Optical Mineralogy 6d6610

ADVANCED OPTICAL MINERALOGY J. Nicholls Department of Geology and Geophysics University of Calgary Calgary, Canada

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

2

Chapter 1: Background to the Notes 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.

3

Chapter 1: Background to the Notes 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 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.

4

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 ing 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

5

Chapter 2: Vector Representation of the Indicatrix 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. The angle between t and s in the plane of the thin section. 2θ Vector sum of t and s. R A unit vector parallel to the normal to the thin section and parallel to the wave w normal in orthoscopic light S, E Angles defining the spindle stage coordinates. Unit vector normal to w and x in the spindle stage reference frame g 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. Vector parallel to n with magnitude N. N X, Y, Z The principal vibration directions of the indicatrix. Refractive indices associated with the principal vibration directions of the α, β, γ 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). 6

Chapter 2: Vector Representation of the Indicatrix 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: w = w1i + w 2 j + w 3k

(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

7

Chapter 2: Vector Representation of the Indicatrix 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: u = sin Vz i + cos Vx k

(2.2)

u = u1 i + u 3 k

(2.3)

where: u1 = sin Vz and u 3 = cos Vz. These vectors are illustrated on Figure 2-2B. The second optic axis will be parallel to a unit vector: v = − sin Vz i + cos Vz k

(2.4)

v = − u1 i + u 3 k

(2.5)

Note that:

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:

8

Chapter 2: Vector Representation of the Indicatrix

t=

(u × w )

s=

(w × v )

( u × w )<( u × w )

(2.8)

( 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:  −u3 w2i + (u3 w1 − u1w3 ) j + u1w2 k  t= 2 1- (u1 w1 + u3 w3 )

(2.10)

u3 w2 i + (u3 w1 + u1 w3 ) j + u1 w2 k  s=  2 1- (u1w1 − u3 w3 )

(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: t < s = cos 2θ = 1 − 2sin 2 θ

(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 sin θ gives: sin θ =

1 2

(1 − 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:

9

Chapter 2: Vector Representation of the Indicatrix n × t = − sin θ w

(2.14)

n
(2.15)

n × s = sin θ w

(2.16)

n< s = cos θ

(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, 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: n=

G G •G

(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)

 m1   n2 w3 − n3 w2  m  =  n w − n w   2  2 1 1 3   m3   n1w2 − n2 w1 

(2.21)

or, in component form:

(See The Appendix, Products of Vectors). 10

Chapter 2: Vector Representation of the Indicatrix 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 1.60

Indices of Refraction

Calculated Values Vz [see Bloss 1961, p. 156; Equation (44)]

β 1.62

γ 1.70

27.618

Optic Axis Vector: u Optic Axis Vector: v

i 0.46357 -0.46357

j 0.0 0.0

k 0.88606 0.88606

t: Eqn. (2.8) s: Eqn. (2.9)

-0.81620 0.52750

0.38918 -0.80348

0.42703 0.27598

G = t + s: Eqn. (2.18) n: Eqn. (2.19) m: Eqn. (2.20)

-0.28871 -0.33354 -0.74526

-0.41430 -0.47865 0.66149

0.70301 0.81219 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

11

Chapter 2: Vector Representation of the Indicatrix 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: t=

(u × w ) 2 1 − (u • w )

(2.22)

s=

(w × v ) 2 1 − (v • w )

(2.23)

Next we substitute these two equations into Equations (2.14) and (2.16) and equate the in sin θ w: n × (w × v ) 1 − (v • w )

2

n × (u × w )

=

(2.24)

1 − (u • w )

2

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): w (n • v ) − v ( n • w ) 1 - (v • w )

2

=

w (n • u ) − u ( n • w ) 1 − (u • w )

2

(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:  n•v n•u w − 2  1 - (v • w )2 1 - (u • w ) 

 =0  

(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:

(n • v )

2

1 − ( u • w )2  − ( n • u )2 1 − ( u • w )2  = 0    

12

(2.27)

Chapter 2: Vector Representation of the Indicatrix 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 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). 13

Chapter 2: Vector Representation of the Indicatrix 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 northsouth 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 w3 = cos S A similar discussion provides us with the components of g:

14

(2.28)

Chapter 2: Vector Representation of the Indicatrix g1 = 0 g2 = cos S g3 = sin S The components of n are now easily calculated from the two criteria: n × g = w sin (π 2 − E ) = w cos E n12 + n22 + n32 = 1

(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: u • v = cos 2V

15

(2.32)

Chapter 2: Vector Representation of the Indicatrix 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: u × v = sin 2Vj

(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: e × v = sin Vj e • v = cos 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: 16

Chapter 2: Vector Representation of the Indicatrix N = Nn

(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: Xi + Yj + Zk = Nn1i + Nn2 j + Nn3k

(2.37)

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

( α)

2

X

( β ) ( γ ) =1

+ Y

2

+ Z

2

(2.38)

gives: 2

2

2

 Nn1   Nn2   Nn3   +  =1   +  α   β   γ 

(2.39)

Hence: N=

αβγ

( βγ n1 ) + (αγ n2 ) + (αβ n3 ) 2

2

2

(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 17

Chapter 2: Vector Representation of the Indicatrix form:  1   1  1  1  n12  2  + n22  2  + n32  2  = 2 α  β  γ  N

(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 1 α 2 , 1 β 2 , and 1 γ 2 . 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: X = a11 x + a12 y + a13 z Y = a21 x + a22 y + a23 z Z = a31 x + a32 y + a33 z

(2.42)

where aij , j = 1, 2, 3, are the direction cosines of the X vibration direction in the spindle stage frame of reference. The a2 j and a3 j 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: n1 = a11n1∗ + a12 n2∗ + a13 n3∗ n2 = a21n1∗ + a22 n2∗ + a23 n3∗ n3 = a31n1∗ + a32 n2∗ + a33 n3∗

18

(2.43)

Chapter 2: Vector Representation of the Indicatrix where ni* , i = 1, 2,3, are the components of n in the spindle stage frame of reference and the ni 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 ( aij , j = 1, 2, 3 ) 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. 1 1 1 6. Using a least squares technique, solve for 2 , 2 , 2 . 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):

19

Chapter 2: Vector Representation of the Indicatrix

cos 2 Vz =

α 2 (γ 2 − β 2 ) β 2 (γ 2 − α 2 )

(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:

20

Chapter 2: Vector Representation of the Indicatrix F = (α − α ' ) + ( β − β ' ) + (γ − γ ') + (2Vz − aVz ' ) + 2

2

2

2

λ α 2 (γ 2 − β 2 ) − 12 β 2 (γ 2 − α 2 ) (1 + cos 2Vz )

(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 doubleangle trigonometric formula. Equation (2.44) is the constraint on the minimization. Carrying through the mathematical procedure provides the following set of equations:

{

}

α 1 + λ γ 2 − β 2 + 12 β 2 (1 + cos 2Vz ) − a ' = 0

{

}

β 1 − λ α 2 + 12 (γ 2 − α 2 ) (1 + cos 2Vz ) − β ' = 0

{

}

α 1 − λ α 2 − 12 β 2 (1 + cos 2Vz ) − γ ' = 0

(2.46)

2Vz + 14 λβ 2 (γ 2 − α 2 ) sin 2Vz − 2Vz ' = 0

α 2 (γ 2 − β 2 ) − 12 β 2 (γ 2 − α 2 ) (1 + cos 2Vz ) = 0 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 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 21

Chapter 2: Vector Representation of the Indicatrix 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? w =  3  (i + j + k )  3

22

Chapter 2: Vector Representation of the Indicatrix 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 1.779 0.001 1.545 0.001 α 1.815 0.002 1.557 0.001 β 1.827 0.001 1.561 0.001 γ 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

23

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 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 of the crystallographic vectors, a, b, and c . For example: d =u a+v b+w 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

24

(3.2)

Chapter 3: Optical Orientation of Biaxial Crystals 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

25

Chapter 3: Optical Orientation of Biaxial Crystals 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 of the indicatrix unit vectors, i, j, and k. These are set down in Table 3-2 and illustrated on Figure 3-1. Table 3-2: Components of the unit cell vectors along the indicatrix axes for the six optical orientations in orthorhombic crystals. Unit Cell Indicatrix Unit Cell Indicatrix Vector Components Vector Components i j k i j k a 0 0 -a 0 0 a a Case 1 b 0 b 0 Case 4 b 0 0 b 0 0 c 0 c 0 c c 0 a 0 0 -a 0 a a Case 2 b 0 0 b Case 5 b b 0 0 c 0 0 0 0 c c c 0 0 a 0 0 -a a a Case 3 b b 0 0 Case 6 b 0 b 0 0 c 0 c 0 0 c c 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 a26

Chapter 3: Optical Orientation of Biaxial Crystals axis and the positive c-axis and is less than 90°. Consequently, the extinction angle will lie in the range: − π2 ≤ θ ≤ π2 . 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 = bj

(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: c • k = c cos θ

(3.4)

c × k = c sin θ j

(3.5)

Next express both equations in component form: c • k = c3

(3.6)

c × k = c2 i − c1 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: 27

Chapter 3: Optical Orientation of Biaxial Crystals c1 = − c sin θ c2 = 0 c3 = c cos θ

(3.8) (3.9) (3.10)

Using these components, the equation for c in the indicatrix frame of reference becomes: c = c ( − sin θ i + cos θ k )

(3.11)

Starting with the equations relating c and a: c • a = ac cos β , c × a = ac sin β , j

(3.12) (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: a = a sin ( β , − θ ) i + cos ( β , − θ ) 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: Y

&

b

Case 2: X & b

a = a sin ( β , − θ ) i + cos ( β , − θ ) k 

a = a  cos ( β , − θ ) j + sin ( β , − θ ) k 

b = bj

b = bi

c = c [− sin θ i + cos θ k ] θ = k^c Case 3: Z & b

c = c [cosθ j − sin θ k ] θ = j^c

a = a  cos ( β , − θ ) i + sin ( β , − θ ) j  b = bk c = c [cosθ i − sin θ j ] θ = i^c

28

Chapter 3: Optical Orientation of Biaxial Crystals 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: a = ax x + a y y + az 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 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: x = x1i + x2 j + x3k

(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 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

29

Chapter 3: Optical Orientation of Biaxial Crystals 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 = cx

(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: c • a = ac cos β ,

(3.18)

c × a = ac sin β , z

(3.19)

Expanding the products in of components along the xyz axes and equating like components in the same way we treated vectors in the monoclinic cases gives: a = a  cos β , x + sin β , 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:

30

Chapter 3: Optical Orientation of Biaxial Crystals b • c = bc cos α , a • b = ab cos γ , b • b = b2

(3.21) (3.22) (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 of components along the xyz frame of reference:  cos γ , − cos α , cos β ,  b = b cos α , x + b   y+ sin β ,    1 − cos 2 α , − cos2 β , − cos2 γ , + 2 cos α , cos β , cos γ ,  b z sin 2 β ,  

(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 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: x • i = cos Θ

(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: E = x × i = sin Θe

31

(3.26)

Chapter 3: Optical Orientation of Biaxial Crystals 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: y × E = E sin Φx

(3.27)

Substitution of the magnitude of E from Equation (3.26) into Equation (3.27) gives: y × E = sin Θ sin Φx

(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: E × j = sin Θ sin Ψi

(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: E × k = sin Θ sin ( Ψ + π2 ) i

(3.30)

E × k = sin Θ cos Ψi

(3.31)

or:

Likewise, the z-axis is related to E by: E × z = sin Θ sin ( π2 − Φ ) x

(3.32)

E × z = sin Θ cos Φx

(3.33)

or:

32

Chapter 3: Optical Orientation of Biaxial Crystals 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)

( x • j ) i - ( i • j ) x = sin Θ sin Ψi

(3.35)

( x • k ) i - ( i • k ) x = sin Θ cos Ψi

(3.36)

to get:

and:

Because the dot product of a pair of normal vectors is zero, the last 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: x2 = sin Θ sin Ψ

(3.37)

x3 = sin Θ cos Ψ

(3.38)

From Equation (3.25) we get immediately: x1 = cos Θ

(3.39)

Thus, in indicatrix components, the equation for x is: x = cos Θi + sin Θ sin Ψj + sin Θ cos Ψk

(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: y × ( x × i ) = sin Θ sin Φx

(3.41)

y • ( x × i ) = sin Θ cos Φ

(3.42)

and:

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

33

Chapter 3: Optical Orientation of Biaxial Crystals z × ( x × i ) = − sin Θ cos Φx

(3.43)

z • ( x × i ) = sin Θ sin Φ

(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 a = a  cos β , x + sin β , y   cos γ , − cos α , cos β ,  b = b cos α , x + b   y+ sin β ,    1 − cos 2 α , − cos2 β , − cos2 γ , + 2 cos α , cos β , cos γ ,  b z sin 2 β ,   c = cx Unit Vectors Along the xyz Axes in the Indicatrix Reference Frame x = cos Θi + sin Θ sin Ψj + sin Θ cos Ψk y = sin Θ sin Φ i + (cos Φ cos Ψ − cos Θ sin Φ sin Ψ ) j -

(cos Φ sin Ψ + cos Θ sin Φ cos Ψ ) k z = − sin Θ cos Φi + (sin Φ cos Ψ + cos Θ cos Φ sin Ψ ) j (sin Φ sin Ψ + cos Θ cos Φ cos Ψ ) 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

34

Chapter 3: Optical Orientation of Biaxial Crystals 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: p • q = cos ε

(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: p × q = sin ε 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): c × a = sin β , p

(3.47)

c • a = cos β ,

(3.48)

35

Chapter 3: Optical Orientation of Biaxial Crystals 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: x • i = cos Θ

(3.49)

, 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: E = x × i = sin Θe

(3.51)

The magnitude of the second Euler angle, Φ, can next be calculated from the expression: y • e = sin Θ cos Φ

(3.52)

The magnitude of the third Euler angle, Ψ is then calculated from the expression: E • j = sin Θ cos Ψ

(3.53)

Again 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.

36

Chapter 3: Optical Orientation of Biaxial Crystals 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 aaxis. 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:

37

Chapter 3: Optical Orientation of Biaxial Crystals p • q = cos ε

(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: p × q = sin ε s

(3.55)

In theory, another vector parallel to s can be constructed from the unit cell properties of the mineral: S=ua+vb+wc

(3.56)

where a, b, and c are vectors along the unit cell axes and u, v, w are given by:  u   k1l2 v = h l    21  w   h1k 2

− k 2 l1  − h1l2  − h 2k1 

(3.57)

The two vectors, s and S are related by: s= The magnitude of S,

S S •S

(3.58)

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)]: s • s = u 2 a 2 + v 2 b 2 + w 2 c2 + 2uvab cos γ , + 2uwac cos β , + 2vwbc cosα ,

(3.59)

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

c×a = cos µ acsin β ,

(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: z•s =

V = cos µ s • s ac sin β ,

(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

38

Chapter 3: Optical Orientation of Biaxial Crystals 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: z • p = cos δ 1

(3.62)

z • q = cos δ 2

(3.63)

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

z ×( p × q) sin ε

(3.64)

This last expression can be expanded to give: z×s =

(z • q ) p − (z • p)q sin ε

(3.65)

Substitute Equations (3.62) and (3.63) into (3.65): z×s =

p cos δ 2 − q cos δ1 sin ε

(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

39

Chapter 3: Optical Orientation of Biaxial Crystals calculate the angle between R and c. This latter value is given by: cos χ =

l1a cos β , − h1c R•R

(3.67)

where: R • R = l12 a 2 + h12 c2 − 2h1l1ac cos β ,

(3.68)

Finally, the components of c are calculated by solving the pair of equations: c × r = sin χ z

(3.69)

c • r = cos χ

(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:

40

Chapter 3: Optical Orientation of Biaxial Crystals p• p =1

(3.72)

q•q =1

(3.73)

p • q = cos ε

(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.

41

Chapter 3: Optical Orientation of Biaxial Crystals

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.

42

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 = w ×d

(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: n • Q = w × d cos θ

(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.

43

Chapter 4: Extinction Angles in Biaxial Crystals 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 of the lattice vectors, a, b, and c: d[uvw ] = ua + vb + wc

(4.3)

d[uvw ] 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:

(

)(

d(hkl) = b − a × c − a k h l h

)

(4.4)

Expanding this expression, we can write it as: d(hkl) =

h (b × c ) + k (c × a ) + l (a × b ) , h, k , l ≠ 0 hkl

44

(4.5)

Chapter 4: Extinction Angles in Biaxial Crystals 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 d•d

(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 aaxis 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, 45

Chapter 4: Extinction Angles in Biaxial Crystals 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), ] 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

46

Chapter 4: Extinction Angles in Biaxial Crystals 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 caxis [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: P = d(hkl ) × d[001]

(4.7)

We next transform this vector into a unit vector: p=

P P•P

(4.8)

and define a set of wave normals by means of the equations: w n × p = d (hkl ) sin θ n w n • p = cos θ n

(4.9) (4.10)

where θ1 = 0, θ n+1 = θ n + ∆θ . By asg 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 47

Chapter 4: Extinction Angles in Biaxial Crystals 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: w n × d [100] = d[100] sin θ n

(4.11)

w n • d[100] = cos θ n

(4.12)

where, as before, θ1 = 0, θ n+1 = θ n + ∆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 Carlsbadalbite 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 48

Chapter 4: Extinction Angles in Biaxial Crystals 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 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

49

Chapter 4: Extinction Angles in Biaxial Crystals 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

50

Chapter 4: Extinction Angles in Biaxial Crystals 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 ) , (1 10 ) , ( 1 10 ) , 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.

51

Chapter 4: Extinction Angles in Biaxial Crystals 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: P = d[uvw ] × d[001]

(4.13)

and convert it to a unit vector: p=

P P•P

(4.14)

We can then define the set of wave normals by the equations: wn × p = d[ uvw ] sin θ n

(4.15)

wn • p = cos θ n

(4.16)

where θ1 = 0, θ n+1 = θ n + ∆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: wn × d (100) = d[001] sin θ n

(4.17)

wn • d (100) = cos θ n

(4.18)

where θ1 = 0, θ n+1 = θ n + ∆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

52

Chapter 4: Extinction Angles in Biaxial Crystals 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: x=

c c•c

(4.19)

y = d (100) , i.e. ⊥ (100 ) z=

b =j b•b

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-

53

Chapter 4: Extinction Angles in Biaxial Crystals 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: n • c = cos δ

(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: c • T = c • u − ( w • u )(c • w )

(4.25)

c • S = c • v − ( w • v )(c • w )

(4.26)

However, c and w are orthogonal vectors, consequently the last 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: T • T = 1 − (w • u )

(4.27)

S • S = 1− (w • v )

(4.28)

2

2

The results are:

54

Chapter 4: Extinction Angles in Biaxial Crystals

(c • u ) (ε + δ ) = 2 1 − ( w • u)

(4.29)

(c • v ) (ε − δ ) = 2 1 − (w • v )

(4.30)

2

cos

2

2

cos

2

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: 1 − ( w • u ) − (c • u ) 2

tan (ε + δ ) = 2

2

(c • u )

2

1 − ( w • v ) − (c • v ) 2

tan (ε − δ ) = 2

(4.31)

2

(c • v )

2

(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: c • u = cos (Vz + θ )

(4.33)

c • v = cos (Vz − θ )

(4.34)

The others are readily evaluated if we write the optic axes vectors in of c and y (Figure 4-11). u = c cos (Vz + θ ) + y sin (Vz + θ )

(4.35)

v = c cos (Vz − θ ) − y sin (Vz − θ )

(4.36)

We can quickly calculate from these last two equations the dot products with the wave normal vector: w • u = cos φ sin (Vz + θ )

(4.37)

w • v = cos φ sin (Vz − θ )

(4.38)

55

Chapter 4: Extinction Angles in Biaxial Crystals Substitution of Equations (4.33), (4.34), (4.37) and (4.38) into (4.31) and (4.32) gives: tan (ε + δ ) = tan (Vz + θ ) sin φ

(4.39)

tan (ε − δ ) = tan (Vz − θ ) sin 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 (α − β ) =

tan α − tan β 1 + tan α tan β

(4.42)

to arrive at Daly’s Equation: tan 2δ =

A sin φ 1 + B sin 2 φ

(4.43)

where: A = tan (Vz + θ ) − tan (Vz − θ )

(4.44)

B = tan (Vz + θ ) tan (Vz − 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:

56

Chapter 4: Extinction Angles in Biaxial Crystals

dδ

dφ

=

A cos φ (1 + B sin 2 φ )

2 2 (1 + B sin 2 φ ) + A2 sin 2 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 (Vz + θ ) − tan (Vz − θ ) = 0

(4.47)

cos φ = 0

(4.48)

1 − tan (Vz + θ ) tan (Vz − θ ) sin 2 φ = 0

(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: cos 2 Vz =

1 − cos 2 θ cos 2 φm 1 + sin 2 φ m

(4.50)

where φm 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 φm from Equation (4.50). Equation (4.50) is an equation in three variables, Vz, θ and φm . If we treat φm as a parameter, we can plot Vz versus θ and produce a family of curves, one curve for each

57

Chapter 4: Extinction Angles in Biaxial Crystals value of φm , 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: Vz = π 4 Vz = π 2 − θ Vz = π 2 + θ

(4.51)

The values of the parameter, φm , 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 φm . 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 φm = 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:

φm = 5.13 scarcely 5° from the pole to (100).

58

Chapter 4: Extinction Angles in Biaxial Crystals The mathematical criteria for a monoclinic crystal (with Y = b, ) 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, > Vz + θ > 90,

(4.52)

135, > θ > 45, Negative crystals:

90, > Vz + θ > 45,

(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: A sin φ (1 − B sin 2 φ + 2 B cos 2 φ ) A cos φ (1 − B sin 2 φ ) dD d 2δ = − − d 2φ D D2 dφ

(4.54)

where D is the denominator on the right hand side of Equation (4.46): 2 D = 2 (1 + B sin 2 φ ) + A2 sin 2 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

59

Chapter 4: Extinction Angles in Biaxial Crystals 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 AB 2 sin φ cos 2 φ cos φ m ≠ 0 1 − B sin 2 φm = 0 − 4 B + A2 A (1 − B ) cos φ = 0 1 − B sin 2 φm ≠ 0 2 2 (1 + B ) + A2    where:

A = tan (Vz + θ ) − tan (Vz − θ )

B = tan (Vz + θ ) tan (Vz − θ ) or:

2sin 2θ cos 2Vz + cos 2θ cos 2Vz − cos 2θ B=− cos 2Vz + cos 2θ A=

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:

60

Chapter 4: Extinction Angles in Biaxial Crystals 0 ≤ δ ≤ 45D

(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 δ m that correspond to φm [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: Vz + θ = π

2

(4.57)

Its reflection across θ equal to zero (Figure 4-14) would satisfy the requirement: Vz − θ = π

2

(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 61

Chapter 4: Extinction Angles in Biaxial Crystals conditions or one can write Daly’s Equation in the form: tan 2δ =

2sin 2θ sin φ cos φ cos 2Vz + cos 2θ (1 + sin 2 φ ) 2

(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 2δ = ±

tan 2Vz sin φ

(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: tan 2δ =

A sin φ B − C sin 2 φ

where:

62

(4.61)

Chapter 4: Extinction Angles in Biaxial Crystals A = cos 2 Vz sin 2θ B = 1 − cos 2 Vz sin 2 θ

(4.62)

C = 1 − cos Vz cos θ 2

2

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 , 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.

63

Chapter 4: Extinction Angles in Biaxial Crystals 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 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 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].

64

Chapter 4: Extinction Angles in Biaxial Crystals Table 4-2: Crystallographic and optical properties of the olivine end 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 1.636 1.827 α 1.650 1.870 β 1.670 1.878 γ 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

65

Chapter 4: Extinction Angles in Biaxial Crystals 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 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 of a crystallographic form, including cleavages, crystal faces and the composition planes of twins. If you , 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

66

Chapter 4: Extinction Angles in Biaxial Crystals the holosymmetric triclinic class, (111) and ( 1 1 1 ) but consists of four faces in the holosymmetric monoclinic class, (111), ( 1 1 1 ), (1 11) and ( 11 1 ). 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 67

Chapter 4: Extinction Angles in Biaxial Crystals 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: qhkl = w × d hkl

(4.63)

qrst = w × d rst

(4.64)

qhkl × qrst = 0

(4.65)

[w × d hkl ]× [w × drst ] = 0

(4.66)

Thus:

if the traces are parallel. Expanding this last equation [i.e. let w × d (hkl ) equal A in Equation (3.34), Chapter 3] gives:

{d • [w × d ]} w − {w • [w × d ]}d rst

hkl

hkl

rst

=0

(4.67)

The last term is zero because w is normal to w × d hkl 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: d rst • [w × d hkl ]

(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: w • [d hkl × d rst ] = 0

(4.69)

Equation (4.69) can be true only if one or more of the following situations occur: 1. Either w or d hkl × d rst 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

68

Chapter 4: Extinction Angles in Biaxial Crystals 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 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 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.

69

Chapter 4: Extinction Angles in Biaxial Crystals 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=ua+vb+wc 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: 70

Chapter 4: Extinction Angles in Biaxial Crystals a = 0.973 nm b = 0.895 nm c = 0.526 nm b° = 108°

b=X 2Vz = 25° θ = Y^c =:49.0°

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

71

Chapter 4: Extinction Angles in Biaxial Crystals c b° Cleavage

≅ 0.525 nm ≅ 105° {110}

≅ 0.525 nm ≅ 105° {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.

72

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. 73

Chapter 5: Location of ray Paths 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

74

Chapter 5: Location of ray Paths 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: ∇×H =

∂D ∂t

∇ × E = − µD

∂H ∂t

(5.2)

(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

75

Chapter 5: Location of ray Paths 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:

(

E = ED cos ω t − ω r • w

v

)

(5.4)

where ED 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 ± ED , 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 ED , 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

76

Chapter 5: Location of ray Paths 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 = Xi + Yj + Zk

(5.5)

r • w = Xw1 + Yw2 + Zw3

(5.6)

Next, we can show that Equation (4) satisfies the classic wave equation, which is a partial differential equation of the form: v 2∇ 2 F =

∂2F ∂t 2

(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: E = φ ED

(5.8)

where φ is a scalar function of the space and time coordinates:

(

φ = cos ω t − ω r • w

v

)

(5.9)

The second derivative of Equation (8) with respect to time is: ∂ 2E ∂ 2φ = E D ∂t 2 ∂t 2

(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: v 2∇ 2 E = ED ( v 2∇ 2φ )

(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

77

Chapter 5: Location of ray Paths 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): µD

∂H = − ( ∇φ ) × ED ∂t

(5.12)

Next calculate the gradient of φ from Equation (9): ∇φ =

ω∇ ( r • w ) v

ω (r • w )   sin  ω t −  v  

(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:  ω (r • w )  ω sin  ω t −  v   φ'= v

(5.15)

then, using Equation (14), Equation (12) becomes: 1 ∂H = − φ ' ( w × ED ) ∂t µD

(5.16)

Equation (16) is important because at a fixed point in space [ 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:

78

Chapter 5: Location of ray Paths H=

φ ( w × ED ) µD v

(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): D=−

φ w × ( w × ED ) µD v

(5.18)

Because there are no derivatives left in Equations (17) and (18), we can combine ED and φ and recover E in Equations (17) and (18): w×E µD v

(5.19)

w × ( w × E) µD v 2

(5.20)

H=

D=−

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: µD v 2 D + ( w • E ) w - E = 0

(5.21)

Equation (21) is a vector equation relating D and E. It has been derived without

79

Chapter 5: Location of ray Paths 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 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: E = µD v 2 D

(5.22)

The 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.

80

Chapter 5: Location of ray Paths 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: V = Vx i rather than:

81

(5.23)

Chapter 5: Location of ray Paths V = Vx' i' + V y' 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: κ 11 E1 + κ 12 E2 + κ 13 E3 = D1 κ 21 E1 + κ 22 E2 + κ 23 E3 = D2 κ 31 E1 + κ 32 E2 + κ 33 E3 = D3

(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: κ 11 κ 12 κ 13   E1   D1  κ      21 κ 22 κ 23   E2  =  D2  κ 31 κ 32 κ 33   E3   D3 

(5.26)

An even simpler notation for the same equations is: KE = D

(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: κ 1 0 0   E1   D1  0 κ 0   E2  =  D2  2   0 0 κ 3   E3   D3 

(5.28)

where the κ i are nonzero components of K in this new reference frame. The restrictions on the tensor and details of obtaining the κ i from the κ ij 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

82

Chapter 5: Location of ray Paths coordinate system. Using the tensor representation in Equations (28), then we can write out Equation (27) as: κ 1 E1i + κ 2 E2 j + κ 3 E3k = D

(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: κE=D

(5.30)

Comparing Equations (22) and (30) we see that: κ=

1 µD v 2

(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:

83

Chapter 5: Location of ray Paths ∇•E = 0

(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): ∇ × ( ∇ × E ) = − µD

∂ [∇ × H ] ∂t

(5.34)

∂2D ∂t 2

(5.35)

∇ × ( ∇ × E ) = − µD

∇ × ( ∇ × E ) = − µDκ D

∂ 2E ∂t 2

(5.36)

Next use identity 1 in Table A-1 and take into Equation (33): ∂ 2E ∇ E = µDκ D 2 ∂t 2

(5.37)

Equation (37) has the form of the classic wave equation [cf. Equation (7)] if we make the following identification: c2 =

1 µDκ D

(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):

84

Chapter 5: Location of ray Paths v12 =

1 µDκ 1

v22 =

1 µDκ 2

v32 =

1 µDκ 3

(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: κ1 = κ D

c2 v12

κ2 = κD

c2 v22

κ3 = κD

c2 v32

(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 Xaxis as the direction along which D1 lies then κ1 is the tensor component of interest. If κ1 2 is assigned the smallest value, then v1 must be the fastest. But we know that c

v12

, 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: κ Dα 2 0 0   E1   D1    κ Dβ 2 0   E2  =  D2   0  0 0 κ Dγ 2   E3   D3  

(5.41)

Finally we note that the relationship between n and D can be expressed: n=

D D•D

If we define a new vector e in the following fashion:

85

(5.42)

Chapter 5: Location of ray Paths 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 n • e n = 0.12922 -0.29112 0.25205 0.02899 em e m • e m = 0.14912 rn: Eqn(42) 0.07222 0.07127 0.07974 e n • wn = 0.01827 0.08440 0.08756 0.08626 rm e m • wm = -0.00582 rn: Unit Vector: Eqn. (43) 0.55966 0.55228 0.61788 n • e n = 0.35901 rm: Unit Vector 0.56606 0.58725 0.57855 m • e m = 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.

e=

κ DE D•D

We can write the set of Equations (41) as:

86

(5.43)

Chapter 5: Location of ray Paths e = Kn

(5.44)

where K is the tensor: 1  α2   0   0 

 0   0   1 2 γ 

0 1

β2

0

(5.45)

as d 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: X 2 Y2 Z2 + + =1 α2 β 2 γ 2

(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: e = Kn =

n1 n n i + 22 j + 32 k 2 α β γ

(5.47)

the ray path is then parallel to a vector: r = e × ( w × e)

(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) w - (e • w ) e then calculate the unit vector from:

87

(5.49)

Chapter 5: Location of ray Paths

r=

r r •r

(5.50)

Explicitly, the two dot products in Equation (49) are given by: n12 n22 n32 + + α4 β 4 γ 4

(5.51)

w1n12 w2 n22 w3 n32 + 2 + 2 α2 β γ

(5.52)

e•e =

e•w =

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 = n •

e e•e

(5.53)

Explicitly, the two dot products are given by Equation (51) and: n•e =

n12 n22 n32 + + α2 β 2 γ 2

(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

88

Chapter 5: Location of ray Paths fashion: f ( X ,Y , Z ) =

X 2 Y2 Z2 + + α2 β2 γ 2

(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: ∇f =

2X 2Y 2Z i+ 2 j+ 2 k 2 α β γ

(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=Nn

(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: n=

X Y Z i+ j+ k N N N

(5.58)

In other words, we have: X N Y n2 = N Z n3 = N n1 =

(5.59)

Substitute for X, Y, and Z in Equation (56) to get: n n n ∇f = 2 N  12 i + 22 j + 32 β γ α

 k 

(5.60)

Thus, we have the result that the gradient, ∇f, is parallel to the electric vector e: ∇ f = 2 Ne

89

(5.61)

Chapter 5: Location of ray Paths Consequently, this parallelism means that the angle between n and e can be calculated from: cosθ = n •

∇f ∇f • ∇ 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: f ( P, Q ) =

P2 Q2 + =1 b2 a 2

(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: ∇f =

2P 2Q p+ 2 q 2 b a

(5.64)

In the plane containing n and ∇fe , n can be written in the form: n = n p p + nq q

90

(5.65)

Chapter 5: Location of ray Paths where: np = nq =

P P2 + Q2 Q P + Q2 2

Next calculate the cosine of the angle between n and ∇f with the dot product. After some algebra the result is: cos θ =

b 2 − n 2p ( b 2 − a 2 )

b 4 − n 2p ( b 4 − a 4 )

(5.66)

The substitution of trig identities into the equation for uniaxial ray paths (Wahlstrom, 1979): tanψ =

ω2 tan φ ε2

(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: n pm =

b a + b2 2

 2ab  θ m =Arccos  2 2   a +b 

(5.67)

(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

91

Chapter 5: Location of ray Paths 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:  2r  θ m = Arccos  , r ≥1 2   1+ r 

(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:  2αγ  θ m = Arccos  2 2  α +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: cos φm = 1 + n 2pm =

a α2 + β2

(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

92

Chapter 5: Location of ray Paths aragonite). Except for the calcum-bearing chain silicates, Fe-end of solidsolution series have larger values for qm than do the Mg-end . This relationship is reversed for the calcium-bearing chain silicates because of the lower birefringences of the Fe-end . 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.

93

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 xaxis, giving a length, Vx. In like manner, Vy is the component of V resolved along the yaxis. Notice particularly that the components are given by: Vx = V cosθ x

(A1)

Vy = V cos θ y

(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: Vx = V cosθ x Vy = V cos θ y

(A3)

Vz = V cos θ z 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 94

Appendix: Mathematical Methods designated i. Unit vectors along the y-axis and z-axis will be given the symbols j, and k, respectively. In of components along the coordinate axes, any vector can be written: V = Vx i + Vy j + Vz k

(A4)

Substituting the set of equations (3) into (4) gives: V = V ( cos θ x i + cos θ y j + cosθ z k )

(A5)

A unit vector parallel to V is thus given by: v=

V = cosθ x i + cos θ y j + cosθ z k V

(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: A • B = AB cos θ

(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.

95

Appendix: Mathematical Methods In of components in a particular component system, the dot product is given by: A • B = Ax Bx + Ay By + Az Bz

(A8)

When the vectors, a and b, are unit vectors, equations (7) and (8) become: a • b = cosθ = cosθ ax cosθ bx + cosθ ay cosθ by + cosθ az cosθ bz

(A9)

where cosθ ax , 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: A × B = AB sin θ 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 d, 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: i Ax Bx

j Ay By

k Az Bz

96

Appendix: Mathematical Methods We next expand this expression as if it were a 3x3 determinant, using cofactors of the top row: A × B = ( Ay Bz − Az By ) i + ( Az Bx − Ax Bz ) j + ( Ax By − Ay Bx ) 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: a × b = sin θ 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 + a y bz − az by = sin θ nx − ax bz + 0 + az bx = sin θ ny

(A13)

ax by − a y bx + 0 = sin θ nz These three equations are linear in three unknowns and can be written in matrix form in the following fashion:  0   −bz  by 

bz 0 −bx

−by   ax   sin θ nx   bx   a y  = sin θ n y  0   az   sin θ nz 

97

(A14)

Appendix: Mathematical Methods 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 this directly or perhaps you 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 that a and b are unit vectors, we get: ax bx + a y by + az bz = cos 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:  bx   −bz  by 

by 0 −bx

bz   ax   cos θ   bx   a y  = sin θ n y  0   az   sin θ nz 

(A16)

The determinant of the coefficient matrix in this equation is bx , 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: bx2 + by2 + bz2 = 1

(A17)

Solving for the three unknowns in equation (16) gives, after considerable manipulation: ax = bx cosθ + by nz sin θ − bz n y sin θ a y = by cosθ + bz nx sin θ − bx nz sin θ az = bz cosθ + bx n y sin θ − by nx sin θ

(A18)

These last three equations provide the components of the unknown unit vector, a, that we are seeking.

98

Appendix: Mathematical Methods 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:

( A '− A) + ( B '− B ) + ( C '− C ) 2

2

2

= S2

(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: F = S 2 + λ ( A + B + C − 180 ) = 0

(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:

99

Appendix: Mathematical Methods ∂F = −2 ( A '− A ) + λ = 0 ∂A ∂F = −2 ( B '− B ) + λ = 0 ∂B ∂F = −2 ( C '− C ) + λ = 0 ∂C

(A21)

In addition we have the constraint equation: A + B + C − 180 = 0

(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 righthanded and Cartesian. Suppose we know the components of a vector, V, in the first reference frame. Then we can express it as: V = Vi i + V j j + Vk 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: V = Vx x + Vy y + Vz 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

100

Appendix: Mathematical Methods 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: x = a11i + a12 j + a13 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: y = a21i + a22 j + a23 k

(A25)

z = a31i + a32 j + a33k

(A26)

We can use the same set of direction cosines to write equations for the unit vectors in the first frame in 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: i = a11 x + a21 y + a31 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: j = a12 x + a22 y + a32 z

(A25)

k = a13 x + a23 y + a33 z

(A26)

In matrix notation, the two sets of equations can be written:  a11 a  21  a31

a12 a22 a32

a13   i   x  a23   j  =  y  a33   k   z 

and:

101

(A27)

Appendix: Mathematical Methods  a11 a  12  a13

a21 a22 a23

a31   x   i  a32   y  =  j  a33   z   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 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:  a11 a  21  a31

a12 a22 a32

a13  Vi  Vx  a23  V j  = Vy  a33  Vk  Vz 

(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.

102

Appendix: Mathematical Methods 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: ∇=

∂ ∂ ∂ j+ k i+ ∂x ∂y ∂z

(A30)

Because it is an operator, ∇ has to do something to a function of the spatial coordinates. Suppose such a function is φ. Then: ∇φ =

∂φ ∂φ ∂φ i+ j+ k ∂x ∂y ∂z

(A31)

where φ is a scalar function, and ∇φ is called the gradient of φ. As an example, suppose φ is given by: φ = x+ y+ z−c

(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: ∇•A =

∂A1 ∂A2 ∂A3 + + ∂x ∂y ∂z

(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: A = xi + y j + zk

103

(A35)

Appendix: Mathematical Methods 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 A ) = ( ∇φ ) • A + φ ( ∇ • A )

5.

∇ × (φ A ) = φ ( ∇ × A ) + ( ∇φ ) × A

6.

∇ × ( ∇φ ) = 0

7. 8.

∇ • AD = 0 ∇ × AD = 0

Then ∇ • A is given by: ∇ • A = 1+1+1 = 3

(A36)

a scalar as promised. The cross product equivalent is (see p. 96):  ∂A ∂A   ∂A ∂A   ∂A ∂A  ∇× A =  3 − 2 i +  1 − 3  j +  2 − 1 k ∂z   ∂z ∂x   ∂x ∂y   ∂y

(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: A = − yi + xj

(A38)

∇ × A = 2k

(A39)

Then ∇ × A is given by:

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.

104

References Cited Bambauer, H. U., Taborszky, F., and Trochim, H. D., 1979, Optical Determination of Rock-Forming Minerals [4th ed.]: Stuttgart, , 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 Birkha, 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.

105

References Cited Hoffmann, B., 1966, About Vectors: New York, Dover Publications, Inc., 134 p. Johannsen, A., 1918, Manual of Petrographic Methods [2nd ed.]: New York, McGrawHill 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, AddisonWesley 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.

106

References Cited 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. 111.

107

OA1, OA2 = Optic Axes Vibration Directions CS1, CS2 = Circular Sections c a CS2

q OA1 CS1

q OA2

b d 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.

Z

w Z w3

Y

Plane of Thin Section

OA2

OA1

w2

Vz v3

-X

X

A

u1

v1 = - u1

w1 v

u

(0,0)

u3

X

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 wi are equal 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 u1, u2, u3 and v1, v2, v3. Constraints on these quantities are: v1 = -u1, u2 = v2 = 0, and v3 = u3. The positive end of the Y axis projects into the plane of the diagram for a right-hand system of coordinates.

m

2q

CS1

t

OA2

n

v u

s

OA1

w

CS2

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, OA1 and OA2 emerge from the projection at the tips of the vectors u and v and are normal to the circular sections CS1 and CS2.

x S g n

E y

w z w

S g

S y

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.

Z

Z

c c 1

-a

b

a

2 -a b

Y

a

Y

X

X Z

Z b

3 b

a

c

a

4 Y

c

Y

-a

-a

X

X Z

Z

c 5 a

-a

Y

c

b X

-a

6

b

Y

a

X

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..

Z

Y

c q

X a

Z a

Y j || b

o

o

b

c q

q

o

b

X

c

Y

b

a X i || b

Z k || b

Figure 3-2: The three optical orientations for monoclinic crystals. Red lines represent unit cell vectors and the black lines represent indicatrix axes.

x||c

Y Y

Y

B

e x||c

Z

X

z

Q X

F>0 Z

Tr(010)

A

Q

y [001]/(010)

C

y [001]/(010) X Z

Q e

x||c

z

Y Y

F< 0 y [001]/(010) 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).

z

q = d (001) e c1 p = d (010)

a

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, c1 and c2 mark the possible locations of the c-axis. The unit vector a will point along 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).

r

c1

s

(01

0)

(h

1

l 2) k2 (h 2

k1 l

1

)

c2

Figure 3-5: Stereographic projection showing the ambiguity in locating the c-axis in a mineral with two cleavages, (h1k1l1) and (h2k2l2). The c-axis will be located at either c1 and c2 (red arrows). The cleavage surface plot as the yellow surfaces. The (010) is the gray surface.

c

c/l

B

c/l

-a/

A

h

c

b/k-

b

a/h b/k-

b a/h -a/

a

c/l

a

b/k

h

h a/

d (hkl)

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).

40 Tr (001)

30

Tr (010)

20 Section cut normal to a-axis

94

10 Trace (010)

0 -10 -20

(001)

90 Tr (010)

q

Section parallel to (001) cleavage

-30 -40 0

10

20

30

40

50

60

70

80

90

100

%An 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.

w19 = -p w10 w9 w7 w5 w3

[001]

(010)

d (hkl)

w1 = p [001]/(010) Figure 4-3: Sterographic projection of a set of evenly spaced wave normals that lie in the plane (hkl). d(hkl) is a unit vector parallel to the pole to the plane. The first wave normal of the set is equivalent to a unit vector p derived from the cross product d(hkl) x d[001].

180

Wave Normal Direction in (010)

160

An0

An50

An100

140 120 100

14.5°

80 60 40 20 0

-20 -10

0

10

20

30 40

50

60

Extinction Angle, fast^Trace (010) 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.

m

B A

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).

-15 -10 -5 0

180

5

10

15

20

25

30

35

40

Wave Normal Angle in Zone (010)

160

a 45

140 50

120 55

100 60

c

80

35

60 40 0

5

10

15

20

30

-5

20 -10

0

25

0

35

10 20 30 40 50 60 70 80 90 100

Mole %An 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.

Wave Normal Angle in Zone (010)

0

Mole %An

10 20 30 40 50 60 70 80 90 100

-15 -10 -5 0

180

5

10

15

20

25

30

35

35

80

40

160

a

60

45

140 50

40 120

55

20 35

100 60

c

0

A

m c

80

35

60 40 20 35

0

0

10 20 30 40 50 60 70 80 90 100

Mole %An

C

B 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).

Angle of Wave Normal in Zone (010)

0 180

-15 -10 -5 0

5

10

15

20

25

30

35

40

20 160

a 45

40 140 50

60 120 55

80 100 60

c

0

10

20

30

40

50

60

70

80

90 100

M ole % A n 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).

[001] [102]

[101] [201]

[101]

[100]

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.

w19 = -p

w10 w9

[001]

^(010)

w7

d [uvw]

w5 w3 w1 = p

^[001]/(010) 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].

z = b = j f w

Vectors in Section perpendicular to [001]

y = ^(100)

v n e

d

s c=x k

z

t u y

w

Vectors in section parallel to (010)

f

k c = xv u

Vz

q

y = ^(100) 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).

90 fm = 0

(-)

15

60

45 90

Vz 30 (+) 0 -90 Pyroxenes Amphiboles

-60

-30

0

.

30

60

90

q

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).

60 4 5

30 1 6

d 0

2 3

-30 -60

0

30

60

90

120

150

180

f w = ^(010)

w = ^(100) Curve 1 2 3

Vz 60 72 60

q 15 -15 -30

Curve 4 5 6

w = ^(-100) Vz 45 30 __

q 45 40 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.

6

90 3

60

2 1 4 -40 -30 -20 -10

10 20 30 40

Vz

-40 -30 -20 -10

10 20 30 40

30

5

0 -90

-60

-30

0

30

60

90

q 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.

OAP

OAP OA Z'

e ac Tr

c

TS

Z c

Z

e ac Tr

OA

1

OAP Tr ac e TS

OAP

f

Z'

c = OA OA

Z'

w OA

4

f

Z'

e ac Tr

c

OA

OA

f

TS

Z OA

5

Z

w OAP

Tra ce TS

c

3

OA

2

OA

Z

Z'

TS

w

w

f

OA

f

w

Figure 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.

OAFa OAFo

Fa

Fo X

Trace (010)

e ac r T

1) 1 (1

OAFa

(111)

OAFo

Z Figure 4-16: Stereographic projection of the optical elements of the olivine end 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.

140 130 120

2Vz

110 100 90 80

0

25

50

75

Mole % Fa Figure 4-17: 2Vz versus composition for olivine solid solutions.

100

21)

Tr(0

116 o

Y ) Tr (021

Tr (021) ) 021 Tr(

102o

Y

Y 132

o

100o

Y

1) (02 Tr

Tr ( 010 )

1) Tr(02

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

Trace 70 (021)^(021) in OA 60 Sections 50 0

25

50

75

100

Mole % Fo 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.

Z Opx =c Opx cx

Vz

Trace (100)

OA

(001)

(100)

Opx

(010)

Z

Opx

(010)

Au git e

(001)

X Opx =b Opx Y x =b x

A

Au git e

x

a

X

B

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.

Section || (010) 90

S5 S8 Di S5

80

Hd S8 Pg

70 En

Fs

En50

60

Vz Opx 50 or w^c in 40 (100) x

j

30

En100 Di Hd Pg

20 10

Section 0 ^ [001] 0

10

20

30

40

50

60

X'^Tr(100) in x 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.

y' y

x

Vx

Vy' V

q

Vx' 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’x and V’y, neither equal to zero. In the unprimed system, the components of V are Vx = |V| and Vy = 0.

x'

D = E1 i + 2E2 j = K E E3

1 0 0 K = 0 2 0 0 0 0 E

E1 E2

D

2E2 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, rn and rm, associated with w. The projection of w is shown with a cross. Red filled circles mark the projections of the vibration directions and yellow filled circles mark the projections of the ray paths.

Figure 5-6: A. Contours of the maximumangle, qm, between ray paths and wave normals as functions of the birefringence (g - a) and the refringence, g. B. Contours of the angle, fm, between the wave normal associated with qm and the Z vibration direction of the indicatrix. Also plotted are values expected for common minerals and end 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.

V

Vy

Y qx qy X

Vx

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.

Z F = f(X,Y,Z) F

Y

X Figure A-2: Illustration of the gradient of a function, f, defined by f = x + y + z - c.

Z

A

A A

A A A

A A

A A

Y

A A

X 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.