PROFESSOR: I think you will get going.
A lot of people are not here, but I think if the MRS meeting
were going on across the river I would be there too if I
didn't have a class to go to.
We had last time finished a discussion strain.
And we introduced strain by defining it as a displacement,
each component of a vector displacement U as being given
by a set of coefficients, eij times x of j, which says that
the displacement in a deformed body varies linearly with the
position of the particular point in the body.
We could also express the same thing in terms
of change of length.
The change in the vector U would be given by-- we show
the same set of coefficients--
eij times delta x of j.
So whether you want the displacement vector or you
want the fractional change of length, you get this by an
expression of the same sort.
We saw though that unless we define the coefficients
correctly we could have a situation where a body is not
only deformed but is rotated as well.
And we saw that in general unless we define these
coefficients carefully we would include in this tensor a
component of pure rotation, rigid body rotation.
And this would not measure deformation.
However, regardless of its suitability or unsuitability,
the positional vector is a vector.
It transforms like a vector.
The displacement U is a vector, and it will transform
like a vector; ergo, the 3 by 3 array eij is a second-rank
tensor and we'll transform like a tensor.
We showed though that we could take eij and get rid of the
pure rotation that might be contained in those
coefficients by writing it as a tensor epsilon ij plus
another tensor omega ij.
And that was a new wrinkle, the concept of adding two
tensors together to get a third tensor.
We did run into that before.
And we define the elements of the two tensors epsilon and
omega; epsilon ij was given by one half of eij plus eji, and
omega ij is given by one half of omega ij.
And notice the order of the subscripts here in this
difference.
That is a matter of definition, but it defines
which term is omega ij and which is omega
ji minus omega ji.
I'm sorry.
This should be eij minus eji.
And we showed that at a tensor of this form which would have
all of the diagonal terms zero and the off-diagonal terms
equal to the negative of one another that a tensor of this
form does indeed correspond to pure body rotation.
We showed that by looking at a point at the end of a
positional vector U and looked at a displacement which would
be exactly at right angles to it.
And the U would be given by the tensor relation.
We showed that U dot r is identically zero when the
displacement vector is given by this tensor with diagonal
term zero and the off-diagonal terms equal to the negative of
one another.
So we have then from a general tensor of the form eij
constructed in the terms epsilon ij something that is a
measure of true deformation.
It is a symmetric tensor by definition.
And therefore in addition to all of the other properties of
second-rank tensor, namely having a law of transformation
and staying symmetric for any arbitrary change of coordinate
systems, the radius normal property also works.
Hence, the representation surface that we construct from
the tensor epsilon ij xi xj equals 1, gives
us the strain quadric.
And it has the property that the radius gives us the value
of strain in a given direction.
And the value of the radius literally is going to be the
vector that's on the left hand side, the magnitude of U times
the component of U that is parallel to the radial vector
per unit radial vector.
And this then is the tensile strain, the fractional change
of length as we look in this direction.
The radius normal is going to give us the direction of the
total displacement U. And the value of the property in that
direction is going to be the product of that displacement
with a unit vector in the direction of interest.
And therefore that is the tensile strain as we said
before from the general properties of the quadric.
So everything we said about second-rank tensor and in
particular symmetric tensor holds for the strain tensor.
We can talk about diagonalizing the strain
tensor into a form epsilon 1, 1, epsilon 22, zero zero,
epsilon 33.
And I'll remind you that we know how to do this to set up
the normal equations and solve for the eigenvalues and then
look for the principal axes of a tensor.
There are special forms of the strain tensor.
Something that 1 diagonal goes into this form is
called plane strain.
And I love the melodic nature of that, the plane strain.
If this were strain induced by a Modernistic transformation,
we could call it the Bain plane strain.
If we diagonalized it, it would be the
main Bain plane strains.
And if that deformation took place in an aircraft window,
we could call it the plane-pane main Bain strains.
And we could continue on indefinitely, but I think
you're finding this tiresome.
Another form of the strain tensor that doesn't look what
it actually represents is epsilon 1 zero minus epsilon 1
zero zero zero zero.
That looks peculiar, very special.
But, again, this is analogous to what we found for stress.
If we rotate this tensor 45 degrees, we would find that
treat tensor was diagonally, had all the
diagonal elements zero.
And we would have the epsilon 11 prime--
let's just call it epsilon prime--
and this epsilon prime zero zero zero.
And this is pure shear
Something that we'll show shortly, and in fact I left it
to you as a problem on a problems set.
The sum of the diagonal elements, epsilon 11 plus
epsilon 22 plus epsilon 33 is something that's defined as
the trace of the tensor.
And for the strain tensor, the trace turns out to be equal to
delta V over the fractional change in volume.
And we'll show that in just a little bit as well.
So a characteristic of pure shear is that it does not
result in any change of volume of the solid.
There's deformation, but the volume stays exactly the same.
Let me then proceed to show that the trace of the strain
tensor is indeed the change in volume.
And want I'll do is to look at an element of volume that is
oriented along the principal elements of strain.
So if this as x1, this is x2, and this is x3, the strain
tensor I'll assume is oriented.
So we have a term epsilon 11 zero zero zero zero; epsilon
22 zero zero zero; epsilon 33.
So let's suppose that the solid originally has edges L1,
L2, and L3 along x1, x2, and x3, respectively.
So the initial volume before any deformation will be simply
L1 times L2 times L3.
Let's suppose now we impose the strain, and the volume
then we'll increase to some value of V plus delta V. And
the length L1 will change to a value L1 times 1
plus epsilon 11.
That is to say it'll be L1 plus epsilon 11 times L1,
which can be factored out in this fashion.
L2 changes to L2 times 1 plus epsilon 22.
And L3 changes to L3 plus epsilon 33 times L3, which
I'll write in this fashion.
So expanding this product, it'll be L1, L2, L3.
And then if I simply multiply out these terms, I'll have 1,
and then I'll have epsilon 11 plus epsilon 22 plus epsilon
11 times epsilon 22.
And this will be times 1 plus epsilon 33.
And if I carry out that multiplication,
that will be 1 plus--
and I'll get exactly these terms again-- epsilon 11 plus
epsilon 22 plus epsilon 11 times epsilon 22.
And then a term epsilon 33 plus epsilon 11 epsilon 33
plus epsilon 22 times epsilon 33.
And then a third-order term, epsilon 11 times epsilon 22
times epsilon 33.
Now having gone through the painful process of expanding
that, I will proceed to say that the epsilons are always
going to be very small.
Elastic strains are typically on the order of 10
to the minus 6.
So a product of two of those strains is 10 to the minus 12.
And a product of three of those terms is going to be on
the order of 10 to the minus 18.
So this last term is going to be minuscule.
And now I use a higher order term intentionally rather than
a higher rank term.
This is really a term of higher order.
The same is true of all the peer-wise projects of this
[INAUDIBLE].
So I now have left if I turn out these higher order terms
to pasture, I'll have 1 plus epsilon 11 plus epsilon 22
plus epsilon 33.
And the rest is negligible.
And this will be the original volume V times 1 plus epsilon
11 plus epsilon 22 plus epsilon 33.
So delta V over V, if I subtract off V on the left,
delta V over V then is simply going to be equal to epsilon
11 plus epsilon 22 plus epsilon 33.
So this is indeed the fractional change of volume,
and it's the trace of epsilon ij.
Now, physically, you wouldn't expect the volume to change if
I change my reference axes because
that's a scalar quantity.
So I've shown by a hand-waving, backdoor argument
that in fact the trace of a tensor does stay invariant for
at least this particular tensor.
But it's fairly straightforward to show that
for any tensor aij the trace of the tensor is invariant
when you change the coordinate system.
And that's not hard to do.
If you're clever, it takes about three lines.
So I'll invite you to the exhilaration of discovering
that for yourself on a problem set.
Any questions or comments?
All right.
Having defined strain as a symmetric second-rank tensor
when we factor out rigid body rotation, we can view strain
as a physical property.
And I think I tried to hoodwink you last time by
saying this is a property of material.
So therefore strain tensor has to also conform to all of the
symmetry restrains that we have derived very
exhaustively.
And you said, no, no, no, no, that can't be; I don't believe
that I can only give a uniform uniaxial deformation to an
orthorhombic crystal.
And I say, yeah, but to get the deformation, you have to
squeeze the crystal.
And that is going to be linked to the deformation by the
elastic constants, and they are physical properties.
And that maybe makes you think a little bit.
But actually even though operationally you impose a
strain by imposing a stress or by doing something else to the
crystal that results in a strain, you can nevertheless
pick any stimulus you wish to achieve a
desired state of strain.
So there is no symmetry constraint on the form of the
strain tensor that can result.
So the stain tensor then is not a property tensor.
It is also a field tensor just like stress.
So it can have any form we choose to create in it.
So we now have the potential of having
stress as a field tensor.
And we have strain as a field tensor.
And within limits, we can create strains of these two
field tensors of any form that we wish.
And we can now regard these as things that we do to crystals
and look at a whole range of properties that result when
the generalized force is not a vector, a tensor of first
rank, but the generalized force is a--
let's say-- a stress tensor.
It's a generalized force which itself is not
a first rank tensor.
It's a second rank tensor.
So we'll see a whole collection of interesting
properties that fall into this category.
But first I'd like to look at one remaining second-rank
tensor property, and that is thermal expansion.
And this is a curious second-rank tensor property,
and we couldn't discuss it until we had
talked about strain.
And let me introduce it by one-dimensional example.
Let's suppose we have a rodlike specimen of length L.
And let's suppose it is in equilibrium at some
temperature T. And then we heat it up to some temperature
T plus delta T.
In response to that particular change in temperature, the
length of the rod will increase to a length L plus
delta L. It expands.
And we would find experimentally that delta L,
first of all, is proportional to L. If you make the rod
twice as long, the amount of expansion in an absolute sense
is twice as large.
And secondly, provided you're well above very, very low
temperatures, close to absolute zero, delta L will
also be proportional to the change in temperature.
Thermal expansion is one of those properties which
thermodynamically must go to zero as
temperature goes to zero.
So the thermal expansion coefficient that we've yet to
define will be a function of temperature as we get down to
very low temperatures on the order of a few degrees Kelvin.
So we can define then the linear thermal expansion
coefficient for this rod as 1 over L, delta L over delta T.
So this is something that will give us the fractional change
of lengths.
And what we can do now in general is to say that if we
create a general strain, not a tensile strain, we have a
general strain epsilon 11 plus epsilon 12, epsilon 13, and so
on, has to be symmetric.
So epsilon 21 is equal to epsilon 12.
We'll say that this tensor epsilon ij will be written in
general for the three-dimensional case as a
second-rank tensor aij times a scalar quantity delta T.
The strain is the field tensor.
That's the response.
The stimulus that creates this response is an incremental
change in temperature.
But as this is a second-rank tensor and this is a scalar,
an array of coefficients in which we multiply every
element of a tensor by a scalar is also a tensor.
So this is called the linear thermal
expansion coefficient tensor.
And it is a symmetric tensor by definition since a measure
of true strain is by definition a symmetric tensor.
So this is the linear thermal expansion coefficient tensor.
AUDIENCE: Quick question.
PROFESSOR: Yes.
AUDIENCE: [INAUDIBLE] the temperature gradient
[INAUDIBLE]
and talk about the directionality say between
here and here because of the temperature [INAUDIBLE]?
PROFESSOR: Oh, yeah.
Sure.
Sure if you wanted to do that.
That assumes that the thermal conductivity would be very
small so that the temperature inside the body would not
attempt to [INAUDIBLE].
Yeah.
In the same way that you could create strains that are in
homogeneous as well.
We'll now for the first time start talking not in terms of
abstractions but in terms of some real physical properties.
What is the range of linear thermal
expansion coefficient tensors?
And unlike many physical properties, that's an easy one
to answer, 10 to the minus 6 degree C per degree C.
Materials have a very, very small range linear thermal
expansion coefficients.
And let me give you some examples for real materials.
They go up to maybe 10 to the minus 5 for a very soft weakly
bonded materials.
But the for metals--
and I'm giving you a single number now because these are
averages for polycrystalline materials--
the value of A times 10 to the 6.
For lead, a low- melting metal, is 28, 2.8 times 10 to
the minus 5.
For copper, 18.
For iron, 12.
Four tungsten, a very refractory metal, 5.
And for diamond, a very strongly bonded materials,
it's 0.89 times 10 minus 6.
If we look at compounds, again, the average linear
thermal expansion coefficient times 10 to the 6 for
polycrystalline material.
For aluminum bromide, hardly a technology material of
commerce, but in light of this because this has one of the
largest linear thermal expansion coefficients of any
material, very weakly bonded to compound.
For a more typical ionic compound, NaCL, 40 times 10 to
the minus 6.
For calcium fluoride, this has a bivalent cation in it, so
the material is stronger, more strongly bonded, 20.
For MgO, both a divalent cation and the divalent anion,
as you would expect, the thermal expansion coefficient
drop still more.
For Al2O3, a trivalent cation that goes down to 8.8, and
that's really an average because this
is a hexagonal material.
And a second-rank property tensor for a hexagonal crystal
has to have two equal diagonal elements and one independent
diagonal element when you diagonalize the tensor.
Spinal, MgAl2O4 this is representative of all of the
ferrites for example, 7.6.
Silicon carbide, a very refractory
covalent compound, 4.7.
For glasses, here's a surprise.
You might think that a glass might have about the same
linear expansion coefficient as the crystalline form of the
same composition.
But a sodium calcium silicon oxide glass, so called
soda-lime silicate glass, is 9, fairly low.
A borosilicate glass, like the Vicor that we love to make
glass apparatus out of, 4.5.
And one of the near record holder is fused silica,
silicio glass, and here a measly value of 0.5.
So glasses have very low linear thermal expansion
coefficients compared to crystallize materials that are
predominantly ionic.
And the reason lies in the nature of the structure.
Something like MgO has ordered ions, and when you heat it up,
it expands isotropically.
A glass is this random network of edge-shared polyhedra.
So when you heat it up, yes, the dimensions of the
polyhedra increase in proportion to the change of
temperature.
But the framework because it's so meandering and open can
buckle to accommodate the increased size of the
tetrahedra that are in the linkage.
So the result is the network buckles, but the overall
microscopic thermal expansion is very, very slight even
though the tetrahedra are changing dimensions in the
same degree that they are in these crystalline inorganic
nonmetallic materials.
Let me give you a handout not that it's up to date, but
there was a very nice, convenient one-page article in
the Journal of the American Ceramic Society a number of
years ago that lists thermal expansion coefficients for low
expansion oxides.
So let me pass that around and let you take off
for copy with that.
And that gives some examples of the value of this property
for, again, for polycrystalline materials,
which if you looked at a single crystal would show
anisotropy.
And here you'll find a few materials in the list near the
top they're arranged in order of increasing linear thermal
expansion, you'll find some that are just about zero when
you have a polycrystalline form of the material.
And you'll have some that have this very unusual behavior
where the linear thermal expansion coefficient is
actually negative.
So let me pass this around for reference.
Now this may seem to be a curiosity that you can fine
for a polycrystalline material for which the individual
grains expand anisotropically a bulk linear thermal
expansion coefficient of zero.
Why should one care other than that being a curious result?
Can anybody guess?
AUDIENCE: Can you say that again?
PROFESSOR: That these are materials which in a
polycrystalline form with random grain orientation have
a linear thermal expansion coefficient this is
essentially zero?
AUDIENCE: [INAUDIBLE]?
PROFESSOR: Yes.
Exactly.
The materials out of which you build furnace linings and
tanks for melting glass or smelting steel have to be made
out of polycrystalline materials.
You can't have a monolithic single crystal that's large
enough to melt a significant amount of
glass in for example.
And being polycrystalline, when the material expands,
there is a great deal of stress between neighboring
grains because they're randomly oriented.
If the bulk microscopic thermal expansion coefficient
is zero, that is going to be refractory that is very, very
resistant to thermal shock.
When you heat it up, you don't find intergranular cracking.
So the materials that have essentially zero spatially
average thermal expansion coefficients are very
attractive for refractories.
You can pick up any issue of the Journal of
the American Ceramics--
not the journal, but the Bulletin of the American
Ceramic Society and you will find refractory companies
hustling materials like Cordierite, which is a
silicate material that on average has one positive
thermal expansion coefficient, one negative thermal expansion
coefficient.
And the volume change of the individual grains is
essentially zero.
So that makes a dandy material for refractories.
Cordierite has another interesting property that is a
really nice example of anisotropy
of a physical property.
If I ask you to rattle off some properties which you know
to be very anisotropic, one of them that you wouldn't think
of is color.
Your t-shirt is pink, so how can that be
a function of direction?
Well your t-shirt is not a single crystal.
And their single crystals which if you hold them up to
light and pass through a plain polarized beam of light the
crystal has a very different color for one direction of
polarization then another.
And these colors, interestingly, very often are
strikingly different.
They're crystals that are green in light polarized in
one direction and red for light polarized in the
opposite direction.
And there is in the old Icelandic sagas--
few people have heard of them.
The Icelandic sagas which were written in about the year 500
are one of the early forms of Western literature that is
really great, enduring world-class literature.
In the sagas, there's one episode where Thor is sleeping
and Olaf comes along and steals his sunstone.
And Thor is upset he takes his battle as and cut
Olaf's head in twain.
And people puzzled for hundreds of
years, what is sunstone.
This doesn't compute in today's age.
And finally a Norwegian archaeologist came up with a
hypothesis that in Iceland there are remarkably large and
remarkably transparent and perfect single crystals of
Cordierite.
And Cordierite is strikingly pleochroic.
That's color that is a function of direction.
And what this archaeologist theorized is that the Vikings
would take a sunstone when they went sailing on the North
Sea, which is notoriously damp and overcast and gray, and
your principal means of navigation was the sun.
But when the sky was overcast, you couldn't see the sun.
But the sun as it comes through the cloud gives you
light that's very strongly polarized.
So the theory is that the Vikings would have very
perfect crystals of transparent Cordierite, would
hold them up to the sky, and turn them around until the
Cordierite crystal lit up a bright, golden sunny yellow.
What better name for the sunstone.
You found the sun from the direction of polarization of
the light scattered from the cloud when you've got the
crystal oriented just so.
In a different orientation, it would look murkier, look sort
of a bluish purple for the other principal direction of
the birefringence.
So there is putting crystal anisotropy hundreds of years
ago to a very useful purpose to navigate on cloudy days.
Unfortunately, the Vikings used it to sail down the
British coast and sacked the next village.
So it shows you that even the most simple of science can be
put to application in war research.
Anyway, won't go there anymore.
So index of refraction and color can also be a strong
function of direction
The numbers that I put on the blackboard suggest that the
magnitude of thermal expansion coefficients are influenced
very strongly by the strength of the interatomic bonding.
And this shows up in a very, very striking way if you group
together classes of comparable materials.
And if you plot the value of the linear thermal expansion
coefficient in units of 10 to the minus 6 per degree C as a
function of the melting point in degrees Kelvin, the numbers
range from about 400 K for materials like sulfur up to
about 3,700 for tungsten.
And the variation is so beautiful it
could make you cry.
It goes as the inverse of the melting point.
And way up here are materials like for sulfur and lithium.
And the way out here are materials like tungsten.
And their relation is given by a equals 0.020 over the
melting point in degrees Kelvin.
So very strong correlation between melting point and
linear expansion for simple elements.
If you do the same thing for compounds, you find a similar
parabolic relation except it is offset.
For oxides and halides, many of them with the rock salt
structure, so they are isotropic.
And again, if you plot here the melting point in degrees
Kelvin, the numbers here range up to 60 times 10
to the minus 6.
And again, a variation with 1 over T, but
offset from the origin.
And the curve here is that a equals 0.038 over T, the
melting point T sub m in Kelvin, minus 7.0 times 10 to
the minus 6.
Finally, let me finish with some data for single crystals
that provide examples of and anisotropy.
These are all hexagonal materials.
And I will give you the--
not all hexagonal.
I take that back.
But they're all uniaxial crystals.
The value of the thermal expansion coefficient along C
and the thermal expansion coefficient that is parallel
to C, so these are the elements that would be in the
diagonal isothermal expansion tensor.
So for Al2O3, which is a hexagonal rhombohedral oxide,
the two expansion coefficients are 8.39 and 9.0 times 10 to
the minus 6, not terribly anisotropic.
But the structure of alumina is a close-packed arrangement
of oxygen with aluminum filling 2/3 of the available
octahedral holes.
So it's hexagonal only because of the sites that are filled
in the array.
For TiO2, which is [INAUDIBLE]
and therefore also uniaxial, 6.8 perpendicular to C; 8.3
parallel to C. For zirconium silicate, 3.7 and 6.2, almost
a 2:1 difference.
For the quartz form of silica, which is hexagonal, 14
perpendicular to C; 9 parallel to C.
For carbon, the graphite form, a layer structure--
you can almost guess how this is going to turn out.
It's going to be humongous perpendicular to the lawyers
and very small within the plane of these tightly bonded
hexagonal nets.
And that is indeed the case, 1 perpendicular to C; 27
parallel to C. So there is an anisotropy of a factor of 27.
A few more.
Aluminum titanate, perpendicular to C minus 2.6;
parallel to C 11.5.
A very well known example of anisotropy, extreme
anisotropy, where one principal coefficient is
positive and two are negative is the calcite forms calcium
carbonate, minus 6 and 25.
For zinc metal, a hexagonal close-packed metal, surprising
degree of anisotropy, 14 and 64.
And for tellurium, which is a chain structure like selenium,
27 perpendicular to C and minus 1.6 in a direction
parallel to C.
Even simple hexagonal close-packed metals can do
strange things at low temperatures.
For zinc, the expansion coefficients perpendicular to
C and parallel to C as a function of temperature go at
300 degrees C from 13 and 64.
At 150, the values have dropped to 8 and 65.
At 60 degrees K perpendicular to C, the thermal expansion is
negative and has not dropped terribly much of at all
parallel to C. So this is a hexagonal close-packed
structure lot.
A lot of action and strange things going on in the plane
at the close-packed layers, but not very much change in a
direction perpendicular to the layers.
Let me raise one more issue.
If we were to look at some of these strangely anisotropic
materials which had negative thermal expansion coefficients
in one direction and positive in another, what would the
representation quadric look like?
Suppose we looked at calcite, for example, which has a large
negative thermal expansion coefficient.
And if we look at the thermal expansion quadric when the
tensor was referred to the principal
axes, parallel to C--
well, the thermal expansion tensor aij along its principal
axes would be along A1 in a direction perpendicular to C,
it would be for the data given here, minus 6, zero, zero,
zero, minus 6, zero, zero, zero.
And A33 is 25.
So that's very, very anisotropic.
But what is this going to be?
This is going to be a quadric that has the shape of an
hyperboloid of two sheets.
So it's going to look like this.
And there'll be an asymptote like this and then
asymptote like this.
Along the asymptote, the radius of
the quadric is infinite.
So therefore along these directions the
strain is equal to zero.
Strain goes as 1 over the radius squared.
In this range of directions here, the radius is imaginary.
But remember that the strain is given by 1
over the radius squared.
So this says that the strain is negative, so the material
has contracted.
And finally, in this range of directions where the radius is
finite and positive, if the value of strain is 1 over
radius squared, this says that the maximum deformation is
along the direction of the C axis, and it's positive.
The material expands along C; the minimum radius of the
quadric corresponds to the maximum thermal expansion.
So let me close with a question.
Along the asymptote of the quadric, the strain is zero.
Does this mean that this direction in a crystal of
calcite does not move, no deformation at all when you
heat it up?
I see faint shakes of the head.
I don't know if that's just in awe of what's going on here or
an opinion on the question.
Do you think it'll be no strain?
AUDIENCE: It would be [INAUDIBLE].
PROFESSOR: Good answer.
Let's remember a property of the representation quadric.
The direction of what happens is normal to the surface of
the quadric.
The thing that is happening here is the displacement; ui
is given by epsilon ij times U sub j.
So when we look in this direction, the displacement,
which would be normal to the surface of the quadric, gives
us this as a displacement.
So along the asymptotes of the quadric, you can't say that
there's no deformation, but that the deformation
corresponds to pure rotation and not any
fractional change of length.
In these directions, the length changes, but it is a
decrease in length within the asymptotes.
For those directions, the strain is an extension.
And again, if you want to know the direction of the
displacement, you find the normal to the surface of the
quadric in the direction of interest, and that's the
direction in which the radius vector points.
All right.
That's I think a good place to quit.
I think I'll say a little bit more after the break about the
atomistic reason for increases in interionic separations,
which we can get some insight into from a simple model.
But before we disperse, I don't know if everybody got a
copy of problem set 13.
If anybody didn't, there's some extra copies up here.
But I will with great pleasure pass out to you problems set
number 14, which has served two purposes.
One is to have you show for yourself that the trace of a
second-rank tensor--
I called it second-order tensor--
second-rank tensor the sum of the diagonal elements is
invariant for a change of reference axes for a general
direction cosine scheme for the change of axes.
And then the other two questions are to give you some
practice in diagonalizing a second-rank property tensor
which is not in diagonal form.
And I asked you to do this in two ways.
One is by direct solution of the secular equation and
finding the eigenvectors.
And then the third example, which is for a completely
general tensor, to do this by the method of successive
approximations.
And you'll find after a couple of iterations you get fairly
close to convergence.
So I shall pass this around.
And again, I think there's something to be
learned from that.
Let's take our stretch then.
I'm sure that lecture involved a lot of stress and strain.
Ha, ha.
Let's resume in 10 minutes.
And we'll then move on to some additional higher rank tensor
properties, which will be very, very interesting.
Go for it.