Crystallography and Mineralogy: Symmetry, Form and Twinning, Projections, Mineral Classification and Optics, Nucleation and Mineral Chemistry
2. Form, Miller indices, zones, spacing and Bragg's law
Steno's law: angles between corresponding faces of a mineral are constant. A face is labelled by Miller indices (hkl): take its intercepts on the axes in units of the cell edges, invert them and clear fractions. Intercepts 2a, 3b and 6c give 1/2, 1/3, 1/6, which times 6 give (321). A face parallel to an axis has intercept ∞ and index 0. (hkl) is one plane, {hkl} a form (all symmetry-equivalent faces: in the cubic system {100} is the cube with 6 faces, {111} the octahedron with 8, {110} the rhombic dodecahedron with 12), and [uvw] a direction. Hexagonal and trigonal crystals use four-index Miller–Bravais (hkil), with i = −(h + k).
Zone law: face (hkl) lies in the zone of axis [uvw] when hu + kv + lw = 0, and the zone axis common to two faces is the cross product [k1l2 − l1k2, l1h2 − h1l2, h1k2 − k1h2]. For (110) and (001) it is [1 −1 0]. Interplanar spacing d is d = a/√(h² + k² + l²) for cubic, 1/d² = h²/a² + k²/b² + l²/c² for orthorhombic and 1/d² = (4/3)(h² + hk + k²)/a² + l²/c² for hexagonal crystals. Bragg's law, nλ = 2 d sin θ, gives the diffraction angle. Worked example: halite, a = 5.64 Å, plane (200): d = 2.82 Å; for λ = 1.54 Å and n = 1, sin θ = 1.54/(2 × 2.82) = 0.273, θ = 15.8° and the detector angle is 2θ = 31.7°.
3. Crystallographic projections
A projection reduces face orientations to a plane diagram. Each face is represented by its pole, where the face normal meets a reference sphere, located by the azimuth φ around the c axis and the polar angle ρ from it. Poles of faces in one zone lie on one great circle. Clinographic and orthographic projections serve for drawing crystal shapes.
| Projection | Distance of pole from centre | Property and use |
|---|---|---|
| Stereographic | r = R tan(ρ/2), R the primitive radius | Preserves angles; circles stay circles; zones are great circles; standard for crystal and structural orientation |
| Gnomonic | r = d tan ρ, d the plane's distance from the sphere's centre | Zones are straight lines, good for indexing; faces at ρ = 90° go to infinity |
| Equal-area | Area-preserving radial scale | Poles keep their density, so used to contour orientation data |
Worked example: on a stereographic net with primitive radius R = 10 cm, a pole at ρ = 60° plots at r = 10 tan 30° = 5.77 cm and one at ρ = 90° on the primitive, r = 10 cm; on a gnomonic plane at d = 10 cm the same ρ = 60° pole is at 10 tan 60° = 17.3 cm. The interfacial angle between two faces is the arc between their poles measured along the great circle through both.
4. Twinning
A twin is an oriented intergrowth of crystals related by a symmetry operation that is not part of the crystal's own point group: a twin plane (reflection), a twin axis (usually a 180° rotation) or a centre. The boundary is the composition plane. Contact twins share a planar boundary (spinel-law twins on (111)); penetration twins interpenetrate (Carlsbad twins of orthoclase, the pyrite iron cross, staurolite crosses at about 60° and 90°); polysynthetic twins repeat many times (albite twinning of plagioclase on (010)); and cyclic twins repeat around an axis (aragonite and chrysoberyl trillings). Microcline's tartan (cross-hatched) twinning combines the albite and pericline laws.
By origin, growth twins form during crystallisation; transformation twins form when a high-symmetry phase inverts to a lower-symmetry one on cooling, as in Dauphiné twins of quartz at the high–low inversion near 573 °C (at atmospheric pressure) and the tartan twins commonly attributed to the monoclinic to triclinic inversion of K-feldspar; and deformation (gliding) twins form under stress, as in calcite, whose twin lamellae serve as a palaeostress and strain gauge. In plagioclase, adjacent albite lamellae extinguish alternately under crossed polars, which separates twinning from exsolution lamellae, which differ in composition rather than in orientation.
5. Classification and physical properties of rock-forming minerals
Minerals are classified by the dominant anion or anionic group (Dana and Strunz): native elements, sulphides, oxides, halides, carbonates, sulphates, phosphates and silicates. Silicates, the bulk of the crust, are classified by how the SiO₄ tetrahedra link.
| Class | Linkage | Si : O | Examples |
|---|---|---|---|
| Nesosilicate | Isolated tetrahedra | 1 : 4 | Olivine, garnet, zircon, kyanite |
| Sorosilicate | Paired tetrahedra | 2 : 7 | Epidote group |
| Cyclosilicate | Rings | 1 : 3 | Beryl, tourmaline, cordierite |
| Inosilicate, single chain | Single chains | 1 : 3 | Pyroxenes |
| Inosilicate, double chain | Double chains | 4 : 11 | Amphiboles |
| Phyllosilicate | Sheets | 2 : 5 | Micas, chlorite, clays, talc |
| Tectosilicate | Frameworks | 1 : 2 | Quartz, feldspars, feldspathoids, zeolites |
Cleavage mirrors structure: micas have one perfect cleavage; pyroxenes two near 90° (87° and 93°) and amphiboles two at about 56° and 124°; feldspars two near 90°; calcite three rhombohedral (about 75° and 105°); halite three at 90°; fluorite four (octahedral); quartz none, with conchoidal fracture. Mohs hardness: talc 1, gypsum 2, calcite 3, fluorite 4, apatite 5, orthoclase 6, quartz 7, topaz 8, corundum 9, diamond 10. Specific gravity, lustre, streak and magnetism complete the hand-specimen description.
6. Optical properties in thin section
In plane-polarised light a mineral shows colour, pleochroism (colour change on rotation, as in biotite, hornblende and tourmaline), relief (contrast with the mounting medium, set by the difference in refractive index; the Becke line moves into the higher-index medium when the stage is lowered), cleavage and form. Isotropic minerals (cubic crystals such as garnet and fluorite, and glass) stay dark in every position between crossed polars. Anisotropic minerals split light into two rays. Uniaxial minerals (tetragonal, hexagonal, trigonal) have indices ω and ε, with optic sign positive if ε > ω (quartz) and negative if ε < ω (calcite). Biaxial minerals (orthorhombic, monoclinic, triclinic) have nα < nβ < nγ and an optic angle 2V, positive when the acute bisectrix is γ.
Birefringence δ = n(max) − n(min), that is |ε − ω| for uniaxial and nγ − nα for biaxial minerals. Retardation Γ = t δ, with t the thickness in the same units as Γ, fixes the interference colour: in a standard 30 μm (30 000 nm) section quartz (δ = 0.009) gives Γ = 270 nm, first-order grey to white, and calcite (δ = 0.172) gives 5160 nm, pale high-order pastel colours. First-order red-violet is at about 550 nm, so the order number is roughly Γ/550.
| Mineral | Relief and birefringence | Diagnostic features |
|---|---|---|
| Quartz | Low; grey, δ about 0.009 | Colourless, no cleavage, undulose extinction; uniaxial positive |
| Plagioclase | Low; grey, δ about 0.007 to 0.013 | Polysynthetic albite twins |
| Olivine | High; upper second- to third-order colours, δ about 0.035 to 0.05 | Colourless, no cleavage, irregular cracks |
| Augite | High; second-order, δ about 0.02 to 0.03 | Two cleavages near 90°, inclined extinction |
| Hornblende | Moderate to high; δ about 0.02 | Green-brown pleochroism, cleavages at 56° and 124° |
| Garnet | Very high; isotropic | Dark between crossed polars, rounded, no cleavage |
7. Mineral nucleation
Crystals nucleate when a melt or solution is undercooled or supersaturated. Classical nucleation theory balances the volume free energy released against the energy of the new interface: ΔG(r) = (4/3)π r³ ΔGv + 4π r² γ, where ΔGv < 0 per unit volume and γ is the interfacial energy. The barrier peaks at the critical radius r* = −2γ/ΔGv, with ΔG* = 16π γ³/(3 ΔGv²); smaller clusters redissolve and larger ones grow. For γ = 0.3 J/m² and ΔGv = −1.5 × 10⁸ J/m³, r* = 2 × 0.3/(1.5 × 10⁸) = 4 nm. Since |ΔGv| rises roughly in proportion to the undercooling ΔT, r* falls as 1/ΔT and ΔG* as 1/ΔT².
Heterogeneous nucleation on existing surfaces (crystals, walls, xenocrysts) lowers the barrier by a factor f(θ) = (2 + cos θ)(1 − cos θ)²/4, below 1, and so occurs at smaller undercooling. Nucleation and growth rates each rise and then fall with undercooling, nucleation peaking at the larger undercooling. Slow cooling gives few nuclei that grow large (coarse, equigranular), faster cooling many small crystals, and quenching gives glass, or dendritic and spherulitic forms when diffusion limits growth. A change in cooling history makes porphyritic texture. Ostwald ripening grows large grains at the expense of small ones, to reduce interfacial energy, and metamorphic minerals often need overstepping of the reaction boundary to nucleate.
8. Mineral chemistry: major, trace and rare earth elements, and cation recalculation
Major-element analyses are given as oxide weight per cent. To get a formula on a stated oxygen basis (olivine 4, pyroxene 6, feldspar 8, garnet 12, anhydrous amphibole 23, anhydrous mica 11 or 22):
- Divide each oxide wt% by its molecular weight to get moles of oxide, then multiply by the number of cations per oxide (SiO₂ 1, Al₂O₃ 2) and the oxygens per oxide (SiO₂ 2, Al₂O₃ 3, FeO and MgO 1).
- Sum the oxygen proportions, divide the oxygen basis by this sum to get the normalising factor, and multiply every cation proportion by it. Check the cation total against the ideal (olivine 3 on 4 O, pyroxene 4 on 6 O, feldspar 5 on 8 O, garnet 8 on 12 O).
- If all iron is analysed as FeO, charge balance estimates Fe³⁺ by Droop's formula, Fe³⁺ = 2X(1 − T/S), with X the ideal oxygens, T the ideal cation total and S the calculated total with all Fe as Fe²⁺.
Worked example, olivine on 4 O (molecular weights SiO₂ 60.08, FeO 71.85, MgO 40.30): SiO₂ 39.0, FeO 15.0, MgO 45.0 wt% give 0.649, 0.209 and 1.117 mol, oxygen sum 2.624, factor 4/2.624 = 1.525, and cations Si 0.990, Fe 0.318, Mg 1.702, total 3.01, so (Mg₁.₇₀Fe₀.₃₂)Si₀.₉₉O₄ and Fo = 100 × Mg/(Mg + Fe) = 84. Solid-solution content is always a mole ratio of cations, not a weight ratio.
Trace elements (below about 0.1 wt%) are described by the partition coefficient D = C(mineral)/C(melt). Compatible elements (D > 1), such as Ni, Cr and Co in olivine and pyroxene, stay in the solid; incompatible ones (D < 1), such as Rb, Ba, Th, U and the light REE, concentrate in the melt. By Goldschmidt's rules ions of the same charge and radius within about 15% substitute freely, and where charges differ the higher-charged ion is preferred by the crystal. The rare earth elements (La to Lu) are trivalent with radius shrinking along the series (lanthanide contraction), so light REE are more incompatible than heavy REE; they are plotted normalised to chondrite. Eu also occurs as Eu²⁺ and substitutes for Ca and Sr in plagioclase, so plagioclase removal leaves a negative Eu anomaly in the melt and accumulation a positive one. Garnet takes heavy REE strongly, so garnet in the source gives HREE-depleted melts, and Ce shows an anomaly under oxidising conditions (Ce⁴⁺), as in seawater.
Key takeaways
- 32 classes fall into 7 systems; only 1-, 2-, 3-, 4- and 6-fold axes exist; class m3m has 9 mirror planes and 48 operations; quartz and calcite are trigonal. Miller indices are inverted intercepts cleared of fractions, and a face lies in a zone when hu + kv + lw = 0.
- Bragg's law nλ = 2 d sin θ with the detector at 2θ; cubic d = a/√(h² + k² + l²). Stereographic projection (r = R tan(ρ/2)) preserves angles, gnomonic (r = d tan ρ) turns zones into straight lines, equal-area is for contouring.
- Twins are related by an operation outside the point group: contact, penetration, polysynthetic and cyclic; growth, transformation (Dauphiné, microcline) and deformation (calcite) in origin. Albite twinning is on (010).
- Silicates are classified by Si : O (olivine 1 : 4, pyroxene 1 : 3, amphibole 4 : 11, micas 2 : 5, framework 1 : 2). Birefringence is δ = nmax − nmin and retardation Γ = t δ. Uniaxial quartz is positive and calcite negative; isotropic minerals stay dark between crossed polars.
- Nucleation has a barrier peaking at r* = −2γ/ΔGv, with ΔG* ∝ γ³/ΔGv². More undercooling means smaller r* and a lower barrier; heterogeneous nucleation lowers it further. Cooling rate controls texture, from coarse to glassy to porphyritic.
- Cation recalculation: moles of oxide, oxygen proportions, factor = oxygen basis / sum, cations = moles × cations per oxide × factor. Solid-solution end members are mole ratios. D > 1 is compatible, D < 1 incompatible; plagioclase removal makes a negative Eu anomaly and garnet retains heavy REE.
Practice questions (18)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
How many mirror planes does the highest-symmetry cubic crystal class, 4/m 3̄ 2/m (m3m), possess?
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Answer: C — 9, three coordinate planes plus six diagonal planes
The class m3m has three mirror planes parallel to the cube faces and six diagonal planes, nine in all, with 48 symmetry operations in total. Three planes is the number for pyrite (2/m 3̄), which is the tempting answer, and the other counts do not correspond to any combination of the cube's symmetry planes.A crystal face cuts the a, b and c axes at 2a, 4b and 1c respectively. Its Miller indices are:
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Answer: A — (214)
Miller indices are the reciprocals of the intercepts cleared of fractions: 1/2, 1/4, 1 multiplied by 4 gives 2, 1, 4, so (214). The set (241) simply lists the intercepts without inverting them, which is the usual slip, and (124) and (421) reverse the order of the axes.Which of the following faces lies in the zone whose axis is [1 −1 0]?
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Answer: D — (112)
A face (hkl) lies in the zone [uvw] when hu + kv + lw = 0. With [u v w] = [1 −1 0] this reduces to h − k = 0, so h = k. Only (112) satisfies it: 1 − 1 + 0 = 0. (121) gives −1, (211) gives +1 and (101) gives +1.Fluorite is cubic with a = 5.46 Å. X-rays of wavelength 1.54 Å are diffracted from the (111) planes. The detector angle 2θ for the second-order reflection (n = 2), in degrees to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 58.5
For cubic crystals d(111) = a/√3 = 5.46/1.732 = 3.152 Å. Bragg's law nλ = 2 d sin θ gives sin θ = 2 × 1.54/(2 × 3.152) = 0.4885, so θ = 29.24° and 2θ = 58.5°. Using n = 1 gives 2θ = 28.3°, and reporting θ instead of 2θ gives 29.2°; both are the usual slips.Quartz can be indexed on hexagonal axes with a = 4.91 Å and c = 5.41 Å. For hexagonal crystals 1/d² = (4/3)(h² + hk + k²)/a² + l²/c². The interplanar spacing d of the (101) planes, in Å to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 3.34
For (101), h = 1, k = 0, l = 1, so h² + hk + k² = 1. Then 1/d² = (4/3)(1)/24.108 + 1/29.268 = 0.05531 + 0.03417 = 0.08948 Å⁻², and d = 3.34 Å. Dropping the factor 4/3, which belongs to the hexagonal metric, would give d = 3.64 Å.In which projection do the poles of all the faces of one zone plot on a straight line?
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Answer: B — Gnomonic projection, because a plane through the sphere's centre cuts the tangent plane in a straight line
In gnomonic projection each pole is projected from the sphere's centre onto a tangent plane, so the zone, being a plane through the centre, appears as a straight line. In stereographic projection the zone is a great circle, a straight diameter only for the special case of a plane through the net's centre, which is the tempting claim. The poles of a zone do lie in one plane through the centre, but other projections turn that plane into a curve.Which of the following statements about twinning are correct?
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Answer: A — Polysynthetic albite twinning in plagioclase is on the plane (010); B — The tartan (cross-hatched) twinning of microcline combines the albite and pericline laws; D — Staurolite cruciform twins are penetration twins whose two individuals cross at about 60° or 90°
Albite twinning is on (010), microcline's tartan pattern is the albite plus pericline combination, and staurolite crosses form at about 60° or 90°, so those three are correct. A twin operation is by definition not part of the crystal's own point group; if it were, it would be a symmetry of the single crystal and produce no twin.A rock-forming silicate has double chains of tetrahedra with Si : O = 4 : 11 and two cleavages meeting at about 56° and 124°. It belongs to the:
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Answer: C — amphibole group, whose double chains give Si : O = 4 : 11 and cleavages at about 56° and 124°
Double-chain silicates with Si : O = 4 : 11 are the amphiboles, whose cleavage angles are about 56° and 124°. Pyroxenes are the tempting alternative since both are chain silicates, but their single chains give 1 : 3 and cleavages near 90°. Micas are sheet silicates and olivine has isolated tetrahedra.In thin section a colourless mineral has high relief, no cleavage, irregular cracks, parallel extinction and interference colours of upper second to third order. It is most likely:
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Answer: A — olivine, which combines high relief and strong birefringence with irregular cracks instead of cleavage
High relief with no cleavage, irregular cracks, parallel extinction and upper second- to third-order colours is the standard set for olivine. Augite would show two cleavages near 90° and inclined extinction, garnet would be isotropic and stay dark, and quartz has low relief and first-order grey, so each of those fails on at least one stated property.A uniaxial mineral has ω = 1.658 and ε = 1.486. Its optic sign and its birefringence are:
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Answer: D — negative, with δ = 0.172
The sign is negative because ε < ω, and the birefringence is the difference |ε − ω| = 1.658 − 1.486 = 0.172, the very high value of calcite. A positive sign would need ε > ω, the birefringence is a difference rather than a sum, and the stem already states the mineral is uniaxial.A mineral grain in a thin section 25 μm thick shows a retardation of 600 nm. Its birefringence, to three decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.024
Retardation Γ = t δ, so δ = Γ/t = 600 nm / 25 000 nm = 0.024. Forgetting to convert 25 μm to 25 000 nm gives 24, which is the usual slip.Which of the following statements about optical properties are correct?
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Answer: A — Isotropic minerals such as garnet remain dark in every position between crossed polars; B — Tetragonal, hexagonal and trigonal minerals are optically uniaxial; D — Pleochroism is observed in plane-polarised light, with the analyser out of the beam
Isotropic minerals stay extinct, the three lower-symmetry non-cubic systems named are uniaxial, and pleochroism needs only polarised light with no analyser, so those three are correct. Quartz has ε (1.553) greater than ω (1.544), so it is optically positive, not negative.In classical nucleation theory a crystal nucleus has interfacial energy γ = 250 mJ/m² and the volume free-energy change on crystallising is ΔGv = −4 × 10⁸ J/m³. The critical radius r* = −2γ/ΔGv, in nm to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 1.25
Convert γ to SI: 250 mJ/m² = 0.25 J/m². Then r* = 2 × 0.25 / (4 × 10⁸) = 1.25 × 10⁻⁹ m = 1.25 nm. Using γ = 250 without the conversion gives 1250 nm, a thousand times too large, which is the usual slip.In classical nucleation theory the barrier is ΔG* = 16π γ³ / (3 ΔGv²). Take γ as constant and |ΔGv| as proportional to the undercooling ΔT. If the undercooling doubles, the homogeneous nucleation barrier changes by a factor of:
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Answer: B — 1/4, because ΔG* varies as 1/ΔGv², and ΔGv doubles
Since ΔG* ∝ 1/ΔGv² and ΔGv doubles, the barrier falls to 1/4. The factor 1/2 forgets the square, which is the tempting answer. The 1/8 argument is wrong because ΔG* = (4π/3) γ r², which varies as the square of r, not the cube (halving r* gives 1/4), and a larger undercooling lowers the barrier, it does not raise it.An orthopyroxene analysis gives SiO₂ 57.0, MgO 40.0 and FeO 3.0 wt% (take molecular weights SiO₂ 60.08, MgO 40.30, FeO 71.85). Recalculated on a basis of 6 oxygens, the number of Si atoms per formula unit, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 1.94
Moles: SiO₂ 57.0/60.08 = 0.9487, MgO 40.0/40.30 = 0.9926, FeO 3.0/71.85 = 0.0418. Oxygen proportions: 2 × 0.9487 + 0.9926 + 0.0418 = 2.9318. Factor = 6/2.9318 = 2.0465, so Si = 0.9487 × 2.0465 = 1.94 (with Mg 2.03 and Fe 0.09, cation total 4.06). Forgetting the factor of 2 oxygens per SiO₂ gives a factor of about 3.03 and Si near 2.87.An olivine contains 38.0 wt% MgO and 20.0 wt% FeO (molecular weights MgO 40.30, FeO 71.85; ignore any other Mg- or Fe-bearing component). Its forsterite content Fo = 100 × Mg/(Mg + Fe) in mole terms, to the nearest whole number, is ____.
Numerical answer — type the value.
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Answer: 77
Moles: MgO 38.0/40.30 = 0.9429 and FeO 20.0/71.85 = 0.2784. Fo = 100 × 0.9429 / (0.9429 + 0.2784) = 77.2, so 77. Using the weight ratio 38/(38 + 20) = 65.5 is the usual slip, because end-member content is a mole ratio.A basaltic magma crystallises and removes a large amount of plagioclase. The chondrite-normalised REE pattern of the residual melt will show:
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Answer: A — a negative Eu anomaly, because Eu²⁺ substitutes for Ca and Sr in plagioclase and is removed with it
In reduced magmas part of the europium is Eu²⁺, which has a radius and charge close to Ca²⁺ and Sr²⁺ and enters plagioclase, so removing plagioclase depletes the residual melt in Eu relative to its neighbours: a negative Eu anomaly. Europium does not behave like the other, strictly trivalent, REE, which rules out the no-anomaly answer. Plagioclase does not strongly retain the heavy REE; garnet does.Which of the following statements about mineral chemistry and cation recalculation are correct?
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Answer: A — A compatible element has a partition coefficient D greater than 1 and is retained in the crystallising solid; B — The forsterite content of olivine is the mole ratio 100 × Mg/(Mg + Fe), not a weight ratio; D — A cation total far from the ideal for the chosen oxygen basis signals a poor analysis, an unanalysed element or ferric iron
Compatible elements have D above 1, end-member contents are mole ratios, and a cation total that misses the ideal value is a standard quality check, so those three are correct. Olivine, (Mg,Fe)₂SiO₄, is recalculated on 4 oxygens; 6 oxygens is the pyroxene basis.