Three places for the two-fold axis

The axes a class pins down separated two things symmetry does to a property tensor. It fixes the tensor’s shape, how many of its principal values are equal, and it fixes some of the tensor’s orientation, how many of its principal directions are the same for every crystal of the class. A monoclinic crystal fixes one direction, the two-fold axis $b$, and leaves the other two free to turn in the plane across it. Because a principal direction is a property like any other, it depends on wavelength, and in white light a monoclinic crystal’s indicatrix is a family of ellipsoids sharing one axis and fanning slightly about it.

That essay stopped at one fixed axis. It did not ask which of the ellipsoid’s three axes that is, and the question has three answers, all of which occur in real crystals. They are what mineralogy textbooks call inclined, horizontal and crossed dispersion, and the textbooks introduce them as three appearances of an interference figure to be recognised and memorised. This essay derives all three from a single computation: the subgroup of the crystal’s Laue group that leaves the line of sight in place.

The computation has a corollary that is easy to state and easy to get wrong. In light of one colour the three are indistinguishable, from each other and from an orthorhombic crystal. A single ellipsoid always has three mirror planes, whatever crystal it belongs to, so the lower symmetry of a monoclinic crystal is invisible until a second wavelength is compared with the first.

Three axes, and the plane of the optic axes #

The vocabulary is fixed first, since the result depends on it. A biaxial crystal has three different principal refractive indices. Call the direction of the least index $X$, of the intermediate index $Y$ and of the greatest index $Z$. The two optic axes, the directions along which both polarisations travel at the same speed, lie in the plane containing $X$ and $Z$. That plane is the optic axial plane, and $Y$, perpendicular to it, is the optic normal. The optic axes sit symmetrically about $Z$, at an angle $V_z$ given by the three indices, and symmetrically about $X$ at $90^\circ - V_z$. Whichever of $X$ and $Z$ makes the smaller angle with them is the acute bisectrix, and the other is the obtuse bisectrix.

An interference figure is what a crystal plate looks like between crossed polars in strongly convergent light. Each point of the figure is a direction through the plate, and the two optic axes appear as two dark points called melatopes. A plate cut across the acute bisectrix shows both melatopes in the field of view, and that is the figure this essay draws throughout. In the figures here the melatopes are placed by a gnomonic projection about the acute bisectrix at 589 nanometres, the sodium line.

In a monoclinic crystal the two-fold axis $b$ must be one of $X$, $Y$ and $Z$, because the indicatrix must be unchanged by the half-turn about it, and an ellipsoid with three different axes is unchanged by a half-turn only about one of its own axes. So $b$ is the optic normal, or it lies in the optic axial plane as one of the two bisectrices, acute or obtuse. Those are the three places it can be.

In every case the turning directions stay in the plane across $b$, and only the identity of the pair differs. That difference is the whole content of the classification. When $b$ is $Y$, the optic axial plane is the plane across $b$; as the colour changes, the bisectrices turn within it, and the optic axes stay in that same plane while their bisector moves. When $b$ is the obtuse bisectrix, the plane across $b$ contains the acute bisectrix and the optic normal, which turn together; the optic axial plane always contains $b$, so it swings about $b$ like a door on a hinge. When $b$ is the acute bisectrix, the optic axial plane contains $b$ and turns about it, and $b$ is the direction the figure is viewed along, so the plane turns about the centre of the figure.

What a figure keeps #

The pictures can be computed directly, and are. Each model crystal here has three principal indices following Cauchy’s formula, $n(\lambda) = A + B/\lambda^2$, with a different $B$ for each direction so that the angle between the optic axes changes slightly with colour. In the monoclinic cases the principal frame turns about $b$ by about nine degrees between 400 and 700 nanometres. That is more than most real crystals show, and it is chosen so the effect is visible at the scale of a figure. The symmetry argument depends on none of these numbers. At thirteen wavelengths across the visible, the optic axes are computed from the indices and the frame, and their positions in the figure plotted.

Those four patterns are the textbook descriptions, and the figure reproduces them without being told what to draw. The orthorhombic crystal’s colours spread symmetrically along the optic axial plane, which is the ordinary dispersion of the optic axes: $2V$ is different in red and in blue, and nothing else changes. In inclined dispersion the optic axes stay in one plane and their two ends move by different amounts, because the acute bisectrix itself is tilting within that plane. In horizontal dispersion the optic axial planes for different colours are parallel and displaced. In crossed dispersion they cross one another at the centre.

The symmetry of each pattern can be read off the tracks. Take the four operations an interference figure could have when its melatopes lie on a horizontal line through the centre: the identity, the half-turn about the centre, the mirror along the line and the mirror across it. Keep those that carry the whole set of coloured melatopes onto itself, colour by colour. The orthorhombic crystal keeps all four. The crystal with $b$ along the optic normal keeps the identity and the mirror along. The crystal with $b$ along the obtuse bisectrix keeps the identity and the mirror across. The crystal with $b$ along the acute bisectrix keeps the identity and the half-turn. The triclinic model, whose principal frame turns about a direction in no special position, keeps only the identity.

The same answer from the class alone #

That reading took the computed tracks as data. The same four answers come from the crystal’s class without any optics at all, and this is the step that explains them rather than describing them.

An optical property of this kind is a symmetric second-rank tensor, unchanged by inversion, so by Neumann’s principle the symmetry that matters for it is the crystal’s Laue group, its point group with the inversion added. For a monoclinic crystal that is $2/m$: the identity, the half-turn about $b$, the mirror across $b$, and the inversion. An interference figure is a picture of directions near a line of sight, and a direction and its opposite are the same ray through the plate. So an operation of the Laue group acts on the figure when it carries the line of sight to itself, in either sense. An operation that reverses the line of sight acts on rays as its negative does, which turns it into one that keeps the line and acts on the plane of the figure.

The figure’s symmetry is therefore the stabiliser of the line of sight in the Laue group, read as a group of operations of a plane. Computed case by case, it is the following. Looking down $b$, every operation of $2/m$ keeps the line, the half-turn acts as a half-turn of the figure, and the mirror across $b$, which reverses the line, acts as its negative, the half-turn again: the figure’s group is $2$. Looking across $b$, along a direction in the mirror plane, the half-turn about $b$ reverses the line and acts as its negative, the mirror across $b$; that mirror contains the line of sight, and on the figure it is a mirror whose line is the trace of the plane across $b$. The figure’s group is $m$, and where its mirror line falls depends on where the optic axial plane is. If $b$ is the optic normal, the plane across $b$ is the optic axial plane, and the mirror runs along it. If $b$ is the obtuse bisectrix, the plane across $b$ is perpendicular to the optic axial plane, and the mirror runs across it.

Both derivations agree in all five cases, and the agreement is required rather than noticed: the stabiliser is computed from the matrices of the Laue group and the tracks from the indices and the frame, and a disagreement in any case stops the calculation. That is the sense in which the names are consequences. Inclined dispersion is a figure whose symmetry is a mirror along the optic axial plane; horizontal dispersion is a mirror across it; crossed dispersion is a half-turn. The three textbook descriptions are what those three groups permit a family of ellipsoids sharing one axis to do.

This is the Curie principle at work in an unusual place. The effect, a figure in white light, has at least the symmetry of its cause, the crystal and the line of sight together. It can have more by accident, and the orthorhombic model shows what more looks like. A monoclinic crystal cannot show more than its stabiliser unless its frame happens not to turn, which is a fact about a material at a wavelength and not about its class.

One colour hides all of it #

The computation has a striking negative half. Repeat it with one wavelength instead of thirteen.

Every figure keeps all four operations. The reason is short: at one wavelength the indicatrix is one ellipsoid, and an ellipsoid with three different axes has the symmetry $mmm$, three mirror planes through its principal sections. So a figure taken down any principal axis in one colour has two mirror lines and a half-turn, whatever the crystal’s class. The class enters only through the way the ellipsoid changes from one colour to the next, and a single colour has no next.

This is the optical form of a fact three optical characters used to count, and the same one twenty-one, thirteen, nine, three meets for the elastic tensor. The dielectric tensor at one frequency is a symmetric second-rank tensor, and such a tensor cannot tell a monoclinic crystal from an orthorhombic one by its shape. It tells them apart by its orientation, and the orientation can be seen only when it moves. The monoclinic crystal’s lower symmetry is real at every wavelength and observable through this property only when more than one wavelength is compared.

There is a practical consequence that petrographers know without stating it this way. Under a sodium lamp a monoclinic crystal’s figure looks like an orthorhombic one. Under white light the colour fringes at the melatopes show the asymmetry, and they are the evidence. The fringes are not a nuisance laid over a symmetric figure; they carry the only symmetry information the figure contains.

A name that changes while the symmetry stays #

The classification depends on which bisectrix is acute, and that depends on the angle between the optic axes, which depends on colour. In most crystals the dependence is small and the acute bisectrix stays the acute bisectrix across the visible. When $2V$ is close to $90^\circ$ it need not.

The crystal is one crystal and its symmetry does not change. What changes is which of its two bisectrices the textbook definition calls acute. Looking down $b$, which is $Z$, the figure has the half-turn of crossed dispersion; looking down $X$, the other bisectrix, the figure has the mirror of horizontal dispersion. Below 575 nanometres the acute bisectrix is $X$, and a plate cut across it shows a horizontal figure. Above, the acute bisectrix is $Z$, and a plate cut across it shows a crossed one. A crystal of this kind does not have a dispersion type. It has two figures, one down each bisectrix, and which of them the name is attached to depends on the colour in which the angle is measured.

That is a small point about nomenclature and a larger one about symmetry. The group computed above is attached to a line of sight, not to a crystal. A classification that names the crystal by the figure down its acute bisectrix inherits a choice of line, and when that choice changes with wavelength, so does the name. The stabiliser computation gives the honest form: down $b$ the figure has $2$, across $b$ along a principal direction it has $m$, and the mirror’s line is fixed by whether that direction is in the optic axial plane or perpendicular to it.

The orientation the Faraday rotation needed #

The three places for $b$ also answer a question left open in a field lets every class rotate light. That essay found a term in the magneto-optical response of monoclinic crystals that rotates light when the field lies across the two-fold axis, and noted that in a uniaxial crystal the corresponding term is all but unobservable, because the birefringence quenches any rotation except along the optic axis and the term vanishes there. In a monoclinic crystal it could be seen in full along an optic axis lying in the plane across $b$, and both optic axes lie in that plane exactly when $b$ is the optic normal.

So the crystals that could show the transverse Faraday term in full are exactly those with inclined dispersion, and the textbook name is a test that can be applied with a microscope. A white-light figure with a single mirror along the optic axial plane identifies a crystal whose two-fold axis is the intermediate index, and in such a crystal a ray along either optic axis can carry the whole transverse term with no birefringence to quench it. Gypsum, whose optic axial plane is the plane across its two-fold axis, is the familiar example of that orientation, and its strong inclined dispersion is how it is usually recognised.

Which other crystals have that orientation is still a question for tables of measured indices, as the earlier essay said. The difference is that the tables already record it, under a heading nobody had connected to magneto-optics: the dispersion type.

What the picture cannot show #

The models here are models. Their indices and their nine degrees of turn are chosen to make the effect visible, and a real crystal’s turn is usually a degree or two, sometimes less. The figures show the shape of each type and not its size, and a real inclined figure can look nearly symmetric to the eye.

The figures also show only the melatopes. A real interference figure has isogyres, the dark brushes through the melatopes, and isochromes, the coloured rings around them, and in white light the colours of all of these carry the dispersion too. The symmetry argument covers every feature at once, since every feature is a function of direction and colour, and the stabiliser acts on all of them. What is drawn is the simplest feature that shows the symmetry, not the whole of what a microscope shows.

Finally, the argument concerns the symmetric part of the dielectric tensor. A crystal in one of the fifteen classes that may rotate light adds a gyration term whose symmetry is not $mmm$, and along an optic axis that term is what a figure shows. For the monoclinic classes that permit it the stabiliser argument still holds, because the gyration tensor obeys the same Laue group; but whether its contribution survives the inversion that the Laue group adds is exactly what the gyration census decided class by class, and the figures here leave it out.

Who named them, and what is computed here #

The three dispersion types appear in mineralogical optics from the nineteenth century onwards, with the names inclined, horizontal and crossed fixed in the German literature and in the standard English texts on optical mineralogy that followed it. Their association with the position of the two-fold axis among the principal directions is part of the classical description. It is given as three cases to recognise, and the connection to the Laue group’s stabiliser is not usually drawn.

What is computed here is that connection, made twice. The tracks come from Cauchy indices and a turning frame, at thirteen wavelengths, for five model crystals. The stabiliser comes from the matrices of the Laue groups $\bar{1}$, $2/m$ and $mmm$ and a line of sight in each case, with no optics in it. The two agree in all five cases. The monochromatic computation runs the same tracks at a single wavelength.

Five tests and one refusal #

The first test is the agreement between the two derivations. The second checks that the three places for $b$ give the three named types, by classifying the tracks geometrically: an optic axial plane whose trace turns, one whose trace stays on a line, and one whose trace stays parallel and moves. The third and fourth are the controls, orthorhombic and triclinic, which must keep everything and nothing. The fifth checks that the switching model really passes $90^\circ$ inside the visible.

The refusal is the claim a careless reading of the holohedry is the ceiling might suggest: that a crystal’s property, being bounded by its class, displays its class. A property can be more symmetric than its class allows, and a single ellipsoid always is.

Still open: the figure down an optic axis #

Every figure here is viewed down a bisectrix. A plate cut perpendicular to one optic axis shows a single melatope at the centre, and the stabiliser of that line in the Laue group is smaller still. In a monoclinic crystal with inclined dispersion the optic axes lie in the mirror plane across $b$, so the stabiliser of either axis is the mirror, and the single-axis figure keeps one mirror line. In the other two cases the optic axes lie in no symmetry plane and the stabiliser is trivial. That predicts that a plate cut across one optic axis of a monoclinic crystal shows a symmetric colour pattern around its melatope only when the crystal has inclined dispersion. The prediction is easy to state; checking it would need the isochromes, which depend on retardation and thickness as well as direction, and they are not computed here.