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STUDENT: Today
we're going to look

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at Hooke's Law in Cubic Solids.

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Hooke's law describes
behavior of springs.

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What we're going to
do is we're going

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to model the solid as
a collection of springs

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connecting a whole bunch of
atoms in a cubic lattice.

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Now, disclaimer-- I'm probably
going to do this all wrong.

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But what's science if we don't
make a few mistakes, right?

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Let's get started.

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The potential that
describes fairly well

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the behavior of the
interaction between two atoms

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is the Lennard-Jones potential.

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It describes the van
der Waals interaction,

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and it looks
something like this.

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It's a potential with
a term which goes 1

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over r to the 12th, and one term
which goes 1 over r to the 6th.

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And as you can see, as
the atoms get very close,

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they repel greatly, and then
they have a sweet spot here,

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which is the distance
they prefer to be at.

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As they get farther away, they
start to repel more again.

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The atom like to sit right there
in that little potential well.

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It's written up in Mathematica.

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You can see the equation
here, what each term means.

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And when I come here as
I take in the derivative.

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What that does is give me the
force on the particle when

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it's in this potential.

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So as we see, the force
is high and positive

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when it's closer
to the other atom,

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then the equilibrium
distance and the force

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is negative when it's farther--

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so it gets pulled
towards that sweet spot

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that I mentioned earlier.

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And here are the
two plus together.

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So with Hooke's Law, we're
modeling the atomic bonds

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as springs.

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And the way that's
going to work is,

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we have Hooke's Law which
is that the force is

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proportional to the
spring constant times

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the displacement.

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So the further you pull
it, the greater the force

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in its linear relationship.

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And the problem is, though,
our Lennard-Jones potential is

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not--

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it's not a normal
potential array.

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It's this weird wobbly shape.

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But for Hooke's Law
to work, you have

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to have something that
behaves like a spring, which

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has a parabolic
potential well, which

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is how you get that linear
force relationship when

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you take the derivative.

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So what I've done is
I've used Mathematica

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and its wonderful math tools
to take the Taylor expansion

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about that equilibrium point.

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And you can see here, this
is the second order Taylor

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expansion, so it's a parabola.

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And if you look at this
little manipulate here,

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I have set it up so you
can adjust the potential

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while depth can choose
the equilibrium distance,

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And you can look at the
different order Taylor

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expansion, so that's the
first order-- it's a line.

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The second order is
a parabola-- that's

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what we're going to be using.

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The other ones are just better
and better approximations

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of the action potential.

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Mathematica supposedly
can do negative powers,

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but I'm not seeing
that here, so could

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be the actual perfect model
would be obviously one over r

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to the 12th, there's r
to the negative 12th.

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So I differentiated that second
order potential like that.

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And you can see here
the force response

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for our generalized for an
approximated spring bond.

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And so you can look at
our force response here.

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Notice like before,
the force is positive

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when the atoms are close
together, and negative when

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they're farther apart--
so it gets pulled out

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of that sweet spot.

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And I've plotted it
here with the potential,

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so you can see how exactly it
interacts with that potential.

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Here is the
approximate potential

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with the actual
Lennard-Jones potential.

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So you can see that it's not
a very good approximation,

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but very small displacements
on the order of about 0.1 times

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that minimum distance
will probably be OK.

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I simplified it here to
get the-- and then I take

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the derivative again, just
so I can get that slope--

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and that'll be useful later.

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So here, we're modeling
our cubic lattice.

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Say we have something
like alpha polonium, which

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has a simple cubic structure--
the only metal that

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has a simple cubic structure
with a single atom motif.

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So we have our stress
over here, and our strain,

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and there's this
fourth ranked tensor

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that connects our second ranked
tensors with stress and strain.

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But, because of the wonders
of mathematics and matrices,

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we can actually break that down
into two first ranked tensors

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and a second ranked tensor.

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Which is pretty
great, because it

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means we don't want to do
some really early math.

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And so we can find that
infinitesimal strain tensor,

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if we're looking at a tiny,
tiny piece of a solid,

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will look something like this.

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It's this-- epsilon
IJ is equal to 1/2

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of the displacements
in each direction

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of the infinitesimal piece.

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And so from that, we
can build that out

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to the second rate
strain tensor,

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which is this tensor here.

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And it looks like this,
where u is the position,

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so we have all
these displacements.

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As you can see along
the principal axes,

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they're just partial
to the displacements

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and the directions
along the other axes,

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they're a little
more complicated,

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because you have things moving
in two directions at once.

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But the coolest part is that
only six of these entries

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are unique--

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because this is
the same as that.

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That's the same as that, and
that is the same as that.

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So what we can do is we
can just break it down

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to these six unique things.

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And then here is our second
ranked tensor, like I

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said before, of our elasticity.

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But, because we're using
Hooke's Law, these are springs

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and this is a simple
cubic lattice.

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So each atom is only attached
to six of its neighbors,

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so it's not going to
have any weird stresses.

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This is where I'm probably
wrong, but at this point,

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we are going to do this.

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So here are our diagonal
elasticity values.

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And if we do anything before and
simplify that spring potential,

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and grab the slope of that line,
we get our spring constant.

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So if this were a
spring, that's what

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the spring constant would be.

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And we just plug that in here
to our matrix, and we get--

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oh look, it's Hooke's Law.

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So if everything is
wonderful and linear,

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which it's probably
not, but if it were,

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we have the Hooke's
Law, and we can

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use that to determine from the
strain, the stress-- or vice

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versa.

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As you can see, it would
look something like this--

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so in the original position
of the spring's the black,

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and then when you drag
the atom over here,

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you get this red,
deformed spring.

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This one here's squished,
these are stretched.

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And yeah, that's about it.