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PHYS216 Practical Astrophysics Lecture 5 – Photometry 1

PHYS216 Practical Astrophysics Lecture 5 – 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)

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PHYS216 Practical Astrophysics Lecture 5 – Photometry 1

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  1. PHYS216 Practical AstrophysicsLecture 5 – Photometry 1 Module Leader: Dr Matt Darnley Course Lecturer: Dr Chris Davis

  2. 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 D2 F therefore: D2 F = R★2 f 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 • Magnitude scale – Hipparchos: • 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. He proposed: • 6thmagnitude star should be precisely 100 times fainter than a 1stmagstar • 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: m1 - m2 = -2.5 log10(F1/F2) i.e. difference in magnitude between two stars is given by the ratio of fluxes. If m2 = 0, then.. Apparent magnitude, m: m = -2.5 log10(F / F0) F0 - flux from zeroth magnitude star, Vega. This equation can also be re-witten: m = -2.5 log10F +Z, Where Zis thezero-point (described later). Betelgeuse m = 0.4 Bellatrix m = 1.6 Rigel m = 0.1 Saiph m = 2.1 Q. If a star if 100x brighter, what’s the magnitude difference? Q. How much brighter is Rigel than Bellatrix?

  5. The Magnitude System Apparent magnitude, m (see previous slide): m = -2.5 log10(F / F0) nb. F0 - flux of “zerothmag” star Vega m1 - m2 = -2.5 log10(F1/F2) Absolute magnitude, M - apparent magnitude a star at a distance of 10 parsecs (pc) 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 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 (in terms of flux ratio) is Sirius than the Sun? • 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 “wavebands” Top-left: RATCam on the Liverpool Telescope Above: RATCam’s filter wheel Bottom-left: CCD detector in RATCam

  7. Filter sets Left –standard optical filter profiles Below – IR filters plotted against atmospheric transmission. 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 • Wavelengths listed above correspond to the centre of the filter’s transmission. • Filter bandwidths typically 20% (i.edl ~ 0.2 l) in the optical; 10% in the IR. • Infrared bands correspond to atmospheric “windows”.

  8. 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 mB, mV, mR … or simply by B, V, R…. For example,Bellatrixhas apparent magnitudes: U = 0.54mag B = 1.42mag V = 1.64 mag K = 2.38mag Q. Is Bellatrix a Red or Blue star?

  9. 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, for example – see right. 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 the Absolute Magnitudes, 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 of maybe Toe Star!

  12. Spectral Energy Distribution • Flux - specific to the waveband, but also the photometric system. This is because the filters have a different central wavelength and band-pass (width), so they let a different amount of light through. • Flux Density – • Flux per unit wavelength, Fl(in W m-2 nm-1 or W m-2 Angstrom-1) • Flux per unit frequency, Fn (in W m-2 Hz-1). • Approximate Flux Density by dividing Flux by the • “width” of the filter (nm, Angstrom, Hz).

  13. Spectral Energy Distribution • Flux - specific to the waveband, but also the photometric system. This is because the filters have a different central wavelength and band-pass (width), so they let a different amount of light through. • Flux Density – • Flux per unit wavelength, Fl(in W m-2 nm-1 or W m-2 Angstrom-1) • Flux per unit frequency, Fn (in W m-2 Hz-1). • Approximate Flux Density by dividing Flux by the • “width” of the filter (nm, Angstrom, Hz). • E.g.Rigel: • - 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

  14. Spectral Energy Distribution • Flux Density vsWavelength 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. 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

  15. 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 - FB,Rigel= 4.6x10-9 W m-2. • 60W - 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: • LB,bulb= 6 x 0.1 x 0.2 = 0.12 W • Finally, if we assume our bulb is 1 km away:. • L = 4 π D2 F • … so … • FB,bulb= 0.12 / (4 π 10002 ) • = 9.5x10-9 W m-2 B filter

  16. 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. mB - mV = (B - V ), where (B - V ) is referred to as a the `colour index'. • [ Remember: m1 - m2 = -2.5 log10(F1/F2) ] Betelgeuse • Any colour index can be constructed, e.g. • 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! Rigel

  17. Colour and Colour Index (B - V ) - most frequently used optical colour index - a measure of the effective temperature, Teff, of a star. (Teff 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: 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

  18. Colour and Colour Index The observed relationship between (B - V) and Teff for Main Sequence stars is given by: • Note: • Teff corresponds to a single value of (B - V) • can’t use B or V (or any other magnitude) alone to measure Teff.

  19. Bolometric Luminosity (and Bolometric Corrections) Teff is related to the total (or bollometric) luminosity, L = 4pR*2s Teff4 - s 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 l). Can often determine total (or bolometric) magnitude (apparent or absolute) via a simple bolometric correction: mbol = mV - BC and hence Mbol = MV - BC Bolometric correction, BC, is a function of (B - V) or Teff. BC is ~zero for a star with a Teff = 5700 K, i.e. like our sun.

  20. Bolometric Luminosity (and Bolometric Corrections) • The Sun: • MV = 4.82 and BC= 0.07 • Hence: • Mbol = 4.82 - 0:07 = 4.75 • (absolute bolometric magnitude). • What about using other colour • indices, e.g. (U - B)? • The relationship between (U - B) and Teff is not monotonic; the spectral energy distribution deviates from a simple black body in the U-band.

  21. Extinction and Reddening • 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 • For small dust grains: • scattering cross-section - sscatl-4 • absorption cross-section - sabsl-1 • Overall extinction cross-section is: • sextl-1 l-2 • Red light (longer wavelengths) is less extinguished than Blue light - hence the term reddening. Tiny interstellar dust particles (image: A Davis, U Chicago)

  22. Extinction and Reddening • Rayleigh scattering occurs in the Earth’s atmosphere too. • Molecules in the air much smaller than the lof light, so • Scattering is more efficient atthe 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. A sunset is red because the “reddening” is most extreme at this time of day. Images from www.teachastronomy.com

  23. Extinction and Reddening • AV = absorption in the V band, in magnitudes – the visual extinction • Absorption in other bands is different because of dependency on wavelength, e.g. • AU = 1.53 AV - absorption in UV is more than in V • AB = 1.32 AV - absorption in BLUE is (a bit) more than in V • AK = 0.11 AV - absorption in IR is much less than in V • Extinction in the IR ~10-times less than at V… B68 - optical B68 - IR

  24. Extinction and Reddening • 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 ) = AB - AV = 1.32AV - AV = (1.32 - 1) AV • E(B - V ) = 0.32 AV • AV = 3.1 E(B - V ) • The greater the extinction, Av , the greater the affect on the B-V colour! Nice blue star… Nice blue star with a cloud in front! 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 looks a bit reddish… 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!

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

  26. Extinction and Reddening • From effective temperature, Teff , can calculate the intrinsic colour, (B - V)0 and thus the colour excess. • E(B - V ) = (B - V)observed - (B - V)0 • From E(B - V ) we can calculate extinction, AV • and correct 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 Remember from a few slides back?

  27. Extinction and Reddening • From effective temperature, Teff , can calculate the intrinsic colour, (B - V)0 and thus the colour excess. • E(B - V ) = (B - V)observed - (B - V)0 • From E(B - V ) we can calculate extinction, AV • and correct 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 the affects the apparent magnitude of the star. : • (mV,observed - AV ) - MV = 5 log d - 5 Remember from a few slides back? i.e. must correct apparent magnitude, m, for extinction before calculate absolute magnitude, M

  28. Atmospheric Absorption (and Airmass corrections) • The altitude of your target affects how much light gets through to the telescope • Distance traveled through the • atmosphere, d: • 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 • 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). • …To be discussed in more detail next time! m0 mz

  29. End.. See you next week….

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