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In this lesson, we're going to turn our attention now to

a topic that we refer to as the Miller Indices.

Illustrated here is a unit q and

let's say that we have two directions that are of interest to us.

First of all, let's say that this cube is being stressed in some way.

And let's say the direction of the applied force is along the arrow that

is describing the y direction.

When we look at the vector that says the direction of interest,

there may be some property in that particular direction like

the elongation or the stress in that direction.

As a result of the applied force that we're putting in to the system.

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We are interested then, in making a distinction between those two.

So we will be interested in describing those directions and

coming up with a consistent way of doing that.

And what we're going to do then is to look at a unit cube.

And what I have described up here is a simple unit cube in which I've labeled O,

that represents the origin, and the other corners, A, B, C, D, F and G.

And what we'll be able to do then is to describe

The co-ordinates of each one of those points because in order for

us to be able to get directions we need to describe the co-ordinates

that tell us something about the distances between the two points of interest.

So first let's look at the position that we're referring to as the origin.

The way we describe this in crystallographic terms is

that the origin is given as 0,0,0.

So we're looking at the origin, and

we're positioning at that origin at that position given as 0.

Now when we come over here and we look at position A.

Position A is along the X axis and we will call that position since it is

at the boundary of the unit cell, it will be the position one comma zero comma zero.

Now we turn our attention to B and what we now have is the position

which it will be described as the 0, 1, 0.

Again, these integers are delineated by commas.

And up here we look at C.

That´s along the Z axis and that position will be 0,0,1.

And we look at D and that position is 1,1,1.

And we look at E, which is on the XY plane.

So that will be 1,1,0.

And we look at F.

That's on that back corner, and what we see here is it is in the plane of Z and Y.

And what we will see is that that is going to be 0,1,1.

And up here we look at G and

we'll see that that's going to be at the position 1,0,1.

The last position, which is indicated as a face position,

position H, what we see is we are coming out in the direction of X and Y.

But in the case of X, we're coming out at a distance of one

half along the x, one alone the Y, and one half along the Z.

And as a result,

what we will see is the positions that are idicated over here on the right.

So the last position of H is 1/2, 1, 1/2.

And remember these are integers, or they're fractions of an integer, and

that puts them inside of the unit cell of interest.

Now that we've come up with a way of identifying the positions of interest.

If, for example, we're interested in talking about the vector that is

described by the position AO or the vector AO.

We can do that by recognizing.

First, we get the two points of interest.

We have the 0,0,0, and the 1,0,0.

And what we're going to do is to note those two positions, and

then what we do is we subtract the tail of

the arrow from the tip, and we clear any fractions.

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And then what we will do then is to write these in integers

in a square bracket and if there happens to be a negative direction,

we're going to put a minus sign on top of it.

So in the case of the vector AO, what we see is it will be written

as [1 0 0], and again it's delineated by the bar.

So when we take a look at all of the vectors that are included here,

the OA vector.

It starts out at position 0,0,0 and it moves to position A, and

what we have is at the tip minus the tail that gives us the orientation of [1 0 0].

And, again, note that the brackets are square brackets,

indicating a specific direction.

If we look at OB, we find that the vector then is

the difference between the coordinates for O and B, and

what we'll see is that that vector becomes [1 1 0].

Now if we go down to the vector OE where OE represents

the vector that's going in the negative Y direction.

If we consider the position, what we will see is, again the tip

of the arrow which is going to be at minus Y and the position O at zero.

What we will get is the vector, and we describe that as [0 1 bar 0].

So when we talk about the negative,

we put the negative sign on top of the number, and we read it as bars.

So that then becomes the [0 1 bar 0], and then when we look at OF,

now we have to pay attention to the fact that F is up and

along the Z direction and it's one unit up in the Z direction.

It's one half of a unit in the X and one half in the Y direction.

And so, when we clear fractions that vector then becomes the [1 1 2].

Because we started out with1/2, 1/2, 1, and

we clear fractions and that gives us the [1, 1, 2] vector.

Now we've described all of the particular vectors in those directions.

Now what I'd like to do is to introduce something that we referred to

as a family of directions.

So, for example, here is our unit cube, again.

And what we'll do is, we'll go around, and

we're going to define all the x, y and z directions.

So, again, the X direction is going to be the [100] direction,

the Y, the [010], And the Z the [001] direction.

Now when we start talking about the negative directions,

again in the negative Z direction it becomes the [1 bar 00].

And in the Y, it becomes the [01 bar 0].

And in the z, it becomes [001 bar].

So we now have a family of directions,

and so what we're looking at is the fact that we have just described six vectors,

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as indicated there on the right hand portion of the screen.

Now, what we can do is we can write those specific directions,

if we are talking about this cube, and we can talk about a family of directions.

In other words, those are all the same type of direction, but

once we've established the coordinate system where we have placed

the origin of our coordinate system, we have,

those are all equivalent vectors, and we've written them in very specific ways.

However, what we can do is,

we can describe all of those similar vectors that are of the type [100].

And we do this by, just like we did with the specific vector,

except rather than talking about the brackets, we now have

those bars which indicate the fact that we're dealing with a family of directions.

So we have the specific directions.

And then according to the visual,

what we've done is we've described all six of those in a very simple way.

So rather than writing all six, we just simply identify that grouping of

equivalent vectors using the nomenclature that is up on the visual.

Thank you.