1 Lecture 5 By Tom Wilson. 2 Dust clouds in Taurus Regions shown in gray are those where very few stars are seen (Becvar atlas)

11

Lecture 5Lecture 5

By Tom Wilson

22

Dust clouds in TaurusDust clouds in Taurus

Regions shown in gray are those where very few stars are seen (Becvar atlas)

33

The Discovery of Molecular The Discovery of Molecular CloudsClouds

-The discussion began in 1960’s when it -The discussion began in 1960’s when it was noticed that dark clouds had a was noticed that dark clouds had a deficiency of HI emission, based on an Adeficiency of HI emission, based on an Avv to HI relation obtained from studies of to HI relation obtained from studies of diffuse clouds. diffuse clouds.

Some thought this could be due to high Some thought this could be due to high optical depth in HI, or self absorption, optical depth in HI, or self absorption, etc. etc.

Barrett et al. found intense OH linesBarrett et al. found intense OH lines

44

HI in Dark CloudsHI in Dark Clouds

The dashed line is the boundary of the visible cloud

55

HI Self Absorption in Dark HI Self Absorption in Dark CloudsClouds

The broader HI emission is from a warm background

66

Molecular CloudsMolecular Clouds

The discovery of thermal OH emission from these The discovery of thermal OH emission from these clouds, and then NHclouds, and then NH33, H, H22O and HO and H22CO were found in CO were found in warmer clouds. This made it likely that the gas in warmer clouds. This made it likely that the gas in these clouds was mostly Hthese clouds was mostly H2 2

The final piece of evidence was the discovery of The final piece of evidence was the discovery of COCO

Still were problems with the abundance (even the Still were problems with the abundance (even the very presence) of complex molecules. very presence) of complex molecules.

77

--Make this molecule on grains, since must Make this molecule on grains, since must conserve energy and momentumconserve energy and momentum

-The exact production rate is known to a -The exact production rate is known to a factor of 2factor of 2

-Once made, H-Once made, H2 2 enters the gas phaseenters the gas phase

- H- H22 is dissociated by 11 eV photons is dissociated by 11 eV photons

- HH2 2 is destroyed by spectral line radiation, is destroyed by spectral line radiation, so is self shielded, and also shielded by so is self shielded, and also shielded by dust dust

- HH2 2 combines with Hcombines with H++ to form H to form H33++ which which

promotes promotes ion-moleculeion-molecule chemistry chemistry

The Life Cycle of HThe Life Cycle of H22

88

TMC-1: A special dust cloud in TaurusTMC-1: A special dust cloud in Taurus

Location of TMC-1

99

Trapeziumstars, which ionize the HII region

Bar, anEdge-onPhotonDominatedRegion

Photo From HST(O’Dell)

The molecular cloudin Orion is extended north-south,behind the HII region

OrionHot CoreA “chemistry factory” in the sky

1010

Methyl alcohol

Methyl formate

Dimethyl ether

Structures of a few interstellar molecules

These are asymmetric

top molecules which have many transitions in the

millimeter range

in warm sources

1111

CLASSIFICATIONS OF PROTOSTARS

1212

Sketch of a Photon Dominated RegionSketch of a Photon Dominated Region

From Hollenbach & Tielens 1999

1313

Lecture5 page 2Lecture5 page 2

For molecules, the electronic states are given analogous assignments:

S

PP

DD

2 electron systems

3S = “triplet S” parallel electron spins

1S = “singlet S” auto-parallel electron spins

1414

MOLECULES

Vastly more complex than hydrogen atoms. The main assumption in a discussion of molecules is Born-Oppenheimer approximation:

W W W WT O T E L V IB R O T

WEL (electronic energies): transitions with energies of a few eV

WVIB (vibrational energies): transitions have energies of 0.1 to 0.01 eV

WROT (rotational energies): transitions have energies of 0.001 eV

To first order there no mixed terms in Hamiltonian, this means that the wavefunctions are product of:

Another way to say this, is that the electrons move much fasten than the rotation of the nuclei. Thus have a negative bowl of potential energy from electrons. In this, the nuclei rotate.

(electronic) * (vibration) * (rotation)

What is the energy equivalent of 1 eV in degrees Kelvin?What is the energy equivalent of 1 eV in degrees Kelvin?

1515

MorseMorse

Potential Potential

P r D eea r re( ) 1

2

rr

P(r)P(r)

1616

Approximations for P(r) P r D eea r re( ) 1

2

Often can use 1st order expansion P r a D r re e( ) 2 2

Since for most radio lines, one is deep in potential well

ROTATIONAL SPECTRA OF DIATOMIC MOLECULES

The Hamiltonian isH O

J

Oee

1

2 22

2

╫

╫

O m r m r m r

r r r mm m

m m

J O

e A A B B e

e A BA B

A B

e

2 2 2

╫

╫

1717

,..1,01 JJJJ

Demand

Assumed perfectly rigid

Calculate the Be for 12C16O and 13C16O using the nuclear masses in

Atomic mass units (AMU), rounded to integers. Compare the values and calculate the line frequencies using h=EROT

)(2

122

1

2

2

definitionO

Bwhere

JJBO

JJ

O

JE

ee

eee

ROT

╫ ╫

╫

1818

Effects of distortion

1919

So E h B J J

h E E JH E J h B J jR O T e

R O T e

( )

( ) ( ) ( ) , , . . .

1

2 1 0 1

This is correct if molecule is perfectly rigid. If stretching, have an extra term

h D J J ( )12

2 1 4 1 3B J D Je e( ) ( )

Where D is > 0 and D / B =10-5

EXAMPLE: For 12C 16O, Be = 57.6360 GHz and De = 0.185 MHz

Use this relation given above to calculate the J = 1 – 0, 2 – 1, 3 – 2 line frequencies

2020

For linear molecules, find J

J

JJ J

2

02 1

2 31

From before A G H z J 1 1 6 5 1 0 11 3 2.

And

For total population of CO, under LTE, get

z J ek T

h Bh B k T

h B J J

k T

J ee

e

2 11

0

( )

n to ta l

z

Jn J e

T

Jn J e

h B J J

k T

h B J J

k Te e

( ) ( ).

( )( ) ( )

2 1

2 2

2 1

1 1 1

Work out the Einstein A coefficients for CS, SiO, HCO+, N2H+.

Then calculate the critical density, n*, where A = n* < v > = 10-10 n*.

Compare to n* for CO J = 1 → 0

exTk

h

l

u

l

u

Vu

l

ul

GHzl

eg

g

N

NbydefinedisT

TkhifdVTg

g

AN

331094.1

2121

SYMMETRIC AND ASYMMETRIC TOPS

is constant, and projection on ax axis

1.- SYMMETRIC TOPS – projection on molecule axis of symmetry is also a constant.

Definition: O O O O Ox y z II

So

HJ

O

J

O

J

O

J

OJ

O O

call J K

x y z

IIz

II

z

2 2 2 22

2

2 2 2 2

1

2

1

2

╫╫ ╫ ╫ ╫

╫ ╫ ╫ ╫ ╫ ╫

J 2

2222

CH3CN: Prolate symmetric Top

NH3: Oblate symmetric Top

Usually one defines

4 41

OB

OC

II╫ ╫

Then for energy eigenvalues

And the energy above the ground state is

2221 KKJJJ

)(1

1 2

JWJWh

BChKJJBhW e

CBABCKJB

BAKJB

)()1(2

)()1(22

2

2323

Ammonia Ammonia

Quantum numbersQuantum numbers

are (J,K),are (J,K),

where where

J is total angularJ is total angular

momentum,momentum,

K is componentK is component

along moleculealong molecule

axis axis

2424

Note that J J JF IN A L IN IT IA L P H O T O N

So 0 → 0 is forbidden, but J J is a llow ed for JF IN IT IA L 1

An extra feature of NH3 is inversion doubling. This is caused by the

Q. M. tunneling of N through the plane of the 3 hydrogen atoms

GHzatKJ

BCKJB

5.572)0,0()0,1(),(

12 2

2525

The inversion doubling gives rise to many lines in the centimeter range from 20 GHz to 40 GHz. These are (J,K) → (J,K) lines

Ammonia is an OblateOblate symmetric top. CH3CN and CH3C2H are ProlateProlate symmetric tops. For prolate, “C” becomes “A”.

Dipole moment along axis of symmetry of molecule. So no radiative transition involving a change in K quantum number

2.- ASYMMETRIC TOPS

Most common type of polyatomic molecule: H2CO, H2O,…

Now, only is a constant. Describe energy levels by the approximation that the asymmetric top is somewhere between a PROLATE and OBLATE symmetric top. That is between K values Ka and Kc. Levels are described by the quantum numbers

Each of the K values is less than or equal to the J value

J 2

J K Ka c

2626

J K Ka cprolateprolate oblateoblate

2727

H C O

A G H z

B G H z

C G H z2

2 8 2 1

3 8 3

3 4 0

.

.

.

Almost a symmetric top molecule, with dipole moment along A axis. In an allowed radiative transition, no charge in Ka. There will be a change in J an Kc.

So 101 → 000 (n = 72 GHz)

Can also have 110 → 111 or 110 ← 111 (6 cm, or 4.8 GHz)

H2CO is nearly a prolate symmetric top

H2O very asymmetric top, with dipole moment along B axis

FormaldehydeFormaldehyde

2828 Almost like a prolate symmetric topAlmost like a prolate symmetric top

2929

EXAMPLE: H2CO

A rotation about the A axis interchanges the 2 hydrogens. This must have an anti-symmetric wavefunction since these are FERMI particles.

So (space) * (spin) = anti symmetric

Definite relation of spin and (-1)Ka : if spins are parallel, ↑↑, then is ortho. If so, the parity of the space wavefunction must be anti-symmetric

↑↓ If spin is para, then the space wavefunction must be symmetric

A rotation about B, A, and C, must leave the molecule in the initial state. So the product of parities of the 3 rotations must be symmetric. Then for ortho: the space part is anti symmetric in B since (-1)Ka is the symmetry. Then have Ka Kc = (even)(odd).

Transitions for ortho: Since J must change in an allowed radiation transition, must connect in upper and lower states, Kc must change must change between states. between states.

Ka Kc==(e)(o)-(e)(e)

(e)(e)-(e)(o)

For para: the space symmetry of A is even, so wavefunctions must be odd. So para transitions for H2CO are (o)(o)-(o)(e)

(o)(e)-(o)(o)

3030

Allowed radiative transitions are dipole. Thus parity of initial and final states must be different:

r rf i

For i and f states the parity must be different

r: over all space r changes sign, but product be positive, so the partiy of the initial and final wavefunctions must be opposite.

Symmetry of spatial part of wavefunctions is crucial: look at H2CO

If proton spins parallel, have OrthoOrtho: the spin wavefunction is symmetric. Rotation of 180° about A axis interchanges protons. These are FERMIFERMI particles as total symmetry must be odd. Thus since the spin wavefunction is even so space wavefunction must be odd. Since the spin wavefunction varies as (-1)Ka. So when Ka = 0, the spin wavefunction is

symmetric. When the spins ↑↓ this is a para state; the Ka = 1. For parallel spins, ↑↑, ortho states, Ka must be 0, 2, 4, 6, etc…

3131

EXAMPLE: H2O

A rotation about the B axis interchanges the 2 hydrogens. This must have an anti-symmetric wavefunction since these are FERMIFERMI particles.

So (space) * (spin) = anti symmetric

Definite relation of spin and (-1)Kb : ↑↑ orthoortho space: anti-symmetric

↑↓ parapara space: symmetric

A rotation about B, A, and C, leave the molecule in the initial state. So the parities of the 3 rotations must be symmetric then orthoortho: space anti symmetric in B since (-1)Ka is the symmetry have Ka Kc = (even)(odd) or (odd)(even)

OrthoOrtho: Since J must change in an allowed radiation transition, must connect in upper and lower states (o)(e)-(e)(o)

(e)(o)-(o)(e)

ParaPara: the symmetry of B is even, so A and C wavefunctions must be both even or both odd. So para para transitions for H2O are (o)(o)-(e)(e)

(e)(e)-(o)(o)

3232

3333

MethanolMethanol Not just more complexNot just more complex Has different notationHas different notation The CHThe CH33 and OH arm can rotate relative to each and OH arm can rotate relative to each

other, giving rise to ‘internal rotation’other, giving rise to ‘internal rotation’

3434

J, TotalJ, Total

Angular Angular

momentummomentum

AngularAngular

MomentumMomentum

AlongAlong

Axis of Axis of

Symmetry Symmetry

3535

OH MoleculeOH Molecule

Not shown areNot shown are

The Far IR The Far IR

TransitionsTransitions

connecting connecting

rotationally rotationally

excited states excited states

to groundto ground

Sketch is to showSketch is to show

The effect of The effect of

Doubling Doubling

84 micron lines84 micron lines

120 micron lines120 micron lines

3636

How convert line intensity into a column density?

Use

ex

ex

GHz

Tk

h

l

u

l

u

ulGHzul

B

Tul

GHz

u

ll

eg

g

N

N

A

dVT

eAg

gN

2311

108.4

3

10165.1

1

1

(sec)5.93 2

General General

Relation Relation

exTk

h

l

u

l

u

Vu

l

ul

GHzl

eg

g

N

NbydefinedisT

TkhifdVTg

g

AN

331094.1

T is TT is Texex

Emission lines:Emission lines:

TTLL=T=Texex(1-e(1-e).).

If If <<1, T<<1, TLL=T=Texex

3737

J

J

J

2

02 1

2 3

JK

JK

J K

J JJ K J K

K

J JJ K

2

02

2 2

2

02

1

1 2 31

10 0

, ,

,

AS u b

JG H z x

1 1 6 5 1 02 1

11 3 2.

( , )

'

Asymmetric Tops: T&S Table IV or Journal of Physical and Chemical Ref. Data for S(u,b)Asymmetric Tops: T&S Table IV or Journal of Physical and Chemical Ref. Data for S(u,b)

In General:In General:

Ammonia:

CO:

3838

LVG MODEL

Assumes close coupling of molecule and radiation field. Also

V Vr

rSobo lev

0

0

This makes the radiative transfer a local problem.

Qualitative result:A

Aue

ue

Instead of an escape from cloud, after a collision the photon is reabsorbed many times. Only happens if > 1

3939

4040

The story of X

The clumps are more dense. Each is assumes to be Virially stable, and then entire cloud is stable Virially.

From LVG,

Virial theorem:

T C O n H n( ) ( )

M m R pc V km s

VM m

R pc

n R

Rn

SU N

SU N

( ) ( ) ( )

( )

( )

2 5 0

2 5 0

43

2 5 0

1 2 1

2

1 2

3

So T dV n n or n H LB ( )2

4141

Need more data to get a secure total column density

1. Could measure lots of lines of a species, make “ Boltzmann Plot”

N J etJ

E

k Tex

2 1

2. For CO, a number of models of excitation. From Mauersberger, one is

Another is

N H X T C O J dV T C O J dVM B M B( ) ( , ) . ( , )22 01 0 2 3 1 0 1 0

based on large velocity gradient model

N(H2): based on “clumpy cloud” model and a virialized cloud

)(1065.2)2( 1821 OdVCTHN MB

4242

THE END !THE END !