Lesson 11. The information lost, continuation. Monday, December 3, 12

Transcription

1 Lesson 11 The information lost, continuation

2 Atmospheric opacity The vertical axis represents opacity From wikipedia

3 Atmospheric layers Exosphere The outermost layer of Earth's atmosphere. It is mainly H and He. Very rarify, almost no atomic/molecular collisions. These freemoving particles follow ballistic trajectories and may migrate into and out of the magnetosphere or the solar wind.

4 Thermosphere Temperature increases with height from the mesopause up to the thermopause, and then it remains constant. T inversion is due to the extremely low mol density. T of this layer can rise to 1,500 C, ISS orbits in this layer, (320 to 380 km). The top of the thermosphere is the bottom of the exosphere, called the exobase. Its height varies with solar activity and ranges from about km.

5 Mesosphere From the stratopause to km; where most meteors burn up. T decreases with height. The mesopause, the minimum T that marks the top of the mesosphere, has an average temperature around 85 C and up to 100 C. Water vapor is frozen, forming ice clouds. A type of lightning referred to as either sprites or ELVES, form many miles above thunderclouds in the troposphere.

6 Stratosphere From the tropopause to about 50 km. T increases with height due to increased absorption of UV radiation by the ozone layer, less turbulence. While T may be 60 C at the tropopause, the top of the stratosphere is much warmer. The stratopause, which is the boundary between the stratosphere and mesosphere, is at 50 to 55 km. The pressure here is 1/1000 sea level.

7 Troposphere It begins at the surface going from 9 km at the poles to 17 km at the equator. Mostly heated by surface energy transfer, so T decreases with altitude. This promotes vertical mixing (from Greek tropos). The troposphere contains roughly 80% of the mass of the atmosphere.the tropopause is the boundary between the troposphere and stratosphere.

8 The atmosphere different layers, at glance Not to scale From wikipedia

9 Photograph prior to STS-130 r docking operations with the ISS. The orange layer is the troposphere. The whitish, the Stratosphere and then above the Mesosphere. From NASA

10 Other layers The ozone layer: It s within the stratosphere. Concentrations are about 2 to 8 ppm. It is located in the lower portion of the stratosphere from about km. Thickness varies seasonally and geographically. About 90% of the ozone in our atmosphere is contained in the stratosphere.

11 The ionosphere, The atmospheric layer that is ionized by solar radiation. From 50 to 1,000 km, and it overlaps both the exosphere and the thermosphere. It forms the inner edge of the magnetosphere. It influences radio propagation on the Earth. It is where auroras occur.

12 The planetary boundary layer is the part of the troposphere that is nearest the Earth's surface. Strong vertical mixing. Most turbulence. At night it becomes stratified, with weak mixing. The depth of it ranges from as little as about 100 m on clear, calm nights to 3000 m or more during the afternoon in dry regions.

13 More layers within layers From wikipedia

14 Sources of extinction From Chromey

15 The previous plot highlights contributions to extinction by four different processes. Rayleigh scattering by molecules. Absorption by ozone. Scattering by aerosols. Molecular-band absorption.

16 Rayleigh scattering by molecules. Photons are scattered by air molecules. The probability is much greater for shortwavelength photons (for pure Rayleigh scattering, extinction is proportional to λ 4 ). Molecular scattering explains why the sky is blue. It is stable over time, and it scales directly with the atmospheric pressure higher altitudes will have more transparent skies.

17 Absorption by ozone Continuous absorption by the O 3 molecule in the UV cuts off λ<320 nm. the Hartley bands nm maximum absorption at 255 nanometres. the Huggins bands, nm. the Chappuis bands, nm. the Wulf bands IR λ>700 nm, centered at 4,700, 9,600 and 14,100 nm, the latter being the most intense.

18 Scattering by aerosols particulates in air (up to 50 μm size). They stay longer in the atmosphere. usually produces gray extinction, like λ 1. A pale-blue sky indicates high aerosol extinction. Aerosols stay below 1,500 m, but volcanic ashes and forest fires can get up to the stratosphere.

19 Molecular-band absorption The main molecular absorbers are water vapor and CO 2. Oxygen has a few relatively narrow features. Water vapor and CO2 bands demarcate the windows in the NIR and MIR. CO 2 stays mixed with altitude. H 2 O vapor remains concentrated near the surface. Sea level: 10 mm in one air mass. Mauna Kea 1 mm (corresponds to 30% humidity).

20 Measuring monochromatic extinction 1. Mean extinction High altitudes, shorter λ Rayleigh scattering is the reason. Stable. Safe to use average or median obtained at the site by previous observers. 2. Standard exoatmospheric magnitudes If the value is available for a known star use Bouguer s law. Better is several stars are known.

21 3. Bouguer s law from several observations. We need to take several observations with different air masses. This, of course, requires waiting for the zenith distance of our site to change. 90 minutes may suffice. From Chromey fig a page 352

22 Recap on air mass It is the optical path length through Earth s atmosphere for light from a celestial source. It indicates relative air mass, the path length relative to that at the zenith at sea level (sealevel air mass at the zenith is 1).

23 4.Variable extinction If the extinction changes and we know several exoatmospheric star magnitudes we can infer k(λ,t). If we have standard magnitudes for two stars we can observe them at different air masses and infer k(λ,t). Take one frame for a standard star near the meridian (the D frame), and then take a frame for a second standard at large air mass in the east (the M frame).

24 We obtain: 4m A MD = m A M (X M ) m A D (X D )=m M m D + k(,t)(x M X D ) 4m A MD = 4m MD + k(,t) 4 X MD For monochromatic cases 4m STD MD = m M m D (11.1) and k(,t)= 4mA MD 4mSTD MD 4X MD (11.2)

25 5. Use all data available Every frame taken during the run of several nights is affected by extinction and therefore contains information about extinction (Chromey). The extinction problem can then be solved by fitting data to a least square fit with constraints imposed by standard stars.

26 Bandpass case Bouguer s law m A λ = m λ + k(λ)x has to be rewritten: k P X = m A P m P = 2.5log R TP f S atm d R TP f d (11.3) or k P X = 2.5log (R TP f exp [ k( )X] d R TP f d ) (11.4) where k P is the extinction coefficient for band P. For wider bands k P =k P (X,S P T).

27 Extinction measured for red stars and blue stars, will differ since the center of the bandpass is different for the two. The Bouguer plot of apparent magnitude versus air mass will give straight lines of different slopes for stars of different spectral shapes. In addition the extinction effect changes the spectrum shape (Forbes effect) for same star at different air masses.

28 The Forbes effect From Chromey

29 (a) Notice the monochromatic energy flux f λ, bandpass transmission T BP, and extinction coefficient. The extinction is due entirely to a strong feature near the blue edge of the band. (b) The flux actually detected at telescope as a function of λ for four different air masses. (c) The Bouguer diagram for the data in (b). Dark circle actual exo magnitude; open circle the corresponding one using linear extrapolation.

30 The Forbes effect, is acute if strong atm absorption affects some segments of the band more than others, i.e. the Johnson U and many of the wider infrared bands. Johnson filter From Chromey Fig 10.3

31 The magnitude change in going from X = 2 to X = 1 can be considerably less than in going from X = 1 to X = 0. In some cases, the width of the band and its effective λ can change a lot at the smaller air masses. Use of exoatmospheric magnitudes then depends on having a good model of the response function, the atmosphere, and the unknown source. It is best to use bands that exclude strong atmospheric absorptions (e.g. the MKO near-infrared system).

32 Second-order extinction coefficients We could solve for k P X in eq. (11.4) in slide 26 if we could make a good approximation of both the monochromatic extinction function k(λ) and the spectrum f λ. A way of doing this is to write: k P = k P + k P. ci (11.5) where ci could be a color index, i.e. (B-V).

33 k P is called the second order extinction coefficient. i.e. for the Johnson V band: k P = k P + k V,BV. (B-V) (11.6) For broadband extinction, therefore, one has for the ith observation of star j: m A P,i,j = m P,j +(k 0 p + k 00 p.(ci))x i,j (11.7)

34 Indices or magnitudes? Typically one reports n-color photometric data with one magnitude and n 1 indices. For coherence the method has also been followed for extinction calculations. i.e. if we have a J-B and J-V observations with the same X, we can write the previous eq for each band and then take the diff: (b v) A =(b v)+ (k 0 B k 0 V )+(k 00 B,BV k 00 V,BV.(B V ) X (b v) A =(b v)+ k 0 B V + k 00 B V.(B V ) X (11.8) (11.9)

35 Change to a standard system How can we compare results with the corresponding ones obtained with other instruments by other people? Instrumental standard system. Depends on: 1) diff. between instrument response and the standard system. 2) shape of the source s spectrum.

36 Monochromatic case Assume two infinitesimally narrow-band filters with λ c = λ 1 & λ 2 and λ S = λ S1 & λ S2 The difference between the first standard magnitude and the instrumental result: m STD S1 m 1 = 2.5[logf S1 logf 1 ]+C STD 1 C 1 (11.10) The quantity would be the same for every source. This looks like a color index. So we can write: m STD S1 m (m 1 m 2 )+ 1 = m (ci) (11.11)

37 α 12 is called the color coefficient for the transformation between systems and α 1 is the zero point constant. Remember that traditionally people work with one magnitude transformation and n-1 index transforms, (m 1 m 2 ) STD m 1 m 2 +( )(m 1 m 2 )+ 1 2 (11.12) With some redefinitions: (ci) STD (ci) (11.13) Assuming in (11.10), (11.13) is fine... See plot next slide.

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39 At narrow band observations...life is cruel. What strategy to follow then? Measure m λ1 and (ci) 12 for many standards that you would like to transform. Then plot (m STD λ1 - m λ1 ) as a function of (ci) 12 as in the figure in the next slide.,then fit the experimental data to what may be the most appropriate curve...linear piecewise or quadratic, or...

40 Empirical determination for standards transform

41 Multicolor case If we want to repeat the previous procedure when doing broadband studies we replace the fluxes with the appropriate integrals: apple m STD P m P = 2.5 log Z T P,STD f d log Z T P f d + C STD P C P (11.14) (11.11) would work better because broadband features smooth out discontinuous features that appear in narrow band. The rest of the procedure is similar. Plot as a function of color the diff between instrumental and standard magnitudes.