WEBVTT
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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
00:08:57.480 --> 00:08:59.760
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
00:09:47.500 --> 00:09:49.640
interpretation.
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If you were indeed to measure
a physical property as a
00:09:53.810 --> 00:09:59.270
function of direction, and you
get a tensor that is a general
00:09:59.270 --> 00:10:02.315
tensor, A11, A12, A13.
00:10:02.315 --> 00:10:07.555
A21, A22, A23.
00:10:07.555 --> 00:10:13.340
A31, A32, A33.
00:10:13.340 --> 00:10:14.590
All the numbers are non-zero.
00:10:16.930 --> 00:10:22.630
You know that the diagonal
values in the tensor are going
00:10:22.630 --> 00:10:25.440
to give you the value of
the property along X1.
00:10:30.650 --> 00:10:33.860
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
00:10:47.310 --> 00:10:51.790
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,
00:11:01.720 --> 00:11:04.890
what is the smallest value
of the property?
00:11:04.890 --> 00:11:09.030
And if the crystal has any
symmetry these directions
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might be the direction of
symmetry axes in the crystal.
00:11:13.480 --> 00:11:15.870
So for many reasons you might
want to know the maximum and
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minimum value of the property.
00:11:18.120 --> 00:11:21.300
And then for some applications
for a real chunk of crystal
00:11:21.300 --> 00:11:25.350
that you've examined relative
to an arbitrary set of axes,
00:11:25.350 --> 00:11:28.700
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
00:11:32.330 --> 00:11:36.730
maximum or minimum thermal
conductivity.
00:11:36.730 --> 00:11:39.710
If you're using a piece of
ceramic let's say, as a probe
00:11:39.710 --> 00:11:43.260
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
00:11:54.430 --> 00:11:57.810
crystal, it'd be a very small
window, but you might want the
00:11:57.810 --> 00:12:02.360
smallest thermal conductivity
normal to the surface, and
00:12:02.360 --> 00:12:04.690
therefore you might want to
make your single crystal
00:12:04.690 --> 00:12:07.280
window oriented in a particular
direction.
00:12:07.280 --> 00:12:10.950
So I hope I've convinced you
with enough straw questions
00:12:10.950 --> 00:12:13.390
that we can knock down easily
that, yes, this would be an
00:12:13.390 --> 00:12:16.340
interesting thing to know.
00:12:16.340 --> 00:12:19.490
So how can we do this?
00:12:19.490 --> 00:12:22.100
Let me show you some simple
geometry that let's us set up
00:12:22.100 --> 00:12:26.990
an equation for finding these
directions right away.
00:12:26.990 --> 00:12:32.820
And it'll be based on the
following observation, that
00:12:32.820 --> 00:12:38.610
when the direction of, let's
say, the applied electric
00:12:38.610 --> 00:12:44.040
field is a long one of the
principal axes, then and only
00:12:44.040 --> 00:12:50.750
then is the direction of the
generalized displacement
00:12:50.750 --> 00:12:55.020
exactly parallel to
the radius vector.
00:12:55.020 --> 00:12:57.390
We go off in any other direction
other than a
00:12:57.390 --> 00:13:00.090
principal axis, the generalized
displacement is
00:13:00.090 --> 00:13:03.190
not parallel to the radius
vector out to the surface of
00:13:03.190 --> 00:13:04.440
the quadric.
00:13:09.970 --> 00:13:22.480
We look at the equation of the
property A11 X1 plus A12 X2
00:13:22.480 --> 00:13:26.030
plus A13 X3.
00:13:26.030 --> 00:13:33.505
This is the X1 component of the
generalized displacement.
00:13:37.840 --> 00:13:46.860
Similarly, A21 X1 plus A22 X2
plus A23 X3 is going to be the
00:13:46.860 --> 00:13:48.110
X2 component.
00:13:51.000 --> 00:13:59.710
And A31 X1 plus A32 X2 plus
A33 times X3, guess what?
00:13:59.710 --> 00:14:04.140
That's going to be the X3
component of the displacement.
00:14:04.140 --> 00:14:09.880
Now we said that these X's give
us the direction of the
00:14:09.880 --> 00:14:14.180
displacement as we let the
coordinates of that point X1,
00:14:14.180 --> 00:14:18.370
X2, X3 roam around the surface
of the quadric.
00:14:18.370 --> 00:14:22.500
When we land at a point on the
surface of the quadric which
00:14:22.500 --> 00:14:27.490
is where a principal axis
emerges, then, and only then,
00:14:27.490 --> 00:14:29.830
is the resulting
vector parallel
00:14:29.830 --> 00:14:31.570
to the applied vector.
00:14:31.570 --> 00:14:40.140
And this means each component of
the resulting vector has to
00:14:40.140 --> 00:14:44.490
be proportional to a vector
out to the point at that
00:14:44.490 --> 00:14:47.260
location, X1, X2, X3.
00:14:47.260 --> 00:14:50.360
So what I'm saying then is
that when we are on a
00:14:50.360 --> 00:14:55.130
principal axis the X1 component
of what happens is
00:14:55.130 --> 00:14:59.490
going to be proportional
to X1.
00:14:59.490 --> 00:15:03.320
And the X2 component of the
vector that happens has to be
00:15:03.320 --> 00:15:06.420
proportional to X2, and this
is the radial vector out to
00:15:06.420 --> 00:15:07.230
that point.
00:15:07.230 --> 00:15:10.940
And, similarly, the X3 component
of what happens has
00:15:10.940 --> 00:15:16.470
to be parallel to the coordinate
X3, the vector out
00:15:16.470 --> 00:15:19.840
to the surface of the
quadric along X3.
00:15:19.840 --> 00:15:24.670
So let me make this an equation,
now, by putting in a
00:15:24.670 --> 00:15:28.430
constant for the unknown
proportionality constant.
00:15:28.430 --> 00:15:32.930
And what I'll do, just for
arbitrary reasons, is call
00:15:32.930 --> 00:15:37.850
that proportionality constant
"lambda." So this, then, is a
00:15:37.850 --> 00:15:43.245
set of equations that will give
me, if I solve for X1,
00:15:43.245 --> 00:15:49.240
X2, and X3, the coordinates of
one of the three points that
00:15:49.240 --> 00:15:53.140
sit out on the surface of the
quadric at the location where
00:15:53.140 --> 00:15:56.840
a principal axis emerges.
00:15:56.840 --> 00:15:59.970
So let me rearrange these
equations to a
00:15:59.970 --> 00:16:02.410
form that I can solve.
00:16:02.410 --> 00:16:07.400
And I'll make the equations
homogeneous by bringing lambda
00:16:07.400 --> 00:16:12.510
X1 over to the left hand side
of the equation, and let me
00:16:12.510 --> 00:16:14.740
point out that this
is the same lambda
00:16:14.740 --> 00:16:16.730
in all three equations.
00:16:16.730 --> 00:16:20.340
There's a proportionality
constant between the radial
00:16:20.340 --> 00:16:28.140
vector out to the surface of the
quadric and each component
00:16:28.140 --> 00:16:30.210
of the vector that results.
00:16:30.210 --> 00:16:38.020
So my set of equations will be
A11 minus lambda X1 plus A12
00:16:38.020 --> 00:16:45.850
times X2 plus A13 times X3, and
that's now equal to zero
00:16:45.850 --> 00:16:46.800
on the right.
00:16:46.800 --> 00:16:51.670
Next equation would
be A21 X1--
00:16:51.670 --> 00:16:56.590
and notice I'm not combining A12
and A21 into numerically
00:16:56.590 --> 00:16:59.520
the same constant, representing
it by the same
00:16:59.520 --> 00:17:01.990
quantity, I'm just leaving
them be separate and
00:17:01.990 --> 00:17:03.830
independent at this point.
00:17:03.830 --> 00:17:07.900
And then the middle term here
is A22 minus lambda times X2
00:17:07.900 --> 00:17:12.599
plus A23 times X3 equals 0.
00:17:12.599 --> 00:17:15.060
And the third equation, you can
anticipate how that turns
00:17:15.060 --> 00:17:24.940
out, it's A31 X1 plus A32
times X2 plus A33 minus
00:17:24.940 --> 00:17:31.560
lambda, and that's
equal to zero.
00:17:31.560 --> 00:17:38.220
So this, then, is a set of
linear equations and they're
00:17:38.220 --> 00:17:39.710
called homogeneous equations.
00:17:47.270 --> 00:17:50.980
And that has solved our problem
for us, all we have to
00:17:50.980 --> 00:17:54.210
do is solve for X1, X2, X3.
00:17:54.210 --> 00:18:00.140
Except that if the nine
coefficients Aij are all
00:18:00.140 --> 00:18:05.360
arbitrary and lambda is a
constant this set of equations
00:18:05.360 --> 00:18:13.850
has only one solution, and that
solution is X1 equals X2
00:18:13.850 --> 00:18:18.530
equals X3 equals 0, and that
indeed will wipe out every one
00:18:18.530 --> 00:18:19.630
of these lines.
00:18:19.630 --> 00:18:21.820
And that is not a very
interesting solution.
00:18:24.380 --> 00:18:31.350
The only case in which this is
not the only solution is that
00:18:31.350 --> 00:18:34.470
if the condition that the
determinant of the
00:18:34.470 --> 00:18:40.710
coefficients is equal to zero,
then we can find a real set of
00:18:40.710 --> 00:18:42.220
X1, X2, X3.
00:18:42.220 --> 00:18:46.070
So the condition that a solution
exists is that A11
00:18:46.070 --> 00:19:01.046
minus lambda A12 A13, and A21
A22 minus lambda A23, A31 A32
00:19:01.046 --> 00:19:05.590
A33 minus lambda, this
determinant has
00:19:05.590 --> 00:19:06.840
to be equal to zero.
00:19:12.090 --> 00:19:14.540
All right, so we know how to
expand the determinant, we'll
00:19:14.540 --> 00:19:17.750
do a little number crunching,
and I'm not going to write it
00:19:17.750 --> 00:19:23.450
down explicitly, but we'll have
a term A11 minus lambda
00:19:23.450 --> 00:19:28.690
and this will be among other
terms in the expansion A22
00:19:28.690 --> 00:19:35.870
minus lambda A23 A32 A33 minus
lambda and then there'll be
00:19:35.870 --> 00:19:39.960
another determinant that
involves this term A12 plus
00:19:39.960 --> 00:19:43.980
its cofactor plus A13
times its cofactor.
00:19:43.980 --> 00:19:46.560
Unfortunately none of these
terms are zero in general, so
00:19:46.560 --> 00:19:48.900
I'm going to have three
terms here.
00:19:48.900 --> 00:19:52.830
If I expand this term, I'm going
to get something in A11
00:19:52.830 --> 00:20:01.590
minus lambda A22 minus lambda
times A33 minus lambda and a
00:20:01.590 --> 00:20:04.640
bunch of other terms that I
won't bother to write down
00:20:04.640 --> 00:20:05.890
explicitly.
00:20:08.070 --> 00:20:14.360
The only point I want to draw at
this particular juncture is
00:20:14.360 --> 00:20:19.300
to note that this is a
third rank equation.
00:20:23.160 --> 00:20:24.410
In lambda.
00:20:26.940 --> 00:20:28.190
It's a cubic equation.
00:20:31.380 --> 00:20:33.880
And what this means
is that there are
00:20:33.880 --> 00:20:35.150
going to be three roots.
00:20:40.080 --> 00:20:44.000
And lets call them lambda 1,
lambda 2, and lambda 3.
00:20:48.180 --> 00:20:49.810
And that's the way
it should be.
00:20:49.810 --> 00:20:55.010
There are three principle axes,
so we should have three
00:20:55.010 --> 00:20:59.050
values of the constant lambda
that we could put into this
00:20:59.050 --> 00:21:02.720
equation that will let us
solve explicitly for the
00:21:02.720 --> 00:21:06.670
coordinates X1, X2, and X3,
which sits on the surface of
00:21:06.670 --> 00:21:09.920
the quadric where our principal
axis pokes out.
00:21:09.920 --> 00:21:13.000
Three different principal axes,
so this should be three
00:21:13.000 --> 00:21:18.770
solutions to the third
rank equation that
00:21:18.770 --> 00:21:20.020
we've defined here.
00:21:23.090 --> 00:21:27.120
OK I think you've all seen this
sort of problem before.
00:21:27.120 --> 00:21:30.435
The lambdas are called
characteristic values.
00:21:38.930 --> 00:21:42.640
Or, on the basis that things
sound more impressive in
00:21:42.640 --> 00:21:52.770
German, these are called
the eigenvalues,
00:21:52.770 --> 00:21:55.210
meaning the same thing.
00:21:55.210 --> 00:21:59.450
So using nothing more than the
properties of the quadric and
00:21:59.450 --> 00:22:03.040
some linear algebra we have
stumbled headlong into an
00:22:03.040 --> 00:22:06.150
eigenvalue problem, which
is a well known sort of
00:22:06.150 --> 00:22:09.700
mathematical problem associated
with a large number
00:22:09.700 --> 00:22:12.980
of physical situations and
a lot to do with physical
00:22:12.980 --> 00:22:16.520
properties in particular.
00:22:16.520 --> 00:22:19.570
OK, now I have a question
for you.
00:22:19.570 --> 00:22:22.180
How we going to solve
this silly thing?
00:22:22.180 --> 00:22:22.950
I know--
00:22:22.950 --> 00:22:27.260
well the answer is I poke it
into my laptop, and out comes
00:22:27.260 --> 00:22:28.210
the answer.
00:22:28.210 --> 00:22:31.450
But that's not satisfying, we
should know how to do this if
00:22:31.450 --> 00:22:35.834
our batteries were dead and we
needed the answer in a hurry.
00:22:35.834 --> 00:22:38.530
Well, I know how to solve
a second rank question.
00:22:38.530 --> 00:22:45.500
If I have equation of the form
ax squared plus bx plus c
00:22:45.500 --> 00:22:49.250
equals 0, I know the
solution to that.
00:22:49.250 --> 00:22:52.800
It's etched indelibly in my
memory and it says that x
00:22:52.800 --> 00:22:57.080
equals minus b plus or minus the
square root of b squared
00:22:57.080 --> 00:23:01.190
minus 4ac all over 2a.
00:23:01.190 --> 00:23:02.450
Impressed you, didn't I?
00:23:02.450 --> 00:23:05.210
You know why I remember that?
00:23:05.210 --> 00:23:11.030
I learned high school algebra
from a throwback--
00:23:11.030 --> 00:23:15.050
a teacher who was a throwback
to the Elizabethan period.
00:23:15.050 --> 00:23:18.770
You know how when people go to
a remote island they suddenly
00:23:18.770 --> 00:23:21.250
find an example of
a species that
00:23:21.250 --> 00:23:23.700
everyone thought was extinct?
00:23:23.700 --> 00:23:30.150
Well, to secondary education
[? Ms. Dourier ?]
00:23:30.150 --> 00:23:34.410
was a species that had long gone
extinct, but happened to
00:23:34.410 --> 00:23:37.820
be preserved at the high
school that I attended.
00:23:37.820 --> 00:23:38.750
[? Ms. Dourier ?]
00:23:38.750 --> 00:23:44.600
was a very tall woman, ramrod
straight, with a pile of snow
00:23:44.600 --> 00:23:46.950
white hair on the
top of her head.
00:23:46.950 --> 00:23:51.090
She invariably dressed in a
long, black skirt, and a white
00:23:51.090 --> 00:23:55.910
blouse that had puffy sleeves,
and a collar with frilly
00:23:55.910 --> 00:23:59.930
things on the top, and,
invariably, a black ribbon
00:23:59.930 --> 00:24:03.820
around the frilly
lace collar top.
00:24:03.820 --> 00:24:09.530
Her mode of enlightened
education was to have all of
00:24:09.530 --> 00:24:14.170
the students in the class have
a notebook that was open.
00:24:14.170 --> 00:24:20.160
And one hapless member of the
class was called upon to solve
00:24:20.160 --> 00:24:24.310
a problem in real time, and as
the student recited we all
00:24:24.310 --> 00:24:27.120
wrote down what the student was
saying in our notebooks.
00:24:27.120 --> 00:24:28.960
All the while [? Ms. Dourier, ?]
00:24:28.960 --> 00:24:32.730
holding a ruler like a swagger
stick, strode up and down the
00:24:32.730 --> 00:24:37.080
aisles, and woe be unto the
poor student who faltered
00:24:37.080 --> 00:24:40.660
slightly, let alone get
something wrong.
00:24:40.660 --> 00:24:45.780
And that would bring a slap of
the ruler down on the desk and
00:24:45.780 --> 00:24:48.450
a sigh of disgust, and
then, all right,
00:24:48.450 --> 00:24:50.100
you take it, Audrey.
00:24:50.100 --> 00:24:53.240
And then Audrey would begin to
recite until she screwed up
00:24:53.240 --> 00:24:56.710
and we would all copy
in our notebooks.
00:24:56.710 --> 00:25:04.920
That tyrant so terrorized and
intimidated a bunch of kids in
00:25:04.920 --> 00:25:09.760
a redneck, working class
high school that we
00:25:09.760 --> 00:25:11.220
really learned algebra.
00:25:11.220 --> 00:25:16.960
So to this very day, as a reflex
action, minus b plus or
00:25:16.960 --> 00:25:21.140
minus the square root of b
squared minus 4ac all over 2a.
00:25:21.140 --> 00:25:24.410
But I don't have the foggiest
idea how to solve a third
00:25:24.410 --> 00:25:27.050
order equation until
I look it up.
00:25:29.560 --> 00:25:33.390
You probably never had to do
it, so let me, for your
00:25:33.390 --> 00:25:36.370
general education and amusement,
pass around a sheet
00:25:36.370 --> 00:25:41.975
that tells you how to solve
a third rank equation.
00:25:46.520 --> 00:25:51.290
OK the first step is to convert
an equation in, let's
00:25:51.290 --> 00:25:57.020
say, y squared plus y cubed plus
py squared plus qy plus r
00:25:57.020 --> 00:26:01.010
equals 0 to a so-called normal
form where you get rid of one
00:26:01.010 --> 00:26:02.770
of the terms.
00:26:02.770 --> 00:26:07.980
So if you make a substitution
of variables and let y be
00:26:07.980 --> 00:26:12.590
replaced by x minus the
coefficient p/3, then you get
00:26:12.590 --> 00:26:17.440
an equation that has a
cubic term, linear
00:26:17.440 --> 00:26:20.040
term x, and a constant.
00:26:20.040 --> 00:26:26.370
And the constants a and b are
combinations of p and q and r
00:26:26.370 --> 00:26:28.670
in the original equation.
00:26:28.670 --> 00:26:32.090
And then the solutions, not
well known to be sure, are
00:26:32.090 --> 00:26:36.470
there's a first solution x is
equal to a plus b, and then,
00:26:36.470 --> 00:26:43.720
something messy, x2 and x3 are
minus 1/2 of a plus b plus or
00:26:43.720 --> 00:26:46.560
minus an imaginary term--
those are the
00:26:46.560 --> 00:26:48.100
second and third roots--
00:26:48.100 --> 00:26:51.610
i times the square root of 3
over 2a minus b where A and B
00:26:51.610 --> 00:26:55.420
are complicated functions
of a and b.
00:26:55.420 --> 00:26:57.275
Capital A and capital B are
functions of little
00:26:57.275 --> 00:26:58.690
a and little b.
00:26:58.690 --> 00:27:02.990
And then if the coefficients in
the original equation are
00:27:02.990 --> 00:27:06.070
real, then you get
one real and two
00:27:06.070 --> 00:27:08.070
conjugated imaginary roots.
00:27:08.070 --> 00:27:11.920
If this combination of terms
b squared and a squared are
00:27:11.920 --> 00:27:18.200
greater than 0, it is b squared
over 4 plus a cubed
00:27:18.200 --> 00:27:19.850
over 27 equals 0.
00:27:19.850 --> 00:27:22.520
If it's that, then you get
three real roots, two
00:27:22.520 --> 00:27:23.820
of which are equal.
00:27:23.820 --> 00:27:26.590
In the most general case that we
would be encountering here
00:27:26.590 --> 00:27:32.260
is b squared over 4 plus a cubed
over 27 is less than 0,
00:27:32.260 --> 00:27:35.300
and then there are three
real unequal roots.
00:27:38.100 --> 00:27:44.570
Unfortunately, if b squared over
4 plus a cubed over 27 is
00:27:44.570 --> 00:27:49.150
negative that term appears
inside of a square root sign
00:27:49.150 --> 00:27:51.870
in the solutions.
00:27:51.870 --> 00:27:55.500
So the only case that we'd
really be interested in
00:27:55.500 --> 00:27:57.430
doesn't work for the solution.
00:27:57.430 --> 00:27:59.760
But fortunately there's another
solution and that's
00:27:59.760 --> 00:28:02.930
given at the bottom of the page,
and if you really wanted
00:28:02.930 --> 00:28:07.910
to solve a third rank equation
by hand, these solutions would
00:28:07.910 --> 00:28:10.560
do it for you.
00:28:10.560 --> 00:28:13.750
But I think you'd much prefer
to have your computer solve
00:28:13.750 --> 00:28:17.890
the equations for you, but let
me show you a way of doing
00:28:17.890 --> 00:28:25.090
this without any computer,
without solving any equations.
00:28:25.090 --> 00:28:31.550
And it is a very clever method
of successive approximations
00:28:31.550 --> 00:28:33.600
that's based on the properties
of the quadric.
00:28:38.470 --> 00:28:43.560
So let's suppose we have a
tensor, and that tensor has
00:28:43.560 --> 00:28:50.640
some set of coefficients Aij,
all of which are non-zero.
00:28:50.640 --> 00:28:53.630
And I'll assume although this
will work for other quadratic
00:28:53.630 --> 00:28:59.690
surfaces that the quadric has
the shape of an ellipsoid.
00:29:05.294 --> 00:29:09.400
All right let me now pick some
direction at random, actually
00:29:09.400 --> 00:29:11.840
I don't have to pick it at
random, but I'll show you what
00:29:11.840 --> 00:29:15.160
the shrewd first
guess would be.
00:29:15.160 --> 00:29:19.380
So let's say we picked
this direction.
00:29:19.380 --> 00:29:25.650
And let us find the direction
if this is--
00:29:25.650 --> 00:29:28.850
let's do it in terms of current
and conductivity.
00:29:28.850 --> 00:29:34.110
Let's let this be the first
guess for the applied field.
00:29:34.110 --> 00:29:37.690
That's clearly not going to be
one of the principal axes, and
00:29:37.690 --> 00:29:40.250
let this be the first resulting
direction of
00:29:40.250 --> 00:29:41.500
current flow J1.
00:29:44.190 --> 00:29:51.600
So what I'm finding then is a
J sub i in terms of an Aij
00:29:51.600 --> 00:29:56.060
times an E sub J, and this is my
first result, and this was
00:29:56.060 --> 00:29:58.590
my first assumption.
00:29:58.590 --> 00:30:05.380
Let me now let the direction of
J be taken as the direction
00:30:05.380 --> 00:30:09.150
of the applied field
for a second guess.
00:30:09.150 --> 00:30:12.130
And we could normalize to a unit
vector, but we don't even
00:30:12.130 --> 00:30:13.020
have to do that.
00:30:13.020 --> 00:30:16.710
So let's simply say that my
second guess for the electric
00:30:16.710 --> 00:30:19.900
field that I hope will point
along one principal axis is
00:30:19.900 --> 00:30:23.390
E2, and E2 I'll take
as identical--
00:30:23.390 --> 00:30:24.460
let's say proportional--
00:30:24.460 --> 00:30:28.370
to J1 from my original guess.
00:30:28.370 --> 00:30:32.516
So this then would be E2,
the second guess.
00:30:32.516 --> 00:30:34.900
And if I find the direction
to the normal--
00:30:34.900 --> 00:30:37.410
to the quadric in that
direction, this would be my
00:30:37.410 --> 00:30:44.070
second result for the current
flow J, I should have put this
00:30:44.070 --> 00:30:46.990
in parentheses to indicate that
these are not components
00:30:46.990 --> 00:30:49.140
of E and J.
00:30:49.140 --> 00:30:51.330
And you can see what's
happening, look at where I am,
00:30:51.330 --> 00:30:55.650
I started out here after just
one iteration I have defined
00:30:55.650 --> 00:30:58.800
this as the potential direction
of one of the
00:30:58.800 --> 00:31:00.480
principal axes.
00:31:00.480 --> 00:31:02.540
So I'm going to find E1--
00:31:02.540 --> 00:31:03.910
the direction of the field--
00:31:03.910 --> 00:31:08.955
E1 that comes from sigma ij
times Ji, I'll take my second
00:31:08.955 --> 00:31:19.250
guess E2 either as identical
to J1 and this is going to
00:31:19.250 --> 00:31:24.920
give me then a new value for
my second iteration, this
00:31:24.920 --> 00:31:27.730
would be J2.
00:31:27.730 --> 00:31:34.330
And this process is going to
converge very, very rapidly on
00:31:34.330 --> 00:31:35.730
the shortest principal axis.
00:31:45.050 --> 00:31:47.940
As you can see in two shots
I'm pretty close to being
00:31:47.940 --> 00:31:51.100
parallel to this direction,
which would be X2 in my
00:31:51.100 --> 00:31:52.350
illustration.
00:31:54.285 --> 00:31:57.410
Now if I wanted to-- if I wanted
to see if I was close
00:31:57.410 --> 00:31:58.420
to convergence--
00:31:58.420 --> 00:32:02.330
this is going to be awkward
because if I don't normalize
00:32:02.330 --> 00:32:06.030
the magnitude of the resulting
vector, J is going to get
00:32:06.030 --> 00:32:07.710
larger and larger and larger.
00:32:07.710 --> 00:32:11.060
So periodically I would
want to normalize--
00:32:11.060 --> 00:32:13.770
take the magnitude of J,
divide that into the
00:32:13.770 --> 00:32:16.060
components, and then I'd
have a unit vector.
00:32:16.060 --> 00:32:20.740
If I wrote a computer program
to do this, I would do the
00:32:20.740 --> 00:32:24.300
normalization each time
to test convergence.
00:32:24.300 --> 00:32:29.580
Now, this is my kind of
procedure, because if I'm
00:32:29.580 --> 00:32:33.470
doing this by hand and I screw
up and I make a mistake and my
00:32:33.470 --> 00:32:37.190
answer is thrown off a little
bit, if I continue to iterate,
00:32:37.190 --> 00:32:42.940
the thing will proceed to
continue to converge to the
00:32:42.940 --> 00:32:46.310
shortest principal axis.
00:32:46.310 --> 00:32:48.990
So I can make a mistake and it
will correct itself and come
00:32:48.990 --> 00:32:51.160
back again, and that's
my kind of solution.
00:32:51.160 --> 00:32:52.020
Yeah, Jason?
00:32:52.020 --> 00:32:53.500
AUDIENCE: It's going to give you
the minimum value of the
00:32:53.500 --> 00:32:54.980
property, right?
00:32:54.980 --> 00:32:55.924
PROFESSOR: No.
00:32:55.924 --> 00:32:59.000
It's going to give you the
minimum principal axis, that's
00:32:59.000 --> 00:33:02.620
going to be the maximum
value of the property.
00:33:02.620 --> 00:33:05.520
So that's one out of three,
hey, that's not bad.
00:33:05.520 --> 00:33:08.010
What do I do for the others?
00:33:08.010 --> 00:33:16.650
Let me tell you without
proof to find the
00:33:16.650 --> 00:33:18.215
maximum principal axis.
00:33:25.760 --> 00:33:28.470
And that would be the minimum
value of the property.
00:33:40.260 --> 00:33:43.530
What you would do is exactly
the same procedure, and you
00:33:43.530 --> 00:33:56.060
would operate on not the tensor,
but on-- using as a
00:33:56.060 --> 00:33:58.270
matrix of coefficients--
00:33:58.270 --> 00:34:02.435
the matrix of the
tensor inverted.
00:34:08.280 --> 00:34:14.040
And I assume you know how to
find the inverse of a matrix,
00:34:14.040 --> 00:34:17.110
except that you don't have to
even find the inverse, the
00:34:17.110 --> 00:34:19.920
inverse is going to be a
collection of functions of the
00:34:19.920 --> 00:34:22.610
original elements divided
by the determinant.
00:34:22.610 --> 00:34:25.000
You don't have to worry about
normalizing, all that's
00:34:25.000 --> 00:34:27.245
important is the relative value
of these coefficients.
00:34:27.245 --> 00:34:28.090
Yes, sir?
00:34:28.090 --> 00:34:30.719
AUDIENCE: If your tensor is
symmetric, can't you say that
00:34:30.719 --> 00:34:32.870
they'll just [INAUDIBLE]
on to another?
00:34:32.870 --> 00:34:34.310
PROFESSOR: That's what
you're going to do.
00:34:34.310 --> 00:34:36.505
But when you know just
one, that won't work.
00:34:36.505 --> 00:34:38.822
The other two are floating
around somewhere and you don't
00:34:38.822 --> 00:34:39.870
know in what direction.
00:34:39.870 --> 00:34:45.340
So if you do this, then you have
two out of the three and
00:34:45.340 --> 00:34:49.590
now very shrewdly, as you point
out, if we have two of
00:34:49.590 --> 00:34:53.969
the principal axes and the
tensor is symmetric, then we
00:34:53.969 --> 00:34:56.910
can automatically get the
direction of the third.
00:34:56.910 --> 00:35:02.370
If it's not symmetric, then
you have to use the set of
00:35:02.370 --> 00:35:06.360
equations, and you have three
principal axes as
00:35:06.360 --> 00:35:10.140
eigenvectors, as they're called
in eigenvalue problems.
00:35:10.140 --> 00:35:13.440
And they do not have to be
orthogonal, but you have the
00:35:13.440 --> 00:35:17.740
components in a Cartesian
coordinate system so you can
00:35:17.740 --> 00:35:20.836
find the angle between
those axes.
00:35:20.836 --> 00:35:21.250
OK?
00:35:21.250 --> 00:35:22.480
Yes, sir.
00:35:22.480 --> 00:35:25.286
AUDIENCE: With respect to
whether the matrix is
00:35:25.286 --> 00:35:27.360
symmetrical or is
not symmetric,
00:35:27.360 --> 00:35:29.520
the quadric is always--
00:35:29.520 --> 00:35:32.400
has three principal axes
at right angles right?
00:35:32.400 --> 00:35:36.720
PROFESSOR: Only if the
tensor is symmetric.
00:35:36.720 --> 00:35:39.130
Only if the tensor
is symmetric.
00:35:39.130 --> 00:35:44.340
And let me put your question
aside for about five minutes,
00:35:44.340 --> 00:35:47.640
and I want to take an overview
of what we've learned about
00:35:47.640 --> 00:35:51.160
the geometry of the quadric and
the symmetry restrictions
00:35:51.160 --> 00:35:54.460
that we can impose on the tensor
in a formal fashion,
00:35:54.460 --> 00:35:57.100
and see how they compare
and how one allows an
00:35:57.100 --> 00:35:59.600
interpretation of the other.
00:35:59.600 --> 00:36:02.110
So I'm going to answer your
question, I'd like to wait for
00:36:02.110 --> 00:36:04.880
about three minutes.
00:36:04.880 --> 00:36:07.520
Other questions?
00:36:07.520 --> 00:36:10.930
OK, again this is all very
symbolic, I'm not giving you
00:36:10.930 --> 00:36:13.370
an explicit answer,
I can't do it.
00:36:13.370 --> 00:36:14.960
But what I can do you--
00:36:14.960 --> 00:36:16.980
do for you is, ha
ha, do to you.
00:36:16.980 --> 00:36:19.870
I almost slipped and said that,
what I can't do to you,
00:36:19.870 --> 00:36:24.800
is give you a problem that
asks you to solve for the
00:36:24.800 --> 00:36:28.560
principal axes, for example
of a property tensor.
00:36:28.560 --> 00:36:31.955
I will be merciful and perhaps
not have all nine coefficients
00:36:31.955 --> 00:36:34.320
non-zero, I will have
one of them zero, or
00:36:34.320 --> 00:36:36.590
maybe two of them zero.
00:36:36.590 --> 00:36:42.190
So again to try this it is very
instructive and it will
00:36:42.190 --> 00:36:45.200
cement what the individual steps
that I performed here
00:36:45.200 --> 00:36:48.820
actually involve.
00:36:48.820 --> 00:36:50.940
Now, a question that I thought
you were going to ask about
00:36:50.940 --> 00:36:57.640
principal axes is that in order
to do what I did here I
00:36:57.640 --> 00:37:02.280
am using the radius-normal
property, and the
00:37:02.280 --> 00:37:08.580
radius-normal property only
works for a symmetric tensor.
00:37:08.580 --> 00:37:12.850
So sounds like I'm swindling
you, except that if I use this
00:37:12.850 --> 00:37:19.910
procedure, the radius-normal
won't actually be identically
00:37:19.910 --> 00:37:24.390
parallel to the generalized
displacement.
00:37:24.390 --> 00:37:27.960
But it's not going to be
terribly different from it,
00:37:27.960 --> 00:37:30.600
and I'm going to get something
that again will converge
00:37:30.600 --> 00:37:32.940
towards the shortest
principal axis.
00:37:32.940 --> 00:37:36.750
And once I'm close to the
principal axis it is true--
00:37:36.750 --> 00:37:38.500
symmetric or not--
00:37:38.500 --> 00:37:43.880
that the only three directions
before which the generalized
00:37:43.880 --> 00:37:46.570
displacement is parallel to
the radius vector are the
00:37:46.570 --> 00:37:48.990
principal axes.
00:37:48.990 --> 00:37:52.280
But anyway this is a dicey
situation if the tensor is not
00:37:52.280 --> 00:37:56.530
symmetric, and it probably comes
as a great relief to
00:37:56.530 --> 00:38:00.710
learn that there's really only
one property tensor of second
00:38:00.710 --> 00:38:04.060
rank that's known for sure
to be non-symmetric.
00:38:04.060 --> 00:38:06.440
So in most cases--
00:38:06.440 --> 00:38:09.560
99 out of 100, if not more--
we will be dealing with
00:38:09.560 --> 00:38:12.350
property tensors that are
symmetric, so we could use
00:38:12.350 --> 00:38:13.600
this procedure.
00:38:16.240 --> 00:38:19.730
So again this property converges
remarkably well, and
00:38:19.730 --> 00:38:23.230
as I say it has the admirable
quality of being
00:38:23.230 --> 00:38:26.040
self-correcting if you make a
mistake, if you're grinding
00:38:26.040 --> 00:38:27.410
this out by hand.
00:38:27.410 --> 00:38:32.110
But a computer program that
you write that just simply
00:38:32.110 --> 00:38:35.600
takes the direction of J,
normalizes it to a unit
00:38:35.600 --> 00:38:40.420
vector, applies that as the
generalized force, finds a
00:38:40.420 --> 00:38:44.020
quote "J" as a second iteration,
normalizes that,
00:38:44.020 --> 00:38:47.400
and then you can check to
whatever degree of convergence
00:38:47.400 --> 00:38:49.330
you wish to have in
your solution.
00:38:49.330 --> 00:38:52.320
And it's a simple matter to set
up a program to do this.
00:38:56.487 --> 00:39:02.390
All right, I think we still have
some time, so let me now
00:39:02.390 --> 00:39:04.650
put everything together.
00:39:04.650 --> 00:39:12.630
And look at what we've seen of
the geometric properties of
00:39:12.630 --> 00:39:20.080
the quadric, and the tensor
arrays that we found for
00:39:20.080 --> 00:39:22.340
different symmetries.
00:39:22.340 --> 00:39:29.520
For triclinic crystals, for
which the recommended
00:39:29.520 --> 00:39:32.940
procedure is to leave by the
nearest exit and work on
00:39:32.940 --> 00:39:40.340
something different instead, you
would have nine elements
00:39:40.340 --> 00:39:46.210
altogether that are necessary
to define the property.
00:39:53.830 --> 00:39:57.690
I'll discuss this in terms
of tensors that
00:39:57.690 --> 00:40:00.870
gives you an ellipsoid.
00:40:00.870 --> 00:40:05.050
The argument is not different
for hyperboloids of one or two
00:40:05.050 --> 00:40:08.640
sheets, and we can't draw
imaginary ellipsoids, but an
00:40:08.640 --> 00:40:12.510
ellipsoid is a much easier
thing to draw.
00:40:12.510 --> 00:40:15.910
OK, triclinic symmetries, either
one or one bar, nine
00:40:15.910 --> 00:40:17.160
elements in the tensor.
00:40:21.700 --> 00:40:28.010
No constraints whatsoever on the
shape of the quadric or on
00:40:28.010 --> 00:40:32.150
its orientation relative to
our coordinate system.
00:40:32.150 --> 00:40:36.960
So, let's total up now looking
at the quadric how many
00:40:36.960 --> 00:40:39.650
degrees of freedom there
are in this situation.
00:40:39.650 --> 00:40:46.710
We have three principal
axes, so those are
00:40:46.710 --> 00:40:48.910
three degrees of freedom.
00:40:48.910 --> 00:40:54.360
The orientation of the quadric
is arbitrary, so there are
00:40:54.360 --> 00:40:57.080
three orientational degrees
of freedom.
00:41:08.790 --> 00:41:11.635
And that comes out to six.
00:41:14.360 --> 00:41:16.320
Supposedly nine degrees
of freedom, what
00:41:16.320 --> 00:41:19.190
are the other three?
00:41:19.190 --> 00:41:22.340
Again, if this is a general
tensor, which is
00:41:22.340 --> 00:41:24.840
non-symmetric, the
eigenvectors--
00:41:27.720 --> 00:41:32.330
the principal axes that you have
to choose to squeeze this
00:41:32.330 --> 00:41:34.550
thing into a diagonal form--
00:41:34.550 --> 00:41:37.360
become non-orthogonal.
00:41:37.360 --> 00:41:38.280
OK?
00:41:38.280 --> 00:41:43.860
So you have another three
degrees of freedom that
00:41:43.860 --> 00:41:49.160
specify the mutual directions
of the eigenvectors--
00:41:53.690 --> 00:41:55.140
the angles between them.
00:41:57.970 --> 00:41:58.380
OK?
00:41:58.380 --> 00:42:02.720
The directions in which you have
to pick your coordinate
00:42:02.720 --> 00:42:10.080
system, X1, X2, and X3, that
force this thing into a
00:42:10.080 --> 00:42:13.520
diagonal form is going to
involve a coordinate system in
00:42:13.520 --> 00:42:18.820
which these three angles are not
90 degrees and are fixed
00:42:18.820 --> 00:42:20.640
if you're going to squeeze
this thing
00:42:20.640 --> 00:42:21.895
into a diagonal form.
00:42:26.690 --> 00:42:27.300
And that's it.
00:42:27.300 --> 00:42:30.595
So we add these three interaxial
angles in, they're
00:42:30.595 --> 00:42:34.150
a total of nine degrees of
freedom for a general non
00:42:34.150 --> 00:42:35.400
symmetric tensor.
00:42:41.860 --> 00:42:45.800
To convince you that these
interaxial angles for the
00:42:45.800 --> 00:42:49.410
eigenvectors really are
variables, let me tell you
00:42:49.410 --> 00:42:53.400
something that we may prove
later as a recreational
00:42:53.400 --> 00:42:56.300
exercise, or I may give it to
you on a problem set, it's not
00:42:56.300 --> 00:42:57.600
difficult to prove.
00:42:57.600 --> 00:43:06.470
And that is the result that
a symmetric tensor remains
00:43:06.470 --> 00:43:13.660
symmetric for any arbitrary
transformation of axes.
00:43:16.720 --> 00:43:22.320
That is from one Cartesian
coordinate system to another.
00:43:48.888 --> 00:43:55.250
Now a very salutary effect of
this is that the tensor has
00:43:55.250 --> 00:43:56.500
nine elements.
00:44:06.900 --> 00:44:08.620
And if the tensor
is symmetric--
00:44:08.620 --> 00:44:10.860
if you want to transform
that tensor--
00:44:10.860 --> 00:44:14.870
you only have to do the
off-diagonal terms, because
00:44:14.870 --> 00:44:17.750
whatever it turns into when
you change the axes, these
00:44:17.750 --> 00:44:20.060
off-diagonal terms are going
to be equal to the
00:44:20.060 --> 00:44:22.746
off-diagonal terms in
the new tensor.
00:44:22.746 --> 00:44:23.170
OK?
00:44:23.170 --> 00:44:25.040
So this means that only six--
00:44:25.040 --> 00:44:27.810
if the tensor was symmetric in
one coordinate system, only
00:44:27.810 --> 00:44:31.750
six elements have to
be transformed.
00:44:31.750 --> 00:44:33.870
Actually, it's better than that,
you don't have to do
00:44:33.870 --> 00:44:36.770
six, you only have to do five.
00:44:36.770 --> 00:44:41.770
And this is the last diagonal
term that can be obtained from
00:44:41.770 --> 00:44:43.020
the trace of the tensor.
00:44:50.460 --> 00:44:55.620
And that would be the trace of
the tensor T prime after
00:44:55.620 --> 00:45:03.040
transformation because A11 prime
plus A22 prime plus A33
00:45:03.040 --> 00:45:07.360
prime has to be equal to the
original trace T, which was
00:45:07.360 --> 00:45:11.680
A11 plus A22 plus A33 in
the original system.
00:45:14.960 --> 00:45:17.480
So if the tensor is symmetric,
you really have to crank
00:45:17.480 --> 00:45:22.200
through a transformation for
only five of the nine terms.
00:45:22.200 --> 00:45:25.150
Now what is the relevance of
this to what I just said?
00:45:25.150 --> 00:45:31.057
If we have the tensor
diagonalized, to a new form
00:45:31.057 --> 00:45:41.380
A11, 0, 0, 0, A22 prime, 0,
0, 0, A33 prime, that is a
00:45:41.380 --> 00:45:45.180
special case of a symmetric
tensor.
00:45:45.180 --> 00:45:48.450
So if you decide to go from
the coordinate system that
00:45:48.450 --> 00:45:52.270
produced this diagonalized form
to some other coordinate
00:45:52.270 --> 00:45:55.130
system, the new tensor that
you're going to get is going
00:45:55.130 --> 00:45:56.530
to be symmetric.
00:45:56.530 --> 00:46:00.225
But the thing was not originally
symmetric, so how
00:46:00.225 --> 00:46:02.010
can that be?
00:46:02.010 --> 00:46:05.140
The answer is you can put it
in diagonal form only in a
00:46:05.140 --> 00:46:07.740
non-Cartesian coordinate
system.
00:46:07.740 --> 00:46:12.150
And, therefore, that
is evidence that
00:46:12.150 --> 00:46:14.540
the reference axes--
00:46:14.540 --> 00:46:17.680
the principal axes-- cannot
be orthogonal.
00:46:17.680 --> 00:46:20.600
Otherwise you could take the
diagonalized tensor and crank
00:46:20.600 --> 00:46:23.690
it back to some arbitrary
Cartesian coordinate, and
00:46:23.690 --> 00:46:26.080
suddenly it would go to a
symmetric tensor when it was
00:46:26.080 --> 00:46:28.674
not symmetric to begin with.
00:46:28.674 --> 00:46:29.500
OK?
00:46:29.500 --> 00:46:30.820
QED.
00:46:30.820 --> 00:46:33.950
So you can diagonalize a general
tensor only if you
00:46:33.950 --> 00:46:37.940
take the axes along the
eigenvectors, which cannot be
00:46:37.940 --> 00:46:39.960
orthogonal.
00:46:39.960 --> 00:46:42.940
OK, I saw you raise your hand,
I wasn't ignoring you.
00:46:42.940 --> 00:46:43.695
That was it?
00:46:43.695 --> 00:46:44.455
Good.
00:46:44.455 --> 00:46:45.425
AUDIENCE: I wanted to make
sure that Cartesian
00:46:45.425 --> 00:46:46.675
[INAUDIBLE]
00:46:49.310 --> 00:46:51.610
PROFESSOR: It's always nice to
see the class one step ahead
00:46:51.610 --> 00:46:52.860
of what you're doing.
00:46:56.786 --> 00:47:01.290
All right, let's look at a
couple of more crystal systems
00:47:01.290 --> 00:47:05.705
in the form of tensors in those
systems and show that
00:47:05.705 --> 00:47:09.680
that in fact does correspond to
the geometric constraints
00:47:09.680 --> 00:47:11.070
on the quadric as well.
00:47:11.070 --> 00:47:12.320
For monoclinic crystals--
00:47:16.530 --> 00:47:22.780
and this is symmetry 2, symmetry
M, and symmetry 2/M.
00:47:22.780 --> 00:47:25.560
The form of the tensor when we
took the coordinate system
00:47:25.560 --> 00:47:31.080
along symmetry elements
was A11, A12, 0--
00:47:31.080 --> 00:47:33.230
terms with a single
three vanished--
00:47:33.230 --> 00:47:39.648
A21, A22, 0, 0, 0, A33.
00:47:42.980 --> 00:47:48.130
So this was the case where X3
was along the two-fold axis
00:47:48.130 --> 00:47:50.420
and or perpendicular to
the mirror point.
00:47:55.690 --> 00:48:01.440
OK, number of independent
variables to describe this
00:48:01.440 --> 00:48:02.810
property is five.
00:48:06.150 --> 00:48:09.600
And if we look at this in terms
of a constraint and
00:48:09.600 --> 00:48:18.530
shape in orientation of an
ellipsoidal quadric, says that
00:48:18.530 --> 00:48:21.540
one of the principal axes, if
the quadric is to remain
00:48:21.540 --> 00:48:28.690
invariant, has to be along
the two-fold axis and or
00:48:28.690 --> 00:48:32.700
perpendicular to a plane that
contains the mirror plane, if
00:48:32.700 --> 00:48:36.220
a mirror plane is
also present.
00:48:36.220 --> 00:48:38.810
So what are the degrees
of freedom here?
00:48:38.810 --> 00:48:42.640
This has to be along one
reference axis, x3.
00:48:42.640 --> 00:48:47.570
And, indeed, this form of the
tensor occurred only when the
00:48:47.570 --> 00:48:52.730
two-fold axes were
parallel, 2 X3.
00:48:52.730 --> 00:48:55.900
And this ellipsoid then
was left with
00:48:55.900 --> 00:48:57.880
one degree of freedom.
00:48:57.880 --> 00:49:02.950
If this is the direction of the
reference axis X2, we have
00:49:02.950 --> 00:49:05.040
a degree of freedom in--
00:49:05.040 --> 00:49:08.330
shouldn't have called these X1
and X2, these need not be the
00:49:08.330 --> 00:49:10.050
coordinate systems.
00:49:10.050 --> 00:49:14.260
Looking down on the quadric
along the two-fold axis, the
00:49:14.260 --> 00:49:17.310
section of the quadric
perpendicular to the two-fold
00:49:17.310 --> 00:49:23.370
axis can have a general shape,
and it can have one degree of
00:49:23.370 --> 00:49:28.670
freedom in the orientation of
the principal axis relative to
00:49:28.670 --> 00:49:30.370
the coordinate system.
00:49:30.370 --> 00:49:34.480
So we have three parameters to
specify the principal axes,
00:49:34.480 --> 00:49:35.960
since this is a general
quadric.
00:49:39.180 --> 00:49:57.590
We have one degree of freedom in
orientation, and that adds
00:49:57.590 --> 00:50:00.425
up to four, but we said five.
00:50:03.180 --> 00:50:04.260
So what's going on here?
00:50:04.260 --> 00:50:07.180
Again this is a question of
where the eigenvectors point.
00:50:10.770 --> 00:50:15.910
If this term is not equal to
this term, in order to force
00:50:15.910 --> 00:50:20.240
the tensor into a diagonal
form, you have to pick
00:50:20.240 --> 00:50:25.350
eigenvectors in the X1, X2
plane, which will force the
00:50:25.350 --> 00:50:27.740
tensor into a symmetric form.
00:50:27.740 --> 00:50:31.810
That is, to make this term
equal to this term.
00:50:31.810 --> 00:50:35.150
It's only a pair of axes
involved, so there's only one
00:50:35.150 --> 00:50:36.420
angle between these two
00:50:36.420 --> 00:50:38.320
eigenvectors, which is a variable.
00:50:38.320 --> 00:50:42.930
So for a nonsymmetric tensor
plus one angle between the
00:50:42.930 --> 00:50:44.180
eigenvectors.
00:50:56.220 --> 00:50:58.150
So that comes out, a-ha, five.
00:51:04.380 --> 00:51:07.880
And again the way to see that is
I will never get the tensor
00:51:07.880 --> 00:51:18.420
into a diagonal form making
all the off-diagonal terms
00:51:18.420 --> 00:51:22.560
zero when I'm starting with a
tensor which is not symmetric,
00:51:22.560 --> 00:51:25.960
and I know that a symmetric
tensor, even in the special
00:51:25.960 --> 00:51:29.080
case where the off-diagonal
terms are zero, is going to go
00:51:29.080 --> 00:51:33.860
to a symmetric tensor in another
coordinate system.
00:51:33.860 --> 00:51:35.270
So I know that there
has to be an
00:51:35.270 --> 00:51:37.680
additional degree of freedom.
00:51:37.680 --> 00:51:41.290
OK, the rest come very, very
easily, so let me take just a
00:51:41.290 --> 00:51:44.270
couple of minutes to finish up
this stage of the discussion,
00:51:44.270 --> 00:51:47.080
and we will shortly go on
to something different.
00:51:51.310 --> 00:51:53.240
The next step up in symmetry
is orthorhombic.
00:52:00.070 --> 00:52:06.890
And if we pick arc-- and this
could be 222, 2MM or 2/M 2/M
00:52:06.890 --> 00:52:19.680
2/M. We found that when we
took X1, X2, and X3 along
00:52:19.680 --> 00:52:29.416
two-fold axes that the tensor
took a diagonal form A11, 0,
00:52:29.416 --> 00:52:35.665
0, 0, A22, 0, 0, 0, A33.
00:52:38.370 --> 00:52:44.310
That says that the quadric
if it's an ellipsoid--
00:52:44.310 --> 00:52:47.690
or any other quadratic form--
00:52:47.690 --> 00:52:52.330
has all three of its principal
axes along the reference
00:52:52.330 --> 00:52:55.230
system X1, X2, X3.
00:52:55.230 --> 00:52:58.120
So the only degree of freedoms
here-- and things are again
00:52:58.120 --> 00:53:02.410
starting to set up like
supercooled water--
00:53:02.410 --> 00:53:06.065
the only degrees of freedom are
the three principal axes.
00:53:11.960 --> 00:53:14.065
And that's it, the orientation
is fixed.
00:53:20.030 --> 00:53:25.400
And, moreover, the tensor is
diagonal in this reference
00:53:25.400 --> 00:53:29.180
system-- it's going to be
symmetric in this reference
00:53:29.180 --> 00:53:32.850
system, it's going to remain
symmetric in any other
00:53:32.850 --> 00:53:34.060
coordinate system.
00:53:34.060 --> 00:53:40.040
So the tensor is always
symmetric, the eigenvectors
00:53:40.040 --> 00:53:48.300
then are always orthogonal, even
if the coordinate system
00:53:48.300 --> 00:53:49.700
stays Cartesian.
00:53:49.700 --> 00:53:52.260
And there are three variables,
three degrees of freedom.
00:54:02.520 --> 00:54:05.210
And finally two remaining
cases can
00:54:05.210 --> 00:54:08.550
be disposed of quickly.
00:54:08.550 --> 00:54:11.510
We saw that for an arbitrary
theta that
00:54:11.510 --> 00:54:13.220
is a symmetry element--
00:54:16.940 --> 00:54:18.110
so let me not say a symmetry--
00:54:18.110 --> 00:54:20.245
for an arbitrary rotation
about X3.
00:54:25.970 --> 00:54:34.560
This was the ingenious little
proof in which we transform
00:54:34.560 --> 00:54:41.470
the tensor by an arbitrary
rotation theta about X3.
00:54:41.470 --> 00:54:48.370
We found that the form of the
tensor was A11, A12, 0, and
00:54:48.370 --> 00:54:52.380
now I get to use tensor notation
by calling this term
00:54:52.380 --> 00:54:57.810
A12 again and not the proper
indices A21, so I'll put
00:54:57.810 --> 00:54:59.190
quotation marks.
00:54:59.190 --> 00:55:07.170
"A22" was equal to "A11."
00:55:07.170 --> 00:55:11.640
So going to an arbitrary
rotation theta this must be
00:55:11.640 --> 00:55:14.670
the form of the tensor we
discovered, and this will
00:55:14.670 --> 00:55:16.810
cover then fourfold
axes, three-fold
00:55:16.810 --> 00:55:19.220
axes, and sixfold axes.
00:55:19.220 --> 00:55:23.570
These two elements are
constrained to have the two
00:55:23.570 --> 00:55:27.410
values, these two are
constrained to have the same
00:55:27.410 --> 00:55:29.650
values, so the tensor
is always symmetric.
00:55:33.080 --> 00:55:35.820
Symmetric relative to these
axes, symmetric in all
00:55:35.820 --> 00:55:40.470
coordinates, and any choice of
coordinate system, then.
00:55:55.280 --> 00:55:59.870
And that means that if the
quadric has this property,
00:55:59.870 --> 00:56:04.080
it's going to have one principal
axis along X3, it's
00:56:04.080 --> 00:56:11.120
going to be circular in the
plane that contains X1 and X2,
00:56:11.120 --> 00:56:20.160
so there are only, count them
up, there are only three
00:56:20.160 --> 00:56:21.410
degrees of freedom.
00:56:29.670 --> 00:56:32.790
In terms of tensor elements,
these degrees of
00:56:32.790 --> 00:56:37.400
freedom are A1, A12--
00:56:37.400 --> 00:56:41.430
in the diagonalized
form-- and A33.
00:56:41.430 --> 00:56:46.540
And there are three degrees
of freedom in terms of the
00:56:46.540 --> 00:56:48.540
independent elements as well.
00:56:48.540 --> 00:56:53.850
So this is for anything that
involves rotational symmetry
00:56:53.850 --> 00:56:56.540
other than 180 degrees.
00:56:56.540 --> 00:57:07.420
And, finally, for cubic
crystals, if a symmetry
00:57:07.420 --> 00:57:11.150
operation other than 180 degrees
gives us this form of
00:57:11.150 --> 00:57:14.960
the tensor, two off-diagonal
terms are zero, two diagonal
00:57:14.960 --> 00:57:16.100
terms equal.
00:57:16.100 --> 00:57:21.140
If we impose this along all
three reference axes, then the
00:57:21.140 --> 00:57:24.540
form of the tensor is going to
go to a diagonal form with all
00:57:24.540 --> 00:57:28.440
diagonal terms equal.
00:57:31.730 --> 00:57:39.720
The quadric that corresponds to
that form is a sphere that
00:57:39.720 --> 00:57:44.640
says that there's one quantity,
one degree of
00:57:44.640 --> 00:57:47.720
freedom in the form
of the tensor.
00:57:54.410 --> 00:57:56.200
There are zero degrees of
00:57:56.200 --> 00:57:58.120
orientational degree of freedom.
00:58:11.430 --> 00:58:14.310
Thus the spherical quadrics
stay spherical for any
00:58:14.310 --> 00:58:15.540
orientation.
00:58:15.540 --> 00:58:19.120
So again, one independent
element in the tensor, just
00:58:19.120 --> 00:58:21.150
one degree of freedom
in the quadric--
00:58:21.150 --> 00:58:22.530
namely its radius--
00:58:22.530 --> 00:58:25.510
and, again, the formal
constraints that we obtained
00:58:25.510 --> 00:58:31.700
by symmetry transformations
agree with those that are the
00:58:31.700 --> 00:58:34.090
same degrees of freedom when
you want the quadric to
00:58:34.090 --> 00:58:36.050
coincide with the axes.
00:58:36.050 --> 00:58:36.844
Yes?
00:58:36.844 --> 00:58:39.925
AUDIENCE: Speaking of the circle
and the exponents to
00:58:39.925 --> 00:58:43.960
play only if A12 equals zero?
00:58:43.960 --> 00:58:48.450
PROFESSOR: These two were
equal, but non-zero.
00:58:48.450 --> 00:58:51.400
These two are equal, this
one was independent.
00:58:51.400 --> 00:58:53.590
And that would be for
the specific case
00:58:53.590 --> 00:58:55.660
of a fourfold axis.
00:58:55.660 --> 00:58:59.300
Now if we put a fourfold axis
along X1, that's going to
00:58:59.300 --> 00:59:02.910
involve these two being
equal and the
00:59:02.910 --> 00:59:07.400
non-zero elements become--
00:59:07.400 --> 00:59:08.350
let's see--
00:59:08.350 --> 00:59:16.718
along X2 it would
be A13 and A23.
00:59:16.718 --> 00:59:17.968
No.
00:59:21.200 --> 00:59:26.680
OK, we put the four-fold axis
along X2, then this one would
00:59:26.680 --> 00:59:40.440
be A22, A11, and A13 would have
to be equal, and 21 and
00:59:40.440 --> 00:59:48.650
12 would have to be zero, 23 and
32 would have to be equal.
00:59:48.650 --> 00:59:51.110
I think that's the way
it'll go, and these
00:59:51.110 --> 00:59:53.440
three have to be zero.
00:59:53.440 --> 00:59:56.210
So when you impose all three
constraints, here we've had
00:59:56.210 --> 01:00:00.060
these two equal, now we have
these two equal, put the
01:00:00.060 --> 01:00:01.860
four-fold access in the next--
01:00:01.860 --> 01:00:03.880
in the remaining direction,
and again it has to be
01:00:03.880 --> 01:00:08.290
diagonal, and pairwise all of
the off-diagonal terms will be
01:00:08.290 --> 01:00:09.748
required to be zero.
01:00:09.748 --> 01:00:15.130
AUDIENCE: I mean, when you
diagonalize your matrix the
01:00:15.130 --> 01:00:20.820
diagonal term are going to be
all different in this case if
01:00:20.820 --> 01:00:25.430
A12 is different from zero.
01:00:25.430 --> 01:00:27.860
PROFESSOR: This is not a final
result, this is an
01:00:27.860 --> 01:00:29.270
intermediate step.
01:00:29.270 --> 01:00:33.230
So we impose this constraint,
plus this constraint, and
01:00:33.230 --> 01:00:35.270
that's actually going to give
us all the qualities we're
01:00:35.270 --> 01:00:38.105
going to get, but we'd want
to do the same thing for a
01:00:38.105 --> 01:00:42.600
four-fold axis along X3.
01:00:42.600 --> 01:00:46.890
You put them all together, the
fact that these are zero will
01:00:46.890 --> 01:00:48.750
wipe out all of the
off-diagonal
01:00:48.750 --> 01:00:51.880
terms, and these things--
01:00:51.880 --> 01:00:53.420
for another choice of
axes, these two
01:00:53.420 --> 01:00:54.880
would have to be zero.
01:00:54.880 --> 01:00:58.790
And for another choice of the
orientation of the four-fold
01:00:58.790 --> 01:01:01.010
axis, this and this would
have to be equal.
01:01:01.010 --> 01:01:03.900
And, finally, this and this
would have to be equal.
01:01:03.900 --> 01:01:06.780
So this is what we found, in
fact, when we cranked through
01:01:06.780 --> 01:01:11.050
the symmetry transformations,
and this is what we would get
01:01:11.050 --> 01:01:14.640
when we said that the quadric
has to have a shape that
01:01:14.640 --> 01:01:17.912
conforms to cubic symmetric.
01:01:17.912 --> 01:01:22.268
AUDIENCE: My question was in
the case of the arbitrary
01:01:22.268 --> 01:01:24.700
relation of those three.
01:01:24.700 --> 01:01:28.859
All three degrees of freedom,
are they, in fact, the three
01:01:28.859 --> 01:01:30.109
principal axes?
01:01:32.840 --> 01:01:34.040
PROFESSOR: For this case here?
01:01:34.040 --> 01:01:36.010
AUDIENCE: Yes.
01:01:36.010 --> 01:01:38.870
PROFESSOR: For a symmetric
tensor, they are
01:01:38.870 --> 01:01:41.190
two principal axes.
01:01:41.190 --> 01:01:41.650
OK?
01:01:41.650 --> 01:01:45.490
If the tensor is nonsymmetric,
then these two don't have to
01:01:45.490 --> 01:01:49.640
be equal any longer, and
they give you the
01:01:49.640 --> 01:01:50.890
extra degree of freedom.
01:01:53.640 --> 01:01:54.890
OK?
01:02:00.640 --> 01:02:01.890
AUDIENCE: What are the
three degrees of
01:02:01.890 --> 01:02:03.140
freedom in this case?
01:02:17.640 --> 01:02:17.990
PROFESSOR: I just want
you to know I
01:02:17.990 --> 01:02:19.240
appreciate your question.
01:02:23.336 --> 01:02:24.797
AUDIENCE: May I make
a suggestion?
01:02:24.797 --> 01:02:25.771
PROFESSOR: Yeah.
01:02:25.771 --> 01:02:27.719
AUDIENCE: I think you need two
degrees of freedom to specify
01:02:27.719 --> 01:02:30.641
the first one, and since it's
symmetric and since it's
01:02:30.641 --> 01:02:33.439
orthogonal you'd only need one
more to find the second one
01:02:33.439 --> 01:02:35.260
and then another to
divide the third.
01:02:35.260 --> 01:02:37.525
So that's three degrees of
freedom to find three
01:02:37.525 --> 01:02:39.430
principal axes in an
orthogonal system.
01:02:39.430 --> 01:02:41.603
PROFESSOR: Not sure I like that,
because this has to be
01:02:41.603 --> 01:02:46.395
the shape of the quadric for an
arbitrary rotation angle.
01:02:56.359 --> 01:02:58.545
AUDIENCE: Does that angle
between X1 and X2-- does it
01:02:58.545 --> 01:03:00.335
have to be 90 degrees if
it's not connected?
01:03:00.335 --> 01:03:03.814
PROFESSOR: That's correct,
that's correct.
01:03:03.814 --> 01:03:09.030
OK, let's not tie up the whole
audience, let me--
01:03:09.030 --> 01:03:11.620
let's talk about this and
I'll think about it too.
01:03:11.620 --> 01:03:15.160
But you make a very, very good
point, but let's not settle it
01:03:15.160 --> 01:03:16.820
in real time.
01:03:16.820 --> 01:03:19.930
We'll throw everybody out, we'll
close the door, and you
01:03:19.930 --> 01:03:25.010
can come back and see who
won the scuffle, OK?
01:03:25.010 --> 01:03:27.050
OK, so let's take our break,
I think you're more
01:03:27.050 --> 01:03:28.300
than ready for it.