We%20are%20now%20ready%20to%20move%20to%20the%20orthorhombic%20system.

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We are now ready to move to the orthorhombic
system. There are 59 orthorhombic space groups.
Orthorhombic crystals belong to crystal classes
222, mm2 or mmm. In this section well consider
many (but not yet all) of the orthorhombic space
groups in crystal class 222, namely, space groups
P222, P2221, P21212, P212121, and I222. Later
versions of the tutorial will complete the 222
groups, and move on to the remaining orthorhombic
groups. Note that all space groups which belong
to crystal class 222 will be enantiomorphous and
noncentrosymmetric.
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Primitive Orthorhombic1
End-Centered Orthorhombic (A- or B- or C- )1
Body-Centered Orthorhombic (I)1
Face-Centered Orthorhombic (F)1
(Click on any green label above to rotate)
In the orthorhombic system, a ? b ? c ? ? ?
90º. We can thus have three possible crystal
classes or point groups 222, mm2 or mmm.
1http//phycomp.technion.ac.il/sshaharr/intro.ht
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In order to derive the primitive orthorhombic
space groups in crystal class 222, we need to
consider how a pair of 2 (or a pair of 21 axes)
parallel to the crystallographic a and b axes may
be combined. Once we've figured out how to
approach this, we'll jump in, do the derivations,
and see what is generated along the
crystallographic c direction. This will require
your careful attention, but the results are very
useful and very interesting!
Martin J. Buerger
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Hmmm.getting started should be fairly
straightforward a pair of twofold axes may
either intersect or they may not. Similarly, a
pair of 21 screw axes may intersect or they may
not. Let's see what happens when we try a
derivation of an orthorhombic space group
beginning with a pair of intersecting 2's. Hey,
I'm really getting into this!
A. Student
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Now for the test do you know what this group is
called?
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That worked, and we got space group P222. What
will we get if we have two nonintersecting 2's?
I'll try it first one will go along a, while
I'll put the second one b, but at c ¼ so
that it doesn't intersect the first twofold axis.
Right now I'm not certain what I'll get, but it
cannot be the same group! Here goes..
Will B. Learning
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So. do you know what this group is called?
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So.when we have two intersecting 2's, we derive
P222, and when the axes do not intersect, we
obtain P2221. Now what we need to do is to try
this for a pair of 21 axes the drill should now
be familiar! First we'll see what we get with a
pair of intersecting 21's, and then with a pair
of non-intersecting 21's. The new group will
likely be P2121X, where X a new generated
element.
Before we begin to look at that, we'll add a
shortcut to our derivations we've noted several
times before that the combinations of symmetry
elements and translations lead to additional
elements that appear every half-unit cell this
is a very general observation. Look back at the
one we just did
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After that "interlude", we need to remind
ourselves where we were! A few slides back, we
said that our next step would be to attempt the
derivation for a pair of 21 axes the drill
should now be familiar! First we'll see what we
get with a pair of intersecting 21's, and then
with a pair of non-intersecting 21's. The new
group will likely be P2121X, where X a new
generated element.
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Yes! We seem to have derived space group P21212,
given the appearance of the diagram, but our
diagram has a different origin from that in the
International Tables for Crystallography, Volume
A. Thus we need to examine the conventions for
choice of origin, as specified on page 21 of that
volume.
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From the International tables for
Crystallography, Volume A, page 21, we read (i)
All centrosymmetric space groups are described
with an inversion centre as origin. A second
description is given if a space group contains
points of high site symmetry that do not coincide
with a centre of symmetry (ii) For
non-centrosymmetric space groups, the origin is
at a point of highest site symmetry... If there
is no site symmetry higher than 1, except for the
cases listed below under (iii), the origin is
placed on a screw axis, or a glide plane, or at
the intersection of several such symmetry
elements (iii) In space group P212121 the
origin is chosen in such a way that it is
surrounded symmetrically by three pairs of 21
axes
Let us look at our putative P21212 diagram again
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Now we'll derive P212121 - recall that the
convention is to place the three axes at
locations equally displaced from the origin.
Andwe've been doing our derivations by starting
with two axes a and b, respectively. So,
let's begin by putting the first 21, a, at b
¼, and the second 21, b, at c ¼. If all
goes as we now may have learned to expect, we
should generate a third 21, c, at a ¼.
J. Monteath Robertson
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Essentially, we've carried out the permutations
and combinations for derivations of space groups
starting with pairs of 2 or 21 axes. We did not
look at combining a 2 and a 21 axis, butin
effect we've done that job too! How?
Our results were the space groups P222, P2221,
P21212 and P212121. From observing these groups,
it should be clear that combining a 2 and a 21
axis will produce either an additional 2 or an
additional 21, depending upon whether the 2 and
21 intersect or not. The result would be a
non-standard setting of one of the four above
groups, e.g., P21221 or P2122, etc. The standard
setting could be readily produced by a simple
transformation of axes. For example, if we say
that P2221 has axes abc, then P2122 will
correspond to an axial transformation with axial
settings cab relative to the standard group P2221.
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However, there is an obvious combination that we
did not try namely, to force three 21 axes to
intersect! To test this case, we shall start
with 21 axes along the crystallographic a, b and
c directions then we will perform the derivation
of the unknown space group as before. After
the Figure is complete, it should be possible to
identify the space group as one of the 9
orthorhombic space groups belonging to crystal
class 222.
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Note that there is a 222 symmetry site in this
"version" of I222. If we look in the
International Tables, Volume A, we can find I222,
but not this diagram. To derive the diagram in
the Tables, we need to place the three 2's at the
origin, and add I-centering operationsand, of
course, then we should get the conventional,
standard diagram, as well as three intersecting
21's displaced by (¼, ¼, ¼).
The take-home message is twofold, no pun
intended! First, if we force three 21's to
intersect, we obtain a non-primitive lattice.
Second, the non-primitive lattice contains pairs
of parallel 2's and 21's along a, b and c. Let's
rederive I222 starting with I-centering and all
the 2's and 21's
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End of Section 5, Pointgroup 222