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Home , Astrochemistry, Molecule

School of and FACULTY OF MATHEMATICS & PHYSICAL and Molecular

Paola Caselli Outline

1. The formation of H2 + 2. The formation of H3 + 3. The initiated by H3 4. Formation and destruction of CO 5. The two-level system 6. Transfer of in spectral lines 7. The rotational diagram 8. The escape probability method Interstellar

Known Interstellar Molecules (Total: > 170)

Amino in SgrB2(N) H (Belloche et al. 2008) H C O C N

H H O H http://www.astrochymist.org/astrochymist_ismGlycine - the.html simplest amino How do molecules form in the ?

The most elementary is the association of A and B to form a AB with internal :

A + B → AB*

The molecule AB* must lose the internal energy. In the Earth , where the number of per cubic centimeter (cc) is very large (~1019), 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 ~108. 1. The formation of H2 The reaction that starts the chemistry in the interstellar medium is the one between two to form molecular hydrogen:

H + H → H2

This reaction happens on the of grains. 1. The formation of H2

-3 -1 The H2 formation rate (cm s ) 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):

1 R = n v A n S " H 2 2 H H g H # 10$17 - 10-16 cm-3s-1

n ≡ H PAHs vH≡ H atoms speed in gas-phase A ≡ grain cross sectional area n dust grain number density ! g≡ SH≡ sticking probability γ ≡ surface reaction probability Once H2 is formed, the fun starts…

H2 is the key to the whole of interstellar chemistry. Some important species that might react with H2 are C, C+, O, N… To decide whether a certain reaction is chemically favored, we need to examine internal energy changes.

Molecule Dissociation energy (eV) H2 4.48 CH 3.47 OH 4.39 CH+ 4.09 OH+ 5.10 Question: Can the following reactions proceed in the cold interstellar medium?

C + H2 → CH + H ?? O + H2 → OH + H ?? + + + + C + H2 → CH + H ?? O + H2 → OH + H ?? Once H2 is formed, the fun starts…

C + H2 → CH + H ??

4.48 eV 3.47 eV

The bond strength of H2 is larger than that of CH the reaction is not energetically favorable.

The reaction is endothermic (by 4.48-3.47 = 1.01 eV) and Dissociation cannot proceed in cold clouds, energy or bond strength where kb T < 0.01 eV ! Once H2 is formed, the fun starts…

Molecule Dissociation energy (eV) H2 4.48 CH 3.47 OH 4.39 CH+ 4.09 OH+ 5.10

C + H2 ✗→ CH + H (endothermic by 1.01 eV) + + (endothermic by 0.39 eV) C + H2 ✗→ CH + H (endothermic by 0.09 eV) O+ H2 →✗ OH + H

+ ✔ + (exothermic by 0.62 eV!) O + H2 → OH + H Rate coefficients and

The rate coefficient k (cm3 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 !

Most reactions possess activation energies Ea (~0.1-1 eV) even if exothermic and k is given by the Arrhenius (Herbst 1990):

k = a(T)exp("E a /kBT)

! -Neutral reactions

AA+ ++ BB →→ CC+ ++ DD

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 e2/2R4

1/2 kLANGEVIN = 2 πe(α/µ) ∼ 10-9 cm3 s-1 independent of T Neutral-Neutral reactions

AA ++ BCBC ABAB ++ CC Energy to break →→ the bond of the 1 eV for endothermic reactions reactant. E ∼ 0.1-1 eV for exothermic reactions k T < 0.01 eV Energy released by b the formation of in molecular clouds the new bond.

Example: O + H2 ✗→ OH + H (does not proceed in cold clouds)

Duley & Williams 1984, Interstellar Chemistry; Bettens et al. 1995, ApJ + 2. The formation of H3

After the formation of molecular hydrogen, cosmic rays ionize H2 initiating fast routes towards the formation of complex molecules in dark clouds:

+ - H2 + c.r. → H2 + e + c.r.

+ Once H2 is formed (in small percentages), it very quickly reacts + with the abundant H2 molecules to form H3 , the most important molecular ion in interstellar chemistry: H H H + + H → H + + H 2 2 3 H + 3. The chemistry initiated by H3

+ + H3 Once H3 is formed, a cascade of reactions greatly enhance the O chemical complexity of the ISM. OH+

+ H2 In fact, H3 can easily donate a and allow larger molecules + H2O to form. H2

+ H3O e e e Example  H2O OH O CHEMISTRY (the formation of in dark clouds) O

O2 + 3. The chemistry initiated by H3

CARBON CHEMISTRY (the formation of hydrocarbons)

The formation of more complicated species from neutral atomic begins with a sequence very similar to that which starts the oxygen chemistry:

CH2 H3+ H2 H2 e C CH+ CH2+ CH3+ e CH

+ A. Proton transfer from H3 to a neutral ; B. Hydrogen abstraction reactions terminating in a molecular ion that does

not react with H2; C. Dissociative recombination with . 4. Formation and destruction of CO

+ + [a] C + H3O → HCO + H2 + + [b] O + CH3 → HCO + H2 [c] HCO+ + e → CO + H is the most important source of CO.

+ CO is very stable and difficult to remove. It reacts with H3 : + + [d] H3 + CO → HCO + H2 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 H2O (but not with H2): [g] C+ + OH → CO+ + H + + [h] CO + H2 → HCO + H + + [i] C + H2O → HCO + H The timescale to form CO

Assume: dark region where all H is in H2 and all atoms more massive than He are in neutral atomic form.

The timescale on which almost all carbon becomes contained in

CO (nO > nC) is at least equal to the timescale for one H2 to be ionized for every C: nC/[ζ n(H2)] = 2 nC/[ζ nH]

-17 -1 -4 For ζ = 3×10 s and nC/nH = 10 , the above expression gives a value of about 2x105 yr.

O e-  HCO+ CO 5. The two-level system

Collisional deexcitation Collisional coefficient coefficient γ γlu ul nu Einstein coefficient for B ΔE Bul lu stimulated emission Einstein coefficient for Aul Einstein coefficient for nl

Problem: find the level populations nu and nl as a of the ambient kinetic temperature Tkin and density ntot.

In order for the populations of both levels to remain constant in time, we need:

" luntot nl + BluJ nl = " ul ntot nu + Bul J nu + Aul nu

Probability per unit time of % Total rate of collisional being excited radiatively excitations per unit time J " & J# $(#)d# and per unit volume 0 ! ! In Local Thermodinamic Equilibrium (i.e. when radiative transitions are negligible): $ ' " lu nu gu #E = = exp& " ) (Tex = Tkin , in LTE) " ul nl gl % kBTkin (

When ntot is so low that that radiative transitions dominate:

A !n A /B !J = ul u = ul ul Blunl Bul nu (gl Blu /guBul )exp(#E /kBTrad ) 1

Under the assumption of thermodynamic equilibrium (Jν=Bν): ! 3 2 2h# 0 /c J " B# 0 = exp($E /kBTrad ) %1  3 2h" 0 Aul = 2 Bul gl Blu = guBul ! c

! " luntot nl + BluJ nl = " ul ntot nu + Bul J nu + Aul nu  $ #E ' $ #E ' $ #E ' exp " = f exp " + (1" f )exp " ! & ) coll & ) coll & ) % kBTex ( % kBTkin ( % kBTrad ( n f * tot fraction of all downward coll 2 3 transitions due to collisions ntot + ncrit (1+ c J /2h+ 0 )

E Δ / 10 ex

T k T /"E B B kin

k 8 ncrit " Aul /# ul ! 6 4 ! 2 kBTrad /"E 0 0 1 2

Excitation Temperature Temperature Excitation Ambient Density log(ntot/n’crit) ! ! Emission in the J=1→0 line of 12C16O

Increasing the density in a cloud can enhance the J=1-0 emission, but only for subcritical values of ntot.  Observations in a given transition are most sensitive to

gas with densities near the corresponding ncrit. Transfer of radiation in spectral lines

The propagation of the specific intensity Iν is governed by the radiative transfer equation: dI " = #$ I + j ds " " "

I (0) I (s) I ( s) ν ! ν ν Δ s=0 s=Δs

j" I" (#s) = I" (0)exp($%" #s) + [1$ exp($%" #s)] %" Stahler & Palla (2004)

! 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: h " 0 2 ln2 j" = nu Aul$(") "(#) = 4# $ %#

h" 0 %" = (nl Blu & nuBul )$(") 4# nu gu = exp("#E /k Tex )  ! nl gl j n A 2h" 3 /c 2 " = u ul = 0 #" nl Blu $ nuBul exp(h" 0 /kBTex ) $1 ! !  2h" 3 /c 2 I (#$ ) = I (0)e%#$ 0 + 0 1% e%#$ 0 ! " 0 0 " 0 ( ) exp(h" 0 /kBTex ) %1

! The quantity of interest to the observer is not Iν itself, but rather the difference between Iν and the background intensity. Accordingly, we define a brightness temperature TB by: c 2 TB " 2 [I# ($%# ) & I# (0)] 2# kB

If we make the final assumption that the background radiation field is Planckian

with an associated temperature Tbg: ! T = T [ f (T ) " f (T )] 1" e"#$ 0 B0 0 ex bg [ ]

T T ( ), T h /k B0 % B & = & 0 0 % & 0 B "1 f (T) % [exp(T0 /T) "1]

TBo is what we measure if we are observing the source with a perfect above the atmosphere, and with an angular resolution much smaller than the source size. ! Antenna and Brightness Temperature

Brightness $S temperature TA " # TB $A

Ratio of solid angles Beam! efficiency subtended by the source and the antenna beams ≡ beam dilution factor

The brightness temperature is the temperature c 2 1 which would result in the given brightness if T B inserted into the Rayleigh-Jeans law (T / T<< 1 B = 2 " 0 2k " or hν<

! Column Density

h# " = 0 (n B % n B )&(#) # 4$ l lu u ul 

3 - ' *0 ! 2 ln2 c Aul (gu /gl ) "E "# 0 = 3 / 2 3 /1 & exp) & ,2 n l"s 8$ % 0"v . ( kBTex +1

For the two-level system, nl~n and nlΔs~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: ! n = nlQ Rotational ) # BhJ(J + 1)& partition Q = (2J + 1)exp%" ( function * j= 0 $ kBTex '

! 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 (Trot) 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) The Rotational Temperature Diagram (RTD)

c 2 1 $S T " # T TB = B" A B 2k 2 $A B "

T = h" /k [ f (T ) # f (T )] 1# e# $% 0 B0 0 B rot bg [ ] ! ! T T ( ) B0 & B " = " 0 #1 f (T) & [exp(h" 0 /kBT) #1]

B Now we assume that f ( T rot ) > > f ( T bg ) . We can then write " 0as:

! ! ! The Rotational Temperature Diagram (RTD)

1# e#$ B = T f (T ) $ (%$ & $) " 0 0 rot $ 0

The optical depth of the transition can be written as: ! Column density in the upper state

h " # N B (eh% / kTrot &1) $v u ul Half-maximum line width in units of velocity 2 c 1  Substituting these two expressions in: TB = 2 B" ! 2kB "

! The Rotational Temperature Diagram (RTD)

3 ' %& * hc Nu Aul 1% e TB = 2 ) , 8"kB# $v( & +

We can now invert the above equation to yield an expression for N . Considering the integrated line intensity, W T dv u! = " B ( " T B # v ) we obtain:

2 8"k # W & $! ) N = B ( + ! u 3 %$ hc Aul ' 1% e *

! The Rotational Temperature Diagram (RTD)

If the molecule is in LTE:

N N = g e" Eu / kT u Q u

Combining this with the previous expression of Nu, assuming optically thin conditions, and taking the log  ! $ 2 ' $ ' 8"kB# W N E u loge log& 3 ) = log& ) * % guhc Aul ( % Q(Trot )( kBTrot

! The Rotational Temperature Diagram (RTD) $ 8 k 2W ' The previous expression shows that " B # is a linear log& 3 ) % guhc Aul (

function of Eu/kB, with slope -(log e)/Trot and intercept log[N/Q(Trot)] at Eu = 0: ! $ 2 ' 8"kB# W log& 3 ) % guhc Aul (

!

White et al. (2010) Solving radiative transfer problems

For the two level system, the statistical equilibrium equations are:

dn l = "n (B J + # ) + n (A + B J + # ) dt l lu lu u ul ul ul dn u = n (B J + # ) " n (A + B J + # ) dt l lu lu u ul ul 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) 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 at some position in the cloud can escape the system (based on Sobolev 1960).

We need to estimate J to calculate the level population. J is the amount of radiation “inside” the source. For a completely opaque source, J = S, the source function.

If β is the chance! that a newly created photon! can escape from the cloud, then ! J = S(1" #)

! Solving radiative transfer problems: escape probability

Now the statistical equilibrium equations can be simplified: dn l = "n # + n # + $n A dt l lu u ul u ul dn u = n # " n # " $n A dt l lu u ul u 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). 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:

1 $ 1# e#$ " =< e#$ >= & e#$ % d$ % = $ 0 $

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). Solving radiative transfer problems: escape probability

For a homogeneous slab:

(1# e#3$ ) " = 3$

For a uniform sphere (Osterbrock 1974): !

1.5+ 2 % 2 2 ( $# . " = -1 $ 2 + ' + 2 * e 0 # , # & # # ) /

This last formula is used in RADEX (on-line code): http://www.sron.rug.nl/~vdtak/radex/radex.php !

SUMMARY

H2 molecules are formed on the surface of dust grains with a rate 1 R = n v A n S " H 2 2 H H g H # 10$17 - 10-16 cm-3s-1

H + H H3 is formed from cosmic-ray of H2, followed by + H2 + H2. This starts interstellar chemistry. H ! The two level system and the brightness temperature γ γlu ul nu of a molecular line at ν = ν0. B ΔE Bul lu A ul T = T [ f (T ) " f (T )] 1" e" #$ 0 B0 0 ex bg [ ]

nl u

/g The rotational temperature diagram allows to derive the u

N  column density and gas temperature when multiple  transitions of the same molecules are observed

Log !  (assuming LTE and optically thin conditions). Eul/kB (K)