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

Luminosity and Flux Luminosity (L) - total Power emitted in all directions over all wavelengths (Joules/sec, or Watts) Flux – Luminosity emitted per unit area of the source (f), or detected per unit area by the observer (F), over all wavelengths (W m -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 D 2 F therefore: D 2 F = R ★ 2 f F / f = R ★ 2 / D 2 This is the Inverse Square Law for radiation Remember! Surface Area of a Sphere = 4 π R 2 ; if R doubles, surface area quadruples… 2

The Magnitude System Magnitude scale – Hipparchos (2 nd Century): the brightest stars - magnitude of 1 the faintest stars - magnitude of 6 However, in terms of the amount of energy received, a sixth magnitude star is approx 100 times fainter, due to the eye's non-linear response to light. Norman Pogson formalize the magnitude system in 1856: 6 th magnitude star should be precisely 100 times fainter than a 1 st mag star each magnitude corresponds to a change in brightness of 100 1/5 = The bottom line: Magnitude is proportional to the log10 of Flux. Remember: The GREATER the magnitude, the FAINTER the object! 3

The Magnitude System Relative magnitudes: m1 - m2 = -2.5 log 10 (F 1 /F 2 ) i.e. difference in magnitude between two stars is given by the ratio of fluxes. If m 2 = 0, then.. Apparent magnitude, m: m = -2.5 log 10 (F / F 0 ) F 0 - flux from zeroth magnitude star, Vega. This equation can also be re-witten: m = -2.5 log 10 F + Z where Z is the zero-point (described later). Q. If star A is 100x brighter than star B, what’s the magnitude difference? Q. How much brighter is Rigel than Bellatrix? Bellatrix m = 1.6 Rigel m = 0.1 Saiph m = 2.1 Betelgeuse m = 0.4 4

The Magnitude System Apparent magnitude, m (from previous slide): m = -2.5 log 10 (F / F 0 ) nb. F 0 - flux of “zeroth mag” star Vega m1 - m2 = -2.5 log 10 (F 1 /F 2 ) Absolute magnitude, M - apparent magnitude a star at a distance of 10 parsecs (pc) Remember, F α 1/d 2 so: m - M = -2.5 log 10 (10 2 / d 2 ) = -2.5 log 10 (d -2 ) log 10 (10 2 ) - where d is in parsecs Distance Modulus: m - M = 5 log d - 5 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… 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 = mag, d = pc Q2. How much brighter would Sirius be if both stars were at a distance of 10 pc?

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

Filter sets U 3600 Å B 4300 Å V 5500 Å R 6500 Å I 8200 Å Z 9000 Å J 1.25  m H 1.65  m K 2.20  m L 3.7  m M 4.7  m N 10.5  m Q 20.9  m NB. 1 Angstrom (Å) = m; 9000 Å = m = 0.9  m Wavelengths listed above correspond to the centre of the filter’s transmission. Filter bandwidths typically 20% (i.e  ) in the optical; 10% in the IR. Infrared bands correspond to atmospheric “windows”. Left –standard optical filter profiles Below – IR filters plotted against atmospheric transmission. 7

Johnson-Morgan-Cousins vs Sloan The most widely used Photometric System: Johnson-Morgan-Cousin UBVRI system Modified by Bessel in the 1990s to better match the performance of CCDs. The magnitude of an object through a given filter, is referred to as m B, m V, m R … or simply by B, V, R…. For example, Bellatrix has apparent magnitudes: U = 0.54 mag B = 1.42 mag V = 1.64 mag K = 2.38 mag Is Bellatrix a Red or Blue star? 8

Johnson-Morgan-Cousins vs Sloan However, in 1996… The Sloan Digital Sky Survey (SDSS) introduced a new set of optical filters: u’ g’ r’ i’ z’ Have: broader bandwidths than J-M-C higher transmission bandwidths don’t overlap in wavelength. Ideal for measuring the red-shifts of galaxies – see right. The SDSS map of Galaxies out to redshift z=0.15 between -1.5 o <  < 1.5 o. Each dot is a galaxy containing perhaps 100 billion stars… 9

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 Vega has U = 0 mag, B = 0 mag, V = 0 mag, J = 0 mag, etc.. 10 Debris disk around Vega (HST image)

The Magnitude System Apparent magnitude: m = -2.5 log 10 (F / F 0 ) where F 0 is the flux of Vega. If you know the Flux of Vega, F 0, 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, m U, m V, m B of Rigel Q2. Calculate the Absolute Magnitudes, M U, M B, M V (assume a distance, d = 250 pc) Rigel (the bright blue star in Orion): F U,Rigel = W m -2, F U,Vega = W m -2 F B,Rigel = W m -2, F B,Vega = W m -2 F V,Rigel = W m -2, F V,Vega = W m In ancient Egypt, Rigel’s name was… Seba-en-Sah, which means Foot Star or maybe Toe Star!

Flux vs Flux Density Flux – Specific to the waveband, but also the photometric system, because the filters have a different central wavelength and band-pass (width) Flux Density – Flux per unit wavelength, F  in W m -2 nm -1 or W m -2 Angstrom -1 ) Flux per unit frequency, F (in W m -2 Hz -1 ). Approximate Flux Density by dividing Flux by the “width” of the filter (nm, Angstrom, Hz). 12

Flux vs Flux Density Flux – Specific to the waveband, but also the photometric system, because the filters have a different central wavelength and band-pass (width) Flux Density – Flux per unit wavelength, F  in W m -2 nm -1 or W m -2 Angstrom -1 ) Flux per unit frequency, F (in W m -2 Hz -1 ). Approximate Flux Density by dividing Flux by the “width” of the filter (nm, Angstrom, Hz). E.g. Rigel: - F B = W m -2 B-band filter:  = 72 nm,  = Hz Therefore: - F  = W m -2 nm -1 - F  = W m -2 Hz Rigel Flux density in the centre of the B-band

Spectral Energy Distribution Flux Density vs Wavelength gives Spectral Energy Distribution (spectrum). The curve is called a Blackbody spectrum and is defined by the Planck function. Spectrum peaks at a different wavelength depending on the temperature of the star. 14 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 F

An aside: Rigel vs. the Lightbulb (!?) How does flux from a BRIGHT STAR compare to flux from a 60 W LIGHT BULB? E.g. Rigel’s B-band flux - F B,Rigel = 4.6x10 -9 W m W - power consumed by the bulb! Incandescent light bulb ~ 10% efficient total flux radiated (across all wavelengths) ≈ 6W 2. Bulb burns at 3,000 K (i.e. like Betelgeuse), most of energy is radiated in the IR! ~ 10% radiated in the optical (between 300 nm and 800 nm) ~ 20% of this optical wavelength range covered by B-band filter. The Luminosity of our bulb in the B-band is therefore: L B,bulb = 6 x 0.1 x 0.2 = 0.12 W 3.Finally, if we assume our bulb is 1 km away:. L = 4 π D 2 F … so … F B,bulb = 0.12 / (4 π ) F B,bulb = 9.5x10 -9 W m B filter

Colour and Colour Index Colour - defined in terms of the ratio of fluxes in different wavebands. - corresponds to a difference in magnitudes in two different bands e.g. m B - m V = (B - V ), where (B - V ) is referred to as a the `colour index'. [ Remember: m1 - m2 = -2.5 log 10 (F 1 /F 2 ) ] 16 Any colour index can be constructed, e.g. Betelgeuse: B – V = 2.7 – 0.42 = (optical) J – K = -3.0 – (-4.4) = +1.4 (near-IR) Rigel: B – V = 0.09 – 0.12 = J – K = – = Note: colours can be negative, or even zero, like Vega! Betelgeuse Rigel

Colour and Colour Index (B - V ) - most frequently used optical colour index - a measure of the effective temperature, T eff, of a star. (T eff is the temperature of a blackbody that would emit the same amount of radiation – typically the temperature near the surface of a star) For example: 17 For the Sun: m B = , m V = hence (B - V) sun = T eff ≈ 5,700 K For Vega: m B = m V = 0.0 (by definition) hence (B - V) vega = T eff ≈ 9,900 K F For Betelgeuse: m B = 2.70, m V = 0.42 hence (B – V) bet = T eff ≈ 3,600 K B - V = -2.5 log 10 (F B / F V ) Sun Betelgeuse  m  m

Empirical Formula for Colour Index The observed relationship between (B - V) and T eff for Main Sequence stars is given by: 18 So Colour gives you the Temperature: T eff corresponds to a single value of (B - V) Can’t use B or V (or any other magnitude) alone to measure T eff.

Bolometric Luminosity (and Bolometric Corrections) T eff is related to the total (or bollometric) luminosity, L = 4  R * 2  T eff 4 -  is the Stephan-Boltzman constant 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 ). Can often determine total (or bolometric) magnitude (apparent or absolute) via a simple bolometric correction: m bol = m V - BC and hence M bol = M V - BC Bolometric correction, BC, is a function of (B - V) or T eff. 19 BC is ~zero for a star with a T eff = 5700 K, i.e. like our sun.

Bolometric Luminosity (and Bolometric Corrections) Most of the sun’s energy is radiated in the optical. 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 The Sun has M V = 4.82 and BC= 0.07, Hence: M bol = 4.82 – 0.07 = 4.75 (absolute bolometric magnitude).

Bolometric Luminosity (and Bolometric Corrections) What about using other colour indices, e.g. (U - B) to measure temperature? The relationship between (U - B) and T eff is not monotonic; the spectral energy distribution deviates from a simple black body in the U- band. The same value of U-B gives multiple values of B-V ! 21

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Affects of interstellar dust Light absorbed and scattered by interstellar dust: absorbed light - re-emitted in the far-IR scattered light - absorbed and re-emitted Both cause extinction - objects appear fainter and “redder” For small dust grains: scattering cross-section -  scat -4 absorption cross-section -  abs -1 Overall extinction cross-section is:  ext Red light (longer wavelengths) is less extinguished than Blue light - hence the term reddening. 23 Tiny interstellar dust particles (image: A Davis, U Chicago)

A blue sky (and red sunset) These processes also occur in the Earth’s atmosphere. Molecules in the air much smaller than the of light, so Scattering is more efficient at the shorter wavelengths Blue sunlight is scattered many times; reaches our eyes with nearly equal intensity from every direction. We therefore see the sky as blue. 24 A sunset is red because the “reddening” is most extreme at this time of day. Images from

Extinction and Reddening A V = absorption in the V band, in magnitudes – the visual extinction Absorption in other bands is different because of dependency on wavelength, e.g. A U = 1.53 A V - absorption in UV is more than in V A B = 1.32 A V - absorption in BLUE is (a bit) more than in V A K = 0.11 A V - absorption in IR is much less than in V Extinction in the IR ~10-times less than at V… 25 B68 - opticalB68 - IR

Colour “excess” Extinction changes the colour of a star! Colour Excess, E(B - V) - also referred to as reddening - is the additional (B - V) colour caused by this wavelength-dependent extinction, so: E(B - V ) = A B - A V = 1.32A V - A V = ( ) A V E(B - V ) = 0.32 A V A V = 3.1 E(B - V ) The greater the extinction, A v, the greater the affect on the B-V colour! 26 Nice blue star… m B = 7.1 mag m V = 7.6 mag (B-V) colour = 7.1 – 7.6 = -0.5 mag A negative number; star is brighter in the blue (B) band! Nice blue star with a cloud in front! A V of cloud = 3.0 mag E(B-V) = 0.32.A V = 0.96 mag (B-V) is now = = 0.46 mag A positive number; star is brighter in V and looks a bit reddish…

Extinction and Reddening From effective temperature, T eff, can calculate the intrinsic colour, (B - V) 0, compare this to the observed colour, and thus calculate the colour excess. E(B - V ) = (B - V) observed - (B - V) 0 From E(B - V ) we can calculate extinction, A V and correct the V-band magnitude. For example: G2V star has T eff = 5520 K; m B = 15.3 (observed); m V = 14.1 (observed). Q. What is the extinction-corrected apparent V-band magnitude of this star? Remember… A V = 3.1. E(B - V ) m V (extinction corrected) = m V (observed) - A V 27 Remember from a few slides back?

Extinction and Reddening From effective temperature, T eff, can calculate the intrinsic colour, (B - V) 0, compare this to the observed colour, and thus calculate the colour excess. E(B - V ) = (B - V) observed - (B - V) 0 From E(B - V ) we can calculate extinction, A V and correct the V-band magnitude. For example: G2V star has T eff = 5520 K; m B = 15.3 (observed); m V = 14.1 (observed). Its intrisic (B-V) colour (based on T eff )  (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, A V = 3.1. E(B - V )  A V = 1.61 mag Extinction-corrected V-band mag of the star  m V = 12.5 mag 28 Remember from a few slides back?

Extinction and Reddening From effective temperature, T eff, can calculate the intrinsic colour, (B - V) 0, compare this to the observed colour, and thus calculate the colour excess. E(B - V ) = (B - V) observed - (B - V) 0 From E(B - V ) we can calculate extinction, A V and correct the V-band magnitude. For example: G2V star has T eff = 5520 K; m B = 15.3 (observed); m V = 14.1 (observed). Its intrisic (B-V) colour (based on T eff )  (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, A V = 3.1. E(B - V )  A V = 1.61 mag Extinction-corrected V-band mag of the star  m V = 12.5 mag Finally, when using the Distance Modulus equation it is important to account for extinction: (m V,observed - A V ) - M V = 5 log d i.e. must correct apparent magnitude, m, for extinction before calculate absolute magnitude, M Remember from a few slides back?

Atmospheric Absorption (and Airmass corrections) As we saw earlier, extinction occurs in the earth’s atmosphere too! The altitude of your target affects how much light gets through to the telescope Distance traveled through the atmosphere, d: d ≈ h/cos z = h sec z ‘sec z’ is known as the airmass. - at an airmass of 1: z=0 o - at an airmass of 2: z= 60 o Attenuation is proportional to sec z, and: m z = m 0 - C sec z C is a constant which depends on (but changes from site to site, and time to time). …To be discussed in more detail next time! 30 m0m0 mzmz

End.. See you next week…. 31