cristallography - hexagonal crystal system i

8/2/2019 Cristallography - Hexagonal Crystal System I

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The Hexagonal Crystal System (PartOne)

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Introduction

For a general introduction to Crystal Systems and Crystal Classes, see the Essay on The Morphologyof Crystals, which the reader should first consult, if he or she is not familiar with Crystal Symmetry andits description.The reader should also consult the previous Essays on The Tetragonal Crystal System in order to fullyunderstand the issues in the present Essay and the next Essays on the Hexagonal System.

The Hexagonal Crystal System is based on four crystallographic axes (although of course any face couldbe described already by three axes, using four axes is more convenient).The system of crystallographic axes of the Hexagonal Crystal System consists of three equivalent (i.e.

interchangeable) horizontal (equatorial) axes of which the positive ends make an angle of 1200. They aredenoted by a

1, a

2and a

3. They all lie in the same (horizontal) plane. When we erect a line perpendicular

to this plane and passing through the intersection point of the three horizontal axes, we have the fourthcrystallographic axis, the c axis. The upper end of the c axis is provided with a + sign, the lower endwith a - sign. Also the other axes are provided with such signs as indicated in the next figure, thatdepicts the configuration and orientation of the three horizontal crystallographic axes.

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Figure 1. The three equatorial (horizontal) equivalent crystallographic axes (red lines) of the HexagonalCrystal System. These axes are sometimes denoted as a, b and d axes.

The next Figure gives an oblique view of the system of crystallographic axes with the c axis vertical.




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Figure 2. The Hexagonal axial system, seen with the c axis vertical.

Indexing of faces

The method of indexing of faces of crystals belonging to the Hexagonal System is not exactly the same asin the case of crystals of the other systems. For the Hexagonal System the method of Bravais-Miller isadopted. In this method the indexing of a face involves four indexes, h, k, i and l.

Because in the Hexagonal System four axis are used the Weiss symbol also contains four symbols,referring to the a1, a2, a3 and c axis respectively. In determining the Bravais-Miller indices we take the

reciprocals of the Weissian derivation coefficients and convert them to whole numbers, and place theresulting set of indices, characterizing a certain crystal face, between round brackets ( ). When we want toindicate a Form (generally consisting of a set of faces) by one of its faces we put the set of Bravais-Millerindices between braces { }.The next figures show how to index faces of the Hexagonal Crystal System (In fact we show it withrespect to the three horizontal crystallographic axes).

Remark : In the Figure we see that the a3

axis is cut at its negative end. To indicate this, one places a

minus sign (-) above the relevant index. Because of typographical reasons we, here on this website, placean asterisk (*) after the relevant index instead.

Figure 3. The indexing of the face A ,characterized by the Weiss symbol

a : ~a : -a : c ( The sign " ~ " stands forinfinity) (and the reciprocal of infinity is0).It is a face of a pyramid. The derivationcoefficients are respectively1 infinity 1(-) 1. The reciprocals ofthe derivation coefficients arerespectively

1/1 1/~ 1/1 (-) 1/1, and these areequal to 1 0 1(-) 1 respectively. So the

Bravais-Miller symbol for this face is( 101*1 ) (For the sign * see the remarkbelow).

Figure 4. The indexing of the faceB , characterized by the Weisssymbol

a : a : -1/2a : c .It is a face of another pyramid. Thederivation coefficients arerespectively1 1 1/2(-) 1. The reciprocals ofthe derivation coefficients arerespectively1/1 1/1 2/1 (-) 1/1, and these areequal to 1 1 2(-) 1 respectively.So the Bravais-Miller symbol for




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Hermann-Maugin notation of symmetry content of hexagonal Classes.

In the Hermann-Maugin notation the first number refers to the principal axis of symmetry coincident withc. The second and third symbols, if present, refer respectively to symmetry elements parallel with andperpendicular to the crystallographic axes a

1, a

2and a

3. Equivalently expressed, the second symbol

refers to symmetry elements each of them parallel with a horizontal crystallographic axis, while the thirdsymbol refers to the bisectors between these axes.

An Oveview of the twelve Classes of the Hexagonal CrystalSystem

The Hexagonal System

this face is ( 112*1 ). (For the sign* see the remark above).

Division Classmirrorplanes

axes cs

HolohedricDihexagonal-bipyramidal

6/m 2/m 2/m1 + (3 + 3)

one [6]three + three [2] +

Hemimorphy of

Holohedric

Dihexagonal-pyramidal

6 m m

3 + 3 one [6] p. -

PyramidalHemihedric

Hexagonal-bipyramidal6/m

1 one [6] +

Hemimorphy ofPyramidal Hemihedric

Hexagonal-pyramidal6 - one [6] p. -

TrapezohedricHemihedric

Hexagonal-trapezohedric6 2 2 -

one [6]three + three [2] -

RhombohedricHemihedric

Ditrigonal-scalenohedric3* 2/m

3one [3]three [2] +




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The Dihexagonal-bipyramidal Class(= Holohedric Division) 6/m 2/m 2/m

The symmetry elements of this Class are :

One main mirror plane, containing the horizontal crystallographic axes.

Three vertical mirror planes, each one containing the main axis and a horizontal axis.

Three vertical mirror planes bisecting the angles of the afore mentioned vertical mirror planes. (So

we have 6 vertical mirror planes in all, making angles of 300

with each other).

One 6-fold rotation axis, coinciding with the vertical crystallographic axis.

Three 2-fold rotation axes, coinciding with the horizontal crystallographic axes.

Three 2-fold rotation axes, lying exactly (in the middle) between the horizontal crystallographicaxes. (So there are six 2-fold rotation axes, lying in the equatorial plane, making angles with each

other of 300).

Center of symmetry.

The stereographic projection of these symmetry elements is depicted in Figure 5.

RhombohedricTetartohedric

Trigonal-rhombohedric3* - one [3] +

TrigonalHemihedric

Ditrigonal-bipyramidal6* m 2

1 + 3one [3]three [2] p. -

Hemimorphy ofTrigonal Hemihedric

Ditrigonal-pyramidal3 m

3 one [3] p. -

TrigonalTetartohedric

Trigonal-bipyramidal6*

1 one [3] -

Hemimorphy ofTrigonal Tetartohedric

(Ogdohedric)

Trigonal-pyramidal3 - one [3] p. -

TrapezohedricTetartohedric

Trigonal-trapezohedric3 2 -

one [3]three [2] p. -




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Figure 5. Stereogram of the symmetry elements of the Dihexagonal-bipyramidal Class, and all the faces(face poles) of the general Form.

The basic Form of the Hexagonal System is the (primary) protopyramid.It is itself based on some existing crystal species. So by definition all the intercepts (i.e. pieces cut-offfrom the crystallographic axes by the faces of the pyramid) are of unit length.

Its basic face, the upper front face, has as its Weiss symbol a : ~a : -a : c (where " ~ " means infinity).From this face we can in turn derive the hexagonal protopyramid by applying the symmetry operationsof the present Class to it, yielding a Form that possesses all the symmetries of the Class. Like in theTetragonal System we will use the stereographic projection of the symmetry elements and face poles toderive all the hexagonal holohedric Forms, of which the most basic one is the mentioned hexagonalprotopyramid (= type I pyramid).

The position of the face a : ~a : -a : c in the stereographic projection of the symmetry elements of theClass is shown in (1) of Figure 6.

When we apply the symmetry elements of our Class to this face, then the following happens : The 6-fold rotation axis generates five new faces, so we now have six faces. Their configuration complies with6-fold rotational symmetry. It is a six-sided monopyramid. Next we reflect this result with respect to theequatorial mirror plane resulting in a bipyramid. The other symmetry elements do not have any furthereffect, because the produced bipyramid already possess these other symmetry elements (they areimplied). So our result is this bipyramid, the hexagonal protopyramid (= type I pyramid).Its stereographic projection is given in (2) of Figure 6.




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Figure 6. (1). Stereogram of the symmetry elements of the Dihexagonal-bipyramidal Crystal Class and of

the facea : ~a : -a : c.(2). Stereogram of the (primary) hexagonal Protopyramid, the first Form of the Class.

The hexagonal protopyramid itself is shown in Figure 7.

The Weiss symbol for this primary protopyramid is (a : ~a : -a : c), and the face a : ~a : -a : c is,besides being the unit face, also one of the basic faces of the Hexagonal Crystal System. When such abasic face is subjected to all the symmetry elements of a certain Crystal Class then one of the Forms ofthat Class is generated.The Miller symbol for the protopyramid can be found as follows.The derivation coefficients can be read off from the Weiss symbol. They are consequently

1 ~ (= infinity) 1(-) 1. Their reciprocals are 1/1 1/~ 1/1(-) 1/1. And they are equal to1 0 1(-) 1. So the Miller symbol for this pyramid is { 101*1 }.The Naumann symbol is P.

The next Form is the (primary) hexagonal deuteropyramid (= type II pyramid).Although it is, like the protopyramid, a hexagonal bipyramid, it is not just a rotated protopyramid, but hasa slightly different shape, it is a little bit more obtuse. I will explain why this is the case. The next Figureshows this deuteropyramid.

Figure 7.HexagonalProtopyramid, and itsorientation with respect to theour crystallographic axes.

From these axes unit pieces arecut off by the upper front face.

Figure 8.HexagonalDeuteropyramid, and itsorientation with respect to theour crystallographic axes (red




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The Weiss symbol for this (primary) deuteropyramid is (2a : 2a : -a : c). So the derivation coefficientsare respectively 2 2 1(-) 1. Their reciprocals are 1/2 1/2 1/1(-) 1/1. To convert these to wholenumbers we multiply each one of them by two resulting in 1 1 2(-) 2. So the Miller symbol for thisForm is { 112*2 }.The Naumann symbol is P2 (i.e. the c-intercept has unit length, expressed by the absence of any number

before P, while the non-unit intercepts with respect to the horizontal axes are of length 2, expressed bythe 2 after the P).

To derive the deuteropyramid from the protopyramid a rotation of the latter (by 300) is not sufficient. It iseven impossible in the world of crystals, because if we just rotate the protopyramid relative to the systemof crystallographic axes, namely rotating it about the c axis, then the derivation coefficient belonging tothe a3 axis becomes irrational (namely half of the square root of three [1/2 (square root)3]). Let us

explain this by considering Figure 7 and 8.In Figure 7, the hexagonal primary protopyramid, OA = OB = 1 (the unit distance). If we rotate this

pyramid by 300 clockwise about the vertical axis and if we imagine that we then get the pyramid ofFigure 8, then OC (in Figure 8) would be equal to OB (in Figure 7), i.e. it would be 1. But then OD

would be 1/2 (square root)3 (= 0.866...). See Figure 9.

lines). From the a3

axisa unit

piece is cut off by the upper frontace (the unit piece refers, with

respect to its length, to theProtopyramid.

Figure 9.Horizontalcrystallographic axes and (thetraces of) three faces (black)(one face and parts of twoadjacent faces) of theDeuteropyramid.




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In Figure 9 the angle COL is 600.

OL = OC.So OCL is an equilateral triangle. This implies that CD = 1/2 CL = 1/2 OC.

So if OC = 1 (= the unit distance OB in Figure 7), then (1/2)2 + (OD)2 = 12, and this is equivalent to 1/4 +

(OD)2 = 1.

So (OD)2 = 3/4.

Thus OD = (square root)3 / (square root)4, which is equal to 1/2(square root)3 (= 0.866...).

Remark :From Figure 9 we can also see that KOD is half an equilateral triangle,which implies that OK = 2OD. In the same way OM = 2OD holds.

So if OD is set equal to one (i.e. equal to OB of Figure 7), in other words if OD is

the unit distance (unit intercept), then OK = 2, and OM = 2, establishing the Weisssymbol for the face CL and giving (2a : 2a : -a : c) for the deuteropyramid.

This means (if OC = 1) that the coefficient for the a3

axis has become irrational. But it is proven that the

derivation coefficients can never become irrational (this impossibility is a direct consequence of theperiodic internal structure of crystals), and so we cannot derive the deuteropyramid from the

protopyramid by just rotating the latter (300). We must enlarge the equatorial plane such that OD (infigure 8) will become equal to 1 again, and thus equal to OB (in Figure 7). So we must enlarge theequatorial plane by a linear factor of 1.1547... (because 1.1547... times 0.866... equals 1). Thus, if theintercept with respect to the c axis remains the same, then this pyramid is not congruent with theprotopyramid but is a little more obtuse. When we would proportionally enlarge the c-intercept, the two

pyramids still would not be congruent but they are then similar pyramids, they have the same shape, butdiffer in size. However in this case the deuteropyramid (with enlarged c-intercept) is not a primarydeuteropyramid anymore, because of the enlargement mentioned (the distance OF in Figure 7 is the unitintercept with respect to the c axis, and the corresponding intercept in the similar deuteropyramid (wherethe intercept was enlarged) is longer, i.e. longer than the unit intercept for this axis).Let's recapitulate and draw some conclusions.A deuteropyramid is not similar to (and thus certainly not congruent with) the primary protopyramid (P)when this deuteropyramid is itself primary (P2), in which "similar" means "having exactly the sameshape". But it can be made similar to the primary protopyramid by an appropriate change of length of thec-intercept, causing the deuteropyramid not to be primary anymore. And even when this deuteropyramid




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is now (made) similar, it still is not congruent with the primary protopyramid. So if a deuteropyramidcombines with a primary protopyramid its faces must have a definite relationship to those of theprotopyramid : they cannot just be anywhere in the combination, but can only occur at certain definitedistances from the faces of the protopyramid with which they combine, because the deuteropyramid isnot congruent with the protopyramid. To fully understand the relationships between the hexagonal proto-and deuteropyramids, and also the relationship between them and the dihexagonal pyramid (a Form yet tobe discussed), it is useful to consider the equatorial planes of those Forms (All such considerations are

equally valid for all the yet to be discussed prismatic Forms, the protoprism, deuteroprism and thedihexagonal prism). See Figure 10, 11 and 12.

Figure 10.Equatorial planes of the hexagonal Protopyramid and Deuteropyramid.When we want to derive the Deuteropyramid from the Protopyramid, we rotate the Protopyramid (red)

300 about the main crystallogaphic axis (the resulting equatorial plane is given in a greenish color) andenlarge that resulted equatorial plane tillthe a1-intercept is equal to 1 again.

In the next Figure I show the relationship of the dihexagon with the hexagons (equatorial planes) of theproto- and deuteropyramids. As the hexagons are the equatorial planes of the proto- and deuteropyramids

as well as of the proto- and deuteroprisms (Forms yet to be discussed), the dihexagon is the equatorialplane of the dihexagonal pyramid as well as of the dihexagonal prism (Forms yet to be discussed).

Figure 11. When we construct adihexagon in the way indicated(black) we get a specialdihexagon, namely a regulardodecagon, i.e. a completelyregular polygon consisting oftwelve equal sides and twelve




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equal angles (corners). This isevident from the fact that thegreen hexagon is just the result ofthe rotation of the red hexagon

( by 300 ), implying congruenttriangles from which the absoluteregularity is easily proved.In the next Figure we willconstruct a general dihexagon.

Figure 12. When we construct adihexagon in the way indicated(black) we get a general dihexagon.It is a kind of irregular dodecagon,i.e. a polygon consisting of twelveequal sides and twelve angles(corners) which are alternately thesame. I.e. it has six angles of thesame size, and six others, differentrom the first group but equal

among each other. The dihexagoncan vary between two extremes, theProtopyramid and theDeuteropyramid, as is evident fromthis and from the next Figure.




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Figure 13.Range of possible dihexagons (the general ones, but also the special (regular) case). Thepossible intercepts related to a representative side of the dihexagon (corresponding to a representative

ace of the associated Dihexagonal Pyramid or Prism) can vary between those of the Protopyramid andthose of the Deuteropyramid.

In the next Figure we show the axial relationships of the general dihexagon (as the equatorial plane of thedihexagonal pyramid and prism, and as such representing those Forms).




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Figure 14.Axial relationships of the sides -- representing faces -- of the general Dihexagon with thehorizontal crystallographic axes of the Hexagonal Crystal System.

In Figure 14 the intercept with respect to the a1

axis is OC. The intercept with respect to the a2

axis is

OE. The intercept with respect to the a3

axis is OB.

OB is equal to 1 (unit intercept).If OE = s, then the length of OC is determined, it in fact is s / (s-1), which we will prove below SeeFigure 15.




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Figure 15.If the intercept with respect to one axis is set to 1, and if the intercept with respect to a secondaxis is set to, say, s, then the intercept with respect to the third axis is determined. It is s / (s-1). (Seebelow).

The axes a1, a

2and a

3are effecting six angles of 600 around O. (See Figure 15). So the angle COB is

600.

AB is a side of a dihexagon. It represents a face of a dihexagonal pyramid or prism.The intercept for AB with respect to the a2 axis is OE. Let its length be s.

The intercept for AB with respect to the a1 axis is OC.The intercept for AB with respect to the a

3axis is OB, which is (set) equal to 1 (unit distance).

Now we draw ED parallel to OC. We then get two similar triangles COB and BED.

The angle DOE = 600. Because the angle enclosed between -a2

and +a1

is equal to 600, the angle OED is

also 600 (recall that ED is parallel to OC).

It follows that now the angle EDO is also 600, and thus the triangle OED is an equilateral triangle. Thismeans that ED = OE = s, and OD is also equal to s.Because OB = 1, BD = s-1.Because the triangles COB and BED are similar, the following proportions hold :OB : BD = 1 : s-1 = OC : ED.

Because ED = s, we get :1 : s-1 = OC : s, which can be written as1 / (s-1) = OC / s, which is equivalent to OC = s / (s-1).

So if the intercept with respect to the a2 axis is equal to s, and the intercept with respect to the a3 axis is

equal to 1, then the intercept with respect to the a1

axis (= OC) is equal to s / (s-1).

So the general Weiss symbol for a dihexagonal pyramid is( [s/(s-1)]a : sa : -a : mc ) , with the derivation coefficient for the first a equal to s / (s-1).




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The derivation coefficient for the a1

axis must lie between 1 and 2.

When it would be 1 we get the protopyramid (P).When it would be 2 we get the deuteropyramid (P2).So when we take (for example) s = 3 and m = 1 (the latter means that the intercept with respect to the caxis has unit length), then we get ( [3/2]a : 3a : -a : c ) (and the coefficient for the first a is indeedbetween 1 and 2), and we could consider the face [3/2]a : 3a : -a : c as one of the basic faces for theHexagonal Crystal System.So we now have already found three such basic faces :

a : ~a : -a : c2a : 2a : -a : c[3/2]a : 3a : -a : c

We will find four more of these, and we will later use them (in our "Facial Approach") to derive theForms of the lower symmetry Classes of the present Crystal System.

The equatorial plane of the dihexagonal pyramid is already given earlier. In the next Figure the pyramid

itself is shown.

The Weiss symbol representing a general dihexagonal pyramid is ( [s/(s-1)]a : sa : -a : mc ), and theNaumann symbol is mPn, where n = s / (s-1). If (for example) s = 3 and m = 1, then the Miller symbol is{213*3}.

The next three Forms can be derived from the pyramids (protopyramid, deuteropyramid, dihexagonalpyramid) by making the intercept with respect to the c axis (i.e. the vertical axis) infinitely long, whichmeans that all the faces are parallel to the c axis and thus are vertical. In this way we get respectively the

Figure 16. The DihexagonalPyramid, is a bipyramidconsisting of 24 faces, each ofwhich is an unequilateraltriangle. Upper faces shaded.Also visible is the system ofcrystallographic axes (red).




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protoprism, deuteroprism and dihexagonal prism.

We begin with the dihexagonal prism.Its equatorial plane is identical to that of the dihexagonal pyramid. It is depicted in Figure 17.

The Weiss symbol for a dihexagonal prism is ( [s/(s-1)]a : sa : -a : ~c ), the Naumann symbol is ~Pn. Ifs = 3, then the Miller symbol is {213*0}.With all this we have also found yet another basic face, compatible with the Hexagonal Crystal System,

namely -- generally -- the face [s/s-1)]a : sa : -a : ~c. If we take s = 3, then we can take as the (newlyfound) basic face [3/2]a : 3a : -a : ~c.

The next Form is the hexagonal protoprism. It can be derived from the protopyramid by steepening thesides till they are vertical. It is depicted in Figure 18.

Figure 17. TheDihexagonal Prism, is aprism consisting of 12vertical faces. Anyhorizontal section of it is asame dihexagon. It is anopen Form, which in thiscase means that it is notclosed by top or bottomaces. Such a prism can

however engage in acombination with anotherForm of this Class (yet to

be discussed), namely theBasic Pinacoid, whichconsists of two horizontalparallel faces which couldclose the prism at thebottom as well as at thetop. Also visible is thesystem of crystallographicaxes (red), and theirbisectors.

Figure 18. The hexagonal

Protoprism, is a prismconsisting of 6 verticalaces. Any horizontal

section of it is a samehexagon. Like theDihexagonal Prism it isan open Form. It too cancombine with a BasicPinacoid to close its topand its bottom. Also




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The Weiss symbol representing the protoprism is (a : ~a : -a : ~c). The Naumann symbol is ~P. TheMiller symbol is {101*0}.With it we have yet another basic face compatible with the Hexagonal Crystal System, namely the face

a : ~a : -a : ~c.

The next Form is the deuteroprism. It can be derived from the deuteropyramid by steepening its sides tillthey are vertical. Like the protoprism any horizontal section of it is a regular hexagon. The deuteroprismis depicted in Figure 19.

visible is the system ofcrystallographic axes(red).

Figure 19. The hexagonalDeuteroprism, is a prismconsisting of 6 vertical faces.Any horizontal section of it isa same hexagon. Like theDihexagonal Prism and theProtoprism it is an openForm. It too can combine

with a Basic Pinacoid toclose its top and its bottom.Also visible is the system ofcrystallographic axes (red)and their bissectors.




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The Weiss symbol for the deuteroprism is (2a : 2a : -a : ~c), and its Naumann symbol is ~P2. Its Millersymbol is {112*0}.With all this we have found yet another basic face compatible with the Hexagonal Crystal System,

namely the face 2a : 2a : -a : ~c.

The last Form to be discussed is the basic pinacoid. It consists of two faces, parallel to each other and tothe equatorial plane of the system of crystallographic axes. This Form is, like the prisms, an open Form, it

can only occur in combinations which result in closed Forms, as is the case when a prism is closed at itsbottom and at its top by this pinacoid. The basic pinacoid is depicted in Figure 20.

Figure 20. Thehexagonal BasicPinacoid, is a Formconsisting of 2horizontal faces.The system ofcrystallographicaxes (red) is alsoshown.




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The Weiss symbol for the hexagonal basic pinacoid is (~a : ~a : ~a : c). The Naumann symbol is 0P,and the Miller symbol is {0001}.With all this we've also found the seventh and last basic face compatible with the Hexagonal Crystal

System, namely the face ~a : ~a : ~a : c.

All these Forms can engage in combinations with each other in real crystals.

This concludes our exposition of the Dihexagonal-bipyramidal Crystal Class.

Besides (and correlated with) the Forms of this Class we found the following seven basic facescompatible with the Hexagonal Crystal System :

a : ~a : -a : c2a : 2a : -a : c[3/2]a : 3a : -a : c

[3/2]a : 3a : -a : ~ca : ~a : -a : ~c2a : 2a : -a :

~c

~a : ~a : ~a : c

As has been said, we'll use these basic faces to derive (in one out of two approaches) the Forms of thelower symmetrical Classes of the Hexagonal Crystal System.

To continue, klickHERE for Part Two (Class 6 m m).




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