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PROFESSOR: OK.
Let's start.
So for several lectures, we've been
talking about the XY model, which
is a collection of two component spins,
for example defined on a square lattice,
but it could be any type of lattice.
And today we are going to compare and contrast
that with the case of a solid.
More specifically for simplicity, an isotropic solid.
And a typical example of an isotropic solid in two
dimension is the triangular lattice.
So something like this.
What both of these systems have in common
is that there is an underlying continuous symmetry that
is broken.
In the case of the XY model, at low temperatures
all of the spins would be pointing more or less
in the same direction.
Potentially with fluctuations, let's say, go
around the direction where they're
supposed to be pointing-- being, let's
say, here in the vertical-- characterized
by some kind of an angle.
Let's call it theta.
And essentially the uniform state at 0 temperature
can be characterized by any theta.
And because of the deformations, it's
captured through these low energy Goldstone modes.
And the probability of some particular deformation
will be proportional at low temperature to the integral,
let's say, in d dimensions of gradient,
the change in theta within different spins squared.
Of course, with some considerations as
we discussed about topological defects, et cetera,
being buried in this description.
Now, similarly, away from 0 temperature
the atoms of a solid will fluctuate around
the position that corresponds to the minimum.
And so the actual picture that you
would see at any finite temperature
would be distorted with respect to the perfect configuration
and there would be some kind of a vector [INAUDIBLE] u
at each location that would describe this fluctuation.
Again, there is no cost in making a uniform distortion
theta that is the same across space.
And similarly here, there is no cost
in uniformly translating things.
So these would both imply that this row can [? continue ?]
[INAUDIBLE] leads to Goldstone modes.
And the corresponding energy cost
in the case of the isotropically distorted system
is typically returned as an integral involving
elastic moduli, as we've seen.
And the traditional way to write that
is in this form where the same reason here
that it wasn't theta that was appearing
but gradient of theta, because a uniform theta
does not make any cross.
These strain fields are related derivatives of the distortion.
But now we remember that this u is a vector
and this strain field uij it was one half of the derivative
of uj in the i direction derivative of ui
in the j direction symmetrized.
Actually, probably better to write this symmetrically
as beta h equals to this quantity.
And then the typical type of calculation
that we did for the XY model was to go
in terms of the independent spin wave modes.
So we go to Fourier space.
This becomes an integral ddq 2 pi to the d.
The derivative here gave q squared.
And then I had something like the Fourier transformed theta
squared.
Now I can also more clearly represent
this in the Fourier description.
I will get something like 1/2 integral vdq 2 pi to the d.
And a little bit of work shows that this form Fourier
transforms to something that is proportional to mu.
Again, it has to be proportionality to q squared
because the energy should go to 0
as q goes to 0 for very long wavelength modes.
And so we'll have something like q squared mu tilde of q
squared.
This is a vector squared.
u is a vector.
q is a vector.
And there is another component that gives you mu plus lambda.
And consistent with rotational symmetry,
we can have a term that is q dotted with u tilde of q,
the whole thing squared.
And again, you can reason that you
can describe the isotropic system in terms of just two
elastic moduli is that the only rotational invariant
quantities are q squared, u squared, and q.u,
that those are the only terms that you can [? find. ?]
Now the next thing.
Once you know this form, you can immediately say something
about the fluctuations.
So you say that expectation value of theta tilde q theta
tilde q prime, if I assume that this is a Gaussian theory
and these are the only fluctuations
that I'm considering, the answer is
going to be 2 pi to the d delta function q plus q prime.
And then I will get a factor of 1 over kq squared.
The corresponding thing here is a bit more complicated.
Because as we said, this quantity u tilde is a vector.
And I can look at the correlation between, say,
the i-th component of that vector along mode
q, the j-th component of that vector along mu 2 prime
and ask what this is.
And again, only the same values of q
are coupled together, so we get the usual formula over here.
If this term was the only term in the story, that
is if we were dealing with vectors q and u that
are orthogonal to each other, then we
would just get the 1 over q squared.
And so there is actually a term that is like that.
But because of this other term, there is the possibility
that q and u are in the same direction, in which case,
these two costs will add up and I
will get something that is proportional to 2
mu plus lambda.
And so then I will get 2 mu plus lambda q squared.
And here I will have qi qj.
This becomes q to the 4th.
Let me make sure that I did not write this incorrectly.
Yes, I did write it incorrectly.
It turns out that there is a mu plus lambda out here too.
But essentially there's just a small complication
that we get because of the fact that you have a vector u.
But overall the scaling is something
that goes like 1 over q squared, which is characteristics
of Goldstone modes here with additional vectorial things
to worry about.
Let me make sure I separate these two.
Now one of the things that we have been concerned with
is whether these fluctuations destroy long range order, so
for long range order the information that we would like
to have is that the entirety of the system
is roughly pointing in the same direction,
meaning that if I know that this beam here
is pointed in the vertical direction,
if I go very far away, it is still more or less
pointed out-- pointing in the same direction.
So for that, we could look, for example,
as the degree which [? with ?] these two spins?
Yes, question?
AUDIENCE: [INAUDIBLE] q squared [INAUDIBLE]?
PROFESSOR: This is 4.
Is that what you're worried about?
AUDIENCE: Yeah.
It should be 2, I guess.
PROFESSOR: No, because I have two powers of q out front.
AUDIENCE: Right.
PROFESSOR: So I think in the notes
what I have is that I have actually
written things this way.
I put the factor of mu q squared out front.
And then I had this as qi qj over q.
And I think--
AUDIENCE: [INAUDIBLE] q squared [INAUDIBLE]?
PROFESSOR: q squared.
Let me one more time check to make sure that that's not--
actually there's a minus sign.
All right.
AUDIENCE: Wouldn't there be another factor of mu
to compensate for the factoring out?
PROFESSOR: Yeah.
So let's check this in the following fashion.
So I can look at modes where u and q are
perpendicular to each other.
When u and q are perpendicular to each other,
I don't have this there.
And so then what I should have is just a cost
that is 1 over mu q squared.
OK.
So let's see if that comes about.
What I can do is let's say imagine
that I'm looking at a q that is oriented along the x direction.
And I look at u's that are in the y direction.
And I correlate two u's in the y direction.
So I will definitely get a delta ij here.
I would have a 1.
Here I would get a factor of nothing,
because this term is absent.
The q's that I'm looking at don't
have any component along the direction of u that I'm having.
So this is absent and I will get a 1
over mu q squared, which is consistent with that.
Now let's look at the case where I look at u's that
are aligned with the vector q.
Let's say again in the x direction.
Then, as far as the energy is concerned,
both terms are present and the cost
should be 2 mu plus lambda.
So the answer that I should ultimately get
should be 1 over 2 mu plus lambda.
So let's check.
So here q squared-- and cancels with this q squared.
All of the q's are in the direction of the indices
that I'm looking at.
So that cancels.
This is 1.
So I will get 1 minus mu plus lambda over 2 mu plus lambda.
So then I will get 2 mu plus lambda minus mu minus lambda.
So I will get a mu in front that cancels this.
And I will be left with just the 2 mu plus lambda.
So we've checked that this is correct.
All right.
So this is a proof, if you like, by induction.
You knew that the only forms that are consistent
are delta ij and qi qj over q squared.
So we can give them two coefficients, a and b,
and then adjust a and b so that these two limits that I
considered are reproduced.
So-- OK.
I didn't have to resort to my notes
if I was willing to spend the corresponding three
minutes here.
So going back here, in order to see the degree of alignment
of the two spins that we have over here, what we need to do
is to do is to look something like cosine of, say,
theta at location x minus theta at location x
prime, which is non other than the real part
of the expectation value of e to the theta at location
x minus theta at location x prime.
And assuming that everything is governed by this Gaussian rate,
the answer is going to obtain by computing this Gaussian
as exponential of minus 1/2 the average
of theta x minus theta x prime squared.
So then my task is to replace theta x
and theta x prime in terms of these theta tilde in Fourier
space.
And then this will give me some expectation
that involves this average.
And when I complete that, eventually I
need to Fourier transform 1 over q squared.
And we've seen that the Fourier transform of the 1 over q
squared is the Coulomb interaction.
And so ultimately this becomes exponential of minus of 1
over k.
The Coulomb interaction as a function
of x minus x prime with the appropriate cutoff
included in this expression.
And then the statement that we always make
is that this entity at large distances,
the Coulomb interaction saying three dimension, falls off as 1
over separation.
So in three dimensions, this at large distances
goes to a constant 1 over k is something that
is proportional to temperature.
So after doing enough temperature,
this entity at large distances goes to a constant.
So there is some knowledge about the correlation
between these things that is left.
Whereas in two dimensions and below,
when I go to sufficiently large distances
the Coulomb interaction diverges.
This whole thing goes to 0 and information is lost.
So this is the in content that Goldstone modes
destroy through long range order in two dimensions and below.
Except that we saw that in two dimensions,
because the Coulomb interaction grows logarithmically,
this correlation fall off according
to this description only as a power law with an exponent
connected to k.
And then the discussion that we had for the XY model
was that at high temperatures you
have exponential decay in two dimensions.
At low temperature you have power law decay.
Although there is no true long range order,
there has to be a phase transition.
And we saw that this transition was
described by the unbinding of these topological defects.
So we would like to state that something similar to that
is going on over here.
The only reason I went through this
was to figure out what the analog of this entity
should be in the case of the solid.
So the analog of our theta is this vector u, the distortion.
What should be the analog of e to the i theta?
And the things that you should keep in mind is that theta,
I can't really tell it's difference
between theta and theta plus 2 pi, theta of 4 pi, et cetera.
So this e to the i theta captures that.
If theta goes to theta plus a multiple of 2 pi, e
to the i theta remains invariant.
So it's a good measure of ordering.
It's better than theta itself.
So similarly, there's a better measure
of deviation from perfect order here
for the case of the lattice.
You say, well, why should I worry about
that since I can measure what u is from the perfect position?
Well, my answer is this point here.
Did it come from here or from there?
Should you use for u this or should you use that?
You don't know.
So the u that you have the to use similar to the theta
that you're using is somewhat arbitrary.
And the arbitrariness of it is given by unit vectors
along the original lattice.
That is, u and any u that is increased or decreased
by some lattice vector of the original lattice
is the same thing.
You can't take them apart in the same sense that theta
and multiple of 2 pi you cannot.
So then that suggest that I can construct some kind of an order
parameter, I call it row G, which is like this.
This e to i something.
It's e to the iu but u is a vector.
And what I do I multiple by a G, which is an inverse lattice
vector.
So G is an inverse lattice vector.
And essentially the inverse lattice vectors
are defined to have the property that when you multiply them
with any of the original lattice vectors-- which are, let's say,
parametrized by two integers, m and n--
the answer is going to be some multiple of 2 pi.
So you can see that irrespective of whether I choose
u as distortion from one lattice position or any other lattice
position that happens to be nearby,
this measure is going to be giving you the same face.
So in the same way as before for the spins where
I asked whether spins at large distances
are correlated long range, I can do the same thing here
and define a long range correlations for a solid
by looking at e to the iG dotted with, let's say, u at position
x minus u at position x prime.
And x and x prime are two points on the lattice.
And again, the u's that I have are Gaussian distributed.
So this I will complete the square.
I will have exponential minus 1/2.
Well, OK, let me be a little bit careful with this.
I have G alpha G beta over 2 because I
have two factors of G dot u.
I will write them using Einstein's summation convention
as G alpha times u alpha component times the u beta
component.
So once more what we can do is to then write
these u's in terms of their Fourier vectors.
So what I will have is exponential minus G alpha G
beta over 2.
I will need to integrate over q's.
Because of these factors, I will get
something like 2 minus 2 cosine of qx minus x prime.
So whenever you go to Fourier space, you will get e to the iq
x minus e to the iq x prime.
Different one for here with different q
primes, but when you take the average,
q prime is set to minus q so then you
get this form as usual.
And then I will have the average Fourier transform use, which
is what I wrote over there.
I will get a 1 over mu q square delta i delta alpha beta minus
q alpha q beta q squared mu plus lambda over mu plus lambda.
There is a little bit of subtlety here.
So I will do something this is not quite kosher
but gives me the right answer.
Certainly I can take this G alpha a beta inside here.
When it multiplies delta alpha beta,
I will simply get G squared.
I really G alpha G alpha.
Sum over alpha.
It will give me G squared.
Here what I will get is a G dot q
squared because I will have two Gq's divided by q squared.
Ultimately I have to integrate over all values of q, including
all angle values.
So there will be, as far as the angle is concerned,
something like a cosine squared here.
And the rather unconventional thing to do
is to angular averages that first
and write this as G squared over 2.
Once you do that, then essentially
you can see that this structure just gives you a constant.
And what you will need to do is the usual Fourier
transformation of 1 over q squared.
So eventually you can see that the answer
is going to be proportional to minus G squared.
I will get a factor of 2 mu u from here.
Once I have written this as G squared over 2,
I will get 1 minus 1/2 of this quantity.
So that becomes 2 mu plus lambda--
or 4 mu plus lambda minus mu minus lambda.
So I will get a 3 mu plus lambda divided by 2 2 mu plus lambda.
And so there's a factor of 4 here.
So that's just the numerics that goes out front.
The key to the whole thing will be the usual integration of 1
over q squared.
Just as we had before, I will get again the same Coulomb
interaction as a function of x minus x
prime with appropriate lattice cut-off
introduced so that the short distance
behavior is controlled.
This whole thing has a name.
It's called a Debye-Waller factor.
Well, almost.
In 3D.
And the point is that if I'm looking
at this whole thing in three dimensions,
at large enough separation this will go to a constant,
just as we had before.
And I will conclude that this entity, as I
look at two points that are very far apart, at least that's
sufficiently low temperature and all of my mus and lambdas
are scaled inversely with temperature,
so this whole thing is proportional to temperature.
Essentially, you can see that this information
at large distance goes to a finite constant,
telling me that there's long range orders preserved.
Again, when I go to two dimensions,
you would say that this, at large distances,
grows logarithmically.
And so the correlations will fall off,
but fall of as a power law.
Now for the case of a solid, usually
what you do in order to probe for the order is
you shine x-rays and you see what comes out.
So essentially, you will see that there
will be some scattering.
And the way that the wave vector of the light has changed
is characterized by a vector q.
And we discussed this already, that the structure factor
s of q is going to be related to expectation value of the phase
that you get from the different points.
So I will get a factor of e to the i q.
the location of the different atoms, let's call them r, of x.
I have to do a sum over x.
I have to scatter from all of the atoms.
There's some overall form factor that if I'm not
really interested, but this whole thing
has to be first squished.
So once I square this, I will be getting averages
between pairs of points.
Because of translational symmetry
I can relate to that sum over pairs of points
to a center of mass that typically
would pick up the number of points that I have.
And then the expectation value of e
to the something like sum over all points, e
to the i q.r r of x minus r of 0 average.
This average I can certainly take out here.
Yes?
STUDENT: Sorry.
Is r what you're calling u before?
PROFESSOR: No.
So r is the actual location of where the particle is location.
So r, let's say for a particle that at zero temperature
was sitting at lattice point-- that is,
let's say, in two dimensions labeled by m and n--
it is shifted by an amount that is u that I
would assign to that point.
So the r's that I put here are the actual physical position
of the atom.
Which is composed of where it was
sitting at 0 temperature plus a little bit more.
So when I do this, indeed I will get a sum over all positions.
This could be a label mn, whatever, of e to the iq.
The difference between the two r's
would be, first of all, the difference between these two
at 0 temperature, which would be some lattice
vector of some variety.
Let's call r 0 of x.
And that quantity has [INAUDIBLE] fluctuations.
The fluctuations go into e to the iq.u of x minus mu of 0.
Now if I'm doing this at 0 temperature or very close to 0
temperature where this is either 1 at 0 temperature or not too
much fluctuating at high temperatures,
I will essentially be adding a lot of e to iqr's.
These are essentially phases.
They can be positive, negative, all over the complex plane.
And typically you would expect that for a randomly chosen q,
the answer from adding all of these things will be 0.
So the only time we're close to 0 temperature,
you expect this to be something that is significant.
Scattering is when q is one of these inverse lattice vectors.
So if I was really doing this at 0 temperature, this would be 1.
And the answer that I would get is essentially
I would get these delta function back peaks
at the locations that correspond to the inverse lattice spacing.
And that's presuming something that you have seen
in crystallography or whatever.
You take a solid and you scatter light from it
and you will back points.
So what I expect is that when I have
a q that is roughly close to G but not necessarily sitting
exactly at that point, what I would get
is a scattering that these proportional to an integral
over the entire lattice of something
like how far I am away from the right position.
And I should really write this as a sum over lattice points,
but I can approximate this by an integral.
The integral again has the property that it will give me 0
unless q is exactly sitting at G.
And then for here, I evaluate this quantity
when q is close to G according to what
I have calculated up there.
So I will get exponential of minus G
squared-- What do I have?
Over 4 mu, three mu plus lambda, 2 mu plus lambda.
The Coulomb interaction as a function of x.
Yes?
STUDENT: Are you missing an overall factor of m?
PROFESSOR: That's why I wrote the proportionality.
Because what I'm really interested is what you see.
And what you see is how things are varying as a function of q.
So I'll tell you what the answer is.
So you do the Fourier transform from this.
And in d equals to 3, what you find at 0 temperature,
as we said, you will get Bragg spots
at the locations of G's that correspond
to the inverse lattice vectors, which actually themselves
form a lattice.
So this is essentially the space of q.
And these are the different values of G.
Now what happens is that the strength of each one
of these delta functions is then modulated
by this factor evaluated at the corresponding G.
And since it is proportional to G squared,
you find that the spots that are close to the origin
are well defined.
As you go further and further away, the become very weaker.
And at some point you cease to see them.
So that's the-- the Debye-Waller factor
describes how these things vanish as you
go further and further away.
And essentially, you find that they go and diminish
as e to the minus G squared.
But clearly, this kind of scattering from a crystal
is very different from the scattering
that you expect from an liquid, which
is essentially you will have a bright spot at q close to 0
and things that would symmetrically fall off,
maybe with a couple of oscillations
due to the size of the particles, at cetera.
Now what happens in d equals to 2 is kind of interesting.
Because in d equals to 2, I'm doing a two
dimensional integral, and this thing
is growing as a power law.
So basically what I end up having
to deal with is something like e to the i q minus G x.
And something that falls off as a over x to some power
that I will simply cause eta G. So related to that combination.
And when you integrate this, just
dimensionally you can see that it
is an integral that scales as x to the power of 2 minus eta.
x is inverse to q minus G. So this is going to scale as 1
over q minus g to the power of 2 minus eta.
So as you go along some particular direction
and ask what you see, you find that will
see peaks close to the locations of the lattice vectors.
So let's say they are equally spaced along the particular
axis that I'm picking.
In three dimension, you would have been seeing delta function
at each one of these positions whose strength is governed
by this Debye-Waller factor.
In two dimensions, you will essentially
see power laws close to each one of them.
But as you go further, the exponent of the power law
changes.
So at some point you even cease to have a divergence.
But you can essentially ignore anything
like that [? happening. ?]
So you can see that because we don't have true long range
order into two dimensions because
of these logarithmic divergences,
et cetera, the traditional picture
that we have of a solid that is characterized by Bragg
scattering and delta function peaks is modified,
but still it is something that looks
very different from the liquid.
So we expect that as we raise temperature
there is a phase transition between scattering
of this form and something that looks like liquid.
And you can roughly guess that essentially it
is going to be related with these etas becoming
larger and larger so that these divergences disappear.
But what is the mechanism?
So let's go back and see what the mechanism was that we
discovered for the XY model.
So we are going to look at d equals to 2 from now on.
That discussion so far was general.
In d equals to 2, we said that we have topological defect.
And the story here was that, as I have been discussing,
the angle is undefined up to 2 pi.
So it would configuration of angles that kind of radiate
away from the origin.
And the idea was that if I take a circuit
and integrate around this circuit--
let's call this the s gradient of theta-- the answer does not
need to come back to 0.
It will be some multiple of 2 pi .
Some integer multiple.
And similarly here, we said that the value of u
is undefined up to a lattice spacing.
That leads to topological defects, which
I find it very difficult to draw for the triangular lattice,
so I'll draw it for the square lattice.
So at each one of these positions
there's a particle sitting.
And you look out here.
You have a perfect square lattice.
You look out here.
You have a perfect square lattice.
But clearly it is not a perfect square lattice.
And the analog of the circuit that I drew over here
that encloses a singlularty-- there's clearly
some kind of a defect sitting over here--
is that I can start with some point on the lattice
and perform what is called a Burgers Circuit.
What Burgers Circuit is is some walk
that you would take on a lattice, which
in a perfect lattice would bring you back
to your starting point.
So for example-- that was not a good starting point.
Let me start from here.
Let's say I take four steps up-- one, two, three, four.
I take five steps to the right.
One, two, three, four, five.
I reverse my four steps.
So I go four steps down.
One, two, three, four.
I reverse my five steps and I go five steps left.
One, two, three, four, five.
And you see that I did not end up where I was.
I ended up here.
And a failure to close my circuit
has to be a lattice vector.
And this is called d and it's called a Burgers Vector
that characterize this defect, which is called a dislocation.
So the uncertainty that I have in assigning a value for u
because of being on a lattice or deforming a lattice
gives rise to these kinds of ambiguity
that are similar to the topological defects
that you had over here.
Whereas the defects here were characterized by an integer,
the defects here are characterized
by a lattice vector.
And here the simplest things were, say, plus or minus.
Clearly here on the square lattice
the simplest things are plus b, minus b along the x direction
and plus b minus b along the y direction.
For example, on the triangular lattice
it would be plus minus at any of the three directions
that would define the lattice.
So calculating the distortion field here was actually
an easy matter because we said that the gradient of theta
was something like 1 over r.
And so the gradient of theta, which was 1 over r,
I could in fact write-- we saw as vector
that is orthogonal to the vector that is going out of the plane.
So we call that a z hat.
It was also orthogonal to the direction of motion
away from the defect, which we could characterize
by looking at the gradient.
The gradient vector for this distortion
is clearly in the radial direction.
And the answer that we had here for one defect
was ni log of r minus ri.
And since I'm taking the gradient it doesn't matter.
I can't put or not put the cut off.
And if I had multiple ones, I would simply sum over it.
So this was the contribution to the distortion
that came from a collection of defects such as this.
Now you can see that essentially in some continuum sense,
all I'm doing here is similar.
I can be taking a big circuit.
And as I go along that big circuit,
I can take a gradient-- So let's-- yeah.
The s.
gradient of the distortion field.
Distortion field is, of course, a vector.
So it has components that I can write as the x component
or the y component.
So this alpha could be the x component of the distortion
or the y component of the distortion.
And once I complete the circuit, here the answer was 2 pi n.
Here the answer is a lattice vector,
which is this Burgers Vector that could be in any direction
that you want.
Now as far as mathematics is concerned,
this line and this line for each component are exactly the same.
So the solution that I can write for gradient of u
I can just copy from here for each component.
So I can write that gradient of the u that
is due to a collection of dislocations
is something like z hat crossed with curl
of a sum over all of the potential locations
off my defect.
Rather than n, I have b over 2.
And it's a vector log of minus ri over [INAUDIBLE].
So if I have a collection of these dislocations
at different places on my lattice-- locations ri,
strengths, b alpha which is a vector--
then the distortion field that they would generate
is given by this.
It's very just following the answers that the [INAUDIBLE].
So then what we did for the case of the overall system--
in order to find how the different topological defects
interact with each was to calculate
beta h, which was an integral of gradient of theta squared.
Yes?
STUDENT: So the b alpha, you have an index i [? also? ?]
PROFESSOR: b alpha have index i also.
Like like the ni have index i.
So maybe I write it in this fashion.
Yeah.
So this is the cost that we have to evaluate,
except that we have to be somewhat
careful with the meaning of this gradient of theta
in that the gradient of theta, as we said,
has a contribution that is from regular spin
waves, the Goldstone modes, that can be deformed back
to everybody pointing in the same direction
without topological defect.
Plus a contribution due to this topological defect that are
categorized in that case by ni.
Now similarly here for the case of the solid,
we have the beta h.
Slightly more complicated integral d 2x.
We have mu uij uij plus lambda over 2 uii ujj.
And what we can do is to say that our strain field
uij has a component that is like the Goldstone modes
that we have been calculating so far,
essentially treating everything as Gaussians.
Except that I will write it as phi ij.
And then a part that would come from the distortion field u
bar that I calculated.
So I take that distortion field and then take derivatives of it
to symmetrize appropriately to construct
the strain that sums from them.
So we substituted this form over here.
And we found that beta h that we got had a part that was simply
relate to the Goldstone modes.
This was the part that could be treating as a Gaussian.
And then we had a part that corresponded
to the interactions among these defects.
And that part we saw had the character
of charges ni and nj plus minus 1
characterizing these topological defects
having a Coulomb interaction [? between them ?].
And then, of course, this i less than j,
we had to worry about what was happening when i was j.
And that we considered to be the contributions
of the core energy that, once exponentiated,
we described as y.
So it's beta h which is log of that would be log of y.
And actually this k bar, by the way, was simply 2 pi k.
And it was 2 pi k because each one of these charges is 2 pi n.
So there's 2 pi 2 pi.
But the Coulomb interaction really
should be log divided by 2 pi.
And so k bar becomes 2 pi k times this.
So I can do the same thing over here.
And what I find happens is that my beta h gets decomposed
as follows.
There will be a part that is simply the original expression
now for the field that is well behaved
and has no dislocation piece.
And then you wouldn't be surprised
because this structure is no different
from the other structure.
There's essentially a gradient of this distortion
field squared.
And you can see that if I take a second derivative
of this distortion field or one derivative of this, effectively
I will have two derivatives of a log, which
will give me a delta function.
So essentially these things do behave
like what you would get from Coulomb type of potential.
[INAUDIBLE] plus [INAUDIBLE] if the potential is 0.
So not surprisingly, I we get a minus k bar sum
over pairs of where these dislocations are located.
And then I will have b at location i, b at location j.
And then I would have a log of ri minus rj with some cut off.
Except that these b's are actually vector.
So this term is a dot product of the two b's.
And it turns out that when you go through the algebra,
there is another term, which is bi.
ri minus rj, bj dotted with ri minus rj
divided by ri minus rj squared.
So the charges that we had in our original theory
were scalar quantities.
These defects were characterized by a scalar value of n.
And they were interacting with something
that was like the ordinary Coulomb potential.
Whereas now we are looking at a system
that is characterized by charges that are vectors.
And it turns out that what we have here
is the vectorial analog of the Coulomb potential.
And again, if you think about an isotropic system and vectors,
this is a vector that you can form,
but b.r is another vector that you can form.
And so both of them do appear.
And once you go through the whole algebra
and through the inverse Fourier transform, et cetera,
you get this additional contribution
to this vector Coulomb interaction.
That's really is the only difference.
And of course, you will again get
a sum over all locations of the core energies
for creating these defects.
And the value of k bar here is related to these parameters
by mu mu plus lambda divided by pi 2u plus lambda.
So this combination controls what you have.
So the next thing that we did was
to construct a perturbation.
We essentially said that if I have these pairs of charges
that can spontaneously appear at an energy
cost but an entropy gain, they will effectively weaken
the overall Coulomb potential.
For example, if I have to test charges
then there will be some polarization
of the medium, some reorientation
of these charges that appear, that weakens
the effective charge here.
And we could perturbatively calculate
what the correction was, except that we
found that that correction, which
involves an integration over the separation of these things.
There was an integration that was potentially
divergent for us no matter how small we made the core
energies here.
And so what we did was eventually to construct an RG.
And the RG was that the value of this parameter k
bar-- actually more usefully its inverse that
is related to temperature-- changed
as a function of integrating out short distance
degrees of freedom.
It was something like 4 pi-- I think it was 4 pi cubed,
it doesn't matter-- y squared.
And that the scaling of to core energy
itself was determined by 2 minus pi k,
which in terms of this k bar becomes 2 minus k bar
over 2 times y.
And we can do exactly the same thing over here,
except for the complication of having
to deal with the charges that are vectorial.
And you get that the strength of the Coulomb interaction,
it's inverse that is related temperature
gets re-normalized by exactly the same process.
That is, there will be charges that will appear.
All the mathematics you can see is clearly similar.
And there will be a reduction, which then you
can think of as an increase in temperature, which will
be proportional to y squared.
I don't know what the constant of proportionality is.
And that dy by dl-- so essentially I
focus on the simplest type of dislocations
that I would have of unit spacing--
has exactly the same [INAUDIBLE],
2 minus k bar over 2 y.
It turns out that there is actually a difference.
And the difference is as follows.
That while we did not calculate the next correction
in this series, I said that this correction came
from essentially what would a single dipole do,
which would be some contribution that is order of y squared.
And I expect that there will be correction
if I look at a configuration that has pairs of dipoles.
And this will change this by order of y
to the fourth and this by order of y cubed.
Whereas if I look at something like a triangular lattice,
you can see that if I have two dislocations out there
as test dislocations pointing out
in opposite directions, which would
have some kind of a interaction such as this.
And I ask how that interaction is
modified by the presence of dipoles of dislocation
in the medium.
I could put a pair of dislocations
integrate over the separation between them.
I will get, essentially, the same type
of mathematical structure that would ultimately give me this.
But then the next order term in this series
is not for dislocations because I
can have a neutral configuration of three dislocations,
such as this.
And so once you do that, you will
find that the next correction here
is order of y cubed and order of y squared here.
But generically, both of them have a phase diagram
for RG flows that involves the inverse of the interaction
of these charges.
That is a temperature-like quantity.
And there is a critical value of that inverse, which is 1/4.
And what happens is that-- let's see, this other axis is y.
Anything smaller than 1/4 y tends to go to 0.
Anything larger, y tends to get larger.
And there is going to be some kind of a separatrix
that he describes the transition between flows that go down here
and the flows that go away.
And presumably, as I heat up my original system-- my lattice,
different types of lattices-- at low temperature,
I can figure out what this combination of mu an lambda is.
That gives me some value of k.
And there is some kind of the core energy for the vertices
that I can calculate.
So there will be some point in this phase diagram, which
at low enough temperature I will go over here, which corresponds
to a system that has an effective logarithmic
interaction between dislocations.
As I change the temperature, I will proceed around
some trajectory such as this because everything will change.
And presumably, at some point I will intersect this
and I will go to the other phase.
Now the characteristic of the other phase
was that this parameter k was going to 0.
It was essentially the logarithmic interaction
was disappeared.
And you can see that that parameter k going to 0
means that mu, the shear modulus, has to go to 0.
So basically this phase out here is definitely characterized
by 0 shear modulus.
So again, the two parameters mu and lambda, mu
is the one that gives you the cost
of trying to shear in a system.
And essentially the presence of the resistance to shear
is what defines a solid for you.
So this is a solid to something that does not have rigidity
transition.
It turns out that there is one subtlety here.
I don't know whether it's good to mention or not.
But so in the case of alpha this superfluidity, for example,
we said that this coupling strength as you
approach the transition rose to a universal value.
In these units it will be 4.
But that it approaches that universal value
with a logarithmic singularity.
Sorry, with a square root singularly.
Sorry.
And we saw that this was experimentally observed
in films of superfluid film.
I also said that on the other side
you have a finite correlations.
The correlations will be decaying exponentially.
And this correlation length has this very unusual signature,
which is that it diverged as a square root of t minus tc
in the exponent.
It was giving rise to these essentially singularities.
While everything kind of looks identical between this RG
and the other RG, it turns out that once you
keep track of these corrections, these corrections actually
are important.
And for the vectorial version of the Coulomb gas,
you find that they shear modulus that is finite in this phase
reaches its final value, which is not
universal because the thing that the universal is
that combination that involves both mu and lambda.
But rather than stc minus t to the 1/2,
stc minus t to some exponent that
is called mu bar, which is 0.36963 [INAUDIBLE].
So this is a kind of subtle thing
that is buried in these recursion relationships
once you go to a higher order.
And accompanying that is a correlation length that
also diverges with some behavior dominated by the same exponent.
So once people understood the original [? Koster ?]
[INAUDIBLE] transition, it was kind
of a very natural next step for them
to think about these locations in a two dimensional
solid and say that in the same manner
that the unbinding of these defects
got rid of the residual long range order that
was left in the XY model.
That this kind of ordering that we said
exists for the two dimensional solid,
this appears as a result of unbinding of dislocations
and you get a liquid.
Turns out that that's not correct.
And it was pointed out by Bert Halperin
I think that if you look at a solid,
there's two things that you.
You have the translational order,
but there's also orientational order.
That is, you can look, for example,
at a collection of spins here and the bonds
are pointing along the x and y directions
and you can look down here and the bonds are pointing
in the x and y directions.
And from the picture that I drew over here,
you can kind of see that once I insert this dislocation which
corresponds to an additional line,
the positions here are distorted.
Once I have many of these inserted,
it is kind of obvious that I will
lose any idea of how the position of this lattice point
and something out there is related.
It's not so obvious that inserting
lots and lots of these lines will
remove the knowledge that I have about
the orientation of the bonds.
And the insight that we will follow up--
and maybe it's probably a good idea to do it next time--
is that actually once the dislocations unbind,
we will lose this kind of ordering.
If I do this in two dimensions, these
are power law decays at low temperatures.
Once the dislocations unbind, this
becomes an exponential decay, as I would expect.
But I can also define locally some kind of an orientation.
Let's again call it theta.
But theta is, let's say, the orientation of the bond that
connects two neighboring spins.
Now what I can do is to look at the correlation of orientations
at different positions.
And it turns out that if I'm doing something
like a triangular lattice, I can't just pick the angle
and correlate the angle from one point to another point
because let's say on the square lattice the angle itself,
I don't know whether it was coming
from a bond that was originally in the x direction
or in the y direction.
So this is unknown up to a factor of 90 degrees.
In the triangular lattice it is an unknown up
to a factor of 60 degrees.
So what I should do is I should define e to the 6 i theta
at each location as a measure of the orientational order.
You can see that, again, if I'm at 0 temperature,
I can reorient the lattice any way that I like.
This phase would be the same across the entire system.
So what I'm interested is to find out
what is the expectation value of this object when
I look at points that are further apart.
And again, clearly if I'm at 0 temperature
these thetas are the same.
This will go to 1 at finite temperature.
Because of the fluctuations it will start to move a bit.
And what we can show-- and you will do it next time--
is that as long as you are in the 0 temperature phase
you will find that this goes to a constant,
even in two dimensions.
But once the dislocations unbind, what we will show
is that it doesn't decay exponentially,
but then it starts to decay as 1 over x to some power,
let's call it eta theta.
And then at some higher temperatures,
it eventually starts to decay exponentially.
So there is the original solid in two
dimension that we started that has true long range
order in the orientations.
And that is reflected in these Fourier
transforms that we make.
So we said that when we do the Fourier transform and look
at the x-ray scattering, let's say from a two
dimensional crystal, you were seeing these kinds of pictures.
And then these were becoming weaker
as we went further and further out.
But now you can see that this picture clearly
has a very, very defined orientational aspect to it.
Now once we go to this intermediate phase,
all of these peaks disappear.
But what you find is that when you look at this, rather
than getting a ring that is uniform,
you will a see a ring that has very well- defined variation.
One, two, three, four, five, six.
Six-fold symmetry to it.
So this phase has no knowledge of where
the particles are located, but knows the orientations.
It's a kind of a liquid crystal.
It actually has a name.
It's called hexatic.
But it's one of the family of different types of materials
that have no translational order but some kind
of orientational order that are hexatics.
So the transition between this hexatic phase
to this fully disordered phase, it turns out in two dimension
to be another one of these dislocation--
well, topology defect unbinding transition.
So maybe we will finish with that next time around.