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1 radiation specific intensity flux density bolometric flux summary of last lecture Js 1 m 2 Hz 1 sr 1 Js 1 m 2 Hz 1 Js 1 m 2 blackbody radiation Planck function(s) Wien s Law λ max T = 2898 µm K Js 1 m 2 m 1 sr 1 Js 1 m 2 m 1 B ν (T ) = 2hν3 c 2 1 e hν/kt 1 B λ(t ) = 2hc2 λ 5 1 e hc/λkt 1 stellar properties luminosity L radius R effective temperature Teff L = 4πR 2 σt 4 eff

2 summary of last lecture flux (Wm -2 ) from blackbody surface (unit area) at temperature T S = σt 4 luminosity (W) of a spherical blackbody surface of radius R L = 4πR 2 σt 4 luminosity (W) of a spherical stellar surface of radius R L = 4πR 2 σt 4 eff (definition of effective temperature Teff)

3 flux (Wm -2 ) from blackbody surface (unit area) at temperature T S = σt 4 more generally, for a real surface... need to consider emissivity ɛ 0 < ɛ < 1 S = ɛσt 4 a perfect blackbody has ɛ = 1 reflectivity r = 1 ɛ a good absorber is a good emitter... (dimensionless) ɛ may be a function of frequency... need to write as integral formula for flux not valid: greybody I ν = ɛ ν B ν ɛ ν

4 dependence of flux on distance d flux at stellar surface (Wm -2 ) S = L 4πR 2 flux at distance d (Wm -2 ) S(d) = L 4πd 2 d flux diminishes 1 d 2 L measure flux S know distance d L = 4πd 2 S(d) (assumptions: (1) no absorption, (2) radiation emitted isotropically)

5 flux at Earth from Sun (Luminosity 3.9 x W, distance = 1AU) flux at distance d (Wm -2 ) S(d) = L 4πd 2 flux ~ 1370 Wm -2 (a.k.a. solar constant ) like many constants in astronomy, not actually constant... varies a bit around this because Earth-Sun distance varies (~ 1320 Wm -2 to ~1410 Wm -2 ) this includes radiation at all wavelengths... only some of this radiation makes it to the Earth s surface

6 flux at Earth from nearest star (Alpha Cen) (Luminosity 5.9 x W, distance = 1.34 pc) flux ~ 2.75 x 10-8 Wm -2 (Sun is factor ~ 5 x brighter) this is the total flux (integrated over wavelength) with any single observation (restricted wavelength range) we see a small fraction of this

7 effective temperatures can assign effective temperatures to other bodies (not just stars) e.g. Earth: receives a flux from the Sun: S = 1370 Wm -2 absorbs this radiation over a projected area of πr 2 E Earth re-radiates this energy in equilibrium: total energy in = total energy out assume that Earth radiates isotropically from surface area 4πRE 2 at an effective temperature T E

8 flux from Sun at Earth effective temperatures S = L 4πd 2 total energy incident on Earth (πr 2 E) ( L 4πd 2 ) total energy re-radiated by Earth at equilibrium temperature TE for balance: (πr 2 E) ( L 4πd 2 ) = 4πR 2 EσT 4 E L E = 4πR 2 EσT 4 E (luminosity of Earth effective temperature of Earth) solve for TE: T E = ( L 16πσd 2 ) 1/4 280 K

9 Earth: T E = ( L 16πσd 2 effective temperatures ) 1/4 280 K note that the equilibrium temperature doesn t depend on Earth s radius for given solar luminosity, it only depends on the Earth-Sun distance, d a planet at distance dp from the Sun should therefore have a temperature given by: T p = ( L 16πσd 2 p ) 1/4 convenient to use Earth result as a reference point: T p = T E ( de d p ) 1/2 (labelled Earth-Sun distance explicitly as de) planetary temperatures should fall proportional to (1/d p ) 1/2

10 writing dp in AU effective temperatures T p = T E ( de d p ) 1/2 T p = 280 K (1/d p,au ) 1/2 Planet dp Predicted Tp Actual Tp Mercury 0.34 AU 480 K 452 K Venus 0.72 AU 329 K 735 K Mars 1.45 AU 232 K 227 K Jupiter 5.2 AU 123 K 112 K (0.1 bar) Saturn 9.5 AU 91 K 84 K (0.1 bar) Uranus 19 AU 64 K 53 K (0.1 bar) Neptune 30 AU 51 K 55 K (0.1 bar) table by Mark McCaughrean

11 Planet dp Predicted Tp Actual Tp Mercury 0.34 AU 480 K 452 K Venus 0.72 AU 329 K 735 K Mars 1.45 AU 232 K 227 K Jupiter 5.2 AU 123 K 112 K (0.1 bar) Saturn 9.5 AU 91 K 84 K (0.1 bar) Uranus 19 AU 64 K 53 K (0.1 bar) Neptune 30 AU 51 K 55 K (0.1 bar) re-radiated light from planets use Wien s Law to predict wavelength where spectrum peaks λ max T = 2898 µm K Earth ~ 10 microns Kuiper Belt objects (~50AU, T~40K) ~ 70 microns not observable from the ground (unless you re observing Earth!) table by Mark McCaughrean

12 not all light is absorbed and re-radiated some is reflected this is why we can see the planets (and other solar system objects) at optical wavelengths) albedo: ratio of reflected em energy to incident em energy... recall: emissivity, reflectivity Earth (overall) ~ 0.3 (varies depending on terrain...) Moon ~ 0.07 Enceladus (icy moon of Saturn) ~ 0.99 Kuiper Belt Objects: down to ~0.05 effect of emissivity/reflectivity on temperatures...?

13 stellar fluxes and magnitudes fluxes we re taking about here are really flux densities not Wm -2 but Wm -2 Hz -1 or Wm -2 micron -1 or Janskys (Jy) (1 Jy = W m 2 Hz 1 ) radiation from star measured over finite solid angle measuring radiation over finite bandwidth will (loosely) use the symbol F for flux in this context where appropriate will use: F ν F λ

14 we have already seen that flux diminishes 1 d 2 this also applies to flux densities flux at stellar surface (Wm -2 ) S = L 4πR 2 flux at distance d (Wm -2 ) S(d) = L 4πd 2 d flux diminishes 1 d 2 L measure flux S know distance d L = 4πd 2 S(d) (assumptions: (1) no absorption, (2) radiation emitted isotropically)

15 stellar fluxes (flux densities!) vary over huge range intrinsic range in luminosity for stars (> ) diminution of fluxes with distance fluxes very small... Alpha Cen total flux ~ 2.75 x 10-8 Wm -2 Alpha Cen flux density in optical ~ 3.5 x Wm -2 Hz -1 (3500 Jy) James Webb Space Telescope (JWST) will detect stars at nanojy level (factor of fainter than Alpha Cen) use logarithmic scale: magnitudes

16 Magnitudes: history Hipparchus (c BCE) Catalogued ~ 850 stars Listed positions and brightnesses Brightnesses in 6 bins Stars of magnitude 1 : brightest visible stars Stars of magnitude 6 : faintest visible stars Norman Pogson ( ) Determined that magnitude 6 stars ~ 100x fainter than magnitude 1 Proposed standardising this Pogson ratio Decrease of 1 magnitude represents decrease in brightness of 5 (100) = mag difference means ~ = 6.25 in brightness, 3 mags means ~ = 15.6, etc. observed magnitude = apparent magnitude 525

17 Magnitudes: formal definition Derivation of flux-magnitude relation Assume two stars: F1 = flux of fainter star F2 = flux of brighter star m1 = magnitude of fainter m2 = magnitude of brighter ( F1 F 2 ) = 100 (m 2 m 1 )/5 m 2 m 1 = 2.5 log 10 ( F1 F 2 ) = 2.5 log 10 ( F2 F 1 ) difference in magnitude ratio in flux a larger magnitude means a FAINTER star 526

18 Magnitudes: formal definition m 2 m 1 = 2.5 log 10 F2 F (m 2 m 1 ) = F2 F 1 For mag difference (m2-m1) = 1, flux ratio = Do not confuse factors of and 2.5: they are different is flux difference for 1 mag = = 10 (1/2.5) 2.5 in magnitude definition is not rounded; it is precise fluxes (flux densities) are always positive magnitudes can be negative or positive (more negative magnitude = brighter!) 527

19 Magnitudes: sample calculations Assume Star1 has a magnitude of 15.4 If Star2 is 750x brighter, calculate its magnitude m 2 m 1 = 2.5 log 10 F2 F 1 = 2.5 log = = 7.2 m 2 = =

20 Magnitudes: sample calculations Assume Star2 has a magnitude of 19.3 mag If Star1 is 17.5 mag, calculate flux ratio m 2 m 1 = 2.5 log 10 F2 F 1 F2 log 10 F 1 F2 F 1 = = 1.8 = = = 0.72 = i.e. Star2 is x the brightness of Star1 529

21 stellar fluxes and magnitudes fluxes we re taking about here are really flux densities not referred to restriction of frequency / wavelength range yet if we did have bolometric (total) flux: bolometric flux bolometric magnitude in general, fluxes (magnitudes) will be measured within a range of wavelengths around a central wavelength

22 Magnitudes: filter systems flux densities from restricted wavelength ranges using filters e.g. Johnson optical-infrared system: central wavelength and wavelength range Filter λ (µm) Δλ Filter λ (µm) Δλ U (µm) H (µm) B K V L R M I N J Q Many variations on basic set, optimised for different sites, detectors, science Harold L. Johnson

23 Typical optical filter profiles McDonald Observatory Sky & Telescope 532

24 Typical near-ir filter profiles 533 JACH

25 Magnitudes: calibration Magnitude systems require calibration Vega (α Lyr) defined to be 0.0 mag in all filters Flux difference with respect to Vega yields magnitude: m = 2.5 log 10 ( F F Vega ) Thus Vega has: mu = 0.0 mb = 0.0 mv = 0.0 mr = 0.0 etc. by definition Features: Simple relative calibration: independent of size of telescope, type of detector system, output units of system In practice, zero point defined as average of many bright stars, of which Vega may be just one Vega often too bright or unavailable: calibration transferred all over sky to systems of secondary standards 534

26 Example secondary standard field: Rubin 149 Landolt (1992) UBVRI photoelectric photometry for 8 stars; RU149, A-G Stetson (2000) CCD imaging photometry for other stars in cluster to faint magnitudes Region provides wide range of colours too Single CCD image provides many standard stars simultaneously 535

27 Magnitudes: calibration Absolute calibration of Vega flux yields physical fluxes Filter λ Δλ Fλ (W m -2 µm -1 ) Fν (W m -2 Hz -1 ) Jansky U 0.36 (µm) 0.15 (µm) 4.19 x x B x x V x x R x x I x x J x x H x x K x x

28 Magnitudes: calibration Absolute calibration (e.g. W m -2 Hz -1 ) difficult Requires careful use of standard radiometric calibration units to compare with stellar light: very rarely done Also, Vega calibration has unsatisfactory physical basis Calibration follows spectral shape of one ~10000K star Flux density for 0.0 mag different at every wavelength AB magnitude system provides good alternative: m AB = 2.5 log 10 (F ν ) ( ± 0.005) (F ν in erg s 1 cm 2 Hz 1 ) Constant (48.585) yields Vega = 0.0 mag at 5480Å (V-band) But then constant flux density at every wavelength, independent of shape of Vega spectrum 537

29 Magnitudes: for reference AB magnitude calibration m AB = 2.5 log 10 (F ν ) ( ± 0.005) (F ν in erg s 1 cm 2 Hz 1 ) Equivalent for flux in SI units m AB = 2.5 log 10 (F ν ) ( ± 0.005) (F ν in W m 2 Hz 1 ) Equivalent for flux in Janskys (1 Jy = W m -2 Hz -1 ) m AB = 2.5 log 10 (F ν ) (8.915 ± 0.005) (F ν in Janskys) 538

30 Magnitudes: colours Magnitude is measure of monochromatic flux i.e. flux at central wavelength of particular filter No indication of slope of stellar spectrum Colour indicates of spectral slope or temperature Colour = (Magnitude at λ1 Magnitude at λ2) (with λ1 < λ2) If +ve, then star red (more flux at longer λ); if -ve, then blue Note: magnitudes are logarithmic, so colour is flux ratio Example: take star with following properties: Magnitudes: mb = 19.5, mv = 19.0, mr = 18.2 Colours: mb - mv (B - V) = +0.5 mag mv - mr (V - R) = +0.8 mag Vega has (B - V) = (V - R) = (R - I) = 0.0 mag by definition Thus star is redder than Vega and thus (likely) cooler 539

31 Stellar spectra, filters, & colours 540 Significant differences in colours for hot (blue) and cool (red) stars when measured at optical wavelengths; in the infrared, colours more subtle, as most stars on Rayleigh-Jeans tail

32 Infrared colours of main sequence stars Sp type Teff (K) (V - K) (J - H) (H - K) B0V B5V A0V A5V F0V F5V G0V G4V K0V K5V M0V M5V Adapted from JACH / Tokunaga (2000), in Allen s Astrophysical Quantities 541

33 Magnitudes: absolute magnitude Stars with same R, T should have same magnitudes But usually different: lie at range of distances (1/r 2 effect) Thus have range of different apparent brightnesses Introduce absolute magnitude Measure of intrinsic luminosity of star Apply inverse square law to same star (same L) at different distances: Link between flux f at distance d to flux F at distance D: ) 2 ( F f ) = L/4πD2 L/4πd 2 = ( d D The magnitude difference is ( ) ( F F M m = 2.5 log 10 m M = 2.5 log f 10 f ) = 2.5 log 10 ( d D ) 2 M is magnitude of source corresponding to flux F, m corresponding to flux f 542

34 Magnitudes: absolute magnitude Definition of absolute magnitude, M: Magnitude of star when distance D = 10 parsecs From apparent magnitude, m, at distance d, via: ( d m M = 2.5 log = 2.5 log 10 (d 2 ) 2.5 log 10 (10 2 ) = 5 log 10 (d) 5 Factor m - M is distance modulus, DM Example: take star with m = 15.0 at 200 parsecs DM = 5 log10(200) - 5 = 6.5 DM = m - M M = m - DM = = 8.5 Makes sense: should be 2.5log10[(200/10) 2 ] mag brighter at 10pc ) 2 543

35 Magnitudes: sample calculation Take star with following characteristics: Absolute KVega magnitude MK = 6.75; distance 2.5 kiloparsecs calculate apparent magnitude, mk Distance modulus: m M = 5 log 10 (d) 5 = 5 log 10 (2500) 5 = (5 3.4) 5 = 12.0 Thus apparent magnitude mk = MK = with calibration standards, can express as flux density can do it the other way: for an observed apparent magnitude, can convert to absolute magnitude if you know the distance

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