1. 1 Lecture 5 By Tom Wilson

2. 2 Dust clouds in Taurus Regions shown in gray are those where very few stars are seen (Becvar atlas)

3. 3 The Discovery of Molecular Clouds The Discovery of Molecular Clouds -The discussion began in 1960’s when it was noticed that dark clouds had a deficiency of HI emission, based on an A v to HI relation obtained from studies of diffuse clouds. Some thought this could be due to high optical depth in HI, or self absorption, etc. Barrett et al. found intense OH lines

4. 4 HI in Dark Clouds HI in Dark Clouds The dashed line is the boundary of the visible cloud

5. 5 HI Self Absorption in Dark Clouds The broader HI emission is from a warm background

6. 6 Molecular Clouds Molecular Clouds The discovery of thermal OH emission from these clouds, and then NH 3, H 2 O and H 2 CO were found in warmer clouds. This made it likely that the gas in these clouds was mostly H 2 The discovery of thermal OH emission from these clouds, and then NH 3, H 2 O and H 2 CO were found in warmer clouds. This made it likely that the gas in these clouds was mostly H 2 The final piece of evidence was the discovery of CO The final piece of evidence was the discovery of CO Still were problems with the abundance (even the very presence) of complex molecules. Still were problems with the abundance (even the very presence) of complex molecules.

7. 7 - Make this molecule on grains, since must conserve energy and momentum -The exact production rate is known to a factor of 2 -Once made, H 2 enters the gas phase - H 2 is dissociated by 11 eV photons -H 2 is destroyed by spectral line radiation, so is self shielded, and also shielded by dust -H 2 combines with H + to form H 3 + which promotes ion-molecule chemistry The Life Cycle of H 2

8. 8 TMC-1: A special dust cloud in Taurus Location of TMC-1

9. 9 Trapezium stars, which ionize the HII region Bar, an Edge-on Photon Dominated Region Photo From HST (O’Dell) The molecular cloud in Orion is extended north-south, behind the HII region Orion Hot Core A “chemistry factory” in the sky

10. 10 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

11. 11  CLASSIFICATIONS OF PROTOSTARS  

12. 12 Sketch of a Photon Dominated Region From Hollenbach & Tielens 1999

13. 13 Lecture5 page 2 For molecules, the electronic states are given analogous assignments: S  P  D  2 electron systems 3 S = “triplet S” parallel electron spins 1 S = “singlet S” auto-parallel electron spins

14. 14 MOLECULES Vastly more complex than hydrogen atoms. The main assumption in a discussion of molecules is Born-Oppenheimer approximation: W EL (electronic energies): transitions with energies of a few eV W VIB (vibrational energies): transitions have energies of 0.1 to 0.01 eV W ROT (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?

15. 15 MorsePotential r P(r)

16. 16 Approximations for P(r) Often can use 1 st order expansion Since for most radio lines, one is deep in potential well ROTATIONAL SPECTRA OF DIATOMIC MOLECULES The Hamiltonian is ╫ ╫ ╫ ╫

17. 17 Demand Assumed perfectly rigid Calculate the B e for 12 C 16 O and 13 C 16 O using the nuclear masses in Atomic mass units (AMU), rounded to integers. Compare the values and calculate the line frequencies using h =E ROT ╫╫ ╫

18. 18 Effects of distortion

19. 19 So This is correct if molecule is perfectly rigid. If stretching, have an extra term Where D is > 0 and D / B =10 -5 EXAMPLE: For 12 C 16 O, B e = 57.6360 GHz and D e = 0.185 MHz Use this relation given above to calculate the J = 1 – 0, 2 – 1, 3 – 2 line frequencies

20. 20 For linear molecules, find From before And For total population of CO, under LTE, get Work out the Einstein A coefficients for CS, SiO, HCO +, N 2 H +. Then calculate the critical density, n*, where A = n* = 10 -10 n*. Compare to n* for CO J = 1 → 0

21. 21 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: So ╫╫╫╫╫ ╫╫╫╫╫╫

22. 22 CH 3 CN: Prolate symmetric Top NH 3 : Oblate symmetric Top Usually one defines ╫╫ Then for energy eigenvalues And the energy above the ground state is

23. 23 Ammonia Quantum numbers are (J,K), where J is total angular momentum, K is component along molecule axis

24. 24 Note that So 0 → 0 is forbidden, but An extra feature of NH 3 is inversion doubling. This is caused by the Q. M. tunneling of N through the plane of the 3 hydrogen atoms

25. 25 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 OblateProlate Ammonia is an Oblate symmetric top. CH 3 CN and CH 3 C 2 H are Prolate 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: H 2 CO, H 2 O,… 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 K a and K c. Levels are described by the quantum numbers Each of the K values is less than or equal to the J value

26. 26 prolate oblate

27. 27 Almost a symmetric top molecule, with dipole moment along A axis. In an allowed radiative transition, no charge in K a. There will be a change in J an K c. So 1 01 → 0 00 (n = 72 GHz) Can also have 1 10 → 1 11 or 1 10 ← 1 11 (6 cm, or 4.8 GHz) H 2 CO is nearly a prolate symmetric top H 2 O very asymmetric top, with dipole moment along B axis Formaldehyde

28. 28 Almost like a prolate symmetric top

29. 29 EXAMPLE: H 2 CO 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 K a K c = (even)(odd). must change between states. Transitions for ortho : Since J must change in an allowed radiation transition, must connect in upper and lower states, K c must change between states. = K a K c =(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 H 2 CO are (o)(o)-(o)(e) (o)(e)-(o)(o)

30. 30 Allowed radiative transitions are dipole. Thus parity of initial and final states must be different: 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 H 2 CO Ortho FERMI If proton spins parallel, have Ortho: the spin wavefunction is symmetric. Rotation of 180° about A axis interchanges protons. These are FERMI 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 K a = 0, the spin wavefunction is symmetric. When the spins ↑ ↓ this is a para state; the K a = 1. For parallel spins, ↑↑, ortho states, K a must be 0, 2, 4, 6, etc…

31. 31 EXAMPLE: H 2 O FERMI A rotation about the B axis interchanges the 2 hydrogens. This must have an anti-symmetric wavefunction since these are FERMI particles. So  (space) *  (spin) = anti symmetric ortho Definite relation of spin and (-1) Kb : ↑↑ ortho space: anti-symmetric para ↑↓ para space: symmetric ortho 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 ortho: space anti symmetric in B since (-1) Ka is the symmetry have K a K c = (even)(odd) or (odd)(even) Ortho Ortho: Since J must change in an allowed radiation transition, must connect in upper and lower states (o)(e)-(e)(o) (e)(o)-(o)(e) Para para Para: the symmetry of B is even, so A and C wavefunctions must be both even or both odd. So para transitions for H 2 O are (o)(o)-(e)(e) (e)(e)-(o)(o)

32. 32

33. 33 Methanol Not just more complex Has different notation Has different notation The CH 3 and OH arm can rotate relative to each other, giving rise to ‘internal rotation’ The CH 3 and OH arm can rotate relative to each other, giving rise to ‘internal rotation’

34. 34 J, Total Angularmomentum AngularMomentumAlong Axis of Symmetry

35. 35 OH Molecule Not shown are The Far IR Transitionsconnectingrotationally excited states to ground Sketch is to show The effect of  Doubling 84 micron lines 120 micron lines

36. 36 How convert line intensity into a column density? Use GeneralRelation  T is T ex Emission lines: T L =T ex (1-e  ). If  <<1, T L =T ex 

37. 37 Asymmetric Tops: T&S Table IV or Journal of Physical and Chemical Ref. Data for S(u,b) In General: Ammonia: CO:

38. 38 LVG MODEL Assumes close coupling of molecule and radiation field. Also This makes the radiative transfer a local problem. Qualitative result: Instead of an escape from cloud, after a collision the photon is reabsorbed many times. Only happens if  > 1

39. 39

40. 40 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: So

41. 41 Need more data to get a secure total column density 1. Could measure lots of lines of a species, make “ Boltzmann Plot” 2. For CO, a number of models of excitation. From Mauersberger, one is Another is based on large velocity gradient model N(H 2 ): based on “clumpy cloud” model and a virialized cloud

42. 42 THE END !