Radiation and the ISM

Question 1

Radians

Definition

Answer 1

θ = l/r radians
l = arc length
r = radius

Question 2

Steradians

Definition

Answer 2

ω = A/r² steradians
A = area on surface of sphere
r = radius

Question 3

What is the energy dEν entering a solid angle dω un time dt?

Answer 3

-the energy dEv in a frequency range v->v+dv, entering a solid angle dω in time dt is:
dEv = IvcosθdAdvdω*dt
-where Iv is the specific intensity

Question 4

Units of Specific Intensity

Answer 4

-the cgs units of specific intensity are erg/(scm²Hzst) or in metric units: W/(m²Hzst)

Question 5

Total Intensity in Terms of Specific Intensity

Answer 5

I = ∫ Iv dv

-integrated between 0 and ∞

Question 6

Intensity and Distance

Answer 6

• intensity is independent of distance as, once inside the beam, radiation stays within it, it is conserved
• as an example, the Sun’s intensity remains unchanged as long as you resolve the radiation that is emitted

Question 7

Flux of Radiation

Definition

Answer 7

-flux of radiation fv is defined as the amount of energy crossing a unit area per unit time per unit frequency interval
fv = 1/dAdvdt ∫ dEv
-where the integral is taken over the area S

Question 8

Total Flux in Terms of Flux of Radiation

Answer 8

F = ∫ fv dv

-where the integral is taken between 0 and ∞

Question 9

Units of Flux

Answer 9

-the units of flux in cgs units are erg cm^(-2) Hz^(-1) or in metric units: J m^(-2) s^(-1) Hz^(-1)

Question 10

Relationship Between Flux and Intesity

Answer 10

-the relationship between flux and intensity is:

fv = ∫ Iv cosθ dω

Question 11

Intensity in Thermodynamic Equilibrium

Answer 11

-for blackbodys in thermodynamic equilibrium, the specific intensity is given by:
Iv=Bv
-where Bv is the Planck function

Question 12

Planck Function

Answer 12

Bv = 2hv³/c² * 1/[exp(hv/kT)-1]

-in units of W m^-2 Hz^-1 st^-1

Question 13

Total Intensity of a Blackbody in Thermodynamic Equilibrium

Answer 13

B = ∫ Bv dv = σT^4/π W/m²
-where σ is the Stefan-Boltzmann constant:
σ = 5.67 * 10^(-8) W m^-2 K^-4

Question 14

Flux of a Blackbody in Thermodynamic Equilibrium

Answer 14

-integrating over radiation travelling from a surface in all outwards directions (i.e. over a solid angle equal to half a sphere)
-total flux is given by:
F = πB

Question 15

Notable Features of the Planck Function on a Log-Log Scale

Answer 15

• straight line portion of the graph at lower frequencies in log-log space
• shift of peak to higher frequency with temperature

Question 16

Rayleigh-Jeans Law

Answer 16

-a low frequency / high temperature simplification of the Planck function
Bv = 2v²kT/c²
-so for the log scale:
log(Bv) = 2log(v) + log(T) + C
-where C = log(2K/c²) is a constant
-this expression for log(Bv) is essentially the equation for a straight line, the straight portion of the Planck curve
- an increase in log(v) by one order of magnitude results in an increase in log(Bv) by two orders of magnitude

Question 17

Wien’s Displacement Law

Peak Frequency

Answer 17

-to find the value of frequency at which the peak intensity is reached
-calculate the derivative of the Planck Function and set equal to 0
-solve the resulting equation using numerical methods
=>
vmax/T = 2.82k/h = 5.88*10^8 Hxz/K

Question 18

Wien’s Displacement Law

Peak Wavelength

Answer 18

-to determine the peak wavelength λmax the law must be derived using the wavelength form of the Planck Function
-the following relation DOES NOT hold:
vmaxλmax=c
-since:
Bλ dλ = Bv dv
-rearrange:
Bλ = Bv * dv/dλ
=>
λmaxT = 0.29cm K

Question 19

Typical Wavelengths of Dust Cloud Emission

Answer 19

-starlight warms the dust grains to T≈ 10-100K
-thus from Wien’s displacement law:
λmax*T = 0.29cm K
=>
λmax ≈ 30-300µm
-these are far-infrared (FIR) wavelengths
-at these wavelengths, interstellar dust radiates more strongly than stars, so a FIR view of the sky is a view of the location of interstellar dust

Question 20

Visible vs FIR vs NIR View of the Milky Way

Answer 20

• at visible wavelengths, light suffers from so much interstellar extinction through absorption by dust that the galactic nucleus is obscured from view
• BUT the amount of interstellar extinction is roughly inversely proportional to wavelength
• a FIR view shows up the cold interstellar dust clouds as they emit FIR more strongly than stars
• a NIR view shows up the stars, stars emit more strongly at NIR than FIR

Question 21

Radio View of the Milky Way

Answer 21

• as radio wavelengths are long, radio waves can penetrate the interstellar medium even more easily than infrared light
• atomic hydrogen, the most abundant element in the universe, emits radio waves even in relatively cold clouds

Question 22

What is the origin of the 21cm hydrogen line?

Answer 22

• particles such as protons and electrons possess spin
• in the case of atomic hydrogen with a single proton and single electron, the state can change from high energy, both spins aligned, to a lower energy arrangement, opposite spins
• this transition releases a photon of 1420MHz or 21cm wavelength, a radio photon
• this is a tiny energy transition, approximately 10^(-6) times the magnitude of that for electronic transitions

Question 23

Radio View of the Galactic Disc

Answer 23

• looking within the plane of the galaxy from out position, hydrogen clouds at different locations along our line of site will be moving at slightly different speeds relative to us
• this is because they are at different distances from the galactic centre than us and speed of orbiting clouds increases closer to the galactic centre
• radio waves from these clouds are subject to slightly different Doppler shifts and this allows us to map the morphology of the milky way

Question 24

Limitations of the Radio View of the Galactic Disc

Answer 24

-gas in the region of the galaxy immediately opposite us is moving perpendicular to our line of sight so does not exhibit a Doppler shift and so we cannot map that part of the galaxy using this technique

Question 25

Why is carbon monoxide an important diagnostic molecule?

Answer 25

• carbon monoxide is a heteronuclear diatomic molecule (unlike H2)
• this means that it has a non-zero electric dipole moment i.e. it is a polar molecule
• when a polar molecule rotates in space it changes energy between rotational states that obey the laws of quantum mechanics
• approximating a diatomic molecule as a linear rigid rotor (a dumbbell) allows computation of its quantised rotational energy states

Question 26

Expression for Rotational Energy

Answer 26

Erot = ℏ²/2I * J(J+1) = BJ(J+1)
-where B is known as the rotational constant, a fundamental property of every heteronuclear molecule, and J is the rotational quantum number:
B = ℏ²/2I

Question 27

Selection Rule for Rotational Transitions

Answer 27

ΔJ = ±1

-i.e. molecules can only switch to neighbouring rotational energy states, they can’t skip over states

Question 28

Energy Difference Between Two Adjacent Rotational Energy States

Answer 28

ΔE = E_J+1 - E_J
= 2B(J+1)
-the energy transition from J=1 from the ground state J=0 is only 0.48meV or 5.5K
-this explains why the first excited rotational state of CO is very easy to populate in dark, quiescent and cold molecular clouds
-this excitation occurs primarily through collisions with ambient H2

Question 29

Energy Distribution of CO

Answer 29

-in thermodynamic equilibrium, the energy level population follows a Boltzmann distribution:
ni/no = gi*exp(-Ei/kT)
-where ni is the number of molecule in energy level i and no is the number in the ground state
-where gi=2J+1 is the degeneracy of the energy level

Question 30

Molecular Gas vs Atomic Gas

Answer 30

molecular gas mass ~ atomic gas mass ≈ 2-4*10^(9)M☉

• dust/gass mass ration ~ 0.01
• most molecular gas is in Giant Molecular Clouds (GMCs) confined in the spiral arms: ≥ 10^5 M☉
• throughout the galactic disc are small clouds and complexes ~ 10^4 M☉

Question 31

Using Other Tracer Molecules

Answer 31

• we can zoom in on smaller regions using other tracer molecules
• an isotope of carbon monoxide C-O18, a less abundant form of CO to find the most dense regions
• we can also trace ammonia
• and the spin transition in molecular nitrogen N2 with a hydrogen ion H+, at low temperatures ~20K other tracers start to stick to dust but this doesn’t happen with N2H+

Question 32

Bok Globules

Answer 32

-small, dense, almost spherical clouds:
T~10K
n ≥ 10^4 cm^(-3)
M ≈ 10-50M☉
L ~ 1pc
-infrared surveys have revealed that many harbour young (usually) low-mass stars

Question 33

How do we detect atomic gas vs molecular gas?

Answer 33

• the atomic gas throughout galaxies is detectable via the spin-flip transition of atomic hydrogen [HI] at a wavelength of 21cm,
• the molecular gas is detectable exploiting the low-lying rotational energy levels of carbon monoxide that are easily excited in cold gas

Question 34

Distribution of Gas Clouds Across the Milky Way

Answer 34

• cold T~10K and dense (n~10^3-10^5cm^(-3)) molecular clouds are the sites of star formation in galaxies
• dust grains (1% by mass) in molecular clouds emit radiation predominantly at FIR wavelengths (30-300 microns)
• most of the molecular gas in the milky way is in Giant Molecular Clouds confined in the spiral arms, that have masses, M ≥ 0^5 M☉
• throughout the galaxy are also found in smaller clouds (~10^2-10^5M☉) and isolated clouds also known as Bok globules (10-50M☉)