PROFESSOR: That I know of that derives two-dimensional
symmetries.
And so, what I will give you instead is a set of notes--
nothing fancy, just handwritten--
but which will carry out each of the derivations in detail,
so you'll have something to fall back on if it seems
overly complicated.
Any questions on what we've done to this point?
We've really set forth the ground rules and components of
our final step.
And that, as I mentioned earlier, is to combine one or
more of these collections of symmetry elements around a
fixed point in space, with one of the five two-dimensional
lattices that can accommodate them.
And when we find a successful, feasible combination, we will
have derived something that we could call the two-dimensional
crystallographic space groups.
These are collections of operations which really pick
up everything-- the whole space--
move it, tumble it end over end, and throw it back down
into coincidence with itself.
So hence the term space group, rather than point group, where
it's only a point that's left invariant at worst.
OK, so what we're going to do then is to take each of the
point groups in turn, and combine each of them, when
possible, with one of the five lattice types.
And I remind you that those are the oblique lattice or
parallelogram lattice.
And this was what was required by two and one,
no symmetry at all.
Then there was the primitive rectangular net, and the
symmetry element that required that was a mirror plane, or we
could have, depending on how the mirror plane was aligned
relative to the translation, we could also have a lattice
that was non-primitive, a double cell.
And that was a centered rectangular net.
And either of those was compatible
with a mirror plane.
Then we had a square cell--
square net or square cell-- and that was what was required
by a 4-fold axis.
And then the hexagonal net, and that could be required by
either a 3-fold or a 6-fold.
Well, if you add up the number of point groups that we've
listed here, there are only six, but there are 10
crystallographic point groups.
So what about the ones that we have not yet considered adding
to a lattice, and these are the nets of the form Cnv.
So let's consider, in turn, each of those point groups and
see what they would require of the lattice that could
accommodate them.
But let's do 2MM first.
2MM, the 2 is compatible with an oblique net.
The mirror plane wants either a rectangular net or a
centered rectangular net.
So all we can do is to add this to a rectangular or
centered rectangular net.
Let's put a 2-fold axis at the lattice point.
The mirror plane has to be aligned along the edge of a
rectangular net.
If there are two of them, you just put one along each of the
edges of the rectangular net.
A mirror plane in a net could either go in like this, or
could go, for the centered net, in like this.
So what we're saying, 2-fold doesn't care, add a mirror
plane, it means that the parallelogram
has to become a rectangle.
You can put the two mirror planes of 2MM in along the two
edges of the rectangle, or do this equally well with a
centered rectangular net, put the mirror plane in parallel
to the edges, just like we did in CM, put the 2-fold axis in
at the lattice points.
So there's a lattice point at the corner, and also a lattice
point in the middle of the cell.
So for the rectangular net, we could put in either M or 2MM
in both of these rectangular nets, the
primitive and the centered.
4-fold access requires that the net become still more
specialized and have the shape of a square.
AUDIENCE: Professor?
PROFESSOR: Hmm?
AUDIENCE: We can't see on this side.
PROFESSOR: Oh, I'm sorry.
I'll pace back and forth, which I usually do.
Then everybody will get equal time.
OK, so I'm glad you spoke up.
4-fold axis requires a square net.
There are two mirror planes, 45 degrees apart, and those
mirror planes have to be along the edges of
a rectangular net.
Put one of them in this orientation, and that surely
is along the edge of a rectangular net, along the
edge of a net that's not only rectangular but square.
But it has a higher symmetry, and that's fine.
Looks as though this diagonal mirror plane might cause
troubles, but let's note that if we look at the diagonal
translations in this lattice, that they will define a
centered rectangular net, so putting the other mirror plane
in this way puts it along the edge of a
centered rectangular net.
So this mirror plane is perfectly happy as well.
So in the square lattice we could pop either
symmetry 4 or 4MM.
3-fold is an interesting one.
That's the odd symmetry that does strange things.
Here is a hexagonal net, and that's what
a 3-fold axis requires.
Two translations, identical in magnitude, and in terms of
what goes on along them, in exactly 120 degrees apart.
Now, I don't readily see any rectangle along the edge of
which I could align the mirror plane in 3M.
But it's the oblique shape of this net that throws you off.
Let's notice that if I draw several unit cells side by
side, here is a double cell that has a rectangular shape.
OK?
Take 2T1 plus T2 as one edge of the cell, and leave the
other one as T2, and we have defined a cell where that is
identically a 90-degree angle.
So one of the mirror planes of 3M could go in this
orientation, and the other one would be 30 degrees away.
So this would be the orientation of
the two mirror planes.
So we're putting in 3M with M perpendicular to the
translations that define the primitive hexagonal
cell, T1 and T2.
But there's another way you can do it.
Let me draw the same net, the same hexagonal net, and look
at a different kind of double cell.
OK, here's my hexagonal net, two translations, T1 and T2,
with a 3-fold axis at the lattice point.
And now I will again draw a couple of extra cells--
double T1, and let's double T2.
And then let me point out that I can put in the mirror plane
this way and along this way, and here is a double cell, and
I can put the mirror plane along this
direction and this direction.
So this is crazy.
This is the same lattice and the same point group, but
there are two different ways I can align the point group in
the lattice.
I can put the mirror plane in such that it's perpendicular
to a cell edge, or put the mirror plane in 30 degrees
away from that orientation such that it is along the edge
of a centric rectangular net.
So I could add 3M as well with the mirror plane aligned
parallel to the translations in the net.
So lots of funny permutations here.
Here I have one symmetry element, two different
lattices that I can combine with it.
Here I have one point group, 3M, but two different ways in
which I can set it relative to the edges of the cell.
Yes?
AUDIENCE: Why are you putting two mirror planes in each of
those figures when you really only could
do one mirror plane?
PROFESSOR: Actually, I need--
well, there are going to be a radiating sheaf of mirror
lines coming out of that 3-fold axis.
AUDIENCE: An angle of 60 degrees.
PROFESSOR: They are at an angle of 60 degrees.
It has to be 60 degrees, and that is
indeed what I have here.
I've just drawn part of 3M, and if I draw the remaining
mirror planes, they would do this.
OK, I think that's fairly intelligible.
Another reason for notes is to draw these diagrams when
they're reasonably complex.
On a blackboard, when you're rushing to get through it, is
something that's easier said than done, even though you may
have done it a number of times.
So there is the job that we've laid out for ourselves.
Five lattices, 10 point groups, and we can drop each
of the point groups into one or more of the lattices of the
five that we have determined.
As soon as we combine things, we're going to have the
situation that we've encountered before.
In general, if I have an operation number one that
produces one transformation of coordinates, and combine it
with an operation number two which has another change of
coordinates, there must automatically
arise some third operation.
And we can write it in terms of the language of operators
by saying operation number one followed by operation number
two is equivalent to an operation number three.
And in terms of the pattern, this says that whenever you
throw two symmetry transformations together, a
third one pops up.
You can't push it down.
It's going to be there once you add the first two
combinations.
You may have to be really clever to see where it is and
what it is, but it's going to be there.
It's going to be there.
It comes in automatically.
So this general class of statement, this equality in
operations, is something that I like to call, for short, a
combination theorem.
And we've seen several examples of this already,
notably the one in the last hour that said whenever you
combine two mirror lines at an angle mu, you automatically
create, like it or not, a rotation operation of a 2 mu.
So as soon as we start making these combinations, we're
going to find a new operation arising.
So let's start with the simplest one.
Add a 1-fold axis to a general oblique net.
So there is T1, there is T2, we don't put anything in it,
it's just a simple array of translations with lattice
points that we can put at the terminal point of these
translations.
There's no symmetry.
If we put in one atom, that atom hangs on every lattice
point, and that's all there is to it.
It's so simple it's not only uninteresting, it's ugly.
It's just a simple atom arranged in a periodic array.
As with everything we've done so far, it's convenient to
have a notation to indicate which particular plane group
one has, and the symbol here is to give side by side the
type of lattice as the first part of the symbol, and then
the point group that you have added.
So what do we have here?
We've taken a primitive oblique lattice, and we've
added no symmetry at all, a 1-fold axis.
And here people who have derived symmetry theory assume
that the reader knows something.
And so you're going to get a lot of respect when you show
your colleagues that you know what this notation means.
All you have to do, if you know the symmetry, is to state
whether the lattice is primitive, as it has to be for
an oblique net, or whether it's centered, in the case of
the rectangular net.
Three dimensions is a lot more complicated.
If the lattice is cubic, it could be primitive, it could
be face-centered, it could be body-centered.
So you need a symbol that tells the type of lattice.
In this case it's primitive, and the point group that we've
added is 1, so this rather boring ugly thing
here is called P1.
Another bit of code.
We're going to do eventually the same thing in three
dimensions.
We're going to take just a general oblique
three-dimensional lattice, and not adorn it with any sort of
symmetry at all, and that will also be called P1.
But to distinguish the two, capital P1 means a
three-dimensional lattice, and we won't be working with those
for a while.
We haven't even enumerated them.
A lowercase p, that is the symbol for a
two-dimensional net.
In other words, a plane group rather than a space group.
So here's the first, P1, and it really doesn't give shivers
running up and down my spine to look at it.
So let's do one that's a little more interesting.
Let's take a 2-fold axis and combine it first with a
primitive rectangular net.
I'm sorry, what am I saying?
I want an oblique net.
So the symbol--
and I'm assuming this is going to be unique, and we haven't
seen anything with a 2-fold axis in it, so it is.
So this would be P for a primitive lattice, 2 for the
symmetry that you've added to the lattice.
And since you are all cognoscenti, I don't have to
tell you it's an oblique lattice, because you know,
don't you, deep down in your heart, that a 2-fold axis
requires only that the lattice be an oblique lattice.
Nothing special is required.
So what we will do is to take this general oblique lattice
with two translations, T1 and T2, and to this-- remember, we
don't deal with symmetry elements.
We have to do this operation by operation.
So what we're doing is taking the operation A pi, the only
non-trivial operation contained in a 2-fold axis,
and putting that in at a lattice point.
OK.
Bells ring and whistles go off.
We don't know what happens when we make this combination.
And let me do this in more general terms, so we can
derive the theorem once and for all.
Here's a translation, and let me add to the end of that
translation a rotation operation A alpha.
In this case, it would be the operation A pi.
What's the net result of rotating and then translating?
That's a non-trivial question.
So if I'm doing this for a general rotation of alpha, let
me take a translation, and I'm going to make my construction
in the following way, because I'm really clever.
And I'm going to put this translation T at one half of
alpha on one side of the perpendicular to T.
And then I'll let that rotation go to work.
And it'll take the translation and move it to alpha over 2 on
the other side of the perpendicular.
And now I'll let the translation move to this
rotated translation, and if it does so, it's going to move
the translation over to here.
Here's a motif hanging on this translation.
It's rotated over to here, and then gets slid over to here.
This one is right-handed, this one is right-handed, this one
is right-handed.
How do I get from this one to this one?
Only two ways, translation or rotation.
If we rotate, about what point are we rotating?
Now I'm going to use a definition
which seemed trivial.
I said, a symmetry element is the locus of points that is
unmoved by the operation.
And if we decided, because these two motifs are not
parallel to one another and have the same corrality, that
the rotation--
has to be a rotation that relates the two.
When asked finally the question, what
point is left unmoved?
Ha.
It's this point where these two translations come together
at a single point.
That's the point that's left unmoved by rotating from here
to here and then sliding over by the translation T.
And I can say exactly where that rotation is.
It's going to be along the perpendicular bisector of my
original translation.
And where is it going to be located?
This line makes an angle of alpha over 2 with respect to
the perpendicular to T. So that angle is
also alpha over 2.
This line and this line both make an angle of alpha over 2
relative to the normal to the translation, so this angle is
also alpha over 2.
And look how this has turned out.
The combination of a rotation with a translation is another
rotation, B, about a different point but through the same net
amount alpha as the original translation.
On top of that, really to nail this down,
where should we look?
This distance in here is T over 2, and the tangent of
alpha is T over 2 over this distance x.
And if I just solve for that distance, x is
equal to T over 2--
I'm sorry.
I got this garbled.
T over 2 over x is equal to the tangent of alpha.
That's what I wanted to do.
So if I solve for x, that distance x is T over 2 over
the tangent of alpha, and I can write that as T over 2
times the cotangent of alpha over 2.
So here it is, done properly at last.
Go up a distance x along the perpendicular bisector of T,
and you go up a distance T over 2 times the cotangent of
alpha over 2.
So this is a general result for any rotation operation.
Combine a rotation operation A alpha with a perpendicular
translation--
perpendicular to the rotation axis A--
and the result is a new rotation operation through the
same angle, but at a different location.
It's at a distance x, which is equal to T over 2 times the
cotangent of alpha over 2 along the perpendicular
bisector of T.
So this is a general theorem.
This is a fact of two-dimensional life that is
always going to be true regardless of alpha.
And the only constraint is that the translation has to be
added perpendicular to the locus of the rotation axis,
and that has to be the case in two dimensions, because the
rotation axis has to always be normal to the space, the
two-dimensional space.
In three dimensions, we'll have to generalize this to
make the translation have arbitrary orientation relative
to the rotation axis, but we've got enough to chew on at
the moment.
Yes.
AUDIENCE: [INAUDIBLE]
PROFESSOR: The distance x is the distance up along the
perpendicular bisector.
I'm sorry.
I said it but I didn't write it in.
OK?
Let me go through it again since we've got
everything on the board.
We start with the original object, rotate it by alpha,
and then we then further map it to a new location by the
translation T.
Both motifs are right-handed, but they're not parallel, so
therefore, they have to be related by a rotation.
If we ask what is the locus about which this location has
occurred, we use the fact that the locus of a rotation axis
is the only point that's left unmoved by the net
transformation, and the place where this translation
intersects the translation that has been shifted over by
T is a point that sits up along the perpendicular
bisector of the translation by a distance T over 2 times the
cotangent of alpha over 2.
So that's in the abstract.
Let's now look at a particular application of this to our
addition of a 2-fold rotation to a translation.
So here's the translation T. We're going to put a 2-fold
axis and the only operation which it
possesses, namely A pi.
And according to our theorem, up along the perpendicular
bisector of T by distance x equals T over 2 times the
cotangent of alpha over 2 should be a
new operation B pi.
So in this case, this would be T over 2 times the cotangent
of pi over 2.
Cotangent of 90 degrees is 0.
So this says that you go 0 of the way along the
perpendicular bisector, which means you stay right at the
midpoint of T. So this says that if you rotate through 180
degrees, follow that by this translation, you should find
that the net effect is a rotation of pi over 2 about
this point.
Boggles the mind, but let's show that it works.
Here is an initial point, initial motif number 1, and it
is right-handed.
We rotate by 180 degrees.
Here's number 2, it stays right-handed, and I pick it up
and I move it by this translation, and it moves from
here to here.
Here's number 3, it stayed right-handed, and the way I
get from 1 to 3 in one shot is to rotate 180 degrees about
the midpoint of the translation.
How about that?
People are so astounded that they're stunned, except for
the people who couldn't see what I was doing, and they're
moving back and forth frantically trying
to see what I wrote.
I'll let you catch your breath, and then let's do
this, use this to derive the two-dimensional space group,
the plane group that results when we add a 2-fold access to
an oblique net.
And it's turned out very simply, because if I've got
the operation B pi here, that implies that I've got a 2-fold
axis, because that's the only operation that exists within a
2-fold axis.
So I'll start by putting in a 2-fold axis here.
I'll combine A pi with this translation T1, I get a new
rotation operation, B pi, in the middle of this
translation, so there's a new 2-fold axis there.
The 2-fold axis here hanging at this lattice point as well,
and at these other lattice points.
Combine the rotation A pi with this translation T2, I get a
new operation, B pi, sitting in the middle of this
translation, so I've got all I need to say that there's a
2-fold axis there.
That 2-fold axis will get translated down here by T1.
This 2-fold axis will be translated over to here by T2.
Then the only other thing that I have to combine this
translation with is the translation T1 plus T2, and
that combined with A pi says I should have B pi in the middle
of that translation as well, so I'll get another 2-fold
axis in the middle of the cell.
And that's it.
We've got our first nontrivial two-dimensional space group.
And what do we call this?
Confusing as hell, some might say, but crystallographers
would say that it's a primitive lattice combined
with a 2-fold axis, this is P2.
I don't have to tell you that it's an oblique net, because
you know that that is all that a 2-fold axis requires, that
the net be oblique.
What does the pattern look like?
I'll say another thing that's so simple, it takes a while
for it to sink in.
The pattern of a plane group is the pattern that is
generated by the point group merely hung at every lattice
point in the net.
So if this is what a 2-fold axis does, relates a pair of
motifs like this, a pattern for P2 is this pair hung at
every lattice point.
So all of the objects are of the same pirality, all either
left-handed or right-handed.
I've got the same pair at every corner of the cell.
And notice that these new symmetry elements that popped
up just like what we found in the point groups, the new
mirror planes that came arose as a consequence of the
combination that we had made.
And they were independent in most cases, 3M being the one
case where they weren't.
These additional three 2-fold axes are
independent 2-fold axes.
They're different kinds of 2-fold axes than the ones at
the corner.
Not only are they not equivalent by translation, but
they do different things relative to the motifs that
are in the pattern.
This 2-fold axis has a different relation relative to
the two closest motifs than does this one,
than does this one.
They relate different objects pairwise in the pattern.
And no two of these 2-fold axes describe the same
pairwise relationship.
So these are independent 2-fold axes.
Independent in the sense that they do different things
relative to the motifs in the pattern, different in the
sense that there is no other operation that throws this one
into this one, and therefore, of necessity, would require
that the pair about this 2-fold axis be the same as the
pair about this 2-fold axis.
OK.
So there is a first nontrivial two-dimensional space group.
I should mention--
probably an obvious fact--
that this also has a cousin in three dimensions.
If I just imagine all of these 2-fold axes extending out
through space perpendicular to the plane of the blackboard,
and take another translation that is also perpendicular to
the blackboard, I would have the three-dimensional space
group which has the same symbol except capital P,
indicating that it's a space lattice.
So almost at no extra charge, we've done all the work to
determine one of the
three-dimensional space groups.
Yes?
AUDIENCE: If you change the order of the combination of
the translation rotation, do we get the same operation?
PROFESSOR: OK.
That is a good question.
I really meant to mention it earlier in connection with the
point groups.
Does interchange of the order of the operations in a
combination change the result, or change, in other words, the
symmetry element that is equivalent to the net
combination of the two?
How many think yes, it does change?
AUDIENCE: Could you say that again?
PROFESSOR: OK, if we have two operations A and B, is
operation A followed by B the same result as the operation B
followed by A?
Does the order make a difference?
And how many say yes--
AUDIENCE: The order makes a difference?
PROFESSOR: The order does make a difference.
How many say no?
Well, this is one of these happy occasions where I can
say you're all right.
Sometimes yes, sometimes no.
So let's ask when does it make a difference and
when does it not.
Let's look at one of our point groups.
Let me take a non-trivial point group.
This would be point group 4MM.
OK?
So those are all mirror planes of two different kinds.
And let me quickly sketch in the pattern.
The pattern is just a pair of things hanging on the mirror
plane, and that pair is rotated by 90 degrees.
OK.
Suppose we look at the operation A pi over 2, a
90-degree rotation in a counterclockwise direction,
and let the other operation be sigma 1, this
reflection plane here.
Suppose we reflect and follow that by A pi over 2.
We reflect to here and then we rotate by 90 degrees, and
that's going to give us the number 3 sitting up here.
So we reflect it and then rotate it.
How about if we first rotate that same object by A pi over
2 and follow it by sigma 1.
So we'll rotate number 1 to here, so this is 2 prime, and
you can see we're already in trouble.
If we now reflect along sigma 1, this is 3 prime.
So in one case we ended up here, in the other case we
ended up here.
So the order did make a difference.
Let us look at another example.
Let's look at 2MM.
So here's an operation sigma 1, here's an operation sigma
2, so let's first do sigma 1 followed by A pi.
So here's object 1, reflect to here, and then we rotate by
180 degrees, and this is number 3.
Now let's do the operation A pi and follow it by sigma 1.
So we'll take number 1 and rotate it up to here.
So here's number 1 prime.
And then we reflect by sigma 1, and 3 and 3 prime are
exactly the same.
So it didn't make a difference.
So the answer is sometimes yes, sometimes no, which is
interesting, but even more interesting
is how can you tell?
Yeah, you have a question?
AUDIENCE: In that first identity you drew--
PROFESSOR: Yep.
AUDIENCE: The right side, it seems like your commas are
drawn backwards and that just affects--
PROFESSOR: Could be.
Mm, no, they're--
ah, these are drawn backwards.
They should all have their tails pointing in.
No, it doesn't work, actually, because even if I screwed up
on the orientation, the positions are different.
Again, first time I reflected, and then
rotated, so this is 3.
If I rotate and then reflect, I'm way over here.
So this is one location, this is another location, and even
if I screwed up by having the tails pointed the wrong
orientation, still not the same.
But here, clearly, reflecting and rotating is the same as
rotating and reflecting.
I end up in the same place.
So how can you tell?
It's a very difficult thing to prove that this is the case,
but you can interchange the order of operations when the
two symmetry operations that are involved leave each other
untransformed.
And if the two symmetry operations move one another,
then changing the order changes the net result.
It doesn't change the nature of the operation, but it does
change the locus about which it operates.
Let me show you what I mean.
Here's 2MM.
The 2-fold axis takes this mirror plane,
turns it upside down.
It takes this mirror plane and flips it over.
The mirror plane is left unchanged in both cases.
The mirror plane has the 2-fold axis sitting right on
it, so it doesn't change the location of the 2-fold axis
when it acts on it.
And similarly, this mirror plane takes this mirror plane
and reflects it, and it doesn't change anything.
So all of these operations--
the three independent operations, sigma 1,
sigma 2, and A pi--
all leave their locus of operation unchanged.
Whereas that, clearly, is not the same for a 4-fold axis and
a mirror plane.
The mirror plane, when it acts on the 4-fold axis, leaves it
unchanged, but the 4-fold axis, when it acts on the
mirror plane, moves it into a new,
entirely different location.
And when that is the case, the order does make a difference.
So I can take that corny joke of the mathematicians, what is
purple and commutes?
An Abelian grape.
I can say, what is purple and commutes?
It's a purple 4-fold axis in a mirror plane, a purple 4MM.
OK, what other aspect of the plane groups and three
dimensional space groups is an analytic representation of the
way the plane group maps atoms around?
And I see that it is five of the hour, so let me just make
one final statement to indicate what I'm going to
talk about next time.
Here's the arrangement of 2-fold axes, and if you have
an atom sitting in here, and this is T1 and T2, and we use
this as the basis of a coordinate system, so this
atom sits at a location x along T1 and y along T2.
And then they ask, what other atoms are related to it?
Well, that 2-fold axis will move x to minus x, and move y
to minus y, and those are the only two atoms I get.
So a characteristic of this plane group, if I were
describing to you a crystal structure which had this
symmetry, I might say something like a lithium in a
position with coordinates xy, and therefore minus x minus y
would also have to occur.
I have an oxygen at a different number, x prime y
prime, and that must then also occur at minus x
prime minus y prime.
And it looks as though any time you throw in an atom at a
location xy, you get a symmetry related companion to
it with negative coordinates.
And this is what is called the general position
of the space group.
There's nothing special about it, and you put in an atom
here, you get another atom here.
This is going to be a nice economical way of describing
an atomic arrangement, particularly if you realize
that there are almost 150 atoms that appear in the unit
cell for some of the cubic symmetries.
Drop in one, wow, all hell breaks loose, 150 other ones.
But what you can do is codify the coordinates of all these
symmetry-related atoms.
And that is a great utility in describing crystal structures,
because I only have to give you the coordinates of one
representative atom.
But this is not the only thing that happens.
This is the general position because it's general relative
to its location with respect to the symmetry elements.
If x and y were both 0, I would have one atom moving to
the origin to coincide with the second one.
And if I put one in at 0, 0, that's all I'm going to get.
The 2-fold axis is just going to twirl it around, and the
translation will repeat it to the other lattice points.
So if I put in an atom at 0, 0, that's all
I'm going to get.
And the same thing would occur for any of these four
independent 2-fold axes.
If I let x and y migrate to 0, 1/2, this atom will join
together with this atom, and I would get just one atom
sitting there.
So there's a position also of the form 0, 1/2, and a
position 1/2, 0, and a position 1/2, 1/2 that are the
location of these four distinct 2-fold axes.
OK, to have a shorthand way of referring to the general
position in these positions, which are called special
positions--
what's special about them is that they're right on top of a
symmetry element, and whenever that happens, the number per
cell is a sub-multiple of the number in the general
position, because coalescence has occurred.
And so they're three independent 2-fold axes, so
there are four locations where you get only one atom per cell
rather than two.
And then to give you a way of referring to these positions
readily, they are labeled starting with the most
specialized, just going through the alphabet, position
a, position b, position c, position d, position e, which
is the general position.
And then another piece of information that's almost
trivial here, but is not for more complicated symmetries,
what's called the rank of the position, and that is the
number that you get per cell, one per cell, one per cell,
and for the general position you get two.
So the rank of the position is just the number per cell.
And then, if it's a special position, the atom has to sit
at a site of some symmetry, and in this case, the atoms
all sit at 2-fold axes for the special position.
The general position always has site symmetry 1 always.
So that's the characteristics of the way in which this
symmetry can move atoms around.
And this is of great utility in interpreting structures.
For example, suppose our motif was not an individual atom,
but was a molecule.
And suppose you had one molecule per cell.
That would tell you the molecule has to fit in a
special position.
And the special positions are all 2-fold axes.
So if the molecule has to sit on a 2-fold axis, the
configuration of that molecule has to have at least a 2-fold
symmetry, or else you can't fit in one per cell in this
particular plane group.
So you can use this information to determine
structures.
I'm going to give you a problem a little bit later on
when we get to three-dimensional
crystallography, and I'm going to give you the magnitude of
the cell edge and the density of rock salt.
And from that, in 10 minutes, you can find out that there's
only one possible structure for rock salt.
The Braggs fiddled around for the better part of a year
doing x-ray experiments and trying to calculate what the
intensities would be for different atomic positions.
You, in another two weeks, can get the same answer in 10
minutes knowing only the lattice constant and the
density and the space group.
So a lot of physical consequences of the nature of
the structure, the confirmation of the molecule,
the extent or absence of solid substitution of one species
for another, can be dredged out of the properties of the
space group.
OK, I've run 2 minutes over, but we took our break two
minutes later, so let's call it a day and I shall see you
on Thursday, hopefully armed with pictures so that I can
personally present each one of you with your corrected
problem sets.