Ay 20 Basic Astronomy and the Galaxy Problem Set 2 October 19, 2008 1 Angular resolutions of radio and other telescopes Angular resolution for a circular aperture is given by the formula, 1.22λ θ = (1) min d Where θmin is the least resolvable angle (in radians), d is the diameter of the aperture, and λ is the wavelength of observation. The factor 1.22 is for circular apertures. In case of synthesized apertures like interferometry, the angular resolution is ill-defined as the least resolvable angle varies as a function of diameter. The resolution is still proportional to λ/d but the numerical coefficient can be different from 1.22. For the purpose of this problem, we use 1.22 as the factor even for interferometric observations. The resolutions of the various telescopes are: Telescope Wavelength Resolution Resolution (radians) (other units) a) Owens Valley 21 cm 6.5 × 10−3 22’ b) CARMA 2.6 mm 1.6 × 10−6 0.33” c) Arecibo 21 cm 8.4 × 10−4 2.9’ d) VLA 21 cm 7.1 × 10−6 1.5” e) VLBAa 21 cm 2.6 × 10−8 5.3 masb f) Hubble 380 nm 1.9 × 10−7 40 mas g) Spitzer 24 µm 3.4 × 10−5 7” h) TMT 1µm 4.1 × 10−8 8.4 mas i) JWST 1.2µm 2.4 × 10−7 50 mas aThe longest baseline for the VLBA is from Virgin Islands to Hawaii, roughly 9500 km. bmilli-arc seconds 2 What we’d like to see The angular size of an object of size l (note that l should be diameter, not a radius, to be consistent with the definitions of problem 1) at a distance d is given by: l θ = (2) d 1 Object Size (cm) Distance (cm) θ (rad) θ (other units) a) Sun-like star 1.4 × 1011 3.1 × 1020 4.5 × 10−10 93 µas b) Red giant star 1.4 × 1014 3.1 × 1020 4.5 × 10−7 93 mas c) Binary star orbit, 5 day period 2.2 × 1012 3.1 × 1020 7.2 × 10−9 1.5 mas d) Binary star orbit, 100 yr period 8 × 1014 3.1 × 1020 2.6 × 10−6 0.56” e) Supernova shell after 1 month 5.2 × 1015 3.1 × 1025 1.7 × 10−10 34 µas f) Region around central BH 5.3 × 1019 3.1 × 1026 1.7 × 10−7 35 mas g) Cluster in interacting galaxies 6.2 × 1018 3.1 × 1027 2 × 10−9 0.4 mas To find the maximum separation a of the binary systems, use Kepler’s Third Law: !1/3 P 2 a = G(M + M ) (3) 2π 1 2 where P is the orbital period and Mi are the masses of the stars. The diameters entered in the table are the relative major axis 2a, though of course projection effects could reduce this somewhat. To find the radius of the region around the central black hole where stars have orbital velocities >500 km/s, use the expression for circular velocity: r GM v = (4) c R 8 using M = 10 M . 3 Fading into the distance (a) P60 The Palomar 60-inch can detect up to mV = 22. (i) Since we know the absolute magnitude of the sun in V -band (4.83), we can use the follow- ing relation between apparent (m) and absolute (M) magnitudes to find the distance at which it our sun would have mV = 22: d m − M = 5 log (5) 10 pc giving d = 27 kpc . To get this so-called distance modulus, recall that the definition of absolute magnitude is the apparent magnitude an object would have at a distance of 10 pc, and that d−2 shows up in flux. (ii) A white dwarf with radius 0.01 R with the same surface temperature as the sun will have a −4 2 luminosity of 10 L , since L ∼ R . And since FWD MWD − M = −2.5 log (6) F and F ∼ L, then MWD = M + 10 = 14.83, using the distance modulus above gives a distance of 270 pc for the white dwarf to have mV = 22. 2 Another way to see this is to notice that in order to have the same magnitude, two objects must have the same flux, and so: −4 L LWD 10 L −2 F = FWD → 2 = 2 = 2 → dWD = 10 d (7) d dWD dWD 46 31 −1 −1 14 (iii) If νLν = 10 , then Lν = 1.83 × 10 erg s Hz at ν = 5.45 × 10 Hz (or equiva- 43 −1 −1 lently, Lλ = 1.82 × 10 erg s nm at 550 nm). Within a bandpass of 100 nm, the luminosity is 45 −1 then roughly LV = 1.82 × 10 erg s . To find the absolute magnitude MV of this quasar then: FV LV MV − M = −2.5 log = −2.5 log . (8) F L ,V (Note that the flat νLν throughout optical and UV is important here; it allows us to not worry about how redshift affects these calculations.) Now, in order to find the luminosity of the sun in the V -band, approximate the sun as a black body at 5800 K, and find the fraction of the sun’s radiation that falls within this band: R 500 nm=6×1014 Hz 600 nm=5×1014 Hz Bνdν R ∞ = 0.13 (9) 0 Bνdν 33 32 Thus, L ,V = 0.13 L ,tot = 0.13 × 4 × 10 erg/s = 5.2 × 10 erg/s. Using this in (7), we obtain MV = −26.5 for this quasar, giving us a distance of 50.5 Gpc . Again, using the equivalent of (7) gives the same result quicker. This is interesting scientifically, as it implies that even the most distant possible luminous quasars are easily detectable with modest instruments, indicating that the observed rarity of very high-z quasars is in fact intrinsic rather than a selection effect. This is also a good place to point out that the “distance” talked about here is what’s known as the “luminosity distance” in cosmology, which is significantly greater than geometric distance due to the redshifting of photon energies. (b) The Heroic Hubble Deep Field The AB magnitude system is defined such that: mAB = −2.5 log Fν − 48.6 (10) We assume throughout that this filter goes from λ = 556-656 nm, or ν = 4.57-5.40 × 1014 Hz. (i) We begin by finding out mAB for the sun, for which we need Fν of the sun at earth at 606 nm: Z R 2 F = B cos θdΩ = B ∆Ω (for small θ) = B · π (11) ν ν ν ν 1 AU Plugging in 10 13 −5 R = 6.96 × 10 cm, 1 AU = 1.5 × 10 cm,Bν=4.96×1014 Hz(5800 K) = 3.01 × 10 cgs 3 −9 −1 −2 −1 gives Fν = 2.04 × 10 erg s cm Hz , which from (9) gives mAB, = −26.9. From this we can find the distance at which the sun would have an AB magnitude of 27.5: d 27.5 + 26.9 = 5 log → d = 1.14 × 1024 cm. (12) 1 AU And so, our sun at a distance of 369 kpc would have mAB = 27.5. −2 (ii) By analogy to part (a), dWD = 10 d , giving 3.69 kpc . (iii) Here again the calculation is essentially the same as in part (a), except the slightly differ- ent band means that LV, = 0.12L instead of 0.13L . Using the same idea as (7), we get 6 dquasar = 1.95×10 d , or 719 Gpc for our quasar to have mAB = 27.5. With current cosmological models that you will learn about next term, this corresponds to a redshift z = 57, so the observed V band is really 100 Angstrom soft X-ray radiation in the quasar rest frame! 4 Spectroscopy is a lot slower than imaging (a) Compute the specific flux Fν and Fλ from a mV = 22 galaxy. Again, we can compare to the sun, since we know MV, = 4.83, and differences in magnitudes are related to logs of ratios of fluxes: F 22 − 4.83 = −2.5 log V,galaxy (13) FV, ,10pc 2 Remember, we already found out that LV, = 0.13L , and F = L/(4πd ), so 0.13 × 4 × 1033 F = = 4.35 × 10−8 erg s−1 cm−2 (14) V, ,10pc 4π(10pc)2 giving −8 −(22−4.83)/2.5 −15 −1 −2 FV,galaxy = (4.35 × 10 )10 = 5.89 × 10 erg s cm . (15) But remember, we’re looking for specific flux Fν, and F = Fν∆ν, where ∆ν is the width of the V-band as defined above (from 500-600 nm: i.e. a width of 1 × 1014 Hz). So, F F = V,galaxy , (16) ν 1014 Hz −29 −1 −2 −1 giving Fν = 5.89 × 10 erg s cm Hz . Similarly, FV,galaxy Fλ = , (17) 1000 A˚ −18 −1 −2 −1 giving Fλ = 5.89 × 10 erg s cm A˚ The V -band photon number flux is given by: F 5.89 × 10−15 F = V = = 1.6 × 10−3photons s−1 cm−2 (18) γ hν (6.626 × 10−27)(5.45 × 1014) giving a total of 330 photons s−1 striking over the area of the P200. 4 (b) The surface brightness of the sky in V -band is 20.4 mag/arcsec2. The effective magnitude of a 2” × 2” patch is thus the equivalent of four mag 20.4 stars, so: mV,sky = 20.4 − 2.5 log 4 = 18.9 mag (19) (c) So the key here (and this is what makes spectroscopy so slow) is the narrowness of the band, which is 1/100 the width of the entire V -band (for one resolution element).