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Sandbox Physics
From planet catalogs to populations

AC24 · L1

Keep the stars with no detected planet

Sandbox Physics ·

A star with no detected planet contributes to an occurrence calculation just as surely as a star with a transit. It tells us where a search was performed. To use that information, we first need to calculate how often an orbit puts a planet in front of its star.

Define the viewing direction

Place a spherical star at the origin. Its radius is R⋆R_\star, measured in metres. A planet has radius RpR_p and moves on a circular orbit of radius aa. The inclination ii is the angle between the orbital angular-momentum axis and our line of sight. An edge-on orbit has i=90∘i=90^\circ.

At the closest projected approach on the foreground half of the orbit, the planet’s centre lies a∣cos⁡i∣a|\cos i| from the star’s centre. We call the distance divided by the stellar radius the impact parameter bb. The two projected disks touch if their separation is no greater than the sum of their radii. A finite planet can therefore transit even when its centre passes just outside the stellar limb.

b=a∣cos⁡i∣R⋆,b≤1+RpR⋆b=\frac{a|\cos i|}{R_\star},\qquad b\le1+\frac{R_p}{R_\star}
The impact parameter and radius ratio are dimensionless. The inequality includes grazing disk overlap.

Reading: About 30 percent of Sun-like Stars Have Kepler-like Planetary Systems

Count directions on a sphere

Imagine drawing the orbital axis in a random direction. Equal patches of a sphere receive equal probability. The area of a thin spherical strip is proportional to its change in cosine, so ∣cos⁡i∣|\cos i| is uniform from zero to one. The angle itself is not uniform. Choosing inclinations uniformly would put too many axes near the pole.

The transit condition accepts the interval from zero to (R⋆+Rp)/a(R_\star+R_p)/a in that uniform cosine. Its length is the geometric transit probability. The maximum is one: a probability cannot exceed all possible directions. Circular orbits outside the star occupy the usual small-probability case.

For a worked example, take a star with radius three quarters of the nominal solar radius and a planet with three nominal equatorial Earth radii on a twenty-day orbit. Give the star eight tenths of the nominal solar gravitational parameter. Kepler’s circular-orbit relation gives the orbital radius; dividing the summed physical radii by it then gives the probability. Every length must use the same units before division. This gives a=2.00×1010 ma=2.00\times10^{10}\,\mathrm m and a transit probability of about 2.70%2.70\%. Only about one orbit in thirty-seven lines up.

pgeom=min⁡ ⁣(1,R⋆+Rpa),a3=GM⋆P24π2p_{\rm geom}=\min\!\left(1,\frac{R_\star+R_p}{a}\right),\qquad a^3=\frac{GM_\star P^2}{4\pi^2}
The period PP is in seconds and GM⋆GM_\star is in cubic metres per second squared. A period of twenty days is converted to seconds before taking the cube root.

Reading: About 30 percent of Sun-like Stars Have Kepler-like Planetary Systems

The denominator includes the unsuccessful searches

Suppose a survey searches one thousand stars. In a chosen period-radius region, imagine that each planet has a two-percent chance of entering the catalog after geometry, observations and recovery are combined. Ten discoveries correspond to ten divided by twenty, or half a planet per searched star. Twenty is the effective number of searched stars: one thousand times two percent.

Now imagine the same ten discoveries came from ten thousand searched stars with the same selection probability. The effective exposure would be two hundred stars and the corresponding estimate would be one twentieth of a planet per star. Keeping only the ten host stars would make these two very different searches look identical.

A real survey has different chances around different stars. Small stars provide deeper transits at a fixed planet radius. Noisy stars can hide the same signal. Add the individual probabilities over all searched stars. An average based only on detected hosts describes a different sample. The hosts have already been selected for successful detection.

E=∑j=1N⋆qj,E[Ndet]=fEE=\sum_{j=1}^{N_\star}q_j,\qquad \mathbb E[N_{\rm det}]=fE
qjq_j is the final probability for a planet around searched star jj under the declared population model. EE is an effective exposure and ff is planets per star.

Reading: A Probabilistic Approach to Kepler Completeness and Reliability for Exoplanet Occurrence Rates

One star can contribute several planets

Planets per star and the fraction of stars with planets answer different questions. A hundred stars could host fifty planets on fifty separate stars, or the same fifty planets on ten rich systems. Both populations contain half a planet per star, but their host fractions differ.

For a useful mathematical example, suppose each star has an independent Poisson number of planets with mean ff. A Poisson model assigns probability e−fe^{-f} to zero planets. The host fraction is then 1−e−f1-e^{-f}. At a generating mean of six tenths of a planet per star, this fraction is about forty-five percent. This relation follows from that particular multiplicity model; clustered planetary systems can obey a different relation.

The synthetic survey keeps all five thousand searched stars and draws their planet counts from this Poisson model. Changing the detector noise keeps the generated planets and orbital directions. Changing the generating mean changes the population itself. This distinction lets you identify whether a catalog changed because the sky changed or because the search improved.

Reading: About 30 percent of Sun-like Stars Have Kepler-like Planetary Systems

Exercises and explanations

  1. A catalog lists twenty planets around twelve hosts. Can you calculate the fraction of searched stars with planetary systems?

    Read the explanation

    You still need the full searched-star count and the selection model. Twelve is a detected-host count. Systems whose planets failed every detection remain outside that number, and planets in the same system can share orbital alignment.

  2. The telescope searches twice as many otherwise identical stars. What changes if the population and selection are unchanged?

    Read the explanation

    The effective exposure doubles, so the expected number of detections doubles. The occurrence in planets per star stays the same. Random counts fluctuate around their new expectation.

Try the prediction

Change the detector and watch the missing planets

  1. Write down the orange planet count and the number of all searched stars.
  2. Reduce noise without changing the generating mean. Check that the orange population stays fixed.
  3. Then change the generating mean. Explain which comparison changes the sky population and which changes the measurement.

Sources and calculation scope

  1. A Probabilistic Approach to Kepler Completeness and Reliability for Exoplanet Occurrence Rates

    Version 1906.03575v4, sections II–VI, Equations 10–12 and Appendix B: externally weighted integer catalogs and occurrence likelihoods.

  2. About 30 percent of Sun-like Stars Have Kepler-like Planetary Systems

    Version 1802.09526v2, section III.1, Equations 1–4, sections III.4–III.5 and Figures 7–9: planets per star, host fraction and system architecture.

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