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hw-num1-sol - Ph 444 Solutions for Numerical Assignment 1 1...

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Ph 444 Solutions for Numerical Assignment 1 1. The particle horizon at the time t is calculated by taking each interval of proper distance, cdt , covered by a photon between t and t + dt , increasing it by the expan- sion of the universe between t and t , a ( t ) /a ( t ), and adding up the intervals between zero and t : d hor ( t ) = a ( t ) integraldisplay t 0 cdt a ( t ) . (1) Changing variables in the integral to a and using da = ˙ a dt , H = ˙ a/a , and Ryden equation 6.6 for H : d hor ( t ) = a ( t ) integraldisplay a ( t ) 0 cda a ˙ a = a ( t ) integraldisplay a ( t ) 0 cda ( a ) 2 H ( a ) (2) = ca ( t ) H 0 integraldisplay a ( t ) 0 da ( a ) 2 r, 0 / ( a ) 4 + Ω m, 0 / ( a ) 3 + Ω Λ , 0 + (1 Ω 0 ) / ( a ) 2 ) 1 / 2 (3) = c (1 + z ) H 0 integraldisplay 1 / (1+ z ) 0 da r, 0 + Ω m, 0 a + Ω Λ , 0 ( a ) 4 + (1 Ω 0 )( a ) 2 ) 1 / 2 (4) The last step uses a ( t ) = 1 / (1 + z ). The dimensionless integral in equation 4 does not have a simple closed form. It can be expressed in terms of exotic special functions, but this is not of much use in obtaining numerical values. I wrote a simple program to evaluate the integral using trapezoidal rule integration with 10 4 equal-size steps in a ( i.e. , multiplying the integrand evaluated at the middle of the interval by the interval width and summing the contribution from all of the intervals). I chose the number of steps by increasing the number until the result stopped changing. The value of the integral, I , for Ω r, 0 = 8 . 4 × 10 5 , Ω m, 0 = 0 . 27, Ω Λ , 0 = 0 . 73, and z = 1090 is I = 6 . 697 × 10 2 . Since c/H 0 = 4220 Mpc for H 0 = 71 km/s/Mpc, d hor ( z = 1090) = 0 . 2592 Mpc. This small size reflects the young age of the universe at the time of decoupling, about 300,000 yrs. The interval of comoving coordinate that corresponds to the size of the horizon at z = 1090 is r = d hor ( z = 1090) /a ( z = 1090). The distance today that corresponds to this interval of comoving coordinate is just a 0 r = d hor ( z = 1090)( a 0 /a ( z = 1090)) = d hor (1 + z ) = 283 Mpc. Thus, any structure imposed on the size of the horizon at the time of decoupling will show up at large physical sizes today, though this size is still much smaller than the particle horizon today. For z = 1090, it is a good approximation to ignore the Ω Λ , 0 ( a ) 4 and (1 Ω 0 )( a ) 2 terms in the denominator of the integrand since they are smaller by a factor of at least 10 3 compared to the Ω m, 0 a term, even at the upper limit of the integration.
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