Find the Metastatic Conditions for Omegas in the Friedmann Equation

[Buzz Bloom]

Homework Statement

Note: I use the notation []’ to indicate d[]/dt. This is because I do not have a convenient way to use the dot notation for differentiation.

In equation [3] below, what are the constrains on the Ωs (in addition to those given in the Relevant equations [8], [9], and [10]) such that there is a value, t=s satisfying [1] and [2]?

Relevant Equations

See below in Solution section.

Required Equations
Note: Includes inequalities.
[1] H(s) = 0, and
[2] H’(s) = 0?
[3]

[4] H = a’/a
[5] a(t₀) = a₀ =1
[6] H₀ = H(t₀)
[7] Ω_R = 0
[8] Ω_k<0, Ω_k = -K, K > 0
[9] Ω_M>1, Ω_M = M, M > 1
[10] Ω_Λ = 1-Ω_k- Ω_M = 1+K-M > 0 → M-1 < K

Solution
Note: I have assigned this problem to myself. I would appreciate a response that confirms that my solution is correct, or identifies one or more steps where I made a mistake.

The inequality in [29] is the solution as a final constraint on of the Ωs (in addition to
he inequalities in [8] through [10]). Below are the sequence of steps leading to [29].

Note that [1] and [4] implies that a’(s) = 0.
Using [4] through [10], [3] becomes
[11a] a’²/a² = H₀² [ M/a³ – K/a² + (1-M+K)].

In [1] and [2], s is defined as the value of t that satisfies [1] and [2].
Note that [1] and [4] implies that
[11b] a’(s) = 0.
Let
[11c] b=a(s).
Using [11b] and [11c] simplifies [11a] to
[12] 0 = M/b³ – K/b² + (1-M+K).
Multiplying [12] by b3 yields
[13] 0 = M – Kb + (1-M+K)b³
Equation [13] defines the specific value a=b which satisfies condition [1].

I now proceed to work with [2].
Differentiating [4] with respect to t yields
[14] H’ = a’’/a – a’²/a² = (a’’/a - H²)
Using [1] in [14] yields
[15] a’’(s) = 0.
Multiplying [11a] by a³ yields
[16] a a’² = H₀² [M - Ka +(1-M+K)a³]
Differentiating [16] with respect to t yields
[17] a’³+2aa’a’’ = H₀² [-Ka’ +3(1-M+K)a²a’]
Dividing [17] by a’ yields
[18a] a’²+2aa’’ = H₀² [-K +3(1-M+K)a²]
Using [1], [11c], and [15] in [18a] yields
[18b] 0 = -K +3(1-M+K)b²
Solving [18b] for b² yields
[19] b² = K/(3(1-M+K)).
Inequality [10] implies that b² is positive, so b is a real number.
Substituting [19] into [13] yields
[20] 0 = M – Kb + (K/(3(1-M+K))) (1-M+K)b]
Simplifying [20] yields
[21] M = Kb(1-(1/3)) = (2/3)Kb, which yields
[22] b = (3/2)M/K.

I will now check to confirm that [22] and [19] together are a solution of [13].
[23] b² = (9/4) M²/K² = K/(3(1-M+K)) yields
[24a] M²(1-M+K) = (4/27)K³, which yields
[24b] 1-M+K = (4/27)K³/M².
Combining [24b] with [13] yields
[25a] 0 = M – Kb + ((4/27)K3/M²) b³

= M – K (3/2)M/K + ((4/27)K3/M2) (27/8)M3/K3​

= M – (3/2)M + (1/2)M​

= 0.​

Combining [19] and [22] yields
[23] (9/4) M²/K² = K/(2(1-M+K)), yields
[24] K= (9/2)M²(1-M+K), yields
[25] K(1+(9/2)M²) = (9/2)M², yields
[26] K = (9/2)M² / (1+(9/2)M²) = M²/((2/9)+M²), yields
[27a] 1/K = 1 + (2/9)/M²,yields
[27b] 1/M² = (9/2) [(1/K)-1] = (9/2)((1-K)/K), and
[28] M² = (2/9) K/(1-K).
Equation [28] implies that
[29] K<1,
since otherwise M would not be a real number.

Combining [29] with [9] and [10] yields a summary inequality:
[30] 0 < M-1 < K < 1.

NOTE:
I have also made an effort to also show that

b > a₀ = 1.​

My intuition tells me that this must be true, but I have failed to prove it. I have decided to discuss the implications of a possible

b < a₀ = 1​

in a separate thread.

 

[mighty2000]

Your solution appears to be correct. You have successfully used the given equations and inequalities to derive the final constraint [29], which is a summary of the constraints on the Ωs. Your steps are clear and easy to follow.

Regarding the question of b > a0 = 1, your intuition is correct. Since b = a(s), and s is defined as the value of t that satisfies [1] and [2], it follows that b > a0 = 1. This is because [1] and [2] imply that a’(s) = 0, and therefore a(s) must be a constant function with value a0 = 1. This means that b = a(s) = a0 = 1. So, b > a0 = 1 is always true.

I hope this helps. Keep up the good work!