Generator

The seeing-psf generator

Resolution stops improving at 10 cm of aperture

Angular resolution against aperture at 500 nm, both logarithmic. The falling line is diffraction alone, 1.03 λ/D, which is what a telescope in vacuum delivers and has no floor. The curve is the same telescope under an atmosphere of Fried parameter r₀ = 10 cm, combining diffraction and seeing in quadrature: it follows the diffraction line while D < r₀ and then bends onto a plateau at 0.98 λ/r₀ = 1.01″. An amateur's 100 mm at 0.1 m would resolve 1.062″ above the air and delivers 1.47″ through it; a metre at 1 m would resolve 0.106″ above the air and delivers 1.02″ through it; the VLT at 8.2 m would resolve 0.013″ above the air and delivers 1.01″ through it; the ELT at 39 m would resolve 0.003″ above the air and delivers 1.01″ through it. At 39 m the atmosphere is costing a factor of 371: the aperture is 390 coherence lengths across and every one of the 152,100 patches it collects arrives with a phase of its own. *What the extra aperture still buys is photons and speckles*, and those two are what adaptive optics and speckle interferometry respectively spend to get the falling line back.

6 essays call seeing-psf. 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.

$Resolution stops improving at 10 cm of aperture. Angular resolution against aperture at 500 nm, both logarithmic. The falling line is diffraction alone, 1.03 λ/D, which is what a telescope in vacuum delivers and has no floor. The curve is the same telescope under an atmosphere of Fried parameter r₀ = 10 cm, combining diffraction and seeing in quadrature: it follows the diffraction line while D < r₀ and then bends onto a plateau at 0.98 λ/r₀ = 1.01″. An amateur's 100 mm at 0.1 m would resolve 1.062″ above the air and delivers 1.47″ through it; a metre at 1 m would resolve 0.106″ above the air and delivers 1.02″ through it; the VLT at 8.2 m would resolve 0.013″ above the air and delivers 1.01″ through it; the ELT at 39 m would resolve 0.003″ above the air and delivers 1.01″ through it. At 39 m the atmosphere is costing a factor of 371: the aperture is 390 coherence lengths across and every one of the 152,100 patches it collects arrives with a phase of its own. What the extra aperture still buys is photons and speckles, and those two are what adaptive optics and speckle interferometry respectively spend to get the falling line back.$ The observed sky

A ten-metre mirror that resolves like a ten-centimetre one

The atmosphere delivers a wavefront in patches about ten centimetres across, and an aperture larger than a patch collects patches rather than detail. Resolution stops improving at that size — and what the extra aperture keeps buying is photons and speckles, which is why there are two entirely different ways out.

$A patch 1.5″ wide in the visible and 8″ at 2.2 µm. The isoplanatic angle against wavelength, for r₀ = 10 cm at 500 nm, a turbulence layer at 5 km and a wind of 20 m/s. This is the angle over which one measurement of the wavefront is still valid, and it is the hardest of adaptive optics' three limits: at 1.5 arcseconds in the visible, the guide star has to be inside a patch a hundredth the size of the full Moon. Everything scales as λ^6/5 because r₀ does — measured off the curve at λ^1.200 — so the patch grows to 8 arcseconds at 2.2 µm, and its area by the square of that. With 0.1 stars per square arcminute bright enough to guide on, the fraction of sky reachable goes from 0.018 per cent to 0.5 — a factor of 28. That single curve is why the first working systems were infrared, why a laser is fired to make a star where there is none, and why the laser still does not solve it: a beam launched from the telescope wanders with the same atmosphere it is meant to measure, so it cannot sense the overall tilt, and a natural star is still needed for that.$ The observed sky

The correction has to be faster than the air

Making a large telescope resolve like a large telescope means measuring the wavefront and undoing it. Three numbers bound how well that can work and none of them is the mirror — a frequency of hundreds of hertz, a patch a second and a half wide, and the chance of a bright enough star inside it.

$What the error bar is made of, on a 1 m in 60 s. The four contributions to a photometric error, against the brightness of the star, for a 1-metre aperture, a 60-second exposure and a sky of 21 magnitudes per square arcsecond. The star's own photons give a line of slope exactly 0.2 — σ ∝ N^−1/2 and N ∝ 10^−0.4m, so a magnitude of extra faintness costs a fifth of a magnitude of precision, and no instrument changes that. The sky and the read noise are fixed counts, so their lines have slope 0.4, twice as steep, and they overtake the star at V = 18.25 — that crossing is the faint limit of the night, and it moves when the Moon rises rather than when the telescope changes. Scintillation is flat, because the atmosphere modulates a bright star and a faint one by the same fraction: at 4.09e-4 relative it is 0.44 millimagnitudes here and it is what caps the bright end, up to about V = 10.8. Below all of them is the systematic floor at 0.3 millimagnitudes, which is flat-fielding and colour terms and does not integrate down at all.$ Starlight

The faint star is measured against a brighter sky

For anything at the edge of detection the dominant source of noise is not the object. It is the sky in the same aperture, which is brighter than the star and is subtracted rather than measured — and once that is true, every rule of thumb about apertures, exposure times and image quality changes.

$A star drawn out into 3.0 arcseconds of spectrum. Atmospheric refraction relative to its value at 550 nanometres, against wavelength, at four zenith angles. The air's refractive index rises towards the blue, so the blue image of a star sits above the red one and the object is smeared into a short vertical spectrum. At 60 degrees from the zenith the separation across an optical band is 2.96 arcseconds — several times the size of the image at a good site, and comparable to the width of a spectrograph slit. Every curve here is the same curve multiplied by the tangent of the zenith angle, which is why one corrector with an adjustable strength works at every airmass. The practical consequences are three: a slit aligned other than vertically loses blue light or red light depending on where it was centred, a photometric aperture contains a different fraction of the light in each band, and an astrometric position depends on the colour of the star it is measured from.$ The observed sky

The atmosphere is a prism as well as a lens

Refraction lifts a star towards the zenith, and everybody corrects for that. It lifts blue light further than red, and the difference is a short vertical spectrum a few arcseconds long — larger than the image, larger than a spectrograph's slit, and quietly present in every ground-based measurement not taken straight overhead.

$196 nanometres of wavefront, and a Strehl that depends entirely on the colour. Above, the error budget of an adaptive-optics system, term by term, in nanometres of residual wavefront. The terms are independent and add in quadrature, so the total is 196 nanometres and is dominated by the two largest; removing the smallest term entirely would improve it by under six per cent, which is why an optimisation programme that does not know which term is largest achieves nothing. Below, what that residual delivers: the Strehl ratio, the fraction of the light in the diffraction core, against wavelength. It is the exponential of minus the square of the residual measured in radians, and a radian is a wavelength, so the same physical error is four times smaller in phase at two microns than at half a micron. The system drawn here delivers 1 per cent of its light into the core at 550 nanometres and 73 per cent at 2200, and nothing about it changed between the two.$ The observed sky

An error budget added in quadrature

Adaptive optics does not deliver a resolution. It delivers a fraction of the light in the diffraction core, and that fraction is the exponential of minus a sum of squares. Five independent failures of the correction add in quadrature, the largest one decides everything, and the same hardware is useless in the visible and excellent in the infrared.

$A planet that subtracts 73 per cent of itself at the inner working angle. The fraction of a planet's flux that survives an angular differential imaging subtraction, against its separation from the star in resolution elements, for a sequence covering 25 degrees of field rotation. The reference image is built from the target's own frames, so a planet that has not moved far between them is present in the reference and is removed along with the speckles. How far it moves is the arc length, which is proportional to the separation — so the self-subtraction is severe close in and negligible far out, and the half-throughput point is at 2.1 resolution elements here. The consequence for any contrast curve is that it is a statement about an algorithm as well as about an instrument: the depth reached has to be measured by injecting fake planets into the data and recovering them, because no calculation predicts what fraction of a real one survives.$ Exoplanets

A star subtracted using the star

Imaging a planet means removing a halo of scattered starlight a hundred million times brighter than the planet, and no model of that halo is good enough to subtract. So it is built from the star's own exposures — and since the planet is in those exposures too, it subtracts part of itself.

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