the d. from the two derivatives, iwill get v to the minus 2.

from the fact that i havereplace two pi lessers with pi primes, i will get afactor of zeta squared. and then i will have thisfactor k 1 plus 1/k id of b. so this is the recursion formulathat we will be dealing with. now there is somesubtleties that go with this formula thatare worth thinking about. our original system had reallyone coupling parameter k that because of the constraintsof the full symmetry of this field, s, part of itbecame the quadratic part that

was the free field theory. but part of it madethe interactions. but because of thisvertical symmetry, to form of thatinteraction was fixed and had to beproportional to k. now if we do our normalization groupcorrectly, to full symmetry that we had has to bemaintained at all levels. which means that the functionalform that i should end up with should have the same propertyin that the higher order

coefficient should berelated to the lower order coefficient, exactly the sameway as we had over there. and at least at this stage, itlooks like that did not happen. that is, we got thecorrection to this term, but we didn't get thecorrection to this term. we shouldn't be worriedabout that right here because we calculatedthings consistently corrections to order of t. andthis was already a term that was orderof t. so the real check

is if you go and calculatethe next order correction, you better get a correctionto this term at next order that matches exactly this. people have done that,and have checked that. and that indeed is the case. so one is consistent with this. there are other kindsof consistency checks that have happenedall over the place, like the fact that thiscame back 1 minus 1

came out to be b to the minusd, so that the density is the same as before,consistent with the fact that you shrunk thelattice after rg so that the pixel size wasthe same as before is a consequence of that. you may worry that that'snot entirely the case because when i dothis, i will have also a factor of zeta squared. but it turns out that zeta is1 plus order of temperature,

as we will shortly see. so i gain consistent--everything's consistent at this levelthat we've calculated things. and the only changeis this factor. now the one thing thatwe haven't calculated is what this zeta is. so to calculate zeta,i note the following that i start witha unit vector that is pointing at 0 temperaturealong this direction.

now because offluctuations, this is going to be kind ofrotating around this. so there is this vectorthat is rotating. if you average it over sometime, what you will see is that the average in allof these direction is 0. the variance is not 0,but the average is 0. but because ofthose fluctuations, the effective length thatyou see in this direction has shrunk.

how much has it shrunk byis related to this rescaling factor that i should chose. and so it's essentiallyaverage of something like 1 minus pi squared. but really it is thepi lesser squares that i'm averaging over. which the lowest order is 1minus 1/2 the average of pi lesser squared, whichis 1 minus 1/2 . now this is an n minus1 component vector.

so each one of the componentswill give you one contribution. the contribution thatyou get for one of these is simply the averageof pi squared, which is 1 over k k squared, whichi have to integrate over k's that lie between lambdaover b and lambda. and you can see that this is1 minus 1/2 n and minus 1. it is inverselyproportional to k. and then the integrationthat i have to do is precisely the sameintegration as here.

so it is, again, this id of b. so let me write the answer,say a couple of words about it, and then we will dealwith it next time. so k prime is goingto be b to the d minus 2-- a newinteraction parameters. it is one factor of1 plus 1/k id of b. and then there's one factorof this square of zeta. so that gives you n minus1 over k id of b times k. so we'll analyze thismore next time around.

but i thought i would give youthe physical reason for how this interactionparameter changes. let's say we arein two dimensions. so let's forgetabout this factor. in two dimensions, we can seethat there is one factor that says at finite temperature,we are going to get weaker. the interaction isgoing to get weaker. and the reason forthat is precisely what i was explaining over here.

that is, you have somekind of a unit vector, but because of itsfluctuations, you will see that itwill loop shorter. and it is lesslikely to be ordered. the more componentsit has to fluctuate, the shorter it will look like. so there is that term. so if this was theonly effect, then k will become weaker and weaker.

and it will have disorder. but this effect saysthat it actually gets stronger becauseof the interactions that you have among the modes. and to show you this to you,we can experiment with it yourself. so this is a sheet of paper. this bend is an exampleof a goldstone mode because i could haverotated this sheet

without any cost of energy. so this bend is a goldstone modethat costs very little energy. now this paper has thekinds of constraints that we have over here. and because ofthose constraints, if i make a modein this direction, i'm not going to be able tobend it in the other directions. so clearly the mode thatyou have this direction, and this direction, are coupled.

that's kind of an exampleof something like this. now while it is easy to do thisbend because of this coupling, if thermal fluctuationshave created modes that are shorter wavelength,and i have already created those modesover here, then you can experiment yourself. you'll see that this is harderto bend compared to this. you can see this already. so that's the effectthat you have over here.

so it's the competitionbetween these two. and depending on whichone wins, and you can see that thatdepends on whether n is larger than 2 or less than 2. so essentially for nthat is larger than 2, you'll find that this term wins. and you get disorderin two dimension. and is less than 2,you will get order like we know the isingmodel can be order.

but there's otherthings that can be captured by this expression,that we will look at next time.