Optical Mineralogy

ADVANCED OPTICAL MINERALOGYJ. NichollsDepartment of Geology and GeophysicsUniversity of CalgaryCalgary, Canada

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Chapter 1: Background to the Text

Optical crystallography and mineralogy are difficult subjects to master. Yet, they

are fundamental to the description and identification of minerals and rocks with a

microscope. In addition to their practical use, optical mineralogy and crystallography

contain as complete and logical part of science as any other subjects in geology.

Consequently, mastery of the subject is also intellectually satisfying. The objective of this

text is to present much of the optical theory used by petrologists in the language of vector

algebra and vector calculus. Vectors are mathematical concepts that some people find

particularly easy to visualize. Perhaps one of the more difficult aspects of optical

crystallography is the visualization of the three dimensional nature of the optical

properties of crystals. The description of these properties with vector algebra provides an

alternative to the word descriptions and perspective drawings commonly used to display

optical properties and can enhance your skill in visualizing the three dimensional aspects

of optical crystallography. Because vectors are mathematical concepts, they also provide

a mechanism for obtaining numerically more precise and accurate descriptions of the

geometry of optical properties than can be had with word descriptions and drawings. A

mastery of the description of optical properties by vector algebra has a practical use. It

will give you, the practitioner, the ability to construct determinative charts and diagrams

relating optical properties to crystallographic and chemical properties that suit your

purposes rather than having to rely on charts and diagrams in the literature that were

perhaps constructed for some other purpose. In the event that new optical data become

available, you will be able to easily revise such charts and diagrams to take advantage of

the new data.

The use of optical properties to estimate mineral compositions has been largely

replaced, at the research level, by electron microprobe techniques. The study of the

crystallographic properties of minerals is now almost the exclusive domain of the x-ray

crystallographer and electron microscopist. These factors have caused a decline in the

importance of optical methods in mineralogy and crystallography. Yet, for the

petrologist, optical techniques remain the prime methods for mineral identification,

textural description, and zoning determination. Vectors can be used to efficiently

Chapter 1: Background to the Notes

3

represent the optical properties of crystals in forms that are easy to manipulate.

Consequently, the techniques by which we study rocks with the microscope are

enhanced. Vector descriptions of the optical properties of several rock-forming mineral

groups, notably olivines, pyroxenes, feldspars, and amphiboles, are included in the text.

The description of optical properties with vector algebra emphasizes the

importance of crystal orientation to the appearance of the mineral in thin section. Random

sections through biaxial crystals can display a variety of properties. In order to correctly

identify minerals and to interpret the fabric of rocks, a petrographer has to relate optical

properties to optical orientations, that is the relative orientation of the indicatrix to the

crystal lattice.

Optical theory as applied to transparent crystals was essentially complete by 1900.

The interested reader can check the validity of this statement by examining the references

cited by Johannsen (1918) in his definitive work on petrographic methods. At the turn of

the century vector calculus had not been invented by J.W. Gibbs, nor were calculators, let

alone computers, available. Consequently, quantitative optical crystallography as used by

mineralogists and petrologists was grounded in 19th century mathematics. The results

were very complicated algebraic and trigonometric equations that yielded quantitative

results for specific cases only after an inordinate amount of hand calculation. The curious

can examine the equations needed to calculate the extinction angle on any face of any

zone of any crystal derived in Johannsen (1918, p. 403). Because of the arithmetical

labor, correlation of crystallographic and optical properties never reached the level of

precision attainable. Today, the availability of computers can be used to relieve the

drudgery of calculation and the language of vector algebra and vector calculus can be

used to efficiently and concisely formulate optical problems.

It is assumed that you, the reader, are conversant with the theory and practice of

optical crystallography, such as covered in Bloss (1961) or Nesse (1991) and can plot

optical properties on stereographic projections and obtain quantitative data from such

projections (e.g. extinction angles). In general, anything that can be plotted and estimated

on a stereonet can be formulated into equations and more precisely calculated with a

computer. In essence, the purpose of this text is to explain how to make such calculations.

Chapter 1: Background to the Notes

4

Most of the mathematics needed to understand this text are discussed in high

school and introductory university level textbooks. You should know what a vector is and

how to represent it in terms of components. You should know how to multiply vectors

and the kinds of products that result: the dot and cross products. You will also need to

know how to solve three simultaneous linear equations and how to manipulate

trigonometric functions. In a couple of instances, use is made of more advanced

techniques involving calculus and statistics such as taught in university level math

courses. Hopefully, these parts have been written in such a way that the problem can be

understood and the concepts underlying its solution followed even if the details and

mechanics of the solution are not part of your background. A review of vector algebra

and the specific features of the calculus used in the text are given in the Appendix.

Several texts that are repeatedly cited; these amplify many parts of the subject and

are recommended as sources of additional information and as review of topics needed as

background. Optical theory is covered in Bloss (1961) and Nesse (1991), crystallography

is covered in Bloss (1971) and spindle stage theory and practice is covered in Bloss

(1981). The texts by Nye (1957) and Lovett (1991) discuss the applications of tensors to

describe the physical properties of crystals and the transformation of coordinates from

one reference frame to another. Hoffmann (1975) discusses vectors and their meaning.

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Chapter 2: Vector Representation of the Indicatrix

Introduction

In this chapter we will derive the equations that relate optical directions within the

indicatrix. The important directions are the principal vibration directions, X, Y, Z, the

optic axes and, in any given section through the indicatrix, the two vibration directions of

light and the wave normal associated with the two vibration directions in the plane of the

section. The material will be organized in the following manner. First, we will use the

Law of Biot-Fresnel to relate the vibration directions in a section cut through a biaxial

crystal to the wave normal associated with these directions. In orthoscopic light the

normal to our thin section is also parallel to the wave normal. Second, we will derive an

equation relating vibration directions in random thin sections to vectors parallel to the

optic axes. If the need should arise, this exercise provides us with the means for locating

the optic axes from extinction angle measurements. Third, we will discuss calculation of

the refractive indices and birefringence of a random section through the indicatrix. We

will end the chapter with a discussion of the need for consistency between the principal

refractive indices and 2V and how this consistency can be obtained from measurements of

the optical properties of crystals.

The names of the vectors and other quantities associated with the optical

properties of biaxial crystals used in this chapter are listed in Table 2-1 for reference.

Quantities of a mathematical nature are described or defined in the Appendix.

The Law of Biot-Fresnel

Because the law of Biot-Fresnel is central to determining the vector representation

of the indicatrix, we first show its meaning with a stereographic projection (Figure 2-1).

Given the stereographic projections of the optic axes (OA1 and OA2, Figure 2-1) of a

biaxial mineral and the stereographic projection of the wave normal (w) of light passing

through a random section of the mineral, the law of Biot-Fresnel can be demonstrated as

follows. The law of Biot-Fresnel states that the vibration directions of the light associated

with w bisect the dihedral angles between the two planes containing the wave normal and


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the optic axes (see Bloss, 1961, Fig. 9-9A, p. 163). To apply this law, the locations of the

wave normal and the optic axes in a frame of reference are required. The indicatrix

provides a convenient frame of reference for the location of vectors representing optical

directions.

Table 2-1: Notation adopted for optical quantities in Chapter 2. Vector notation is described in the Appendix. u, v Unit vectors parallel to the optic axes. Vz Optic axial angle. 2Vz is the angle between u and v. T, S Vectors parallel to the lines of intersection between the circular sections and

the plane of the thin section. t, s Unit vectors parallel to T and S. 2θ The angle between t and s in the plane of the thin section. R Vector sum of t and s. w A unit vector parallel to the normal to the thin section and parallel to the wave

normal in orthoscopic light S, E Angles defining the spindle stage coordinates. g Unit vector normal to w and x in the spindle stage reference frame e, f Unit vectors parallel to the acute and obtuse bisectrices. n, m Unit vectors parallel to the vibration directions in the plane of the thin section. N, M Refractive indices associated with the vibration directions, n and m. N Vector parallel to n with magnitude N. X, Y, Z The principal vibration directions of the indicatrix. α, β, γ Refractive indices associated with the principal vibration directions of the

indicatrix

Construct the circular sections, CS1 and CS2, normal to OA1 and OA2,

respectively. The projections of the diameters of the circular sections of the indicatrix

intersect the primitive circle of the projection along the lines a-b and c-d, Figure 2-1. The

wave normal is plotted at the pole of the primitive circle; consequently, the section

through the indicatrix, normal to w, lies in the plane of the projection. In general, this

section through the indicatrix will be an ellipse with major and minor axes equal to γ’ and

α’, respectively, where γ’ and α’ are the slow and fast indices of refraction in the section

through the crystal. Since the radii of the circular sections of the indicatrix are equal to β,

the four radii of the elliptical section along the lines a-b and c-d will also equal β. The

properties of an ellipse require that the bisectors of the angles between equal radii be

parallel to the semi-axes of the ellipse. Hence, these are the vibration directions

associated with the wave normal, w (see Bloss, 1961, p. 229. Fig. 11-9).


7

Construct the projections of the planes containing w and the optic axes (Dashed in

Figure 2-1). Measure the angles labeled θ. The law of Biot-Fresnel states that these

angles are equal.

Location of Vibration Directions

The location of the vibration directions in a section identified by a given wave

normal is the necessary first step in describing the optical properties of crystals with

vector algebra. Consequently, this section will be referred to again in the book and the

reader should be certain that this section is thoroughly mastered before continuing.

As we know from our study of stereographic projections, we must be given a

certain amount of data before we can determine the vibration directions of a section

through the indicatrix with a stereonet. Transferring the calculations to a set of equations

does not lessen the need for data. To make the calculations we will ultimately need to

know the direction of the normal to the thin section, 2V, and the optical sign of the

crystal. We are guided by the Law of Biot-Fresnel in our search for the vibration

directions in a section through a biaxial crystal. The law of Biot-Fresnel states that a

vibration direction bisects the angle between the two planes formed by the wave normal

and each of the optic axes. From this law we see that we need to know the location of the

wave normal and the optic axes in order to find the vibration directions. Consequently,

we take as given the direction of the wave normal relative to the indicatrix axes. The

directions of the optic axes, relative to the indicatrix axes, are determined if we know the

sign and 2V for the substance.

If we know the direction of the wave normal, we can write down an equation for a

unit vector parallel to this direction:

1 2 3w w w= + +w i j k (2.1)

where i, j, and k are unit vectors parallel to the axes of the indicatrix, X, Y, and Z,

respectively. The wi, i = 1, 2, 3, are the components of w parallel to X, Y, and Z, in that

order. Because the components of a unit vector are the same as the direction cosines of


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the vector in the frame of reference, it is relatively simple to locate the wave normal

vector by the angles it makes with the axes of the frame of reference (Figure 2-2A).

A unit vector parallel to the optic axis that falls between positive X and positive Z

is given by:

sin cos Vz Vx= +u i k (2.2)

1 3u uu = i + k (2.3)

where: 1 3u sin and u cos .Vz Vz= = These vectors are illustrated on Figure 2-2B. The

second optic axis will be parallel to a unit vector:

sin cos Vz Vz= − +v i k (2.4)

Note that:

1 3u u = − +v i k (2.5)

A stereographic projection of the relationships between w, u, v and the sections

(planes) normal to each vector is shown in Figure 2-3. A vector parallel to the line of

intersection of the circular section normal to OA1 and the plane of the thin section will be

normal to both OA1 and the wave normal. Let’s label this vector t. The same situation

will be true for another vector parallel to the intersection of the second circular section

and the plane of the thin section; label it s. In other words, the vector t lies at 90° to u and

w; s lies at 90° to v and w.

We first want to calculate the components of the vectors parallel to the lines of

intersection. The cross product was designed to find such a vector as it produces a new

vector normal to two original ones. A vector parallel to the first intersection is:

= ×T u w (2.6)

and a vector parallel to the second intersection is:

×S = w v (2.7)

Unit vectors parallel to the two lines of intersection are obtained by dividing by

the magnitudes of the T and S:


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( )( ) ( )

×=× ×

u wt

u w u w� (2.8)

( )( ) ( )

×=× ×

w vs

w v w v� (2.9)

Substitution of Equations (2.3) and (2.5) into Equations (2.8) and (2.9), plus the

fact that the sum of squares of a set of direction cosines is one, provides after a bit of

algebraic manipulation:

( )

( )3 2 3 1 1 3 1 2

2

1 1 3 31-

u w u w u w u w

u w u w

− + − + =+

i j kt (2.10)

( )

( )3 2 3 1 1 3 1 2

2

1 1 3 31-

u w u w u w u w

u w u w

+ + + =−

i j ks (2.11)

Notice here that Equations (2.10) and (2.11) are formulae for calculation. The wi

are given quantities and the ui can be calculated from 2Vz [Equations (2.2) and (2.4)].

Consequently, t and s can be calculated.

According to the Law of Biot-Fresnel, one of the vibration directions in the plane

of the thin section bisects the angle between t and s, labeled 2θ on Figure 2-3. The value

of this angle can be calculated by using the dot product of t and s:

2cos 2 1 2sinθ θ= = −t s� (2.12)

Note that t and s have magnitudes of one. Consequently, the product of their

magnitudes is also unity and does not appear explicitly in Equation (2.12). Solving for

sinθ gives:

( )12sin 1θ = − t s� (2.13)

We can now find θ easily with the Arcsin function. Our next task is to find a unit

vector parallel to the bisector of the angle 2θ. This unit vector will be normal to w, hence

the bisector is related to t, s, and w by:


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sin θ× = −n t w (2.14)

cosθ=n t� (2.15)

sin θ×n s = w (2.16)

cosθn s =� (2.17)

Either pair of Equations (2.14) and (2.15) or (2.16) and (2.17) are sufficient to

locate n. The details of calculating a vector, such as n, given two other unit vectors and

the angle between n and t are described in The Appendix (Solution of the Product

Equations). In short, we can calculate the components of one vibration direction, n, using

either the pair of Equations (2.14) and (2.15) or the pair (2.16) and (2.17). The minus sign

is required in Equation (2.14) in order for the triple of vectors, n, t, and: -w to form a right

hand set.

As an alternative to the sets of Equations (2.14) and (2.15) or (2.16) and (2.17), n

can be calculated in the following fashion. Because n bisects the angle between t and s

and because t and s are of equal magnitude, remember they are unit vectors, the

parallelogram law of vector addition requires that:

= +G t s (2.18)

where G is a vector parallel to n. It is then a simple matter to convert G into a unit vector

by dividing by the square root of the dot product of G with itself:

•

G

G Gn = (2.19)

To find the second vibration direction, we note that it is normal to both w and n.

As a result they are simply related by the cross product:

= ×m n w (2.20)

or, in component form:

1 2 3 3 2

2 2 1 1 3

3 1 2 2 1

m n w n w

m n w n w

m n w n w

− = − −

(2.21)

(See The Appendix, Products of Vectors).


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The foregoing contains the information we need to find the vibration directions in

any section through a biaxial mineral if we are given the wave normal vector, w, 2V and

optic sign. In fact, all we need to calculate the vibration directions for a given wave

normal defined in the frame of reference describing the indicatrix are the indices of

refraction. 2V can be calculated from the indices of refraction. An example of the

calculations is given in Table 2-2.

Table 2-2: Example of the calculations. The components of the unit vectors, w, u, v, n, and m, are equal to the direction cosines of the angles between the vector and the axes of the frame of reference, the indicatrix. The wave normal vector, w, was chosen to make equal angles with the axes of the indicatrix.

Given Values w1 w2 w3 Wave normal vector: w 0.57735 0.57735 0.57735 α β γ Indices of Refraction 1.60 1.62 1.70

Calculated Values Vz [see Bloss 1961, p. 156; Equation (44)] 27.618 i j k Optic Axis Vector: u 0.46357 0.0 0.88606 Optic Axis Vector: v -0.46357 0.0 0.88606 t: Eqn. (2.8) -0.81620 0.38918 0.42703 s: Eqn. (2.9) 0.52750 -0.80348 0.27598 G = t + s: Eqn. (2.18) -0.28871 -0.41430 0.70301 n: Eqn. (2.19) -0.33354 -0.47865 0.81219 m: Eqn. (2.20) -0.74526 0.66149 0.08378

Location of the Optic Axes

In the last section we derived the equations for finding the vibration directions in

random sections if we are given the vectors parallel to the wave normal and the optic

axes. In this section we will try to find out what we need to know in order to determine

the vectors parallel to the optic axes, u and v, from extinction angle measurements. In

other words, how many wave normals and their associated vibration directions must we


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know in order to locate the optic axes.

We can write Equations (2.8) and (2.9) in the following way by manipulating the

vectors under the radical:

( )

( )21

×

− •

u wt =

u w (2.22)

( )

( )21

×

− •

w vs =

v w (2.23)

Next we substitute these two equations into Equations (2.14) and (2.16) and

equate the terms in sin θ w:

( )( )

( )( )2 2

1 1

× × × ×=

− • − •

n w v n u w

v w u w (2.24)

Notice that we can remove the negative sign from Equation (2.14) by reversing the order

of w and u in the cross product. There is a vector identity that can be used to expand the

vector triple products in the following manner (Hoffmann, 1975, p. 80):

( ) ( )

( )( ) ( )

( )2 21 1

• − • • − •=

• − •

w n v v n w w n u u n w

- v w u w (2.25)

But because n and w are normal vectors, the angle between them is 90°. As a result the

cosine of the angle between them is zero causing their dot product to be zero.

Consequently, Equation (2.25) simplifies to:

( ) ( )2 2

1 1

• • − = • •

0n v n u

w- v w - u w

(2.26)

The wave normal has a definite direction that cannot be represented by a zero vector;

therefore, Equation (2.26) can equal zero only if the scalar coefficient of w (i.e. the term

in square brackets) is zero. Setting this term equal to zero and squaring gives:

( ) ( ) ( ) ( )2 2 2 21 1 0 • − • − • − • = n v u w n u u w (2.27)


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Equation (2.27) is the required expression that relates n and w to u and v. If we

treat u and v (the optic axes vectors) as unknowns, then for any one wave normal, w, and

vibration direction, n, we have one equation in four unknowns: the independent

components of u and v. At first we might suspect that there are six unknowns; the vectors

have three components each. However, the components of each vector are not

independent because the sum of squares of the components of a unit vector must equal

one. The location of the second vibration vector in the section through the crystal, m,

does not provide an additional constraint on the location of the optic axes because m is

not independent of n. [See Equation (2.20)]. Consequently, we are left with one equation

and four unknowns.

The significance of Equation (2.27) lies in the fact that it tells us we must

determine the vibration directions in at least four randomly chosen sections if we want to

find vectors parallel to the optic axes by extinction angle measurements, a fact first

discovered by Weber (1921). Further, the vibration directions located by positions of

extinction, must all refer to a common coordinate system. This can be done by mounting

the crystal on a rotational device such as a spindle stage (Bloss, 1981) or, much less

accurately, a universal stage (Slemmons, 1962). If we have vector representations of

more than four different extinction positions, we can use the method of least squares (see

The Appendix) to calculate the components of the optic axial vectors, u and v, that best

agree with all the data.

Equation (2.27) also tells us that in thin section, the extinction position of a

randomly chosen section only gives us 25% of the information we need to locate the optic

axes. Even if we change to conoscopic illumination and look at an interference figure,

just any old section will not allow us to locate the optic axis, determine the sign or

possibly even establish whether the crystal is biaxial. The point to remember is that we

have been discussing arbitrarily chosen sections and that the secret to optical mineralogy

is to decrease the arbitrariness. You can do this by searching for grains with low

interference colors and that, ideally, remain at extinction on rotation of the stage. These

sections are not chosen at arbitrarily and will give approximately centered optic axis

figures. In this instance, we can locate optic axes with just one section, not four (see,

however, page 19).


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Spindle Stage Coordinates

This section will outline the method for finding vectors parallel to the optic axes

from spindle stage measurements. The material is more thoroughly covered in the text by

Bloss (1981) and is summarized here as background for later discussions.

Spindle stage data can be plotted on a stereographic projection using a particularly

convenient reference frame (Bloss, 1981) as shown on Figure 2-4. The reference frame is

Cartesian with axes labeled x, y, and z. The axis of the microscope is parallel to z and the

x-y plane is parallel to the microscope stage. The east-west axis is labeled x, the north-

south axis is labeled y and the system is a right-handed one. Two angles are needed to

locate a vector in this reference frame and are labeled E and S on Figure 2-4. E is

measured by rotating the microscope stage and S is measured by rotating the spindle

stage axis. The details of making the measurements and the precautions to follow to

ensure the data are precise and accurate are described in Bloss (1981). It is sufficient here

to know that the primary data from the spindle stage are several pairs of E and S angles

that serve to locate w and n for each pair.

To find these locations we calculate the components of w and n in the spindle

stage reference frame. To begin, note that w lies in the y-z plane and consequently w1, the

component of w along the x-axis, is zero. The other components of w are easily deduced

from a section through the stereographic sphere parallel to the y-z plane (Figure 2-4B).

An auxiliary vector, g, is drawn along the intersection of the y-z plane and the plane

containing n and the x-axis. Consequently, g lies at an angle S to the y-axis. The wave

normal vector, w, is, by definition, normal to g and consequently makes an angle S to the

z-axis. The projection of w on the y-axis lies along the negative portion of y and is equal

in magnitude to sin S. The projection of w along the z-axis is equal to cos S and we have

as components of w in the spindle stage reference frame:

w1 = 0

w2 = sin S (2.28)

w3 = cos S

A similar discussion provides us with the components of g:


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g1 = 0

g2 = cos S

g3 = sin S

The components of n are now easily calculated from the two criteria:

( )2

2 2 21 2 3

sin cos

1

E E

n n n

π× − =

+ + =

n g = w w (2.30)

The results give the components of n in the spindle stage reference system:

n1 = cos E

n2 = sin E cos S (2.31)

n3 = sin E sin S

The components of n are calculated for each pair of E and S angles measured with the

spindle stage.

The components of each resulting n are entered into Equation (2.27). This

provides a number of simultaneous equations that can be solved, by least squares

techniques, for the components of u and v in the spindle stage reference frame.

Unfortunately, Equation (2.27) is nonlinear in these unknowns, the optic axial vectors,

and standard linear least-squares techniques cannot be used. Nonlinear least-squares

methods, such as described by Meyer (1975) are required instead. The essence of these

procedures is to find a first approximate location for u and v from a stereographic

projection of the extinction angle data (see Bloss, 1981 for details). Next, Equation (2.27)

is evaluated at each extinction position by entering the components of n, u, and v, the

latter obtained from the stereographic projection. New positions of u and v are calculated

by nonlinear least-squares methods and the exercise of evaluating Equation (2.27) is

repeated, using the new values of u and v. Iteration continues until the corrections to u

and v are smaller than some small number.

If all goes well, the nonlinear least-squares procedure produces the best values for

u and v. Their dot product immediately gives 2V:

cos 2V•u v = (2.32)


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The location of the unit vector parallel to the y-axis, j, in the spindle stage reference

frame, is easily found with the cross product:

sin 2V×u v = j (2.33)

A second application of the cross product, in conjunction with the dot product (Appendix,

Solution of the Product Equations) serves to define a unit vector, e, parallel to the acute

bisectrix:

sin

cos

V

V

×•

e v = j

e v = (2.34)

Finally, we can once more apply the cross product and calculate the unit vector, f, parallel

to the obtuse bisectrix:

×e j = f (2.35)

We cannot assign the labels i and k, as defined earlier, to e and f unless we know

the sign of the crystal. If the crystal is positive then e equals k and f equals i. Otherwise

the reverse is true. The locations of j, e and f, as calculated above, are with respect to the

spindle stage reference axes. To convert u and v to the indicatrix reference frame, we

return to Equations (2.2) and (2.4).

Refractive Indices in Arbitrary Sections

If we know the sign and 2V of the crystal, then given a wave normal, we can

calculate the vibration directions associated with the wave normal. But what else can we

find out about the optical properties of the section we are looking at? Some obvious

properties are the refractive indices associated with the vibration directions of the section

and the birefringence of the section.

Having determined the vibration directions we next calculate the refractive

indices of the light vibrating in these directions. As usual, before we can calculate

anything, there must be some data available to enter into the equations. We will assume

therefore, that we know the principal refractive indices, α, β, and γ.

A vector parallel to the vibration vector that stretches from the center of the

indicatrix to a point (X,Y,Z) on the indicatrix is given by:


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N=N n (2.36)

where N is a scalar that multiplies n and gives the magnitude of N. From the definition of

the indicatrix we conclude that its value is equal to the refractive index of the light

vibrating parallel to n. In component form we have:

1 2 3X Y Z Nn Nn Nn+ + = + +i j k i j k (2.37)

Equating components and substituting the results into the equation for the indicatrix:

( ) ( ) ( )2 22

1X Y Zα β γ+ + = (2.38)

gives:

2 22

31 2 1NnNn Nn

α β γ + + =

(2.39)

Hence:

( ) ( ) ( )2 2 2

1 2 3

Nn n n

αβγ

βγ αγ αβ=

+ + (2.40)

Equation (2.40) is important because it is a formula for calculating the refractive

index of light vibrating parallel to a vibration vector, n, in a section identified by a given

wave normal, w. Associated with n and w will be a second vibration direction, m. The

refractive index, M, of this second vibration direction can be calculated from the same

kind of equation by substituting mi for ni in Equation (2.40) [See equations 2.20 and

2.21]. The birefringence of the section is easily calculated by taking the absolute value of

the difference between M and N: M N− .

Principal Refractive Indices by Extrapolation of Spindle Stage Data

A succession of vibration directions through a crystal mounted on a spindle stage

can be brought into the plane of the microscope stage. At each instance, the refractive

indices associated with the two vibration vectors in the plane of the stage can be

measured. If the components of three different vibration vectors and their associated

refractive indices are substituted into Equation (39) we produce three equations of the


18

form:

2 2 21 2 32 2 2 2

1 1 1 1n n n

Nα β γ + + =

(2.41)

The resulting system of equations can be solved for the principal refractive indices, α, β,

and γ.

If more than three refractive indices are measured, the system of equations will be

over determined and the principal refractive indices can be determined by least-squares

analysis. In contrast to the least squares estimation of optic axis vectors, the actual least

squares calculation of the principal refractive indices is linear in 2 2 21 , 1 , and 1 .α β γ

Before the least squares analysis can be done, the reference frames assigned to the

indicatrix and the spindle stage must be related. The components of the vibration vectors

that appear in Equation (2.37) are those in the indicatrix reference frame. However, the

vectors are measured in the spindle stage reference frame. The equations needed to

transform the components of a vector from one reference frame to another are given by

Nye (1957, p. 9-10) [See also, The Appendix, Transformation of Vector Components]. If

a point in the indicatrix frame of reference is represented by the numbers (X,Y,Z) and the

same point in the spindle stage frame of reference is represented by (x,y,z) then the two

sets of numbers are related by:

11 12 13

21 22 23

31 32 33

X a x a y a z

Y a x a y a z

Z a x a y a z

= + += + += + +

(2.42)

where , 1, 2, 3,ija j = are the direction cosines of the X vibration direction in the spindle

stage frame of reference. The 2 3 and j ja a are the analogous direction cosines for Y and Z,

respectively. Consequently (see The Appendix, Transformation of Vector Components),

the components of a vibration vector in the two frames of reference are related by:

1 11 1 12 2 13 3

2 21 1 22 2 23 3

3 31 1 32 2 33 3

n a n a n a n

n a n a n a n

n a n a n a n

∗ ∗ ∗

∗ ∗ ∗

∗ ∗ ∗

= + += + += + +

(2.43)


19

where *, 1, 2,3,in i = are the components of n in the spindle stage frame of reference and

the in are the components of n in the indicatrix reference frame.

An outline of the procedure for the determination of the principal refractive

indices is:

1. On a spindle stage, locate X, Y, and Z.

2. Calculate their direction cosines ( , 1, 2, 3ija j = ) with respect to the spindle stage

reference frame.

3. Locate several (more than three) vibration vectors (n) and measure their refractive indices (N).

4. Calculate the components of each n in the indicatrix reference frame with Equation (2.43).

5. Form the over-determined system of equations using Equation (2.41) as a model.

6. Using a least squares technique, solve for 2 2 2

1 1 1, ,

α β γ.

7. Calculate α, β and γ.

Earlier, we showed that extinction angles in four arbitrary sections were required

to locate the optic axes of a biaxial crystal. If, however, extinction angles AND indices of

refraction in three sections are measured on a crystal mounted on a spindle stage, then the

optic axes can be calculated from the data. Julian and Bloss (1987) describe methods for

making the required calculations with eigenvalues and eigenvectors.

Consistency of 2V and the Refractive Indices

If we are given the principal refractive indices, sign and 2V of a mineral, we can

calculate the optical properties in any section through a crystal. Mineralogists and

crystallographers for have collected these data for over a century. Therefore, we would

expect that the data for doing these exercises or having a computer do them for us is

available. Unfortunately, different workers measured different properties on different

crystal fragments. Because the optical properties are not all independent, the data are

often (usually) inconsistent. For example, the principal refractive indices and 2V are not

independent properties because they are related by the equation (e.g. Bloss, 1961, p. 156):


20

( )( )

2 2 2

2

2 2 2cos Vz

α γ β

β γ α

−=

− (2.44)

Because of experimental errors, tabulated values of 2V and the principal refractive indices

seldom exactly satisfy Equation (2.44) and in some cases calculated and measured values

of 2V differ by several degrees. This is illustrated in Table 2-3 for some plagioclase data

compiled by Phillips and Griffen (1981).

Table 2-3: Optic axial angles for the plagioclase series. Tabulated values compared with values calculated from the principal refractive indices. Data from Phillips and Griffen (1981). Mineral An Content Calc 2V Tab 2V Diff Low Albite An 0 74.47 77.0 -2.53 Oligoclase An20 90.22 93.0 -2.78 Andesine An40 90.22 83.0 7.22 Labradorite An60 90.22 80.0 10.22 Bytownite An80 90.27 95.0 -4.73 Anorthite An100 103.69 102.0 1.69

It is worth noting that consistency is a necessary requirement for accurate data but

it is not a sufficient one. If data are inconsistent we know that some or all of them are not

correct. However, consistent data may also be inaccurate; they just agree among

themselves.

If we are going to use tabulated data to construct determinative charts for

minerals, we obviously would like the data to be theoretically consistent. To enforce this

consistency we will calculate new values of 2V and the refractive indices that are

consistent with Equation (2.44) and as close to the tabulated values as possible.

The mathematical device for doing this is called a Lagrange multiplier. The

technique of constrained minimization is outlined in The Appendix, (Constrained

Minimization and Lagrange Multipliers) and the theoretical development of the technique

is explained in many math texts (Thomas, 1968, p. 528; Kaplan, 1973, p. 184; Marsden

and Tromba, 1981, p. 217).

The mathematical procedure leads to our wanting to find values of α, β, γ, and 2V

such that the following function is a minimum:


21

( ) ( ) ( ) ( )

( ) ( )( )

2 2 2 2

2 2 2 2 2 212

' ' ' 2 '

1 cos 2

F Vz aVz

Vz

α α β β γ γ

λ α γ β β γ α

= − + − + − + − +

− − − + (2.45)

where the primes indicate measured or tabulated values and λ is the Lagrange multiplier.

Note that the term in square brackets is derived from Equation (2.44) by using a double-

angle trigonometric formula. Equation (2.44) is the constraint on the minimization.

Carrying through the mathematical procedure provides the following set of equations:

( ){ }( )( ){ }

( ){ }( )

( ) ( )( )

2 2 212

2 2 212

2 212

2 2 214

2 2 2 2 2 212

1 1 cos 2 ' 0

1 1 cos 2 ' 0

1 1 cos 2 ' 0

2 sin 2 2 ' 0

1 cos 2 0

Vz a

Vz

Vz

Vz Vz Vz

Vz

α λ γ β β

β λ α γ α β

α λ α β γ

λβ γ α

α γ β β γ α

+ − + + − =

− + − + − =

− − + − =

+ − − =

− − − + =

(2.46)

Equations (2.46) are a set of five equations in five unknowns, α, β, γ, 2Vz and λ,

that can be solved simultaneously. Unfortunately, the equations are nonlinear (i.e.

squares, higher powers and trigonometric functions of the unknowns are present in the

expressions) and their solution requires iteration techniques. As a result, their solution is

only feasible on a computer. Methods for achieving this are described by Burden, et al.

(1981, Chapter 9). The important point to remember is that we can, in theory and

practice, obtain consistent values of the optical parameters for entering into our

equations. The program, Optics.exe, used to calculate the optical properties in sections

through biaxial crystals contains a subroutine that ensures consistency of the primary

optical data, 2Vz, α, β, and γ.

Summary

We need the following data to locate the vibration directions in a random section

through a biaxial crystal: the direction cosines of the wave normal, w, 2V and the sign of

the crystal. The last two bits of data are equivalent to 2Vz. We first calculate the

components of u and v from Equations (2.2) and (2.4). Next we calculate the components

of t and s from Equations (2.8) and (2.9). Equation (2.13) is used to find sin θ. The


22

components of the first vibration vector are found by solving the pair of Equations (2.14)

and (2.15) or the pair (2.16) and (2.17). The components of the second vibration vector

are found by solving Equation (2.20). Next, we can calculate the refractive indices in a

particular section, N and M, with Equation (2.40) if we know the principal refractive

indices, α, β, and γ. The problem of locating the optic axes from extinction angles

illustrates the importance of removing the random element from optical observations. To

locate the optic axes from extinction angle measurements, data collected from at least

four different randomly chosen sections are needed unless the sections are special ones.

For example, optic axis or bisectrix sections are sufficient in themselves, but such

sections cannot be randomly chosen every time. Least squares methods can be applied to

extinction vector and refractive index data collected on a spindle stage. Best fit locations

of the optic axes and of the principal refractive indices can be extracted with these

methods. Theoretically consistent optical parameters can be objectively obtained from

optical measurements with constrained minimization using Lagrange multipliers.

Problems

1. The reported value of 2V for an orthopyroxene having a composition of En50 is 56°. The sign of the mineral is negative. Calculate unit vectors along the optic axes. Use the indicatrix as the primary frame of reference.

2. Jadeite has an extinction angle of Z^c = 40° in a section parallel to (010). In a reference frame with the following unit vectors:

z parallel to the c-axis y parallel to the b-axis x perpendicular to (100)

The optic axial vectors, u and v, are given by: u = 0.9588 x + 0.2840 z v = 0.1309 x + 0.9914 z What is the sign and optic angle of jadeite?

3. A positive biaxial mineral has an interaxial angle, 2V, equal to 60°. What are the vibration vectors associated with the following wave normal vector?

( )33

= w i + j + k


23

4. The principal refractive indices for the crystal in problem 3 are: α = 1.6000 β = 1.6061 γ = 1.6250

Are the refractive indices consistent with a 2V of 60°?

5. What are the refractive indices associated with each vibration vector in problem 3? What interference color would by produced by a crystal plate 0.03 mm thick?

6. The following refractive indices are reported for olivine and plagioclase. What are the values of 2V that agree with these data? What is the standard error on each optic angle? Assume each refractive index was obtained independently of the others.

RI Olivine Std. Error Plagioclase Std. Error α 1.779 0.001 1.545 0.001 β 1.815 0.002 1.557 0.001 γ 1.827 0.001 1.561 0.001

Explain why there is a difference in the standard errors on the optic angles.

7. The following refractive indices were measured at three orientations of a crystal mounted on a spindle stage. What are the principal refractive indices of the crystal?

S E RI 0.00 30.00 1.556

30.00 50.00 1.568 110.00 20.00 1.602

The spindle stage coordinates of the principal vibration vectors are: S E

X 30.00 10.00 Y 20.00 99.85 Z 110.30 91.70

24

Chapter 3: Optical Orientation of Biaxial Crystals

Introduction

Our objective in this chapter is to obtain equations that relate the indicatrix to the

crystal lattice; this relationship is called the optical orientation of the crystal. The reason

for specifying the optical orientation is that mineral identification, estimation of

compositions of solid solutions and characterization of structural states by optical

methods often depend on measuring an optical property in a particular section cut through

a crystal. This optical property, for example, could be an extinction angle to a cleavage

trace or it could be a refractive index in a plane identified by its crystallographic

orientation such as a plane parallel to the lattice plane (hkl). In order to measure the

optical property with respect to the crystallographic feature, the relationship between the

crystal’s lattice and its indicatrix must be known. The complexity of the relationship

depends on the crystal system; the lower the symmetry of the crystal system, the more

complex the relationship can be. What we have done so far towards specifying the optical

orientations is to find equations that describe directions in the indicatrix without any

attempt to take into account crystallographic directions. Because the optical properties are

our main concern, the indicatrix will remain our primary frame of reference. This means

that the unit vectors, i, j, k, will continue to be parallel to X, Y, and Z of the indicatrix. To

obtain the optical orientation for a crystal we will have to mesh two frames of reference,

the indicatrix and the crystal lattice. Ultimately we will have to find the components of

the crystallographic vectors along the indicatrix axes. Suppose we know a vector that is

described in terms of the crystallographic vectors, , , and a b c . For example:

u v w = + +d a b c (3.1)

where u, v, w, are the indices for a crystallographic direction, [uvw], and a, b, and c, are

vectors parallel to the unit cell edges. We want to write an expression for d in the

following form:

d = d1 i + d2 j + d3 k (3.2)


25

where the di are the components of d along the axes of the indicatrix and d is a unit

vector. The vector d, described by Equation (3.1) need not be a unit vector. It is much

more convenient to mesh two frames of reference if both of them are right hand Cartesian

systems. Because monoclinic and triclinic lattices are not Cartesian, this convenience is

not immediately available. To obtain this convenience, we will find expressions for the

components of the unit cell vectors, a, b, c, along the axes of a secondary Cartesian

system that we will label with axes x, y, and z. The specification of the optical orientation

of a crystal will then be a matter of stating the relationship between the frame of

reference described by x, y, z and the frame of reference described by X, Y, and Z, the

indicatrix.

Orthorhombic Orientations

Symmetry requirements limit the variants on the optical orientation of

orthorhombic crystals to six. The principal vibration directions must parallel the

crystallographic axes (Nye, 1957, Chapter 13). Complications arise only when we label

the positive and negative ends of the axes. Otherwise the two frames of reference, the

indicatrix and the crystal lattice, coincide. In three of the six possible cases, the positive

end of an axis in one reference frame, say the indicatrix, must coincide with a negative

end of an axis in the other reference frame, the crystal lattice, in order to ensure that both

reference frames maintain right hand orientations. Only one axis in one reference frame

need be given a negative sign in order to maintain a right hand orientation; which axis we

choose to label with the negative sign is arbitrary. The details of the six orientations we

will use are given in Table 3-1; the scheme was chosen because of its apparent symmetry.

Note that apart from sign, the secondary frame of reference coincides with the primary

one.

Table 3-1: Orthorhombic optical orientations. The notation was chosen to preserve right hand orientations in both frames of reference.

Vibration Parallel set of crystallographic axes Direction Case 1 Case 2 Case 3 Case 4 Case 5 Case 6

X a c b -a b c Y b a c c -a b Z c b a b c -a


26

The concern for maintaining a right hand orientation in our frames of reference is

forced on us by the cross product. If we change the hand of our frame of reference, we

change the sign of the cross product (Hoffmann, 1975, p. 103). Because we use the cross

product in calculations in the indicatrix and, as we shall see, in the crystal lattice it makes

it worth our while to maintain right hand reference systems.

It remains to express the components of the unit cell vectors in terms of the

indicatrix unit vectors, i, j, and k. These are set down in Table 3-2 and illustrated on

Figure 3-1.

Monoclinic Orientations

The symmetry of the monoclinic system requires that one principal vibration

direction coincide with the two-fold axis of symmetry (see Nye, 1957, chapter 13).

Mineralogists almost exclusively use the second setting for monoclinic crystals and thus

label the two-fold symmetry axis the b-axis (an exception is Smith, 1982, Chapter 4). In

conformity with nearly everyone, we will also label the symmetry axis the b-axis. The

most common optical orientation in minerals is shown in Figure 3-2 where Y is parallel

to the b-axis. To complete the specification of an optical orientation in monoclinic

crystals, an extinction angle in (010) must be given. For the first case illustrated on

Figure 3-2, with Y parallel to b, the extinction angle we will recognize is between Z and

the c-axis of the crystal. By convention, this extinction angle, labeled θ on Figure 3-2,

will be assigned a positive sign if Z lies in the obtuse angle, β°, between the positive a-

Table 3-2: Components of the unit cell vectors along the indicatrix axes for the six optical orientations in orthorhombic crystals. Unit Cell

Vector Indicatrix Components

Unit Cell Vector

Indicatrix Components

i j k i j k a a 0 0 a -a 0 0 Case 1 b 0 b 0 Case 4 b 0 0 b c 0 0 c c 0 c 0 a 0 a 0 a 0 -a 0 Case 2 b 0 0 b Case 5 b b 0 0 c c 0 0 c 0 0 c a 0 0 a a 0 0 -a Case 3 b b 0 0 Case 6 b 0 b 0 c 0 c 0 c c 0 0


27

axis and the positive c-axis and is less than 90°. Consequently, the extinction angle will

lie in the range: 2 2π πθ− ≤ ≤ . The convention regarding the sign of the extinction angle, θ,

in (010) is not consistent in the literature. Bloss (1961, p. 227), for example, describes the

extinction angle as we do here. Phillips and Griffen (1981), on the other hand, assign a

negative value where we use a positive one.

The other two orientations, X parallel to the b-axis and Z parallel to the b-axis, are

also illustrated on Figure 3-2. The extinction angles in (010), labeled θ in Figure 3-2, are

measured positive if the appropriate principal vibration direction lies in the interaxial

angle, β°. Again, we have ensured the frames of reference maintain right hand

orientations, this time by the appropriate definition of the extinction angles.

We next define vectors parallel to the crystallographic axes, each with a

magnitude equal to the corresponding length of the unit cell edge. We want to find the

components of these vectors that are along the indicatrix axes. For the case, Y parallel to

b, shown on Figure 3-1, the crystallographic vector along the Y-axis will be given by:

b=b j (3.3)

We can use the following systematic procedure to find the components of a and c.

First, from the sketch on Figure 3-2, we note that the following product equations are

valid:

cosc θ• =c k (3.4)

sinc θ× =c k j (3.5)

Next express both equations in component form:

3c• =c k (3.6)

2 1c c× = −c k i j (3.7)

Equation (3.7) can be derived from the expansion of the cross product as a determinant

(see The Appendix). Equating components in Equations (3.4) and (3.6) with the like

components of Equations (3.5) and (3.7) provides the expressions for the components of

c:


28

1 sinc c θ= − (3.8)

2 0c = (3.9)

3 cosc c θ= (3.10)

Using these components, the equation for c in the indicatrix frame of reference becomes:

( )sin cosc θ θ= − +c i k (3.11)

Starting with the equations relating c and a:

cosac β• =c a � (3.12)

sinac β× =c a j� (3.13)

we can substitute the components of c from Equations (3.8)-(3.10) and arrive at the

following result after some algebra with trigonometric identities:

( ) ( )sin cosa β θ β θ = − + − a i k� � (3.14)

These results and the representation of the monoclinic lattice vectors in the other two

monoclinic frames of reference are set down in Table 3-3.

Table 3-3: Components of the lattice vectors along the indicatrix axes for the three possible monoclinic optical orientations. The extinction angle in (010), θ , is positive if the indicated principal vibration direction lies in the interaxial angle, β � .

( ) ( )

[ ]

Case 1:

sin cos

sin cos

Y

a

b

c

β θ β θ

θ θθ

= − + − == −=

b

a

b

c

c

i k

j

i + k

k ^

� �

�

( ) ( )

[ ]

Case 2:

cos sin

cos sin

X

a

b

c

β θ β θ

θ θθ

= − − == −

b

a

b

c

c

j + k

i

j k

= j^

� �

�

( ) ( )

[ ]

Case 3:

cos sin

cos sin

Z

a

b

c

β θ β θ

θ θθ

= − + − == −=

b

a

b

c

c

i j

k

i j

i ^

� �

�


29

Triclinic Orientations: Euler Angles

The optical orientations of triclinic crystals are the most complex, both to

visualize and to describe mathematically. The complexity arises because triclinic crystals

have insufficient symmetry to require a coincidence of vibration direction and lattice

direction and because the triclinic lattice is not even partly Cartesian. Consequently, there

is only one case for the optical orientation of triclinic crystals as compared to six for

orthorhombic and three for monoclinic crystals. The angles between the indicatrix axes

and the lattice vectors, however, can be any value.

There are a large number of ways to mathematically describe the optical

orientation of triclinic crystals. We will use a scheme based on three angles, called Euler

angles. To define these angles, we proceed in the following fashion. First we will define a

right hand frame of reference to which we can refer the lattice vectors, a, b, c. The axes

of this frame of reference we will label x, y, and z, and call it the xyz reference system.

Note the use of lower case letters for this secondary frame of reference. The primary

frame of reference, the indicatrix, has capital letters for labels on the axes. Unit vectors

along the axes (x,y,z) of this frame of reference will carry the labels x, y, and z. For

example, the lattice vector parallel to the a-axis will be expressed as:

x y za a a= + +a x y z (3.15)

Second, we will relate the xyz frame of reference to the frame of reference defined by the

indicatrix. This will result in our being able to express each of the unit vectors, x, y, z, in

terms of the unit vectors along the primary frame of reference, i, j, k. For example, the

unit vector along the x-axis will be expressed in an equation of the form:

1 2 3x x x= + +x i j k (3.16)

where the components of x will consist of trigonometric functions of the three Euler

angles. We can then substitute Equation (3.16) and two more like it, one for y and one for

z, into Equation (3.15). On doing so we will have specified the lattice vector, a, in terms

of components along the indicatrix axes. If we do the same thing for b and c, then we will

have specified the optical orientation of the triclinic crystal.

The choice of the xyz frame of reference is arbitrary and limited only by the


30

requirements of a right hand orientation and Cartesian axes. If we have so much

discretion in choosing the xyz system, we might as well make it as convenient as

possible. It would be satisfying if the xyz frame of reference would reduce to the

monoclinic and orthorhombic frames of reference with an increase in symmetry of the

crystal lattice. Such a frame of reference is one with its axes parallel to [001], normal to

(010) and normal to [001] in the (010) plane. These axes are the ones chosen by Burri

(1956) to describe the optical orientations in the plagioclase feldspars. Notice particularly

that these axes are at right angles to each other, even in triclinic crystals, and that they

coincide with three important twin axes in the feldspars: the Carlsbad twin axis, the

Albite twin axis, and the Roc Tourne (or Carlsbad-Albite) twin axis. Having followed the

lead of Burri (1956) this far, we will continue to do so and label the axes the way he did:

x = [001]

y = Normal to [001] in (010)

z = Normal to (010)

With this choice of labels, we have a right hand frame of reference and can see

immediately that:

c=c x (3.17)

The a-axis lies in the (010) plane and, by definition, lies in the x-y plane of our coordinate

system. Hence, a has no component along the z-axis which is normal to (010). The lattice

vectors, a and c, are related by:

cosac β• =c a � (3.18)

sinac β× =c a z� (3.19)

Expanding the products in terms of components along the xyz axes and equating like

components in the same way we treated vectors in the monoclinic cases gives:

cos sina β β = + a x y� � (3.20)

In order to find the components of b along the axes of the xyz frame of reference, we can

use the following equations, treating the components of b as unknowns:


31

cosbc α• =b c � (3.21) cosab γ• =a b � (3.22)

2b• =b b (3.23)

Solving for the components of b by substituting the components of a and c calculated

above, we finally arrive at the expression for b in terms of components along the xyz

frame of reference:

2 2 2

2

cos cos coscos

sin

1 cos cos cos 2cos cos cos

sin

b b

b

γ α βαβ

α β γ α β γβ

−= + +

− − − +

b x y

z

� � �

�

�

� � � � � �

�

(3.24)

Equations (3.17), (3.20) and (3.24) complete the first half of our program of

writing equations for the lattice vectors in terms of components along the indicatrix axes.

What remains is to relate the unit vectors, x, y, z, to the indicatrix axes.

To complete the set of equations needed to specify the optical orientation of

triclinic crystals we first look in detail at Euler angles. Euler angles are usually described

as the angles through which one set of axes must be rotated to bring it into coincidence

with a second frame of reference (Bloss, 1981, Chapter 7). If we are trying visualize how

the indicatrix is sitting in the crystal lattice, rotations may not be concepts of primary

interest. Consequently, as an alternative representation, we will present Euler angles as

elements of vector equations and try to visualize the orientation of the two frames of

reference without recourse to thinking about rotations. The relationships between our two

frames of reference and the Euler angles are shown in stereographic projection in

Figure 3-3A.

We can define the first Euler angle, Θ, by the equation:

cos• = Θx i (3.25)

That is, Θ is the angle between the positive ends of the x-axis and the X-axis. Next we

can define a vector by:

sin= × ΘE x i = e (3.26)


32

Note that the magnitude of E is sin Θ. The vector E is parallel to the line of intersection

of the Y-Z and y-z planes of the two coordinate systems. Bloss (1981) refers to this line

as the nodal line. We will in like fashion call E the nodal vector. Being a vector, E has a

definite direction and is not the same vector as that produced by reversing x and i in

Equation (3.26) nor is it equal to the vector produced by replacing Θ by: -Θ.

Consequently Θ is positive if measured in a counter clockwise direction from x towards

X.

The second Euler angle, Φ, will be defined by the equation:

sinE× = ΦEy x (3.27)

Substitution of the magnitude of E from Equation (3.26) into Equation (3.27) gives:

sin sin× = Θ ΦEy x (3.28)

The cross product is used to define Φ because a sign is given to the angle. Φ is positive

when measured in a counter clockwise direction from y (see Figure 3-3).

The third Euler angle, Ψ, is defined by the equation:

sin sin× = Θ ΨE j i (3.29)

Ψ is positive when measured from E towards the Y-axis in a counter clockwise

direction.The Z-axis is located another 90° beyond the Y-axis and, as a result, is related

to E by:

( )2sin sin π× = Θ Ψ +E k i (3.30)

or:

sin cos× = Θ ΨE k i (3.31)

Likewise, the z-axis is related to E by:

( )2sin sin π× = Θ − ΦE z x (3.32)

or:

sin cos× = Θ ΦE z x (3.33)


33

Equations (3.26)-(3.33) are the raw material for constructing the components of the unit

vectors, x, y, z, along the axes of the indicatrix. First, substitute ×x i for E in Equations

(3.29) and (3.31) and use the vector identity (Hoffmann, 1975, p. 80):

( ) ( ) ( )× × = • − •A B C A C B A B C (3.34)

to get:

( ) ( ) sin sin• • = Θ Ψx j i - i j x i (3.35)

and:

( ) ( ) sin cos• • = Θ Ψx k i - i k x i (3.36)

Because the dot product of a pair of normal vectors is zero, the last terms before the equal

signs in Equations (3.35) and (3.36) vanish. Writing x in component form, taking the

indicated dot products and equating the coefficients of i in the last two equations gives:

2 sin sinx = Θ Ψ (3.37)

3 sin cosx = Θ Ψ (3.38)

From Equation (3.25) we get immediately:

1 cosx = Θ (3.39)

Thus, in indicatrix components, the equation for x is:

cos sin sin sin cos= Θ + Θ Ψ + Θ Ψx i j k (3.40)

To calculate the components of y and z we have to solve two pairs of vector

equations. The components of y can be obtained from:

( ) sin sin× × = Θ Φy x i x (3.41)

and:

( ) sin cos• × = Θ Φy x i (3.42)

The components of z can be found by solving the pair of equations:


34

( ) sin cos× × = − Θ Φz x i x (3.43)

( ) sin sin• × = Θ Φz x i (3.44)

The solutions of the last four equations for the components of y and z involve a

considerable amount of algebra and the details are left as exercises. The results are

included in the complete set of equations for specifying the optical orientation of triclinic

crystals set down in Table 3-4.

Table 3-4: Equations for representing the lattice vectors of triclinic crystals in the frame of reference defined by the indicatrix. As an intermediate step, the lattice vectors are represented by components in a frame of reference with the x-axis parallel to [001], the y-axis normal to [001] in the plane (010) and the z-axis normal to (010). Equations of the unit vectors along the xyz axes as functions of the Euler angles complete the specification.

Lattice Vectors in the xyz Frame of Reference

cos sina β β = + a x y� �

2 2 2

2

cos cos coscos

sin

1 cos cos cos 2cos cos cos

sin

b b

b

γ α βαβ

α β γ α β γβ

−= + +

− − − +

b x y

z

� � �

�

�

� � � � � �

�

c=c x

Unit Vectors Along the xyz Axes in the Indicatrix Reference Frame

cos sin sin sin cos= Θ + Θ Ψ + Θ Ψx i j k

( )( )sin sin cos cos cos sin sin

cos sin cos sin cos

= Θ Φ + Φ Ψ − Θ Φ Ψ

Φ Ψ + Θ Φ Ψ

y i j -

k

( )( )sin cos sin cos cos cos sin

sin sin cos cos cos

= − Θ Φ + Φ Ψ + Θ Φ Ψ

Φ Ψ + Θ Φ Ψ

z i j -

k

Euler Angles from Spindle Stage Measurements

Generally, the optical orientations of monoclinic and triclinic crystals can be

precisely determined only with a combination of single crystal x-ray and spindle stage

studies (see Bloss, 1981, chapter. 7). In special cases, the optical orientations of crystals

with two optically recognizable crystal surfaces (cleavages, composition planes, crystal


35

faces) can be determined with a combination of spindle stage and x-ray powder

diffraction data. Specifically, we need to know the unit cell constants and the locations, in

the spindle stage reference frame, of the poles to two lattice planes. Because the Euler

angles we use to describe the optical orientation of triclinic crystals refer to a reference

frame parallel to three important twin axes in feldspars, feldspars are a convenient

mineral group on which to demonstrate the method. Equations for a more general case are

given subsequently.

Given the sign of the mineral, spindle stage studies locate unit vectors along the

indicatrix axes (see Chapter 2: Location of the Optic Axes, page 5). To locate the unit

vectors in the auxiliary reference frame (xyz), we need the locations of the poles to the

feldspar cleavage forms {010} and {001}. We will designate unit vectors normal to the

(010) and (001) cleavages by p and q. The location of these vectors in the spindle stage

reference frame can be determined optically. The first step is to locate the a-axis of the

crystal. We begin by determining the interfacial angle between the cleavages from the dot

product of p and q:

cosε• =p q (3.45)

where ε is the angle between the cleavages. The cross product of p and q can now be used

to provide an expression from which a unit vector along the a-axis can be calculated:

sinε× =p q a (3.46)

where a is a unit vector parallel to the a-axis of the unit cell. Next we locate the c-axis.

The interaxial angle, β°, between the a-axis and the c-axis of the crystal lattice is needed

for the calculation. This angle can be obtained from powder diffraction data that has been

refined for the unit cell parameters. The (010) plane is normal to p and contains both the

a-axis and the c-axis. Consequently, we can locate the c-axis by solving the set of

equations (see Appendix: Solution of the Product Equations, page 96):

sin β× =c a p� (3.47)

cos β=c • a � (3.48)


36

By definition, the x-axis of the auxiliary reference frame is parallel to c so we

immediately know x and can calculate the first Euler angle, Θ, from:

cos• = Θx i (3.49)

Remember, i is a unit vector we locate from the spindle stage data, given the optic sign of

the crystal (see Chapter 2: Location of the Optic Axes, page 5).

The y-axis of the auxiliary reference frame is normal to the c-axis (i.e. x) and to

the pole to (010), p. The x-axis and the pole to (010) are also perpendicular.

Consequently a unit vector parallel to the y-axis can be calculated from:

×p x = y (3.50)

The nodal vector, E, can be located because we know the locations of i, x and the value

of the first Euler angle, Θ:

sin= × = ΘE x i e (3.51)

The magnitude of the second Euler angle, Φ, can next be calculated from the expression:

sin cos• = Θ Φy e (3.52)

The magnitude of the third Euler angle, Ψ is then calculated from the expression:

sin cos• = Θ ΨE j (3.53)

Again remember that j is a unit vector that results from the spindle stage data (see

Chapter 2: Location of the Optic Axes, page 5).

In summary, the calculation of the Euler angles for minerals with (010) and (001)

cleavages proceeds as follows:

1. Calculate the unit vectors normal to the cleavages (010) and (001), p and q, from their spindle stage angles, S and E.

2. Calculate the interfacial angle, e, with the dot product of p and q [Equation (3.45)].

3. Locate the unit vector parallel to the a-axis, a, with the dot and cross products of p and q [Equations (3.45) and (3.46)].

4. Calculate the unit vector parallel to the c-axis, c, using the interaxial angle β° in the dot and cross products of c and a. The x-axis of the auxiliary reference frame is parallel to [001], consequently c is parallel to x.


37

5. The first Euler angle, Θ, is calculated with the dot product of i and x; unit vectors parallel to the X vibration direction and the x-axis.

6. Calculate the unit vector parallel to the y-axis of the auxiliary reference frame with the cross products of p and x [Equation (3.50)].

7. Calculate the nodal vector, E, from the cross product of x and i [Equation (3.51)].

8. Calculate the second Euler angle, Φ, with the expression for the dot product of E and y [Equation (3.52)].

9. The last Euler angle, Ψ, is calculated with the expression for the dot product of E and j [Equation (3.53)].

This procedure for obtaining Euler angles does not produce unique results. The

reason for this will be apparent to those familiar with the problem of positive and

negative extinction angles on determinative curves. Because of the centrosymmetric

properties of the usual optical observations, we cannot always distinguish the positive and

negative sides of a crystallographic plane. Although the intersection of (010) and (001)

defines the a-axis we cannot be sure that Equation (3.46) actually defines the positive a-

axis. Another way of expressing the same dilemma is illustrated on Figure 3-4. The cross

product of p and q define a unit vector normal to the plane that contains them, a. Without

additional data, however, we do not know in which direction along the great circle

representing the trace of (010) to go to locate c. Is it at c1, or c2? The procedure which we

have outlined cannot distinguish between these possibilities and we must turn to

mineralogy texts for information that designates the correct orientation. The monograph

by Burri, et al. (1967) will give sufficient information to choose the correct orientation

for plagioclase. For other minerals, Deer, et al. (1966) or Bambauer, et al. (1979) are

likely sources of information.

The optically recognizable crystal surfaces in minerals other than feldspars may

not be parallel to (010) and (001). However the Euler angles refer to a Cartesian reference

frame in which c and x coincide and in which z is normal to (010) (i.e. z equals p in the

feldspar example). Consequently, the Euler angles can be calculated only after c and z are

located [see Equations (3.49)-(3.53) with z substituted for p and c for x]. It remains then,

to locate c and z, given the locations of two unit vectors, p and q, normal to two crystal

surfaces, (h1k1l1) and (h2k2l2).

Again, we begin by calculating the interfacial angle between p and q:


38

cosε• =p q (3.54)

Next, locate a vector along the line of intersection of (h1k1l1) and (h2k2l2). Such a vector

can be obtained from p and q:

sinε× =p q s (3.55)

In theory, another vector parallel to s can be constructed from the unit cell properties of

the mineral:

S = u a + v b + w c (3.56)

where a, b, and c are vectors along the unit cell axes and u, v, w are given by:

1 2 2 1

2 1 1 2

1 2 2 1

u k l k l

v h l h l

w h k h k

− = − −

(3.57)

The two vectors, s and S are related by:

=•

S

S Ss (3.58)

The magnitude of S, •S S , can be calculated from powder diffraction data. To do this,

form the dot product of S with itself [i.e. use Equation (3.56)]:

2 2 2 2 2 2u a v b w c 2uvab cos 2uwaccos 2vwbc cosγ β α• = + + + + +s s � � � (3.59)

Next, we note that our unknown vector, z, is related to the unit cell vectors by:

cosacsin

µβ

×= =c az

�

(3.60)

At this point we are in a position to calculate the angle, µ, between z and s by using

Equations (3.56), (3.58) and (3.60). After some algebra, we get:

cos acsin

V µβ

• = =•

z ss s �

(3.61)

where V is the unit cell volume [see Bloss, 1971, p. 501]. Notice particularly that

everything between the equal signs in Equation (3.61) can be evaluated from powder


39

diffraction data that has been refined for the unit cell parameters. Consequently, µ can be

calculated.

The next step is to locate z in the spindle stage reference frame. To do this we will

need the angles between z and p and between z and q. Equations for calculating

interfacial angles, which are equal to the angles between face normals, are given in

crystallography texts (see Bloss, 1971, p. 64). The data required for the calculations are

the unit cell parameters and the Miller indices of the faces, (h1k1l1), (h2k2l2) and (010).

Alternatively, you can read ahead in Chapter 4 and find out how to construct vectors

normal to a crystallographic plane. The dot product will then produce the required angles.

In any case, given the unit cell parameters, we can calculate the angles, δ1 and δ2, that

satisfy the relationships:

1cosδ• =z p (3.62)

2cosδ• =z q (3.63)

Substitution of Equation (3.55) into the cross product of z and s gives:

( )sinε

× ×× =

z p qz s (3.64)

This last expression can be expanded to give:

( ) ( )

sinε• − •

× =z q p z p q

z s (3.65)

Substitute Equations (3.62) and (3.63) into (3.65):

2 1cos cos

sin

δ δε

−× p qz s = (3.66)

Equation (3.66) plus the dot product of z and s [Equation (3.61)] are sufficient for

locating z. The procedure is a slight modification of that described in The Appendix

(Solution of the Product Equations, page 96). Hence, we have located z.

To locate c, which is parallel to x, we form an auxiliary vector R along the

intersection of (010), the plane normal to z, and (h1k1l1), the plane normal to p. By

treating R in the same way as we did S [Equations (3.56)-(3.59)] we can locate R and


40

calculate the angle between R and c. This latter value is given by:

1 1l a cos h ccos

βχ −=•R R

�

(3.67)

where:

2 2 2 21 1 1 1l a h c 2h l accos β• = + −R R � (3.68)

Finally, the components of c are calculated by solving the pair of equations:

sin χ× =c r z (3.69)

cos χ• =c r (3.70)

After locating c and z, the Euler angles are calculated as described earlier for the

feldspars. Again, there is an ambiguity in the results; on which side of r does c lie in the

(010) plane? Recourse to a mineralogy text must be made to determine which is correct.

A stereographic projection of the general case is shown in Figure 3-5. The location of c

can be either parallel to c1 or c2.

The detent spindle stage (Bloss, 1981) is not constructed to provide accurate

measures of the S angles except at even 5° intervals. Consequently, accurate locations of

p and q are difficult to attain with the spindle stage. Powder diffraction data, on the other

hand, give precise estimates of the unit cell parameters and from which a precise value of

the interfacial angle between (h1k1l1) and (h2k2l2) can be calculated. In other words, the

value of ε in Equation (3.46) or (3.54) is more precisely determined from the x-ray data

than from the spindle stage data. To take advantage of this precision, one can calculate

new locations for p and q that are as close as possible to the spindle stage locations but

satisfy the interfacial angle determined by the x-ray data. This is done by employing the

Lagrange multiplier technique (Appendix, page 98). To do this, minimize the expression:

( ) ( ) ( ) ( )F = − • − + − • −p p' p p' q q' q q' (3.71)

where p and q are the new, relocated, vectors and p' and q' are their locations from the

spindle stage data, subject to the constraints:


41

1• =p p (3.72)

1• =q q (3.73)

cosε• =p q (3.74)

Summary

Tables 3-2, 3-3, and 3-4 contain the expressions for writing down the lattice

vectors as functions of the unit vectors along the indicatrix axes. To use the resulting

equations we must have recourse to some primary optical data such as compiled by Deer,

et al. (1966) or by Bambauer, et al. (1979). If the mineral is orthorhombic all we need is

information as to which of the six possible orientations is correct for the mineral under

consideration. For monoclinic minerals, we need to find out which of the principal

vibration directions parallels the b-axis and what is the appropriate extinction angle in

(010). Often the extinction angle between the c-axis and the principal vibration direction

listed in Table 3-3 is not specifically given in the compilations of data. In such case, you

will need to calculate it from the extinction angle given and, if necessary, the interaxial

angle, β°. To date, Euler angles are available only for the plagioclase feldspars. The

primary reference to these data is Burri et al. (1967). Diagrams constructed with these

data are in Bambauer et al. (1979). It’s likely that future compilations will take advantage

of the precision offered by Euler angles to specify the optical orientations of triclinic

crystals.

Euler angles can be obtained from spindle stage and powder diffraction data if the

crystal has two optically recognizable crystal surfaces (cleavages, composition planes,

crystal faces). The Lagrange multiplier technique allows one to specify the locations of

unit vectors normal to the crystal surfaces that are consistent with the x-ray data.


42

Problems

1. An orthorhombic mineral with axial ratios 0.5:1:0.3 shows an optic axis figure with a negative sign on a section cut parallel to (110). What is the optic angle? What are the components of the wave normal vector in the indicatrix reference frame?

2. A monoclinic mineral is known to have the following properties: a:b:c = 0.5:1:0.6, β° = 120° (101), ( 101) and (010) cleavages. On one cleavage fragment a centered optic axis figure is displayed with a negative sign. On a second cleavage fragment a Bxo figure is displayed and on a third cleavage fragment, an optic normal figure is shown. What is the optical orientation of the mineral? Is your answer unique? If not, what measurements could you make to remove the ambiguities?

3. Derive Equation (3.24) from Equations (3.21)-(3.23).

4. Use the expression for x given by Equation (3.40) to find an expression for y from Equations (3.41) and (3.42).

5. Use the expression for x given by Equation (3.40) to solve Equations (3.43) and (3.44) for z.

6. Estimate values for the Euler angles of kyanite from the description of its optical properties in Deer, et al. (1992), p.59.

7. The spindle stage coordinates of the poles to the cleavages of a plagioclase are: S E (010) 114.0 56.0 (001) 96.0 141.0 If the spindle stage coordinates of the principal vibration directions are: S E X 16.1 55.7 Y 20.7 145.6 Z 107.5 87.8 what are the Euler angles for this crystal? A stereographic projection is one method for the obtaining the required values.

43

Chapter 4 - Extinction Angles in Biaxial Crystals

Introduction

An extinction angle is often one of the most diagnostic measurements we can

make to identify minerals in thin section. Witness the plagioclase feldspars whose

compositions we estimate with extinction angle measurements. Invariably, an extinction

angle is measured between a vibration direction and the trace of a crystallographic

surface in the plane of the thin section. Such features are commonly produced by

cleavage planes or the composition planes of twins intersecting the thin section. In this

chapter we will describe how the extinction angles in sections defined by a given

(known) wave normal can be calculated. The results of the calculations can then be used

to plot determinative curves of extinction angle versus composition or extinction angle

versus structural state. Several of the more common types of calculations have been

incorporated into the computer program, Optics.exe.

We begin by mathematically defining a representation of the trace of the crystal

surface in thin section, against which the extinction angle is measured. This trace can be

described as the line of intersection between a lattice plane, representing the

crystallographic surface, and a plane parallel to the thin section. If we represent the

normals to the thin section by the wave normal vector, w, and the lattice plane by d, then

a vector parallel to the trace of the crystal surface and parallel to the thin section is:

= ×Q dw (4.1)

If n is a unit vector parallel to a vibration direction in the plane of the thin section, then

the extinction angle, θ, will be implicitly defined by:

cosθ• = ×Q dn w (4.2)

Equation (4.2) can provide us with a value of the extinction angle if we know n,

w, and d. The relationship between n and w was covered in Chapter 2 and d can be

calculated if we can relate the normal to a lattice plane to the optical orientation of the

crystal. We covered optical orientations in the last chapter. The equations for representing

w and d as functions of the lattice vectors are the first subjects of this chapter.

Chapter 4: Extinction Angles in Biaxial Crystals

44

Our immediate problem is to find a representation for the wave normal as a

function of the lattice vectors. As far as the optics of crystals are concerned, the only

definitely recognizable sections in thin section are those showing an optic axis, a bisectrix

or an optic normal. Of these, only the optic axis sections are quickly found. Given the

limitations of locating a wave normal vector by optical criteria alone, it follows that we

generally identify a wave normal vector according to some set of crystallographic criteria

rather than optical ones. Judging by how most petrographers, both novice and

experienced, examine thin sections, this identification is often a subconscious activity. On

reflection, we identify sections, and what is the same thing, wave normal vectors, by

whether the section is normal to a lattice plane or normal to a lattice direction. So our first

task is to represent these directions as vectors.

Crystallographic Vectors

A vector parallel to a particular lattice direction, [uvw], is particularly easy to

write in terms of the lattice vectors, a, b, and c:

[ ]uvw u v w= + +d a b c (4.3)

[ ]uvwd can be reduced to a unit vector by dividing it by its magnitude.

To find a unit vector normal to a lattice plane with Miller indices (hkl) we proceed

as follows. The intercepts of the plane and the lattice vectors determine new vectors

parallel to the lattice vectors: a/h, b/k, and c/l. From these we can construct two more

vectors that lie in the lattice plane, such as (b/k - a/h) and (c/l - a/h). The relationships

between these vectors are shown on Figure 4-1. A vector normal to the lattice plane will

be parallel to the cross product of the two vectors:

( ) ( ) ( )hkl k h l h= − × −b a c ad (4.4)

Expanding this expression, we can write it as:

( )( ) ( ) ( )

hkl , , , 0h k l

h k lhkl

× + × + ×= ≠

b c c a a bd (4.5)


45

If either the vector defined by Equation (4.3) or by Equation (4.5) is to represent a wave

normal vector, we will need to change it into a unit vector in conformity with our usage

in Chapter 2. To do this, simply divide d by the square root of the dot product of the

vector with itself:

=•

d

d dd (4.6)

Substitution of the appropriate expression for d [Equation (4.3) or Equation (4.5)]

into Equation (4.6) produces the required expression for defining a wave normal parallel

to [uvw] or normal to (hkl).

In thin section, wave normals parallel to [uvw] are recognized by the intersection

of two crystal surfaces, (h1k1l1) and (h2k2l2). The Equation Set (3.57), Chapter 3, relates

[uvw] to the Miller indices of the two planes. Those interested can form two vectors,

d(h1k1l1) and d(h2k2l2), with Equation (4.5) as a model, take the cross product and show

that it is a vector parallel to that defined by Equation (4.3). This exercise is also a

derivation of the Equation set (3.57) in Chapter 3.

Examples of sections, whose wave normals in orthoscopic light are parallel to a

lattice direction, [uvw], are sections normal to the c-axes of amphiboles and pyroxenes or

the a-axes of feldspars. The wave normal in sections cut perpendicular to the c-axis of

monoclinic amphiboles is parallel to [001], a direction defined by the cleavage faces in

the form {110}. A determinative curve for plagioclase feldspars showing extinction

angles between the fast directions and the trace of (010) in sections cut normal to the a-

axis is shown on Figure 4-2. The data for drawing Figure 4-2 were produced with the

computer program Optics.exe, using 2Vz, α, β and γ with Euler angles from Burri, et al.

(1967) as primary data.

As a practical matter, wave normal vectors perpendicular to a crystal surface,

(hkl), occur most frequently in grain mounts containing minerals with good cleavages.

The most obvious example is mica. Sprinkle some mica fragments on a slide and,

because of the perfect (001) cleavage, they will all lie with the cleavage parallel to the

stage. Consequently, the wave normal vectors, with orthoscopic illumination, will be

perpendicular to (001). Plagioclase feldspars have perfect (001) cleavage. Consequently,


46

grain mounts of plagioclase fragments contain several grains with wave normals

perpendicular to (001). Such grains can be recognized by sharp albite twin composition

planes and (010) cleavage traces. The extinction angle between the fast direction and the

a-axis [which lies in (001), remember] can be used to estimate the An content. A

determinative curve for this purpose is plotted in Figure 4-2 from data calculated with the

program Optics.exe.

Extinction Angles in Sections Normal to (hkl)

Most of the methods that use extinction angles to estimate plagioclase

compositions have as one of their first requirements that the thin section be cut normal to

the (010) plane. Because the albite twin is a normal twin, the composition planes, (010),

between lamellae are parallel to macroscopic twin planes. Consequently, any directional

property in one twin individual is related by mirror plane symmetry to the same

directional property in all individuals related to the first by the albite twin law. This

means that extinction angles in sections cut normal to (010) are of equal magnitude but

opposite direction (symmetrical) in lamellae related by the albite twin law. If the

extinction angles are not symmetrical either the twin is not an albite twin, or the thin

section is not normal to (010), or a mistake was made in measuring the extinction angles.

Symmetrical extinction angles, sharp composition planes and parallel (010) cleavage

traces that remain in focus on raising and lowering the stage ensure that the thin section is

cut normal to (010). None of these criteria are capable of specifying which direction in

(010) is parallel to our wave normal; rather, it is only one of an infinite set, all of which

lie in (010). The various extinction angle methods for estimating plagioclase

compositions complete the specification of the wave normal vector in different ways. The

Michel-Levy or maximum extinction angle method assumes the largest measured

extinction angle in the thin section is equal to the largest extinction angle possible in

(010). The a-normal method completes the specification by insisting that the thin section

be cut normal to both (010) and (001); hence, the wave normal is parallel to [100]. The

third commonly used method of estimating plagioclase compositions with extinction

angles measured in the zone normal to (010), the Carlsbad-albite method, locates the

particular wave normal in (010) by measuring the extinction angles in twin individuals


47

related by the Carlsbad twin law. In this section we will develop a systematic method for

calculating extinction angles in sections with wave normals that are evenly spaced within

a given lattice plane, (hkl). From these kinds of data one can construct determinative

curves for extinction angle measurements made in sections normal to (hkl).

First of all, we suppose that the lattice plane (hkl) is not perpendicular to the c-

axis [001]. Note that this supposition can be wrong only if the crystal is orthorhombic. In

monoclinic and triclinic crystals, [001] cannot be normal to a lattice plane (hkl);

otherwise it would no longer be monoclinic or triclinic. If our supposition is true, then the

plane normal to the c-axis and the plane (hkl) will intersect. A vector that lies in the plane

(hkl) and is normal to the c-axis, [001], is given by:

( ) [ ]hkl 001= ×P d d (4.7)

We next transform this vector into a unit vector:

=•

P

P Pp (4.8)

and define a set of wave normals by means of the equations:

( )hkl sinn nθ× =w p d (4.9)

cosn nθ• =w p (4.10)

where 1 n+10, nθ θ θ θ= = + ∆ .

By assigning a value to ∆θ we form a set of evenly spaced wave normals. Notice

that w1 is equal to p (i.e. θ1 = 0 so w and p are parallel vectors). If we let ∆θ be 10° and n

run from 1 to 19, the wave normals will be 10° apart over a range of 180°. The last wave

normal in the set, w19, will be equal to: − p . The set of wave normals all lie in the plane

(hkl) by virtue of the cross product, thus ensuring they are normal to d(hkl) which is itself

normal to the plane. One other convenience that results from defining the set of wave

normals in this way is that they plot along a great circle on an upper hemisphere

stereographic projection of a crystal in standard orientation (Figure 4-3. See also Bloss,

1971, p. 87-91). Finally, we point out that Equations (4.9) and (4.10) are another pair of

product equations whose method of solution is described in The Appendix (Solution of


48

the Product Equations, page 96).

If the plane in question is (001) and if the crystal is orthorhombic, our supposition

that (hkl) is not normal to [001] is false. In this case, we define a set of wave normals in

(001) by the equations:

[ ] [ ]100 100 sinn nθ× =w d d (4.11)

[ ]100 cosn nθ• =w d (4.12)

where, as before, 1 n+10, nθ θ θ θ= = + ∆ .

The Carlsbad-albite method (Tobi, 1963; Tobi and Kroll, 1975) of determining

plagioclase compositions is based on measurements of extinction angles for wave

normals that lie in (010). The method is arguably the most widely applicable, precise

method of determining plagioclase compositions in thin section by optical methods. It is

also the most complicated of the commonly used methods of determining plagioclase

compositions and requires the most insight into the theoretical basis for the method. The

a-normal method is just as precise as the Carlsbad-albite method but finding sections cut

normal to [100] is more difficult than finding sections cut normal to (010). The Carlsbad-

albite method is superior to the Michel-Levy method because no assumptions are made

about finding the maximum extinction angle and about uniform plagioclase compositions

in the thin section.

The Carlsbad-albite method could easily be called the Carlsbad method because it

is the Carlsbad twin that is required for the method to work. The albite twins provide

redundant information that can be used to check the reliability of the Carlsbad data but

extinction angles in adjacent albite twins are not required for the method to work.

Redundant data or information is seldom, if ever, useless. Redundant data or information

is always, or nearly always, necessary for estimates of reliability of an experimental or

analytical result.

Extinction angle curves in sections normal to (010) for three compositions of high

structural state plagioclase series are plotted in Figure 4-4. The plots were made with

results calculated by the program Optics.exe. Burri, et al. (1967) measured the Euler

angles required for the calculations. The signs (±) of the extinction angles in Figure 4-4


49

were assigned according to the location of the fast direction in a stereographic plot of the

crystal lattice in standard orientation (see Bloss, 1971, p. 87-91). If the fast extinction

direction falls between (010) and: − y , the extinction angle is positive. If the fast

direction falls between (010) and y, then the extinction angle is negative. This convention

conforms to the designation of signs of extinction angles in plots for determining

plagioclase compositions with the Carlsbad-albite method, in most instances (see Deer, et

al., 1992, p. 449). These curves are of no practical use, however, unless one knows the

angle of the wave normal in (010).

The curves plotted in Figure 4-4 provide insight into how extinction angles in

Carlsbad twins are related. If a plagioclase grain displays a Carlsbad twin, the individuals

forming the simple twin are related as shown in Figure 4-5. The twin axis [001] is normal

to the twin plane, a macroscopic mirror plane. The composition plane belongs to the form

{hk0} and is most commonly (010). If the c-axis is oriented vertically, then the twin

individuals are side-by-side in a horizontal arrangement (Figure 4-5). A thin section cut

perpendicular to (010) in such a grain will pass through both parts of the twin. One

individual will have a wave normal that falls between 90° (wave normal parallel to [001])

and 180° [wave normal ⊥[001]/(010)] on Figure 4-4, whereas the other individual will

have a wave normal between 0° [wave normal ⊥[001]/(010)] and 90° (wave normal

parallel to [001]). Consequently, the two individuals forming the Carlsbad twin will have

different extinction angles between the fast direction and the trace of (010). There are two

particular orientations for which this conclusion does not follow. First, if the thin section

is cut parallel to [001], but normal to (010), then the extinction angles in the individuals

will be symmetrical (Figure 4-5). Second, in thin sections cut perpendicular to the c-axis,

thus parallel to the Carlsbad twin plane, the extinction angles will be identical and the

trace of the composition plane will be invisible.

If extinction angles are measured in the two individuals of a Carlsbad twin, the

positions of the wave normals will be symmetrically displaced from the Carlsbad twin

plane (90° position, Figure 4-4). The lines marking the two extinction angles both cross

the same compositional contour at equal distances from the 90° position on the vertical

axis. An example is shown on Figure 4-4. Figure 4-4 would obviously be difficult to use


50

as a determinative diagram because interpolating between compositional contours and

finding points symmetrically placed is difficult when the extinction angles are the

available data.

Figure 4-6 has composition plotted against the position of the wave normal in

(010). This diagram shows the extinction angles, rather than compositions, as contours.

To use this diagram as a determinative chart, the measured extinction angles are

interpolated in the parts of the diagram above and below 90°. Again, the wave normal

positions must be symmetrical about 90° and single valued in composition. Finding the

appropriate composition on Figure 4-6 is still no walk in the park. If the diagram is

reflected at the position of the Carlsbad twin plane so that upper and lower parts of the

diagram are superimposed, then finding symmetrical position is simplified. The

geometrical relationship between the reflected diagram and the Carlsbad twin is shown

on Figure 4-7.

The modified diagram for disordered plagioclase is shown on Figure 4-8. To use

the diagram, the dashed contour representing the larger extinction angle is followed until

it crosses the solid contour that represents the smaller extinction angle. The point of

intersection locates the positions of the wave normals in (010). Points on the solid

contours lie between 0° and 90°, whereas points on the dashed contours fall between 90°

and 180°.

Construction of Figure 4-6 and Figure 4-7 requires extinction angles at fixed,

integer values. Numerical methods can supply the required data. Iterative methods can

provide the needed positions of the wave normals for the fixed values of the extinction

angles if initial estimates of the positions can be found. The routine to calculate extinction

angles at evenly spaced positions of the wave normals supplies the required estimates. A

routine exists in Optics.exe for doing the iterations to provide the wave normal position

at a fixed value for the extinction angle. Another method for obtaining the required

extinction angles and wave normal positions is to calculate the extinction angles for wave

normals parallel to lattice directions, [u0v], in (010). Some examples are shown on

Figure 4-9. An extinction angle is calculated for a wave normal for a particular lattice

direction and particular plagioclase composition. The wave normal will have a fixed


51

position in (010) if the mineral has a fixed composition and degree of order. The position

of the wave normal changes very little across the compositional range of disordered

plagioclase.

Extinction Angles in the Zone [uvw]

The characteristic extinction angle between the c-axis and a principal vibration

direction in the (010) plane of monoclinic crystals can be a diagnostic property if either

the c-axis or the a-axis can be optically recognized in sections cut parallel to (010) and if

sections cut parallel to (010) can be found in the microscope. In amphiboles and

pyroxenes, the cleavage form {110} contains the faces ( ) ( ) ( ) ( )110 , 110 , 110 , and 110 .

Their intersections are parallel to the c-axis. Hence, sections cut parallel to the c-axis will

display parallel traces of intersecting cleavage planes. Unfortunately, this is not a

sufficient criterion for ensuring that the plane of the thin sections is parallel to (010)

because all sections in the zone [001] will display these features. The identification of

sections parallel to (010) is usually made with an interference figure that shows a

centered bisectrix if the b-axis is parallel to X or Z or a centered optic normal figure if the

b-axis is parallel to Y. To use the characteristic extinction angle as a diagnostic property,

one must ensure that it is measured in (010). In the zone [001] the extinction angle in

sections parallel to (100) will be zero whereas in a section parallel to (010) it will be the

characteristic value, θ [Table 3-3]. Although it would be convenient if the maximum

extinction angle in the zone [001] occurred on (010), such need not be the case. This has

been known since Daly (1899) proved it by deriving a simple equation relating extinction

angles to the angle between the thin section and the pole to (100) in the zone [001]. Su

and Bloss (1984) provide a recent discussion of the problem.

Extinction angles to the c-axis in sections parallel to [001] are given by relatively

simple equations that are particular cases of the more general problem of locating

vibration directions in thin sections cut parallel to a zone axis [uvw]. Because a general

procedure for locating such vibration directions is very similar to that just described for

sections normal to (hkl), we will discuss the general case first. A detailed look at the

special monoclinic calculations will be presented later.


52

We want to find a set of wave normals that lie in a common plane. The plane

itself is identified as one perpendicular to a crystallographic direction [uvw]. If [uvw] is

not [001], the c-axis, then one way of defining the set is to first calculate the vector:

[ ] [ ]uvw 001= ×P d d (4.13)

and convert it to a unit vector:

=•

PP P

p (4.14)

We can then define the set of wave normals by the equations:

[ ]uvw sinn nθ× =w p d (4.15)

cosn nθ• =w p (4.16)

where 1 n+10, nθ θ θ θ= = + ∆ .

A schematic representation of this set of wave normals is shown in stereographic

projection in Figure 4-10. Again, the poles representing the wave normals all plot on a

great circle of the upper hemisphere projection of a crystal in standard orientation. Notice

again, that the method for finding solutions to Equations (4.15) and (4.16) for the

components of the wave normals is that outlined in The Appendix (Solution of the

Product Equations, page 96).

If [uvw] is [001] then an analogous result can be obtained by defining the set of

wave normals with the equations:

( ) [ ]100 001 sinn nθ× =w d d (4.17)

( )100 cosn nθ• =w d (4.18)

where 1 n+10, nθ θ θ θ= = + ∆ .

By definition, d(100) is normal to the (100) lattice plane. The c-axis, [001], must lie

in this plane, hence d[001] is normal to d(100). The reciprocal lattice vector, a*, is also

normal to the lattice plane (100) by virtue of the cross product. Consequently, d(100) and

a* are parallel. Because the standard stereographic projection has the a*-b* plane parallel


53

to the plane of projection, the poles representing the wave normals in Equations (4.17)

and (4.18) plot on the primitive (See Bloss, 1971, Chapter 13 for a description and

discussion of the nature of the reciprocal lattice).

Extinction in Sections Parallel to [001] for Monoclinic Crystals

We now take up the special problem of calculating extinction angles to the c-axis

in sections parallel to [001] of monoclinic crystals. We will first look at the problem for

the orientation Y = b (Case 1, Table 3-3). Until and unless explicitly stated otherwise, the

following discussion only applies to crystals with the Case 1 orientation. As mentioned,

Daly published a paper in 1899 relating the extinction angle to the c-axis in a section

parallel to [001] to its orientation with respect to (100). For this special case, the

machinery encapsulated in Equations (4.13)-(4.18) is not required. Because Daly’s

Equation has been lost or forgotten in the later literature, we will derive it anew even

though it is like reinventing the wheel. Daly used a geometrical construction to provide a

simple and elegant derivation of his formula. In keeping with the vector methods used in

this book, we will forgo the geometry even if our derivation is not so elegant. All of

which goes to show that vectors are not always the most economical mathematical tools

to use.

The most convenient reference frame in which to derive Daly’s Equation is the

auxiliary frame used to describe triclinic orientations with Euler angles (Chapter 3).

Monoclinic geometry in this reference frame produces the following equivalences:

=•

c

c cx (4.19)

( ) ( )100 , 100i.e.= ⊥y d

= =•

b

b bz j

These elements are shown in stereographic and orthographic projection in Figure 4-11.

The other elements used in the derivation are shown also.

The stereographic projection shows two great circles representing planes defined

by the wave normal, w, and optic axes vectors, u and v. According to the law of Biot-


54

Fresnel, the vibration vector, n, in the section normal to w bisects the angle between these

planes. The angle between the vibration vector, n, and the c-axis vector, c, is the

extinction angle δ:

cosδ• =n c (4.20)

In a similar fashion, the principal vibration vector, k, bisects the angle between the optic

axes vectors and lies at an angle, θ, to c (see Chapter 3). Our aim is to find a relationship

between δ, the optic angle (Vz) and the orientation angles, θ and φ (Figure 4-11).

As a first step, we find vectors in the plane normal to w that also lie in the planes

defined by w and the optic axes vectors. These are given by:

( )= × ×T w u w (4.21)

( )= × ×S w v w (4.22)

These triple products can be expanded to give:

( )= − •T u w u w (4.23)

( )= − •S v w v w (4.24)

Next, take the dot products with the unit vector along the c-axis:

( )( )• = • − • •Tc c u w u c w (4.25)

( )( )• = • − • •Sc c v w v c w (4.26)

However, c and w are orthogonal vectors, consequently the last terms in Equations (4.25)

and (4.26) are zero. To find the angles between c and T and between c and S, we divide

Equations (4.25) and (4.26) by the magnitudes of T and S, respectively:

( )21• = − •T T w u (4.27)

( )21• = − •S S w v (4.28)

The results are:


55

( ) ( )( )

2

22cos

1ε δ

•+ =

− •

c u

w u (4.29)

( ) ( )( )

2

22cos

1ε δ

•− =

− •

c v

w v (4.30)

where ε is the angle between the vibration vector, n, and the unit vector t or between n

and s, Figure 4-11. Equations (4.29) and (4.30) can be rearranged to give:

( ) ( ) ( )( )

2 2

22

1tan ε δ

− • − •+ =

•

w u c u

c u (4.31)

( ) ( ) ( )( )

2 2

22

1tan ε δ

− • − •− =

•

w v c v

c v (4.32)

This last transformation to the tangent function is a natural step in Daly’s derivation. Our

vector methods, on the other hand, provide no direction in this regard. Although our use

of the tangent function seems arbitrary at this point, it does lead to a simple result.

The next step is to evaluate the dot products. We have immediately from the

definitions:

( )cos Vz θ• = +c u (4.33)

( )cos Vz θ• = −c v (4.34)

The others are readily evaluated if we write the optic axes vectors in terms of c and y

(Figure 4-11).

( ) ( )cos sinVz Vzθ θ= + + +u c y (4.35)

( ) ( )cos sinVz Vzθ θ= − − −v c y (4.36)

We can quickly calculate from these last two equations the dot products with the wave

normal vector:

( )cos sin Vzφ θ• = +w u (4.37)

( )cos sin Vzφ θ• = −w v (4.38)


56

Substitution of Equations (4.33), (4.34), (4.37) and (4.38) into (4.31) and (4.32) gives:

( ) ( )tan tan sinVzε δ θ φ+ = + (4.39)

( ) ( )tan tan sinVzε δ θ φ− = − (4.40)

If we have the good fortune to notice that:

( ) ( )2δ ε δ ε δ= + − − (4.41)

then it is fairly obvious that we can use the trig identity:

( ) tan tantan

1 tan tan

α βα βα β−− =

+ (4.42)

to arrive at Daly’s Equation:

2

sintan 2

1 sin

A

B

φδφ

=+

(4.43)

where:

( ) ( )tan tanA Vz Vzθ θ= + − − (4.44)

( ) ( )tan tanB Vz Vzθ θ= + − (4.45)

Daly’s Equation, equation (4-43) shows that the extinction angle, �, to the c-axis

is a function of three variables: the position of the thin section in the zone [001], φ , the

optic axial angle of the crystal, Vz, and the extinction angle in (010), �. The optic axial

angle and the extinction angle in (010) are fixed quantities in a given monoclinic crystal

with a uniform composition and structural state. Consequently, we can use Daly’s

Equation to classify and characterize extinction angle curves for monoclinic crystals with

Case 1 orientations and to predict extinction angles on {hk0} cleavage forms.

In order to locate the maximum or minimum extinction angles in the zone [001]

for a crystal with known values for Vz and the extinction angle in (010), θ, we calculate

the derivative of δ with respect to φ and set the result equal to zero. This operation

requires some algebra but the required derivative is given by:


57

( )

( )2

22 2 2

cos 1 sin

2 1 sin sin

A Bd

dB A

φ φδφ φ φ

+=

+ +

(4.46)

This expression will be zero whenever at least one of the expressions in the numerator is

zero, subject to the proviso that the denominator is not simultaneously zero. The factors

of the numerator are:

( ) ( )tan tan 0Vz Vzθ θ+ − − = (4.47)

cos 0φ = (4.48)

( ) ( ) 21 tan tan sin 0Vz Vzθ θ φ− + − = (4.49)

The points at which these expressions are zero may represent maxima, minima or points

of inflection. Additional work will be needed to decide which is the case.

It is relatively easy to show that Equation (4.47) is true only for θ equal to zero. If

θ is zero, the Z vibration direction will be parallel to the c-axis of the crystal.

Consequently, the extinction angle will be zero for all sections in the zone [001] a value

that we will call a minimum.

Equation (4.48) is zero when φ is 90°. In other words, the extinction angle in

(010) represents a maximum or minimum (possibly an inflection point).

The most interesting requirement for a minimum or maximum on the extinction

angle curve is Equation (4.49). It is relatively straightforward to determine the conditions

for which Equation (49) holds. First, rewrite it in the form:

2 2

22

1 cos coscos

1 sinm

m

Vzθ φ

φ−=

+ (4.50)

where mφ is the angle between the wave normal and the pole to (100) where the

extinction angle takes an extreme value. As we shall see, not all combinations of Vz and θ

can produce a valid value of mφ from Equation (4.50).

Equation (4.50) is an equation in three variables, Vz, θ and mφ . If we treat mφ as a

parameter, we can plot Vz versus θ and produce a family of curves, one curve for each


58

value of mφ , that relates values of the three variables that satisfy Equation (4.49) and thus

defines the location of an extreme value on the extinction angle curve. Some examples

are shown in Figure 4-12. Solution of Equation (4.50) lies within regions of Vz- θ space

bounded by the curves:

4

2

2

Vz

Vz

Vz

ππ θπ θ

== −= +

(4.51)

The values of the parameter, mφ , along these boundary curves are 0 and π/2 (0 and 90

degrees, Figure 4-12). The regions which contain Vz- θ values that produce a maximum

or minimum on the extinction angle curves are contoured in values of mφ . The result is a

chart that provides a rapid and convenient way of determining whether a mineral has the

orientation and optic angle for a maximum extinction angle in a section not parallel to

(010). Further it gives the approximate location of the section in the zone [001].

A few amphibole and pyroxene combinations are plotted on Figure 4-12 (Data

from Deer, et al., 1966; Su and Bloss, 1984). Many, but not all, of the optically negative

amphiboles have their maximum extinction angles on sections in the zone [001] that are

not coincident with (010). On the other hand, most of the optically positive chain silicates

lack this feature. The location of the section with the maximum extinction angle can be

anywhere from a section nearly parallel to (100) [i.e. at φm = 0)] to parallel to (010) [i.e.

at mφ = π/2]. Amphiboles practically cover the range; the most extreme value is for an

amphibole culled from the literature by Su and Bloss (1984) [the point labeled ∆ on

Figure 4-12], which has the values:

Vz = 68.5

θ = 21.0

The value of φm corresponding to these data can be calculated with the aid of Equation

(4.49). The result is:

mφ = 5.13

scarcely 5° from the pole to (100).


59

The mathematical criteria for a monoclinic crystal (with Y = b, remember) to

have a maximum or minimum extinction angle on a section not parallel to (010) is easily

deduced from Figure 4-12. These are:

Positive crystals: 135 90Vz θ> + >� � (4.52)

135 45θ> >� �

Negative crystals: 90 45Vz θ> + >� � (4.53)

45 θ>� 45°

These same criteria were obtained by Su and Bloss (1984) with a trigonometric inequality

and combined into the statement that a maximum extinction angle occurs on a section not

parallel to (010) if the obtuse bisectrix is within 45° of the c-axis.

In spite of our labeling the points at which the derivative of δ with respect to φ

[Equation (4.46)] is zero as maxima or minima, all we have shown is that they are critical

points on the extinction angle curves. They may well be inflection points. In order to

determine whether any of the critical points are inflection points we need to calculate the

second derivative of δ with respect to φ. From Equation (4.46) we get:

( ) ( )2 2 22

2 2

sin 1 sin 2 cos cos 1 sinA B B A Bd dD

d D D d

φ φ φ φ φδφ φ

− + −= − − (4.54)

where D is the denominator on the right hand side of Equation (4.46):

( )22 2 22 1 sin sinD B Aφ φ = + + (4.55)

To determine what is the nature of the critical points, we need to know the sign of

the second derivative when the first derivative is zero. If the second derivative is negative

when the first derivative is zero, then the point on the curve is a maximum. If the second

derivative is positive, then the point is a minimum. If both derivatives are zero, then the

point is an inflection point. The only equation among the set of three, Equations (4.47)-

(4.49), that causes Equation (4.54) to be zero is Equation (4.47). As already noted,

Equation (4.47) is true only if θ is zero. In this case, however, extinction angles in the


60

zone [001] are all zero. The extinction angle curve corresponding to this special case is

plotted on Figure 4-13 [Curve 6]. The curve has neither a maximum or minimum nor has

it a normal inflection point. It is just plain degenerate. Consequently, extinction angle

curves, for Case 1 monoclinic crystals, have no inflection points, only maxima or

minima.

Table 4-1: Nature of the second derivative at the critical points on the extinction angle curve of sections in [001] for monoclinic crystals with Y = b.

Criteria for Critical Points Value of Second Derivative

21 sin 0mB φ− = cos 0mφ ≠ 2 2

2

sin cos

4AB

B A

φ φ−+

21 sin 0mB φ− ≠ cos 0φ = ( )

( )2 2

1

2 1

A B

B A

− + +

where: ( ) ( )tan tanA Vz Vzθ θ= + − −

( ) ( )tan tanB Vz Vzθ θ= + −

or: 2sin 2

cos 2 cos 2A

Vz

θθ

=+

cos 2 cos 2

cos 2 cos 2Vz

BVz

θθ

−= −+

When either Equation (4.48) or (4.49) holds, then the second term in Equation

(4.54) is zero and has no influence on the sign of the second derivative. We need only

look at the first term. The criteria for locating the critical points and the value of the

second derivative at these points are listed in Table 4-1. By evaluating the second

derivative at several points in Vz-θ space, one can quickly see that both maxima and

minima exist. Both occur because we distinguish positive from negative extinction angles

in the (010) plane. If the slow vibration direction, Z’, lies within 45° of the c-axis and also

falls between the positive ends of the a-axis and the c-axis, then a maximum will occur. If

the slow direction, Z’, falls between the negative end of the a-axis and the positive end of

the c-axis, then a minimum will occur on the extinction angle curves. An analogous

situation holds when the fast vibration direction, X’, is within 45° of the c-axis. As a

result, Daly’s Equation [Equation (4.43)] only predicts extinction angles within the range:


61

0 45δ≤ ≤ (4.56)

Some typical extinction curves are plotted in Figure 4-13. They were calculated

with Daly’s Equation [Equation (43)] using the values of Vz and θ listed in the caption to

the Figure.

The crystals which have their Bxo within 45° of the c-axis also have an extremum

on their extinction angle curves at φ equal to 90° [i.e. in the sections parallel to (010)].

The extremum at this location is always of the opposite kind than that located at φm [see

Equation (4.50), Figure 4-11]. The complete extinction angle curve ( 0 φ π≤ ≤ ) for

crystals with their Bxo within 45° of the c-axis are characterized by two maxima and a

minimum or vice-versa. Case 1 monoclinic crystals with their Bxo at angles greater than

45° from the c-axis possess extinction angle curves with only one extremum. This occurs

at φ equal to 90°, that is, in sections parallel to (010).

The values of mδ that correspond to mφ [see Equation (4.50), Figure 4-11] are

contoured on Figure 4-14. This chart can be used to obtain a rapid estimate of the size of

the extremum on an extinction angle curve, given θ and Vz. Figure 4-12 will quickly give

its location in the zone [001]. If Vz is greater than 45° (i.e. the crystal is negative), then

the maximum extinction angle will be Z’^c on Figure 4-14. If the crystal is positive, the

plotted angles are X’^c.

The numbered points on Figure 4-14 correspond to the numbered curves on

Figure 4-13. Points 3 and 4 represent special cases. Points like number 3 are

characterized by the condition:

2Vz πθ+ = (4.57)

Its reflection across θ equal to zero (Figure 4-14) would satisfy the requirement:

2Vz πθ− = (4.58)

When either condition (4.57) or (4.58) [but not both, at least for the moment]

hold, Daly’s Equation (4.43) takes on an indeterminate form, ∞ ∞ . One can apply

L’Hospital’s rule to examine the nature of the extinction angle curve under these


62

conditions or one can write Daly’s Equation in the form:

( )2 2

2sin 2 sintan 2

cos cos 2 cos 2 1 sinVz

θ φδφ θ φ

=+ +

(4.59)

and examine its behavior when either Equation (4.57) or Equation (4.58) is true.

Substitution for θ in Equation (4.59) from either (4.57) or (4.58) gives:

tan 2

tan 2sin

Vzδφ

= ± (4.60)

The sign is chosen according to which substitution is made. If Equation (4.57) is used, the

negative sign applies. Except at φ equal zero, Equation (4.60) produces a perfectly

acceptable extinction angle curve. At φ equal to zero, the wave normal is parallel to an

optic axis and the thin section is parallel to a circular section for which there is no defined

extinction position. Because of the unbounded nature of the tangent function at π/2

radians, a limiting value of 45° is obtained as φ approaches zero. An extinction angle

curve for this special case is shown on Figure 4-13 (Curve 3). A stereographic

representation is shown on Figure 4-15.

The last special case (Curve 4, Figure 4-13) occurs when both Vz and θ are 45° In

this case, the wave normals all lie in one circular section and the c axis is parallel to the

other. If 2Vz is 90°, then Equation (4.60) indicates that δ will equal 45°, regardless of the

value of φ. At φ equal to zero, there is no defined extinction angle.

The various extinction positions for an arbitrary value of φ for each kind of

extinction angle are shown in stereographic projection in Figure 4-15.

In the other monoclinic orientations that are possible, either X or Z will parallel

the b-axis of the unit cell. For the case when X = b (Case 2, Table 3-3), an equation

analogous to Daly’s equation for Case 1 monoclinic crystals can be derived. It is:

2

sintan 2

sin

A

B C

φδφ

=−

(4.61)

where:


63

2

2 2

2 2

cos sin 2

1 cos sin

1 cos cos

A Vz

B Vz

C Vz

θθθ

== −= −

(4.62)

It is an easy exercise to show that the only critical point occurs when φ is π/2 and that the

critical point is a maximum or minimum, depending on the value of θ.

The third kind of monoclinic orientation, Z = b, produces an extinction angle

equation identical in form to Equation (4.61). Consequently, no new features result.

Measurable Angles in Sections Normal to an Optic Axis

Petrographers frequently have the problem of estimating the composition of an

orthorhombic solid solution in thin section. Typical examples are olivines and

orthopyroxenes. Because the principal vibration directions are tied to the lattice vectors

by the symmetry of orthorhombic crystals, extinction angles do not vary significantly

with composition in orthorhombic solid solutions. Extinction angles, when measured in a

section containing at least one principal vibration direction, against the trace of a crystal

plane that itself contains two principal vibration directions, cannot vary with composition

at all. In other cases, extinction angles can vary with composition, but not by much. There

are two sources for the variation in extinction angle; the first is the compositional

dependence of the unit cell parameters and the second is the change in 2V with

composition. Of the two, the second produces the much larger effect. We will illustrate

these statements with some calculations employing data for the olivine series end

members, forsterite and fayalite. These data and the results are set down in Table 4-2.

To illustrate the effect of changing unit cell dimensions we will first look at the

difference in the extinction angles, Y^Tr(101) measured in (010). For forsterite this angle

is 38.5° and for fayalite is it 38.3°, a difference of only 0.2°. It is unusual to find an

olivine crystal with a zoning range greater than 30%. A zoning range of this extent would

produce a change in the extinction angle of approximately 0.06° - hardly significant, let

alone measurable in thin section.

In a section described by a general set of Miller indices, such as (111), both

changes in lattice parameters and changes in 2V cause a variation in extinction angles.


64

The direction cosines to the pole of (111) in forsterite are 0.7352, 0.3430 and 0.5846. In

fayalite the corresponding cosines are 0.7384, 0.3396 and 0.5826. The differences

between these numbers correspond to an angle of approximately 0.7°. Compared to the

change in position of an optic axis, 26° (δ2Vz/2), this is an insignificant change. The

difference in the extinction angle between the fast direction and the trace of (010),

measured on (111), for the two end members can be obtained from the data in Table 4-2

and is only 7.6°. The extinction angles are shown in stereographic projection for the two

end members on Figure 4-16. Assuming the variation in this extinction angle is

approximately linear across the olivine series, a change in composition of 10% Fo will

cause a variation of only 0.76° in the extinction angle - a change that would be difficult to

detect in thin section.

The point of this discussion is that measurements of conventional extinction

angles are not feasible methods for estimating the compositions of orthorhombic solid

solutions. Optical estimates of the composition of orthorhombic solid solutions depend on

a measurement that is a function of the refractive indices. For example, the zoning pattern

in olivines, normal or reverse, can be detected by changes in the interference colors if the

compositional range is sufficiently large (see Deer, et al., 1966, p. 7). The most

commonly measured property for estimating compositions of orthorhombic minerals is

2V which is related to the principal refractive indices [see Equation (40), Chapter 2]. The

optic axial angle can be estimated for minerals in thin section with Kamb’s method (see

Bloss, 1961, p. 180) or Tobi’s method (Bloss, 1961, p. 205). Both methods have their

limitations; an important one is that a centered bisectrix (or at least a near centered one) is

required. Orthopyroxene crystals that will exhibit a centered bisectrix figure in thin

section are relatively easy to find because Z is parallel to the c-axis. Consequently,

sections of orthopyroxene that show sharp cleavage traces at angles of approximately 88°

and 92° will be cut normal to the c-axis and will provide estimates of 2V with Kamb’s

method. Kamb’s method, if repeatedly applied to the same mineral section, will provide

an estimate of 2V that is precise to approximately ±10°. This translates into a

compositional precision of approximately ±5 mole percent En on a determinative curve

of 2V versus orthopyroxene composition [see, for example, Deer, et al., 1966, p. 112].


65

Table 4-2: Crystallographic and optical properties of the olivine end members and some calculated angles between optical and crystallographic directions. Fo Fa a (nm) 0.4756 0.4817 b (nm) 1.0195 1.0477 c (nm) 0.5981 0.6105 Optical Orientation: a = Z, b = X, c = Y

2Vz 82.0 134.0 α 1.636 1.827 β 1.650 1.870 γ 1.670 1.878

X’^Tr(010) on (111) 9.4° 1.8°

Tr(021)^Tr(021) in

Optic Axis Section 83.1° 49.0°

Olivines, on the other hand, have no crystallographic guides for finding bisectrix

figures in thin section. Even experienced petrographers can become frustrated trying to

find appropriate sections for use with Kamb’s method. Such sections have intermediate

interference colors but not all sections with the same intermediate color will display a

bisectrix. In addition, the variation of 2V with olivine composition is such that a precision

of ±10° in 2V translates into a precision of approximately ±25 mole percent Fo

(Figure 4-17). Ideally, we would like a method of estimating olivine compositions that

utilizes easily discovered sections and is relatively precise. The method to be described

has approximately the same precision as the optic axial angle method but for which

appropriate sections are considerably easier to find. In short, the method to be described

is not very precise but is easier to use than one requiring an estimate of 2V. I think it fair

to say that a precise and efficient method of estimating olivine compositions in thin

section by optical methods does not exist and probably cannot be created.

Sections cut normal to an optic axis are much easier than bisectrix sections to

find. They are ideally isotropic and even with white light as the illuminator, the

dispersion of the optic axis is usually sufficiently small that such sections remain nearly

dark and evenly illuminated on rotation of the stage under crossed polarizers. Because

such sections are permanently at extinction, the angle cannot be an extinction angle but


66

rather it must be an angle between the traces of crystallographic surfaces. Obviously one

limitation of the method is that the crystal you are examining must display intersecting,

extended if need be, traces of the crystal planes. Phenocrysts of olivine in volcanic rocks

commonly develop as euhedral crystals showing the form {021}. Some illustrations of

these are shown in Figure 4-18. Looking down an optic axis in a crystal bounded by this

form, the optic axis will emerge on a line bisecting the angle between (021) and ( 021)

[depending on which end of which optic axis is emerging from the section, the signs on

the Miller indices may have to be permuted].

Consequently, by rotating the stage of the microscope, we can measure the angle

between the traces of (021) and ( 021). This angle for the two olivine end members is set

down in Table 4-2 and the difference is 34.1°. Assuming the variation is linear across the

compositional range, a variation of one degree corresponds to approximately 3 percent

variation in composition. A calculated determinative curve for the olivine series is shown

on Figure 4-19.

Crystallographic Forms in Thin Section

The location of vibration directions, calculation of extinction angles and other

properties that we have outlined in these notes are extensive and time consuming if done

by hand. As a practical matter, a computer is required if charts for mineral identification

and compositional estimates are going to be constructed. If one is going to have the

computer calculate extinction angles, it would be an advantage if we had a systematic

method for identifying the traces of the crystallographic surfaces we see in thin section.

All of the surfaces that outcrop in the plane of the thin section are members of a

crystallographic form, including cleavages, crystal faces and the composition planes of

twins. If you remember, a form includes all the faces that can be generated by the

symmetry elements of the lattice operating on the Miller indices of one face in the form.

A form can contain 1, 2 or more faces depending on the initial set of Miller indices and

the symmetry of the crystal.

For example, in all crystals with a center of symmetry, the form {100} consists of

two faces, (100) and ( 100 ). On the other hand, the form {111} consists of two faces in


67

the holosymmetric triclinic class, (111) and ( 111 ) but consists of four faces in the

holosymmetric monoclinic class, (111), ( 111 ), (111) and ( 111 ). The question is, given

the form symbol and the symmetry of the crystal, how many nonparallel sets of traces can

we see in thin section? To answer the question, we need a list of the number of faces in

the various forms of the three crystal systems. Because of the symmetry center operation,

crystals with this property produce parallel faces on their forms. For example, the form

{hkl} in a crystal with a center of symmetry contains, at least, the faces (hkl) and ( hk l ).

In thin section, the trace of (hkl) and ( hk l ) will be parallel and, if contained within the

crystal as cleavage planes for example, indistinguishable. Consequently, we can ignore

faces in a form if their Miller indices can be obtained by multiplying each index of

another face by minus one. This is equivalent to ignoring the center of symmetry which is

what one does, in part, to obtain the other crystal classes from the holosymmetric one in

the biaxial systems. The list of faces that can produce distinguishable traces in thin

section is considerably shortened and is given in Table 4-3.

Table 4-3: Faces in forms that can create distinguishable sets of parallel traces in thin section. Form Triclinic Monoclinic Orthorhombic {100} (100) (100) (100) {010} (010) (010) (010) {001} (001) (001) (001) {hk0} (hk0) (hk0)( hk0 ) (hk0)( hk0 ) {h0l} (h0l) (h0l) (h0l)( h0l ) {0kl} (0kl) (0kl)( 0kl ) (0kl)( 0kl ) {hkl} (hkl) (hkl)( hkl ) (hkl)( hkl )( hkl )( hkl )

Besides the crystallographic limitations on the number of sets of parallel traces we

can see in thin section, it may happen that nonparallel faces will give rise to parallel

traces in thin section. To use an analogy from field geology, suppose two planes have the

same strike but opposing dips and the thin section is cut parallel to the strike then the

traces of the two planes will be parallel. We next examine the conditions that will

produce this situation: parallel traces from nonparallel faces.

Suppose we are given the Miller indices of two crystal planes, (hkl) and (rst). We

then form two vectors normal to these planes [Equation (4.5)]. Next, we find vectors


68

parallel to the traces of their intersection with the plane of the thin section [Equation

(4.1)]. Next, we impose the condition that these two vectors must be parallel (or

antiparallel) if the traces are to be parallel:

hkl hkl= ×q w d (4.63)

rst rst= ×q w d (4.64)

hkl rst× = 0q q (4.65)

Thus:

[ ] [ ]hkl rst× × × = 0w d w d (4.66)

if the traces are parallel. Expanding this last equation [i.e. let ( )hkl×w d equal A in

Equation (3.34), Chapter 3] gives:

[ ]{ } [ ]{ }rst hkl hkl rst• × − • × = 0d w d w w w d d (4.67)

The last term is zero because w is normal to hkl×w d and the cosine of 90° is zero.

Because the wave normal vector is not a zero length vector, its coefficient must be zero if

the traces of the two planes are parallel:

[ ]rst hkl• ×d w d (4.68)

The position in which the vectors appear in a scalar triple product can be changed as long

as the cyclic order is preserved (Hoffmann, 1975, p. 78-79). As a result, we can write

Equation (4.68) in the form:

[ ] 0• × =hkl rstw d d (4.69)

Equation (4.69) can be true only if one or more of the following situations occur:

1. Either w or ×hkl rstd d is a zero vector.

2. The line of intersection of the two planes and the wave normal vector are at right

angles.

In other words, the wave normal and the zone axis for the two faces are


69

perpendicular. Because the wave normal vector is a nonzero vector, the only remaining

possibility, which arises from the first situation, is for the cross product of the vectors

normal to the faces (hkl) and (rst) to be zero. In other words, the normals are parallel or

antiparallel. If the normals are parallel, then the faces themselves are parallel. This

situation was taken into account when we collected the data in Table 4-3.

Table 4-3 lists the Miller indices of the faces that will contribute a distinguishable

set of parallel planes belonging to the forms in biaxial crystals. If the wave normal, by

chance, happens to be normal to the zone axis of faces in the form, the number of sets of

traces will be reduced by the number of traces generated by the faces parallel to the zone

axis minus one. Against each of the sets of traces we can measure extinction angles.

Application to Pyroxene Epitaxy

Several Mauna Loa tholeiite basalts contain overgrowths of augite on

orthopyroxene. A sketch of one such overgrown orthopyroxene is shown on Figure 4-20.

The augite overgrowths appear to form only on faces parallel to the optic axial plane of

orthopyroxene. Two optical orientations for orthopyroxene are in the literature. Deer, et

al. (1966) describe the orientation of the optic axial plane of orthopyroxene as parallel to

(100) whereas Phillips and Griffen (1981) describe it as parallel to (010). The orientation

described by Deer, et al. (1966) is the better choice because the b-axes and c-axes of the

two minerals, augite and orthopyroxene, are approximately the same size and the atomic

arrangements in the (100) plane would be similar; both factors would promote epitaxic

growth. The suggested correlation of orientations is shown in Figure 4-20.

If the overgrowths are epitaxic, then we should be able to correctly predict the

wave normal in the augite given a wave normal in orthopyroxene. An easy wave normal

to locate in the orthopyroxene is one parallel to an optic axis. The corresponding wave

normal in augite should then lie in (100) and at an angle to the common c-axis direction

equal to Vz for orthopyroxene. Consequently we can construct curves of extinction

angles to the trace of (100) versus the angle between the wave normal and the c-axis for

several members of the augite series (Figure 4-21). The extinction angle we measure

should then fall on the appropriate curve for an augite if the overgrowth is epitaxic.


70

It is perhaps worth noting that a correct prediction does not prove epitaxy - rather

we can only say that the available data are consistent with such an interpretation or that

epitaxic growth did not take place because the predicted extinction angle does not agree

with the measured one.

Nicholls and Stout (1997) studied several overgrowths from the 1881 lava flow

that erupted from Mauna Loa, Hawaii. The results of more detailed study are consistent

with epitaxic growth (Figure 4-21). Pigeonite, however, was found to also be part of the

overgrowths and the sequence of crystallization was orthopyroxene followed by pigeonite

or augite. Thermodynamic models of the crystallization paths are consistent with

pyroxene crystallization at pressures greater than 0.18 GPa followed by magma transport

to shallower depths and pressures less than approximately 0.1 GPa where olivine

crystallized.

Problems

1. Derive an equation of the form: d = u a + v b + w c for a vector along the line of intersection of two lattice planes, (h1k1l1) and (h2k2l2).

2. Typical unit cell parameters for a monoclinic amphibole are: a = 0.985 nm b = 1.805 nm c = 0.525 nm b° = 105° What is the 2V of an amphibole that has an extinction angle, Z^c, of 21 degrees on (010) and an extinction angle of 16.6 degrees on (110)? What is the maximum extinction angle in the zone [001]?

3. A titanian augite shows a centered optic axis figure in a section normal to [001] and an extinction angle, Z^c, on (110) equal to 27.5 degrees. If its unit cell parameters are:

a = 0.975 nm b = 0.895 nm c = 0.525 nm b° = 105° What is its optic axial angle? What is its maximum extinction angle in the zone[001]? On what section does this maximum occur?

4. Derive Equation (61) and show that the only critical point occurs at f equal to p/2.

5. Construct an extinction angle curve for sections cut parallel to [001] of a pigeonite with the following properties:


71

a = 0.973 nm b = X b = 0.895 nm 2Vz = 25° c = 0.526 nm θ = Y^c =:49.0° b° = 108°

6. Plagioclase zoning patterns are often complex. They can be normal, reverse, oscillatory, patchy or a combination of any or all of these patterns. The most convenient tool for describing the zoning pattern is the petrographic microscope. Explain how extinction angles in sections normal to (010) can be used to determine the zoning pattern. Describe any limitations or precautions that must be taken.

7. A mineral of the epidote group has the following properties: Y = b θ = 90° Cleavage: {001} b° = 115° 2Vz = 116° Describe the extinction behavior in sections in the zone [001]. Do the same for sections cut normal to (001).

8. Is it possible to use sections cut normal to an optic axis to determine plagioclase compositions? Explain.

9. Construct determinative curves for estimating compositions of low structural state plagioclase in sections cut:

Normal to [100] Parallel to (001) Parallel to (010). In each case explain the situations (e.g. grain mount or thin section) where the curve would be most useful or applicable (see Figure 4-2).

10. Construct extinction angle curves for sections cut normal to (010) in low structural state plagioclases (see Figure 4-8).

11. Plot a curve in stereographic projection that shows the position of the vibration vector closest to the c-axis in sections in the zone [001] for a monoclinic crystal with the following properties:

Y = b 2V = 90° θ = 45 How does this curve compare with analogous curves for monoclinic crystals that have the properties listed in Figure 4-12? How do the analogous curves for minerals with either X = b or Z = b compare?

12. Deer, et al. (1966) provide the following data for the hastingsite-ferrohastingsite series:

Pargasite Ferrohastingsite 2Vα 120° 10° θ (Y = b) 26° 12° a ≅ 0.985 nm ≅ 0.985 nm b ≅ 1.805 nm ≅ 1.805 nm


72

c ≅ 0.525 nm ≅ 0.525 nm b° ≅ 105° ≅ 105° Cleavage {110} {110}

Assuming the variations in 2V and q vary linearly with composition, plot: (1) The extinction angle on (110) against composition. (2) The maximum extinction angle against composition. (3) The location of the maximum extinction angle, dm, against composition.

Could any of these curves be practical tool for estimating composition? Explain.

73

Chapter 5: Location of Ray Paths

Introduction

If asked, most of us would define a ray path as the direction along which light

travels. On reflection however, such a definition is unsatisfactory because the wave

normal could be defined as the direction of travel and all optics texts emphasize that the

ray path and wave normal need not coincide in anisotropic media. Perhaps ray path would

be easier to define if we first answer the question: How do we know light travels from

here to there? The answer is, of course, that some action creates light at a source and

there is a reaction to the light at a receiver (i.e. we see something). The reaction takes

place because energy is transferred from the source to the receiver. In other words, the

ray path is the path along which energy is transferred. By and large ray paths do not play

a prominent role in optical mineralogy (e.g. Bloss, 1961; Nesse, 1991). Rather, it is the

vibration directions that are important and, as we have seen, these are associated with the

wave normal not the ray path. In spite of the relative importance of ray paths and wave

normals, students of optical mineralogy seem to have an instinctive need to know where

the ray paths are and students of optical mineralogy do ask about the location of ray

paths. In addition, the location of ray paths is required for a complete description of

interference figures. A completely rigorous and general mathematical description of the

phenomena leading to interference figures has not been developed, even though research

into the matter dates back to the late 1800s (Bethke & Birnie 1980). It is much easier to

determine the wave normals than the ray paths. The latest model for describing

interference figures (Bethke & Birnie 1980) assumes that wave normals are adequate

approximations to the ray paths. This chapter describes a method for finding the ray paths

associated with a given wave normal. Consequently, a quantitative evaluation of the wave

normal approximation follows. Bloss (1961, p. 77-78, 160-161), Wahlstrom (1979,

Appendix B), and Nesse (1991, p. 56-57, 80-81) describe graphical procedures for

finding ray paths. Graphical procedures, however, are not always practical methods for

finding numerical solutions to real problems. The vector-based method outlined in this

chapter results in an analytical expression for the angles between the wave normal and

ray path.

Chapter 5: Location of ray Paths

74

In this chapter, we will discuss the location of ray paths in anisotropic media.

Along the way, we will derive the indicatrix from a set of more fundamental equations,

called Maxwell’s Equations, and gain a deeper understanding of the manner in which

light and matter interact. Again, vectors provide a concise and elegant means of

describing the relationship between wave normals and ray paths but it is vector calculus

rather than vector algebra that is required. If you are not familiar with the operators of

differential vector calculus, div, grad, and curl, then you can either skip this section or,

more profitably, you can read the excellent book by Schey (1973) during an evening and

pick up enough to follow through this chapter.

Maxwell’s Equations

To start with, we need to define the vectors that describe the behavior of light in

crystals. Because light is a form of electromagnetic radiation, it seems natural that the

vectors will have to do with electricity and magnetism. There are actually two vectors

associated with each kind of phenomenon. The vectors having to do with electricity will

be labeled E and D. The nature of E and D are described by Bloss (1971, p.361-371) and

need not be elaborated here. Sufficient to say that if a crystal is placed in an electric field

described by the vector function E, then a current will tend to flow in the crystal parallel

to D. If the crystal is isometric or if an anisotropic crystal is given a particular orientation

with respect to E, then D will be parallel to E. In general, however, D and E are not

parallel and this is one way of defining optical anisotropy; the cause, E, and the response,

D, are not parallel in anisotropic media. When light is the source of the electric field, E,

the field oscillates, causing D to oscillate also. As it turns out, D is parallel to a vibration

direction in the crystal. Inside the crystal, E presumably still exists and it is the normal to

E that constitutes the ray path. In other words, energy is carried by the electric field and

its direction of transfer is normal to E. In summary then, two vectors D and E represent

the electric part of light. D is parallel to the polarization or vibration direction we have

been calculating. In fact a unit vector parallel to D is n. The normal to E is the ray path

and is not, in general, parallel to w, the wave normal. This latter vector, w, is normal to D.

Finally, we should note that even though D is the direction along which an electric

current tends to flow in response to E, there will be no net current if E is oscillating; as


75

many electrons will try to flow along +D as -D, thereby canceling the flow of electricity.

Similar to the vectors describing the electrical properties of crystals, there are two

vectors associated with their magnetic properties, B and H. If we place a crystal in a

sufficiently strong magnetic field, B, it will acquire a magnetic direction parallel to H.

The crystals that transmit light are found by experiment to be effectively isotropic in their

magnetic behavior. Consequently, B and H will be parallel and we can write:

B = µoH (5.1)

where µo is a scalar constant.

The equations relating B, H, D and E are known as Maxwell’s Equations. They

are partial differential vector equations and contain a complete macroscopic description

of electromagnetic phenomena (Schey, 1973). Maxwell’s Equations are derived from

experimental results and a large measure of their value resides in their ability to predict

the behavior of matter under conditions that are experimentally difficult or impossible to

attain. Several branches of science are founded on mathematical generalizations of

experimental or empirical findings. For example, thermodynamics is based on its three

laws and the mathematical consequences of these laws provide the equations we use to

describe mineral equilibria in rocks. Likewise, Newton’s laws underwrite the science of

mechanics. Maxwell’s Equations, although of a more complex mathematical nature than

the laws of thermodynamics or Newton’s laws, have an analogous position in

electromagnetism. Our ultimate goal is to start with Maxwell’s Equations and arrive at a

representation of the indicatrix.

The essentials of Maxwell’s Equations, when applied to crystals that transmit

light, are:

t

∂× =

∂DH∇ (5.2)

t

µ∂

× = −∂HE∇ (5.3)

Notice that B does not appear in these equations. This is because of Equation (1), which

is valid only for small magnetic fields and for substances that appear isotropic in their


76

magnetic properties. When these conditions are not met, then the more extensive and

complete set of Maxwell’s equations must be used. Equations (2) and (3) are specialized

and do not represent a complete description of electromagnetic phenomena. Schey (1973,

p. 109) provides a complete statement of Maxwell’s Equations and with the help of Nye

(1957, Appendix H) you can easily obtain Equations (2) and (3) for the special case of

transparent crystals.

Electromagnetic Waves and Maxwell’s Equations

Our immediate goal is to find expressions for D and E that satisfy Equations (2)

and (3) and that also describe an electromagnetic plane-polarized wave. We will employ

that time-honored method of guessing what the solution must be and then check that the

solution has the required properties. An equation that describes a wave produced by an

oscillating vector is:

( )cos t vω ω= − •E E r w (5.4)

where E is a constant vector, ω is the frequency of oscillation with units of reciprocal

time (i.e. 1/second), t is the time, v is the speed with which the wave travels (units of

length/time), w is our familiar wave normal unit vector and r is a position vector. It

stretches from the origin of the coordinate system to the point in space labeled (X,Y,Z).

For a fixed wave normal, w, the variable quantities are t and r whereas ω and v will be

constants.

You should convince yourself that Equation (4) describes a plane-polarized wave

in both space and time. For example, the cosine term is a scalar function of position and

time. Because it is a scalar, it can only modify the magnitude of E; it cannot change its

direction except by 180° when the cosine function switches from positive to negative.

That is, the cosine can take positive or negative values and E will be parallel to ± E ,

depending on the sign of the cosine term, or E will be zero if the cosine term is zero.

Hence, E is forced to oscillate in the plane containing E , forming a plane-polarized

wave. To show that Equation (4) represents a wave, plot the magnitude of E as a function

of time for a fixed value of r. Next, take t as constant and investigate the effect of the dot


77

product, •r w , on the magnitude of E.

In the indicatrix reference frame, the triple of numbers (X,Y,Z) represents a point

on the indicatrix and the equations for r and w can be written explicitly as:

=r X Y Zi + j + k (5.5)

1 2 3Xw Yw Zw• + +r w = (5.6)

Next, we can show that Equation (4) satisfies the classic wave equation, which is a partial

differential equation of the form:

2

2 22

FFvt

∂=

∂∇ (5.7)

where v is the speed of the wave and F represents whatever is doing the waving. In our

problem, this would be E. If our guess as to the form of E is correct [Equation (4)], then

when we substitute E for F in Equation (7), the result should be an identity. This can be

shown, perhaps most elegantly, by first writing E in the form:

φ=E E (5.8)

where φ is a scalar function of the space and time coordinates:

( )cos t vφ ω ω= − •r w (5.9)

The second derivative of Equation (8) with respect to time is:

2 2

2 2t tφ∂ ∂

=∂ ∂

E E (5.10)

In order to get an identity out of Equations (7) and (8) we have to show that by

substituting Equation (8) into the left hand side of Equation (7) we can obtain:

( )2 2 2 2v v φ∇ = ∇E E (5.11)

Equation (8) can be reduced in such a fashion by applying some vector algebra and the

identities listed in Table A-1 in the Appendix. This is left as an exercise for the reader.

As a result of these machinations we have shown that Equation (4) represents a


78

plane-polarized wave in space and time and that it satisfies the classic wave equation. We

next want to obtain an equation for H by substituting our expression for E [Equation (8)]

into Equation (3).

We take the curl of E, as given by Equation (8) and use Identities 2 and 7, Table

A-1, to get from Equation (3):

( )t

µ φ∂

= − ∇ ×∂H E (5.12)

Next calculate the gradient of φ from Equation (9):

( ) ( )sin tv v

ω ωφ ω

∇ • • ∇ = −

r rw w (5.13)

But you can easily check from Equation (6) that the gradient of the dot product, •r w , is

equal to:

( )∇ • =r w w (5.14)

If we define a new scalar function, φ’, as:

( )sin'

tv

v

ωω ω

φ

• −

=

r w

(5.15)

then, using Equation (14), Equation (12) becomes:

( )1 't

φµ

∂= − ×

∂H Ew (5.16)

Equation (16) is important because at a fixed point in space [remember the partial

derivative with respect to time is evaluated at a fixed point where (X,Y,Z) are constant] it

says that the magnetic field vector, H, is normal to the plane containing w and E.

Figure 5-1 shows in a schematic fashion, the relationship between E, w and H that

Equation (16) requires. An explicit expression for H is obtained by integrating Equation

(16) with respect to time at constant r:


79

( )v

φµ

= ×H Ew (5.17)

where we have arbitrarily set the constant of integration equal to zero. This is our

expression for H and all that is left is to take the curl of H, substitute the result into

Equation (2) and solve for D. The details follow almost exactly the steps that lead to

Equation (17):

( )v

φµ

= − × ×D Ew w (5.18)

Because there are no derivatives left in Equations (17) and (18), we can combine E and

φ and recover E in Equations (17) and (18):

vµ

×=

EH w (5.19)

( )2vµ

× ×= −

ED

w w (5.20)

according to Equations (19) and (20), D is normal to the plane of H and w. The

plane normal to H, according to Equation (19), contains E. Consequently, E, D and w

must be coplanar. D is shown schematically on Figure 5-1 in relation to E, w and H. It

can be shown that the direction of energy transfer is proportional to ×E H (see for

example Lipson and Lipson, 1969, p. 73) by employing the theorems of integral vector

calculus. Consequently, the ray path, R, can be inserted into Figure 5-1 and we reach the

important result that the wave normal, a vibration direction and the ray path associated

with that vibration direction are all coplanar. This statement is included in texts on optical

crystallography. We have just shown that it is a consequence of Maxwell’s Equations and

the wave properties of light.

We next expand the triple product in Equation (20) [see Equation (3.34), Chapter

3, p. 33] with the result:

( )2vµ + • =D E E 0w w - (5.21)

Equation (21) is a vector equation relating D and E. It has been derived without


80

any reference as to whether the crystal is isotropic or otherwise. In this sense it is

completely general. However, we cannot go any further without taking into account how

the crystal itself relates the two vectors D and E.

Relationship Between D and E

The relationship between D and E is most simple in isotropic media where D and

E are parallel. If D and E are parallel, then E is normal to w and the dot product in

Equation (21) vanishes, leaving us with the result:

2vµ=E D (5.22)

The terms before D constitute a scalar, hence, Equation (22) says the two vectors, E and

D, are parallel as they must be in isotropic media.

In anisotropic substances, E and D are not parallel and we need a mathematical

method for relating nonparallel vectors. This mathematical method is a tensor. An

excellent introduction to the use of tensors to describe the physical properties of crystals

is in Bloss (1971, p. 361-372). Nye (1957, Ch. 1) presents an extensive description of

physical properties represented by tensors but still at a mathematically straightforward

level. If you are not familiar with tensor properties, it is strongly recommended that you

consult one or both of these references. Actually, if you know indicatrix theory, you are

familiar with a tensor. The indicatrix is the geometrical representation of a tensor

property, the reciprocal of the square of the index of refraction. In other words, the

optical property, 1/N2, associated with a known vibration direction, n, is a tensor property

of the crystal that relates two vectors.

Before we describe these two vectors and their relationship, we need to fix in our

minds the following facts about tensors.

1. A tensor is a mathematical method that can be used to describe or represent a part of

the universe in which we live. A vector can describe some property and, like a tensor,

it is also a mathematical method.

2. Tensors come in various sizes and the size of a particular tensor depends on the

number of dimensions involved and the complexity of the property being described.


81

For example, in three dimensions, the world we live in, the size of a tensor is given by

3n, where n is a measure of the complexity of the property being represented. In the

space-time world of relativity the number of dimensions is four and the size of a

tensor would be given by 4n. If n is equal to zero, then the tensor has only one

component and is equivalent to a scalar. In fact a scalar can be described as a zero

rank tensor. In our everyday world of three dimensions, a tensor for which n is equal

to one has three components, just like a vector. A vector then, is the same as a first

rank tensor. If n is equal to two, the tensor is of second rank and can be used to

describe the optical properties associated with the indicatrix. Because indicatrix space

spans three dimensions, such tensors will have nine components (32).

3. Tensors are designed to be invariant with respect to the coordinate system used to

describe the property. For example, if the temperature in Calgary is 25øC, then the

zero rank tensor that describes temperature has to give the same result whether we

locate Calgary by latitude and longitude or whether we locate Calgary with respect to

a coordinate system with an origin at the center of the moon. In fact, mathematicians

rigorously define tensors by how they behave under a change of coordinate system. If

we look at a fixed vector with components V1, V2, V3 in one coordinate system, then

it is easy to show that in a different coordinate system the components will change but

that the vector itself doesn’t. A two dimensional example of this fact is shown in

Figure 5-2. What is true for vectors is true for tensors in general. A change in

coordinate systems produces a change in the components of a tensor but their total

representation of a physical property does not change.

4. If tensors are invariant under a change of coordinate system although their

components do change, then it is reasonable to suppose that we can make the

components easier to manipulate if we make the proper choice of coordinate system.

Figure 5-2 illustrates this point. If we choose the unprimed coordinate system, (x,y)

rather than the primed one, (x’, y’), then the vector V can be written:

xV=V i (5.23)

rather than:


82

'xV=V '

yi' +V j' (5.24)

In an analogous manner, by a proper choice of reference frame, several components of a

second rank tensor can be reduced to zero.

We now return to the description of the tensor relating D and E. In an arbitrary

reference frame, the equations that relate D and E are:

11 1 12 2 13 3 1

21 1 22 2 23 3 2

31 1 32 2 33 3 3

E E E DE E E DE E E D

κ κ κκ κ κκ κ κ

+ + =+ + =+ + =

(5.25)

where Di and Ei are the components of D and E in our arbitrary reference frame and the

κij are the components, in this reference frame, of the second rank tensor that relates D

and E. If we write the set of Equations (25) in matrix form we get:

11 12 13 1 1

21 22 23 2 2

31 32 33 3 3

E DE DE D

κ κ κκ κ κκ κ κ

=

(5.26)

An even simpler notation for the same equations is:

=E DK (5.27)

where K represents the second order tensor and E and D have their usual meaning as

vectors.

The tensors that represent many physical properties can be referred to a special set

of axes such that Equations (26) can be written:

1 1 1

2 2 2

3 3 3

0 00 00 0

E DE DE D

κκ

κ

=

(5.28)

where the iκ are nonzero components of K in this new reference frame. The restrictions

on the tensor and details of obtaining the iκ from the ijκ are given in Nye (1957). What

we need to note is that K is equally well represented by either the square matrix in

Equations (26) or in Equations (28). The differences are solely due to a change in


83

coordinate system. Using the tensor representation in Equations (28), then we can write

out Equation (27) as:

1 1 2 2 3 3E E Eκ κ κ Di + j + k = (5.29)

Hence, another way of writing a vector is KE, the matrix product of a second rank tensor

and another vector.

You should convince yourself that, in general, the tensor K operates on E in such

a way that a new vector results which is not parallel to E (Figure 5-3).

In order for Equation (29) to reduce to the isotropic case and look something like

Equation (22), all the κi’s would have to be equal, say to κ:

κ E = D (5.30)

Comparing Equations (22) and (30) we see that:

2

1v

κµ

= (5.31)

for isotropic substances. Optically isotropic substances are characterized by a uniform

index of refraction, regardless of vibration direction. Because the index of refraction is

inversely proportional to the speed of light in a substance, we expect the components of

K to be functions of the refractive indices [cf. Equation (31)]. Our next task is to find

these functions.

An index of refraction is the ratio of the speed of light in a vacuum to the speed in

a substance. If we are going to relate speeds and refractive indices, we will need an

expression for the speed of light in a vacuum. Presumably, a vacuum is isotropic and

uniform throughout the universe. Consequently, in a vacuum, D and E should be related

by:

κο E = D (5.32)

where κo is a universal scalar constant. In order to relate κo to the speed of light, c, we

need to manipulate the two Maxwell’s Equations [Equations (2) and (3)] in conjunction

with one more:


84

0∇ • =E (5.33)

Equation (33) is another of Maxwell’s Equations that has been simplified because we are,

at present, dealing with a vacuum [see Schey (1973), p. 146]. First take the curl of

Equation (3) and substitute Equations (2) and (32):

( ) [ ]t

µ∂ ∇×

∇× ∇× = −∂

HE (5.34)

( )2

2tµ

∂∇× ∇× = −

∂DE (5.35)

( )2

2tµ κ

∂∇× ∇× = −

∂EE (5.36)

Next use identity 1 in Table A-1 and take into account Equation (33):

2

22t

µ κ∂

∇ =∂

EE (5.37)

Equation (37) has the form of the classic wave equation [cf. Equation (7)] if we make the

following identification:

2 1cµ κ

= (5.38)

We have retraced Maxwell’s most outstanding achievement, the identification of light as

an electromagnetic phenomenon. The two constants, µo and κo, can be experimentally

determined. The reciprocal of their product turns out to be the same as the measured

speed of light in a vacuum.

Equations (30), (31) and (38) suggest that the velocity of light vibrating in a

particular direction parallel to D would have as its components in the reference frame

defined by Equation (28):


85

21

1

22

2

23

3

1

1

1

v

v

v

µ κ

µ κ

µ κ

=

=

=

(5.39)

The nonzero components of K, when referred to the special coordinate system that

led to Equations (28), can be written with Equations (38) and (39) as:

2

1 212

2 222

3 23

cvcvcv

κ κ

κ κ

κ κ

=

=

=

(5.40)

The set of Equations (40) relate the components of our tensor K to three velocities

of light through the crystal. When written as they were for Equation (28), there is one

component that is the largest possible, one that is the smallest possible. If we label the X-

axis as the direction along which D1 lies then κ1 is the tensor component of interest. If κ1

is assigned the smallest value, then v1 must be the fastest. But we know that 2

21

cv , if v1 is

the fastest, is equal to α2, and κ1 is equal to κoα2. In a like manner, we find κ2 equal to

κoβ2 and κ3 equal to κoγ2. Consequently, we can write equations for E in the form:

21 1

22 2

23 3

0 00 00 0

E DE DE D

κ ακ β

κ γ

=

(5.41)

Finally we note that the relationship between n and D can be expressed:

•

DD D

n = (5.42)

If we define a new vector e in the following fashion:


86

κ•Ee

D D= (5.43)

We can write the set of Equations (41) as:

Table 5-1. Example of the calculations. Given Values

w1 w2 w3 Wave normal vector: w 0.57735 0.57735 0.57735

α β γ Indices of Refraction 1.60 1.62 1.70

Calculated Values Vz (see Bloss 1961, p. 156) 27.618 I j k Optic Axis Vector: u 0.46357 0.0 0.88606 Optic Axis Vector: v -0.46357 0.0 0.88606 t: Eqn. (15) -0.81620 0.38918 0.42703 s: Eqn. (16) 0.52750 -0.80348 0.27598 G = t + s: Eqn. (17) -0.28871 -0.41430 0.70301 n: Eqn. (18) -0.33354 -0.47865 0.81219 m: Eqn. (20) -0.74526 0.66149 0.08378 en: Eqn. (40) -0.13029 -0.18238 0.28103

•e en n = 0.12922 em -0.29112 0.25205 0.02899

•e em m = 0.14912 rn: Eqn(42) 0.07222 0.07127 0.07974

•en nw = 0.01827 rm 0.08440 0.08756 0.08626

• =em mw -0.00582 rn: Unit Vector: Eqn. (43) 0.55966 0.55228 0.61788

• =enn 0.35901 rm: Unit Vector 0.56606 0.58725 0.57855

• =emm 0.38612 cos θn: Eqn. (46) = 0.99871 cos θm = 0.99989 θn = 2.91° θm = 0.86° The components of the unit vectors, w, u, v, n, m, and r, are equal to the direction cosines of the angles between the vector and the axes of the frame of reference, the indicatrix. The wave normal vector, w, was chosen to make equal angles with the axes of the indicatrix. The m and n subscripts refer to values calculated for the two vibration directions.


87

=e Kn (5.44)

where K is the tensor:

2

2

2

1 0 0

10 0

10 0

α

β

γ

(5.45)

as advertised earlier. Notice, in particular, that e and E are parallel vectors because of

Equation (43).

Nye (1957) shows that a tensor with positive principal components, such as K,

can be geometrically represented by the ellipsoid:

2 2 2

2 2 2 1X Y Zα β γ

+ + = (5.46)

Hence, we have obtained our indicatrix from Maxwell’s Equations and the tensor

relationship between D and E.

The Indicatrix and Ray Paths

In order to calculate the ray path, we first calculate e from the expression:

31 22 2 2

nn nα β γ

=e Kn = i + j + k (5.47)

the ray path is then parallel to a vector:

( )= × ×r e ew (5.48)

The location of the ray path with respect to the axes of the indicatrix is most

easily calculated if a unit vector parallel to r is first calculated. First, expand the triple

product in Equation (48) into a form that is easier to express in a numerical format:

( ) ( )= • •r e e e ew - w (5.49)

then calculate the unit vector from:


88

•

rr r

r = (5.50)

Explicitly, the two dot products in Equation (49) are given by:

22 231 2

4 4 4

nn nα β γ

• = + +e e (5.51)

22 2

3 31 1 2 22 2 2

w nw n w nα β γ

• = + +e w (5.52)

Equation (50) has the components of the wave normal vector and the principal indices of

refraction as its only variables; the ni are functions of the wi and the indices of refraction

(see Chapter 2). It is the analytical expression for calculating the ray path. An example of

the calculations is shown in Table 5-1.

In summary, to find the ray paths associated with the wave normal w, first

calculate the two vibration vectors, m and n. Next calculate the two e vectors, Km and

Kn. Finally calculate the two ray paths with Equation (50).

Magnitude of the Angle Between r and w

Optical crystallography texts state that the angle between the ray path and wave

normal is small. We now have the capability of calculating this angle and determining

how small it is. The angle between w and r will equal the angle between n and e because

of the cross product relationship shown in Equation (48) [see Figure 5-1]. The angle

between e and n is most easily obtained with the dot product:

cosθ = ••

ee e

n (5.53)

Explicitly, the two dot products are given by Equation (51) and:

22 231 2

2 2 2

nn nα β γ

• = + +en (5.54)

Equation (50) can be obtained by a second method and from which we can derive

some additional results. Consider the equation for the indicatrix written in the following


89

fashion:

( )2 2 2

2 2 2, , X Y Zf X Y Zα β γ

= + + (5.55)

where f(X,Y,Z) is a scalar function of position. The gradient of a scalar function of

position is normal to the surfaces on which the function, f(X,Y,Z) is constant (Schey,

1973, p. 138). Hence, a vector normal to the indicatrix is given by:

2 2 2

2 2 2X Y Zfα β γ

∇ = i + j + k (5.56)

A vector parallel to the vibration vector that stretches from the centre of the indicatrix to

a point (X,Y,Z) on the indicatrix is given by:

N = N n (5.57)

where N is a scalar that multiples n and gives the magnitude of N. From the definition of

the indicatrix one concludes that the value of N is equal to the refractive index of the light

vibrating parallel to n. In component form Equation (57) becomes:

X Y ZN N N

n = i + j + k (5.58)

In other words, we have:

1

2

3

XnNYnNZnN

=

=

=

(5.59)

Substitute for X, Y, and Z in Equation (56) to get:

31 22 2 22 nn nf N

α β γ

∇ =

i + j + k (5.60)

Thus, we have the result that the gradient, ∇f, is parallel to the electric vector e:

2f N∇ = e (5.61)


90

Consequently, this parallelism means that the angle between n and e can be calculated

from:

cosθ∇

= •∇ •∇

fnf f

(5.62)

Note that because ∇f is normal to the indicatrix surface, the electric vector is also.

These relationships are illustrated in Figure 5-4. The shaded plane is the plane of

the thin section and contains the vibration vector n. In this particular example, the thin

section is parallel to the X-axis. Consequently, the wave normal, the vibration vector and

the electric vector lie in the Y-Z plane, as does the ray path, parallel to r, in Figure 5-4.

Because ∇f is normal to the indicatrix surface, the ray path will be parallel to its tangent.

As a result, the vibration vector and the ray path vector, r, are parallel to conjugate radii

of the indicatrix, a fact described by Bloss (1961).

We can next examine the factors that control the size of the angle between the

vibration and electric vectors, θ. Any central section through the indicatrix is an ellipse,

including the section that contains the vibration, electric and wave normal vectors n, ∇f,

and w (Figure 5-5A). This plane is not the plane of the thin section; rather, the plane of

the thin section, the plane normal to w, is perpendicular to the ellipse under discussion

(Figure 5-5A). Label the major and minor axes of the ellipse containing n, ∇f, and w as P

and Q with unit vectors, p and q, along these axes:

( )2 2

2 2, 1P Qf P Qb a

= + = (5.63)

where a and b are the lengths of the minor and major axes of the ellipse. The gradient to

( ),f P Q is:

2 2

2 2P Qfb a

∇ = p + q (5.64)

In the plane containing n and f∇ e , n can be written in the form:

n np qn = p + q (5.65)


91

where:

2 2

2 2

PnP Q

QnP Q

=+

=+

p

q

Next calculate the cosine of the angle between n and ∇f with the dot product. After some

algebra the result is:

( )( )

2 2 2 2

4 2 4 4cos

b n b a

b n b aθ

− −=

− −

p

p

(5.66)

The substitution of trig identities into the equation for uniaxial ray paths

(Wahlstrom, 1979):

2

2tan tanωψ φ

ε= (5.67)

will transform it into Equation (66) with ω replacing a and ε replacing b. ψ is the angle

between the ray path and the major axis of the ellipse (Figure 5-5A). The results of the

calculations listed in Table 5-1 are plotted on Figure 5-5B. The wave normal vector, w, is

coplanar (in two planes) with the vibration vectors, n and m, and with the two ray paths,

rn and rm.

To find the maximum value of θ for our given ellipse, we calculate the derivative

of θ with respect to np and set the result to zero. After some more algebraic manipulation,

the result is:

2 2pmbn

a b=

+ (5.67)

2 2

2=Arccosmab

a bθ

+ (5.68)

where θm is the maximum value of θ for the particular ellipse, and npm is the value of np

for which θ is a maximum. You can check that θm is a maximum rather than a minimum


92

by substituting one and zero for np in Equation (66). In both cases, θ will be zero.

Because θm is greater than zero for positive values of a and b in Equation (68), θm must be

a maximum. If r represents the ratio of b to a with r greater than one, Equation (68)

becomes:

2

2Arccos , 11m

r rr

θ = ≥ + (5.69)

One can show by several techniques that θm increases with increasing values of r. For

any given crystal, r is largest when a is equal to α and b is equal to γ. Hence, the

maximum value for the angle between the ray path and wave normal for any given crystal

is:

2 2

2Arccosmαγ

θα γ

= +

(5.70)

Direction of the Wave Normal Farthest from the Ray Path

Because the largest angle between the ray path and the wave normal lies in the

optic axial plane where the largest and smallest indices of refraction are γ and α, the thin

section that shows this largest divergence will be normal to the optic axial plane and

parallel to the Y vibration direction. In the optic axial plane, the unit vectors

corresponding to p and q are k and i. The angle, φm, between the Z vibration direction and

the wave normal associated with the ray path most divergent from the wave normal can

be calculated with the dot product, with the result:

2

2 2cos 1m pm

anφα β

= + =+

(5.71)

Examples and Applications

A plot of θm as a function of γ and birefringence ( )γ α− is shown in Figure 5-6A.

The common rock-forming minerals, olivine, pyroxene, amphibole, and plagioclase, have

small angles between ray paths and wave normals (approximately 0.5° to 2°). Minerals

with higher birefringences have larger angles (3° - 6°, titanite; 5.5°, strontianite and


93

aragonite). Except for the calcum-bearing chain silicates, Fe-end members of solid-

solution series have larger values for qm than do the Mg-end members. This relationship

is reversed for the calcium-bearing chain silicates because of the lower birefringences of

the Fe-end members.

A plot of φm, the angle between the Z vibration direction and the wave normal that

diverges most from its associated ray path, as a function of γ and birefringence ( )γ α− is

shown on Figure 5-6B. The wave normal and ray path will diverge most in thin sections

cut parallel to the Y vibration direction and with wave normals between 45° and 50° of

the Z vibration direction. Crystals with larger birefringences show maximum divergence

between wave normals and ray paths in sections closer to the Y-Z plane of the indicatrix.

For most rock-forming minerals, the ray path and wave normal will diverge by

less than 2°. Consequently, the approximation that the wave normal and ray path are

parallel is a good. Vector algebra provides a convenient way to calculate optical

directions in crystals.

94

Appendix: Mathematical Methods

Coordinate Systems and Vector Components

It is often stated in mathematics texts that vectors are independent of a coordinate

system. Certainly the algebraic manipulation of vectors can be done without twitching

over the coordinate system. At the end of the algebra, however, we want to know a

direction or an angle. This usually means that a coordinate system must enter the

calculations at one time or another. To treat this occurrence we need to be able to

decompose a vector into its components along the coordinate axes of the frame of

reference. In this appendix we will be concerned only with coordinate systems that have

axes at right angles and which have the same scale along all three axes. The frames of

reference that we will use will all be right handed ones. Figure A-1 illustrates how a

vector in the x-y plane can be resolved into components along these axes.

The magnitude of a vector V we will denote by V. In Figure A-1, the length of the

arrow represents the magnitude; hence, V is the length of the arrow. The component of V

along the x-axis is obtained by projecting V onto the x-axis along a line normal to the x-

axis, giving a length, Vx. In like manner, Vy is the component of V resolved along the y-

axis. Notice particularly that the components are given by:

cosx xV V θ= (A1)

cosy yV V θ= (A2)

where θx and θy are the angles between the vector and the x-axis and y-axis, respectively.

In three dimensions, the components of a vector of magnitude V are:

cosx xV V θ=

cosy yV V θ= (A3)

cosz zV V θ=

Practically all the vectors that we will encounter in describing the optical

properties of crystals are unit vectors, that is, vectors of magnitude one. For example, a

unit vector parallel to the x-axis of the coordinate system in use at the time will be


95

designated i. Unit vectors along the y-axis and z-axis will be given the symbols j, and k,

respectively. In terms of components along the coordinate axes, any vector can be

written:

x y zV V V= + +V i j k (A4)

Substituting the set of equations (3) into (4) gives:

( )cos cos cosx y zV θ θ θ= + +V i j k (A5)

A unit vector parallel to V is thus given by:

cos cos cosx y zVθ θ θ= = + +

Vv i j k (A6)

This last equation makes it obvious that the cosines of the angles between the coordinate

axes and a unit vector are the components of that unit vector. These quantities, the

cosines of the angles, are used so frequently they are given the appropriate name,

direction cosines. Any time we are given the task of finding the direction cosines of a

line, we are looking for the components of a unit vector parallel to the line. Direction

cosines and the components of a unit vector are one and the same.

Products of Vectors

We frequently need to find the angle between vectors, for example, the angle

between vectors parallel to the optic axes, 2V. This angle appears in the expressions for

the products of the two vectors. There are two kinds of products, the dot or scalar product

and the cross or vector product. If you multiply two vectors according to one set of rules,

the result is simply a number or, as it is called, a scalar. If you multiply vectors according

to a second set of rules the result or product is a vector.

One way of defining the dot product of two vectors, A and B, is:

cosAB θ• =A B (A7)

where θ (0 ≤ θ ≤ 180°) is the angle between the positive ends of A and B. A and B are, as

usual, the magnitudes of the vectors. Notice that as the term scalar product implies, the

result is a number or scalar.


96

In terms of components in a particular component system, the dot product is given

by:

x x y y z zA B A B A B• = + +A B (A8)

When the vectors, a and b, are unit vectors, equations (7) and (8) become:

cos cos cos cos cos cos cosax bx ay by az bzθ θ θ θ θ θ θ• = = + +a b (A9)

where cos axθ , etc., are the direction cosines for a and b, respectively. The significance of

equation (9) lies in the fact that if we know the components of the two unit vectors we

can calculate the angle between them. For example, if we know the components of the

unit vectors parallel to the optic axes, we can use equation (9) to find 2V.

The cross or vector product can be defined by:

sinAB θ× =A B n (A10)

where θ (0 ≤ θ ≤ 180°) is the angle between the positive ends of A and B and n is a unit

vector normal to the plane of A and B. The question immediately arises, to which side of

the plane containing A and B does n point? By convention A, B and n are oriented

according to the right hand rule. Bloss (1971, p. 386-387) describes a rule for orienting n:

point the right thumb along A, the right forefinger along B, then bend the middle finger

of the right hand. It can only be bent towards n. Note that as advertised, ×A B , is a

vector. Its direction is parallel to n and its magnitude is AB sin θ.

We often need the components of a vector that is formed from a cross product. If

we are given the components of two vectors that are to be multiplied together, then the

components of the cross product are given by a rule that looks like the expansion of a 3x3

determinant. Suppose we want the components of ×A B . Then write down the unit

vectors along the coordinate axes and the components of A and B in the following

manner and in the following order:

x y z

x y z

A A AB B B

i j k


97

We next expand this expression as if it were a 3x3 determinant, using cofactors of the top

row:

( ) ( ) ( )y z z y z x x z x y y xA B A B A B A B A B A B× = − + − + −A B i j k (A11)

Depending on what information we have, either equation (10) or equation (11)

can be used to evaluate ×A B . In one case we have to know the magnitudes of the

vectors, the angle between them and the direction of n. In the other case, we need to

know the components of A and B.

Solution of the Product Equations

In the course of describing optical properties with vector algebra, we sometimes

find ourselves in the position of knowing that three vectors are related by a cross product

and of having the components of two vectors and the angle between another two vectors

given to us. Specifically, in the equation:

sin θ× =a b n (A12)

we know the components of the unit vectors, b and n, as well as the angle, θ. The

components of a are unknown. Our problem is to find the components of the third unit

vector, a. The first step towards the solution is to use the fact that if two vectors are equal,

their components referred to the same coordinate axes must be separately equal.

Combining the components of the two equations (10) and (11) provides three scalar

equations:

0 sin

0 sin 0 sin

y z z y x

x z z x y

x y y x z

a b a b na b a b n

a b a b n

θθθ

+ − =− + + =

− + = (A13)

These three equations are linear in three unknowns and can be written in matrix form in

the following fashion:

0 sin

0 sin 0 sin

z y x x

z x y y

y x z z

b b a nb b a n

b b a n

θθθ

− − = −

(A14)


98

where a is the unknown unit vector whose components we wish to find. At first glance it

appears that we have essentially solved our problem; we have three equations in three

unknowns that we should be able to solve simultaneously. However, the matrix of

coefficients (i.e. the 3x3 matrix) has a determinant equal to zero. You can verify this

directly or perhaps you remember that a skew-symmetric matrix of odd order has a zero

determinant. Consequently, the three equations are not independent and we must cast

about for a replacement for at least one of them.

If we combine equations (7) and (8), after taking into account that a and b are unit

vectors, we get:

cosx x y y z za b a b a b θ+ + = (A15)

If we replace one equation in the set (13) with equation (15) there are three possible

matrix equations, all equivalent to each other. Replacing the first equation of the set (13)

by equation (15) gives:

cos

0 sin 0 sin

x y z x

z x y y

y x z z

b b b ab b a n

b b a n

θθθ

− = −

(A16)

The determinant of the coefficient matrix in this equation is xb , which you can easily

show for yourself. To do this you will need to use the fact that the sum of squares of a

unit vector is one:

2 2 2 1x y zb b b+ + = (A17)

Solving for the three unknowns in equation (16) gives, after considerable manipulation:

cos sin sincos sin sincos sin sin

x x y z z y

y y z x x z

z z x y y x

a b b n b na b b n b na b b n b n

θ θ θθ θ θθ θ θ

= + −= + −= + −

(A18)

These last three equations provide the components of the unknown unit vector, a, that we

are seeking.


99

Constrained Minimization and Lagrange Multipliers

The measurement of optical properties can easily provide estimates of values that

are theoretically related. For example, if we measure the principal refractive indices, a, b,

and g, we can calculate 2V. We can also measure 2V as well and compare it to the

calculated value. It would be fortuitous of the two values agreed exactly because of

experimental error in determining both the refractive indices and 2V. The question arises,

what are the best estimates of the physical properties? One method of answering this

question is to recalculate both 2V and the refractive indices in such a way that they are

consistent with an equation relating them and still be as close to the measured values as

possible. An example may make these ideas more precise. The classic example of this

problem is that of a surveyor who measures the angles of a triangle. Let’s call the

measured values A’, B’, and C’. The surveyor believes Euclid and knows that their sum

must equal 180°. Let’s label the best estimates of the angles that also obey Euclid’s ideas

about triangles as A, B, and C (i.e. without primes). We next have to attach a meaning to

best values. In science generally, it has become the practice to define best such that the

sum of the squares of the errors is smallest. The errors in measuring the angles are, by

definition, (A’-A), (B’-B), and (C’-C). Thus we want the following sum to be a minimum:

( ) ( ) ( )2 2 2 2' ' 'A A B B C C S− + − + − = (A19)

Notice that if the angles did not have to sum to 180°, then the measured values are best,

according to our criterion, because the sum (19) would then be zero. But we do want the

angles to sum to 180° so we add the constraint in the following way:

( )2 180 0F S A B Cλ= + + + − = (A20)

where λ is, for the moment, an unknown number, called a Lagrange multiplier. We next

find the values of A, B, and C that make equation (20) a minimum. It is shown in calculus

books that in order for a function, such as (20), to achieve a minimum, then the partial

derivatives must be zero:


100

( )

( )

( )

2 ' 0

2 ' 0

2 ' 0

F A AAF B BBF C CC

λ

λ

λ

∂= − − + =

∂∂

= − − + =∂∂

= − − + =∂

(A21)

In addition we have the constraint equation:

180 0A B C+ + − = (A22)

Equations (21) and (22) provide four equations in four unknowns (A, B, C, and λ) that

can be solved to provide the best estimates of A, B, and C. In addition we get the bonus of

knowing λ. However, no one ever seems to put it to much use.

Transformation of Components

The solution of optical problems frequently requires that we calculate the

components of a vector in second reference frame when we know the components in an

original frame. Usually the two reference frames are the indicatrix and another tied to the

microscope. Suppose we label one reference frame by calling unit vectors along the three

axes i, j, and k. The second reference frame will be labeled with unit vectors, x, y, and z,

along its three axes. In both instances, the reference frames will be assumed to be right-

handed and Cartesian. Suppose we know the components of a vector, V, in the first

reference frame. Then we can express it as:

i j kV V V= + +V i j k (A23)

where the Vi are the known components of V. What we desire are the components of V in

the second reference frame. In other words, we would like to be able to write:

x y zV V V= + +V x y z (A24)

Unless some or all of the axes of the first reference frame parallel the axes of the second,

the components of V in the first frame will not equal any of the components in the

second.

In order to calculate the components of V in the second reference frame we have


101

to know how the two frames are related. The most common method of relating two

reference frames with a common origin is to specify the direction cosines of one set of

axes in the other reference frame. For example, the unit vector x, along the x-axis of the

second reference frame, can be written as a function of the unit vectors in the first

reference frame:

11 12 13a a a= + +x i j k (A25)

where the aij are the direction cosines of x in the i, j, k frame of reference (i.e. a11 is the

cosine of the angle between x and i, a12 is the cosine of the angle between x and j, and a13

is the cosine of the angle between x and k). The same type of relationship will obtain for y

and z:

21 22 23a a a= + +y i j k (A25)

31 32 33a a a= + +z i j k (A26)

We can use the same set of direction cosines to write equations for the unit vectors in the

first frame in terms of the unit vectors in the second frame. For example, the unit vector i,

parallel to the x-axis of the second frame, can be expressed as:

11 21 31a a a= + +i x y z (A27)

Note the relationship between the aij in the equations for x and i [equations (25) and

(27)]. The remaining two equations for the first frame unit vectors are:

12 22 32a a a= + +j x y z (A25)

13 23 33a a a= + +k x y z (A26)

In matrix notation, the two sets of equations can be written:

11 12 13

21 22 23

31 32 33

a a aa a aa a a

=

i xj yk z

(A27)

and:


102

11 21 31

12 22 32

13 23 33

a a aa a aa a a

=

x iy jz k

(A28)

The matrices of direction cosines are related by an interchange of rows and

columns. In the language of matrix algebra, the second matrix is the transpose of the first.

The two matrix equations are actually all we need to transform the components of a

vector from one reference frame to another. If the components of the vector V in the first

reference frame are known, then calculate the components of the unit vectors, i, j, and k,

in the second reference frame, equation (28), and substitute the results into equation (23).

On the other hand, if we know the components of V in the second reference frame (Vx,

Vy, Vz), then we can calculate x, y, and z in terms of i, j, and k, equation (27), and

substitute the result into equation (24).

The results can be put into a more elegant form with matrix notation:

11 12 13

21 22 23

31 32 33

i x

j y

k z

a a a V Va a a V Va a a V V

=

(A29)

This last matrix equation can be easily derived by substituting the three equations from

the set (28) into equation (23).

In summary, the matrix of direction cosines in equation (27) and its transpose can

be multiplied by the appropriate set of vector components to produce the components in

the alternate reference frame. The matrix method of transforming components can be

conveniently programmed for a computer but is not the simplest method available.

Because the sum of squares of the direction cosines is equal to one, there are only three

independent quantities in the direction cosine matrix (i.e. both the sum of squares of the

elements of each row and column of the cosine matrix must equal one). The three

independent quantities can be chosen in several ways. One way to specify the three

independent quantities is with Euler angles as described in Chapter 3.


103

Vector Identities Involving ∇

In Chapter 5, we used the ∇ notation from vector calculus to derive the

relationship between ray path and wave normal. The symbol ∇ is a differential operator

that can be treated like a vector. Specifically it is defined as:

x y z

∂ ∂ ∂∇ = +

∂ ∂ ∂i + j k (A30)

Because it is an operator, ∇ has to do something to a function of the spatial coordinates.

Suppose such a function is φ. Then:

x y zφ φ φ

φ∂ ∂ ∂

∇ = +∂ ∂ ∂

i + j k (A31)

where φ is a scalar function, and ∇φ is called the gradient of φ. As an example, suppose φ

is given by:

x y z cφ = + + − (A32)

A drawing of the surface represented by φ ιs shown on Figure A-2. The gradient of this

example function is:

φ∇ = i + j + k (A33)

which represents a vector, 3 units long, normal to the surface represented by φ

(Figure A-2).

When operating on a vector function of position, say A, ∇ can produce the

equivalent of a dot product or a cross product. The dot product equivalent is:

31 2 AA Ax y z

∂∂ ∂∇ • = + +

∂ ∂ ∂A (A34)

where A1, A2, and A3 are the components of A along the coordinate axes. ∇ • A is called

the divergence and is a scalar. As a simple example, suppose A is given by:

x y z= + +A i j k (A35)


104

Table A-1 Vector identities. φ is a scalar function of position. A and B represent vector functions. Ao represents a constant vector (i.e. it does not vary with position). 1. ( ) ( ) ( ) ( ) ( )∇ • = •∇ + •∇ + × ∇× + × ∇×A B B A A B A B B A

2. ( ) ( ) ( ) ( ) ( )∇× × = •∇ − •∇ + ∇ • − ∇ •A B B A A B A B B A

3. ( ) ( )2∇ = ∇ ∇ • − ∇× ∇×A A A

4. ( ) ( ) ( )φ φ φ∇ • = ∇ • + ∇ •A A A

5. ( ) ( ) ( )φ φ φ∇× = ∇× + ∇ ×A A A

6. ( )φ∇× ∇ = 0 7. 0∇ • =A 8. ∇× =A 0

Then ∇ • A is given by:

1 1 1 3∇ • = + + =A (A36)

a scalar as promised.

The cross product equivalent is (see p. 96):

3 32 1 2 1A AA A A Ay z z x x y

∂ ∂∂ ∂ ∂ ∂ ∇× = − + − + − ∂ ∂ ∂ ∂ ∂ ∂ A i j k (A37)

where, as before, the Ai are the components of A along the coordinate axes. Again, a

simple example to show the properties of ∇× A will suffice. Suppose A is given by:

y x= −A i + j (A38)

Then ∇× A is given by:

2∇× =A k (A39)

The vector A is shown schematically in Figure A-3, along with ∇× A . The last quantity,

∇× A , is called the curl of A.

Operations with ∇ can be combined in several ways to produce equations that are

commonly called identities, analogous to trig identities. A short list of the identities

needed in the text are listed in Table A-1.

105

References Cited

Bambauer, H. U., Taborszky, F., and Trochim, H. D., 1979, Optical Determination of

Rock-Forming Minerals [4th ed.]: Stuttgart, Germany, E. Schweizerbart'sche

Verlagsbuchhandlung (Naglele und Obermiller), v. 1, 188 p.

Bloss, F. D., 1961, An Introduction to the Methods of Optical Crystallography: New

York, Holt, Rinehart and Winston, 294 p.

---, 1971, Crystallography and Crystal Chemistry [1st ed.]: New York, Holt, Rinehart and

Winston, Inc., 545 p.

---, 1981, The Spindle Stage. Principles and Practice: Cambridge, Cambridge University

Press, 340 p.

Burden, R. L., Faires, J. D., and Reynolds, A. C., 1981, Numerical Analysis [2nd ed.]:

Boston, Prindle, Weber & Schmidt, 598 p.

Burri, C., 1956, Charakterisierung der Plagioklasoptik durch drei Winkel und Neuentwurf

des Stereogramms der optischen Orientierung fur konstante Anorthit-Intervalle:

Schweiz Mineralogische un Petrographische Mitteilungen, v. 36, p. 539-592.

Burri, C., Parker, R. L., and Wenk, E., 1967, Die optische orientierung der plagioklase:

Basel, Switzerland, Verlag Birkhauser, 333 p.

Daly, R. A., 1899, On the optical characters of the vertical zone of amphiboles and

pyroxenes; and on a new method of determining extinction angles in these minerals

by means of cleavage pieces: Proceedings of the American Academy of Arts and

Sciences, v. 34, p. 311-323.

Deer, W. A., Howie, R. A., and Zussman, J., 1992, An Introduction to the Rock-Forming

Minerals [2nd ed.]: Essex, England, Longman Group UK Limited, 696 p.

Gunter, M., and Bloss, F. D., 1982, Andalusite-kanonaite series: lattice and optical

parameters: American Mineralogist, v. 67, p. 1218-1228.

References Cited

106

Hoffmann, B., 1966, About Vectors: New York, Dover Publications, Inc., 134 p.

Johannsen, A., 1918, Manual of Petrographic Methods [2nd ed.]: New York, McGraw-

Hill Book Company, Inc.

Julian, M. M., and Bloss, F. D., 1987, Matrix calculation of optical indicatrix parameters

from central cross sections through the index ellipsoid: American Mineralogist, v.

72, p. 612-616.

Kaplan, W., 1973, Advanced Calculus [2nd ed.]: Reading, Massachusetts, Addison-

Wesley Publishing Company, 709 p.

Lovett, D. R., 1989, Tensor Properties of Crystals: Bristol, England, IOP Publishing Ltd,

139 p.

Marsden, J. E., and Tromba, A. J., 1981, Vector Calculus [2nd ed.]: San Francisco, W.H.

Freeman and Company, 591 p.

Meyer, S. L., 1975, Data Analysis for Scientists and Engineers: New York, John Wiley

and Sons, Inc., 513 p.

Nesse, W. D., 1991, Introduction to optical mineralogy [2nd ed.]: New York, Oxford

University Press, 335 p.

Nicholls, J., and Stout, M. Z., 1997, Epitactic overgrowths and intergrowths of

clinopyroxene on orthopyroxene: Implications for paths of crystallization, 1881

lava flow, Mauna Loa Volcano, Hawaii: Canadian Mineralogist, v. 35, p. 909-922.

Nye, J. F., 1957, Physical Properties of Crystals, Their Representation by Tensors and

Matrices: Oxford, England, Oxford University Press, 322 p.

Phillips, W. R., and Griffen, D. T., 1981, Optical Mineralogy, The Nonopaque Minerals:

San Francisco, W.H. Freeman and Company, 677 p.

References Cited

107

Press, W. H., Teukolsky, S. A., Vetterling, W. T., and Flannery, B. P., 1992, Numerical

Recipes in FORTRAN, The Art of Scientific Computing [2nd ed.]: Cambridge,

Cambridge University Press, 963 p.

Schey, H. M., 1973, Div, Grad, Curl and All That: New York, W.W. Norton & Company,

Inc., 163 p.

Slemmons, D. B., 1962, Determination of volcanic and plutonic plagioclases using a

three- or four-axis universal stage: Geological Society of America Special Paper, v.

69, p. 1-64.

Smith, J. V., 1982, Geometrical and structural crystallography [1st ed.]: New York, John

Wiley and Sons, Inc., 450 p.

Su, S. C., and Bloss, F. D., 1984, Extinction angles for monoclinic amphiboles or

pyroxenes: A cautionary note: American Mineralogist, v. 69, p. 309-433.

Thomas, G. B., Jr., 1968, Calculus and Analytic Geometry, 4th Edition: Reading, Mass.,

Addison-Wesley Publishing Co., 818 p.

Tobi, A. C., 1963, Plagioclase determination with the aid of the extinction angles in

sections normal to (010). A critical comparison of current Albite-Carlsbad charts:

American Journal of Science, v. 261, p. 157-167.

Tobi, A. C., and Kroll, H., 1975, Optical determination of the An-content of plagioclases

twinned by Carlsbad-Law: A revised chart: American Journal of Science, v. 275, p.

731-736.

Weber, L., 1921, Ist durch die ausloschungsschiefe von vier kristallplatten der winkel der

optischen achsen ein deutig bestimmt?: Zeitschrift fur Kristallographie, v. 56, p. 1-

11.

Vibration Directions

OA , OA = Optic Axes1 2

CS , CS = Circular Sections1 2

OA1

OA2

CS1

CS2

a

b

c

d

q

q

Figure 2-1: Stereographic projection of a random plane through a biaxial indicatrix. The wave normal for the vibration directions is the pole of the projection.

-X X

Z

Vz

OA1OA2

(0,0)

uv u3v3

u1v = - u1 1

A B

Figure 2-2. A. Schematic illustration of a general wave normal in the frame of reference defined by the axes of the indicatrix. The projections of the unit vector parallel to the wave normal, w,on the axes are the components of w in this frame of reference. The w are equal i

to the cosines of the angles w makes with the axes of the indicatrix. B. Sketch of the optic axial plane through an indicatrix showing the unit vectors, u and v, parallel to the optic axes. The components of these vectors are u , u , u and v , v , v Constraints on these 1 2 3 1 2 3.

quantities are: v = -u , u = v = 0, and v = u . The positive end of the Y axis projects into 1 1 2 2 3 3

the plane of the diagram for a right-hand system of coordinates.

wZ

X

YPlane of ThinSection

w3

w2

w1

w

t

OA1

OA2

CS1

CS2

q2 n

m

u

v

Figure 2-3. Stereographic projection of the optical indicatrix and a Biot-Fresnel construction showing the relationship of the unit vector parallel to the wave normal, w, to unit vectors parallel to the optic axes, u and v, and to unit vectors parallel to the intersections of the circular sections with the plane of the thin section, t and s. Unit vectors parallel to the vibration directions in the plane of the thin section, n and m, bisect the angles between t and s. The optic axes, OA and OA emerge from the projection at the tips of the vectors u and v 1 2

and are normal to the circular sections CS and CS .1 2

s

x

yng

w

S

y

z

wg S

S

E

Figure 2-4: Spindle stage coordinate system. The reference frame is defined by unit vectors parallel to the x, y, z axes. E and S are angles measured by rotating the microscope stage and spindle stage, respectively.

X

Z

Ya-a

c

b1

X

Z

Y-ab

a

c

2

X

Z

Yc

b

-a

a4

X

Z

Y

-a

a

b

c3

X

Z

Yc

b

a

-a6

X

Z

Ya -a

b

c

5

Figure 3-1: The six optical orientations for orthorhombic crystals. The red lines represent unit cell vectors and the black lines represent the axes of the indicatrix. Numbers correspond to the orientations listed in Table 3-1..

X

a

Y

q

bo

||bj

Z c

X

a

Z

q

bo

||bi

Y c

Y

a

Z

q

bo

||bk

X c

Figure 3-2: The three optical orientations for monoclinic crystals. Red lines represent unit cell vectors and the black lines represent indicatrix axes.

X

Y

Z

e

x||c

y [001]/(010)

zQ

Y

F > 0

A

z

XZ

Ye

<F 0

Q

Y

x||c

y [001]/(010)

CT

r(010)

x||c

z

XY

Z

y [001]/(010)

Q

B

Figure 3-3: Stereographic projections and block diagram showing the indicatrix axes, the frame of reference defined by the twin axes of the Carlsbad, Roc Tourne, and Albite twins, and the Euler angles relating the indicatrix axes to the frame of reference. A. Defining projection for Euler angles. B. Block diagram corresponding to A. C. Euler angles for plagioclase feldspars (schematic).

p = d ( 0 1 0 )

q = d ( 0 0 1 )

a

ec1

c2

Figure 3-4: Stereographic projection showing the ambiguity in locating the c-axis in feldspars from the locations of the (010) cleavage vector, p, and the (001) cleavage vector, q. Unit vectors, c and c mark the possible locations of the c-axis. The unit vector a will point along 1 2

either the positive or negative a-axis. The yellow surface contains p and q. The gray surface is (010) and is normal to p. The c-axis and the a-axis lie in (010).

( 01

0)

( hk

l)

11

1

( hk

l)

22

2

sr

c2

c1

Figure 3-5: Stereographic projection showing the ambiguity in locating the c-axis in a mineral with two cleavages, (h k l ) and (h k l ). The c-axis will be located at either c and c1 1 1 2 2 2 1 2

(red arrows). The cleavage surface plot as the yellow surfaces. The (010) is the gray surface.

Figure 4-1: A. Sketch of the lattice plane (hkl) in relation to the lattice vectors a, b, and c. The two vectors in the a-b and a-c planes, (b/k – a/h) and (c/l – a/h) also lie in the lattice plane (hkl). B. The two vectors (b/k – a/h) and (c/l – a/h) are translated to the origin. Their cross product, d(hkl), is normal to (hkl).

a

b

c

b / k

c/

l

a / h

b / k - a / h

c/

l-

a/

h

A

b / k - a / h

c/

l-

a/

ha

b

c

d ( h k l )

B

40

30

20

10

0

-10

-20

-30

-40

q

10 200 30 40 50 60 70 80 90 100

%An

Tr

(010

)

(001

)

90

Section parallel to (001) cleavage

Trace (010)

Section cut normal to a-axis 94

Tr (001)

Tr

(010)

Figure 4-2: Crystallographically defined wave normals and extinction angles in disordered plagioclase. Sections parallel to a cleavage, (001), lower curve, have a wave normal perpendicular to (001), a prominent cleavage surface. In sections cut normal to a lattice direction defined by the intersection of the (001) and (010) cleavage planes, upper curve, the wave normal is parallel to [001], the a-axis.

w = -p19

w = p1

d (hkl)

[001] (010)

[001]/(010)

w3

w5

w7

w9

w10

Figure 4-3: Sterographic projection of a set of evenly spaced wave normals that lie in the plane (hkl). d is a unit vector parallel to the pole to the plane. The first wave normal of (hkl)

the set is equivalent to a unit vector p derived from the cross product d x d .(hkl) [001]

Figure 4-4: Plot of extinction angle between the fast direction and the trace of (010) for three compositions in the disordered plagioclase series. The wave normals all lie in (010) [i.e. thin sections cut normal to (010)]. Red lines mark two hypothetical extinction angles in two individuals of a Carlsbad twins. The Carlsbad twin forces the location of the wave normals to be symmetric about the Carlsbad twin plane (90° position) andthe composition to fall on the same compositional contour.

-20 -10 0 10 20 30 40 50 60Extinction Angle, fast^Trace (010)

0

20

40

60

80

100

120

140

160

180

Wav

eN

orm

alD

irec

tion

in(0

10)

An100An50An0

14.5°

Figure 4-5: A. Carlsbad twin relations. The Carlsbad twin axis is [001] and the twin plane is normal to the twin axis. The twin plane is represented by the yellow plane and acts to reflect the upper crystal to the lower crystal. The individual parts of the twinned crystal meet along the composition plane, in this drawing (010). Ellipses (Black) represent the intersection of the indicatrix with the crystal face (schematic). B. Schematic depiction of extinction positions in an arbitrary section normal to (010).

m

A

B

0 10 20 30 40 50 60 70 80 90 100

Mole %An

0

20

40

60

80

100

120

140

160

180-15 -10 -5 0 5 10 15 20 25 30 35

40

45

50

55

60

35

35-10

-5

0 5 10 15 20 3025

aW

ave

Nor

mal

Ang

le i

n Z

one

(0

10)

c

Figure 4-6: Contours of plagioclase extinction angles between the fast direction and the trace of (010) in sections cut normal to (010).The locations of the a-axis, c-axis, and the normal to the (001) cleavage plane are indicated on the left side of the diagram. The origin of the y-axis has a wave normal parallel to [001]\(010).

The location of the trace of the Carlsbad twin plane is located at 90 . Contours were drawn with a spline interpolation algorithm (Price, et al, 1992) through the points marked with squares. Points were calculated with Optics.exe.

Figure 4-7:The relationship between the extinction angle diagram and the Carlsbad twin. The lower part of Figure 4-6 is reflected across the line at 90° on the y-axis of the chart (B). The reflected part is then moved to overlap the upper part of the diagram (A). The twin is reflected across the Carlsbad twin plane and then placed alongside the original (C).

Wav

e N

orm

al A

ngle

in

Zon

e

(010

)

0 10 20 30 40 50 60 70 80 90 100Mole %An

0

20

40

60

8035

35

c

A

100

120

140

160

180-15 -10 -5 0 5 10 15 20 25 30 35

40

45

50

55

60

a

c

0 10 20 30 40 50 60 70 80 90 100

Mole %An

0

20

40

60

80 35

35

B

m

C

0 10 20 30 40 50 60 70 80 90 100

M ole % A n

0

20

40

60

80 100

120

140

160

180-15 -10 -5 0 5 10 15 20 25 30 35

40

45

50

55

60

a

Ang

le o

f W

ave

Nor

mal

in

Zon

e

(010

)

c

Figure 4-8: Determinative chart for disordered plagioclase that display a Carlsbad twin. The contours all meet at point that represents the position and composition of the plagioclase that has an optic axis in (010).

[101]

[102] [101]

[201]

[100]

[001]

Figure 4-9: Lattice directions in (010) that can be usedas wave normal directionsto calculate extinction anglesin (010). The calculated extinction angles can be entered in an interpolation algorithm to obtain fixed, interger values for extinction angles.

d [uvw]

[001]^(010)

^[001]/(010)

w = p1

w3

w5

w7

w9

w10

w = -p19

Figure 4-10: Stereographic projectionof a set of evenly spaced wave normals that are themselves normal to the crystallographic direction [uvw]. d[uvw] is a unit vector parallel to the crystallographic direction [uvw]. The first wave normal of the set is equivalent to a unit vector p derived from the cross pproduct, d[uvw]xd[001].

Figure 4-11:Stereographic projection of the optical elements and crystallographic directions of a monoclinic crystal with Y = b. The vibration directions and wave normal are drawn for a thin section cut parallel to the c-axis, [001]. f is the angle between the wave normal and the pole to (100).

c = x

kn

sv

t

u

z

w

y

d

e

w

z = b = j

f Vectors in Sectionperpendicular to [001]

vc = xk

u

y = ^(100)

Vzq

Vectorsin section

parallel to (010)f

y = ^(100)

-90 -60 -30 0 30 60 90

q

0

30

60

90

(+)

(-)

Vz

f = 0m

15

4590

Amphiboles

Pyroxenes

Figure 4-12: Vz versus q, the extinction angle in (010) for monoclinic crystals with Y = b. Contours in fm, the angle between the pole to (100) and the wave normal that produces the maximum extinction angle in the zone [001]. Points outside the shaded areas have maximum extinction angles at f = 90°, which are sections parallell to (010). Several points are shown for amphiboles and pyroxenes.Circles - Data from Deer, et al. (1966).Triangle - Data from Su and Bloss (1984).

.

0 30 60 90 120 150 180-60

-30

0

30

604

5

1

6

2

3

f

d

w = (100)^ w = (010)^ w = (-100)^

Curve Vz q Curve Vz q1 60 15 4 45 452 72 -15 5 30 403 60 -30 6 __ 0

Figure 4-13: Examples of extinction angle curves in the zone [001] for monoclinic crystals with Y = b. : Point where the curve is mathematically undefined.

q

Vz10 20 30 40

-40 -30 -20 -10 10 20 30 40

-40 -30 -20 -10

60

90

30

0-90 -60 -30 0 30 60 90

6

3 21

4

5

Figure 4-14: Vz versus q, the characteristic extinction angle in (010) for monoclinic crystals with Y = b. Contours are in degrees for the maximum extinction angle in the zone [001]. Numbered points correspond to numbered extinction angle curves on Figure 4-13.

OA

OA

ZZ'

w

Trace

TS

OA

P

f

c

f

Z'Z

OA

OA

w

c

OA

P

Tra

ce T

S

OA

ZZ'

OA

c = OA

Trac

e TS

w

OA

P

f

f

OA

OA

Z

Z'c

w

Trace

TS

OA

P

f

Z'Zc

w

Trace

TS

OA

P

OA

OA

1 2

3 4

5Figure 4-15: Stereographic projections of crystals with the types of extinction angle curves shown on Figure 4-13. Numbers correspond to the curves labeled with the same number on Figure 4-13.

OAFo OAFo

OAFa OAFa(111)

Trace

(111)

Tra

ce

(010

)X

Z

FaFo

Figure 4-16: Stereographic projection of the optical elements of the olivine end members forsterite (Fo) and fayalite (Fa). The calculated extinction angle between the fast (X’) direction and the trace of (010) in the plane (111) is shown for each end member.

2Vz

Mole % Fa

140

130

120

110

100

90

800 25 50 75 100

Figure 4-17: 2Vz versus composition for olivine solid solutions.

Y

Tr (021)

Tr (010)T

r(021)

Y

Tr(021)

o116

o132

Tr (021)

Tr(021)

o102

o100

Y

Y

Tr (021)

Figure 4-18: Tracings of partly corroded olivine phenocrysts from a porphyritic basaltic rock (oceanite). The sections are cut nearly normal to an optic axis, consequently the Y-vibration direction lies in the plane of the thin section. Its orientation is marked. Such sections remain dark on rotation and are easily identified in thin section. An interference figure should show a single, centered isogyre. A tangent to the isogyre at the optic axis will parallel the Y-vibration direction.

80

70

60

50

0 25 50 75 100

Mole % Fo

Trace(021)^(021)

in OASections

Figure 4-19: Plot of the angle between the traces of the faces in the form {021} as seen in sections cut normal to an optic axis of a member of the olivine solid solution series.

Cpx

Opx

Y =bCpx Cpx

X =bOpx Opx

X

Z

Z =cOpx Opx

cCpx

a

OAVz

B

(001)

(100)

(01

0)

(01

0)

(001)

Aug

ite

Aug

ite

OpxOpx

A

Trace (100)

Figure 4-20: A. Sketch of clinopyroxene overgrowth on orthopyroxene from the 1881 Mauna Loa tholeiite basalt. B. Orientation of orthopyroxene and augite crystals in position of epitaxic growth. The (100) faces of the two crystals coincide.

0 10 20 30 40 50 600

10

20

30

40

50

60

70

80

90

Section^ [001]

Section|| (010)

En50

En100

X'^Tr(100) in Cpx

V Opxz

orw^c in

(100) Cpx

S8S5

En

Di Hd

Fs

S5 S8

Pg

DiHdPg

j

Figure 4-21: Extinction angles for clinopyroxenes in epitaxic overgrowths on orthopyroxene in sections cut normal to the common (100) plane. The gray area delineates values that can be measured in sections cut normal to an optic axis of orthopyroxene. The star marks the location of the coordinates on the graph from an overgrowth found in the 1881 lava flow from Mauna Loa. Inset shows the compositions of the clinopyroxenes for which the curves were drawn (Nicholls and Stout, 1997).

Figure 5-1: Schematic diagram showing the relationships between electromagnetic vectors. E is the vector representing the electric field, D is parallel to a vibration direction, w is aunit vector parallel to the wave normal. H is normal to the plane defined by E and w. R is parallel to the ray path and is defined by ExH.E, D, and w are coplanar.

V

y'

x'

xy

Vx'

Vy'

Vx

q

Figure 5-2: Diagram showing that the values of the components of a vector change with change in coordinate system but the direction and magnitude of the vector do not. In the primed coordinate system, the vector, V, has the components V’ and V’ , neither equal to x y

zero. In the unprimed system, the components of V are V = |V| and V = 0.x y

E

D

E3

E1

E2

2E2

D = E i + 2E j = K E1 2

K =1 0 00 2 00 0 0

Figure 5-3: Diagram illustrating how a vector D can be represented as the product of a second rank tensor, K, and a nonparallel vector, E.

Figure 5-4: Diagram showing the relationship between the gradient to the indicatrix, Ñf, a vibration direction, n, and their associated wave normall, w. The angle between the ray path, r, and the wave normal is equal to the angle between n and Ñf (see text).

Figure 5-5: A. Diagram of an arbitrary section through the indicatrix. Major and minor axes of the elliptical section are P and Q with lengths a and b. Ñf is the gradient of the equation of the ellipse, p and q are unit vectors parallel to the P and Q axes, w is a unit vector parallel to the wave normal, n is a unit vector parallel to the vibration direction in the plane of the ellipse, and r is parallel to the ray path. B. Stereographic projection of the wave normal, w, the vibration vectors, n and m, two ray path vectors, r and r , associated with w. The projection of w is shown with a cross. Red filled circles mark the n m

projections of the vibration directions and yellow filled circles mark the projections of the ray paths.

Figure 5-6: A. Contours of the maximumangle, q , between ray paths and wave normals as functions of the m

birefringence (g - a) and the refringence, g. B. Contours of the angle, f , between the wave normal associated m

with q and the Z vibration direction of the indicatrix. Also plotted are values expected for common minerals m

and end members of mineral solutions. Ttn = titanite, Fa = fayalite, Fo = forsterite, Fs = ferrosilite, En = enstatite, Hd = hedenbergite, Di = diopside, Act = actinolite, Tr = tremolite, An = anorthite, Ab = albite, Arg = aragonite, Str = strontianite.

X

Y

VVy

Vx

qy

qx

Figure A-1: Decomposition of a vector into its components along the x-axis and y-axis. The vector, V, lies in the x-y plane. Hence, its component along the z-axis is zero.

X

Y

Z

F

F = f(X,Y,Z)

Figure A-2: Illustration of the gradient of a function, f, defined by f = x + y + z - c.

X

Y

Z

A

A

A

A

A

AA

AA

A

A

A

Figure A-3: Diagram showing the form of the curl of a vector function, A, and the function itself. The vector function is: A = - y i + x j.

Optical Mineralogy

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