The mirror a plate keeps in one colour

Three places for the two-fold axis ended on a prediction it could not check. Its figures looked down a bisectrix, with both melatopes in the field, and they drew only where those melatopes go as the colour changes. A plate cut across a single optic axis shows one melatope at the centre, surrounded by rings. The prediction was that this pattern is symmetric in white light only for one of the three monoclinic arrangements, the one mineralogy calls inclined dispersion, and that the check would need the rings themselves, the isochromes, which depend on retardation and thickness as well as direction. They were not computed there. They are computed here.

The prediction holds, in the sense that the numbers come out as the group said they would. But the rings make the statement sharper than the tracks could, and they correct one way of reading it. It is not quite true that a plate in one colour shows no sign of its crystal’s class. What is true is narrower and more exact. In the single colour a plate was cut for, every crystal shows a perfect mirror, and in every other colour only a crystal with a mirror keeps it. A monoclinic plate in crossed dispersion looked at with a sodium lamp is indistinguishable by symmetry from an orthorhombic one, and looked at with a blue filter it is not, with no change at all to the plate.

What a plate across an optic axis looks like #

The vocabulary comes from the earlier essay, and it needs only a short reminder. A biaxial crystal’s refractive indices at a given wavelength are the three semi-axes of an ellipsoid, the indicatrix, and those semi-axes run along three orthogonal directions called $X$, $Y$ and $Z$ in order of increasing index. Two directions through the crystal have both polarisations travelling at the same speed. Those are the optic axes, and they lie in the $XZ$ plane, the optic axial plane.

Take a plate of a fixed thickness, its faces cut perpendicular to one optic axis at one chosen wavelength, here the sodium line at 589 nanometres. The convention for the rest of this essay is that the plate is fixed once cut. Changing the colour of the light moves the optic axes, because dispersion moves them, and it does not move the plate. Put the plate between crossed polarisers and look through it in strongly convergent light, so that each point of the field is one direction through the plate. For a ray in direction $s$, the crystal offers two polarisations with two indices, $n_1$ and $n_2$, along two directions in the plane across $s$. If $\chi$ is the angle from the polariser to one of those directions, the transmitted intensity is

$$I = \sin^2 2\chi \; \sin^2\!\left(\frac{\pi L\,|n_1 - n_2|}{\lambda \cos\theta}\right),$$

with $L$ the thickness and $\theta$ the angle of the ray from the plate’s normal. Two families of dark curve appear. Where $\sin 2\chi$ vanishes, the eigen-directions line up with the polarisers, and that gives the dark brushes, the isogyres. Where the second factor vanishes, the retardation is a whole number of wavelengths, and that gives the rings, the isochromes, which are nearly circular close to the melatope and swell into the familiar shape further out.

The computation is direct. For each direction in a square field the impermeability tensor is projected onto the plane across the ray, the two-by-two matrix that results has two eigenvalues whose inverse square roots are the indices, and $\chi$ comes from its eigenvectors. Refraction at the plate’s face is left out, on the grounds that the indices barely change across the field, and leaving it out moves the rings a little and changes no symmetry. The thickness is a quarter of a millimetre, and the model crystals are those of the earlier essay, so that every number can be compared: three Cauchy indices near 1.52, with the principal frame in the monoclinic cases turning by about nine degrees across the visible. That is more than most real crystals show. It is chosen so the effect can be seen at this scale, and the symmetry argument uses none of those numbers.

Five plates, one colour #

Start with the colour each plate was cut for.

All five have a mirror, and the mirror is exact. The residual used to measure it is the mean of $|I(x, y) - I(x, -y)|$ over the field, with $y$ measured across the optic axial plane’s trace, and at 589 nanometres it is zero to rounding error for every crystal, including the triclinic one, whose group has nothing in it but the identity and the inversion.

The reason is the single ellipsoid. At one wavelength the indicatrix has three mirror planes, because every ellipsoid does, and the optic axial plane is one of them. The plate’s normal is an optic axis at that wavelength, so it lies in that plane. Reflection in that plane carries the ellipsoid to itself and the normal to itself, and so it carries the whole figure to itself. Nothing about the crystal’s class enters. This is Neumann’s principle working in reverse: the property has more symmetry than the crystal is required to give it, and in this one colour the extra symmetry is all that is visible. A coincidence the group did not ask for described the same thing for energy levels, a symmetry the structure has and the group does not explain. Here it comes from the fact that a symmetric second-rank tensor has few enough components to be an ellipsoid at all.

The same plates, another colour #

Now change the light and leave the plates where they are.

Two of the five keep the mirror. The others lose it, and the loss is large. The residual for the crystal in horizontal dispersion goes from zero to 0.26, in crossed dispersion to 0.21, and in the triclinic crystal to 0.24, on a scale where the intensity runs between zero and one.

Each row is what the class said it should be, and the class’s answer was worked out first. A figure’s symmetry at a chosen line of sight is the part of the Laue group that keeps that line, as the earlier essay derived. A monoclinic crystal’s Laue group is $2/m$. Its mirror is the plane across the two-fold axis $b$. So the figure keeps a mirror exactly when the line of sight lies in that plane, and it has to lie there at every wavelength, which needs a line of sight fixed in the crystal.

In inclined dispersion $b$ is the optic normal, so the optic axial plane is the plane across $b$. It is the crystal’s mirror. The optic axis at sodium lies in it, so the plate’s normal lies in the crystal’s own mirror, and at every wavelength the crystal really does have that reflection as a symmetry. Hence an exact zero in every column. The orthorhombic crystal has the same plane as a mirror for a different reason, and the same zero.

In horizontal and crossed dispersion $b$ lies in the optic axial plane, so the plate’s normal, in that plane and away from $b$, is not in the plane across $b$ and not along $b$. The Laue group keeps that line only with its identity. So the class promises no symmetry at all, and the mirror that appears at sodium belongs to the ellipsoid at sodium rather than to the crystal.

That is the prediction, and the isochromes confirm it. The argument that gives it is short, and it is striking how much of the optics it avoids. It needs no value of an index, no thickness and no intensity formula.

One plate at three colours #

The contrast is clearest side by side.

Reading the figure takes a moment, because a figure’s mirror is not as noticeable as a figure’s centre. The rings around the melatope are nearly round whatever happens, so the loss of symmetry is not in the innermost rings. It is in the outer rings and in how the two lobes of each ring compare above and below the line. A change of colour also shrinks or spreads all the rings together, because the retardation is divided by the wavelength. That change is large, and it is what an eye sees first, and it has nothing to do with symmetry. Only the difference between the two sides of the line is the signal, and in the crossed crystal at 450 nanometres it is plainly there.

In real white light all the colours arrive together. The rings of different colours fall at different radii, because of the factor $\lambda$, which is the source of the colour in an interference figure in the first place. An inclined crystal’s coloured rings are displaced along the line, by different amounts for different colours, and so a white-light figure is balanced across the line and unbalanced along it. A crossed crystal’s are displaced across the line, and the pattern loses its balance altogether.

That account of what white light does is an inference from the single-colour figures rather than a computed colour image, which a sum over wavelengths and a model of the eye would be needed for. The statement about symmetry holds regardless. If each colour’s figure has a mirror across one line, then any weighted sum of them has it too, and if none of the coloured figures has it, a mirror in the sum could occur only by cancellation of a kind nothing here suggests.

Where the melatope goes #

The tracks that the earlier essay drew for bisectrix figures can also be drawn on these plates. They are the simplest way to see the difference and they carry a surprise.

The surprise is the first panel against the second. The orthorhombic melatope barely moves: its total excursion is about a hundredth, in units of the tangent of the angle. The inclined melatope moves more than twenty times as far, about 0.23, and it does so without leaving the line. So the crystal with the larger movement has the same mirror, and a plate of an orthorhombic crystal and a plate of an inclined monoclinic one are told apart by how far the melatope goes and not by any symmetry. The axes a class pins down is why. In the orthorhombic crystal all three principal directions are pinned, and the only thing dispersion can move is the angle between the axes; in the inclined crystal the principal frame itself turns in the mirror plane.

The horizontal and crossed melatopes leave the line by 0.195 and 0.104, nearly straight across it. Those two are not the same as each other, since one is a shift and the other is a rotation in the bisectrix picture, yet on a single-axis plate both look like movement across the line. That is a real loss compared with the bisectrix view. A single-axis plate does not distinguish horizontal from crossed dispersion by symmetry at all. Both have the trivial stabiliser, both figures have no mirror. Only the bisectrix figure, whose line of sight is special for one of the two and not the other, tells them apart.

The triclinic melatope moves mostly along the line in this model, with only 0.021 across it, because of the particular axis about which that model’s frame was chosen to turn. Its figure still has no mirror. This is a useful warning about reading the tracks alone. The track of one point can stay close to a line for accidental reasons, and only the whole figure decides.

The line the eye would move #

A careful observer who sees the melatope leave the line will not expect the mirror to be about the old line any more. The natural correction is to move the line sideways to pass through the melatope, and to ask whether the figure is symmetric about that. This can be measured too, and it is the part of the test most worth doing, because the earlier argument does not obviously rule it out.

The moved line helps. At 450 nanometres the residual falls from 0.21 to 0.11; at 650, from 0.14 to 0.02. So most of the visible asymmetry at long wavelengths is the melatope’s displacement. But it does not reach zero anywhere except at sodium, and the reason is exact. At 450 nanometres the ellipsoid does have a mirror plane through that colour’s melatope, its own optic axial plane, and the rings around that melatope would be symmetric if the plate had been cut for 450 nanometres. It was not. The plate’s normal is not in that plane, so the reflection does not carry the plate to itself. The thickness along a ray, and the angle $\theta$, differ between the ray and its reflection, and so does the angle of the eigen-directions against the fixed polarisers. Those small differences are what the residual still measures.

This is the sense in which the plate belongs to the experiment. The crystal’s symmetry and the plate’s symmetry are separate things, and the figure has only what they share. A mirror of the ellipsoid that the plate does not respect is not a mirror of the figure.

What the computation leaves out #

The figures here are level curves at one intensity, and a real interference figure is a continuous pattern of intensities. The level curve was chosen because it shows the rings and the brushes as lines that can be compared by eye. The measured residuals use the intensity itself, over a square grid, so they are not affected by the choice of level. What can be lost is the appearance: no figure here looks quite like a photograph of an interference figure.

The field is a square of about 23 degrees to each side of the normal, with the rays’ angles inside the crystal rather than outside it. A real microscope objective sets the field, and refraction at the plate’s face compresses it. That changes where the rings fall, by a little, and it leaves every exact zero in place, since a mirror in the crystal survives refraction at a face whose normal lies in the mirror.

Absorption is left out, and with it the difference between polarisations that makes many real minerals pleochroic. Optical activity, which fifteen classes may show, is also left out, and it has a direct effect on what happens near the melatope, where the birefringence vanishes and rotation is no longer small beside it. In a chiral crystal the dark cross at the melatope is partly filled in. That belongs to the same computation with a complex dielectric tensor, and it has not been done here.

And the model crystals are models. No mineral has these particular Cauchy coefficients, and few have a principal frame that turns by nine degrees across the visible. The symmetry statements do not depend on that; the sizes of the residuals do.

What had to come out exactly #

The checks here are of a kind that floating point rarely passes by accident. An exact zero of the residual means that the intensities above and below the line agree to the last digits at every one of a large grid of points, in a calculation that involves two eigenvalue problems and two trigonometric functions per point. That cannot happen unless reflection really does carry one ray’s whole calculation onto the other’s. A mistaken frame, or a plate normal slightly out of plane, would show up as a small number rather than a zero. In the same way, a birefringence that vanishes along the optic axis at sodium ties the plate to the same optic axes that the three optical characters computed from the indices, by a route that uses none of the projection.

The refusals are the other half. Claims that the crossed crystal keeps its mirror in blue, that a horizontal crystal keeps it at 450 nanometres, and that a triclinic one does, each fail with a residual far from zero, so the test does say no when no is the right answer.

Still open: a chiral crystal’s melatope #

The figures have treated the plate as if the two polarisations were linear, and in a crystal of one of the eleven chiral classes they are not, close to an optic axis. The gyration tensor mixes them into elliptical polarisations, and the effect is largest exactly where these figures are centred, where the birefringence falls to zero and whatever else there is takes over. The crossed polars no longer give a dark point at the melatope, and the size and sign of what appears there measure a component of gyration directly. A field lets every class rotate light found the transverse part of that effect in the bulk. Whether a chiral monoclinic crystal shows a mirror in the bright patch at its melatope when the rest of its figure has one is a question about a symmetry the ellipsoid does not have, since an axial tensor changes sign under a mirror. A monochromatic figure of a crystal in class $2$ might then reveal its class where a monochromatic figure of a crystal in class $2/m$ cannot. Answering that needs the gyration tensor added to the projected matrix, with the complex eigenvalues that follow, and it has not been done here.