PHYS216 Practical Astrophysics Lecture 4 – Photometry 1

1. PHYS216 Practical AstrophysicsLecture 4 – Photometry 1 Module Leader: Dr Matt Darnley Course Lecturer: Dr Chris Davis

2. Luminosity and Flux Luminosity (L) - the total Power from a source of radiation (in units of energy/second=Watts) emitted in all directions over all wavelengths. Flux - the luminosity, or Energy per second, emitted per unit area of the source (f), or detected per unit area by the observer (F), over all wavelengths (Wm-2) Total luminosity, L, given by: L = 4 π R★2 f where f = surface flux and R★ = stellar radius. At a distance D from the source, if the measured Flux (power received per unit detector area) = F then: L = 4 π D2 F therefore: D2 F = R★2 f or F / f = R★2 / D2 This is the Inverse Square Law for radiation Remember! Surface Area of a Sphere = 4 π R2 ; if R doubles, surface area quadruples…

3. The Magnitude System The Greek astronomer Hipparchos is usually credited with the origin of the magnitude scale. He assigned the brightest stars he could see with his eye a magnitude of 1 and the faintest a magnitude of 6. However, in terms of the amount of energy received, a sixth magnitude star is not 6 times fainter than a first magnitude star; its is approx 100 times fainter, due to the eye's non-linear response to light. This led the English astronomer Norman Pogsonto formalize the magnitude system in 1856. He proposed that a sixth magnitude star should be precisely 100 times fainter than a first magnitude star, so that each magnitude corresponds to a change in brightness of 1001/5 = 2.512 • The bottom line: • Magnitude is proportional to the log10 of Flux. • Remember: • The GREATER the magnitude, the FAINTER the object!

4. The Magnitude System Relative magnitudes are given by: m1 - m2 = -2.5 log10(F1/F2) i.e. difference in magnitude between two stars is given by the ratio of fluxes. If one star has a magnitude of zero, then the above equation defines the Apparent magnitude, m, e.g.: m = -2.5 log10(F / F0) Here F0 is the flux from the zeroth magnitude star, Vega, the “primary” standard. This equation can also be re-witten: m = -2.5 log10F + Z where Z is the zero-point (described later). Betelgeuse m = 0.4 Bellatrix m = 1.6 Rigel m = 0.1 Saiph m = 2.1 Q. If star A is 100x brighter than star B, what’s the magnitude difference? Q. How much brighter is Rigel than Bellatrix? In other words, if we can measure the Flux of Vega, and the Flux of another star, we can calculate the apparent mag of that star

5. The Magnitude System Apparent magnitude, m, is given by (see previous slide): m = -2.5 log10(F / F0) - where F0 is the flux of the “zerothmag” star, Vega m1 - m2 = -2.5 log10(F1/F2) Absolute magnitude, M,is the apparent magnitude a star would have if it was at a distance of 10 parsecs (pc). It is a measure of the intrinsic brightness of a star Remember, F α 1/d2 so: m - M = -2.5 log10(102 / d2) = -2.5 log10(d-2) - 2.5 log10(102) - where d is in parsecs Therefore the Distance Modulus:m - M = 5 log d - 5 • Q1. What is the Absolute magnitude, M, of • Sirius, the brightest star in the sky: • m = -1.5 mag, d=2.6 pc? • The Sun, which is actually the brightest star in the sky: • m = -26.7 mag, d = 5.10-6 pc • Q2. How much brighter would Sirius be if both stars were at a distance of 10 pc? • For example: • if the distance modulus, m - M = 0, d = 10 pc • if the distance modulus, m - M = 5, d = 100 pc • if the distance modulus, m - M = 10, d = 1000 pc • etc…

6. Optical imaging through Filters Stars have different magnitudes at different wavelengths, i.e. when viewed through different filters/in different wavelength bands, or “wavebands” Top-left: RATCam on the Liverpool Telescope Above: RATCam’s filter wheel Bottom-left: the CCD detector mounted in RATCam

7. Filter sets Left – the standard optical filter profiles (Johnson system). Below – IR filters plotted against atmospheric transmission in the near-IR. Two narrow-band filters are also plotted (in green). U 3600 Å B 4300 Å V 5500 Å R 6500 Å I 8200 Å Z 9000 Å J 1.25 mm H 1.65 mm K 2.20 mm L 3.7 mm M 4.7 mm N 10.5 mm Q 20.9 mm NB. 1 Angstrom (Å) = 10-10 m; 9000 Å = 0.9 10-6 m = 0.9 mm The wavelengths listed above correspond to the centre of the filter’s transmission. Filter bandwidths are typically 20% (i.edl ~ 0.2 l) in the optical; 10% in the IR. The near- and mid-IR bands roughly correspond to atmospheric “windows”. H was defined after J and K – its not true that IR astronomers struggle with their alphabet!

8. Johnson-Morgan-Cousins vs Sloan The most widely used Photometric System(i.e set of filter definitions) is the Johnson-Morgan-Cousin UBVRI system, which is an extension of the Ultraviolet-Blue-Visible system defined by Johnson and Morgan in the 1950s to include redder filters, Red and Infrared, by Cousins a decade later. The central wavelengths and bandwidths have been modified slightly over the years – most notably by Bessel in the 1990s to better match the performance of CCDs. Bessel optical filters are used today in many telescopes – including the LT! The magnitude of an object, when seen through a given filter, is referred to as mB, mV, mR, etc., or simply by B, V, R. For example,Bellatrixhas apparent magnitudes U = 0.54mag B = 1.42mag V = 1.64 mag K = 2.38mag Bellatrix is a red star; its brighter at longer wavelengths

9. Johnson-Morgan-Cousins vs Sloan However, in 1996… The Sloan Digital Sky Survey (SDSS) introduced a new set of optical filters, which are referred to as u’ g’ r’ i’ z’. These filters have broader bandwidths than the J-M-C set, they have higher transmission (they let through more light – so fainter objects like distant galaxies can be seen), and their bandwidths don’t overlap in wavelength. They are ideal for measuring the red-shifts of galaxies, for example – see right. Sloan optical filters are now used in many other telescopes – including the LT! The SDSS map of Galaxies out to redshiftz=0.15 between -1.5o < d < 1.5o. Each dot is a galaxy containing perhaps 100 billion stars…

10. The Magnitude System The Apparent magnitude, m - specific to the waveband through which it is observed. For example: Betelgeuse has U = 4.3 mag, B = 2.7, V = 0.42 mag, J = -3.0 mag, K = -4.4 mag Betelgeuse is a very red star! The higher the magnitude, the fainter the star. Vega has U = 0 mag, B = 0 mag, V = 0 mag, J = 0 mag, etc.. Debris disk around Vega (HST image)

11. The Magnitude System • Apparent magnitude: • m = -2.5 log10(F / F0) where F0 is the flux of Vega. • If you know the Flux of Vega, F0 , in each filter • If you measure the Flux of a star on your CCD, through the same filters • ….. You can work out the apparent magnitude of that star. • Q1. Calculate the Apparent U,B and V mags, mU, mV, mB of Rigel • Q2. Calculate its Absolute Magnitude, MU, MB, MV(assume a distance, d = 250 pc) • Rigel(the bright blue star in Orion): • FU,Rigel = 3.47.10-9 W m-2 , FU,Vega = 2.09.10-9 W m-2 • FB,Rigel = 4.58.10-9 W m-2 , FB,Vega = 4.98.10-9 W m-2 • FV,Rigel= 4.30.10-9 W m-2 , FV,Vega = 4.80.10-9 W m-2 In ancient Egypt, Rigel’s name was… Seba-en-Sah,which means Foot Star or maybe Toe Star!

12. Flux vs. Flux Density • The Flux is specific to the waveband, but also the photometric system, used. For example, the Johnson-Morgan-Cousin’s U-band flux of a star is slightly different from the Sloan u’-band flux. This is because the filters have a different central wavelength and band-pass (width), so they let a different amount of light through. • Therefore, its often better to quote the Flux Density, i.e. the Flux per unit wavelength, Fl(in W m-2 nm-1 or W m-2 Angstrom-1) or Flux per unit frequency, Fn (in W m-2 Hz-1). • You can approximate the Flux Density • by simply dividing the flux by the • “width” of the filter, in nm, Angstrom, • or even Hz. • E.g.Rigel(the bright blue star in Orion): • - FB = 4.58.10-9 W m-2 • B-band filter: Dl = 72 nm, Dn = 1.17.1014 Hz • Therefore: • - Fl,B = 6.36.10-11 W m-2 nm-1 • - Fl,B = 3.91.10-23 W m-2 Hz-1 Rigel Flux density in the centre of the B-band www.astropixels.com

13. Spectral Energy Distribution • Plot Flux Density against Wavelength to get a Spectral Energy Distribution (or simply ‘spectrum’) for a star. The curve is called a Blackbody spectrum and is defined by the Planck function; the spectrum peaks at a different wavelength depending on the temperature of the star. Cool stars are brighter in the IR (l > 1 mm) than in the optical (l < 1 mm). Fl The sun has a surface temperature of 5,800 K; it’s a yellow star. Rigel (blue) : 11,000 K Betelgeuse (red) : 3,500 K

14. An aside: Rigel vs. the Lightbulb (!?) You may have noticed (from the last few slides) that even bright stars seem to produce very little Flux. For example, Rigel’s B-band flux is FB,Rigel= 4.6x10-9 W m-2. How does this compare to a 60W light-bulb? 1. First, 60W is the power consumed by the bulb! An incandescent light bulb is perhaps 10% efficient, so the total flux radiated (across all wavelengths) may be only 6W! 2. This energy is radiated over a broad spectral (wavelength) range. In fact, burning at 3,000 K (i.e. like Betelgeuse), most of an incandescent bulb’s energy is radiated in the IR! Only about 10% is radiated in the optical (between 300 nm and 800 nm), and only ~20% of this optical wavelength range is covered by our B-band filter. The Luminosity of our bulb in the B-band is therefore: LB,bulb= 6 x 0.1 x 0.2 = 0.12 W Finally, if we assume our bulb is 1 km away (so that our bulb looks like a star). L = 4 π D2 F … so … FB,bulb= 0.12 / (4 π 10002 ) = 9.5x10-9 W m-2 Our bulb is now almost as faint as Rigel, when viewed through a B-band filter at a distance of ~1 km B filter

15. Colour and Colour Index • The colour of an object is defined in terms of the ratio of fluxes in different wavebands. This corresponds to a difference in magnitudes in two different bands, • e.g. mB - mV = (B - V ), where (B - V ) is referred to as a the `colour index'. • [ Remember: m1 - m2 = -2.5 log10(F1/F2) ] • Any colour index can be constructed; (U - R), (V - I), (J - K) etc. For example: • Betelgeuse: • B – V = 2.7 – 0.42 = +2.28 (optical) • J – K = -3.0 – (-4.4) = +1.4 (near-IR) • Rigel: • B – V = 0.09 – 0.12 = -0.03 • J – K = 0.206 – 0.213 = -0.07 • Note: colours can be negative, or even zero, like Vega! Betelgeuse Rigel

16. Colour and Colour Index (B - V ) is the most frequently used optical colour index, and is a measure of the effective temperature, Teff, of a star. (The effective temperature is the temperature of a blackbody that would emit the same amount of radiation – typically this is similar to the temperature near the surface of a star; the temperature at the core is usually much higher!) For example: Fl • For Vega: mB = mV = 0.0 (by definition) • hence • (B - V)vega= 0.0 - Teff ≈ 9,900 K B - V = -2.5 log10(FB / FV) B@0.43mm V@0.55mm Sun • For the Sun: mB = -26.14, mV = -26.78 • hence • (B - V)sun= 0.64 - Teff ≈ 5,700 K • For Betelgeuse: mB = 2.70, mV = 0.42 • hence • (B – V)bet= 2.28 - Teff ≈ 3,600 K Betelgeuse

17. Empirical Formula for Colour Index The observed relationship between (B - V) and Teff for Main Sequence stars (stars in the main/stable part of their evolution, like the sun) is given by: Note: For Main Sequence stars (like the Sun) each Teff corresponds to a single value of (B - V) . We can’t use B or V (or any other magnitude) alone to measure Teff. However, the stars colourdoes give you its temperature.

18. Bolometric Luminosity (and Bolometric Corrections) Teff is related to the total (or bollometric) luminosity, L = 4pR*2s Teff4 - s is the Stephan-Boltzman constant Where L is the intrinsic or absolute (not apparent!) brightness of the star: it represents the total outflow of radiation per second from the star (at all l). We can determine the total (or bolometric) magnitude (apparent or absolute) via a bolometric correction. We define: mbol = mV - BC and hence Mbol = MV - BC Where the bolometric correction, BC, is a function of (B - V) or, equivalently, Teff. (We assume there is no extinction, see next section.) Note that very hot and very cool objects have large BCs, because their spectral energy distribution peaks a long way from the V band (at 5500 Angstrom).… BC is ~zero for a star with a Teff = 5700 K, i.e. like our sun.

19. Bolometric Luminosity (and Bolometric Corrections) The Sun has MV = 4.82 and BC= 0.07, Hence: Mbol = 4.82 – 0.07 = 4.75 (absolute bolometric magnitude). Most of the sun’s energy is radiated in the optical. Our eyes have evolved to be sensitive over the same wavelength range. The sun doesn’t radiate much light in the ultraviolet or infrared, so your eyes aren’t sensitive in these parts of the electromagnetic spectrum.

20. Bolometric Luminosity (and Bolometric Corrections) What about using other colour indices, e.g. (U - B)? The relationship between (U - B) and Teff is not monotonic, which suggests that the spectral energy distribution (or spectrum, for short) deviates from a simple black body in the U-band or between U and B. (This is the Balmer discontinuity at 364 nm, which is due to the ionisation of hydrogen out of the n=2 level: this affects A0-F0 stars in particular). A single value of U-B can be equivalent to multiple values of B-V !

22. Affects of interstellar dust Light from distant stars is both absorbed and scattered by interstellar dust (absorbed light is re-emitted in the far-IR; scattered light is actually absorbed and re-emitted in a different direction!). Both affects cause extinction, and mean that objects appear fainter than they should. For small dust grains: scattering cross-section, sscat prop to l-4 absorption cross-section, sabs prop to l-1 The overall effect is that the observed extinction cross-section is: sext prop to l-a , where a in the range 1-2. Red light (longer wavelengths) is less extinguished than Blue light, so objects appear redder than they should when viewed through dust - hence the term reddening (see e.g. the example to the right). Tiny interstellar dust particles (image: A Davis, U Chicago)

23. A blue sky (and red sunset) Scattering of light (Rayleigh scattering) occurs in the Earth’s atmosphere too. Many of the molecules and dust particles in the air are much smaller than the wavelength of light, so scattering is more efficient atthe shorter wavelengths (see bottom-right). Light from the Sun must passes through these particles before it can reach our eyes. Blue light is scattered many times until it reaches our eyes with nearly equal intensity from every direction. We therefore see the sky as blue. A sunset is red because the “reddening” is most extreme at this time of day. Images from www.teachastronomy.com

24. Extinction and Reddening • AV = absorption in the V (visible) band, in magnitudes – the visual extinction • Absorption in other bands is different because of the dependency on wavelength, e.g. • AU = 1.53 AV - absorption in the UV is more than in the visible • AB = 1.32 AV - absorption in the BLUE is (a bit) more than in the visible • AK = 0.11 AV - absorption in the IR is much less than in the visible! • The variation in extinction with wavelength means that, in the IR, we can often see through clouds of dust that are opaque in the optical! B68 - optical B68 - IR

25. Colour “excess” • Extinction changes the colour of a star! The Colour Excess, E(B - V), also referred to as the reddening, is the additional (B - V) colour caused by this wavelength-dependent extinction, so: • E(B - V) = AB - AV = 1.32AV - AV = (1.32 - 1) AV • E(B - V) = 0.32 AV • AV = 3.1 E(B - V) • In other words, the greater the extinction, Av (the more dust between you and the star), the greater the affect on the B-V colour! Nice blue star… Nice blue star with a cloud in front! mB = 7.1 mag mV = 7.6 mag (B-V) colour = 7.1 – 7.6 = -0.5 mag a negative number; star is brighter in the blue (B) band! AV of cloud = 3.0 mag E(B-V) = 0.32.AV = 0.96 mag (B-V) is now = -0.5 + 0.96 = 0.46 mag A positive number; star is brighter in V and is looking a bit reddish…

26. Extinction and Reddening • If we know the effective temperature, Teff , in advance (e.g. from the star’s spectrum), we can calculate the intrinsiccolour, (B - V)0, compare this to the observed colour, and thus calculate the colour excess. • Then: E(B - V ) = (B - V)observed - (B - V)0 • From E(B - V ) we can calculate the extinction, AV , • and make the appropriate correction to the V-band • magnitude. • For example: • G2V star has Teff = 5520 K; mB = 15.3; mV = 14.1 (observed). • Its intrisic (B-V) colour (based on Teff) (B – V)0 = 0.68 mag • Its observed colour is (B - V)observed = 1.2 mag • Colour excess, E(B - V ) = 1.2 – 0.68 E(B – V) = 0.52 mag • Visual extinction, AV = 3.1. E(B - V ) AV = 1.61 mag • Extinction-corrected V-band mag of the star mV = 12.5 mag • Finally, when using the Distance Modulus equation it is important to account for extinction, because it affects the apparent magnitude of the star. The modified equation becomes: • (mV,observed - AV ) - MV = 5 log d - 5 Remember from a few slides back? i.e. you have to correct the apparent magnitude, m, for extinction before you calculate the absolute magnitude, M

27. Atmospheric Absorption (and Airmass corrections) • At night the sky ain’t blue – however, it still absorbs and scatters star-light! • The altitude of your target therefore affects how much light gets through to the telescope at a given wavelength… • The distance traveled through the • atmosphere is given by: • d ≈ h/cosz = h sec z • ‘sec z’ is known as the airmass. • - at an airmass of 1: z=0o • - at an airmass of 2: z= 60o • Its preferable not to observe at airmass > 2. • Attenuation is proportional to sec z, and: mz = m0 - C sec z • C is a constant which depends on l(but changes from site to site, and time to time). • Before magnitudes can be properly used in data analysis etc, they must all be `corrected' to the same airmass - generally, an airmass of 1.0, which corresponds to an observation made directly overhead. m0 mz