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PROFESSOR: OK.
Let's resume.
I cut things off at a time when we had the final answer.
And I left you hanging because we don't know what the final
answer is telling us.
This says that as we change the orientation of the normal
to the plate, relative to x3 and, secondarily, as we change
the angle between the threefold axis and x2, we get
this strange third-rank trigonometric function.
Let's convert these cosines into the appropriate angles.
This was the twofold axis.
And this was x1.
This was the direction of the threefold axis.
This was x3.
And this is the direction x2.
And that comes out in between a pair of twofold axes.
And we may want to look at this from up above, relative
to these twofold axes that occur.
This thing here looks like a trigonometric identity,
doesn't it?
Let's let this angle here be theta.
And l1, then, is cosine of theta.
And so our geometry is that this is l1 cosine of theta.
This is l2, which is the cosine of the angle between
our direction and x2.
And that is the cosine of pi over 2, minus theta.
And that, then, is equal to sine of theta.
So this identity, as you all know--
I'm not telling you anything that you don't already know--
this is d is equal to d1,1 1, times the
cosine of 3 theta, right?
You knew that.
Believe it or not, it is.
It's one of these obscure trigonometric identities,
which wouldn't occur to you in a million years unless you go
digging through some handbooks.
So this is rather astonishingly simple.
It's simply a cosine function.
But the interesting thing is it goes as a
function of 3 theta.
So this goes through one cycle between a pair of adjacent
twofold axes.
So it starts out as d1,1.
And it finishes up at d1,1.
And in between it is minus d1,1.
And so it goes down.
And now we come to something that I do differently than
most people do.
Most people will say it goes as minus d1,1.
But minus d1,1 I like to show as a lobe going off in this
direction, with a minus sign.
AUDIENCE: How do you get that cosine of 3 theta?
PROFESSOR: Hmm?
AUDIENCE: How do you get d equals d1 cosine 3 theta?
PROFESSOR: How do I get that?
AUDIENCE: Yeah.
PROFESSOR: That is just a trigonometric identify,
believe it or not, a well-known
trigonometric identify.
Actually, it's an exceedingly obscure
trigonometric identity.
So the way this is going to go from x1 is it's going to be a
positive lobe around x1.
There's going to be a negative lobe along x2 and then a
positive lobe again about the twofold axis that's 120
degrees away.
And then a negative lobe, and then a positive lobe opposite
this negative lobe, and a negative lobe opposite this
positive lobe.
So it's a six-membered--
six lobes.
There is always a positive lobe opposite a negative lobe.
And what this means is that the charge is of opposite sign
on opposite ends of the twofold axes.
So the response peaks up on twofold axes.
Now the thing that I don't like is that what Nye does is
to say, OK, this is a negative value.
So you should plot the radius in a negative direction.
And that puts it over here, right on top of
this positive lobe.
So what Nye shows is the polar plot of this result, in the
plane of the twofold axes, is simply this.
This is x1.
And he shows a lobe here.
And if you interpret that as a polar plot you say, well, I
know what the value is in here.
It's decreasing.
And there's nothing going on in here.
So the response must be 0.
And then it starts coming up again.
And there's a response in different directions here.
And then it goes back down to 0.
And in this range, there's nothing going on.
And that's not true.
What's going on is a negative value of the modulus.
And to me, that becomes abundantly clear if you just
put a sign that labels the sign of the modulus within
those lobes.
Now I submit, that's pretty anisotropic, isn't it?
Yes, Steve?
AUDIENCE: Did you just arbitrarily choose where your
positive and negatives go, is it [INAUDIBLE]
out here?
PROFESSOR: No.
This comes out here.
When l1 is 0, that's--
excuse me.
When the angle is 0, l1 is plus 1.
Remember that l1 is the cosine of this angle.
When that angle is 0, then the value is plus 1.
So I was careful to put the label x1 on this.
Now the other thing that we should examine is how these
lobes vary in a direction perpendicular to x, in a
direction that includes x3.
So let us look at how the function varies in a direction
perpendicular to--
that includes x1 and x3.
OK.
Call this angle phi.
And in the x1, x3 plane, d1,1,1 prime turns out to be
l1 cubed d1,1.
Before you go to our general expression.
And l2 is cosine of 90.
This thing drops out.
We're left with simply d1,1 prime equals
l cubed times d1,1.
So this then goes as cosine cubed of phi, which is a lobe
that starts out at plus 1.
And then because it's cosine cubed, this dies out very,
very rapidly.
So these lobes are very flat in the x1, x2 plane, die out
very rapidly as the cube of phi, where phi is the angle
between the twofold axis and x3.
So the interesting thing about this surface is that if you
decided to pick up the random fragment of crystal and
determine what its Piezoelectric Modulus is.
One way of doing this, a poor man's test for
piezoelectricity is to just clamp a fragment of crystal
between two electrodes and then hook this up to a
variable frequency generator that sweeps through a range of
frequencies, changes the frequency of a voltage across
these plates, and then comes back and sweeps again.
Or alternatively, have a knob one on that
lets you change frequency.
Then put a pair of earphones across the crystal.
And if we would do that, by having this variable frequency
generator, as we turned it, when we hit a resonant
frequency that set up either a full wavelength of a
vibrational wave in the crystal--
or half wavelength, or one wavelength, or 3/2
wavelength--
there'd be a resonance.
And the capacitance does something crazy.
It does like this as you go through
the resonant frequency.
And then your simple detector--
a pair of earphones--
as you tuned the frequency, you would hear static as you
went through this discontinuity in capacitance.
Or alternatively, you could put this on a cathode ray tube
and just have a sweep frequency and put the voltage
across the electrodes on the oscilloscope screen.
And you would see something like this, and then this
discontinuity, and then maybe second harmonic.
And it would work just fine.
This is a poor man's way of detecting piezoelectricity.
But what if--
what if--
you happen to have your piece of quartz with the electrodes
directly on the c-axis?
The modulus would be 0, and you wouldn't detect anything.
Or if you put on the probes such that they were in a
direction that was in between these lobes where, again, the
Piezoelectric Modulus has gone to 0.
You wouldn't find anything.
So measuring the Piezoelectric Modulus for a random chunk of
material is dangerous, if you just look at one direction and
say nothing's going on; it's not piezoelectric.
The other thing is the material could have a
piezoelectric response that's so weak you just
can't measure it.
Yeah, OK?
AUDIENCE: This stuff's only for single crystals, right?
Because if you [INAUDIBLE] polycrystal material, you're
getting [INAUDIBLE] averaging.
PROFESSOR: You're averaging over all directions.
So this is for a single crystal.
And that's the only time you get these exotic, very
anisotropic surfaces.
Another method that I've seen in a rather old
book is really nice.
What you do is you cut your crystal into a little plate.
And then you drive the plate by means of putting electrodes
on it and hook this up to a variable frequency.
And if the material is transparent, you will set up--
depending on the velocity of sound in the material--
you will set up standing waves when you
hit the right frequency.
And these lines that I've drawn are places where the
displacement is always 0.
And these things that I've indicated at little squares is
a region where the displacement goes up and down
between its maximum and minimum extreme.
Now if this crystal is transparent and you shine a
light through it, these little regions bounded by lines of
zero displacement act like little lenses.
So if you pass a beam of light through it, you get a bunch of
right maxima, little focused spots on the sheet.
And you can determine for a particular frequency if this
is the 1 and 1/2 wavelengths along a dimension that you
know, you can again find the Piezoelectric Moduli.
So there are lots of ways of detecting this effect.
But when you have a single crystal, you've got to really
look at this as a function of direction, to
be absolutely sure.
One other Piezoelectric Modulus is in the problem set.
4 bar 3m, this is the structure of sphalerite and a
lot of the compound semiconductors.
This is not in the notes.
But if you try the problem, what you find is, again, a
highly, highly anisotropic behavior.
And this is the direction of the unit cell of the crystal.
You find that the piezoelectric response is a
series of very sharp lobes, positive and negative, going
along the directions of the body diagonal, namely the 1,
1, 1 directions in the crystal.
And then there's another lobe that goes like
this, negative, positive.
Another lobe that goes up here, positive and negative.
So very sharp lobes along the body diagonal, alternately
positive, negative, positive, negative as you go around 0,
0, 1 plane.
OK.
That's what the longitudinal piezoelectric effect is like.
You can invent single-crystal devices for weighing fish.
And then you have an interesting question.
If you hang the weight on the crystal, and it's a flat
plate, how would you orient the crystal to get the optimum
sensitivity?
So you define a scalar modulus for the crystal in the
particular direction you're exerting a uniaxial stress.
And then you express this module in terms of theta, the
angle of rotation, within the crystographic plane.
In the case of the sup cell if you looked at that geometry,
that is a flat plate subject to compression, where you
measure the surface charge perpendicular to the direction
of the uniaxial stress, not parallel to it
as we've done here.
And then there's a question if you rotate the plane of the
crystal about the imposed stress, what orientation gives
you optimum, maximum response?
So there are fun things that you could do with that.
We have a quiz next time.
There's a lame-duck period after the quiz will be over.
And I want to say a little bit about the tensor aspects of
elastic moduli.
Mechanical behavior is something that you have or
will cover in great detail in a graduate-level class on
mechanics, but I think very few people who deal with
things other than cubic materials or with
single-crystal materials where the elastic constants are
functions of crystal symmetry and direction.
So let me, at least at the time available, set up a
definition of stress in terms of strain.
Stress is a second-rank tensor.
There are six unique components.
So we could write an expression for sigma 1,1;
sigma 2,2; sigma 3,3; sigma 4,4; and sigma 5,5.
And sigma--
AUDIENCE: --1,2.
PROFESSOR: No, I want to go down like this.
So this would be sigma 2,3; sigma 1,3; and sigma 1,2;
eventually to be known as sigma 1, sigma 2, sigma 3,
sigma 4, sigma 5, and sigma six.
But actually, if we're writing out a full tensor relation, we
should write down all six elements here.
So let me do that and put in a 3,2.
Put in all nine of them, because before we condense to
matrix form, that is what we're going to have.
So we'll have a sigma 2,3; we'll have a sigma 3,2; we'll
have a sigma 1,3; a sigma 3,1; a sigma 1,2; and a sigma 2,1.
And we can express each of the stresses in terms of the
elements of strain.
So I'll have an epsilon 1,1; an epsilon 1,2; an epsilon
2,2; an epsilon 3,3; an epsilon 2,3; an epsilon 3,2;
an epsilon 1,3; an epsilon 3,1; an epsilon 1,2; and an
epsilon 2,1.
Nine terms, nine terms.
And in between is going to be a tensor
consisting of 9 by 9 elements.
There will be 81 different coefficients in here.
If we want to derive symmetry constraints, we're going to
have to, for every one of those 81 elements, do a
transformation.
And each transformed element is going to be a sum of four
direction cosines times one tensor element, repeated 81
different times.
Not something to be undertaken on a short afternoon.
OK.
What are the tensor elements in here?
They are represented by the symbol c.
This would be c1,1,1,1.
This would be c1,1,2,2 times epsilon 2,2; a c1,1,3,3; a
c1,1,2,3; and so on.
And these c's, in one of the great perversions of
scientific notation, are called stiffnesses.
The other thing we could do is to write
strain in terms of stress.
And really, stress is something you
can do to the crystal.
So stress is an independent variable.
And this is something that I feel more at home with.
So I'll have an epsilon 1,1; 2,2; 3,3; epsilon 3,2; 2,3;
epsilon 1,3; epsilon 3,1; epsilon 2,1;
and an epsilon 1,2.
Again, nine elements of strain.
And these will be given by coefficients times each of the
nine elements of stress, sigma 1,1; sigma 2,2; sigma 3,3;
sigma 2,3; sigma 3,2; and so on for nine different
components.
Andy the coefficients here are designated with the symbol s.
These are the tensor elements.
So this is s1,1,1.
This is s1,1,2,2.
And the s's stand for compliances.
Now English is a very strange language.
But to call compliance s and to call stiffness c is surely
a perversion that can be designed for no other purpose
than to confuse the introductory student and give
the instructor some feeling of superiority.
I don't feel superior, I feel embarrassed that I have to
explain this.
Stiffness, c; compliance, s.
It's got be one of the atrocities
of scientific notation.
To remember which goes with which, I'll tell you what
works for me.
If it doesn't work for you, forget it.
The s goes with sigma.
And the little c goes with the epsilon.
That's a c with a bar in it.
That works for me.
If it doesn't work for you, use your own mnemonic device.
English does a lot of things strangely, but nothing is as
perverse as this.
Now what's some examples?
Consider this.
You park your car in your driveway, but you drive your
car on a parkway.
Why don't you drive your car on a driveway and park your
car in a parkway?
No, it's the other way around.
It's almost as bad as this.
What are some other ones?
In my basement, I have something that's called a hot
water heater.
You don't heat hot water, you heat cold water.
Why isn't it called a cold water heater?
Even better, I once worked for a year and a half in
Switzerland.
And in the kitchen of my home I had, in German, an
elektrowarmwasse rsheissungsapparat.
That's one word that's 11 syllables.
The Germans have a knack with the language that we have
never approached in English.
But there's another one you've probably heard.
You know what the prefix "pro" means.
That means for, and "con" means against.
And so progress is moving forward, and I leave it to you
to decide what Congress means.
OK.
So I could go on and on.
I probably shouldn't, unless you egg me on.
No, I won't.
But anyway, there are--
OK, I'll give another one.
Why is what a doctor does called "practice"?
This is not reassuring at all.
Let me invite you over to my practice.
And why is it when you take an airplane your flight ends with
a terminal?
That's doesn't sound very encouraging either.
So anyway, language is silly.
And I guess if you want to call c stiffnesses and s
compliances, it's OK.
The verbal description of these terms means something.
Because if the stiffness is very high, that says that a
little bit of strain requires that you haven't posed a very
large stress.
So stiff is when the material is resistant to stress.
So you have to have a very, very large strain to get
yourself a given level of stress.
Conversely, if the material is very compliant, a very small
stress for a compliant material should
give you a big strain.
And that's exactly what a large value of s will do.
It will give you a large strain for a
relatively small stress.
So the words describe what's going on.
OK.
There is great utility in reducing these equations to a
matrix form, which is going to cut us down from a 9 by 9, or
a tensor with 81 elements, to a smaller number of subscripts
and a smaller number of matrix elements, so if we let sigma 1
be sigma 1,1; if we let sigma 2 be equal to sigma 2,2 and
sigma 3 be equal to sigma 3.
And then if we write the inequalities we've had before,
and we let epsilon 1,1 to be replaced by epsilon 1, then we
would have out in front here a term simply c1,1, a 1 for the
sigma and a 1 for the epsilon.
The next term would be c1,2 times epsilon 2, where
obviously epsilon 2 has replaced epsilon 2,2, and 1
stands for the two indices, 1,1.
So this will go c1,3; epsilon 3.
And then you hit these messy factors of two.
You'll have a pair of terms.
You'll have c1,4.
And that will stand for a term c1,1,2,3.
And then there'll be another term c1,1,3,2.
And if you want to write this as a matrix, c1,4 has to be
equal to the sum of these two things, times epsilon 4.
That's the only way you can write it as a matrix.
And if you think it was bad worrying about a factor of 2
in front of half of the terms on the right-hand side of the
equation, we're going to be dealing with factors of 2 for
terms in the upper 3 by 3 array.
We're going to be worrying about a factor of 4, somehow
or other, down in the lower 3 by 3 array.
So the way you handle the factor of 2 really is a
nightmare for relations in elasticity.
But continuing on with the first line, c1,5 times epsilon
5 would be a combination of c1,1,3,1 plus 1,1,1,3.
And then finally, when you get to c1,6 times epsilon 6, that
would represent a combination of c1,1,2,1 plus c1,1,1,2.
You have to make a decision where to swallow the 2.
And probably the best thing I can do is to give you
conventions for relabeling stress, strain, stiffness, and
compliance, which I'll pass around to you.
No big deal, except that this is a case where you can eat
the 2's early on, but you can't even break even.
So very, very briefly, what you do is you convert the
tensor elements of stress to matrix elements in exactly the
same way we did with piezoelectricity.
For strains, we do exactly the same thing that we did with
the converse piezoelectric effect.
The factor of 2 that you swallowed in definition of
stress pops up to haunt you when you
deal with the strains.
And you cannot write a nice matrix relation unless you
divide the off-diagonal elements of strain by 1/2.
And that was exactly the same thing we encountered with the
converse piezoelectric effect, which related strain to an
applied electric field.
In defining the matrix stiffnesses, the c's, you
forge straight ahead, and you let ci,j,k,l, which is
identical to ci,j,l,k--
that's the term that relates the two equal shear stresses--
and you define that as a matrix term with two
subscripts.
No factors of 2 or 4 are involved.
Then when you hit the s's, there's a nightmare.
si,j,k,l is sm,n if m and n are 1, 2, or 3; that is to
say, not 4, 5, or 6.
si,j,k,l, which is the same as s,i,j,l,k, has to be defined
as one half of sm,n, where m or n is 4, 5, or 6.
And then finally, you have to throw in a factor of 4 when
both m and n are 4, 5, or six or, in other words, m and n
are not 1, 2, or 3.
So it is a mess.
And these are rules that you have to bear in mind if you're
ever going to go from matrix form, which works absolutely
lovely in a fixed coordinate system.
But if you have a single crystal and you want to refer
it to different axes, you have to be prepared to resurrect
the full tensor form.
And then and only then can you work symmetry transformations.
I don't want to go through the simple algebra of expanding
and contracting these terms.
So the next two sheets show you how you condense from
tensor to matrix notation, and then go back from matrix
notation to tensor notation.
This is for the compliances si,j,k.
And the next page does the same thing, if you care to
write stress in terms of strain.
And there are a lot more factors of
2 that appear there.
But again, it shows you how you go from tensor notation,
for subscripts on the c's, down to matrix notation with 2
subscripts, and then go back up again to tensor notation.
So it's a tedious business.
And you've got to keep careful tracks of your factors of 2
and your factors of 4.
How about symmetry restrictions?
Holy mackerel.
Transformation of 81 different elements.
Well, in order to do the tensor transformations, you
have to go to the full-force subscript notation.
Only then is a law of transformation defined.
Fourth-rank properties are even tensors.
So mercifully, there are not as many different
possibilities.
All point groups that differ by presence or absence of
inversion have exactly the same form of the tensor.
So symmetry 2, m, and 2/m look alike.
Symmetry 2, 2mm, and 2/m, 2/m, 2/m look alike.
So there's only one orthorhombic tensor, only one
kind of monoclinic tensor, only one kind of triclinic
tensor, and so on.
And the only place that you have more than one form of the
tensor for a particular set of point groups is looking at the
point groups that are based on single-rotation axes,
something like 4, 4 bar 4m, and those that are based on
the axial arrangements 4, 2, 2.
So they're different, just as they were for second-rank
tensor properties.
For cubic, a rather remarkable result.
There are three independent compliances.
But cubic crystals are not elastically isotropic.
They are anisotropic.
And it's just the nature of the tensor that requires that.
So rather than letting you hang by your thumbs wondering
what's on these pages, let me pass around the summary of
symmetry constraints.
If you understood what we did for third-rank tensors or even
second-rank tensors, you how to do this.
And fortunately, the terms almost over.
So I can't make you do it, which is probably an enormous
relief to you.
Some additional bits of information.
The transformations for hexagonal crystals are
complicated because threefold and sixfold axes do not change
the reference axes x1, x2, x3 into one another.
And the equations that relate the individual tensor elements
are more complex.
For the threefold axis, for example, 2 s1,1 minus 2 s1,2
is equal to s4,4.
So they're not simple relations between them.
And that's simply because you're not changing one
reference access into another.
You're changing x1, for example, into a linear
combination of x1 and x2.
Two things down here of interest.
I said that cubic crystals are not isotropic.
What would have to be the case if the
material were to be isotropic?
Well, it turns out that if you do this for the compliances,
s, if 1/2 of s4,4 is equal to s1,1 minus s1,2, then the
material is elastically isotropic.
If you do this in terms stiffnesses, the c's, then 2
c4,4 is equal to c1,1 minus c1,2 if the material is
supposed to be isotropic or going to be isotropic.
Then there is one other inequality
that depends on structure.
And this is something known as the Cauchy relation.
And it depends on the interatomic forces.
The first condition that has to be met is that the forces
between the atoms should be central forces.
What is a central force?
Well, this is a case where the attractive force between the
atoms is directly along the line joining their centers.
Isn't that always the case?
Don't crystals hold together because there's an attractive
force between atoms?
Well actually, for metallic crystals and ionic crystals,
maybe that's not a bad assumption, particularly for
ionic crystals.
But if you had a covalent material, where the bonding
was due to overlap of orbitals like this, then the thing that
holds the crystal together is overlap
between these orbitals.
And the force holding the atoms together is a force that
goes through this shared electron pair.
And that's not a central force.
And so the Cauchy relation generally fails rather badly
for covalently bonded materials.
Second assumption is that each atom is
at a center of symmetry.
And the reason for that is so the force in the plus-x
direction is the same as the force
in the minus-x direction.
And that is not true for any material that has tetrahedral
coordination.
So it's not true for any of the forms of SiO2.
It's not true for any of the compound semiconductors that
are based on tetrahedral units.
So there, the atom inside of these tetrahedra is decidedly
not at an inversion center.
And finally, you assume that the crystal is under no state
of initial stress.
Because if it is under initial stress, you've squished it,
you've stretched the bonds.
And the forces--
not to a major degree, to be sure, but-- the forces will
not truly be perfect central forces.
So if all three of these assumptions are satisfied,
then you have an additional equality
c1,2 is equal to c4,4.
What I'll do next time is I'll bring in some examples of data
for stiffnesses and compliances and, in
particular, show you how some of these elastic tensor
elements change with temperature.
And we can examine them to see how well the Cauchy equality
or the isotropy condition is satisfied.
All right.
That's about all that we'll do about the definition of
fourth-rank tensor properties.
On next Tuesday to wrap the term up, we'll look at the
variation of some elastic moduli, such as Young's
Modulus or the shear modulus, for different symmetries as a
function of direction in the crystal, OK?