extragalactic astronomy (astro 504) spring 2020extragalactic astronomy (astro 504) spring 2020...
TRANSCRIPT
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.