VI. 21cm Radiation. 1 AY230-21cm. A. History

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1 1 AY230-21cm VI. 21cm Radiation A. History B. Physics van der Hulst 1941 Collq in a bunker! Ground state of HI: l = 0, s = 1/2, j = 1/2 Degeneracy of the G.S. is removed by the interaction of the magnetic moments of the electron and proton Proton has a spin and therefore a magnetic moment I is the nuclear spin g p 5.5 Hyperfine interaction with the electron µ p = g pe 2m p c I 1) H = µ p B 2) Spin aligned has highest energy To estimate B, model the electron orbit as a magnetized sphere Substituting Introduce F = I + s ala spin-orbit coupling H = 16π 3 2 B = 8π 3 M = 8π 3 µ eψ 2 0 3) I s = 1 2 ) 2 e m e g p Ψ 2 0 I 2m e c m s 4) p F 2 I 2 S 2 ) 5) I = 1 2 = s F = 0, 1 for spins aligned, anti-aligned For a 1s state Ψ 2 0 = 1 πa 3 0 Transition Energy 6)

2 2 AY230-21cm Energy Frequency: ν 21cm = MHz Wavelength: λ = 21.1 cm Approx energy: E = 10 5 ev Transition probability Spontaneous emission coefficient Lifetime hν = U + U 7) = 8π ) 2 e m e 2I + 1 g p 8) 3 2m e c m p πa 3 0 Excited state is primarily de-excited by collisions C. Spin Temperature = 8 3 α2 g p m e m p Ryd 9) A 10 = s 1 10) τ 21cm 10 7 yr 11) < σv > cm 3 s 1 for T 100K 12) τ coll yr 13) T S : Quantifies the level popuplations of the two hyperfine levels of H Level population Boltzmann n 1 = g 1 exp hν n 0 g 0 kt ) S T = 3 exp T [ S 3 1 T T S + O ) T T S ) 2 ] 14) 15) 16) T = 0.068K This generally implies T T S The excitation temperature T S is referred to as the Spin Temperature G. Field 1959, ApJ, 129, 551: T S = T kin even if the gas is NOT in thermal equilibrium Majority of H gas will have 3/4 of the atoms in the excited state and only 1/4 in the ground state

3 3 AY230-21cm HI gas is always emitting 21cm radiation Not a dominant coolant, but a very important diagnostic Furthermore, one cannot neglect stimulated emission!! D. Brightness Temperature Transfer equation Optical depth di ν ds = κ νi ν + j ν 17) dτ ν = κ ν ds 18) Consider radiation from a single cloud with optical depth τ νr radiation I ν 0) I ν = I ν 0)e τνr + τ νr 0 and incident j ν κ ν e τν dτ ν 19) If one has strict thermodynamic equilibrium, Kirchoff s law applies E. Optical Depth j ν = B ν T ) = 2hν3 1 κ ν c 2 e hν/kt 1 In the low frequency regime, hν kt 20) B ν T ) = 2ν2 kt c 2 21) Define brightness temperature, T b is the temperature where I ν = B ν T ) Radiation transfer equation becomes Opacity of H I gas to 21cm radiation 2nd term is due to stimulated emission Without this term, we could not probe the Galaxy Again, T T T b = T b0 e τνr + T 1 e τνr ) 22) κ ν = κ ν 1 e T /T ) 23) κ ν κ ν T T T = hν 10B 01 n 0 φ ν 4π T = hν 10 g 1 A10 c 2 4π g 0 2hν ) 25) ) 1 4 n H T T φ ν 26)

4 4 AY230-21cm Hot regions are less opaque φ ν is the line profile we will return to this when discussing Lyα absorption) Optical depth Frequency τ ν = L 0 = const) T κ ν ds 27) L 0 n H ds 28) Velocity space δν = ν 10 v c φv) τ v N HI T φv) is in km/s N HI is in cm 2 For an isolated cloud with a Gaussian velocity distribution φv) = e v2 /2σ 2 σ 2π 29) 30) 31) Peak optical depth τ 0 = N HI T 1 σ 2π 32) For a single cloud with T 80K, σ 3km/s, and N HI cm 2, we find τ F. Temperatures of HI Gas Observe a cloud in 21cm emission Characterize the cloud by T S and τ ν Radiative transfer low frequency limit) T b = T S 1 e τ ν ) 33) Observe toward the Galactic center Cartoon

5 5 AY230-21cm. Sun τ 0 1 because of the lack of differential rotation Line saturates at T b T S T S 80K, the Harmonic mean of all gas T b Tb 80K!v Random motions + saturation v

6 6 AY230-21cm Observe off the Galactic center emission) Cartoon Sun. Now, τ ν 1 T b = T S τ ν 34) It appears we have sensitivity to T S But recall τ ν 1/T S T b is independent of T S Thin limit κ ν negligible): I ν = j ν ds 35) A 10 n 1 ds 36) Observation is independent of T S Can accurately measure N HI Fig

7 7 AY230-21cm Require absorption spectra to determine T S Sun. Quasar Absorption-line analysis of a continuum source e.g. quasar) with intensity T C Sun. Cloud TS,!") Quasar TC) a) Radiative transfer T b = T S 1 e τ ν ) + TC e τν 37) b) Determine the unattenuated brightness of the QSO Observe the quasar near but not at the 21cm frequency off-line ) τ ν 1 T off b = T C by definition) c) Measure the brightness Tb on at the 21cm frequency on-line ) [Eq. 37] d) Take the difference spectrum T b = Tb on T off b 38) = T S 1 e τ ν ) + TC e τν T C 39) = T S T C ) 1 e τν ) 40)

8 8 AY230-21cm e) Subtle complication: The telescope has a finite beam size Beam size Θ beam = λ D 3 for Aracebo) 41) Source quasar) size Θ S Θ beam True intensity is related to the apparent intensity by Therefore, Updating our expression I true = Ω beam = I app Ω S Θbeam Θ S ν 42) ) 2 Iν app 43) ) 2 Tb true Θbeam = T C T A 44) Θ S T b = T S T A ) 1 e τν ) 45) Given T S or τ ν, one can infer the other f) Measuring the 21cm absorption τ ν ) Observe the cloud slightly off the quasar Assume physical properties of the cloud vary slowly Observe T off b = T S 1 e τ ν ) 46) g) Solve for τ ν e τν = T A T off b T b 47) T A Solving for T S follows trivially Observations through the plane There is gas detected in emission that is not seen in absorption Fig Furthermore, the absorption components are more narrow than those in emission

9 9 AY230-21cm Implication: Some gas detected in emission has low τ ν due to high T S warm clouds) WNM: Warm Neutral Medium Surveys of the Milky Way Many areas with emission detected without absorption Large filling factor of warm gas 40% of gas detected in emissio is WNM Low density Large filling factor implies low density This means the cooling time is long: Λ n 2 Overall characteristics Wide velocity components σ 5 17km/s) Large filling factor Scale height of the gas in our Galaxy) nz) = n 0 exp z /h) 48) h 250 pc 49) Surface density vs. stars which have Σ 70M pc 2 ) Σ W NM = ρ dz = 3.8M pc 2 50) CNM: Cold Neutral Medium Dense clouds Λ CNM Λ W NM Low filling factor: 1/3 of radio sources do NOT show 21cm absorption Dispersion: σ CNM 2 3km/s σ CNM < σ W NM Dispersion is due to macroscopic velocities within the cloud, not thermal broadening Cloud-cloud dispersion: σ cc 7km/s Surface density Σ CNM = 4.5M pc 2 51) Actual temperature observations: Harmonic mean of the WNM and CNM Consider two clouds, one warm and one cold Temperatures T S1, T S2 Optical depths τ 1, τ 2 Cloud 1 is in front of cloud 2 Off-source temperature T off b = T S1 1 e τ 1 ) + TS2 1 e τ 2 ) e τ 1 52)

10 10 AY230-21cm If we assigned an average temperature to the clouds T naive = T S 1 1 e τ 1 ) + T S2 1 e τ 2 ) e τ 1 1 e τ 1+τ 2 ) 53) Consider several limiting cases a) τ 1 1, τ 2 1 b) τ 1 1, τ 2 1 T naive = T S1 54) T naive = T S1 τ 1 + T S2 55) c) τ 1, τ 2 1 T naive = T S 1 τ 1 + T S2 τ 2 τ 1 + τ 2 56) Tend to overestimate T S of the CNM and underestimate T S of the WNM Careful treatment 5000 K < T W NM < K 57) 20 K < T CNM < 250 K 58)

11 11 AY230-21cm G. Galactic rotation curve Diagram!0 Sun. R0 l " r R! Θ 0 = ω 0 R 0 = rotational velocity at our position Θ = ωr = rotational velocity at radius R r is the distance to a given cloud l is angle off the center of the galaxy Observationally, we measure the velocity projected along the sightline [ π ] vr) = Θ cos 2 l + θ ) Θ 0 sin l 59) Constraints = ωr [ sin θ cos l + sin l cos θ ] Θ 0 cos l 60) R sin θ = r sin l 61) R cos θ + r cos l = R 0 62)

12 12 AY230-21cm Rewrite [ R vr) = ωr 0 sin θ cos l + R ] sin l cos θ Θ 0 sin l 63) R 0 R 0 = ωr 0 sin l Θ 0 sin l 64) = R 0 sin l [ ωr) ω 0 ] 65) If the Galaxy were rotating as a solid body i.e. ωr) = ω 0 ), then every parcel of gas would have zero velocity on the los The optical depth would be extremely large We would be unable to see through the Galaxy! vr) Only unique value of vr) vt Double valued) Rc r Observations of vr) Velocity peaks when R r Define this R = R c Define this velocity as v T Finally, v T = R c ωr c ) R 0 ω 0 sin l 66) R c = R 0 sin l 67) Θ c = ωr c )R c = v T + Θ 0 sin l 68)

13 13 AY230-21cm Knowing R 0 and Θ 0 these are the hardest parts!), we can observe at a range of l to map out the rotation curve Fig! 220 km/s Rkpc) Solid body in the interior Flat at R > 2 kpc H. Electron Density in HI Regions of the ISM General: Radio observations but not 21cm Examine attenuation of Pulsar radiation by electrons Measure Dispersion Measure and distance to determine n e Faraday rotation Consider the e response to a plane wave Treat the electrons as a fluid EM radiation drives a current which modifies the incident radiation field Ignoring a magnetic field, the frequency of the plane wave becomes ω 2 = ω 2 p + c 2 k 2 69) ω 2 p 4πn ee 2 m e 70) Index of refraction: m m c v p = [ ] 1 1 ω2 2 p ω 2 71) Can measure radiation over a frequency range to determine ω p and then n e Observe pulsars

14 14 AY230-21cm a) ω < ω p Rem) = 0 No propagation of the radiation! For n e = cm 3, ω p = 10 4 Hz Why doesn t the wave propagate? Debye Radius Let v 2 T = kt/m e 4πn e e 2 R 2 D kt 72) ωpm 2 e RD 2 kt 73) ωpr 2 D 2 kt m e 74) 1 ω p R D v T 75) If ω < ω p, then e can travel fast enough to cancel out the propagating wave! b) ω > ω p Each pulse has a wide spectrum of frequencies Pulse delays will occur as a function of ω Group velocity v g = dω [ dk = c 1 ω2 p ω 2 c Pulse arrival time Lower ω longer time ] 1 2 ωp 2 ω 2 ) 76) 77) T ω) = L 0 ds v g ω) L c + 1 2c Take measurements at a wide range of ω Evaluate T ω) d1/ω 2 ) = 1 2 L 0 L ds ω2 p c = 1 ) 4πe 2 L 2 m e c 0 ds ω2 p ω 2 78) 0 n e ds 79) Define DM, the Dispersion measure DM = n e ds pc cm 3 ) 80) T ω) d1/ω 2 ) = DM 81)

15 15 AY230-21cm Answer Difficult part Measure L for the pulsar Use 21cm + spiral arms Compare with Spitzer s estimate Assumed all e s came from C Puzzle What leads to the extra source of electrons? Is it related to missing heating? I. Extragalactic Rotation Curves <n e >= 0.03 cm 3 82) n e = n C = C H n H 83) = 10 4 n H = cm 3 84) As gas falls into a potential well e.g. dark matter halo), it may dissipate and settle into a rotating disk Assume non-negligible angular momentum For galaxies, this angular momentum is believed to result from tidal torques HI gas Excellent tracer of the potential well Generally extends far beyond the detection limit of stars 21cm telescope allow for excellent spectral resolution kinematics) Challenge: Spatial resolution Either study nearby galaxies Or build bigger telescopes Or use interferometric techniques Or use optical/molecular lines First studies Roberts & Rots 1973, A& A, 26, 483

16 16 AY230-21cm Results a) The HI gas is extended far beyond the Holberg Radius b) The rotation curve flattens at large radii c) The surface density of gas decreases with radii power-law or even exponential) Implications a) HI gas exists beyond where stars form

17 17 AY230-21cm b) Dark matter! Current observational research 38 Large 21cm surveys: HIPASS, ALFALA, THINGS Example Walter et al. 2008) Fig. 8. NGC 628. Top left: integrated HI map moment 0). Greyscale range from Jy km s 1. Top right: Optical image from the digitized sky survey DSS). Bottom left: Velocity field moment 1). Black contours lighter greyscale) indicate approaching emission, white contours darker greyscale) receding emission. The thick black contour is the systemic velocity v sys =659.1 km s 1 ), the iso velocity contours are spaced by v=12.5 km s 1. Bottom right: Velocity dispersion map moment 2). Contours are plotted at 5, 10 and 20 km s 1.

18 18 AY230-21cm Fig. 55. ISO and NFW rotation curve fits for NGC 7793 using the entire observed extent. Lines and symbols as in Fig. 25. Open questions How does not) star-formation trace HI gas? How does the rotation curve behave in the very inner regions of low surface brightness galaxies? 94 J. HI Mass Function Concept Mass function of galaxies in HI gas

19 19 AY230-21cm Number of objects per solar mass interval per comoving Mpc 3 Akin to a luminosity function for galaxies Observations Large sky surveys in 21cm Aracebo, Parkes telescopes Measure position, total flux, and velocity Analysis Convert velocity to distance v = H 0 r) Convert 21cm flux to HI mass Assess the survey volume and calculate number densities Example: Zwaan et al HIPASS)

20 2005MNRAS.359L..30Z 20 AY230-21cm K. Reionization Definition: Epoch when the ionizing radiation from quasars+galaxies ionized the majority of baryons in the Universe Research area of strong, current interest What are the exact sources? How are they distributed? Challenges Very few if any) emitters to study Very distant universe radiation is highly redshifted + dimmed Very simple universe prior to reionization Neutral hydrogen Neutral helium Not much else A solution: Probe the hydrogen gas via 21cm emission Spin temperature at Reionization

21 21 AY230-21cm Three processes will determine T S a) Absorption of CMB photons including stimulated emission) b) Spin-exchange collisions with other hydrogen atoms, free electrons and protons c) Scattering of ambient Lyα photons the Wouthuysen-Field effect) The third process may be neglected at very early times, when there is no source of Lyα photons Level populations 0=ground, 1=excited) dn 0 dt = n 0 C01 + B 01 I CMB ν ) ) + n1 C10 + A 10 + B 10 Iν CMB Definitions A 10 is the spontaneous emission coefficient Iν CMB = 2kT CMB /λ 2 10 is the intensity of the CMB at 21cm B 10, B 01 are the Einstein coefficients 85) B 10 = λ3 10 2hc A 10 86) B 01 = g 1 g 0 B 10 = 3B 10 87) C 10 is the de-excitation rate per atom) from collisions Rewriting B 10 I CMB ν Detailed balance in thermal equilibrium tells us = A 10T CMB T 88) n 0 C 01 = n 1 C 10 89) C 01 = n 1 = g 1 exp T /T K ) 3 1 T /T K ) C 10 n 0 g 0 90) T K is the gas kinetic temperature In the early universe, it is set by adiabatic expansion In steady state, dn 0 /dt = 0, giving n 1 = 3C 10 1 T /T K ) + 3A 10 T CMB /T n 0 C 10 + A T CMB /T ) 91) Solving for T S with n 1 /n 0 31 T /T S ) T S = T CMB + yt K + T 1 + y 92) T S is the weighted mean between T K and T CMB

22 22 AY230-21cm The collisional coupling coefficient y is the sum of three terms, y T A 10 T K C 10 = T A 10 T K C H + C e + C p ), 93) where C H, C e, and C p are the de-excitation rates of the triplet due to collisions with neutral atoms, electrons, and protons. The H-H collision term can be written as C H = n H κ 10, κ 10 = T K exp 32/T K ) cm 3 s 1 94) This is valid in the range 10 < T K < 10 3 K Diagnosing the gas during Reionization with 21cm Key: The CMB drives the temperature of the gas toward T CMB Need the gas to have yt K > T CMB or lots of collisions to have T S T CMB Consider the universe at z = 100 Characteristics of the gas T K = 168K n H = 0.19 cm 3 C H > A 10 T CMB T 95) yt K = 730K T CMB = 275K Therefore, before the Reionzation and the appearance of the first emitting objects, when the IGM is cooling adiabatically faster than the CMB, spin-exchange collisions between hydrogen atoms couple T S to the kinetic temperature T K of the cold gas Cosmic hydrogen can be observed in absorption since T K < T CMB ) against the CMB! Current+Future experiments LOFAR MWS Simple dipole experiments

23 23 AY230-21cm S.R. Furlanetto et al. / Physics Reports ) a) b) M temperature evolution if only adiabatic cooling and Compton heating are involved. The spin temperature T S includes only coll ) Differential Figure brightness 1: a) temperature IGM temperature against theevolution CMB forif T S only shown adiabatic in panelcooling a. and Compton heating are involved. The spin temperature T S includes only collisional coupling. b) Differential brightness temperature against the CMB for T S shown in panel a. From Furlanetto, Oh, & Briggs 2006, PhR, 433, 181. have assumed that T K 1 + z) i.e., coupling to the CMB is still strong). Freeze-out occurs when t nity; at later times x i remains roughly constant because the recombination time then exceeds the expa. Numerical calculations with RECFAST yield x i = at z = 200 which is past freeze-out). Inse 63), we find 1 z)t K dt K dt 10 7 Ω b h 2 T γ/t K 1)1 + z) 5/2. mal decoupling occurs when z dec 150Ω b h 2 /0.023) 2/5.