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5. Astronomy Through the Atmosphere

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5. Astronomy Through the Atmosphere 5. Astronomy Through the Atmosphere Ground based observations are affected by: • absorption • refraction • scattering • scintillation In this section we will briefly consider some of the effects of these four phenomena. 1 Absorption The Earth’s atmosphere is opaque to E-M radiation, apart from two windows: in the optical and radio regions of the E-M spectrum. X-rays UV Visible Infrared sub mm Radio mm & microwave hard soft EUV (FIR) Completely opaque Optical window Radio window Molecular transitions (inc. H2O) 0 Ionisation of air molecules (starting with ozone) Ionosphere reflective Fraction of energy absorbed 1 1nm 10nm 100nm 1mm 10mm 100mm 1mm 1cm 10cm 1m 10m2 100m Completely transparent Wavelength Absorption Between the optical and radio windows there are numerous absorption bands due to molecular transitions (mainly of water). It is possible to get above the clouds containing this water vapour because of the 4 temperature structure (km) of the atmosphere. Above about 2 km there can be a thin inversion Altitude 2 layer, where the temperature Inversion increases with height. Clouds layer form at the base of the inversion layer, leaving generally clear, dry 020 Temperature (C) air above. 3 Absorption The world’s best observatories (e.g. La Palma, Hawaii, La Silla, Paranal) are all at altitudes and locations which place them above inversion layers. 4 Absorption How does absorption in the atmosphere affect the apparent brightness of objects? We model the atmosphere as a series of plane-parallel slabs. Consider a thin slab of thickness d l and radiation of intensity I incident perpendicular to the slab: We model the change in intensity due to the thin I slab as: dI = −I κ dl (5.1) dl I + dI Absorption coefficient, which is not in general constant, but depends on depth in the atmosphere 5 Absorption Re-arranging, and integrating from the source (at distance D , emitting radiation of intensity I 0 ) to the Earth’s surface I obs dI D Although we are thinking mainly = − κ d (5.2) about atmospheric absorption in this ∫ ∫ l section, the same formula can I describe interstellar absorption. I 0 0 We define the right hand integral as the optical depth, denoted by τ , so or −τ ⎛ I obs ⎞ I = I e ln⎜ ⎟ = −τ obs 0 I 0 ⎝ ⎠ (5.4) 6 (5.3) Absorption • If τ = 0 we describe the atmosphere as transparent : I obs = I 0 • If τ << 1 we describe the atmosphere as optically thin : I obs ≈ I 0 • If τ ≥ 1 we describe the atmosphere as optically thick : I obs << I 0 Expressing in terms of apparent magnitude, we can write ⎛ F ⎞ ⎛ I ⎞ ⎜ obs ⎟ ⎜ obs ⎟ ()mobs − m0 = −2.5log10 ⎜ ⎟ = −2.5log10 ⎜ ⎟ (5.5) ⎝ F0 ⎠ ⎝ I 0 ⎠ ⎛ I ⎞ and, from (5.4), ⎜ obs ⎟ (5.6) log10 ⎜ ⎟ = −τ log10 e ⎝ I 0 ⎠ so (5.7) ∆m = ()m obs − m 0 = 2.5τ log10 e ≈ 1.09 τ 7 Absorption Suppose we observe a star at zenith angle θ Ë Path length through slab of thickness d l is d ds = l = d secθ (5.8) θ cosθ l So this introduces an extra factor of dl ds secθ in eqs. 5.1 – 5.3. (5.9) Hence ∆m θ ≈ 1.09 τ sec θ = ∆m sec θ Zenith extinction This treatment is only an approximation, as the light ray is also refracted by the atmosphere, thus changing θ along its path. 8 Refraction We model the atmosphere as plane-parallel and consider a light ray incident at zenith angle θ on the top of the atmosphere (valid for θ ≤ 60 o . Larger zenith angles need a treatment that includes the curvature of the Earth.) θ Suppose we break the atmosphere into four A parallel slabs, each with uniform refractive Top of atmosphere index: η 0 Lη 3 η B θ3 3 Applying Snell’s Law, at level A: sin θ = η sin θ η 3 3 C θ2 2 at level B: η 3 sin θ 3 = η 2 sin θ 2 η sin θ = η sin θ η at level C: 2 2 1 1 D θ1 1 Finally, at level D: η1 sin θ1 = η 0 sin θ 0 η θ0 0 Putting these together: sin θ = η sin θ GROUND 0 0 (5.10) Extends to an arbitrary number of slabs: to correct the observed zenith 9 angle for refraction, we only need only to know η 0 at ground level! Scattering Air molecules, dust and water vapour all scatter light. However, their different sizes cause different effects on light. 3 regimes: 1. Particle size, a >> λopt: particles scatter all wavelengths equally. Atmospheric water droplets fall into this category, and this is why cloud and mist appear white. 10 Scattering 2. Particle size, a ~ λopt Scattering strength ∝ 1 λ Examples: fine dust, cigarette smoke This is why e.g. cigarette smoke, illuminated from the side, has a bluish tinge: blue light is scattered by the smoke particles slightly more than red light. 11 Scattering 3. Particle size, a << λopt scattering power ∝ 1 λ4 so blue light is strongly scattered (e.g., the blue sky is due to scattering by air molecules) This is Rayleigh scattering This explains why the daytime sky is blue, and why the sun appears red at sunset (blue light scattered out of line of sight)12 Scattering Rayleigh scattering is anisotropic ⇒ sky light is polarised Only this polarisation is scattered strongly 13 Scattering We can analyse the loss of intensity due to scattering in the same way as for absorption. D The optical dI = −I κ d ⇒ τ = ∫κ(l) dl depth is larger l 0 for blue light Scattering coefficient: −4 κ ∝ λ for Rayleigh scattering. ⇒ Stars appear reddened by the atmosphere (including the Sun!) 14 Scattering • At visible wavelengths, the scattering of sunlight makes the sky so bright that we must observe at night. • In the Far Infra-Red (and beyond) on the other hand, scattering makes a small enough contribution that we can observe during the day too. • (As we saw in previous sections, however, the thermal emission from the daytime sky may be a problem at FIR wavelengths). 15 Incoming parallel rays Scintillation from point source Scintillation, or ‘twinkling’ of Turbulent ‘cells’ r0 ~ 10cm starlight is caused by turbulence v ≈ 10 m/s in the atmosphere. Air ‘cells’ of varying density – and hence h ~ 7km Overdense cells scatter rays and smear-out light refractive index – are continually from point source to form passing across the line of sight to a seeing disk a star, and changing the pattern of Darker patch illumination from the star which reaches ground level. GROUND Brighter patch Typical scale length for the cells is r0~10 cm, at a height of ~7 km. Hence, illumination from the star at any instant will not be uniform, but will consist of brighter and darker patches, typically also ~10 cm across. Cells deflect the light from a point source over seeing disk: Angular radius of seeing disk ~10 cm = 3 arcsec. 16 7 km Scintillation The cells are continually moving across the line of sight, with a transverse speed of about 10m/s. Scintillation timescale of variations ~ 10 cm = 0.01 sec. 10 m/s If the telescope aperture, D ~ r0 We see rapid variations in position and brightness of the image as individual cells cross the line of sight. If the telescope aperture, D >> r0 We see an image formed from many cells added together, ⇒ Average brightness of the image is ~ constant, but there are rapid variations in the position, size and shape of the seeing disk This can be corrected using Adaptive Optics (see Honours Astronomy) 17 ESO 3.6m telescope at La Silla, Chile 18 Gamma Perseus 0.2 arcsecond binary 250 frames Wavelength 570nm 25 ms per frame 12.5 arsec/pixel Data from the William Herschel 4.2 metre Telescope in La Palma. 19 Scintillation Radio observations are also affected by scintillation – this time not from the Earth’s atmosphere, but from turbulence in the interstellar medium and the interplanetary medium. The typical size of turbulent cells is much larger (as is the wavelength of the radiation). The scale-size for interplanetary scintillation at ~100 MHz is ~100 km, and the scintillation timescale is about 1 s. Interstellar scintillation causes variations in the brightness of pulsars (small, bright) on timescales of minutes to hours. 20 Example At 700 nm the zenith extinction (∆m700) is 0.08 magnitudes. Estimate the extinction (∆m400) at 400 nm. After correction for the atmosphere, a star is found to have a true colour index (B-V)0 = -0.13. At a particular observatory, the zenith extinction (∆m) for the B band is 0.29 and for the V band is 0.17. At what zenith distance would the star have the same apparent magnitude in the two bands? 21.
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