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PROFESSOR: Having looked at the
quizzes in a preliminary
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fashion, I come to the
conclusion that some notes on
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tensors would be useful for
the remainder of the term.
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So at great pain and personal
sacrifice I will endeavor to
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[INAUDIBLE]
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All right, let me remind you of
where we were a week ago--
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before a slight unpleasantness
intervened--
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and we had just begun to look
at the properties of a very
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useful surface, the
representation quadric.
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And we said that to define it we
would take the elements of
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a second rank tensor, something
of the form A11,
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A12, A13, A21, A22, A23,
A31, A32, A33.
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And we'd use those elements as
coefficients in a second rank,
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a quadratic equation, of the
form Aij Xi Xj equals 1.
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OK, so the Xi and Xj are
the coordinates in our
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three-dimensional space.
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And the set of coordinates that
satisfy this equation--
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setting the right hand side
equal to a constant--
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will define some surface
in space.
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And it's going to be a surface
that is a quadratic form, so
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it is going to be one of four
different surfaces that can be
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defined by such an equation.
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One would be an ellipsoid, and
in general this would be an
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ellipsoid oriented in an
arbitrary fashion with respect
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to the reference axes.
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And it would have a general
shape, so the three principle
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axes of the ellipsoid would
be of different lengths.
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Then we saw we could also get
a hyperboloid of one sheet.
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One sheet means one surface--
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continuous surface.
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This was the surface that had
an hourglass like shape that
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pinched down in the middle,
and the cross-section, in
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general, would be
an ellipsoid.
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And the asymptotes to this
surface would define a
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boundary between directions in
which the radius was real,
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which meant a positive value
of the property, and a
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direction in which the radius
was imaginary, even though
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that may boggle the mind.
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And if you take an imaginary
quantity and square it, you're
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going to get a negative
number.
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So this surface would describe
a property that was positive
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in some directions and negative
in magnitude over
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another range of directions.
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Then the third surface was a
hyperboloid of two sheets.
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And this was a surface that
looked like two [INAUDIBLE]
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with an elliptical cross-section
that were nose
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to nose, but not touching
the origin.
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And in this range of directions
between the
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asymptote to those two lobes the
radius was imaginary, the
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property negative and then
there were a range of
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directions that would intersect
the surface of
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either of these two sheets in
there in those directions the
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property would be positive.
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Then the fourth one--
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which is encountered only
rarely, but as we pointed out
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does exist in the
case of thermal
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expansion as a property--
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and this would be an imaginary
ellipsoid.
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This would be an ellipsoid in
which the distance from the
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origin out to the surface was
everywhere imaginary.
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And this would be a property
then that was negative in all
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directions in space.
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And that doesn't seem to be a
physically realizable thing,
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we pointed out that for the
linear thermal expansion
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coefficient there are a small
number of very unusual
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materials that when you heat
them up contract uniformly in
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all directions.
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Very remarkable property, but
that does indeed, thankfully,
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provide me with an example of an
imaginary ellipsoid quadric
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that does represent, indeed, a
realizable physical property.
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OK, this surface, we said,
had two rather remarkable
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properties.
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The first property was that if
we looked at the radius from
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the origin out to the surface of
the quadric, a property of
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the quadric is that the radius
out in some direction--
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that would be defined
in terms of the
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three direction cosines--
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the radius has the property that
it is given by 1 over the
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square root of the value of the
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property in that direction.
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Or, alternatively, the value
of the property in the
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direction is 1 over
the magnitude
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of the radius squared.
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OK, so this is what gives us the
meaning of the imaginary
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radii, the imaginary radii 1
squared would give you a
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negative property.
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But again, I point out, it's
obvious here-- but we have to
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keep it in mind--
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that the value of the property
and the magnitude of the
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radius are related in a
reciprocal fashion, not only
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reciprocal, but reciprocal
of the square.
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So if this is the quadric A, a
polar quad of the value of the
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property A as a function of
direction would have its
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maximum value along the minimum
principal axis and,
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correspondingly, the minimum
value of the property along
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the maximum radius
of the quadric.
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So that is sort of
counter-intuitive, you think
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big radius, big value of the
property, no, goes not only as
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the reciprocal, but as the
reciprocal of the square.
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So the value of the property
with direction is not a
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quadratic form any longer,
it's based on a quadratic
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equation, but when you
square the radius
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it's no longer quadratic.
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Looks as though we have the
value of the property as a
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function of direction, but looks
as though we've lost
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information about
the direction of
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the resulting vector.
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The direction that we're talking
about is always the
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direction in which we're
applying the generalized
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force, the electric field, or
the temperature gradient, or
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the magnetic field.
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But the direction of the thing
that happens is defined by the
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basic equation that gives
us the tensor.
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But as I demonstrated with you
last time, that information is
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in the quadric also, and
this is the so-called
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radius-normal property.
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Which doesn't involve exactly
what it sounds like, but what
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it tells us to do is a way of
determining the direction of
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what happens is to go out in a
particular direction, along
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some radius that will intersect
the surface of the
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quadric at some point.
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And then at the point where
the directional vector
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emerges, construct a vector that
is normal to the surface.
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And so if this then were the
direction of an applied
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electric field, the
direction to the--
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normal to the surface of the
quadric where that direction
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intersected--
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the surface of the
representation quadric--
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this would be in the case of
conductivity the direction in
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which the current flows.
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So everything that you care to
know about the anisotropy of
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the property that's described
by a given tensor, and about
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the direction of
what happens--
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the generalized displacement
as you apply a vector--
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is contained within
the quadric.
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But one caveat, this
holds only if
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the tensor is symmetric.
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And I think this is where we
finished up just before the
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quiz, only if Aij equals Aji
does the radius-normal
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property work.
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OK, any comments or
questions on this?
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OK, let me now turn to a
practical question of
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interpretation.
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If you were indeed to measure
a physical property as a
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function of direction, and you
get a tensor that is a general
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tensor, A11, A12, A13.
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A21, A22, A23.
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A31, A32, A33.
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All the numbers are non-zero.
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You know that the diagonal
values in the tensor are going
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to give you the value of
the property along X1.
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Or, in general, the diagonal
terms give you the value of
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the property along Xi.
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Aii gives you the property
along Xi.
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You have not the foggiest idea,
if the quadric has a
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general orientation, what the
maximum and minimum values of
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the property might be, and
these might concern you.
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Given the tensor that you've
determined, what are the
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largest values, what is the
largest value of the property,
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what is the smallest value
of the property?
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And if the crystal has any
symmetry these directions
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might be the direction of
symmetry axes in the crystal.
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So for many reasons you might
want to know the maximum and
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minimum value of the property.
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And then for some applications
for a real chunk of crystal
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that you've examined relative
to an arbitrary set of axes,
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you might want to ask how should
I orient that crystal
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so that I can cut a rod from
it that has, let's say, the
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maximum or minimum thermal
conductivity.
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If you're using a piece of
ceramic let's say, as a probe
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into a furnace, you wouldn't
want that probe to conduct
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much heat, so you'd like the
minimum thermal expansion
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coefficient, for example.
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If you were making a window out
of a transparent single
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crystal, it'd be a very small
window, but you might want the
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smallest thermal conductivity
normal to the surface, and
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therefore you might want to
make your single crystal
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window oriented in a particular
direction.
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So I hope I've convinced you
with enough straw questions
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that we can knock down easily
that, yes, this would be an
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interesting thing to know.
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So how can we do this?
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Let me show you some simple
geometry that let's us set up
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an equation for finding these
directions right away.
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And it'll be based on the
following observation, that
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when the direction of, let's
say, the applied electric
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00:12:38,610 --> 00:12:44,040
field is a long one of the
principal axes, then and only
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00:12:44,040 --> 00:12:50,750
then is the direction of the
generalized displacement
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00:12:50,750 --> 00:12:55,020
exactly parallel to
the radius vector.
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00:12:55,020 --> 00:12:57,390
We go off in any other direction
other than a
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principal axis, the generalized
displacement is
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00:13:00,090 --> 00:13:03,190
not parallel to the radius
vector out to the surface of
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00:13:03,190 --> 00:13:04,440
the quadric.
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00:13:04,440 --> 00:13:09,970
212
00:13:09,970 --> 00:13:22,480
We look at the equation of the
property A11 X1 plus A12 X2
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00:13:22,480 --> 00:13:26,030
plus A13 X3.
214
00:13:26,030 --> 00:13:33,505
This is the X1 component of the
generalized displacement.
215
00:13:33,505 --> 00:13:37,840
216
00:13:37,840 --> 00:13:46,860
Similarly, A21 X1 plus A22 X2
plus A23 X3 is going to be the
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00:13:46,860 --> 00:13:48,110
X2 component.
218
00:13:48,110 --> 00:13:51,000
219
00:13:51,000 --> 00:13:59,710
And A31 X1 plus A32 X2 plus
A33 times X3, guess what?
220
00:13:59,710 --> 00:14:04,140
That's going to be the X3
component of the displacement.
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00:14:04,140 --> 00:14:09,880
Now we said that these X's give
us the direction of the
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00:14:09,880 --> 00:14:14,180
displacement as we let the
coordinates of that point X1,
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00:14:14,180 --> 00:14:18,370
X2, X3 roam around the surface
of the quadric.
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00:14:18,370 --> 00:14:22,500
When we land at a point on the
surface of the quadric which
225
00:14:22,500 --> 00:14:27,490
is where a principal axis
emerges, then, and only then,
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00:14:27,490 --> 00:14:29,830
is the resulting
vector parallel
227
00:14:29,830 --> 00:14:31,570
to the applied vector.
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00:14:31,570 --> 00:14:40,140
And this means each component of
the resulting vector has to
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00:14:40,140 --> 00:14:44,490
be proportional to a vector
out to the point at that
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00:14:44,490 --> 00:14:47,260
location, X1, X2, X3.
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00:14:47,260 --> 00:14:50,360
So what I'm saying then is
that when we are on a
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00:14:50,360 --> 00:14:55,130
principal axis the X1 component
of what happens is
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00:14:55,130 --> 00:14:59,490
going to be proportional
to X1.
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00:14:59,490 --> 00:15:03,320
And the X2 component of the
vector that happens has to be
235
00:15:03,320 --> 00:15:06,420
proportional to X2, and this
is the radial vector out to
236
00:15:06,420 --> 00:15:07,230
that point.
237
00:15:07,230 --> 00:15:10,940
And, similarly, the X3 component
of what happens has
238
00:15:10,940 --> 00:15:16,470
to be parallel to the coordinate
X3, the vector out
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00:15:16,470 --> 00:15:19,840
to the surface of the
quadric along X3.
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00:15:19,840 --> 00:15:24,670
So let me make this an equation,
now, by putting in a
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00:15:24,670 --> 00:15:28,430
constant for the unknown
proportionality constant.
242
00:15:28,430 --> 00:15:32,930
And what I'll do, just for
arbitrary reasons, is call
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00:15:32,930 --> 00:15:37,850
that proportionality constant
"lambda." So this, then, is a
244
00:15:37,850 --> 00:15:43,245
set of equations that will give
me, if I solve for X1,
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00:15:43,245 --> 00:15:49,240
X2, and X3, the coordinates of
one of the three points that
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00:15:49,240 --> 00:15:53,140
sit out on the surface of the
quadric at the location where
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a principal axis emerges.
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00:15:56,840 --> 00:15:59,970
So let me rearrange these
equations to a
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00:15:59,970 --> 00:16:02,410
form that I can solve.
250
00:16:02,410 --> 00:16:07,400
And I'll make the equations
homogeneous by bringing lambda
251
00:16:07,400 --> 00:16:12,510
X1 over to the left hand side
of the equation, and let me
252
00:16:12,510 --> 00:16:14,740
point out that this
is the same lambda
253
00:16:14,740 --> 00:16:16,730
in all three equations.
254
00:16:16,730 --> 00:16:20,340
There's a proportionality
constant between the radial
255
00:16:20,340 --> 00:16:28,140
vector out to the surface of the
quadric and each component
256
00:16:28,140 --> 00:16:30,210
of the vector that results.
257
00:16:30,210 --> 00:16:38,020
So my set of equations will be
A11 minus lambda X1 plus A12
258
00:16:38,020 --> 00:16:45,850
times X2 plus A13 times X3, and
that's now equal to zero
259
00:16:45,850 --> 00:16:46,800
on the right.
260
00:16:46,800 --> 00:16:51,670
Next equation would
be A21 X1--
261
00:16:51,670 --> 00:16:56,590
and notice I'm not combining A12
and A21 into numerically
262
00:16:56,590 --> 00:16:59,520
the same constant, representing
it by the same
263
00:16:59,520 --> 00:17:01,990
quantity, I'm just leaving
them be separate and
264
00:17:01,990 --> 00:17:03,830
independent at this point.
265
00:17:03,830 --> 00:17:07,900
And then the middle term here
is A22 minus lambda times X2
266
00:17:07,900 --> 00:17:12,599
plus A23 times X3 equals 0.
267
00:17:12,599 --> 00:17:15,060
And the third equation, you can
anticipate how that turns
268
00:17:15,060 --> 00:17:24,940
out, it's A31 X1 plus A32
times X2 plus A33 minus
269
00:17:24,940 --> 00:17:31,560
lambda, and that's
equal to zero.
270
00:17:31,560 --> 00:17:38,220
So this, then, is a set of
linear equations and they're
271
00:17:38,220 --> 00:17:39,710
called homogeneous equations.
272
00:17:39,710 --> 00:17:47,270
273
00:17:47,270 --> 00:17:50,980
And that has solved our problem
for us, all we have to
274
00:17:50,980 --> 00:17:54,210
do is solve for X1, X2, X3.
275
00:17:54,210 --> 00:18:00,140
Except that if the nine
coefficients Aij are all
276
00:18:00,140 --> 00:18:05,360
arbitrary and lambda is a
constant this set of equations
277
00:18:05,360 --> 00:18:13,850
has only one solution, and that
solution is X1 equals X2
278
00:18:13,850 --> 00:18:18,530
equals X3 equals 0, and that
indeed will wipe out every one
279
00:18:18,530 --> 00:18:19,630
of these lines.
280
00:18:19,630 --> 00:18:21,820
And that is not a very
interesting solution.
281
00:18:21,820 --> 00:18:24,380
282
00:18:24,380 --> 00:18:31,350
The only case in which this is
not the only solution is that
283
00:18:31,350 --> 00:18:34,470
if the condition that the
determinant of the
284
00:18:34,470 --> 00:18:40,710
coefficients is equal to zero,
then we can find a real set of
285
00:18:40,710 --> 00:18:42,220
X1, X2, X3.
286
00:18:42,220 --> 00:18:46,070
So the condition that a solution
exists is that A11
287
00:18:46,070 --> 00:19:01,046
minus lambda A12 A13, and A21
A22 minus lambda A23, A31 A32
288
00:19:01,046 --> 00:19:05,590
A33 minus lambda, this
determinant has
289
00:19:05,590 --> 00:19:06,840
to be equal to zero.
290
00:19:06,840 --> 00:19:12,090
291
00:19:12,090 --> 00:19:14,540
All right, so we know how to
expand the determinant, we'll
292
00:19:14,540 --> 00:19:17,750
do a little number crunching,
and I'm not going to write it
293
00:19:17,750 --> 00:19:23,450
down explicitly, but we'll have
a term A11 minus lambda
294
00:19:23,450 --> 00:19:28,690
and this will be among other
terms in the expansion A22
295
00:19:28,690 --> 00:19:35,870
minus lambda A23 A32 A33 minus
lambda and then there'll be
296
00:19:35,870 --> 00:19:39,960
another determinant that
involves this term A12 plus
297
00:19:39,960 --> 00:19:43,980
its cofactor plus A13
times its cofactor.
298
00:19:43,980 --> 00:19:46,560
Unfortunately none of these
terms are zero in general, so
299
00:19:46,560 --> 00:19:48,900
I'm going to have three
terms here.
300
00:19:48,900 --> 00:19:52,830
If I expand this term, I'm going
to get something in A11
301
00:19:52,830 --> 00:20:01,590
minus lambda A22 minus lambda
times A33 minus lambda and a
302
00:20:01,590 --> 00:20:04,640
bunch of other terms that I
won't bother to write down
303
00:20:04,640 --> 00:20:05,890
explicitly.
304
00:20:05,890 --> 00:20:08,070
305
00:20:08,070 --> 00:20:14,360
The only point I want to draw at
this particular juncture is
306
00:20:14,360 --> 00:20:19,300
to note that this is a
third rank equation.
307
00:20:19,300 --> 00:20:23,160
308
00:20:23,160 --> 00:20:24,410
In lambda.
309
00:20:24,410 --> 00:20:26,940
310
00:20:26,940 --> 00:20:28,190
It's a cubic equation.
311
00:20:28,190 --> 00:20:31,380
312
00:20:31,380 --> 00:20:33,880
And what this means
is that there are
313
00:20:33,880 --> 00:20:35,150
going to be three roots.
314
00:20:35,150 --> 00:20:40,080
315
00:20:40,080 --> 00:20:44,000
And lets call them lambda 1,
lambda 2, and lambda 3.
316
00:20:44,000 --> 00:20:48,180
317
00:20:48,180 --> 00:20:49,810
And that's the way
it should be.
318
00:20:49,810 --> 00:20:55,010
There are three principle axes,
so we should have three
319
00:20:55,010 --> 00:20:59,050
values of the constant lambda
that we could put into this
320
00:20:59,050 --> 00:21:02,720
equation that will let us
solve explicitly for the
321
00:21:02,720 --> 00:21:06,670
coordinates X1, X2, and X3,
which sits on the surface of
322
00:21:06,670 --> 00:21:09,920
the quadric where our principal
axis pokes out.
323
00:21:09,920 --> 00:21:13,000
Three different principal axes,
so this should be three
324
00:21:13,000 --> 00:21:18,770
solutions to the third
rank equation that
325
00:21:18,770 --> 00:21:20,020
we've defined here.
326
00:21:20,020 --> 00:21:23,090
327
00:21:23,090 --> 00:21:27,120
OK I think you've all seen this
sort of problem before.
328
00:21:27,120 --> 00:21:30,435
The lambdas are called
characteristic values.
329
00:21:30,435 --> 00:21:38,930
330
00:21:38,930 --> 00:21:42,640
Or, on the basis that things
sound more impressive in
331
00:21:42,640 --> 00:21:52,770
German, these are called
the eigenvalues,
332
00:21:52,770 --> 00:21:55,210
meaning the same thing.
333
00:21:55,210 --> 00:21:59,450
So using nothing more than the
properties of the quadric and
334
00:21:59,450 --> 00:22:03,040
some linear algebra we have
stumbled headlong into an
335
00:22:03,040 --> 00:22:06,150
eigenvalue problem, which
is a well known sort of
336
00:22:06,150 --> 00:22:09,700
mathematical problem associated
with a large number
337
00:22:09,700 --> 00:22:12,980
of physical situations and
a lot to do with physical
338
00:22:12,980 --> 00:22:16,520
properties in particular.
339
00:22:16,520 --> 00:22:19,570
OK, now I have a question
for you.
340
00:22:19,570 --> 00:22:22,180
How we going to solve
this silly thing?
341
00:22:22,180 --> 00:22:22,950
I know--
342
00:22:22,950 --> 00:22:27,260
well the answer is I poke it
into my laptop, and out comes
343
00:22:27,260 --> 00:22:28,210
the answer.
344
00:22:28,210 --> 00:22:31,450
But that's not satisfying, we
should know how to do this if
345
00:22:31,450 --> 00:22:35,834
our batteries were dead and we
needed the answer in a hurry.
346
00:22:35,834 --> 00:22:38,530
Well, I know how to solve
a second rank question.
347
00:22:38,530 --> 00:22:45,500
If I have equation of the form
ax squared plus bx plus c
348
00:22:45,500 --> 00:22:49,250
equals 0, I know the
solution to that.
349
00:22:49,250 --> 00:22:52,800
It's etched indelibly in my
memory and it says that x
350
00:22:52,800 --> 00:22:57,080
equals minus b plus or minus the
square root of b squared
351
00:22:57,080 --> 00:23:01,190
minus 4ac all over 2a.
352
00:23:01,190 --> 00:23:02,450
Impressed you, didn't I?
353
00:23:02,450 --> 00:23:05,210
You know why I remember that?
354
00:23:05,210 --> 00:23:11,030
I learned high school algebra
from a throwback--
355
00:23:11,030 --> 00:23:15,050
a teacher who was a throwback
to the Elizabethan period.
356
00:23:15,050 --> 00:23:18,770
You know how when people go to
a remote island they suddenly
357
00:23:18,770 --> 00:23:21,250
find an example of
a species that
358
00:23:21,250 --> 00:23:23,700
everyone thought was extinct?
359
00:23:23,700 --> 00:23:30,150
Well, to secondary education
[? Ms. Dourier ?]
360
00:23:30,150 --> 00:23:34,410
was a species that had long gone
extinct, but happened to
361
00:23:34,410 --> 00:23:37,820
be preserved at the high
school that I attended.
362
00:23:37,820 --> 00:23:38,750
[? Ms. Dourier ?]
363
00:23:38,750 --> 00:23:44,600
was a very tall woman, ramrod
straight, with a pile of snow
364
00:23:44,600 --> 00:23:46,950
white hair on the
top of her head.
365
00:23:46,950 --> 00:23:51,090
She invariably dressed in a
long, black skirt, and a white
366
00:23:51,090 --> 00:23:55,910
blouse that had puffy sleeves,
and a collar with frilly
367
00:23:55,910 --> 00:23:59,930
things on the top, and,
invariably, a black ribbon
368
00:23:59,930 --> 00:24:03,820
around the frilly
lace collar top.
369
00:24:03,820 --> 00:24:09,530
Her mode of enlightened
education was to have all of
370
00:24:09,530 --> 00:24:14,170
the students in the class have
a notebook that was open.
371
00:24:14,170 --> 00:24:20,160
And one hapless member of the
class was called upon to solve
372
00:24:20,160 --> 00:24:24,310
a problem in real time, and as
the student recited we all
373
00:24:24,310 --> 00:24:27,120
wrote down what the student was
saying in our notebooks.
374
00:24:27,120 --> 00:24:28,960
All the while [? Ms. Dourier, ?]
375
00:24:28,960 --> 00:24:32,730
holding a ruler like a swagger
stick, strode up and down the
376
00:24:32,730 --> 00:24:37,080
aisles, and woe be unto the
poor student who faltered
377
00:24:37,080 --> 00:24:40,660
slightly, let alone get
something wrong.
378
00:24:40,660 --> 00:24:45,780
And that would bring a slap of
the ruler down on the desk and
379
00:24:45,780 --> 00:24:48,450
a sigh of disgust, and
then, all right,
380
00:24:48,450 --> 00:24:50,100
you take it, Audrey.
381
00:24:50,100 --> 00:24:53,240
And then Audrey would begin to
recite until she screwed up
382
00:24:53,240 --> 00:24:56,710
and we would all copy
in our notebooks.
383
00:24:56,710 --> 00:25:04,920
That tyrant so terrorized and
intimidated a bunch of kids in
384
00:25:04,920 --> 00:25:09,760
a redneck, working class
high school that we
385
00:25:09,760 --> 00:25:11,220
really learned algebra.
386
00:25:11,220 --> 00:25:16,960
So to this very day, as a reflex
action, minus b plus or
387
00:25:16,960 --> 00:25:21,140
minus the square root of b
squared minus 4ac all over 2a.
388
00:25:21,140 --> 00:25:24,410
But I don't have the foggiest
idea how to solve a third
389
00:25:24,410 --> 00:25:27,050
order equation until
I look it up.
390
00:25:27,050 --> 00:25:29,560
391
00:25:29,560 --> 00:25:33,390
You probably never had to do
it, so let me, for your
392
00:25:33,390 --> 00:25:36,370
general education and amusement,
pass around a sheet
393
00:25:36,370 --> 00:25:41,975
that tells you how to solve
a third rank equation.
394
00:25:41,975 --> 00:25:46,520
395
00:25:46,520 --> 00:25:51,290
OK the first step is to convert
an equation in, let's
396
00:25:51,290 --> 00:25:57,020
say, y squared plus y cubed plus
py squared plus qy plus r
397
00:25:57,020 --> 00:26:01,010
equals 0 to a so-called normal
form where you get rid of one
398
00:26:01,010 --> 00:26:02,770
of the terms.
399
00:26:02,770 --> 00:26:07,980
So if you make a substitution
of variables and let y be
400
00:26:07,980 --> 00:26:12,590
replaced by x minus the
coefficient p/3, then you get
401
00:26:12,590 --> 00:26:17,440
an equation that has a
cubic term, linear
402
00:26:17,440 --> 00:26:20,040
term x, and a constant.
403
00:26:20,040 --> 00:26:26,370
And the constants a and b are
combinations of p and q and r
404
00:26:26,370 --> 00:26:28,670
in the original equation.
405
00:26:28,670 --> 00:26:32,090
And then the solutions, not
well known to be sure, are
406
00:26:32,090 --> 00:26:36,470
there's a first solution x is
equal to a plus b, and then,
407
00:26:36,470 --> 00:26:43,720
something messy, x2 and x3 are
minus 1/2 of a plus b plus or
408
00:26:43,720 --> 00:26:46,560
minus an imaginary term--
those are the
409
00:26:46,560 --> 00:26:48,100
second and third roots--
410
00:26:48,100 --> 00:26:51,610
i times the square root of 3
over 2a minus b where A and B
411
00:26:51,610 --> 00:26:55,420
are complicated functions
of a and b.
412
00:26:55,420 --> 00:26:57,275
Capital A and capital B are
functions of little
413
00:26:57,275 --> 00:26:58,690
a and little b.
414
00:26:58,690 --> 00:27:02,990
And then if the coefficients in
the original equation are
415
00:27:02,990 --> 00:27:06,070
real, then you get
one real and two
416
00:27:06,070 --> 00:27:08,070
conjugated imaginary roots.
417
00:27:08,070 --> 00:27:11,920
If this combination of terms
b squared and a squared are
418
00:27:11,920 --> 00:27:18,200
greater than 0, it is b squared
over 4 plus a cubed
419
00:27:18,200 --> 00:27:19,850
over 27 equals 0.
420
00:27:19,850 --> 00:27:22,520
If it's that, then you get
three real roots, two
421
00:27:22,520 --> 00:27:23,820
of which are equal.
422
00:27:23,820 --> 00:27:26,590
In the most general case that we
would be encountering here
423
00:27:26,590 --> 00:27:32,260
is b squared over 4 plus a cubed
over 27 is less than 0,
424
00:27:32,260 --> 00:27:35,300
and then there are three
real unequal roots.
425
00:27:35,300 --> 00:27:38,100
426
00:27:38,100 --> 00:27:44,570
Unfortunately, if b squared over
4 plus a cubed over 27 is
427
00:27:44,570 --> 00:27:49,150
negative that term appears
inside of a square root sign
428
00:27:49,150 --> 00:27:51,870
in the solutions.
429
00:27:51,870 --> 00:27:55,500
So the only case that we'd
really be interested in
430
00:27:55,500 --> 00:27:57,430
doesn't work for the solution.
431
00:27:57,430 --> 00:27:59,760
But fortunately there's another
solution and that's
432
00:27:59,760 --> 00:28:02,930
given at the bottom of the page,
and if you really wanted
433
00:28:02,930 --> 00:28:07,910
to solve a third rank equation
by hand, these solutions would
434
00:28:07,910 --> 00:28:10,560
do it for you.
435
00:28:10,560 --> 00:28:13,750
But I think you'd much prefer
to have your computer solve
436
00:28:13,750 --> 00:28:17,890
the equations for you, but let
me show you a way of doing
437
00:28:17,890 --> 00:28:25,090
this without any computer,
without solving any equations.
438
00:28:25,090 --> 00:28:31,550
And it is a very clever method
of successive approximations
439
00:28:31,550 --> 00:28:33,600
that's based on the properties
of the quadric.
440
00:28:33,600 --> 00:28:38,470
441
00:28:38,470 --> 00:28:43,560
So let's suppose we have a
tensor, and that tensor has
442
00:28:43,560 --> 00:28:50,640
some set of coefficients Aij,
all of which are non-zero.
443
00:28:50,640 --> 00:28:53,630
And I'll assume although this
will work for other quadratic
444
00:28:53,630 --> 00:28:59,690
surfaces that the quadric has
the shape of an ellipsoid.
445
00:28:59,690 --> 00:29:05,294
446
00:29:05,294 --> 00:29:09,400
All right let me now pick some
direction at random, actually
447
00:29:09,400 --> 00:29:11,840
I don't have to pick it at
random, but I'll show you what
448
00:29:11,840 --> 00:29:15,160
the shrewd first
guess would be.
449
00:29:15,160 --> 00:29:19,380
So let's say we picked
this direction.
450
00:29:19,380 --> 00:29:25,650
And let us find the direction
if this is--
451
00:29:25,650 --> 00:29:28,850
let's do it in terms of current
and conductivity.
452
00:29:28,850 --> 00:29:34,110
Let's let this be the first
guess for the applied field.
453
00:29:34,110 --> 00:29:37,690
That's clearly not going to be
one of the principal axes, and
454
00:29:37,690 --> 00:29:40,250
let this be the first resulting
direction of
455
00:29:40,250 --> 00:29:41,500
current flow J1.
456
00:29:41,500 --> 00:29:44,190
457
00:29:44,190 --> 00:29:51,600
So what I'm finding then is a
J sub i in terms of an Aij
458
00:29:51,600 --> 00:29:56,060
times an E sub J, and this is my
first result, and this was
459
00:29:56,060 --> 00:29:58,590
my first assumption.
460
00:29:58,590 --> 00:30:05,380
Let me now let the direction of
J be taken as the direction
461
00:30:05,380 --> 00:30:09,150
of the applied field
for a second guess.
462
00:30:09,150 --> 00:30:12,130
And we could normalize to a unit
vector, but we don't even
463
00:30:12,130 --> 00:30:13,020
have to do that.
464
00:30:13,020 --> 00:30:16,710
So let's simply say that my
second guess for the electric
465
00:30:16,710 --> 00:30:19,900
field that I hope will point
along one principal axis is
466
00:30:19,900 --> 00:30:23,390
E2, and E2 I'll take
as identical--
467
00:30:23,390 --> 00:30:24,460
let's say proportional--
468
00:30:24,460 --> 00:30:28,370
to J1 from my original guess.
469
00:30:28,370 --> 00:30:32,516
So this then would be E2,
the second guess.
470
00:30:32,516 --> 00:30:34,900
And if I find the direction
to the normal--
471
00:30:34,900 --> 00:30:37,410
to the quadric in that
direction, this would be my
472
00:30:37,410 --> 00:30:44,070
second result for the current
flow J, I should have put this
473
00:30:44,070 --> 00:30:46,990
in parentheses to indicate that
these are not components
474
00:30:46,990 --> 00:30:49,140
of E and J.
475
00:30:49,140 --> 00:30:51,330
And you can see what's
happening, look at where I am,
476
00:30:51,330 --> 00:30:55,650
I started out here after just
one iteration I have defined
477
00:30:55,650 --> 00:30:58,800
this as the potential direction
of one of the
478
00:30:58,800 --> 00:31:00,480
principal axes.
479
00:31:00,480 --> 00:31:02,540
So I'm going to find E1--
480
00:31:02,540 --> 00:31:03,910
the direction of the field--
481
00:31:03,910 --> 00:31:08,955
E1 that comes from sigma ij
times Ji, I'll take my second
482
00:31:08,955 --> 00:31:19,250
guess E2 either as identical
to J1 and this is going to
483
00:31:19,250 --> 00:31:24,920
give me then a new value for
my second iteration, this
484
00:31:24,920 --> 00:31:27,730
would be J2.
485
00:31:27,730 --> 00:31:34,330
And this process is going to
converge very, very rapidly on
486
00:31:34,330 --> 00:31:35,730
the shortest principal axis.
487
00:31:35,730 --> 00:31:45,050
488
00:31:45,050 --> 00:31:47,940
As you can see in two shots
I'm pretty close to being
489
00:31:47,940 --> 00:31:51,100
parallel to this direction,
which would be X2 in my
490
00:31:51,100 --> 00:31:52,350
illustration.
491
00:31:52,350 --> 00:31:54,285
492
00:31:54,285 --> 00:31:57,410
Now if I wanted to-- if I wanted
to see if I was close
493
00:31:57,410 --> 00:31:58,420
to convergence--
494
00:31:58,420 --> 00:32:02,330
this is going to be awkward
because if I don't normalize
495
00:32:02,330 --> 00:32:06,030
the magnitude of the resulting
vector, J is going to get
496
00:32:06,030 --> 00:32:07,710
larger and larger and larger.
497
00:32:07,710 --> 00:32:11,060
So periodically I would
want to normalize--
498
00:32:11,060 --> 00:32:13,770
take the magnitude of J,
divide that into the
499
00:32:13,770 --> 00:32:16,060
components, and then I'd
have a unit vector.
500
00:32:16,060 --> 00:32:20,740
If I wrote a computer program
to do this, I would do the
501
00:32:20,740 --> 00:32:24,300
normalization each time
to test convergence.
502
00:32:24,300 --> 00:32:29,580
Now, this is my kind of
procedure, because if I'm
503
00:32:29,580 --> 00:32:33,470
doing this by hand and I screw
up and I make a mistake and my
504
00:32:33,470 --> 00:32:37,190
answer is thrown off a little
bit, if I continue to iterate,
505
00:32:37,190 --> 00:32:42,940
the thing will proceed to
continue to converge to the
506
00:32:42,940 --> 00:32:46,310
shortest principal axis.
507
00:32:46,310 --> 00:32:48,990
So I can make a mistake and it
will correct itself and come
508
00:32:48,990 --> 00:32:51,160
back again, and that's
my kind of solution.
509
00:32:51,160 --> 00:32:52,020
Yeah, Jason?
510
00:32:52,020 --> 00:32:53,500
AUDIENCE: It's going to give you
the minimum value of the
511
00:32:53,500 --> 00:32:54,980
property, right?
512
00:32:54,980 --> 00:32:55,924
PROFESSOR: No.
513
00:32:55,924 --> 00:32:59,000
It's going to give you the
minimum principal axis, that's
514
00:32:59,000 --> 00:33:02,620
going to be the maximum
value of the property.
515
00:33:02,620 --> 00:33:05,520
So that's one out of three,
hey, that's not bad.
516
00:33:05,520 --> 00:33:08,010
What do I do for the others?
517
00:33:08,010 --> 00:33:16,650
Let me tell you without
proof to find the
518
00:33:16,650 --> 00:33:18,215
maximum principal axis.
519
00:33:18,215 --> 00:33:25,760
520
00:33:25,760 --> 00:33:28,470
And that would be the minimum
value of the property.
521
00:33:28,470 --> 00:33:40,260
522
00:33:40,260 --> 00:33:43,530
What you would do is exactly
the same procedure, and you
523
00:33:43,530 --> 00:33:56,060
would operate on not the tensor,
but on-- using as a
524
00:33:56,060 --> 00:33:58,270
matrix of coefficients--
525
00:33:58,270 --> 00:34:02,435
the matrix of the
tensor inverted.
526
00:34:02,435 --> 00:34:08,280
527
00:34:08,280 --> 00:34:14,040
And I assume you know how to
find the inverse of a matrix,
528
00:34:14,040 --> 00:34:17,110
except that you don't have to
even find the inverse, the
529
00:34:17,110 --> 00:34:19,920
inverse is going to be a
collection of functions of the
530
00:34:19,920 --> 00:34:22,610
original elements divided
by the determinant.
531
00:34:22,610 --> 00:34:25,000
You don't have to worry about
normalizing, all that's
532
00:34:25,000 --> 00:34:27,245
important is the relative value
of these coefficients.
533
00:34:27,245 --> 00:34:28,090
Yes, sir?
534
00:34:28,090 --> 00:34:30,719
AUDIENCE: If your tensor is
symmetric, can't you say that
535
00:34:30,719 --> 00:34:32,870
they'll just [INAUDIBLE]
on to another?
536
00:34:32,870 --> 00:34:34,310
PROFESSOR: That's what
you're going to do.
537
00:34:34,310 --> 00:34:36,505
But when you know just
one, that won't work.
538
00:34:36,505 --> 00:34:38,822
The other two are floating
around somewhere and you don't
539
00:34:38,822 --> 00:34:39,870
know in what direction.
540
00:34:39,870 --> 00:34:45,340
So if you do this, then you have
two out of the three and
541
00:34:45,340 --> 00:34:49,590
now very shrewdly, as you point
out, if we have two of
542
00:34:49,590 --> 00:34:53,969
the principal axes and the
tensor is symmetric, then we
543
00:34:53,969 --> 00:34:56,910
can automatically get the
direction of the third.
544
00:34:56,910 --> 00:35:02,370
If it's not symmetric, then
you have to use the set of
545
00:35:02,370 --> 00:35:06,360
equations, and you have three
principal axes as
546
00:35:06,360 --> 00:35:10,140
eigenvectors, as they're called
in eigenvalue problems.
547
00:35:10,140 --> 00:35:13,440
And they do not have to be
orthogonal, but you have the
548
00:35:13,440 --> 00:35:17,740
components in a Cartesian
coordinate system so you can
549
00:35:17,740 --> 00:35:20,836
find the angle between
those axes.
550
00:35:20,836 --> 00:35:21,250
OK?
551
00:35:21,250 --> 00:35:22,480
Yes, sir.
552
00:35:22,480 --> 00:35:25,286
AUDIENCE: With respect to
whether the matrix is
553
00:35:25,286 --> 00:35:27,360
symmetrical or is
not symmetric,
554
00:35:27,360 --> 00:35:29,520
the quadric is always--
555
00:35:29,520 --> 00:35:32,400
has three principal axes
at right angles right?
556
00:35:32,400 --> 00:35:36,720
PROFESSOR: Only if the
tensor is symmetric.
557
00:35:36,720 --> 00:35:39,130
Only if the tensor
is symmetric.
558
00:35:39,130 --> 00:35:44,340
And let me put your question
aside for about five minutes,
559
00:35:44,340 --> 00:35:47,640
and I want to take an overview
of what we've learned about
560
00:35:47,640 --> 00:35:51,160
the geometry of the quadric and
the symmetry restrictions
561
00:35:51,160 --> 00:35:54,460
that we can impose on the tensor
in a formal fashion,
562
00:35:54,460 --> 00:35:57,100
and see how they compare
and how one allows an
563
00:35:57,100 --> 00:35:59,600
interpretation of the other.
564
00:35:59,600 --> 00:36:02,110
So I'm going to answer your
question, I'd like to wait for
565
00:36:02,110 --> 00:36:04,880
about three minutes.
566
00:36:04,880 --> 00:36:07,520
Other questions?
567
00:36:07,520 --> 00:36:10,930
OK, again this is all very
symbolic, I'm not giving you
568
00:36:10,930 --> 00:36:13,370
an explicit answer,
I can't do it.
569
00:36:13,370 --> 00:36:14,960
But what I can do you--
570
00:36:14,960 --> 00:36:16,980
do for you is, ha
ha, do to you.
571
00:36:16,980 --> 00:36:19,870
I almost slipped and said that,
what I can't do to you,
572
00:36:19,870 --> 00:36:24,800
is give you a problem that
asks you to solve for the
573
00:36:24,800 --> 00:36:28,560
principal axes, for example
of a property tensor.
574
00:36:28,560 --> 00:36:31,955
I will be merciful and perhaps
not have all nine coefficients
575
00:36:31,955 --> 00:36:34,320
non-zero, I will have
one of them zero, or
576
00:36:34,320 --> 00:36:36,590
maybe two of them zero.
577
00:36:36,590 --> 00:36:42,190
So again to try this it is very
instructive and it will
578
00:36:42,190 --> 00:36:45,200
cement what the individual steps
that I performed here
579
00:36:45,200 --> 00:36:48,820
actually involve.
580
00:36:48,820 --> 00:36:50,940
Now, a question that I thought
you were going to ask about
581
00:36:50,940 --> 00:36:57,640
principal axes is that in order
to do what I did here I
582
00:36:57,640 --> 00:37:02,280
am using the radius-normal
property, and the
583
00:37:02,280 --> 00:37:08,580
radius-normal property only
works for a symmetric tensor.
584
00:37:08,580 --> 00:37:12,850
So sounds like I'm swindling
you, except that if I use this
585
00:37:12,850 --> 00:37:19,910
procedure, the radius-normal
won't actually be identically
586
00:37:19,910 --> 00:37:24,390
parallel to the generalized
displacement.
587
00:37:24,390 --> 00:37:27,960
But it's not going to be
terribly different from it,
588
00:37:27,960 --> 00:37:30,600
and I'm going to get something
that again will converge
589
00:37:30,600 --> 00:37:32,940
towards the shortest
principal axis.
590
00:37:32,940 --> 00:37:36,750
And once I'm close to the
principal axis it is true--
591
00:37:36,750 --> 00:37:38,500
symmetric or not--
592
00:37:38,500 --> 00:37:43,880
that the only three directions
before which the generalized
593
00:37:43,880 --> 00:37:46,570
displacement is parallel to
the radius vector are the
594
00:37:46,570 --> 00:37:48,990
principal axes.
595
00:37:48,990 --> 00:37:52,280
But anyway this is a dicey
situation if the tensor is not
596
00:37:52,280 --> 00:37:56,530
symmetric, and it probably comes
as a great relief to
597
00:37:56,530 --> 00:38:00,710
learn that there's really only
one property tensor of second
598
00:38:00,710 --> 00:38:04,060
rank that's known for sure
to be non-symmetric.
599
00:38:04,060 --> 00:38:06,440
So in most cases--
600
00:38:06,440 --> 00:38:09,560
99 out of 100, if not more--
we will be dealing with
601
00:38:09,560 --> 00:38:12,350
property tensors that are
symmetric, so we could use
602
00:38:12,350 --> 00:38:13,600
this procedure.
603
00:38:13,600 --> 00:38:16,240
604
00:38:16,240 --> 00:38:19,730
So again this property converges
remarkably well, and
605
00:38:19,730 --> 00:38:23,230
as I say it has the admirable
quality of being
606
00:38:23,230 --> 00:38:26,040
self-correcting if you make a
mistake, if you're grinding
607
00:38:26,040 --> 00:38:27,410
this out by hand.
608
00:38:27,410 --> 00:38:32,110
But a computer program that
you write that just simply
609
00:38:32,110 --> 00:38:35,600
takes the direction of J,
normalizes it to a unit
610
00:38:35,600 --> 00:38:40,420
vector, applies that as the
generalized force, finds a
611
00:38:40,420 --> 00:38:44,020
quote "J" as a second iteration,
normalizes that,
612
00:38:44,020 --> 00:38:47,400
and then you can check to
whatever degree of convergence
613
00:38:47,400 --> 00:38:49,330
you wish to have in
your solution.
614
00:38:49,330 --> 00:38:52,320
And it's a simple matter to set
up a program to do this.
615
00:38:52,320 --> 00:38:56,487
616
00:38:56,487 --> 00:39:02,390
All right, I think we still have
some time, so let me now
617
00:39:02,390 --> 00:39:04,650
put everything together.
618
00:39:04,650 --> 00:39:12,630
And look at what we've seen of
the geometric properties of
619
00:39:12,630 --> 00:39:20,080
the quadric, and the tensor
arrays that we found for
620
00:39:20,080 --> 00:39:22,340
different symmetries.
621
00:39:22,340 --> 00:39:29,520
For triclinic crystals, for
which the recommended
622
00:39:29,520 --> 00:39:32,940
procedure is to leave by the
nearest exit and work on
623
00:39:32,940 --> 00:39:40,340
something different instead, you
would have nine elements
624
00:39:40,340 --> 00:39:46,210
altogether that are necessary
to define the property.
625
00:39:46,210 --> 00:39:53,830
626
00:39:53,830 --> 00:39:57,690
I'll discuss this in terms
of tensors that
627
00:39:57,690 --> 00:40:00,870
gives you an ellipsoid.
628
00:40:00,870 --> 00:40:05,050
The argument is not different
for hyperboloids of one or two
629
00:40:05,050 --> 00:40:08,640
sheets, and we can't draw
imaginary ellipsoids, but an
630
00:40:08,640 --> 00:40:12,510
ellipsoid is a much easier
thing to draw.
631
00:40:12,510 --> 00:40:15,910
OK, triclinic symmetries, either
one or one bar, nine
632
00:40:15,910 --> 00:40:17,160
elements in the tensor.
633
00:40:17,160 --> 00:40:21,700
634
00:40:21,700 --> 00:40:28,010
No constraints whatsoever on the
shape of the quadric or on
635
00:40:28,010 --> 00:40:32,150
its orientation relative to
our coordinate system.
636
00:40:32,150 --> 00:40:36,960
So, let's total up now looking
at the quadric how many
637
00:40:36,960 --> 00:40:39,650
degrees of freedom there
are in this situation.
638
00:40:39,650 --> 00:40:46,710
We have three principal
axes, so those are
639
00:40:46,710 --> 00:40:48,910
three degrees of freedom.
640
00:40:48,910 --> 00:40:54,360
The orientation of the quadric
is arbitrary, so there are
641
00:40:54,360 --> 00:40:57,080
three orientational degrees
of freedom.
642
00:40:57,080 --> 00:41:08,790
643
00:41:08,790 --> 00:41:11,635
And that comes out to six.
644
00:41:11,635 --> 00:41:14,360
645
00:41:14,360 --> 00:41:16,320
Supposedly nine degrees
of freedom, what
646
00:41:16,320 --> 00:41:19,190
are the other three?
647
00:41:19,190 --> 00:41:22,340
Again, if this is a general
tensor, which is
648
00:41:22,340 --> 00:41:24,840
non-symmetric, the
eigenvectors--
649
00:41:24,840 --> 00:41:27,720
650
00:41:27,720 --> 00:41:32,330
the principal axes that you have
to choose to squeeze this
651
00:41:32,330 --> 00:41:34,550
thing into a diagonal form--
652
00:41:34,550 --> 00:41:37,360
become non-orthogonal.
653
00:41:37,360 --> 00:41:38,280
OK?
654
00:41:38,280 --> 00:41:43,860
So you have another three
degrees of freedom that
655
00:41:43,860 --> 00:41:49,160
specify the mutual directions
of the eigenvectors--
656
00:41:49,160 --> 00:41:53,690
657
00:41:53,690 --> 00:41:55,140
the angles between them.
658
00:41:55,140 --> 00:41:57,970
659
00:41:57,970 --> 00:41:58,380
OK?
660
00:41:58,380 --> 00:42:02,720
The directions in which you have
to pick your coordinate
661
00:42:02,720 --> 00:42:10,080
system, X1, X2, and X3, that
force this thing into a
662
00:42:10,080 --> 00:42:13,520
diagonal form is going to
involve a coordinate system in
663
00:42:13,520 --> 00:42:18,820
which these three angles are not
90 degrees and are fixed
664
00:42:18,820 --> 00:42:20,640
if you're going to squeeze
this thing
665
00:42:20,640 --> 00:42:21,895
into a diagonal form.
666
00:42:21,895 --> 00:42:26,690
667
00:42:26,690 --> 00:42:27,300
And that's it.
668
00:42:27,300 --> 00:42:30,595
So we add these three interaxial
angles in, they're
669
00:42:30,595 --> 00:42:34,150
a total of nine degrees of
freedom for a general non
670
00:42:34,150 --> 00:42:35,400
symmetric tensor.
671
00:42:35,400 --> 00:42:41,860
672
00:42:41,860 --> 00:42:45,800
To convince you that these
interaxial angles for the
673
00:42:45,800 --> 00:42:49,410
eigenvectors really are
variables, let me tell you
674
00:42:49,410 --> 00:42:53,400
something that we may prove
later as a recreational
675
00:42:53,400 --> 00:42:56,300
exercise, or I may give it to
you on a problem set, it's not
676
00:42:56,300 --> 00:42:57,600
difficult to prove.
677
00:42:57,600 --> 00:43:06,470
And that is the result that
a symmetric tensor remains
678
00:43:06,470 --> 00:43:13,660
symmetric for any arbitrary
transformation of axes.
679
00:43:13,660 --> 00:43:16,720
680
00:43:16,720 --> 00:43:22,320
That is from one Cartesian
coordinate system to another.
681
00:43:22,320 --> 00:43:48,888
682
00:43:48,888 --> 00:43:55,250
Now a very salutary effect of
this is that the tensor has
683
00:43:55,250 --> 00:43:56,500
nine elements.
684
00:43:56,500 --> 00:44:06,900
685
00:44:06,900 --> 00:44:08,620
And if the tensor
is symmetric--
686
00:44:08,620 --> 00:44:10,860
if you want to transform
that tensor--
687
00:44:10,860 --> 00:44:14,870
you only have to do the
off-diagonal terms, because
688
00:44:14,870 --> 00:44:17,750
whatever it turns into when
you change the axes, these
689
00:44:17,750 --> 00:44:20,060
off-diagonal terms are going
to be equal to the
690
00:44:20,060 --> 00:44:22,746
off-diagonal terms in
the new tensor.
691
00:44:22,746 --> 00:44:23,170
OK?
692
00:44:23,170 --> 00:44:25,040
So this means that only six--
693
00:44:25,040 --> 00:44:27,810
if the tensor was symmetric in
one coordinate system, only
694
00:44:27,810 --> 00:44:31,750
six elements have to
be transformed.
695
00:44:31,750 --> 00:44:33,870
Actually, it's better than that,
you don't have to do
696
00:44:33,870 --> 00:44:36,770
six, you only have to do five.
697
00:44:36,770 --> 00:44:41,770
And this is the last diagonal
term that can be obtained from
698
00:44:41,770 --> 00:44:43,020
the trace of the tensor.
699
00:44:43,020 --> 00:44:50,460
700
00:44:50,460 --> 00:44:55,620
And that would be the trace of
the tensor T prime after
701
00:44:55,620 --> 00:45:03,040
transformation because A11 prime
plus A22 prime plus A33
702
00:45:03,040 --> 00:45:07,360
prime has to be equal to the
original trace T, which was
703
00:45:07,360 --> 00:45:11,680
A11 plus A22 plus A33 in
the original system.
704
00:45:11,680 --> 00:45:14,960
705
00:45:14,960 --> 00:45:17,480
So if the tensor is symmetric,
you really have to crank
706
00:45:17,480 --> 00:45:22,200
through a transformation for
only five of the nine terms.
707
00:45:22,200 --> 00:45:25,150
Now what is the relevance of
this to what I just said?
708
00:45:25,150 --> 00:45:31,057
If we have the tensor
diagonalized, to a new form
709
00:45:31,057 --> 00:45:41,380
A11, 0, 0, 0, A22 prime, 0,
0, 0, A33 prime, that is a
710
00:45:41,380 --> 00:45:45,180
special case of a symmetric
tensor.
711
00:45:45,180 --> 00:45:48,450
So if you decide to go from
the coordinate system that
712
00:45:48,450 --> 00:45:52,270
produced this diagonalized form
to some other coordinate
713
00:45:52,270 --> 00:45:55,130
system, the new tensor that
you're going to get is going
714
00:45:55,130 --> 00:45:56,530
to be symmetric.
715
00:45:56,530 --> 00:46:00,225
But the thing was not originally
symmetric, so how
716
00:46:00,225 --> 00:46:02,010
can that be?
717
00:46:02,010 --> 00:46:05,140
The answer is you can put it
in diagonal form only in a
718
00:46:05,140 --> 00:46:07,740
non-Cartesian coordinate
system.
719
00:46:07,740 --> 00:46:12,150
And, therefore, that
is evidence that
720
00:46:12,150 --> 00:46:14,540
the reference axes--
721
00:46:14,540 --> 00:46:17,680
the principal axes-- cannot
be orthogonal.
722
00:46:17,680 --> 00:46:20,600
Otherwise you could take the
diagonalized tensor and crank
723
00:46:20,600 --> 00:46:23,690
it back to some arbitrary
Cartesian coordinate, and
724
00:46:23,690 --> 00:46:26,080
suddenly it would go to a
symmetric tensor when it was
725
00:46:26,080 --> 00:46:28,674
not symmetric to begin with.
726
00:46:28,674 --> 00:46:29,500
OK?
727
00:46:29,500 --> 00:46:30,820
QED.
728
00:46:30,820 --> 00:46:33,950
So you can diagonalize a general
tensor only if you
729
00:46:33,950 --> 00:46:37,940
take the axes along the
eigenvectors, which cannot be
730
00:46:37,940 --> 00:46:39,960
orthogonal.
731
00:46:39,960 --> 00:46:42,940
OK, I saw you raise your hand,
I wasn't ignoring you.
732
00:46:42,940 --> 00:46:43,695
That was it?
733
00:46:43,695 --> 00:46:44,455
Good.
734
00:46:44,455 --> 00:46:45,425
AUDIENCE: I wanted to make
sure that Cartesian
735
00:46:45,425 --> 00:46:46,675
[INAUDIBLE]
736
00:46:46,675 --> 00:46:49,310
737
00:46:49,310 --> 00:46:51,610
PROFESSOR: It's always nice to
see the class one step ahead
738
00:46:51,610 --> 00:46:52,860
of what you're doing.
739
00:46:52,860 --> 00:46:56,786
740
00:46:56,786 --> 00:47:01,290
All right, let's look at a
couple of more crystal systems
741
00:47:01,290 --> 00:47:05,705
in the form of tensors in those
systems and show that
742
00:47:05,705 --> 00:47:09,680
that in fact does correspond to
the geometric constraints
743
00:47:09,680 --> 00:47:11,070
on the quadric as well.
744
00:47:11,070 --> 00:47:12,320
For monoclinic crystals--
745
00:47:12,320 --> 00:47:16,530
746
00:47:16,530 --> 00:47:22,780
and this is symmetry 2, symmetry
M, and symmetry 2/M.
747
00:47:22,780 --> 00:47:25,560
The form of the tensor when we
took the coordinate system
748
00:47:25,560 --> 00:47:31,080
along symmetry elements
was A11, A12, 0--
749
00:47:31,080 --> 00:47:33,230
terms with a single
three vanished--
750
00:47:33,230 --> 00:47:39,648
A21, A22, 0, 0, 0, A33.
751
00:47:39,648 --> 00:47:42,980
752
00:47:42,980 --> 00:47:48,130
So this was the case where X3
was along the two-fold axis
753
00:47:48,130 --> 00:47:50,420
and or perpendicular to
the mirror point.
754
00:47:50,420 --> 00:47:55,690
755
00:47:55,690 --> 00:48:01,440
OK, number of independent
variables to describe this
756
00:48:01,440 --> 00:48:02,810
property is five.
757
00:48:02,810 --> 00:48:06,150
758
00:48:06,150 --> 00:48:09,600
And if we look at this in terms
of a constraint and
759
00:48:09,600 --> 00:48:18,530
shape in orientation of an
ellipsoidal quadric, says that
760
00:48:18,530 --> 00:48:21,540
one of the principal axes, if
the quadric is to remain
761
00:48:21,540 --> 00:48:28,690
invariant, has to be along
the two-fold axis and or
762
00:48:28,690 --> 00:48:32,700
perpendicular to a plane that
contains the mirror plane, if
763
00:48:32,700 --> 00:48:36,220
a mirror plane is
also present.
764
00:48:36,220 --> 00:48:38,810
So what are the degrees
of freedom here?
765
00:48:38,810 --> 00:48:42,640
This has to be along one
reference axis, x3.
766
00:48:42,640 --> 00:48:47,570
And, indeed, this form of the
tensor occurred only when the
767
00:48:47,570 --> 00:48:52,730
two-fold axes were
parallel, 2 X3.
768
00:48:52,730 --> 00:48:55,900
And this ellipsoid then
was left with
769
00:48:55,900 --> 00:48:57,880
one degree of freedom.
770
00:48:57,880 --> 00:49:02,950
If this is the direction of the
reference axis X2, we have
771
00:49:02,950 --> 00:49:05,040
a degree of freedom in--
772
00:49:05,040 --> 00:49:08,330
shouldn't have called these X1
and X2, these need not be the
773
00:49:08,330 --> 00:49:10,050
coordinate systems.
774
00:49:10,050 --> 00:49:14,260
Looking down on the quadric
along the two-fold axis, the
775
00:49:14,260 --> 00:49:17,310
section of the quadric
perpendicular to the two-fold
776
00:49:17,310 --> 00:49:23,370
axis can have a general shape,
and it can have one degree of
777
00:49:23,370 --> 00:49:28,670
freedom in the orientation of
the principal axis relative to
778
00:49:28,670 --> 00:49:30,370
the coordinate system.
779
00:49:30,370 --> 00:49:34,480
So we have three parameters to
specify the principal axes,
780
00:49:34,480 --> 00:49:35,960
since this is a general
quadric.
781
00:49:35,960 --> 00:49:39,180
782
00:49:39,180 --> 00:49:57,590
We have one degree of freedom in
orientation, and that adds
783
00:49:57,590 --> 00:50:00,425
up to four, but we said five.
784
00:50:00,425 --> 00:50:03,180
785
00:50:03,180 --> 00:50:04,260
So what's going on here?
786
00:50:04,260 --> 00:50:07,180
Again this is a question of
where the eigenvectors point.
787
00:50:07,180 --> 00:50:10,770
788
00:50:10,770 --> 00:50:15,910
If this term is not equal to
this term, in order to force
789
00:50:15,910 --> 00:50:20,240
the tensor into a diagonal
form, you have to pick
790
00:50:20,240 --> 00:50:25,350
eigenvectors in the X1, X2
plane, which will force the
791
00:50:25,350 --> 00:50:27,740
tensor into a symmetric form.
792
00:50:27,740 --> 00:50:31,810
That is, to make this term
equal to this term.
793
00:50:31,810 --> 00:50:35,150
It's only a pair of axes
involved, so there's only one
794
00:50:35,150 --> 00:50:36,420
angle between these two
795
00:50:36,420 --> 00:50:38,320
eigenvectors, which is a variable.
796
00:50:38,320 --> 00:50:42,930
So for a nonsymmetric tensor
plus one angle between the
797
00:50:42,930 --> 00:50:44,180
eigenvectors.
798
00:50:44,180 --> 00:50:56,220
799
00:50:56,220 --> 00:50:58,150
So that comes out, a-ha, five.
800
00:50:58,150 --> 00:51:04,380
801
00:51:04,380 --> 00:51:07,880
And again the way to see that is
I will never get the tensor
802
00:51:07,880 --> 00:51:18,420
into a diagonal form making
all the off-diagonal terms
803
00:51:18,420 --> 00:51:22,560
zero when I'm starting with a
tensor which is not symmetric,
804
00:51:22,560 --> 00:51:25,960
and I know that a symmetric
tensor, even in the special
805
00:51:25,960 --> 00:51:29,080
case where the off-diagonal
terms are zero, is going to go
806
00:51:29,080 --> 00:51:33,860
to a symmetric tensor in another
coordinate system.
807
00:51:33,860 --> 00:51:35,270
So I know that there
has to be an
808
00:51:35,270 --> 00:51:37,680
additional degree of freedom.
809
00:51:37,680 --> 00:51:41,290
OK, the rest come very, very
easily, so let me take just a
810
00:51:41,290 --> 00:51:44,270
couple of minutes to finish up
this stage of the discussion,
811
00:51:44,270 --> 00:51:47,080
and we will shortly go on
to something different.
812
00:51:47,080 --> 00:51:51,310
813
00:51:51,310 --> 00:51:53,240
The next step up in symmetry
is orthorhombic.
814
00:51:53,240 --> 00:52:00,070
815
00:52:00,070 --> 00:52:06,890
And if we pick arc-- and this
could be 222, 2MM or 2/M 2/M
816
00:52:06,890 --> 00:52:19,680
2/M. We found that when we
took X1, X2, and X3 along
817
00:52:19,680 --> 00:52:29,416
two-fold axes that the tensor
took a diagonal form A11, 0,
818
00:52:29,416 --> 00:52:35,665
0, 0, A22, 0, 0, 0, A33.
819
00:52:35,665 --> 00:52:38,370
820
00:52:38,370 --> 00:52:44,310
That says that the quadric
if it's an ellipsoid--
821
00:52:44,310 --> 00:52:47,690
or any other quadratic form--
822
00:52:47,690 --> 00:52:52,330
has all three of its principal
axes along the reference
823
00:52:52,330 --> 00:52:55,230
system X1, X2, X3.
824
00:52:55,230 --> 00:52:58,120
So the only degree of freedoms
here-- and things are again
825
00:52:58,120 --> 00:53:02,410
starting to set up like
supercooled water--
826
00:53:02,410 --> 00:53:06,065
the only degrees of freedom are
the three principal axes.
827
00:53:06,065 --> 00:53:11,960
828
00:53:11,960 --> 00:53:14,065
And that's it, the orientation
is fixed.
829
00:53:14,065 --> 00:53:20,030
830
00:53:20,030 --> 00:53:25,400
And, moreover, the tensor is
diagonal in this reference
831
00:53:25,400 --> 00:53:29,180
system-- it's going to be
symmetric in this reference
832
00:53:29,180 --> 00:53:32,850
system, it's going to remain
symmetric in any other
833
00:53:32,850 --> 00:53:34,060
coordinate system.
834
00:53:34,060 --> 00:53:40,040
So the tensor is always
symmetric, the eigenvectors
835
00:53:40,040 --> 00:53:48,300
then are always orthogonal, even
if the coordinate system
836
00:53:48,300 --> 00:53:49,700
stays Cartesian.
837
00:53:49,700 --> 00:53:52,260
And there are three variables,
three degrees of freedom.
838
00:53:52,260 --> 00:54:02,520
839
00:54:02,520 --> 00:54:05,210
And finally two remaining
cases can
840
00:54:05,210 --> 00:54:08,550
be disposed of quickly.
841
00:54:08,550 --> 00:54:11,510
We saw that for an arbitrary
theta that
842
00:54:11,510 --> 00:54:13,220
is a symmetry element--
843
00:54:13,220 --> 00:54:16,940
844
00:54:16,940 --> 00:54:18,110
so let me not say a symmetry--
845
00:54:18,110 --> 00:54:20,245
for an arbitrary rotation
about X3.
846
00:54:20,245 --> 00:54:25,970
847
00:54:25,970 --> 00:54:34,560
This was the ingenious little
proof in which we transform
848
00:54:34,560 --> 00:54:41,470
the tensor by an arbitrary
rotation theta about X3.
849
00:54:41,470 --> 00:54:48,370
We found that the form of the
tensor was A11, A12, 0, and
850
00:54:48,370 --> 00:54:52,380
now I get to use tensor notation
by calling this term
851
00:54:52,380 --> 00:54:57,810
A12 again and not the proper
indices A21, so I'll put
852
00:54:57,810 --> 00:54:59,190
quotation marks.
853
00:54:59,190 --> 00:55:07,170
"A22" was equal to "A11."
854
00:55:07,170 --> 00:55:11,640
So going to an arbitrary
rotation theta this must be
855
00:55:11,640 --> 00:55:14,670
the form of the tensor we
discovered, and this will
856
00:55:14,670 --> 00:55:16,810
cover then fourfold
axes, three-fold
857
00:55:16,810 --> 00:55:19,220
axes, and sixfold axes.
858
00:55:19,220 --> 00:55:23,570
These two elements are
constrained to have the two
859
00:55:23,570 --> 00:55:27,410
values, these two are
constrained to have the same
860
00:55:27,410 --> 00:55:29,650
values, so the tensor
is always symmetric.
861
00:55:29,650 --> 00:55:33,080
862
00:55:33,080 --> 00:55:35,820
Symmetric relative to these
axes, symmetric in all
863
00:55:35,820 --> 00:55:40,470
coordinates, and any choice of
coordinate system, then.
864
00:55:40,470 --> 00:55:55,280
865
00:55:55,280 --> 00:55:59,870
And that means that if the
quadric has this property,
866
00:55:59,870 --> 00:56:04,080
it's going to have one principal
axis along X3, it's
867
00:56:04,080 --> 00:56:11,120
going to be circular in the
plane that contains X1 and X2,
868
00:56:11,120 --> 00:56:20,160
so there are only, count them
up, there are only three
869
00:56:20,160 --> 00:56:21,410
degrees of freedom.
870
00:56:21,410 --> 00:56:29,670
871
00:56:29,670 --> 00:56:32,790
In terms of tensor elements,
these degrees of
872
00:56:32,790 --> 00:56:37,400
freedom are A1, A12--
873
00:56:37,400 --> 00:56:41,430
in the diagonalized
form-- and A33.
874
00:56:41,430 --> 00:56:46,540
And there are three degrees
of freedom in terms of the
875
00:56:46,540 --> 00:56:48,540
independent elements as well.
876
00:56:48,540 --> 00:56:53,850
So this is for anything that
involves rotational symmetry
877
00:56:53,850 --> 00:56:56,540
other than 180 degrees.
878
00:56:56,540 --> 00:57:07,420
And, finally, for cubic
crystals, if a symmetry
879
00:57:07,420 --> 00:57:11,150
operation other than 180 degrees
gives us this form of
880
00:57:11,150 --> 00:57:14,960
the tensor, two off-diagonal
terms are zero, two diagonal
881
00:57:14,960 --> 00:57:16,100
terms equal.
882
00:57:16,100 --> 00:57:21,140
If we impose this along all
three reference axes, then the
883
00:57:21,140 --> 00:57:24,540
form of the tensor is going to
go to a diagonal form with all
884
00:57:24,540 --> 00:57:28,440
diagonal terms equal.
885
00:57:28,440 --> 00:57:31,730
886
00:57:31,730 --> 00:57:39,720
The quadric that corresponds to
that form is a sphere that
887
00:57:39,720 --> 00:57:44,640
says that there's one quantity,
one degree of
888
00:57:44,640 --> 00:57:47,720
freedom in the form
of the tensor.
889
00:57:47,720 --> 00:57:54,410
890
00:57:54,410 --> 00:57:56,200
There are zero degrees of
891
00:57:56,200 --> 00:57:58,120
orientational degree of freedom.
892
00:57:58,120 --> 00:58:11,430
893
00:58:11,430 --> 00:58:14,310
Thus the spherical quadrics
stay spherical for any
894
00:58:14,310 --> 00:58:15,540
orientation.
895
00:58:15,540 --> 00:58:19,120
So again, one independent
element in the tensor, just
896
00:58:19,120 --> 00:58:21,150
one degree of freedom
in the quadric--
897
00:58:21,150 --> 00:58:22,530
namely its radius--
898
00:58:22,530 --> 00:58:25,510
and, again, the formal
constraints that we obtained
899
00:58:25,510 --> 00:58:31,700
by symmetry transformations
agree with those that are the
900
00:58:31,700 --> 00:58:34,090
same degrees of freedom when
you want the quadric to
901
00:58:34,090 --> 00:58:36,050
coincide with the axes.
902
00:58:36,050 --> 00:58:36,844
Yes?
903
00:58:36,844 --> 00:58:39,925
AUDIENCE: Speaking of the circle
and the exponents to
904
00:58:39,925 --> 00:58:43,960
play only if A12 equals zero?
905
00:58:43,960 --> 00:58:48,450
PROFESSOR: These two were
equal, but non-zero.
906
00:58:48,450 --> 00:58:51,400
These two are equal, this
one was independent.
907
00:58:51,400 --> 00:58:53,590
And that would be for
the specific case
908
00:58:53,590 --> 00:58:55,660
of a fourfold axis.
909
00:58:55,660 --> 00:58:59,300
Now if we put a fourfold axis
along X1, that's going to
910
00:58:59,300 --> 00:59:02,910
involve these two being
equal and the
911
00:59:02,910 --> 00:59:07,400
non-zero elements become--
912
00:59:07,400 --> 00:59:08,350
let's see--
913
00:59:08,350 --> 00:59:16,718
along X2 it would
be A13 and A23.
914
00:59:16,718 --> 00:59:17,968
No.
915
00:59:17,968 --> 00:59:21,200
916
00:59:21,200 --> 00:59:26,680
OK, we put the four-fold axis
along X2, then this one would
917
00:59:26,680 --> 00:59:40,440
be A22, A11, and A13 would have
to be equal, and 21 and
918
00:59:40,440 --> 00:59:48,650
12 would have to be zero, 23 and
32 would have to be equal.
919
00:59:48,650 --> 00:59:51,110
I think that's the way
it'll go, and these
920
00:59:51,110 --> 00:59:53,440
three have to be zero.
921
00:59:53,440 --> 00:59:56,210
So when you impose all three
constraints, here we've had
922
00:59:56,210 --> 01:00:00,060
these two equal, now we have
these two equal, put the
923
01:00:00,060 --> 01:00:01,860
four-fold access in the next--
924
01:00:01,860 --> 01:00:03,880
in the remaining direction,
and again it has to be
925
01:00:03,880 --> 01:00:08,290
diagonal, and pairwise all of
the off-diagonal terms will be
926
01:00:08,290 --> 01:00:09,748
required to be zero.
927
01:00:09,748 --> 01:00:15,130
AUDIENCE: I mean, when you
diagonalize your matrix the
928
01:00:15,130 --> 01:00:20,820
diagonal term are going to be
all different in this case if
929
01:00:20,820 --> 01:00:25,430
A12 is different from zero.
930
01:00:25,430 --> 01:00:27,860
PROFESSOR: This is not a final
result, this is an
931
01:00:27,860 --> 01:00:29,270
intermediate step.
932
01:00:29,270 --> 01:00:33,230
So we impose this constraint,
plus this constraint, and
933
01:00:33,230 --> 01:00:35,270
that's actually going to give
us all the qualities we're
934
01:00:35,270 --> 01:00:38,105
going to get, but we'd want
to do the same thing for a
935
01:00:38,105 --> 01:00:42,600
four-fold axis along X3.
936
01:00:42,600 --> 01:00:46,890
You put them all together, the
fact that these are zero will
937
01:00:46,890 --> 01:00:48,750
wipe out all of the
off-diagonal
938
01:00:48,750 --> 01:00:51,880
terms, and these things--
939
01:00:51,880 --> 01:00:53,420
for another choice of
axes, these two
940
01:00:53,420 --> 01:00:54,880
would have to be zero.
941
01:00:54,880 --> 01:00:58,790
And for another choice of the
orientation of the four-fold
942
01:00:58,790 --> 01:01:01,010
axis, this and this would
have to be equal.
943
01:01:01,010 --> 01:01:03,900
And, finally, this and this
would have to be equal.
944
01:01:03,900 --> 01:01:06,780
So this is what we found, in
fact, when we cranked through
945
01:01:06,780 --> 01:01:11,050
the symmetry transformations,
and this is what we would get
946
01:01:11,050 --> 01:01:14,640
when we said that the quadric
has to have a shape that
947
01:01:14,640 --> 01:01:17,912
conforms to cubic symmetric.
948
01:01:17,912 --> 01:01:22,268
AUDIENCE: My question was in
the case of the arbitrary
949
01:01:22,268 --> 01:01:24,700
relation of those three.
950
01:01:24,700 --> 01:01:28,859
All three degrees of freedom,
are they, in fact, the three
951
01:01:28,859 --> 01:01:30,109
principal axes?
952
01:01:30,109 --> 01:01:32,840
953
01:01:32,840 --> 01:01:34,040
PROFESSOR: For this case here?
954
01:01:34,040 --> 01:01:36,010
AUDIENCE: Yes.
955
01:01:36,010 --> 01:01:38,870
PROFESSOR: For a symmetric
tensor, they are
956
01:01:38,870 --> 01:01:41,190
two principal axes.
957
01:01:41,190 --> 01:01:41,650
OK?
958
01:01:41,650 --> 01:01:45,490
If the tensor is nonsymmetric,
then these two don't have to
959
01:01:45,490 --> 01:01:49,640
be equal any longer, and
they give you the
960
01:01:49,640 --> 01:01:50,890
extra degree of freedom.
961
01:01:50,890 --> 01:01:53,640
962
01:01:53,640 --> 01:01:54,890
OK?
963
01:01:54,890 --> 01:02:00,640
964
01:02:00,640 --> 01:02:01,890
AUDIENCE: What are the
three degrees of
965
01:02:01,890 --> 01:02:03,140
freedom in this case?
966
01:02:03,140 --> 01:02:17,640
967
01:02:17,640 --> 01:02:17,990
PROFESSOR: I just want
you to know I
968
01:02:17,990 --> 01:02:19,240
appreciate your question.
969
01:02:19,240 --> 01:02:23,336
970
01:02:23,336 --> 01:02:24,797
AUDIENCE: May I make
a suggestion?
971
01:02:24,797 --> 01:02:25,771
PROFESSOR: Yeah.
972
01:02:25,771 --> 01:02:27,719
AUDIENCE: I think you need two
degrees of freedom to specify
973
01:02:27,719 --> 01:02:30,641
the first one, and since it's
symmetric and since it's
974
01:02:30,641 --> 01:02:33,439
orthogonal you'd only need one
more to find the second one
975
01:02:33,439 --> 01:02:35,260
and then another to
divide the third.
976
01:02:35,260 --> 01:02:37,525
So that's three degrees of
freedom to find three
977
01:02:37,525 --> 01:02:39,430
principal axes in an
orthogonal system.
978
01:02:39,430 --> 01:02:41,603
PROFESSOR: Not sure I like that,
because this has to be
979
01:02:41,603 --> 01:02:46,395
the shape of the quadric for an
arbitrary rotation angle.
980
01:02:46,395 --> 01:02:56,359
981
01:02:56,359 --> 01:02:58,545
AUDIENCE: Does that angle
between X1 and X2-- does it
982
01:02:58,545 --> 01:03:00,335
have to be 90 degrees if
it's not connected?
983
01:03:00,335 --> 01:03:03,814
PROFESSOR: That's correct,
that's correct.
984
01:03:03,814 --> 01:03:09,030
OK, let's not tie up the whole
audience, let me--
985
01:03:09,030 --> 01:03:11,620
let's talk about this and
I'll think about it too.
986
01:03:11,620 --> 01:03:15,160
But you make a very, very good
point, but let's not settle it
987
01:03:15,160 --> 01:03:16,820
in real time.
988
01:03:16,820 --> 01:03:19,930
We'll throw everybody out, we'll
close the door, and you
989
01:03:19,930 --> 01:03:25,010
can come back and see who
won the scuffle, OK?
990
01:03:25,010 --> 01:03:27,050
OK, so let's take our break,
I think you're more
991
01:03:27,050 --> 01:03:28,300
than ready for it.
992
01:03:28,300 --> 01:03:29,373