the interstellar medium – ism extragalactic nguido/extra_06.pdf1 the interstellar medium – ism...

1

The Interstellar Medium – ISMExtragalactic N.6

Before we pass to the study of external galaxies and Clusters of Galaxies we should probably consider in some details the Interstellar Medium. This because much of what we know about it is related to the observations within our Galaxy and because what we learn about it will also be useful in the study of external Galaxies.I first make sure, however, that some general and very simplified concepts are known to everyone.

2

Spectroscopic Notationn 1 2 3l 0, 1, 2, ....n-1 => 2 (2 l+1) closed shelll s, p, d, fs ≤ ½& with several electrons:L = S l S = Ss |L-S| § J § L+SL ¥ S Term splits in 2 S+1 levels with different J

QuartetTripletDoubletSing let43212S+1

Etc1/2, 3/2, 5/2, 7/21,2,33/2, 5/22D21/2, 3/2, 5/20,1,21/2, 3/21P13/211/20S0J (S=3/2)J (S=1)J (S=1/2)J (S=0)TermL

3

More Details – Ground Configuration

H 1s 1 proton – 1 electron He 1s2 2Li 1s2 2s 3Be 1s2 2s2 etcB 1s2 2s2 2pC 1s2 2s2 2p2

N 1s2 2s2 2p3

O 1s2 2s2 2p4

F 1s2 2s2 2p5

Ne 1s2 2s2 2p6

The letters S, P, D, F are related to the appearance the series had on a photographic plate. Principal, Sharp, Diffuse and Fundamental.

= See Transition rules in Physics or a summary could be prepared.

4

Spectroscopic Notation - HydrogenMultiplicity=2 s + 1

232

2 PPrincipal quantum Number

Orbital Angular MomentumSpin Angular Momentum

l = 0 1 2 3 4 to n-1S P D F G

Total Angular Momentum

n2

32

52

12

21

2

1 1s for l 0 on

3l 1 ; J l s 2 P2

n 2 Fine Structure 5 10 eV1l 0; J l s 2 S21l 1; J l

y

P

2 2

s 22

l−

= = + = ⇒

= ⇒ = = + = ⇒

= = + = ⇒

= ± =

21

21 S

32

6n 1 Hyperfine St. l 0 E 5.8 10 eV∆ −↑ ↑= = =

↑ ↓1

5

Transitions n=3 to n=2 => Hα 6563 A

5 3 3 1 12 22 2 2

3 1 12 22

n 33 3 51 1l 0,1,2 ; l s 2 2 2 2 2

D D P S P

n 2P S P

=

⇒ = + ⇒

=

P1/2S1/2

P3/2D3/2

D5/2

The student discuss whether we can observe the fine structure in celestial objects.

P3/2

P1/2

S1/2

6

Ortho and Para Helium

Proton

Neutron

He3He4 0.00013%

Ortho-Helium Electron spins parallel

S=1 J=L≤S Triplet L-1, L, L+1

Para-Helium Electron spins anti-parallel

S=0 J=L Sing lets

7

Result

• The fine structure shown in the previous slide has been confirmed with the detection of the triplets.

• Very low transition probability between sing lets and triplets. The “spin flip” transition is a very low probability process compared to the electric dipole transition through which almost all optical spectra are produced.

• Level diagrams:– Hydrogen and series.– He I– He II

• Transition in the X ray domain – innermost shells

8

Einstein Transition Prob. & Black Body

• Uν Density of Energy (per unit volume and frequency)• N2 Number of atoms at level 2• N1 Number of atoms at level 1• A21 Spontaneous Emission• Uν B21 Induced Emission• Uν B21 Induced Absorption

( )( )2 1E E h

1 kT kT21 21 2 12 1

2

21 21h

1 kT12 21 12 21

2

Equilibrium :

NA U B N U B N e eN

A AU NB B B e BN

ν

ν ν

ν ν

−

+ = = =

= =− −

9

( )

( )

x

221

0 3

12 12 21

321

12 21 312

3

h3kT

3

h2kT

Case 0 Rayleigh & Jeans RJ approximation e 1 x ...

A 8 kTU RJh cB B BkT

I must have :A 8 hB B &B c

8 h 1U Recov er Planck equation for Uc e 1

8 h 1Bc e 1

ν

ν νν

ν ν

ν

πνν

π ν

π ν

π ν

→

→ = + +

= = =+ −

= =

= ⇒−

=−

10

A nicer way Using Transfer equation

τν=0• Radiation passing through a slab

ds

( ) ( ) ( ) [ ] ( )

dI j ds k ds IdI j ds I define optical depth d k ds

k ds k dsdI j Id k

dIDefine source function S I B in Therm. eq.d

solution

I 0 I e S 1 e for S 0 I eν ν ν

ν ν ν ν

ν νν ν

ν ν

ν νν

ν

νν ν ν

τ τ τν ν ν ν ν ν ν

ρ ρρ τ ρ

ρ ρ

τ

τ

τ τ− − −

= −

= − ⇒ = −

= − +

= − − =

= + − = =

s=0

11

The slab and two atomic states

Balance in Energy

2 21 2 21 1 12

Below the balance equation

Emission Stimulated Emission AbsorptionRadiation Fi

h h hn A n B I n B I4 4 4ν νν ν νπ π π

( )2 21 1 12 2 21

eldI

dI h n A n B n B Ids 4

Transfer EquationdI I Sd

ν

νν

νν ν

ν

νπ

τ

= − −

− = −

12

( ) ( )

( )

( )

2 21

1 12 2 211 12 2 21

1 12 2 21

2 21

1 12 2 21

h 32 2 kT

h21 1 kT

dI n AIh n B n Bds n B n B4

d ds n B n Bn AS & Ther mod inamic Equil.

n B n B

n g 2h 1e ; S Bn g c

e 1

νν

ν

ν

ν

ν ν ν

νπ

τ

ν−

+ =−−

= − −

=−

= = =−

13

( )3

2 21 21h2

1 12 2 21 1kT12 21

2

321

h h21kT kT

12 212

112 21

23

21h2 hk

3

T kT21

21 212

n A A2h 1B Sc n B n B ne 1 B B

n

A2h 1 must bec ge 1 e B B

gg B B and I haveg

A2h 1c

e 1 B e 1

2hA Bc

ν νν

ν ν

ν ν

ν

ν

ν ν

= = = =−

− −

= ⇒− −

= ⇒

=

− −

=

⇒

14

Note

• In some text books you might find a different derivation and relation between the Einstein coefficients.

• Instead of using Iν in the derivation you may use Uν as you mught find in various textbooks and in slide 7. Here we prefer to use the notation as given in Rybicki and Lightman.

• If I use density of radiation instead of using Iν we should use 4πIν/c.

• That is pay attention also on the units of the Einstein coefficients.

15

21 cm 1420.4 MHz

S

N

S

N

p

e

Spin flips

N

S

N

S

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Characteristics

• High Energy State 1/2+1/2 F=1• # of states 2 F +1 = 3 g=3

• Low Energy State 1/2 + -1/2 =0 F=0• # of states 2 F+1 = 1 g=1

• Separation between levels:• 6 10-6 eV ª T=0.07 K ª 1420.4 MHz ª 21.105 cm• Transition probability• 2.869 10-15 s-1

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( ) ( )

( ) ( ) ( )

0 0

0

0

dI I S *ed

dI e I e S e I e A & S e Bd

dA B dA Bdd

A 0 A Bd

I I e S e d for S const

I 0 S 1 e I e

ν

ν ν ν ν ν

ν ν

ν

νν ν

ν ν

τνν ν

ν

τ τ τ τ τνν ν ν ν

ν

ντ τν

τ

ν ν

ττ τν ν ν ν ν

τ τν ν ν ν

τ

τ

ττ

τ τ

τ

τ

−

− − − − −

− −

− −

= −

= − = =

= − ⇒ = −

− =

= + ⇒ =

= − +

∫ ∫

∫∫

I detect

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Optical Thin & Thick

( ) ( ) ( ) ( )( ) ( ) ( )

( )

0 0lim I 0 lim S 1 e I e

S 1 1 I 1

S I

ν ν

ν ν

τ τν ν ν ντ τ

ν ν ν ν ν

ν ν ν ν

τ

τ τ τ

τ τ

− −

→ →= − + =

= − − + − =

+

( ) ( ) ( ) lim I 0 lim S 1 e I e Sν ν

ν ν

τ τν ν ν ν ντ τ

τ− −

→∞ →∞= − + =

19

Brightness Temperature

( ) ( ) ( )

( )

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

3 2

h2 2kT

2

2

2

B 2

B B

I 0 S 1 e I e S B

2h 1 2kTB for h kTc c

e 12kTI 0 B 1 e I e 1 e I e

ccDefine T I

2kT 0 T e 1T e

ν ν

ν ν ν ν

ν ν

τ τν ν ν ν ν ν

ν ν

τ τ τ τν ν ν ν ν ν

ν

τ τν

τ

ν νν

ντ τ

ντ

− −

− − − −

− −

= − + ≡

= ⇒ ≈−

= − + = − +

≡

= + −

20

Reading the Equation

• We have a gas at a certain Temperature at the optical depth τ corresponding to a distance from the observer (or from the surface of the star) –s.

• Assume the radiation emitted is similar to that of a Black Body I can substitute I with T simply because I define as I did. Eventually I interpret later on.

• Likewise for the source function I define a temperature that might, but may be not, similar to the temperature we are talking about.

• The advantage of the notation is related to Radio Telescopes since here the Rayleigh Jeans approximation is valid and I measure energy of radiation received on the focal plane of the Radio Telescope.

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Example

Ts

( )

( )( ) ( ) ( )

( )( )

s

h 2kT s

s 2

B

B s

B s

B s

2kTh h1 e 1 B TkT kT cT 0 since there is no extra source

T 0 1 e T 1 e

optically Thick T 0 T

optically

T

Thin T 0 T

ν ν

ν

ν

ν

τ τ

ν

νν ν

τ

τ

− −

+ =

=

= − = −

⇒ =

⇒ ≈

∼

Black Body

Bνν(T(Tss))

22

Back to the 21 cm

( )

( )

1 10

0

The column density is N n ds

The line profile is generally broadened

we use : d 1

ϕ ν

ϕ ν ν

∞

∞

=

=

∫

∫sn1

( ) ( )

( ) ( )

2 21 1 12 2 21

1 12 2 21

That is we should writedI h n A n B n B Ids 4

and for :hn B n B ds4

νν

ν

ν

ν ϕ νπ

τντ ϕ νπ

= − −

= −∫

N1

23

( ) ( ) ( )

( )

( )

( ) ( )

s

s

s

11 12 2 21 1 12 2 12

2

hkT2 1

1 12 1 121 2

hkT

1 12

hkT

1 12 2 21 1 12

gh hn B n B I n B n B I4 4 g

g gh n B n e B I4 g g

h n B 1 e I4

h hn B n B ds n B 1 e4 4

ν ν

ν

ν

ν

ν

ν

ν

ν νϕ ν ϕ νπ π

ν ϕ νπ

ν ϕ νπ

ν ντ ϕ νπ π

−

−

−

− = − =

= − = = −

= − = −

∫ ( )I dsν ϕ ν

∫

24

( ) ( )

( ) ( )

s

s

hkT

s

hkT

1 12 1 12s

2 2 2 212 12

1 10s s

he 1kT

h h hn B 1 e I ds n B I ds4 4 kT

B h B hn ds N4 kT 4 kT

ν

ν

ν ν ν

ν

ν ν ντ ϕ ν ϕ νπ π

ν νϕ ν ϕ νπ π

−

−

∞

≈ −

= − = =

= =

∫ ∫

∫

( )2 2 2H 1

1 1 12

212 21 21 21 3

1

N n g 3 N 3 1 NN n g 1

g cB B B Ag 2hν

= = ⇒ = +

= =

∼3

1

25

( ) ( )

( ) ( )

2 2 2 2 212 2 H

1 21 3s s 1

221 2 H H

1 s s

15 2 1

B h g Nh cN A4 kT 4 kT g 2h 4

hc A g N NC32 k g T T

C 2.57 10 cm K

ν

ν

ν ντ ϕ ν ϕ νπ π ν

τ ϕ ν ϕ νπ ν

− −

= =

= =

=

26

Line width of HI 21 cm

• The natural width of the 21 cm emission line, as for any other atomic line, is extremely small. However the lines are broadened because of thermal motion and turbulence.

• For the interstellar medium where HI is present and emitting we assume a temperature of about 100 K.

• In this case the thermal motion will be of about 1 km/s so that turbolence, estimated to be about 5 to 10 km/s will dominate. Below we will use 10 km/s for simplicity.

( )

( )

20

D 0

HD

6D 5

4 50

D

1 vel 2kTe ; ; vel 1 km / sc m

11420 10 * 4.74 kHz3 10

1With Turbolence 5 10 Hz ; Line center 1.1 10

ν ν∆ν ν νϕ ν

νπ∆ν

∆ν

ϕ νπ∆ν

−−

−

−= = =

≡ = =

∼

∼

∼

27

Sun

Galactic Center

If the width of the line is about 10 km/s a cloud moving with velocity larger than that is missed by the beam of the telescope or in any case is well separated by the previous one.

l

( )

( ) ( )( )rad

21

v A d Sin 2l ; A 15km / s / kpc10 km / s 10d for Sin 2l 1 700 pc 2 10 cmA Sin 2l 15

Max optical depth

=

≈ =∼ ∼ ∼

28

( )

3H

21 2H

15 21 5H

s

Assume N density 0.5 atoms cm

N column 10 atoms cmC N 2.65 10 10 1.1 10 0.3 1

T 100ν

ϕ ντ

−

−

− −

= = <

∼∼

∼

• Along most of the line of sight the Galaxy is optically thin. However at l=0 and l=180 degrees we have no radial velocity and the argument does not work.

• The temperature we used is somewhat smaller than the actual value which is Ts ~ 135.

29

Problem

• I observe with a Radio Telescope a cloud that I assume to be optically thin. The collecting area is S. What is the rate of Photons I receive?

• Apply the result to the characteristic of the Leiden early observations in 21 cm.

• Leiden radio Telescope: S=180 m2 ; ∆ν = 12 kHz ; Ω ~ 30 sqdegrees.

( ) ( )( )

( )

2

s 2

s2

16 4 38

227 6

I 0 ~ 2kT 1 ec

S I 0 kT SN 2h h

2 0.3 1.38 10 100 180 10 12 10 0.0091 3.9 10 photons / s6.62 10 1420 10 21.105

ντν

νν ν

ν

Ω ∆ντ ∆νΩ

ν ν λ

−

−

−

−

⋅ ⋅ ⋅ = = =

⋅ ⋅ ⋅ ⋅

=⋅ ⋅

30

In conclusion

• The Energy we receive therefore, since each photon carry 6 10-6 eV, is of :

3.9 108 photons/s or 2400 eV• Modern Aperture Synthesis Telescopes have much higher

space resolution and spectroscopic (km/s) resolution.• When we consider a thin cloud we should also keep in

mind that a large amount of photons is absorbed an re-emitted before they leave the cloud. The absorption

(n1B12 – n2B21) ϕ(ν)• Is a small difference between large numbers.

31

The Column Density

( )

( ) ( )

[ ]

( ) ( )

( ) ( )

H

s

H s H s0 0 0 0

14H s0

614 18

5

18H s B s0

18H B0

C NT

C N d T d C N d T d

N 3.88 10 T d

1420 10vel c 3.88 10 1.8 103 10

N l,b 1.8 10 vel T dvel ;Thin Cloud T T

N l,b 1.8 10 T l,b,v dvel

ν

ν ν

ν

ν ν

ϕ ντ

ϕ ν ν τ ν ϕ ν ν τ ν

τ ν

∆νν

τ τ

∞ ∞ ∞ ∞

∞

∞

∞

=

= ⇒ =

=

= = =

=

=

∫ ∫ ∫ ∫∫

∫∫

∼

32

External Galaxy

dΩ

D

dS = D2 dΩHere the whole surface of the Galaxy within the beam will contribute to the signal.

( ) ( )( )

18 2H BGalaxy 0

N Total 1.8 10 D d T l,b,v dvelΩ

Ω∞

= ∫ ∫

33

Units

1 Jansky = 10-23 ergs cm-2 s-1 Hz-1 = 10-26 W m-2 Hz-1

∆vel = ∆ν c/νI = 2 k ν2/c2 TB

( ) ( ) ( )( )

( )

( ) ( )( ) ( )

( )

218H BGalaxy 0

2 218 24 10 23 245

216 9 33

25HI

N Total # atoms 1.8 10 D cm d T Degrees dvel km / s

in solar Masses, Mpc, Jansky, km / s

1.82 10 3.086 10 2.9979 10 10 1.67 102.35 10

2 1.38 10 1.42 10 1.99 10

S velD2.35 10Mpc Jansky km / s

ΩΩ

∞

− −

−

=

=

=

∫ ∫

M

M 0dvel

∞

∫

34

Problem

• Estimate the HI Mass of a cloud at a distance of 30 pc and whose flux is 4.5 10-15 erg cm-2 s-1.

2 26 1HI

15 121 cm H 21 21

21 cm 58 H H HH

21

L 4 d f 4.85 10 erg s3L N A h A 2.869 10 s4

L m NN 2.39 10 ; 203 A h4

π

ν

ν

−

− −

= =

= =

= = = =

MM M