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PROFESSOR: Do them individually
so I can continue
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to put names and
faces together.
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I'm happy to announce that
the registrar has now got
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everybody's photograph
online for
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registration in the course.
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So anonymity is a thing of the
past, so you have to watch
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your step from now on.
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I handed back, to those who
didn't get it, problem set
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number four, which asked you
to tackle some patterns,
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nontrivial patterns.
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And actually, that was a dirty
trick, because we hadn't, at
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that point, derived the plane
groups, and you really didn't
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know what to do or
what to look for.
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But nevertheless, it got you
thinking about patterns and
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some of the symmetry elements
which we had discussed up to
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that point.
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At this point we have derived
exhaustively every last one of
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the 17 plane groups.
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So now you are armed with this
new-found power, and when
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faced with a pattern, you should
know exactly what to
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look for and how to go
about deciding what
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plane group it is.
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At the very least, you'll
have the drawings of the
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arrangement of symmetry elements
in the plane groups
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before you, and you can work by
the process of elimination.
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For example, high symmetry
usually hits you right between
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the eyes, and if something is
square-ish, you can pretty
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quickly guess that it's based
on a square lattice.
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And if it has a square lattice,
there jolly well
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better be a 4-fold axis in there
that makes it square.
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If you can find the 4-fold axis,
then you have to ask
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yourself only three questions.
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So 4-fold axis, fine.
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Is there a mirror
line in there?
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Yeah.
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Does the mirror line go through
the 4-fold axis?
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Then it is P 4 MM.
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And you know just where to
look for everything else,
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including these very
subtle glide planes
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that are hard to spot.
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If there is a mirror plane
there, but it doesn't go
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through the 4-fold axis,
then it's P 4 MG.
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And if there is no mirror
plane, then it's P 4.
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So just by asking one or two
simple questions, you can
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narrow it down to what the
plane group has to be.
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This is another indication that
the informed intellect is
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always more than a match
for sheer, raw native
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intelligence.
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If you know what to look
for, it's a lot easier.
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Because you really didn't have
much practice with patterns,
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we're having a quiz, as you
know, next Thursday, that will
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cover up through completion of
the plane groups and not the
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material we've been doing now.
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So I think it might be of use
to you to have some practice
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analyzing a few more patterns.
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So there are four additional
patterns in this problem set.
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As always, it's optional, but if
you would like to try them,
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and you want to see if you've
got them right, come in and
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see me tomorrow or on
Thursday morning.
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I'd be happy to go over them
with you on the spot.
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So this is for practice.
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And some of you did
extraordinarily well
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on the first try.
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Others, I think, could use
the additional practice.
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There were a few people who
identified the plane group
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correctly, but got the name
that's assigned to it wrong.
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And one other very confusing
thing is that there is one
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plane group that has the symbols
G and M, and another
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plane group which has
the symbols M and G.
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So if you found a mirror plane
and a glide plane is an
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independent symmetry plane,
when is it MG,
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and when is it GM?
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I have a little mnemonic
device.
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GM, general manager, is the
guy who sits on top of the
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organization.
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So GM should be the
plane group that
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has the highest symmetry.
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P 4 GM, P 4 general manager.
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Now that is--
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hey, it works for me.
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But another way of saying it
is that there are two plane
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groups, one has MG and the other
has GM, and I like cars,
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and I think an MG is much
classier than anything that
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General Motors, GM, puts out,
so MG should be the one of
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highest quality, highest
symmetry.
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And that's just the reverse.
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But whatever works for you.
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You could keep them straight
through that simple algorithm.
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And as they say, if it works for
me, but if it doesn't work
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for you, don't use it.
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All right.
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What I will bring in during
intermission, for those of you
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had trouble identifying
translations in the patterns
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that we handed out earlier, I've
taken these and put them
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on overhead transparencies.
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And I'll have two of each.
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So if you don't see the symmetry
or translations that
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are present, you can actually
take one pattern and
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physically move it and lay it
on top of the other one, and
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that's a good way to convince
yourself what a translation
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looks like when it occurs
in a pattern.
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So I'll bring those in at
our break between class.
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All right, any questions
before we move on?
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Any questions that have arisen
as you have gotten
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ready for the quiz?
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You haven't gotten ready
for the quiz yet, so
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there are no questions.
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That's OK.
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I know how things work at MIT.
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You deal with one crisis
at a time.
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Any questions?
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Anything you want to go over?
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There was one interesting
wrinkle in a problem that I
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had not encountered before,
and this was the one that
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asked you to look at, in two
dimensions, a plane with
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indices h and k.
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And then, when h and k were
mutually prime, to move that
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plane by the translations plus
T1 and plus and minus T2, and
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then show that the number of
intervals between the origin
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and the intercept plane, the one
that hit lattice points on
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both translations, was equal
to h times k if they were
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mutually prime.
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That is true only if the
lattice is primitive.
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And what the problem said was to
pick one of the cells that
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you used in the first
problem, number one.
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Well, that problem asked you
to begin with identifying
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different primitive cells.
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If you take a multiple cell,
this operation of going plus
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and minus T1 does not put a
lattice line through each of
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the lattice points.
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If you picked a double cell,
that process decorated only
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half of the lattice points with
planes, and the other
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half sat there with nothing
hanging on them at all.
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The key to the difference, if
you looked at a double cell,
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was that if h plus k was even,
then you automatically got a
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plane on every lattice point.
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If h plus k was odd, as it would
have been for the plane
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2,1 for example, 2 plus 1 is 3--
hey, this isn't even one
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of my good days--
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then half of the lattice points
did not have planes
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hanging on them.
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Now, there's great relevance
of this observation to
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diffraction, and you probably
are all familiar, if only
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vaguely, with the magic rules
that say, if h plus k is equal
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to 3 pi plus 4, then the
intensity is identically 0.
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Well, for a double cell, the
lattice planes that are
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repeated by translation diffract
x-rays, and there is
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no reason why the intensity
should be
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something other than 0.
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So here comes, a la Bragg, an
x-ray beam coming in at angle
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theta, and then you say you
get diffraction when
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scattering from this lattice
plane is exactly in
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phase with this one.
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And this gives the familiar
relation that an integral
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number of wavelengths is equal
to 2 d sine of theta when the
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crystal diffracts.
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So there is exactly-- this
says there's exactly 2 pi
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phase difference or n lambda
path difference between these
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two planes.
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Now, if the lattice would be a
double cell, then there is an
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additional lattice point in here
that does not get a plane
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hung on it.
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So if the lattice is a double
cell, there's another plane
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that has to hang on this lattice
point, and that one is
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exactly out of phase with this
plane, and the intensity is 0.
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So this observation that a
non-primitive lattice has a
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interplanar spacing that is a
sub-multiple of that of a
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primitive lattice gives some
insight into why certain
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reflections--
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certain diffraction maxima-- are
identically 0 in intensity
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for a crystal that has a
non-primitive lattice.
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So that is something I had
not noticed before.
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I should have, but I will phrase
the problem a little
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bit more precisely
in the future.
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All right.
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So to conclude my preamble, I
hope you'll try playing with
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some of the four additional
patterns that I handed out,
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just to give yourself
some practice.
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And the implication of this is
that you're going to see a
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pattern on the quiz, and I will
tell you that you will.
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So if you want to see how you
did on the patterns that I
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distributed, please come in
and talk to me about them.
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All right then.
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Let me remind you where
we were last time.
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We started to begin to build a
framework of symmetry elements
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in three dimensions.
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And we asked the question, what
would happen if we take a
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first rotation axis, A alpha,
combine it with a second
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rotation axis, B beta, in
such a way that they
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intersect at a point.
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This means that their operation
and reproducing
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atoms or motifs is going to
leave at least one point in
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space unchanged, and that will
be the point of intersection.
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We ask ourselves, what will
be the combined effect--
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we have two operations
in space--
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00:11:07,580 --> 00:11:11,160
what would be the combined
effect of rotating alpha
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degrees about A followed
immediately by beta degrees
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00:11:15,910 --> 00:11:16,710
about B.
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00:11:16,710 --> 00:11:20,785
So what we're going to do then
is to take a first motif--
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and let's say it's
left handed.
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Being a left-handed person, I
like to give right handed
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00:11:26,600 --> 00:11:29,080
motifs and left handed
motifs equal time.
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If we rotate that through an
angle alpha, and this is
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00:11:34,760 --> 00:11:37,830
number 2, it will stay
left handed.
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00:11:37,830 --> 00:11:42,100
Then if we rotate that by B
beta, it'll move it over here
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to number 3 and it will stay
left handed as well.
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00:11:45,682 --> 00:11:49,980
And the question is then, what
net operation is equivalent to
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the combined operation of these
two transformations?
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00:11:55,258 --> 00:11:59,070
And to specify the type of
operation is really a
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no-brainer.
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00:11:59,730 --> 00:12:03,500
All of these motifs are of the
same corality so the only
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00:12:03,500 --> 00:12:04,950
thing that can relate them is
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00:12:04,950 --> 00:12:08,260
translation or another rotation.
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00:12:08,260 --> 00:12:12,420
And clearly the first and the
third have no reason to be
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00:12:12,420 --> 00:12:15,450
parallel to one another, and the
distance between them is
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00:12:15,450 --> 00:12:19,120
going to depend on how far they
are away from the axis,
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00:12:19,120 --> 00:12:23,260
so translation won't do the
job, a and the only thing
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00:12:23,260 --> 00:12:29,270
that's left as a net operation
that's equivalent to those two
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00:12:29,270 --> 00:12:35,050
steps is rotation about
a third axis C.
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00:12:35,050 --> 00:12:39,240
And what we're going to do today
is answer the question,
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00:12:39,240 --> 00:12:48,670
where is axis C located, and
what is the angle of rotation,
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given the value of alpha and
beta and the angle between
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these two axes?
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00:12:55,010 --> 00:12:59,540
And let's define that
as a lowercase c.
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So clearly the location of the
axis and the amount of the
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rotation is going to be a
function of alpha, beta, and
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the angle between them.
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00:13:08,350 --> 00:13:11,970
We want this to be a combination
of operations that
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exists in a symmetry
operation.
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00:13:16,470 --> 00:13:19,260
And if this is to be a
crystallographic symmetry,
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these will be restricted to
the angular throws of a
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1-fold, 2-fold, 3-fold,
4-fold, a 6-fold axis.
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We can take these two at a time,
ask what the net effect
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is if we combine at
a given angle c.
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00:13:33,020 --> 00:13:38,890
And what comes out here must
be a rotation which is also
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crystallographic.
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So there are going to be severe
constraints on this
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combination.
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00:13:48,730 --> 00:13:51,440
Two rotations about an
intersecting point will always
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00:13:51,440 --> 00:13:56,060
be a third rotation, but if
this is to be a set of
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00:13:56,060 --> 00:14:00,060
operations in a symmetry group,
the result must be
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00:14:00,060 --> 00:14:02,880
crystallographic.
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00:14:02,880 --> 00:14:06,320
That's a tough problem, and how
will we undertake it is
244
00:14:06,320 --> 00:14:07,570
going to be non-intuitive.
245
00:14:11,370 --> 00:14:14,870
OK, the problem is most readily
treated with spherical
246
00:14:14,870 --> 00:14:16,610
trigonometry.
247
00:14:16,610 --> 00:14:23,530
So on the surface of a sphere,
I'm going to map the point at
248
00:14:23,530 --> 00:14:25,300
which A alpha protrudes--
249
00:14:25,300 --> 00:14:28,190
and I'll call this point A--
250
00:14:28,190 --> 00:14:32,510
and then I'll mark out the point
where axis B beta exits
251
00:14:32,510 --> 00:14:41,210
the sphere, and I'll mark that
point B. And this is the angle
252
00:14:41,210 --> 00:14:43,120
C.
253
00:14:43,120 --> 00:14:47,650
And we said that in spherical
trigonometry, the measure of
254
00:14:47,650 --> 00:14:51,910
the length of the arc separating
A and B is given by
255
00:14:51,910 --> 00:14:55,630
the angle subtended at the
center, so the length of this
256
00:14:55,630 --> 00:14:58,530
distance between A and
B is the angle c.
257
00:14:58,530 --> 00:15:01,500
Again, it sort of boggles the
mind when you measure lengths
258
00:15:01,500 --> 00:15:04,500
in terms of degrees rather
than some metric unit.
259
00:15:07,657 --> 00:15:08,110
All right.
260
00:15:08,110 --> 00:15:14,350
So I will now not bother to show
the sphere on which the
261
00:15:14,350 --> 00:15:16,080
geometry is taking place.
262
00:15:16,080 --> 00:15:20,710
I'll just draw A and B,
and this is the arc
263
00:15:20,710 --> 00:15:22,850
between them, c.
264
00:15:22,850 --> 00:15:26,590
And somewhere or other there
will be some third axis, C,
265
00:15:26,590 --> 00:15:29,900
which is going to be the
combined effect of the
266
00:15:29,900 --> 00:15:34,990
rotation about axis A and axis
B. So what I would like to do
267
00:15:34,990 --> 00:15:39,530
is to locate the position
of this axis C.
268
00:15:39,530 --> 00:15:43,880
In order to do that, I'll have
to know what the angle between
269
00:15:43,880 --> 00:15:47,600
A and C is, and I'll call that,
by analogy to what I've
270
00:15:47,600 --> 00:15:49,400
done here, I'll call that b.
271
00:15:49,400 --> 00:15:53,250
And I'll want to know what the
angle between B and C is, and
272
00:15:53,250 --> 00:15:56,270
I'll call that angle a.
273
00:15:56,270 --> 00:16:00,190
So again, going back to three
dimensions momentarily, if
274
00:16:00,190 --> 00:16:04,920
this is the rotation operation
C gamma, and this is A alpha,
275
00:16:04,920 --> 00:16:11,200
and this is B beta, the axis c
is this, the angle b is this,
276
00:16:11,200 --> 00:16:12,940
and the angle a is this.
277
00:16:19,240 --> 00:16:19,480
OK.
278
00:16:19,480 --> 00:16:22,870
It's a non-trivial problem and
it is not by accident that the
279
00:16:22,870 --> 00:16:27,080
solution to this problem was
first given by a very, very
280
00:16:27,080 --> 00:16:32,130
famous mathematician, Leonhard
Euler, and this construction
281
00:16:32,130 --> 00:16:34,950
that we're about to go through
is called Euler's
282
00:16:34,950 --> 00:16:36,200
construction.
283
00:16:41,719 --> 00:16:42,200
All right.
284
00:16:42,200 --> 00:16:50,290
Let me find where these
different locations
285
00:16:50,290 --> 00:16:51,850
are going to be.
286
00:16:51,850 --> 00:16:55,230
We've specified the location of
point A and the location of
287
00:16:55,230 --> 00:17:00,100
point B, and we know that the
angle between them is the
288
00:17:00,100 --> 00:17:04,460
length of the arc ab, which is
the angle between A and C. So
289
00:17:04,460 --> 00:17:06,690
let me now do some
constructions.
290
00:17:06,690 --> 00:17:15,270
Let me find a great circle that
by design is alpha over 2
291
00:17:15,270 --> 00:17:20,619
away from the arc ab, and
that is by construction.
292
00:17:20,619 --> 00:17:24,859
And I'm going to say, then, that
if A alpha works in this
293
00:17:24,859 --> 00:17:29,900
direction, the operation of A
alpha is going to take this
294
00:17:29,900 --> 00:17:34,020
great circle and move it over
to a great circle which is
295
00:17:34,020 --> 00:17:39,730
alpha over 2 on the other
side of the arc ab.
296
00:17:39,730 --> 00:17:41,100
Fine, you say, so what?
297
00:17:41,100 --> 00:17:45,120
Well, just going to leave
those there for now.
298
00:17:45,120 --> 00:17:49,300
I'll have B beta work
in the same sense.
299
00:17:49,300 --> 00:17:54,660
And I'm now going to create a
line here that by construction
300
00:17:54,660 --> 00:18:00,250
is beta over 2 on one side of
the arc ab, and if I let B
301
00:18:00,250 --> 00:18:06,130
beta go to work, that will map
this great circle over to a
302
00:18:06,130 --> 00:18:09,660
new location beta over
2 on the other side.
303
00:18:13,190 --> 00:18:17,010
What has this done for me, other
than perhaps confuse me
304
00:18:17,010 --> 00:18:19,530
and clutter the diagram?
305
00:18:19,530 --> 00:18:23,710
Well, now I'm going to determine
unequivocally the
306
00:18:23,710 --> 00:18:27,630
location of the axis C, and
where it emerges from the
307
00:18:27,630 --> 00:18:29,750
reference here.
308
00:18:29,750 --> 00:18:32,880
And how will I do that?
309
00:18:32,880 --> 00:18:37,680
I'm going to use a definition
that may have seemed trivial
310
00:18:37,680 --> 00:18:40,550
the first time we made
the observation.
311
00:18:40,550 --> 00:18:45,610
I said that a symmetry element
is the locus of points that is
312
00:18:45,610 --> 00:18:49,832
left unmoved by an operation.
313
00:18:49,832 --> 00:18:52,730
OK?
314
00:18:52,730 --> 00:18:57,540
I rotated by A alpha from here
to here, that took everything
315
00:18:57,540 --> 00:19:00,800
along this line and mapped it
to a new location here.
316
00:19:00,800 --> 00:19:07,290
I took this line and rotated
it by B beta, and that took
317
00:19:07,290 --> 00:19:10,760
everything along this line and
moved it to a new location.
318
00:19:10,760 --> 00:19:15,010
So my question now is if I
rotate by A alpha and then
319
00:19:15,010 --> 00:19:19,120
rotate in the same direction
by B beta, what
320
00:19:19,120 --> 00:19:20,405
point is left unmoved?
321
00:19:24,110 --> 00:19:26,940
It's only one point that can
make that claim, and that is
322
00:19:26,940 --> 00:19:30,320
where these two great
circles intersect.
323
00:19:30,320 --> 00:19:33,510
The rotation A alpha will
take this location--
324
00:19:33,510 --> 00:19:36,250
and I'm going to call it C
because I've identified now
325
00:19:36,250 --> 00:19:38,760
what it is-- it's going to take
C and move it over to
326
00:19:38,760 --> 00:19:42,470
here, call that C prime, and
then B beta takes that point
327
00:19:42,470 --> 00:19:45,350
and only that point, and
restores it back to its
328
00:19:45,350 --> 00:19:47,410
original location.
329
00:19:47,410 --> 00:19:50,770
So this, then, ladies and
gentlemen, is where the
330
00:19:50,770 --> 00:19:54,420
rotation axis C gamma pokes
out of the sphere of
331
00:19:54,420 --> 00:19:55,670
reflection.
332
00:20:00,680 --> 00:20:04,940
Still don't know what this angle
is in here, and I would
333
00:20:04,940 --> 00:20:11,820
dearly love to know what these
arcs b and a are, and then I
334
00:20:11,820 --> 00:20:15,170
will have specified all three
of the interaxial angles
335
00:20:15,170 --> 00:20:22,490
between A, B and C.
336
00:20:22,490 --> 00:20:28,420
OK, let me do something quite
similar to what I did before.
337
00:20:28,420 --> 00:20:32,290
I'm going to again let A alpha
work on a particular point,
338
00:20:32,290 --> 00:20:35,090
and then let B beta map it.
339
00:20:35,090 --> 00:20:37,830
So here's A alpha,
here's B beta.
340
00:20:37,830 --> 00:20:41,580
And now I'm going to look
specifically at how these two
341
00:20:41,580 --> 00:20:46,820
rotations transform point A,
where A is the point at which
342
00:20:46,820 --> 00:20:50,000
axis A alpha pokes out
of the sphere.
343
00:20:50,000 --> 00:20:53,640
A alpha, when it acts on this
point, does nothing to it.
344
00:20:53,640 --> 00:20:55,240
It leaves it alone.
345
00:20:55,240 --> 00:21:00,820
B beta is going to map A to
a new location, A prime.
346
00:21:07,110 --> 00:21:11,750
Now, doing A alpha and following
up by B beta is
347
00:21:11,750 --> 00:21:17,230
supposed to be equal to
the rotation C gamma.
348
00:21:17,230 --> 00:21:24,770
So that says that this point and
this point must be related
349
00:21:24,770 --> 00:21:26,170
by the rotation gamma.
350
00:21:28,740 --> 00:21:29,720
So say that again.
351
00:21:29,720 --> 00:21:32,360
We're doing exactly what we did
here except we're starting
352
00:21:32,360 --> 00:21:35,720
with an initial point, not this
arc, but we're starting
353
00:21:35,720 --> 00:21:39,480
with the specific point A,
operate on it by A alpha, it
354
00:21:39,480 --> 00:21:42,120
twirled around but stays put.
355
00:21:42,120 --> 00:21:47,480
Rotate that by B beta, it goes
through a total angle beta to
356
00:21:47,480 --> 00:21:49,030
this location here.
357
00:21:49,030 --> 00:21:52,340
The net effect of getting from A
to A prime is supposed to be
358
00:21:52,340 --> 00:21:56,530
the rotation C gamma, so this
angle is then gamma and this
359
00:21:56,530 --> 00:21:59,835
is the location of C. OK?
360
00:22:04,585 --> 00:22:08,620
OK, one other step that's
a fairly easy one.
361
00:22:08,620 --> 00:22:14,700
This length is equal to this
length, because they were
362
00:22:14,700 --> 00:22:18,010
produced by rotation.
363
00:22:18,010 --> 00:22:29,490
This side is common to these two
triangles, and this angle
364
00:22:29,490 --> 00:22:33,110
then is beta over 2, this
is beta over 2.
365
00:22:33,110 --> 00:22:39,470
And if these two triangles, A,
B, and C, that triangle is
366
00:22:39,470 --> 00:22:47,110
similar to A prime BC, and
therefore I can say that angle
367
00:22:47,110 --> 00:22:58,780
A prime CA is identical to ACB,
so therefore this angle
368
00:22:58,780 --> 00:23:02,540
has to equal this angle, and if
the total angle is gamma,
369
00:23:02,540 --> 00:23:05,410
this is gamma over 2, and
this is gamma over 2.
370
00:23:11,090 --> 00:23:14,020
So now let me extract from this
the information that I
371
00:23:14,020 --> 00:23:15,550
would like to use.
372
00:23:15,550 --> 00:23:20,560
Here are three axes, A alpha,
B beta, and C gamma.
373
00:23:20,560 --> 00:23:23,820
This angle in here
is gamma over 2.
374
00:23:23,820 --> 00:23:27,930
This angle in here is beta over
2, and this angle in here
375
00:23:27,930 --> 00:23:29,180
is alpha over 2.
376
00:23:32,200 --> 00:23:36,030
And let me emphasize that in
this magic triangle, out of
377
00:23:36,030 --> 00:23:40,600
which we're going to extract
some dazzlingly profound
378
00:23:40,600 --> 00:23:45,920
stuff, it is half the angular
throw of the rotation axes
379
00:23:45,920 --> 00:23:49,430
that appear in here as these
spherical angles, and not the
380
00:23:49,430 --> 00:23:51,720
entire angle of rotation.
381
00:23:56,780 --> 00:24:01,020
So here's how properties of the
three rotation axes are
382
00:24:01,020 --> 00:24:02,400
related one to another.
383
00:24:05,620 --> 00:24:11,340
And now, we introduced without
proof last time something
384
00:24:11,340 --> 00:24:16,140
called the law of cosines in
spherical trigonometry.
385
00:24:16,140 --> 00:24:20,120
And I not only do not want to
prove it, but I have no idea
386
00:24:20,120 --> 00:24:23,010
how I would go about doing so,
but that doesn't prevent me
387
00:24:23,010 --> 00:24:24,480
from using it.
388
00:24:24,480 --> 00:24:29,460
So if here are three edges, a,
b, c, and three angles in
389
00:24:29,460 --> 00:24:33,810
there, A, B, and C, we said
that the law of cosines in
390
00:24:33,810 --> 00:24:37,350
spherical trigonometry,
analogous in a way to the law
391
00:24:37,350 --> 00:24:41,070
of cosines and plane geometry,
except since the lengths of
392
00:24:41,070 --> 00:24:43,500
the triangles are measured
in degrees, there are
393
00:24:43,500 --> 00:24:47,460
trigonometric functions of these
angles that appear in
394
00:24:47,460 --> 00:24:48,880
the law of cosines.
395
00:24:48,880 --> 00:24:54,750
This says that cosine of b
cosine of c plus sine b sine
396
00:24:54,750 --> 00:25:00,745
of c times cosine of a is
equal to cosine of a.
397
00:25:03,880 --> 00:25:09,630
So this now is an interesting
relation that we can apply to
398
00:25:09,630 --> 00:25:13,620
this spherical triangle, which
connects together the three
399
00:25:13,620 --> 00:25:16,080
rotation axes.
400
00:25:16,080 --> 00:25:24,770
Let me apply it to find the
angle c which we have picked
401
00:25:24,770 --> 00:25:30,850
as the angle between the initial
two axes a and b.
402
00:25:30,850 --> 00:25:34,650
That says that this should be
equal to cosine of b cosine of
403
00:25:34,650 --> 00:25:40,240
c, the angle between the other
two axes, plus sine of b sine
404
00:25:40,240 --> 00:25:54,340
of c times the cosine of angle
a, and angle a is cosine of
405
00:25:54,340 --> 00:25:55,738
alpha over 2.
406
00:26:00,260 --> 00:26:02,940
So all these quantities that
we'd like to determine are
407
00:26:02,940 --> 00:26:07,640
hooked together by the
law of cosines.
408
00:26:07,640 --> 00:26:13,060
And this is a lovely relation,
but it doesn't do us a bit of
409
00:26:13,060 --> 00:26:20,220
good, because in this relation
we know only one quantity, and
410
00:26:20,220 --> 00:26:23,530
that is the rotation
angle of a.
411
00:26:23,530 --> 00:26:28,170
We can pick the angle between
a and b, that's this, but I
412
00:26:28,170 --> 00:26:30,130
have no idea what these
other angles are.
413
00:26:30,130 --> 00:26:31,600
That's what I'd like
to find out.
414
00:26:31,600 --> 00:26:33,980
I'd like to find out the
angles at which three
415
00:26:33,980 --> 00:26:38,870
rotations have to be combined
in order that rotation about
416
00:26:38,870 --> 00:26:42,150
one followed by rotation about
the second be the third.
417
00:26:42,150 --> 00:26:45,650
So this equation is a beautiful
equation, but it
418
00:26:45,650 --> 00:26:48,860
involves everything that I
don't know and only one
419
00:26:48,860 --> 00:26:50,470
quantity that I do know.
420
00:26:50,470 --> 00:26:52,130
So it looks as though
we're up the creek.
421
00:26:52,130 --> 00:26:53,910
Yes, sir?
422
00:26:53,910 --> 00:26:59,637
AUDIENCE: So you're looking for
cosine c, so shouldn't it
423
00:26:59,637 --> 00:27:01,700
be cosine b cosine a?
424
00:27:01,700 --> 00:27:02,950
PROFESSOR: Oh, I'm sorry.
425
00:27:02,950 --> 00:27:04,200
Yeah, I did that wrong.
426
00:27:09,840 --> 00:27:10,340
Yeah.
427
00:27:10,340 --> 00:27:12,030
You're absolutely right.
428
00:27:12,030 --> 00:27:17,070
This should be cosine of a,
and this a goes with this
429
00:27:17,070 --> 00:27:18,205
alpha over 2.
430
00:27:18,205 --> 00:27:18,890
Absolutely.
431
00:27:18,890 --> 00:27:20,140
Sorry about that.
432
00:27:25,380 --> 00:27:29,000
So anyway, the point still
stands that what this equation
433
00:27:29,000 --> 00:27:32,230
involves is the three interaxial
angles, and I would
434
00:27:32,230 --> 00:27:35,460
like to know how I could combine
a and b to get it to
435
00:27:35,460 --> 00:27:39,170
come out to a crystallographic
rotation c, and where that
436
00:27:39,170 --> 00:27:42,460
location is relative to
the first two axes.
437
00:27:42,460 --> 00:27:45,710
So it involves everything I
don't know, and only one
438
00:27:45,710 --> 00:27:47,370
thing that I do.
439
00:27:47,370 --> 00:27:53,340
But now we introduce another
curious aspect of spherical
440
00:27:53,340 --> 00:27:55,830
triangles, which I mentioned
last time.
441
00:27:55,830 --> 00:27:57,450
You may have thought
that that's
442
00:27:57,450 --> 00:28:00,230
interesting, but who cares?
443
00:28:00,230 --> 00:28:06,640
Here are the three points, A,
B, and C, and these are the
444
00:28:06,640 --> 00:28:11,290
three arcs little c, little
a and little b.
445
00:28:11,290 --> 00:28:14,210
And then we said we could
construct something called the
446
00:28:14,210 --> 00:28:17,350
polar triangle of ABC.
447
00:28:17,350 --> 00:28:23,480
And what we would do, we would
find the pole of arc b, and
448
00:28:23,480 --> 00:28:27,340
that will be some
point B prime.
449
00:28:27,340 --> 00:28:31,680
We'll find the pole of arc
a, and that will be
450
00:28:31,680 --> 00:28:33,360
some point A prime.
451
00:28:33,360 --> 00:28:37,060
We'll find, similarly, the pole
of arc c, and that will
452
00:28:37,060 --> 00:28:39,210
be some point C prime.
453
00:28:39,210 --> 00:28:43,250
And now we can connect together
A prime, B prime, and
454
00:28:43,250 --> 00:28:45,440
C prime, and get something
that's
455
00:28:45,440 --> 00:28:47,005
called the polar triangle.
456
00:28:49,540 --> 00:28:51,400
And now comes the useful part.
457
00:28:51,400 --> 00:28:57,990
We said that a curious property
of the polar triangle
458
00:28:57,990 --> 00:29:05,290
is that the side of the polar
triangle plus the angle
459
00:29:05,290 --> 00:29:10,680
opposite it add up
to 180 degrees.
460
00:29:10,680 --> 00:29:14,790
In my original triangle, this is
beta over 2, this is gamma
461
00:29:14,790 --> 00:29:17,960
over 2, and this is
alpha over 2.
462
00:29:17,960 --> 00:29:23,370
So the length of this side is
going to be 180 degrees minus
463
00:29:23,370 --> 00:29:28,540
beta over 2, the length of this
side is going to be 180
464
00:29:28,540 --> 00:29:32,490
degrees minus alpha over 2, and
the length of this side is
465
00:29:32,490 --> 00:29:38,270
going to be 180 degrees
minus gamma over 2.
466
00:29:38,270 --> 00:29:43,630
And now let's use these angles
and these lengths in the law
467
00:29:43,630 --> 00:29:46,310
of cosines, and I'll leave
out the little bit
468
00:29:46,310 --> 00:29:49,210
of intervening algebra.
469
00:29:49,210 --> 00:29:55,760
And what we will get out of
this is that cosine of c--
470
00:29:55,760 --> 00:29:57,720
and I'll solve for that--
471
00:29:57,720 --> 00:30:03,940
is equal to cosine of alpha over
2 cosine of beta over 2
472
00:30:03,940 --> 00:30:10,140
plus cosine of gamma over 2
divided by sine alpha over 2
473
00:30:10,140 --> 00:30:11,860
sine beta over 2.
474
00:30:15,820 --> 00:30:18,820
And that is something we can
sink our teeth into and run
475
00:30:18,820 --> 00:30:23,570
with, because now I can ask the
question, suppose I want a
476
00:30:23,570 --> 00:30:28,010
to be a 2-fold rotation axis,
b to be a 3-fold rotation
477
00:30:28,010 --> 00:30:31,830
axis, and c be a 4-fold
rotation axis?
478
00:30:31,830 --> 00:30:39,040
Then the value of alpha over 2
is half of 180 degrees or 90.
479
00:30:39,040 --> 00:30:41,470
Well, you can see I put
in half the value of
480
00:30:41,470 --> 00:30:42,970
the rotation axes.
481
00:30:42,970 --> 00:30:46,990
And then I solve for c, and that
is the angle at which I
482
00:30:46,990 --> 00:30:51,950
have to put axis a and b
together to get c to turn out
483
00:30:51,950 --> 00:30:55,050
to be whatever angle
gamma over 2 is.
484
00:30:59,440 --> 00:31:00,220
So I can do this
485
00:31:00,220 --> 00:31:02,850
systematically now without thinking.
486
00:31:02,850 --> 00:31:11,090
And I can set up the problem by
taking the crystallographic
487
00:31:11,090 --> 00:31:15,140
rotation axes and combining them
together three at a time
488
00:31:15,140 --> 00:31:18,060
in all possible combinations.
489
00:31:18,060 --> 00:31:20,070
Right?
490
00:31:20,070 --> 00:31:23,400
In addition to this relation,
I have two other relations.
491
00:31:23,400 --> 00:31:27,120
And let me assemble them off to
the left, because we have
492
00:31:27,120 --> 00:31:30,820
to solve three equations to
find out the nature of the
493
00:31:30,820 --> 00:31:34,460
combination that is required.
494
00:31:34,460 --> 00:31:40,010
So just permuting terms, the
angle between A and B, c, has
495
00:31:40,010 --> 00:31:45,330
to follow from cosine of c
equals cosine of alpha over 2
496
00:31:45,330 --> 00:31:51,050
cosine of beta over 2 plus
cosine of gamma over 2.
497
00:31:51,050 --> 00:31:56,470
Notice that the single term by
itself is the cosine of half
498
00:31:56,470 --> 00:32:02,200
the angle of the opposite
rotation axis c.
499
00:32:02,200 --> 00:32:05,340
Then in the denominator is the
sine of these two angles.
500
00:32:08,130 --> 00:32:12,690
And so just permuting terms, one
can see that cosine of b
501
00:32:12,690 --> 00:32:16,850
is going to turn out to be
cosine of alpha over 2 cosine
502
00:32:16,850 --> 00:32:23,280
of gamma over 2 plus cosine of
theta over 2 divided by sine
503
00:32:23,280 --> 00:32:27,570
of alpha over 2 sine
of gamma over 2.
504
00:32:27,570 --> 00:32:31,600
And a third analogous expression
will give me the
505
00:32:31,600 --> 00:32:37,230
angle that will be the one
between B and C. And this will
506
00:32:37,230 --> 00:32:42,590
be cosine of beta over 2 cosine
gamma over 2 plus
507
00:32:42,590 --> 00:32:48,330
cosine of alpha over 2 divided
by sine beta over 2
508
00:32:48,330 --> 00:32:52,690
sine gamma over 2.
509
00:32:52,690 --> 00:32:54,590
OK?
510
00:32:54,590 --> 00:32:56,090
So now we don't have
to think anymore.
511
00:32:56,090 --> 00:32:57,990
It's just plug and chug.
512
00:32:57,990 --> 00:33:01,100
And I'll pause to suck in air
and let you catch up, and then
513
00:33:01,100 --> 00:33:03,515
we'll set up the problem and
look at a few solutions.
514
00:33:25,320 --> 00:33:28,210
And all this, in the event
that you're thoroughly
515
00:33:28,210 --> 00:33:31,240
bewildered, is in the
set of notes that I
516
00:33:31,240 --> 00:33:32,260
handed out last time.
517
00:33:32,260 --> 00:33:37,060
So you can read it over
at your leisure.
518
00:33:37,060 --> 00:33:39,540
AUDIENCE: Will this stuff
be on the quiz?
519
00:33:39,540 --> 00:33:39,980
PROFESSOR: No.
520
00:33:39,980 --> 00:33:42,940
Quiz will go up to the end of
the two-dimensional plane
521
00:33:42,940 --> 00:33:45,100
groups and stop.
522
00:33:45,100 --> 00:33:46,350
We won't say anything
three dimensional.
523
00:34:00,180 --> 00:34:06,030
OK, let's, then, if there's no
objection or complaint, look
524
00:34:06,030 --> 00:34:08,449
at possible values for--
525
00:34:08,449 --> 00:34:11,179
let me do it the same way that
I did it in the notes so that
526
00:34:11,179 --> 00:34:12,340
it's consistent--
527
00:34:12,340 --> 00:34:17,690
let's put down the value for
axis b, the rank of axis b and
528
00:34:17,690 --> 00:34:19,120
the rank of axis a.
529
00:34:25,370 --> 00:34:32,460
And A could be a 1-fold axis,
B could be a 1-fold access,
530
00:34:32,460 --> 00:34:36,280
and we could take a 1 with a 1
with a 1, a 1 with a 1 with a
531
00:34:36,280 --> 00:34:39,810
2, a 1 with a 1 with a 3, a 1
with a 1 with a 4, and a 1
532
00:34:39,810 --> 00:34:41,060
with a 1 with a 6.
533
00:34:44,440 --> 00:34:46,739
This is clearly impossible.
534
00:34:46,739 --> 00:34:50,920
If I did nothing, and followed
it by doing nothing, and
535
00:34:50,920 --> 00:34:54,989
wanted it to come out to be a
6-fold rotation, you'd all be
536
00:34:54,989 --> 00:34:57,580
spinning on your axes
like tops right now.
537
00:34:57,580 --> 00:35:00,270
So you can't do nothing and
follow it by doing nothing and
538
00:35:00,270 --> 00:35:02,200
have it come out to
be a net rotation.
539
00:35:02,200 --> 00:35:05,630
So these are impossible.
540
00:35:05,630 --> 00:35:08,590
So we don't have to
consider those.
541
00:35:08,590 --> 00:35:13,220
A could be a 2, though, and I
don't want to do 2, 1, 1,
542
00:35:13,220 --> 00:35:14,930
because I've got
a 1, 1, 2 here.
543
00:35:14,930 --> 00:35:16,310
The order doesn't make
any difference.
544
00:35:16,310 --> 00:35:22,440
So I'll start with a 2, 1,
2, a 2, 1, 3, a 2, 1,
545
00:35:22,440 --> 00:35:25,050
4, and a 2, 1, 6.
546
00:35:25,050 --> 00:35:31,020
So those are four combinations
that I should be examining.
547
00:35:31,020 --> 00:35:36,070
I could look at a 3 with a--
548
00:35:39,380 --> 00:35:43,200
2 with a 1, I have here in the
form of 2, 1, 3, so the next
549
00:35:43,200 --> 00:35:55,250
one I would want to look at
is a 3, 1, 3, a 3, 1,
550
00:35:55,250 --> 00:35:59,850
4, and a 3, 1, 6.
551
00:35:59,850 --> 00:36:03,310
And let me put in a
couple more here.
552
00:36:03,310 --> 00:36:10,480
If B were a 2, I should look at
a 2 with a 2 with a 2, a 2
553
00:36:10,480 --> 00:36:15,040
with a 2 with a 3, a 2
with a 2 with a 4, 2
554
00:36:15,040 --> 00:36:17,230
with a 2 with a 6.
555
00:36:17,230 --> 00:36:20,530
And then A 3 with a 2--
556
00:36:20,530 --> 00:36:25,192
and I've got 3, 2, 2 up here, so
I'll start with 3, 2, 3, 3,
557
00:36:25,192 --> 00:36:28,980
2, 4, 3, 2, 6, run out of room
here, but there should be a
558
00:36:28,980 --> 00:36:33,700
similar entry with
a 4 and a 6.
559
00:36:33,700 --> 00:36:35,000
So this sets up the problem.
560
00:36:35,000 --> 00:36:39,170
If you count up the number of
ways one can do this, we only
561
00:36:39,170 --> 00:36:42,600
have to consider the
off-diagonal boxes here,
562
00:36:42,600 --> 00:36:48,160
because interchanging a and b,
for example, looking at 3, 1,
563
00:36:48,160 --> 00:36:53,030
3, that's going to be the
same as 1, 3, 3 up here.
564
00:36:53,030 --> 00:36:55,110
So it's just the off-diagonal
boxes
565
00:36:55,110 --> 00:36:56,680
that we have to consider.
566
00:36:56,680 --> 00:37:06,330
So there should be a 3,
3, 3 in here, a 3, 3,
567
00:37:06,330 --> 00:37:09,065
4, and a 3, 3, 6.
568
00:37:12,720 --> 00:37:16,890
So what we would have to do in
order to determine the unique
569
00:37:16,890 --> 00:37:21,950
combinations is to look at all
of these combinations in turn,
570
00:37:21,950 --> 00:37:28,190
and I'm going to not
try all of them.
571
00:37:28,190 --> 00:37:34,740
I will do one that's going
to be clearly impossible.
572
00:37:34,740 --> 00:37:36,730
So let's look at 2, 1, 4.
573
00:37:36,730 --> 00:37:44,390
So here A corresponds to a
2-fold axis, B corresponds to
574
00:37:44,390 --> 00:37:53,830
a 1-fold axis, and C would
correspond to a 4-fold axis.
575
00:37:53,830 --> 00:37:58,500
So could we combine these three
axes at appropriate
576
00:37:58,500 --> 00:38:01,620
angles such that a 2-fold
followed by a 1-fold is
577
00:38:01,620 --> 00:38:04,420
equivalent to a 4-fold?
578
00:38:04,420 --> 00:38:06,310
This clearly isn't
going to work.
579
00:38:06,310 --> 00:38:10,820
If I do a 180-degree rotation
then don't do anything and ask
580
00:38:10,820 --> 00:38:15,850
is that equivalent to a 4-fold
rotation, that is saying that
581
00:38:15,850 --> 00:38:19,080
the 2-fold axis should be
identical to the 4-fold axis,
582
00:38:19,080 --> 00:38:21,460
and that is not going to work.
583
00:38:24,580 --> 00:38:28,580
So let me do now a generic
family that I know turns out
584
00:38:28,580 --> 00:38:31,220
to be possible.
585
00:38:31,220 --> 00:38:36,120
Let me look at an n-fold axis
with a 2-fold axis with a
586
00:38:36,120 --> 00:38:41,590
2-fold axis, and I can show
that this combination is
587
00:38:41,590 --> 00:38:45,260
possible for any integer
n whatsoever.
588
00:38:45,260 --> 00:38:48,080
So this will include a lot
of non-crystallographic
589
00:38:48,080 --> 00:38:50,980
symmetries.
590
00:38:50,980 --> 00:38:58,130
So let's say that this is C,
this is A, and this is B. But
591
00:38:58,130 --> 00:38:59,995
make it C, B, A if you'd like.
592
00:39:03,960 --> 00:39:11,770
OK, so my first equation says
that cosine of c, the angle
593
00:39:11,770 --> 00:39:17,610
between A and B, should be equal
to the cosine of alpha
594
00:39:17,610 --> 00:39:22,640
over 2 times the cosine of beta
over 2 plus the cosine of
595
00:39:22,640 --> 00:39:28,290
gamma over 2 divided by sine
of alpha over 2 sine
596
00:39:28,290 --> 00:39:31,370
of beta over 2.
597
00:39:31,370 --> 00:39:39,390
B is a 2-fold axis, so alpha
is equal to 180 degrees.
598
00:39:39,390 --> 00:39:40,780
A is a 2-fold--
599
00:39:40,780 --> 00:39:41,970
I'm sorry.
600
00:39:41,970 --> 00:39:44,530
B beta, A alpha.
601
00:39:44,530 --> 00:39:47,510
Beta is equal to 180 degrees,
the angular
602
00:39:47,510 --> 00:39:50,510
throw of a 2-fold axis.
603
00:39:50,510 --> 00:39:53,570
Alpha is equal to 180
degrees, the angular
604
00:39:53,570 --> 00:39:56,080
throw of a 2-fold axis.
605
00:39:56,080 --> 00:40:04,140
And gamma is equal to whatever
2 pi over n would be.
606
00:40:06,730 --> 00:40:10,300
That's the throw of the n-fold
axis than I'm letting be equal
607
00:40:10,300 --> 00:40:11,520
to C.
608
00:40:11,520 --> 00:40:17,940
So the cosine of A and B, to get
the result of rotation A
609
00:40:17,940 --> 00:40:21,360
followed by rotation B being
equal to the net rotation of
610
00:40:21,360 --> 00:40:25,340
an n-fold axis is that the
cosine of c should be the
611
00:40:25,340 --> 00:40:33,210
cosine of 180 degrees over 2
times the cosine of 180 over 2
612
00:40:33,210 --> 00:40:39,340
plus the cosine of gamma over
2, and gamma is whatever the
613
00:40:39,340 --> 00:40:43,600
rank of the axis determines, and
that's divided by sine of
614
00:40:43,600 --> 00:40:50,920
alpha over 2, and alpha is 180
and sine beta over 2, and
615
00:40:50,920 --> 00:40:53,870
that's 180 over 2.
616
00:40:53,870 --> 00:40:59,550
So the cosine of c is going to
be the cosine of 90, which is
617
00:40:59,550 --> 00:41:04,600
0, times the cosine of 90, which
is 0, plus the cosine of
618
00:41:04,600 --> 00:41:08,540
gamma over 2, whatever that
might be, over the sine of 90
619
00:41:08,540 --> 00:41:12,020
which is 1, sine of
90 which is 1.
620
00:41:12,020 --> 00:41:18,405
So this says that cosine
of c is equal to cosine
621
00:41:18,405 --> 00:41:20,350
of gamma over 2.
622
00:41:20,350 --> 00:41:25,520
So the angle between axis A and
axis B ought to be equal
623
00:41:25,520 --> 00:41:32,710
to one half the angular throw
of rotation axis C. So let's
624
00:41:32,710 --> 00:41:34,680
start putting down some
of this information.
625
00:41:34,680 --> 00:41:43,470
This says that if this is axis
C gamma, then A pi and B pi,
626
00:41:43,470 --> 00:41:47,180
the 2 180-degree rotations,
should be at an
627
00:41:47,180 --> 00:41:49,000
angle gamma over 2.
628
00:41:52,310 --> 00:41:57,530
We still need values
for b, and we still
629
00:41:57,530 --> 00:41:59,470
need a value for a.
630
00:41:59,470 --> 00:42:03,400
So let's find out
what those are.
631
00:42:03,400 --> 00:42:08,620
And let me start over here at
the left again, because I've
632
00:42:08,620 --> 00:42:12,430
got the relation that I need
for b sitting here.
633
00:42:12,430 --> 00:42:18,160
The cosine of b is equal to the
cosine of alpha over 2,
634
00:42:18,160 --> 00:42:26,185
and that is the cosine of pi
over 2, plus alpha is a
635
00:42:26,185 --> 00:42:27,850
90-degree rotation.
636
00:42:27,850 --> 00:42:32,990
Then cosine of gamma over 2,
whatever gamma happens to be,
637
00:42:32,990 --> 00:42:36,290
plus the cosine of beta over
2, and that's cosine of pi
638
00:42:36,290 --> 00:42:41,760
over 2, and this is all over
sine of pi over 2 times the
639
00:42:41,760 --> 00:42:50,110
sine of gamma over 2.
640
00:42:50,110 --> 00:42:55,360
So this is going to be cosine of
pi over 2, which is 0, plus
641
00:42:55,360 --> 00:43:05,620
cosine of pi over 2 divided by
sine of pi over 2, which is 1.
642
00:43:05,620 --> 00:43:09,510
This gamma over 2--
643
00:43:09,510 --> 00:43:13,800
no, cosine of pi over 2 is 0.
644
00:43:13,800 --> 00:43:20,530
So this is 0 plus 0 times sine
of gamma over 2, whatever that
645
00:43:20,530 --> 00:43:22,070
turns out to be.
646
00:43:22,070 --> 00:43:27,090
So cosine of b is 0, and that
says that the angle b between
647
00:43:27,090 --> 00:43:32,680
axis A and axis C turns
out to be 90 degrees.
648
00:43:32,680 --> 00:43:40,780
So in order to get pi followed
by c gamma to be equal to b
649
00:43:40,780 --> 00:43:44,400
pi, I've got to make b be
equal to pi over 2.
650
00:43:47,420 --> 00:43:50,420
And if I put it in this
orientation, then a pi
651
00:43:50,420 --> 00:43:54,960
followed by c gamma is going
to be equal to b pi.
652
00:43:54,960 --> 00:43:59,870
One final angle, and that's
the value for a.
653
00:44:07,300 --> 00:44:12,890
OK, cosine of a should be equal
to the cosine of beta
654
00:44:12,890 --> 00:44:18,410
over 2, that's pi over 2 times
the cosine of gamma over 2,
655
00:44:18,410 --> 00:44:26,700
whatever it is, plus the cosine
of beta over 2, and
656
00:44:26,700 --> 00:44:34,200
that's cosine of pi over 2, over
sine pi over 2 sine of
657
00:44:34,200 --> 00:44:35,930
gamma over 2.
658
00:44:35,930 --> 00:44:43,880
And this, as for b, turns out
to be 0 plus 0 over sine pi
659
00:44:43,880 --> 00:44:50,810
over 2, which is 1 times
sine of gamma over 2.
660
00:44:50,810 --> 00:44:54,820
So cosine of a turns out to be
0, and this says that the
661
00:44:54,820 --> 00:44:58,750
angle a is also pi over 2.
662
00:44:58,750 --> 00:44:59,600
So we've got a whole--
663
00:44:59,600 --> 00:45:00,502
Yeah.
664
00:45:00,502 --> 00:45:01,704
AUDIENCE: Did you really
need to go
665
00:45:01,704 --> 00:45:03,210
through all three equations?
666
00:45:03,210 --> 00:45:06,760
PROFESSOR: Yeah, because I had
to show that all three work.
667
00:45:06,760 --> 00:45:10,300
And in general, if I combine,
let's say, a 4-fold with a
668
00:45:10,300 --> 00:45:14,350
3-fold with a 2-fold, which is
something I want to do, all
669
00:45:14,350 --> 00:45:18,238
three angles a, b, and
c, will be different.
670
00:45:18,238 --> 00:45:20,700
OK?
671
00:45:20,700 --> 00:45:22,510
So the answer is yes.
672
00:45:22,510 --> 00:45:26,270
And there will be a few cases
where a value for one angle
673
00:45:26,270 --> 00:45:31,600
will exist and the value for the
one or two others will be
674
00:45:31,600 --> 00:45:32,880
impossible.
675
00:45:32,880 --> 00:45:36,240
And that's also something
that I have to know.
676
00:45:36,240 --> 00:45:40,830
So repetitious as the exercise
might be, the answer is yeah,
677
00:45:40,830 --> 00:45:42,822
you do have to do all three.
678
00:45:42,822 --> 00:45:44,072
AUDIENCE: [INAUDIBLE]?
679
00:45:48,234 --> 00:45:50,460
PROFESSOR: Oh, you don't have to
do different permutations.
680
00:45:50,460 --> 00:45:52,830
That's just a question
of labeling.
681
00:45:52,830 --> 00:45:53,080
OK?
682
00:45:53,080 --> 00:45:56,490
So n with a 2 with a 2 is the
same as a 2 with an n with a 2
683
00:45:56,490 --> 00:45:58,080
is the same as a 2
with a 2 with an
684
00:45:58,080 --> 00:46:00,000
n, that's just labeling.
685
00:46:00,000 --> 00:46:03,500
So that's why my boxes, when I
filled them out, the list got
686
00:46:03,500 --> 00:46:06,940
shorter and shorter, until
finally for a 6-fold axis, it
687
00:46:06,940 --> 00:46:12,780
would be just 6, 1, 1, 6, 1,
2, 6, 1, 3, and so on.
688
00:46:12,780 --> 00:46:15,450
Just that one entry in the box
where I enumerated what should
689
00:46:15,450 --> 00:46:16,700
be considered.
690
00:46:19,840 --> 00:46:28,440
Well, here is a whole slew of
possible solutions and a lot
691
00:46:28,440 --> 00:46:29,990
of them are
non-crystallographic, but
692
00:46:29,990 --> 00:46:31,450
still possible.
693
00:46:31,450 --> 00:46:37,450
This says that a combination
of a 2-fold axis--
694
00:46:37,450 --> 00:46:40,540
but remember now that these are
equations and operations.
695
00:46:40,540 --> 00:46:44,910
But the only operation that's
present for a 2-fold axis is a
696
00:46:44,910 --> 00:46:47,415
rotation a pi, b pi or c pi.
697
00:46:47,415 --> 00:46:52,940
So I could combine three 2-fold
axes that are mutually
698
00:46:52,940 --> 00:46:58,850
orthogonal, and that is an
allowable combination.
699
00:46:58,850 --> 00:47:03,650
And what we are obtaining here
is a sort of scaffolding, a
700
00:47:03,650 --> 00:47:08,170
framework, based on pure
rotation operations, that by
701
00:47:08,170 --> 00:47:11,270
themselves will be an allowable
3-dimensional point
702
00:47:11,270 --> 00:47:16,630
group, but which also provides a
framework, a Christmas tree,
703
00:47:16,630 --> 00:47:19,230
that we can decorate with mirror
planes and inversion
704
00:47:19,230 --> 00:47:21,500
centers to get still additional
705
00:47:21,500 --> 00:47:23,760
groups of higher symmetry.
706
00:47:23,760 --> 00:47:26,470
So here's one possible
crystallographic o combination
707
00:47:26,470 --> 00:47:28,030
of rotation axes.
708
00:47:28,030 --> 00:47:31,860
What we will use to denote this
combination is the same
709
00:47:31,860 --> 00:47:35,200
rule that we use for
our other notation.
710
00:47:35,200 --> 00:47:38,880
We will make a running list of
the independent operations
711
00:47:38,880 --> 00:47:43,030
that are present, and what we
have combined here are three
712
00:47:43,030 --> 00:47:45,390
distinct independent
2-fold axes.
713
00:47:48,390 --> 00:47:54,150
A solid that would have this
symmetry plus some other
714
00:47:54,150 --> 00:48:00,790
symmetry would be an orthogonal
brick with one
715
00:48:00,790 --> 00:48:03,050
2-fold axis coming out here.
716
00:48:03,050 --> 00:48:10,130
This is the operation c pi,
another 2-fold axis coming out
717
00:48:10,130 --> 00:48:15,740
the front, and this would be
the operation a pi, and
718
00:48:15,740 --> 00:48:18,510
another 2-fold axis coming out
of this face, and this would
719
00:48:18,510 --> 00:48:20,040
be the operation b pi.
720
00:48:23,410 --> 00:48:26,430
Now let me show you-- it's
rather amusing-- that what we
721
00:48:26,430 --> 00:48:28,740
have done really works.
722
00:48:28,740 --> 00:48:34,990
We've shown supposedly that a pi
followed by b pi should be
723
00:48:34,990 --> 00:48:40,390
equal to a net rotation c
pi about an axis that's
724
00:48:40,390 --> 00:48:42,070
orthogonal to the first two.
725
00:48:42,070 --> 00:48:45,680
So let's pick a motif, and for
convenience I'll put it at one
726
00:48:45,680 --> 00:48:47,340
corner of this brick.
727
00:48:47,340 --> 00:48:52,420
Here's object 1, I rotate it
by 180 degrees about a.
728
00:48:52,420 --> 00:48:57,090
Here sits object number 2,
same corality, and then I
729
00:48:57,090 --> 00:49:02,900
rotate it 180 degrees about B
beta, and that's going to give
730
00:49:02,900 --> 00:49:05,240
me number 3.
731
00:49:05,240 --> 00:49:09,990
What is the net way of
getting from 1 to 3?
732
00:49:09,990 --> 00:49:10,860
Holy mackerel.
733
00:49:10,860 --> 00:49:14,720
It's a net 180-degree
rotation about c pi.
734
00:49:14,720 --> 00:49:15,970
It really works.
735
00:49:18,270 --> 00:49:21,360
Or I could do the operations
in a different order.
736
00:49:21,360 --> 00:49:25,580
I could rotate by d, rotate by
c, and the way I get from the
737
00:49:25,580 --> 00:49:30,160
first to the third is
a rotation a pi.
738
00:49:30,160 --> 00:49:33,260
So that is a self consistent
set of rotation axes.
739
00:49:33,260 --> 00:49:36,610
That is 2, 2, 2.
740
00:49:36,610 --> 00:49:38,722
Let me do one more.
741
00:49:38,722 --> 00:49:41,730
Well, no, let me take a break
here and let you absorb all
742
00:49:41,730 --> 00:49:45,140
this, and then we'll look at
some remaining ones, and this
743
00:49:45,140 --> 00:49:47,340
will include some that are
non- crystallographic.
744
00:49:47,340 --> 00:49:48,800
And that's perfectly OK.
745
00:49:48,800 --> 00:49:51,980
But they're lovely groups, they
constitute groups, but
746
00:49:51,980 --> 00:49:54,250
they won't be groups
that can occur in
747
00:49:54,250 --> 00:49:55,760
combination with a lattice.