Purpose of these lectures

To prepare you for observing at the Isaac Newton Telescope What the telescope + instrument can do What it can’t do What the atmospheric and lunar conditions are likely to be and their effect on the observations To choose what we’ll observe Overview of lectures

Lecture 1: an overview of observing – telescope, instrument, coordinate systems, atmosphere, and moon Lecture 2: an overview of photometry and photometric data reduction Lecture 3: the TAC – “Time Allocation Committee” La Palma observing trip: preparation Lecture 1 S.C. Trager Outline

Astronomical coordinate systems Telescope properties: INT plate scale, speed, type Instrument: the WFC A short recap of noise Effects of the atmosphere and background sources Coordinate systems The Celestial Sphere

First we need the concept of the celestial sphere. It would be nice if we knew the distance to every object we’re interested in — but we don’t. And it’s actually unnecessary in order to observe them! The Celestial Sphere

Instead, we assume that all astronomical sources are infinitely far away and live on the surface of a sphere at infinite distance. This is the celestial sphere. If we define a coordinate system on this sphere, we know where to point! Further, stars (and galaxies) move with respect to each other. The motion normal to the line of sight — i.e., on the celestial sphere — is called proper motion (which we’ll return to shortly) Astronomical coordinate systems

A bit of terminology: great circle: a circle on the surface of a sphere intercepting a plane that intersects the origin of the sphere Horizon coodinates

A natural coordinate system for an Earth- bound observer is the “horizon” or “Alt-Az” coordinate system The great circle of the horizon projected on the celestial sphere is the equator of this system. Horizon coodinates

Altitude (or elevation) is the angle from the horizon up to our object — the zenith, the point directly above the observer, is at +90º Horizon coodinates

We need another coordinate: define a great circle perpendicular to the equator (horizon) passing through the zenith and, for convenience, due north This line of constant longitude is called a meridian Horizon coodinates

The azimuth is the angle measured along the horizon from north towards east to the great circle that intercepts our object (star) and the zenith. Horizon coodinates

The origin of these angles (coordinates) is the observer Note that this is a left- handed coordinate system! Horizon coodinates

Nearly all big telescopes (diameter ≥ 4m, telescopes built after ~1990, most radio telescopes) are in alt-az mounts This is the natural coordinate system for these telescopes But this system is dependent on the location of the observer and time of the observation: makes consistent cataloguing of objects difficult! Equatorial coordinates +90º

Let’s consider a coordinate system that is tied to the astronomical objects themselves — and preferably those that don’t move! ♈

–90º Equatorial coordinates +90º

In equatorial coordinates, the celestial equator is the great circle that intersects both the celestial sphere and the Earth’s equator: it’s the projection of the equator onto the celestial ♈ sphere

–90º Equatorial coordinates +90º The declination δ is the celestial latitude and is measured in degrees, with 0º at the equator, +90º at the North Celestial Pole (NCP) — the intersection of the Earth’s north (rotational) pole with the celestial sphere — and ♈ –90º at the South Celestial Pole –90º Equatorial coordinates +90º The right ascension (RA) α is the celestial longitude and is measured in units of time, 0–24 hours, from west to east, with 0h at the Sun’s position when it crosses the equator from south to north, ♈ approximately at noon on 21 March. –90º Equatorial coordinates +90º The position α=0h, δ=0º is called the vernal equinox ♈ this is the sign of the constellation Aries, where the vernal equinox happened 2500 years ago ♈ The equatorial system is a right-handed system –90º Equatorial coordinates +90º

Because the Earth precesses around an average direction perpendicular to the ecliptic (the plane of the Earth’s orbit around the Sun) due to the torques exerted on by the Moon, Sun, and Jupiter (more ♈ later!), the equatorial system slowly changes with time.

–90º Equatorial coordinates +90º

This means that the vernal equinox and the celestial equator move with respect to the distant background objects (galaxies, quasars). There we need to assign an epoch — a date — to any equatorial coordinate. ♈ (We’ll return to this shortly!)

–90º The local equatorial system

The local equatorial system is used to point polar-axis (or “equatorial”) mount telescopes These telescopes rotate around an axis parallel to the Earth’s rotation axis In the Northern Hemisphere, this means that the primary mount axis always points north The local equatorial system

These telescopes track a star by rotations around only one axis Note that this means that the field of the image does not rotate, like it does for an alt-az telescope The local equatorial system

In the local equatorial system, the hour angle HA replaces the right ascension: HA=LST–α Here LST is the local sidereal time (which we’ll define shortly!) So knowing the time of day (the LST) and the α,δ of an object, it’s very easy to locate your object with a polar-axis telescope. Note that the minus sign makes this a left-handed coordinate system! Julian Date

Julian Date (JD) is an extremely useful way of keeping track of observations made over long time periods. Julian Dates are defined as the number of (Julian) days since noon on 1 January 4713 BCE (really!) Roughly now — 27 March 2014 at 08h 30m UT1 — is JD 2456743.854167 The beginning of each Julian day is defined to be at noon in Greenwich, 12h UT1 Julian Date

The Modified Julian Date (MJD) is often used (because it’s shorter!): MJD=JD–2400000.5 Note that it starts at midnight in Greenwich rather than at noon (in fact, it started precisely at 00h 00m UT on Wednesday 17 November 1858!) Note also that J2000.0 is defined on the Julian day/year (=365.2500d exactly)/century (36525d) system and began at 12h (TDB) 1 January 2000 exactly, i.e., JD2451545.0 (TDB) Telescope plate scales, speeds and types y

h′ θ z O D

f nebula at ∞

Let’s replace our telescope with a thin lens with an equivalent focal length f Note that rays that pass through the center of a thin lens are undeviated, so that h0 = f tan ✓ f✓ ⇡ y

h′ θ z O D

f nebula at ∞

So the angle imaged onto a unit length on the focal plane is ✓ 1 S = = ! h0 f 206265 [00/rad] or S(00/mm) = f [mm] is the plate scale y

h′ θ z O D

f nebula at ∞

The Shane 3m telescope at Lick Observatory has f=15.3 m at prime focus, so S=14″/mm The full moon then images onto ≈13 cm at prime focus This is big for an optical system! Now let’s consider the size on the detector of our nebula with angular size θ

For pixels (or photographic grains) of fixed size, smaller diameter h′ will put more energy into a single pixel: –2 –2 Epixel∝h′ ∝f

Of course, a larger aperture (mirror) will collect more 2 photons, so Epixel∝D

2 Therefore, for diffuse (resolved) sources, Epixel∝(D/f)

The ratio F=(f/D) is called the focal ratio We often speak of the f-ratio of an optical system, f/F The speed of an optical system is proportional to the energy deposited in a unit area (like a pixel), so –2 speed∝Epixel∝F , because an exposure to a given depth (the sensitivity) can be carried out more quickly in a faster system The speed of an optical system is only dependent on the focal ratio F and is independent of the diameter D A 1m telescope at f/6 is as fast as an 8m telescope at f/6, because the 1m telescope has a shorter focal length than the 8m telescope more light is concentrated into a smaller area at the focal plane Of course, sometimes we want to use a larger F — because it results in a higher magnification — and so big telescopes are useful after all! Telescope types and configurations

One-mirror telescopes: A single mirror directs the light to a detector at the prime focus Usually the primary is not the appropriate shape for this configuration Need a prime focus corrector to eliminate aberrations Classical radio telescopes are often this design, but most modern (radio) telescopes don’t have a prime focus corrector Two-mirror telescopes

Cassegrain, with a convex secondary A Gregorian telescope: Gregorian, with a concave secondary The INT

The INT is a 2.54 m Cassegrain telescope typically used as a single- mirror telescope with an imager, the Wide-Field Camera (WFC), at the prime focus The INT has a prime-focus corrector that corrects the aberrations induced by the paraboloid primary The WFC

The WFC is a mosaic of 4 thinned 2048x4096 CCDs with 13.5 µm pixels mounted at the prime focus of the INT The WFC

Together with the inter- chip gaps, the mosaic is ~34’x34’ The WFC

Together with the inter- chip gaps, the mosaic is ~34’x34’ The WFC

Filters: six-position filter wheel with many, many filters available – see http://catserver.ing.iac.es/filter/list.php? instrument=WFC for a full list Quantum efficiency: ~60% at 380nm, ~80% at 400– 650 nm, ~15% at 950nm The INT: questions

The primary mirror of the INT has a diameter of 2.54 m The corrected focal ratio of the prime focus corrector (behind which the WFC sits) is f/3.29 What’s the focal length at prime focus? What’s the plate scale of the INT at prime focus? The WFC has 13.5 µm/pixel; what is the pixel scale of the WFC? Signal-to-noise

Our confidence in detecting an object or feature in astronomy is often expressed as the signal-to-noise ratio If we consider an astrophysical source from which we collect N photons in some amount of time from a detector that has its own noise — caused, say, by its own electronics — we can use the Poisson pdf to determine the total variance of the measurement If we assume that N≈⟨N⟩=µ, then the variance of the sources is also µ≈N

If we call the detector’s own noise the “read noise” RN, its variance is (RN)2

Then the total variance is

! 2 = N + (RN)2 and the “noise” — the error bar or standard deviation — is = N + (RN)2 p More generally, we also have a background of photons — from the night sky, contaminating sources, etc. — so we really should write

! 2 = NS + NB + (RN)

where NS is the numberp of photons from the source and NB is the number of photons from the background

Then our signal-to-noise ratio is

S NS NS ! = = 2 N NS + NB + (RN) This is effectively the significance of the detection of p our source photons in units of the error bar σ Be careful with this formula!

“Counts” from a CCD image are actually in ADU, analogue-to-digital units, where a multiple of the number of photons collected:

! ADU = N /G

where G [e–/ADU] is the “gain” of the system

ALWAYS convert from ADU to e– (photons) before computing your statistics! An example: the WFC on the Isaac Newton Telescope

Gain chip RN ( (

1 4,6 1,26

2 4,8 1,7

3 4,3 1,2

4 4,3 1,67 Seeing and background Seeing

All ground-based observatories suffer from a major problem: light from distant objects must pass through the Earth’s atmosphere recall from our discussion of optics that the atmosphere has refractive power Seeing

The refraction itself isn’t the problem: it’s the variation in the index of refraction that causes problems Wind, convection, and land masses cause turbulence, mixing the layers of different indices of refraction in non-uniform and continuously varying ways Seeing top of atmosphere

turbulent atmosphere

near ground

This non-uniform, continuous variation in the index of refraction cause the initially plane-parallel wavefronts to tilt, bend, and corrugate This causes the twinkling we see when we look at stars with our eyes Any degradation of images by the atmosphere is called seeing Seeing

Modern observatories are built at very dry sites, so the overall effect of humidity is typically unimportant for seeing (but important for differential refraction) For an ideal gas and adiabatic conditions, dn p ! dT / T 2 Therefore the primary source of index of refraction variations in the atmosphere is thermal variations We care (mostly) about small-scale variations in T: thermal turbulence Thermal turbulence

Thermal turbulence in the atmosphere is created on a variety scales by

convection: air heated undisturbed turbulence by conduction with the wind flow warm surface of the Earth becomes buoyant and first rises, displacing cooler air mountain ridge humid (wet) air also rises and displaces dry ocean air Thermal turbulence

wind shear: high winds — in particular the jet stream — generate wind shear undisturbed turbulence and eddies at various wind flow

scales, creating a first turbulent interface mountain between other layers in ridge laminar (non-turbulent) ocean flow Thermal turbulence

disturbances: large undisturbed turbulence land-form variations — wind flow

like mountains — can first create turbulence mountain ridge

ocean Seeing effects

Image blurring (smearing) dominates for D>r0 Each isoplanatic patch gives rise to its own Airy disk (FWHM ~ λ/D): these are called speckles

Seen together, they have FWHM ~ λ/r0 and give rise to a shimmering blur If integrating for long times, they will have this size for the long- exposure PSF In general, the larger the telescope, the more photons it gathers — but the PSF is limited to λ/r0 need adaptive optics (AO) or speckle interferometry Remember that FWHM ~ λ–1/5, so seeing is always better in the red Backgrounds, absorption, scattering, and extinction

In order to get accurate results, we need to correct for the background and for light removed by the atmosphere through absorption and scattering Many background sources come from the Earth (its inhabitants and its atmosphere) But perhaps the most important background source is the sunlight reflected from the moon: moonlight Sky brightness at CTIO

days from new moon U B V R I

0 22 22,7 21,8 20,9 19,9

3 21,5 22,4 21,7 20,8 19,9

7 19,9 21,6 21,4 20,6 19,7

10 18,5 20,7 20,7 20,3 19,5 from NOAO newsletter #37 from

Surface brightness in mag/sq.arcsec, Surface brightness in mag/sq.arcsec, 14 17 19,5 20 19,9 19,2 Airglow is emission from molecules and some atoms (like O, Na, H) in the atmosphere

this is the source of “sky (emission) lines” we see in spectra taken from ground-based sites 1992PASP..104...76O 1992PASP..104...76O 1992PASP..104...76O

Night sky at a “light- polluted” site 1992PASP..104...76O Scattered sunlight during twilight is also a problem at the beginning and end of the night

although very useful for flat fields!

We define three “official” twilights according to the Sun’s position with respect to the horizon:

6 degree twilight: “civil twilight” — when you need to turn your headlights on (at least in the USA!)

12 degree twilight: “nautical twilight” — sailors can still see the horizon but also see enough stars to fix a position with a sextant

18 degree twilight: “astronomical twilight” — fully dark sky Zodiacal light is sunlight scattered off of Solar System dust, very near the plane of the ecliptic

this is the limiting background on moonless nights at a very dark site away from the Galactic plane

Unresolved galaxy and starlight — like in the plane of the Milky Way — can also cause a significant background

In the near- to mid-IR (>2.4 µm), thermal emission from the Earth and the telescope dominate the background

Clouds will reflect light back from the Earth, Sun, and Moon

Finally, light pollution from poorly-regulated human settlements is a serious — and growing — problem for astronomy Absorption and scattering

The atmosphere not only adds background but also removes flux, decreasing signal-to-noise (or even removing signal altogether) Absorption and scattering

Scattering arises from two major regimes of particle-photon scattering Absorption and scattering

Molecular scattering, where the scattering radius a≪λ

Rayleigh scattering is dominant, with a cross- section 3 2 2 8⇡ (n 1) 2 4 ()=! N R 3 N 24 / where N is the particle density and is a function of temperature and pressure, and therefore altitude, etc. Absorption and scattering

Rayleigh scattering is the main cause of blue extinction and is responsible for the blue sky! Absorption and scattering

Aerosol scattering, where the scattering radius a & /10 !

This is Mie scattering, which has (roughly)

2 1 ! a / and so is the main cause of red extinction Absorption and scattering

Aerosols are highly variable from night to night, as they come from air pollution “haze” (dust) volcanic ash dust storms etc. Absorption and scattering

In the summer at La Palma, wind-blown dust from Saharan dust storms are particularly annoying, causing significant increase in optical extinction Annual variation at ORM (La Palma)

Long-term variation at ORM Airmass

In order to determine extinction corrections, top of we need to define the atmosphere airmass of our observations X We quote magnitudes at z

“the top of the one airmass

atmosphere” top of telescope Airmass

We define the airmass top of between the top of the atmosphere telescope and the top of the atmosphere at the zenith as one airmass or X X=1 z one airmass

top of telescope Airmass

Define z as the zenith angle (or zenith distance), where top of z=90º–altitude atmosphere Then in a plane-parallel approximation to the X Earth’s atmosphere, z one airmass X =1/ cos z =secz top of telescope Airmass

The plane-parallel approximation is good for z=60º, X=2 z<60º

X=1 For larger zenith angles, a z=90º, X=38 spherical-shell description is required: X =secz 0.0018167(sec z 1) 0.002875(sec z 1)2 0.0008083(sec z 1)3 Airmass

Note that the cosine rule of spherical trigonometry z=60º, X=2 gives

X=1 cos z =sin! sin + cos cos cos h z=90º, X=38 where ϕ is your At 60º, you look through 2 latitude, δ is the airmasses object’s declination, At 71º, you look through 3 and h is its hour angle airmasses (all in radians or At 90º, you look through 38 degrees) airmasses Atmospheric extinction and photometry

Although this topic actually belongs with our discussion of photometry, we can now discuss the effect of atmospheric extinction on photometry, the measurement of magnitudes Atmospheric extinction and photometry

Because atmospheric absorption and scattering is a radiative transfer problem, we can characterize as an optical depth problem Atmospheric extinction and photometry Then we can write that the flux remaining after passing through the atmosphere is

X/⌧ ! f = f0e

where f0 is the original flux of the object above the atmosphere, X is the airmass, and τ is the optical depth Atmospheric extinction and photometry Then, in magnitudes, m = 2.5 log10 f ! = 2.5 log10 f0 + kX Therefore, in magnitudes, the loss of light per airmass is linear, with a proportionality constant

! k = 2.5(log e)/⌧ 10 called the extinction coefficient Atmospheric extinction and photometry

We will return to the process filter k of determining kλ later... In the meantime, for the sky U 0,55 over La Palma, on average, the extinction per unit B 0,25 airmass is given to the right... V 0,15 Clearly, we want to observe objects in the U R 0,09 band as close to overhead as possible! I 0,06