A shared axis hides at the angle where it does no harm

A rapidly rotating star looks different from different directions. Its poles are hotter and brighter than its equator, so an observer looking down on a pole sees a brighter, hotter star than one looking at the equator. For a population of such stars the differences average out exactly, provided the axes point at random. Every photon goes somewhere, so the mean apparent luminosity over random orientations is the true luminosity, and even a survey limited by brightness tilts that average only slightly.

The proviso carries the whole result. Field stars come from many clouds and have been stirred around the Galaxy for billions of years, so random axes are the natural assumption for them. The stars of a cluster formed together, out of one cloud, at one time. If that cloud was rotating as a whole when it fragmented, the angular momentum of the pieces might have kept some memory of the common rotation. A cluster’s stars might then share an axis, and the averaging that protects population studies would not apply to them.

That can be tested, because each star’s inclination can be measured, poorly, one star at a time. This essay asks how many stars the test needs. The answer depends sharply on where the common axis points, and the orientation at which it is hardest to detect is not the one usually expected. It is also the orientation at which a shared axis would do the least damage.

What random axes look like through a sine #

A star’s inclination $i$ is the angle between its rotation axis and the line of sight. The cheapest way to measure it combines three quantities: the projected rotation speed $v\sin i$ from the width of the spectral lines, the rotation period from the brightness modulation of spots carried round the star, and the radius from luminosity and temperature. The equatorial speed is $2\pi R/P$, and

$$\sin i = \frac{v\sin i}{2\pi R/P}.$$

For axes pointing at random, every direction on the sphere is equally likely, which makes $\cos i$ uniform between 0 and 1. The inclination itself is not uniform: its density is $\sin i$, because there are many more ways to point an axis near the sky plane than near the line of sight. Only 13 per cent of random axes lie within 30 degrees of the line of sight. In terms of the measured quantity, the fraction of stars with $\sin i$ below some value $s$ is $1 - \sqrt{1-s^2}$, and 60 per cent of random stars have $\sin i$ above 0.8.

The figure sets three cluster models against that curve. A cluster whose axes are all exactly parallel has a single value of $\sin i$, so its distribution is a vertical step. The step sits at $\sin i = 0.5$ if the axis is 30 degrees from the line of sight and at $0.996$ if it is 85 degrees. Either step is very far from the smooth curve of random axes. The largest vertical gap, $D$, is near 0.9 in both cases, the most a cumulative distribution can differ from another.

Real alignment would not be perfect. A convenient model scatters each star’s axis about the common direction with a Fisher distribution, the spherical version of a Gaussian, with concentration $\kappa$. At $\kappa = 5$ a typical star’s axis lies about 37 degrees from the common direction, which is a weak alignment but a definite one. About a common axis at 55 degrees, that cluster’s curve is almost on top of the random-axes curve, with $D = 0.04$. The model is strongly aligned in physical terms, and statistically it is close to random.

The same test with real errors #

The ratio that gives $\sin i$ is not measured perfectly. The line width carries errors of a few kilometres per second and the radius several per cent. For slow rotators, the dominant population among the cool stars whose periods spots reveal, the line width is barely resolved. An error of 0.1 in $\sin i$ is a fair round number for a good cluster sample. Because the ratio is noisy it can exceed one, and with random axes piled up near $\sin i = 1$ it often does.

The errors turn each step into a ramp of width 0.1. The perfectly aligned cluster at 30 degrees still stands well apart. At 85 degrees the ramp is centred near one, where random axes also crowd, and the gap falls to 0.45. That is smaller but still large. A cluster of stars that all present their equators would not have a spread-out distribution of measured $\sin i$; it would have a narrow one, with nothing below about 0.8, and random axes would put two-fifths of their stars there.

The loose alignment about 55 degrees has nearly disappeared. With a gap of 0.02, its curve differs from random axes by less than the width of the line drawing it.

How many stars it takes #

The vertical gap is a rough guide. A sharper measure is the expected log-likelihood ratio between the two hypotheses, which is how a sample of $N$ stars actually accumulates evidence. Each star contributes, on average, the Kullback–Leibler divergence between the aligned model’s distribution of measured $\sin i$ and that of random axes. A three-standard-deviation preference corresponds to a likelihood ratio of about 370 to one, so the number of stars needed is $\ln 370$ divided by that divergence.

Perfect alignment can be found with a handful of stars at any orientation: four about an axis at 30 degrees, ten about one at 85. Even a tight alignment, $\kappa = 20$, with axes typically 18 degrees from the common direction, needs no more than about forty. These are small samples by the standard of a modern cluster survey, which measures periods for hundreds of stars and line widths for many of them.

The loose alignment is a different matter, and where the difficulty peaks is unexpected. The difficulty does not lie near the sky plane, where the sine is flat and where individual inclinations are worst determined. About an axis at 85 degrees, 38 stars suffice. It lies in a sharp peak near 55 degrees, where the same alignment needs almost a thousand stars. That is more than most open clusters hold with measurable rotation. The orientation that hides an alignment is an oblique one, not the edge-on one.

Why 54.7 degrees #

The peak is at the angle where $3\cos^2 i_0 = 1$, $i_0 = 54.7°$, which crystallographers and spectroscopists call the magic angle. The reason it appears here is a theorem about averages on a sphere.

A sample of measured $\sin i$ fixes its low moments much better than its full shape. The most informative is the mean of $\sin^2 i$, or equivalently the mean of the second Legendre polynomial,

$$P_2(\cos i) = \tfrac12(3\cos^2 i - 1), \qquad \langle\sin^2 i\rangle = \tfrac23\big(1 - \langle P_2\rangle\big).$$

For random axes $\langle P_2 \rangle = 0$ and $\langle \sin^2 i\rangle = 2/3$. For axes scattered about a common direction at $i_0$, the addition theorem for spherical harmonics splits the average into two factors:

$$\langle P_2(\cos i)\rangle = P_2(\cos i_0)\,\langle P_2(\cos\theta)\rangle,$$

where $\theta$ is each star’s angle from the common axis. The second factor describes how tight the alignment is: one for perfect alignment, zero for none, and $1 - 3\coth\kappa/\kappa + 3/\kappa^2$ for a Fisher cone. The first factor describes only where the axis points, and it is zero at 54.7 degrees.

So every alignment, however tight, has the random-axes value of $\langle\sin^2 i\rangle$ when its common axis lies at 54.7 degrees. The test is not completely blind there. Higher moments, $P_4$ and above, are not zero at that angle, and a perfect alignment is a spike that no moment-based view misses. But a loose alignment has small higher moments to begin with. With its second moment cancelled by the geometry, all that remains is a faint difference in the shape of a distribution smeared by measurement errors, and the figure prices that at a thousand stars.

The cancellation is exact for any distribution symmetric about its axis, not only the Fisher form. It depends on two things only: the cone is symmetric, and the axis lies at the angle where $P_2$ vanishes.

How often a cluster would sit in the blind band #

A single cluster’s common axis, if it has one, has no reason to point anywhere in particular relative to the Sun. The cluster formed with its axis set by its parent cloud, and the Sun’s position was settled independently, a few hundred parsecs away. So across many clusters the common axes should themselves be randomly oriented relative to the line of sight, with $\cos i_0$ uniform like any other random direction.

That makes it possible to say how often the blind spot matters. The band in which a loose alignment needs more than about three hundred stars runs roughly from 48 to 62 degrees. The fraction of random directions inside it is $\cos 48° - \cos 62° = 0.20$, one cluster in five. The band is not narrow on the sphere, because 55 degrees is a long way from the pole and the solid angle at those inclinations is large. For comparison, only 13 per cent of axes lie within 30 degrees of the line of sight, the region where a loose alignment shows most easily. A survey of twenty clusters, each with an alignment as loose as $\kappa = 5$, would expect to detect it in fifteen or sixteen with modest samples and to miss it in four, whatever the size of each sample within reason.

This gives a survey of many clusters a way round the blind spot that no single cluster has. If loose alignments are common, most clusters will show them and a fifth will not, and the fraction that do sets the population rate. If none do, the absence is significant across the four-fifths of clusters whose axes would have been visible. The clusters in the blind band then contribute nothing either way, but they do not bias the answer: they are a known, random fifth, set by the geometry and not by the stars.

What a population of clusters cannot do is identify which of them are the hidden fifth. That would need a measurement of the common axis’s direction that does not go through the second moment.

A second clock that reads the same angle differently #

The spot modulation that gives each star its period also carries information about the inclination. A spot near a star’s equator, seen equator-on, rotates fully into and out of view and makes a large brightness swing. Seen pole-on, the same spot stays in view all the time and barely modulates the light. The amplitude of the periodic signal is therefore a second function of $i$, independent of line widths and radii.

It is a poorer measure star by star, because the amplitude depends as much on how large the spots are and at what latitudes they sit as on the angle from which they are seen. As a population statistic, though, it depends on the inclination through a different function from $\sin i$. A test that combined the two would not have its blind spot at the same place. That combination has been used to compare planet-hosting stars with others, as one of the ways a whole population’s orientation is measured when no single star’s can be. Applied to clusters, it is one of the few ways to look into the oblique band, and its success depends on modelling spot distributions well enough that their scatter does not swamp the signal.

Where a shared axis would do harm #

The reason for asking was the brightness of rapid rotators. A gravity-darkened star seen pole-on looks brighter than its true luminosity, and seen equator-on fainter. With random axes the cluster’s rapid rotators scatter both ways about the truth. With a shared axis they would all be displaced together, and a cluster’s whole upper main sequence could sit too bright or too faint by one amount.

The figure gives the size of the displacement. A star turning at 0.98 of its critical rate looks 0.13 magnitudes brighter than its true luminosity from the pole and 0.06 fainter from the equator. A cluster whose rapid rotators all shared a pole-on axis would have its upper main sequence shifted by a tenth of a magnitude. That is comparable to the corrections that cluster distances and ages are fitted to, and it would be systematic, not random, so averaging over stars would not remove it.

The curves cross zero at about 55 degrees. This is the same angle as before and the same reason. To lowest order the anisotropy of a rotating star’s light, flattened at the equator and bright at the poles, is a quadrupole: it goes as $P_2$ of the angle between the line of sight and the spin axis. Its mean over a cluster aligned about $i_0$ is therefore proportional to $P_2(\cos i_0)$, by the same addition theorem, and it vanishes at 54.7 degrees. The cancellation is not perfect, because the darkening has higher-order terms that do not vanish there, but the residual at 54.7 degrees is 0.004 magnitudes for the faster star and 0.001 for the slower, against offsets of a tenth at the pole.

The orientation at which an alignment is hardest to detect is the orientation at which it does the least harm to the cluster’s photometry. The second moment carries both the bias and the evidence. A cluster whose shared axis is invisible to the inclination test also has rapid rotators that, on average, look like the truth. A cluster whose alignment would distort its photometry badly, with an axis near the pole or the sky plane, has an alignment that a few dozen stars reveal.

What has been measured #

Inclinations measured from the splitting of stellar oscillation frequencies do not depend on line widths or spots. They come from the relative heights of a mode’s components, which depend on the angle from which the star is seen. In 2017 an analysis of red giants in two old open clusters observed by Kepler, NGC 6791 and NGC 6819, reported that most of their axes lay close to a common direction. A reanalysis of the same stars found the individual inclinations far less certain than the first fits implied, and once they were treated with more care the evidence for a shared axis became much weaker.

Measurements from spectra and spot periods in younger, closer clusters, where hundreds of stars have periods, have reported both random and partially aligned axes. The figures here explain why the answers can differ without anyone erring. A loose alignment about an axis anywhere between about 45 and 65 degrees needs a sample of several hundred stars before it can be told from random axes. A cluster sample of fifty stars that finds random axes has excluded tight alignments at any angle and loose ones at most angles. It has not excluded a loose alignment about an oblique axis, and it cannot.

The theoretical expectation is not settled either. Simulations of cluster formation produce shared axes only when ordered rotation carries a substantial share of the cloud’s kinetic energy at the moment stars form. In the turbulent clouds that make most clusters, random motions usually dominate, and the fragments’ spins should come out close to random. A loose alignment, a cone with a typical spread of thirty or forty degrees, is what an intermediate case would produce. It is exactly the case the oblique blind spot can hide.

What the figures leave out #

The measurement error is a single number, 0.1 in $\sin i$, applied equally to every star. In practice the error is larger for slow rotators, whose line widths are hard to resolve, and smaller for fast ones. A real sample is a mixture, and the counts above shift towards higher numbers wherever the slow rotators dominate. The error is also taken as Gaussian and unbiased. The ratio method has a known bias near $\sin i = 1$, where the true equatorial speed appears in both measurements, and a careful analysis integrates over it instead of dividing. That affects the edge-on cases most and the oblique ones little.

The Fisher model is one choice of how axes scatter about a common one. A distribution with a sharp core and wide wings, or one made of two populations, aligned and random, behaves differently at large angles. But the cancellation at 54.7 degrees depends only on symmetry about the axis, so it applies to all of them. The star counts are averages; any particular cluster’s sample could reach its threshold sooner or later.

Still open: whether the oblique clusters are hiding anything #

The population argument for gravity-darkened stars needs random axes, and the evidence that cluster stars have them is good for tight alignments and weak for loose ones at oblique angles. The cancellation means that the gap is less dangerous than it looks, since an oblique shared axis biases the photometry by only a few thousandths of a magnitude. But it is a gap in what is known about how clusters form. A cloud that leaves a loose memory of its rotation in its stars’ spins would be telling something real about the balance of turbulence and rotation when it collapsed, and at 54.7 degrees that record cannot be read.

Breaking the degeneracy needs a quantity the second moment does not carry. One candidate is the position angle of each axis on the sky, which interferometry and spectroastrometry have measured for a few bright stars and which, combined with the inclination, gives the full direction instead of one angle of it. Another is the large rotating sample that cluster surveys are now assembling, hundreds of stars with periods and line widths in each of dozens of clusters. Whether that sample shows an oblique alignment in any cluster, and whether it is common, is a question its size can answer and no individual cluster could.