Hey, hi.
Let's take a look at crystallographic directions,
just the facts, just really quickly.
So the process for determining crystallographic direction
is first to appreciate you can translate
the vector if necessary-- I'm actually
going to call that point 0-- translate
if it simplifies things.
And you'll see some examples of that.
So then once you've got that translated vector-- translated
means it's just shifted in space parallel to the original--
then you determine projection onto the axes.
We'll start with cubic, x, y, and z.
That's projection.
Next step is reduce to the lowest integer values.
And then finally, we enclose in square brackets plus
do a little bit of housekeeping, which means negative signs are
moved above and no commas.
So those are the steps.
Let's take a look at some examples.
In our steps, we're going to go through projection, reduction,
and then enclosure.
So how about we start with a vector
that originates here and travels down to there.
So if we proceed systematically, we
do projection onto x, y, and z.
We're going to reduction if necessary
and then an enclosure.
So the projection onto the x-axis-- remember,
this is the x-axis-- the projection of that
is negative 1.
It's 1 in the negative direction, negative 1.
y, there's no component in the y-axis.
it's in the xz plane.
This is y over here.
And zed has a value of negative 1, a projection of negative 1.
Again, you could start with-- if I
wanted to I could define the origin here, o prime.
And then I could say, well, the point coordinates
of this point, if that's the origin are negative 1 comma
0 comma negative 1.
So you see that the point coordinates
of the head of the vector that originates at the origin are
the same as the projection.
Reduction, in this case, there's no reduction necessary.
These are already lowest integer values.
So the enclosure with our housekeeping steps
just becomes 1 bar, 0, 1 bar.
How about another example, perhaps one
that originates here and then travels down halfway
along the y-axis there.
I'm just going to draw that one in yellow actually,
so we don't confuse it with the other one that I drew.
So again, we'll do projection, reduction, and enclosure, x, y,
z.
I encourage you to do it systematically.
With this case here, we could start
with defining a new origin over here, o double prime.
So if that's the origin, the point coordinates
of the head of the vector here are going to be 0 in the x,
negative 1/2 in the y, and negative 1 in the zed.
That's as well what the projection is.
The projection onto the x-axis is 0,
onto the y is negative 1/2, and onto the zed is negative 1.
The reduction we're going to do here
then is going to be clear that fraction.
So we'll multiply across by 2.
We get 0, negative 1, and negative 2.
And the enclosure then becomes 0, 1 bar, 2 bar.
And that's what that vector is.
Last topic is families of directions.
And families of directions are directions
that have the same atoms in the scene positioning,
and for example, the face diagonals.
If you picked a crystal structure,
say FCC or BCC or simple cubic, and you do all the face
diagonals, they would all be in the same family directions.
And so the way to describe a family of directions
is the same as-- we had the same process, except we close it
in angle brackets, 0, 1, 1.
And usually, we move the 0 out to the left.
Non-zeros increasing towards the right.
And expressing it in the simplest way.
That is we wouldn't express this as 0,
1, 1 family of directions.
Rather we'd just call it-- sorry,
the 0, 1 bar, 1, something like that.
We'd just drop the negative signs.
And that includes all of those vectors both positive and
negative direction of each and so on for all the faces.
So this includes the specific 0, 1, 1 direction;
the 0, 1, 1 bar; 0, 1 bar, 0; et cetera.
Thank you.