Extragalactic Astronomy (ASTRO 504)Spring 2020

Problem Set 8

Due 27 March 2020

Solving problem sets is one of the most efficient ways of learning the subject. You are encouraged tocollaborate with fellow students and/or to consult senior students, local postdocs and me. But, pleasewrite the solution by yourself. No homework will be accepted after I post the solution on thecourse webpage.

1. (25 pts) FRB150418.The dispersion relation of electromagnetic waves passing through an ionized medium (such asIGM) is given by

k2c

2 =!2 �!2p

(1)

where the plasma frequency is given by !2p⌘ 4⇡nee

2/me with electron number density ne. Showthat the dispersion relation implies that the arrival time of a radio burst at different frequency islagged by

�tDM = 4.146⇣ ⌫

GHz

⌘�2 Å DMcm�3pc

ãms (2)

and the IGM contribution of the dispersion measure (DM) is given by

DMIGM =3cH0⌦IGM

8⇡Gmp

Z z

0dz0 (1+ z

0)Xe(z0)p⌦M(1+ z0)3 +⌦⇤

, (3)

where z is the redshift of the radio burst, Xe is the ionization fraction of the IGM along the lineof sight, ⌦M, ⌦⇤, ⌦IGM are the density parameters of, respectively, matter, cosmological constant,and IGM.(b) FRB150418 is at redshift z = 0.492± 0.008, and DM = 776.2 cm�3pc. When DMMilky Way =189cm�3pc, DMhalo = 30 cm�3pc, DMhost galaxy = 37 cm�3pc (in the rest-frame), and the uncer-tainty due to the line-of-sight inhomogeneitiese in the IGM is�DMinhomo. = 100cm�3pc, estimatethe baryon fraction⌦baryon with the uncertainty range due to�DMinhomo.. Assume that the Heliumabundance is Yp = 0.24 and He is neutral in the IGM. About 90 % of baryons are in IGM.

2. (25 pts) Gunn-Peterson troughThe Gunn-Peterson effect is the absorption trough produced in the spectra of high-redshift quasarsby the absorption of Lyman-↵ photons due to neutral hydrogen in the foreground intergalacticmedium. The task of this problem is to compute and plot the resultant spectrum of a high-redshiftquasar.

(a) First, compute the Lyman-↵ optical depth that we would detect as a function of observedwavelength for observed wavelengths between the Lyman limit (=912 Å) redshifted by zem andLyman-↵ (=1216 Å) redshifted by zem. Assume the Lyman-↵ scattering cross-section at rest fre-quency ⌫ is given by

�⌫ =⇡e

2

mecf12�

D(⌫� ⌫12) (4)

1

where f12 = 0.4164 is the oscillator strength of the Lyman-↵ transition and �D(x) is the Dirac-delta function. Plot your result for the neutral fraction fH = 10�6, 10�5, · · · , 100 and zem = 3.5.(b) Suppose the quasar has a flat continuum spectrum with intensity I⌫ = 4⇥10�7erg/cm2/s/Hz/str.Add to this a Lyman alpha emission line modeled as a Gaussian with emission redshift, zem = 3.5,Doppler velocity dispersion� = 1000km/s, and central intensity I

Ly↵⌫ = 4⇥10�6erg/cm2/s/Hz/str.

(i) Use the radiative transfer in the expanding Universe, plot the resultant spectrum as a functionof wavelength for the range discussed above in the absence of foreground neutral Hydrogen;(ii) Now do this in the presence of foreground neutral hydrogen for the range of fH discussedabove;(c) Now repeat your calculation assuming zem = 6.5. What is the minimum fH required to explainthe strong absorption trough that we see in the class?(d) Suppose you were to use an absorptio-line profile that is more realistic than the delta-functionapproximation used above. Specifically, suppose

�⌫ =⇡e

2

mecf12

A21/4⇡2

(⌫� ⌫12)2 + (A21/4⇡)2(5)

with the Einstein coefficient A21 = 6⇥ 108 s�1. How would your results change from the abovecalculation?

2

:a) dispersion relation : KZ?_ wz.wp'

Ee→ zokdk - zwdw ⇒ no - date =kw±=-: ftp.T

radio burst at distance L away

⇒ arrival time at fey.ve v =

feta,

=f¥l(tfffyknfootflitztfty= to + ¥ 4te÷e[ nedl

me'

W -ZEV -.

. K + zhu' DM

ln Cosmological setting , we need to include the change in the time intend at Z & now.

fat÷ . sate → He .

-

ttflo." time interval at 2- is related to the current value by FG = CHafez

@cosmological redshift of the frequency Vo - ( 1+2-5'Ve ⇒ Vz .

- (Ht) Do

: DM = fdhneatzstadx. fitcttnzeftthgndz'

Nea) -mgncz)Xect) : Mgm (take a) :kitchen (Hzpxecz )

- 3←tg÷EmEa+⇒3XeH

:m=:¥Ek÷f:#Intimateot,"= under DM = 4.146K¥)→(aiB?p-) ms .

(b) 2=0.492

DM = 776.2 t 189 + 30 +374+2-5'tDMKM → DMKM ÷ (b32.41=100) Cni 3- Pc

He is neutral → me- : Ned- net'e - Neall -2mg

neall = Ny +2ha. i TT =Yp, Ny .

. ( 1-Y. )nb

= ( I-Yptkk ) Nb = ( l - ¥12 ) Nb

→ Xe - 1- ¥12 = 0.88

os - mama , .'

÷ytE¥H÷tm#dir!ft¥3"

: Scjlt.

= (Dhl )x(1.1232×6' 5h'mylar)t@S×hwkx¥H÷)mp<

116

= ( 99.68 aitlypc ) DM

= 0.053 ± 0.01

:a) E .

- T÷ fnf " ( w ( Hz ) .

k.)

Tu = fol Ntnidl=fFhµHEat Isth, =Ik÷fzf?Th*a) 84mHz) -k)k€+2,= MTIEMMETT: zaikb - 1

NH±Ha) = fynycza) .

. ¥, ( 1 . Yp ) Nba = fy ( 1 . Yp ) k¥6 Rbfmp C

H±: a:(FEE)f£ct¥s3±E÷rm÷ CHH')[w#t⇒]

±auohifr(o?i÷)e!thIdjµfa. : za . Yu

:HI

:ne= lot

C b)

Zen = 3.5

(C) Zen = 6.5

(d) line profile width ± Azika ~ 50MHz ⇒ ME ~ hosting ~ lot ! ! "

extremely narrow ! "

the result should be the same.