1. Mellinger Lesson 6 molecular line & clouds Toshihiro Handa Dept. of Phys. & Astron., Kagoshima University Kagoshima Univ./ Ehime Univ. Galactic radio astronomy

2. Mellinger Physical state of Int.-stellar gas categorytemperaturedensitymajor objects molecular gas20K>100 cm -3 molecular clouds atomic gas100K1 cm -3 WNM, CNM ionized gas6000-10000K100 cm -3 HII regions ionized gas10 6 K<0.01 cm -3 coronal gas expansion Gravitational collapse Phase change Pressure equilibrium Radiative cooling (very slow) SNR heating

3. Mellinger Molecular clouds ▶ It is a “interstellar molecular gas cloud”. ■ Condensation of IS gas mainly composed by H 2 ■ The most dense part of the ISM ▶ Non-equilibrium with surrounding gas ■ Self gravity works efficiently. ■ Self-gravity ＞ gas pressure gives contraction(?)

4. Mellinger Prove of molecular clouds(1) ▶ Hydrogen molecules H 2 ■ Symmetric 2 atom molecule is  =0 ■ No electric-dipole emission! ■ No radio emission for rotational transition ▶ Q:What’s the next? A ： CO ■ Abundance is not well fixed. ■ CO/H 2 ~10 -4

5. Mellinger Prove of molecular clouds(2) ▶ Problem on CO ■ Saturate with even a moderate column density ▶ Isotope molecules of 12 C 16 O ■ 13 CO, C 18 O ▶ Other molecules ■ CS, HCO +, HCN, etc.

6. Mellinger Mol. lines : rotational transition ▶ diatomic molecule ▶ Electronegativity difference ■ C=2.55, O=3.44 Pauling electronegativity ▶ Mass difference ■ C=12, O=16 ▶ Cnt of electr. distr. ≠ Mass center ■ Electric dipole moment ■ Rot. gives charge vib.→radio emission C O axis ： mass center

7. Mellinger Rot. trans. of diatomic molecule(1) ▶ Model: rotation of an electric dipole ■ Rotation quantum number J ■ Energy level E J =hBJ(J+1) ■ Allowed transition is only  J=±1 ■ Line frequency n J+1,J =E J+1 -E J =2B(J+1) A J+1,J =(64  4 n 3 )/(3hc 3 )|  J+1,J | 2 =(64  4 n 3 )/(3hc 3 ) (J+1)/(2J+3)  2

8. Mellinger Rot. trans. of diatomic molecule(2) ▶ Absorption coefficent   =(8  3 )/(3hc) (J+1)/(2J+1)  2 n J {1-exp[(-h )/(kT ex )]}  ( ) ▶ Assume LTE, total number of molecules are n J =n g J exp{-E J /(kT ex )}/ Q ▶ Dist. funct.: Q=  (2J+1) exp{-E J /(kT ex )} =∫(2J+1)exp{-hBJ(J+1)/kT ex } dJ = kT ex /(hB)

9. Mellinger Rot. trans. of diatomic molecule(3) ▶ Absorption coefficient   =(8  3 B )/(3kT ex c) (J+1)  2 n exp[(-hBJ(J+1)/(kT ex )] {1-exp[(-h )/(kT ex )]}  ( ) ▶ If Tex=const on the line-of-sight,   =N(4  3 2  2 )/(3kT ex c) exp[(-h  J)/(2kT ex )] {1-exp[(-h )/(kT ex )]}  ( ) where we used h =2hB(J+1).

10. Mellinger Rot. trans. of diatomic molecule(4) ▶ Use d  =(  /c) d v (Doppler effect)   =N(4  3 2  2 )/(3kT ex  v ) exp[(-h  J)/(2kT ex )] {1-exp[(-h )/(kT ex )]} ■ solve for N, N  =   (3kT ex  v ) /(4  3 2  2 ) exp[(h  J)/(2kT ex )] {1-exp[(-h )/(kT ex )]} -1

11. Mellinger Rot. trans. of diatomic molecule(5) ▶ Complex calc due to nonlinearity, Invalid RJ appr. h /k =5.5 [K] ～ T ex =20 [K](for CO) Use R-J equiv. temp. J(T)=h /k{exp(h /kT)-1} -1 We cannot neglect CMB ▶ Line intensity  T =[J(T ex )-J(T BG )][1-exp(-  )] ▶ From this we got , which gives N. N 13CO(1-0) [cm -2 ] =2.5x10 14  13CO T ex  v [K km s -1 ] {1-exp(- 5.29/T ex [K]} -1 for 13 CO(1-0) line

12. Mellinger Molecular line in a mol. cloud ▶ Collisional excitation ■ Equilibrium between collision and line emission ▶ Two level mode dn 1 =n 2 A 21 -n 1 B 12 I+n 2 B 21 I-n 1 C 12 +n 2 C 21 n=n 1 +n 2 total number is const. ▶ Solve under steady state with dn 1 =0 n2n2 n1n1 C 12 C 21 B 21 B 12 A 21

13. Mellinger Consider the extreme case ▶ When collision term neglect (C 12 = C 21 =0) I=(A 21 /B 21 )/[(n 1 /n 2 ) (B 12 /B 21 )-1] ■ The same as derivation of Einstein coefficients. ▶ When radi. part neglect A 12 = B 12 = B 21 =0 ） n 2 /n 1 =C 12 /C 21 ■ Due to frequent collision, thermal equiv. under T k ■ T k : kinetic temperature n 2 /n 1 =C 12 /C 21 =( g 2 / g 1 )exp[-(h )/(kT k )] ■ With Einstein coefficients it gives…

14. Mellinger Reduction of coefficients ▶ Derived equation n 2 /n 1 =( g 2 / g 1 ){(c 2 /(2h  3 )I A 21 +C 21 exp[-(h )/(kT k )]} /{A 21 [1+ c 2 /(2h  3 ) I]+C 21 } ■ Show I with T r using Planck function formally ■ Show n 2 /n 1 with T ex using Boltzmann distr. ▶ They give… exp[-(h )/(kT ex )]=[A 21 /{exp(-h /kT r )-1}+C 21 exp(-h /kT k ) /{A 21 exp(h /kT r )/[exp(h /kT r )-1]+C 21 }

15. Mellinger Consider the extreme case Equation given on the previous page exp[-(h )/(kT ex )]=[A 21 /{exp(-h /kT r )-1}+C 21 exp(-h /kT k ) /{A 21 exp(h /kT r )/[exp(h /kT r )-1]+C 21 } ■ When radiation dominant (A 21 ≫ C 21 ) T ex →T r ■ When collision dominant (A 21 ≪ C 21 ) T ex →T k ■ Weak radiation approximation (I=0) n 2 /n 1 =[n 2 /n 1 ] Bol (A 21 /C 21 +1) -1 [n 2 /n 1 ] Bol : Boltzmann distr. with T k

16. Mellinger Critical density(1) ▶ Classical collision model: C 21 C 21 =n(H 2 )  ▶ no line with small n 2 /n 1 n 2 /n 1 =[n 2 /n 1 ] Bol (A 21 /C 21 +1) -1 ■ Critical value is given by A 21 <C 21 ■ n(H 2 )>A 21 /(  )=n(H 2 ) crit : critical density ▶ Easy misunderstanding ■ “strong line with large A coefficient” is false.

17. Mellinger Critical density(2) ▶ In the case of CO(J=1-0) ■ A 10 =7.203×10 -8 s -1,  ~10 -15 cm 2 ■ If T k ~20K, ~0.5 km s -1 ■ They gives n(H 2 ) crit, CO(1-0) ~10 3 cm -3 :crit. density ▶ In the case of CO(J=4-3) ■ With A 43 =( 43 / 10 ) 3 A 10 =6.4×10 -6 s -1, we get ■ n(H 2 ) crit, CO(4-3) ~10 5 cm -3

18. Mellinger Critical density(3) ▶ Appropriate line to address typical density ■ high density tracer CS, HCN, HCO + CO(4-3), CO(3-2) NH 3 ▶ Molecular gas without any line emission ■ Very less dense gas may exist. ■ Candidates of baryonic dark matter=dark gas

19. Mellinger Multi-line observations(1) ▶ LTE approximation ■ T ex is constant between any two levels ■ Line intensities differ due to  T B =T ex (1-e -  ) ■ Compare lines with  ≫ 1 and  ≪ 1 T B,thick =T ex, T B,thin =T ex , ▶ Optical depth from intensity→column density ▶ Optically thick line→excitation temperature

20. Mellinger Multi-line observations(2) ▶ Multi-levels (allow  j=±1: diatomic mol.) dn j =n j+1 A j+1,j -n j B j, j+1 I j+1,j +n j+1 B j+1,j I j+1,j -n j C j,j+1 +n j+1 C j+1,j n=  n j total number is const. ■ Solve it under steady state dn j =0 ▶ Change of I j+1,j :simliar to the 2 level model   = (h  )/(4  )  ( ) n j A j+1,j   = (h  )/(4  )  ( ) (n j B j,j+1 -n j+1 B j+1,j ) ▶ Change of intensity dI =(   –   I )dx ■ Depend on the large scale structure of the cloud