Astrochemistry and Molecular Astrophysics Paola Caselli

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1 School of Physics and Astronomy FACULTY OF MATHEMATICS & PHYSICAL SCIENCES Astrochemistry and Molecular Astrophysics Paola Caselli

2 Outline 1. The formation of H 2 2. The formation of H The chemistry initiated by H Formation and destruction of CO 5. The two-level system 6. Transfer of radiation in spectral lines 7. The rotational temperature diagram 8. The escape probability method

3 Interstellar Molecules Known Interstellar Molecules (Total: > 170) Amino acetonitrile in SgrB2(N) (Belloche et al. 2008) H C H O N C H H O H Glycine - the simplest amino acid

4 How do molecules form in the interstellar medium? The most elementary chemical reaction is the association of A and B to form a molecule AB with internal energy: A + B AB * The molecule AB * must lose the internal energy. In the Earth atmosphere, where the number of particles per cubic centimeter (cc) is very large (~10 19 ), the molecule looses its energy via three-body reactions: AB * + M AB But this is not an efficient process in interstellar clouds, where the number of particles per cc ranges between a few hundred and ~10 8.

5 1. The formation of H 2 The reaction that starts the chemistry in the interstellar medium is the one between two hydrogen atoms to form molecular hydrogen: H + H H 2 This reaction happens on the surface of dust grains.

6 1. The formation of H 2 The H 2 formation rate (cm -3 s -1 ) is given by (Gould & Salpeter 1963; Hollenbach & Salpeter 1970; Jura 1974; Pirronello et al. 1999; Cazaux & Tielens 2002; Bergin et al. 2004; Cuppen & Herbst 2005; Cazaux et al. 2008): R H 2 = 1 2 n Hv H A n g S H " # 10 $ cm -3 s -1 n H gas number density v H H atoms speed in gas-phase A grain cross sectional area n g dust grain number density S H sticking probability γ surface reaction probability PAHs

7 Once H2 is formed, the fun starts H 2 is the key to the whole of interstellar chemistry. Some important species that might react with H 2 are C, C+, O, N To decide whether a certain reaction is chemically favored, we need to examine internal energy changes. Molecule H 2 CH C + H 2 CH + H?? C + + H 2 CH + + H?? Dissociation energy (ev) OH 4.39 CH OH Question: Can the following reactions proceed in the cold interstellar medium? O + H 2 OH + H?? O + + H 2 OH + + H??

8 Once H2 is formed, the fun starts C + H 2 CH + H?? 4.48 ev 3.47 ev The bond strength of H 2 is larger than that of CH the reaction is not energetically favorable. The reaction is endothermic (by = 1.01 ev) and cannot proceed in cold clouds, where k b T < 0.01 ev! Dissociation energy or bond strength

9 Once H2 is formed, the fun starts Molecule H 2 CH Dissociation energy (ev) OH 4.39 CH OH C + H 2 CH + H C + + H 2 CH + + H O+ H 2 OH + H (endothermic by 1.01 ev) (endothermic by 0.39 ev) (endothermic by 0.09 ev) O + + H 2 OH + + H (exothermic by 0.62 ev!)

10 Rate coefficients and activation energies The rate coefficient k (cm 3 s -1 ) of a generic reaction A + B -> C + D is given by: k =< " v > σ total cross section of the reactants v relative velocity <The average is performed over the thermal distribution> Most reactions possess activation energies E a (~0.1-1 ev) even if exothermic and k is given by the Arrhenius formula (Herbst 1990): k = a(t)exp("e a /k B T)

11 Ion-Neutral reactions A + + B C + + D Exothermic ion-molecule reactions do not possess activation energy because of the strong long-range attractive force (Herbst & Klemperer 1973; Anicich & Huntress 1986): R V(R) = - α e 2 /2R 4 k LANGEVIN = 2 πe(α/µ) 1/ cm 3 s -1 independent of T

12 Neutral-Neutral reactions Energy to break the bond of the reactant. A + BC AB + C 1 ev for endothermic reactions E ev for exothermic reactions Energy released by the formation of the new bond. k b T < 0.01 ev in molecular clouds Example: O + H 2 OH + H (does not proceed in cold clouds) Duley & Williams 1984, Interstellar Chemistry; Bettens et al. 1995, ApJ

13 2. The formation of H 3 + After the formation of molecular hydrogen, cosmic rays ionize H 2 initiating fast routes towards the formation of complex molecules in dark clouds: H 2 + c.r. H e - + c.r. Once H 2 + is formed (in small percentages), it very quickly reacts with the abundant H 2 molecules to form H 3+, the most important molecular ion in interstellar chemistry: H H 2 H H H H H

14 3. The chemistry initiated by H 3 + Once H 3 + is formed, a cascade of reactions greatly enhance the chemical complexity of the ISM. In fact, H 3 + can easily donate a proton and allow larger molecules to form. Example OXYGEN CHEMISTRY (the formation of water in dark clouds) H 3 + O OH + H 2 H 2 O + H 2 H 3 O + e e e H 2 O OH O O O 2

15 3. The chemistry initiated by H 3 + CARBON CHEMISTRY (the formation of hydrocarbons) The formation of more complicated species from neutral atomic carbon begins with a sequence very similar to that which starts the oxygen chemistry: H 3 + H 2 H 2 e C CH+ CH 2 + CH 3 + e CH 2 CH A. Proton transfer from H 3 + to a neutral atom; B. Hydrogen abstraction reactions terminating in a molecular ion that does not react with H 2 ; C. Dissociative recombination with electrons.

16 4. Formation and destruction of CO [a] C + H 3 O + HCO + + H 2 [b] O + CH 3 + HCO + + H 2 [c] HCO + + e CO + H is the most important source of CO. CO is very stable and difficult to remove. It reacts with H 3+ : [d] H CO HCO + + H 2 but reaction (c) immediately reform CO. The main mechanisms for removing CO are: [e] He + + CO He + C + + O [f] hν + CO C + O Some of C + can react with OH and H 2 O (but not with H 2 ): [g] C + + OH CO + + H [h] CO + + H 2 HCO + + H [i] C + + H 2 O HCO + + H

17 The timescale to form CO Assume: dark region where all H is in H 2 and all atoms more massive than He are in neutral atomic form. The timescale on which almost all carbon becomes contained in CO (n O > n C ) is at least equal to the timescale for one H 2 to be ionized for every C: n C /[ζ n(h 2 )] = 2 n C /[ζ n H ] For ζ = s -1 and n C /n H = 10-4, the above expression gives a value of about 2x10 5 yr. O e - HCO + CO

18 5. The two-level system Collisional coefficient Einstein coefficient for absorption γ lu B lu n u ΔE n l γ ul B ul A ul Collisional deexcitation coefficient Einstein coefficient for stimulated emission Einstein coefficient for spontaneous emission Problem: find the level populations n u and n l as a function of the ambient kinetic temperature T kin and density n tot. In order for the populations of both levels to remain constant in time, we need: " lu n tot n l + B lu J n l = " ul n tot n u + B ul J n u + A ul n u Total rate of collisional excitations per unit time and per unit volume Probability per unit time of being excited radiatively J " % & 0 J # $(#)d#

19 In Local Thermodinamic Equilibrium (i.e. when radiative transitions are negligible): " lu " ul = n u n l = g $ u exp " #E ' & ) (T ex = T kin, in LTE) g l % k B T kin ( When n tot is so low that that radiative transitions dominate: J = A ul n u B lu n l B ul n u = A ul /B ul (g l B lu /g u B ul )exp(#e /k B T rad ) 1 Under the assumption of thermodynamic equilibrium (J ν =B ν ): J " B # 0 = 2h# 0 3 /c 2 exp($e /k B T rad ) %1 A ul = 2h" 3 0 B c 2 ul g l B lu = g u B ul

20 " lu n tot n l + B lu J n l = " ul n tot n u + B ul J n u + A ul n u $ exp &" #E % k B T ex ' $ ) = f coll exp " #E ' $ & ) + (1" f coll )exp&" #E ( % k B T kin ( % k B T rad ' ) ( f coll * n tot n tot + n crit (1+ c 2 J /2h+ 0 3 ) fraction of all downward transitions due to collisions n crit " A ul /# ul Excitation Temperature k B T ex /ΔE k B T kin /"E k B T rad /"E Ambient Density log(n tot /n crit )

21 Emission in the J=1 0 line of 12 C 16 O Increasing the density in a cloud can enhance the J=1-0 emission, but only for subcritical values of n tot. Observations in a given transition are most sensitive to gas with densities near the corresponding n crit.

22 Transfer of radiation in spectral lines The propagation of the specific intensity Iν is governed by the radiative transfer equation: di" = #$ " I" + j" ds Iν(0)! Iν(s) s=0 Iν(Δs) s=δs j" I" (#s) = I" (0)exp($% " #s) + [1$ exp($% " #s)] %" Stahler & Palla (2004)

23 We consider the case in which both absorption and emission are due to transitions between two discrete levels in an atom or molecule. The macroscopic emission and absorption coefficients can be written in terms of the microscopic Einstein coefficients: j " # " = j " = h" 0 4# n u A ul$(") % " = h" 0 4# (n lb lu & n u B ul )$(") n u A ul n l B lu $ n u B ul = 2h" 0 3 /c 2 exp(h" 0 /k B T ex ) $1 n u n l "(#) = 2 ln2 $ %# = g u exp ("#E /k T ex ) g l I " 0 (#$ 0 ) = I " 0 (0)e %#$ 0 + 2h" 0 3 /c 2 ( ) exp(h" 0 /k B T ex ) %1 1% e%#$ 0

24 The quantity of interest to the observer is not I ν itself, but rather the difference between I ν and the background intensity. Accordingly, we define a brightness temperature T B by: T B " c 2 2# 2 k B [I # ($% # ) & I # (0)] If we make the final assumption that the background radiation field is Planckian with an associated temperature T bg : T B0 = T 0 [ f (T ex ) " f (T bg )] [ 1" e "#$ ] 0 T B0 % T B (& = & 0 ), T 0 % h& 0 /k B f (T) % [exp(t 0 /T) "1] "1 T Bo is what we measure if we are observing the source with a perfect telescope above the atmosphere, and with an angular resolution much smaller than the source size.

25 Antenna and Brightness Temperature T A " # $ S $ A T B Brightness temperature Beam efficiency Ratio of solid angles subtended by the source and the antenna beams beam dilution factor The brightness temperature is the temperature which would result in the given brightness if inserted into the Rayleigh-Jeans law (T 0 / T<< 1 or hν<<kt, valid in radioastronomy) T B = c 2 2k 1 " 2 B "

26 Column Density " # = h# 0 4$ (n l B lu % n u B ul )&(#) "# 0 = 2 ln2 c 3 A ul (g u /g l ) 8$ 3 / 2 % 0 3 "v - ' / 1& exp )& "E. ( k B T ex * 0, 2 n l "s + 1 For the two-level system, n l ~n and n l Δs~N, the column density. Such an approximation is however not adequate for a system in which the two levels are closely spaced rungs in the ladder (e.g. CO). In these cases: Rotational partition function n = n l Q ) # Q = *(2J + 1)exp %" $ j= 0 BhJ(J + 1) k B T ex & ( '

27 The Rotational Temperature Diagram (RTD) The RTD is a customary technique to analyze the data of an individual molecule with many transitions. This method assumes LTE conditions so that a single excitation temperature (T rot ) characterizes all transitions. If the transitions are all optically thin, the RTD provides a good estimate of the column density of the molecule towards the source. Goldsmith & Langer (1999)

28 The Rotational Temperature Diagram (RTD) T A " $ # S T B T B = c 2 1 $ A 2k B " B 2 " T B0 = h" 0 /k B [ f (T rot ) # f (T bg )] [ 1# e # $% ] 0 T B0 & T B (" = " 0 ) f (T) & [exp(h" 0 /k B T) #1] #1 Now we assume that f (T rot ) >> f (T bg ). We can then write B " 0 as:

29 The Rotational Temperature Diagram (RTD) B " 0 1# e#$ = T 0 f (T rot ) $ $ (%$ 0 & $) The optical depth of the transition can be written as: Column density in the upper state " # h $v N u B ul (eh% / kt rot &1) Half-maximum line width in units of velocity Substituting these two expressions in: T B = c 2 2k B 1 " 2 B "

30 The Rotational Temperature Diagram (RTD) T B = hc 3 N u A ul 8"k B # 2 $v ' 1% e %& * ), ( & + We can now invert the above equation to yield an expression for N u. Considering the integrated line intensity, W = " T B dv ( " T B #v ) we obtain: N u = 8"k B# 2 W & $ ) ( + hc 3 A ul ' 1% e %$ *

31 The Rotational Temperature Diagram (RTD) If the molecule is in LTE: N u = N Q g u e" E u / kt Combining this with the previous expression of N u, assuming optically thin conditions, and taking the log $ log 8"k B# 2 W & % g u hc 3 A ul ' $ ) = log& ( % N ' ) * E u loge Q(T rot )( k B T rot

32 The Rotational Temperature Diagram (RTD) The previous expression shows that $ log 8"k B# 2 W ' & ) % g u hc 3 A ul ( function of E u /k B, with slope -(log e)/t rot and intercept log[n/q(t rot )] at E u = 0: is a linear $ log 8"k B# 2 W & % g u hc 3 A ul ' ) ( White et al. (2010)

33 Solving radiative transfer problems For the two level system, the statistical equilibrium equations are: dn l dt dn u dt = "n l (B lu J + # lu ) + n u (A ul + B ul J + # ul ) = n l (B lu J + # lu ) " n u (A ul + B ul J + # ul ) To solve this, we need to know the radiation field which was what we were after in the first place. This problem can be solved with some simplifying assumptions. van der Tak et al. (2007)

34 Solving radiative transfer problems: escape probability The problem is how to decouple the radiative transfer calculations from the calculations of the level populations. The escape probability β is a factor which allows us to determine the chance that a photon at some position in the cloud can escape the system (based on Sobolev 1960). We need to estimate to calculate the level population. is the amount of radiation inside the source. For a completely opaque source, = S, the source function. J J If β is the chance that a newly created photon can escape from the cloud, then J = S(1" #) J

35 Solving radiative transfer problems: escape probability Now the statistical equilibrium equations can be simplified: dn l dt dn u dt = "n l # lu + n u # ul + $n u A ul = n l # lu " n u # ul " $n u A ul And we can solve the level populations and the radiation field separately; they are decoupled. The contribution from background radiation can easily be added: - take the background intensity. The average chance that it penetrates into the source is (1 - β).

36 Solving radiative transfer problems: escape probability How to estimate β? Several forms have been proposed that depend on geometry (we have to find a form that estimates the average local escape probability over all directions). A very crude form of β in a one-dimensional case can be estimated as: " =< e #$ >= 1 $ $ & e # $ % d % 0 $ = 1# e#$ $ The form of β for a radially expanding sphere is equal to the above result. This is called the Sobolev or large velocity gradient (LVG) approximation (see also Elitzur 1992).

37 Solving radiative transfer problems: escape probability For a homogeneous slab: " = (1# e#3$ ) 3$ For a uniform sphere (Osterbrock 1974): " = # 1$ 2 # + % - ' 2 2 # + 2, & # 2 (. * e $# 0 ) / This last formula is used in RADEX (on-line code):

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39 SUMMARY H 2 molecules are formed on the surface of dust grains with a rate R H 2 = 1 2 n H v H A n g S H" # 10 $ cm -3 s -1 H H H H 3 + is formed from cosmic-ray ionization of H 2, followed by H H 2. This starts interstellar chemistry. γ lu B lu n u ΔE γ ul B ul A ul The two level system and the brightness temperature of a molecular line at frequency ν = ν 0. T B0 = T 0 [ f (T ex ) " f (T bg )] [ 1" e " #$ ] 0 n l Log N u /g u E ul /k B (K) The rotational temperature diagram allows to derive the column density and gas temperature when multiple transitions of the same molecules are observed (assuming LTE and optically thin conditions).