Generator

The olbers-sky generator

Two reasons the sky is dark, and only one of them is the expansion

The surface brightness of the sky accumulated out to a given distance, in units of the surface brightness of a star, both axes logarithmic. Every shell of thickness dr contributes the same amount, because the number of stars in it grows as r² and each one's flux falls as r⁻², so the straight rising line is the paradox: in an infinite static universe the sky reaches stellar surface brightness at the mean free path to a stellar surface, λ = 1/nσ, which for 10⁹ stars per cubic megaparsec of radius 0.6 R☉ is 1.74·10¹⁸ Mpc. The two suppressions are then computed separately, and they are wildly unequal. The finite age cuts the integral off at the particle horizon, 1.41·10⁴ Mpc, which is a factor of 1.23·10¹⁴ short of λ. The expansion then dims what is left by a further factor of 6.03, computed as the mean of (1+z)⁻² over the comoving distance to the horizon. The horizon does 14 orders of magnitude and the redshift does less than one. The common answer that the sky is dark because the universe is expanding is therefore very nearly the wrong answer: the sky is dark because the universe is young, and the expansion is a small correction on top.

8 essays call olbers-sky. The drawing above is what it returns with no arguments at all; every call below passes it something, because a placement that passes nothing draws whichever member of the family the generator happens to default to rather than the one its essay argues about.

Where it is called

Every figure listed here is the same construction drawn at different numbers, so a correction to one is a correction to all of them.

Every shell contributes the same. Concentric shells of equal thickness around an observer, with stars scattered uniformly in volume. The number of stars in a shell grows as its radius squared and the flux from each falls as the radius squared, so the two cancel exactly and every shell delivers the same total light — the drawn counts are 6, 15, 31, 55, 88, in proportion to r³ − r₀³, and the drawn sizes fall as 1/r. That cancellation is the paradox, and it is why no amount of dust helps: dust absorbs the light and then re-radiates it, and in a universe old enough for the sum to converge it would come to the same temperature as the stars. The sum diverges linearly with radius, so something has to stop it — and the only two candidates are that the shells eventually overlap, or that there are no shells beyond a certain distance because there has not been time for their light to arrive.

Why the sky is dark

In an infinite universe of stars every sight line ends on a stellar surface, so the whole sky should be as bright as the Sun's disc. It is not, and the usual answer — that the expansion redshifts the light away — accounts for a factor of six out of a hundred trillion.

The darkness has a number in it

The night sky is not black. It carries about sixty nanowatts per square metre per steradian, and that number is the sum of every photon every star has ever emitted, redshifted and added up over thirteen billion years.

A background weighed by what it stops

The faintest light in the universe cannot be photographed from inside the Solar System, because the zodiacal foreground is a hundred times brighter. It can be weighed instead, by the bite it takes out of a blazar at a trillion electronvolts.

Three messengers, and the three walls they end on. The redshift of the last opaque surface, for the three things that cross the universe, on one logarithmic axis spanning thirty-one decades. Olbers' argument compares how far a sight line runs before it ends on something against how far anything has had time to come — and for starlight the first is 1.7·10¹⁸ Mpc against a horizon of 1.4·10⁴ Mpc, which is why the optical sky is dark. That comparison is not what decides the other two. A relic neutrino's mean free path against ordinary matter works out at 7.7·10³⁷ Mpc — 4·10¹⁹ times starlight's, so on the mean-free-path argument alone the neutrino sky should be darker still. It is not, because the quantity that terminates a sight line is the WALL, and the walls are at z = 1,090, z ≈ 6×10⁹ and nowhere at all. Every neutrino sight line ends on a surface from one second after the beginning; every gravitational-wave sight line runs to the beginning, or to a binary. Both of those skies are saturated — which is what Olbers' argument predicted, and what the optical sky refuses to do. The three are also wildly unequal in brightness: the microwave sky is 996 nW m⁻² sr⁻¹, and the gravitational-wave background at Ω = 10⁻⁹ is 0.018 — a saturated sky five decades fainter than a dark one.

Two skies where the paradox comes out right

Olbers argued that every sight line should end on a source and the sky should blaze. In neutrinos and in gravitational waves it does — the walls are at one second and at no time at all — and both of those skies have now been detected.

$How much of a background a survey can resolve, against sources per beam. The fraction of a background resolved into individual sources before confusion stops the count, against X, the number of sources brighter than the break in the counts that fall inside one beam, on a logarithmic axis. The curve is computed for counts falling as S^−2.5 above the break and S^−1.5 below it, with a source counted as recoverable while there is less than one source per 30 beams brighter than it. It is a single curve: telescope, wavelength and integration time enter only through X. Half the background is resolved at X ≈ 0.033, and the fall from nearly all to nearly none takes about four decades. The surveys are placed on it at their own beams and counts: Hubble, optical at X = 8.7·10⁻⁵, 100 per cent; Chandra, soft X-ray at X = 6.4·10⁻⁶, 100 per cent; Spitzer, 24 µm at X = 0.012, 54 per cent; Herschel, 250 µm at X = 0.043, 24 per cent; Herschel, 500 µm at X = 0.18, 11 per cent. The points do not sit exactly on the line because each band has its own slopes.$ Cosmology

A background that stops breaking into sources

A deep enough image turns a glowing sky into a list of galaxies. Past a certain depth it stops doing so, and the depth is set by how many sources share one beam — which a longer exposure cannot change.

The histogram of a sky too crowded to count

When every beam holds more than one faint source, a map stops being a list. It remains a measurement — the histogram of its pixels carries the number of sources nobody can pick out, down to the depth where there are so many that their sum turns Gaussian.

40 galaxies detected, and a map of all of them: the summed light traces the structure the catalogue misses. A 160 by 160 Mpc/h slice of a simulated universe at redshift 6, its galaxies clustered on a density field and given luminosities from a Schechter function with a faint-end slope of -1.5. Left: the 40 galaxies bright enough to detect individually, above 1.5 times the characteristic luminosity. Right: the same slice as a map of summed line intensity in 32 by 32 cells, shaded linearly between the 5th and 97th percentiles of the cells, with no galaxy detected at all. The catalogue is a sparse scatter in which the structure is barely visible; the map shows it, and its cells correlate with the underlying density at 0.65. The map's light comes overwhelmingly from galaxies below the detection limit.

A map of galaxies none of which was seen

A survey that lists galaxies one by one stops at the faintest it can detect, and at high redshift most of the light comes from galaxies fainter than that. Map one spectral line in summed intensity instead, pixel by pixel and frequency by frequency, and every galaxy contributes whether it is detected or not. The fluctuations of that map have two parts — clustering, which weighs the total light, and graininess, which weighs how unevenly it is shared — so one power spectrum measures two moments of a luminosity function nobody has counted.

[C II] power is round in the plane of k; CO(4–3) mapped at the wrong distance is elongated 5.1 times along k∥. Contours of the clustering power spectrum in the plane of wavenumber across the sky, k⊥, and along the line of sight, k∥, in h/Mpc, at 10², 10²·⁵, 10³, 10³·⁵ (Mpc/h)³ in arbitrary units, for the target line, [C II] from redshift 6.60 (black), and for the CO(4–3) interloper from redshift 0.84 placed at the target's distance (colour). The target's contours are quarter-circles, because its structure has no preferred direction. The interloper's are ellipses, drawn in towards the k∥ axis and elongated along it by the ratio of the two rescalings, α∥/α⊥ = 5.1: a structure that is round where it really is looks magnified across the sky and flattened along the line of sight where the map puts it. That difference of shape, not of amplitude, is what separates the two in a map where both have been summed.

A line from the wrong redshift lands at the wrong distance

A map of ionised carbon from redshift six is built by assuming every photon at a given frequency came from that line. Carbon monoxide from galaxies at redshift one sends photons to the same frequency, and the map places them five billion light-years too far away. Their structure arrives magnified across the sky and squashed along the line of sight — round where it really is, flattened in the map — and that distortion of shape, which no amount of brightness can hide, is how the impostor is found.

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