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I have the following formula: $$t(a) = \frac{1}{H_0}\int_0^a \frac{da'}{\sqrt{\Omega_{r,0}a^{-2} + \Omega_{m,0}a^{-1}+\Omega_{\Lambda,0}a^2 + (1-\Omega_0)}}$$ I want to go in the opposite direction. That is, I want to know what a is given t. $$a(t) = ??$$ Can I use Solve to find this value?

EDIT: Here's the Mathematica form of the equation:

t[scaleFactor_] := (Mpc/Subscript[H, 0])*
   NIntegrate[
1/Sqrt[Subscript[Ω, r]/a^2 + 
   Subscript[Ω, m]/a + 
   Subscript[Ω, k] + 
           Subscript[Ω, Λ]^2*a], {a, 
 0, scaleFactor}];
Bob Hanlon
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Quark Soup
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    Your equation makes no mathematical sense. You cannot have the same variable as the dummy integration one and as the limit of the integral. Which $a$ in the integrand are integrated over and which ones are not? – SuperCiocia Apr 27 '20 at 22:50
  • There are several ways to determine numerically the inverse of a function. Please provide your function in Mathematica code. – bbgodfrey Apr 27 '20 at 23:06
  • Without specific values of $\Omega_{r,0}, \Omega_{m,0},\Omega_{\Lambda,0},\Omega_{0}$ your task is a bit involved because of many cases to be considered, nevertheless I can say that in general $t(a)$ is an elliptic integral and $a(t)$ can be cast in terms of the Weierstrass elliptic function. I would put some effort in order to provide an answer if you prescribed appropriate values of $\Omega$ constants as well as $H_0$. – Artes Apr 27 '20 at 23:06
  • Solve will not solve your equation, you should examine these answers: 1, 2, 3, then you should understand well the underlying issues. – Artes Apr 27 '20 at 23:15
  • @Artes - For all practical purposes, you can drop the $\Omega_r$ and $(1-\Omega_0)$ terms. $\Omega_m=0.3$ and $\Omega_{\Lambda}=0.7$, $H_0=67.3/Mpc=2.18E-18$. – Quark Soup Apr 28 '20 at 00:59

1 Answers1

2

Easy, differentiate your equation with respect to $a$:

$$\frac{\mathrm{d} t}{\mathrm{d} a} = \frac{1}{H_0} \frac{1}{\sqrt{\Omega_{r,0}a^{-2} + \Omega_{m,0}a^{-1}+\Omega_{\Lambda,0}a^2 + (1-\Omega_0)}}.$$

Now you have an autonomous differential equation for $a(t)$:

$$ \frac{1}{H_0}\frac{\mathrm{d} a(t)}{\mathrm{d} t} = \sqrt{\Omega_{r,0}a^{-2} + \Omega_{m,0}a^{-1}+\Omega_{\Lambda,0}a^2 + (1-\Omega_0)}.$$

eq=1/Subscript[H, 0] a'[t]==Sqrt[Subscript[Ω, m]/a[t]+Subscript[Ω, Λ] a[t]^2];
DSolve[eq,a[t],t]
(*{{a[t]->((-(1/2))^(2/3) E^(-(C[1]+t Subscript[H, 0]) Sqrt[Subscript[Ω, Λ]]) (E^(3 (C[1]+t Subscript[H, 0]) Sqrt[Subscript[Ω, Λ]])-Subscript[Ω, m] Subscript[Ω, Λ])^(2/3))/Subsuperscript[Ω, Λ, 2/3]},

   {a[t]->(E^(-(C[1]+t Subscript[H, 0]) Sqrt[Subscript[Ω, Λ]]) (E^(3 (C[1]+t Subscript[H, 0]) Sqrt[Subscript[Ω, Λ]])-Subscript[Ω, m] Subscript[Ω, Λ])^(2/3))/(2^(2/3) Subsuperscript[Ω, Λ, 2/3])},

   {a[t]->-(((-1)^(1/3) E^(-(C[1]+t Subscript[H, 0]) Sqrt[Subscript[Ω, Λ]]) (E^(3 (C[1]+t Subscript[H, 0]) Sqrt[Subscript[Ω, Λ]])-Subscript[Ω, m] Subscript[Ω, Λ])^(2/3))/(2^(2/3) Subsuperscript[Ω, Λ, 2/3]))}}*)

Out of 3 solutions this one seems to be physical one:

$$\frac{e^{-\left(C[1]+t H_0\right) \sqrt{\Omega _{\Lambda }}} \left(e^{3 \left(C[1]+t H_0\right) \sqrt{\Omega _{\Lambda }}}-\Omega _m \Omega _{\Lambda }\right){}^{2/3}}{2^{2/3} \Omega _{\Lambda }^{2/3}}$$

yarchik
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