Book a Demo!
CoCalc Logo Icon
StoreFeaturesDocsShareSupportNewsAboutPoliciesSign UpSign In
Download

Assignment 8

32 views
Tweet
Share
Share

ASSIGNMENT 8

BY Andrej Njegus u5808872

# Conversions

mpc2km=3.086e19
mpc2m=3.086e22
solm2kg=1.989e30
mas2rad=RR(pi/1e3/648000)
deg2rad=RR(pi/180)
yr2sec=3.154e+7

# Constants

c=3.0e8 #m/s
h= 6.626e-34 # m^2*kg/s
sigma=5.670e-8 #W/(m^2*K^4) = J/(s*m^2*K^4) = (kg*m^2/s^2) Stefan Boltzmann constant
H0=70 #km/(s*Mps)
H0_s=70/mpc2km
G=6.673e-11 #Grav constant N * m^2 / (kg)^2 = kg*m/(s^2) * m^2 / (kg)^2
k=1.38e-23
m_H=1.67e-27 #kg
CMB= 2.73 #K

#Units

%var m, kg, s, K, W, J, Mpc, rho_rad, rho_c, Omega_M, Omega_gamma, a, a_0, Omega, Omega_Lambda, Omega_0, H, H_0, z, theta

Questions 1 and 2.1 done solely on paper.

QUESTION 2.2

NOTE: R=aa0R=\frac{a}{a_{0}} and Rrec=T0Trec.R_{rec} = \frac{T_0}{T_{rec}}.

As per discussion with Brian via video-link, in Q2 we are allowed to pre-packaged integrators. SAGE uses QAG adaptive quadrature method for numerical integration. For more details see: https://en.wikipedia.org/wiki/QUADPACK.

Initially I approached question 2.2 by assuming Ωγ\Omega_\gamma is zero (since ΩM\Omega_M and ΩΛ\Omega_\Lambda add up to 1). Calculation by this assumption is given bellow. Two known formulas for ΩM+ΩΛ=1\Omega_M + \Omega_\Lambda = 1 , k=0k=0 are used. Answer given in unit yearsyears.

RR(2*(1+1097.9)^(-3/2) / (3*H0_s*sqrt(0.3)))/yr2sec
2/(H0_s*3*sqrt(1-0.3)) * arcsinh(sqrt((1-0.3)/0.3)*(1+1097.9)^(-3/2))/yr2sec
467032.438481317 467032.438344450

Here I calculate trect_{rec}, with Ωγ\Omega_\gamma as per resut in Q1.1 in previous assignment. Integral range is R∈[0,Rrec]R\in[0,R_{rec}]. Answer given in unit yearsyears.

numerical_integral(1/(x*sqrt(0.3*x^(-3) + 5.07e-5*x^(-4) + 0.7)), 0, 2.73/3000)[0]/H0_s/yr2sec
394417.741080156

QUESTION 2.3

Answer given in units ms\frac ms

RR(c/sqrt(3))
1.73205080756888e8

QUESTION 2.4

Co-moving distance (ie. acoustic scale) as: Dcomac=cs∫0RrecdR′1R′2ΩMR′−3+ΩγR′−4+ΩΛ D_{com_{ac}} = c_s \int_{0}^{R_{rec}} dR' \frac{1}{R'^{2}\sqrt{\Omega_{M}R'^{-3} + \Omega_{\gamma}R'^{-4} + \Omega_\Lambda}} Answer given in units MpcMpc.

pretty_print(numerical_integral(1/((x^2)*sqrt(0.3*x^(-3) + 5.07e-5*x^(-4) + 0.7)), 0, 2.73/3000)[0]/H0_s/RR(sqrt(3)) * c /mpc2m, '$Mpc$')
179.329471996106\displaystyle 179.329471996106 MpcMpc

Angular-size distance is: Dθac=Rrec cs∫Rrec1dR′1R′2ΩMR′−3+ΩγR′−4+ΩΛ D_{\theta_{ac}} = R_{rec} \ c_s \int_{R_{rec}}^{1} dR' \frac{1}{R'^{2}\sqrt{\Omega_{M}R'^{-3} + \Omega_{\gamma}R'^{-4} + \Omega_\Lambda}} Answer given in units MpcMpc.

pretty_print(numerical_integral(1/((x^2)*sqrt(0.3*x^(-3) + 5.07e-5*x^(-4) + 0.7)), 2.73/3000, 1)[0] / H0_s / RR(sqrt(3)) * c /mpc2m *2.73/3000, '$Mpc$')
7.17278597256324\displaystyle 7.17278597256324 MpcMpc

How large this appears today: Dhorac=Rrec Dcomac D_{hor_{ac}} = R_{rec} \ D_{com_{ac}} Answer given in units MpcMpc.

pretty_print(numerical_integral(1/((x^2)*sqrt(0.3*x^(-3) + 5.07e-5*x^(-4) + 0.7)), 0, 2.73/3000)[0]/H0_s/RR(sqrt(3)) *2.73/3000* c /mpc2m, '$Mpc$')
0.163189819516456\displaystyle 0.163189819516456 MpcMpc

Finally we have θac\theta_{ac}:

pretty_print(theta, '$=$', (2 * arctan(0.163189819516456/2/7.17278597256324))/deg2rad, '$^\circ$')
θ\displaystyle \theta == 1.30349416104823\displaystyle 1.30349416104823 ∘^\circ

Small angle approximation works also:

0.163189819516456/7.17278597256324/deg2rad
1.30355038524219