- #1

Buzz Bloom

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- TL;DR Summary
- I am seeking help in finding my error. I am integrating the Friedmann equation (with just H_0, Ω_m, and Ω_Λ) expecting a result of about 13.8 x 10^9 years. My result and how I calculated it is in the main body of this post.

NOTE: I am attempting to convey the equations in this post into LaTerX format in Post #19.

My result is way off. It is about 7.44 x 10^9 years. The values I use are:

dt = (1/H_0) (1/(M/a^3 + L)^(1/2)) (da/a)

dt = (1/3) (1/H_0) (1/L^1/2)) dx / (x^2 + 2Q x)^(1/2)

Integrating produces

Calculating the value of t uses the integration limits between x=a=0 and x=a=1. The result is the following.

My result is way off. It is about 7.44 x 10^9 years. The values I use are:

1/H_0 = 14.4 X 10^9 years,

M = Ω_m = 0.3103, and

L = Ω_Λ = 1 - Ω_m = 0.6897.

The equation I start with is the following.dt = (1/H_0) (1/(M/a^3 + L)^(1/2)) (da/a)

I make the following substitutions.x = a^3 and a = x^(1/3),

da = (1/3) dx / x^(2/3)

Q = (1/2) M/L and M/L = 2Q

This leads to the following equation.dt = (1/3) (1/H_0) (1/L^1/2)) dx / (x^2 + 2Q x)^(1/2)

Integrating produces

t = (1/3) (1/H_0) (1/L^(1/2) [log ( x+Q + (x^2 + 2Qx)^(1/2) )].

Note: I did differentiate the above integration result to confirm it leads to the previous differential equation.Calculating the value of t uses the integration limits between x=a=0 and x=a=1. The result is the following.

t = (1/3) (1/H_0) (1/L^(1/2) log (W(1) / W(0)), where

W(1) = 1+Q + (1+2Q)^(1/2), and

W(0) = (Q+2Q)(1/2).

Now comes the numerical calculations.(1/3) (1/L^(1/2) (1/H_0) = (1/3) (1/0.6897)^1/2 14.4 x 10^9 years

~= 5,780,000,000 years.

W(1) ~= ~2.43

W(0) ~= 0.896

W(1)/W(0) = ~2.71

log(W(1)/W(0)) =~1.00

t = (1/3) (1/L^(1/2) (1/H_0) log (W(1) / W(0))

~=5,780,000,000 years X ~1.00 ~=5,770,000,000 years

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