Astronomy & Celestial Mechanics
the sky, computed
The collection

Every essay — page 22

One idea per essay, ordered so that the earlier ones set up the later ones — but nothing here depends on being read in sequence. Essays 421–440 of 534.

Spaceflight

Celestial mechanics used forwards — where to burn, and what it costs.

A 5/5/1 constellation at one instant: where no satellite is above 0°. The points beneath the 5 satellites of a 5/5/1 Walker delta pattern at 43.5°, at 11,605 km, at one instant, with each satellite's 0° footprint — a circle 69.2° in radius at the Earth's centre — drawn round it, on an equirectangular map that swells the circles towards the poles. The shaded cells are ground with no satellite above the mask: none at this instant. Everywhere else is seen by between 1 and 3 at once. The largest empty circle at this instant is 67.6° in radius, computed exactly from the satellites' directions, inside the footprint, which is why there are none.

Five satellites and not four

No single orbit keeps a satellite above every point on the Earth. How many are needed is a question about the largest empty circle among their directions at the worst instant, and it has sharp answers. Four can never do it, five can from 11,605 kilometres up, and past that the arrangement of the orbits matters as much as their number.

8 figures · Ground tracks

The observed sky

Coordinates, seasons, phases and shadows — geometry seen from inside it.

When the Orion Nebula can be observed from latitude 52°, through a year. Every night of 2027, from local noon to the following noon, for a station at latitude 52°. The shaded cells are the times at which the Orion Nebula (right ascension 5.59 h, declination −5.4°) stands at least 20° above the horizon while the Sun is more than 18° below it. The two outer curves are the start and end of astronomical darkness, which never comes on 64 nights around midsummer; the diagonal line is the object's transit, which arrives four minutes earlier each night and wraps through the whole day once a year. It crosses local midnight on 15 Dec, when the object stands opposite the Sun. The shaded season is the stretch of the year either side of that date in which the transit falls inside the dark hours. From this latitude the object transits at 32.6° altitude; it is observable on 181 nights for at least an hour, for 6.3 hours on the best of them (3 Jan), and for 860 hours in the year.

A right ascension is a date

An object can be observed only when two clocks agree — the sidereal clock that brings it high in the sky and the solar clock that makes the sky dark. The two drift apart by one turn a year, so every right ascension has a season, every latitude gives that season a different length, and an object that never sets is best observed at the opposite time of year from the one its right ascension names.

7 figures · Sidereal time The zenith sky after sunset, as single scattering predicts it. The brightness of the zenith sky at 550 nm, in magnitudes per square arcsecond with brighter upward, against the Sun's depression below the horizon, computed by single scattering in a spherical atmosphere: US Standard Atmosphere densities, Rayleigh scattering, a 300 Dobson-unit ozone layer, and every photon scattered exactly once. The Earth's shadow climbs the zenith as R(sec d − 1) — 8.7 km at 3°, 35.1 at 6°, 142 at 12° — through air that thins by a factor of e every eight kilometres or so, so the brightness does not fall steadily: it falls 1.4 magnitudes a degree between 3° and 6°, and 2.7 a degree between 7° and 10°. The model gives 14.3 at the end of civil twilight and 27.0 at the end of nautical. The dashed line is the natural night sky, 21.9 magnitudes per square arcsecond. Single scattering reaches it at 9.25° of depression — 8.8 degrees before the 18° at which astronomical twilight is observed to end. The difference is not an error in the arithmetic; it is the light this model leaves out, scattered more than once.

The shadow that climbs the zenith

After sunset the sky overhead is lit only above the Earth's own shadow, and the shadow climbs as the square of the Sun's depression. Scattering computed once from that geometry predicts a sky that dims faster and faster and is as dark as night by nine degrees. The real sky takes eighteen, and the difference is light that has been scattered more than once.

8 figures · Twilight

Cosmology

One object, seen once, from inside — and every number in it the output of a model.

How much a galaxy's redshift changes in 10 years, and which way. The change in a galaxy's apparent recession velocity over 10 years of the observer's time, c ż/(1 + z), against the galaxy's redshift, for three universes with the same present expansion rate of 67.36 km/s/Mpc. The drift is (1 + z) H₀ − H(z): positive if the expansion rate at the galaxy's epoch was less than (1 + z) times today's, which is to say if the expansion has been accelerating since. In the empty universe the expansion rate is exactly (1 + z) H₀ at every epoch and nothing drifts at all. In the matter-only universe the expansion has only ever slowed, and every redshift falls: −8.6 cm/s over 10 years at z = 1 and −25.5 at z = 4. In ΛCDM the drift is positive nearby, largest at z = 0.63 where it reaches 2.51 cm/s, changes sign at z = 1.91, and is −5.5 cm/s at z = 4. The whole signal is a few centimetres per second in a decade, against the thirty kilometres per second of the Earth's own orbital motion that has to be removed from every spectrum first.

A redshift that changes while it is watched

A galaxy's redshift is a ratio of two sizes of the universe, and the second one is still growing while the light is being collected. So every redshift drifts, by a few centimetres per second in a decade, and the direction of the drift says whether the expansion has been speeding up since the light left — the one test of that question that needs no distance and no model of any source.

6 figures · Expansion

Galaxies

Where the unknown stops being a number and becomes a profile — and most of it is not light.

The horizon and the orbit that cannot come back, against the hole's spin. Two radii round a rotating black hole, in gravitational radii GM/c², against its spin a from −1 (an orbit against the rotation) to +1 (with it), for orbits in the equatorial plane. The lower curve is the horizon, 1 + √(1 − a²), which is the same whichever way the orbit goes. The upper one is the marginally bound orbit, 2 − a + 2√(1 − a): the closest a body falling in from far away can pass and still escape back out. For a hole with no spin the horizon is at 2 and the marginally bound orbit at 4 — twice as far out. With maximal spin the marginally bound orbit comes in to 1.09 for a prograde orbit (a = 0.998) and moves out to 5.83 for a retrograde one. A star whose tidal radius lies inside this curve is swallowed whole, so the curve, not the horizon, is the line a disruption flare is measured against — and it moves by a factor of 5.3 with the spin.

The line a star is swallowed at is not the horizon

A star torn apart by a black hole makes a flare, and a star swallowed whole makes nothing, so the heaviest hole that can produce a flare is a measurement. That ceiling is set not at the horizon but at the closest orbit from which infalling matter can still come back out — twice as far out for a hole that does not spin, and moved by a factor of five by the hole's rotation.

6 figures · Tidal disruption

Exoplanets

Planets nobody has seen, weighed and measured from a dip, a wobble and a delay.

How long a planet's hydrogen lasts, against how much of it there is. The time a young Sun-like star's saturated X-ray and ultraviolet output would take to remove a planet's whole hydrogen envelope, at 100 times the Earth's insolation, against the envelope's share of the planet's mass, for cores of 3, 5, 8 Earth masses. It is computed with energy-limited escape and an interior fit for the envelope's thickness, and it is not monotonic. A heavy envelope takes long to remove because there is a lot of it. A very light one takes long because the planet is small and intercepts little light. In between the time peaks, and it peaks where the envelope has swollen the planet most for its mass: 3 Earth masses at an envelope of 1.9 per cent, 129 Myr, where the envelope is 1.31 times as thick as the core's radius; 5 Earth masses at an envelope of 2.8 per cent, 364 Myr, where the envelope is 1.30 times as thick as the core's radius; 8 Earth masses at an envelope of 4.0 per cent, 946 Myr, where the envelope is 1.30 times as thick as the core's radius. The peak is what makes a valley. A planet above it losing gas moves up the curve, its remaining envelope lasting longer and longer, and settles; a planet below it moves down, lasting less and less, and loses everything. The horizontal lines are the saturated phase, 100 Myr, and the 300 Myr of saturated-equivalent exposure the whole history delivers: a core whose peak lies below the second line cannot keep any envelope at all.

The envelope that doubles a planet lasts longest

The gap in the radii of small planets is not merely a place where planets are rare — it is nearly empty, and the reason is a peak. The time a young star needs to strip a planet's hydrogen is longest for the envelope that swells the planet to a little over twice its core's size. Anything thinner runs away to nothing, and anything thicker settles back towards the peak.

7 figures · Radius valley

Cosmology

One object, seen once, from inside — and every number in it the output of a model.

The expansion rate and the acceleration, and which of them reaches inside an orbit. Two quantities per unit distance through cosmic time, both in units of today's H₀². The square of the expansion rate, H², falls steeply from the big bang and is 1.00 today by definition. The acceleration of the expansion, ä/a, is the sum of a matter-and-radiation part, −Ωₘ/(2a³) − Ωᵣ/a⁴, drawn dashed, and the cosmological constant's part, ΩΛ = 0.685, which is the same at every time. The sum was negative — the expansion decelerating — until the universe was 7.7 Gyr old, at a = 0.614 or redshift 0.63, and is 0.527 today. The equation of motion of anything orbiting inside a bound system carries ä/a and never H: the rate at which distant galaxies recede does not appear in it at all. Of ä/a, the matter part is the mean density of the universe, which inside a galaxy or a planetary system is already counted in the mass that is doing the holding, many million times over. What is left is the constant: a fixed outward acceleration per unit distance, ΩΛ H₀², that does not grow with time and does not care how fast the universe is expanding.

An orbit feels the acceleration and never the rate

If space expands, it is natural to ask why the Earth's orbit does not. The answer is not that gravity resists the stretching. It is that the expansion rate never appears in the equation of motion of a bound orbit at all — only the acceleration does, and of that only the cosmological constant's part survives, as a fixed outward push that moves the Earth's orbit once, by twelve picometres, and never again.

5 figures · Expansion Two galaxies that stopped moving apart, and the mass that did it. The separation of two galaxies on a radial orbit that began together at the big bang, against cosmic time, solved so that after 13.797 Gyr they are 770 kpc apart and approaching at 110 km/s — the present separation and approach speed of the Milky Way and the Andromeda galaxy. The curve is a cycloid, r = A(1 − cos θ) and t = B(θ − sin θ), and only one cycloid passes through that point with that slope. It rose to 1037 kpc, turned round when the universe was 8.5 Gyr old, and on this purely radial orbit the two meet 3.3 Gyr from now. Its period fixes the mass: A³/(GB²) = 4.2·10¹² solar masses. The dashed curve is the same calculation with the cosmological constant's outward push included, integrated rather than solved; to arrive at the same place at the same speed against that push it needs 4.76·10¹² solar masses, 13 per cent more. Far more than the stars of the two galaxies, it is the timing argument's measurement of the Local Group's dark matter.

The age of the universe weighs the Local Group

The Andromeda galaxy is approaching the Milky Way, and in an expanding universe that means the two once moved apart, stopped and turned round. One radial orbit passes through their present separation with their present speed after exactly the age of the universe, and its period fixes the mass that turned them — four trillion suns, twenty-five times what their stars can account for.

5 figures · Expansion

Galaxies

Where the unknown stops being a number and becomes a profile — and most of it is not light.

How far the returning debris swings round, against the hole's mass. The relativistic advance of pericentre, per orbit, of the most bound debris from a disrupted star of 1 solar radius and 1 solar mass, passing at the tidal radius, against the hole's mass. Δω = 6π GM / c² a(1 − e²), and for these nearly parabolic orbits a(1 − e²) is about twice the pericentre, so the angle is set by how many gravitational radii the pericentre is — and that falls as the mass to the minus two thirds, because the tidal radius grows as the cube root of the mass while the gravitational radius grows as the mass. At 10⁵ solar masses the pericentre is 219 gravitational radii and the stream swings round by 2.5°; at 10⁶ solar masses the pericentre is 47 gravitational radii and the stream swings round by 11.6°; at 10⁷ solar masses the pericentre is 10 gravitational radii and the stream swings round by 53.4°. A light hole barely bends its debris's orbit, and a heavy one bends it by tens of degrees.

The debris that returns fastest lights up last

The debris of a star torn apart by a light black hole comes back within two weeks, in greater excess of what the hole can swallow than for any heavier hole. It should make the promptest flare, and it may make the slowest — because to shine, the returning stream has to crash into itself, and where it does so is decided by how far relativity swings its orbit round. Round a light hole the swing is a few degrees, and the streams meet only near the far end of their orbit, moving slowly.

6 figures · Tidal disruption Where the stars that are torn apart come from, round a 10⁶ solar-mass hole. The rate at which stars are delivered onto orbits reaching the tidal radius, per logarithmic interval of their orbital radius, for a hole of 10⁶ solar masses at the centre of a nucleus that is isothermal outside the hole's influence and a Bahcall–Wolf cusp within it, with a velocity dispersion of 54 km/s set by the black hole mass–dispersion relation. The radius is in units of the hole's influence radius, GM/σ² = 1.47 pc. Close in, a star's orbit is so short and its angular momentum changes so slowly that the loss cone empties every orbit, and the rate is limited by relaxation diffusing stars into it; far out, the angular momentum wanders across the whole cone in one orbit, the cone is full, and the rate is limited by how many stars there are and how long they take to come in. The flux peaks at 0.27 influence radii, near where the two regimes meet (0.19), and the total is 7.4·10⁻⁵ disruptions a year. The nucleus model is the simplest there is and real nuclei differ from it by factors of several; the shape — a narrow band of radii supplying most of the flares — is what survives.

The stars a black hole eats come from a narrow band

A star is torn apart only if its orbit happens to point almost exactly at the hole — within a millionth of the possible directions of its angular momentum. Stars on those orbits are gone within one orbit, and the only way new ones arrive is by the slow random walk of encounters with other stars. Close to the hole that walk is too slow; far from it the orbits take too long; and most of the stars a hole destroys come from a narrow band between.

6 figures · Tidal disruption

The observed sky

Coordinates, seasons, phases and shadows — geometry seen from inside it.

The colour of the zenith at twilight, with and without ozone. The colour of the zenith sky relative to sunlight — the 450 nm brightness against the 650 nm brightness, in magnitudes, bluer upward — against the Sun's depression, computed by single scattering with a 300 Dobson-unit ozone layer and again with none. Scattering alone favours blue by λ⁻⁴, which would make the sky 1.60 magnitudes bluer than sunlight if nothing were removed on the way; that is the dotted line. But after sunset every ray has travelled a long grazing path, and Rayleigh scattering removes blue from that path faster than red, so the two effects fight. Without ozone they very nearly cancel: the zenith is 0.01 magnitudes from sunlight's own colour at sunset and 0.06 at 6° — a pale, colourless sky — and only turns bluer, −0.13 at 10°, once the lit layer has climbed above most of the air the grazing ray used to cross. With ozone the zenith is −0.34 at sunset, −0.67 at 6° and −0.76 at 10°: bluer than without by 0.72 magnitudes at 6°, because the grazing ray also crosses the ozone layer near its tangent point, and ozone's Chappuis band absorbs orange and red rather than blue. The ozone cross-sections are approximate, and the conclusion does not depend on them to better than a factor of two.

Ozone keeps the twilight zenith blue

After sunset the sky overhead turns a deep blue, and scattering alone cannot explain it. The light that reaches the zenith has first grazed hundreds of kilometres of air, which strips blue out of the sunlight as fast as scattering puts it back, and the two very nearly cancel. What tips the balance is a gas that makes up a few parts in ten million of the atmosphere and absorbs the orange and red end of the spectrum.

6 figures · Twilight How far the Sun has to sink before the sky darkens, on four worlds. The dimming of the zenith sky relative to sunset, in magnitudes, against the Sun's depression, for single scattering in an isothermal atmosphere with the stated scale height and vertical optical depth, scattering isotropically: the Earth (R 6371 km, H 8.5 km, τ 0.097); Mars (dust) (R 3389.5 km, H 11.1 km, τ 0.5); Titan (haze) (R 2574.7 km, H 50 km, τ 4); Pluto (haze) (R 1188 km, H 50 km, τ 0.02). These are representative values for the layer that does the scattering — air on the Earth, dust on Mars, haze on Titan and Pluto — not full models of those atmospheres. The angle over which a sky darkens is set by the height of the scatterers against the size of the planet, √(2H/R), because that is the depression at which the shadow over the zenith has risen one scale height. The Earth dims by ten magnitudes at 8.6° of depression, which at the equator of a world whose solar day is 24 hours takes the Sun 35 minutes. Mars (dust) dims by ten magnitudes at 13.4° of depression, which at the equator of a world whose solar day is 24.66 hours takes the Sun 55 minutes. Titan (haze) dims by ten magnitudes at 29.8° of depression, which at the equator of a world whose solar day is 382.7 hours takes the Sun 31.7 hours. Pluto (haze) dims by ten magnitudes at 43.9° of depression, which at the equator of a world whose solar day is 153.3 hours takes the Sun 18.7 hours.

The air's height against the planet sets the twilight

On the Earth the sky overhead fades over about nine degrees of the Sun's descent. That angle is not a property of air or of sunlight. It is the square root of twice the height of the scattering layer divided by the radius of the planet, and on a small world with a tall haze it grows to tens of degrees — so that twilight on Titan lasts more than a day.

5 figures · Twilight

Exoplanets

Planets nobody has seen, weighed and measured from a dip, a wobble and a delay.

When the stripping happens, in a model population. For the same 3,000 model planets, the share that have lost their whole envelope by each age, scaled to the share bare at 5.0 Gyr (35 per cent of the population), beside the share of the star's lifetime XUV energy delivered by then. Half of all the stripping in this model is finished by 72 Myr and nine tenths by 457 Myr, while the star has delivered 27 and 76 per cent of its XUV energy. That is the clock photoevaporation keeps: the valley is essentially finished within the first few hundred million years, because the planets near the boundary are the ones that run away, and they run away early. A mechanism powered instead by the slow cooling of the planets' own cores would keep moving planets across the valley for billions of years. The difference is in when, not where, and it is why the ages of the stars hosting planets on either side of the valley are the measurement that can separate the two.

The stripping runs ahead of the starlight that drives it

If young stars carve the radius valley with their X-ray light, the valley should be finished early — half of it before the star has delivered a third of that light, and nine tenths of it within a few hundred million years. If the planets' own cooling cores carve it instead, planets should still be crossing it billions of years later. The two accounts put the valley in the same place, and they are separated by the one thing a histogram cannot show — when.

6 figures · Radius valley Where photoevaporation puts the valley, round stars of different mass. The radius of the largest core stripped bare, against orbital period, round stars of 0.5 M☉, 0.75 M☉, 1 M☉, 1.25 M☉, in the same energy-limited model, with each star's luminosity taken as its mass to the fourth power and its saturated phase lengthened for smaller stars as the mass to the −1.5, 100 Myr for the Sun. At ten days the valley is at 1.22 Earth radii round the 0.5 M☉ star, 1.38 Earth radii round the 0.75 M☉ star, 1.50 Earth radii round the 1 M☉ star, 1.60 Earth radii round the 1.25 M☉ star, and every one of them tilts as the −0.176 power of the orbital period, because the tilt comes from the exponents of the escape law and the interior fit, which do not depend on the star. A lower-mass star is far fainter, so at a given period its planets receive much less light, and even its longer active phase does not make up the difference: this model puts the valley at smaller radii round smaller stars while keeping the same sign of tilt. The stellar scalings are rough, and they are the least certain part of the calculation; what is robust is that photoevaporation ties the valley to the XUV energy a planet received, which falls with the star's mass at fixed period, and a mechanism tied to something else would tie it differently.

A smaller star puts the valley lower

Round stars of half the Sun's mass, the gap between bare rocky cores and sub-Neptunes should sit at a smaller radius than round the Sun, because at the same orbital period their planets receive a tenth of the light. Their stars also stay young and active far longer, which pushes the other way. Photoevaporation weighs those two against each other in a definite proportion, and the answer is a valley that scales as the star's mass to about the power three tenths.

5 figures · Radius valley The XUV energy that reached one astronomical unit, for three young Suns. The cumulative X-ray and extreme-ultraviolet energy delivered per square metre at the Earth's distance from the Sun, against age, for three histories that differ only in how long the young Sun stayed magnetically saturated: 20 Myr for a slow rotator, 100 Myr for a medium rotator, 300 Myr for a fast rotator. After saturation each declines to a common track by 1 Gyr, as rotation histories are observed to converge. By 4.5 Gyr the totals are 2.06·10¹⁵ J/m² for the slow rotator, 3.66·10¹⁵ J/m² for the medium rotator, 6.47·10¹⁵ J/m² for the fast rotator: the fast rotator delivered 3.1 times as much as the slow rotator, nearly all of it in the first few hundred million years. The Sun's own rotation at that age is not measured; it is inferred from the spread of rotation periods in young clusters, and all three histories are consistent with a Sun that ends up rotating as it does now.

The young Sun's spin decides what the Earth kept

The calculation that strips sub-Neptunes applies just as well to a planet with a trace of hydrogen. An Earth that captured a few hundredths of a per cent of its mass from the gas it formed in — several times the hydrogen now in its oceans — would have lost all of it, or kept most of it, depending on something nobody has measured — how fast the Sun was spinning in its first few hundred million years.

5 figures · Radius valley

Cosmology

One object, seen once, from inside — and every number in it the output of a model.

The dipole a 370 km/s motion has to put into counts of distant sources. The amplitude of the dipole in the number of sources per unit solid angle expected from the Sun's motion at 369.82 km/s relative to the microwave background, β = 1.234e-3, split into its two parts: aberration, 2β, which crowds sources towards the direction of motion, and the Doppler boost, x(1 + α)β, which brightens sources there and lifts fainter ones above the flux limit. For radio sources, with counts steepening as S^(−1) and spectra falling as ν^(−0.75), the expected dipole is 0.0046; for mid-infrared quasars, with counts steepening as S^(−1.7) and spectra falling as ν^(−1.26), the expected dipole is 0.0072, and the measured one is 0.0155 — 2.16 times larger, which would need a speed of 797 km/s. A dipole of a few parts in a thousand needs a catalogue of more than a million sources to see at all. The disagreement is not with the direction, which lies close to the microwave background's, but with the size, and it is not yet explained: either the samples carry a systematic nobody has found, or the matter and the radiation do not share one rest frame on these scales — in which case the assumption that the universe looks the same from everywhere is wrong in a way it has never been caught being wrong before.

The Sun's speed counted in quasars comes out twice too large

The microwave background is warmer in one direction by a part in a thousand, and that dipole is read as the Sun's motion through it at 370 kilometres per second. The same motion must crowd the counts of distant galaxies and quasars towards the same direction by an amount that can be calculated exactly. When a million quasars were counted, the direction agreed and the size came out more than twice too large — as though the Sun were moving at 800 kilometres per second relative to the matter.

5 figures · Expansion m²φ²: where the observed scales left, and where inflation ends. The m²φ² potential, drawn as a shape with its height divided out, against the field in reduced Planck masses. The field rolls downhill towards zero and inflation ends where the slow-roll parameter ε reaches one, at φ = 1.41. The shaded band is the stretch of field the scales now seen on the sky left the Hubble radius from: 60 e-folds before the end at φ = 15.56 and 50 before it at φ = 14.21. Nothing about the sky depends on the rest of the curve. Across that band the two numbers the tilt is made of are ε = 9.01e-3 and η = 0.0090, which give a spectral index of 0.9640 and a tensor-to-scalar ratio of 0.1441 at 55 e-folds. The height is not in either: it is fixed separately by the amplitude of the fluctuations, which puts the potential at (2.0 × 10¹⁶ GeV)⁴ there — and multiplying the whole curve by any constant leaves the band, the tilt and the ratio exactly where they are, because every slow-roll quantity is a ratio of the potential to its own derivatives. The field travels 13.49 Planck masses from the middle of the band to the end.

The tilt knows the slope and not the height

The measured spectral index, 0.965, is quoted as the strongest evidence for inflation, and it is a statement about two dimensionless numbers — how steeply the potential fell and how sharply that slope was changing, over the few e-folds the sky can see. The height of the potential is not in it at all, which is why potentials that look nothing alike reproduce it.

7 figures · Inflation 6 potentials against the tilt and the tensor bound. Predictions in the plane of spectral index and tensor-to-scalar ratio, each drawn as a short track from 50 e-folds to 60, the range usually allowed for the pivot scale to have left before the end of inflation. The vertical band is the measured index, 0.9649 ± 0.0042 with its two-sigma extent, and the shaded region above 0.036 is excluded at 95 per cent by the B-mode polarisation limit — drawn as two independent limits, which the published constraint is not quite: the real likelihood is a correlated contour, and it is somewhat tighter than this box in the corner where the tilt is high. λφ⁴: nₛ 0.9412, r 0.3137 at 50 e-folds; 0.9508, 0.2623 at 60 — outside; m²φ²: nₛ 0.9604, r 0.1584 at 50 e-folds; 0.9669, 0.1322 at 60 — outside; linear φ: nₛ 0.9701, r 0.0796 at 50 e-folds; 0.9751, 0.0664 at 60 — outside; φ^⅔: nₛ 0.9734, r 0.0532 at 50 e-folds; 0.9778, 0.0443 at 60 — outside; natural, f = 7: nₛ 0.9569, r 0.0906 at 50 e-folds; 0.9628, 0.0670 at 60 — outside; Starobinsky: nₛ 0.9616, r 0.0042 at 50 e-folds; 0.9678, 0.0030 at 60 — inside. The simplest potential of all, a mass term, is excluded over its whole range of e-folds, and not by the tilt, which it matches — by the tensors.

A ratio that is an energy and a distance

The tensor-to-scalar ratio is the one inflationary observable that measures the height of the potential rather than its shape, and the quantity it fixes is an energy — a ratio of 0.01 means inflation happened at 10¹⁶ GeV. It fixes a second thing as well, how far the field travelled, and near the present bound that distance is several Planck masses, which is where the theory stops being able to vouch for itself.

5 figures · Inflation Starobinsky: e-folds before the end, against how reheating went. N, the number of e-folds between the pivot scale leaving the Hubble radius and the end of inflation, for the Starobinsky potential, against the temperature at which reheating finished, for 3 equations of state during it. All the lines meet on the right at instant reheating, 2.6 × 10¹⁵ GeV, where N = 55.6. The left edge is 5 MeV, below which nucleosynthesis would not have happened. w = 0, oscillating field: 42.0 at 5 MeV; w = ⅓, like radiation: 55.6 at 5 MeV; w = 1, kination: 69.0 at 5 MeV. The reason is how far the universe stretches while the energy density falls: an oscillating field dilutes like matter, as a⁻³, so for the same fall in density it expands further than radiation would, more of the growth of today's scales happens after inflation, and fewer e-folds of inflation are needed to put them where they are. A stiff epoch with w = 1 dilutes as a⁻⁶, stretches less, and needs more. A radiation-like epoch changes nothing. None of this epoch has been observed; the lines are the arithmetic of energy and entropy, and the spread between them is how much an unobserved history moves a quantity the spectral index depends on.

The epoch nobody saw moves the tilt

Between the end of inflation and the hot universe that made the light elements lies an interval nothing has observed, in which the energy of the inflaton became radiation. How long that took changes how many e-folds before the end the observed scales left — by as many as fourteen — and that moves every model's predicted spectral index by more than the measurement's uncertainty. A potential is never tested by the tilt alone; a potential and a reheating history are tested together.

7 figures · Inflation One sky with and without a local non-Gaussianity of fNL·σ = 0.3. The same scale-invariant random field drawn twice, from one seed: on the left as it is, Gaussian, and on the right after the local transformation Φ → Φ + fNL(Φ² − ⟨Φ²⟩) with fNL·σ = 0.3. Solid contours are one and two standard deviations above the mean, dashed ones below, and each map is measured against its own mean and spread. The transformation adds to every value in proportion to its square, so peaks are pushed up and troughs are pulled back towards the mean: the area above +2σ goes from 1.8 to 4.7 per cent of the map and the area below −2σ from 2.4 to 0.0, and the skewness of the values rises from −0.073 to 1.484. This is exaggerated by a factor of about 2200. The primordial potential varies by about 3 × 10⁻⁵, so even fNL = 5 — the size of the current uncertainty — makes fNL·σ about 10⁻⁴.

A test that can only fail one way

Single-field inflation predicts a local non-Gaussianity of 0.015, a skewness in the primordial potential of a few parts in a million. The measurement is −0.9 ± 5.1. A detection at the level of one would eliminate every model with a single clock at once; a null result at any reachable precision confirms nothing, because the prediction lies below anything the sky has enough independent modes to measure.

6 figures · Inflation

Every sequence · Every named object · Search