PROFESSOR: Having looked at the quizzes in a preliminary
fashion, I come to the conclusion that some notes on
tensors would be useful for the remainder of the term.
So at great pain and personal sacrifice I will endeavor to
[INAUDIBLE]
All right, let me remind you of where we were a week ago--
before a slight unpleasantness intervened--
and we had just begun to look at the properties of a very
useful surface, the representation quadric.
And we said that to define it we would take the elements of
a second rank tensor, something of the form A11,
A12, A13, A21, A22, A23, A31, A32, A33.
And we'd use those elements as coefficients in a second rank,
a quadratic equation, of the form Aij Xi Xj equals 1.
OK, so the Xi and Xj are the coordinates in our
three-dimensional space.
And the set of coordinates that satisfy this equation--
setting the right hand side equal to a constant--
will define some surface in space.
And it's going to be a surface that is a quadratic form, so
it is going to be one of four different surfaces that can be
defined by such an equation.
One would be an ellipsoid, and in general this would be an
ellipsoid oriented in an arbitrary fashion with respect
to the reference axes.
And it would have a general shape, so the three principle
axes of the ellipsoid would be of different lengths.
Then we saw we could also get a hyperboloid of one sheet.
One sheet means one surface--
continuous surface.
This was the surface that had an hourglass like shape that
pinched down in the middle, and the cross-section, in
general, would be an ellipsoid.
And the asymptotes to this surface would define a
boundary between directions in which the radius was real,
which meant a positive value of the property, and a
direction in which the radius was imaginary, even though
that may boggle the mind.
And if you take an imaginary quantity and square it, you're
going to get a negative number.
So this surface would describe a property that was positive
in some directions and negative in magnitude over
another range of directions.
Then the third surface was a hyperboloid of two sheets.
And this was a surface that looked like two [INAUDIBLE]
with an elliptical cross-section that were nose
to nose, but not touching the origin.
And in this range of directions between the
asymptote to those two lobes the radius was imaginary, the
property negative and then there were a range of
directions that would intersect the surface of
either of these two sheets in there in those directions the
property would be positive.
Then the fourth one--
which is encountered only rarely, but as we pointed out
does exist in the case of thermal
expansion as a property--
and this would be an imaginary ellipsoid.
This would be an ellipsoid in which the distance from the
origin out to the surface was everywhere imaginary.
And this would be a property then that was negative in all
directions in space.
And that doesn't seem to be a physically realizable thing,
we pointed out that for the linear thermal expansion
coefficient there are a small number of very unusual
materials that when you heat them up contract uniformly in
all directions.
Very remarkable property, but that does indeed, thankfully,
provide me with an example of an imaginary ellipsoid quadric
that does represent, indeed, a realizable physical property.
OK, this surface, we said, had two rather remarkable
properties.
The first property was that if we looked at the radius from
the origin out to the surface of the quadric, a property of
the quadric is that the radius out in some direction--
that would be defined in terms of the
three direction cosines--
the radius has the property that it is given by 1 over the
square root of the value of the
property in that direction.
Or, alternatively, the value of the property in the
direction is 1 over the magnitude
of the radius squared.
OK, so this is what gives us the meaning of the imaginary
radii, the imaginary radii 1 squared would give you a
negative property.
But again, I point out, it's obvious here-- but we have to
keep it in mind--
that the value of the property and the magnitude of the
radius are related in a reciprocal fashion, not only
reciprocal, but reciprocal of the square.
So if this is the quadric A, a polar quad of the value of the
property A as a function of direction would have its
maximum value along the minimum principal axis and,
correspondingly, the minimum value of the property along
the maximum radius of the quadric.
So that is sort of counter-intuitive, you think
big radius, big value of the property, no, goes not only as
the reciprocal, but as the reciprocal of the square.
So the value of the property with direction is not a
quadratic form any longer, it's based on a quadratic
equation, but when you square the radius
it's no longer quadratic.
Looks as though we have the value of the property as a
function of direction, but looks as though we've lost
information about the direction of
the resulting vector.
The direction that we're talking about is always the
direction in which we're applying the generalized
force, the electric field, or the temperature gradient, or
the magnetic field.
But the direction of the thing that happens is defined by the
basic equation that gives us the tensor.
But as I demonstrated with you last time, that information is
in the quadric also, and this is the so-called
radius-normal property.
Which doesn't involve exactly what it sounds like, but what
it tells us to do is a way of determining the direction of
what happens is to go out in a particular direction, along
some radius that will intersect the surface of the
quadric at some point.
And then at the point where the directional vector
emerges, construct a vector that is normal to the surface.
And so if this then were the direction of an applied
electric field, the direction to the--
normal to the surface of the quadric where that direction
intersected--
the surface of the representation quadric--
this would be in the case of conductivity the direction in
which the current flows.
So everything that you care to know about the anisotropy of
the property that's described by a given tensor, and about
the direction of what happens--
the generalized displacement as you apply a vector--
is contained within the quadric.
But one caveat, this holds only if
the tensor is symmetric.
And I think this is where we finished up just before the
quiz, only if Aij equals Aji does the radius-normal
property work.
OK, any comments or questions on this?
OK, let me now turn to a practical question of
interpretation.
If you were indeed to measure a physical property as a
function of direction, and you get a tensor that is a general
tensor, A11, A12, A13.
A21, A22, A23.
A31, A32, A33.
All the numbers are non-zero.
You know that the diagonal values in the tensor are going
to give you the value of the property along X1.
Or, in general, the diagonal terms give you the value of
the property along Xi.
Aii gives you the property along Xi.
You have not the foggiest idea, if the quadric has a
general orientation, what the maximum and minimum values of
the property might be, and these might concern you.
Given the tensor that you've determined, what are the
largest values, what is the largest value of the property,
what is the smallest value of the property?
And if the crystal has any symmetry these directions
might be the direction of symmetry axes in the crystal.
So for many reasons you might want to know the maximum and
minimum value of the property.
And then for some applications for a real chunk of crystal
that you've examined relative to an arbitrary set of axes,
you might want to ask how should I orient that crystal
so that I can cut a rod from it that has, let's say, the
maximum or minimum thermal conductivity.
If you're using a piece of ceramic let's say, as a probe
into a furnace, you wouldn't want that probe to conduct
much heat, so you'd like the minimum thermal expansion
coefficient, for example.
If you were making a window out of a transparent single
crystal, it'd be a very small window, but you might want the
smallest thermal conductivity normal to the surface, and
therefore you might want to make your single crystal
window oriented in a particular direction.
So I hope I've convinced you with enough straw questions
that we can knock down easily that, yes, this would be an
interesting thing to know.
So how can we do this?
Let me show you some simple geometry that let's us set up
an equation for finding these directions right away.
And it'll be based on the following observation, that
when the direction of, let's say, the applied electric
field is a long one of the principal axes, then and only
then is the direction of the generalized displacement
exactly parallel to the radius vector.
We go off in any other direction other than a
principal axis, the generalized displacement is
not parallel to the radius vector out to the surface of
the quadric.
We look at the equation of the property A11 X1 plus A12 X2
plus A13 X3.
This is the X1 component of the generalized displacement.
Similarly, A21 X1 plus A22 X2 plus A23 X3 is going to be the
X2 component.
And A31 X1 plus A32 X2 plus A33 times X3, guess what?
That's going to be the X3 component of the displacement.
Now we said that these X's give us the direction of the
displacement as we let the coordinates of that point X1,
X2, X3 roam around the surface of the quadric.
When we land at a point on the surface of the quadric which
is where a principal axis emerges, then, and only then,
is the resulting vector parallel
to the applied vector.
And this means each component of the resulting vector has to
be proportional to a vector out to the point at that
location, X1, X2, X3.
So what I'm saying then is that when we are on a
principal axis the X1 component of what happens is
going to be proportional to X1.
And the X2 component of the vector that happens has to be
proportional to X2, and this is the radial vector out to
that point.
And, similarly, the X3 component of what happens has
to be parallel to the coordinate X3, the vector out
to the surface of the quadric along X3.
So let me make this an equation, now, by putting in a
constant for the unknown proportionality constant.
And what I'll do, just for arbitrary reasons, is call
that proportionality constant "lambda." So this, then, is a
set of equations that will give me, if I solve for X1,
X2, and X3, the coordinates of one of the three points that
sit out on the surface of the quadric at the location where
a principal axis emerges.
So let me rearrange these equations to a
form that I can solve.
And I'll make the equations homogeneous by bringing lambda
X1 over to the left hand side of the equation, and let me
point out that this is the same lambda
in all three equations.
There's a proportionality constant between the radial
vector out to the surface of the quadric and each component
of the vector that results.
So my set of equations will be A11 minus lambda X1 plus A12
times X2 plus A13 times X3, and that's now equal to zero
on the right.
Next equation would be A21 X1--
and notice I'm not combining A12 and A21 into numerically
the same constant, representing it by the same
quantity, I'm just leaving them be separate and
independent at this point.
And then the middle term here is A22 minus lambda times X2
plus A23 times X3 equals 0.
And the third equation, you can anticipate how that turns
out, it's A31 X1 plus A32 times X2 plus A33 minus
lambda, and that's equal to zero.
So this, then, is a set of linear equations and they're
called homogeneous equations.
And that has solved our problem for us, all we have to
do is solve for X1, X2, X3.
Except that if the nine coefficients Aij are all
arbitrary and lambda is a constant this set of equations
has only one solution, and that solution is X1 equals X2
equals X3 equals 0, and that indeed will wipe out every one
of these lines.
And that is not a very interesting solution.
The only case in which this is not the only solution is that
if the condition that the determinant of the
coefficients is equal to zero, then we can find a real set of
X1, X2, X3.
So the condition that a solution exists is that A11
minus lambda A12 A13, and A21 A22 minus lambda A23, A31 A32
A33 minus lambda, this determinant has
to be equal to zero.
All right, so we know how to expand the determinant, we'll
do a little number crunching, and I'm not going to write it
down explicitly, but we'll have a term A11 minus lambda
and this will be among other terms in the expansion A22
minus lambda A23 A32 A33 minus lambda and then there'll be
another determinant that involves this term A12 plus
its cofactor plus A13 times its cofactor.
Unfortunately none of these terms are zero in general, so
I'm going to have three terms here.
If I expand this term, I'm going to get something in A11
minus lambda A22 minus lambda times A33 minus lambda and a
bunch of other terms that I won't bother to write down
explicitly.
The only point I want to draw at this particular juncture is
to note that this is a third rank equation.
In lambda.
It's a cubic equation.
And what this means is that there are
going to be three roots.
And lets call them lambda 1, lambda 2, and lambda 3.
And that's the way it should be.
There are three principle axes, so we should have three
values of the constant lambda that we could put into this
equation that will let us solve explicitly for the
coordinates X1, X2, and X3, which sits on the surface of
the quadric where our principal axis pokes out.
Three different principal axes, so this should be three
solutions to the third rank equation that
we've defined here.
OK I think you've all seen this sort of problem before.
The lambdas are called characteristic values.
Or, on the basis that things sound more impressive in
German, these are called the eigenvalues,
meaning the same thing.
So using nothing more than the properties of the quadric and
some linear algebra we have stumbled headlong into an
eigenvalue problem, which is a well known sort of
mathematical problem associated with a large number
of physical situations and a lot to do with physical
properties in particular.
OK, now I have a question for you.
How we going to solve this silly thing?
I know--
well the answer is I poke it into my laptop, and out comes
the answer.
But that's not satisfying, we should know how to do this if
our batteries were dead and we needed the answer in a hurry.
Well, I know how to solve a second rank question.
If I have equation of the form ax squared plus bx plus c
equals 0, I know the solution to that.
It's etched indelibly in my memory and it says that x
equals minus b plus or minus the square root of b squared
minus 4ac all over 2a.
Impressed you, didn't I?
You know why I remember that?
I learned high school algebra from a throwback--
a teacher who was a throwback to the Elizabethan period.
You know how when people go to a remote island they suddenly
find an example of a species that
everyone thought was extinct?
Well, to secondary education [? Ms. Dourier ?]
was a species that had long gone extinct, but happened to
be preserved at the high school that I attended.
[? Ms. Dourier ?]
was a very tall woman, ramrod straight, with a pile of snow
white hair on the top of her head.
She invariably dressed in a long, black skirt, and a white
blouse that had puffy sleeves, and a collar with frilly
things on the top, and, invariably, a black ribbon
around the frilly lace collar top.
Her mode of enlightened education was to have all of
the students in the class have a notebook that was open.
And one hapless member of the class was called upon to solve
a problem in real time, and as the student recited we all
wrote down what the student was saying in our notebooks.
All the while [? Ms. Dourier, ?]
holding a ruler like a swagger stick, strode up and down the
aisles, and woe be unto the poor student who faltered
slightly, let alone get something wrong.
And that would bring a slap of the ruler down on the desk and
a sigh of disgust, and then, all right,
you take it, Audrey.
And then Audrey would begin to recite until she screwed up
and we would all copy in our notebooks.
That tyrant so terrorized and intimidated a bunch of kids in
a redneck, working class high school that we
really learned algebra.
So to this very day, as a reflex action, minus b plus or
minus the square root of b squared minus 4ac all over 2a.
But I don't have the foggiest idea how to solve a third
order equation until I look it up.
You probably never had to do it, so let me, for your
general education and amusement, pass around a sheet
that tells you how to solve a third rank equation.
OK the first step is to convert an equation in, let's
say, y squared plus y cubed plus py squared plus qy plus r
equals 0 to a so-called normal form where you get rid of one
of the terms.
So if you make a substitution of variables and let y be
replaced by x minus the coefficient p/3, then you get
an equation that has a cubic term, linear
term x, and a constant.
And the constants a and b are combinations of p and q and r
in the original equation.
And then the solutions, not well known to be sure, are
there's a first solution x is equal to a plus b, and then,
something messy, x2 and x3 are minus 1/2 of a plus b plus or
minus an imaginary term-- those are the
second and third roots--
i times the square root of 3 over 2a minus b where A and B
are complicated functions of a and b.
Capital A and capital B are functions of little
a and little b.
And then if the coefficients in the original equation are
real, then you get one real and two
conjugated imaginary roots.
If this combination of terms b squared and a squared are
greater than 0, it is b squared over 4 plus a cubed
over 27 equals 0.
If it's that, then you get three real roots, two
of which are equal.
In the most general case that we would be encountering here
is b squared over 4 plus a cubed over 27 is less than 0,
and then there are three real unequal roots.
Unfortunately, if b squared over 4 plus a cubed over 27 is
negative that term appears inside of a square root sign
in the solutions.
So the only case that we'd really be interested in
doesn't work for the solution.
But fortunately there's another solution and that's
given at the bottom of the page, and if you really wanted
to solve a third rank equation by hand, these solutions would
do it for you.
But I think you'd much prefer to have your computer solve
the equations for you, but let me show you a way of doing
this without any computer, without solving any equations.
And it is a very clever method of successive approximations
that's based on the properties of the quadric.
So let's suppose we have a tensor, and that tensor has
some set of coefficients Aij, all of which are non-zero.
And I'll assume although this will work for other quadratic
surfaces that the quadric has the shape of an ellipsoid.
All right let me now pick some direction at random, actually
I don't have to pick it at random, but I'll show you what
the shrewd first guess would be.
So let's say we picked this direction.
And let us find the direction if this is--
let's do it in terms of current and conductivity.
Let's let this be the first guess for the applied field.
That's clearly not going to be one of the principal axes, and
let this be the first resulting direction of
current flow J1.
So what I'm finding then is a J sub i in terms of an Aij
times an E sub J, and this is my first result, and this was
my first assumption.
Let me now let the direction of J be taken as the direction
of the applied field for a second guess.
And we could normalize to a unit vector, but we don't even
have to do that.
So let's simply say that my second guess for the electric
field that I hope will point along one principal axis is
E2, and E2 I'll take as identical--
let's say proportional--
to J1 from my original guess.
So this then would be E2, the second guess.
And if I find the direction to the normal--
to the quadric in that direction, this would be my
second result for the current flow J, I should have put this
in parentheses to indicate that these are not components
of E and J.
And you can see what's happening, look at where I am,
I started out here after just one iteration I have defined
this as the potential direction of one of the
principal axes.
So I'm going to find E1--
the direction of the field--
E1 that comes from sigma ij times Ji, I'll take my second
guess E2 either as identical to J1 and this is going to
give me then a new value for my second iteration, this
would be J2.
And this process is going to converge very, very rapidly on
the shortest principal axis.
As you can see in two shots I'm pretty close to being
parallel to this direction, which would be X2 in my
illustration.
Now if I wanted to-- if I wanted to see if I was close
to convergence--
this is going to be awkward because if I don't normalize
the magnitude of the resulting vector, J is going to get
larger and larger and larger.
So periodically I would want to normalize--
take the magnitude of J, divide that into the
components, and then I'd have a unit vector.
If I wrote a computer program to do this, I would do the
normalization each time to test convergence.
Now, this is my kind of procedure, because if I'm
doing this by hand and I screw up and I make a mistake and my
answer is thrown off a little bit, if I continue to iterate,
the thing will proceed to continue to converge to the
shortest principal axis.
So I can make a mistake and it will correct itself and come
back again, and that's my kind of solution.
Yeah, Jason?
AUDIENCE: It's going to give you the minimum value of the
property, right?
PROFESSOR: No.
It's going to give you the minimum principal axis, that's
going to be the maximum value of the property.
So that's one out of three, hey, that's not bad.
What do I do for the others?
Let me tell you without proof to find the
maximum principal axis.
And that would be the minimum value of the property.
What you would do is exactly the same procedure, and you
would operate on not the tensor, but on-- using as a
matrix of coefficients--
the matrix of the tensor inverted.
And I assume you know how to find the inverse of a matrix,
except that you don't have to even find the inverse, the
inverse is going to be a collection of functions of the
original elements divided by the determinant.
You don't have to worry about normalizing, all that's
important is the relative value of these coefficients.
Yes, sir?
AUDIENCE: If your tensor is symmetric, can't you say that
they'll just [INAUDIBLE] on to another?
PROFESSOR: That's what you're going to do.
But when you know just one, that won't work.
The other two are floating around somewhere and you don't
know in what direction.
So if you do this, then you have two out of the three and
now very shrewdly, as you point out, if we have two of
the principal axes and the tensor is symmetric, then we
can automatically get the direction of the third.
If it's not symmetric, then you have to use the set of
equations, and you have three principal axes as
eigenvectors, as they're called in eigenvalue problems.
And they do not have to be orthogonal, but you have the
components in a Cartesian coordinate system so you can
find the angle between those axes.
OK?
Yes, sir.
AUDIENCE: With respect to whether the matrix is
symmetrical or is not symmetric,
the quadric is always--
has three principal axes at right angles right?
PROFESSOR: Only if the tensor is symmetric.
Only if the tensor is symmetric.
And let me put your question aside for about five minutes,
and I want to take an overview of what we've learned about
the geometry of the quadric and the symmetry restrictions
that we can impose on the tensor in a formal fashion,
and see how they compare and how one allows an
interpretation of the other.
So I'm going to answer your question, I'd like to wait for
about three minutes.
Other questions?
OK, again this is all very symbolic, I'm not giving you
an explicit answer, I can't do it.
But what I can do you--
do for you is, ha ha, do to you.
I almost slipped and said that, what I can't do to you,
is give you a problem that asks you to solve for the
principal axes, for example of a property tensor.
I will be merciful and perhaps not have all nine coefficients
non-zero, I will have one of them zero, or
maybe two of them zero.
So again to try this it is very instructive and it will
cement what the individual steps that I performed here
actually involve.
Now, a question that I thought you were going to ask about
principal axes is that in order to do what I did here I
am using the radius-normal property, and the
radius-normal property only works for a symmetric tensor.
So sounds like I'm swindling you, except that if I use this
procedure, the radius-normal won't actually be identically
parallel to the generalized displacement.
But it's not going to be terribly different from it,
and I'm going to get something that again will converge
towards the shortest principal axis.
And once I'm close to the principal axis it is true--
symmetric or not--
that the only three directions before which the generalized
displacement is parallel to the radius vector are the
principal axes.
But anyway this is a dicey situation if the tensor is not
symmetric, and it probably comes as a great relief to
learn that there's really only one property tensor of second
rank that's known for sure to be non-symmetric.
So in most cases--
99 out of 100, if not more-- we will be dealing with
property tensors that are symmetric, so we could use
this procedure.
So again this property converges remarkably well, and
as I say it has the admirable quality of being
self-correcting if you make a mistake, if you're grinding
this out by hand.
But a computer program that you write that just simply
takes the direction of J, normalizes it to a unit
vector, applies that as the generalized force, finds a
quote "J" as a second iteration, normalizes that,
and then you can check to whatever degree of convergence
you wish to have in your solution.
And it's a simple matter to set up a program to do this.
All right, I think we still have some time, so let me now
put everything together.
And look at what we've seen of the geometric properties of
the quadric, and the tensor arrays that we found for
different symmetries.
For triclinic crystals, for which the recommended
procedure is to leave by the nearest exit and work on
something different instead, you would have nine elements
altogether that are necessary to define the property.
I'll discuss this in terms of tensors that
gives you an ellipsoid.
The argument is not different for hyperboloids of one or two
sheets, and we can't draw imaginary ellipsoids, but an
ellipsoid is a much easier thing to draw.
OK, triclinic symmetries, either one or one bar, nine
elements in the tensor.
No constraints whatsoever on the shape of the quadric or on
its orientation relative to our coordinate system.
So, let's total up now looking at the quadric how many
degrees of freedom there are in this situation.
We have three principal axes, so those are
three degrees of freedom.
The orientation of the quadric is arbitrary, so there are
three orientational degrees of freedom.
And that comes out to six.
Supposedly nine degrees of freedom, what
are the other three?
Again, if this is a general tensor, which is
non-symmetric, the eigenvectors--
the principal axes that you have to choose to squeeze this
thing into a diagonal form--
become non-orthogonal.
OK?
So you have another three degrees of freedom that
specify the mutual directions of the eigenvectors--
the angles between them.
OK?
The directions in which you have to pick your coordinate
system, X1, X2, and X3, that force this thing into a
diagonal form is going to involve a coordinate system in
which these three angles are not 90 degrees and are fixed
if you're going to squeeze this thing
into a diagonal form.
And that's it.
So we add these three interaxial angles in, they're
a total of nine degrees of freedom for a general non
symmetric tensor.
To convince you that these interaxial angles for the
eigenvectors really are variables, let me tell you
something that we may prove later as a recreational
exercise, or I may give it to you on a problem set, it's not
difficult to prove.
And that is the result that a symmetric tensor remains
symmetric for any arbitrary transformation of axes.
That is from one Cartesian coordinate system to another.
Now a very salutary effect of this is that the tensor has
nine elements.
And if the tensor is symmetric--
if you want to transform that tensor--
you only have to do the off-diagonal terms, because
whatever it turns into when you change the axes, these
off-diagonal terms are going to be equal to the
off-diagonal terms in the new tensor.
OK?
So this means that only six--
if the tensor was symmetric in one coordinate system, only
six elements have to be transformed.
Actually, it's better than that, you don't have to do
six, you only have to do five.
And this is the last diagonal term that can be obtained from
the trace of the tensor.
And that would be the trace of the tensor T prime after
transformation because A11 prime plus A22 prime plus A33
prime has to be equal to the original trace T, which was
A11 plus A22 plus A33 in the original system.
So if the tensor is symmetric, you really have to crank
through a transformation for only five of the nine terms.
Now what is the relevance of this to what I just said?
If we have the tensor diagonalized, to a new form
A11, 0, 0, 0, A22 prime, 0, 0, 0, A33 prime, that is a
special case of a symmetric tensor.
So if you decide to go from the coordinate system that
produced this diagonalized form to some other coordinate
system, the new tensor that you're going to get is going
to be symmetric.
But the thing was not originally symmetric, so how
can that be?
The answer is you can put it in diagonal form only in a
non-Cartesian coordinate system.
And, therefore, that is evidence that
the reference axes--
the principal axes-- cannot be orthogonal.
Otherwise you could take the diagonalized tensor and crank
it back to some arbitrary Cartesian coordinate, and
suddenly it would go to a symmetric tensor when it was
not symmetric to begin with.
OK?
QED.
So you can diagonalize a general tensor only if you
take the axes along the eigenvectors, which cannot be
orthogonal.
OK, I saw you raise your hand, I wasn't ignoring you.
That was it?
Good.
AUDIENCE: I wanted to make sure that Cartesian
[INAUDIBLE]
PROFESSOR: It's always nice to see the class one step ahead
of what you're doing.
All right, let's look at a couple of more crystal systems
in the form of tensors in those systems and show that
that in fact does correspond to the geometric constraints
on the quadric as well.
For monoclinic crystals--
and this is symmetry 2, symmetry M, and symmetry 2/M.
The form of the tensor when we took the coordinate system
along symmetry elements was A11, A12, 0--
terms with a single three vanished--
A21, A22, 0, 0, 0, A33.
So this was the case where X3 was along the two-fold axis
and or perpendicular to the mirror point.
OK, number of independent variables to describe this
property is five.
And if we look at this in terms of a constraint and
shape in orientation of an ellipsoidal quadric, says that
one of the principal axes, if the quadric is to remain
invariant, has to be along the two-fold axis and or
perpendicular to a plane that contains the mirror plane, if
a mirror plane is also present.
So what are the degrees of freedom here?
This has to be along one reference axis, x3.
And, indeed, this form of the tensor occurred only when the
two-fold axes were parallel, 2 X3.
And this ellipsoid then was left with
one degree of freedom.
If this is the direction of the reference axis X2, we have
a degree of freedom in--
shouldn't have called these X1 and X2, these need not be the
coordinate systems.
Looking down on the quadric along the two-fold axis, the
section of the quadric perpendicular to the two-fold
axis can have a general shape, and it can have one degree of
freedom in the orientation of the principal axis relative to
the coordinate system.
So we have three parameters to specify the principal axes,
since this is a general quadric.
We have one degree of freedom in orientation, and that adds
up to four, but we said five.
So what's going on here?
Again this is a question of where the eigenvectors point.
If this term is not equal to this term, in order to force
the tensor into a diagonal form, you have to pick
eigenvectors in the X1, X2 plane, which will force the
tensor into a symmetric form.
That is, to make this term equal to this term.
It's only a pair of axes involved, so there's only one
angle between these two
eigenvectors, which is a variable.
So for a nonsymmetric tensor plus one angle between the
eigenvectors.
So that comes out, a-ha, five.
And again the way to see that is I will never get the tensor
into a diagonal form making all the off-diagonal terms
zero when I'm starting with a tensor which is not symmetric,
and I know that a symmetric tensor, even in the special
case where the off-diagonal terms are zero, is going to go
to a symmetric tensor in another coordinate system.
So I know that there has to be an
additional degree of freedom.
OK, the rest come very, very easily, so let me take just a
couple of minutes to finish up this stage of the discussion,
and we will shortly go on to something different.
The next step up in symmetry is orthorhombic.
And if we pick arc-- and this could be 222, 2MM or 2/M 2/M
2/M. We found that when we took X1, X2, and X3 along
two-fold axes that the tensor took a diagonal form A11, 0,
0, 0, A22, 0, 0, 0, A33.
That says that the quadric if it's an ellipsoid--
or any other quadratic form--
has all three of its principal axes along the reference
system X1, X2, X3.
So the only degree of freedoms here-- and things are again
starting to set up like supercooled water--
the only degrees of freedom are the three principal axes.
And that's it, the orientation is fixed.
And, moreover, the tensor is diagonal in this reference
system-- it's going to be symmetric in this reference
system, it's going to remain symmetric in any other
coordinate system.
So the tensor is always symmetric, the eigenvectors
then are always orthogonal, even if the coordinate system
stays Cartesian.
And there are three variables, three degrees of freedom.
And finally two remaining cases can
be disposed of quickly.
We saw that for an arbitrary theta that
is a symmetry element--
so let me not say a symmetry--
for an arbitrary rotation about X3.
This was the ingenious little proof in which we transform
the tensor by an arbitrary rotation theta about X3.
We found that the form of the tensor was A11, A12, 0, and
now I get to use tensor notation by calling this term
A12 again and not the proper indices A21, so I'll put
quotation marks.
"A22" was equal to "A11."
So going to an arbitrary rotation theta this must be
the form of the tensor we discovered, and this will
cover then fourfold axes, three-fold
axes, and sixfold axes.
These two elements are constrained to have the two
values, these two are constrained to have the same
values, so the tensor is always symmetric.
Symmetric relative to these axes, symmetric in all
coordinates, and any choice of coordinate system, then.
And that means that if the quadric has this property,
it's going to have one principal axis along X3, it's
going to be circular in the plane that contains X1 and X2,
so there are only, count them up, there are only three
degrees of freedom.
In terms of tensor elements, these degrees of
freedom are A1, A12--
in the diagonalized form-- and A33.
And there are three degrees of freedom in terms of the
independent elements as well.
So this is for anything that involves rotational symmetry
other than 180 degrees.
And, finally, for cubic crystals, if a symmetry
operation other than 180 degrees gives us this form of
the tensor, two off-diagonal terms are zero, two diagonal
terms equal.
If we impose this along all three reference axes, then the
form of the tensor is going to go to a diagonal form with all
diagonal terms equal.
The quadric that corresponds to that form is a sphere that
says that there's one quantity, one degree of
freedom in the form of the tensor.
There are zero degrees of
orientational degree of freedom.
Thus the spherical quadrics stay spherical for any
orientation.
So again, one independent element in the tensor, just
one degree of freedom in the quadric--
namely its radius--
and, again, the formal constraints that we obtained
by symmetry transformations agree with those that are the
same degrees of freedom when you want the quadric to
coincide with the axes.
Yes?
AUDIENCE: Speaking of the circle and the exponents to
play only if A12 equals zero?
PROFESSOR: These two were equal, but non-zero.
These two are equal, this one was independent.
And that would be for the specific case
of a fourfold axis.
Now if we put a fourfold axis along X1, that's going to
involve these two being equal and the
non-zero elements become--
let's see--
along X2 it would be A13 and A23.
No.
OK, we put the four-fold axis along X2, then this one would
be A22, A11, and A13 would have to be equal, and 21 and
12 would have to be zero, 23 and 32 would have to be equal.
I think that's the way it'll go, and these
three have to be zero.
So when you impose all three constraints, here we've had
these two equal, now we have these two equal, put the
four-fold access in the next--
in the remaining direction, and again it has to be
diagonal, and pairwise all of the off-diagonal terms will be
required to be zero.
AUDIENCE: I mean, when you diagonalize your matrix the
diagonal term are going to be all different in this case if
A12 is different from zero.
PROFESSOR: This is not a final result, this is an
intermediate step.
So we impose this constraint, plus this constraint, and
that's actually going to give us all the qualities we're
going to get, but we'd want to do the same thing for a
four-fold axis along X3.
You put them all together, the fact that these are zero will
wipe out all of the off-diagonal
terms, and these things--
for another choice of axes, these two
would have to be zero.
And for another choice of the orientation of the four-fold
axis, this and this would have to be equal.
And, finally, this and this would have to be equal.
So this is what we found, in fact, when we cranked through
the symmetry transformations, and this is what we would get
when we said that the quadric has to have a shape that
conforms to cubic symmetric.
AUDIENCE: My question was in the case of the arbitrary
relation of those three.
All three degrees of freedom, are they, in fact, the three
principal axes?
PROFESSOR: For this case here?
AUDIENCE: Yes.
PROFESSOR: For a symmetric tensor, they are
two principal axes.
OK?
If the tensor is nonsymmetric, then these two don't have to
be equal any longer, and they give you the
extra degree of freedom.
OK?
AUDIENCE: What are the three degrees of
freedom in this case?
PROFESSOR: I just want you to know I
appreciate your question.
AUDIENCE: May I make a suggestion?
PROFESSOR: Yeah.
AUDIENCE: I think you need two degrees of freedom to specify
the first one, and since it's symmetric and since it's
orthogonal you'd only need one more to find the second one
and then another to divide the third.
So that's three degrees of freedom to find three
principal axes in an orthogonal system.
PROFESSOR: Not sure I like that, because this has to be
the shape of the quadric for an arbitrary rotation angle.
AUDIENCE: Does that angle between X1 and X2-- does it
have to be 90 degrees if it's not connected?
PROFESSOR: That's correct, that's correct.
OK, let's not tie up the whole audience, let me--
let's talk about this and I'll think about it too.
But you make a very, very good point, but let's not settle it
in real time.
We'll throw everybody out, we'll close the door, and you
can come back and see who won the scuffle, OK?
OK, so let's take our break, I think you're more
than ready for it.