HIA Summer SchoolMolecular Line ObservationsPage 1 Molecular Line Observations or “What are molecules good for anyways?” René Plume Univ. of Calgary Department

HIA Summer School Molecular Line Observations Page 1

Molecular Line Observations or

“What are molecules good for anyways?”René Plume

Univ. of CalgaryDepartment of Physics & Astronomy

more detailed notes can be found athttp://www.ism.ucalgary.ca/courses/asph50

3/notes.html


Molecules in the ISM

http://www.cv.nrao.edu/~awootten/allmols.html129 molecules as of 2005

• Molecular line observations are not just stamp collecting.• Can be used to determine:- gas density- gas temperature- molecular abundances- cloud kinematics

• HOW???- for that we need to look at properties of molecules & radiation


Molecular Excitation

Molecules can rotate Can also vibrate but we will focus on rotationKinetic energy of rotation is:

€

H rot =12Iω2 =

J 2

2I

€

J = Iωsince

Angular momentum

But this is quantum mechanics so rotation is quantized Energies of rotational levels are low enough that collisions

between particles can cause excitation & de-excitationThus molecular rotation lines can be used to find physical conditions (densities & temps) in interstellar gas


A Perfectly Rigid Rotor

Solution of Schroedinger’s equation results in eigenvalues for the rotational energy

€

Erot =h2

2IeJ (J +1) = hBeJ (J +1)

where

J in the QN for the total ang momJ = 0, 1, 2, ……

€

ν(J )=1h

(Erot(J +1)−Erot(J ))=2Be(J +1)

€

Be =h

4πIe

= rotation constant (often expressed in Hz)


NON-linear Molecules

In general All 3 MoI are different called an ASYMMETRIC ROTOR or ASYMMETRIC TOPMoI are labeled as IA, IB, IC

IA < IB < IC

In some cases 2 MoI are equalcalled a SYMMETRIC ROTOR or SYMMETRIC TOP

If 2 largest MoI are equal (IB = IC) Have a prolate symmetric top Foot ball shaped molecule Linear molecule is a special case of a prolate symmetric top CH3CN

N

H

HH

H

HH

C

N

C

If 2 smallest MoI are equal (IA = IB) Have an oblate symmetric topFrisbee shaped moleculeNH3


Symmetric Top Molecules

Much of the behavior can be deduced from Classical mechanicsMolecule rotates about molecular axis with ang mom JZ

And there is a precession of this axis about the total ang mom (J)Energy of rotation is:

JZ

J

H =12

IAωA2 +

12

IBωB2 +

12

ICωC2

=J A2

2IA

+J B2

2IB

+J C2

2IC

For a PROLATE symmetric topIA < IB = IC

Using the fact that J2 = JA2 + JB

2 + JC2

H =J 2

2IB

+ J A2 1

2IA

−12IB

⎛

⎝⎜⎞

⎠⎟


Prolate Symmetric Top Molecules

Now we make the transition to quantum mechanicsJ2 and JA

2 both have eigenvalues

JZ

J

J2 =J (J +1)h2

JA2 =K 2h2

€

A =h

4πIA

€

B =h

4πIB

€

C =h

4πIC

€

E = hBJ(J +1) + (A − B)hK 2Energies =

Where

H =J 2

2IB

+ J A2 1

2IA

−12IB

⎛

⎝⎜⎞

⎠⎟

Since IA < IB A > BSecond term is positiveSo each J value corresponds to a series of 2J+1 levels lying progressively higher in energy

JA is the projection of the total J onto the molecular axisK is the quantum number associated with this projection2J + 1 values of K K = 0, ±1, ±2,…,±J

012

3

5

4

12

3

5

4

2

3

5

4J

J

J

K=0K=1

K=2

Prolate


Oblate Symmetric Top Molecules

For Oblate symmetric tops:IA = IB < IC

And J2 = JA2 + JB

2 + JC2

€

E = BhJ(J +1) + (C − B)hK 2Energies =

Where

€

W =J 2

2IB

+J C2 1

2IC−

12IB

⎛

⎝ ⎜

⎞

⎠ ⎟

Since IB < IC C < BSecond term is negativeSo each J value corresponds to a series of 2J+1 levels lying progressively lower in energy

K ladders are RADIATIVELY DECOUPLEDSo populations across K-ladders are controlled by collisionsMore on this later…..

012

3

5

4

12

3

5

4

2

3

5

4

J JJ

K=0 K=1 K=2

Oblate

€

A =h

4πIA

€

B =h

4πIB

€

C =h

4πIC

See “Townes & Schaalow”


So What? What does this give us?

Consider an Interstellar cloud…….

0

S So0

I

orB(TB)

I(bg)or

B(Tbg) TK

Basics of Radiative Transfer

For a photon traveling in a straight line….

0

S So0

dIνds

=−Κν Iν + jν

I = Specific Intensity (erg s-1 cm-2 sr-1 Hz-1)

K = Absorption coefficient

j = Emission coefficient

d = –Kds = Opacity = 0 at observer and increases toward source (if K > 0)

A measure of how far we see into the source€

= Κso

s

∫ ds

Basics of Radiative Transfer

0

S So0

The radiative transfer equation can be solved and, in LTE, can be re-written as:

B (TB) =B (Tbg)e− + B (TK ) 1−e−( )

€

Bν (T)=2hν3

c2e

hνkT −1

⎛

⎝ ⎜

⎞

⎠ ⎟ −1

Planck functionWhere:


Special cases

If << 1A Taylor expansion gives:

e- 1; therefore:

Absorption of background radiation by foreground cloud

Emission of foreground cloud at a temperature T into the beam

B (TB) =B (Tbg) + B (TK )

All background radiation is absorbed by the intervening cloud and there is only emission of foreground cloud at a temperature T into the beam

Virtually no foreground emission and no absorption of background radiationIf >> 1

e- 0; therefore: B (TB) =B (TK )

€

e−τ ≈1−τ+...

B (TB) =B (Tbg)e− + B (TK ) 1−e−( )


Radiative Transfer using Einstein A Coefficients

As before, pass radiation through a slab of thickness dsIntensity changes

3 different processes to consider:

€

dEe(ν)=hνon2A21ϕ(ν)dσdsdΩ4π

dνdtTotal amount of energy emitted spontaneously over the full solid angle 4.

€

dEa(ν)=hνon1B124πc

Iνϕ(ν)dσdsdΩ4π

dνdt Total amount of energy absorbed

€

dEs(ν) =hνon2B214πc

Iνϕ(ν)dσdsdΩ4π

dνdt Total amount of stimulated emission

Total amount of energy emitted is:

€

dIνdΩdσdνdt=dEe(ν)+dEs(ν)−dEa(ν)

=hν4π

n2A21+n2B214πc

Iν −n1B124πc

Iν⎡ ⎣ ⎢

⎤ ⎦ ⎥ ϕ(ν)dΩdσdsdνdt


Radiative Transfer using Einstein A Coefficients

So the radiative transfer equation is:

€

dIνdΩdσdνdt=dEe(ν)+dEs(ν)−dEa(ν)

=hν4π

n2A21+n2B214πc

Iν −n1B124πc

Iν⎡ ⎣ ⎢

⎤ ⎦ ⎥ ϕ(ν)dΩdσdsdνdt

€

dIνds

= −hν

cn1B12 − n2B21[ ]Iνϕ (ν ) +

hν

4πn2A21ϕ (ν )

€

dIνds

=−Κ νIν +jνRemember:

€

Κ =hν

cn1B12 − n2B21[ ]ϕ (ν )

Therefore:

€

Kν = n1

g2

g1

A21

c 2

8πν 21− e

−hν kT ⎡ ⎣ ⎢

⎤ ⎦ ⎥ϕ (ν )

After some algebra….


Column Densities from ObservationsSo, if << 1 becomes

N2 is the column density (cm-2) of particles in the upper state

€

B(TB)=τB(TK )

= Kν∫ dsB(TK )

= c2

8πν2g2

g1n1A211−e

−hνkT

⎡

⎣ ⎢

⎤

⎦ ⎥ ϕ(ν)

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ ∫ dsB(TK )

= c2

8πν2g2

g1n1A211−e

−hνkT

⎡

⎣ ⎢

⎤

⎦ ⎥ ϕ(ν)

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ s

2hν3

c21

ehν

kTK −1

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

=hν4π

N1A21ϕ(ν)g2

g1e

−hνkT e

hνkTK −1

ehν

kTK −1

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

=hν4π

N2A21ϕ(ν)

€

B(TB)=B(Tbg)e−τ +Bν (TK ) 1−e−τ

( )

If B(Tbg) = 0 of course we can always subtract B(Tbg) from the observations by chopping on the sky


Column Densities from Observations

Now to incorporate the line profile, we measure the frequency or velocity integrated intensity:

€

B(TB)=2kν2

c2TB =

hν4π

N2A21ϕ(ν)

€

TB =hc2

8kπνN2A21ϕ(ν)

So,

And therefore,

€

TB∫ dν =νc

TB∫ dV=hc2

8kπνN2A21 ϕ(ν)∫ dν

€

TB∫ dV=hc3

8kπν2N2A21 ϕ(ν)∫ dν

€

N2 =8kπν2

hc31

A21TB∫ dV

Normalized to unity

€

Bν (T ) =2hν 3

c2

1

ehν

kT −1≈

2hν 3

c2

1

1+hνkT

−1≈

2kν 2

c2TBIf h << kT



For optically thin emission (all photons created escape cloud) for a single transition:

∫TBdv = integrated intensity of the line

Aul = spontaneous emission coefficient

= frequency of the given transition

Nu = fuNtot fu = fraction in upper state

Ntot = total column density

€

Nu (cm−2 ) =8πkν 2

Aulhc3

TBdv∫directly from observables!

want this



Fraction in the upper state given by the partition function.

€

f(T)= gie−

EikT

i=0

∞∑

€

ntot= nii=0

∞∑ =

no

go

gie−

EikT

i=0

∞∑ =

no

go

f(T)

€

nu

no=

gu

goe

−EuokTBut….

€

ntot=nu

gu

gie−

EikT

i=0

∞∑

e−

EuokT

So….

So to calculate the total column density of say 13CO J = 2-1:

€

Ntot=N2

5(1+3e

−5.29T +5e

−15.88

T +7e−

31.72T +...)

e−15.88

T

€

N2 =8kπν2

hc31

A21TB∫ dV

See the file: example_column_den.doc for a worked example


Why do we give a #$&^%!?

Dickens et al. 2000, ApJ 542, 870L134N

So for every molecule we observe we can make a map of the column density (abundance)

But we often find that the distribution of different chemical species is different…

WHY?Not all related to excitationMany species/line have similar excitation requirementsTex and ncrit

Only way to understand the complex distribution of molecular species is to apply both physics & chemistryto understand why the abundances vary with positionAge, temp, dust properties, etc.

So let’s look at interstellar chemistry for a moment….


Formation of Molecules

In any gas, atoms can chemically interact to form molecules

Given table above and the fact that He & Ne are chemically inert

Expect molecules with H, C, N & O to be most abundantMost common is H2

Element Atomic Number

Abundance

H 1 1

He 2 0.1

C 6 4x10-4

N 7 10-4

O 8 9x10-4

Ne 10 10-4

Cosmic elemental abundances


Gas-Phase Chemical Reactions

€

A + B → M + N

€

M + X →Y + Z

Formation of molecule M

Destruction of molecule M

Reaction rate = kf (cm3 s-1)

Reaction rate = kd (cm3 s-1)

€

M +CR → P +QCR-destruction of molecule MReaction rate = cr (s-1) (related to CR flux)

Photo-destruction of molecule M

Reaction rate = uv (s-1) (related to UV flux)

M +γ → P +Q


Gas-Phase Chemical Reactions

So the rate of change of the Abundance of molecule M is given by:

• In reality there will be a reaction network of 1000’s of reactions and you need to solve for the abundance change in each simultaneously • via a series of stiff differential equations• I.e. UMIST data base

- 4000 reactions coupling 400 species

d

dtn(M ) =kfn(A)n(B)−n(M ) CR(M ) + UV (M ) + kdn(X)[ ] cm−3 s−1


Time dependent chemical evolution

Bergin, Langer & Goldsmith 1995, ApJ, 441, 222

Log time

So you can begin to see why different species have different abundance distributions. It depends on the chemical history of the core


Gas - Grain Interactions

Dust grains are also important in chemistry

Primarily responsible for the formation of H2

In addition, all species can interact with grains gas-phase species can accrete (adsorb or freeze-out) onto dust grains

And thus be removed from gas phase

species attached to grain surfaces may react with one another

Forming new speciesWhich can later be ejected back into the gas phase (desorption or evaporation)


Bergin, Langer & Goldsmith 1995, ApJ, 441, 222

Log time


Example of grain freeze out - H2O & O2

Steady state chemical models predict high H2O and O2 abundances (since the 70’s)

Maréchal et al. 1997,A&A,324,221

ObservedAbundanceranges


Example of freeze out - O2 & H2O Abundances

Why are O2 and H2O abundances so low?

Freeze out onto dust grainsFollowed by chemical reactions on the surface of grains

Roberts & Herbst 2002, A&A, 395, 233


Surface Migration & surface reactions

Once adsorbed onto a grain surface the species does not just sit in one spot

It can migrate from binding site to binding site via quantum tunneling through the potential wells that separate each binding site for light species (H, D, etc.)

Timescale for H tunneling is 1.5x10-10 s via thermal hopping for heavier species

In a typical grain, H will visit all 2x106 binding sites in ~ 10-4 seconds thus it will visit each binding site many times before evaporating

heavier atoms will visit each binding site in ~ 100 hours (again less than the evaporation timescale)

So during this migration, species can encounter one another at various binding sites and react Forming new species which can eventually evaporate back into gas phase the best way to produce observed abundances of certain species like H2CO and CH3OH


Hot Cores - an example of Grain surface reactions

Rodgers & Charnley 2003, ApJ, 585, 355

Chemical abundances in a collapsing envelope as a function of distance from a protostar (after 105 years)


No Really…Why Do We Care??

Distribution and abundance of molecules critical to understand chemistry

Understanding chemistry is important in understanding: ionization fraction

Controls magnetic support of cloud against collapse

thermal balance Controls thermal support of cloud against collapse


Molecular Cloud Cooling

Goldsmith & Langer 1978, ApJ, 222, 881

Speaking of temperature…how do we measure the temperature in a molecular cloud?


Temperatures from Observations

One way is to take a transition from a linear molecule that is:

Optically thickLow energyIn LTEApply radiative transfer and use the R-J limit

B (T ) =2h 3

c2

1

eh

kT −1≈2h 3

c2

1

1+hkT

−1≈2k 2

c2 TSince:

And if >> 1: B (TB) =B (TK )

So:TB =TK


Temperatures from Observations (Better Way)

Remember, for a symmetric top molecule

2 principal moments of inertia are equalRotational energy levels described by 2 QNJ - the total angular momentum

K - the projection of J along axis of symmetry

We get K-ladders

012

3

5

4

12

3

5

4

2

3

5

4J

J

J

K=0K=1

K=2012

3

5

4

12

3

5

4

2

3

5

4

J JJ

K=0 K=1 K=2

Prolate Oblate

Since there is no dipole moment perpendicular to symmetry axis

There are no radiative transitions across K laddersK ladders connected ONLY through collisionsPopn of one K ladder wrt another should reflect a thermal distribution at the kinetic temperature


Methyl Acetylene

Can only decay radiatively within a given “K” ladder

populations across “K” ladder will reflect gas temperature


Temperature of Molecular Hydrogen Gas

K = 0

K = 1

K = 2



Again assuming optically thin emissionIntegrated intensity is proportional to column density in upper state

€

Nu =8kπν2

hc31

AulTB∫ dV

In LTE

€

Nu

Nl=

gu

gle−Eul

kTex

In general, Tex will be different for different pairs of levelsBut if populations of ALL levels are in LTE we get our old friend:

Ntot

hc3

f 8k 2 guAule−Eu

kT = TB∫ dV

NtotCe−Eu

kT = TB∫ dV



NtotCe−Eu

kT = TB∫ dV

elnCNtot e−Eu

kT =e−EukT + lnCNtot =e−Eu

kT +b = TB∫ dV

ln e−Eu

kT +b⎛⎝

⎞⎠ =ln TB∫ dV( )

−EukT + b = ln TB∫ dV( )

m x y


Temperature of Molecular Gas

Observations!

See the file: example_temp.doc for a worked example


Temperature Distributions

Observations!

Temperature Map of the

Orion Molecular Ridge

Documents

HIA Summer SchoolMolecular Line ObservationsPage 1 Molecular Line Observations or “What are molecules good for anyways?” René Plume Univ. of Calgary Department