Turn changing starlight into competing periods

First calculate what the telescope receives

A surface patch contributes light only when it faces us. Let $i$ be the angle between the spin axis and the line of sight, $\lambda$ its latitude and $\phi$ its longitude measured from the central meridian. The cosine of its viewing angle is $\mu$. Positive values lie on the visible side.

For a small patch of surface area $dA$, the projected area is $\mu dA$. If it removes a fraction $c$ of the local continuum intensity $I(\mu)$, its flux deficit is approximately the expression below. The unspotted disk flux is $F_0$. The real acquisition integrates a finite spherical Gaussian region across the visible surface; the small-patch expression explains its limiting behaviour.

At zero inclination the longitude drops out of the viewing cosine. At a nonzero inclination, rotation changes the deficit. The light curve need not resemble a sine wave: a patch can pass behind the limb for part of each circuit. That shape motivates including a few harmonics in the observation model.

Reading: Curtis et al. (2020), When Do Stalled Stars Resume Spinning Down? Advancing Gyrochronology with Ruprecht147, arXiv2010.02272v1

A sinusoid provides a useful first question

Write the saved data as times $t_j$, relative fluxes $y_j$ and stated standard deviations $\sigma_j$. At trial frequency $g$, fit a constant plus sine and cosine. Their coefficients are linear, so weighted least squares supplies a best fit without guessing an amplitude or starting phase.

A familiar periodogram compares the squared residual of this fit with that of a constant. The fraction of variance removed is a descriptive fit score. It answers which trial sinusoid reduces those residuals most. Converting that score into a false-alarm probability requires additional assumptions about noise and the search; our variance-fraction demonstration makes no such conversion.

A finite observing window broadens the response to a periodic signal. Gaps add secondary peaks. Taking the grid point with the largest score discards both the width and the alternatives. Our inference refines every local maximum found on its coarse search and then integrates a positive likelihood measure around and between them.

Put a probability model around the repeating signal

The interactive analysis uses four harmonics, a baseline offset and a linear trend. The data vector is the flux minus one. Its design matrix $X_g$ contains the baseline, the centred time divided by the observation duration, and four sine/cosine pairs. A rectangular exposure of length $\Delta t_j$ multiplies harmonic $k$ by its exact averaging response.

A harmonic has an integer multiple of the trial frequency. Four sine/cosine pairs describe detail within a single repeating pattern.

Each linear coefficient has a proper, finite Gaussian prior. The baseline standard deviation is 0.1, the trend coefficient has standard deviation 0.03, and harmonic $k$ has sine and cosine standard deviations $0.05/k$. These are declared observation-model choices. They can favour smoother explanations, so the numerical package records them.

Let $\Lambda$ be that diagonal prior covariance and $D$ the diagonal covariance of the saved measurement errors. Integrating the linear coefficients gives a normalized Gaussian likelihood for each frequency.

The implementation solves a ten-dimensional positive normal matrix instead of inverting the full data covariance. An independent full-covariance calculation checks the result. It also compares the periodic model with a proper constant-plus-trend model. Their Bayes factor compares these declared models and priors; it is a different quantity from a frequentist false-alarm probability.

Reading: Bouma, Palumbo & Hillenbrand (2023), The Empirical Limits of Gyrochronology, arXiv2303.08830v1

One observing day can hide many frequencies

Suppose observations occur at integer days $n$. Adding an integer number of cycles per day leaves every sampled complex phase unchanged. In particular, frequencies 0.1 and 1.1 per day produce the same phases on that window.

The instantaneous daily synthetic test gives equal marginal evidence at 0.1,1.1,2.1 and 3.1 per day to floating-point precision. A prior on period can assign different integrated weights to their neighbourhoods. The data equality survives that prior choice. Irregular observing times usually weaken the equality, but do not guarantee a single remaining solution.

A second ambiguity concerns what repeats. We define two pattern-clock hypotheses: a surface clock can span one or two repeats of the fitted pattern. With multiplicity $m$, the surface period is $P=m/g$. Equal model prior weights are explicit. The supported surface-period interval is 0.5 to 60 days, wider than the four-to-forty-day simulation control, so changing a control cannot secretly delete a short daily alias.

A lasting ten-day synthetic patch consequently retains narrow ten-day and twenty-day alternatives under these hypotheses. Reporting their median alone would conceal the gap between them. The displayed modes and complete mixture carry the useful answer.

Ask whether a stable clock describes these points

Before using a period in an age calculation, inspect the saved-data checks. A weak Bayes factor against the constant-plus-trend model leaves the rotation signal unsupported. A residual chi-square per degree of freedom above three rejects our stated stationary harmonic noise model. Most probability on periods longer than half the observing span means that fewer than two rotations dominate.

These thresholds are declared screening rules for the teaching model. The pole-on and short-life presets fail for different data-based reasons. Longer or more precise observations can improve a weak signal. A changing pattern calls for a time-dependent surface model. The next step depends on the failure you actually observed.