1
00:00:00,000 --> 00:00:10,248
2
00:00:10,248 --> 00:00:12,680
PROFESSOR: Do them individually
so I can continue
3
00:00:12,680 --> 00:00:16,090
to put names and
faces together.
4
00:00:16,090 --> 00:00:19,150
I'm happy to announce that
the registrar has now got
5
00:00:19,150 --> 00:00:21,740
everybody's photograph
online for
6
00:00:21,740 --> 00:00:23,250
registration in the course.
7
00:00:23,250 --> 00:00:28,580
So anonymity is a thing of the
past, so you have to watch
8
00:00:28,580 --> 00:00:31,060
your step from now on.
9
00:00:31,060 --> 00:00:34,940
I handed back, to those who
didn't get it, problem set
10
00:00:34,940 --> 00:00:40,660
number four, which asked you
to tackle some patterns,
11
00:00:40,660 --> 00:00:42,290
nontrivial patterns.
12
00:00:42,290 --> 00:00:45,410
And actually, that was a dirty
trick, because we hadn't, at
13
00:00:45,410 --> 00:00:47,950
that point, derived the plane
groups, and you really didn't
14
00:00:47,950 --> 00:00:49,800
know what to do or
what to look for.
15
00:00:49,800 --> 00:00:52,710
But nevertheless, it got you
thinking about patterns and
16
00:00:52,710 --> 00:00:56,500
some of the symmetry elements
which we had discussed up to
17
00:00:56,500 --> 00:00:58,150
that point.
18
00:00:58,150 --> 00:01:02,400
At this point we have derived
exhaustively every last one of
19
00:01:02,400 --> 00:01:04,330
the 17 plane groups.
20
00:01:04,330 --> 00:01:08,130
So now you are armed with this
new-found power, and when
21
00:01:08,130 --> 00:01:11,600
faced with a pattern, you should
know exactly what to
22
00:01:11,600 --> 00:01:15,080
look for and how to go
about deciding what
23
00:01:15,080 --> 00:01:16,180
plane group it is.
24
00:01:16,180 --> 00:01:20,290
At the very least, you'll
have the drawings of the
25
00:01:20,290 --> 00:01:22,600
arrangement of symmetry elements
in the plane groups
26
00:01:22,600 --> 00:01:26,400
before you, and you can work by
the process of elimination.
27
00:01:26,400 --> 00:01:30,390
For example, high symmetry
usually hits you right between
28
00:01:30,390 --> 00:01:34,090
the eyes, and if something is
square-ish, you can pretty
29
00:01:34,090 --> 00:01:36,980
quickly guess that it's based
on a square lattice.
30
00:01:36,980 --> 00:01:40,590
And if it has a square lattice,
there jolly well
31
00:01:40,590 --> 00:01:43,510
better be a 4-fold axis in there
that makes it square.
32
00:01:43,510 --> 00:01:46,300
If you can find the 4-fold axis,
then you have to ask
33
00:01:46,300 --> 00:01:48,060
yourself only three questions.
34
00:01:48,060 --> 00:01:49,510
So 4-fold axis, fine.
35
00:01:49,510 --> 00:01:51,650
Is there a mirror
line in there?
36
00:01:51,650 --> 00:01:52,160
Yeah.
37
00:01:52,160 --> 00:01:54,730
Does the mirror line go through
the 4-fold axis?
38
00:01:54,730 --> 00:01:57,560
Then it is P 4 MM.
39
00:01:57,560 --> 00:01:59,690
And you know just where to
look for everything else,
40
00:01:59,690 --> 00:02:02,050
including these very
subtle glide planes
41
00:02:02,050 --> 00:02:03,640
that are hard to spot.
42
00:02:03,640 --> 00:02:05,940
If there is a mirror plane
there, but it doesn't go
43
00:02:05,940 --> 00:02:11,290
through the 4-fold axis,
then it's P 4 MG.
44
00:02:11,290 --> 00:02:13,350
And if there is no mirror
plane, then it's P 4.
45
00:02:13,350 --> 00:02:16,550
So just by asking one or two
simple questions, you can
46
00:02:16,550 --> 00:02:20,560
narrow it down to what the
plane group has to be.
47
00:02:20,560 --> 00:02:26,330
This is another indication that
the informed intellect is
48
00:02:26,330 --> 00:02:29,440
always more than a match
for sheer, raw native
49
00:02:29,440 --> 00:02:30,330
intelligence.
50
00:02:30,330 --> 00:02:34,520
If you know what to look
for, it's a lot easier.
51
00:02:34,520 --> 00:02:40,810
Because you really didn't have
much practice with patterns,
52
00:02:40,810 --> 00:02:44,510
we're having a quiz, as you
know, next Thursday, that will
53
00:02:44,510 --> 00:02:50,060
cover up through completion of
the plane groups and not the
54
00:02:50,060 --> 00:02:52,290
material we've been doing now.
55
00:02:52,290 --> 00:02:55,900
So I think it might be of use
to you to have some practice
56
00:02:55,900 --> 00:02:57,970
analyzing a few more patterns.
57
00:02:57,970 --> 00:03:01,640
So there are four additional
patterns in this problem set.
58
00:03:01,640 --> 00:03:04,670
As always, it's optional, but if
you would like to try them,
59
00:03:04,670 --> 00:03:07,310
and you want to see if you've
got them right, come in and
60
00:03:07,310 --> 00:03:09,360
see me tomorrow or on
Thursday morning.
61
00:03:09,360 --> 00:03:12,170
I'd be happy to go over them
with you on the spot.
62
00:03:12,170 --> 00:03:14,110
So this is for practice.
63
00:03:14,110 --> 00:03:17,130
And some of you did
extraordinarily well
64
00:03:17,130 --> 00:03:19,050
on the first try.
65
00:03:19,050 --> 00:03:21,625
Others, I think, could use
the additional practice.
66
00:03:21,625 --> 00:03:32,110
67
00:03:32,110 --> 00:03:35,460
There were a few people who
identified the plane group
68
00:03:35,460 --> 00:03:39,620
correctly, but got the name
that's assigned to it wrong.
69
00:03:39,620 --> 00:03:42,610
And one other very confusing
thing is that there is one
70
00:03:42,610 --> 00:03:47,470
plane group that has the symbols
G and M, and another
71
00:03:47,470 --> 00:03:51,200
plane group which has
the symbols M and G.
72
00:03:51,200 --> 00:03:54,090
So if you found a mirror plane
and a glide plane is an
73
00:03:54,090 --> 00:03:56,800
independent symmetry plane,
when is it MG,
74
00:03:56,800 --> 00:03:59,070
and when is it GM?
75
00:03:59,070 --> 00:04:02,790
I have a little mnemonic
device.
76
00:04:02,790 --> 00:04:07,220
GM, general manager, is the
guy who sits on top of the
77
00:04:07,220 --> 00:04:08,500
organization.
78
00:04:08,500 --> 00:04:12,350
So GM should be the
plane group that
79
00:04:12,350 --> 00:04:13,650
has the highest symmetry.
80
00:04:13,650 --> 00:04:18,320
P 4 GM, P 4 general manager.
81
00:04:18,320 --> 00:04:20,910
Now that is--
82
00:04:20,910 --> 00:04:22,690
hey, it works for me.
83
00:04:22,690 --> 00:04:26,750
But another way of saying it
is that there are two plane
84
00:04:26,750 --> 00:04:30,760
groups, one has MG and the other
has GM, and I like cars,
85
00:04:30,760 --> 00:04:35,340
and I think an MG is much
classier than anything that
86
00:04:35,340 --> 00:04:38,880
General Motors, GM, puts out,
so MG should be the one of
87
00:04:38,880 --> 00:04:40,530
highest quality, highest
symmetry.
88
00:04:40,530 --> 00:04:41,820
And that's just the reverse.
89
00:04:41,820 --> 00:04:43,010
But whatever works for you.
90
00:04:43,010 --> 00:04:46,720
You could keep them straight
through that simple algorithm.
91
00:04:46,720 --> 00:04:49,310
And as they say, if it works for
me, but if it doesn't work
92
00:04:49,310 --> 00:04:53,170
for you, don't use it.
93
00:04:53,170 --> 00:04:54,420
All right.
94
00:04:54,420 --> 00:04:59,470
95
00:04:59,470 --> 00:05:03,170
What I will bring in during
intermission, for those of you
96
00:05:03,170 --> 00:05:06,650
had trouble identifying
translations in the patterns
97
00:05:06,650 --> 00:05:11,300
that we handed out earlier, I've
taken these and put them
98
00:05:11,300 --> 00:05:13,780
on overhead transparencies.
99
00:05:13,780 --> 00:05:15,540
And I'll have two of each.
100
00:05:15,540 --> 00:05:18,540
So if you don't see the symmetry
or translations that
101
00:05:18,540 --> 00:05:21,100
are present, you can actually
take one pattern and
102
00:05:21,100 --> 00:05:24,470
physically move it and lay it
on top of the other one, and
103
00:05:24,470 --> 00:05:27,960
that's a good way to convince
yourself what a translation
104
00:05:27,960 --> 00:05:30,010
looks like when it occurs
in a pattern.
105
00:05:30,010 --> 00:05:32,250
So I'll bring those in at
our break between class.
106
00:05:32,250 --> 00:05:35,150
107
00:05:35,150 --> 00:05:38,340
All right, any questions
before we move on?
108
00:05:38,340 --> 00:05:42,380
Any questions that have arisen
as you have gotten
109
00:05:42,380 --> 00:05:45,290
ready for the quiz?
110
00:05:45,290 --> 00:05:46,956
You haven't gotten ready
for the quiz yet, so
111
00:05:46,956 --> 00:05:48,170
there are no questions.
112
00:05:48,170 --> 00:05:48,800
That's OK.
113
00:05:48,800 --> 00:05:50,750
I know how things work at MIT.
114
00:05:50,750 --> 00:05:54,560
You deal with one crisis
at a time.
115
00:05:54,560 --> 00:05:55,210
Any questions?
116
00:05:55,210 --> 00:05:58,280
Anything you want to go over?
117
00:05:58,280 --> 00:06:03,620
There was one interesting
wrinkle in a problem that I
118
00:06:03,620 --> 00:06:06,590
had not encountered before,
and this was the one that
119
00:06:06,590 --> 00:06:09,990
asked you to look at, in two
dimensions, a plane with
120
00:06:09,990 --> 00:06:12,160
indices h and k.
121
00:06:12,160 --> 00:06:17,260
And then, when h and k were
mutually prime, to move that
122
00:06:17,260 --> 00:06:26,670
plane by the translations plus
T1 and plus and minus T2, and
123
00:06:26,670 --> 00:06:30,850
then show that the number of
intervals between the origin
124
00:06:30,850 --> 00:06:33,710
and the intercept plane, the one
that hit lattice points on
125
00:06:33,710 --> 00:06:36,860
both translations, was equal
to h times k if they were
126
00:06:36,860 --> 00:06:39,790
mutually prime.
127
00:06:39,790 --> 00:06:42,800
That is true only if the
lattice is primitive.
128
00:06:42,800 --> 00:06:46,870
And what the problem said was to
pick one of the cells that
129
00:06:46,870 --> 00:06:51,130
you used in the first
problem, number one.
130
00:06:51,130 --> 00:06:54,590
Well, that problem asked you
to begin with identifying
131
00:06:54,590 --> 00:06:56,230
different primitive cells.
132
00:06:56,230 --> 00:07:00,920
If you take a multiple cell,
this operation of going plus
133
00:07:00,920 --> 00:07:09,010
and minus T1 does not put a
lattice line through each of
134
00:07:09,010 --> 00:07:10,670
the lattice points.
135
00:07:10,670 --> 00:07:14,420
If you picked a double cell,
that process decorated only
136
00:07:14,420 --> 00:07:18,440
half of the lattice points with
planes, and the other
137
00:07:18,440 --> 00:07:21,150
half sat there with nothing
hanging on them at all.
138
00:07:21,150 --> 00:07:23,785
139
00:07:23,785 --> 00:07:27,560
The key to the difference, if
you looked at a double cell,
140
00:07:27,560 --> 00:07:32,070
was that if h plus k was even,
then you automatically got a
141
00:07:32,070 --> 00:07:34,020
plane on every lattice point.
142
00:07:34,020 --> 00:07:38,110
If h plus k was odd, as it would
have been for the plane
143
00:07:38,110 --> 00:07:42,450
2,1 for example, 2 plus 1 is 3--
hey, this isn't even one
144
00:07:42,450 --> 00:07:44,830
of my good days--
145
00:07:44,830 --> 00:07:48,010
then half of the lattice points
did not have planes
146
00:07:48,010 --> 00:07:49,850
hanging on them.
147
00:07:49,850 --> 00:07:52,880
Now, there's great relevance
of this observation to
148
00:07:52,880 --> 00:07:56,970
diffraction, and you probably
are all familiar, if only
149
00:07:56,970 --> 00:08:01,060
vaguely, with the magic rules
that say, if h plus k is equal
150
00:08:01,060 --> 00:08:05,480
to 3 pi plus 4, then the
intensity is identically 0.
151
00:08:05,480 --> 00:08:10,340
Well, for a double cell, the
lattice planes that are
152
00:08:10,340 --> 00:08:16,160
repeated by translation diffract
x-rays, and there is
153
00:08:16,160 --> 00:08:19,750
no reason why the intensity
should be
154
00:08:19,750 --> 00:08:21,240
something other than 0.
155
00:08:21,240 --> 00:08:24,810
So here comes, a la Bragg, an
x-ray beam coming in at angle
156
00:08:24,810 --> 00:08:27,850
theta, and then you say you
get diffraction when
157
00:08:27,850 --> 00:08:30,640
scattering from this lattice
plane is exactly in
158
00:08:30,640 --> 00:08:31,700
phase with this one.
159
00:08:31,700 --> 00:08:35,100
And this gives the familiar
relation that an integral
160
00:08:35,100 --> 00:08:39,360
number of wavelengths is equal
to 2 d sine of theta when the
161
00:08:39,360 --> 00:08:41,799
crystal diffracts.
162
00:08:41,799 --> 00:08:46,380
So there is exactly-- this
says there's exactly 2 pi
163
00:08:46,380 --> 00:08:51,430
phase difference or n lambda
path difference between these
164
00:08:51,430 --> 00:08:52,680
two planes.
165
00:08:52,680 --> 00:08:54,970
166
00:08:54,970 --> 00:08:58,690
Now, if the lattice would be a
double cell, then there is an
167
00:08:58,690 --> 00:09:02,710
additional lattice point in here
that does not get a plane
168
00:09:02,710 --> 00:09:03,300
hung on it.
169
00:09:03,300 --> 00:09:06,070
So if the lattice is a double
cell, there's another plane
170
00:09:06,070 --> 00:09:08,650
that has to hang on this lattice
point, and that one is
171
00:09:08,650 --> 00:09:13,890
exactly out of phase with this
plane, and the intensity is 0.
172
00:09:13,890 --> 00:09:20,260
So this observation that a
non-primitive lattice has a
173
00:09:20,260 --> 00:09:24,000
interplanar spacing that is a
sub-multiple of that of a
174
00:09:24,000 --> 00:09:28,190
primitive lattice gives some
insight into why certain
175
00:09:28,190 --> 00:09:29,100
reflections--
176
00:09:29,100 --> 00:09:33,520
certain diffraction maxima-- are
identically 0 in intensity
177
00:09:33,520 --> 00:09:36,480
for a crystal that has a
non-primitive lattice.
178
00:09:36,480 --> 00:09:38,750
So that is something I had
not noticed before.
179
00:09:38,750 --> 00:09:41,600
I should have, but I will phrase
the problem a little
180
00:09:41,600 --> 00:09:43,300
bit more precisely
in the future.
181
00:09:43,300 --> 00:09:48,810
182
00:09:48,810 --> 00:09:49,110
All right.
183
00:09:49,110 --> 00:09:53,410
So to conclude my preamble, I
hope you'll try playing with
184
00:09:53,410 --> 00:09:58,300
some of the four additional
patterns that I handed out,
185
00:09:58,300 --> 00:10:00,370
just to give yourself
some practice.
186
00:10:00,370 --> 00:10:04,010
And the implication of this is
that you're going to see a
187
00:10:04,010 --> 00:10:06,680
pattern on the quiz, and I will
tell you that you will.
188
00:10:06,680 --> 00:10:09,300
So if you want to see how you
did on the patterns that I
189
00:10:09,300 --> 00:10:13,120
distributed, please come in
and talk to me about them.
190
00:10:13,120 --> 00:10:15,760
191
00:10:15,760 --> 00:10:16,330
All right then.
192
00:10:16,330 --> 00:10:20,150
Let me remind you where
we were last time.
193
00:10:20,150 --> 00:10:29,530
We started to begin to build a
framework of symmetry elements
194
00:10:29,530 --> 00:10:31,410
in three dimensions.
195
00:10:31,410 --> 00:10:37,680
And we asked the question, what
would happen if we take a
196
00:10:37,680 --> 00:10:42,230
first rotation axis, A alpha,
combine it with a second
197
00:10:42,230 --> 00:10:46,440
rotation axis, B beta, in
such a way that they
198
00:10:46,440 --> 00:10:48,270
intersect at a point.
199
00:10:48,270 --> 00:10:51,500
This means that their operation
and reproducing
200
00:10:51,500 --> 00:10:55,630
atoms or motifs is going to
leave at least one point in
201
00:10:55,630 --> 00:10:59,480
space unchanged, and that will
be the point of intersection.
202
00:10:59,480 --> 00:11:03,270
We ask ourselves, what will
be the combined effect--
203
00:11:03,270 --> 00:11:07,580
we have two operations
in space--
204
00:11:07,580 --> 00:11:11,160
what would be the combined
effect of rotating alpha
205
00:11:11,160 --> 00:11:15,910
degrees about A followed
immediately by beta degrees
206
00:11:15,910 --> 00:11:16,710
about B.
207
00:11:16,710 --> 00:11:20,785
So what we're going to do then
is to take a first motif--
208
00:11:20,785 --> 00:11:23,670
and let's say it's
left handed.
209
00:11:23,670 --> 00:11:26,600
Being a left-handed person, I
like to give right handed
210
00:11:26,600 --> 00:11:29,080
motifs and left handed
motifs equal time.
211
00:11:29,080 --> 00:11:31,790
212
00:11:31,790 --> 00:11:34,760
If we rotate that through an
angle alpha, and this is
213
00:11:34,760 --> 00:11:37,830
number 2, it will stay
left handed.
214
00:11:37,830 --> 00:11:42,100
Then if we rotate that by B
beta, it'll move it over here
215
00:11:42,100 --> 00:11:45,682
to number 3 and it will stay
left handed as well.
216
00:11:45,682 --> 00:11:49,980
And the question is then, what
net operation is equivalent to
217
00:11:49,980 --> 00:11:55,258
the combined operation of these
two transformations?
218
00:11:55,258 --> 00:11:59,070
And to specify the type of
operation is really a
219
00:11:59,070 --> 00:11:59,730
no-brainer.
220
00:11:59,730 --> 00:12:03,500
All of these motifs are of the
same corality so the only
221
00:12:03,500 --> 00:12:04,950
thing that can relate them is
222
00:12:04,950 --> 00:12:08,260
translation or another rotation.
223
00:12:08,260 --> 00:12:12,420
And clearly the first and the
third have no reason to be
224
00:12:12,420 --> 00:12:15,450
parallel to one another, and the
distance between them is
225
00:12:15,450 --> 00:12:19,120
going to depend on how far they
are away from the axis,
226
00:12:19,120 --> 00:12:23,260
so translation won't do the
job, a and the only thing
227
00:12:23,260 --> 00:12:29,270
that's left as a net operation
that's equivalent to those two
228
00:12:29,270 --> 00:12:35,050
steps is rotation about
a third axis C.
229
00:12:35,050 --> 00:12:39,240
And what we're going to do today
is answer the question,
230
00:12:39,240 --> 00:12:48,670
where is axis C located, and
what is the angle of rotation,
231
00:12:48,670 --> 00:12:53,670
given the value of alpha and
beta and the angle between
232
00:12:53,670 --> 00:12:55,010
these two axes?
233
00:12:55,010 --> 00:12:59,540
And let's define that
as a lowercase c.
234
00:12:59,540 --> 00:13:02,460
So clearly the location of the
axis and the amount of the
235
00:13:02,460 --> 00:13:06,260
rotation is going to be a
function of alpha, beta, and
236
00:13:06,260 --> 00:13:08,350
the angle between them.
237
00:13:08,350 --> 00:13:11,970
We want this to be a combination
of operations that
238
00:13:11,970 --> 00:13:16,470
exists in a symmetry
operation.
239
00:13:16,470 --> 00:13:19,260
And if this is to be a
crystallographic symmetry,
240
00:13:19,260 --> 00:13:22,380
these will be restricted to
the angular throws of a
241
00:13:22,380 --> 00:13:26,180
1-fold, 2-fold, 3-fold,
4-fold, a 6-fold axis.
242
00:13:26,180 --> 00:13:30,110
We can take these two at a time,
ask what the net effect
243
00:13:30,110 --> 00:13:33,020
is if we combine at
a given angle c.
244
00:13:33,020 --> 00:13:38,890
And what comes out here must
be a rotation which is also
245
00:13:38,890 --> 00:13:40,140
crystallographic.
246
00:13:40,140 --> 00:13:42,540
247
00:13:42,540 --> 00:13:46,270
So there are going to be severe
constraints on this
248
00:13:46,270 --> 00:13:48,730
combination.
249
00:13:48,730 --> 00:13:51,440
Two rotations about an
intersecting point will always
250
00:13:51,440 --> 00:13:56,060
be a third rotation, but if
this is to be a set of
251
00:13:56,060 --> 00:14:00,060
operations in a symmetry group,
the result must be
252
00:14:00,060 --> 00:14:02,880
crystallographic.
253
00:14:02,880 --> 00:14:06,320
That's a tough problem, and how
will we undertake it is
254
00:14:06,320 --> 00:14:07,570
going to be non-intuitive.
255
00:14:07,570 --> 00:14:11,370
256
00:14:11,370 --> 00:14:14,870
OK, the problem is most readily
treated with spherical
257
00:14:14,870 --> 00:14:16,610
trigonometry.
258
00:14:16,610 --> 00:14:23,530
So on the surface of a sphere,
I'm going to map the point at
259
00:14:23,530 --> 00:14:25,300
which A alpha protrudes--
260
00:14:25,300 --> 00:14:28,190
and I'll call this point A--
261
00:14:28,190 --> 00:14:32,510
and then I'll mark out the point
where axis B beta exits
262
00:14:32,510 --> 00:14:41,210
the sphere, and I'll mark that
point B. And this is the angle
263
00:14:41,210 --> 00:14:43,120
C.
264
00:14:43,120 --> 00:14:47,650
And we said that in spherical
trigonometry, the measure of
265
00:14:47,650 --> 00:14:51,910
the length of the arc separating
A and B is given by
266
00:14:51,910 --> 00:14:55,630
the angle subtended at the
center, so the length of this
267
00:14:55,630 --> 00:14:58,530
distance between A and
B is the angle c.
268
00:14:58,530 --> 00:15:01,500
Again, it sort of boggles the
mind when you measure lengths
269
00:15:01,500 --> 00:15:04,500
in terms of degrees rather
than some metric unit.
270
00:15:04,500 --> 00:15:07,657
271
00:15:07,657 --> 00:15:08,110
All right.
272
00:15:08,110 --> 00:15:14,350
So I will now not bother to show
the sphere on which the
273
00:15:14,350 --> 00:15:16,080
geometry is taking place.
274
00:15:16,080 --> 00:15:20,710
I'll just draw A and B,
and this is the arc
275
00:15:20,710 --> 00:15:22,850
between them, c.
276
00:15:22,850 --> 00:15:26,590
And somewhere or other there
will be some third axis, C,
277
00:15:26,590 --> 00:15:29,900
which is going to be the
combined effect of the
278
00:15:29,900 --> 00:15:34,990
rotation about axis A and axis
B. So what I would like to do
279
00:15:34,990 --> 00:15:39,530
is to locate the position
of this axis C.
280
00:15:39,530 --> 00:15:43,880
In order to do that, I'll have
to know what the angle between
281
00:15:43,880 --> 00:15:47,600
A and C is, and I'll call that,
by analogy to what I've
282
00:15:47,600 --> 00:15:49,400
done here, I'll call that b.
283
00:15:49,400 --> 00:15:53,250
And I'll want to know what the
angle between B and C is, and
284
00:15:53,250 --> 00:15:56,270
I'll call that angle a.
285
00:15:56,270 --> 00:16:00,190
So again, going back to three
dimensions momentarily, if
286
00:16:00,190 --> 00:16:04,920
this is the rotation operation
C gamma, and this is A alpha,
287
00:16:04,920 --> 00:16:11,200
and this is B beta, the axis c
is this, the angle b is this,
288
00:16:11,200 --> 00:16:12,940
and the angle a is this.
289
00:16:12,940 --> 00:16:19,240
290
00:16:19,240 --> 00:16:19,480
OK.
291
00:16:19,480 --> 00:16:22,870
It's a non-trivial problem and
it is not by accident that the
292
00:16:22,870 --> 00:16:27,080
solution to this problem was
first given by a very, very
293
00:16:27,080 --> 00:16:32,130
famous mathematician, Leonhard
Euler, and this construction
294
00:16:32,130 --> 00:16:34,950
that we're about to go through
is called Euler's
295
00:16:34,950 --> 00:16:36,200
construction.
296
00:16:36,200 --> 00:16:41,719
297
00:16:41,719 --> 00:16:42,200
All right.
298
00:16:42,200 --> 00:16:50,290
Let me find where these
different locations
299
00:16:50,290 --> 00:16:51,850
are going to be.
300
00:16:51,850 --> 00:16:55,230
We've specified the location of
point A and the location of
301
00:16:55,230 --> 00:17:00,100
point B, and we know that the
angle between them is the
302
00:17:00,100 --> 00:17:04,460
length of the arc ab, which is
the angle between A and C. So
303
00:17:04,460 --> 00:17:06,690
let me now do some
constructions.
304
00:17:06,690 --> 00:17:15,270
Let me find a great circle that
by design is alpha over 2
305
00:17:15,270 --> 00:17:20,619
away from the arc ab, and
that is by construction.
306
00:17:20,619 --> 00:17:24,859
And I'm going to say, then, that
if A alpha works in this
307
00:17:24,859 --> 00:17:29,900
direction, the operation of A
alpha is going to take this
308
00:17:29,900 --> 00:17:34,020
great circle and move it over
to a great circle which is
309
00:17:34,020 --> 00:17:39,730
alpha over 2 on the other
side of the arc ab.
310
00:17:39,730 --> 00:17:41,100
Fine, you say, so what?
311
00:17:41,100 --> 00:17:45,120
Well, just going to leave
those there for now.
312
00:17:45,120 --> 00:17:49,300
I'll have B beta work
in the same sense.
313
00:17:49,300 --> 00:17:54,660
And I'm now going to create a
line here that by construction
314
00:17:54,660 --> 00:18:00,250
is beta over 2 on one side of
the arc ab, and if I let B
315
00:18:00,250 --> 00:18:06,130
beta go to work, that will map
this great circle over to a
316
00:18:06,130 --> 00:18:09,660
new location beta over
2 on the other side.
317
00:18:09,660 --> 00:18:13,190
318
00:18:13,190 --> 00:18:17,010
What has this done for me, other
than perhaps confuse me
319
00:18:17,010 --> 00:18:19,530
and clutter the diagram?
320
00:18:19,530 --> 00:18:23,710
Well, now I'm going to determine
unequivocally the
321
00:18:23,710 --> 00:18:27,630
location of the axis C, and
where it emerges from the
322
00:18:27,630 --> 00:18:29,750
reference here.
323
00:18:29,750 --> 00:18:32,880
And how will I do that?
324
00:18:32,880 --> 00:18:37,680
I'm going to use a definition
that may have seemed trivial
325
00:18:37,680 --> 00:18:40,550
the first time we made
the observation.
326
00:18:40,550 --> 00:18:45,610
I said that a symmetry element
is the locus of points that is
327
00:18:45,610 --> 00:18:49,832
left unmoved by an operation.
328
00:18:49,832 --> 00:18:52,730
OK?
329
00:18:52,730 --> 00:18:57,540
I rotated by A alpha from here
to here, that took everything
330
00:18:57,540 --> 00:19:00,800
along this line and mapped it
to a new location here.
331
00:19:00,800 --> 00:19:07,290
I took this line and rotated
it by B beta, and that took
332
00:19:07,290 --> 00:19:10,760
everything along this line and
moved it to a new location.
333
00:19:10,760 --> 00:19:15,010
So my question now is if I
rotate by A alpha and then
334
00:19:15,010 --> 00:19:19,120
rotate in the same direction
by B beta, what
335
00:19:19,120 --> 00:19:20,405
point is left unmoved?
336
00:19:20,405 --> 00:19:24,110
337
00:19:24,110 --> 00:19:26,940
It's only one point that can
make that claim, and that is
338
00:19:26,940 --> 00:19:30,320
where these two great
circles intersect.
339
00:19:30,320 --> 00:19:33,510
The rotation A alpha will
take this location--
340
00:19:33,510 --> 00:19:36,250
and I'm going to call it C
because I've identified now
341
00:19:36,250 --> 00:19:38,760
what it is-- it's going to take
C and move it over to
342
00:19:38,760 --> 00:19:42,470
here, call that C prime, and
then B beta takes that point
343
00:19:42,470 --> 00:19:45,350
and only that point, and
restores it back to its
344
00:19:45,350 --> 00:19:47,410
original location.
345
00:19:47,410 --> 00:19:50,770
So this, then, ladies and
gentlemen, is where the
346
00:19:50,770 --> 00:19:54,420
rotation axis C gamma pokes
out of the sphere of
347
00:19:54,420 --> 00:19:55,670
reflection.
348
00:19:55,670 --> 00:20:00,680
349
00:20:00,680 --> 00:20:04,940
Still don't know what this angle
is in here, and I would
350
00:20:04,940 --> 00:20:11,820
dearly love to know what these
arcs b and a are, and then I
351
00:20:11,820 --> 00:20:15,170
will have specified all three
of the interaxial angles
352
00:20:15,170 --> 00:20:22,490
between A, B and C.
353
00:20:22,490 --> 00:20:28,420
OK, let me do something quite
similar to what I did before.
354
00:20:28,420 --> 00:20:32,290
I'm going to again let A alpha
work on a particular point,
355
00:20:32,290 --> 00:20:35,090
and then let B beta map it.
356
00:20:35,090 --> 00:20:37,830
So here's A alpha,
here's B beta.
357
00:20:37,830 --> 00:20:41,580
And now I'm going to look
specifically at how these two
358
00:20:41,580 --> 00:20:46,820
rotations transform point A,
where A is the point at which
359
00:20:46,820 --> 00:20:50,000
axis A alpha pokes out
of the sphere.
360
00:20:50,000 --> 00:20:53,640
A alpha, when it acts on this
point, does nothing to it.
361
00:20:53,640 --> 00:20:55,240
It leaves it alone.
362
00:20:55,240 --> 00:21:00,820
B beta is going to map A to
a new location, A prime.
363
00:21:00,820 --> 00:21:07,110
364
00:21:07,110 --> 00:21:11,750
Now, doing A alpha and following
up by B beta is
365
00:21:11,750 --> 00:21:17,230
supposed to be equal to
the rotation C gamma.
366
00:21:17,230 --> 00:21:24,770
So that says that this point and
this point must be related
367
00:21:24,770 --> 00:21:26,170
by the rotation gamma.
368
00:21:26,170 --> 00:21:28,740
369
00:21:28,740 --> 00:21:29,720
So say that again.
370
00:21:29,720 --> 00:21:32,360
We're doing exactly what we did
here except we're starting
371
00:21:32,360 --> 00:21:35,720
with an initial point, not this
arc, but we're starting
372
00:21:35,720 --> 00:21:39,480
with the specific point A,
operate on it by A alpha, it
373
00:21:39,480 --> 00:21:42,120
twirled around but stays put.
374
00:21:42,120 --> 00:21:47,480
Rotate that by B beta, it goes
through a total angle beta to
375
00:21:47,480 --> 00:21:49,030
this location here.
376
00:21:49,030 --> 00:21:52,340
The net effect of getting from A
to A prime is supposed to be
377
00:21:52,340 --> 00:21:56,530
the rotation C gamma, so this
angle is then gamma and this
378
00:21:56,530 --> 00:21:59,835
is the location of C. OK?
379
00:21:59,835 --> 00:22:04,585
380
00:22:04,585 --> 00:22:08,620
OK, one other step that's
a fairly easy one.
381
00:22:08,620 --> 00:22:14,700
This length is equal to this
length, because they were
382
00:22:14,700 --> 00:22:18,010
produced by rotation.
383
00:22:18,010 --> 00:22:29,490
This side is common to these two
triangles, and this angle
384
00:22:29,490 --> 00:22:33,110
then is beta over 2, this
is beta over 2.
385
00:22:33,110 --> 00:22:39,470
And if these two triangles, A,
B, and C, that triangle is
386
00:22:39,470 --> 00:22:47,110
similar to A prime BC, and
therefore I can say that angle
387
00:22:47,110 --> 00:22:58,780
A prime CA is identical to ACB,
so therefore this angle
388
00:22:58,780 --> 00:23:02,540
has to equal this angle, and if
the total angle is gamma,
389
00:23:02,540 --> 00:23:05,410
this is gamma over 2, and
this is gamma over 2.
390
00:23:05,410 --> 00:23:11,090
391
00:23:11,090 --> 00:23:14,020
So now let me extract from this
the information that I
392
00:23:14,020 --> 00:23:15,550
would like to use.
393
00:23:15,550 --> 00:23:20,560
Here are three axes, A alpha,
B beta, and C gamma.
394
00:23:20,560 --> 00:23:23,820
This angle in here
is gamma over 2.
395
00:23:23,820 --> 00:23:27,930
This angle in here is beta over
2, and this angle in here
396
00:23:27,930 --> 00:23:29,180
is alpha over 2.
397
00:23:29,180 --> 00:23:32,200
398
00:23:32,200 --> 00:23:36,030
And let me emphasize that in
this magic triangle, out of
399
00:23:36,030 --> 00:23:40,600
which we're going to extract
some dazzlingly profound
400
00:23:40,600 --> 00:23:45,920
stuff, it is half the angular
throw of the rotation axes
401
00:23:45,920 --> 00:23:49,430
that appear in here as these
spherical angles, and not the
402
00:23:49,430 --> 00:23:51,720
entire angle of rotation.
403
00:23:51,720 --> 00:23:56,780
404
00:23:56,780 --> 00:24:01,020
So here's how properties of the
three rotation axes are
405
00:24:01,020 --> 00:24:02,400
related one to another.
406
00:24:02,400 --> 00:24:05,620
407
00:24:05,620 --> 00:24:11,340
And now, we introduced without
proof last time something
408
00:24:11,340 --> 00:24:16,140
called the law of cosines in
spherical trigonometry.
409
00:24:16,140 --> 00:24:20,120
And I not only do not want to
prove it, but I have no idea
410
00:24:20,120 --> 00:24:23,010
how I would go about doing so,
but that doesn't prevent me
411
00:24:23,010 --> 00:24:24,480
from using it.
412
00:24:24,480 --> 00:24:29,460
So if here are three edges, a,
b, c, and three angles in
413
00:24:29,460 --> 00:24:33,810
there, A, B, and C, we said
that the law of cosines in
414
00:24:33,810 --> 00:24:37,350
spherical trigonometry,
analogous in a way to the law
415
00:24:37,350 --> 00:24:41,070
of cosines and plane geometry,
except since the lengths of
416
00:24:41,070 --> 00:24:43,500
the triangles are measured
in degrees, there are
417
00:24:43,500 --> 00:24:47,460
trigonometric functions of these
angles that appear in
418
00:24:47,460 --> 00:24:48,880
the law of cosines.
419
00:24:48,880 --> 00:24:54,750
This says that cosine of b
cosine of c plus sine b sine
420
00:24:54,750 --> 00:25:00,745
of c times cosine of a is
equal to cosine of a.
421
00:25:00,745 --> 00:25:03,880
422
00:25:03,880 --> 00:25:09,630
So this now is an interesting
relation that we can apply to
423
00:25:09,630 --> 00:25:13,620
this spherical triangle, which
connects together the three
424
00:25:13,620 --> 00:25:16,080
rotation axes.
425
00:25:16,080 --> 00:25:24,770
Let me apply it to find the
angle c which we have picked
426
00:25:24,770 --> 00:25:30,850
as the angle between the initial
two axes a and b.
427
00:25:30,850 --> 00:25:34,650
That says that this should be
equal to cosine of b cosine of
428
00:25:34,650 --> 00:25:40,240
c, the angle between the other
two axes, plus sine of b sine
429
00:25:40,240 --> 00:25:54,340
of c times the cosine of angle
a, and angle a is cosine of
430
00:25:54,340 --> 00:25:55,738
alpha over 2.
431
00:25:55,738 --> 00:26:00,260
432
00:26:00,260 --> 00:26:02,940
So all these quantities that
we'd like to determine are
433
00:26:02,940 --> 00:26:07,640
hooked together by the
law of cosines.
434
00:26:07,640 --> 00:26:13,060
And this is a lovely relation,
but it doesn't do us a bit of
435
00:26:13,060 --> 00:26:20,220
good, because in this relation
we know only one quantity, and
436
00:26:20,220 --> 00:26:23,530
that is the rotation
angle of a.
437
00:26:23,530 --> 00:26:28,170
We can pick the angle between
a and b, that's this, but I
438
00:26:28,170 --> 00:26:30,130
have no idea what these
other angles are.
439
00:26:30,130 --> 00:26:31,600
That's what I'd like
to find out.
440
00:26:31,600 --> 00:26:33,980
I'd like to find out the
angles at which three
441
00:26:33,980 --> 00:26:38,870
rotations have to be combined
in order that rotation about
442
00:26:38,870 --> 00:26:42,150
one followed by rotation about
the second be the third.
443
00:26:42,150 --> 00:26:45,650
So this equation is a beautiful
equation, but it
444
00:26:45,650 --> 00:26:48,860
involves everything that I
don't know and only one
445
00:26:48,860 --> 00:26:50,470
quantity that I do know.
446
00:26:50,470 --> 00:26:52,130
So it looks as though
we're up the creek.
447
00:26:52,130 --> 00:26:53,910
Yes, sir?
448
00:26:53,910 --> 00:26:59,637
AUDIENCE: So you're looking for
cosine c, so shouldn't it
449
00:26:59,637 --> 00:27:01,700
be cosine b cosine a?
450
00:27:01,700 --> 00:27:02,950
PROFESSOR: Oh, I'm sorry.
451
00:27:02,950 --> 00:27:04,200
Yeah, I did that wrong.
452
00:27:04,200 --> 00:27:09,840
453
00:27:09,840 --> 00:27:10,340
Yeah.
454
00:27:10,340 --> 00:27:12,030
You're absolutely right.
455
00:27:12,030 --> 00:27:17,070
This should be cosine of a,
and this a goes with this
456
00:27:17,070 --> 00:27:18,205
alpha over 2.
457
00:27:18,205 --> 00:27:18,890
Absolutely.
458
00:27:18,890 --> 00:27:20,140
Sorry about that.
459
00:27:20,140 --> 00:27:25,380
460
00:27:25,380 --> 00:27:29,000
So anyway, the point still
stands that what this equation
461
00:27:29,000 --> 00:27:32,230
involves is the three interaxial
angles, and I would
462
00:27:32,230 --> 00:27:35,460
like to know how I could combine
a and b to get it to
463
00:27:35,460 --> 00:27:39,170
come out to a crystallographic
rotation c, and where that
464
00:27:39,170 --> 00:27:42,460
location is relative to
the first two axes.
465
00:27:42,460 --> 00:27:45,710
So it involves everything I
don't know, and only one
466
00:27:45,710 --> 00:27:47,370
thing that I do.
467
00:27:47,370 --> 00:27:53,340
But now we introduce another
curious aspect of spherical
468
00:27:53,340 --> 00:27:55,830
triangles, which I mentioned
last time.
469
00:27:55,830 --> 00:27:57,450
You may have thought
that that's
470
00:27:57,450 --> 00:28:00,230
interesting, but who cares?
471
00:28:00,230 --> 00:28:06,640
Here are the three points, A,
B, and C, and these are the
472
00:28:06,640 --> 00:28:11,290
three arcs little c, little
a and little b.
473
00:28:11,290 --> 00:28:14,210
And then we said we could
construct something called the
474
00:28:14,210 --> 00:28:17,350
polar triangle of ABC.
475
00:28:17,350 --> 00:28:23,480
And what we would do, we would
find the pole of arc b, and
476
00:28:23,480 --> 00:28:27,340
that will be some
point B prime.
477
00:28:27,340 --> 00:28:31,680
We'll find the pole of arc
a, and that will be
478
00:28:31,680 --> 00:28:33,360
some point A prime.
479
00:28:33,360 --> 00:28:37,060
We'll find, similarly, the pole
of arc c, and that will
480
00:28:37,060 --> 00:28:39,210
be some point C prime.
481
00:28:39,210 --> 00:28:43,250
And now we can connect together
A prime, B prime, and
482
00:28:43,250 --> 00:28:45,440
C prime, and get something
that's
483
00:28:45,440 --> 00:28:47,005
called the polar triangle.
484
00:28:47,005 --> 00:28:49,540
485
00:28:49,540 --> 00:28:51,400
And now comes the useful part.
486
00:28:51,400 --> 00:28:57,990
We said that a curious property
of the polar triangle
487
00:28:57,990 --> 00:29:05,290
is that the side of the polar
triangle plus the angle
488
00:29:05,290 --> 00:29:10,680
opposite it add up
to 180 degrees.
489
00:29:10,680 --> 00:29:14,790
In my original triangle, this is
beta over 2, this is gamma
490
00:29:14,790 --> 00:29:17,960
over 2, and this is
alpha over 2.
491
00:29:17,960 --> 00:29:23,370
So the length of this side is
going to be 180 degrees minus
492
00:29:23,370 --> 00:29:28,540
beta over 2, the length of this
side is going to be 180
493
00:29:28,540 --> 00:29:32,490
degrees minus alpha over 2, and
the length of this side is
494
00:29:32,490 --> 00:29:38,270
going to be 180 degrees
minus gamma over 2.
495
00:29:38,270 --> 00:29:43,630
And now let's use these angles
and these lengths in the law
496
00:29:43,630 --> 00:29:46,310
of cosines, and I'll leave
out the little bit
497
00:29:46,310 --> 00:29:49,210
of intervening algebra.
498
00:29:49,210 --> 00:29:55,760
And what we will get out of
this is that cosine of c--
499
00:29:55,760 --> 00:29:57,720
and I'll solve for that--
500
00:29:57,720 --> 00:30:03,940
is equal to cosine of alpha over
2 cosine of beta over 2
501
00:30:03,940 --> 00:30:10,140
plus cosine of gamma over 2
divided by sine alpha over 2
502
00:30:10,140 --> 00:30:11,860
sine beta over 2.
503
00:30:11,860 --> 00:30:15,820
504
00:30:15,820 --> 00:30:18,820
And that is something we can
sink our teeth into and run
505
00:30:18,820 --> 00:30:23,570
with, because now I can ask the
question, suppose I want a
506
00:30:23,570 --> 00:30:28,010
to be a 2-fold rotation axis,
b to be a 3-fold rotation
507
00:30:28,010 --> 00:30:31,830
axis, and c be a 4-fold
rotation axis?
508
00:30:31,830 --> 00:30:39,040
Then the value of alpha over 2
is half of 180 degrees or 90.
509
00:30:39,040 --> 00:30:41,470
Well, you can see I put
in half the value of
510
00:30:41,470 --> 00:30:42,970
the rotation axes.
511
00:30:42,970 --> 00:30:46,990
And then I solve for c, and that
is the angle at which I
512
00:30:46,990 --> 00:30:51,950
have to put axis a and b
together to get c to turn out
513
00:30:51,950 --> 00:30:55,050
to be whatever angle
gamma over 2 is.
514
00:30:55,050 --> 00:30:59,440
515
00:30:59,440 --> 00:31:00,220
So I can do this
516
00:31:00,220 --> 00:31:02,850
systematically now without thinking.
517
00:31:02,850 --> 00:31:11,090
And I can set up the problem by
taking the crystallographic
518
00:31:11,090 --> 00:31:15,140
rotation axes and combining them
together three at a time
519
00:31:15,140 --> 00:31:18,060
in all possible combinations.
520
00:31:18,060 --> 00:31:20,070
Right?
521
00:31:20,070 --> 00:31:23,400
In addition to this relation,
I have two other relations.
522
00:31:23,400 --> 00:31:27,120
And let me assemble them off to
the left, because we have
523
00:31:27,120 --> 00:31:30,820
to solve three equations to
find out the nature of the
524
00:31:30,820 --> 00:31:34,460
combination that is required.
525
00:31:34,460 --> 00:31:40,010
So just permuting terms, the
angle between A and B, c, has
526
00:31:40,010 --> 00:31:45,330
to follow from cosine of c
equals cosine of alpha over 2
527
00:31:45,330 --> 00:31:51,050
cosine of beta over 2 plus
cosine of gamma over 2.
528
00:31:51,050 --> 00:31:56,470
Notice that the single term by
itself is the cosine of half
529
00:31:56,470 --> 00:32:02,200
the angle of the opposite
rotation axis c.
530
00:32:02,200 --> 00:32:05,340
Then in the denominator is the
sine of these two angles.
531
00:32:05,340 --> 00:32:08,130
532
00:32:08,130 --> 00:32:12,690
And so just permuting terms, one
can see that cosine of b
533
00:32:12,690 --> 00:32:16,850
is going to turn out to be
cosine of alpha over 2 cosine
534
00:32:16,850 --> 00:32:23,280
of gamma over 2 plus cosine of
theta over 2 divided by sine
535
00:32:23,280 --> 00:32:27,570
of alpha over 2 sine
of gamma over 2.
536
00:32:27,570 --> 00:32:31,600
And a third analogous expression
will give me the
537
00:32:31,600 --> 00:32:37,230
angle that will be the one
between B and C. And this will
538
00:32:37,230 --> 00:32:42,590
be cosine of beta over 2 cosine
gamma over 2 plus
539
00:32:42,590 --> 00:32:48,330
cosine of alpha over 2 divided
by sine beta over 2
540
00:32:48,330 --> 00:32:52,690
sine gamma over 2.
541
00:32:52,690 --> 00:32:54,590
OK?
542
00:32:54,590 --> 00:32:56,090
So now we don't have
to think anymore.
543
00:32:56,090 --> 00:32:57,990
It's just plug and chug.
544
00:32:57,990 --> 00:33:01,100
And I'll pause to suck in air
and let you catch up, and then
545
00:33:01,100 --> 00:33:03,515
we'll set up the problem and
look at a few solutions.
546
00:33:03,515 --> 00:33:25,320
547
00:33:25,320 --> 00:33:28,210
And all this, in the event
that you're thoroughly
548
00:33:28,210 --> 00:33:31,240
bewildered, is in the
set of notes that I
549
00:33:31,240 --> 00:33:32,260
handed out last time.
550
00:33:32,260 --> 00:33:37,060
So you can read it over
at your leisure.
551
00:33:37,060 --> 00:33:39,540
AUDIENCE: Will this stuff
be on the quiz?
552
00:33:39,540 --> 00:33:39,980
PROFESSOR: No.
553
00:33:39,980 --> 00:33:42,940
Quiz will go up to the end of
the two-dimensional plane
554
00:33:42,940 --> 00:33:45,100
groups and stop.
555
00:33:45,100 --> 00:33:46,350
We won't say anything
three dimensional.
556
00:33:46,350 --> 00:34:00,180
557
00:34:00,180 --> 00:34:06,030
OK, let's, then, if there's no
objection or complaint, look
558
00:34:06,030 --> 00:34:08,449
at possible values for--
559
00:34:08,449 --> 00:34:11,179
let me do it the same way that
I did it in the notes so that
560
00:34:11,179 --> 00:34:12,340
it's consistent--
561
00:34:12,340 --> 00:34:17,690
let's put down the value for
axis b, the rank of axis b and
562
00:34:17,690 --> 00:34:19,120
the rank of axis a.
563
00:34:19,120 --> 00:34:25,370
564
00:34:25,370 --> 00:34:32,460
And A could be a 1-fold axis,
B could be a 1-fold access,
565
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
566
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
567
00:34:39,810 --> 00:34:41,060
with a 1 with a 6.
568
00:34:41,060 --> 00:34:44,440
569
00:34:44,440 --> 00:34:46,739
This is clearly impossible.
570
00:34:46,739 --> 00:34:50,920
If I did nothing, and followed
it by doing nothing, and
571
00:34:50,920 --> 00:34:54,989
wanted it to come out to be a
6-fold rotation, you'd all be
572
00:34:54,989 --> 00:34:57,580
spinning on your axes
like tops right now.
573
00:34:57,580 --> 00:35:00,270
So you can't do nothing and
follow it by doing nothing and
574
00:35:00,270 --> 00:35:02,200
have it come out to
be a net rotation.
575
00:35:02,200 --> 00:35:05,630
So these are impossible.
576
00:35:05,630 --> 00:35:08,590
So we don't have to
consider those.
577
00:35:08,590 --> 00:35:13,220
A could be a 2, though, and I
don't want to do 2, 1, 1,
578
00:35:13,220 --> 00:35:14,930
because I've got
a 1, 1, 2 here.
579
00:35:14,930 --> 00:35:16,310
The order doesn't make
any difference.
580
00:35:16,310 --> 00:35:22,440
So I'll start with a 2, 1,
2, a 2, 1, 3, a 2, 1,
581
00:35:22,440 --> 00:35:25,050
4, and a 2, 1, 6.
582
00:35:25,050 --> 00:35:31,020
So those are four combinations
that I should be examining.
583
00:35:31,020 --> 00:35:36,070
I could look at a 3 with a--
584
00:35:36,070 --> 00:35:39,380
585
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
586
00:35:43,200 --> 00:35:55,250
one I would want to look at
is a 3, 1, 3, a 3, 1,
587
00:35:55,250 --> 00:35:59,850
4, and a 3, 1, 6.
588
00:35:59,850 --> 00:36:03,310
And let me put in a
couple more here.
589
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
590
00:36:10,480 --> 00:36:15,040
with a 2 with a 3, a 2
with a 2 with a 4, 2
591
00:36:15,040 --> 00:36:17,230
with a 2 with a 6.
592
00:36:17,230 --> 00:36:20,530
And then A 3 with a 2--
593
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,
594
00:36:25,192 --> 00:36:28,980
2, 4, 3, 2, 6, run out of room
here, but there should be a
595
00:36:28,980 --> 00:36:33,700
similar entry with
a 4 and a 6.
596
00:36:33,700 --> 00:36:35,000
So this sets up the problem.
597
00:36:35,000 --> 00:36:39,170
If you count up the number of
ways one can do this, we only
598
00:36:39,170 --> 00:36:42,600
have to consider the
off-diagonal boxes here,
599
00:36:42,600 --> 00:36:48,160
because interchanging a and b,
for example, looking at 3, 1,
600
00:36:48,160 --> 00:36:53,030
3, that's going to be the
same as 1, 3, 3 up here.
601
00:36:53,030 --> 00:36:55,110
So it's just the off-diagonal
boxes
602
00:36:55,110 --> 00:36:56,680
that we have to consider.
603
00:36:56,680 --> 00:37:06,330
So there should be a 3,
3, 3 in here, a 3, 3,
604
00:37:06,330 --> 00:37:09,065
4, and a 3, 3, 6.
605
00:37:09,065 --> 00:37:12,720
606
00:37:12,720 --> 00:37:16,890
So what we would have to do in
order to determine the unique
607
00:37:16,890 --> 00:37:21,950
combinations is to look at all
of these combinations in turn,
608
00:37:21,950 --> 00:37:28,190
and I'm going to not
try all of them.
609
00:37:28,190 --> 00:37:34,740
I will do one that's going
to be clearly impossible.
610
00:37:34,740 --> 00:37:36,730
So let's look at 2, 1, 4.
611
00:37:36,730 --> 00:37:44,390
So here A corresponds to a
2-fold axis, B corresponds to
612
00:37:44,390 --> 00:37:53,830
a 1-fold axis, and C would
correspond to a 4-fold axis.
613
00:37:53,830 --> 00:37:58,500
So could we combine these three
axes at appropriate
614
00:37:58,500 --> 00:38:01,620
angles such that a 2-fold
followed by a 1-fold is
615
00:38:01,620 --> 00:38:04,420
equivalent to a 4-fold?
616
00:38:04,420 --> 00:38:06,310
This clearly isn't
going to work.
617
00:38:06,310 --> 00:38:10,820
If I do a 180-degree rotation
then don't do anything and ask
618
00:38:10,820 --> 00:38:15,850
is that equivalent to a 4-fold
rotation, that is saying that
619
00:38:15,850 --> 00:38:19,080
the 2-fold axis should be
identical to the 4-fold axis,
620
00:38:19,080 --> 00:38:21,460
and that is not going to work.
621
00:38:21,460 --> 00:38:24,580
622
00:38:24,580 --> 00:38:28,580
So let me do now a generic
family that I know turns out
623
00:38:28,580 --> 00:38:31,220
to be possible.
624
00:38:31,220 --> 00:38:36,120
Let me look at an n-fold axis
with a 2-fold axis with a
625
00:38:36,120 --> 00:38:41,590
2-fold axis, and I can show
that this combination is
626
00:38:41,590 --> 00:38:45,260
possible for any integer
n whatsoever.
627
00:38:45,260 --> 00:38:48,080
So this will include a lot
of non-crystallographic
628
00:38:48,080 --> 00:38:50,980
symmetries.
629
00:38:50,980 --> 00:38:58,130
So let's say that this is C,
this is A, and this is B. But
630
00:38:58,130 --> 00:38:59,995
make it C, B, A if you'd like.
631
00:38:59,995 --> 00:39:03,960
632
00:39:03,960 --> 00:39:11,770
OK, so my first equation says
that cosine of c, the angle
633
00:39:11,770 --> 00:39:17,610
between A and B, should be equal
to the cosine of alpha
634
00:39:17,610 --> 00:39:22,640
over 2 times the cosine of beta
over 2 plus the cosine of
635
00:39:22,640 --> 00:39:28,290
gamma over 2 divided by sine
of alpha over 2 sine
636
00:39:28,290 --> 00:39:31,370
of beta over 2.
637
00:39:31,370 --> 00:39:39,390
B is a 2-fold axis, so alpha
is equal to 180 degrees.
638
00:39:39,390 --> 00:39:40,780
A is a 2-fold--
639
00:39:40,780 --> 00:39:41,970
I'm sorry.
640
00:39:41,970 --> 00:39:44,530
B beta, A alpha.
641
00:39:44,530 --> 00:39:47,510
Beta is equal to 180 degrees,
the angular
642
00:39:47,510 --> 00:39:50,510
throw of a 2-fold axis.
643
00:39:50,510 --> 00:39:53,570
Alpha is equal to 180
degrees, the angular
644
00:39:53,570 --> 00:39:56,080
throw of a 2-fold axis.
645
00:39:56,080 --> 00:40:04,140
And gamma is equal to whatever
2 pi over n would be.
646
00:40:04,140 --> 00:40:06,730
647
00:40:06,730 --> 00:40:10,300
That's the throw of the n-fold
axis than I'm letting be equal
648
00:40:10,300 --> 00:40:11,520
to C.
649
00:40:11,520 --> 00:40:17,940
So the cosine of A and B, to get
the result of rotation A
650
00:40:17,940 --> 00:40:21,360
followed by rotation B being
equal to the net rotation of
651
00:40:21,360 --> 00:40:25,340
an n-fold axis is that the
cosine of c should be the
652
00:40:25,340 --> 00:40:33,210
cosine of 180 degrees over 2
times the cosine of 180 over 2
653
00:40:33,210 --> 00:40:39,340
plus the cosine of gamma over
2, and gamma is whatever the
654
00:40:39,340 --> 00:40:43,600
rank of the axis determines, and
that's divided by sine of
655
00:40:43,600 --> 00:40:50,920
alpha over 2, and alpha is 180
and sine beta over 2, and
656
00:40:50,920 --> 00:40:53,870
that's 180 over 2.
657
00:40:53,870 --> 00:40:59,550
So the cosine of c is going to
be the cosine of 90, which is
658
00:40:59,550 --> 00:41:04,600
0, times the cosine of 90, which
is 0, plus the cosine of
659
00:41:04,600 --> 00:41:08,540
gamma over 2, whatever that
might be, over the sine of 90
660
00:41:08,540 --> 00:41:12,020
which is 1, sine of
90 which is 1.
661
00:41:12,020 --> 00:41:18,405
So this says that cosine
of c is equal to cosine
662
00:41:18,405 --> 00:41:20,350
of gamma over 2.
663
00:41:20,350 --> 00:41:25,520
So the angle between axis A and
axis B ought to be equal
664
00:41:25,520 --> 00:41:32,710
to one half the angular throw
of rotation axis C. So let's
665
00:41:32,710 --> 00:41:34,680
start putting down some
of this information.
666
00:41:34,680 --> 00:41:43,470
This says that if this is axis
C gamma, then A pi and B pi,
667
00:41:43,470 --> 00:41:47,180
the 2 180-degree rotations,
should be at an
668
00:41:47,180 --> 00:41:49,000
angle gamma over 2.
669
00:41:49,000 --> 00:41:52,310
670
00:41:52,310 --> 00:41:57,530
We still need values
for b, and we still
671
00:41:57,530 --> 00:41:59,470
need a value for a.
672
00:41:59,470 --> 00:42:03,400
So let's find out
what those are.
673
00:42:03,400 --> 00:42:08,620
And let me start over here at
the left again, because I've
674
00:42:08,620 --> 00:42:12,430
got the relation that I need
for b sitting here.
675
00:42:12,430 --> 00:42:18,160
The cosine of b is equal to the
cosine of alpha over 2,
676
00:42:18,160 --> 00:42:26,185
and that is the cosine of pi
over 2, plus alpha is a
677
00:42:26,185 --> 00:42:27,850
90-degree rotation.
678
00:42:27,850 --> 00:42:32,990
Then cosine of gamma over 2,
whatever gamma happens to be,
679
00:42:32,990 --> 00:42:36,290
plus the cosine of beta over
2, and that's cosine of pi
680
00:42:36,290 --> 00:42:41,760
over 2, and this is all over
sine of pi over 2 times the
681
00:42:41,760 --> 00:42:50,110
sine of gamma over 2.
682
00:42:50,110 --> 00:42:55,360
So this is going to be cosine of
pi over 2, which is 0, plus
683
00:42:55,360 --> 00:43:05,620
cosine of pi over 2 divided by
sine of pi over 2, which is 1.
684
00:43:05,620 --> 00:43:09,510
This gamma over 2--
685
00:43:09,510 --> 00:43:13,800
no, cosine of pi over 2 is 0.
686
00:43:13,800 --> 00:43:20,530
So this is 0 plus 0 times sine
of gamma over 2, whatever that
687
00:43:20,530 --> 00:43:22,070
turns out to be.
688
00:43:22,070 --> 00:43:27,090
So cosine of b is 0, and that
says that the angle b between
689
00:43:27,090 --> 00:43:32,680
axis A and axis C turns
out to be 90 degrees.
690
00:43:32,680 --> 00:43:40,780
So in order to get pi followed
by c gamma to be equal to b
691
00:43:40,780 --> 00:43:44,400
pi, I've got to make b be
equal to pi over 2.
692
00:43:44,400 --> 00:43:47,420
693
00:43:47,420 --> 00:43:50,420
And if I put it in this
orientation, then a pi
694
00:43:50,420 --> 00:43:54,960
followed by c gamma is going
to be equal to b pi.
695
00:43:54,960 --> 00:43:59,870
One final angle, and that's
the value for a.
696
00:43:59,870 --> 00:44:07,300
697
00:44:07,300 --> 00:44:12,890
OK, cosine of a should be equal
to the cosine of beta
698
00:44:12,890 --> 00:44:18,410
over 2, that's pi over 2 times
the cosine of gamma over 2,
699
00:44:18,410 --> 00:44:26,700
whatever it is, plus the cosine
of beta over 2, and
700
00:44:26,700 --> 00:44:34,200
that's cosine of pi over 2, over
sine pi over 2 sine of
701
00:44:34,200 --> 00:44:35,930
gamma over 2.
702
00:44:35,930 --> 00:44:43,880
And this, as for b, turns out
to be 0 plus 0 over sine pi
703
00:44:43,880 --> 00:44:50,810
over 2, which is 1 times
sine of gamma over 2.
704
00:44:50,810 --> 00:44:54,820
So cosine of a turns out to be
0, and this says that the
705
00:44:54,820 --> 00:44:58,750
angle a is also pi over 2.
706
00:44:58,750 --> 00:44:59,600
So we've got a whole--
707
00:44:59,600 --> 00:45:00,502
Yeah.
708
00:45:00,502 --> 00:45:01,704
AUDIENCE: Did you really
need to go
709
00:45:01,704 --> 00:45:03,210
through all three equations?
710
00:45:03,210 --> 00:45:06,760
PROFESSOR: Yeah, because I had
to show that all three work.
711
00:45:06,760 --> 00:45:10,300
And in general, if I combine,
let's say, a 4-fold with a
712
00:45:10,300 --> 00:45:14,350
3-fold with a 2-fold, which is
something I want to do, all
713
00:45:14,350 --> 00:45:18,238
three angles a, b, and
c, will be different.
714
00:45:18,238 --> 00:45:20,700
OK?
715
00:45:20,700 --> 00:45:22,510
So the answer is yes.
716
00:45:22,510 --> 00:45:26,270
And there will be a few cases
where a value for one angle
717
00:45:26,270 --> 00:45:31,600
will exist and the value for the
one or two others will be
718
00:45:31,600 --> 00:45:32,880
impossible.
719
00:45:32,880 --> 00:45:36,240
And that's also something
that I have to know.
720
00:45:36,240 --> 00:45:40,830
So repetitious as the exercise
might be, the answer is yeah,
721
00:45:40,830 --> 00:45:42,822
you do have to do all three.
722
00:45:42,822 --> 00:45:44,072
AUDIENCE: [INAUDIBLE]?
723
00:45:44,072 --> 00:45:48,234
724
00:45:48,234 --> 00:45:50,460
PROFESSOR: Oh, you don't have to
do different permutations.
725
00:45:50,460 --> 00:45:52,830
That's just a question
of labeling.
726
00:45:52,830 --> 00:45:53,080
OK?
727
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
728
00:45:56,490 --> 00:45:58,080
is the same as a 2
with a 2 with an
729
00:45:58,080 --> 00:46:00,000
n, that's just labeling.
730
00:46:00,000 --> 00:46:03,500
So that's why my boxes, when I
filled them out, the list got
731
00:46:03,500 --> 00:46:06,940
shorter and shorter, until
finally for a 6-fold axis, it
732
00:46:06,940 --> 00:46:12,780
would be just 6, 1, 1, 6, 1,
2, 6, 1, 3, and so on.
733
00:46:12,780 --> 00:46:15,450
Just that one entry in the box
where I enumerated what should
734
00:46:15,450 --> 00:46:16,700
be considered.
735
00:46:16,700 --> 00:46:19,840
736
00:46:19,840 --> 00:46:28,440
Well, here is a whole slew of
possible solutions and a lot
737
00:46:28,440 --> 00:46:29,990
of them are
non-crystallographic, but
738
00:46:29,990 --> 00:46:31,450
still possible.
739
00:46:31,450 --> 00:46:37,450
This says that a combination
of a 2-fold axis--
740
00:46:37,450 --> 00:46:40,540
but remember now that these are
equations and operations.
741
00:46:40,540 --> 00:46:44,910
But the only operation that's
present for a 2-fold axis is a
742
00:46:44,910 --> 00:46:47,415
rotation a pi, b pi or c pi.
743
00:46:47,415 --> 00:46:52,940
So I could combine three 2-fold
axes that are mutually
744
00:46:52,940 --> 00:46:58,850
orthogonal, and that is an
allowable combination.
745
00:46:58,850 --> 00:47:03,650
And what we are obtaining here
is a sort of scaffolding, a
746
00:47:03,650 --> 00:47:08,170
framework, based on pure
rotation operations, that by
747
00:47:08,170 --> 00:47:11,270
themselves will be an allowable
3-dimensional point
748
00:47:11,270 --> 00:47:16,630
group, but which also provides a
framework, a Christmas tree,
749
00:47:16,630 --> 00:47:19,230
that we can decorate with mirror
planes and inversion
750
00:47:19,230 --> 00:47:21,500
centers to get still additional
751
00:47:21,500 --> 00:47:23,760
groups of higher symmetry.
752
00:47:23,760 --> 00:47:26,470
So here's one possible
crystallographic o combination
753
00:47:26,470 --> 00:47:28,030
of rotation axes.
754
00:47:28,030 --> 00:47:31,860
What we will use to denote this
combination is the same
755
00:47:31,860 --> 00:47:35,200
rule that we use for
our other notation.
756
00:47:35,200 --> 00:47:38,880
We will make a running list of
the independent operations
757
00:47:38,880 --> 00:47:43,030
that are present, and what we
have combined here are three
758
00:47:43,030 --> 00:47:45,390
distinct independent
2-fold axes.
759
00:47:45,390 --> 00:47:48,390
760
00:47:48,390 --> 00:47:54,150
A solid that would have this
symmetry plus some other
761
00:47:54,150 --> 00:48:00,790
symmetry would be an orthogonal
brick with one
762
00:48:00,790 --> 00:48:03,050
2-fold axis coming out here.
763
00:48:03,050 --> 00:48:10,130
This is the operation c pi,
another 2-fold axis coming out
764
00:48:10,130 --> 00:48:15,740
the front, and this would be
the operation a pi, and
765
00:48:15,740 --> 00:48:18,510
another 2-fold axis coming out
of this face, and this would
766
00:48:18,510 --> 00:48:20,040
be the operation b pi.
767
00:48:20,040 --> 00:48:23,410
768
00:48:23,410 --> 00:48:26,430
Now let me show you-- it's
rather amusing-- that what we
769
00:48:26,430 --> 00:48:28,740
have done really works.
770
00:48:28,740 --> 00:48:34,990
We've shown supposedly that a pi
followed by b pi should be
771
00:48:34,990 --> 00:48:40,390
equal to a net rotation c
pi about an axis that's
772
00:48:40,390 --> 00:48:42,070
orthogonal to the first two.
773
00:48:42,070 --> 00:48:45,680
So let's pick a motif, and for
convenience I'll put it at one
774
00:48:45,680 --> 00:48:47,340
corner of this brick.
775
00:48:47,340 --> 00:48:52,420
Here's object 1, I rotate it
by 180 degrees about a.
776
00:48:52,420 --> 00:48:57,090
Here sits object number 2,
same corality, and then I
777
00:48:57,090 --> 00:49:02,900
rotate it 180 degrees about B
beta, and that's going to give
778
00:49:02,900 --> 00:49:05,240
me number 3.
779
00:49:05,240 --> 00:49:09,990
What is the net way of
getting from 1 to 3?
780
00:49:09,990 --> 00:49:10,860
Holy mackerel.
781
00:49:10,860 --> 00:49:14,720
It's a net 180-degree
rotation about c pi.
782
00:49:14,720 --> 00:49:15,970
It really works.
783
00:49:15,970 --> 00:49:18,270
784
00:49:18,270 --> 00:49:21,360
Or I could do the operations
in a different order.
785
00:49:21,360 --> 00:49:25,580
I could rotate by d, rotate by
c, and the way I get from the
786
00:49:25,580 --> 00:49:30,160
first to the third is
a rotation a pi.
787
00:49:30,160 --> 00:49:33,260
So that is a self consistent
set of rotation axes.
788
00:49:33,260 --> 00:49:36,610
That is 2, 2, 2.
789
00:49:36,610 --> 00:49:38,722
Let me do one more.
790
00:49:38,722 --> 00:49:41,730
Well, no, let me take a break
here and let you absorb all
791
00:49:41,730 --> 00:49:45,140
this, and then we'll look at
some remaining ones, and this
792
00:49:45,140 --> 00:49:47,340
will include some that are
non- crystallographic.
793
00:49:47,340 --> 00:49:48,800
And that's perfectly OK.
794
00:49:48,800 --> 00:49:51,980
But they're lovely groups, they
constitute groups, but
795
00:49:51,980 --> 00:49:54,250
they won't be groups
that can occur in
796
00:49:54,250 --> 00:49:55,760
combination with a lattice.
797
00:49:55,760 --> 00:49:57,975