W. J. (William James) Lewis.

A treatise on crystallography online

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Similarly, any other zone-axis 0A>
which is neither parallel nor perpen-
dicular to 0A and which is not in
the plane 0B&, is repeated in another
zone-axis 0A t \ and the plane AOkA t F « g

also contains 0A. As the plane A0^A t
contains two zone-axes, it is parallel to a possible face.

Hence, 0A is parallel to two possible faces, and is, therefore, a
possible zone-axis.

Cor. 1. In the case of a tetrad axis, we can turn the crystal
about it through 90° four times in succession until the crystal
returns to the original position. Hence, the edge LM t of Fig. 84
takes up three new positions, which are obtained by turning

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the crystal about the tetrad axis through 90°, through 2x90°= 180°,

and through 3x90° = 270°. One of these

positions, when the crystal has been turned

through 1 80°, is clearly LM '. But the face

parallel to the edges LM ti LM\ is also parallel

to the tetrad axis OL. Similarly, the plane

OLM is parallel to a possible face. Hence,

a tetrad axis is a possible zone-axis.

Cor. 2. The same is also true in the case of a hexad axis ; for if
the crystal is turned thrice in the same
direction, each time through 60°, a
position is attained which is the same
as that given by a single rotation of
180° about the same axis. Thus the
edges VA t , VA\ Fig. 85, are coplanar
with the hexad axis OV. Similarly,
the planes OA F, OA tl V contain two
edges and are parallel to possible faces.
Hence, a hexad axis is, also, a possible zone-axis.

The same argument would apply to any axis of symmetry of
even degree, but we shall see in Prop. 6 that no axis can have
a degree higher than six.

Cob. 3. Suppose an axis of symmetry of odd degree, n, to exist
in a crystal, n being > 3. Let the
continuous lines VA 9 VA\ <fec. of
Fig. 86 represent the polar edges 1
of a pyramid formed by the homo-
logous and interchangeable faces of
such a crystal when it is projected
on a plane perpendicular to the
axis of symmetry OV.

Produce the lines A A 1 , A X A\ <fec.,
in which the faces meet the plane
of projection; and let alternate pairs

1 Several classes of crystals have a single axis of symmetry of degree higher
than 2, which may, or may not, be associated with planes of symmetry and with
dyad axes perpendicular to it. Such an axis will be called a principal axis.
Thus, in Figs. 84, 86, OL and OV are principal axes. We shall also denote the
edges of a pyramid, formed by the homologous faces, which meet at an apex on
the principal axis, such as V of Fig. 85, polar edges.

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of these lines meet in the points M, M\ <fcc. The lines VM, VM\
<fcc., clearly bisect the angles A*VA\ A 4 VA*, <fcc. But VM, VM 1 , <fcc,
are the orthogonal projections on the paper of the possible edges
in which alternate faces of the pyramid would meet, if produced.
Hence, the plane containing the opposite possible edges VA and
VM also contains OV, for VA and VM are co-linear. Similarly,
the planes containing other opposite pairs VA 1 and VM\ <fec,
also contain OV. If then axes of odd degree, higher than three,
were possible, they would be possible zone-axes.

The above proofs are not applicable to the case of a triad axis.

5. Prop. 3. To prove that a plane perpendicular to an axis of
symmetry is a possible face.

Let, in Fig. 87, 0A be a dyad axis, and let a face EFG of the
crystal, not parallel or perpendicular to the axis, meet it at A.
There must be a second face
JSFH, which is interchangeable
with EFG when the crystal is
turned about 0A through 180°.
Let EF be the line of intersection
of the two planes. When the
crystal has been turned through
180° about 0A, the edge EF re- Pia 8?

tains its direction, for the two

faces have changed places : the only difference is that points E and
F equally distant from A have changed places. Hence the edge
EF is perpendicular to 0A.

Similarly, any other face whatever, not parallel or perpendicular
to 0A, and not in a zone with EFG and EFH, must be associated
with a second like face, so that the two faces meet in an edge
perpendicular to 0A.

Hence 0A is perpendicular to two edges which are not parallel.
It is, therefore, perpendicular to the possible face which is parallel
to these edges.

Cor. 1. In a manner similar to that employed in Cors. 1 and 2
of Art. 4, it can be shown that tetrad and hexad axes are perpen-
dicular to possible faces. For two successive rotations in the same
direction, each of 90°, about a tetrad axis interchange a pair of
faces intersecting in a line perpendicular to the axis; and similarly,
three successive rotations, each of 60°, about a hexad axis inter-
change a similar pair of faces which intersect in an edge

L. c. 8

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perpendicular to the axis. Hence a number of non-parallel edges
can be perpendicular to tetrad and hexad axes. These axes must
therefore be perpendicular to possible faces.

Cor. 2. On the supposition of an axis of symmetry of odd
degree, w>3, it has been seen in Prop. 2, Cor. 3, that VM, VM 1 , &c.,
would be possible edges. Then VMM 1 , <fec., would be possible faces
belonging to a second pyramid. But, from Fig. 86, it is clear that
the pair of opposite possible faces VAA\ VMM 1 intersect one
another in a straight line parallel to AA 1 and MM 1 and perpen-
dicular to OF. The same is true of all the opposite pairs of
possible faces. Hence, OV is perpendicular to a number of
possible edges, and is therefore perpendicular to a possible face.

6. It is not possible to establish, as a consequence of the law of
rational indices, that a triad axis is, when alone, a possible zone-
axis, or that it is perpendicular to a possible face.

Let, in Fig. 88, VO be a, triad axis, and VM a zone-axis of the
crystal which is not parallel or
perpendicular to OV. Let MO ^Jr^

be drawn perpendicular to the s^f ^v

triad axis and meet it at 0. S^ / : fi\^

Rotation of the crystal about Jfj<^-^-/— -^
VO through 120° must bring ^sZszr-^u^ ~~*~"

VM to VM, and 0M to 0M t , M

Fig 88

so that the triangle V0M t is

not coplanar with VOM. A further rotation of 60° is required to
bring 0M t into the continuation Ofi of OM. But this is not a
possible angle of rotation about a triad axis, nor is there an edge V\l
associated with VM. A second rotation of 120° in the same
direction as the first brings the triangle into the position VOM so
that 0M tl lies as far beyond 0/jl as 0M t was short of it. A third
rotation of 120° clearly brings the line back to VM.

The proof of Prop. 2 depends, in the case of an axis of sym-
metry of even degree, on the fact that two homologous zone-axes
are coplanar with the axis; and in the case of an axis of odd
degree, n > 3, the proof depends on the fact that two non-homo-
logous edges, such as VA and VM, are coplanar with the axis.
Here the homologous zone-axes form the edges of a trigonal pyramid,
and no pair of them lies in a plane containing the triad axis. Nor
can an auxiliary pyramid be formed by the edges of alternate faces.

Again, the proof that a possible face is perpendicular to an axis

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of symmetry depends on the fact that pairs of possible faces intersect
in lines perpendicular to the axis. The homologous faces, such as
MVM /y M § VM„ in Fig. 88, intersect in VM t which is not at 90° to
the triad axis, for 0M t is at 90° to VO. Hence, we cannot prove
that the triad axis is perpendicular to a possible face.

But, although it is not possible to prove that a triad axis, when it
is alone or associated only with a centre of symmetry, is a zone-axis
perpendicular to a possible face, we can prove that it is so when it
is associated with a plane, or with axes, of symmetry. It is also
found to be an actual zone-axis, and perpendicular to faces, in all
the known crystals in which triad axes exist. Hence, we shall
assume that, where a triad axis exists, it is a possible zone-axis and
perpendicular to a possible face.

Hence, generally, an axis of symmetry is (1) parallel to a possible
zone-axis, and (2) perpendicular to a possible face.

7. Prop. 4. If a crystal has any two of the three elements
of symmetry — a plane, a centre, and an axis of even degree perpen-
dicular to the plane — then it must also have the third element.

(a) Let, in Pig. 89, ABCD be a plane of symmetry, 2; and
let 0, a point in 2, be the centre of symmetry. Let OP be the
normal on any face ABE y and 0P X the
normal on the corresponding homologous
face ABF on the opposite side of the
plane 2; and let 0A be the normal on
the plane ABCD.

The plane containing OP and 0P 1 is
perpendicular to 2, and must contain OA.
Now, since the crystal has a centre of sym-
metry there must be faces at both ends of
a normal continued through the centre.

Hence, 0P % and 0P t9 the continuations of OP and 0P 19 are
perpendicular to faces FCD and ECD which are parallel to ABE
and ABF respectively (Euclid xi. 14). The four faces form a four-
sided prism, like the roof of a long building and its reflection in a
horizontal plane passing through the eaves.

It is clear that rotation through 180° about the normal 0A
interchanges OP and 0P 9 , 0P t and 0P %9 and the corresponding
faces. For 0A is the external bisector of the angle P0P 19 and,
consequently, bisects the angles P0P s and P^P^

The same relations must hold for each set of four faces


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symmetrical with respect to ABCD and the centre 0. They
must be interchangeable in pairs about the normal 0A on ABCD.
Hence rotation through 180° about the line 0A interchanges
homologous faces. But 0A, being the normal on a plane of symmetry,
is a zone-axis perpendicular to a possible face (Prop. 1). Hence,
0A is an axis of symmetry of at least twofold rotation. The
possibility of rotations through 60°, or 90°, is not excluded. For
three rotations in the same direction, each of 60°, about 0A bring
OP to the position 0P 9 ; and similarly, two successive rotations,
each of 90°, are equivalent to a single rotation of 180° about the
same axis. The axis may, therefore, be one of dyad, tetrad, or
hexad symmetry; i.e. it is an axis of symmetry of even degree.

(b) When a crystal has a centre of symmetry 0, and an axis
of symmetry of even degree 0A, it can, by the aid of the same
diagram, be shown that the plane ABCD, perpendicular to 0A, is
a plane of symmetry. For rotations through 180° about 0A are
always possible and interchange pairs of normals OP and OP s ,
where all three lines are co-planar and 0A bisects the angle P0P s .
Owing to the centre of symmetry each normal is diplohedral, and
the faces at opposite ends of a normal are necessarily parallel. But
parallel planes intersect a third in parallel lines (Euclid xi. 16).
Hence the edges AB and CD are each parallel to the edge EA, in
which the faces ABU and DCE meet. The edges are therefore all
perpendicular to the plane containing OA, OP, 0P 3 , and therefore
to 0A. The plane, containing the edge AB and the line through
in the plane P0P % which is perpendicular to 0A, bisects the angle
P0P lt and therefore the angle between the faces of which OP and
0P X are the normals.

A similar relation can be found for every other set of two
pairs of parallel faces, which are also interchangeable, in pairs, by
rotation through 180° about 0A. Hence, the plane ABCD is a
plane of symmetry; for it bisects the angles between homologous
pairs of faces, and is parallel to a possible face and perpendicular to
a zone-axis.

(c) If a plane of symmetry, ABCD, and an axis of symmetry of
even degree, 0A, perpendicular to ABCD coexist* in a crystal, then
parallel faces must be invariably present, i.e. the crystal is centro-
8ymmetrical. The proof is obtained from the diagram, already
used, by taking the relations of the lines and faces in a different
order. OP and 0P X are symmetrical with respect to ABCD which

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bisects the angle between them and is perpendicular to their plane.
Hence, the plane P x 0P contains 0A. Again, 0A bisects the angle
POP z , and 0P S lies in the plane containing OP and 0A. Hence,
0P X , OP, 0P S all lie in one plane, and POP x + POP % = 180°.
Therefore 0P X and 0P S are opposite directions on the same straight
line. The faces ABF and DCE at the opposite ends P l and P %
must therefore be parallel; and similarly, ABE and DGF are
parallel faces. The same is true of every other set of faces which
are homologous with respect to 2 and to 0A. The crystal is
therefore centro-symmetrical.

Hence, the presence of any two of the three elements of
symmetry — a centre, a plane and an axis of even degree perpen-
dicular to the plane of symmetry — always involves the presence
of the third element.

8. Cor. The proof given above also establishes a very im-
portant relation of the triad axis ; viz. that in a centro-symmetrical
crystal having an axis of triad symmetry, the plane perpendicular to
the axis cannot be a plane of symmetry. For the axis perpendicular
to a plane of symmetry in a centro-symmetrical crystal must be an
axis of symmetry of even degree. There is nothing to exclude the
possibility of a triad axis and a plane of symmetry perpendicular to
it occurring together in a crystal; but in such a case the crystal
cannot be centro-symmetrical, and parallel faces, if they occur, are
to be regarded as belonging to different forms and will often be
distinguished by a difference of character; i.e. such parallel faces
are not homologous.

9. We shall now consider the limitations to the number
of planes of symmetry possible in a zone. We shall, thereby, see
that the arrangements possible are such that the least angles
between planes of symmetry are 90°, 60°, 45° and 30°.

Prop. 5. To prove that the only angles, less than 60°, possible
between planes of symmetry are 45° and 30°.

Let, in Fig. 90, P and Q be the poles
of planes of symmetry such that no pair of
planes of symmetry in their zone make with
one another a less angle than PQ = <j> (say).
Now, P must be repeated on the other side
of Q in a similar pole P x where P X Q = QP ;
and similarly, Q must be repeated in a similar
pole Q 2 . Suppose the planes of symmetry

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to be perpendicular to the paper, so that the poles P, Q f P u
Ac, all lie in the primitive, where PQ = QP X = P x Qi = Ac. = <£.
But, since P and Q are the poles of planes of symmetry, so must
also the poles P u Q i9 <fec., be poles of planes of symmetry.
Let the successive poles be inserted in the zone until R 9 the last
pole before P, is reached, where P is the opposite pole to P.
Now the next step gives either P, or a pole S beyond it But
P is also the pole of a plane of symmetry — that first taken — and
RP, SP f are both less than RS = <£. This contravenes the limi-
tation on the poles P and Q first taken; viz. that no planes
of symmetry in the zone make with one another a less angle than <f>.
Hence, in proceeding beyond R the next pole must be P, and RP = <f>.
The angle </> is therefore an exact submultiple of 180°. Call the
number w, then w^ = 180°.

If, now, n is greater than 3, there must be at least three other
poles between P and P, and we can apply the anharmonic ratio
of four tautozonal poles to investigate the possible values of n and </>.
But, if <f> is 60° or 90°, the anharmonic ratio is inapplicable, since
pairs of poles are at 180° to one another. Supposing </><60°, we

sinPP! sin^Px . . ... . .

— — wA -r . -7T7T = w, where m is some positive rational number,
sin PQ Bin Q&

sin PPi sin Q^Q sin 8 2<£ 4 sin 2 <f> cos 9 <f>

Then, m = — — nrk . — -^—^ = . g , = . „—

sin PQ sin Q % P X sin 2 </> sin 2 <£

= 4 cos 2 <£ = 2 (1 + cos 2<£) (1).

Let2<£ = 0, .'. n0 = 36O°; andco80 = cos2<£ = (m-r2) - l...(2).

The problem involves, therefore, the determination of the angles
less than 120°, which have rational cosines and are exact sub-
multiples of 360°. These angles are 90° and 60°, as is proved in
Art. 10. In these cases, cos 6 = cos 90° = 0, and cos = cos 60° = 1+2.
Henoe, when = 90°, <£ = 45° and n-4; and when = 60°, <£ = 30°
and n = 6. The value of the anharmonic ratio is, in the former case,
m=2; and in the latter, m = 3.

10. Proof. Series (3) (Todhunter's Trig,, p. 226, 1859), and
the properties of equations having integral coefficients, enable us to
prove that the only angles <120°, which are exact submultiples of
360° and have rational cosines, are 90° and 60°.

2 cos nO = (2 cos 6)* - y (2 cos 0)*-* + —^ (2 cos 0)»-«-<fcc. ... (3).

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The rth term is

(- iy n(n-r-l)( W -r-2)...(n-2r + l) (2 ^ ^

the coefficient of (2 cos 0) n ~ ar being integral

But, when n0 = 360°, cos n$ = 1 ; and writing x for 2 cos 0, we
have the equation

a ^_ na ^-» + n ^"" 3 ) g n-^ < fec. = 2 (4).

Further, the commensurable roots of an equation of the nth
degree, which has integral coefficients and unity for that of the
highest term, are integral and exactly divide the constant term.
It follows that the possible values of x are limited to + 2, + 1, and 0.

If 2 cos = ±2, is 0° or 180°. The former value is absurd,
and the latter gives <£ = 90°. This is a possible angle between
planes of symmetry, but it is not admissible as a solution of
equation (1), for it gives a meaningless anharmonic ratio.

If 2cos0 = ±l, is 60° or 120°, and <f> is 30° or 60°. The
former establishes that the least angle possible between planes of
symmetry is 30°, and that we cannot have more than six such
planes in a zone. The latter value (60°) is a possible angle between
planes of symmetry, but it is not admissible as a solution of (1) ;
for, if four successive poles are taken at 60° to one another, the
extreme pair are at 180°, and the anharmonic ratio is meaningless.

If 2cos0 = O, = 90°, <£ = 45° andw = 4. This is the only
other solution possible, and is that given when n is even and the
last term of series (3) is 2.

It follows that the greatest number of planes of symmetry in a
zone is six in one case and four in the other 1 . Crystals may, also,
have only two, or three, tautozonal planes of symmetry.

1 The proof of this important proposition given in the text is due to Professor
Story- Maskelyne, who gave it in a course of lectures attended by the author in
1869. Professor von Lang (KrysU p. 75, 1866) gives an expression, equivalent
to (1), limiting the angles of Uogonal zones, i.e. of zones in which equal angles
recur. He points out that 30°, 45°, 60° and 90°, satisfy the equation ; but he
does not prove them to be the only angles which do so. In a classical memoir,
published in Acta Soc. Sci. Fennica, rx. p. 1, 1871, but read on 19 March, 1867,
Professor Axel Gadolin finds that the angles of rotation about axes of symmetry
are subject to equation (2) ; and he proves, by the method used in Art. 10, that
60° and 90° are the least possible angles. His method of arriving at equation (2)
is, however, not altogether satisfactory; for it is not applicable to the case of a
tetrad axis. A different method has, therefore, been adopted in Art 11.

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Cor. 1. No crystal can have five tautozonal planes of symmetry ;
for the least angle between them is 36°, and this angle does not
satisfy equation (2).

Cor. 2. Planes of symmetry inclined to one another at angles
of 45°, and also at 30°, cannot coexist in one and the same zone.
It is clear that a zone-circle cannot be divided by poles at distances
of 30° to one another and also at distances of 45° without some of
the poles making angles of less than 30° with one another.

11. Prop. 6. To prove that tetrad and hexad axes are the
only axes of symmetry of degree higher than three which are
possible in crystals.

Suppose A n to be an axis of symmetry of degree n (> 3) occur-
ring alone in a crystal, and let it be placed vertically. By Props.
2 and 3, A n is a possible zone-axis perpendicular to a possible face.
If then a face meets the axis at a point F, we have a pyramid of n
similar faces all passing through the apex V and meeting the hori-
zontal base in possible edges.

We proceed to show that n possible faces can be found parallel
to A n and inclined to one another in succession at an angle

When n is even, n-r-2 faces through opposite polar edges are
possible, each of which contains the axis of symmetry. Thus, in

Fig. 92.

Fig. 91, the planes M t LM\ MLO, both containing the tetrad axis
OL y are possible faces inclined to one another at an angle of 90°.
Similarly, in Fig. 92, the planes AJA\ A V0 f A„V0, containing the
hexad axis of symmetry 0V 9 are possible faces inclined to one
another at angles of 60°. Generally, therefore, we have n-s-2
possible tautozonal faces inclined to one another in successive pairs
at an angle $ = 360° -r w.

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If now the alternate faces of the pyramid are extended to meet
one another, we obtain in Fig. 91 horizontal edges through Z,
parallel to OX and 0Y 9 which bisect the angles M t OM, MOM'.
Through these edges and the axis OL possible faces can be drawn ;
and we have four faces through the axis OL inclined to one another
in succession at angles of 45°. In the general case, when n is greater
than 4, we obtain new polar edges, such as VM t of Fig. 92, through
which and the axis n~2 possible faces can be drawn bisecting the
angles between adjacent faces of the first set. Thus, for instance,
the face OVM t bisects the angle
between OVA and OVA n . We
thus have, altogether, n tautozonal
faces through A n inclined to one
another in succession at an angle

The same can be shown to be
the case, when n is odd and >3.
Thus, in Fig. 93, the face through
the axis and the possible edge VM
bisects the angle between the pos-
sible faces- through VA* and VA*
parallel to the axis.

Fig. 90 may now be taken to represent the poles of the n tauto-
zonal faces through the axis A ni inclined to one another in succession
at angles <£, where n<£=180°. Forming the anharmonic ratio,
a.r. {PQPxQz}', of four consecutive poles, we obtain equations (1)
and (2). Hence, as proved in Art. 10, the possible values of
0=2$ are restricted to 60° and 90°; and the axis can only be a
hexad or a tetrad axis. Similarly, 72° not being a possible value of
6, a pentad axis is inadmissible.

The only axes of symmetry possible in crystals are therefore
those of two-, three-, four-, and six-, fold symmetry.

12. Isogonal zones. Zones in which one of the crystallometric
angles, 60°, 45° or 30°, recur are common in cubic crystals and in
crystals having a principal axis. We can, however, show that, even
when the faces are not planes of symmetry, no pair of faces can
include an angle of 30°, or 60°, in a zone in which 45° recurs in
succession ; and, vice versa, that 45° cannot occur in a zone in which
either 30°, or 60°, recurs in succession.

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Thus, in Fig. 94, let PQ-QP=P t Q=&c=<t>, where <f> is either 46° or
30°. Take any possible pole R in the zone, and let /\PR=<*. Also,
let S be a point such that RS =#,.

Then the A.R. {PRQP} and the A.R. {PSQP} give the equations

sin PA sinP,^ sin a , .. , , x /rN

- — u?^ - — 5^= . , a . \— m ( a rational number) (5),

amPQ BUiP f Q sin(2<£-a) x / \ />

sin PS . sin P,S sin(a+<k) , . /a .

^pQ-^prQ^^^^^r n{a&J) ()>

(a) If <f> is 46°, then (5) gives

8m « i. /*x

8in(90°-a) =tana = W (7)J

and (6) gives

, / . » n tana+tan A

tan (>+<£,)=»=- — i T—-1

™ 1 - tan a tan 0,

=(by substitution from (7)) m+tan< ^ ,..( 8 ).
x * x " l-mtan<£, v '

If 0, is 45°, then tan^,=l, and n is rationaL
Hence, in a zone having 46° recurring, we can have Fl °* 9*.

a pole 8 at 45° to any other known pole R

If, however, <£,=30°, then tan <£,= 1-^-^3 ; and from (8)

n= V3-m 3-m* 3-m* * '*

Online LibraryW. J. (William James) LewisA treatise on crystallography → online text (page 11 of 53)