Radiative Transfer and Molecular Lines Sagan Workshop PDF Free Download

Radiative Transfer and Molecular Lines Sagan Workshop 2009 Sara Seager Trent Schindler Trent Schindler MIT

Lecture Contents Overview of Equations for Planetary Atmospheres Radiative Transfer Thermal Inversions Molecular Line Primer

Planet Atmosphere Equations di(s,",µ,t) ds = #(s,",µ,t) $%(s,",t)i(s,",µ,t) Energy transport E out = E in,* + E in,planet dp(r) dr P = nkt = " Gm(r)#(r) r 2 No simple equation Consv. of Energy (in each layer) Hydrostatic Eq. Ideal Gas Law Chemical Equilibrium Want to derive: Flux, T, P, ρ, chemical composition

Atmosphere Model Uncertainties Only known inputs Surface gravity Radiation from star Unknowns: Atmospheric circulation Clouds Particle size distribution, composition, and shape Fraction of gas condensed Vertical extent of cloud Chemistry Elemental abundances Nonequilibrium Photochemistry Opacities Internal Luminosities Lodders 2005

Lecture Contents Overview of Equations for Planetary Atmospheres Radiative Transfer Thermal Inversions Molecular Line Primer

Definition of Intensity Intensity is a ray of infinitesimally small area, frequency interval, solid angle, and time interval. The intensity is traveling in the direction of photon propagation.

1D Radiative Transfer Equation Change in Intensity = gains to the beam - losses from the beam di(s,",t) ds = #(s,",t) $%(s,",t)i(s,",t) I = intensity, ε = emissivity, k = absorption coefficient

Optical Depth τ d" = #$ds A measure of transparency Think of an object in a fog When the object is immediately in front of you, the intervening fog has τ = 0 As the object moves away, τ increases τ is frequency-dependent We use τ as a vertical distance scale

1D Radiative Transfer Equation Equation from before di(z,",t) dz = #(z,",t) $%(z,",t)i(z,",t) Divide by κ di(z,",t) %(z,",t)dz = #(z,",t) %(z,",t) $ I(z,",t) Substitute τ and use Kirchoff s Law $ di(&,",t) d& = B(&,",t) $ I(&,",t)

1D Radiative Transfer Solution di(",#,t)e $" $ I(",#,t)e $" = $B(",#,t)e $" d" % di(",#,t)e $" (, + ' $ I(",#,t)e $" * d" =, + $B(",#,t)e $" d" 0 & d" 0 ) I(" +,#,t)e $" =$+ $ I(" 0,#,t)e $" = 0 =, + $B(",#,t)e $" d" 0 I(" 0,#,t) =, + B(",#,t)e $" d" 0 Emergent intensity is like a black body with bites taken out of it.

Earth s Thermal Emission Spectrum Pearl and Christensen 1997 I(" 0,#,t) = & % B(",#,t)e $" d" 0 Emergent intensity is like a black body with bites taken out of it.

Lecture Contents Overview of Equations for Planetary Atmospheres Radiative Transfer Thermal Inversions Molecular Line Primer

Origin of Absorption and Emission Lines

Constant Temperature Assume that the deepest layer of a planetary atmosphere is a black body. For an isothermal planet atmosphere, the spectrum emerging from the top of the atmosphere will also be a black body

Constant Temperature Read going up I(0,",t) = B(T,",t)... I 3 (z,",t) = B 5 (z,",t) di 3 (z,",t) = 0 dz di 3 (z,",t) = # 3 (z,",t) $% 3 (z,",t)i 4 (z,",t) dz di 4 (z,",t) = 0,I 4 (z,",t) = B 5 (z,",t) "(z,#,t) = $(z,#,t)b(z,#,t) dz di 4 (z,",t) = # 4 (z,",t) $% 4 (z,",t)i 5 (z,",t) dz I 5 (z,",t) = B 5 (z,",t)

Constant Temperature I(",#,t) = & % B(",#,t)e $" d" 0 0 No absorption or emission lines! Mathematically, constant T means B(τ,ν,t) = constant and can be removed from integrand $ % e "# d# =1 0 I(# 0,&,t) = B(T,&)

Origin of Absorption and Emission Lines

Decreasing T with Increasing Altitude Think of this situation as cooler layers of gas overlying hotter layers. According to the conceptual picture law, an absorption spectrum will result.

Decreasing T with Increasing Altitude & % I(" 0,#,t) = B(",#,t)e $" d" 0 Take two values of κ at neighboring frequencies. κ 1 is in the band center and κ 2 is in the neighboring continuum ; κ 1 is >> κ 2. Consider an optical depth τ From optical depth definition τ = κs, s 1 is shallower than s 2. "(z,#,t) = $(z,#,t)b(z,#,t) And T 1 < T 2, and B 1 < B 2. Therefore, I 1 < I 2 namely the intensity is lower in the band center than the continuum, generating an absorption line.

Decreasing T with Increasing Altitude Do you expect emission or absorption lines if the temperature is increasing with altitude? Think of the simple conceptual picture of where emission lines come from.

Earth Hanel et al. 2003

Lecture Contents Overview of Equations for Planetary Atmospheres Radiative Transfer Thermal Inversions Molecular Line Primer

Atmosphere Processes Kiehl and Trenberth, 1997

Earth s Thermal Emission Spectrum Turnbull et al. 2007 Pearl and Christensen 1997 Molecular absorption controls energy balance. Which molecule is Earth s strongest greenhouse gas? CO 2, N 2, H 2 O, N 2 O, or CH 4

Line-By-Line Opacities Burrows and Sharp 2007

Absorption Coefficient k " = Sf (" #" 0 ) For a given molecule at a given frequency k is in units of [m 2 /molecule] S is the line strength f is the (normalized) broadening function ν 0 is is the wavenumber of an ideal, monochromatic line $ % k " dv =S Normalized #$ *Note: κ ν = nk ν in units of [m -1 ], where n is the number density for a given molecule

k " = Sf (" #" 0 ) $ ( ) % f " #" 0 d" =1 #$ S ~ R ij 2 B fi = 1 8" 3 # fi 4" 3hc Summary We want to find the absorption coefficient for a given transition N j N = g e "E j / KT j # i g i e "E i / KT % Line broadening includes Doppler broadening and collisional broadening. Line strength is related to the square of transition probability, itself originating from quantum mechanics $ * f ( µ $ i )dv 2 % & f * & i d' Selection rules for permitted lines The energy level population ratio in LTE is defined by Boltzmann statistics. g is the degeneracy. " # = k # B # Emission coefficient in LTE (Kirchoff s Law) 2

H Atom Energy Levels

Molecular Energy Levels Electrons travel much faster than nuclei We may assume that the electronic energy depends only on the positiions of the nuclei and not on their own velocities The potential energy distribution is then a function of the internuclear distances alone The potential energy has a minimum at locations where the attractive binding forces balance repulsive internuclear forces.

Molecular Energy Levels Which molecules have rotational-vibrational transitions? Must have an electric dipole to interact with radiation Electric dipole: a difference between the center of charge and the center of mass Homonuclear molecules do not have a dipole moment Brent Iverson http://www.chem.tamu.edu/organic/spring99/dipolemoments.html

Rotational Energy Levels Rotation of a diatomic molecule approximate the molecule as two masses m 1 and m 2 at a fixed separation r 1 and r 2 the molecule rotations about an axis perpendicular to the line joining the nuclei Linear diatomic molecules can be approximated by a classical rigid rotor. The classical expression for energy is Total energy = Kinetic energy + Potential energy. A classical rigid rotor has no potential energy. E = 1 " 2 m i v i = 1 2 2 I# 2 = L2 2I i ω is the rotation frequency I is the moment of inertia = Σ i m i r i 2 L is the angular momentum L = Σ i m i ω r i

Rotational Energy Levels In classical theory E and ω can assume any value In quantum mechanics, the rigid rotor can only exist in discrete energy states. To find the energy states, we would use the timeindependent Schrodinger equation (with the Hamiltonian expressed in terms of the angular momentum operator). A pure rotational spectrum has energies at microwave frequencies. H" = L2 2I " = E" H" = J ( J +1 )h 2 " 2I E J = h2 2I J =0, 1, 2, J(J +1) B = h2 2I N J = (2J +1)e "BJ(J +1)/ KT N 0

Vibrational Energy Levels The potential energy can be approximated by the parabolic form of the simple harmonic oscillator. Similar to a simple harmonic oscillator but with quantized energy levels at equally spaced values. In a linear diatomic molecule all vibratory motion takes place along the line joining the atoms. Think of the atoms in periodic motion with respect to the center of mass. As the molecule vibrates, rotational modes are also excited. Vibration and rotation occur together.

Vibrational Energy Levels E = 1 2 m v 2 1 1 + 1 2 m v 2 2 2 +V (x) V (x) = 1 2 Cx 2 E = p 2µ + 1 2 Cx 2...Associate with the QM operator... # E v = ( v +1/2)h% 1 $ 2" E vk = h) 0 v k +1/2 v k = 0, 1, 2K ( ) C & ( µ ' µ is the reduced mass k denotes the normal modes C is the bond force constant Energies ~ infrared " 0 = C µ

Rotational Vibrational Transitions Simulated intensity of rotational-vibrational transitions of the CO molecule for the v = 1-0 band. ΔJ = -1 P branch ΔJ = 1 R branch The vibrational transitions are always accompanied by rotational transitions.

Rotational Vibrational Cross- Sections for CO v = 2-0 v = 1-0 From the VPL website

Non-Linear Molecules 3N-6 vibrational modes, where N is the number of atoms http://en.wikipedia.org/wiki/water_absorption

Water Vapor Cross Sections From the VPL website

Summary Overview of Equations for Planetary Atmospheres Radiative Transfer 1D equation: change of intensity is the sum of gains to and losses from the beam Thermal Inversions Absorption lines for no thermal inversion Emission lines indicate thermal inversion Molecular Line Primer Molecular vibrational and rotational lines (which appear in the IR spectrum) were outlined for a linear molecule

Summary Overview of Equations for Planetary Atmospheres Radiative Transfer 1D equation is change of intensity is the sum of gains to and losses from the beam Thermal Inversions Absorption lines for no thermal inversion Emission lines indicate thermal inversion Molecular Line Primer Molecular vibrational and rotational lines (which appear in the IR spectrum) were outlined

A New Text Book Exoplanet Atmospheres: Physical Processes Sara Seager Princeton University Press Should appear by end of 2009