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PROFESSOR: All right.
The quiz on Thursday will cover up through
piezoelectricity.
You've had a set of notes in your hands that cover just
about everything that I wanted to say about piezoelectricity.
There are other modulae that one could talk about.
But these follow quite directly from the one or two
that we will do.
So we will not have anything to say about elasticity until
the last lecture of the term, which is a nice outcome
because you certainly don't want to take a quiz on
forthright tensors.
You'll spend the entire hour just writing out all these
cumbersome equations.
All right.
So the quiz will cover up through piezoelectricity,
including a few of the representation surfaces that
we'll examine today.
The thing that we'll be looking at is--
I'm sorry.
You have a question?
AUDIENCE: Yes.
A question.
Well, you said [INAUDIBLE]
example.
PROFESSOR: I'm sorry.
Third-rank tensors.
AUDIENCE: OK.
PROFESSOR: Also, we had just a little bit of that going into
the second quiz.
And we didn't really ask anything about that on the
second quiz.
So it'd be third-rank tensors.
But you have to use second-rank tensors to define
third-rank tensors.
So it really will be not the emphasis, but certainly you
should be familiar with the early part of what we did with
second-rank tensors.
All right.
So today we'll look at some representation surfaces to
your wonder and delight at how incredibly anisotropic the
variation of these third-rank properties are with direction.
One of the things that I love to do for problems when we get
to these different piezoelectric effects is make
up hypothetical devices just for fun.
So among those that I've invented, the first one is a
earthquake sensing device because it's well known that
California is going to split in half, and half will fall
into the sea any day now.
So if this is the San Andreas fault, I have developed large,
prismatic monoclinic crystals that I embed into the San
Andreas fault at regular intervals.
The bottom of these crystals is grounded.
There's an electrode at the top.
And if the San Andreas fault starts to move, and there is
shear on these crystals, there will be a charge
developed on the top.
And I ask you to relate the charge on the top in terms of
the piezoelectric modulae to the shear along the San
Andreas fault.
So there's a very clever little device that clearly is
going to be lucrative because these crystals will have to be
huge in size.
And they'll cost more even than silicon.
Another device that I've invented is used down at the
Boston fish pier.
This is a crystal on which I hang a pan, and
you put fish in it.
That creates a tensile stress in this direction.
We measure the charge on this face.
And we put a little meter that measures the charge that's
accumulated.
So this is how the fishermen can weigh their fish after
they haul them in off the boat.
So you see this is really practical material we're
dealing with.
This has applications in all realms of life.
So today I'd like to show you, also, another
problem that sets up.
And I pass this out just so you can, again, have a look at
some of the questions that you should be equipped to answer.
So here--
not fully expecting anybody to do it, but you can see the
sorts of problems one might ask.
Here is problem set number 16, which asks you, should you be
so inclined, to think about manipulations
of third-rank tensors.
But then the second question is an idea for a device which
the Center for Material Science and Engineering is
considering employing here in building 13.
And you can read all about that..
This is the famous soup cell.
I think probably you will see some sort of for fun modulus
for a particular device on the quiz.
I haven't made one up yet.
But I think you'll have a look at something like that.
And when we do the longitudinal piezoelectric
modulus for quartz, which we'll do momentarily, this
will give you an idea of how to set up these expressions.
The problem, basically, is that the direct piezoelectric
effect measures a polarization in terms of all of the
elements of applied stress.
So there is a modulus dijk times all of the elements of
stress, sigma jk.
So this, until you use the condensation of subscripts,
contains nine terms going this way and three
equations going this way.
As we discussed earlier, since only six of the nine elements
of stress are independent, you can condense this down into a
3 by 6 array of terms.
The problem in doing that, though, even though one has
only 18 different modulae to work with, instead of 27--
that's a considerable economy in notation, and an
elimination of a great deal of redundancy--
the matrix form of this relation-- and I emphasize
that it is a matrix, and it's no longer a tensor--
the matrix form cannot be transformed.
Again, there are only six terms in the
matrix elements of stress.
And there are three equations, again, one for each component
of polarization.
So instead of having 27, one has only 18.
But if you are considering changing the reference axes--
and that is one of the things that it's interesting to do
for these various modulae that one can define--
change the orientation of a particular rod-shaped specimen
that you cut out of a crystal to different crystallographic
orientations and then ask how the scalar modulus changes as
you change the direction in which you've sliced out the
wafer or the rod of material.
In order to do that, you want to transform the piezoelectric
tensor to a new set of reference axes.
And you cannot transform the dij's because they are a
matrix and not a tensor.
And no law of transformation is defined.
So what you have to do in any generic problem of this sort
is, for a particular single crystal, look up the form of
the dij matrix that will have the equalities between tensor
elements written in and the 0's, those modulae which are
identically 0, entered into the array.
And then you have to work your way backwards to get to the
full tensor notation.
So we'll see this when we look at the modulae for
symmetry three two.
You'll have to write in the exact matrix subscripts
without absorbing the equalities in the notation.
And then you'll have to expand the matrix terms into full
three-subscript tensor terms.
Then if you want to transform the axes, which is to say you
want to cut out your specimen, be it a plate or a rod, in a
different orientation, you have to transform the full
three subscript tensor elements to a new setting.
And then if you want to continue to work in that
setting in the compact matrix form, collapse it back down to
matrix form, insert the equalities, and then you're
back to where you started from.
So this problem of seeing how modulae that describe
different phenomena vary with crystal symmetry, you have to
go through this problem of expanding the compact form and
then collapsing back down when you've got it as a function of
some orientational angle or in terms of a coordinate system
that you want to work in.
The problem is exactly the same for elasticity.
And we'll look at some of these modulae next term.
You've heard these names before, I'm sure--
Young's modulus, shear modulus, and so on.
We'll take a look on next Tuesday, a week from today, at
Young's modulus, which is one of the more important ones.
And probably are used to seeing this in the form of
information for a polycrystalline material,
which is essentially isotropic.
The modulus is much more interesting for
single crystal materials.
And then the surfaces that are defined particularly for the
lower symmetries are absolutely wild things with
lumps and wiggles and lobes and things of that sort,
nothing like the dumb old, uninteresting ellipsoids that
we encountered for second-rank properties.
So let's, then, take a look at how we had set up and defined
the direct piezoelectric effect.
We set this up as a proper tensor relation, saying that
P1 is equal to d1.
V1 And then, you'll recall, we have nine elements of strain--
sigma 11, sigma 12, sigma 13, sigma 21, sigma 22, sigma 23,
sigm a 31, sigma 32, and sigma 33.
But the tensor is symmetric, so we really only need to
enter into our relation six of these nine terms explicitly.
And what we did was to replace the two subscripts on the
elements of strain, which, again, you need if you want to
refer to those elements of stress through a different
coordinate system.
You have to know the elements of stress in tensor form.
But we convert it to six terms by going and replacing the
pairs of subscripts with a single one, two, and three,
marching down the diagonal of the tensor this way and then
marching up the right-hand side, calling two, three,
four, and calling one, three, five, and finally ending up in
this slot here.
And we call that six.
So that was the notation we used to get to a matrix
representation.
But the place where all this started is--
and I'll write just one line of this to be merciful--
d 111 times sigma 11, so this pair of subscripts goes with
this pair, plus d 122.
I'm putting that one in next because we're going to number
the terms for stress in this order, one through six.
Times sigma 22 plus d 133 times sigma 33 plus d 123
times sigma 23 plus d 132 times sigma 32 plus d113 times
sigma 23 plus d 132 times sigma 32.
And finally we end up in slot number six.
And we have a d 112 times sigma 12 plus a d
121 times sigma 21.
Look at that.
There's an equation that covers two whole blackboards.
So if we now condense this down to matrix form we would
say that P1 is d 11 times sigma 1 plus d 12 times sigma
2 times d 13 times sigma 3 plus--
and now we have this messy problem with the 2's--
we have a d 14 times sigma 4 plus, again, a d 14
times a sigma --
2 3 is equal to 32, so I can call this 4--
and then these terms become 15 sigma 5 and, again, a 15 times
sigma 5 plus a 16 times sigma 6 plus d 16 times sigma 6.
Now we have to make a choice.
Either we are going to have, in a general matrix relation,
that P sub i is equal to dij times sigma j, if j is equal
to 1, 2, or 3.
But it's equal to 2 dij times sigma j if j is
equal to 4, 5, or 6.
And that is something we like to avoid, if possible.
That's ugly.
And it's going to be a hell of a matrix if we have factors of
two in front of some of the matrix elements but not in
terms of others.
So this is something we could do.
Hey, it's our ballgame.
We make up the rules.
But that's going to be an ugly thing to have to deal with.
So instead, as we mentioned last time, what we will do is
to lump together these terms and define d 14 not as these
individual tensor elements, but define those matrix
elements as the sum of these two elements.
And then we saw before that--
we saw last time in our earlier meeting--
that from the converse piezoelectric effect, which
expresses strain in terms of an applied field, where the
elements of strain 1, 23, and 32
appear in separate equations.
And knowing that the same array of piezoelectric
coefficients amazingly describes the converse
piezoelectric effect as well as the direct piezoelectric
effect, we know that d 123 equals d 132.
And that is from the converse effect.
So equivalent to saying this is to define d 14 as twice d
123 because these two elements are equal.
So we're eating the factor of two here so that we can write
a nice matrix relation that doesn't
involve a factor of two.
So making this combination of terms for the shear stresses,
we would have simply d 14 sigma 4 plus d 15 times sigma
5 plus d 16 times sigma 6.
And we can say, in general, for the other two equations,
by analogy to this one, that P sub i is dij times sigma j--
a nice, neat matrix relation but one for which,
unfortunately, there's no law of transformation for the
matrix modulae dij.
If you want to change to another coordinate system, we
have to be prepared to resurrect this full three
subscript notation on the piezoelectric modulae.
OK.
Comments or questions at this point?
All right.
If not, let me remind you that in the notes which I
distributed last time, there are summarized all of the
constraints imposed on the piezoelectric modulae for
single crystals.
And, again, these constraints, these requirements that the
tensors remain invariant for the change of axes produced by
a symmetry element that the crystal possesses, these
transformations show that no third-rank tensor property can
exist in a crystal that has inversion.
So the 11 [INAUDIBLE] group, so-called, that possess
inversion have absolutely no property and can be not
considered further for third-rank properties.
And then one must consider all of the 32 minus 11 21 point
groups that lack conversion separately.
There's no reason why they should behave the same way.
Remember that for second-rank tensor properties we pulled
the argument that inversion imposes no restrictions or
constraints whatsoever on second-rank properties so,
therefore, two different symmetries that differ only by
the presence or absence of an inversion center.
That is to say, you change 2 to 2 over
m if you add inversion.
But the argument was inversion requires nothing, so the
constraints or symmetry, too, look exactly
like those for symmetry.
And here you've got to plod through every single one of
the non-centrosymmetric point groups separately.
And they all have tensors that have different forms.
So what I'd like to do is look at one specific one.
And that is the matrix for symmetry 32.
And that is a point group that you'll recall has a threefold
axis and twofold axes at intervals of 60 degrees.
And in your list of the qualities and absences, 32,
where the 3 is parallel to the axis x3, and the twofold axis
is parallel to x1.
So we're defining this as x1, this as x3, and x2 comes out
halfway between the two.
So let me draw this looking down along the threefold axis.
These are all twofold axes.
And we'll take x1 in this direction.
x2 pokes out in between two twofold axes.
And x3 comes straight up along the threefold axis.
With that coordinate system, the form of the piezoelectric
modulus matrix has this form-- d 11 minus d 11 0 d 14 0 0 0 0
0 d 15 minus d 14 0 and in the bottom row d 31
d 31 d 33 0 0 0.
So this matrix with the equalities put in is obtained
by looking at the full three-subscript tensor and
requiring that it look the same before and after any of
the rotations involved by these three distinct axes--
the threefold and the pair of twofolds.
OK.
Now, there are many different scalar modulae that one could
define, some of them serious and worth the consideration
because of the application in devices or other practical
situations.
The modulus that I'd like to examine is something called
the longitudinal piezoelectric effect.
And let's emphasize, again, that it is impossible to come
up with one representation surface that fits every need
because, again, the direct piezoelectric effect relates
the components of a vector to a tensor, sigma ij.
There are six independent tensor elements.
So how can you describe how this vector is going to change
as you change orientation of a crystal whose behavior is
described by all of these modulae?
But I would point out, however, that there are only
two distinct--
oops, I'm sorry.
This is d 14.
I don't know how I made that d 15.
There are only, in this array--
and I slipped a notch.
Excuse me.
Last couple of days are such that I am not able to even
read from my notes.
So these bottom lines, my apologies, are all 0.
And there are two modulae, d 11 and d 14.
So there are two independent numbers, but they appear as
different matrix elements and, therefore, different tensor
elements, as well.
And this should be minus 2 d 14.
I'm sorry.
I slipped down a notch, and I got some for 32
and some for 6.
And I'm glad I found it at this point, or I'd really be
in deep trouble.
OK.
So two numbers and that is, in fact, the form of the matrix
for symmetry 32.
So the longitudinal piezoelectric effect is one of
the representation surfaces that gives you the way in
which the polarization will change for one very, very
specific type of stress.
In particular, what we'll do is set this up as
a coordinate system.
And we will look at a coordinate system where this
is the reference axis, x1.
We'll come down with a compressive stress, sigma 11,
along that axis.
And, therefore, we're applying a uni-axial stress, which has
just one component of stress.
So that's very specialized.
In general, there would be six different
components of stress.
But we're looking at one specific stimulus applied to
this crystal plate.
In response to that sigma 11 there are charges induced on
all of these surfaces.
And these charges, this charge per unit area, is proportional
to the component of polarization P1, the component
of polarization P2 along the surface out of which x 2
comes, and the charge per unit area or the
polarization along x3.
So this would be P3.
And this would be the direction of x3.
Now I deliberately tried to show this sample as a thin
wafer, which has a much larger surface area here than it does
on the other two surfaces-- a much smaller area there.
Therefore, since polarization is charge per unit area, if
this area normal to x1 is a very large area, there's a lot
of charge accumulated there.
It's going to be easy to measure.
If we make the wafer vanishingly thin, then the
charge per unit area is high, but the total area is small.
So there's going to be a negligible accumulation of
charge on these two side surfaces.
So the longitudinal piezoelectric effect and the
longitudinal piezoelectric electric modulus is an effect,
where we look at the component of polarization, P1, in
response to an applied stress, sigma 11.
So it's that simple.
Look at all the terms we've thrown out.
We've thrown out a whole bunch of elements of stress, which
we could impose if we wanted to.
And we've thrown away two of the three components of
polarization by designing a specialized sample.
So all that's left then is that P1 equals sigma 11.
And the relation between those two parameters is the 111.
So all this is going to hinge on one single piezoelectric
electric modulus, d 111, and how that
changes with direction.
So this is for one orientation of a plate.
And I had not specified how the orientation of this plate
is related to the symmetry axes.
So let's do that now.
What I'm going to assume is that this is a crystal.
It doesn't look like it's hexagonal.
But imagine that this is a crystal of quartz.
And we could look at an x1 that's in this direction.
And imagine that we have cut out of this crystal a wafer
that has a normal along x1.
And then relative to this coordinate system, if this is
x1, the modulus d 111 would tell us what charges
accumulated on these two surfaces.
But now, what we could do if we wanted to know how this
modulus changed with direction would be to cut out a plate, a
thin plate, in another orientation,
where this is x1 prime.
And this has changed relative to the orientation of the cell
edges in the crystal.
The crystal is fixed.
We're just cutting a wafer out in a different orientation.
So this is x1 prime.
We're going to, again, squeeze it with a tensile
stress sigma 11 prime.
And we'll ask how the polarization P1 prime is
related to sigma 11 prime.
And the answer is that P1 prime will be a tensor element
d 11 prime times sigma 11 prime.
So what we are asking, essentially, is how does d 11
transform when we take the direction of x1 in a different
orientation and, thus, change the value of d 11 prime?
It's going to change all of the
piezoelectric matrix elements.
But we're looking at an effect in a sample that is
deliberately prepared such that we will measure only the
surface charge given by P1 prime.
And, therefore, the way in which the properties of this
plate change as we vary the way in which we've sliced it
out of the single crystal is going to be simply the
variation of d 11 prime with direction.
So this is the general nature of what we will do when we
define any of the scalar modulae related to the
piezoelectric modulus tensor.
We can change our notation a little bit in that we have a
modulus which I'll define as a scalar modules d.
And that d is going to be d 111 prime.
And I know how to evaluate that.
d 111 priime will be C 1l, C1m, C 1n, where these are
direction cosines, times all of the elements in the
original tensors, dlmn, in the tensor referred to the
original coordinate system.
So even though this looks simple--
it's just one modulus--
when we transform it, we've got a product of three
direction cosines out in front at every single one of the 27
tensor elements in the original tensor.
So it's not as trivial as it seems.
So this is how this modulus that relates compressive
stress to induced surface charge will change with
orientation.
But what are these direction cosines?
These are the direction cosines--
not the full direction cosine matrix.
These are the direction cosines for x1 prime.
OK?
So we can get rid of this two-subscript notation if it's
understood that these are the direction cosines of x1 and
simply call these l l, l m, and l n, just as we did for
the direction cosines of a vector because we're only
concerned about the orientation of one of the
axes, namely x1 prime.
We don't care diddly-bop about x2 prime or x3 prime because
these don't enter into the modulus that we have defined.
And this will be times dlmn.
OK.
Is what we're doing clear?
So we have defined this particular effect in terms of
those of the 27 piezoelectric modulae which are necessary to
describe it.
And then we've established how they will change with a change
of the direction of one particular direction.
And we don't care anything about x2 or x3.
All right.
Now we go through this process of inserting for matrix
notation with the equalities built in.
The proper matrix notation in the first term is d 11 in
matrix notation.
That's this term up here in the upper left-hand corner.
The second term, the term that we've written as minus d 11,
that's not d 11 at all.
This is, by definition, d 12.
And we need the true subscripts, if we're going to
transform this.
So this really is not even a matrix because the subscripts
have lost meaning, and we're just using them to identify
equalities.
Then comes a 0.
And next comes something that we've labeled d 14.
And the subscripts there are correct.
That is, indeed, the fourth term in the first row.
But we're going to want to convert d 14 into a tensor
element momentarily.
Now let's get the rest of the terms that are non-zero.
This is really d 25.
So the next term that is non-zero is d 25, which just
happens, because of symmetry, to be equal to minus d 14.
But this is the true matrix subscripts, and this is the
true matrix subscript here.
The next term over to the right is the fifth and final
non-zero term.
This is minus 2 d 11.
And this is really d 26.
Those are the true matrix elements.
So we put in the proper matrix subscripts.
And now the next, final, step in the expansion is to convert
these terms into actual tensor elements.
So this is d 111.
And these are tensor subscripts, so this is
something we can transform.
This is d 12.
In tensor notation this is d 122.
And that's something that has a law of transformation.
d 14 is really d 123 plus d 132.
We lumped two tensor elements together
to define this modulus.
Minus d 14 that appears in the next to the last
non-zero spot is d 25.
d 25 is really d 231 plus d 213.
Then, finally, d 26 is d 121 plus d 112.
AUDIENCE: Shouldn't that be d221?
PROFESSOR: Sorry.
d 26, you're right.
That's down in the second row. d 221 and d 212.
Now we've got something we can transform.
The law for transformation is l sub l, l
sub m, l sub n, dlmn.
So And these are the direction cosines of x1.
So this term will transform as l 1, l 1, l1 times d 111 .
1 The next term will transform as l 1, l 2, l 2 times d 122.
And that will be d 122 prime for different
orientation of x1.
This will be two terms.
This will be l 123.
And I can write them in any order.
So this is l 123 times d 123 plus d 132.
And these terms prime, when we change axes, are going to be
equal to l 2, l 1, l 3 times d 231 plus d 213.
And this last term will be l 1, l 2 squared times d 221
plus d 212.
All right.
So this now is our new tensor element, d 11 prime.
And that's given by this sum of terms.
So we'll have a first term l 1 cubed times d 111.
And if I look through these other terms, that's the only
term in l 1 cubed that I'll have.
The next term will involve the product of three cosines--
l 1 and l 2 squared times d 122.
And if I go down here, here's an l 1, l 2 squared again.
So I have plus d 221 plus d 212.
And then, finally, the other coefficient that I have is
plus l 1, l 2, l 3--
which is what this should be.
And that will be times the sum of terms d 123 plus d 132.
Up here, you've got the same thing again--
plus d 231 plus d 213.
And I have a total of 1, 2, 3, 4, 5--
1, 2, 3, 4, 5 terms.
OK, that is how the longitudinal piezoelectric
modulus will change as we change the direction of the
normal to the plate that we have cut out of the crystal.
So these are direction cosines relative to the
crystallographic axes.
l 3 is the angle between the normal to the plate and the
threefold axis.
l 1 is the angle cosine to the angle between the normal to
the plate and one of the twofold axes.
And l 2 is the direction cosine for the normal to the
threefold axis and the twofold axis.
OK.
So we've got it now in terms of tensor elements.
And now--
yeah?
AUDIENCE: Is it at all reasonable to assume instead
of taking those sums in d 123, d 122, just saying 2 d 123?
Is that OK in assuming?
Or not necessarily?
PROFESSOR: Well, we could do that.
But I did it the long way to not obscure what we're doing.
OK?
This is a well-defined summation over subscripts.
And we're going to collapse immediately down to the sums.
And we're going to replace the equalities.
So let's see what comes out of this, if we now, having
reached the zenith, having transformed the tensor
elements, go down and replace this with a consolidation of
terms and an insertion of the equalities
between the matrix elements.
OK The first term is d 11.
So I will have l 1 cubed times d 11.
Notice I'm getting third powers of direction cosines,
which is going to be what causes the exotic nature of
these anisotropies.
And then I have a product of l 1 and l 2 squared.
And this is d 12.
And this second term is--
where did it go?
This is d 21 plus d 212.
And that is what we call d 26.
And then, finally, this product of three different
direction cosines--
l 1, l 2, l 3.
And we have d 231 plus d 213.
And this is d 25.
And, again, an l 1, l 2, l 3 times d 14.
And the second term here is d 25, if I've done it correctly.
d 25 --
this is d 24.
And this one is d 24.
2/4 OK.
Let's now insert the equalities--
back to where we came from.
d 11 is d 11.
d 12, however, for symmetry 32--
I'm going to my handy-dandy chart of symmetry
restrictions.
I don't want to do that.
That's fourth rank.
AUDIENCE: It's still on the board.
PROFESSOR: It's still on the board?
Yes.
Thank you.
When your nose is in it, it's hard to see.
d 12 is minus d 11.
1 And d 26 is minus 2 d 11.
So these two terms can be consolidated.
l 1 cubed plus l 1, l 2 squared times d 11 minus d 11.
So these two terms die.
And I have a minus 2 d 11 that's left.
If I insert the equalities here, I'll have l 1, l 2, l 3.
d 25 is minus d 14.
And here's a d 14 itself.
So these two terms die.
And then I had d 25.
And that is--
AUDIENCE: You don't have d25.
PROFESSOR: I don't have d 25.
Where did I get the extra one?
AUDIENCE: [INAUDIBLE].
PROFESSOR: OK.
I'll take your word for it.
And I know how it has to turn out.
OK.
So these two terms kill each other.
And I'm left with, then, an l 1, l 2, l 3.
Or have I left something out?
This is d 1--
this is d 15 and d 231 and d 23 --
uh, this is d 2 --
and the combination of 13 and 31 is d 25.
Right?
And if I look at my equalities, this is l 1, l 2,
l 3 minus d 14 plus d 14 plus d 25.
And d 25 is minus d 14.
AUDIENCE: Why would you add this d25?
PROFESSOR: Let me check my notes and see
what I've got here.
AUDIENCE: [INAUDIBLE].
PROFESSOR: OK.
See what I --
d 25 is minus d 14.
And then I have just a d 14.
I don't know --
I see what I did.
I put it in the wrong slot.
This is d 14.
And d 25 is the one that's minus d 14.
so this term dies, which is nice because that
cross term is messy.
So what I'm left with, then, if I check against my notes,
is d l 1 cubed plus l 1, l 2 squared times d 11.
And then I have minus d 1 minus--
this doesn't belong in here.
This wants to end up being l 1 cubed minus 3 l 1, l 2 squared
times d 11.
The first term is l 1 cubed.
That's correct.
The second term is l 1, l 2 squared.
We've got a minus d 1 plus minus 2 d 11.
So I have minus 3.
It should be a minus.
Yeah, that carries down to a minus.
So I have minus 3 l 1, l 2 squared all time d 11.
OK.
So what we have ended up with is an expression for the
longitudinal piezoelectric modulus as a function of
orientation.
The surprising thing is that l 3 does not appear here at all.
It doesn't depend on the angle between the normal to the
plate and the threefold axis.
It depends only on one modulus, and that is a
remarkable thing.
This says that the shape of this surface is independent,
essentially, of the property, any property that relates the
one one prime to a uni-axial stimulus, sigma 11.
And you measure a vectory component in the same
direction is always going to have this universal surface.
And it involves just a geometric term and then one
modulus that changes the magnitude of the longitudinal
piezoelectric modulus but does not change the asymmetry.
So let's see what this function looks like as a
function of direction.
Maybe we better wait for that until we come back because
that's going to take a few minutes.
So this is what we found.
That is correct.
And we have to now decide what this looks like, which will
take a few more minutes.
But let's stop here rather than run late.
All right.
Let's take our 10-minute break as usual.