Assume that the evolution equation for the density contrast δ is given by (found in image 1 below) where α and β are positive constants, k is the comoving wave number, and ρ¯ is the matter energy density in the background. i) Show that in the large scale limit k = 0, equation (1) can be written as where θ is a constant and Ωm is the density parameter for matter. Determine θ. ii)Solve Equ. (2) in order to find the time evolution for δ in the case of a flat universe dominated by radiation, neglecting the term containing Ωm.
Assume that the evolution equation for the density contrast δ is given by (found in image 1 below) where α and β are positive constants, k is the comoving wave number, and ρ¯ is the matter energy density in the background. i) Show that in the large scale limit k = 0, equation (1) can be written as where θ is a constant and Ωm is the density parameter for matter. Determine θ. ii)Solve Equ. (2) in order to find the time evolution for δ in the case of a flat universe dominated by radiation, neglecting the term containing Ωm.
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Assume that the evolution equation for the density contrast δ is given by (found in image 1 below)
where α and β are positive constants, k is the comoving wave number, and ρ¯ is the matter
energy density in the background.
i) Show that in the large scale limit k = 0, equation (1) can be written as
where θ is a constant and Ωm is the density parameter for matter. Determine θ.
ii)Solve Equ. (2) in order to find the time evolution for δ in the case of a flat universe
dominated by radiation, neglecting the term containing Ωm.
![2
8+26+ 0-6 – 86 = 0,
-
5+2=
(1)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3b765a29-fcb0-4895-9cef-5ef60c256f5b%2F482445f9-7e89-46ed-9193-e2e9305715ef%2Fcd7z4td_processed.png&w=3840&q=75)
Transcribed Image Text:2
8+26+ 0-6 – 86 = 0,
-
5+2=
(1)
![5+2 6+0H²m6= 0,
a
(2)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3b765a29-fcb0-4895-9cef-5ef60c256f5b%2F482445f9-7e89-46ed-9193-e2e9305715ef%2F0hqg1j7_processed.png&w=3840&q=75)
Transcribed Image Text:5+2 6+0H²m6= 0,
a
(2)
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