PROFESSOR: OK, its always disappointing when, just as
you're ready to tie everything up in a nice, neat little
package and leave the room to a round of applause, you screw
up the last little detail.
Anyway, what's wrong here is, I said for an arbitrary
rotation around X3, and I'm comforted only by the fact
that nobody else caught this little glitch either, the
answer elements are a11, a11, a33, a12, not 0, and
-a12, 0, 0, 0, 0.
So there's a skew-symmetric form when there's just a
single rotation axis.
So this would be asymmetry n and this would be one of a
family of two different symmetries, the dihedral
groups n22, the two-fold axes would require that these be
equal, so, 0, 0, 0, a11, 0, 0, 0, a33.
So this is not the form of the tensor for a four-fold axis,
it's the form for a crystal with symmetry 422.
And same for a three-fold axis.
This is the form for 32 and 622.
For just the single rotation axis of any sort other than 2,
these two terms are equal in magnitude, but non-0.
Now if you look at the equation for the tensor, the
term in a12, x1, x2, and a21, x1, x2 are opposite in sign
and disappear.
So the quadric is, in fact a surface of revolution
So the question is why are there two degrees of freedom
rather than three?
And I think the answer to that is going to lie in what is
required of the eigenvectors when the off diagonal turns
are skew-symmetric, and that is something I've not thought
about before.
All right, so that one minor loose end sticking out of the
lid as we try to close the box on this segment of our
discussion.
I'd like t move on to a different set of
considerations.
One of the things that we can subject a crystal to as a
generalized force is not merely a vector force, but
stimulations that are more complicated.
And I would like to in particular now, as a prelude
to get into getting into property
tensors of higher rank.
I'd like to . first discuss stress, and then define
strain, and you've all heard of these
questions in other contexts.
I'm going to introduce it, though, in terms of the tensor
notation that we've used for other sorts of tensors, just
so that we have everything in terms of the same language.
So let me introduce stress by supposing I have a volume
element in a solid, and it will not surprise you that I'm
going to define the coordinate system in the solid by three
axes, X1, X2, X3 in a Cartesian.
system, and I will look at a surface on the exterior of the
solid that cuts axis X1 at A, axis X2 at B, and axis X3 at
C.
Then I'll assume that I have a force density, force per unit
area applied to the surface.
And if this is force per unit area, why
don't I call it a pressure?
Well, a special case of what we normally think of as a
pressure is a hydrostatic pressure, and that is
something that's always exactly normal to the surface.
So I don't want to call it a pressure to mislead you into
the assumption that this force per unit area, which is a
vector, has to be normal to the surface.
So I'll call it a force density, which becomes a
hydrostatic pressure when that force vector is
normal to the surface.
And I'll call that force density vector K, and it will
have three components, KI.
Now, the solid is in equilibrium, I will assume.
And if that external force is not balance on the other side
by balancing forced, the volume element in the solid
would undergo a linear acceleration, or an angular
acceleration at the very least.
So I will assume that this force on the outside, KI,
which is a force per unit area.
So to make that a force, I will have to multiply that
force density by the area of the surface
ABC, on which it acts.
And I'll say that this has to be balanced for each of the
three directions, X1 and X2 and X3 by a force per unit
area acting on these internal surfaces.
So I'll assume that there is a force acting in the x1
direction, and I'll call those a sigma IJ.
And I'll say that there's a first force acting in the X1
direction on--
now I need to somehow specify the orientation to the
surface, because on this internal surface there's also
a force acting in this direction.
Now on this back surface there's also a force acting in
this direction.
So the forces acting on all three internal surfaces in the
X1 direction have to balance exactly the X1 direction, the
X1 component of the force density.
So let me turn this to K1, and I'll say this has to be a
force per unit area acting on area.
First of all, I'll call the origin O on area AOC.
Plus a force acting in the X1 direction on area A--
whoops, what did I do?
I wanted to do this first on area BOC, excuse me.
And a force in the X1 direction acting on area AOC.
And finally a force acting on area AOB.
And these are all acting in the X1 direction, and I now
would like to introduce some notation that indicates the
direction of the direction orientation of the surface on
which that force acts.
And I will use the normal to these internal surfaces to
define that.
So the force per unit area on area AOB, that is a surface
whose normal is X3.
And I will call this component sigma 13.
And I'll call this one area AOC is normal to X2.
That's why I did that little sleight of hand as I began
this, and finally area BOC has as its normal X1.
So my use of subscripts here is that this first subscript
is the direction in which the force acts.
And the second subscript will be the normal to
the internal surface.
So I've set up algebraically a system of
subscripts that works.
We'll see that these have some physical
significance in just a moment.
So what will I call these?
Well, let me get this in a slightly different form.
Let me take the left hand side and divide area ABC into all
of these other areas.
So this then will have sigma 11 times area
BOC over area ABC.
And not surprisingly, sigma 12 will then have multiplying it
a term area AOC divided by area ABC.
And this is all looking very cumbersome, but you'll be
amazed at how this cleans up in just a moment.
And this will be area AOB over finally area AOC.
ABC again.
OK, so how can we tidy this up?
Well, let me imagine that I have some surface, a planar
surface, and its normal points in this direction.
And I'm going to take that area and project it onto
another surface whose normal points in this direction.
And let's suppose that the angle between these two
directions is phi, and the angle between the
two surfaces is phi.
The area of this original surface is A, and its normal
points in this direction.
And I project it onto a surface whose normal is an
angle phi away.
This new area, A prime is equal to A times
the cosine of phi.
So if we accept that, then the first line of this equation,
which will be eventually a set of three equations, this is
the ratio of area BOC over area ABC, and what is that?
That is the angle between the direction of the force density
and area ABC.
And here is area ABC.
That's the normal here.
And the first term involves area BOC, and that is the
direction of X, the normal to that is X1, and therefore the
ratio of these two areas is the same thing as the cosine
of the angle between K and X1.
And this second ratio of areas is going to be sigma 12 times
the cosine of the angle between the force density
vector K and X2.
And the third term is going to be sigma 13 times the cosine
of the angle between K and X3.
What these cosines are are just the cosines of the angle
between K and the three reference
axes, X1, X2, and X3.
So this is simply the direction cosine L1 of K. And
this is the direction cosine L2 of K. And this is the
direction cosine L3.
So the X1 component of the force density vector is going
to be equal to a term sigma 11 times L1, plus the term sigma
12 times L2, the term sigma 13 L3.
And if I would write down similar balances, forces for
the X2 component of the force density vector, not
surprisingly these things are going to give me coefficients
that I'll label sigma 121, sigma 22 times L2, plus sigma
23 times L3.
And in general then the requirement that this body be
in equilibrium states that each component of the force
density vector should be given by a coefficient sigma IJ
times L sub J. These are the components of the force
density vector.
This is the direction cosines of the direction of K.
This is a unit vector in the direction of K. This is the
direction of a vector.
These are the components of K. And therefore these
coefficients sigma IJ relate to vectors, and they must be a
second rank tensor.
And these are called, as you all know, it comes as no
surprise as the components of stress.
So the reason that these are really tensors, and not just a
vector force per unit area essentially comes down to the
fact that the behavior of the body, dynamically and
mechanically, is going to depend not only on the
direction of the force, but on the direction of the surface
on which it acts.
Quite clear here, if I shove, is a force density,
and that's a vector.
But the way in which this desk responds to my pushing on it
with a force vector of that magnitude is going to depend
whether I do this, which is going to compress it, or
whether it takes the same force and direct it this way,
which if I don't move the thing is going
to result in a shear.
So the behavior of the body will depend on the direction
of the force per unit area that I apply, and the
direction of the surface on which I apply it.
In this case, that would be the surface ABC.
Yes, sir?
AUDIENCE: You say that the force density is not parallel
to the normal of the surface?
PROFESSOR: That's correct.
AUDIENCE: So when you do the ratio of [INAUDIBLE]
isn't it supposed to be [INAUDIBLE]
of the normal to the [INAUDIBLE] and X1?
PROFESSOR: Another way of defining it.
This is a vector, and what I'm doing is splitting that vector
up into three internal forces that act on surfaces whose
normal are here.
So actually, what I'm going to do is to take the way in which
the internal force pushes back.
I see, that's what's confusing.
If the force that I'm putting on the body, I didn't mean to
indicate any particular direction, but if this
component of K points in this direction, this force that
balances it has to go in the opposite direction.
AUDIENCE: So you're saying that there were surface ABC is
defined as normal to K?
PROFESSOR: ABC is not normal.
No.
ABC is not normal to K. All I'm saying is that it if it
were normal to K, then the balancing force would be just
a function of this orientation of the surface.
But if I allow this to be a general force per unit area,
then the balancing force on the inside depends on the
angle between K and the angle between this internal surface.
AUDIENCE: I don't see the relation between K and the
ratio between other ends.
BOC and ABC [INAUDIBLE].
PROFESSOR: OK.
The reason that there are three areas, is that what's
troubling you?
Because I'm really now splitting the balancing force
up into a force per unit area on this surface, and a force
per unit area on this surface, and a force per unit area on
this surface.
And that's so that I can look at these different components
of resistance in terms of a Cartesian system.
Clear that the ratio of these two areas does go as the--
I erased it now, that is going to go as the--
This is A prime, and this is the original A, the relation
between those areas is given by the cosine
of this angle phi.
If I [INAUDIBLE] this down in this projection, and this is
the area of A.
This is the direction of the normal to the surface ABC.
And I'm going to let this be the direction of the normal to
one of the internal surfaces, and these internal surfaces
are going to have as their normal either X1, X2, or X3.
AUDIENCE: So [INAUDIBLE]
between the normal to ABC and X1.
PROFESSOR: And X1, yes.
And that's what I'm saying its, and that is going to be
the same thing.
As the cosine of that angle is just going to be the direction
cosine of the normal to ABC relative to X1, X2, X3.
OK?
AUDIENCE: So in the end, you're saying that K is
parallel to the normal of ABC?
PROFESSOR: No, I'm not.
No, I'm not.
This is a relation between the areas.
And these are the forces acting on
those internal surfaces.
And I'm saying there are three of them, and those three
components that are all acting in the X1 direction will
depend on the ratio of these areas.
The ratio of this area is the cosine of the angle
between X1 and K. OK.
This you're OK with?
AUDIENCE: Yeah.
PROFESSOR: OK.
And what I'm saying now is that there is acting on this
internal surface something that has to balance the
component of K, which points in the direction of X1.
And the component of K before we divided out the area, the
component of K1 times the area gave me the net force.
So this K1 now is a force per unit area, the component of K
per unit area that points along the X1 direction.
Now as a force per unit area I multiply it by an area that
gave me force.
There are three forces internally that balance this
force, and they all act in the X1 direction.
AUDIENCE: You're just [INAUDIBLE].
Shouldn't it be the cosine of the angle between the normals
of [INAUDIBLE]?
PROFESSOR: Yeah.
That's what I'm saying here.
That for X, for this surface, yeah, that's what I'm saying.
One of these surfaces here, if we look at the first term,
this is the direction of X1, and the force here will be the
force along K. And if I take this force and put it down in
X1, this is going to be the force times the
cosine of the angle.
And the two areas are going to go as the cosine of the angle.
AUDIENCE: [INAUDIBLE]
cosine of the normal [INAUDIBLE]?
PROFESSOR: OK.
This is indeed the ratio between the internal area and
the external area.
Since the internal area has a normal that's either X1, X2,
or X3 the angle between the normal to these internal
surfaces and K is going to be the direction cosine of K.
I suggest since we're getting out of time quickly we either
resolve this after the end of class or leave
it until next time.
That's a bad note on which to end, too.
The point I want to leave, still subject to resolution
and debate is that the coefficient sigma IJ I would
like to claim relate to vectors.
One is the direction of the external surface, and if it
relates to vectors, then this set of coefficients qualifies
as a second rank tensor.
And if you accept that, however grudgingly, everything
that we've said about second rank tensors holds for the
quantity sigma IJ, which are called the elements of stress.
In particular, we can talk about a stress quadrant.
We can talk about the value of a stress in
a particular direction.
If we have a stress tensor in which show all the terms are
non-0, it can be diagonalized.
We can also say that if the crystal has symmetry, for
example, if the crystal is cubic, you can say that the
form of a second rank tensor for a cubic
crystal has to be diagonal.
And these diagonal elements represent compressive
stresses, so you can say that you can only subject a cubic
crystal to compressive stress.
On the other hand, if the crystal is a triclinic
crystal, you can say that there's a sigma 11, there's a
sigma 12, a sigma 13, a sigma 21, a sigma 22, sigma 23,
sigma 31, sigma 32, sigma 33.
So for a triclinic crystal, there appear to be nine
elements, but next time we will show that the nature of
stress is the body has to be in equilibrium requires that
the stress tensor be by definition symmetric if
there's to be no net couple to force on it.
Going to buy this?
You didn't buy the things that I said that perhaps were
slightly incorrect.
This is massively incorrect.
Jason?
AUDIENCE: You can rotate a cubic crystal [INAUDIBLE].
PROFESSOR: No.
Because say a tensor of this form acts
just like the scalar.
Just like a cubic crystal, you don't suddenly get off
diagonal properties if you transfer it
to a different axis.
So unless you tell me what's wrong here, I'm going to leave
this on your plate until next week.
And I don't want you fretting over the weekend.
This is clearly nonsense, and it is a contradiction that
arises only if you carry too much of what we've been doing
for the last month with you into a discussion of stress
and strain.
Stress is something that you impose on a crystal.
It has nothing to do with what kind of crystal it is and
what's going on in the interior of the crystal.
There may be constraints on the coefficients that relate
stress and strains, and there surely are.
But I can take a cubic crystal, and I can squeeze it,
and I can twist it and share it anyway I like.
So this is an important distinction between the
tensors that represent physical properties, which
we've been discussing up to this point.
And the properties such as conductivity, diffusivity,
susceptibility and so on and all their many varieties,
these are something which we'll refer
to as property tensors.
And they are tensors demonstrably, but they are
tensors which describe the physical
behavior of a crystal.
Nye uses a term that I don't think is self explanatory, so
I don't care for it.
He calls these field tensors.
That sounds like some description of a corn patch
that requires a tensor.
But no.
We probably don't really think of what we mean
when we say a field.
When we say an electric field, you think of an e-vector.
You think of a vector.
But what a field is is just a set of coordinates, XYZ, and
that describes an area or a volume.
And just like the things that corn grow in are referred to
as corn fields, it's an area that has corn on a lattice
very often also.
So a field is just a set of coordinates in space to which
a value of something is a sign.
So if we have an electric field, and we express that
electric field as a function of X, Y, and Z, that is a
field of vectors.
If we take something that has intrinsically, that
intrinsically needs to be described as a tensor, then we
have to every point in space XYZ a tensor sigma IJ, whose
value and whose components value vary with position
within the space.
So talking about the field as assigning values of something
to every coordinate in the space that could be a tensor.
What we refer to as an electric field really is a
field vector.
So that's Nye's term, but the fact that it took me three
minutes to explain why he uses it explains as well why I
don't care for it.
So we'll talk about these as property tensors.
And property tensors are something that's
innate to the material.
Something like stress, strain, and other second rank tensors
are external stimuli, just like an electric field.
And they have value that is imposed on the crystal, and
the stress field is defined as a function of position.
So that's the distinction between a property tensor,
which is subject to symmetry constraints, and a tensor that
represents a stimulus, an external stimulus applied to
the crystal, and it's defined as a function of position.
And that is a field tensor.
OK, so that's the difference.
Since this is something that's imposed externally, it can
have any form of the tensor that you wish to oppose.
And there are no symmetry restrictions on it.
The only restriction is, as we'll show next time, that it
must be symmetric if the body is to be in equilibrium.
All right.
It is time to quit, and that I think is a good
place to leave things.
And we'll say more about stress next time.